Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Rationalize $\left(\sqrt{3x+5}-\sqrt{5x+11} -\sqrt{x+9}\right)^{-1}$ I was trying to find if there a method similar to multiplying and dividing by the conjugate $$\frac{1}{\sqrt{3x+5}-\sqrt{5x+11} - \sqrt{x+9}},$$ but that doesn't seem to work here. Also, is there a method of multiplying by a conjugate for roots other ... | You can apply it twice:
$$\frac1{\sqrt a-\sqrt b-\sqrt c}=\frac{(\sqrt a-\sqrt b)+\sqrt c}{(\sqrt a-\sqrt b)^2-c}=\frac{\sqrt a-\sqrt b+\sqrt c}{a+b-c-2\sqrt{ab}},$$
then
$$\frac{\sqrt a-\sqrt b+\sqrt c}{a+b-c-2\sqrt{ab}}=\frac{(\sqrt a-\sqrt b+\sqrt c)(a+b-c+2\sqrt{ab})}{(a+b-c)^2-4ab}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/934774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Two methods to integrate? Are both methods to solve this equation correct?
$$\int \frac{x}{\sqrt{1 + 2x^2}} dx$$
Method One:
$$u=2x^2$$
$$\frac{1}{4}\int \frac{1}{\sqrt{1^2 + \sqrt{u^2}}} du$$
$$\frac{1}{4}log(\sqrt{u}+\sqrt{{u} +1})+C$$
$$\frac{1}{4}log(\sqrt{2}x+\sqrt{2x^2+1})+C$$
Method Two
$$u=1+2x^2$$
$$\frac{... | Another one just for fun ;-)
Set $x=\frac{1}{\sqrt 2}\sinh(u)$ then,
$$\int^t\frac{x}{\sqrt{1+2x^2}}dx=\frac{1}{ 2}\int^{\text{argsinh}(\sqrt 2 t)}\frac{\cosh(x)\sinh(u)}{\sqrt{1+\sinh^2(u)}}du=\frac{1}{ 2}\int^{\text{argsinh}(\sqrt2 t)}\frac{\cosh(u)\sinh(u)}{\cosh u}du=\frac{1}{ 2}\int^{\text{argsinh}(\sqrt2 t)}\sinh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
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Calculate the binomial sum $ I_n=\sum_{i=0}^n (-1)^i { 2n+1-i \choose i} $ I need any hint with calculating of the sum
$$
I_n=\sum_{i=0}^n (-1)^i { 2n+1-i \choose i}.
$$
Maple give the strange unsimplified result
$$
I_n={\frac {1/12\,i\sqrt {3} \left( - \left( \left( 1+i\sqrt {3} \right)
^{2\,{\it n}+2} \right) ^... | Generating Functions
Let's compute the generating function of $\displaystyle\sum_{k=0}^n(-1)^k\binom{n-k}{k}$:
$$
\begin{align}
\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k\binom{n-k}{k}x^n
&=\sum_{k=0}^\infty(-1)^kx^k\sum_{n=k}^\infty\binom{n-k}{k}x^{n-k}\\
&=\sum_{k=0}^\infty(-1)^kx^k\frac{x^k}{(1-x)^{k+1}}\\
&=\frac1{1-x}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
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commutative matrix multiplication of nxn matrices? If there are two matrices A and B that are both nxn matrices, will AB = BA always?
Is there a way to have those two matrices so that AB = 0 but BA ≠ 0?
| No. I can give a simple counterexample:
$\begin{pmatrix}1 & 2 \\ 3 &4\end{pmatrix}$ * $\begin{pmatrix}4 & 3\\2 & 1\end{pmatrix}$ = $\begin{pmatrix}8 & 5\\20 & 13\end{pmatrix}$
while
$\begin{pmatrix}4 & 3\\2 & 1\end{pmatrix}$ * $\begin{pmatrix}1 & 2\\3&4\end{pmatrix}$ = $\begin{pmatrix}16 & 20\\8 & 8\end{pmatrix}$.
Two ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/938020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Sums of binomial coefficients Does anyone know something about the following sums?
$$
S_m(n)=\sum\limits_{k=o}^n(-1)^k{mn\choose mk}
$$
Notice that $S_m(n)=0$ for odd $n$, so we only consider $S_m(2n)$. It holds that $S_0(2n)=1$, $S_1(2n)=0$, $S_2(2n)=(-4)^n$, $S_3(2n)=\frac{2}{3}(-27)^n$, but $S_4(2n)$ is no longer a ... | Start by restating the problem: we seek to evaluate
$$S_m(n) = \sum_{k=0}^n (-1)^k {nm\choose km}
= \sum_{k=0}^n (-1)^k {nm\choose nm-km}.$$
Introduce the integral representation
$${nm\choose nm-km}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{nm}}{z^{nm-km+1}} \; dz.$$
This gives the following for the sum:
$$\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/940028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Convergent or divergent $\sum\limits_{n=0}^{\infty }{\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}}$ \begin{align}
& \sum\limits_{n=0}^{\infty }{\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}} \\
& \text{ordering} \\
& a_{n}=\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}=\frac{1\... | An explicit computation is even better. We have:
$$\frac{(2n-1)!!}{(2n+2)!!}=\frac{1}{(2n+2)\,4^n}\binom{2n}{n}=\frac{1}{4^n}\binom{2n}{n}-\frac{1}{4^{n+1}}\binom{2n+2}{n+1}$$
hence your series is telescopic and we have:
$$\sum_{n=1}^{N}\frac{(2n-1)!!}{(2n+2)!!}=\frac{1}{2}-\frac{1}{4^{N+1}}\binom{2N+2}{N+1}=\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/940087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Which one is true? (CSIR) Let $a,b,c$ be a positive real number such that $b^2+c^2<a<1$. Let
$A=\begin{bmatrix} 1&b&c\\ b&a & 0\\ c & 0 & 1\end{bmatrix}$. Then
(1) all eigen values of $A$ are positive
(2) all eigenvalues of $A$ are negative
(3) all eigenvalues of $A$ are either positive or negative
(4) all eigenvalue... | Using submatrices:
$$\Delta_1=\det\{[1]\}=1>0$$
$$\Delta_2=\det\left\{\left[\begin{array}{cc}1&b\\b&a\end{array}\right]\right\}=a-b^2>a-(b^2+c^2)>0$$
$$\begin{array}{rcl}
\Delta_3&=&\det\left\{\left[\begin{array}{ccc}1&b&c\\b&a&0\\c&0&1\end{array}\right]\right\}\\
&=&(a+0+0)-(ac^2+b^2+0)\\
&=&a-ac^2-b^2>a-c^2-b^2>0
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/945594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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finding a function from given function here is a function for:
$f(x-\frac{\pi}{2})=\sin(x)-2f(\frac{\pi}{3})$
what is the $f(x)$?
I calculate $f(x)$ as follows:
$$\begin{align}
x-\frac{\pi}{2} &= \frac{\pi}{3} \Rightarrow x= \frac{5\pi}{6} \\
f(\frac{\pi}{3}) &=\sin\frac{5\pi}{6}-2f(\frac{\pi}{3}) \\
3f(\frac{\pi}{3}) ... | Shift the argument by $\pi/2$:
$$f(x)=\sin(x+\frac\pi2)-2f(\frac{\pi}3).$$
Then plug $\pi/3$:
$$f(\frac{\pi}3)=\sin(\frac{5\pi}6)-2f(\frac{\pi}3).$$
Solve for $f(\pi/3)$:
$$f(x)=\sin(x+\frac\pi2)-\frac23\sin(\frac{5\pi}6).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/947010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How do I show that $Var(Y) = n\frac{(\theta _{1} - \theta _{2})^{2}}{(n+1)^{2} (n+2)}$? My original pdf is $f(y) = \frac{n (y_{n} - \theta_{1})^{n-1}}{(\theta_{2} - \theta_{1})^{n}}$ for $\theta_{1} < y < \theta_{2}$.
After using U-substitution, I obtain $E(Y) = \frac{n \theta_{2} + \theta_{1}}{(n+1)}$.
For variance o... | Instead of $\mathbb{E}(Y^2)$ consider
$$
c_2 =\mathbb{E}\left(\left(Y-\theta_1\right)^2\right) = \int_{\theta_1}^{\theta_2} \frac{n}{\left(\theta_2-\theta_1\right)^n} \left(y-\theta_1\right)^{n+1} \mathrm{d} y = \frac{n}{n+2} \left(\theta_2-\theta_1\right)^2
$$
Similarly
$$
c_1 = \mathbb{E}\left(\left(Y-\theta_1\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/947692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Trig limit without L'Hospital Rule: $\lim\limits_{x\to0}\frac{\tan x-\sin x}{x^3}$ I'm really getting stuck on this and would appreciate some help:
$$
\lim_{x\ \to\ 0}\left[\,\tan\left(\,x\,\right) - \sin\left(\,x\,\right) \over x^{3}\,\right]
$$
I know I need to change $\tan\left(\,x\,\right)$ into $\sin\left(\,x\,\... | $$\begin{align}\lim_{x\to 0}\frac{\tan x-\sin x}{x^3}&=\lim_{x\to 0}\frac{\sin x-\sin x\cos x}{x^3\cos x}\\&=\lim_{x\to 0}\frac{\sin x}{x}\cdot \frac{1-\cos x}{x^2}\cdot \frac{1}{\cos x}\\&=\lim_{x\to 0}\frac{\sin x}{x}\cdot\frac{\sin^2x}{x^2(1+\cos x)}\cdot\frac{1}{\cos x}\\&=\lim_{x\to 0}\frac{\sin x}{x}\cdot \left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/950236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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If $a_{n+1}=a_n+\frac1{a_n}$, then $a_n/n$ converges to $0$ Let $a_{n+1}=a_n+\dfrac1{a_n}$, with $a_n=1$.
Prove $\lim \limits_{n\to \infty }\left(\dfrac{a_n}{n}\right)=0$.
Now I already know that it is monotonically increasing and that $a_n\to \infty$ as $n\to \infty$.
I thought of using Cauchy here, but I don't know ... | If you know that $a_n \to \infty,$ you can apply Stolz theorem to obtain
$$\lim_{n \to \infty} \frac{a_n}{n} = \lim_{n \to \infty} \frac{a_{n+1} - a_n}{(n+1)-n} = \lim_{n \to \infty} \frac{1}{a_n} = 0.$$
Edit: I'm adding solution which doesn't use Stolz theorem.
