Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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On the biconditional $I(n^2) = 2 - \frac{5}{3q} \iff (k = 1 \land q = 5)$, where $q^k n^2$ is an odd perfect number MOTIVATION
Let $N$ be an odd perfect number given in the so-called Eulerian form
$$N = q^k n^2,$$
i.e., $q$ is the special / Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
In what... | Let $q^k n^2$ be an odd perfect number given in Eulerian form.
Since
$$I(n^2) \leq 2 - \frac{5}{3q}$$
holds in general, and since the biconditional
$$\bigg(I(n^2) = 2 - \frac{5}{3q}\bigg) \iff \bigg(k=1 \land q=5\bigg)$$
holds, the simplest biconditional that I could come up with is
$$\bigg(I(n^2) < 2 - \frac{5}{3q}\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\frac{a}{7a+b}+\frac{b}{7b+c}+\frac{c}{7c+a}\le\frac38$ Suppose that $a,b,c>0$. How to prove $$\frac{a}{7a+b}+\frac{b}{7b+c}+\frac{c}{7c+a}\le\frac38$$
?
My first idea: By AM-GM, $$7a+b\geq \sqrt{7ab}$$ so $$\sum_{cyc} \frac{a}{7a+b}\le\sum_{cyc}\sqrt{\frac{a}{7b}}$$ but I am not sure if we can continue ... | By AM-GM we have $$a^2b+ac^2+b^2c\geq3abc$$ and $$a^2c+ab^2+bc^2\geq 3abc$$ so that $$35(a^2b+ac^2+b^2c)+13(a^2c+ab^2+bc^2)\geq 3(35+13)abc=144abc.$$
Now, note that $$\frac38-\sum_{\text{cyc}} \frac{a}{7a+b}=\frac{35(a^2b+ac^2+b^2c)+13(a^2c+ab^2+bc^2)-144abc}{8 (7 a+b) (a+7
c) (7 b+c)},$$
which is non-negative by th... | {
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"url": "https://math.stackexchange.com/questions/3483200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve $\frac{y+1}{x-y}+\frac{x+2}{x+y}=\frac{x^2+y^2+10}{x^2-y^2}$ and $ 2x+5y=1$
Solve
$$\begin{array}{|l}\dfrac{y+1}{x-y}+\dfrac{x+2}{x+y}=\dfrac{x^2+y^2+10}{x^2-y^2}
\\ 2x+5y=1\end{array}$$
As always:
$$\begin{array}{|l} x-y \ne0 \\ x+y \ne 0 \\ x^2-y^2\ne0\end{array} \Leftrightarrow \begin{array}{|l} x \ne y \... | $$\dfrac{y+1}{x-y}+\dfrac{x+2}{x+y}=\dfrac{x^2+y^2+10}{x^2-y^2}$$
$$\dfrac{xy+x+y+y^2+x^2+2x-xy-2y}{x^2-y^2} = \dfrac{x^2+y^2+10}{x^2-y^2}$$
Since $(x^2-y^2) \ne 0$ , we have :
$$x+y+2x-2y = 10\implies 3x-y =10 \quad \color{#2d0}{\text{(1.)}}$$
Also $$2x+5y=1 \quad \color{#d05}{\text{(2.)}}$$
Multiplying $(1.)$ by $5$... | {
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"timestamp": "2023-03-29T00:00:00",
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The number of solutions to $y^2\equiv x^3+nx\pmod p$ is exactly $p$
Question
Suppose $p\equiv3\pmod4$ and $n\in\mathbb Z$, is the number of solutions to $y^2\equiv x^3+nx\pmod p$ exactly $p$? I have numerically confirmed it when $|n|\le10$ and $p\le229$.
Thoughts
Fix an $x$, the number of solutions to $y^2\equiv x^3+... | Note that $$\sum_{x = 0}^{p -1} \left(\frac{x^3 + nx}{p}\right) = \sum_{x = 1}^{(p -1)/2} \left(\frac{x^3 + nx}{p}\right) + \left(\frac{(-x)^3 -nx}{p}\right)$$ and $$\left(\frac{(-x)^3 -nx}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{x^3 +nx}{p}\right) = - \left(\frac{x^3 + nx}{p}\right),$$ since $p \equiv 3 \pmod... | {
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"timestamp": "2023-03-29T00:00:00",
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Double integral with change of coordinates Evaluate the following double integral:
$$\iint_D \sqrt{x^2+y^2}\,dx\,dy$$
where $D=\{(x,y)\in\mathbb{R^2},\ x^2+y^2-4x<0\}$. I'm trying to use the polar coordinates by making the substitution $x=\rho\cos\theta$ and $y=\rho\sin\theta$.
After replacing in the domain, I find out... | First, let's look at the region $D$: $$x^2+y^2-4x < 0 \iff (x-2)^2 + y^2 < 4$$ Hence, $D$ is the interior of the circle with centre $(2,0)$ and radius $2$. In polar coordinates, this is the region $0<\rho<4\cos\theta$ and $-\pi/2<\theta<\pi/2$. (See here for explanation).
Therefore, the integral is:
\begin{align}
\ii... | {
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"timestamp": "2023-03-29T00:00:00",
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How to use other method to prove $2^x>2x-1$. let $x\in R$,show $$2^x>2x-1$$
My try:
take a function $f(x)=2^x-2x+1$, then we have
$$f'(x)=2^x\ln{2}-2,~~~~f''(x)=2^x\ln^2{2}>0$$
if let $x_{0}$ such $f'(x_{0})=0\Longrightarrow 2^{x_{0}}=\dfrac{2}{\ln{2}}$ is minimum the point (since $f''(x)>0$).
so we have
$$f(x)\ge... | $\require{begingroup} \begingroup$
$\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
\begin{align}
2^x&>2x-1
\tag{1}\label{1}
\end{align}
Let's consider
\begin{align}
2^x&=2x-1
\tag{2}\label{2}
.
\end{align}
It is known, that solution to equation \eqref{2}
can be found in terms of t... | {
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"timestamp": "2023-03-29T00:00:00",
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write 33 as sum of three cubes to solve 3 as sum of three cubes I am interested in writing numbers as the sum of three positive or negative cubed integers. This article describes how a supercomputer was used to find that
$$33 = 8{,}866{,}128{,}975{,}287{,}528^3 + (−8{,}778{,}405{,}442{,}862{,}239)^3 + (−2{,}736{,}111{,... | The resolution for $33$ didn't help directly for the resolution for $3$. The initial author rather designed an efficient method to tackle the general problem.
In the paper (https://people.maths.bris.ac.uk/~maarb/papers/cubesv1.pdf), he describes an alternative method based on factorization: if
$$k-z^3=(x+y)(x^2-xy+y^2)... | {
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Express $\sin (2x)$ in the form $\dfrac{a\pi^2+b\pi + c}{d},$ where $a,b,c,d$ are integers
Let $0<x<\dfrac{\pi}{2}.$ If $x$ is such that $\cos\left(\dfrac{3}{2}\cos x\right) = \sin\left(\dfrac{3}{2}\sin x\right),$ then express $\sin\, (2x)$ in the form $\dfrac{a\pi^2+b\pi + c}{d},$ where $a,b,c,d$ are integers.
Here ... | $(\sin(x)+\cos(x))^2-2\sin(x)\cos(x)=\sin^2(x)+\cos^2(x)$
$\sin(2x)=\left(\frac\pi3\right)^2-1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Can we find $x,y \in \mathbb{N} $ such that $x^y -y^x=2020$? Happy new year.
Can we find $x,y \in \mathbb{N} $ such that : $$x^y -y^x=2020$$
One solution is $(x,y) = (2021,1)$.
It is clear that both of $x,y$ are odd or even .
Note that $2020=2^2 \times 5 \times 101 $
| If $x$ or $y$ is $1$, then $y=1$ and $x=2021$.
If $x$ or $y$ is even then both are even. Since $2020$ is not divisible by $8$, either $x$ or $y$ is $2$. For $ t\in \mathbb{N}$,$2^t-t^2\ge 0 $ if $t\ne 3$ and is an increasing function of $t$ for $t\ge 4$. Therefore $x=2$ and $y\approx 11$ but $2^{11}-11^2=1927\ne 2020. ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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The calculation of the series $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n^4}$ Happy New Year 2020, Romania
In this recent post, Evaluate $\int_0^1 \frac{\arctan x\ln^2 x}{1+x^2}\,dx$, the proposed integral reduces to the calculation of $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n^3}$ which is known in the literatu... | Consider
$$\psi \left( -z \right)+\gamma \underset{z\to n}{\mathop{=}}\,\frac{1}{z-n}+{{H}_{n}}+\sum\limits_{k=1}^{\infty }{\left( {{\left( -1 \right)}^{k}}H_{n}^{k+1}-\zeta \left( k+1 \right) \right){{\left( z-n \right)}^{k}}}, n\ge 0$$
Then
$$\left( \psi \left( -2z \right)+\gamma \right)\frac{\pi f\left( z \right)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaluate $\int \sin^{-1}\frac{2x}{1+x^2}dx$ $\int \sin^{-1}\dfrac{2x}{1+x^2}dx$
My attempt is as follows:-
$$x=\tan\theta$$
$$dx=\sec^2\theta d\theta$$
$$\int \sin^{-1}(\sin2\theta) \cdot\sec^2\theta d\theta$$
So here should we make cases on the basis of values of $\theta$ or can we write $\sin^{-1}(\sin2\theta)$ as $2... | Let $ \displaystyle x=\tan \frac{\theta}{2} \text{ then }\displaystyle d x=\frac{1}{2} \sec ^{2} \frac{\theta}{2} d \theta$ and
$\displaystyle \frac{2 x}{1+x^{2}}=\frac{2 \tan \frac{\theta}{2}}{1+\tan ^2\frac{\theta }{2}}=\sin \theta .$
$\therefore \displaystyle \int \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d \displ... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_{0}^{\frac{\pi}{6}}\frac{\sqrt{3\cos2x-1}}{\cos x}dx$ $$\int_{0}^{\frac{\pi}{6}}\dfrac{\sqrt{3\cos2x-1}}{\cos x}dx$$
$$\int_{0}^{\frac{\pi}{6}}\dfrac{\sqrt{3(1-\tan^2x)-(1+\tan^2x)}}{\sec x\cos x}dx$$
$$\int_{0}^{\frac{\pi}{6}}\sqrt{2-4\tan^2x}\quad dx$$
$$\sqrt{2}\int_{0}^{\frac{\pi}{6}}\sqrt{1-2\tan^2x... | Hint
$$3\cos2x-1-=3(1-2\sin^2x)-1$$
Use $\sin x=u$ to find $$I=\int\sqrt2\dfrac{\sqrt{1-3u^2}}{1-u^2}du$$
$$\dfrac{2I}{\sqrt2}=\int\dfrac{(1+u)\sqrt{1-3u^2}}{1-u^2}du+\int\dfrac{\sqrt{1-3u^2}(1-u)}{1-u^2}$$
For the first integral, set $1-u=y$ and $1+u=z$ in the second
| {
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"timestamp": "2023-03-29T00:00:00",
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If $z = \frac{(\sqrt{3} + i)^n}{(\sqrt{3}-i)^m}$, find the relation between $m$ and $n$ such that $z$ is a real number. I am given the following number $z$:
$$z = \dfrac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^m}$$
with $n, m \in \mathbb{N}$. I have to find a relation between the natural numbers $n$ and $m$ such that the numb... | Hint:
Use the complex exponential notation and congruences: your final fraction is none other than
$$z=2^{n-m}\frac{\mathrm e^{\tfrac{ni\pi}6}}{\mathrm e^{\tfrac{-mi\pi}6}}=2^{n-m} \mathrm e^{\tfrac{(n+m)i\pi}6},$$
anf it is a real number if and only if
$$\frac{(n+m) \pi}6\equiv 0\mod \pi\iff (n+m)\pi \equiv 0\mod 6\pi... | {
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"url": "https://math.stackexchange.com/questions/3498784",
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Derivative of sum of two functions I have to find $\frac{dy}{dx}\left[(x\sqrt{x})+\frac{1}{x^2\sqrt{x}}\right]$ but would like to find where I made a mistake in my solution.
