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If $f(1)=1\;,f(2)=3\;,f(3)=5\;,f(4)=7\;,f(5)=9$ and $f'(2)=2,$ Then sum of all digits of $f(6)$
$(1):$ If $P(x)$ is a polynomial of Degree $4$ such that $P(-1) = P(1) = 5$ and
$P(-2)=P(0)=P(2)=2\;,$Then Max. value of $P(x).$
$(2):$ If $f(x)$ is a polynomial of degree $6$ with leading Coefficient $2009.$ Suppose
further that $f(1)=1\;,f(2)=3\;,f(3)=5\;,f(4)=7\;,f(5)=9$ and $f'(2)=2,$
Then sum of all digits of $f(6)$ is
$\bf{My\; Try\; For\; (1):}$ Given $x=-2\;,x=0\;,x=+2$ are the roots of $P(x)=0.$
So $(x+2)\;,(x-0)\;,(x-2)$ are factors of $P(x)=0$. So we can write $P(x)$ as
$P(x) = A\cdot x\cdot (x-2)\cdot (x+2)(x-r)\;,$ So we have calculate value of $A$ and $r$
Now put $x=-1\;,$ we get $P(-1)=-3A\cdot (1+r)\Rightarrow -3A\cdot (1+r)=5............................(1)$
Now put $x=1\;,$ we get $P(1)=-3A\cdot (1-r)\Rightarrow -3A\cdot (1-r)=5..................................(2)$
So from $(1)$ and $(2)\;,$ We get $r=0$ and $\displaystyle A=-\frac{5}{3}.$
So Polynomial $\boxed{\boxed{\displaystyle P(x)=-\frac{5}{3}\cdot x^2\cdot (x^2-4)}}$
$\bf{My\; Try\; For \; (2):}$Let $f(x)=2x-1\;\forall\; x=1\;,2\;,3\;,4\;,5.$
So we can say that $(x-1)\;,(x-2)\;,(x-3)\;,(x-4)\;,(x-5)$ are the roots of $f(x)-2x+1=0$
So $f(x)-2x+1=2009\cdot \underbrace{(x-1)\cdot(x-2)\cdot (x-3)\cdot (x-4)\cdot (x-5)}\cdot\underbrace{(x-r)}$
Now How can i solve after that
Help me and plz explain me, is my $(1)$ Try is right or not
Thanks
| Question 1 directly states that $P(-2)=P(0)=P(2)=2$. Why on earth are you then saying that they're roots of the polynomial?
What you can say is that $x=-2,0,2$ are roots of $P(x)=2$, or equivalently $P(x)-2=0$. So we can say that, if $P(x)$ is 4th degree, then $P(x)-2$ is as well, and hence $P(x)-2=A(x-\alpha)(x-2)x(x+2)$ for some $A,\alpha$. Then you can insert your values for $P(1)$ and $P(-1)$.
Also, you haven't answered the question. Read it carefully, see what it's asking, and make sure you answer it.
As for 2, you're on the right track. The next step would probably be to look at what the derivative of that expression will look like at $x=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Domain, range and zeros of $f(x,y)=\frac{\sqrt{4x-x^2-y^2}}{x^2y^2-4xy^2+3y^2}$ Given the following function with two variables:
\begin{equation}
\frac{\sqrt{4x-x^2-y^2}}{x^2y^2-4xy^2+3y^2}
\end{equation}
I need to find a) the domain for the above function.
Can anyone give me a hint on how to find the domain in f?
I already know that:
\begin{equation}
dom f = \{ f(x,y) ∈ \mathbb{R}^2\backslash(x^2y^2-4xy^2+3y^2=0) : \sqrt{4x-x^2-y^2} \geq 0 \}
\end{equation}
But of course this needs to be written in a more simpler form.
During class we solve simpler functions like without fractions and roots, so I don't have anything that can help me.
After this I also need to find:
*
*b) zeros of the function
*c) Calculating Algebraically, the range of the function:
\begin{equation}
T(x,y) = \sqrt{4x-x^2-y^2}
\end{equation}
*d) Extreme values of the function T
I'm of course not expecting the complete solution but something like a kick start.
| You factored your first equation correctly. Now determine which values of $y$, $x$, yield a zero denominator: $$y^2(x-3)(x-1) = 0 \implies y = 0, \text{ or } x = 3, \text{ or } x = 1$$
So all ordered pairs of the form $(x, 0), (3, y), (1, y), \; x, y \in \mathbb R$ must be omitted from the domain. One way to see this is to note that the horizontal line $y = 0$ (i.e., the $x$-axis), and the vertical lines $x = 1, x = 3$ are thus excluded from $\mathbb R^2$.
Also, the second restriction does not require $\sqrt{4x-x^2-y^2} \geq 0$ (though it certainly will be true, given the proper restriction on the argument). Rather, we need the argument of the square root to be greater than or equal to zero, in order for the square root of the argument to be defined. $$\begin{align} 4x - x^2 - y^2 \geq 0 & \iff -4x + x^2 + y^2\leq 0\\ &\iff x^2 - 4x + 4 + y^2 \leq 4 \\ &\iff (x - 2)^2 + y^2\leq (2)^2\end{align}$$
Note that $$ (x - 2)^2 + y^2= (2)^2$$ is the equation of a circle of radius 2, centered at the point $(2, 0)$. So $$(x - 2)^2 + y^2\leq (2)^2$$ is the subset of $\mathbb R^2$ consisting of all points on and inside the circle.
So the domain will be equal to the portion of the plane on and inside circle (all points $(x, y)$ such that $(x - 2)^2 + y^2\leq (2)^2)$, and excluding the lines $y = 0, x=1, x=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim_{n\rightarrow\infty}(1+\frac{1}{n})(1+\frac{2}{n})^{\frac{1}{2}}\cdots(1+\frac{n}{n})^{\frac{1}{n}}$ Evaluate
$$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)^{\frac{1}{2}}\cdots \left(1+\frac{n}{n}\right)^{\frac{1}{n}}$$
solve:
$$ \exp\left\{\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{k=1}^{n} \frac{\ln(\frac{k}{n}+1)}{\frac{k}{n}}\right\}= \exp\left\{\int_0^1 \frac{\ln(1+x)}{x}\,dx\right\}$$
how to evaluate
$$\int_0^1\frac{\ln(1+x)}{x}\,dx$$
| $\dfrac{\ln (1+x)}{x} = 1 -\dfrac{x}{2} + \dfrac{x^2}{3} - \dfrac{x^3}{4} + ....$, thus:
$\displaystyle \int_{0}^1 \dfrac{\ln (1+x)}{x}dx = 1 - \dfrac{1}{2^2} + \dfrac{1}{3^2} - \dfrac{1}{4^2} - ... = \displaystyle \sum_{k=1}^\infty \dfrac{1}{k^2} - 2\cdot \dfrac{1}{2^2}\displaystyle \sum_{k=1}^\infty \dfrac{1}{k^2} = \dfrac{\pi^2}{12}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\lim_{(x,y)\rightarrow (0,0)} \frac{(xy)^3}{x^2+y^6}$ $$\lim_{(x,y)\rightarrow (0,0)} \frac{(xy)^3}{x^2+y^6}$$
I don't really know how to do, but I was trying to do like that:
$a=x$,
$b=y^2$
then I was trying to do this
$$\lim_{(x,y)\rightarrow (0,0)} \frac{ab}{a^2+b^2}$$
then I don't know no more how to do...
| If the correct wording is
$$\lim_{(x,y)\rightarrow (0,0)} \frac{x(y^3)}{x^2+y^6}$$
it is a good idea to change into :
$a=x$,
$b=y^3$
then is there a "limit" to :
$$\lim_{(a,b)\rightarrow (0,0)} \frac{ab}{a^2+b^2}$$
$$\frac{ab}{a^2+b^2}=\frac{1}{\frac{a}{b}+\frac{b}{a}}$$
if $a$ and $b$ both tend to $0$ then $\frac{a}{b}$ and $\frac{b}{a}$ are undetermined such as $\frac{0}{0}$. So a limit doesn't exists.
$\frac{ab}{a^2+b^2}$ is undeterminated in $a$ and $b$ tending to $0$.
You obtain as many different values as you want, depending the manner of making $a$ and $b$ tend to $0$. For examples :
If $a=b$ tend to 0, then $\lim_{(a,b)\rightarrow (0,0)} \frac{ab}{a^2+b^2}=\frac{1}{2}$
If $a=2b$ tend to 0, then $\lim_{(a,b)\rightarrow (0,0)} \frac{ab}{a^2+b^2}=\frac{2}{5}$
If $a=kb$ tend to 0, then $\lim_{(a,b)\rightarrow (0,0)} \frac{ab}{a^2+b^2}=\frac{k}{1+k^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
if $6^m+2^{m+n}\cdot 3^w+2^n = 332\;,$ Then $m^2+mn+n^2 = \;,$ where $m,n,w \in \mathbb{Z^{+}}$ Suppose $m.n$ are positive integers such that $6^m+2^{m+n}\cdot 3^w+2^n = 332\;,$ Then $m^2+mn+n^2 = $
$\bf{My\; Try::}$ Given $6^m+2^{m+n}\cdot 3^w+2^n=332\Rightarrow 2^m\cdot 3^n+2^{m}\cdot 3^n\cdot 3^w+2^n=332=2^2\cdot 133$
$\Rightarrow 2^m\cdot \left(3^n+3^{n}\cdot 3^{w}+2^{n-m}\right)=2^2\cdot 133\;,$ Using Camparasion, we get $m=2$
$\Rightarrow (3^n+3^m\cdot 9+2^{n-2}) = 133$
Now How Can i solve after that, Help me
| You made some arithmetic errors in what you did (the initial factorization is incorrect, plus $332 = 4\cdot 83$, not $4\cdot 133$) , but you are on the right track.
$$
6^m + 2^{m+n}\cdot 3^w + 2^n = 332 \Rightarrow
2^m\cdot 3^m + 2^m\cdot 2^n \cdot 3^w + 2^n = 332.
$$
Now assuming $m\le n$ (if it doesn't work, you must also try $n\le m$), this is equal to
$$2^m(3^m + 2^n\cdot 3^w + 2^{n-m}) = 2^2\cdot 83,$$
so that as you say $m=2$, and then
$$9 + 2^n\cdot 3^w + 2^{n-2} = 83 \Rightarrow 2^{n-2}(4\cdot 3^w + 1) = 74.$$
Thus $n=3$ (and $w=2$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $I_{\alpha}=\int_{0}^{\infty} \frac{\ln(1+x^2)}{x^\alpha}dx$ using complex analysis Again, improper integrals involving $\ln(1+x^2)$
I am trying to get a result for the integral $I_{\alpha}=\int_{0}^{\infty} \frac{\ln(1+x^2)}{x^\alpha}dx$ - asked above link- using some complex analysis, however, I couldn't find an appropriate solution. (Of course, for $\alpha \in (0,3)$ as stated above link (o.w. it is divergent))
Any ideas?
| Let $\alpha = 1 + p$. Then we have
$$
\int_0^{\infty}\frac{\ln(1+x^2)}{x^{1+p}}dx
$$
where $0<p<2$. Let $u = \ln(1+x^2)$ and $dv = x^{-1-p}dx$. Then $du = \frac{2x}{1+x^2}$ and $v = \frac{-1}{px^p}$.
$$
uv\big|_0^{\infty}\to 0
$$
for $p\in(0,2)$.
$$
\int_0^{\infty}\frac{\ln(1+x^2)}{x^{1+p}}dx=\frac{2}{p}\int_0^{\infty}\frac{x^{1-p}}{x^2+1}dx
$$
Let $\beta -1 = 1-p$ so $p=2-\beta$. Let $x=t^2$ so $dx=2tdt$.
$$
\frac{4}{2-\beta}\int_0^{\infty}\frac{t^{2\beta - 1}}{t^4+1}dt
$$
The contour we desire is
The integrals over the semicircles go to zero as $R\to\infty$ and $\delta\to 0$ so we have
\begin{align}
\frac{4}{2-\beta}\int_0^{\infty}\frac{t^{2\beta - 1}}{t^4+1}dt &=
\frac{4}{2-\beta}\int_{-\infty}^{\infty}\frac{z^{2\beta - 1}}{z^4+1}dz\\
&= \frac{4}{2-\beta}\int_0^{\infty}\frac{z^{2\beta - 1}+(-z)^{2\beta - 1}}{z^4+1}dz
\end{align}
Now, $(-z)^{2\beta} = -e^{2\pi i\beta}z^{2\beta}$.
$$
\frac{4}{2-\beta}\int_0^{\infty}\frac{z^{2\beta - 1}+(-z)^{2\beta - 1}}{z^4+1}dz
= \frac{4(1-e^{2\pi i\beta})}{2-\beta}\int_0^{\infty}\frac{z^{2\beta - 1}}{z^4+1}dz\tag{1}
$$
Then $(1)$ is equal to $2\pi i$ times sum of the residue is the upper half plane.
\begin{align}
\frac{8\pi i}{(1-e^{2\pi i\beta})(2-\beta)}\Bigl[\lim_{z\to e^{\frac{i \pi }{4}}}\frac{(z-e^{\frac{i \pi }{4}}) z^{2\beta -1}}{z^4+1}+\lim_{z\to e^{\frac{3i\pi}{4}}}\frac{(z-e^{\frac{3i\pi}{4}})z^{2\beta-1}}{z^4+1}\Bigr] &=
\frac{\pi\csc\bigl(\frac{\pi\beta}{2}\bigr)}{2-\beta}\\
&= \frac{\pi\csc\bigl(\frac{\pi p}{2}\bigr)}{p}\tag{2}\\
&= \frac{\pi\sec\bigl(\frac{\pi\alpha}{2}\bigr)}{1-\alpha}\tag{3}
\end{align}
Equations $(2)$ and $(3)$ occur by back substituting for $\beta$ and $p$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Trigonometric substitution in the integral $\int x^2 (1-x^2)^{\frac{9}{2}} \ \mathrm dx$ I'm trying to solve
$$\int_{-1}^{1} x^2(1-x^2)^{\frac{9}{2}} \, dx$$
The hint said to use the substitution $x=\sin y$
I got $$\int_{-\pi/2}^{\pi/2} \sin^2y \cos^{\frac{11}{2}} y \,dy$$
|
$$x=\sin y$$
$$\int x^2 ( 1-x^2)^{\frac{9}{2}}\,dx = \int x^2 (\sqrt{1-x^2})^9 \,dx= \int \sin^2 y (\sqrt{1-sin^2 y})^9 \cos y\, dy $$
$$= \int \sin^2 y (\sqrt\cos^2 y)^9 \cos y\, dy= \int \sin^2 y (\cos y)^9 \cos y\, dy$$
$$\int \sin^2 y \cos^{10} y \,dy$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1028779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_0^\infty \frac{(\log x)^2}{1+x^2} dx$ using complex analysis How do I compute
$$\int_0^\infty \frac{(\log x)^2}{1+x^2} dx$$
What I am doing is take
$$f(z)=\frac{(\log z)^2}{1+z^2}$$
and calculating
$\text{Res}(f,z=i) = \dfrac{d}{dz} \dfrac{(\log z)^2}{1+z^2}$
which came out to be $\dfrac{\pi}{2}-\dfrac{i\pi^2}{8}+\dfrac{i\pi}{2}$
Im not too sure how to move on from here. the given answer is $\dfrac{\pi^3}{8}$
Any help will be appreciated. thank you in advanced.
| We have
$$\int_1^{\infty} \dfrac{\ln^2(x)}{1+x^2}dx = \int_1^0 \dfrac{\ln^2(1/x)}{1+1/x^2} \dfrac{-dx}{x^2} = \int_0^1 \dfrac{\ln^2(x)}{1+x^2}dx$$
Hence, our integral is
$$I = 2\int_0^1 \dfrac{\ln^2(x)}{1+x^2}dx$$
From complex analysis, since $\dfrac1{1+z^2}$ is analytic in the open unit disk, we have $$\dfrac1{1+z^2} = \sum_{k=0}^{\infty}(-1)^kz^{2k}$$
Hence,
$$I = 2\sum_{k=0}^{\infty}(-1)^k \int_0^1 x^{2k}\ln^2(x)dx = 2\sum_{k=0}^{\infty}(-1)^k \dfrac2{(2k+1)^3} = 2 \cdot 2 \cdot \dfrac{\pi^3}{32} = \dfrac{\pi^3}8$$
where the last equality follows from the fact that
$$\text{Li}_n(e^{2\pi ix}) + (-1)^n \text{Li}_n(e^{-2\pi ix}) = -\dfrac{(2 \pi i)^n}{n!} B_n(x)$$
taking $n=3$, $x=1/4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $ \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+\cdots + \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}$ for $n\in \mathbb N$ I want to prove that if $n \in \mathbb N$ then $$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+ \cdots+ \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}.$$
I think I am stuck on two fronts. First, I don't know how to express the leading terms on the left hand side before the $\dfrac{n}{(n+1)!}$ (or if doing so is even necessary to solve the problem). I am also assuming that the right high side should initially be expressed $1 - \dfrac{1}{(n+2)!}$. But where to go from there.
I'm actually not sure if I'm even thinking about it the right way.
| We'll use induction to prove this
given statement is true for $n=1$
then we assume that it's true for $n=k$
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+ \cdots+ \frac{k}{(k+1)!} = 1 - \frac{1}{(k+1)!}$$
and then by adding $\dfrac{k+1}{(k+2)!}$ on both sides we get
$$\begin{align}
\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+ \cdots+ \frac{k}{(k+1)!} +\frac{k+1}{(k+2)!}&= 1 - \frac{1}{(k+1)!}+\frac{k+1}{(k+2)!}\\
&=1 - \frac{k+2}{(k+2)!}+\frac{k+1}{(k+2)!}\\
&=1 +\frac{k+1-k-2}{(k+2)!}\\
&=1-\frac{1}{(k+2)!}\end{align}$$
it's also true for $n=k+1$
And then using Proncipal of mathematical induction...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proving and Finding a limit I need to find the following limit and prove using the definition of limits.
$$\lim_{x\to1} {x \over x+1} = \frac 1 2$$.
Following the definition:
$$\forall \epsilon \exists \delta : \lvert x - c \rvert < \delta \Rightarrow \lvert F(x) - L \rvert < \epsilon$$
$$\left\lvert \frac{x}{x+1} - \frac{1}{2} \right\rvert < \epsilon = \left\lvert 2x-x-1 \over 2x+2 \right\rvert = \left\lvert x-1 \over 2x+2 \right\rvert < \epsilon$$
I have trouble around here. I don't know how to reach $\left\lvert x - c \right\rvert < \delta$
I tried:
$$ \frac{x-1}{2x+2} < \frac{x}{2x} = \frac{1}{2} < \epsilon$$
But something about that doesn't seem right to me. Can I get any hints?
| Showing the versatility of this problem.
$$\begin{align}|x - 1| < \delta \leq \frac{1}{2} &\Leftrightarrow -\frac{1}{2} < x - 1 < \frac{1}{2} \Leftrightarrow \frac{1}{2} < x < \frac{3}{2} \\ &\Leftrightarrow \frac{3}{2} < x + 1 < \frac{5}{2} \Rightarrow \frac{2}{5} < \frac{1}{x+1} < \frac{2}{3} \end{align}$$
In particular $\Big|\frac{1}{x+1}\Big| < \frac{2}{3}$.
Then as you have already reached
$$\left\lvert \frac{x}{x+1} - \frac{1}{2} \right\rvert = \left\lvert 2x-x-1 \over 2x+2 \right\rvert = \left\lvert x-1 \over 2x+2 \right\rvert = \frac{1}{2}\left\lvert x-1 \over x+1 \right\rvert < \frac{\delta}{3}$$
Now take $\delta = \min \lbrace \epsilon , \frac{1}{2}\rbrace$. And you will have for every $\epsilon > 0$ given, there exits $\delta = \min \lbrace \epsilon, \frac{1}{2}\rbrace$ such that
$$|x - 1| < \delta \Rightarrow \Bigg|\frac{x}{x + 1} - \frac{1}{2}\Bigg| < \epsilon$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1032193",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Maximum GCD of two polynomials Consider $f(n) = \gcd(1 + 3 n + 3 n^2, 1 + n^3)$
I don't know why but $f(n)$ appears to be periodic. Also $f(n)$ appears to attain a maximum value of $7$ when $n = 5 + 7*k $ for any $k \in \Bbb{Z}$.
Why?
And how would one find this out?
Given two polynomials $p_1(n)$ and $p_2(n)$, how does one calculate the maximum value of $\gcd( p_1(n), p_2(n) ) $ and where this maximum is found?
| If integer $d$ divides both $1+n^3,1+3n+3n^2$
$d$ must divide $n(1+3n+3n^2)-3(1+n^3)=3n^2+n-3$
$d$ must divide $1+3n+3n^2-(3n^2+n-3)=2n+4$
$d$ must divide $3n(2n+4)-2(1+3n+3n^2)=6n-2$
$d$ must divide $3(2n+4)-(6n-2)=14$
So, $(1+n^3,1+3n+3n^2)$ must divide $14$
Now, $3n^2+3n+1=6\dfrac{n(n+1)}2+1\equiv1\pmod2\implies 3n^2+3n+1$ is odd
So, $(1+n^3,1+3n+3n^2)$ must be $1$ or $7$
Again, $3n^2+3n+1\equiv0\pmod7\iff3(n^2+n-2)\equiv0\iff n^2+n-2\equiv0$
$\iff(n+2)(n-1)\equiv0\iff n\equiv1,-2$
Now, $n\equiv1\implies n^3+1\not\equiv0\pmod7$
and $n\equiv-2\pmod7\implies n^3+1\equiv0$
So, $(1+n^3,1+3n+3n^2)=7$ if $n\equiv-2\pmod7$ else they are co-prime
| {
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"timestamp": "2023-03-29T00:00:00",
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Double Integral $\int\limits_0^a\int\limits_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$ How to solve this integral?
$$\int_0^a\!\!\!\int_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$$
my attempt
$$
\int_0^a\!\!\!\int_0^a\frac{dx \, dy}{(x^2+y^2+a^2)^\frac{3}{2}}=
\int_0^a\!\!\!\int_0^a\frac{dx}{(x^2+\rho^2)^\frac{3}{2}}dy\\
\rho^2=y^2+a^2\\
x=\rho\tan\theta\\
dx=\rho\sec^2\theta \, d\theta\\
x^2+\rho^2=\rho^2\sec^2\theta\\
\int_0^a\!\!\!\int_0^{\arctan\frac{a}{\rho}}\frac{\rho\sec\theta}{\rho^3\sec^3\theta}d\theta \, dy=
\int_0^a\!\!\!\frac{1}{\rho^2}\!\!\!\int_0^{\arctan\frac{a}{\rho}}\cos\theta \, d\theta \, dy=\\
\int_0^a\frac{1}{\rho^2}\sin\theta\bigg|_0^{\arctan\frac{a}{\rho}} d\theta \, dy=
\int_0^a\frac{1}{\rho^2}\frac{x}{\sqrt{x^2+\rho^2}}\bigg|_0^ady=\\
\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy$$
Update:
$$\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy=\frac{\pi}{6a}$$
| The natural course of action whenever you see $x^2 + y^2$ in a multiple integration problem is to convert to polar coordinates.
Set $x = r\cos\theta$, $y = r\sin\theta$, $0 \leq \theta <2\pi$. Then $dx~dy = r~dr~d\theta$, and the integrand becomes
$$
\frac{r}{(r^2 + a^2)^{3/2}}~dr~d\theta.
$$
This integral doesn't depend on $\theta$, so in fact when you integrate in $r$ it is a one-variable integral, and from single-variable calculus we know that the substitution $u = r^2 + a^2$ will do the trick. The only thing left to do is to find the limits of integration.
The region $0 \leq x \leq a$, $0 \leq y \leq a$ is a square of side length $a$ in the first quadrant of the plane. This can be parametrized by $0 \leq \theta \leq \frac{\pi}{4}$, $0 \leq r \leq \sqrt{a^2 + (a\sin\theta)^2} = a\sqrt{1 + \sin^2 \theta}$. (Draw a picture! This is the meat of the problem.) Therefore the integral becomes
$$
\int_{\theta = 0}^{\pi/4}\int_{r=0}^{a\sqrt{1+\sin^2\theta}} \frac{r}{(r^2 + a^2)^{3/2}}~dr~d\theta.
$$
Can you take it from here?
| {
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Sinha's Theorem for Equal Sums of Like Powers $x_1^7+x_2^7+x_3^7+\dots$ Sinha’s theorem can be stated as, excluding the trivial case $c = 0$, if,
$$(a+3c)^k + (b+3c)^k + (a+b-2c)^k = (c+d)^k + (c+e)^k + (-2c+d+e)^k\tag{1} $$
for $\color{blue}{\text{both}}$ $k = 2,4$ then,
$$a^k + b^k + (a+2c)^k + (b+2c)^k + (-c+d+e)^k = \\(a+b-c)^k + (a+b+c)^k + d^k + e^k + (3c)^k
\tag{2}$$
for $k = 1,3,5,7$.
The system $(1)$ be equivalently expressed as,
$$\begin{align}
x_1^k+x_2^k+x_3^k\, &= y_1^k+y_2^k+y_3^k,\quad \color{blue}{\text{both}}\; k = 2,4\\
x_1+x_2-x_3\, &= 2(y_1+y_2-y_3)\\
x_1+x_2-x_3\, &\ne 0\tag{3}
\end{align}$$
There are only two quadratic parameterizations known so far to $(3)$, namely,
$$(-5x+2y+z)^k + (-5x+2y-z)^k + (6x-4y)^k = \\(9x-y)^k + (-x+3y)^k + (16x-2y)^k\tag{4}$$
where $126x^2-5y^2 = z^2$ and,
$$(6x+3y)^k + (4x+9y)^k + (2x-12y)^k = \\(-x+3y+3z)^k + (-x+3y-3z)^k + (-6x-6y)^k\tag{5}$$
where $x^2+10y^2 = z^2$ found by Sinha and (yours truly). The square-free discriminants are $D = 70, -10$, respectively.