Let $\varepsilon > 0$. Since $a_n \to \infty$, there is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/951274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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$ \cos {A} \cos {B} \cos {C} \leq \frac{1}{8} $ In an acute triangle with angles $ A, B $ and $ C $, show that
$ \cos {A} \cdot \cos {B} \cdot \cos {C} \leq \dfrac{1}{8} $
I could start a semi-proof by using limits: as $ A \to 0 , \; \cos {A} \cos {B} \cos {C} $ becomes big (as we want), but $ A+B+C $ becomes too small... | $$y=2\cos A\cos B\cos C=[\cos(A-B)+\cos(A+B)]\cos C=[\cos(A-B)-\cos C]\cos C$$
$$\implies\cos^2C-\cos(A-B)\cos C+y=0$$ which is Quadratic Equation in $\cos C$
As $C$ is real $\implies\cos C$ is real,
the discriminant $\cos^2(A-B)-4y\ge0\iff y\le\dfrac{\cos^2(A-B)}4\le\dfrac14$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Simplest way to integrate $\int \frac{1}{1+\tan x}dx,$ $$\int \frac{1}{1+\tan x}dx,$$
A substitution like $t = \tan x, \;dt = (1+t^2)dx$ etc. immediately comes to mind, but I find this method a bit lengthy with the partial fractions. Is there a more concise solution to this?
| Generalization:
For $\int\dfrac{a\cos x+b\sin x}{c\sin x+d\cos x}dx,$ where at least one of $a,b$ is non-zero
write $a\cos x+b\sin x=A(c\sin x+d\cos x)+B\dfrac{d(c\sin x+d\cos x)}{dx}\ \ \ \ (1)$
So, $\int\dfrac{a\cos x+b\sin x}{c\sin x+d\cos x}dx=A\int\ dx+B\dfrac{d(c\sin x+d\cos x)}{(c\sin x+d\cos x)dx}dx$
$=Ax+\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/953691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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For which natural n is $3^n+5^n$ divisible by $3^{n-1}+5^{n-1}$? For which natural n is $3^n+5^n$ divisible by $3^{n-1}+5^{n-1}$? I only got that $3^n+5^n=8k$ when $n$ is odd. How to solve this one?
| Let $A_n=5^n+3^n,$
If $d$ divides $A_m,A_{m-1}$
$d$ will divide $A_m-3A_{m-1}=2\cdot5^{m-1}$
$d$ will divide $5A_{m-1}-A_m=2\cdot3^{m-1}$
So, $d$ will divide $(2\cdot3^{m-1},2\cdot5^{m-1})=2(3,5)^{m-1}=2$
So, the necessary condition for $A_{m-1}\mid A_m$ is $: A_{m-1}$ must divide $2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/955040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Solve for $x$ when $\sin 2x = \cos x$ where $ x$ is in the domain $ [0, 2\pi]$ Quick question on trig (which I haven't dealt with in a long time):
since $\sin 2x = 2\sin x\cos x $
$2\sin x\cos x = \cos x$
$2\sin x\cos x/\cos x = 1$
$\sin x = 1/2$
since $\sin x = 1/2$ in quadrants $1$ and $2$, $x = \pi/6$ and $x = 5\pi... | Generalized Solution :
$$\sin ax=\cos bx$$ for example, $\sin\dfrac{3x}2=\cos5x$ can also be addressed this way
Method $\#1:$
$$\cos bx=\sin\left(\frac\pi2-bx\right)\implies\sin ax=\sin\left(\frac\pi2-bx\right)$$
$$\implies ax=n\pi+(-1)^n\left(\frac\pi2-bx\right)$$ where $n$ is any integer
If $n$ is odd, $=2m+1$(say)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/955971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Intuition for Euler's Partition Theorem Euler's Partition Theorem states the following:
Every number has as many integer partitions into odd parts as into distinct parts.
I played around with small examples (I wrote out the partitions into odd/distinct parts for integers 1 through 10), but can't develop an intuition ... | You can set up a one-to-one correspondence (aka a bijection) between the partitions of $N$ into odd parts and the partitions of $N$ into distinct parts. I will just illustrate with an example; I leave it to you to formalize it and verify that it works.
Here is one way to partition the number $N=863$ into odd parts: par... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $(ac+bd)^2 + (ad-bc)^2 \geq 144$ , if $a+b=4, c+d=6$. I got that $(ac)^{2}+(bd)^{2}+(ad)^{2}+(bc)^{2} \geq 144$. How is this one solved?
| $$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$$
Method $\#1:$
Assuming $a,b$ are real,
$(a-b)^2\ge0\iff a^2+b^2\ge2ab\iff2(a^2+b^2)\ge(a+b)^2=4^2$
Similarly, for $c^2+d^2$
Method $\#2:$
Assuming $a,c$ are real,
$a^2+b^2=a^2+(4-a)^2=2(a^2-4a+8)=2[(a-2)^2+4]\ge2\cdot4$
Similarly, for $c^2+d^2$
See also: Brahmagupta-Fibonacc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculation of $\int\frac1{\tan \frac{x}{2}+1}dx$ Calculation of $\displaystyle \int\frac{1}{\tan \frac{x}{2}+1}dx$
$\bf{My\; Try}::$ Let $\displaystyle I = \displaystyle \int\frac{1}{\tan \frac{x}{2}+1}dx$, Now let $\displaystyle \tan \frac{x}{2}=t\;,$ Then $\displaystyle dx=\frac{2}{1+t^2}dt$
So $\displaystyle I = 2\... | Set $t=\dfrac{x}{2}$, then multiply the integrand by $\cos t$, we get
\begin{equation}
\frac{2\cos t}{\cos t+\sin t}
\end{equation}
then let
\begin{equation}
I=\int\frac{2\cos t}{\cos t+\sin t}dt
\end{equation}
and
\begin{equation}
J=\int\frac{2\sin t}{\cos t +\sin t}dt
\end{equation}
Find $I+J$ and $I-J$, where $I-J$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/958679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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The convergence of an infinite seqeunce Suppose that
$$
a_n = \prod_{k=n}^{\infty}\left(1 - \frac{1}{k^2}\right),
$$
for $n \geq 2$. How can we show that
$$
\lim_{n \to \infty} a_n = \lim_{n \to \infty}\prod_{k=n}^{\infty}\left(1 - \frac{1}{k^2}\right) = 1?
$$
Thanks very much.
| Consider the following.
Method 1
The product is
\begin{align}
\prod_{k=n}^{\infty} \left(1 - \frac{1}{k^2}\right) = \frac{ \prod_{k=2}^{\infty}\left(1 - \frac{1}{k^2}\right) }{ \prod_{k=2}^{n-1} \left(1 - \frac{1}{k^2}\right) }
\end{align}
for which
\begin{align}
\lim_{n \rightarrow \infty} \, \prod_{k=n}^{\infty}\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/962679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove using induction principles $$\forall{n,a>1}:\;\sum\limits_{k=1}^{2^n-1}\frac{1}{k^a}\;\leq\left(\frac{1-2^{n(1-a)}}{1-2^{1-a}}\right)$$
For any fixed value of $a > 1$.
Induction step:
$$\sum_{k=1}^{2^{n+1} - 1} \frac{1}{k^a} = (\sum\limits_{k=1}^{2^n-1}\frac{1}{k^a}) + \frac{1}{(2^n)^a} + \frac{1}{(2^n + 1)^a} + ... | Ellaborating on the hint:
Notice that $$\frac{1 - 2^{n(1-a)}}{1 - 2^{1-a}} = 1 + 2^{1-a} + \ldots + 2^{(n-1)(1-a)}$$
and that
$$\frac{1}{(2^n)^a} + \frac{1}{(2^n + 1)^a} + \ldots + \frac{1}{(2^{n+1} - 1)^a} \leq 2^n \cdot \frac{1}{(2^n)^a}.$$
Now
$$\sum_{k=1}^{2^{n+1}-1}\frac{1}{k^a} = \sum_{k=1}^{2^{n}-1}\frac{1}{k^a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/962912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this exponential equation? $2^{2x}3^x=4^{3x+1}$. I haven't been able to find the correct answer to this exponential equation:
$$\eqalign{
2^{2x}3^x&=4^{3x+1}\\
2^{2x} 3^x &= 2^2 \times 2^x \times 3^x\\
4^{3x+1} &= 4^3 \times 4^x \times 4\\
6^x \times 4 &= 4^x \times 256\\
x\log_6 6 + \log_6 4 &= x\log_64 ... | I think the problem with your solution might be here:
$2^2x * 3^x => 2^2 * 2*x * 3^x$
$4^{3x+1} => 4^3 * 4^x * 4$
since $a^{bc} = (a^b)^c$ and not $a^b * a^c$. So for instance:
$4^{3x+1} => 4^x * 4^x * 4^x * 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/964962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Inverse Laplace of $\frac{\sinh{x\sqrt{s}}}{s^2\sinh{\sqrt{s}}}$ What is the inverse Laplace of $\frac{\sinh{x\sqrt{s}}}{s^2\sinh{\sqrt{s}}}$?
Using the residues, I can calculate the residues at $s_n=2n\pi i$, but I have problem in calculating residue at $s=0$.
The final answer should be:
$f(t)=\frac{1}{6}x(x^2-1)+xt+... | First of all, the poles you are thinking of are at $s=-n^2 \pi^2$. ($s=0$ is a removable pole of the sinh terms.)
Also note that, even though there are square roots in the sinh terms, we need not worry about taking branches because any odd phase behavior will cancel in the fraction. Therefore, what we have is a strai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/965257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof of $\sqrt{n^2-4}, n\ge 3$ being irrational Is the proof of $n\ge 3$, $\sqrt{n^2-4} \notin \mathbb{Q} \ \text{correct}$?
$\sqrt{n^2-4} \in \mathbb{Q}
\\
\sqrt{n^2-4} = \frac{p}{q}
\\
(\sqrt{n^2-4})^2 = \left(\frac{p}{q}\right)^2
\\
q^2\left( n^2-4\right)=p^2
\\
\text{p is divisible by} \left (n^2-4 \right)
\Rig... | The fault is when you deduce that $p$ is divisible by $n^2-4$ from the fact that
$$
q^2(n^2-4)=p^2
$$
which only implies that $p^2$ is divisible by $n^2-4$.
