Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to find the sum: $\sum_{n=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right) $ $$S_n=\sum_{n=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right) = \frac{1}{r-s}\left( \frac{1}{s+1}+\frac{1}{s+2}\cdots+\frac{1}{r}\right)$$
I fail to see how this:
$$S_n = \frac{1}{r-s}\left( \frac... | Let $r=s+k$. Then:
$$\begin{align}S_n&=\sum_{n=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right) = \\
&=\frac1{r-s}(\ \boxed{\color{black}{\frac1{s+1}}}-\color{red}{\frac1{s+k+1}}+\color{black}{\boxed{\frac1{s+2}}}-\color{blue}{\frac1{s+k+2}}\color{black}{+\cdots +\boxed{\frac1{s+k}}}-\color{green}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Simplifying $\frac{1}{x^4+4x^2}$ I'm trying to solve this problem on my own and it involves simplifying the expression in the title.
In the solutions it says it's this:
$$\frac{1}{x^4+4x^2} = \frac{1}{4}\Biggl[\frac{1}{x^2}-\frac{1}{x^2+4}\Biggr]$$
But I can't for the life of me figure out where the 1/4 came from. Or e... | We can write the fraction as,
$$\frac{1}{x^4 +4x^2} = \frac{1}{4}\cdot \frac{4}{x^2(x^2+4)}$$
The fraction $\frac{4}{x^2(x^2+4)}$ can be written as..
\begin{align}
\frac{1}{4}\cdot \frac{4}{x^2(x^2+4)}=\frac{1}{4}\Biggl[\frac{(x^2+4)-x^2}{x^2(x^2+4)}\Biggr]\\
=\frac{1}{4}\Biggl[\frac{1}{x^2}-\frac{1}{x^2+4}\Biggr]
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3231405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to prove De Moivre's theorem inductively It is given that $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ where $n\in Z^+$.
I can show it works for $n=1$ but I am stuck in showing it inductively. I have got as far as below but are stuck in the rearranging:
$$(\cos\theta + i\sin\theta)^{k+1}= (\cos k\t... | $$(\cos \theta + i\sin \theta)^k = \cos k\theta + i\sin k\theta$$
$$\text{So for } (\cos \theta + i\sin \theta)^{k+1} = (\cos \theta + i\sin \theta)^k \cdot (\cos \theta + i\sin \theta) = (\cos k\theta + i\sin k\theta)\cdot (\cos \theta + i\sin \theta)$$
$$=\cos k\theta\cos\theta +i\sin\theta\cos k\theta +i\sin k\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3231547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Eliminate $\theta$ from from $\frac{x-k\sin \theta \cot \alpha}{k \cos \theta}=\frac{y-k\cos \theta \tan \alpha}{k \sin \theta}=\frac{z}{c}$ I am stuck with the following problem :
Eliminate $\theta$ from
$$\frac{x-k\sin \theta \cot \alpha}{k \cos \theta}=\frac{y-k\cos \theta \tan \alpha}{k \sin \theta}=\frac{z}{c}... | Write your line as
$$
\frac{x-k\sin\theta\cot\alpha}{k\cos\theta}=\frac{z}{c}\quad\text{and}\quad\frac{y-k\cos\theta\tan\alpha}{k\sin\theta}=\frac{z}{c}.
$$
Manipulate the two equations into
\begin{align*}
cx/k&=c\cot\alpha\sin\theta+z\cos\theta\\
cy/k&=z\sin\theta+c\tan\alpha\cos\theta
\end{align*}
so solving for $\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Pythagorean Quadruples and Stereographic Projection I am trying to solve Diophantine equation $a^2+b^2+c^2=d^2$ by transforming this equation, assuming $d \neq 0$, into a sphere $ (\frac{a}{d})^2 + (\frac{b}{d})^2 +(\frac{c}{d})^2 = x^2 + y^2 +z^2 =1$ and using stereographic projection of this sphere from $N = (0,0,1)$... | You are on the right track; indeed you have $\gcd(b,c)=2nq\gcd(np,mq)$, and also
$$\gcd(a,d)=\gcd(d,d-a)=\gcd(n^2p^2+m^2q^2+n^2q^2,2n^2q^2),$$
which shows that $\gcd(a,b,c,d)$ divides $2n^2q^2$. It follows that
\begin{eqnarray*}
\gcd(a,b,c,d)&=&\gcd(d,d-a,\gcd(b,c))=\gcd(d,2n^2q^2,2nq\gcd(np,mq)),
\end{eqnarray*}
where... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Linear Algebra - How to find the axis of a parabola? Given the parabola $x^2+y^2-2xy+4x=0$ defined in $\mathbb{R}^2$, how can I find the axis?
The matrices associated to the curve are:
$$A=\begin{bmatrix}1 & -1\\ -1 & 1\end{bmatrix}, B=\begin{bmatrix}2 \\ 0\end{bmatrix}, C=\begin{bmatrix}1 & -1 & 2\\ -1 & 1 & 0\\ 2 & 0... | Equation of a parabola can be written as $L_1^2= 4 A L_2$, provided $L_1$ and $ L_2$ are equations of lines which are non-parallel.
Further if $L_1$ and $L_2$ are perpendicular and normalized, then $4A$ is the length of
latus rectum. Equation of axis is $L_1=0$, Equation of tangent at vertex is $L_2=0$. The focus is g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3237377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Limit of powers of $3\times3$ matrix Consider the matrix
$$A = \begin{bmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{bmatrix}$$
What is $\lim_{n→\infty}$$A^n$ ?
A)$\begin{bmatrix} 0 & 0 & 0\\ 0& 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$
B)$\begin{bmatrix} \frac{1}{4} &\f... | If you are in $1$, you have same probability to stay there or to pass to $2$, but no way to get back from there. Thus you are finally drifting to $2$.
States $2$ and $3$ are symmetrical: at long they will tend to be equally populated, independently of the starting conditions.
Therefore also starting from $1$ you wil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Compute convolution $f*g(x)$ such that : $f,g\in L^{p}$ Question :
Compute convolution : $f*g(x)$
$f(x)=\begin{cases}3x^2 & \text{ if } |x|\leq4 \\0 & \text{ otherwise}\end{cases}$
$g(x)=\begin{cases}1 &\text{ if }|x|\leq 2 \\0 & \text{ otherwise}\end{cases}$
My try :
\begin{align}
f*g(x)&=\int f(x-y)g(y)dy\\
&=\int... | Your bounds aren't quite right. We have $$f(x-y) = 3(x-y)^2\cdot\mathsf 1_{[-4,4]}(x-y) $$
and $g(y) = \mathsf 1_{[-2,2]}(y)$, so we have the following inequalities for $y$:
\begin{align}
x-4&\leqslant y\leqslant x+4\\
-2&\leqslant y\leqslant 2.
\end{align}
For $-6\leqslant x\leqslant -2$ we have
$$
f\star g(x) = \int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does dividing a common factor out from numerator and denominator of a rational function create a new function with different domain? Suppose a function is defined as $ f(x) =\frac{x^2 - 9}{x-3}. $
If we divide the common factor $ x-3 $ from both the numerator and denominator :-
$$ \frac{x^2 - 9}{x - 3} \\ = \frac{(x+3... | $x+3$ is different from $\frac{x^2-9}{x-3}$, but only in the domain. All the simplification does is to add $x=3$ to the domain.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Why do we take the positive square root only from $\sqrt{a^2-x^2}$ when integrating using trig substitutions? $$\int\frac{\mathrm dx}{x^2 \sqrt{16-x^2}}$$
when substituting $\,x=4\sin\theta ,\;\mathrm dx=4\cos\theta\, \mathrm d\theta,\,$ it becomes
$$\int\frac{4\cos\theta\, \mathrm d\theta}{4^2\sin^2\theta\sqrt{16-4^2\... | With $x=4\sin\theta$ your integral becomes $\int\frac{4\cos\theta d\theta}{64\sin^2\theta|\cos\theta|}=\frac{1}{16}\int\frac{\operatorname{sgn}(\cos\theta)d\theta}{\sin^2\theta}$ with $\operatorname{sgn}y:=\frac{y}{|y|}$. Luckily, choosing $\theta$ to obtain $\sin\theta=\frac{x}{4}$ lets us arbitrarily choose whether $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing the matrix powers of a non-diagonalizable matrix Define
\begin{equation}
A = \begin{pmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{pmatrix}.
\end{equation}
Note that the sum of the dimensions of the eigenspaces of $A$ is only two. $A$ is thus not diag... | Note that your matrix $A$ has the generalized eigenvectors
\begin{equation}v_1=\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}, v_2=\begin{pmatrix}0 \\ 2 \\ -2\end{pmatrix}, v_3=\begin{pmatrix}0\\ 0 \\ 1\end{pmatrix}.\end{equation}
Thus, by Jordan decomposition, $A=\big(v_1,v_2,v_3\big)J\big(v_1,v_2,v_3\big)^{-1}$, where
\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to find probability of having at least 2 out of 3 students selected to sit together? I have the problem. 25 students are seated in a circle at the campfire night. 3 students are selected (the probability of being selected for each student is the same) and asked to join a game. Find the probability of having at leas... | Yes, your solution is correct.
Refer to the diagram:
$\hspace{4cm}$
You considered two cases:
Case 1: triple students (order does not matter):
$$(\color{red}1,2,3),(\color{red}2,3,4),(\color{red}3,4,5),...,(\color{red}{23},24,25),(\color{red}{24},25,1),(\color{red}{25},1,2) \Rightarrow 25$$
Case 2: double students (or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Modular Arithmetic. Integer Solutions How can you show that $|a^2 -10b^2|=2$ has no integer solutions for a and b using modular arithmetic?
Thank you.
| This is a case where the principal form also represents $-1,$ as $3^2 - 10 = -1.$ So, if $a^2 - 10 b^2 = n,$ we find
$$ (3a+10b)^2 - 10 (a+3b)^2 = -n $$
Meanwhile, $x^2 - 10 y^2$ represents primes $p \equiv 1, 9, 31, 39 \pmod {40}$
The other class of this discriminant $2x^2 - 5 y^2,$ represents $2$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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studying the series $\sum_\limits{n=0}^\infty \tan (\frac{n}{1+n^3})$ studying the series $\sum_\limits{n=0}^\infty \tan (\frac{n}{1+n^3})$.
The series has got a sense for each natural number.
