Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Is it correct? $\tan x + \cot x \ge 2$ proof The question is to prove $\tan x + \cot x \ge 2$ when $x$ is an acute angel.
This is what I did
$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &... | Even if your implications are correct, your logic is flawed.
Logically, what you have done is similar to
$$\begin{align}
+1&=-1
\\ \Rightarrow (+1)^2&=(-1)^2
\\ \Rightarrow 1&=1.
\end{align}$$
Now $1=1$ so $+1=-1$ must be true also.
Deriving a true statement says nothing about the truth of the hypothesis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 6
} |
Proving formula for sum of squares with binomial coefficient $$\sum_{k=0}^{n-1}(k^2)= \binom{n}{3} + \binom{n+1}{3}$$
How should I prove that it is the correct formula for sum of squares?
Should I use induction to prove the basis? Any help is appreciated.
| Using repeated differences we get
$$
\begin{array}{llll}
0 & 1 & 5 & 14 & \\
1 & 4 & 9 & \\
3 & 5 & \\
2& \\
\end{array}
$$
The last difference repeats forever, because $k^2$ has degree $2$.
Newton's interpolation formula then gives us this formula for the sequence of sums:
$$
0 \binom{n-1}{0} + 1 \binom{n-1}{1} + 3 \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Closed-form of $\int_0^{\pi/2}\frac{\sin^2x\arctan\left(\cos^2x\right)}{\sin^4x+\cos^4x}\,dx$ I have just seen two active posts about integrals of inverse trigonometric function, $\arctan(x)$, here on MSE. So I decide to post this question. This integral comes from a friend of mine (it's not a homework problem) and we ... | $\newcommand{\al}[1]{\begin{align}#1\end{align}} \renewcommand{\Im}{\operatorname{Im}}$Result:
$$I = \frac{\pi \left[2 \pi +\log \left(-4 \sqrt{17+13 \sqrt{2}}+8 \sqrt{2}+13\right)-4 \tan ^{-1}\left(\sqrt{\frac{1}{\sqrt{2}}-\frac{1}{2}}+\frac{1}{\sqrt{2}}+1\right)\right]}{8 \sqrt{2}} \\\approx 0.299397.$$
The evalua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
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.
| By indeterminate coefficients:
$$(x-1)(ax^5+bx^4+cx^3+dx^2+ex+f)\\=ax^6+(b-a)x^5+(c-b)x^4+(d-c)x^3+(e-d)x^2+(f-e)x-f.$$
After identification, $$a=b=c=d=e=f=1.$$
| {
"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": 0
} |
Integral of Rational Functions $$\int \frac{dx}{ax^2 + bx + c} \quad \text{for} \quad 4ac-b^2 >0$$
then
$$\begin{align}
ax^2 + bx + c
&= a\biggl(x+\frac{b}{2a}\biggr)^2 + \frac{4ac-b^2}{4a} \\
&= \Biggl(\sqrt{a}\biggl(x+\frac{b}{2a}\biggr)\Biggr)^2 + \Biggl(\sqrt{\frac{4ac-b^2}{4a}}\Biggr)^2
\end{align}$$
implies
$$ \i... | Hard to debug, since some detail is missing. To make things less messy, I would rewrite the integral as
$$ \int\frac{4a\,dx}{(2ax+b)^2+4ac-b^2}.$$
Let $4ac-b^2=K^2$. Then let $2ax+b=Ku$. We have $2a \,dx=K\,du$. So our integral is
$$\int \frac{2K\,du}{K^2u^2+K^2},$$
which is
$$\frac{2}{K}\arctan u+C.$$
Remark: I am ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
finding the jordan canonical with one eigenvalue Let $$A= \left[
\begin{array}{ c c }
-2 & -3 & 1 \\
0 & -2 & 0 \\
0 & 0 & -2
\end{array} \right]
$$
Determine the Jordan canonical form of A.
The only eigenvalue I found was $\lambda=-2$
So finding the eigenvector of this is
$$ \left[
\begin{array}{... | Hint: You can find a third generalized and linearly independent eigenvector by solving (use Gaussian Elimination):
$$[A-\lambda I]v_3 = [A + 2I]v_3 = v_1$$
You should get:
$$v_3 = \left(0, -\dfrac 13, 0 \right)$$
You can then use the eigenvectors to find (this uses your current eigenvectors) and $v_3$ as:
$$J = P^{-1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1058241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Series representation
I have started this problem by expanding it so that i can get some cancellation term, but couldn't reach on the correct result.I got the result like -ln4 -ln5-ln6..... Please have a look on this.
| Consider the sum:
$$S=\sum_{n=1}^m \left(1-\frac{2n+5}{2}\ln\left(\frac{n+3}{n+2}\right)\right)=m-\frac{1}{2}\sum_{n=1}^m \ln\left(\frac{n+3}{n+2}\right)^{2n+5}$$
Since,
$$\sum_{n=1}^m \ln\left(\frac{n+3}{n+2}\right)^{2n+5}=\ln\left(\frac{4^7}{3^7}\cdot \frac{5^9}{4^9}\cdot \frac{6^{11}}{5^{11}}\cdots\frac{(m+3)^{2m+5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1058714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
evaluation of the integral of a certain logarithm I come across the following integral in my work
$$\int_a^\infty \log\left(\frac{x^2-1}{x^2+1}\right)\textrm{d}x,$$
with $a>1$.
Does this integral converge ? what is its value depending on $a$ ?
| Using
$$
-i\log\left(\frac{x+i}{x-i}\right)=\pi-2\tan^{-1}(x)
$$
we get
$$
\begin{align}
\int_a^\infty\log\left(\frac{x^2-1}{x^2+1}\right)\mathrm{d}x
&=\int_a^\infty\left[\vphantom{\sum}\log(x+1)+\log(x-1)-\log(x+i)-\log(x-i)\right]\mathrm{d}x\\
&{=}+(x+1)\log(x+1)-(x+1)\\
&\hphantom{=}+(x-1)\log(x-1)-(x-1)\\
&\hphanto... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sqrt{n^2 + 2}$ is irrational
Suppose $n$ is a natural number. Prove that $\sqrt{n^2 + 2}$ is irrational.
From looking at the expression, it seems quite obvious to me that $\sqrt{n^2 + 2}$ will be irrational, since $n^2$ will be a natural number, and after adding $2$ to it, $n^2 + 2$ will no longer be a p... | Without loss of generality we can assume $x,y\in \mathbb{N}\ (?)$.
$$n^2+2=\dfrac{x^2}{y^2} \implies x^2=y^2\left(n^2+2\right)$$
For odd or even $n$, $x,y$ can be both odd, or exactly one of them should be even.
Case 1 ($n$ odd)
Case 1.1 (both odd)
$\left(x^2\equiv1\pmod4, y^2\equiv1\pmod4\right)\land \left(x^2=y^2\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Help with a tricky limit $\lim_{n\to\infty} \sum\limits_{i=1}^n (i/n)(\sqrt[2]{(i+1)/n}-\sqrt[2]{i/n})$ I have been attempting to follow the answer to a question previously asked on the site (Lebesgue integral basics), but am lost on how one might go about evaluating $$\lim_{n\to\infty} \sum\limits_{i=1}^n \frac in\lef... | You can manipulate the sums to get this working:
\begin{align}
\sum\limits_{i=1}^n\frac{i}{n}(\sqrt{\frac{i+1}{n}}-\sqrt{\frac{i}{n}})&=\sum\limits_{i=1}^n\frac{i}{n}\sqrt{\frac{i+1}{n}}-\sum\limits_{i=1}^n\frac{i}{n}\sqrt{\frac{i}{n}}\\&=\sum\limits_{i=1}^n\frac{i+1}{n}\sqrt{\frac{i+1}{n}}-\sum\limits_{i=1}^n\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Extremely difficult log integral, real methods only $$\int_{0}^{1}\frac{x^2 + x\log(1-x)- \log(1-x) - x}{(1-x)x^2} dx$$
I tried this:
$$M_1 = \int_{0}^{1} \frac{1}{1-x} \cdot \left(\frac{x^2 + x\log(1-x) - \log(1-x) - x)}{x^2}\right) dx$$
$$M_1 = \int_{0}^{1} \frac{x^2 + x\log(1-x) - \log(1-x) - x}{(1-x)x^2} dx$$
$$M_1... | Using Paul_I's answer and the classic expansion
$$\log\left(1-x\right)=-\sum_{n=1}^\infty \frac{x^n}{n}, \quad -1<x<1,$$
we get
$$-\int_0^1\frac{\log(1-x)+x}{x^2}\mathrm dx =\sum_{n=2}^\infty\int_0^1 \frac{x^{n-2}}{n}\mathrm dx = \sum_{n=1}^\infty\frac{1}{n(n+1)}=1$$
in accordance with Mathematica.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate this series? I need to evaluate this series.Without using derivative.
$$A=\frac{2} {2^2} + \frac{4} {2^5}+ \frac{6} {2^8} + \cdots $$
Where the $i$ th member is calculated with the the formula below:
$$A_i=\frac{i} {2^{3i-2}}$$
Feel free to edit the tags please.Thanks for you help.
| $$a=\frac{2}{2^2}+\frac{4}{2^5}+\frac{6}{2^8}+\frac{8}{2^11}...+ $$multiply a by $$\frac{1}{2^3} $$ then find $$a -\frac{1}{2^3}a $$
$$a -\frac{1}{2^3}a=\\\frac{2}{2^2}+\frac{4}{2^5}+\frac{6}{2^8}+\frac{8}{2^{11}}...\\-(\frac{2}{2^5}+\frac{4}{2^8}+\frac{6}{2^11}+\frac{8}{2^{14}}...)\\a -\frac{1}{2^3}a=\frac{2}{2^2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Properties of prime mod $3$ We know that if $p$ is a prime congruent to $3 \mod 4$, we cannot represent it as sum of two squares. Is there a positive property of such $p$? That is, do we have any statements that say "$p$ is a prime congruent to $3 \mod 4$ iff $\underline{\mbox{a positive statement}}$ is TRUE in an uniq... | For example, we can say, a prime is $1,3 \pmod 8$ if and only if there is just one expression $p = x^2 + 2 y^2.$
You get a little flexibility by throwing in indefinite forms: a prime is $1,7 \pmod 8$ if and only if there are just two infinite sequences of expressions $p = x^2 - 2 y^2,$ under the action (and its inverse... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Suppose that two polar curves are given by: $R_1 = \cos(2\theta)$ and $R_2 = \sin(3\theta)$. Find the smallest positive solution exactly. Suppose that two polar curves are given by: $R_1 = \cos(2\theta)$ and $R_2 = \sin(3\theta)$. Find the smallest positive solution exactly.
