Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Sum of the form $r+r^2+r^4+\dots+r^{2^k} = \sum_{i=1}^k r^{2^i}$ I am wondering if there exists any formula for the following power series :
$$S = r + r^2 + r^4 + r^8 + r^{16} + r^{32} + ...... + r^{2^k}$$
Is there any way to calculate the sum of above series (if $k$ is given) ?
| I haven’t been able to obtain a closed form expression for the sum, but maybe you or someone else can do something with what follows.
In Blackburn's paper (reference below) there are some manipulations involving the geometric series
$$1 \; + \; r^{2^n} \; + \; r^{2 \cdot 2^n} \; + \; r^{3 \cdot 2^n} \; + \; \ldots \; +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/276892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 5,
"answer_id": 1
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How to show that $\frac{x^2}{x-1}$ simplifies to $x + \frac{1}{x-1} +1$ How does $\frac{x^2}{(x-1)}$ simplify to $x + \frac{1}{x-1} +1$?
The second expression would be much easier to work with, but I cant figure out how to get there.
Thanks
| Very clever trick: If you have to show that two expressions are equivalent, you work backwards. $$\begin{align}=& x +\frac{1}{x-1} + 1 \\ \\ \\ =& \frac{x^2 - x}{x-1} +\frac{1}{x - 1} + \frac{x-1}{x-1} \\ \\ \\ =& \frac{x^2 - x + 1 + x - 1}{x - 1} \\ \\ \\ =&\frac{x^2 }{x - 1}\end{align}$$Now, write the steps backwards... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/278481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
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Compute $\lim_{n\to\infty} \left(\sum_{k=1}^n \frac{H_k}{k}-\frac{1}{2}(\ln n+\gamma)^2\right) $ Compute
$$\lim_{n\to\infty} \left(\sum_{k=1}^n \frac{H_k}{k}-\frac{1}{2}(\ln n+\gamma)^2\right) $$
where $\gamma$ - Euler's constant.
| We have
\begin{align}
2\sum_{k=1}^n \frac{H_k}{k} &= 2\sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} \\
&= \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} + \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} \\
&= \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} + \sum_{j=1}^n \sum_{k=j}^n \frac{1}{jk}, \text{ swapping the order of summation on the second ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/279380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 1,
"answer_id": 0
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prove the divergence of cauchy product of convergent series $a_{n}:=b_{n}:=\dfrac{(-1)^n}{\sqrt{n+1}}$ i am given these series which converge. $a_{n}:=b_{n}:=\dfrac{(-1)^n}{\sqrt{n+1}}$ i solved this with quotient test and came to $-1$, which is obviously wrong. because it must be $0<\theta<1$ so that the series conver... | $\sum_{n=0}^\infty\dfrac{(-1)^n}{\sqrt{n+1}}$ is convergent by Leibniz's test, but it is not absolutely convergente (i.e. it is conditionally convergent.)
To show that the Cauchy product does not converge use the inequality
$$
x\,y\le\frac{x^2+y^2}{2}\quad x,y\in\mathbb{R}.
$$
Then
$$
\sqrt{n-k+1}\,\sqrt{k+1}\le\frac{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/279890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What is limit of: $\displaystyle\lim_{x\to 0}$$\tan x - \sin x\over x$ I want to search limit of this trigonometric function:
$$\displaystyle\lim_{x\to 0}\frac{\tan x - \sin x}{x^n}$$
Note: $n \geq 1$
| Checking separatedly the cases for $\,n=1,2,3\,$, we find:
$$n=1:\;\;\;\;\;\;\frac{\tan x-\sin x}{x}=\frac{1}{\cos x}\frac{\sin x}{x}(1-\cos x)\xrightarrow [x\to 0]{}1\cdot 1\cdot 0=0$$
$$n=2:\;\;\frac{\tan x-\sin x}{x^2}=\frac{1}{\cos x}\frac{\sin x}{x}\frac{1-\cos x}{x}\xrightarrow [x\to 0]{}1\cdot 1\cdot 0=0\;\;(\te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/279967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}} \geq \frac{n^2+3n}{2n+2}$ How to prove the following inequalities without using Bernoulli's inequality?
*
*$$\prod_{k=1}^{n}{\sqrt[k+1]{k}} \leq \frac{2^n}{n+1},$$
*$$\sum_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}} \geq \frac{n^2+3n}{2n+2}.$$
My proof:
*
... | The inequality to be shown is
$$(n+1)^{n+1}\geqslant(n+3)^n,
$$
for every positive integer $n$.
For $n = 1$ it is easy. For $n \ge 2$, apply AM-GM inequality to $(n-2)$-many $(n+3)$, 2 $\frac{n+3}{2}$, and $4$, we get
$$(n+3)^n < \left(\frac{(n-2)(n+3) + \frac{n+3}{2} + \frac{n+3}{2} + 4}{n+1}\right)^{n+1} = \left(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/280360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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how to find inverse of a matrix in $\Bbb Z_5$ how to find inverse of a matrix in $\Bbb Z_5$
please help me explicitly how to find the inverse of matrix below, what I was thinking that to find inverses separately of the each term in $\Bbb Z_5$ and then form the matrix?
$$\begin{pmatrix}1&2&0\\0&2&4\\0&0&3\end{pmatrix}$$... | Hint: Use the adjugate matrix.
Answer: The cofactor matrix of $A$ comes
$\color{grey}{C_A=
\begin{pmatrix}
+\begin{vmatrix} 2 & 4 \\ 0 & 3 \end{vmatrix} & -\begin{vmatrix} 0 & 4 \\ 0 & 3 \end{vmatrix} & +\begin{vmatrix} 0 & 2 \\ 0 & 0 \end{vmatrix} \\
-\begin{vmatrix} 2 & 0 \\ 0 & 3 \end{vmatrix} & +\begin{vmatrix} 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/280522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Showing that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left (\sqrt{a^2+1}-1\right)$. How can I show that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left(\sqrt{a^2+1}-1\right)$?
| Let $x = a \sin(y)$. Then we have
$$\dfrac{\sqrt{a^2-x^2}}{1+x^2} dx = \dfrac{a^2 \cos^2(y)}{1+a^2 \sin^2(y)} dy $$
Hence,
$$I = \int_{-a}^{a}\dfrac{\sqrt{a^2-x^2}}{1+x^2} dx = \int_{-\pi/2}^{\pi/2} \dfrac{a^2 \cos^2(y)}{1+a^2 \sin^2(y)} dy $$
Hence,
$$I + \pi = \int_{-\pi/2}^{\pi/2} \dfrac{a^2 \cos^2(y)}{1+a^2 \sin^2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/281587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Let $a,b$ and $c$ be real numbers.evaluate the following determinant: |$b^2c^2 ,bc, b+c;c^2a^2,ca,c+a;a^2b^2,ab,a+b$| Let $a,b$ and $c$ be real numbers. Evaluate the following determinant:
$$\begin{vmatrix}b^2c^2 &bc& b+c\cr c^2a^2&ca&c+a\cr a^2b^2&ab&a+b\cr\end{vmatrix}$$
after long calculation I get that the answer w... | Imagine expanding along the first column. Note that the cofactor of $b^2c^2$ is $$(a+b)ac-(a+c)ab=a^2(c-b)$$ which is a multiple of $a^2$. The other two terms in the expansion along the first column are certainly multiples of $a^2$, so the determinant is a multiple of $a^2$. By symmetry, it's also a multiple of $b^2$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/282655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How I study these two sequence?
Let $a_1=1$ , $a_{n+1}=a_n+(-1)^n \cdot 2^{-n}$ , $b_n=\frac{2 a_{n+1}-a_n}{a_n}$
(1) $\{\ {a_n\}}$ converges to $0$ and $\{\ {b_n\}}$ is a cauchy sequence .
(2) $\{\ {a_n\}}$ converges to non-zero number and $\{\ {b_n\}}$ is a cauchy sequence .
(3) $\{\ {a_n\}}$ converges to $0$ and $\... | We have $b_n=\frac{2 a_{n+1}-a_n}{a_n}=2\frac{a_{n+1}}{a_n}-1$. For very large values of $n$, since $a_n\to2/3$ we have $a_{n+1}\sim a_n$. So $b_n\to 2-1=1$ so it is Cauchy as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/283294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Question about theta of $T(n)=4T(n/5)+n$ I have this recurrence relation $T(n)=4T(\frac{n}{5})+n$ with the base case $T(x)=1$ when $x\leq5$. I want to solve it and find it's $\theta$. I think i have solved it correctly but I can't get the theta because of this term $\frac{5}{5^{log_{4}n}}$ . Any help?
$T(n)=4(4T(\fra... | An alternative approach is to prove that $T(n)\leqslant5n$ for every $n$. This holds for every $n\leqslant5$ and, if $T(n/5)\leqslant5(n/5)=n$, then $T(n)\leqslant4n+n=5n$. By induction, the claim holds.
On the other hand, $T(n)\geqslant n$ for every $n\gt5$, hence $T(n)=\Theta(n)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/284848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to find finite trigonometric products I wonder how to prove ?
$$\prod_{k=1}^{n}\left(1+2\cos\frac{2\pi 3^k}{3^n+1} \right)=1$$
give me a tip
| Let $S_n = \sum_{k=0}^n 3^k = \frac{3^{n+1}-1}{2}$. Then
$$3^{n}- S_{n-1} = 3^{n} - \frac{3^{n}-1}{2} = \frac{3^{n}+1}{2} = S_{n-1}+1.
$$
Now by induction we have the following product identity for $n \geq 0$:
$$
\begin{eqnarray}
\prod_{k=0}^{n}\left(z^{3^k}+1+z^{-3^k}\right) &=& \left(z^{3^{n}}+1+z^{-3^{n}}\right)\pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/284971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Find infinitely many pairs of integers $a$ and $b$ with $1 < a < b$, so that $ab$ exactly divides $a^2 +b^2 −1$. So I came up with $b= a+1$ $\Rightarrow$ $ab=a(a+1) = a^2 + a$
So that:
$a^2+b^2 -1$ = $a^2 + (a+1)^2 -1$ = $2a^2 + 2a$ = $2(a^2 + a)$ $\Rightarrow$
$(a,b) = (a,a+1)$ are solutions.
