Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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The Decision of three methods of the solutions $dx/P=dy/Q=dz/R$ Question:
(A) $$\frac{adx}{(b-c)yz}= \frac{bdy}{(c-a)xz}=\frac{cdz}{(a-b)xy}$$
(B) $$\frac{dx}{xz-y}=\frac{dy}{yz-x}=\frac{dz}{1-z^2}$$
These are simultaneous diff eq. of the first order and the first degree in three variables
And I know that three metho... | A.
$$\frac{a dx}{(b -c ) yz} = \frac{b dy}{(c -a ) xz} = \frac{c dx}{(a -b ) xy}$$
Now multiply the numerator and denominator first fraction by $x$ for second fraction by $y$ and third fraction by $z$ and add all of them and get it
$$\frac{ax dx}{(b-c)xyz} = \frac{by dy}{(c-a)xz} = \frac{cz dz}{(a-b)xy} = \frac{ax dx ... | {
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"timestamp": "2023-03-29T00:00:00",
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integer ordered pair,s $(x,y)$ in $1!+2!+3!+............+x! =y^3$ (1) Total no. of integer ordered pair,s $(x,y)$ in $1!+2!+3!+............+x! =y^2$
(2) Total no. of integer ordered pair,s $(x,y)$ in $1!+2!+3!+............+x! =y^3$
(3) Total no. of integer ordered pair,s $(x,y)$ in $1!+2!+3!+............+x! =y^4$
where... | For Q1, you have already noticed that for $x = 4$, the LHS ends in $3$, and for $x > 4$, the terms that are added have last digit $0$. As no square ends in $3$, we do not have any solution for $x \ge 4$.
For Q2, note that if $x \ge 9$, the LHS is divisible by $3$, so if this is the cube of an integer, that integer mus... | {
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real analysis The e number How can show
$$\frac{{{{\left( {1 + \frac{1}{n}} \right)}^{{n^2}}}}}{{{{\left( {1 + \frac{1}{{n + 1}}} \right)}^{{{(n + 1)}^2}}}}} < \frac{1}{{{{\left( {1 + \frac{1}{n}} \right)}^n}}}$$
| Using Bernoulli's Inequality,
$$
\begin{align}
\frac{\left(1+\frac1{n+1}\right)^{(n+1)^2}}{\left(1+\frac1n\right)^{n^2}}
&=\left(\frac{n(n+2)}{(n+1)^2}\right)^{(n+1)^2}\left(1+\frac1n\right)^{2n+1}\\
&=\left(1-\frac1{(n+1)^2}\right)^{(n+1)^2}\left(1+\frac1n\right)^{n+1}\left(1+\frac1n\right)^n\\
&\ge\left(1-\frac1{n+1}... | {
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"timestamp": "2023-03-29T00:00:00",
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system of equations $\sqrt{x}+y = 11$ and $x+\sqrt{y} = 7$. If $x,y\in \mathbb{R}$ and $\sqrt{x}+y = 11\;$ and $x+\sqrt{y} = 7$. Then $(x,y) = $
$\underline{\bf{My\;\; Try::}}$ Let $x=a^2$ and $y=b^2$, Then equation is $a+b^2 = 11$ and $a^2+b = 7$.
$(a+b)+(a+b)^2-2ab = 18$ and Now Let $a+b=S$ and $ab=P$, we get $S+S^2-... | You were on the right track, all you need to do is subtract rather than add.
Doing that will give you $(a+b^2)-(a^2+b)=4$
Which can become
$(b^2-a^2)-(b-a)=4=(b-a)(b+a)-(b-a)=(b-a)(b+a-1)$
Since the only way to can obtain 4 by multiplying is either $2*2$ or $4*1$, either
$b-a=2$ and $b+a-1=2$
or
$b-a=1$ and $b+a-1=4$
i... | {
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Integral $ \int _0^6 \lfloor x \rfloor \sin( \frac {6x}{\pi}) \ \mathrm dx $ Question
$ \int _0^6 \lfloor x \rfloor \sin( \frac {6x}{\pi}) \ \mathrm dx $ = ?
we tried to bound it from both sides using $x$ and $(x-1)$, which yield nice estimation ($\frac {24}\pi$) - ($\frac {36}\pi$) but not a precise one.
we also trie... | x is real number, m is integer, and $\mathbb{Z}$ is the set of integers (positive, negative, and zero).
$$\lfloor x \rfloor=\max\, \{m\in\mathbb{Z}\mid m\le x\}$$
So
$$\begin{alignat*}{1}
\int_{0}^{6}\lfloor x\rfloor\sin(\frac{6x}{\pi})\ \mathrm{d}x= & \int_{0}^{1}0.\sin(\frac{6x}{\pi})\ \mathrm{d}x+\int_{1}^{2}1.\sin(... | {
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Prove that $1<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}$ Prove that $1<\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n+1}$.
By using the Mathematical induction. Suppose the statement holds for $n=k$.
Then for $n=k+1$. We have $\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}+\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{... | By cauchy-schwarz inequality
$$\left(\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{3n+1}\right)(n+1+n+2+\cdots+3n+1)>(1+1+\cdots+1)^2=(2n+1)^2$$
note$$ (n+1+n+2+\cdots+3n+1)=\dfrac{(n+1+3n+1)(2n+1)}{2}=(2n+1)^2$$
| {
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If $a^3 + b^3 +3ab = 1$, find $a+b$
Given that the real numbers $a,b$ satisfy $a^3 + b^3 +3ab = 1$, find $a+b$.
I tried to factorize it but unable to do it.
| Hint: \begin{align} x^3+y^3+z^3-3xyz& =(x+y+z)(x^2+y^2+z^2-xy-xz-yz) \\ &=(x+y+z)\left(\frac{(x-y)^2+(x-z)^2+(y-z)^2}{2}\right) \end{align}
Solution:
$$0=a^3+b^3+(-1)^3-3(a)(b)(-1)=(a+b-1)\left(\frac{(a-b)^2+(a+1)^2+(b+1)^2}{2}\right)$$
so $a+b=1$ or $a=b=-1$. The latter gives $a+b=-2$.
| {
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Prove that for $\cos (\alpha ) = \frac{1}{3}$, $\alpha < \frac{\pi}{2} - \frac{1}{3}$ I have the following question in a mock exam:
$\beta = \frac{\pi}{2} - \alpha$, show that $\beta > \frac{1}{3}$
From the earlier part of the question we know that $\alpha$ is an angle between two vectors (so it's between $0$ and $\pi$... | You mean that: if $0<\alpha<\frac{\pi}{2}$ and $\cos\alpha=\frac{1}{3}$, then $\alpha<\frac{\pi}{2}-\frac{1}{3}$?
Let $F(x)=\cos x-\frac{1}{3}$, then $F(0)=\frac{2}{3}$ and $F(\frac{\pi}{2}-\frac{1}{2})=\sin\frac{1}{3}-\frac{1}{3}<0$ since $\sin x<x (0<x<\frac{\pi}{2})$.
This shows that $\alpha<\frac{\pi}{2}-\frac{1}{3... | {
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Show $g(z+1) = zg(z)$ This is for homework, and I am in need of a hint. Given the product
$$ g(z) = \prod_{k=1}^{\infty} \frac{k}{z+k}\left( 1 + \frac{1}{k} \right)^z, $$
I am trying to show that $g(z+1) = zg(z)$. Here is what I have so far.
I tried to get a better sense of $g$, and wrote
$$ g(z) = \frac{1}{z+1}\left... | Notice that $g(z+1)$ is the product for $g(z)$ "shifted" by one: that is,
$$ g(z+1) = \prod_{k=2}^\infty \frac{k}{z+k}\left(1+\frac{1}{k-1}\right)^z.$$
We can try to rewrite this in terms of the product $\prod\frac{1}{z+k}(1+\frac{1}{k})^z$:
$$\prod_{k=2}^\infty \frac{k}{z+k}\left(1+\frac{1}{k-1}\right)^z = \frac{z+1}... | {
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Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$
Question : Is the following true for any $m\in\mathbb N$?
$$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$
Motivation : I reached $(\star)$ by using computer. It seems true, but I can't prove it... | Note that
$$\frac{1}{\sin^2(x)}=\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}$$
using this identity we can write
$$\begin{align}\sum_{k=0}^{m-1}\frac{1}{\sin^2(\frac{x+k\pi}{m})}&=\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{(\frac{x+k\pi}{m}+n\pi)^2}\\
&=\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{\frac{(x+k\pi+mn\pi)^... | {
"language": "en",
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perfect squares possible? If we let a, b, c, d, and x be integers is it possible that $$x^2+a^2 = (x+1)^2 + b^2 = (x+2)^2 + c^2 = (x+3)^2 + d^2$$
My initial thought is no way! I tried expanding and simplifying, getting $$a^2 = 2x+1 + b^2 = 4x+4 + c^2 = 6x+9 + d^2$$.
It seems that the difference between these perfect sq... | I get it. Mod 4 suffices. Well, mod 8 anyway. One of $x,x+1,x+2,x+3$ is divisible by 4. If its matching partner out of $a,b,c,d$ is even, the common sum is divisible by 4, and we actually cannot have any odd numbers involved, which cannot occur with consecutive $x,x+1,x+2,x+3.$ Required detail: the sum of two odd squar... | {
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What is the last digit of $\operatorname{lcm}(3^{2003}-1,3^{2003}+1)$? What is the last digit of $\operatorname{lcm}(3^{2003}-1,3^{2003}+1)$?
I am able to find out that LCM is $\dfrac{3^{4006}-1}2$. Since $3^{4006}$ has last digit as $8$, now second last digit can be anything from $0-9$. Based on that second last digit... | The two numbers have GCD equal to $2$. So, you're basically asking for the last digit of $(3^{2003}-1)(3^{2003}+1)/2 = (3^{4006}-1)/2$.