This would be true if $n^2-4$ is prime, but it isn't in general. For instance, if $n=4$ and $p=6$, $p^2=36$ is divisible by $n^2-4=12$, but $6$ is not divisible by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/965390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $\int \frac{x^2 - x}{x^2 +x + 1}dx$ I know the answer is $x - \ln|x^2 + x + 1|$ but I don't understand how to get it. Its in the partial fraction decomposition section of homework. The way the homework worked it is like this...
$$
\int \frac{2x+1}{x^2 + x + 1} \, dx
$$
I see they split it up and got the derivative... | Suppose you have
$$
\frac{\text{some polynomial}}{x^2 + x + 1}.
$$
Do long division and get
$$
\text{some polynomial} + \frac{\text{some first-degree polynomial}}{x^2+x+1}.
$$
In this case you get
$$
1 + \frac{-2x-1}{x^2+x+1}.
$$
First degree is the derivative of second degree, so $\dfrac{d}{dx}(x^2+x+1) = 2x+1$. So
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/966080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding the oblique asymptote of: Given $$f(x)=\frac{x^2+1}{(x+1)^{\frac{1}{2}}}$$ how would you find the oblique asymptote of that?
| When $x$ is large, $x^2+1 \approx x^2$ and $\sqrt{x+1} \approx \sqrt{x}$. So, for large values of $x$, $$f(x)=\frac{x^2+1}{(x+1)^{\frac{1}{2}}}\approx x^{3/2}$$ I suppose that you consider "oblique" in a very extended way.
More precise answers could be obtained writing $$f(x)=\frac{x^2+1}{(x+1)^{\frac{1}{2}}}=\frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/967700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Trigonometry equation with arctan Solve the following equation: $\arctan x + \arctan (x^2-1) = \frac{3\pi}{4}$.
What I did
Let $\arctan x = \alpha, \arctan(x^2-1) = \beta$, $\qquad\alpha+\beta = \frac{3\pi}{4}$
$\tan(\alpha+\beta) = \tan(\frac{3\pi}{4}) = -1$
$$\frac{\tan\alpha + tan\beta}{1-\tan\alpha\tan\beta} = \fr... | Liek Show that $2\tan^{-1}(2) = \pi - \cos^{-1}(\frac{3}{5})$,
$$\arctan x+\arctan(x^2-1)=\begin{cases} \arctan\left(\dfrac{x+x^2-1}{1-x(x^2-1)}\right) &\mbox{if } x(x^2-1)<1 \\\pi+ \arctan\left(\dfrac{x+x^2-1}{1-x(x^2-1)}\right) & \mbox{if } x(x^2-1)> 1. \end{cases} $$
Now, $-\dfrac\pi2\le\arctan(z)\le\dfrac\pi2$
$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/970578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_{0}^{1}\left(\frac{\ln{(1+x)}}{1+x}\right)^n dx $ I wonder if this integral $$\int_{0}^{1}\left(\frac{\ln{(1+x)}}{1+x}\right)^n dx \quad n=1,2,3,...$$ admits a general formula for integers $n$. I've found $$\int_{0}^{1}\frac{\ln{(1+x)}}{1+x} dx = \left[ \frac{1}{2}\left(\ln(1+x)\right)^2 \right]_0^1=\f... | Result:
$$\int^1_0\frac{\ln^n(1+x)}{(1+x)^n}{\rm d}x=\frac{n!}{(n-1)^{n+1}}-\frac{n!}{2^{n-1}}\sum^n_{j=0}\frac{\ln^{j}{2}}{j!(n-1)^{n-j+1}}$$
$\text{for $n\in \mathbb{N}$, $n\geq2$}.$
Derivation:
\begin{align}
\small{\int^1_0\frac{\ln^n(1+x)}{(1+x)^n}{\rm d}x}
&\small{=\int^2_1\frac{\ln^n{x}}{x^n}{\rm d}x\tag1}\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/971699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How to get tangent of inverse of curve?? Ok so my question is. Let $ f(x)=(1/7)x^3+21x-1.$ and let y=g(x) be the inverse function of f. Determine all points on the graph of the inverse function g so that the tangent line is perpendicular to the straight line $ y=-42x+4. $There are two point $P(x1,y1)$ and $Q(x2,y2)$ wh... | The curve $y=g(x)$ has equation that can be written in implicit form as
$$\frac{y^3}{7}+21y-1=x.$$
Differentiate. We get
$$\frac{dy}{dx}\left(\frac{3y^2}{7}+21\right)=1.\tag{1}$$
We want the tangent line to $y=g(x)$ to be perpendicular to a line with slope $-42$. So we want the tangent line to $y=g(x)$ to have slope ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/972658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Given $\sqrt{1+x} < 1 +0.5x$ for all $x>0$, prove that $\sqrt{1+x} > 1 + 0.5x - 0.125x^2$ Given $\sqrt{1+x} < 1 +0.5x$ for all $x>0$,
prove that $\sqrt{1+x} > 1 + 0.5x - 0.125x^2$
Im thinking along the lines of binominal expansion as $\sqrt{1+x} = 1 + 0.5x - 0.125x^2 + ...$
But how im not sure how to continue
| You do not have a calculus tag, so the first solution below may be inappropriate.
First Solution: Let
$$f(x)=\sqrt{1+x}-(1+0.5x-0.125x^2).$$
We want to show that $f(x)\gt 0$ if $x\gt 0$. Note that $f(0)=0$. We will show that $f(x)$ is increasing.
We have
$$f'(x)=\frac{1}{2\sqrt{1+x}}-(0.5-0.25 x).$$
We will show that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/972878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Property of greatest integer function I came across the following mathematical statement in a proof. Can somebody tell me which property of greatest integer function makes it possible?
$x + y - \lfloor x + y \rfloor + z - \lfloor x + y - \lfloor x + y \rfloor + z \rfloor = x + y - \lfloor x + y \rfloor + z - \lfloor x... | In general, if $N$ is an integer, then
$$\lfloor N+\alpha\rfloor=N+\lfloor\alpha\rfloor.$$
Proof : Let $N+\alpha=M+\beta$ where $M$ is an integer and $0\le \beta\lt 1$. Then, since we have
$$\alpha=M-N+\beta\Rightarrow \lfloor\alpha\rfloor=M-N,$$
we have
$$\begin{align}\lfloor N+\alpha\rfloor&=\lfloor M+\beta\rfloor\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/973861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that a random variable has a Poisson distribution? If $X$ is a random variable whose PMF, $p_X(x)$, is positive on and only on non-negative integers. How can we show that $X$ has a Poisson distribution if
$$\begin{align*}
p_X(x) = \frac{3}{x} p_x(x-1) && x = 1, 2, 3, \ldots
\end{align*}$$
| $$
p_X(4) = \frac 3 4 p_X(3) = \frac 3 4 \cdot \frac 3 3 p_X(2) = \frac 3 4 \cdot \frac 3 3\cdot \frac 3 2p_X(1) = \frac 3 4 \cdot \frac 3 3 \cdot \frac 3 2 \cdot\frac 3 1p_X(0) = \frac{3^4 p_X(0)}{4!}.
$$
As with $4$, so with other positive integers (or a bit less informally: prove by induction that $p_X(x)= \dfrac{3^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/977489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show $\sum_n \frac{z^{2^n}}{1-z^{2^{n+1}}} = \frac{z}{1-z}$ Show $\displaystyle\sum_{n=0}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}} = \frac{z}{1-z}$ for $|z|<1$.
This is an additional problem for my complex analysis class and I've attempted it for a few hours but ended up taking wrong routes. All of my attempts I haven't us... | $$\sum_{n=0}^{\infty}\frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\cdots$$
Add $\displaystyle \frac{-z}{1-z}$ to both sides. It's Telescoping series:
$$\frac{-z}{1-z}+\sum_{n=0}^{\infty}\frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{-z}{1-z}+\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/978266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
Hint: solve for $b^2$ in terms of $a^2$ and then solve for $a$
I've attempted the question but I don't think I've done it correctly:
$$
\begin{align*}
b^2 &= 4 - a^2\\
b &= \sqrt{4-a^2}
\end{align*}
$$
Therefore,
$$
\... | $$a^3+3a^2(ib)-3ab^2+(ib)^3=8$$
Equating the real & the imaginary parts,
$$a^3-3ab^2=8;3a^2b-b^3=0\iff b(3a^2-b^2)=0$$
Case $\#1:b=0\implies a^3=8\implies a=2$
Case $\#2:3a^2-b^2=0\iff b^2=3a^2$
and we have $a(a^2-3b^2)=8\implies a[a^2-3(3a^2)]=8\iff a^3=-1\implies a=-1$ as $a$ is real
$\implies b^2=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/979252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 3
} |
Proof of Nesbitt's Inequality: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$? I just thought of this proof but I can't seem to get it to work.
Let $a,b,c>0$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$
Proof: Since the inequality is homogeneous, WLOG $a+b+c=1$. So $b+c=1-a$, and... | I'm sorry but AM-GM Inequality is all that I know,
So, By using AM-GM Inequality for $ (a+b) , (b+c) , (c+a)$ and
$\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$
We get
$$\frac{(a+b)+(b+c)+(c+a)}{3}
\ge\sqrt[3]{(a+b)(b+c)(c+a)}\tag{1}$$
$$\frac{\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}}{3}
\ge\sqrt[3]{\frac{1}{(b+c)(c+a)(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/980751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Hermite Polynomials and Brownian motion I am asked to prove the following :
Let $B_t$ be a standard brownian motion.
The $n$th Hermite polynomial is $\displaystyle H_n(t,x)=\frac{(−t)^n}{n!} e^{x^2/(2t)} \frac{d^n}{dx^n}e^{-x^2/(2t)}$.
Show that the $H_n$ play the role that the monomials $\dfrac{x_n}{n!}$ play in ord... | First of all, the statement
$$dH_{n+1}(t,B_t) = H_n(t,B_t)$$
doesn't make sense. Or can you explain what you mean by it? I guess, it should read
$$dH_{n+1}(t,B_t) = H_n(t,B_t) \, dB_t;$$
at least that's what I prove in this answer.
Lemma 1 $$H_{n+1}(t,x) = \frac{x}{n+1} H_n(t,x) - \frac{t}{n+1} H_{n-1}(t,x). \tag{0}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/981606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
solve indefinite integral I have this indefinite integral $\int 3 \sqrt{x}\,dx$ to solve.