I know that we are looking for the smallest ... | Using formulas:
$$\sin{(3\theta)}=3\sin{(\theta)}-4\sin^3{(\theta)}$$
$$\cos{(2\theta)}=1-2\sin^2{(\theta)}$$
We can find a cubic equation for $\sin{(\theta)}$.
We have;
\begin{equation}
\begin{aligned}
\cos{(2\theta)}&=\sin{(3\theta)} \\ \cos{(2\theta)}-\sin{(3\theta)} &= 0 \\
1-2\sin^2{(\theta)} - 3\sin{(\theta)}+4\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1064268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve cubic equation $x^3-9 x^2-15x-6 =0$ without going Cardano
Solve the cubic equation for $x\in\mathbb{R}$ $$x^3-9 x^2-15x-6 =0$$
Note that the only real solution is $x=3+2\sqrt[3]{7}+\sqrt[3]{7^2}$. Given the regularity of this solution, can we solve for it constructively, without going full Cardano?.
Also, can w... | As an alternative to the Cardano’s method, substituting $x=3+2\sqrt{14}\cosh t$ into $x^3-9 x^2-15x-6=0$ to get
$$4\cosh^3t-3\cosh t = \cosh 3t = \frac{15}{4\sqrt{14}}$$
which leads to $t=\frac13\cosh^{-1}\frac{15}{4\sqrt{14}}$ and, then, the solution
$$x=3+2\sqrt{14}\cosh\left(\frac13\cosh^{-1}\frac{15}{4\sqrt{14}}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
A problem with proving using definition that $\lim_{n\to\infty}\frac {n^2-1}{n^2+1}=1$
Prove using the definition that: $$\displaystyle\lim_{n\to\infty}\frac {n^2-1}{n^2+1}=1 $$
What I did:
Let $\epsilon >0$, finding $N$: $\mid\frac {n^2-1}{n^2+1}-1\mid=\mid\frac {-2}{n^2+1}\mid\le\mid\frac {-2}{n^2}\mid=\frac {2}{n^... | $$\left|\frac{n^2-1}{n^2+1}-1\right|=2\frac1{n^2+1}<\epsilon\iff n^2+1>\frac2\epsilon\iff n>\sqrt{\frac2\epsilon-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Can this interesting property be proven? $$2^2+3^2+5^2+7^2+9^2+11^2=(17)^2$$
$$22^2+33^2+55^2+77^2+99^2+11^2=(143)^2$$
Also:
$$22^2+33^2+55^2+77^2+99^2+121^2=(187)^2$$
$$222^2+333^2+555^2+777^2+999^2+1221^2=(1887)^2$$
$$2222^2+3333^2+5555^2+7777^2+9999^2+12221^2=(18887)^2$$
$$22222^2+33333^2+55555^2+77777^2+99999^2+122... | Hint: What happens if you multiply your first identity with $111\cdots1^2$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Give the following linear transformation find values of parameter Find values of parameter t for which transformation is epimorphic:
$\psi([x_1,x_2,x_3,x_4])=x_1+x_2+x_3+2x_4,x_1+tx_2+x_3+3x_4,2x_1+x_2+tx_3+3x_4 $
When this transformation is epimorphic i.e. what should i look for in the reduced form of matrix of this l... | It seems the following.
The transformation $\psi$ is ephimorphic iff rank of its matrix is maximal, that is $3$.
Buy elementary transormations, which does not change the rank, we can transform the matrix as follows.
$\begin{pmatrix}
1 & 1 & 1 & 2 \\
0 & t-1 & 0 & 1 \\
0 & -1 & t-2 & -1
\end{pmatrix} $
$\begin{pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the solid angle of the intersection loop between a cone and an off-axis sphere?
An upright (green) cone with opening angle $2a < \pi/10$ has its vertex at point O with cartesian xyz coordinates $(0,0,0)$. The cone axis (dotted line) lies in the plane $y=0$ and is parallel to the z-axis.
A (blue) sphere of radi... | Let's move everything so that the center of the sphere is in the origin. Then the sphere is simply
$$x^2+y^2+z^2=r^2$$
and the cone becomes
$$\left(x+P_x\right)^2+y^2=c^2\left(z+P_z\right)^2$$
and since you don't want a double cone, you also want
$$z+P_z>0\quad.$$
The $y=0$ plane intersects the cone in two lines, namel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $ \int \frac {\mathrm{d}x}{(4x^2-1)^{3/2}}$ I have trouble using trig sub. After I get that x = 2x+1, should I substitute back into the original problem's $4x^2$ with $(4(2x+1)^2)$?
| $y=2x-1 \implies \dfrac{1}{2}dy=dx$
$\therefore\displaystyle\int\dfrac{dx}{\left(4x^2-1\right)^{\frac{3}{2}}}=\dfrac{1}{2}\displaystyle\int\dfrac{dy}{y^{\frac{3}{2}}\left(y+2\right)^{\frac{3}{2}}}$
$y=2\tan^2\theta \implies dy=4\sec^2\theta\tan\theta\ d\theta$
$\therefore\dfrac{1}{2}\displaystyle\int\dfrac{dy}{y^{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1073018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Partial Fraction Decomposition of Exponential Generating Functions I want to see if it is possible to write
$$ \left(\frac{x}{e^x-1}\right) \left(\frac{x^2/2! }{e^x-1-x}\right) \left(\frac{x^3/3!}{e^x-1-x-x^2/2}\right)$$
as a linear combination of the factors
$$p(x)\frac{x}{e^x-1}+q(x)\frac{x^2/2}{e^x-1-x}+r(x)\frac{x^... | I think that your idea to decompose in simple fractions is good. You can put $K=\mathbb{Q}(x)$, replace $\exp(x)$ by $y$, and decompose the following fraction in $K(y)$:
$$\left(\frac{x}{y-1}\right) \left(\frac{x^2/2! }{y-1-x}\right) \left(\frac{x^3/3!}{y-1-x-x^2/2}\right)=\frac{A}{y-1}+\frac{B}{y-1-x}+\frac{C}{y-1-x-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1073287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find value of f(2013)? Given a function $f(x)$ such that: $f(1) + f(2) + f(3)+\cdots+f(n) = n^2f(n)$
Find the value of $f(2013)$. It is given that $f(1) = 2014$.
I tried attempting the question as a bottom-up DP, but soon realized the numbers are too irregular and large to deal with in a mechanical fashion.
| Calculating a few of the first $n$ by hand reveals a pattern:
$$f(n) = \frac{f(1)}{T_n},$$
where $T_n$ is the $n$th triangular number.
Proof by induction:
The relationship holds for $n=1$ since $T_1 = 1.$
We assume that
$$f(x) = \sum_{k=1}^n f(k) = \frac{2}{n(n+1)} f(1).$$
This implies that
$$\frac{2}{n(n+1)} f(1) = n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1073713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How to prove $ \lim_{n\to\infty}\sum_{k=1}^{n}\frac{n+k}{n^2+k}=\frac{3}{2}$? How to prove that
$\displaystyle \lim_{n\longrightarrow\infty}\sum_{k=1}^{n}\frac{n+k}{n^2+k}=\frac{3}{2}$?
I suppose some bounds are nedded, but the ones I have found are not sharp enough (changing $k$ for $1$ or $n$ leads to the limit being... | Squeeze (without integral calculus).
I. $\displaystyle\sum_{k=1}^{n}\dfrac{n+k}{n^2+k} \leq \sum_{k=1}^{n}\dfrac{n+k}{n^2+1}=\frac{1}{n^2+1}\sum_{k=1}^{n}(n+k) = \frac{1}{n^2+1}(n^2+\frac{n(n+1)}{2})\rightarrow\frac{3}{2} $.
II.$\displaystyle\sum_{k=1}^{n}\dfrac{n+k}{n^2+k} \geq\sum_{k=1}^{n}\dfrac{n+k}{n^2+n}=\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove without induction $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $
Prove $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $ for all $n\ge 1$ and $b,c$ are integers.
Is it possible to prove this without induction?
| We want to show$$2^n\vert\left(\left(b+\sqrt{b^2-4c}\right)^n+\left(b-\sqrt{b^2-4c}\right)^n\right).$$
Using $x-y=\frac{x^2-y^2}{x+y}$ indeed we have
$$ 2^n\vert\left(\left(b+\sqrt{b^2-4c}\right)^n+\left(\frac{4c}{b+\sqrt{b^2-4c}}\right)^n\right) \\ 2^n\vert\frac{\left(b+\sqrt{b^2-4c}\right)^{2n}+4^nc^n}{\left(b+\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
How to express the equations as the Square root Like? $$2\sin \frac{\pi}{16}= \sqrt{2-\sqrt{2+\sqrt{2}}}$$
What law is need to be applied here? Do I have to make the $\frac{\pi}{16}$ in a form that will be give us $\sqrt{2}$ like sin 45 degree?
| Call $x = \sin \frac{\pi}{16} \to 1-2x^2 = \cos \frac{\pi}{8} \to \left(1-2x^2\right)^2 = \cos^2 \frac{\pi}{8} = \dfrac{1+\cos \frac{\pi}{4}}{2} = \dfrac{2+\sqrt{2}}{4} \to 1-2x^2 = \dfrac{\sqrt{2+\sqrt{2}}}{2} \to x^2 = \dfrac{2-\sqrt{2+\sqrt{2}}}{4} \to 2x = \sqrt{2-\sqrt{2+\sqrt{2}}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Maximum $\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx$ where $y\in \left[0,1\right]$? How find maximum this integral $$\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx$$ where $y\in \left[0,1\right]$?
| By differentiation under the integral sign we get
\begin{align*}\left(\int_0^y\sqrt{x^4+(y-y^2)^2}\,dx\right)'&=\int_0^y\frac{2(y-y^2)(1-2y)}{2\sqrt{x^4+(y-y^2)^2}}dx+\sqrt{y^4+(y-y^2)^2}\\&=\sqrt{y^4+(y-y^2)^2}+\int_0^y\frac{y(1-y)(1-2y)}{\sqrt{x^4+(y-y^2)^2}}dx,\end{align*}
which is positive for $y\in(0,\frac12)$. El... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
How to show $\lim_{n \to \infty}a_n=\frac{5^{3\cdot n}}{2^{\left(n+1\right)^2}}$? $$\lim_{n \to \infty}a_n=\dfrac{5^{3\cdot n}}{2^{\left(n+1\right)^2}}$$
I am trying to solve it using the squeeze theorem.
