Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the partial sum of the series and the limit of it $\sum_{n=1}^{\infty}\frac{1}{2^n}\tan\frac{1}{2^n}$ Find the partial sum of the series and the limit of it:
$\sum_{n=1}^{\infty}\frac{1}{2^n}\tan\frac{1}{2^n}$
I did $\lim_{n\to\infty}\frac{1}{2^n}\tan\frac{1}{2^n} = 0$ so we can not say that the series is divergen... | Hint :
$$\cot x-\tan x =2\cot 2x\iff\dfrac12\tan x=? $$
Put $x=1/2^n, n=1,2,3,\cdots m$ and add to find the partial sum
Finally set $m\to\infty $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2966450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Suppose $(a, b, c)\in\Bbb R^3$, $a,b,c$ are all nonzero, and we have $
\sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$. Find the value of $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$$
Here is my attempt:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}.$$
I am having tr... | Solving your equation $$\sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$$ for $c$ we get
$$c=-\frac{ab}{a+b}$$ then we get
$$\frac{1}{a}+\frac{1}{b}-\frac{a+b}{ab}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2969596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Integral $\int_0^\infty \frac{x-\sin x}{x^3(x^2+4)} dx$ The following integral appeared on the $8$th Open Mathematical Olympiad of the Belarusian-Russian University. $$I=\int_0^\infty \frac{x-\sin x}{x^3(x^2+4)} dx$$ I used power series: $$x-\sin x = \sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n+1}}{(2n+1)!}\rightarrow I=\su... | Let $J(a)=\displaystyle\int_{0}^{\infty}\frac{ax-\sin ax}{x^3(x^2+1)}dx$. Differentiating by $a$ three times (which is admissible under the integral sign, because the resulting integrals converge uniformly for $a$ in any finite interval), we get $$J'''(a)=\displaystyle\int_{0}^{\infty}\frac{\cos ax}{x^2+1}dx=\frac{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Two different color rooks are placed on a chessboard so they don't attack each other. What is a number of ways we can do that?
Of course we have $64$ choices for the first one and then $49$ choices for the second one. So we have $64\cdot 49$ ways to put them. But some of the configurations are essentially the same if ... | Hmm, I think I found a proper number.
We can (essentialy) put white rook only on one of a marked cells in pictures (A) or (B):
(A):
\begin{array} {|r|r|r||r|r|r||r|r|}
\hline
\color{white}{*}& * &*& * & \color{white}{*} &\color{white}{*} & \color{white}{*} & \color{white}{*} \\
\hline
& & *& * & & & & \\
\hline... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2976385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help Factoring A Difficult Expression Given A = $\begin{bmatrix}a & b & c \\c & a & b \\b & c & a \\\end{bmatrix}$, B =$\begin{bmatrix}a & b & c \\a^2 & b^2 & c^2 \\a^4 & b^4 & c^4 \\\end{bmatrix}$
Suppose $C =AB=\begin{bmatrix}
a^2 + a^2b + a^4c & ab + b^3 + b^4c & ac +bc^2 + c^5 \\
\\
ca + a^3 +a^4b & cb + ab^2 + b... | Actually, we have
$$
\det(A)=(a^2 - ab - ac + b^2 - bc + c^2)(a + b + c),
$$
$$
\det(B)=(a + b + c)(a - b)(a - c)(b - c)abc.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Showing that $2^{2^{2^x}}<100^{100^x}$ for large $x$ I wish to show that $2^{2^{2^x}}<100^{100^x}$ for $x$ sufficiently large. I have taken logs (base 10) of both sides to get $2^{(2^x-1)}\log_{10} 2$ and $100^x$. It is not immediately clear how I can prove that the second term here is larger than the first.
Any help a... | HINT
Since $\log$ function is strictly increasing we have
$$2^{2^{2^x}}<100^{100^x}\iff \log_{10}\left(2^{2^{2^x}}\right)<\log_{10}\left(100^{100^x}\right) \iff 2^{2^x}\log_{10}2<2\cdot 100^x$$
then again
$$2^{2^x}\log_{10}2<2\cdot 100^x \iff \log_{10}\left(2^{2^x}\log_{10}2\right)<\log_{10}\left(2\cdot 100^x\right)$$
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove that $f$ is additive if $f(x)f(x-y)+f(y)f(x+y)= f(x)^2+f(y)^2$
Say $f:\mathbb{R}\to \mathbb{R}$ is non-constant such that $$f(x)f(x-y)+f(y)f(x+y)= f(x)^2+f(y)^2$$
Prove that $f(x+y)=f(x)+f(y)$.
If we put $a=f(0)$ and $y=0$ we get $$f(x)^2+af(x)= f(x)^2+a^2 \implies af(x) = a^2$$
If $a\ne 0$ then $f(x)=a$ is... | This is only part of an answer, as I can't prove that $f$ is even or odd.
First, we want to show that $f$ is either even or odd, given $f(x)^2 = f(-x)^2$.
Say there are two points $x$ and $y$ such that $f(x) = f(-x)$ and $f(y) = -f(-y)$. Then, $f(x)f(x-y) + f(x+y) = f(x)^2+f(y)^2$, and, by plugging in $-x,y$, we get $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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On integer solutions of $2\sqrt{\sqrt{2x}+\sqrt{y}}=\sqrt{x}+\sqrt{2y}$
$\textsf{Background}$
From the double-angle formula $\cos2\alpha=2\cos^2\alpha-1$, we can get $$\cos15^\circ=\sqrt{\frac{1+\cos30^\circ}2}=\frac{\sqrt{2+\sqrt3}}2$$ but we also know that it is equivalent to $\dfrac{\sqrt6+\sqrt2}4$.
This is an ex... | The stated condition leads easily to
$$(\sqrt x + \sqrt{2y} - 2\sqrt2)^2 = 8 - 4\sqrt y.$$
Hence $0 \le y \le 4.$ Trying each value of $y$ in turn, one finds that the only solutions with $x$ also an integer are the four already stated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Integral of $\int \frac{-x^2+2x-3}{x^3-x^2+x-1}dx$ I have this simple integral: $\int \frac{-x^2+2x-3}{x^3-x^2+x-1}dx$ and I can't come up with the correct answer.
Here's what I did:
I found the roots for $x^3-x^2+x-1$ so I could do partial fractions:
$\frac{-x^2+2x-3}{x^3-x^2+x-1}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}=\fra... | The $\arctan$ comes as a result of the antiderivative
$$\int \frac{1}{1 + x^2} dx = \tan^{-1}(x) + C$$
You calculated that antiderivative wrong: the antiderivatives of fractions are a little more nuanced than just throwing the denominator in a natural logarithm.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2984961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $5 \mid (n^5-n)$ for all $n \in \mathbb{Z}^+$
Prove for all $n \in \mathbb{Z}^+$ that $5 \mid (n^5-n)$
My proof
Basis step: Since $5 \mid (1^5-1) \iff 5 \mid 0$ and $5 \mid 0$ is true, the statement is true for $n=1$.
Inductive step: Assume the statement is true for $n = k$; that is, assume that $5 \mid (k^... | Here is another way. For every $n \in \mathbb{Z}^{+}$, we have
(here we use $a^2 -b^2 = (a-b)(a+b)$)
\begin{align}
n^5 - n = n(n^4 -1) & = n(n^2-1)(n^2+1) \\
&= n(n-1)(n+1)(n^2 -4 +5) \\
& = n(n-1)(n+1)(n^2 -4) + 5n(n-1)(n+1) \\
& = n(n-1)(n+1)(n-2)(n+2) + 5n(n-1)(n+1) \\
& = (n-2)(n-1)n(n+1)(n+2) + 5n(n-1)(n+1).
\end{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Integrating a 2D Gaussian over a linear strip How do I show that
$$\int_{-\infty}^{\infty} \frac{e^{-\frac{x^2}{2 \sigma ^2}} \left(\text{erf}\left(\frac{2 d-\sqrt{2} x}{2 \sigma
}\right)+\text{erf}\left(\frac{2 d+\sqrt{2} x}{2 \sigma }\right)\right)}{2 \sqrt{2
\pi } \sigma }dx=\text{erf}\left( \frac{d}{\sqrt{2}... | Direct $x$-integration is not possible. However, expressing the erf functions by (direct) indefinite integrals does the job, after exchanging integration orders:
Notice$$
\text{erf}\left(\frac{2 d\pm\sqrt{2} x}{2 \sigma }\right)= \int
\frac{2 e^{-\frac{( \sqrt{2}d \pm x)^2}{2 \sigma^2} }}{\sqrt{\pi} \sigma} \text{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Area bounded by $y=\sqrt{\frac{1+\sin{x}}{\cos{x}}}$ and $y=\sqrt{\frac{1-\sin{x}}{\cos{x}}}$
The area of the region between the curves
$y=\sqrt{\frac{1+\sin{x}}{\cos{x}}}$ and
$y=\sqrt{\frac{1-\sin{x}}{\cos{x}}}$ bounded by the lines $x=0$ and
$x=\frac{\pi}{4}$ is:
a) $\int_{0}^{\sqrt{2}-1}\frac{t}{(1+t^2)\sqrt... | HINT
We could use Tangent half-angle substitution to obtain
$$\sin x=\frac{2t}{1+t^2}\quad \cos x=\frac{1-t^2}{1+t^2}\quad t=\tan \frac x 2 \implies dt=\frac12(1+t^2)dx$$
then
$$\int_{0}^{\frac{\pi}{4}}\sqrt{\frac{1+\sin{x}}{\cos{x}}}-\sqrt{\frac{1-\sin{x}}{\cos{x}}}dx=\int_0^{\sqrt2 -1} \left(\sqrt{\frac{(1+t)^2}{1-t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Expectation of total scores when rolling a die until the score is not 6 An unbiased six-sided die 1, 2, ..., 6 is used during a game, where if one player scores a 6, then he rolls the die again. The player continues playing the game until he scores another value which is NOT 6.