My attempt:
$$\int 3 \sqrt{x}\,dx = 3 \cdot \frac {x^{\frac {1}{2} + \frac {2}{2}}}{\frac {1}{2} + \frac {2}{2}}$$
$$\int 3 \sqrt{x}\,dx = 3 \frac{x^{\frac {3}{2}}}{\frac {3}{2}} = \frac{2}{3} \cdot \frac{9}{3} x^{\frac {3}{2}}$$
... | You failed in $\displaystyle\frac{2\cdot9}{3\cdot3} = \frac{18}{3}$, actually $\displaystyle\frac{2\cdot9}{3\cdot3} = \frac{2}{3}\cdot3 = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/982826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Without actually calculating the value of cubes find the value of $(1)^3+(2)^3+2(4)^3+(-5)^3+(-6)^3$. Also write the identity used Without actually calculating the value of cubes find the value of $(1)^3+(2)^3+2(4)^3+(-5)^3+(-6)^3$.
Also write the identity used
| use the sum of cubes of continuous integers: $\dfrac{n^2 (n+1)^2}{ 4}$ , this is the same as $1^3 + 2^3 +\cdots+n^3$. You can use this. But you need to have coefficients of 1 in front of your cubes, so split the $(2)4^3=4^3+4^3$. Leave the negatives at the end and do the same thing
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/985039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Understanding the Cholesky decomposition I'm attempting to understand the Cholesky decomposition via the following site: http://en.wikipedia.org/wiki/Cholesky_decomposition
If I have a matrix, say
$$A = \begin{bmatrix}
2 & -1 & 0\\
-1 & 2 & -1\\
0 & -1 & 2\end{bmatrix},$$ then I'd like to use the Cholesky Algorithm... | Wikipedia’s probably not the ideal place to learn about the Cholesky decomposition or of the various algorithms that can be used to generate the decomposition. I’d recommend reading a textbook on numerical linear algebra or general book on numerical mathematics. That being said, I’ve gone ahead and fixed up your step... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/986642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the roots of $(w−1)^4 +(w−1)^3 +(w−1)^2 +w=0$
Write down, in any form, all the roots of the equation $z^5 − 1 = 0$
Hence find all the roots of the equation
$$(w−1)^4 +(w−1)^3 +(w−1)^2 +w=0$$
and deduce that none of them is real
My Try:
I know how to do the first part:
$$z^5=1=cos 2\pi k + i sin 2\pi k$$
$... | Your approach looks good: $z^5-1=0$ has five solutions that are evenly distributed around the unit circle, as defined by your trigonometric solution. These are the black points in the following figure:
As you say, the left hand side factors into
$$z^5-1 = (z-1)(z^4+z^3+z^2+z+1).$$
If we divide off the first order ter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/988877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the coefficient of $x^{24}$ in $(1 + x + x^2 + x^3 + x^4 + x^5)^8$ I'm not sure how to go about doing this. Do I find the ways to add up to 24 using the exponents with repetition? Is the multinomial theorem useful here? I also have a feeling that generating functions might be useful here, but I can't see how. Any ... | $\bf{My\; Solution::}$ Let $S = 1+x+x^2+x^3+........+x^5......(1)$
Multiply both side by $x\;,$ We get
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;xS = x+x^2+x^3+..............x^6........(2)$
Now Subtract $(1)$ and $(2)\;,$ we get
$\Rightarrow \displaystyle S(1-x) = 1-x^6\Rightarrow S = \frac{(1-x^6)}{(1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/989862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result R when tested w... | Here I have taken the compact answer by /u/Quanto (to whom all credit is due) and simply filled in some intermediate steps. His trick was to (twice) differentiate inside the integral ( a special case of the Leibniz integral rule ).
Using the known definite integral:- $\int_0^{2\pi}\frac{1}{E+F\cos x}\mathrm{d}x= \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/990813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
Beautiful triangle problem Circle, inscribed in $ABC$, touches $BC, CA, AB$ in points $A', B', C'$. $AA' BB', CC'$ intersect at $G$. Circumcircle of $GA'B'$ crosses the second time lines $AC$ and $BC$ at $C_A$ and $C_B$. Points $ A_B, A_C, B_C,B_A, C_A, C_B$ are concyclic.
Looks straightforward, but I'm struggling to... | This is a coordinate-based approach, making heavy use of tools from projective geometry.
Without loss of generality, you can choose the coordinate system in such a way that the inscribed circle is the unit circle. On that you can use a rational parametrization, i.e. choose $a,b,c\in\mathbb R$ such that $A'=(1-a^2,2a)/(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/991368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Double Integration with change of variables I am having trouble with the following double integral:
$$\iint\limits_D(x^2+y^2) \;dA$$
where $D$ is given by the region enclosed by the curves
*
*$xy=1$
*$xy=2$
*$x^2-y^2 =1$
*$x^2-y^2 =2$
I have tried changing variables to $u=x^2+y^2$, $v=x^2-y^2$ (and other chang... | Consider the variables
\begin{align*}
u = xy, && v = x^2-y^2.
\end{align*}
We obtain that $x=\frac{u}{y}$. Therefore:
$$y^2=x^2-v,$$
$$y^2 -\frac{u^2}{y^2}=-v,$$
$$\frac{y^4-u^2}{y^2} =-v,$$
$$y^4+vy^2-u^2=0,$$
$$y^2 = \frac{-v\pm \sqrt{v^2-4(-u^2)}}{2} = \frac{-v \pm \sqrt{4u^2+v^2}}{2}.$$
From which we obtain:
$$x^2=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find the sum of the series $\sum_{k=2}^\infty \frac{1}{k^2-1}$? I have this problem :
$$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1}$$
My solution
$$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1} = -\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k+1} -\frac{1}{k-1} = -\frac{1}{2}[(\frac{1}{3}-1)+(\frac{1}{4}-\frac{1}{2})+(\frac{1}{5}-\f... | $$-\frac{1}{2}[(\frac{1}{3} - 1) + (\frac{1}{4} - \frac{1}{2}) + (\frac{1}{5} - \frac{1}{3}) + (\frac{1}{6} - \frac{1}{4})\ldots $$
$$-\frac{1}{2}[-1 - \frac{1}{2} + (\frac{1}{3} - \frac{1}{3}) + (\frac{1}{4} - \frac{1}{4}) + (\frac{1}{5} - \frac{1}{5}) \ldots $$
$$-\frac{1}{2}[-1 - \frac{1}{2}] = \frac{3}{4} $$
Notice... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/993890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Convergent series and comparison test Show that the series converges, but not absolutely: $\displaystyle \sum_{n=1}^\infty\Bigg(\exp\Bigg(\frac{(-1)^n}{n}\Bigg)-1\Bigg)$.
This is what I did so far:
$\exp\Bigg(\dfrac{(-1)^n}{n}\Bigg)-1=\exp(\frac{1}{n})-1>0$ when $n$ is even
$\exp\Bigg(\dfrac{(-1)^n}{n}\Bigg)-1=\exp(\fr... | Use the fact that $\exp(x) = 1 + x + x^2/2! + x^3/3!+ \cdots$ for any $x$, so
$$
\exp(\frac{(-1)^n}{n}) = 1 + \frac{(-1)^n}{n} + \frac{1}{2!}\left(\frac{(-1)^n}{n}\right)^2 + \dots.
$$
Then
$$
\Bigg|\exp\Bigg(\dfrac{(-1)^n}{n}\Bigg)-1\Bigg|\leq\frac{2}{n} \to 0.
$$
Indeed
$$
\Bigg|\exp\Bigg(\dfrac{(-1)^n}{n}\Bigg)-1\B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the minimum distance between the curves $y^2-xy-2x^2 =0$ and $y^2=x-2$ How to find the minimum distance between the curves $y^2-xy-2x^2 =0$ and $y^2=x-2$
Let $y^2-xy-2x^2 =0...(1)$ and $y^2=x-2...(2)$
In equation (1) coefficient of $x^2 =-2; y^2=1, 2xy =\frac{-1}{2}$
We know that a second degree equation where $a... | When the shortest distance is achieved, the normal to the two curves coincide.
We are lucky that the first curve degenerates in $x+y=0$ and $2x-y=0$, giving two normal directions $(1,1)$ and $(2,-1)$.
Then, taking the gradient of the second expression, we express parallelism:
$$(-1,2y)\times(1,1)=-1-2y=0,$$
$$y=-\frac1... | {
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"answer_id": 2
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Evaluate the sum. Evalute the following sum: $ 1 + 2 + 2 + 3 + 4 + 4 + 5 + 6 + 6 + . . . .+ (n-1) + n + n$.
I tried doing it but I keep getting the wrong answer. I've used known sums to solve it.
| Assuming $n$ is even
$$S=1 + 2 + 2 + 3 + 4 + 4 + 5 + 6 + 6 + . . . .+ (n-1) + n + n$$
$$S=(1+2+3+\cdots+(n-1)+n)+(2+4+6+\cdots+n)$$
$$S=\sum_{k=1}^{n} k +\sum_{k=1}^{n/2} 2k$$
$$S=\frac{n(n+1)}{2}+\frac{n}{2}\left(\frac{n}{2}+1\right)=
\frac{2n(n+1)}{4}+\frac{n(n+2)}{4}
=\frac{3n^2+4n}{4}$$
| {
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Find a matrix transformation mapping $\{(1,1,1),(0,1,0),(1,0,2)\}$ to $\{(1,1,1),(0,1,0),(1,0,1)\}$
Find a matrix transformation mapping $\{(1,1,1),(0,1,0),(1,0,2)\}$ to $\{(1,1,1),(0,1,0),(1,0,1)\}$.
Is the answer
$$
\begin{bmatrix}1& 0& -1\\0& 1& 1\\0& 0& 1\end{bmatrix}?
$$
I understand the concept of Matrix Tran... | The columns of the matrix tell you where it sends the standard basis vectors. For instance if I am interested in the third column then I need to determine what the action of our linear operator is on the column vector ,
$$\left[ \begin{array}{c}0 \\ 0 \\ 1 \end{array}\right].$$
This vector can be written as a linear co... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
constructive counting math problem about checkers on a checkerboard In how many ways can we place anywhere from 0 to 9 indistinguishable checkers on a 3x3 checkerboard (no more than one checker per square), such that no row or column contains exactly 1 checker?
I tried reducing the checkerboard to a 2x2 checkerboard an... | We can also do this using inclusion-exclusion. There are $\binom3j\binom3k$ ways to choose $j$ particular rows and $k$ particular columns for which the restriction is violated. Due to the symmetry with respect to $j$ and $k$, we have $4(4+1)/2=10$ cases to consider:
$j=0$, $k=0$, $2^9=512$ arrangements
$j=0$, $k=1$: $3... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
find common ratio of $\sum_{k=1}^\infty \frac{1}{k(k+1)}$ I have this problem, I need to find the sum.
$$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\cdots+\frac{1}{k(k+1)}$$
The problem is that the ratio is not conclusive, Any idea how to find the ratio?