Here is my work:
\begin{align}
(f+g)'=& \ f'+g', f = x \sqrt{x}, g= \frac{1}{x^2 \sqrt x}\\ f=& \ x \cdot x^{1/2} = x^{3/2}
\\ f'=& \ \frac{3}{... | In general if $f(x)=x^{n}$ for some constant $n$ then $f'(x)=nx^{n-1}$ which is the power rule.
| {
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"timestamp": "2023-03-29T00:00:00",
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A Diophantine equation: solve $(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$ (without using Fermat's last theorem) Solve this Diophantine equation: $(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$
My attempt (use Fermat's last theorem)
$$(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$$
$$\Leftrighta... | $$(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$$
If we put $A=3x^2+y^2-4y-17$ and $B=2x^2+2y^2-4y-6$, observe that $A-B=x^2-y^2-11$ (the factor on the right).
In other words, we have $a^3-b^3=(a-b)^3$
Expand this to get $a^3-b^3=a^3-3a^2b+3ab^2-b^3$
So $3ab(a-b)=0$ is the simplified equation. You can then show... | {
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"timestamp": "2023-03-29T00:00:00",
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Number of ways to pair people but with restrictions I have a question regarding combinatorics that I hope you guys can help me clarify:
There are 14 people. 3 of them are TAs. The professor wants to group them into pairs such that no TAs are paired with each other. Find the number of ways to have such arrangement.
So w... | Method 1: Line up the TA's in alphabetical order. Do the same for the students.
There are $11$ ways to select a student to pair with the first TA, $10$ ways to select a student to pair with the second TA, and $9$ ways to select a student to pair with the third TA. Remove those students from the line. That leaves us ... | {
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Find all natural solutions $(a, b)$ such that $(ab - 1) \mid (a^2 + a - 1)^2$.
Find all natural solutions $(a, b)$ such that $$\large (ab - 1) \mid (a^2 + a - 1)^2$$
We have that $$(ab - 1) \mid (a^2 + a - 1)^2 \implies (ab - 1) \mid [(ab)^2 - ab^2 - b^2]^2$$
$$\iff (ab - 1) \mid (ab^2 + b^2 + 1)^2 \iff (ab - 1) \mid... |
Let $(a,b)$ be a solution of $$ (ab - 1) \mid (a^2 \pm a - 1)^2$$
then $(b,c)$ is a solution of $$ (bc-1) \mid (b^2 \mp b - 1)^2,$$ where $$c=\frac{(b^2\mp b-1)^2+ab-1}{b(ab-1)}.$$
Proof
Let $N=ab-1$. Then $0\equiv (a^2b^2\pm ab^2-b^2)^2\equiv (b^2\mp b-1)^2$ (mod $N$). Therefore $(b^2\mp b -1)^2=NM$ for some na... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Denesting $\sqrt{20+ \sqrt{96}+\sqrt{12}}$ into four possible radicals Denesting $\sqrt{20+ \sqrt{96}+\sqrt{12}}$ into possible radicals. This is an answer to an obscure closed question, here on the site. While there is an answer posted, it isn't the complete solution.
| $$\begin{align}
\sqrt{20+ \sqrt{96}+\sqrt{12}} &= \sqrt{5\cdot4+ 4\sqrt{6}+2\sqrt{3}} \\
&= \sqrt{2}\sqrt{5\cdot2+ 2\sqrt{6}+\sqrt{3}} \\
&= \sqrt{2}\sqrt{5\cdot2+\sqrt{3}\left(2\sqrt{2}+1\right)} \\
\end{align}$$
By Using the identity of denesting $\sqrt{a\pm b\sqrt{c}}=\sqrt{(a+d)/2} \pm\sqrt{(a-d)/2}$, where $d=\... | {
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"url": "https://math.stackexchange.com/questions/3511485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Irreducibility of Polynomials without Eisenstein In order to do a different task, I need to show that $f:=X^4-10X^2+1$ is irreducible over $\mathbb{Q}$ and that $g:=X^2Y^2-2Y^2-3X+6$ is irreducible over $\mathbb{Q}[X]$. Since I don't think that Eisenstein criterion works here, I was wondering if there is another simpl... | For the first one, there is a simplification due to https://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial)
We see quickly that there are no rational roots. We are left with showing that the quartic is not the product of two quadratics with integer polynomials. The choices are
$$ (x^2 + Ax + 1)(x^2 + Bx + 1) $$ or
$... | {
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"timestamp": "2023-03-29T00:00:00",
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Suppose that $z\in\mathbb C$ with $|z^2+1|\le 1$. How to prove $|z+1|\geq\frac12$. Let $z\in\mathbb C$ with $|z^2+1|\le 1$. I want to prove $|z+1|\geq\frac12$.
I noticed that $|z^2+1|\le 1$ means that $z$ lies in a cassini oval. I tried with the substitution $z=r\exp(i\theta)$ where $r\geq0, \theta\in[0,2\pi]$. The co... | Let $z=a+ib$, then $$|1+z^2|^2=1+2a^2+a^4+b^2(2a^2-2)+b^4\le1.$$ We want to minimize $|1+z|^2=\color{orange}{(1+a)^2+b^2}$. Let me complete the square in the constraint: $$b^4+b^2(2a^2-2)\color{blue}{+(a^2-1)^2}\le\color{blue}{(a^2-1)^2}-a^4-2a^2=1-4a^2$$ which is the same as $$(b^2+(a^2-1))^2\le1-4a^2.$$ In particular... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $f(1)= 4, f'(x)= \sum_{k=0}^∞ \frac{(-1)^{k} (x-1)^{k}}{(k+1)!}$ then find $f''(1)$. Let $f(1)= 4, f'(x)= \sum_{k=0}^∞ \frac{(-1)^{k} (x-1)^{k}}{(k+1)!}$ then find $f''(1)$.
I directly differentiated $f'(x)$ and put $x=1$, then I got $0$.
If I expand the series $f'(x)$ then then coefficient of $(x-1)$ is $f"(1)$, w... | Your sum is
$$f'(x) = \sum_{k=0}^{\infty} \frac{(-1)^{k} (x-1)^{k}}{(k+1)!} \tag{1}\label{eq1A}$$
Differentiating this term by term gives
$$f''(x) = \sum_{k=0}^{\infty} \frac{k(-1)^{k} (x-1)^{k-1}}{(k+1)!} \tag{2}\label{eq2A}$$
The first term, i.e., for $k = 0$, is $0$. The second term, i.e., for $k = 1$, is $\frac{(-1... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve inverse trigonometric equation $\frac{\pi}{6}=\tan^{-1} \frac{11}{x} -\tan^{-1} \frac{1}{x}$ How do I go about solving for $x$ when I have:
$\frac{\pi}{6}=\tan^{-1} \left( \frac{11}{x} \right)-\tan^{-1}\left( \frac{1}{x} \right)$.
| Let $y = 1/x$. Then solve
$$\pi/6 = \tan^{-1} (11 y) - \tan^{-1} y$$
or
$$\pi/6 + \tan^{-1} y = \tan^{-1} (11 y)$$
to find $y =\frac{5 \sqrt{3}}{11}-\frac{8}{11}$, making $x$ easy to determine.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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probability that two of these boxes contain exactly $2$ and $3$ balls is
If $10$ different balls are to be placed in $4$ distinct boxes at random, then the probability that two of these boxes contain exactly $2$ and $3$ balls is
What I tried:
Total number of ways $\displaystyle 4^{10}$
probability that two of these... | The given answer is wrong! The correct one has been found by NCh.
If the question was: find the probability that box 1 contains exactly 3 balls and box 2 contains exactly 2 balls then result is
$$p:=\frac{\binom{10}{3}\cdot\binom{7}{2}\cdot 2^5}{4^{10}}=\frac{315}{2^{12}}$$
$\binom{10}{3}$ ways to choose the balls to ... | {
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"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find area of ellipse $5x^2 -6xy +5y^2=8$
Find the area of ellipse whose equation in the $xy$- plane is given by $5x^2 -6xy +5y^2=8$
My attempt : I know that area of ellipse $ = \pi a b$ ,where $a$ is semi-major axis and $b$ is semi minor axis
Now if we make matrix $\begin{bmatrix} 5 & -3 \\-3& 5\end{bmat... | Under the variable changes $x = \frac{u+v}{\sqrt2}$ and $y = \frac{u-v}{\sqrt2}$, the equation $5x^2 -6xy +5y^2=8$ is of the standard ellipse form
$$\frac{u^2}4+v^2=1$$
with the major and minor axes 2 and 1, respectively. Thus, the area is $2\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How can I factor out $x^2+ax-2 (3a+1)(2a+1)$? I have tried it this way but could not factor it out :
$$x^2+ax-2 (3a+1)(2a+1)$$
Here, let
$$3a+1=m\tag{i}$$
$$2a+1=n\tag{ii}$$
So, (i)-(ii) gives us $$m-n=a$$ Then
$$\begin{align}
x^2+ax-2(3a+1)(2a+1) &= x^2+(m-n)x-2mn \\
&= x^2+mx-nx-mn-mn \\
&= x(x+m)-n(x+m)-mn \... | Writing out the two equations
$$
x^2+ax-2 (3a+1)(2a+1)=(x-b)(x-c)=x^2-(b+c)x+bc
$$
and comparing coefficients gives the equations
$$
a+b+c=0,\quad 12a^2 +10a +bc + 2=0.
$$
Substituting $c=-b-a$ and solving for $b$ we obtain
$$
b=\frac{\pm \sqrt{49a^2 + 40a + 8} - a}{2}
$$
Edit: Following the suggestion of lab bhattach... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove $\frac{x}{\sqrt{1+x^2}}<\arctan\left(x\right)$ for every $x>0$
Prove $\frac{x}{\sqrt{1+x^2}}<\arctan\left(x\right)$ for every $x>0$
I thought about MVT , what i did is :
let $g(x) = \arctan\left(x\right) $
so by MVT there is $c\in (0,x)$ such that :
$g'(c) = \frac{g\left(x\right)-g\left(0\right)}{x-0} \impli... | Try comparing derivatives and integrating the difference from 0 to $x$.
$$\frac{\sqrt{x^2+1}-\frac{x^2}{\sqrt{1+x^2}}}{1+x^2}=\frac{1}{(1+x^2)^{\frac{3}{2}}}<\frac{1}{(1+x^2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3520272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $\frac{a+b}{\csc x}=\frac{a-b}{\cot x}$, show that $\csc x \cot x=\frac{a^2-b^2}{4ab}$
If $$\frac{a+b}{\csc x}=\frac{a-b}{\cot x}$$
show that
$$\csc x \cot x=\frac{a^2-b^2}{4ab}$$
(original problem image)
I tried getting rid of the denominator and then expanding the given equation, but couldn't get to the an... | From $1+\cot^2 x= \csc^2 x$ we have
\begin{eqnarray*}
(\color{red}{\csc x} +\cot x)(\csc x -\color{blue}{\cot x}) = 1
\end{eqnarray*}
Now sub
\begin{eqnarray*}
\color{red}{\csc x} = \frac{a+b}{a-b} \cot x
\end{eqnarray*}
and
\begin{eqnarray*}
\color{blue}{\cot x} = \frac{a-b}{a+b} \csc x .
