Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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I need the proving of $x\log(x)=(\frac{x-1}{x})+\frac{3}{2!}(\frac{x-1}{x})^2+\frac{11}{3!}(\frac{x-1}{x})^3+...\frac{S_{n}}{n!}(\frac{x-1}{x})^n$ I could find the role of Striling Numbers in the natural logarithm function as follows
$$x\log(x)=(\frac{x-1}{x})+\frac{3}{2!}(\frac{x-1}{x})^2+\frac{11}{3!}(\frac{x-1}{x}... | $$\log(1-x)=-\sum_{n=1}^\infty \frac{1}{n}x^n$$
so
$$\log(x)=-\log\frac{1}{x}=-\log(1-\frac{x-1}{x})=\sum_{n=1}^\infty \frac{1}{n}(\frac{x-1}{x})^n$$
And
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$
so
$$x=\frac{1}{1-\frac{x-1}{x}}=\sum_{n=0}^\infty (\frac{x-1}{x})^n$$
and thus
$$x\log(x)=(\sum_{n=1}^\infty \frac{1}{n}(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Does $\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} =0$? $$\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} \stackrel{?}{=} 0$$
My calculations (usage of L'Hôpital's rule will be denoted by L under the equal sign =):
(Sorry for the small font, but you can zoom in to see better with Firefox)
$$
\b... | Take the logarithm of your limit then use L'Hopital and Taylor series. Let the expression in the limit be $L$:
$$\begin{align}
\ln L&=\ln\left[ \left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} \right] \\[2ex]
&= \frac{\ln\left(\frac{\sin(x)}{x}\right)}{x^2} \\[2ex]
&\to\frac{\frac{x}{\sin x}\cdot \frac{x\cos x-\sin x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Integral proof $I_n=\int\frac{x^n}{\sqrt{x^2+a}} \, dx$ If $$I_n=\int\frac{x^n}{\sqrt{x^2+a}}\,dx$$ prove that $$I_n=\frac{1}{n}x^{n-1}\sqrt{x^2+a}-\frac{n-1}{n}aI_{n-2}$$
I tried
$$I_n=\int\frac{x\times x^{n-1}}{\sqrt{x^2+a}} \,dx =\int(\sqrt{x^2+a})' \times x^{n-1} \,dx \\ =x^{n-1}\sqrt{x^2+a} -(n-1)\int x^{n-2}\sqr... | Note that
\begin{align}
\int\frac{x^n}{\sqrt{x^2+a}}\,dx
= &\int \frac{x^{n-1}}{n(x^2+a)^{\frac{n-1}2}}\ d\left[(x^2+a)^{\frac n2}\right]
\end{align}
Then, integrate by parts to immediately establish
$$I_n= \frac{x^{n-1}}{n}\sqrt{x^2+a}-\frac{(n-1)a}{n}I_{n-2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that $ABCD$ has an incircle if and only if $\frac {1}{PE} + \frac {1}{PG} = \frac {1}{PF} +\frac {1}{PH}$ Let $ABCD$ be a convex quadrilateral.Let the diagonals $AC$ and $BD$ intersect in $P$.Let $PE,PF,PG,PH$ be the altitudes from $P$ onto the sides $AB,BC,CD$ and $DA$ respectively.Show that $ABCD$ has an incircl... | Source: http://www.artofproblemsolving.com/Forum/blog.php?u=166324&b=103587
Let me denote the distances $PE,PF,PG,PH$ as $h_1,h_2,h_3,h_4$. Let $AB=a,BC=b,CD,c,DA=d$
$\frac{h_1}{h_4}=\frac{d(C,AB)}{d(C,AD)}=\frac{b \sin B}{c \sin D}$
$\frac{h_1}{h_2}=\frac{d(D,AB)}{d(D,BC)}=\frac{d \sin A}{c \sin C}$
$\frac{h_1}{h_3}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1133077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $(1+\cos\frac{\pi}{8})(1+\cos\frac{3\pi}{8}) (1+\cos\frac{5\pi}{8})(1+\cos\frac{7\pi}{8})$
How to find the value of $$\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right) \left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)$$
I used: $ 1+ \cos \theta=2\sin^2(\theta/2)$ to get $$... | First notice that $\dfrac{7\pi }{8}=\pi-\dfrac{\pi}{8}$ and also $\dfrac{5\pi }{8}=\pi-\dfrac{3\pi}{8}$.
Substituting these values in the original expression, you'll get:
$$\left(1+\cos {\pi \over 8}\right)\left(1+\cos {3\pi \over 8}\right)\left(1-\cos {3\pi \over 8}\right)\left(1-\cos {\pi \over 8}\right)$$
Using the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1133590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
limit of $ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $ Hello I am trying to find the limit of
$ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $
I've tried applying L'H rule but it ends up getting really messy.
The answer is $ \frac {1}{e} $ so I assume it must simplify into something which I ca... | $$\dfrac {x(x+1)^{x+1}}{(x+2)^{x+2}}=\dfrac{(x+1)^{x+2}-(x+1)^{x+1}}{(x+2)^{x+2}}$$
$$\dfrac {x(x+1)^{x+1}}{(x+2)^{x+2}}=\Big(1-\dfrac{1}{x+2}\Big)^{x+2}\Big(\dfrac{x+2}{x+1}\Big)-\Big(1-\dfrac{1}{x+2}\Big)^{x+1}\dfrac{1}{x+2}\to e^{-1}+0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Suppose that $a,b$ are reals such that the roots of $ax^3-x^2+bx-1=0$ are all positive real numbers. Prove that... Suppose that $a,b$ are reals such that the roots of $ax^3-x^2+bx-1=0$ are all positive real numbers. Prove that:
$(i)~~0\le 3ab\le 1$
$(ii)~~b\ge \sqrt3$.
My attempt:
I could solve the first part by ... | Let $x, y, z > 0$ be the three roots. Then, $x+y+z = xyz = \dfrac1a$ and $xy+yz+zx = \dfrac{b}a$
$(i),\quad $ Clearly, $a, b > 0$. Also $(x+y+z)^2 \ge 3(xy+yz+zx) \implies \dfrac1{a^2} \ge 3\dfrac{b}a \implies 1 \ge 3ab$.
For $(ii),\quad (xy+yz+zx)^2 \ge 3xyz(x+y+z) \implies \dfrac{b^2}{a^2} \ge 3\dfrac{1}{a^2} \imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1136109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Having problems finding $x$ in terms of $a$ and $b$. I have attempted this question but have not found a solution. I am currently stuck. Hints on how I may go further would be helpful. Thank You in advance.
The Question:
$$\frac{a^2b}{x^2} + \left(1+\frac{b}{x}\right)a = 2b+ \frac{a^2}{x}$$
What I have done so far that... | Take the equation:
$$\frac{a^2b}{x^2} + \frac{ab}{x} + a = 2b+ \frac{a^2}{x}$$
Multiply each side by $x^2$:
$$a^2b+abx+ax^2=2bx^2+a^2x$$
Subtract $2bx^2+a^2x$ from each side:
$$a^2b+abx+ax^2-2bx^2-a^2x=0$$
Rearrange it as a quadratic equation:
$$(a-2b)x^2+(ab-a^2)x+(a^2b)=0$$
And solve as you would any quadratic equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
differentiate $(2x^3 + 3x)(x − 2)(x + 4)$ So, I'm really stuck on this problem.
Differentiate $(2x^3 + 3x)(x − 2)(x + 4)$
This is what I come up with $10x^4+16x^3+39x^2+6x-18$.
But, the answer in the book has $16x^4$ as the leading term
Here's my work:
$(2x^3+3x)d/dx(x^2+2x-8)+(x^2+2x-8)d/dx(2x^3+3x)$
$(2x^3+3x)(2x+2)+... | Let $y=(2x^3+3x)(x-2)(x+4)$. Then if we are not too fussy about whether this makes sense in the reals, we get
$$\ln y=\ln(2x^3+3x)+\ln(x-2)+\ln(x+4).$$
Differentiating, we get
$$\frac{1}{y}\frac{dy}{dx}=\frac{6x^2+3x}{2x^3+3x}+\frac{1}{x-2}+\frac{1}{x+4},$$
and now we know $\frac{dy}{dx}$.
In this problem, the implicit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
$\int \frac{x^3+2}{(x-1)^2}dx$ In order to integrate
$$\int \frac{x^3+2}{(x-1)^2}dx$$
I did:
$$\frac{x^3+2}{(x-1)^2} = x+2+3\frac{x}{(x-1)^2}\implies$$
$$\int \frac{x^3+2}{(x-1)^2} dx= \int x+2+3\frac{x}{(x-1)^2}dx$$
But I'm having trouble integrating the last part:
$$\int \frac{x}{(x-1)^2}dx$$
Wolfram alpra said me ... | change of variable gets you there faster. let $u = x - 1, x = 1 + u.$ then
$\begin{align}\int\dfrac{x^3 + 2}{(x-1)^2}\,dx &= \int\frac{(1+u)^3+2}{u^2} \,du\\
&=\int\frac{3+3u+3u^2+u^3}{u^2}\,du\\
& = \int3u^{-2}+3u^{-1}+ 3+u \,du\\
&=-\frac{3}{u} + 3\ln u + 3u + \frac{u^2}{2} +C
\end{align}$
note that this substituti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1139221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Given that $m^2+n^2=1$, $p^2+q^2=1$, and $mp+nq=0$, how much is $mn+pq$? without using
$(m,n)\perp(p,q)$
| The answer is obtained using trigonometry.
The condition $x^2 + y^2 = 1$ implies $x = \cos\theta$ and $y = \sin\theta$. Write
$$ m = \cos \alpha \ , \quad n = \sin\alpha $$
$$ p = \cos \beta \ , \quad q = \sin\beta $$
Then
$$ mp + nq = \cos\alpha cos\beta + \sin\alpha \sin \beta = \cos(\alpha - \beta) $$
$$ mn + pq = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
finding the max of $f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$ I need to find the max of $$f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$$
When $x$ is a real number.
What i did is to simplify: $$f(x)=\sqrt{x^4-7x^2-10x+41}-\sqrt{x^4-3x^2-2x+5}$$.
Then i compute: $$f'(x)=\frac{-5-7x+2x^3}{\sqrt{41-... | Hint:
Consider the points $A=(4,5)$ and $B=(1,2)$ and a point $P$ on the line $x=y$ parallel to the line AB. We have for the distances $PA$, $PB$ $AB$ the triangle inequality
$$PA-PB \le AB$$ with equality if an only if $A$,$B$, $P$ lie on a line in this order. This will only happen for points $P$ chosen on $x=y$ at... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
When is $2^x+3^y$ a perfect square?