Thanks!
| $$\sum_{k=1}^n\frac{1}{k(k+1)} = (1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\cdots+(\frac{1}{n}-\frac{1}{n+1})=\frac{n}{n+1}$$
So
$$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \lim_{n\to \infty} \sum_{k=1}^n\frac{1}{k(k+1)}= \lim_{n\to \infty} \frac{n}{n+1}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/997525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Simplify $f(x) = \arctan(2x) + \arctan(3x)$
Simplify $f(x) = \arctan(2x) + \arctan(3x)$
I had a go at it and this is what I got to :
We have: $-\pi<\arctan(2x)+\arctan(3x)<\pi$
Let $a=\arctan(2x)$ and $b=\arctan(3x)$
Then I cut it into different intervals :
$-\pi<a+b<-\pi/2$
$-\pi/2<a+b<\pi/2$
$\pi/2<a+b<\pi$
In the... | Drawing 2 right triangles with common side of length 1 and having adjacent sides of lengths 2x and 3x, respectively, the Law of Cosines gives
$25x^2=(1+4x^2)+(1+9x^2)-2\sqrt{1+4x^2}\sqrt{1+9x^2}\cos\theta\;\;\;$ where $\theta=\arctan 2x+\arctan 3x$,
so $\displaystyle\cos\theta=\frac{1-6x^2}{\sqrt{(1+4x^2)(1+9x^2)}}$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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continuous functional calculus; spectrum of an self adjoint element in a c*algebra Let A be a C$^*$-Algebra, $a\in A$ selfadjoint and $\|a^2-a\|<\frac{1}{4}$. The claim is: $\sigma(a)\subseteq (-\frac{1}{2},\frac{1}{2}) \cup (\frac{1}{2},\frac{3}{2})$ and there is a projection $p\in A$ such that $\|a-p\|<\frac{1}{2}$.... | What you did is correct, $||a^2 - a || < \frac{1}{4}$ means $\sigma(a^2 - a) \subset (-1,1)$, that is the function $t^2 - t$ takes $\sigma(a)$ to the subset $(-1,1)$, and so, $\sigma(a) \subset (\frac{1-\sqrt{2}}{2},\frac{1}{2}) \cup (\frac{1}{2} , \frac{1+\sqrt{2}}{2}) \subset (-\frac{1}{2},\frac{1}{2})\cup (\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Inequality proof by induction, what to do next in the step I have to prove that for $n = 1, 2...$ it holds: $2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}}$
Base: For $n = 1$ holds, because $2\sqrt{2}-2 < 1$
Step: assume holds for $n_0$.
$2\sqrt{n+2} - 2 < 1 + \frac{1}{\sqrt2} + \f... | Let's prove this first: x>0, : $ 2*\sqrt{x} +\frac{1}{\sqrt{x}} > 2*\sqrt{x+1} $
This is true iif : $ (2*\sqrt{x} +\frac{1}{\sqrt{x}})^2 > (2*\sqrt{x+1})^2 $
You get : $ 2*\sqrt{x} +\frac{1}{\sqrt{x}} > 2*\sqrt{x+1} $ <=> $ 4x + \frac{1}{x} +4 > 4*(x+1) $ <=> $ \frac{1}{x} > 0 $ which is true, so the inequality is tru... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Trigonometric equation, missing some solutions I'm missing part of the answer, and I'm not quite sure why. The given answer doesn't even seem to hold...
Solve for x: $$\tan 2x = 3 \tan x $$
First some simplifications:
$$\tan 2x = 3 \tan x $$
$$\tan 2x - 3 \tan x = 0$$
$$\frac{\sin 2x}{\cos 2x} - \frac{3 \sin x}{\cos ... | By your method, $$\frac{\sin x(2 \cos^2x - 3 \cos 2x)}{\cos 2x \cos x} = 0$$
Either $\sin x=0\implies x=n\pi$ where $n$ is any integer
else $2 \cos^2x - 3 \cos 2x=0\iff 1+\cos2x-3\cos2x=0\iff\cos2x=\dfrac12=\cos\dfrac\pi3$
The rest is like my other answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Getting $(1-3x^6 + 3x^{12} - x^{18}) \sum_{i=0}^{\infty} \binom{i+2}{2} x^{i}$ from $(\frac{1-x^{6}}{1-x})^3$ using generating functions I'm not sure how to get $(1-3x^6 + 3x^{12} - x^{18}) \sum_{i=0}^{\infty} \binom{i+2}{2} x^{i}$ from $(\frac{1-x^{6}}{1-x})^3$.
I know the following series.
$$\frac{1}{1-x}=(1+x+x^2 + ... | Hint. From
$$
\frac{1}{1-x}=\sum_{i=0}^{\infty}x^i,\quad |x|<1,
$$
by differentiating, you get
$$
\frac{1}{(1-x)^2}=\sum_{i=1}^{\infty}ix^{i-1},\quad |x|<1,
$$
or, changing $i-1$ to $i$,
$$
\frac{1}{(1-x)^2}=\sum_{i=0}^{\infty}(i+1)x^{i},\quad |x|<1,
$$
differentiating again and changing $i-1$ to $i$ again :
$$
\frac{... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to integrate the dilogarithms? $\def\Li{\operatorname{Li}}$
How can you integrate $\Li_2$? I tried from $0 \to 1$
$\displaystyle \int_{0}^{1} \Li_2(z) \,dz = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)}$
$$\frac{An + B}{n^2} + \frac{D}{n+1} = \frac{1}{n^2(n+1)}$$
$$(An + B)(n+1) + D(n^2) = 1$$
Let $n = -1, \implies D = 1... | $$
\begin{align}
\int_0^1\mathrm{Li}_2(x)\,\mathrm{d}x
&=\int_0^1\sum_{k=1}^\infty\frac{x^k}{k^2}\,\mathrm{d}x\\
&=\sum_{k=1}^\infty\frac1{(k+1)k^2}\\
&=\sum_{k=1}^\infty\left(\frac1{k^2}-\frac1k+\frac1{k+1}\right)\\
&=\zeta(2)-\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)\\
&=\frac{\pi^2}{6}-1
\end{align}
$$
where... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
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Volume of rotated region (integration) Let $T$ be a right-angled triangular region with vertices $(0,−b)$,$(1,0)$ and $(0,a)$ where $a$ and $b$ are positive numbers. When $T$ is rotated about the line $x=2$, it generates a solid with volume $V=\dfrac{410\pi}{27}$
Find $a$ and $b$.
Really having trouble with this one. I... | Using Pappus's Theorem, $V=A(2\pi\rho)=\frac{1}{2}(a+b)(2\pi(2-\frac{1}{3}))=\pi(a+b)\frac{5}{3}=\frac{410}{27}\pi$,
$\hspace{.3 in}$ so $a+b=\frac{82}{9}$.
By the Pythagorean Theorem,
$a^2+1+b^2+1=(a+b)^2,\;\;\;$ so $2=2ab\implies ab=1\implies b=\frac{1}{a}$.
Then $a+\frac{1}{a}=\frac{82}{9}\implies \frac{a^2+1}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to show for $\alpha\in (0,1)$, any $f\in C^\alpha([0,1]/{\sim})$ has a Fourier series $S_nf$ uniformly converging to $f$ Technically homework(a midterm) but its over and I'm itching to know the solution. I know how to show it for $\alpha>1/2$ (the Fourier series will converge absolutely), but apparently its true fo... | Suppose that $|f(x)|\le C$ and $|f(x)-f(y)|\le C|x-y|^\alpha$.
Express the Difference Using the Dirichlet Kernel
Using the Dirichlet Kernel, we get
$$
\begin{align}
|S_nf(x)-f(x)|
&=\left|\,\int_{-1/2}^{1/2}\frac{\sin((2n+1)\pi y)}{\sin(\pi y)}[f(x-y)-f(x)]\,\mathrm{d}y\,\right|\\
&=\left|\,\sum_{k=-n}^n\int_{\frac{2k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Product series general formula? For a series like :
$$1^3 + 2^3 + 3^3 + 4^3 + \cdots + n^3$$
There is a general formula : $(n(n+1)/2)^2$
My question: Is there any general formula possible for following series :
$$1^1 \cdot 2^2 \cdot 3^3 \cdot 4^4 \cdots n^n$$
| let $$s_n=\prod_{r=1}^nr^r$$
Write$$s_n=1^1 \times2^2 \times 3^3 \times 4^4 \times\cdots \times n^n$$
Take $\log$ on both sides
$$\ln (s_n)=\ln(1^1 \times2^2 \times 3^3 \times 4^4 \times\cdots \times n^n)$$
$$\ln (s_n)=\ln(1^1)+\ln(2^2)+\ln(3^3)+\ln(4^4)+\cdots +\ln(n^n)$$
$$\ln (s_n)=1\ln(1)+2\ln(2)+3\ln(3)+4\ln(4)+\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Parametric solutions to $(4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2=\square$ Let $a,b,c$ and $d$ be rational.Find a rational parametric solutions for $a,b,c$ and $d$ so that
$$(4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2=\square.$$
| For the equation:
$$4c^2b^2+4d^2a^2-c^2a^2-4d^2b^2=3t^2$$
Can specify any number : $d,c$ . Then decisions will be.
$$a=(d^2-c^2)p^2+3s^2$$
$$b=(d^2-c^2)p^2-3cps-3s^2$$
$$t=c(d^2-c^2)p^2+4(d^2-c^2)ps-3cs^2$$
$p,s$ - any integer asked us.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to evaluate this improper integral? I got stuck when evaluating these two improper integrals:$$
\int_a^b\frac{dx}{\sqrt{(b-x)(x-a)}}
$$
and$$
\int_0^1\frac{dx}{\sqrt{x-x^3}}
$$
How to evaluate them? Thank you!
| \begin{align*}
\int_a^b{\frac{\mathrm dx}{\sqrt{(b-x)(x-a)}}}&=\int_a^b\frac{\mathrm d x}{\sqrt{-ab+(a+b)x-x^2}}\\
&=\int_a^b\frac{\mathrm dx}{\sqrt{-ab+\left(\frac{a+b}{2}\right)^2-\left(x-\frac{a+b}{2}\right)^2}}\\
&=\int_a^b\frac{\mathrm dx }{\sqrt{\left(\frac{b-a}{2}\right)^2-\left(x-\frac{a+b}{2}\right)^2}}
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Positive integers $x$,$y$,$z$ If $x$ , $y$ and $z$ are positive integers and $3x = 4y = 7z$, then calculate the smallest possible value for $x+y+z$.
