Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
$R$ be the row reduced echelon form of a $4 \times 4$ real matrix $A$ Let $R$ be the row reduced echelon form of a $4 \times 4$ real matrix $A$ and
let the
third column of $R$ be $\left[\begin{array}{l}0 \\ 1 \\ 0 \\ 0\end{array}\right]$. Then which is true?
P): If $\left[\begin{array}{l}\alpha \\ \beta \\ \gamma \\ 0\end{array}\right]$ is a solution of $A \mathrm{x}=0$, then $\boldsymbol{\gamma}=0$.
Q): For all $\mathrm{b} \in \mathbb{R}^{4}, \operatorname{rank}[A \mid \mathrm{b}]=\operatorname{rank}[R \mid \mathbf{b}]$.
For $P$ \begin{aligned}
&{\left[\begin{array}{llll}
a & b & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & \phi \\
0 & 0 & 0 & 0
\end{array}\right]\left[\begin{array}{l}
\alpha \\
\beta \\
\gamma \\
0
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0\\
0
\end{array}\right]} \\
&a \alpha+b \beta+0+0=0 \\
&0+0+\gamma+0=0 \\
&\Rightarrow \gamma=0
\end{aligned}
For $Q$
Rank of a matrix and rank of Row echelon matrix is same . So Q is also correct.
So both the statements are true. Is my approach correct?
Thanks in advance.
| For $P$ \begin{aligned}
&{\left[\begin{array}{llll}
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{array}\right]\left[\begin{array}{l}
0 \\
-2 \\
2 \\
0
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0\\
0
\end{array}\right]} \\
\\
\\
&\Rightarrow \gamma=2
\end{aligned}
The statement $P$ is also false.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding $\int^1_0 \frac{x\,\mathrm dx}{((2x-1)\sqrt{x^2+x+1} + (2x+1)\sqrt{x^2-x+1})\sqrt{x^4+x^2+1}}$ $$\int^1_0 \frac{x\,\mathrm dx}{((2x-1)\sqrt{x^2+x+1} + (2x+1)\sqrt{x^2-x+1})\sqrt{x^4+x^2+1}}$$
I've tried :
*
*Substitution:
*
*$t = x^2$
*$x = \cos{t}$
*Definite Integral Properties like $a+b-x$ (which messes up the integral symmetry)
*Partial Integration by breaking integral into into $\frac{x}{\sqrt{\left(x^4+x^2+1\right)}}$
*Partial Fractions which are very ugly
As per wolframAlpha, a beautiful elementary closed form exists
$$
\frac{\sqrt{\left(x^4+x^2+1\right)}}{3}\left(\frac{1}{\sqrt{\left(x^2-x+1\right)}}- \frac{1}{\sqrt{\left(x^2+x+1\right)}}\right)
$$
I'll be happy with solution of just definite integral, bonus points if you get closed form solution too.
|
Let $\mathcal{I}$ denote the value of the following definite integral:
$$\mathcal{I}:=\int_{0}^{1}\mathrm{d}x\,\frac{x}{\left[\left(2x+1\right)\sqrt{x^{2}-x+1}+\left(2x-1\right)\sqrt{x^{2}+x+1}\right]\sqrt{x^{4}+x^{2}+1}}\approx0.2440169.$$
Observe that $\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)=x^{4}+x^{2}+1$ and
$\left[\left(2x+1\right)\sqrt{x^{2}-x+1}-\left(2x-1\right)\sqrt{x^{2}+x+1}\right]\left[\left(2x+1\right)\sqrt{x^{2}-x+1}+\left(2x-1\right)\sqrt{x^{2}+x+1}\right]=6x$.
Then,
$$\begin{align}
\mathcal{I}
&=\int_{0}^{1}\mathrm{d}x\,\frac{x}{\left[\left(2x+1\right)\sqrt{x^{2}-x+1}+\left(2x-1\right)\sqrt{x^{2}+x+1}\right]\sqrt{x^{4}+x^{2}+1}}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{x}{\left(2x+1\right)\sqrt{x^{2}-x+1}+\left(2x-1\right)\sqrt{x^{2}+x+1}}\\
&~~~~~\times\frac{1}{\sqrt{\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)}}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{\left(2x+1\right)\sqrt{x^{2}-x+1}-\left(2x-1\right)\sqrt{x^{2}+x+1}}{6}\\
&~~~~~\times\frac{1}{\sqrt{x^{2}-x+1}\sqrt{x^{2}+x+1}}\\
&=\frac13\int_{0}^{1}\mathrm{d}x\,\frac{\left(2x+1\right)\sqrt{x^{2}-x+1}-\left(2x-1\right)\sqrt{x^{2}+x+1}}{2\sqrt{x^{2}-x+1}\sqrt{x^{2}+x+1}}\\
&=\frac13\int_{0}^{1}\mathrm{d}x\,\left[\frac{\left(2x+1\right)}{2\sqrt{x^{2}+x+1}}-\frac{\left(2x-1\right)}{2\sqrt{x^{2}-x+1}}\right]\\
&=\frac13\int_{0}^{1}\mathrm{d}x\,\frac{d}{dx}\left[\sqrt{x^{2}+x+1}-\sqrt{x^{2}-x+1}\right]\\
&=\frac{\sqrt{3}-1}{3}.\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4171475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Expected number of leaves of a planar tree with $n$ nodes I want to find out the expected number of leaves of a planar tree with $n$ nodes using the bivariate generating function for the number of leaves in a planar tree.
For the (univariate) generating function we have $$P(z)=\frac{z}{1-P(z)}$$Solving the quadratic equation gives us $$P(z)=\frac{1-\sqrt{1-4z}}{2}$$
Does this imply that the bivariate generating function is given by $$P(z,u)=\frac{z}{1-u\frac{1-\sqrt{1-4z}}{2}}$$?
And for the expected number of leave I have to evaluate $$\frac{[z^n](\frac{\partial}{\partial u}P(z,u))\big |_{u=1}}{[z^n]P(z,1)}$$?
| The combinatorial class for plane trees with leaves marked is
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{P} = \mathcal{Z} \times \mathcal{U} +
\mathcal{Z}\times \textsc{SEQ}_{\ge 1}(\mathcal{P})$$
This gives the functional equation
$$P(z) = uz + z \frac{P(z)}{1-P(z)}$$
or
$$z = \frac{P(z)(P(z)-1)}{u(P(z)-1)-P(z)}.$$
Now applying the residue operator for $n\ge 1$ we get
(the constant term of $P(z)$ is zero)
$$[z^n] P(z) = \frac{1}{n} [z^{n-1}] P'(z)
= \frac{1}{n}
\; \underset{z}{\mathrm{res}} \;
\frac{1}{z^n} P'(z).$$
We put $w=P(z)$ and find
$$\frac{1}{n}
\; \underset{w}{\mathrm{res}} \;
\frac{(u(w-1)-w)^{n}}{w^{n} (w-1)^{n}}.$$
Extracting the coefficient on $[u^k]$ where we must have $1\le k\le n-1$
we obtain
$$\frac{1}{n} {n\choose k}
\; \underset{w}{\mathrm{res}} \;
\frac{(w-1)^k (-1)^{n-k} w^{n-k}}
{w^{n} (w-1)^{n}}
\\ = \frac{1}{n} {n\choose k}
\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{k} (1-w)^{n-k}}
\\ = \frac{1}{n} {n\choose k} {k-1+n-1-k\choose k-1}
= \frac{1}{n} {n\choose k} {n-2\choose k-1}
= \frac{1}{k} {n-1\choose k-1} {n-2\choose k-1}.$$
As a sanity check we get for the sequence of leaf counts with $k$
increasing from $k=1$ to $n-1$ the following values
$$1, 1, 1, 1, 3, 1, 1, 6, 6, 1, 1, 10, 20, 10, 1,
1, 15, 50, 50, 15, 1, \ldots $$
which is OEIS A001263, the Narayana numbers.
What we have is $N(n-1,k).$ When we look this up in turn at Wikipedia
on Narayana
numbers
we find that the number of ordered rooted trees with $n+1$ vertices and
$k$ leaves is $N(n,k)$ which means we have the correct value.
We continue with the expectation where we first need the total count of
these trees which is
$$\frac{1}{n} \sum_{k=1}^{n-1} {n\choose k} {n-2\choose k-1}
= \frac{1}{n} \sum_{k=1}^{n-1} {n\choose k} {n-2\choose n-k-1}
= \frac{1}{n} {2n-2\choose n-1}$$
by Vandermonde. These are the familiar Catalan numbers $C_{n-1}.$ We
get for the sum total of leaves
$$\sum_{k=1}^{n-1} {n-1\choose k-1} {n-2\choose k-1}
= \sum_{k=0}^{n-2} {n-1\choose k} {n-2\choose k}
\\ = \sum_{k=0}^{n-2} {n-1\choose k} {n-2\choose n-2-k}
= {2n-3\choose n-2} = {2n-3\choose n-1}.$$
again by Vandermonde. This will produce for $n\ge 2$
$$\frac{n-1}{2n-2} {2n-2\choose n-1}
= \frac{1}{2} {2n-2\choose n-1}.$$
Looking it up we find OEIS A088218
where it says "total number of leaves in all rooted ordered trees with
$n$ edges" is $\frac{1}{2} {2n\choose n}.$ As before this is $n+1$
vertices so we definitely have the right answer.
We are ready at last to compute the queried expectation of the
number of leaves for $n\ge 2$
$$\frac{(2n-3)!}{(n-2)! \times (n-1)!}
\times n \times \frac{(n-1)! \times (n-1)!}{(2n-2)!}
\\ = \frac{1}{2n-2} \times n \times (n-1).$$
This is
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2} n.}$$
Remark. In the above we have used OP's functional equation for
$P(z)$ to deduce the type of tree being sought (ordered rooted or
rooted plane tree).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\displaystyle I=\iiint_V(x+y+z)^2 \, dx \, dy \, dz$, where $V$ is the region bounded by $x^2+y^2+z^2=3a^2$ and $x^2+y^2=2az$ Compute $\displaystyle I=\iiint_V(x+y+z)^2 \,dx \,dy \,dz$, where $V$ is the region bounded by $x^2+y^2+z^2=3a^2$ and $x^2+y^2=2az$.
What I did was I found the intersection of the two surfaces, I got that it is the circle $x^2+y^2=2a^2$ in the plane $z=a$, so I think that $D=\{(x, y)\mid x^2+y^2\le 2a^2\}$ is the projection of $V$ on the $xy$ plane. Then everything is simple because I can easily see that $0\le z\le a\sqrt{3}$, so I should have $$I=\iint_D\left(\int_0^{a\sqrt 3}(x+y+z)^2\,dz\right)\,dx\,dy$$ which is easy to compute by polar coordinates.
However, my book does the following: after finding that the intersection is the circle $x^2+y^2=2a^2$ in the plane $z=a$, it says that $0\le z\le a\sqrt 3$ and it goes on to find the section with the plane $z=\text{constant}$, finding this section to be $x^2+y^2\le 2az$ for $z\in [0,a]$ and $x^2+y^2\le 3a^2-z^2$ for $z\in [a, a\sqrt 3]$. It basically goes on to use the method that I asked about here A question about the cross section method for triple integrals.
I am not really sure if my approach is correct. I kind of understand their approach, but I don't understand why mine would be wrong.
| As Ninad Munshi put in his comments, the way question reads, it could mean two different regions. I am considering the region as,
$x^2+y^2+z^2 \leq 3a^2$ and $2az \geq x^2+y^2$.
First find the radius at the intersection of both surfaces.
$x^2+y^2+z^2=3a^2$ and $2az = x^2+y^2$.
$z^2+2az = 3a^2 \implies z = a$. So using cylindrical coordinates, at intersection,
$r^2 = 2az = 2a^2 \implies r = a\sqrt2$.
So integral should be,
$\displaystyle \int_0^{2\pi} \int_0^{a\sqrt2} \int_{r^2/(2a)}^{\sqrt{3a^2-r^2}} f(x,y,z) \ r \ dz \ dr \ d\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4174060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How many non-congruent triangles can be formed by the vertices of a regular polygon of $n$ sides On a circle there are $n$ evenly spaced points (i.e., they form a regular polygon). How many triangles can be formed when congruent triangles are considered the same?
This question is closely related to
How many triangles can be formed by the vertices of a regular polygon of $n$ sides?, but here congruent triangles are considered to be the same.
| For each triangle $A_iA_jA_k$, the angles $\angle A_iOA_j$, $\angle A_jOA_k$ and $\angle A_kOA_i$ are respectively equal to $|j-i|\frac{2\pi}{n}$, $|j-k|\frac{2\pi}{n}$ and $|k-i|\frac{2\pi}{n}$ with $1\le i\ne j\ne k \le (n-1)$.
Let denote
$$
\begin{cases}
x=|j-i|\\
y=|k-j|\\
z=|k-i|
\end{cases}
$$
Then
$$
\begin{cases}
x+y+z = n \\
x,y,z \in \Bbb N^* \\
\end{cases} \tag{1}
$$
The number of triangles when congruent triangles are considered the same is equal to the number of solutions of $(1)$ (Two solutions that differ only in the order of their summands are considered the same, for example, with $n = 4$: $1+1+2$ is the same as $2 +1 +1$).
The problem $(1)$ is solved here and the number of solutions of $(1)$ is the number of partitions $p(3,n)$ of $n$ into $3$ non-zero parts. The number of partitions $p(k,n)$ satisfies
$$
\begin{cases}
p(0,0) &= 0 \\
p(k,n) &= p(k,n-k)+ p(k-1,n-1) & \text{otherwise}. \\
\end{cases} \tag{2}
$$
Remark: It seems that the recurrent formula in the answer is not correct, the recurrent formula $(2)$ is taken from the wikipedia here.
From $(2)$, we can deduce the general formula of $p(3,n)$ as follows
*
*For $k = 1$, it's obvious that
$$p(1,n) = 1$$
*For $k =1$, we have
$$p(2,n) = p(2,n-2)+1 =... \implies p(2,n) = \left\lfloor \frac{n}{2} \right\rfloor$$
*For $k = 3$, we have
$$p(3,n) =p(3,n-3)+\left\lfloor \frac{n-1}{2} \right\rfloor=...$$
$$
\begin{align}
\implies p(3,n) &=\left\lfloor \frac{n-1}{2} \right\rfloor+\left\lfloor \frac{n-4}{2} \right\rfloor+...+\left\lfloor \frac{n-1-3i}{2} \right\rfloor+... \\
& =\sum_{0 \le i \le \left\lfloor \frac{n-1}{3}\right\rfloor}\left\lfloor \frac{n-1-3i}{2} \right\rfloor
\end{align}
$$
Hence, the number of triangles is equal to
$$p(3,n)=\sum_{0 \le i \le \left\lfloor \frac{n-1}{3} \right\rfloor}\left\lfloor \frac{n-1-3i}{2} \right\rfloor$$
Remark: The last steps have a lot of calculation, please feel free to correct if you find any mistake.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4175040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding the range of $a$ for which line $y=2x+a$ lies between circles $(x-1)^2+(y-1)^2=1$ and $(x-8)^2+(y-1)^2=4$ without intersecting either
Find the range of parameter $a$ for which the variable line $y = 2x + a$ lies between the circles $(x-1)^2+(y-1)^2=1$ and $(x-8)^2+(y-1)^2=4$ without intersecting or touching either circle.
Now, how I solved it was realising the line has a positive slope of $+2$, and thus if it's a tangent to circle #1 then its intercept should be negative. And so, using the condition of tangency,
$$a^2= m^2(r_1)^2+(r_1)^2= 4+1 $$
And thus, as $a$ can only be negative (otherwise a positive value of $a$ will make the line intersect the circle). Thus, making the lower bound of $a> - \sqrt 5$.
Similarly for the bigger circle $a<-\sqrt{20}=-\sqrt{(2^2)(2^2)+(2^2)}$
Hence I find that the solution should be $(-\sqrt5,-\sqrt{20})$, but the actual solution is $\left(2\sqrt 5-15,-\sqrt 5-1\right)$.
| Your condition of tangency is assuming your circles are centered at the origin.
It derives for setting $y = mx + a$ and $x^2 +y^2 = r^2$ and substituting $x^2 + (mx+a)^2 = r^2$. Solving for $x$ and getting two values based an the discriminate $4a^2m^2 -4(a^2 - r^2)(1+m^2)$ and knowing if this is a tangent line the two points of intersection coincide so the discriminate is $0$. Solving we get $a^2 = (m^2 + 1)r^2$ as you did.
But our circles are not centered at the origin.
So our work is $y = 2x + a$ and $(x-1)^2 + (y-1)^2 = 1$
$(x-1)^2 + (2x+(a-1))^2 = 1$ so $
$(4+1)x^2 +(4(a-1)-2)x +(1 + (a-1)^2 -1) = 0$ so
$5x^2 + (4(a-1)-2)x + (a-1)^2 = 0$
So $x = \frac {-4(a-1)-2 \pm \sqrt {[4(a-1)-2]^2 - 20(a-1)^2}}{10}$
and we must have $[4(a-1)-2]^2 - 20(a-1)^2=0$
So $-4(a-1)^2 - 16(a-1) + 4 =0$
$(a-1)^2 + 4(a-1) - 1 = 0$ so $a-1 = \frac {-4\pm \sqrt{20}}2 = -2 \pm \sqrt 5$.
And $a = -1\pm \sqrt 5$. As we are passing between the circles and the other circle is centered to the right of this circle we wan $a < -1 -\sqrt 5$.
You can do the same for the other circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4175558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Let $x^2 −(m−3)x+m=0,(m\in R)$ be a quadratic equation. Find the value of $m$ for which at least one root is greater than $2$
Let $x^2 −(m−3)x+m=0,(m\in R)$ be a quadratic equation. Find the value of $m$ for which at least one root is greater than $2$.
Discriminant$=m^2-10m+9\ge0\implies m\in(-\infty,1]\cup[9,\infty)$
And putting $m=0$, we get both roots less than $2$. Does that mean we can eliminate all the values of $m\le1?$ Why?
Putting $m=9,$ we get double root at $3$. Does that mean all values of $m\ge9$ can be accepted?
| The quadratic polynomial $ \ x^2 − (m−3)x + m \ \ $ represents an "upward-opening" parabola, for which the "vertex form" is $ \ \left(x \ − \ \frac{ m−3}{2} \right)^2 \ + \ \left( m \ - \ \frac{ [m−3]^2}{4} \right) \ \ . $ It may also be noted that $ \ f(1) \ = \ 4 \ \ , $ independent of the value of $ \ m \ \ . $ So there can be no zeroes of the polynomial greater than $ \ 2 \ $ at the very least for when the vertex is at $ \ x \le 1 \ \Rightarrow \ \frac{ m−3}{2} \ \le \ 1 \ \Rightarrow \ m \ \le \ 5 \ \ . $
The vertex is tangent to or "below" the $ \ x-$axis for
$$ m \ - \ \frac{ [m−3]^2}{4} \ \le \ 0 \ \ \Rightarrow \ \ 0 \ \le \ -4m \ + \ m^2 \ -6m \ + \ 9 \ \ = \ \ (m - 1)·(m - 9) $$ $$ \Rightarrow \ \ m \ \le \ 1 \ \ , \ \ m \ \ge \ 9 $$
(as also found by other respondents). We may thus discard the first inequality. As for the second, $ \ m \ = \ 9 \ $ gives us the polynomial $ \ \left(x \ − \ \frac{ 9−3}{2} \right)^2 \ = \ (x \ − \ 3 )^2 \ \ , $ which has a "double zero" at $ \ x = 3 \ \ $ and is therefore an admissible result. All larger values of $ \ m \ $ correspond to parabolas with vertices at $ \ x \ \ge \ 3 \ $ and $ \ y \ < \ 0 \ \ , $ so they always have an $ \ x-$intercept "to the right" of $ \ x = 3 \ \ . $ Hence, all of the quadratic polynomials $ \ x^2 − (m−3)x + m \ \ $ have at least one real zero greater than $ \ 2 \ $ for $ \ \mathbf{m \ \ge \ 9} \ \ . $
[In fact, we may note that for $ \ m = 10 \ \ , $ the polynomial becomes $ \ x^2 - 7x + 10 \ = \ (x - 2)·(x - 5) \ \ , $ so both zeroes are greater than $ \ 2 \ $ for $ \ 9 \ \le \ m \ < \ 10 \ \ . \ $ ]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4176361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Finding a general solution of pde $(x^2-y^2-u^2)\cdot u_x+(2xy)\cdot u_y=2xu$ how can I solve this partial dif. equation. I try to use Langarange methods
This is my solution;
$$\frac {dx}{x^2-y^2-u^2}=\frac {dy}{2xy}=\frac {du}{2xu}= λ$$
$$\frac {dy}{2xy}=\frac {du}{2xu}$$
$$\frac {du}{u}-\frac {dy}{y}=0$$
$$\ln u-\ln y=\ln c_1$$
$$\frac {u}{y}=c_1=w$$
then i know that we now need to find another function. but i am stuck.
$$F(w,v)=0$$
$$w=f(v)$$
can u help me to find $v$ with steps, please? i would appreciate it if you help.
| There is a another way and this
$$\frac{du}{2xu}=\frac{xdx+ydy+udu}{x.(x^2-y^2-u^2)+y.2xy+u.2xu}$$
$$\frac{du}{2xu}=\frac{xdx+ydy+udu}{x^3-x.y^2-x.u^2+2xy^2+2xu^2}$$
$$\frac{du}{2xu}=\frac{xdx+ydy+udu}{x^3+x.y^2+x.u^2}$$ and if we simplify the equation
$$\frac{du}{2x.(u)}=\frac{xdx+ydy+udu}{x(x^2+y^2+u^2)}$$
$$\frac{du}{2x.(u)}=\frac{2xdx+2ydy+2udu}{2x(x^2+y^2+u^2)}$$ and cancled the $2x$ and use $d(x^2+y^2+u^2)=2dx+2dy+2du$ then we get
$$\frac{du}{u}=\frac{d(x^2+y^2+z^2)}{x^2+y^2+u^2}$$ and
$$ln|u|=ln|x^2+y^2+u^2|+lnd$$
$$ln|\frac{u}{x^2+y^2+u^2}|=lnd$$ then $$d=\frac{u}{x^2+y^2+u^2}=v$$ so $F(w,v)=F(\frac{u}{y},\frac{u}{x^2+y^2+u^2})=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Asymptotic expansion of $\sum_{k=1}^n {n \choose k} (-1)^k \frac 1 {1-x^k} $ I'm trying to find the limit (and asymptotic expansion as $n\to\infty $) of
$$\sum_{k=1}^n {n \choose k} (-1)^k \frac 1 {1-x^k} $$
for $0<x<1$.
So far, I have no idea...
I found this question when dealing with the expected value of $\max (X_1, \dots X_n)$ when $X_1, \dots, X_n$ are geometric random variables with same success probability $p$...
Thus, the expected value is this sum for $x=1-p$ (in absolute value)...
Could you help me please ?
Thanks in advance.
| Not a complete answer, but an idea of where to look and some lower bounds.
Let $$p_k(x)=\frac{1-x^k}{1-x}=1+x+\dots+x^{k-1}$$
Then:
$$\frac{1}{1-x^k}=\frac{1}{k(1-x)}+\frac{(k-1)+(k-2)x+\cdots+x^{k-2}}{kp(x)}$$
Also, $$\begin{align}\sum_{k=1}^{n}(-1)^k\binom{n}k\frac1k&=\int_0^1\frac{(1-t)^n-1}{t}\,dt\\
&=\int_0^1 \frac{t^n-1}{1-t}\,dt\\
&=-\int_0^1(1+t+t^2+\cdots+t^{k-1})\,dt\\
&=-\left(1+\frac12+\cdots+\frac{1}{n-1}\right) =-H_{n-1}
\end{align}$$
So if $f_n(x)$ is your term, then $$f_n(x)=\frac{-H_n}{1-x}+\sum_{k=1}^{n}(-1)^k\binom nkq_k(x)$$
When $$q_k(x)=\frac{(k-1)+(k-2)x+\cdots+x^{k-2}}{kp(x)}$$
Notice that $$q_k(x)=\frac{x^{k-2}p_k’(1/x)}{kx^{k-1}p_k(1/x)}=\frac{1}{x}\frac{p_k’(1/x)}{kp_k(1/x)}$$
For any polynomial $p$ with no repeating roots, $$\frac{p’(y)}{p(y)}=\sum_j \frac{1}{y-r_i}$$ where the $r_i$ are the roots of the polynomial.
So:
$$\begin{align}q_k(x)&=\frac{1}{k}\sum_{j=1}^{k-1}\frac{1}{1-xe^{2\pi ij/k}}\\
&= \frac{1}{k}\sum_{j=1}^{k-1}\frac{1-xe^{-2\pi ij/k}}{1+x^2-2x\cos(2\pi ij/k)}\\
&=\frac1k\sum_{j=1}^{k-1}\frac{1-x\cos(2\pi j/k)} {1+x^2-2x\cos(2\pi j/k)}
\end{align}
$$
The last step because we know the imaginary values cancel.
Now:
$$\begin{align}\frac{1}{1+x^2-2x\cos\theta}&=\frac{1}{1+x^2}\sum_{p=0}^{\infty}\left(\frac{2x\cos\theta}{1+x^2}\right)^p\\
&\geq\frac{1}{1+x^2}+\frac{2x\cos\theta}{(1+x^2)^2}
\end{align}$$
So:
$$\frac{1-x\cos\theta}{1+x^2-2x\cos\theta}\geq\\ \frac{1}{1+x^2}+\frac{2x\cos\theta}{(1+x^2)^2} -\frac{x(1+x^2)\cos\theta}{(1+x^2)^2}-\frac{2x^2\cos^2\theta}{(1+x^2)^2}\\=
\frac{1}{1+x^2}+\frac{x(1-x^2)\cos\theta}{(1+x^2)^2}-\frac{2x^2\cos^2\theta}{(1+x^2)^2}
$$
Now:
$$\sum_{j=1}^k\cos(2\pi j/k)=-1$$
and $$\sum_{j=1}^k\cos^{2}(2\pi j/k)=\frac k2$$
So: $$q_k(x)\geq\frac1k\left(\frac{1}{1+x^2}-\frac{x(1-x^2)}{(1+x^2)^2}-\frac{kx^2} {(1+x^2)^2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4179268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Number of integer solutions of $a^2+b^2=10c^2$ Find the number of integer solutions of the equation $a^2+b^2=10c^2$.
I can only get by inspection that $a=3m, b=m,c=m$ satisfies for any $m \in Z$.
Is there a formal logic to find all possible solutions? Any hint?
Also i tried taking $a=p^2-q^2$, $b=2pq$ and $10c^2=(p^2+q^2)^2$
which gives $$\frac{p^2+q^2}{c}=\sqrt{10}$$ which is invalid, since a rational can never be an irrational.
| We first transform the given equation $a^2+b^2=10c^2$ as follows:
Consider $x=\frac{a}{c}, y=\frac{b}{c}$ (assuming we are interested in $c \neq 0$ case). Then we have to find rational points on the circle
$$x^2+y^2=10.$$
Now $(x,y)=(3,1)$ is an obvious solution. To find other rational solutions, consider the line that passes through $(3,1)$ and has slope $m$. It is given by
$$y=mx+(1-3m).$$
Consider the intersection of this line with the circle $x^2+y^2=10$. We can find the intersection from
$$x^2+(mx+(1-3m))^2=10 \implies (1+m^2)x^2+2m(1-3m)x+(1-3m)^2-10=0.$$
But instead of solving the quadratic, we can argue that if two roots are $x_1$ and $x_2$, then we have $x_1=3$, so by means of Viete formula etc. we have
$$x_1+x_2=-\frac{2m(1-3m)}{1+m^2} \implies x_2=\frac{3m^2-2m-3}{1+m^2}.$$
Now if $m$ is rational then $x_2$ is also rational and so will $y_2$ be, because
$$y_2=\frac{1-m^2-6m}{1+m^2}.$$
Now for $m=\frac{p}{q}$, where $p,q \in \Bbb{Z}$ and $q \neq 0$, we can have
$$\color{blue}{a=3p^2-2pq-3q^2, \quad b=q^2-p^2-6pq, \quad c=p^2+q^2}$$
Now the main question is: this combined with your answer (which can be obtained if we allow $q=0$) is that the totality of all solutions?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Comparing numbers of the form $c+\sqrt{b}$ (eg, $3+3\sqrt{3}$ and $4+2\sqrt{5}$) without a calculator It is easy to compare to numbers of the form $a\sqrt{b}$, simply by comparing their squares, for example $3\sqrt{3}$ and $2\sqrt{5}$.
But what if we have $a=3+3\sqrt{3}$ and $b=4+2\sqrt{5}$ for example?
How to compare them without using a calculator? Is there a method that works for all numbers of the form $c+\sqrt{b}$?
I found a similar question to this on the site, but that one is a bit different and I am looking for an answer suitable for elementary students to understand. (but I would love to know any kind of answer even if it does not fit the elementary level).
I was thinking about subtracting $b-a$ and checking wether it is negative or positive,
I got $b-a=1+2\sqrt{5}-3\sqrt{3}$ but $2\sqrt{5}-3\sqrt{3}<0$
(I don't want to continue from here by approximating $\sqrt{5}=2.something$, because we will be somehow using the calculator in our mind; so I am stuck here)
| I think you can use Maclaurin series
For example, $3+3\sqrt{1+2}\approx3+3\cdot(1+2/2)=9$ and $4+2\sqrt{1+4}\approx4+2\cdot (1+4/2)=10$
Clearly, $10>9$ so $4+2\sqrt{5}>3+3\sqrt{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Computing the character table of $SL_2(\mathbb{Z}/3)$ I'm self studying some group representation theory and I've hit a wall with Exercise 3.10 (Computing the character table of $G = SL_2(\mathbb{Z}/3)$) in Fulton and Harris' book. Any advice or hints would be really appreciated!
Here's what I know so far:
(1) The representatives of the conjugacy classes can be chosen to be:
$I, -I,
A_1 = \left(\begin{array}{cc}
1 & 1 \\
0 & 1
\end{array} \right),
A_2 = \left(\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array} \right),
B_1 = \left(\begin{array}{cc}
-1 & 1 \\
0 & -1
\end{array} \right),
B_2 = \left(\begin{array}{cc}
-1 & -1 \\
0 & -1
\end{array} \right), \text{ and }
C = \left(\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array} \right)$
Thus there are 7 irreducible representations of $G$.
There are 1, 1, 4, 4, 4, 4, and 6 conjugates for each of these representatives respectively (see this post).
(2) $G/\{\pm I\}$ has a faithful action on the lines in the $\mathbb{Z}/3$ plane, of which there are 4, thus $G/\{\pm I\}$ is a subgroup of $S_4$ of size 12, thus it must be $A_4$. This allows us to pullback the irreducible representations of $A_4$ to get irreducible representations of $G$. By doing this we can obtain 4 rows of the character table:
\begin{equation*}
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
& 1 & 1 & 4 & 4 & 4 & 4 & 6 \\
\hline
SL_2(\mathbb{Z}/3) & I & -I & A_1 & A_2 & B_1 & B_2 & C \\
\hline
U & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline
U' & 1 & 1 & \omega & \omega^2 & \omega^2 & \omega & 1 \\
\hline
U'' & 1 & 1 & \omega^2 & \omega & \omega & \omega^2 & 1 \\
\hline
V & 3 & 3 & 0 & 0 & 0 & 0 & -1 \\
\hline
W & d_1 & ? & ? & ? & ? & ? & ? \\
\hline
W' & d_2 & ? & ? & ? & ? & ? & ? \\
\hline
W'' & d_2 & ? & ? & ? & ? & ? & ? \\
\hline
\end{array}
\end{equation*}
(3) For the remaining three, using the fact that the sum of the squared dimensions of the irreducible representations is $|G| = 24$, we can deduce that $d_1^2+d_2^2+d_3^2 = 12$, implying that we must have $d_1 = d_2 = d_3 = 2$.
My question is, how can we determine the remaining three characters of the 2 dimensional representations? I've tried looking at tensor products of the first 4 representations among other things and nothing seems to work.
Note:
*
*This is over $\mathbb{C}$.
*The person in this post was considering a similar problem, but ran into the same issue.
Thanks in advance!
| I'm rusty in representation theory, but I'll give it a try.
The characters $W,W',W''$ at $A_1,A_2,B_1,B_2$ can't all be zero, since this would break the orthogonality relation (not sure but this feels true).
Therefore one of them must be non-zero, without loss of generality a character $W$. This implies that $W'=W\otimes U'$ and $W''=W \otimes U''$ (modulo permutations).
Therefore it is enough to find one irreducible representation of degree 2.