Find the expected value of the total scor... | Let the expected value be X. If you throw a six, your expected value is 6+X, otherwise the score is whatever you threw. Therefore
X = (1 + 2 + 3 + 4 + 5 + (6+X)) / 6
(5/6) X = 21/6
X = 21/5
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2989179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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$a,b,c,d,e$ are zeroes of $6x^5+5x^4+4x^3+3x^2+2x+1$ find $(a+1)(b+1)(c+1)(d+1)(e+1)$
If $a,b,c,d,e$ are zeroes of the polynomial
$$6x^5+5x^4+4x^3+3x^2+2x+1$$
find the value of $(a+1)(b+1)(c+1)(d+1)(e+1)$.
According to me in order to solve this problem one should first factorize the given polynomial in the form o... | It is just $$-{1\over 6}p(-1)= -(-6+5-4+3-2+1) = {1\over 2}$$
Proof: $$p(-1) = 6(-1-a)(-1-b)(-1-c)(-1-d)(-1-e)$$
$$ = -6(1+a)(1+b)(1+c)(1+d)(1+c)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$
Prove the identity
$$8\cos^4 \theta -4\cos^3 \theta-8\cos^2 \theta+3\cos \theta +1=\cos4\theta-\cos3\theta$$
If $7\theta $ is a multiple of $2\pi,$ Show that $\cos4\theta=\cos3\theta$ and deduce,
$$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{... | $$
\begin{align}
&1+\color{#C00}{2\cos\left(\frac{2\pi}7\right)}+\color{#090}{2\cos\left(\frac{4\pi}7\right)}+\color{#00F}{2\cos\left(\frac{6\pi}7\right)}\\
&=1+\color{#C00}{\cos\left(\frac{2\pi}7\right)}+\color{#090}{\cos\left(\frac{4\pi}7\right)}+\color{#00F}{\cos\left(\frac{6\pi}7\right)}+\color{#00F}{\cos\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Factoring the polynomial $3(2x+3)^2 + 7(2x+3) - 6$
Factor $3(2x+3)^2 + 7(2x+3) - 6$
What I did:
With the substitution $X=2x+3$:
\begin{align}
3(2x+3)^2 + 7(2x+3) - 6
&= 3X^2-9X+2X-6 \\
&= 3X(X-3)+2(X-3) \\
&= (X-3)(3X+2) \\
&=((2x+3)-3)(3(2x+3)+2) \\
&=(2x+0)(6x+9+2) \\
&=(2x)(6x+11)
\end{align}
| Your method is correct. You made a sign error:
$$...=(X+3)(3X-2)=...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Show that the following inequality holds when $x>0$ $\require{cancel}$
Show that the following inequality holds for $x>0$
$$1+\frac{x}{2}-\frac{x^2}{8}<\sqrt{x+1}<1+\frac{x}{2}.$$
I proceeded as follows
$$\sqrt{x+1}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16 (\xi+1)^{5/2}},\quad \xi\in[0,x]$$
and substituing in the in... | At $x=0$, all three members equal $1$. Then we can take the derivative and
$$\frac12-\frac x4<\frac1{2\sqrt{x+1}}<\frac12.$$
Again, we have equality at $x=0$ and
$$-\frac14<-\frac1{4(x+1)^{3/2}}<0.$$
This final bracketing is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Prove that number $\underbrace{11 \ldots1}_{100} \underbrace{22 \ldots2}_{100}$ is product of two consecutive numbers Prove that number $\underbrace{11 \ldots1}_{100}$$\underbrace{22 \ldots2}_{100}$ is product of two consecutive numbers
$\begin{align}\underbrace{11 \ldots1}_{100} \underbrace{22 \ldots2}_{100}&=10^{199... | Insert $1=\frac{1}{3}*3$
$$
(10^{100}+2)\frac{1}{3} 3\frac{10^{100}-1}{10-1}\\
x= \frac{10^{100}+2}{3}\\
y=3\frac{10^{100}-1}{10-1} = \frac{10^{100}-1}{3}\\\\
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Residue of order 3 - Find the Laurent Series for the function
\begin{align}
f(z) = \frac{1}{(z^2+4)^3}
\end{align}
about the isolated singular pole $z = 2i$. What is the pole order? What is the residue at the pole?
My attempt:
\begin{align}
f(z) &= \frac{1}{(z^2+4)^3}\\
&= \frac{1}{(z+2i)^3(z-2i)^... | Integration is completely unnecessary here. More generally, never integrate if you can differentiate. And most of the times, geometric series is more than enough.
$$
\begin{aligned}
\frac{1}{(z+2i)^3}&=\frac{1}{2}\frac{\mathrm d^2}{\mathrm dz^2}\frac{1}{(z-2i)+4i}\\
&=\frac12\frac{\mathrm d^2}{\mathrm dz^2}\left[-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculate $\sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k}$ Calculate $\sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k}$, I do not know hot get rid of that $k$, for me it is similar like $\binom{n}{k}=\frac{k}{n} \binom{n-1}{k-1}$, do you have some idea?
| We can write your sum as
$$
\eqalign{
& f(n) = \sum\limits_{k = 1}^n {\left( { - 1} \right)^{\,k + 1} \left( \matrix{
n \cr
k \cr} \right){1 \over k}} = \cr
& = \sum\limits_{k = 0}^{n - 1} {\left( { - 1} \right)^{\,k} \left( \matrix{
n \cr
k + 1 \cr} \right){1 \over {k + 1}}} = \sum\limits_{k = 0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Study the convergence of a series I am studying the convergence of the series $$\sum\limits_{n>=0} \left(\sqrt[n]{n} - 1\right)^n$$
I just know that the limit tend to $0$, but I don't know how to prove the convergence.
| I assume you meant $n \geq 1$. Using $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x^{2}+x+1)$ we have
$$n-1=
\left(n^{\frac{1}{n}}\right)^{n}-1=\\
\left(n^{\frac{1}{n}}-1\right)\left(n^{\frac{1}{n}(n-1)}+n^{\frac{1}{n}(n-2)}+...+n^{\frac{1}{n}2}+n^{\frac{1}{n}}+1\right)\geq ...$$
using AM-GM
$$... \geq\left(n^{\frac{1}{n}}-1\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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What am I doing wrong finding $\lim_{x\to 0} \left( \frac{1+x\cdot2^x}{1+x\cdot3^x} \right)^{1/x^2}$? It has an answer here, but I'd like to know where my solution went wrong.
$$\lim_{x\to 0} \left( \frac{1+x\cdot2^x}{1+x\cdot3^x} \right)^{\frac{1}{x^2}} $$
$$\lim_{x\to 0} \left( \frac{1+x\cdot2^x +x\cdot 3^x-x\cdot 3^... | You have used binomial theorem to expand $2^x$ and $3^x$ and this is not valid as the binomial theorem for a general exponent is valid only under certain conditions. To be precise you have $$(1+t)^x=\sum_{k=0}^{\infty}\binom{x}{k}t^k$$ for $|t|<1$. The result is valid for $t=\pm 1$ with certain restrictions on $x$. For... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3003184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2$ If $a, b, c$ are distinct real numbers then you demonstrate that:
$$ S=\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2.$$
Using inequality $ |x-y|\leq |x|+|y|$ we showed that $ S >\frac{3}{2}.$
For $b = 2a, c = 3a, S=5,$ that is, the ... | Notice that $|x-y|\leq |x|+|y|$ implies that:
$$\frac{|z|}{|x-y|} \geq \frac{|z|}{|x|+|y|}.$$
Then:
$$S \geq \frac{|a|}{|b|+|c|} + \frac{|b|}{|a|+|c|} + \frac{|c|}{|a|+|b|} = \\
=\frac{A}{B+C} + \frac{B}{A+C} + \frac{C}{A+B},$$
where I set $A = |a|, B = |b|$ and $C= |c|$ for the sake of simplicity.
Multiplying and div... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
An AMM-like integral $\int_0^1\frac{\arctan x}x\ln\frac{(1+x^2)^3}{(1+x)^2}dx$
How can we evaluate $$I=\int_0^1\frac{\arctan x}x\ln\frac{(1+x^2)^3}{(1+x)^2}dx=0?$$
I tried substitution $x=\frac{1-t}{1+t}$ and got
$$I=\int_0^1\frac{2 \ln \frac{2 (t^2+1)^3}{(t+1)^4} \arctan \frac{t-1}{t+1}}{t^2-1}dt\\
=\int_0^1\frac{2 ... | @Kemono Chen proved here
$$\int_0^y\frac{\ln(1+yx)}{1+x^2}dx=\frac12 \tan^{-1}(y)\ln(1+y^2)$$
Divide both sides by $y$ then integrate between $0$ and $1$ we get
$$\color{red}{\frac12\mathcal{I}}=\frac12\int_0^1\frac{\tan^{-1}(y)\ln(1+y^2)}{y}dy=\int_0^1\int_0^y\frac{\ln(1+yx)}{y(1+x^2)}dxdy$$
$$=\int_0^1\frac{1}{1+x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
Distance from focus to nearest point in ellipse Consider an ellipse with semi-axes $a$ (major) and $b$ (minor). For such an ellipse the distance of focus to the centre is:
$f = \sqrt{a^2-b^2}$
Now, the distance from the focus to the nearest point on the ellipse is along the major semi-axis a, thus this distance is:
$r_... | For 1) I wanted to do a very simple thing. A top part of ellipse has the equation
$y = b\sqrt{1-\frac{x^2}{a^2}}$
The squared distance from focus to any point is then given by
$d^2 = \left(x - \sqrt{a^2-b^2}\right)^2 + b^2\left(1-\frac{x^2}{a^2}\right)$.
If now we compute the derivative:
$\frac{\mathrm{d}}{\mathrm{d}x... | {
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"url": "https://math.stackexchange.com/questions/3010449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solve without a calculator: If $x+\sqrt{x}=13$ then $x+\frac{13}{\sqrt{x}}=?$
$$x+\sqrt{x}=13$$
$$x+\frac{13}{\sqrt{x}}=?$$
I tried to square(and also triple in another attempt) both sides of both of the equations hoping that I would find some expression to plug in. It didn't help.