$1+n^3>n, \forall n \in N \Rightarrow \frac{n}{1+n^3}<1,\forall n \in N\Rightarrow \tan \frac{n}{1+n^3} $ tends to 0 as $n \in ... | From $$x\leq\tan x\leq \frac{4x}{\pi} \quad \left(0\leq x \leq \frac{\pi}{4}\right)$$
We get $$0\leq \frac{n}{1+n^3}\leq \tan \left(\frac{n}{1+n^3}\right) \leq \frac{4n}{\pi(1+n^3)} \leq \frac{4}{\pi n^2}$$
So original series converges by comparison test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x + y + z = 2$, then show $\frac{x y z}{(1-x)(1-y)(1-z)} \geq 8$, added a second question(Problem 2). Problem number 1:
The problem is that $x, y, z$ are proper fractions, and each one of them is greater than zero.
Given $x + y + z = 2$, prove
$$\frac{x y z}{(1-x)(1-y)(1-z)} \geq 8$$
I have tried to solve this... | The first inequality:
We need to prove that:
$$\frac{8xyz}{\prod\limits_{cyc}(x+y+z-2x)}\geq8$$ or
$$xyz\geq\prod_{cyc}(y+z-x)$$ or
$$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is Schur.
The second inequality it's just Holder for three sequences:
$$\prod_{cyc}(1+x^3)\geq\left(\sqrt[3]{1\cdot1\cdot1}+\sqrt[3]{x^3y^3z^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find a matrix $C \in M_{3\times 3} (\mathbb C)$ such that $A_4=C^{T}A_3C$
Let $$A_3=\begin{bmatrix} -1 & 2 & -1 \\ 2 & 0 & 0 \\ -1 & 0 & -1 \end{bmatrix},$$ $$A_4=\begin{bmatrix} 4 & 4 & 4 \\ 4 & 0 & 2 \\ 4 & 2 & 5 \end{bmatrix}.$$
a) Show that $A_3$ and $A_4$ are congruent over $\mathbb C$, but not over $\mathbb R$... | Note that if you write $A_{4}=C^{T}A_{3}C$, then you are essentially applying simultaneous row and column operations. $C^{T}$ represent some operations applied to the rows of $A_{3}$ and $C$ represent the same operations applied to the columns. Now you can start doing calculations
$$A_{3}=\left(\begin{matrix}-1&2&-1\\2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3248664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\frac{x}{4} = \frac{y}{x} = \frac{7}{y}$ Could someone please explain to me what's wrong here?
$\cfrac{x}{4} = \cfrac{y}{x} = \cfrac{7}{y}$
So $$\cfrac{x^2}{4} = y \ \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \ y = \sqrt{7x}$$
Hence $$\cfrac{x^2}{4} =\sqrt{7x}$$
With a solution of $x = 2.57$
But $$\cfrac{2.57^2}{4} \neq \sqr... | Hint $ $ Multiplyimg $\,\overbrace{ \dfrac{\color{#c00}x}{4} = \dfrac{\color{#0a0}y}{\color{#c00}x} = \dfrac{7}{\color{#0a0}y}}^{\Large a}\,\Rightarrow\,a^{\large 3} = \dfrac{7}4$
hence $\, x = 4a,\ \ y = xa = 4a^2,\ \ 7 = ya = 4a^3\ \ \checkmark$
Remark $ $ This telescoping product view works generally
$\,\overbrace... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Calculate the minimum value of $\sum_\mathrm{cyc}\frac{a^2}{b + c}$ where $a, b, c > 0$ and $\sum_\mathrm{cyc}\sqrt{a^2 + b^2} = 1$.
$a$, $b$ and $c$ are positives such that $\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{c^2 + a^2} = 1$. Calculate the minimum value of $$\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}... | By rearrangement-inequality
$$\frac{a^{\,2}}{b+ c}+ \frac{b^{\,2}}{c+ a}+ \frac{c^{\,2}}{a+ b}\geqq \frac{a^{\,2}}{a+ b}+ \frac{b^{\,2}}{b+ c}+ \frac{c^{\,2}}{c+ a}$$
We have to prove
$$\frac{a^{\,2}}{a+ b}+ \frac{b^{\,2}}{b+ c}+ \frac{c^{\,2}}{c+ a}\geqq \frac{1}{2}(\,\sqrt{\frac{a^{\,2}+ b^{\,2}}{2}}+ \sqrt{\frac{b^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Discrepancy in finding the coefficient of a polynomial After asking a question here let us consider on the following case which is simpler, this might help me to make some progress:
suppose we have the following polynomial:
$$f(x)=(x^0+x^1+x^2+x^3)^3$$ expanding this gives:
$$f(x)=1 + 3 x + 6 x^2 + 10 x^3 + 12 x^4 +... | $$f(x)=\frac{(1-x^4)^3}{(1-x)^3}$$
We have that $$\frac{1}{(1-x)^3}=\sum_{k=0}^{\infty}\binom{k+2}{2}x^k.$$
And $(1-x^4)^3=1-3x^4+3x^8-x^{12}.$
So $f(x)=\sum_{k=0}^{\infty} a_kx^k$ where:
$$a_k = \binom{k+2}{2}-3\binom{k-2}{2}+3\binom{k-6}{2}-\binom{k-10}{2}.$$
More general, if $$\begin{align}f_{n,m}(x)&=\left(1+x+x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250882",
"timestamp": "2023-03-29T00:00:00",
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Solve integer equation $2^m.m^2=9n^2-12n+19$
Problem: Find $m,n\in \mathbb{N}^*$ satisfies: $2^m.m^2=9n^2-12n+19$.
This is my attempt:
We have $9n^2-12n+19\equiv 1 \pmod 3$, so: $2^{m}m^2\equiv 1 \pmod 3\tag{1}$
In addition, we have: $$m^2\equiv 0\text{ or }1 \pmod 3 $$
So $(1)\implies m\equiv \pm 1\pmod 3$.
Suppose ... | The case left open by the above answer is for $m$ odd. If $m$ is odd, then $m^2\times 2^m = 2A^2$ for some integer $A$; namely $A = m\times 2^{\frac{m-1}{2}}$. Thus we are left with the equation
$$2A^2 = (3n-2)^2+15$$
Note that the LHS of the equation is either 2 or 0 mod 3, whereas the RHS is always 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Double Integration Problem $\int_{0}^{1} \int_0^1 \frac{1}{1+y(x^2-x)}dydx$
Compute
$$I = \int_{0}^{1} \int_0^1 \frac{1}{1+y(x^2-x)}dydx$$
Here are my steps:
$$\begin{split}
I &=\int_{0}^{1} \left(\int_0^1 \frac{dy}{1+y(x^2-x)}\right)dx\\
&=\int_{0}^{1} \left[\frac{\ln(1+y(x^2-x))}{x^2-x}\right]_0^1dx\\
&=\int_... | Here is an alternate way that avoids your integral.
Writing the integrand in terms of a geometric sum we have:
$$\frac{1}{1 + y(x^2 - x)} = \sum_{n = 0}^\infty y^n (x - x^2)^n, \quad |x|, |y| < 1,$$
then
\begin{align}
I &= \int_0^1 \int_0^1 \frac{1}{1 + y(x^2 - x)} \, dy dx\\
&= \sum_{n = 0}^\infty \int_0^1 y^n \, dy \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the sum of $\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$ $$S(x)=\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$$
The question is divided into three parts:
1. Determine its radius of convergence
2. By using the power series of $\frac{1}{1-x}$, show that for all x $\in$ ]-1,1[ , we get $\ln(1-x)=-\sum_{n}^\infty \frac{1}{n}x^n$
3.... | Another approach is to start with the geometric series $\sum_{n=1} x^{n-1}=\frac{1}{1-x}$ (assuming you are in appropriate interval for convergence etc.).
\begin{align*}
\frac{1}{1-x} & = \sum_{n=1}x^{n-1}\\
\int \frac{1}{1-x} \, dx& =\sum_{n=1}\frac{x^n}{n}\\
-\ln(1-x)+c & = \sum_{n=1}\frac{x^n}{n}\\
-x\ln(1-x)+xc & =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3253257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplifying $\sum_{cyc}\tan^{-1}\left(\sqrt{\frac{x(x+y+z)}{yz}}\right)$. I get $0$, but the answer is $\pi$. So the question is
$$ \tan^{-1}\left(\sqrt{\frac{x(x+y+z)}{yz}}\right)+\tan^{-1}\left(\sqrt{\frac{y(x+y+z)}{xz}}\right)+\tan^{-1}\left(\sqrt{\frac{z(x+y+z)}{yx}}\right) =\ ? $$
So my take on the question is to... | Hint:-
$$ \tan^{-1}a + \tan^{-1}b + \tan^{-1}c= \pi$$
Only and only if
$$a+b+c=abc$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3255546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Easier method of finding the equation of the circle circumscribing the triangle formed by 3 lines? The equation of the circle circumscribing the triangle formed by the lines $y = 0, y = x$ and $2x + 3y = 10$ is?
I know this can be done by solving two equations at a time and finding the vertex. Then forming 3 different ... | Easy to see that $(0,0)$, $(2,2)$ and $(5,0)$ are the vertices of the triangle.
Let $M\left(\frac{5}{2},b\right)$ be the center of the circle.
Thus, $$\left(\frac{5}{2}\right)^2+b^2=\left(\frac{1}{2}\right)^2+(b-2)^2,$$ which gives $$b=-\frac{1}{2}$$ and $$M\left(\frac{5}{2},-\frac{1}{2}\right).$$
Now, for the radius o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to tackle this squaring of inequality problem
If the roots of quadratic equation $$x^2 − 2ax + a^2 + a – 3 = 0$$
are real and less than $3$, find the range of $a$.
The roots are $a \pm \sqrt {3 – a}$
For the roots to be real, we must have a < 3.
Also, for the roots to be less than 3, we must have $\pm \... | Since $x_1,x_2<3$ we have $3-x_i>0$ so their product is positive:$$ 0<9-3(x_1+x_2)+x_1x_2$$
thus $$0<9-6a+a^2+a-3 = a^2-5a+6=(a-3)(a-2)$$
So $a\in (-\infty ,2)\cup (3,\infty)$. But since the discriminat must be $\geq 0$ we get $a\leq 3$ so we have $a\in (-\infty ,2)$
Now let us prove that all $a<2$ are good.