We must have something like this:
\begin{equation*}
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
& 1 & 1 & 4 & 4 & 4 & 4 & 6 \\
\hline
SL_2(\mathbb{Z}/3) & I & -I & A_1 & A_2 & B_1 & B_2 & C \\
\hline
U & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline
U' & 1 & 1 & \omega & \omega^2 & \omega^2 & \omega & 1 \\
\hline
U'' & 1 & 1 & \omega^2 & \omega & \omega & \omega^2 & 1 \\
\hline
V & 3 & 3 & 0 & 0 & 0 & 0 & -1 \\
\hline
W & 2 & a & b & c & d & e & f \\
\hline
W' & 2 & a & \omega b & \omega^2 c & \omega^2 d & \omega e & f\\
\hline
W'' & 2 & a & \omega^2 b & \omega c & \omega d & \omega^2 e & f\\
\hline
\end{array}
\end{equation*}
Consider the following orthogonality relations:
*
*Column $I$ and column $-I$ give $a=-2$.
*Column $I$ and column $C$ give $f=0$.
*Column $A_1$ and column $A_1$ give $|b|=1$.
*Column $A_2$ and column $A_2$ give $|c|=1$.
*Column $B_1$ and column $B_1$ give $|d|=1$.
*Column $B_2$ and column $B_2$ give $|e|=1$.
*Column $A_1$ and column $B_2$ give $b=-e$.
*Column $A_2$ and column $B_1$ give $c=-d$.
Since $A_1$ has order 3, the eigenvalues of its representations must be $1,\omega$ or $\omega^2$. Therefore the trace (sum of eigenvalues) is in $\{2,2\omega,2\omega^2,-1,-\omega,-\omega^2\}$, but since $|b|=1$ we are only constrained to $\{-1,-\omega,-\omega^2\}$. Up to permuting the $W$'s, we may assume $b=-1$.
\begin{equation*}
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
& 1 & 1 & 4 & 4 & 4 & 4 & 6 \\
\hline
SL_2(\mathbb{Z}/3) & I & -I & A_1 & A_2 & B_1 & B_2 & C \\
\hline
U & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline
U' & 1 & 1 & \omega & \omega^2 & \omega^2 & \omega & 1 \\
\hline
U'' & 1 & 1 & \omega^2 & \omega & \omega & \omega^2 & 1 \\
\hline
V & 3 & 3 & 0 & 0 & 0 & 0 & -1 \\
\hline
W & 2 & -2 & -1 & c & -c & 1 & 0 \\
\hline
W' & 2 & -2 & -\omega & \omega^2 c & -\omega^2 c & \omega & 0\\
\hline
W'' & 2 & -2 & -\omega^2 & \omega c & -\omega c & \omega^2 & 0\\
\hline
\end{array}
\end{equation*}
Since $A_2$ also has order 3, there are three possibilities for the character table of $SL(\mathbb{F}_3)$, depending on the value of $c\in\{-1,-\omega,-\omega^2\}$. If we can show that there is a rational second degree irreducible representation, then $c=1$ and we're finished. But since $W$ is an irreducible representation then $\overline{W}$ is too. Since $\overline{W}$ can't be equal to $W'$ nor $W''$ (look at column of $A_1$), we must have $\overline{W}=W$ and therefore $c$ is real which implies that $c=-1$ thus finishing the proof.
The final table is
\begin{equation*}
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
& 1 & 1 & 4 & 4 & 4 & 4 & 6 \\
\hline
SL_2(\mathbb{Z}/3) & I & -I & A_1 & A_2 & B_1 & B_2 & C \\
\hline
U & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline
U' & 1 & 1 & \omega & \omega^2 & \omega^2 & \omega & 1 \\
\hline
U'' & 1 & 1 & \omega^2 & \omega & \omega & \omega^2 & 1 \\
\hline
V & 3 & 3 & 0 & 0 & 0 & 0 & -1 \\
\hline
W & 2 & -2 & -1 & -1 & 1 & 1 & 0 \\
\hline
W' & 2 & -2 & -\omega & -\omega^2 & \omega^2 & \omega & 0\\
\hline
W'' & 2 & -2 & -\omega^2 & -\omega & \omega & \omega^2 & 0\\
\hline
\end{array}
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How do I calculate the sum of sum of triangular numbers? As we know, triangular numbers are a sequence defined by $\frac{n(n+1)}{2}$. And it's first few terms are $1,3,6,10,15...$. Now I want to calculate the sum of the sum of triangular numbers. Let's define
$$a_n=\frac{n(n+1)}{2}$$
$$b_n=\sum_{x=1}^na_x$$
$$c_n=\sum_{x=1}^nb_x$$
And I want an explicit formula for $c_n$. After some research, I found the explicit formula for $b_n=\frac{n(n+1)(n+2)}{6}$. Seeing the patterns from $a_n$ and $b_n$, I figured the explicit formula for $c_n$ would be $\frac{n(n+1)(n+2)(n+3)}{24}$ or $\frac{n(n+1)(n+2)(n+3)}{12}$.
Then I tried to plug in those two potential equations,
If $n=1$, $c_n=1$, $\frac{n(n+1)(n+2)(n+3)}{24}=1$, $\frac{n(n+1)(n+2)(n+3)}{12}=2$. Thus we can know for sure that the second equation is wrong.
If $n=2$, $c_n=1+4=5$, $\frac{n(n+1)(n+2)(n+3)}{24}=5$. Seems correct so far.
If $n=3$, $c_n=1+4+10=15$, $\frac{n(n+1)(n+2)(n+3)}{24}=\frac{360}{24}=15$.
Overall, from the terms that I tried, the formula above seems to have worked. However, I cannot prove, or explain, why that is. Can someone prove (or disprove) my result above?
| Notice that after $k$ summations, the formula is
$$\binom{n+k-1}{n-1}.$$
As we can check, by the Pascal identity
$$\binom{n+k-1}{n-1}-\binom{n-1+k-1}{n-2}=\binom{n-1+k-1}{n-1},$$
which shows that the last term of a sum (sum up to $n$ minus sum up to $n-1$) is the sum of the previous stage ($k-1$) up to $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Let $n$ and $m$ be integers such that $5$ divides $1+2n^2+3m^2.$ Then show that $5$ divides $n^2-1.$
Let $n$ and $m$ be integers such that $5$ divides $1+2n^2+3m^2.$ Then show that $5$ divides $n^2-1.$
$\textbf{My attempts} :$
From the condition, we can write $$1+2n^2+3m^2\equiv 0(\mod 5)\tag 1$$
Now, since $5$ is prime so by Fermat's little theorem, we can write $$n^4\equiv 1(\mod 5)\quad\text{and} \quad m^4\equiv 1(\mod 5).$$
So, we get $n^4-m^4\equiv 0(\mod 5)$.
Since $5$ prime so, either $5|(m^2+n^2)$ or $5|(m^2-n^2).$
Now if $5|(m^2+n^2)$ then from $(1)$ we get $$1-n^2+3n^2+3m^2\equiv 0(\mod 5)$$ So, we are done.
Now, if $5|(m^2-n^2)$ then from $(1)$ we get $$1+5n^2-3n^2+3m^2\equiv 0(\mod 5).$$ So, we shall arrive at a contradiction that $1\equiv 0(\mod 5).$
In this way, I have tried to solve this problem. I will be highly obliged if you kindly check this or correct me.
Thanks in advance.
| You can make it a bit easier. The cool thing working $\mod 5$ is that squares are always $\equiv 0$ or $\equiv \pm 1$ (you see this by just entering all possibilities). If you know that
$$
1+2n^2+3m^2 \equiv 0 \mod 5
$$
test out the possibilities of $n,m \mod 5$ to see what works:
\begin{align}
n^2\equiv 0, m^2\equiv 0 \Rightarrow 1+2n^2+3m^2\equiv 1 \not\equiv 0 \\
n^2\equiv 1, m^2\equiv 0 \Rightarrow 1+2n^2+3m^2\equiv 3 \not\equiv 0 \\
n^2\equiv -1, m^2\equiv 0 \Rightarrow 1+2n^2+3m^2\equiv -1 \not\equiv 0 \\ \\
n^2\equiv 0, m^2\equiv 1 \Rightarrow 1+2n^2+3m^2\equiv 4 \not\equiv 0 \\
n^2\equiv 1, m^2\equiv 1 \Rightarrow 1+2n^2+3m^2\equiv 6 \not\equiv 0 \\
n^2\equiv -1, m^2\equiv 1 \Rightarrow 1+2n^2+3m^2\equiv 2 \not\equiv 0 \\ \\
n^2\equiv 0, m^2\equiv -1 \Rightarrow 1+2n^2+3m^2\equiv -2 \not\equiv 0 \\
n^2\equiv 1, m^2\equiv -1 \Rightarrow 1+2n^2+3m^2\equiv 0 \\
n^2\equiv -1, m^2\equiv -1 \Rightarrow 1+2n^2+3m^2\equiv -4\not\equiv 0 \\
\end{align}
By the table above you (you could also get this by systematic thinking without trying out all possibilities), you see that $n^2\equiv 1$ which means that $n^2-1\equiv 0 \mod 5$, so $n^2-1$ is divisible by $5$ if $1+2n^2+3m^2$ is (and in that case you also know that $m^2+1$ is divisible by $5$ as well)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding the discriminant of a quadratic equation from the given information on the roots of a quadratic equation I recently came accross an old question that I solved during my school days. Which is
If $\alpha, \beta$ are two real roots of a quadratic equation $ ax^2+bx+c=0 $ and $\alpha+\beta, \alpha^2+\beta^2, \alpha^3+\beta^3$ are in GP,
then which of the following is correct?
a) $\Delta\neq0$
b) $b\Delta=0$
c) $c\Delta=0$
d) $\Delta=0$
After seeing the question, I immediately realised that $\alpha+\beta=\frac{-b}{a}$ and $\alpha\beta=\frac{c}{a}$.
And since $\alpha+\beta, \alpha^2+\beta^2, \alpha^3+\beta^3$ are in GP, I got the following equation, I wrote them as $(\alpha^2+\beta^2)^2=(\alpha+\beta)(\alpha^3+\beta^3)$ -> call eqn $i$.
I rewrote the above equation as
$$[(\alpha+\beta)^2-2\alpha\beta]^2=(\alpha+\beta)[(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)]$$
I combined the above two and simplified further which resulted in $ac(b^2-4ac)=0$. And since $a$ can not be zero and $\Delta=b^2-4ac$, I concluded that $c\Delta=0$ and option c is correct.
However I later realised that eqn $i$ can expanded as follows,
$$\alpha^4+\beta^4+2\alpha^2\beta^2=\alpha^4+\beta^4+\alpha\beta^3+\beta\alpha^3$$.
Which will ultimately result in saying that $(\alpha-\beta)^2=0$ and therefore $\alpha=\beta$. If the roots are equal, the discriminant ($\Delta$) has to be zero which means option d is more correct. But most online websites only marked option c as the correct answer.
So which one is the correct answer really? Are they both correct? Or Am I missing something here?
| There is another approach that might be taken that largely avoids the issue of factorization. While it is tempting upon seeing the expressions $ \ \alpha + \beta \ , \ \alpha^2 + \beta^2 \ , \ \alpha^3 + \beta^3 \ $ to consider an argument based on Newton's identities, that leads to a great deal of writing (and a bit of interpretative complication). Instead, we might deal with this as a "logic puzzle" by working with the coefficients of the quadratic equation.
The only requirement on the coefficients is that $ \ a \ \neq \ 0 \ \ . $ We can start with the case of $ \ c = 0 \ \ $ only, keeping in mind that $ \ \alpha + \beta \ = \ -\frac{b}{a} \ $ and $ \ \alpha · \beta \ = \ \frac{c}{a} \ \ . $ This tells us that one or both of $ \ \alpha \ $ and $ \ \beta \ $ must equal zero, but the sum of the zeroes requires that one of these be equal to $ \ -\frac{b}{a} \ \ . $ (We also see this by observing that the quadratic equation in this case is $ \ ax^2 + bx = 0 \ \ . $ ) The expressions under discussion then have one term equal to zero, making them fall into the geometric progression $ \ -\frac{b}{a} \ , \ (-\frac{b}{a})^2 \ , \ (-\frac{b}{a})^3 \ \ . $
If we have $ \ b = 0 \ \ $ , then $ \ \alpha + \beta \ = \ 0 \ \Rightarrow \ \alpha \ = \ -\beta \ \ , $ in which case $ \ \alpha^3 + \beta^3 \ = \ 0 \ \ , $ but $ \ \alpha^2 + \beta^2 \ \neq \ 0 \ \ $ generally. The only exception is to have also $ \ c = 0 \ \ , $ which leads to $ \ \alpha \ = \ -\beta \ = \ 0 \ $ and the trivial geometric progression $ \ 0 \ , \ 0 \ , \ 0 \ \ . $ (The quadratic equation has "collapsed" to $ \ ax^2 \ = \ 0 \ \ . ) \ $ So $ \ b = c = 0 \ $ can be considered a subsidiary case of $ \ c = 0 \ \ . $
If we have for the discriminant $ \ \Delta = 0 \ \ , $ then the quadratic equation has the "double root" $ \ \alpha \ = \ \beta \ = \ -\frac{b}{2a} \ \ . $ We again obtain a geometric progression
$$ \alpha + \beta \ = \ 2 · \alpha \ = \ -\frac{b}{a} \ \ \ , \ \ \ \alpha^2 + \beta^2 \ = \ 2 · \alpha^2 \ = \ 2· \left(-\frac{b}{2a} \right)^2 \ = \ \frac{b^2}{2a^2} \ \ \ , $$
$$ \alpha^3 + \beta^3 \ = \ 2 · \alpha^3 \ = \ 2· \left(-\frac{b}{2a} \right)^3 \ = \ -\frac{b^3}{4a^3} \ \ . $$
It remains to determine whether $ \ \Delta \neq 0 \ $ (two distinct, non-zero roots) can produce a geometric progression. To save a bit of writing, we will identify the roots by their "components" $ \ \alpha \ = \ B + D \ \ , \ \ \beta \ = \ B - D \ \ . $ The expressions are then
$$ \alpha + \beta \ = \ 2 · B \ \ \ , \ \ \ \alpha^2 + \beta^2 \ \ = \ \ (B + D)^2 \ + \ (B - D)^2 \ \ = \ \ 2· (B^2 + D^2) \ \ \ , $$
$$ \alpha^3 + \beta^3 \ \ = \ \ (B + D)^3 \ + \ (B - D)^3 \ \ = \ \ 2 B^3 \ + \ 6 B D^2 \ \ . $$
In order for these to be in geometric progression, we require
$$ r \ \ = \ \ \frac{2· (B^2 + D^2)}{2 · B} \ \ = \ \ \frac{2 B^3 \ + \ 6 B D^2}{2· (B^2 + D^2)} $$
$$ \Rightarrow \ \ (B^2 + D^2)^2 \ \ = \ \ B · (B^3 \ + \ 3 B D^2) \ \ \Rightarrow \ \ B^4 \ + \ 2·B^2D^2 \ + \ D^4 \ \ = \ \ B^4 \ + \ 3·B^2D^2$$
$$ \Rightarrow \ \ B^2D^2 \ - \ D^4 \ \ = \ \ D^2 \ · \ (B^2 \ - \ D^2) \ \ = \ \ 0 \ \ . $$
(This is along the lines of the equation you developed.)
This produces no new zeroes: we have already rejected $ \ D \ = \ \frac{\sqrt{\Delta}}{2a} \ = \ 0 \ \ $ in our assumption, and $ \ B \ = \ \pm D \ $ gives us the zeroes $ \ 0 \ $ and $ \ 2·B \ = \ -\frac{b}{a} \ \ $ (and we will have the associated geometric progression). This does appear permit the condition $ \ \Delta \ = \ b^2 \ \neq \ 0 \ \ \ , $ but that is equivalent to $ \ b^2 \ = \ b^2 - 4ac \ \Rightarrow \ 4ac \ = \ 0 \ \Rightarrow \ c = 0 \ \ , $ since $ \ a \neq 0 \ \ . $
The proposition can thus be stated as:
If $ \ c = 0 \ \ $ (or $ \ b = 0 \ \ \mathbf{and} \ c = 0 \ ) \ $ or $ \ \Delta = 0 \ \ $ (or $ \ \Delta = b^2 \ ) \ , $ then $ \ \alpha + \beta \ , \ \alpha^2 + \beta^2 \ , \ \alpha^3 + \beta^3 \ $ are in geometric progression for the real roots $ \ \alpha \ $ and $ \ \beta \ $ of the quadratic equation $ \ ax^2 + bx + c = 0 \ \ . $
This makes choices $ \ \mathbf{(a)} \ \ \text{and} \ \ \mathbf{(b)} \ $ provisionally correct, since they can be true under specific conditions . These are really "gotcha" choices, and shouldn't properly be accepted (or even offered as choices) on an exam. Choice $ \ \mathbf{(d)} \ $ is technically correct, but it clearly doesn't "give the whole story". Choice $ \ \mathbf{(c)} \ $ is the only one that covers the requisite conditions fully.
My feeling is that this is a poorly-constructed set of choices, and perhaps this shouldn't have been posed as a multiple-choice question. (I am picturing the minor riot that erupted when students contested the instructor's answer.) I agree with Yathiraj Sharma in that the problem could be "saved" if it instead asks which choice is always correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$\sqrt{x+1}, \sqrt{x+2}, \sqrt{x+3}$ cannot be all three rational numbers. $x$ is a rational number. Then,
$\sqrt{x+1}, \sqrt{x+2}, \sqrt{x+3}$ cannot be all three rational numbers. Any idea? Thanks!
I tried to solve this problem and I arrived to prove that in natural set (positive integers) we cannot have:
$m^2 + n^2 = k^2$ and
$m^2 + 2n^2 = l^2$.
Also is rather easy to see that $m$ is odd, $n$ is even, also $n$ is multiple of $3$.
But I'm not sure that is the way.
| Note that your question is equivalent to showing that for a rational $x$, all of $\sqrt{x-1}$, $\sqrt{x}$ and $\sqrt{x+1}$ cannot be rational. If possible, on the contrary, let that be true.
Let $x=\frac{a^2}{b^2}$ with $(a,b)=1$ so that
\begin{align}
\sqrt{x-1}&=\frac{\sqrt{a^2-b^2}}{b}\\
\sqrt{x}&=\frac{a}{b}\\
\sqrt{x+1}&=\frac{\sqrt{a^2+b^2}}{b}
\end{align}
Since, $\sqrt{x+1}$ is rational, then $\sqrt{a^2+b^2}$ must be an integer. So,
$$a^2+b^2=c^2$$
which implies
\begin{align*}
&a=p^2-q^2\\
&b=2pq\\
&c=p^2+q^2
\end{align*}
for some integers $p,q,r$.
Similarly, $\sqrt{a^2-b^2}$ should also be an integer. So,
$$(p^2-q^2)-4p^2q^2=x^2$$
for some integer $x$, which is not possible because of this.
This completes the proof.
Is that okay?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4184424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Determine which of the following are metrics on $X$ Let $d$ and $d'$ be metrics on a non empty set $X$. Then which of the following are metrics on $X$.
*
*$\rho_1 (x,y)=d(x,y)d'(x,y)$ for all $x,y \in X$
*$\rho_2 (x,y)=max\{5,d(x,y)\}$ for all $x,y \in X$
*$\rho_3 (x,y)$=max$\{d(x,y),d'(x,y)\}$ for all $x,y \in X$
*$\rho_4 (x,y)$=min$\{d(x,y),d'(x,y)\}$ for all $x,y \in X$
My Attempt:
If $d$ satisfy the following conditions then it becomes a metric:
*
*$d(x,y) \geq 0$ for all $x$, $y \in X$.
*$d(x,y) = 0$ if and only if $x=y$.
*$d(x,y) = d(y,x)$ for all $x$, $y\in X$.
*$d(x,z) \leq d(x,y) + d(y,z)$ for all $x$, $y$, $z \in X$ (the triangle inequality)
I have seen that $\rho (x,y)=min\{5,d(x,y)\}$ for all $x,y \in X$ form a metric.
Option 3 is option that is true and options 1,2,4 are false.
But in my calculation (not written here), all the four options are correct. First three properties are holds trivially. I think there is a mistake made by me in last property I.e. triangle inequality fails for options 1,2,4. Please provide hints or solutions. Thanks
| Option 1 is false. For a counterexample, take $X = \mathbb R$ and let $d = d'$ be the usual metric $d(x, y) = |x - y|$. Then observe that:
$$
\rho(4, 3) + \rho(3, 2)
= |4 - 3|^2 + |3 - 2|^2
= 2
< 4
= |4 - 2|^2
= \rho(4, 2)
$$
Option 2 is false, since $\rho(7, 7) = 5$, instead of $0$.
Option 4 is false. For a counterexample, take $X = \mathbb R^2$ and let:
\begin{align*}
d(x, y) &= \sqrt{4(x_1 - y_1)^2 + (x_2 - y_2)^2} \\
d'(x, y) &= \sqrt{(x_1 - y_1)^2 + 4(x_2 - y_2)^2}
\end{align*}
Now consider $x = (1, 0)$ and $y = (0, 0)$ and $z = (0, 1)$. Observe that:
\begin{align*}
\rho(x, y) + \rho(y, z)
&= \min\left\{\sqrt{4 \cdot 1^2 + 0^2}, \sqrt{1^2 + 4 \cdot 0^2}\right\} + \min\left\{\sqrt{4 \cdot 0^2 + 1^2}, \sqrt{0^2 + 4 \cdot 1^2}\right\} \\
&= 2\\
&< \sqrt 5 \\
&= \min\left\{\sqrt{4 \cdot 1^2 + 1^2}, \sqrt{1^2 + 4 \cdot 1^2}\right\} \\
&= \rho(x, z)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4186413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The product of two roots of $x^4-18x^3+kx^2+200x-1984$ is $-32$; what is $k$?
In the quartic equation, $x^4-18x^3+kx^2+200x-1984$, The product of two roots is $-32$. What is the value of $k$?
This is my effort: Without losing generality, let's say $ab=-32$. Then since $abcd=-1984$, $cd=62$. $bcd+acd+abd+abc=ab(c+d)+cd(a+b)$, so $-32(c+d)+62(a+b)=-200$. Since $a+b+c+d=18$, let's say $a+b=X$, and $c+d=18-X$. $62X-32(18-X)=-200$, $94X+576=-200$, $94X=-776$, $a+b=\frac{776}{94}$
| Good attempt!
$$94X\color{red}-576=-200$$
$$X=\frac{376}{94}=4$$
That is now, we know $a+b=4$ and $c+d=18-4=14$.
Also $ab=-32$ and $cd=62$.
While we can solve for the roots, we can observe that we can make use of some tricks:
\begin{align}k&=ab+ac+ad+bc+bd+cd \\&=(ab+cd) + (ac+ad+bc+bd)
\\
&=(ab+cd) + (a+b)(c+d)\\&=(-32+62)+4(14)\\
&=30+56\\&=86 \end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that: $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}+5\ge (a+b)(b+c)(c+a)$ Let $a,b,c>0$ satisfy $a+b+c=3$
Prove that: $$\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}+5\ge (a+b)(b+c)(c+a)$$I have my solution, we need to prove:
$$3\sqrt[3]{a}+3\sqrt[3]{b}+3\sqrt[3]{c}+15\ge 3(a+b)(b+c)(c+a)$$
We know that: $$3(a+b)(b+c)(c+a)=(a+b+c)^3-a^3-b^3-c^3$$
So the problem is:
$$a^3+3\sqrt[3]{a}+b^3+3\sqrt[3]{b}+c^3+3\sqrt[3]{c}+15\ge 27$$
By AM-GM, we have: $$a^3+3\sqrt[3]{a}+b^3+3\sqrt[3]{b}+c^3+3\sqrt[3]{c}\ge4a+4b+4c=12$$ and we done.
Is there any other way? please help me
| Another way.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $v^2\leq u^2=1$ and by Schur $w^3\geq4uv^2-3u^3=4v^2-3$.
Id est, by AM-GM and C-S we obtain:
$$5-\prod_{cyc}(a+b)+\sum_{cyc}\sqrt[3]a=5-(9uv^2-w^3)+\sum_{cyc}\frac{3a}{3\sqrt[3]{a^2}}\geq$$
$$\geq5-9v^2+4v^2-3+3\sum_{cyc}\frac{a}{2a+1}=2-5v^2+3\sum_{cyc}\frac{a^2}{2a^2+a}\geq$$
$$\geq2-5v^2+\frac{3(a+b+c)^2}{\sum\limits_{cyc}(2a^2+a)}=2-5v^2+\frac{27}{2(9u^2-6v^2)+3}=$$
$$=2-5v^2+\frac{9}{7-4v^2}=\frac{(1-v^2)(23-20v^2)}{7-4v^2}\geq0.$$
The following stronger inequality is also true.
Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3$. Prove that:
$$\sqrt[3]a+\sqrt[3]b+\sqrt[3]c+2\geq\frac{5}{8}(a+b)(a+c)(b+c).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4188877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Eliminating $x,y,z$ from given set of equalities
Given that: $x-2y+z=a$, $x^2-2y^2+z^2=b$, $x^3-2y^3+z^3=c$ and $\frac{1}{x}-\frac{2}{y}+\frac{1}{z}=0$. Eliminate $x,y,z$ from given set of equations. [Hint: Use $x^2+z^2=(x+z)^2-2xz$ and $x^3+z^3=(x+z)(x^2+z^2)-xz(x+z)$]
Now using hints as given, I can solve $y$ in terms of $a,b,c$, but now sure how to do it for $x,z$. So for example, for $y$:
$$\begin{aligned}b&=(x^2+z^2)-2y^2 \\ &=(x+z)^2-2xz-2y^2\\ &=(a+2y)^2-y(a+2y)-2y^2\end{aligned}$$
If we plug in the solution of $y$ from this into the third equation, one involving third powers of $x,y,z$, then we would be left with an expression only in $a,b,c$, but doing so is quite laborious. Is there any alternate method?
How to proceed? Any hints are appreciated. Thanks.
| We leave it to the OP to show that if $a = 0$ then $b = 0$ and $c =0$ and $x=y=z$.
Else
$\tag 1 \Large{y = \frac{b-a^2}{3a}}$
and by the 4th equation $b \ne a^2$.
and since
$x^3+z^3=(x+z)(x^2+z^2)-xz(x+z)= (x+z)\big((x+z)^2-2xz\big)-xz(x+z)$
and
$xz = \frac{1}{2} (x+z)y$
we have
$x^3+z^3= (x+z)\big((x+z)^2-(x+z)y\big)-\frac{1}{2} (x+z)y(x+z)$
and therefore
$c + 2y^3 = (x+z)\big((x+z)^2-(x+z)y\big)-\frac{1}{2} (x+z)y(x+z)$
And so
$\tag 2 c + 2y^3 = (a+2y)^3 - \frac{3}{2} (a+2y)^2 y$
Work check
Let $a = 1$ and $b = 4$ so that $y = 1$.
Solving $\text{(2)}$ we get $c = 11.5$
Returning to the original system we find
$x = \frac{1}{2}(3 + \sqrt3) \land z = \frac{1}{2} (3 - \sqrt3)$
or
$x = \frac{1}{2}(3 - \sqrt3) \land z = \frac{1}{2} (3 + \sqrt3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to make use of angle sum and difference identities to find the value of sine and cosine?
Calculate: $\cos\left({5\pi\over12}\right)$ and $\cos\left({\pi\over12}\right)$
What is the easiest way to find $\cos\left({5\pi\over12}\right)$ and $\cos\left({\pi\over12}\right)$ (without a calculator)? If I know that $\frac{5\pi }{12}=\frac{\pi }{4}+\frac{\pi }{6}$ and $\frac{\pi }{12}=\frac{\pi }{3}-\frac{\pi }{4}$, then I can apply angle sum and difference identities. But how do I know $\frac{5\pi }{12}= \frac{\pi }{4}+\frac{\pi }{6}$ and $\frac{\pi }{12}= \frac{\pi }{3}-\frac{\pi }{4}$ in the first place. I know $ \frac{\pi }{4}+\frac{\pi }{6} = \frac{5\pi }{12}$ and $ \frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}$ but I can't go the other way round.
I gave $\frac{5\pi}{12}$ and $\frac{\pi}{12}$ as an example, I want the general solution for any value in pi rational form $\frac{\pi p}{q}$.
| 1 equation and 2 unknows is, generally, an overdetermined system.
In this case, pick any solution for $\dfrac{5\pi}{12} = \dfrac{p_1\pi}{q_1} + \dfrac{p_2\pi}{q_2}$, and sum $\pm \dfrac{p_3\pi}{q_3}$ in a smart way
$$\dfrac{5\pi}{12} = \left( \dfrac{p_1\pi}{q_1} + \dfrac{p_3\pi}{q_3} \right) + \left( \dfrac{p_2\pi}{q_2} - \dfrac{p_3\pi}{q_3} \right)$$
$$\dfrac{5\pi}{12} = \dfrac{\pi}{6} + \dfrac{\pi}{4} = \dfrac{\pi}{3} + \dfrac{\pi}{12} = \dfrac{3\pi}{10} + \dfrac{7\pi}{60} = \dots $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find the value of $\lim_{ n \to \infty} \left(\frac{1^2+1}{1-n^3}+\frac{2^2+2}{2-n^3}+\frac{3^2+3}{3-n^3}+...+\frac{n^2+n}{n-n^3}\right)$ The following question is taken from the practice set of JEE Main exam.
Find the value of $$\lim_{ n \to \infty} \left(\frac{1^2+1}{1-n^3}+\frac{2^2+2}{2-n^3}+\frac{3^2+3}{3-n^3}+...+\frac{n^2+n}{n-n^3}\right)$$
I wrote it as$$\lim_{ n \to \infty} \sum_{r=1}^n\frac{r^2+r}{r-n^3}\\=\lim_{ n \to \infty} \sum_{r=1}^n\frac{r(r+1)}{r-n^3}\\=\lim_{ n \to \infty} \sum_{r=1}^n\frac{r(\dfrac rn+\dfrac1n)}{\dfrac rn-n^2}\\=\lim_{ n \to \infty} \sum_{r=1}^n\frac{\dfrac rn(\dfrac rn+\dfrac1n)n}{\dfrac rn-n^2}$$
To convert it into into integration, $\dfrac rn$ can be written as $x$. $\dfrac1n$ is written as $dx$, but instead we have $n$. How to tackle that? Or any other approach for the question? Just a hint would suffice. Thanks.
| $R=\lim_{ n \to \infty} \left(\frac{1^2+1}{-n^3}+\frac{2^2+2}{-n^3}+\frac{3^2+3}{-n^3}+...+\frac{n^2+n}{-n^3}\right)=\lim_{ n \to \infty} \left(\frac{n(n+1)(2n+1)+3n(n+1)}{-6n^3}\right)=\frac{-1}{3}$
$P=\lim_{ n \to \infty} \left(\frac{1^2+1}{n-n^3}+\frac{2^2+2}{n-n^3}+\frac{3^2+3}{n-n^3}+...+\frac{n^2+n}{n-n^3}\right)=\lim_{ n \to \infty} \left(\frac{n(n+1)(2n+1)+3n(n+1)}{6(n-n^3)}\right)=\frac{-1}{3}$
$$R\ge\lim_{ n \to \infty} \left(\frac{1^2+1}{1-n^3}+\frac{2^2+2}{2-n^3}+\frac{3^2+3}{3-n^3}+...+\frac{n^2+n}{n-n^3}\right)\ge P$$
So $~\lim_{ n \to \infty} \left(\frac{1^2+1}{1-n^3}+\frac{2^2+2}{2-n^3}+\frac{3^2+3}{3-n^3}+...+\frac{n^2+n}{n-n^3}\right)=\frac{-1}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4193228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Probability two blocks have exactly 2 out of 4 properties the same?
I have 120 blocks. Each block is one of 2 different materials, 3 different colors, 4 different sizes, and 5 different shapes. No two blocks have exactly the same of all four properties. I take two blocks at random. What is the probability the two blocks have exactly two of these four properties the same?
I ended up getting:
$${{\binom{2}{1}\binom{3}{1}\binom{4}{2}\binom{5}{2} + \binom{2}{1}\binom{3}{2}\binom{4}{1}\binom{5}{2} + \binom{2}{1}\binom{3}{2}\binom{4}{2}\binom{5}{1} + \binom{2}{2}\binom{3}{1}\binom{4}{1}\binom{5}{2} + \binom{2}{2}\binom{3}{1}\binom{4}{2}\binom{5}{1} + \binom{2}{2}\binom{3}{2}\binom{4}{1}\binom{5}{1}}\over{\binom{120}{2}}} = {{41}\over{238}}$$Is this correct?
EDIT: The answer given here is ${{35}\over{119}} = {5\over{17}}$:
https://web2.0calc.com/questions/probability-question-blocks
Who is correct?
| Yes.
Since no two blocks have all four properties identical, and $2\cdot 3\cdot 4\cdot 5=120$, there are exactly enough blocks for every combination of properties; thus each arrangement has identical probability. Hence you can use this method.