I have opened the expression to $$a_n=\dfrac{5^3\cdot 5^n}{2^{n^2}\cdot2^{2n}\cdot2)}$$
I think that the LHS should... | $$ \frac{a_{n+1}}{a_{n}}=\\\frac{\frac{5^{3(n+1)}}{2^{(n+2)^2}}}{\frac{5^{3(n)}}{2^{(n+1)^2}}}=\\\frac{5^{3n+3}}{5^{3n}}\frac{2^{(n+1)^2}}{2^{(n+2)^2}}=\\125\frac{2^{n^2+2n+1}}{2^{n^2+4n+4}}=\\125\frac{1}{2^{2n+3}}=\\\frac{125}{8}\frac{1}{4^n}\\$$when n become large $$n\rightarrow \infty\\\frac{a_{n+1}}{a_{n}}=\frac{12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1084355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find prime factorization of $2^{22} + 1$ Find the prime factorization of $2^{22} + 1$.
How can I approach this with a subtle way?
I know that $a^n -b^n = (a-b) (a^{n-1}+a^{n-2}b + \cdots + ab^{n-2}+b^{n-1})$ and $a^{2n+1} + b^{2n+1} = (a+b) (a^{2n}-a^{2n-1}b + \cdots - ab^{2n-1}+b^{2n})$. So far I have tried to apply t... | $2^{22}+1=2^2\cdot2^{20}+1=4\cdot\Big(2^5\Big)^4+1^4.\quad$ But $4x^4+1=\big(2x^2-2x+1\big)\big(2x^2+2x+1\big)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1084440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to calculate : $\frac{1+i}{1-i}$ and $(\frac{1+\sqrt{3i}}{1-\sqrt{3i}})^{10}$ I have these problems :
How to calculate : $\frac{1+i}{1-i}$ and $(\frac{1+\sqrt{3i}}{1-\sqrt{3i}})^{10}$
For some reason this is incorrect I'll be glad to understand why, This is what I done :
I used this formula : $(\alpha,\beta)*(\gamm... | What you calculated in the first part is not the same quantity as what you asked in the original question: is the quantity to be computed $$\frac{1-i}{1+i},$$ or is it $$\frac{1+i}{1-i}?$$ Even if it is the latter, you have made a sign error in the fourth line: it should be $$(1+i)(\tfrac{1}{2} + \tfrac{i}{2}) = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1084725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How prove this diophantine equation $(x^2-y)(y^2-x)=(x+y)^2$ have only three integer solution? HAPPY NEW YEAR To Everyone! (Now Beijing time 00:00 (2015))
Let $x,y$ are integer numbers,and such $xy\neq 0$,
Find this diophantine equation all solution
$$(x^2-y)(y^2-x)=(x+y)^2$$
I use Wolf found this equation only have... | On the LHS the sum of the two factors is nonnegative, so $x^2-y$ and $y^2-x$ cannot be negative at the same time. Hence, if $x+y\ne0$ then both $x^2-y$ and $y^2-x$ must be positive.
If $y^2-x\ge8$ and $x^2-y\ge8$ then
$$
(x+y)^2 = (x^2-y)(y^2-x) = \\
= \frac{x^2-y}2(y^2-x) +\frac{y^2-x}2(x^2-y) +0 \ge \\
\ge
4(y^2-x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Integral of an exponential I have the following: $$ I(a,b) \equiv\int_{-\infty}^\infty e^{\frac{-1}{2}\left(ax^2+\frac{b}{x^2}\right)}dx$$ where $a,b>0$. And I have the following substitution as a hint: $$y=\frac{1}{2}\left(\sqrt{a}x-\frac{\sqrt{b}}{x}\right)$$ And the integral must be evaluated. But I seem to be havin... | I too got stuck on your hint. Here's how I did your integral.
First letting $x = \frac{t}{\sqrt{a}}, \; dx = \frac{dt}{\sqrt{a}}$, we get
$$ \int_{-\infty}^\infty e^{-\frac{1}{2}\left(a x^2 + \frac{b}{x^2}\right)} \, dx = \frac{1}{\sqrt{a}} I(a b), $$
where
$$ I(\omega) = \int_{-\infty}^\infty e^{-\frac{1}{2}\left(t^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Is there a lower bound for the following rational expression, in terms of $b^2$? In the following, $b$ and $r$ are both positive integers.
Is there a lower bound for the following rational expression, in terms of $b^2$?
$$L = \frac{b^2\left(2^{r+1} b^2(2^r-1) + 3\cdot 2^r - 2\right)}{\left(2^{r+1} - 1\right)(2^r b^2 + ... | Taking off from Matt's answer, I get:
$$\frac{1}{2}\left(1 + \frac{2^r}{2^r - 1}\right)(b^2 + 2^{-r}).$$
We need an upper bound for $\frac{1}{2}\left(1 + \frac{2^r}{2^r - 1}\right)$.
We have the upper bound:
$$\frac{1}{2}\left(1 + \frac{2^r}{2^r - 1}\right) = \frac{1}{2}\left(2 + \frac{1}{2^r - 1}\right) \leq \frac{3}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding an integer $n$ such that $\sin(n)$ is close to 1 Given some $\epsilon>0$, is there an efficient way to find an integer $n$ such that
$$1-\sin(n)<\epsilon$$
We all know there is always one (and many), and so I can test all $n$ from $0$ until I find a good candidate, but I ask for some efficient algorithm that g... | Consider the Taylor series of $\sin(x)$ around $\left(2k + \frac{1}{2}\right) \pi$,
\begin{align*}
\sin\left(\left(2k + \frac{1}{2}\right)\pi + x\right)
&= 1 - \frac{x^2}{2} + \frac{x^4}{24} - \ldots \\
&\le 1 - \frac{x^2}{2}
\end{align*}
as it is reasonably clear that $\frac{x^4}{24} - \frac{x^6}{120} + \ldots > 0$ f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Factor the polynomial $z^5 + 32$ in real factors The question that I have trouble solving is the following:
Factor the polynomial $z^5 + 32$ in real factors. The answer should not use trigonometric functions. (Hint: you are allowed to use the fact that: $cos(\pi/5) = \frac{1+\sqrt{5}}{4}$ and $cos(3\pi/5) = \frac{1-\sq... | Set $t=2z$, for the moment. Then
\begin{align}
z^5+32=32(t^5+1)
&=32(t+1)(t^4-t^3+t^2-t+1)\\
&=32(t+1)t^2\left(t^2+\frac{1}{t^2}-\left(t+\frac{1}{t}\right)+1\right)\\
&=32(t+1)t^2\left(\left(t+\frac{1}{t}\right)^{\!2}-\left(t+\frac{1}{t}\right)-1\right)
\end{align}
Consider $u^2-u-1=(u-\alpha)(u-\beta)$, where
$$
\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
sketch points given by condition on complezx number z the condition is
$|z+i| \le 3$ so its a circle with radius three centered at -i
treating $z=x+iy$
$-\sqrt{9} \le \sqrt{(x^2 )+(y+1)^2} \le \sqrt{9}$
squaring everything $-9 \le x^2 + (y+1)^2 \le 9$
$-9 \le x^2 + y^2 + 2y +1 \le 9$
I feel like I am complicating more ... | You're really close.
Well $|z+i|\le 3$ with $z=x+iy$ we have then simply $0\le\sqrt{x^2+(y+1)^2}\le3$ ($\forall Z\in\mathbb{C},\,|Z|\ge 0$).
Thus $x^2+(y+1)^2\le3^2$ which represents the disk of radius $3$ and center $(0,-1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Showing $\binom{2n}{n} = (-4)^n \binom{-1/2}{n}$ Is there a proof for the following identity that only uses the definition of the (generalized) binomial coefficient and basic transformations?
Let $n$ be a non-negative integer.
$$\binom{2n}{n} = (-4)^n \binom{-\frac{1}{2}}{n}$$
| $$\begin{align}
\binom{-1/2}{ n} &= (-1/2)(-3/2)(-5/2)\ldots(-(2n-1)/2)
/n! \\
&= (-1)^n 1\cdot 3\cdot 5\cdot\ldots\cdot(2n-1)/ (2^n n!) \\
&= (-1)^n (2n)!/ (2^n \cdot n! \cdot 2 \cdot 4 \cdot 6\cdot \ldots \cdot (2n)) \\
&= (-1)^n (2n)!/ (2^n n!)^2 \\
&= (-1/4)^n \binom{2n}{n}.
\end{align}$$
Hence:
$$ (1-4X)^{-1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Find the residue at $z=-2$ for $g(z) = \frac{\psi(-z)}{(z+1)(z+2)^3}$ Find the residue at $z=-2$ for
$$g(z) = \frac{\psi(-z)}{(z+1)(z+2)^3}$$
I know that:
$$\psi(z+1) = -\gamma - \sum_{k=1}^{\infty} (-1)^k\zeta(k+1)z^k$$
Let $z \to -1 - z$ to get:
$$\psi(-z) = -\gamma - \sum_{k=1}^{\infty} \zeta(k+1)(z+1)^k$$
therefor... | $\psi(z)$ is regular over $\Re(z)>0$, hence $z=-2$ is a triple pole for $g(z)$. This gives:
$$\operatorname{Res}\left(g(z),z=-2\right)= \frac{1}{2}\left.\frac{d^2}{dz^2} (z+2)^3 g(z)\right|_{z=-2}=\frac{1}{2}\left.\frac{d^2}{dz^2} \frac{\psi(-z)}{z+1}\right|_{z=-2}\tag{1}$$
so:
$$\operatorname{Res}\left(g(z),z=-2\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Describe geometry of complex plane. Let $a \in \Bbb R$ and $c>0$ to be fixed. Describe the set of points $z$ such that $|z-a|-|z+a|=2c$ for every possible choice of $a$ and $c$.
Then let $a$ be a complex number using the rotation of the plane describe the locus of points satisfying the above equation.