Question: Any other solution for $(3)$ in terms of quadratic forms?
P.S. There are a whole bunch of elliptic curves that can solve $(3)$.
| (Too long for a comment.)
I simplified your expression and found they are ternary quadratic forms. (Why didn't you just simplify them? Maple and Mathematica can do it easily.) So,
$$R^n+Q^n+T^n = X^n+Y^n+Z^n,\quad for\; n =2,4\tag{1}$$
$$\begin{align}R =& -2 k^2 - 2 k s + s^2 + 3 k t - t^2\\
Q =&\; k^2 - 2 k s - 2 s^2 + 3 s t - t^2\\
T =&\; k^2 + 4 k s + s^2 - 3 k t - 3 s t + 2 t^2\\
X =&\; k^2 + k s + s^2 - t^2\\
Y =&\; k^2 + k s + s^2 - 3 k t - 3 s t + 2 t^2\\
Z =& -2 k^2 - 2 k s - 2 s^2 + 3 k t + 3 s t - t^2
\end{align}$$
Unfortunately, this also satisfies,
$$R+Q+T =X+Y+Z = 0$$
or equivalently, since $(1)$ involves even powers,
$$R+Q-(-T) =X+Y-(-Z) = 0$$
a case prohibited by Sinha's theorem.
| {
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Irrational number inequality : $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$ it is easy and simple I know but still do not know how to show it (obviously without simply calculating the sum but manipulation on numbers is key here.
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$$
| In general, for $n \ge 3$,
$$\sum_{k=1}^{n} \frac{1}{\sqrt{k}} \gt 2(\sqrt{n+1}-1) \gt \sqrt{n}$$
by using
$$\frac{1}{\sqrt{n}} = \frac{2}{2\sqrt{n}} \gt \frac{2}{\sqrt{n+1} + \sqrt{n}} = 2(\sqrt{n+1} - \sqrt{n})$$
and having a telescopic sum.
For $n \ge 3$ we also have that
$$ 2\sqrt{n+1} - 2 = \sqrt{n+1} + (\sqrt{n+1} -2) \gt \sqrt{n} $$
| {
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Find equation of Tangent line at $(4, 1)$ on $5y^3 + x^2 = y + 5x$ Can someone help me find equation of tangent line at $(4, 1)$ on
$5y^3 + x^2 = y + 5x$
$Y=f(x)$
I dont know how to isolate the $Y$
| $$\dfrac{d}{dx}5y^3 + x^2 = y + 5x$$
$$=y'15y^2 + 2x = y' + 5$$
$$=y'15y^2-y' = -2x + 5$$
$$=y'(15y^2-1) = -2x + 5$$
$$=y' = \dfrac{-2x + 5}{(15y^2-1)}$$
Plugging in the point $(4,1)$ gives
$$y'=-\dfrac{3}{14}$$
| {
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Proving of $\frac{\pi }{24}(\sqrt{6}-\sqrt{2})=\sum_{n=1}^{\infty }\frac{14}{576n^2-576n+95}-\frac{1}{144n^2-144n+35}$ This is a homework for my son, he needs the proving.I tried to solve it by residue theory but I couldn't.
$$\frac{\pi }{24}(\sqrt{6}-\sqrt{2})=\sum_{n=1}^{\infty }\frac{14}{576n^2-576n+95}-\frac{1}{144n^2-144n+35}$$
| Another way to evaluate the sum is to use this. Note that
$$ 576n^2 - 576n + 95=24^2(n^2-n+\frac{95}{24^2})=24^2[(n-\frac{1}{2})^2+(\frac{7i}{24})^2] $$
and
$$ 144n^2 - 144n + 35=12^2(n^2-n+\frac{35}{12^2})=12^2[(n-\frac{1}{2})^2+(\frac{i}{12})^2] $$
and hence we have
\begin{eqnarray*}
&&\sum_{n=1}^\infty \left(\frac{14}{576n^2 - 576 + 95}-\frac{1}{144n^2 - 144n + 35}\right)\\
&=&\sum_{n=1}^\infty\left(\frac{7}{288}\frac{1}{(n-\frac{1}{2})^2+(\frac{7i}{24})^2}-\frac{1}{12^2}\frac{1}{(n-\frac{1}{2})^2+(\frac{i}{12})^2}\right) \\
&=& \frac{1}{2}\sum_{n=-\infty}^\infty\left(\frac{7}{288}\frac{1}{(n-\frac{1}{2})^2+(\frac{7i}{24})^2}-\frac{1}{12^2}\frac{1}{(n-\frac{1}{2})^2+(\frac{i}{12})^2}\right) \\
&=& \frac{1}{2} \left(\frac{\pi\sinh(2\pi b)}{b(\cosh(2\pi b)-\cos(2\pi a)}\bigg|_{a=\frac{1}{2},b=\frac{7i}{24}} - \frac{\pi\sinh(2\pi b)}{b(\cosh(2\pi b)-\cos(2\pi a)}\bigg|_{a=\frac{1}{2},b=\frac{i}{12}}\right)\\
&=&\frac{1}{2}\left(\frac{\pi(1+\sqrt3)}{12+24\sqrt2-12\sqrt3}-\frac{\pi}{24+12\sqrt3}\right)\\
&=&\frac{\pi}{24}(\sqrt6-\sqrt2).
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculus - Functions How do I go about this question? Also how exactly will its graph be?
$$ f(x) = 1 + 4x -x^2 $$
$$g(x) = \begin{cases}
\max f(t) & x \le t \le (x+1) ;\quad 0 \le x < 3 \\
\min (x+3) & 3 \le x \le 5
\end{cases}
$$
Verify continuity of $g(x)$ for all $x$ in $[0,5]$
| here is a way to think of solving this problem geometrically. note that $1 + 4x - x^2$ is a parabola symmetric about $x = 2,$ opens downward, a has a local max at $x = 2, y = 5.$ think of the function $g$ as the global max on the moving window $x, x+1]$ of length $1.$ for $0 \le x \le 1,$ this max occurs at the right end point
$x + 1$ and the value is $f(x+1).$ for $1 \le x \le 2,$ the global max is at $x = 2$ and the value is $f(2)$ and finally for $2 \le x \le 3,$ the global max occurs at the left end point $x$ and the value is $f(x).$ we can write this as piecewise function
$g(x) = \left\{ \begin{array}{ll} f(x+1) & if 0 \le x \le 1 \cr
f(2) & if 1 \le x \le 2 \cr
f(x) & if 2 \le x \le 3 \cr
x + 1 & if 3 \le x \end{array} \right.$
you can verify the continuity of $g$ at the boundary points $1, 2, $ and $3.$
| {
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Find a closed form for the equations $1^3 = 1$, $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$ This is the assignment I have:
Find a closed form for the equations
$1^3 = 1$
$2^3 = 3+5$
$3^3 = 7+9+11$
$4^3 = 13+15+17+19$
$5^3 = 21+23+25+27+29$
$...$
Hints. The equations are of the form $n^3 = a1 +a2 +···+an$, where
$a_{i+1} = a_i +2$ and $a_0 =n(n−1)+1$.
My reasoning:
We have to find a formula that give us $n^3$ summing operands. (why is this useful?)
We know that the first operand (or term) of the sum is $a_0 =n(n−1)+1$.
In fact, if you put $n = 3$, then $a_0 = 3(3 − 1) + 1 = 3*2 + 1 = 7$, which is exactly the first number of sum.
Then I notice that each $n$ sum has $n$ operands, and each operand differs from one another of 2.
Thus I came out with this formula:
$$
\sum\limits_{i=0}^{n-1} a_0 + 2 \cdot i
$$
where $a_0 =n(n−1)+1$
For example, if $n = 3$, then we have
$(n(n−1)+1 + 2 \cdot 0) + (n(n−1)+1 + 2 \cdot 1) + (n(n−1)+1 + 2 \cdot 2) \equiv$
$\equiv (7 + 0) + (7 + 2) + (7 + 4) \equiv$
$\equiv 7 + 9 + 11$
Which is what is written as third example.
I don't know if this is correct form or even if this is a closed form, that's why I am asking...
| Let us derive the $r$ term of $1,3,7,13,21$
Let $S_r=1+3+7+13+21+\cdots+T_r$
$S_r-S_r=1+(3-1)+(7-3)+(13-7)+(21-13)+\cdots+T_r-T_{r-1}-T_r$
$\implies T_r=1+(2+4+6+8+$ up to $r-1$th term $)$
$=1+\dfrac{(r-1)}2\{2\cdot2+(r-2)2\}=1+r^2-r$
Now the $m$th row will be, $$\sum_{r=0}^{m-1}\{(m^2-m+1)+2r\}=(m^2-m+1)\sum_{r=0}^{m-1}1+2\sum_{r=0}^{m-1}r$$
$$=m(m^2-m+1)+2\cdot\frac{m(m-1)}2=m^3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Evaluation of $\int \frac{x^4}{(x-1)(x^2+1)}dx$ Evaluation of $\displaystyle \int \frac{x^4}{(x-1)(x^2+1)}dx$
$\bf{My\; Try::}$ Let $$\displaystyle I = \int\frac{x^4}{(x-1)(x^2+1)}dx = \int \frac{(x^4-1)+1}{(x-1)(x^2+1)}dx = \int\frac{(x-1)\cdot (x+1)\cdot (x^2+1)}{(x-1)(x^2+1)}+\int\frac{1}{(x-1)(x^2+1)}dx$$
So $\displaystyle I = \int (x+1)dx+J\;\,\;,$ Where $\displaystyle J = \int\frac{1}{(x-1)(x^2+1)}dx$
Now can we solve $J$ without using Partial fraction.
If yes then plz explain me, Thanks
| Hint
Using partial fraction decomposition from the very beginning, we have $$\frac{x^4}{(x-1)(x^2+1)}=\frac{-x-1}{2 \left(x^2+1\right)}+x+\frac{1}{2 (x-1)}+1$$ $$\frac{x^4}{(x-1)(x^2+1)}=\frac{-x}{2 \left(x^2+1\right)}+\frac{-1}{2 \left(x^2+1\right)}+x+\frac{1}{2 (x-1)}+1$$
| {
"language": "en",
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If $\tan\theta$, $2\tan\theta+2$, $3\tan\theta +3$ are in geometric progression then find … Problem:
If $\tan\theta$, $2\tan\theta+2$, $3\tan\theta +3$ are in geometric progression then find the value of $$\frac{7-5\cot\theta}{9+4\sqrt{\sec^2\theta -1}}$$
Solution :
Since $\tan\theta$, $2\tan\theta+2$, $3\tan\theta +3$ are in geometric progression
$\therefore (2\tan\theta +2)^2 = \tan\theta (3\tan\theta +3)$
After simplifying we will get the equation $\tan^2\theta +5\tan\theta +4=0$
Which gives two values of $\tan\theta$ viz. $\tan\theta =-1, -4$
However, it is given that $\tan\theta \neq -1$ , my question is why as nothing is defined for the domain of $\theta$ please suggest thanks.
| Simply because if $\tan \theta = -1$, then the "progression" becomes $-1, 0, 0$. So we have to rule it out.
| {
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Can this series be put in a generalized form? I asked a similar question here But this one seems to not work out so nicely...
I started looking at the series,
$$
S = \frac{1}{2}+\frac{2}{3}-\frac{3}{4}-\frac{4}{5}+\frac{5}{6}+\frac{6}{7}-\frac{7}{8}-\frac{8}{9}+\cdots
$$
Which is equivilent to
$$
S = \sum_{k=0}^{\infty}[\frac{4k+1}{4k+2}+\frac{4k+2}{4k+3}-\frac{4k+3}{4k+4}-\frac{4k+4}{4k+5}]
$$
I found $S = \frac{\pi}{4}-\frac{\ln(4)}{4}-1$
Then i put the series in a different form such that
$$
S_n = \sum_{k=0}^\infty[\frac{4k+n}{4k+n+1}+\frac{4k+n+1}{4k+n+2}-\frac{4k+n+2}{4k+n+3}-\frac{4k+n+3}{4k+n+4}]
$$
$$
S_n = -\sum_{k=0}^{\infty}\frac{2(2n^2 + 16nk+10n+32k^2+40k+11)}{(4k+n+1)(4k+n+2)(4k+n+3)(4k+n+4)}
$$
Going back to the question i had previously asked i learned that $\psi(1+x)=\sum_{k=1}^\infty \frac{x}{k(k+x)}$ which does not seem to apply here. The series yields some interesting results
$$
S_1 = \frac{\pi}{4}-\frac{\ln(4)}{4}-1
$$
$$
S_2 = \frac{\pi}{4}+\frac{\ln(4)}{4}-\frac{3}{2}
$$
$$
S_3=\frac{\ln(4)}{4}+\frac{1}{6}-\frac{\pi}{4}
$$
$$
S_4 = \frac{11}{12}-\frac{\pi}{4} - \frac{\ln(4)}{4}
$$
$$
S_5 = \frac{\pi}{4}-\frac{\ln(2)}{2}-\frac{37}{60}
$$
It seems like $\frac{pi}{4}$ occurs for every $n$ value, I would like to know if there is a generalized form of this series like the result i found here and how i would go about identifying this form (if one exists).
| This is nowhere near a complete answer, so I've marked it community wiki.
$$\frac{n-1}{n} + \frac{n}{n+1} = \frac{n^2 -1}{n(n+1)} + \frac{n^2}{n(n+1)} = \frac{2n^2-1}{n^2+n}$$
This means every pair of consecutive terms in your original series that have the same sign sum to $$\pm \frac{2n^2 -1}{n^2+n}$$ for some $n \in \mathbb{Z}$. This gives an alternate way to write your sum:
$$S = \left(\frac{1}{2} + \frac{2}{3}\right) - \left(\frac{3}{4} + \frac{4}{5}\right) + \left(\frac{5}{6} + \frac{6}{7}\right) - \cdots $$
$$S = \frac{2(2^2)-1}{2^2+2} - \frac{2(4^2)-1}{4^2+4}+\frac{2(6^2)-1}{6^2+6}-\cdots$$
$$S = \sum\limits_{k=1}^{\infty}(-1)^{k+1}\frac{8k^2-1}{4k^2+2k} $$
$$S = \sum\limits_{k=1}^{\infty}(-1)^{k+1} \frac{(4k-1)(4k+1)}{2k(2k+1)}$$
| {
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Integral of $\frac{1}{\sqrt{x^2-x}}dx$ For a differential equation I have to solve the integral $\frac{dx}{\sqrt{x^2-x}}$. I eventually have to write the solution in the form $ x = ...$ It doesn't matter if I solve the integral myself or if I use a table to find the integral. However, the only helpful integral in an integral table I could find was:
$$\frac{dx}{\sqrt{ax^2+bx+c}} = \frac{1}{\sqrt{a}} \ln \left|2ax + b +2\sqrt{a\left({ax^2+bx+c}\right)}\right|$$
Which would in my case give:
$$\frac{dx}{\sqrt{x^2-x}} = \ln \left|2x -1 + 2\sqrt{x^2-x}\right|$$
Which has me struggling with the absolute value signs as I need to extract x from the solution. All I know is that $x<0$ which does not seem to help me either (the square root will only be real if $x<-1$).
Is there some other formula for solving this integral which does not involve absolute value signs or which makes extracting $x$ from the solution somewhat easier? Thanks!
| $\sqrt{x}=z \implies\dfrac{dx}{2\sqrt{x}}=dz$
$\displaystyle\int\dfrac{dx}{\sqrt{x^2-x}}=\displaystyle\int\dfrac{dx}{\sqrt{x(x-1)}}=2\displaystyle\int\dfrac{ dz}{\sqrt{z^2-1}}$
$z=\sec \theta \implies dz=\sec \theta \tan \theta\ d\theta$
$\therefore \displaystyle\int\dfrac{ dz}{\sqrt{z^2-1}}=\displaystyle\int{\sec \theta \ d\theta}=\ln \left\lvert \sec\theta+\tan \theta\right\rvert+C$
$\therefore \displaystyle\int\dfrac{dx}{\sqrt{x^2-x}}=2\ln \left\lvert \sqrt{x}+\sqrt{x-1}\right\rvert+C'$
$$\color{blue}{\forall x \geq 1,\ \ln \left\lvert \sqrt{x}+\sqrt{x-1}\right\rvert=\ln \left( \sqrt{x}+\sqrt{x-1}\right)}$$
| {
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Prove $g(a) = b, g(b) = c, g(c) = a$
Let
$$f(x) = x^3 - 3x^2 + 1$$
$$g(x) = 1 - \frac{1}{x}$$
Suppose $a>b>c$ are the roots of $f(x) = 0$. Show that $g(a) = b, g(b) = c, g(c) = a$.
(Singapore-Cambridge GCSE A Level 2014, 9824/01/Q2)
I was able to prove that
$$fg(x) = -f\left(\frac{1}{x}\right)$$
after which I have completely no clue how to continue. It is possible to numerically validate the relationships, but I can't find a complete analytical solution.
| Rearranging $g(x) = 1 - \frac{1}{x},$ we obtain $x = \frac{1}{1-g(x)}$. Substitute into $f$ and we obtain $$\left(\frac{1}{1-g(x)}\right)^3 - 3\cdot\left(\frac{1}{1-g(x)}\right)^2+1.$$ The numerator of this turns out to be: $$1 - 3(1-g(x)) + (1-g(x))^3 = -(g^3(x) - 3g^2(x) +1) \equiv -f.$$ for $x\equiv g(x).$ Hence we can say that the roots of $f$ are equivalent to cyclic evaluations of $g$ for those roots, noting that any one of these roots cannot be substituted into $g$ to produce itself again since $a \neq \frac{1}{1-a} \Rightarrow a^2-a+1 \neq 0 \, $ for real $a\, (\bigtriangleup\,=-3).$
| {
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Show that $x^a+x-b=0$ must have only one positive real root and not exceed the $\sqrt[a]{b-1}$ If we take the equation $$x^3+x-3=0$$ and solve it to find the real roots, we will get only one positive real roots which is $(x=1.213411662)$. If we comparison this with $\sqrt[3]{3-1}=1.259921$, we will find that $x$ is less than $\sqrt[3]{3-1}$.This always happens with any value of $a$ so that $a$ any positive real number and $b$ is a positive real number.
So we can ask the following:
1-prove that $x^a+x-b=0$ must have only one positive real root if $a$ positive real number and $b$ is a positive real number greater than $1$.
2-The value of this roots must be less than $\sqrt[a]{b-1}$
3-What happens when $a$ is a complex value.I mean this thing stays right or not?
| If $a > 0$, you can use the Intermediate Value Theorem.
If $a < 0$, let $v$ be the minimum value of $x^a + x$ on $(0,\infty)$. Then $x^a + x - b$ has no positive real roots if $b < v$, one if $b = v$ and two if $b > v$.
If $a$ is complex, $x^a + x - b$ is real only for a discrete set of positive real $x$'s, so there will usually be no positive real roots.
| {
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Compute $\bar z - iz^2 = 0$ $\bar z - iz^2 = 0, i = $ complex unit.
I've found 2 solutions to this, like this:
$x - iy - i(x+iy)^2 = 0$
$i(-x^2+y^2-y)+2xy+x=0$
$2xy + x = 0$ --> $y=-\frac{1}{2}$
$x^2 - y^2 + y = 0$ --> $x_1=\sqrt\frac34$ $x_2=-\sqrt\frac{3}{4}$
Solution 1: $z = \sqrt\frac34 - \frac12i $
Solution 2: $z = -\sqrt\frac34 - \frac12i $
That's great and everything, but Wolfram gives me another solution, which is $z = i$. How do I get that? LINK to wolfram.
| Did's suggestion is excellent: setting $z = re^{i\theta}$ we have $\bar z = re^{-i\theta}$ and $z^2 = r^2e^{2i\theta}$ whence
$re^{-i\theta} = i r^2 e^{2i\theta}; \tag{1}$
using
$i = e^{i \pi/2} \tag{2}$
yields
$r e^{-i\theta} = r^2 e^{i(2\theta + \pi/2)}; \tag{3}$
taking moduli,
$r = r^2 \Rightarrow r = 0, 1; \tag{4}$
when $r = 0$, $z = 0$; for $r = 1$,
$e^{-\theta} = e^{2\theta+ \pi/2}͵ \tag{5}$
whence
$e^{i(3\theta + \pi/2)} = e^{\pi i/2}(e^{i\theta})^3 = i(e^{i\theta})^3 = 1, \tag{6}$
or
$(e^{i\theta})^3 = -i. \tag{7}$
When $r = 1$, $z = re^{i\theta} = e^{i\theta}$, so (7) tells us that $z$ must be a cube root of $-i$; furthermore, we can work backwards from (7): if
$z^3 = -i, \tag{8}$
then, taking moduli,
$\vert z \vert^3 = 1, \tag{9}$
so that
$\vert z \vert = 1; \tag{10}$
then (8) implies
$i z^2 = i \vert z \vert^2 z^2 = i \bar z z z ^2 = i \bar z z^3 = i (-i) \bar z = \bar z; \tag{11}$
we see the non-zero solutions of
$\bar z = i z^2. \tag{12}$
are precisely the three cube roots of $-i = e^{3\pi i/2}$, viz. $e^{3 \pi i/6} = e^{\pi i / 2} = e^{-3 \pi i / 2} = i$, $e^{(1/2 + 2/3) \pi i} = e^{7 \pi i / 6} = e^{-5 \pi i / 6}$, $e^{(1/2 + 4/3) \pi i} = e^{11 \pi i / 6} = e^{- \pi i / 6}$.
We check this result for $z = e^{7 \pi i / 6}$:
$z = e^{7 \pi i / 6} = -\dfrac{\sqrt{3}}{2} - \dfrac{1}{2} i \tag{13}$
$z^2 = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i; \tag{14}$
$iz^2 = -\dfrac{\sqrt{3}}{2} + \dfrac{1}{2} i = \bar z; \tag{15}$
a similar calculation validates $z = e^{11 \pi i / 6}$.
It is easy to see that $i(i)^2 = -i$ and the solution $z = 0$ "checks itself", as it were. The complete solution set is thus
$\{ 0, i, e^{7 \pi i / 6}, e^{11 \pi i / 6} \}. \tag{16}$
Note: Not to put too fine a point on it, but for the sake of accuracy it should be observed that $2$ of heropup's proposed solutions, namely $e^{2 \pi i / 3}$ and $e^{-2 \pi i / 3} = e^{4 \pi i / 3}$, do not satisfy the given equation (12):
$e^{2 \pi i / 3} = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i; \tag{17}$
$(e^{2 \pi i / 3})^2 = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2} i; \tag{18}$
$\overline{e^{2 \pi i / 3}} = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2} i \ne \dfrac{\sqrt{3}}{2} - \dfrac{1}{2} i = i (e^{2 \pi i / 3})^2, \tag{19}$
with a similar calculation for $e^{4 \pi i / 3}$. One can also simply write
$(e^{2 \pi / 3})^3 = e^{2 \pi i} = 1 \ne -i, \tag{20}$
etc. I do not know exactly where heropup's error lies, but we can see from the above his proposed solutions do not all satisfy $\bar z = i z^2$. End of Note.
Hope this helps. Cheers,
and as always,
Fiat Lux!!!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$1 + 4 + 7 + · · · + 3n − 2 =
\frac{n(3n − 1)}
2$$
for all positive integers $n$.
Proof: $$1+4+7+\ldots +3(k+1)-2= \frac{(k + 1)[3(k+1)+1]}2$$
$$\frac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$$
Along my proof I am stuck at the above section where it would be shown that:
$\dfrac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$ is equivalent to $\dfrac{(k + 1)[3(k+1)+1]}2$
Any assistance would be appreciated.
| Non-inductive derivation:
\begin{align}
\sum_{k=1}^n(3k-2) &= \sum_{k=1}^n3k -\sum_{k=1}^n2\\
&= 3\left(\sum_{k=1}^n k\right) -2n\\
&= \frac{3(n)(n+1)}{2} - \frac{4n}{2}\\
&=\frac{3n^2-n}{2}\\
&= \frac{n(3n-1)}{2}\\
\end{align}
This, of course, relies on one knowing the sum of the first $n$ natural numbers, but that's a well-known identity.
| {
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"url": "https://math.stackexchange.com/questions/1050814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 2
} |
Evaluation of $ \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ Evaluation of $\displaystyle \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ where $\lfloor x \rfloor $ represent floor function of $x$.
My Try:: Here $\displaystyle f(x) = \frac{x^2}{\sin x\cdot \tan x}$ is an even function.
