Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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What's the sum of this series? I would like to know how to find out the sum of this series:
$$1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2} - \frac{1}{6^2} + \cdots$$
The answer is that it converges to a sum between $\frac 34$ and $1$, but how should we go about estimating this sum?
Thanks!
| The sum of this series is $\frac{\pi^2}{12}$.
Explantion
We already know that:
$$1+ \frac{1}{2^2} + \frac{1}{3^2} + \text{...} = \frac{\pi^2}{6}$$
HINT
Note that:
$$\large \frac{-1}{2^2} = \frac{1}{2^2} - \frac{1}{2^2} - \frac{1}{2^2}$$
Now,
$$1- \frac{1}{2^2} + \frac{1}{3^2} - \text{...} = \left(\frac{1}{1} + \frac{1}... | {
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"url": "https://math.stackexchange.com/questions/911508",
"timestamp": "2023-03-29T00:00:00",
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Find $\lim\limits_{x \to \infty} \left(\sqrt{x^2+x+1} - \sqrt{x^2-x} \right)$ I am having a tough time with these TYPE of problems
looking forward ideas, All ideas will be appreciated
| Here is another approach:
$\sqrt{x^2+x+1} - \sqrt{x^2-x} = x (\sqrt{1+{1 \over x}+{1 \over x^2}} - \sqrt{1-{1 \over x}} )$.
For $\delta$ sufficiently small, we have $|\sqrt{1 + \delta}-(1+{\delta \over 2}) | \le |\delta|^2 $ (this follows from the Taylor expansion of $\delta \mapsto \sqrt{1+\delta}$, which is $1+{1 \ov... | {
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"timestamp": "2023-03-29T00:00:00",
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Sum of a series problem involving cubes For any odd integer $n$, evaluate $n^3 - (n-1)^3 + \cdots + (-1)^{n-1} \cdot 1^3 = \sum_{k=1}^n (-1)^{k+1}\cdot k^3$
How would you go about solving such a problem?
Any help would be appreciated.
| Just split the problem up into "odd" and "even" cases and use Faulhaber's formula (this is perhaps the most straightforward way); e.g.:
$$\begin{align}
7^3-6^3+5^3-4^3+3^3-2^3+1^3 \\
&= (7^3+6^3+5^3+4^3+3^3+2^3+1^3)-2(6^3+4^3+2^3) \\
&=(7^3+6^3+5^3+4^3+3^3+2^3+1^3)-2(6^3+4^3+2^3) \\
&=(7^3+6^3+5^3+4^3+3^3+2^3+1^3)-2((2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/914621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Checking a solution to an indefinite integral I recently did some work to try to find $\int{\frac{dx}{Ax^3 - B}}$, but I'm always paranoid that my solution has some minor trivial error in the middle of the process that screwed up the end result entirely, so could someone please help check my solution?
The first step to... | I think that we could make this simpler without involving complex numbers.
Just as rogerl did using partial fractions, we have $$\frac{1}{u^3-1} =\frac{1}{3}\frac{1}{u-1}-\frac{1}{3}\frac{u+2}{u^2+u+1}$$ Now $$\frac{u+2}{u^2+u+1}=\frac{1}{2}\frac{2u+4}{u^2+u+1}=\frac{1}{2}\Big(\frac{2u+1}{u^2+u+1}+\frac{3}{u^2+u+1}\Bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/915414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Getting the correct answer for a limit with a cube in the denominator Given:
$$
\lim_{x \to 1} f(x) = \frac3{1-x^3} - \frac1{1-x}
$$
I first used the difference of cubes to get
$$
\lim_{x \to 1} f(x) = \frac3{(1-x)(1+x+x^2)} - \frac1{1-x}
$$
Then multiplied each term by $(1-x)$, cancelling $(1-x)$ in the first term, an... | You can't multiply each term by $(1-x)$- rather you should factor it out to get:
$\frac{1}{1-x}\frac{3}{(1+x+x^2)-1)}=\frac{1}{1-x}\frac{(2-x-x^2)}{(1+x+x^2)}=\frac{1}{1-x}\frac{(1-x+1-x^2)}{(1+x+x^2)}=\frac{1}{1-x}\frac{(1-x)(2+x)}{(1+x+x^2)}$.
Now, cancel, and take the limit- the answer shall be $1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Nested Sum Encountered in Maclaurin Expansion of $e^{-x^2}$ In the course of working out the Maclaurin expansions of $e^{-x^2}$ and $cos(x^2)$, I ran into the following nested sum:
$$ \underbrace{ \sum_{a=0}^1 \left( a \sum_{b=0}^{a+1} b \left( \sum_{c=0}^{b+1} c \cdots d \sum_{e=0}^{d+1} \left( e \sum_{f=0}^{e+1} f \r... | Returned to the problem and things became clear. Thank you to almagest for the inspiration to consider his recursive function.
To keep everything in one spot, I'll start from the beginning defintions:
Let $S \colon \mathbb{N}^{2} \to \mathbb{N}$ and define as follows:
$$ \begin{align}
S(0,0) & \equiv 0 \\
S(0,k) & \equ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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What is the correct way to compare to algebraic quantities? If I have two algebraic quantities, what is the correct way to determine if they are equal, or if not which is greater than the other?
For example, if I have $\frac{2d}{\sqrt{s^2-c^2}}$ and $\frac{2ds}{s^2-c^2}$, and I want to prove $\frac{2d}{\sqrt{s^2-c^2}} ... | After messing around by doing scratch work like what you've done, I like to prove what I originally wanted by using a single chain of operations:
\begin{align*}
\frac{2d}{\sqrt{s^2-c^2}}
&= \frac{2d}{\sqrt{s^2-c^2}} \cdot \frac{\sqrt{s^2-c^2}}{\sqrt{s^2-c^2}} \\
&= \frac{2d\sqrt{s^2-c^2}}{s^2-c^2} \\
&< \frac{2d\sqrt{s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find coefficient of $x^3$ in (2+x) ^(3/2)/(1-x) I can expand $\dfrac{(2+x)^{3/2}}{1-x}=(1+x+x^2+\ldots)\left({3/2\choose0}+{3/2\choose1}(x+1)+{3/2\choose2}(x+1)^2+\ldots\right)$, but that doesn't seem to lead anywhere.
| Your idea is right but it helps to factor out the $2$ in the numerator so the binomial expansion is in $x$ instead of $x+1$:
\begin{eqnarray*}
\dfrac{(2+x)^{3/2}}{1-x} &=& \dfrac{2^{3/2}(1+\frac{x}{2})^{3/2}}{1-x} \\
&=& 2^{3/2}(1+x+x^2+\ldots)\left({3/2\choose0}+{3/2\choose1}\frac{x}{2}+{3/2\choose2}\left(\frac{x}{2}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $\int\frac{x^3}{(x^3+1)^2}dx$ any ideas on how I could continue this integral
$$\int\frac{x^3}{(x^3+1)^2}dx$$
I am half way done, by entirely using fraction decomposition
$\int\frac{x^3}{(x^3+1)^2}dx=\int\frac{1}{x^3+1}dx-\int\frac{1}{(x^3+1)^2}dx$
$\frac{1}{x^3+1}=\frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}$ from whi... | The polynomial $x^3+1$ is not irreducible (as you have seen),
$$x^3+1=(x+1)(x^2-x+1).$$
You need to continue decomposing in order to compute the integral.
| {
"language": "en",
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Calculus - Derivatives and Tangent Lines I am trying to find the equation of the lines (there's 2) tangent to the graph of $f(x) = x^3$ and parallel to the line $12x − y + 1 = 0$.
So far obviously the slope of the line must be $12$.
The equation of the line tangent to $f(x) = x^3$ is the derivative and must be equal t... | Yes, the slope of our line(s) needs to be $12$ to ensure they are parallel to the line
$\,12x − y + 1 = 0.$
Next, finding the derivative of $y = x^3$ gives us $\,y' = 3x^2$.
To find the points on the curve at which the tangent lines at those points have slope $12$, we solve $$3x^2 = 12 \iff x^2 = 4 \iff x =\pm 2.$$
Fo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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can you show the solution for this in finding the value of a? $$\frac{a^2+4^2-5^2}{8a} = \frac{\sqrt{3}}{2}\frac{a^2+4^2-6^2}{8a}+\frac{1}{2} \sqrt{1- \Big(\frac{a^2+4^2-6^2}{8a}\Big)^2}$$
show the solution in finding the value of A
| Let
$$\sin(\theta) = \frac{a^2 + 4^2 - 6^2}{8a}.$$
Then we have:
$$\sin(\theta) + \frac{11}{8a} = \frac{\sqrt{3}}{2}\sin(\theta) + \frac{1}{2}\cos(\theta).$$
Simplifying, we have:
$$\sin(\theta) + \frac{11}{8a} = \sin\left(\theta + \frac{\pi}{6}\right).$$
Consequently:
$$\sin\left(\theta + \frac{\pi}{6}\right) - \sin(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/924957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Verifying proof of $\lim_{x \to\sqrt{2}}\frac{x^2-2}{x^2+\sqrt{2}x-4} = \frac 2 3$ $$\lim_{x \to\sqrt{2}} \dfrac{x^2-2}{x^2+\sqrt{2}x-4} = \lim_{w \to2} \dfrac{w^2-4}{w^2+2w-8} =\lim_{w \to2} \dfrac{(w-2)(w+2)}{(w+4)(w-2)} = \frac 2 3$$
Change of variable:
$$w=\sqrt{2}x \Rightarrow x=\frac{w}{\sqrt{2}}\Rightarrow x^2=\... | Your work is correct and helps in removing square roots. But you can dispense with the change of variables.
