Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Evaluate this power series Evaluate the sum
$$x+\frac{2}{3}x^3+\frac{2}{3}\cdot\frac{4}{5}x^5+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}x^7+\dots$$
Totally no idea. I think this series may related to the $\sin x$ series because of those missing even powers. Another way of writing this series:
$$\sum_{k=0}^{\infty}\fra... | In this answer, I mention this identity, which can be proven by repeated integration by parts:
$$
\int_0^{\pi/2}\sin^{2k+1}(x)\;\mathrm{d}x=\frac{2k}{2k+1}\frac{2k-2}{2k-1}\cdots\frac{2}{3}=\frac{1}{2k+1}\frac{4^k}{\binom{2k}{k}}\tag{1}
$$
Your sum can be rewritten as
$$
f(x)=\sum_{k=0}^\infty\frac1{(2k+1)}\frac{4^k}{\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Maximum N that will hold this true Find the largest positive integer $N$ such that
$$\sqrt{64 + 32^{403} + 4^{N+3}}$$
is an integer
Is $N = 1003$?
| Note that with $N=2008$ we have $ (2^{N+3}+8)^2=4^{N+3}+2\cdot 8\cdot 2^{N+3}+64=4^{N+3}+2^{2015}+64=64+32^{403}+4^N,$ so we conjecture that the maximal value is $2008$.
If $2^{2015}+2^6+2^{2N+6}$ is a perfect square then also $\frac1{64}$ of it, i.e. $2^{2009}+1+2^{2N}=m^2$ for some $m\in\mathbb N$.
But if $N> 2008$... | {
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"source": "stackexchange",
"question_score": "2",
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absolute value inequalities When answer this kind of inequality
$|2x^2-5x+2| < |x+1|$
I am testing the four combinations when both side are +, one is + and the other is - and the opposite and when they are both -.
When I check the negative options, I need to flip the inequality sign?
Thanks
| Another way - no tricks, just systematically looking at all cases, where we can write the inequality without the absolute value sign, to convince you that all possibilities are covered..
Following the definition of the absolute value function, RHS is easy to rewrite as,
$$|x+1| = \begin{cases} x+1 & x \ge -1\\ -x-1 &x ... | {
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Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$
Question : Is the following true for any $m\in\mathbb N$?
$$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$
Motivation : I reached $(\star)$ by using computer. It seems true, but I can't prove it... | (First time I write in a math blog, so forgive me if my contribution ends up being useless)
I recently bumped into this same identity while working on Fourier transforms. By these means you can show in fact that
$$\sum_{k=-\infty}^{+\infty}\frac{1}{(x-k)^2}=\frac{\pi^2}{\sin^2 (\pi x)}.\tag{*}\label{*}$$
Letting $x=\fr... | {
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"source": "stackexchange",
"question_score": "49",
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Three Variables-Inequality with $a+b+c=abc$ $a$,$b$,$c$ are positive numbers such that $~a+b+c=abc$
Find the maximum value of $~\dfrac{1}{\sqrt{1+a^{2}}}+\dfrac{1}{\sqrt{1+b^{2}}}+\dfrac{1}{\sqrt{1+c^{2}}}$
| Trigonometric substitution looks good for this, especially if you know sum of cosines of angles in a triangle are $\le \frac32$. However if you want an alternate way...
Let $a = \frac1x, b = \frac1y, c = \frac1z$. Then we need to find the maximum of
$$F = \sum \frac{x}{\sqrt{x^2+1}}$$
with the condition now as $xy + ... | {
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"question_score": "2",
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evaluating $\sum_{n=1}^\infty \frac{n^2}{3^n}$ I am trying to compute the sum $$\sum_{n=1}^\infty \frac{n^2}{3^n}.$$
I would prefer a nice method without differentiation, but if differentiation makes it easier, then that's fine.
Can anyone help me?
Thanks.
| We have
\begin{gather*}
S=\sum_{n=1}^{\infty}\frac{n^2}{3^n}=\frac{1}{3}\sum_{n=0}^{\infty}\frac{(n+1)^2}{3^{n-1}}=\frac{1}{3}\left(\sum_{n=0}^{\infty}\frac{n^2}{3^{n-1}}+2\sum_{n=0}^{\infty}\frac{n}{3^{n-1}}+\sum_{n=0}^{\infty}\frac{1}{3^{n-1}}\right)=\frac{1}{3}\left(S+2 S_1+S_2\right).
\end{gather*}
Then
$$
S_2=\s... | {
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How prove this inequality $\frac{x}{x^3+y^2+z}+\frac{y}{y^3+z^2+x}+\frac{z}{z^3+x^2+y}\le 1$ for $x+y+z=3$ let $x,y,z$ be positive numbers,and such $x+y+z=3$,show that
$$\dfrac{x}{x^3+y^2+z}+\dfrac{y}{y^3+z^2+x}+\dfrac{z}{z^3+x^2+y}\le 1$$
My try:$$(x^3+y^2+z)(\dfrac{1}{x}+1+z)\ge 9$$
so
$$\dfrac{x}{x^3+y^2+z}+\dfrac{y... | Note that $2+x^3=x^3+1+1\geqslant 3x$, $y^2+1\geqslant 2y$, thus$$\dfrac{x}{x^3+y^2+z}=\frac{x}{3+x^3+y^2-x-y}\leqslant\frac{x}{3x+2y-x-y}=\frac{x}{2x+y}.$$Similarly, we can get $$\dfrac{y}{y^3+z^2+x}\leqslant\frac{y}{2y+z},\,\,\,\,\,\dfrac{z}{z^3+x^2+y}\leqslant\frac{z}{2z+x}.$$It suffices to show$$\frac{x}{2x+y}+\fra... | {
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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... | In case of $(w,a)$ or $(a,w)$ you need to consider that one of these two balls is chosen (randomly as by coin tossing, we should assume). Therefore these cases have to be weighted by a factor of $\frac 12$.
| {
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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 = \... | Substitute $x=\frac{t}{\sqrt{\cos a+t^2}}$. Then
$dx= \frac{\cos a\ dt}{(\cos a+t^2)^{3/2}}$ and
\begin{align}\int_{0}^{1}\frac{\sqrt{1-x^2}}{1-x^2\sin^2 a}dx
=& \int_{0}^{\infty}\frac{\sqrt{\cos a}\ \frac1{t^2} }{(t^2+ \frac1{t^2} )\cos a +(1+\cos^2a)}\overset{t\to 1/t}{dt}\\
=& \ \frac{\sqrt{\cos a}}2\int_{0}^{\inft... | {
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"question_score": "6",
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"answer_id": 1
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How find this sum $\sum_{n=1}^{\infty}n\sum_{k=2^{n-1}}^{2^n-1}\frac{1}{k(2k+1)(2k+2)}$ Find the sum
$$\sum_{n=1}^{\infty}n\sum_{k=2^{n - 1}}^{2^{n}\ -\ 1}\dfrac{1}{k(2k+1)(2k+2)}$$
My try:
note
$$\dfrac{1}{k(2k+1)(2k+2)}=\dfrac{2}{(2k)(2k+1)(2k+2)}=\left(\dfrac{1}{(2k)(2k+1)}-\dfrac{1}{(2k+1)(2k+2)}\right)$$
Then I ca... | Are you really sure about the outer $n$? If not, Wolfram Alpha gives a sum that telescopes nicely
$$\sum_{k=2^{n-1}}^{2^n-1} \frac{1}{k (2 k+1) (2 k+2)}=-2^{-n-1}-\psi^{(0)}(2^{n-1})+\psi^{(0)}(2^n)-\psi^{(0)}\left(\frac{1}{2}+2^n\right) +\psi^{(0)}\left(\frac{1}{2}+2^{n-1}\right)$$
Update:
It is not hard to prove user... | {
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"question_score": "6",
"answer_count": 3,
"answer_id": 2
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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.
| Let $f(x) = 1+x + x^2+x^3+ \cdots$. Then the radius of convergence of $f$ is $1$, and inside this disc we have $f(x) = \frac{1}{1-x}$, and $f'(x) = 1+2x+3x^2+\cdots = \frac{1}{(1-x)^2}$.
Suppose $|x|<1$, then we have $xf'(x) = x+2x^2+3x^3+\cdots = \frac{x}{(1-x)^2}$.