So we will calculate for $\displaystyle \lim_{x\to 0^{+}}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$
Put $x=0+h$ where $h$ is a small positive quantity, and using series expansion
So limit convert into $\displaystyle \lim_{h\to 0}\left\lfloor \frac{h^2}{\sin h\cdot \tan h}\right\rfloor = \lim_{h\to 0}\left\lfloor \dfrac{h^2}{\left(h-\dfrac{h^3}{3!}+\dfrac{h^5}{5!}- \cdots\right)\cdot \left(h+\dfrac{h^3}{3}+\dfrac{2}{15}h^5+ \cdot\right)}\right\rfloor$
Now how can i solve after that, Help me
Thanks
| Note that the function under consideration is even and hence it is sufficient to consider $x \to 0^{+}$. Now we need to compare $x^{2}$ and $\sin x \tan x$ for $x > 0$. Let $$f(x) = x^{2} - \sin x\tan x$$ Then $f(0) = 0$ and $$f'(x) = 2x - \sin x\sec^{2}x - \sin x$$ Then we have $f'(0) = 0$. Further $$f''(x) = 2 - \sec x - 2\tan^{2}x - \cos x$$ and $f''(0) = 0$. Next we have $$f'''(x) = -\sec x\tan x - 4\tan x\sec^{2}x + \sin x = \sin x(1 - \sec^{2}x - 4\sec^{3}x)$$ As $x \to 0^{+}$ we can clearly see that $f'''(x) < 0$. This means that $f''(x)$ is decreasing and considering that $f''(0) = 0$ we have $f''(x) < 0$ when $x \to 0^{+}$. Continuing further in the same manner we see that $f(x) < 0$ when $x \to 0^{+}$. It follows that $\sin x\tan x > x^{2} > 0$ when $x \to 0^{+}$ so that $$0 < \dfrac{x^{2}}{\sin x\tan x} < 1$$ when $x \to 0^{+}$. Hence we have $$\left\lfloor\dfrac{x^{2}}{\sin x\tan x}\right\rfloor = 0$$ when $x \to 0^{+}$. Now it is obvious that the desired limit is $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Find limit of $\frac {1}{x^2}- \frac {1}{\sin^2(x)}$ as x goes to 0 I need to use a taylor expansion to find the limit.
I combine the two terms into one, but I get limit of $\dfrac{\sin^2(x)-x^2}{x^2\sin^2(x)}$ as $x$ goes to $0$. I know what the taylor polynomial of $\sin(x)$ centered around $0$ is… but now what do I do?
| Outline: If we can write down the Taylor series for $\sin^2 x$, or just the first couple of terms, we will be finished.
This can be done from $x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$ by ordinary squaring, treating the series for $\sin x$ casually as a "long" polynomial. Maybe it will feel better, however, to use the trigonometric identity
$$\cos 2x=1-2\sin^2 x, \quad\text{rewritten as}\quad \sin^2 x=\frac{1}{2}(1-\cos 2x).$$
Now from the series for $\cos t$ we get that the series for $\sin^2 x$ is
$$\frac{2x^2}{2!}-\frac{8x^4}{4!}+\frac{32x^6}{6!}-\cdots.$$
Now it's over. The leading term of $\sin^2 x-x^2$ is $-\frac{8x^4}{4!}$ and the leading term of $x^2\sin^2 x$ is $\frac{2x^4}{2!}$. Cancel.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Four cards are drawn without replacement. What is the probability of drawing at least two kings? Four cards are drawn without replacement. What is the probability of drawing at least two kings?
is the below my answer correct or not?!
since cards are drawn without replacement, (4/52)(3/51)(48/50)*(47/49)=1/240=0.0042
1-0.0042=0.9958
Probability of drawing AT LEAST 2 kings is 0.9958.
| There are $\binom{4 \cdot 13}{4} = \binom{52}{4} = 270725$ cases of "drawing four cards", the ways to get $2$ out of $4$ kings is $\binom{4}{2}$, to get $3$ is $\binom{4}{3}$, for $4$ it is $\binom{4}{4}$. In all:
$$\binom{4}{2} + \binom{4}{3} + \binom{4}{4}
= 11$$
The probability is thus $\frac{11}{270725} = 4.06316 \cdot 10^{-5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $g(x) = \max(y^2-xy)(0 \leq y\leq 1)\;,$ Then minimum value of $g(x)$ If $g(x) = \max\limits_{0 \leq y\leq 1}(y^2-xy)$, then minimum value of $g(x)$
$\bf{My\; try::}$ We can write $\displaystyle f(y) = y^2-xy = y^2-xy+\frac{x^2}{4}-\frac{x^2}{4} = \left(y-\frac{x}{2}\right)^2-\frac{x^2}{4}$
Now when $y=0\;,$ Then expression $f(0) = y^2-xy=0$
Now when $y=1\;,$ Then expression $f(1) = (1-x).$
So $\displaystyle g(x) = \max\left(f(0)\;,f(1),f\left(\frac{x}{2}\right)\right)$.
Now How can i solve after that, Help me
Thanks
| It seems like you picked $y=x/2$ as a possible maximum value of $\left(y-\frac{x}{2}\right)^2-\frac{x^2}{4}$ instead of a possible minimum. You're almost there. Here's a full argument:
To evaluate $\max\limits_{0 \leq y\leq 1}(y^2-xy)$, you need to first find its stationary point by taking $f(y) = y^2-xy$ and solving $df/dy = 0$ (there's only one, since $f$ is a parabola): $2y - x = 0 \Rightarrow y = x/2$. That might be a local maximum or a local minimum, so to calculate the global maximum for $y\in [0,1]$, you need to compare this value with the extremities $f(0)$ and $f(1)$.
$$\begin{align}
g(x) &= \max\left(f(0)\;,f(1),f\left(\frac{x}{2}\right)\right)\\
&= \max\left(0\;,1-x,\frac{-x^2}{4}\right)\\
&= \max\left(0\;,1-x\right)\\
&= \begin{cases}
0 & x>1 \\
1-x & x\le 1
\end{cases}
\end{align}$$
Note that $g(x)\ge0$ if $x\le1$. So the minimum value of $g(x)$ is $0$.
MATLAB code to check the answer (for $-1000<x<1000$):
N = 1000;
X = 1000;
M = X*1000;
y = 0:1/N:1-1/N;
x = -X:2*X/M:X-2*X/M;
g=zeros(M,1);
for k=1:M
g(k) = max(y.^2-x(k)*y);
end
plot(x,g);
min(g)
Plot:
Output:
ans =
0
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find the value of the sum $$\arctan\left(\dfrac{1}{2}\right)+\arctan\left(\dfrac{1}{3}\right)$$
We were also given a hint of using the trigonometric identity of $\tan(x + y)$
Hint
$$\tan\left(x+y\right)\:=\:\dfrac{\tan x\:+\tan y}{1-\left(\tan x\right)\left(\tan y\right)}$$
| Let $x = \arctan \dfrac{1}{2}$ and $y = \arctan \dfrac{1}{3}$. Then, $\tan x = \dfrac{1}{2}$ and $\tan y = \dfrac{1}{3}$.
Using that formula, you can easily compute $\tan(x+y)$. Do you see how to get $x+y$ from that?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differential equation, substitution By means of the substitute $y = v(x)Y (x)$, where $Y (x)$ is to be specified, solve the differential equation:
$$\dfrac{dy}{dx}+\dfrac{y}{x}=\dfrac{y^2}{x}$$
with $y=2$ at $x=1$
Anyone can solve it for me with explaining the steps please, I have no idea how to do it. Thank you very much
| $$
y' = v'Y + vY' = -\frac{vY}{x} + \frac{v^2Y^2}{x}
$$
if we re-write
$$
v'Y = -vY' -\frac{vY}{x} + \frac{v^2Y^2}{x} = -\left(Y' + \frac{Y}{x}\right)v + \frac{v^2Y^2}{x}
$$
set the brackets term to zero i.e.
$$
Y' = -\frac{Y}{x}\implies Y(x) = \frac{C}{x}
$$
then we obtain
$$
v' = \frac{1}{Y}\frac{v^2Y^2}{x} = \frac{Y}{x}v^2 = \frac{C}{x^2} v^2
$$
which is separable.
Thus
$$
v = \frac{x}{C+C_1x}\implies y(x) = \frac{C}{C+C_1x}
$$
put it back into the equation we find
$$
y' +\frac{y}{x} = \frac{-C C_1}{\left(C+C_1x\right)^2} + \frac{C}{\left(C+C_1x\right)x} = \\
\frac{-C}{\left(C+C_1x\right)^2x}\left[C_1 x -\left(C+C_1x\right)\right] = \frac{C^2}{\left(C+C_1x\right)^2x} = \left(\frac{C}{\left(C+C_1x\right)}\right)^2\frac{1}{x} = \frac{y^2}{x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using the half/double angle formulas to solve an equation I am completely stumped by this problem: $$\cos\theta - \sin\theta =\sqrt{2} \sin\frac{\theta}{2} $$
I know that I should start by isolating $\sin\dfrac{\theta}{2}$
and end up with $$\frac {\cos\theta - \sin\theta}{\sqrt{2}} = \sin\frac{\theta}{ 2}$$
From here on out, I have no idea what steps to take.
| Given : $\dfrac{\cos\theta - \sin\theta}{\sqrt{2}} = \sin\dfrac{\theta}{ 2}$
using $\sin(\frac{\theta}{2})=\sqrt{\frac{1-\cos(\theta)}{2}}$ and squaring both sides,
\begin{align*}
\frac{\sin^2(\theta)+\cos^2(\theta)-2\sin(\theta)\cos(\theta)}{2} & = \frac{1-\cos(\theta)}{2}\\
1-2\sin(\theta)\cos(\theta) & = 1-\cos(\theta) && \text{because $\sin^2(\theta)+\cos^2(\theta) = 1$}\\
-2\sin(\theta)\cos(\theta) & = -\cos(\theta)\\
-\sin(\theta) & = -\frac{1}{2}\\
\sin(\theta) & = \frac{1}{2} = \sin(\frac{\pi}{3})
\end{align*}
General Solutions $\theta$ = $n\pi +(-1)^{n}\frac{\pi}{3}$ , $n\in Z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Integrate a periodic absolute value function \begin{equation}
\int_{0}^t \left|\cos(t)\right|dt = \sin\left(t-\pi\left\lfloor{\frac{t}{\pi}+\frac{1}{2}}\right\rfloor\right)+2\left\lfloor{\frac{t}{\pi}+\frac{1}{2}}\right\rfloor
\end{equation}
I got the above integral from https://www.physicsforums.com/threads/closed-form-integral-of-abs-cos-x.761872/. It seems to hold and the way I approached it was to see that the integrand was periodic and $\int_{\frac{\pi}{2}}^\frac{3\pi}{2} -\cos(t)dt=\int_{\frac{3\pi}{2}}^{\frac{5\pi}{2}} \cos(t)dt=\ldots=2$.
I need to evaluate a similar integral.
\begin{equation}
\int_{0}^t \sin\left(\frac{1}{2}(s-t)\right)\left|\sin\left(\frac{1}{2}s\right)\right|ds
\end{equation}
Here too the integrand is periodic but I am unable to get the closed form. Can someone help me out?
| We can use the Fourier series of the square wave. We have:
$$\operatorname{sign}\left(\sin\frac{s}{2}\right)=\frac{4}{\pi}\sum_{k=0}^{+\infty}\frac{1}{2k+1}\sin\frac{(2k+1)s}{2},$$
and since:
$$\frac{1}{2k+1}\int_{0}^{t}\sin\frac{t-s}{2}\sin\frac{s}{2}\sin\frac{(2k+1)s}{2}\,ds=\\ = \frac{4\cos(t/2)}{(2k-1)(2k+1)^2(2k+3)}-\frac{\cos(kt)}{(2k-1)(2k+1)^2}+\frac{\cos((k+1)t)}{(2k+1)^2(2k+3)}$$
we have:
$$\color{red}{\int_{0}^{t}\sin\frac{t-s}{2}\left|\sin\frac{s}{2}\right|\,dt} =\\= -\frac{\pi}{2}\cos\frac{t}{2}-\frac{4}{\pi}\sum_{k=0}^{+\infty}\left(\frac{\cos(kt)}{(2k-1)(2k+1)^2}-\frac{\cos((k+1)t)}{(2k+1)^2(2k+3)}\right)=\\=-\frac{\pi}{2}\cos\frac{t}{2}-\frac{4}{\pi}\left(-1-2\sum_{k=1}^{+\infty}\frac{\cos(kt)}{(2k-1)^2(2k+1)^2}\right)=\\=\color{red}{\frac{4}{\pi}-\frac{\pi}{2}\cos\frac{t}{2}+\frac{8}{\pi}\sum_{k=1}^{+\infty}\frac{\cos(kt)}{(2k-1)^2(2k+1)^2}}.$$
The last sum gives a $2\pi$-periodic function whose absolute value is always $\leq\frac{\pi^2-8}{2\pi}=0.297556782\ldots<0.3.$ It is also interesting to notice that the integral, as a function of $t$, is pretty well gaussian-shaped on the interval $[0,4\pi]$:
| {
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"question_score": "5",
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Prove that : $f(\sin x)+f(\cos x) \ge 196, \forall x\in\left(0;\frac{\pi}{2}\right)$ Given: $$f(\tan2x)=\tan^{4}x+\frac{1}{\tan^{4}x}, \forall x\in\left(0;\frac{\pi}{4}\right)$$
Prove that :$f(\sin x)+f(\cos x) \ge 196, \forall x\in\left(0;\frac{\pi}{2}\right)$
Could someone help me ?
| since
$$t=\tan{2x}=\dfrac{2\tan{x}}{1-\tan^2{x}}=-\dfrac{2}{\tan{x}-\dfrac{1}{\tan{x}}},t>0$$
so
$$\tan{x}-\dfrac{1}{\tan{x}}=-\dfrac{2}{t}$$
and
$$\tan^4{x}+\dfrac{1}{\tan^4{x}}=\left[\left(\tan{x}-\dfrac{1}{\tan{x}}\right)^2+2\right]^2-2$$
so
$$f(t)=\left[\dfrac{4}{t^2}+2\right]^2-2=\dfrac{16}{t^4}+\dfrac{16}{t^2}+2$$
so
$$f(\sin{x})+f(\cos{x})=16\left(\dfrac{1}{\sin^4{x}}+\dfrac{1}{\cos^4{x}}\right)+16\left(\dfrac{1}{\sin^2{x}}+\dfrac{1}{\cos^2{x}}\right)+4$$
since use Holder inequality we have
$$\left(\dfrac{1}{\sin^4{x}}+\dfrac{1}{\cos^4{x}}\right)(\sin^2{x}+\cos^2{x})^2\ge (1+1)^3$$
$$\Longrightarrow \dfrac{1}{\sin^4{x}}+\dfrac{1}{\cos^4{x}}\ge 8$$
$$\left(\dfrac{1}{\sin^2{x}}+\dfrac{1}{\cos^2{x}}\right)(\sin^2{x}+\cos^2{x})\ge (1+1)^2$$
$$\Longrightarrow \dfrac{1}{\sin^2{x}}+\dfrac{1}{\cos^2{x}}\ge 4$$
so
$$f(\sin{x})+f(\cos{x})\ge 196$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How find $a,b$ if $\int_{0}^{1}\frac{x^{n-1}}{1+x}dx=\frac{a}{n}+\frac{b}{n^2}+o(\frac{1}{n^2}),n\to \infty$ let
$$\int_{0}^{1}\dfrac{x^{n-1}}{1+x}dx=\dfrac{a}{n}+\dfrac{b}{n^2}+o(\dfrac{1}{n^2}),n\to \infty$$
Find the $a,b$
$$\dfrac{x^{n-1}}{1+x}=x^{n-1}(1-x+x^2-x^3+\cdots)=x^{n-1}-x^n+\cdots$$
so
$$\int_{0}^{1}\dfrac{x^{n-1}}{1+x}=\dfrac{1}{n}-\dfrac{1}{n+1}+\dfrac{1}{n+2}-\cdots$$
and note
$$\dfrac{1}{n+1}=\dfrac{1}{n}\left(\frac{1}{1+\dfrac{1}{n}}\right)=\dfrac{1}{n}-\dfrac{1}{n^2}+\dfrac{1}{n^3}+o(1/n^3)$$
and simaler
$$\dfrac{1}{n+2}=\dfrac{1}{n}-\dfrac{2}{n^2}+o(1/n^2)$$
$$\dfrac{1}{n+3}=\dfrac{1}{n}-\dfrac{3}{n^2}+o(1/n^2)$$
then which term end?
| Here's a possible to find the coefficients:
From
$$
n\int_0^1 \frac{x^{n-1}}{1+x} d\;x =\left[\frac{x^n}{1+x} \right]^1_0 + \int_0^1 \frac{x^n}{(1+x)^2} d\;x =1/2 + \int_0^1 \frac{x^n}{(1+x)^2} d\;x
$$
Since the last term tends to zero as $n$ increases, we get $a =1/2$.
The same process allows to determine $b$.
| {
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"url": "https://math.stackexchange.com/questions/1060893",
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Prove by induction that an expression is divisible by 11
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.
I am rather confused by this question. This is my attempt so far:
For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such that $2^{3k-1} + 5\cdot 3^k$ is divisible by $11$.
We show that it is also divisible by $11$ when $n = k + 2$
$2^{3k+5} + 5\cdot 3^{k+2}$
$32\cdot 2^3k + 5\cdot 9 \cdot3^k$
$32\cdot 2^3k + 45\cdot 3^k$
$64\cdot 2^{3k-1} + 45\cdot 3^k$ (Making both polynomials the same as when $n = k$)
$(2^{3k-1} + 5\cdot 3^k) + (63\cdot 2^{3k-1} + 40\cdot 3^k)$
The first group of terms $(2^{3k-1} + 5\cdot 3^k)$ is divisible by $11$ because we have made an assumption that the term is divisible by $11$ when $n=k$. However, the second group is not divisible by $11$. Where did I go wrong?
| hint: You may want to reconsider the way you split the terms at the end.
Note that $64(2^{3k - 1}) + 45(3^k) = 9(2^{3k - 1} + 5(3^k)) + 55(2^{3k - 1})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 0
} |
Generic rotation to remove Quadratic Cross-product Show that if $b\neq 0$, then the cross-product term can be eliminated from the quadratic $ax^2 + 2bxy + cy^2$ by rotating the coordinate axes through an angle $\theta$ that satisfies the equation $$
\cot{2\theta}=\frac{a-c}{2b}.
$$
I know that $ax^2 + 2bxy + cy^2$ can be rewritten as $$
\begin{bmatrix}x&y\end{bmatrix}
\begin{bmatrix}a&b\\b&c\end{bmatrix}
\begin{bmatrix}x\\y\end{bmatrix},
$$ and I am familiar with how to rotate it using unitary matrices. However, it is very tedious to find the eigenvectorss of $$
\begin{bmatrix}a&b\\b&c\end{bmatrix}.
$$ Is there a method for solving this problem that does not require finding the eigenvectors?
| Let $u$ and $v$ be the rotated coordinates. That is,
$$
\begin{pmatrix} u \\ v \end{pmatrix} =
\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix}.
$$
You want to be able to substitute a quadratic with no $uv$ term for the quadratic
$ax^2 + bxy + cy^2.$ That is, you want there to be fixed coefficients $p$ and $q$ such that
$$ pu^2 + qv^2 = ax^2 + bxy + cy^2 \tag{1}$$
for every pair $(x,y).$
It's slightly tedious, but not difficult, to write $u$ and $v$ each in terms of
$x,$ $y,$ and $\theta,$ and to plug these substitutions into $pu^2 + qv^2.$
If you collect the terms in $x^2,$ in $xy,$ and in $y^2,$
knowing that the result must equal $ax^2 + bxy + cy^2,$ you can deduce
the values of $a$, $b$, and $c$ in terms of these other parameters.
Now see if you can take $a - c$ in those terms and make it look like
$(\mathit{something})\times \cos 2\theta.$
If you can then make $2b$ look like
$(\mathit{the\ same\ thing})\times \sin 2\theta,$
you will have proved your desired result.
You can avoid carrying around all those factors of $x^2,$ $xy,$ and $y^2$
if you rewrite Equation $(1)$ using the matrices
$\begin{pmatrix} p&0 \\ q&0 \end{pmatrix}$ and $\begin{pmatrix} a&b \\ x&d \end{pmatrix}$
and the vectors $\begin{pmatrix} u \\ v \end{pmatrix}$ and
$\begin{pmatrix} x \\ y \end{pmatrix},$ and peform the obvious substitution
for $\begin{pmatrix} x \\ y \end{pmatrix}$ immediately; you can then derive a
matrix equation that doesn't involve $x$ or $y$ at all.
That notation may be a bit more agreeable with your initial thoughts on the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Question about sines of angles in an acute triangle
Let $\triangle ABC$ be a triangle such that each angle is less than $ 90^\circ $.
I want to prove that $\sin A + \sin B + \sin C > 2$.
Here is what I have done:
Since $A+B+C=180^{\circ}$ and $0 < A,B,C < 90^\circ$, at least two of $A,B,C$ are in the range 45 < x < 90, without loss of generality, let these angles be $A$ and $B$.
$\sin A + \sin B + \sin C = \sin A + \sin B + \sin(180^\circ-A-B) = \sin A + \sin B + \sin(A+B)$
Since $45^\circ < A,B < 90^\circ$ it follows that $2^{0.5} < \sin A + \sin B < 2.$ Am I near the answer?
| I have found a simpler solution.
Observing the graph of $y = \sin x$ and $y = \frac{2}{\pi}x$, $x\in[0,\frac{\pi}{2}]$.
We can see that $\sin x\ge\frac{2}{\pi}x$.
Let $\bigtriangleup ABC$ be a triangle such that each angle is less than $90^{\circ}$. So we have:
$A\in(0,\frac{\pi}{2}),B\in(0,\frac{\pi}{2}),C\in(0,\frac{\pi}{2})$.
Because $\sin x>\frac{2}{\pi}x$, when $x\in(0,\frac{\pi}{2})$,
$$\sin A+\sin B+\sin C > \frac{2}{\pi}(A+B+C)=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Avoiding extraneous solutions When solving quadratic equations like $\sqrt{x+1} + \sqrt{x-1} = \sqrt{2x + 1}$ we are told to solve naively, for example we would get $x \in \{\frac{-\sqrt{5}}{2},\frac{\sqrt{5}}{2}\}$, even though the first solution doesn't work, and then try all the solutions and eliminate the extraneous ones. This is not a very elegant algorithm! How would one use the fact that $\sqrt{x}^2= |x|$ to avoid having to check answers?
| Try to square both sides.
$$ \sqrt{x+1} + \sqrt{x-1} = \sqrt{2x+1} \Leftrightarrow $$
$$ \Leftrightarrow \left ( \sqrt{x+1} + \sqrt{x-1} \right )^{2} = \left ( \sqrt{2x+1} \right )^{2} \Leftrightarrow $$
$$ \Leftrightarrow \left ( \sqrt{x+1} \right )^2 + 2\sqrt{x+1}\sqrt{x-1} + \left ( \sqrt{x-1} \right )^2 = 2x+1 $$
$$ \Leftrightarrow (x+1) + 2\sqrt{(x+1)(x-1)} + (x-1) = 2x+1 \Leftrightarrow $$
$$ \Leftrightarrow \sqrt{(x+1)(x-1)} = \frac{1}{2} \Leftrightarrow $$
$$ \Leftrightarrow (x+1)(x-1) = \frac{1}{4} \Leftrightarrow $$
$$ \Leftrightarrow x^{2} - 1 - \frac{1}{4} = 0 \Leftrightarrow $$
$$ \Leftrightarrow x^{2} - \frac{5}{4} = 0 $$
And now you can just use the quadratic formula.
$$ x = \frac{-(0)\pm \sqrt{(0)^2-4(1)(-\frac{5}{4}})}{2(1)} \Leftrightarrow $$
$$\Leftrightarrow x = \frac{\pm \sqrt{4\times \frac{5}{4}}}{2} \Leftrightarrow $$
$$\Leftrightarrow x = \frac{\pm \sqrt{5}}{2} $$
Also, $ -\frac{\sqrt{5}}{2} $ is not included in the final solution, try to see why...
I hope I have helped. Saclyr.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Rules for whether an $n$ degree polynomial is an $n$ degree power Given an $n$ degree equation in 2 variables ($n$ is a natural number)
$$a_0x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_{n-1}x+a_n=y^n$$
If all values of $a$ are given rational numbers, are there any known minimum or sufficient conditions for $x$ and $y$ to have:
*
*Real number
*Rational number
*Integer
solutions and how many of them would exist. If it is not known/possible (or too hard) for an $n$ degree polynomial, do such conditions exist for quadratic ($n=2$) and cubic ($n=3$) polynomials.
| (The OP suggested a connection to this post.)
Because this question is too broad, it spreads thin and may be vague. I suggest it be limited so coefficients $a_i$ are rational, and $x,y$ are also rational.
Having said that, two nice results are discussed in Kevin Brown's website.
I. Deg 2: The sum of $24$ consecutive squares.
$$F(x) = x^2+(x+1)^2+(x+2)^2+\dots+(x+23)^2=y^2\tag1$$
$$F(x) = 24x^2+552x+4324=y^2$$
which has solution,
$$x=p^2+70pq+144q^2,\quad\quad y =10(7p^2+30pq+42q^2)$$
where $p,q$ solve the Pell equation $p^2-6q^2=1$. This has an infinite number of integer solutions with the case $p,q = 1,0$ yielding the famous cannonball stacking problem,
$$1^2+2^2+3^2+\dots+24^2 =70^2$$
II. Deg 3: The sum of $n$ consecutive cubes.
$$G(x) = x^3+(x+1)^3+(x+2)^3+\dots+(x+n-1)^3=y^3\tag2$$
$$G(x) = n x^3 + \tfrac{1}{4}n(n - 1)\big(6x^2 + 4 n x - 2x + n(n - 1)\big) =y^3$$
a solution of which (by Dave Rusin) is,
$$x=\tfrac{1}{6}(v^4 - 3v^3 - 2v^2 + 4),\quad\quad n=v^3$$
hence for $2^3=8$ and $4^3=64$ cubes,
$$(-2)^3+(-1)^3+\dots+3^3+4^3+5^3 = 6^3$$
$$6^3+7^3+8^3+9^3+10^3+\dots+69^3 = 180^3$$
and so on.
