Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove $\displaystyle\frac{H(x^2)}{H(x)}$ increases. For $x\in[0,1]$, let $f(x):=-x\ln x$ and the two-sample entropy function $H(x)=f(x)+f(1-x)$. Prove $h(x):=\displaystyle\frac{H(x^2)}{H(x)}$ increases.
Here is my proof which is a bit cumbersome. I am seeking a much more elegant approach.
The numerator of the derivati... | Partial Hints :
We broke the fraction into two part :
Show that :
$$f(x)=x^{-1}\left(-x^{2}\ln\left(x^{2}\right)-\left(1-x^{2}\right)\ln\left(1-x^{2}\right)\right)$$
Is increasing on $[0,2/5]$
And :
$$g(x)=x^{-1}\left(-x\ln\left(x\right)-\left(1-x\right)\ln\left(1-x\right)\right)$$
Is decreasing on $(0,1)$
Obviously a ... | {
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"url": "https://math.stackexchange.com/questions/4613282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Given the curve $y=\frac{5x}{x-3}$. Find its asymptotes, if any. Given the curve $y=\frac{5x}{x-3}$. To examine its asymptotes if any.
We are only taking rectilinear asymptotes in our consideration
My solution goes like this:
We know that, a straight line $x=a$, parallel to $y$ axis can be a vertical asymptote of a br... | You have correctly found the asymptotes of the function. Observe that we can express the function in the form
$$f(x) = \frac{5x}{x - 3} = \frac{5x - 15 + 12}{x - 3} = \frac{5x - 15}{x - 3} + \frac{12}{x - 3} = \frac{12}{x - 3} + 5$$
which tells us that the graph of $f$ can be obtained by dilating the graph of $y = 1/x... | {
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"url": "https://math.stackexchange.com/questions/4614269",
"timestamp": "2023-03-29T00:00:00",
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Is there any other simpler method to evaluate $\int_0^{\infty} \frac{x^n-2 x+1}{x^{2 n}-1} d x,$ where $n\geq 2?$ Though $n\geq 2$ is a real number, which is not necessarily an integer, we can still resolve the integrand into two fractions,
$$\displaystyle I=\int_0^{\infty} \frac{x^n-2 x+1}{\left(1+x^n\right)\left(1-x^... | It is debatable whether the following method is simpler, but it leads to a stronger result:
For any $z \in \mathbb{C}$ with $\operatorname{Re}(z) > 0$ we have
\begin{align}
\int \limits_0^1 \frac{1 - 2 x + x^z}{1 - x^{2z}} \, \mathrm{d} x &\stackrel{x=t^{1/2z}}{=} \frac{1}{2z} \int \limits_0^1 \frac{t^{\frac{1}{2z} - 1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Is this logical deduction regarding some modular restrictions on odd perfect numbers valid? - Part II Let $p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
He... | This is not a comprehensive proof, it is a mere observation that there are other prime components that do not satisfy the condition you stated. $13^1$ is not the only prime component that violates this rule. Let us take $p^kn^2$ to be an odd perfect number and k=1.
We know that $\sigma (p^k)/2\times gcd(n^2,\sigma (n^... | {
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"timestamp": "2023-03-29T00:00:00",
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Limit of $\frac 12$, $\frac{1\cdot 4}{2 \cdot 3}$, $\frac{1\cdot 4\cdot 5}{2\cdot 3\cdot 6}$, ... The following is not a homework, just curiosity.
Consider the integers grouped by consecutive pairs : $(1,2)$, $(3,4)$, ...
What is the limit of the "switching fractions" where we alternatively use the largest number in a ... | A more simpler, yet heavier way to look at this is,
A more common series is given by $\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\ldots=\prod\limits^{n}_{k=1}{\dfrac{2k-1}{2k}}=\dfrac{\left(n-\frac{1}{2}\right)!}{n!\cdot\left(\frac{-1}{2}\right)!}$.
Your problem is slight altered with every alternate term inver... | {
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"timestamp": "2023-03-29T00:00:00",
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Maximize $x^2 + y^2$ subject to $x^2 + y^2 = 1 - xy$ The equation $x^2 + y^2 = 1 - xy$ represents an ellipse. I am trying to show that its major axis is along $y=-x$ and find the vertex. To find the vertex I tried to find the vector in the ellipse with the greatest norm, which is equivalent to maximizing $x^2 + y^2$ su... | Using polar coordinates, then
$ x = r \cos \theta $
$ y = r \sin \theta $
So we want to maximize $ r^2 = x^2 + y^2 $ subject to $ r^2 = 1 - \dfrac{1}{2} r^2 \sin(2 \theta) $
Hence,
$ r^2 = \dfrac{1}{1 + \frac{1}{2} \sin(2 \theta) } $
Clearly the maximum is when $ \sin(2 \theta) = -1 $, which corresponds to $ \theta = \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What does $\sin(4\theta)$ equal? I want to write $\sin(4\theta)$ in terms of $\cos(\theta)$ and $\sin(\theta)$. My work:
\begin{align*}\sin(4\theta) & = 2\sin(2\theta) \cos(2\theta)
\\ &= 2[(2\sin(\theta) \cos(\theta)(2\cos^2(\theta)-1)]
\\ &= 2(4\sin(\theta) \cos^3(\theta) - 2\sin(\theta) \cos(\theta))
\\ &= 8\sin(... | Those two equations are the same. Using $\cos^2(\theta) = 1 - \sin^2(\theta)$ you get your equation:
$= 4\cos(\theta)\sin(\theta) - 8\sin(\theta)^3\cos(\theta) \\
= 4\cos(\theta)\sin(\theta) - 8\sin(\theta)\cos(\theta) \cdot (1-\cos^2(\theta)) \\
= -4\cos(\theta)\sin(\theta) + 8\sin(\theta)\cos^3(\theta) \\
= 8\sin(\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4617244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Arithmetic Progression with a Sum Property Question I had a question in a exam, and I had no idea how to do it.
The question is:
Let $(a_k)_{k\geq1}$ a sequence of not null real numbers such that:
$$\displaystyle \sum_{k=1}^{n-1}\frac{1}{a_k\cdot a_{k+1}}=\frac{n-1}{a_1\cdot a_n} \, \forall \,n\geq 2$$
Show that this s... | Let $r = a_2 - a_1$. To prove that $(a_n)$ is arithmetic is equivalent to prove that:
$$\forall n \geq 1, a_n = (n - 1) r + a_1$$
Let's proceed by strong induction.
*
* It's clear, it's true for $n = 1$.
* Assume it's true for $k \in \{1, \ldots, n - 1\}$ then :
$$\begin{array}{lcl}
\displaystyle \dfrac{n - 1}{a_1 a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4619860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Studying $\lim_{x \rightarrow 0^+} \frac{xe^{-2x^2}-\sin(x)+\beta x^3}{x^{\alpha}\cos(x^2)}$ Study the following limit with $\alpha$ and $\beta$ parameters:
$$\lim_{x \rightarrow 0^+} \frac{xe^{-2x^2}-\sin(x)+\beta x^3}{x^{\alpha}\cos(x^2)}$$
My attempt:
$$\frac{xe^{-2x^2}-\sin(x)+\beta x^3}{x^{\alpha}\cos(x^2)} \;\sim... | $$xe^{-2x^2}-\sin x+\beta x^3$$$$=x\left(1-2x^2+\frac{(-2x^2)^2}2\right)-\left(x-\frac{x^3}6+\frac{x^5}{120}\right)+\beta x^3+o(x^5)$$$$\sim_{(x\to0)}\begin{cases}\left(\beta-\frac{11}6\right)x^3&\text{if }\beta\ne\frac{11}6\\\frac{239}{120}x^5&\text{if }\beta=\frac{11}6.\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Finding The Square mean of A Gaussian Function. I've been trying to find the square mean of a gaussian function using the limits of $+/-$ infinity.
$$\int_{-\infty}^{\infty} x^2e^{-2x^2}\mathrm{d}x$$
Why does splitting the function into a $$ u = x$$ and $$v'=xe^{-2x^2}$$ and integrating by parts give a different answer... | If you integrate by parts it doesn't matter in which functions you split your original integral. Both ways will resolve to the same solution but one way might be easier to evaluate. Lets look at both of your proposals and compare the result:
$$
\begin{align}
\int \underbrace{x}_u\cdot \underbrace{xe^{-2x^2}}_{v'}\text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Change of variables example in an ODE I do not understand how this was obtained
Why the $(y')^2$ becomes $y'$ and the $y'$ becomes $(y')^2$ and why $y^2$ becomes $x^2 (y')^3$ in the new ode?
How does $\frac{d y(x)}{d x}$ change under this change of variables?
| IMO Leibniz's notation is a bit easier to follow.
$$\frac{d^2y}{dx^2} - \frac1y \left(\frac{dy}{dx}\right)^2 + \sin(x) y \frac{dy}{dx} + \cos(x) y^2 = 0$$
By the chain rule,
$$\frac{dy}{dx} = \frac1{\frac{dx}{dy}} \implies \frac{d^2y}{dx^2} = -\frac1{\left(\frac{dx}{dy}\right)^2} \frac{d\frac{dx}{dy}}{dx} = -\frac1{\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4624886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\sum_{k=0}^n\frac{3^{k+4}\binom{n}{k}}{\binom{k+4}4}+\sum_{m=0}^3\frac{\binom{n+4}{m}3^m}{\binom{n+4}4}=\frac{4^{n+4}}{\binom{n+4}4}$
Prove that
$$\sum_{k=0}^n\frac{3^{k+4}\binom{n}{k}}{\binom{k+4}{4}}
+\sum_{m=0}^3\frac{\binom{n+4}{m}3^m}{\binom{n+4}{4}}
=\frac{4^{n+4}}{\binom{n+4}{4}}$$
Wolfram Alpha shows... | As Bruno B suggested in a comment, multiply by $\binom{n+4}4$ and simplify the terms in the first sum:
$$
\binom{n+4}4\frac{\binom nk}{\binom{k+4}4}=\frac{(n+4)!}{n!4!}\frac{\frac{n!}{k!(n-k)!}}{\frac{(k+4)!}{k!4!}}=\frac{(n+4)!}{(k+4)!(n-k)!}=\binom{n+4}{k+4}\;.
$$
Then merge the two sums, with $m=k+4$, to get the sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4626102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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On factoring $2+128(a+b)^3$ Factor $2+128(a+b)^3$
$=2[1+64(a+b)^3]= 2[1+64(a^3+3a^2b+3ab^2+b^3)]$
Then I'm at a loss for what to do.