If $x$ and $y$ are positive integers, then when is $2^x+3^y$ a perfect square?
I tried this question a lot but failed. I tried dividing cases into when $x,y$ are even/odd, but still have no idea what to do when they are both odd. I tried looking into when is $2^x+3^y$ is of the fo... | Let $2^x+3^y=k^2$.
mod $3$ gives $x=2m$ for some $m\in\mathbb Z^+$, since $(-1)^x\equiv k^2\pmod {3}$ and since $2$ (i.e. $-1$) is not a quadratic residue mod $3$.
More generally, $\left(\frac{-1}{p}\right)=1\iff p\equiv 1\pmod {4}$, where $p$ is an odd prime and $\left(\frac{a}{b}\right)$ is the Legendre symbol.
Thus ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
$\int \sqrt{\frac{x}{x+1}}dx$ In order to integrate
$$\int \sqrt{\frac{x}{x+1}}dx$$
I did:
$$x = \tan^2\theta $$
$$\int \sqrt{\frac{x}{x+1}}dx = \int\sqrt{\frac{\tan^2(\theta)}{\tan^2(\theta)+1}} \ 2\tan(\theta)\sec^2(\theta)d\theta = \int \frac{|\tan(\theta)|}{|\sec^2(\theta)|}2\tan(\theta)\sec^2(\theta)d\theta = \int... | Consider $$I=\int\sqrt{\frac{x}{1+x}}dx.$$ We proced by the change of variable $u=\sqrt{1+x}$, $du=\frac{1}{2\sqrt{1+x}}$dx and $x=u^2-1,$ which gives $$I=2\int \sqrt{u^2-1}\ du=u\sqrt{u^2-1}-\log\left(u+\sqrt{u^2-1}\right)+C,$$ see https://owlcation.com/stem/How-to-Integrate-Sqrtx2-1-and-Sqrt1-x2 for more informations... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
reduction of order method for 3rd order DE Use the Reduction of Order method to solve $$(2-x)y'''+(2x-3)y''-xy'+y=0$$ (such that $x<2$) using $u(x)=e^x$.
How do you use the method for $3$rd order... I have seen it been used on $2$nd but not higher. Please help.
| Don't forget $y=x$ is also a particular solution of the ODE.
Let $y=xu$ ,
Then $y'=xu'+u$
$y''=xu''+u'+u'=xu''+2u'$
$y'''=xu'''+u''+2u''=xu'''+3u''$
$\therefore(2-x)(xu'''+3u'')+(2x-3)(xu''+2u')-x(xu'+u)+xu=0$
$x(2-x)u'''+3(2-x)u''+x(2x-3)u''+2(2x-3)u'-x^2u'-xu+xu=0$
$x(2-x)u'''+2(x^2-3x+3)u''-(x^2-4x+6)u'=0$
Let $v=u'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find two numbers whose sum is 20 and LCM is 24 With some guess work I found the answers to be 8 and 12.
But is there any general formula for this?
Note that the question is asked to my nephew who is at 4th grade.
| A general algorithm
If $a+b = n$ and $\mathrm{lcm}(a,b) = c$ we let $\gcd(a,b) = d$ to get
$$ab = cd\\
n = a+b$$
Solving $b = n-a$ gives us
$$a(n-a) = cd \\
\Leftrightarrow a^2-na + cd = 0 \\
\Leftrightarrow a = \frac n2 \pm \sqrt{\frac{n^2}4 - cd}$$
So for even $n$ we must find $d$ such that $\frac{n^2}4 - cd$ is a pe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1143716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 5
} |
Another method of proving $x^3+y^3=z^3$ has no integral solutions? The equation $$ax^2+by^2=z^3$$ has the following parametrization:
$$y=q(3ap^2-bq^2)$$
$$x=p(ap^2-3bq^2)$$
$$z=ap^2+bq^2$$ can we deduct from that the Diophantine equation $$x^3+y^3=z^3$$ has no nontrivial solutions by choosing $x=a$ and $y=b$?
| Your parametrization is almost right, you flipped the second factors for $x$, $y$.
In fact, from
$$\sqrt{a}\, x + i \sqrt{b}\, y = (\sqrt{a}\, p + i \sqrt{b}\, q)^3$$
we have with
\begin{eqnarray}
x &=& p (a p^2 - 3 b q^2) \\
y &=& q( 3 a p^2 - b q^2) \\
z &=& ap^2 + b q^2
\end{eqnarray}
($\pm$ if you wish) the equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to factorize $n(n+1)(n+2)(n+3)+1$? How to factorize $n(n+1)(n+2)(n+3)+1$ ?
It's turned into the lowest power of $n$ ever possible but the question wants me to factorize it.
How can I do that ?
| Set $\dfrac{n+n+1+n+2+n+3}4=m\iff 2n+3=2m$
$\implies n(n+1)(n+2)(n+3)+1=\dfrac{2n(2n+2)(2n+4)(2n+6)}{2^4}+1$
$=\dfrac{(2m-3)(2m-1)(2m+1)(2m+3)}{2^4}+1$
$=\dfrac{(4m^2-9)(4m^2-1)}{2^4}+1$
$=\dfrac{(4m^2)^2-2\cdot4m^2\cdot5+5^2}{2^4}=\dfrac{(4m^2-5)^2}{(2^2)^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
prove that : $\cos x \cdot \cos(x-60^{\circ}) \cdot \cos(x+60^{\circ})= \frac14 \cos3x$ I should prove this trigonometric identity.
I think I should get to this point :
$\cos(3x) = 4\cos^3 x - 3\cos x $
But I don't have any idea how to do it (I tried solving $\cos(x+60^{\circ})\cos(x-60^{\circ})$ but I got nothing)
| Using sum of angles identity,
\begin{align*}
\cos x\cos (x-60^{\circ})&\cos(x+60^{\circ})\\
& = \cos x(\cos x\cos 60^{\circ}+\sin x\sin 60^{\circ})(\cos x\cos 60^{\circ}-\sin x\sin 60^{\circ})\\ \\
& = \cos x(\cos^2x\cos^260^{\circ}-\sin^2x\sin^260^{\circ})\\\\
& = \cos x\left(\frac{1}{4}\cos^2x-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1145406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Strong induction with Fibonacci numbers I have two equations that I have been trying to prove. The first of which is:F(n + 3) = 2F(n + 1) + F(n) for n ≥ 1.For this equation the answer is in the back of my book and the proof is as follows:1) n = 1: F(4) = 2F(2) + F(1) or 3 = 2(1) + 1, true.2) n = 2: F(5) = 2F(3) + F(2)... | the following relation on the Fibonacci numbers is sometimes useful: (for $n \ge k \ge 0$):
$$
F(n+k) = \sum_{j=0}^k \binom{k}{j}F(n-j)
$$
this gives:
$$
F(n+6) = \sum_{j=0}^3 \binom{3}{j}F(n+3-j) \\
= F(n+3) +\color{blue}{3F(n+2)+3F(n+1)} +F(n) \\
=4 F(n+3)+F(n)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Inequality in Hanoi Open Mathematics Competition $\frac{1}{bc} + \frac{1}{ab} + \frac{1}{ca} \geq 1$.
Prove $\frac{a}{bc} + \frac{c}{ab} + \frac{b}{ca} \geq 1$, where
$a,b,c$ are positive real numbers
| Let $x=\frac{1}{bc},y=\frac{1}{ab},z=\frac{1}{ca}$. The question becomes given $x+y+z\geq1$ , prove $\frac{x\sqrt{x}}{\sqrt{yz}}+\frac{y\sqrt{y}}{\sqrt{xz}}+\frac{z\sqrt{z}}{\sqrt{xy}}=\frac{x^2+y^2+z^2}{\sqrt{xyz}}\geq 1$.
From Cauchy-Schwarz $3(x^2+y^2+z^2)\geq(x+y+z)^2$
From AM-GM $x^2+y^2+z^2\geq 3(xyz)^{2/3}$
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving $\sum_{n =1,3,5..}^{\infty }\frac{4k \ \sin^2\left(\frac{n}{k}\right)}{n^2}=\pi$ Proving
$$\sum_{n =1,3,5..}^{\infty }\frac{4k \sin^2\left(\frac{n}{k}\right)}{n^2}=\pi$$
Where $k$ any number greater than $0$
I tried to prove it by using the Fourier series but I couldnt find any form likes the above form... | Using (in the interval $(0,\pi)$
) $$\frac{2\pi x-x^{2}}{8}=\underset{n\geq1}{\sum}\frac{\sin\left(n\frac{x}{2}\right)^{2}}{n^{2}}\,\,(1)$$
we have$$\frac{2\pi x-x^{2}}{8}=\underset{n\geq1}{\sum}\frac{\sin\left(\left(2n-1\right)\frac{x}{2}\right)^{2}}{\left(2n-1\right)^{2}}+\frac{1}{4}\underset{n\geq1}{\sum}\frac{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1153068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Complete the character table of group of order $21$ You are given the incomplete character table of a group $G$ with order $21$ which has $5$ conjugacy classes, $C_1,\dots,C_5$, which have sizes $1,7,7,3,3$.
$$ \begin{array}{|c|c|c|c|c|}
\hline
& C_1 & C_2 & C_3 & C_4 & C_5 \\ \hline
& & & & & \\ \hline
& & && & ... | The remaining two characters are algebraic conjugates of $\chi_2$ and $\chi_4$. You get them by replacing $\zeta_3$ by $\zeta_3^2$, and $\zeta_7$ by $\zeta_7^3$, respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1153190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Euler Mascheroni Constant is Zero $$
H_n - \ln n = \int_0^1 \frac{1 - x^n}{1 - x} dx - \int_1^n \frac{dy}{y}
$$
Let $x = \frac{y - 1}{n - 1}$, or $y = (n-1)x + 1$. Then, $dy = (n - 1) dx$.
$$
H_n - \ln n = \int_0^1 \frac{1 - x^n}{1 - x} dx - \int_0^1 \frac{n - 1}{(n - 1) x + 1} dx = \int_0^1 (\frac{1 - x^n}{1 - x} - \f... | You obviously took a wrong turn along the way to conclude
$$\gamma = \int_0^1 \frac{dx}{1 - x} - \int_0^1 \frac{dx}{x},$$
since each integral is improper and divergent. The limit/integral switch cannot be justified by uniform convergence, monotone convergence, etc.
As an example where the switch is valid, change varia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1155144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the last digit of $77777^{77777}$ Find the last digit of $$77777^{77777}$$
I got a pattern going for $77777^n$ for $n=1, 2, ....$ to be:
$$7, 9, 3, 1$$ for $n = 1, 2, 3, 4$ respectively.