If we choose $|a| >1$, then letting $x = \frac{1}{a... | {
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$\left( \frac{1 \cdot 2}{p} \right) + \left( \frac{2 \cdot 3}{p} \right) + \cdots + \left( \frac{(p-2)(p-1)}{p} \right) = -1$ Let $p$ be an odd prime number. Prove that $$\left( \frac{1 \cdot 2}{p} \right) + \left( \frac{2 \cdot 3}{p} \right) + \left( \frac{3 \cdot 4}{p} \right) + \cdots + \left( \frac{(p-2)(p-1)}{p} \... | Let $a^\ast$ be the inverse of $a$ modulo $p$. Then
$$\left(\frac{a(a+1)}{p}\right)=\left(\frac{a(a+aa^\ast)}{p}\right)=\left(\frac{a^2(1+a^\ast)}{p}\right)=\left(\frac{1+a^\ast}{p}\right).$$
As $a$ ranges from $1$ to $p-2$, the number $1+a^\ast$ ranges, modulo $p$, through the integers from $2$ to $p-1$. But the su... | {
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"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
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Show that 2 surfaces are tangent in a given point Show that the surfaces $ \Large\frac{x^2}{a^2} + \Large\frac{y^2}{b^2} = \Large\frac{z^2}{c^2}$ and $ x^2 + y^2+ \left(z - \Large\frac{b^2 + c^2}{c} \right)^2 = \Large\frac{b^2}{c^2} \small(b^2 + c^2)$ are tangent at the point $(0, ±b,c)$
To show that 2 surfaces are tan... | The respective gradients of the surfaces are locally perpendicular to them:
$$
\nabla f_1 = 2\left(\frac{x}{a^2}, \frac{y}{b^2}, -\frac{z}{c^2} \right) \\
\nabla f_2 = 2\left(x, y, z - \frac{b^2+c^2}{c}\right)
$$
At any point of common tangency, the gradients are proportional. Therefore $\nabla f_1 = \lambda \nabla f_2... | {
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"question_score": "4",
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Integral $\int_0^\infty\frac{1}{x\,\sqrt{2}+\sqrt{2\,x^2+1}}\cdot\frac{\log x}{\sqrt{x^2+1}}\mathrm dx$ I need your assistance with evaluating the integral
$$\int_0^\infty\frac{1}{x\,\sqrt{2}+\sqrt{2\,x^2+1}}\cdot\frac{\log x}{\sqrt{x^2+1}}dx$$
I tried manual integration by parts, but it seemed to only complicate the i... | This is not a full answer but how far I got, maybe someone can complete
$\frac{1}{x\,\sqrt{2}+\sqrt{2\,x^2+1}}=\frac{1}{\sqrt{2\,x^2}+\sqrt{2\,x^2+1}}=\frac{\sqrt{2\,x^2+1}-\sqrt{2\,x^2}}{\sqrt{2\,x^2+1}^2-\sqrt{2\,x^2}^2}=\sqrt{2\,x^2+1}-\sqrt{2\,x^2}$
So you are looking at
$$\int_0^\infty(\sqrt{\frac{2\,x^2+1}{x^2+1}... | {
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"question_score": "34",
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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 $x+y = x^2 + y^2 - xy$
| Added: The approach below is ugly: It would be most comfortable to delete.
We look at your second equation. Look first at the case $x\ge 0$, $y\ge 0$. We have $x^2+y^2-xy=(x-y)^2+xy$. Thus $x^2+y^2-xy\ge xy$. So if the equation is to hold, we need $xy\le x+y$.
Note that $xy-x-y=(x-1)(y-1)-1$. The only way we can have... | {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Understanding an Approximation I am reading the paper A Group-theoretic Approach to Fast Matrix Multiplication and there is an approximation in the paper I don't fully understand.
In the proof of Theorem 3.3. it is stated that
$$
\frac{\ln (n(n+1)/2)!)}{\ln (1!2!\dots n!)}= 2+ \frac{2-\ln 2}{\ln n}+O\left( \frac{1}{\l... | The highest order term in $\log \prod\limits_{k=1}^n k!$ is $\frac{n^2}{2}\log n$, so in the desired
$$\log \left(\frac{n(n+1)}{2}\right){\Large !} = \left(2 + \frac{2-\log 2}{\log n} + O\left(\frac{1}{\log^2 n}\right)\right)\log \prod_{k=1}^n k!,\tag{1}$$
we get on the right hand side a term $O\left(\frac{n^2}{\log n}... | {
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Radius and Interval of Convergence of $\sum (-1)^n \frac{ (x+2)^n }{n2^n}$ I have found the radius of convergence $R=2$ and the interval of convergence $I =[-2,2)$ for the following infinite series:
$\sum_{n=1}^\infty (-1)^n \frac{ (x+2)^n }{n2^n}$
Approach:
let
$a_n = (-1)^n \frac{ (x+2)^n }{n2^n}$
Take the ratio test... | The radius of convergence is right. The interval of convergence started out OK, you wrote that for sure we have convergence if $-2\lt x+2\lt 2$. But this should become $-4\lt x\lt 0$.
The endpoint testing was inevitably wrong, since the incorrect endpoints were being tested.
We have convergence at $x=0$ (alternating s... | {
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"question_score": "3",
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In a triangle $\angle A = 2\angle B$ iff $a^2 = b(b+c)$ Prove that in a triangle $ABC$, $\angle A = \angle 2B$, if and only if:
$$a^2 = b(b+c)$$
where $a, b, c$ are the sides opposite to $A, B, C$ respectively.
I attacked the problem using the Law of Sines, and tried to prove that if $\angle A$ was indeed equal to $2\... | $$a^2-b^2=bc\implies \sin^2A-\sin^2B=\sin B\sin C\text{ as }R\ne0$$
Now, $\displaystyle\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)=\sin(\pi-C)\sin(A-B)=\sin C\sin(A-B)\ \ \ \ (1)$
$$\implies \sin B\sin C=\sin C\sin(A-B)$$
$$\implies \sin B=\sin(A-B)\text{ as }\sin C\ne0$$
$$\implies B=n\pi+(-1)^n(A-B)\text{ where }n\text... | {
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Find the maximum and minimum values of $A \cos t + B \sin t$ Let $A$ and $B$ be constants. Find the maximum and minimum values of $A \cos t + B \sin t$.
I differentiated the function and found the solution to it as follows:
$f'(x)= B \cos t - A \sin t$
$B \cos t - A \sin t = 0 $
$t = \cot^{-1}(\frac{A}{B})+\pi n$
Howe... | $A\cos t+ B \sin t = \sqrt{A^2+B^2} ( \frac{A}{\sqrt{A^2+B^2}} \cos t + \frac{B}{\sqrt{A^2+B^2}} \sin t)$. Choose $\theta$ such that $e^{i \theta} = \frac{A}{\sqrt{A^2+B^2}} + i\frac{B}{\sqrt{A^2+B^2}} $. Then
$A\cos t+ B \sin t = \sqrt{A^2+B^2} ( \cos \theta \cos t + \sin \theta \sin t) = \sqrt{A^2+B^2} \cos(\theta-t... | {
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Complex numbers problem I have to solve
where n is equal to n=80996.
| $$\frac{\sqrt3+5i}{4+2\sqrt3i}=\frac{\sqrt3+5i}{4+2\sqrt3i}\cdot\frac{4-2\sqrt3i}{4-2\sqrt3i}=\frac{14\sqrt3+14i}{28}=\frac{\sqrt3+i}2=\cos\frac{\pi}6+i\sin\frac{\pi}6;$$$$\left(\frac{\sqrt3+5i}{4+2\sqrt3i}\right)^n=\left(\cos\frac{\pi}6+i\sin\frac{\pi}6\right)^n.$$
| {
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"question_score": "1",
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How prove this inequality $\sum\limits_{k=1}^{n}\frac{1}{k!}-\frac{3}{2n}<\left(1+\frac{1}{n}\right)^n<\sum\limits_{k=1}^{n}\frac{1}{k!}$ show that
$$\sum_{k=1}^{n}\dfrac{1}{k!}-\dfrac{3}{2n}<\left(1+\dfrac{1}{n}\right)^n<\sum_{k=0}^{n}\dfrac{1}{k!}(n\ge 3)$$
Mu try: I konw
$$\sum_{k=0}^{\infty}\dfrac{1}{k!}=e$$
and I ... | I think you mean this inequality$$\sum_{k=0}^{n}\dfrac{1}{k!}-\dfrac{3}{2n}<\left(1+\dfrac{1}{n}\right)^n.$$In fact, we can prove a sharper one $$\left(1+\frac{1}{n}\right)^n+\frac{3}{2n}>e.$$Let $f(x)=\left(1+\dfrac{1}{x}\right)^x+\dfrac{3}{2x}$, then$$f'(x)=\left(1+\dfrac{1}{x}\right)^x\left(\ln\left(1+\frac{1}{x}\ri... | {
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"source": "stackexchange",
"question_score": "1",
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consecutive convergents Problem: Let $\phi=\frac{1+\sqrt{5}}{2}$ be the golden ratio and let $a$, $b$, $c$, $d$ be positive integers so that $\frac{a}{b}>\phi>\frac{c}{d}$. It is also known that $ad-bc=1$. Prove that $a/b$ and $c/d$ are consecutive convergents of $\phi$.
Numerical experimentations point towards the va... | Note that $\phi=1+\frac{1}{1+\frac{1}{1+\ldots}}$ has convergents $\frac{f_{n+1}}{f_n}$, i.e. ratios of consecutive Fibonacci numbers. Note that I have used lower case for the Fibonacci numbers so as to avoid confusion with the Farey sequence $F_n$.
The main idea is to appeal to the properties of Farey sequences.
*
... | {
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$1 + \frac{1}{1+2}+ \frac{1}{1+2+3}+ ... + \frac{1}{1+2+3+...+n} = ?$ How do I simplify the following series
$$1 + \frac{1}{1+2}+ \frac{1}{1+2+3}+ \frac{1}{1+2+3+4} + \frac{1}{1+2+3+4+5} + \cdot\cdot\cdot + \frac{1}{1+2+3+\cdot\cdot\cdot+n}$$
From what I see, each term is the inverse of the sum of $n$ natural numbers.
... | HINT: $$\frac {2} {n(n+1)} = \frac 2 n - \frac 2 {n+1}$$
| {
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"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Solving for the zero of a multivariate
How does one go about solving the roots for the following equation
$$x+y+z=xyz$$
There simply to many variables. Anyone have an idea ?
| If we fix one of the variables, we get a hyperbola in that plane. So, for example, fixing any $z = z_0,$ this is your relationship:
$$ \left(x - \frac{1}{z_0} \right) \left(y - \frac{1}{z_0} \right) = \; 1 + \frac{1}{z_0^2} $$
Makes me think the surface could be connected.
