Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Does $\int_0^\infty\frac{\ln(1+x)}{x(1+x^n)}dx$ have a general form?
Does $$I_n=\int_0^\infty\frac{\ln(1+x)}{x(1+x^n)}dx$$ have a general form?
I tried to evaluate some small $n$s.
For $n=1$, $I_1$ is obviously $\frac16\pi^2$.
For $n=2$, see here.
$I_2=\frac5{48}\pi^2$.
For $n=3$, I put it in Mathematica and get $$\... | $$\begin{aligned}
I_n &= \int_0^1 \frac{\ln(1+x)}{x(1+x^n)}dx + \int_1^\infty \frac{\ln(1+x)}{x(1+x^n)}dx \\
&= \int_0^1 \frac{\ln(1+x)}{x(1+x^n)}dx + \int_0^1 \frac{x^n \ln(1+x)}{x(1+x^n)}dx - \int_0^1 \frac{x^{n-1} \ln x}{1+x^n}dx \\
&= \int_0^1 \frac{\ln(1+x)}{x}dx + \frac{1}{n}\int_0^1 \frac{\ln(1+x^n)}{x}dx \\
&= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the limit of the sequence: $\lim_{n_\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}$ Evaluate the limit of the sequence:
$$\lim_{n\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}$$
My try:
Stolz-cesaro: The li... | Note that
$$
\frac{\sqrt{(n-1)!}}{\prod_{k=1}^n\big(1+\sqrt{k}\big)}=\frac{1}{1+\sqrt{n}}
\prod_{k=1}^{n-1}\frac{\sqrt{k}}{1+\sqrt{k}}<\frac{1}{1+\sqrt{n}}\to 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Limit $\lim_{(x, y) \to (\infty, \infty)} \frac{x+\sqrt{y}}{x^2+y}$ Show whether the limit exists and find it, or prove that it does not.
$$\lim_{(x, y) \to(\infty,\infty)}\frac{x+\sqrt{y}}{x^2+y}$$
WolframAlpha shows that limit does not exist, however, I do fail to conclude so.
$$\lim_{(x, y) \to(\infty,\infty)}\frac... | Since $(x,y)\rightarrow (\infty,\infty),$ assume $x,y>0.$ $$0<\frac{x+\sqrt{y}}{x^2+y}=\frac{x\sqrt {y}\left(\frac{1}{\sqrt y}+\frac 1x\right)}{x\sqrt {y}\left(\frac{x}{\sqrt y}+\frac {\sqrt y}{x}\right)}\leq
\frac{\frac{1}{\sqrt y}+\frac 1x}{2} \rightarrow 0$$
as $\frac ab + \frac ba \geq 2$ for $a,b>0.$ Thus by the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3040482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Calculate the limit $\lim_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)}$. We could use L'Hospital here, because both numerator as well as denominator tend towards 0, I guess. The derivative of the numerator is $$x^2\cdot \left(-\sin\left(\frac{1}{x}\right)\right) \cdot \left( -\frac{1}{x^2}\right) + ... | With the Taylor power series, $\sin x= x+o(x)$
$$\lim_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)}=\lim_{x\to0}{x \cos\left(\frac{1}{x}\right)}=0$$
Because $x\to0$ and $\cos(1/x)$ is bounded from $-1$ to $1$ as $x\to0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integral$\int_{\pi/6}^{\pi/2} \frac{\sin^{1/3} (x)}{\sin^{1/3} (x)+\cos^{1/3} (x)}$ $$\int_{\pi/6}^{\pi/2} \frac{\sin^{1/3} (x)}{\sin^{1/3} (x)+\cos^{1/3} (x)}$$
I tried using the substitution $t^3=\tan x$. Which gives me
$$\int\frac{3t^3}{(t+1)(t^6+1)}$$
How should I proceed?
| I will put you on the path:
\begin{align}
\frac{3x^3}{\left(x + 1\right)\left(x^6 + 1\right)} &= \frac{3x^3}{\left(x + 1\right)\left(x^2 + 1\right)\left(x^2 + \sqrt{3}x + 1\right)\left(x^2 - \sqrt{3}x + 1\right)} \\
&= -\frac{3}{2}\frac{1}{x + 1} - \frac{1}{2}\left[\frac{1}{x^2 + 1} +\frac{x}{x^2 + 1}\right] + \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find the other asymptote of $y=\sqrt{x^2+x}$ The task is to find the asymptotes of $y=\sqrt{x^2+x}$.
I first calculated the limits to infinity and found that $\lim_{x \to \pm}y= \infty$.
Next, to find $m_{1,2}$: $$m_1=\lim_{x \to +\infty}\frac{y}{x}=\frac{\sqrt{x^2+x}}{x}=\sqrt{1+ \frac{1}{x}}=1=m_1$$
and $$m_2=... | You may try also in this way. By definition $y=mx+q$ is an asymptote of $f$ as $x\to \pm\infty$ if $f(x)=mx+q+o(1)$. Now, as $|x|\to +\infty$,
$$\sqrt{x^2+x}=|x|\sqrt{1+1/x}=|x|\left(1+\frac{1}{2x}+o(1/x)\right)\\
=|x|+\frac{|x|}{2x}+o(1).$$
Hence if $x>0$ then
$$\sqrt{x^2+x}=x+\frac{1}{2}+o(1)$$
and for $x<0$
$$\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Divisor of $x^2+x+1$ can be square number? $$1^2+1+1=3$$
$$2^2+2+1=7$$
$$8^2+8+1=73$$
$$10^2+10+1=111=3\cdot37$$
There is no divisor which is square number.
Is it just coincidence? Or can be proved?
*I'm not english user, so my grammer might be wrong
| What about $x=653$, where
$$x^2+x+1=427063=7\cdot (13\cdot 19)^2\,?$$
How did I find this $x$? I first suppose that $x^2+x+1=ay^2$ for some positive integers $a$ and $y$. This means
$$(2x+1)^2-a(2y)^2=-3\,.$$
Let $u:=2x+1$ and $v:=2y$. Then, we are to solve the Pell-type equation $$u^2-av^2=-3\,,$$ where $u,v\in\ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to solve equations of this type My math teacher gave us some questions to practice for the midterm exam tomorrow, and I noticed some of them have the same wired pattern that I don't know if it has a name:
Q1. If: $x - \frac{1}{x} = 3$ then what is $x^2 + \frac{1}{x^2}$ equal to?
The answer for this question is 11, ... | For the first one :
$$\left(x-\frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2} \Leftrightarrow x^2 + \frac{1}{x^2} = \left(x-\frac{1}{x}\right)^2 + 2 \implies x^2+ \frac{1}{x^2} = 11$$
For the second one, observe that :
$$\frac{x}{x+y} + \frac{y}{x+y} = 1 \Rightarrow 5 + \frac{y}{x+y} = 1 \Leftrightarrow \frac{y}{x+y} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Radius of convergence of $\sum\limits_{n=1}^{\infty} \frac{n+2}{2n^2+2} x^n$ I want to determine the convergence of the following series in dependency of $x$:
$\sum\limits_{n=1}^{\infty} \frac{n+2}{2n^2+2} x^n=\frac{3}{4}x+\frac{2}{5}x^2+\frac{1}{4}x^3+\frac{3}{17}x^4+ ... $
How can I solve this?