By Fermat's theorem, $3^{10 \cdot 4} \equiv 1 \pmod{100}$, since: $$\phi(100) = \phi(5^2 \cdot 2^2) = 5 \cdot 2 \cdot (5-1) \cdot (2-1) = 40 .$$
Thus, the two last digits of $(3^{4006}... | {
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Complex number: Roots
Solve all the roots of the following equation: $$(z-i)^2(z+i)^2=\frac{1}{4}.$$ Find the set of complex numbers $z$ such that $$\left|\frac{z-3}{z+3}\right|=2.$$
Would anyone mind telling me how to solve the above problems? I really have no idea.
| The first one:
$$(z-i)^2(z+i)^2 = \left( (z-i)(z+i) \right)^2 = (z^2 - i^2)^2 = (z^2+1)^2,$$
so
$$z^2+1 = \pm\frac{1}{2}.$$
For the second one, write $z = x + iy$, so $|z-3| = 2|z+3|$, which is equivalent to $|z-3|^2 = 4|z+3|^2$, gives you
$$(x-3)^2 + y^2 = 4(x+3)^2 + 4y^2.$$
| {
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Probability task (Find probability that the chosen ball is white.) I have this task in my book:
First box contains $10$ balls, from which $8$ are white. Second box contains $20$ from which $4$ are white. From each box one ball is chosen. Then from previously chosen two balls, one is chosen. Find probability that the ch... | The desired probability is
$$\frac{8}{10}\frac{16}{20}\frac{1}{2}+\frac{2}{10}\frac{4}{20}\frac{1}{2}+\frac{8}{10}\frac{4}{20}$$
You will notice that in $(a, w)$ or $(w, a)$ cases, there is only $\frac12$ chance of selecting the white ball.
| {
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How find this $\lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac{1}{2n}\right)$ Find this limit
$$\lim_{n\to\infty}n^2\left(\dfrac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\dfrac{1}{k+1}-\dfrac{1}{2n}\right)\tag{1}$$
I can only solve this limit
$$I=\lim_{n\to\infty}\left(\dfrac{1^k+2^k+\cdots+n^k... | We know that $\sum_{i=1}^n i^k=s_k(n)$ for some polynomial $s_k(x)=a_0+a_1x+\ldots +a_{k+1}x^{k+1}$. (We know this because the map $\Delta\colon f\mapsto f(X)-f(X-1)$ is a linear map from the $\mathbb Q$-vector space of polynomials of degree $\le k+1$ to the vector space of polynomials of degree $\le k$ and the kernel ... | {
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Prove that $ \int_{0}^{1}\frac{\sqrt{1-x^2}}{1-x^2\sin^2 \alpha}dx = \frac{\pi}{4\cos^2 \frac{\alpha}{2}}$. Prove that $\displaystyle \int_{0}^{1}\frac{\sqrt{1-x^2}}{1-x^2\sin^2 \alpha}dx = \frac{\pi}{4\cos^2 \frac{\alpha}{2}}$.
$\bf{My\; Try}::$ Let $x = \sin \theta$, Then $dx = \cos \theta d\theta$
$\displaystyle = \... | Letting $x=\sin \theta$ yields
$$
\begin{aligned}
I &=\int_0^{\frac{\pi}{2}} \frac{d t}{\left(1+t^2\right)^2-\left(1+t^2\right) t^2 \sin ^2 \alpha} \\
&=\int_0^{\infty} \frac{d x}{t^4 \cos ^2 \alpha+\left(1+\cos ^2 \alpha\right) t^2+1}
\end{aligned}
$$
Dividing both denominator and numerator of the integrand gives
$$
\... | {
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Limit of $\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$ What is the limit of this sequence $\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$?
Where $a$ is a constant and $n \to \infty$.
If answered with proofs, it will be best.
| The limit does not exist if $|a|\le 1$, and does if $|a|\gt 1$. The fact that it exists if $|a|\gt 1$ can be shown using the Ratio Test.
So we concentrate on the value of the limit, when it exists.
Let $S_n$ be our sum, and let $S$ be its limit. Then
$$aS_{n+1}-S_n=1+\frac{1}{a}+\frac{1}{a^2}+\cdots+\frac{1}{a^n}.$$
... | {
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Calculation of integers $b,c,d,e,f,g$ such that $\frac{5}{7} = \frac{b}{2!}+\frac{c}{3!}+\frac{d}{4!}+\frac{e}{5!}+\frac{f}{6!}+\frac{g}{7!}$ There are unique integers $b,c,d,e,f,g$ such that $\displaystyle \frac{5}{7} = \frac{b}{2!}+\frac{c}{3!}+\frac{d}{4!}+\frac{e}{5!}+\frac{f}{6!}+\frac{g}{7!}$
Where $0\leq b,c,d,e... | Look at this equality that you've got: $$ 2520 \cdot b + 840 \cdot c + 210 \cdot d + 42 \cdot e + 7 \cdot f + g = 3600.$$
Note that if you consider everything modulo $7$, then most of the summands disappear, because $2520,840,210,42$ and $7$ are all multiples of $7$. So, taking remainders modulo $7$, we get $g \equiv 2... | {
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Infinite Series $\sum\limits_{n=1}^\infty\left(\frac{H_n}n\right)^2$ How can I find a closed form for the following sum?
$$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$
($H_n=\sum_{k=1}^n\frac{1}{k}$).
| Starting with $\displaystyle \dfrac{H_n}{n} = \sum_{k=1}^{\infty} \dfrac{1}{k(k+n)}$ we have,
\begin{align*}\sum_{n=1}^{\infty} \dfrac{H_n^2}{n^2} &= \sum_{n=1}^{\infty} \left(\sum_{k=1}^{\infty}\dfrac{1}{k(k+n)}\right)^2\\&= \sum_{n=1}^{\infty}\sum_{k,j=1}^{\infty} \dfrac{1}{jk(j+n)(k+n)} \\&= \sum_{n=1}^{\infty} \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/554003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
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Number of ways a sum of r can occur if a die is rolled a number of times. Show that $(1 - x - x^2 - x^3 - x^4 - x^5 -x^6)^{-1}$ is the generating function for the umber of ways a sum of r can occur if a die is rolled a number of times.
| Consider the product
$$
\left(x+x^2+x^3+x^4+x^5+x^6\right)^n\tag{1}
$$
The choice of a term in each factor of $(1)$ corresponds to the choice of a face on each of $n$ dice. The coefficient of $x^k$ in $(1)$ is the number of ways for the faces of $n$ dice to sum to $k$.
The coefficient of $x^k$ in the sum of $(1)$ over ... | {
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Proof: $2^{n-1}(a^n+b^n)>(a+b)^n$ If $n \in \mathbb{N}$ with $n \geq 2$ and $a,b \in \mathbb{R}$ with $a+b >0$ and $a \neq b$, then $$2^{n-1}(a^n+b^n)>(a+b)^n.$$
I tried to do it with induction. The induction basis was no problem but I got stuck in the induction step: $n \to n+1$
$2^n(a^{n+1}+b^{n+1})>(a+b)^{n+1} $
$ \... | Start from the other side:
$$(a+b)^{n+1} < (a+b)(2^{n-1}(a^n+b^n))= 2^{n-1}a^{n+1}+2^{n-1}a^{n}b+2^{n-1}ab^{n}+2^{n-1}b^{n+1}$$
Now, prove that $(a^{n}-b^{n})(a-b)>0$
which implies
$$2^{n-1}a^{n}b+2^{n-1}ab^{n}< 2^{n-1}a^{n+1}+2^{n-1}b^{n+1} \,.$$
P.S. If you are familiar, the last inequality:
$$a^{n}b+ab^{n}< a^{n+1}... | {
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The number of integral solutions for the equation $x-y = x^2 + y^2 - xy$ Find the number of integral solutions for the equation $x-y = x^2 + y^2 - xy$ and the equation of type $x+y = x^2 + y^2 - xy$
| This equation can be rewritten just in another form: $\frac{m+1}{n}+\frac{n+1}{m}=a$
Can be solved using the equation Pell:
$p^2-(a^2-4)s^2=1$
Solutions have the form:
$n=2(p-(a+2)s)s$
$m=-2(p+(a+2)s)s$
And more:
$n=\frac{2p(p+(a-2)s)}{a-2}$
$m=\frac{2p(p-(a-2)s)}{a-2}$
If this equation has a solution: $p^2-(a^2-4)... | {
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Test for convergence/divergence of $\sum_{k=1}^\infty \frac{k^2-1}{k^3+4}$ Given the series
$$\sum_{k=1}^\infty \frac{k^2-1}{k^3+4}.$$
I need to test for convergence/divergence. I can compare this to the series $\sum_{k=1}^\infty \frac{1}{k}$, which diverges.
To use the comparison test, won't I need to show that $... | To show that the given series is (ultimately) greater than the harmonic series for some k and beyond, We could consider this limit:
$$\lim_{k\rightarrow\infty}\frac{\frac{1}{k}}{\frac{k^2-1}{k^3+4}} = \lim_{k\rightarrow\infty}\frac{k^3+4}{k^3-k}= 1$$
Hence, by the definition of limit,
$$\forall\varepsilon\gt0,\exists N... | {
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How prove this inequality $\sum_{i=1}^{n}\frac{\ln{i}}{i^4}<\frac{1}{14}$ I need to show that
$$\sum_{i=2}^{n}\dfrac{\ln{i}}{i^4}<\dfrac{1}{14}$$
This problem from high school competition, so usage of integrals and infinite serries is forbidden.
My try: let $x\in (n-1,n)$,then
$$\dfrac{ln{x}}{x^4}<\dfrac{\ln{(n-1)}}{(n... | Since
$$\frac{d}{dx}\left[\frac{\log x}{x}\right] = \frac{1 - \log x}{x^2} < 0
\quad\text{ when } x > e$$
We have
$$\frac{\log k}{k} \le \frac{\log 4}{4} = \frac{\log 2}{2}\quad\text{ for any } k \ge 4.$$ As a result,
$$\sum_{k=2}^\infty\frac{\log k}{k^4}
= \sum_{k=2}^\infty\left(\frac{\log k}{k}\right)\frac{1}{k^3}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/556737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Divisibility by $10^6$? Let $p_k$ be the $k^{th}$ prime number.
Find the least $n$ for which $(p_1^2+1)(p_2^2+1) \cdots (p_n^2+1)$ is divisible by $10^6$.
I have no idea where to start on this problem. Any help would be appreciated.
| Let's evaluate each $(p^2 + 1)$ expression, counting up the factors of $2$ and $5$, stopping when they both are at least $6$:
$n = 1$
$p = 2$, and $(p^2 + 1) = 5$. 2s so far = $0$, 5s so far = $1$.
$n = 2$
$p = 3$, and $(p^2 + 1) = 10$. 2s so far = $1$, 5s so far = $2$.
$n = 3$
$p = 5$, and $(p^2 + 1) = 26$. 2s so f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/557255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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Non isolated minimum Consider the $C^2$ function $F:\mathbb{R}^k \rightarrow \mathbb{R}^k$
is it possible for it to be such that $x_n$ is a strict local minimizer for all $n$ and $x_n \rightarrow x$ where $x$ is a strict local minimizer itself.