The denominator counts the ways to select 2 among these 120 combinations, of course. This is the size of the sample space.
Your numerator counts the ways to select one of two properties and two of the other for each of the six such arrangements of properties. Thus properly counting the favoured outcomes.
So... everything checks out okay.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4196967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If $f(x) = x^2 - 2x + 5$ what is $f^{-1}(x)$?
If $f(x) = x^2 - 2x + 5$ what is $f^{-1}(x)$?
with the condition : $x > 1$
I solved this problem in this way:
$f(x) = x^2 - 2x + 5 -1 +1 \longrightarrow (x-2)^2 + 1 = f(x) $
$f^{-1}(x) = \sqrt{x-1} + 2$
But I saw someone else solved it in this way:
$f(x) = x^2 - 2x + 1 + 4 \longrightarrow (x+1)^2 + 4 = f(x) $
$f^{-1}(x) = \sqrt{x-4} + 1$
Which one is correct? If the second one is correct why mine is wrong?
| After the edit, in the given interval $f(x)$ is bijective and hence invertible. But your method is incorrect because:
$$(x-2)^2+1=x^2-4x+5\neq f(x)$$
Your friend is right, but I believe in the second-last step you meant $f(x)=(x-1)^2+4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4198095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove by limit definition $\lim _{x\to 3}\left(\frac{x^2-9}{x^3-3x^2-2x+6}\right)=\frac{6}{7}$. Prove, using the limit definition, that
$\lim _{x\to 3}\left(\frac{x^2-9}{x^3-3x^2-2x+6}\right)=\frac{6}{7}$.
I tried to do it this way, but I can’t move forward. Can anyone help me how to continue?
By definition we have
$\forall \epsilon >0,\exists \delta >0\::\:0<\left|x-a\right|<\delta \:\Rightarrow \left|f\left(x\right)-L\right|<\epsilon \Leftrightarrow a-\delta<x<a+\delta \Rightarrow L-\epsilon<f(x)<L+\epsilon.$
Proof:
\begin{eqnarray*}
3-\delta <x<3+\delta \:\Rightarrow \:\frac{6}{7}-\epsilon <\:f\left(x\right)<\frac{6}{7}+\epsilon &\:\Leftrightarrow& \:\frac{6}{7}-\epsilon <\frac{\left(x-3\right)\left(x+3\right)}{x^2\left(x-3\right)+2\left(x-3\right)}<\frac{6}{7}+\epsilon\\ &\:\Leftrightarrow& \:\frac{6}{7}-\epsilon <\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x^2-2\right)}<\frac{6}{7}+\epsilon\\ &\:\Leftrightarrow& \:\frac{6}{7}-\epsilon <\frac{\left(x+3\right)}{\left(x^2-2\right)}<\frac{6}{7}+\epsilon \\
&\:\Leftrightarrow& \:\:?
\end{eqnarray*}
| Note that\begin{align}\frac{x^2-9}{x^3-3x^2-2x+6}-\frac67&=\frac{-6 x^2+7 x+33}{7 \left(x^2-2\right)}\\&=\frac{-6x-11}{7(x^2-2)}(x-3).\end{align}So, take $\delta_1>0$ such that $|x-3|<\delta_1\implies|x-3|<1\iff2<x<4$. Then$$7(x^2-2)>7(2^2-2)=14\quad\text{and}\quad|-6x-11|=|6x+11|<35.$$ Therefore$$|x-3|<\delta_1\implies\left|\frac{x^2-9}{x^3-3x^2-2x+6}-\frac67\right|<\frac{35}{14}|x-3|.$$Now, take $\delta>0$ such that $\delta\leqslant\delta_1$ and also that $\delta<\frac{14}{35}\varepsilon$. Then$$|x-3|<\delta\implies\left|\frac{x^2-9}{x^3-3x^2-2x+6}-\frac67\right|<\frac{35}{14}\times\frac{14}{35}\varepsilon=\varepsilon.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $k^7/7 + k^5/5 + 2k^3/3 - k/105$ is an integer I tried to prove this using induction.
Let $k=1$; then the equation gives
$$1/7 + 1/5 +2/3 – 1/105 = 105/105 = 1,$$
which is an integer. So it is true for $k=1$. Now let it be true for $n>k$. This gives
$$105|(15n^7 + 21n^5 + 70n^3 – n).$$
For $(n+1)$ we have
$$105|(15(n+1)^7 + 21(n+1)^5 + 70(n+1)^3 – (n+1)),$$
which is the same as
$$105|(15n^7 + 105n^6 + 336n^5 + 630n^4 + 875n^3 + 945n^2 + 629n + 175)$$
since $105|(15n^7 + 21n^5 + 70n^3 – n)$.
If $105|(105n^6 + 315n^5 + 630n^4 + 805n^3 + 945n^2 + 630n + 175),$
which is the same as if $$105|((105n^6 + 315n^5 + 630n^4 +945n^2 + 630n) + 805n^3 + 175),$$
which is the same as if $$105|((\text{multiple of } 105) + 805n^3 + 175).$$
I’m stuck at this part, because neither $805$ nor $175$ is a multiple of $105$, so how to prove that they are multiples of $105$?
| The generating function is $$\sum_{k\ge 0} f(k)x^k=\frac{320}{(x-1)^3}+\frac{3672}{(x-1)^5}+\frac{1540}{(x-1)^4}+\frac{4584}{(x-1)^6}+\frac{1}{(x-1)}+\frac{2880}{(x-1)^7}+\frac{720}{(x-1)^8}+\frac{29}{(x-1)^2}$$ with $$f(k)=\frac12 k^7+\frac15 k^5+\frac23 k^3-\frac{1}{105}k$$. Since all the factors of this polynomial of $1/(x-1)$ are integers, all $f(k)$ are integers.
Another method of proof is that the polynomial has the recurrence $$f(k) = 8f(k-1)-28f(k-2)+56f(k-3)-70f(k-4)+56f(k-5)-28f(k-6)+8f(k-7)-f(k-8)$$ and to start an induction on $k$ by showing that the first 8 terms of $f(k)$ are integers. [Of course the recurrence is valid for any polynomial $f(k)$ of order 7 and the signed binomial coefficients are a well-known property of the recurrence.]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\lim\limits_{n\to\infty}\frac{1\cdot3+2\cdot4+\dots+n(n+2)}{n^3}.$
Evaluate: $$\lim\limits_{n\to\infty}\frac{1\cdot3+2\cdot4+\dots+n(n+2)}{n^3}.$$
I'm learning limits for the first time and this is an exercise problem from my book. Here is my solution to the problem:
Let $S=1\cdot3+2\cdot4+\dots+n(n+2)\\ =(1^2+2)+(2^2+4)+\dots+(n^2+2n)\\ =(1^2+2^2+\dots+n^2)+2(1+2+\dots+n)\\ =\frac{n(n+1)(2n+1)}{6}+2\cdot\frac{n(n+1)}2\\ =\frac13n^3+\frac32n^2+\frac76n$
Hence, $\lim\limits_{n\to\infty}\frac{1\cdot3+2\cdot4+\dots+n(n+2)}{n^3}\\ =\lim\limits_{n\to\infty}\frac13+\frac3{2n}+\frac7{6n^2}\\ =\frac13.$
I'm quite sure about the solution. But my book says the answer is $\frac16$. So, is the answer in the book wrong, or am I missing something? And can the problem be solved with L'Hôpital's rule? (I've just started learning the rule and I don't know how to solve this using this). Some other methods to solve the problem are also welcome.
| Another way using summation notation.
$\begin{array}\\
s(n)
&=\dfrac{\sum_{k=1}^n k(k+2)}{n^3}\\
&=\dfrac{\sum_{k=1}^n (k^2+2k)}{n^3}\\
&=\dfrac{\sum_{k=1}^n (k^2+2k+1-1)}{n^3}\\
&=\dfrac{\sum_{k=1}^n ((k+1)^2-1)}{n^3}\\
&=\dfrac{\sum_{k=1}^n (k+1)^2}{n^3}-\dfrac{\sum_{k=1}^n 1}{n^3}\\
&=\dfrac{\sum_{k=2}^{n+1} k^2}{n^3}-\dfrac{n}{n^3}\\
&=\dfrac{\sum_{k=1}^{n+1} k^2-1}{n^3}-\dfrac1{n^2}\\
&=\dfrac{\sum_{k=1}^{n+1} k^2}{n^3}-\dfrac{1}{n^3}-\dfrac1{n^2}\\
&=\dfrac{(n+1)(n+2)(2n+3)}{6n^3}-\dfrac{1}{n^3}-\dfrac1{n^2}\\
&\to \dfrac13\\
\end{array}
$
Probably too involved,
but I enjoy this kind
of playing around.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Computing $\sum_{k = 1}^{n} \dfrac{k}{2^k} $ in different ways I am trying to compute the following sum \begin{align} S = \sum_{k = 1}^{n} \dfrac{k}{2^k} \end{align}
I have computed this sum and found that
$$ \bbox[5px,border:2px solid red]
{
S = 2 - \dfrac{n+2}{2^{n}}
}
$$
For prooving this let's rewrite the sum in the following way:
\begin{align} S = \dfrac{1}{2} + \dfrac{1 + 1}{2^2} + \dfrac{1+1+1}{2^3} + \ldots + \dfrac{1+ \ldots+1}{2^n} = \bbox[5px,border:2px solid yellow]
{\dfrac{1}{2} + \dfrac{1}{2^2} + \ldots + \dfrac{1}{2^n}} + \bbox[5px,border:2px solid yellow]
{ \dfrac{1}{2^2} + \dfrac{1}{2^3} + \ldots+\dfrac{1}{2^n}} + \ldots + \bbox[5px,border:2px solid yellow]{
\sum_{j = k}^{n}\dfrac{1}{2^j}} + \ldots +\bbox[5px,border:2px solid yellow]{
\dfrac{1}{2^n}} = \sum_{j = 1}^{n}\left( \dfrac{1}{2^{k-1}} - \dfrac{1}{2^n} \right) = 2 - \dfrac{1}{2^{n-1}} - \dfrac{n}{2^n} =2 - \dfrac{n+2}{2^{n}} \end{align}
This computations are rather awfull and i made them just to find the answer, and i am looking for another interesting ways how to compute this sum (especcialy not elementary, using calculus, or number theory, or theory of functions of complex variable, or something else), any ways will be very appriciated!
| Here is a generating function approach. We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series.
We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^n}\color{blue}{\frac{k}{2^k}}
&=[z^n]\sum_{q=0}^{\infty}\left(\sum_{k=0}^q\frac{k}{2^k}\right)z^q\tag{1}\\
&=[z^n]\frac{1}{1-z}\,\sum_{q=0}^\infty \frac{q}{2^ q}z^q\tag{2}\\
&=[z^n]\frac{z}{1-z}\,\frac{d}{dz}\sum_{q=0}^\infty \left(\frac{z}{2}\right)^q\\
&=[z^{n-1}]\frac{1}{1-z}\,\frac{d}{dz}\left(\frac{1}{1-\frac{z}{2}}\right)\tag{3}\\
&=\frac{1}{2}[z^{n-1}]\frac{1}{1-z}\,\frac{1}{\left(1-\frac{z}{2}\right)^2}\\
&=[z^{n-1}]\left(\frac{2}{1-z}-\frac{1}{1-\frac{z}{2}}-\frac{1}{2\left(1-\frac{z}{2}\right)^2}\right)\tag{4}\\
&=[z^{n-1}]\left(2\sum_{j=0}^\infty z^j-\sum_{j=0}^\infty \left(\frac{z}{2}\right)^j\right.\\
&\qquad\qquad\qquad\left.-\frac{1}{2}\sum_{j=0}^\infty(j+1)\left(\frac{z}{2}\right)^j\right)\\
&=2-\frac{1}{2^{n-1}}-\frac{n}{2^n}\tag{5}\\
&\,\,\color{blue}{=2-\frac{n+2}{2^n}}
\end{align*}
in accordance with OPs result.
Comment:
*
*In (1) we write the sum as coefficient of $z^n$ of a generating function.
*In (2) we use that multiplication with $\frac{1}{1-z}$ transforms a coefficient $a_q$ to the sum of the first $q$ coefficients: $\sum_{k=0}^q a_k$.
*In (3) we apply the rule $[z^p]z^qA(z)=[z^{p-q}]A(z)$ and use the differential operator $\frac{d}{dz}$.
*In (4) we make a partial fraction expansion to ease the following series expansion.
*In (5) we select the coefficient of $z^{n-1}$.
Btw. I think OPs approach is smart. Using sigma notation it can be written as
\begin{align*}
\color{blue}{\sum_{k=1}^n}\color{blue}{\frac{k}{2^k}}
&=\sum_{k=1}^n\frac{1}{2^k}\sum_{j=1}^k1\\
&=\sum_{j=1}^n\sum_{k=j}^n\frac{1}{2^k}\\
&=\sum_{j=1}^n\left(\frac{1-\frac{1}{2^{n+1}}}{1-\frac{1}{2}}-\frac{1-\frac{1}{2^j}}{1-\frac{1}{2}}\right)\\
&=\sum_{j=1}^n\left(\frac{1}{2^{j-1}}-\frac{1}{2^n}\right)\\
&=\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}-\frac{n}{2^n}\\
&\,\,\color{blue}{=2-\frac{n+2}{2^n}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Using Rearrangement Inequality .
Let $a,b,c\in\mathbf R^+$, such that $a+b+c=3$. Prove that $$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a^2+b^2+c^2}{2}$$
$Hint$ : Use Rearrangement Inequality
My Work :-$\\$
Without Loss of Generality let's assume that $0\le a\le b\le c$ then we can infer that $c^2\ge b^2\ge a^2$ also $b+c\ge c+a \ge a+b$. Thus $\frac{1}{b+c}\le \frac{1}{c+a}\le \frac{1}{a+b}$. Hence by Rearrangement Inequality $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$ is the greatest permutation however I am not able to express $\frac{a^2+b^2+c^2}{2}$ as another permutation of the same, hence I require assistance.
Thank You
| Since triples $(a^2,b^2,c^2)$ and $\left(\frac{1}{b+c},\frac{1}{a+c},\frac{1}{a+b}\right)$ have the same ordering,
by Rearrangement and C-S we obtain: $$\sum_{cyc}\frac{a^2}{b+c}\geq\frac{1}{3}\sum_{cyc}a^2\sum_{cyc}\frac{1}{b+c}\geq\frac{1}{3}\sum_{cyc}a^2\cdot\frac{(1+1+1)^2}{\sum\limits_{cyc}(b+c)}=\frac{a^2+b^2+c^2}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
$(1+A)^n \ge (1+a_1)\cdot(1+a_2)\cdots(1+a_n) \ge (1+G)^n$ $A$ and $G$ are arithmetic mean and the geometric mean respectively of $n$ positive real numbers $a_1$,$a_2$,$\ldots$,$a_n$ . Prove that
*
*$(1+A)^n$$\ge$$(1+a_1)\cdot(1+a_2)\cdots(1+a_n)$$\ge$$(1+G)^n$
*if $k$$\gt$$0$, $(k+A)^n$$\ge$$(k+a_1)\cdot(k+a_2)\cdots(k+a_n)$$\ge$$(k+G)^n$
My try (Ques -1): To prove $(1+a_1)\cdot(1+a_2)\cdots(1+a_n)$$\ge$$(1+G)^n$
$(1+a_1)\cdot(1+a_2)\cdots(1+a_n)=1+\sum{a_1}+\sum{a_1}\cdot{a_2}+\sum{a_1}\cdot{a_2}\cdot{a_3}+\cdots+(a_1\cdot{a_2}\cdots{a_n})$
Consider $a_1,a_2,\ldots,a_n$ be $n$ positive real numbers and then apply AM$\ge$GM
$\frac{(a_1+a_2+\cdots+a_n)}{n}\ge(a_1\cdot{a_2}\cdots{a_n})^\frac{1}{n}$ imples $\sum{a_1}\ge n\cdot{G}$ because $G=(a_1\cdot{a_2}\cdots{a_n})^\frac{1}{n}$
again consider $(a_1\cdot{a_2}),(a_1\cdot{a_3}),\ldots({a_1}\cdot{a_n}),(a_2\cdot{a_3}),(a_2\cdot{a_4})\ldots,(a_2\cdot{a_n}),(a_3\cdot{a_4})\ldots(a_{n-1}\cdot{a_n})$ be $\frac{n(n-1)}{2!}$ positive real numbers.
Then Applying AM$\ge$GM , we get,
$\frac{\sum{a_1}\cdot{a_2}}{\frac{n(n-1)}{2!}}$ $\ge$ $\bigl(a_{1}^{n-1}\cdot{a_{2}^{n-1}}\cdots{a_{n}^{n-1}}\bigl)^\frac{2!}{n(n-1)}$ implies $\sum{a_1}\cdot{a_2}\ge \frac{n(n-1)}{2!}G^2$
Similary , $\sum{a_1}\cdot{a_2}\cdot{a_3}\ge \frac{n(n-1)(n-2)}{3!}G^3$ and so on.
Therefore, $(1+a_1)\cdot(1+a_2)\cdots(1+a_n)\ge 1+nG+\frac{n(n-1)}{2!}G^2+\frac{n(n-1)(n-2)}{3!}G^3+\cdots+G^n$
Therefore, $(1+a_1)\cdot(1+a_2)\cdots(1+a_n)\ge (1+G)^n$
To prove $(1+A)^n$$\ge$$(1+a_1)\cdot(1+a_2)\cdots(1+a_n)$
Consider $(1+a_1),(1+a_2),\ldots,(1+a_n)$ be $n$ positive real numbers and then applying AM$\ge$GM
$\frac{(1+a_1)+(1+a_2)+\cdots+(1+a_n)}{n} \ge [(1+a_1)\cdot(1+a_2)\cdot\cdots(1+a_n)]^\frac{1}{n}$
$\frac{n+a_1+a_2+\cdots{a_n}}{n}\ge [(1+a_1)\cdot(1+a_2)\cdot\cdots(1+a_n)]^\frac{1}{n}$
$(1+A)^n\ge(1+a_1)\cdot(1+a_2)\cdot\cdots(1+a_n)$
My try (Ques -2): To prove $(k+A)^n$$\ge$$(k+a_1)\cdot(k+a_2)\cdots(k+a_n)$
Consider $(k+a_1),(k+a_2)\ldots(k+a_n)$ be $n$ positive real numbers and applying AM$\ge$GM
$\frac{(k+a_1)+(k+a_2)+\cdots(k+a_n)}{n}\ge[(k+a_1)\cdot(k+a_2)\cdots(k+a_n)]^\frac{1}{n}$
$(k+A)^n\ge (k+a_1)\cdot(k+a_2)\cdots(k+a_n)$
To prove $(k+a_1)\cdot(k+a_2)\cdots(k+a_n)$$\ge$$(k+G)^n$
$(k+a_1)\cdot(k+a_2)\cdots(k+a_n)=k^n+k^{n-1}\sum{a_1}+k^{n-2}\sum{a_1\cdot{a_2}}+\cdots+(a_1\cdot{a_2}\cdots{a_n})$
now
$\sum{a_1}\ge n\cdot{G}$,
$\sum{a_1}\cdot{a_2}\ge \frac{n(n-1)}{2!}G^2$
$\sum{a_1}\cdot{a_2}\cdot{a_3}\ge \frac{n(n-1)(n-2)}{3!}G^3$ and so on.
Therefore
$(k+a_1)\cdot(k+a_2)\cdots(k+a_n)\ge k^n+nk^{n-1}G+\frac{n(n-1)}{2!}k^{n-2}G^2+\cdots+G^n$
Therefore, $(k+a_1)\cdot(k+a_2)\cdots(k+a_n)\ge (k+G)^n$
My Ques :
*
*Have I solved the questions correctly?
*How to prove $\sum{a_1}\cdot{a_2}\cdot{a_3}\ge \frac{n(n-1)(n-2)}{3!}G^3$
Thanks
| Adding on to @notallwrong's answer:
For the case $n = 3$, you want to know how many times say $a_1$ appears. We have to make triples. Choose $a_1$. The remaining two can be chosen in $\binom{n-1}{2}$ ways.
For the case $n = 4$, you want to know how many times say $a_1$ appears. We have to make quadruples. Choose $a_1$. The remaining three can be chosen in $\binom{n-1}{3}$ ways. and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
For which value of $t \in \mathbb R$ the equation has exactly one solution : $x^2 + \frac{1}{\sqrt{\cos t}}2x + \frac{1}{\sin t} = 2\sqrt{2}$ For which value of $t \in R $ the equation has exactly one solution : $x^2 + \frac{1}{\sqrt{\cos t}}2x + \frac{1}{\sin t} = 2\sqrt{2}$
Here $t \neq n\pi, t \neq (2n+1)\frac{\pi}{2}$
Therefore , for the given equation to have exactly one solution we should have :
$(\frac{2}{\sqrt{\cos t}})^2 -4.(\frac{1}{\sin t} - 2\sqrt{2}) = 0 $
$\Rightarrow \frac{4}{\cos t} - 4 (\frac{1}{\sin t} - 2\sqrt{2}) = 0 $
$\Rightarrow \sin t -\cos t +2 \sqrt{2}\sin t\cos t = 0 $
$\Rightarrow \sqrt{2}( \frac{1}{\sqrt{2}}\sin t - \frac{1}{\sqrt{2}}\cos t) = -2\sqrt{2} \sin t\cos t$
$\Rightarrow \sqrt{2}(\cos(\pi/4)\sin t -\sin(\pi/4)cos t = -\sqrt{2}\sin2t $ [Using $\sin x\cos y -\cos x\sin y = \sin(x-y)$]
$\Rightarrow \sqrt{2}\sin(\frac{\pi}{4}-t) =-\sqrt{2}\sin2t$
$\Rightarrow \sin(\frac{\pi}{4}-t) =-\sin2t $ [ Using $-\sin x = \sin(-x)$ and comparing R.H.S. with L.H.S. ]
$\Rightarrow \frac{\pi}{4}-t = -2t $
$\Rightarrow t = - \frac{\pi}{4}$
Is it correct answer, please suggest.. thanks
| $$x^2 + \frac{1}{\sqrt{\cos t}}2x + \frac{1}{\sin t} = 2\sqrt{2}\\
\implies x=\frac{-\frac{2}{\sqrt{\cos t}} \pm\sqrt{\frac{4}{\cos t}
-4(1)\big(\frac{1}{\sin t}-2\sqrt{2}\big)}}{2}\\
=\frac{-1}{\sqrt{\cos t}}\pm \sqrt{\sec {t}-\csc{t} +2\sqrt{2}}$$
Now, the problem is finding what values of $t$ yield
$\quad
\sec t - \csc t + 2\sqrt{2}=0
\quad$
Wolfram Alpha shows infinite $t$-values
here where the equation has only one solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to solve this complicated trigonometric equation for the solution of this triangle.
In $\triangle ABC$, if $AB=AC$ and internal bisector of angle $B$ meets $AC$ at $D$ such that $BD+AD=BC=4$, then what is $R$ (circumradius)?
My Approach:- I first drew the diagram and considered $\angle ABD=\angle DBC=\theta$ and as $AB=AC$, $\angle C=2\theta$. Therefore $\angle A=180-4\theta$. Also as $AE$ is the angular bisector and $AB=AC$, then $BE=EC=2.$ Now applying the sine theorem to $\triangle ADB$ and $ \triangle BDC$ gives $$\frac{BD}{\sin {(180-4\theta)}}=\frac{AD}{\sin \theta}$$
$$\frac{BC}{\sin(180-3\theta)}=\frac{BD}{\sin 2\theta}$$
Now we know that $BC=4$ and then solving both the equations by substituting in $BD+AD=4$, we get $$\sin 2\theta .\sin4\theta+\sin2\theta.\sin\theta=\sin3\theta.\sin4\theta$$
$$\sin4\theta+\sin\theta=\frac{\sin3\theta.\sin4\theta}{\sin2\theta}$$
Now I have no clue on how to proceed further from here. Though I tried solving the whole equation into one variable ($\sin\theta$), but it's getting very troublesome as power of $4$ occurs. Can anyone please help further or else if there is any alternative method to solving this problem more efficiently or quickly?
Thank You
| Indeed you are proceeding correctly. The equation can be solved as follows:
$$\sin 2\theta \sin \theta=\sin 4\theta(\sin 3\theta-\sin 2\theta) {\tag 1}$$
Now $\sin 3\theta-\sin 2\theta=2\cos \frac {5\theta}{2} \sin \frac {\theta}{2}$.
Also $\sin \theta=2\cos \frac {\theta}{2} \sin \frac {\theta}{2}$, and $\sin 4\theta=2\sin 2\theta \cos 2\theta$.
So, $(1)$ simplifies to:
$$\cos \frac {\theta}{2}=2\cos \frac {5\theta}{2}\cos 2\theta {\tag 2}$$
Since $2\cos \frac {5\theta}{2} \cos 2\theta=\cos \frac {9\theta}{2}+\cos \frac {\theta}{2}$, we have, from $(2)$:
$$\cos \frac {9\theta}{2}=0$$
This means that $\frac {9\theta}{2}=\frac {\pi}{2}$, hence $\theta=\frac {\pi}{9}$.
This means that all angles of triangle are known, and we know $BC=4$. Thus using sine law, $$2R=\frac {BC}{\sin A}$$ is easy to calculate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4216355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Question on partial answer of the generalized Catalan's conjecture (case $n =2$) Edited question.
Let $n,m\in\Bbb N$ be integers greater or equal to $2$. If $3\leq m<n$ then there is no $m,n$ such that $m^n -n^m =2$. If $n<m$ and $m$ is prime then there is no $m,n$ such that $m^n-n^m =2$. If $m = 2$ then there is no $n$ such that $m^n- n^m = 2$.
Original question.
Let $n\geq 2$ be an integer and $p$ be a prime number. Then there is no $p,n$ such that $p^n - n^p =2$.
False Proof : Consider the case when $p$ is odd prime. Then
by FLT, $n^{p-1}\equiv 1\bmod p$ so $n^p = p^n-2 \equiv n\bmod p$. $\color{#C00}{\text{Then $n^{p+1} = (p+1)^n -2\equiv n\bmod p+1$ so $n^p\equiv 1\bmod p+1$}}$. $p^n - 2 = (p+1)m+1$ for some $m$ so that $p^n =(p+1)m+3 = pm+m+3$. Hence, $p(p^{n-1}-m) = m+3 = pk_0$ for some integer $k_0$ so that $m = pk_0-3$. Then $p(p^{n-1}-pk_0+3) = pk_0$ so $p^{n-1}-pk_0= k_0-3$. Since, $p(p^{n-2}-k_0) = k_0-3$ and letting $k_0 = pk_1+3$, we get $p(p^{n-2}-pk_1-3) = pk_1$ i.e. $p^{n-2}-pk_1 = k_1+3$. Hence inductively, we conclude that $p(1-k_N) = k_N\pm 3$ for some large $N$. This gives the contradiction.
Rough idea given by @robjohn ♦: We want to show that $p^n-n^p=2$ is impossible, so if $n=kp-2$, consider $p^{kp-2}-(kp-2)^p$
Dividing both sides by $(kp)^p$, we have $\frac{p^{(k-1)p}}{k^pp^2}-\left(1-\frac2{kp}\right)^p$.
If $k=1$, this is $\frac1{p^2}-\left(1-\frac2{p}\right)^p$ which is less than $0$. If $k\ge2$, then $\left(\frac{p^{k-1}}{kp^{2/p}}\right)^p-\left(1-\frac2{kp}\right)^p$ is greater than $0$ and multiplied by $(kp)^p$ is much bigger than $2$.
| This is a particular case of $m^n = n^m+2$ with $m,n\geq 2$. First note if both numbers are even then $m^n$ and $n^m$ are multiples of $4$, so we have $m,n \geq 3$. Let $n = m+k$ with $k\geq 1$. ( You can see that if $m$ is the larger one it wont work because the left side will be smaller than the right).
Taking logs gives $n\log(m) = m\log(n) + \epsilon$ where $\epsilon$ is the error term $\log(n^m+2) - \log(n^m) = \int \limits_{n^m}^{n^m+2} \frac{1}{x} dx \leq \frac{2}{n^m}$.
$n \log(m) - m\log(n) = (m+k)\log(m) - m(\log(m+k))$.
Differentiating with respect to $k$ we get $\log(m) - \frac{m}{m+k}$ so the function is increasing in $k$ in $(1,\infty)$.
Setting $k=1$ yields $(m+1)\log(m) - m\log(m+1) = \log(m) - m(\log(m+1) - \log(m))$ where $\log(m+1) - \log(m) = \int\limits_{m}^{m+1} \frac{1}{x}dx \leq \frac{1}{m}$. It follows that $\epsilon$ is at least $\log(m) - 1 \geq \log(3) - 1$.
Hence we must have $\log(3) -1 \leq \frac{2}{n^m}$ or $n^m \leq \frac{2}{\log(3)-1}<21$.
$4^3$ is already $64$ so there is no candidate we must check.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4216663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integration and summation: prove $399< \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots+\frac{1}{\sqrt{40000}}<400$ is false. I need help proving that the following statement is false:
$$
399 < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}}
+ \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{40000}} < 400.
$$
I tried to bound the summation between two integrals
$$
\text{1)} \int_1^{40001} \frac{2}{\sqrt{x}}
\qquad \text{and} \qquad
\text{2)} \int_0^{40000} \frac{2}{\sqrt{x}}
$$
and have convinced myself that the statement is true, but this is not the correct answer as per my textbook.
| Your sum is equal to
$$1 + \sum_{n=2}^{40000} \frac{1}{\sqrt{n}} < 1 + \int_1^{40000} \frac{1}{\sqrt{x}} \; dx = 1+ \left.2\sqrt{x}\right|_{x=1}^{40000} = 1+(2\sqrt{40000} - 2) = 399.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $\csc^2(x)=\csc^2(\frac x 2)+\csc^2 (\frac x 4)$
Solve $\displaystyle\csc^2(x)=\csc^2\left(\frac x 2\right)+\csc^2\left(\frac x 4\right)$ , $\forall
x\in[0,\pi/2] $.
My attempt:
$$\frac{1}{\sin^2\theta}=\frac{1}{\sin^2\frac \theta 2}+\frac{1}{\sin^2\frac \theta 4}$$
then $$1=\frac{\sin^2\theta}{\sin^2\frac \theta 2}+\frac{\sin^2\theta}{\sin^2\frac \theta 4}$$
we have $\theta=2\times\frac \theta 2$ and $\frac \theta 2=2\times\frac \theta4$ then
$$1= 4\cos^2 \frac \theta 2+16\cos^2 \frac \theta 2 \cos^2 \frac \theta 4 $$
$$1= \left(4+16\cos^2 \frac \theta 4\right)\cos^2 \frac \theta 2.$$
I'm stuck here. Any help
| Starting from your last equation
$$1=\left(4+16 \cos ^2\left(\frac{\theta }{4}\right)\right) \cos ^2\left(\frac{\theta
}{2}\right)$$
$$\cos ^2\left(\frac{\theta }{4}\right)=x \implies \theta=4 \cos ^{-1}\left(\sqrt{x}\right)\implies 1=4 (1-2 x)^2 (4 x+1)$$ So, we need to solve the cubic equation
$$64 x^3-48 x^2+3=0$$ Since $\Delta=331776$, this equation has three real roots.
Using the trigonometric method, they are given by
$$x_k=\frac{1}{4}+\frac{1}{2} \sin \left(2k\frac{\pi }{3}+\frac{\pi }{18}\right)\qquad \text{for}\qquad k=0,1,2$$ The problem is that the trigonometric functions of angles which are multiples of $\frac \pi{18}$ are given by infinite nested radicals (see here). So, numerically
$$x_0=-0.219846\cdots\qquad x_1=0.336824\cdots\qquad x_2=0.633022\cdots$$ The first root is negative, then discarded. Back to $\theta$, we have
$$\theta_1=4 \cos ^{-1}\left(\sqrt{\frac{1}{4}+\frac{1}{2} \sin \left(\frac{\pi
}{18}\right)}\right)=3.80648\cdots$$
$$\theta_2=4 \cos ^{-1}\left(\sqrt{\frac{1}{4}+\frac{1}{2} \cos \left(\frac{2 \pi
}{9}\right)}\right)=2.60302\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is my logic valid in showing that any two consecutive terms of the Fibonacci sequence are coprime Assume terms $a_n$ and $a_{n+1}$ share some common factor $x$ so that $a_n = xm$ for some integer $m$ and $a_{n+1} = xk$ for some integer $k$.
Since $a_{n-1}$ = $a_{n+1} - a_n = xk - xm = x(k-m)$, $a_{n-1}$ is a multiple of $x$ as well. Then $a_{n-2}$ must also be a multiple of $x$ because $a_{n-2} = a_{n} - a_{n-1} = xm - (xk - xm) = x(2m-k)$. Since each lower or upper term can be found by subtracting or adding one multiple of $x$ to another, we can say that every term in the sequence must then be a multiple of $x$.
