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Sum of the form $r+r^2+r^4+\dots+r^{2^k} = \sum_{i=1}^k r^{2^i}$ I am wondering if there exists any formula for the following power series : $$S = r + r^2 + r^4 + r^8 + r^{16} + r^{32} + ...... + r^{2^k}$$ Is there any way to calculate the sum of above series (if $k$ is given) ?

I haven’t been able to obtain a closed form expression for the sum, but maybe you or someone else can do something with what follows. In Blackburn's paper (reference below) there are some manipulations involving the geometric series $$1 \; + \; r^{2^n} \; + \; r^{2 \cdot 2^n} \; + \; r^{3 \cdot 2^n} \; + \; \ldots \; + \; r^{(m-1) \cdot 2^n}$$ that might give some useful ideas to someone. However, thus far I haven't found the identities or the manipulations in Blackburn's paper to be of any help. Charles Blackburn, Analytical theorems relating to geometrical series, Philosophical Magazine (3) 6 #33 (March 1835), 196-201. As for your series, I tried exploiting the factorization of $r^m – 1$ as the product of $r-1$ and $1 + r + r^2 + \ldots + r^{m-1}:$ First, replace each of the terms $r^m$ with $\left(r^{m} - 1 \right) + 1.$ $$S \;\; = \;\; \left(r – 1 \right) + 1 + \left(r^2 – 1 \right) + 1 + \left(r^4 – 1 \right) + 1 + \left(r^8 – 1 \right) + 1 + \ldots + \left(r^{2^k} – 1 \right) + 1$$ Next, replace the $(k+1)$-many additions of $1$ with a single addition of $k+1.$ $$S \;\; = \;\; (k+1) + \left(r – 1 \right) + \left(r^2 – 1 \right) + \left(r^4 – 1 \right) + \left(r^8 – 1 \right) + \ldots + \left(r^{2^k} – 1 \right)$$ Now use the fact that for each $m$ we have $r^m - 1 \; = \; \left(r-1\right) \left(1 + r + r^2 + \ldots + r^{m-1}\right).$ $$S \;\; = \;\; (k+1) + \left(r – 1 \right)\left[1 + \left(1 + r \right) + \left(1 + r + r^2 + r^3 \right) + \ldots + \left(1 + r + \ldots + r^{2^{k} - 1} \right) \right]$$ At this point, let's focus on the expression in square brackets. This expression is equal to $$\left(k+1\right) \cdot 1 + kr + \left(k-1\right)r^2 + \left(k-1\right)r^3 + \left(k-2\right)r^4 + \ldots + \left(k-2\right)r^7 + \left(k-3\right)r^8 + \ldots + \left(k-3\right)r^{15} + \ldots + \left(k-n\right)r^{2^n} + \dots + \left(k-n\right)r^{2^{n+1}-1} + \ldots + \left(1\right) r^{2^{k-1}} + \ldots + \left(1\right) r^{2^{k} - 1}$$ I'm now at a loss. We can slightly compress this by factoring out common factors for groups of terms such as $(k-2)r^4 + \ldots + (k-2)r^7$ to get $(k-2)r^4\left(1 + r + r^2 + r^3\right).$ Doing this gives the following for the expression in square brackets. $$\left(k+1\right) + \left(k\right)r + \left(k-1\right)r^2 \left(1+r\right) + \left(k-2\right)r^4\left(1+r+r^2+r^3\right) + \ldots + \;\; \left(1\right)r^{2^{k-1}} \left(1 + r + \ldots + r^{2^{k-1} -1} \right)$$

How to show that $\frac{x^2}{x-1}$ simplifies to $x + \frac{1}{x-1} +1$ How does $\frac{x^2}{(x-1)}$ simplify to $x + \frac{1}{x-1} +1$? The second expression would be much easier to work with, but I cant figure out how to get there. Thanks

Very clever trick: If you have to show that two expressions are equivalent, you work backwards. $$\begin{align}=& x +\frac{1}{x-1} + 1 \\ \\ \\ =& \frac{x^2 - x}{x-1} +\frac{1}{x - 1} + \frac{x-1}{x-1} \\ \\ \\ =& \frac{x^2 - x + 1 + x - 1}{x - 1} \\ \\ \\ =&\frac{x^2 }{x - 1}\end{align}$$Now, write the steps backwards (if you're going to your grandmommy's place, you turn backwards and then you again turn backwards, you're on the right way!) and act like a know-it-all. $$\begin{align}=&\frac{x^2 }{x - 1} \\ \\ \\ =& \frac{x^2 - x}{x-1} +\frac{1}{x - 1} + \frac{x-1}{x-1} \\ \\ \\ =& x +\frac{1}{x-1} + 1 \end{align}$$ Q.E.Doodly dee! This trick works and you can impress your friends with such elegant proofs produced by this trick.

Compute $\lim_{n\to\infty} \left(\sum_{k=1}^n \frac{H_k}{k}-\frac{1}{2}(\ln n+\gamma)^2\right) $ Compute $$\lim_{n\to\infty} \left(\sum_{k=1}^n \frac{H_k}{k}-\frac{1}{2}(\ln n+\gamma)^2\right) $$ where $\gamma$ - Euler's constant.

We have \begin{align} 2\sum_{k=1}^n \frac{H_k}{k} &= 2\sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} \\ &= \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} + \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} \\ &= \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} + \sum_{j=1}^n \sum_{k=j}^n \frac{1}{jk}, \text{ swapping the order of summation on the second sum}\\ &= \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} + \sum_{k=1}^n \sum_{j=k}^n \frac{1}{jk}, \text{ changing variables on the second sum}\\ &= \sum_{k=1}^n \sum_{j=1}^n \frac{1}{jk} + \sum_{k=1}^n \frac{1}{k^2} \\ &= \left(\sum_{k=1}^n \frac{1}{k} \right)^2 + \sum_{k=1}^n \frac{1}{k^2} \\ &= H_n^2+ H^{(2)}_n. \\ \end{align} Thus \begin{align*} \lim_{n\to\infty} \left(\sum_{k=1}^n \frac{H_k}{k}-\frac{1}{2}(\log n+\gamma)^2\right) &= \lim_{n\to\infty} \frac{1}{2}\left(H_n^2+ H^{(2)}_n-(\log n+\gamma)^2\right) \\ &= \lim_{n\to\infty} \frac{1}{2}\left((\log n + \gamma)^2 + O(\log n/n) + H^{(2)}_n-(\log n+\gamma)^2\right) \\ &= \frac{1}{2}\lim_{n\to\infty} \left( H^{(2)}_n + O(\log n/n) \right) \\ &= \frac{1}{2}\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{k^2}\\ &= \frac{\pi^2}{12}. \end{align*}

prove the divergence of cauchy product of convergent series $a_{n}:=b_{n}:=\dfrac{(-1)^n}{\sqrt{n+1}}$ i am given these series which converge. $a_{n}:=b_{n}:=\dfrac{(-1)^n}{\sqrt{n+1}}$ i solved this with quotient test and came to $-1$, which is obviously wrong. because it must be $0<\theta<1$ so that the series converges. my steps: $\dfrac{(-1)^{n+1}}{\sqrt{n+2}}\cdot \dfrac{\sqrt{n+1}}{(-1)^{n}} = - \dfrac{\sqrt{n+1}}{\sqrt{n+2}} = - \dfrac{\sqrt{n+1}}{\sqrt{n+2}}\cdot \dfrac{\sqrt{n+2}}{\sqrt{n+2}} = - \dfrac{(n+1)\cdot (n+2)}{(n+2)\cdot (n+2)} = - \dfrac{n^2+3n+2}{n^2+4n+4} = -1 $ did i do something wrong somewhere? and i tried to know whether the cauchy produkt diverges as task says: $\sum_{k=0}^{n}\dfrac{(-1)^{n-k}}{\sqrt{n-k+1}}\cdot \dfrac{(-1)^{k}}{\sqrt{k+1}} = \dfrac{(-1)^n}{nk+n-k^2+1} = ..help.. = diverging $ i am stuck here how to show that the produkt diverges, thanks for any help!

$\sum_{n=0}^\infty\dfrac{(-1)^n}{\sqrt{n+1}}$ is convergent by Leibniz's test, but it is not absolutely convergente (i.e. it is conditionally convergent.) To show that the Cauchy product does not converge use the inequality $$ x\,y\le\frac{x^2+y^2}{2}\quad x,y\in\mathbb{R}. $$ Then $$ \sqrt{n-k+1}\,\sqrt{k+1}\le\frac{n+2}{2} $$ and $$ \sum_{k=0}^n\frac{1}{\sqrt{n-k+1}\,\sqrt{k+1}}\ge\frac{2(n+1)}{n+2}. $$ This shows that the the terms of the Cauchy product do not converge to $0$, and the series diverges.

What is limit of: $\displaystyle\lim_{x\to 0}$$\tan x - \sin x\over x$ I want to search limit of this trigonometric function: $$\displaystyle\lim_{x\to 0}\frac{\tan x - \sin x}{x^n}$$ Note: $n \geq 1$

Checking separatedly the cases for $\,n=1,2,3\,$, we find: $$n=1:\;\;\;\;\;\;\frac{\tan x-\sin x}{x}=\frac{1}{\cos x}\frac{\sin x}{x}(1-\cos x)\xrightarrow [x\to 0]{}1\cdot 1\cdot 0=0$$ $$n=2:\;\;\frac{\tan x-\sin x}{x^2}=\frac{1}{\cos x}\frac{\sin x}{x}\frac{1-\cos x}{x}\xrightarrow [x\to 0]{}1\cdot 1\cdot 0=0\;\;(\text{Applying L'Hospital})$$ $$n=3:\;\;\;\frac{\tan x-\sin x}{x^3}=\frac{1}{\cos x}\frac{\sin x}{x}\frac{1-\cos x}{x^2}\xrightarrow[x\to 0]{}1\cdot 1\cdot \frac{1}{2}=\frac{1}{2}\;\;(\text{Again L'H})$$ $$n\geq 4:\;\;\;\;\frac{\tan x-\sin x}{x^4}=\frac{1}{\cos x}\frac{\sin x}{x}\frac{1-\cos x}{x^2}\frac{1}{x^{n-3}}\xrightarrow[x\to 0]{}1\cdot 1\cdot \frac{1}{2}\cdot\frac{1}{\pm 0}=$$ and the above either doesn't exists (if $\,n-3\,$ is odd), or it is $\,\infty\,$ , so in any case $\,n\geq 4\,$ the limit doesn't exist in a finite form.

Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}} \geq \frac{n^2+3n}{2n+2}$ How to prove the following inequalities without using Bernoulli's inequality? * *$$\prod_{k=1}^{n}{\sqrt[k+1]{k}} \leq \frac{2^n}{n+1},$$ *$$\sum_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}} \geq \frac{n^2+3n}{2n+2}.$$ My proof: * * \begin{align*} \prod_{k=1}^{n}{\sqrt[k+1]{k}} &= \prod_{k=1}^{n}{\sqrt[k+1]{k\cdot 1 \cdot 1 \cdots 1}} \leq \prod^{n}_{k=1}{\frac{k+1+1+\cdots +1}{k+1}}\\ &=\prod^{n}_{k=1}{\frac{2k}{k+1}}=2^n \cdot \prod^{n}_{k=1}{\frac{k}{k+1}}=\frac{2^n}{n+1}.\end{align*} * $$\sum_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}} \geq n \cdot \sqrt[n]{\prod_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}}} \geq n \cdot \sqrt[n]{\frac{n+1}{2^n}}=\frac{n}{2}\cdot \sqrt[n]{n+1}.$$ It remains to prove that $$\frac{n}{2}\cdot \sqrt[n]{n+1} \geq \frac{n^2+3n}{2n+2}=\frac{n(n+3)}{2(n+1)},$$ or $$\sqrt[n]{n+1} \geq \frac{n+3}{n+1},$$ or $$(n+1) \cdot (n+1)^{\frac{1}{n}} \geq n+3.$$ We apply Bernoulli's Inequality and we have: $$(n+1)\cdot (1+n)^{\frac{1}{n}}\geq (n+1) \cdot \left(1+n\cdot \frac{1}{n}\right)=(n+1)\cdot 2 \geq n+3,$$ which is equivalent with: $$2n+2 \geq n+3,$$ or $$n\geq 1,$$ and this is true becaue $n \neq 0$, $n$ is a natural number. Can you give another solution without using Bernoulli's inequality? Thanks :-)

The inequality to be shown is $$(n+1)^{n+1}\geqslant(n+3)^n, $$ for every positive integer $n$. For $n = 1$ it is easy. For $n \ge 2$, apply AM-GM inequality to $(n-2)$-many $(n+3)$, 2 $\frac{n+3}{2}$, and $4$, we get $$(n+3)^n < \left(\frac{(n-2)(n+3) + \frac{n+3}{2} + \frac{n+3}{2} + 4}{n+1}\right)^{n+1} = \left(n+1\right)^{n+1}$$

how to find inverse of a matrix in $\Bbb Z_5$ how to find inverse of a matrix in $\Bbb Z_5$ please help me explicitly how to find the inverse of matrix below, what I was thinking that to find inverses separately of the each term in $\Bbb Z_5$ and then form the matrix? $$\begin{pmatrix}1&2&0\\0&2&4\\0&0&3\end{pmatrix}$$ Thank you.

Hint: Use the adjugate matrix. Answer: The cofactor matrix of $A$ comes $\color{grey}{C_A= \begin{pmatrix} +\begin{vmatrix} 2 & 4 \\ 0 & 3 \end{vmatrix} & -\begin{vmatrix} 0 & 4 \\ 0 & 3 \end{vmatrix} & +\begin{vmatrix} 0 & 2 \\ 0 & 0 \end{vmatrix} \\ -\begin{vmatrix} 2 & 0 \\ 0 & 3 \end{vmatrix} & +\begin{vmatrix} 1 & 0 \\ 0 & 3 \end{vmatrix} & -\begin{vmatrix} 1 & 2 \\ 0 & 0 \end{vmatrix} \\ +\begin{vmatrix} 2 & 0 \\ 2 & 4 \end{vmatrix} & -\begin{vmatrix} 1 & 0 \\ 0 & 4 \end{vmatrix} & +\begin{vmatrix} 1 & 2 \\ 0 & 2 \end{vmatrix} \end{pmatrix}= \begin{pmatrix} 6 & 0 & 0 \\ -6 & 3 & 0 \\ 8 & -4 & 2 \end{pmatrix}=} \begin{pmatrix} 1 & 0 & 0 \\ -1 & 3 & 0 \\ 3 & 1 & 2 \end{pmatrix}.$ Therefore the adjugate matrix of $A$ is $\color{grey}{\text{adj}(A)=C_A^T= \begin{pmatrix} 1 & 0 & 0 \\ -1 & 3 & 0 \\ 3 & 1 & 2 \end{pmatrix}^T=} \begin{pmatrix} 1 & -1 & 3 \\ 0 & 3 & 1 \\ 0 & 0 & 2 \end{pmatrix}$. Since $\det{(A)}=1$, it follows that $A^{-1}=\text{adj}(A)= \begin{pmatrix} 1 & -1 & 3 \\ 0 & 3 & 1 \\ 0 & 0 & 2 \end{pmatrix}$. And we confirm this by multiplying them matrices: $\begin{pmatrix} 1 & -1 & 3 \\ 0 & 3 & 1 \\ 0 & 0 & 2\end{pmatrix} \begin{pmatrix} 1 & 2 & 0 \\ 0 & 2 & 4 \\ 0 & 0 & 3\end{pmatrix}= \begin{pmatrix} 1 & 0 & 5 \\ 0 & 6 & 15 \\ 0 & 0 & 6\end{pmatrix}= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$.

Showing that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left (\sqrt{a^2+1}-1\right)$. How can I show that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left(\sqrt{a^2+1}-1\right)$?

Let $x = a \sin(y)$. Then we have $$\dfrac{\sqrt{a^2-x^2}}{1+x^2} dx = \dfrac{a^2 \cos^2(y)}{1+a^2 \sin^2(y)} dy $$ Hence, $$I = \int_{-a}^{a}\dfrac{\sqrt{a^2-x^2}}{1+x^2} dx = \int_{-\pi/2}^{\pi/2} \dfrac{a^2 \cos^2(y)}{1+a^2 \sin^2(y)} dy $$ Hence, $$I + \pi = \int_{-\pi/2}^{\pi/2} \dfrac{a^2 \cos^2(y)}{1+a^2 \sin^2(y)} dy + \int_{-\pi/2}^{\pi/2} dy = \int_{-\pi/2}^{\pi/2} \dfrac{1+a^2}{1+a^2 \sin^2(y)} dy\\ = \dfrac{1+a^2}2 \int_0^{2 \pi} \dfrac{dy}{1+a^2 \sin^2(y)}$$ Now $$\int_0^{2 \pi} \dfrac{dy}{1+a^2 \sin^2(y)} = \oint_{|z| = 1} \dfrac{dz}{iz \left(1 + a^2 \left(\dfrac{z-\dfrac1z}{2i}\right)^2 \right)} = \oint_{|z| = 1} \dfrac{4z^2 dz}{iz \left(4z^2 - a^2 \left(z^2-1\right)^2 \right)}$$ $$\oint_{|z| = 1} \dfrac{4z^2 dz}{iz \left(4z^2 - a^2 \left(z^2-1\right)^2 \right)} = \oint_{|z| = 1} \dfrac{4z dz}{i(2z + a(z^2-1))(2z - a(z^2-1))}$$ Now $$ \dfrac{4z}{(2z + a(z^2-1))(2z - a(z^2-1))} = \dfrac1{az^2 - a + 2z} - \dfrac1{az^2 - a - 2z}$$ $$\oint_{\vert z \vert = 1} \dfrac{dz}{az^2 - a + 2z} = \oint_{\vert z \vert = 1} \dfrac{dz}{a \left(z + \dfrac{1 + \sqrt{1+a^2}}a\right) \left(z + \dfrac{1 - \sqrt{1+a^2}}a\right)} = \dfrac{2 \pi i}{2 \sqrt{1+a^2}}$$ $$\oint_{\vert z \vert = 1} \dfrac{dz}{az^2 - a - 2z} = \oint_{\vert z \vert = 1} \dfrac{dz}{a \left(z - \dfrac{1 + \sqrt{1+a^2}}a\right) \left(z - \dfrac{1 - \sqrt{1+a^2}}a\right)} = -\dfrac{2 \pi i}{2 \sqrt{1+a^2}}$$ Hence, $$\oint_{|z| = 1} \dfrac{4z dz}{i(2z + a(z^2-1))(2z - a(z^2-1))} = \dfrac{2 \pi i}i \dfrac1{\sqrt{1+a^2}} = \dfrac{2 \pi}{\sqrt{1+a^2}}$$ Hence, we get that $$I + \pi = \left(\dfrac{1+a^2}2\right) \dfrac{2 \pi}{\sqrt{1+a^2}} = \pi \sqrt{1+a^2}$$ Hence, we get that $$I = \pi \left(\sqrt{1+a^2} - 1 \right)$$

Let $a,b$ and $c$ be real numbers.evaluate the following determinant: |$b^2c^2 ,bc, b+c;c^2a^2,ca,c+a;a^2b^2,ab,a+b$| Let $a,b$ and $c$ be real numbers. Evaluate the following determinant: $$\begin{vmatrix}b^2c^2 &bc& b+c\cr c^2a^2&ca&c+a\cr a^2b^2&ab&a+b\cr\end{vmatrix}$$ after long calculation I get that the answer will be $0$. Is there any short processs? Please help someone thank you.

Imagine expanding along the first column. Note that the cofactor of $b^2c^2$ is $$(a+b)ac-(a+c)ab=a^2(c-b)$$ which is a multiple of $a^2$. The other two terms in the expansion along the first column are certainly multiples of $a^2$, so the determinant is a multiple of $a^2$. By symmetry, it's also a multiple of $b^2$ and of $c^2$. If $a=b$ then the first two rows are equal, so the determinant's zero, so the determinant is divisible by $a-b$. By symmetry, it's also divisible by $a-c$ and by $b-c$. So, the determinant is divisible by $a^2b^2c^2(a-b)(a-c)(b-c)$, a poynomial of degree $9$. But the detrminant is a polynomial of degree $7$, so it must be identically zero.

How I study these two sequence? Let $a_1=1$ , $a_{n+1}=a_n+(-1)^n \cdot 2^{-n}$ , $b_n=\frac{2 a_{n+1}-a_n}{a_n}$ (1) $\{\ {a_n\}}$ converges to $0$ and $\{\ {b_n\}}$ is a cauchy sequence . (2) $\{\ {a_n\}}$ converges to non-zero number and $\{\ {b_n\}}$ is a cauchy sequence . (3) $\{\ {a_n\}}$ converges to $0$ and $\{\ {b_n\}}$ is not a cauchy sequence . (4) $\{\ {a_n\}}$ converges to non-zero number and $\{\ {b_n\}}$ is not a cauchy sequence . Trial: Here $$\begin{align} a_1 &=1\\ a_2 &=a_1 -\frac{1}{2} =1 -\frac{1}{2} \\ a_3 &= 1 -\frac{1}{2} + \frac{1}{2^2} \\ \vdots \\ a_n &= 1 -\frac{1}{2} + \frac{1}{2^2} -\cdots +(-1)^{n-1} \frac{1}{2^{n-1}}\end{align}$$ $$\lim_{n \to \infty}a_n=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}$$ Here I conclude $\{\ {a_n\}}$ converges to non-zero number. Am I right? I know the definition of cauchy sequence but here I am stuck to check. Please help.

We have $b_n=\frac{2 a_{n+1}-a_n}{a_n}=2\frac{a_{n+1}}{a_n}-1$. For very large values of $n$, since $a_n\to2/3$ we have $a_{n+1}\sim a_n$. So $b_n\to 2-1=1$ so it is Cauchy as well.

Question about theta of $T(n)=4T(n/5)+n$ I have this recurrence relation $T(n)=4T(\frac{n}{5})+n$ with the base case $T(x)=1$ when $x\leq5$. I want to solve it and find it's $\theta$. I think i have solved it correctly but I can't get the theta because of this term $\frac{5}{5^{log_{4}n}}$ . Any help? $T(n)=4(4T(\frac{n}{5^{2}})+\frac{n}{5})+n$ $=4^{2}(4T(\frac{n}{5^{3}})+\frac{n}{5^{2}})+4\frac{n}{5}+n$ $=...$ $=4^{k}T(\frac{n}{5^{k}})+4^{k-1}\frac{n}{5^{k-1}}+...+4\frac{n}{5}+n$ $=...$ $=4^{m}T(\frac{n}{5^{m}})+4^{m-1}\frac{n}{5^{m-1}}+...+4\frac{n}{5}+n$ Assuming $n=4^{m}$ $=4^{m}T(\lceil(\frac{4}{5})^{m}\rceil)+((\frac{4}{5})^{m-1}+...+1)n$ $=n+\frac{1-(\frac{4}{5})^{m}}{1-\frac{4}{5}}n=n+5n-n^{2}\frac{5}{5^{log_{4}n}}$ $=6n-n^{2}\frac{5}{5^{log_{4}n}}$

An alternative approach is to prove that $T(n)\leqslant5n$ for every $n$. This holds for every $n\leqslant5$ and, if $T(n/5)\leqslant5(n/5)=n$, then $T(n)\leqslant4n+n=5n$. By induction, the claim holds. On the other hand, $T(n)\geqslant n$ for every $n\gt5$, hence $T(n)=\Theta(n)$.