Let $f(x)=x^2+x\sqrt{2}-4$; then $f(\sqrt{2})=0$, so you know $\sqrt{2}$ is a root of the polynomial. Then the usual scheme (you may be used to a different one)
$$
\begin{array}{r|rr|r}
& 1 & \sqrt{2} & -4 \\
\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/926507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of a product I need to find the value of $$L=\lim_{n \rightarrow \infty}\displaystyle\prod_{r=1}^{n} \left(1+\dfrac{r^2}{n^2}\right)^{1/n}$$
Is doing this OK?--
$$\begin{align} L &=\lim_{n \rightarrow \infty}\left(1+\dfrac{1^2}{n^2}\right)^{1/n}\left(1+\dfrac{2^2}{n^2}\right)^{1/n} \ ... \ \left(1+\dfrac{n^2}{n^2... | $$L=\lim_{n \rightarrow \infty}\displaystyle\prod_{r=1}^{n} \left(1+\dfrac{r^2}{n^2}\right)^{1/n}$$
$$\ln L=\lim_{n \rightarrow \infty}\frac1n\sum_{r=1}^n\ln\left(1+\dfrac{r^2}{n^2}\right)=\int_0^1\ln(1+x^2)dx$$
$$\int\ln(1+x^2)dx=\ln(1+x^2)\int dx-\int\left(\frac{d[\ln(1+x^2)]}{dx}\int dx\right)dx$$
$$=x\ln(1+x^2)-\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rational Expression Simplification The function:
$$f(x) = \frac{3x - 4}{x^2 - 2x}$$
is simplified to:
$$f(x) = \frac{2}{x} + \frac{1}{x - 2}$$
How? And in what way?
| $$f(x)=\frac {3x-4}{x^2-2x}\\
=\frac {3x-4}{x(x-2)}\\
=\frac Ax +\frac B{x-2}\\
=\frac {A(x-2)+Bx}{x(x-2)}$$
Equation numerators:
$$3x-4=A(x-2)+Bx\\
=(A+B)x-2A$$
gives $A=2,B=1$, i.e.
$$f(x)=\frac {3x-4}{x^2-2x}=\frac 2x +\frac 1{x-2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/930525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the particular solution of $u_x+2u_y-4u=e^{x+y}$ satisfying the following side condition $u(x,-x) = x$ 2.Find the particular solution of $u_x+2u_y-4u=e^{x+y}$ satisfying the following side condition $u(x,-x) = x$
I know that under that condition $y = -x$ which is the reflection of the $x$ graph. I have problems wi... | I spotted the following mistake, which invalidates all steps after it: from
$$e^{-4z}(v) = \frac{-2}{5}e^{\frac{w-5z}{2}}+C(w)$$
you should have gotten
$$v = \frac{-2}{5}e^{\frac{w-5z}{2}}e^{4z}+C(w)e^{4z}$$
Instead, you omitted $C$, then multiplied by $e^{4z}$, then brought $C$ back in, un-multiplied.
Why don't yo... | {
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"url": "https://math.stackexchange.com/questions/931921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Evaluating $\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x}$ I did this:
$$\begin{align}
\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x} &\sim \lim_{x \to 0} \frac{(1 + x + x^2)^{1/x} - (1 + x)^{1/x}}{x} = \\
&= \lim_{x \to 0} \left [ (1+x)^{1/x} \frac{\left ( \fr... | I think that, in the same spirit as you show in the post, we could do a little faster considering the development of $$1+\sin (x)+a \sin ^2(x)=1+x+a x^2+O\left(x^3\right)$$ Then going to logarithms, series, followed by exponentiation of series and series again $$f(a)=\Big(1+\sin (x)+a \sin ^2(x)\Big)^{1/x}=e+\left(e a-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/933388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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What is $\frac{2x}{1-x^2}$ when $x=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$? If
$$x=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$$
Find
$$\frac{2x}{1-x^2}$$
I got till here by simplification by taking the previous value of x, ie,
$$x={\frac{\sqrt{1-\cos\theta}}{1+\cos\theta}}$$
$$\frac{2\tan\theta\sqrt{1+\cos\theta}}{\cos... | $$x=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$$
But $\cos2a=2\cos^2a-1=1-2\sin^2a$, so
$$1-\cos\theta=2\sin^2 \frac{\theta}{2}$$
$$1+\cos\theta=2\cos^2 \frac{\theta}{2}$$
Hence
$$x=\left|\tan \frac{\theta}2\right|$$
Therefore
$$\frac{2x}{1-x^2}=\frac{2\left|\tan\frac{\theta}2\right|}{1-\tan^2\frac{\theta}2}=\frac{\left|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/933564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Closed-form of $\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx$ Does the following series or integral have a closed-form
\begin{equation}
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx
\end{equation}
where $\Psi_3(x)$ is the polygam... | \begin{align}
\sum^\infty_{n=1}\frac{(-1)^{n-1}\psi_3(n+1)}{n}
&=-12\zeta(5)+\frac{45}{4}\zeta(4)\ln{2}+\frac{9}{4}\zeta(2)\zeta(3)
\end{align}
Let $\displaystyle f(z)=\frac{\pi\csc(\pi z)\psi_3(-z)}{z}$. Then at the positive integers,
\begin{align}
\sum^\infty_{n=1}{\rm Res}(f,n)
&=\sum^\infty_{n=1}\operatorname*{Res... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/934981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
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"answer_id": 2
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Solve the following integration $$\int \sqrt{cot\theta} d\theta$$
I tried to set $t=\sqrt{cot\theta},t^2=cot\theta$ and substitute into the original integration and get$$-\int\frac{2t^2}{1+t^4}dt$$, but then what can I do?
| Consider that $$t^4+1 = (t^4+2t^2+1)-2t^2 = (t^2+1)^2-(\sqrt{2}\,t)^2 = (t^2+\sqrt{2}\,t+1)(t^2-\sqrt{2}\, t+1)$$
hence:
$$\frac{2t^2}{t^4+1}=\frac{t}{\sqrt{2}\, t^2 -2t+\sqrt{2}}-\frac{t}{\sqrt{2}\, t^2 +2t+\sqrt{2}}\tag{1}$$
and the integral of both terms in the RHS is given by:
$$\frac{1}{2\sqrt{2}}\left(\log(\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/935198",
"timestamp": "2023-03-29T00:00:00",
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How to integrate $\frac{y^2-x^2}{(y^2+x^2)^2}$ with respect to $y$? In dealing with the integration,
$$\int\frac{y^2-x^2}{(y^2+x^2)^2}dy$$
I have tried to transform it to polar form, which yields
$$\int\frac{\sin^2\theta-\cos^2\theta}{r^2}d(r\cos\theta)$$
But, what should I do now to continue?
I am sticking on it now.
| Another approach is
$\int \frac{y^2-x^2}{(x^2+y^2)^2} dy = \int\frac{y^2+x^2}{(x^2+y^2)^2} dy - \int \frac{2x^2}{(x^2+y^2)^2} dy = \int\frac{1}{x^2+y^2} dy - 2x^2\int \frac{1}{(x^2+y^2)^2} dy$
Once in this form, simple variable substitution for the first one and integration by parts or by $y = tan(u)$ substitution shou... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$P(X^2+Y^2<1)$ of two independent n(0,1) random variables Suppose that X and Y are independent n(0,1) random variables.
a) Find $P(X^2+Y^2<1)$
Attempt:
a) Let $U = X^2 + Y^2$, $V = Y$.
Then $X = \sqrt{V^2 -U}$, $Y = V$.
$J = \left| \begin{array}{ccc}
\frac{-1}{\sqrt{V^2-U}} & \frac{V}{V^2-U} \\
0 & 1\\
\end{array} \r... | Although the following method is more convoluted than necessary, I think it provides a fun way to see the convolution method for computing the density of a sum of independent random variables in action. It also requires a few fun integration tricks.
First we will find the distribution for $X^2$ (and, by symmetry, the... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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In $\triangle ABC$, $D$ is a point on side $BC$ that $\angle BAD = \angle CAD =\angle ABC$. If $BD=1$ and $DC=2$, what would be the length of $AB$?
In $\triangle ABC$, $D$ is a point on side $\overline{BC}$ that $\angle BAD = \angle CAD =\angle ABC$. If $\overline{BD}=1$ and $\overline{DC}=2$, what would be the length... | You are correct that the Angle Bisector Theorem gives you that $|AC| = 2|AB|$. Let's take that as our jumping-off point, assigning $x$ and $y$ to lengths as shown:
(I'm using "$x$" instead of "$1$", so that we can follow the value through the formulas better. "$1$"s are so easy to lose!)
By the Law of Cosines in $\tri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/938801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is this proof by induction for a sum of odd squares correct? Statement: $1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = (n/3)*(2n-1)*(2n+1)$
Proof by induction
-Base case: when $n = 1$
$1^2 = 1/3 * (2 * 1 -1) * (2 * 1 +1) = 1$
$1=1$ hence statement holds for $n = 1$
-Inductive step
assume $n = k$ is true
then $1^2 + 3^2 + 5^2 + ... | Your line "hence something complicated is equal to itself" is unconvincing and suggests that you might do better with a different order, perhaps like $$1^2 + 3^2 + 5^2 + ... (2k - 1)^2 + (2(k+1)-1)^2$$ $$=(k/3) * (2k-1)*(2k+1) + (2(k+1) -1)^2$$ $$=(4k^3+12k^2 + 11k + 3)/3$$ $$= ((k+1)/3) * (2(k+1) - 1) * (2(k+1) +1)$$ ... | {
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"url": "https://math.stackexchange.com/questions/939004",
"timestamp": "2023-03-29T00:00:00",
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} |
Integrate $\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$
Integrate $$I=\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$$
Let $$\begin{align}u^2=\frac{\sin(x-a)}{\sin(x+a)}\implies 2udu&=\frac{\sin(x+a)\cos(x-a)-\sin(x-a)\cos(x+a)}{\sin^2(x+a)}dx\\2udu&=\frac{\sin((x+a)-(x-a))}{\sin^2(x+a)}dx\\
2udu&=\frac{\sin(2a)}{\sin^2(x+a)}d... | Since $\sin(x-a)\sin(x+a)=\cos^2(a)-\cos^2(x)=\sin^2(x)-\sin^2(a)$, if we let
$$
u=\frac{\cos(x)}{\cos(a)}\quad\text{and}\quad v=\frac{\sin(x)}{\sin(a)}\tag{1}
$$
then
$$
\begin{align}
&\int\sqrt{\frac{\sin(x-a)}{\sin(x+a)}}\,\mathrm{d}x\\
&=\int\frac{\sin(x-a)}{\sqrt{\sin(x+a)\sin(x-a)}}\,\mathrm{d}x\\
&=\int\left[\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/940037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 5,
"answer_id": 2
} |
Find side of an equilateral triangle inscribed in a rhombus The lengths of the diagonals of a rhombus are 6 and 8. An equilateral triangle inscribed in this
rhombus has one vertex at an end-point of the shorter diagonal and one side parallel to the longer
diagonal. Determine the length of a side of this triangle.
Expr... |
Hopefully one can see anything in this drawing...