\end{eqnarray*}
This giv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3520401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How many spanning trees does this graph contains? How many spanning tree does below graph contains?
I tried to calculate by using edge deletion / contraction approach and I am getting 48:
I tried to calculate it by Kirchoffs method as a determinant of Laplacian matrix:
$$ \left(\begin{matrix}2&0&0&0&0&0&0&0\\0&3&0&0... | If you delete BF and CG then you need to delete one of the remaining $8$ edges.
If you delete BF but not CG then you need to delete one of the $5$ edges of the LH cycle and one of the $3$ edges of the RH cycle. Similarly for deleting CF but not BF.
If you delete neither BF nor CG then you need to delete one of the $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3522409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate matrix by using Cayley-Hamilton theorem Calculate matrix $B = A^{10}-3A^9-A^2+4A$ using Cayley-Hamilton theorem on $A$.
$$A = \begin{pmatrix}
2 & 2 & 2 & 5 \\
-1 & -1 & -1 & -5 \\
-2 & -2 & -1 & 0 \\
1 & 1 & 3 & 3
\end{pmatrix}$$
Now, I've calculated the characteristic polynomial of $A$:
$P_A(\lambda... | We know that $A^4-3A^3=3A-A^2$. Now,
\begin{align*}
B&=A^{10}-3A^9-A^2+4A\\
&=A^6(A^4-3A^3)-A^2+4A\\
&=A^6(3A-A^2)-A^2+4A\\
&=3A^7-A^8-A^2+4A\\
&=-A^4(A^4-3A^3)-A^2+4A\\
&=-A^4(3A-A^2)-A^2+4A\\
&=-3A^5+A^6-A^2+4A\\
&=A^2(A^4-3A^3)-A^2+4A\\
&=A^2(3A-A^2)-A^2+4A\\
&=-A^4+3A^3-A^2+4A\\
&=A
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3522975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Non-homogeneous cyclic $\frac{x+1}{\sqrt{x+y}}+\frac{y+1}{\sqrt{y+z}}+\frac{z+1}{\sqrt{z+x}} \geq 3\sqrt{2}$
Consider three non-negative real numbers $x$, $y$ and $z$, no two of
which are zero at the same time. Prove that the following inequality
holds:
$$\frac{x+1}{\sqrt{x+y}}+\frac{y+1}{\sqrt{y+z}}+\frac{z+1}{\s... | Remarks: The inequality (1) is @arqady's idea. Although the original inequality is not homogeneous, (1) is. Hope to see a nice proof of (1).
A proof utilizing arqady@AoPS's nice idea.
Since the desired inequality is cyclic,
assume that $z = \min(x, y, z)$.
The desired inequality is written as
$$\sum_{\mathrm{cyc}} \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3523354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Given that $\alpha + \beta - \gamma = \pi$, prove that $\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = 2 \sin \alpha \sin \beta \cos \gamma$. I am told:
$$\alpha + \beta - \gamma = \pi$$
And I have to prove:
$$\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = 2 \sin \alpha \sin \beta \cos \gamma$$
What should I be looking... | $$\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma-2 \sin \alpha \sin \beta \cos \gamma=$$
$$=\sin^2 \alpha + \sin^2 \beta -1+\cos^2(\alpha+\beta)+2 \sin \alpha \sin \beta \cos (\alpha+\beta)=$$
$$=\sin^2 \alpha-\cos^2 \beta+\cos(\alpha+\beta)\cos(\alpha-\beta)=$$
$$=\sin^2 \alpha-\cos^2 \beta+\cos^2\alpha\cos^2\beta-\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3523461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Derivative of $\dfrac{\sqrt{3-x^2}}{3+x}$ I am trying to find the derivative of this function
$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}$
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-2x(3+x)}{2\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-x(3+x)}{\sqrt{3-... | Your derivative of $\sqrt{3 - x^2}$ is incorrect.
$$\begin{align}\dfrac{\mathrm d}{\mathrm dx}\dfrac{\sqrt{3 - x^2}}{3 + x^2} &= \dfrac{\frac{\mathrm d}{\mathrm dx}\sqrt{3 - x^2}\cdot(3 + x) - \sqrt{3 - x^2}\cdot\frac{\mathrm d}{\mathrm dx}(3 + x)}{(3 + x)^2} \\ &= \dfrac{\frac{-2x}{2\sqrt{3 - x^2}}(3 + x) - \sqrt{3 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
Minimum value of $\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$ If $$x^2+y^2=1$$
Prove that Minimum value of $$f(x,y)=\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$$ is
$$\frac{2\sqrt{ab}}{a+b}$$
My try:
I used basic Trigonometry:
Let $x=\cos t$ and $y=\sin t$
Then we get $$f(x,y)=g(t)=\frac{a\cos^2 t+b\sin^2 t}{\sqrt{a^2\cos^2 t+b... | Since you have
$$x^2 + y^2 = 1 \implies y^2 = 1 - x^2 \tag{1}\label{eq1A}$$
you can substitute this into your $f(x,y)$ to have a function $g(x)$ for $0 \le x \le 1$ with
$$g(x) = \frac{ax^2 + b(1-x^2)}{\sqrt{a^2x^2 + b^2(1-x^2)}} = \frac{(a-b)x^2 + b}{\sqrt{(a^2-b^2)x^2 + b^2}} \tag{2}\label{eq2A}$$
Since $x$ only appe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\lim\limits_{n\to\infty}\left(\sqrt [n+1]{\frac {a_{n+1}}{b_{n+1}}}-\sqrt[n]{\frac {a_n}{b_n}}\ \right)$
Consider two sequences $(a_n)_{n\ge 1}$ and $(b_n)_{n\ge 1}$ of
positive real numbers such that
$$\lim_{n\to\infty} \frac {a_{n+1}}{n^2\cdot a_n}=x>0$$
and
$$\lim_{n\to\infty} \frac {b_{n+1}}{n\cdot b... | Some observations. Define
$$
c_n : = \frac{1}{n!}\frac{a_n}{b_n}.
$$
Then by the assumptions on $a_n$ and $b_n$, we have
$$
\mathop {\lim }\limits_{n \to + \infty } \frac{c_{n + 1}}{c_n} = \frac{x}{y}\quad \text{ and hence }\quad \mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{c_n} = \frac{x}{y}.
$$
Note that
$$
\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3528462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solve a system of equation: $\cos(2x) + \cos(y) = 1$, $\sin(2x) + \sin(y) = 1$ Solve a system of equation:
$$\cos(2x) + \cos(y) = 1$$
$$\sin(2x) + \sin(y) = 1$$
My idea:
Let's see what is product of this two equations.
$$\cos(2x)\sin(2x) + \cos(2x)\sin(y) + \cos(y)\sin(2x) + \cos(y)\sin(y) = 1$$
$$\cos(2x)\sin(2x) + \s... | Let $z=2x$. Then
$$1=\cos z + \cos y = 2\cos\frac{z-y}2\cos\frac{x+y}2\tag 1$$
$$1=\sin z + \sin y = 2\cos\frac{z-y}2\sin\frac{x+y}2\tag 2$$
Take (1) - (2) to have a factorized equation,
$$\cos\frac{z-y}2 \sin\left( \frac{z+y}2 - \frac\pi4 \right) =0$$
So, two cases to consider:
Case 1) $\cos\frac{z-y}2 = 0$ leads to $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Prove that $x + \frac{2x^3}{3} + \cdots + \frac{2\cdot 4 \cdot \cdots 2nx^{2n+1}}{3\cdot 5 \cdot (2n+1)}+\cdots = \frac{\arcsin(x)}{\sqrt{1-x^2}}$
Prove that $x + \dfrac{2x^3}{3} + \cdots + \dfrac{2\cdot 4 \cdot \cdots 2nx^{2n+1}}{3\cdot 5 \cdot (2n+1)}+\cdots = \dfrac{\arcsin(x)}{\sqrt{1-x^2}}.$
Let $f(x) = x + \dfr... | In this proof i will use some proprieties of Wallis'integral (Which easy to show by simple induction),
We have :
$$W_{2n+1}=\prod_{k=1}^n\frac{2k}{2k+1}$$
So your series is equal to :
$$f(x)=\sum_{n=0}^\infty W_{2n+1}x^{2n+1}=\sum_{n=0}^\infty \biggl(\int_0^{\pi/2}\sin^{2n+1}t\ dt\biggr)x^{2n+1}$$
Then we put $t=\arcsi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find whether $\sum_{n=1}^\infty \frac{2^{3n-3}3^{n+2}}{5^{2n-2}} $ is geometric and convergent and if so, state the value of the sum $$\sum_{n=1}^\infty \frac{2^{3n-3}3^{n+2}}{5^{2n-2}} $$
First I tried to find the ratio by doing $\frac{a_{n+1}}{a_n}$. Many calculations were made, and the result was $24/25$, so it is ... | $$\sum_{n=1}^\infty \frac{2^{3n-3}3^{n+2}}{5^{2n-2}} =\frac{9.25}{8}\sum_{n=1}^{\infty } \frac{8^n.3^n}{25^n}=\frac{225}{8}\frac{\frac{24}{25}}{1-\frac{24}{25}}=675$$
So your answer is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Limit of $\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$ $$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$$
I tried to used $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ but it did not worked out so I tried to use the squeeze theorem.
$$0=\sqrt[3]{x^3}-\sqrt{x^2}\leq \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}\leq\sqrt[3]{8x^3}-\sqrt{4x^2}=... | Moving factors of $x^3$ and $x^2$ out, we transform the limit to
$$\lim_{x\to\infty}x\left(\sqrt[3]{1+2/x^2}-\sqrt{1-2/x}\right)$$
Then applying the binomial series yields
$$=\lim_{x\to\infty}x\left(\left(1+O(x^{-2})\right)-\left(1-\frac12\cdot\frac2x+O(x^{-2})\right)\right)$$
$$=\lim_{x\to\infty}x\left(1-1+\frac12\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
remainder when $a_{1000}$ is divided by $1000$
If $a_{1}=7,a_{2}=7^7,a_{3}=7^{7^{7}}.$ Then the remainder when
$a_{1000}$ is divided by $1000$
what i try
$a_{1}=7=1\mod(1000)$ and
$a_{2}=7^7=1^7\mod(1000)=1\mod(1000)$
from using modulo theorem
$a_{3}=7^{7^{7}}=1^7\mod(1000)=1\mod(7)$
can we say that $a_{1000}= 1\mod... | $$7^4=(50-1)^2=2401$$
$$7^{4n}=(1+2400)^n\equiv1+2400n\pmod{1000}$$
So, $n$ must be divisible by $5$ to make residue $\equiv1$
$\implies$ord$_{1000}7=20$
We need $a_{999}\pmod{20}$
Again, $a_r,r\ge2$ are of the form $7^{4n+3}$
$7^{4n+3}=7^3(50-1)^{2n}\equiv3(1-50)^{2n}\equiv3(1-\binom{2n}150)\pmod{20}\equiv3$
So for $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How many pairs of $(x,y,z)$ for $x+y+2z=n$
How many pairs of $(x,y,z)$ for $x+y+2z=n$ when $n$ is an odd integer $\geq 5$
For $n=5$ there are 2 pairs. $n=7$ there are 6 pairs. $n=9$ there are 12 pairs. $n=11$ there are 20 pairs.