$=2[1+64(a^3+b^3)+192ab(a+b)]= 2[1+64(a+b)(a^2-ab+b^2)+192ab(a+b)]=2[1+64(a+b)(a^2-ab+b^2+3ab)]=2[1+64(a+b)^3]$
Back to where I began...
But the answer is given as $2[1+4a+4b][1-4(a+b)+16... | You're almost there, in fact consider
$$2[1+64(a+b)^3] $$
and recall that $x^3+y^3=(x+y)(x^2-xy+y^2)$. In your case, take $x=1$ and $y=4(a+b)$.
Hence you have:
$$2[1+64(a+b)^3]=2[1+4(a+b)][1-4(a+b)+16(a+b)^2] $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4627574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Showing that $\sum_{n \geq 1} \frac{a}{n^2 + a^2} $ is bounded by integrals. For $ a > 0$, we have: $ S(a) = \sum_{n \geq 0} \frac{a}{n^2 + a^2} $
Prove that $ S(a) $ is bounded between 2 integrals.
If we take: $ k \leq x \leq k + 1 $, we will have:
$ S(k+1) \leq \frac{a}{n^2 + a^2} \leq S(k) $
$ \implies \int_{k}^{k+... | What is $x$ supposed to be in the first line ? What you want to have is
$$ \int_k^{k+1}\frac{a}{x^2+a^2}dx\leqslant\frac{a}{k^2+a^2}\leqslant\int_{k-1}^k\frac{a}{x^2+a^2}dx $$
for all $k\geqslant 1$, which follows from the fact that $x\mapsto\frac{a}{x^2+a^2}$ is decreasing on $[0,+\infty)$. Summing this gives
$$ \int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4630619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $ y^2(y^2-6)+x^2-8x+24=0$ and the minimum value of $x^2+y^4$ is $m$ and maximum value is $M$, then find the value of $m$ and $M$ If $ y^2(y^2-6)+x^2-8x+24=0$ and the minimum value of $x^2+y^4$ is $m$ and maximum value is $M$, then find the value of $m$ and $M$
My Attempt:
I converted them to perfect square i.e., $(x... | Your work seems fine to me.
A geometric approach for the second half (since the bounds on $8 \cos \theta + 6 \sin \theta$ were not obvious to me): Note that if you let $u=y^2$, then you are basically finding the points of the circle defined by $(x-4)^2 + (u-3)^2 = 1$ that are closest/farthest from the origin, which are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Line integral of $\int_\gamma\left(\frac{1}{x+1}-\frac{y}{(x+y)^2}\right)dx+\left(\frac{1}{y+1}-\frac{x}{(x+y)^2}\right)dy$ Determine the line integral $\int_\gamma\left(\frac{1}{x+1}-\frac{y}{(x+y)^2}\right)dx+\left(\frac{1}{y+1}-\frac{x}{(x+y)^2}\right)dy$ where $\gamma$ follows the circle from $(1,0)$ to $(0,1)$.
Le... | As it is stated the result is 0. To obtain $1$ it should be
$$\int_\gamma\left(\frac{1}{x+1}-\frac{y}{(x+y)^2}\right)dx+\left(\frac{1}{y+1}+\frac{x}{(x+y)^2}\right)dy.$$
Then
$$
\begin{align}\int_0^{\pi/2}&\left(\left(\frac{1}{\cos\theta+1}-\frac{\sin\theta}{(\cos\theta+\sin\theta)^2}\right)(-\sin\theta)+\left(\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4632079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve for all $x$ such that $x^3 = 2x + 1, x^4 = 3x + 2, x^5 = 5x + 3, x^6 = 8x +5 \cdots$ Question:
Solve for all $x$ such that $\begin{cases}&{x}^{3}=2x+1\\&{x}^{4}=3x+2\\&{x}^{5}=5x+3\\&{x}^{6}=8x+5\\&\vdots\end{cases}$.
My attempt:
I sum up everything.
$$\begin{aligned}\sum_{i=1}^n x^{i+2} &= (2 + 3 + 5 + 8 + \cdo... | $$x^4=x^3x=2x^2+x\implies3x+2=2x^2+x\implies-2x^2+2x+2=0\implies x^2-x-1=0$$
So $x=\frac{1\pm\sqrt{5}}{2}=\varphi,\varphi^{\dagger}$. You could use induction to prove that they are solutions to all other equations.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve for $x$: $\frac{x-a+b}{x-a}+\frac{x-b}{x-2b}=\frac{x}{x-b}+\frac{x-a}{x-a-b}$ Solve for $x$: $\dfrac{x-a+b}{x-a}+\dfrac{x-b}{x-2b}=\dfrac{x}{x-b}+\dfrac{x-a}{x-a-b}$
$\Rightarrow \dfrac{(x-a)+b}{x-a}+\dfrac{(x-2b)+b}{x-2b}=\dfrac{(x-b)+b}{x-b}+\dfrac{(x-a-b)+b}{x-a-b} \ \ \ ...(1)$
$\Rightarrow 1+\dfrac{b}{x-a}+1... | From (6) to (7) you lost the solution $x=(2b+a)/2.$ And from (8) to (9) you lost $b=0.$ You must present the set of solutions $x$ in the case $b=0$ or $a$, and in the case $b\ne0,a.$ And in each case, exclude the values which would make some of the 4 denominators equal to $0.$
So you need to split in three cases:
*
*... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int \sqrt{ a^2 - x^2} dx$ Problem:
Evaluate the integral
$$
\int \sqrt{ a^2 - x^2 } dx.
$$
Solution:
Let us put
$$
x = a \sin \theta . \tag{0}
$$
Then we have
$$
dx = a \cos \theta d \theta, \tag{1}
$$
and
$$
a^2 - x^2 = a^2 \left( 1 - \sin^2 \theta \right) = a^2 \cos^2 \theta. \tag{2}
$$
So we have
$... | You obtained the correct answer. However, your assumption that $0 < \theta < \pi/2$ is not necessary.
Since you let $x = a\sin\theta$, where $a > 0$, then
$$\theta = \arcsin\left(\frac{x}{a}\right)$$
is defined to be the unique angle in the interval $[-\pi/2, \pi/2]$ such that $\sin\theta = x/a$. Since $\theta \in [-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4641251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solve the differential equation: $(x^2-y^2)dx+2xydy=0$. Given $(x^2-y^2)dx+2xydy=0$
My solution-
Divide the differential equation by $dx$
$\Rightarrow x^2-y^2+2xy\frac{dy}{dx}=0$
$\Rightarrow 2xy\frac{dy}{dx}=y^2-x^2$
Divide both sides by $2xy$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2}[\frac{y}{x}-\frac{x}{y}]$
This is a ... | Your solution it is correct. Answering your second question, here is another slightly different solution.
If we can write the ODE as a Bernoulli equation: $y'+a(x)y=b(x)y^{\alpha}$ for some real $\alpha$ so the substitution $v=y^{1-\alpha}$ transforms the Bernoulli equation into a linear equation $\frac{1}{1-\alpha}v'+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4642892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Evaluate: $\int \frac{2}{(1-x)(1+x^2)}dx$ Given $\int \frac{2}{(1-x)(1+x^2)}dx$
The most obvious approach is to use Partial fractions
Let $\frac{2}{(1-x)(1+x^2)}=\frac{A}{1-x}+\frac{Bx+C}{1+x^2}$
$\Rightarrow \frac{2}{(1-x)(1+x^2)}=\frac{A+Ax^2+Bx-Bx^2+C-Cx}{(1-x)(1+x^2)}$
We get $A=1, B=1, C=1$
The integral now become... | $$\Im \left(\frac{1}{x-i}\right)=\frac{1}{1+x^2}$$
$$\int \frac{2}{(1-x)(1+x^2)}dx=\Im\int\frac{2}{(1-x)(x-i)}dx$$$$=\Im\int\frac{(1+i)((1-x)+(x-i))}{(1-x)(x-i)}dx=\Im\left[\int\frac{1+i}{x-i}dx+\int\frac{1+i}{1-x}dx\right]$$
$$=\Im\left[(1+i)\ln(x-i)-(1+i)\ln(1-x)\right]+c$$$$=\frac{1}{2}\ln(x^2+1)+\arctan x-\ln(1-x)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4643054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Compute the integral $\int_{-\infty}^\infty \frac{\sin(x) (1+\cos(x))}{x(2+\cos(x))} dx$ Background
I recently found out about Lobachevsky's integral formula, so I tried to create a problem on my own for which I'd be able to apply this formula. The problem is presented below.
Problem
Compute the integral $\int_{-\inft... | There is a generalization of Lobachevsky's integral formula that is applicable.