The idea is:
$$77777^{77777} \pmod{10}$$
I see that:
$$77777^n \equiv 77777^{n + 4} \pmod{10}$$
Using the spotted pattern, but... | $\varphi(10)
= \varphi(2 \times 5) = \varphi(2) \times \varphi(5) = (2-1)(5-1) = 4
$
Since $\gcd(7,10) = 1,$ then $7^4 \equiv 1 \pmod{10}$
Note that $77777 \equiv 777 \times 100 + 76 + 1 \equiv 1 \pmod 4$
So
\begin{align}
77777^{77777} \pmod{10}
&\equiv 7^{77777} \pmod{10}\\
&\equiv 7^{4(\text{something... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1155705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Un-Simplifying a fraction, i.e. computing partial fraction decomposition $\frac{3x^2+17x}{x^3+3x^2+-6x-8}$
I need to find the value of C in the form of
$\frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+4}$
which is based on the fraction give at the top.
I can get so far to do the following:
$A(x^2+2x-8) + B(x^2+5x+4) + C(x^... | if you want to find just $C$ not other two, then look at the behavior of $$ \frac{3x^2 + 17x}{(x+1)(x-2)(x+4)} = \frac{3*(-4)^2+17*(-4)}{(-4+1)(-4-2)(x+4)} +\cdots=-\frac{10/9}{(x-4)}+\cdots$$ therefore $$C = -\frac{10}9. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1156046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find the equation of a circle, given a point on it and a point where it is tangent to a given line The given question is:
Find the equation of the circle that passes through point $(-3,-4)$ and touches the line $x-y+7=0$ at the point $(-5,2).$
What I did was:
Took the given points $(-5,2)$ and $(-3,-4)$ as the diameter... | the mid point of the chord connecting the points $(-5, 2)$ and $(-3, -4)$ is $(-4, -1)$ and the chord has slope $\frac{-4-2}{-3 + 5} = -3.$ the parametric equation of a point on the bisector of this chord is $$x = -4 + 3t, y= -1 + t$$ the slope of the line connection this point and point of contact $(-5, 2)$ is $$\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1157135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
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Integral $\int_{-\infty}^{\infty} \frac{\sin^2x}{x^2} e^{ix}dx$ How do I determine the value of this integral?
$$\int_{-\infty}^{\infty} \frac{\sin^2x}{x^2} e^{ix}dx$$
Plugging in Euler's identity gives
$$\int_{-\infty}^{\infty} \frac{i\sin^3x}{x^2}dx + \int_{-\infty}^{\infty} \frac{\sin^2x \cos x}{x^2}dx$$
and since $... | The integral can be evaluate by converting it to a double integral first.
Let
$$I = \int_{-\infty}^\infty \frac{\sin^2 x \cos x}{x^2} \, dx = 2 \int_0^\infty \frac{\sin^2 x \cos x}{x^2} \, dx.$$
Noting that
$$\sin^2 x \cos x = \frac{1}{4} (\cos x - \cos 3x),$$
yields
$$I = \frac{1}{2} \int_0^\infty \frac{\cos x - \cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1157772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 5
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Showing that $f_n$ is the number closest to $\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^n$ I want to prove that the the $n$th Fibonacci number $f_n$ is the integer closest to $\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^n$. What would be a rigorous way to go about this? I assume I'll have to use a recurrence relation and ma... | Copy-paste from Wikipedia:
Like every sequence defined by a linear recurrence with constant coefficients, the Fibonacci numbers have a closed-form solution. It has become known as Binet's formula, even though it was already known by Abraham de Moivre:
$$F_n = \frac{\varphi^n-\psi^n}{\varphi-\psi} = \frac{\varphi^n-\psi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1159514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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how does this expression cancel out How does $$\frac{2t-1-i}{2t^2-2t+1}=\frac{1+i}{-1+(1+i)t}$$ I just can't see how this works...
I typed the LHS in WFA and it gave the RHS but I don't know how anything can cancel.
| By using quadratic formula we get the roots of $2t^2-2t+1$, they are $\frac{1+i}{2}$ and $\frac{1-i}{2}$, then, we can factor out $2t^2-2t+1$ as
\begin{align*}
2t^2-2t+1&=2\left(t-\frac{1+i}{2}\right)\left(t-\frac{1-i}{2}\right)\\
&=(2t-1-i)\left(t-\frac{1-i}{2}\right)
\end{align*}
Hence
\begin{align*}
\frac{2t-1-i}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1162241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Partial fraction decomposition of $\frac{x^3+x+2}{x(x^2+1)^2}$ Help me solve this
$$\dfrac{x^3+x+2}{x(x^2+1)^2}$$
It looked like a simple one, but became complicated in my hands because i tried it like this:
$$\dfrac{x^3+x+2}{x(x^2+1)^2}=\dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2} $$
multiply all sides by ... | You're working with an incorrect equation.
After multiplying both sides by $x(x^2 + 1)^2$, you should have $$\begin{align} x^3+x+2 & = A(x^2+1)^2+Bx(x^3+x)+C(x^3+x)+Dx^2+Ex\\
&= Ax^4+ 2Ax^2 + A +Bx^4 + Bx^2 +Cx^3 + Cx +Dx^2 + E x
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1163864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Introduction to probability Dice We roll a fair die repeatedly until we see the number four appear and then we stop.
What is the probability that we needed an even number of die rolls?
For this problem I said that it would be the Sum of (5/36)*(25/36)^j and I got an answer of 5/9? Does that seem correct.
Thanks
| We want the probability of having an even number of rolls. That means we want an odd number of non-fours, and then a four. So, we could have 1, 3, 5, 7, etc. non-fours, followed by a four. Given that the die is fair, we will assume that $P(4) = \frac{1}{6}$ and that the rolls are independent, so we can multiply when ap... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1164163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve this limit, normal methods result in zero. Online sources say otherwise. in one part of my homework, we are asked to solve this limit:
$$
\lim_{x\to\infty}\sqrt{\frac{x^3}{x-3}} - x
$$
The result should be $3/2$, but when I try it, I always get $0$ instead.
Please, do not post the full solution, I would on... | We have
\begin{align} \sqrt{\frac{x^3}{x - 3}} - x &= x\sqrt{\frac{x}{x - 3}} - x\\
&= x\left(\sqrt{\frac{1}{1 - \frac{3}{x}}} - 1\right)\\
&= x\left(\frac{1}{\sqrt{1 - \frac{3}{x}}} - 1\right)\\
&= \frac{x}{\sqrt{1 - \frac{3}{x}}}\left(1 - \sqrt{1 - \frac{3}{x}}\right)\\
&= \frac{x}{\sqrt{1 - \frac{3}{x}}}\frac{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Expansion of cumulant transform
Verify the following expansion for a cumulant generating function of a random variable $X$.
\begin{align}
\kappa(t) & = \mu t + \frac{1}{2}\sigma^2t^2+\frac{1}{6}\rho_3\sigma^3t^3 + \frac{1}{24}\rho_4\sigma^4t^4 + \cdots \\
& = \mu t + \frac{1}{2}\sigma^2t^2+\frac{1}{6}\kappa_3 t^3 + ... | The Taylor expansion of $\log(1+x)$ is $x-x^2/2+x^3/3-\cdots$. Therefore up to second order we get
$$
\newcommand{\EE}{\mathbb{E}}
\newcommand{\VV}{\mathbb{V}}
\log\left(1+t\EE[X]+\frac{t^2}{2}\EE[X^2]+O(t^3)\right) \\ =
\left(t\EE[X]+\frac{t^2}{2}\EE[X^2]+O(t^3)\right)-\frac{1}{2}\left(t\EE[X]+O(t^2)\right)^2 \\ =
t\E... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Product of Sums: Show that the following is a Polynomial by converting it into standard form. $$\prod_{k=0}^n (1+x^{2^k})$$
The given expression simplifies to $(1+x)(1 + x^2)...(1 + x^{2^n})$
I am not able to proceed further. How do I express this in Summation form?
| We have
$$\prod_{k = 0}^n (1 + x^{2^k}) = \prod_{k = 0}^n \frac{1 - (x^{2^k})^2}{1 - x^{2^k}} = \prod_{k = 0}^n \frac{1 - x^{2^{k+1}}}{1 - x^{2^k}} = \frac{1 - x^{2^{n+1}}}{1 - x},$$
and
$$\frac{1 - x^{2^{n+1}}}{1 - x} = 1 + x + x^2 + \cdots + x^{2^{n+1} - 1}.$$
Thus
$$\prod_{k = 0}^n (1 + x^{2^k}) = 1 + x + x^2 + \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\int_{0}^{\pi} \int_{0}^{\pi/3} \int_{\sec \phi}^{2} 5\rho^2 \sin(\phi) \mathrm d\rho \mathrm d\phi \mathrm d\theta$
Evaluate the following \begin{align*} \int_{0}^{\pi} \int_{0}^{\pi/3} \int_{\sec \phi}^{2} 5\rho^2 \sin(\phi) \mathrm d\rho \mathrm d\phi \mathrm d\theta \end{align*}
Attempt at solution: We ... | $\displaystyle \int_{0}^{\pi} \int_{0}^{\pi/3} \int_{\sec \phi}^{2} 5\rho^2 \sin(\phi) \ d\rho \ d\phi \ d\theta=\frac{5}{3}\int_0^{\pi}\int_0^{\pi/3}(8-\sec^3\phi)\sin\phi\;d\phi d\theta$
$\displaystyle=\frac{5}{3}\int_0^{\pi}\int_0^{\pi/3}(8\sin\phi-\tan\phi\sec^2\phi)\;d\phi d\theta=\frac{5}{3}\pi\left[-8\cos\phi-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1169179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that $ \int_{0}^{\pi/2} \frac{\mathrm{d}{x}}{\sqrt{a^{2} {\cos^{2}}(x) + b^{2} {\sin^{2}}(x)}} = \frac{\pi}{2 \cdot \text{AGM}(a,b)} $. I know Neumann’s solution of this famous definite integral that is totally based on substitution, but is there any solution using complex analysis? Assuming that $ a > b $, sho... | Let
$$I(a,b)=\int_{0}^{\pi/2}\frac{dx}{\sqrt{a^2\sin^2 x+b^2\cos^2 x}}=\int_{0}^{1}\frac{dt}{\sqrt{(1-t^2)(a^2 t^2 + b^2(1-t^2))}}.\tag{1}$$
Obviously $I(a,a)=\frac{\pi}{2a}$, hence in order to prove our identity we just need to prove that:
$$ I(a,b)=I\left(\sqrt{ab},\frac{a+b}{2}\right).\tag{2}$$
By replacing $b\tan x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1169877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Show the series is convergent and find the limit Sequence of real numbers $a_n$ defined recursively with $a_1=1/2$ and $a_{n+1}=\frac{a_n^2}{a_n^2-a_n+1}$ for all $n \geq 1$.