Indeed, as $|z| \rightarrow \infty,$ the cu... | {
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Proving that if these quadratics are equal for some $\alpha$, then their coefficients are equal Let
$$P_1(x) = ax^2 -bx - c \tag{1}$$
$$P_2(x) = bx^2 -cx -a \tag{2}$$
$$P_3(x) = cx^2 -ax - b \tag{3}$$
Suppose there exists a real $\alpha$ such that
$$P_1(\alpha) = P_2(\alpha) = P_3(\alpha)$$
Prove
$$a=b=c$$
Equating ... | Denote
$$Q_1(x)=P_1(x)-P_2(x)=(a-b)x^2-(b-c)x-(c-a);$$
$$Q_2(x)=P_2(x)-P_3(x)=(b-c)x^2-(c-a)x-(a-b);$$
$$Q_3(x)=P_3(x)-P_1(x)=(c-a)x^2-(a-b)x-(b-c).$$
Then $\alpha$ is a real root of the equations $Q_i(x);$ so that $\Delta_{Q_i(x)}\geq 0 \ \ \forall i=1,2,3;$ where $\Delta_{f(x)}$ denoted the discriminant of a quadrati... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Rudin Example 3.35B Why the $n$th root of $a_n$ is $1/2$? For Baby Rudin Example 3.35(b), I understand how the $\liminf$ and $\limsup$ of the ratio test were found, but I am not clear why $\ \lim \sqrt[n]{a_n } = \frac{1}{2} $.
Please help.
| The sequence in question is
$$\frac{1}{2} + 1 + \frac{1}{8} + \frac{1}{4}+ \frac{1}{32}+ \frac{1}{16}+\frac{1}{128}+\frac{1}{64}+\cdots$$
In case the pattern is not clear, we double the first term, the divide the next by $8$, the double, then divide by $8$, and so on.
The general formula for an odd term is $a_{2k-1}=\... | {
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"source": "stackexchange",
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Value of sum of telescoping series $$\sum_{n\geqslant1}\frac{1}{\sqrt{n}} -\frac{ 1}{\sqrt {n+2}}$$
In looking at the first five partial sums, I am not convinced the series is telescopic (the middle terms don't cancel out).
Thanks in advance!
| $\sum_{k=1} ^n \frac{1}{\sqrt{k}}- \frac{1}{\sqrt{k+2}}$
$= \frac{1}{\sqrt{1}}- \frac{1}{\sqrt{3}}+ \frac{1}{\sqrt{2}}- \frac{1}{\sqrt{4}} +\frac{1}{\sqrt{3}}- \frac{1}{\sqrt{5}}+ \frac{1}{\sqrt{4}}- \frac{1}{\sqrt{6}}+ \frac{1}{\sqrt{5}}- \frac{1}{\sqrt{7}}+...+ \frac{1}{\sqrt{n-2}}- \frac{1}{\sqrt{n}}+ \frac{1}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/568227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How prove this $a_{n}>1$ let $0<t<1$, and $a_{1}=1+t$, and such
$$a_{n}=t+\dfrac{1}{a_{n-1}}$$
show that $a_{n}>1$
My try: since
$$a_{1}=1+t>1$$
$$a_{2}=t+\dfrac{1}{a_{1}}=t+1+\dfrac{1}{1+t}-1>2\sqrt{(t+1)\cdot\dfrac{1}{1+t}}-1=2-1=1$$
$$a_{3}=t+\dfrac{1}{a_{2}}=t+\dfrac{1}{t+\dfrac{1}{t+1}}=t+\dfrac{t+1}{t^2+t+1}=1+\... | Let $\displaystyle \mu = \frac{t + \sqrt{t^2+4}}{2}$, we have
$$\mu > 1\quad\text{ and }\quad\mu(t - \mu) = \left(\frac{t + \sqrt{t^2+4}}{2}\right)\left(\frac{t - \sqrt{t^2+4}}{2}\right) = -1$$
From this, we get
$$a_{n+1} - \mu = t - \mu + \frac{1}{a_n} = \frac{1}{a_n} - \frac{1}{\mu} = \frac{\mu - \alpha_n}{\mu a_n}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/570116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Getting angles for rotating $3$D vector to point in direction of another $3$D vector I've been trying to solve this in Mathematica for $2$ hours, but got the wrong result.
I have a vector, in my case $\{0, 0, -1\}$. I want a function that, given a different vector, gives me angles DX and DY, so if I rotate the original... | So you need a 3×3 rotation matrix $E$ such that
$$ E\,\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ -1 \end{pmatrix} = -\hat{k} $$
This rotation matrix consists of two elementary rotations
$$ \begin{aligned}
E & = {\rm Rot}(\hat{i},\varphi_x){\rm Rot}(\hat{j},\varphi_y) \\
& = \begin{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/575472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solution to the limit of a series I'm strugling with the following problem:
$$\lim_{n\to \infty}(n(\sqrt{n^2+3}-\sqrt{n^2-1})), n \in \mathbb{N}$$
Wolfram Alpha says the answer is 2, but I don't know to calculate the answer.
Any help is appreciated.
| For the limit: We take advantage of obtaining a difference of squares.
We have a factor of the form $a - b$, so we multiply it by $\dfrac{a+b}{a+b}$ to get $\dfrac{a^2 - b^2}{a+b}.$
Here, we multiply by $$\dfrac{\sqrt{n^2+3}+ \sqrt{n^2 - 1}}{\sqrt{n^2+3}+ \sqrt{n^2 - 1}}$$
$$n(\sqrt{n^2+3}-\sqrt{n^2-1})\cdot\dfrac{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/576550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Prove $13|19^n-6^n$ by congruences I am trying to prove $13|19^n-6^n$. With induction its not so bad but by congruences its quite difficult to know how to get started.
Any hints?
| Because $x^n - y^n$ is divisible by $x-y$ as
$$x^n - y^n = (x-y)\sum_{i=0}^{n-1}x^iy^{n-1-i}$$
Substitute $x=19$ and $y = 6$.
$$\begin{align*}
x^n-y^n =& x^n\left[1-\left(\frac yx\right)^n\right]\\
=& x^n \left[1+\left(\frac yx\right)+\left(\frac yx\right)^2+\cdots+\left(\frac yx\right)^{n-1}\right]\left[1-\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/582670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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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... | $$t+\frac1t=18\sqrt3\iff t^2-(2\cdot9\sqrt3)t+1=0\iff t_{1,2}=\frac{9\sqrt3\pm\sqrt{(9\sqrt3)^2-1\cdot1}}1=$$
$$=9\sqrt3\pm\sqrt{81\cdot3-1}\quad=\quad9\sqrt3\pm\sqrt{243-1}\quad=9\sqrt3\pm\sqrt{242}\quad=\quad9\sqrt3\pm\sqrt{2\cdot121}=$$
$$=9\sqrt3\pm\sqrt{2\cdot11^2}\quad=\quad9\sqrt3\pm11\sqrt2\quad\iff\quad x_{1,2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/583062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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How to prove $\frac{y^2-x^2}{x+y+1}=\pm1$ is a hyperbola? How to prove $\frac{y^2-x^2}{x+y+1}=\pm1$ is a hyperbola, knowing the canonical form is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=\pm1$ where $a$ and $b$ are constants? Thanks !
| Let
$$
\frac{y^2-x^2}{x+y+1}=1\\
\Rightarrow y^2-x^2=x+y+1\\
\Rightarrow y^2-x^2-x-y=1
$$
Complete the squares for x and y . You will get rectangular hyperbola. Similar will be the case if
$$
\frac{y^2-x^2}{x+y+1}=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/585301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Checking on some convergent series I need some verification on the following 2 problems I attemped:
I have to show that the following series
is convergent: $$1-\frac{1}{3 \cdot 4}+\frac{1}{ 5 \cdot 4^2 }-\frac{1}{7 \cdot 4^3}+ \ldots$$ .
My Attempt: I notice that the general term is given by $$\,\,a_n=(-1)^{n}{1 ... | Hint: For all sufficiently large $n$ (in fact, $n \ge 1$ suffices for this), we have $\ln{n} \le \sqrt{n}$; thus
$$\sum\limits_{n = 2}^{\infty} \frac{\ln n}{n^2} \le \sum\limits_{n = 2}^{\infty} \frac{1}{n^{3/2}}$$
which is a $p$-series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/587782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $f$ is continuous, why is $f$ with the property $f\left(\frac{x+y}{2}\right)\le \frac{1}{2}f(x)+\frac{1}{2}f(y)$ is convex?
If $f$ is continuous, why is $f$ with the property
$$f\left(\frac{x+y}{2}\right)\le \frac{1}{2}f(x)+\frac{1}{2}f(y),$$ where $0\le x,y\le 1$
is convex?
| By induction we can prove that: if $k,m, l\in\mathbb{N} , k+m=2^l , x,y\in \mbox{domain} f $ then $$f\left( \frac{k}{2^l} \cdot x +\frac{m}{2^l} \cdot y \right)\leq \frac{k}{2^l} \cdot f(x) +\frac{m}{2^l} \cdot f(y). $$
Indeed the asertion is true when $l=1 .$ Suppose that it is true for some $l\geq 1 .$ And let $k+m=2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Different ways of proving a polynomial is irreducible in finite field. Is there a general characterization of irreducible polynomials over a finite field?
I was going through a problem in finding whether $p(x):=x^7+x^5+1$ is irreducible over $\mathbb F_2[x]$ or not.
If the polynomial is of degree less than or equal ... | The given polynomial is in fact not irreducible. There is at least one decomposition:
$$
(x^2+x+1)\cdot(x^5+x^4+x^3+x+1) = x^7+x^5+1
$$
This can be found by resolving the equality for coefficients:
$$
(x^2+ax+1)\cdot(x^5+bx^4+cx^3+dx^2+ex+1) = x^7+x^5+1
$$
which, equating term by term, and ignoring the terms of degree ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/589625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Question regarding 3 x 3 matrices If $A$ is a $3 \times 3$ matrix with real elements and $\det(A)=1$, then are these affirmations equivalent:
$$
\det(A^2-A+I_3)=0 \leftrightarrow \det(A+I_3)=6 \text{ and } \det(A-I_3)=0?
$$
| $\Leftarrow)$ From $\det (A-I)=0$ we know that $1$ is an eigenvalue. Let $x,y$ be the two others (possibly complex, and counting multiplicities). From $\det A=1$ we know that $xy=1$. And $6=\det(A+I)=2(x+1)(y+1)$, so
$$
3=xy+x+y+1=2+x+y,
$$
so $x+y=1$. We get a system of two equations on $x,y$, namely
$$
x+y=1,\ \ xy=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/589971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $A_k=(-1)^k\binom nk$?