Then, I tried to simply bring the s... | If $x + \frac{13}{\sqrt x} = a$, then:
$$13a = (x+\sqrt x)(x + \frac{13}{\sqrt x})$$
$$= x^2 + 13\sqrt x + x\sqrt x + 13$$
Now use the equality $x + \sqrt x = 13$:
$$13a = x^2 + (x + \sqrt x)\sqrt x + x \sqrt x + (x +\sqrt x) $$
$$(x + \sqrt x)a = x^2 + x\sqrt x + x + x \sqrt x + x + \sqrt x $$
$$(x + \sqrt x)a = (x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find all real numbers $x,y,z\in [0,1]^3$ such that $(x^2+y^2)\sqrt{1-z^2}\ge z$.... Such that:
$$(x^2+y^2)\sqrt{1-z^2}\ge z$$
and
$$(z^2+y^2)\sqrt{1-x^2}\ge x$$
and
$$(x^2+z^2)\sqrt{1-y^2}\ge y$$
Since $x,y,z$ $\in ]0,1[^3$
then , there are some real numbers $a,b,c$ such that
$\cos a=x, \cos b=y , \cos c=z$
After some ... | The inequality is symetric.
so we can suppose that $x\ge y \ge z$
the second inequality becomes
$$2x^2\sqrt{1-x^2}\ge x$$
$$2x\sqrt{1-x^2}\ge 1$$
$$4x^2-4x^4-1\ge 0$$
$$-(2x^2-1)^2\ge 0$$
$$2x^2-1=0$$
$$x=\frac{1}{\sqrt{2}}$$
By the same way , after remplacing $x$ by its value
we will find that $x=y=z=\frac{1}{\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Limit of $\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$ (No L'Hôpital)
$\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$
I can't get to the end of this limit. Here is what I worked out:
\begin{align*}
& \lim_{x \to 0} \frac{\cos 2x}{\sin 2x}\cdot\frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
\lim_{x \... | $\displaystyle \cot 2x\cot \left(\frac{\pi }{2}-x\right)=\cot (2x)\tan (x)=\dfrac{\cos 2x\cdot \sin x}{2\sin x\cos^2 x}=\dfrac{\cos 2x}{2\cos^2x}$
$\displaystyle\lim_{x\to0} \cot 2x\cot \left(\frac{\pi }{2}-x\right)=\displaystyle\lim_{x\to0}\dfrac{\cos 2x}{2\cos^2x}=\dfrac12 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Is it true that there aren't any three different numbers $x,y,z$ such that $x^3+x \equiv y^3+y \equiv z^3+z \pmod p $? Let $p$ be a prime number. Is it true that there aren't any three different numbers $x,y,z$ such that $$x^3+x \equiv y^3+y \equiv z^3+z \pmod p $$
with $x -y, y-z, z-x$, each of them cannot be divided... | If $p=n^2+1$ (such as $5,17,37\dots$), then $x=0,\ y=n,\ z=(p-n)$ will solve your equivalence with all three terms congruent to $0$.
$x^3+x=0;\ y^3+y=n(n^2+1)=np\equiv 0 \mod{p}; z^3+z=(p-n)(p^2-2np+n^2+1)=(p-n)(p^2-2np+p)=p(p-n)(p-2n+1)\equiv 0\mod{p}$
$p$ cannot be of the form $n^2+1$ if $p\equiv 3\mod{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to show $x^{p-2}+\cdots+x^2+x+1\equiv0\pmod{p}$ has $p-2$ incongruent solutions: $2, 3,..., p-1.$ when $p$ is an odd prime? Let $p$ be an odd prime. How would I prove that the congruence $x^{p-2}+\cdots+x^2+x+1\equiv0\pmod{p}$ has exactly $p-2$ incongruent solutions, and they are the integers $2, 3,..., p-1?$
I fou... | To phrase the yellow block differently; the equation $x^{p-1}-1\equiv0\pmod{p}$ has precisely $p-1$ incongruent solutions by Lagrange's theorem. These are precisely the nonzero residues modulo $p$, i.e. the congruence classes $1$, $2$,..., $p-1$. We can factor the left hand side of this equation as follows:
$$x^{p-1}-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3013542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Another Limit Conundrum
For what values of $a$ and $b$ is the following limit true?
$$\lim_{x\to0}\left(\frac{\tan2x}{x^3}+\frac{a}{x^2}+\frac{\sin bx}{x}\right)=0$$
This question is really confusing me. I know that $\tan(2x)/x^3$ approaches infinity as $x$ goes to $0$ (L'Hôpital's). I also understand that $\sin(bx... | Using the usual Taylor series for
$$y=\frac{\tan(2x)}{x^3}+\frac{a}{x^2}+\frac{\sin (bx)}{x}=\frac{\tan(2x)+ax+x^2\sin (bx)}{x^3}$$ The numerator write
$$\left(2 x+\frac{8 x^3}{3}+\frac{64 x^5}{15}+O\left(x^7\right) \right)+a x+x^2\left(b x-\frac{b^3 x^3}{6}+\frac{b^5 x^5}{120}+O\left(x^7\right) \right)$$ that is to sa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3013664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Number of Non negative integer solutions of $x+2y+5z=100$ Find Number of Non negative integer solutions of $x+2y+5z=100$
My attempt:
we have $x+2y=100-5z$
Considering the polynomial $$f(u)=(1-u)^{-1}\times (1-u^2)^{-1}$$
$\implies$
$$f(u)=\frac{1}{(1-u)(1+u)}\times \frac{1}{1-u}=\frac{1}{2} \left(\frac{1}{1-u}+\frac{... | An alternative way.
Given $x+2y+5z=100$ and it is clear that $0\le z\le20$.
For any possible values of $z$, $x+2y=100-5z$.
Let us take $p=100-5z\ge0$. Solving the equation $x+2y=p$,
$(-p,p)$ is a solution. The general solution of $(x,y)$ is $$x=-p+2q,\ \ y=p-q,\ \ q\in\mathbb{Z}$$
If $p=2k$, then $k=\dfrac p2\le q\le p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3014438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
An Inconsistency in Numerical Approximation Consider the expression
$$
10^5 - \frac{10^{10}}{1+10^5}.
$$
Using the elementary properties of fractions we can evaluate the expression as
$$
10^5 - \frac{10^{10}}{1+10^5} = \frac{10^5 + 10^{10} - 10^{10}}{1+10^5} = \frac{10^5}{1+10^5}\approx 1.
$$
Note that the approximat... | The first approximation is fine. The second on is not, because, $10^5$ and $\dfrac{10^{10}}{1+10^5}$ are large numbers with approximately the same size. You are saying that since $10\,001$ is close to $10\,000$, then $1$ is close to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3017585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $ how can I find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $?
It is so weird for me because I put this to Mathematica and it tells me that sum does not converge...
Let consider sum no to infinity, but to n
$$ \sum_... | If I'm not mistaken, the actual sum is
$$ \frac{1 - \ln(2)}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3019050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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Show that the equation system $I:x^2-y^2=a;II: 2xy= b$ has always a solution $(x,y)\in \mathbb {R}^2$ The statement must be false because if $b=0$ then $x$ or $y$ must be $0$. If $x$ is $0$ we get from $I$ that $-y^2=a\iff y^2 = -a$. Wouldn't that be a contradiction to the statement? Because if $a>0$ then there is no s... | Actually, the statement is true even if $b=0$: just take $(x,y)=\left(\pm\sqrt a,0\right)$ if $a\geqslant0$ and $(x,y)=\left(0,\pm\sqrt{-a}\right)$.
Otherwise, your approach is fine, but not your computations. You should have obtained, when $b\neq0$,$$\pm\left(\sqrt{\frac{a+\sqrt{a^2+b^2}}2},\frac b{|b|}\sqrt{\frac{-a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3022475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Seeking Methods to solve $ I = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx$ I was wondering what methods people knew of to solve the following definite integral? I have found a method using Feynman's Trick (see below) but am curious as to whether there are other Feynman's Tricks and/or Meth... | $$ I = \int_{0}^{1}\frac{\arctan x}{x\sqrt{1-x^2}}\,dx =\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\int_{0}^{1}\frac{x^{2n}}{\sqrt{1-x^2}}\,dx=\frac{\pi}{2}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)}\cdot\frac{\binom{2n}{n}}{4^n}$$
is a fairly simple hypergeometric series, namely $\frac{\pi}{2}\cdot\phantom{}_2 F_1\left(\tfrac{1}{2},\tf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 1
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Finding $\lim\limits_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)$ What is$$\lim_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)?$$So it is$$\lim_{n→∞}\frac{n^3(\sqrt{n^2+\sqrt{n^4+1}})^2-(n\sqrt{2})^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}=\lim_{n→∞}\frac{n^3(n^2+\sqrt{n^4+1}-2n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}.$$
I do not ... | Let $1/n=h$
$$\lim_{h\to0^+}\dfrac{\sqrt{1+\sqrt{1+h^4}}-\sqrt2}{h^4}$$
$$=\lim_{h\to0^+}\dfrac{1+\sqrt{1+h^4}-2}{h^4}\cdot\lim_{h\to0^+}\dfrac1{\sqrt{1+\sqrt{1+h^4}}+\sqrt2}$$
$$=\lim_{h\to0^+}\dfrac{1+h^4-1}{h^4}\cdot\lim_{h\to0^+}\dfrac1{\sqrt{1+h^4}+1}\cdot\lim_{h\to0^+}\dfrac1{\sqrt{1+\sqrt{1+h^4}}+\sqrt2}$$
$$=\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3025375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Show $\sqrt{x (1-y)} - \sqrt{y (1-x)} + y\sqrt{x} - x\sqrt{y} - \frac{x - y}{3} > 0$ Consider the inequality
$$
\sqrt{x (1-y)} - \sqrt{y (1-x)} + y\sqrt{x} - x\sqrt{y} - \frac{x - y}{3} > 0
$$
where $0<y<x<1$ and $x+y\leq 1$.