Show that $\sum_{n=1}^\infty a_n$ is convergent and find its limit.
I have tried to convert the recursive form to explicit form but it's too dif... | Define the following sequence $b_n = \frac{1}{a_n}$. Then
$$b_{n+1} =
\frac{a_n^2-a_n+1}{a_n^2} = 1 - \frac{1}{a_n}+ \frac{1}{a_n^2} = b_n^2-b_n+1
$$
so the sequence $b_n$ is an increasing sequence of natural numbers whose first terms are
$$2, \ 3, \ 7, \ 43, \ 1807, \ \dots$$
Partial sums have a curious pattern: if yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Bounds for $\log(1-x)$ I would like to show the following
$$-x-x^2 \le \log(1-x) \le -x, \quad x \in [0,1/2].$$
I know that for $|x|<1$, we have $\log(1-x)=-\left(x+\frac{x^2}{2}+\cdots\right)$. The inequality on the right follows because the difference is $\frac{x^2}{2}+ \frac{x^3}{3} + \cdots \ge 0$.
For the inequa... | $\begin{array}\\
-\ln(1-x)
&=\sum_{k=1}^{\infty} \frac{x^k}{k}\\
&=x+\sum_{k=2}^{\infty} \frac{x^k}{k}\\
&=x+x^2\sum_{k=2}^{\infty} \frac{x^{k-2}}{k}\\
&=x+x^2\sum_{k=0}^{\infty} \frac{x^{k}}{k+2}\\
&\le x+x^2\sum_{k=0}^{\infty} \frac{x^{k}}{2}\\
&\le x+\frac{x^2}{2}\sum_{k=0}^{\infty} x^{k}\\
&= x+\frac{x^2}{2}\frac1{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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quadratic Gauss sum over a power of 2 Is there a general formula for the generalized quadratic Gauss sum defined by
$$
G(a,b,c)=\frac{1}{c}\sum_{n=0}^{c-1}e\left(\frac{an^2+bn}{c}\right)
$$
where $e(x)=\exp(2\pi ix)$ and $c$ is a power of 2?
| You could probably use the generalized quadratic Gauss sum reciprocity formula, defined for integers $a, b, c$ satisfying $ac \neq 0$ and $ac+b$ even. In this case, we set $\displaystyle S(a,b,c) = \sum_{x=0}^{|c|-1} e\left(\frac{ax^2+bx}{2c}\right)$, and the theorem states
\begin{eqnarray*}
S(a,b,c) = \left|\frac{c}a\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1172541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the derivative of $\arccos\frac{b+a\cos x}{a+b\cos x}$ Find the derivative of $\arccos\dfrac{b+a\cos x}{a+b\cos x}$
is there a smart way to find this derivative
i tried by the conventional chain rule way, and it got very complicated
| It is not too complicated if one goes through the chain rule of derivatives.
$$\begin{split}\arccos \left( {\frac{{b + a\cos x}}{{a + b\cos x}}} \right) &= \arccos \left( u \right) = F\left( {u\left( x \right)} \right)\\
u &= \frac{{b + a\cos x}}{{a + b\cos x}}\\
\frac{{dF}}{{dx}} &= \frac{{dF}}{{du}}\frac{{du}}{{dx}}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1173050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to prove that $x^9-9x^7+27x^5-30x^3+9x-1$ is irreducible in $\mathbb{Q}[x]$? The problem says:
Given the irreducible polynomial $x^3-3x-1$ with root $2\cos(\pi/9)$, prove that $2\cos(\pi/27)$ is a root of a monic irreducible polynomial of degree 9 over $\mathbb{Q}$, and hence $[\mathbb{Q}(2\cos(\pi/27)):\mathbb{Q}]... | The computation is a little horrendous (I used WolframAlpha), but you can show that
$$(x+1)^9-9(x+1)^7+27(x+1)^5-30(x+1)^3+9(x+1)-1$$
expands to be
$$x^9+9 x^8+27 x^7+21 x^6-36 x^5-54 x^4+9 x^3+27 x^2-3$$
which is irreducible by Eisenstein's criterion with $p=3$, which means that
$$x^9-9x^7+27x^5-30x^3+9x-1$$
is also i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1174827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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sums of squares and Pythagorean Triples I'm told that if $r^2 + s^2 = z^2$ then $(r+s)^2 + (r-s)^2 = 2z^2$, which is obvious. But i'm trying to show that every integer solution to $x^2 + y^2 = 2z^2$ arises this way from a Pythagorean Triple (r,s,z) and then go on to find the general solution of $x^2 + y^2 = 2z^2$ where... | From $a^2+b^2=2z^2,$ assuming $a\ge b,$ if you want $a=a_1+b_1,\ b=a_1-b_1$ then $a_1=(a+b)/2$ and $b_1=(a-b)/2.$ Here you need then to require that $a,b$ have the same parity to be able to do this manipulation.
Note: If one already has $a^2+b^2=2z^2$ then the requirement mentioned above, that the parity of $a,b$ are t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1177661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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writing $pq$ as a sum of squares for primes $p,q$ Let $p$ and $q$ be distinct primes congruent to $1$ mod $4$. How many ways are there to write $pq$ as a sum of squares?
I know that any prime $p\equiv 1\pmod 4$ can be written uniquely as a sum of squares and $pq\equiv 1\pmod 4$ but $pq$ is not a prime so I'm stuck.
| Since $n$ can be written as a sum of two squares in a number of ways that depends on how many ways there are to split $n$ as:
$$ n = z\bar{z} = (a-bi)(a+bi) $$
over $\mathbb{Z}[i]$, and the latter is an Euclidean domain hence a UFD, it follows that for any $n\in\mathbb{N}$:
$$\#\{(a,b)\in\mathbb{Z}^2:a^2+b^2= n\} = 4(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$n^2(n^2-1)(n^2-4)$ is always divisible by 360 $(n>2,n\in \mathbb{N})$ How does one prove that $n^2(n^2-1)(n^2-4)$ is always divisible by 360? $(n>2,n\in \mathbb{N})$
I explain my own way:
You can factorize it and get $n^2(n-1)(n+1)(n-2)(n+2)$.
Then change the condition $(n>2,n\in \mathbb{N})$ into $(n>0,n\in \mathbb{N... | $$n^2(n^2-1)(n^2-4)=(n-2)(n-1)n(n+1)(n+2)\cdot n \\
=(n-2)(n-1)n(n+1)(n+2)(n+3)-3(n-2)(n-1)n(n+1)(n+2)$$
And
$$(n-2)(n-1)n(n+1)(n+2)(n+3)=720 \cdot \binom{n+3}{6}$$
$$3(n-2)(n-1)n(n+1)(n+2)=3\cdot 120 \binom{n+2}{5}=360 \binom{n+2}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\lim_{x\to0}\frac{e^x-1-x}{x^2}$ using only rules of algebra of limits. I would like to solve that limit solved using only rules of algebra of limits.
$$\lim_{x\to0}\frac{e^x-1-x}{x^2}$$
All the answers in How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion? do... | $$\left(\frac{e^x-1}x\right)^2=\frac{e^{2x}-2e^x+1}{x^2}=2^2\frac{e^{2x}-1-2x}{(2x)^2}-2\frac{e^x-1-x}{x^2}.$$
Then, taking the limit,
$$1=4L-2L.$$
UPDATE: the same approach can be used for the next order,
$$\left(\frac{e^x-1}x\right)^3=\frac{e^{3x}-3e^{2x}+3e^x-1}{x^3}=3^3f(3x)-3\cdot2^3(2x)+3f(x),$$
where $f(x)=\dfr... | {
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Integral fraction of polynomials I have this problem:
$$\int \frac{-2x^2+6x+8}{x^4-4x+3}dx$$
I have tried using partial fractions, but I can't get solution. Thank you for any advice.
| A quick check shows that $x = 1$ is a root of $x^4 - 4x + 3$. By synthetic division we find $x^4 - 4x + 3 = (x - 1)(x^3 + x^2 + x - 3)$. By similar reasoning, $x^3 + x^2 + x - 3 = (x - 1)(x^2 + 2x + 3)$. So $x^4 - 4x + 3 = (x - 1)^2(x^2 + 2x + 3)$. Now you have the setup to do partial fraction decomposition. The result... | {
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"url": "https://math.stackexchange.com/questions/1182910",
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"source": "stackexchange",
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How to solve equations of the form $(a+x)^2 + (b+x)^2 +(c+x)^2 + (d+x)^2-e=0$ I have an equation of the form $(a+x)^2 + (b+x)^2 +(c+x)^2 + (d+x)^2-e=0$ where $a$, $b$, $c$, $d$ and $e$ are known.
My question is how would I derive $x$.
| HINT:
$(a+x)^2+(b+x)^2+(c+x)^2+(d+x)^2-e=0\implies$
$\color{red}{4}x^2+\color{green}{2(a+b+c+d)}x+\color{blue}{a^2+b^2+c^2+d^2-e}=0\implies$
$x_{1,2}=\frac{-\color{green}{2(a+b+c+d)}\pm\sqrt{(\color{green}{2(a+b+c+d)})^2-4\cdot\color{red}{4}\cdot(\color{blue}{a^2+b^2+c^2+d^2-e})}}{2\cdot\color{red}{4}}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A sum involving a ratio of two binomial factors. Let $a\ge 0$, $a_1\ge 0$ ,$b \ge 0$ and $b_1\ge0$ be real numbers subject to $1+b+a_1-b_1-a >0$. Let $m$ be a positive integer. Then using methods similar to those in Another sum involving binomial coefficients. I have shown that the following identity holds:
\begin{eqn... | We need to analyze a certain rational function of $m$. We decompose that rational function into simple fractions and we have:
\begin{equation}
\frac{(1+a+m)^{(n)}}{(1+a_1+m)^{(n)}} = 1 + \sum\limits_{l=1}^n \frac{{\mathcal A}_l}{m+a_1+l}
\end{equation}
where ${\mathcal A}_l = (a-a_1-l+1)^{(n)} (-1)^{l-1}/((l-1)!(n-l)!)... | {
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How am I supposed to know that $\frac{1}{(1-x)^2} = \sum_{n=0}^{\infty } \binom{n+1}{n}x^n$? I'm currently reading through the solution to a problem that involves finding generating functions. In some of the intermediary steps, it is written that
$$\frac{1}{(1-x)^2} = \sum_{n=0}^{\infty } \binom{n+1}{n}x^n$$
without an... | $$\dfrac{1}{1-x}=1+x+x^2+x^3+\cdots\,\,\,\,\,\forall|x|\lt1$$ Then $$\dfrac{1}{(1-x)^2}=(1+x+x^2+x^3+\cdots)(1+x+x^2+x^3+\cdots)$$ $x^k$ in the expansion of right hand side is produced by $$1.x^k+x.x^{k-1}+x^2.x^{k-2}+\cdots+x^k.1=(k+1)x^k.$$ Hence $$\dfrac{1}{(1-x)^2}=1+2x+3x^2+4x^3+\cdots.$$ Similarly we can find $\d... | {
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How many bags would need to be bought to have all the red noses? There are 9 types of Red Noses, for comic relief this year:
Each one is sold in an opaque packet, so it is lucky dip which one you will get.