How to find finite trigonometric products I wonder how to prove ? $$\prod_{k=1}^{n}\left(1+2\cos\frac{2\pi 3^k}{3^n+1} \right)=1$$ give me a tip

Let $S_n = \sum_{k=0}^n 3^k = \frac{3^{n+1}-1}{2}$. Then $$3^{n}- S_{n-1} = 3^{n} - \frac{3^{n}-1}{2} = \frac{3^{n}+1}{2} = S_{n-1}+1. $$ Now by induction we have the following product identity for $n \geq 0$: $$ \begin{eqnarray} \prod_{k=0}^{n}\left(z^{3^k}+1+z^{-3^k}\right) &=& \left(z^{3^{n}}+1+z^{-3^{n}}\right)\prod_{k=0}^{n-1}\left(z^{3^k}+1+z^{-3^k}\right) \\ &=& \left(z^{3^{n}}+1+z^{-3^{n}}\right) \left(\sum_{k=-S_{n-1}}^{S_{n-1}} z^k\right) \\ &=&\sum_{k=S_{n-1}+1}^{S_n}z^k + \sum_{k=-S_{n-1}}^{S_{n-1}}z^k+\sum_{k=-S_n}^{-S_{n-1}-1} z^k \\ &=& \sum_{k=-S_n}^{S_n} z^k \end{eqnarray} $$ Now take $z = \exp\left(\frac{\pi \, i}{3^n + 1}\right)$ and use that $z^{3^n+1}=-1$ to get $$\begin{eqnarray} \prod_{k=0}^n\left(1 + 2 \cos \left(\frac{2 \pi \,3^k}{3^n+1}\right)\right) &=& \sum_{k=-S_n}^{S_n}z^{2k} = \frac{z^{2S_n+1}-z^{-2S_n-1}}{z-z^{-1}} = \frac{z^{3^{n+1}}-z^{-3^{n+1}}}{z-z^{-1}} \\ &=& \frac{z^{3(3^n+1)-3} - z^{-3(3^n+1)+3}}{z-z^{-1}} = \frac{z^3-z^{-3}}{z-z^{-1}} = z^2 + 1 + z^{-2} \\ &=& 1 + 2\cos\left(\frac{2\pi}{3^n+1}\right) \end{eqnarray}$$ and your identity follows.

Find infinitely many pairs of integers $a$ and $b$ with $1 < a < b$, so that $ab$ exactly divides $a^2 +b^2 −1$. So I came up with $b= a+1$ $\Rightarrow$ $ab=a(a+1) = a^2 + a$ So that: $a^2+b^2 -1$ = $a^2 + (a+1)^2 -1$ = $2a^2 + 2a$ = $2(a^2 + a)$ $\Rightarrow$ $(a,b) = (a,a+1)$ are solutions. My motivation is for this follow up question: (b) With $a$ and $b$ as above, what are the possible values of: $$ \frac{a^2 +b^2 −1}{ab} $$ Update With Will Jagy's computations, it seems that now I must show that the ratio can be any natural number $m\ge 2$, by the proof technique of vieta jumping. Update Via Coffeemath's answer, the proof is rather elementary and does not require such technique.

$(3,8)$ is a possible solution. This gives us 24 divides 72, and a value of 3 for (b). Have you considered that if $ab$ divides $a^2+b^2-1$, then we $ab$ divides $a^2 + b^2 -1 + 2ab$? This gives us $ab$ divides $(a+b+1)(a+b-1)$. Subsequently, the question might become easier to work with.

Show that $(a+b+c)^3 = a^3 + b^3 + c^3+ (a+b+c)(ab+ac+bc)$ As stated in the title, I'm supposed to show that $(a+b+c)^3 = a^3 + b^3 + c^3 + (a+b+c)(ab+ac+bc)$. My reasoning: $$(a + b + c)^3 = [(a + b) + c]^3 = (a + b)^3 + 3(a + b)^2c + 3(a + b)c^2 + c^3$$ $$(a + b + c)^3 = (a^3 + 3a^2b + 3ab^2 + b^3) + 3(a^2 + 2ab + b^2)c + 3(a + b)c^2+ c^3$$ $$(a + b + c)^3 = a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3ab^2 + 3b^2c + 3ac^2 + 3bc^2 + 6abc$$ $$(a + b + c)^3 = (a^3 + b^3 + c^3) + (3a^2b + 3a^2c + 3abc) + (3ab^2 + 3b^2c + 3abc) + (3ac^2 + 3bc^2 + 3abc) - 3abc$$ $$(a + b + c)^3 = (a^3 + b^3 + c^3) + 3a(ab + ac + bc) + 3b(ab + bc + ac) + 3c(ac + bc + ab) - 3abc$$ $$(a + b + c)^3 = (a^3 + b^3 + c^3) + 3(a + b + c)(ab + ac + bc) - 3abc$$ $$(a + b + c)^3 = (a^3 + b^3 + c^3) + 3[(a + b + c)(ab + ac + bc) - abc]$$ It doesn't look like I made careless mistakes, so I'm wondering if the statement asked is correct at all.

In general, $$a^n+b^n+c^n = \sum_{i+2j+3k=n} \frac{n}{i+j+k}\binom {i+j+k}{i,j,k} s_1^i(-s_2)^js_3^k$$ where $s_1=a+b+c$, $s_2=ab+ac+bc$ and $s_3=abc$ are the elementary symmetric polynomials. In the case that $n=3$, the triples possible are $(i,j,k)=(3,0,0),(1,1,0),$ and $(0,0,1)$ yielding the formula: $$a^3+b^3+c^3 = s_1^3 - 3s_2s_1 + 3s_3$$ which is the result you got. In general, any symmetric homogeneous polynomial $p(a,b,c)$ of degree $n$ can be written in the form: $$p(a,b,c)=\sum_{i+2j+3k=n} a_{i,j,k} s_1^i s_2^j s_3^k$$ for some constants $a_{i,j,k}$. If you don’t know the first statement, you can deduce the values $a_{i,j,k}$ by solving linear equations. This is because the triples $(i,j,k)$ are limited to $(3,0,0), (1,1,0),(0,0,1).$. So if: $$a^3+b^3+c^3=a_{3,0,0}(a+b+c)^3+a_{1,1,0}(ab+ac+bc)(a+b+c)+a_{0,0,1}abc$$ Then try it for specific values of $(a,b,c).$ For example, when $(a,b,c)=(1,0,0),$ you get: $$1=a_{3,0,0}\cdot 1+a_{1,1,0}\cdot 0+a_{0,0,1}\cdot 0.$$ Try $(a,b,c)=(1,1,0)$ and $(1,1,1).$ I've often thought Fermat's Last Theorem was most interesting when stated as a question about these polynomials. One statement of Fermat can be written as: If $p$ is an odd prime, then $a^p+b^p+c^p=0$ if and only if $a+b+c=0$ and $abc=0$.

Another two hard integrals Evaluate : $$\begin{align} & \int_{0}^{\frac{\pi }{2}}{\frac{{{\ln }^{2}}\left( 2\cos x \right)}{{{\ln }^{2}}\left( 2\cos x \right)+{{x}^{2}}}}\text{d}x \\ & \int_{0}^{1}{\frac{\arctan \left( {{x}^{3+\sqrt{8}}} \right)}{1+{{x}^{2}}}}\text{d}x \\ \end{align}$$

For the second integral, consider the more general form $$\int_0^1 dx \: \frac{\arctan{x^{\alpha}}}{1+x^2}$$ (I do not understand what is special about $3+\sqrt{8}$.) Taylor expand the denominator and get $$\begin{align} &=\int_0^1 dx \: \arctan{x^{\alpha}} \sum_{k=0}^{\infty} (-1)^k x^{2 k} \\ &= \sum_{k=0}^{\infty} (-1)^k \int_0^1 dx \: x^{2 k} \arctan{x^{\alpha}} \end{align}$$ Now we can simply evaluate these integrals in terms of polygamma functions: $$\int_0^1 dx \: x^{2 k} \arctan{x^{\alpha}} = \frac{\psi\left(\frac{a+2 k+1}{4 a}\right)-\psi\left(\frac{3 a+2 k+1}{4 a}\right)+\pi }{8 k+4}$$ where $$\psi(z) = \frac{d}{dz} \log{\Gamma{(z)}}$$ and we get that $$\int_0^1 dx \: \frac{ \arctan{x^{\alpha}}}{1+x^2} = \frac{\pi^2}{16} - \frac{1}{4} \sum_{k=0}^{\infty} (-1)^k \frac{\psi\left(\frac{3 \alpha+2 k+1}{4 \alpha}\right)-\psi\left(\frac{\alpha+2 k+1}{4 \alpha}\right) }{2 k+1} $$ This is about as close as I can get. The sum agrees with the numerical integration out to 6 sig figs at about $10,000$ terms for $\alpha = 3+\sqrt{8}$.

Factorization problem Find $m + n$ if $m$ and $n$ are natural numbers such that: $$\frac {m+n} {m^2+mn+n^2} = \frac {4} {49}\;.$$ My reasoning: Say: $$m+n = 4k$$ $$m^2+mn+n^2 = 49k$$ It follows:$$(m+n)^2 = (4k)^2 = 16k^2 \Rightarrow m^2+mn+n^2 + mn = 16k^2 \Rightarrow mn = 16k^2 - 49k$$ Since: $$mn\gt0 \Rightarrow 16k^2 - 49k\gt0 \Rightarrow k\gt3$$ Then no more progress.

Observe that $k$ must be a non-zero integer. We know that $m, n$ are the roots of the quadratic equation $$X^2 - 4kX + (16k^2 - 49k)$$ The roots, from the quadratic equation, are $$ \frac { 4k \pm \sqrt{(4k)^2 - 4(16k^2 - 49k) }} {2} = 2k \pm \sqrt{ 49k - 12k^2}$$ The expression in the square root must be a perfect square. Try $k = 1$, $49 k - 12k^2 = 37$ is not a perfect square. Try $k = 2$, $49k - 12k^2 = 50$ is not a perfect square. Try $k=3$, $49k-12k^2 = 39$ is not a perfect square. Try $k=4$, $49k-12k^2 = 4$ is a perfect square. This leads to roots 6, 10, which have sum 16. For $k\geq 5$, $49k - 12k^2 < 0$ has no solution. For $k \leq -1$, $49k - 12k^2 < 0$ has no solution.

$\gcd(m,n) = 1$ and $\gcd (mn,a)=1$ implies $a \cong 1 \pmod{ mn}$ I have $m$ and $n$ which are relatively prime to one another and $a$ is relatively prime to $mn$ and after alot of tinkering with my problem i came to this equality: $a \cong 1 \pmod m \cong 1 \pmod n$ why is it safe to say that $a \cong 1 \pmod {mn}$?..

It looks as if you are asking the following. Suppose that $m$ and $n$ are relatively prime. Show that if $a\equiv 1\pmod{m}$ and $a\equiv 1\pmod{n}$, then $a\equiv 1\pmod{mn}$. So we know that $m$ divides $a-1$, and that $n$ divides $a-1$. We want to show that $mn$ divides $a-1$. Let $a-1=mk$. Since $n$ divides $a-1$, it follows that $n$ divides $mk$. But $m$ and $n$ are relatively prime, and therefore $n$ divides $k$. so $k=nl$ for some $l$, and therefore $a-1=mnl$. Remark: $1.$ There are many ways to show that if $m$ and $n$ are relatively prime, and $n$ divides $mk$, then $n$ divides $k$. One of them is to use Bezout's Theorem: If $m$ and $n$ are relatively prime, there exist integers $x$ and $y$ such that $mx+ny=1$. Multiply through by $k$. We get $mkx+nky=k$. By assumption, $n$ divides $mk$, so $n$ divides $mkx$. Clearly, $n$ divides $nky$. So $n$ divides $mkx+nky$, that is, $n$ divides $k$. $2.$ Note that there was nothing special about $1$. Let $m$ and $n$ be relatively prime. If $a\equiv c\pmod{m}$ and $a\equiv c\pmod{n}$, then $a\equiv c\pmod{mn}$.

Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Let $s_n$ be a sequence defined as given below for $n \geq 1$. Then find out $\lim\limits_{n \to \infty} s_n$. \begin{align} s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx \end{align} I have written a solution of my own, but I would like to know it is completely correct, and also I would like if people post more alternative solutions.

Notice (1) $\frac{s_n}{n} + \frac{s_{n+1}}{n+1} = \int_0^1 x^{n-1} dx = \frac{1}{n} \implies s_n + s_{n+1} = 1 + \frac{s_{n+1}}{n+1}$. (2) $s_n = n\int_0^1 \frac{x^{n-1}}{1+x} dx < n\int_0^1 x^{n-1} dx = 1$ (3) $s_{n+1} - s_n = \int_0^1 \frac{d (x^{n+1}-x^n)}{1+x} = \int_0^1 x^n \frac{1-x}{(1+x)^2} dx > 0$ (2+3) $\implies s = \lim_{n\to\infty} s_n$ exists and (1+2) $\implies s+s = 1 + 0 \implies s = \frac{1}{2}$. In any event, $s_n$ can be evaluated exactly to $n (\psi(n) - \psi(\frac{n}{2}) - \ln{2})$ where $\psi(x)$ is the diagamma function. Since $\psi(x) \approx \ln(x) - \frac{1}{2x} - \frac{1}{12x^2} + \frac{1}{120x^4} + ... $ as $x \to \infty$, we know: $$s_n \approx \frac{1}{2} + \frac{1}{4 n} - \frac{1}{8 n^3} + ...$$

half-angle trig identity clarification I am working on the following trig half angle problem. I seem to be on the the right track except that my book answer shows -1/2 and I didn't get that in my answer. Where did I go wrong? $$\sin{15^{\circ}} = $$ $$\sin \frac { 30^{\circ} }{ 2 } = \pm \sqrt { \frac { 1 - \cos{30^{\circ}} }{ 2 } } $$ $$\sin \frac { 30^{\circ} }{ 2 } = \pm \sqrt { \frac { 1 - \frac {\sqrt 3 }{ 2 } }{ 2 } } $$ $$\sin \frac { 30^{\circ} }{ 2 } = \pm \sqrt { \frac { 1 - \frac {\sqrt 3 }{ 2 } }{ 2 } (\frac {2} {2}) } $$ $$\sin \frac { 30^{\circ} }{ 2 } = \pm \sqrt { 2 - \sqrt {3} } $$ Book Answer $$\sin \frac { 30^{\circ} }{ 2 } = -\frac {1} {2} \sqrt { 2 - \sqrt {3} } $$

$$\sqrt { \dfrac { 1 - \dfrac {\sqrt 3 }{ 2 } }{ 2 } \times \dfrac22} = \sqrt{\dfrac{2-\sqrt3}{\color{red}4}} = \dfrac{\sqrt{2-\sqrt3}}{\color{red}2}$$

Solving Recurrence $T(n) = T(n − 3) + 1/2$; I have to solve the following recurrence. $$\begin{gather} T(n) = T(n − 3) + 1/2\\ T(0) = T(1) = T(2) = 1. \end{gather}$$ I tried solving it using the forward iteration. $$\begin{align} T(3) &= 1 + 1/2\\ T(4) &= 1 + 1/2\\ T(5) &= 1 + 1/2\\ T(6) &= 1 + 1/2 + 1/2 = 2\\ T(7) &= 1 + 1/2 + 1/2 = 2\\ T(8) &= 1 + 1/2 + 1/2 = 2\\ T(9) &= 2 + 1/2 \end{align}$$ I couldnt find any sequence here. can anyone help!

The generating function is $$g(x)=\sum_{n\ge 0}T(n)x^n = \frac{2-x^3}{2(1+x+x^2)(1-x)^2}$$, which has the partial fraction representation $$g = \frac{2}{3(1-x)} + \frac{1}{6(1-x)^2}+\frac{x+1}{6(1+x+x^2)}$$. The first term contributes $$\frac{2}{3}(1+x+x^2+x^3+\ldots)$$, equivalent to $T(n)=2/3$ the second term contributes $$\frac{1}{6}(1+2x+3x^2+4x^3+\ldots)$$ equivalent to $T(n) = (n+1)/6$, and the third term contributes $$\frac{1}{6}(1-x^2+x^3-x^5+x^6-\ldots)$$ equivalent to $T(n) = 1/6, 0, -1/6$ depending on $n\mod 3$ being 0 or 1 or 2. $$T(n) = \frac{2}{3}+\frac{n+1}{6}+\left\{\begin{array}{ll} 1/6,& n \mod 3=0\\ 0,& n \mod 3=1 \\ -1/6,&n \mod 3 =2\end{array}\right.$$

Solve $\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$ for $x$ Is there any smart way to solve the equation: $$\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$$ Use Maple I can find $x \in \{1;ab+bc+ca\}$

I have a partial solution, as follows: Note that $\frac{(b-c)(1+a^2)}{x+a^2}=\frac{(b-c)\left((x+a^2)+(1-x)\right)}{x+a^2}=(b-c)+\frac{{(b-c)}(1-x)}{x+a^2}$. Likewise, $\frac{(c-a)(1+b^2)}{x+b^2}=(c-a)+\frac{(c-a)(1-x)}{x+b^2}$ and $\frac{(a-b)(1+c^2)}{x+c^2}=(c-a)+\frac{(a-b)(1-x)}{x+c^2}$. Now, $\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=\frac{(b-c)(1-x)}{x+a^2}+\frac{(c-a)(1-x)}{x+b^2}+\frac{(a-b)(1-x)}{x+c^2}$ as $(b-c)+(c-a)+(a-b)=0$. Hence $(1-x)\left(\frac{b-c}{x+a^2}+\frac{c-a}{x+b^2}+\frac{a-b}{x+c^2}\right)=0$ and so $x=1$ or $\frac{b-c}{x+a^2}+\frac{c-a}{x+b^2}+\frac{a-b}{x+c^2}=0$

Show that $(m^2 - n^2, 2mn, m^2 + n^2)$ is a primitive Pythagorean triplet Show that $(m^2 - n^2, 2mn, m^2 + n^2)$ is a primitive Pythagorean triplet First, I showed that $(m^2 - n^2, 2mn, m^2 + n^2)$ is in fact a Pythagorean triplet. $$\begin{align*} (m^2 - n^2)^2 + (2mn)^2 &= (m^2 + n^2)^2 \\ &= m^4 -2m^2n^2 + n^4 + 4m^2n^2 \\ &= m^4 + 2m^2n^2 + n^4 \\ &= 1\end{align*}$$ which shows that it respect $a^2+b^2 = c^2$ let p be a prime number, $ p|(m^2 + n^2) \text { and } p|(m^2 - n^2) $ if $gcd(m^2 + n^2, (m^2 - n^2)) = 1$ $p | (m^2 + n^2) , \text { so, } p |m^2 \text { and } p |n^2$ that means $ (m^2 + n^2) \text { and } (m^2 - n^2) $ are prime together I'm kind of lost when I begin to show the gcd = 1... I think I know what to do, just not sure how to do it correctly. Thanks

To show $(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2$ is equivalent to showing $(m^2 - n^2)^2 + (2mn)^2 - (m^2 + n^2)^2 = 0$ so \begin{align*} && (m^2 - n^2)^2 + (2mn)^2 - (m^2 + n^2)^2 \\ &=& m^4 - 2m^2n^2 + n^4 + 4m^2n^2 - m^4 - 2m^2n^2 - n^4 \\ &=& m^4 + n^4 - m^4 - n^4 \\ &=& 0\end{align*}

Find $\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}.$ Find $$\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}\;.$$

Here is another sketch of proof. Let $$J_n = \{(j, k) : 0 \leq j, k < n \text{ and } (j, k) \neq (0, 0) \}.$$ Then for each $(j, k) \in J_n$ and $(x_0, y_0) = (j/n, k/n)$, we have $$ \frac{x_0 + y_0}{(x_0+\frac{1}{n})^{3} + (y_0 + \frac{1}{n})^3} \leq \frac{x+y}{x^3 + y^3} \leq \frac{x_0 + y_0 + \frac{2}{n}}{x_0^3 + y_0^3} $$ for $(x, y) \in [x_0, y_0] \times [x_0 + 1/n, y_0 + 1/n]$. Thus if we let $D_n$ be the closure of the set $[0, 1]^2 - [0, 1/n]^2$, then $$ \sum_{(j,k) \in J_n} \frac{j+k}{(j+1)^3 + (k+1)^3} \leq \int_{D_n} \frac{x+y}{x^3 + y^3} \, dxdy \leq \sum_{(j,k) \in J_n} \frac{j+k+2}{j^3 + k^3}. $$ It is not hard to establish the relation that $$ \sum_{(j,k) \in J_n} \frac{j+k}{(j+1)^3 + (k+1)^3} = \sum_{j,k=1}^{n} \frac{j+k}{j^3 + k^3} + O(1) $$ and that $$ \sum_{(j,k) \in J_n} \frac{j+k+2}{j^3 + k^3} = \sum_{j,k=1}^{n} \frac{j+k}{j^3 + k^3} + O(1). $$ By noting that \begin{align*} \int_{D_n} \frac{x+y}{x^3 + y^3} \, dxdy &= 2\int_{\frac{1}{n}}^{1} \int_{0}^{y} \frac{x+y}{x^3 + y^3} \, dxdy \\ &= (2 \log n) \int_{0}^{1} \frac{1}{x^2 - x + 1} \, dx \\ &= \frac{4 \pi \log n}{3\sqrt{3}}, \end{align*} we obtain the asymptotic formula $$ \frac{1}{\log n} \sum_{j,k=1}^{n} \frac{j+k}{j^3 + k^3} = \frac{4 \pi}{3\sqrt{3}} + O\left( \frac{1}{\log n} \right) $$ and hence the answer is $\displaystyle \frac{4 \pi}{3\sqrt{3}} $.

Computing $\lim_{n\to\infty}n\sum_{k=1}^n\left( \frac{1}{(2k-1)^2} - \frac{3}{4k^2}\right)$ What ways would you propose for the limit below? $$\lim_{n\to\infty}n\sum_{k=1}^n\left( \frac{1}{(2k-1)^2} - \frac{3}{4k^2}\right)$$ Thanks in advance for your suggestions, hints! Sis.