The formula states that if $f(x)$ is an odd periodic function of period $a$, then $$\int_{0}^{\infty} \frac{f(x)}{x} \, \mathrm dx = \frac{\pi}{a} \int_{0}^{a/2} f(x) \cot \left(\frac{\pi x}{a} \right) \, \mathrm dx \tag{1}$$ if the integra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4645216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to solve this series $f(n) = f(n/2) + n$? Can I solve this as:
$f(n) = f(n/2) + n$
let find, $$f(n/2) = f(n/2/2) + n/2\\
f(n/2) = f(n/4) + n/2$$
Now,
$$\begin{split}
f(n) &= f(n/4) + n/2 + n\\
f(n) &= f(n/8) + n/4 + n/2 + n
\end{split}$$
hence so on.
$$\vdots$$
Now, $n = 2^i$.
$$\begin{split}
f(2^i) &= f(2^i/2^i)... | At some point you need to have the value of say $f(1)$ or something. The classical way of solving this kind of questions would be to do as you did until:
$$ f(n) = f(1) + \sum_{i=0}^{k} \frac{n}{2^i}$$
where $2^k = n$. Then you should be able to solve the problem by computing $\sum_{i=0}^{k} \frac{n}{2^i}$ using geomet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Variation of the Polya Urn Model An urn contains ${b \ge 1}$ blue balls and one red ball. At each stage a ball is randomly chosen from the urn and then replaced along with a new ball of the same color. Let $T$ denote the first step when a red ball is chosen. What is ${P[T \ge i]}$ and what is $E[T]$?
I calculated $1- P... | For $T \ge i$ you must draw a blue ball for each of the first $i-1$ steps.
The probability of this occuring is
$$\frac{b}{b+1}\times\frac{b+1}{b+2}\times\frac{b+2}{b+3}\ \times\ ... \ \times\frac{b+i-2}{b+i-1}$$
Hence, $P[T \ge i]$ is equal to $\frac{b}{b+i-1}$
The expected value $E[T]$ is given by
$$(1\times\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $|z-1|=1$, where $z$ is a point on the argand plane, show that $\dfrac{z-2}{z}=i\cdot\tan(\arg z)$, where $i=\sqrt{-1}$ If $|z-1|=1$, where $z$ is a point on the argand plane, show that $\dfrac{z-2}{z}=i\cdot\tan(\arg z)$, where $i=\sqrt{-1}$
M y Approach: Let $z=x+iy$
$|z-1|^2=1\implies (z-1)(\bar z-1)=1\implies x^... | You've already got $\dfrac{z-2}{z}=i\cdot \dfrac{2y}{x^2+y^2}$ (for $z\not=0$).
Now you can use $x^2+y^2=2x$ again to have $\dfrac{z-2}{z}=i\cdot \dfrac{2y}{2x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4650758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\int\frac{dx}{1+\tan x}$ How do we solve $$\int\frac{dx}{1+\tan x}$$? I found two ways to solve it, one with the Weierstrass substitution ($t=\tan(\dfrac x2)$) and one with simply substituting $u=\tan x$ (which is better in my opinion). To solve it with the Weierstrass substitution, we get $$\int\dfrac{\frac{2... | Solution using Euler substitution, provided $\cos(x)>0$:
$$\begin{align*}
& \int \frac{dx}{1+\tan(x)} \\
&= \int \frac{\cos(x)}{\sin(x)+\cos(x)} \, dx \\
&= \int \frac{du}{u+\sqrt{1-u^2}} & u=\sin(x) \\
&= -2 \int \frac{1-t^2}{(1+t^2) \left(1-2t -t^2\right)}\,dt & u=-\frac{2t}{1+t^2}\\
&= \int \left(\frac{-1-t}{1-2t-t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4650904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Recurrence relation for $a_{n}$ where $6a_{n}$ and $10a_{n}$ are both triangular According A180926, the elements of the set
{$a:\exists m,n|60a=5n^2+5n=3m^2+3m$} satisfy the following recurrence relation:
$$a_{n}=\frac{62a_{n-1}+1+\sqrt{(48a_{n-1}+1)(80a_{n-1}+1)}}{2}$$
$$a_1=0$$
How this can be derived?
| Here's a complete solution, including a derivation of the recurrence relation in A180926. First, notice
$\rm\ 60\:a \:=\: 5\:(n^2+n)\ =\ 3\:(m^2+m)\ \ $ is easily transformed into the following Pell equation
$\rm\quad\quad\quad\quad\ \: 5\:x^2-3\:y^2 \:= \ 2\ \ \ $ for $\rm\ \ \ (x,\:y)\: =\: (2\:n+1,\:2\:m+1)$
Hence... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/5577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$? I would like to find the apothem of a regular pentagon. It follows from
$$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$
But how can this be proved (geometrically or trigonometrically)?
|
Consider a $\triangle ABC$ with $AB=1$, $\mathrm{m}\angle A=\frac{\pi}{5}$ and $\mathrm{m}\angle B=\mathrm{m}\angle C=\frac{2\pi}{5}$, and point $D$ on $\overline{AC}$ such that $\overline{BD}$ bisects $\angle ABC$. Now, $\mathrm{m}\angle CBD=\frac{\pi}{5}$ and $\mathrm{m}\angle BDC=\frac{2\pi}{5}$, so $\triangle ABC... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/7695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 11,
"answer_id": 2
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If $2 x^4 + x^3 - 11 x^2 + x + 2 = 0$, then find the value of $x + \frac{1}{x}$?
If $2 x^4 + x^3 - 11 x^2 + x + 2 = 0$, then find the value of $x + \frac{1}{x}$ ?
I would be very grateful if somebody show me how to factor this polynomial by hand, as of now I have used to Mathematica to get $(x-2) (2x - 1) (1 + 3x + ... | You don't need to factor the polynomial to find the value of $x + \frac{1}{x}$. Note that the given condition implies $\left( 2x^2 + \frac{2}{x^2} \right) + \left( x + \frac{1}{x} \right) - 11 = 0$ by dividing by $x^2$ and collecting symmetric terms. Now observe that $\left( x + \frac{1}{x} \right)^2 = x^2 + \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$ I wanna know how to solve this equation: $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
I have some trouble to do that and I'd glad with any help I may get.
| Cube both sides and collect terms. You should get
\begin{eqnarray}
512 - 3264 x + 8856 x^2 - 13457 x^3 + 12702 x^4 - 7794 x^5 +
3136 x^6 - 844 x^7 + 120 x^8 = 0
\end{eqnarray}
which factorizes into
\begin{eqnarray}
(8 - 17 x + 6 x^2) (64 - 272 x + 481 x^2 - 456 x^3 + 258 x^4 -
84 x^5 + 20 x^6) = 0.
\end{eqnarray}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $ How to find the value of
$$\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ})$$
manually ?
| For $\sin x\ne0,$
$$2\sin3x\cot x-4\cos x$$
$$=\dfrac{2\sin3x\cos x-2(2\cos x\sin x)}{\sin x}$$
$$=\dfrac{\sin4x+\sin2x-2\sin2x}{\sin x}$$
$$=\dfrac{\sin4x-\sin2x}{\sin x}$$
$$=2\cos3x$$
Now set $\ 2\sin3x=\dfrac{\sqrt3}2=\sin60^\circ\implies 3x=n180^\circ+(-1)^n60^\circ\text{ where } n \text{ is any integer}$
So, $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/10661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Partial Derivative proof For
$$f(x,y) = \begin{cases}
\frac{xy}{x^2+y^2} &\text{if }(x,y)\neq (0,0);\\
0 &\text{if }(x,y)=(0,0).\end{cases}$$
I'm trying to prove that $\frac{\partial f}{\partial x}$, the partial derivative with respect to $x$, exists.
Taking the partial derivative freehand I get
$$\frac{y(x^2+y^2)-x... | I have no idea what you did with the limit. For $y\neq 0$ or $c\neq 0$, we have:
\begin{align*}
\lim{x\to c}\frac{f(x,y)-f(c,y)}{x-c} &= \lim_{x\to c}\frac{\frac{xy}{x^2+y^2} - \frac{cy}{c^2+y^2}}{x-c}\\
&= \lim_{x\to c}\frac{xy(c^2+y^2)-cy(x^2+y^2)}{(c-y)(x^2+y^2)(c^2+y^2)}\\
&= \lim_{x\to c}\frac{(x-c)(y^3-xyc)}{(x-c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/11657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Quadratic Congruence and Sum of Two Squares How would one go about showing how many solutions the following congruence has?
$$x^2 + y^2 \equiv 23 \pmod{93}.$$
| If you are working modulo a prime $p$, then you can obtain the result using the Legendre symbol or by simple trial and error if it is small. For instance, working modulo $3$, as CJost suggests, you are trying to solve $x^2 + y^2 \equiv 23 \equiv 2 \pmod{3}$. then you want to solve $x^2 \equiv 2-y^2\pmod{3}$. Since $y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/11983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Geometry problem on the incentre and circumcenter of a triangle I have the following problem:
In a triangle $ABC$ the line joining incentre and circumcentre is parallel to side $BC$. Prove that $\cos B + \cos C=1$.
Could someone help me solve it?
| This one has been untouched for a few days now, so here's a solution. It does the job but it feels like I beat it into submission with a stick.
I would be interested to know if anyone has a more elegant approach, as it was tougher than I was expecting.
In the above diagram $I$ is the incentre and $K$ is the circumcen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/20246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Probability of 3 of a kind with 7 dice Similar questions:
Chance of 7 of a kind with 10 dice
Probability of getting exactly $k$ of a kind in $n$ rolls of $m$-sided dice, where $k\leq n/2$
Probability was never my thing, so please bear with me.
I've reviewed the threads above to the best of my ability, but I still wonde... | Carrying around the $6^7$ is just complicating your life. Instead, just count how many distinct rolls have exactly one triple; the end probability will be that count, divided by the total number of possible rolls (namely, $6^7$).