Therefore, if any two consecutive terms share a common factor, then all terms must share a common factor. However, if $a_1=1$ we can see the counterexample of $a_3 = 2, a_4=3$ where $2$ and $3$ are consecutive terms with no common factors. Since not all terms then share a common factor, no two consecutive terms can share a common factor.
| Alternative:
One has $\begin{pmatrix} 1 & 1 \\ 1 & 0\end{pmatrix}^n = \begin{pmatrix} f_{n+1} & f_{n} \\ f_{n} & f_{n-1} \end{pmatrix}$ and so taking the determinant $(-1)^n = f_{n-1}f_{n+1} - f_n^2$. This shows $1$ is a linear combination of $f_{n+1}$ and $f_n$ and they are therefore coprime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding $a+b+c+d$, where $ab+c+d=15$, $bc+d+a=24$, $cd+a+b=42$, $da+b+c=13$ Let $a,b,c,d \in \mathbb{R}$. Consider the following constraints:
\begin{cases} ab+c+d=15 \\ bc+d+a=24 \\ cd+a+b=42 \\da+b+c=13 \end{cases}
Calculate the value of $a+b+c+d$.
It is easy to use the Gröbner basis to get the value:
\begin{cases}
10849-4501 d+380d^2,-39409+2320c+3420d,-20+29b-9d,1801+2320 a-380 d\}
\end{cases}
so the value of $a+b+c+d$ is $\frac{169}{10}$.
What I am curious about is how to use high schools mathematics to get an answer without too much complicated mathematical calculations ?
| Playing around with the four LHSs, I tried to obtain polynomials with some symmetry. I first noted that
$$\left((ab+c+d)+(cd+a+b)\right)-\left((bc+d+a)+(da+b+c)\right)=(b-d)(a-c)$$
then after replacing the $+$ sign with $\cdot$, we get
$$(ab+c+d)\cdot(cd+a+b)-(bc+d+a)\cdot(da+b+c)=(b-d)(a-c)(a+b+d+c-1).$$
Putting all together,
$$\begin{align}
&\frac{(ab+c+d)(cd+a+b)-(bc+d+a)(da+b+c)}{(ab+c+d)+(cd+a+b)-(bc+d+a)-(da+b+c)}\\&\qquad=
\frac{(b-d)(a-c)(a+b+d+c-1)}{(b-d)(a-c)}=a+b+c+d-1.
\end{align}$$
Hence with the given values we find
$$a+b+c+d=1+\frac{15\cdot 42-24\cdot 13}{15+42-24-13}=1+\frac{318}{20}=\frac{169}{10}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Integrate via substitution: $x^2\sqrt{x^2+1}\;dx$ I need to integrate the following using substitution:
$$
\int x^2\sqrt{x^2+1}\;dx
$$
My textbook has a similar example:
$$
\int \sqrt{x^2+1}\;x^5\;dx
$$
They integrate by splitting $x^5$ into $x^4\cdot x$ and then substituting with $u=x^2+1$:
$$
\int \sqrt{x^2+1}\;x^4\cdot x\;dx\\
=\frac{1}{2}\int \sqrt{u}\;(u-1)^2\;du\\
=\frac{1}{2}\int u^{\frac{1}{2}}(u^2-2u+1)\;du\\
=\frac{1}{2}\int u^{\frac{5}{2}}-2u^{\frac{3}{2}}+u^{\frac{1}{2}}\;du\\
=\frac{1}{7}u^{\frac{7}{2}}-\frac{2}{5}u^{\frac{5}{2}}+\frac{1}{3}u^{\frac{3}{2}}+C
$$
So far so good. But when I try this method on the given integral, I get the following:
$$
\int x^2\sqrt{x^2+1}\;dx\\
=\frac{1}{2}\int \sqrt{x^2+1}\;x\cdot x\;dx\\
=\frac{1}{2}\int \sqrt{u}\;\sqrt{u-1}\;du\;(u=x^2+1)\\
=\frac{1}{2}\int u^{\frac{1}{2}}(u-1)^\frac{1}{2}\;du
$$
Here is where it falls down. I can't expand the $(u-1)^\frac{1}{2}$ factor like the $(u-1)^2$ factor above was, because it results in an infinite series. I couldn't prove, but I think any even exponent for the $x$ factor outside the square root will cause an infinite series to result. Odd exponents for $x$ will work, since it will cause the $(u-1)$ term to have a positive integer exponent.
How should I proceed? I don't necessarily want an answer. I just want to know if I'm missing something obvious or if it is indeed above first year calculus level and probably a typo on the question.
| Proceeding with your method
$\begin{align} \int x^2\sqrt{x^2+1}\;dx &=\int \sqrt{x^2+1}\;x\cdot x\;dx\\
&=\frac1{2}\int \sqrt{u}\;\sqrt{u-1}\;du\;\left( \text{ let $\begin{align} u&=x^2+1 \\ du&=2xdx\end{align}$}\right)\\
&=\frac1{2}\int \sqrt{u^2-u} \;du\\
&=\frac1{2} \int \sqrt{\left(u-\frac1{2}\right)^2-\left(\frac1{2}\right)^2} \;du \;\text{ (completing the square)} \\
&=\frac1{2}\left({\left(u-\frac1{2}\right)\over 2}{\sqrt{u^2-u}}-{\left(\frac1{2}\right)^2 \over 2}{ \ln \left|\left(u-\frac1{2}\right) +\sqrt{u^2-u}\right|} +C\right)\\
&\text{ (using $\int \sqrt{x^2-a^2} dx={x\over 2}{\sqrt{x^2-a^2}}-{a^2 \over 2}{ \ln |x+\sqrt{x^2-a^2}|} +C)$}\\
&={(2u-1)\over 8}{\sqrt{u^2-u}}-{\frac{ \ln |(2u-1) +2 \sqrt{u^2-u}|}{16}} +C'\\
&\text{ substituting $\left(\begin{align} u&=x^2+1 \\ 2u-1 &=2x^2 + 1 \\ \sqrt{u^2-u}&=\sqrt{(x^2 +1)^2-(x^2+1)}=\sqrt{x^4+x^2}\end{align}\right)$}
\\
&=\frac{{(2x^2 + 1)}{\sqrt{x^4+x^2}}}{8}-{\frac{ \ln |(2x^2+1) +2 \sqrt{x^4+x^2}|}{16}} +C'\\
&=\frac{{(2x^3 + x)}{\sqrt{x^2+1}}}{8}-{\frac{ \ln |(x^2+(x^2+1) +2x \sqrt{x^2+1}|}{16}} +C'\\
&=\frac{{(2x^3 + x)}{\sqrt{x^2+1}}}{8}-{\frac{ \ln |(x+\sqrt{x^2+1})^2|}{16}} +C'\\
&=\frac{{(2x^3 + x)}{\sqrt{x^2+1}}-\ln |x+\sqrt{x^2+1}|}{8} +C'
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Limit of a function in terms of exponents
Let $f(n)$ denote the $n$th term of the sequence $2, 5, 10, 17, 26,\cdots$ and $g(n)$ denote the $n$th term of the
sequence $2, 6, 12, 20,30,\cdots$.
Let $F(n)$ and $G(n)$ denote respectively the sum of n terms of the above sequences.
$$\lim_{n\to\infty}\left(\frac{F(n)}{G(n)}\right)^n-\left(\frac{f(n)}{g(n)}\right)^n$$
I found
*
*$G(n)=\frac {n(n+1)(n+2)}{3}$
*$F(n)=\frac {n(2n^2+3n+7)}{6}$
*$f(n)=n^2+1$
*$g(n)=n^2+n$
I am not able to further calculate the limit. The answer is $e^{-3/2}-e^{-1}$.
| The expressions you got for $F(n),G(n),f(n),g(n)$ are correct.
Now,
$\displaystyle \begin{align} \lim_{n \to\infty}\left(\frac{F(n)}{G(n)}\right)^n-\left(\frac{f(n)}{g(n)}\right)^n &= \lim_{n \to\infty}\left(\frac1{2} \frac{2n^2+3n+7}{n^2+3n+2}\right)^n-\left(\frac{n^2+1}{n^2+n}\right)^n\\
&= \lim_{n \to\infty}\left(\frac{n^2+\frac{3n}{2}+\frac{7}{2}}{n^2+3n+2}\right)^n- \lim_{n \to \infty}\left(\frac{1+\frac1{n^2}}{1+\frac1{n}}\right)^n\\
&=\frac{ \lim\limits_{n \to\infty}\left({1+\frac{3}{2n}+\frac{7}{2n^2}}\right)^n}{\lim\limits_{n \to\infty}\left({1+\frac{3}{n}+\frac{2}{n^2}}\right)^n} - \frac{ \lim\limits_{n \to\infty}\left({1+\frac{1}{n^2}}\right)^n}{\lim\limits_{n \to\infty}\left({1+\frac{1}{n}}\right)^n} \\
& \text{(All the limits are now in $1^\infty$ form)} \\
&=\frac{e^{ \lim\limits_{n \to\infty}\left({\frac{3}{2n}+\frac{7}{2n^2}}\right)n}}{e^{ \lim\limits_{n \to\infty}\left({\frac{3}{n}+\frac{2}{n^2}}\right)n}} - \frac{e^{ \lim\limits_{n \to\infty}\left({\frac1{n^2}}\right)n}}{e^{ \lim\limits_{n \to\infty}\left({\frac{1}{n}}\right)n}} \tag{1} \label{1}\\
&=\frac{e^{ \lim\limits_{n \to\infty}\left({\frac{3}{2}+\frac{7}{2n}}\right)}}{e^{ \lim\limits_{n \to\infty}\left({3+\frac{2}{n}}\right)}} - \frac{e^{ \lim\limits_{n \to\infty}\left({\frac1{n}}\right)}}{e^{ \lim\limits_{n \to\infty} 1}} \\
&=\frac{e^{\frac{3}{2}}}{e^3} - \frac{e^0}{e}\\
&=e^{\frac{-3}{2}} -e^{-1} \end{align}$
where in (\ref{1}) we used $\lim\limits_{n\to \infty} {f_1(n)}^{f_2(n)} = e^{\lim\limits_{n\to \infty}({f_1(n)-1}){f_2(n)}}$ since $f_1(n) \to 1,\; f_2(n) \to \infty$ as $n \to \infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Two ways of computing $\lim_{a \to 0^+} \left( \frac{1}{(x+ia)^2} - \frac{1}{(x-ia)^2} \right)$ gives contradictory result In this question, we assume that $\lim_{a\to 0^+}$ is implicit.
It is well-known that
$$\frac{1}{x-ia} - \frac{1}{x+ia} = 2\pi i \delta(x),$$
where $\delta(x)$ is the Dirac delta distribution. Differentiating this equation gives
$$\frac{1}{(x+ia)^2} - \frac{1}{(x-ia)^2} = 2\pi i\nabla \delta(x).$$
However, we can also evaluate LHS as
$$\frac{1}{(x+ia)^2} - \frac{1}{(x-ia)^2} = \left( \frac{1}{x+ia} - \frac{1}{x-ia}\right) \left( \frac{1}{x+ia} + \frac{1}{x-ia}\right) \\ = -2\pi i\delta(x) \left( \frac{1}{x+ia} + \frac{1}{x-ia}\right) =0,$$
since substituting $x=0$ to the last factor $\left( \frac{1}{x+ia} + \frac{1}{x-ia}\right)$ gives 0. Since all calculations are done formally, I am not sure which one is correct. Which one is correct, and why?
Finally, the same problem occurs for higher powers. For example, differentiating again gives
$$\frac{1}{(x-ia)^3} - \frac{1}{(x+ia)^3} = \pi i\nabla^2\delta(x),$$
but evaluating LHS as
$$\frac{1}{(x-ia)^3} - \frac{1}{(x+ia)^3} = \left( \frac{1}{x-ia} - \frac{1}{x+ia} \right) \left( \frac{1}{(x-ia)^2} + \frac{1}{(x-ia)(x+ia)}+\frac{1}{(x+ia)^2} \right)$$
gives undefined result $2\pi i\delta(x)/(ia)^2$. Which one is correct?
| Your observation is not wrong, but you have to keep in mind that you take the limit $a\rightarrow 0$ after all. I addition your product is in indeterminate form, since plugging in $x=0$ gives something of the form $\infty\cdot 0$.
If you really want to follow the route of factorization, rather than considering $$\frac{1}{(x+ia)^2} - \frac{1}{(x-ia)^2} = \frac{-4ia x}{(x^2+a^2)^2}$$ directly, you have to find out how
$$- \left( \frac{1}{x+ia} + \frac{1}{x-ia} \right) = \frac{-2x}{a^2+x^2}$$ behaves, i.e. how the RHS acts via integration on test-functions and not just point-wise considerations for $x=0$ and $x\neq 0$. Indeed the RHS seems to vanish for $x=0$, but at the same time it has maxima at $x=\pm a$ with the values $\pm \frac{1}{a}$ when $a \rightarrow 0$. So really the RHS as a distribution is indeterminate at $x=0$.
However, you can easily check how this expression behaves on some test-function $f(x)$ (assuming integrals are defined) $$\int \frac{-2x}{a^2+x^2} \, f(x) \, {\rm d}x = \log\left(\frac{a/\pi}{x^2+a^2}\right) f(x) - \int \log\left(\frac{a/\pi}{x^2+a^2}\right) f'(x) \, {\rm d}x \\
\stackrel{a\rightarrow 0}{=}\log(\delta(x)) f(x) - \int \log(\delta(x)) f'(x) \, {\rm d}x = \int \frac{\delta'(x)}{\delta(x)} f(x) \, {\rm d}x$$
which is what you would expect.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given the function $f(x,y)=\frac{5xy^2}{x^2+y^2}$, argue analytically, if the discontinuity of $f$ at $(0,0)$ is avoidable or unavoidable. It is clear that $f(0,0)$ does not exist, and therefore the function is discontinuous. Now to know if the discontinuity is avoidable or inevitable, we must see that the limit at that point exists or not. To see the above we will use polars.
\begin{align*}
\lim_{(x,y)\rightarrow(0,0)}\frac{5xy^2}{x^2+y^2}&=\lim_{r\rightarrow0}\frac{5(r\cos(\theta))(r\sin(\theta))^2}{(r\cos(\theta))^2+(r\sin(\theta))^2}\\
& =\lim_{r\rightarrow0}\frac{5r^3\cos(\theta)\sin^2(\theta)}{r^2(\cos^2(\theta)+\sin^2(\theta))}\\
& =\lim_{r\rightarrow0}\frac{5r^3\cos(\theta)\sin^2(\theta)}{r^2}\\
& =\lim_{r\rightarrow0}5r\cos(\theta)\sin^2(\theta)=0.
\end{align*}
Now let's test for the definition $\epsilon-\delta$, to see that the limit exists and is zero.
\begin{align*}
\left|\frac{5xy^2}{x^2+y^2}-0\right| & =\left|\frac{5xy^2}{x^2+y^2}\right|\\
& =\frac{5|x|y^2}{x^2+y^2}\\
& \leq 5|x| \text{ since $y^2\leq x^2+y^2$}\\
& \leq 5\sqrt{x^2+y^2} \text{ since $|x|\leq\sqrt{x^2+y^2}$}\\
& <5\delta = \epsilon.
\end{align*}
Therefore we have:
\begin{align*}
\forall\epsilon>0, \exists\delta>0 \text{ such that, if } 0<\sqrt{x^2+y^2}<\delta \Rightarrow |f(x,y)-0| &\leq5 \sqrt{x^2+y^2}\\
&<5 \delta = 5\left(\frac{\epsilon}{5}\right)=\epsilon.
\end{align*}
Thus the limit exists and is zero, therefore the discontinuity of $f$ is avoidable.
I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.
| Your first sentence is wrong; since $f$ is undefined at $(0,0)$, it is neither continuous nor discontinuous at that point.
Otherwise, you proved correctly that $\lim_{(x,y)\to(0,0)}f(x,y)=0$ and that therefore you can extend $f$ in such a way that it becomes continuous at $(0,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4227337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Summation of square root fractional part Is there a way to calculate the following sum for large $c>K$, closed-form or approximation?
$$\sum_{n=1}^{c}\left\{\sqrt{K+n^2}\right\}$$Where K and n are positive integers and $\{\}$ indicates the fractional part.
A similar question: Fractional part summation
| The given sum can be written as
$$
\Sigma = \Sigma_1 - \Sigma_2
$$
where
$$
\Sigma_1 = \sum_{n=1}^{c} \sqrt{n^2 + K}
$$
and
$$
\Sigma_2 = \sum_{n=1}^{c} \lfloor\sqrt{n^2 + K}\rfloor.
$$
Now using $H_n = \log n + \gamma + \mathcal{O}(1/n)$ where $H_n$ is the $n$-th harmonic number,
\begin{align*}
\sum_{n=1}^{c} \sqrt{n^2 + K} - n &= \sum_{n=1}^c \frac{K}{\sqrt{n^2 + K} + n} \\
&\le \frac{K}{2} \sum_{n=1}^{c} \frac{1}{n} = \frac{K}{2}(\log c + \gamma + \mathcal{O}(1/c)).
\end{align*}
Hence
$$
\Sigma_{1} = \sum_{n=1}^{c} \sqrt{n^2 + K} = \frac{1}{2}c(c+1) + \frac{K}{2}(\log c + \gamma) + \mathcal{O}(1/c)
$$
Let $A = \lfloor\sqrt{K+1} - 1\rfloor$. Then,
\begin{align*}
\Sigma_{2} = \sum_{n=1}^{c} \lfloor\sqrt{n^2 + K}\rfloor &= \sum_{j=1}^{A}\sum_{\frac{K-(j+1)^2}{2(j+1)} < n \le \frac{K-j^2}{2j}}(n+j) + \sum_{\frac{K-1}{2} < n \le c}n \\
&= \sum_{n=1}^c n + \sum_{j=1}^{A}\sum_{\frac{K-(j+1)^2}{2(j+1)} < n \le \frac{K-j^2}{2j}}j\\
&= \frac{1}{2}c(c+1) + \sum_{j=1}^A j\left(\frac{3}{2} + \frac{K}{2j(j+1)}+ \mathcal{O}(1)\right)\\
&= \frac{1}{2}c(c+1) + \frac{3}{4}A(A+1) + \frac{K}{2}(H_A - 1) + \mathcal{O}\left(\sum_{j=1}^A j\right)\\
&= \frac{1}{2}c(c+1) + \frac{3}{4}A(A+1) + \frac{K}{2}(H_A - 1) + \mathcal{O}(1)
\end{align*}
since $\mathcal{O}\left(\sum_{j=1}^A j\right) = \mathcal{O}(K) = \mathcal{O}(1)$.
Thus,
\begin{align*}
\Sigma &= \frac{K}{2}(\log c + \gamma) - \frac{3}{4}A(A+1) - \frac{K}{2}(H_A - 1) + \mathcal{O}(1).\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Complex number related problem Let $z_1,z_2,z_3$ be complex numbers such that $|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2$ and $|z_1–z_2| =|z_1–z_3|$,$(z_2 \ne z_3)$, then the value of $|z_1+z_2||z_1+z_3|$ is_______
My solution is as follow
${z_1} = 2{e^{i{\theta _1}}};{z_2} = 2{e^{i{\theta _2}}};{z_3} = 2{e^{i{\theta _3}}}$ & $Z = {z_1} + {z_2} + {z_3} = 2\left( {{e^{i{\theta _1}}} + {e^{i{\theta _2}}} + {e^{i{\theta _3}}}} \right)$
$\left| {{z_1} - {z_2}} \right| = \left| {{z_1} - {z_3}} \right| \Rightarrow \left| {{e^{i{\theta _1}}} - {e^{i{\theta _2}}}} \right| = \left| {{e^{i{\theta _1}}} - {e^{i{\theta _3}}}} \right|$
Let ${\theta _1} = 0$
$\left| {{z_1} - {z_2}} \right| = \left| {{z_1} - {z_3}} \right| \Rightarrow \left| {1 - \left( {\cos {\theta _2} + i\sin {\theta _2}} \right)} \right| = \left| {1 - \left( {\cos {\theta _3} + i\sin {\theta _3}} \right)} \right|$
$ \Rightarrow \left| {1 - \cos {\theta _2} - i\sin {\theta _2}} \right| = \left| {1 - \cos {\theta _3} - i\sin {\theta _3}} \right| \Rightarrow \left| {2{{\sin }^2}\frac{{{\theta _2}}}{2} - 2i\sin \frac{{{\theta _2}}}{2}\cos \frac{{{\theta _2}}}{2}} \right| = \left| {2{{\sin }^2}\frac{{{\theta _3}}}{2} - 2i\sin \frac{{{\theta _3}}}{2}\cos \frac{{{\theta _3}}}{2}} \right|$
$\Rightarrow \left| { - 2{i^2}{{\sin }^2}\frac{{{\theta _2}}}{2} - 2i\sin \frac{{{\theta _2}}}{2}\cos \frac{{{\theta _2}}}{2}} \right| = \left| { - 2{i^2}{{\sin }^2}\frac{{{\theta _3}}}{2} - 2i\sin \frac{{{\theta _3}}}{2}\cos \frac{{{\theta _3}}}{2}} \right| \Rightarrow \left| { - 2i\sin \frac{{{\theta _2}}}{2}\left( {\cos \frac{{{\theta _2}}}{2} + i\sin \frac{{{\theta _2}}}{2}} \right)} \right| = \left| { - 2i\sin \frac{{{\theta _3}}}{2}\left( {\cos \frac{{{\theta _3}}}{2} + i\sin \frac{{{\theta _3}}}{2}} \right)} \right|$
$ \Rightarrow \left| { - 2i\sin \frac{{{\theta _2}}}{2}\left( {{e^{i\frac{{{\theta _2}}}{2}}}} \right)} \right| = \left| { - 2i\sin \frac{{{\theta _3}}}{2}\left( {{e^{i\frac{{{\theta _3}}}{2}}}} \right)} \right|$
$ \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}\left( {{e^{ - i\frac{\pi }{2}}}} \right)\left( {{e^{i\frac{{{\theta _2}}}{2}}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}\left( {{e^{ - i\frac{\pi }{2}}}} \right)\left( {{e^{i\frac{{{\theta _3}}}{2}}}} \right)} \right| \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}\left( {{e^{i\left( {\frac{{{\theta _2}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}\left( {{e^{i\left( {\frac{{{\theta _3}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right|$
$\Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}} \right|\left| {\left( {{e^{i\left( {\frac{{{\theta _2}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}} \right|\left| {\left( {{e^{i\left( {\frac{{{\theta _3}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}} \right|$
${\theta _2} \ne {\theta _3}$
How do I proceed further?
| Consider $$u=z_2/z_1,v=z_3/z_1$$ and then we have $$|u-1|=|v-1|\tag{1}$$ and $$u\neq v, |u|=|v|=1\tag{2}$$ and $$|1+u+v|=1\tag{3}$$ Then from first equation we get $$(u-1)(\bar{u} - 1)=(v-1)(\bar{v}-1)$$ or $$u+\bar{u}=v+\bar{v}\tag{4}$$ This means that $u, v$ have same real part and using $(2)$ we can see that they are conjugates. If $u=x+iy, v=x-iy$ then from $(3)$ we get $$|1+2x|=1$$ so that either $x=0$ or $x=-1$. But $x=-1$ gives $y=0$ so that $u=v$ which is not allowed.
Thus $u, v=\pm i$ and the expression whose value is to be found is $$|z_1|^2|1+u||1+v|=4|1+i||1-i|=8$$
The key to this problem is to understand that the individual values of $z_1,z_2,z_3$ don't matter, but what matters is their ratios. Also the choice of dividing by $z_1$ to get $u, v$ is because all the constraints in question are symmetric in $z_2,z_3$ and $z_1$ plays a special role as being equidistant from both $z_2,z_3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Pairs of integers $ (x,m)$ for which $\sqrt[3]{\sqrt[3]{x-2}+m}+\sqrt[3]{-\sqrt[3]{x-2}+m}=2$ hold?
Find all pairs of integers $(x,m)$ for which $$\sqrt[3]{\sqrt[3]{x-2}+m}+\sqrt[3]{-\sqrt[3]{x-2}+m}=2$$ hold.
I have used this property :
Property:
if $$a+b+c=0 \implies a^3+b^3+c^3=3abc, $$ I come up to the following equation:
$(2m-8)^3=-216(m^2-(x-2)^{2/3})$ , such that $a=-2, b=(\sqrt[3]{x-2}+m)^{1/3}, c=(-\sqrt[3]{x-2}+m)^{1/3}$, I can't solve the last equation however i tried $x$ as a paramater instead of $m$ , The solution from wolfram alpha are $(x,m)=(2,1),(66,4)$, Any Help ?
| Generally solutions $(x,m)$ can be split into three cases:
*
*$(x,m)=\left(\frac{1}{243} \left(486-\sqrt{3} (m-1)^{3/2} (m+8)^3\right),m\right)$
*$(x,m)=\left(\frac{1}{243} \left(486+\sqrt{3} (m+8)^3 (m-1)^{3/2}\right)),m\right)$
*$(x,m)=(2,1)$
In the first two cases $m\ge2$ and we choose $m$ such that the resulting term for $x$ is an integer. For example if we choose $m=4$, the first case yields the solution $(x,m)=(-62,4)$ and from the second case we obtain $(x,m)=(66,4)$.
You can generate infinitely such solution pairs by choosing a $m$ where $m-1$ is an odd cubic, e.g. $m=3^1+1,3^3+1,3^5+1,\ldots$ or, more precisely, as shown by @Aqua where $m=3c^2+1$ for natural $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Calculating $\int\frac{a^4-x^4}{x^4+a^2x^2+a^4}dx$
$$\int\frac{a^4-x^4}{x^4+a^2x^2+a^4}dx$$
I tried factoring as $${(a^2-x^2)(a^2+x^2)}$$ $$\text{and}$$ $$(x^2+a^2+ax)(x^2+a^2-ax)$$ but didn't get much after this. I sew that the expression is symmetric in a,x but I don't know how to use that.
| Hint:
First,
$$\dfrac{a^4 - x^4}{x^4 + a^2x^2 + a^4} = \dfrac{a^2x^2 + 2a^4}{x^4 + a^2x^2 + a^4} - 1 = a^2\dfrac{x^2 + 2a^2}{x^4 + a^2x^2 + a^4} - 1$$
Second,
$$\dfrac{x^2 + 2a^2}{x^4 + a^2x^2 + a^4} = \dfrac{x^2 + 2a^2}{\left(x^2 -ax + a^2\right)\left(x^2 + ax + a^2\right)}$$
Now, perform partial-fraction decomposition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$ Let $a,b,c>0$:
Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$
My solution:
We have: $\left[\begin{matrix}\frac{1}{x}+\frac{1}{y} \geq \frac{4}{x+y} \\\frac{1}{x}+\frac{1}{y} +\frac{1}{z} \geq \frac{9}{x+y+z}\end{matrix}\right.$
$=>S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)}$
$\leq \frac{a^2}{4}.(\frac{1}{2a+b}+\frac{1}{2a+c})+\frac{b^2}{4}.(\frac{1}{2b+c}+\frac{1}{2b+a})+\frac{c^2}{4}.(\frac{1}{2c+a}+\frac{1}{2c+b})$
$=\frac{1}{4}.[a^2.(\frac{1}{2a+b}+\frac{1}{2a+c})+b^2.(\frac{1}{2b+a}+\frac{1}{2b+c})+c^2.(\frac{1}{2c+a}+\frac{1}{2c+b})]$
$\leq \frac{1}{4}.[\frac{a^2}{9}.(\frac{2}{a}+\frac{1}{b}+\frac{2}{a}+\frac{1}{c}) +\frac{b^2}{9}.(\frac{2}{b}+\frac{1}{c}+\frac{2}{b}+\frac{1}{a}) +\frac{c^2}{9}.(\frac{2}{c}+\frac{1}{a}+\frac{2}{c}+\frac{1}{b})]$
$=\frac{1}{36}.[a^2.(\frac{4}{a}+\frac{1}{b}+\frac{1}{c})+b^2.(\frac{1}{a}+\frac{4}{b}+\frac{1}{c})+c^2.(\frac{1}{a}+\frac{1}{b}+\frac{4}{c})]$
$= \frac{1}{36}.(4a+4b+4c+\frac{a^2}{b}+\frac{a^2}{c}+\frac{b^2}{a}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{c^2}{b}) $
We need prove that: $S \le \frac{1}{3}$
$=> S \le \frac{1}{12}.(4a+4b+4c+\frac{a^2}{b}+\frac{a^2}{c}+\frac{b^2}{a}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{c^2}{b}) \le 1 $ <- in there, I don't know how to do that :<
I'm trying to find a solution to continue for me or another solution, Can you help me?
| Another way.
$$\frac{1}{3}-\sum_{cyc}\frac{a^2}{(2a+b)(2a+c)}=\sum_{cyc}\left(\frac{1}{9}-\frac{a^2}{(2a+b)(2a+c)}\right)=$$
$$=\frac{1}{9}\sum_{cyc}\frac{2ab+2ac+bc-5a^2}{(2a+b)(2a+c)}=\frac{1}{18}\sum_{cyc}\frac{(c-a)(5a+b)-(a-b)(5a+c)}{(2a+b)(2a+c)}=$$
$$=\frac{1}{18}\sum_{cyc}(a-b)\left(\frac{5b+c}{(2b+c)(2b+a)}-\frac{5a+c}{(2a+b)(2a+c)}\right)=$$
$$=\frac{1}{18}\sum_{cyc}\frac{(a-b)^2(c^2+10ab-ac-bc)}{(2a+b)(2a+c)(2b+a)(2b+c)}\geq$$
$$\geq\frac{1}{18}\sum_{cyc}\frac{(a-b)^2(c^2-ac-bc+ab)}{(2a+b)(2a+c)(2b+a)(2b+c)}=$$
$$=\frac{1}{18\prod\limits_{cyc}((2a+b)(2a+c))}\sum_{cyc}(a-b)^2(c-a)(c-b)(2c+a)(2c+b)=$$
$$=\frac{(a-b)(b-c)(c-a)\sum\limits_{cyc}(b-a)(4c^2+2ac+2bc+ab)}{18\prod\limits_{cyc}((2a+b)(2a+c))}=$$
$$=\frac{(a-b)(b-c)(c-a)\sum\limits_{cyc}(4a^2c-4a^2b+2a^2b-2a^2c+a^2c-a^2b)}{18\prod\limits_{cyc}((2a+b)(2a+c))}=$$
$$=\frac{(a-b)(b-c)(c-a)\sum\limits_{cyc}(a^2c-a^2b)}{6\prod\limits_{cyc}((2a+b)(2a+c))}=\frac{\prod\limits_{cyc}(a-b)^2}{6\prod\limits_{cyc}((2a+b)(2a+c))}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Calculating value of $\delta$ for a function $y=\frac1{(1-x)^2}$ unbounded at $x\rightarrow 1$ The function $y=\frac 1{(1-x)^2}$ is unbounded at $x\rightarrow 1$.
How can I calculate the value of $\delta$ such that $y> 10^6$ if $\left | x-1 \right |< \delta$ ?
Taking $y=f(x)=10^6$, putting in ${y=\frac1{(1-x)^2}}$, we get
$10^6 = \frac1{(1-x)^2}\\
\frac1{(1-x)^2} = \frac1{10^6}\\
x=1-\sqrt{(10^{-6})}$ meaning $x= 0.999$.
As, $\left |x-1\right |< \delta,$ hence $ \left|0.999 -1\right| < \delta,$ or
$10^{-3} < \delta$.
It can imply that,
$ \delta$ must be at least $10^{-3}$ or greater $ \delta \geq 10^{-3}$
(After correction by user:505767)
| We have that for $A,B>0$ (i.e. dividing both sides by $AB$)
$$A>B \iff \frac 1 A < \frac 1 B$$
therefore
$$\frac{1}{(1-x)^2}>10^6 \iff (1-x)^2<\frac1{10^6}=10^{-6}$$
form which, using that $\sqrt{A^2}=|A|$ and (taking the square roots both sides)
$$A^2<B \iff |A|<\sqrt B$$
we find
$$(1-x)^2<10^{-6} \iff |1-x|<10^{-3}$$
and $|1-x|=|x-1|$.
In your evaluation it should be
$$x=1-\sqrt{10^{-6}}=1-10^{-3}=0.999$$
which is correct but not much effective and clear to obtain the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Let $x^3+\frac{1}{x^3}$ and $x^4+\frac{1}{x^4}$ are rational numbers. Show that $x+\frac{1}{x}$ is rational.
$x^3+\dfrac{1}{x^3}$ and $x^4+\dfrac{1}{x^4}$ are rational numbers. Prove that $x+\dfrac{1}{x}$ is rational number.