My motivation is for th... | $(3,8)$ is a possible solution.
This gives us 24 divides 72, and a value of 3 for (b).
Have you considered that if $ab$ divides $a^2+b^2-1$, then we $ab$ divides $a^2 + b^2 -1 + 2ab$?
This gives us $ab$ divides $(a+b+1)(a+b-1)$.
Subsequently, the question might become easier to work with.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/288672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Show that $(a+b+c)^3 = a^3 + b^3 + c^3+ (a+b+c)(ab+ac+bc)$ As stated in the title, I'm supposed to show that $(a+b+c)^3 = a^3 + b^3 + c^3 + (a+b+c)(ab+ac+bc)$.
My reasoning:
$$(a + b + c)^3 = [(a + b) + c]^3 = (a + b)^3 + 3(a + b)^2c + 3(a + b)c^2 + c^3$$
$$(a + b + c)^3 = (a^3 + 3a^2b + 3ab^2 + b^3) + 3(a^2 + 2ab + ... | In general, $$a^n+b^n+c^n = \sum_{i+2j+3k=n} \frac{n}{i+j+k}\binom {i+j+k}{i,j,k} s_1^i(-s_2)^js_3^k$$
where $s_1=a+b+c$, $s_2=ab+ac+bc$ and $s_3=abc$ are the elementary symmetric polynomials.
In the case that $n=3$, the triples possible are $(i,j,k)=(3,0,0),(1,1,0),$ and $(0,0,1)$ yielding the formula:
$$a^3+b^3+c^3 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/288965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Another two hard integrals Evaluate :
$$\begin{align}
& \int_{0}^{\frac{\pi }{2}}{\frac{{{\ln }^{2}}\left( 2\cos x \right)}{{{\ln }^{2}}\left( 2\cos x \right)+{{x}^{2}}}}\text{d}x \\
& \int_{0}^{1}{\frac{\arctan \left( {{x}^{3+\sqrt{8}}} \right)}{1+{{x}^{2}}}}\text{d}x \\
\end{align}$$
| For the second integral, consider the more general form
$$\int_0^1 dx \: \frac{\arctan{x^{\alpha}}}{1+x^2}$$
(I do not understand what is special about $3+\sqrt{8}$.)
Taylor expand the denominator and get
$$\begin{align} &=\int_0^1 dx \: \arctan{x^{\alpha}} \sum_{k=0}^{\infty} (-1)^k x^{2 k} \\ &= \sum_{k=0}^{\infty} (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/289421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Factorization problem Find $m + n$ if $m$ and $n$ are natural numbers such that: $$\frac {m+n} {m^2+mn+n^2} = \frac {4} {49}\;.$$
My reasoning:
Say: $$m+n = 4k$$ $$m^2+mn+n^2 = 49k$$
It follows:$$(m+n)^2 = (4k)^2 = 16k^2 \Rightarrow m^2+mn+n^2 + mn = 16k^2 \Rightarrow mn = 16k^2 - 49k$$
Since: $$mn\gt0 \Rightarrow 16k^... | Observe that $k$ must be a non-zero integer.
We know that $m, n$ are the roots of the quadratic equation
$$X^2 - 4kX + (16k^2 - 49k)$$
The roots, from the quadratic equation, are
$$ \frac { 4k \pm \sqrt{(4k)^2 - 4(16k^2 - 49k) }} {2} = 2k \pm \sqrt{ 49k - 12k^2}$$
The expression in the square root must be a perfect s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/290166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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$\gcd(m,n) = 1$ and $\gcd (mn,a)=1$ implies $a \cong 1 \pmod{ mn}$ I have $m$ and $n$ which are relatively prime to one another and $a$ is relatively prime to $mn$
and after alot of tinkering with my problem i came to this equality:
$a \cong 1 \pmod m \cong 1 \pmod n$
why is it safe to say that $a \cong 1 \pmod {mn}$?.... | It looks as if you are asking the following. Suppose that $m$ and $n$ are relatively prime. Show that if $a\equiv 1\pmod{m}$ and $a\equiv 1\pmod{n}$, then $a\equiv 1\pmod{mn}$.
So we know that $m$ divides $a-1$, and that $n$ divides $a-1$. We want to show that $mn$ divides $a-1$.
Let $a-1=mk$. Since $n$ divides $a-1$, ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Let $s_n$ be a sequence defined as given below for $n \geq 1$. Then find out $\lim\limits_{n \to
\infty} s_n$.
\begin{align}
s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx
\end{align}
I have written a solution of my own, but I would like to ... | Notice
(1) $\frac{s_n}{n} + \frac{s_{n+1}}{n+1} = \int_0^1 x^{n-1} dx = \frac{1}{n} \implies s_n + s_{n+1} = 1 + \frac{s_{n+1}}{n+1}$.
(2) $s_n = n\int_0^1 \frac{x^{n-1}}{1+x} dx < n\int_0^1 x^{n-1} dx = 1$
(3) $s_{n+1} - s_n = \int_0^1 \frac{d (x^{n+1}-x^n)}{1+x} = \int_0^1 x^n \frac{1-x}{(1+x)^2} dx > 0$
(2+3) $\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/292251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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half-angle trig identity clarification I am working on the following trig half angle problem. I seem to be on the the right track except that my book answer shows -1/2 and I didn't get that in my answer. Where did I go wrong?
$$\sin{15^{\circ}} = $$
$$\sin \frac { 30^{\circ} }{ 2 } = \pm \sqrt { \frac { 1 - \cos{30^{\c... | $$\sqrt { \dfrac { 1 - \dfrac {\sqrt 3 }{ 2 } }{ 2 } \times \dfrac22} = \sqrt{\dfrac{2-\sqrt3}{\color{red}4}} = \dfrac{\sqrt{2-\sqrt3}}{\color{red}2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/295038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solving Recurrence $T(n) = T(n − 3) + 1/2$; I have to solve the following recurrence.
$$\begin{gather}
T(n) = T(n − 3) + 1/2\\
T(0) = T(1) = T(2) = 1.
\end{gather}$$
I tried solving it using the forward iteration.
$$\begin{align}
T(3) &= 1 + 1/2\\
T(4) &= 1 + 1/2\\
T(5) &= 1 + 1/2\\
T(6) &= 1 + 1/2 + 1/2 = 2\\
T(7) ... | The generating function is $$g(x)=\sum_{n\ge 0}T(n)x^n = \frac{2-x^3}{2(1+x+x^2)(1-x)^2}$$, which has the partial fraction representation
$$g = \frac{2}{3(1-x)} + \frac{1}{6(1-x)^2}+\frac{x+1}{6(1+x+x^2)}$$. The first
term contributes $$\frac{2}{3}(1+x+x^2+x^3+\ldots)$$, equivalent to $T(n)=2/3$ the second term contrib... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/300934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Solve $\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$ for $x$ Is there any smart way to solve the equation: $$\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$$
Use Maple I can find $x \in \{1;ab+bc+ca\}$
| I have a partial solution, as follows:
Note that $\frac{(b-c)(1+a^2)}{x+a^2}=\frac{(b-c)\left((x+a^2)+(1-x)\right)}{x+a^2}=(b-c)+\frac{{(b-c)}(1-x)}{x+a^2}$. Likewise, $\frac{(c-a)(1+b^2)}{x+b^2}=(c-a)+\frac{(c-a)(1-x)}{x+b^2}$ and $\frac{(a-b)(1+c^2)}{x+c^2}=(c-a)+\frac{(a-b)(1-x)}{x+c^2}$.
Now, $\frac{(b-c)(1+a^2)}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/306154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
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Show that $(m^2 - n^2, 2mn, m^2 + n^2)$ is a primitive Pythagorean triplet Show that $(m^2 - n^2, 2mn, m^2 + n^2)$ is a primitive Pythagorean triplet
First, I showed that $(m^2 - n^2, 2mn, m^2 + n^2)$ is in fact a Pythagorean triplet.
$$\begin{align*} (m^2 - n^2)^2 + (2mn)^2 &= (m^2 + n^2)^2 \\
&= m^4 -2m^2n^2 + n^4 + ... | To show $(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2$ is equivalent to showing $(m^2 - n^2)^2 + (2mn)^2 - (m^2 + n^2)^2 = 0$ so \begin{align*} && (m^2 - n^2)^2 + (2mn)^2 - (m^2 + n^2)^2 \\ &=& m^4 - 2m^2n^2 + n^4 + 4m^2n^2 - m^4 - 2m^2n^2 - n^4 \\ &=& m^4 + n^4 - m^4 - n^4 \\ &=& 0\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}.$ Find
$$\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}\;.$$
| Here is another sketch of proof.
Let
$$J_n = \{(j, k) : 0 \leq j, k < n \text{ and } (j, k) \neq (0, 0) \}.$$
Then for each $(j, k) \in J_n$ and $(x_0, y_0) = (j/n, k/n)$, we have
$$ \frac{x_0 + y_0}{(x_0+\frac{1}{n})^{3} + (y_0 + \frac{1}{n})^3} \leq \frac{x+y}{x^3 + y^3} \leq \frac{x_0 + y_0 + \frac{2}{n}}{x_0^3 + y_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/306831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Computing $\lim_{n\to\infty}n\sum_{k=1}^n\left( \frac{1}{(2k-1)^2} - \frac{3}{4k^2}\right)$ What ways would you propose for the limit below?
$$\lim_{n\to\infty}n\sum_{k=1}^n\left( \frac{1}{(2k-1)^2} - \frac{3}{4k^2}\right)$$
Thanks in advance for your suggestions, hints!