There is also the issue of distinguishable and non-distinguishable dice. But let's assume ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/23262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Dot product of two vectors How does one show that the dot product of two vectors is $ A \cdot B = |A| |B| \cos (\Theta) $?
| Think about a triangle with sidelengths $|\textbf{a}|,|\textbf{b}|,|\textbf{c}|$. Then we can use the law of cosines.
$$
\begin{align}
|\textbf{c}|^2&=|\textbf{a}|^2+|\textbf{b}|^2-2|\textbf{a}||\textbf{b}| \cos \theta \\
\implies 2|\textbf{a}||\textbf{b}| \cos \theta &= |\textbf{a}|^2+|\textbf{b}|^2-|\textbf{c}| =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/25266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ This is being asked in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions, and here: List of abstract duplicates.
What methods can be used to evaluate the limit $$\lim_{x\righta... | Possibly more elementary proof based on $\frac{c^{n}-d^{n}}{c-d} = \sum_{k=0}^{n-1} c^{n-1-k} d^k$. Using this for $c = \sqrt[n]{ x^n+ \sum_{m=0}^{n-1} a_m x^m }$ and $d=x$.
$$
c - d = \frac{c^{n}-d^{n}}{ \sum_{k=0}^{n-1} c^{n-1-k} d^k } = \frac{ a_{n-1} x^{n-1} + \ldots + a_1 x + a_0}{ x^{n-1} \sum_{k=0}^{n-1} (\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/30040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 6,
"answer_id": 0
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Sum of series $\sum_{k=1}^\infty{-\frac{1}{2^k-1}}$ What is the sum of the series $\displaystyle\sum_{k=1}^\infty{\frac{-1}{2^k-1}}$?
Also, more generally, can we find $\displaystyle\sum_{k=1}^\infty{\frac{-1}{c^k-1}}$ for some $c$?
| Ignoring the negative sign, you're looking for
$$ {1 \over 2^1 - 1} + {1 \over 2^2 - 1} + {1 \over 2^3 - 1} + \cdots $$
but we have
$$ {1 \over 2^{jk} - 1} = {1 \over 2^{jk}} + {1 \over 2^{2jk}} + {1 \over 2^{3jk}} + \cdots. $$
If we rewrite the first expression using the second one, then your sum is
$$ \left( {1 \over... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/31042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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How can I sum the infinite series $\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad$ How can I find the sum of the infinite series
$$\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad ?$$
My attempt at a solution - I saw that I co... | \begin{align*}(-1)^{n-1}\frac{1\cdot 4 \dots (3n-2)}{5\cdot 10 \dots 5n}&=
(-1)^{n-1}\frac{3^n}{5^n}(-1)^n\frac{(-\frac{1}{3})\cdot (-\frac{4}{3}) \dots (-\frac{3n-2}{3})}{1\cdot 2 \dots n}\\ &= -(3/5)^n\binom{-1/3}{n}\end{align*}
Therefore, you can obtain
$$\sum_{n=1}^{\infty} -(3/5)^n\binom{-1/3}{n} = 1 - \sum_{n=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/34671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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Writing a percent as a decimal and a fraction I am having a problem understanding some manipulations with recurring decimals. The exercise is
Write each of the following as a
decimal and a fraction:
(iii) $66\frac{2}{3}$%
(iv) $16\frac{2}{3}$%
For (iii), I write a decimal $66\frac{2}{3}\% = 66.\bar{6}\% = 0.66\bar{... | The model answer for this exercise is too complicated: $66 {2\over 3} = 66 + {2\over 3} = {200 \over 3}$. So $66 {2\over 3} \% = {200 \over 300} = {2 \over 3}$. Similarly, $16 {2\over 3} = {50 \over 3}$ and $16 {2\over 3}\% = {50 \over 300} = {1 \over 6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/35631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Real and imaginary parts of the Möbius transformation Given that the Möbius transformation is:
$f(z) = \dfrac{az+b}{cz+d} ,\, (a d-b c) \neq 0$ and with $a,b,c$ and $d$ complex numbers written $a= a_1 + a_2i$ etc.
I think I must be missing something because when separating the Möbius transformation in to its real and... | Edit: Sorry - this only holds for $a,b,c,d \in \mathbb{R}$!
Thanks to some help in a question I asked:
\begin{align*}
\textrm{Re} (f(z))
&=\textrm{Re} \left(\frac{az+b}{cz+d}\right) \\
&=\textrm{Re} \left( \frac{(az+b)(c \overline{z}+d)}{(cz+d)(c \overline{z}+d)}\right) \\
&=\frac{1}{|cz+d|^2} \textrm{Re} \left( acz\o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/36542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Finding double root. An easier way? Given the polynomial $f = X^4 - 6X^3 + 13X^2 + aX + b$ you have to find the values of $a$ and $b$ such that $f$ has two double roots.
I went about this by writing the polynomial as:
$$f = X^4 - 6X^3 + 13X^2 + aX + b = (X - r)^2(X - s)^2$$
... and from there I arrived at a system of f... | HINT $\ $ Simpler, first solve $\rm\ f\: =\: (x^2 + c\ x + d)^2\:.\:$ This yields
$$\rm x^4 - 6\ x^3 +13\ x^2 +\cdots\ =\ x^4 + 2\:c\ x^3 + (2\:d+c^2)\ x^2\:+\cdots$$
hence $\rm\ 2\:c = -6\ $ so $\rm\ c = -3\:.\ $ And $\rm\ 2\:d+c^2 = 2\:d+9 = 13\ $ so $\rm\ d = 2\:.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/38267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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How to factor quadratic $ax^2+bx+c$? How do I shorten this? How do I have to think?
$$ x^2 + x - 2$$
The answer is
$$(x+2)(x-1)$$
I don't know how to get to the answer systematically. Could someone explain?
Does anyone have a link to a site that teaches basic stuff like this? My book does not explain anything and I h... | Still one more way: if you don't want to remember the quadratic formula, then just remember how it is proved. Let's do an example: Factor $3x^2 + 4x - 5$.
1) What gives $3x^2$ when squared? Answer: $\sqrt{3}x$.
2) What number $a$ will I have to add to this to get $4x$ as one of the terms in $(\sqrt{3}x + a)^2$? Answer:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/39917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 8,
"answer_id": 4
} |
Evaluate $\sum\limits_{k=1}^n k^2$ and $\sum\limits_{k=1}^n k(k+1)$ combinatorially
$$\text{Evaluate } \sum_{k=1}^n k^2 \text{ and } \sum_{k=1}^{n}k(k+1) \text{ combinatorially.}$$
For the first one, I was able to express $k^2$ in terms of the binomial coefficients by considering a set $X$ of cardinality $2k$ and par... | We show bijectively that
$$2^2+4^2+6^2+\cdots +(2n)^2=\binom{2n+2}{3}.$$
That does not quite give a purely combinatorial expression for $1^2+2^2+ \cdots +n^2$, since we still need to divide by $4$, which is "algebra." But the sin of multiplying or dividing by a constant seems to be a small one in this game. And the r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/43317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 0
} |
How do I get the square root of a complex number? If I'm given a complex number (say $9 + 4i$), how do I calculate its square root?
| Claim 1. Suppose $b\neq 0$. Then the two roots to the equation $x^2 = a +bi$ are: $$\pm\frac{\sqrt{2}}{2}\left(\sqrt{\sqrt{a^{2}+b^{2}}+a}+\mathrm{i}\frac{b}{\left|b\right|}\sqrt{\sqrt{a^{2}+b^{2}}-a}\right).$$
Claim 2. Suppose $b>0$. Then:
(a) The two roots to the equation $x^2 = a +bi$ are $$x = \pm\frac{\sqrt{2}}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/44406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "121",
"answer_count": 12,
"answer_id": 0
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Prove that $6|2n^3+3n^2+n$ My attempt at it: $\displaystyle 2n^3+3n^2+n= n(n+1)(2n+1) = 6\sum_nn^2$
This however reduces to proving the summation result by induction, which I am trying to avoid as it provides little insight.
| to check for divisibility by 6 a number must be divisible by both 2 and 3 so we will prove that
$2n^3 + 3n^2 + n
= n (2n^2 + 3n +1)
= n (n+1) (2n+1)$
If $n$ is even then 2 divides $n$ and $n+1$ will be odd so $n+1$ can be $3k+2$ or $3k$ where $k$ is some integer.
If $3k+1 = n+1$ as it would make $n$ itself a multiple... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/47734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 12,
"answer_id": 11
} |
Seriously: What is the inverse of this 2-by-2 matrix? Okay, I have a hangover and it must be a stupid error, but I just don't get it:
The inverse of a 2-by-2 matrix $A=\left( \matrix{a & b \\ c& d} \right )$ is $\frac{1}{det A}\left( \matrix{d & -b \\ -c& a} \right ) = \frac{1}{ad-bc}\left( \matrix{d & -b \\ -c& a} \r... | Here is another way to solve the problem altogether. Notice that $$\det\left(\begin{array}{cc}
1+y^{2} & -xy\\
-xy & 1+x^{2}\end{array}\right)=1+x^{2}+y^{2},$$ and that $$\det\left(\begin{array}{cc}
d & -b\\
-c & a\end{array}\right)=\det\left(\begin{array}{cc}
a & b\\
c & d\end{array}\right).$$
This means $G$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/49216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to add compound fractions? How to add two compound fractions with fractions in numerator like this one:
$$\frac{\ \frac{1}{x}\ }{2} + \frac{\ \frac{2}{3x}\ }{x}$$
or fractions with fractions in denominator like this one:
$$\frac{x}{\ \frac{2}{x}\ } + \frac{\ \frac{1}{x}\ }{x}$$
| One easy way to figure this out is that dividing by $x$ is the same as multiplying by $1/x$ (but all bets are off when $x=0$, as division by $0$ is undefined). So
$$
\begin{align*}
\frac{ \frac{a}{b} }{c} &= \frac{1}{c} \frac{a}{b} = \frac{a}{bc} \\ \\
\frac{a}{\frac{b}{c}} &= a \frac{1}{\frac{b}{c}} = a \frac{c}{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/51410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
number of combinations without repetition with limited supply Say, I got $1$ red ball, $1$ blue, $2$ yellow, $3$ green, totally $7$ balls.