You know that all the green and all the blue lines are equal. Additionally, you know that the top green line is parallel to one of the diagonals and that the diagonals are perpendicular. Therefore, the top green line is perpendicular to the short diagonal as well.
From... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/940104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Integral $\int_0^\pi \frac{x\,\operatorname dx}{a^2\cos^2x+b^2\sin^2x}$ Integrate:
$$
\int_0^\pi \frac{x\,\operatorname dx}{a^2\cos^2x+b^2\sin^2x}
$$
| Hint:
We can find the anti-derivative,
$$\begin{align}
\int\frac{\mathrm{d}x}{a^2\cos^2{x}+b^2\sin^2{x}}
&=\int\frac{\sec^2{x}\,\mathrm{d}x}{a^2+b^2\tan^2{x}}\\
&=\int\frac{\mathrm{d}u}{a^2+b^2u^2}\\
&=\frac{\arctan{\frac{bu}{a}}}{ab}+\color{grey}{constant}\\
&=\frac{\arctan{\left(\frac{b\tan{x}}{a}\right)}}{ab}+\color... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/941172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why is $\frac 25$ the real part of $\frac{1}{2+i}$? According to Wolfram Alpha, Re(1/(2+i))=2/5.
How did it calculate that?
| Multiply numerator and denominator by $2-i$:
$$\frac 1{2+i}\cdot \frac{2-i}{2-i} = \frac{2-i}{4-(-1)} = \frac{2-i}{5} = \frac 25 - \frac 15\cdot i$$
Edit for more explanation as to why this strategy works.
We multiply numerator and denominator by the conjugate of $2+ i$, which is $2+i$, to remove the imaginary number ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/941617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 2
} |
Problem with Equilateral Triangle Randomly choose a point $P$ in the interior of an equilateral triangle $ABC$, which has a side of length $a$. Let $D,E,F$ be the feet of the perpendiculars from $P$ to $BC,CA,AB$ respectively. Show that $(PDC)+(PFB)+(PEA)=\frac{a^2\sqrt3}{8}$.
Here are my thoughts so far:
We want to... | Unfortunately, getting a proof from the two equations reported in the OP seems difficult, since the heights of the three triangles are all different.
The proof can be achieved using an alternative method. Orient an equilateral triangle with side $s$ in a way that $A$ is the vertex at the top, and the base $BC$ is at t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/947511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Derivative of $f(x)=-6\sin^4 x$
$f(x)=-6\sin^4x$
$f(x)=-6\sin x^4$
$f'(x)=-6\cos x^4(4x^3)=-24x^3\cos x^4$
What am I doing wrong? Please show the steps.
| Whether $f(x) = -6\sin^4 x = -6(\sin x)^4$ or $f(x) = -6 \sin (x^4)$, the chain rule is the way to go.
If
$f(x) = -6(\sin x)^4, \tag{1}$
set
$u = \sin x \tag{2}$
so that (1) becomes
$f(u(x)) = -6(u(x))^4; \tag{3}$
now we apply the chain rule, which says that $df(u(x))/dx$ is given by
$\dfrac{df(u(x))}{dx} = \dfrac{df}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/952037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Infinite product: $(1-0.5^2)(1-0.5^3)(1-0.5^4)...$ Find a closed form for the value of the infinite product $(1-0.5^2)(1-0.5^3)(1-0.5^4)...$
I know it converges. At first I thought it was the Euler–Mascheroni constant, but it's only accurate to about 3 sig fig. I haven't been able to to solve it yet.
| $$A_n=\prod_{i=2}^n(1-(\frac{1}{2})^{k})=2 \left(\frac{1}{2};\frac{1}{2}\right)_n$$ where appears the Pochhammer symbol. For $$A_{\infty}=\prod_{i=2}^{\infty}(1-(\frac{1}{2})^{k})=2 \left(\frac{1}{2};\frac{1}{2}\right)_{\infty }\approx 0.5775761901732048425577994$$ while $$\gamma \approx 0.5772156649015328606065121$$ I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/952262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Complete the square to change into standard form Here is the equation:
$x^2 + y^2 + 4x - 6y - 3 = 0$
Here are the instructions:
Complete the square to change the equation info standard form. Then graph the equation.
Because both $y^2$ and $x^2$ are present, I do not know what to do. Thank you!
| $$\begin{align}x^2+y^2+4x-6y-3=0&\iff x^2+2\cdot 2x+2^2-2^2+y^2-2\cdot 3y+3^2-3^2-3=0\\&\iff (x^2+4x+4)+(y^2-6y+9)=3+4+9\\&\iff (x+2)^2+(y-3)^2=4^2.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to integrate $\int_{0}^{\infty} \frac{dx}{\sqrt{x}(x^{2}+1)}$ using the residue theorem. He was doing this integral using the formula
$$\int_{0}^{\infty} \frac{dx}{\sqrt{x}(x^{2}+1)}= \frac{2\pi i}{1-e^{-2\pi i\alpha}}(\sum(Res(\frac{F(z)}{z^{\alpha}};z_{k})))$$ where $F(z)=\frac{1}{(x^{2}+1)}$, $\alpha=\frac{1}{2... | Here are some relevant hints including basic complex number arithmetic.
Put $$\sqrt{z} = \exp(1/2\log z)$$
where the logarithm has the branch cut on the positive real axis
(argument from zero to $2\pi$.)
Use a keyhole contour with the slot of the key on the positive real
axis. On the circular part ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/953364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Number of ways the letters of two k-letter words can match $M(k) :=$ the number of ways the letters of two (ordered) k-letter words can match.
$M(1) = 2$: the single letters of the two words are either matching or not.
$M(2) = 12$:
$$\begin{array}{ c|c c }
& A & A \\\hline
B & \circ & \circ \\
B & \circ ... | After some time trying to solve definitively the problem @Tamas finally found a paper with the "solution".
Based on the calculated values of $M$ given by a computer program that checked all possible matrices, $M(n)$ gives the https://oeis.org/A014235 series. It's formula is therfore is $\sum_{k=0}^{n} k! * S(n+1, k+1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/953703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integration without substitution of $\frac{x^2+3}{x^6\left(x^2+1\right)}$ This is a repost of a question i had written incorrectly earlier.
How do I integrate this without substitutions ?
$$
\frac{x^2+3}{x^6\left(x^2+1\right)}
$$
I got:
$$
\frac{1}{x^6}+\frac{2}{x^6\left(x^2+1\right)},
$$
but wasn't able to eliminat... | $$\frac{x^2+3}{x^6(x^2+1)}=\frac{2x^4-2x^2+3}{x^6}-\frac{2}{x^2+1}=2x^{-2}-2x^{-4}+3x^{-6}-2\cdot\frac{1}{x^2+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/954101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Deriving $\frac{8}{\sqrt{x-2}}$ I'm not sure how to derive this:
$$\frac{8}{\sqrt{x-2}}$$
I tried
$$8 \cdot \frac{1}{\sqrt{x-2}}$$
$$8 \cdot (\sqrt{x-2})^{-1}$$
Differentiating w.r.t. $x$,
$$8 \cdot -1 \cdot (\sqrt{x-2})^{-2}$$
$$8 \cdot -1 \cdot \frac{1}{(\sqrt{x-2})^{2}}$$
$$\frac{-8}{(\sqrt{x-2})^{2}}$$
$$\frac{-8}... | $$\frac{d(f(g(x)))}{dx}\ne f'(g(x))\quad\text{instead}\quad \frac{d(f(g(x)))}{dx}= f'(g(x))g'(x)$$
So
$$(f(\sqrt{x-2}))'=f'(\sqrt{x-2})(\sqrt{x-2})'=f'(\sqrt{x-2})\frac1{2(\sqrt{x-2})}$$
When $f(x)=8/x$ then $f'(x)=-8/x^2$
$$f'(\sqrt{x-2})=\frac{-8}{(\sqrt{x-2})^2}\frac1{2(\sqrt{x-2})}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/955066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Proof for $\log\frac{2n+1}{n+1}<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}<\log 2$ How can I prove that
$\displaystyle \log\frac{2n+1}{n+1}<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}<\log2$
using $\displaystyle \frac{x}{1+x}<\log(1+x)<x$
| We have
$$\frac{1}{n+k}>\log\left(1+\frac{1}{n+k}\right) $$
hence:
$$\sum_{k=1}^{n}\frac{1}{n+k}>\log\prod_{k=1}^{n}\frac{n+k+1}{n+k}=\log\frac{2n+1}{n+1}$$
while:
$$\frac{1}{n+k}=\frac{\frac{1}{n+k-1}}{1+\frac{1}{n+k-1}}<\log\left(1+\frac{1}{n+k-1}\right)$$
gives:
$$\sum_{k=1}^{n}\frac{1}{n+k}<\log\prod_{k=1}^{n}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/957977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Can someone explain how to solve linear inequality?
Can someone explain how to solve this linear inequality?
| We have $$\frac{6+(13-x)(x-3)-3(x-3)-2(x+3)}{x^2-9}\le0$$
$$\iff-\frac{(x-6)(x-5)}{(x-3)(x+3)}\le0$$
For the equality $x=6$ or $5$
For the inequality,
$$-\frac{(x-6)(x-5)}{(x-3)(x+3)}<0\iff\frac{(x-6)(x-5)}{(x-3)(x+3)}>0$$
$$\iff[x-(-3)](x-3)(x-5)(x-6)>0 (\text{multiplying either sides by }$(x-3)(x+3)^2>0)$$
Check fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/959421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $4^{7 x - 10} = 10^{7 x - 6}$ for $x$ Solve for $x$ of the following equation.
$$4^{7 x - 10} = 10^{7 x - 6}$$
I tried to make $4$ and $10$ have a common base but I could not find one so I don't know where to go from here.
| $$
\begin{align}
& 4^{7 x - 10} = 10^{7 x - 6} \\[6pt]
\implies & \log_4(4^{7 x - 10}) = \log_4(10^{7 x - 6}) \\[6pt]
\implies & 7x-10 = (7x-6)\log_4(10) \\[6pt]
\implies & 7x-7x\log_4(10) = 10-6\log_4(10) \\[6pt]
\implies & 7x[1-\log_4(10)] = 10-6\log_4(10) \\[6pt]
\implies & 7x = \frac{10-6\log_4(10)}{1-\log_4(10)} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/959918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Prove the given condition from given two quadratic equation
Question: If the quadratic equations $x^2+bx+c=0$ and $bx^2+cx+1=0$ have a
common root then prove that either $b + c + 1 = 0$ or $b^2 + c^2 + 1 =bc + b + c$
Till yet,
I had figured the common root of the given two quadratic equation. i.e.
