Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Showing two things are equal by Fourier series Given the Fourier series for the function:
$$f(x) = x+\frac14x^2 \quad -\pi\leq x \lt \pi$$
$$f(x)=f(x+2\pi) \quad -\infty \leq x \lt \infty$$
is
$$\frac{\pi^2}{12}+\sum \limits_{n=1}^\infty (-1)^n \left(\frac{\cos(nx)}{n^2}-\frac{2\sin(nx)}{n}\right)$$
show that$$\sum \li... | Try $x = \pi$. When $x=\pi$, by Dirichlet Theorem (will look up which one),
$$
f(\pi) = \frac{f(-\pi)+f(\pi)}{2} = \frac{\pi^2}{4}
$$
since you have a jump discontinuity at $x=\pi$ (see theorem).
This gives you
$$
\sum_{n=1}^{\infty}\frac{1}{n^2} + \frac{\pi^2}{12}=\frac{\pi^2}{4}
$$
Theorem: Let $f$ be periodic and... | {
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"url": "https://math.stackexchange.com/questions/1023743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factorizing Given Problem I have searched through various site's and forums but couldn't find the answer to my problem,
$$z^2-\frac{1}{2}z-\frac{1}{4}=0$$
How will you factorize this
As I can't find $2$ numbers that give me $ac = -\frac{1}{4}$ and $b = -\frac{1}{2}$ when added
Is there a different approach for such sum... | The quadratic formula can be used to find the roots of a quadratic equation.
$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$ for a quadratic of the form $ax^2+bx+c=0$.
So for $a=1,b=-\frac{1}{2},c=-\frac{1}{4}$, we have that
$x=\dfrac{-\left(\frac{1}{2}\right)\pm\sqrt{(-\frac{1}{2})^2-4(1)(-\frac{1}{4})}}{2(1)}=\dfrac{-\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I evaluate $\lim_{n \to\infty}\left(1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)\right)/\left(1^2+2^2+3^2+\dots+n^2\right)^2$? How can I evaluate this limit? Give me a hint, please.
$$\lim_{n \to\infty}\frac{1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)}{\left(1^2+2^2+3^2+\dots+n^2\right)^2}$$
| The numerator is bounded above by $n.n(n+1)(n+2)$, fourth degree. The denominator is exactly $(2n^3+3n^2+n)^2/36$ (from the square pyramidal number formula), sixth degree. So the denominator "wins".
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Sum of $\sum_{n=0}^{\infty }\frac{1}{4^{(n/3)+1}}$ Find the sum of $$\sum_{n=0}^{\infty }\frac{1}{4^{(n/3)+1}}$$
| geometric sequence with $$q=\frac{1}{4^{\frac{1}{3}}}\\$$
$$\frac{1}{4^{1}}+\frac{1}{4^{\frac{4}{3}}}++\frac{1}{4^{\frac{7}{3}}}++\frac{1}{4^{\frac{10}{3}}}+ ...\\=\frac{a_{1}}{1-q}=\\=\frac{\frac{1}{4^{1}}}{1-\frac{1}{4^{\frac{1}{3}}}}=\\$$
| {
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How do I express the sum $(1+k)+(1+k)^2+\ldots+(1+k)^N$ for $|k|\ll1$ as a series? Wolfram Alpha provides the following exact solution
$$ \sum_{i=1}^N (1+k)^i = \frac{(1+k)\,((1+k)^N-1)}{k}.$$
I wish to solve for $N$ of the order of several thousand and $|k|$ very small (c. $10^{-12}).$ When I do this on a computer in... | An alternative approach using binomial coefficients:
$$\begin{align}
\sum_{i=1}^{N}(1+k)^i&=\sum_{i=1}^N\sum_{r=0}^i \binom ir k^r\\
&=k^0\sum_{i=1}^N\binom i0+\sum_{r=1}^Nk^r\sum_{i=r}^N\binom ir\\
&=N+\sum_{r=1}^Nk^r\binom{N+1}{r+1}
\end{align}$$
NB - the result is the same as
$$\qquad N+\sum_{r=2}^{N+1}k^{r-1}\bi... | {
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Prove that $ \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+\cdots + \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}$ for $n\in \mathbb N$ I want to prove that if $n \in \mathbb N$ then $$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+ \cdots+ \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}.$$
I think I am stuck on two fronts. First, I don't... | Hint. If, for somespecific $n$, we have
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+ \cdots+ \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}\ ,$$
then
$$\eqalign{\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{n+1}{(n+2)!}
&=1-\frac{1}{(n+1)!}+\frac{n+1}{(n+2)!}\cr
&=1-\frac{n+2}{(n+2)!}+\frac{n+1}{(n+2)!}\cr
&=1-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Double Integral $\int\limits_0^a\int\limits_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$ How to solve this integral?
$$\int_0^a\!\!\!\int_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$$
my attempt
$$
\int_0^a\!\!\!\int_0^a\frac{dx \, dy}{(x^2+y^2+a^2)^\frac{3}{2}}=
\int_0^a\!\!\!\int_0^a\frac{dx}{(x^2+\rho^2)^\frac{3}{2}}dy\\
\r... | Let $y=\sqrt{a^2+x^2} \tan \theta$, $dy=\sqrt{a^2+x^2}~ \mbox{sec}^2 \theta~ d\theta$ Then
$$\int_{0}^{a} \int_{0}^{a} \frac{dx ~ dy} {(a^2+x^2+y^2)^{3/2}} =\int_{0}^{a}\frac{dx}{a^2+x^2}\int_{0}^{\tan^{-1}(a/\sqrt{a^2+x^2})} \cos \theta ~d\theta$$ $$ = a \int_{0}^{a} \frac{dx}{(x^2+a^2)\sqrt{(2a^2+x^2)}}.$$
$$ \Right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Integrate $\frac{\lambda y^{2}}{\sqrt{2\pi}} e^{-(\frac{1}{2}+ \lambda x)y^{2}}$ with respect to $y$ $$F(x,y)= \frac{\lambda y^{2}}{\sqrt{2\pi}} e^{-(\frac{1}{2}+ \lambda x)y^{2}}$$
Please show that the function when integrated with respect to $y$ is
$F_X(x)= \frac{\lambda}{(2\sqrt{2}(\frac{1}{2}+ \lambda x)^\frac{3}{... | So, the two main passages are the change of variable and the gamma function. I'm supposing you want to integrate it between $0$ and $+\infty$.Therefore:
\begin{equation}
F_x(x) = \int_0^{+\infty} \frac{\lambda y^2}{\sqrt{2\pi}} e^{-(\frac{1}{2} + \lambda x)y^2} dy= \int_0^{+\infty} \frac{\lambda}{\sqrt{2\pi}} \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Characteristics method applied to the PDE $u_x^2 + u_y^2=u$ I am trying to solve: $u_x^2 + u_y^2=u$ with boundary conditions: $u(x,0)=x^2$. Unfortunately it leads to equations that makes no sense (sum of squares is $0$ and all constants are $0$). I would be grateful for a reasonable explanation how to solve that with h... | Let $u=v^2$ ,
Then $u_x=2vv_x$
$u_y=2vv_y$
$\therefore(2vv_x)^2+(2vv_y)^2=v^2$ with $v(x,0)=x$
$4v^2(v_x)^2+4v^2(v_y)^2=v^2$ with $v(x,0)=x$
$v_x^2+v_y^2=\dfrac{1}{4}$ with $v(x,0)=x$
$v_y^2=\dfrac{1}{4}-v_x^2$ with $v(x,0)=x$
$v_y=\pm\sqrt{\dfrac{1}{4}-v_x^2}$ with $v(x,0)=x$
$v_{xy}=\mp\dfrac{v_xv_{xx}}{\sqrt{\dfrac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $(b-a)\cdot f(\frac{a+b}{2})\le \int_{a}^{b}f(x)dx$ Let $f$ be continuously differentiable on $[a,b]$. If $f$ is concave up, prove that
$$(b-a)\cdot f\left(\frac{a+b}{2}\right)\le \int_{a}^{b}f(x)dx.$$
I know that (and have proved) $$(b-a)\cdot f\left(\frac{a+b}{2}\right)= \int_{a}^{b}f(x)dx$$ for any linear func... | Let $h(x)=f(x)-m x$ where $m=f'(\frac{b+a}2)$. By the Mean Value Theorem for integrals (also related to average value of a function), there is $c\in(a,b)$ such that $h(c)(b-a)=\int_{a}^{b}h(x)dx$. Note that $h$ is concave up and $h'(\frac{b+a}2)=0$ hence $h$ has a global (over $[a,b]$) minimum at $\frac{b+a}2$. Hence ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$. Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$.
My solutions: the homogeneous portion is $a_n=c3^n$, and the inhomogeneous portion is $a^*_n=-1/2n^2-3/4n+9/8$.
This results in a final recurrence relation of
$$a_n=-\frac{1}{8}3^n-\frac{... | Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply the shifted recurrence by $ẓ^n$ and sum over $n \ge 0$, recognize some sums:
$\begin{align*}
\sum_{n \ge 0} a_{n + 1} z^n
&= 3 \sum_{n \ge 0} a_n z^n + \sum_{n \ge 0} (n + 1)^2 - 3 \sum_{n \ge 0} z^n \\
\frac{A(z) -a_0}{z}
&= 3 A(z) + \sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Sinha's Theorem for Equal Sums of Like Powers $x_1^7+x_2^7+x_3^7+\dots$ Sinha’s theorem can be stated as, excluding the trivial case $c = 0$, if,
$$(a+3c)^k + (b+3c)^k + (a+b-2c)^k = (c+d)^k + (c+e)^k + (-2c+d+e)^k\tag{1} $$
for $\color{blue}{\text{both}}$ $k = 2,4$ then,
$$a^k + b^k + (a+2c)^k + (b+2c)^k + (-c+d+e)^k ... | I think not to introduce additional equations, and directly solve the system of equations.
$$\left\{\begin{aligned}&R^2+Q^2+T^2=X^2+Y^2+Z^2\\&R^4+Q^4+T^4=X^4+Y^4+Z^4\end{aligned}\right.$$
Using integer parameters $k,s,t$ - Will make a replacement.
$$a=3(k+s-t)^2+k(k-t)$$
$$b=3(k+s-t)^2+s(s-t)$$
$$c=3(k+s-t)^2-t^2+(k+s)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1037013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Irrational number inequality : $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$ it is easy and simple I know but still do not know how to show it (obviously without simply calculating the sum but manipulation on numbers is key here.