OK, it turns out that $$\sum_{k=1}^n\left( \frac{1}{(2k-1)^2} - \frac{3}{4k^2}\right) = \sum_{k=1}^{n-1} \frac{1}{(k+n)^2}$$ This may be shown by observing that $$\sum_{k=1}^n \frac{1}{(2k-1)^2} = \sum_{k=1}^{2 n-1} \frac{1}{k^2} - \frac{1}{2^2} \sum_{k=1}^n \frac{1}{k^2}$$ The desired limit may then be rewritten as $$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n-1} \frac{1}{(1 + (k/n))^2}$$ which you may recognize as a Riemann sum, equal to $$\int_0^1 dx \: \frac{1}{(1+x)^2} = \frac{1}{2}$$

Really Stuck on Partial derivatives question Ok so im really stuck on a question. It goes: Consider $$u(x,y) = xy \frac {x^2-y^2}{x^2+y^2} $$ for $(x,y)$ $ \neq $ $(0,0)$ and $u(0,0) = 0$. calculate $\frac{\partial u} {\partial x} (x,y)$ and $\frac{\partial u} {\partial y} (x,y)$ for all $ (x,y) \in \Bbb R^2. $ show that $ \frac {\partial^2 u} {\partial x \partial y} (0,0) \neq \frac {\partial^2 u} {\partial y \partial x} (0,0) $. Check, using polar coordinates, that $ \frac {\partial u}{\partial x} and \frac {\partial u}{\partial y} $ are continuous at $(0,0)$ Any help really appreciated. Cheers

We are given: $$u(x, y)=\begin{cases} xy \frac {x^2-y^2}{x^2+y^2}, ~(x, y) \ne (0,0)\\\\ ~~~0, ~~~~~~~~~~~~~~~(x, y) = (0,0)\;. \end{cases}$$ I am going to multiply out the numerator for ease in calculations, so we have: $$\tag 1 u(x, y)=\begin{cases} \frac {x^3y - xy^3}{x^2+y^2}, ~(x, y) \ne (0,0)\\\\ ~~~0, ~~~~~~~~~~~~~~(x, y) = (0,0)\;. \end{cases}$$ We are asked to: * *(a) Find: $\displaystyle \frac{\partial u} {\partial x} (x,y) ~\forall x~ \in \Bbb R^2$ *(b) Find: $\displaystyle \frac{\partial u} {\partial y} (x,y) ~\forall x~ \in \Bbb R^2$ *(c) Show that $ \displaystyle \frac {\partial^2 u} {\partial x \partial y} (0,0) \neq \frac {\partial^2 u} {\partial y \partial x} (0,0)$ *(d) Check, using polar coordinates, that $\displaystyle \frac {\partial u}{\partial x} \text{and} \frac {\partial u}{\partial y} $ are continuous at $(0,0)$ Using $(1)$, for part $(a)$, we get: $\tag 2 \displaystyle \frac{\partial u} {\partial x} (x,y) = \frac{(3x^2y- y^3)(x^2+y^2) - 2x(x^3y - xy^3)}{(x^2 + y^2)^2} = \frac{x^4y + 4x^2y^3-y^5}{(x^2+y^2)^2}$ Using $(1)$, for part $(b)$, we get: $\tag 3 \displaystyle \frac{\partial u} {\partial y} (x,y) = \frac{(x^3-3xy^2)(x^2+y^2) - 2y(x^3y - xy^3)}{(x^2+y^2)^2} = \frac{x^5 - 4x^3y^2-xy^4}{(x^2+y^2)^2}$ Next, we need mixed partials, so using $(2)$, we have: $ \tag 4 \displaystyle \frac {\partial^2 u} {\partial x \partial y} (x, y) = \frac{(x^4+ 12x^2y^2-5y^4)(x^2+y^2)^2 - 2(x^2+y^2)(2y)(x^4y + 4x^2y^3-y^5)}{(x^2+y^2)^4} = \frac{x^6 + 9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3} = \frac {\partial^2 u} {\partial y \partial x} (x, y)$ Thus, we get: $$\tag 5 \displaystyle \frac {\partial^2 u} {\partial x \partial y} (x, y) = \frac {\partial^2 u} {\partial y \partial x} (x, y) = \begin{cases} \frac{x^6 + 9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}, ~(x, y) \ne (0,0)\\\\ ~~~~~~~~~~~~~~~~~~0, ~~~~~~~~~~~~~~~~~~~~~~~~(x, y) = (0,0)\;. \end{cases}$$ Now, for part $(c)$, we want to show that $ \frac {\partial^2 u} {\partial x \partial y} (0,0) \neq \frac {\partial^2 u} {\partial y \partial x} (0,0)$, so we need to find the limits of each mixed partial. We have: $ \tag 6 \displaystyle \frac{\partial^2 u} {\partial x \partial y} (0,0) = \lim\limits_{h \to 0} \frac{\frac{\partial u}{\partial x} (0, h) - \frac{\partial u}{\partial x} (0, 0)}{h} = \lim\limits_{h \to 0} \frac{-h^5/h^4}{h} = \lim\limits_{h \to 0} \frac{-h}{h} = -1$, and $ \tag 7 \displaystyle \frac{\partial^2 u} {\partial y \partial x} (0,0) = \lim\limits_{h \to 0} \frac{\frac{\partial u}{\partial y} (h, 0) - \frac{\partial u}{\partial y} (0, 0)}{h} = \lim\limits_{h \to 0} \frac{h^5/h^4}{h} = \lim\limits_{h \to 0} \frac{h}{h} = +1$ $\therefore$, for part $(c)$, we have shown that: $$\frac {\partial^2 u} {\partial x \partial y} (0,0) \neq \frac {\partial^2 u} {\partial y \partial x} (0,0)$$ as desired. Can you handle part $(d)$? Regards

If $P(A \cup B \cup C) = 1$, $P(B) = 2P(A) $, $P(C) = 3P(A) $, $P(A \cap B) = P(A \cap C) = P(B \cap C) $, then $P(A) \le \frac14$ We have ($P$ is probability): $P(A \cup B \cup C) = 1$ ; $P(B) = 2P(A) $ ; $P(C) = 3P(A) $ and $P(A \cap B) = P(A \cap C) = P(B \cap C) $. Prove that $P(A) \le \frac{1}{4} $. Well, I tried with the fact that $ 1 = P(A \cup B \cup C) = 6P(A) - 3P(A \cap B) + P(A \cap B \cap C) $ but I got stuck... Could anyone help me, please?

This solution is quite possibly messier than necessary, but it’s what first occurred to me. For convenience let $x=P\big((A\cap B)\setminus C\big)$ and $y=P(A\cap B\cap C)$. Let $a=P(A)-2x-y$, $b=P(B)-2x-y$, and $c=P(C)-2x-y$; these are the probabilities of $A\setminus(B\cup C)$, $B\setminus(A\cup C)$, and $C\setminus(A\cup B)$, respectively. (A Venn diagram is helpful here.) Now $$b+2x+y=2(a+2x+y)=2a+4x+2y\;,$$ so $b=2a+2x+y$. Similarly, $$c+2x+y=3(a+2x+y)=3a+6x+3y\;,$$ so $c=3a+4x+2y$. Then $$\begin{align*} 1&=P(A\cup B\cup C)\\ &=a+b+c+3x+y\\ &=6a+9x+4y\\ &=4(a+2x+y)+2a+x\\ &=4P(A)+2a+x\;. \end{align*}$$ Since $2a+x\ge 0$, we must have $P(A)\le \frac{1}{4}$.

$T:P_3 → P_3$ be the linear transformation such that... Let $T:P_3 \to P_3$ be the linear transformation such that $T(2 x^2)= -2 x^2 - 4 x$, $T(-0.5 x - 5)= 2 x^2 + 4 x + 3$, and $T(2 x^2 - 1) = 4 x - 4.$ Find $T(1)$, $T(x)$, $T(x^2)$, and $T(a x^2 + b x + c)$, where $a$, $b$, and $c$ are arbitrary real numbers.

hint:T is linear transformation $T(a+bx+cx^2)=aT(1)+bT(x) +cT(x^2) $ $$T(2 x^2)= -2 x^2 - 4 x \to T( x^2)= - x^2 - 2 x$$$$T(-0.5 x - 5)= 2 x^2 + 4 x + 3\to-0.5T( x) - 5T(1)= 2 x^2 + 4 x + 3\to 16x^2+72x-46$$$$T(2 x^2 - 1) = 4 x - 4.\to T( x^2) - \frac12T(1) = 2 x - 2\to T(1)=-2 x^2 - 8x +4.$$

Probability of choosing two equal bits from three random bits Given three random bits, pick two (without replacement). What is the probability that the two you pick are equal? I would like to know if the following analysis is correct and/or if there is a better way to think about it. $$\Pr[\text{choose two equal bits}] = \Pr[\text{2nd bit} = 0 \mid \text{1st bit} = 0] + \Pr[\text{2nd bit} = 1 \mid \text{1st bit} = 1]$$ Given three random bits, once you remove the first bit the other two bits can be: 00, 01, 11, each of which occurring with probability $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$. Thus, $$\Pr[\text{2nd bit} = 0] = 1\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{4} + 0\cdot\frac{1}{4} = \frac{3}{8}$$ And $\Pr[\text{2nd bit} = 1] = \Pr[\text{2nd bit} = 0]$ by the same analysis. Therefore, $$\Pr[\text{2nd bit}=0 \mid \text{1st bit} = 0] = \frac{\Pr[\text{1st and 2nd bits are 0}]}{\Pr[\text{1st bit}=0]} = \frac{1/2\cdot3/8}{1/2} = \frac{3}{8}$$ and by the same analysis, $\Pr[\text{2nd bit} = 1 \mid \text{1st bit} = 1] = \frac{3}{8}$. Thus, $$\Pr[\text{choose two equal bits}] = 2\cdot\frac{3}{8} = \frac{3}{4}$$

Whatever the first bit picked, the probability the second bit matches it is $1/2$. Remark: We are assuming what was not explicitly stated, that $0$'s and $1$'s are equally likely. One can very well have "random" bits where the probability of $0$ is not the same as the probability of $1$.

Help with writing the following as a partial fraction $\frac{4x+5}{x^3+1}$. I need help with writing the following as a partial fraction: $$\frac{4x+5}{x^3+1}$$ My attempts so far are to factor $x^3$ into $(x+1)$ and $(x^2-x+1)$ This gives me: $A(x^2-x+1)+B(x+1)$. But I have problems with solving the equation system that this gives: $A = 0$ (since there are no $x^2$ terms in $4x+5$) $-A+B =4$ (since there are $4$ $x$ terms in $4x+5$) $A+B = 5$ (since the constant is $5$ in $4x+5$) this gives me $A=0.5$ and $B=4.5$ and $\frac{1/2}{x+1}, \frac{9/2}{x^2-x+1}$ This is appearantly wrong. Where is my reasoning faulty? Thank you!

You need to use one less exponent per factor in the numerator after your factorization. This leads to: $$\frac{Ax+B}{x^2-x+1} + \frac{C}{x+1} = \frac{4x+5}{x^3+1}$$ This gives us: $$Ax^2 + Ax + Bx + B + Cx^2 - Cx + C = 4x + 5$$ This leads to: $A + C = 0$ $A + B - C = 4$ $B + C = 5$ yielding: $$A = -\frac{1}{3}, B = \frac{14}{3}, C = \frac{1}{3}$$ Writing the expansion out yields: $$\frac{4x+5}{x^3+1} = \frac{14 - x}{3(x^2-x+1)} + \frac{1}{3(x+1)}$$

Prove that $ \left(1+\frac a b \right) \left(1+\frac b c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$. Given $a,b,c>0$, prove that $\displaystyle \left(1+\frac a b \right) \left(1+\frac b c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$. I expanded the LHS, and realized I have to prove $\displaystyle\frac a b +\frac a c +\frac b c +\frac b a +\frac c a +\frac c b \geq \frac{2(a+b+c)}{\sqrt[3]{abc}}$, but I don't know how. Please help. Thank you.

We can begin by clearing denominators as follows $$a^2c+a^2b+b^2a+b^2c+c^2a+c^2b\geq 2a^{5/3}b^{2/3}c^{2/3}+2a^{2/3}b^{5/3}c^{2/3}+2a^{2/3}b^{2/3}c^{5/3}$$ Now by the Arithmetic Mean - Geometric Mean Inequality, $$\frac{2a^2c+2a^2b+b^2a+c^2a}{6} \geq a^{5/3}b^{2/3}c^{2/3}$$ That is, $$\frac{2}{3}a^2c+\frac{2}{3}a^2b+\frac{1}{3}b^2a+\frac{1}{3}c^2a \geq 2a^{5/3}b^{2/3}c^{2/3}$$ Similarly, we have $$\frac{2}{3}b^2a+\frac{2}{3}b^2c+\frac{1}{3}c^2b+\frac{1}{3}a^2b \geq 2a^{2/3}b^{5/3}c^{2/3}$$ $$\frac{2}{3}c^2b+\frac{2}{3}c^2a+\frac{1}{3}a^2c+\frac{1}{3}b^2c \geq 2a^{2/3}b^{2/3}c^{5/3}$$ Summing these three inequalities together, we obtain the desired result.

Prove $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots+\frac{1}{\sqrt{n}} > 2\:(\sqrt{n+1} − 1)$ Basically, I'm trying to prove (by induction) that: $$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots+\frac{1}{\sqrt{n}} > 2\:(\sqrt{n+1} − 1)$$ I know to begin, we should use a base case. In this case, I'll use $1$. So we have: $$1 > 2\:(1+1-1) = 1>-2$$ Which works out. My problem is the next step. What comes after this? Thanks!

Mean Value Theorem can also be used, Let $\displaystyle f(x)=\sqrt{x}$ $\displaystyle f'(x)=\frac{1}{2}\frac{1}{\sqrt{x}}$ Using mean value theorem we have: $\displaystyle \frac{f(n+1)-f(n)}{(n+1)-n}=f'(c)$ for some $c\in(n,n+1)$ $\displaystyle \Rightarrow \frac{\sqrt{n+1}-\sqrt{n}}{1}=\frac{1}{2}\frac{1}{\sqrt{c}}$....(1) $\displaystyle \frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{c}}<\frac{1}{\sqrt{n}}$ Using the above ineq. in $(1)$ we have, $\displaystyle \frac{1}{2\sqrt{n+1}}<\sqrt{n+1}-\sqrt{n}<\frac{1}{2\sqrt{n}}$ Adding the left part of the inequality we have,$\displaystyle\sum_{k=2}^{n}\frac{1}{2\sqrt{k}}<\sum_{k=2}^{n}(\sqrt{k}-\sqrt{k-1})=\sqrt{n}-1$ $\Rightarrow \displaystyle\sum_{k=2}^{n}\frac{1}{\sqrt{k}}<2\sum_{k=2}^{n}(\sqrt{k}-\sqrt{k-1})=2(\sqrt{n}-1)$ $\Rightarrow \displaystyle1+\sum_{k=2}^{n}\frac{1}{\sqrt{k}}<1+2\sum_{k=2}^{n}(\sqrt{k}-\sqrt{k-1})=2\sqrt{n}-2+1=2\sqrt{n}-1$ $\Rightarrow \displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt{k}}<2\sqrt{n}-1$ Similarly adding the right side of the inequality we have, $\displaystyle\sum_{k=1}^{n}\frac{1}{2\sqrt{k}}>\sum_{k=1}^{n}(\sqrt{k+1}-\sqrt{k})=\sqrt{n+1}-1$ $\Rightarrow \displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt{k}}>2(\sqrt{n+1}-1)$ This completes the proof. $\displaystyle 2\sqrt{n+1}-2<\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}<2\sqrt{n}-1.$ This is a much better proof than proving by induction(Ofcourse if someone knows elementary calculus).

$\sum \limits_{k=1}^{\infty} \frac{6^k}{\left(3^{k+1}-2^{k+1}\right)\left(3^k-2^k\right)} $ as a rational number. $$\sum \limits_{k=1}^{\infty} \frac{6^k}{\left(3^{k+1}-2^{k+1}\right)\left(3^k-2^k\right)} $$ I know from the ratio test it convergest, and I graph it on wolfram alpha and I suspect the sum is 2; however, I am having trouble with the manipulation of the fraction to show the rational number. ps. When it says write as a rational number it means to write the value of $S_{\infty}$ or to rewrite the fraction?

That denominator should suggest the possibility of splitting the general term into partial fractions and getting a telescoping series of the form $$\sum_{k\ge 1}\left(\frac{A_k}{3^k-2^k}-\frac{A_{k+1}}{3^{k+1}-2^{k+1}}\right)\;,$$ where $A_k$ very likely depends on $k$. Note that if this works, the sum of the series will be $$\frac{A_1}{3^1-2^1}=A_1\;.$$ Now $$\frac{A_k}{3^k-2^k}-\frac{A_{k+1}}{3^{k+1}-2^{k+1}}=\frac{3^{k+1}A_k-3^kA_{k+1}-2^{k+1}A_k+2^kA_{k+1}}{(3^k-2^k)(3^{k+1}-2^{k+1})}\;,$$ so you want to choose $A_k$ and $A_{k+1}$ so that $$3^{k+1}A_k-3^kA_{k+1}-2^{k+1}A_k+2^kA_{k+1}=6^k\;.$$ The obvious things to try are $A_k=2^k$, which makes the last two terms cancel out to leave $3^{k+1}2^k-3^k2^{k+1}=6^k(3-2)=6^k$, and $A_k=3^k$, which makes the first two terms cancel out and leaves $6^k(3-2)=6^k$; both work. However, summing $$\sum_{k\ge 1}\left(\frac{A_k}{3^k-2^k}-\frac{A_{k+1}}{3^{k+1}-2^{k+1}}\right)\tag{1}$$ to $$\frac{A_1}{3^1-2^1}=A_1$$ is valid only if $$\lim_{k\to\infty}\frac{A_k}{3^k-2^k}=0\;,$$ since the $n$-th partial sum of $(1)$ is $$A_1-\frac{A_{n+1}}{3^{n+1}-2^{n+1}}\;.$$ Checking the two possibilities, we see that $$\lim_{k\to\infty}\frac{2^k}{3^k-2^k}=0,\quad\text{but}\quad\lim_{k\to\infty}\frac{3^k}{3^k-2^k}=1\;,$$ so we must choose $A_k=2^k$, and the sum of the series is indeed $A_1=2$.

Show that $x$, $y$, and $z$ are not distinct if $x^2(z-y) + y^2(x-z) + z^2(y-x) = 0$. Suppose that $x^2(z-y) + y^2(x-z) + z^2(y-x) = 0$. How can I show that $x$, $y$, and $z$ are not all distinct, that is, either $x=y$, $y=z$, or $x=z$?

$x^2(z-y)+y^2(x-z)+z^2(y-x)=x^2(z-y)+x(y^2-z^2)+z^2y-y^2z=(z-y)(x^2-x(y+z)+zy)=(z-y)(x-y)(x-z)$.

Homogeneous equation I am trying to solve the following homogeneous equation: Thanks for your tips $xy^3y′=2(y^4+x^4)$ I think this isHomogeneous of order4 => $xy^3dy/dx=2(y^4+x^4/1)$ => $xy^3dy=2(x^4+y^4)dx$ => $xy^3dy-2(x^4+y^4)dx=0$ I do not know how to continued

Make the substitution $v=y^4$. Then by the chain rule we have $v'=4y^3 y'$. Now your DE turns into: $$x \frac{v'}{4}=2(v+x^4)$$ Then can be simplified to: $$v'-8\frac{v}{x}=8x^3$$ We first solve the homogeneous part: $$v_h' -8\frac{v_h}{x}=0$$ This leads to $v_h=c\cdot x^8$ so that $v_h'=c\cdot 8x^7$. This is our homogeneous solution. Now to find a particular solution we guess it will look something like $$v_p=ax^4+bx^3+dx^2+3x+f$$ Filling this into the DE we get: $$(4ax^3 +3bx^2+2dx+3)-\frac{8}{x}(ax^4+bx^3+dx^2+3x+f)=8x^3$$ the only terms with $x^3$ in it are the $4ax^3$ and the $-8ax^3$. Conclusion: $-4a=8$ so $a=-2$. All the other terms are zero. We now have our particular solution $v_p=-2x^4$. The general solution is the sum of the particular and homogeneous solution so $$v_g=c\cdot x^8 -2x^4$$ Now backsubstitute $v=y^4$ or $y=v^{\frac{1}{4}}$ to obtain the final answer.

How do we integrate, $\int \frac{1}{x+\frac{1}{x^2}}dx$? How do we integrate the following integral? $$\int \frac{1}{x+\large\frac{1}{x^2}}\,dx\quad\text{where}\;\;x\ne-1$$

The integral is equivalent to $$\int dx \frac{x^2}{1+x^3} = \frac{1}{3} \int \frac{d(x^3)}{1+x^3} = \frac{1}{3} \log{(1+x^3)} + C$$

How to choose the starting row when computing the reduced row echelon form? I'm having hell of a time going around solving matrices to reduced row echelon form. My main issue is which row to start simplifying values and based on what? I have this example so again, the questions are: 1.Which row to start simplifying values? 2.Based on what criteria? Our professor solved it in the class with no fractions but I could not do it. Even though I know the 3 operations performed on matrices

Where you start is not really a problem. My tip: * *Always first make sure you make the first column: 1,0,0 *Then proceed making the second one: 0,1,0 *And lastly, 0,0,1 Step one: $$\begin{pmatrix} 1&2&3&9 \\ 2&-1&1&8 \\ 3&0&-1&3\end{pmatrix}$$ row 3 - 3 times row 1 $$\begin{pmatrix} 1&2&3&9 \\ 2&-1&1&8 \\ 0&-6&-10&-24\end{pmatrix}$$ row 2 - 2 times row 1 $$\begin{pmatrix} 1&2&3&9 \\ 0&-5&-5&-10 \\ 0&-6&-10&-24\end{pmatrix}$$ Which simplifies to $$\begin{pmatrix} 1&2&3&9 \\ 0&1&1&2 \\ 0&3&5&12\end{pmatrix}$$ Now you can proceed with step 2, and 3. row 1 - 2 times row 2 and row 3 - 3 times row 2 $$\begin{pmatrix} 1&0&1&5 \\ 0&1&1&2 \\ 0&0&2&6\end{pmatrix}$$ Simplifies to $$\begin{pmatrix} 1&0&1&5 \\ 0&1&1&2 \\ 0&0&1&3\end{pmatrix}$$ row 2 - row 3 $$\begin{pmatrix} 1&0&1&5 \\ 0&1&0&-1 \\ 0&0&1&3\end{pmatrix}$$ row 1 - row 3 $$\begin{pmatrix} 1&0&0&2 \\ 0&1&0&-1 \\ 0&0&1&3\end{pmatrix}$$

Intersection points of a Triangle and a Circle How can I find all intersection points of the following circle and triangle? Triangle $$A:=\begin{pmatrix}22\\-1.5\\1 \end{pmatrix} B:=\begin{pmatrix}27\\-2.25\\4 \end{pmatrix} C:=\begin{pmatrix}25.2\\-2\\4.7 \end{pmatrix}$$ Circle $$\frac{9}{16}=(x-25)^2 + (y+2)^2 + (z-3)^2$$ What I did so far was to determine the line equations of the triangle (a, b and c): $a : \overrightarrow {OX} = \begin{pmatrix}27\\-2.25\\4 \end{pmatrix}+ \lambda_1*\begin{pmatrix}-1.8\\0.25\\0.7 \end{pmatrix} $ $b : \overrightarrow {OX} = \begin{pmatrix}22\\-1.5\\1 \end{pmatrix}+ \lambda_2*\begin{pmatrix}3.2\\-0.5\\3.7 \end{pmatrix} $ $c : \overrightarrow {OX} = \begin{pmatrix}22\\-1.5\\1 \end{pmatrix}+ \lambda_3*\begin{pmatrix}5\\-0.75\\3 \end{pmatrix} $ But I am not sure what I have to do next...

The side $AB$ of the triangle has equation $P(t) = (1-t)A + tB$ for $0 \le t \le 1$. The $0 \le t \le 1$ part is important. If $t$ lies outside $[0,1]$, the point $P(t)$ will lie on the infinite line through $A$ and $B$, but not on the edge $AB$ of the triangle. Substitute $(1-t)A + tB$ into the circle equation, as others have suggested. This will give you a quadratic equation in $t$ that you can solve using the well-known formula. But, a solution $t$ will give you a circle/triangle intersection point only if it lies in the range $0 \le t \le 1$. Solutions outside this interval can be ignored. Do the same thing with sides $BC$ and $AC$.

Find $x$ such that $\sum_{k=1}^{2014} k^k \equiv x \pmod {10}$ Find $x$ such that $$\sum_{k=1}^{2014} k^k \equiv x \pmod {10}$$ I knew the answer was $3$.