$an^2 + bn + c = f(n)$
$25a + 5b + c = 2$
$49a + 7b + c = 6$
$81a + 8b + c = 12$
$f(n) = ... | For $x,y,z\in \mathbb{N}, x,y,z\geq 0$,
When $n$ is odd, the number is:
$$ (n+1)+(n-1)+(n-3)+...+2=\frac{(n+1)(n+3)}{4} $$
when $n$ is even, the number is:
$$ (n+1)+(n-1)+(n-3)+...+1=(\frac{n+2}{2})^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that $\sum_{k=1}^{89} \frac{1}{\tan^3(k^\circ)+1}=44.5$ So I've been stuck on this problem for a while now, and to be honest I have no idea how to start. I stumbled upon it while practicing for an Olympiad, and it is a past Olympiad problem (for teens) itself.
The problem is as follows:
Given is the function $f(x... | We can write $\tan(90^\circ - k^\circ) = \frac{1}{\tan(k^\circ)}$.
\begin{align*}
\sum_{k=1}^{89} \frac{1}{\tan^3(k^\circ) + 1} &= \sum_{k=1}^{44} \frac{1}{\tan^3(k^\circ) + 1} + \sum_{k=46}^{89} \frac{1}{\tan^3(k^\circ) + 1} + \frac{1}{\tan^3(45^\circ) + 1} \\
&= \sum_{k=1}^{44} \frac{1}{\tan^3(k^\circ) + 1} + \sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is $x^3 + x + 1$ for $x=\frac{\sqrt 5 - 1}{2}$
What is $x^3 + x + 1$ for $x=\frac{\sqrt 5 - 1}{2}$
Of course we can subtitute and expand it
${(\frac{\sqrt 5 -1}{2})}^3 + \frac{\sqrt 5 -1}{2} + 1$
Is there a better way to do it?
| Since $x^2=\frac{3-\sqrt{5}}{2}=1-x$, $x^3=x-x^2=2x-1$, so $x^3+x+1=3x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Help with "Smallest Euclidean Norm problem". I'm trying to find the smallest euclidian norm of the general solution of an algebraic equation $Ax=y,$ where $A$ is an $n\times m$ matrix and the gral solution is:
$x=x_{p}+\alpha_{1}n_{1}+\alpha_{2}n_{2}=\begin{pmatrix}
0\\
-4\\
0\\
0
\end{pmatrix}+\alpha_{1}\begin{pmatrix... | setting the gradient to the zero vector gives system
$$ 6 \alpha_1 + 4 \alpha_2 = 8 \; , $$
$$ 4 \alpha_1 + 10 \alpha_2 = 16 \; . $$
Or augmented matrix
$$
\left(
\begin{array}{cc|c}
6 & 4 & 8 \\
4 & 10 & 16
\end{array}
\right)
$$
$$
\left(
\begin{array}{cc|c}
3 & 2 & 4 \\
2 & 5 & 8
\end{array}
\right)
$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Nice integral involving omega constant, Lambert's function and logarithms
$$\int_1^e \Big(\operatorname{W}(x)^2 \ln(x)0.25-\operatorname{W}(x) 0.5 + \operatorname{W}(x) \ln(x) \Big) \, dx=\operatorname{W}(1)0.5 + \operatorname{W}(1)^2 0.25$$
I can express the antiderivative wich involves exponential integral and La... | I unfortunately do not have a smart way to evaluate this interesting integral. My approach is more just "plug and chug".
Let $u = \operatorname{W}_0 (x)$, then $x = ue^u, dx = e^u(1 + u) \, du$ while the limits of integration become: $u = \operatorname{W}_0 (1) = \Omega$ and $u = \operatorname{W}_0 (e) = 1$. Here $\Om... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Common integrating factor of two differential equations. The following differential equations have a common integrating factor.
$$(3y+4xy^2)dx+(4x+5x^2y)dy=0$$
$$(6y+x^2y^2)dx+(8x+x^3y)dy=0$$
I have to find this integrating factor.
I have arrived to the solution of this problem assuming that the integrating factor I ... | This may help
Let us re-write the first one as:
$$\frac{xdy}{ydx}=-\frac{3+6xy}{4+5xy}\implies\frac{xdy}{ydx}+1=-\frac{3+6xy}{4+5xy}+1. $$
$$\implies \frac{xdy+ydx}{ydx}=\frac{1-xy}{4+5xy}$$
$$\implies \frac{dv}{ydx}=\frac{1-v}{4+5v}, xy=v \implies \frac{dv}{dx}=\frac{v-v^2}{4+5v} \frac{1}{x}.$$
$$\implies \int \frac{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3547454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
coefficient of bivariate generating functions how can I find/what is the coefficient of the $x^4y^4$ and $x^6y^6$ term for the following generating function
$$\frac{1}{1-x-y-x^2y}$$
I've been informed I can do this by using Pascal's triangle, by adding an extra value of 2 back and 1 down but am unsure of how to? (I'm n... | A method to find the coefficient of $x^m y^n$ in $\frac{1}{1-x-y-x^2y}$ is described as follows.
First, let
$$\frac{1}{1 - x - y - x^2y} = a_0(x) + a_1(x) y + a_2(x) y^2 + a_3(x) y^3 + \cdots .$$
We need to determine $a_n(x) y^n$.
Clearly,
$$\frac{\partial^n \tfrac{1}{1 - x - y - x^2y}}{\partial y^n} \Big\vert_{y=0} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3547566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof that combined matrices are idempotent. Suppose that $X_{nxp} (n>p)$ is a matrix such that $X'X$ is invertible.
*
*Prove that both $X(X'X)^{-1}X'$ and $I_n-X(X^TX)^{-1}X^T$ are idempotent.
*Prove that the tr $X(X'X)^{-1}X' =p$ and tr $(I_n - X(X'X)^{-1}X')=n-p$.
| Let us put
$$ A \colon= X \left(X^\prime X\right)^{-1} X^\prime. $$
Then we note that
$$
\begin{align}
A^2 &= AA \\
&= \left[ X \left(X^\prime X\right)^{-1} X^\prime \right] \left[ X \left(X^\prime X\right)^{-1} X^\prime \right] \\
&= X \left[ \left(X^\prime X\right)^{-1} \left( X^\prime X \right) \right] \left(X^\pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3547987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to calculate this determinant of a $2\times 2$matrix? This matrix arises from a homework problem which our professor gave.
We need to find the determinant of this matrix.
Does there exist any simple way to find the determinant of this matrix?
$\begin{pmatrix}
x-pq-p+3-(q-1)(\frac{x+2-n}{x-n+2-l}) && (1-p)(2+\fra... | It helps to rewrite your matrix as $A+B$ with$$A:=\left(\begin{array}{cc}
x-pq-p+3 & 1-p\\
1-q & x-pq-q+3
\end{array}\right),\,B:=\frac{x+2-n}{x+2-n-l}\left(\begin{array}{cc}
1-q & 1-p\\
1-q & 1-p
\end{array}\right).$$Write $A=\left(\begin{array}{cc}
a & c\\
b & d
\end{array}\right),\,B=\left(\begin{array}{cc}
e & g\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $\int_a^b \frac{1}{x}\sqrt{-(x-a)(x-b)}dx = (\frac{a+b}{2}-\sqrt{ab})\pi$ Does anyone know how to solve $$\int_a^b \frac{1}{x}\sqrt{-(x-a)(x-b)}dx$$
After trying Wolfram Alpha, I conclude that it possibly equals $(\frac{a+b}{2}-\sqrt{ab})\pi = \frac{(\sqrt{a}-\sqrt{b})^2}{2}\pi$.
The book I am reading says one ca... | Let $x=a\cos^2 t+b \sin^2 t \implies dx=(b-a) \sin 2t dt$
Then $$I=2\int_{0}^{\pi/2} \frac{(b-a)^2 \sin^2 t \cos^2 tdt}{a \cos^2 t+b \sin^2t}= 2 (b-a)^2 \int_{0}^{\pi/2}\frac{ \sin^2 t ~dt}{a+b\tan^2 t}$$
Let $\tan t=u$, then
$$I=2\int_{0}^{\infty} \frac{u^2 du}{(1+u^2)^2(a+bu^2)}=2\int_{0}^{\infty} du \left(\frac{b-a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3549527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
Linearization of $C^r$ ODE system Consider the following ODE system
\begin{align*}
\dot{x} &= x^5 + y^3 = f(x\,,y) \\
\dot{y} &= x^3 - y^5 = g(x\,,y)
\end{align*}
Clearly, $(x\,,y) = (0\,,0)$ is the only fixed point. Linearizing
$$
J =
\begin{bmatrix}
5x^4 & 3y^2 \\
3x^2 & -5y^4
\end{bmatrix}
$$
evaluating at the fixe... | From
\begin{align*}
\dot{x} &= x^5 + y^3 \\
\dot{y} &= x^3 - y^5
\end{align*}
we have
\begin{align*}
x^3\dot{x} &= x^8 + x^3 y^3 \\
y^3\dot{y} &= x^3y^3 - y^8
\end{align*}
and then
$$
\frac 14\frac{d}{dt}(x^4-y^4) = x^8+y^8
$$
analyzing the behavior along a line $y(t) = \alpha x(t)$ we conclude
$$
\frac 14(1-\alpha^4)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3549658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to prove that $\tan \left( \frac{\pi}{2} - \theta \right) = \cot \theta$ I am asked to simplify the following expression:
$$\tan \left( \frac{\pi}{2} - \theta \right)$$
The book gives me the answer $\cot \theta$ but, when I try to derive that formula using
$$\tan \left( A \pm B \right) = \frac{\tan A \pm \tan B}{1... | There are several ways one can play with this type of problems.