In the identity $$\frac{n!}{x(x+1)(x+2)\cdots(x+n)}=\sum_{k=0}^n\frac{A_k}{x+k},$$prove that $$A_k=(-1)^k\binom nk.$$
My try: The given identity implies $$\frac{1\cdot2\cdots n}{x(x+1)(x+2)\cdots(x+n)}=\frac{A_0}{x}+\frac{A_1}{x+1}+\dots+\frac{A_n}{x+n}.$$
Now putting $A_k=(-1)... | HINT: Multiply both side by $x+k$ and then put $x=-k.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this reccurence relation? Let a,b,c be real numbers. Find the explicit formula for $f_n=af_{n-1}+b$ for $n \ge 1$ and $f_0 = c$
So I rewrote it as $f_n-af_{n-1}-b=0$ which gives the characteristic equation as $x^2-ax-b=0$. The quadratic formula gives roots $x= \frac{a+\sqrt{a^2+4b}}{-2}, \frac{a-\sqrt{a^2+... | Why not consider this?
$f_n + m = a(f_{n-1} + m) \Longrightarrow (a-1)m=b$
1) $a=1$, simple recurrence $f_n = f_{n-1} + b$, $f_n = bn+c$
2) $a\neq 1$, $m=\frac{b}{a-1}$, $f_n+m = a(f_{n-1}+m)$, geometric sequence $f_n+m=a^n(c+m)$
Hope it is helpful!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/592215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$? I'd like to find out why
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6
\end{align}
I tried to rewrite it into a geometric series
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = \sum_{n=0}^{\infty} \Big(\frac{1}{2}\Big)^nn^2
\end{align}
But I don't know... | Let me show you a slightly different approach; this approach is very powerful, and can be used to compute values for a large number of series.
Let's think of this in terms of power series. You noticed that you can write
$$
\sum_{n=0}^{\infty}\frac{n^2}{2^n}=\sum_{n=0}^{\infty}n^2\left(\frac{1}{2}\right)^n;
$$
so, let'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/593996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 8,
"answer_id": 4
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How to put $2x^2 + 4xy + 6y^2 + 6x + 2y = 6$ in canonical form We are given the equation $2x^2 + 4xy + 6y^2 + 6x + 2y = 6$
We did an example of this in class but the equation had less terms.
I took a note in class that says : if there are linear terms, I have to rotate...
This is what I think I have to do.
*
*Put t... | Not knowing what exactly "canonical form" is, here is what I get.
Translating to get rid of the linear terms:
$$
2(x+2)^2+4(x+2)(y-1/2)+6(y-1/2)^2=\frac{23}{2}\tag{1}
$$
With $P=\dfrac{\sqrt{2+\sqrt2}}{2}\begin{bmatrix}1&1-\sqrt2\\-1+\sqrt2&1\end{bmatrix}=\begin{bmatrix}\cos(\pi/8)&-\sin(\pi/8)\\\sin(\pi/8)&\hphantom{+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/594693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the value of $\cos(2\pi /5)$ using radicals This is homework so if there is another example that can illustrate the technique I would happily accept that as guidance. The only thing I have been able to find is a question asking about $\cos(2\pi/7)$, which I think is a much harder problem.
I dont have the faintest ... | Note that
$2\cdot \dfrac{2\pi}{5} + 3\cdot \dfrac{2\pi}{5} = 2\pi,$
therefore
$\cos(2⋅\dfrac{2\pi}{5})=\cos(3⋅\dfrac{2\pi}{5})$.
Put $\cos(\dfrac{2\pi}{5})=x$. Using the formulas
$\cos2x=2\cos2x−1,\cos 3x=4\cos 3x−3\cos x$,
we have
$4x^3−2x^2−3x+1=0⇔(x−1)(4x^2+2x−1)=0$.
Because $\cos(\dfrac{2\pi}{5})≠1$, we get
$4x^2+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/595162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding a 2x2 Matrix raised to the power of 1000 Let $A= \pmatrix{1&4\\ 3&2}$. Find $A^{1000}$.
Does this problem have to do with eigenvalues or is there another formula that is specific to 2x2 matrices?
| Perform an eigenvalue decomposition of $A$, we then get
$$A =
\begin{bmatrix}
-4/5 & -1/\sqrt2\\
3/5 & -1/\sqrt2
\end{bmatrix}
\begin{bmatrix}
-2 & 0\\
0 & 5
\end{bmatrix}
\begin{bmatrix}
-4/5 & -1/\sqrt2\\
3/5 & -1/\sqrt2
\end{bmatrix}^{-1}
=VDV^{-1}
$$
where $V = \begin{bmatrix}
-4/5 & -1/\sqrt2\\
3/5 & -1/\sqrt2
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/597602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
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Equations and inequalities as parameters: proving that an equation holds. I have $x-y=3$ and $y\le1$ and $x\ge\frac12$.
I proved that $\sqrt{(2x-1)^2}+\sqrt{(2y-2)^2}=7$ and that $-\frac52\le y\le 1$ and $\frac12\le x\le4$.
How can I prove that $|x+y-5|+|x+y+2|=7$?
| ok let us consider following cases :
first side is just $-x-y+5$,second case is $x+y+2$,so sum is $5+2=7$
to be more deeply,let us take such situations
*
*$y=1$
2.$x=4$
we have
$|x+y-5]+|x+y+2|=x+y-5+x+y+2=2*x+2*y-3=2*(x+y)-3
=2*(5)-3=7$
can you continue from this?
just consider this situation when $x<0$ th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/598192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Concentric and Tangent Ellipse from 2 Hyperbolas Find the equation of the ellipse that is concentric and tangent to the following hyperbolas:
$$\begin{align}
-2x^2 + 9y^2 - 20x - 108y + 256 &= 0 \\
x^2 - 4y^2 + 10x + 48y - 219 &= 0
\end{align}$$
I did the math for both equations and the center is the same: $(-5,6)$.
... | Not a complete solution, but this approach will work:
An ellipse with centre at the point $(-5,6)$ would be
$$\frac{(x+5)^2}{a^2}+\frac{(y-6)^2}{b^2} = 1$$
Now change to a new set of axes ($u, v$) parallel to the $x,y$ axes, but with origin at the point $(-5,6)$. In other words, put $u = x+5$ and $v = y-6$.
Referred t... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of $\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)$ I have to determine the following:
$\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)$
$\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)=\lim\limits_{x \rightarrow \infty}(\sqrt{x^8(1+\frac{4}{x^8})}-x^4 = \lim\limits_{x \rightarrow \infty}(x^4\sqrt... | A short way to (non-rigorously) find the limit is to observe that for large $x$,
$$
\sqrt{x^8+4} \approx \sqrt{x^8}=x^4
$$
so that for large $x$ (especially in $\lim_{x \to \infty}$)
$$
\sqrt{x^8+4}-x^4 \approx x^4-x^4=0
$$
So the limit must be $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/598928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral involving a confluent hypergeometric function I have the following integral involving a confluent hypergeometric function:
$$\int_{0}^{\infty}x^3e^{-ax^2}{}_1F_1(1+n,1,bx^2)dx$$
where $a>b>0$ are real constants, and $n\geq 0$ is an integer.
Wolfram Mathematica returns the following solution: $\frac{a^{n-1}(a+b... | Let's start with the hypergeometric function. We have:
\begin{eqnarray}
F_{2,1}[1+n,1;b x^2] &=&
\sum\limits_{m=0}^\infty \frac{(1+n)^{(m)}}{m!} \cdot \frac{(b x^2)^m}{m!} \\
&=& \sum\limits_{m=0}^\infty \frac{(m+1)^{(n)}}{n!} \cdot \frac{(b x^2)^m}{m!} \\
&=&\left. \frac{1}{n!} \frac{d^n}{d t^n} \left( t^n \cdot e^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $f$ if $f(f(x))=\sqrt{1-x^2}$ Find $f$ if $f(f(x))=\sqrt{1-x^2} \land [-1; 1] \subseteq Dom(f)$
$$$$Please give both real and complex functions. Can it be continuous or not (if f is real)
| I guessed that $f$ is of the form $\sqrt{ax^2+b}$. Then, $f^2$ is $\sqrt{a^2x^2 + \frac{a^2-1}{a-1}b}$. From here on in, it is algebra:
$$
a^2 =-1 \implies a = i ~~~~\text{and}~~~~\frac{a^2-1}{a-1}b = 1 \implies b = \frac{1-i}{2}
$$
So we get $f(x) = \sqrt{ix^2 + \frac{1-i}{2}}$. I checked using Wolfram, and $f^2$ appe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/607234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$ How can I prove that
$$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\frac{11}{4}\zeta(3)+\zeta(2)+4\log(2)-4$$
I think this post can help me, but I'm not sure.