My solution:
$x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}-1\right)$
$x^4+\dfrac{1}{x^4}=x^4+2+\dfrac{1}{x^4}-2=\left(x^2+\dfrac{1}{x^2}\right)^2-1-1=\left(x^2+\dfrac{1}{x^2}-1\right)\left(x^2+\dfrac{1}{x^2}+1\right)-1$
Because $x^4+\dfrac{1}{x^4}$ is rational number so
$\left(x^2+\dfrac{1}{x^2}-1\right)$ is rational number too and
$x^3+\dfrac{1}{x^3}$ is rational number so
$\left(x+\dfrac{1}{x}\right)$ is rational number.
Am I wrong? Please check my solution, thank you.
| I wanted to write this answer with the thought that it might be cleaner than my previous answer.
We have,
$$\begin{align}&\begin{cases}x+\frac 1x=u,\thinspace u \in \mathbb R\\
x^2+\frac{1}{x^2}=v,\thinspace v\in\mathbb R\\
x^3+\frac{1}{x^3}=p,\thinspace p\in\mathbb Q\\
x^4+\frac{1}{x^4}=q,\thinspace q\in\mathbb Q\end{cases}\\ \implies &vq=p^2-2+v\\
\implies &vq-v=p^2-2\\
\implies &v=\frac{p^2-2}{q-1}\in\mathbb Q\\
\implies &up=q+v\\
\implies &u=\frac{q+v}{p}\in\mathbb Q.\end{align}$$
Final answer:
$$u=\frac{p^2q-p^2-q+2}{p}\in\mathbb Q.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 2
} |
Proving divisibility by $9$ for $10^n + 3 \times 4^{n+2} + 5$. I am trying to prove that for all $n$, $9$ divides $10^n + 3 \times 4^{n+2} + 5$. I first tried induction on $n$, but couldn't get anywhere with the induction step. I then tried to use modular arithmetic. The furthest I could get was:
As $10 \equiv 1 \mod 9$, $10^n \equiv 1^n = 1$ for all $n$, so modulo $9$, we have
\begin{align*}
10^n + 3 \cdot 4^{n+2} + 5 & = 3 \cdot 4^{n+2} + 6 \\
& = 3\left(4^{n+2} + 2\right) \\
& = 3\left(\left(2^2\right)^{n+2} + 2 \right) \\
& = 3\left(2^{2(n+2)} + 2 \right) \\
& = 3\left(2\left(2^{2(n+2) - 1} + 1\right) \right)
\end{align*}
I need to something get a factor of $3$ to show that the entire expression is divisible by $9$ and hence equal to $0$, mod $9$. But, with a sum of even terms, this does not appear possible.
Any hints on how to proceed would be appreciated. Is induction the standard way to prove something like this?
| We can construct one of the possible proofs as follows:
$$\begin{align}&10^n + 3 \times 4^{n+2} + 5 \mod 9
\\
\iff &10^n + 3 \times 4^{n+2} -4\mod 9\\
\iff &10^n-1^n+3 \times 4^{n+2} -3\mod 9\\
\iff &3\left(4^{n+2}-1\right)\mod 9\\ \iff
&4^{n+2}-1^{n+2} \mod 3 =0.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Why is this approach wrong?
If $\alpha$ and $\beta$ are the roots of $x^{2} - 4 x - 3$, then find $$\frac{1}{(\alpha-4)^{2}}+\frac{1}{(\beta-4)^{2}}$$
Solution
This is the approach that gives wrong answer.
\begin{array}{l}
x^{2}-4 x-3=0 \\
(x-4)^{2}=19 \\
So, {(\alpha-4)^{2}} {\&(\beta-4)^{2}}=19 \\
\text { Hence, } \frac{1}{(\alpha-4)^{2}}+\frac{1}{(\beta-4)^{2}}=\frac{2}{19}
\end{array}
This is how it is solved :
\begin{array}{l}
x^{2}-4 x-3=0 \\
\alpha^{2}-4 \alpha-3=0 \\
\alpha(\alpha-4)=3 \\
\alpha-4=\frac{3}{\alpha} \\
\& \beta-4=\frac{3}{\beta}
\end{array}
So , $$
\frac{1}{(\alpha-4)^{2}}+\frac{1}{(\beta-4)^{2}}
$$
\begin{array}{l}
=\frac{\alpha^{2}}{9}+\frac{\beta^{2}}{9} \\
=\frac{1}{9}\left(\alpha^{2}+\beta^{2}\right) \\
=\frac{22}{9}
\end{array}
Please tell me why the first approach is wrong .
| You did not factorise $x^2-4x-3$ correctly. Actually, $x^2-4x-3=(x-2)^2-4-3=(x-2)^2-7$. We find that $(x-2)^2=7$, and so $x-2=\pm\sqrt{7}$. Therefore, $x-4=-2\pm\sqrt{7}$, and $(x-4)^2=11\pm4\sqrt{7}$. Hence,
$$
\frac{1}{(\alpha-4)^2}+\frac{1}{(\beta-4)^2}=\frac{1}{11+4\sqrt{7}}+\frac{1}{11-4\sqrt{7}}=\frac{11-4\sqrt{7}+11+4\sqrt{7}}{121-112}=\frac{22}{9} \, .
$$
An alternative method (which would be slower in this case), would be to make the substitution $w=x-4$. The resulting quadratic in $w$ will have roots $\alpha-4$, and $\beta-4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4239918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What happens to $w=\frac{x^2y^3}{z^4}$ , if $x,y,z$ increase by $\%1$ , $\%2$ , $\%3$ respectively? If $x,y,z$ increase by $\%1$ , $\%2$ , $\%3$ respectively , then $w=\dfrac{x^2y^3}{z^4}$ ............. approximately.
$1)\ \%\ 3\text{ decrease}$
$2)\ \%\ 4\text{ decrease}$
$3)\ \%\ 3\text{ increase}$
$4)\ \%\ 4\text{ increase}$
I denote the new $w$ by $w'$,
$$w'-w=\left( \frac{1.01^2.1.02^3}{1.03^4}-1\right)\times w=\left(\frac{101^2\times102^3}{103^4\times100}-1\right)\times w$$
From here should I evaluate $\dfrac{101^2\times102^3}{103^4\times100}-1$ by hand (calculator is not allowed) or there is a quicker method to get to the correct answer?
| You can use partial elasticities, also know as condition numbers:
\begin{align*}
\varepsilon_w \approx & \frac{x w'_x}{w} \varepsilon_x+\frac{y w'_y}{w} \varepsilon_y + \frac{z w'_z}{w} \varepsilon_z\\
= & 2\varepsilon_x + 3 \varepsilon_y - 4 \varepsilon_z= 2 + 3\times 2-4 \times 3 = -4
\end{align*}
So, the answer is an approximate decreasing by $\text{4%}$.
This comes from Taylor's formula:
\begin{align*}
f(\tilde x) \approx f(x) + f'(x)(\tilde x -x) \Rightarrow \\
f(\tilde x) - f(x) \approx f'(x)(\tilde x -x) \Rightarrow \\
\frac{f(\tilde x) - f(x)}{f(x)} \approx \frac{f'(x)(\tilde x -x)}{f(x)} \Rightarrow\\
\frac{f(\tilde x) - f(x)}{f(x)} \approx \frac{xf'(x)}{f(x)} \frac{\tilde x -x}{x}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4240956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Largest possible side of a triangle when $\cos(3P)+ \cos(3Q)+ \cos(3R) = 1$.
In triangle $PQR$, $\cos(3P)+ \cos(3Q)+ \cos(3R) = 1$. Two sides of the
triangle have lengths of $15 cm$ and $18 cm$. If the length of the
third side of the triangle PQR is $\sqrt{m}$ cm, then the largest
possible value that $'m'$ can take is?
At first I thought the information given i.e. $\cos(3P)+ \cos(3Q)+ \cos(3R) = 1$ to be a distraction and I went with the usual way to solve this problem that sum of two sides is always greater than the third side. So;
$15+18>\sqrt{m}$ $\Rightarrow m<1089$
So highest value that $m$ can take is $1088$ but this was the wrong answer. Then I used the cosine rule as follows :-
$\cos P=\frac{15^2+18^2-m}{2.15.18}$
$\Rightarrow -1 \le \frac{15^2+18^2-m}{2.15.18} \le1 $
$\Rightarrow9 \le m \le1089$
So from here I got the highest value of $m$ to be 1089 but this was also wrong. Now I am stuck as I am not able to think how to solve this question. I think I need to use the given condition some how to get to the answer but how? Please help me on this !!!
Thanks in advance !!!
| Let $p$, $q$ and $r$ be sides-lengths of the triangle.
Thus, $$\sum_{cyc}\left(4\left(\frac{p^2+q^2-r^2}{2pq}\right)^3-3\cdot\frac{p^2+q^2-r^2}{2pq}\right)=1$$ or
$$\prod_{cyc}((p+q-r)(p^2+q^2+pq-r^2))=0,$$ which says that one of the measured angles of the triangle is equal to $120^{\circ}$.
Can you end it now?
I got $m=819$.
Another way:
$$\cos3P+\cos3Q+\cos3R-1=2\cos\frac{3P+3Q}{2}\cos\frac{3P-3Q}{2}-2\sin^2\frac{3R}{2}=$$
$$=2\cos\left(270^{\circ}-\frac{3R}{2}\right)\cos\frac{3P-3Q}{2}-2\sin^2\frac{3R}{2}=$$
$$=-2\sin\frac{3R}{2}\left(\cos\frac{3P-3Q}{2}+\sin\frac{3R}{2}\right)=$$
$$=-2\sin\frac{3R}{2}\left(\cos\frac{3P-3Q}{2}+\sin\frac{3(180^{\circ}-P-R)}{2}\right)=$$
$$=-2\sin\frac{3R}{2}\left(\cos\frac{3P-3Q}{2}-\cos\frac{3P+3Q}{2}\right)=-4\sin\frac{3R}{2}\sin\frac{3P}{2}\sin\frac{3Q}{2}$$ and we get the same result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding smooth behaviour of infinite sum Define
$$E(z) = \sum_{n,m=-\infty}^\infty \frac{z^2}{((n^2 + m^2)z^2 + 1)^{3/2}} = \sum_{k = 0}^\infty \frac{r_2(k) z^2}{(kz^2 + 1)^{3/2}} \text{ for } z \neq 0$$
$$E(0) = \lim_{z \to 0} E(z) = 2 \pi$$
where $r_2(k)$ is the number of ways of writing $k$ as a sum of two squares of integers. Is $E(z)$ smooth at $z = 0$, and can we evaluate its derivatives $(\partial_z)^n E(z)|_{z = 0}$?
Footnote 1: There is a physical motivation for these equations, $E(z)$ is the electric field generated by a $2d$ lattice of point charges where the lattice spacing is $z$ and the charge density is held fixed. $E(0)$ here is the continuum limit.
Footnote 2: Here is a plot of the behavior of the sum in the second representation I wrote above from $k = 0$ to $k = N$, as a function of $z \in [-2,2]$. The black line is $2 \pi$. It looks like the limit will be very flat at $z = 0$ (possibly all derivatives vanish? That would be cool)
| §1. Basic results
We have to investigate the sum
$$E_s(z) = \sum_{n,m=-\infty}^\infty \frac{z^2}{((n^2 + m^2)z^2 + 1)^{3/2}}\tag{1.1}$$
Writing
$$\frac{z^2}{((n^2 + m^2)z^2 + 1)^{3/2}}=\int_{0}^{\infty} z^2 \sqrt{\frac{4 t}{\pi} }e^{-t(z^2 (m^2 + n^2) + 1)}\,dt\tag{1.2}$$
and doing the sums under the integral the double sum becomes this integral
$$E_i(z) = z^2 \int_0^{\infty}\sqrt{\frac{4 t}{\pi} }e^{-t} \vartheta _3\left(0,e^{-t z^2}\right)^2 \,dt\tag{1.3}$$
where $\vartheta _3$ is a Jacobi theta function.
This expression is better suitable to be plotted (using the NIntegrate command of Mathematica) than the sum.
Here is a picture of your function $E(z)$ for $z>0$ (for $z<0$ we have $E(-z)=E(z)$).
Notice that it is rather different from the plot in the OP.
Further results are:
For $z \to 0$ we have $E(z) = 2 \pi$, for $z>>1$ we have $E(z) \simeq z^2$.
All derivatives of $E(z)$ with respect to $z$ vanish for $z\to 0$ like $z^p e^{-1/z^2}$ (with some power $p$).
§2. Asymptotic expansions
Expanding the integrand $i(z,t)$ with respect to $z$ we obtain
*
*for $z\simeq 0$
$$\begin{align}i = \frac{2\sqrt{\pi }}{\sqrt{t}} e^{-\frac{8 \pi ^2}{t z^2}-t} \left(2 e^{\frac{3 \pi ^2}{t z^2}}+e^{\frac{4 \pi ^2}{t z^2}}+2\right) \times\left(e^{\frac{3 \pi ^2}{t z^2}} \left(2-t z^2\right)+e^{\frac{4 \pi ^2}{t z^2}}+2\right)\end{align}\tag{2.1}$$
integrating over $t$ results in
$$\begin{align}E(0<z\simeq 0)=\pi \left(-e^{-\frac{2 \pi }{z}} \left(z^2+2 \pi z-8\right)-2 e^{-\frac{2 \sqrt{2} \pi }{z}} \\
\left(z^2+2 \sqrt{2} \pi z-4\right)-2 e^{-\frac{2 \sqrt{5} \pi }{z}} \left(z^2+2 \sqrt{5} \pi z-8\right)\\
+8 e^{-\frac{4 \pi }{z}}+8 e^{-\frac{4 \sqrt{2} \pi }{z}}+2\right)\end{align}\tag{2.2}$$
the leading terms of which are
$$\begin{align}E(0<z\simeq 0)=\pi \left(2 + 8 e^{-\frac{2 \pi }z}+ 8 e^{-\frac{4 \pi }z}\right)\end{align}\tag{2.3}$$
*for $z\to \infty$
Developing first $\vartheta _3\left(0,e^{-u}\right)^2$ for $u\to \infty$ and then replacing $u \to t z^2$ we find for the integrand
$$i(z\to +\infty) \simeq \frac{2 e^{-t} \sqrt{t} z^2 \left(2 \left(e^{-4 t z^2}+e^{-t z^2}\right)+1\right)^2}{\sqrt{\pi }}\tag{2.4}$$
After integration over $t$ this leads to the interesting result
$$\begin{align}E(z\to \infty) \simeq z^2 \left(1+\frac{4}{\left(1+z^2\right)^{3/2}}+\frac{4}{\left(1+2 z^2\right)^{3/2}}\\
+\frac{4}{\left(1+4 z^2\right)^{3/2}}+\frac{8}{\left(1+5 z^2\right)^{3/2}}+\frac{4}{\left(1+8 z^2\right)^{3/2}}\right) \end{align}\tag{2.5}$$
Here the following two sequence of numbers appear: $\{0,1,2,4,5,8,...\}$ as a factor before $z^2$ in the denominator and $\{1,4,4,4,8,4,...\}$ in the numerator.
The first series (if prolonged by taking into account higher order terms) is easily identified in OEIS as
A001481 Numbers that are the sum of 2 squares.
The second series is A004018 Theta series of square lattice (or number of ways of writing n as a sum of 2 squares). Often denoted by r(n) or r_2(n).
Hence the asymptotic form $(2.5)$ leads automatically to the second sum in the OP (the one containing $r_2(k)$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
Integration of $\sin^2x$ without using double angle identity of $\cos 2x$ I want to integrate $\sin^2x$ without using the double angle identity of $\cos2x$.
Here's what I tried:
$$
\int \sin^2x dx
= \int \sin x \tan x \cos x dx
$$
Let $u = \sin x$ => $du = \cos x dx$
And if $u = \sin x$, $\tan x = \frac{u}{\sqrt{1-u^2}}$, therefore
$$
=\int u × \frac{u}{\sqrt{1-u^2}}du = \int \frac{u^2}{\sqrt{1-u^2}}du
$$
Now if $t = \sqrt{1 - u^2}$, $2dt = \frac{du}{\sqrt{1-u^2}}$ and $u^2 = 1 - t^2$
$$
= 2\int (1-t^2)dt = 2t - \frac{2t^3}{3}$$
Substituting back $u$ and $\sin x$
$$
= 2\sqrt{1-u^2} - \frac{2}{3}(\sqrt{1 - u^2})^3
$$
$$
= 2\cos x - \frac{2}{3}\cos^3x
$$
But when you differentiate it you get $-2sin^3x$
All the steps seem right to me, why is the answer wrong or what I did is wrong, and is using the double angle formula the only way to integrate it?
| If $t=\sqrt{1-u^2}$, then$$\mathrm dt=-\frac u{\sqrt{1-u^2}}\,\mathrm du\quad\text{and}\quad u^2=1-t^2.$$So,\begin{align}\int\frac{u^2}{\sqrt{1-u^2}}\,\mathrm du&=-\int u\frac{-u}{\sqrt{1-u^2}}\,\mathrm du\\&=-\int\sqrt{1-t^2}\,\mathrm dt\\&=-\frac12\left(t\sqrt{1-t^2}+\arcsin (t)\right).\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Solve for $\sqrt{x}+\sqrt[4]{y}=\sqrt{p}$ with $x$, $y$ are positive integers and $p$ is a prime number For that given problem, I have constructed a solution to it:
Squaring both sides, we get $x+\sqrt{y}+2\sqrt[4]{x^2y}=p$. Suppose $\sqrt{y}$ and $\sqrt[4]{x^2y}$ are integers, we can let $y=y'^2$, $y'$ being a positive integer. Subbing to the equation, we get $x+y'+2\sqrt{xy'}=p$, so $(\sqrt{x}+\sqrt{y'})^2=p$. We can see that $\sqrt{x}+\sqrt{y'}$ is a positive integer, and $x$ and $y'$ are positive, so $\sqrt{x}+\sqrt{y'}>1$, so $(\sqrt{x}+\sqrt{y'})^2$ has at least 3 positive divisors, contradiction.
The only problem here is that I can't show that $\sqrt{y}$ and $\sqrt[4]{x^2y}$ are integers, because $\sqrt[4]{x^2y}=\sqrt{x\sqrt{y}}$, and I cannot guarantee that $x\sqrt{y}$ is an integer, so I can't show that they are both integers.
I would really appreciate it if I can get a hint on this problem, thank you a lot in advance.
| See that $\sqrt[4]y = (\sqrt p - \sqrt x)$ so
$$ y = (\sqrt p -\sqrt x)^4 = p^{2} - 4 \, p \sqrt{px} + 6 \, p x - 4x \, \sqrt{px} + x^{2} =p^2 +6px - \sqrt{px}(4x+4p) +x^2 $$ So you can easily see that $\sqrt{xp}$ is an integer.
Now $\sqrt y = (\sqrt p -\sqrt x)^2 = p + x - 2\sqrt{px}$ so $\sqrt y$ is an integer.
Also $\sqrt{x}\sqrt[4]y = x - \sqrt{px}$ which is an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Q: Epsilon-delta proof for $\lim_{x\to1} \frac{x}{x^2+1}=\frac{1}{2}$ In proving $\lim_{x\to1} \frac{x}{x^2+1}=\frac{1}{2}$, I have done this working:
Let $ϵ>0$.
$0<|x-1|< \Rightarrow|\frac{x}{(x^2+1)}-\frac{1}{2}|<$
$ |\frac{x}{(x^2+1)}-\frac{1}{2}|< = |\frac{2x-x^2+1}{2(x^2+1)}| = |\frac{-2(x-1)^2}{x(x-1)^2+4x}| = |\frac{-(x-1)^2}{(x-1)^2+2(x-1)+2}| = \frac{(x-1)^2}{|(x-1)^2+2(x-1)+2|}|$
after which I get stuck at trying to implement triangle inequality to push in the modulo in the denominator to get $|x-1|$ terms to substitute .
| You have\begin{align}\frac x{x^2+1}-\frac12&=\frac{-x^2+2x-1}{2x^2+2}\\&=\frac{-(x-1)^2}{2x^2+2}.\end{align}Now, you alwas have $2x^2+2\geqslant2>1$. So$$\left|\frac{-(x-1)^2}{2x^2+2}\right|<(x-1)^2\tag1$$and therefore, given $\varepsilon>0$, if you take $|x-1|<\sqrt{\varepsilon}$, it will follow from $(1)$ that$$\left|\frac x{x^2+1}-\frac12\right|<\varepsilon.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to show that $\displaystyle\lim_{x\rightarrow0}\dfrac{a^{2x}-2}{x^x}=-1$ I tried like this:
Let $y=a^{2x}-2\Rightarrow a^{2x}=y+2\Rightarrow 2x\ln a=\ln\left(y+2\right)\Rightarrow x=\dfrac{\ln\left(y+2\right)}{2\ln a}$
Also if $x\longrightarrow0,$ then $y\longrightarrow a^{2(0)}-2=-1.$
But we I put each and every this assumption in the given expression, then I get hanged due to $x^x.$ How to use algebra or any other easy procedure to show that $\displaystyle\lim_{x\rightarrow0}\dfrac{a^{2x}-2}{x^x}=-1$.
| The secret here is to express $\dfrac{a^{2x}-2}{x^x}$ a function of the $\dfrac{e^{y}-1}{y}$, with $y=(2\cdot \ln a)\cdot x$, and function the $x\cdot \ln x$.
Note that
\begin{align}
\dfrac{a^{2x}-2}{x^x}
=&
\dfrac{a^{2x}-1}{x^x}-\dfrac{1}{x^x}
\\
=&
\dfrac{(e^{\ln a})^{2x}-1}{2x}\cdot\dfrac{2x}{x^x}-\dfrac{1}{x^x}
\\
=&
\dfrac{e^{2(\ln a)x}-1}{2x}
\cdot
\dfrac{2x}{e^{x\ln x}}
-
\dfrac{1}{e^{x\ln x}}
\\
=&
\left(
(\ln a)
\cdot\dfrac{e^{2(\ln a)x}-1}{2(\ln a)x}
\cdot
2x
-
1
\right)
\cdot
\frac{1}{e^{x\cdot \ln x}}
\\
\end{align}
We have
*
*$\displaystyle\lim_{x\to 0}\dfrac{e^{2(\ln a)x}-1}{2(\ln a)x}=1$ implies
$\displaystyle\lim_{x\to 0}(\ln a)
\cdot\dfrac{e^{2(\ln a)x}-1}{2(\ln a)x}
\cdot
2x=0$
*$\displaystyle\lim_{x\to 0}x\cdot \ln x=\displaystyle\lim_{x\to 0}\dfrac{\ln x}{\left(\dfrac{1}{x} \right)}=\displaystyle\lim_{x\to 0}\dfrac{D(\ln x)}{D\left(\dfrac{1}{x} \right)}=\displaystyle\lim_{x\to 0}\dfrac{\left(\frac{1}{x} \right)}{\left(-\frac{1}{x^2}\right)}=\displaystyle\lim_{x\to 0}\dfrac{1}{-\left(\frac{1}{x}\right)}=\displaystyle\lim_{x\to 0}-x=0$
implies $\displaystyle\lim_{x\to 0}\dfrac{1}{e^{x\cdot \ln x}}=\dfrac{1}{e^{(\; \lim_{x\to 0}x\cdot \ln x)}}=\dfrac{1}{e^{0}}=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Does Diophantine equation $1+n+n^2+\dots+n^k=2m^2$ have a solution for $n,k \geq 2$? When studying properties of perfect numbers (specifically this post), I ran into the Diophantine equation
$$
1+n+n^2+\dots+n^k=2m^2, n\geq 2, k \geq 2.
$$
Searching in range $n \leq 10^6$, $k \leq 10^2$ yields no solution. So I wonder if there are any solutions, and if not, is there some elementary reason for that? Or if it can be converted to some known open problem, that would do too...
Some thoughts: Of course if we allow $n=1$ or $k=1$ we could get trivial solutions. We can also see by mod $2$ that both $n$ and $k$ must be odd. Now we could try to solve some small cases such as $k=3$ or $k=5$... So set $k=3$ and try to solve $$1+n+n^2+n^3=2m^2.$$ Left side factors and hence we want to solve $(n+1)(n^2+1)=2m^2$. Now this imples $4 \mid 2m^2$ and so $m$ is even. Also we can see that $4 \nmid n^2+1$ for any integer $n$ (becase $n^2\equiv 0,1 \pmod 4 $). So all powers of $2$ in $2m^2$ except one will divide $n+1$. So let $m=2^t r$ with $2 \nmid r$, then $2m^2=2^{2t+1}r^2$, $2^{2t} \mid n+1$, $2 \mid n^2+1$. So we can put $n=2^{2t}s-1$ with $2 \nmid s$, substitute it back, divide all powers of $2$ and we have the equation
$$
s(2^{4t-1}s^2-2^{2t}s+1)=r^2.
$$
Now the two expressions in the product are coprime, so they both have to be square, and that is farthest I got so far. Also the expression $2^{4t-1}s^2-2^{2t}s+1$ being square has a solution $s=15,t=2$, and I am yet to find another (but of course $s$ is not a square in this case so it does nothing for the original problem).
| Here is an elementary proof that $1+n+n^2+n^3=2m^2$ has only integer solution $(n,m)=(-1,0)$. This is the minimal case referred in the general proof and this post gives an alternative without using the elliptic curves.
Since $(n+1)(n^2+1)=2m^2$ and gcd of the expressions on the left is $2$ (by parity argument $n$ must be odd, and $4 \nmid n^2+1$), we thus have $n^2+1$ either square of twice a square. It cannot be a square unless $n=0$ (which does not solve the original equation), hence $n^2+1=2a^2$, and so $n+1=b^2$. As $n$ is odd $b$ is even, say $b=2c$. Thus $n=4c^2-1$ and plugging to the previous equation we get $(4c^2-1)^2+1=2a^2$. However with some manipulation this can be written as:
$$
(2c^2-1)^2+(2c^2)^2=a^2.
$$
This is a primitive Pythagorean triple ($2c^2$ and $2c^2-1$ are coprime). It implies $2c^2=2\alpha \beta$, $2c^2-1=\alpha^2-\beta^2$ for some coprime integers $\alpha, \beta$. But then $c^2=\alpha \beta$ and $\alpha=\gamma^2$, $\beta=\delta^2$ for some integers $\gamma$, $\delta$. Plugging to the other equation we get $2\gamma^2 \delta^2-1=\gamma^4-\delta^4$. We rewrite this as $(\delta^2+\gamma^2)^2=2\gamma^4+1$, now we are ready to use the following lemma:
Lemma. Only integer solutions to $x^2=2y^4+1$ are $(x,y)=(\pm 1,0)$.
Proof. There is an elementary proof on this site already, see Integer solutions to $x^2=2y^4+1$.. It was also the Problem 4 on China IMO Team Selection Test in 1993, see this AoPS thread. Here is a proof based on ideas in linked posts/comments:
Clearly $y=0$ is a solution with $x=\pm 1$, so assume $y \neq 0$. Since $x$ is odd, put $x=2k+1$. The equation becomes $2k(k+1)=y^4$. Hence $y=2z$ and $k(k+1)=8z^4$. Since $\gcd(k,k+1)=1$, one of $k,k+1$ is fourth power and the other is $8$ times a fourth power.
If $k+1 = 8a^4, k = b^4$, then $b^4=8a^4-1 \equiv -1 \pmod 8$, impossible.
If $k+1 = b^4, k = 8a^4$, we can write $8a^4=(b^4-1)=(b^2-1)(b^2+1)$. Since $4 \nmid b^2+1$, it's clear that $\gcd(b^2-1,b^2+1) = 2$ and we have $a^4=\frac{b^2-1}{4} \frac{b^2+1}{2}$. Then $b^2-1=4t^2=(2t)^2$ and so $b=\pm 1$. But then $a=0$, impossible since $y \neq 0$.
So by the Lemma with $x=\delta^2+\gamma^2$ and $y=\gamma$, we have $\gamma=0$ and $\delta^2=1$. By a backwards substitution $\alpha=0, \beta=1$, then $c^2=0$ and finally $n=-1$ is the only solution (with $m=0$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Solving two equations in polar coordinates Problem:
Find the points of intersection of the following two curves.
\begin{align*}
r &= 1 - \cos \theta \\
r^2 &= \cos \theta
\end{align*}
Answer:
Note: While I do not show a plot, I did produce one. It indicated that there were only two solutions.
Now we solve for the points of intersection:
\begin{align*}
r^2 &= r^2 \\
\left( 1 - \cos \theta \right) ^2 &= \cos \theta \\
\left( \cos \theta - 1 \right) ^2 &= \cos \theta \\
\cos^2 \theta - 2 \cos \theta + 1 &= \cos \theta \\
\cos^2 \theta - 3 \cos \theta + 1 &= 0
\end{align*}
\begin{align*}
\cos \theta &= \dfrac{ 3 \pm \sqrt{ 9 - 4(1)(1) } } { 2(1) }
= \dfrac{ 3 \pm \sqrt{ 5 } } { 2 }
\end{align*}
Here are the two values for $\theta$. Notice they are in the first and
fourth quadrant.
\begin{align*}
\theta &= \cos^{-1} \left( \dfrac{ 3 + \sqrt{ 5 } } { 2 } \right) \\
\theta &= \cos^{-1} \left( \dfrac{ 3 - \sqrt{ 5 } } { 2 } \right)
\end{align*}
Since $\cos(-x) = \cos(x)$ it would appear to me that there are four valid
solutions. However, the graph indicates that there are only two valid solutions. Did I do something wrong?
The book's answer is:
$$ \left( \dfrac{-1 + \sqrt{5}}{2},
\cos^{-1}\left( \dfrac{3 - \sqrt{5}}{2} \right) \right),
\left( \dfrac{-1 + \sqrt{5}}{2},
2\pi - \cos^{-1}\left( \dfrac{3 - \sqrt{5}}{2} \right) \right) $$
I am feeling confused and I am hoping somebody can shed some light
on the problem.
| There are indeed only two valid solutions (up to $2\pi$ periodicity). Note that $|\cos(\theta)|\leq 1$ so we discard the case $\cos(\theta)=\frac{3+\sqrt{5}}{2}$. Then $\cos(-x)=\cos(x)$ does not introduce new solutions, since we have
$$\cos(\theta)=\frac{3-\sqrt{5}}{2}\implies\theta=\pm\cos^{-1}\left(\frac{3-\sqrt{5}}{2}\right)+2n\pi,\space n\in\mathbb{Z}$$
$$\cos(-\theta)=\frac{3-\sqrt{5}}{2}\implies\theta=\mp\cos^{-1}\left(\frac{3-\sqrt{5}}{2}\right)+2k\pi,\space k\in\mathbb{Z}$$
and they are the same set of solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Converting sets into intervals with variables I am trying to convert the set to interval form where center is $c$ and the radius is $r$. I have not had any problem doing the first example but I cannot find a way to get the $c$ value for the second question. How would I be able to solve the $c$ for the second question as well as turning it into interval notation?
*
*$ \left|x^2 - 1\right| > 2 $
*$ \left|x^2 - 2x\right| > 2 $
Work
*
*$ \left|x^2 - 1\right| > 2 $
$c= 1$ and $r = 2$
$$(-\infty, \sqrt{c-r})\cup(\sqrt{c+r}, +\infty)$$
$$(-\infty, \sqrt{1-2})\cup(\sqrt{1+2}, +\infty)$$
$$(-\infty, \sqrt{-1})\cup(\sqrt3, +\infty)$$
So the intervals are between $(-\infty, -\sqrt3)\cup(\sqrt3, +\infty)$ for example 1.
*$ \left|x^2 - 2x\right| > 2 $
$c = ?$ and $r = 2$
| It is ideal to solve these problems directly.
First problem
Note that for the first problem, you have to find all $x\in\mathbb{R}$ such that
$$ |x^2 -1| > 2 \tag{1}$$
Let $x\in\mathbb{R}$ be a real number that satisfies (1). By definition of the modulus sign, this condition is equivalent to the statement that
$$
\begin{eqnarray}
&x^2 - 1 > 2 \tag{1a}\\
\text{or } &x^2 - 1 < -2. \tag{1b}
\end{eqnarray}
$$
What remains is for us to solve each of these inequalities, and find all the real values that satisfy (1a) or (1b). Indeed, all the $x$ values that satisfy (1a) are such that
$$
x^2 > 3 \quad \iff \quad x > \sqrt{3} \text{ or }x<-\sqrt{3}.
$$
Regarding the second inequality (1b) $ x^2 < -1 $, no real number satisfies this constraint, and thus the only real values that satisfy the initial inequality $|x^2 - 1| > 2$ are such that $x\in(-\infty,-\sqrt{3})\cup(\sqrt{3},\infty)$, i.e.
$$
|x^2 - 1| > 2 \quad \iff \quad x\in(-\infty,-\sqrt{3})\cup(\sqrt{3},\infty).
$$
Second problem:
For the second problem, one needs simply to apply the same procedure. We have to find all $x\in\mathbb{R}$ such that
$$
|x^2 - 2x| > 2. \tag{2}
$$
Assume that $x\in\mathbb{R}$ is a real number that satisfies (2). Then by definition of the modulus sign, (2) is equivalent that $x$ satisfies either of the following conditions:
$$
\begin{eqnarray}
&x^2 - 2x > 2 \tag{2a} \\
\text{or } &x^2 - 2x < -2. \tag{2b}
\end{eqnarray}
$$
Solving the first inequality (2a), we find that
$$
x^2 - 2x > 2 \quad \iff \quad (x - 1)^2 > 3 \quad \iff \quad x-1 > \sqrt{3} \text{ or } x - 1 < -\sqrt{3}
$$
That is, $x > 1 + \sqrt{3}$ or $x < 1 - \sqrt{3}$.