I know that $\sin(\frac{1}{x})$ has a somewhat similar property, but it is ... | Here is an example: $F(x) = x^6 (\sin \frac{1}{x})^2 + x^8$.
Establishing that this satisfies your criteria takes a little work:
First, let us establish the relevant smoothness properties:
Note that $F$ is smooth for $x \ne 0$, and near $x=0$ we have some $K_0$ such that $F(x) \le K_0x^6$, we have $|F(x)-F(0) - 0.x| \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/557322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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homework - Show a matrix as a combination of other matrices and long division The topic we are dealing with here is polynomial division.
The question is:
We are given a polynomial: $f(x) = (x+1)(x-1)^2$, and a matrix $D \in R^{nxn}$ such that $f(D)=0$
Using only $I, D, D^2$ find:
1) $D^5+3D^2-D+I$
2) $D^{2013}$
3) $D^{... | Hint: Suppose $x^{2013} = (x+1)(x-1)^2q(x) + r(x)$ with $\operatorname{deg}(r)\le2$. Then $r(1)=1,\ r(-1)=-1$ and $r'(1)=2013$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Express $\cos2\theta$ in terms of $\cos$ and $\sin$ (De Moivre's Theorem) Use De Moivre's to express $\cos2\theta$ in terms of powers of $\sin$ and $\cos$.
What I have is:
$\cos2\theta + i\sin2\theta\\
= (\cos\theta + i \sin\theta)^2\\
= \cos^2\theta + 2 \cos\theta ~i \sin\theta + (i \sin)^2\theta\\
= \cos^2\theta + i(... | Yes,
$$\cos^2\theta - \sin^2\theta = (\cos\theta)(\cos\theta) - (\sin\theta)(\sin\theta)$$
$$=\cos(\theta + \theta) = \cos(2\theta)$$
| {
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"source": "stackexchange",
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"answer_count": 4,
"answer_id": 0
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In $\mathbb{Z_2[x]}/(x^2 + x +1)$, what is $\overline{x} \cdot \overline{x}$? In $\mathbb{Z_2[x]}/(x^2 + x +1)$, what is $\overline{x} \cdot \overline{x}$?
$\overline{x} \cdot \overline{x} = \overline{x^2}$
If I divide $\overline{x^2}$ by $(x^2 + x +1)$ I get an answer of $1$ with remainder $(-x-1)$.
$\overline{-1}$ in... | Observe that $$\overline{x^2 + x + 1} = \overline 0$$ since $x^2 + x + 1 \in (x^2 + x + 1)$, which means that $$\overline{x^2} =\overline{ - (x + 1)},$$ but since $-1 \equiv 1 \pmod 2,$ we get $$\overline{x^2} = \overline{x + 1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/565505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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how to find the barycentric coordinates of the orthocenter $A = (0,0),B = (4,0),C = (1,2)$
How can I find the barycentric coordinates of the orthocenter of $\triangle ABC$?
| This Wikipedia page describes the orthocentre in barycentric coordinates of the triangle $ABC$ as
$$
\left((a^2+b^2-c^2)(a^2-b^2+c^2):(a^2+b^2-c^2)(-a^2+b^2+c^2):(a^2-b^2+c^2)(-a^2+b^2+c^2)\right),
$$ where $a=|BC|,b=|CA|$ and $c=|AB|$ are the side lengths of the triangle.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the sum of $\sum_{n=1}^\infty (-1)^{n+1} (2n-1)x^{2n-1}$ Find the sum of $S(x) = \sum_{n=1}^\infty (-1)^{n+1} (2n-1)x^{2n-1}$.
I know convergence radius is $1$ because $\frac{1}{\sqrt[n] {(2n-1)}} = \frac{1}{1} = 1.$
Then:
$$x^{-1} S(x) = \sum_{n=1}^\infty (-1)^{n+1}(2n-1)x^{2n+2}. $$
$$\int x^{-1}S(x) = \sum_{n=1... | $$S=\sum_{n=0}^\infty (-1)^n (2n+1)x^{2n+1}$$
Thus, $$\begin{align}
S&=\sum_{n=0}^\infty (-1)^n(2n)x^{2n+1}+\sum_{n=0}^\infty (-1)^nx^{2n+1}\\
&=2x\sum_{n=0}^{\infty}n(-x^2)^{n}+\frac{x}{1+x^2}\\
&=2x\left(\sum_{n=0}^\infty(n+1)(-x^2)^n-\sum_{n=0}^\infty (-x^2)^n\right)+\frac{x}{1+x^2}\\
&=2x\left(\frac{1}{(1+x^2)^2}-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/572929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $x_1, \ldots, x_6$ are positive real numbers that add up to $2$. Show that:
If $x_1,x_2,x_3,x_4,x_5$ and $x_6$ are positive real numbers that add up to $2$, then:
$$2^{12} \leq \left(1+\dfrac{1}{x_1}\right) \left(1+\dfrac{1}{x_2}\right)\left(1+\dfrac{1}{x_3}\right)\left(1+\dfrac{1}{x_4}\right)\left(1+\dfrac{1}{x_5}... | The inequality is equivalent to
$$ \frac{1}{ \left(1+\dfrac{1}{x_1}\right) \left(1+\dfrac{1}{x_2}\right)\left(1+\dfrac{1}{x_3}\right)\left(1+\dfrac{1}{x_4}\right)\left(1+\dfrac{1}{x_5}\right)\left(1+\dfrac{1}{x_6}\right)} \leq \frac{1}{2^{12}}$$
or
$$\sqrt[6]{\prod \frac{x_i}{x_i+1}} \leq \frac{1}{4}$$
Now, by AM-GM we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/573625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If 4n+1 and 3n+1 are both perfect sqares, then 56|n. How can I prove this?
Prove that if $n$ is a natural number and $(3n+1)$ & $(4n+1)$ are both perfect squares, then $56$ will divide $n$.
Clearly we have to show that $7$ and $8$ both will divide $n$.
I considered first $3n+1=a^2$ and $4n+1=b^2$. $4n+1$ is a odd per... | $3n+1=x^2$
$4n+1=y^2$
$4y^2-3x^2=1$
Put $z=2y $
$z^2-3x^2=1$
One of solution is: $(z_0,x_0)=(2,1) $
Other solutions are given by:
$(z_n+ \sqrt3 y_n)=(2+ \sqrt 3)^{n} $
From above we get ,
$z_{n+1}=7z_n+12y_n $
And
$y_{n+1}=4z_n+7y_n $
Or (in terms of $x_n $);
$x_{n+1}=7x_n+6y_n $
$y_{n+1}=8x_n+7y_n $
Now,
$x_{n+1} \equ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Indefinite Integral of $(x^2+1)\over (x^3 + 8)$ I need help in solving this indefinite integral: $$\int {(x^2+1)\over (x^3 + 8)} $$
I know it needs to be reduced to the form $A\over (x+2)$ + $B\over (x+2)^2$ where A and B are constants, but I cannot seem to solve this integral.
Thanks in advance for any help.
| $$(x^3 + 8) = (x + 2)(x^2 - 2x + 4) \neq (x+2)^3$$
To break the integrand into partial fractions, use $$\int {x^2+1\over x^3 + 8}\,dx = \int \left(\frac{A}{x+2}+\frac{Bx+C}{x^2-2x+4}\right)\,dx$$
The first term will integrate as $\;A\log|x + 2| + \text{constant}$.
Once you solve for $A, B, C$, you'll get a better idea... | {
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"source": "stackexchange",
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Proof of Heron's Formula for the area of a triangle
Let $a,b,c$ be the lengths of the sides of a triangle. The area is given by Heron's formula:
$$A = \sqrt{p(p-a)(p-b)(p-c)},$$
where $p$ is half the perimeter, or $p=\frac{a+b+c}{2}$.
Could you please provide the proof of this formula?
Thank you in advance.
| Consider a triangle in the 3D cartesian system whose side lengths are a, b and c. Assume the three vertices of the triangle are on three coordinate axes. The area of this triangle can be represented as the sum of three area vectors where each vector corresponds to the projections of the given triangle in each of the th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 4
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Limit Proof Check: $\lim _{x \to a} x^4 = a^4$ Reviewing limits and I'm afraid I may be making mistakes, just looking for a quick proof check.
$f(x)=x^4$, prove that $\lim _{x \rightarrow a}f(x)=a^4$ by showing how to find $\delta$ .
This is my work. $|x^4 - a^4 | < \epsilon$ and $0<|x-a|<\delta$
Factoring: $|x^2 +a^... | Alternate Solution.
Need to find a bound for $|x^2 + a^2||x + a|$. Take for instance, $|x - a| < 1$ then
$$|x^2 + a^2||x + a| = |x^2 - a^2 + 2a^2 ||x - a + 2a| = (|x^2 - a^2 + 2a^2 |)(|x - a + 2a|)$$
This does not change the expression since $a^2 - a^2 = 0$ and $a - a = 0$
Using the triangle inequality $|x + y| \le |x|... | {
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If $x^3+\frac{1}{x^3}=18\sqrt{3}$ then to prove $x=\sqrt{3}+\sqrt{2}$
If $x^3+\frac{1}{x^3}=18\sqrt{3}$ then we have to prove $x=\sqrt{3}+\sqrt{2}$
The question would have been simple if it asked us to prove the other way round.
We can multiply by $x^3$ and solve the quadratic to get $x^3$ but that would be unnecessa... | $x^3+x^{-3}=18\sqrt{3}\Rightarrow x^6+1=18\sqrt{3}x^3\Rightarrow x^6-18\sqrt{3}x^3+1=0$
From there if you look at it like a quadratic equation, you can find 2 solutions for $x^3$ So x would simply be the cube roots of that. and $x^3=a$ has only 1 real root for any real-valued a.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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Find the residue of $\frac{e^{iz}}{(z^2+1)^5}$ at $z = i$ and evaluate $\int_0^{\infty} \cos x/(x^2+1)^5 dx$ I know the evaluation of $\int_0^{\infty} \cos x/(x^2+1)^5 dx$ requires that I solve the first part, but for some reason I'm stumped. I get that I should use $\lim_{z \to i}\frac{1}{24}\frac{d^4}{dz^4}((z-i)^5\f... | When you have such a high order pole, it is likely better to evaluate the residue directly from the Laurent series about the pole. That is, rewrite
$$\frac{e^{i z}}{(z^2+1)^5} = \frac{1}{e} \frac{e^{i (z-i)}}{(z-i)^5 [(z-i)+2 i]^5} = \frac{1}{i 2^5 e}\frac{e^{i (z-i)}}{(z-i)^5 [1-i(z-i)/2]^5}$$
So we want to find the ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Math induction sum of even numbers I need to prove by induction this thing:
$2+4+6+........+2n = n(n+1)$
so, this thing is composed by sum of pair numbers, so its what I do, but I'm stucked.