Here is one:
$\tan\left(\frac{\pi}{2}-\theta \right) =\frac{\sin\left(\frac{\pi}{2}-\theta \right) }{ \cos\left(\frac{\pi}{2}-\theta \right) }$
But
$\sin\left(\frac{\pi}{2}-\theta \right)=\sin\left(\frac{\pi}{2}\right)\cos\left(\theta \right)-\cos\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
A sequence $(a_n)_{n\ge 1}$ such that $a_1>0$ and $a_{n+1}=a_n-\ln(1+a_n)$ Let $(a_n)_{n\ge 1}$ be a sequence such that $a_1>0$ and $$a_{n+1}=a_n-\ln(1+a_n)$$
a) Prove that $a_{n+1}<\frac{a_n^2}{2}, \forall n\in \mathbb{N}$ and $\lim\limits_{n\to \infty} (n^{2019}a_n)=0$.
b) If $a_1<2$, prove that $a_{12} \in (0,10^{-... | Hint: to finish $(a)$. You already proved that $a_{n+1}<\frac{a_n^2}{2}$, use it further, i.e.
$$0<a_{n+1}<\frac{a_n^2}{2}<\frac{1}{2}\cdot\left(\frac{a_{n-1}^2}{2}\right)^{2}=
\frac{a_{n-1}^4}{2^3}<\\
\frac{1}{2^3}\left(\frac{a_{n-2}^2}{2}\right)^4=
\frac{a_{n-2}^8}{2^7}<\\
\frac{1}{2^7}\left(\frac{a_{n-3}^2}{2}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proving $(\sin^2 \alpha+\sin\alpha \cos \alpha)^{\sin \alpha}(\cos^2 \alpha+\sin \alpha \cos \alpha)^{\cos \alpha}\leq 1$
If $\alpha \in \left(0, \frac{\pi}{2}\right)$, prove that:
$$(\sin^2 \alpha+\sin\alpha \cos \alpha)^{\sin \alpha}(\cos^2 \alpha+\sin \alpha \cos \alpha)^{\cos \alpha}\leq 1$$
I know $\sin \alpha$ ... | Rewrite the last line of your proof as: $a^{\frac{a}{a+b}}\cdot b^{\frac{b}{a+b}}\cdot (a+b) \le 1$. Apply the weighted AM-GM inequality you have: $LHS \le (\dfrac{a^2}{a+b}+\dfrac{b^2}{a+b})\cdot (a+b) = \dfrac{1}{a+b}\cdot (a+b) = 1 = RHS$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3551745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Barnard and Child inequality exercise Prove that,
$$3(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)≥abc(a+b+c)^3$$
For positive $a,b,c$
The exercises in this book are making me crazy.
Any help would be appreciated.
My attempts:
I opened the LHS brackets and used the cube identity on RHS
I get expressions which are somewhat similar b... | Another way.
By your work, AM-GM and Muirhead we obtain:
$$3\prod_{cyc}a^2b\prod_{cyc}a^2c-abc(a+b+c)^3=\frac{1}{2}\sum_{sym}(3a^3b^3+2a^4bc+a^2b^2c^2-6a^3b^2c)\geq$$
$$\geq \frac{1}{2}\sum_{sym}\left(6\sqrt[6]{(a^3b^3)^3(a^4bc)^2a^2b^2c^2}-6a^3b^2c\right)=3\sum_{sym}\left(a^{\frac{19}{6}}b^{\frac{13}{6}}c^{\frac{2}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to solve a differential equation $(x^2y+y^5)dx+(x^3-xy^4)dy=0$?
Solve the following differential equation:
$$(x^2y+y^5)dx+(x^3-xy^4)dy=0$$
I noticed that it is not exact, since:
$$
(x^2y+y^5)'_y=x^2+5y^4\ne3x^2-y^4=(x^3-xy^4)'_x
$$
Then I tried to express $y'$:
$$
y'=\frac{x^2y+y^5}{xy^4-x^3}
$$
And now I don't... | Try the Ansatz $df=0$ with $f_x=x^ay^b(x^2y+y^5),\,f_y=x^ay^b(x^3-xy^4)$ so
$$\begin{align}0&=\frac{(f_x)_y-(f_y)_x}{x^ay^b}\\&=(b-a-2)x^2+(a+b+6)y^4\\\implies(a,\,b)&=(-4,\,-2).\end{align}$$The equations $f_x=\frac{1}{x^2y}+\frac{y^3}{x^4},\,f_y=\frac{1}{xy^2}-\frac{y^2}{x^3}$ are consistent with $f=-\frac{1}{xy}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to prove $\sum _{k=1}^{\infty } (-1)^k H_{\frac{2 k}{3}} = -\frac{\pi }{2 \sqrt{3}}+\frac{3 \pi }{8}-\frac{3}{4} \log (2)$? I stumbled on this problem in the wake of the discussion https://math.stackexchange.com/a/3553902/198592
Can you make sense of the equation in the question involving the harmonic number with ... | Start with using the integral representation of the harmonic number $H_n=\int_0^1\frac{1-x^n}{1-x}\ dx$ we have
$$\sum_{k=0}^\infty(-1)^k H_{\frac{2k}{3}}=\int_0^1\frac{1}{1-x}\sum_{k=0}^\infty((-1)^k-(-x^{\frac23})^n)\ dx$$
$$\int_0^1\frac{1}{1-x}\left(\frac12-\frac{1}{1+x^{\frac23}}\right)\ dx\overset{x\to x^3}{=}-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3554285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $4^{1/3}+10^{1/2}$ is irrational Show that $4^{1/3}+10^{1/2}$ is irrational.
I start by assuming it to be rational and want to come to a contradiction
$$
4^{1/3}+10^{1/2} = r \\
\Rightarrow 4^{1/3} = r - 10^{1/2} \\
\Rightarrow 4 = (r-10^{1/2})^{3} \\
\Rightarrow 4 = r^{1/3} - 10^{1/3} + 3r^2*10^{1/2} - 30r
... | Since $\sqrt[3]{4}$ is a root of the polynomial $f(x) = x^3 - 4$, $r \stackrel{def}{=}\sqrt[3]{4} + \sqrt{10}$ is a root of
$$g(x) = f(x-\sqrt{10})f(x+\sqrt{10}) = x^6-30x^4-8x^3+300x^2-240x-984$$
By rational root theorem,
if $r = \frac{p}{q}$ is a rational root of $g(x)$ for coprime integers $p$ and $q$, then $q$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove $\binom{n}{0} + \frac{1}{2} \binom{n}{1} + \frac{1}{3} \binom{n}{2} + ... + \frac{1}{n+1} \binom{n}{n} = \frac{2^{n + 1} - 1}{n + 1}$. I have to prove the identity:
$$\binom{n}{0} + \dfrac{1}{2} \binom{n}{1} + \dfrac{1}{3} \binom{n}{2}
+ ... + \dfrac{1}{n+1} \binom{n}{n} = \dfrac{2^{n + 1} - 1}{n + 1}$$
First ... | We can find the probabilistic interpretation of both sides of the equation: Imagine having $n+1$ people. We then select every possible nonempty subset of them (there are exactly $2^{n+1}-1$ such subsets) and then select one person from it uniformly at random. Both sides are equal to the expected value of the number of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3557675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Primes in the form $n^2+4$ For what numbers $n$ is $n^2+4$ prime?
Obviously $n$ has to be odd number (if $n$ is even, then $n^2+4>2$ is also even, so it couldn't be a prime number).
So far I recognize that the latest digit of $n$ is not 9 and 1, because then $n^2\equiv 1 \pmod{5}$, so $n^2+4\equiv 0 \pmod{5}$.
Unfortu... | A partial result is that if $n>1$ is itself a square, then $n^2+4$ will not be a prime. For if $n=m^2$ then:
$$n^2+4=m^4+4=(m^2+2m+2)(m^2-2m+2)$$
and since $m=\sqrt{n}>1$ both factors will be $>1$.
Also, if $n=ax+b$ with $a>1$ and $b^2+4=ka$ for some $k$, then $n^2+4$ will not be a prime since:
$$n^2+4=(ax+b)^2+4=a^2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
An interesting question of algebraic manipulation If $a+b+c+d+e=0$ then prove that $a^3+b^3+c^3+d^3+e^3=3(abc + abd + abe + acd + ace + ade + bcd + bce + bde + cde)$ Extend this argument to n integers such that if $a_1+a_2+a_3+\cdots a_n=0$ then $$a_1^3+a_2^3+a_3^3\cdots a_n^3=3\left(\sum_{i>j>k} a_ia_ja_k\right)$$
My... | Because $$\left(\sum_{k=1}^na_k\right)^3=\sum_{k=1}^na_k^3+3\sum_{k=1}^na_k\sum_{1\leq i<j\leq n}a_ia_j-3\sum_{1\leq i<j<k\leq n}a_ia_ja_k.$$
We can get it by the following way.
Firstly, it's obvious that $$\left(\sum_{k=1}^na_k\right)^3=\sum_{k=1}^na_k^3+3\sum_{k=1}^na_k\sum_{1\leq i<j\leq n}a_ia_j+K\sum_{1\leq i<j<k\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3567992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Divisibility Trick for 11 I am trying to prove the 11 alternating sum divisibility trick. I know that $10\equiv -1\pmod{11}$ so for every power of $10$ in a number, we should be able to substitute in $(-1)$ like so:
$$ a(10^n) + b(10^{(n-1)}+\cdots+ c(10) + d \equiv a(-1)^{n} + b(-1)^{n-1}+\cdots+c(-1)+d\pmod{11}$$
Wha... | Why are you shifting the variable names. You can but its confusing. It=n $a(-1)^2 + b(-1)^1+c=a-b+c$ you have $a,c$ in the odd positions and $b$ in the even position.
But with $a(-1)^3+b(-1)^2+c(-1)^1 + d= -a+b-c+d$ you have $a,c$ switching to the even position, while $b$ switches to the odd. And $d$ is new in the o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3570028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding the MacLaurin series of $\frac{x+3}{2-x}$ I did:
$$\frac{x+3}{2-x} = \frac{x}{2-x}+\frac{3}{2-x} = x(\frac{1}{2-x})+3(\frac{1}{2-x}) = \\
= \frac{x}{2}(\frac{1}{1-\frac{x}{2}})+\frac{3}{2}(\frac{1}{1-\frac{x}{2}}) = \\
= \frac{x+3}{2}\sum(\frac{x}{2})^n = ...?$$
The answer my professor got was $\frac{(3 + x)}{(... | Use $\frac{1}{1-t} = 1+t+t^2+t^3+\>...$ to get,
$$\frac{3 + x}{2 - x} = -1 + \frac{\frac52}{1 - \frac x2}
= -1+\frac52\left(1+ \frac x2 + \frac {x^2}{2^2} + \frac {x^3}{2^3}+\> …\right)= \frac{3}{2} + \sum_{n=1}^∞ \frac{5x^n}{ 2^{n+1}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3570936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Given positives $a, b, c$ such that $a + b + c = 3$, prove that $\sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} \le \frac{1}{2}$.