III. Deg 4: The sum of 4th powers in arithmetic progression.
No analogous results known so far. See linked post in first line.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\int_{0}^{\pi/2}\frac{x\sin x\cos x\;dx}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}$ How to evaluate the following integral
$$\int_{0}^{\pi/2}\frac{x\sin x\cos x}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}dx$$
For integrating I took $\cos^{2}x$ outside and applied integration by parts.
Given answer is $\dfrac{\pi}{4ab^{2}(a+b)}$.
But I am not getting the answer.
| The case $a^2=b^2$ being simple, let's just consider, by symmetry, the case $a>b>0$.
Observe that
$$
\partial _x \left(\frac{1}{a^2 \cos^2x+b^2 \sin^2 x}\right)=2(a^2-b^2)\frac{\cos x\sin x}{(a^2 \cos^2x+b^2 \sin^2 x)^2}
$$
then, integrating by parts, you may write
$$
\begin{align}
I(a,b)&=\int_0^{\pi/2}\frac{x\cos x\sin x}{(a^2 \cos^2x+b^2 \sin^2 x)^2} dx\\\\
&=\frac{x}{2(a^2-b^2)}\left.\frac{1}{a^2 \cos^2x+b^2 \sin^2 x}\right|_{0}^{\pi/2}-\frac{1}{2(a^2-b^2)}\int_0^{\pi/2}\frac{dx}{a^2 \cos^2x+b^2 \sin^2 x}\\\\
&= \frac{\pi}{4(a^2-b^2)b^2}-\frac{1}{2(a^2-b^2)}\int_0^{\pi/2}\frac{1}{\left(a^2 +b^2 \large{\frac{\sin^2 x}{\cos^2x}}\right)}\frac{1}{\cos^2 x}dx\\\\
&= \frac{\pi}{4(a^2-b^2)b^2}-\frac{1}{2(a^2-b^2)}\int_0^{\pi/2}\frac{1}{\left(a^2 +b^2 \tan ^2x\right)}(\tan x)'dx\\\\
&= \frac{\pi}{4(a^2-b^2)b^2}-\frac{1}{2(a^2-b^2)}\frac{1}{ab}\left.\arctan \left(\frac ba \tan x \right)\right|_{0}^{\pi/2}\\\\
&= \frac{\pi}{4(a^2-b^2)b^2}-\frac{\pi}{4(a^2-b^2)}\frac{1}{ab}\\\\
&= \frac{\pi}{4(a+b)ab^2}.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 0
} |
Limits using Maclaurins expansion for $\lim_{x\rightarrow 0}\frac{e^{x^2}-\ln(1+x^2)-1}{\cos2x+2x\sin x-1}$ $$\lim_{x\rightarrow 0}\frac{e^{x^2}-\ln(1+x^2)-1}{\cos2x+2x\sin x-1}$$
Using Maclaurin's expansion for the numerator gives:
$$\left(1+x^2\cdots\right)-\left(x^2-\frac{x^4}{2}\cdots\right)-1$$
And the denominator:
$$\left(1-2x^2\cdots\right) + \left(2x^2-\frac{x^4}{3}\cdots\right)-1$$
$$\therefore \lim_{x\rightarrow 0} f(x) = \frac{-\dfrac{x^4}{2}}{-\dfrac{x^4}{3}} = \frac{3}{2}$$
But Wolfram gives that the limit is $3$.
I thought, maybe I used too few terms. What is a thumb rule for how many terms in expansion to use to calculate limits?
Using three terms yielded the answer $\lim_{x\rightarrow 0}f(x) = -4$.
What did I do wrong?
| Let's begin by writing out some of the expansions.
$$ \begin{align}
e^{x^2} &= 1 + x^2 + \frac{x^4}{2} + \frac{x^6}{6!} + \cdots \\
\ln(1 + x^2) &= x^2 - \frac{x^4}{2} + \frac{x^6}{3} + \cdots \\
\cos(2x) &= 1 - \frac{4x^2}{2} + \frac{16x^4}{4!} - \frac{64x^6}{6!} + \cdots \\
2x\sin x &= 2x^2 - \frac{2x^4}{3!} + \frac{2x^6}{5!} + \ldots
\end{align}$$
We will see in a moment if we need more or fewer terms. Then the numerator looks like
$$\begin{align}
e^{x^2} - \ln(1 + x^2) - 1 &\approx \left( 1 + x^2 + \frac{x^4}{2} + \color{#BB0000}{\frac{x^6}{4!}}\right) - \left( 1 + x^2 - \frac{x^4}{2} + \color{#BB0000}{\frac{x^6}{3}}\right) \\
&\approx x^4 +\color{#BB0000}{\left( \frac{1}{4!} - \frac{1}{3}\right)x^6},
\end{align}$$
which is slightly different than what I said in the comments because I unknowingly dropped a negative sign. In black is the lowest number of coefficients that will matter, but I'm keeping extras in red so that we can better understand how many are necessary.
Similarly, the denominator will look like
$$\begin{align}
\cos 2x + 2x\sin x - 1 &\approx \left(\frac{-4x^2}{2} + \frac{16x^4}{4!} - \color{#BB0000}{\frac{64x^6}{6!}}\right) + \left(2x^2 - \frac{2x^4}{3!} + \color{#BB0000}{\frac{2x^6}{5!}}\right) \\
&\approx \frac{1}{3}x^4 + \color{#BB0000}{\left(\frac{2}{5!} - \frac{64}{6!}\right)x^6}.
\end{align}$$
Putting this all together, your fraction looks a lot like
$$ \lim_{x \to 0}\frac{x^4 +\color{#BB0000}{\left( \frac{1}{4!} - \frac{1}{3}\right)x^6}}{\frac{1}{3}x^4 + \color{#BB0000}{\left(\frac{2}{5!} - \frac{64}{6!}\right)x^6}}.$$
We can factor out $x^4$ from the numerator and the denominator, leaving us with
$$ \lim_{x \to 0} \frac{1 + \color{#BB0000}{\left( \frac{1}{4!} - \frac{1}{3}\right)x^2}}{\frac{1}{3} + \color{#BB0000}{\left(\frac{2}{5!} - \frac{64}{6!}\right)x^2}} = 3.$$
In this last form, you can see that everything in $\color{#BB0000}{\text{red}}$ doesn't contribute, since it all goes to $0$ as $x \to 0$. Similarly, if we included many more terms, they would all have even higher powers of $x$, and thus go to $0$ as well.
So we see that we needed to go up to degree $4$ approximations everywhere to get the answer. Why degree $4$? Because that is how many contribute to the final answer. A general rule is that you can't go wrong by including too many terms, but you can go wrong if you include too few. $\diamondsuit$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Sum of $k$-th powers Given:
$$
P_k(n)=\sum_{i=1}^n i^k
$$
and $P_k(0)=0$, $P_k(x)-P_k(x-1) = x^k$ show that:
$$
P_{k+1}(x)=(k+1) \int^x_0P_k(t) \, dt + C_{k+1} \cdot x
$$
For $C_{k+1}$ constant.
I believe a proof by induction is the way to go here, and have shown the case for $k=0$. This is where I'm stuck. I have looked at the right hand side for the k+1 case:
$$
(k+2)\int^x_0P_{k+1}(t) \, dt + C_{k+2} \cdot x
$$
and I don't see how this reduces to $P_{k+2}(x)$. Even if we are assuming the kth case, replacing $P_{k+1}$ in the integrand of the $(k+1)$-st case just makes it more messy. I am not looking for the answer just a push in the right direction. I can see that each sum ends up as a polynomial since expressions like $P_1(x) = 1+2+\cdots+x=\frac{x(x+1)}{2}$, but I don't know how to do that for arbitrary powers, and I believe I don't need to in order to solve this problem.
| From the definition of $P_k(n)$ we get
$$
P_k(n) = \sum_{i=1}^n i^k \Rightarrow
P_0(n) = \sum_{i=1}^n 1 = n.
$$
We now use complete induction over $k$ to proof the statement $S(k)$
$$
P_k(x) = k \int\limits_0^x P_{k-1}(t) \, dt + C_k \, x \quad (*)
$$
Base case
For $k=1$ we have $S(1)$:
$$
1 \int\limits_0^x P_0(t) \, dt + C_1 \, x
=
\int\limits_0^x t \, dt + C_1 \, x
=
\frac{1}{2} x^2 + C_1 \, x
$$
which corresponds to the well known Gauss summation formula $P_1(n)$, if $C_1 = 1/2$.
Inductive step
Assuming equation $(*)$ is true for $\{ 1, \ldots, k \}$ we perform the recursion by starting with $S(k)$ and then applying $S(k-1), S(k-2), \ldots$ until we hit the bottom with $S(1)$ and get $P_0 = \mbox{id}$. That integrand is then simple enough to perform the the $k$ iterated integrations:
\begin{align}
P_k(x)
&=
k \int\limits_0^x P_{k-1}(t_k) \, dt_k + C_k \, x \\
&=
k \int\limits_0^x \left( (k-1) \int\limits_0^{t_k}
P_{k-2}(t_{k-1}) \, dt_{k-1} + C_{k-1} \, t_k \right) \, dt_k + C_k \, x \\
&=
k (k-1) \int\limits_0^x \int\limits_0^{t_k}
P_{k-2}(t_{k-1}) \, dt_{k-1} \, dt_k + \frac{k}{2} C_{k-1} \, x^2 + C_k \, x \\
&=
k! \int\limits_0^x \cdots \int\limits_0^{t_2}
P_0(t_1) \, dt_1 \cdots \,dt_k
+ \sum_{j=1}^k C_{k-j+1} \frac{k!}{(k-j+1)!j!} x^j \\
&=
k! \int\limits_0^x \cdots \int\limits_0^{t_2}
t_1 \, dt_1 \cdots \,dt_k
+ \sum_{j=1}^k \binom{k}{j} \frac{C_{k-j+1}}{k-j+1} x^j \\
&=
k! \int\limits_0^x \cdots \int\limits_0^{t_3}
\frac{1}{2}t_2^2 \, dt_2 \cdots \,dt_k
+ \sum_{j=1}^k \binom{k}{j} \frac{C_{k-j+1}}{k-j+1} x^j \\
&=
k! \int\limits_0^x
\frac{1}{k!}t_k^k \,dt_k
+ \sum_{j=1}^k \binom{k}{j} \frac{C_{k-j+1}}{k-j+1} x^j \\
&=
\frac{1}{k+1}x^{k+1}
+ \sum_{j=1}^k \binom{k}{j} \frac{C_{k-j+1}}{k-j+1} x^j \\
\end{align}
Then we try to arrive at $S(k+1)$:
\begin{align}
(k+1) \int\limits_0^x P_k(t) \, dt + C_{k+1} \, x
&=
(k+1) \int\limits_0^x \left(
\frac{1}{k+1}t^{k+1} +
\sum_{j=1}^k \binom{k}{j} \frac{C_{k-j+1}}{k-j+1} t^j \right)
\, dt \\
& + C_{k+1} \, x \\
&=
(k+1) \left(
\frac{1}{(k+1)(k+2)}x^{k+2}
+ \right. \\
& \left. \sum_{j=1}^k \binom{k}{j} \frac{C_{k-j+1}}{(k-j+1)(j+1)} x^{j+1}
\right) + C_{k+1} \, x \\
&=
\frac{1}{k+2}x^{k+2}
+ \sum_{j=1}^k \binom{k}{j} \frac{(k+1)C_{k-j+1}}{(k-j+1)(j+1)} x^{j+1}
+ C_{k+1} \, x \\
&=
\frac{1}{k+2}x^{k+2}
+ \sum_{j=1}^k \binom{k+1}{j+1} \frac{C_{k-j+1}}{k-j+1} x^{j+1}
+ C_{k+1} \, x \\
&=
\frac{1}{k+2}x^{k+2}
+ \sum_{j=2}^{k+1} \binom{k+1}{j} \frac{C_{k+1-j+1}}{k+1-j+1} x^j
+ C_{k+1} \, x \\
&=
\frac{1}{k+2}x^{k+2}
+ \sum_{j=1}^{k+1} \binom{k+1}{j} \frac{C_{k+1-j+1}}{k+1-j+1} x^j \\
&=
P_{k+1}(x)
\end{align}
Thus $S(k+1)$ follows.
By the principle of induction $(*)$ holds for all $k \in \mathbb{N} \setminus \{ 0\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
In triangle ABC, Find $\tan(A)$.
In triangle ABC, if $(b+c)^2=a^2+16\triangle$, then find $\tan(A)$ . Where $\triangle$ is the area and a, b , c are the sides of the triangle.
$\implies b^2+c^2-a^2=16\triangle-2bc$
In triangle ABC, $\sin(A)=\frac{2\triangle}{bc}$, and $\cos(A)=\frac{b^2+c^2-a^2}{2bc}$,
$\implies \tan(A)=\frac{4\triangle}{b^2+c^2-a^2}$
$\implies \tan(A)=\frac{4\triangle}{16\triangle-2bc}$. But the answer is in the form, $\frac{x}{y}$ where $x$ and $y$ are integers.
Any help is appreciated. Thanks in advance.
| $$b^2+c^2-a^2=16\Delta - 2bc \Rightarrow \frac{b^2+c^2-a^2}{2bc}=\frac{8\Delta}{bc}-1 \Rightarrow \cos A=4\sin A-1$$
$$\Rightarrow 2\cos^2\frac{A}{2}-1=8\sin\frac{A}{2}\cos\frac{A}{2}-1 \Rightarrow \tan\frac{A}{2}=\frac{1}{4}$$
Hence,
$$\tan A=\frac{2\cdot \frac{1}{4}}{1-\frac{1}{4^2}}=\boxed{\dfrac{8}{15}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Advanced Integration techniques: Quadratic Expressions and U-Substitution Find $$\int \frac{2x-1}{x^2-6x+13}dx $$
In the final steps after a u-substitution, one arrives at $$\int \frac{2u}{u^2+4}du + \int\frac{5}{ u^2+4}du$$
The next step is arriving at $$\ln(u^2+4) + 5\arctan(\frac{u}{2}) + C$$
How does $\int$ $dx(2u)\over(u^2+4)$ yield $\ln(u^2+4)$? The power of the denominator is two. The denominator is not equivalent to a variable to the first power, so I do not think ln(...) can be the answer.
| Letting $u = x - 3$ we have that $du = dx$ and $2u + 5 = 2x -1$.
$$\begin{align}\int \frac{2x- 1}{x^2-6x + 13}dx &= \int \frac{2x- 1}{(x-3)^2 + 4}dx\\&=\int \frac{2u + 5}{u^2 + 4}du\\&=\int \frac{2u}{u^2 + 4}du + \int \frac{5}{u^2 + 4}du\\&=\ln |u^2 + 4| + \frac{5}{2}\arctan\Big(\frac{u}{2}\Big) \end{align}$$
Because $\frac{1}{u^2 + 4} = \frac{1}{4}\frac{1}{\frac{u^2}{4}+1} = \frac{1}{2}\frac{1}{2}\frac{1}{\frac{u^2}{4}+1} = \frac{d}{du}\arctan (\frac{u}{2})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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No. of different real values of $x$ which satisfy $17^x+9^{x^2} = 23^x+3^{x^2}.$ Number of different real values of $x$ which satisfy $17^x+9^{x^2} = 23^x+3^{x^2}.$
$\bf{My\; Try::}$Using Hit and trial $x=0$ and $x=1$ are solution of above exponential equation.
Now we will calculate any other solution exists or not.
If $x\geq 2\;,$ Then $17^x+9^{x^2}>9^{x^2} = (6+3)^{x^2}>6^{x^2}+3^{x^2} = (6^x)^x+3^{x^2}>23^x+3^{x^2}\;,$
bcz $(6^x>23)\; \forall x\geq 2.$
So no solution in $x\in \left[2,\infty\right)$
Now i did not understand how can i calculate in $x<0$ and $0<x<1$.
Help me, Thanks
|
$$17^x+9^{x^2} = 23^x+3^{x^2}$$
Clearly $0,1$ are the two roots.I would prefer rough sketching the graph:
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Power series for the rational function $(1+x)^3/(1-x)^3$
Show that $$\dfrac{(1+x)^3}{(1-x)^3} =1 + \displaystyle\sum_{n=1}^{\infty} (4n^2+2)x^n$$
I tried with the partial frationaising the expression that gives me
$\dfrac{-6}{(x-1)} - \dfrac{12}{(x-1)^2} - \dfrac{8}{(x-1)^3} -1$
how to proceed further on this having doubt with square and third power term in denominator.
| $\dfrac{(1+x)}{(1-x)^3}^3 =\dfrac{6}{(1-x)} - \dfrac{12}{(1-x)^2} + \dfrac{8}{(1-x)^3} -1=6\sum_{n=0}^\infty x^n-12\sum_{n=0}^\infty (n+1)x^n+4\sum_{n=0}^\infty (n+1)(n+2)x^n=1+\sum_{n=1}^\infty (6-12n-12+4n^2+12n+8)x^n=1 + \displaystyle\sum_{n=1}^{\infty} (4n^2+2)x^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1078092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
AlgebraII factoring polynomials equation: $2x^2 - 11x - 6$
Using the quadratic formula, I have found the zeros: $x_1 = 6, x_2 = -\frac{1}{2}$
Plug the zeros in: $2x^2 + \frac{1}{2}x - 6x - 6$
This is where I get lost. I factor $-6x - 6$ to: $-6(x + 1)$, but the answer says otherwise. I am also having trouble factoring the left side.
Could someone please explain to me why the answer to the question was: $(x - 6)(2x + 1)$. How does $-\frac{1}{2}$ become $1$?
| $$
2x^2-11x - 6 = 2(x -\bullet)(x-\bullet)
$$
The two "$\bullet$"s are the two roots, $6$ and $\dfrac{-1}2$. So you have
$$
2(x-6)\left(x-\frac {-1} 2 \right) = (x-6)\ \underbrace{2\left(x+\frac 1 2 \right)} = (x-6)(2x+1).
$$
So $2\left(x + \dfrac 1 2\right)$ becomes $2x+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Number of $ 6 $ Digit Numbers with Alphabet $ \left\{ 1, 2, 3, 4 \right\} $ with Each Digit of the Alphabet Appearing at Least Once Find the number of 6 digit numbers that can be made with the digits 1,2,3,4 if all the digits are to appear in the number at least once.
This is what I did -
I fixed four of the digits to be 1,2,3,4 .
Now remaining 2 places can be filled with 4 digits each. Number of 6 digit numbers if two places are filled with same digit are 4 * 6!/3! and if filled by different digits are 12 * 6!/(2!*2!). Therefore, total such numbers are 2880.
But the correct answer is 1560.
Any hint would be appreciated.
| Instead of dividing into cases, we use the Principle of Inclusion/Exclusion.
There are $4^6$ strings of length $6$ over our $4$-letter alphabet. Now we count the bad strings, in which one or more digits are missing.
There are $3^6$ strings with the digit $1$ mising, and also $3^6$ with $2$ missing, with $3$ missing, with $4$ missing.
So our first estimate for the number of bad strings is $4\cdot 3^6$.
However, when we added, we multiply counted the bad strings that have more than one missing digit. For example, there are $2^6$ strings with $1$ and $2$ missing, and the same for all $6$ ways to choose $2$ digits from $4$.
So our next estimate for the number of bad strings is $4\cdot 3^6-6\cdot 2^6$.
However, we have subtracted too much, for we have subtracted one too many times the $4$ strings with $3$ digits missing. So the number of bads is $4\cdot 3^6-6\cdot 2^6+4$.
More neatly, we can write the number of good strings as
$$\binom{4}{0}4^6-\binom{4}{1}3^6+\binom{4}{2}2^6-\binom{4}{3}1^6.$$
The method readily generalizes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
Finding singularities in circle of convergence of $f(z)$ and showing taylor series diverges there $$f(x)=\arctan(x)$$
I know that $$\dfrac{1}{1+x^2}=1-x+x^2-x^3+\cdots=\sum_{i=0}^\infty (-x)^i$$
Also: $$\arctan(x)=\int \dfrac{1}{1+x^2}dx = \int \sum_{i=0}^\infty (-x)^i dx = x - 0.5x^2+1/3x^3+\cdots=\sum_{i=0}^\infty \frac{(-1)^i(x)^{i+1}}{2i+1}$$
Radius of convergence $R = 1$ using ratio test $\implies |z|\le1$ is the circle of convergence
Singularities are at $\pm i$ and both lie in $|z|$ however how do i show that the series diverges there?
EDIT:
$$\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 +\cdots + (-1)^kx^{2k} + \cdots\to \int \dfrac{dx}{1+x^2} = \sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{2k+1}.$$
| Observe: $$\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 +\cdots + (-1)^kx^{2k} + \cdots\to \int \dfrac{dx}{1+x^2} = \sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{2k+1}.$$ When $ x = \pm i$, the series equals $\pm i\cdot \displaystyle \sum_{k=0}^\infty \dfrac{1}{2k+1} = \infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Show without differentiation that $\frac {\ln{n}}{\sqrt{n+1}}$ is decreasing
Show that the function $\displaystyle \frac {\ln{n}}{\sqrt{n+1}}$ is decreasing from some $n_0$
My try: $\displaystyle a_{n+1}=\frac{\ln{(n+1)}}{\sqrt{n+2}}\le \frac{\ln{(n)+\frac{1}{n}}}{\sqrt{n+2}}$
so we want to show that $\ln{n}\cdot(\sqrt{n+1}-\sqrt{n+2})+\frac{\sqrt{n+1}}{n}\le 0$ or equivalently $n\cdot \ln{n} \cdot (\sqrt{\frac{n+2}{n+1}}-1)\ge1$ and I'm stuck here.
| First, you can show very easily without calculus that $k!>4^k$ whenever $k\ge18$: we then have
$$\eqalign{k!
&=(1\times2\times3)\times(4\times5\times\cdots\times15)\times(16\times17\times18)\times19\times\cdots\times k\cr
&>(1\times1\times1)\times(4\times4\times\cdots\times4)\times(4^2\times4^2\times4^2)\times4\times\cdots\times4\cr
&=4^k\ .\cr}$$
(In fact, this is true for $k\ge9$, but the proof takes a little more work.) Now for any $n$ we have
$$\eqalign{\Bigl(1+\frac1n\Bigr)^{2n+3}
&=\sum_{k=0}^{2n+3}\binom{2n+3}k\frac1{n^k}\cr
&\le\sum_{k=0}^{2n+3}\frac{(3n)^k}{k!}\frac1{n^k}\cr
&\le\sum_{k=0}^{17}\frac{3^k}{k!}+\sum_{k=18}^\infty\Bigl(\frac34\Bigr)^k\cr
&=\Bigl(\sum_{k=0}^{17}\frac{3^k}{k!}\Bigr)+3\Bigl(\frac34\Bigr)^{17}\ .\cr}$$
This last quantity is a specific, fixed number. So if $n$ is large enough we have
$$\eqalign{
n>\Bigl(1+\frac1n\Bigr)^{2n+3}
&\Rightarrow n^{2n+4}>(n+1)^{2n+3}\cr
&\Rightarrow n^{n+2}>(n+1)^{n+\frac32}\cr
&\Rightarrow n^{n+2}>(n+1)^{\sqrt{(n+1)(n+2)}}\cr
&\Rightarrow n^{1/\sqrt{n+1}}>(n+1)^{1/\sqrt{n+2}}\cr
&\Rightarrow \frac{\log n}{\sqrt{n+1}}>\frac{\log(n+1)}{\sqrt{n+2}}\ .\cr}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplification a trigonometric equation $$16 \cos \frac{2 \pi}{15} \cos\frac{4 \pi}{15} \cos\frac{8 \pi}{15} \cos\frac{14 \pi}{15}$$
$$=4\times 2 \cos \frac{2 \pi}{15} \cos\frac{4 \pi}{15} \times2 \cos\frac{8 \pi}{15} \cos\frac{14 \pi}{15}$$
I am intending in this way and then tried to apply the formula, $2\cos A \cos B$ but i think I might not get the answer. What to do now? the result will be 1.
| use this well know identity
$$2\sin{x}\cos{x}=\sin{2x}$$
so
\begin{align*}\cos{x}\cos{2x}\cos{4x}\cos{8x}&=\dfrac{2\sin{x}\cos{x}\cos{2x}\cos{4x}\cos{8x}}{2\sin{x}}\\
&=\dfrac{\sin{2x}\cos{2x}\cos{4x}\cos{8x}}{2\sin{x}}\\
&=\dfrac{2\sin{2x}\cos{2x}\cos{4x}\cos{8x}}{4\sin{x}}\\
&=\dfrac{\sin{4x}\cos{4x}\cos{8x}}{4\sin{x}}\\
&=\dfrac{2\sin{4x}\cos{4x}\cos{8x}}{8\sin{x}}\\
&=\dfrac{\sin{8x}\cos{8x}}{8\sin{x}}\\
&=\dfrac{2\sin{8x}\cos{8x}}{16\sin{x}}\\
&=\dfrac{\sin{16x}}{16\sin{x}}
\end{align*}
let
$x=\dfrac{\pi}{15}$
since
$$\cos{\dfrac{14\pi}{15}}=-\cos{\dfrac{\pi}{15}},\sin{\dfrac{16\pi}{15}}=-\sin{\dfrac{\pi}{15}}$$
so\begin{align*}
&16\cos{\dfrac{2\pi}{15}}\cos{\dfrac{4\pi}{15}}\cos{\dfrac{8\pi}{15}}\cos{\dfrac{14\pi}{15}}\\
&=-16\cos{\dfrac{\pi}{15}}\cos{\dfrac{2\pi}{15}}\cos{\dfrac{4\pi}{15}}\cos{\dfrac{8\pi}{15}}\\
&=-\dfrac{\sin{\dfrac{16\pi}{15}}}{\sin{\dfrac{\pi}{15}}}\\
&=1
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The locus of points $z$ which satisfy $|z - k^2c| = k|z - c|$, for $k \neq 1$, is a circle Use algebra to prove that the locus of points z which satisfy $|z - k^2c| = k|z - c|$, for $k \neq 1$ and $c = a + bi$ any fixed complex number, is a circle centre $O$.