Here is how I u... | with $z=x+iy$ it is equivalent to
$$\sqrt{(x-a)^2+y^2}-\sqrt{(x+a)^2+y^2}=2c$$
Does this help you?
squaring the equation $$\sqrt{(x-a)^2+y^2}=2c+\sqrt{(x+a)^2+y^2}$$
we get $$-xa-c^2=c\sqrt{(x+a)^2+y^2}$$
squaring again we get $$x^2a^2+c^4+2xac^2=c^2((x+a)^2+y^2)$$
this is equivalent to
$$x^2(a^2-c^2)+c^2(c^2-a^2)=y^2c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1104414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
extracting the middle term of $ (z \cos \theta + w\sin \theta )^m(- z\sin \theta + w\cos \theta )^m $ Is there a systematic way to extract the middle term of the following expression?
$$ (z \cos \theta + w\sin \theta )^m(- z\sin \theta + w\cos \theta )^m $$
This is homogeneous polynomial of degree $2m$, so I am looki... | We use the binomial theorem, to see that
$$
(z\cos\theta+w\sin\theta)^m = \sum_{k=0}^m {m\choose k}z^{m-k}\cos^{m-k}\theta \,w^k\sin^k\theta
$$
and
$$
(-z\sin\theta+w\cos\theta)^m=\sum_{k=0}^m {m\choose k}(-z)^k\sin^k\theta\,w^{m-k}\cos^{m-k}\theta.
$$
Written this way, I think it is clear that when we multiply these f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1106814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Hypergeometric 2F1 with negative c I've got this hypergeometric series
$_2F_1 \left[ \begin{array}{ll}
a &-n \\
-a-n+1 &
\end{array} ; 1\right]$
where $a,n>0$ and $a,n\in \mathbb{N}$
The problem is that $-a-n+1$ is negative in this case. So when I try to use Gauss's identity
$_2F_1 \left[ \begin{array}{ll}
a & b \\
... | We may try to keep it simple.
Suppose we seek to evaluate
$$\sum_{k=0}^n {a-1+k\choose k} {a-1+n-k\choose n-k}.$$
It is immediately apparent that this is a convolution of two ordinary generating functions.
To see what they are re-write the sum as
$$\sum_{k=0}^n {a-1+k\choose a-1} {a-1+n-k\choose a-1}.$$
By the Newton b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Showing ${n + a - 1 \choose a - 1} = \sum_{k = 0}^{\left\lfloor n/2 \right\rfloor} {a \choose n-2k}{k+a-1 \choose a-1}$ Prove that for integers $n \geq 0$ and $a \geq 1$, $${n + a - 1 \choose a - 1} = \sum_{k = 0}^{\left\lfloor n/2 \right\rfloor} {a \choose n-2k}{k+a-1 \choose a-1}.$$
I figured I'd post this question, ... | This one can also be done using complex variables.
Suppose we seek to evaluate
$$\sum_{k=0}^{\lfloor n/2\rfloor}
{a\choose n-2k} {k+a-1\choose a-1}.$$
Introduce the integral representation
$${a\choose n-2k}
=\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{a}}{z^{n-2k+1}} \; dz.$$
Note that this integral is zero whe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
If $x^2+x+1=0 $, then find the value of $(x^3+1/x^3)^3$?
If $x^2+x+1=0 $, then find the value of $(x^3+1/x^3)^3$
This thing doesn't make sense how should I use first identity to find the second one.
| $x^2+x+1=0 \to x+1/x=-1 \to (x+1/x)^3 = -1\to x^3 + 3(x+1/x) + 1/x^3 = -1\to x^3+1/x^3 = 2 \to (x^3+1/x^3)^3 = 2^3 = 8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Solutions of the functional equation $f(2x) = \frac{f(x)+x}{2}$ How can I solve the following functional equation?
$$f(2x) = \frac{f(x)+x}{2},$$
for $x \in \mathbb{R}$ with $f$ being a continuous function.
| Here's an alternative proof. Notice that by substituting in $\frac{1}{2^{n+1}}x$ for $x$, we can obtain
$$f\left(\frac{1}{2^n}x\right) = \frac{1}{2}f\left(\frac{1}{2^{n+1}}x\right) + \frac{x}{2^{n+2}}$$
Using this expression, we find that
\begin{align*}f(x) &= \frac{1}{2}f\left(\frac{1}{2}x\right) + \frac{x}{4}\\
&= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1111361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Is this a correct way to prove this? I've just looked at this question and sketched a way to do it my head. When I looked at the answer it looked slightly more complicated than the way I did it so I just wanted to check whether this is a correct way:
The question states:
Let $$f(x)=ax-\frac{x^3}{1+x^2}$$ where $a$ is... | Not very clear. If I were you I will show in this order:
first,
$$f'(x)=a-\frac{x^2 (x^2+3)}{(x^2+1)^2}$$
Secondly,
$$ \forall x, \quad \frac{9}{8}-\frac{x^2 (x^2+3)}{(x^2+1)^2} = \frac{1}{8}\left( \frac{x^2-3}{x^2+1} \right)^2 \geq 0 $$
Thus if $a \geq \frac98$, then $\forall x$ we have
$$f'(x)=a-\frac{x^2 (x^2+3)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int^{1}_{-1} \frac{\ln(ax^2+2bx+a)}{x^2+1}dx$ if $a>b>0$ I am trying to evaluate the following integral:
$$\int^{1}_{-1} \frac{\ln(ax^2+2bx+a)}{x^2+1}dx,$$ where $a>b>0$.
I can't really think of a way to find it. So, please give me a hint.
| With $\sin\theta =\frac ba$\begin{align}
&\int^{1}_{-1} \frac{\ln(ax^2+2bx+a)}{x^2+1}dx\\
=& \int^{1}_{-1} \frac{\ln a}{x^2+1}dx+
\int^{1}_{-1} \frac{\ln(x^2+2x \sin \theta+1)}{x^2+1}\overset{x\to1/x}{dx}\\
=& \ \frac\pi2\ln a+\frac12 \int^{\infty}_{-\infty} \frac{\ln(x^2+2x \sin \theta+1)}{x^2+1}dx-2\int_1^\infty \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Limit $\lim_\limits{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)}$ Evaluate the given limit:
$$\lim_{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)} .$$
I've tried to evaluate it but I always get stuck... Obviously I need L'Hôpital's Rule here, but still get confused on the way. May someone show m... | This solution is based on no l'Hospital rule nor Taylor expansion. The following standard limits only are used:
\begin{eqnarray*}
\lim_{x\rightarrow 0}e^{x} &=&1. \\
\lim_{x\rightarrow 0}\frac{\tan x}{x} &=&1. \\
\lim_{x\rightarrow 0}\frac{1+x+\frac{1}{2}x^{2}-e^{x}}{x^{3}} &=&-\frac{1}{6}%
. \\
\lim_{x\rightarrow 0}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 2
} |
Inverse of $3$ by $3$ matrix with non-constant entries. I'm solving a question in nonhomogenous ordinary differential equation system $x'=Px+q$, and to solve my question I need to compute the inverse of the matrix
$A=\begin{pmatrix}e^{-2t} & e^{-t} & 0 \\ -\frac{5}{4}e^{-2t} & -\frac{4}{3}e^{-t} & e^{2t} \\ -\frac{7}{4... | We can factor out the variable dependence in this particular case:
$$ A=\begin{pmatrix}e^{-2t} & e^{-t} & 0 \\ -\frac{5}{4}e^{-2t} & -\frac{4}{3}e^{-t} & e^{2t} \\ -\frac{7}{4}e^{-2t} & -\frac{2}{3}e^{-t} & -e^{2t} \end{pmatrix} $$
$$ = \begin{pmatrix} 1 & 1 & 0 \\ -\frac{5}{4} & -\frac{4}{3} & 1 \\ -\frac{7}{4} & -\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determining the best possible substitution for an integrand What substitution is best used to calculate $$\int \frac{1}{1 + \sqrt{x^2 -1}}dx$$
| @Pp. suggests an Euler substitution of the form
\begin{align*}
x &= \frac{t^2+1}{2t} = \frac{t}{2} + \frac{1}{2t} \\
dx &= \left(\frac{1}{2} - \frac{1}{2t^2}\right)\,dt = \frac{t^2-1}{2t^2}\,dt
\end{align*}
By this method you gain being able to solve all integrals of this type: Additions, multiplications and divisio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Infinitely many primes of the form $6n - 1$ Prove there are infinitely many primes of the form $6n - 1$ with the following: (i) Prove that the product of two numbers of the form $6n + 1$ is also of that form. That is, show that $(6j + 1)(6k + 1) = 6m + 1$, for some choice of $m$. (ii) Show that every prime $p$ greater ... | (i) $(6m+1)\times(6n+1) = 36mn + 6m + 6n + 1 = 6(6mn + m + n) + 1$
(ii) To prove all primes are of the form $6n + 1$ or $6n - 1$, consider:
*
*$6x+0 = 6x$
*$6x+2 = 2(3x+1)$
*$6x+3 = 3(2x+1)$
*$6x+4 = 2(3x+2)$
*$6x+5 = 6(x+1) - 1$
(iii) To prove that there are infinitely many primes of the form $6n-1$ consider... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that if $({x+\sqrt{x^2+1}})({y+\sqrt{y^2+1}})=1$ then $x+y=0$ Let
$$\left({x+\sqrt{x^2+1}}\right)\left({y+\sqrt{y^2+1}}\right)=1$$
Prove that $x+y=0$.