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$$
| Or, one can consider the sum $\sqrt{3} + \sqrt{3/2} + 1$; we have
$\sqrt{3} > 1, \tag{1}$
$\sqrt{\dfrac{3}{2}} > 1, \tag{2}$
$1 \ge 1, \tag{3}$
(this last inequality being particularly subtle!) and adding (1)-(3) we have
$\sqrt{3} + \sqrt{\dfrac{3}{2}} + 1 > 3; \tag{4}$
now we divide by $\sqrt{3}$, and voila!, we obtai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1037112",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find equation of Tangent line at $(4, 1)$ on $5y^3 + x^2 = y + 5x$ Can someone help me find equation of tangent line at $(4, 1)$ on
$5y^3 + x^2 = y + 5x$
$Y=f(x)$
I dont know how to isolate the $Y$
| knowing
$$
y-y_0 = y'(x-x_0)
$$
$$
\frac{d}{dx}[3y^3+x^2] = \frac{d}{dx}[y+5x]
$$
$$
15y^2 y' + 2x = y' + 5
$$
$$
2x-5 = y'[1-15y^2] \therefore
$$
$$
y' = \frac{2x-5}{1-15y^2}
$$
When at the point (4,1) you get
$$
y-1 = \frac{2(4)-5}{1-15}(x-4)
$$
$$
y = \frac{12-3x}{14}+1 = \frac{13}{7}-\frac{3x}{14}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving of $\frac{\pi }{24}(\sqrt{6}-\sqrt{2})=\sum_{n=1}^{\infty }\frac{14}{576n^2-576n+95}-\frac{1}{144n^2-144n+35}$ This is a homework for my son, he needs the proving.I tried to solve it by residue theory but I couldn't.
$$\frac{\pi }{24}(\sqrt{6}-\sqrt{2})=\sum_{n=1}^{\infty }\frac{14}{576n^2-576n+95}-\frac{1}{14... | I wouldn't say it's very illuminating, but here's an answer:
We have
\begin{align*}
\sum_{n=1}^\infty \frac{14}{576n^2 - 576 + 95}
&= \sum_{n=1}^\infty \frac{1}{24n - 19} - \frac{1}{24n - 5} \\
&= \frac{1}{24} \sum_{n=1}^\infty \frac{1}{n - 19/24} - \frac{1}{n - 5/24} \\
&= \frac{1}{24} \left(\psi\left(\frac{19}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $(1+x+x^2)^{25} =\sum^{50}_{r=0} a_r x^r$ then .... If $$(1+x+x^2)^{25} =\sum^{50}_{r=0} a_r x^r$$
then find :
$\sum^{16}_{r=0} a_{3r} =$
My approach :
let (1+x) =t therefore,
$(1+x+x^2)^{25} =\sum^{50}_{r=0} a_r x^r$
=$(1+x+x^2)^{25} = (t+x^2)^{25}$
$= ^{25}C_0t^{25} +^{25}C_1 t^{24}x^2 +^{25}C_2 t^{23}x^4 +\cd... | Hint: Evaluate at the two roots of the equation $1+x+x^2=0$ and at $x=1$, that is, at the three cube roots of unity, and add. The key fact is that if $\omega$ is one of the roots of $1+x+x^2=0$, then $\omega^2$ is the other, and $1^k+\omega^k+\omega^{2k}=0$ unless $k$ is divisible by $3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find a closed form for the equations $1^3 = 1$, $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$ This is the assignment I have:
Find a closed form for the equations
$1^3 = 1$
$2^3 = 3+5$
$3^3 = 7+9+11$
$4^3 = 13+15+17+19$
$5^3 = 21+23+25+27+29$
$...$
Hints. The equations are of the form $n^3 = a1 +a2 +···+an$, where
$a_{i+1} = a_... | $$\begin{align}
1^3&=1\\
2^3&=3+5\\
3^3&=7+9+11\\
4^3&=13+15+17+19\\
5^3&=21+23+25+27+29\\
\vdots &= \vdots\\
n^3&=[(n^2-n+1)]+[(n^2-n+1)+2]+[(n^2-n+1)+4]+\cdots+[(n^2-n+1)+2(n-1)]\\
&=\sum_{r=1}^n(n^2-n+1)+2(r-1)\qquad \blacksquare
\end{align}$$
That appears to be the formula required for the "series".
The formula ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $\int \frac{x^4}{(x-1)(x^2+1)}dx$ Evaluation of $\displaystyle \int \frac{x^4}{(x-1)(x^2+1)}dx$
$\bf{My\; Try::}$ Let $$\displaystyle I = \int\frac{x^4}{(x-1)(x^2+1)}dx = \int \frac{(x^4-1)+1}{(x-1)(x^2+1)}dx = \int\frac{(x-1)\cdot (x+1)\cdot (x^2+1)}{(x-1)(x^2+1)}+\int\frac{1}{(x-1)(x^2+1)}dx$$
So $\disp... | Since we aren't permitted to use Partial Fraction, therefore undergo the following method,
$x=\tan \theta \implies dx=\sec^2 \theta\ d\theta$
$\therefore\displaystyle\int \dfrac{dx}{(x-1)\left(x^2+1\right)}=\displaystyle\int \dfrac{\sec^2 \theta\ d\theta}{(\tan \theta-1)\sec^2 \theta}=\displaystyle\int\dfrac{d\theta}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} \ge \frac{1}{a + 2b}$
Prove that $$\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} \ge \frac{1}{a + 2b}$$
I tried to to prove the above statement using the AM-HM inequality:
$$\begin{align}\frac{1}{2^n - 2^{n-1}}\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} &\ge \f... | For $a,b,x>0$, the function $f(x)=\frac{1}{a+bx}$ is convex. So by Jensen's inequality
$$
\frac{1}{2^{n-1}}\sum_{i=2^{n-1}+1}^{2^n}f(i)\geq f\left(\frac{1}{2^{n-1}}\sum_{i=2^{n-1}+1}^{2^n}i\right)=f(0.5+3\times2^{n-2})=\frac{1}{a+(0.5+3\times2^{n-2})b}.
$$
It remains to show $2^{n-1}$ times the rightmost expression abo... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving formula for sum of squares with binomial coefficient $$\sum_{k=0}^{n-1}(k^2)= \binom{n}{3} + \binom{n+1}{3}$$
How should I prove that it is the correct formula for sum of squares?
Should I use induction to prove the basis? Any help is appreciated.
| It's easy to prove this directly from the definition of the binomial coefficient. The advantage is you don't need to remember any binomial identities, or try to rewrite a fraction to look like a binomial.
The left-hand side is $$\sum_{k=0}^{n-1}k^2=\frac{n(n-1)(2n - 1)}{6}$$ by a well known formula (you can also prove ... | {
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"url": "https://math.stackexchange.com/questions/1046693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Evaluating $ \int_{0}^{\infty}\frac{\ln (1+16x^2)}{1+25x^2}\mathrm d x$ how to solve such type of definite integration?
I would like to see various methods to evaluate following integral
$$
\int_{0}^{\infty}\frac{\ln (1+16x^2)}{1+25x^2}\mathrm d x$$
| Rewrite the integral as
$$ 4 \log{2}\int_0^{\infty} dx \frac{1}{1+25 x^2} + \int_0^{\infty} dx \frac{\log {\left(\frac1{16}+x^2 \right)}}{1+25 x^2} $$
Consider
$$I(a) = \int_0^{\infty} dx \frac{\log{(a+x^2)}}{1+25 x^2} $$
$$\begin{align}I'(a) &= \int_0^{\infty} dx \frac{1}{(a+x^2)(1+25 x^2)}\\ &= \frac12 \int_{-\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1049545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the linear homogeneous recurrence relation with constant coefficients $$9a_{n} = 6a_{n-1}-a_{n-2}, a_{0}=6, a_{1}=5$$
So
$$x^n = (6x^{n-1}-x^{n-2})\div9$$
thus
$$[x^2 = (6x-1)\div9] \equiv [x^2 - \frac{2}{3}x + \frac{1}{9} = 0], x=\frac{1}{3}$$
also
$$a_{2}=\frac{8}{3}, a_{3}=\frac{31}{27}$$
How do I plug in that... | Here are the steps
$$
a_{0}=6
$$
$$
a_{1}=5
$$
$$
9a_{n} = 6a_{n-1}-a_{n-2}
$$
Lets rewrite this recurrence as
$$
a_{n} = \frac69a_{n-1}-\frac19a_{n-2}
$$
$$
a_{n} = \frac23a_{n-1}-\frac1{3^2}a_{n-2}
$$
Now let's multiply by $3^n$
$$
3^0a_{0}=3^06=6
$$
$$
3^1a_{1}=3^15=15
$$
$$
3^na_{n} = 2\cdot 3^{n-1}a_{n-1}-3^{n-2}a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$1 + 4 + 7 + · · · + 3n − 2 =
\frac{n(3n − 1)}
2$$
for all positive integers $n$.
Proof: $$1+4+7+\ldots +3(k+1)-2= \frac{(k + 1)[3(k+1)+1]}2$$
$$\frac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$$
Along my proof I am stuck at the above section where it w... | The formula must be a quadratic polynomial (because its first order difference is a linear polynomial) and has three independent coefficients. It suffices to identify for three different values of $n$:
$$\begin{align}1=\frac{1(3\cdot1-1)}2
\\1+4=\frac{2(3\cdot2-1)}2
\\1+4+7=\frac{3(3\cdot3-1)}2\end{align}$$
This comple... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 1
} |
Show that $\prod_{i=2}^n \left(1-\frac{1}{i^2}\right) = \frac{n+1}{2n}$for $n \in \Bbb{N}$, $ n \ge 2$ Use mathematical induction to shoe that fpr any $n\in N$, if $n\ge2$, then
$$\prod_{i=2}^{n}\left(1-\frac{1}{i^2}\right)=\frac{n+1}{2n}$$
So I understand what's happening up until the first red circle. Why does 1 - b... | $$\color{blue}{1}-\frac{1}{(n+1)^2}=\color{blue}{\frac{(n+1)^2}{(n+1)^2}}-\frac{1}{(n+1)^2}=\frac{(n+1)^2-1}{(n+1)^2}$$
Now use $$(a+b)^2=(a+b)(a+b)=a^2+ab+ab+b^2=a^2+2ab+b^2$$
to get $(n+1)^2=n^2+2n+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1051587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $ \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ Evaluation of $\displaystyle \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ where $\lfloor x \rfloor $ represent floor function of $x$.