EDIT:
@Winther said, I... | If the ratio test works, there's no need to check with other tests; in your case you want to compute, for $x\ne0$,
$$
\lim_{n\to\infty}
\left|\,\frac{\dfrac{(n+1)+2}{2(n+1)^2+2}x^{n+1}}{\dfrac{n+2}{2n^2+2}x^n}\,\right|
=\lim_{n\to\infty}\frac{n+3}{n+2}\frac{2(n+1)^2+2}{2n^2+2}|x|=|x|
$$
This limit is $<1$ if and only i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solve the system $x^2(y+z)=1$ ,$y^2(z+x)=8$ and $z^2(x+y)=13$ Solve the system of equations in real numbers
\begin{cases} x^2(y+z)=1 \\ y^2(z+x)=8 \\z^2(x+y)=13 \end{cases}
My try:
Equations can be written as:
\begin{cases}\frac{1}{x}=xyz\left(\frac{1}{y}+\frac{1}{z}\right)\\
\frac{8}{y}=xyz\left(\frac{1}{x}+\frac{1}... | I would substitute $$y+z=a,z+x=b,x+y=c$$ then $$z=\frac{a+b-c}{2}$$ and so on.
Eliminating the variables $$y,z$$ we get for $x$:
$$25 x^9-3828 x^6+109 x^3=-42$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove that $a_n \in [0,2)$
Let $(a_n)_{n \in \mathbb{N}}$ be a sequence, with $a_0=0$, $a_{n+1}=\frac{6+a_n}{6-a_n}$.
Prove that $a_n \in [0,2)$ $\forall n \in \mathbb{N}$
Here's what I did:
I tried to prove this by induction:
Base case:
$0 \leq a_0 (=0) < 2$.
Inductive step:
Suppose that $0 \leq a_n < 2$
So $$\beg... | First, let's rewrite the recurrence formula:
$$a_{n+1}=\frac{6+a_n}{6-a_n}=\frac{a_n-6+12}{6-a_n}=-1+\frac{12}{6-a_n}$$
Now, we have:
$$0\leq a_n<2$$
$$6\geq 6-a_n > 4$$
$$2\leq \frac{12}{6-a_n} < 3$$
$$1\leq -1+\frac{12}{6-a_n}=a_{n+1} < 2$$
Thus, $a_{n+1} \in [1, 2)$ and since $[1, 2) \subset [0, 2)$, $a_{n+1} \in [0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Finding $B^*$, the dual basis Find a basis $B$ for
$$V = \left\{ \left[
\begin{array}{cc}
x\\
y\\
z
\end{array}
\right] \in \mathbb{R}^3 \vert x+y+z = 0\right\}$$ and then find $B^*$, the dual basis for $B$.
The way we learned it was that given a basis, we build a matrix A whose columns is the vectors, find $A^{... | By definition, the dual basis functionals $f_1, f_2$ are given on your basis $\{v_1, v_2\}$ as $$f_1(\alpha_1v_1 + \alpha_2v_2) = \alpha_1, \quad f_2(\alpha_1v_1 + \alpha_2v_2) = \alpha_2$$
Now $$f_1\left(\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right) = f_1\left(\begin{bmatrix} -y-z \\ y \\ z\end{bmatrix}\right) = f_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Limlit of $\frac{x-x\ln(1+x)-\ln(1+x)}{x^3+x^2}$ at $0$ without l'Hôpital or Taylor series I have $f : x \mapsto \frac{\ln(1+x)}{x}$ which derivative is $$\frac{1}{x(1+x)}-\frac{\ln(1+x)}{x^2}$$
I want to find the limit as $x$ goes to $0$ of this derivative. I've tried simplifying the expression to $$\frac{x - x\ln(1+x... | First, you can decompose the first term in your derivative as:
$\frac{1}{x(1+x)} = \frac{1}{x}-\frac{1}{1+x}$.
Second, you can expand the second term using Taylor series:
$\frac{ln(1+x)}{x^2}$
$ = \frac{1}{x^2} \left( x - \frac{x^2}{2} + \frac{x^3}{3} - .. \right)$
$ = \frac{1}{x} - \frac{1}{2} + \frac{x}{3} - $
Puttin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Prove that $ \int\limits_0^\infty\frac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2}\,dx=\dfrac{3\pi}{8} $ Is this conjecture true?
Conjecture:
$$
\int_0^\infty\frac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2}\,dx=\dfrac{3\pi}{8}
$$
I found it myself based on numerical evidence. Need help in analytical proof. Thanks.
| Well, we can just rewrite the integral as: $$\int_0^\infty\frac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2}\,dx=\int_0^\infty \frac{x^2}{e^{2x}\left(\frac{1-x}{1+x}\right)^2+e^{-2x}}\frac{1}{(1+x)^2}dx$$
$$=\int_0^\infty \frac{1}{e^{4x}\left(\frac{1-x}{1+x}\right)^2+1}\frac{1}{e^{-2x}}\frac{x^2}{(1+x)^2}dx =\int_0^\infty \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
calculate flux through surface I need to calculate the flux of the vector field $\vec{F}$ through the surface $D$, where
$$\vec{F} = \left<z, \, y \sqrt{x^2 + z^2}, \, -x \right> \\
D = \{x^2+6x+z^2\le 0 \,| -1\le y \le 0\}.$$
So there should be a cylinder (height on $y$ axis) shifted by 3 units (center $x=-3$), with c... | I'm not exactly sure where the $3\sqrt{3}$ comes from in your result, but there is indeed more than one way to evaluate this problem.
(1) Direct method
Here is some technical information about this method from MIT's open notes, and some visualization for what the flux of a vector field through a surface is. Let the flu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3071218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Seeking methods to solve: $\int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt$ Seeking Methods to solve the following two definite integrals:
\begin{equation}
I_(n) = \int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt \qquad J(n) = \int_0^\infty \frac{\ln^2(t)}{t^n + 1}\:dt
\end{equation}
For $n \in \mathbb{R},\:n \gt 1$
The method I took... | You can use the pole expansions
\begin{align}
\csc^2 (z) &= \sum_{m \in \mathbb{Z}} \frac{1}{[m \pi + z]^2} \, , \, z \in \mathbb{C}\setminus\pi\mathbb{Z} \, , \\
\sec^2 (z) &= \sum_{m \in \mathbb{Z}} \frac{1}{\left[\left(m+\frac{1}{2}\right) \pi + z\right]^2} \, , \, z \in \mathbb{C}\setminus\pi \left(\mathbb{Z} + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3073077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Pythagorean triples where the sum of the two cubes is also a square Are there any Primitive Pythagorean triple solutions $(a,b,c)$ where the sum of the two cubes is also a square? In other words are there coprime $a,b>0 \in \mathbb{N} \;, (a,b)=1$ where $a^2+b^2=c^2$ and $a^3+b^3=d^2$ for some $c,d \in \mathbb{N}$
| Disclaimer: These are some unfinished thoughts I will leave here to work on later, or for others to continue.
Given that $a$ and $b$ are coprime, it follows that $\gcd(a+b,a^2-ab+b^2)$ divides $3$ because
$$\gcd(a+b,a^2-ab+b^2)=\gcd(a+b,3b^2)=\gcd(a+b,3).$$
Suppose towards a contradiction that the gcd equals $3$: Then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3077097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Householder transformations to upper triangular form Let $A=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix}$.