I wanna select $3$ balls from them. How many ways I can do this?I counted manually
$$123, 124, 133, 134, 144, 233, 234, 244, 334, 344, 444,$$ so $11$ combinations.Is there a formul... | Your problem is equivalent with this one.
$x_1+x_2+x_3+x_4=3$ where
$0\le x_1\le 1,~0\le x_2\le 1, 0\le x_3\le 2,~0\le x_4\le 3$
$0+0+0+3=3$
$0+0+1+2=3$
$0+1+0+2=3$
$1+0+0+2=3$
$0+0+2+1=3$
$0+1+2+0=3$
$1+0+2+0=3$
$0+1+1+1=3$
$1+0+1+1=3$
$1+1+0+1=3$
$1+1+1+0=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/52135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Limit of This Complicated Formula My brain had twisted because of this nasty problem.
let
$$r_{n}=\sqrt{n^2+n+\frac{3}{4}}$$
$$x_{n}=\left \lfloor \frac{r_{n}}{\sqrt{2}}-\frac{1}{2} \right \rfloor$$
$$a_{n}=\sum_{k=1}^{\left \lfloor r_{n}-x_{n} \right \rfloor} \left \lfloor \sqrt{n^2+n-k^2-k+\frac{1}{2}-(x_{n})^2-(2k+1... | Let's use some little-O notation and hand-waving:
$$r_n = n + o(n)$$
$$x_n = \frac{n}{\sqrt{2}} + o(n)$$
$$\lfloor r_n - x_n\rfloor = \frac{2 - \sqrt{2}}{2}n + o(n)$$
$$a_n = \sum_{k = 1}^{\frac{2 - \sqrt{2}}{2}n + o(n)} \Big(\sqrt{1 - \frac{1}{\sqrt{2}}}\Big) n + o(n) - k + o(k) - \sqrt{2kn} $$
Thus,
$$ a_n = \Big(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/53257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Is this Batman equation for real? HardOCP has an image with an equation which apparently draws the Batman logo. Is this for real?
Batman Equation in text form:
\begin{align}
&\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{... | Sorry but this is not the answer but too long for a comment:
Probably the easiest verification is to type the equation on Google you'l be surprised :
The easiest way is to Google :2 sqrt(-abs(abs(x)-1)abs(3-abs(x))/((abs(x)-1)(3-abs(x))))(1+abs(abs(x)-3)/(abs(x)-3))sqrt(1-(x/7)^2)+(5+0.97(abs(x-.5)+abs(x+.5))-3(abs(x-.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/54506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "466",
"answer_count": 10,
"answer_id": 3
} |
Inequality involving sides of a triangle How can one show that for triangles of sides $a,b,c$ that
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} < 2$$
My proof is long winded, which is why I am posting the problem here.
Step 1: let $a=x+y$, $b=y+z$, $c=x+z$, and let $x+y+z=1$ to get
$\frac{1-x}{1+x}+\frac{1-y}{1+y}+\f... | Let $c$ be the largest of $a$, $b$, and $c$. Then
\begin{align}
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}&\le\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{a+b}\\
&=1+\frac{c}{a+b}\\
&< 2
\end{align}
since $c<a+b$.
For a degenerate triangle with $a=0$, we have $c=a+b$ and the sum equals $2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Surface area of revolution Given the function $y=\frac{1}{3}x^3$ on the interval [0, 3], find the surface area of the revolution on the x-axis.
The derivative is $y'=x^2$, so plugging everything in the integral gives
$$2\pi\int_0^3\frac{x^3}{3}\sqrt{1+(x^2)^2}dx$$
$$2\pi\int_0^3\frac{x^3}{3}\sqrt{1+x^4}dx$$
I got a lit... | $$
2\pi \int_0^3 {\frac{{x^3 }}{3}\sqrt {1 + x^4 } dx} = \frac{{2\pi }}{3}\int_0^3 {x^3 (1 + x^4 )^{1/2} dx} = \frac{{2\pi }}{{3 \cdot 4}}\int_0^3 {4x^3 (1 + x^4 )^{1/2} dx}$$
$$
= \frac{\pi }{6}\int_0^3 {(1 + x^4 )^{1/2} 4x^3 dx} .
$$
Hope this helps.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the perimeter of any triangle given the three altitude lengths The altitude lengths are 12, 15 and 20. I would like a process rather than just a single solution.
| Consider the triangle
Triangle with angles $A,B,C$ and opposite sides $a,b,c$
Its area is given e.g. by $$S=\dfrac{bc\sin A}{2}.\tag{1}$$ From the following trigonometric relations valid for a triangle
$$\sin A=2\sin \frac{A}{2}\cos \frac{A}{2}\tag{2}$$
and
$$\sin \frac{A}{2}=\sqrt{\dfrac{(p-b)(p-c)}{bc}}\tag{3}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/55440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
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How to prove the following identities?
Prove:
\begin{align}
\tan(A) + \cot(A) & = 2 \text{cosec}(2A)\\
\tan(45^{\circ}+A^{\circ}) - \tan(45^{\circ}-A^{\circ}) & = 2 \tan(2A^{\circ})\\
\text{cosec}(2A) + \cot(2A) & = \cot(A)
\end{align}
I have got all the formulas that I need but I just couldn't solve these. ... | For the second one:
*
*$2)$ Call $\alpha = \frac{\pi}{4} +A$ and $\beta= \frac{\pi}{4}-A$. So you have $\alpha + \beta = \frac{\pi}{2} \Rightarrow \cot(\alpha+\beta) = 0$. From this you have $$\frac{1}{\tan(\alpha+\beta)} = \frac{1 - \tan{\alpha}\tan{\beta}}{\tan{\alpha} + \tan{\beta}}=0 \Rightarrow \tan{\alpha} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/56122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Finding the minimal polynomial of a $A$, given a diagonal matrix equivalent to $XI_{10}-A$ Let $A\in M_{10}(\mathbb{C})$. We know that $XI_{10}-A $ is equivalent(By finite number of operations on the rows) to the diagonal matrix:
$$C=\begin{pmatrix}
1& & & & & & & & & \\
& 1 & & & & & & & & \\
& ... | Let $K$ be a field, $X$ an indeterminate, and $V$ a finite dimensional $K[X]$-module.
Recall that the multiset of elementary divisors of $V$ is the multiset
$$f_1^{n(1,1)},\dots,f_1^{n(1,k(1))},$$
$$\vdots$$
$$f_r^{n(r,1)},\dots,f_r^{n(r,k(r))},$$
defined by the following conditions: the $f_i$ are distinct monic ir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/57435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The rank of skew-symmetric matrix is even I know that the rank of a skew-symmetric matrix is even. I just need to find a published proof for it. Could anyone direct me to a source that could help me?
| $\newcommand{\rank}{\mathrm{rank}}$
$\newcommand{\diag}{\mathrm{diag}}$
If $K$ is a real skew symmetric matrix, typically it is proven by showing $K$ has even non-zero eigenvalues. Below is a proof using the Hermitian form of $K$.
Since $\rank(K) = r$, there exist order $n$ invertible matrices $P$ and $Q$, such that $K... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/57696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Tangent of an ellipse to an outside point Let $C$ be a curve that is given by the equation:
$$
2x^2 + y^2 = 1
$$
and let P be a point $(1,1)$, which lies outside of the curve.
We want to find all lines that are tangent to $C$ and intersect $P$, and have found $y=1$, but are not sure how to find the other line.
| Let the equation of the line be $y = mx+c$. Since the line passes through $(1,1)$, we have $1 = m + c$ i.e. $c = 1 - m$. Hence, the equation of the line is $y = mx + 1 - m$.
Now we want the above line to be a tangent to $2x^2 + y^2 = 1$. This means the line should intersect the ellipse at exactly one point. Plug in $y ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/58055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Elementary central binomial coefficient estimates
*
*How to prove that $\quad\displaystyle\frac{4^{n}}{\sqrt{4n}}<\binom{2n}{n}<\frac{4^{n}}{\sqrt{3n+1}}\quad$ for all $n$ > 1 ?
*Does anyone know any better elementary estimates?
Attempt. We have
$$\frac1{2^n}\binom{2n}{n}=\prod_{k=0}^{n-1}\frac{2n-k}{2(n-k)}... | You can get an even more precise answer than those already provided by using more terms in the Stirling series. Doing so yields, to a relative error of $O(n^{-5})$,
$$\binom{2n}{n} = \frac{4^n}{\sqrt{\pi n}} \left(1 - \frac{1}{8n} + \frac{1}{128n^2} + \frac{5}{1024n^3} - \frac{21}{32768 n^4} + O(n^{-5})\right).$$
To t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/58560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 7,
"answer_id": 3
} |
tough integral involving $\sin(x^2)$ and $\sinh^2 (x)$ I ran across this integral I get no where with. Can someone suggest a method of attack?.
$$\int_0^{\infty}\frac{\sin(\pi x^2)}{\sinh^2 (\pi x)}\mathrm dx=\frac{2-\sqrt{2}}{4}$$
I tried series, imaginary parts, and so forth, but have made no progress.
Thanks very ... | Although this question is two years old, the integral was mentioned in chat recently, I evaluated it, and then found this question. Since there is no complete solution, although Hans Lundmark's suggestion is excellent and similar in nature, I am posting what I have done.
Contours
Since the integrand is even,
$$
\begin{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 4,
"answer_id": 1
} |
Probability of balls in the box In a box there are 12 balls; 4 defective, 8 not defective.
What is the probability that when 3 balls are drawn, at least two of them are defective.