My Try:: Here $\displaystyle f(x) = \frac{x^2}{\sin x\cdot \tan x}$ is an... | Hint?
$$
\lim_{x\to 0} \frac{x^2}{\sin x \cdot \tan x}
= \lim_{x\to 0} \cos x \left( \frac{x}{\sin x} \right)^2
$$
Now we can split $\lim ab=\lim a \cdot \lim b$, since both limits exists. Alternatively one can use L'ôpitals rule twice. $\lim_{x\to a} f/g = \lim_{x\to a} f'/g'$. Somewhat of a hazzle but it works
$$
\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$x^3-3x^2+4x-2$ cannot be factored over $\mathbb R$ I'm new to the site, and I need a bit of help from you.
How can I prove that the polynomial: $f(x)=x^3-3x^2+4x-2$
cannot be factored as a product of polynomials of degree 1 with real coefficients?
Thanks.
| Observe that$$\begin{align*}
f(x)=x^3-3x^2+4x-2&=(x^3-3x^2+3x-1)+(x-1)\\\\
&=(x-1)^3+(x-1)\\\\
&=(x-1)((x-1)^2+1)\\\\
&=(x-1)(x^2-2x+2)
\end{align*}$$
By the quadratic formula, the roots of $x^2-2x+2$ are complex, so $f$ cannot be factored any further into polynomials with real coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Find $\lim_{x \to 0^-}\frac{\sqrt{1 - \cos4x}}{\sin3x - \sin x}$ Find $ \lim_{x \to 0^-}\frac{\sqrt{1 - \cos4x}}{\sin3x - \sin x}$
What I did
$\frac{\sqrt{1 - \cos4x}}{\sin3x - \sin x}$
$\frac{\sqrt{2}\sin2x}{\sin3x - \sin x}$
$\frac{\sqrt{2}\sin2x}{2\cos 2x \sin x}$
$\frac{\sqrt{2}(2 \sin x \cos x)}{2\cos 2x \sin x}$
... | $$\begin{gathered}
\mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt {1 - \cos \left( {4x} \right)} }}
{{\sin \left( {3x} \right) - \sin \left( x \right)}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt {2{{\sin }^2}\left( {2x} \right)} }}
{{2\cos \left( {2x} \right)\sin \left( x \right)}} = \mathop {\lim }\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show $\sum_{n=1}^\infty\frac{1}{n^2+3n+1}=\frac{\pi\sqrt{5}}{5}\tan\frac{\pi\sqrt{5}}{2}$. How to show that
$$\sum_{n=1}^\infty\frac{1}{n^2+3n+1}=\frac{\pi\sqrt{5}}{5}\tan\frac{\pi\sqrt{5}}{2}$$
?
My try:
We have
$$n+3n+1=\left(n+\frac{3+\sqrt{5}}{2}\right)\left(n+\frac{3-\sqrt{5}}{2}\right),$$
so
$$\frac{1}{n^2+3n+1}=... | Note
$$ n^2+3n+1=(n+\frac{3}{2})^2+\left(\frac{\sqrt 5i}{2}\right)^2 $$
and
and hence
\begin{eqnarray}
\sum_{n=0}^\infty\frac{1}{n^2+3n+1}&=&\sum_{n=0}^\infty\frac1{(n+\frac{3}{2})^2+\left(\frac{\sqrt 5i}{2}\right)^2}\\
&=&\frac12\sum_{n=-\infty}^\infty\frac1{(n+\frac{3}{2})^2+\left(\frac{\sqrt 5i}{2}\right)^2}.
\end{e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Find limit of $\frac {1}{x^2}- \frac {1}{\sin^2(x)}$ as x goes to 0 I need to use a taylor expansion to find the limit.
I combine the two terms into one, but I get limit of $\dfrac{\sin^2(x)-x^2}{x^2\sin^2(x)}$ as $x$ goes to $0$. I know what the taylor polynomial of $\sin(x)$ centered around $0$ is… but now what do I... | Using Taylor series
$$\lim_{x\rightarrow0} \frac{\sin^2 x - x^2}{x^2\sin^2 x} = \lim_{x\rightarrow0} \frac{\Big(x - \frac{x^3}{6} + O(x^5)\Big)^2 - x^2}{x^2\Big(x - \frac{x^3}{6} + O(x^5)\Big)^2} = \lim_{x\rightarrow 0} \frac{\Big(x^2 - \frac{x^4}{3} + O(x^6)\Big) - x^2}{x^2\Big(x^2 - \frac{x^4}{3} + O(x^6)\Big)} \\= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What is the value of the expression $2x^2 + 3xy – 4y^2$ when $x = 2$ and $y = - 4$? What is the value of the expression $2x^2 + 3xy – 4y^2$ when $x = 2$ and $y = - 4$?
I'm not good at algebra so please explain in easy to understand steps.
Thanks
| $2x^2 + 3xy - 4y^2$ when $x = 2$ and $y = -4$,
$2(2)^2 + 3(2)(-4) - 4(-4)^2 = 2 *4 - 24 - 4*16 = 8 - 24 - 64 = -80$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.
| You can easily see that $x^6 - 1$ has a root at $1$, so you know that
$$x^6 - 1 = (x-1) \cdot p(x)$$
Where $p(x)$ is a polynomial of degree $5$. Perform polynomial division to find $p$:
$$\begin{align*} (x^6 - 1) \div (x-1) & = x^5 + (x^5 - 1) \div (x-1) \\
& = x^5 + x^4 + (x^4 - 1) \div (x-1) \\
& = x^5 + x^4 + x^3 + ... | {
"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": 3
} |
Find value of sum of reciprocals of powers of a number Is there a simple way to find the value of the following expression?
$$\frac1x+\frac1{x^2}+\frac1{x^3}+\cdots$$
On trial and error, I was getting $\frac1{x-1}$, but I'm looking for a mathematical proof to it.
Please don't use complicated notation like summation unl... | Define the $n$th partial sum by
$$
S_n = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots + \frac{1}{x^n}
$$
Then
\begin{align*}
x S_n &= 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots + \frac{1}{x^{n-1}} \\
\implies x S_n - S_n &= 1 - \frac{1}{x^n} \iff S_n = \frac{1- \frac{1}{x^n}}{x-1}, \; x \neq 1
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Using the half/double angle formulas to solve an equation I am completely stumped by this problem: $$\cos\theta - \sin\theta =\sqrt{2} \sin\frac{\theta}{2} $$
I know that I should start by isolating $\sin\dfrac{\theta}{2}$
and end up with $$\frac {\cos\theta - \sin\theta}{\sqrt{2}} = \sin\frac{\theta}{ 2}$$
From here o... | $$\cos \theta . \sin \frac{\pi}{4} - \cos \frac{\pi}{4} . \sin \theta = \sin \frac{\theta}{2}$$
$$\sin (\frac{\pi}{4} - \theta) = \sin \frac{\theta}{2}$$
$$\frac{\pi}{4} - \theta = n\pi + (-1)^n(\frac{\theta}{2})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Integral with series How do I represent this integral
$$\int_{0}^{1} \frac{10}{10+x^4} dx$$
as a series so that I can calculate with an error of less than $10^{-5}$.
| \begin{align}
\frac{10}{10+x^4} = \frac{1}{1-\left( \frac{-x^4}{10} \right)} = \frac 1 {1-r} & = 1+r+r^2+r^3+\cdots \\[6pt]
& = 1-\frac{x^4}{10} + \frac{x^8}{100} - \cdots
\end{align}
Integrating term by term from $0$ to $1$ gives
$$
1 - \frac 1 {50} + \frac 1 {900} -\cdots
$$
Since the terms alternate in sign and get ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to sum $\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2}$? Does anyone know the general strategy for summing a series of the form:
$$\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2},$$
where $a$ is a positive integer?
Any hints or ideas would be great!
| Certain series of the form $\sum_{n = -\infty}^\infty f(n)$ can be evaluated by means of residue calculus. One important result which is useful to this problem states that if $f$ is holomorphic on $\Bbb C \setminus \{z_1,\ldots, z_k\}$ (where the $z_i$ are the isolated singularities of $f$) and $|zf(z)|$ is bounded for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1064217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"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... | When you say "full Cardano" are you referring to using a formula? Because there is a process for solving a cubic that is constructive. But maybe you do not want this because it is essentially how to construct Cardano's formula.
Substituting $x=y+3$ leads to:
$$\begin{align}
x^3-9x^2-15x-6&=0\\
(y+3)^3-9(y+3)^2-15(y+3)-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Converting a polar equation to a rectangular one $$r=\frac { 4 }{ 1+2\sin\theta } $$
Steps I took:
$$(1+2\sin\theta )r=\frac { 4 }{ 1+2\sin\theta } (1+2\sin\theta )$$
$$r+2r\sin\theta =4$$
$$r+2y=4$$
$$(r+2y)^2=16$$
$$(r+2y)(r+2y)=r^2+4yr+4y^2$$
$$r^2+4yr+4y^2-16=0$$
My outcome doesn't seem to match the correct answe... | use that $$r=\sqrt{x^2+y^2}$$ and $$\sin(\theta)=\frac{y}{\sqrt{x^2+y^2}}$$ and you will get
$$\sqrt{x^2+y^2}=\frac{4}{1+\frac{2y}{\sqrt{x^2+y^2}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $c_n = \frac1n \bigl(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} \bigr)$ converges I want to show that $c_n$ converges to a value $L$ where:
$$c_n = \frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}}}{n}$$
First, it's obvious that $c_n > 0$.
I was abl... | Just wondering that the following elementary way is not among the presented solutions.