Give the radius of the circle in terms of $k$ and $|c|$.
I squared both sides and got this:
$$(k^2−1)x^2+(k^2−1)y^2+(a^2+b^2-k^2a^2-k^2b^2)k^2=0$$
I might have gone wrong somewhere though.
Edit. Never mind, I didn't go wrong.
$$(k^2-1)x^2+(k^2-1)y^2-(k^2-1)k^2a^2-(k^2-1)k^2b^2=0$$
$$x^2+y^2=k^2(a^2+b^2)$$
$$r^2=k^2(a^2+b^2)$$
$$r=k|c|$$
| Here are the steps
$$ \left|z - k^2c\right| = k\left|z - c\right| $$
$$ \left|z - k^2c\right|^2 = k^2\left|z - c\right|^2 $$
$$ \left(z - k^2c\right)\left(\overline{z - k^2c}\right) = k^2(z - c)\left(\overline{z-c}\right) $$
$$ \left(z - k^2c\right)\left(\overline{z} - k^2\overline{c}\right) = k^2(z - c)\left(\overline{z}-\overline{c}\right) $$
$$ \left|z\right|^2 - k^2\overline{c}z-k^2c\overline{z}+k^4\left|c\right|^2= k^2\left(\left|z\right|^2-\overline{c}z-c\overline{z} + \left|c\right|^2\right) $$
$$ \left|z\right|^2 - k^2\overline{c}z-k^2c\overline{z}+k^4\left|c\right|^2= k^2\left|z\right|^2-k^2\overline{c}z-k^2c\overline{z} + k^2\left|c\right|^2 $$
$$ \left|z\right|^2 -k^2c\overline{z}+k^4\left|c\right|^2= k^2\left|z\right|^2-k^2c\overline{z} + k^2\left|c\right|^2 $$
$$ \left|z\right|^2 +k^4\left|c\right|^2= k^2\left|z\right|^2 + k^2\left|c\right|^2 $$
$$ \left|z\right|^2 -k^2\left|z\right|^2= k^2\left|c\right|^2 -k^4\left|c\right|^2$$
$$ \left|z\right|^2\left(1 -k^2\right)= k^2\left|c\right|^2\left(1 -k^2\right)$$
$$ \left|z\right|^2= k^2\left|c\right|^2$$
$$ \left|z\right|= k\left|c\right|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is $\ln(1+\frac{1}{x-1}) \ge \frac{1}{x}$ for all $x \ge 2$? Plotting both functions $\ln(1+\frac{1}{x-1})$ and $\frac{1}{x}$ in $[2,\infty)$ gives the impression that $\ln(1+\frac{1}{x-1}) \ge \frac{1}{x}$ for all $x \ge 2$.
Is it possible to prove it?
| Expand $\ln(1+\frac{1}{x-1})=\ln(\frac{x}{x-1})$ using Taylor series. You will get:
$$\ln(\frac{x}{x-1})=\frac{1}{x-1}-\frac12\left(\frac{1}{x-1}\right)^2+O(x^3)$$
Now you have to show that:
$$\frac{1}{x-1}-\frac12\left(\frac{1}{x-1}\right)^2\geq\frac1x$$
$$\frac{2x-3}{2\cdot(x-1)^2}\geq\frac1x$$
Which is equivalent to:
$$x>1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Compute the Jacobi bracket $R(x,y,z)=(x,y,z)$ and $\Theta(x,y,z)=(xz,yz,-(x^2+y^2))$
Show $[R,\Theta]=\Theta$
$[R,\Theta]=(R\cdot\nabla)\Theta-(\Theta\cdot\nabla)R$
$[R,\Theta]=(\frac{d}{dx}x+\frac{d}{dy}y+\frac{d}{dz}z)(xz,yz,-(x^2+y^2))-(\frac{d}{dx}xz+\frac{d}{dy}yz-\frac{d}{dz}(x^2+y^2)(x,y,z)$
$=((\frac{d}{dx}x+\frac{d}{dy}y+\frac{d}{dz}z)(xz),(\frac{d}{dx}x+\frac{d}{dy}y+\frac{d}{dz}z)(yz),(\frac{d}{dx}x+\frac{d}{dy}y+\frac{d}{dz}z)(-(x^2+y^2)))-((\frac{d}{dx}xz+\frac{d}{dy}yz-\frac{d}{dz}(x^2+y^2))x,(\frac{d}{dx}xz+\frac{d}{dy}yz-\frac{d}{dz}(x^2+y^2))y,(\frac{d}{dx}xz+\frac{d}{dy}yz-\frac{d}{dz}(x^2+y^2))z)$
$=(5xz,5yz,-5(x^2+y^2))-(3xz,3yz,2z^2-x^2-y^2)$
$=(2xz,2yz,-4x^2-4y^2-2z^2)\neq(xz,yz,-(x^2+y^2))$
Have tried this computation many times and keep getting the same wrong answer, must have a problem with my method, please help, it would be greatly appreciated. Not a hard computation, just long.
If you arrive at $(R\cdot\nabla)\Theta-(\Theta\cdot\nabla)R=\Theta$, please post working.
| Thanks to Ted Shifrin's post, I have found the mistakes I was making. Here is the correct working:
$R(x,y,z)=(x,y,z)$, $\Theta(x,y,z)=(xz,yz,-(x^2+y^2))$
$[R,\Theta]=(R\cdot\nabla)\Theta-(\Theta\cdot\nabla)R$
$[R,\Theta]=(x\frac{d}{dx}+y\frac{d}{dy}y+z\frac{d}{dz})(xz,yz,-(x^2+y^2))-((xz\frac{d}{dx}+yz\frac{d}{dy}-(x^2+y^2)\frac{d}{dz})(x,y,z))$
$=((x\frac{d}{dx}+y\frac{d}{dy}+z\frac{d}{dz})(xz),(x\frac{d}{dx}+y\frac{d}{dy}+z\frac{d}{dz})(yz),(x\frac{d}{dx}+y\frac{d}{dy}+z\frac{d}{dz})(-(x^2+y^2)))-((xz\frac{d}{dx}+yz\frac{d}{dy}-(x^2+y^2)\frac{d}{dz})x,(xz\frac{d}{dx}+yz\frac{d}{dy}-(x^2+y^2)\frac{d}{dz})y,(xz\frac{d}{dx}+yz\frac{d}{dy}-(x^2+y^2)\frac{d}{dz})z)$
$=(xz+xz,yz+yz,-2x^2-2y^2)-(xz,yz,-(x^2+y^2))$
$=2\Theta-\Theta=\Theta$
| {
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"url": "https://math.stackexchange.com/questions/1083947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int \frac{ax^2-b}{x\sqrt{c^2x^2-(ax^2+b)^2}} dx$ Integrate:
$$\int \frac{ax^2-b}{x\sqrt{c^2x^2-(ax^2+b)^2}} dx$$
What should be the substitution here so that it becomes simpler??? Please help
| This question follows a series of substitution, I'd say it's very tedious. However, yet simple.
Here's a hint:
First substitute. $x^2=t$, giving $dx=\dfrac{dt}{2x}$
Giving us, the integral as :
$$\large\int\dfrac{(at-b)}{2t\sqrt{c^2t-(at+b)^2}}.dt$$
The same can be re written as,
$$\large\int\dfrac{a}{2\sqrt{c^2t-(at+b)^2}}.dt-\large\int\dfrac{b}{2t\sqrt{c^2t-(at+b)^2}}.dt$$
Considering the first integral.
Put $at+b=y$ , so that $adt=dy$ and $t=\dfrac{y-b}{a}$.
After all the substitution, you should get.
$I_1=\dfrac{1}{2}\large\int\dfrac{1}{\sqrt{c^2\left(\dfrac{y-b}{a}\right)-(y)^2}}.dy$
Which is a rational function, and can be easily solved by completing the square.(try!)
End result should be(hoping no careless mistake!)
$~~~~~~~~~~~~~~~~~$$\begin{equation}I_1=\dfrac{1}{2}\sin^{-1}\left(\dfrac{(ax^2+b)-\dfrac{c^2}{2a}}{\left(\left(\dfrac{c^2}{2a}\right)^2-\dfrac{c^2b}{a}\right)^{\dfrac{1}{2}}}\right) \end{equation}$ $~~~~~~~~~~~~~~~~~~~$ Iff, $\left(\dfrac{c^2}{2a}\right)^2>\dfrac{c^2b}{a}$
For integral $I_2$ or the second integral. Procedure is fairly similar.
The only difference is that first you have to put $t=\dfrac{\large{1}}{\large{z}}$. Then the procedure is same as first integral giving you.
$~~~~~~~~~~~~~~~~~$$\begin{equation}I_2=\dfrac{1}{2}\sin^{-1}\left(\dfrac{(a+\dfrac{b}{x^2})-\dfrac{c^2}{2b}}{\left(\left(\dfrac{c^2}{2b}\right)^2-\dfrac{c^2a}{b}\right)^{\dfrac{1}{2}}}\right) \end{equation}$ $~~~~~~~~~~~~~~~~~~~$ Iff, $\left(\dfrac{c^2}{2b}\right)^2>\dfrac{c^2a}{b}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How find the maximum possible length of OC, where ABCD is a square, and AD is the chord of the circle? Given a circle $o(O(0,0), r=1)$. How to find the maximum possible length of $OC$, where $ABCD$ is a square, and $AD$ is the chord of the circle?
I have no idea how to do this, can this be proved with simple geometry?
| This might not exclusively be using simple geometry, but here is a solution.
Without loss of generality, we can assume that the point $D$ is located at $(1,0)$, and that the point $A$ is located with a nonnegative $y$ coordinate.
We can construct a function which gives the $y$ coordinate of the $A$ point given the $x$ coordinate, it will simply be a function following a half-circle.
The equation for the circle is: $y^2+x^2=1$, and to construct a function outlining this circle, we simply rearrange it to $y=\pm\sqrt{1-x^2}$. Since we want the upper half of the circle, we want the function $y=\sqrt{1-x^2}$.
We now have 2 points, one fixed at $(1,0)$ and one located somewhere on the upper half of the circle perimeter.
We assume that the square is pointing away from center of the circle. And to find $C$ point of the square, we can do the following:
For the line $AD$ the change in $x$ is $1-x$ and the change in $y$ is $\sqrt{1-x^2}$
Since the line $DC$ is ortogonal to $AD$, the change in $x$ and $y$ have ben exchanged, so the coordinate of $C$ is
$$C=\left(1+\sqrt{1-x^2},1-x\right)$$
For finding the distance from $(0,0)$ we use the pythagorean theorem and get
$$
\begin{align}
|OC|&=\sqrt{\left(1+\sqrt{1-x^2}\right)^2+(1-x)^2}\\
&=\sqrt{2\sqrt{1-x^2}-2x+3}
\end{align}
$$
The maximum distance is simply the maximum of this function. Since the function is exclusively positive in the domain the $x$ coordinate of the maximum does not change by taking the square
$$2\sqrt{1-x^2}-2x+3$$
For finding the maximum, take the derivative and put it equal to zero, since that gives the point where the curve turns around.
$$\frac d{dx}2\sqrt{1-x^2}-2x+3 = \frac{-2x}{\sqrt{1-x^2}}-2$$
$$
\begin{align}
\frac{-2x}{\sqrt{1-x^2}}-2&=0\\
\frac{-2x}{\sqrt{1-x^2}}&=2\\
-2x&=2\sqrt{1-x^2}\\
4x^2&=4(1-x^2)\\
x^2&=1-x^2\\
2x^2&=1\\
x&=\pm\frac1{\sqrt{2}}
\end{align}
$$
We introduced a extra solution when we took the square, check both solutions and find that the $x$ coordinate is
$$x=-\frac1{\sqrt2}$$
And plug that into the formula and find the distance is
$$
\begin{align}
&\sqrt{2\sqrt{1-\frac12}+2\frac1{\sqrt2}+3}\\
=&1+\sqrt2
\end{align}
$$
Therefore the solution is $1+\sqrt2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Closed form for integral of inverse hyperbolic function in terms of ${_4F_3}$ While attempting to evaluate the integral $\int_{0}^{\frac{\pi}{2}}\sinh^{-1}{\left(\sqrt{\sin{x}}\right)}\,\mathrm{d}x$, I stumbled upon the following representation for a related integral in terms of hypergeometric functions:
$$\small{\int_{0}^{1}\frac{x\sinh^{-1}{x}}{\sqrt{1-x^4}}\,\mathrm{d}x\stackrel{?}{=}\frac{\Gamma{\left(\frac34\right)}^2}{\sqrt{2\pi}}\,{_4F_3}{\left(\frac14,\frac14,\frac34,\frac34;\frac12,\frac54,\frac54;1\right)}-\frac{\Gamma{\left(\frac14\right)}^2}{72\sqrt{2\pi}}{_4F_3}{\left(\frac34,\frac34,\frac54,\frac54;\frac32,\frac74,\frac74;1\right)}}.$$
I'm having some trouble wading through the algebraic muckity-muck, so I'd like help confirming the above conjectured identity. More importantly, can these hypergeometrics be simplified in any significant way? The "niceness" of the parameters really makes me suspect it can be...
Any thoughts or suggestions would be appreciated. Cheers!
| $$
\newcommand{\as}{\sinh^{-1}}
\newcommand{\at}{\tan^{-1}}
\begin{align}
I &:= 2 \int_0^{\pi/2} \frac{x \as x}{\sqrt{1-x^4}}
\\&= \int_0^{\pi/2} \as\sqrt{ \sin x }
\\&= \frac 1 2\int_0^{\pi} \as\sqrt{ \sin x }
\\&= \frac 1 2\int_0^{\infty} \frac{2}{1+t^2}\as\sqrt{ \frac{2t}{1+t^2}}
\\&= \left.\at x \as\sqrt{ \frac{2t}{1+t^2}}\right\lvert _0^\infty - \int_0^{\infty} \at t \frac{1-t^2}{\sqrt{2t}(1+t^2)^{3/2}} \frac{1}{\sqrt{1+\frac{2t}{1+t^2}}}
\\&= \int_0^{\infty} \at t \frac{t-1}{\sqrt{2t}(1+t^2)}
\\&= \sqrt 2\int_0^{\infty} \frac{x^2-1}{1+x^4} \at x^2
\end{align}
$$
Let
$$
J(a) = \sqrt 2\int_0^{\infty} \frac{x^2-1}{1+x^4} \log(1+ a^2 x^2),
$$
so $J(0) = 0$ and
$$
\begin{align}
J'(a) &= \sqrt 2\int_0^{\infty} \frac{x^2-1}{1+x^4} \frac{2a x^2}{1+a^2x^2}
\\&= \sqrt 2\int_0^{\infty} \frac{2 \left(a^3 x^2-a^3-a x^2-a\right)}{\left(a^4+1\right) \left(x^4+1\right)}+\frac{2 a \left(a^2+1\right)}{\left(a^4+1\right) \left(a^2 x^2+1\right)}
\\&= \frac{ \pi\sqrt{2} }{a^2+\sqrt{2} a+1}
\\&= i \pi \left[\frac{1}{a+ \frac{1+i}{\sqrt 2}}-\frac{1}{a+\frac{1-i}{\sqrt 2}} \right].
\end{align}
$$
This implies
$$
J\left(\frac{1+i}{\sqrt 2}\right) = i \pi\left[\log\left(\sqrt 2 (1+i) \right) - \log \sqrt 2 \right],
$$
whence
$$
\boxed{
I = \operatorname{Im} J\left(\frac{1+i}{\sqrt 2}\right) = \frac \pi 2 \log 2.}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
How do I integrate: $\int\sqrt{\frac{x-3}{2-x}} dx$? I need to solve:
$$\int\sqrt{\frac{x-3}{2-x}}~{\rm d}x$$
What I did is:
Substitute: $x=2\cos^2 \theta + 3\sin^2 \theta$. Now:
$$\begin{align}
x &= 2 - 2\sin^2 \theta + 3 \sin^2 \theta \\
x &= 2+ \sin^2 \theta \\
\sin \theta &= \sqrt{x-2} \\
\theta &=\sin^{-1}\sqrt{x-2}
\end{align}$$
and, $ cos \theta = \sqrt{(3-x)} $
$ \theta=\cos^{-1}\sqrt{(3-x)}$
The integral becomes:
$$\begin{align}
&= \int{\sqrt[]{\frac{2\cos^2 \theta + 3\sin^2 \theta-3}{2-2\cos^2 \theta - 3\sin^2 \theta}} ~~(2 \cos \theta\sin\theta)}~{\rm d}{\theta}\\
%
&= \int{\sqrt[]{\frac{2\cos^2 \theta + 3(\sin^2 \theta-1)}{2(1-\cos^2 \theta) - 3\sin^2 \theta}}~~(2 \cos \theta\sin\theta)}~{\rm d}{\theta} \\
%
&= \int\sqrt[]{\frac{2\cos^2 \theta - 3\cos^2 \theta}{2\sin^2 \theta - 3\sin^2 \theta}}~~(2 \cos \theta\sin\theta) ~{\rm d}\theta \\
%
&= \int\sqrt[]{\frac{-\cos^2 \theta }{- \sin^2 \theta}}~~(2 \cos \theta\sin\theta) ~{\rm d}\theta \\
%
&= \int \frac{\cos \theta}{\sin\theta}~~(2 \cos \theta\sin\theta)~{\rm d}\theta \\
%
&= \int 2\cos^2 \theta~{\rm d}\theta \\
%
&= \int (1- \sin 2\theta)~{\rm d}\theta \\
%
&= \theta - \frac {\cos 2\theta}{2} + c \\
%
&= \sin^{-1}\sqrt{x-2} - \frac {\cos 2(\sin^{-1}\sqrt{x-2})}{2} + c
\end{align}$$
But, The right answer is :
$$\sqrt{\frac{3-x}{x-2}} - \sin^{-1}\sqrt{3-x} + c $$
Where am I doing it wrong?
How do I get it to the correct answer??
UPDATE:
I am so sorry I wrote:
= $\int 2\cos^2 \theta .d\theta$
= $\int (1- \sin 2\theta) .d\theta$
It should be:
= $\int 2\cos^2 \theta .d\theta$
= $\int (1+ \cos2\theta) .d\theta$
= $ \theta + \frac{\sin 2\theta}{2} +c$
What do I do next??
UPDATE 2:
= $ \theta + \sin \theta \cos\theta +c$
= $ \theta + \sin \sin^{-1}\sqrt{(x-2)}. \cos\cos^{-1}\sqrt{(3-x)}+c$
= $ \sin^{-1}\sqrt{(x-2)}+ \sqrt{(x-2)}.\sqrt{(3-x)}+c$
Is this the right answer or I have done something wrong?
| Hint:
For solving the integral, notice that
$$t^2 = \frac{x-3}{2-x} \Rightarrow dx = -\frac{2t}{(t^2+1)^2}dt.$$ Hence,
$$\int \sqrt{\frac{x-3}{2-x}}dx = -2\int\frac{t^2}{(t^2+1)^2}dt. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
A simple Partial Differential Equation? I thought... I was looking at the Partial Differential Equation involving function:
$$ z(x,y)$$
$$ \frac{\partial z}{\partial x} + c \frac{\partial z}{\partial y} = 0 $$
Which fairly intuitively has a solution:
$$ z = f\left( x - \frac{1}{c} y \right)$$
Now I was considering the generalization of this equation to:
$$ \frac{\partial z}{\partial x} + (c_1x + c_2) \frac{\partial z}{\partial y} = 0 $$
But instead of starting too large I would focus first on:
$$ \frac{\partial z}{\partial x} + c_1 x\frac{\partial z}{\partial y} = 0 $$
Which itself i focused on:
$$ \frac{\partial z}{\partial x} + x\frac{\partial z}{\partial y} = 0 $$
I did some algebraic debauchery and have concluded that:
$$ g = a_0 + a_1 x - 2 a_1 x\sum_{i=1}^{\infty} \left[\frac{1}{i}\begin{pmatrix} 2i-2 \\ i-1 \end{pmatrix} \left( \frac{y}{2x^2} \right)^n \right] $$
Is a solution to this. But I do not know its closed form.
Work So Far:
One notes that the solution $ z = f\left( x - \frac{1}{c} y \right)$ to the first problem can be derived by noting that:
$$ z = 1, x - \frac{1}{c}y, \left(x - \frac{1}{c}y\right)^2 ... $$
Are all solutions and the transform is linear so any function of the form:
$$ F = a_0(1) + a_1 \left(x - \frac{1}{c}y \right) + a_2 \left(x - \frac{1}{c}y\right)^2 ... $$
Is a solution, which is the general power series of all functions of the form
$$ f \left( x - \frac{1}{c}y\right) $$
Using that same intuition note:
$$ g_0 = a_0 $$
Is a solution... then we can consider $$ g_1 = a_0 + a_1 x $$ and ask what term should be added to $g_1$ to make $g_2$ such that $$ \frac{\partial g_1}{\partial x} + x \frac{\partial g_2}{\partial y} = 0 $$. Then recursively generate $g_3, g_4 ... $
By doing so we conclude that one solution is
$$ g = a_0 + a_1 x - 2 a_1 x\sum_{i=1}^{\infty} \left[\frac{1}{i}\begin{pmatrix} 2i-2 \\ i-1 \end{pmatrix} \left( \frac{y}{2x^2} \right)^n \right] $$
What is the closed form for this guy? From here it becomes clear that if we consider $g^r$
$$ \frac{\partial g^r}{\partial x} + x \frac{\partial g^r}{\partial y} = $$
$$ r g^{r-1} \left( \frac{\partial g}{\partial x} + x \frac{\partial g}{\partial y} \right) = r g^{r-1}(0) = 0 $$
And since our transform is linear:
$$ z = e_0 + e_1g + e_2 g^2 + e_3 g^3 ... $$
Is a solution to the equation meaning that the general solution is
$$ z = f(g) $$
| After playing around with wolfram alpha I found that
$$ \sum_{i=1}^{\infty} \left[ \frac{1}{i} \begin{pmatrix} 2i-2 \\ i-1 \end{pmatrix} u^i \right] = \frac{1}{2} (1 - \sqrt{1 - 4u} )$$
Therefore
$$g = a_0 + a_1 x - 2a_1 x \frac{1}{2} \left(1 - \sqrt{1 - 4\frac{y}{2x^2}} \right) = $$
$$ a_0 + a_1x - a_1x \left(1 - \sqrt{1 - 2 \frac{y}{x^2}} \right) = $$
$$ a_0 + a_1x - a_1x \left(1 - \frac{\sqrt{x^2 - 2 y}}{x} \right) = $$
$$ a_0 + a_1x - a_1 \left(x - \sqrt{x^2 - 2 y} \right) = $$
$$ a_0 + a_1 \sqrt{x^2 - 2y}$$
We verify:
$$ \frac{\partial g}{\partial x} = a_1 \frac{x}{\sqrt{x^2 - 2y}}$$
$$ x\frac{\partial g}{\partial y} = -a_1 \frac{x}{\sqrt{x^2 - 2y}}$$
$$\frac{\partial g}{\partial x} + x\frac{\partial g}{\partial y} = 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Calculate the limit of $(1+x2^x)/(1+x3^x)$ to the power $1/x^2$ when $x\to 0$ I have a problem with this: $\displaystyle \lim_{x \rightarrow 0}{\left(\frac{1+x2^x}{1+x3^x}\right)^\frac{1}{x^2}}$.
I have tried to modify it like this: $\displaystyle\lim_{x\rightarrow 0}{e^{\frac{1}{x^2}\ln{\frac{1+x2^x}{1+x3^x}}}}$ and then calculate the limit of the exponent: $\displaystyle \lim_{x\rightarrow 0}{\frac{1}{x^2}\ln{\frac{1+x2^x}{1+x3^x}}}$.
But I don't know what to do next. Any ideas?
| Since
$$
2^x=1+\log(2)x+O\!\left(x^2\right)\\
\Downarrow\\
\log\left(1+x2^x\right)=x+\left(\log(2)-\tfrac12\right)x^2+O\!\left(x^3\right)
$$
and
$$
3^x=1+\log(3)x+O\!\left(x^2\right)\\
\Downarrow\\
\log\left(1+x3^x\right)=x+\left(\log(3)-\tfrac12\right)x^2+O\!\left(x^3\right)
$$
Therefore,
$$
\frac1{x^2}\log\left(\frac{1+x2^x}{1+x3^x}\right)=\log\left(\frac23\right)+O\!\left(x\right)
$$
Applying $e^x$ gives
$$
\left(\frac{1+x2^x}{1+x3^x}\right)^{1/x^2}=\frac23\left(1+O(x)\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Find $\sin^3 a + \cos^3 a$, if $\sin a + \cos a$ is known
Given that $\sin \phi +\cos \phi =1.2$, find $\sin^3\phi + \cos^3\phi$.
My work so far:
(I am replacing $\phi$ with the variable a for this)
$\sin^3 a + 3\sin^2 a *\cos a + 3\sin a *\cos^2 a + \cos^3 a = 1.728$. (This comes from cubing the already given statement with 1.2 in it.)