How do you do this? Can someone please give me a hint?
| $k=3x=4y=7z\implies x=\dfrac{k}{3},y=\dfrac{k}{4},z=\dfrac{k}{7}\implies x+y+z=k\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{7}\right)=k\left(\dfrac{4\cdot7+7\cdot3+3\cdot4}{84}\right)\geq \dfrac{61}{84}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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Need explain how to find sum of series Can someone explain me how to find sum of next series:
$\sum_{n=1}^\infty n^4 \tan^{n-1}(x)$
Thanks for answers in advance.
| We can write $n^4$ as a sum of combinatorial polynomials of degree $4$ or less:
$$
n^4=24\binom{n}{4}+36\binom{n}{3}+14\binom{n}{2}+\binom{n}{1}\tag{1}
$$
Next, for $|x|\lt1$, we can use the Generalized Binomial Theorem and negative binomial coefficients to show
$$
\begin{align}
\sum_{n=k}^\infty\binom{n}{k}x^{n-1}
&=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proof By Induction Fibonacci Numbers How do I prove that
$$ f_{ 2n+1 } = 3f_{ 2n } + 1 - f_{ 2n-3 } $$
I'm not sure how to prove it using the defining recurrence of Fibonacci numbers.
| Let's do this the most basic way:
$f_{2n+1} = 3f_{2n} + 1 - f_{2n-3}$
$f_{2n} + f_{2n-1} = 3f_{2n} + 1 - f_{2n-3}$
$f_{2n-1} = 2f_{2n} + 1 - f_{2n-3}$
$f_{2n-1} + f_{2n-3} = 2f_{2n} + 1$
$f_{2n-1} + f_{2n-3} = f_{2n} + f_{2n-1} + f_{2n-2} + 1$
$f_{2n-3} = f_{2n} + f_{2n-2} + 1$
$f_{2n-3} = f_{2n} + f_{2n-3} + f_{2n-4} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding roots in marginally stable system modeled by complex number A system can be modeled by $(z + 3)(z + 2)(z + 1) + C = 0$, where $C > 0$, and $z = x + iy$. When it is marginally stable $Re(z) = 0$.
What are the values of the roots in marginally stable condition? At what values of $C$ do they occur?
Please give me ... | If $(z + 3)(z + 2)(z + 1) + C = 0$ is the characteristic equation for a transfer function (usually denoted in powers of $s$), we can apply the Routh-Hurwitz criterion. Unless there is more to the question, I will assume this is the case. We have that
$$
(z + 3)(z + 2)(z + 1) + C = z^3 + 6z^2 + 11z + 6 + C
$$
Then
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Domain, range and zeros of $f(x,y)=\frac{\sqrt{4x-x^2-y^2}}{x^2y^2-4xy^2+3y^2}$ Given the following function with two variables:
\begin{equation}
\frac{\sqrt{4x-x^2-y^2}}{x^2y^2-4xy^2+3y^2}
\end{equation}
I need to find a) the domain for the above function.
Can anyone give me a hint on how to find the domain in f?
... | we have for the domian two cases
$4x-x^2-y^2\geq 0$ and $y^2(x^2-4x+3)>0$
or
$4x-x^2-y^2\le 0$ and $y^2(x^2-4x+3)<0$
note that $x^2-4x+3=0$ gives $x_1=1$ or $x_2=3$ and $4x-x^2-y^2=4-(y^2+(x-2)^2)$ thus we get for the first case
$4>y^2+(x-2)^2$ and $x>3$ and $y\neq 0$
or
$4>y^2+(x-2)^2$ and $x<1$ and $y \neq 0$
and for... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Showing that $a_n$ is a contracting sequence I have a sequence defined by $a_1=\frac{1}{2}$ and $a_{n+1}=\frac{1}{2}(2+a_n-a_n^2)=\frac{1}{2}(\frac{9}{4}-(a_n-\frac{1}{2})^2)$
I have shown that $\frac{1}{2}\le a_n\le\frac{9}{8}$ for all $n$ and now need to show that $a_n$ is a contracting sequence. I know the definitio... | So simplify further and factor the difference:
$a_{n+2}-a_{n+1}= \frac{1}{2}(a_n^2-a_n-a_{n+1}^2+a_{n+1}) =
\frac{1}{2}(a_{n+1}-a_n-(a_{n+1}^2-a_n^2)) =
\frac{1}{2}(a_{n+1}-a_n-(a_{n+1}-a_n)(a_{n+1}+a_n)) =
\frac{1}{2}(a_{n+1}-a_n)(1-(a_{n+1}+a_n))$.
Note $a_{n+1}-1=\frac{a_n(1-a_n)}2$. Also, if $a>1$ then $a-a^2<... | {
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"url": "https://math.stackexchange.com/questions/1023011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\int_0^{\pi/4}\sqrt{1+\left( \tan x\right)^2}dx$ I would like to understand all the steps to find out this integral $$ \int_0^{\pi/4} \sqrt{1+\left( \tan x\right)^2} dx$$ Wolfram Alpha returns: $$ \frac12 \log(3+2 \sqrt2) = 0.881373587019543...$$ Why does it give this value? Thank you.
| Here are different steps.
$$ \int_0^{\pi/4} \sqrt{1+\left( \tan x\right)^2} \:dx = \ln\left(1+\sqrt{2}\right).$$
Proof.
For $0<x<\pi/4$, set $\displaystyle t:=\tan\left(\frac{x}{2}\right)$ giving $$\cos x = \frac{1-t^2}{1+t^2},\quad x=2 \arctan t, \quad dx=\frac{2}{1+t^2}dt.$$
Then
$$
\begin{align}
\int_0^{\pi/4} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Derivation of formula involving Gamma function? I'm trying to prove that:
$$\prod_{n=1}^{\infty}\frac{n(n+a+b)}{(n+a)(n+b)} = \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)}$$
whenever $a$ and $b$ are positive.
I know that
$$\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)} = \frac{\int_0^{\infty}e^{-s}s^ads \int_0^{\infty}e^{... | Another hint (bit more rigorous perhaps): The Beta Function can be written as
\begin{equation}
B(x,y) = \frac{x+y}{x y} \prod_{n=1}^\infty \left( 1+ \dfrac{x y}{n (x+y+n)}\right)^{-1},
\end{equation}
the right hand side of the above formula can be expanded
\begin{eqnarray}
B(x,y) &=& \frac{x+y}{x y} \prod_{n=1}^\infty ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$a+b+c+d+e$ divides $a^5+b^5+c^5+d^5+e^5-5abcde$ Let $a,b,c,d,e$ be integers such that $a(b+c)+b(c+d)+c(d+e)+d(e+a)+e(a+b)=0$. Prove that $a+b+c+d+e$ divides $a^5+b^5+c^5+d^5+e^5-5abcde$.
I'm reminded of the factorization $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. But for $5$th degree, how can I find a factoriza... | Hint: firts note that; if $a(b+c)+b(c+d)+c(d+e)+d(e+a)+e(a+b)=0$, then $$(a+b+c+d+e)^2=a^2+b^2+c^2+d^2+e^2$$
Now take $P(x)=x^5+kx^4+rx^3+sx^2+tx+u$, with roots $a,b,c,d,e$ then from Viète’s
Relations;
$\boxed{ u=-abcde}$,
$\boxed{k=-(a+b+c+d+e)}$,
$\boxed{r=ae+be+ce+de+ab+ac+ ad+bc+bd+cd=0}$
Take $m=a^4+b^4+c^4+d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1027093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Proving the inequality $4\ge a^2b+b^2c+c^2a+abc$ So, a,b,c are non-negative real numbers for which holds that $a+b+c=3$.
Prove the following inequality: $$4\ge a^2b+b^2c+c^2a+abc$$
For now I have only tried to write the inequality as $$4\left(\frac{a+b+c}3\right)^3\ge a^2b+b^2c+c^2a+abc$$ but I don't know what to do af... | Let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.
Hence, by Rearrangement and AM-GM we obtain:
$a^2b+b^2c+c^2a+abc=a\cdot ab+b\cdot bc+c\cdot ca+abc\leq x\cdot xy+y\cdot xz+z\cdot yz+xyz=$
$=y(x+z)^2=4y\left(\frac{x+z}{2}\right)^2\leq4\left(\frac{y+\frac{x+z}{2}+\frac{x+z}{2}}{3}\right)^3=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1028027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
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Series convergence and limit I am trying to solve an exercise that asks me this:
Prove using the $ε–N$ method that the sequence
$a(n) = \frac{n^2 + n - 1}{n^2 + n}$
converges and state the limit.
My attempt is the following:
First, I guess that the $\lim \limits_{n\to\infty} a(n) = 1$.
Then we require
\begin{equation}... | $2n^2 > 1/\epsilon$ will not guarantee $n^2+n > 1/\epsilon$.
You need to find a nice function of $n$, that lies between $n^2+n$ and $1/\epsilon$, which will allow you to solve for $n$ in terms of $\epsilon$. We can in fact choose $n>1/\epsilon$, since this will ensure that $n^2+n > 1/\epsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Two roots of the polynomial $x^4+x^3-7x^2-x+6$ are $2$ and $-3$. Find two other roots. I have divided this polynomial first with $(x-2)$ and then divided with $(x+3)$ the quotient. The other quotient I have set equal to $0$ and have found the other two roots. Can you explain to me if these actions are correct and why?
| $(x-2)(x+3)=x^2+x-6$
$x^4+x^3-7x^2-x+6\\=(x^4+x^3-6x^2)-(x^2+x-6)\\= x^2(x^2+x-6)-1(x^2+x-6)\\=(x^2-1)(x^2+x-6)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1036803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Help me, a doubt $f(x)=\cot^{-1} \frac{1-x}{1+x}$ I have a doubt
$$f(x)=\cot^{-1} \frac{1-x}{1+x}$$
$$f´(x)=\frac{1}{(\frac{1-x}{1+x})^2}\cdot\frac{(-1)(1+x)-(1-x)}{1+\frac{(1-x)^2}{(1+x)^2}}$$
mm this could to be really easy but I do not understand in the first denominator gives one, someone who can explain,
| Hint: Let $f(x) = \cot^{-1} x$ then $f'(x) = -\frac{1}{1+x^2}$ and if $g(x) = \frac{1-x}{1+x}$ then $g'(x) = -\frac{2}{(x+1)^2}$
Now use the chain rule $$[f(g( x ))]' = f'(g(x))\ g'(x) $$
Notice that $$-\frac{1}{1+\Big(\frac{1-x}{1+x}\Big)^2} = -\frac{(1+x)^2}{(1+x)^2+(1-x)^2} = -\frac{(1+x)^2}{1+2x + x^2 +1-2x+ x^2} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
$ (3+\sqrt{5})^n+(3-\sqrt{5})^n\equiv\; 0 \; [2^n] $ Proof that for all $n\in \mathbb{N}$ :
$$
(3+\sqrt{5})^n+(3-\sqrt{5})^n\equiv\; 0 \; [2^n]
$$
| Another view on this:
$\left(\dfrac{3+\sqrt 5}{2}\right)^n +\left(\dfrac{3-\sqrt 5}{2} \right)^n$
Is the trace of the $n$-th power of
$\left(\begin{array}{cc} 2 & 1 \\ 1 & 1\end{array}\right)$
And another:
$\dfrac{3+\sqrt 5}{2}$ and $\dfrac{3-\sqrt 5}{2}$ are algebraic integers (they're roots of $x^2 - 3x + 1$). Then s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Compute$\int\limits_{0}^{2} \sqrt{x^2-2x+2}\ln(2+x)dx$. Compute: $I=\displaystyle \int\limits_{0}^{2} \sqrt{x^2-2x+2}\ln(2+x)dx$.