We have $$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to evaluate $\int_{0}^{1}\frac{\arctan x}{x} \log{\left(\frac{ 1+ x}{\sqrt{1+x^2}}\right)}\mathrm dx$
How to evaluate $$\int_{0}^{1}\frac{\arctan x}{x} \log{\left(\frac{1+ x}{\sqrt{1+x^2}}\right)}\mathrm dx$$
I tried to integrate by parts, but no way so far, help me, thanks.
| From here , we have $\ \displaystyle \ 3\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx-2\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx=0$
or $\ I=\displaystyle\int_{0}^{1}\frac{\arctan x}{x} \ln{\left(\frac{ 1+ x}{\sqrt{1+x^2}}\right)}\ dx=\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx$
using $\ \displaystyle\arctan x\ln(1+x^2)=-2\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
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Taylor expansion of $\sqrt{n-k}$ I am reading a paper which casually assumes the asymptotic $\sqrt{n-k} \simeq \sqrt{n}-\frac{k}{2\sqrt{n}}$. This expression is what Wolfram calls Taylor expansion at infinity and from what I understand we work with $\sqrt{x-k}=\sqrt{x}\cdot \sqrt{1-k/x}$ and then we proceed on doing th... | If you want to show that $\sqrt{x-1/2}\geq \sqrt{x}\left(1-\frac{4}{x}\right),$
the following argument is simpler than bounding Taylor series terms:
(a) for $\frac12\le x<4$: $\sqrt{x-1/2}\geq0\gt \sqrt{x}\left(1-\frac{4}{x}\right);$
(b) for $x>\frac{32}{15}, \frac{15}2>\frac{16}x,$ so $x-\frac12>x-8+\frac{16}x=x(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Ordinary Differential Equation getting two different answers Solve $y(y^2-2x^2)dx+x(2y^2-x^2)dy=0$ and find a particular curve passing through $(1,2)$
My attempt:
1st Solution: Rewrite as
$y(y^2dx+x2ydy)-x(x^2dy+y2xdx)=0$
$\implies xy^2d(xy^2)-x^2yd(x^2y)=0$ (multiplying $xy$)
$\implies (xy^2)^2-(x^2y)^2 = c\,\,\,$ ... | In your second solution, you work out your constant too soon. It is the case that if $|x| = c$, then you can conclude that $ x = c$ because the arbitrary constant can absorb the $\pm$. In the first equation, $c$ nonzero. In the second, it's allowed to be negative.
So I think you should have written
$$x^2y^2(y^2-x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $(\sqrt{3}+i)^{14}+(\sqrt{3}-i)^{14}$ Evaluate $(\sqrt{3}+i)^{14}+(\sqrt{3}-i)^{14}$
I tried by using the De'moivers theorem but I didn't get proper value I get a mess value 4..but I am not sure about the answer can anyone please tell me
| First of all, try to factorise by $2^{14}$ :).
$(\sqrt{3} + i)^{14} + (\sqrt{3} - i)^{14} = 2^{14}(\dfrac{\sqrt{3}}{2} + \dfrac{i}{2})^{14} + 2^{14}(\dfrac{\sqrt{3}}{2} - \dfrac{i}{2})^{14} = 2^{14}\left((\dfrac{\sqrt{3}}{2} + \dfrac{i}{2})^{14} + (\dfrac{\sqrt{3}}{2} - \dfrac{i}{2})^{14}\right)$
Now, use the fact that... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\sqrt{2} x^2 - \sqrt{3} x +k=0$ with solutions $\sin\theta$ and $\cos\theta$, find k
If the equation $\sqrt{2} x^2 - \sqrt{3} x +k=0$ with $k$ a constant has two solutions $\sin\theta$ and $\cos\theta$ $(0\leq\theta\leq\frac{\pi}{2})$, then $k=$……
My approach is suggested below but I am not sure how to continue.
Sin... | $\sin\theta+\cos\theta=\dfrac{\sqrt3}{\sqrt2}$ and $\sin\theta\cos\theta=\dfrac k{\sqrt2}$.
$(\sin\theta+\cos\theta)^2-2\sin\theta\cos\theta=1$
$\dfrac32-\sqrt2 k=1$
$k=\dfrac 1{2\sqrt2}$
| {
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"url": "https://math.stackexchange.com/questions/3264274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $x\mathcal{R}y\iff x^2-y^2=2(y-x)$ is an equivalence relation Prove that $$x\mathcal{R}y\iff x^2-y^2=2(y-x)$$ is an equivalence relation.
Reflexive. For all $x$ we have $x^2-x^2=2(x-x)$, so $x\mathcal{R}x$.
Symmetric. For all $x,y$ we have \begin{align}x\mathcal{R}y&\implies x^2-y^2=2(y-x)\\&\implies(-1)(x^... | Your proof is correct and well-presented. Good job!
Mostly posting this so this question can be finally considered to have an answer. Made it Community Wiki since I have nothing of substance to add.
| {
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Find the solution set of $\frac{3\sqrt{2-x}}{x-1}<2$ Find the solution set of $\frac{3\sqrt{2-x}}{x-1}<2$
Start by squaring both sides
$$\frac{-4x^2-x+14}{(x-1)^2}<0$$
Factoring and multiplied both sides with -1
$$\frac{(4x-7)(x+2)}{(x-1)^2}>0$$
I got
$$(-\infty,-2)\cup \left(\frac{7}{4},\infty\right)$$
Since $x\leq2$ ... | For $\dfrac{3\sqrt{2-x}}{x-1}$ to be defined, $x\le2$ and $x\ne1$.
If $x<1,$ then the expression is negative (i.e., $<0$), so of course it is $< 2$.
If $x>1,$ then, as you showed, the inequality holds when $x>\dfrac74$.
Therefore, the solution set is $x<1$ or $\dfrac74<x\le2$.
| {
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"url": "https://math.stackexchange.com/questions/3266367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find remainder of division of $x^3$ by $x^2-x+1$ I am stuck at my exam practice here.
The remainder of the division of $x^3$ by $x^2-x+1$ is ..... and that of $x^{2007}$ by $x^2-x+1$ is .....
I tried the polynomial remainder theorem but I am not sure if I did it correctly.
By factor theorem definition, provided by W... | Since $$x^3+1 = (x+1)(x^2-x+1)$$ so $$x^3 = (x+1)(x^2-x+1)-1$$ the answer is $-1$.
Similarly for \begin{eqnarray}x^{3n}+1 &=& (x^3+1)\underbrace{\Big((x^3)^{n-1}-(x^3)^{n-2}+...-(x^3)+1\Big)}_{q(x)}\\
&=& (x+1)(x^2-x+1)q(x)\\
\end{eqnarray}
so the answer is again $-1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given three positive numbers $a,\,b,\,c$ . Prove that $(\!abc+ a+ b+ c\!)^{3}\geqq 8\,abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!)$ . Given three positive numbers $a,\,b,\,c$ . Prove that $$(\!abc+ a+ b+ c\!)^{3}\geqslant 8\,abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!).$$
My own problem is given a solution, and I'm looking forward to seeing... | After replacing $a\rightarrow\frac{1}{a},$ $b\rightarrow\frac{1}{b}$ and $c\rightarrow\frac{1}{c}$ we need to prove that
$$(1+ab+ac+bc)^3\geq8abc(1+a)(1+b)(1+c).$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$ and $abc=w^3$.
Thus, we need to prove that:
$$(1+3v^2)^3\geq8w^3(1+w^3+3v^2+3u)$$ or
$$(1+3v^2)\geq8(w^3+w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is the complexity of $\binom{2n}{n} = O(2^n)$? How to find the complexity of $f(n)=\binom{2n}{n}$?
We know that $f(n)=\binom{2n}{n} = \frac{(2n)!}{(n!)^2}$.
Is this $O(n^2)$? What concerns me is $n!$
| First off, you know that:
$\begin{align*}
2^{2 n}
&= (1 + 1)^{2 n} \\
&= \sum_{0 \le k \le 2 n} \binom{2 n}{k} \\
&\ge \binom{2 n}{n}
\end{align*}$
so that $\binom{2 n}{n} = O(2^{2 n})$.
More precise estimates are from Stirling's approximation, in the variant given by Robbins ("A Remark on Stirling's Formu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Using triangle inequality to find $\lim_{(x,y)\to(0,0)}\frac{x^3-x^2y}{x^2+y^2+xy}$ This is an exercise from my textbook where the problem is to find the limit of the function $\frac{x^3-x^2y}{x^2+y^2+xy}$ when $(x,y) \to (0,0)$.
So after changing to polar coordinates and simplifying I get the equivalent function $$r\c... | Because by the triangle inequality we obtain:
$$\left|\frac{x^3-x^2y}{x^2+xy+y^2}\right|=\frac{x^2|x-y|}{x^2+xy+y^2}\leq\frac{x^2(|x|+|y|)}{x^2+xy+y^2}\leq$$
$$\leq\frac{x^2(|x|+|y|)}{x^2-|xy|+y^2}\leq\frac{x^2(|x|+|y|)}{\left(\frac{|x|+|y|}{2}\right)^2}=\frac{4x^2}{|x|+|y|}=4|x|\cdot\frac{|x|}{|x|+|y|}\leq4|x|\rightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3272681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding general solution of $3{\times}3$ system I am given the following:
$$
x'=
\begin{bmatrix}
2 &0 &0 \\
-7&9 &7 \\
0&0 &2
\end{bmatrix}
x
$$
Solving $\det(A-\lambda I)$, I get $\lambda = 2,2,9$. Solving $\det(A-2\lambda)$, I get
\begin{bmatrix}
0&0 &0 \\
-7&7 &7 \\
0&0 &0
\end{bmatrix}
So we have g... | For $\lambda=2$, $(A-2I)v=0$ gives $-x+y+z=0$.
Letting $y=k_1$ and $z=k_2$ you get $x=k_1+k_2$.
Then $v=\begin{pmatrix}k_1+k_2\\k_1\\k_2\end{pmatrix}=k_1\begin{pmatrix}1\\1\\0\end{pmatrix}=k_2\begin{pmatrix}1\\0\\1\end{pmatrix}$.