I know the answer is $$\frac{{4 \choose 2}{8 \choose 1} + {4 \choose 3}}{{12 \choose 3}}$$
But why isn't the answer also $$\frac{{4 \choose ... | Another way is to look at the possible sequences. Let's say $D$ is a defect ball and $G$ is a nondefect (good ball). If you sample 3 balls without replacement the event 'to get at least 2D' is $\{DDD,GDD,DGD,DDG\}$. The probability of the first event is
$P(DDD)= \frac{4 \cdot 3 \cdot 2}{12 \cdot 11 \cdot 10}$
It seems... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/62648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Simplifying the expression $\sqrt[4]{\frac{g^3h^4}{4r^{14}}}$? How would I simplify this expression?
$$\sqrt[4]{\frac{g^3h^4}{4r^{14}}}\ ?$$
I did this
$$\begin{align*}
\sqrt[4]{g^3h^3h^4}\\
h\sqrt[4]{g^3h^3}\\
\sqrt[4]{4r^{14}}\\
\sqrt[4]{2r^2r^{12}}\\
r^3\sqrt[4]{2r^2}\\
\end{align*}$$
But I am stuck?
Yes that ... | If the initial expression is $$\sqrt[4]{\dfrac{g^3 h^4}{4 r^{14} } }$$ then you have made slight errors as $\sqrt[4]{g^3 h^4} = h\sqrt[4]{g^3}$ not $h\sqrt[4]{g^3 h^3}$, while $\sqrt[4]{4 r^{14}} = r^3 \sqrt[4]{4 r^2}$ not $r^3 \sqrt[4]{2 r^2}$.
But otherwise you seem to have done sensible things.
So you could end up... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/62962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need help solving a particular system of non-linear equations analytically How would one go about analytically solving a system of non-linear equations of the form:
$a + b + c = 4$
$a^2 + b^2 + c^2 = 6$
$a^3 + b^3 + c^3 = 10$
Thanks!
| Hint: Newton's identities.
$$\begin{align*}a+b+c&=4\\ ab+bc+ca=\frac12((a+b+c)^2-(a^2+b^2+c^2))&=\frac12(4^2-6)=5\\abc=\frac13((ab+bc+ca-a^2-b^2-c^2)(a+b+c)+(a^3+b^3+c^3))&=\frac13((5-6)\cdot 4+10)=2\end{align*}$$
Remember that $a,\ b,\ c$ are the three roots of a polynomial $P(t)=(t-a)(t-b)(t-c)=t^3-(a+b+c)t^2+(ab+bc+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proving this identity $\sum_k\frac{1}{k}\binom{2k-2}{k-1}\binom{2n-2k+1}{n-k}=\binom{2n}{n-1}$ using lattice paths How can I prove the identity $\sum_k\frac{1}{k}\binom{2k-2}{k-1}\binom{2n-2k+1}{n-k}=\binom{2n}{n-1}$?
I have to prove it using lattice paths, it should be related to Catalan numbers
The $n$th Catalan numb... | This one can also be done using complex variables.
Suppose we seek to evaluate
$$\sum_{k=1}^n \frac{1}{k} {2k-2\choose k-1}
{2n-2k+1\choose n-k}.$$
Introduce the integral representation
$${2n-2k+1\choose n-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n-2k+1}}{z^{n-k+1}} \; dz.$$
This has the property that i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/65944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
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evaluating $ \int_0^{\sqrt3} \arcsin(\frac{2t}{1+t^2}) \,dt$ $$\begin{align*}
\int \arcsin\left(\frac{2t}{1+t^2}\right)\,dt&=t\arcsin\left(\frac{2t}{1+t^2}\right)+\int\frac{2t}{1+t^2}\,dt\\
&=t\arcsin\left(\frac{2t}{1+t^2}\right) + \ln(1+t^2)+C
\end{align*}$$
So
$$ \int\nolimits_0^{\sqrt3} \arcsin\left(\frac{2t}{1+t^... | part of integral solution is $\ln(1+t^2)$. When you insert integral bounds you get $\ln(1+(\sqrt{3})^2)-\ln(1+(0)^2)$$=\ln(4)-ln(1)$$=2\ln(2)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Proof of dividing fractions $\frac{a/b}{c/d}=\frac{ad}{bc}$ For dividing two fractional expressions, how does the division sign turns into multiplication? Is there a step by step proof which proves
$$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}=\frac{ad}{bc}?$$
| There is something assumed about the order of operations in the notation
$a/b \div c/d$. I see this as a problem in pedagogy when fractions are first introduced in schools. Somehow we have to assume that the slash bars are to be done before the division symbol, even though each of those symbols stands for "divide". ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 6,
"answer_id": 2
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Partial Fractions of form $\frac{1}{(ax+b)(cx+d)^2}$ When asked to convert something like $\frac{1}{(ax+b)(cx+d)}$ to partial fractions, I can say
$$\frac{1}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}$$
Then why can't I split $(cx+d)^2$ into $(cx+d)(cx+d)$ then do
$$\frac{1}{(ax+b)(cx+d)^2} = \frac{A}{ax+b} + \fr... | HINT $\rm\displaystyle\quad \frac{f(x)}{g(x)\ (c\:x+d)^2}\: =\: \frac{h(x)}{(a\:x+b)\:(c\:x+d)}\ \ \Rightarrow\ \ f(x)\:(a\:x+b)\: =\: g(x)\:h(x)\:(c\:x+d)\:$
hence, evaluating at $\rm\:x = -d/c\:$ yields that either $\rm\:f(-d/c) = 0\:$ (so the LHS isn't in lowest terms) or $\rm\ a\:x+b\ $ has root $\rm\:x = -d/c\:$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = n(n+1)(2n+7)/6$ by mathematical induction I am doing mathematical induction. I am stuck with the question below. The left hand side is not getting equal to the right hand side.
Please guide me how to do it further.
$1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = ... | HINT $\: $ First trivially inductively prove the Fundamental Theorem of Difference Calculus
$$\rm\ F(n)\ =\ \sum_{i\: =\: 1}^n\:\ f(i)\ \ \iff\ \ \ F(n) - F(n-1)\ =\ f(n),\quad\ F(0) = 0$$
Your special case now follows immediately by noting that
$$\rm\ F(n)\ =\ \dfrac{n\ (n+1)\ (2\:n+7)}{6}\ \ \Rightarrow\ \ F(n)-F(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/72660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Find $x$ if $(7+4\sqrt 3)^{x^2-8}+(7-4\sqrt 3)^{x^2-8}=14$
Known that:
$$(7+4\sqrt 3)^{x^2-8}+(7-4\sqrt 3)^{x^2-8}=14$$
What is the value of $x$
| Note that $$(7+4 \sqrt{3}) \cdot (7-4\sqrt{3}) = 49-48 =1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/74272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Pythagorean quadruples Another Project Euler problem has me checking the internet again. Among other conditions, four of my variables satisfy:
$$a^2+b^2+c^2=d^2 .$$
According to Wikipedia, this is known as a Pythagorean Quadruple. It goes on to say all quadruples can be generated from an odd value of $a$ and an even ... | Sum3Squares
I thought of extremely simple derivation of the parametrization of three squares which sum to square.
Suppose $a^2 + b^2 + c^2 = d^2$
then $a^2 + b^2 = d^2 - c^2$
As is well known, any number which is the sum of two squares is the product of only primes which are the sum of two squares.
We can thus easily d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/76892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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Decomposition of a nonsquare affine matrix I have a $2\times 3$ affine matrix
$$
M = \pmatrix{a &b &c\\ d &e &f}
$$
which transforms a point $(x,y)$ into $x' = a x + by + c, y' = d x + e y + f$
Is there a way to decompose such matrix into shear, rotation, translation,and scale ? I know there's something for $4\time... | If $(x, y, 1)$ is a vector in homogeneous coordinates, we have, by decomposing $M$ into blocks, that
$$M \left[\begin{array}{c}x\\y\\1\end{array}\right] = \left[\begin{array}{cc} a& b\\ d&e\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right] + \left[\begin{array}{c}c\\f\end{array}\right].$$
Here $(c,f)$ is th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/78137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
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How can I prove the inequality $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z}$? For $x > 0$, $y > 0$, $z > 0$, prove:
$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z} .$$
I can see that this is true, I also checked it with a few numbers. But I guess that is not enough to prove it. So how ... | Since $w+\frac{1}{w} \geq 2$ for all $w>0$, we have
$$
\begin{split}
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)
&= 3 + \frac{x}{y}+\frac{y}{x} + \frac{x}{z}+ \frac{z}{x} +\frac{y}{z}+\frac{z}{y} \\&\geq 3+2+2+2 = 9
\end{split}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/78406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
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Asymptotic formula of $\sum_{n \le x} \frac{d(n)}{n^a}$ As the title says, I'm trying to prove
$$\sum_{n \le x} \frac{d(n)}{n^a}= \frac{x^{1-a} \log x}{1-a} + \zeta(a)^2+O(x^{1-a}),$$
for $x \ge 2$ and $a>0,a \ne 1$, where $d(n)$ is the number of divisors of $n$. There is a post here dealing with the case $a=1$. This ... | What you're trying to show isn't true. You should have $\zeta(a)^2$ rather than $\zeta(a)$. (See Exercise 3.3 in Apostol's number theory book in the link above.)
With that correction, you're almost there! Picking up where you left off we have, and using two of the formulas in Apostol's text in the link above,
$$
... | {
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Probability that a $(n, \frac12)$-binomial random variable is even Let $X$ be a $(n, \frac12)$-Binomial RV. Show that $X$ is even with probability $\frac12$.
| A more general question might be "What is the probability that a $(n,p)$-Binomial random variable $X$ is even?" Generalizing on my hint on the main question,
$$
\begin{align*}
\left((1-p) + p\right)^n &= \sum_{k=0}^n\binom{n}{k}p^k(1-p)^{n-k}\\
&= \sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}p^{2k}(1-p)^{n-2k} +
... | {
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How to prove whether a polynomial function is even or odd We know that a function is even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$. With this reasoning is it possible to prove that a polynomial function such as $f(x) = a_{2n}x^{2n} + a_{2n-2}x^{2n-2} + ...+a_{2}x^2 + a_{0}$ is even or odd?