How to transform this matrix with Householder transformations to an upper triangular matrix?
I started like this:
$\alpha_1=$sgn$(0)(0^2+0^2+1^2+0^2)^{\frac{1}{2}}=1$.
So $v_1=\b... | You are full on track towards
$\,\left(\begin{smallmatrix}\ddot\smile& *& *\\ 0& \ddot\smile& *\\ 0& 0& \ddot\smile\\ 0& 0& 0\end{smallmatrix}\right)\;$ just go ahead:
$$v_3\:=\: \begin{pmatrix}0\\ 0\\ \sqrt 2+1\\ 1\end{pmatrix}\quad\Longrightarrow\;
-2\frac{v_3\,v_3^T}{v_3^Tv_3}\:=\: -\frac 2{\left(\sqrt 2+1\right)^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3078096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the formula of the linear transformation Consider the linear transformation $f:R^3\rightarrow R^3$ with $f(0,1,2)=(2,1,1)$ and $\ker f=\{(0,2,1),(1,0,1)\}.$ Find:
(i) $f(0,1,0)$ (ii) the general formula of $f$.
Any help, please.
| Now,
$$
(0,1,0)=\frac{1}{3}\big[2(0,2,1)-(0,1,2)\big],
$$
hence
$$f(0,1,0)
=\frac{1}{3}\big[2 \cdot f(0,2,1)-f(0,1,2)\big]
=\frac{1}{3}\big[2 \cdot 0 -(2,1,1)\big]
=\left(-\frac{2}{3},-\frac{1}{3},-\frac{1}{3}\right).
$$
Now,
$$
(0,0,1)
=\frac{1}{2}\big[(0,1,2)-(0,1,0)\big],
$$
hence
$$
f(0,0,1)
=\frac{1}{2}\left[(2,1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i}$ To prove $$\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i},$$
I demonstrated that the equality is true for any $n,$ for $m = 0, 1,$ and for any $r < n + m$ simply by fixing $n$ and $r$ and inserting $0,1$ for $m.$ Then, I procee... | You can think of a combinatoric proof as follows.
Suppose we have $n$ men and $m$ women, and we wish to form a committee of $r$ people. We can count in two different ways.
Case 1 $\binom{n+m}{r}$ is the number of r-subsets of a set with n+m elements.
Case 2 First we pick $i$ males. That leaves $r-i$ female to choose. S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
$f(x) = (x-2)(x-4)(x-6) +2$ then $f$ has all real roots between $0$ and $6$. True or false? $f(x) = (x-2)(x-4)(x-6) +2$ then $f$ has all real roots between $0$ and $6$
$($ true or false$)?$
Here
$f(0) = -46$ and $f(6) = 2$ since function is continuous so it must have at least one root between $0$ and $6$, but how to ch... | Consider $F(x):=f(x)-2= (x-2)(x-4)(x-6).$
1)$F(x), f(x)$ polynomials of degree $3.$
2) $x \rightarrow \infty$ $F(x), f(x) \rightarrow \infty.$
3) $x \rightarrow -\infty$ $F(x), f(x) \rightarrow -\infty.$
4) Roots of $F(x)$ at $x=2,4,6$. That's it.
5) $x> 6:$ $F(x) > 0,$ $f(x)= F(x)+2$.
No roots of $f$ for $x>6$.
6) Fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Functional equation involving $f(x^4)+f(x^2)+f(x)$
Find all increasing functions $f$ from positive reals to positive reals satisfying $f(x^4) + f(x^2) + f(x) = x^4 + x^2 + x$.
It's easy to show that $f(1)=1$, and I was also able to show that
$$f(x)-x = f(x^{(8^k)}) - x^{(8^k)}$$
for all integers $k$.
But where to go ... | First we show that $\lim_{x \to 1} f(x) = f(1) = 1$. Note that the limit of an increasing function from below or from above always exists. Let the limit from below be $a$ and the limit from above be $b$. If we take the limit from below of $f(x) + f(x^2) + f(x^4) = x + x^2 + x^4$, we get $a + a + a = 3$, so $a = 1$ (sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Calculus : Plug in value into a derivative? $$ f(x) = x^3-2x^2-6x $$
after derivative
$$ 3x^2-4x-6 $$
If u plugin x=1,2,3,4,5
as a result:
$$ 3(1)^2-4(1)-6=-7 $$
$$ 3(2)^2-4(2)-6=-2 $$
$$ 3(3)^2-4(3)-6=9 $$
I know it means the slope of the current point on the curve or instantaneous rate of change
Look at the answer ... | For $x=1$ :
$$f(1)=(1)^3-2(1)^2-6(1)=-7$$
$$f'(1)=3(1)^2-4(1)-6=-7$$
This means that at point $(1\:,\:-7)$ the slope of the tangent is $=-7$ .
For $x=2$ :
$$f(2)=(2)^3-2(2)^2-6(2)=-12$$
$$f'(2)=3(2)^2-4(2)-6=-2$$
This means that at point $(2\:,\:-12)$ the slope of the tangent is $=-2$ .
For $x=3$ :
$$f(3)=(3)^3-2(3)^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3088150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding sum of none arithmetic series I have a question to find the sum of the following sum:
$$
S = \small{1*1+2*3+3*5+4*7+...+100*199}
$$
I figured out that for each element in this series the following holds:
$$
a_n = a_{n-1} + 4n - 3
$$
But I don't know where to go from here, I tried subtracting some other series b... | Via generating functions we first find the generating function for each term and then sum them up. I'll build from the ground up. It may look overcomplicated but it also doesn't require remembering too many special identities. Our goal is to find $S_{99}$ where
\begin{align*}
S_m = \sum_{n=0}^{m} a_n
\end{align*}
where... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3088304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
$| \sin x| > \frac{\sqrt{2-\sqrt{2}}}{2}$ iff $1/8< \{ \frac{x}{\pi}\} < 7/8$ where $\{x\}$ is the fractional part of $x$ This is a problem that arose while reading the book "Putnam and Beyond":
Why is $| \sin x| > \frac{\sqrt{2-\sqrt{2}}}{2}$ iff $\frac18 < \{ \frac{x}{\pi}\} < \frac78$ where $\{x\}$ is the fractiona... | Because we need to prove that:
$$|\sin x|>\frac{\sqrt{2-\sqrt2}}{2},$$ which is
$$\frac{\pi}{8}+2\pi k<x<\frac{7\pi}{8}+2\pi k$$ or
$$\frac{9\pi}{8}+2\pi k<x<\frac{15\pi}{8}+2\pi k,$$ which is
$$\frac{\pi}{8}+\pi k<x<\frac{7\pi}{8}+\pi k,$$
where $k$ is an integer numbers, which is true because
$$\frac{1}{8}<\left\{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3089031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Problem when convert $\sqrt{A}+\sqrt{B}=\sqrt{C}+\sqrt{D}$ to A+B=C+D. This is what my lecturer taught me.
If you have $\sqrt{A}+\sqrt{B}=\sqrt{C}+\sqrt{D}$
You can easily convert to $A+B=C+D$ or $AB=CD$
And then he gave me an example.
$\sqrt{8x+1}+\sqrt{3x-5}=\sqrt{7x+4}+\sqrt{2x-2}$ Find x.
Which is totally work.
Af... | Try $A=25$, $B=1$ and $C=D=9$.