Assuming there is the same amount of each type (${1\over9}$th). On average, how many bags would need to be bought... |
I'm going assume that there are, effectively, infinite bags of noses—if we only had 9 bags... well, you get it. ;)
Say we have $n-1$ of the $9$ unique noses. The probability of getting a new style of nose on the next bag is $p_n = 1 - \frac{n-1}{9}$. Then, on average, it takes $\frac{1}{p_n}$ bags to get the next uniq... | {
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Find trig derivative of $y=4x(7x+\cot{7x})^6$ Find trig derivative of $y=4x(7x+\cot{7x})^6$.
I got $y'= 4(7x+\cot{7x})^6 + 168x(7x+\cot{7x})^5 (\cot^2 {7x})$ but I'm not sure I did it right.
Your help is appreciated (:
| use the logarithmic derivative. here is how it works.
let $$y = 4x(7x + \cot 7x)^6, \ln y = \ln 4 + \ln x + 6 \ln(7x \sin 7x + \cos 7x) - 6 \ln \sin 7x $$ taking the derivative, we get
$$\begin{align}\frac 1y\frac{dy}{dx} &=\frac 1x+6\left(\frac{7\sin 7x+49x\cos7 x-7\sin7 x}{7x\sin 7x + \cos 7x}- \frac{7\cos 7x}{\sin... | {
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Finding the sum of $\sum_{n=0}^{\infty}\frac {(x+1)^{n+2}}{3^n}$
Find the sum of the series and for which values of $x$ does it converge:
$$\sum_{n=0}^{\infty}\frac {(x+1)^{n+2}}{3^n} $$
My attempt:
$$\begin{align}&S_n=(x+1)^2+(x+1)^3/3+...+(x+1)^{n+1}/3^{n} \\
-(x+1)^2&S_n=-(x+1)^3/3-...-(x+1)^{n+2}/3^{n} \\
&S_... | The series is of the form
\begin{align}
f(t) = \sum_{n=0}^{\infty} t^{n} = \frac{1}{1-t}.
\end{align}
When $t = (x+1)/3$ the value becomes
\begin{align}
f\left(\frac{x+1}{3}\right) = \sum_{n=0}^{\infty} \frac{(x+1)^{n}}{3^{n}} = \frac{3}{2-x}.
\end{align}
Now multiply by $(x+1)^{2}$ to obtain
\begin{align}
\sum_{n=0}^{... | {
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The limit : $ \lim _{x \to \infty } \sqrt{x^2 +x} - \sqrt{x^2 +1} $ The limit:
$ \lim _{x \to \infty } \sqrt{x^2 +x} - \sqrt{x^2 +1} $
(A) is 0 (B) is $\frac 1 2 $ (C) is 2 (D) does not exist
Is doing it with Binomial expansion and cancelling the terms only way?
| $$ \lim \limits_{x \to \infty } \sqrt{x^2 +x} - \sqrt{x^2 +1}=\lim \limits_{x \to \infty }\dfrac{(\sqrt{x^2 +x} - \sqrt{x^2 +1})(\sqrt{x^2 +x} + \sqrt{x^2 +1})}{\sqrt{x^2+x}+\sqrt{x^2+1}}=\lim \limits_{x \to \infty } \dfrac{x-1}{\sqrt{x^2+x}+\sqrt{x^2+1}}=\lim \limits_{x \to \infty } \dfrac{1-\dfrac{1}{x}}{\sqrt{1+\dfr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Why General Leibniz rule and Newton's Binomial are so similar? The binomial expansion:
$$(x+y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k}$$
The General Leibniz rule (used as a generalization of the product rule for derivatives):
$$(fg)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} f^{(k)} g^{(n-k)}$$
Both formulas can be obta... | Taylor series expansion of $f(x+h)$
$$
f(x+h)=f(x)+h\frac{d}{dx} \left(f(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x) \right)+....
$$
Taylor series expansion of $g(x+h)$
$$
g(x+h)=g(x)+h\frac{d}{dx} \left(g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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calculating a limit of sequence Can some one help to show that for $a>0$
$$ \lim_{n\to \infty} \frac{a^\frac{1}{n}}{n+1}+\frac{a^\frac{2}{n}}{n+\frac{1}{2}}+...+\frac{a^\frac{n}{n}}{n+\frac{1}{n}}=\frac{a-1}{\log a}$$
My tries : I was not able to use riemann sum for calculating.
For the special case $a=1$ the RHS has ... | Let
$$S(n) = \frac{a^{\frac{1}{n}}}{n + 1} + \frac{a^{\frac{2}{n}}}{n + \frac{1}{2}} + \cdots + \frac{a^{\frac{n}{n}}}{n + \frac{1}{n}}.$$
Then
$$\frac{a^{\frac{1}{n}}}{n + 1} + \frac{a^{\frac{2}{n}}}{n + 1} + \cdots + \frac{a^{\frac{n}{n}}}{n + 1} < S(n) < \frac{a^{\frac{1}{n}}}{n} + \frac{a^{\frac{2}{n}}}{n} +\cdot... | {
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"source": "stackexchange",
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Find a formula for $\sin(3a)$ and use to calculate $\sin(π/3)$ and $\cos(π/3)$? Problem:
Find a formula for $\sin(3a)$ in terms of $\sin(a)$ and $\cos(a)$. Use this to calculate $\sin(π/3)$ and $\cos(π/3)$.
My attempt:
\begin{align}
\sin(3a) &= \sin(2a + a) = \sin(2a)\cos(a) + \cos(2a)\sin(a) \\
&= \sin(a + a)\cos(... | $$\sin 3a = \sin (2a+a) = \sin(2a)\cos(a)+ \cos(2a)\sin(a)= 2\sin(a)\cos^2(a)+\sin(a)(1-2\sin^2(a))=2\sin(a)(1-\sin^2(a))+\sin(a)(1-2\sin^2(a)) =3\sin(a)-4\sin^3(a) $$
So...
$$\sin(\pi) = 0 = 3\sin(\dfrac{\pi}{3})-4\sin^3(\dfrac{\pi}{3})\implies \dfrac{3}{4} =\sin^2(\dfrac{\pi}{3})\implies \sin(\dfrac{\pi}{3})=\dfrac{\... | {
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Help integrate $\frac{1}{x^3+x^8}$ Could you help me to integrate
$$
\int{\frac{dx}{x^3+x^8}}
$$
I've tried partial fraction decomposition but got $(x^4-x^3+x^2-x+1)$ as the last term when factored the denominator.
Thank you.
| The best way to get the answer given by Wolfram Alpha is to write:
$$\frac{1}{x^3+x^8} = \frac{1}{x^3} -\frac{x^2}{x^5+1}$$
Then solve:
$$\frac{x^2}{1+x^5}=\frac{a}{x+1} + \frac{bx+c}{x^2-2\cos(\pi/5)x+1}+\frac{dx+e}{x^2-2\cos(3\pi/5)x+1}$$
Since $\sin\pi/5$ and $\cos\pi/5$ are in terms of $\sqrt{5}$, you can rewrite t... | {
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Why Does $ \sum\limits_{k=0}^n \begin{pmatrix} n+1 \\ k+1 \end{pmatrix} p^{k+1} (1-p)^{n-k} $ sum to $ (1-(1-p)^{n+1}) $? I was browsing around when I found this question: Find the expected value of $\frac{1}{X+1}$ where $X$ is binomial.
I understood the solution until I hit this portion where $ \sum\limits_{k=0}^n \be... | By using $t = k + 1$:
$$ \sum\limits_{k=0}^n \begin{pmatrix} n+1 \\ k+1 \end{pmatrix} p^{k+1} (1-p)^{n-k} = \sum\limits_{t = 1}^{n+1} \begin{pmatrix} n+1 \\ t \end{pmatrix} p^{t} (1-p)^{n-t +1}$$
$$ = \sum\limits_{t = 0}^{n+1} \begin{pmatrix} n+1 \\ t \end{pmatrix} p^{t} (1-p)^{n-t +1} - (1 - p)^{n+1} $$
$$... | {
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If $x,y$ are integers greater than $1$ and $n$ is a positive integer such that $2^n + 1=xy$ , $\exists 1< aIf $x,y$ are integers greater than $1$ and $n$ is a positive integer such that $2^n + 1=xy$ , then is it true that either $2^n|x-1$ or $2^n|y-1$ ? I have only been able to observe that both $x,y$ are odd . Please... | Let $x=2^{a+p}c+1,y=2^ad+1$ where $p\ge0$ and $c,d$ are add
$2^n+1=xy=2^{2a+p}cd+2^{a+p}c+2^ad+1$
$\iff2^n=2^a(2^{a+p}cd+2^pc+d)$
$\iff2^{n-a}-[2^{a+p}cd+2^pc]=d$ which is odd
If $p>0,2^{n-a}-[2^{a+p}cd+2^pc]$ is even
$\implies p=0\implies$ the highest power of $2$ that divides $x-1$ and $y-1$ will be the same
| {
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Implicit line equation Trying to understand parametric and implicit line equations but I'm at a complete halt now. I have a line that goes through P(0,1) and Q(3,2), I need to find the implicit equation N((x,y)- Z) = 0 such that Z is a point on the line and N is a vector perpendicular to the line. Please point me in th... | Given the points:
$$\begin{gathered}
P(0,1) \hfill \\
Q(3,2) \hfill \\
\end{gathered}$$
Build the vector pointing from $P$ to $Q$:
$$\overrightarrow {PQ} = Q - P = \left( {\begin{array}{*{20}{c}}
3 \\
1
\end{array}} \right)$$
Set parametrization:
$$\left( {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \... | {
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Proving bounds of a harmonic series Let $p>1$. Prove that the series:
$\sum_{n=1}^\infty \frac {(-1)^{n+1}}{n^p}$
is between $\frac {1}{2}$ and 1.
Any help is appreciated.