Given positives $a, b, c$ such that $a + b + c = 3$, prove that $$\frac{1}{c^2 + 4a^2 + b^2} + \frac{1}{a^2 + 4b^2 + c^2} + \frac{1}{b^2 + 4c^2 + a^2} \le \frac{1}{2}$$
We have that $$a^2 + 4b^2 + c... | We need to prove that
$$\sum_{cyc}\frac{1}{b^2+c^2+4a^2}\leq\frac{9}{2(a+b+c)^2}$$ or
$$\sum_{cyc}(2a^6-4a^5b-4a^5c+13a^4b^2+13a^4c^2-4a^4bc-12a^3b^3-12a^3b^2c-12a^3c^2b+20a^2b^2c^2)\geq0$$ or
$$\sum_{cyc}(a-b)^2(2c^4+2(a^2-4ab+b^2)c^2+a^4-2a^3b+4a^2b^2-2ab^3+b^4)\geq0$$ for which it's enough to prove that:
$$(a^2-4ab+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$k\cos\beta \sin\alpha-\sin\beta+d\sqrt{1+k^2}=0$ ，Find $\beta$ $k\cos\beta \sin\alpha-\sin\beta+d\sqrt{1+k^2}=0$. Find $\beta$.
$K$ and $d$ are constant.
$d∈[0,\pm1],k∈[0,\pm\infty)$.
This is an equation in a spherical coordinate system. Where $\alpha$ is longitude and $\beta$ is latitude.
| \begin{align}
& A\cos\beta + B\sin\beta \\[10pt]
= {} & \sqrt{A^2+B^2} \left( \frac A{\sqrt{A^2+B^s}} \cos\beta + \frac B {\sqrt{A^2+B^2}} \sin\beta \right) \\[10pt]
= {} & \sqrt{A^2+B^2} \Big( \sin\gamma \cos\beta + \cos\gamma\sin\beta\Big) \\
& \quad \text{where } \tan\gamma = A/B \\[10pt]
= {} & \sqrt{A^2+B^2} \, \s... | {
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"url": "https://math.stackexchange.com/questions/3574345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove $\cos x +\cos y = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$ Prove that $$\cos(x) + \cos(y) = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$$ holds true for any $x, y \in \mathbb{R}$.
Even though I managed to prove its brother $\sin(x) + \sin(y)$, I haven't been able to tackle this one.
Important identities needed for th... | $$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$$
$$\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$$
Adding the two,
$$\cos(a+b)+\cos(a-b) = 2\cos(a)\cos(b)$$
Setting $a = \frac{x+y}{2}$ and $b = \frac{x-y}{2}$, we get
$$\cos(x) + \cos(y) = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to find the range of $x^{T}Ax$
Let $A=
\begin{pmatrix}
1 & -2 & -1 \\
-2 & 1 & 1 \\
-1 & 1 & 4
\end{pmatrix}, x =
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}$ with $x^{T}x = 1$, what's the range of $x^TAx$ subject to $(1,1,1)x = 0$?
We can diagonize $A$ to $\begin{pmatrix}
-1 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 ... | The intersection of the sphere $x^T x = 1$ with the plane $(1,1,1) \; x = 0$ is a circle. It may be parameterized as
$$ \pmatrix{x_1\cr x_2\cr x_3\cr} = \cos(\theta) \pmatrix{1/\sqrt{2}\cr 0 \cr -1/\sqrt{2}\cr} + \sin(\theta) \pmatrix{1/\sqrt{6}\cr -2/\sqrt{6}\cr 1/\sqrt{6}\cr} $$
Compute $x^T A x$ as a function of $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3577333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Use epsilon-delta definition of limit to establish the following: $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$ I understand that my solution here is probably not the most efficient (My professor's solution is "cleaner") but it is how my mind attacked the problem. I have been losing lots of points for m... | You need to relate $\displaystyle \left| \frac{1}{2+\sqrt{x}} - \frac{1}{3} \right|$ with $|x-1|$ in a proper way:
$\displaystyle \left| \frac{1}{2+\sqrt{x}} - \frac{1}{3} \right| = \left| \frac{1 - \sqrt{x}}{3(2+\sqrt{x})} \right| = \left| \frac{1-x}{3(1+\sqrt{x})(2+\sqrt{x})} \right| \leq \frac{|x-1|}{3\cdot1\cdot2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Confused about positive and negative signs: Find the value of $\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$. Without tables or a calculator, find the value of $\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$.
I do not understand how the positive/negative signs are obtained as shown in the book; is there... | Hint
$a-b=4$
$a+b=2\sqrt5$
$ab=1$
$a^3-b^3=(a-b)^3+3ab(a-b)=?$
$a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)((a-b)^2+ab)=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Show $\sum_{k=0}^{n}2^{n-k}\binom{a+k}{k}\frac{a-k}{a+k}=\binom{a+n}{n}$ How it can be shown that:
$$\sum_{k=0}^{n}2^{n-k}\binom{a+k}{k}\frac{a-k}{a+k}=\binom{a+n}{n}$$
Where $a \ne 0$
My try:
$$\sum_{k=0}^{n}2^{n-k}\binom{a+k}{k}\frac{a-k}{a+k}=\sum_{k=0}^{n}2^{n-k}\binom{a+k}{a}\frac{a-k}{a+k}$$
$$=\frac{1}{a}\sum... | I know you would prefer a combinatorial proof and thinks that induction does not tell us anything, but since you would like to know an algebraic proof, I'll just include an inductive proof for completion's sake.
Base case $n = 0$:
$$
\sum_{k=0}^0 2^{-k}{a + k \choose k}\frac{a - k}{a + k} = 2^0 {a \choose 0}\frac{a}{a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3579670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Find center of circle, which tangent to $y=2$ at $(3,2)$ and $y=-x\sqrt 3 +2$ Using $$r=\frac{|Ax + By + c|}{\sqrt{A^2+B^2}}$$
I get $$b=2+a\sqrt3$$ and stuck,because the options are on numbers. What do i do next?
| Suppose $C=(a,b)$ is the center of the circle.
Then $a=3$ because the circle is tangent to $y=3$ at point $(3,2)$.
Also the distance of $C$ from both lines $y=2$ and $y=-x\sqrt{3}+2$ is same:
$$|b-2|=\frac{|b+3\sqrt{3}-2|}{2}$$
and note that we have two solutions for $b<2$ and $b>2$.
can you proceed?
Edit: If $b>2$,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can I solve this other than using Newton's sums or Newton's identities?
$\begin{cases}
&x + y + z = 12\\& x^2 + y^2 + z^2 = 12
\\ & x^3 +y^3 + z^3 = 12
\end{cases}$
If $x,$ $y,$ and $z$ satisfy the system of equations above, what is the value of $x^4+y^4+z^4?$
People told me that this can be solved by using Newton... |
People told me that this can be solved by using Newton's sums or Newton's identities which I don't know how. Does anybody know how to do that or any other method to solve this?
We can solve this by doing this method:
$\begin{align} (x+y+z)^2 &\rightarrow x^2 + y^2 + z^2 + 2(xy+xz+yz) = 12^2 \\
&\rightarrow 12 + 2(xy+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Interpreting an olympiad inequality as a convex function We need to prove the following inequality:
$$\frac{y^3xz}{x^3(xy+z^2)}+\frac{z^3xy}{y^3(x^2+yz)}+\frac{x^3yz}{z^3(xz+y^2)}\geq \frac{3}{2}$$
This equation is convex in each of the variables $x,y,z$. Moreover, its minimum seems to be at the origin (the equation ... | $$\Longleftrightarrow \sum\dfrac{y^3z}{x^2(xy+z^2)}\ge\dfrac{3}{2}$$
take
$$a=\dfrac{x}{z},b=\dfrac{y}{x},c=\dfrac{z}{y}$$where $abc=1$
$$\Longleftrightarrow \sum \dfrac{b^2}{a+b}\ge \dfrac{3}{2}$$
use Cauchy-Schwarz inequality we have
$$\sum\dfrac{b^2}{a+b}\ge\dfrac{(a+b+c)^2}{\sum(a+b)}=\dfrac{a+b+c}{2}\ge\dfrac{3}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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The value of the determinant
Calculate the value of determinant:
$$D = \begin{vmatrix}
0 & 1 & 2 & ... & 2020 \\
1 & 0 & 1 & ... & 2019 \\
2 & 1 & 0 & ... & 2018 \\
... & ... & ... & ... & ... \\
... & ... & ... & ... & ... \\
2019 & 2018 & 2017 & ... & 1 \\
2020 & 2019 & 2018 & ... & ... | Let's do this for a $5 \times 5$ matrix and hopefully it would make sense how to generalize this. The key idea is that when you apply operations of Gaussian Elimination to the matrix,
*
*flipping rows multiplies the result by
*rescaling a row by $k$ rescales the determinant by $k$ as well
*adding multiples of row... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that: $\sin \,\cot^{-1} \cos \,\tan^{-1}x = \sqrt{\dfrac{x^2+1}{x^2+2}} $. Prove that: $\sin \,\cot^{-1} \cos \,\tan^{-1}x = \sqrt{\dfrac{x^2+1}{x^2+2}} $.
This question was taken from Miscellaneous Example of S.L. Loney's Trigonometry.
Since, the question involves 2 parts, LHS and RHS, LHS is completely Trigonom... | Let $y=\tan^{-1}x$. Then $\tan y=x$ and so drawing a right triangle with angle $y$ and sides $1,x,\sqrt{1+x^2}$, we find $\cos y=\dfrac{1}{\sqrt{1+x^2}}$.
Now let $t=\cot^{-1}(\cos y)$. Then $\cot t=\cos y=\dfrac{1}{\sqrt{1+x^2}}$, so drawing a right triangle with angle $t$, we find the hypotenuse is $\sqrt{2+x^2}$, so... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluate $\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p}{n^{p+1}}$
Evaluate
$\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p}{n^{p+1}}$ using Stolz-Cesaro theorem .
Now this is my attempt :$\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p+(2n)^p-1^p-3^p-\dots-(2n-1)^p}{(n+1)^{p+1}-n^{p+1}}$=$\l... | Let $ n $ be a positive integer greater than $ 1 \cdot $
Since $ \sum\limits_{k=0}^{n-1}{\left(2k+1\right)^{p}}=\sum\limits_{k=1}^{2n}{k^{p}}-\sum\limits_{k=1}^{n}{\left(2k\right)^{p}}=\sum\limits_{k=1}^{n}{k^{p}}+\sum\limits_{k=n+1}^{2n}{k^{p}}-2^{p}\sum\limits_{k=1}^{n}{k^{p}}$ $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve this equation : $\tan^{-1} \frac{x+1}{x-1} + \tan^{-1} \frac{x-1}{x} = \tan^{-1} (-7)$ Solve this equation : $\tan^{-1} \frac{x+1}{x-1} + \tan^{-1} \frac{x-1}{x} = \tan^{-1} (-7)$
This was an exam question, my try was as follows:
$$ \tan^{-1} \frac{x+1}{x-1} = \tan^{-1} (-7) - \tan^{-1} \frac{x-1}{x} $$
Now, assu... | Take tangents on the both sides, you'll get:
$$ \dfrac{\tan \left [ \tan^{-1} \frac{x+1}{x-1} \right] + \tan \left [ \tan^{-1} \frac{x-1}{x} \right]}{ 1 - \tan \left [ \tan^{-1} \frac{x+1}{x-1} \right] \tan \left [ \tan^{-1} \frac{x-1}{x} \right]} = -7 $$
$$ \frac{2x^2 - x + 1}{1-x} = -7 $$
Upon solving that equation,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If the equation of the curve on the reflection of the ellipse $\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1$ about the line $x-y-2=0$ is ...