Sis.
| OK, it turns out that
$$\sum_{k=1}^n\left( \frac{1}{(2k-1)^2} - \frac{3}{4k^2}\right) = \sum_{k=1}^{n-1} \frac{1}{(k+n)^2}$$
This may be shown by observing that
$$\sum_{k=1}^n \frac{1}{(2k-1)^2} = \sum_{k=1}^{2 n-1} \frac{1}{k^2} - \frac{1}{2^2} \sum_{k=1}^n \frac{1}{k^2}$$
The desired limit may then be rewritten as
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/307117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 5,
"answer_id": 2
} |
Really Stuck on Partial derivatives question Ok so im really stuck on a question. It goes:
Consider $$u(x,y) = xy \frac {x^2-y^2}{x^2+y^2} $$ for $(x,y)$ $ \neq $ $(0,0)$ and $u(0,0) = 0$.
calculate $\frac{\partial u} {\partial x} (x,y)$ and $\frac{\partial u} {\partial y} (x,y)$ for all $ (x,y) \in \Bbb R^2. $ show ... | We are given:
$$u(x, y)=\begin{cases}
xy \frac {x^2-y^2}{x^2+y^2}, ~(x, y) \ne (0,0)\\\\
~~~0, ~~~~~~~~~~~~~~~(x, y) = (0,0)\;.
\end{cases}$$
I am going to multiply out the numerator for ease in calculations, so we have:
$$\tag 1 u(x, y)=\begin{cases}
\frac {x^3y - xy^3}{x^2+y^2}, ~(x, y) \ne (0,0)\\\\
~~~0, ~~~~~~~~~~... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/307710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $P(A \cup B \cup C) = 1$, $P(B) = 2P(A) $, $P(C) = 3P(A) $, $P(A \cap B) = P(A \cap C) = P(B \cap C) $, then $P(A) \le \frac14$ We have ($P$ is probability):
$P(A \cup B \cup C) = 1$ ; $P(B) = 2P(A) $ ; $P(C) = 3P(A) $ and $P(A \cap B) = P(A \cap C) = P(B \cap C) $. Prove that $P(A) \le \frac{1}{4} $.
Well, I tried ... | This solution is quite possibly messier than necessary, but it’s what first occurred to me. For convenience let $x=P\big((A\cap B)\setminus C\big)$ and $y=P(A\cap B\cap C)$. Let $a=P(A)-2x-y$, $b=P(B)-2x-y$, and $c=P(C)-2x-y$; these are the probabilities of $A\setminus(B\cup C)$, $B\setminus(A\cup C)$, and $C\setminus(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/308386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$T:P_3 → P_3$ be the linear transformation such that... Let $T:P_3 \to P_3$ be the linear transformation such that $T(2 x^2)= -2 x^2 - 4 x$, $T(-0.5 x - 5)= 2 x^2 + 4 x + 3$, and $T(2 x^2 - 1) = 4 x - 4.$
Find $T(1)$, $T(x)$, $T(x^2)$, and $T(a x^2 + b x + c)$, where $a$, $b$, and $c$ are arbitrary real numbers.
| hint:T is linear transformation $T(a+bx+cx^2)=aT(1)+bT(x) +cT(x^2) $ $$T(2 x^2)= -2 x^2 - 4 x \to T( x^2)= - x^2 - 2 x$$$$T(-0.5 x - 5)= 2 x^2 + 4 x + 3\to-0.5T( x) - 5T(1)= 2 x^2 + 4 x + 3\to 16x^2+72x-46$$$$T(2 x^2 - 1) = 4 x - 4.\to T( x^2) - \frac12T(1) = 2 x - 2\to T(1)=-2 x^2 - 8x +4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Probability of choosing two equal bits from three random bits Given three random bits, pick two (without replacement). What is the probability that the two you pick are equal?
I would like to know if the following analysis is correct and/or if there is a better way to think about it.
$$\Pr[\text{choose two equal bits}]... | Whatever the first bit picked, the probability the second bit matches it is $1/2$.
Remark: We are assuming what was not explicitly stated, that $0$'s and $1$'s are equally likely. One can very well have "random" bits where the probability of $0$ is not the same as the probability of $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/310508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Help with writing the following as a partial fraction $\frac{4x+5}{x^3+1}$. I need help with writing the following as a partial fraction:
$$\frac{4x+5}{x^3+1}$$
My attempts so far are to factor $x^3$ into $(x+1)$ and $(x^2-x+1)$
This gives me: $A(x^2-x+1)+B(x+1)$.
But I have problems with solving the equation system t... | You need to use one less exponent per factor in the numerator after your factorization.
This leads to:
$$\frac{Ax+B}{x^2-x+1} + \frac{C}{x+1} = \frac{4x+5}{x^3+1}$$
This gives us:
$$Ax^2 + Ax + Bx + B + Cx^2 - Cx + C = 4x + 5$$
This leads to:
$A + C = 0$
$A + B - C = 4$
$B + C = 5$
yielding:
$$A = -\frac{1}{3}, B = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/313151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $ \left(1+\frac a b \right) \left(1+\frac b c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$.
Given $a,b,c>0$, prove that $\displaystyle \left(1+\frac a b \right) \left(1+\frac b c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$.
I e... | We can begin by clearing denominators as follows
$$a^2c+a^2b+b^2a+b^2c+c^2a+c^2b\geq 2a^{5/3}b^{2/3}c^{2/3}+2a^{2/3}b^{5/3}c^{2/3}+2a^{2/3}b^{2/3}c^{5/3}$$
Now by the Arithmetic Mean - Geometric Mean Inequality,
$$\frac{2a^2c+2a^2b+b^2a+c^2a}{6} \geq a^{5/3}b^{2/3}c^{2/3}$$
That is,
$$\frac{2}{3}a^2c+\frac{2}{3}a^2b+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/313605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Prove $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots+\frac{1}{\sqrt{n}} > 2\:(\sqrt{n+1} − 1)$ Basically, I'm trying to prove (by induction) that:
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots+\frac{1}{\sqrt{n}} > 2\:(\sqrt{n+1} − 1)$$
I know to begin, we should use a base case. In this case, I'll use $1$. So we ha... | Mean Value Theorem can also be used,
Let $\displaystyle f(x)=\sqrt{x}$
$\displaystyle f'(x)=\frac{1}{2}\frac{1}{\sqrt{x}}$
Using mean value theorem we have:
$\displaystyle \frac{f(n+1)-f(n)}{(n+1)-n}=f'(c)$ for some $c\in(n,n+1)$
$\displaystyle \Rightarrow \frac{\sqrt{n+1}-\sqrt{n}}{1}=\frac{1}{2}\frac{1}{\sqrt{c}}$..... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/313834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
$\sum \limits_{k=1}^{\infty} \frac{6^k}{\left(3^{k+1}-2^{k+1}\right)\left(3^k-2^k\right)} $ as a rational number. $$\sum \limits_{k=1}^{\infty} \frac{6^k}{\left(3^{k+1}-2^{k+1}\right)\left(3^k-2^k\right)} $$
I know from the ratio test it convergest, and I graph it on wolfram alpha and I suspect the sum is 2; however, I... | That denominator should suggest the possibility of splitting the general term into partial fractions and getting a telescoping series of the form
$$\sum_{k\ge 1}\left(\frac{A_k}{3^k-2^k}-\frac{A_{k+1}}{3^{k+1}-2^{k+1}}\right)\;,$$
where $A_k$ very likely depends on $k$. Note that if this works, the sum of the series wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/319400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Show that $x$, $y$, and $z$ are not distinct if $x^2(z-y) + y^2(x-z) + z^2(y-x) = 0$. Suppose that $x^2(z-y) + y^2(x-z) + z^2(y-x) = 0$.
How can I show that $x$, $y$, and $z$ are not all distinct, that is, either $x=y$, $y=z$, or $x=z$?
| $x^2(z-y)+y^2(x-z)+z^2(y-x)=x^2(z-y)+x(y^2-z^2)+z^2y-y^2z=(z-y)(x^2-x(y+z)+zy)=(z-y)(x-y)(x-z)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Homogeneous equation I am trying to solve the following homogeneous equation:
Thanks for your tips
$xy^3y′=2(y^4+x^4)$
I think this isHomogeneous of order4
=> $xy^3dy/dx=2(y^4+x^4/1)$
=> $xy^3dy=2(x^4+y^4)dx$
=> $xy^3dy-2(x^4+y^4)dx=0$
I do not know how to continued
| Make the substitution $v=y^4$. Then by the chain rule we have $v'=4y^3 y'$. Now your DE turns into:
$$x \frac{v'}{4}=2(v+x^4)$$
Then can be simplified to:
$$v'-8\frac{v}{x}=8x^3$$
We first solve the homogeneous part:
$$v_h' -8\frac{v_h}{x}=0$$
This leads to $v_h=c\cdot x^8$ so that $v_h'=c\cdot 8x^7$. This is our homog... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/320328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do we integrate, $\int \frac{1}{x+\frac{1}{x^2}}dx$? How do we integrate the following integral?
$$\int \frac{1}{x+\large\frac{1}{x^2}}\,dx\quad\text{where}\;\;x\ne-1$$
| The integral is equivalent to
$$\int dx \frac{x^2}{1+x^3} = \frac{1}{3} \int \frac{d(x^3)}{1+x^3} = \frac{1}{3} \log{(1+x^3)} + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to choose the starting row when computing the reduced row echelon form? I'm having hell of a time going around solving matrices to reduced row echelon form. My main issue is which row to start simplifying values and based on what? I have this example
so again, the questions are:
1.Which row to start simplifying ... | Where you start is not really a problem.
My tip:
*
*Always first make sure you make the first column: 1,0,0
*Then proceed making the second one: 0,1,0
*And lastly, 0,0,1
Step one:
$$\begin{pmatrix} 1&2&3&9 \\ 2&-1&1&8 \\ 3&0&-1&3\end{pmatrix}$$
row 3 - 3 times row 1
$$\begin{pmatrix} 1&2&3&9 \\ 2&-1&1&8 \\ 0&-6&-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/324798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
Intersection points of a Triangle and a Circle How can I find all intersection points of the following circle and triangle?