This is my solution:
Let
$$a=x+\sqrt{x^2+1}$$
and
$$b=y+\sqrt{y^2+1}$$
Then $x=\dfrac{a^2-1}{2a}$ and $y=\dfrac{b^2-1}{2b}$. Now $ab=1\implies b=\dfrac1a$. Then ... | Add other solution
since
$$\ln{(x+\sqrt{x^2+1})}+\ln{(y+\sqrt{y^2+1})}=0$$
note $f(x)=\ln{(x+\sqrt{x^2+1})}$ ia odd function and strictly increasing on real numbers,
if $$f(x)+f(y)=0\Longleftrightarrow x+y=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 3
} |
Recurrence equation solution? Can you help me with the solution of this recurrence equation?
$$
f(n+2) = -2f(n) +3f(n+1) +n \quad\mid\quad f(1)=4 \quad\mid\quad f(2)=5
$$
Thank you.
| suppose we define $u(n) = f(n+1) - f(n).$ then we can rewrite
$f(n+2) = -2f(n) + 3f(n+1) + n$ as $$u(n+1) = 2u(n) + n, u(1) = 1$$
we will loo for a particular solution in the form of $u(n) =an+ b $ we need
$an + a + b = 2an+ 2b + n$ is satisfied if $a = 2a+1, a+b = 2b$ which is satified if $a = b = -1$ and the hom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $\sum_{n=1}^{\infty} \log~ ( n ~\sin \frac {1 }{ n })$ Convergence of $$\sum_{n=1}^{\infty} \log~ ( n ~\sin \dfrac {1 }{ n })$$
Attempt:
Initial Check : $\lim_{n \rightarrow \infty } \log~ ( n ~\sin \dfrac {1 }{ n }) = 0$
$\log~ ( n ~\sin \dfrac {1 }{ n }) < n ~\sin \dfrac {1 }{ n }$
$\implies \... | A slightly different way than the two current answers, dealing mostly with intuition:
$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots$$
or alternately,
$$\sin\left(\frac 1x \right) = \frac{1}{x} - \frac{1}{3!x^3} + \frac{1}{5!x^5} + \ldots$$
Multiplying by $x$ gives that
$$x\sin\left( \frac 1x \right) = 1 - \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Finding an order of a coset in $A/B$ where $A$ is a free abelian group and $B$ is a subgroup. Let $A$ be a free abelian group with basis $x_1,x_2,x_3$ and let $B$ be a subgroup of A generated by $x_1+x_2+4x_3, 2x_1-x_1+2x_3$. In the group $A/B$ find the order of the coset $(x_1+2x_3)+B$.
How can I find this order? Jus... | If you stick to the "$QAP^{-1}$ convention" and do things horizontally, then $(Q^{-1})^{\top}$ is what gives you the basis and $P^{-1})^{\top}$ is what gives you the generators. To avoid transposing, take $\begin{pmatrix}1&1&4\\2&-1&-2\end{pmatrix}$.
Using elementary row and column operations and recording them in mat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Limit of non-linear multi-variable function I'm trying to prove the limit of the following function is $0$:
$\lim_{(x,y) \to (1,-1)} {x^3} - {2xy^2} + 1$
I know that I'm trying to find a $\delta$ s.t $ 0 < \sqrt{(x - 1)^2 + (y + 1)^2} < \delta $ which implies $|{x^3} - {2xy^2} + 1| < \epsilon$
I tried factoring $x$ so... | We have $$\begin{align}|x^3-2xy^2+1|&=|(x-1+1)^3-2(x-1+1)(y+1-1)^2+1|\\&=\left|(x-1)^3+3(x-1)^2+3(x-1)-2(x-1)\left[(y+1)^2+2(x+1)\right]-2\left[(y+1)^2-2(y+1)\right]\right|\end{align}$$
Idea: We have those $x-1$ and $y+1$ in the condition $\sqrt{(x-1)^2+(y+1)^2}<\delta$, we make them appear in the expression $x^3-2xy^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show that if $m^2 + n^2 $ is divisible by $4$, then $mn$ is also divisible by $4$. Show that if $m$ and $n$ are integers such that $m^2 + n^2 $ is divisible by $4$, then $mn$ is also divisible by $4$.
I am not sure where to begin.
| $(m+n)^2=m^2+n^2+2mn$, let $m^2+n^2=4k$ then $(m+n)^2=4k+2mn$ , now as RHS is divisible by $2$ LHS will also be divisible by $2$ but LHS is square of some quantity thus if $2$ divides it $4$ will also divide now consider $(m+n)^2-4k=2mn$. Now LHS is divisible by $4$ so $2mn$ is divisible by $4$ thus $mn$ is divisible ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Inequality $x^4+y^4+(x^2+1)(y^2+1)\ge x^3(1+y) +y^3(1+x)+x+y$ for $x,y \in\mathbb{R}$ Prove for $x,y \in\mathbb{R}$ that such inequality exists ;
$x^4+y^4+(x^2+1)(y^2+1)\ge x^3(1+y) +y^3(1+x)+x+y$
And here is what I realised ;
because $(x^2+1)(y^2+1) >=1$
and $x^4+y^4 \ge 0$
$1\ge x^3(1+y) +y^3(1+x)+x+y$
I'd prefer a h... | Hint: First expand the inequality. Then expand $(x-y)^4$ and use it to get rid of the $x^3y+xy^3$ terms. Beat the rest with repeated usages of AM-GM in the form $a^2+b^2\ge 2ab$ with various choices of $a,b$.
A short solution: It's equivalent to
$$\frac14(x-y)^4+\frac14(x^2-y^2)^2+\frac12(x^2+1)(x-1)^2+\frac12(y^2+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Inequality $(a+b)^2 + (a+b+4c)^2\ge \frac{kabc}{a+b+c}$ for $a,b,c \in\mathbb{R}$ Find biggest constans k such that $(a+b)^2 + (a+b+4c)^2\ge \frac{kabc}{a+b+c}$ is true for any $a,b,c \in\mathbb{R}$
Could you check up my solution? I'm not sure it's ok -
$(a+b)^2 + (a+b+4c)^2 \ge 0$
and
$0\ge \frac{kabc}{a+b+c}$
so
$0... | We need to worry only about $a+b+c> 0^\dagger$ and the inequality is homogeneous, so we may set $a+b+c=1$ and $c \in (0, 1]$
Then we have $ab \le \frac14(1-c)^2$ and we need to find largest $k$ satisfying
$$(1-c)^2+(1+3c)^2 \ge \frac{k}4 c(1-c)^2 \implies \frac{k}4 \le \frac1c+\frac1c\left(\frac{1+3c}{1-c}\right)^2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Given that a,b,c are distinct positive real numbers, prove that (a + b +c)( 1/a + 1/b + 1/c)>9
Given that $a,b,c$ are distinct positive real numbers, prove that $(a + b +c)\big( \frac1{a}+ \frac1{b} + \frac1{c}\big)>9$
This is how I tried doing it:
Let $p= a + b + c,$ and $q=\frac1{a}+ \frac1{b} + \frac1{c}$.
Using A... | By the AM-GM inequality,
$$\frac{a}{b}+\frac{b}{a}\geq 2 $$
and so on, so:
$$ (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) = 3+\sum_{cyc}\left(\frac{b}{a}+\frac{a}{b}\right) \geq 3+6 = \color{red}{9}$$
as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Volume of figure between $x^2+y^2+z^2=16$ and $ x^2+y^2=6z$ if $z\geq 0$ I have a problem where I have to find volume of figure formed, when $x^2+y^2+z^2=16$ and $ x^2+y^2=6z$ intersects if $z\geq 0$.
Here is a graphic for clarity:
So far I have transformed the problem to cylindrical coordinates (given the fact that w... | Thank to @Lucian I realized, that, the problem could be solved easily, by calculating volume when rotating $x=\sqrt{16-z^2}$ and $x=\sqrt{6z}$ about the $Oz$ axis. Therefore my volume would be:
$$\int_{2 \sqrt{3}}^4 \pi \left(16-z^2\right) \, dz+\int_0^{2 \sqrt{3}} \pi 6 z \, dz=\frac{236 \pi }{3}-24 \sqrt{3} \pi$$
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Where am I going wrong in my linear Diophantine solution? Let $-2x + -7y = 9$. We find integer solutions $x, y$.
These solutions exist iff $\gcd(x, y) \mid 9$. So,
$-7 = -2(4) + 1$
then
$-2 = 1(-2)$
so the gcd is 1, and $1\mid9$. OK.
In other words,
$(-7)(1) + (-2)(-4) = 1$
Multiply by 9 on each side to get
$(-7)(9) + ... | $x_0$ should be $-36$ and $y_0$ should be $9$. Somehow they got switched.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Why does solving the spherical Bessel equation using Frobenius series produce two quadratic equations for the exponents at the singularity? The spherical Bessel equation is:
$$x^2y'' + 2xy' + (x^2 - \frac{5}{16})y = 0$$
If I seek a Frobenius series solution, I will have:
\begin{align*}
&\quad y = \sum_{... | The main indicial is the the first equation. You get two roots from the first equation so you should set $a_1=0$. If there is a common root between the first and second equations, then you can consider that root and set both $a_1$ and $a_0$ nonzero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that the equation $a^2+b^2=c^2+3$ has infinitely many integer solutions $(a,b,c)$. Prove that the equation $a^2+b^2=c^2+3$ has infinitely many integer solutions $(a,b,c)$.
My attempt:
$(a+1)(a-1)+(b+1)(b-1)=c^2+1$
This form didn't help so I thought of $\mod 3$, but that didn't help either. Please help. Thank you.... | Note that the requirement is simply to demonstrate infinitely many solutions - we are not required to find all solutions.
Setting $c=b+1$, we see that $a^2 = c^2-b^2+3 = 2b+4$
Therefore for any even $a>2$, we can choose $b=\frac{a^2-4}{2}$ and $c=b+1$. This gives infinitely many solutions as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1127860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Relationship between trigonometric and hyperbolic sine Why is the following identity true?
$$ \sin(ix) = i\sinh(x)$$
When I do the calculation, I get this:$$\sin(ix) = \frac{{e^{i(ix)}}-e^{-i(ix)}}{2i}=\frac{e^{-x}-e^x}{2i}=-\frac{e^x-e^{-x}}{2i}=-\left(\frac{\sinh(x)}{i}\right)$$
| Another derivation:
$$\displaystyle \sin(ix) = ix - \frac {(ix)^3}{3!} + \frac {(ix)^5}{5!} - \frac {(ix)^7}{7!}+ \cdots$$
$$ = ix + \frac {ix^3} {3!} + \frac {ix^5}{5!} + \frac {ix^7}{7!} + \cdots $$
$$ = i \left({x + \frac {x^3}{3!} + \frac {x^5}{5!} + \frac {x^7}{7!} + \cdots }\right)= i \sinh x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
find the length of rectangle based on area of frame A rectangular picture is $3$ cm longer than its width, $(x+3)$. A frame $1$ cm wide is placed around the picture. The area of the frame alone is $42 \text{cm}^2$. Find the length of the picture.