We see that $$A+B\neq C+D$$ and $$AB\neq CD,$$
but $$\sqrt{A}+\sqrt{B}=\sqrt{C}+\sqrt{D}.$$
We can try to understand, when it happens.
Firstly, $A$, $B$, $C$ and $D$ are non-negatives.
If $A=B$ and $C=D$ we obtain $A+B=C+D$ and $AB=CD.$
Now, let $A>B$, $C>D$, but $A+B=C+D$.
Thus, after sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3093188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Integral $\int\frac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$
Integrate $\displaystyle\int\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$
I tried dividing by $\cos^2(x)$ and then substituting $\tan(x)=t$.
| After substituting $\cos(2x)=\cos^2x-\sin^2x$ and $\sin(2x)=2\sin x\cos x$ and doing a bit of algebra, we find that
$$2x\cos(2x)+(x^2-1)\sin(2x)=2(x\sin x+\cos x)(x\cos x-\sin x)$$
and thus
$$\begin{align}{2x^2\over 2x\cos(2x)+(x^2-1)\sin(2x)}
&={x^2\over(x\sin x+\cos x)(x\cos x-\sin x)}\\
&={x\cos x\over x\sin x+\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3097078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Find the density of $Y = a/(1 + X^{2})$, where $X$ has the Cauchy distribution. Find the density of $Y = a/(1 + X^{2})$, where $X$ has the Cauchy distribution.
MY SOLUTION
To begin with, let us remember that the Cauchy probability density function is given by
\begin{align*}
f_{X}(x) = \frac{1}{\pi(1+x^{2})}\quad\text{f... | Your approach is fine.
Perhaps you are aware of this result (very useful to generate a Cauchy variable): if $Z \sim U(-\frac{\pi}{2},\frac{\pi}{2})$ then $X = \tan(Z)$ follows a canonical Cauchy distribution.
Now, we have $Y = \frac{a}{1+X^2}= a \cos^2(Z)$ which is slightly easier.
Then, assuming $a>0$ $$\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
I integrated $\int_0^1 \sqrt{x-x^2} dx$ after a $u$-substitution and trig substitution, but I'm not sure what to do with the limits I'm a bit lost on the technique in this problem and could use some insight. When do we have to change the limits?
$$\int_0^1 \sqrt{x-x^2} dx$$
completing the square gives me:
$$\int \sqrt... | Just so you are aware, this can be handled with the Beta and by extension the Gamma Function.
Here you have:
\begin{align}
I &= \int_0^1 \sqrt{x - x^2}\:dx = \int_0^1 \sqrt{x\left(1 - x\right)}\:dx = \int_0^1 x^{\frac{1}{2}}\left(1 - x\right)^{\frac{1}{2}}\:dx\\& = B\left(\frac{1}{2} + 1, \frac{1}{2} + 1 \right)= B\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How do I rearrange this? Hi please could some one explain the steps taken to rearange the below. I've had a dabble but don't seem to be getting it.
$$\frac{C}{\sqrt{C^2 - V^2}} $$
To
$$\frac{1}{\sqrt{1 - \frac{V^2}{C^2}}}$$
Any help would be greatly appreciated.
Thanks
| Assuming $C > 0$ then
$\frac C{\sqrt{C^2 - V^2}}=$
$\frac {C*\frac 1C}{(\sqrt{C^2 - V^2})\frac 1C}=$
$\frac {1}{(\sqrt{C^2 - V^2})\frac 1{\sqrt{C^2}})}=$
$\frac {1}{\sqrt{(C^2 - V^2)\frac 1{C^2}}}=$
$\frac {1}{\sqrt{\frac {C^2}{C^2} - \frac {V^2}{C^2}}}=$
$\frac 1{\sqrt{1 - \frac {V^2}{C^2}}}$
.... or ...
$\frac {C}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3102159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving Trig Function with both $\sin$ and $\cos$ I am trying to understand the function $h(t) = 65 + 36\sin\left(\frac{3t}{2}\right) - 15\cos\left(\frac{3t}{2}\right)$, where $t$ is the time in seconds and $\frac{3t}{2}$ is expressed in radians.
A question asks that I find the times in the first revolution when the he... | You are looking for $t$ such that
$$
\begin{align}
&\quad h(t) = 65\\
\Rightarrow &\quad 36\sin\left(\frac{3t}{2}\right) = 15\cos\left(\frac{3t}{2}\right)\\
\Rightarrow &\quad 12\sin\left(\frac{3t}{2}\right) = 5\cos\left(\frac{3t}{2}\right)\\
\Rightarrow &\quad 12^2\sin^2\left(\frac{3t}{2}\right) = 5^2\cos^2\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Probability that the sum of three dice is not greater than 9
What is the probability that the sum of 3 indistinguishable dice is less than or equal to 9?
I have tried counting the pairs $X_1, X_2$ such that their sum is less than or equal to 8, but I seem to be overcounting because I get that for each pair of dice I ... | Consider the sum of the first two dice. If the sum is lower than or equal to 3, the third die can have any value. As the sum increases from 3 to 8, fewer possible values can be thrown by the third die. Since the number of ways to throw a sum of $n$ with two dice equals $6 - |n - 7|$, the number of ways to throw a total... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
how to find the radius of convergence ? $\sum_{n=1}^{\infty} (\frac{x^n}{n} - \frac{x^{n+1}}{n+1})$ How can i find the radius of convergence ? i dont know where to start i cant use $\frac{a_n}{a_{n+1}}$ test here.
$\sum_{n=1}^{\infty} (\frac{x^n}{n} - \frac{x^{n+1}}{n+1})$
the question looks simple but i dont know how ... | Let us consider the partial sums.
\begin{align}
s_N &= \sum_{n=1}^N\left(\frac{x^n}n - \frac{x^{n+1}}{n+1}\right)\\
&= \left(x-\frac{x^2}{2}\right) + \left(\frac{x^2}{2} - \frac{x^3}{3}\right) + \left(\frac{x^3}{3} - \frac{x^4}{4}\right) + \text{ ... } + \left(\frac{x^{N}}{N} - \frac{x^{N+1}}{N+1}\right)\\
&= x - \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\frac{2a}{a^2-4} - \frac{1}{a-2}-\frac{1}{a^2+2a}$
Evaluate
$$\dfrac{2a}{a^2-4} - \dfrac{1}{a-2}-\dfrac{1}{a^2+2a}$$
We have to see what their common term is. Therefrom, we can evaluate the simplified expression by canceling out.
$$a^2 - 4 = (a)^2 - (2)^2 = (a-2)(a+2) \tag {1}$$
$$a^2 +2a = a(a+2)\tag{2}$$... | Hint: You can write $$\frac{2a^2}{a(a-2)(a+2)}-\frac{a(a+2)}{a(a-2)(a+2)}-\frac{a-2}{a(a+2)(a-2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If $a_{n} = ((-1)^{n})/(\sqrt{1+n})$ then is $\sum_{k=0}^{n} (a_{n-k})(a_{k})$ convergent? If $a_{n} = ((-1)^{n})/(\sqrt{1+n})$ then is $C_{n} = \sum_{k=0}^{n} (a_{n-k})(a_{k})$ convergent?
$a_{n}$ is conditionally convergent, but how do I check the convergence of $C_{n}$?