$2+4+6+\cdots+2n = n(n+1)$
$(2+4+6+\cdots+2n)+(2n+2) = n(n+1) + (2n+2) $
$n(n+1)+(2n+2) = n(n+1)+(2n+2) $
$n^2 + 3n + 2$
$n(n+2+1... | $2+4+6+........+2n = n(n+1)$
$2+4+6+........+2n+2(n+1) = n(n+1)+2(n+1)=(n+1)(n+2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/583884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Find $a,b \in \mathbb{Z}^+$ such that : $\frac{a^{2}-2}{ab+2}\in \mathbb{Z}$
$1$. Find $a;b\in \mathbb{Z}^+$ such that : $\frac{a^{2}-2}{ab+2}\in \mathbb{Z}$
$2$. Find $m;n>1$ such that : $2^m+3^n=k^2$ $(k\in \mathbb{Z})$
Problem 1. I thought :
$\frac{a^{2}-2}{ab+2}\in \mathbb{Z}\Rightarrow a^{2}-2\vdots ab+2\Rightar... | 1
Case 1 : $\frac{a^2-2}{ab+2} =0$ no solution
Case 2 : $\frac{a^2-2}{ab+2} >0 $ and $a < b$ : So $$
\frac{a^2-2}{ab+2} < 1$$
no solution.
Case 3 : $\frac{a^2-2}{ab+2} >0 $ and $a \geq b\ (a>1,\ a>b)$ : So $$
\frac{a^2-2}{ab+2} < \frac{a^2}{ab} = \frac{a}{b},\ ba^2-2b < a^2b +2a$$
So $$ ab\leq ab(a-b)< 2(a+b),\ (a-2)(b... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $a = b = c$, given $P_1(x) = ax^2-bx-c$ , $P_2(x) = bx^2-cx-a$, $P_3(x)=cx^2-ax-b$ and $P_1(v)=P_2(v)=P_3(v)$ Prove $a = b = c$, given $P_1(x) = ax^2-bx-c$, $P_2(x) = bx^2-cx-a$, $P_3(x)=cx^2-ax-b$ and $P_1(v)=P_2(v)=P_3(v)$
where $v$ is a real number.
$a,b,c$ are non zero real numbers.
| Let us put $w=P_1(v)=P_2(v)=P_3(v)$. We have
$$
\begin{array}{lcl}
P_1(v)=av^2-bv-c, & -(v+v^2)P_1(v) &=& a(-v^4-v^3)+b(v^3+v^2)+c(v^2+v) \\
P_2(v)=bv^2-cv-a, & (v^2+1)P_2(v) &=& a(-v^2-1)+b(v^4+v^2)+c(-v^3-v) \\
P_3(v)=cv^2-av-b, & (v-1)P_3(v) &=& a(-v^2+v)+b(-v+1)+c(v^3-v^2) \\
\end{array}
$$
So if we put $z_1=-(v+v... | {
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"source": "stackexchange",
"question_score": "1",
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Doubts about a nested exponents modulo n (homework) As part of my homework I am supposed to find the remainder of the division of $2^{{14}^{45231}}$ by $31$.
Using the ideas explained in calculating nested exponents modulo n I tried the following:
$\phi(31)=30$ since $31$ is prime. Then:
$2^{{14}^{45231}}$ (mod $31$) =... | Note that $14 = 2 \cdot 7$. We have $2^5 \equiv 2 \pmod{30}$. Similarly, $7^4 \equiv 1 \pmod{30}$. Hence, $14^{45231} = 2^{45231} \cdot 7^{45231}$.
$$2^{45231} \equiv \pmod{30} \equiv 2^{5(9046)+1} \equiv \pmod{30} \equiv 2^{9047} \cdot 2^{5(1809)+2} \pmod{30} \equiv 2^{1811} \pmod{30} \equiv 2^{5(362)+1} \pmod{30} \eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/587652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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struggle simplifying $\sqrt{9+\sqrt{5}}$ I need to simplify $\sqrt{9+\sqrt{5}}$
I already do this (proven it) $\sqrt{9-4\sqrt{5}}=2-
\sqrt{5}$
But I couldn't when apply to $\sqrt{9+\sqrt{5}}=\sqrt{9-4\sqrt{5}+5\sqrt{5}}=\sqrt{(2+\sqrt{5})^2+5\sqrt{5}}$
PLEASE help me out
| Suppose
$\sqrt{9+\sqrt{5}}
= a+\sqrt{b}
$.
Squaring both sides,
$9+\sqrt{5}
= a^2+b+2a\sqrt{b}
= a^2+b+\sqrt{4a^2b}
$.
Equating the parts,
$9 = a^2+b$
and
$5 = 4a^2b$.
From the second,
$a^2 = 5/(4b)$,
so, from the first,
$9 = 5/(4b)+b$,
or $4b^2-36b+5 = 0$.
The discriminant is
$d = 36^2-4\cdot 4\cdot 5
=16(9^2-5)
=16\c... | {
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"source": "stackexchange",
"question_score": "6",
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How to solve this reccurence relation? Let a,b,c be real numbers. Find the explicit formula for $f_n=af_{n-1}+b$ for $n \ge 1$ and $f_0 = c$
So I rewrote it as $f_n-af_{n-1}-b=0$ which gives the characteristic equation as $x^2-ax-b=0$. The quadratic formula gives roots $x= \frac{a+\sqrt{a^2+4b}}{-2}, \frac{a-\sqrt{a^2+... | Define $F(z) = \sum_{n \ge 0} f_z z^n$, and write your recurrence as:
$$
f_{n + 1} = a f_n + b \quad f_0 = c
$$
Multiply by $z^n$, sum over $n \ge 0$, and recognize:
\begin{align}
\sum_{n \ge 0} f_{n + 1} z^n &= \frac{F(z) - f_0}{z} \\
\sum_{n \ge 0} b z^n &= \frac{b}{1 - z}
\end{align}
to get:
$$
\frac{F(z) - ... | {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $y = \frac{1}{x-1} +\frac {1}{x-5}$ for $x$ I'm stuck on this one (too long ago for me I guess).
I expanded the fractions coming to $y = \frac{2x-6}{x^2-6x+5}$ and even tried to apply a polynom division (translation?) but this came to nothing.
What's the proper approach on this one?
| You have $y (x^2 - 6 x + 5) = 2x - 6$, or $y x^2 - (6y + 2)x + (5y + 6) = 0$, so applying the quadratic formula yields
$$
x = \frac{(6y + 2) \pm \sqrt{(6y +2)^2 - 4 y (5y + 6)}}{2y}.
$$
This simplifies to
$$
x = \frac{(3y + 1) \pm \sqrt{4y^2 + 1}}{y}.
$$
| {
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"url": "https://math.stackexchange.com/questions/592467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $x$ for $\left(\frac1{1\times101} + \frac1{2\times102} + \dots +\frac1{10\times110}\right)x = \frac1{1\times11} + \frac1{2\times12}...$ $$\left(\frac1{1\times101} + \frac1{2\times102} + \dots +\frac1{10\times110}\right)x = \frac1{1\times11} + \frac1{2\times12} + \dots +\frac1{100\times110}$$
Find x
My younger sist... | One can group the $100$ summands on the right side by the ending digit of the denominators. For example, taking the numbers ending in 2:
$$\frac{1}{2\times 12}+\frac{1}{12 \times 22}+ ... +\frac{1}{92 \times 102}=\\
= \frac{1}{10}\left(\frac{1}{2} -\frac{1}{12}\right) +\frac{1}{10}\left(\frac{1}{12} -\frac{1}{22}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/593818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$? I'd like to find out why
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6
\end{align}
I tried to rewrite it into a geometric series
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = \sum_{n=0}^{\infty} \Big(\frac{1}{2}\Big)^nn^2
\end{align}
But I don't know... | We will use the same technique found here to show that $\sum_{n=1}^{\infty}\frac{n}{2^n} = 2$. But we'll need the following theory extension (see this link) to that result. For $k \ge 1$,
$\tag 1 \sum_{n=k}^{\infty}\frac{n}{2^n} = 2^{1 - k} \, (k + 1)$
We decompose the summands of $\sum_{n=1}^{\infty}\frac{n^2}{2^n}$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/593996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 8,
"answer_id": 6
} |
How do you simplify this expression? $$\lim_{h\to0}(\frac{x}{h(x+h+1)} + \frac{1}{x+h+1} - \frac{x}{h(x+1)})$$
I know the answer is $$\frac{1}{(1+x)^2}$$
But I can't get there
| Using fraction addition rules, we have
$$\lim_{h\to0}\left(\frac{x}{h(x+h+1)} + \frac{1}{x+h+1} - \frac{x}{h(x+1)}\right)\\
=\lim_{h\to0}\left(\frac{x(x+1)}{h(x+h+1)(x+1)} + \frac{h(x+1)}{h(x+h+1)(x+1)} - \frac{x(x+h+1)}{h(x+1)(x+h+1)}\right)\\
=\lim_{h\to0}\frac{(x+h)(x+1)-x(x+h+1)}{h(x+1)(x+h+1)}\\
=\lim_{h\to0}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/596253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\triangle ABC;AB=c;AC=b;BC=a$ such that $a\geq b\geq c$. Prove : $\frac{a^2-b^2}{c}+\frac{b^2-c^2}{a}+\frac{c^2+2a^2}{b}\geq \frac{2ab-2bc+3ca}{b}$ $\triangle ABC;AB=c;AC=b;BC=a$ such that $a\geq b\geq c$. Prove :
$$\frac{a^2-b^2}{c}+\frac{b^2-c^2}{a}+\frac{c^2+2a^2}{b}\geq \frac{2ab-2bc+3ca}{b}$$
I have tried that :
... | Another way to look at it would be:
\begin{align}
LHS &= \frac{(a - b)(a + b)}{c} + \frac{(b - c)(b + c)}{a} + \frac{c^2 + 2a^2}{b} \\
&> (a - b) + (b - c) + \frac{c^2 + 2a^2}{b} \\
&= a - c + \frac{c^2 + 2a^2}{b}.