If the equation of the curve on the reflection of the ellipse $\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1$ about the line $x-y-2=0$ is $16x^2+9y^2+k_1x-36y+k_2=0$, then find $k_1$ and $k_2... | Consider the line $y=x-2$;
New coordinates : $X=x$; and $Y=y+2$;
Reflect about $Y=X.$
The given curve expressed in $X,Y:$
$(X-4)^2/4^2+(Y-5)^2/3^2=1$;
The reflected curve:
$(Y-4)^2/4^2+(X-5)^2/3^2=1$;
Back to $x,y$:
$(y-2)^2/4^2+(x-5)^2/3^2=1$;
$9(y-2)^2+16(x-5)^2=16\cdot 9$;
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $ \int_{\pi/4}^{\pi/2} \frac{2\sin x+\cos x}{\sin x+2\cos x}\,dx$
Evaluate
$$ \int_{\pi/4}^{\pi/2} \frac{2\sin x+\cos x}{\sin x+2\cos x}\,dx,$$
My attempt : $u=\tan\frac{x}{2} \rightarrow x=2\arctan(u) \rightarrow \frac{2}{1+u^2}du=dx$
$$\sin x= \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}=\frac{2u}{1+u^2... |
\begin{align} I&=\int_{\pi/4}^{\pi/2} \frac{2\sin x+\cos x}{\sin x+2\cos x}\,dx\tag{1}\label{1}.\end{align}
\begin{align}
I&=
\int_{\pi/4}^{\pi/2}
\frac{2\cot x+1}{\cot x+2}\,dx
\overset{\color{blue}{t=\cot x}}
{=}
\int_0^1
\frac{2+t}{(1+2t)(1+t^2)}
\, dt
\tag{2}\label{2}
\\
&=
\int_0^1
\frac{2+t}{(1+2t)(1+t^2)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find the function $f(x)$ whose graph passes through the point $(0,\frac{4}{3})$ and whose derivative is: $f'(x) = x \sqrt{16-x^2}$ Section 5.2
Can somebody verify this solution for me?
Find the function $f(x)$ whose graph passes through the point $(0,\frac{4}{3})$ and whose derivative is:
$f'(x) = x \sqrt{16-x^2}$
So ... | Your mistake is introduced here:
$\rightarrow \frac{4}{3} = \frac{-1}{3} (16)^{\frac{3}{2}}+C$
$\rightarrow 4= -(16)^{\frac{3}{2}}+C$
What you did here is multiply by $3$ on both sides, I believe. You also went ahead and let the factor of $3$ be "absorbed" into the constant $C$ (after all, $C$ is a constant, and ther... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Derivative of quadratic expression involving outer product of estimate I was trying to calculate the derivative with respect to $x$ of the following expression:
$\frac{1}{2}\left|\left|\left(\frac{Axx^TA^T}{x^TA^TAx}-I\right)b\right|\right|^2_2$
Here, $x$, $b$ are vectors and $A$ is a matrix and $I$ the identity matrix... | We can rewrite the function as
\begin{align*}
f(x) ={}& \frac{1}{2}b^\top\left( \frac{Axx^\top A^\top}{x^\top A^\top Ax} - I \right)^\top\left( \frac{Axx^\top A^\top}{x^\top A^\top Ax} - I \right) b \\
={}& \frac{1}{2}b^\top \left( \frac{Axx^\top A^\top}{x^\top A^\top Ax} - I \right)^2 b \\
={}& \frac{1}{2}b^\top \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3593094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding the $2008$th derivative of $\sin^6(\frac{x}{4}) + \cos^6\left(\frac{x}{4}\right)$ at $x = 0$?
Let $f(x) = \sin^6 \frac{x}{4} + \cos^6 \frac{x}{4}$ for all real numbers $x$. Determine $f^{(2008)}(0)$ (ie. $f$ differentiated $2008$ times and then evaluated at $x=0$)
Hi,
I have tried to find the $2008$th deriva... | Several among the list of trigonometric identities to consider are
$$\sin^2(x) = 1 - \cos^2(x) \tag{1}\label{eq1A}$$
$$\sin(2y) = 2\sin(y)\cos(y) \implies \sin(y)\cos(y) = \frac{\sin(2y)}{2} \tag{2}\label{eq2A}$$
$$\cos(2y) = 1 - 2\sin^2(y) \implies \sin^2(y) = \frac{1 - \cos(2y)}{2} \tag{3}\label{eq3A}$$
Using these i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3597173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find positive integer solutions for $n^4(n+1)+1=7^m$ This is the statement:
Find all pairs of natural numbers $(n,m)$ that satisfies equation $$n^4(n+1)+1=7^m$$
I found one solution (2,2) and believe there is no more solutions. I'm beginner in number theory and don't know how to even start this. Can anyone help?
| Let $d=\gcd(n^3-n+1,n^2+n+1)$.
Since $d\mid n^3-n+1$ and $d\mid n^2+n+1 \implies d\mid n^3-1$, so $$d\mid (n^3-1)-(n^3-n+1)= n-2$$
So $$d\mid (n^2+n+1) -(n-2)- (n^2-4)=7$$ So $d=1$ or $d=7$.
Case 1 $\boxed{d=1}$:
*
*$n^2+n+1= 7^m$ and $n^3-n+1=1$ (so $n=1$ which does not work) or
*$n^2+n+1= 1$ (impossible) and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3597979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve quadratic matrix equations of the form $A^T B A=C$? I want to solve the following matrix equation
$$A^T
\begin{pmatrix}
10 & 0 & 0 \\
0 & 20 & 0 \\
0 & 0 & 25 \\
\end{pmatrix} A =
\begin{pmatrix}
\frac{35}{2} & \frac{5 \sqrt{3}}{2} & 0 \\
\frac{5 \sqrt{3}}{2} & \frac{25}{2} & 0 \\
0 & 0 & 25 \\
\en... | The matrix $C$ is real symmetric so that you can diagonalize it as
$$C=P^TDP$$ where $P$ is orthogonal (hence its transpose is its inverse).
If $D$ matches the given $B$, you have the solution. If it does not, there is no solution, as the Eigenvalues are uniquely defined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3601237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Evaluate $\int_{0}^{\frac{\pi}{4}} \tan^{-1}\left(\sqrt{\frac{\cos 2x}{2\cos^2 x}}\right)\:dx$ Evaluate $$I=\int_{0}^{\frac{\pi}{4}} \tan^{-1}\left(\sqrt{\frac{\cos 2x}{2\cos^2 x}}\right)\:dx$$
I tried with substitution:
$$\frac{\cos 2x}{2\cos^2 x}=\tan^2 y$$ So we have
$$\sec^2 x=2-2\tan^2 y$$ Differentiating we get:
... | \begin{align}J&=\int_{0}^{\frac{\pi}{4}} \tan^{-1}\left(\sqrt{\frac{\cos 2x}{2\cos^2 x}}\right)\,dx\\
&=\int_{0}^{\frac{\pi}{4}}\left(\int_0^1 \frac{\sqrt{\frac{2\cos^2 x}{\cos(2x)}}}{a^2+\frac{2\cos^2 x}{\cos(2x)}}\,da\right)\,dx\\
&\overset{u=\tan x}=\int_0^1 \int_0^1 \frac{\sqrt{2}\sqrt{1-u^2}}{\left(a^2(1-u^2)+2\ri... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $(x,y,z)$ is a Pythagorean triple such that each of $x,y,z$ can be written as sum of two squares then prove that $180|xyz$ This is a problem based on Pythagorean Triples.
If $(x,y,z)$ is a Pythagorean triple such that each of $x,y,z$ can be written as sum of two squares then prove that $180|xyz$
Any ideas of how to... | $$x^2= y^2+z^2\;\;\;(*)$$
*
*Divisibilty with $4$:
*
*If $y$ and $z$ are even then we are done.
*If $y$ and $z$ are odd then $x $ is even so $x=2a$ and $y=2b+1$ and $z=2c+1$ so $$4a^2 = 4b^2+4b+4c^2+4c+2$$ which is impossible
*If $y=2b+1$ and $z=2c$ then $x=2a+1$ so $$4a^2+4a+1= 4b^2+4b+1+4c^2$$ so $$\unde... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Calculate $ \int_0^1{\frac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx$
Given a function $f(x)$ that is differentiable on $\left[0; 1\right]$
satsifies: $$ f(1) = 1 $$ $$ f(x)f(1-x) = e^{x^2 - x} $$
Calculate: $$ \int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx $$
Attempt number 1:
Using integration by parts... | Denoting $ I=\int\limits_{0}^{1}{x\left(1-x\right)\ln{\left(f\left(x\right)\right)}\,\mathrm{d}x} $, we have :
\begin{aligned} \int_{0}^{1}{\frac{\left(2x^{3}-3x^{2}\right)f'\left(x\right)}{f\left(x\right)}\,\mathrm{d}x}&=\left[\left(2x^{3}-3x^{2}\right)\ln{\left(f\left(x\right)\right)}\right]_{0}^{1}-6\int_{0}^{1}{\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3605371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\ x^3+px-q=0 $ has three roots $a$, $b$, $c$ then find an equation with roots $a+b$, $b+c$ and $c+a$ The roots are not $a$, $b$, $c$ but $\alpha$, $\beta$, and $\gamma$. I wrote $a$, $b$, $c$, due to space constraints.
I know that I have to start with
$$\ \alpha + \beta + \gamma = 0 $$
$$\ \alpha + \beta = -\gamma,... | Well, you rather need to compute $$\begin{align}(a+b)+(b+c)+(a+c)&=2(a+b+c)\\&=0\\(a+b)(b+c)+(a+b)(c+a)+(b+c)(c+a)&=a^2 + 3 a b + 3 a c + b^2 + 3 b c + c^2\\&=(a+b+c)^2+(ab+bc+ca)\\&=p\\(a+b)(b+c)(c+a)&=a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2\\&=(ab+bc+ca)(a+b+c)-abc\\&=q\end{align}$$
which, by Vieta's ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find delta for the limit I'm having difficulty to solve this problem:
I know that ${\displaystyle \lim_{x\to a} f(x) = L}$ means for every $\varepsilon > 0$, there exists $\delta > 0$ such that $|f(x) - L| < \varepsilon$ whenever $0 <|x-a| < \delta$.