I could transform $C_{n}$ in following way
$C... | $C_{n}
= \sum_{k=0}^{n} \frac{(-1)^{n}}{(1+k)^{1/2}(1+n-k)^{1/2}}
$
$\begin{array}\\
(1+k)(1+n-k)
&=1+n+kn-k^2\\
&=1+n-(k^2-kn)\\
&=1+n-(k^2-kn+n^2/4)+n^2/4\\
&=1+n-(k-n/2)^2+n^2/4\\
&= (n+2)^2/4-(k-n/2)^2\\
\text{so}\\
\dfrac1{\sqrt{(1+k)(1+n-k)}}
&= \dfrac1{\sqrt{(n+2)^2/4-(k-n/2)^2}}\\
\end{array}
$
so
$\begin{arra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving that |z|=1. I am trying to prove that
If $z\in \mathbb{C}-\mathbb{R}$ such that $\frac{z^2+z+1}{z^2-z+1}\in \mathbb{R}$. Show that $|z|=1$.
1 method , through which I approached this problem is to assume $z=a+ib$ and to see that $$\frac{z^2+z+1}{z^2-z+1}=1+\frac{2z}{z^2-z+1}$$.
So problem reduces to show that $... | Let $w=\frac{z^2+z+1}{z^2-z+1}=1+\frac{2z}{z^2-z+1}$
Since $\frac{z^2+z+1}{z^2-z+1}\in \mathbb{R}$ , so $\operatorname{Im}(w)=0$
$$\iff w-\bar{w}=0$$
Now, let's solve $1+\frac{2z}{z^2-z+1}=\overline{1+\frac{2z}{z^2-z+1}}$
$$\implies \frac{z}{z^2-z+1}=\overline{\frac{z}{z^2-z+1}}$$
Since $z^2-z+1=0 \ when\ z= \frac{1 \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3108847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Proof Explanation - Odd prime division
Prove that if $p$ is an odd prime, then $$p\mid \lfloor(2+\sqrt5)^p\rfloor - 2^{p+1}$$
The solution posted by another user is as follows:
Let $=(2+\sqrt5)^p + (2-\sqrt5)^$.
Note that $N$ is an integer. There are various ways to see this. One can, for example, expand using the b... | 1) We have
$(2+\sqrt 5)^p=2^p+\binom p12^{p-1}\sqrt 5+\binom p22^{p-2}\sqrt 5^2+\binom p23^{p-3}\sqrt 5^3+...+\binom pp\sqrt 5^p$, and
$(2-\sqrt 5)^p=2^p-\binom p12^{p-1}\sqrt 5+\binom p22^{p-2}\sqrt 5^2-\binom p23^{p-3}\sqrt 5^3+...-\binom pp\sqrt 5^p$.
When you add these together, all the odd terms are equal and oppo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3110101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$ prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$ .
I think that must I use from $\dfrac{a^2}{2}+\dfrac{b^2}{2} \geq ab$ but no result please help me .!
| $$\frac{a^2}{2}+\frac{b^3}{3}+\frac{c^6}{6}=\frac{a^2}{6}+\frac{a^2}{6}+\frac{a^2}{6}+\frac{b^3}{6}+\frac{b^3}{6}+\frac{c^6}{6}.$$
Then use AM-GM with these $6$ variables.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3110691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Find closed form of composition. Consider $f(z) = \frac{1-a}{1-az}$, where $a \in(0,1)$
. Now we want to find $f(f\dots(f(z)\dots)$. It will begood if there is a closed form of it.
I've consider first step : $f(f(z)) = \frac{1 - a}{1 -a\frac{1-a}{1-az}} = \frac{(1-a)(1-az)}{1-az-a+a^2}$ , but it looks non-simplified.
I... | If we identify linear fractional transformation with the matrix (up to multiplicative constant), i.e.
$$
f(z) = \frac{az+b}{cz+d}\ \ \ \Longleftrightarrow \ \ \ \left[\begin{array}{cc}a&b\\c&d\end{array}\right],
$$ then composition corresponds to matrix multiplication. Thus the problem boils down to calculating
$$
A^n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that $\cot{142\frac{1}{2}^\circ} = \sqrt2 + \sqrt3 - 2 - \sqrt6$.
Show that $\cot{142\frac{1}{2} ^\circ} = \sqrt2 + \sqrt3 - 2 - \sqrt6$.
What I have tried:
Let $\theta = 142\frac{1}{2}^\circ \text{ and } 2\theta = 285^\circ$.
$$\cos 285^\circ = \cos 75^\circ$$
$$\cos 75^\circ = \frac{\sqrt3 - 1}{2\sqrt2}$$
$$\... | First render $\cot(142\frac{1}{2}°)=-\cot(37\frac{1}{2}°)$. Then for the latter angle put in:
$\cot x= \dfrac{2\cos^2 x}{2\sin x \cos x}=\dfrac{1+\cos 2x}{\sin2x}$
with $x=37\frac{1}{2}°$ so $2x=75°$. If you know the sine and cosine of the latter the rest is algebra.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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A formula for tan(2x) Help with solving...
Suppose that $\tan^2x=\tan(x-a)·\tan(x-b)$, show that $$\tan(2x)=\frac{2\sin(a)·\sin(b)}{\sin(a+b)}$$
As far as I know, so far the $\tan2x$ can be converted to $\frac{2\tan x}{1-\tan^2x}$ using the double angle formula and the $\tan 2x$ can be further be substituted to the f... | Variation of lab bhattacharjee's answer:
$$\tan^2x=\tan(x-a)·\tan(x-b) \Rightarrow \\
\tan^2x+1=\tan(x-a)·\tan(x-b)+1 \Rightarrow \\
\frac{1}{\cos^2 x}=\tan(x-a)·\tan(x-b)+1 \Rightarrow \\
\cos^2x =\frac1{\tan(x-a)·\tan(x-b)+1} \Rightarrow \\
\frac{1+\cos 2x}{2}=\frac1{\tan(x-a)·\tan(x-b)+1} \Rightarrow \\\\
\cos 2x=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Proving an equality using given ones; no need for differentiation Prove that $(\frac ab+\frac bc+\frac ca)(\frac ba+\frac cb+\frac ac)\geqslant9$
The formulas given were
$$\frac{a+b} {2}\geqslant\sqrt {ab}$$
$$a^2+b^2\geqslant2ab$$
$$\frac{a+b+c} {3}\geqslant\root3\of{abc}$$
$$a^3+b^3+c^3\geqslant3abc$$
for all $a\gt0,... | Method 1: AM-GM
Apply the last given inequality, but with the fractions instead.
$$\frac ab+\frac bc+\frac ca\ge 3 \sqrt[3]{\frac ab\,\frac bc\,\frac ca} = 3.$$
Do the same for $\frac ba+\frac cb+\frac ac$
Method 2: Cauchy Schwarz
$$9 = 3^2 = \left(\frac{\sqrt{a}}{\sqrt{b}} \, \frac{\sqrt{b}}{\sqrt{a}} + \frac{\sqrt{b}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Evaluate $\int_{0}^{\pi} x\sin\big(\frac{1}{x}\big)\cos x \,dx$ I wonder if an integral of the form $$\int_{0}^{\pi} x\sin\Bigl(\frac{1}{x}\Bigr)\cos x \,dx$$ which can be further simplified to
$$\int_{0}^{\pi} \frac{\sin (x^{-1})}{(x^{-1})}\,\cos x\, dx=\cos x\biggl(\int_{0}^{\pi} \frac{\sin (x^{-1})}{(x^{-1})}\,dx\bi... | NOT THE SOLUTION:
DAMN, I just got to the of the solution as below and realised I had the bounds wrong the whole time - Thought I would leave in case you're interested!