\end{align}
And
RHS $= 2a - 2c + \dfrac{3ca}{b}$.
\begin{align}
\text{So, }LHS > RHS &\iff \frac{c^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/597664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Limit of $\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)$ I have to determine the following:
$\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)$
$\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)=\lim\limits_{x \rightarrow \infty}(\sqrt{x^8(1+\frac{4}{x^8})}-x^4 = \lim\limits_{x \rightarrow \infty}(x^4\sqrt... | We have
$$
x^4 (\sqrt{1 + \frac{4}{x^8}}-1)=
x^4 (\frac{2}{x^8} + O(x^{-8}))
=\frac{2}{x^4} + O (x^{-4}) \rightarrow 0
$$
when $x \rightarrow \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/598928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Is it a Fermat polynomial? A Fermat polynomial is a polynomial which can be written as the sum of squares of two polynomials with integer coefficients. Let $f(x)$ be a Fermat polynomial such that $f(0)=1000$. Prove that $f(x)+2x$ is not a Fermat polynomial.
| $1000$ can be expressed as a sum of two squares in two different ways: either $1000 = 10^2 + 30^2$ or $1000 = 18^2 + 26^2$. Thus if $f(x)$ is a Fermat polynomial with $f(0) = 1000$ and $f(x) = [a(x)]^2 + [b(x)]^2$, then $a(0)^2 + b(0)^2 = 1000$ so that (wlog) either $a(0) = 10$ and $b(0) = 30$, or $a(0) = 18$ and $b(0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/602781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
For a prime $p$ find integers $n,m$ s.t $ p > n>m>0$ and $n^3 \equiv m^3 \pmod p$ Sitting at home on daddy leave i have decided to try to learn more number theory. I was playing around with some equations and got to this, which i don't know how to proceed with:
"
For a prime $p$, how to find (if there is) the smallest ... | $$n^3\equiv m^3\pmod p\implies p|(n^3-m^3)=(n-m)(n^2+mn+m^2)$$
Since $p$ is prime, $p|(n-m)$ or $p|(n^2+mn+m^2)$ but $n\ne m$ and $0<m<n<p$ so we can see $p\not\mid(n-m)$.
Now I'm just going to kind of pull something out of a hat and say that $m^2+mn+n^2$ is the norm on the Euclidean domain $\Bbb Z[\omega]$ where $\o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/608939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If $a+b+c \ge 1$ and $a,b,c>0$, prove that $\frac{1}{2a+s}+\frac{1}{2b+s}+\frac{1}{2c+s} \ge \frac{1}{s}$, where $s=ab+bc+ca$. What I know is that $$s \le \frac{(a+b+c)^2}{3} \ge \frac{1}{3}$$But as you can see the sign is pointing to different sides. So I can't see how this could be helpful. Just a small observation. ... | first we prove $a+b+c=1$ case:
$\dfrac{1}{2a+s}+\dfrac{1}{2b+s}+\dfrac{1}{2c+s} \ge \dfrac{1}{s} \iff s^3+s^2-4abc\ge 0$
let $x=3a,y=3b,z=3c,3u=x+y+z=3,3v^2=xy+yz+xz=9s,w^3=xyz=27abc$
$s^3+s^2-4abc\ge 0 \iff v^6+3v^4-4w^3 \ge 0 \iff v^6+3v^4\ge 4w^3$
accoding to $uvw$ method, $w^3\le 3v^2-2+2\sqrt{(1-v^2)^3}$
$\iff v^6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/609068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate the lim $\lim_{x\to 1} \frac{ \sqrt{x} - 1}{\sqrt[3]{x} - 1}$ $$
\lim_{x\to 1} \frac{ \sqrt{x} - 1}{\sqrt[3]{x} - 1}
$$
I multiply it with:
$$
\frac{ \sqrt{x} + 1}{\sqrt{x} + 1}
$$
And I get :
$$
\lim_{x\to 1} \frac{ \sqrt{x}^2 - 1^2}{(\sqrt[3]{x} - 1) * (\sqrt{x} + 1)}
$$
But the solution is still division b... | You basically want to insert $1$ written as a “conjugate” that removes the radical divided by itself. For $\sqrt{x}-1$ it is $\sqrt{x}+1$, for $\sqrt[3]{x}-1$ it is $\sqrt[3]{x^2}+\sqrt[3]{x}+1$. So you get
\begin{align}
\lim_{x\to1}\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}&=
\lim_{x\to1}\left((\sqrt{x}-1)\frac{\sqrt{x}+1}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/609146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Diagonalization of Matrix with Trigonometric Functions Problem statement:
Diagonalize the following matrix:
$$
\begin{pmatrix}
\cos \theta & \sin \theta \\
\sin \theta & -\cos \theta\\
\end{pmatrix}
$$
My attempt:
Ie found that the eigenvalues of this matrix are $\lambda = -1$ or $\lambda = 1$, so I plugged in $\lambd... | For $\;\lambda=-1\;$ :
$$\begin{align*}I&\;\;(\cos\theta+1)x+\sin\theta y=0\iff y=-\frac{\cos\theta+1}{\sin\theta}x\\II&\;\;\sin\theta x-(\cos\theta-1)y=0\end{align*}$$
We need only one equation as these two are linearly dependent (proof?), so
$$II\longrightarrow x=\frac{\cos\theta-1}{\sin\theta}y\;\implies\binom {\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/610310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
solving an expression based on sin $\theta$ If $\sin^2 \theta = \frac{x^2 + y^2 + 1}{2x}$, then $x$ must be equal to what?
What does the following solution mean?
$0 \le \sin^2 \le 1$
This implies $0 \le \frac{x^2 + y^2 + 1 }{ 2x } \le 1$
This implies $\frac{(x - 1)^2 + y^2 }{2x} \le 0 $
This implies $x = 1$.
Can you ... | On rearrangement we have $$x^2-2x\sin^2\theta+y^2+1=0\ \ \ \ (1)$$
which is clearly a Quadratic Equation in $x$
As $x$ is real, the discriminant must be $\ge0$
But actually the discriminant,
$\displaystyle (2\sin^2\theta)^2-4(y^2+1)=-4(y^2+1-\sin^4\theta)$
$\displaystyle=-4\{y^2+(1-\sin^2\theta)(1+\sin^2\theta)\}=-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/610399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to integrate $\int{ dx \over \sqrt{1 + x^2}}$ How to integrate $dx \over \sqrt{1 + x^2}$?
Answer should be $\log ( x + \sqrt{1 + x^2})$
Please help as possible...
Thank you
| We have
$$\int \frac{1}{\sqrt{1+x^2}}\, dx.$$
We make the following substitution. Let
$$
\begin{align*}
x&=\tan \theta \\
dx &= \sec^2 \theta \, d \theta\\
1+x^2&=1+\tan^2 \theta \\
&=\sec^2 \theta.
\end{align*}
$$
Hence our integral becomes
$$
\begin{align*}
\int \frac{1}{\sqrt{1+x^2}}\, dx &= \int\frac{\sec^2 \theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/610733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Motivation for $y(x - y) = \frac{1}{2}(x^2 - y^2) - \frac{1}{2}(x-y)^2$ It's a really simple identity and according to the text, I am reading, this is an obvious identity
$$y(x - y) = \frac{1}{2}(x^2 - y^2) - \frac{1}{2}(x-y)^2$$
Without expanding the RHS, I don't know how they came up with this.
| Factor the first term on the right hand side: $\frac{1}{2} (x^2-y^2) = \frac{1}{2} (x+y)(x-y).$ Factor the second term: $(x-y)^2=(x-y)(x-y)$ since squaring a number is defined to be the product of that number with itself.
Then factor out $(x-y)$ from $\frac{1}{2} (x+y)(x-y)-\frac{1}{2}(x-y)(x-y)$ since it is a common t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all postive integer numbers $x,y$,such $x+y+1$ divides $2xy$ and $x+y-1$ divides $x^2+y^2-1$ Find all postive integer $x$ and $y$ such that
$x+y+1$ divides $2xy$ and $x+y-1$ divides $x^2+y^2-1$
My try: since
$$(x+y)^2-2xy=x^2+y^2$$
I know this well know reslut:
$$xy|(x^2+y^2+1),\Longrightarrow \dfrac{x^2+y^2+1}{xy... | Noting that
$\left(x+y+1\right)\left(x+y-1\right)=x^2+y^2-1+2xy\quad * $
Use the above expression to prove that $x+y-1$ divides $2xy$ and $x+y+1$ divides $x^2+y^2-1$
Note that $x+y-1$ and $x+y+1$ have the same parity.
Let both of them be odd, then they are coprime
$\implies \left(x+y+1\right)\left(x+y-1\right)$ divi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$, where $x$, $y$ and $z$ are positive integers Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$ , where $x,y,z$ are positive integers.
Found ten solutions $(x,y,z)$ as ${(3,3,3),(2,4,4),(4,2,4),(4,4,2),(2,3,6),(... | Hint: Deduce that none of $x, y, z \in \mathbb{N}$ exceeds $7$. This can be done by mathlove's hint above.
Additionally, you can show that there is only one $(a, 2, 2)$-tuple and thus use $(a, 2, 3)$ to bound the solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/616639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
Find coefficient of $x^8$ in $(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)^6$ Find coefficient of $x^8$ in $(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)^6$
how to do it? I think it should be $3^6$ since $(3x^2)^6=3^6x^8$. (this is false)
Is this true?
| Let $S:=(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)$. Then $$\begin{align}S(1+x)&=1-x+x^2-x^3+x^4-x^5+x^6+7x^7\\
&=\frac{1+x^7}{1+x}+7x^7\end{align}$$
So $$\begin{align} S&=\frac{1+x^7}{(1+x)^2}+\frac{7x^7}{1+x} \end{align}$$
Then we have $$ S^6=\sum_{r=0}^6{n\choose r}\frac{(1+x^7)^r(7x^7)^{n-r}}{(1+x)^{2r+n-r}} $$
The observant... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Dealing with Generating Functions accurately I'm currently working through Iven Niven's "Mathematics of Choice." In the chapter on Generating Functions, the exercises include problems like:
How many solutions in non-negative integers does the equation $2x+3y+7z+9r=20$ have?