$\sin^3 a + 3\sin a * \cos a (\sin a + \cos a) + \cos^3 a = 1.728$
$\sin^3 a + 3\sin a * \cos a (1.2) + \cos^3 a = 1.728$
$\sin^3 a + \cos^3 a = 1.728 - 3\sin a * \cos (a) *(1.2)$
Now I am stuck. What do I do next? Any hints?
| Note
$$
x^3+y^3 = (x+y)\left(x^2-xy +y^2\right)
$$
Now we can see that if we set
$$
x =\sin a\\
y = \cos a.
$$
Then we get
$$
\sin^3a+\cos^3a = (\sin a +\cos a)\left(1-\sin a\cos a\right)
$$
We can utilise
$$
(\sin a+ \cos a)^2 = 1 +2\sin a\cos a
$$
Combining all the above will yield
$$
\sin^3a+\cos^3a = (\sin a +\cos a)\left(1-\frac{(\sin a+ \cos a)^2-1}{2}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 0
} |
Finding value of 1 variable in a 3-variable $2^{nd}$ degree equation The question is: If $a,b,\space (a^2+b^2)/(ab-1)=q$ are positive integers, then prove that $q=5$. Also prove that for $q=5$ there are infinitely many solutions in $\mathbf N$ for $a$ and $b$. I simplified the equation as follows:-$$\frac {a^2+b^2}{ab-1}=q\\\begin{align}\\&=>\frac {2a^2+2b^2}{ab-1}=2q\\&=>\frac{a^2+b^2+2ab+a^2+b^2-2ab}{ab-1}=2q\\&=>(a+b)^2+(a-b)^2=2q(ab-1)\\&=>2(a+b)^2+2(a-b)^2=q(4ab-4)\\&=>2(a+b)^2+2(a-b)^2=q((a+b)^2-(a-b)^2-4)\end{align}$$Substituting $a+b=X$ and $a-b=Y$, we get $$2X^2+2Y^2=q(X^2-Y^2-4)\\\begin{align}&=>(q-2)X^2=(q+2)Y^2+4q\end{align}$$Now using the quadratic residues modulo $5$, I know that $X^2,Y^2\equiv0, \pm1(mod\space 5)$. But using this directly doesn't give the answer. So what to do after this? An answer without the use of co-ordinate geometry would be greatly appreciated as it seems there is a very good resemblance of the equation to a pair of hyperbolas which are symmetric with respect to the line $y=x$ but I don't understand co-ordinate geometry very well.
| I take it that you have looked into the details of Vieta Jumping wiki link, so if we know the curve $x^2+y^2 - qxy + q = 0$ has a single integer solution then there are infinitely many, generated by the pair of iterations:
$$(x,y) \mapsto (qx-y,x)$$ $$(x,y) \mapsto (y,x)$$
on the graph of the curve.
Further observe that for a lattice point $(x,y)$ with $y > x >0$, the orbit of the iteration $(x,y) \mapsto (qx-y,x)$ does not change the branch of hyperbola, i.e., remains in the branch in the first quadrant.
Geometrically this is equivalent to moving from a lattice point $(x_0,y_0)$ to the point $(x_0,x_0)$ on the line $y = x$ and then join and extend the line joining $(x_0,x_0)$ and $(y_0,x_0)$ to meet the curve again at $(qx_0 - y_0,x_0)$.A sequence of iterations that will look like:
Now, $x^2+y^2 - qxy + q = 0$ has no solution in natural numbers if $q = 1,2$
since,$x^2+y^2 - qxy + q > 0$ for $q = 1,2$.Thus, $q \ge 3$.
We note that the vertex of the branch in concern $\color{blue}{A} = \left(\sqrt{\frac{q}{q-2}},\sqrt{\frac{q}{q-2}}\right)$ lies on the line joining $\color{black}{J} = (1,1)$ and $\color{black}{H} = (2,2)$ for $q \ge 3$.
If $I = (x_i,y_i)$ is the lattice point in the orbit of the iteration closest to the vertex $\color{blue}{A}$, then the next point in the orbit $K$ no longer stays in the upper half (region $y>x$) part of the hyperbola, $K = (x_k,y_k)$ must lie in the lower half region ($x > y$).
But in order to go from $I$ to $K$ we find that the point on the line $x = y$, i.e., $(y_k,y_k)$ must be a lattice point as well. The only lattice point on this line between the origin and vertex $\color{blue}{A}$ is $J = (1,1)$, hence $y_k = 1$.
Therefore, both the roots of the quadratic $x^2+1 - qx + q = 0$ must be positive integers, viz $x_k$ and $qx_k - 1$.
The discriminant $\Delta = q^2 - 4(q+1)$ must be a perfect square. But $\Delta$ differs from $(q-2)^2$ by $8$, and the only perfect squares that differ by $8$ are $1$ & $9$.
$$\implies (q-2)^2 = 9 \implies q = 5 \textrm{ since } q \ge 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1093297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Help on proving a trigonometric identity involving cot and half angles Prove: $\cot\frac{x+y}{2}=-\left(\frac{\sin x-\sin y}{\cos x-\cos y}\right)$.
My original idea was to do this:
$\cot\frac{x+y}{2}$ = $\frac{\cos\frac{x+y}{2}}{\sin\frac{x+y}{2}}$, then substitute in the formulas for $\cos\frac{x+y}{2}$ and $\sin\frac{x+y}{2}$, but that became messy very quickly.
Did I have the correct original idea, but overthink it, or is there any easier way? Hints only, please.
| You should know the following factorisation formulae:
*
*$\sin p +\sin q= 2\sin\dfrac{p+q}2 \cos\dfrac{p-q}2$
*$\sin p-\sin q= 2\sin\dfrac{p-q}2 \cos\dfrac{p+q}2$
*$\cos p +\cos q= 2\cos\dfrac{p+q}2 \cos\dfrac{p-q}2$
*$\cos p-\cos q= -2\sin\dfrac{p+q}2 \sin\dfrac{p-q}2$
(They're derived from : \begin{align*}
&\sin(a+b)+\sin(a-b)= 2\sin a \cos b &&
\sin(a+b)-\sin(a-b)= 2\sin b \cos a\\
&\cos(a+b)+\cos(a-b)= 2\cos a \cos b&&
\cos(a+b)-\cos(a-b)= -2\sin a \sin b
\end{align*} by setting $p=a+b$, $q=a-b$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1093646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show that: $\frac{1}{1+\log_ab+\log_bc}+\frac{1}{1+\log_bc+\log_ca}+\frac{1}{1+\log_ca+\log_ab}\leq1$
Let $a, b, c>1$. Show that:
$$\frac{1}{1+\log_ab+log_bc}+\frac{1}{1+\log_bc+\log_ca}+\frac{1}{1+\log_ca+\log_ab}\leq1.$$
My attempt:
We noted $\log_bc=x, \log_ca=y, \log_ab=z$ with $xyz=1$ and we have reduced inequality at this time:$$2(x+y+z)\leq xy^2+x^2y+yz^2+y^2z+xz^2+x^2z.$$
Does anyone have idea how it goes? Thank You!
I found a simple continuation:
$$2(x+y+z)\leq xy^2+x^2y+yz^2+y^2z+xz^2+x^2z\iff$$
$$\iff2(x+y+z)\leq xy(x+y)+yz(y+z)+zx(z+x)\iff$$
$$\iff2(x+y+z)\leq (xy+yz+zx)(x+y+z)-3xyz\iff$$
$$\iff3\leq(x+y+z)(xy+yz+zx-2).$$
The last inequality follows easily using:
$$3\leq x+y+z$$
$$1=3(xyz)^{\frac{2}{3}} -2\leq xy+yz+zx-2.$$
These last inequality is obtained by applying AM-GM inequality.
| Following OP's simplification of the inequality:
$\displaystyle \begin{align} xy^2+x^2y+yz^2+y^2z+xz^2+x^2z &= \sum\limits_{cyc} (x^2y + xy^2) \\ &=\sum\limits_{cyc} (x^{6/3}y^{3/3} + x^{3/3}y^{6/3}) \\ & \ge \sum\limits_{cyc} (x^{5/3}y^{4/3} + x^{4/3}y^{5/3}) \tag{1}\\ &=\sum\limits_{cyc} x^{5/3}(y^{4/3} + z^{4/3})\tag{2} \\ & \ge 2\sum\limits_{cyc} x^{5/3}y^{2/3}z^{2/3} \tag{3}\\ &= 2\sum\limits_{cyc} x \end{align}$
Justification of steps:
$(1) \, (x^{6/3}y^{3/3} + x^{3/3}y^{6/3}) - (x^{5/3}y^{4/3} + x^{4/3}y^{5/3}) = xy(x^{1/3}+y^{1/3})(x^{1/3}-y^{1/3})^2 \ge 0$
$(2)$ Rearranging the cyclic summation.
$(3) \, y^{4/3} + z^{4/3} \ge 2y^{2/3}z^{2/3}$ and put $xyz = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $x,y,z \geq 1/2, xyz=1$, showing that $2(1/x+1/y+1/z) \geq 3+x+y+z$ If $x,y,z \geq 1/2, xyz=1$, showing that $2(1/x+1/y+1/z) \geq 3+x+y+z$
I tried Schturm's method for quite some time, and Cauchy Schwarz for numerators because of the given product condition.
| $f(x,y,z) = 2xy+2yz+2zx -x-y-z-3$ subject to $xyz=1$.
$f_x = \lambda zy \to 2xy+2xz - x = \lambda$.
$f_y = \lambda xz \to 2xy + 2yz - y = \lambda$.
$\Rightarrow (2z-1)(x-y) = 0$.
If $z=\dfrac{1}{2} \to f(x,y,z) = f(x,y,\frac{1}{2}) = 2xy + y+x-x-y-\dfrac{7}{2}=2(2)-\dfrac{7}{2} = \dfrac{1}{2}>0$
Thus assume $x=y$, and other cases gives: $x=y=z \to x =y=z = 1 \to f(1,1,1) = 0$. Thus $f(x,y,z) \geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
The roots of a certain recursively-defined family of polynomials are all real Let $P_0=1 \,\text{and}\,P_1=x+1$ and we have $$P_{n+2}=P_{n+1}+xP_n\,\,n=0,1,2,...$$
Show that for all $n\in \mathbb{N}$, $P_n(x)$ has no complex root?
| For $x = u+iv$ define the pair of sequences $u_n,v_n$ such that $P_n(x) = u_n+iv_n$ are the real and imaginary parts of $P_n(x)$.
Then, $u_{n+1}+iv_{n+1} = u_n+iv_n + (u+iv)(u_{n-1}+iv_{n-1})$
$\implies u_{n+1} = u_n + uu_{n-1} - vv_{n-1}$ and $v_{n+1} = v_n + uv_{n-1} + vu_{n-1}$
Assume $a+ib$ (where, $a,b \in \mathbb{R}$ and $b \neq 0$) is a root of $P_n(x)$ i.e., $P_n(a+ib) = 0$ and say $m$ is the smallest index $n$ such that $P_m(a+ib) = 0$.
Since, the coefficients of $P_n$ are all real numbers, $P_m(a-ib) = 0$.
Thus, $u_m = 0 = u_{m-1}+au_{m-2}-bv_{m-2} = u_{m-1}+au_{m-2}+bv_{m-2}$
and $v_{m} = 0 = v_{m-1} + av_{m-2} + bu_{m-2} = v_{m-1} + av_{m-2} - bu_{m-2}$
That is $-bv_{m-2} = bv_{m-2}$ and $-bu_{m-2} = bu_{m-2}$ and since $b \neq 0$ we must also have
$u_{m-2} = 0 = v_{m-2}$,i.e., $a+ib$ is a root of $P_{m-2}$ which contradicts the minimality of $m$.
Hence, $b = 0$, i.e., all the roots of $P_n$ are real.
Note: The roots of $P_n(x)$ are infact: $-\dfrac{1}{4}\sec^2 \frac{k\pi}{n+2}$ for $k = 1,2,\cdots \left[\frac{n+1}{2}\right]$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Mean value theorem with trigonometric functions Let $f(x) = 2\arctan(x) + \arcsin\left(\frac{2x}{1+x^2}\right)$
*
*Show that $f(x)$ is defined for every $ x\ge 1$
*Calculate $f'(x)$ within this range
*Conclude that $f(x) = \pi$ for every $ x\ge 1$
Can I get some hints how to start? I don't know how to start proving that $f(x)$ is defined for every $ x\ge 1$ and I even had problems calculating $f'(x)$
Thanks everyone!
| The expression for $f(x)$ is defined whenever
$$
-1\le\frac{2x}{1+x^2}\le1
$$
that is
\begin{cases}
2x\le 1+x^2\\
2x\ge -(1+x^2)
\end{cases}
or
\begin{cases}
x^2-2x+1\ge0\\
x^2+2x+1\ge0
\end{cases}
which is satisfied for every $x$.
Computing the derivative of
$$
g(x)=\arcsin\frac{2x}{1+x^2}
$$
just requires some patience; consider $h(x)=2x/(1+x^2)$; then
$$
g'(x)=\frac{1}{\sqrt{1-h(x)^2}}h'(x)
$$
by the chain rule. Now
$$
1-h(x)^2=1-\frac{4x^2}{(1+x^2)^2}=\frac{1+2x^2+x^4-4x^2}{(1+x^2)^2}=
\frac{(1-x^2)^2}{(1+x^2)^2}
$$
so
$$
\frac{1}{\sqrt{1-h(x)^2}}=\frac{1+x^2}{|1-x^2|}
$$
Note the absolute value! Always remember that $\sqrt{a^2}=|a|$. On the other hand, we can write $\sqrt{(1+x^2)^2}=1+x^2$, because this expression is always positive.
Now we tackle $h'(x)$:
$$
h'(x)=2\frac{1\cdot(1+x^2)-x\cdot 2x}{(1+x^2)^2}=\frac{2(1-x^2)}{(1+x^2)^2}
$$
and so, putting all together, we have
$$
g'(x)=\frac{1+x^2}{|1-x^2|}\frac{2(1-x^2)}{(1+x^2)^2}=
\frac{2}{1+x^2}\frac{1-x^2}{|1-x^2|}
$$
Note that this is not defined for $x=1$ or $x=-1$.
Thus you have
$$
f'(x)=\frac{2}{1+x^2}+\frac{2}{1+x^2}\frac{1-x^2}{|1-x^2|}
$$
For $x>1$ we have $|1-x^2|=x^2-1$, so $\frac{1-x^2}{|1-x^2|}=-1$
and therefore
$$
f'(x)=0
$$
for $x>1$ and the function is constant in the interval $(1,\infty)$. Since it is continuous at $1$ it is constant also on $[1,\infty)$. The constant value is
$$
f(1)=2\arctan1+\arcsin1=2\frac{\pi}{4}+\frac{\pi}{2}=\pi
$$
Note also that, for $x<-1$, we can draw the same conclusion and
$$
f(x)=-\pi \qquad(x\le -1)
$$
On the other hand, for $-1<x<1$ we have
$$
f'(x)=\frac{4}{1+x^2}
$$
so
$$
f(x)=k+4\arctan x
$$
(because in $(-1,1)$ the two functions have the same derivative), where $k$ is constant; this can be evaluated by seeing that $f(0)=0$, so $k=0$.
Thus your function $f$ can also be represented by
$$
f(x)=\begin{cases}
-\pi & \text{for $x\le-1$}\\
4\arctan x & \text{for $-1<x<1$}\\
\pi & \text{for $x\ge 1$}
\end{cases}
$$
Note. I showed the full exercise because it uses just basic facts. I wanted to underline how splitting the computation into pieces can lead to the result in an easier way than by having gigantic expressions to deal with.
Here's a picture of the graph.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
$\lim_{n \rightarrow \infty} \frac{1-(1-1/n)^4}{1-(1-1/n)^3}$ Find
$$\lim_{n \rightarrow \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}$$
I can't figure out why the limit is equal to $\dfrac{4}{3}$ because I take the limit of a quotient to be the quotient of their limits.
I'm taking that $\lim_{n \rightarrow \infty}1-\left(1-\frac{1}{n}\right)^4 = 0$ and likewise that $\lim_{n \rightarrow \infty}1-\left(1-\frac{1}{n}\right)^3 = 0$, which still gives me that the limit should be 0.
| If you expand the binomials, you get $\dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}=\dfrac{1-\left(1-\dfrac{4}{n}+\dfrac{6}{n^2}-\dfrac{4}{n^3}+\dfrac{1}{n^4}\right)}{1-\left(1-\dfrac{3}{n}+\dfrac{3}{n^2}-\dfrac{1}{n^3}\right)}=\dfrac{\dfrac{4}{n}-\dfrac{6}{n^2}+\dfrac{4}{n^3}-\dfrac{1}{n^4}}{\dfrac{3}{n}-\dfrac{3}{n^2}+\dfrac{1}{n^3}}$. Multiplying by $n$ on the top and bottom yields $\dfrac{4-\dfrac{6}{n}+\dfrac{4}{n^2}-\dfrac{1}{n^3}}{3-\dfrac{3}{n}+\dfrac{1}{n^2}}$. Then the limit is easier to evaluate. Any term with a division by $n$ will drop out as $n\to \infty$.
$\lim_{n \to \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}=\dfrac{4-0+0-0}{3-0+0}=\dfrac{4}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
} |
How do I solve this, first I have to factor $ 2x\over x-1$ + $ 3x +1\over x-1$ - $ 1 + 9x + 2x^2\over x^2-1$? I am doing calculus exercises but I'm in trouble with this
$$\frac{ 2x}{x-1} + \frac{3x +1}{ x-1} - \frac{1 + 9x + 2x^2}{x^2-1}$$
the solution is
$$ 3x\over x+1$$
The only advance that I have done is factor $ x^2-1$ = $( x-1)$ $ (x+1)$.
I do not know how can I factor $1 + 9x + 2x^2$, can someone please guide me in how to solve this exercise.
| Notice that $$\frac{A}{B} + \frac{A'}{BC} = \frac{C}{C}\frac{A}{B} + \frac{A'}{BC} = \frac{AC + A'}{BC}$$
Then
$$\begin{align}\frac{(x+1)}{(x+1)}\frac{2x}{(x-1)}+&\frac{(x+1)}{(x+1)}\frac{3x+1}{(x-1)}+\frac{1 + 9x + 2x^2}{(x-1)(x+1)}\\&=\frac{2x(x+1) + (3x+1)(x+1) - (1+9x + 2x^2)}{(x-1)(x+1)} \\&= \frac{2x^2 + 2x + 3x^2+3x + x + 1 - 1-9x - 2x^2}{(x-1)(x+1)}\\&=\frac{3x^2 - 3x}{(x-1)(x+1)}\end{align}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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} |
Why is $\sum_{m=0}^k \frac{k^2+k}{2} = \frac{(k^2+k)(k+1)}{2}$? I recieved an answer like this :
$S= \displaystyle \sum_{m=0}^k \dfrac{k^2+k}{2} - k\displaystyle \sum_{m=0}^k m + \displaystyle \sum_{m=0}^k m^2=\dfrac{(k^2+k)(k+1)}{2}-\dfrac{k\cdot k(k+1)}{2}+\dfrac{k(k+1)(2k+1)}{6}=k(k+1)\left(\dfrac{k+1}{2} - \dfrac{k}{2}+\dfrac{2k+1}{6}\right)= \dfrac{k(k+1)(k+2)}{3}$
But I don't understand this equation $\sum_{m=0}^k \dfrac{k^2+k}{2} = \dfrac{(k^2+k)(k+1)}{2}.$
Could anyone explain me please?
| Since $\displaystyle \sum_{x=0}^n C =(n+1) C$. you are summing constant $n+1$ times.
Note that you are summing over $m$ and $\dfrac {k^2+k}{2}$ is independent of $m$.
Note that $\displaystyle \sum_{x=0}^n C = \underbrace{C+C+ \dots +C}_{k+1 \text{times}} = (n+1) C$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding determinant of $n \times n$ matrix I need to find a determinant of the matrix:
$$
A = \begin{pmatrix}
1 & 2 & 3 & \cdot & \cdot & \cdot & n \\
x & 1 & 2 & 3 & \cdot & \cdot & n-1 \\
x & x & 1 & 2 & 3 & \cdot & n-2 \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
x & x & \cdot & \cdot & x & 1 & 2 \\
x & x & \cdot & \cdot & \cdot & x & 1 \\
\end{pmatrix}
$$
We know that $x \in R$
So far I managed to transform it to the form:
$$
\begin{pmatrix}
1-x & 1 & 1 & \cdot & \cdot & \cdot & 1 \\
0 & 1-x & 1 & 1 & \cdot & \cdot & 1 \\
0 & 0 & 1-x & 1 & 1 & \cdot & 1 \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
0 & 0 & \cdot & \cdot & 0 & 1-x & 1 \\
x & x & \cdot & \cdot & \cdot & x & 1 \\
\end{pmatrix}
$$
by the operations: (Let's say $r_i$ is the ith row)
$$r_1 = r_1 - r_n,r_2 = r_2-r_n, r_3 = r_3 - r_n, ..., r_{n-1} = r_{n-1} - r_n$$
and then $$r_1 = r_1 - r_2, r_2 = r_2 - r_3, r_3 = r_3 - r_4,...,r_{n-2} = r_{n-2} - r_{n-1}$$
Unfortunately, I have no idea how to eliminate the last row. Any hints?
| If $c_i$ is $i$th column of your second determinant, do $c_n= c_n-c_{n-1}$, $c_{n-1}=c_{n-1}-c_{n-2}$, ..., $c_2=c_2-c_1$ to get: $$\left|\begin{array}{ccccccc}
1-x & x & 0 & 0 & \cdots & 0 & 0\\
0 & 1-x & x & 0 & \cdots & 0 & 0\\
0 & 0 & 1-x & x & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & 0 & \cdots & 1-x & x\\
x & 0 & 0 & 0 & \cdots & 0 & 1-x\\
\end{array}\right|$$
This determinant is obviously equal $(1-x)^n+(-1)^{n+1}x^n$ (expand it by the first column).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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complex algebra Calculate: \begin{align*}
&\left( 1+\frac{1+i}{2} \right) \left( 1+ \left( \frac{1+i}{2} \right)^2 \right) \left( 1+{\left( \frac{1+i}{2} \right)^2}^2 \right) \ldots \\
&\left( 1+{\left( \frac{1+i}{2} \right)^2}^k \right) \ldots \left( 1+{\left( \frac{1+i}{2} \right)^2}^{2001} \right) \text{.}
\end{align*}
Can someone help?Whatever I trying i can't find some rule to calculate this.
| Multiply by $$\left(1-\frac{1+i}{2}\right)$$ compute, and divide by that number at the end.
The thing is that $$\left(1-\frac{1+i}{2}\right)\left(1+\frac{1+i}{2}\right)=\left(1-\left(\frac{1+i}{2}\right)^2\right)$$ then $$\left(1-\left(\frac{1+i}{2}\right)^2\right)\left(1+\left(\frac{1+i}{2}\right)^2\right)=\left(1-\left(\frac{1+i}{2}\right)^4\right)$$
and so on.
The whole product collapses to $$1-\left(\frac{1+i}{2}\right)^{2^{2002}}$$
Then we divide by $1-\frac{1+i}{2}$ to get back the original product:
$$\begin{align}\frac{1-\left(\frac{1+i}{2}\right)^{2^{2002}}}{1-\frac{1+i}{2}}&=2\left(1-\left(\frac{1+i}{2}\right)^{2^{2002}}\right)\cdot\left(1-\frac{1-i}{2}\right)\\&=2\left[1-\left(\frac{1+i}{2}\right)^{2^{2002}}-\left(\frac{1+i}{2}\right)+\left(\frac{1+i}{2}\right)^{2^{2003}}\right]=2\left[1-\left(2^{-1/2}e^{\pi i/4}\right)^{2^{2002}}-\left(\frac{1+i}{2}\right)+\left(2^{-1/2}e^{\pi i/4}\right)^{2^{2003}}\right]\\&=2\left[1-\left(2^{-1/2}e^{\pi i/4}\right)^{2^{2002}}-\left(\frac{1+i}{2}\right)+\left(2^{-1/2}e^{\pi i/4}\right)^{2^{2003}}\right]\\&=2\left[1-2^{-2^{2000}}-\frac{1+i}{2}+2^{-2^{2001}}\right]\\&=\left[\frac{3}{2}-2^{1-2^{2000}}+2^{-2^{2001}}\right]-i\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Number theory: prove that if $a,b,c$ odd then $2\gcd(a,b,c) = \gcd(a+b,b+c, c+a)$ Please help! Am lost with the following:
Prove that if $a,b,c$ are odd integers, then $2 \gcd(a,b,c) = \gcd( a+b, b+c, c+a)$
Thanks a lot!!
| Let $d=\gcd(a+b,b+c,c+a)$.
$$d\mid a+b,b+c\implies d\mid (a+b)-(b+c)\implies d\mid a-c$$
$$\begin{cases}d\mid a-c\\d\mid a+c\end{cases}\implies d\mid (a+c)-(a-c)\implies d\mid 2c$$
Similarly, we have $d\mid 2a$, $d\mid 2b$.
$$\begin{cases}d\mid 2a\\d\mid 2b\\d\mid 2c\end{cases}\implies d\mid \gcd(2a,2b,2c)\implies d\mid 2\gcd(a,b,c)$$
So we've proved that $\gcd(a+b,b+c,c+a)\mid 2\gcd(a,b,c)$. What is left to prove is that $2\gcd(a,b,c)\mid \gcd(a+b,b+c,c+a)$.
$$\begin{cases}\gcd(a,b,c)\mid a\\\gcd(a,b,c)\mid b\\\gcd(a,b,c)\mid c\end{cases}\implies \begin{cases}\gcd(a,b,c)\mid a+b\\\gcd(a,b,c)\mid b+c\\\gcd(a,b,c)\mid c+a\end{cases}\implies \gcd(a,b,c)\mid \gcd(a+b,b+c,c+a)\implies $$
$$\implies 2\gcd(a,b,c)\mid \gcd(a+b,b+c,c+a)$$
(since $\gcd(a+b,b+c,c+a)$ is even and $\gcd(a,b,c)$ is odd). $\ \ \ \square$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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why is $\lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}}$=-3? Exercise taken from here: https://mooculus.osu.edu/textbook/mooculus.pdf (page 42, "Exercises for Section 2.2", exercise 4).