$\implies v_1=\begin{pmatrix}1\\1\\0\end{pmatrix}$ and $v_2=\begin{pmatrix}1\\0\\1\end{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $ \int_{0}^2 x^{26} (x-1)^{17} (5x-3)dx ~~ ?$ How to evaluate $$ \int_{0}^2 x^{26} (x-1)^{17} (5x-3)dx ~~ ?$$
I have tried to evaluate, $$ I = \int_{0}^2 x^{26} (x-1)^{17} (5x-3)dx = 5\int_{0}^2 x^{27} (x-1)^{17} dx - 3\int_{0}^2 x^{26} (x-1)^{17} dx$$ by parts but I am getting very lengthy expressions ... | $$\frac{d}{dx} x^{a+1} (x-1)^{b+1} = x^a (x-1)^b \left\{(a+b+2)x - (a+1)\right\}$$
So
\begin{align*}
\int x^{26} (x-1)^{17} \left\{(26+17+2)x - (26+1)\right\} dx
&= \int x^{26} (x-1)^{17} (45x - 27) dx \\
&= 9\int x^{26} (x-1)^{17} (5x - 3) dx \\
&= x^{27} (x-1)^{18}
\end{align*}
Thus $$\int x^{26} (x-1)^{17} (5x - 3) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3274580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Integral $\int_{-\pi/2}^{\pi/2}\cos^2(x)\cos(a+b\tan(x))dx$
Evaluate
$$\int_{-\pi/2}^{\pi/2}\cos^2(x)\cos(a+b\tan(x))\mathrm dx$$
I tried the following ideas:
$u=a+b\tan(x)$, $du=b\sec^2(x)dx$
$$\frac{1}{b}\int \cos^4(x)\cos(u)du$$
using $1+\tan^2(x)=\sec^2(x)$
$$b^3\int \frac{\cos(u)du}{(u-a)^4}$$
I am stuck at th... | $$I=\int_{-\pi/2}^{\pi/2}\cos^2(x)\cos(a+b\tan(x))\mathrm dx\overset{x\to -x}=\int_{-\pi/2}^{\pi/2}\cos^2(x)\cos(a-b\tan(x))\mathrm dx$$
Summing up the two integrals from above gives us:
$$ 2I= 2\cos a \int_{-\pi/2}^{\pi/2} \cos^2 x \cos(b\tan x) dx\Rightarrow I=2\cos a\int_0^{\pi/2}\cos^2 x\cos(b\tan x) dx$$
$$\overse... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3274705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can I formally prove that $3\mid 2^n+1\iff n=2m+1,m \in \mathbb N$ How can I formally prove that two to the power of some n all plus one is divisible by three when n is odd (1,3,5,7,...)? Or, another words,$$2^n+1=3k \Leftrightarrow (n=2m+1,m \in \mathbb N)\bigwedge(k \ge 1,k \in \mathbb N)$$It is actually true, fo... | 1) Let $n$ be odd.
Polynomial.$y:= x^n+1^n$, has a zero at $x=-1$.
$(x-(-1)) =(x+1)$ is a factor.
$x^n+1^n= (x+1)p_{n-1}(x)$, where $p_{n-1}$ is a polynomial of degree $n-1$.
$x=2$:
$2^n +1= (2+1)p_{n-1}(2)$, and we are done.
2) Le n be even.
$y= x^n+1^n$;
Is $x=-1$ a zero, i.e $(x+1)$ a factor?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prime factor inversion Define a function $f(n)$ for $n \in \mathbb{N}$ to "invert" the prime
factorization of $n$ in the following sense. Let me start with an example.
If $n= 3564 = 2^2 \cdot 3^4 \cdot 11^1$,
then $f(n) = 2^2 \cdot 4^3 \cdot 1^{11} = 256$,
inverting the base primes and their exponents.
In general, if t... | Your function is OEIS sequence A008477, where I find the comment
For any n, the sequence n, a(n), a(a(n)), a(a(a(n))), ... is eventually periodic with period <= 2 [Farrokhi]. - N. J. A. Sloane, Apr 25 2009
The reference is to
M. Farrokhi, The Prime Exponentiation of an Integer: Problem 11315, Amer. Math. Monthly, 11... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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find the $\cos 40(2\cos 80-1)=?$ find the $\cos 40(2\cos 80-1)=?$
My try :
$\cos 40(2\cos 80-1)=2\cos 40 \cos 80 -\cos40=2\cos40 (2\cos ^240-1)-\cos 40\\=4\cos^340-3\cos 40=\cos 3(40)=\cos (120)=-1/2$
I do not want to use the formula $4\cos^3x -3\cos x=\cos 3x$ .
Now How to solve ?
| $$ \cos(40°)\cdot(2\cdot \cos(80°)-1)\\
= \cos(40°)\cdot(4\cos^2(40°)-3)\\
= 4\cos^3(40°)-3\cos(40°)\\
= \cos(120°)\\
= -\frac{1}{2} $$
Just look how simple the solution is. Why would you want to not use this approach.
Unless you want to prove the triple angle identity from scratch.
Or
$$ \cos(40°)\cdot(2\cdot \cos(80°... | {
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"source": "stackexchange",
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Recurrence relation for $\int_{0}^{\infty} \frac{1}{(1+x^2/a)^n}dx$
Let$$I_{n,a} = \int_{0}^{\infty} \frac{1}{(1+x^2/a)^n}dx$$ where $a>0$.
Show that $$I_{n+1,a} = \frac{2n-1}{2n}I_{n,a}$$
I have tried integrating by parts but it didn't work for me, and I don't know what else can I try. Can anyone please help?
| Just in case you're interested you can solve this for any $n \in \mathbb{R}^+$, $n \geq 1$ using the Beta and by extension the Gamma Function.
Here we will address your integral:
\begin{equation}
I(a,n) = \int_0^\infty \frac{1}{\left(1 + \frac{x^2}{a}\right)^n}\:dx
\end{equation}
We begin by making the substitution... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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What is the radius of convergence for the Taylor expansion of $\frac{e^x\sin(x)}{x^2+25}$ about $x=0$? I would like to know how I could go about finding the radius of convergence for the Maclaurin series of $\frac{e^x\sin(x)}{x^2+25}$.
I am familiar with how to find the radius of convergence of more simple series such... | $$\frac{e^x\sin x }{x^2+25}= e^x \cdot \sin x \cdot f(x)$$
where
$$f(x) = \frac{1}{25} \frac{1}{1 + \left( \frac{x}{5} \right)^2}$$
Since $e^x$ and $\sin x$ have Maclaurin series with infinite radius of convergence, we need look only at $f(x)$ which has radius of convergence $r=5$.
$$\frac{1}{1-x} = 1 + x + x^2 + x^3... | {
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"url": "https://math.stackexchange.com/questions/3279413",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Evaluate $\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx$
How to prove $\ \displaystyle\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx=\frac{3\pi^3}{64}+\frac{\pi}{16}\ln^22-\Im\operatorname{Li}_3(1+i)$
Where $\ \displaystyle\operatorname{Li}_3(x)=\sum_{n=1}^\infty\frac{x^n}{n^3}\ $
is the trilogarithm.
I managed to prove the abov... | Using $\displaystyle\int_0^1\frac{u\ln^nx}{1-ux}\ dx=(-1)^nn!\operatorname{Li}_{n+1}(u)$ .
which can be found in Cornel's book, (Almost) impossible integrals, sums and series.
and with $n=1$, we get $\displaystyle\int_0^1\frac{u\ln x}{1-ux}\ dx=-\operatorname{Li}_2(u)$
Divide both sides by $1+u^2$ then integrate fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Integrating quadratics in denominator I'm following a book on Calculus that introduces partial fraction expansion. They discuss common outcomes of the partial fraction expansion, for example that we are left with an integral of the form:
$$
\int \frac{dx}{x^2+bx+c}
$$
And then we can use complete the square and $u$-sub... | The answer is that this method doesn't always work. In this particular case, we can use the following: Since
$$x^2-8x+1=0 \quad \Leftrightarrow \quad x=\frac{8\pm \sqrt{60}}{2}=4\pm \sqrt{15}$$
this polynomial can be factored as $$\big(x-(4+\sqrt{15})\big)\big(x-(4-\sqrt{15})\big)$$
use partial fraction like this
$$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Designate set $ \lbrace n \in\mathbb N : 2^{n-1} | n! \rbrace $ Actually I found out that if $n = 2^k$, $k\in\mathbb Z_{+}$ we can say that it's true.
Here's the proof:
Let $n = 2^k$, $k\in\mathbb Z_{+}$
Then $\nu _{2}(n!) = \nu _{2}((2^k)!) = \lfloor \frac{2^k}{2} \rfloor + \lfloor \frac{2^k}{4} \rfloor +\dots+ \lfloo... | The power of $2$ in $n$ is given by $\left\lfloor\frac n2\right\rfloor+\left\lfloor\frac n4\right\rfloor+\cdots\left\lfloor\frac n{2^{p}}\right\rfloor$ where $2^p\leq n<2^{p+1}$
For the given condition to be satisfied, we have
$$n-1\leq \left\lfloor\frac n2\right\rfloor+\left\lfloor\frac n4\right\rfloor+\cdots+\left\lf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that the diophantine equation $2x^2-5y^2=7$ has no integer solutions. My attempt: I rewrote it as $2x^2=5y^2+7. 2x^2$ is always even, so in order for the RHS to be even, this means that $5y^2$ must be odd since an odd number plus $7$ is even.
If I evaluate when y is odd, so if $y=2k+1$ for some integer $k$, I ge... | Modulo $7$, $2x^2-5y^2=7$ would mean $2x^2+2y^2\equiv0$ or $x^2+y^2\equiv0$ or $x^2\equiv-y^2$.
Now $x^2, y^2\equiv 0, 1, 2, $ or $4 \pmod 7$, so the only solution would be $x^2\equiv y^2\equiv0\pmod7$.