This of course ends up being the same as find... | Dealing with coefficients of generating functions is either synthetically (by learning patterns from experience what the formula for coefficients might be) or directly by manipulating the coefficients of constituent gfs (which helps with the first one). Usually the latter ends up by resulting in the multiplication of p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/617907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Help with finding tangent to curve at a point
Find an equation for the tangent to the curve at $P\left( \dfrac{\pi}{2},3 \right )$ and the horizontal tangent to the curve at $Q.$
$$y=5+\cot x-2\csc x$$
$y\prime=-\csc ^2 x -2(-\csc x \cot x)$
$y\prime= 2\csc x \cot x - \csc ^2 x\implies$ This is the equation of the... | This means that there can be infinitely many horizontal tangents. As simple as that .
Just pick up the one you want.
$x = 2n \pm \dfrac{\pi}3$ is one solution.
Corresponding $y$ value
$y = 5 + \cot\left(2n \pm \dfrac{\pi}3\right) -2\csc\left(2n \pm \dfrac{\pi}3\right)$
$y = 5 \pm \dfrac1{\sqrt3} \pm \dfrac4{\sqrt3} $
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
If $a,b,c\in(0;+\infty)$, prove that $\frac{a^2+b^2+c^2}{ab+bc+ca}+a+b+c+2\ge2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$. If $a,b,c\in(0;+\infty)$ and $$\frac{c}{1+a+b}+\frac{a}{1+b+c}+\frac{b}{1+c+a}\ge\frac{ab}{1+a+b}+\frac{bc}{1+b+c}+\frac{ca}{1+c+a}$$Prove that $$\frac{a^2+b^2+c^2}{ab+bc+ca}+a+b+c+2\ge2(\sqrt{ab}+\sqrt{bc}+\... | The given condition is equivalent to:
$$\sum_{cyc}\frac{c-ab}{1+a+b} \ge 0$$
$$\implies \sum_{cyc}\frac{c(1+a+b)-ab}{1+a+b} \ge \sum_{cyc}\frac{c(a+b)}{1+a+b} \qquad \text{adding to both sides}$$
$$\implies \sum_{cyc}c \ge \sum_{cyc}\frac{c(a+b)+ab}{1+a+b} = (ab+bc+ca)\sum_{cyc}\frac{1}{1+a+b} \tag{1}$$
By Cauchy-Sc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/620003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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proving the inequality $\triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$ If $\triangle$ be the area of $\triangle ABC$ with side lengths $a,b,c$. Then show that $\displaystyle \triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$
and also show that equality hold if $a=b=c$.
$\bf{My\; Try}::$ Here we have to prove $4\tria... | Thanks mathlove,phira,and user for solution.
My prove for $(b+c-a)\cdot(c+a-b)\cdot(a+b-c)\leq abc$, where $a,b,c$ are the sides of a $\triangle$.
Using $\;\;\;\; \{b+(c-a)\}\cdot \{b-(c-a)\} = b^2-(c-a)^2\leq b^2$
similarly $ \{c+(a-b)\}\cdot \{c-(a-b)\} = c^2-(a-b)^2\leq c^2$
similarly $\{a+(b-c)\}\cdot \{a-(b-c)\} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/621182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Proving that if $a^2+b^2=c^2$, then $a+b\ge c$. Hello, I'm trying to prove this statement.
Let a,b & c be three positive real numbers and if $a^2+b^2=c^2$ then $a+b\ge c$
Any help, please?
| $(a+b)^2=a^2+b^2+2ab\geq a^2+b^2= c^2\rightarrow (a+b)^2\geq c^2\rightarrow a+b\geq c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/621514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
Closed form of $\arccos\left(\frac{2 \pi}{2^N}\right)$ Is there a closed form for $\arccos\left(\dfrac{2 \pi}{2^N}\right)$ in terms of $N \in \mathbb{Z}, N \ge 3$?
I'm not super optimistic, but I'm not sure how to really start exploring the problem, either.
| $$\arccos z= \frac {\pi} {2} - \left( z + \left( \frac {1} {2} \right) \frac {z^3} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {z^5} {5} + \cdots\ \right)
= $$
$$=\frac {\pi} {2} - \sum_{n=0}^\infty \frac {\binom{2n} n z^{2n+1}} {4^n (2n+1)}; \qquad | z | \le 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/623679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Solution verification: solving $\sqrt{x-4}-\sqrt{x-5}+1=0$ I solved the following equation, and I just want to be sure I did it right.
This is the procedure:
$$
\sqrt{x-4}-\sqrt{x-5}+1=0\\
\sqrt{x-4}=\sqrt{x-5}-1\\
\text{squaring both sides gives me:}\\
(\sqrt{x-4})^2=(\sqrt{x-5}-1)²\\
x-4=(\sqrt{x-5})²-\sqrt{x-5}+1\\
... | It is easily seen that the equation $\sqrt{x-4}-\sqrt{x-5}=-1$ has no solutions since $\sqrt{x-4}>\sqrt{x-5}$ for all $x>=5$.
Your method is correct but you made a mistake in going from step $3$ to step $4$. After correcting that, you will still reach the same conclusion that there are no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
} |
calculus limit question: another difficult limit problem I have posted previously on a problem in a similar vein here:
Limit evaluation: very tough question, cannot use L'hopitals rule
I believe this problem is very similar, but it has stumped me.
$$\lim_{x \to 0}\frac{1-\frac12 x^2 - \cos\left(\frac{x}{1-x^2}\right)}{... | Like what Emanuele said, asymptotic expansions are useful for this kind of limits, and in fact better than L'Hopital (which fails miserably for some limits):
$\frac{x}{1-x^2} \in x + x^3 + O(x^5) \to 0$ as $x \to 0$
[We keep the error term so that at the end we know the error of the final approximation.]
$\cos( \frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/625574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find solutions of this equation If $a+b+c+d = 30$ and $a,b,c,d$ lie between $0$ and $9$. How to find number of solutions of this equation.
| Hint: Consider the coefficient of $x^{30}$ in
$$
\left(\frac{x^{10}-1}{x-1}\right)^4
$$
Since
$$
(1-x)^{-4}=\sum_{n=0}^\infty\binom{n+3}{3}x^n
$$
and
$$
\left(1-x^{10}\right)^4=\sum_{k=0}^4(-1)^k\binom{4}{k}x^{10k}
$$
we get the coefficient of $x^n$ to be
$$
\sum_{k=0}^4(-1)^k\binom{4}{k}\binom{n-10k+3}{n-10k}
$$
Note ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/626837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Why is it that $2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2$? Why is it that $$2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2?$$
| $$2^{10} + 2^{9} + 2^{8} +2^7+2^6+2^5+2^4+ 2^{3} + 2^2 + 2^1 = 2^{11} - 2$$
Add $(2-2)$ to the left-hand side, obtaining:
$$\begin{align}
2^{10} + 2^{9} + 2^{8} + 2^7+2^6+2^5+2^4 + 2^{3} + 2^2 + 2^1 \color{green}{+ 2 - 2} & = 2^{11} -2 \\
2^{10} + 2^{9} + 2^{8} + 2^7+2^6+2^5+2^4 + 2^{3} + 2^2 + \color{maroon}{2^1 + 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/628501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 0
} |
How to memorize the families that are $\sin$, $\cos$, and $\tan$ of $\pi$ over something? Is there any way to easily memorize these? Such as $\sin \frac{\pi}{6} = 1/2$. Any help needed!!!
| Here's a nice trick:
$$
\sin\left(0\right) = \sqrt{\frac{0}{4}}\\
\sin\left(\frac{\pi}{6}\right) = \sqrt{\frac{1}{4}}\\
\sin\left(\frac{\pi}{4}\right) = \sqrt{\frac{2}{4}}\\
\sin\left(\frac{\pi}{3}\right) = \sqrt{\frac{3}{4}}\\
\sin\left(\frac{\pi}{2}\right) = \sqrt{\frac{4}{4}}
$$
where the angles are in increasing or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/628644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Basis of Kernel of a matrix Given $\theta>0$. Let $H$ be $5 \times 6$ matrix
$$\left[\begin{matrix}
1 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & -1 & 0 & 0 & 0 \\
1 & 0 & 0 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & -1 \\
\theta & 0 & 0 & 0 & 0 & -1
\end{matrix}\right]$$
Consider the subspace $S=\{x\in\mathbb{R}^6:Hx=0$}. I know... | We perform the row operations:
$$\begin{bmatrix}
1 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & -1 & 0 & 0 & 0 \\
1 & 0 & 0 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & -1 \\
\theta & 0 & 0 & 0 & 0 & -1 \\
\end{bmatrix}
\xrightarrow{R_5 \gets \tfrac{1}{\theta} R_5}
\begin{bmatrix}
1 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & -1 & 0 & 0 & 0 \\
1 & 0 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/630065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Calculation of $\bf{Max.}$ value of $\sqrt{10x-x^2}-\sqrt{18x-x^2-77}\;\;\forall x\in \mathbb{R}$ (1) Calculation of Max. and Min. value of $\sqrt{x^2-3x+2}+\sqrt{2+3x-x^2}\;\; \forall x\in \mathbb{R}$
(2) Calculation of Max. value of $\sqrt{10x-x^2}-\sqrt{18x-x^2-77}\;\;\forall x\in \mathbb{R}$
My Try:: for $(1)$ one:... | (1):the approach is not right ,you can't make sure $a\ge b,b\le c \to a \ge c $
the easy way is let$ u=x^2-3x, f(x)=f(u)=\sqrt{2+u}+\sqrt{2-u},f^2=4+2\sqrt{4-u^2}$
$0\le\sqrt{4-u^2} \le 2 \implies 4 \le f^2 \le 8 \implies 2 \le f \le 2\sqrt{2}$ when $u=0$ and $u=2$ get Max and Min.
(2): $\sqrt{10x-x^2}$ domain is [$0,... | {
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How to prove $(b-2)^2 > 12a(5c + 2)$ provided $(3a + b + 5c)(5c + 2) < 0$? $a$, $b$, $c$ are rational numbers. It is known that $(3a + b + 5c)(5c + 2) < 0$.
How do I prove that $(b-2)^2 > 12a(5c + 2)$?
According to Bernoulli's inequality, I got this:
$(1 + b - 3)^2 \ge 1 + 2(b - 3) > 12a(5c + 2)$
$2b - 5 > 12a(5c + 2)$... | It seems the following.