| Different approach:
$$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\sum_{n=1}^\infty\frac{H_{2n}+\frac{1}{2n+1}}{n^2}$$
$$=\sum_{n=1}^\infty\frac{H_{2n}}{n^2}+\sum_{n=1}^\infty\frac{1}{n^2(2n+1)}$$
where
$$\sum_{n=1}^\infty\frac{H_{2n}}{n^2}=4\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^2}$$
$$=4\sum_{n=1}^\infty\frac{1}{2n}\left(-\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/609056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 1
} |
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 ... | We have $$(x-1)^2=x^2+1-2x\ge0\iff x^2+1\ge 2x$$
and since $\sin^2\theta\ge0$ hence we have $x>0$ and then
$$1\ge\sin^2\theta =\frac{x^2+y^2+1}{2x}\ge1\quad (>1\;\text{if}\; y\ne 0\;\text{which gives a contradiction})$$
hence $y=0$ and
$$\sin^2\theta=\frac{x^2+1}{2x}=1\iff x=1$$
| {
"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": 1
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A fair dice is tossed until a number greater than $4$ appears. The probability that an even number of tosses will be required is A fair dice is tossed until a number greater than $4$ appears. The probability that an even number of tosses will be required is:
$A. 1/2$
$B. 3/5$
$C. 1/5$
$D. 2/3$
What I did: The probabili... | You didn't go wrong. The probability of "success" is 1/3, so the probability of succeding in $2k$ tosses is
$$\left(\frac{2}{3}\right)^{2k-1} \frac{1}{3}$$
And
$$\sum_{k=1}^\infty\left(\frac{2}{3}\right)^{2k-1} \frac{1}{3}=\frac{1}{3} \frac{3}{2}\sum_{k=1}^\infty\left(\frac{4}{9}\right)^k =\frac{1}{2} \frac{4/9}{1-4/9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/610731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$\cos^2\frac{1}{2}(\alpha-\beta)=\frac{3}{4}$ if........... Help please: If $\sin\alpha+\sin\beta= \sqrt{3} (\cos\beta-\cos\alpha)$ then show that $\cos^2\frac{1}{2}(\alpha-\beta)=\frac{3}{4}$ please tell me how can... | Apart from the Prosthaphaeresis Formulas already mentioned with the unmentioned assumption that $\displaystyle \sin\frac{\alpha+\beta}2\ne0$
we can try as follows :
Rearranging we have $\displaystyle\sin\alpha+\sqrt3\cos\alpha=\sqrt3\cos\beta-\sin\beta$
As $\displaystyle 1^2+(\sqrt3)^2=4,$ we write $1=2\sin30^\circ,\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/614789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Doubts in Trigonometrical Inequalities I'm now studying Trigonometrical Inequalities, and I've just got struck when I have modified arguments to my trigonometrical functions, for example:
$\sqrt{2} - 2\sin\left(x - \dfrac{\pi}{3} \right) < 0$ when $-\pi < x < \pi$
With some work I've got: $\sin\left(x - \dfrac{\pi}{3} ... | The inequality
$$
\sin \left( x - \frac{\pi}{3} \right) > \frac{\sqrt{2}}{2}
$$
implies that
$$
\frac{\pi}{4} < x - \frac{\pi}{3} < \frac{3\pi}{4}.
$$
Adding $\frac{\pi}{3}$ to all three expressions yields
$$
\frac{7\pi}{12} < x < \frac{13\pi}{12}.
$$
If you impose the initial restriction, then the upper bound is $\pi$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/615272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Help finding the $\lim\limits_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$ I need help finding the $$\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$$
I did the following:
$$\begin{align*}
\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \... | $$\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}=\lim_{x \to \infty} \frac{x^{1/3}-x^{1/5}}{x^{1/3} + x^{1/5}}=$$
$$=\lim_{x \to \infty} \frac{1-x^{1/5-1/3}}{1 + x^{1/5-1/3}}=\lim_{x \to \infty} \frac{1-x^{-2/15}}{1 + x^{-2/15}}=\lim_{x \to \infty} \frac{1-\frac{1}{x^{2/15}}}{1 + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/615505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$, where $x$, $y$ and $z$ are positive integers Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$ , where $x,y,z$ are positive integers.
Found ten solutions $(x,y,z)$ as ${(3,3,3),(2,4,4),(4,2,4),(4,4,2),(2,3,6),(... | We may as well assume $x\le y\le z$ (and then count rearrangements of the variables as appropriate). The smallest variable, $x$, cannot be greater than $3$ (or else $1/x+1/y+1/z\lt1/3+1/3+1/3=1$), nor can it be equal to $1$ (or else $1/x+1/y+1/z=1+1/y+1/z\gt1$). So either $x=2$ or $x=3$.
If $x=3$, then $y=z=3$ as wel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
Double Euler sum $ \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} $ I proved the following result
$$\displaystyle \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} =- \frac{97}{12} \zeta(6)+\frac{7}{4}\zeta(4)\zeta(2) + \frac{5}{2}\zeta(3)^2+\frac{2}{3}\zeta(2)^3$$
After consideration of powers of polylogarithms.
You can refer to t... | The following new solution is proposed by Cornel Ioan Valean. Based on a few ideas presented in the book, (Almost) Impossible Integrals, Sums, and Series, like the Cauchy product of $(\operatorname{Li}_2(x))^2$, that is $\displaystyle (\operatorname{Li}_2(x))^2=4\sum_{n=1}^{\infty}x^n\frac{H_n}{n^3}+2\sum_{n=1}^{\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Help with finding tangent to curve at a point
Find an equation for the tangent to the curve at $P\left( \dfrac{\pi}{2},3 \right )$ and the horizontal tangent to the curve at $Q.$
$$y=5+\cot x-2\csc x$$
$y\prime=-\csc ^2 x -2(-\csc x \cot x)$
$y\prime= 2\csc x \cot x - \csc ^2 x\implies$ This is the equation of the... | The first part looks good. For the second part as you see the solution to the equation gives
$$2 \cot x \csc x - \csc^2 x = 0 \Rightarrow x=2n \pi \pm\frac{\pi}{3}.$$
Choose $x=\frac{\pi}{3}$ as it is one such solution. We find the point on $y$ to be
$$5+\cot\left( \frac{\pi}{3} \right)-2\csc\left( \frac{\pi}{3} \right... | {
"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": 0
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Plane Geometry problem I came across this problem in a mathematics-related facebook group. Could anyone advise on the solution to it(i.e. hints only)? Thank you.
|
Since triangle ABC is equilateral, we know that angle BAC is 60º. Therefore, angle PAQ is also 60º; we have a circle theorem that tells us that angle POQ is thus twice that measure or 120º. The diameter AD bisects that angle, so angle POD is 60º.
Diameter AD is also an altitude of equilateral triangle ABC: thus, if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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evaluation of $\lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} = $ (1) $\displaystyle \lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} = $, where $n\in \mathbb{N}$ and $\lfloor x \rfloor = $ floor function of $x$
(2)$\displaystyle \lim_{x\rightarrow \infty}\l... | No. You cannot write so. In this case, going back to the definition is the best. So, in order to solove your questions, use the followings :
$$x-1\lt \lfloor x\rfloor \le x\Rightarrow \frac{x-1}{x}\le \frac{\lfloor x\rfloor}{x}\lt \frac{x}{x}\Rightarrow \lim_{x\to\infty}\frac{\lfloor x\rfloor}{x}=1$$
where $x\gt0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How prove this stronger than Weitzenbock's inequality:$(ab+bc+ac)(a+b+c)^2\ge 12\sqrt{3}\cdot S\cdot(a^2+b^2+c^2)$ In $\Delta ABC$,$$AB=c,BC=a,AC=b,S_{ABC}=S$$
show that
$$(ab+bc+ac)(a+b+c)^2\ge 12\sqrt{3}\cdot S\cdot(a^2+b^2+c^2)$$
I know this Weitzenböck's_inequality
$$a^2+b^2+c^2\ge 4\sqrt{3}S$$
But my inequality i... | Square both sides, put$S^2=\dfrac{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}{16}$ in, we have:
$(ab+bc+ac)^2(a+b+c)^4\ge 27\cdot (a+b+c)(a+b-c)(a+c-b)(b+c-a)(a^2+b^2+c^2)^2$
with brutal force method(BW method), WOLG let $a=$Min{$a,b,c$},$b=a+u,c=a+v,u\ge0,v\ge0$ , put in the inequality and rearrange them, we have:
$ \iff k_6a^6+k_5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/621099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
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... | For a triangle
$\Delta = \frac{abc}{4R} = rs$
Now in your inequality
you can put in the values to get
$R \ge 2r$
This is known to be true since the distance between incentre and circumcentre $d^2 = R(R-2r)$
Thus your inequality is proved
| {
"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": 0
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if range of $f(x) = \frac{x^2+ax+b}{x^2+2x+3}$ is $[-5,4]$. Then $a$ and $b$ are If Range of $\displaystyle f(x) = \frac{x^2+ax+b}{x^2+2x+3}$ is $\left[-5,4\; \right]$ for all $\bf{x\in \mathbb{R}}$. Then values of $a$ and $b$.
$\bf{My\; Try}::$ Let $\displaystyle y=f(x) = \frac{x^2+ax+b}{x^2+2x+3} = k$,where $k\in \ma... | To find the places where $f(x)$ is minimal and maximal, differentiate $f$ wrt $x$. Then solve $f'(x)=0$. Call the solution $x_0$ an $x_1$ (and so on if there are move). Now, you know for which $x$ $f(x)$ is minimal/maximal. Calculate $f(x_0)$ and $f(x_1)$. These should be equal to $-5$ and $4$. You only have to know wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/621512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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If real number x and y satisfy $(x+5)^2 +(y-12)^2=14^2$ then find the minimum value of $x^2 +y^2$ Problem :
If real number x and y satisfy $(x+5)^2 +(y-12)^2=14^2$ then find the minimum value of $x^2 +y^2$
Please suggest how to proceed on this question... I got this problem from [1]: http://www.mathstudy.in/
| As mathlove has already identified the curve to be circle $\displaystyle (x+5)^2+(y-12)^2=14^2$
But I'm not sure how to finish from where he has left of without calculus.