Triangle
$$A:=\begin{pmatrix}22\\-1.5\\1 \end{pmatrix} B:=\begin{pmatrix}27\\-2.25\\4 \end{pmatrix} C:=\begin{pmatrix}25.2\\-2\\4.7 \end{pmatrix}$$
Circle
$$\frac{9}{16}=(x-25)^2 + (y+2)^2 + (z-3)... | The side $AB$ of the triangle has equation $P(t) = (1-t)A + tB$ for $0 \le t \le 1$. The $0 \le t \le 1$ part is important. If $t$ lies outside $[0,1]$, the point $P(t)$ will lie on the infinite line through $A$ and $B$, but not on the edge $AB$ of the triangle. Substitute $(1-t)A + tB$ into the circle equation, as ot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/325411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $x$ such that $\sum_{k=1}^{2014} k^k \equiv x \pmod {10}$ Find $x$ such that $$\sum_{k=1}^{2014} k^k \equiv x \pmod {10}$$
I knew the answer was $3$.
| We are going to compute the sum mod $2$ and mod $5$. The Chinese Remainder Theorem then gives us the result mod $10$.
Mod $2$, obviously $$k^k \equiv \begin{cases}0 & \text{if }k\text{ even,}\\1 & \text{if }k\text{ odd,}\end{cases}$$
so
$$\sum_{k = 0}^{2014} \equiv \frac{2014}{2} = 1007 \equiv 1 \mod 2.$$
By Fermat, $k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/325529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
A probability question that involves $5$ dice For five dice that are thrown, I am struggling to find the probability of one number showing exactly three times and a second number showing twice.
For the one number showing exactly three times, the probability is:
$$
{5 \choose 3} \times \left ( \frac{1}{6} \right )^{3} \... | First, I assume they wil all come out in neat order, first three in a row, then two in a row of a different number. The probability of that happening is
$$
\frac{1}{6^2}\cdot \frac{5}{6}\cdot\frac{1}{6} = \frac{5}{6^4}
$$
The first die can be anything, but the next two have to be equal to that, so the $\frac{1}{6^2}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/325703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
$\lim_{n\to\infty}(\sqrt{n^2+n}-\sqrt{n^2+1})$ How to evaluate $$\lim_{n\to\infty}(\sqrt{n^2+n}-\sqrt{n^2+1})$$
I'm completely stuck into it.
| A useful general approach to limits is, in your scratch work, to take every complicated term and replace it with a similar approximate term.
As $n$ grows large, $\sqrt{n^2 + n}$ looks like $\sqrt{n^2} = n$. More precisely,
$$ \sqrt{n^2 + n} = n + o(n) $$
where I've used little-o notation. In terms of limits, this means... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/326253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove $a\sqrt[3]{a+b}+b\sqrt[3]{b+c}+c\sqrt[3]{c+a} \ge 3 \sqrt[3]2$ Prove $a\sqrt[3]{a+b}+b\sqrt[3]{b+c}+c\sqrt[3]{c+a} \ge 3 \sqrt[3]2$ with $a + b+c=3 \land a,b,c\in \mathbb{R^+}$
I tried power mean inequalities but I still can't prove it.
| Here is my proof
by AM-GM inequality,we have
$$ a\sqrt[3]{a+b}=\frac{3\sqrt[3]{2}a(a+b)}{3\sqrt[3]{2(a+b)(a+b)}}\geq 3\sqrt[3]{2}\cdot \frac{a(a+b)}{2+2a+2b} $$
Thus,it's suffice to prove that
$$ \frac{a(a+b)}{a+b+1}+\frac{b(b+c)}{b+c+1}+\frac{c(c+a)}{c+a+1}\geq 2 $$
Or
$$ \frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/326500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Limit of definite sum equals $\ln(2)$ I have to show the following equality:
$$\lim_{n\to\infty}\sum_{i=\frac{n}{2}}^{n}\frac{1}{i}=\log(2)$$
I've been playing with it for almost an hour, mainly with the taylor expansion of $\ln(2)$. It looks very similar to what I need, but it has an alternating sign which sits in ... | Truncate the Maclaurin series for $\log(1+x)$ at the $2m$-th term, and evaluate at $x=1$. Take for example $m=10$. We get
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots+\frac{1}{19}-\frac{1}{20}.$$
Add $2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{20}\right)$, and subtract the sa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/326545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
How to calculate the asymptotic expansion of $\sum \sqrt{k}$? Denote $u_n:=\sum_{k=1}^n \sqrt{k}$. We can easily see that
$$ k^{1/2} = \frac{2}{3} (k^{3/2} - (k-1)^{3/2}) + O(k^{-1/2}),$$
hence $\sum_1^n \sqrt{k} = \frac{2}{3}n^{3/2} + O(n^{1/2})$, because $\sum_1^n O(k^{-1/2}) =O(n^{1/2})$.
With some more calculations... | Let us substitute into the sum
$$\sqrt k=\frac{1}{\sqrt \pi }\int_0^{\infty}\frac{k e^{-kx}dx}{\sqrt x}. $$
Exchanging the order of summation and integration and summing the derivative of geometric series, we get
\begin{align*}
\mathcal S_N:=
\sum_{k=1}^{N}\sqrt k&=\frac{1}{\sqrt \pi }\int_0^{\infty}\frac{\left(e^x-e^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/326617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Need help proving this integration If $a>b>0$, prove that :
$$\int_0^{2\pi} \frac{\sin^2\theta}{a+b\cos\theta}\ d\theta = \frac{2\pi}{b^2} \left(a-\sqrt{a^2-b^2} \right) $$
| I'll do this one $$\int_{0}^{2\pi}\frac{cos(2\theta)}{a+bcos(\theta)}d\theta$$if we know how to do tis one you can replace $sin^{2}(\theta)$ by $\frac{1}{2}(1-cos(2\theta))$ and do the same thing. if we ae on the unit circle we know that $cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$ so by letting $z=e^{i\theta}$ we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/326714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Math question functions help me? I have to find find $f(x,y)$ that satisfies
\begin{align}
f(x+y,x-y) &= xy + y^2 \\
f(x+y, \frac{y}x ) &= x^2 - y^2
\end{align}
So I first though about replacing $x+y=X$ and $x-y=Y$ in the first one but then what?
| Hint:
If $x+y=X$ and $x-y=Y$ then
\begin{align*}
x&=X-y \implies Y= X-y-y \implies 2y=X-Y \dots
\end{align*}
And be careful with the second for $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/328481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to create a generating function / closed form from this recurrence? Let $f_n$ = $f_{n-1} + n + 6$ where $f_0 = 0$.
I know $f_n = \frac{n^2+13n}{2}$ but I want to pretend I don't know this. How do I correctly turn this into a generating function / derive the closed form?
| In two answers, it is derived that the generating function is
$$
\begin{align}
\frac{7x-6x^2}{(1-x)^3}
&=x\left(\frac1{(1-x)^3}+\frac6{(1-x)^2}\right)\\
&=\sum_{k=0}^\infty(-1)^k\binom{-3}{k}x^{k+1}+6\sum_{k=0}^\infty(-1)^k\binom{-2}{k}x^{k+1}\\
&=\sum_{k=1}^\infty(-1)^{k-1}\left(\binom{-3}{k-1}+6\binom{-2}{k-1}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/329424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Integral of irrational function $$
\int \frac{\sqrt{\frac{x+1}{x-2}}}{x-2}dx
$$
I tried:
$$
t =x-2
$$
$$
dt = dx
$$
but it didn't work.
Do you have any other ideas?
| Let $y=x-2$ and integrate by parts to get
$$\int dx \: \frac{\sqrt{\frac{x+1}{x-2}}}{x-2} = -2 (x-2)^{-1/2} (x+1)^{1/2} + \int \frac{dy}{\sqrt{y (y+3)}}$$
In the second integral, complete the square in the denominator to get
$$\int \frac{dy}{\sqrt{y (y+3)}} = \int \frac{dy}{\sqrt{(y+3/2)^2-9/4}}$$
This integral may be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/329513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to simplify a square root How can the following:
$$
\sqrt{27-10\sqrt{2}}
$$
Be simplified to:
$$
5 - \sqrt{2}
$$
Thanks
| Set the nested radical as the difference of two square roots so that $$\sqrt{27-10\sqrt{2}}=(\sqrt{a}-\sqrt{b})$$ Then square both sides so that $$27-10\sqrt{2}=a-2\sqrt{a}\sqrt{b}+b$$ Set (1) $$a+b=27$$ and set (2) $$-2\sqrt{a}\sqrt{b}=-10\sqrt{2}$$ Square both sides of (2) to get $$4ab= 200$$ and solve for $b$ to g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/329838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 4
} |
How to find $f:\mathbb{Q}^+\to \mathbb{Q}^+$ if $f(x)+f\left(\frac1x\right)=1$ and $f(2x)=2f\bigl(f(x)\bigr)$
Let $f:\mathbb{Q}^+\to \mathbb{Q}^+$ be a function such that
$$f(x)+f\left(\frac1x\right)=1$$
and
$$f(2x)=f\bigl(f(x)\bigr)$$
for all $x\in \mathbb{Q}^+$. Prove that
$$f(x)=\frac{x}{x+1}$$
for all $x\in \mathb... | Some ideas:
$$\text{I}\;\;\;\;x=1\Longrightarrow f(1)+f\left(\frac{1}{1}\right)=2f(1)=1\Longrightarrow \color{red}{f(1)=\frac{1}{2}}$$
$$\text{II}\;\;\;\;\;\;\;\;f(2)=2f(f(1))=2f\left(\frac{1}{2}\right)$$
But we also know that
$$f(2)+f\left(\frac{1}{2}\right) =1$$
so from II we get
$$3f\left(\frac{1}{2}\right)=1\Longri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/329894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How do I transform the left side into the right side of this equation? How does one transform the left side into the right side?
$$
(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2
$$
| Expand the left hand side, you get
$$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2$$
Add and substract $2abcd$
$$a^2c^2+a^2d^2+b^2c^2+b^2d^2=(a^2c^2-2abcd+b^2d^2)+(a^2d^2+2abcd+b^2c^2)$$
Complete the square, you can get
$$(a^2c^2-2abcd+b^2d^2)+(a^2d^2+2abcd+b^2c^2)=(ac-bd)^2+(ad+bc)^2$$
Therefore,
$$(a^2+b^2)(c^2+d^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/330863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 5
} |
Prove inequality: $28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4$ Prove: $28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4$ with $a, b, c \ge0$
I can do this by: $EAT^2$ (expand all of the thing)
*
*$(x+y+z)^4={x}^{4}+{y}^{4}+{z}^{4}+4\,{x}^{3}y+4\,{x}^{3}z+6\,{x}^{2}{y}^{2}+6\,{
x}^{2}{z}^{2... | A nice way of tackling the calculations might be as follows:$$~$$
Let $x=b+c-a,y=c+a-b,z=a+b-c.$ Then the original inequality is just equivalent with
$$\frac74\Bigl((x+y)^4+(y+z)^4+(z+x)^4\Bigr)\geq x^4+y^4+z^4+(x+y+z)^4.$$
Now we can use the identity
$$\sum_{cyc}(x+y)^4=x^4+y^4+z^4+(x+y+z)^4-12xyz(x+y+z),$$
So that it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/331954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find lim sup $x_n$ Let $x_n = n(\sqrt{n^2+1} - n)\sin\dfrac{n\pi}8$ , $n\in\Bbb{N}$
Find $\limsup x_n$.