Put $x=3a$, $y=b-2$, and $z=5c+2$. Given $(x+y+z)z<0$ we need to show that $y^2>4xz.$ Assume the converse. Then $(x+z)^2\ge 4xz>y^2$. So $|x+z|>|y|$ and the sign of $x+y+z$ is equal to the sign of $x+z$. Then $(x+z)z<0$, but $xz>y^2/4\ge 0$ and $z^2\ge 0$, a contradiction.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A claim that a is a square We have integers $a,b,c,d$ such that $a<b\le c<d$ and $ad=bc$ and $\sqrt{d}-\sqrt{a}\le 1$.Show that $a$ is a perfect square.This question is pretty unbelievable for me.anyway I don't know if I am reposting this here.
| Let us show that $a = n^2 ,\; b = n^2+n = c,\; d = (n+1)^2$ is the only solution.
The condition $\sqrt{d}-\sqrt{a} \leq 1$ shows that $$ d\leq a+2\sqrt{a}+1.$$
Let $b = a+x,\; c = a+y,\; d = a+z$ with $x,y,z \in \Bbb N$. Then $0<x\leq y < z \leq 2\sqrt{a}+1$, showing $0<x\leq y \leq 2 \sqrt{a}$. Furthermore:
$$ad = b... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Prove that $\lim\limits_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right)=0$
Prove that:
$$\lim_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1}
\,\right)=0$$
Epsilon>0. According to the Archimedean Property of reals, we have n1 element N with epsilon*n1>11. (Why 11? Seems so random...)... | The numerator was made bigger (or equal, if $n=3$) and the denominator was made smaller, so the fraction becomes bigger. This only works, when $n\geqslant3$ is assumed somewhere.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/632553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Inverse Function (and WolframAlpha gives different Result) I wanted to calculate the inverse function of
$$
f(x) = \frac{1}{x} + \frac{1}{x-1}
$$
Quite simple I thought, put
$$
y = \frac{1}{x} + \frac{1}{x-1} = \frac{2x-1}{x(x-1)}
$$
rearrange and solve
$$
y(x(x-1)) - 2x + 1 = 0
$$
which give the quadratic equation... | Your error is in the solution formula. You have $(y+2)^2 - 4\cdot y\cdot 1 = y^2+4 \neq (y+2)(y-2)$. It would be $y^2-4 = (y+2)(y-2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/632854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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prove that $a^{nb}-1=(a^n-1)((a^n)^{b-1}+...+1)$ Prove that $a^{nb}-1=(a^n-1)\cdot ((a^n)^{b-1}+(a^n)^{b-2}+...+1)$
We can simplify it, like this:
$$(a^{n})^b-1=(a^n-1)\cdot \sum_{i=1}^{b}(a^{n})^{b-i}$$
How can we prove this?
| For $\ x = a^b\ $ it is $\ f_n \,:=\, \dfrac{x^n-1}{x-1}\, =\, x^{n-1}+ x^{n-2} +\cdots + x + 1.\,$ Here is a proof by telescopy.
Notice that $\ \color{#0a0}{f_{n} - f_{n-1} = x^{n-1}}\ $ since $\ \dfrac{x^n-1}{x-1} - \dfrac{x^{n-1}-1}{x-1}\, =\, \dfrac{x^n-x^{n-1}}{x-1\quad }\, =\, x^{n-1}$
$\begin{eqnarray}{\rm\! He... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/633035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Linear algebra question on rank and null space This is an exercise from linear algebra and optimization by Gill, I do exercises to be prepared for my final exam and these are not homework questions!
Exercise $\mathbf{6.1.}\,$ Consider the following matrix $A$ and vector $b$:
$$
A=\begin{pmatrix}
2 & 4 \\
1 & 2 \\
... | If you do row reduction, you find
$$
\begin{pmatrix}
2 & 4 \\
1 & 2 \\
1 & 2
\end{pmatrix}
\to
\begin{pmatrix}
1 & 2 \\
0 & 0 \\
0 & 0
\end{pmatrix}
$$
which means that the first column is a basis for the column space of $A$ (which is better terminology than “range of $A$”, in my opinion). So the general form of the ve... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Help with local extrema of $f(x)=x^4-5x^2$
Find the coordinates of any local extreme points and inflection points of the function $f(x)=x^4-5x^2$
My try:
Find critical points: $f^{\prime}(x)=4x^3-10x=0$
$f^{\prime}(x)=2x(2x^2-5)=0 \implies x=0, x=\pm\sqrt{\dfrac{5}{2}}$
I would then use the critical points to deter... | From where you left off with the critical points you correctly found, we determine their $y$-coordinates as followed
$$\begin{aligned}
f(0) &= 0\\
f\left(-\dfrac{\sqrt{10}}{2} \right) &= -\dfrac{25}{4}\\
f\left(\dfrac{\sqrt{10}}{2} \right) &= -\dfrac{25}{4}
\end{aligned}$$
To determine the nature of those critical poin... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $\lim_{(x,y) \to (0,0)} \frac{x^3y}{x^4+y^2} = 0?$
How to prove that $\lim_{(x,y) \to (0,0)} \dfrac{x^3y}{x^4+y^2} = 0?$
First I tried to contradict by using $y = mx$ , but I found that the limit exists.
Secondly I tried to use polar coordinates, $x = \cos\theta $ and $y = \sin\theta$,
And failed .... | With polar coordinates we're cool, too:
$$x=r\cos\theta\;\;,\;\;y=r\sin\theta$$
$$\frac{x^3y}{x^4+y^2}=\frac{r^4\cos^3\theta\sin\theta}{r^4\cos^4\theta+r^2\sin^2\theta}=\frac{r^2\cos^3\theta\sin\theta}{r^2\cos^4\theta+\sin^2\theta}\xrightarrow[r\to 0]{}\frac0{0+\sin^2\theta}=0$$
But what if $\;\sin^2\theta=0\;$ ? Well... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
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mixed numbers subtraction vertically In the following subtraction we are subtracting $2$ mixed numbers vertically. I know how it works except the last step.
$$ 7 \frac{1}{3} - 4 \frac{1}{2} = 3 + \frac{-1}{6} = 2 + \frac{5}{6} = 2 \frac{5}{6}$$
I am confused about this step: $ 3 + \frac{-1}{6} = 2 + \frac{5}{6}$
How do... | Write it as
$$ 3 - \frac{1}{6}= 2 + 1 - \frac{1}{6}= 2 + \frac{6}{6} -\frac{1}{6} = 2 + \frac{6-1}{6} = 2+\frac{5}{6} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/638931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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The sum of areas of 2 squares is 400and the difference between their perimeters is 16cm. Find the sides of both squares. The sum of areas of 2 squares is 400and the difference between their perimeters is 16cm. Find the sides of both squares.
I HAVE TRIED IT AS BELOW BUT ANSWER IS NOT CORRECT.......CHECK - HELP!
Let sid... | $x^2+y^2=400$
$4*x-4*y=16$
could you continue?
ok we have $x^2+y^2=400$ and $x-y=4$
from there $x=4+y$
$(4+y)^2+y^2=400$
$16+8*y+y^2+y^2=400$
$2*y^2+8*y-384=0$
$y^2+4*y-192=0$
$D=4+192=196$
$y_1=(-2+\sqrt{196})$
$y_2=(-2-\sqrt{196})$
now calculate $x_1$ and $x_2$
$y=12$
$x=16$
now check $16*16+12*12=256+144=400$
$6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/639271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing $\lim_{n\to \infty}{\frac{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)}{7\cdot11\cdot15\cdot\dots.\cdot(4n+3)}}$ Let $\{a_n\}_{n\ge1}^{\infty}=\bigg\{\cfrac{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)}{7\cdot11\cdot15\cdot\dots.\cdot(4n+3)}\bigg\}$. Find $\lim_{n\to \infty}{a_n}$.
I.) In the first step I studied monotony:
$... | One can check that
$$
\bigg( \frac{4n+1}{4n+3} \bigg)^2 < \frac{n+1}{n+2}
$$
for all $n\ge0$. Therefore
$$
0 < a_n = \frac57 \frac9{11} \cdots \frac{4n+1}{4n+3} < \bigg( \frac23 \frac34 \cdots \frac{n+1}{n+2} \bigg)^{1/2} = \sqrt{\frac2{n+2}},
$$
and so $a_n\to0$ by the squeeze theorem.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
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Pre-cal trigonometric equation problem am i correct to factor out a 2 first?
$$2\sin^2 2x =1$$
$$2(\sin^2 x-1) =1$$
$$2\cos^2 x =1$$
$$\cos^2 x ={1\over2}$$
$$\cos x =\pm{\sqrt{2}\over 2 }$$
i'm only looking for solutions from $$0≤ x ≤ 2\pi $$
$$x = {\frac{\pi}{4}},{\frac{7\pi}{4}},{\frac{3\pi}{4}},{\frac{5\pi}{4}}$$
t... | $$
\begin{align*}
& 2\sin^2(2x)=1 \\
\Rightarrow & 2\left( \frac{1-\cos(4x)}{2} \right)=1\\
\Rightarrow&1-\cos(4x)=1\\
\Rightarrow & \cos(4x)=0 \\
\Rightarrow & 4x=\frac{n\pi}{2}, \, n \in \{ 1,3,5,7,\ldots \}\\
\Rightarrow & x=\frac{n\pi}{8}, n \in\{ 1,3,5,7,\ldots \}
\end{align*}
$$
Hence the set is
$$x\in\left\{\fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Why $\lim_{n\to \infty}{\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+\dots+\frac{1}{n\cdot(n+2)}}=\frac{3}{4}$? Let $\{a_n\}_{n\ge1}^{\infty}=\bigg\{\cfrac{1}{1\cdot3}+\cfrac{1}{2\cdot4}+\dots+\cfrac{1}{n\cdot(n+2)}\bigg\}$. Find $\lim_{n\to \infty}{a_n}$.
I write:
$$\lim_{n\to \infty}{a_n}=\sum_{n=1}^{\infty}{\frac{1}{n\cdot(n... | Telescope! Note that $$ \frac1{n(n+2)}=\frac12\cdot\left(\frac1n-\frac1{n+2}\right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Complex numbers - Exponential numbers - Proof Let $z$ be a complex number, and let $n$ be a positive integer such that $z^n = (z + 1)^n = 1$. Prove that $n$ is divisible by 6.
For this problem I am stumped...how should I begin?