Here is one of the ways:
Using Parametric equation, any point $P(x,y)$ on the circle can be represented as $\displaystyle (14\cos\phi-5, 14\sin\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/623444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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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:
$$
\sqrt{x-4}-\sqrt{x-5}+1=0\\
\sqrt{x-4}=\sqrt{x-5}-1\\
\text{squaring both sides gives me:}\\
(\sqrt{x-4})^2=(\sqrt{x-5}-1)²\\
x-4=(\sqrt{x-5})²-\sqrt{x-5}+1\\
... | Easiest way to see it, is take both square roors to the other side and square, to get
$$
\begin{split}
1 &= (x-5)-(x-4) - 2\sqrt{(x-5)(x-4)}\\
2 + 2\sqrt{(x-5)(x-4)} &= 0\\
1 + \sqrt{(x-5)(x-4)} &= 0\\
\end{split}
$$
but $\sqrt{\ldots} \geq 0$ so this is impossible...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Finding the improper integral $\int^\infty_0\frac{1}{x^3+1}\,dx$ $$\int^\infty_0\frac{1}{x^3+1}\,dx$$
The answer is $\frac{2\pi}{3\sqrt{3}}$.
How can I evaluate this integral?
| $$x^3+1 = (x+1)(x^2-x+1)$$
Logic: Do partial fraction decomposition.Find $A,B,C$.
$$\frac{1}{x^3+1} = \frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}$$
By comparing corresponding co-efficients of different powers of $x$, you will end up with equations in A,B,C.After solving you get :
$$A=\frac{1}{3},B=\frac{-1}{3},C=\frac{2}{3}$$
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/625821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving an irrational equation Solve for $x$ in:
$$\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}=\sqrt{5}$$
I used the property of proportions ($a=\sqrt{3+x}$, $b=\sqrt{3-x})$:
$$\frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{2a}{2b}=\frac{a}{b}$$
I'm not sure if that's correct.
Or maybe the notations $a^3=3+x$, $b^3=3-x$... | Here is another simple way, exploiting the innate symmetry.
Let $\ \bar c = \sqrt{3\!+\!x}+\sqrt{3\!-\!x},\,\ c = \sqrt{3\!+\!x}-\sqrt{3\!-\!x}.\,$ Then $\,\color{#0a0}{\bar c c} = 3\!+\!x-(3\!-\!x) = \color{#0a0}{2x},\ $ so
$\,\displaystyle\sqrt{5} = \frac{\bar c}c\, \Rightarrow \color{#c00}{\frac{6}{\sqrt{5}}} = {\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/628314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solve $7x^3+2=y^3$ over integers I need to solve the following
solve $7 x^3 + 2 = y^3$ over integers.
How can I do that?
| To solve this kind of equations, we have several 'tools' such as
using mod, using inequalities, using factorization...
In your question, using mod will help you.
Since we have
$$y^3-2=7x^3,$$
the following has to be satisfied :
$$y^3\equiv 2\ \ \ (\text{mod $7$}).$$
However, in mod $7$,
$$0^3\equiv 0,$$
$$1^3\equiv 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/628711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Algebra 2-Factoring sum of cubes by grouping Factor the sum of cubes: $81x^3+192$
After finding the prime factorization of both numbers I found that $81$ is $3^4$
and $192$ is $2^6 \cdot 3$.
The problem is I tried grouping and found $3$ is the LCM so it would outside in parenthesis. The formula for the sum of cubes i... | $$81x^3+192 = 3 (27 x^3 + 64) = 3 ((3x)^3+4^3) \\= 3 (3x + 4) ((3x)^2 - 3x\cdot 4 + 4^2) = 3 (3x+4)(9x^2-12x+16)$$
Since $12^2-4\cdot9\cdot16$ does not have a nice square root further factorization is not possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/629765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What's the explanation for why n^2+1 is never divisible by 3? What's the explanation for why $n^2+1$ is never divisible by $3$?
There are proofs on this site, but they are either wrong or overcomplicated.
It can be proved very easily by imagining 3 consecutive numbers, $n-1$, $n$, and $n+1$. We know that exactly one of... | one of $n-1,n$ or $n+1$ is divisible by $3$.
If it is $n$ then so is $n^2$.
If it is not $n$, then one of $n-1$ or $n+1$ is divisible by $3$, and hence so is their product $n^2-1$.
Thus, either $n^2$ or $n^2-1$ is a multiple of $3$. If$n^2+1$ would be a multiple of three, then one of $2=(n^2+1)-(n^2-1)$ or $1=(n^2+1)-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/630742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 1
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Prove that $\lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4} $. My attempt:
We prove that $$\displaystyle \lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4} $$
It is sufficient to show that for an arbitrary real number $\epsilon\gt0$, there is a $K$
... |
Is this proof correct? What are some other ways of proving this?
Thanks!
Your proof is correct with the caveat that you are a bit more precise about what $\epsilon$ and $K$ mean. Another way to prove this is using l'Hôpital's rule. Let $f(n)=23n+2$, $g(n)=4n+1$, then we can see that
$$
\lim_{n\rightarrow\infty} f(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/631462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How to compute $\lim_{n\to \infty}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{n^2+n-k^2}}$ Find this follow limit
$$I=\lim_{n\to \infty}\sum_{k=1}^{n}\dfrac{1}{\sqrt{n^2+n-k^2}}$$
since
$$I=\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{\sqrt{1+\dfrac{1}{n}-\left(\dfrac{k}{n}\right)^2}}$$
I guess we have
$$I=\lim_{n\to... | The difference term-wise is
\begin{align}
\frac{1}{ \sqrt {1+({\frac{k}{n}})^2}} - \frac{1}{ \sqrt {1 + \frac{1}{n} - {(\frac{k}{n}})^2}}&= \frac{\frac{1}{n}}{ \sqrt {1+({\frac{k}{n}})^2}\cdot \sqrt {1 + \frac{1}{n} - ({\frac{k}{n}})^2} ( \sqrt {1+({\frac{k}{n}})^2}+ \sqrt {1 + \frac{1}{n} - ({\frac{k}{n}})^2}) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/633371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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A second order recurrence relation problem I was asked by a friend with this problem but I can't solve it. Can anyone help?
We have a sequence $\left\{a_n\right\}$ that satisfies $a_1=1$, $a_2=2$,
$$a_n+\frac{1}{a_n} =\frac{a_{n+1}^2+1}{a_{n+2}}$$
where $n$ is a positive integer.
Prove that
*
*$a_{n+1}=a_n+\frac{1}... | (a)
By induction, assume $a_{k+1} = a_k + \frac1{a_k}$. Then to get the result for $n=k+1$, consider
$$\begin{align*}
a_k+\frac1{a_k} =& \frac{a_{k+1}^2+1}{a_{k+2}}\\
a_{k+1}=&\frac{a_{k+1}^2+1}{a_{k+2}}\\
a_{k+2} =& a_{k+1}+\frac1{a_{k+1}}
\end{align*}$$
Also prove the base case for $n=1$ holds.
(b)
Using $a_{n+1}^2 ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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 .... | Observe that $x^4 + y^2 \geq |x^2y|$ (for instance, because $x^4+y^2+2x^2y = (x^2+y)^2\geq0$ and $x^4+y^2-2x^2y = (x^2-y)^2 \geq0$). Hence $\displaystyle \left|\frac{x^2y}{x^4+y^2}\right| \leq 1$ when $(x,y)\neq (0,0)$ and thus $$\lim_{(x,y)\rightarrow (0,0)} \left|\frac{x^3y}{x^4+y^2}\right| \leq \lim_{(x,y)\rightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/637987",
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"source": "stackexchange",
"question_score": "10",
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"answer_id": 2
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If $n,k\in\mathbb N$, solve $2^8+2^{11}+2^n=k^2$. If $n,k\in\mathbb N$, solve $$2^8+2^{11}+2^n=k^2$$
It's hard for me to find an idea. Some help would be great. Thanks.
| HINT:
We have $\displaystyle2^8+2^{11}=2304$
Now, $2^8+2^{11}+2^0=2304+1\ne k^2$
$\displaystyle 2^8+2^{11}+2^1=2304+2\equiv2\pmod8$, but $a^2\equiv0,1,4\pmod8$
So, $n\ge2$ let $n=m+2$ where $m\ge0$
$\displaystyle 2^8+2^{11}+2^{m+2}=4(576+2^m)\implies 576+2^m$ must be perfect square
Like either method $m\ge2$ let $m=r+... | {
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Calculate the pseudo inverse of the matrix The subject is to calculate the pseudo inverse if matrix $\begin{equation*}
\mathbf{A} = \left(
\begin{array}{ccc}
1 & 0 \\
2 & 1 \\
0 & 1 \\
\end{array}
\right)
\end{equation*}$
My answer is as follows: (SVD decomposition)
First, $\begin{equat... | Recall, for $\mathbf{\Sigma}$ we take the square roots of the non-zero eigenvalues and populate the diagonal with them, putting the largest in $\mathbf{\Sigma}_{11}$, the next largest in $\mathbf{\Sigma}_{22}$ and so on until the smallest value
ends up in $\mathbf{\Sigma}_{mm}$.
$$\begin{equation*}
\mathbf{\Sigma} =... | {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Find the volume of the solid bounded by $z=x^2+y^2+1$ and $z=2-x^2-y^2$. Question: Find the volume of the solid bounded by $z=x^2+y^2+1$ and $z=2-x^2-y^2$.
Setting the 2 equations equal w.r.t. $z$, $x^2+y^2+1=2-x^2-y^2 \rightarrow x=\pm\sqrt{\frac 12-y^2}$
Therefore the boundary of $y=\pm\frac {1}{\sqrt2}$.
So to find ... | Your setup is right. Here is the method you could have done to compute the volume.
Assume the density is $f(x,y,z) = 1$, so
$$V = \iiint_D \,dx\,dy\,dz$$
We are given that the solid is bounded by $z = x^2 + y^2 + 1$ and $z = 2 - x^2 - y^2$. As I commented under your question, you need to use cylindrical coordinates ... | {
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"source": "stackexchange",
"question_score": "2",
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Convergence of the integral $\int\limits_{1}^{\infty} \left( \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x+3}} \right) \, dx$ Would someone please help me prove that the integral
$$
\int\limits_{1}^{\infty} \left( \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x+3}} \right) \, dx
$$
is convergent?