Here we are addressing the integral:
\begin{equation}
I = \int_0^\infty x \sin\left(\frac{1}{x}\right)\cos(x)\:dx
\end{equation}
Here we will employ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3120759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Applying comparison theorem to limits when inequality is known I somehow confuse myself whenever I apply the comparison lemma. I know there is the comparison lemma that says the following:
Let the sequence $\{a_{n}\}$ converge to the number $a$. Then the sequence $\{b_{n}\}$ converges to the number $b$ if there is a ... | The problem is that $$\sqrt{1+\frac{1}{n}} + 1 > 2$$ for all $n\in\mathbb N$, and therefore
$$\frac{1}{\sqrt{1+\frac{1}{n}}+1}-\frac{1}{2} < 0$$
When you compare the absolute values of both sides of your last inequality, the "$<$" becomes "$>$", and the theorem does not apply.
Namely, because the values inside the abso... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3120860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Can anybody help with the integral: $\int \frac{\sqrt{1-x^2}}{\sqrt{1+x^2}} dx$ please? I have searched online. I have got a solution from Wolfram but I don't understand it, or how to reach it. If anyone has a method that would be fantastic, thanks!
| If the integral is a definite integral with the interval of integration being from $0$ to $1$, here is an approach that makes use of the following form for the Beta function $\operatorname{B} (x,y)$:
$$\operatorname{B} (x,y) = \int_0^1 t^{x - 1} (1 - t)^{y - 1} \, dt, \qquad x,y > 0. \tag1$$
Let
$$I = \int_0^1 \sqrt{\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3121075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Distributing pairs of coloured socks to people Say you have $100$ socks, where $50$ are black, $30$ are white, $20$ are blue. You want to distribute them to $10$ different people such that none of them end up with an odd number of any colour, but there is no requirement that all $10$ people get socks. We are assuming t... | I was looking into the Generating Function approach to this question.
The 25 black pairs of socks could be represented by;
$$(1+x+x^2+x^3+x^4+...+x^{25})$$
and the ten boxes by raising this to the power of 10;
$$(1+x+x^2+x^3+x^4+...+x^{25})^{10}$$
So, essentially, the answer could be obtained from expanding the bracket... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Trisected Triangle Perimeter ICTM State 2017 Division AA Precalculus Individual Question 17 reads:
$C$ and $D$ lie on $BE$ such that $AC$ and $AD$ trisect $\angle$BAE in $△ABE$. $BC = 2$, $CD = 3$, and $DE = 6$. Then the perimeter of $△ABC$ may be expressed as $P = f + k \sqrt{w} + p \sqrt{q}$ in simplified and reduced... | Let $AB=x$.
Thus, since $$\frac{AD}{AB}=\frac{CD}{BC},$$ we obtain
$$\frac{AD}{x}=\frac{3}{2},$$ which gives
$$AD=\frac{3}{2}x.$$
Also, let $AC=y$.
Thus, since $$\frac{AE}{AC}=\frac{DE}{DC},$$ we obtain $AE=2y$.
But, $$AC^2=AB\cdot AD-BC\cdot CD$$ and
$$AD^2=AC\cdot AE-CD\cdot DE,$$ which gives
$$y^2=\frac{3}{2}x^2-6$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $2018^{2019}> 2019^{2018}$ without induction, without Newton's binomial formula and without Calculus. Prove that $2018^{2019}> 2019^{2018}$ without induction, without Newton's binomial formula and without Calculus. This inequality is equivalent to
$$
2018^{1/2018}>2019^{1/2019}
$$
One of my 'High school' stu... | Compared to my first answer, this answer makes a simpler use of the same familiar identity, which is proved without explicit use of induction, thus:
\begin{align*}
x^m - y^m & = x^m - (x^{m-1}y - x^{m-1}y) - (x^{m-2}y^2 - x^{m-2}y^2) - \cdots - (xy^{m-1} - xy^{m-1}) - y^m \\
& = (x^m - x^{m-1}y) + (x^{m-1}y - x^{m-2}y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 13,
"answer_id": 5
} |
Find the number of ways to express 1050 as sum of consecutive integers I have to solve this task:
Find the number of ways to present $1050$ as sum of consecutive
positive integers.
I was thinking if factorization can help there:
$$1050 = 2 \cdot 3 \cdot 5^2 \cdot 7 $$
but I am not sure how to use that information... | We want to find the number of solutions of
$$n+(n+1)+\ldots + (n+k) = 1050,\ n\in\mathbb Z_{>0},\ k\in\mathbb Z_{\geq 0}.\tag{1}$$
Rewrite the sum as $$n(k+1) + 0 + 1 +\ldots + k = n(k+1) + \frac{k(k+1)}{2}= \frac 12(2n+k)(k+1).$$
Thus, the number of solutions to $(1)$ is the same as the number of solutions of
$$(2n+k)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
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Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2$ where a, b, c and d are positive real numbers
I have to prove the following inequality using the Cauchy-Schwarz inequality:
$$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2$$
where a, b, c and d are positive real numbers.