But this means $7|x,y$, which means $49|2x^2-5y^2=7,$ a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
A finite product : $\prod_{k=0}^{n-1}(1-\frac{1}{n-1+k})$ Find the maximum and minimum of the following products :
$A)$ $\prod_{k=0}^{n-1}(1-\frac{1}{n-1+k})$
$B)$ $\prod_{k=0}^{n-1}(1-\frac{1}{n+1-k})$
My idea is :
$n-1+k>k$ then : $\frac{1}{n-1+k}<\frac{1}{k}$
We obtain :
$\prod_{k=0}^{n-1}(1-\frac{1}{k})$
But I... | HINT: (Assuming $ n \geq 2 $)
$\prod_{k=0}^{n-1}(1-\frac{1}{n+k-1}) = \prod_{k=0}^{n-1}\frac{n+k-2}{n+k-1} = \frac{n-2}{n+(n-1)-1} = \frac{1}{2} \cdot \frac{n-2}{n-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
The integral $\int_{0}^{\infty} \frac{\cot^{-1}\sqrt{1+x^2}}{\sqrt{1+x^2}}dx =\frac{\pi}{2}\ln (1+\sqrt{2})$ At Mathematica the numerical value of the integral
$$\int_{0}^{\infty} \frac{\cot^{-1}\sqrt{1+x^2}}{\sqrt{1+x^2}} dx$$
equals 1.3844.., which is nothing but $\frac{\pi}{2}\ln (1+\sqrt{2})=z$. Also, one of its t... | Here we will address your integral:
\begin{equation}
I = \int_0^\infty \frac{\cot^{-1}\left(\sqrt{1 + x^2} \right)}{\sqrt{1 + x^2}}\:dx
\end{equation}
We first let $x = \tan(s)$:
\begin{align}
I &= \int_0^\frac{\pi}{2} \frac{\cot^{-1}\left(\sqrt{1 + \tan^2(s)} \right)}{\sqrt{1 + \tan^2(s)}}\cdot \sec^2(s)\:ds = \int_0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3286507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solve $x^{3} = 6+ 3xy - 3 ( \sqrt{2}+2 )^{{1}/{3}} , y^{3} = 9 + 3xy(\sqrt{2}+2)^{{1}/{3}} - 3(\sqrt{2}+2)^{{2}/{3}}$ Solve the system of equations for $x,y \in \mathbb{R}$
$x^{3} = 6+ 3xy - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}} $
$ y^{3} = 9 + 3xy(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}}$
I j... | Write the equations as $$
x^{3} = 6+ 3xy - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}}\\
\frac{(xy)^{3}}{x^3} = 9 + 3xy(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}}
$$
Now let $z = xy$ which gives you
$$
f(z) = z^3 - (9 + 3z(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}})(6+ 3z - 3\left ( \sqrt{2}+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3286730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
$\int_{0}^{\infty }{{{x}^{n}}\sin \left( {{x}^{1/4}} \right)\exp \left( -{{x}^{1/4}} \right)dx}=0$ I'd like to show that for all positive integers $n$ we have
$$I\left( n \right)=\int_{0}^{\infty }{{{x}^{n}}\sin \left( {{x}^{1/4}} \right)\exp \left( -{{x}^{1/4}} \right)dx}=0.$$
This is true after some computer experime... | Let
$$I_n = \int_0^\infty x^n \sin (\sqrt[4]{x}) \exp (\sqrt[4]{x}) \, dx, \qquad n \in \mathbb{N}.$$
After enforcing a substitution of $x \mapsto \sqrt[4]{x}$ one has
$$I_n = 4 \int_0^\infty x^{4n + 3} e^{-x} \sin x \, dx.$$
The following useful property for the Laplace transform will now be employed to evalaute the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3289346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Show $\int_{\max\{0,1-c/x\}}^1(1-v)\,dv=\frac{1}{2}\min\{c/x,1\}^2$ Let $c,x\in [0,\infty)$. Using $-\max\{a,b\}=\min\{-a,-b\}$ we get
\begin{align}
\int_{\max\{0,1-c/x\}}^1(1-v)\,dv&= 1-\max\{0,1-c/x\}-\left(\frac{1}{2}-\frac{1}{2}\max\{0,1-c/x\}^2\right)\\
&=\frac{1}{2}-\max\{0,1-c/x\}+\frac{1}{2}\max\{0,1-c/x\}^2\\
... | \begin{align*}
&=\frac{1}{2}\left(1+2\min(0,c/x-1)+\min(0,c/x-1)^2\right) \\
&=\frac{1}{2}\left(1+\min(0,c/x-1)\right)^2 \\
&=\frac{1}{2}\min(1,c/x)^2
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Radius of convergence of $\sum_{n=0}^\infty a_n x^n$, with $a_{n+2} = \frac{n(n+1) a_{n+1} - a_n}{(n+2)(n+1)}, a_2 = -a_0/2$ Problem
Find the radius of convergence of the power series
$$
\sum_{n=0}^\infty a_n x^n
$$
where $a_n$'s are defined by the following recurrence relation
$$
\begin{aligned}
a_{n+2} &= \frac{n(... | The trick is to notice $a_{n+2}=-\frac{f_n}{(n+2)!}(a_0+a_1)$ for $n\ge1$ with $f_1=f_2=1,\,f_n=nf_{n-1}-f_{n-2}$. Asymptotically $f_n\sim nf_{n-1}$, so $\frac{f_n}{(n+2)!}$ is a sequence in which the ratio of consecutive terms $\to1$. Thus the radius of convergence is also $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Given three positive numbers $a,b,c$. Prove that $\sum\limits_{cyc}\sqrt{\frac{a+b}{b+1}}\geqq3\sqrt[3]{\frac{4\,abc}{3\,abc+1}}$ .
Ji Chen. Given three positive numbers $a, b, c$. Prove that
$$\sum\limits_{cyc}\sqrt{\frac{a+ b}{b+ 1}}\geqq 3\sqrt[3]{\frac{4\,abc}{3\,abc+ 1}}$$
Of course, we've to solve it by $uvw$... | By AM-GM $$\sum_{cyc}\sqrt{\frac{a+b}{b+1}}\geq3\sqrt[6]{\prod\limits_{cyc}\frac{a+b}{a+1}}.$$
Thus, it's enough to prove that
$$(a+b)(a+c)(b+c)(3abc+1)^2\geq16a^2b^2c^2(a+1)(b+1)(c+1).$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$(9uv^2-w^3)(3w^3+1)^2\geq16w^6(w^3+3v^2+3u+1)$$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Solving the following limit without L'Hospital's rule: $\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x} $
I have been trying to solve the following limit $$\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x}.$$
I came across the right answer as shown by the steps below, but I would to check if the steps are correct or if so... | The second limit is not really hard: leaving out the $\sin2$ factor, we have
$$
\lim_{x\to0}\frac{\cos(x^2)-\cos x}{x}=
\lim_{x\to0}\frac{\cos(x^2)-1+1-\cos x}{x}
$$
It's easy to show that
$$
\lim_{x\to0}\frac{1-\cos x}{x}=0
$$
so we remain with
$$
\lim_{x\to0}\frac{\cos(x^2)-1}{x}=\lim_{x\to0}\frac{\cos(x^2)-1}{x^2}x
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Find all extrema of a complicated trigonometric function Problem
Find all local extrema for
$$f(x) = \frac{\sin{3x}}{1+\frac{1}{2}\cos{3x}}$$
Attempted solution
My basic approach is to take the derivative, set the derivative equal to zero and solve for x.
Taking the derivative with the quotient rule and a few cases of ... | This function has period $\frac{2\pi}3$ and it is an odd function, so we need to determine its variations only on $\bigl[0,\frac\pi 3\bigr]$.
Now simplifying the derivative, you get
$$f'(x)=\frac{3\bigl(\frac12+\cos 3x\bigr)}{\bigl(1+\frac{1}{2}\cos{3x}\bigr)^2},$$
which has the sign of $\;\frac12+\cos 3x$, so we have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Calculate the maximum value of $x^3y + y^3z + z^3x$ where $x + y + z = 4$ and $x, y, z \ge 0$.
Given non-negatives $x, y, z$ such that $x + y + z = 4$. Calculate the maximum value of $$\large x^3y + y^3z + z^3x$$
As an assumption, the maximum value is $27$, occured when $(x, y, z) = (0, 1, 3)$.
I have a guess about a... | I like the following way.
Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.
Thus, by Rearrangement and AM-GM we obtain:
$$x^3y+y^3z+z^3x=x^2\cdot xy+y^2\cdot yz+z^2\cdot zx\leq a^2\cdot ab+b^2\cdot ac+c^2\cdot bc=$$
$$=b(a^3+c^3+abc)\leq b(a+c)^3=27\cdot b\left(\frac{a+c}{3}\right)^3\leq27\left(\frac{b+3\cdot\frac{a+c}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Matrices commuting with a given $3\times 3$ complex matrix.
Let $A$ be a $3\times 3$ complex matrix. Let $C(A)$ be the vector space of complex matrices that commute with $A$. Show that the complex dimension of $C(A)$ is at least $3$.
I know that this kind of questions has been asked many times on this site. And ther... | By change of basis we can assume that $A$ is in Jordan normal form.
Case 1: $A = \begin{pmatrix}a&0&0\\0&b&0\\0&0&c\end{pmatrix}$.
Then the matrices $\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}, \begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix}, \begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}$ are linearly independen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How can one integrate $\int\frac{1}{(x+1)^4(x^2+1)} dx$?
How can one integrate $\displaystyle\int\frac{1}{(x+1)^4(x^2+1)}\ dx$?
Attempt:
I tried partial fraction decomposition (PFD) and got lost. The method of u-substitution didn't work for me either.
What else can I do? Can one calculate the integral without PFD?
| We use a variant of the Heaviside method. Shift by one and consider
$$\frac{1}{z^4(z^2-2z+2)}\text{.}$$
Develop $1/(z^2-2z+2)$ in series about $z=0$, keeping the remainder exactly as you go:
$$\frac{1}{z^2-2z+2}=\frac{1}{2}+\frac{z}{2}+\frac{z^2}{4}-\frac{z^4}{4(z^2-2z+2)}\text{.}$$
Then
$$\frac{1}{z^4(z^2-2z+2)}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to show the matrix has Rank $\le 5$
I want to show that the following matrix has Rank $\le 5$.