Also there's a hint for it:
From $z^n = 1$, prove that $|z| = 1$. What does the equation $(z... | A problem way too cool and cute to pass up, so check this out:
$(1.) \; z^n = 1 \Rightarrow \vert z \vert^n = 1, \tag{1}$
$(2.) \; \vert z \vert^n = 1 \Rightarrow \vert z \vert = 1 \Rightarrow \exists \theta \in \Bbb R \;\text{such that} \; z = e^{i\theta}, \tag{2}$
$(3.) \; \vert z \vert = 1 \Rightarrow z \bar z = 1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/643024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
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Numerical Analysis - Richardson Extrapolation Question: Suppose that N(h) is an approximation to $M$ for every $h > 0$ and that
$M = N(h) + K_1 h + K_2 h^2 + K_3 h^3 +\cdots$, for some constants $K_1, K_2, K_3,\cdots$. Use the values $N(h), N( h/3),$ and $N (h/9)$ to produce an $O(h^3)$
approximation to $M$.
My work:
$... | It is not true that $N_1(h)=O(h^2), N_2(h)=O(h^4), \cdots$ for the formula
$$
M=N(h)+K_1 h+K_2 h^2+K_3 h^3+\cdots
$$
but it is true for
$$
M=N(h)+K_1 h^2+K_2 h^4+K_3 h^6+\cdots
$$
It's easy to get an $O(h^3)$ formula. Subtracting the first formula from $3$ times the second formula eliminates the $h$ term
$$
2M=3N(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/644446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove this inequality: $\frac{x^3+1}{\sqrt{x^4+y+z}}+\frac{y^3+1}{\sqrt{y^4+z+x}}+\frac{z^3+1}{\sqrt{z^4+x+y}}\geq2\sqrt{xy+yz+zx}$ Let $x,y,z>0$ such that $xyz=1$. Show that: $$\dfrac{x^3+1}{\sqrt{x^4+y+z}}+\dfrac{y^3+1}{\sqrt{y^4+z+x}}+\dfrac{z^3+1}{\sqrt{z^4+x+y}}\geq2\sqrt{xy+yz+zx}$$
I've tried many things but... | proof: since
\begin{align*}2\sqrt{(x^4+y+z)(xy+zx+yz)}&=2\sqrt{[x^4+xyz(y+z)][xy+yz+xz]}\\
&=2\sqrt{(x^3+y^2z+yz^2)(x^2y+
x^2z+xyz)}
\end{align*}
Use AM-GM inequality we have
$$2\sqrt{(x^3+y^2z+yz^2)(x^2y+x^2z+xyz)}\le x^3+y^2z+yz^2+x^2y+x^2z+xyz=(x+y+z)(x^2+yz)=\dfrac{(x+y+z)(x^3+1)}{x}$$
so
$$\dfrac{x^3+1}{\sqrt{x^4... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve the error in the simultaneous equation. The question is:
$$\begin{align}
\tag{1} 2x - 3y &= 3\\
\tag{2} 4x^2 - 9y^2 &= 3
\end{align}$$
From equation (1):
$$\tag{3} 2x = 3 - 3y.$$
Substitute equation (3) in (2):
$$\begin{align}
4x^2 - 9y^2 &= 3\\
(2x)^2 - 9y^2 &= 3\\
(3 - 3y)^2 - 9y^2 &= 3\\
[(3)^2 - 2\cdot 3\c... | There is other way:
$\ 2x - 3y = 3 \tag{1}$
$(2x - 3y)(2x + 3y) = 3 \tag{2} $
Using $(1)$ we have
$$ 2x - 3y = 3 $$
$$ 2x + 3y = 1 $$
$$(1)+(2)$$
$$4x=4$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to prove all $c_{n},d_{n}$ to be integers if $(n+1)c_{n}=nc_{n-1}+2nd_{n-1}$ and $d_{n}=2c_{n-1}+d_{n-1}$? Let sequences $(c_n)$ and $(d_n)$ be given by
$$c_0=0,\:d_0=1$$
and recursively for $n\ge 1$ by
$$\begin{align}
c_n & =\frac{n}{n+1}c_{n-1}+\frac{2n}{n+1}d_{n-1} \\[2ex]
d_n & =2c_{n-1}+d_{n-1}
\end{align}$$
I... | Consider the polynomials $(1+x+x^2)^n$; the only coefficients that will be non zero are terms whose $x$ exponent is between $0$ and $2n$. Note two other things; firstly they will all be whole numbers and secondly they satisfy $a_{i}= a_{2n-i}$ (i.e they will be symmetrical around the central term).
Now we have define s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/653089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Proving a property of about the Fermat numbers
Show that the last digit in the decimal expansion of $F_n=2^{2^n}+1$ is $7$ for $n \geq 2$.
For our base step we let $n=2$. Now we have $2^{2^2}=16$. So the assertion holds for our base case. Then we assume it holds for $n$. For the $n+1$ case, is there a way to de... | Continuing from where you left, we just need to prove that $F_{n+1}$ holds true.
$$F_{n+1} = 2^{2^{n+1}}+1$$
$$F_{n+1} = 2^{2^n.2}+1$$
$$F_{n+1} = (2^{2^n})^2 + 1$$
$$F_{n+1} = (2^{2^n} + 1)^2 - 2.2^{2^n}$$
First term on the right hand side is $F_n^2$ which has a last digit of $7^2=9$.
Second term on the right hand sid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/655528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Legendre polynomial problem (please help!)
Problem: Show that $P^{'}_{2n+1}(0)= \frac{(-1)^n (2n+1){^{2n}}C_n}{2^n}$.
My attempt:
Given: $P_{n}(x)=\frac{1}{2^n}\sum_{k=0}^{\frac{1}{2}n}(-1)^{k} {{^n}}C_k{^{2n-2k}C_n}x^{n-2k}$
Let $n = 2n+1$.
We have $P_{2n+1}(x)=\frac{1}{2^{2n+1}}\sum_{k=0}^{{n}+\frac{1}{2}}(-1)^{k} ... | Your result is correct although it is difficult to read the derivation: I strongly suggest you use the binomial coefficient notation $\binom{a}{b}$ instead of $^aC_b$.
My own derivation is as follows: The recurrence relation
$$(x^2-1)P_n'=n(xP_n-P_{n-1})$$
implies that
$$P_n'(0)=nP_{n-1}(0)$$
so it remains to find the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/655808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Intuitively, why is the Euler-Mascheroni constant near $\sqrt{1/3}$? Questions that ask for "intuitive" reasons are admittedly subjective, but I suspect some people will find this interesting.
Some time ago, I was struck by the coincidence that the Euler-Mascheroni constant $\gamma$ is close to the square root of $1/3$... | From the continuous fraction expansion, the seventh convergent is
$$\gamma \approx \frac{15}{26}$$
From the limit definition
$$\begin{align}
\gamma
&=
\lim_{n \to \infty} {\left(2H_n-\frac{1}{6}H_{n^2+n-1}-\frac{5}{6}H_{n^2+n}\right)} \\
&=
\frac{7}{12}+\sum_{n=1}^{\infty}\left(\frac{2}{n+1}-\frac{1}{3n(n+1)(n+2)}-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/656283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\int_{0}^{\infty} \frac{x^{3}- \sin^{3}(x)}{x^{5}} \ dx $ using contour integration EDIT: Instead of expressing the integral as the imaginary part of another integral, I instead expanded $\sin^{3}(x)$ in terms of complex exponentials and I don't run into problems anymore.
\begin{align} \int_{0}^{\infty} \fr... | Another approach :
Sorry Random Variable, this is not using contour integration technique since I don't know how to approach the integral using that way. $\ddot\smile$
Consider
$$
\mathcal{I}(\alpha)=\int_0^\infty\frac{(\alpha x)^3-\sin^3\alpha x}{x^5}dx.\tag1
$$
Differentiating $(1)$ four times yields
\begin{align}
\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/656757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 4,
"answer_id": 1
} |
Prove that $2xy\mid x^2+y^2-x$ implies $x$ is a perfect square. Prove that $2xy\mid x^2+y^2-x$ implies $x$ is a perfect square.
My work:
$2xy\mid x^2+y^2-x \implies x^2+y^2-x=2xy\cdot k$
So,$x^2+y^2+2xy-x=(x+y)^2-x=2xy \cdot (k+1)$
And,$x^2+y^2-2xy-x=(x-y)^2-x=2xy \cdot (k-1)$
I found that for $x,y$ both odd, no soluti... | We use the following
Fact: A non-zero integer is a perfect square (by that I mean a number of the form $k^2$ or $-k^2$) if and only if in its prime factorization, the exponent of every prime factor is even.
Now let $p$ be any prime factor of $x$ and $k$ the exponent of $p$ in the prime factorization of $x$. If $k$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/663283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Method of moments on uniform distributions I need help on how to find the estimates $a$ and $b$ in the uniform distribution $\mathcal U[a,b]$ using the method of moments.
This is where I am at:
I have found $U_1=\overline X$ and $m_1=\frac{a+b}2$ Also, $m_2=\frac1{n(E(X_i^2)}$ and $u_2=E(X_i^2)$ when I equate $m_1$ to ... | The population mean is $\dfrac{a+b}{2}$. The population variance is $\dfrac{(b-a)^2}{12}$. If you have already found that the popuation variance of the $\mathcal U[0,1]$ distribution is $1/12$, just notice that the length of the interval has been multiplied by $b-a$, and that is a scale factor, so you multiply the va... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/664811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\left(\frac12(x+y)\right)^2 \le \frac12(x^2 + y^2)$ Prove that $$\left(\frac12(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
I've gotten that
$$\left(\frac12(x+y)\right)^2 \ge 0 $$
but stumped on where to go from here...
| $\frac12 (x^2+y^2)-(\frac12 (x+y))^2$
$=\frac12 (x^2+y^2)-\frac14 (x^2+y^2+2xy)$
$=\frac14 (2x^2+2y^2-x^2-y^2-2xy)$
$=\frac14 (x^2+y^2-2xy)$
$=\frac14 (x-y)^2\ge 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/665206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 0
} |
Show if $a^2+b^2 \le 2$ then $a+b \le 2$ If $a^2+b^2 \le 2$ then show that $a+b \le2$
I tried to transform the first inequality to $(a+b)^2\le 2+2ab$ then $\frac{a+b}{2} \le \sqrt{1+ab}$ and I thought about applying $AM-GM$ here but without result
| $(a+b)^2+(a-b)^2=2(a^2+b^2)\leq 4$, so $|a+b|\leq 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/666217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.