Multiplying first... | Let $y$ be the common root
So, we have $$y^2+by+c=0, by^2+cy+1=0$$
Solving for $y,y^2$ we get $y^2=\dfrac{b-c^2}{c-b^2},y=\dfrac{bc-1}{c-b^2}$
and using the identity $y^2=(y)^2$ we get $$b^3+c^3+1^3-3\cdot b\cdot c\cdot1=0$$
Now use Factorize the polynomial $x^3+y^3+z^3-3xyz$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/960972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving congruence equations Solve: $7x^6\equiv 11 \pmod{23}$ and $5^x\equiv 19 \pmod{23}$
I can solve simple congruence equations but how do I go about solving these?
| Fermat's little theorem says that $n^{22} \equiv 1 \pmod{23}$, which means that the last equation can be rewritten to an equation of the form $x \equiv a \pmod{22}$. To figure out $a$, we just need all powers of $5$ that is congruent to $19 \pmod{23}$. A quick search reveals that $15$ is the only one.
Also, the first ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/962099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
how to prove that $k+1 \ge (1+\frac{1}{k})^{k}$? How to prove that
$$k+1\ge \bigg(1+\frac{1}{k}\bigg)^{k}
$$ when $k>2$
| We will prove this with induction.
Consider the base case of $k=2$. Clearly, $3\ge 2.25$.
Now assume that $k+1\ge \left(1+\frac{1}{k}\right)^k$. We want to show that $\left(k+1\right)+1\ge\left(1+\frac{1}{k+1}\right)^{k+1}$. Consider
$$\begin{align}
\left(1+\frac{1}{k+1}\right)^{k+1}&=\left(1+\frac{1}{k+1}\right)^k\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/962500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Evaluate $\gcd(ab,p^4)$ and $\gcd(a+b,p^4)$, given that $\gcd(a,p^2)=p$ and $\gcd(b,p^3)=p^2$, where $p$ is a prime. Evaluate $\gcd(ab,p^4)$ and $\gcd(a+b,p^4)$, given that $\gcd(a,p^2)=p$ and $\gcd(b,p^3)=p^2$, where $p$ is a prime. Is it true if $\gcd(a,b)=\gcd(a,c)$, then
$\gcd(a^2,b^2)=\gcd(a^2,c^2)$?
We have $\g... | By assumption, we have that there exist $x,y,z,w\in\mathbb{Z}$ so that
$$
ax+p^2y=p\tag{1}
$$
and
$$
bz+p^3w=p^2\tag{2}
$$
Separating $ax$ and $bz$ and multiplying, then moving the multiple of $p^4$ back to the left, we get
$$
abxz+p^4(w+y-pwy)=p^3\tag{3}
$$
Thus, $\gcd(ab,p^4)\mid p^3$. However, since $p\mid a$ and $p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/964251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Evaluate $ \int_0^\pi \left( \frac{2 + 2\cos (x) - \cos((k-1)x) - 2\cos (kx) - \cos((k+1)x)}{1-\cos (2x)}\right) \mathrm{d}x $ Evaluate the following definite integral:
$$ \int_0^\pi \left( \frac{2 + 2\cos (x) - \cos((k-1)x) - 2\cos (kx) - \cos((k+1)x)}{1-\cos(2x)}\right) \mathrm{d}x, $$ where $k \in \mathbb{N}_{>0}$... | I think we can get the answer into a relatively elementary form. Observe that
$$\cos{(k-1) x} + 2 \cos{k x} + \cos{(k+1) x} = 2 \cos{k x} (1+ \cos{x})$$
Thus the integrand simplifies into
$$4 \int_0^{\pi} dx \cos^2{\frac{x}{2}} \frac{\sin^2{k \frac{x}{2}}}{\sin^2{x}} $$
which is equal to
$$\frac12 \int_0^{2 \pi} du \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/965283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\gcd (n^3-1,n+1)=1$ for all even $n$. Prove that if $n$ is even, then
$$\gcd(n^3-1,n+1)=1.$$
I really don't have a clue with this one. Any help would be appreciated.
| If you polynomial long divide $n+1$ into $n^3-1$ (or use the Euclidean Algorithm), you obtain the expression:
\begin{align*}
n^3-1&=(n+1)(n^2-n+1)-2 \\
\implies 2&=(n+1)(n^2-n+1)-(n^3-1)
\end{align*}
The greatest common divisor of $n^3-1$ and $n+1$ divides both terms on the right, so it must divide their difference, 2,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/966202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
differentiation of the following equation 3 i already done the differentiation, just wanna confirm either i got it right or wrong. Can someone verify this for me.
1) f(x) = $ -3\over x^{5/2}$
f '(x)
= $ 3({ 5\over 2}x^{3/2})$ . $\frac{1}{x^5}$
= $ { 15\over 2}x^{(3/2)-5}$
= ${ 15\over 2}x^{-7/2}$
2) f(x) = $\frac{2x... | I haven't been through all of them, but I suggest you redo 1 and 2.
In 1, it should be $x^{-6}$, not $x^{-5}$
In 2, it should be $-9x^{2}(2x^{2}+3)$, not $-3(2x^{2}+3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/966460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solving for $x$ using exponential log laws For
$\log_2(x) + 2\log_2(x-1) = 2 + \log_2(2x+1)$
I moved all the $x$ to left side, used got rid of log and got
$x-(x-1)^2 - (x+1) = 4$
Simplyifing I get
$x^2-2x=4$
The answer should be $x = 4$ (I checked on wolfram alpha)
Help please?
| Here are the steps I would use:
$\log_2(x) + 2\log_2(x-1) = 2 + \log_2(2x+1)$
$\log_2(x) + \log_2\left((x-1)^2\right) = \log_2\left(2^2\right) + \log_2(2x+1)$
That step turns each term into a base $2$ logarithm. Now they can be combined using the product rule for logs:
$\log_2\left(x(x-1)^2\right) = \log_2\left(4(2x+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/967501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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For which values of $a$ is the solution for $x^2 - y^2 = a^3$ unique?
For which values of $a$ is the solution unique? $$x^2-y^2=a^3$$
I'm not sure how to do this, so I've been looking at this guy's solution.
$x^2 - y^2 = a^3$ is factored into $(x-y)(x+y) = a^3$. He says that the parity of both $x-y$ and $x+y$ is the ... | Lets focus on the case when $a>0$.
Let $a=p^{k_1}_{1}...p^{k_r}_{r}$ be the prime factorisation of $a$ then the number of distinct positive divisors of $a$ is equal to $\tau(a)=(k_1+1)...(k_r+1)$.
Therefore the number of distinct positive divisors of $a^3$ is $\tau(a^3)=(3k_1+1)...(3k_r+1)$. As said above $$x^2-y^2=a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/967943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_0^1\frac{\ln(1-x)}{x}\text{Li}_3\left(\frac{1+x}{2}\right)dx$ , $\int_0^1\frac{\ln^2(1-x)}{x}\text{Li}_2\left(\frac{1+x}{2} \right)dx$ How can we evaluate the following integrals:
$$\int_0^1\frac{\ln(1-x)}{x}\text{Li}_3\left(\frac{1 + x}{2} \right)\,dx\\
.\\
\int_0^1\frac{\ln^2(1-x)}{x}\text{Li}_2\left(\... | $$\begin{align*}{\large\int}_0^1\frac{\ln(1-x)\,\operatorname{Li}_3\left(\frac{1+x}2\right)}xdx&=\frac{29\,\zeta(5)}{16}-\frac{19\pi^2}{96}\zeta(3)+\frac{5\,\zeta(3)}{16}\ln^22+\frac{\ln^52}{40}\\&-\frac{5\pi^2}{72}\ln^32+\frac{11\pi^4}{1440}\ln2-3\operatorname{Li}_5\left(\tfrac12\right).\\
\\
{\large\int}_0^1\frac{\ln... | {
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"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
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Proof by induction of a sum? Let $n ∈ N$. Prove by induction that there are $n$ ways to write the number $n$ as a sum $n=x_1+x_2+...+x_k$ where the $x_i$ are natural numbers and $x_1 ≤x_2 ≤...≤x_k ≤x_1+1$.
For example, $5 = 5$, $5 = 1 + 1 + 1 + 1 + 1$, $5 = 1 + 1 + 1 + 2$, $5 = 1 + 2 + 2$, and $5 = 2 + 3$.
| Here's a rough outline of a strategy: With $n=x_1+\cdots+x_k$, you can write $n+1=x_2+\cdots+x_k+(x_1+1)$. There is one allowed way to write $n+1$ that you can't get in this way, however.
Edit: To consider an example, here are the five ways to write $5$, with $x_1$ marked in red:
$5=\color{red}{5}$, $5=\color{red}{1}+1... | {
"language": "en",
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"source": "stackexchange",
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If $(a,b)=1$, prove that $(a^2+b^2,a+b)=1$ or $2$. If $(a,b)=1$, prove that $(a^2+b^2,a+b)=1$ or $2$.
So far, I let
$d=(a^2+b^2,a+b)$
$\implies d|(a^2+b^2-(a+b)^2)$
$\implies d|(a^2+b^2-(a^2+2ab+b^2))$
$\implies d|(-2ab)$
I have heard from other people that this somehow leads to a conclusion, but I am not seeing it. Al... | Many gcd questions can be answered by finding a way to use the Euclidean algorithm. By the Euclidean algorithm, I mean $(a, aq+r) = (a,r)$.
Here, $a$ is played by $a+b$, $q$ is played by $a-b$, and $r$ is played by $2b^2$. The Euclidean algorithm then implies that
$$(a+b, (a+b)(a-b) + 2b^2) = (a+b, 2b^2).$$
This reads... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/972759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Log integrals II By considering the integral
\begin{align}
I_{\mu} = \int_{0}^{\pi/4} \sin(2\theta) \, \left( \cos(\theta) - \sin(\theta) \right)^{\mu} \, d\theta
\end{align}
derivatives can be taken with respect to $\mu$ to obtain integrals involving logarithms. Let these integrals be
\begin{align}
J_{\mu}^{m} = \pa... | Another possible closed-form
$$I_\mu=\frac{2\cdot{_2F_1}\left(\begin{array}c\tfrac12,2\\\tfrac{\mu+5}{2}\end{array}\middle|\,-1\right)}{(\mu+3)(\mu+1)},$$
for $\Re(\mu)>-1$.