I have tries:
$(x+6)(x+3) = 42 \\
x^2+3x+6x+12 = 42 \\
x^2+9x+12x-42 = 0... | The area of the frame is the difference between the area of the rectangle bordered by the outside of the frame and the area of the picture. Since the frame adds $1~\text{cm}$ on each side of the picture, the outer rectangle has length $x + 8~\text{cm}$ and width $x + 5~\text{cm}$. Thus, the area of the frame is
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
The inequality. Regional olympiad 2015 Let $a, b, c$ - the positive real numbers, and $ab+bc+ca=1$
Prove that $\sqrt{a+\frac{1}{a}}+\sqrt{b+\frac{1}{b}}+\sqrt{c+\frac{1}{c}} \geqslant 2(\sqrt{a}+\sqrt{b}+\sqrt{c})$
Probably, we should use these facts:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{abc}$
$(a+b+c)^2 = a... | By Cauchy-Schwarz we have
$$
\sqrt{a + \frac{1}{a}} = \sqrt{a + b + c + \frac{bc}{a}} \geq \frac{1}{2}\left(\sqrt{a} + \sqrt{b} + \sqrt{b} + \sqrt{\frac{bc}{a}}\right)
$$
and similarly
$$
\sqrt{b + \frac{1}{b}} \geq \frac{1}{2}\left(\sqrt{a} + \sqrt{b} + \sqrt{b} + \sqrt{\frac{ca}{b}}\right)
$$
and
$$
\sqrt{c + \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How to prove $p(z-y)(zy)(z^2-yz+y^2) \mid x^p-(z-y)^p \Rightarrow x=z-y$? Assume $p>2$ prime and $1<x<y<z$ coprime. How to prove the following: $$p(z-y)(zy)(z^2-yz+y^2) \mid x^p-(z-y)^p \Rightarrow x=z-y$$
I remember it as an extra exercise which I couldn't solve when I was a student. It still bothers me...
I can only ... | The following reasoning is derived from exercise 1.19.13 in the excellent book about polynomials with Barbeau. This exercise is based on problem A2 from the 37th Putnam Competition in 1976. All lower case symbols used are representing non-negative integers unless specified differently.
Define $F_n=(x+y)^n-x^n-y^n$, $G... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sequence $a_{n+1}=\sqrt{1+\frac{1}{2}a_n^2}$ I am trying but cant figure out anything.
$a_{n+1}=\sqrt{1+\frac{1}{2}a_n^2}$
I am trying to proove that $a_n^2-2<0$.
Getting $$a_{n+1} -a_n=\dots=\frac{2-a_n^2}{2\left(\sqrt{1+\frac{1}{2}a_n^2} +a_n\right)}$$
Then I have no clue how to proove it since I am not given $a_1$.I... | Hints:
1) $\dfrac{1}{2}a_{n}^2-a_{n+1}^2=\dfrac{1}{2}a_{n+1}^2-a_{n+2}^2=\ldots=\dfrac{1}{2}a_{n+i}^2-a_{n+i+1}^2=-1$
Therefore...
2) Proving that $a_n^2-2<0$ is the same as proving that $a_{n+1}^2-2<0$, and
3) Proving that $a_n^2-2<0$ is the same as proving that $a_{n-1}^2-2<0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Limit of $(\cos{xe^x} - \ln(1-x) -x)^{\frac{1}{x^3}}$ So I had the task to evaluate this limit
$$ \lim_{x \to 0} (\cos{(xe^x)} - \ln(1-x) -x)^{\frac{1}{x^3}}$$
I tried transforming it to:
$$ e^{\lim_{x \to 0} \frac{ \ln{(\cos{xe^x} - \ln(1-x) -x)}}{x^3}}$$
So I could use L'hospital's rule, but this would just be imposs... | First notice that $$\cos (xe^x) = \sum_{n=0}^{\infty}(-1)^n \frac{(xe^x)^{2n}}{2n!}$$
and $$\ln (1 - x) = -\sum_{n=0}^{\infty} \frac{x^n}{n}$$
Thus $$\cos (xe^x) - \ln (1 - x) = \sum_{n=0}^{\infty}(-1)^n \frac{(xe^x)^{2n}}{2n!} +\sum_{n=0}^{\infty}\frac{x^n}{n} = 1 + x - \frac{2x^3}{3} - O(x^4) $$
Therefore we have
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1145687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
finding the minimum value of $\frac{x^4+x^2+1}{x^2+x+1}$ given $f(x)=\frac{x^4+x^2+1}{x^2+x+1}$.
Need to find the min value of $f(x)$.
I know it can be easily done by polynomial division but my question is if there's another way
(more elegant maybe) to find the min?
About my way: $f(x)=\frac{x^4+x^2+1}{x^2+x+1}=x^2-x+... | Multiply the fraction by $\frac{x^2-1}{x^2-1}$:$$\frac{(x^2-1)(x^4+x^2+1)}{(x+1)(x-1)(x^2+x+1)}=\frac{x^6-1}{(x+1)(x^3-1)}=\frac{x^3+1}{x+1}=x^2-x+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Question about Lagrange multiplier and maximum point Find the maximum of $\log{x}+\log{y}+3\log{z}$ on portion of the sphere $x^2 + y^2 + z^2 =5r^2$
where $x,y,z>o $
I found that maximum is $5\log{r} + 3\log{\sqrt{3}}$ at $(r,r,3\sqrt{3})$
And Use this result to prove that for real positive $a, b, c$,
$abc^3$ is less... | There are two typos:
The maximum happens at $(r,r,\sqrt{3}r)$, and the right hand side of the inequality should be $27(\frac{a+b+c}{5})^5$.
Suppose $a=x^2,b=y^2,c=z^2$. Use the fact that $\log{x}+\log{y}+3\log{z}\leq 5\log{r} + 3\log{\sqrt{3}}$ when $x,y,z$ satisfies the sphere equation.
So
$$\log{xyz^3}\leq\log{3\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
By viewing the polynomials as a difference of two squares, factorise the following polynomials. By viewing the polynomials as a difference of two squares, factorise the following polynomial:
$$x^4+x^2+1.$$
I searched but couldn't find a way to solve this
Edit: By using Hans Lundmark hint, I get:
$$(x^2+1)^2-x^2$$
Is i... | We don't factorization:
$$\dfrac2{x^4+ax^2+1}=\dfrac{x^2+1}{x^4+ax^2+1}-\dfrac{x^2-1}{x^4+ax^2+1}=\dfrac{1+\dfrac1{x^2}}{x^2+a+\dfrac1{x^2}}+\dfrac{1-\dfrac1{x^2}}{x^2+a+\dfrac1{x^2}}$$
Now as $\displaystyle\int\left(1\pm\dfrac1{x^2}\right)dx=x\mp\dfrac1x,$
for the first integral, choose $x-\dfrac1x=u$ and $x+\dfrac1x=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Implicit differentiation of $e^{x^2+y^2} = xy$ I just want to reconfirm the steps needed to answer this question. Thank you
Find $\dfrac{dy}{dx}$ in the followng:
$$e^{\large x^2 + y^2}= xy$$
I got this so far.
$\newcommand{\dd}{\mathrm{d}}\frac{\dd x}{\dd y} e^{x^2}\cdot e^{y^2} = \frac{\dd x}{\dd y} xy$
$u=e^{x^2}$:... | $$ e^{x^2+y^2} =xy
\\ e^{x^2+y^2}\; [x^2+y^2]' =[x]'\, y+x\,[y]'
\\ e^{x^2+y^2}\; [2x+2y\,\color{blue}{y'}] =[1]\, y+x\,[\color{blue}{y'}]
\\ 2xe^{x^2+y^2}+2ye^{x^2+y^2}\,\color{blue}{y'} =y+x\,\color{blue}{y'}
\\ 2ye^{x^2+y^2}\,\color{blue}{y'}-x\,\color{blue}{y'} =y-2xe^{x^2+y^2}
\\ (2ye^{x^2+y^2}-x)\,\color{blue}{y'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Summation inductional proof: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$ Having the following inequality
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$$
To prove it for all natural numbers is it enough to show that:
$\frac{1}{(n+1)^2}-\frac{1}{n^2} <2$ or $\frac{1}{(n+1)^2}... | $$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}+...\uparrow\frac{\pi^2}{6}=1,6449340668482...\lt2.$$
See: http://mathworld.wolfram.com/PiFormulas.html
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
For odd $m\ge3$, does it follow: $\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$ Unless I am making a mistake, I am calculating that:
$$\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$$
Here's my reasoning:
*
*$\dfrac{x^m + y^m}{x+y} = x^{m-1} - x^{m-2}y ... | Well, I would do it that way :
$$ x^m = x\ x^{m-1} $$
So :
$$\begin{align}\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} &= \frac{x^m + y^m+ (xy)\ x^{m-2} + (xy)\ y^{m-2}}{x+y} \\
&=\frac{x^m + y^m+ y\ x^{m-1} + x\ y^{m-1}}{x+y} \\
&=\frac{ x\ x^{m-1} + y\ y^{m-1}+ y\ x^{m-1} + x\ y^{m-1}}{x+y}\\
&=\frac{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int{\frac{\sqrt[3]{x+8}}{x}}dx$ So I've tried solving the equation below by using $u=x+8$, and I get $\int{\frac{\sqrt[3]{u}}{u-8}}du$ which doesn't seem to lead anywhere, I've also tried taking the $ln$ top and bottom, but I don't know how to proceed. Any hints?
$$\int{\frac{\sqrt[3]{x+8}}{x}}dx$$
Update:... | Let $x=u^3-8$, then $\mathrm dx =3u^2\mathrm du$ and we get the integral
$$\int\frac{u}{u^3-8}3u^2\mathrm du$$
Last integral can be solved by partial fractions since
\begin{align*}
\frac{3u^3}{u^3-8}&=3+\frac{24}{u^3-8}\\
&=3+\frac{A}{u-2}+\frac{Bu+C}{u^2+2u+4}
\end{align*}
where $A$, $B$ and $C$ are constants we must ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1153991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$.
Let $a$, $b$ and $c$ be the three sides of a triangle.