So, I add it here as a late answer:
The following inequality for $k \in \mathbb{N}$ will be used:
$$\sqrt{k+1}-\sqrt k = \frac 1{\sqrt{k+1}+\sqrt k} > \frac 1{2\sqrt{k+1}}$$
Hence,
\begin{eqnarray*} \frac 1n\sum_{k=1}^n \frac 1{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 5
} |
Prove that $\int_0^\infty\frac{x^n}{1+e^{x-t}}\mathrm{d}x = \frac{t^{n+1}}{n+1} + o(t^n)$, when $t \to \infty,\,n\in\Bbb{R}^+$ I hae to prove that
$$\int_0^\infty\frac{x^n}{1+e^{x-t}}\mathrm{d}x = \frac{t^{n+1}}{n+1} + o(t^n), \quad\text{ when } t \to \infty,\,n\in\Bbb{R}^+$$
where $o(\cdot)$ is the Little-o notation.... | First, let's split up the integral:
$$
\begin{align}
\int_0^\infty\frac{x^n}{1+e^{x-t}}\mathrm{d}x
&=\int_{-t}^\infty\frac{(x+t)^n}{1+e^x}\mathrm{d}x\\
&=\color{#C00000}{\int_{-t}^0\frac{(x+t)^n}{1+e^x}\mathrm{d}x}+\color{#00A000}{\int_0^\infty\frac{(x+t)^n}{1+e^x}\mathrm{d}x}\tag{1}
\end{align}
$$
Note that the first ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation $2x^2+5y^2+6xy-2x-4y+1=0$ in real numbers
Solve the equation $2x^2+5y^2+6xy-2x-4y+1=0$
The problem does not say it but I think solutions should be from $\mathbb{R}$. I tried to express the left sum as a sum of squares but that does not work out. Any suggestions?
| $$2x^2 + 5y^2 + 6xy -2x -4y+1=0$$
$$(1+1)x^2 + (4+1)y^2 + (4+2)xy - 2x -4y + 1=0$$
$$(x^2 +4y^2 +4xy -2x -4y + 1) + (x^2 +2xy + y^2)=0$$
$$(x+2y-1)^2 + (x+y)^2=0$$
Your idea to write the expression as a sum of squares is good intuition. Now, when can a sum of squares be zero? Exactly when both of the squares are zero.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Prove $4^k - 1$ is divisible by $3$ for $k = 1, 2, 3, \dots$ For example:
$$\begin{align}
4^{1} - 1 \mod 3 &=
\\
4 -1 \mod 3 &=
\\
3 \mod 3 &=
\\3*1 \mod 3 &=0
\\
\\
4^{2} - 1 \mod 3 &=
\\
16 -1 \mod 3 &=
\\
15 \mod 3 &=
\\3*5 \mod 3 &= 0
\\
\\
4^{3} - 1 \mod 3 &=
\\
64 -1 \mod 3 &=
\\
21 \mod 3 &=
\\3*7 \mod 3 &=
... | You may also go about this easily with induction:
$$4^{n+1}-1 = (3+1)\cdot 4^n-1 = 3\cdot 4^n + (4^{n}-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 0
} |
Least squares with matrix in $GF(2)$? Here's an example of a problem I'm working on involving finding combination of bit vectors that yield a certain sum (in the $GF(2)$ sense):
$
\begin{pmatrix}
1 & 1 & 1 & 1 & 0 & 0 & 1 & 0 & 0 \\
1 & 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 \\
1 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 1 \\
1 & 0 & 0 & 1... | The norm you're talking about is the Hamming weight of an element and you want the element of minimum Hamming weight in your solution space. As far as I know there does not exist a good general algorithm for computing it.
By computing the nullspace of that matrix and then a reduced basis of that nullspace, you can pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Why the differentiation of $e^x$ is $e^x?$ $$\frac{d}{dx} e^x=e^x$$
Please explain simply as I haven't studied the first principle of differentiation yet, but I know the basics of differentiation.
| \begin{align}
\frac{d}{dx}e^x &= \frac{d}{dx}\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \right)\\
&=0+\frac{1}{1!}+\frac{2x}{2!}+\frac{3x^2}{3!}+\dots\\
&=0+1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots\\
&=0 + e^x\\
&=e^x
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 5
} |
Integration by nonobvious substitutions The standard technique for evaluating the integral
$$\int \sec x \,dx$$
is making the nonobvious substitution
$$u = \sec x + \tan x, \qquad du = (\sec x \tan x + \sec^2 x) dx,$$ which transforms the integral to
$$\int \frac{du}{u} .$$
What are other integrals that are nicely hand... | If you're unaware of a substitution, which usually happens when you have'nt attempted a problem of that particular type, the substitution usually seems non-obvious(happened to me a lot when beginning to learn integration), some of the examples which I liked the most are:
From here:$$\int \frac{\mathrm{dx}}{x^4[x(x^5-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Number of Solutions of $y^2-6y+2x^2+8x=367$? Find the number of solutions in integers to the equation
$$y^2-6y+2x^2+8x=367$$
How should I go about solving this?
Thanks!
| Use the completing the square technique to get $(y-3)^2+2(x+2)^2 = 384$.
Make the substitution $a = y-3$ and $b = x+2$ to get $a^2+2b^2 = 384$.
Since $2b^2$ and $384$ are even, $a^2$ must be even, and hence $a$ is even. So let $a = 2a_1$.
Then, we get $4a_1^2+2b^2 = 384$, i.e. $2a_1^2+b^2 = 192$.
Since $2a_1^2$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Inequality involving $\frac{\sin x}{x}$ Can anybody explain me, why the following inequality is true?
$$\sum_{k=0}^{\infty} \int_{k \pi + \frac{\pi}{4}}^{(k+1)\pi-\frac{\pi}{4}} \left| \frac{\sin \xi}{\xi} \right| \, \text{d} \xi \geq \sum_{k=0}^{\infty} \frac{\left| \sin\left( (k+1) \pi - \frac{\pi}{4} \right)\right|... | Hint
For $x\in[k\pi+\pi/4,(k+1)\pi-\pi/4]$,
$$|\sin x|\ge|\sin\big((k+1)\pi-\pi/4\big)|$$
and
$$x\le(k+1)\pi-\pi/4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the greatest common divisor of $2003^4 + 1$ and $2003^3 + 1$ Find the greatest common divisor of $2003^4 + 1$ and $2003^3 + 1$ without the use of a calculator. It is clear that $2003^4+1$ has a $082$ at the end of its number so $2003^4+1$ only has one factor of 2, while $2003^3+1$ has a $028$ at the end of its num... | $d = \text{gcd}(2003^4+1,2003^3+1) \Rightarrow d \mid 2003^4-2003^3 = 2003^3\cdot 2002 = 2003^3\cdot 2\cdot 7\cdot 11\cdot 13$, and by checking case by case, we have $2$ is the only number such that $d$ divides into evenly. So $d = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Is $\ln(1+\frac{1}{x-1}) \ge \frac{1}{x}$ for all $x \ge 2$? Plotting both functions $\ln(1+\frac{1}{x-1})$ and $\frac{1}{x}$ in $[2,\infty)$ gives the impression that $\ln(1+\frac{1}{x-1}) \ge \frac{1}{x}$ for all $x \ge 2$.
Is it possible to prove it?
| We know that for $|x| < 1$, $$\log(1-x) = -x-\frac{1}{2}x^2 - \frac{1}{3}x^3 - \dotsb.$$
Since $$1 + \frac{1}{x-1} = \frac{x-1+1}{x-1} = \frac{x}{x-1},$$
we see that
$$\log \left(1+\frac{1}{x-1}\right) = \log \left(\frac{x}{x-1}\right) = -\log\left(\frac{x-1}{x}\right) = -\log\left(1-\frac{1}{x}\right).$$
So if $x \ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the zeros of $f(x)=x^3+64$ $$f(x)=x^3+64$$
Again, I am really not sure how to do this I tried to factor but it clearly was not the right answer
| $x^3 + 64 = x^3+4^3 = (x+4)(x^2-4x+16) = (x+4)((x-2)^2+12) = 0 \to x+4 = 0 \to x= -4$. I suppose you want real roots. If you want complex roots, then $(x-2)^2 = -12 = (2\sqrt{3}\cdot i)^2 \to x = 2 \pm 2\sqrt{3}\cdot i$.
In general, if you are interested in finding the roots of $p(x) = x^3 + q$, then you first write: $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1084023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Getting wrong answer trying to evaluate $\int \frac {\sin(2x)dx}{(1+\cos(2x))^2}$ I'm trying to evaluate $$\int \frac {\sin(2x)dx}{(1+\cos(2x))^2} = I$$
Here's what I've got:
$$t=1+\cos(2x)$$
$$dt=-2\sin(2x)dx \implies dx = \frac {dt}{-2}$$
$$I = \int \frac 1{t^2} \cdot \frac {-1}2 dt = -\frac 12 \int t^{-2}dt = -\frac... | Probably, you saw the solution as
$$\frac{\sec^2(x)}{4}+C$$
But note that
$$\begin{array}{rclcc}
\frac{1}{2}\frac{1}{1+\cos(2x)}&=&\frac{1}{2}\frac{1}{1+\cos^2(x)-\sin^2(x)}&/\,\,\cos(2x)=\cos^2(x)-\sin^2(x)\\
&=&\frac{1}{2}\frac{1}{1+\cos^2(x)-\sin^2(x)}&/\,\,1=\cos^2(x)+\sin^2(x)\\
&=&\frac{1}{2}\frac{1}{2\cos^2(x)}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1084586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Minimum of $\newcommand{\b}[1]{\bigl(#1\bigr)} \newcommand{\f}{\frac} \b{\f3a-1}^2+\b{\f ab-1}^2+\b{\f bc-1}^2+(3c-1)^2$ The minimum of :
$$\newcommand{\b}[1]{\left(#1\right)}
\newcommand{\f}{\frac}
\b{\f3a-1}^2+\b{\f ab-1}^2+\b{\f bc-1}^2+(3c-1)^2$$ where $0<a,b,c\le9$, is $p-q\sqrt{r}$; $p,q,r\in\mathbb Z$ and $q,r$ ... | USe Cauchy-Schwarz inequality we have
$$x^2+y^2+z^2+w^2\ge\dfrac{1}{4}(x+y+z+w)^2$$
so
$$LHS\ge \dfrac{1}{4}\left(\dfrac{3}{a}+\dfrac{a}{b}+\dfrac{b}{c}+3c-4\right)^2$$
and Use AM-GM inequality we have
$$\dfrac{3}{a}+\dfrac{a}{b}+\dfrac{b}{c}+3c\ge 4\sqrt[4]{\dfrac{3}{a}\cdot\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot 3c}= 4\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to define an affine transformation using 2 triangles? I have $2$ triangles ($6$ dots) on a $2D$ plane.