Why is $\lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}}$=-3*? I always find 3 as the solution. I tried two approaches:
Approach 1:
$$
\lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}} = \lim_{x\to -\infty} \frac{3x+7}{x} = \lim_{x\to -\infty} \frac{3x}{x}+\lim_{x\to -\infty} \frac{7}{x}\\
= \lim_{x\to -\infty} 3+\lim_{x\to -\infty} \frac{7}{x} = 3 + 0 = 3
$$
Approach 2:
$$
\lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}} = \lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}} \times \frac{\frac{1}{x}}{\frac{1}{x}} = \lim_{x\to -\infty} \frac{3+\frac{7}{x}}{\frac{\sqrt{x^2}}{x}}\\
= \lim_{x\to -\infty} \frac{3+\frac{7}{x}}{\frac{\sqrt{x^2}}{\sqrt{x^2}}}
= \lim_{x\to -\infty} \frac{3+\frac{7}{x}}{\sqrt{\frac{x^2}{x^2}}}\\
= \frac{\lim_{x\to -\infty}3+\lim_{x\to -\infty}\frac{7}{x}}{\lim_{x\to -\infty}\sqrt{1}}\\
= \frac{3 + 0}{1} = 3
$$
* -3 is given as the answer by the textbook (cf. page 247) as well as wolfram|alpha
EDIT: I just re-read page 40 of the textbook and realized that I made a mistake in my approach 2. Instead of multiplying with $\frac{1}{x}$ I should have multiplied with $\frac{-1}{x}$ which is positive because as $x\to -\infty$, x is a negative number.
It thus reads:
$$
\lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}} = \lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}} \times \frac{\frac{-1}{x}}{\frac{-1}{x}} = \lim_{x\to -\infty} \frac{-3+\frac{-7}{x}}{\frac{\sqrt{x^2}}{-x}}\\
= \lim_{x\to -\infty} \frac{-3+\frac{-7}{x}}{\frac{\sqrt{x^2}}{\sqrt{x^2}}}
= \lim_{x\to -\infty} \frac{-3+\frac{-7}{x}}{\sqrt{\frac{x^2}{x^2}}}\\
= \frac{\lim_{x\to -\infty}-3+\lim_{x\to -\infty}\frac{-7}{x}}{\lim_{x\to -\infty}\sqrt{1}}\\
= \frac{-3 + 0}{1} = -3
$$
| $\sqrt{x^2}=|x|$ for real $x$
For $x\le0, |x|=-x$
| {
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"timestamp": "2023-03-29T00:00:00",
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Summation of an infinite series The sum is as follows:
$$
\sum_{n=1}^{\infty} n \left ( \frac{1}{6}\right ) \left ( \frac{5}{6} \right )^{n-1}\\
$$
This is how I started:
$$
= \frac{1}{6}\sum_{n=1}^{\infty} n \left ( \frac{5}{6} \right )^{n-1} \\
= \frac{1}{5}\sum_{n=1}^{\infty} n \left ( \frac{5}{6} \right )^{n}\\\\
= \frac{1}{5}S\\
S = \frac{5}{6} + 2\left (\frac{5}{6}\right)^2 + 3\left (\frac{5}{6}\right)^3 + ...
$$
I don't know how to group these in to partial sums and get the result. I also tried considering it as a finite sum (sum from 1 to n) and applying the limit, but that it didn't get me anywhere!
PS: I am not looking for the calculus method.
I tried to do it directly in the form of the accepted answer,
$$
\textrm{if} \ x= \frac{5}{6},\\
S = x + 2x^2 + 3x^3 + ...\\
Sx = x^2 + 2x^3 + 3x^4 + ...\\
S(1-x) = x + x^2 + x^3 + ...\\
\textrm{for x < 1},\ \ \sum_{n=1}^{\infty}x^n = -\frac{x}{x-1}\ (\textrm{I looked up this eqn})\\
S = \frac{x}{(1-x)^2}\\
\therefore S = 30\\
\textrm{Hence the sum} \sum_{n=1}^{\infty} n \left ( \frac{1}{6}\right ) \left ( \frac{5}{6} \right )^{n-1} = \frac{30}{5} = 6
$$
| Letting $a = d = 1/6$ and $r = 5/6$, our sum is:
$$
S = a + (a + d)r + (a + 2d)r^2 + (a + 3d)r^3 + \cdots
$$
Scaling by $r$, we find that:
$$
rS = ar + (a + d)r^2 + (a + 2d)r^3 + \cdots
$$
Subtracting the two equations (by collecting like powers of $r$), we obtain:
$$
(1 - r)S = a + dr + dr^2 + dr^3 + \cdots = a + dr(1 + r + r^2 + \cdots) = a + \frac{dr}{1 - r}
$$
Hence, we conclude that:
$$
S = \frac{a}{1 - r} + \frac{dr}{(1 - r)^2} = \frac{1/6}{1 - 5/6} + \frac{(1/6)(5/6)}{(1 - 5/6)^2} = \frac{1}{6 - 5} + \frac{(1)(5)}{(6 - 5)^2} = 6
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral $\int \frac{x+2}{x^3-x} dx$ I need to solve this integral but I get stuck, let me show what I did:
$$\int \frac{x+2}{x^3-x} dx$$
then:
$$\int \frac{x}{x^3-x} + \int \frac{2}{x^3-x}$$
$$\int \frac{x}{x(x^2-1)} + 2\int \frac{1}{x^3-x}$$
$$\int \frac{1}{x^2-1} + 2\int \frac{1}{x^3-x}$$
now I need to resolve one integral at the time so:
$$\int \frac{1}{x^2-1}$$ with x = t I have:
$$\int \frac{1}{t^2-1}$$
Now I have no idea about how to procede with this...any help?
| HINT:
$$\int\frac{x+2}{x^3-x}=\int\frac{-2}{x}+\int\frac{\frac{3}{2}}{x-1}+\int\frac{\frac{1}{2}}{x+1}$$
Can u do it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the real and imaginary part of the following I'm having trouble finding the real and imaginary part of $z/(z+1)$ given that z=x+iy. I tried substituting that in but its seems to get really complicated and I'm not so sure how to reduce it down. Can anyone give me some advice?
| Just multiply the fraction by the complex conjugate of $z+1$, that is,
\begin{equation*}
\frac{z}{z+1} = \frac{x+iy}{1+x+iy} = \frac{1+x-iy}{1+x-iy} \frac{x+iy}{1+x+iy} = \frac{(1+x)x+y^2+iy }{(1+x)^2+y^2} =
\end{equation*}
\begin{equation*}
= \frac{(1+x)x+y^2}{(1+x)^2+y^2} + i \frac{y}{(1+x)^2+y^2}
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
The maximum of $\binom{n}{x+1}-\binom{n}{x}$ The following question comes from an American Olympiad problem. The reason why I am posting it here is that, although it seems really easy, it allows for some different and really interesting solutions. Do you want to give a try?
Let $n$ be one million. Find the maximum value attained by $\binom{n}{x+1}-\binom{n}{x}$, where $0<x<n$ is an integer.
Edit: I saw the answers below, and I can say that I posted this one because there is at least one really nice solution, which is not so "mechanical" :) The value of $n$ has no particular meaning, it stands just for a "sufficiently large integer"..
| Note that
$$
\begin{align}
\binom{n}{x+1}-\binom{n}{x}
&=\left[\frac{n-x}{x+1}-1\right]\binom{n}{x}\\
&=\frac{n-2x-1}{x+1}\binom{n}{x}\tag{1}
\end{align}
$$
$(1)$ is positive for $x\lt\frac{n-1}2$ and negative for $x\gt\frac{n-1}2$.
Furthermore,
$$
\begin{align}
&\hphantom{}\left[\binom{n}{x+1}-\binom{n}{x}\right]-\left[\binom{n}{x}-\binom{n}{x-1}\right]\\
&=\left(\left[\frac{n-x}{x+1}-1\right]-\left[1-\frac{x}{n-x+1}\right]\right)\binom{n}{x}\\
&=\frac{4x^2-4nx+(n+1)(n-2)}{(x+1)(n-x+1)}\binom{n}{x}\tag{2}
\end{align}
$$
$(2)$ is $0$ when $x=\frac{n\pm\sqrt{n+2}}2$. If this $x$ is an integer, then $\binom{n}{x+1}-\binom{n}{x}=\binom{n}{x}-\binom{n}{x-1}$. Otherwise, the maximum occurs somewhere between $x$ and $x-1$.
Thus, the maximum value for $(1)$ is where $x=\left\lfloor\dfrac{n-\sqrt{n+2}}2\right\rfloor$.
For $n=1000000$, $x=499499$.
$$
\binom{1000000}{499500}-\binom{1000000}{499499}=9.582658295\times10^{301023}
$$
whereas
$$
\binom{1000000}{499499}-\binom{1000000}{499498}=9.582581749\times10^{301023}
$$
and
$$
\binom{1000000}{499501}-\binom{1000000}{499500}=9.582658257\times10^{301023}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find a parametrization of the intersection curve between two surfaces in $\mathbb{R^3}$ $x^2+y^2+z^2=1$ and $x^2+y^2=x$. Find a parametrization of the intersection curve between two surfaces in $\mathbb{R}^3$
$$x^2+y^2+z^2=1$$ and $$x^2+y^2=x.$$
I know that $x^2+y^2+z^2=1$ is a sphere and that $x^2+y^2=x$ is a circular cylinder. Any help is greatly appreciated, thank you.
| +1) The polar equation of $x^2+y^2=x$ is $r=cos\theta$ and so we have $x=rcos\theta$ and $y=rsin\theta$ as parametrics (with $r$ known) Substituting into the first equation, you can solve for $z$. Note that $1-cos^2\theta=sin^2\theta$ Hope this helps, if not or incorrect, I will take it off.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $\frac{a+1}{b}+\frac{b}{a}$ is an integer then it is $3$. If $\frac{a+1}{b}+\frac{b}{a}$ is an integer for positive integers $a,b$ then prove that this integer is $3$.
I reduced the to prove that if $\frac{c^2+d^2+1}{cd}$ is an integer then it is $3$ where $c,d\in{\mathbb{N}}$. And this is equivalent to prove that the Pell's equation $x^2-(k^2-4)y^2=-4$ has solutions for $k=3$ only. But I dont know how to prove it.
| Let
$$\dfrac{a^2+a+b^2}{ab}=k,k\in N^{+}$$
so
$$b^2-ka\cdot b+a^2+a=0$$
case 1
if $a\le b$ and $b$ is minimum
Now Assmue that $b>a$,
then $b_{1}=ka-b\in Z$,and
$$b\cdot b_{1}=a^2+a$$
so $b_{1}$ is postive integer number.
then
$$b_{1}=\dfrac{a^2+a}{b}<\dfrac{a^2+a}{a}=a+1$$
since $b_{1}>b\ge a+1$ impossible.
so $a=b$
then
$$a^2|(2a^2+a)\Longrightarrow a^2|a$$ then we have
$a=1,b=1$
so
$$k=3$$
case 2: if $a\ge b$,then we consider
$$a^2+(1-kb)a+b^2=0$$
Now Assmue that $a>b$,and $a$ is minimum
and
$$a_{1}+a=kb-1\in Z,aa_{1}=b^2$$
then
$$a_{1}=\dfrac{b^2}{a}<b$$
impossible.
so $a=b$,then following is same as case 1
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Use integration by substitution I'm trying to evaluate integrals using substitution. I had
$$\int (x+1)(3x+1)^9 dx$$
My solution: Let $u=3x+1$ then $du/dx=3$
$$u=3x+1 \implies 3x=u-1 \implies x=\frac{1}{3}(u-1) \implies x+1=\frac{1}{3}(u+2) $$
Now I get $$\frac{1}{3} \int (x+1)(3x+1)^9 (3 \,dx) = \frac{1}{3} \int \frac{1}{3}(u+2)u^9 du = \frac{1}{9} \int (u+2)u^9 du \\
= \frac{1}{9} \int (u^{10}+2u^9)\,du = \frac{1}{9}\left( \frac{u^{11}}{11}+\frac{2u^{10}}{10} \right) + C$$
But then I get to this one
$$\int (x^2+2)(x-1)^7 dx$$
and the $x^2$ in brackets is throwing me off.
I put $u=x-1\implies x=u+1,$ hence $x^2+2 =(u+1)^2 +2 = u^2+3$. So
$$ \int(x^2+2)(x-1)^7\, dx = \int (u+1)u^7 du = \int (u+u^7) du = \frac{u^2}{2}+\frac{u^8}{8} + C $$
Is this correct or have I completely missed the point?
| I know the question was about integration by substitution, but if you want another way, you can also do it using tabular integration by parts
$$x^2 + 2 \quad \quad \quad (x-1)^7$$
$$2x \quad \quad \quad \frac 1 8 (x-1)^8$$
$$2 \quad \quad \quad \frac 1 {72} (x-1)^9$$
$$0 \quad \quad \quad \frac 1 {720} (x-1)^{10}$$
So you get
$$(x^2 + 2)\left({\frac 1 8}\right)(x-1)^8 - (2x)\left({\frac 1 {72}}\right)(x-1)^9 + (2)\left({\frac 1 {720}}\right)(x-1)^{10} + C$$
If anyone can make the LaTeX look cool with arrows or an actual table I'd be much obliged :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Number of real roots of $2 \cos\left(\frac{x^2+x}{6}\right)=2^x+2^{-x}$ Find the number of real roots of
$ \cos \,\left(\dfrac{x^2+x}{6}\right)= \dfrac{2^x+2^{-x}}{2}$
1) 0
2) 1
3) 2
4) None of these
My guess is to approach it in graphical way. But equation seems little difficult.
| Rearranging we get $$(2^x)^2-2\cos\dfrac{x^2+x}6(2^x)+1=0\ \ \ \ (0)$$ which is a Quadratic equation in $2^x$
For real $2^x,$ the discriminant must be $\ge0$
i.e., $$\left(2\cos\dfrac{x^2+x}6\right)^2-4=-4\sin^2\dfrac{x^2+x}6\ge0\ \ \ \ (1)$$
As $\sin\dfrac{x^2+x}6$ is real, $$\sin^2\dfrac{x^2+x}6\ge0\iff-4\sin^2\dfrac{x^2+x}6\le0\ \ \ \ (2)$$
By $(1),(2);$ $$\sin^2\dfrac{x^2+x}6=0\implies\cos\dfrac{x^2+x}6=\pm1$$
What happens to $(0)$ if $(i):\cos\dfrac{x^2+x}6=1$
Or if $(ii):\cos\dfrac{x^2+x}6=-1$
Remember for real $x, 2^x>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Let a,b,c be positive real numbers numbers such that $ a^2 + b^2 + c^2 = 3 $ Let $a,b,c\in\mathbb{R^+}$ such that $ a^2 + b^2 + c^2 = 3 $. Prove that
$$
(a+b+c)(a/b + b/c + c/a) \geq 9.
$$
My Attempt
I tried AM-GM on the symmetric expression so the $a+b+c \geq 3$, but I found $a+b+c \leq 3$.
| We'll prove that $\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\geq\frac{3}{2}\sum\limits_{cyc}\left(\frac{a}{b}+\frac{a}{c}\right)$.
Let $\frac{a}{b}=x$, $\frac{b}{c}=y$ and $\frac{c}{a}=z$.
Thus, we need to prove that $2(x+y+z)^2\geq3(x+y+z+xy+xz+yz)$,
where $x>0$, $y>0$ and $z>0$ such that $xyz=1$
or $\sum\limits_{cyc}(2x^2-3x+xy)\geq0$ or $\sum\limits_{cyc}\frac{(2x+1)(x-1)^2}{x}\geq0$, which is obvious.
Hence, it remains to prove that $(a+b+c)^2\sum\limits_{cyc}(a^2b+a^2c)\geq18abc(a^2+b^2+c^2)$.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $9u^2(9uv^2-3w^3)\geq18w^3(9u^2-6v^2)$ or $(7u^2-4v^2)w^3\leq3u^3v^2$.
But $w^3$ gets a maximal value, when two numbers from $\{a,b,c\}$ are equal.
Since the last inequality is homogeneous, it remains to check one case only: $b=c=1$,
which gives $(a-1)^2(a^2-2a+4)\geq0$. Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Independence of path in a closed curve line integral
Let $f(t)$ be a continuous function. Let $C$ be a smooth closed curve. Show that $$\oint\limits_C xf(x^2 + y^2)\,dx + y f(x^2 + y^2)\,dy = 0$$ Hint: Remember that $f(t)$ has a primitive function $F(t)$. Use this fact to construct a potential function for the vector field.
If we prove that the vector field $F(x,y) = (x \cdot f(x^2 + y^2),y \cdot f(x^2 + y^2))$ is conservative in its domain, then we can use the fact $$\int\limits_C {F \cdot dr} = \varphi (r(b)) - \varphi (r(a)) = 0$$ to express the initial line integral as a potential function difference between two points on a closed smooth curve, which would imply that any path taken between these two points would yield the same integral value.
Now in order to show that the given vector field is conservative we need to find its potential, which means we have to calculate $$\frac{d\varphi}{dx} = x \cdot f(x^2 + y^2)\text{ and }\frac{d\varphi}{dy} = y \cdot f(x^2 + y^2)$$ How can we find a potential function $\varphi (x,y)$ if we do not know the form of the function $f(x^2 + y^2)$? If we replace the argument with, say, $t = {x^2} + {y^2}$ we would need to calculate the integrals $$\frac{d\varphi}{dx} = x \cdot f(t)\text{ and }\frac{d\varphi}{dy} = y \cdot f(t)$$ but we cannot treat $f(t)$ as constant since its dependent on both variables $x$ and $y$.
What i had in mind is to show that due to symmetry $F(x,y) = F(y,x)$, which implies that a closed loop is cut evenly by two curves ${C_1}$ and ${C_2}$, their sum defines the closed loop region. $F$ is conservative vector of the field if $$\frac{d}{dy} (x \cdot f(x^2 + y^2)) = 2xyf' = \frac{d}{dx} ( y \cdot f(x^2 + y^2))$$
Now we know that the field is conservative, which implies $F = \nabla \varphi $ for some scalar potential function $\varphi $ defined over the closed loop. Therefore, $$F \cdot dr = \left( \left( \frac{d\varphi}{dx} \right)i + \left( \frac{d\varphi}{dy} \right)j \right) \cdot \left( {dxi + dyj} \right) = \frac{{d\varphi }}{{dx}}dx + \frac{{d\varphi }}{{dy}}dy = d\varphi $$ Since $C$ is a continuous smooth, closed curve, parametrized, say, by $r = r(t),\,\,\,a \le t \le b$ then $r(a) = r(b)$ and $$\int\limits_C F \cdot dr = \int\limits_a^b \frac{d\varphi (r(t))}{dt} \, dt = \varphi (r(b)) - \varphi (r(a)) = 0$$
However, i don't think that it is sufficient to claim that the given vector field is conservative based on the equality of partial derivative of its components. We still need to find a potential function i think. For example, if instead of $F(x,y) = (x \cdot f(x^2 + y^2),y \cdot f(x^2 + y^2))$ we were given a vector field, say, $F(x,y) = (x \cdot ({x^2} + {y^2}),y \cdot (x^2 + y^2))$ we could easily find its potential to be $\varphi (x,y) = \frac{x^4}{4} + \frac{x^2 y^2}{2} + \frac{y^4}{4} = \frac{1}{4} (x^2 + y^2 )^2$. Now we can take any smooth closed curve, parametrize it and show that the potential at any two points on the curve is the same, hence its difference is zero. For example, in our case we can take a unit circle and parametrize it as $x = \cos t \text{ and }y = \sin t$. Now we just chose two random points on the curve, say, $P_1(1,0) \text{ and } P_2 (0,1)$ and show that $$\oint\limits_C x(x^2 + y^2)\,dx + y(x^2 + y^2)\,dy = \left[ \frac{1}{4} ( x^2 + y^2)^2 \right]_{(1,0)}^{(0,1)} = \frac{1}{4} - \frac{1}{4} = 0$$
How to deal when we are given $f(x^2 + y^2)$ instead of a function itself?
| Let $g(u) := \frac{1}{2}\int_0^u f(t)\, dt$. Then $$\frac{\partial}{\partial x} g(x^2 + y^2) = \frac{1}{2}f(x^2 + y^2) \frac{\partial}{\partial x}(x^2 + y^2) = xf(x^2 + y^2)$$
and similarly $$\frac{\partial}{\partial y}g(x^2 + y^2) = yf(x^2 + y^2).$$ Therefore, the vector field $xf(x^2 + y^2)\vec{i} + yf(x^2 + y^2)\vec{j}$ is the gradient of $g(x^2 + y^2)$. Hence, $$\oint_C xf(x^2 + y^2)\, dx + yf(x^2 + y^2)\, dy = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solution of 2nd order linear ODE with regular singular points, and complex exponents at singularity The steady state temperature distribution of a rod given by:
\begin{equation}
\frac{\textrm{d}p(x)y'}{\textrm{d}x} - y = 0,\; 0 \leq x \leq 1,\; \text{and} \;y(0) = 0,
\end{equation}
where $y(x)$ is the steady state temperature distribution, and $p(x) = x^s$ is the spatially dependent conductivity of the rod, for some $s \in [0, 1)$. For $0 \leq s < 1$ what is the form of the solution close to $x = 0$ (i.e. the first the term of the series solution)? Does a solution exist for $s = 1$? $s > 1$?
First, simplify the equation into a recognizable form:
\begin{align*}
&\quad \frac{\textrm{d}p(x)y'}{\textrm{d}x} - y = 0 \\
&\equiv p(x)y'' + p'(x)y' - y = 0 \\
&\equiv y'' + \frac{p'(x)}{p(x)}y' - \frac{1}{p(x)} y = 0
\end{align*}
Since $p(x) = x^s$, thus $p'(x) = sx^{s-1}$ and $p''(x) = s(s-1)x^{s-2}$. Substituting:
\begin{align*}
&\quad y'' + \frac{p'(x)}{p(x)}y' - \frac{1}{p(x)} y = 0 \\
&\equiv y'' + \frac{s}{x}y - \frac{s^2 - s}{x^2}y = 0
\end{align*}
We can immediately recognize that $x = 0$ is a regular singular point since:
\begin{align*}
&\lim_{x\rightarrow 0} \frac{sx}{x} = s
&\lim_{x\rightarrow 0} -\frac{(s^2 - s)x^2}{x^2} = s - s^2
\end{align*}
It is worth noting that these limits exist for all $s$.
So, we look for solutions of the form:
\begin{align*}
y = \sum_{n=0}^{\infty} a_nx^{n+r}
\end{align*}
where $r(r-1) + sr + (s - s^2) = 0 = r^2 + (s-1)r + (s - s^2)$ is the corresponding indical equation. Solving for $r$, we obtain:
\begin{align}
\nonumber &\quad r = \frac{-(s-1) \pm \sqrt{(s-1)^2 - 4(s - s^2)}}{2} \\
\nonumber &\equiv r = \frac{-(s-1) \pm \sqrt{s^2 - 2s + 1 - 4s + 4s^2}}{2} \\
&\equiv r = \frac{-(s-1) \pm \sqrt{5s^2 - 6s + 1}}{2}
\end{align}
For what values of $s$ do we get imaginary $r$? That would be when the discriminant of the quadratic formula is less than zero:
\begin{align*}
5s^2 - 6s + 1 < 0
\end{align*}
Consider when $s = 0$:
\begin{align*}
&\quad 5s^2 - 6s + 1 = 0 \\
&\equiv s = \frac{6 \pm \sqrt{36 - 20}}{10} \\
&\equiv s = \frac{6 \pm \sqrt{16}}{10} \\
&\equiv s = \frac{6 \pm 4}{10} \\
&\equiv s = 1 \vee s = \frac{1}{5}\\
\end{align*}
Since for $(s=0 \wedge d = 5s^2 - 6s + 1) \implies d = 1$, then we know that the discriminant is positive for all values of $s \leq \frac{1}{5}$ and $s \geq 1$, and we have:
\begin{align*}
r_1 &= \frac{-(s-1) + \sqrt{(s - 1)(s - \frac{1}{5})}}{2} \\
r_2 &= \frac{-(s-1) - \sqrt{(s + 1)(s - \frac{1}{5})}}{2}
\end{align*}
The general solution is of the form:
\begin{equation*}
y = c_1\sum_{n = 0}^{\infty} a_nx^{n + r_1} + c_2\sum_{n = 0}^{\infty} a_nx^{n + r_2}
\end{equation*}
where $a_n$ might be complex if $\frac{1}{5} s < 1$
We know that $y(0) = 0$, therefore:
\begin{align*}
&\quad y(0) = 0 = c_1\sum_{n = 0}^{\infty} a_n0^{n + r_1} + c_2\sum_{n = 0}^{\infty} a_n0^{n + r_2} \\
&\equiv 0 = c_1\sum_{n = 0}^{\infty} a_n + c_2\sum_{n = 0}^{\infty} a_n \\
&\equiv 0 = (c_1 + c_2) \\
&\equiv c1 = -c_2
\end{align*}
The first term of this series is:
\begin{align*}
\nonumber &\quad c_1\left(a_0x^{r_1} - a_0x^{r_2}\right) \\
\nonumber &\equiv c_1a_0\left(x^{\frac{-(s-1) + \sqrt{(s - 1)(s - \frac{1}{5})}}{2}} - x^{\frac{-(s-1) - \sqrt{(s - 1)(s - \frac{1}{5})}}{2}}\right) \\
\end{align*}
Let $\frac{1 - s}{2} = \alpha$ and $\frac{\sqrt{(s - 1)(s - \frac{1}{5})}}{2} = \beta$
If $0 \leq s \leq \frac{1}{5}$:
\begin{equation*}
y \approx c_1a_0\left(x^{\alpha + \beta} - x^{\alpha - \beta}\right)
\end{equation*}
If $\frac{1}{5} < s < 1$:
\begin{equation*}
y \approx c_1a_0\left(x^{\alpha + i\beta} - x^{\alpha - i\beta}\right)
\end{equation*}
Recall that:
\begin{align*}
&\quad x^r = e^{rln(x)} \\
&\implies x^{\lambda + i\mu} = e^{\lambda\ln(x)}e^{i\mu\ln(x)} \\
&\equiv x^{\lambda + i\mu} = e^{\lambda\ln(x)}(\cos(\mu\ln(x)) + i\sin(\mu\ln(x)))
\end{align*}
So we have oscillatory solutions.