Just a challenge problem I was presented and curious on the solution
Thanks
| Since $\sum_{n = 1}^\infty (-1)^{n+1}/n^p$ is absolutely convergent for $p > 1$, we may rearrange the terms without affecting the sum. Now
$$\sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n^p} = \left(1 - \frac{1}{2^p}\right) + \left(\frac{1}{3^p} - \frac{1}{4^p}\right) + \cdots$$
and each term in parentheses is positive, so th... | {
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A quick way to prove the inequality $\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$ Can anyone suggest a quick way to prove this inequality?
$$\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$$
| assuming $$x\geq0 \land y\geq0$$
$$\dfrac{\sqrt{x}+\sqrt{y}}{2} \leq \sqrt{\dfrac{x+y}{2}}$$
Squaring:
$$\dfrac{x+y+2\sqrt{xy}}{4} \leq \dfrac{x+y}{2} \implies \dfrac{\sqrt{xy}}{2} \leq \dfrac{x+y}{4}$$
Squaring again
$$\dfrac{\sqrt{xy}}{2} \leq \dfrac{x+y}{4} \implies \dfrac{xy}{4} \leq \dfrac{x^2+2xy+y^2}{16}\implies... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$. Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$.
I've started by letting $P(n) = n^3+11n$
$P(1)=12$ (divisible by 6, so $P(1)$ is true.)
Assume $P(k)=k^3+11k$ is divisible by 6.
$P(k+1)=(k+1)^... | First, show that this is true for $n=1$:
$1^3+11=6\cdot2$
Second, assume that this is true for $n$:
$n^3+11n=6k$
Third, prove that this is true for $n+1$:
$(n+1)^3+11(n+1)=$
$n^3+3n^2+14n+12=$
$\color{red}{n^3+11n}+3n^2+3n+12=$
$\color{red}{6k}+3n^2+3n+12=$
$6k+3n(n+1)+12=$
*
*$2 |n\implies6|3n \implies6|3n(n+1)\... | {
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If $x^2 +px +1$ is a factor of $ ax^3 +bx+c$ then relate $a,b,c$ Suppose If $x^2 +px +1$ is a factor of $ax^3 +bx+c$ then relate $a,b,c$ such that $a,b,c \in R$
I can write $$ax^3 +bx+c=(x^2 +px +1)(\lambda x +D)$$
$$\implies ax^3 +bx+c =\lambda x^3 + x^2.p\lambda + x(\lambda+pD)+D $$
and then compare coefficient to fi... | Just observe that the product of two roots of the quadratic is $1$. So the third root has to be $\frac{-c}{a}$. Now substitute this root instead of $x$ in the cubic.
| {
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the sum of the squares I think it is interesting, if we have the formula
$$\frac{n (n + 1) (2 n + 1)}{6} = 1^2 + 2^2 + \cdots + n^2 .$$
If the difference between the closest numbers is smaller (let's call is a) we obtain, for example, if a=0.1
$$\frac{n (n + 0.1) (2 n + 0.1) }{6 \cdot 0.1} = 0.1^2 + 0.2^2 + \cdots ... | The quantity on the right of your last equation can be expressed as
$$ R(n, \epsilon) =
\sum_{k=1}^{n/\epsilon} k^2 \epsilon^3 $$ where $\epsilon = 0.00\ldots 01$ approaches zero.
Yore conclusion is correct in the following sense:
$$\lim_{\epsilon \to 0^+} \frac{R(n, \epsilon)}{n^3/3} = 1 $$
| {
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Set A = {1,2,3,4,5} Pick randomly one digit and remove it. What is the prob. that we pick an odd digit the 2nd time. The probability that we pick any number for the first time is $\dfrac{1}{5}$
the sample space of sample spaces after the first event is then
{2,3,4,5}
{1,3,4,5}
{1,2,4,5}
{1,2,3,5}
{1,2,3,4}
prob. to pic... | The probability we pick an odd number first is $\frac{3}{5}$. Then the probability of another odd number being chosen from the remaining numbers is $\frac{2}{4}=\frac{1}{2}$. Put together, the probability is $\frac{3}{5}\times\frac{1}{2}=\frac{3}{10}$.
The probability we pick an even number first is $\frac{2}{5}$. Then... | {
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Roots to the quartic equation, $(x+1)^2+(x+2)^3+(x+3)^4=2$ Solving with Mathematica gives me the four roots, $$x=-4,-2,\dfrac{-7\pm\sqrt5}{2}$$ Is there some trick to solving this that doesn't involve expanding and/or factoring by grouping?
| Although the substitution $y=x+2$ works well, I would first try $y=x+3$, which avoids having to expand the highest power (quartic) term and leads to $$y^4+y^3-2y^2-y+1=0$$This yields two roots $y=\pm 1$ to the rational root theorem. Or alternatively to the observation that $$y^4+y^3-2y^2-y+1=(y^4+y^3-y^2)-(y^2+y-1)=(y^... | {
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How to find all positive integers $m,n$ such that $3^m+4^n$ is a perfect square? How to find all positive integers $m$, $n$ such that $3^m+4^n$ is a perfect square? I have found $m=n=2$ is a solution, but cannot find any other and cannot prove whether there is any other solution or not. Please help. Thanks in advance.
| Since $3=-1$ modulo $4$ and $3^2=1$ modulo $4$ the sum $3^m+4^n$ can only be a square when $m$ is even. Therefore we want that
$$(3^j)^2+ (2^n)^2$$
is a square. Using the formula for Pythagorean triples we see that necessarily
$$3^j=p^2-q^2,\qquad 2^n=2pq\ .$$
It follows that $p=2^r$, $q=2^s$ with $r\geq s\geq0$ and th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Maximum and minimum value of $f(z)= |1+z|+|1-z+z^2|$ Let $f(z)= |1+z|+|1-z+z^2|$. Then what is the maximum and minimum value of $f(z)$, if $|z|=1$, where $z$ is a complex number.
Its very difficult to go through the process when we put $z=x+iy$ in the function and then quantify the maximum value. Is there any other met... | $$\begin{align}z = e^{2it}, f &= |1 + z| + |1-z+z^2|\\ &= |1+z| + \frac{|1 + z^3|}{|1+z|}\\
& = 2\cos t+ \frac{\cos 3t}{\cos t} \\
&= 2|\cos t|+|4\cos^2 t - 3|\\
& = 2|x| + |4x^2-3|, x = \cos t \end{align}$$
we can break $$f(x) = 2|x| + |4x^2 - 3| = \begin{cases} 2x+3-4x^2 & if\, 0 < x < \sqrt 3/2\\ 2x + 4x^2 - 3 & if ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1211249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\sum\limits_{k=1}^{2n} {(-1)^k \cdot k^2}=(2n+1)\cdot n$ for all $n\geq 1$ by induction How prove the following equality:
$a_n$:=$\sum\limits_{k=1}^{2n} {(-1)^k \cdot k^2}=(2n+1)\cdot n$
$1$.presumption: $(-1)^1 \cdot 1^2+(-1)^2\cdot2^2=(2 \cdot 1+1) \cdot 1=3$ that seems legit
$2$.precondition:
$a_{n-1}$= $(2... | The equality is true for $n=1$, because
$$
(-1)^1\cdot 1^2+(-1)^2\cdot 2^2=-1+4=3
$$
and
$$
(2\cdot 1+1)\cdot 1=3.
$$
Now suppose the assert is true for $a_{n-1}$. Then
\begin{align}
a_{n}&=\sum_{k=1}^{2n}(-1)^k\cdot k^2\\
&=\biggl(\sum_{k=1}^{2n-2}(-1)^k\cdot k^2\biggr)+
(-1)^{2n-1}\cdot(2n-1)^2+(-1)^{2n}(2n)^2\\
&=a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1215840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof that $(2^m - 1, 2^n - 1) = 2^d - 1$ where $d = (m,n)$ $(2^m - 1, 2^n - 1) = 2^d - 1$ where $d = (m,n)$
my work:
I assumed $m = da$ , $n = db$ for $a,b \in \mathbb{Z}$.
Now, $2^m - 1$ = $2^{da} - 1$ = $(2^d)^a - 1$ = $x^a - 1$ where $x = 2^d$.
similarly
$2^n - 1$ = $x^b - 1$
Now,
using $x^a - 1 = (x - 1)(x^{a-1... | If $a=b$ this means you have $m=n=d$ so this is trivial.
Assume now that $a\neq b$, well you can exchange both so that we can always assume $a>b$ :
$$x^{a-b}(x^{b-1}+...+x+1)=x^{a-1}+...+x^{a-b+1}+x^{a-b} $$
Hence :
$$x^{a-1}+...+x+1-x^{a-b}(x^{b-1}+...+x+1)=x^{a-b-1}+...+x+1 $$
It implies that $\gcd(x^{a-1}+...+x+1,x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1216998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to factor $\frac{27}{125}a^6b^9-\frac{1}{64}c^{12}$ I'm stuck with the following:
$\frac{27}{125}a^6b^9-\frac{1}{64}c^{12}$
My idea was/is the following:
$\frac{3^3a^6b^9}{5^3}-\frac{c^{12}}{8^2}$
Trouble is that I don't know where to go from here. If I go via LCD for $5^3$ and $8^2$ I'll get 8000 in the denomina... | Use
$x^3-y^3=(x-y)(x^2+xy+y^2)$
$\frac{27}{125}a^6b^9-\frac{1}{64}c^{12}
=\left(\frac{3a^2b^3}{5}\right)^3-\left(\frac{c^4}{4}\right)^3$.
Now,
let $x=\frac{3a^2b^3}{5}$
and
$y=\frac{c^4}{4}$
and use that factorization to get
$[\frac{3a^2b^3}{5}-\frac{c^4}{4}][\frac{9a^4b^6}{25}+\frac{3a^2b^3c^4}{20}+\frac{c^8}{16}]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1218208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Integrate with substitution Use the substitution $x=2\sin(\theta)$ to find the value of
$$\int_0^{\sqrt{3}} \frac{1}{(4+x^2)^{3/2}}dx$$
I got to $$\int_{0}^{\pi/3} \frac{2\cos(\theta)}{(4+4\cos^2(\theta))^{3/2}}d\theta.$$
However, I don't know how to further integrate this!
Thanks!
| For now let's perform the indefinite integral: $$\int \frac{1}{(4+x^2)^{3/2}}dx.$$
If we use the substitution $x=2\tan(\theta)$ this gives $dx=2\sec^2(\theta)d\theta$.