Show that $$\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3\,.$$
A full expanding results in:
$$\sum_{cyc}a(a+b-c)(a+c-b)\g... | By C-S $\sum\limits_{cyc}\frac{a}{b+c-a}=\sum\limits_{cyc}\frac{a^2}{ab+ac-a^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2ab-a^2)}\geq3$, where the last inequality it's just
$$\sum\limits_{cyc}(a-b)^2\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1155955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
} |
Why is $\lim_{n\to\infty} n(e - (1+\frac{1}{n})^n) = \frac{e}{2}$ I'm having trouble understanding why
$$\lim_{n\to\infty} n(e - (1+\tfrac{1}{n})^n) = \frac{e}{2}$$
Can someone offer me a proof for this?
| Using: $\ln(1 + \frac{1}{n}) \sim \frac{1}{n} - \frac{1}{2n^2}$
And: $e^{-\frac{1}{2n}} \sim 1 - \frac{1}{2n} $
$$n(e - (1 + \frac{1}{n})^n) = n(e - e^{n\ln(1 + \frac{1}{n})}) = ne(1 - e^{n\ln(1 + \frac{1}{n}) - 1}) \sim ne(1 - e^{1 - \frac{1}{2n} - 1}) = ne(1 - e^{-\frac{1}{2n}}) = ne(1 - (1 - \frac{1}{2n}) = ne(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1156356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Proof of a limit of a sequence I want to prove that $$\lim_{n\to\infty} \frac{2n^2+1}{n^2+3n} = 2.$$
Is the following proof valid?
Proof
$\left|\frac{2n^2+1}{n^2+3n} - 2\right|=\left|\frac{1-6n}{n^2+3n}\right| =\frac{6n-1}{n(n+3)} $ (because $n \in \mathbb N^+)$. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \... | Your proof is incorrect: you want to show that
$$\frac{6n-1}{n^2-3n}<\varepsilon$$
so what you need is
$$\frac{6n-1}{n^2-3n}<{\rm something}$$ or $6n-1<$something, that is an estimate of $6n-1$ from above, not from below.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1158558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Example of contour integration Could someone help me evaluate the following integral with contour integration ?
$$\int_{0}^{2\pi}\frac{d\theta}{(a+b\cos\theta)^2}.$$
Constraints are: $a>b>0$.
| Using the parametrization $z = e^{i\theta}$, $0 \le \theta \le 2\pi$ for the unit circle, we can write the integral as the contour integral
$$\oint_{|z| = 1} \dfrac{1}{\left(a + b\frac{z + z^{-1}}{2}\right)^2} \frac{dz}{iz},$$
which can be rewritten
$$\oint_{|z| = 1} \dfrac{4z}{i(bz^2 + 2az + b)^2}\, dz,$$
or
$$\frac{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1160331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the sum $\sum_{n=1}^{50}\frac{1}{n^4+n^2+1}$ Find the sum $\sum_{n=1}^{50}\frac{1}{n^4+n^2+1}$
$$\begin{align}\frac{1}{n^4+n^2+1}&
=\frac{1}{n^4+2n^2+1-n^2}\\
&=\frac{1}{(n^2+1)^2-n^2}\\
&=\frac{1}{(n^2+n+1)(n^2-n+1)}\\
&=\frac{1-n}{2(n^2-n+1)}+\frac{1+n}{2(n^2+n+1)}\\
\end{align}$$
For $n={1,2,3}$ it is not givi... | Partial answer.
Rewrite the terms as:
$$\frac{1}{2}\left(\frac{n+1}{n^2+n+1} - \frac{n-1}{1+(n-1)+(n-1)^2}\right) = \\\frac{1}{2}\left(\frac{1}{n^2+n+1}+f(n)- f(n-1)\right)$$ where $f(n)=\frac{n}{1+n+n^2}$. In particular $\sum_{n=1}^\infty \left(f(n)-f(n-1)\right)=-f(0)=0$.
So we've reduced it to:
$$\frac12\sum \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1163703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
geometric proof of $2\cos{A}\cos{B}=\cos{(A+B)}+\cos{(A-B)}$ I have seen geometric proof of identities
$$\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}$$
and
$$\cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}$$
By adding two equation, $$2\cos{A}\cos{B}=\cos{(A+B)}+\cos{(A-B)}$$.
But how to prove this by geometry?
Thank you.
|
$$\begin{align}
2 \cos A \cos B &= \cos(A-B)+\cos(A+B) \\[6pt]
2 \sin A \,\sin B &= \cos(A-B)-\cos(A+B)
\end{align}$$
Note. Although not labeled (yet), these identities are also evident:
$$\begin{align}
2 \,\sin A \cos B &= \sin(A+B)+\sin(A-B) \\[6pt]
2 \cos A \,\sin B &= \sin(A+B)-\sin(A-B)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Simplification of $\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$ I was trying to simplify $\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$. Numerical evaluation suggested that the answer is $\sqrt{2}$ and it checked out when I substituted $\sqrt{2}$ in the equation $x= \sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$.
But I still cannot simplify the ini... | Suppose that
$$
\left(\sqrt{a}\pm\sqrt{b}\right)^2
=\overbrace{a+b\vphantom{\sqrt4}}^c\pm\overbrace{\sqrt{4ab}}^{\sqrt{d}}
$$
Then
$$
\begin{align}
c^2-d
&=a^2+2ab+b^2-4ab\\
&=(a-b)^2
\end{align}
$$
Thus,
$$
a=\frac{c+\sqrt{c^2-d}}2
\qquad\text{and}\qquad
b=\frac{c-\sqrt{c^2-d}}2
$$
If $c=4$ and $d=7$, we get
$$
4-\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1167576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Four numbers drawn from 1 to 100000 randomly, the same number may be chosen more than once. Four numbers drawn from 1 to 100000 randomly, the same number may be chosen more than once. Determine probability the last digit from multiplying that four numbers is 1 or 9.
I have tried with many cases with it but it come a bi... | As mentioned in the comments, we may assume we only draw numbers from the set $\{0,1,2,\ldots,9\}$. This is the case because only the last digit of the number we picked matters. Let us call the random numbers $A$, $B$, $C$, $D$. If one of $A$, $B$, $C$, $D$ is even or divisible by $5$, then $ABCD$ is as well and then i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1167687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
If $a$ is a quadratic residue of odd prime $p$, then is it the case that $a^{-1}$ is also a quadratic residue? If $a$ is a quadratic residue of odd prime $p$, then is it the case that $a^{-1}$ is also a quadratic residue ?
I notice quadratic residues of odd prime $13$ :
$$1^2\equiv 12^2\equiv 1\\2^2\equiv 11^2\equiv ... | If $a^2 = b$ in a ring $R$ with unit, $a$ is invertible if and only if $b$ is and then $b^{-1} = (a^{-1})^2$. (Tell if this is clear to you, if not, I'll detail in an edit.) Apply this to $R = \mathbf{Z} / n \mathbf{Z}$. Apply the latter to the case $n = p$ is prime. ;-)
Detail. If $a$ is (for instance left) invertible... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What am i doing wrong when solving this differential equation $$
f(x) = \frac{\frac{d^2}{dx^2}[e^y]}{\frac{d}{dx}[e^y]}
$$
Given that $f(x) = cx$
$$
\frac{c}{2}x^2 + k_1 = \ln(e^y y')
$$
$$
k_2\int e^{\frac{c}{2}x^2} dx = e^y
$$
$$
y = \ln(k_2\int e^{\frac{c}{2}x^2} dx)
$$
Therefore
$$
y'^2 - cxy' + y'' = 0
$$
$$
y' = ... | Using the very long way, assuming that $y$ is a function of $x$,$$f(x) = \frac{\frac{d^2}{dx^2}[e^y]}{\frac{d}{dx}[e^y]}$$ finally write $$y''-y'^2-y' f(x)=0$$ which eventually allows reduction order setting $z=y'$. So, if $f(x)=c x$, the solution of $z'-z^2-z f(x)=0$ is $$z=\frac{2 \sqrt{c} e^{\frac{c x^2}{2}}}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Explaining that $1 \cdot 3 \cdot 5 \dotsm (2n+1) = 1 \cdot 3 \cdot 5 \dotsm (2n-1)(2n+1)$ I have a few students that are having trouble understanding that
$$1 \cdot 3 \cdot 5 \dotsm (2n+1) = 1 \cdot 3 \cdot 5 \dotsm (2n-1)(2n+1),$$
specifically that
$$\frac{1 \cdot 3 \cdot 5 \dotsm (2n+1)}{1 \cdot 3 \cdot 5 \dotsm (2n-... | I would suggest put some various number
for example n=3 , and force them to write numbers in two rows
then simplify
$$n=3 \to 2n+1=2(3)+1=7 \to1.3.5.7 \\2n-1=2(3)-1=5 \to 1.3.5\\ \frac{1.3.5...(2n+1)}{1.3.5...(2n-1)}=\frac{1.3.5.7}{1.3.5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
How to prove $2^{n+1} * 2^{n+1} = (2^n*2^n)+(2^n*2^n)+(2^n*2^n)+(2^n*2^n)$ Below diagram is used as part of a proof of induction to prove that $E$ a way to tile a $2^n * 2^n$ region with square missing :
What is the proof that $2^{n+1} * 2^{n+1}$ = $(2^n*2^n)+(2^n*2^n)+(2^n*2^n)+(2^n*2^n)$ ?
| In general, $a^m\cdot a^n=a^{m+n}$, so $a^n\cdot a^n=a^{n+n}=a^{2n}$. Thus
$$
a^n\cdot a^n+a^n\cdot a^n+a^n\cdot a^n+a^n\cdot a^n=4a^n\cdot a^n=4a^{2n}
$$
Similarly, $a^{n+1}\cdot a^{n+1}=a^{2n+2}$. Now let $a=2$; the former expression is
$$
4\cdot2^{2n}=2^2\cdot2^{2n}=2^{2n+2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1174146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find all integral solutions to $a+b+c=abc$. Find all integral solutions of the equation $a+b+c=abc$.
Is $\{a,b,c\}=\{1,2,3\}$ the only solution? I've tried by taking $a,b,c=1,2,3$.
| The solutions with $abc=0$ obviously are of form $0,b,-b$. And since $a,b,c$ is a solution iff $-a,-b,-c$ is, we find all solutions with $abc\neq 0$ by finding all with $abc>0$. We show the only such solution is $1,2,3$.