We are going to compute the sum mod $2$ and mod $5$. The Chinese Remainder Theorem then gives us the result mod $10$. Mod $2$, obviously $$k^k \equiv \begin{cases}0 & \text{if }k\text{ even,}\\1 & \text{if }k\text{ odd,}\end{cases}$$ so $$\sum_{k = 0}^{2014} \equiv \frac{2014}{2} = 1007 \equiv 1 \mod 2.$$ By Fermat, $k^k$ mod $5$ only depends on the remainder of $k$ mod $\operatorname{lcm}(5, 4) = 20$. So $$\sum_{k = 1}^{2014}k^k \equiv \underbrace{100}_{\equiv 0} \cdot \sum_{k=1}^{20} k^k + \sum_{k=1}^{14} k^k \\ \equiv 1 + 4 + 2 + 1 + 0 + 1 + 3 + 1 + 4 + 0 + 1 + 1 + 3 + 1 \\\equiv 3 \mod 5.$$ Combining the results mod $2$ and mod $5$, $$\sum_{k=1}^{2014} k^k \equiv 3\mod 10.$$

A probability question that involves $5$ dice For five dice that are thrown, I am struggling to find the probability of one number showing exactly three times and a second number showing twice. For the one number showing exactly three times, the probability is: $$ {5 \choose 3} \times \left ( \frac{1}{6} \right )^{3} \times \left ( \frac{5}{6}\right )^{2} $$ However, I understand I cannot just multiply this by $$ {5 \choose 2} \times \left ( \frac{1}{6} \right )^{2} \times \left ( \frac{5}{6}\right )^{3} $$ as this includes the probability of picking the original number twice which allows the possibility of the same number being shown $5$ times. I am unsure of what to do next, I tried to write down all the combinations manually and got $10$ possible outcomes so for example if a was the value found $3$ times and $b$ was the value obtained $2$ times one arrangement would be '$aaabb$'. However I still am unsure of what to do after I get $10$ different possibilities and I am not sure how I could even get the $10$ different combinations mathematically. Any hints or advice on what to do next would be much appreciated.

First, I assume they wil all come out in neat order, first three in a row, then two in a row of a different number. The probability of that happening is $$ \frac{1}{6^2}\cdot \frac{5}{6}\cdot\frac{1}{6} = \frac{5}{6^4} $$ The first die can be anything, but the next two have to be equal to that, so the $\frac{1}{6^2}$ comes from there. Then the fourth die has to be different, and the odds of that happening is the $\frac{5}{6}$ above, and lastly, the last die has to be the same as the fourth. Now, we assumed that the three equal dice would come out first. There are other orders, a total of $\binom{5}{3}$. Multiply them, and you get the final answer $$ \binom{5}{3}\frac{5}{6^4} $$ You might reason another way. Let the event $A$ be "there are exactly 3 of one number" and $B$ be "There are exactly 2 of one number". Then we have that $$ P(A\cap B) = P(A) \cdot P(B|A) = {5 \choose 3} \cdot \frac{1}{6^3} \cdot\frac{5^2}{6^2} \cdot \frac{1}{5} = \binom{5}{3}\frac{5}{6^4} $$ Where $P(A)$ you allready calculated on your own, and $P(B|A)$ is the probability that there is an exact pair given that there is an exact triple, which again is the probability that the two last dice are equal, and that is $\frac{1}{5}$.

$\lim_{n\to\infty}(\sqrt{n^2+n}-\sqrt{n^2+1})$ How to evaluate $$\lim_{n\to\infty}(\sqrt{n^2+n}-\sqrt{n^2+1})$$ I'm completely stuck into it.

A useful general approach to limits is, in your scratch work, to take every complicated term and replace it with a similar approximate term. As $n$ grows large, $\sqrt{n^2 + n}$ looks like $\sqrt{n^2} = n$. More precisely, $$ \sqrt{n^2 + n} = n + o(n) $$ where I've used little-o notation. In terms of limits, this means $$ \lim_{n \to \infty} \frac{\sqrt{n^2 + n} - n}{n} = 0 $$ but little-o notation makes it much easier to express the intuitive idea being used. Unfortunately, $\sqrt{n^2 + 1}$ also looks like $n$. Combining these estimates, $$ \sqrt{n^2 + n} - \sqrt{n^2 + 1} = (n + o(n)) - (n + o(n)) = o(n) $$ Unfortunately, this cancellation has clobbered all of the precision of our estimates! All this analysis reveals is $$ \lim_{n \to \infty} \frac{\sqrt{n^2 + n} - \sqrt{n^2+1}}{n} = 0 $$ which isn't good enough to answer the problem. So, we need a better estimate. A standard way to get better estimates is differential approximation. While the situation at hand is a little awkward, there is fortunately a standard trick to deal with square roots, or any power: $$ \sqrt{n^2 + n} = n \sqrt{1 + \frac{1}{n}} $$ and now we can invoke differential approximation (or Taylor series) $$ f(x+h) = f(x) + h f'(x) + o(h) $$ with $f(x) = 1 + \frac{1}{x}$ at $x=1$ to get $$ \sqrt{n^2 + n} = n \left( 1 + \frac{1}{2n} + o\left(\frac{1}{n} \right)\right) = n + \frac{1}{2} + o(1) $$ or equivalently in limit terms, $$ \lim_{n \to \infty} \sqrt{n^2 + n} - n - \frac{1}{2} = 0$$ similarly, $$ \sqrt{n^2 + 1} = n + o(1)$$ and we get $$ \lim_{n \to \infty} \sqrt{n^2 + n} - \sqrt{n^2 + 1} = \lim_{n \to \infty} (n + \frac{1}{2} + o(1)) - (n + o(1)) = \lim_{n \to \infty} \frac{1}{2} + o(1) = \frac{1}{2} $$ If we didn't realize that trick, there are a few other tricks to do, but there is actually a straightforward way to proceed too. Initially, simply taking the Taylor series for $g(x) = \sqrt{n^2 + x}$ around $x=0$ doesn't help, because that gives $$ g(x) = n + \frac{1}{2} \frac{x}{n} + o(x^2) $$ the Taylor series for $h(x) = \sqrt{x^2 + x}$ doesn't help either. But this is why we pay attention to the remainder term! One form of the Taylor remainder says that: $$ g(x) = n + \frac{1}{2} \frac{x}{n} - \frac{1}{8} \left( n^2 + c \right)^{-3/2} x^2$$ for some $c$ between $0$ and $x$. It's easy to bound this error term for $x > 0$. $$ \left| \frac{1}{8} \left( n^2 + c \right)^{-3/2} x^2 \right| \leq \left| \frac{1}{8} \left( n^2 \right)^{-3/2} x^2 \right| = \left| \frac{x^2}{8 n^3} \right| $$ So, for $x > 0$, $$ g(x) = n + \frac{1}{2} + O\left( \frac{x^2}{n^3} \right) $$ (note I've switched to big-O). Plugging in $n$ gives $$ g(n) = n + \frac{1}{2} + O\left( \frac{1}{n} \right) $$ which gives the approximation we need (better than we need, actually). (One could, of course, simply stick to limits rather than use big-O notation) This is not the simplest way to solve the problem, but I wanted to demonstrate a straightforward application of the tools you have learned (or will soon learn) to solve a problem in the case that you can't find a 'clever' approach.

Prove $a\sqrt[3]{a+b}+b\sqrt[3]{b+c}+c\sqrt[3]{c+a} \ge 3 \sqrt[3]2$ Prove $a\sqrt[3]{a+b}+b\sqrt[3]{b+c}+c\sqrt[3]{c+a} \ge 3 \sqrt[3]2$ with $a + b+c=3 \land a,b,c\in \mathbb{R^+}$ I tried power mean inequalities but I still can't prove it.

Here is my proof by AM-GM inequality,we have $$ a\sqrt[3]{a+b}=\frac{3\sqrt[3]{2}a(a+b)}{3\sqrt[3]{2(a+b)(a+b)}}\geq 3\sqrt[3]{2}\cdot \frac{a(a+b)}{2+2a+2b} $$ Thus,it's suffice to prove that $$ \frac{a(a+b)}{a+b+1}+\frac{b(b+c)}{b+c+1}+\frac{c(c+a)}{c+a+1}\geq 2 $$ Or $$ \frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\leq 1 $$ After homogenous,it's $$\frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b}\leq \frac{1}{3} $$ Now,multiply $4a+4b+4c$ to each sides.we can rewrite the inequality into $$ \frac{9ca}{4a+4b+c}+\frac{9ab}{4b+4c+a}+\frac{9bc}{4c+4a+b}\leq a+b+c $$ Using Cauchy-Schwarz inequality,we have $$ \frac{9}{4a+4b+c}=\frac{(2+1)^2}{2(2a+b)+(2b+c)}\le \frac{2}{2a+b}+\frac{1}{2b+c} $$ Therefore \begin{align} \sum{\frac{9ca}{4a+4b+c}}&\leq \sum{\left(\frac{2ca}{2a+b}+\frac{ca}{2b+c}\right)}\\ &=a+b+c \end{align} Hence we are done!

Limit of definite sum equals $\ln(2)$ I have to show the following equality: $$\lim_{n\to\infty}\sum_{i=\frac{n}{2}}^{n}\frac{1}{i}=\log(2)$$ I've been playing with it for almost an hour, mainly with the taylor expansion of $\ln(2)$. It looks very similar to what I need, but it has an alternating sign which sits in my way. Can anyone point me in the right direction?

Truncate the Maclaurin series for $\log(1+x)$ at the $2m$-th term, and evaluate at $x=1$. Take for example $m=10$. We get $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots+\frac{1}{19}-\frac{1}{20}.$$ Add $2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{20}\right)$, and subtract the same thing, but this time noting that $$ 2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{20}\right)=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{10}.$$ We get $$\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{10}+\frac{1}{11}+\cdots+\frac{1}{20}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{10}\right).$$ There is nice cancellation, and we get $$\frac{1}{11}+\frac{1}{12}+\cdots+\frac{1}{20}.$$

How to calculate the asymptotic expansion of $\sum \sqrt{k}$? Denote $u_n:=\sum_{k=1}^n \sqrt{k}$. We can easily see that $$ k^{1/2} = \frac{2}{3} (k^{3/2} - (k-1)^{3/2}) + O(k^{-1/2}),$$ hence $\sum_1^n \sqrt{k} = \frac{2}{3}n^{3/2} + O(n^{1/2})$, because $\sum_1^n O(k^{-1/2}) =O(n^{1/2})$. With some more calculations, we get $$ k^{1/2} = \frac{2}{3} (k^{3/2} - (k-1)^{3/2}) + \frac{1}{2} (k^{1/2}-(k-1)^{-3/2}) + O(k^{-1/2}),$$ hence $\sum_1^n \sqrt{k} = \frac{2}{3}n^{3/2} + \frac{1}{2} n^{1/2} + C + O(n^{1/2})$ for some constant $C$, because $\sum_n^\infty O(k^{-3/2}) = O(n^{-1/2})$. Now let's go further. I have made the following calculation $$k^{1/2} = \frac{3}{2} \Delta_{3/2}(k) + \frac{1}{2} \Delta_{1/2}(k) + \frac{1}{24} \Delta_{-1/2}(k) + O(k^{-5/2}),$$ where $\Delta_\alpha(k) = k^\alpha-(k-1)^{\alpha}$. Hence : $$\sum_{k=1}^n \sqrt{k} = \frac{2}{3} n^{3/2} + \frac{1}{2} n^{1/2} + C + \frac{1}{24} n^{-1/2} + O(n^{-3/2}).$$ And one can continue ad vitam aeternam, but the only term I don't know how to compute is the constant term. How do we find $C$ ?

Let us substitute into the sum $$\sqrt k=\frac{1}{\sqrt \pi }\int_0^{\infty}\frac{k e^{-kx}dx}{\sqrt x}. $$ Exchanging the order of summation and integration and summing the derivative of geometric series, we get \begin{align*} \mathcal S_N:= \sum_{k=1}^{N}\sqrt k&=\frac{1}{\sqrt \pi }\int_0^{\infty}\frac{\left(e^x-e^{-(N-1)x}\right)-N\left(e^{-(N-1)x}-e^{-Nx}\right)}{\left(e^x-1\right)^2}\frac{dx}{\sqrt x}=\\&=\frac{1}{2\sqrt\pi}\int_0^{\infty} \left(N-\frac{1-e^{-Nx}}{e^x-1}\right)\frac{dx}{x\sqrt x}=\\ &=\frac{1}{2\sqrt\pi}\int_0^{\infty} \left(N-\frac{1-e^{-Nx}}{e^x-1}\right)\frac{dx}{x\sqrt x}. \end{align*} To extract the asymptotics of the above integral it suffices to slightly elaborate the method used to answer this question. Namely \begin{align*} \mathcal S_N&=\frac{1}{2\sqrt\pi}\int_0^{\infty} \left(N-\frac{1-e^{-Nx}}{e^x-1}+\left(1-e^{-Nx}\right)\left(\frac1x-\frac12\right)-\left(1-e^{-Nx}\right)\left(\frac1x-\frac12\right)\right)\frac{dx}{x\sqrt x}=\\ &={\color{red}{\frac{1}{2\sqrt\pi}\int_0^{\infty}\left(1-e^{-Nx}\right)\left(\frac1x-\frac12-\frac{1}{e^x-1}\right)\frac{dx}{x\sqrt x}}}+\\&+ {\color{blue}{\frac{1}{2\sqrt\pi}\int_0^{\infty} \left(N-\left(1-e^{-Nx}\right)\left(\frac1x-\frac12\right)\right)\frac{dx}{x\sqrt x}}}. \end{align*} The reason to decompose $\mathcal S_N$ in this way is that * *the red integral has an easily computable finite limit: since $\frac1x-\frac12-\frac{1}{e^x-1}=O(x)$ as $x\to 0$, we can simply neglect the exponential $e^{-Nx}$. *the blue integral can be computed exactly. Therefore, as $N\to \infty$, we have $$\mathcal S_N={\color{blue}{\frac{\left(4n+3\right)\sqrt n}{6}}}+ {\color{red}{\frac{1}{2\sqrt\pi}\int_0^{\infty}\left(\frac1x-\frac12-\frac{1}{e^x-1}\right)\frac{dx}{x\sqrt x}+o(1)}},$$ and the finite part you are looking for is given by $$C=\frac{1}{2\sqrt\pi}\int_0^{\infty}\left(\frac1x-\frac12-\frac{1}{e^x-1}\right)\frac{dx}{x\sqrt x}=\zeta\left(-\frac12\right).$$

Need help proving this integration If $a>b>0$, prove that : $$\int_0^{2\pi} \frac{\sin^2\theta}{a+b\cos\theta}\ d\theta = \frac{2\pi}{b^2} \left(a-\sqrt{a^2-b^2} \right) $$

I'll do this one $$\int_{0}^{2\pi}\frac{cos(2\theta)}{a+bcos(\theta)}d\theta$$if we know how to do tis one you can replace $sin^{2}(\theta)$ by $\frac{1}{2}(1-cos(2\theta))$ and do the same thing. if we ae on the unit circle we know that $cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$ so by letting $z=e^{i\theta}$ we wil get$$cos(\theta)=\frac{z+\frac{1}{z}}{2}=\frac{z^2+1}{2z}$$and$$cos(2\theta)=\frac{z^2+\frac{1}{z^2}}{2}=\frac{z^4+1}{2z^2}$$ thus, if$\gamma: |z|=1$ the integral becomes $$\int_{\gamma}\frac{\frac{z^4+1}{2z^2}}{a+b\frac{z^2+1}{2z}}\frac{1}{iz}dz=\int_{\gamma}\frac{-i(z^4+1)}{2z^2(bz^2+2az+b)}dz$$now the roots of $bz^2+2az+b$ are $z=\frac{-2a\pm \sqrt{4a^2-4b^2}}{2b}=\frac{-a}{b}\pm\frac{\sqrt{a^2-b^2}}{b}$, you can check that the only root inside $|z|=1$ is $z_1=\frac{-a}{b}+\frac{\sqrt{a^2-b^2}}{b}$ so the only singularities of the function that we want to integrate inside $\gamma$ are $z_0=0$ and $z_1$ both are poles. find the resudies and sum them to get the answer.Notice that this will only give youe "half" the answer you still have to do$$\frac{1}{2}\int_{\gamma}\frac{d\theta}{a+bcos(\theta)}$$ in the same way to get the full answer.

Math question functions help me? I have to find find $f(x,y)$ that satisfies \begin{align} f(x+y,x-y) &= xy + y^2 \\ f(x+y, \frac{y}x ) &= x^2 - y^2 \end{align} So I first though about replacing $x+y=X$ and $x-y=Y$ in the first one but then what?

Hint: If $x+y=X$ and $x-y=Y$ then \begin{align*} x&=X-y \implies Y= X-y-y \implies 2y=X-Y \dots \end{align*} And be careful with the second for $x=0$.

How to create a generating function / closed form from this recurrence? Let $f_n$ = $f_{n-1} + n + 6$ where $f_0 = 0$. I know $f_n = \frac{n^2+13n}{2}$ but I want to pretend I don't know this. How do I correctly turn this into a generating function / derive the closed form?

In two answers, it is derived that the generating function is $$ \begin{align} \frac{7x-6x^2}{(1-x)^3} &=x\left(\frac1{(1-x)^3}+\frac6{(1-x)^2}\right)\\ &=\sum_{k=0}^\infty(-1)^k\binom{-3}{k}x^{k+1}+6\sum_{k=0}^\infty(-1)^k\binom{-2}{k}x^{k+1}\\ &=\sum_{k=1}^\infty(-1)^{k-1}\left(\binom{-3}{k-1}+6\binom{-2}{k-1}\right)x^k\\ &=\sum_{k=1}^\infty\left(\binom{k+1}{k-1}+6\binom{k}{k-1}\right)x^k\\ &=\sum_{k=1}^\infty\left(\binom{k+1}{2}+6\binom{k}{1}\right)x^k\\ &=\sum_{k=1}^\infty\frac{k^2+13k}{2}x^k\\ \end{align} $$ Therefore, we get the general term is $f_k=\dfrac{k^2+13k}{2}$

Integral of irrational function $$ \int \frac{\sqrt{\frac{x+1}{x-2}}}{x-2}dx $$ I tried: $$ t =x-2 $$ $$ dt = dx $$ but it didn't work. Do you have any other ideas?

Let $y=x-2$ and integrate by parts to get $$\int dx \: \frac{\sqrt{\frac{x+1}{x-2}}}{x-2} = -2 (x-2)^{-1/2} (x+1)^{1/2} + \int \frac{dy}{\sqrt{y (y+3)}}$$ In the second integral, complete the square in the denominator to get $$\int \frac{dy}{\sqrt{y (y+3)}} = \int \frac{dy}{\sqrt{(y+3/2)^2-9/4}}$$ This integral may be solved using a substitution $y+3/2=3/2 \sec{\theta}$, $dy = 3/2 \sec{\theta} \tan{\theta}$. Using the fact that $$\int d\theta \sec{\theta} = \log{(\sec{\theta}+\tan{\theta})}+C$$ we may evaluate the integral exactly. I leave the intervening steps to the reader; I get $$\int dx \: \frac{\sqrt{\frac{x+1}{x-2}}}{x-2} = -2 (x-2)^{-1/2} (x+1)^{1/2} + \log{\left[\frac{2}{3}\left(x-\frac{1}{2}\right)+\sqrt{\frac{4}{9}\left(x-\frac{1}{2}\right)^2-1}\right]}+C$$

How to simplify a square root How can the following: $$ \sqrt{27-10\sqrt{2}} $$ Be simplified to: $$ 5 - \sqrt{2} $$ Thanks

Set the nested radical as the difference of two square roots so that $$\sqrt{27-10\sqrt{2}}=(\sqrt{a}-\sqrt{b})$$ Then square both sides so that $$27-10\sqrt{2}=a-2\sqrt{a}\sqrt{b}+b$$ Set (1) $$a+b=27$$ and set (2) $$-2\sqrt{a}\sqrt{b}=-10\sqrt{2}$$ Square both sides of (2) to get $$4ab= 200$$ and solve for $b$ to get $$b=\frac{50}{a}$$ Replacing $b$ in (1) gives $$a+\frac{50}{a}=27$$ Multiply all terms by $a$ and convert to the quadratic equation $$a^2-27a+50=0$$ Solving the quadratic gives $a=25$ or $a=2$. Replacing $a$ and $b$ in the first difference of square roots formula above with $25$ and $2$ the solutions to the quadratic we have $$\sqrt{25}-\sqrt{2}$$ or $$5-\sqrt{2}$$

How to find $f:\mathbb{Q}^+\to \mathbb{Q}^+$ if $f(x)+f\left(\frac1x\right)=1$ and $f(2x)=2f\bigl(f(x)\bigr)$ Let $f:\mathbb{Q}^+\to \mathbb{Q}^+$ be a function such that $$f(x)+f\left(\frac1x\right)=1$$ and $$f(2x)=f\bigl(f(x)\bigr)$$ for all $x\in \mathbb{Q}^+$. Prove that $$f(x)=\frac{x}{x+1}$$ for all $x\in \mathbb{Q}^+$. This problem is from my student.

Some ideas: $$\text{I}\;\;\;\;x=1\Longrightarrow f(1)+f\left(\frac{1}{1}\right)=2f(1)=1\Longrightarrow \color{red}{f(1)=\frac{1}{2}}$$ $$\text{II}\;\;\;\;\;\;\;\;f(2)=2f(f(1))=2f\left(\frac{1}{2}\right)$$ But we also know that $$f(2)+f\left(\frac{1}{2}\right) =1$$ so from II we get $$3f\left(\frac{1}{2}\right)=1\Longrightarrow \color{red}{f\left(\frac{1}{2}\right)=\frac{1}{3}}\;,\;\;\color{red}{f(2)=\frac{2}{3}}$$ and also: $$ \frac{1}{2}=f(1)=f\left(2\cdot \frac{1}{2}\right)=2f\left(f\left(\frac{1}{2}\right)\right)=2f\left(\frac{1}{3}\right)\Longrightarrow \color{red}{f\left(\frac{1}{3}\right)=\frac{1}{4}}\;,\;\color{red}{f(3)=\frac{3}{4}}$$ One more step: $$f(4)=f(2\cdot2)=2f(f(2))=2f\left(\frac{2}{3}\right)=2\cdot 2f\left(f\left(\frac{1}{3}\right)\right)=4f\left(\frac{1}{4}\right)$$ and thus: $$1=f(4)+f\left(\frac{1}{4}\right)=5f\left(\frac{1}{4}\right)\Longrightarrow \color{red}{f\left(\frac{1}{4}\right)=\frac{1}{5}}\;,\;\;\color{red}{f(4)=\frac{4}{5}}$$ ...and etc.