The matrix is
\begin{bmatrix}
2&1&1&1&0&1&1&1\\
1&2&1&1&1&0&1&1\\
1&1&2&1&1&1&0&1\\
1&1&1&2&1&1&1&0\\
0&1&1&1&2&1&1&1\\
1&0&1&1&1&2&1&1\\
1&1&0&1&1&1&2&1\\
1&1&1&0&1&1&1&2
\end{bmatrix}
I found that there is a submatri... | Let
$$
A = \begin{bmatrix} 2 & 1 & 1 & 1 \\ 1 & 2 & 1 &1 \\ 1 &1&2&1\\1&1&1&2\end{bmatrix}, \quad B = \begin{bmatrix} 1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1 \end{bmatrix} = vv^{T}
$$
where $v = [1, 1, 1 ,1]^{T}$. It is easy to see that $rank(B) = 1$. Now our original matrix is
$$
X = \begin{bmatrix} A & 2B - A\\ 2B -A & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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an interesting inequality with Muirhead If $x,y,z>0$ I have to prove that
$\sum\limits_{cyc}^{} \frac { x(x^3 yz+x^2-x y^3 z-yz) }{(1+x y^2)(1+xyz)} \ge 0$ holds. My approach is that from Muirhead's inequality the inequality is true since $(4,1,1)≻(2,3,1)$ and $(2,1)≻(1,1)$. Am I right?
| We need to prove that
$$\sum_{cyc}\frac{x(x^2(1+xyz)-yz(1+xy^2)}{(1+xy^2)(1+xyz)}\geq0$$ or
$$\sum_{cyc}\frac{x^3}{1+xy^2}\geq\frac{3xyz}{1+xyz}.$$
Now, by C-S
$$\sum_{cyc}\frac{x^3}{1+xy^2}=\sum_{cyc}\frac{x^4}{x+x^2y^2}\geq\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(x+x^2y^2)}.$$
Id est, it's enough to prove that
$$(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove equality - Necessity and sufficiency I'm trying to figure out an example from a book with math problems. So here is an example (the equality should be proven)
$$ \sqrt[3]{38 + \sqrt{1445}} + \sqrt[3]{38 - \sqrt{1445}} = 4 $$
The author is suggesting to have $ \sqrt[3]{38 + \sqrt{1445}} + \sqrt[3]{38 - \sqrt{1445}... | Because the substitution $\sqrt[3]{38 + \sqrt{1445}} + \sqrt[3]{38 - \sqrt{1445}}=x$ in the equation
$$38 + \sqrt{1445} + 38 - \sqrt{1445}+3 \sqrt[3]{(38 + \sqrt{1445})(38 - \sqrt{1445})}\left( \sqrt[3]{38 + \sqrt{1445}} + \sqrt[3]{38 - \sqrt{1445}}\right)=x^3$$
is not an equivalent transformation of the last equation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3307439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find all possible values of $\gcd(n^2+3, (n+1)^2+3)$. Let $n \in \mathbb{N}$. Find all possible values of $\gcd(n^2+3, (n+1)^2+3)$.
I began this problem giving some values for $n$ and I found that $\gcd(n^2+3, (n+1)^2+3)=1$ for most of $n$ I tried, but if $n=6$, then $\gcd=13$.
Then I tried to prove that only for $n=... | $$\gcd(n^2+3,(n+1)^2+3)=\gcd(n^2+3,2n+1)=\gcd(4n^2+12,2n+1)=$$
$$=\gcd(4n^2+4n+1-4n+11,2n+1)=\gcd(4n-11,2n+1)=$$
$$\gcd(4n+2-13,2n+1)=\gcd(13,2n+1).$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Natural number solutions of $x^3+y^3=p^2$, $x$ and $y$ are integers, $p$ is prime number. Natural number solutions of $x^3+y^3=p^2$, $x$ and $y$ are integers, $p$ is prime number.
I have found $(1,2,3)$ is a solution and there seem to be no other solution.
Can anyone prove it?
| Hint: Since $p$ is a prime number, we must have : $x+y = p, x ^2 - xy + y^2 = p$ or $x + y = 1, x^2 - xy + y^2 = p^2$ or $x+y = p^2, x^2-xy+y^2 = 1$. Either case is quite simple enough to handle. But for the last case which is the case the OP pointed out as a missing case in the comment.We claim that this case can't h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Evaluating $\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx$ My first instinct was to evaluate the indefinite form of the integral, which I did by substituting $x=\tan t$, therefore yielding
\begin{align}
\int \frac{x\ln x}{(1+x^2)^2} \,dx
&= \int \frac{\tan t \sec^2 t \ln\tan t}{(1+\tan^2 t)^2} \,dt
&& \text{by substi... | Using $x\mapsto\frac1x$ for $x>1$,$$\int_{0}^{\infty}\frac{x^{a}\ln^{b}x}{\left(1+x^{2}\right)^{c}}dx=\int_{0}^{1}\left(\frac{x^{a}\ln^{b}x}{\left(1+x^{2}\right)^{c}}+\left(-1\right)^{b}\frac{x^{2c-a-2}\ln^{b}x}{\left(1+x^{2}\right)^{c}}\right)dx$$vanishes provided $b$ is odd with $c=a+1$, as in your problem viz. $a=b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3311259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
Calculate $\int \frac{dx}{1-\sin^4x}$
Calculate $$\int \frac{dx}{1-\sin^4x}$$
My try:
\begin{align}
\int \frac{dx}{1-\sin^4x}&=\int \frac{(1-\sin^2x)+(1+\sin^2x)dx}{1-\sin^4x}
\\&=\int \frac{dx}{1-\sin^2x}+\int \frac{dx}{1+\sin^2x}
\\&=\tan x+\int \frac{dx}{1+\sin^2x}
\end{align}
How to deal with the second one?
| For the 2nd integral, rewrite the integrand,
$$\frac{1}{1+\sin^2x}=\frac{\sec^2x}{2\sec^2x-1}=\frac{(\tan x)’}{2\tan^2x+1}$$
Then, integrate,
$$\int \frac{d\tan x}{2\tan^2x+1}
=\frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\tan x) + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3313468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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$\epsilon - N$ proof of $\sqrt{4n^2+n} - 2n \rightarrow \frac{1}{4}$ I have the following proof for $\lim_{n\rightarrow\infty} \sqrt{4n^2+n} - 2n = \frac{1}{4}$ and was wondering if it was correct. Note that $\sqrt{4n^2+n} - 2n = \frac{n}{\sqrt{4n^2+n} + 2n}$.
$$\left|\frac{n}{\sqrt{4n^2+n} + 2n} - \frac{1}{4}\right| ... | Like the other commenters have mentioned, the implication is in the wrong direction. Since you want $|\frac{1}{64n}| < \epsilon$, you are required to have $n > \frac{1}{64 \epsilon}$. In other words, $n > \frac{1}{64 \epsilon}$ implies $|\frac{1}{64n}| < \epsilon$. What you have written is the other way around.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Help differentating $f(x) = \sqrt\frac{x^2-1}{x^2+1}$ The equation I'm trying to differentiate is, $ f(x) = \sqrt\frac{x^2-1}{x^2+1}$ and I know the answer is meant to be
$$=\frac{\frac{x\sqrt {x^2+1}}{\sqrt {x^2-1}}-\frac{x\sqrt {x^2-1}}{\sqrt {x^2+1}}}{x^2+1}$$
But when I do the working out I get this
$$=\frac{(x^2-... | You have erroneously transposed the two lone $x$'s in the second-last line into $1/x$'s in the last. (That is, they incongrously turn from being mulitplied with the radicals to dividing them.)
$$=\frac{\color{blue}x(x^2-1)^\frac{-1}{2}\cdot(x^2+1)^\frac{1}{2}-(x^2-1)^\frac{1}{2}\cdot \color{blue}x(x^2+1)^\frac{-1}{2}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}$ Problem:
$$\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}$$
$$\lim_{x \to -\infty} \sqrt{\frac{1}{x^6}}=0$$ so...
$$\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}=\frac{1}{0}$$
The answer is $-1$ and I kn... | $\frac 1 {-\sqrt {\frac 1 {x^{6}}} \sqrt {x^{6}+4}}$ is nothing but $\frac 1 {- \sqrt {\frac 4 {x^{6}}+1}}$ so the limit is $\frac 1 {-1}=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Sum of products of combinatorials In the proof of some proposition, it appears that the following statement should hold:
$$ \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} C^{n+1}_r C^n_{\beta} =2^{2n}. $$
However, using the definition of combinatorials does not help:
$$ \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} C^{n+1}_r C^n_{\be... | Let the sum be $S$. By using the definition $C^n_r=C^n_{n-r}$, we’ll change the sum into:
$\quad S \\= \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} C^{n+1}_r C^{n}_\beta \\ = \sum_{0 \le \beta < r \le n+1} C^{n+1}_r C^{n}_\beta \\ = \sum_{0 \le \beta < r \le n+1} C^{n+1}_{n-r+1} C^{n}_{n-\beta}$
Let $s=n-r+1$ and $\alpha=n-\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3324706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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If $a+b+c = 4, a^2+b^2+c^2=7, a^3+b^3+c^3=28$ find $a^4+b^4+c^4$ and $a^5+b^5+c^5$ I have tried to solve it but cannot find any approach which would lead me to the answer
| Hint
Use Newton’s identities and Vieta's formulas to find $a,b,c$ as roots of a degree 3 polynomial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing a circle from a given set with a tangent line condition I was asked to compute the circles' equations from the set
$$x^2 + y^2 -3x + (k-6)y + (9-3k)=0$$
that fulfill the following request: the circles must be tangent to the line
$$x+y-3=0$$
I started to compute
$$
\begin{cases}
x^2 + y^2 -3x + (k-6)y + (9-3k)... | Once you spotted the obvious solution
$k=3$, you can factor
$$(k-3)^2-8(9-3k)=(k-3)\left[k-3+24\right]=(k-3)(k+21)$$
and so the second solution is $21$, not $24$.
There is another mistake earlier when you develop $x^2=(y-3)^2$, it looks like you implicitly wrote $(y-3)^2=y^2+3^2$, forgetting the rectangular term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can we not use Muirhead's Inequality for proving the following inequality? There was a question in the problem set in my math team training homework:
Show that $∀a, b, c ∈ \mathbb{R}_{≥0}$ s.t. $a + b + c = 1, 7(ab + bc + ca) ≤ 2 + 9abc.$
I used Muirhead's inequality to do the question (you can try out yourself):... | Schur's inequality:
$$a^3+b^3+c^3+3abc \ge a^2(b+c)+b^2(c+a)+c^2(a+b) \iff \\
(a+b+c)^3+9abc\ge 4(a+b+c)(ab+bc+ca) \Rightarrow \\
1+9abc\ge 4(ab+bc+ca) \quad (1)$$
Rearrangement:
$$a+b+c=1 \Rightarrow a^2+b^2+c^2=1-2(ab+bc+ca)\ge ab+bc+ca \Rightarrow \\
1\ge 3(ab+bc+ca) \quad (2)$$
Now add $(1)$ and $(2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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What is the general solution of this equation :$2^x 3^y+1=7^z$ with $x, y , z$ are integers? I have got these triplet solution $(x,y,z)=(1,1,1),(4,1,2)$ for this equation:
$$2^x 3^y+1=7^z$$
with $x, y , z$ are integers, But i can't get general solution of it, I have attempted to use Gausse theorem for the solution... | Clearly $z>0$, since $2^x3^y+1>1$, so $7^z-1\in \mathbb{Z}$ so $x,y\in \mathbb{N}$.