Plugging in the subsitution yields: $$\int \frac{2\sec^2(\theta)}{(4+4\tan^2(\theta))^{3/2}} d\theta=\frac{1}{4} \int \frac{\sec^2(\theta)}{\sec^3(\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1222449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Easy inequality going wrong
Question to solve: $$\frac{3}{x+1} + \frac{7}{x+2} \leq \frac{6}{x-1}$$
My method:
$$\implies \frac{10x + 13}{(x+1)(x+2)} - \frac{6}{x-1} \leq 0$$
$$\implies \frac{4x^2 -15x-25}{(x-1)(x+1)(x+2)} \leq 0$$
$$\implies (x-5)(4x+5)(x-1)(x+1)(x+2) \leq 0$$
Using method of intervals, I get:
For $... | After you reached the stage
$$ (x-5)(4x+5)(x-1)(x+1)(x+2)\le 0\stackrel{\div 4}\iff$$
$$ (x+2)\left(x+\frac54\right)(x+1) (x-1)(x-5)\le0$$
and since all the factors are with odd exponent, you can apply the snake method, and get the solution
$$x<-2\;\;,\;\;\;or\;\;\;-\frac54\le x<-1\;\;,\;\;\;or\;\;\;1<x\le 5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1228475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int \frac{dx}{1+\sin x+\cos x}$ Evaluate $$\int \frac{1}{1+\sin x+\cos x}\:dx$$
I tried several ways but all of them didn't work
I tried to use Integration-By-Parts method but it's going to give me a more complicated integral
I also tried u-substitution but all of my choices of u didn't work
Any suggestions?... | $$
\begin{aligned}
\int \frac{d x}{1+\sin x+\cos x}
= & \int \frac{d x}{2 \cos ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}\\ &=\frac{1}{2} \int \frac{\sec ^2 \frac{x}{2}}{1+\tan \frac{x}{2}} d x\\&=\int \frac{d\left(\tan \frac{x}{2}\right)}{1+\tan \frac{x}{2}}
\\&=\ln \left|1+\tan \frac{x}{2}\right|+C
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1230164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
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How to simplify expression with Fibonacci numbers I have to simplify the expression $\sum_{n=0}^\infty \sum_{k=0}^n \frac{F_{2k}F_{n-k}}{10^n}$. I only noticed that $\sum_{n=0}^\infty \sum_{k=0}^n \frac{F_{2k}F_{n-k}}{10^n} = \sum_{n=0}^\infty \frac{1}{10^n} \sum_{k=0}^n F_{2k}F_{n-k}$. What to do next?
| The generating function of the Fibonacci numbers is well-known:
$$ \frac{x}{1-x-x^2} = \sum_{n=0}^\infty F_n x^n. $$
We also want to isolate only the even Fibonacci numbers:
$$ \frac{x}{1-x-x^2} + \frac{-x}{1+x-x^2} = \sum_{n=0}^\infty F_n (x^n + (-x)^n) = \sum_{n=0}^\infty 2F_{2n} x^{2n}. $$
Since
$$
\frac{x}{1-x-x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1232400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})=\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$? I understand that $\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$
But I am struggling to algebraically show that $\sqrt{2}$,$\sqrt[3]{5}\in\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$ to conclude that $\mathbb{Q}(\sqrt{2},\sqrt[3]{... | I'm afraid I can't think of any methods other than brute force. If we set $\alpha = \sqrt{2} + \sqrt[3]{5}$, then all powers of $\alpha$ lie in the span over $\Bbb Q$ of the set $\{1, \sqrt{2}, \sqrt[3]{5}, \sqrt[3]{25}, \sqrt{2}\sqrt[3]{5}, \sqrt{2}\sqrt[3]{25}\}.$ What we essentially need to show is that the set sp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1234930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
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Evaluating: $I_1 = \int\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) dx$ $$I_1 =\int \sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) dx= ?$$
I tried substitution: $\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) = \Xi$, but then I'm not able to do anything after the resulting integral.
Could someone help? There must be a ... | Hint. Assume $a>0,\,x+a>0$. Integrating by parts gives
$$\int \sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) dx=x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) -\int x\left( \frac1{2x} \frac{\sqrt{ax}}{x+a}\right)dx \tag1 $$ and the last integral is easy to evaluate
$$\int \frac{\sqrt{x}}{x+a}\:dx=2\int \frac{u^2}{u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1235849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Finding the inverse of $2+\sqrt{5}+2\sqrt{7}$ Finding the inverse of $2+\sqrt{5}+2\sqrt{7}$ in the field $\mathbb{Q}(\sqrt{5},\sqrt{7})$.
I know that all the elements of $\mathbb{Q}(\sqrt{5},\sqrt{7})$ are of the form:
$a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35}$, where $a,b,c,d \in \mathbb{Q} $
So I could simply solve: $(2+\sqr... | \begin{align}
\frac{1}{2+\sqrt{5}+2\sqrt{7}}
&=\frac{2+\sqrt{5}-2\sqrt{7}}{(2+\sqrt{5})^2-28}\\
&=\frac{2+\sqrt{5}-2\sqrt{7}}{-19+4\sqrt{5}}\\
&=\frac{(2+\sqrt{5}-\sqrt{7})(-19-4\sqrt{5})}{361-80}\\
&=\dots
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1236171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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What is the value of $1^2 + 2^2 + 3^2 + \cdots + (p-1)^2\pmod{p}$? What is the value of $1^2 + 2^2 + 3^2 + \cdots + (p-1)^2\pmod{p}$?
Let's try a several primes greater than 3...
If $p=5$, then we have $1^2 + 2^2 + 3^2 + 4^2 = 30$, so that $30\pmod{5} = 0$
If $p=7$, then we have $1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2= 91$... | HINT: You don't need $a$: consider separately the cases $p\equiv1\pmod3$ and $p\equiv2\pmod3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1236894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find multivariable limit $\frac{x^2y}{x^2+y^3}$ Find multivariable limit of: $$\lim_{ \left( x,y\right) \rightarrow \left(0,0 \right)}\frac{x^2y}{x^2+y^3}$$
How to find that limit? I was trying to do the following, but i am not able to find a proper inequality:
$$| \frac{x^2y}{x^2+y^3} | = |y-\frac{y^4}{x^2+y^3}| \le$$... | Let $a>b>0$, $x=a\cdot t$ and $b\cdot t$. Then we have
$$
\lim_{t^+\to 0}\frac{a^2t^2\cdot bt}{a^2t^2 + b^3t^3}
=
\lim_{t^+\to 0}\frac{a^2bt^3}{t^3(a^2\frac{1}{t} + b^3)}
=
\lim_{t^+\to 0}\frac{a^2b}{(a^2\frac{1}{t} + b^3)}=+\infty
$$
and
$$
\lim_{t^-\to 0}\frac{a^2t^2\cdot bt}{a^2t^2 + b^3t^3}
=
\lim_{t^-\to 0}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1237309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Compute $\int_{0}^{\infty}\frac{x \log(x)}{(1+x^2)^2}dx$ Given
$$\int_{0}^{\infty}\frac{x \log(x)}{(1+x^2)^2}dx$$
I couldn't evaluate this integral. My only idea here was evaluating this as integration by parts.
\begin{align}
\int\frac{x \log(x)}{(1+x^2)^2}dx & = \frac{1}{2} \int\frac{ \log(x)}{(x^2+1)^2}d(x^2+1)\\
& =... | We have
$$\int_1^{\infty} \dfrac{x\log(x)}{(1+x^2)^2}dx = \int_1^0 \dfrac{1/x\log(1/x)}{(1+1/x^2)^2}\dfrac{-dx}{x^2} = \int_1^0 \dfrac{x\log(x)}{(1+x^2)^2} = -\int_0^1 \dfrac{x\log(x)}{(1+x^2)^2}dx $$
Hence,
$$\int_0^{\infty} \dfrac{x\log(x)}{(1+x^2)^2}dx = \int_0^1 \dfrac{x\log(x)}{(1+x^2)^2}dx + \int_1^{\infty} \dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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What is the tip for this exact differential equation? $$ xdx + ydy = \frac{xdy - ydx}{x^2 + y^2} $$
I have multiplied the left part $x^2+y^2$ for $x dx + y dy$ getting
$$(x^3+xy^2+y)dx+(x^2y+y^3-x)dy=0$$
And the derivative test give me:
$\frac{dM}{dy}= 0+2xy+1$ and $\frac{dN}{dx} = 2xy+0-1$.
Where´s my mistake?
| $\frac{1}{2}d(x^2+y^2) = xdx+ydy$
$d(y/x) = \frac{xdy-ydx}{x^2}$
Let $u = x^2+y^2$ and $v = \frac{y}{x}$
$LHS = \frac{1}{2}d(x^2+y^2) = \frac{1}{2}du$
$RHS = \frac{xdy-ydx}{x^2+y^2} = x^2 \frac{dv}{u} = \frac{u}{1+v^2}\frac{dv}{u} = \frac{dv}{1+v^2}$
Thus,
$$
\frac{du}{dv} = \frac{1}{1+v^2}\
$$
Now integrate and subst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How is $\frac{(10^{4})^{6}-1}{10^4-1} = 1 + 10^{4} + 10^{8} + 10^{12} + 10^{16} + 10^{20}$? As the title states, how is: $$\frac{(10^{4})^{6}-1}{10^4-1} = 1 + 10^{4} + 10^{8} + 10^{12} + 10^{16} + 10^{20}$$
I can't see the pattern. Can someone please help? Thanks.
| Note that for $x\not=1$,
$$(x-1)(x^5+x^4+x^3+x^2+x+1)=x^6-1$$
$$\Rightarrow \frac{x^6-1}{x-1}=x^5+x^4+x^3+x^2+x+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Floor inequality: $\lfloor x+y\rfloor\ge \lfloor x\rfloor+\lfloor y\rfloor$ I remember seeing the inequality $\lfloor x+y\rfloor\ge \lfloor x\rfloor+\lfloor y\rfloor$ somewhere which is true for all reals.
So I was wondering what's wrong with this proof?
For all reals $a,b$ with $|b|>1$ we can use the inequality to get... | $$ \left\lfloor -\dfrac{1}{b} \right\rfloor = -1 $$
so you should get
$$ \left\lfloor \dfrac{a-1}{b} \right\rfloor \ge \left\lfloor \dfrac{a}{b} \right\rfloor + \left\lfloor \dfrac{-1}{b} \right\rfloor = \left\lfloor \dfrac{a}{b} \right\rfloor - 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Given $r>0$, find $k>0$ such that $\sqrt{(x-2)^2+(y-1)^2}Using the axioms, theorem, definitions of high school algebra concerning the real numbers, then prove the following:
Given $r>0$, find a $k>0$ such that:
$$\text{for all }x, y: \sqrt{(x-2)^2+(y-1)^2}<k\implies|xy-2|<r $$
I tried with several values given to $k$ a... | what you are looking for are the two hyperbolas that either touch at two place or touch and go through the end point of a diameter of the semicircle. take the upper part of the semicircle $$(x-2)^2 + (y-1)^2 = k^2.$$ let the hyperbola $$xy = 2+r, \frac{dy}{dx} = -\frac y x$$ touch at $$x = 2 + \cos t, y = 1 + \sin t$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int\frac{x^2}{\sqrt{1+x+x^2}}\,dx$ The task is to evaluate $$\int\frac{x^2}{\sqrt{1+x+x^2}}\,dx$$
My best approach has been substitution of $u=x+\frac{1}{2}$, and from there onto (some terrible) trig sub - finally arriving at a messy answer.