I don't know how to simplify it further, but I think it's a good idea for integer $\mu$ values to separate the even and the odd $\mu$ cases, since... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/972837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Solving the differential equation: $f(x)yy'=(y')^2-0.5$ I am trying to solve this equation:
$f(x)yy'=(y')^2-0.5$
I have already tried traditional methods...
Any ideas?
| We have
\begin{align*}
&y y^\prime = (y^\prime)^2 - 0.5 \\
& (y^\prime)^2 - y y^\prime = 0.5 \\
& (y^\prime)^2 - y y^\prime + \frac{1}{4}y^2 = 0.5 + \frac{1}{4} y^2 \\
& \left(y^\prime - \frac{1}{2} y \right)^2 = 0.5 + \frac{1}{4} y^2 \\
& y^\prime = \frac{1}{2} \left(y \pm \sqrt{1 + y^2} \right)
\end{align*}
You can... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Calc I limit question involing trig functions Find the limit
$$\lim_{x\to2}\frac{\sin^2(x^2-4)\sec^2(3x-6)}{(x^3-8)\tan(2x-4)}$$
i have been having trouble finding this limit, i have tried using having trig identities and making all terms sin and cos but i cant figure it out, keep in mind i am not allowed to use L'hopi... | A little prep work gives us
$${\sin^2(x^2-4)\sec^2(3x-6)\over(x^3-8)\tan(2x-4)}={\sin^2(x^2-4)\over(x-2)\sin(2x-4)}\cdot{\cos(2x-4)\over(x^2+2x+4)\cos^2(3x-6)}$$
Now the second term causes no trouble as $x\to2$. It gives
$$\lim_{x\to2}{\cos(2x-4)\over(x^2+2x+4)\cos^2(3x-6)}={\cos0\over12\cos^20}={1\over12}$$
If we rew... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solving the system $a^2-6=2\sqrt{2c+6}, \, b^2-6=2\sqrt{2a+6}, \, c^2-6=2\sqrt{2b+6}$ Question:
Solve the following system for $a,b,c\in \mathbb{R}$:
$$\begin{cases}
b^2-6=2\sqrt{2a+6}\\
c^2-6=2\sqrt{2b+6}\\
a^2-6=2\sqrt{2c+6}
\end{cases}$$
I found the following:$$ (b^2-6)^2=4(2a+6)$$
$$(c^2-6)^2=4(2b+6)$$
$$(a^2-6... | Let $$ 3T:=a^2+b^2+c^2-18 = 2\sqrt{2a+6} +
2\sqrt{2b+6} + 2\sqrt{2c+6} $$
If $$ a= {\rm max}\ \{ a,b,c\},\ b
= {\rm min}\ \{ a,b,c\},\ a>b$$ then
$$T>b^2-6=2\sqrt{2a+6} > T$$
Contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/975207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to find $\int_0^{\pi/4}\sec(x)^3dx$ How do I find $$\int_0^{\pi/4}\sec(x)^3dx$$
I arrive at $\sec x\tan x + 1/3 \ln| (\cos^3(x))| dx$
where $u = \sec x$ and $v' = \sec^2(x)$
What's my error? Could I get a step by step solution?
| So you're integrating by parts with $u = \sec x$ and $dv = \sec^2 x = (\tan x)'$. Then:
$$
I := \int_{0}^{\frac{\pi}{4}} \sec^3 x dx = \sec \left(\frac{\pi}{4}\right)\tan \left(\frac{\pi}{4}\right) - \int_{0}^{\frac{\pi}{4}}\tan^2 x \sec x dx
$$
Where the last integral is
$$
\int_{0}^{\frac{\pi}{4}}\tan^2 x \sec x dx ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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When $a\to \infty$, $\sqrt{a^2+4}$ behaves as $a+\frac{2}{a}$? What does it mean that $\,\,f(a)=\sqrt{a^2+4}\,\,$ behaves as $\,a+\dfrac{2}{a},\,$
as $a\to \infty$?
How can this be justified?
Thanks.
| This goes back to Newton: The binomial theorem doesn't only work for natural number exponents! If $a > 2$, the following formula holds:
$$\sqrt{a^2+ 4} = (a^2 + 4)^{1/2} = \sum_{k=0}^\infty {{1/2}\choose k}(a^2)^{1/2-k}4^k$$ $$= a + (1/2)\frac{4}{a} + \frac{(1/2)(-1/2)}{2!} \frac{4^2}{a^3} + \cdots$$ $$= a + \frac{2}{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Prove that for all positive integers $x$, $\left\lfloor \frac{x^2 +2x + 2}{4}\right\rfloor =\left\lfloor \frac{x^2 + 2x + 1}{4}\right\rfloor$. Title says it all, basically. I believe it to be true that
$$\left\lfloor \dfrac{x^2 + 2x + 2}{4} \right\rfloor=\left\lfloor \dfrac{x^2 + 2x + 1}{4} \right\rfloor$$ for all posi... | $x$ is either odd or even. If $x$ is even, let $x=2r+1$. Then the right hand side is
$$
\lfloor \frac{(x+1)^2}{4} \rfloor = \lfloor 4 \frac{(r+1)^2}{4} \rfloor
= \lfloor (r+1)^2 \rfloor = (r+1)^2
$$
and the left hand side is
$$
\lfloor \frac{(x+1)^2+1}{4} \rfloor = \lfloor 4 \frac{(r+1)^2+1}{4} \rfloor
= \lfloor ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/977003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Determine whether or not a binary number is divisible by $3$ Let $K$ be a natural number with $n$ binary digits.
Is there an $O(n)$ method for deciding whether or not $K$ is divisible by $3$?
$3|K \iff d_1-d_2+d_3-d_4\dots\pm d_n=0$ works correctly up to $20$, but fails for $21$.
| Another way to prove the accepted answer without using the congruence relation. First, note that we have:
$$
\frac{1}{3} \text{ = } 0+\frac{1}{3} \\
\frac{2}{3} \text{ = } 1-\frac{1}{3} \\
\frac{4}{3} \text{ = } 1+\frac{1}{3} \\
\frac{8}{3} \text{ = } 3-\frac{1}{3} \\
\\ \vdots
$$ And, generally:
$$
\frac{2^k}{3}= \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/979274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Triplets of distinct integers > 1 that return integer values. If $(A, B, C)$ are distinct integers $> 1$, and $$f(A, B, C) = \frac{\frac{A^2-1}{A} + \frac{B^2-1}{B}}{\frac{C^2-1}{C}},$$ then for what (if any) triplets $(A, B, C)$ is $f(A, B, C)$ an integer?
UPDATE: I've done some more work and have come up with the fol... | We can write $f(A,B,C)=\frac {(A^2-1)BC+(B^2-1)AC}{AB(C^2-1)}$ then $f(A,B,C)$ is an integer if $AB(C^2-1)|C((A^2-1)B+(B^2-1)A)$ $\Rightarrow$ because $GCD(C,C^2-1)=1$ then $A|C \ or\ B|C$ and $ (C^2-1)|(A^2-1)B+(B^2-1)A $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/979947",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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minimize a function using AM-GM inequality I want to minimize the function
$$ \frac{x}{1-x^2} + \frac{y}{1-y^2} + \frac{z}{1-z^2} $$ subject to the constraint $$x^2 + y^2 + z^2 = 1 \space\text{and} \space x,y,z > 0$$
Wolfram Alpha tells me that the minimum occurs at $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\s... | We can write the inequality to prove as
$$\frac{x}{1-x^2}+\frac{y}{1-y^2}+\frac{z}{1-z^2}\ge \frac{3\sqrt3}2 \tag{$\star$}$$
As equality is achieved for $x=y=z=\frac1{\sqrt3}$, if $(\star)$ holds then we have established the minimum.
Consider function $f(t) = \dfrac{t}{1-t^2}-\frac{\sqrt 3}2 -\frac{\sqrt 3}2(3t^2-1)$. ... | {
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"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$ How can I calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$? I know that $1+2+\cdots+n=\dfrac{n+1}{2}\dot\ n$. But what should I do next?
| Hint: use also that
$$
1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}6
$$
$$
1 + (1+2) + \dots + (1 +2+\dots +n) =
\frac{1(1+1)}2 + \frac{2(2+1)}2 + \dots + \frac{n(n+1)}2
\\=\frac 12 \left[
(1^2 + 1) + (2^2 + 2 ) + \dots + (n^2 + n)
\right]
\\=\frac 12 \left[
(1^2 + 2^2 + \dots + n^2) + (1 + 2 + \dots + n)
\right]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/980941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 3
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Which of the following numbers does not divide $2^{1650}-1$? I'm practicing for a math competition that is coming up, and I got stuck on this question:
Which of the following numbers does not divide $2^{1650}-1$?
$3$, $7$, $31$, $127$, $2047$
I've seen a link on Yahoo Answers that says this:
Note that $3 = 2^2 - 1$,... | Here is an easier way of understanding that theorem.
For any positive integer $m$ we have: $x^m-1 = (x-1)(x^{m-1}+x^{m-2}+\cdots+x+1)$.
Now set $m = b$ and $x = 2^a$, to get $(2^a)^b - 1 = (2^a-1)[(2^a)^{b-1}+(2^a)^{b-1}+\cdots+(2^a)+1]$.
This simplifies as $2^{ab}-1 = (2^a-1)[\text{some integer}]$.