One can, for instance, consider optimizing the expression using the Karush-Kuhn-Tucker condi... | If you let $x = \sin^2 a, y = \sin^2 b$ with $0<b<a<\pi/2$, then your inequality is equivalent to:
$$\sin a\cos b - \sin b\cos a +\sin^2b\sin a - \sin^2a\sin b >\dfrac{\sin^2 a-\sin^2 b}{3}$$ or $$\sin(a-b)>(\sin a-\sin b)\left(\frac 13+\sin a\sin b\right)$$
or $$\dfrac{\cos\left(\frac {a-b}2\right)}{\cos\left(\frac{a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x^2\equiv 5 \pmod{1331p^3}$ Let $p$ be given by $p=2^{89}-1$ and note that it is a Mersenne Prime. The problem is to find the number of incongruent solutions to
$$
x^2\equiv 5 \pmod{1331p^3}
$$
I began the problem by splitting it up into the congruences
$$
x^2\equiv 5 \pmod{1331} $$and$$
x^2\equiv 5 \pmod{p^3}
$$
I f... | Consider Legendere symbol $\left(\frac{5}{p}\right)$. By quadratic reciprocity
$$\left(\frac{5}{p}\right)\Big(\frac{p}{5}\Big)=(-1)^{\left(\frac{5-1}{2}\right)\left(\frac{p-1}{2}\right)}=1 \implies \left(\frac{5}{p}\right)=\left(\frac{p}{5}\right).$$
But $p=2^{89}-1 \equiv 2(2^2)^{44}-1 \equiv 1 \pmod{5}$. Thus
$$\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3033409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Strange sum that always end up with 9 If we have any number, example 4896, and sum all digits
sum = 4+8+9+6 = 27
and than substract this number from the original number, we always get a number that is divisible by 9:
4896-27=4869 -> 4869/9 = 541.
Moreover now everytime i sum the digits from new number (4869), i will ge... | I will give a formal proof for numbers with $4$ digits. You can think about how to generalize it (Edit: or refer to @fleablood's answer who gave a formal proof for arbitrarily many digits).
Let $n=abcd$ be a number where $0\leq a,b,c,d\leq 9$ are it's digits. For example if $n=4821$ then $a=4,b=8,c=2,d=1$.
Another way ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to factor $-x^3+x^2-2$ Is there an easy way to factor $$f(x)=-x^3+x^2-2\;?$$
I have checked step-by-step calculators that all use theorems I am not very familiar with.
It doesn't seem like a sum or difference of cubes, and grouping is not an option. I tried factoring out the $-x^2$ $$f(x)=-x^2(x-1)-2$$ but that d... | $$-x^3+x^2-2=-1-x^3+x^2-1$$
$$=-(1+x^3)+(x^2-1)$$
$$=-(1+x)(1-x+x^2)+(x-1)(x+1)$$
$$=(x+1)(x-1-(1-x+x^2))$$
$$=(x+1)(-x^2+2x-2)$$
Edit: At the fourth step, there is a factor of $x+1$ in both terms. So I factor this out of the expression as follows:
$$-(1+x)(1-x+x^2)+(x-1)(x+1)=(x+1)[-(1-x+x^2)]+(x+1)(x-1)$$
$$=(x+1)[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Finding Bounds in proof of Stirling's Formula In a proof of Stirling's Formula, my professor claims that
$\frac 12 $($ln$k + $ln$(k+1)) $\le$ $\int_k^{k+1}$$ln$x $dx$ $\le$ $\frac 12$($ln$k + $ln$(k+1)) + $\frac 1{k^2}$.
I can see that the first inequality is true by concavity of the natural log function (hence the ar... | My attempt:
\begin{align}
\int_{\ln(k)}^{\ln(k+1)} &= \int_0^1 \ln(k+x) dx = \int_0^{1/2} \ln(k+x) + \int_{1/2}^{1} \ln(k+1 - x) dx \\
&= \int_0^{1/2} \ln(k) + \frac{x}{k} - \frac{x^2}{2k^2} + o(\frac{1}{k^2}) dx \\
&\quad + \int_{1/2}^1 \ln(k+1) - \frac{x}{k+1} - \frac{x^2}{(k+1)^2} + o(\frac{1}{(k+1)^2}) dx \\
&= \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3035076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find all integers x for which $ |x^3 + 6x^2 + 2x − 6| $ is prime. I think that this means that $|x^3 + 6x^2 + 2^x − 6| \equiv 0 \bmod p$. Ignoring the absolute value first, I get $
x(x^2+2)+6(x^2−1) \equiv 0 \mod p \Rightarrow x(x^2+2) \equiv -6(x^2−1) \bmod p$. But I'm not sure how to proceed.
| Following your thoughts of grouping the terms into:
$$x(x^2+2)+6(x^2−1)$$
consider the following:
*
*The second term, $6(x^2-1)$ is always divisible by $6$ and consequently is divisible by $3$.
*For the first term $x(x^2+2)$, it is also divisible by 3. Why? $x$ is either $0,-1,1 \pmod 3$. Hence, $x^2 \equiv 0,1 \pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Half of Vandermonde's Identity We know Vandermonde's Identity, which states
$\sum_{k=0}^{r}{m \choose k}{n \choose r-k}={m+n \choose r}$.
Does anyone know what happens if we walk bigger steps with k? Say we skip all the odd ks, is something like
$\sum_{k=0}^{r/2}{m \choose 2k}{n \choose r-2k}=\frac{1}{2} {m+n \choose ... | In general, having the ogf (z-Transform)
$$
F(z) = \sum\limits_{0\, \le \;n} {a_{\,n} \,z^{\,n} }
$$
then
$$
{1 \over m}\sum\limits_{0 \le \,k\, \le \,m - 1} {\left( {z^{\,{1 \over m}} \;e^{\,i\,{{2k\pi } \over m}} } \right)^{\,j}
F(z^{\,{1 \over m}} \;e^{\,i\,{{2k\pi } \over m}} )}
= \sum\limits_{0\, \le \;n} {\,a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Prove that if $a,b,c \in \mathbb{R^+}\text{ and } abc=8\text{ then } {ab+4\over a+2}+{bc+4\over b+2}+{ca+4\over c+2}\ge6$ Question:
Prove that if $a,b,c \in \mathbb{R^+}\text{ and } abc=8\text{ then } {ab+4\over a+2}+{bc+4\over b+2}+{ca+4\over c+2}\ge6$
My Approach:
Now we have:
$${ab+4\over a+2}={2\times (ab+4)\over... | Because you wrote $$\tfrac{ab(2+c)}{2(a+2)}+\tfrac{bc(2+a)}{b+2)}+\tfrac{ca(2+b)}{2(c+2)}
\ge\sqrt[3]{{ab(2+c)\over2(a+2)}\cdot{bc(2+a)\over2(b+2)}\cdot{ca(2+b)\over2(c+2)}},$$ but it should be
$$\tfrac{ab(2+c)}{2(a+2)}+\tfrac{bc(2+a)}{b+2)}+\tfrac{ca(2+b)}{2(c+2)}
\ge3\sqrt[3]{{ab(2+c)\over2(a+2)}\cdot{bc(2+a)\over2(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Prove the inequality $\ln {(1+\frac{1}{x})}> \frac{2}{2x+1}$ Prove the inequality $$\ln {(1+\frac{1}{x})}> \frac{2}{2x+1}$$
for $x>0$.
My attempt: Let $$f(x)=\ln {(1+\frac{1}{x})}-\frac{2}{2x+1}$$
Then $$f'(x)=-\frac{1}{x(x+1)}+\frac{4}{(2x+1)^2}$$
$$f''(x)=\frac{1}{x^2}-\frac{1}{(x+1)^2}-\frac{8}{(2x+1)^3}>0$$
Then th... | Here is a more natural way to prove this. Note that:
$$\ln\bigg(1+\frac{1}{u}\bigg) = \int\limits_u^{u+1} \frac{1}{x} dx$$
i.e. the area under the graph of $\frac{1}{x}$ between $u$ and $(u+1)$. This explains why $\frac{2}{2u+1}$ is such a good approximation in the first place -- it comes from approximating the area as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How would I go about solving for $x$ in $\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=a-b$? The question
This is a homework question. Given the following, I am to solve for $x$ in terms of $a$ and $b$:
$$\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=a-b;a>b.$$
My attempt
Although I see ... | Another way is as follows:
*
*Set $\boxed{x = a + t(a-b)}$ for $t \geq 0$
$$\begin{eqnarray*}
\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}} & = & a-b \\
& \Leftrightarrow & \\
\frac{t(a-b)\sqrt{t(a-b)}+(t+1)(a-b)\sqrt{(t+1)(a-b)}}{\sqrt{t(a-b)}+\sqrt{(t+1)(a-b)}} & = & a-b \\
& \Leftrightarrow & \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solve similar right triangles, given one's hypotenuse, the other's base, and the sum of their heights. I encountered this problem while trying to determine a generic equation for entasis, but this question is not about entasis.
$\theta$ is wanted—given this lovely figure
given that the two triangles are similar, and g... | It is pretty easy to see from the geometry of the figure that
$b \cot \theta + a \cos \theta = h, \tag 1$
whence,
$b \dfrac{\cos \theta}{\sin \theta} + a\cos \theta = h; \tag 2$
now using $\cos^2 \theta + \sin^2 \theta = 1$, i.e. $\cos \theta = \pm \sqrt{1 - \sin^2 \theta}$,
$b \dfrac{\sqrt{1 - \sin^2 \theta}}{\sin \th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Computing $\sqrt[4]{28+16 \sqrt 3}$ I want to compute following radical
$$\sqrt[4]{28+16 \sqrt 3}$$
For that, I first tried to rewrite this in terms of exponential.
$$(28+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
We know that $ 28 = 2 \cdot 7^{\frac{1}{2}}$
$$(2 \cdot 7^{\frac{1}{2}}+16\cdot 3^{\frac{1}{2}})^{\frac{1}{... | $$\sqrt[4] {28+16\sqrt 3}=\sqrt[4] {(\sqrt {12})^2 +(\sqrt {16})^2 +2\sqrt {16\cdot 12}}=\sqrt {4+2\sqrt 3}=\sqrt {(\sqrt 3)^2 +(1)^2 +2\sqrt {3\cdot 1}}=\sqrt 3 +1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:
If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:
I have not been able to find a better method than to calculate $a$ and $b$ then substitute ... | Hint:
As $a,b$ are the roots of $x^2-2x+3=0$
$a^3-3a^2+5a-3=(a^2-2a+3)(a-1)+1=1$
Similarly for $b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Mistake in solving an equation involving a square root I want to solve $2x = \sqrt{x+3}$, which I have tried as below:
$$\begin{equation}
4x^2 - x -3 = 0 \\
x^2 - \frac14 x - \frac34 = 0 \\
x^2 - \frac14x = \frac34 \\
\left(x - \frac12 \right)^2 = 1 \\
x = \frac32 , -\frac12
\end{equation}$$
This, however, is incorrect... | You made a mistake when completing the square.
$$x^2-\frac{1}{4}x = \frac{3}{4} \color{red}{\implies\left(x-\frac{1}{2}\right)^2 = 1}$$
This is easy to spot since $(a\pm b)^2 = a^2\pm2ab+b^2$, which means the coefficient of the linear term becomes $-2\left(\frac{1}{2}\right) = -1 \color{red}{\neq -\frac{1}{4}}$. This m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3052746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 5,
"answer_id": 1
} |
Evaluating $\int_0^1\arctan x\ln(1+x)\left(\frac2x-\frac3{1+x}\right)dx$
How can we find the value of $$\int_0^1\arctan x\ln(1+x)\left(\frac2x-\frac3{1+x}\right)dx$$ using elementary methods?