Thank you.
| Use
$$\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+3}}=\frac{\sqrt{x+3}-\sqrt{x}}{\sqrt{x^2+3x}}=\frac{3}{(\sqrt{x+3}+\sqrt{x})\sqrt{x^2+3x}}$$
So, the integrand is positive and $\le \frac{3}{2x\sqrt{x}}$.
Here I use $\sqrt{x^2+3x}>x$ and $\sqrt{x+3}>\sqrt{x}$
So the integral converges by the comparison criterium.
| {
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"source": "stackexchange",
"question_score": "2",
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how to find $ \lim\limits_{x \to \infty} \left(\sqrt{x^2 +1} +\sqrt{4x^2 + 1} - \sqrt{9x^2 + 1}\right)$ How can I find this?
$ \lim\limits_{x \to \infty} \left(\sqrt{x^2 +1} +\sqrt{4x^2 + 1} - \sqrt{9x^2 + 1}\right)$
| Since for any $A>0$
$$\sqrt{A^2 x^2+1}-A|x| = \frac{1}{A|x|+\sqrt{A^2 x^2+1}}<\frac{1}{2A|x|}$$
holds, we have:
$$\left|\sqrt{x^2+1}+\sqrt{4x^2+1}-\sqrt{9x^2+1}\right|=\left|\sqrt{x^2+1}-|x|+\sqrt{4x^2+1}-2|x|-\sqrt{9x^2+1}+3|x|\right|\leq \left|\sqrt{x^2+1}-|x|\right|+\left|\sqrt{4x^2+1}-2|x|\right|+\left|\sqrt{9x^2+1... | {
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"url": "https://math.stackexchange.com/questions/651148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is the Series : $\sum_{n=1}^{\infty} \left( \frac 1 {(n+1)^2} + ..........+\frac 1 {(n+n)^2} \right) \sin^2 n\theta $ convergent?
Is the Series :
$$\sum_{n=1}^{\infty} \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right) \sin^2 n\theta $$ convergent?
Attempt:
$$\sum_{n=0}^{\infty} \left( \frac 1 {(n+1)^2} + ... | Notice that
$$\sum_{k=1}^n\frac 1{(n+k)^2}=\frac 1{n}\frac 1n\sum_{k=1}^nf\left(\frac kn\right),$$
with $f(x):=\frac 1{1+x^2}$, a continuous positive function. Hence the convergence of the initial series reduces to the convergence of
$$\sum_{n\geqslant 1}\frac{\sin^2(n\theta)}n.$$
Write $\sin^2(A)=\frac{1-\cos(2A)}2$... | {
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"source": "stackexchange",
"question_score": "3",
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solve equation in positive integers Can anybody help me with this equation?
Solve in $\mathbb{N}$:
$$3x^2 - 7y^2 +1=0$$
One solution is the pair $(3,2)$, and i think this is the only pair of positive integers that can be a solution. Any idea?
| There are infinitely many solutions in positive integers. $7y^2-3x^2=1$ is an example of a "Pell equation", and there are standard methods for finding solutions to Pell equations.
For example, the fact that $(x,y)=(2,3)$ is a solution to $7y^2-3x^2=1$ is equivalent to noting that $(2\sqrt7+3\sqrt3)(2\sqrt7-3\sqrt3)=1$.... | {
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"source": "stackexchange",
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Prove that $\frac{1}{1*3}+\frac{1}{3*5}+\frac{1}{5*7}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$ Trying to prove that above stated question for $n \geq 1$. A hint given is that you should use $\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1})$. Using this, I think I reduced it to $\frac{1}{2}(\frac{1}{n... | $$\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1})
$$This implies that the sum is
$$\frac{1}{2}\big\{\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{2n-1}+\frac{1}{2n-1}-\frac{1}{2n+1}\big\}$$ cancel terms and complete
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to tackle a recurrence that contains the sum of all previous elements? Say I have the following recurrence:
$$T(n) = n + T\left(\frac{n}{2}\right) + n + T\left(\frac{n}{4}\right) + n + T\left(\frac{n}{8}\right) + \cdots +n + T\left(\frac{n}{n}\right) $$
where $n = 2^k$, $k \in \mathbb{N} $ and $T(1) = 1$.
simplifie... | I'd just start with $T(1)$ and look for a pattern:
$$T(2^1) = 1 \cdot 2^1 + 2^{1-1}T(1)$$
$$T(2^2) = 2\cdot 2^2 + 1\cdot 2^1 + (2^{2-1}) T(1)$$
$$T(2^3) = 3\cdot 2^3 + 2\cdot 2^2 + 2 \cdot 1 \cdot 2^1 + 2^{3-1} T(1)$$
$$T(2^4) = 4\cdot 2^4 + 3\cdot 2^3 + 2 \cdot 2 \cdot 2^2 + 4 \cdot 1 \cdot 2^1 + 2^{4-1} T(1)$$
so tha... | {
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"url": "https://math.stackexchange.com/questions/667929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Center of Mass via integration for ellipsoid I need some help with the following calculation:
I have to calculate the coordinates of the center of mass for the ellipsoid
$$\left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 + \left( \frac{z}{c} \right)^2 \le 1, \quad z \ge 0$$
with mass-density $\mu(x,y,z)=z^... | The mass density is invariant under $x\rightarrow -x$ and $y\rightarrow -y$, so the center of mass must have $x=y=0$. You do still need to find its $z$-coordinate, but since the mass density is only a function of $z$, you can reduce this to a one-dimensional integral. At a given value of $z$, the cross-section is an ... | {
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"source": "stackexchange",
"question_score": "4",
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Suppose $f$ is a thrice differentiable function on $\mathbb {R}$ . Showing an identity using taylor's theorem Suppose $f$ is a thrice differentiable function on $\mathbb {R}$ such that $f'''(x) \gt 0$ for all $x \in \mathbb {R}$. Using Taylor's theorem show that
$f(x_2)-f(x_1) \gt (x_2-x_1)f'(\frac{x_1+x_2}{2})$ for a... | Using the Taylor expansion to third order, for all $y$ there exists $\zeta$ between $(x_1+x_2)/2$ and $y$ such that
$$
f(y) = f \left( \frac{x_1+x_2}2 \right) + f'\left( \frac{x_1+x_2}2 \right)\left(y - \frac{x_1 + x_2}2 \right) \\
+ \frac{f''(\frac{x_1+x_2}2)}2 \left( y - \frac{x_1 + x_2}2 \right)^2 + \frac{f'''(\zeta... | {
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Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$? Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$?
So far I have:
$a^2+(a+2)^2+(a+4)^2+1$
$=a^2+a^2+4a+4+a^2+8a+16+1 $
$=3a^2+12a+21$
$=3(a^2+4a+7) $
where do I go from here.. the solution I ha... | If $a$ is odd, then $a = 2b+1$ for some integer $b$.
Then $a^2 + 4a + 7 = 4b^2 + 4b + 1 + 8b + 4 + 7 = 4(b^2 + 3b + 3)$, which is evenly divisible by $4$.
Combine this with the divisibility by $3$ that you already have, and you're done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $ It is known that
\begin{align}
\arcsin x + \arcsin y =\begin{cases}
\arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 \le 1 &\text{or} &(x^2+y^2 > 1 &\text{and} &xy< 0);\\
\pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\t... | Using this, $\displaystyle-\frac\pi2\leq \arcsin z\le\frac\pi2 $ for $-1\le z\le1$
So, $\displaystyle-\pi\le\arcsin x+\arcsin y\le\pi$
Again, $\displaystyle\arcsin x+\arcsin y=
\begin{cases}
\\-\pi- \arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})& \mbox{if } -\pi\le\arcsin x+\arcsin y<-\frac\pi2\\
\arcsin(x\sqrt{1-y^2}+y\sqrt{1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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How many distinct factors can be made from the number $2^5*3^4*5^3*7^2*11^1$? How many distinct factors can be made from the number $2^5*3^4*5^3*7^2*11^1$?
Hmmm... So I didn't know what to do here so I tested some cases for a rule.
If a number had the factors $3^2$ and $2^1$, you can make $5$ distinct factors: $2^1$, ... | If $\begin{equation}x = a^p \cdot b^q\cdot c^r+...\end{equation}$
then there are $(p+1)(q+1)(r+1)...$ numbers that divde $x$.
Any number that divides $x$ will be of the form $a^\alpha\cdot b^\beta\cdot c^\gamma$ .
So we have p p+1 options for $\alpha$ because we need to consider $\alpha = 0$ also. Similarly, we have $q... | {
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"source": "stackexchange",
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evaluation of $\int\frac{1}{\sin^3 x-\cos^3 x}dx$ Evaluation of $\displaystyle \int\frac{1}{\sin^3 x-\cos^3 x}dx$
$\bf{My\; Try::}$ Given $\displaystyle \int\frac{1}{\sin^3 x-\cos^3 x}dx = \int\frac{1}{(\sin x-\cos x)\cdot (\sin^2 x-\sin x\cos x+\cos^2 x)}dx$
$\displaystyle = 2\int\frac{(\sin x-\cos x)}{(\sin x-\cos x)... | You're missing a minus sign at one point, but other than that I think you're OK. Next, use partial fractions:
$$
\frac{1}{(2-t^2)(3-t^2)} = \frac{A}{\sqrt{2}-t} + \frac{B}{\sqrt{2}+t} + \frac{C}{\sqrt{3}-t} + \frac{D}{\sqrt{3}+t}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/674019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Find the value of $\lim_{x \to - \infty} \left( \sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} \right)$ I am stuck on this. I would like the algebraic explanation or trick(s) that shows that the equation below has limit of $-2$ (per the book). The wmaxima code of the equation below.
$$
\lim_{x \to - \infty} \left( \sqrt{x^2 + 2x} ... | For $x>0$: For brevity let $A=\sqrt {x^2+2 x}.$ We have $(x+1)^2=A^2 +1>A^2>0$ so $x+1>A>0 . $ ..... So we have $$0<x+1-A=$$ $$=(x+1-A)\frac {x+1+A}{x+1+A}=\frac {(x+1)^2-A^2}{x+1+A}=\frac {1}{x+1+A}<1/x.$$ Therefore $$(i)\quad \lim_{x\to \infty} (x+1-A)=0.$$ For $x>2$: For brevity let $B=\sqrt {x^2-2 x}.$ We have $(x-... | {
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"url": "https://math.stackexchange.com/questions/675516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Incongruent solutions to $7x \equiv 3$ (mod $15$) I'm supposed to find all the incongruent solutions to the congruency $7x \equiv 3$ (mod $15$)
\begin{align*}
7x &\equiv 3 \mod{15} \\
7x - 3 &= 15k \hspace{1in} (k \in \mathbb{Z}) \\
7x &= 15k+3\\
x &= \dfrac{15k+3}{7}\\
\end{align*}
Since $x$ must be an integer, we mus... | We can solve this congruence equation in elementary way also.