Hence, $2^{ab}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/982073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $a^2\cos B\cos C+b^2\cos C\cos A+c^2\cos A\cos B\leq2S.$ Prove that in any triangle inequality holds:
$$a^2\cos B\cos C+b^2\cos C\cos A+c^2\cos A\cos B\leq2S.$$
Is gender inequality that occurs right triangle, not an equilateral triangle. For this reason I suspect you have used other ways different from the usual... | Let $H$ be the orthocenter of $ABC$. Then
$$ d(H,BC) = 2R\cos B\cos C $$
hence:
$$ \sum_{cyc} 2R a\cos B\cos C = 2S. $$
If $ABC$ is an acute triangle, then $\cos A,\cos B,\cos C>0$ and $a,b,c < 2R$, hence:
$$ \sum_{cyc} a^2\cos B\cos C < \sum_{cyc} 2R a\cos B\cos C = 2S. $$
However, if $ABC$ is an obtuse triangle, the ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral: $\int_0^{\pi/12} \ln(\tan x)\,dx$ I am trying to evaluate:
$$\int_0^{\pi/12} \ln(\tan x)\,dx$$
I think the integral is quite simple but I am having a hard time evaluating it. I started with the result:
$$\int_0^{\pi/4} \ln(\tan x)\,dx= -G$$
where $G$ is the Catalan's constant. With the change of variables $x\... | Using the Fourier series of $\ln(\tan{x})$,
\begin{align}
&\int^\frac{\pi}{12}_0\ln(\tan{x})\ {\rm d}x\\
=&-2\sum^\infty_{n=0}\frac{1}{2n+1}\int^\frac{\pi}{12}_0\cos\Big{[}(4n+2)x\Big{]}\ {\rm d}x\\
=&-\sum^\infty_{n=0}\frac{\sin\Big[(2n+1)\tfrac{\pi}{6}\Big{]}}{(2n+1)^2}\\
=&\color{#E2062C}{-\frac{1}{2}\sum^\infty_{n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/983044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 8,
"answer_id": 4
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Finding inverse of the rational fucntion I am trying to find the inverse of $f(x)=\frac{2x-1}{x(x-1)}$. So far I figured out that the domain should be (0,1) and the range is R itself. I found that by drawing a graph after I turned the function $f(x)=\frac{1}{x}+\frac{1}{x-1}$. since function is 1-1 and onto between thi... | If
$y = \frac{2x-1}{x(x-1)}
$,
$yx(x-1) = 2x-1$
or
$yx^2-yx-(2x-1) = 0$
or
$yx^2-(y+2)x+1 = 0
$.
Applying the quadratic formula,
the discriminant
$D^2
=(y+2)^2-4y
=y^2+4y+4-4y
=y^2+4
$
so the roots are
$x
= \frac{y+2\pm D}{2y}
= \frac{y+2\pm \sqrt{y^2+4}}{2y}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/983122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Evaluating $(\frac{\cos x}{1-\sin x})^2$ $(\dfrac{\cos x}{1-\sin x})^2$
$f\;'(x)= 2(\dfrac{\cos x}{1-\sin x}) \times (\dfrac{-\sin x+\sin^2x-\cos^2x}{(1-\sin x)^2})$
Does $\sin^2x-\cos^2=1$? or $-1$? Then it could factor with the bottom and the answer would be $\dfrac{2\cos x}{(1-\sin x)^2}$. Is this right?
| Using $\sin^2 x + \cos^2 x=1$,
$$f\;'(x)= 2(\dfrac{\cos x}{1-\sin x}) \times (\dfrac{-\sin x(1-\sin x)+\cos^2x}{(1-\sin x)^2})= 2(\dfrac{\cos x}{1-\sin x}) \times (\dfrac{1-\sin x}{(1-\sin x)^2})=\frac{2\cos x}{(1-\sin x)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/983530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How can I calculate $\lim_{x \to 0}\left(\frac{\sin{x}-\ln({\text{e}^{x}}\cos{x})}{x\sin{x}}\right)$ limit without using L'Hopital's rule? $$\lim_{x \to 0}\left(\frac{\sin{x}-\ln({\text{e}^{x}}\cos{x})}{x\sin{x}}\right)$$
Can this limit be calculated without using L'Hopital's rule?
| Clearly $$\begin{aligned}L &= \lim_{x \to 0}\frac{\sin x - \log(e^{x}\cos x)}{x\sin x}\\
&= \lim_{x \to 0}\frac{\sin x - x - \log \cos x}{x^{2}}\cdot\frac{x}{\sin x}\\
&= \lim_{x \to 0}\frac{\sin x - x - \log \cos x}{x^{2}}\\
&= \lim_{x \to 0}\frac{\sin x - x}{x^{2}} - \frac{\log \cos x}{x^{2}}\\
&= A - B\end{aligned}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/983893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Simplifying radicals I am stuck in the following puzzle and couldn't find a way to approach this.
$\sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{14 + \sqrt{180}}}}$
Please help.
| Hint:
$$\eqalign{
& \sqrt{14 + \sqrt {180}} = 3 + \sqrt 5 \cr
& \sqrt {3 + \sqrt 5 + 3 + \sqrt 5 } = \sqrt {6 + 2\sqrt 5 } = 1 + \sqrt 5 \cr
& \sqrt {5 + \sqrt 5 + \sqrt {3 + \sqrt 5 + 3 + \sqrt 5 } } = \sqrt {6 + 2\sqrt 5 } = 1 + \sqrt 5 \cr} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/984040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Fourier series representation of $\sin^4 x$ I tried solving for fourier coefficients of Fourier series for the multiples of fundamental frequency $\omega_0=2$. So $F_n=\int_0^{\pi} \sin^4 x \, e^{-i2nx} dx$. And my calculator says answer should be $3e^{-i2nx}/8$.
But for some reasons I get $i(2+1/(4-2n)-1/(4+2n))/(16\p... | Simply rewrite $\sin^4 x$ in terms of multiple angle trigonometric functions. You will get a finite series.
In fact
$$
1 - 2 \sin^2 x = \cos(2x) \quad \text{and} \quad \cos(2x) = 2\cos^2(x)-1\\
\Rightarrow \sin^2 x = \frac{1-\cos(2x)}{2} \\
\Rightarrow \sin^4 x = \frac{1+\cos^2(2x)-2\cos(2x)}{4} \\
\Rightarrow \sin^4 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/985920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find derivative of $f(x)=\frac{1}{\sqrt{x+2}}$ by definition Use the definition of a derivative to find the derivative of:
$$f(x)=\frac{1}{\sqrt{x+2}}+2x$$
my work:
$$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
$$\lim_{h\to0}\frac{\frac{1}{\sqrt{x+h+2}}+2(x+h)-\frac{1}{\sqrt{x+2}}-2x}{h}$$
$$\lim_{h\to0}\frac{\frac{1}{\sqrt{x+... | Hint: Note that
$$\lim_{h\to0}\frac{\frac{1}{\sqrt{x+h+2}}+2h-\frac{1}{\sqrt{x+2}}}{h}=\lim_{h\to 0}\frac{1}{h}\left[\frac{1}{\sqrt{x+h+2}}-\frac{1}{\sqrt{x+2}}\right]+\lim_{h\to0}\frac{2h}{h}=\lim_{h\to 0}\frac{1}{h}\left[\frac{1}{\sqrt{x+h+2}}-\frac{1}{\sqrt{x+2}}\right]+2$$
Now look that:
$$\lim_{h\to 0}\frac{1}{h}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/986541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solve complex equation $z^3 = i$ I have this $z^3 = i$ complex equation to solve.
I begin with rewriting the complex equation to $a+bi$ format.
1 $z^3 = i = 0 + i$
2 Calculate the distance $r = \sqrt{0^2 + 1^2} = 1$
3 The angle is $\cos \frac{0}{1}$ and $\sin \frac{1}{1}$, that equals to $\frac {\pi}{2}$.
4 The complex... | Way easier way;
$$z^3=i \\
\iff z^3-i=0 \\
\stackrel{-i=i^3}{\iff}z^3+i^3=0 \\
\iff (z+i)(z^2-iz-1) = 0 \\
\iff z_1=-i,\; z_2=\frac12 (i-\sqrt 3), \; z_3=\frac12 (i+\sqrt 3)$$
Disregard this answer if your exercises restrict you to trigonometric/polar form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/987222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Find the area of triangle APB, where P is a point $(a\cos\theta, b\sin\theta)$ on an ellipse and $A, B$ are its radii points $(a,0) (0,b)$
A point $P(a\cos\theta, b\sin\theta)$ sits on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The points $A$ and $B$ have coordinates $(a,0)$ and $(0,b)$ respectively. Show th... | A triangle with two sides given by vectors ${\bf x}, {\bf y}$ has area
$$\frac{1}{2}|{\bf x} \times {\bf y}|.$$ In this case we may take our vectors to be
\begin{align}
{\bf x} &= (a \cos \theta, b \sin \theta) - (a, 0) = (a(\cos \theta - 1), b \sin \theta)\\
{\bf y} &= (0, b) - (a, 0) = (-a, b).
\end{align}
Su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/991641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding a function with certain properties I ran into a problem, and I'm not sure how to continue.
Problem: Let $f$ be a function such that $\sqrt {x - \sqrt { x + f(x) } } = f(x)$, for $x > 1$. In that domain, $f(x)$ has the form $\dfrac{a+\sqrt{cx+d}}{b}$, where $a$, $b$, $c$, $d$ are integers and $a$, $b$ are relati... | We have $f(x)=O(\sqrt x)$. Note that $\lim_{x\to\infty}\frac{f(x)}{\sqrt x}=\frac{\sqrt c}{b}$, whereas
$\lim_{x\to\infty}\frac{\sqrt{x-\sqrt{x+f(x)}}}{\sqrt x}=1$. So we conclude $c=b^2$.
Then $$ \sqrt{x+f(x)}=x-f(x)^2=x-\frac{a^2+2a\sqrt{b^2x+d}+b^2x+d}{b^2}=-\frac{(a^2+d)+2a|b|\sqrt{x+d/b^2}}{b^2}$$
Repeat the trick... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Use modular arithmetic to show that a number is divisible by 11 iff the sum of its alternating digits is divisible by 11 We can expand the number $n = n_0 + 10n_1 + ... + (10^s)n_s$
Then we have $10^k ≡ (-1)^k \mod11$.
How do we go from here to here:
$n ≡ n_0 - n_1 + ... + (-1)^s n_s \mod 11$
I do not understand how t... | We know that $10^k \equiv (-1)^k \pmod{11}$. If $k$ is even, then $(-1)^k \equiv 1 \pmod{11}$. If $k$ is odd, then $(-1)^k \equiv -1 \pmod{11}$. Hence,
\begin{align*}
n & \equiv n_0 + 10n_1 + 10^2n_2 + 10^3n_3 + \cdots + 10^sn_s \pmod{11}\\
& \equiv n_0 + (-1)^1n_1 + (-1)^2n_2 + (-1)^3n_3 + \ldots + (-1)^sn_s \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to find the roots of $x^4-i=0$ I need the manually analysis to calculate the roots without using the numerical methods
| $$x^4-i=0$$
$$x^4=i=e^{i(2k+1)\pi/2}=\cos\left(\frac{(2k+1)\pi}{2}\right)+i\sin\left(\frac{(2k+1)\pi}{2}\right)$$
$$x=i^{1/4}=e^{i(2k+1)\pi/8}=\left[\cos\left(\frac{(2k+1)\pi}{2}\right)+i\sin\left(\frac{(2k+1)\pi}{2}\right)\right]^{1/4}$$
Using De Moivre's formula
$$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin (n\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Determining unknown coefficients of cubic splines The problem : Find $c$ in the following cubic spline.