But I am ... | By C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c+d)^2}{\sum\limits_{cyc}(ab+ac)}=2+\frac{(a+b+c+d)^2-2\sum\limits_{cyc}(ab+ac)}{\sum\limits_{cyc}(ab+ac)}=$$
$$=2+\frac{a^2+c^2+b^2+d^2-2ac-2bd}{\sum\limits_{cyc}(ab+ac)}\geq2+\frac{2\sqrt{a^2c^2}+2\sqrt{b^2d^2}-2ac-2bd}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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How to solve these trigonometry equations? I have to work with the following 5 equations:
*
*$(1-\tan^2x)(1-\tan^22x)(1-\tan^24x)=8$
*$(2\cos 2x+1)(2\cos 6x+1)(2\cos 18x+1)=1$
*$\dfrac{\cos 2x}{\sin 3x}+\dfrac{\cos 6x}{\sin 9x}+\dfrac{\cos 18x}{\sin 27x}=0$
*$\dfrac{\cos x}{\sin 3x}+\dfrac{\cos 3x}{\sin 9x}+\dfra... | Using that
$$\tan(2x)=2\,{\frac {\tan \left( x \right) }{1- \left( \tan \left( x \right)
\right) ^{2}}}
$$
$$\tan(4x)={\frac {4\,\tan \left( x \right) -4\, \left( \tan \left( x \right)
\right) ^{3}}{1-6\, \left( \tan \left( x \right) \right) ^{2}+
\left( \tan \left( x \right) \right) ^{4}}}
$$
we get for your fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3140794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Minimum value of $(x^2+y^2)^2$ if $x,y$ are real number such that $x^2+2xy-y^2=6$ Then find minimum value of $(x^2+y^2)^2$
what i try : $x^2+2xy+y^2-2y^2=6$ or $(x+y)^2-\bigg(\sqrt{2} y\bigg)^2=6$
put $\displaystyle (x+y)=\sqrt{6}\cos \alpha$ and $\displaystyle \sqrt{2}y=\sqrt{6}\sin \alpha$
$\displaystyle x=\sqrt{6}\c... | $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=(6-2xy)^2+4x^2y^2=2(2xy-3)^2+18\ge18$$
The equality occurs if $2xy=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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$I_1=\int_0^1 \frac 1 {1+\frac 1 {\sqrt x}} dx$ If we let $I_2= \int_0^1 \frac 1 {1+\frac 1 {1+\sqrt x}} dx$ and so on, we’re tasked with solving $I_n$. We can show that $I_2 = \int \frac {1+\sqrt x} {1-\sqrt x + 1} = \int \frac {1+\sqrt x} {2+\sqrt x }$
And that $I_3=\int \frac {2+\sqrt x}{3+2\sqrt x}$, and that in g... | We define $$j=j(a,b)=\int_0^1\frac{a+\sqrt x}{b+\sqrt x}dx$$
So we see that
$$j^*(a_1,a_2,a_3,a_4)=\int_0^1\frac{a_1+a_2\sqrt x}{a_3+a_4\sqrt x}dx=\frac{a_2}{a_4}j\left(\frac{a_1}{a_2},\frac{a_3}{a_4}\right)$$
So then we see that
$$\begin{align}
j&=\int_0^1\frac{a-b+b+\sqrt x}{b+\sqrt x}dx\\&=(a-b)\int_0^1\frac{dx}{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3148168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all positive integer $n$ such that there exist $m\in\mathbb{Z}$ with $2^n-1|m^2+9$ I have an problem with elementary number theory:
Find all positive integer $n$ such that there exist $m\in\mathbb{Z}$ with $2^n-1|m^2+9$
It's look like the problem in this link, but there some differences: Find all positive integer ... | If $2^n-1$ has a prime divisor $p\equiv3\pmod4$, then $m^2+3^2\equiv0\pmod p$. If $p\neq 3$, let $a$ be an inverse of $3$ modulo $p$. Then $(am)^2\equiv-1\pmod p$, contradicting $p\equiv3\pmod4$.
Thus $3$ must be the only prime divisor of $2^n-1$ that is $3$ mod $4$. First, $n=1$ clearly works. Suppose $n \geq 2$. Then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Value of $(p,q)$ in indefinite integration Finding value of $(p,q)$ in $\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx=\frac{px^3+qx^8}{x^{10}-2x^5+1}+c$
what i try
$\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx$
put $x=1/t$ and $dx=-1/t^2dt$
$\displaystyle -\frac{2t^5+3x^{10}}{t^{10}-2t^5+1}dt$
How do i so... | \begin{equation}
\left(\frac{px^3+qx^8}{x^{10}-2x^5+1}\right)^\prime=\frac{(8qx^7+3px^2)(x^5-1)^2-10x^7(qx^5+p)(x^5-1)}{(x^5-1)^4}
\end{equation}
Let $p=1,\quad q=-1$ and we get
\begin{eqnarray}
&&\frac{(-8x^7+3x^2)(x^5-1)^2-10x^7(-x^5+1)(x^5-1)}{(x^5-1)^4}\\
&=&\frac{(-8x^7+3x^2)(x^5-1)^2+10x^7(x^5-1)^2}{(x^5-1)^4}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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How to simplify $\sqrt{2+\sqrt{3}}$ $?$
Simplify $\dfrac{2\left(\sqrt2 + \sqrt6\right)}{3\sqrt{2+\sqrt3}}$
The answer to this question is $\frac{4}{3}$ in a workbook.
How would I simplify $\sqrt{2+\sqrt3}$ $?$ If it was something like $\sqrt{3 + 2\sqrt2}$ , I would have simplified it as follows:
$\sqrt{3 + 2\sqrt2... | $$\left(\frac{2\left(\sqrt2+\sqrt6\right)}{3\sqrt{2+\sqrt3}}\right)= \frac{(2√2+2√6)\sqrt{2+\sqrt{3}}}{3(2+√3)}=\frac{8+4√3}{2+√3}=\frac{4}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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$\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}=K(a-b)(b-c)(c-a)$, solve for $K$ Qestion: $\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}=K(a-b)(b-c)(c-a)$, solve for $K$
Answer: $K=(ab+bc+ca)$
My attempt: $$\begin{align}\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}&=\begin{vmatr... | The mistake is here
$$\begin{vmatrix}b+a& c+a\\b^2+ba+a^2&c^2+ca+a^2\end{vmatrix}=(c-b)(a-b)(b+a)(c+a)\begin{vmatrix}1&1\\(b+a)-ba&(c+a)-ca\end{vmatrix}$$
These aren't equal to each other. If you want to continue row reduction methods, you should continue with
$$\begin{vmatrix}b+a& c+a\\b^2+ba+a^2&c^2+ca+a^2\end{vmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\dfrac{3}{4}$ where $a+b+c=3$
If $a$, $b$, $c$ are non-negative numbers such that $a + b + c = 3$ then prove
$$(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\frac{3}{4}$$
Here's what I did.
Let $c \ge a \ge b$.
We have that
\begin{align*}
(c - 1)^3 + (a - 1)^3 &= (c + a ... | Perhaps a different approach than you are using. Using the Lagrange multipliers, we define the auxiliary problem
$$ F = (x-1)^3 + (y-1)^3 + (z-1)^3 - \lambda(x + y + z - 3). $$
Differentiating and setting to $0$ we obtain the conditions
\begin{align}
3(x-1)^2 = \lambda \\
3(y-1)^2 = \lambda \\
3(z-1)^2 = \lambda
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Prove an inequality with positives $a$, $b$ and $c$.
If $a$, $b$ and $c$ are positives such that $(a + b + c)\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right) = x$ ($x \ge 9$) then prove that $$\large(a^2 + b^2 + c^2)\left(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}\right) \le x(\sqrt x - 2)^2$$
The equali... | Let $a+b+c=3u$, $ab+ac+bc=v^2$ and $abc=w^3$.
Thus, $a$, $b$ and $c$ are roots of the equation:
$$t^3-3ut^2+3v^2t-w^3=0.$$
Now, by your work the equality in your inequality occurs for
$$(ab+ac+bc)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=(a+b+c)^2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)$$ or
$$\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3154291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove without induction that $2×7^n+3×5^n-5$ is divisible by $24$. I proved this by induction. But I want to show it using modular arithmetic. I tried for sometime as follows
$$2×7^n-2+3×5^n-3\\
2(7^n-1)+3(5^n-1)\\
2×6a+3×4b\\
12(a+b)$$
In this way I just proved that it is divisible by 12 but it is not enough. Am I mis... | You may split it up by calculating $\mod 8$ and $\mod 3$:
*
*$\mod 8$:
\begin{eqnarray*} 2×7^n+3×5^n-5
& \equiv_8 & 2\times (-1)^n + 3\times (-3)^n +3 \\
& \equiv_8 & 2\times (-1)^n + 3((-3)^n + 1)\\
& \stackrel{3^2 \equiv_8 1}{\equiv_8}& \begin{cases} 2+3\times (1+1) & n = 2k \\ -2 +3 (-3 + 1) & n= 2k+1\end{cases}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3155303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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Three dice: conditional probability After throwing $3$ dice, we know that on every die there is a different number. What is the probability that there is a $6$ on exactly one dice?