Hint: lim sup $x_n = \sup C(x_n)$.
How to make it into a fraction to find the cluster point of $x_n$?
| Expand: $n · \left( \sqrt{n²+1} - n \right) = n · \tfrac{1}{\sqrt{n²+1} + n}$ for all $n ∈ ℕ$.
Since $\tfrac{\sqrt{n²+1} + n}{n} = \sqrt{1+\tfrac{1}{n²}} + 1 \overset{n → ∞}{\longrightarrow} 2$, you have $n · \left( \sqrt{n²+1} - n \right) \overset{n → ∞}{\longrightarrow} \tfrac{1}{2}$.
Now $x_n = n · \left( \sqrt{n²+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/332803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Need to prove the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges I need to prove that the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges. I do not have to find the limit. I have tried to prove it by proving that the sequence is monotone and bounded, but I am hav... | Notice that $ 2k^2 \geq k(k+1) \implies \frac{1}{k^2} \leq \frac{2}{k(k+1)}$.
$$ \sum_{k=1}^{\infty} \frac{2}{k(k+1)} = \frac{2}{1 \times 2} + \frac{2}{2 \times 3} + \frac{2}{3 \times 4} + \ldots $$
$$ \sum_{k=1}^{\infty} \frac{2}{k(k+1)} = 2\Big(\, \Big(1 - \frac{1}{2}\Big) + \Big(\frac{1}{2} - \frac{1}{3} \Big) + \Bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/333417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 8,
"answer_id": 5
} |
Evaluating $\int \frac{1}{{x^4+1}} dx$ I am trying to evaluate the integral
$$\int \frac{1}{1+x^4} \mathrm dx.$$
The integrand $\frac{1}{1+x^4}$ is a rational function (quotient of two polynomials), so I could solve the integral if I can find the partial fraction of $\frac{1}{1+x^4}$. But I failed to factorize $1+x^4$.... | Without using fractional decomposition:
$$\begin{align}\int\dfrac{1}{x^4+1}~dx&=\dfrac{1}{2}\int\dfrac{2}{x^4+1}~dx
\\&=\dfrac{1}{2}\int\dfrac{(x^2+1)-(x^2-1)}{x^4+1}~dx
\\&=\dfrac{1}{2}\int\dfrac{x^2+1}{x^4+1}~dx-\dfrac{1}{2}\int\dfrac{x^2-1}{x^4+1}~dx
\\&=\dfrac{1}{2}\int\dfrac{1+\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}}~d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/333611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Need to prove the sequence $a_n=(1+\frac{1}{n})^n$ converges by proving it is a Cauchy sequence
I am trying to prove that the sequence $a_n=(1+\frac{1}{n})^n$ converges by proving that it is a Cauchy sequence.
I don't get very far, see: for $\epsilon>0$ there must exist $N$ such that $|a_m-a_n|<\epsilon$, for $ m,n>... | We have the following inequalities:
$\left(1+\dfrac{1}{n}\right)^n = 1 + 1 + \dfrac{1}{2!}\left(1-\dfrac{1}{n}\right)+\dfrac{1}{3!}\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right) + \ldots \leq 2 + \dfrac{1}{2} + \dfrac{1}{2^2} + \dots =3$
Similarly,
$
\begin{align*}
\left(1-\dfrac{1}{n^2}\right)^n &= 1 - {n \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/334382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Variation of Pythagorean triplets: $x^2+y^2 = z^3$ I need to prove that the equation $x^2 + y^2 = z^3$ has infinitely many solutions for positive $x, y$ and $z$.
I got to as far as $4^3 = 8^2$ but that seems to be of no help.
Can some one help me with it?
| the equation:
$X^2+Y^2=Z^3$
Has the solutions:
$X=2k^6+8tk^5+2(7t^2+8qt-9q^2)k^4+16(t^3+2qt^2-tq^2-2q^3)k^3+$
$+2(7t^4+12qt^3+6q^2t^2-28tq^3-9q^4)k^2+8(t^5+2qt^4-2q^3t^2-5tq^4)k+$
$+2(q^6-4tq^5-5q^4t^2-5q^2t^4+4qt^5+t^6)$
..................................................................................................... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/334839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 8,
"answer_id": 5
} |
halfspaces question How do I find the supporting halfspace inequality to epigraph of
$$f(x) = \frac{x^2}{|x|+1}$$
at point $(1,0.5)$
| For $x>0$, we have
$$
f(x)=\frac{x^2}{x+1}\quad\Rightarrow\quad f'(x)=\frac{x^2+2x}{(x+1)^2}.
$$
Hence $f'(1)=\frac{3}{4}$ and an equation of the tangent to the graph of $f$ at $(1,f(1))$ is
$$y=f'(1)(x-1)+f(1)=\frac{3}{4}(x-1)+\frac{1}{2}=\frac{3}{4}x-\frac{1}{4}.$$
An inequality defining the halfspace above this ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/335490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the property that allows $5^{2x+1} = 5^2$ to become $2x+1 = 2$? What is the property that allows $5^{2x+1} = 5^2$ to become $2x+1 = 2$? We just covered this in class, but the teacher didn't explain why we're allowed to do it.
| $5^{(2x+1)} = 5^2$
Multiplying by $1/5^2$ om both sides we get,
$\frac{5^{(2x+1)}}{5^2} = 1$
$\Rightarrow 5^{(2x+1)-2} = 1$
Taking log to the base 5 on both sides we get $2x+1-2=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Prove $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$ Let $a,b,c$ are non-negative numbers, such that $a+b+c = 3$.
Prove that $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$
Here's my idea:
$\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(ab + bc + ca)$
$2(\sqrt{a} + \sqrt... | I will use the following lemma (the proof below):
$$2x \geq x^2(3-x^2)\ \ \ \ \text{ for any }\ x \geq 0. \tag{$\clubsuit$}$$
Start by multiplying our inequality by two
$$2\sqrt{a} +2\sqrt{b} + 2\sqrt{c} \geq 2ab +2bc +2ca, \tag{$\spadesuit$}$$
and observe that
$$2ab + 2bc + 2ca = a(b+c) + b(c+a) + c(b+c) = a(3-a) + b(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/336362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 2
} |
Calculate:$\lim_{x \rightarrow (-1)^{+}}\left(\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}} \right)$ How to calculate following with out using L'Hospital rule
$$\lim_{x \rightarrow (-1)^{+}}\left(\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}} \right)$$
| Let $\sqrt{\arccos(x)} = t$. We then have $x = \cos(t^2)$. Since $x \to (-1)^+$, we have $t^2 \to \pi^-$. Hence, we have
$$\lim_{x \to (-1)^+} \dfrac{\sqrt{\pi} - \sqrt{\arccos(x)}}{\sqrt{1+x}} = \overbrace{\lim_{t \to \sqrt{\pi}^-} \dfrac{\sqrt{\pi} - t}{\sqrt{1+\cos(t^2)}}}^{t = \sqrt{\arccos(x)}} = \underbrace{\lim_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/337603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Surface integral over ellipsoid I've problem with this surface integral:
$$
\iint\limits_S {\sqrt{ \left(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\right)}}{dS}
$$, where
$$
S = \{(x,y,z)\in\mathbb{R}^3: \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}= 1\}
$$
| Let the ellipsoid $S$ be given by
$${\bf x}(\theta,\phi)=(a\cos\theta\cos\phi,b\cos\theta\sin\phi,c\sin\theta)\ .$$
Then for all points $(x,y,z)\in S$ one has
$$Q^2:={x^2\over a^4}+{y^2\over b^4}+{z^2\over c^4}={1\over a^2b^2c^2}\left(\cos^2\theta(b^2c^2\cos^2\phi+a^2c^2\sin^2\phi)+a^2b^2\sin^2\theta\right)\ .$$
On the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/338155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
Find a closed form of the series $\sum_{n=0}^{\infty} n^2x^n$ The question I've been given is this:
Using both sides of this equation:
$$\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$$
Find an expression for $$\sum_{n=0}^{\infty} n^2x^n$$
Then use that to find an expression for
$$\sum_{n=0}^{\infty}\frac{n^2}{2^n}$$
This is a... | You've got $\sum_{n\geq 0} n(n-1)x^n$, modulo multiplication by $x^2$. Differentiate just once your initial power series and you'll be able to find $\sum_{n\geq 0} nx^n$. Then take the sum of $\sum_{n\geq 0} n(n-1)x^n$ and $\sum_{n\geq 0} nx^n$. What are the coefficients?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
} |
approximation of law sines from spherical case to planar case we know for plane triangle cosine rule is $\cos C=\frac{a^+b^2-c^2}{2ab}$ and on spherical triangle is $ \cos C=\frac{\cos c - \cos a \cos b} {\sin a\sin b}$ suppose $a,b,c<\epsilon$ which are sides of a spherical triangle, and $$|\frac{a^2 +b^2-c^2}{2ab}-... | Note that for $x$ close to $0$,
$$1-\frac{x^2}{2!} \le \cos x\le 1-\frac{x^2}{2!}+\frac{x^4}{4!}$$
and
$$x-\frac{x^3}{3!} \le \sin x\le x.$$
(We used the Maclaurin series expansion of $\cos x$ and $\sin x$.)