I need to find $\delta$ when $\varepsilon = 0.001$ for ${\displayst... | $$| \frac{1}{\sqrt{x^2+1}} - \frac{1}{\sqrt 2}|$$ $$=\frac{|\sqrt{x^2+1}-\sqrt 2|}{\sqrt2 \sqrt{x^2+1}}.$$ Let $0<x-(-1)< \delta<1.$ We shall look at the numeraor and denominator of the expression above. First we deal with the denominator: $\sqrt{x^2+1}>1,$ so $$\frac{1}{\sqrt2 \sqrt{x^2+1}}<\frac{1}{\sqrt 2}$$ Now loo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let $A,B,C$ be square matrices. Calculate $(A+B+C)^3$
Let $A,B,C$ be square matrices. Calculate $(A+B+C)^3$
I used the fact that
$$(A+B+C)^2=A^2+AB+AC+BA+B^2+BC+CA+CB+C^2$$
and multiply this with $(A+B+C)$ and I got
$$A^3+A^2B+A^2C
+ABA+AB^2+ABC
+ACA+ACB+AC^2
+BA^2+BAB+BAC
+B^2A+B^3+B^2C
+BCA+BCB+BC^2
+CA^2+CAB+CAC
+... | The Binomial and multinomials work as usual only for commutative matrices. So this trinomial can be simplifies only if $AB=BA$,~ $BC=CB$ and $CA=AC$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}$ Let $1>a>0$, $1>b>0$, $1>c>0$ and $a+b+c=1$. Prove that
$$
\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}.
$$
I saw the following solution. Let $x=\frac{2}{1-a^2}$, $y=\frac{2}{1-b^2}$, $z=\frac{2}{1-c^2}$, ... | No it’s wrong. Hint : use the CS inequality but the “ Angel “ form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to evaluate a sum of square roots rounded to the nearest integer? This is (adapted from) Problem 11 from the 2007 AIME :
For each positive integer $p$, let $b(p)$ denote the unique positive integer $k$ such that $|k-\sqrt{p}|<\frac{1}{2}$. For example, $b(6)=2$ and $b(23)=5$. Find $S=\sum_{p=1}^{2007} b(p)$.
I st... | As you notice, for positive integers $p$ and $k$, $b(p)=k$ if and only if $k-\dfrac12< \sqrt{p} < k+\dfrac12$, which is equivalent to $$(k-1)^2+(k-1)=k^2-k<p\leq k^2+k\,.$$
Now, $44\cdot 45=1980<2007<2070=45\cdot 46$, we see that $b(P)=K$ with $P:=2007$ and $K:=45$.
Rewrite the required sum $S:=\sum\limits_{p=1}^P\,b(p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Seeking alternative methods for $\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$ I've solved this one by first tackling,
$$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$
But i'd like to know other ways to solve it since the way i did it was a bit lengthy and not that straight... | $$\int_0^1\frac{\ln(x^2-x+1)}{x(1-x)}\ dx=\underbrace{\int_0^1\frac{\ln(x^2-x+1)}{1-x}\ dx}_{x\to 1-x}+\int_0^1\frac{\ln(x^2-x+1)}{x}\ dx$$
$$=2\int_0^1\frac{\ln(x^2-x+1)}{x}\ dx=2\underbrace{\int_0^1\frac{\ln(x^3+1)}{x}\ dx}_{x^3\to x}-2\int_0^1\frac{\ln(1+x)}{x}\ dx$$
$$=-\frac43\int_0^1\frac{\ln(1+x)}{x}\ dx=-\frac4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 0
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Prove that $DE\perp EF$.
Question: Point $D$ lies inside $\Delta ABC$ such that $\angle DAC=\angle DCA=30^\circ$ and $\angle DBA = 60^\circ$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF=2FC$. Prove that $DE\perp EF$.
My approach: Let $\angle CDF=\alpha$ and $\angle EDC=\beta$.
... |
Let $M$ and $G$ be the midpoints of $AC$ and $AF$, respectively. If $h$ is the homothety about $C$ with dilatation ratio $2$, then we see that
$$h(F)=G\,,\,\,h(M)=A\,,\text{ and }h(E)=B\,.$$
Let $\omega$ and $\Omega$ denote the circumcircles of the triangles $DFM$ and $DGA$, respectively.
First, if $D'$ is the refl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Expressions for system of equations in a neighborhood of the origin, $x' = y+y^2 - 2xy + x^2$, $y'=x+y^2 - 2xy + x^2.$ Do you guys agree with my solution to the following problem? Please provide feedback if possible, thanks!
Find expressions for the local stable and local unstable manifolds for the following system of ... | $$x'=y+(x-y)^2 $$
$$y'=x+(x-y)^2$$
HINT :
$$y'-x'=x-y\quad\implies\quad \frac{y'-x'}{y-x}=-1\quad\implies\quad y-x=c_1e^{-t}$$
$$y=x+c_1e^{-t}$$
$y'=x+(c_1e^{-t})^2=x'-c_1e^{-t}$
$$x'-x=c_1e^{-t}+c_1^2e^{-2t}$$
$$x(t)=c_2e^t-\frac{c_1}{2}e^{-t}-\frac{c_1^2}{3}e^{-2t}$$
$$y(t)=c_2e^t+\frac{c_1}{2}e^{-t}-\frac{c_1^2}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3616630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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A simpler non-calculator proof for $17^{69}<10^{85}$ I have proved that $17^{69}<10^{85}$ by using the following inequalities:
$x<\exp\left(\dfrac{2(x-1)}{x+1}\right)$ for all $x\in \left]-1,1\right[$
and $x<{\mathrm e}^{x-1}$ for all $x\in \left] 1,+\infty \right[$, but I am looking for a simpler non-calculator proof.... | Proof that $17^{69} < 10^{85}$ without using calculator, there are many ways this can be done you'll need to understand number theory, so I'll highlight a few examples
$$17^{69} < 10^{85}$$
$$17^{69} < 10^{17×5}$$
$$17^{1/17} < 10^{5/69}$$
Remember $$\lim_{n→∞} \sqrt[n]{n} = 1$$
Meaning the value of 17^{1/17} is betwee... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3623860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Find $\min$ for$ f(x) = (x + a + b)(x + a - b)(x - a + b)(x - a - b)$ I'm trying to find $minf(x)$ for $f(x) := (x + a + b)(x + a - b)(x - a + b)(x - a - b)$, where $a, b \in \mathbb{R},$ using inequalities.
For example, i can find $maxf(x)$, using AM-GM ineq:
$$\sqrt[4]{(x + a + b)(x + a - b)(x - a + b)(x - a - b)})^... | For a fixed nonzero $x,$ there is no minimum and no maximum.
To exceed any upper bound, take $a = t$ and $b=0$ for some large $t.$ The original polynomial becomes $t^4 - 2 x^2 t^2 = t^2 (t^2 - 2 x^2) \; . \;$ As soon as $t^2 > 2 x^2 + 1,$ the polynomial is larger than $t^2,$ and we may take $t$ as large as we like. In... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3629941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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For $a,b,c\ge0$, show $a^2 + b^2 + c^2 = 3$ implies $(a^3 + b^3 + c^3)^2 \geq 3 + 2(a^4 + b^4 + c^4)$. $a,b,c$ are nonnegative real numbers. If $a^2 + b^2 + c^2 = 3$,
prove that $$(a^3 + b^3 + c^3)^2 \geq 3 + 2(a^4 + b^4 + c^4).$$
I tried using $(a^3 + b^3 + c^3)^2 >= 3(a^3 + b^3 + c^3)abc$ to make it maybe better.
I ... | We need to prove that:
$$(a^3+b^3+c^3)^2\geq\frac{(a^2+b^2+c^2)^3}{9}+\frac{2(a^4+b^4+c^4)(a^2+b^2+c^2)}{3}$$ or
$$\sum_{cyc}(2a^6-9a^4b^2-9a^4c^2+18a^3b^3-2a^2b^2c^2)\geq0,$$ which is true by AM-GM:
$$\sum_{cyc}(2a^6-9a^4b^2-9a^4c^2+18a^3b^3-2a^2b^2c^2)\geq$$
$$\geq\sum_{cyc}(2a^6-9a^4b^2-9a^4c^2+16a^3b^3)=$$
$$=\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3631024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculus distance maximizing
Point $(x, y)$ satisfies the inequality $x^4 + y^2 \leqslant 1$. Determine the largest possible distance from the origin for $(x, y)$.
So the largest distance will result when $x^4+x^2=1$.
The distance from the origin for point $(x,y)$ is $d=\sqrt{x^2+y^2}$, but from the given inequality ... | Attempt:
Constraint: $x^4+y^2 \le 1$; $y^2\le 1-x^4;$
Maximize : $x^2+y^2\le x^2+1-x^4=$
$-(x^4-x^2)+1=$
$-[(x^2-1/2)^2-1/4]+1=-(x^2-1/2)^2+5/4\le 5/4$.
Maximum attained at $x^2=1/2$;
$y^2=5/4 -1/2=3/4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3631719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Show that $\lim _{x\to \infty }\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right) =1/2$ I don't know how to start. Is it simple algebraic manipulation where,
if, let $a=\sqrt{x+\sqrt{x+\sqrt{x}}} $
and, $b=\sqrt{x}$
the above equation can be manipulated as
$\implies a-b$$.\:\frac{a+b}{a+b}=\frac{a^2-b^2}{a+b}$
giving, $\... | $a=\sqrt{x+\sqrt{x+\sqrt{x}}}
$.
$a^2
=x+\sqrt{x+\sqrt{x}}
$
and
$(\sqrt{x}+\frac12)^2
=x+\sqrt{x}+\frac14
$.
$\sqrt{x+\sqrt{x}}^2
=x+\sqrt{x}
$
and
$(\sqrt{x}+\frac14)^2
=x+\frac12\sqrt{x}+\frac1{16}
\lt x+\sqrt{x}
$
so
$a > \sqrt{x}+\frac12$.
Numerically,
it looks like
$a < \sqrt{x}+\frac12+\frac1{8\sqrt{x}}
$,
so le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3632164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Minimize $|a-1|^3+|b-1|^3$ with constant product $ab=s$ Let $0<s$, and define
$$ F(s):=\min_{a,b \in \mathbb{R}^+,ab=s} \left(|a-1|^3+|b-1|^3\right). $$
I would like to find proofs for the claim
$$
F(s)=\begin{cases}
1 - 3 s - 2s^{3/2}=F\big(a(s),b(s)\big), &\text{ if } 0<s\le1/9, \\
2 + 6 s - 2(3 + s)s^{1/2}=F(\sqrt ... | Since you have established that the signs of $a - 1$ and $b - 1$ are identical, one approach would be to find the extrema of $(a - 1)^3 + (b-1)^3$ and then remove those where $a - 1$ and $b - 1$ are of opposite signs.
Plugging in $b = {s \over a}$ directly, you are finding an extremum of the expression
$$f(a) = (a - 1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3633097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.