We shall write $[15]$ to denote the word mod 15. Fine?
Note that \begin{align*}
&7x\equiv 3[15]\\
-&15x +7x\equiv 3-15[15] ~~\text{because}~~ 15x\equiv 0\equiv 15[15]\\
-&8x\equiv -12[15]\\
&2x\equiv 3\left[\frac{15}{\gcd(15, -4)}\right]~~\text{since}~~ax\e... | {
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"timestamp": "2023-03-29T00:00:00",
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Why aren't these two integration methods yielding the same answer? I'm trying to solve this (not homework, if it matters), and both u-substitution and integration by parts are both yielding two different answers. Where am I going wrong?
Equation: $$\int \frac{(4x^3)}{(x^4+7)}dx$$
u-substitution answer: $$=\ln\big|(x^4+... | I don't understand it. For $u = x^4+7, du = 4x^3dx$ so
$$
\int \frac{4x^3}{x^4+7} dx = \int du/u = \ln |x^4+7| + C.
$$
Show work for your by parts results and it will become clear where the error is...
| {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find value of $r$ and the limit For some $r \in \mathbb Q$, the limit
$$\lim_{x \rightarrow \infty}x^r.\frac{1}2.\frac{3}4.\frac{5}6......\frac{2x-1}{2x}$$ exists and is non zero
What is that value of $r$ and what is that limit equal to?
I rewrote the product $\frac{1}2.\frac{3}4.\frac{5}6......\frac{2x-1}{2x}$ =... | Let $x^r.\frac{1}{2}.\frac{3}{4} \dots \frac{2x-1}{2x} = A \tag{1}$.
Clearly $A < x^r. \frac{2}{3} \tag{2}. \frac{4}{5}.\frac{6}{7} \dots \frac{2x}{2x+1}$ and $A > x^r. \frac{1}{2}.\frac{2}{3}.\frac{4}{5} \dots \frac{2x-2}{2x-1} \tag{3}$
Multiplying $(1)$ and $(3)$,
$A^2 > x^{2r}.\frac{1}{2(2x)} \text{ or }, A > \frac{... | {
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"url": "https://math.stackexchange.com/questions/676574",
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"source": "stackexchange",
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How do I put $\sqrt{x+1}$ into exponential notation? I think $\sqrt{x+1} = x^{1/2} + 1^{1/2}$. Is this incorrect? Why or why not?
| Remember the formula for fractional exponents:
$$x^\frac{m}{n} = \sqrt[n]{x^m}$$
It can also be written as:
$$x^\frac{m}{n} = (\sqrt[n]{x})^m$$
$\sqrt{x+1}$ can be rewritten as $\sqrt[2]{(x+1)^1}$
Using our formula, we can say that:
$$\sqrt[2]{(x+1)^1} = (x+1)^\frac{1}{2}$$
Also remember that:
$$x^m + y^m \neq (x+y)^m$... | {
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nonlinear diophantine equation $x^2+y^2=z^2$ how to solve a diophantine equation $x^2+y^2=z^2$ for integers $x,y,z$
i strongly believe there is a geometric solution ,since this is a pythagoras theorem form
or a circle with radius $z$
$x^2+y^2=z^2$
$(\frac{x}{z})^2+(\frac{y}{z})^2=1\implies x=y=\pm z$ or $0$
so we consi... | Euclid's Formula says that in essence, $(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2$ for all positive integers $m > n$.
This is basically a parametrization of Pythagorean Triplets with two parameters.
| {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometric Series Proof I am posed with the following question:
Prove that for even powers of $\sin$:
$$ \int_0^{\pi/2} \sin^{2n}(x) dx = \dfrac{1 \cdot 3 \cdot 5\cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n} \times \dfrac{\pi}{2} $$
Here is my work so far:
*
*Proof by induction
$P(1) \Rightarrow n = 2 \Rightarrow... | For $k=1$, it's straightforward to verify$$\int_0^{\pi/2}\sin^2x~dx=\int_0^{\pi/2}\frac{1-\cos 2x}2dx=\frac\pi4$$
Assume $k=n$ we have
$$I_n=\int_0^{\pi/2}\sin^{2n}x~dx=\frac{(2n-1)!!}{(2n)!!}\frac\pi2$$
Then for $k=n+1$,
$$\begin{align}I_{n+1}&=\int_0^{\pi/2}\sin^{2n}x(1-\cos^2x)dx\\
&=I_n-\int_0^{\pi/2}\sin^{2n}x\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/680340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
An inequality for sides of a triangle Let $ a, b, c $ be sides of a triangle and $ ab+bc+ca=1 $. Show
$$(a+1)(b+1)(c+1)<4 $$
I tried Ravi substitution and got a close bound, but don't know how to make it all the way to $4 $.
I am looking for a non-calculus solution (no Lagrange multipliers).
Do you know how to do it?
| Solving $ab+bc+ca=1$ for $c$ gives
$$
c=\frac{1-ab}{a+b}\tag{1}
$$
The triangle inequality says that for non-degenerate triangles
$$
|a-b|\lt c\lt(a+b)\tag{2}
$$
Multiply $(2)$ by $a+b$ to get
$$
|a^2-b^2|\lt1-ab\lt(a+b)^2\tag{3}
$$
By $(3)$, we have $(a+b)^2-1+ab\gt0$; therefore,
$$
\begin{align}
(a+b+1)(a+b+ab-1)
&=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/681433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
If $a$ and $b$ are odd then $a^2+b^2$ is not a perfect square
Prove if $a$ and $b$ are odd then $a^2+b^2$ is not a perfect square.
We have been learning proof by contradiction and were told to use the Euclidean Algorithm.
I have tried it both as written and by contradiction and can't seem to get anywhere.
| Let $a=2m+1$ and $b=2n+1$.
Assume $a^2+b^2=k^2$. Then:
$$(2m+1)^2+(2n+1)^2=k^2 \iff \\
4(m^2+m+n^2+n)+2=k^2 \iff \\
4(m^2+m+n^2+n)+2=(2r)^2 \iff \\
2(m^2+m+n^2+n)+1=2r^2 \iff \\
2s+1=2r^2,$$
which is a contradiction. Hence, the assumption $a^2+b^2=k^2$ is false.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/683070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
What steps are taken to make this complex expression equal this? How would you show that $$\sum_{n=1}^{\infty}p^n\cos(nx)=\frac{1}{2}\left(\frac{1-p^2}{1-2p\cos(x)+p^2}-1\right)$$ when $p$ is positive, real, and $p<1$?
| Since $0<p<1$, we have
\begin{eqnarray}
\sum_{n=1}^\infty p^n\cos(nx)&=&\Re\sum_{n=1}^\infty p^ne^{inx}=\Re\sum_{n=1}^\infty(pe^{ix})^n=\Re\frac{pe^{ix}}{1-pe^{ix}}=\Re\frac{pe^{ix}(1-pe^{-ix})}{|1-pe^{ix}|^2}\\
&=&p\Re\frac{-p+\cos x+i\sin x}{|1-p\cos x-ip\sin x|^2}=p\frac{-p+\cos x}{(1-p\cos x)^2+p^2\sin^2x}\\
&=&p\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/685705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$f(x) = \arccos {\frac{1-x^2}{1+x^2}}$; f'(0+), f'(0-)? $f(x) = \arccos {\frac{1-x^2}{1+x^2}}$
$f'(x) = 2/(1+x^2)$,
but I see graphic, and it is true only for x>=0.
For x<=0 => $f'(x) = -2/(1+x^2)$
How can I deduce the second formula or proof that it is.
| Method $\#1:$
Let $\displaystyle\arccos\frac{1-x^2}{1+x^2}=y$
$\displaystyle\implies(1) \cos y=\frac{1-x^2}{1+x^2}\ \ \ \ (i)$
Using the definition of Principal values,
$\displaystyle \implies(2)0\le y\le\pi \implies 0\le\frac y2\le\frac\pi2\implies \tan\frac y2\ge0$
Applying Componendo and dividendo on $(i),$
$\displa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/685900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding system with infinitely many solutions The question asks to find equation for which the system has infinitely many solutions.
The system is:
\begin{cases}
-cx + 3y + 2z = 8\\
x + z = 2\\
3x + 3y + az = b
\end{cases}
How should I approach questions like this?
I tried taking it to row... | You can do row reduction; consider the matrix
\begin{align}
\left[\begin{array}{ccc|c}
-c & 3 & 2 & 8 \\
1 & 0 & 1 & 2 \\
3 & 3 & a & b
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & 0 & 1 & 2 \\
-c & 3 & 2 & 8 \\
3 & 3 & a & b
\end{array}\right]\quad\text{swap 1 and 2}\\
&\to
\left[\begin{array}{ccc|c}
1 & 0 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/686489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"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.