How do I transform the left side into the right side of this equation? How does one transform the left side into the right side? $$ (a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2 $$

Expand the left hand side, you get $$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2$$ Add and substract $2abcd$ $$a^2c^2+a^2d^2+b^2c^2+b^2d^2=(a^2c^2-2abcd+b^2d^2)+(a^2d^2+2abcd+b^2c^2)$$ Complete the square, you can get $$(a^2c^2-2abcd+b^2d^2)+(a^2d^2+2abcd+b^2c^2)=(ac-bd)^2+(ad+bc)^2$$ Therefore, $$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2.$$

Prove inequality: $28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4$ Prove: $28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4$ with $a, b, c \ge0$ I can do this by: $EAT^2$ (expand all of the thing) * *$(x+y+z)^4={x}^{4}+{y}^{4}+{z}^{4}+4\,{x}^{3}y+4\,{x}^{3}z+6\,{x}^{2}{y}^{2}+6\,{ x}^{2}{z}^{2}+4\,x{y}^{3}+4\,x{z}^{3}+4\,{y}^{3}z+6\,{y}^{2}{z}^{2}+4 \,y{z}^{3}+12\,x{y}^{2}z+12\,xy{z}^{2}+12\,{x}^{2}yz$ *$(x+y-z)^4={x}^{4}+{y}^{4}+{z}^{4}+4\,{x}^{3}y-4\,{x}^{3}z+6\,{x}^{2}{y}^{2}+6\,{ x}^{2}{z}^{2}+4\,x{y}^{3}-4\,x{z}^{3}-4\,{y}^{3}z+6\,{y}^{2}{z}^{2}-4 \,y{z}^{3}-12\,x{y}^{2}z+12\,xy{z}^{2}-12\,{x}^{2}yz$ ... $$28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4\\ \iff a^4 + b^4 + c^4 \ge a^2b^2+c^2a^2+b^2c^2 \text{(clearly hold by AM-GM)}$$ but any other ways that smarter ?

A nice way of tackling the calculations might be as follows:$$~$$ Let $x=b+c-a,y=c+a-b,z=a+b-c.$ Then the original inequality is just equivalent with $$\frac74\Bigl((x+y)^4+(y+z)^4+(z+x)^4\Bigr)\geq x^4+y^4+z^4+(x+y+z)^4.$$ Now we can use the identity $$\sum_{cyc}(x+y)^4=x^4+y^4+z^4+(x+y+z)^4-12xyz(x+y+z),$$ So that it suffices to check that $$\frac37\Bigl(x^4+y^4+z^4+(x+y+z)^4\Bigr)\geq 12xyz(x+y+z),$$ Which obviously follows from the AM-GM inequality: $(x+y+z)^4\geq \Bigl(3(xy+yz+zx)\Bigr)^2\geq 27xyz(x+y+z)$ and $x^4+y^4+z^4\geq xyz(x+y+z).$ Equality holds in the original inequality iff $x=y=z\iff a=b=c.$ $\Box$

Find lim sup $x_n$ Let $x_n = n(\sqrt{n^2+1} - n)\sin\dfrac{n\pi}8$ , $n\in\Bbb{N}$ Find $\limsup x_n$. Hint: lim sup $x_n = \sup C(x_n)$. How to make it into a fraction to find the cluster point of $x_n$?

Expand: $n · \left( \sqrt{n²+1} - n \right) = n · \tfrac{1}{\sqrt{n²+1} + n}$ for all $n ∈ ℕ$. Since $\tfrac{\sqrt{n²+1} + n}{n} = \sqrt{1+\tfrac{1}{n²}} + 1 \overset{n → ∞}{\longrightarrow} 2$, you have $n · \left( \sqrt{n²+1} - n \right) \overset{n → ∞}{\longrightarrow} \tfrac{1}{2}$. Now $x_n = n · \left( \sqrt{n²+1} - n \right) · \sin \left(\tfrac{n · π}{8}\right) ≤ n · \left( \sqrt{n²+1} - n \right)$ for all $n ∈ ℕ$, so: \begin{align*} \limsup_{n→∞} x_n &= \limsup_{n→∞} \left( n · \left( \sqrt{n²+1} - n \right) · \sin \left(\tfrac{n · π}{8}\right) \right)\\ &≤ \limsup_{n→∞} \left( n · \left( \sqrt{n²+1} - n \right) \right)= \tfrac{1}{2} \end{align*} But $\sin \left(\tfrac{(16n+4)· π}{8}\right) = \sin \left(n·2π + \tfrac{π}{2}\right) = \sin \left(\tfrac{π}{2}\right)= 1$ for all $n ∈ ℕ$. This means $x_{16n+4} = (16n+4) · \left( \sqrt{(16n+4)²+1} - (16n+4) \right) · \sin \left(\tfrac{(16n+4)· π}{8}\right) \overset{n → ∞}{\longrightarrow} \tfrac{1}{2}$. Therefore $\tfrac{1}{2}$ is a limit point of the sequence $(x_n)_{n ∈ ℕ}$ as well as an upper bound of the limit points of that sequence. So it’s the upper limit of that sequence.

Need to prove the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges I need to prove that the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges. I do not have to find the limit. I have tried to prove it by proving that the sequence is monotone and bounded, but I am having some trouble: Monotonic: The sequence seems to be monotone and increasing. This can be proved by induction: Claim that $a_n\leq a_{n+1}$ $$a_1=1\leq 1+\frac{1}{2^2}=a_2$$ Need to show that $a_{n+1}\leq a_{n+2}$ $$a_{n+1}=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}\leq 1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}=a_{n+2}$$ Thus the sequence is monotone and increasing. Boundedness: Since the sequence is increasing it is bounded below by $a_1=1$. Upper bound is where I am having trouble. All the examples I have dealt with in class have to do with decreasing functions, but I don't know what my thinking process should be to find an upper bound. Can anyone enlighten me as to how I should approach this, and can anyone confirm my work thus far? Also, although I prove this using monotonicity and boundedness, could I have approached this by showing the sequence was a Cauchy sequence? Thanks so much in advance!

Notice that $ 2k^2 \geq k(k+1) \implies \frac{1}{k^2} \leq \frac{2}{k(k+1)}$. $$ \sum_{k=1}^{\infty} \frac{2}{k(k+1)} = \frac{2}{1 \times 2} + \frac{2}{2 \times 3} + \frac{2}{3 \times 4} + \ldots $$ $$ \sum_{k=1}^{\infty} \frac{2}{k(k+1)} = 2\Big(\, \Big(1 - \frac{1}{2}\Big) + \Big(\frac{1}{2} - \frac{1}{3} \Big) + \Big(\frac{1}{3} - \frac{1}{4} \Big) + \ldots \Big)$$ $$ \sum_{k=1}^{\infty} \frac{2}{k(k+1)} = 2 (1) = 2 $$. Therefore $ \sum_{k=1}^{\infty} \frac{1}{k^2} \leq 2$.

Evaluating $\int \frac{1}{{x^4+1}} dx$ I am trying to evaluate the integral $$\int \frac{1}{1+x^4} \mathrm dx.$$ The integrand $\frac{1}{1+x^4}$ is a rational function (quotient of two polynomials), so I could solve the integral if I can find the partial fraction of $\frac{1}{1+x^4}$. But I failed to factorize $1+x^4$. Any other methods are also wellcome.

Without using fractional decomposition: $$\begin{align}\int\dfrac{1}{x^4+1}~dx&=\dfrac{1}{2}\int\dfrac{2}{x^4+1}~dx \\&=\dfrac{1}{2}\int\dfrac{(x^2+1)-(x^2-1)}{x^4+1}~dx \\&=\dfrac{1}{2}\int\dfrac{x^2+1}{x^4+1}~dx-\dfrac{1}{2}\int\dfrac{x^2-1}{x^4+1}~dx \\&=\dfrac{1}{2}\int\dfrac{1+\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}}~dx-\dfrac{1}{2}\int\dfrac{1-\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}}~dx \\&=\dfrac{1}{2}\left(\int\dfrac{1+\dfrac{1}{x^2}}{\left(x-\dfrac{1}{x}\right)^2+2}~dx-\int\dfrac{1-\dfrac{1}{x^2}}{\left(x+\dfrac{1}{x}\right)^2-2}~dx\right) \\&=\dfrac{1}{2}\left(\int\dfrac{d\left(x-\dfrac{1}{x}\right)}{\left(x-\dfrac{1}{x}\right)^2+2}-\int\dfrac{d\left(x+\dfrac{1}{x}\right)}{\left(x+\dfrac{1}{x}\right)^2-2}\right)\end{align}$$ So, finally solution is $$\int\dfrac{1}{x^4+1}~dx=\dfrac{1}{4\sqrt2}\left(2\arctan\left(\dfrac{x^2-1}{\sqrt2x}\right)+\log\left(\dfrac{x^2+\sqrt2x+1}{x^2-\sqrt2x+1}\right)\right)+C$$

Need to prove the sequence $a_n=(1+\frac{1}{n})^n$ converges by proving it is a Cauchy sequence I am trying to prove that the sequence $a_n=(1+\frac{1}{n})^n$ converges by proving that it is a Cauchy sequence. I don't get very far, see: for $\epsilon>0$ there must exist $N$ such that $|a_m-a_n|<\epsilon$, for $ m,n>N$ $$|a_m-a_n|=\bigg|\bigg(1+\frac{1}{m}\bigg)^m-\bigg(1+\frac{1}{n}\bigg)^n\bigg|\leq \bigg|\bigg(1+\frac{1}{m}\bigg)^m\bigg|+\bigg|\bigg(1+\frac{1}{n}\bigg)^n\bigg|\leq\bigg(1+\frac{1}{m}\bigg)^m+\bigg(1+\frac{1}{n}\bigg)^n\leq \quad?$$ I know I am supposed to keep going, but I just can't figure out the next step. Can anyone offer me a hint please? Or if there is another question that has been answered (I couldn't find any) I would gladly look at it. Thanks so much!

We have the following inequalities: $\left(1+\dfrac{1}{n}\right)^n = 1 + 1 + \dfrac{1}{2!}\left(1-\dfrac{1}{n}\right)+\dfrac{1}{3!}\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right) + \ldots \leq 2 + \dfrac{1}{2} + \dfrac{1}{2^2} + \dots =3$ Similarly, $ \begin{align*} \left(1-\dfrac{1}{n^2}\right)^n &= 1 - {n \choose 1}\frac{1}{n^2} + {n \choose 2}\frac{1}{n^4} + \dots\\ &= 1 - \frac{1}{n} + \dfrac{1}{2!n^2}\left(1-\dfrac{1}{n}\right) - \dfrac{1}{3!n^3}\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right) + \ldots \end{align*} $ So, $$ | \left(1-\dfrac{1}{n^2}\right)^{n} -\left(1-\frac{1}{n} \right)| \leq \dfrac{1}{2n^2} + \dfrac{1}{2^2n^2} + \dfrac{1}{2^3n^2} + \ldots = \dfrac{1}{n^2} . $$ Now, $ \begin{align*} \left(1+\frac{1}{n+1}\right)^{n+1} - \left(1+\frac{1}{n}\right)^n &= \left(1+\frac{1}{n}-\frac{1}{n(n+1)}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n \\ &=\left(1+\frac{1}{n}\right)^{n+1}\left\{ \left( 1- \frac{1}{(n+1)^2}\right)^{n+1} - \frac{n}{n+1}\right\}\\ &= \left(1+\frac{1}{n}\right)^{n+1}(1 - \frac{1}{n+1} + \text{O}(\frac{1}{n^2}) - \frac{n}{n+1})\\ & = \left(1+\frac{1}{n}\right)^{n+1}\text{O}(\frac{1}{n^2}) \\ &= \text{O}(\frac{1}{n^2}) \text{ (since $(1+1/n)^{n+1}$ is bounded) }. \end{align*} $ So, letting $a_k = (1+1/k)^k$ we have, $|a_{k+1}-a_k| \leq C/k^2$ for some $C$ and hence, $\sum_{ k \geq n } | a_{k+1} - a_k | \to 0$ as $n \to \infty$. Since $|a_n - a_m| \leq \sum_{ k \geq \min\{m,n\}} |a_{k+1} - a_k|$. So given $\epsilon > 0$ chose $N$ such that $\sum_{ k \geq N} |a_k - a_{k+1}| < \epsilon$ and $|a_n - a_m| < \epsilon $ for $n,m \geq N$.

Variation of Pythagorean triplets: $x^2+y^2 = z^3$ I need to prove that the equation $x^2 + y^2 = z^3$ has infinitely many solutions for positive $x, y$ and $z$. I got to as far as $4^3 = 8^2$ but that seems to be of no help. Can some one help me with it?

the equation: $X^2+Y^2=Z^3$ Has the solutions: $X=2k^6+8tk^5+2(7t^2+8qt-9q^2)k^4+16(t^3+2qt^2-tq^2-2q^3)k^3+$ $+2(7t^4+12qt^3+6q^2t^2-28tq^3-9q^4)k^2+8(t^5+2qt^4-2q^3t^2-5tq^4)k+$ $+2(q^6-4tq^5-5q^4t^2-5q^2t^4+4qt^5+t^6)$ ................................................................................................................................................. $Y=2k^6+4(3q+t)k^5+2(9q^2+16qt+t^2)k^4+32qt(2q+t)k^3+$ $+2(-9q^4+20tq^3+30q^2t^2+12qt^3-t^4)k^2+4(-3q^5-tq^4+10q^3t^2+6q^2t^3+5qt^4-t^5)k-$ $-2(q^6+4tq^5-5q^4t^2-5q^2t^4-4qt^5+t^6)$ ................................................................................................................................................. $Z=2k^4+4(q+t)k^3+4(q+t)^2k^2+4(q^3+tq^2+qt^2+t^3)k+2(q^2+t^2)^2$ $q,t,k$ - What are some integers any sign. After substituting the numbers and get a result it will be necessary to divide by the greatest common divisor. This is to obtain the primitive solutions.

halfspaces question How do I find the supporting halfspace inequality to epigraph of $$f(x) = \frac{x^2}{|x|+1}$$ at point $(1,0.5)$

For $x>0$, we have $$ f(x)=\frac{x^2}{x+1}\quad\Rightarrow\quad f'(x)=\frac{x^2+2x}{(x+1)^2}. $$ Hence $f'(1)=\frac{3}{4}$ and an equation of the tangent to the graph of $f$ at $(1,f(1))$ is $$y=f'(1)(x-1)+f(1)=\frac{3}{4}(x-1)+\frac{1}{2}=\frac{3}{4}x-\frac{1}{4}.$$ An inequality defining the halfspace above this tangent is therefore $$ y\geq \frac{3}{4}x-\frac{1}{4}. $$ See here for a picture of the graph and the tangent.

What is the property that allows $5^{2x+1} = 5^2$ to become $2x+1 = 2$? What is the property that allows $5^{2x+1} = 5^2$ to become $2x+1 = 2$? We just covered this in class, but the teacher didn't explain why we're allowed to do it.

$5^{(2x+1)} = 5^2$ Multiplying by $1/5^2$ om both sides we get, $\frac{5^{(2x+1)}}{5^2} = 1$ $\Rightarrow 5^{(2x+1)-2} = 1$ Taking log to the base 5 on both sides we get $2x+1-2=0$.

Prove $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$ Let $a,b,c$ are non-negative numbers, such that $a+b+c = 3$. Prove that $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$ Here's my idea: $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$ $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(ab + bc + ca)$ $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - 2(ab + bc + ca) \ge 0$ $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - ((a+b+c)^2 - (a^2 + b^2 + c^2) \ge 0$ $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) - (a+b+c)^2 \ge 0$ $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$ $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$ And I'm stuck here. I need to prove that: $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$ or $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a+b+c)$, because $a+b+c = 3$ In the first case using Cauchy-Schwarz Inequality I prove that: $(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$ $3(a^2 + b^2 + c^2) \ge (a+b+c)^2$ Now I need to prove that: $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a^2 + b^2 + c^2)$ $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a^2 + b^2 + c^2)$ $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge a^2 + b^2 + c^2$ I need I don't know how to continue. In the second case I tried proving: $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$ and $a^2 + b^2 + c^2 \ge a+b+c$ Using Cauchy-Schwarz Inequality I proved: $(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$ $(a^2 + b^2 + c^2)(a+b+c) \ge (a+b+c)^2$ $a^2 + b^2 + c^2 \ge a+b+c$ But I can't find a way to prove that $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$ So please help me with this problem. P.S My initial idea, which is proving: $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$ maybe isn't the right way to prove this inequality.

I will use the following lemma (the proof below): $$2x \geq x^2(3-x^2)\ \ \ \ \text{ for any }\ x \geq 0. \tag{$\clubsuit$}$$ Start by multiplying our inequality by two $$2\sqrt{a} +2\sqrt{b} + 2\sqrt{c} \geq 2ab +2bc +2ca, \tag{$\spadesuit$}$$ and observe that $$2ab + 2bc + 2ca = a(b+c) + b(c+a) + c(b+c) = a(3-a) + b(3-b) + c(3-c)$$ and thus $(\spadesuit)$ is equivalent to $$2\sqrt{a} +2\sqrt{b} + 2\sqrt{c} \geq a(3-a) + b(3-b) + c(3-c)$$ which can be obtained by summing up three applications of $(\clubsuit)$ for $x$ equal to $\sqrt{a}$, $\sqrt{b}$ and $\sqrt{c}$ respectively: \begin{align} 2\sqrt{a} &\geq a(3-a), \\ 2\sqrt{b} &\geq b(3-b), \\ 2\sqrt{c} &\geq c(3-c). \\ \end{align} $$\tag*{$\square$}$$ The lemma $$2x \geq x^2(3-x^2) \tag{$\clubsuit$}$$ is true for any $x \geq 0$ (and also any $x \leq -2$) because $$2x - x^2(3-x^2) = (x-1)^2x(x+2)$$ is a polynomial with roots at $0$ and $-2$, a double root at $1$ and a positive coefficient at the largest degree, $x^4$. $\hspace{60pt}$ I hope this helps ;-)

Calculate:$\lim_{x \rightarrow (-1)^{+}}\left(\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}} \right)$ How to calculate following with out using L'Hospital rule $$\lim_{x \rightarrow (-1)^{+}}\left(\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}} \right)$$

Let $\sqrt{\arccos(x)} = t$. We then have $x = \cos(t^2)$. Since $x \to (-1)^+$, we have $t^2 \to \pi^-$. Hence, we have $$\lim_{x \to (-1)^+} \dfrac{\sqrt{\pi} - \sqrt{\arccos(x)}}{\sqrt{1+x}} = \overbrace{\lim_{t \to \sqrt{\pi}^-} \dfrac{\sqrt{\pi} - t}{\sqrt{1+\cos(t^2)}}}^{t = \sqrt{\arccos(x)}} = \underbrace{\lim_{y \to 0^+} \dfrac{y}{\sqrt{1+\cos((\sqrt{\pi}-y)^2)}}}_{y = \sqrt{\pi}-t}$$ $$1+\cos((\sqrt{\pi}-y)^2) = 1+\cos(\pi -2 \sqrt{\pi}y + y^2) = 1-\cos(2 \sqrt{\pi}y - y^2) = 2 \sin^2 \left(\sqrt{\pi} y - \dfrac{y^2}2\right)$$ Hence, \begin{align} \lim_{y \to 0^+} \dfrac{y}{\sqrt{1+\cos((\sqrt{\pi}-y)^2)}} & = \dfrac1{\sqrt2} \lim_{y \to 0^+} \dfrac{y}{\sin \left(\sqrt{\pi}y - \dfrac{y^2}2\right)}\\ & = \dfrac1{\sqrt2} \lim_{y \to 0^+} \dfrac{\left(\sqrt{\pi}y - \dfrac{y^2}2\right)}{\sin \left(\sqrt{\pi}y - \dfrac{y^2}2\right)} \dfrac{y}{\left(\sqrt{\pi}y - \dfrac{y^2}2\right)} = \dfrac1{\sqrt{2 \pi}} \end{align}

Surface integral over ellipsoid I've problem with this surface integral: $$ \iint\limits_S {\sqrt{ \left(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\right)}}{dS} $$, where $$ S = \{(x,y,z)\in\mathbb{R}^3: \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}= 1\} $$

Let the ellipsoid $S$ be given by $${\bf x}(\theta,\phi)=(a\cos\theta\cos\phi,b\cos\theta\sin\phi,c\sin\theta)\ .$$ Then for all points $(x,y,z)\in S$ one has $$Q^2:={x^2\over a^4}+{y^2\over b^4}+{z^2\over c^4}={1\over a^2b^2c^2}\left(\cos^2\theta(b^2c^2\cos^2\phi+a^2c^2\sin^2\phi)+a^2b^2\sin^2\theta\right)\ .$$ On the other hand $${\rm d}S=|{\bf x}_\theta\times{\rm x}_\phi|\>{\rm d}(\theta,\phi)\ ,$$ and one computes $$\eqalign{|{\bf x}_\theta\times{\rm x}_\phi|^2&=\cos^4\theta(b^2c^2\cos^2\phi+a^2c^2\sin^2\phi)+a^2b^2\cos^2\theta\sin^2\theta\cr &=\cos^2\theta\ (a^2b^2c^2\ \ Q^2)\ \cr}$$ It follows that your integral ($=:J$) is given by $$\eqalign{J&=\int\nolimits_{\hat S} Q\ |{\bf x}_\theta\times{\rm x}_\phi|\>{\rm d}(\theta,\phi)=\int\nolimits_{\hat S}abc\ Q^2\ \cos\theta\ {\rm d}(\theta,\phi) \cr &={1\over abc}\int\nolimits_{\hat S}\cos\theta\left(\cos^2\theta(b^2c^2\cos^2\phi+a^2c^2\sin^2\phi)+a^2b^2\sin^2\theta\right)\ {\rm d}(\theta,\phi)\ ,\cr}$$ where $\hat S=[-{\pi\over2},{\pi\over2}]\times[0,2\pi]$. Using $$\int_{-\pi/2}^{\pi/2}\cos^3\theta\ d\theta={4\over3},\quad \int_{-\pi/2}^{\pi/2}\cos\theta\sin^2\theta\ d\theta={2\over3},\quad \int_0^{2\pi}\cos^2\phi\ d\phi=\int_0^{2\pi}\sin^2\phi\ d\phi=\pi$$ we finally obtain $$J={4\pi\over3}\left({ab\over c}+{bc\over a}+{ca\over b}\right)\ .$$

Find a closed form of the series $\sum_{n=0}^{\infty} n^2x^n$ The question I've been given is this: Using both sides of this equation: $$\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$$ Find an expression for $$\sum_{n=0}^{\infty} n^2x^n$$ Then use that to find an expression for $$\sum_{n=0}^{\infty}\frac{n^2}{2^n}$$ This is as close as I've gotten: \begin{align*} \frac{1}{1-x} & = \sum_{n=0}^{\infty} x^n \\ \frac{-2}{(x-1)^3} & = \frac{d^2}{dx^2} \sum_{n=0}^{\infty} x^n \\ \frac{-2}{(x-1)^3} & = \sum_{n=2}^{\infty} n(n-1)x^{n-2} \\ \frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} n(n-1)\frac{x^n}{x}(x+1) \\ \frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} (n^2x + n^2 - nx - n)\frac{x^n}{x} \\ \frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} n^2x^n + n^2\frac{x^n}{x} - nx^n - n\frac{x^n}{x} \\ \end{align*} Any help is appreciated, thanks :)

You've got $\sum_{n\geq 0} n(n-1)x^n$, modulo multiplication by $x^2$. Differentiate just once your initial power series and you'll be able to find $\sum_{n\geq 0} nx^n$. Then take the sum of $\sum_{n\geq 0} n(n-1)x^n$ and $\sum_{n\geq 0} nx^n$. What are the coefficients?

approximation of law sines from spherical case to planar case we know for plane triangle cosine rule is $\cos C=\frac{a^+b^2-c^2}{2ab}$ and on spherical triangle is $ \cos C=\frac{\cos c - \cos a \cos b} {\sin a\sin b}$ suppose $a,b,c<\epsilon$ which are sides of a spherical triangle, and $$|\frac{a^2 +b^2-c^2}{2ab}- \frac{\cos c - \cos a \cos b} {\sin a\sin b}|<Ke^m$$ could any one tell me what will be $m$ and $K$?