A similar approach can be done for the second inequality (2b). Indeed, we need to find all the $x$ reals such that
$$
x^2 - 2x < -2.
$$
Indeed, we have that
$$
x^2 - 2x < -2 \quad \iff \quad x^2 - 2x + 2 < 0 \quad \iff \quad (x-1)^2 + 1 < 0,
$$
but this is a sum of two positive real numbers, and so no real numbers that satisfy this second inequality exist.
In conclusion, we have that
$$
|x^2 - 2x| > 2 \quad \iff \quad x\in(-\infty,1-\sqrt{3})\cup(1+\sqrt{3},\infty).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to color a cube with exactly 6 colors using Polya enumeration theory I have seen in other questions (Painting the faces of a cube with distinct colours) , and found for myself there are 30 colorings of the cube with exactly 6 colors. My issue is when I use Polya enumeration theory I take the number of colorings up to 6 to be $\frac{1}{24}(6^6+3*6^4+12*6^3+8*6^2)$ (https://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem) and subtract the number of colorings up to 5, $\frac{1}{24}(5^6+3*5^4+12*5^3+8*5^2)$ to get 1426. What is wrong with my application of Polya enumeration theory? How could I fix it?
| To find the number of ways to color the faces of a cube with exactly $6$ colors using Polya's Enumeration Theorem:
First, as you already know, the cycle index of the symmetries of the face permutations of a cube is
$$Z = \frac{1}{24}(x_1^6 + 6x_1^2x_4+3x_1^2x_2^2 + 8x_3^2 + 6x_2^3)$$
The figure inventory for $6$ colors represented by $a,b,c,d,e,f$ is
$$a+b+c+d+e+f$$
"Substituting" the figure inventory into the cycle index, we have
$$\frac{1}{24} \left((a+b+c+d+e+f)^6 \\ \\+6 (a+b+c+d+e+f)^2
\left(a^4+b^4+c^4+d^4+e^4+f^4\right) \\+3 (a+b+c+d+e+f)^2
\left(a^2+b^2+c^2+d^2+e^2+f^2\right)^2\\+8
\left(a^3+b^3+c^3+d^3+e^3+f^3\right)^2 \\ +6 \left(a^2+b^2+c^2+d^2+e^2+f^2\right)^3 \right) $$
The number of ways to color the cube using each of the six colors exactly once is the coefficient of $abcdef$ when this polynomial is expanded. In difficult cases I would use a computer algebra to find the coefficient, but in this example it's easy to see that what we want is $1/24$ the coefficient of $abcdef$ in $(a+b+c+d+e+f)^6$, i.e.
$$\frac{6!}{24} = 30$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find value of $x^{5} + y^{5} + z^{5}$ given the values of $x + y + z$, $x^{2} + y^{2} + z^{2}$ and $x^{3} + y^{3} + z^{3}$ If$$x+y+z=1$$
$$x^2+y^2+z^2=2$$
$$x^3+y^3+z^3=3$$
Then find the value of $$x^5+y^5+z^5$$
Is there any simple way to solve this problem ? I have tried all my tricks tried to multiply two equations , substitute $z=1-x-y$ , but things got messy nothing seems to work out .
| This form of equations can be solved systematically using Newton's identities.
The problem at hand is a special case of $3$ variables. Let $e_1,e_2,e_3$ be the three elementary symmetric polynomial associated with $x,y,z$:
$$\begin{align}
e_1 &= x + y + z\\
e_2 &= xy+ yz + zx\\
e_3 &= xyz
\end{align}$$
This is equivalent to $\displaystyle\;(\lambda - x)(\lambda -y)(\lambda -z) = \lambda^3 - e_1\lambda^2 + e_2\lambda - e_3$.
For any integer $k > 0$, let $p_k = x^k + y^k + z^k$.
For 3 variables, the Newton's identities are:
$$\begin{align}
p_1 &= e_1\\
p_2 &= e_1 p_1 - 2 e_2\\
p_3 &= e_1 p_2 - e_2 p_1 + 3e_3\\
p_n &= e_1 p_{n-1} - e_2 p_{n-2} + e_3 p_{n-3} \quad\text{ for }\; n > 3
\end{align}$$
Given the known values of $(p_1,p_2,p_3) = (1,2,3)$, we have
$$\begin{align}
e_1 &= p_1 = 1;\\
e_2 &= -\frac12(p_2 - e_1p_1) = -\frac12\left(2 - 1^2\right) = -\frac12\\
e_3 &= \frac13(p_3 - e_1p_2 + e_2p_1) = \frac13\left(3 - 2 - \frac12\right) = \frac16
\end{align}$$
Substitute this into formula of $p_4$ and $p_5$, we have
$$\begin{align}
p_4 &= e_1 p_3 - e_2 p_2 + e_3 p_1 = 3 + \frac12\cdot 2 + \frac16\cdot 1 = \frac{25}{6}\\
p_5 &= e_1 p_4 - e_2 p_3 + e_3 p_2 = \frac{25}{6} + \frac12\cdot 3 + \frac16\cdot 2 = 6
\end{align}$$
The answer is $x^5 + y^5 + z^5 = 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4256912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Probability of One Event Less Than Probability of Second Event I am having a bit of trouble proving some cases when one probability is smaller than the other probability for all positive integers $a, b$, so some suggestions would be appreciated.
Here is the problem:
For all $a, b \in \mathbb{Z}^+$, if $P(A) = \dfrac{a^2 - a + b^2 - b}{a^2 + 2ab + b^2 - a - b}$ and $P(B) = \dfrac{a^2 + b^2}{a^2 + 2ab + b^2}$, prove that $P(A) < P(B)$.
So I attempted using cases.
Case 1: If $a = b \neq 0$, then we want to show that $P(A) < P(B)$
\begin{align*}
P(A) = \dfrac{a^2 - a + b^2 - b}{a^2 + 2ab + b^2 - a - b} = \dfrac{2a^2 - 2a}{4a^2 - 2a} < \dfrac{a^2 + b^2}{a^2 + 2ab + b^2} = \dfrac{2a^2}{4a^2} = \dfrac{1}{2} = P(B)
\end{align*}
Thus, this implies that $P(A) < P(B)$ for all $a, b \in \mathbb{Z}^+$
The next two cases are the cases I am having trouble with.
Case 2: If $a > b > 0$, then we want to show that $P(A) < P(B)$.
Case 3: If $b > a > 0$, then we want to show that $P(A) < P(B)$.
I am not sure how to approach cases 2 and 3. But case 3 should follow from case 2. So some assistance would be appreciated. Thanks
| Incomplete solution:
$$
P(A)=P(B)-\left[1-P(B)\right]\cdot\frac{a+b}{a^{2}+2ab+b^{2}-a-b}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solving or estimating $\int_{-\infty}^\infty \frac{\exp(-a/(x^2+b))}{x^2+c}dx$ Can the integral
$$\int_{-\infty}^\infty \frac{\exp(-a/(x^2+b))}{x^2+c}dx$$
be made explicit ($a,b,c>0$)? I'm also asking those of you who have access to CAS which can solve it.
In case it can't, is there a very good upper bound? I need a better one than the obvious bound
$$\int_{-\infty}^\infty \frac{1}{x^2+c}dx = \frac{\pi}{\sqrt{c}}$$
Thanks a lot!
| Hint:
$$\begin{align}\int_{-\infty}^\infty\dfrac{e^{-\frac{a}{x^2+b}}}{x^2+c}~dx&=2\int_0^\infty\dfrac{e^{-\frac{a}{x^2+b}}}{x^2+c}~dx
\\&=2\int_0^\frac{\pi}{2}\dfrac{e^{-\frac{a}{b\tan^2x+b}}}{b\tan^2x+c}~d(\sqrt b\tan x)
\\&=2\sqrt b\int_0^\frac{\pi}{2}\dfrac{e^{-\frac{a}{b\sec^2x}}\sec^2x}{b\sec^2x+c-b}~dx
\\&=2\sqrt b\int_0^\frac{\pi}{2}\dfrac{e^{-\frac{a\cos^2x}{b}}}{b+(c-b)\cos^2x}~dx
\\&=2\sqrt b\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^na^n\cos^{2n}x}{b^n(b+(c-b)\cos^2x)}~dx\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4261677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find the sequence $a_k$ for generating function $\left(\frac{1-x^3}{1-x}\right)^n$ We know that $\frac{1}{(1-x)^n} = \sum_{k=0}^\infty \binom{k+n-1}{n-1} x^k$
I also worked out $(1-x^3)^n$ using the binomial theorem and got $(1-x^3)^n = \sum_{i=0}^\infty (-1)^i \binom{n}{n-i} x^{3i}$
I'm not sure what to do with these to get $a_k$ from $\sum_{k=0}^\infty a_k x^k$ or if these are even what I need to solve the problem
Any help is appreciated
| Hint:
\begin{align*}
\left(\frac{1-x^3}{1-x}\right)^n=\left(\frac{(1-x)\left(1+x+x^2\right)}{1-x}\right)^n=\left(1+x+x^2\right)^n
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4269464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What did I do wrong in my proof of $4\nmid (n^2+3)\implies 2\nmid (n^4-3)$? A homework problem is asking me to prove the following statement:
For any $n\in\mathbb{Z}$, $2\mid (n^4-3)$ if and only if $4\mid (n^2+3)$
Quick note: we're not allowed to use the theory of congruences in our solution.
I figured the best way to approach this was to prove that $4\mid (n^2+3)$ implies $2\mid (n^4-3)$ and $4\nmid (n^2+3)$ implies $2\nmid (n^4-3)$.
The proof of "$4\mid (n^2+3)$ implies $2\mid (n^4-3)$" was not an issue. It's proving "$4\nmid (n^2+3)$ implies $2\nmid (n^4-3)$" where I had some trouble. Here's my argument so far:
If $4\nmid (n^2+3)$, then $n^2+3=4k+r$ for some integer $k$, where $r$ is either $1$, $2$, or $3$. This is equivalent to
$$n^2=4k+r-3$$
so
\begin{align}
n^4 &= (4k+r-3)^2\\
&= 16k^2+8k(r-3)+(r-3)^2\\
\end{align}
which implies that
$$n^4-3=16k^2+8k(r-3)+(r-3)^2-3$$
Since $r$ is either $1$, $2$, or $3$, we have that
$$n^4-3=16k^2+8k(1-3)+(1-3)^2-3=16k^2-16k+1\text{,}$$
$$n^4-3=16k^2+8k(2-3)+(2-3)^2-3=\color{red}{16k^2-8k-2}\text{,}$$
$$\text{or}$$
$$\vdots$$
I stopped here because the part in red is problematic. Factoring out a $2$, it can be seen that $n^4-3$ is even in this case. I can't have this because I'm trying to show that $2\nmid (n^4-3)$, so what did I do wrong? I checked my algebra over and over again, and I can't seem to find a mistake anywhere. I also graphed $(4k+r-3)^2-3$ with $16k^2+8k(r-3)+(r-3)^2-3$ on Desmos for $r=1,2,3$ and the graphs seem identical.
In light of this, it seems like either (1) I'm blind, (2) I made a subtle mistake in my assumptions, or (3) there's something deeper going on that I'm failing to recognize. Am I seeing everything? Are my assumptions for the proof correct? If the answer to both of these is yes, does this mean that dividing $n^2+3$ by $4$ will never yield a remainder $r=2$? If not, what's going on here? Any help is appreciated.
| You already noted that if $n^2+3= 4k+2$ then it's a problem. So we should show that $r=2$ is not possible.
Now that $2|n^2+3\implies n$ is odd. But odd squares are of the form $4l+1.$
Hence $n^2+3=4l+4.$
So if $2|n^2+3 $ then $4|n^2+3$ too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4269779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to calculate arccos of complex number I want to know what can I do to calculate the acos of a complex number. I looked at this answer on this site but that only work for imaginary numbers.
Given the complex number:
$$ \arccos(2 + 3i) $$
How can I calculate it?
Thanks in advance!
| $z=\arccos(2+3i)$
$2+3i=\cos z=(e^{iz}+e^{-iz})/2$; $e^{iz}+e^{-iz}=4+6i$; $u^2-(4+6i)u+1=0$ where $u=e^{iz}$; $$u={4+6i\pm\sqrt{(4+6i)^2-4}\over2}={4+6i\pm\sqrt{-24+48i}\over2}=2+3i\pm\sqrt{-6+12i}$$ so we need to find $\sqrt{-6+12i}$.
Let $\sqrt{-6+12i}=r+si$. Then $-6+12i=r^2-s^2+2rsi$, so $$r^2-s^2=-6,\qquad 2rs=12$$ Then $$(r^2+s^2)^2=(r^2-s^2)^2+(2rs)^2=36+144=180$$ so $r^2+s^2=\sqrt{180}=6\sqrt5$. Then $2r^2=6\sqrt5-6$ and $2s^2=6\sqrt5+6$, so $$r=\sqrt{3\sqrt5-3},\qquad s=\sqrt{3\sqrt5+3}$$ So $e^{iz}=u=2+3i\pm(\sqrt{3\sqrt5-3}+\sqrt{3\sqrt5-3}i)$, and $z=-i\log u$.
But how do we find $\log u$, when $u$ is a complex number? Let's do it in general. Write $a+bi=\sqrt{a^2+b^2}e^{i\arctan(b/a)}$. Then $\log(a+bi)=(1/2)\log(a^2+b^2)+i\arctan(b/a)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4270736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the partial fraction: $\frac{x^2}{(x-1)^2(x-2)^2(x-3)}$ I'm trying to find the partial fraction for the following $\frac{x^2}{(x-1)^2(x-2)^2(x-3)}$ by the process of long-division, here's what I have tried:
Leaving $(x-3)$ as it is in the denominator and aiming for $(x-1)$ and $(x-2)$.
First step:
$\frac{1^2}{(x-1)^2(1-2)^2(1-3)} = \frac{1}{-2(x-1)^2}, \frac{2^2}{(2-1)^2(x-2)^2(2-3)} = \frac{4}{-(x-2)^2}$
We should then substract this from the first equation to get:
$\frac{x^2}{(x-1)^2(x-2)^2(x-3)} - \frac{1}{-2(x-1)^2} - \frac{4}{-(x-2)^2} = \frac{9x^3-36-45x^2+72x}{2\left(x-1\right)^2\left(x-2\right)^2\left(x-3\right)}$
Taking synthetic division on the numerator I find that 2 and 3 is not a root as there exists remainders. Have I approached this step in the wrong way?
| You've made some error in your simplification.
$$
\frac{x^2}{(x-1)^2(x-2)^2(x-3)} - \frac{1}{-2(x-1)^2} - \frac{4}{-(x-2)^2}$$
$$
\begin{align*}
&= \frac{x^2+\frac12(x-2)^2(x-3)+4(x-1)^2(x-3)}{(x-1)^2(x-2)^2(x-3)} \\
&= \frac{9x^3-45x^2+72x-36}{2(x-1)^2(x-2)^2(x-3)} \\[1ex]
&= \frac{9(x^3-5x^2+8x-4)}{2(x-1)^2(x-2)^2(x-3)} \\[1ex]
&= \frac{9(x-1)(x-2)^2}{2(x-1)^2(x-2)^2(x-3)} \\[1ex]
&= \frac{9}{2(x-1)(x-3)}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4271461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Expanding $(1-i)^\frac{1}{3}$ using De Moivre's formula I want to rewrite $(1-i)^ \frac 1 3$ using de Moivre's formula.
I defined $z := 1 - i$, then $r_z = \sqrt{2}$ and
$1 = \sqrt2\cos\theta$ and $-1 = \sqrt2 \sin\theta \Rightarrow \theta = -\frac \pi 4$
So:
$$1 - i = \sqrt{2}(\cos(\frac \pi 4) + i\sin(\frac \pi 4))$$
$$(1-i)^\frac 1 3 = 2^ \frac 1 6 (\cos(\frac \pi {12} + i\sin(\frac \pi {12}))$$
Am I correct with this derivation?
| You have only take the principal root, indeed we have (there is also a typo for the angle in the $\sin$ term):
$$1 - i = \sqrt{2}\left(\cos\left(\frac \pi 4+2n\pi\right) + i\sin\left(-\frac \pi 4+2n\pi\right)\right)$$
and therefore the three possible roots are
$$2^ \frac 1 6 \left(\cos\left(\frac \pi {12}\right) + i\sin\left(-\frac \pi {12}\right)\right)$$
$$2^ \frac 1 6 \left(\cos\left(\frac \pi {12}+\frac 2 3 n\right) + i\sin\left(-\frac \pi {12}+\frac 2 3 n\right)\right)$$
$$2^ \frac 1 6 \left(\cos\left(\frac \pi {12}+\frac 4 3 n\right) + i\sin\left(-\frac \pi {12}+\frac 4 3 n\right)\right)$$
with $n=0,1,2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4273796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show that $1998 < \sum_1^{10^6}\frac{1}{\sqrt{n}}<1999$ Show that $1998 < \sum_1^{10^6}\frac{1}{\sqrt{n}}<1999$ by first recalling the inequality:
$$2(\sqrt{n+1}-\sqrt{n})<\frac{1}{\sqrt{n}}<2(\sqrt{n}-\sqrt{n-1})$$ when $n=1, 2, 3...$
What I have tried to show using the inequality:
By the first step of induction to see if the inequality holds :
$2(\sqrt{1+1}-\sqrt{1})<\frac{1}{\sqrt{1}}<2(\sqrt{1}-\sqrt{1-1})$ is true
Then by induction we have:
$2(\sqrt{n+1+1}-\sqrt{n+1})<\frac{1}{\sqrt{n+1}}<2(\sqrt{n+1}-\sqrt{n+1-1})$
Then by using the fact that if $a>b$ and $c>0$ then $ac>bc$
The following should be true if the induction is right.
$\frac{2(\sqrt{n+1}-\sqrt{n})}{\sqrt{n+1}}>2(\sqrt{n+1+1}-\sqrt{n+1})$ by taking the square of both sides we have after expanding:
$$\frac{8n-8\sqrt{n^2+n}+4}{n+1}>8n+12-8\sqrt{n^2+3n+2}$$
Then placing everything onto one side:
$$\frac{-8n^2+8n\sqrt{n^2+3n+2}-12n+8\sqrt{n^2+3n+2}-8\sqrt{n^2+n}-8}{n+1}>0$$
Thouigh I'm not sure whether I'm going down the right direction. I had assumed that the LHS would cancel to provide something more exact. Any detailed hint son how to proceed?
| You don't need induction here. We have
$$n(n+1)=n^2+n<n^2+n+\frac14=\left(n+\frac12\right)^2$$
Take square roots:
$$\sqrt n\sqrt{n+1}<n+\frac12$$
Multiply by $2/\sqrt n$ and rearrange:
$$2(\sqrt{n+1}-\sqrt n)<\frac{1}{\sqrt n}$$
Do the same with $n(n-1)$ to get the right-hand side of the inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving $2\tan^{-1}\left(\sqrt{\frac{a}{b}}\tan\frac{x}{2}\right)=\sin^{-1}\frac{2\sqrt{ab}\sin x}{\left(b+a\right)+\left(b-a\right)\cos x}$ $$
2\tan ^{-1}\left(\sqrt{\frac{a}{b}}\tan \frac{x}{2}\right)=\sin ^{-1}\frac{2\sqrt{ab}\sin x}{\left(b+a\right)+\left(b-a\right)\cos x}
$$
I know within inverse of trigonometric function we have the value.
How Do I solve this.
my approach for this solution was:
1st:
$$
\sin ^{-1}\dfrac{2\sqrt{\dfrac{a}{b}}\tan \dfrac{x}{2}}{1+\left( \sqrt{\dfrac{a}{b}}\tan \dfrac{x}{2}\right) ^{2}}
$$
$$
\sin ^{-1}\dfrac{2\dfrac{\sqrt{a}}{\sqrt{b}}\tan \dfrac{x}{2}}{1+\dfrac{a}{b}\tan ^{2}\dfrac{n}{2}}
$$
$$
\sin \dfrac{2\sqrt{a}\tan \dfrac{x}{2}}{\dfrac{\sqrt{b}\left( b+atan ^{2}\dfrac{x}{2}\right) }{b}}
$$
$$
\sin ^{-1}\dfrac{2\sqrt{a}\tan \dfrac{x}{2}\sqrt{b}}{b+atan ^{2}\dfrac{x}{2}}
$$
$$
\sin ^{-1}\dfrac{2\sqrt{ab}\tan \dfrac{x}{2}}{b+atan ^{2}\dfrac{x}{2}}
$$
$$
\sin ^{-1}\dfrac{2\sqrt{ab}\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}}{b+a\dfrac{\sin ^{2}\dfrac{x}{2}}{\cos ^{2}\dfrac{x}{2}}}
$$
$$
=\sin ^{-1}\dfrac{2\sqrt{ab}\sin \dfrac{x}{2}\cos \dfrac{x}{2}}{\cos ^{2}\dfrac{x}{2}b+asin ^{2}\dfrac{x}{2}}
$$
After this I have tried to approach using
$$
\sin ^{2}\theta +\cos ^{2}\theta =1
$$
formula but could not reach the Right Hand Side.
| This should be the next step of your attempt: $$\sin ^{-1}\dfrac{\sqrt{ab}\sin x}{b\left(\frac{1+\cos x}{2}\right)+a\left(\frac{1-\cos x}2\right)}$$ using half-angle formula.
The result follows immediately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve $\int_{0}^{1} \frac{(x-1)\ln x}{(x+1)(x^2+2\cosh \alpha+1)} \,dx$ I have this unpleasant integral that appear in my last exam which i was not able solve
For $\alpha \in \mathbb{R}$. Evaluate $$I=\int_{0}^{1} \frac{(x-1)\ln x}{(x+1)(x^2+2\cosh \alpha+1)} \,dx$$
Anyway my attempt is that I try to do some partial fraction on the integral
$$\frac{(x-1)}{(x+1)(x^2+2\cosh \alpha+1)} = \frac{(x-1)}{(x+1)(x+\cosh \alpha-\sinh \alpha)(x+\cosh \alpha+\sinh \alpha)}$$
$$\frac{(x-1)}{(x+1)(x+\cosh \alpha-\sinh \alpha)(x+\cosh \alpha+\sinh \alpha)}=\frac{a}{x+1}+\frac{b}{x+\cosh \alpha-\sinh \alpha}+\frac{c}{x+\cosh \alpha+\sinh \alpha}$$
And I find that
$$a=\frac{1}{\cosh \alpha-1}, b=c=\frac{1}{2-2\cosh \alpha}$$
That's all I could do. I still don't know what to do next
| It is not so unpleasant. To make things easier (at least to me), consider
$$\frac{x-1}{(x+1)(x^2+k)}=\frac{x-1}{(x+1) (x-a) (x-b)}$$ and, as you did, using partial fraction decomposition, this becomes
$$\frac{a-1}{(a+1) (a-b) (x-a)}-\frac{2}{(a+1) (b+1) (x+1)}+\frac{b-1}{(b+1) (b-a) (x-b)}$$ One integration by parts leads to
$$\int\frac {\log(x)}{x+c} dx=\text{Li}_2\left(-\frac{x}{c}\right)+\log (x) \log \left(1+\frac{x}{c}\right)$$
$$\int_0^1\frac {\log(x)}{x+c} dx=\text{Li}_2\left(-\frac{1}{c}\right)$$ Combining all of the above
$$I=\int_0^1\frac{(x-1)\log(x)}{(x+1) (x-a) (x-b)}=$$
$$\frac{6 (a-1) (b+1) \text{Li}_2\left(\frac{1}{a}\right)-6 (a+1) (b-1)
\text{Li}_2\left(\frac{1}{b}\right)+\pi ^2 (a-b)}{6 (a+1) (b+1) (a-b)}$$
As you found
$$a=-e^{-\alpha} \qquad \text{and} \qquad b=-e^{\alpha}$$ Replacing leads to
$$I=-\frac{e^{\alpha } \left(6 \text{Li}_2\left(-e^{-\alpha }\right)+6
\text{Li}_2\left(-e^{\alpha }\right)+\pi ^2\right)}{6 \left(e^{\alpha
}-1\right)^2}$$ which looks like a gaussian.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4277210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to get the maximum value of the sum in two sides in two right triangles? The problem is as follows:
The figure from below shows two triangles intersected on point $E$. Assume $AE=3\,m$ and $ED=1\,m$. Find the maximum value of $AB+EC$
The given choices in my book are as follows:
$\begin{array}{ll}
1.&\sqrt{10}\,m\\
2.&\4\,m\\
3.&\sqrt{5}\,m\\
4.&\4\,m\\
\end{array}$
The official solution for this problem according to my book is shown below:
Let $M=AB+EC$
Notice on the figure:
$AB=3\sin x$
and $EC=\cos x$
$M=3\sin x+\cos x$
$M=\sqrt{10}\left(\frac{3}{\sqrt{10}}\sin x+\frac{1}{\sqrt{10}}\cos x\right)$
$\tan \alpha =\frac{1}{3}$
$M=\sqrt{10}\left(\sin (x+\alpha)\right)$
Also notice that $x$ and $\alpha$ are acute angles.
Therefore:
$0<x+\alpha<\pi$
$\sin (x+\alpha)\leq 1$
Thus the maximum $M$ is $\sqrt{10}$.
There is where it ends the official solution.
But I am confused, where exactly is that alpha that is being talking about?, how did the author came up with the idea of $\sqrt{10}$ and the use of $\frac{1}{3}$.
Could someone please explain me, the logic that the author used and offer an alternate solution, perhaps easier to understand and less to guess what it was meant?.
I appreciate a step by step solution so I can understand.
| Let $AM=CD$ and $AN=DE$
Since $\angle AMN=\angle CDE$,(cyclic quadrilateral), $\triangle AMN \cong \triangle DCE$, then, $AN=ED=1$
Let $AB=x$, and, $EC=MN=y$
Since $\triangle ABE \sim \triangle AMN$, (AAA Similarity), $BE=3y$
FIRST ALTERNATIVE SOLUTION:
$x^2+9y^2=9$, in $\triangle ABE$
$x=\sqrt{9-9y^2}$
To find max$(x+y)$
$f(y)=\sqrt{9-9y^2}+y$
$f'(y)=\frac{-9y}{\sqrt{9-9y^2}}+1=0$
There is a maximum when $y=\frac{1}{\sqrt{10}}$ ,and, $x=\frac{9}{\sqrt{10}}$
$\text{max}(AB+EC)=\text{max}(x+y)=\frac{10}{\sqrt{10}}=\sqrt{10}$
SECOND ALTERNATIVE SOLUTION:
$x^2+9y^2=9$, in $\triangle ABE$
To find max$(x+y)$
Applying Cauchy-Schwarz inequality
$$(x+y)^2=(x+\frac{1}{3}\times 3y)^2\leq (1^2+(\frac{1}{3})^2)\times (x^2+(3y)^2)=$$
$$=(1+\frac{1}{9})\times (x^2+9y^2)=$$
$$=(1+\frac{1}{9})\times 9=10$$
$$x+y\leq\sqrt{10}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4283209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Children's Fruit Division How many ways can $11$ apples and $9$ pears be divided between 4 children so that each child receives five fruits? (Apples are identical. just like pears).
Solution: $f\left(x,y\right)=\left(x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5\right)^4$
$f\left(x,y\right)=\left(\frac{x^6-y^6}{x-y}\right)^4$
$f\left(x,y\right)=\left(x^6-y^6\right)^4{\cdot\left(x-y\right)}^{-4}$
$f\left( x,y \right)={{\left( {{x}^{6}}-{{y}^{6}} \right)}^{4}}\cdot {{x}^{-4}}{{\left( 1-\frac{y}{x} \right)}^{-4}}$
$f\left(x,y\right)=\left(x^{24}-4x^{18}y^6+{6x}^{12}y^{12}-{4x}^6y^{18}+y^{24}\right)\cdot\sum_{k=0}^{\infty}\binom{3+k}{k}x^{-4-k}y^k$
The coefficient of $x^{11}y^9$ in $f\left(x,y\right)=\binom{3+9}{9}-4\binom{3+3}{3}=140.$
I'm right?
| Yes that is correct. But a simpler solution exists given each kid must get equal number of fruits. We can first distribute $9$ pears (or $11$ apples) to four children such that none of them get more than $5$ pears. We then note that as each kid must have five fruits, once we distribute pears, the distributions of apples are fixed.
Unrestricted number of ways to distribute $9$ pears using stars and bars method: $ \displaystyle {{9+4-1} \choose {4-1}} = {12 \choose 3}$
Now we subtract distributions where a kid would have received more than five pears. We choose a kid, assign $6$ pears and then distribute rest $3$ pears among them.
That is given by, $ \displaystyle 4 \cdot {3 + 4 - 1 \choose 4 - 1} = 4 \cdot {6 \choose 3}$
So number of ways to distribute $9$ pears such that no kid receives more than $5$ pears,
$ \displaystyle = {12 \choose 3} - 4 \cdot {6 \choose 3} = 140$
So, number of ways to distribute $11$ apples and $9$ pears such that each kid receives five fruits is also $140$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4284738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 1,
"answer_id": 0
} |
Is there a natural number $n$ for which $\sqrt[n]{22-10\sqrt7}=1-\sqrt7$ Is there a natural number $n$ for which $$\sqrt[n]{22-10\sqrt7}=1-\sqrt7$$
My idea was to try to express $22-10\sqrt7$ as something to the power of $2$, but it didn't work $$22-10\sqrt7=22-2\times5\times\sqrt7$$ Since $5^2=25, \sqrt7^2=7$ and $25+7\ne22$. What else can we try?
| To paraphrase the excellent answer given in the comments:
Because $\sqrt{7}$ is a pure quadratic surd, if there is such a natural number $n$, then also
$$\sqrt[n]{22+10\sqrt{7}}=1+\sqrt{7}.$$
As $(1+\sqrt{7})(1-\sqrt{7})=1^2-\sqrt{7}^2=-6$ it follows that
\begin{eqnarray*}
-6&=&(1+\sqrt{7})(1-\sqrt{7})\\
&=&\sqrt[n]{22+10\sqrt{7}}\sqrt[n]{22-10\sqrt{7}}\\
&=&\sqrt[n]{(22+10\sqrt{7})(22-10\sqrt{7})}\\
&=&\sqrt[n]{22^2-10^2\sqrt{7}^2}\\
&=&\sqrt[n]{-216}.
\end{eqnarray*}
It is then clear that the only remaining candidate is $n=3$, and a routine verification shows that
$$(1-\sqrt{7})^3=22-10\sqrt{7}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4285350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Possible division by zero in integration by u-substitution When I was solving the indefinite integral $\int x^2(x^3-7)^3\ dx$, my working was as follows:
Let $u=x^3-7$, then $dx=\frac{1}{3x^2}du$.
Hence, $\int x^2u^3\frac{1}{3x^2}\ du=\frac{1}{3}\int u^3\ du=\frac{u^4}{12}+C=\frac{(x^3-7)^4}{12}+C$.
However, during the u-substitution we get $dx=\frac{1}{3x^2}du$. If $x=0$, then wouldn't that mean we would be diving by zero?
So when we go ahead to find a definite integral with $x=0$ in the limit, why is it okay to ignore the fact that we are dividing by zero?
| You may avoid dividing by zero if instead of $$dx=\frac{1}{3x^2}du$$ consider $$ du = 3x^2 dx $$ and substitute $$ x^2dx = \frac {1}{3} du$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4286234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
When does $\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=2mx+4$ have $2$ real solutions? $$\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=2mx+4$$
My solution:
I consider only real numbers, so $|x|\geq1$. After adding the fractions I get the following quadratic equation:
$$2x^2-mx-3=0,$$ and it has 2 real solutions if $D>0$, so $m^2+24>0$.