I tried to : $I=\displaystyle \int \limits_{-1}^{1}\sqrt{t^2+1}\ln(3+t)dt$
set $t=\tan u\Rightarrow dt=(1+\tan^2u)du$
and $I=\displaystyle \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \dfra... | Hint:
*
*${x+2=e^z \implies dx=e^zdz}$
$\displaystyle \int \sqrt{x^2-2x+2}\ln(2+x)dx\\=\displaystyle \int ze^z\sqrt{e^{2z}-6e^z+10}\ dz\\=z\displaystyle \int e^z\sqrt{e^{2z}-6e^z+10}\ dz-\displaystyle \int \left(\displaystyle\int e^z\sqrt{e^{2z}-6e^z+10}\ dz\right)dz$
$\\$
*$e^z=y \implies e^zdz = dy$
$\theref... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1042539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to get $\sqrt {k} + \frac{1}{\sqrt{k+1}}$ in the form $\frac{\sqrt{k^2} + 1}{\sqrt{k+1}}$? I was wondering if it is possible to get $\sqrt {k} + \dfrac{1}{\sqrt{k+1}}$ in the form $\dfrac{\sqrt{k^2} + 1}{\sqrt{k+1}}$, and if so, how?
I ask this, because I'm following this answer, and I get lost at how they arrive a... | It cannot hold obviously. When $k=\color{red}1$, $\sqrt {k} + \frac{1}{\sqrt{k+1}}=\color{red}{1+\frac{\sqrt2}{2}}$, however, $\frac{\sqrt{k^2} + 1}{\sqrt{k+1}}=\color{red}{\sqrt2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1045867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Factor $55 - 88 \sqrt{-2}$ as a product of primes in $\mathbb{Z}[\sqrt{-2}]$ To solve this problem, I let $K = \mathbb{Q}(\sqrt{-2})$, and I thought to take the norm $$N(55 - 88 \sqrt{-2}) = 55^2 + 2 \cdot 88^2 = 18513 = 3^2\cdot11^2 \cdot 17$$ If $a \in \mathbb{Z}[\sqrt{-2}]$ is irreducible, then $N(a) = p^f$, where $... | I think that only after doing several problems like this one do certain "obvious" basic facts really begin to feel truly obvious and basic.
The first such fact, which Zoe already mentioned, is that if for $a + b \sqrt d$ with $a$ and $b \in \mathbb Z$ we have $\gcd(a, b) > 1$, then $a + b \sqrt d$ is divisible by some ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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What am I doing wrong? (Trigonometric Identity) $$\frac { \cos\theta }{ 1-\sin\theta } =\frac { \sin\theta -\csc\theta }{ \cos\theta -\cot\theta } $$
Steps I took:
$$\frac { \sin\theta -\frac { 1 }{ \sin\theta } }{ \cos\theta -\frac { \cos\theta }{ \sin\theta } } $$
$$\frac { \frac { \sin^{ 2 }\theta -1 }{ \si... | $\dfrac{\sin^2\theta-1}{\cos \theta(\sin\theta-1)}=\dfrac{-\cos^2\theta}{\cos \theta(\sin\theta-1)}=\dfrac{-\cos\theta}{\sin\theta-1}=\dfrac{\cos\theta}{1-\sin\theta}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How prove this inequality $\prod_{1\le i
let $z_{1},z_{2},z_{3},z_{4},z_{5}$ are complex numbers,and such
$$|z_{1}|^2+|z_{2}|^2+|z_{3}|^2+|z_{4}|^2+|z_{5}|^2=5$$
$$A=\begin{bmatrix}1&1&1&1&1\\z_1&z_2&z_3&z_4&z_5\\z_1^2&z_2^2&z_3^2&z_4^2&z_5^2\\z_1^3&z_2^3&z_3^3&z_4^3&z_5^3\\z_1^4&z_2^4&z_3^4&z_4^4&z_5^4\\\end{bmatrix... | Not the exact solution, but maybe a way to go: From the AM-GM inequality we have that
$$
\prod_{i<j}|z_i-z_j|^2\leq\frac{1}{5^5}\left(\sum_{i<j}|z_i-z_j|^2\right)^5.
$$
By the parallelogram identity we have $|z-w|^2=2(|z|^2+|w|^2)-|z+w|^2\leq2|z|^2+2|w|^2$ for each $z,w\in\mathbb C$. Thus,
$$
\prod_{i<j}|z_i-z_j|^2\leq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1049570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Riemann sum with circles! My friend and I were talking about Riemann sum and friend suggested that all rectangles would be replaced by circles this is the formula I came up with:
$$
\lim _{n\ \to\ \infty}\sum_{i\ =\ 1}^{n}\pi\,{\rm f}^{2}\left(\, i\,\right)
$$
Is this formula right and if so does this formula work for ... | I've just written a paper on this subject for class. My answer is a bit different from yours, because I am allowing circles to stack on each other.
Rectangular Riemann sums take the form $$\sum_{i=0}^n f(a+i\Delta x)\Delta x $$ when given a function $f(x)$, divided into $n$ partitions, bounded between $ a \text{ and } ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$1 + 4 + 7 + · · · + 3n − 2 =
\frac{n(3n − 1)}
2$$
for all positive integers $n$.
Proof: $$1+4+7+\ldots +3(k+1)-2= \frac{(k + 1)[3(k+1)+1]}2$$
$$\frac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$$
Along my proof I am stuck at the above section where it w... | Sum of the first and last terms = $1 + (3n-2) = 3n-1$
Sum of 2nd and (n-1)th terms = $4 + (3n-5) = 3n-1$
Sum of 3rd and (n-2)th terms = $7 + (3n-8) = 3n-1$
$...$
Sum of (n-1)th and 2nd terms = $(3n-5) + 4 = 3n-1$
Sum of n-th (last) and 1st terms = $(3n-2) + 1 = 3n-1$
Add both sides up .
$(1+4+...+(3n-2)) + (1+4+..... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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"answer_id": 5
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Inequality problem involving Cauchy Schwarz inequality If $a+b+c=3$, prove that $\sum \frac{a^2}{b^2-2b+3} \geq 3/2$. How to prove it using the Cauchy-Schwarz inequality? Denote the expression with $P$. What I got was $P$ is greater than or equal to $\frac{9}{\sum a^2-2a-2b-2c+9}$. Is it correct if I found the minimum ... | By C-S $\sum\limits_{cyc}\frac{a^2}{b^2-2b+3}=\sum\limits_{cyc}\frac{a^4}{a^2b^2-2a^2b+3a^2}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^2b^2-2a^2b+3a^2)}$.
Hence, it remains to prove that $2(a^2+b^2+c^2)^2\geq3\sum\limits_{cyc}(a^2b^2-2a^2b+3a^2)$, which after homogenization gives
$\sum\limits_{cyc}(a^4+a^2b^2-2a^3c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find orthogonal matrices Let $A=\begin{bmatrix} 1 & -1/2&-1/2 \\ -1/2 & 1& -1/2\\ -1/2&-1/2 &1 \end{bmatrix}$. Is it possible to find explicitly orthogonal matrices $P, Q$ such that $2A=PJP^t+QJQ^t$? Here $J$ is a matrix of all entries one.
| Note that $A$ has eigenvalues $0$, $3/2$, $3/2$, with null space spanned by
$[1,1,1]^t$. On the other hand, $J$ has eigenvalues $0,0,3$ with $[1,1,1]^t$ its eigenvector for $3$. Consider an orthonormal basis
$$u = \dfrac{1}{\sqrt{3}}\pmatrix{1\cr 1\cr 1\cr}, \ v = \dfrac{1}{\sqrt{2}} \pmatrix{1\cr -1\cr 0\cr},\ w =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculate a lim $\lim_{x\to \infty } \left ( \frac{x^2+2}{x^2-4} \right)^{9x^2} $ $$
\lim_{x\to \infty } \left ( \frac{x^2+2}{x^2-4} \right)^{9x^2}
$$
Can you help with it?
| Using the polynomial division algorithm, rewrite the base as
$$\frac {x^2 + 2}{x^2 - 4} = 1 + \frac 6{x^2 - 4} = 1 + \frac 1 {\frac {x^2 - 4}{6}}$$
We will make use of the standard limit:
$$\lim_{x \to +\infty} \left(1 + \frac 1x \right)^x = e$$
So:
$$\begin{align}
\lim_{x \to +\infty} \left(\frac {x^2 + 2}{x^2 - 4}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.
| here is how i explain this: look at the numbers $9, 99, 999, 9999, \cdots$ in base ten. they are $9 = 10 -1, 99 = (10-1)*11 = 10^2 - 1, 999 = (10-1)*111 = 10^3 - 1, 9999 = (10-1)*1111 = 10^4 - 1$ and the left hand side has the factor $9 = (10-1)$. now you can rewrite the string of equations in the form $$(10 -1) = 1(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 12,
"answer_id": 7
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Find the minimum possible value of $x(1-z)+y(1-x)+z(1-y)$ It is given that $$xyz=(1-x)(1-y)(1-z)$$ and $$x, y, z \in (0,1)$$
Find the minimum possible value of the expression: $$x(1-z)+y(1-x)+z(1-y)$$
Using the AM-GM inequality concepts, I can write that the value is minimum when
$$x(1-z)=y(1-x)=z(1-y)$$
What else can... | Making the substitutions $\displaystyle \frac{1-x}{x} = a, \frac{1-y}{y} = b$ and $\displaystyle \frac{1-z}{z} = c$,
Then, $abc = 1$ and
$\begin{align}\displaystyle \sum\limits_{cyc} x(1-z) &= \sum\limits_{cyc} \frac{c}{(1+c)(1+a)} \\&= \frac{\sum\limits_{cyc} c(1+b)}{(1+a)(1+b)(1+c)} \\ &= \frac{\sum\limits_{cyc} c(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove that, $(2\cdot 4 \cdot 6 \cdot ... \cdot 4000)-(1\cdot 3 \cdot 5 \cdot ...\cdot 3999)$ is a multiple of $2001$ Prove that the difference between the product of the first 2000 even numbers and the first $2000$ odd numbers is a multiple of $2001$. Please show the method.