Note that for $x$ close to $0$, $$1-\frac{x^2}{2!} \le \cos x\le 1-\frac{x^2}{2!}+\frac{x^4}{4!}$$ and $$x-\frac{x^3}{3!} \le \sin x\le x.$$ (We used the Maclaurin series expansion of $\cos x$ and $\sin x$.) Using these facts on the small angles $a$, $b$, and $c$, we can estimate your difference.

Irreducible components of the variety $V(X^2+Y^2-1,X^2-Z^2-1)\subset \mathbb{C}^3.$ I want to find the irreducible components of the variety $V(X^2+Y^2-1, \ X^2-Z^2-1)\subset \mathbb{C}^3$ but I am completely stuck on how to do this. I have some useful results that can help me decompose $V(F)$ when $F$ is a single polynomial, but the problem seems much harder even with just two polynomials. Can someone please help me? EDIT: In trying to answer this question, I knew it would be useful to know if the ideal $I=(X^2+Y^2-1, X^2-Z^2-1)$ was a prime ideal of $\mathbb{C}[X,Y,Z]$ but I'm finding it hard to describe the quotient ring. Is it a prime ideal?

$\newcommand{\rad}{\text{rad}\hspace{1mm}} $ The ideal $(x^2 + y^2 - 1,x^2 - z^2 - 1)$ is equal to the ideal $(y^2 + z^2 ,x^2 - z^2 - 1)$. This is because \begin{eqnarray*} (y^2 + z^2) + (x^2 - z^2 - 1) &=& y^2 + x^2 - 1\\ (x^2 + y^2 - 1) - (x^2 - z^2 - 1) &=& y^2 + z^2. \end{eqnarray*} Thus we get \begin{eqnarray*} V(x^2 + y^2 - 1,x^2 - z^2 - 1) &=& V( y^2 + z^2,x^2 - z^2 - 1) \\ &=& V\left( (y+zi)(y- zi),x^2 - z^2 - 1\right) \\ &=& V(y+zi,x^2 - z^2 - 1) \cup V(y-zi,x^2 - z^2 - 1).\end{eqnarray*} Now we claim that the ideals $(y+zi,x^2 - z^2 - 1)$ and $(y-zi,x^2 - z^2 - 1)$ are prime ideals. I will only show that the former is prime because the proof for the latter is similar. By the Third Isomorphism Theorem we have \begin{eqnarray*} \Bbb{C}[x,y,z]/(y+ zi,x^2 - z^2 - 1) &\cong& \Bbb{C}[x,y,z]/(y+zi) \bigg/ (y+ zi,x^2 - z^2 - 1)/ (y + zi) \\ &\cong& \Bbb{C}[x,z]/(x^2 - z^2 - 1)\end{eqnarray*} because $\Bbb{C}[x,y,z]/(y + zi) \cong \Bbb{C}[x,z]$. At this point there are two ways to proceed, one of which is showing that $x^2 - z^2 - 1$ is irreducible. However there is a nicer approach which is the following. Writing \begin{eqnarray*} x &=& \frac{x+z}{2} + \frac{x-z}{2} \\ z &=& \frac{z + x}{2} + \frac{z-x}{2}\end{eqnarray*} this shows that $\Bbb{C}[x,z] = \Bbb{C}[x+z,x-z]$. Then upon factoring $x^2 - z^2 - 1$ as $(x+z)(x-z) - 1$ the quotient $\Bbb{C}[x,z]/(x^2 - z^2 - 1)$ is isomorphic to $\Bbb{C}[u][v]/(uv - 1)$ where $u = x+z, v = x-z$. Now recall that $$\left(\Bbb{C}[u] \right)[v]/(uv - 1) \cong \left(\Bbb{C}[u]\right)_{u} $$ where the subscript denotes localisation at the multiplicative set $\{1,u,u^2,u^3 \ldots \}$. Since the localisation of an integral domain is an integral domain, this completes the proof that $(y+zi,x^2 - z^2 - 1)$ is prime and hence a radical ideal. Now use Hilbert's Nullstellensatz to complete the proof that your algebraic set decomposes into irreducibles as claimed in Andrew's answer.

Funny integral inequality Assume $f(x) \in C^1([0,1])$,and $\int_0^{\frac{1}{2}}f(x)\text{d}x=0$,show that: $$\left(\int_0^1f(x)\text{d}x\right)^2 \leq \frac{1}{12}\int_0^1[f'(x)]^2\text{d}x$$ and how to find the smallest constant $C$ which satisfies $$\left(\int_0^1f(x)\text{d}x\right)^2 \leq C\int_0^1[f'(x)]^2\text{d}x$$

solutin 2: by Schwarz,we have $$\int_{0}^{\frac{1}{2}}[f'(x)]^2dx\int_{0}^{\frac{1}{2}}x^2dx\ge\left(\int_{0}^{\frac{1}{2}}xf'(x)dx\right)^2=\left[\dfrac{1}{2}f(\dfrac{1}{2})-\int_{0}^{\frac{1}{2}}f(x)dx\right]^2$$ so $$\int_{0}^{\frac{1}{2}}[f'(x)]^2dx\ge 24\left[\dfrac{1}{2}f(\dfrac{1}{2})-\int_{0}^{\frac{1}{2}}f(x)dx\right]^2$$ the same methods,we have $$\int_{\frac{1}{2}}^{1}[f'(x)]^2dx\ge 24\left[\dfrac{1}{2}f(\dfrac{1}{2})-\int_{0}^{\frac{1}{2}}f(x)dx\right]^2$$ and use $2(a^2+b^2)\ge (a+b)^2$ then we have $$\int_{0}^{1}[f'(x)]^2dx\ge 12\left(\int_{0}^{1}f(x)dx-2\int_{0}^{\frac{1}{2}}f(x)dx\right)^2$$

Find the value of : $\lim_{x\to\infty}x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)=-1$ How can I show/explain the following limit? $$\lim_{x\to\infty} \;x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)=-1$$ Some trivial transformation seems to be eluding me.

The expression can be multiplied with its conjugate and then: $$\begin{align} \lim_{x\to\infty} x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right) &= \lim_{x\to\infty} x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)\left(\frac{\sqrt{x^2-1}+\sqrt{x^2+1}}{\sqrt{x^2-1}+\sqrt{x^2+1}}\right) \cr &=\lim_{x\to\infty} x\left(\frac{x^2-1-x^2-1}{\sqrt{x^2-1}+\sqrt{x^2+1}}\right) \cr &=\lim_{x\to\infty} x\left(\frac{-2}{\sqrt{x^2-1}+\sqrt{x^2+1}}\right) \cr &=\lim_{x\to\infty} \frac{-2}{\frac{\sqrt{x^2-1}}{x} + \frac{\sqrt{x^2+1}}{x}} \cr &=\lim_{x\to\infty} \frac{-2}{\sqrt{\frac{x^2}{x^2}-\frac{1}{x^2}} + \sqrt{\frac{x^2}{x^2}+\frac{1}{x^2}}} \cr &=\lim_{x\to\infty} \frac{-2}{\sqrt{1-0} + \sqrt{1-0}} \cr &=\lim_{x\to\infty} \frac{-2}{1+1} \cr &= -1\end{align}$$

Solve $\frac{1}{2x}+\frac{1}{2}\left(\frac{1}{2x}+\cdots\right)$ If $$\displaystyle \frac{1}{2x}+\frac{1}{2}\left(\frac{1}{2x}+ \frac{1}{2}\left(\frac{1}{2x} +\cdots\right) \right) = y$$ then what is $x$? I was thinking of expanding the brackets and trying to notice a pattern but as it effectively goes to infinity. I don't think I can expand it properly, can I?

Expand. The first term is $\frac{1}{2x}$. The sum of the first two terms is $\frac{1}{2x}+\frac{1}{4x}$. The sum of the first three terms is $\frac{1}{2x}+\frac{1}{4x}+\frac{1}{8x}$. And so on. The sum of the first $n$ terms is $$\frac{1}{2x}\left(1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n-1}}\right).$$ As $n\to\infty$, the inner sum approaches $2$. So the whole thing approaches $\frac{1}{x}$. So we are solving the equation $\frac{1}{x}=y$.

Can the matrix $A=\begin{bmatrix} 0 & 1\\ 3 & 3 \end{bmatrix}$ be diagonalized over $\mathbb{Z}_5$? Im stuck on finding eigenvalues that are in the field please help. Given matrix: $$ A= \left[\begin{matrix} 0 & 1\\ 3 & 3 \end{matrix}\right] $$ whose entries are from $\mathbb{Z}_5 = \{0, 1, 2, 3, 4\}$, find, if possible, matrices $P$ and $D$ over $\mathbb{Z}_5$ such that $P^{−1} AP = D$. I have found the characteristic polynomial: $x^2-3x-3=0$ Since its over $\mathbb{Z}_5$, $x^2-3x-3=x^2+2x+2=0$. But from there I'm not sure how to find the eigenvalues, once I get the eigenvalues that are in the field it will be easy to find the eigenvectors and create the matrix $P$.

yes over $\Bbb Z_5$ because: $\lambda^2 -3\lambda-3=o$ at Z_5 we will have $\Delta=9+12=4+2=6$ (9~4 and 12~2 at Z_5) so $\Delta=1$ and so $\lambda_1=\frac{3+1}{2}=2$ and $\lambda_2=\frac{3-1}{2}=1$ about: $\lambda_1$ we have :$ ( \left[\begin{matrix} 0 & 1\\ 3 &3 \end{matrix}\right]-\left[\begin{matrix} 2 & \\ 0 &2 \end{matrix}\right] )\left[\begin{matrix} x\\ y \end{matrix}\right]=0$ $$-2x+y=0 $$ & $$( 3x+y=0 ~ -2x+y=0 ) $$ and so $$ y=2x $$ is our space of eigen value of $ \lambda_1 =\{(2,4),(0,0)(1,2)\} $ => (dim =1) base={(1,2)} about $\lambda_2$: $ ( \left[\begin{matrix} 0 & 1\\ 3 &3 \end{matrix}\right]-\left[\begin{matrix} 1 & 0\\ 0 &1 \end{matrix}\right] )\left[\begin{matrix} x\\ y \end{matrix}\right]=0$ and so $y=x$ is our answer and eigenvector space of $\lambda_2=\{(0,0)(1,1)(2,2)(3,3)(4,4)\} \implies $ $(\dim=1)$ base ={(1,1)} matrix at base of$ \{(1,1),(1,2)\}$ will be diagonalizable $\left[\begin{matrix} 2 & 0\\ 0 &1 \end{matrix}\right] $

Separating $\frac{1}{1-x^2}$ into multiple terms I'm working through an example that contains the following steps: $$\int\frac{1}{1-x^2}dx$$ $$=\frac{1}{2}\int\frac{1}{1+x} - \frac{1}{1-x}dx$$ $$\ldots$$ $$=\frac{1}{2}\ln{\frac{1+x}{1-x}}$$ I don't understand why the separation works. If I attempt to re-combine the terms, I get this: $$\frac{1}{1+x} \frac{1}{1-x}$$ $$=\frac{1-x}{1-x}\frac{1}{1+x} - \frac{1+x}{1+x}\frac{1}{1-x}$$ $$=\frac{1-x - (1+x)}{1-x^2}$$ $$=\frac{-2x}{1-x^2} \ne \frac{2}{1-x^2}$$ Or just try an example, and plug in $x = 2$: $$2\frac{1}{1-2^2} = \frac{-2}{3}$$ $$\frac{1}{1+2} -\frac{1}{1-2} = \frac{1}{3} + 1 = \frac{4}{3} \ne \frac{-2}{3}$$ Why can $\frac{1}{1-x^2}$ be split up in this integral, when the new terms do not equal the old term?

The thing is $$\frac{1}{1-x}\color{red}{+}\frac 1 {1+x}=\frac{2}{1-x^2}$$ What you might have seen is $$\frac{1}{x-1}\color{red}{-}\frac 1 {x+1}=\frac{2}{1-x^2}$$ Note the denominator is reversed in the sense $1-x=-(x-1)$.

Reasoning about the gamma function using the digamma function I am working on evaluating the following equation: $\log\Gamma(\frac{1}{2}x) - \log\Gamma(\frac{1}{3}x)$ If I'm understanding correctly, the above is an increasing function which can be demonstrated by the following argument using the digamma function $\frac{\Gamma'}{\Gamma}(x) = \int_0^\infty(\frac{e^{-t}}{t} - \frac{e^{-xt}}{1-e^{-t}})$: $\frac{\Gamma'}{\Gamma}(\frac{1}{2}x) - \frac{\Gamma'}{\Gamma'}(\frac{1}{3}x) = \int_0^\infty\frac{1}{1-e^{-t}}(e^{-\frac{1}{3}xt} - e^{-\frac{1}{2}xt})dt > 0 (x > 1)$ Please let me know if this reasoning is incorrect or if you have any corrections. Thanks very much! -Larry

This answer is provided with help from J.M. $\log\Gamma(\frac{1}{2}x) - \log\Gamma(\frac{1}{3}x)$ is an increasing function. This can be shown using this series for $\psi$: The function is increasing if we can show: $\frac{d}{dx}(\log\Gamma(\frac{1}{2}x) - \log\Gamma(\frac{1}{3}x)) > 0$ We can show this using the digamma function $\psi(x)$: $$\frac{d}{dx}(\log\Gamma(\frac{1}{2}x) - \log\Gamma(\frac{1}{3}x)) = \frac{\psi(\frac{1}{2}x)}{2} - \frac{\psi(\frac{1}{3}x)}{3}$$ $$\frac{\psi(\frac{1}{2}x)}{2} - \frac{\psi(\frac{1}{3}x)}{3} = -\gamma + \sum_{k=0}^\infty(\frac{1}{k+1} - \frac{1}{k + {\frac{1}{2}}}) + \gamma - \sum_{k=0}^\infty(\frac{1}{k+1} - \frac{1}{k+\frac{1}{3}})$$ $$= \sum_{k=0}^\infty(\frac{1}{k+\frac{1}{3}} - \frac{1}{k+\frac{1}{2}})$$ Since for all $k\ge 0$: $k + \frac{1}{3} < k + \frac{1}{2}$, it follows that for all $k\ge0$: $\frac{1}{k+\frac{1}{3}} > \frac{1}{k+\frac{1}{2}}$ and therefore: $$\sum_{k=0}^\infty(\frac{1}{k+\frac{1}{3}} - \frac{1}{k+\frac{1}{2}}) > 0.$$

Solve recursive equation $ f_n = \frac{2n-1}{n}f_{n-1}-\frac{n-1}{n}f_{n-2} + 1$ Solve recursive equation: $$ f_n = \frac{2n-1}{n}f_{n-1}-\frac{n-1}{n}f_{n-2} + 1$$ $f_0 = 0, f_1 = 1$ What I have done so far: $$ f_n = \frac{2n-1}{n}f_{n-1}-\frac{n-1}{n}f_{n-2} + 1- [n=0]$$ I multiplied it by $n$ and I have obtained: $$ nf_n = (2n-1)f_{n-1}-(n-1)f_{n-2} + n- n[n=0]$$ $$ \sum nf_n x^n = \sum(2n-1)f_{n-1}x^n-\sum (n-1)f_{n-2}x^n + \sum n x^n $$ $$ \sum nf_n x^n = \sum(2n-1)f_{n-1}x^n-\sum (n-1)f_{n-2}x^n + \frac{1}{(1-z)^2} - \frac{1}{1-z} $$ But I do not know what to do with parts with $n$. I suppose that there can be useful derivation or integration, but I am not sure. Any HINTS?

Let's take a shot at this: $$ f_n - f_{n - 1} = \frac{n - 1}{n} (f_{n - 1} - f_{n - 2}) + 1 $$ This immediately suggests the substitution $g_n = f_n - f_{n - 1}$, so $g_1 = f_1 - f_0 = 1$: $$ g_n - \frac{n - 1}{n} g_{n - 1} = 1 $$ First order linear non-homogeneous recurrence, the summing factor $n$ is simple to see here: $$ n g_n - (n - 1) g_{n - 1} = n $$ Summing: $$ \begin{align*} \sum_{2 \le k \le n} (k g_k - (k - 1) g_{k - 1}) &= \sum_{2 \le k \le n} k \\ n g_n - 1 \cdot g_1 &= \frac{n (n + 1)}{2} - 1 \\ g_n &= \frac{n + 1}{2} \\ f_n - f_{n - 1} &= \frac{n + 1}{2} \\ \sum_{1 \le k \le n} (f_n - f_{n - 1}) &= \sum_{1 \le k \le n} \frac{k + 1}{2} \\ f_n - f_0 &= \frac{1}{2} \left( \frac{n (n + 1)}{2} + n \right) \\ f_n &= \frac{n (n + 3)}{4} \end{align*} $$ Maxima tells me this checks out. Pretty!

Help in proving that $\nabla\cdot (r^n \hat r)=(n+2)r^{n-1}$ Show that$$\nabla \cdot (r^n \hat r)=(n+2)r^{n-1}$$ where $\hat r$ is the unit vector along $\bar r$. Please give me some hint. I am clueless as of now.

You can also use Cartesian coordinates and using the fact that $r \hat{r} = \vec{r} = (x,y,z)$. \begin{align} r^n \hat{r} &= r^{n-1} (x,y,z) \\ \nabla \cdot r^n \hat{r} & = \partial_x(r^{n-1}x) + \partial_y(r^{n-1}y) + \partial_z(r^{n-1}z) \end{align} Each term can be calculated: $\partial_x(r^{n-1}x) = r^{n-1} + x (n-1) r^{n-2} \partial_x r$ $\partial_x r = \frac{x}{r}$. (Here, I used the fact that $r = \sqrt{x^2+y^2+z^2}$.) The terms involving $y$ and $z$ are exactly the same with $y$ and $z$ replacing $x$. And so, \begin{align} \nabla \cdot r^n \hat{r}& = r^{n-1} + x(n-1)r^{n-2}\frac{x}{r} + r^{n-1} + y(n-1)r^{n-2}\frac{y}{r} + r^{n-1} + z(n-1)r^{n-2}\frac{z}{r}\\ & = 3r^{n-1}+(n-1)(x^2 + y^2 + z^2)r^{n-3} \\ & = 3r^{n-1} + (n-1)r^2 r^{n-3} \\ & = (n+2)r^{n-1} \end{align} Of course, polar is the easiest :)

If $J$ is the $n×n$ matrix of all ones, and $A = (l−b)I +bJ$, then $\det(A) = (l − b)^{n−1}(l + (n − 1)b)$ I am stuck on how to prove this by induction. Let $J$ be the $n×n$ matrix of all ones, and let $A = (l−b)I +bJ$. Show that $$\det(A) = (l − b)^{n−1}(l + (n − 1)b).$$ I have shown that it holds for $n=2$, and I'm assuming that it holds for the $n=k$ case, $$(l-b)^{k-1}(a+(k-1)b)$$ but I'm having trouble proving that it holds for the $k+1$ case. Please help.

I think that it would be better to use $J_n$ for the $n \times n$ matrix of all ones, (and similarly $A_n, I_n$) so it is clear what the dimensions of the matrices are. Proof by induction on $n$ that $\det(A_n)=(l-b)^n+nb(l-b)^{n-1}$: When $n=1, 2$, this is easy to verify. We have $\det(A_1)=\det(l)=l=(l-b)^1+b(l-b)^0$ and $\det(A_2)=\det(\begin{array}{ccc} l & b \\ b & l \end{array})=l^2-b^2=(l-b)^2+2b(l-b)$. Suppose that the statement holds for $n=k$. Consider $$A_{k+1}=(l-b)I_{k+1}+bJ_{k+1}=\left(\begin{array}{ccccc} l & b & b & \ldots & b \\ b & l & b & \ldots & b \\ \ldots & \ldots & \ldots &\ldots &\ldots \\ b & b & b & \ldots & l \end{array}\right)$$ Now subtracting the second row from the first gives \begin{align} \det(A_{k+1})& =\det\left(\begin{array}{ccccc} l & b & b & \ldots & b \\ b & l & b & \ldots & b \\ \ldots & \ldots & \ldots &\ldots &\ldots \\ b & b & b & \ldots & l \end{array}\right) \\ & =\det\left(\begin{array}{ccccc} l-b & b-l & 0 & \ldots & 0 \\ b & l & b & \ldots & b \\ \ldots & \ldots & \ldots &\ldots &\ldots \\ b & b & b & \ldots & l \end{array}\right) \\ & =(l-b)\det(A_k)-(b-l)\det\left(\begin{array}{cccccc} b & b & b & b & \ldots & b \\ b & l & b & b & \ldots & b \\ b & b & l & b & \ldots & b \\ \ldots & \ldots & \ldots &\ldots &\ldots & \ldots \\ b & b & b & b & \ldots & l \end{array}\right) \end{align} Now taking the matrix in the last line above and subtracting the first row from all other rows gives an upper triangular matrix: \begin{align} \det\left(\begin{array}{cccccc} b & b & b & b & \ldots & b \\ b & l & b & b & \ldots & b \\ b & b & l & b & \ldots & b \\ \ldots & \ldots & \ldots &\ldots &\ldots & \ldots \\ b & b & b & b & \ldots & l \end{array}\right) & =\det\left(\begin{array}{cccccc} b & b & b & b & \ldots & b \\ 0 & l-b & 0 & 0 & \ldots & 0 \\ 0 & 0 & l-b & 0 & \ldots & 0 \\ \ldots & \ldots & \ldots &\ldots &\ldots & \ldots \\ 0 & 0 & 0 & 0 & \ldots & l-b \end{array}\right) \\ & =b(l-b)^k \end{align} Therefore we have (using the induction hypothesis) \begin{align} \det(A_{k+1}) & =(l-b)\det(A_k)-(b-l)(b(l-b)^k) \\ & =(l-b)((l-b)^k+kb(l-b)^{k-1})+(l-b)(b(l-b)^{k-1}) \\ & =(l-b)^{k+1}+(k+1)b(l-b)^k \end{align} We are thus done by induction.