It comes down to every real number, but the answer in the book states that there is no real m value for which the main equation has 2 real solutions, and no detailed explanation has been included. I kindly ask for explanation.
|
I consider only real numbers, so $|x|\geqslant1.$ After adding the fractions I get the following quadratic equation: $$4x^2-mx+1=0$$
Can you show us the steps of how you got to that equation? Because that is not the equation I got. In general, when calculating $$\frac{a+b}{a-b}+\frac{a-b}{a+b},$$
you have $$\begin{align}
\frac{a+b}{a-b}+\frac{a-b}{a+b} &= \frac{(a+b)(a+b)}{(a-b)(a+b)}+\frac{(a-b)(a-b)}{(a-b)(a+b)} \\&= \frac{a^2+2ab+b^2+a^2-2ab+b^2}{a^2-b^2} \\&= \frac{2(a^2+b^2)}{a^2-b^2}\end{align}$$
and this expression, in your case, when $a=x$ and $b=\sqrt{x^2-1}$, simplifies to something rather simple.
Edit:
Yes, the equation is indeed simplified down to $2x^2-2mx-3=0$, your solution looks correct to me. Also, the book's solution is clearly wrong, because for $m=0$, the equation clearly does have two solutions, and they are $x_1=\sqrt{\frac32},x_2=-\sqrt{\frac32}$.
You can see for yourself (or use Wolfram alpha, see here and here) that when $x$ is any of those two values, then the left side of your expression becomes $4$, and thus, it solves your equation if $m=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4286679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Given triangle ABC with $\alpha=2\beta$ and $b, c, a$ is forming an arithmetic sequence. Find $\alpha, \beta, \gamma$. Given $\triangle ABC$ with $\alpha=2\beta$ and $b, c, a$ is forming an arithmetic sequence ($b$ is the $1$'st term, and $a$ is the last term), find $\alpha, \beta, \gamma$.
My attempts so far:
Let $c=b+k$ and $a=b+2k$. Then, from $\alpha=2\beta$, I got $\gamma = 180^\circ - 3\beta$.
With the Law of Sines , I can write
$$\dfrac{b}{\sin \beta}=\dfrac{b+k}{\sin 3\beta}=\frac{b+2k}{\text{sin } 2\beta}$$
From $\dfrac{b}{\sin\beta}=\dfrac{b+k}{\sin 3\beta}$, it gives $\cos 2\beta = \dfrac{k}{2b}$.
From $\dfrac{b}{\sin \beta}=\dfrac{b+2k}{\sin 2\beta}$, it gives $\cos \beta = \dfrac{b+2k}{2b}$.
After this, I don't know what should I do. Is there any other theorem that can be used to solve this problem?
| From $b,c,a$ forming an arithmetic sequence,
$$c-b=a-c$$ $$2c=a+b$$ Applying sine rule, $$2\sin C=\sin A+\sin B$$
Using the given data $A=2B$, we can rewrite the above as, $$2\sin 3B=\sin2B+\sin B$$ $$\require{cancel}2\cdot \cancel{2\sin\frac{3B}{2}}\cos\frac{3B}2=\cancel{2\sin\frac{3B}{2}}\cos\frac B2\qquad\small{(\because\sin\frac{3B}{2}\neq0})$$
Now letting $\cos\frac B2=x$, we get, $$2(4x^3-3x)=x$$ $$x^2=\frac78$$ Therefore, $$\cos B=2\cos^2\frac B2-1=\frac34$$ $$B=\cos^{-1}\frac34$$
So we have found $B$ alias $\beta$ which leads to find other angles as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4287641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
In equation involving absolute value one solution might get rejected, why is that? For example the equation $|2x-1|+3=4x$,
We can get two equations out of that, $2x-1=4x-3$ and $2x-1=-(4x-3)$. Solve the first one we get $x=1$ and for the second one $x=2/3$. But when you substitute 2/3 into the original equation, you get 10/3=8/3, which is clearly wrong. My answer sheet says the answer is rejected, but if we follow all steps correctly and get to a solution, how can it still be wrong? It's like solving $5=3+x$ and x somehow doesn't equal to two.
| The absolute value of a number cannot be negative.
Observe that $2x - 1 \geq 0 \implies x \geq \dfrac{1}{2}$.
Hence, $|2x - 1| = 2x - 1$ if $x \geq 1/2$. On the other hand, if $x < 1/2$, then $2x - 1 < 0$, so $|2x - 1| = -(2x - 1) = -2x + 1 > 0$. Therefore, we obtain the piecewise definition
$$
|2x - 1| = \begin{cases}
2x - 1 & \text{if $x \geq \dfrac{1}{2}$}\\[2 mm]
-2x + 1 & \text{if $x < \dfrac{1}{2}$}
\end{cases}
$$
Thus, we must consider cases, depending on whether $x \geq 1/2$.
Case 1: If $x \geq 1/2$,
\begin{align*}
|2x - 1| + 3 & = 4x\\
|2x - 1| & = 4x - 3\\
2x - 1 & = 4x - 3\\
2 & = 2x\\
1 & = x
\end{align*}
Since $1 \geq 1/2$, $x = 1$ is a solution, as you can verify by direct substitution.
Case 2: If $x < 1/2$,
\begin{align*}
|2x - 1| + 3 & = 4x\\
|2x - 1| & = 4x - 3\\
-2x + 1 & = 4x - 3\\
4 & = 6x\\
\frac{2}{3} & = x
\end{align*}
However, $2/3 \geq 1/2$, so it is not satisfy the restriction that $x < 1/2$. Therefore, $x = 2/3$ is not a valid solution, as you can verify by direct substitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4288385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Compute the series $\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}$ Compute the series $$\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}.$$
How do I go about with the index notation, for example to arrange the series instead as $\sum_{n=1}^{\infty}a_n $?
I have tried to simplify the expression as:
$$\begin{align}&\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}=\sum_{n=3}^{\infty} \frac{(n-1)(n^3+n+1)}{n^6}\\
\implies&\sum_{n=3}^{\infty} \frac{n^4}{n^6} - \sum_{n=3}^{\infty} \frac{1}{n^6} = \sum_{n=3}^{\infty} \frac{(n-1)(n^3+n+1)}{n^6}\\
\implies& \sum_{n=3}^{\infty} \frac{1}{n^2} - \sum_{n=3}^{\infty} \frac{1}{n^6}\end{align}$$
I'm not sure what to do with the index $n=3$ as I know that I can simplify it otherwise as $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ and $\sum_{n=1}^{\infty} \frac{1}{n^6} = \frac{\pi^6}{945}$.
| $$
\begin{align}
\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}&=\sum_{n=3}^{\infty} \frac{n^4}{n^4}\frac{1-\frac{1}{n^4}}{n^2}\\
&=\sum_{n=3}^{\infty} \frac{1-\frac{1}{n^4}}{n^2}\\
&=\sum_{n=3}^{\infty} \frac{1}{n^2}-\frac{1}{n^6}\\
\end{align}$$
Now note, since each limit exists (see below, for the value)
$$\sum_{n=3}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}-\frac{5}{4}$$
$$\sum_{n=3}^{\infty} \frac{1}{n^6}=\frac{\pi^6}{945}-\frac{65}{64}$$
You can split up the sum
$$
\begin{align}
\sum_{n=3}^{\infty} \frac{1}{n^2}-\frac{1}{n^6}&=\sum_{n=3}^{\infty} \frac{1}{n^2}-\sum_{n=3}^{\infty}\frac{1}{n^6}\\
&=\frac{\pi^2}{6}-\frac{5}{4}-\frac{\pi^6}{945}+\frac{65}{64}\\
&=\frac{\pi^2}{6}-\frac{\pi^6}{945}-\frac{15}{64}\\
\end{align}$$
The calculation of the limits above is done in the following manner:
$$
\begin{align}
\frac{\pi^2}{6}&=\sum_{n=1}^{\infty} \frac{1}{n^2}\\
&=\sum_{n=3}^{\infty} \frac{1}{n^2}+\frac{1}{1^2}+\frac{1}{2^2}\\
&=\sum_{n=3}^{\infty} \frac{1}{n^2}+\frac{5}{4}\\
\end{align}$$
Subtracting $\frac{5}{4}$ on both sides, yields
$$
\begin{align}
\frac{\pi^2}{6}-\frac{5}{4}&=\sum_{n=3}^{\infty} \frac{1}{n^2}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove uniqueness of solution in ODE I need to find function $f(x)$, such that $x > 0$ and $f(x) > 0$, so that the area limited by the function, the $y$ and $x$ axes and $x = a$ ($a > 0$) is $f^3(x)$ for every $a$, and prove that there is only one such function.
Let $y' = f(x)$ and $y = \int^{x}{f(t)dt}$ I get the ODE $(y')^3 = y \Longrightarrow y'=\sqrt[3]{y}$ which is separable thus I get $y = \sqrt[3]{(\frac{2}{3}(C+x))^2}$. But I'm not sure how I'm supposed to prove uniqueness. The existence and uniqueness theorem is defined for $D = \{(x, y)| \alpha < x <\beta, \gamma < y < \delta \}$ where $y$ is continuous and the $y$ partial derivative exists if $(x_0, y_0)\in D$ then there is a solution for the initial value problem with $(x_0, y_0)$ and that solution is unique, but can I use this here? and if so, how?
| From the question we get that $y = \sqrt{\frac{8}{27}}(x+C)^{3/2}$ so $f(x) = y'= \sqrt[3]{y}=\sqrt{\frac{2}{3}(x+C)}$ we know that $y(a) = \int_{0}^{a} f(x)dx = (y'(a))^3$ hence
$$
\int_{0}^{a} f(x)dx = \int_{0}^{a} \sqrt{\frac{2}{3}(x+C)}dx = \sqrt{\frac{8}{27}}(x + C)^{3/2}|_0^a = \sqrt{\frac{8}{27}}(a + C)^{3/2} - \sqrt{\frac{8}{27}}(C)^{3/2} = \sqrt{\frac{8}{27}}(a + C)^{3/2} = \sqrt{\frac{2}{3}(a+C)}^3 = f^3(a) \Longrightarrow \sqrt{\frac{8}{27}}(C)^{3/2}=0 \Longrightarrow C=0
$$
Therefore there is only one unique f(x) in the first quadrant when $C=0$ and $f(x) = \sqrt{\frac{2}{3}(x)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4291760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is there any formula for $\operatorname{arctan}{(a+b)}$ Is there any formula for $$\operatorname{arctan}{(a+b)}$$
I know couple of formulas for trigonometric functions. But I don't know if such formulas exists for inverse trigonometric functions. I don't have a clue where to start with. Any hint will be appreciated.
| The following formula is well known and you can find it here.
$$\arctan(a) + \arctan(b) = \arctan\left(\frac{a+b}{1-ab}\right)$$
(of course up to a multiple of $\pi$ with $ab\neq 1$). Then
$$\arctan(a+b) + \arctan(a-b) = \arctan\left(\frac{2 a}{1-a^2+b^2}\right)$$
$$\arctan(a+b) - \arctan(a-b) = \arctan\left(\frac{2 b}{1+a^2-b^2}\right)$$
Hence we have
\begin{align*}
2 \arctan(a+b) &= \arctan\left(\frac{2 a}{1-a^2+b^2}\right)+\arctan\left(\frac{2 b}{1+a^2-b^2}\right)\\[2mm]
\arctan(a+b) &= \frac{1}{2}\arctan\left(\frac{2 a}{1-a^2+b^2}\right)+\frac{1}{2}\arctan\left(\frac{2 b}{1+a^2-b^2}\right)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4294501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How many negative real roots does the equation $x^3-x^2-3x-9=0$ have? How many negative real roots does the equation $x^3-x^2-3x-9=0$ have ?
My approach :-
f(x)= $x^3-x^2-3x-9$
Using rules of signs, there is 1 sign change , so there can be at most 1 positive real root
f(-x)= $-x^3-x^2+3x-9$
2 sign changes here, indicating at most 2 negative real roots
I end up with following 2 possibilities:-
1)1 positive, 2 negative real roots
2)1 positive, 2 imaginary roots
how to progress further ?
| Actually, in working through my Calculus oriented answer, I accidentally thought of a somewhat convoluted precalculus approach to solving the problem.
$f(x)= x^3-x^2-3x-9.$
How many negative real roots does the equation $f(x) = 0$ have ?
$$\text{Let} ~~g(x) = x^3 - x^2 - 3x + 3 = (x - 1) \times (x^2 - 3).\tag1$$
Part of the inspiration for this method, was to construct $g(x)$ such that $f(x) - g(x)$ equals a constant and where $g(x)$ is easy to factor. I noted that the two leftmost coefficients of $f(x)$ followed the pattern $[(+1), (-1)],$ so I constructed $g(x)$ so that the next two coefficients of $g(x)$ would follow the pattern $(-3) \times [(+1), (-1)].$
The other part of the inspiration for this method, was my observation in my other posted answer that the absolute value of the rightmost coefficient of $f(x)$, namely $|-9| = 9$ seemed inordinately large.
Then, with $g(x)$ specified by (1) above, you have that $f(x) = g(x) - 12.$
Therefore, it is sufficient to show that on the interval $-\infty < x < 0$,
the maximum value achieved by $g(x)$ is less than $12$.
Given how $g(x)$ is factored in (1) above, the only values of $x < 0$ for which $g(x)$ will be $ > 0$ will be $-\sqrt{3} < x < 0.$
However, for $-\sqrt{3} < x < 0$, a routine examination of
$|g(x)| = |x - 1| \times |x^2 - 3|$ shows that
*
*$|x - 1| = 1 + (-x) < 1 + (\sqrt{3}) ~< ~3 ~< ~4.$
*$|x^2 - 3| = 3 - x^2 \leq 3.$
Therefore, the product of the two factors above must be strictly less than $(4 \times 3)$. Therefore, the maximum value for $g(x)$ on the interval $-\sqrt{3} < x < 0$ must be less than $(12)$. Since this is the only interval for which $x < 0$ and $g(x) > 0$, the maximum value of $g(x)$ on the interval $-\infty < x < 0$ must be less than $(12)$.
Therefore, since $f(x) = g(x) - 12$, you must have that the maximum value of $f(x)$ on the interval $-\infty < x < 0$ must be less than $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4296303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
If $X_{1},\ldots,X_{n}$ are independent Bernoulli random variables, calculate $E[S_{\tau_{2}}]$ Let $X_{1},\ldots,X_{n}$ be independent Bernoulli random variables. $P(X_{i}=1)=p$, $P(X_{i}=0)=q=1-p$. We denote $S_{n}=X_{1}+\cdots+X_{n}$. There are $\tau_{1}=\min \{n: X_{n}=1\}$ and $\tau_{2}=\min \{n: X_{n}=X_{n-1}=1\}$. How to calculate $E[S_{\tau_{2}}]$?
I've tried through the total expectation formula. I got
$$E[S_{\tau_{2}}]=\frac2p+\frac{E[S_{\tau_{2}-\tau_{1}}]+1}{1-p}$$
But it is not clear how to proceed further.
A similar question for $\tau_{3}=\min \{n: X_{n}=X_{n-1}=X_{n-2}=1\}$. How to calculate $E[S_{\tau_{3}}]$?
| Sketch (almost solution):
Put $$A_n = \{ \text{ in a set $\{ X_1, X_2, \ldots, X_n\}$ there's no two consecutive units \}} \}$$ $$= \{ \not \exists 1 \le i \le n-1: X_i = X_{i+1} \} = \{ \tau_2 > n\}.$$
Put $a_n = P(A_n, X_n = 0 )$ and $b_n = P(B_n, X_n=1)$. We have
$$b_{n+1} = P(A_{n+1}, X_{n+1} = 1) = P(A_{n+1}, X_n = 0, X_{n+1} = 1) + P(A_{n+1}, X_n = 1, X_{n+1} = 1)$$
$$ = P(A_{n}, X_n = 0, X_{n+1} = 1) + 0 = P(A_{n}, X_n = 0) P(X_{n+1} = 1) = \frac{a_n}2,$$
$$a_{n+1} = P(A_{n+1}, X_{n+1} = 0) = P(A_{n+1}, X_n = 0, X_{n+1} = 0) + P(A_{n+1}, X_n = 1, X_{n+1} = 0)$$
$$ = P(A_{n}, X_n = 0, X_{n+1} = 0) + P(A_{n}, X_n = 1, X_{n+1} = 0)=$$
$$ = P(A_{n}, X_n = 0)P(X_{n+1} = 0) + P(A_{n}, X_n = 1)P( X_{n+1} = 0)=$$
$$ = a_n \frac12 + b_n \frac12 = \frac{a_n + b_n}2.$$
Thus
\begin{cases} b_{n+1} = \frac{a_n}2, \\ a_{n+1} = \frac{a_n + b_n}2. \end{cases}
We have $a_{n+1} = \frac{a_n + b_n}2 = \frac{a_n + \frac12 a_{n-1}}2$, i.e. $4a_{n+1} = 2a_n + a_{n-1}$. Hence $a_n = \frac{C_1}{2^n} \cos(\frac{\pi n}{3}) + \frac{C_2}{2^n} \sin(\frac{\pi n}{3})$ and $b_{n+1} = a_n$. It's easy to find $a_1, a_2, b_1, b_2$ and hence to find $C_1$, $C_2$ and $a_n$, $b_n$. Finally $$E \tau_2 = \sum_{n \ge 0} P(\tau_2 > n) = \sum_{n \ge 0} P(A_n) = \sum_{n \ge 0} \big( P(A_n, X_n = 0) + P(A_n, X_n = 1) \big) = \sum_{n \ge 0} (a_n + b_n). $$
As $\frac{a_n}{2^{-n}}$ and $\frac{b_n}{2^{-n}}$ are $O(1)$ we have $E \tau_2 < \infty$.
You can find $E \tau_3$ in the same way but with bigger amount of calculations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4296644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
A non standard way to a probability problem We have teams $A$ and $B$.Every team has equal chance to win. A team wins after they have won $4$ matches. Team $A$ won the first match, what is the probability, that team $B$ won the whole game.
My try:
So team $A$ needs $3$ more wins to win the game. Team $B$ needs $4$ more wins to win the game. So the winner will be determined in maximum of $7$ games. So now I tried to do it my own way (I dont know if it is usual to do it this way)
Let's say that we are looking at the wins of team $A$. (They need $3$ more wins).
$$...A_i,...A_j,...A_n$$,
where $3$ dots symbolize the wins of team $B$. So $$(i-1) + (j-i-1) + (n-j-i-1) \leq 3 $$
(Because there must not be more than 3 losses, because at 4 team $B$ wins the game )
we get:
$$n-i \leq 6 \quad \text{and} \quad 0<i<n $$
If we look what integer tuples can satisfy the upper inequalities we get: $$N = \{(1,6),(2,6), (3,6),(4,6),(5,6),\\(1,5),(2,5),(3,5),(4,5),(1,4),(2,4),(3,4),(1,3),(2,3),(1,2)\}$$
Where $N$...possible winnings of team $A$
So we can count the number of elements ($15$) an that would be our number in the numerator, where in the denominator is the number of elements in our sample space.
To calculate the sample space I would do:
$$|\omega| = 1 + (\binom{4}{3} - 1)\\ + (\binom{5}{3} - \binom{4}{3} - 1)\\ + (\binom{6}{3} - (1 + (\binom{4}{3} - 1) + (\binom{5}{3} - \binom{4}{3} - 1)))\\ + (\binom{7}{3} -(1 + (\binom{4}{3} - 1)+ (\binom{5}{3} - \binom{4}{3} - 1) + (\binom{6}{3} - (1 + (\binom{4}{3} - 1)+ (\binom{5}{3} - \binom{4}{3} - 1)))) )$$
I guess we could also write this with a recursive formula. However we get $|\omega|=35$.
(What I did here is that $1$ is for $3$ consecutive wins, $\binom{4}{3} - 1$ are other possible matches in 4 games and so on.)
So if my procedure is correct the probability is $$P(\text{team B wins the whole game}) = 1 - \frac{|N|}{|\omega|} = 1 - \frac{15}{35} = \frac{20}{35}$$
However is my process atleas partially correct (I do not know the solution).
| To more simply calculate the odds of this problem, we can relate it to flipping a coin. A heads gives team $A$ the victory, while a tails gives team $B$ the victory. To solve the problem, we are looking for the probability that at least $4$ of $6$ flips are tails. This is not difficult to calculate.
Total possible outcomes: $2^6=64$
6 tails: $_6C_6=1$ way; $\dfrac1{64}$
5 tails: $_6C_5=6$ ways; $\dfrac{6}{64}$
4 tails: $_6C_4=15$ ways; $\dfrac{15}{64}$
Total: $\dfrac{1+6+15}{64}=\dfrac{22}{64}=\dfrac{11}{32}$. This gives a $34.375\%$ chance that team B manages to win the tournament.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4300190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
In Desmos, function fills entire screen unless $y$ is used in some denominator In Desmos, I am graphing the equation
$$(g^2 - 4x)y^2 + (2g - 4x - 4)y + 1 = 0$$
which uses the helper function
$$g = x + 1 - \frac{x^2 - 2x + 1}{x + 1 + 2\sqrt x}$$
which fills up the entire screen, even though it should be a smooth curve. Strangely, when I add $y$ in any denominator of the main function, say setting the function equal to $0/y$ instead of $0$, the function works just fine and is displayed correctly.
Buggy Curve:
Intended Curve:
I would think that having $y$ in the denominator like this would change things when $y=0$, but without it, the whole screen is filled up (every $y$ value.) I can't figure out the reason that this might be.
Here are the equation sources to paste into desmos:
Helper Function: g=x+1-\frac{x^{2}-2x+1}{x+1+2\sqrt{x}}
Buggy Equation: \left(g^{2}-4x\right)\cdot y^{2}+\left(2g-4x-4\right)\cdot y+1=0
Fixed Equation: \left(g^{2}-4x\right)\cdot y^{2}+\left(2g-4x-4\right)\cdot y+1=\frac{0}{y}
| There are quite some redundancy in the way your implicit function was formulated. Note that the rational function in $g$ can be simplified.
$$
\frac{x^2 - 2x + 1}{x + 1 + 2\sqrt x}= \left(\frac{x-1}{\sqrt{x} + 1} \right)^2 = \left(\frac{(\sqrt{x} + 1)(\sqrt{x} - 1)}{\sqrt{x} + 1} \right)^2 = (\sqrt{x} - 1)^2
$$
Put this back into $g$ we have
$$g=x+1-(x-2\sqrt{x} + 1) = 2 \sqrt{x}$$
The curve $(g^2 - 4x)y^2 + (2g - 4x - 4)y + 1 = 0$ therefore goes from an implicit function to an explicit function.
$$\begin{align}
&\implies 0 \cdot y^2 +(4\sqrt{x}-4x-4)y+1=0 \\
&\implies y = \frac14 \frac1{x-\sqrt{x}+1}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4302885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $x ,y$ satisfying : $x,y$ are $2$ positive integers and $(xy+x+2)(xy+1)$ is a perfect square. Find $x ,y$ satisfying : $x,y$ are $2$ positive integers and $(xy+x+2)(xy+1)$ is a perfect square.
I have solved this problem and will post the solution as soon as possible. Hope everyone can check my solution! Thanks very much !
| Disclaimer: This is only a partial answer. I may complete this later.
Let $x$, $y$ and $z$ be three positive integers such that
$$(xy+x+2)(xy+1)=z^2,\tag{1}$$
The greatest common divisor $d$ of the two factors divides $1+x$ and $1-y$, so we have
$$x=ud-1\qquad\text{ and }\qquad y=vd+1,$$
for coprime integers $u$ and $v$, and $z=wd$. Plugging this into $(1)$ then yields
$$w^2=\big(uvd+2u-v\big)\big(uvd+u-v\big).$$
Now the two factors on the right hand side are coprime, and hence they are both perfect squares, so
$$uvd+2u-v=a^2\qquad\text{ and }\qquad uvd+u-v=b^2,$$
for some positive integers $a$ and $b$ with $a>b$. Then $u=a^2-b^2$ and so
$$v=\frac{2b^2-a^2}{d(a^2-b^2)-1}.$$
As $v$ is positive this shows that $a^2<2b^2$.
Conversely, if $a$ and $b$ are two coprime positive integers with $b<a<\sqrt{2}b$ and $d$ is any positive integer such that
$$\frac{2b^2-a^2}{d(a^2-b^2)-1},$$
is also an integer, then the integer $x$ and $y$ given by
$$x=d(a^2-b^2)-1\qquad\text{ and }\qquad y=d\frac{2b^2-a^2}{d(a^2-b^2)-1}+1=\frac{db^2-1}{d(a^2-b^2)-1},$$
satisfy
$$(xy+x+2)(xy+1)=(dab)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Let the roots of the polynomial $P(x) = x^3 − x + 1$ be a, b, c. Construct the polynomial with roots $a^4, b^4, c^4$ and degree 3 I know that $P(x) = x^3 − x + 1$ has a single real root. But I guess it's irrelevant to find it.
What I thought was to change $x$ into ${x}^{\frac{1}{4}}$ so it would satisfy the condition. However, the resulting $Q(x) = x^\frac{3}{4} − x^\frac{1}{4} + 1$ is neither a polynomial nor has a real root.
| Maybe I was wrong in calculations, one needs to check.
Vieta: $a+b+c=0$, $ab+ac+bc=-1$, $abc=-1$.
Let's do some algebraic manipulations to obtain the same for $a^4$, $b^4$, $c^4$.
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac) \Rightarrow a^2+b^2+c^2=2$$
$$(ab+ac+bc)^2=a^2b^2+a^2c^2+b^2c^2+2abc(a+b+c)\Rightarrow a^2b^2+a^2c^2+b^2c^2=1$$
$$(a^2+b^2+c^2)^2=a^4+b^4+c^4+2(a^2b^2+a^2c^2+b^2c^2)\Rightarrow a^4+b^4+c^4=2$$
$$a^2b^2c^2=(abc)^2=1$$
$$(a^2b^2+a^2c^2+b^2c^2)^2=a^4b^4+a^4c^4+b^4c^4+2a^2b^2c^2(a^2+b^2+c^2)\Rightarrow a^4b^4+a^4c^4+b^4c^4=-3$$
It's ok, because there are complex roots in $a,b,c$.
$$a^4b^4c^4=(abc)^4=1$$
Reverse Vieta: $x^3-2x^2+3x-1$ is polynomial with roots $a^4$, $b^4$, $c^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4306647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Showing that $\sqrt{2+\sqrt3}=\frac{\sqrt2+\sqrt6}{2}$ So I was playing around on the calculator, and it turns out that $$\sqrt{2+\sqrt3}=\frac{\sqrt2+\sqrt6}{2}$$
Does anyone know a process to derive this?
| Both sides are clearly positive, and squaring the left hand side shows that
$$\begin{align} \left(\frac{\sqrt{2}+\sqrt{6}}{2}\right)^2&=\frac{\sqrt{2}^2+2\sqrt{2}\sqrt{6}+\sqrt{6}^2}{2^2}\\&=\frac{2+4\sqrt{3}+6}{4}\\&=2+\sqrt{3}.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4307201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Proof verification: $\lim_{x\to 2} \frac{\sqrt{x^2 + 5} - 3}{x - 2} = \frac23$ The question is as follows:
Prove that $\displaystyle\lim_{x\to 2} \dfrac{\sqrt{x^2 + 5} - 3}{x - 2} = \dfrac23$.
My proof is:
Fix $\varepsilon > 0$.
Note that $$\begin{array}{rcl}\left|\dfrac{\sqrt{x^2 + 5} - 3}{x - 2} - \dfrac23\right| &=& \left|\dfrac{x + 2}{\sqrt{x^2 + 5} + 3} - \dfrac23\right|\\&=& \dfrac13\left|\dfrac{3x - 2\sqrt{x^2 + 5}}{3 + \sqrt{x^2 + 5}}\right|\\&=& \dfrac13 \left|\dfrac{5x^2 - 20}{\left(3 + \sqrt{x^2 + 5}\right)\left(3x + 2\sqrt{x^2 + 5}\right)}\right|\\&=& \dfrac{5|x - 2|\cdot|x + 2|}{3\left|3 + \sqrt{x^2 + 5}\right|\cdot \left|3x + 2\sqrt{x^2 + 5}\right|}\end{array}$$
Pick $\delta = \min\left\{1, \dfrac{21\varepsilon}5\right\}$. Suppose that $0 < |x - 2| < \delta < 1$. Then $1 < x < 3$. Therefore, $|x + 2| < 5$, $\left|3 + \sqrt{x^2 + 5}\right| > 3 + \sqrt 6 > 5$, and $\left|3x + 2 \sqrt{x^2 + 5}\right| > 3 + 2\sqrt 6 > 7$.
Therefore,
$$\left|\dfrac{\sqrt{x^2 + 5} - 3}{x - 2} - \dfrac23\right| < \dfrac{5 \cdot 5}{3 \cdot 5 \cdot 7} |x - 2| < \dfrac5{21} \delta < \varepsilon$$
Hence, $\displaystyle\lim_{x\to 2} \dfrac{\sqrt{x^2 + 5} - 3}{x - 2} =\dfrac23$.
Is my proof correct?
| Yes, you have the main idea correct.
Though, there seems to be too much formalism without explaining in words, which may obscure your proof.
One key step you make is to rewrite the function as
$$
\frac{\sqrt{x^2+5}-3}{x-2}=\frac{x^2+5-9}{(x-2)(\sqrt{x^2+5}+3)}
=\frac{x+2}{\sqrt{x^2+5}+3}\;.
$$
At this point, you can simply apply one of the limit laws and continuity of the functions to conclude that
$$
\lim_{x\to 2}\frac{x+2}{\sqrt{x^2+5}+3}=\frac{4}{6}=\frac23\;.
$$
Students usually confuse "formal/rigorous proof" with proofs written in terms of $\epsilon$-$\delta$.
If one does want to proceed with an $\epsilon$-$\delta$ proof, then one needs to estimate the quantity
$$
\left|\frac{x+2}{\sqrt{x^2+5}+3}-\frac23\right|\;\tag{1}
$$
Instead of handling this on an ad hoc way, one can follow the idea of proving the quotient law for limits to estimate (1) as
$$
\begin{align}
\left|\frac{x+2}{\sqrt{x^2+5}+3}-\frac46\right|
=& \frac{|6f(x)-4g(x)|}{|6g(x)|} \\
=& \frac{|6f(x)-6\cdot 4+6\cdot 4- 4g(x)|}{|6g(x)|}\\
\le&\frac{6|f(x)-4|+4|g(x)-6|}{|6g(x)|}\tag{2}
\end{align}
$$
where $f(x)=x+2$ and $g(x)=\sqrt{x^2+5}+3$. It is then clear that one should use the following facts to get a desired $\delta$:
*
*$f$ and $g$ are continuous at $x=2$;
*$g$ is bounded away from $0$ near $x=2$.
To spell out the formalism, one can take $\delta=\min(\delta_1,\delta_2,\delta_3)$ such that
\begin{align}
|f(x)-4|<\epsilon\qquad\textrm{whenever } |x-2|<\delta_1\\
|g(x)-6|<\epsilon\qquad\textrm{whenever } |x-2|<\delta_2\\
|g(x)|\ge |g(x)-6+6|\ge 6-|g(x)-6|\ge 1\qquad\textrm{whenever } |x-2|<\delta_3\tag{3}
\end{align}
Combining with (2) one has
$$
\left|\frac{x+2}{\sqrt{x^2+5}+3}-\frac46\right| \le
\frac{10\epsilon}{6}\;.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4308750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Extracting coefficients I stuck at the following problem:
Let
\begin{equation}
f(z) = \frac{1 - \sqrt{1 - 4z}}{2}.
\end{equation}
Find $[z^n]$ of $f(z)^r$ for $r \in \mathbb{N}$. Where $[z^n]f(z)$ is the $n$-th coefficient of the power series $f(z) = \sum_{n \geq 0}{a_n z^n}$ therefore $[z^n]f(z) = a_n$.
So far I got
\begin{equation}
\sqrt{1 - 4z} = \sum_{n \geq 0}\binom{1/2}{n}(-4 x)^n.
\end{equation}
and therefore
\begin{align*}
f(z) = \frac{1 - \sqrt{1 - 4z}}{2} &= \frac{\sum_{n \geq 1}\binom{1/2}{n}(-4)^n z^n}{2} \\
&= \sum_{n \geq 1}\binom{1/2}{n}(-1)^n 2^{n} z^n \\
&= \sum_{n \geq 1}\binom{1/2}{n}(-2)^n z^n \\
&= \sum_{n \geq 0}a_n z^n
\end{align*}
with coefficients $a_0 = 0$ and $a_n = \binom{1/2}{n}(-2)^n$.