I have started with the following process:
$... | We have $$2001=3\times 23 \times 29.$$
Now note that $$\begin{align}2\cdot 4 \cdot 6 \cdot ... \cdot 4000&=(2\times 1)\cdot (2 \times 2) \cdot (2 \times 3) \cdot ... \cdot (2\times 2000)\\
&=2^{2000}\times 1\cdot2\cdot3\cdot... \cdot23 \cdot ... \cdot 29 \cdot....\cdot2000
\end{align}$$
Therefore $2\cdot 4 \cdot 6 \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1058993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Question on the sum $\sum_{n=1}^{\infty}\frac{x^n}{n} = -\ln(1-x)$ $f(x) = \displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n} = x + \frac{x^2}{2} + \frac{x^3}{3} + ... = -\ln(1-x)$ for $|x| < 1$.
$f'(x) = \displaystyle\sum_{n=1}^{\infty}x^{n-1} = 1 + x + x^2 + x^3 +... = \frac{1}{1-x}$ for $|x| < 1$
If $f'(x) = \displaystyl... | There are various things going on here. One is the issue of the $+C.$ What you wrote might better be put
If $f'(x) = \displaystyle\frac{1}{1-x}$, then by definition of the indefinite integral $\displaystyle\int\frac{1}{1-x}dx=f(x)+C$
Also , by definition of definite integrals
If $f'(x) = \displaystyle\frac{1}{1-x}$, ... | {
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"url": "https://math.stackexchange.com/questions/1060082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove by induction that an expression is divisible by 11
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.
I am rather confused by this question. This is my attempt so far:
For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such th... | First, show that this is true for $n=2$:
*
*$\frac{2^{3\cdot2-1}+5\cdot3^1}{11}=7\in\mathbb{N}$
Second, assume that this is true for $n$:
*
*$\frac{2^{3n-1}+5\cdot3^n}{11}=k\in\mathbb{N}$
Third, prove that this is true for $n+2$:
*
*$\frac{2^{3(n+2)-1}+5\cdot3^{n+2}}{11}=\frac{2^{3n+5}+5\cdot3^{n+2}}{11}$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 3
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ordered pairs $(A,B)$ of subsets of $X$ >such that $A\neq \phi\;,B\neq \phi$ and $A\cap B = \phi\;,$ is
Let $X$ be a set of $5$ elements. Then the number of ordered pairs $(A,B)$ of subsets of $X$
such that $A\neq \phi\;,B\neq \phi$ and $A\cap B = \phi\;,$ is
$\bf{My\; Try::}$ Let $X = \left\{1,2,3,4,5\right\}\;,$
th... |
Hi. You might want to look about Stirling numbers of the second kind. You want the number of ways you can take a surjective function from $[5]=\{1,2,3,4,5\}$ to $[2]=\{1,2\}$ plus the number of surjective functions from $[5]=\{1,2,3,4,5\}$ to $[3]=\{1,2,3\}$.(in $3$ to represent elements that are not taken) In your ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How does Wolfram Alpha come up with the substitution $x = 2\sin u$? Integration/Analysis I have to integrate
$$
\int_0^2 \sqrt{4-x^2} \, dx
$$
I looked at the Wolfram Alpha step by step solution, and first thing it does is it substitutes
$x = 2\sin(u)\text{ and } \,dx = 2\cos(u)\,du$
How does it know to substitute $2\... | Theóphile's comment is a good answer. I will give another.
The $\sqrt{4-x^2}$ and indeed anything of the form $\sqrt{a^2-b^2}$ --- think $\sqrt{x^2-b^2}$ or $\sqrt{a^2-x^2}$ --- or $x^2+b^2$ can be considered --- via Pythagoras --- as sides of a right-angled triangle.
In this example we have $\sqrt{2^2-x^2}=b\Rightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1063477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Consider the linear transformation $T: \mathbb{R}^{2\times 2} \rightarrow \mathbb{R}^{2\times 2}$ such that $T(A)=A-A^T$
*
*Find a basis for $\ker T$.
*Find a basis for $\operatorname{im} T$.
*Prove that $T$ is diagonalizable. Find an eigenbasis $\mathcal{B}$ for $T$ and find the $\mathcal{B}$-matrix $[T]_\mathcal{... | You already know three eigenvectors. Since your space is four-dimensional, you just need to find a fourth. Here's a hint: if $Tv = \lambda v$ and $\lambda \neq 0$, this means that $v$ must be in $T$'s....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1064281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Help with inverse matrix problem? (Specific problem in description) \begin{equation}
\text{If}
\begin{vmatrix}A\end{vmatrix}
\text{=}\frac{1}{24}
\text{, solve }
\begin{vmatrix}
\begin{pmatrix}\frac{1}{3}A\end{pmatrix}^{-1} - 120 \text{ }A^*
\end{vmatrix}
\end{equation}
My attempt so far comes from the definition of a... | Use the $A^{-1}$ only in second part or first part of the problem not both
\begin{equation}
A^{*} = \frac{1}{24}{A^{-1}}
\end{equation}
sub this where $A^*$ is in second part and don't change $A^{-1}$ until fully simplified
$=|\frac13 A^{-1} - 120 (\frac{1}{24} A^{-1})|$
$=|\frac13 A^{-1} - 5 A^{-1})|$
$=(\frac{-14}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1064836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Question about converting a polar equation to a rectangular equation $$\sec\theta =2$$
So I went through all the steps and got:
$$\cos\theta =\frac { 1 }{ 2 } $$
$$\sin\theta =\pm \sqrt { 1-\frac { 1 }{ 4 } } $$
$$\sin\theta =\pm \frac { \sqrt { 3 } }{ 2 } $$
$$y=\pm {\sqrt { 3 } }$$
Now why is it that the correct a... | From $\sec\theta=2$, draw a triangle in either the first or fourth quadrant with angle $\theta$, adjacent side $1$, opposite side $\sqrt 3$, hypotenuse $2$. Then
\begin{align}
\sec\theta=2\implies \tan\theta&=\pm\sqrt{3}\\
{r\sin\theta\over r\cos\theta}&=\pm \sqrt{3}\\
{y\over x}&=\pm \sqrt{3}\\
y&=\pm \sqrt{3}\,x.
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
using power expansion to find limit I am preparing for my final exam, and stuck on this question.
Using power series expansion, evaluate $$\lim_{x\to 0} \frac{x\cos(x)
-\sin(x)}{x^2-x\ln(1+x)}$$
I have no idea how to proceed. Any help would be highly appreciated!
| We have: $\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} +\cdots $,
$\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} +\cdots $,
$\log(1+x) = x -\dfrac{x^2}{2} +\dfrac{x^3}{3} + \cdots $
Thus:
\begin{align}
x\cos x - \sin x
&= \left(x - \dfrac{x^3}{2!} + \dfrac{x^5}{4!}+\cdots \right) - \left(x -
\dfrac{x^3}{3!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Closed form of $\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$ Today I discussed the following integral in the chat room
$$\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$$
where $0\leq a, b\leq \pi$ and $k>0$.
... | \begin{align}
\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}&=\int_0^\infty\frac{1}{x}\int_a^b \frac{\mathrm d}{\mathrm dy}\ln \left(x^2+2kx\cos y+k^2\right) \;\mathrm dy\,\mathrm dx\\[10pt]
&=-\int_0^\infty\frac{1}{x}\int_a^b \frac{2kx\sin y}{x^2+2kx\cos y+k^2} \;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 2
} |
What is the required radius of the smaller circles around a larger circle so they touch? I am trying to determine how to calculate the required radius of the smaller circles so they touch each other around the larger circle. (red box)
I would like to be able to adjust the number of smaller circles and the radius of th... | Suppose that the center $c_k,k=1,\ldots,n$ of the small circles are placed equidistantly on the bigger circle.
Then we have $c_k=\left(R\sin(2\pi\frac{k}{n}),R\cos(2\pi\frac{k}{n})\right), k = 1,\ldots,n$. So for $k=1,\ldots,n$ we must have $r=\frac{\|c_{k}-c_{k+1}\|_2}{2}$, since two circles with same radius are tange... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
In triangle ABC, Find $\tan(A)$.
In triangle ABC, if $(b+c)^2=a^2+16\triangle$, then find $\tan(A)$ . Where $\triangle$ is the area and a, b , c are the sides of the triangle.
$\implies b^2+c^2-a^2=16\triangle-2bc$
In triangle ABC, $\sin(A)=\frac{2\triangle}{bc}$, and $\cos(A)=\frac{b^2+c^2-a^2}{2bc}$,
$\implies \ta... | We have $16\triangle=(b+c+a)(b+c-a)$
$\iff16r\cdot s=2s\cdot2(s-a)$ where $r$ is the in-radius & $2s=a+b+c$
Using this, $\tan\dfrac A2=\dfrac r{s-a}=\dfrac14$
Finally use $\tan2x$ formula
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Putnam definite integral evaluation $\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$
Evaluate $$\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$$ Source : Putnam
By the property $\displaystyle \int_0^af(x)\,dx=\int_0^af(a-x)\,dx$:
$$=\int_0^{\pi/2}\frac{(\pi/2-x)\sin x\cos x}{\sin^4 x+\cos^4 x}dx=\fr... | $1 - \dfrac{\sin^2(2x)}{2} = \dfrac{1+\cos^2(2x)}{2}$, and $\sin x\cos x = \dfrac{\sin (2x)}{2} \Rightarrow \displaystyle \int \dfrac{\sin x\cos x}{\cos^4x+\sin^4x}dx = \displaystyle \int -\dfrac{1}{2}\dfrac{d(\cos(2x))}{1+\cos^2(2x)}dx = -\dfrac{1}{2}\arctan(\cos (2x)) + C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 2
} |
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.