If $y\geq 2$ then $1\equiv _9 7^z$. Since ord$_9(7) = 3$ we have $3\mid z$ so $z=3t$.
Now we can write: $$2^x3^y = (7^3-1)\Big((7^3)^{t-1}+\ldots+7^3+1\Big)$$
Since $7^3-1 = 19\cdot 9\cdot 2$ we see this is impossible, so $y\leq 1$ or ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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tangent inequality in triangle Let $a$, $b$ and $c$ be the measures of angles of a triangle (in radians).
It is asked to prove that
$$\tan^2\left(\dfrac{\pi-a}{4}\right)+\tan^2\left(\dfrac{\pi-b}{4}\right)+\tan^2\left(\dfrac{\pi-c}{4}\right) \ge 1$$
When does equality occur ?
My try :
Letting $u:= \tan\left(\dfrac{\pi... | Let $\dfrac{\pi-A}4=x$ etc.
$\implies4(x+y+z)=3\pi-\pi\iff x+y+z=\dfrac\pi2$
Now as $\tan x,\tan y,\tan z$ are real,
$$(\tan x-\tan y)^2+(\tan y-\tan z)^2+(\tan z-\tan x)^2\ge0$$
$$\implies\tan^2x+\tan^2y+\tan^2z\ge\tan x\tan y+\tan y\tan z+\tan z\tan x$$
Finally $$\tan(x+y)=\tan\left(\dfrac\pi2-z\right)$$
$$\iff\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Finding the integral $\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}$ One may take $x= \cos t$ and get
$$I=\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}= -\frac{1}{4}\int \csc^4(t/2)~ dt=-\frac{1}{4} \int [\csc^2(t/2) +\csc^2(t/2) \cot^2(t/2)]~ dt.$$
$$\Rightarrow I=\frac{1}{2} \left [\cot (t/2)] +\frac{1}{3}\cot^3(t/2)\right]=\frac{(2-x)}... | Let us use $$1-x=\frac{1}{u} \Rightarrow x=1-\frac{1}{u} \Rightarrow dx=\frac{du}{u^2}.$$ Then
$$I=\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}= \int \frac{u du}{\sqrt{2u-1}}.$$
Next use $$2u-1=v^2 \Rightarrow du =v dv.$$ Then
$$I=\frac{1}{2} \int (v^2+1) dv= \frac{v}{2}[\frac{v^2}{3}+1]=\frac{\sqrt{2u-1}}{3} (u+1)=\frac{(2-x)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Show that $\left\vert\frac{\pi}{4} - \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)\right\vert < 0.1$
Show that $$\left\vert\frac{\pi}{4} - \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)\right\vert < 0.1 .$$
I know that $\arctan 1 = \frac{\pi}{4}$ and that the sequence ... | Simple answers above. The alternative is as follows:
*
*We know that $$1-\frac13+\frac15-\frac17+\frac19=\frac{263}{315}$$ which can be calculated by hand.
*We also know that $$\frac\pi4<\frac{3.2}4=0.8\quad\text{whereas}\quad\frac{263}{315}>\frac{252}{315}=0.8$$ so the inequality is equivalent to $$\left(1 - \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Locus of mid point of $AB$
If the family of lines $tx+3y-6=0.$ where $t$ is variable intersect the lines $x-2y+3=0$ and $x-y+1=0$ at point $A$ and $B.$ Then locus of mid point of $AB$ is
what i try
Intersection of line $tx+3y-6=0$ and $x-2y+3=0$ is
$\displaystyle A\bigg(\frac{3}{3+2t},\frac{6+3t}{3+2t}\bigg)$
and ... | Let $M(x,y).$
Thus, $$y-x=\frac{1}{2}\left(\frac{3t+3}{2t+3}+1\right)$$ or
$$t=\frac{6x-6y+6}{4y-4x-5}.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Convergence of recurrence sequence Let $a_1=1$ define $a_{n+1}=\frac{1}{2}(a_n+\frac{2}{a_n})$ . Show that the sequence converges to $\sqrt{2}$. The idea is to show the sequence is bounded and monotone . But how should I do it?
| *
*$a_2=\frac{3}{2}>\sqrt{2}$
*Show if $a_n>\sqrt{2}, \sqrt{2}<a_{n+1}<a_n$
$a_{n+1}-\sqrt{2}=\frac{1}{2}(a_n-\sqrt{2}+\frac{2-\sqrt{2}a_n}{a_n})=(a_n-\sqrt{2})\frac{1-\frac{\sqrt{2}}{a_n}}{2}$
so $0<a_{n+1}-\sqrt{2}<\frac{a_n-\sqrt{2}}2$
So $\lim_{n\to\infty}a_n=\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3331520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find value for $k$ such that $(x^2-k)$ is a factor for $f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6$ Find value for $k$ such that $(x^2-k)$ is a factor for, $$f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6$$
My Try
Since $x^2-k=0$ when we substitute $x=\pm k$ to $f(x)$ it should be equal to $0.$
But ... | Use that $$\frac{f(x)}{x^2-k}=2\,{k}^{2}+3\,kx+2\,{x}^{2}-3\,k-4\,x-5+{\frac {2\,{k}^{3}x+2\,{k}^{3}
+{k}^{2}x-3\,{k}^{2}-7\,kx-5\,k-6\,x+6}{{x}^{2}-k}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How to factorize $\frac{4x^3+4x^2-7x+2}{4x^4-17x^2+4}$? How to factorize and simplify the following?
$$\frac{4x^3+4x^2-7x+2}{4x^4-17x^2+4}$$
I've tried everything I know. Trying to factorize the numerator first then denominator, but I get no where. Usual identities like $(x+y)^2=x^2+2xy+y^2$ don't work either, and ne... | Just above the line that says GCD it shows the quotients, which you would want because of wishing to reduce the fraction
$$ \left( 4 x^{3} + 4 x^{2} - 7 x + 2 \right) $$
$$ \left( 4 x^{4} - 17 x^{2} + 4 \right) $$
$$ \left( 4 x^{3} + 4 x^{2} - 7 x + 2 \right) = \left( 4 x^{4} - 17 x^{2} + 4 \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Evaluate $\tan 195^{\circ}$ without using the calculator How to evaluate $\tan 195^{\circ}$ without using the calculator, and how to give the answer in the form $a+b \sqrt{3}$, where $a$ and $b$ are integers?
| The Tangent Identities:
$$\tan (2\theta) = \frac{2\tan\theta}{1-\tan^2\theta} $$
$$\tan (\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1-\tan\alpha\tan\beta} $$
$$\tan (\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1+\tan\alpha\tan\beta} $$
$$\tan 195^\circ = \tan (180^\circ + 15^\circ)$$
$$\tan(180^\circ) = 0$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3337434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
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Proving the equation using binomial theorem I want to prove this theorem using Binomial theorem and I've got trouble in understanding 3rd step if anyone knows why please explain :) Prove that sum:
$\sum_{r=0}^{k}\binom{m}{r}\binom{n}{k-r}=\binom{m+n}{k}$
1st step:
$(1+y)^{m+n}=(1+y)^m(1+y)... | By the equation of 2nd step,
Left:
$$\binom{m+n}{0}y^{0}+\binom{m+n}{1}y^{1}+\cdots+\binom{m+n}{m+n-1}y^{m+n-1}+\binom{m+n}{m+n}y^{m+n}$$
The coefficient of $y^{k}$ is $\binom{m+n}{k}$
Right:
$$(\binom{m}{0}y^{0}+\cdots+\binom{m}{m}y^{m})(\binom{n}{0}y^{0}+\cdots+\binom{n}{n}y^{n}) $$
The term $y^{k}$ come from
$$\bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3338885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Inequality $\left(1-\frac{x(1-x)+y(1-y)}{1-x+1-y}\right)^2+(1-\frac{x+y}{2})^2\geq (1-x)^2+(1-y)^2$ it's a little problem found by myself
let $x,y\neq 1$ .be real numbers then we have :
$$\left(1-\frac{x(1-x)+y(1-y)}{1-x+1-y}\right)^2+\left(1-\frac{x+y}{2}\right)^2\geq (1-x)^2+(1-y)^2$$
I tried Jensen's inequalit... | For $x+y\neq2$ by AM-GM we obtain:$$\left(1-\frac{x(1-x)+y(1-y)}{1-x+1-y}\right)^2+\left(1-\frac{x+y}{2}\right)^2=$$
$$=\left(\frac{(1-x)^2+(1-y)^2}{2-x-y}\right)^2+\left(\frac{2-x-y}{2}\right)^2\geq$$
$$\geq2\sqrt{\left(\frac{(1-x)^2+(1-y)^2}{2-x-y}\right)^2\left(\frac{2-x-y}{2}\right)^2}= (1-x)^2+(1-y)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Volume of a solid created by an extended tetrahedron
Every edge of a tetrahedron with length $p$ is extended through the vertices by $p$.
Now all 12 points create a new solid $J$ of which I seek the volume dependent on the volume of the tetrahedron in the centre.
With some help the solution becomes clear:
The whole ... | There are only three types of faces for the new solid. They are either equilateral triangles of side lenghth $p$ that are $\dfrac{7}{4} h$ away from the center, where $h = p \sqrt{\dfrac{2}{3}} $, or equilateral triangles of side $2p$ that are $ \frac{5}{4} h $ from the centroid, or rectangles of sides $ p $ and $ 2 p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3344522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Evaluating $\lim_{x\to -\infty} \frac{4x^3+1}{2x^3 + \sqrt{16x^6+1}}$ In finding this limit:
$$\lim_{x\to -\infty} \frac{4x^3+1}{2x^3 + \sqrt{16x^6+1}}$$
I've been told to divide all the terms by $-x^3$ (as opposed to $x^3$ if we take the limit as $x \to \infty$), and go from there. Dividing by a negative $x^3$ doesn'... | You can do it either way. If you divide by $x^3$ you have to remember that $x^3$ is a negative number and $x^3 =-\sqrt{x^6}$. you get
$\frac {4+\frac 1{x^3}}{2 +\frac {\sqrt{16x^6 + 1}}{x^3}}=$
$\frac {4+\frac 1{x^3}}{2+\frac {\sqrt{16x^6+1}}{-\sqrt{x^6}} }=$
$\frac {4+\frac 1{x^3}}{2-\sqrt{16+\frac 1{x^6}}}$
It's eas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3344902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Eigenvalue of $A-aI_3$ Question: Let $A=\begin{pmatrix} a+1 & 1 & 1 \\ 1 & a+1 & 1 \\ 1 & 1 & a+1\end{pmatrix}$. Show that $A-aI_3$ has eigenvalue of 3. Also find eigenvector.