I wanted to use cauchys integral formula for
$g(z) = f(z)^r$ to extract the $n$ coefficient
but then I get
\begin{align}
\left( \frac{1 - \sqrt{1 - 4z}}{2}\right)^r &= \frac{1}{2^r} \left(1 - \sqrt{ 1 - 4z} \right)^r \\
&= \frac{1}{2^r} \sum^{r}_{m=0}\binom{r}{m}(-1)^m \sqrt{1 - 4z}^m \\
&= \frac{1}{2^r} \sum^{r}_{m=0}\binom{r}{m}(-1)^m (1 - 4z)^{m/2} \\
&= \frac{1}{2^r} \sum^{r}_{m=0}\binom{r}{m}(-1)^m \sum_{k \geq 0}\binom{m/2}{k}(-1)^k 4^k z^k \\
&= \frac{1}{2^r}\sum^{r}_{m=0} \sum_{k \geq 0}\binom{r}{m} \binom{m/2}{k} (-1)^{m+k} 4^k z^k.
\end{align}
Therefore I should have
\begin{align*}
[z^n] (f(z))^r &= [z^n] \sum^{r}_{m=0} \sum_{k \geq 0}\frac{1}{2^r} \binom{r}{m} \binom{m/2}{k} (-1)^{m+k} 4^k z^k \\
&= \sum^{r}_{m=0}[z^n] \sum_{k \geq 0}{\frac{1}{2^r} \binom{r}{m} \binom{m/2}{k} (-1)^{m+k} 4^k z^k} \\
&= \sum^{r}_{m=0}\frac{1}{2^r}\binom{r}{m} \binom{m/2}{n} (-1)^{m + n} 4^n \\
&= \sum^{r}_{m=0}\frac{1}{2^r}\binom{r}{m} \binom{m/2}{n} (-1)^m (-4)^n \\
&= \frac{1}{2^r} \sum^{r}_{m=0} \binom{r}{m} \binom{m/2}{n} (-1)^m (-4)^n \\
&= \frac{1}{2^r} \sum^{r}_{m=2 n} \binom{r}{m} \binom{m/2}{n} (-1)^m (-4)^n
\end{align*}
For $r \geq 2n$ otherwise the coefficient vanishes.
I would be thankful if anyone can give me a hint.
| Too long for a comment, but not a full flesh answer.
$f(z)$ can be considered as the root of this quadratic equation:
$$f(z)^2-f(z)+z=0,\tag{1}$$
Let us multiply (1) by $f(z)$, giving:
$$f(z)^3-f(z)^2+zf(z)=0\tag{2}$$
Adding (1) and (2) gives an expression for $f(z)^3$.
More generally, iterating $r-2$ times the process of multiplication by $f(z)$ the previous equation gives by successive cancellations, and using the formula for the sum of a finite geometric series, the following recurrence formula:
$$f(z)^r=f(z)-z\dfrac{1-f(z)^{r-1}}{1-f(z)},$$
which can be transformed, due to relationship $1-f(z)=\dfrac{4z}{1-\sqrt{1-4z}}$ into
$$f(z)^r=f(z)-(1-\sqrt{1-4z})\frac14 (1-f(z)^{r-1}),$$
a formula that looks tractable for a recursive computation of the coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What is wrong with the application of L'Hopital's rule to $\lim_{x\rightarrow 2}\frac{x^2 - 7x + 10}{x^2 + x - 6}$? The question is
$$\lim_{x\rightarrow 2}\frac{x^2 - 7x + 10}{x^2 + x - 6}$$ If we plug in $x = 2$ we see that both quadratrics have a zero there. "Logically" this is a L'Hopital question, so if I take the derivative of top and bottom and find the limit I get:
$$\lim_{x\rightarrow2}\frac{2x-7}{2x+1} = \frac{-3}{5}$$
However, if I plot the function as close to $x = 2$ that I wish,, I see that the right handed limit and the left handed limit are positive and negative infinity, respectively, and of course my plotting software cannot plot the $x = 2$ case.
This concerns me. On the one hand it seems that the limit should not exist, but on the other hand L'Hopital pumps out an answer.
What did I miss?
| It holds $x^2-7x+10=(x-2)(x-5)$ and $x^2+x-6=(x-2)(x+3)$. Hence, we have $$\frac{x^2-7x+10}{x^2+x-6}=\frac{(x-5)(x-2)}{(x+3)(x-2)}=\frac{x-5}{x+3}$$
for all $x\notin\{-3,2\}$. Hence, the pole $x=2$ is removable, because we can extend the domain of the lefthandside by $x=2$, such that the function is still continuous. So your result $-\frac 35$ is correct, because $\frac{2-5}{2+3}=-\frac 35$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the number of different colorings of the faces of the cube with 2 white, 1 black, 3 red faces? I tried using Polya theorem same in this https://nosarthur.github.io/math%20and%20physics/2017/08/10/polya-enumeration.html guide, using S4 group for cube faces. But i Have a $\frac{1}{24}12w^2*b*r^3$.
Please help to me how count coloring faces of the cube and other platonic solids.
| The cycle index polynomial for rotational symmetries of the faces of a cube is
$$
\tfrac1{24}(z_1^6+12z_2^3+8z_3^2+3z_1^2z_2^2)
$$
Therefore, in order to count the number of colorings with one black, three red, and two white faces, we need to find the coefficient $b^1r^3w^2$ in
$$
\tfrac1{24}[(b+w+r)^6+12(b^2+r^2+w^2)^3+8(b^3+r^3+w^3)^2+3(b+r+w)^2(b^2+r^2+w^2)^2]
$$
Going term by term...
*
*The coeficient of $b^1r^3w^2$ in $(b+r+w)^6$ is given by the multinomial coefficient $\binom{6!}{1!3!2!}=60$.
*$(b^2+r^2+w^2)^3$ contributes nothing to $b^1r^3w^2$.
*$(b^3+r^3+w^3)^2$ also contributes nothing.
*The trickiest term is $(b+r+w)^2(b^2+r^2+w^2)^2$. I will write this as
$$
(b+r+w)(b+r+w)(b^2+r^2+w^2)(b^2+r^2+w^2)
$$
In order to contribute to $b^1r^3w^2$, the $(b^2+r^2+w^2)$ factors need to contribute an $r^2$ and a $w^2$. This can be done in either order, for $2$ choices. The remaining $b^1r^1$ comes from the $(b+r+w)$ factors; again, either one can give the $b$ while the other gives an $r$, for another $2$ choices. Therefore, the contribution from this term is $2\times 2=4$.
All in all, the coefficient is
$$
\frac1{24}[1\cdot 60+12\cdot 0+8\cdot 0+3\cdot 4]=\boxed{3}
$$
Here is a Wolfram Alpha computation which confirms this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4313417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Signature of $q(x,y)=5x^2-6xy+y^2$ a quadratic form Let $q:R^2\rightarrow R$ a quadratic form such that $q(x,y)=5x^2-6xy+y^2$
find the signature of $q$
I know that $q_A=$ $\begin{pmatrix}
5 & -3\\
-3 & 1
\end{pmatrix}$
and how $q(1,0)=5\neq0$ then $(1,0)$ is an element of a orthogonal basis $\beta=\{(1,0),v\}$
and then how $\beta$ is orthogonal then the matrix associated to $q$ respect to $\beta$
should be
$\begin{pmatrix}
5 & 0\\
0 & 1
\end{pmatrix}$ is right?
| $$ Q^T D Q = H $$
$$\left(
\begin{array}{rr}
1 & 0 \\
- \frac{ 3 }{ 5 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rr}
5 & 0 \\
0 & - \frac{ 4 }{ 5 } \\
\end{array}
\right)
\left(
\begin{array}{rr}
1 & - \frac{ 3 }{ 5 } \\
0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rr}
5 & - 3 \\
- 3 & 1 \\
\end{array}
\right)
$$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$ P^T H P = D $$
$$\left(
\begin{array}{rr}
1 & 0 \\
\frac{ 3 }{ 5 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rr}
5 & - 3 \\
- 3 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rr}
1 & \frac{ 3 }{ 5 } \\
0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rr}
5 & 0 \\
0 & - \frac{ 4 }{ 5 } \\
\end{array}
\right)
$$
$$ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4313649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to find first integral in a given quasilinear partial differential equation? I have to solve PDE
$$y(x+y)\dfrac{\partial z}{\partial x}+x(x+y)\dfrac{\partial z}{\partial y}=2(x^2-y^2)+xz-yz.$$
From this equation I get the system of differential equations:
$$\dfrac{dx}{y(x+y)}=\dfrac{dy}{x(x+y)}=\dfrac{dz}{2(x^2-y^2)+xz-yz}.$$
I have $$\dfrac{dx}{y(x+y)}=\dfrac{dy}{x(x+y)},$$ from which I get $$xdx=ydy$$ and by integrating I get the first integral $$x^2-y^2=C_1.$$
I don't know how to get another first integral.
| $$y(x+y)\dfrac{\partial z}{\partial x}+x(x+y)\dfrac{\partial z}{\partial y}=2(x^2-y^2)+xz-yz.$$
You correctly wrote :
$$\dfrac{dx}{y(x+y)}=\dfrac{dy}{x(x+y)}=\dfrac{dz}{2(x^2-y^2)+xz-yz}.$$
And you correctly found a first characteristic equation
$$x^2-y^2=C_1$$
For a second characteristic equation :
$$\dfrac{dx}{y(x+y)}=\dfrac{dy}{x(x+y)}=\dfrac{dx-dy}{y(x+y)-x(x+y)}=\dfrac{dz}{2(x^2-y^2)+xz-yz}$$
$$\dfrac{dx-dy}{-(x^2-y^2)}=\dfrac{dz}{2(x^2-y^2)+(x-y)z}$$
$$\dfrac{dx-dy}{-C_1}=\dfrac{dz}{2C_1+(x-y)z}$$
With $t=x-y$
$$-\dfrac{dt}{C_1}=\dfrac{dz}{2C_1+t\,z}$$
$$C_1\dfrac{dz}{dt}+t\,z+2C_1=0$$
This is a linear first order ODE which solution is :
$$e^{t^2/(2C_1)}z+\sqrt{2\pi C_1}\text{ erfi}\left(\frac{t}{\pm\sqrt{2C_1}}\right)=C_2$$
Fonction erfi : https://mathworld.wolfram.com/Erfi.html
$$e^{(x-y)^2/(2(x^2-y^2))}z+\sqrt{2\pi (x^2-y^2)}\text{ erfi}\left(\frac{x-y}{\pm\sqrt{2(x^2-y^2)}}\right)=C_2$$
$$e^{(x-y)/(2(x+y))}z\pm\sqrt{2\pi (x^2-y^2)}\text{ erfi}\left(\sqrt{\frac{x-y}{2(x+y)}}\right)=C_2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Legendre relations for elliptic integrals with null imaginary part I am trying to compute the transformation of $K(k)$ as $k \to 1/k$, where $K(k)$ is just the Legendre integral
$$K(k) := F\left(\frac{\pi}{2}, k\right).$$
The Digital Library of Mathematical Functions [1] gives a relation for this transformation that relies on the imaginary part of $k$:
$$K\left(1 / k^{\prime}\right)=k^{\prime}\left(K\left(k^{\prime}\right) \mp \mathrm{i} K(k)\right),$$
where $k^{\prime}$ is the complementary modulus of $k$, and the $\pm$ sign is for $\Im(k) \lessgtr 0$. My question is, how do I find the transformation in the case where $\Im(k)=0$? I understand that there is a branch cut for this formula for $\Im(k)=0$, but what do I do to get around that?
| Let $0<k<1$, then
$$ K\left(\frac{1}{k}\right) = \int_{0}^{1}\frac{dt}{\sqrt{\left(1-t^2\right)\left(1-\frac{1}{k^2}t^2\right)}}dt $$
Make the following change of variable: $\displaystyle w^2 \mapsto \frac{1}{k^2}t^2$
$$K\left(\frac{1}{k}\right) = k\int_{0}^{\frac{1}{k}}\frac{dt}{\sqrt{\left(1-k^2w^2\right)\left(1-w^2\right)}}dw$$
Note that, since $0<k<1$ then $\displaystyle \frac{1}{k}>1$. Hence,
$$K\left(\frac{1}{k}\right) = k\int_{0}^{1}\frac{dt}{\sqrt{\left(1-k^2w^2\right)\left(1-w^2\right)}}dw + k\int_{1}^{\frac{1}{k}}\frac{dt}{\sqrt{\left(1-k^2w^2\right)\left(1-w^2\right)}}dw$$
The firts integral is just $K(k)$
Then
$$K\left(\frac{1}{k}\right) = kK(k) + k\int_{1}^{\frac{1}{k}}\frac{dt}{\sqrt{\left(1-k^2w^2\right)\left(1-w^2\right)}}dw$$
For the second, make the substitution:
$$ w = \frac{1}{\sqrt{1-k'^2z^2}}$$
where $k'$ is the complementary modulus and $k' = \sqrt{1-k^2}$
Hence
$$ dw = \frac{k'^2zdz}{\sqrt{1-k'^2z^2}}$$
$$\frac{1}{\sqrt{1-k^2w^2}} = \frac{\sqrt{1-k'^2z^2}}{k'\sqrt{1-z^2}}$$
$$ \frac{1}{\sqrt{1-w^2}} = \frac{i\sqrt{1-k'^2z^2}}{k'z}$$
$$ z \to 0 \textrm{ if } x \to 1$$
$$ z \to 1 \textrm{ if } x \to \frac{1}{k}$$
Therfore
$$\int_{1}^{\frac{1}{k}}\frac{dt}{\sqrt{\left(1-k^2w^2\right)\left(1-w^2\right)}}dw = i\int_{0}^{1}\frac{dt}{\sqrt{\left(1-z^2\right)\left(1-k'^2z^2\right)}}dz= iK(k') $$
Hence, we can conclude
$$ \boxed{K\left(\frac{1}{k}\right) = k\left[K(k)+iK(k')\right] \quad 0<k<1 }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$100$ numbers are written around the circle
$100$ positive numbers are written in a circle. The sum of any two
neighbors is equal to the square of the number following them
clockwise. Find all such sets of numbers.
Let us introduce the notation $n=100$, not forgetting that $n$ is even. Let the numbers start in clockwise order $a_{1}, a_{2}, \ldots, a_{n}$. The sum of the conditions of the problem gives $(a_{1}-1)^{2}+(a_{2}-1)^{2}+\ldots+(a_{n}-1)^{2}=n$.
Suppose that two adjacent numbers can simultaneously be at least $2$, and at least one of them is strictly greater than $2$. Then all subsequent numbers in the circle are strictly greater than $2$, which contradicts the obtained equality. The case is considered similarly when two adjacent numbers can simultaneously be no more than $2$, and at least one of them is strictly less than $2$.
Thus, the numbers "alternate": no less than two and no more than two. Suppose WLOG that $a_{1}+a_{2} \geq 4$. Then
$$ \begin{gathered} a_{1}+a_{2} \geq 4, a_{3}+a_{4} \geq 4, \ldots, a_{n-1}+a_{n} \geq 4, \\ 4 \geq a_{2}+a_{3}, 4 \geq a_{4}+a_{5}, \ldots, 4 \geq a_{n}+a_{1}. \end{gathered} $$
Adding all the inequalities, we get the inequality $0≥0$. Therefore,
$$ a_{1}+a_{2}=a_{3}+a_{4}=\ldots, a_{n-1}+a_{n}=a_{2}+a_{3}=a_{4}+a_{5}=\ldots=a_{n}+a_{1}=4. $$
So all numbers are equal to two.
QUESTION: If we add up all the inequalities and get the identity, then we cannot draw any conclusions beyond the fact that this situation is possible. How was the consequence obtained that the sums of all pairs are equal to $4$?
| Here is my explanation of the solution.
To me, it's a case where "minor details are skipped", as opposed to Paul's "big jump / stopped trying to explain / logic leap".
However, the exposition could be significantly improved on by stating where the condition $a_i+a_{i+1} = a_{i+2}^2$ was repeatedly used.
Let us introduce the notation $n=100$, not forgetting that $n$ is even. Let the numbers start in clockwise order $a_{1}, a_{2}, \ldots, a_{n}$. The sum of the conditions of the problem gives $(a_{1}-1)^{2}+(a_{2}-1)^{2}+\ldots+(a_{n}-1)^{2}=n$.
The conditions are $ a_i + a_{i+1} = a_{i+2}^2$. Summing all of them and completing the squares gives us the equation.
Suppose that two adjacent numbers can simultaneously be at least $2$, and at least one of them is strictly greater than $2$. Then all subsequent numbers in the circle are strictly greater than $2$, which contradicts the obtained equality.
The case is considered similarly when two adjacent numbers can simultaneously be no more than $2$, and at least one of them is strictly less than $2$.
With the assumptions, $ a_{i+2} = \sqrt{ a_i + a_{i+1} } > 2$.
Proceed via induction to conclude that $ a_j > 2 \forall j$.
Likewise for the other case.
Thus, the numbers "alternate": no less than two and no more than two. Suppose WLOG that $a_{1}+a_{2} \geq 4$. Then
Hence if $ a_i \geq 2$, then $ a_{i+1} \not > 2 $ so $ a_{i+1} \leq 2$.
Likewise, $ a_{i+2} \not < 2$ so $ a_{i+2} \geq 2$.
Suppose WLOG that $a_3 \geq 2$, then $a_1 + a_ 2 = a_3 ^2 \geq 4$.
From the previous statement, $a_{2i+1 } \geq 2, a_{2i} \leq 2$.
$$ \begin{gathered} a_{1}+a_{2} \geq 4, a_{3}+a_{4} \geq 4, \ldots, a_{n-1}+a_{n} \geq 4, \\ 4 \geq a_{2}+a_{3}, 4 \geq a_{4}+a_{5}, \ldots, 4 \geq a_{n}+a_{1}. \end{gathered} $$
This follows because $ a_{2i+1} + a_{2i+2} = a_{2i+3} ^2 \geq 2^2 = 4$ and $ a_{2i} + a_{2i+1} = a_{2i+2} ^2 \leq 2^2 = 4$.
Adding all the inequalities, we get the inequality $0≥0$. Therefore,
$$ a_{1}+a_{2}=a_{3}+a_{4}=\ldots, a_{n-1}+a_{n}=a_{2}+a_{3}=a_{4}+a_{5}=\ldots=a_{n}+a_{1}=4. $$
Adding all of the inequalities, we get that $ 2n + \sum a_j \geq 2n + \sum a_j$, which means that equality holds throughout. (If equality doesn't hold throughout, then at least one of the inequalities is strict, which means that the summed inequality is strict, which is a contradiction.)
Hence, $ a_i + a_{i+1} = 4$.
Note: The solution writer might have intended to sum $a_{2i+1} + a_{2i+2} - 4 \geq 0$ in order to get the $ 0 \geq 0$.
So all numbers are equal to two.
Hence $ a_{i+2} = \sqrt{ a_i + a_{i+1} } = \sqrt{4} = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show that $2Q(x)$ can be written as the sum of three perfect squares (a) Show that $x^3-3xbc+b^3+c^3$ can be written in the form $(x+b+c)Q(x)$, where $Q(x)$ is a qudaratic expression.
(b) Show that $2Q(x)$ can be written as the sum of three expressions, each of which is a perfect square.
My attempt at solution: (a) Through long division, I arrived at $Q(x) = x^2-(b+c)x+b^2-bc+c^2$
(b) The only way that I know to rewrite a quadratic expression as perfect squares is through completing the square. So $$(x-\frac{b+c}{2})^2-(\frac{b+c}{2})^2+b^2-bc+c^2$$ $\therefore (x-\frac{b+c}{2})^2 + (b-c)^2 - \frac{1}{4}(b-c)^2$
$\therefore 2Q(x) = 2(x-\frac{b+c}{2})^2 + (b-c)^2 + \frac{1}{2}(b-c)^2$
This is the best that I've managed to come up with, but it doesn't seem to be right as the question mentioned it can be written as the sum of three perfect squares.
| I will put the Hessian matrix of $x^2 + y^2 + z^2 - yz - zx -xy$ and display , with square matrices $PQ = QP = I,$ $P^T HP = D$ is diagonal, so $Q^T D Q = H$
I will type in the final outcome first. The $Q^T DQ$ version reads
$$ \left( x - \frac{y}{2} - \frac{z}{2} \right)^2 + \frac{3}{4} (y-z)^2 = x^2 + y^2 + z^2 - yz - zx - xy $$
which does show that the quadratic form is rank two, not full rank. Furthermore, the form is zero when $x=y=z$
and only then. This does suggest that we might try to write the form using $(u-v)^2$ and $(v-w)^2.$ It will work out better if $(w-u)^2$ is included somehow
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$ P^T H P = D $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
\frac{ 1 }{ 2 } & 1 & 0 \\
1 & 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
2 & - 1 & - 1 \\
- 1 & 2 & - 1 \\
- 1 & - 1 & 2 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & \frac{ 1 }{ 2 } & 1 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
2 & 0 & 0 \\
0 & \frac{ 3 }{ 2 } & 0 \\
0 & 0 & 0 \\
\end{array}
\right)
$$
$$ $$
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
- \frac{ 1 }{ 2 } & 1 & 0 \\
- \frac{ 1 }{ 2 } & - 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
2 & 0 & 0 \\
0 & \frac{ 3 }{ 2 } & 0 \\
0 & 0 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } \\
0 & 1 & - 1 \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
2 & - 1 & - 1 \\
- 1 & 2 & - 1 \\
- 1 & - 1 & 2 \\
\end{array}
\right)
$$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4317837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Formula for distance between Incenter and Orthocenter. On pg. 205 of Johnson's Advanced Euclidean Geometry, he gives three formulas, preparatory to a proof of Feuerbach's theorem.
$$OI^2=R^2-2R\rho$$
$$IH^2=2\rho^2-2Rr$$
$$OH^2=R^2-4Rr$$
where $O$ is the circumcenter, $I$ the incenter, $H$ the orthocenter; $R$ the radius of the circumcircle, $\rho$, that of the inscribed circle, and $r$, the in-radius of the pedal triangle, the triangle formed by the feet of the altitudes.
The first equation is the theorem of Euler. The third follows from the first, by applying it to the pedal triangle, whose incenter is $H$, and circumcenter is $N$, the midpoint of $HO$.
Is there a correspondingly simple proof of the second formula? A non-trigonometric solution is preferred.
| Here is a try to get the formula quickly.
Disclaimer: I will use more or less standard notations in a triangle $\Delta ABC$. The side lengths are $a,b,c$, the incenter is $I$, the inradius is $r$ (not $\rho$), the circumradius is $R$, the area is $S$. Let $AA'$, $B'$, $CC'$ be the heights intersecting in $H$. The inradius of the $H$-pedal triangle $\Delta A'B'C'$ will be denoted by $r_H$.
The barycentric coordinates of $I$ are $I=[a:b:c]$, explicitly we have (identifying with affixes)
$$
(a+b+c)I=aA+bB+cC\ .
$$
Then considering vectors:
$$
\begin{aligned}
a(a+b+c)\overrightarrow{IA} &=
a^2\overrightarrow{AA} +
ab\overrightarrow{BA} +
ac\overrightarrow{CA}
\ ,\\
b(a+b+c)\overrightarrow{IB} &=
ba\overrightarrow{AB} +
b^2\overrightarrow{BB} +
bc\overrightarrow{CB}
\ ,\\
c(a+b+c)\overrightarrow{IC} &=
ca\overrightarrow{AC} +
cb\overrightarrow{BC} +
c^2\overrightarrow{CC}
\ ,
\end{aligned}
$$
so after adding:
$$
\tag{$1$}
a\overrightarrow{IA}+
b\overrightarrow{IB}+
c\overrightarrow{IC} =0\ .
$$
Write now $\overrightarrow{HA}=\overrightarrow{HI}+\overrightarrow{IA}$, take norms, and multiply with $a$. Write the similar equalities for $B$, $C$ instead of $A$ to get:
$$
\begin{aligned}
aHA^2 &=aHI^2 + aIA^2 + 2a\overrightarrow{HI}\cdot \overrightarrow{IA}\ ,\\
bHB^2 &=bHI^2 + bIB^2 + 2b\overrightarrow{HI}\cdot \overrightarrow{IB}\ ,\\
cHC^2 &=cHI^2 + cIC^2 + 2c\overrightarrow{HI}\cdot \overrightarrow{IC}\ .
\end{aligned}
$$
Adding, we get rid of the scalar products by $(1)$, and obtain:
$$
\tag{$2$}
aHA^2+bHB^2+cHC^2 = (a+b+c)HI^2
+ \underbrace{aIA^2+bIB^2+cIC^2}_{\text{some constant }K^3}\ .
$$
Note: We did not use any special property of $H$, so $(2)$ is valid for any point $X$ instead of $H$, explicitly
$$
\tag{$2_X$}
aXA^2+bXB^2+cXC^2 = (a+b+c)XI^2 + K^3\ .
$$
(Key-word: Leibnitz formula.)
(For $I$ we have used only $I=[a:b:c]$. So we can also replace $I$ by a general point $P=[u:v:w]$, using the corresponding barycentric coordinates $u,v,w$ instead.)
Now compute the relation $a(2_A)+b(2_B)+c(2_C)$ divided by $(a+b+c)$, weights being used to make appear on the R.H.S. the term $\sum a AI^2=K^3$.
So after the sum we have on the R.H.S. $K^3+K^3$.
On the L.H.S. we get $\frac 1{a+b+c}\sum(abc^2+acb^2)=2abc$.
This gives the value for $K^3$,
$$
K^3 = abc = 4R\;S=4R\; rp=2Rr\;(a+b+c)\ .
$$
Here $p=\frac 12(a+b+c)$ is the half perimeter. Inserting this into $(2)$ we get
$$
\tag{$3$}
\frac1{a+b+c}(aHA^2+bHB^2+cHC^2) = HI^2
+ \underbrace{2Rr}_{abc/(a+b+c)}\ .
$$
We want now to show $(!)$ the following similar relation:
$$
\tag{$4$}
2r^2 \overset != HI^2
+ 2Rr_H\ .
$$
Showing it is equivalent to:
$$
\tag{$4'$}
\frac1{2p}(aHA^2+bHB^2+cHC^2) - \frac{abc}{2p}
\overset !=
2r^2 - 2Rr_H\ .
$$
It is time to write down some trigonometric formulas.
$$
\begin{aligned}
BC &= a\ ,\\
B'C' &= BC\cos A = a\cos A &&\text{ from }\Delta ABC\sim \Delta AC'B'\ ,\\
AH &= AB'\sin\widehat{AHB'}=c\cos A\sin C= 2R\cos A\ ,\\
HA'&=HB\cos\widehat{AHB'} =2R\cos B\cos C\ ,\\
r_H &=HA'\cdot \frac{B'C'}{BC}=2R\cos A\cos B\cos C
&&\text{ from }\Delta HB'C'\sim \Delta HCB\ ,\text{ so}\\
2Rr_H &= HA\cdot HA'=HB\cdot HB'=HC\cdot HC'\ .
\end{aligned}
$$
I understand that trigonometry should be avoided, but then computations
are hard to write. Above, there are also some alternative relations showing how to avoid - if really needed - trigonometric formulas. At any rate, no trigonometric relations will be used below. (E.g. like writing $\cos^2A$ in terms of $\cos 2A$ or so.)
We want to show $(4')$ and are in position to show this as an algebraic formula in the variables $a,b,c$. For this, we use for $S^2$ Heron, and $R$ is eliminated via $abc=4RS$.
The version of $(4')$ suited for this polynomial check is:
$$
\tag{$4''$}
\frac1{2p}4R^2(a\cos^2A + b\cos^2B + c\cos^2 C) - \frac{abc}{2p}
\overset !=
\frac{2S^2}{p^2} - 4R^2\cos A\cos B\cos C\ .
$$
Equivalently:
$$
\tag{$4'''$}
\frac1{2p}4R^2(a\cos^2A + b\cos^2B + c\cos^2 C)
+
\frac1{2p}4R^2(2p\; \cos A\cos B\cos C)
\overset !=
\frac{abc+4(p-a)(p-b)(p-c)}{2p} \ .
$$
And we multiply with $2p$.
I have to finish, so a computer check is done for the above.
var('a,b,c')
p = (a + b + c)/2
SS = p*(p - a)*(p - b)*(p - c)
RR = (a*b*c/4)^2 / SS
def f(a, b, c): return (b^2 + c^2 - a^2)/2/b/c
cosA, cosB, cosC = f(a,b,c), f(b,c,a), f(c,a,b)
RHS = a*b*c + 4*(p - a)*(p - b)*(p - c)
LHS = 4*RR*(a*cosA^2 + b*cosB^2 + c*cosC^2
+ 2*p*cosA*cosB*cosC)
print(bool(LHS == RHS))
And the computer checks the relation. Explicitly:
sage: print( bool(LHS == RHS) )
True
sage: 2*RHS.expand()
-a^3 + a^2*b + a*b^2 - b^3 + a^2*c + b^2*c + a*c^2 + b*c^2 - c^3
sage: 2*LHS.factor().expand()
-a^3 + a^2*b + a*b^2 - b^3 + a^2*c + b^2*c + a*c^2 + b*c^2 - c^3
(I will search for a better finish
based on geometric relations.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $\frac{1+2x}{1+\sqrt{1+2x}}+\frac{1-2x}{1-\sqrt{1-2x}}$ for $x=\frac{\sqrt3}{4}$. Find the value of $$\dfrac{1+2x}{1+\sqrt{1+2x}}+\dfrac{1-2x}{1-\sqrt{1-2x}}$$ for $x=\dfrac{\sqrt3}{4}$.
I have no idea why I can't solve this problem. I tried to simplify the given expression, but I was able to reach only $$\dfrac{2-\sqrt{1+2x}\left(\sqrt{(1-2x)(1+2x)}-1\right)}{2x}$$ I also tried to plug in $x=\dfrac{\sqrt3}{4}$ directly, but I wasn't able to get anything. Thank you!
| You can dodge a certain amount of work by re-writing the given expression using $ \ u \ = \ \sqrt{1 + 2x} \ $ and $ \ v \ = \ \sqrt{1 - 2x} \ \ $ to produce
$$ \frac{1 \ + \ 2x}{1 \ + \ \sqrt{1 + 2x}} \ + \ \frac{1 \ - \ 2x}{1-\sqrt{1-2x}} \ \ \rightarrow \ \ \frac{u^2}{1 \ + \ u} \ + \ \frac{v^2}{1 \ - \ v} \ \ = \ \ \frac{u^2 \ - \ u^2v \ + \ v^2 \ + \ uv^2 }{1 \ + \ u \ - \ v \ - \ uv} $$ $$ = \ \ \frac{u^2 \ + \ v^2 \ - \ uv · (u \ - \ v) }{1 \ + \ (u \ - \ v) \ - \ uv} \ \ . $$
We can now make a good deal of headway with $ \ u^2 \ = \ 1 + \frac{\sqrt3}{2} \ \ , \ \ v^2 \ = \ 1 - \frac{\sqrt3}{2} \ \ , $ $ uv \ = \ \sqrt{1 - 4x^2} \ = \ \sqrt{1 - 4·\frac{3}{16}} \ = \ \sqrt{\frac14} \ = \ \frac12 \ \ . $
But we finally have to "buckle down" and deal with the difference $ \ u - v \ \ . $ We can make use of the fact that the irrational square-root under the radical remains part of the resulting value. Thus, $ \ u \ = \ \sqrt{1 + \frac{\sqrt3}{2}} \ $ can be treated as $ \ \sqrt2·u \ = \ \sqrt{2 + \sqrt3 } \ = \ a + b·\sqrt3 \ \ . $ This requires that $ \ a^2 + 3b^2 \ = \ 2 \ \ , \ \ 2ab \ = \ 1 \ \ , \ $ for which we can obtain a solution "by inspection" (or more formally) as $ \ a \ = \ b \ = \ \frac{1}{\sqrt2} \ \ . $ From this, we obtain
$$ u \ \ = \ \ \sqrt{1 + \frac{\sqrt3}{2}} \ \ = \ \ \frac{\frac{1}{\sqrt2} + \frac{1}{\sqrt2}·\sqrt3 }{\sqrt2} \ \ = \ \ \frac12 + \frac{\sqrt3}{2} \ \ \ , \ \ \ v \ \ = \ \ \sqrt{1 - \frac{\sqrt3}{2}} \ \ = \ \ -\frac12 + \frac{\sqrt3}{2} $$
$$ \Rightarrow \ \ u \ - \ v \ \ = \ \ 1 \ \ . $$
(This sort of question comes up a lot on MSE: as just one example, here .)
Hence, for $ \ x \ = \ \frac{\sqrt3}{4} \ \ , $
$$ \frac{1 \ + \ 2x}{1 \ + \ \sqrt{1 + 2x}} \ + \ \frac{1 \ - \ 2x}{1-\sqrt{1-2x}} \ \ = \ \ \frac{\left(1 + \frac{\sqrt3}{2} \right) \ + \ \left(1 - \frac{\sqrt3}{2} \right) \ - \ \frac12 · 1 }{1 \ + \ 1 \ - \ \frac12} \ \ $$ $$ = \ \ \frac{1 \ + \ 1 \ - \ \frac12 }{\frac32} \ \ = \ \ 1 \ \ . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.