The points of the triangles are: a, b, c and x, y, z
I would like to find a matrix, using I can transform every point in the 2D space.
If I transform a, then the result is x. For b the result is y, and for c the r... | The transformation you're looking for has this form:
$$\left[
\begin{array}{ccc}
t_1 & t_2 & t_3\\
t_4 & t_5 & t_6\\
0 & 0 & 1\\
\end{array}
\right]
\left[
\begin{array}{ccc}
a_1 & b_1 & c_1\\
a_2 & b_2 & c_2 \\
1 & 1 & 1 \\
\end{array}
\right]
= \left[\begin{array}{ccc}
x_1 & y_1 & z_1\\
x_2 & y_2 & z_2\\
1 & 1 & 1
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Show that: $\frac{1}{1+\log_ab+\log_bc}+\frac{1}{1+\log_bc+\log_ca}+\frac{1}{1+\log_ca+\log_ab}\leq1$
Let $a, b, c>1$. Show that:
$$\frac{1}{1+\log_ab+log_bc}+\frac{1}{1+\log_bc+\log_ca}+\frac{1}{1+\log_ca+\log_ab}\leq1.$$
My attempt:
We noted $\log_bc=x, \log_ca=y, \log_ab=z$ with $xyz=1$ and we have reduced ineq... | other solution:
since $xyz=1$,we only prove
$$\dfrac{1}{1+x+y}+\dfrac{1}{1+y+z}+\dfrac{1}{1+x+z}\le 1$$
let
$$x=\dfrac{a^2}{bc},y=\dfrac{b^2}{ac},z=\dfrac{c^2}{ab}$$
so
$$\Longrightarrow \dfrac{1}{1+y+z}=\dfrac{1}{1+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}}=\dfrac{abc}{abc+b^3+c^3}$$
so this inequality we only prove
$$\Longle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
find the limit of the following expression Let $f\left(x\right)=\left(x^{2}+1\right)e^{x}$.
Find the following:$$\lim_{n\to+\infty}n\int_{0}^{1}\left(f\left(\frac{x^{2}}{n}\right)-1\right)\,dx$$
I dont know if L'Hopital's Rule may be used and tried using Integral Mean Value Theorem, but I cannot solve it! Any help?
| Explicitly,
$$
\lim_{n\to\infty} n \int_0^1 \left[f\left(\frac{x^2}{n}\right) -1\right]\,dx
= \lim_{n\to\infty} n \int_0^1 \left[\left(\frac{x^4}{n^2} + 1\right)e^{x^2/n} - 1\right]\,dx
$$
You can expand the exponential as a Taylor series:
\begin{align*}
n\int_0^1 \left[\left(\frac{x^4}{n^2} + 1\right)e^{x^2/n} - 1\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Trouble with definite integral calculating probabilities I cannot solve this:
$$\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}} \tan ^{-1}(a+\tan (x)) \, dx$$
it apeared when trying to find out the probability:
$$P\{\tan a - \tan b \leq 2x\},\ \ 0 < x < 1\sqrt{3}$$
Knowing that the joint distribution $f(a,b)$ is
$f(a,b) = \f... | An approach that is less elegant but more elementary than residues.
Let
$$
J(b) = \int_{-\pi/2}^{\pi/2} \tan^{-1} \left( 2b + \tan x \right) dx.
$$
Then $J(0) = 0$ and using the substitution $\tan x = t$,
$$
\begin{align}
J'(b) &= \int_{-\infty}^\infty \frac{2}{(1+t^2)(1+(t+2b)^2)} dt
= \int_{-\infty}^\infty \frac{dt}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
If $x,y,z \geq 1/2, xyz=1$, showing that $2(1/x+1/y+1/z) \geq 3+x+y+z$ If $x,y,z \geq 1/2, xyz=1$, showing that $2(1/x+1/y+1/z) \geq 3+x+y+z$
I tried Schturm's method for quite some time, and Cauchy Schwarz for numerators because of the given product condition.
| We make the usual substitution $x = \dfrac{b}{a},y = \dfrac{c}{b}$ and $z = \dfrac{a}{c}$, then the inequality translates into proving:
$$\displaystyle 2\left(\sum\limits_{cyc} \frac{a}{b} - 3\right) \ge \sum\limits_{cyc} \frac{b}{a} - 3$$
Wlog assume $c$ lies between $a$ and $b$ (that is $a \le c \le b$ or $b \le c \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"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... | $z^2 + 32 = (z+2)(x - \alpha)(z - \overline {\alpha})(z - \beta)(z- \overline {\beta})$ where $\overline {\alpha}, \overline {\beta}$ are complex conjugate of $\alpha, \beta$ respectively. Also $(z - \alpha)(z - \overline {\alpha})$ and $(z - \beta)(z- \overline {\beta})$ are real polynomials (why?). Now choose $\alpha... | {
"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": 3
} |
Sum of consecutive numbers I was wondering if there a way to figure out the number of ways to express an integer by adding up consecutive natural numbers.
For example for $N=3$ there is one way to express it
$1+2 = 3$
I have no idea where to start so any help would be appreciated
| The sum of the integers from $1$ to $n$ is $\dfrac{n(n+1)}{2}$. Hence, the sum of the integers from $m+1$ to $n$ is simply $\dfrac{n(n+1)}{2}-\dfrac{m(m+1)}{2}$. So, if the sum of the integers from $m+1$ to $n$ is $N$, then
$\dfrac{n(n+1)}{2}-\dfrac{m(m+1)}{2} = N$
$(n^2+n) - (m^2+m) = 2N$
$(n-m)(n+m+1) = 2N$
Hence, $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
Prove the following inequality without using differentiation Let $a,b,c$ be real numbers that satisfy $0\le a,b,c\le 1$. Show that
$$\frac a{b+c+1} + \frac b{a+c+1} + \frac c{a+b+1} + (1-a)(1-b)(1-c) \le 1.$$
I don't know where to start. Multiplying everything by the denominators creates extreme mess.
| Let $f(a, b, c)$ denote the left hand side of the inequality. Since $$\frac{\partial^2}{\partial a^2}f=\frac{2b}{(a+c+1)^3}+\frac{2c}{(a+b+1)^3}\ge0$$we have that $f$ is convex in each of the three variables; hence,
the maximum must occur where $a, b, c \in \{0, 1\}$. Since $f$ is $1$ at each of these $8$ points,
the i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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How to prove $\int_0^1 \frac1{1+x^2}\arctan\sqrt{\frac{1-x^2}2}d x=\pi^2/24$? Since I'm stuck at this final step of the solution here. I wished to try contour integral, taking the contour a quadrant with centre ($0$) and two finite end points of arc at $(1),(i)$:
Then:
$$\operatorname{Res}\limits_{x=i}\frac1{1+x^2}\arc... | Disclaimer: now I have a working real-analytic technique, but it is quite a tour-de-force.
We have:
$$\begin{eqnarray*} I &=& \int_{0}^{1}\frac{1}{1+x^2}\arctan\sqrt{\frac{1-x^2}{2}}\,dx = \frac{1}{2}\int_{0}^{1}\frac{1}{(1+x)\sqrt{x}}\arctan\sqrt{\frac{1-x}{2}}\,dx\\&=&\int_{0}^{1/2}\frac{1}{(2-2x)\sqrt{1-2x}}\arctan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
How do I solve this, first I have to factor $ 2x\over x-1$ + $ 3x +1\over x-1$ - $ 1 + 9x + 2x^2\over x^2-1$? I am doing calculus exercises but I'm in trouble with this
$$\frac{ 2x}{x-1} + \frac{3x +1}{ x-1} - \frac{1 + 9x + 2x^2}{x^2-1}$$
the solution is
$$ 3x\over x+1$$
The only advance that I have done is fact... | The first two terms are already over a common denominator.
First simplify $$\frac {2x}{x-1}+\frac {3x+1}{x-1}-\frac {1+9x+2x^2}{x^2-1}=\frac {5x+1}{x-1}-\frac {1+9x+2x^2}{x^2-1}$$
Then put the first fraction over the same denominator as the second $$\frac {(5x+1)(x+1)}{x^2-1}-\frac {1+9x+2x^2}{x^2-1}=\frac {(5x+1)(x+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding a particular solution to a differential equation what is the particular solution for the following differential equation?
$$D^3 (D^2+D+1)(D^2+1)(D^2-3D+2)y=x^3+\cos\left(\frac{\sqrt{3}}2x \right)+xe^{2x}+\cos(x)$$
I tried Undetermined Coefficients and it took so long to solve it,not to mention it was on an exam... | Because this is an equation with constant coefficients, this in principle can be attacked using a Laplace transform. Here, for the sake of simplicity, I will assume that $y(0)$ and the first eight derivatives of $y$ at $x=0$ are zero. Then the Laplace transform of $y(x)$, $Y(p)$, is
$$\begin{align} Y(p) &= \left [ \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Integral $\int \frac{x+2}{x^3-x} dx$ I need to solve this integral but I get stuck, let me show what I did:
$$\int \frac{x+2}{x^3-x} dx$$
then:
$$\int \frac{x}{x^3-x} + \int \frac{2}{x^3-x}$$
$$\int \frac{x}{x(x^2-1)} + 2\int \frac{1}{x^3-x}$$
$$\int \frac{1}{x^2-1} + 2\int \frac{1}{x^3-x}$$
now I need to resolve one... | Use partial fractions
$$\frac{x+2}{x^3-x}=\frac{x+2}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}.$$
Solve for $A,B$ and $C$ and then integrate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Does $\int_0^\infty \sin(x^{2/3}) dx$ converges? My Try:
We substitute $y = x^{2/3}$. Therefore, $x = y^{3/2}$ and $\frac{dx}{dy} = \frac{2}{3}\frac{dy}{y^{1/3}}$
Hence, the integral after substitution is:
$$ \frac{3}{2} \int_0^\infty \sin(y)\sqrt{y} dy$$
Let's look at:
$$\int_0^\infty \left|\sin(y)\sqrt{y} \right| d... | By $y = x^{2/3}$, we get $$\int_0^\infty \sin(x^{2/3})dx = \frac{3}{2}\int_0^\infty \sin(y)\sqrt{y}dy$$
Consider the integral over the interval $[2n\pi, 2(n+1)\pi]$, we have
$$
\begin{align}
\int_{2n\pi}^{2(n+1)\pi} \sin(y)\sqrt{y}dy &= \int_{2n\pi}^{(2n+1)\pi}\sin(y)\sqrt{y}dy + \int_{(2n+1)\pi}^{(2n+2)\pi}\sin(y)\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1111952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Prove that for any positve real Prove that for any positive real numbers $x,y,z$ such that $xyz \geq 1$
$$\frac{x^5-x^2}{x^5+y^2+z^2} + \frac{y^5-y^2}{y^5+z^2+x^2} +\frac{z^5-z^2}{z^5+x^2+y^2} \geq 0.$$
This problem is from the $2005$ IMO competition.