$S \scriptstyle{1}$$(x)$ = $\large4 - \large\frac{11}{4}x + \large\frac{3}{4}x^3$, on $[0,1]$
$S \scriptstyle{2}$$(x)$ = $\large2 - \frac{1}{2}(x-1) + c(x-1)^2 -\frac{3}{4}(x-1)^3$, on $[1,2]$
My attempt:
1.)... | A cubic spline matches second derivatives as well. So we get, at $x=1$
$$
\frac{3}{4} 6x = 2c
$$
from which $c = \frac{9}{4}$.
The best explanation of cubic splines in my opinion is in Numerical Recipes, pages 120-124 or so.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/995781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluating $\int\frac{x^3}{\left(\sqrt{4x^2+9}\right)^3}\,dx$ using a trigonometric substitution I have this integral:
$$\int\frac{x^3}{\left(\sqrt{4x^2+9}\right)^3}\,dx$$
I tried to solve it with a trigonometric substitituon but I can't get any result.
I would appreciate if somebody could help me.
| Hint:
Put $x = \frac{3}{2}\tan\theta \Rightarrow dx = \frac{3}{2} \sec^2 \theta \ d\theta$, we have
\begin{align}
\int\frac{x^3}{\left(\sqrt{4x^2+9}\right)^3}\,dx&=\frac{3}{16}\int\frac{\tan^3\theta}{\sec^3\theta}\cdot\sec^2\theta\,\,d\theta\\
&=\frac{3}{16}\int\frac{\sin^3\theta}{\cos^3\theta}\cdot\cos\theta\,\,d\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/997462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find all numbers $c$ that satisfy Mean Value Theorem Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all numbers $c$ that satisfy the conclusion of the Mean Value Theorem.
My function is a simple one: $x^{1/3}$ on the interval $[0,1]$.
Where I'm screwing up (... | $$\dfrac{1}{3}x^{-2/3} = 1 \\ \implies \dfrac{1}{3}=x^{2/3} \\ \implies \left(\dfrac{1}{3}\right)^{3}=x^2 \\ \implies \sqrt{x^2}= \pm \sqrt{ \left(\dfrac{1}{3}\right)^{3}}$$ And for what it's worth, $$\sqrt{ \left(\dfrac{1}{3}\right)^{3}} = \sqrt{ \left(\dfrac{1}{3}\right)^{2}\left(\dfrac{1}{3}\right)} \\ = \left(\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/997775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$a^2\equiv 10\pmod b$ and $a^3\equiv 33\pmod b$ Let $a,b$ be positive integers such that $a^2\equiv 10\pmod b$ and $a^3\equiv 33\pmod b$. What are all possible values of $b$?
We have that $10a\equiv 33\pmod b$, but how does that determine the value of $b$?
[Source: Russian competition problem]
| For integers c, d, n, and k we have $c\equiv d \mod n \Rightarrow c^k\equiv d^k \mod n$. Then, using Mark Bennet's hint, we have $a^6\equiv 1000\mod b$ and $a^6\equiv 1089\mod b \Rightarrow 1000+bn=1089+bm\Rightarrow 89=b(n-m)$. But 89 is a prime number, so $b=1$ or $b=89$. $a=30$ and $b=89$ works. Lots of solutions ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/998697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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What is the remainder on division of $z^{400} + z^{303} + 1$ by $z^4-1$? I am asked to determine the remainder when the polynomial $f(z)=z^{400}+z^{303}+1$ is divided by the polynomial $g(z)=z^4-1$.
I expressed f as $f(z) = h(z)(z^4-1) + r(z)$ where $r(z)$ is a polynomial
Realising that $z^4-1 = (z^2-1)(z^2+1) = (z-1)(... | Note that we can write $f(z)=h(z)(z^4-1)+r(z)$ where $r(z)$ is a polynomial of degree at most three (one less than the degree of $z^4-1$).
An unknown polynomial of degree $3$ has four coefficients to be determined. If we successively use $z=1, z=-1, z=i, z=-i$ - the four roots of $z^4-1$ the term involving $h(z)$ will ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Second order cone with quadratic interpretation Could you please help me to understand how the second part of the equation (quadratic form) derived form the first one?
The basic definition of the second-order cone is:
$C = \big\{(x,t) \in \mathbb{R}^{n+1} | \|x\|_2 \leq t \big\}= \big\{(x,t) \in \mathbb{R}^{n+1} | x^Tx... | Hint: Why don't you try expanding that quadratic form in terms of $x$ and $t$.
\begin{align}
\begin{pmatrix}
x\\
t
\end{pmatrix}^T
\begin{pmatrix}
I & 0\\
0 & -1
\end{pmatrix}\begin{pmatrix}
x\\
t
\end{pmatrix}
\end{align}
First expand
\begin{align}
\begin{pmatrix}
I_{n\times n} & 0_{n\time... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simple first order differential equation I've got another first order differential that has me stumped. Please see my work and let me know where I'm going wrong:
$$y' = \frac{6x^2}{y(1+x^3)}$$
$$y'y = \frac{6x^2}{1+x^3}$$
$$ y.\frac{dy}{ dx} = \frac{6x^2}{1+x^3} $$
$$\int y dy = \int 6x^2\frac{1}{1+x^3} dx$$
$$\frac{1... | Unlike the integral of a sum or difference of functions , where $$\int (f(x) \pm g(x))\,dx = \int f(x) \,dx \pm \int g(x)\,dx,$$ the integral of the product of functions is not the product of the integrals of each function. I.e., $$\int f(x)g(x)\,dx \neq \left(\int f(x)\,dx\right)\cdot \left(\int g(x) \,dx\right)$$
An... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding $\int_{0}^{\pi/2} \frac{\tan x}{1+m^2\tan^2{x}} \mathrm{d}x$ How do we prove that $$I(m)=\int_{0}^{\pi/2} \frac{\tan x}{1+m^2\tan^2{x}} \mathrm{d}x=\frac{\log{m}}{m^2-1}$$
I see that $$I(m)=\frac{\partial}{\partial m} \int_{0}^{\pi/2} \arctan({m\tan x}) \ \mathrm{d}x$$
But I don't see how to use this fact. Can ... | $\def\artanh{{\rm{artanh}}\;}$Denote the evaluated integral as $I$ and rewrite it as follows
\begin{align}
I&=\frac{1}{2}\int_0^{\Large\frac{\pi}{2}}\frac{\sin2x}{\cos^2x+m^2\sin^2x}\,dx\\
&=\int_0^{\Large\frac{\pi}{2}}\frac{\sin2x}{m^2+1-(m^2-1)\cos2x}\,dx\\
&=\frac{1}{2}\int_0^{\large\pi}\frac{\sin x}{m^2+1-(m^2-1)\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
How to solve $\left|\frac{x+4}{ax+2}\right| > \frac1x$ How to solve:
$$\left|\frac{x+4}{ax+2}\right| > \frac{1}{x}$$
What I have done:
I) $x < 0$:
Obviously this part of the inequation is $x\in(-\infty, 0), x\neq \frac{-2}{a}$
II) $x > 0$:
$$\left|\frac{x+4}{ax+2}\right| > \frac1x$$
$$\frac{|x+4|}{|ax+2|} > \frac1x$$
b... | The answer depends on the discriminants of the two quadratic polynomials:
$D_1= (a-4)^2+8$ and $D_2= (a+4)^2-8$. This shows that you have up to 4 roots.
Can you go on from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Unable to find the sum of a series I am trying to find the sum of the following series:
$$\sum_{n=1}^{\infty} {\frac{1+7^n}{9^n}}$$
which I rewrote as
$$\sum_{n=1}^{\infty} \left(\frac{1}{9^n}+
\left(\frac{7}{9}\right)^n\right)$$
I am assuming that it is a geometric series and the initial value is
$$a_1=\frac{1}{9} + ... | This is not a geometric series.
But it is the sum of two geometric series:
$$\sum_{n=1}^{\infty} {\frac{1+7^n}{9^n}}
= \sum_{n=1}^{\infty} {\frac{1}{9^n}}+
\sum_{n=1}^{\infty} {\frac{7^n}{9^n}}
=\frac 19 \frac 1{1-\frac 19} +\frac 79\frac 1{1-\frac 79}$$
because
$$
\left|\frac 19\right|<1
\\\left|\frac 79\right|<1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Where is the error in this proof : Prove that:
$$\frac {2\Gamma'(2z)}{\Gamma(2z)}-\frac {\Gamma'(z)}{\Gamma(z)}-\frac {\Gamma \prime(z+\frac{1}{2})}{\Gamma(z+\frac{1}{2})} =2 \log 2$$
But I obtain this equal zero.
My proof:
From Weierstrass definition of Gamma we have
$$\frac{1}{\Gamma(z)}=z.e^{\gamma z}.\prod_{n=1}^... | Let's rewrite your equation :
$$\tag{1}\frac {2\Gamma'(2z)}{\Gamma(2z)}-\frac {\Gamma'(z)}{\Gamma(z)}-\frac {\Gamma \prime(z+\frac{1}{2})}{\Gamma(z+\frac{1}{2})} =2 \log 2$$
as
$$\left[\log\Gamma(2z)-\log\Gamma(z)-\log \Gamma\left(z+\frac 12\right)\right]'=2 \log 2$$
or
$$\tag{2}\left[\log\frac{\Gamma(2z)}{\Gamma(z)\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A thinking problem of limit from my teacher. Please find the limit:$$\mathop {\lim }\limits_{n \to \infty } n\left[ {{{\left( {\frac{1}{\pi }\left( {\sin \left( {\frac{\pi }{{\sqrt {{n^2} + 1} }}} \right) + \sin \left( {\frac{\pi }{{\sqrt {{n^2} + 2} }}} \right) + \cdots+ \sin \left( {\frac{\pi }{{\sqrt {{n^2} + n} }}... | Step 1. Since for small $x$ we have $\frac{1}{\sqrt{1+x}}=1-\frac{x}{2}+\frac{3x^2}{8}+O(x^3),$
$$\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{1+\frac{k}{n^2}}}=\frac{1}{n}\sum_{k=1}^{n}\left(1-\frac{k}{2n^2}+\frac{3k^2}{8n^4}\right)+O\left(\frac{1}{n^3}\right)$$
hence:
$$\sum_{k=1}^{n}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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prove that $\frac{ab}{a+b}+\frac{ac}{a+c}+\frac{ad}{a+d}+\frac{bc}{b+c}+\frac{bd}{b+d}+\frac{cd}{c+d}\le \frac{3}{4}$ Let a,b,c,d be positive real numbers such that$\ a+b+c+d=1$, than prove that $$\frac{ab}{a+b}+\frac{ac}{a+c}+\frac{ad}{a+d}+\frac{bc}{b+c}+\frac{bd}{b+d}+\frac{cd}{c+d}\le \frac{3}{4}$$
This question is... | By AM-GM:
$ab\le \left(\frac{a+b}{2}\right)^2=\frac{(a+b)^2}{4}$
Thus
$$\sum_{\text{sym}}\frac{ab}{a+b}\le \sum_{\text{sym}}\frac{a+b}{4}=\frac{3}{4}$$
Equality is achieved iff $a=b=c=d$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
running time of an algorithm I am trying to prove an algorithm with input size $n$ satisties the recurence relation (for $n>=1$)
$T(n) = T(n-1)+n$ and an initial condition of $T(1)=1$ has running time in $Θ(n^2$).