With some help of calculator I get the result: $\displaystyle{\frac3{128}\pi^3-\frac9{32}\pi\ln^22}$.
Thoughts of this integra... | $$I=2\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx-3\int_0^1\frac{\arctan x\ln(1+x)}{1+x}\ dx$$
Apply IBP for the second integral, we get,
$$I=2\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx-\frac{3\pi}{8}\ln^22+\frac32\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx$$
The first integral was calculated here :
$$\int_0^1 \frac{\arctan x\ln(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3054741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Given $\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1$, prove $\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right) = 1$.
Let $x$ and $y$ be real numbers such that
$$\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1$$
Prove that
$$\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt... | Write $x=\tan a$ and $y = \tan b$ for some (angles) $a,b$. They exsist since $\tan $ is surjective function. Pluging in starting equation we get:
$$ {\sin a+1\over \cos a}\cdot {\sin b+1\over \cos b}=1$$
and after rearranging we get $$\sin(a)+ \sin(b) = \cos (a+b) - \cos 0$$
which is equivalent to $$2\sin{a+b\over 2}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Find the residue, state the nature of the singularity, find the constant term in $1/\sin(ze^z)$ at $z=0$
Find the residue, state the nature of the singularity, find the constant term in series $1/\sin(ze^z)$ at $z=0$.
We can rewrite the function $\frac{1}{\sin(ze^z)}$ as $\frac{ze^z}{\sin(ze^z)}\cdot\frac{1}{ze^z}$. ... | You can find the first few terms of the series, step-by-step
$$ ze^z = z\left(1 + z + \frac{z^2}{2} + \dots \right) = z + z^2 + \frac{z^3}{2} + \dots $$
\begin{align} \sin(ze^z) &= \sin\left(z + z^2 + \frac{z^3}{2} + \dots \right) \\
&= \left(z + z^2 + \frac{z^3}{2} \dots \right) - \frac{1}{3!}\left(z + z^2 + \frac{z^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $ B_{n}(x)=\sum^{\infty}_{k=0} \dfrac{(-1)^k}{k!(n+k)!} \left(\dfrac{x}{2}\right)^{n+2k}$ converges uniformly. Let $n\geq 0,$ be a fixed. The Bessel function of order $n$ is the function defined by
\begin{align} B_{n}(x):=\sum^{\infty}_{k=0} \dfrac{(-1)^k}{k!(n+k)!} \left(\dfrac{x}{2}\right)^{n+2k} .\end{ali... | Happy New Year! Credits to Jakobian
\begin{align} \left| \dfrac{(-1)^k}{k!(n+k)!} \left(\dfrac{x}{2}\right)^{n+2k}\right|\leq \dfrac{1} {k!}\left( \dfrac{\left| x\right|} {2}\right)^{n+2k}\leq \dfrac{1} {k!}\left(\dfrac{c}{2}\right)^{n+2k}.\end{align}
Since,
\begin{align} \lim\limits_{k\to\infty}\left|\dfrac{k!} {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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On Yanqi Xu's 2016 joint undergraduate math research project with Dr. Judy Holdener at Kenyon College In what follows, we let $\sigma(X)$ denote the sum of the divisors of the positive integer $X$. Denote the abundancy index of $X$ by $I(X)=\sigma(X)/X$, and the deficiency of $X$ by $D(X)=2X-\sigma(X)$. Finally, let ... | The following answer is based on a theorem in a paper (to appear) communicated to Dris by Holdener (co-authored with Rachfal).
Let $N = q^k n^2$ be an odd perfect number with special/Euler prime $q$. Suppose that $k>1$.
Consider the proper factor
$$q^{\frac{k-1}{2}} n^2.$$
This has abundancy index
$$I\bigg(q^{\frac{k-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
how to show $ \sum_{10}^{\infty} \frac{\sin{\frac{1}{n}}}{\ln(n)}(e^{\frac{1}{n^2}} - 1)(\sqrt{n^4 - 8})$ converge/diverge $$ \sum_{10}^{\infty} \frac{\sin{\frac{1}{n}}}{\ln(n)}\left(e^{\frac{1}{n^2}} - 1\right)\left(\sqrt{n^4 - 8}\right) $$
I have tried a lot of stuff, didn't work at all . A hint will be good too.
I ... | Hint. One may observe that,
$$
\lim_{ n\to \infty}(e^{\frac{1}{n^2}} - 1)(\sqrt{n^4 - 8})= 1
$$giving, for a certain $n_0\ge10$, $n\ge n_0$,
$$
\frac12 \le(e^{\frac{1}{n^2}} - 1)(\sqrt{n^4 - 8}) \tag1
$$ then, as $n \ge 10$,
$$
\frac{1}{n}-\frac{1}{6n^3}\le\sin{\frac{1}{n}}\tag2
$$ yielding, for $N\ge n_0$,
$$
\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Prove $\lim_{n\to\infty}\frac{2\cdot4\cdot6\cdot...\cdot(2n-2)(2n)}{1\cdot3\cdot5\cdot...\cdot(2n-1)}\frac{1}{\sqrt{2n+1}}=\sqrt\frac{\pi}{2}$ I have to prove that the following limit is equal to $\sqrt{\pi/2}$:
$$\lim_{n\to\infty}\frac{2\cdot4\cdot6\cdot...\cdot(2n-2)(2n)}{1\cdot3\cdot5\cdot...\cdot(2n-1)}\frac{1}{\sq... | Using the double factorial notation we need to find $$\lim_{n\to\infty} \frac {(2n)!!}{(2n-1)!!\sqrt {2n+1}}$$
Now using the relation between double factorial and the factorial, the limit changes to $$\lim_{n\to\infty} \frac {2^{2n}(n!)^2}{(2n)!\sqrt {2n+1}}$$
Using Stirling's approximation for factorials we get $$\lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Trying to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$ to be $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ I am asked to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$. The solution is provided as $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ and I am unable to arrive at this. I was able to arrive at $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$
Here is my wor... | Observe that
$$
\sqrt{8}\times (1+\sqrt{3x})=\sqrt{8}+\sqrt{8}\times \sqrt{3x}
$$ and
$$
\sqrt{8}\times \sqrt{3}=\sqrt{24}=\sqrt{4\times 6}=2\sqrt{6}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Find the dimension of the subspace of $\mathbb{R}^4$ spanned by $x_1= (1, 2, -1, 0)$, $x_2=(2, 5, -3, 2)$, $x_3=(2, 4, -2, 0)$, $x_4=(3, 8, -5, 4)$
Find the dimension of the subspace of $\mathbb{R}^4$ spanned by the vectors
$$x_1= \begin{bmatrix}
1 & 2 & -1& 0
\end{bmatrix}^T, \qquad
x_2= \begin{bmatrix}
2 ... | The given vectors are the columns of
$$
A =
\left[\begin{array}{rrrr}
1 & 2 & 2 & 3 \\
2 & 5 & 4 & 8 \\
-1 & -3 & -2 & -5 \\
0 & 2 & 0 & 4
\end{array}\right]
$$
The reduced row echelon form of $A$ is
$$
\operatorname{rref}(A)=\left[\begin{array}{rrrr}
1 & 0 & 2 & -1 \\
0 & 1 & 0 & 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Probability of coins in a bag Two bags contain $10$ coins each, and the coins in each bag are numbered from $1$ to $10$. One coin is drawn at random from each bag. The probability that one coin has the value $1,2,3$ or $4$ and the other coin has the value $7,8,9$ or $10$ is? I think the answer is $4/25$ but, my friend ... | Let the probability that coin drawn from the first bag is $1,2,3,4$ be $p_1$. We can quickly see that
$$p_1 = \frac{4}{10}$$
Now the probability that coin drawn from the second bag is $7,8,9,10$ be $p_2$. We can see again that this probability is
$$p_2 = \frac{4}{10}$$
So net probability will be a product of $p_1$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluating $\sum_{n=1}^\infty \frac{1}{(n^5)!} \approx 1$ and proving that is irrational. Define $\delta = \sum_{n=1}^\infty \frac{1}{(n^5)!}$. Wolfram says it converges by the ratio test. Trying to prove that $\delta$ is irrational, begin defining $S_n$ as:
\begin{align}
S_n = (n^5)!\delta \: - (n^5)!\sum_{k=1}^n\fra... |
A Liouville number is a real number x with the property that, for
every positive integer n, there exist integers p and q with q > 1 and
such that $$ 0<\left|x-{\frac {p}{q}}\right|<{\frac {1}{q^{n}}}. $$ A
Liouville number can thus be approximated "quite closely" by a
sequence of rational numbers. In 1844, Jo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Possible to prove that a particular trigonometric expression is always positive? This is a continuation of the an earlier post where the geometric motivation was presented. Here I'd like to ask: is it possible to prove $\Delta > 0$ always?
$$\begin{align}
\Delta &\equiv \sin(t) \sin\left(r+ (2 \pi -2 r - t)\frac{\epsi... | Given
$$0<r<\frac{\pi}{4} \qquad 0<t<\frac{\pi}{4} \qquad 0<\varepsilon <1\tag1$$
Easily to see that
$$t<\dfrac\pi2,\quad \dfrac{\varepsilon t}4 <\dfrac\pi{16}.$$
At the same time, sine increases in $\left(0,\dfrac\pi2\right).$
Therefore,
$$\begin{align}
&\sin t >\sin\left(t-\dfrac{\varepsilon t}4\right),\\[4pt]
&\Del... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Number of integral values of $c$ in solution set
Let the quadratic equation $(c-5)x^2-2cx+c-4=0$
has one root in $(0,2)$ and other root in $(2,3).$
Then the number of integral values of $c$ in solution set
Try: writing quadratic equation as $$f(x)=x^2-\bigg(\frac{2c}{c-5}\bigg)x+\frac{c-4}{c-5}=0\;\;, c\neq 5$$
$f(x)... | You have only used that $2$ lies between the two roots. But we are given more information than that. We also know that both roots lie between $0$ and $3$ (i.e. $f(0)$ and $f(3)$ are both positive). That will exclude seven of your $18$ solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Probabillity - rolling a dice 20 times, probability of a result gets only once
A dice is rolled $20$ times, with the possible results $\left\{1,2,3,4,5,6\right\}$.