Using these facts on the small angles $a$, $b$, and $c$, we can estimate your difference.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/340786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Irreducible components of the variety $V(X^2+Y^2-1,X^2-Z^2-1)\subset \mathbb{C}^3.$ I want to find the irreducible components of the variety $V(X^2+Y^2-1, \ X^2-Z^2-1)\subset \mathbb{C}^3$ but I am completely stuck on how to do this. I have some useful results that can help me decompose $V(F)$ when $F$ is a single poly... | $\newcommand{\rad}{\text{rad}\hspace{1mm}}
$
The ideal $(x^2 + y^2 - 1,x^2 - z^2 - 1)$ is equal to the ideal $(y^2 + z^2 ,x^2 - z^2 - 1)$. This is because
\begin{eqnarray*} (y^2 + z^2) + (x^2 - z^2 - 1) &=& y^2 + x^2 - 1\\
(x^2 + y^2 - 1) - (x^2 - z^2 - 1) &=& y^2 + z^2. \end{eqnarray*}
Thus we get
\begin{eqnarray*} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/347325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 1
} |
Funny integral inequality Assume $f(x) \in C^1([0,1])$,and $\int_0^{\frac{1}{2}}f(x)\text{d}x=0$,show that:
$$\left(\int_0^1f(x)\text{d}x\right)^2 \leq \frac{1}{12}\int_0^1[f'(x)]^2\text{d}x$$
and how to find the smallest constant $C$ which satisfies
$$\left(\int_0^1f(x)\text{d}x\right)^2 \leq C\int_0^1[f'(x)]^2\text{... | solutin 2:
by Schwarz,we have
$$\int_{0}^{\frac{1}{2}}[f'(x)]^2dx\int_{0}^{\frac{1}{2}}x^2dx\ge\left(\int_{0}^{\frac{1}{2}}xf'(x)dx\right)^2=\left[\dfrac{1}{2}f(\dfrac{1}{2})-\int_{0}^{\frac{1}{2}}f(x)dx\right]^2$$
so
$$\int_{0}^{\frac{1}{2}}[f'(x)]^2dx\ge 24\left[\dfrac{1}{2}f(\dfrac{1}{2})-\int_{0}^{\frac{1}{2}}f(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/347385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Find the value of : $\lim_{x\to\infty}x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)=-1$ How can I show/explain the following limit?
$$\lim_{x\to\infty} \;x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)=-1$$
Some trivial transformation seems to be eluding me.
| The expression can be multiplied with its conjugate and then:
$$\begin{align}
\lim_{x\to\infty} x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)
&= \lim_{x\to\infty} x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)\left(\frac{\sqrt{x^2-1}+\sqrt{x^2+1}}{\sqrt{x^2-1}+\sqrt{x^2+1}}\right) \cr
&=\lim_{x\to\infty} x\left(\frac{x^2-1-x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/348071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Solve $\frac{1}{2x}+\frac{1}{2}\left(\frac{1}{2x}+\cdots\right)$ If
$$\displaystyle \frac{1}{2x}+\frac{1}{2}\left(\frac{1}{2x}+ \frac{1}{2}\left(\frac{1}{2x} +\cdots\right) \right) = y$$
then what is $x$?
I was thinking of expanding the brackets and trying to notice a pattern but as it effectively goes to infinity. I d... | Expand. The first term is $\frac{1}{2x}$.
The sum of the first two terms is $\frac{1}{2x}+\frac{1}{4x}$.
The sum of the first three terms is $\frac{1}{2x}+\frac{1}{4x}+\frac{1}{8x}$.
And so on.
The sum of the first $n$ terms is
$$\frac{1}{2x}\left(1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n-1}}\right).$$
As $n\to\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/349548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Can the matrix $A=\begin{bmatrix} 0 & 1\\ 3 & 3 \end{bmatrix}$ be diagonalized over $\mathbb{Z}_5$? Im stuck on finding eigenvalues that are in the field please help.
Given matrix:
$$
A= \left[\begin{matrix}
0 & 1\\
3 & 3
\end{matrix}\right]
$$
whose entries are from $\mathbb{Z}_5 = \{0, 1, 2, 3, 4\}$, find, if possi... | yes over $\Bbb Z_5$ because:
$\lambda^2 -3\lambda-3=o$ at Z_5 we will have $\Delta=9+12=4+2=6$ (9~4 and 12~2 at Z_5)
so $\Delta=1$
and so $\lambda_1=\frac{3+1}{2}=2$ and $\lambda_2=\frac{3-1}{2}=1$
about:
$\lambda_1$ we have :$ ( \left[\begin{matrix}
0 & 1\\
3 &3
\end{matrix}\right]-\left[\begin{matrix}
2 & \\
0 &2
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/350470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Separating $\frac{1}{1-x^2}$ into multiple terms I'm working through an example that contains the following steps:
$$\int\frac{1}{1-x^2}dx$$
$$=\frac{1}{2}\int\frac{1}{1+x} - \frac{1}{1-x}dx$$
$$\ldots$$
$$=\frac{1}{2}\ln{\frac{1+x}{1-x}}$$
I don't understand why the separation works. If I attempt to re-combine the ter... | The thing is $$\frac{1}{1-x}\color{red}{+}\frac 1 {1+x}=\frac{2}{1-x^2}$$
What you might have seen is
$$\frac{1}{x-1}\color{red}{-}\frac 1 {x+1}=\frac{2}{1-x^2}$$
Note the denominator is reversed in the sense $1-x=-(x-1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/350564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Reasoning about the gamma function using the digamma function I am working on evaluating the following equation:
$\log\Gamma(\frac{1}{2}x) - \log\Gamma(\frac{1}{3}x)$
If I'm understanding correctly, the above is an increasing function which can be demonstrated by the following argument using the digamma function $\frac... | This answer is provided with help from J.M.
$\log\Gamma(\frac{1}{2}x) - \log\Gamma(\frac{1}{3}x)$ is an increasing function. This can be shown using this series for $\psi$:
The function is increasing if we can show: $\frac{d}{dx}(\log\Gamma(\frac{1}{2}x) - \log\Gamma(\frac{1}{3}x)) > 0$
We can show this using the diga... | {
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"source": "stackexchange",
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Solve recursive equation $ f_n = \frac{2n-1}{n}f_{n-1}-\frac{n-1}{n}f_{n-2} + 1$ Solve recursive equation:
$$ f_n = \frac{2n-1}{n}f_{n-1}-\frac{n-1}{n}f_{n-2} + 1$$
$f_0 = 0, f_1 = 1$
What I have done so far:
$$ f_n = \frac{2n-1}{n}f_{n-1}-\frac{n-1}{n}f_{n-2} + 1- [n=0]$$
I multiplied it by $n$ and I have obtained:
$$... | Let's take a shot at this:
$$
f_n - f_{n - 1} = \frac{n - 1}{n} (f_{n - 1} - f_{n - 2}) + 1
$$
This immediately suggests the substitution $g_n = f_n - f_{n - 1}$, so $g_1 = f_1 - f_0 = 1$:
$$
g_n - \frac{n - 1}{n} g_{n - 1} = 1
$$
First order linear non-homogeneous recurrence, the summing factor $n$ is simple to see he... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Help in proving that $\nabla\cdot (r^n \hat r)=(n+2)r^{n-1}$
Show that$$\nabla \cdot (r^n \hat r)=(n+2)r^{n-1}$$ where $\hat r$ is the unit vector along $\bar r$.
Please give me some hint. I am clueless as of now.
| You can also use Cartesian coordinates and using the fact that $r \hat{r} = \vec{r} = (x,y,z)$.
\begin{align}
r^n \hat{r} &= r^{n-1} (x,y,z) \\
\nabla \cdot r^n \hat{r} & = \partial_x(r^{n-1}x) + \partial_y(r^{n-1}y) + \partial_z(r^{n-1}z)
\end{align}
Each term can be calculated:
$\partial_x(r^{n-1}x) = r^{n-1} + x (n... | {
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If $J$ is the $n×n$ matrix of all ones, and $A = (l−b)I +bJ$, then $\det(A) = (l − b)^{n−1}(l + (n − 1)b)$ I am stuck on how to prove this by induction.
Let $J$ be the $n×n$ matrix of all ones, and let $A = (l−b)I +bJ$. Show that $$\det(A) = (l − b)^{n−1}(l + (n − 1)b).$$
I have shown that it holds for $n=2$, and I'... | I think that it would be better to use $J_n$ for the $n \times n$ matrix of all ones, (and similarly $A_n, I_n$) so it is clear what the dimensions of the matrices are.
Proof by induction on $n$ that $\det(A_n)=(l-b)^n+nb(l-b)^{n-1}$:
When $n=1, 2$, this is easy to verify. We have $\det(A_1)=\det(l)=l=(l-b)^1+b(l-b)^0$... | {
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Find minimum in a constrained two-variable inequation I would appreciate if somebody could help me with the following problem:
Q: find minimum
$$9a^2+9b^2+c^2$$
where $a^2+b^2\leq 9, c=\sqrt{9-a^2}\sqrt{9-b^2}-2ab$
| Maybe this comes to your rescue.
Consider $b \ge a \ge 0$
When you expansion of $(\sqrt{9-a^2}\sqrt{9-b^2}-2ab)^2=(9-a^2)(9-b^2)+4a^2b^2-4ab \sqrt{(9-a^2)(9-b^2)}$
This attains minimum when $4ab \sqrt{(9-a^2)(9-b^2)}$ is maximum.
Applying AM-GM :
$\dfrac{9-a^2+9-b^2}{2} \ge \sqrt{(9-a^2)(9-b^2)} \implies 9- \dfrac{9... | {
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"source": "stackexchange",
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Prove that $\tan(75^\circ) = 2 + \sqrt{3}$ My (very simple) question to a friend was how do I prove the following using basic trig principles:
$\tan75^\circ = 2 + \sqrt{3}$
He gave this proof (via a text message!)