By using telescope, I've got up to the point where I got
$T(n) = n+(n-1)+(n-2)+......+ 1$
But I can not... | First note that
$$ n+(n-1)+(n-2)+\cdots +1$$
Can be rewritten as
$$ 1+2+\dots +n$$
Now let's use mathematical induction
Hypothesis
$$ \forall n\in\mathbb{N}:n\ge 1$$
$$ 1+2+\dots +n=\frac{n(n+1)}{2} $$
Basis Step
Let $n=1$,
$$ 1=\frac{1\cdot (1+1)}{2} $$
$$ 1=\frac{1\cdot 2}{2} $$
$$ 1=1 $$
Inductive Step
Let $n=k+1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Subtracting Quarters of Squares Equals Multiply?! Can anyone explain to me how/why this works (hopefully in mostly layman's terms)?
It seems pretty magical to me at the moment.
$${{(a+b)^2\over4} - {(a-b)^2\over4}} = a b.$$
| In addition to the direct derivations already shown,
your magical equation is closely related to the formula
$$x^2 - y^2 = (x + y)(x - y).$$
Just set $a = x + y$ and $b = x - y.$
Then $\frac{a+b}2 = x$ and $\frac{a-b}2 = y,$
so $x^2 = \frac{(a+b)^2}4$ and $y^2 = \frac{(a-b)^2}4.$
Use these facts to replace $x^2,\ y^2,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 7,
"answer_id": 4
} |
Find minima of the multivariable function $\frac{4}{x^2+y^2+1}+2 xy$ $$f(x,y)=\frac{4}{x^2+y^2+1}+2 xy \\ \text{within the domain: }1/5\leq x^2+y^2\leq 4$$
I am able to find the maximum of the function at $x^2+y^2=4$ by substituting x,y for $\cos(t)$ and $\sin(t)$ and I am able to start working for a solution in the l... | $f_x(x,y) = 0 \to \dfrac{-8x}{(x^2+y^2+1)^2} = 2y$
$f_y(x,y) = 0 \to \dfrac{-8y}{(x^2+y^2+1)^2} = 2x$.
Thus $(x,y) = (0,0)$ is a critical point, also $\dfrac{x}{y} = \dfrac{y}{x} \to x = \pm y$
Consider $g(x) = f(x,x) = \dfrac{4}{2x^2+1} + 2x^2$ with $1/5 \leq 2x^2 \leq 4$. let $u = 2x^2$, then $h(u) = g(x,x) = \dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation $\lvert z\rvert^2z-3\overline z=0$ I was trying to solve the equation using the identities $z=x+iy$; $\overline z=x-iy$ and $\lvert z\rvert^2=x^2+y^2$ so as to get
$(x^2+y^2)(x+iy)-3(x-iy)=0$, that is $x^3+x^2iy+xy^2+iy^3-3x+3iy=0$
Now, a complex number is null $\iff Re(z)=Im(z)=0$, therefore
$x^3+xy... | HINT:
As $|z|^2=z\cdot\bar z,$
we have $\bar z(z^2-3)=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Lesser known derivations of well-known formulas and theorems What are some lesser known derivations of well-known formulas and theorems?
I ask because I recently found a new way to derive the quadratic formula which didn't involve completing the square as is commonly taught. Doing so I was wondering what other proofs a... | The proof in
N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58
(1951), 563
that
$(1+1/n)^n
\to e$
which uses the arithmetic-geometric mean
inequality (AGMI),
which we will use in the
form
$((v_1+v_2+...v_n)/n)^n > v_1v_2...v_n$
(all $v_i$ positive)
with equality if and only if all the $v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1017602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
"answer_id": 5
} |
How can I prove $\pi ^2=\sum_{n=0}^{\infty }\frac{1}{(2n+1+\frac{a}{3})^2}+\frac{1}{(2n+1-\frac{a}{3})^2}$ Proving this formula
$$
\pi^{2}
=\sum_{n\ =\ 0}^{\infty}\left[\,{1 \over \left(\,2n + 1 + a/3\,\right)^{2}}
+{1 \over \left(\, 2n + 1 - a/3\,\right)^{2}}\,\right]
$$
if $a$ an even integer number so that
$$
a \ge... | Alternatively, let's consider
$$f(a)=\sum_{n=0}^{\infty }\left(\frac{3}{2n+1-\frac{a}{3}}-\frac{3}{2n+1+\frac{a}{3}}\right)=9\sum_{n=0}^{\infty }\left(\frac{1}{6n+3-a}-\frac{1}{6n+3+a}\right)$$
so that our original sum is $f'(a)$.
$$\begin{align}
f(a)&=9\sum_{n=0}^{\infty }\int_0^1 \left(x^{6n+2-a}-x^{6n+2+a}\right)\,d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Find the interval of convergence of $x + \frac{1}{2} x^2 + 3x^3 + \frac{1}{4}x^4 +...$ How to find the interval of convergence of the following series: $x + \frac{1}{2} x^2 + 3x^3 + \frac{1}{4}x^4 +...$
I have no idea what to proceed. Any help? Thanks!
| I would recommend splitting the $x^k*k$ and the $\frac{x^k}{k}$ terms and seeing where the two intervals intersect.
$(∑(2k+1)x^{2k+1}) + (∑\frac{x^{2k}}{2k})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Why can we square both sides when LHS and RHS is square rooted? When we have something in this form
$$\sqrt{x + a} = \sqrt{y + b},$$
a common technique to solve is to square both side so that:
$$(\sqrt{x + a})^2 = (\sqrt{y + b})^2 \implies x + a = y + b.$$
I'm an engineer and not a mathematician. As I understand it eng... | Travelling down the alternative road, we see that
\begin{align*}
\sqrt{x+a} &= \sqrt{y+b} & \\
\sqrt{x+a}\times\sqrt{y+b} &= \sqrt{y+b}\times\sqrt{y+b} &(\text{multiplying both sides by}\ \sqrt{y+b})\\
\sqrt{x+a}\times\sqrt{y+b} &= y + b &(\text{simplifying})\\
\sqrt{x+a}\times\sqrt{x+a} &= y + b &(\text{using the equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1025671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integration of $\exp[f(x,y)]$ Here is the question i want to solve.
$$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \exp\left[{-2\over3}(y^2-yz+z^2)\right]\,dy\,dz$$
I know that $\exp$ is $e^{f(x)}$ and i can find $\int e^{f(x)}\,dx$ but with two variables makes me confusing.
| \begin{align}
\text{Let }u & = y+z, \\
\text{and }v & = y-z. \\[15pt]
\text{It follows that }y & = \frac{u+v} 2, \\[6pt]
\text{and }z & = \frac{u-v} 2.
\end{align}
Then
\begin{align}
y^2-yz+z^2 & = \left( \frac{u+v} 2 \right)^2-\frac{u+v} 2 \cdot \frac{u-v} 2 + \left(\frac{u-v} 2\right)^2 \\[6pt]
& = \frac{u^2+3v^2} 4.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral with branch cut ( Problem while calculating residue) While calculating this integral $\int_{-1}^{1}\frac{dx}{\sqrt{1-x^2}(1+x^2)}$ , I am really struggling to calculate the residue at (-i), I am getting the value of residue as $\frac{-1}{2\sqrt{2}i}$, but for the value of residue is $\frac{1}{2\sqrt{2}i}$ with... | We will do the residue calculation since it is necessary to conclude on
this problem.
Suppose we seek to compute
$$Q = \int_{-1}^{+1} \frac{dx}{\sqrt{1-x^2}(1+x^2)}.$$
Re-write this as
$$\int_{-1}^{+1}
\exp(-1/2\mathrm{LogA}(1+z))
\exp(-1/2\mathrm{LogB}(1-z))
\frac{1}{1+z^2} dz$$
and call the function $f(z).$
We mus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1027015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Problem involving cube roots of unity Given that $$\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}=2\omega^2\;\;\;\;\;(1)$$ $$\frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}=2\omega\;\;\;\;\;(2)$$
Find $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=?\;\;\;\;\;\;(3)$$
I tried adding the given two eq... | I would like to add to the above answer, the symmetry put a spark in my brain and i came up with this.
As in answer stated by peiter.
Assume
$X=\large\frac{1}{a+1}+\large\frac{1}{b+1}+\large\frac{1}{c+1}$ $\space\space\space\space\space\space\space\space$ and
$\large\frac{1}{a+\omega}+\large\frac{1}{b+\omega}+\large\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1030747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Does this system of simultaneous Pell-like equations have any non-trivial positive integer solutions? Let $a,b,c$ be positive integers satisfying
\begin{align}
2a^2-1 &= b^2, \\
2a^2+1 &= 3c^2.
\end{align}
The trivial solution is $(a,b,c)=(1,1,1)$. Are there others?
|
NOTE: As pointed out by Erick Wong, there was a flaw in the original proof. It may be fixable, though I haven't found a fix yet; leaving here for history (and just in case anyone else can take this method to the goal line).
Theorem. The Diophantine system of equations
\begin{align}
2a^2-1 &= b^2, \\
2a^2+1 &= 3c^2
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 1,
"answer_id": 0
} |
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.