My thinking:
I know that we have to apply characteristic polynomial $|A- \lambda I|$ to find the eigenvalue. I don't understand the part $A-a... | Alternative answer: $A-aI_3=\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{pmatrix}$ clearly has eigenvalue $0$
with independent eigenvectors $\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}.$
The sum of the eigenvalues is the trace of the matrix, which i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Note: $x$ and $y$ are obtuse angles.
My attempt that is not simple is as follows.
Expand both known constraints, so we have
\be... | As $90<x,y<180$ and $\cos(x+y)>0$
$$270<x+y<360\implies\sin(x+y)=-\sqrt{1-(1/3)^2}$$
Again, $-90<x-y<90^\circ\implies\sin(x-y)=\pm\sqrt{1-(1/5)^2}$
Finally $$\sin2y=\sin(x+y+(x-y))=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3346306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Problems with understanding a given Sum Identity In a textbook I found the following equation without any explaination:
\begin{align}
\sum_{\substack{j, k=1,\\ j\neq k}}^n\frac{1}{(x-x_j)(x-x_k)} - \frac{3}{2}\left( \sum_{j=1}^n \frac{1}{x-x_j}\right)^2 \\
= -\frac{1}{2}\sum_{j=1}^n \left( \frac{1}{x-x_j}\right)^2 - ... | Consider the following expression
\begin{eqnarray*}
\left( \sum_{j=1}^{n} \frac{1}{x-x_j} \right)^2 = \left( \sum_{j=1}^{n} \frac{1}{x-x_j} \right) \left( \sum_{k=1}^{n} \frac{1}{x-x_k} \right).
\end{eqnarray*}
Now when you calculate the product we have either $j \neq k$ or $j=k$, so we have
\begin{eqnarray*}
\left( \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3347222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Odd or even function. Determine whether the following function is odd or even :
$$f(x) =(\cos{x} + \sin{x} +1)(\cos{x} + \sin{x} -1)$$
My turn :
$$f(x) = (\cos{x}+ \sin{x})^2 -1= \sin{2x}$$
Then f is an odd function
Another turn :
$$f(-x) = (\cos{x}- \sin{x} +1)(\cos{x} -\sin{x} -1)$$
But the last form seems neither od... | While it seems that everything in the original post is correct, you want to show that:
$f(x) = -f(-x)$
$(\cos x + \sin x + 1)(\cos x +\sin x - 1) = -(\cos (-x) +\sin (-x) + 1)(\cos (-x) + \sin (-x) -1)\\
(\cos x + \sin x + 1)(\cos x +\sin x - 1) = -(\cos x -\sin x + 1)(\cos x - \sin x -1)\\
(\cos x + \sin x)^2 - 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3347610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Eigen Decomposition Check I am following the wiki entry on eigen dicomposition with the following matrix:
$$A = \begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix}$$
I wish to find a diagonalizing matrix T.S.
$$T^{-1}AT=\Lambda$$
$$AT=T\Lambda$$
where,
$$\Lambda = \begin{pmatrix}
x & 0 \\
0 & y \\
\end{pmatri... | Your answer is correct except that you forgot to switch the elements on the diagonal of $T$ when finding its inverse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $xu_x+yu_y+zu_z=4u$ \begin{cases}
xu_x+yu_y+zu_z=4u\\
u(x,y,1)=xy\\
\end{cases}
Using Lagrange method we get:
$$\frac{x}{y}=c_1, \frac{y}{z}=c_2, \frac{z}{\sqrt[4]{u}}=c_3$$
So the general solution is $$u=\frac{z^4}{c_3}$$?
| Converting to spherical coordinates, we get
$$ru_r = 4u \implies u = f(\theta,\phi)r^4$$
Then plugging in our boundary condition at $r\cos\theta = 1$ and $xy=r^2\sin^2\theta\sin\phi\cos\phi$, we can get
$$f(\theta,\phi)r^4 = r^2\sin^2\theta\sin\phi\cos\phi\cdot(1)=r^2\sin^2\theta\sin\phi\cos\phi\cdot (r^2\cos^2\theta)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\displaystyle \lim_{x \to 2}\frac{\sqrt{x^3+1}-\sqrt{4x+1}}{\sqrt{x^3-2x}-\sqrt{x+2}}$ I came across this question.
Evaluate the limit $$ \lim_{x \to 2}\frac{\sqrt{x^3+1}-\sqrt{4x+1}}{\sqrt{x^3-2x}-\sqrt{x+2}}$$
I tried rationalizing the denominator, substitution, yet nothing seems to cancel out with the d... | Using a small trick that I enjoy.
Let $x=t+2$ to make
$$y=\frac{\sqrt{x^3+1}-\sqrt{4x+1}}{\sqrt{x^3-2x}-\sqrt{x+2}}=\frac{\sqrt{4 t+9}-\sqrt{t^3+6 t^2+12 t+9}}{\sqrt{t+4}-\sqrt{t^3+6 t^2+10 t+4}}$$ and now use the binomial expansion or Taylor series around $t=0$.
We have
$$\sqrt{4 t+9}=3+\frac{2 t}{3}-\frac{2 t^2}{27}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Problem with approximating function I want to approximate the expression $$ \frac{a z +1/2 b z^2 + c z^3}{a + b z + d z^2}.$$
This should be approximately equal to $z - \frac{b}{2a}z^2.$ But no matter the approach, I do not get this answer. Anyone who sees what I should have done?
Edit: some users had a fair point, in ... | $$\begin{array}{rcl}\dfrac{az + \dfrac{1}{2} bz^2 + cz^3}{a + bz + dz^2} &=& \dfrac{az + bz^2 +dz^3-\dfrac{1}{2}bz^2-(d-c)z^3}{a +bz+dz^2}\\
&=&\dfrac{az + bz^2 +dz^3}{a+bz+dz^2}-\dfrac{\dfrac{1}{2}bz^2 +(d-c)z^3}{a+bz+dz^2}\\
&=&z -\dfrac{\dfrac{1}2b\left(z^2 +\dfrac{2(d-c)}{b}z^3\right)}{a+bz+dz^2}\\
&=& z - \dfrac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3352667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Formulas for Sequences Removing Multiples of 2, 3, and 5 First off, I am a programmer so please excuse if some of the terms I use are not the correct mathematical terms. I was working on devising a function to improve one of my prime number generation algorithms. With this in mind, I first set out to find the formulas ... | my way is with an 8 value addition sequence 2, 6, 4, 2, 4, 2, 4, 6. and Repeat.
Start @ -1
(-1) [+ 2] = 1 [+6] = 7 [+4] = 11
thus (-1) 1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 77, 79, 83...
and so on
Repeating this addition sequence from start position of -1
[2, 6, 4, 2, 4, 2, 4, 6]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3355398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\frac{x^3 - x + 10}{x^2-9}$ is surjective I'm looking for an elegant way to prove that this function is surjective. One way to solve this would be to set $\frac{x^3-x+10}{x^2-9} = y$ and solve for $x$ in terms of $y$ but this would require using the cubic formula to solve. Is there any other nice way?
| One can also approach the problem by way of Descartes' Rule of Signs.
render $\frac{x^3-x+10}{x^2-9} = y$ as $x^3-x^2y-x+(10+9y)=0$ for $|x|\ne 3$ and let
\begin{eqnarray}
f(x)&=&x^3-x^2y-x+(10+9y)\\
f(-x)&=&-x^3-x^2y+x+(10+9y)
\end{eqnarray}
Then for $y$ in the interval $\left(\infty, -\frac{10}{9}\right)$ Descartes' ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Equation of circle touching three circles, two of which are intersecting Find the equation of the circle which is tangentially touching three given circles: $x^2+y^2=49$, $x^2+(y-3.5)^2=49/4$, and $y^2+(x-3.5)^2=49/4$.
By tangentially i mean, it touches the smaller two circle externally and the larger one internally.
T... |
Let $a$ be the radius of the inscribed circle and apply the cosine rule to the triangle formed by the centers of the large circle, one of the small circles and the inscribed circle,
$$(a+3.5)^2=(7-a)^2+3.5^2-2\cdot 3.5 \cdot (7-a)\cos 135^\circ$$
which yields
$$a= \frac{7(\sqrt{2}+1)}{3\sqrt{2}+1}$$
Therefore, the eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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If $a_{k}=2^{2^k}+2^{-2^k}$ then evaluate $\prod_{k=1}^\infty\left(1-\frac{1}{a_{k}}\right)$ If $$a_{k}=2^{2^k}+2^{-2^k}$$ then evaluate $$\prod_{k=1}^\infty\left(1-\frac{1}{a_{k}}\right)$$
I tried using Sophie-Germaine Identity about factorisation for $x^4+4$ but it did not work
| Note that
$$1 - \frac{1}{a_k} = \frac{1 - 2^{-2^k} + 4^{-2^k}}{1 + 4^{-2^k}} = \frac{1 + 8^{-2^k}}{(1 + 2^{-2^k})(1 + 4^{-2^k})}$$
and for $|x| < 1$, we have
$$\prod_{k=1}^{\infty} (1 + x^{2^k}) = \lim_{n \to \infty} \prod_{k=1}^n (1 + x^{2^k}) = \lim_{n \to \infty} \frac{1 - x^{2^{n+1}}}{1 - x^2} = \frac{1}{1-x^2}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Integration of a rational function involving a quadratic
Evaluate the integral: $$\int\frac{3x}{(1-4x-2x^2)^2}\ dx$$
Here is my work:
*
*Complete the square on the denominator: $$(1-4x-2x^2)=(1-2(2x+x^2+1-1))=(3-2(x+1)^2)$$
*Insert back into denominator. Use substitution $u=x+1$; $x=u-1$; $dx=du$. $$\int\frac{3(u... | Here is how to carry out the second integral. With the substitution $u=\frac{\sqrt3}{\sqrt2}\sin\theta$,
$$I=-\int\frac3{(3-2u^2)^2}\ du$$
$$=-\frac{\sqrt3}{\sqrt2}\int\frac{3\cos\theta}{(3-2(\frac{3}{2}\sin^2\theta))^2}\ d\theta$$
$$=-\frac{\sqrt3}{\sqrt2}\int\frac{3\cos\theta}{(3-3\sin^2\theta)^2}\ d\theta
=-\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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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.