$1. \tan75^\circ$
$2. = \tan(60^\circ + (30/2)^\circ)$
$3. = (\tan60^\circ + \tan(30/2)^\circ) / (1 - \ta... | You can rather use $\tan (75)=\tan(45+30)$ and plug into the formula by Metin. Cause: Your $15^\circ$ is not so trivial.
| {
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"question_score": "9",
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Find the Sum $1\cdot2+2\cdot3+\cdots + (n-1)\cdot n$ Find the sum $$1\cdot2 + 2\cdot3 + \cdot \cdot \cdot + (n-1)\cdot n.$$
This is related to the binomial theorem. My guess is we use the combination formula . . .
$C(n, k) = n!/k!\cdot(n-k)!$
so . . . for the first term $2 = C(2,1) = 2/1!(2-1)! = 2$
but I can't figure... | As I have been directed to teach how to fish... this is a bit clunky, but works.
Define rising factorial powers:
$$
x^{\overline{m}} = \prod_{0 \le k < m} (x + k) = x (x + 1) \ldots (x + m - 1)
$$
Prove by induction over $n$ that:
$$
\sum_{0 \le k \le n} k^{\overline{m}} = \frac{n^{\overline{m + 1}}}{m + 1}
$$
When $n ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating a trigonometric integral using residues Finding the trigonometric integral using the method for residues:
$$\int_0^{2\pi} \frac{d\theta}{ a^2\sin^2 \theta + b^2\cos^2 \theta} = \frac{2\pi}{ab}$$ where $a, b > 0$.
I can't seem to factor this question
I got up to $4/i (z) / ((b^2)(z^2 + 1)^2 - a^2(z^2 - 1)^2 ... | Letting $z = e^{i\theta},$ we get
$$\int_0^{2\pi} \frac{1}{a^2\sin^2\theta + b^2\cos^2\theta} d\theta =
\int_{|z|=1} \frac{1}{iz} \frac{4}{-a^2(z-1/z)^2+b^2(z+1/z)^2} dz \\=
\int_{|z|=1} \frac{1}{iz} \frac{4z^2}{-a^2(z^2-1)^2+b^2(z^2+1)^2} dz =
-i\int_{|z|=1} \frac{4z}{-a^2(z^2-1)^2+b^2(z^2+1)^2} dz.$$
Now the location... | {
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"source": "stackexchange",
"question_score": "2",
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Rank of matrix of order $2 \times 2$ and $3 \times 3$ How Can I calculate Rank of matrix Using echlon Method::
$(a)\;\; \begin{pmatrix}
1 & -1\\
2 & 3
\end{pmatrix}$
$(b)\;\; \begin{pmatrix}
2 & 1\\
7 & 4
\end{pmatrix}$
$(c)\;\; \begin{pmatrix}
2 & 1\\
4 & 2
\end{pmatrix}$
$(d)\;\; \begin{pmatrix}
2 & -3 & 3\\
2 & ... | Follow this link to find your answer
If you are left with any doubt after reading this this, feel free to discuss.
| {
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Finding the remainder when $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by $7$
Find the remainder when $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by $7$.
Please brief about the concept behind this to solve such problems. Thanks.
| Using Euler-Fermat's theorem.
$\phi(7)=6$
$2^{6} \equiv 1 (\mod 7) \implies2^4.2^{96} \equiv 2(\mod7)$
$3^{6} \equiv 1 (\mod 7) \implies3^4.3^{96} \equiv 4(\mod7)$
$4^{6} \equiv 1 (\mod 7) \implies4^4.4^{96} \equiv 4(\mod7)$
$5^{6} \equiv 1 (\mod 7) \implies5^4.5^{96} \equiv 2(\mod7)$
$2^{100}+3^{100}+4^{100}+5^{100} \... | {
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"timestamp": "2023-03-29T00:00:00",
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Graph of $\quad\frac{x^3-8}{x^2-4}$. I was using google graphs to find the graph of $$\frac{x^3-8}{x^2-4}$$ and it gave me:
Why is $x=2$ defined as $3$? I know that it is supposed to tend to 3. But where is the asymptote???
| Because there is a removable singularity at $x = 2$, there will be no asymptote.
You're correct that the function is not defined at $x = 2$. Consider the point $(2, 3)$ to be a hole in the graph.
Note that in the numerator, $$(x-2)(x^2 + 2x + 4) = x^3 - 8,$$ and in the denominator $$(x-2)(x+ 2) = x^2 - 4$$
When we si... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find a polar representation for a curve. I have the following curve:
$(x^2 + y^2)^2 - 4x(x^2 + y^2) = 4y^2$ and I have to find its polar representation.
I don't know how. I'd like to get help .. thanks in advance.
| Just as the Cartesian has two variables, we will have two variables in polar form:
$$x = r\cos \theta,\;\;y = r \sin \theta$$
We can also use the fact that $x^2 + y^2 = (r\cos \theta)^2 + (r\sin\theta)^2 = r^2 \cos^2\theta + r^2\sin^2 \theta = r^2\underbrace{(\sin^2 \theta + \cos^2 \theta)}_{= 1} =r^2$
This gives us $$... | {
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Evaluating Complex Integral. I am trying to evaluate the following integrals:
$$\int\limits_{-\infty}^\infty \frac{x^2}{1+x^2+x^4}dx $$
$$\int\limits_{0}^\pi \frac{d\theta}{a\cos\theta+ b} \text{ where }0<a<b$$
My very limited text has the following substitution:
$$\int\limits_0^\infty \frac{\sin x}{x}dx = \frac{1}{2i}... | For the first one, write $\dfrac{x^2}{1+x^2+x^4}$ as $\dfrac{x}{2(1-x+x^2)} - \dfrac{x}{2(1+x+x^2)}$. Now
$$\dfrac{x}{(1-x+x^2)} = \dfrac{x-1/2}{\left(x-\dfrac12\right)^2 + \left(\dfrac{\sqrt3}2 \right)^2} + \dfrac{1/2}{\left(x-\dfrac12\right)^2 + \left(\dfrac{\sqrt3}2 \right)^2}$$
and
$$\dfrac{x}{(1-x+x^2)} = \dfrac{x... | {
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Can't prove this elementary algebra problem $x^2 + 8x + 16 - y^2$
First proof:
$(x^2 + 8x + 16) – y^2$
$(x + 4)^2 – y^2$
$[(x + 4) + y][(x + 4) – y]$
2nd proof where I mess up:
$(x^2 + 8x) + (16 - y^2)$
$x(x + 8) + (4 + y)(4 - y)$
$x + 1(x + 8)(4 + y)(4 - y)$ ????
I think I'm breaking one of algebra's golden rules, but... | Let us fix your second approach.
$$
\begin{align}
&x^2+8x+16-y^2\\
=&x^2+8x+(4-y)(4+y)\\
=&x^2+x\big[(4-y)+(4+y)\big]+(4-y)(4+y)\\
=&(x+4-y)(x+4+y).
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Integrating a school homework question. Show that $$\int_0^1\frac{4x-5}{\sqrt{3+2x-x^2}}dx = \frac{a\sqrt{3}+b-\pi}{6},$$ where $a$ and $b$ are constants to be found.
Answer is: $$\frac{24\sqrt3-48-\pi}{6}$$
Thank you in advance!
| On solving we will find that it is equal to -$$-4\sqrt{3+2x-x^{2}}-\sin^{-1}(\frac{x-1}{2})$$
Now if you put the appropriate limits I guess you'll get your answer.
First of all write $$4x-5 = \mu \frac{d(3+2x-x^{2})}{dx}+\tau(3+2x-x^{2})$$
We will find that $\mu=-2$ and $\tau=-1$.
$$\int\frac{4x-5}{\sqrt{3+2x-x^{2}}}=\... | {
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How to prove $n$ is prime? Let $n \gt 1$ and
$$\left\lfloor\frac n 1\right\rfloor + \left\lfloor\frac n2\right\rfloor + \ldots + \left\lfloor\frac n n\right\rfloor = \left\lfloor\frac{n-1}{1}\right\rfloor + \left\lfloor\frac{n-1}{2}\right\rfloor + \ldots + \left\lfloor\frac{n-1}{n-1}\right\rfloor + 2$$
and $\lfloor \... | You know that
$$\left( \left\lfloor\frac n 1\right\rfloor - \left\lfloor\frac{n-1}{1}\right\rfloor \right)+\left( \left\lfloor\frac n2\right\rfloor - \left\lfloor\frac{n-1}{2}\right\rfloor\right) + \ldots + \left( \left\lfloor\frac{n}{n-1}\right\rfloor - \left\lfloor\frac{n-1}{n-1}\right\rfloor\right) + \left\lfloor\... | {
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Determining power series for $\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}$ I'm looking for the power series for $f(x)=\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}$
My approach: the given function is a combination of two problems. first i made some transformations, so the function looks easier.
$$\frac{3x^{2}-4x+9}{(x-1)^2(x+3)})=\frac{3x^{2... | You can use the partial fraction decomposition:
$$
\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}=
\frac{A}{1-x}+\frac{B}{(1-x)^2}+\frac{C}{1+\frac{1}{3}x}
$$
and sum up the series you get, which are known.
If you do the computation, you find $A=0$, $B=2$ and $C=1$, so
$$
\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}=
\frac{2}{(1-x)^2}+\frac{1}{... | {
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Arc length of logarithm function I need to find the length of $y = \ln(x)$ (natural logarithm) from $x=\sqrt3$ to $x=\sqrt8$.