If $abc>0$ then at least one of $a,b,c$ is $>0$ and the other two are alike in sign. But if tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 1
} |
If $\frac {1}{2+a} + \frac {1}{2+b} + \frac {1}{2+c} = 1$, prove $\sqrt{ab} + \sqrt{ac} + \sqrt{bc} \leq 3$ Let $a,b,c$ be non-negative numbers such that $$\frac {1}{2+a} + \frac {1}{2+b} + \frac {1}{2+c} = 1.$$
Prove that $ \sqrt{ab} + \sqrt{ac} + \sqrt{bc} \leq 3 $.
| The condition gives that there are $\alpha\geq0$, $\beta\geq0$ and $\gamma\geq0$ such that $\alpha+\beta+\gamma=\pi$ for which $\sqrt{ab}=2\cos\gamma$, $\sqrt{ac}=2\cos\beta$ and $\sqrt{bc}=2\cos\alpha$.
Hence, we need to prove that $\cos\alpha+\cos\beta+\cos\gamma\leq\frac{3}{2}$, which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Show $x^2+4x+18\equiv 0\pmod{49}$ has no solution My method was to just complete the square:
$x^2+4x+18\equiv 0\pmod{49}$
$(x+2)^2\equiv -14\pmod{49}$
$x+2\equiv \sqrt{35}\pmod{49}$
So $x\equiv\sqrt{35}-2\pmod{49}$, which has no real solutions.
I feel that this may be too elementary, is this the correct way to solve t... | Indeed, it is wrong. $49 \mid (x+2)^2+14$ implies $7\mid (x+2)^2$, and in particular $7\mid x+2$. But this means that $49\mid (x+2)^2$, and by difference $49$ should divide $14$ too, which is false.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Is it possible to write Catalan's product for e as a product of a sequence? Catalan found a product for $e$:
$$e=\dfrac{2}{1}\left(\dfrac{4}{3}\right)^{\frac{1}{2}}\left(\dfrac{6\cdot 8}{5\cdot 7}\right)^{\frac{1}{4}}\left(\dfrac{10\cdot 12\cdot 14\cdot 16}{9\cdot 11\cdot 13\cdot 15}\right)^{\frac{1}{8}}\cdots$$
Is it ... | Look at that fourth term:
$$\left(\frac{10\cdot 12\cdot 14\cdot 16}{9\cdot 11\cdot 13\cdot 15}\right)^{1/8}$$
Factor out the $2$s in the numerator:
$$10\cdot 12\cdot 14\cdot 16=2^4\cdot5\cdot6\cdot7\cdot 8=2^4\frac{8!}{4!}$$
The denominator is $$\begin{align}9\cdot 11\cdot 13\cdot 15 &= \frac{16!}{2^8\cdot 8!\cdot (1\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1185697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find all possible values of $ a^3 + b^3$ if $a^2+b^2=ab=4$. Find all possible values of $a^3 + b^3$ if $a^2+b^2=ab=4$.
From $a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)(4-4)=(a+b)0$. Then we know $a^3+b^3=0$.
If $a=b=0$, it is conflict with $a^2+b^2=ab=4$.
If $a\neq0$ and $b\neq0$, then $a$ and $b$ should be one positive and one ... | There are no solutions in the real numbers.
Plotting $a^2+b^2=4$ on a graph, we get a circle radius 2. Plotting $ab=4$ on the same graph, we get a hyperbola which goes through $(2,2)$ (with the negative side going through $(-2,-2)$) which does not intersect the circle.
A tag of "recreational mathematics" wouldn't usu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1186290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Solve $\int\frac{x}{\sqrt{x^2-6x}}dx$ I need to solve the following integral
$$\int\frac{x}{\sqrt{x^2-6x}}dx$$
I started by completing the square,
$$x^2-6x=(x-3)^2-9$$
Then I defined the substitution variables..
$$(x-3)^2=9\sec^2\theta$$
$$(x-3)=3\sec\theta$$
$$dx=3\sec\theta\tan\theta$$
$$\theta=arcsec(\frac{x-3}{3})$... | I'll give you an answer that is completely free from all this trigonometric nonsense, that I personally think is cleaner and easier. We wish to compute
$$\int \frac{x}{\sqrt{x^2-6x}}\,\mathrm{d}x.$$
Consider first rewriting it as
$$\int \frac{x}{\sqrt{x^2-6x}}\,\mathrm{d}x=\frac{1}{2}\int \frac{2x-6}{\sqrt{x^2-6x}}\,\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1189943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
How to solve $\int\sqrt{1+x\sqrt{x^2+2}}dx$ I need to solve
$$\int\sqrt{1+x\sqrt{x^2+2}}dx$$
I've chosen the substitution variables
$$u=\sqrt{x^2+2}$$
$$du=\frac{x}{\sqrt{x^2+2}}$$
However, I am completly stuck at
$$\int\sqrt{1+xu} dx$$
Which let me believe I've chosen wrong substitution variables.
I've then tried lett... | $x=\sqrt{2}\tan{u},dx=\sqrt{2}\sec^2{u}du,\sqrt{1+x\sqrt{x^2+2}}dx=\sqrt{2+4\tan{u}|\sec{u}|}\sec^2{u}du=\dfrac{\sqrt{2\cos^2{u} \pm 4\sin{u}}}{|\cos{u}|\cos^2{u}}du=\pm\dfrac{\sqrt{2\cos^2{u} \pm 4\sin{u}}}{cos^4{u}}d\sin{u}=\pm\dfrac{\sqrt{2(1-v^2) \pm 4v}}{(1-v^2)^2}dv ,v=\sin{u}$
consider case "+",$2(1-v^2)+4v=2(2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
$\sup$ and $\inf$ of this set These are exercises from my textbook, and I am not sure if the solutions are correct or not.
Given a set $B = \{\frac{n}{2n+1} : n \in \mathbb{N} \}$
Find the $\sup$ and $\inf$ of $B$, and maxima and minima if they exist.
The solutions give $\sup B = \frac{1}{2}$ and $\inf B = \frac{1}{3}$... | First $\frac{1}{3} < \frac{1}{2}$.
Second $B$ is bounded above by $\lim \frac{n}{2n+1} = \frac{1}{2}$. Then $\sup B \leq \frac{1}{2}$. Take any $\epsilon > 0$ then there exists $N \in \mathbb N$ such that
$$n > N \implies |x_n - \frac{1}{2}| < \epsilon $$
then there exists $x_k = \frac{k}{2k + 1} \in B$, such that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$ \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$ How do we compute this integral ?
$$ \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$$
I have tried partial fraction but it is quite hard to factorize the denominator. Any help is appreciated.
| This is not a final answer but it is too long for a comment.
Beside the elegant solution given by Dr.MV using the residue theorem, there is something I found interesting in the problem. $$\frac{1+x^6}{1-x^2+x^4-x^6+x^8}=(1+x^2+x^6+x^8)\sum_{k=0}^{\infty}(-1)^k x^{10k}$$ So $$I=\int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find the Derivative and simplify (write w/o negative exponents and factor if possible) y=cos^3(x^2+2) I am currently practicing on an old college calculus final. Would like to know if my work is correct.
Find the derivative for f(x) = x^1/2 + (3/x^2)-2 and y=cos^3(x^2+2) . Thanks for your help.
| You got one part right, but there are a few mistakes. For example, the function $\frac{3}{x^2}-2$ is equivalently written as $3x^{-2}-2$. By the power rule, the derivative would be $$(-2)3x^{-2-1} = -6x^{-3} = \frac{-6}{x^3}$$ Your derivative of $x^{1/2}$ is correct.
The derivative of $\cos^3(x^3+2)$ is wrong for a fe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solution of $y''+xy=0$ The differential equation $y''+xy=0$ is given.
Find the solution of the differential equation, using the power series method.
That's what I have tried:
We are looking for a solution of the form $y(x)= \sum_{n=0}^{\infty} a_n x^n$ with radius of convergence of the power series $R>0$.
Then:
$$y'(x)... | $$y''+xy=0$$
$$y=\sum_{n=0}^{\infty} c_n x^n, y'=\sum_{n=1}^{\infty} nc_n x^{n-1}, y''=\sum_{n=2}^{\infty} n(n-1)c_n x^{n-2}$$
$$\therefore y''+xy=0=\underbrace{\sum_{n=2}^{\infty} n(n-1)c_n x^{n-2}}_{k=n-2\Rightarrow n=k+2}+\underbrace{\sum_{n=0}^{\infty} c_n x^{n+1}}_{k=n+1\Rightarrow n=k-1}=\sum_{k=0}^{\infty} (k+2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1197528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
Derivative Chain Rule $f(x) = (3x^2+2)^2 (x^2 -5x)^3$ I'm learning chain rule in derivative,
and I don't understand this in the example.
$$f(x) = (3x^2+2)^2 (x^2 -5x)^3\\$$
\begin{align}
f'(x)&= (3x^2 +2)^2[3(x^2 -5x)^2(2x-5)] + (x^2-5x)^3[2(3x^2+2)(6x)]
\\&=3(3x^2 +2)(x^2-5x)^2[(3x^2+2)(2x-5)+4x(x^2-5x)]
\\&= 3(3x^2 ... | \begin{align}
f'(x) &= (3x^2 +2)^2[3(x^2 -5x)^2(2x-5)] + (x^2-5x)^3[2(3x+2)(3)]\\
&=3(3x^2+2)(x^2-5x)^2[(2x-5)-(x^2-5x)(2)] \\
&=3(3x^2+2)(x^2-5x)^2[2x-5-4x^2+10x] \\
&=3(3x^2+2)(x^2-5x)^2(-4x^2+12x-5)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the Inverse Laplace Transforms Find the inverse Laplace transform of:
$$\frac{3s+5}{s(s^2+9)}$$
Workings:
$\frac{3s+5}{s(s^2+9)}$
$= \frac{3s}{s(s^2+9} + \frac{5}{s(s^2+9)}$
$ = \frac{3}{s^2+9} + \frac{5}{s}\frac{1}{s^2+9}$
$ = \sin(3t) + \frac{5}{s}\frac{1}{s^2+9}$
Now I'm not to sure on what to do.
Any help will... | The convolution theorem is, found here,
\begin{align}
\mathcal{L}^{-1} \{f(s)g(s)\} = \int_{0}^{t} f(t-u) g(u) \, du
\end{align}
In the case here $f(s) = 1/s$ which is the transform of $1$ and $g(s)$ being the transform of $\sin$. Now
\begin{align}
\mathcal{L}^{-1}\{ \frac{1}{s ( s^{2} + a^{2})} \} &= \frac{1}{a} \, \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"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.