Let $X$ be the number of results, out of the possible 6, which were chosen only once during the 20 rolls.
Calculate $P\left\{X\right\}$
I find it hard to ... | Observe that since there are $20$ rolls, $P(X=6)=0$. So we need only check the probabilities that $X=1,2,3,4,5$. These can be done on a case by case basis.
For $X=5$, the probability is
$$\frac{\binom{6}{5}\cdot\binom{20}{5}\cdot 5!}{6^{20}}$$
since we must choose the the $5$ results which will occur only once, and ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3070013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the sum $\sum _{i=1}^4\:\sum _{j=1}^i\:\left(i\cdot \:j-1\right)$
Is it correct if this sum is solved this way?
| $$\begin{align}
\sum_{i=1}^n\sum_{j=1}^i (ij-1)
&=\sum_{i=1}^ni\left(\sum_{j=1}^ij-1\right)\\
&=\sum_{i=1}^ni\left(\frac {i(i+1)}2-1\right)\\
&=\frac 12\sum_{i=1}^ni(i^2+i-2)\\
&=\frac 12 \sum_{i=1}^n i(i-1)(i+2)\\
&=\frac 12 \sum_{i=1}^n(i+1)i(i-1)+i(i-1)\\
&=\frac 12 \sum_{i=1}^n 6\binom {i+1}3+2\binom i2\\
&=3\binom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Eigenvalue of (some) $ 4 \times 4 $ symmetric matrices $$A=\pmatrix{
0 & 3 & 2 & 0 \\
3 & 0 & 0 & 2 \\
2 & 0 & 0 & 3 \\
0 & 2 & 3 & 0 \\
}$$
Is there a quicker way to compute eigenvalues of this matrix other than to do it the long way? And what are the strategies for similar matrices?
| Observe
\begin{align}
M=
\begin{pmatrix}
A & B\\
B & A
\end{pmatrix}
\end{align}
where $B = 2I_2$ and $A=\begin{pmatrix}0 & 3\\ 3 & 0 \end{pmatrix}$ and $A$ & $B$ commute. Then we see that
\begin{align}
\det\left(M-\lambda I_4 \right) = \det ((A-\lambda I_2)^2-B^2)
\end{align}
where we used the determinant formula for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3074092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the sum $\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{41+\cdots}}}}}$ Okay so this can be written as $$\sqrt{5+\sqrt{(5+6)+\sqrt{(5+6+8)+\sqrt{(5+6+8+10)+\sqrt{(5+6+8+10+12)\cdots}}}}}$$
Putting it as $y$ and squaring both sides doesn't seem to help, and I don't know what else can be done.
| This is a slightly more rigorous form of @Pablo_'s excellent insight. @Sangchul Lee covers the full, analytic answer.
Set $a_n = n^2 + 5n + 5$ for $n \geq 0$. This sequence gives the coefficients of the "infinite radical." Rather than consider the full infinite radical, consider the "partial radicals," defined as $$r_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
Is it true that $\forall n \in \Bbb{N} : (\sum_{i=1}^{n} a_{i} ) (\sum_{i=1}^{n} \frac{1}{a_{i}} ) \ge n^2$ , if all $a_{i}$ are positive?
If $\forall i \in \Bbb{N}: a_{i} \in \Bbb{R}^+$ , is it true that $\forall n \in \Bbb{N} : \big(\sum_{i=1}^{n}a_{i}\big) \big(\sum_{i=1}^{n}
\frac{1}{a_{i}}\big) \ge n^2$ ?
I ha... | Just apply
$$AM\geq GM\geq HM$$
$$\Longrightarrow \frac{\sum_{i=1}^n a_i}{n}\geq \frac{n}{\sum_{i=1}^n\frac{1}{a_i}}$$
Now the result is immediate.
Here equality holds iff all $a_i's$ are equal.
Hope it helps:)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Cauchy Schwarz - Finding minimum of a function For $x, y , z$ in real numbers, If $2x+y+z=5$, then what is the min value of $x^2 + y^2 + z^2 - 2x + 4y + 6$. This is a weekend brain teaser in the 2nd week of Calc 3.
| Here is an alternate solution using the Cauchy-Schwarz Inequality:-
We have to minimize $f(x,y,z)=x^2+y^2+z^2-2x+4y+6=(x-1)^2+(y+2)^2+z^2+1$ subject to $2x+y+z=5$
From the Cauchy-Schwarz Inequality , we have $$(2\cdot(x-1)+1 \cdot (y+2) +1\cdot z)^2 \leq (2^2+1^2+1^2)((x-1)^2+(y+2)^2+z^2)$$ $$ \therefore (2x+y+z)^2 \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Any thoughts on this integral? $\int \cos^2(x)\cdot\sin^4(x)dx$
I tried the usual trigonometric identities but they don't seem helpful
| Write $c$ for $\cos x$ and $s$ for $\sin x$. Then the integrand is
\begin{align}
c^2 s^4
&= c^2 s^2 (s^2) \\
&= \frac{1}{4} (2cs)^2 (1 - c^2) \\
&= \frac{1}{2} \frac{1}{4} (2cs)^2 (2 - 2c^2) \\
&= \frac{1}{2} \left( \frac{1}{4} (2cs)^2 (1 - 2c^2) \right)
+ \frac{1}{2} \left( \frac{1}{4} (2cs)^2 \right) \\
\end{... | {
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"source": "stackexchange",
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I can't see what is wrong with my answer to this P&C problem:
In how many different ways can 8 different books be distributed among 3 students if each receiver gets at least 2 books?
Here's what I did:
First take 6 books from the lot; that can be done in $\binom{8}{6}$ ways.
Then pick two from the 6 books, which can ... | As already explained you are overcounting the correct number: according to your procedure, the same distribution can be obtained in different orders.
Split the enumeration into different cases according to the possible distributions of the books among the three students: say $n_1$ books to the first student, $n_2$ boo... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Inner product induced by the polarisation identitiy I am required to show that given a Banach space $(X,\|\cdot\|)$ on $\mathbb{R}$ that if the parallelogram identity holds then $X$ is a Hilbert space with the inner product given by the polarisation identity.
Proof:
Let $x+z=\frac{x+y}{2}+z+\frac{x-y}{2}$ and $y+z=\fra... | If you take your equality with $y=0$, you have shown that
$$
\left(\tfrac12\,x|z\right)=\tfrac12\,(x|z).
$$
And that's exactly what you need, now you can remove the $1/2$ you wanted.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\left\lfloor \frac{a-b}{2} \right\rfloor + \left\lceil \frac{a+b}{2} \right\rceil = a$ when $a,b$ are integers? Let $a$ and $b$ be positive integers.
If $b$ is even, then we have $$\left\lfloor \frac{a-b}{2} \right\rfloor + \left\lceil \frac{a+b}{2} \right\rceil = a$$
I think the equality also hold when $b$ is odd. Wh... | If $a$ and $b$ are both odd (or both even), then $a-b$ and $a+b$ are both even, and thus $$\left\lfloor \frac{a-b}{2} \right\rfloor = \frac{a-b}{2} \quad\text{and}\quad \left\lceil \frac{a+b}{2} \right\rceil = \frac{a+b}{2}.$$
Otherwise, if exactly one of $a$ and $b$ is odd, then $a-b$ and $a+b$ are both odd, and thus ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$ If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$
My attempt:
$$\cos^6 (x) + \sin^4 (x)=1$$
$$\cos^6 (x) + (1-\cos^2 (x))^{2}=1$$
$$\cos^6 (x) + 1 - 2\cos^2 (x) + \cos^4 (x) = 1$$
$$\cos^6 (x) + \cos^4 (x) - 2\cos^2 (x)=0$$
| $$1=\cos^6x+\sin^4x\leq\cos^2x+\sin^2x=1,$$ which gives the following system.
$$\cos^6x=\cos^2x$$ and $$\sin^4x=\sin^2x.$$
Can you end it now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a_{n+1}=\frac{3+a_n^2}{a_n+1}$ and $a_1=1$, then what is $\lim\limits_{n\to\infty}\left(\frac{4}{3}\right)^n(3-a_n)$? Question:
If $a_{n+1}=\frac{3+a_n^2}{a_n+1}$ and $a_1=1$, then what is $\lim\limits_{n\to\infty}\left(\frac{4}{3}\right)^n(3-a_n)$ ?
My approach:
I am able to prove separately that the sequence $a... | EDIT: I misunderstood the question. This answer only proves convergence, but doesn't find the limit.