| Since,$$\frac{x^5-x^2}{x^5+y^2+z^2} - \frac{x^5-x^2}{x^3(x^2+y^2+z^2)} = \frac{x^2(x^3-1)^2(y^2+z^2)}{x^3(x^2+y^2+z^2)(x^5+y^2+z^2)} \ge 0$$
Hence, it suffices to prove $\displaystyle \sum\limits_{cyc} \frac{x^5-x^2}{x^3(x^2+y^2+z^2)} \ge 0$
$$\sum\limits_{cyc} \frac{x^5-x^2}{x^3(x^2+y^2+z^2)} = \frac{\sum\limits_{cyc}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What Did I Do Wrong When Solving For This 2nd Order Differential Equation? (answered myself) $$
\frac{y''}{y'}+y' = f(x)
$$
I set the following to be true:
$$
y = \sum_{n=0}^{\infty} a_n x^n
$$
$$
f(x) = \sum_{n=0}^{\infty} b_nx^n
$$
Therefore:
$$
y'' = y'(f(x)-y')
$$
$$
\sum_{n=0}^{\infty} (n+1)(n+2)a_{n+2} x^n = \sum... | Your problem is that
you took the terms involving $b_n$
out of the summation.
Here is my take.
I'll start at
$\sum_{n=0}^{\infty} (n+1)(n+2)a_{n+2} x^n = \sum_{n=0}^{\infty} [(n+1)a_{n+1} x^n] \times \sum_{n=0}^{\infty} [(b_n - (n+1)a_{n+1})x^n]$
but use different indices for the summations
on the right.
$\begin{array}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
computation $ \prod_{n=1}^{\infty}(1+q^{2n})(1+q^{2(n-1)}) = 2 \prod_{n=1}^{\infty} (1+q^{2n})^2 $ I want to compute the following identity
$ \prod_{n=1}^{\infty}(1+q^{2n})(1+q^{2(n-1)}) = 2 \prod_{n=1}^{\infty} (1+q^{2n})^2 = \frac{1}{2} \prod_{n=1}^{\infty} (1+q^{2(n-1)})^2 $
Can anyone gives some explict procedure o... | $\prod_{n=1}^\infty{(1+q^{2 \cdot n})\cdot(1+q^{2\cdot(n-1)})} = $
$[\prod_{n=1}^\infty{(1+q^{2 \cdot n})}]\cdot [\prod_{n=1}^\infty{(1+q^{2 \cdot (n - 1)})}] = $
$[\prod_{n=1}^\infty{(1+q^{2 \cdot n})}] \cdot [\prod_{n=0}^\infty{(1+q^{2 \cdot n})}] = $
$[\prod_{n=1}^\infty{(1+q^{2 \cdot n})}] \cdot (1+q^0) \cdot [\pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
factor the following expression $25x^2 +5xy -6y^2$ How to factor
$$25x^2 +5xy -6y^2$$
I tried with $5x(5x+y)-6y^2$. I'm stuck here.
I can't continue.
| Forget about the $y$ for a second.
If you want to factor $25x^2 + 5x - 6$, look for two numbers whose product is $25 \cdot -6 = -150$ and whose sum is $5$. It takes just a moment to see that the two numbers are $15$ and $-10$. Then:
\begin{align*}
25 x^2 + 5x - 6 &= 25 x^2 + (15x - 10x) - 6 \\ &= (25 x^2 + 15x) - (10x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 0
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Find a parametrization of the intersection curve between two surfaces in $\mathbb{R^3}$ $x^2+y^2+z^2=1$ and $x^2+y^2=x$. Find a parametrization of the intersection curve between two surfaces in $\mathbb{R}^3$
$$x^2+y^2+z^2=1$$ and $$x^2+y^2=x.$$
I know that $x^2+y^2+z^2=1$ is a sphere and that $x^2+y^2=x$ is a circular... | A point of the intersection belongs to the cylinder. So $(x-1/2)^2+y^2=1/4$. Take for parameter the angle $\theta$ such that $x-1/2=1/2 \cos \theta$ and $y=1/2 \sin \theta$. This is just the angle of the cylindrical coordinates.
Then you have $x+z^2=1$ which implies $z^2=1/2-1/2 \cos \theta=\sin^2(\theta/2)$. You then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Transformation of an equation How do you get from the left side to the right side in this equation?
$$\frac{1+\sqrt{5}}{2} + 1 =\left(\frac{1+\sqrt{5}}{2}\right)^2$$
| We have $$\left(\frac{1+\sqrt{5}}{2}\right)^2=\frac{1}{4}(1+\sqrt{5})^2=\frac{1}{4}(6+2\sqrt{5})=\frac{3+\sqrt{5}}{2}= \frac{1+\sqrt{5}}{2}+1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of the Determinant If $a^2+b^2+c^2+ab+bc+ca \le 0\quad \forall a, b, c\in\mathbb{R}$,
then find the value of the determinant
$$ \begin{vmatrix}
(a+b+2)^2 & a^2+b^2 & 1 \\
1 & (b+c+2)^2 & b^2+c^2 \\
c^2+a^2 & 1 & (c+a+2)^2 \\
\end{vmatrix}$$
I tried expanding the whole sq... | Hint
$$\dfrac{1}{2}[(a+b)^2+(b+c)^2+(a+c)^2]=a^2+b^2+c^2+ab+ac+bc\le 0$$
so we have
$$a=-b,b=-c,c=-a\Longrightarrow a=b=c=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Use integration by substitution I'm trying to evaluate integrals using substitution. I had
$$\int (x+1)(3x+1)^9 dx$$
My solution: Let $u=3x+1$ then $du/dx=3$
$$u=3x+1 \implies 3x=u-1 \implies x=\frac{1}{3}(u-1) \implies x+1=\frac{1}{3}(u+2) $$
Now I get $$\frac{1}{3} \int (x+1)(3x+1)^9 (3 \,dx) = \frac{1}{3} \int \fr... | Nice work, (though a wee bit hard to read).
Let's look a bit more closely at the second integral.
Your substitution is correct: $$u = x-1 \implies x = u+1,\;dx = du, \\ x^2 + 2 =(u+1)^2 + 2$$
You got a bit mixed up when substituting into the factor $x^2 + 2$, to express it as a function of $u$.
Substituting, we get $$... | {
"language": "en",
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"source": "stackexchange",
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Let a,b,c be positive real numbers numbers such that $ a^2 + b^2 + c^2 = 3 $ Let $a,b,c\in\mathbb{R^+}$ such that $ a^2 + b^2 + c^2 = 3 $. Prove that
$$
(a+b+c)(a/b + b/c + c/a) \geq 9.
$$
My Attempt
I tried AM-GM on the symmetric expression so the $a+b+c \geq 3$, but I found $a+b+c \leq 3$.
| Consider the following optimization problem
$$\min_{x,y,z}\{(x+y+z)(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})\}$$
subject to
$$x^2+y^2+z^2=3$$
All we need to show is that
$$\min_{x,y,z}\{(x+y+z)(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})\}=9$$
under the condition
$$x^2+y^2+z^2=3$$
Set up the Lagrangian
$$\mathcal{L}(x,y,z,\l... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many 20-digit numbers are there which are formed using only the digits 5 and 7 and divisible by both 5 and 7. now i realised last digit has to be $5$ and position of $7$ wont affect divisibility.
$$x_{1}x_{2}x_{3}x_{4}x_{5}x_{6}x_{7}x_{8}x_{9}x_{10}x_{11}x_{12}x_{13}x_{14}x_{15}x_{16}x_{17}x_{18}x_{19}x_{20}\equiv ... | The question boils down to the following: How many $x_i \in \{ 5,7 \}$ can you find such that
$$x_1...x_{19}5 \equiv 0 \pmod{7}$$
Since $7$ is divisible by $7$, we can replace all $7's$ by $0$s.
Define $y_i =1$ if $x_i=5$ and $0$ otherwise. So the problem boils to
$$5\cdot y_1...y_{19}1 \equiv 0 \pmod{7}$$
or
$$ y_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Weird integration issue: $\ln(x+1)=\ln(2x+2)$ ?! Weird integration issue: Using $(\ln[f(x)])'=\frac {f'(x)}{f(x)}$ we get that $\int \frac{2\,dx}{2x+2}=\ln(2x+2)$. Yet, $\int \frac{2\,dx}{2x+2}= \int\frac{dx}{x+1}=\ln(x+1)$ using the same rule as earlier.