So, if I am not mistake, the length should be $$\int^\sqrt8_\sqrt3\sqrt{1+\frac{1}{x^2}}dx$$
I am having trouble calculating the integral. I tried to do substitution, but I still fail to think o... | $\int^{\sqrt{8}}_{\sqrt{3}}\sqrt{1+\frac{1}{x^{2}}}dx$=$\int^{\sqrt{8}}_{\sqrt{3}}\frac{\sqrt{1+x^{2}}}{x}dx$=$\int^{\sqrt{8}}_{\sqrt{3}}\frac{1+x^{2}}{x\sqrt{1+x^{2}}}$=$\int^{\sqrt{8}}_{\sqrt{3}}\frac{x}{\sqrt{1+x^{2}}}dx$+
+$\int^{\sqrt{8}}_{\sqrt{3}}\frac{1}{x\sqrt{1+x^{2}}}dx$=
$=\frac{1}{2}\int^{\sqrt{8}}_{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/389781",
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Bernoulli differential equation help? We have the equation $$3xy' -2y=\frac{x^3}{y^2}$$ It is a type of Bernouli differential equation. So, since B. diff equation type is
$$y'+P(x)y=Q(x)y^n$$
I modify it a little to:
$$y'- \frac{2y}{3x} = \frac{x^2}{3y^2}$$
$$y'-\frac{2y}{3x}=\frac{1}{3}x^2y^{-2}$$
Now I divide both s... | $$\text{We have $3xy^2 y'-2y^3 = x^3 \implies x (y^3)' - 2y^3 = x^3 \implies \dfrac{(y^3)'}{x^2} + y^3 \times \left(-\dfrac2{x^3}\right) = 1$}$$
$$\text{Now note that }\left(\dfrac1{x^2}\right)' = -\dfrac2{x^3}. \text{ Hence, we have }\dfrac{d}{dx}\left(\dfrac{y^3}{x^2}\right) = 1\implies \dfrac{y^3}{x^2} = x + c$$
$$\... | {
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Solve $\sqrt{2x-5} - \sqrt{x-1} = 1$ Although this is a simple question I for the life of me can not figure out why one would get a 2 in front of the second square root when expanding. Can someone please explain that to me?
Example: solve $\sqrt{(2x-5)} - \sqrt{(x-1)} = 1$
Isolate one of the square roots: $\sqrt{(2x-5... | To get rid of the square root, denote: $\sqrt{x-1}=t\Rightarrow x=t^2+1$. Then:
$$\sqrt{2x-5} - \sqrt{x-1} = 1 \Rightarrow \\
\sqrt{2t^2-3}=t+1\Rightarrow \\
2t^2-3=t^2+2t+1\Rightarrow \\
t^2-2t-4=0 \Rightarrow \\
t_1=1-\sqrt{5} \text{ (ignored, because $t>0$)},t_2=1+\sqrt{5}.$$
Now we can return to $x$:
$$x=t^2+1=(1+\... | {
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positive Integer value of $n$ for which $2005$ divides $n^2+n+1$
How Can I calculate positive Integer value of $n$ for which $2005$
divides $n^2+n+1$
My try:: $2005 = 5 \times 401$
means $n^2+n+1$ must be a multiple of $5$ or multiple of $401$
because $2005 = 5 \times 401$
now $n^2+n+1 = n(n+1)+1$
now $n(n+1)+1$ co... | A number of the form $n^2+n+1$ has divisors of the form 3, or any number of $6n+1$, and has a three-place period in base n.
On the other hand, there are values where 2005 divides some $n^2+n-1$, for which the divisors are of the form n, 10n+1, 10n+9. This happens when n is 512 or 1492 mod 2005.
| {
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Help me prove this inequality : How would I go about proving this?
$$ \displaystyle\sum_{r=1}^{n} \left( 1 + \dfrac{1}{2r} \right)^{2r} \leq n \displaystyle\sum_{r=0}^{n+1} \displaystyle\binom{n+1}{r} \left( \dfrac{1}{n+1} \right)^{r}$$
Thank you! I've tried so many things. I've tried finding a series I could compare... | Here is the proof that $(1+1/x)^x$ is concave for $x\ge 1$.
The second derivative of $(1+1/x)^x$ is $(1+1/x)^x$ times $$p(x)=\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)^2-\frac{1}{x(1+x)^2}$$
Now for $x\ge 1$, we have $$\ln(1+1/x)-\frac{2}{1+x}\le \frac{1}{x}-\frac{2}{1+x}=\frac{1-x}{x(1+x)}\le 0$$ and $$... | {
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find out the value of $\dfrac {x^2}{9}+\dfrac {y^2}{25}+\dfrac {z^2}{16}$ If $(x-3)^2+(y-5)^2+(z-4)^2=0$,then find out the value of $$\dfrac {x^2}{9}+\dfrac {y^2}{25}+\dfrac {z^2}{16}$$
just give hint to start solution.
| Hint: What values does the function $x^2$ acquire(positive/negarive)? What is the solution of the equation $x^2=0$?
Can you find the solution of the equation $x^2+y^2=0$?
Now, what can you say about the equation $(x-3)^2+(y-5)^2+(z-4)^2=0
$? Can you find the values of $x,y,z?$
| {
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How prove this $\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$ Prove that
$$\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$$
I have some question. Using this, find this integral is not converge, I'm wrong?
Thank you... | First make the substitution $x=u^2$ to get:
$\displaystyle \int _{0}^{\infty }\!\sin \left( x \right) \sin \left( \sqrt {x}
\right) {dx}=\int _{0}^{\infty }\!2\,\sin \left( {u}^{2} \right) \sin
\left( u \right) u{du}$,
$\displaystyle=-\int _{0}^{\infty }\!u\cos \left( u \left( u+1 \right) \right) {du}+
\int _{0}^{\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/397990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Show that no number of the form 8k + 3 or 8k + 7 can be written in the form $a^2 +5b^2$ I'm studying for a number theory exam, and have got stuck on this question.
Show that no number of the form $8k + 3$ or $8k + 7$ can be written in the form $a^2 +5b^2$
I know that there is a theorem which tells us that $p$ is exp... | $8k+3,8k+7$ can be merged into $4c+3$ where $k,c$ are integers
Now, $a^2+5b^2=4c+3\implies a^2+b^2=4c+3-4b^2=4(c-b^2)+3\equiv3\pmod 4,$
But as $(2c)^2\equiv0\pmod 4,(2d+1)^2\equiv1\pmod 4,$
$a^2+b^2\equiv0,1,2\pmod 4\not\equiv3$
Clearly, $a^2+5b^2$ in the question can be generalized $(4m+1)x^2+(4n+1)y^2$ where $m,n$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/398210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solve the equation $\sqrt{3x-2} +2-x=0$
Solve the equation: $$\sqrt{3x-2} +2-x=0$$
I squared both equations $$(\sqrt{3x-2})^2 (+2-x)^2= 0$$
I got $$3x-2 + 4 -4x + x^2$$
I then combined like terms $x^2 -1x +2$
However, that can not be right since I get a negative radicand when I use the quadratic equation.
$x = 1/2... | $$\sqrt{3x-2} +2-x=0$$
Isolating the radical:$$\sqrt{3x-2} =-2+x$$
Squaring both sides:$$\bigg(\sqrt{3x-2}\bigg)^2 =\bigg(-2+x\bigg)^2$$
Expanding $(-2+x)^2$ and gathering like terms: $$3x-2=-2(-2+x)+x(-2+x)$$
$$3x-2=4-2x-2x+x^2$$
Set x equal to zero:$$3x-2=4-4x+x^2$$
Gather like terms:$$0=4+2-3x-4x+x^2$$
Factor the q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/398984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
A problem on matrices: Find the value of $k$
If $
\begin{bmatrix}
\cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\
\sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7} \\
\end{bmatrix}^k
=
\begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix}
$, then the least positiv... | Powers of matrices should always be attacked with diagonalization, if feasible. Forget $2\pi/7$, for the moment, and look at
$$
A=\begin{bmatrix}
\cos\alpha & -\sin\alpha\\
\sin\alpha & \cos\alpha
\end{bmatrix}
$$
whose characteristic polynomial is, easily, $p_A(X)=1-2X\cos\alpha+X^2$. The discriminant is $4(\cos^2\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/401158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Listing subgroups of a group I made a program to list all the subgroups of any group and I came up with satisfactory result for $\operatorname{Symmetric Group}[3]$ as
$\left\{\{\text{Cycles}[\{\}]\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
1 & 2 \\
\end{array}
\right)\right]\right\},\left... | I have the impression that you only list the cyclic subgroups.
For $S_3$, the full group $S_3$ ist missing as a subgroup (you are mentioning that in your question).
For $S_4$, several subgroups are missing. In total, there should be $30$ of them. $14$ of them are cyclic, which are exactly the ones you listed. To give y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/401971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving $\sqrt{3x^2-2}+\sqrt[3]{x^2-1}= 3x-2 $ How can I solve the equation $$\sqrt{3x^2-2}+\sqrt[3]{x^2-1}= 3x-2$$
I know that it has two roots: $x=1$ and $x=3$.
| Substituting $x = \sqrt{t^3+1}$ and twice squaring, we arrive to the equation
$$ 36t^6-24t^5-95t^4+8t^3+4t^2-48t=0.$$ Its real roots are $t=0$ and $t=2$ (the latter root is found in the form $\pm \text{divisor}(48)/\text{divisor}(36)$), therefore $x=1$ and $x=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/402965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find the integral of implicitly defined function? Let $a$ and $b$ be real numbers such that $ 0<a<b$. The decreasing continuous function
$y:[0,1] \to [0,1]$ is implicitly defined by the equation $y^a-y^b=x^a-x^b.$
Prove
$$\int_0^1 \frac {\ln (y)} x \, dx=- \frac {\pi^2} {3ab}.
$$
| OK, at long last, I have a solution. Thanks to @Occupy Gezi and my colleague Robert Varley for getting me on the right track. As @Occupy Gezi noted, some care is required to work with convergent integrals.
Consider the curve $x^a-x^b=y^a-y^b$ (with $y(0)=1$ and $y(1)=0$). We want to exploit the symmetry of the curve a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/406847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 1
} |
Find all positive integers $x$ such that $13 \mid (x^2 + 1)$ I was able to solve this by hand to get $x = 5$ and $x =8$. I didn't know if there were more solutions, so I just verified it by WolframAlpha. I set up the congruence relation $x^2 \equiv -1 \mod13$ and just literally just multiplied out. This lead me to two ... | Starting with $2,$ the minimum natural number $>1$ co-prime with $13,$
$2^1=2,2^2=4,2^3=8,2^4=16\equiv3,2^5=32\equiv6,2^6=64\equiv-1\pmod{13}$
As $2^6=(2^3)^2,$ so $2^3=8$ is a solution of $x^2\equiv-1\pmod{13}$
Now, observe that $x^2\equiv a\pmod m\iff (-x)^2\equiv a$
So, $8^2\equiv-1\pmod {13}\iff(-8)^2\equiv-1$
Now,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/409005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"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.