Let's try looking at rations again:
$$
\begin{align}
R_n = \frac{(\frac{4}{3})^{n+1}(3-a_{n+1})}{(\frac{4}{3})^{n}(3-a_{n})} &=
\frac{4}{3} \times \frac{3-\frac{3+a_n^2}{a_n+1}}{3-a_{n}} =
\frac{4}{3} \times \frac{\frac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can $2\cos(x-\frac{\pi}2) = -2\sin(x-\frac{\pi}2)$ How can $2\cos(x-\dfrac{\pi}2) = -2\sin(x-\dfrac{\pi}2)$
I know that $\cos(-x) = \cos(x)$ and that $\cos(\dfrac{\pi}2-x) = \sin(x)$
From these two formulas I can get
$1.$ $2\cos(x-\dfrac{\pi}2) = 2\cos(\dfrac{\pi}2-x)$
$2.$ $2\cos(\dfrac{\pi}2-x) = 2 \sin(x)$
$... | $$2\cos\left(x-\dfrac\pi2\right)=-2\sin\left(x-\dfrac\pi2\right)$$
$$\implies\tan\left(x-\dfrac\pi2\right)=-1=\tan\left(-\dfrac\pi4\right)$$
$$\implies x-\dfrac\pi2=n\pi-\dfrac\pi4$$ where $n$ is any integer
Clearly, the given relationship is not an identity
| {
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"source": "stackexchange",
"question_score": "3",
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$X \sim Exp(\frac{1}{3}) $ If $Y= \max(X,2)$ Find $E(Y)$ $X \sim Exp(\frac{1}{3}) $
If $Y= \max(X,2)$ Find $E(Y)$
I did in two ways can anyone point out my mistake ?
First one
$f(y)=
\begin{cases}
\ 2, & \text{if $0<x\le2$ } \\[2ex]
X, & \text{if $2<x<\infty$ }
\end{cases}$
$E(Y)=E(Y|0<x<2)P(0<x\le2)+E(Y|2<x<\infty... | This is how I would proceed (similar to your second method):
$E(Y)=\int_0^22\times\frac{1}{3}e^{\frac{-x}{3}}dx + \int_2^\infty x\times\frac{1}{3}e^{\frac{-x}{3}}dx$
$=2(1-e^{\frac{-2}{3}}) +3\int_\frac{2}{3}^\infty t\times e^{-t}dt$
$=2(1-e^{\frac{-2}{3}}) +3\Gamma(2,\frac{2}{3})$
$=2+3e^{\frac{-2}{3}}$
Here is an eas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"Milk" the integral $\int_0^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2(x^s+1)}\mathrm dx$ I found the following integral in chapter $13$ of Irresistible Integrals, and I would like to see which conclusions you can reach from it. My goal in asking this question is to see which methods I can employ in t... | First I will like to give some steps and maybe some more insight of this integral.$$I=\int_0^1\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2(x^s+1)}\mathrm dx+\int_1^\infty\left(\frac{x^2}{x^4+2ax^2+1}\right)^r\frac{x^2+1}{x^2(x^s+1)}\mathrm dx$$
With $x\rightarrow \frac{1}{x}$ in the second one we get:
$$\int... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Summation of combination of binomial coefficient Is there any way to find:
$$\sum_{i=0}^n {\binom{n}{i}i^k}$$
I know that we can find it for small k by using binomial theorem by differentiating both sides and then multiplying both sides by x and repeating till the form presents itself and then putting x =1, but is ther... | I think that the general form of $$S_k=\sum_{i=0}^n {\binom{n}{i}i^k}$$ is "just" an hypergeometric function
$$S_k=n \, _kF_{k-1}(2,\cdots,2,1-n;1,\cdots,1;-1)$$ in which the arguments appear $(k-1)$ times.
Looking for cases, the result seems to be of the form
$$S_k=2^{n-k}\,n \, P_{k-1}(n)$$ where $P$ is a polynomial.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3091196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find limit of function of two variables (if exist) Find $$
\lim_{(x,0)\to (\pi,0)}\frac{3sin^3(x+y)}{y^2+(x-\pi)^2}.
$$
My attempt was to show that that limit does not exist. But for any pairs(checked by me) of $(x_n,y_n)\to(\pi,0)$ this function tends to $0$ (when $n\to \infty$) (for example I checked $(x_n,y_n)=(\fra... | I would expect the limit to exist and be zero.
*
*Note that $\sin (t\pm\pi)=\pm\sin t$. So $\sin (x+y)=-\sin (x-\pi+y)$.
*We also have, for small $t$, $|\sin t|\leq t$.
*And, $(a+b)^3\leq 4a^3+4b^3$.
Thus, for $(x,y)$ close to $(\pi,0)$,
\begin{align}
\left|\frac{3\sin^3(x+y)}{y^2+(x-\pi)^2}\right|
&=\left|\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3091505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate ${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$
Evaluate $${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$$
It should be $$\frac{1}{12}n(n+1)(2n+1)$$ but I don't know how to prove that. I am also aware that $\lim\limits_{\theta \to 0} \dfrac{1-\cos \theta}{\theta ^2}=... | Multiply by conjugate:
$$\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}=\\
\lim_{x \rightarrow 0} \frac{1-\cos^2 x\cos^2 2x\cdots \cos^2 nx}{x^2\cdot(1+\cos x\cos 2x\cdots \cos nx)}=\\
\frac12 \lim_{x \rightarrow 0} \frac{1-(1-\sin^2 x)(1- \sin^2 2x)\cdots (1-\sin^2 nx)}{x^2}=\\
\frac12 \lim_{x \right... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Hyperparameter gradients for Matérn covariance Similarly to this question I am seeking the derivative of a covariance function with respect to its parameters. However, that question is specifically about the squared exponential kernel.
How can I calculate the hyperparameter gradients for the Matérn 5/2 kernel?
$$k_{\m... |
How can I calculate the hyperparameter gradients for the Matérn 5/2 kernel?
If the general Matern kernel is written
$$ \mathcal{K}_M(r) = \sigma^2\frac{2^{1-\nu}}{\Gamma(\nu)}
\left( \frac{r\sqrt{2\nu}}{\ell} \right)^\nu K_\nu\left( \frac{r\sqrt{2\nu}}{\ell} \right) $$
then the $5/2$ kernel is given by
$$ \mathcal... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find limit (type 0/0) I'm struggling to find the limit
$$I = \lim_{x\to 0}\frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}$$
What I was trying:
$$ I = \lim_{x\to 0}\frac{1-x + 2\sqrt{1-x} + 1 - (1-x) - 1 - \sqrt[3]{8-x}}{x} $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2 - (2-x)- \sqrt[3]{8-x}}{x} \qquad \quad $$
$$ = \lim_{x\to 0}... | Until the last step your attempt is correct. However, the last step (splitting to two limits) is not allowed, because it becomes the indetermined form $\infty-\infty$. Here is an alternative way with splitting:
$$\lim_{x\to 0}\frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}=\\\lim_{x\to 0}\frac{(2\sqrt{1-x}-2)+(2-\sqrt[3]{8-x})}{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve $2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$ for $x$? I'm struggling to figure out how to solve $2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$ for $x$.
The answer is $x = \frac{12}{5}$, but I am getting $x = \frac{-35}{9}$
My steps are:
$2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$
*
*make exponent postive
$2x\frac{1}{\s... | If we put $x=4t$, the equation becomes
$$\frac{4t}{\sqrt{1-t}}=6\sqrt{1-t}$$
and
$$2t=3(1-t)$$
which gives
$$t=\frac 35 \; ,\; x=\frac{12}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Writing a number in polar form (help converting $\theta$ to $\pi$)
Write $w = \sqrt{3} - i$ in polar form.
How is $\theta = \frac{-1}{\sqrt3} \textrm{ converted into } \frac{-\pi}{6}$? I understand that $w$ lies in the fourth quadrant of the unit circle, but that's about all I understand. I tried to convert degr... | The first thing to realize is that if you ever have $x^2 + y^2 = 1$ then there is a $\theta$ so that $x = \cos \theta; y = \sin \theta$ and $\tan \theta = \frac yx$.
And second thing to realise is that if you have any $a,b$ where they don't both equal $0$, you can rewrite $a = (\sqrt{a^2+b^2})a'$ and $b = (\sqrt{a^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3102699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Formula for partial sum of binoms: $\sum_{k=0}^{\lfloor n/2 \rfloor} \binom nk \binom mk$ I am dealing with some expressions containing combinatoric numbers. Does anybody know a formula for this?
$$\displaystyle\sum_{k=0}^{\left\lfloor \dfrac{n}{2} \right\rfloor} \binom{n}{k}\binom{m}{k}$$
| We can find a closed form expression in case $m\leq \left\lfloor\frac{n}{2}\right\rfloor$. We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write for instance
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n
\end{align*}
We obtain for $m\leq \left\lfloor\f... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that the equation represents a sphere, find center and radius $2x^2 + 2y^2 + 2z^2 = 8x - 24z + 1$
So here's my guess:
$2x^2 - 8x + 2y^2 + 2z^2 + 24z = 1$
Get all variables on one side
$2x^2 - 8x + 16 + 2y^2 + 2z^2 + 24z + 144 = 1 + 16 + 144$ Complete square
$(2x - 8)^2 + 2y^2 + (2z^2 + 24)^2 = 161$ Stopped h... | Note that $(2x)^2 \ne 2x^2$. The answer is:
$$2(x-2)^2 + 2(y-0)^2 + 2(z+6)^2 = 8 + 0 + 72 + 1$$
Thus, center $(2,0,-6)$, and radius $\sqrt{81/2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Linear recursion with constant coefficients that fullfil $a_n = 3^n + 7^n$ for $n \in \mathbb{N}_0$ Let $a_n = 3^n + 7^n$ for $n \in \mathbb{N}_0$
I know that the generating function of $a_n= 3^n+7^n = A(x) = \frac{1}{1 - 3x} + \frac{1}{1 - 7x}.$
The exponential generating function
$$A(x) = \sum_{n=0}^{\infty}\frac{a_... | $$a_n-3^n = 7^n$$
so $$ a_{n+1}-3^{n+1}= 7^{n+1}$$
so $$a_{n+1}-3^{n+1}=7\cdot 7^n = 7(a_n-3^n )$$
so $$a_{n+1}-7a_n = -4\cdot 3^n$$
so $$a_{n+2}-7a_{n+1} = -4\cdot 3^{n+1}$$ $$=-12\cdot 3^n$$ $$= 3(-4\cdot 3^n)$$
$$ =3 (a_{n+1}-7a_n )$$
So $$a_{n+2}-10a_{n+1} +21a_n =0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Precalculus algebra exercise Hi I need to solve this problem and I don’t know how so I’d appreciate a hint.