What is wrong here?
| Remember that:
$$\int \frac{2}{2x+2}dx \neq \ln(2x+2)$$
But:
$$\int \frac{2}{2x+2}dx = \ln(2x+2)+C$$
where $C$ is constant, so:
$$\int \frac{2}{2x+2}dx = \ln(2x+2)+C=\ln(2(x+1))+C=\ln(x+1)+\ln 2+C=\ln(x+1)+(\ln 2+C)=\ln(x+1)+C_1$$
So both calculations are almost correct, but you forgot about constant.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of $\lfloor k^{1/3} \rfloor$ I am faced with the following sum:
$$\sum_{k=0}^m \lfloor k^{1/3} \rfloor$$
Where $m$ is a positive integer. I have determined a formula for the last couple of terms such that $\lfloor n^{1/3} \rfloor^3 = \lfloor m^{1/3} \rfloor^3$. For example if the sum is from 0 to 11 I can find the ... | You have
$$
\begin{align*}
\sum_{k=0}^m \lfloor k^{1/3} \rfloor &= \sum_{i=0}^{\lfloor m^{1/3} \rfloor} i |\{0 \leq k \leq m : \lfloor k^{1/3} \rfloor = i\}| \\ &=
\sum_{i=0}^{\lfloor m^{1/3} \rfloor} i |\{0 \leq k \leq m : i^3 \leq k < (i+1)^3\}| \\ &=
\sum_{i=0}^{\lfloor m^{1/3} \rfloor-1} i((i+1)^3-i^3) + \lfloor m^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Determine the value of k for which a matrix system is consistent and the values for which it is inconsistent The non-homogenous system is as follows:
$$3x+2y+5z=10\\
3x-2y=7\\
6x+4y-10z=k$$
I have determined that:
$$z=1-\frac{k}{20}\\
y=\frac{k}{16}-\frac{1}{2}\\
x=2+\frac{k}{24}$$
What are the values of $k$ to form a ... | Let $K$ be a field and $k\in K$. Then we can write the system of linear equations in matrix form $Av=b$ with
$$
A=\begin{pmatrix} 3 & 2 & 5 \\ 3 & -2 & 0 \\ 6 & 4 & -10 \end{pmatrix},\quad v=\begin{pmatrix}
x \\ y \\ z \end{pmatrix},\quad b=\begin{pmatrix}
10 \\ 7 \\ k \end{pmatrix}
$$
The system has a unique solution ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof: $n^p < \frac{(n+1)^{p+1}-n^{p+1}}{p+1} < (n+1)^p$ I've edited the post in order to add at the end what, I think, is the complete proof of these inequalities. I want to apologize by not having given a reply as early as those given by the users who gave me hints about this exercise. It took me a little of time but... | The inequality $\dfrac{(n+1)^{p+1} - n^{p+1}}{p+1} < (n+1)^{p+1}$ is trivial. You obtain the left-hand-side by subtracting from the right and dividing by a number larger than $1$.
On the other hand, if you use part (a) you have
$$(n+1)^{p+1} - n^{p+1} = (1) \bigg((n+1)^p + (n+1)^{p-1}n + \cdots + (n+1)n^{p-1} + n^p\big... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Integral $\int\frac{dx}{x^5+1}$ Calculate $\displaystyle\int\dfrac{dx}{x^5+1}$
| One might as well explain how to derive the factorization in Troy Woo's answer. (This is by hand, not software.) Yes, he's right that any polynomial with real coefficients can be factorized into linear and quadratic factors with real coefficients; this fact can be derived from the fundamental theorem of algebra, by fac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\sum_{n=0}^{\infty }\frac{\sin^4(4n+2)}{(2n+1)^2}=\frac{5\pi ^2}{16}-\frac{3\pi }{4}$ I dont have an idea to prove it because of exist $\sin(4n+2)^4$
$$\sum_{n=0}^{\infty }\frac{\sin^4(4n+2)}{(2n+1)^2}=\frac{5\pi ^2}{16}-\frac{3\pi }{4}$$
| Since:
$$\sin x = \frac{1}{2i}\left(e^{ix}-e^{-ix}\right)\tag{1}$$
we have:
$$\sin^4 x = \frac{1}{16}\left(e^{4ix}+e^{-4ix}-4e^{2ix}-4e^{2ix}+6\right)=\frac{1}{8}\left(\cos(4x)-4\cos(2x)+3\right).\tag{2} $$
It follows that:
$$\begin{eqnarray*} S &=& \sum_{n\geq 0}\frac{\sin^4(4n+2)}{(2n+1)^2}\\&=&\frac{1}{8}\sum_{n\geq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149367",
"timestamp": "2023-03-29T00:00:00",
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$\sqrt{4 -2 \sqrt{3}} = a + b\sqrt{3}$, where numbers $a$ and $b$ are rational If $a$ and $b$ are rational numbers such that $\sqrt{4 -2 \sqrt{3}} = a + b\sqrt{3}$
Then what is the value of $a$? The answer is $-1$.
$$\sqrt{4 - 2\sqrt{3}} = a + b\sqrt{3}$$
$$4 - 2\sqrt{3} = 2^2 - 2\sqrt{3}$$ Let $u =2$ hence,
$$\sqrt{u... | Such square roots can be computed by a Simple Denesting Rule:
Here $\ 4-2\sqrt 3\ $ has norm $= 4.\:$ $\rm\ \color{blue}{subtracting\ out}\,\ \sqrt{norm}\ = 2\,\ $ yields $\,\ 2-2\sqrt 3\:$
which has $\, {\rm\ \sqrt{trace}}\, =\, \sqrt{4}\, =\, 2.\ \ \rm \color{brown}{Dividing\ it\ out}\ $ of the above yields $\ \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150497",
"timestamp": "2023-03-29T00:00:00",
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Exponential Generating Functions (EGF) (a) Find the exponential generating function (in closed form) for the number of ways to place n students into 3 classes (one MATH, one BIOLOGY and one PHYSICS), such that the MATH class does not have exactly one student. (Classes are allowed to be empty).
(b) Using the e.g.f. foun... | You were doing fine until you tried to extract the coefficient of $\frac{x^n}{n!}$ from $xe^{2x}$. The coefficient is just that: it cannot involve $x$. Thus, you need to incorporate that extra factor of $x$ in the summation:
$$x\sum_{n\ge 0}2^n\frac{x^n}{n!}=\sum_{n\ge 0}2^n\frac{x^{n+1}}{n!}\;.$$
The problem now is th... | {
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"timestamp": "2023-03-29T00:00:00",
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Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$.
Let $a$, $b$ and $c$ be the three sides of a triangle.
Show that $$\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... | Suppose that $S>0$. Then for $x\in(0,S/2)$, the function $f(x)=\frac{x}{S-2x}$ is convex. Thus, by Jensen's inequality and with $S=a+b+c$, we have
$$
\frac{1}{3}f(a)+\frac{1}{3}f(b)+\frac{1}{3}f(c)\geq f\left(\frac{1}{3}(a+b+c)\right)=\frac{S/3}{S-2S/3}=1.
$$
This is equivalent to $f(a)+f(b)+f(c)\geq 3$, which is your ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to calculate derivative of $f(x) = \frac{1}{1-2\cos^2x}$? $$f(x) = \frac{1}{1-2\cos^2x}$$
The result of $f'(x)$ should be equals
$$f'(x) = \frac{-4\cos x\sin x}{(1-2\cos^2x)^2}$$
I'm trying to do it in this way but my result is wrong.
$$f'(x) = \frac {1'(1-2\cos x)-1(1-2\cos^2x)'}{(1-2\cos^2x)^2} =
\frac {1-2\cos... | $f(x)=\dfrac{1}{1-2\cos^2 x}=-\dfrac{1}{2\cos^2 x-1}=-\dfrac{1}{\cos 2x}=-\sec 2x$
$\therefore f'(x)=-2\sec(2x)\tan(2x) \quad\blacksquare$
Check the list of formulas for finding derivatives of trigonometric functions here
| {
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"timestamp": "2023-03-29T00:00:00",
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Discrete-time sinusoids with same frequency I've read that sine waves of the form $x_n = \sin(w_{0}n)$, with frequencies $w_{0}$ and $w_{0} + 2\pi$, are indistinguishable from each other when considering discrete time.
The book gives $\cos\big(\frac{\pi}{4}n\big)$ as an example. Adding $2\pi$ to the frequency, we get $... | For $n$ an integer,
$$\cos\frac{7\pi n}{4}= \cos\frac{9\pi n}{4} $$
because for $n$ even both are simply zero, and for $n$ odd, $\frac{7\pi n}{4}$ is $-\left( \frac{9\pi n}{4} \right) + 2\pi$ but
$$
\cos\left[-\left( \frac{9\pi n}{4} \right) + 2\pi \right]= \cos[-\left( \frac{9\pi n}{4} \right)] = \cos\left[+\left( \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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An $\varepsilon-\delta$ proof of $\lim_{x\rightarrow c} (2x^ 2 − 3x + 4) = 2c ^2 − 3c + 4$.
Give an $\varepsilon-\delta$ proof using the “$\delta\leq 1 $ trick” to prove that $\lim_{x\rightarrow c} (2x^ 2 − 3x + 4) = 2c ^2 − 3c + 4$.
My attempt: Let $ε>0$ and there exist $\delta>0$ such that whenever $|x-c|<δ$,
$$|2... | Note that
$$|x+c|\leq |x-c|+|2c|$$
which means
$$2|x-c||x+c|+3|x-c|\leq 2|x-c|^2+4|c||x-c|+3|x-c|=|x-c|(2|x-c|+4|c|+3)$$
Choose
$$\delta:=\min\left\{\frac{\epsilon}{8|c|+6}, 2|c|+\frac{3}{2}\right\}$$
Then for $|x-c|<\delta$, we have
$$|x-c|(2|x-c|+4|c|+3)<\delta(2\delta+4|c|+3)\leq\frac{\varepsilon}{8|c|+6}\left(2\le... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify $\frac{\cos \theta}{\sec \theta} + \frac{\sin \theta}{\csc \theta}$ Please show how to simplify this type of expression without using a calculator. I'm new to trigonometry and I don't know how to simplify this expressions: $\frac{\cos \theta}{\sec \theta} + \frac{\sin \theta}{\csc \theta}$.
| First, we need to make a common denominator of $\frac{\sin\theta}{1-\cos\theta}-\frac{\sin\theta}{1+\cos\theta}$.