What do you suggest? H... | Two polynomials have the same value at every real number if and only if they are identical (exact same coefficients in each and every power of $x$); this follows because a polynomial of degree $n\gt 0$ can have at most $n$ roots. If $f(x)$ and $g(x)$ are the same at every value of $x$, then $f-g$ has infinitely many ro... | {
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Characteristic function of a standard normal random variable The characteristic function of a random variable $X$ is given by
$$\Phi_X(\omega) = \mathbb{E}e^{j\omega X}=\int_{-\infty}^\infty e^{j\omega x}f_X(x) dx.$$
One can easily capture the similarity between this integral and the Fourier transform.
For a standard n... | I will give two answers:
Do it without complex numbers, notice that
$$ \begin{eqnarray}
\mathcal{F}(\omega) = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x &=& \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mat... | {
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Determine if the equation is valid/true The equation is:
$$\log_b \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} = 2\log_b(\sqrt{3}+\sqrt{2}).$$
I can get as far as:
$$\log_b(\sqrt{3}+\sqrt{2}) - \log_b(\sqrt{3}-\sqrt{2}) = 2\log_b(\sqrt{3}+\sqrt{2})$$
Which looks almost too simple, but I can't get the signs to match up r... | HINT 1. $(a-b)(a+b) = a^2-b^2$ for any numbers $a$ and $b$.
HINT 2. Note that
$$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right)\left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\right) = \frac{(\sqrt{3}+\sqrt{2})^2}{3-2}.$$
| {
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$\int \cos^{-1} x \; dx$; trying to salvage an unsuccessful attempt $$
\begin{align}
\int \cos^{-1} x \; dx &= \int \cos^{-1} x \times 1 \; dx
\end{align}
$$
Then, setting
$$\begin{array}{l l}
u=\cos^{-1} x & v=x \\
u' = -\frac{1}{\sqrt{1-x^2}} & v'=1\\
\end{array}$$
Then by the IBP technique, we have:
$$\begin... | You might have created more work for yourself with the second integration by parts, but have you tried using the substitution you mentioned in the previous step? Assuming your work is correct to that point, the substitution $u=1-x^2$ does appear to lead to something more manageable, even after the second integration b... | {
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"source": "stackexchange",
"question_score": "2",
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Is this a proper use of induction? ($(n^2+5)n$ is divisible by 6) Just want to get input on my use of induction in this problem:
Question. Use mathematical induction to prove that $(n^2+5)n$ is divisible by $6$ for all integers $n \geqslant 1$.
Proof by mathematical induction.
(1) show base case ($n=1$) is true:
$$
... | You can just say that it is always divisible by 6 since it is divisible by 3 (from the expression) and it is divisible by 2 since it is a multiplication of two consecutive integers- one of which is even.
Also the whole problem could be solved very easily using divisibility rather than induction, but I guess your probl... | {
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complex stereographic projection inverse
Wikipedia gives this form of the stereographic Projection from $S^{2} \rightarrow \hat{\mathbb C}$:$$ (1) : z=\frac{x_{1}+ix_{2}}{1-x_{3}}$$ and for the inverse projections the points are supposedly:$$(2): x_{1}=\frac{\overline{z}+z}{z\overline{z}+1}, x_{2}=\frac{z-\overline{z}... | If $z=\dfrac{x_{1}+ix_{2}}{1-x_{3}}$, then $$|z|^2=z\overline{z}=\frac{x_{1}+ix_{2}}{1-x_{3}}\cdot\frac{x_{1}-ix_{2}}{1-x_{3}}=\frac{x_{1}^2+x_{2}^2}{(1-x_{3})^2}=\frac{1-x_{3}^2}{(1-x_{3})^2}=\frac{1+x_{3}}{1-x_{3}},$$
since $x_{1}^2+x_{2}^2+x_{3})^2=1$ for $(x_1,x_2,x_3)\in S^2$. From the above equality, we have $|z|... | {
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Solving $z^4 + 2z^3 + 6z - 9 = 0$ I'm trying to solve $z^4 + 2z^3 + 6z - 9 = 0$.
$z$ is a complex number.
I usually can solve those equations when they are of second degree.
I don't know what to do, breaking out $z$ doesn't help...
EDIT: Sorry I forgot to mention that $z$ has a solution where $\Re(z) = 0$.
| The Quartic Formula can be very useful to solve this.
But there is an alternate way, which my lecturer suggested me : Summing and Subtracting some powers of $z$ to make expression factorable.
For example $z^4+2z^3+6z-9=0$ can be factorized as:
$z^4+(3z^3-z^3)+(3z^2-3z^2)+(9z-3z)-9=0$
$(z^4+3z^3+3z^2+9z)-(z^3+3z^2+3z... | {
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Need help deriving recurrence relation for even-valued Fibonacci numbers. That would be every third Fibonacci number, e.g. $0, 2, 8, 34, 144, 610, 2584, 10946,...$
Empirically one can check that:
$a(n) = 4a(n-1) + a(n-2)$ where $a(-1) = 2, a(0) = 0$.
If $f(n)$ is $\operatorname{Fibonacci}(n)$ (to make it short), then i... | Let $\alpha$ and $\beta$ be the two roots of the equation $x^2-x-1=0$. Then the $n$-th Fibonacci number is equal to
$$\frac{\alpha^n-\beta^n}{\sqrt{5}}.$$
We are interested in the recurrence satisfied by the numbers
$$\frac{\alpha^{3n}-\beta^{3n}}{\sqrt{5}}.$$
If $x$ is either of $\alpha$ or $\beta$, then $x^2=x+1$. ... | {
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Difference Equation $y_{n+3} − 3 y_{n+1} + 2 y_n = (−2)^n$ I get the solution to be $y_n = A(-2)^n + Bn + C + \frac{1}{9}n(-2)^{n-1}$ but wolfram alpha gets $y(n) = c_1 (-2)^n+c_2+c_3 n+\frac{1}{27} 2^{n-1} e^{i \pi n} (4-3 n)$
Explanation required.
Help is much appreciated.
| Your answers are the same.
Observe that $e^{i\pi n}=\cos(\pi n)+i\sin(\pi n)=\cos(\pi n)=(-1)^n$. Hence, you have:
$$\begin{align*}
y(n)& = c_1 (-2)^n+c_2+c_3 n+\frac{1}{27} 2^{n-1} e^{i \pi n} (4-3 n) \\
& = c_1 (-2)^n+c_2+c_3 n+\frac{1}{27} (-2)^{n-1} (3n-4) \\
& = c_1 (-2)^n+c_2+c_3 n+\frac{1}{9} n (-2... | {
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If $n\ge 3$, $4^n \nmid 9^n-1$ Could anyone give me a hint to prove the following?
If $n\ge 3$, $4^n \nmid 9^n-1$
| Possibly I’m missing something obvious, but I don’t see any straightforward way to make pedja’s induction work, and ehsanmo’s answer amounts to waving a magic wand unless you sit down and go through the proof of the lifting-the-exponent lemma in the PDF to which he linked, so I’m going to offer an argument using only t... | {
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The inequality $b^n - a^n < (b - a)nb^{n-1}$ I'm trying to figure out why $b^n - a^n < (b - a)nb^{n-1}$.
Using just algebra, we can calculate
$ (b - a)(b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1}) $
$ = (b^n + b^{n-1}a + \ldots + b^{2}a^{n-2} + ba^{n-1}) - (b^{n-1}a + b^{n-2}a^2 + \ldots + ba^{n-1} + a^{n-1}) $
$ ... | You ask why it is necessarily true that
$$(b - a)(b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1}) < (b - a)nb^{n-1}.$$ A quick answer is that it is not necessarily true. We cannot have strict inequality if $b=a$. And there are other issues. For example, if $n=1$, we always have equality. And the inequality sometim... | {
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"source": "stackexchange",
"question_score": "4",
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Remember trig identities like $\cos(\pi/3) = 1/2$ I have started doing complex analysis and I keep having to switch between rectangular co-ordinates and polar form and I keep running into stuff like - $\cos(\pi/3) = 1/2$.
I keep having to look these up. Am I expected to memorize these or what, its getting to be a seri... | All you really need to know are the values of $\sin(x)$, where $x$ is $0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}$, and $\frac{\pi}{2}$. All of the other values come directly from these:
$$
\begin{align}
\sin(0) &= \frac{\sqrt{0}}{2} = 0
\\
\\
\sin(\pi/6) &= \frac{\sqrt{1}}{2} = \frac{1}{2}
\\
\\
\sin(\pi/... | {
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${a_n}$ series of Fibonacci numbers. $f(x)=\sum_{0}^{\infty}a_nx^n$, show that in the convergence radius: $f(x)= \frac{1}{1-x-x^2}$ I'd really like your help with this following problem:
Let ${a_n}$ be a Fibonacci series $a_1=a_0=1$ and $a_{n+2}=a_n+a_{n+1}$ for every $n \geq 0$.
Let $f(x)=\sum_{0}^{\infty}a_nx^n$, I n... | You can find the radius of convergence by noting the well-known identity:
$$
\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \varphi,
$$
where $\varphi = \frac{\sqrt{5}+1}{2}$ is the golden ratio.