Find minimum in a constrained two-variable inequation I would appreciate if somebody could help me with the following problem: Q: find minimum $$9a^2+9b^2+c^2$$ where $a^2+b^2\leq 9, c=\sqrt{9-a^2}\sqrt{9-b^2}-2ab$

Maybe this comes to your rescue. Consider $b \ge a \ge 0$ When you expansion of $(\sqrt{9-a^2}\sqrt{9-b^2}-2ab)^2=(9-a^2)(9-b^2)+4a^2b^2-4ab \sqrt{(9-a^2)(9-b^2)}$ This attains minimum when $4ab \sqrt{(9-a^2)(9-b^2)}$ is maximum. Applying AM-GM : $\dfrac{9-a^2+9-b^2}{2} \ge \sqrt{(9-a^2)(9-b^2)} \implies 9- \dfrac{9}{2} \ge \sqrt{(9-a^2)(9-b^2)}$ $\dfrac{a^2+b^2}{2} \ge ab \implies 18 \ge 4ab$

Prove that $\tan(75^\circ) = 2 + \sqrt{3}$ My (very simple) question to a friend was how do I prove the following using basic trig principles: $\tan75^\circ = 2 + \sqrt{3}$ He gave this proof (via a text message!) $1. \tan75^\circ$ $2. = \tan(60^\circ + (30/2)^\circ)$ $3. = (\tan60^\circ + \tan(30/2)^\circ) / (1 - \tan60^\circ \tan(30/2)^\circ) $ $4. \tan (30/2)^\circ = \dfrac{(1 - \cos30^\circ)}{ \sin30^\circ}$ Can this be explained more succinctly as I'm new to trigonometry and a little lost after (2.) ? EDIT Using the answers given I'm almost there: * *$\tan75^\circ$ *$\tan(45^\circ + 30^\circ)$ *$\sin(45^\circ + 30^\circ) / \cos(45^\circ + 30^\circ)$ *$(\sin30^\circ.\cos45^\circ + \sin45^\circ.\cos30^\circ) / (\cos30^\circ.\cos45^\circ - \sin45^\circ.\sin30^\circ)$ *$\dfrac{(1/2\sqrt{2}) + (3/2\sqrt{2})}{(3/2\sqrt{2}) - (1/2\sqrt{2})}$ *$\dfrac{(1 + \sqrt{3})}{(\sqrt{3}) - 1}$ *multiply throughout by $(\sqrt{3}) + 1)$ Another alternative approach: * *$\tan75^\circ$ *$\tan(45^\circ + 30^\circ)$ *$\dfrac{\tan45^\circ + \tan30^\circ}{1-\tan45^\circ.\tan30^\circ}$ *$\dfrac{1 + 1/\sqrt{3}}{1-1/\sqrt{3}}$ *at point 6 in above alternative

You can rather use $\tan (75)=\tan(45+30)$ and plug into the formula by Metin. Cause: Your $15^\circ$ is not so trivial.

Find the Sum $1\cdot2+2\cdot3+\cdots + (n-1)\cdot n$ Find the sum $$1\cdot2 + 2\cdot3 + \cdot \cdot \cdot + (n-1)\cdot n.$$ This is related to the binomial theorem. My guess is we use the combination formula . . . $C(n, k) = n!/k!\cdot(n-k)!$ so . . . for the first term $2 = C(2,1) = 2/1!(2-1)! = 2$ but I can't figure out the second term $3 \cdot 2 = 6$ . . . $C(3,2) = 3$ and $C(3,1) = 3$ I can't get it to be 6. Right now i have something like . . . $$ C(2,1) + C(3,2) + \cdot \cdot \cdot + C(n, n-1) $$ The 2nd term doesn't seem to equal 6. What should I do?

As I have been directed to teach how to fish... this is a bit clunky, but works. Define rising factorial powers: $$ x^{\overline{m}} = \prod_{0 \le k < m} (x + k) = x (x + 1) \ldots (x + m - 1) $$ Prove by induction over $n$ that: $$ \sum_{0 \le k \le n} k^{\overline{m}} = \frac{n^{\overline{m + 1}}}{m + 1} $$ When $n = 0$, it reduces to $0 = 0$. Assume the formula is valid for $n$, and: $$ \begin{align*} \sum_{0 \le k \le n + 1} k^{\overline{m}} &= \sum_{0 \le n} k^{\overline{m}} + (n + 1)^{\overline{m}} \\ &= \frac{n^{\overline{m + 1}}}{m + 1} + (n + 1)^{\overline{m}} \\ &= \frac{n \cdot (n + 1)^{\overline{m}} + (m + 1) (n + 1)^{\overline{m}}} {m + 1} \\ &= \frac{(n + m + 1) \cdot (n + 1)^{\overline{m}}}{m + 1} \\ &= \frac{(n + 1)^{\overline{m + 1}}}{m + 1} \end{align*} $$ By induction, it is valid for all $n$. Defining falling factorial powers: $$ x^{\underline{m}} = \prod_{0 \le k < m} (x - k) = x (x - 1) \ldots (x - m + 1) $$ you get a similar formula for the sum: $$ \sum_{0 \le k \le n} k^{\underline{m}} $$ You can see that $x^{\overline{m}}$ (respectively $x^{\underline{m}}$) is a monic polynomial of degree $m$, so any integral power of $x$ can be expressed as a combination of appropiate factorial powers, and so sums of polynomials in $k$ can also be computed with some work. By the way, the binomial coefficient: $$ \binom{\alpha}{k} = \frac{\alpha^{\underline{k}}}{k!} $$

Evaluating a trigonometric integral using residues Finding the trigonometric integral using the method for residues: $$\int_0^{2\pi} \frac{d\theta}{ a^2\sin^2 \theta + b^2\cos^2 \theta} = \frac{2\pi}{ab}$$ where $a, b > 0$. I can't seem to factor this question I got up to $4/i (z) / ((b^2)(z^2 + 1)^2 - a^2(z^2 - 1)^2 $ I think I should be pulling out $a^2$ and $b^2$ out earlier but not too sure how.

Letting $z = e^{i\theta},$ we get $$\int_0^{2\pi} \frac{1}{a^2\sin^2\theta + b^2\cos^2\theta} d\theta = \int_{|z|=1} \frac{1}{iz} \frac{4}{-a^2(z-1/z)^2+b^2(z+1/z)^2} dz \\= \int_{|z|=1} \frac{1}{iz} \frac{4z^2}{-a^2(z^2-1)^2+b^2(z^2+1)^2} dz = -i\int_{|z|=1} \frac{4z}{-a^2(z^2-1)^2+b^2(z^2+1)^2} dz.$$ Now the location of the four simple poles of the new integrand is given by $$z_{0, 1} = \pm \sqrt{\frac{a+b}{a-b}} \quad \text{and} \quad z_{2, 3} = \pm \sqrt{\frac{a-b}{a+b}}.$$ We now restrict ourselves to the case $a > b > 0,$ so that only $z_{2,3}$ are inside the contour. The residues $w_{2,3}$ are given by $$ w_{2,3} =\lim_{z\to z_{2,3}} \frac{4z}{-2a^2(2z)(z^2-1)+2b^2(2z)(z^2+1)} = \lim_{z\to z_{2,3}} \frac{1}{-a^2(z^2-1)+b^2(z^2+1)} \\ = \lim_{z\to z_{2,3}} \frac{1}{z^2(b^2-a^2)+b^2+a^2}= \frac{1}{-(a-b)^2+b^2+a^2} = \frac{1}{2ab}.$$ It follows that the value of the integral is given by $$-i \times 2\pi i \times 2 \times \frac{1}{2ab} = \frac{2\pi}{ab}.$$ The case $b > a > 0$ is left to the reader. (In fact it is not difficult to see that even though in this case the poles are complex, it is once more the poles $z_{2,3}$ that are inside the unit circle.)

Rank of matrix of order $2 \times 2$ and $3 \times 3$ How Can I calculate Rank of matrix Using echlon Method:: $(a)\;\; \begin{pmatrix} 1 & -1\\ 2 & 3 \end{pmatrix}$ $(b)\;\; \begin{pmatrix} 2 & 1\\ 7 & 4 \end{pmatrix}$ $(c)\;\; \begin{pmatrix} 2 & 1\\ 4 & 2 \end{pmatrix}$ $(d)\;\; \begin{pmatrix} 2 & -3 & 3\\ 2 & 2 & 3\\ 3 & -2 & 2 \end{pmatrix}$ $(e)\;\; \begin{pmatrix} 1 & 2 & 3\\ 3 & 6 & 9\\ 1 & 2 & 3 \end{pmatrix}$ although I have a knowledge of Using Determinant Method to calculate rank of Given matrix. But in exercise it is calculate using echlon form plz explain me in detail Thanks

Follow this link to find your answer If you are left with any doubt after reading this this, feel free to discuss.

Finding the remainder when $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by $7$ Find the remainder when $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by $7$. Please brief about the concept behind this to solve such problems. Thanks.

Using Euler-Fermat's theorem. $\phi(7)=6$ $2^{6} \equiv 1 (\mod 7) \implies2^4.2^{96} \equiv 2(\mod7)$ $3^{6} \equiv 1 (\mod 7) \implies3^4.3^{96} \equiv 4(\mod7)$ $4^{6} \equiv 1 (\mod 7) \implies4^4.4^{96} \equiv 4(\mod7)$ $5^{6} \equiv 1 (\mod 7) \implies5^4.5^{96} \equiv 2(\mod7)$ $2^{100}+3^{100}+4^{100}+5^{100} \equiv 5(\mod 7)$

Graph of $\quad\frac{x^3-8}{x^2-4}$. I was using google graphs to find the graph of $$\frac{x^3-8}{x^2-4}$$ and it gave me: Why is $x=2$ defined as $3$? I know that it is supposed to tend to 3. But where is the asymptote???

Because there is a removable singularity at $x = 2$, there will be no asymptote. You're correct that the function is not defined at $x = 2$. Consider the point $(2, 3)$ to be a hole in the graph. Note that in the numerator, $$(x-2)(x^2 + 2x + 4) = x^3 - 8,$$ and in the denominator $$(x-2)(x+ 2) = x^2 - 4$$ When we simplify by canceling (while recognizing $x\neq 2$), we end with the rational function $$\frac{x^2 + 2x + 4}{x+2}$$ We can confirm that the "hole" at $x = 2$ is a removable singularity by confirming that its limit exists: $$\lim_{x \to 2} \frac{x^2 + 2x + 4}{x+2} = 3$$ In contrast, however, we do see, that there is an asymptote at $x = -2$. We can know this without graphing by evaluating the limit of the function as $x$ approaches $-2$ from the left and from the right: $$\lim_{x \to -2^-} \frac{x^2 + 2x + 4}{x+2} \to -\infty$$ $$\lim_{x \to -2^+} \frac{x^2 + 2x + 4}{x+2} \to +\infty$$ Hence, there exists a vertical asymptote at $x = -2$.

Find a polar representation for a curve. I have the following curve: $(x^2 + y^2)^2 - 4x(x^2 + y^2) = 4y^2$ and I have to find its polar representation. I don't know how. I'd like to get help .. thanks in advance.

Just as the Cartesian has two variables, we will have two variables in polar form: $$x = r\cos \theta,\;\;y = r \sin \theta$$ We can also use the fact that $x^2 + y^2 = (r\cos \theta)^2 + (r\sin\theta)^2 = r^2 \cos^2\theta + r^2\sin^2 \theta = r^2\underbrace{(\sin^2 \theta + \cos^2 \theta)}_{= 1} =r^2$ This gives us $$r^4 - 4r^3\cos \theta - 4r^2 \sin^2\theta = 0 \iff r^2 - 4r\cos\theta - 4\sin^2\theta = 0$$

Evaluating Complex Integral. I am trying to evaluate the following integrals: $$\int\limits_{-\infty}^\infty \frac{x^2}{1+x^2+x^4}dx $$ $$\int\limits_{0}^\pi \frac{d\theta}{a\cos\theta+ b} \text{ where }0<a<b$$ My very limited text has the following substitution: $$\int\limits_0^\infty \frac{\sin x}{x}dx = \frac{1}{2i}\int\limits_{\delta}^R \frac{e^{ix}-e^{-ix}}{x}dx \cdots $$ Is the same of substitution available for the polynomial? Thanks for any help. I apologize in advance for slow responses, I have a disability that limits me to an on-screen keyboard.

For the first one, write $\dfrac{x^2}{1+x^2+x^4}$ as $\dfrac{x}{2(1-x+x^2)} - \dfrac{x}{2(1+x+x^2)}$. Now $$\dfrac{x}{(1-x+x^2)} = \dfrac{x-1/2}{\left(x-\dfrac12\right)^2 + \left(\dfrac{\sqrt3}2 \right)^2} + \dfrac{1/2}{\left(x-\dfrac12\right)^2 + \left(\dfrac{\sqrt3}2 \right)^2}$$ and $$\dfrac{x}{(1-x+x^2)} = \dfrac{x+1/2}{\left(x+\dfrac12\right)^2 + \left(\dfrac{\sqrt3}2 \right)^2} - \dfrac{1/2}{\left(x+\dfrac12\right)^2 + \left(\dfrac{\sqrt3}2 \right)^2}$$ I trust you can take it from here. For the second one, from Taylor series of $\dfrac1{(b+ax)}$, we have $$\dfrac1{(b+a \cos(t))} = \sum_{k=0}^{\infty} \dfrac{(-a)^k}{b^{k+1}} \cos^{k}(t)$$ Now $$\int_0^{\pi} \cos^k(t) dt = 0 \text{ if $k$ is odd}$$ We also have that $$\color{red}{\int_0^{\pi}\cos^{2k}(t) dt = \dfrac{(2k-1)!!}{(2k)!!} \times \pi = \pi \dfrac{\dbinom{2k}k}{4^k}}$$ Hence, $$I=\int_0^{\pi}\dfrac{dt}{(b+a \cos(t))} = \sum_{k=0}^{\infty} \dfrac{a^{2k}}{b^{2k+1}} \int_0^{\pi}\cos^{2k}(t) dt = \dfrac{\pi}{b} \sum_{k=0}^{\infty}\left(\dfrac{a}{2b}\right)^{2k} \dbinom{2k}k$$ Now from Taylor series, we have $$\color{blue}{\sum_{k=0}^{\infty} x^{2k} \dbinom{2k}k = (1-4x^2)^{-1/2}}$$ Hence, $$\color{green}{I = \dfrac{\pi}{b} \cdot \left(1-\left(\dfrac{a}b\right)^2 \right)^{-1/2} = \dfrac{\pi}{\sqrt{b^2-a^2}}}$$

Can't prove this elementary algebra problem $x^2 + 8x + 16 - y^2$ First proof: $(x^2 + 8x + 16) – y^2$ $(x + 4)^2 – y^2$ $[(x + 4) + y][(x + 4) – y]$ 2nd proof where I mess up: $(x^2 + 8x) + (16 - y^2)$ $x(x + 8) + (4 + y)(4 - y)$ $x + 1(x + 8)(4 + y)(4 - y)$ ???? I think I'm breaking one of algebra's golden rules, but I can't find it.

Let us fix your second approach. $$ \begin{align} &x^2+8x+16-y^2\\ =&x^2+8x+(4-y)(4+y)\\ =&x^2+x\big[(4-y)+(4+y)\big]+(4-y)(4+y)\\ =&(x+4-y)(x+4+y). \end{align} $$

Integrating a school homework question. Show that $$\int_0^1\frac{4x-5}{\sqrt{3+2x-x^2}}dx = \frac{a\sqrt{3}+b-\pi}{6},$$ where $a$ and $b$ are constants to be found. Answer is: $$\frac{24\sqrt3-48-\pi}{6}$$ Thank you in advance!

On solving we will find that it is equal to -$$-4\sqrt{3+2x-x^{2}}-\sin^{-1}(\frac{x-1}{2})$$ Now if you put the appropriate limits I guess you'll get your answer. First of all write $$4x-5 = \mu \frac{d(3+2x-x^{2})}{dx}+\tau(3+2x-x^{2})$$ We will find that $\mu=-2$ and $\tau=-1$. $$\int\frac{4x-5}{\sqrt{3+2x-x^{2}}}=\mu\int\frac{d(3+2x-x^{2})}{\sqrt{3+2x-x^{2}}}+\tau\int\frac{dx}{\sqrt{3+2x-x^{2}}}$$$$= -2(2\sqrt{(3+2x-x^{2}})+-1\left(\int\frac{d(x-1)}{\sqrt{2^{2}-(x-1)^{2}}}\right) $$$$=-4\sqrt{3+2x-x^{2}}-\sin^{-1}(\frac{x-1}{2})$$

How to prove $n$ is prime? Let $n \gt 1$ and $$\left\lfloor\frac n 1\right\rfloor + \left\lfloor\frac n2\right\rfloor + \ldots + \left\lfloor\frac n n\right\rfloor = \left\lfloor\frac{n-1}{1}\right\rfloor + \left\lfloor\frac{n-1}{2}\right\rfloor + \ldots + \left\lfloor\frac{n-1}{n-1}\right\rfloor + 2$$ and $\lfloor \cdot \rfloor$ is the floor function. How to prove that $n$ is a prime? Thanks in advance.

You know that $$\left( \left\lfloor\frac n 1\right\rfloor - \left\lfloor\frac{n-1}{1}\right\rfloor \right)+\left( \left\lfloor\frac n2\right\rfloor - \left\lfloor\frac{n-1}{2}\right\rfloor\right) + \ldots + \left( \left\lfloor\frac{n}{n-1}\right\rfloor - \left\lfloor\frac{n-1}{n-1}\right\rfloor\right) + \left\lfloor\frac n n\right\rfloor=+2 \,.$$ You know that $$\left( \left\lfloor\frac n 1\right\rfloor - \left\lfloor\frac{n-1}{1}\right\rfloor \right)=1$$ $$\left\lfloor\frac n n\right\rfloor =1$$ $$\left( \left\lfloor\frac n k\right\rfloor - \left\lfloor\frac{n-1}{k}\right\rfloor\right) \geq 0, \qquad \forall 2 \leq k \leq n-1 \,.$$ Since they add to 2, the last ones must be equal, thus for all $2 \leq k \leq n-1$ we have $$ \left\lfloor\frac n k\right\rfloor - \left\lfloor\frac{n-1}{k}\right\rfloor = 0 \Rightarrow \left\lfloor\frac n k\right\rfloor = \left\lfloor\frac{n-1}{k}\right\rfloor $$ It is easy to prove that this means that $k \nmid n$. Since this is true for all $2 \leq k \leq n-1$, you are done.

Determining power series for $\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}$ I'm looking for the power series for $f(x)=\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}$ My approach: the given function is a combination of two problems. first i made some transformations, so the function looks easier. $$\frac{3x^{2}-4x+9}{(x-1)^2(x+3)})=\frac{3x^{2}-4x+9}{x^3+x^2-5x+3}$$ Now i have two polynomials. i thought the Problem might be even easier, if thinking about the function as: $$\frac{3x^{2}-4x+9}{x^3+x^2-5x+3)}= (3x^{2}-4x+9)\cdot \frac{1}{(x^3+x^2-5x+3)}$$ Assuming the power series of $3x^{2}-4x+9$ is just $3x^{2}-4x+9$ itself. I hoped, i could find the power series by multiplying the series of the both easier functions.. yeah, i am stuck. $\sum\limits_{n=0}^{\infty}a_{n}\cdot x^{n}=(3x^{2}-4x+9)\cdot \ ...?... =$ Solution

You can use the partial fraction decomposition: $$ \frac{3x^{2}-4x+9}{(x-1)^2(x+3)}= \frac{A}{1-x}+\frac{B}{(1-x)^2}+\frac{C}{1+\frac{1}{3}x} $$ and sum up the series you get, which are known. If you do the computation, you find $A=0$, $B=2$ and $C=1$, so $$ \frac{3x^{2}-4x+9}{(x-1)^2(x+3)}= \frac{2}{(1-x)^2}+\frac{1}{1+\frac{1}{3}x} $$ The development of $(1-x)^{-2}$ can be deduced from the fact that $$ \frac{1}{1-x}=\sum_{n\ge0}x^n $$ so, by deriving, we get $$ \frac{1}{(1-x)^2}=\sum_{n\ge1}nx^{n-1}=\sum_{n\ge0}(n+1)x^n $$ The power series for the other term is again easy: $$ \frac{1}{1+\frac{1}{3}x}=\sum_{n\ge0}\frac{(-1)^n}{3^n}x^n $$ so your power series development is $$ \frac{3x^{2}-4x+9}{(x-1)^2(x+3)}= \sum_{n\ge0}\left(2n+2+\frac{(-1)^n}{3^n}\right)x^n $$

Arc length of logarithm function I need to find the length of $y = \ln(x)$ (natural logarithm) from $x=\sqrt3$ to $x=\sqrt8$. So, if I am not mistake, the length should be $$\int^\sqrt8_\sqrt3\sqrt{1+\frac{1}{x^2}}dx$$ I am having trouble calculating the integral. I tried to do substitution, but I still fail to think of a way to integrate it. This is what I have done so far: $$\sqrt{1+\frac{1}{x^2}}=u-\frac{1}{x}$$ $$x=\frac{2u}{u-1}$$ $$dx=\frac{2}{(u-1)^2}du$$ $$\sqrt{1+\frac{1}{x^2}}=u-\frac{1}{x}=u-\frac{1}{2}+\frac{1}{2u}$$ $$\int\sqrt{1+\frac{1}{x^2}}dx=2\int\frac{u-\frac{1}{2}+\frac{1}{2u}}{(u-1)^2}du$$ And I am stuck.

$\int^{\sqrt{8}}_{\sqrt{3}}\sqrt{1+\frac{1}{x^{2}}}dx$=$\int^{\sqrt{8}}_{\sqrt{3}}\frac{\sqrt{1+x^{2}}}{x}dx$=$\int^{\sqrt{8}}_{\sqrt{3}}\frac{1+x^{2}}{x\sqrt{1+x^{2}}}$=$\int^{\sqrt{8}}_{\sqrt{3}}\frac{x}{\sqrt{1+x^{2}}}dx$+ +$\int^{\sqrt{8}}_{\sqrt{3}}\frac{1}{x\sqrt{1+x^{2}}}dx$= $=\frac{1}{2}\int^{\sqrt{8}}_{\sqrt{3}}\frac{(1+x^{2})'}{\sqrt{1+x^{2}}}dx$ $-\int^{\sqrt{8}}_{\sqrt{3}}\frac{(\frac{1}{x})'}{\sqrt{1+\frac{1}{x^{2}}}}dx =$ $\sqrt{1+x^{2}}|^{\sqrt{8}}_{\sqrt{3}}-ln(\frac{1}{x}+ \sqrt{1+\frac{1}{x^{2}}})$ $|^{\sqrt{8}}_{\sqrt{3}}$ $=1+\frac{1}{2}$$ln\frac{3}{2}$

Bernoulli differential equation help? We have the equation $$3xy' -2y=\frac{x^3}{y^2}$$ It is a type of Bernouli differential equation. So, since B. diff equation type is $$y'+P(x)y=Q(x)y^n$$ I modify it a little to: $$y'- \frac{2y}{3x} = \frac{x^2}{3y^2}$$ $$y'-\frac{2y}{3x}=\frac{1}{3}x^2y^{-2}$$ Now I divide both sides by $y^{-2}$. What should I do now?

$$\text{We have $3xy^2 y'-2y^3 = x^3 \implies x (y^3)' - 2y^3 = x^3 \implies \dfrac{(y^3)'}{x^2} + y^3 \times \left(-\dfrac2{x^3}\right) = 1$}$$ $$\text{Now note that }\left(\dfrac1{x^2}\right)' = -\dfrac2{x^3}. \text{ Hence, we have }\dfrac{d}{dx}\left(\dfrac{y^3}{x^2}\right) = 1\implies \dfrac{y^3}{x^2} = x + c$$ $$\text{Hence, the solution to the differential equation is }\boxed{\color{blue}{y^3 = x^3 + cx^2}}$$

Solve $\sqrt{2x-5} - \sqrt{x-1} = 1$ Although this is a simple question I for the life of me can not figure out why one would get a 2 in front of the second square root when expanding. Can someone please explain that to me? Example: solve $\sqrt{(2x-5)} - \sqrt{(x-1)} = 1$ Isolate one of the square roots: $\sqrt{(2x-5)} = 1 + \sqrt{(x-1)}$ Square both sides: $2x-5 = (1 + \sqrt{(x-1)})^{2}$ We have removed one square root. Expand right hand side: $2x-5 = 1 + 2\sqrt{(x-1)} + (x-1)$-- I don't understand? Simplify: $2x-5 = 2\sqrt{(x-1)} + x$ Simplify more: $x-5 = 2\sqrt{(x-1)}$ Now do the "square root" thing again: Isolate the square root: $\sqrt{(x-1)} = \frac{(x-5)}{2}$ Square both sides: $x-1 = (\frac{(x-5)}{2})^{2}$ Square root removed Thank you in advance for your help

To get rid of the square root, denote: $\sqrt{x-1}=t\Rightarrow x=t^2+1$. Then: $$\sqrt{2x-5} - \sqrt{x-1} = 1 \Rightarrow \\ \sqrt{2t^2-3}=t+1\Rightarrow \\ 2t^2-3=t^2+2t+1\Rightarrow \\ t^2-2t-4=0 \Rightarrow \\ t_1=1-\sqrt{5} \text{ (ignored, because $t>0$)},t_2=1+\sqrt{5}.$$ Now we can return to $x$: $$x=t^2+1=(1+\sqrt5)^2+1=7+2\sqrt5.$$

positive Integer value of $n$ for which $2005$ divides $n^2+n+1$ How Can I calculate positive Integer value of $n$ for which $2005$ divides $n^2+n+1$ My try:: $2005 = 5 \times 401$ means $n^2+n+1$ must be a multiple of $5$ or multiple of $401$ because $2005 = 5 \times 401$ now $n^2+n+1 = n(n+1)+1$ now $n(n+1)+1$ contain last digit $1$ or $3$ or $7$ $\bullet $ if last digit of $n(n+1)+1$ not contain $5$. So it is not divisible by $5$ Now how can I calculate it? please explain it to me.