Let's go!
$$\require{cancel}\begin{aligned}\frac{\sin\theta}{1-\cos\theta}-\frac{\sin\theta}{1+\cos\theta}&=\frac{\sin\theta(1+\cos\theta)-\sin\theta(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}\\&=\frac{\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trick with modular exponentiation For example,
$$123^{25} \pmod{10}$$
$$ 123 \equiv 3 \pmod{10}$$
$$123^{2} \equiv 9 \pmod{10}$$
$$123^{3} \equiv 7 \pmod{10}$$
$$123^{4} \equiv 1 \pmod{10}$$
$$123^{5} \equiv 3 \pmod{10}$$
It is easy to see:
$$123^n \equiv 123^{4(k) + n} \pmod{10}$$
$$4(6) = 4(k) = 24$$ Hence,
$$123^{1}... | First of all $123^n = (120+3)^n = (12\times10+3)^n = \sum_1^n \binom{n}{k}\times(12\times10)^k\times3^{n-k} \equiv 3^n \pmod {10}$, because all multiples of $10 \equiv 0 \pmod {10}$. So it's only needed to solve $3^{25} \pmod {10}$
This reduces it to $3^n \equiv 3^{n+4k} \pmod{10}$. Why $4$? As @André Nicolas stated, t... | {
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"url": "https://math.stackexchange.com/questions/1166558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that the limit of the following function of two variables is zero I need to prove the following:
$$\lim_{(x,y)\to (1 ,2)} \frac{x^2+2xy-6x-2y+5}{\sqrt{(x-1)^2+(y-2)^2}}=0$$
I've tried to solve it by substituting $y=mx$ but I can't get the solution that way. Any help would be appreciated thanks.
| Factor the numerator as $$x^2+2xy-6x-2y+5=(x-1)[(x-1)+2(y-2)]$$Then, use the triangle inequality$$|(x-1)[(x-1)+2(y-2)]|\le|(x-1)^2|+2|x-1||y-2|$$to show that$$\frac{|x^2+2xy-6x-2y+5|}{\sqrt{(x-1)^2+(y-2)^2}}\le\frac{|(x-1)^2|+2|x-1||y-2|}{\sqrt{(x-1)^2+(y-2)^2}}$$Noting that $\sqrt{(x-1)^2+(y-2)^2}\ge|x-1|$ $$\frac{|x^... | {
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Find $\lim \limits_{(x,y) \rightarrow (0,0)}\frac{x^3 y^3 }{x^2+y^2}$ (epsilon delta proof) Suppose we have the following function:
$f(x,y)=\dfrac{x^3 y^3 }{x^2+y^2}$.
Determine $\lim \limits_{(x,y) \rightarrow (0,0)}f(x,y)$.
I did the following, but I cannot continue.
Suppose $0 < \sqrt{x^2+y^2} < \delta$.
$$\left|\f... | Hint: $\forall (x,y)\in \mathbb R^2\left(2|xy|\leq x^2+y^2\right)$.
This inequality helps not only for finding $\delta$, but also to get $$\forall (x,y)\in \mathbb R^2\left((x,y)\neq (0,0)\implies \left|\dfrac{xyx^2y^2}{x^2+y^2}\right|= \dfrac{x^2y^2}2\left|\dfrac{2xy}{x^2+y^2}\right|\leq \dfrac{x^2y^2}{2}\right),$$
wh... | {
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"timestamp": "2023-03-29T00:00:00",
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Proof of an Limit Using the formal definition of convergence, Prove that $\lim\limits_{n \to \infty} \frac{3n^2+5n}{4n^2 +2} = \frac{3}{4}$.
Workings:
If $n$ is large enough, $3n^2 + 5n$ behaves like $3n^2$
If $n$ is large enough $4n^2 + 2$ behaves like $4n^2$
More formally we can find $a,b$ such that $\frac{3n^2+5n}{... | Note
$$\left|\frac{3n^2 + 5n}{4n^2 + 2} - \frac{3}{4}\right| = \left|\frac{20n - 6}{4(4n^2 + 2)}\right| = \frac{|10n - 3|}{4(2n^2 + 1)} < \frac{10n + 3}{8n^2} < \frac{20n}{8n^2} = \frac{5}{2n}.\tag{1}$$
Hence, given $\epsilon > 0$, setting $N > 2\epsilon/5$ will make the left hand side of $(1)$ less than $\epsilon$ whe... | {
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Prove that $ \cos x - \cos y = -2 \sin ( \frac{x-y}{2} ) \sin ( \frac{x+y}{2} ) $ Prove that $ \cos x - \cos y = -2 \sin \left( \frac{x-y}{2} \right) \sin \left( \frac{x+y}{2} \right) $ without knowing cos identity
We don't know that $ \cos0 = 1 $
We don't know that $ \cos^2 x + \sin^2 x = 1 $
I have managed to prove i... | Remember this trick for life: write $x = a+b$ and $y = a-b$. So: $$\begin{cases} x&= a+b \\ y &= a-b \end{cases} \implies \begin{cases} a &= \frac{x+y}{2}\\ b &= \frac{x-y}{2} \end{cases}.$$
Then: $$\begin{align} \cos x - \cos y &= \cos(a+b)-\cos(a-b) \\ &= \cos a \cos b - \sin a \sin b - (\cos a \cos b + \sin a \sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1170630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Integral of square root. We have
$$\int \sqrt{x} \text{ }dx=\int x^{1/2} \text{ }dx=\frac{2}{3}x^{3/2}+C.$$
But, let $u=\sqrt{x}$. Then $x=u^2$, and $dx=2udu$. Substituting, we have
$$\int \sqrt{x} \text{ }dx=\int 2u^2 \text{ }du=\frac{2}{3}x^3+C.$$
Which one is correct, and why?
| You have $u= \sqrt{x}$, hence $\int 2u^2 du = \frac{2u^3}{3}+C= \frac{2x^{3/2}}{3}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the following conditional divisibility If $gcd(a,b)=1$ and $n$ is a prime number,then prove that $\frac{(a^n + b^n)}{(a+b)}$ and $(a+b)$ have no factors in common unless $(a+b)$ is a multiple of $n$.
I don't know how to establish the relation between $n$ and $(a+b)$. This is how much I have been able to derive :
... | Because $n$ is odd (note: for $n=2$, $\frac{a^2+b^2}{a+b}$ need not be an integer) we have $$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-a^{n-4}b^3+\cdots-a^1b^{n-2}+b^{n-1})$$
We compute $$gcd(a^{n-1}-a^{n-2}b+a^{n-3}b^2-a^{n-4}b^3+\cdots-a^1b^{n-2}+b^{n-1},a+b)=$$
$$gcd(a^{n-1}-a^{n-2}b+a^{n-3}b^2-a^{n-4}b^3+\cdots-a^1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Proof that every number ≥ $8$ can be represented by a sum of fives and threes. Can you check if my proof is right?
Theorem. $\forall x\geq8, x$ can be represented by $5a + 3b$ where $a,b \in \mathbb{N}$.
Base case(s): $x=8 = 3\cdot1 + 5\cdot1 \quad \checkmark\\
x=9 = 3\cdot3 + 5\cdot0 \quad \checkmark\\
x=10 = 3\cdot... | Write $x=8n+k$ for $0\leq k<8$ and $n\geq 1$. Because $8n=(3+5)n$, the problem is clear if $k=0,3,5$. Let's consider the remaining 5 cases:
\begin{align*}
k=1 & \implies 8n+1=5(n-1)+3(n+2),\\
k=2 & \implies 8n+2=5(n+1)+3(n-1),\\
k=4 & \implies 8n+4=5(n-1)+3(n+3),\\
k=6 & \implies 8n+6=5n+3(n+2),\\
k=7 & \implies 8n+7=5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 13,
"answer_id": 6
} |
Prove $7$ divides $13^n- 6^n$ for any positive integer I need to prove $7|13^n-6^n$ for $n$ being any positive integer.
Using induction I have the following:
Base case:
$n=0$: $13^0-6^0 = 1-1 = 0, 7|0$
so, generally you could say:
$7|13^k-6^k , n = k \ge 1$
so, prove the $(k+1)$ situation:
$13^{(k+1)}-6^{(k+1)}$
$... | There's always this:
$$\begin{split}
x^2-y^2 &= (x-y)(x+y)\\
x^3-y^3 &= (x-y)(x^2+xy+y^2)\\
x^4-y^4 &= (x-y)(x^3 +x^2y+xy^2+y^3)\\
&\vdots\\
x^n-y^n &= (x-y)\left(\sum_{i=1}^{n}x^{n-i}y^{i-1}\right) \mid n\in \mathbb{N} \end{split}$$
In the above, let $x=13$ and $y=6$. Then you can visually see that no matter what $n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"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... | $$(x+2)^2 \equiv -14 \equiv 35 \pmod{49} \text{ is solvable}$$
$$\iff (-14)^{\phi(49)/2} \equiv 35^{\phi(49)/2} \equiv 1 \pmod{49} \text { (But this is false.)}$$
Theorem: If $m$ is has a primitive root (2, 4, and odd prime powers), $x^k \equiv a \pmod{m}$ has a solution if and only if $$\Large a^{\frac{\phi(m)}{gcd(\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Parity of a Permutation Determine the parity of the following.
$$\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 & 7 & 4 & 1 & 6 & 5 & 3 \end{pmatrix}$$
Workings:
$\sigma = \begin{pmatrix} 1 & 2 & 7 & 3 & 4 \end{pmatrix}\begin{pmatrix} 5 & 6 \end{pmatrix}$
$\sigma = \begin{pmatrix} 1 & 4 \end{pmatrix} \begin{pm... | If $\tau=\beta_{1}\cdots \beta_{k}$ where each $\beta_{i}$, for $1\le i \le k$, is a transposition, then we have $$sgn(\tau)=(-1)^{k}$$
But, for $\sigma$ we have $k=5$. Hence, $$sgn(\sigma)=(-1)^{5}=-1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1184352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
how to show $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ I know that $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ is correct. But having some hard time proofing it using trig relations. Some of the relations I used are
$$
\sin x \cos x= \frac{1}{2} \sin(2 x)\\
\sin^2 x + \... | Rearrange and Cross multiply
it is required to prove that
$$\dfrac{\cos x +1 +\sin x}{1-\sin x +\cos x} = \dfrac{1+\sin x}{\cos x}$$ or
$$ \cos^2 x + \cos x (1 +\sin x ) = (1-\sin^2 x) + \cos x (1 +\sin x ) $$ or
$$ \cos x (1 +\sin x ) = \cos x (1 +\sin x ). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1185058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
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 ... | It is given that $a^{2} + b^{2} = ab =4$ and the goal is to determine all values of $a^{3} + b^{3}$. Now, it is seen that $a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2}) = 0$. So far the set of equations becomes
\begin{align}
a^{3} + b^{3} &= 0 \\
a^{2} + b^{2} &= 4 \\
ab &= 4.
\end{align}
From the first, multiply by $a^{3}... | {
"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": 3
} |
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.