Once you know the radius of convergence, to find the partial sum $S_N$, just write:
$$\begin{align}
S_N := \sum_{n=0}^N a_n x^... | {
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Prove that $x_1^2+x_2^2+x_3^2=1$ yields $ \sum_{i=1}^{3}\frac{x_i}{1+x_i^2} \le \frac{3\sqrt{3}}{4} $ Prove this inequality, if $x_1^2+x_2^2+x_3^2=1$:
$$ \sum_{i=1}^{3}\frac{x_i}{1+x_i^2} \le \frac{3\sqrt{3}}{4} $$
So far I got to $x_1^4+x_2^4+x_3^4\ge\frac{1}3$ by using QM-AM for $(2x_1^2+x_2^2, 2x_2^2+x_3^2, 2x_3^2+x... | Another solution. By the AM-GM inequality, we have
$$
\frac{ 2 \cdot \dfrac{1}{\sqrt{3}} \cdot x_i } {1+x_i^2}
\le
\frac{\dfrac{1}{3} + x_i^2 }{1+x_i^2} = 1 - \frac{2/3}{1+x_i^2},
$$
But $-1/(1+z)$ is a concave function of $z = x^2$, so
\begin{align}
\frac{ \dfrac{2}{\sqrt{3}} x_i } {1+x_i^2}
&\le
\sum_{i=1}^3 \left[... | {
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"source": "stackexchange",
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zeta of three, question about closed form If $\sum\limits_{n=2}^\infty \frac1{(n^2-n)^3}=10-\pi^2$, then what is the limit in closed form of $\sum\limits_{n=1}^\infty \frac1{n^3}$?
| It's called Apéry's constant, $\zeta(3)$, and it doesn't have any known closed form.
Allow me to demonstrate how the first sum doesn't help. Use $(a-b)^3=a^3-3ab(a-b)-b^3$ below:
$$\sum_{n=2}^\infty \frac{1}{(n^2-n)^3}=\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+1}\right)^3=\sum_{n=1}^\infty \color{DarkOrange}{\fra... | {
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"source": "stackexchange",
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The sum of the coefficients of $x^3$ in $(1-\frac{x}{2}+\frac{1}{\sqrt x})^8$ I know how to solve such questions when it's like $(x+y)^n$ but I'm not sure about this one:
In $(1-\frac{x}{2}+\frac{1}{\sqrt x})^8$, What's the sum of the
coefficients of $x^3$?
| The formula you're looking for is
$$
(x+y+z)^8 = \sum_{i+j+k = 8} \begin{pmatrix} 8 \\\ i,j,k \end{pmatrix} x^i y^j z^k
$$
with
$$
\begin{pmatrix} 8 \\\ i,j,k \end{pmatrix} = \frac{8!}{i!j!k!}.
$$
This is known as the multinomial expansion (it works for more than $3$ variables too, you just have to add more indice... | {
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"source": "stackexchange",
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Consider the sequence $\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\ldots$ Consider the sequence
$$\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\ldots$$ For which numbe... | Your sequence includes every rational number in $(0,1)$ infinitely many times. For any $b\in[0,1]$, there exists a sequence of rational numbers in $(0,1)$ converging to $b$. That sequence necessarily occurs as a subsequence of your sequence. Conversely, any $b\notin[0,1]$ cannot have a subsequence of your sequence conv... | {
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Solving $ xy = a + b\cdot\operatorname{lcm}(x,y) + c\cdot\gcd(x,y)$ given $a,b,c$ Given $a$, $b$, and $c$, find the number of pairs of positive integers $(x, y)$ satisfying this equation:
$$ xy = a + b\cdot\operatorname{lcm}(x,y) + c\cdot\gcd(x,y).$$
If $a=2, b= 1, c= 1$, then the answer is 2.
If $a=160, b= 0, c= 90$, ... | First, note that $xy=lcm(x,y)\gcd(x,y)$. So you first want to find solutions to $LG-bL-cG-a = (L-c)(G-b) - (a+bc) = 0$.
So we first need to know the factorizations of $a+bc$.
Even then, we need the conclusion that $G|L$, and then we can get many different $x,y$ for that pair $(L,G)$. Specifically, if $k$ is the numbe... | {
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"source": "stackexchange",
"question_score": "1",
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modulo question I thought you could only mod positive numbers but then I saw this
and became confused...How does this even work ? how can you have negatives?
$$\begin{align*}
-8 &\equiv 7 \pmod{5}\\
2 &\equiv -3 \pmod{5}\\
-3 &\equiv -8\pmod{5}
\end{align*}$$
Actually....What I thought was that mod just means how ... | In simple terms mod gives you remainder. Also if $a\equiv b\pmod{n}$, then $b\equiv a\pmod{n}$, because $n$ divides $(a-b)$, then $n$ also divides $(b-a)$. With the examples you had, for instance
$$
\begin{align*}
-8 &\equiv 7 \pmod{5}\\
2 &\equiv -3 \pmod{5}\\
-3 &\equiv -8\pmod{5}
\end{align*}
$$
$-8 \equiv ... | {
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"source": "stackexchange",
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Finding a limit of a series How would you calculate this limit it just blew me off on my midterms i seem to have calculated the limit correctly but my process is bougus < what my friend said.
$$
\lim_{n\to\infty}\frac{n \sqrt{n} +n}{\sqrt{n^3}+2}
$$
How i calculated the limit:
$$
\lim_{n\to\infty}\frac{n \sqrt{n^2 ... | A more readable form would be
$$
\large {
\lim_{n\to\infty}\frac{n \sqrt{n} +n}{\sqrt{n^3}+2} \hspace{4pt} = \hspace{4pt} \lim_{n\to\infty}\frac{n^{\frac{3}{2}}+n}{n^{\frac{3}{2}}+2}
\hspace{4pt} = \hspace{4pt} \lim_{n\to\infty}
\frac
{1+
\frac{1}{\sqrt{n}}
}
{1+\frac{2}
{n^{\frac{3}{2}}}
}
} \hspace{4pt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/112463",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Resolve this double integral $\iint_D x(y+x^2)e^{y^2-x^4}dxdy$ over $D=\{ x \geq 0, x^2\leq y \leq x^2+1,2-x^2\leq y \leq 3-x^2 \}$ We can rewrite
$$
I=\iint_D x(y+x^2)e^{(y-x^2)(y+x^2)} dx dy
$$
$$
D= \{ (x,y) \in \mathbb{R}^2:x \geq 0 \land 0 \leq y-x^2 \leq 1 \land 2 \leq y+x^2 \leq 3 \}.
$$
With this new notat... | You considered condition $x\geq 0$ implicitly when you canceled out $x$ and $\frac{1}{|4x|}$. Also condition $x\geq 0$ is neccesasry because it guarantees that $\Phi$ is bijective. As for computations - everything is correct.
| {
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Integers and fractions How would I write this as an integer or a fraction in lowest terms?
$(1-\frac12)(1+\frac 12)(1-\frac13)(1+\frac13)(1-\frac14)(1+\frac14).....(1-\frac1{99})(1+\frac1{99})$
I really need to understand where to start and the process if anyone can help me.
| Rewrite it instead as
$$\color{Green}{\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\cdots\left(1-\frac{1}{98}\right)\left(1-\frac{1}{99}\right)}\times \color{Blue}{\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots\left(1+\frac{1}{98}\right)\left(1+\frac{1}{99}\right)}$$
Evaluate left and right parts sep... | {
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Show that $\binom{2n}{ n}$ is divisible by 2?
Possible Duplicate:
prove that $(2n)!/(n!)^2$ is even if $n$ is a positive integer
Show that $\binom{2n}{ n}$ is divisible by 2?
Any help would be appreciated..
| An algebraic approach:
$$ { 2n \choose n } = \frac{(2n)!}{n! \cdot n!} $$
$$ = \frac{ 1 \cdot 2 \cdot 3 \cdots n \cdot (n+1) \cdots (2n-1)\cdot (2n) }{n! \cdot n!}$$
$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [2 \cdot 4 \cdot 6 \cdots (2n)]}{n! \cdot n!} $$
$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/114330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Salem numbers and Lehmer's decic Given,
$$x^{12}-x^7-x^6-x^5+1 = 0\tag1$$
This has Lehmer’s decic polynomial as a factor,
$$x^{10} + x^9 - x^7 - x^6 - x^5 - x^4 - x^3 + x + 1=0\tag2$$
hence one of its roots is the smallest known Salem number. All ten roots obey the beautiful cyclotomic relation,
$$x^{630}-1=\frac{(x^{... | Revisiting this old question, now armed with Mathematica's "Integer Relations", I find that it is quite easy to look for similar polynomial relations. For example, let $x$ be a root of Lehmer's decic, then it is also the case that,
$$x^{630}-1 = \frac{(x^{315}-1)(x^{210}-1)(x^{126}-1)^2(x^{90}-1)(x^{10}-1)(x^{9}-1)}{(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/114909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to eliminate the repeated case from polynomial counting? How to eliminate the repeated case from polynomial counting?
assume a die throw $3$ times, do not allow repeated number appear
$$(x+x^2+x^3+x^4+x^5+x^6)^3 - y$$
how to count the repeated case, that should be minused in above polynomial counting?
what is $y$ i... | Instead of subtracting out the repeated cases, it's easier to only generate the non-repeated cases in the first place:
$$(1+zx)(1+zx^2)(1+zx^3)(1+zx^4)(1+zx^5)(1+zx^6)\;.$$
Then the coefficient of $z^3x^n$ counts the number of partitions of $n$ into $3$ distinct parts from $1$ to $6$, and the number of ways of getting ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/115910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there something like Cardano's method for a SOLVABLE quintic. So there is no quadratic formula equivalent for a GENERAL fifth degree equation, but is there an equivalent formula for a SOLVABLE fifth degree equation.
| For any solvable polynomial equation $a_px^p+a_{p-1}x^{p-1}+\cdots+a_1x+a_0=0$ of odd prime degree $p$, the roots have the form:
$x=-\frac{a_{p-1}}{pa_p}+\omega\sqrt[p]{r_1}+\omega^2\sqrt[p]{r_2}+\cdots+\omega^{p-1}\sqrt[p]{r_{p-1}}$
Where $r_1,\cdots,r_{p-1}$ are solutions to a polynomial of degree $p-1$ with coeffici... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/117849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
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.