Is there a slicker way of doing this?
| As you suggest, complete the square,
$$ 1+x+x^2 = (x+1/2)^2 + 3/4 $$
Now, the key is to substitute to make the square root something nice. I'll do a couple of substitutions to make it clearer. Set $y=x+1/2$, so we have
$$ \int \frac{(y-1/2)^2}{\sqrt{y^2+3/4}} \, dy $$
Now the denominator looks in the form appropriate f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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How can I integrate $\frac{1}{x^2-x-1}$? I need to find $$\int\frac{1}{x^2-x-1}dx$$ but I don't know what to do. I've thought about substitution or partial fractions but neither has worked.
| As stated above, complete the square of the integrand to get
\begin{equation*}
\int\frac{1}{(x-\frac{1}{2})^2-\frac{5}{4}}.
\end{equation*}
Now substitute $u=x-\frac{1}{2}$ & $du=dx:$
\begin{equation*}
\int\frac{1}{u^2-\frac{5}{4}}=-\frac{4}{5}\int \frac{1}{1-\frac{4u^2}{5}}.
\end{equation*}
Substitute again using $s=... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How prove that $0 < a_{10} - \sqrt{2}< 10^{-370}$ for $a_n = \frac{1}{2}(a_{n-1} + \frac{2}{a_{n-1}})$? Let $a_1=1,$ , $a_n = \frac{1}{2}(a_{n-1} + \frac{2}{a_{n-1}})$. How prove that
$0 < a_{10} - \sqrt{2}< 10^{-370}$?
$a_n - a_{n-1} = \frac{1}{2a_{n-1}} \left(2 - a_{n-1}^2\right) < 0 \Rightarrow a_{n-1} > \sqrt{2}$ a... | We have
$$a_{n+1} - \sqrt2 = \dfrac12\left(a_n-\sqrt2 + \dfrac2{a_n} - \sqrt2\right) = \dfrac12\left(a_n-\sqrt2 + \dfrac{2-\sqrt2a_n}{a_n}\right) = \dfrac{(a_n-\sqrt2)^2}{2a_n}$$
Clearly, $a_n > \sqrt2$ for all $n >1$. Hence, we obtain that
$$a_{n+1} -\sqrt2 \leq \dfrac{(a_n-\sqrt2)^2}{2\sqrt2} \leq \dfrac{(a_{n-1}-\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Square Roots: Variables with Exponents. Alright, so let me get this straight:
$\sqrt{x^2} = |x|$
$\sqrt{x^3} = x\sqrt{x}$
$\sqrt{x^4} = x^2$
$\sqrt{x^6} = |x^3|$
Are these correct?
| Well The first and last one are more tricky than the other two.
you know that $x^2$ is always positive and so when we take the square root it will be also positive, This is why $$\sqrt{x^2} = |x|$$
And the last one follows from the first one, But here you have to know that the an even power is always positive $x^{2n}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1253664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find vectors when added up equal (1, 1, 1) Question:
Let $V$ be the 2-dim subspace of $\mathbb R^3$ spanned by $(1, 2, -3)$ and $(-2, 0, 1)$. Write the vector $u = (1,1,1)$ in the form $u = v + w$, where $v$ is in $V$ and $w$ is in $V^\perp$, which is the subspace orthogonal to $V$.
What I tried:
So I found a vector th... | You have found a basis. Let these vectors be the columns of a matrix $A$, and solve $Ax=(1,1,1)^T $.
Let $A = \begin{pmatrix} 2 & 1 & -2 \\ 5 & 2 & 0 \\ 4 & -3 & 1 \end{pmatrix}$.
Compute $$Ax = \begin{pmatrix} 1 \\ 1\\ 1\end{pmatrix}.$$
Solving, we get $x = \begin{pmatrix} \frac{11}{45} \\ -\frac{1}{9} \\ -\frac{14}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1253760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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problem based on LCM and HCF Find the least number which when divided by 7,9,11 produce 1,2,3 as reminders .
I tried a method which works only when the difference between the elements is same. But in this case 7-1=6 , 9-2=7, 11-3=8 . So that method is not working.
Please, provide the way by which I can solve the proble... | This may be a bit longer, but it is a bit more mechanical of a solution.
To solve
$$
\begin{align}
x&\equiv1\pmod{7}\\
x&\equiv2\pmod{9}\\
x&\equiv3\pmod{11}
\end{align}
$$
we can solve three simpler problems. To solve the simpler equations, we can use the Extended Euclidean Algorithm as implemented in this answer.
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1257957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Determine the degree of the extension over Q Determine the degree of the extension $Q(\sqrt{3+2 \sqrt{2}})$ over Q.
I can see that $$3+2 \sqrt{2} = (1+ \sqrt2)(1+ \sqrt2) =(1+ \sqrt2)^2$$ does that mean $$x^2 -(1+ \sqrt2)^2)$$ has a degree $2$. Is this correct
| Example: Let $p(x) = \sqrt{1+\sqrt{2}}$.
Let $u = \sqrt{1+\sqrt{2}} \Rightarrow u^2 = (1+\sqrt{2})^2 = 2+2\sqrt{2} \Rightarrow (u^2-2)^2 = 8 \Rightarrow u^4-4u^2+4=8 \Rightarrow u^4-4u2-4=0$
Hence $u$ is the root of $x^4-4u^2-4$. Try this techniques with your adjoined root and see if you can say anything about the poly... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find all $c\in\mathbb Z^+$ for which $\exists a,b\in\mathbb Z^+, a\neq b$ with $a+c\mid ab$ and $b+c\mid ab$
Find all $c\in\mathbb Z^+$ for which $\exists a,b\in\mathbb Z^+, a\neq b$ with $\begin{cases}a+c\mid ab\\b+c\mid ab\end{cases}$
For those $c$, prove only finitely many $(a,b)$ exist.
My attempt:
Let $(a+... | Your findings so far are fine, but not (yet) fully conclusive towards the main problem.
Let $$C=\{\,c\in\mathbb Z^+\mid \exists a,b\in\mathbb Z^+\colon a\ne b\land a+c| ab\land b+c| ab\,\}$$
be the set we are looking for.
For $c\ge 2$ we can let $$\tag{$\star$} a=c^3+c^2-c,\qquad b=c(a-1)=c(c+1)(c^2-1)$$Then certainly... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $n$ is a product of two consecutive integers, its units digits must be $0,2, $or $6$
Prove that if $n$ is a product of two consecutive integers, its units digits must be $0,2,$ or $6$.
I'm having a hard time with the $0,2,6$ and part but here is what I have so far.
Since $n$ is the product of two consec... | You can also work over all the cases of $k$ modulo $5$.
*
*If $k=5t$, then
$$(2(5t))^2+2(5t)=100t^2+10t=10(10t^2+t)$$
so our number ends with a $0$.
*If $k=5t+1$, then
$$(2(5t+1))^2+2(5t+1)=100t^2+40t+4+10t+2=10(10t^2+5t)+6$$
so our number ends with a $6$.
*If $k=5t+2$, then
$$(2(5t+2))^2+2(5t+2)=100t^2+80t+16+10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
How can I finish integrating $\int {\sqrt{x^2-49} \over x} $ using trig substitution? $$\int {\sqrt{x^2-49} \over x}\,dx $$
$$ x = 7\sec\theta$$
$$ dx = 7\tan\theta \sec\theta \,d\theta$$
$$\int {\sqrt{7^2\sec^2\theta - 7^2} \over 7\sec\theta}\left(7\tan\theta \sec\theta \,d\theta\right) = \int \sqrt{7^2\sec^2\theta - ... | This is the correct answer. Unless the original problem asked for a definite integral, in which case all you would need to do is plug in your bounds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Help solve for length $PQ$ how do I approach this question using simultaneous equations with trig and or pythag??? Solve for length $PQ$
Cheers bob
| Let $PQ=x$ and $QS=y$
Then for triangle $PQS$:
$$x^2 +y^2 = 12^2$$
$$x^2+y^2 = 144$$
For triangle PTR:
$$(x-2)^2+y^2=11^2$$
$$x^2-4x+4+y^2=121$$
So adding $23$ to both sides:
$$x^2-4x+4+y^2+23=121+23$$
$$x^2-4x+27+y^2=144$$
So
$$x^2-4x+27+y^2 = x^2+y^2$$
Take $x^2+y^2$ from both sides
$$-4x+27=0$$
$$4x=27$$
$$x=\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1266041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Represent the transformation with respect to the standard basis
Consider a linear transformation from $R^2$ to $R^2$ defined via:
$$\left(\begin{matrix} 1 \\ 3\end{matrix} \right) \mapsto \left(\begin{matrix} 3 \\ 1\end{matrix} \right)$$ and $$\left(\begin{matrix} -1 \\ 3\end{matrix} \right) \mapsto \left(\begin{matri... | We can represent a linear map: $F:\mathbb{R}^{2} \to \mathbb{R}^{2}$ using a matrix $\mathbf{A}\in\mathbb{R}^{2\times2}$, generically, we can write:
$$\mathbf{A}=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$$
Applying this to the vector $\begin{pmatrix}1 & 3\end{pmatrix}^{T}$, we have:
$$\begin{pmatrix}a & b \\ c & d\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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PDE Manipulation - Calculus I need help for this question, its a lot of calculus but I'm confuse.
let $$ u= \dfrac{(x-b)^{2}+y^{2}-q^{2}}{(x-b-1)^{2}+y^{2}-q^{2}-1} $$
I need show that
$$ u_{x}^{2}+u_y^{2}= \dfrac{1}{(x-b)^{2}}\left(u-1 \right)^{2}\left( u^{2}+q^{2}(u-1)^{2}\right) $$
I begin with this
\begin{eqnarray... | It is a bit easier to write
$$\begin{align}
u&=\frac{(x-b)^2+y^2-q^2}{(x-b-1)^2+y^2-q^2-1}&\\\\
&=\frac{(x-b-1)^2+y^2-q^2-1+2(x-b)}{(x-b-1)^2+y^2-q^2-1}\\\\&=1+2\frac{(x-b)}{(x-b-1)^2+y^2-q^2-1}\\\\
&=1+2\frac{s}{(s-1)^2+t^2}
\end{align}$$
where $s=x-b$ and $t^2=y^2-q^2-1$.
Using the chain rule, we have
$$\frac{\parti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1269159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.