A number of the form $n^2+n+1$ has divisors of the form 3, or any number of $6n+1$, and has a three-place period in base n. On the other hand, there are values where 2005 divides some $n^2+n-1$, for which the divisors are of the form n, 10n+1, 10n+9. This happens when n is 512 or 1492 mod 2005.

Help me prove this inequality : How would I go about proving this? $$ \displaystyle\sum_{r=1}^{n} \left( 1 + \dfrac{1}{2r} \right)^{2r} \leq n \displaystyle\sum_{r=0}^{n+1} \displaystyle\binom{n+1}{r} \left( \dfrac{1}{n+1} \right)^{r}$$ Thank you! I've tried so many things. I've tried finding a series I could compare one of the series to but nada, I tried to change the LHS to a geometric series but that didn't work out, please could someone give me a little hint? Thank you!

Here is the proof that $(1+1/x)^x$ is concave for $x\ge 1$. The second derivative of $(1+1/x)^x$ is $(1+1/x)^x$ times $$p(x)=\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)^2-\frac{1}{x(1+x)^2}$$ Now for $x\ge 1$, we have $$\ln(1+1/x)-\frac{2}{1+x}\le \frac{1}{x}-\frac{2}{1+x}=\frac{1-x}{x(1+x)}\le 0$$ and $$\ln(1+1/x)\ge \frac{1}{x}-\frac{1}{2x^2}\ge 0,$$ so \begin{align*}p(x)&= \ln^2(1+1/x)-\frac{2\ln(1+1/x)}{1+x}+\frac{1}{(1+x)^2}-\frac{1}{x(1+x)^2}\\ &=\ln(1+1/x)(\ln(1+1/x)-2/(1+x))+\frac{x-1}{x(1+x)^2}\\ &\le \left(\frac{1}{x}-\frac{1}{2x^2}\right)\left(\frac{1}{x}-\frac{2}{1+x}\right)+\frac{x-1}{x(1+x)^2}\\ &=-\frac{(x-1)^2}{2x^3(1+x)^2}\le 0 \end{align*} proving that $(1+1/x)^x$ is concave for $x\ge 1$.

find out the value of $\dfrac {x^2}{9}+\dfrac {y^2}{25}+\dfrac {z^2}{16}$ If $(x-3)^2+(y-5)^2+(z-4)^2=0$,then find out the value of $$\dfrac {x^2}{9}+\dfrac {y^2}{25}+\dfrac {z^2}{16}$$ just give hint to start solution.

Hint: What values does the function $x^2$ acquire(positive/negarive)? What is the solution of the equation $x^2=0$? Can you find the solution of the equation $x^2+y^2=0$? Now, what can you say about the equation $(x-3)^2+(y-5)^2+(z-4)^2=0 $? Can you find the values of $x,y,z?$

How prove this $\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$ Prove that $$\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$$ I have some question. Using this, find this integral is not converge, I'm wrong? Thank you everyone

First make the substitution $x=u^2$ to get: $\displaystyle \int _{0}^{\infty }\!\sin \left( x \right) \sin \left( \sqrt {x} \right) {dx}=\int _{0}^{\infty }\!2\,\sin \left( {u}^{2} \right) \sin \left( u \right) u{du}$, $\displaystyle=-\int _{0}^{\infty }\!u\cos \left( u \left( u+1 \right) \right) {du}+ \int _{0}^{\infty }\!u\cos \left( u \left( u-1 \right) \right) {du}$, and changing variable again in the second integral on the R.H.S such that $u\rightarrow u+1$ this becomes: $=\displaystyle\int _{0}^{\infty }\!-u\cos \left( u \left( u+1 \right) \right) {du}+ \int _{-1}^{\infty }\!\left(u+1\right)\cos \left( u \left( u+1 \right) \right) {du} $, $\displaystyle=\int _{0}^{\infty }\!\cos \left( u \left( u+1 \right) \right) {du}+ \int _{-1}^{0}\! \left( u+1 \right) \cos \left( u \left( u+1 \right) \right) {du} $. Now we write $u=v-1/2$ and this becomes: $\displaystyle\int _{1/2}^{\infty }\!\cos \left( {v}^{2}-1/4 \right) {dv}+\int _{-1/ 2}^{1/2}\! \left( v+1/2 \right) \cos \left( {v}^{2}-1/4 \right) {dv}=$ $\displaystyle \left\{\int _{0}^{\infty }\!\cos \left( {v}^{2}-1/4 \right) {dv}\right\}$ $\displaystyle +\left\{\int _{-1/2} ^{1/2}\!v\cos \left( {v}^{2}-1/4 \right) {dv}+\int _{-1/2}^{0}\!1/2\, \cos \left( {v}^{2}-1/4 \right) {dv}-1/2\,\int _{0}^{1/2}\!\cos \left( {v}^{2}-1/4 \right) {dv}\right\},$ but the second curly bracket is zero by symmetry and so: $\displaystyle \int _{0}^{\infty }\!\sin \left( x \right) \sin \left( \sqrt {x} \right) {dx}=\displaystyle \int _{0}^{\infty }\!\cos \left( {v}^{2}-1/4 \right) {dv}$, $\displaystyle =\int _{0}^{ \infty }\!\cos \left( {v}^{2} \right) {dv}\cos \left( 1/4 \right) + \int _{0}^{\infty }\!\sin \left( {v}^{2} \right) {dv}\sin \left( 1/4 \right) $. We now quote the limit of Fresnel integrals: $\displaystyle\int _{0}^{\infty }\!\cos \left( {v}^{2} \right) {dv}=\int _{0}^{ \infty }\!\sin \left( {v}^{2} \right) {dv}=\dfrac{\sqrt{2\pi}}{4}$, to obtain: $\displaystyle \int _{0}^{\infty }\!\sin \left( x \right) \sin \left( \sqrt {x} \right) {dx}=\dfrac{\sqrt{2\pi}}{4}\left(\cos\left(\dfrac{1}{4}\right)+\sin\left(\dfrac{1}{4}\right)\right)=\dfrac{\sqrt{\pi}}{2}\sin{\left(\dfrac{3\pi-1}{4}\right)}$.

Show that no number of the form 8k + 3 or 8k + 7 can be written in the form $a^2 +5b^2$ I'm studying for a number theory exam, and have got stuck on this question. Show that no number of the form $8k + 3$ or $8k + 7$ can be written in the form $a^2 +5b^2$ I know that there is a theorem which tells us that $p$ is expressible as a sum of $2$ squares if $p\equiv$ $1\pmod 4$. This is really all I have found to work with so far, and I'm not really sure how/if it relates. Many thanks!

$8k+3,8k+7$ can be merged into $4c+3$ where $k,c$ are integers Now, $a^2+5b^2=4c+3\implies a^2+b^2=4c+3-4b^2=4(c-b^2)+3\equiv3\pmod 4,$ But as $(2c)^2\equiv0\pmod 4,(2d+1)^2\equiv1\pmod 4,$ $a^2+b^2\equiv0,1,2\pmod 4\not\equiv3$ Clearly, $a^2+5b^2$ in the question can be generalized $(4m+1)x^2+(4n+1)y^2$ where $m,n$ are any integers

Solve the equation $\sqrt{3x-2} +2-x=0$ Solve the equation: $$\sqrt{3x-2} +2-x=0$$ I squared both equations $$(\sqrt{3x-2})^2 (+2-x)^2= 0$$ I got $$3x-2 + 4 -4x + x^2$$ I then combined like terms $x^2 -1x +2$ However, that can not be right since I get a negative radicand when I use the quadratic equation. $x = 1/2 \pm \sqrt{((-1)/2)^2 -2}$ The answer is 6

$$\sqrt{3x-2} +2-x=0$$ Isolating the radical:$$\sqrt{3x-2} =-2+x$$ Squaring both sides:$$\bigg(\sqrt{3x-2}\bigg)^2 =\bigg(-2+x\bigg)^2$$ Expanding $(-2+x)^2$ and gathering like terms: $$3x-2=-2(-2+x)+x(-2+x)$$ $$3x-2=4-2x-2x+x^2$$ Set x equal to zero:$$3x-2=4-4x+x^2$$ Gather like terms:$$0=4+2-3x-4x+x^2$$ Factor the quadratic and find the solutions:$$0=x^2-7x+6$$ $$0=(x-6)(x-1)$$ $$0=x-6\implies\boxed{6=x}$$ $$0=x-1\implies\boxed{1=x}$$ Checking 6 as a solution: $$\sqrt{3(6)-2} +2-(6)=0$$ $$\sqrt{16} +2-6=0$$ $$4+2-6=0$$ $$6-6=0$$ Checking 1 as a solution: $$\sqrt{3(1)-2} +2-(1)=0$$ $$\sqrt{1} +2-1=0$$ $$2\neq0$$ The solution $x=1$ does not equal zero and therefore is not a solution. The solution $x=6$ does equal zero and therefore is our only solution to this equation.

A problem on matrices: Find the value of $k$ If $ \begin{bmatrix} \cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\ \sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7} \\ \end{bmatrix}^k = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} $, then the least positive integral value of $k$ is? Actually, I got no idea how to solve this. I did trial and error method and got 7 as the answer. how do i solve this? Can you please offer your assistance? Thank you

Powers of matrices should always be attacked with diagonalization, if feasible. Forget $2\pi/7$, for the moment, and look at $$ A=\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix} $$ whose characteristic polynomial is, easily, $p_A(X)=1-2X\cos\alpha+X^2$. The discriminant is $4(\cos^2\alpha-1)=4(i\sin\alpha)^2$, so the eigenvalues of $A$ are \begin{align} \lambda&=\cos\alpha+i\sin\alpha\\ \bar{\lambda}&=\cos\alpha-i\sin\alpha \end{align} Finding the eigenvectors is easy: $$ A-\lambda I= \begin{bmatrix} -i\sin\alpha & -\sin\alpha\\ i\sin\alpha & \sin\alpha \end{bmatrix} $$ and an eigenvector is $v=[-i\quad 1]^T$. Similarly, an eigenvector for $\bar{\lambda}$ is $w=[i\quad 1]^T$. If $$ S=\begin{bmatrix}-i & i\\1 & 1\end{bmatrix} $$ you get immediately that $$ S^{-1}=\frac{i}{2}\begin{bmatrix}1 & -i\\-1 & -i\end{bmatrix} $$ so, by well known rules, $$ A=SDS^{-1} $$ where $$ D= \begin{bmatrix} \cos\alpha+i\sin\alpha & 0 \\ 0 & \cos\alpha-i\sin\alpha \end{bmatrix}. $$ By De Moivre's formulas, you have $$ D^k= \begin{bmatrix} \cos(k\alpha)+i\sin(k\alpha) & 0 \\ 0 & \cos(k\alpha)-i\sin(k\alpha) \end{bmatrix}. $$ Since $A^k=S D^k S^{-1}$ your problem is now to find the minimum $k$ such that $\cos(k\alpha)+i\sin(k\alpha)=1$, that is, for $\alpha=2\pi/7$, $$ \begin{cases} \cos k(2\pi/7)=1\\ \sin k(2\pi/7)=0 \end{cases} $$ and you get $k=7$. This should not be a surprise, after all: the effect of $A$ on vectors is exactly rotating them by the angle $\alpha$. If you think to the vector $v=[x\quad y]^T$ as the complex number $z=x+iy$, when you do $Av$ you get $$ Av= \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} =\begin{bmatrix} x\cos\alpha-y\sin\alpha\\x\sin\alpha+y\cos\alpha \end{bmatrix} $$ and $$ (x\cos\alpha-y\sin\alpha)+i(x\sin\alpha+y\cos\alpha)= (x+iy)(\cos\alpha+i\sin\alpha)=\lambda z $$ (notation as above). Thus $z$ is mapped to $\lambda z$, which is just $z$ rotated by an angle $\alpha$.

Listing subgroups of a group I made a program to list all the subgroups of any group and I came up with satisfactory result for $\operatorname{Symmetric Group}[3]$ as $\left\{\{\text{Cycles}[\{\}]\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left( \begin{array}{cc} 1 & 2 \\ \end{array} \right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left( \begin{array}{cc} 1 & 3 \\ \end{array} \right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left( \begin{array}{cc} 2 & 3 \\ \end{array} \right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left( \begin{array}{ccc} 1 & 2 & 3 \\ \end{array} \right)\right],\text{Cycles}\left[\left( \begin{array}{ccc} 1 & 3 & 2 \\ \end{array} \right)\right]\right\}\right\}$ It excludes the whole set itself though it can be added seperately. But in case of $SymmetricGroup[4]$ I am getting following $\left\{\{\text{Cycles}[\{\}]\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left( \begin{array}{cc} 1 & 2 \\ \end{array} \right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left( \begin{array}{cc} 1 & 3 \\ \end{array} \right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left( \begin{array}{cc} 1 & 4 \\ \end{array} \right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left( \begin{array}{cc} 2 & 3 \\ \end{array} \right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left( \begin{array}{cc} 2 & 4 \\ \end{array} \right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left( \begin{array}{cc} 3 & 4 \\ \end{array} \right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left( \begin{array}{cc} 1 & 2 \\ 3 & 4 \\ \end{array} \right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left( \begin{array}{cc} 1 & 3 \\ 2 & 4 \\ \end{array} \right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left( \begin{array}{cc} 1 & 4 \\ 2 & 3 \\ \end{array} \right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left( \begin{array}{ccc} 1 & 2 & 3 \\ \end{array} \right)\right],\text{Cycles}\left[\left( \begin{array}{ccc} 1 & 3 & 2 \\ \end{array} \right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left( \begin{array}{ccc} 1 & 2 & 4 \\ \end{array} \right)\right],\text{Cycles}\left[\left( \begin{array}{ccc} 1 & 4 & 2 \\ \end{array} \right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left( \begin{array}{ccc} 1 & 3 & 4 \\ \end{array} \right)\right],\text{Cycles}\left[\left( \begin{array}{ccc} 1 & 4 & 3 \\ \end{array} \right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left( \begin{array}{ccc} 2 & 3 & 4 \\ \end{array} \right)\right],\text{Cycles}\left[\left( \begin{array}{ccc} 2 & 4 & 3 \\ \end{array} \right)\right]\right\}\right\}$ The matrix form shows double transposition. Can someone please check for me if I am getting appropriate results? I doubt I am!!

I have the impression that you only list the cyclic subgroups. For $S_3$, the full group $S_3$ ist missing as a subgroup (you are mentioning that in your question). For $S_4$, several subgroups are missing. In total, there should be $30$ of them. $14$ of them are cyclic, which are exactly the ones you listed. To give you a concrete example, the famous Klein Four subgroup $$\{\operatorname{id},(12)(34),(13)(24),(14)(23)\}$$ is not contained in your list.

Solving $\sqrt{3x^2-2}+\sqrt[3]{x^2-1}= 3x-2 $ How can I solve the equation $$\sqrt{3x^2-2}+\sqrt[3]{x^2-1}= 3x-2$$ I know that it has two roots: $x=1$ and $x=3$.

Substituting $x = \sqrt{t^3+1}$ and twice squaring, we arrive to the equation $$ 36t^6-24t^5-95t^4+8t^3+4t^2-48t=0.$$ Its real roots are $t=0$ and $t=2$ (the latter root is found in the form $\pm \text{divisor}(48)/\text{divisor}(36)$), therefore $x=1$ and $x=3$.

How to find the integral of implicitly defined function? Let $a$ and $b$ be real numbers such that $ 0<a<b$. The decreasing continuous function $y:[0,1] \to [0,1]$ is implicitly defined by the equation $y^a-y^b=x^a-x^b.$ Prove $$\int_0^1 \frac {\ln (y)} x \, dx=- \frac {\pi^2} {3ab}. $$

OK, at long last, I have a solution. Thanks to @Occupy Gezi and my colleague Robert Varley for getting me on the right track. As @Occupy Gezi noted, some care is required to work with convergent integrals. Consider the curve $x^a-x^b=y^a-y^b$ (with $y(0)=1$ and $y(1)=0$). We want to exploit the symmetry of the curve about the line $y=x$. Let $x=y=\tau$ be the point on the curve where $x=y$, and let's write $$\int_0^1 \ln y \frac{dx}x = \int_0^\tau \ln y \frac{dx}x + \int_\tau^1 \ln y \frac{dx}x\,.$$ We make a change of coordinates $x=yu$ to do the first integral: Since $\dfrac{dx}x = \dfrac{dy}y+\dfrac{du}u$, we get (noting that $u$ goes from $0$ to $1$ as $x$ goes from $0$ to $\tau$) \begin{align*} \int_0^\tau \ln y \frac{dx}x &= -\int_\tau^1 \ln y \frac{dy}y + \int_0^1 \ln y \frac{du}u \\ &= -\frac12(\ln y)^2\Big]_\tau^1 + \int_0^1 \ln y \frac{du}u = \frac12(\ln\tau)^2 + \int_0^1 \ln y \frac{du}u\,. \end{align*} Next, note that as $(x,y)\to (1,0)$ along the curve, $(\ln x)(\ln y)\to 0$ because, using $(\ln x)\ln(1-x^{b-a}) = (\ln y)\ln(1-y^{b-a})$, we have $$(\ln x)(\ln y) \sim \frac{(\ln y)^2\ln(1-y^{b-a})}{\ln (1-x^{b-a})} \sim \frac{(\ln y)^2 y^{b-a}}{a\ln y} = \frac1a y^{b-a}\ln y\to 0 \text{ as } y\to 0.$$ We now can make the "inverse" change of coordinates $y=xv$ to do the second integral. This time we must do an integration by parts first. \begin{align*} \int_\tau^1 \ln y \frac{dx}x &= (\ln x)(\ln y)\Big]_{(x,y)=(\tau,\tau)}^{(x,y)=(1,0)} + \int_0^\tau \ln x \frac{dy}y \\ & = -(\ln\tau)^2 + \int_0^\tau \ln x \frac{dy}y \\ &= -(\ln\tau)^2 - \int_\tau^1 \ln x \frac{dx}x + \int_0^1 \ln x \frac{dz}z \\ &= -\frac12(\ln\tau)^2 + \int_0^1 \ln x \frac{dz}z\,. \end{align*} Thus, exploiting the inherent symmetry, we have $$\int_0^1 \ln y\frac{dx}x = \int_0^1 \ln y \frac{du}u + \int_0^1 \ln x \frac{dz}z = 2\int_0^1 \ln x \frac{dz}z\,.$$ Now observe that \begin{multline*} x^a-x^b=y^a-y^b \implies x^a(1-x^{b-a}) = x^az^a(1-x^{b-a}z^{b-a}) \\ \implies x^{b-a} = \frac{1-z^a}{1-z^b}\,, \end{multline*} and so, doing easy substitutions, \begin{align*} \int_0^1 \ln x \frac{dz}z &= \frac1{b-a}\left(\int_0^1 \ln(1-z^a)\frac{dz}z - \int_0^1 \ln(1-z^b)\frac{dz}z\right) \\ &=\frac1{b-a}\left(\frac1a\int_0^1 \ln(1-w)\frac{dw}w - \frac1b\int_0^1 \ln(1-w)\frac{dw}w\right) \\ &= \frac1{ab}\int_0^1 \ln(1-w)\frac{dw}w\,. \end{align*} By expansion in power series, one recognizes that this dilogarithm integral gives us, at long last, $$\int_0^1 \ln y\frac{dx}x = \frac2{ab}\int_0^1 \ln(1-w)\frac{dw}w = \frac2{ab}\left(-\frac{\pi^2}6\right) = -\frac{\pi^2}{3ab}\,.$$ (Whew!!)

Find all positive integers $x$ such that $13 \mid (x^2 + 1)$ I was able to solve this by hand to get $x = 5$ and $x =8$. I didn't know if there were more solutions, so I just verified it by WolframAlpha. I set up the congruence relation $x^2 \equiv -1 \mod13$ and just literally just multiplied out. This lead me to two questions: * *But I was wondering how would I do this if the $x$'s were really large? It doesn't seem like multiplying out by hand could be the only possible method. *Further, what if there were 15 or 100 of these $x$'s? How do I know when to stop?

Starting with $2,$ the minimum natural number $>1$ co-prime with $13,$ $2^1=2,2^2=4,2^3=8,2^4=16\equiv3,2^5=32\equiv6,2^6=64\equiv-1\pmod{13}$ As $2^6=(2^3)^2,$ so $2^3=8$ is a solution of $x^2\equiv-1\pmod{13}$ Now, observe that $x^2\equiv a\pmod m\iff (-x)^2\equiv a$ So, $8^2\equiv-1\pmod {13}\iff(-8)^2\equiv-1$ Now, $-8\equiv5\pmod{13}$ If we need $x^2\equiv-1\pmod m$ where integer $m=\prod p_i^{r_i}$ where $p_i$s are distinct primes and $p_i\equiv1\pmod 4$ for each $i$ (Proof) $\implies x^2\equiv-1\pmod {p_i^{r_i}}$ Applying Discrete logarithm with respect to any primitive root $g\pmod {p_i^{r_i}},$ $2ind_gx\equiv \frac{\phi(p_i^{r_i})}2 \pmod {\phi(p_i^{r_i})}$ as if $y\equiv-1\pmod {p_i^{r_i}}\implies y^2\equiv1 $ $\implies 2ind_gy\equiv0 \pmod {\phi(p_i^{r_i})}\implies ind_gy\equiv \frac{\phi(p_i^{r_i})}2 \pmod {\phi(p_i^{r_i})}$ as $y\not\equiv0\pmod { {\phi(p_i^{r_i})}}$ Now apply CRT, for relatively prime moduli $p_i^{r_i}$ For example, if $m=13, \phi(13)=12$ and $2$ is a primitive root of $13$ So, $2ind_2x\equiv 6\pmod {12}\implies ind_2x=3\pmod 6$ $\implies x=2^3\equiv8\pmod{13}$ and $x=2^9=2^6\cdot2^3\equiv(-1)8\equiv-8\equiv5\pmod{13}$