Thus:
\begin{equation*}
y \approx c_1a_0x^{\alpha}\left(2i\sin(\beta\ln(x))\right)
\end{equation*}
Since the limits determining if $x = 0$ is a regular point exist for all $s$, we can write a (non-trivial) Frobenius series solution for all values of $s$.
Questions:
1) I have not used information about $0 \leq s < 1$, which is bothersome. Did I miss an opportunity to use it to simplify the form of the first term?
2) Is my reasoning for why solutions exist for $s = 1$ and $s > 1$ sound?
| $$y''+\frac{s}{x}y'-x^{-s}y=0$$
This is an ODE of the Bessel kind, but not on standard form. In order to make it standard, let $y(x)=x^aF(b\:x^c)=x^aF(X)$. The goal is to transform it to :
$$F''(X)+\frac{1}{X}F'(X)-\left(1+\frac{\nu^2}{X^2}\right)F(X)=0$$
The calculus is arduous. The result is :
$$a=\frac{1-s}{2} \: ; \: b=\frac{2}{2-s} \: ; \: c=\frac{2-s}{2} \: ; \: \nu=\frac{s-1}{s-2}$$
The general solution of the Bessel function is :
$$F(X)=c_1 I_\nu(X)+c_2 K_\nu(X)$$
$I_\nu(X)$ and $K_\nu(X)$ are the mofified Bessel functions of first and second kind. So, the general solution is :
$$y(x)=C_1 x^{\frac{1-s}{2}}I_{\frac{s-1}{s-2}}\left(\frac{2}{2-s}x^{\frac{2-s}{2}}\right) + C_2 x^{\frac{1-s}{2}}K_{\frac{s-1}{s-2}}\left(\frac{2}{2-s}x^{\frac{2-s}{2}}\right)$$
The formula is valid any $s$, but some particular cases arise sometimes with simplification of the Bessel functions to functions of lower level :
Case $s=1$ then $y=C_1 I_0\left(2 x^{1/2}\right)+C_2 K_0\left(2 x^{1/2}\right)$
Case $s=0$ then $y=C_1 \sinh(x) +C_2 \cosh(x)$
Case $0<s<1$ then :
In case of $0\leq x\leq 1$ , the condition $y(0)=0$ implies $C_2=0$
$$y(x)=C x^{\frac{1-s}{2}}I_{\frac{s-1}{s-2}}\left(\frac{2}{2-s}x^{\frac{2-s}{2}}\right)$$
The first term of the series expansion around $x=0$ of the Bessel function is :
$$I_\nu(X)\simeq\frac{1}{2^\nu \Gamma(\nu+1)}X^\nu$$
$$y(x)\simeq C \frac{1}{(2-s)^{\frac{s-1}{s-2}} \Gamma(1+\frac{1-s}{2-s})}x^{1-s} $$
This is consistent with a direct approach :
Suppose that the first term be on the form $y\simeq cx^p$ then $y'\simeq cpx^{p-1}$ and $y''\simeq cp(p-1)x^{p-2}$
Puting them into the ODE : $cp(p-1)x^{p-2}+scpx^{p-2}-cx^{-s+p}\simeq 0$
$p(p-1+s)-x^{2-s}\simeq 0$
This implies $p-1+s=0$ hense $p=1-s$ and $y\simeq cx^{1-s}$ already obtained above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 1,
"answer_id": 0
} |
Chinese Remainder Theorem for non prime-numbers. Let's say I want to find x such that x leaves remainder 2 when divided by 3 and x leaves remainder 3 when divided by 5.
x % 3 = 2
x % 5 = 3
We break down the problem to:
x % 3 = 1
x % 5 = 0
Therefore,
5k % 3 = 1
2k % 3 = 1
k = 2
10, when remainder = 1
20, when remainder = 2
Now x % 3 = 0
and x % 5 = 1
3k % 5 = 1
k = 2
6, when remainder = 1
18, when remainder = 3
Therefore, final solution is 20 + 18 = 38.
38 is a solution
LCM of 3,5 = 15.
38 - 15 - 15 = 8.
8 is the least number, that is the solution.
But now if I have
x % 7 = 3
x % 4 = 2
How do I solve the question ?
| It has nothing to see with primeness of the moduli, but to their coprimeness.
The systematic method consists in finding Bézout's coefficients for $7$ and $3$: $\enspace-1\cdot 7+2\cdot 4=1$
(For bigger numbers you can use the Extended Euclidean Algorithm).
The solution to the system of congruences \begin{equation*}\begin{cases}x\equiv 3 \pmod 7\\x\equiv 2 \pmod 4
\end{cases}
\end{equation*}
are then simply:$$x\equiv -2\cdot 7+3\cdot 2\cdot 4=10 \pmod{\operatorname{lcm(7,4)}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How to prove this: $a^4+b^4+2 \ge 4ab$? How to prove this: $a^4+b^4+2 \ge 4ab$?
$a$ and $b$ are reals.
| Notice that:
$$(a^2+b^2)^2 = a^4 + b^4 + 2a^2b^2$$
So, then we have that: $(a^2 + b^2)^2 - 2a^2b^2 + 2 \ge 4ab$
It follows that:
$$(a^2 + b^2)^2 \ge 2a^2b^2 + 4ab +2 -4,\ (a^2 + b^2)^2 \ge 2(ab+1)^2 -2^2 = \left(\sqrt{2}(ab+1) + 2\right)\left(\sqrt{2}(ab+1) - 2\right)$$
Our inequality now becomes:
$$(a^2 + b^2)^2 \ge\left(\sqrt{2}ab+\sqrt{2} + 2\right)\left(\sqrt{2}ab+\sqrt{2} - 2\right)$$
NOTE This manipulation may or may not assist you in proving the inequality. I only provided it as a possible way to help you in your proof.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that the arithmetic mean is less or equal than the quadratic mean I tried to solve this for hours but no success.
Prove that the arithmetic mean is less or equal than the quadratic mean.
I am in front of this form:
$$
\left(\frac{a_1 + ... + a_n} { n}\right)^2 \le \frac{a_1^2 + ... + a_n^2}{n}
$$
With rewriting the inequality in other forms I had no luck.
I think maybe induction would be OK, but I have no idea, how to do it in this case. Do you know a good proof for this?
Thanks!
| for sake of contradiction suppose:
$\sqrt{\frac{a_1^2+ \cdots +a_n^2}{n}}<\frac{a_1+ \cdots +a_n}{n}\\
\frac{a_1^2+ \cdots +a_n^2}{n}<\frac{(a_1+ \cdots +a_n)^2}{n^2}\\
\frac{a_1^2+ \cdots +a_n^2}{n}<\frac{a_1^2+ \cdots +a_n^2}{n^2}+2\frac{\sum_{sym}a_1a_2}{n^2}\\
\\
(n-1)(a_1^2+\cdots+a_n^2)<2\sum_{sym}a_1a_2 \quad \quad (1)$
At this point I show by induction that the last line is wrong:
conjecture:
$(n-1)(a_1^2+\cdots+a_n^2)\geq2\sum_{sym}a_1a_2 \quad \forall \:n\geq2, n\in \mathrm{Z^+}\\
\text{let} \quad n=2\\
(a_1-a_2)^2\geq0\\
a_1^2+a_2^2-2a_1a_2>0\\
\Rightarrow a_1^2+a_2^2\geq 2a_1a_2$
Now assume for $n=k$
$(k-1)(a_1^2+\cdots +a_n^2)\geq 2\sum_{sym}a_1a_2 \quad \quad (2)\\
\text{I have from the case $n=2$}\\
a_{k+1}^2+a_1^2\geq 2a_{k+1}a_1\\
\Rightarrow 2a_{k+1}^2+a_1^2+a_2^2\geq2a_{k+1}a_1+a_{k+1}^2+a_2^2\geq2a_{k+1}a_1+2a_{k+1}a_2\\\text{proceding as above}:\\
ka_{k+1}^2+a_1^2+\cdots+a_k^2 \geq 2 a_{k+1}(a_1+\cdots+a_k) \quad \quad (3)\\
\text{now adding (2) and (3):}\\
(k-1)(a_1^2+\cdots +a_k^2)+ka_{k+1}^2+a_1^2+\cdots +a_k^2\geq2\sum_{sym}a_1a_2+2a_{k+1}(a_1+\cdots +a_k)\\
\Rightarrow k(a_1^2+\cdots +a_{k+1}^2)\geq 2\sum_{sym}a_1a_2\\
\therefore (n-1)(a_1^2+\cdots+a_n^2)\geq2\sum_{sym}a_1a_2 \quad \forall \:n\geq2, n\in \mathrm{Z^+} \quad\text{by induction}$
This clearly contradicts (1), so I have:
$\sqrt{\frac{a_1^2+ \cdots +a_n^2}{n}}\geq\frac{a_1+ \cdots +a_n}{n}$
by contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Does $\sum_{n\ge0} \cos (\pi \sqrt{n^2+n+1}) $ converge/diverge? How would you prove convergence/divergence of the following series?
$$\sum_{n\ge0} \cos (\pi \sqrt{n^2+n+1}) $$
I'm interested in more ways of proving convergence/divergence for this series.
My thoughts
Let
$$u_{n}= \cos (\pi \sqrt{n^2+n+1})$$
Let's first check A necessary condition for the convergence of that seires is that the limit $$\lim\limits_{n \to \infty} u_n$$ should exist and be equal $0$
but i have answer in my old book but i can't remember how they got that
$$u_n=\cos \left(n\pi+\frac{\pi}{2}+\frac{3\pi}{8n}+o(\frac{1}{n^2})\right)=\frac{(-1)^{n+1}.3\pi}{8n}+o(\frac{1}{n^2})$$
since $|\frac{(-1)^{n+1}.3\pi}{8n}|$ decreasing. and converge $0$
then by compraison $u_n$ is convergent
any help would be appreciated
| From the Taylor expansion
$$
(1+x)^{1/2}=1+\frac{x}{2}-\frac{x^2}{8}+O(x^3)
$$
we get
$$\begin{align}
\sqrt{n^2+n+1}&=n\Bigl(1+\frac{1}{n}+\frac{1}{n^2}\Bigr)^{1/2}\\
&=n\Bigl(1+\frac{1}{2}\Bigl(\frac{1}{n}+\frac{1}{n^2}\Bigr)-\frac{1}{8}\Bigl(\frac{1}{n}+\frac{1}{n^2}\Bigr)^2+O\Bigl(\frac{1}{n}+\frac{1}{n^2}\Bigr)^3\Bigr)\\
&=n+\frac12+\frac{3}{8\,n}+O\Bigl(\frac{1}{n^2}\Bigr).
\end{align}$$
Thus
$$
\cos(\pi\sqrt{n^2+n+1})=\cos\Bigl(\Bigl(n+\frac12\Bigr)\pi+\Bigl(\frac{3\,\pi}{8\,n}+O\Bigl(\frac{1}{n^2}\Bigr)\Bigr)
$$
Now use the formula for the cosine of a sum and take into account that
$$
\cos\Bigl(\Bigl(n+\frac12\Bigr)\pi\Bigr)=0,\quad\sin\Bigl(\Bigl(n+\frac12\Bigr)\pi\Bigr)=(-1)^n.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
need help on this fraction equation $2/5 = 2/3 - r/5$ $$\frac{2}{5} = \frac{2}{3} - \frac{r}{5}$$
I'm trying to find $r$. Can anyone give me a step by step?
| First note that
$$ \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm b}{c}$$
And
$$ \frac{a}{b}\pm c=\frac{a\pm bc}{c} $$
So now we have
$$\frac{2}{5} = \frac{2}{3} - \frac{r}{5} $$
$$\frac{2}{5} + \frac{r}{5} = \frac{2}{3} $$
$$\frac{2+r}{5} = \frac{2}{3} $$
$$ 2+r = \frac{2\times 5}{3} $$
$$ r = \frac{10}{3}-2 =\frac{10-(2\times 3)}{3}$$
$$r=\frac{10-6}{3}=\frac43$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Representation of Quaternion group in $GL(2,3)$ I am working with the representation of the quaternion group in $GL(2,3)$ generated by $A=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}, B=\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}, C=\begin{pmatrix} -1 & 1\\ 1 & 1 \end{pmatrix}$.
Now $C=PBP^{-1}$ where $P= \begin{pmatrix} 0 & 1\\ 1& 0 \end{pmatrix}$. I am trying to show that $A$ and $B$ are similar by an element in $GL(2,3)$.
They are obviously similar in $GL(2,9)$ because they are both diagonalizable with the same eigenvalues: Their characteristic polynomials are both $x^2+1$, and since its irreducible over $F_3$ and $F_3$ is perfect, the roots are distinct.
I feel that because of the relations $A=CB, B=AC, C=BA$, $A$, $B$, $C$ should all "be on the same footing": $B$ and $C$ similar by P should mean that $A,B$ are similar too, without me having to worry about whether the similarity matrix has entries in $F_3$. After all I should be able to use jacobi's identity and the matrix P to achieve this. It has not been working out for some time!
edit:
another thing I am trying -
Probably the best way to do this is to look at all the embeddings of $Q_8 \hookrightarrow GL(2,3)$.
| Let's suppose they are conjugate by
$$M = \left(\begin{array}{rr}
x & y\\
z & w \end{array} \right).$$
Then, we have
$$MB = AM \Rightarrow \left(\begin{array}{rr}
x + y & x - y\\
z+w & z-w \end{array} \right) = \left(\begin{array}{rr}
-z & -w\\
x & y \end{array} \right).$$
So, $z = -x- y, w = y - x$ and putting these into the last two equations gives
$$x = x, y = y$$
and hence we can choose arbitrary values of $x$ and $y$. So define $M$ as
$$M = \left(\begin{array}{rr}
1 & 0\\
-1 & -1 \end{array} \right).$$
Just for confirmation,
$$MB = \left(\begin{array}{rr}
1 & 1\\
1 & 0 \end{array} \right)$$
and
$$AM =\left(\begin{array}{rr}
1 & 1\\
1 & 0 \end{array} \right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find a solution: $3(x^2+y^2+z^2)=10(xy+yz+zx)$ I'm something like 90% sure that this diophantine equation has nontrivial solutions:
$3(x^2+y^2+z^2)=10(xy+yz+zx)$
However, I have not been able to find a solution using my calculator. I would greatly appreciate if someone could try to find one using a program. Or maybe you can just guess one that happens to work?
Thanks!
EDIT: By nontrivial I mean no $0$'s. (Credits to Slade for reminding me to define this)
EDIT2: In fact, you are free to find a nontrivial solution to $(3n-3)(x^2+y^2+z^2)=(9n+1)(xy+yz+zx)$ where $n\equiv 1\pmod 5$ is a positive integer. The one I posted above is the case $n=5(2)+1$, but you will make my day if you can find a nontrivial solution for any $n=5k+1$.
| Because the equation is homogenous, the integer solutions can be derived from the rational solutions, in other words swapping between projective and affine form.
I prove below that the set of non-zero rational solutions are common rational multiples of the following (which, conversely, satisfies the equation identically) for any rational parameter t:
$x,\ y,\ z = 2 t - 1,\ 3 t^2 - 8 t + 5,\ t^2 - 6 t + 8$
So, explicitly, the complete set of integer solutions with GCD(x, y, z) = 1 can be expressed as follows, as $m,\ n$ range over coprime integer pairs
$x,\ y,\ z = (2 m - n) n,\ 3 m^2 - 8 m n + 5 n^2,\ m^2 - 6 m n + 8 n^2$
Proof:
Let $p,\ q,\ r = - x + y + z,\ x - y + z,\ x + y - z$
<=> $2 x,\ 2 y,\ 2 z = q + r,\ r + p,\ p + q$
Then $p^2 + q^2 + r^2 = 3 (x^2 + y^2 + z^2) - 2 (x y + y z + z x)$
and $p q + q r + r p = - (x^2 + y^2 + z^2) + 2 (x y + y z + z x)$
So if $a (p^2 + q^2 + r^2) = b (p q + q r + r p)$
then $(3 a + b) (x^2 + y^2 + z^2) = 2 (a + b) (x y + y z + z x)$
So the required equation is obtained with $3 a + b,\ a + b = 3,\ 5$, i.e. $a,\ b = -1,\ 6$ and the original is equivalent to
$p^2 + q^2 + r^2 + 6 (p q + q r + r p) = 0$
If r = 0 then this becomes $(p + 3 q)^2 = 8 q^2$, which for rational $p, q$ has only the solution p = q = 0
Otherwise, we can replace $\frac{p}{r},\ \frac{q}{r}$ by $p,\ q$ respectively and the equation becomes
$q^2 + 6 (p + 1) q + (p^2 + 6 p + 1) = 0$
which for rational $q$ (assuming rational $p$) requires rational $s$ with
$q = 2 s - 3 (p + 1)$
$9 (p + 1)^2 - (p^2 + 6 p + 1) = 4 s^2$
The latter is equivalent to $8 s^2 - (4 p + 3)^2 = 7$
So in view of the obvious rational solution $4 p + 3,\ s = 1, 1$ we can replace in this $4 p + 3,\ s = 2 t u + 1,\ u + 1$
which implies either $u = 0$, which recovers the solution already observed, or $u = \frac{4 - t}{t^2 - 2}$
which gives successively
$p = \frac{- t^2 + 2 t + 1}{t^2 - 2}$
$s = \frac{t^2 - t + 2}{t^2 - 2}$
$q = \frac{2 v^2 - 8 v + 7}{t^2 - 2}$
So the original $p,\ q,\ r$ can be taken as follows, and the result follows
$p,\ q,\ r = - t^2 + 2 t + 1,\ 2 t^2 - 8 t + 7,\ t^2 - 2$
Sanity check: In the expressions for $x, y, z$ take $t = 0$ to give $x, y, z = 1, 5, 8$ and the equation becomes 2.3^3.5 = 2.3^3.5
Regards
John R Ramsden
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
If $x = 3^{1/3} + 3^{2/3} + 3$, find the value of $x^3 - 9x^2 + 18x - 12$
If $x = 3^{1/3} + 3^{2/3} + 3$, find the value of $$x^3 - 9x^2 + 18x - 12.$$
This is not a homework problem. I'm not even a student. I'm going through an old textbook. I know this is a simple problem. Can't seem to crack it though.
| Let $\,a = \sqrt[3]3.\, $ $\, (\!\overbrace{x\!-\!3}^{\Large a^2+a}\!)^2 = \overbrace{a^4}^{\Large 3a}\!+\overbrace{2a^3}^{\Large 6}+a^2\,$
Therefore $\ \color{#0a0}{x^2\!-6x\!+\!3\, =\, a^2\!+\!3a}.\ $ By very simple arithmetic we have:
$\begin{align} f(x)\, & =\ (x-3)\,(\color{#0a0}{x^2\!-6x+3})\, -\, 3(x+1)\ \ {\rm\ via\ \ Division\ Algorithm}\\
& = (a^2\!+\!a)\,\color{#0a0}{(a^2\!+\!3a)}\, -\, 3(a^2+a+4)\\
&= \underbrace{a^4}_{\Large 3a}\!\!+\underbrace{4a^3}_{\Large\! 4\,\cdot\, 3}\!+3a^2\, -\, 3(a^2+a+4)\, =\, 0
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1135322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to integrate $\frac{1}{x\sqrt{1+x^2}}$ using substitution? How you integrate
$$\frac{1}{x\sqrt{1+x^2}}$$
using following substitution? $u=\sqrt{1+x^2} \implies du=\dfrac{x}{\sqrt{1+x^2}}\, dx$
And now I don't know how to proceed using substitution rule.
| If $u=\sqrt{1+x^2}$ then $u^2 = 1+x^2$, so $x^2= u^2-1$. Then you have
\begin{align}
& \int \frac 1 {x\sqrt{1+x^2}} \,dx = \int \frac {x} {x^2\sqrt{1+x^2}} \,dx \\[8pt]
= {} & \int\frac{du}{u^2-1} = \int\frac{du}{(u-1)(u+1)}.
\end{align}
Then use partial fractions.
| {
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"url": "https://math.stackexchange.com/questions/1137842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 0
} |
Sum $\sum_{n=2}^{\infty} \frac{n^4+3n^2+10n+10}{2^n(n^4+4)}$ I want to evaluate the sum $$\large\sum_{n=2}^{\infty} \frac{n^4+3n^2+10n+10}{2^n(n^4+4)}.$$ I did partial fraction decomposition to get $$\frac{1}{2^n}\left(\frac{-1}{n^2+2n+2}+\frac{4}{n^2-2n+2}+1\right)$$ I am absolutely stuck after this.
| You are almost there.
So once you have obtained $$\frac{1}{2^n}\left(\frac{-1}{n^2+2n+2}+\frac{4}{n^2-2n+2}+1\right)$$ then observe the following:
$n^2+2n+2=(n+1)^2+1$ and $n^2-2n+2=(n-1)^2+1$. Which means the given sum becomes
$$\sum\limits_{n=2}^\infty \left[\frac{-1}{2^n\{(n+1)^2+1\}}+\frac{1}{2^{n-2}\{(n-1)^2+1\}} \right] +\sum\limits_{n=2}^\infty \frac{1}{2^n}=:\sum\limits_{n=2}^\infty [u_n-u_{n+2}]+\frac 12 $$ where $u_n:=\frac{1}{2^{n-2}\{(n-1)^2+1\}}$.
If you calculate $\sum\limits_{n=2}^m[u_n-n_{n+2}]$, you will find $u_2+u_3-(u_{m+1}+u_{m+2})$. hence taking $m\rightarrow \infty$ we shall get
$$\sum\limits_{n=2}^\infty [u_n-u_{n+2}]=u_2+u_3=\frac{1}{2}+\frac{1}{10}.$$
Hence the final sum is $2\times \frac 12+\frac{1}{10}=\cdots.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find all real solutions for $x$ in $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 .$ Find all real solutions for $x$ in $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 .$
I have found out that the answers were 0,1, and -1. But I used sort of a guess-and check way.
$2(2^x-1)x^2+(2^{x^2}-2)x=2^{x+1}-2$
I expanded it into:
$(x^2-1)(2^{x+1}-2)+x(2^{x^2}-2)=0$
I just found all the possibilities to make each group of equation 0 resulting in 0,1, and -1.
How can I prove this is correct instead of trial/error?
| Mistake when you divide by 2. You get:
$$2^{x^2-1} = 2^{(x+1)(x-1)} = 2^{{x+1}^{x-1}}$$
Now maybe it gets easier?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1143714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $(z^3-z)(z+2)$ is divisible by $12$ for all integers $z$ I am a student and this question is part of my homework.
May you tell me if my proof is correct?
Thanks for your help!
Prove that $(z^3-z)(z+2)$ is divisible by $12$ for all integers $z$.
$(z^3-z)(z+2)=z(z^2-1)(z+2)=z(z-1)(z+1)(z+2)=(z-1)(z)(z+1)(z+2)$
$(z-1)(z)(z+1)(z+2)$ means the product of $4$ consecutive numbers.
Any set of $4$ consecutive numbers has $2$ even numbers, then $(z-1)(z)(z+1)(z+2)$ is divisible by $4$.
Any set of $4$ consecutive number has at least one number that is multiple of $3$, then $(z-1)(z)(z+1)(z+2)$ is divisible by 3.
Therefore $(z-1)(z)(z+1)(z+2)$ is divisible by $12$. Q.E.D.
| That's correct. Alternatively it is divisible by $\,24\,$ by integrality of binomial coefficients
$$\,(z+2)(z+1)z(z-1)\, =\, 4!\ \dfrac{(z+2)(z+1)z(z-1)}{4!}\, =\, 24{ {z+2\choose 4}}\qquad\qquad$$
Similarly $\,n!\,$ divides the product of $\,n\,$ consecutive naturals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int\frac{dx}{x^5+1}$ Calculate $\displaystyle\int\dfrac{dx}{x^5+1}$
| Hint:
$$
\begin{split}
x^5+1&=(1 + x)(x^4 - x^3 + x^2 - x + 1) \\
&=(1 + x)(x^2 + ((-1 + \sqrt 5) / 2)x + 1)(x^2 + ((-1 - \sqrt 5) / 2)x + 1)
\end{split}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
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