If
$a^2x^2 + b^2y^2 + c^2z^2 = 0$
$a^2x^3 + b^2y^3 + c^2z^3 = 0$
$\frac 1x - a^2 = \frac 1y - b^2 = \frac 1z - c^2$
Then $a^4x^3 + b^4y^3 + c^4z^3 = 0$
I think that $a^4x^3 + b^4y^3 + c^4z^3 = 0$ is a factor in... | Suppose that
$$
\begin{cases}
(ax)^2 + (by)^2 + (cz)^2 = 0\\
a^2x^3 + b^2y^3 + c^2z^3 = 0\\
\frac 1x - a^2 = \frac 1y - b^2 = \frac 1z - c^2
\end{cases}
$$
However, for any $a, x \in \mathbb R$, $(ax)^2 \geq 0$. So we have
$$
0 = (ax)^2 + (by)^2 + (cz)^2 \geq 0 + 0 + 0 = 0
$$
Therefore in each inequality, there's equal... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How to evaluate $\lim_{n \to \infty} \sqrt[n]{n^{(4n)}+(4n)^n}\left[\left(2+\frac{1}{n^2}\right)^{18}-\left(4+\frac{4}{n^2}\right)^9\right]$
$$\lim_{n \to \infty} \sqrt[n]{n^{(4n)}+(4n)^n}\left[\left(2+\frac{1}{n^2}\right)^{18}-\left(4+\frac{4}{n^2}\right)^9\right]$$
How can I deal with n-th root and the second part ... | You may also proceed using
*
*$(\star)$: $a^n-b^n = (a-b)\sum_{k=0}^{n-1}a^kb^{n-k}$
\begin{eqnarray*} \sqrt[n]{n^{(4n)}+(4n)^n}\left[\left(2+\frac{1}{n^2}\right)^{18}-\left(4+\frac{4}{n^2}\right)^9\right]
& = & \color{blue}{n^4} \sqrt[n]{1+\frac{4^n\cdot n^n}{n^{4n}}}\left[\left( 4+\frac{4}{n^2} + \frac{1}{n^4... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find if terms are terms of the same arithmetic progression
Is it possible that numbers $\frac{1}{2}, \frac{1}{3}, \frac{1}{5}$ are (not necessarily adjacent) terms of the same arithmetic progression?
Hint: Yes. Try $\frac{1}{30}$ as a difference.
I was going back and forth how they found out that difference.
My id... | The key is to find the greatest common divisor of the two differences $\frac{1}{6}$ and $\frac{2}{15}$. How? Exactly like we would for integers.
If we divide $\frac{1}{6}$ by $\frac{2}{15}$, we get $1$ with a remainder of $\frac{1}{6}-\cdot\frac{2}{15} = \frac{5}{30}-\frac{4}{30}=\frac{1}{30}$.
Next, we have the pair $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Geometry problem (proving a relation between sides of a triangle) Let $ABC$ be a triangle with unequal sides. The medians of $ABC$,
when extended, intersect its circumcircle in points $L, M, N$. If $L$ lies
on the median through $A$ and $LM = LN$, prove that: $2BC^{2}=CA^2+AB^2$.
So I first wrote midpoint theorem $AB^2... | Below is a brute force solution. Though there should be more elegant solutions, I have decided to present this one due to the beauty of some resulting algebraic expressions.
First, it should be clarified that (as will be seen) the real meaning of the condition "a triangle with unequal sides" is "a triangle with $AC\ne ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find integral of this fraction with a radical. Is my process right? I think this involves a u-sub and then partial fractions? Is this process right? Basically I learned any integral can be solved with using usub and integration by parts... and using partial fractions is just some algebraic manipulation that then relies... | Notice how $\int \frac{1}{x} dx = \ln{(abs(x))} + C$ and not $\ln{(x)} + C$, for instance, the term $\frac{1}{2}\ln{(u-1)}$ is actually $\frac{1}{2} \ln{(abs(u-1))}$. The final expression becomes:
$\int \frac{dx}{2 \sqrt{x+3} + x} = \frac{3}{2} \ln {(\sqrt{x+3} + 3)} + \frac{1}{2} \ln {(abs(\sqrt{x+3} - 1))} + C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int \left(x^2(1+x^3)^{\frac{2}{3}}\right)^{-1} dx$ I'm stuck on evaluating this indefinite integral.
$$\int\frac{dx}{x^2(1+x^3)^{\frac{2}{3}}}$$
I tried doing a u-substitution on the $1+x^3$ term inside the two-thirds power but didn't get anywhere. Any help?
| There is a "dirty" trick having in mind that
*
*$\frac{d}{dx}\left( \sqrt[3]{1+x^3}\right)= \frac{x^2}{\left(1+x^3\right)^{\frac{2}{3}}}$
Then partial integration gives:
\begin{eqnarray*} \int\frac{dx}{x^2(1+x^3)^{\frac{2}{3}}}\; dx
& = & \int\frac{1+x^3 - x^3}{x^2\left(1+x^3\right)^{\frac{2}{3}}}\; dx \\
& = & \int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3118512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral check. Is partial fractions the only way? I'm getting a different answer from wolfram and I have no idea where. I have to integrate:
$$\int_0^1 \frac{xdx}{(2x+1)^3}$$
Is partial fractions the only way?
So evaluating the fraction first:
$$\frac{x}{(2x+1)^3} = \frac{A}{2x+1} + \frac{B}{(2x+1)^2} + \frac{C}{(2x+1... | Hint: Substitute $x=\tfrac12 u -\tfrac12$
Addendum: Your original problem was $\int_0^1\frac{x\; dx}{(2x+1)^3}$. For this substitution, compute:
Integrand: We have $x=\tfrac12 u -\tfrac12=\tfrac12(u-1)$, so
$$\frac{x}{(2x+1)^3}=\frac{\tfrac12(u-1)}{(2\cdot\frac12(u-1)+1)^3}
=\tfrac12\frac{u-1}{u^3}=\boxed{\tfrac12\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3120889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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calculation of $\int^{1}_{0}\frac{1}{\sqrt{x-x^2}(x^2+3x+2)}dx$
Calculation of $$\int^{1}_{0}\frac{1}{\sqrt{x-x^2}(x^2+3x+2)}dx$$
My Try: Let $$I = \int^{1}_{0}\frac{1}{\sqrt{x-x^2}(x+1)(x+2)}dx$$
Put $x=\sin^2 \theta$ and $dx = 2\sin \theta \cos \theta$ and changing limits
So $$I = \int^{\frac{\pi}{2}}_{0}\frac{2}{(... | We have $$\frac{I}{2}=\int_0^{\pi/2}\left(\frac{1}{1+\sin^2\theta}-\frac{1}{2+\sin^2\theta}\right)d\theta,$$and with $t=\tan\theta$ we have $\sin^2\theta=\frac{t^2}{1+t^2}$ so $$\int_0^{\pi/2}\frac{d\theta}{k+\sin^2\theta}=\frac{1}{k+1}\int_0^\infty\frac{dt}{\frac{k}{k+1}+t^2}=\frac{\pi}{2\sqrt{k(k+1)}},$$so$$I=\pi\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the minimum value of $\dfrac{a_1^2+a_2^2+\dots+a_n^2}{(a_1+a_2+\dots+a_n)^2}$? Motivation: To find the maximum value of $c$ in 2.4-1 Lemma (Linear combinations) of the book Kreyszig - Introductory Functional Analysis with Applications for $X=\mathbb{R^n}$ which reduces to solving the following question:
Wha... | Let $\vec{a} = \begin{bmatrix}a_1 & a_2 & \cdots & a_n\end{bmatrix}^T$ and $\vec{1} = \begin{bmatrix}1 & 1 & \cdots & 1\end{bmatrix}^T$. Using the Cauchy-Schwarz Inequality, we have
\begin{align*}
\|\vec{a}\|^2 \|\vec{1}\|^2 &\ge (\vec{a} \cdot \vec{1})^2
\\
(a_1^2+a_2^2+\cdots+a_n^2)(1^2+1^2+\cdots+1^2) &\ge (a_1\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Why is $\sum_{i=1}^6 2^i = 2^7-2$?
Why is $$\sum_{i=1}^6 2^i = 2^7-2$$
| $$\sum_{i=1}^6 2^i =2+2^2+2^3+2^4+2^5+2^6$$ $$= 2(1+2+2^2+..+2^5) = 2{2^6-1\over 2-1} =2^7-2$$
We have generaly $$1+x+x^2+...+x^n = {x^{n+1}-1\over x-1}$$
| {
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"url": "https://math.stackexchange.com/questions/3128022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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find the maximum of the value $\left(\sum_{i=1}^{n}ia_{i}\right)\left(\sum_{i=1}^{n}\frac{a_{i}}{i}\right)^2$ let $a_{i}\ge 0$ and such $$a_{1}+a_{2}+\cdots+a_{n}=1$$
find the maximum of the value
$$\left(\sum_{i=1}^{n}ia_{i}\right)\left(\sum_{i=1}^{n}\dfrac{a_{i}}{i}\right)^2$$
I try to use
From Pólya-Szegö’s inequa... | Alternative solution
Problem: Let $a_i\ge 0, \forall i$ with $a_1 + a_2 + \cdots + a_n = 1$. Prove that the maximum of
$$\left(\sum_{i=1}^n i a_i\right)\left(\sum_{i=1}^n \frac{a_i}{i}\right)^2$$
is $\frac{4(n+1)^3}{27n^2}$.
We may use the approach in my answer to this equation:
An upper bound of product of two inner p... | {
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"url": "https://math.stackexchange.com/questions/3128289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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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.