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then
$$P(B)=\dfrac{6∗5∗4}{6^3}\text{ and }
P(A\cap B)... | If I understood right from question and also from the discussion before. That every throw it is ensured that you get a different number (so all three different numbers). So you can get 6 only in one throw.
Probability of getting 6 in first throw is $\dfrac{1}{6}$, so it is guaranteed that next two throws there will no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Create generating function: $(1, 1, -2, -2, 10, 3, -4, -4....)$
Find generating function (without using infinite series):
a) (0, 1, 4, 9, 16, 25, 36...)
b) (1, 1, -2, -2, 10, 3, -4, -4, 5, 5, -6, -6, 7...) (Only irregularity is the 10)
Here's what I got:
a) $(0, 1, 4, 9, 16...) = \frac{x(1+x)}{(1-x)^3}$
(Begin with (... | The function is
$$
7x^4+(1+x)(1-2x^2+3x^4-4x^6+\dots)
$$
Can you find a closed form for $1-2x^2+3x^4-4x^6+\dots$? Perhaps you can find a closed form for $1+2y+3y^2+4y^3+\dots$, then substitute $y\gets -x^2$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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can not find root of the equation $(0,0)$ with semi-major axis a and semi-minor axis $b$">
I want find out for each point $(x,y)$ in the white region, the nearest point $(a \cos \theta, b \sin \theta )$ on the ellipse curve.I tried with the following approach.
$$
\text {distance} = \sqrt{(x-a \cos \theta)^2+(y-b \sin \... | You can get it by solving a constrained optimization problem. You want the point $(u,v)$ that minimizes $f(u,v)=(u-x)^2+(v-y)^2$ subject to the constraint $\frac{u^2}{a^2} + \frac{v^2}{b^2} = 1$. The point $(x,y)$ doesn't need to be inside the ellipse.
The Lagrangian is given by
$$
L(u,v,\lambda) = (u-x)^2+(v-y)^2-\la... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3165762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Sum of Infinite series $\frac{1.3}{2}+\frac{3.5}{2^2}+\frac{5.7}{2^3}+\frac{7.9}{2^4}+......$ Prove that the sum of the infinite series $\frac{1.3}{2}+\frac{3.5}{2^2}+\frac{5.7}{2^3}+\frac{7.9}{2^4}+......$ is 23.
My approach
I got the following term
$S_n=\sum_1^\infty\frac{4n^2}{2^n}-\sum_1^\infty\frac{1}{2^n}$.
For ... | $$\frac{1.3}{2}+\frac{3.5}{2^2}+\frac{5.7}{2^3}+\frac{7.9}{2^4}+…=\sum_{n=1}^{\infty }\frac{(2n-1)(2n+1)}{2^n}=\sum_{n=1}^{\infty }\frac{(4n^2-1)}{2^n}$$
depending on the geometric series
$$\frac{1}{1-x}=\sum_{n=0}^{\infty }x^n$$
$$(\frac{1}{1-x})'=\sum_{n=1}^{\infty }nx^{n-1}$$
$$x(\frac{1}{1-x})'=\sum_{n=1}^{\infty }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3165878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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exist infinitely many positive integers $n$ such $\omega (n)+\omega (n+1)\equiv 0\pmod 3$ For integer $n>1$, $\omega (n)$ denotes number of distinct prime factors of $n$, and $\omega (1)=0$. Prove that:there exist infinitely many positive integers $n$ satisfying
$$\omega (n)+\omega (n+1)\equiv 0\pmod 3$$
Positive inte... | With $\Omega$ instead of $\omega$ I have a solution.
For any $n$ then $a = \Omega(2n) \bmod 3,b=\Omega(2n+1)\bmod 3,c=\Omega(2n+2)\bmod 3$, $\Omega((2n+1)^2) = 2b\bmod 3, \Omega((2n+1)^2-1) = a+c\bmod 3,\Omega(n)+\Omega(n+1) = a+c-2 \bmod 3$
Assume we chose $n$ such that $a=2\bmod 3$
The claim $\Omega(l)+\Omega(l+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3166601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why do these laurent series approaches conflict? I was working on a problem of finding the Laurent series of $\frac{1}{z-3}$ that converges where $|z-4| > 1$
So I had one approach, let $u=z-4$ then:
$$\frac{1}{z-3} = \frac{1}{1+u} $$
$$ = \frac{1}{u} - \frac{1}{u^2} + \frac{1}{u^3}...$$
$$ = \frac{1}{z-4} - \frac{1}... | ${1\over 1+x} = 1 - x + x^2 - x^3 + \cdots$ when $|x| < 1$.
This is derived from ${1 \over 1-x} = 1 + x + x^2 + \cdots$ with $|x| < 1$.
| {
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"url": "https://math.stackexchange.com/questions/3175400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Expand $\frac{1}{z^2}$ as a series on $|z - 2| < 2$ I am asked to show that
$$\frac{1}{z^2} = \frac{1}{4} + \frac{1}{4}\sum_{n = 1}^{\infty}(-1)^{n+1}(n+1)\left(\frac{z-2}{2}\right)^n$$
on $|z - 2| < 2$.
My plan is to consider a series expansion of $\frac{-1}{z}$ in $|z - 2| < 2$ and differentiate this series term by ... | Perhaps a simpler approach would be to write
$$
{1 \over z^2} = {1 \over 1 - \left(1 - z^2\right)}.
$$
Letting $a = 1 - z^2$, we see that the above expression has (at least formally) the geometric series expansion:
$$
{1 \over 1 - \left(1 - z^2\right)} = {1 \over 1 - a} = \sum_{n \geq 0} a^{n}.
$$
Now expand each power... | {
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"url": "https://math.stackexchange.com/questions/3180164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Alternate complex binomial series sum
Calculation of $\displaystyle \sum^{2n-1}_{r=1}(-1)^{r-1}\cdot r\cdot \frac{1}{\binom{2n}{r}}$ is
My Try: Using $$\int^{1}_{0}x^m(1-x)^ndx = \frac{1}{(m+n+1)}\cdot \frac{1}{\binom{m+n}{n}}$$
So $\displaystyle \int^{1}x^{2n-r}(1-x)^r=\frac{1}{2n}\cdot \frac{1}{\binom{2n}{r}}$
Sum ... | Here is a more elementary method to find the sum which gives a more general result. Let $\displaystyle f(n)=\sum^{n-1}_{r=1}(-1)^{r-1}\frac{r}{\binom{n}{r}}$. We have been asked to find $\displaystyle f(2n)$. Note that $\displaystyle \frac{1}{\binom{n+1}{k+1}}+\frac{1}{\binom{n+1}{k}}=\frac{n+2}{n+1}\frac{1}{\binom{n}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Proving $\frac{1}{6}a+\frac{1}{3}b+\frac{1}{2}c \geq \frac{6abc}{3ab+bc+2ca}$ for positive $a$, $b$, $c$ I'm at the end of an inequality proof that started out complex and I was able to simplify it to:
$$\frac{1}{6}a+\frac{1}{3}b+\frac{1}{2}c \geq \frac{6abc}{3ab+bc+2ca} \quad\text{where}\quad a, b, c > 0$$
I'm able ... | Since $\frac{1}{6}+\frac{1}{3}+\frac{1}{2}=1,$ by AM-GM we obtain:
$$\left(\frac{1}{6}a+\frac{1}{3}b+\frac{1}{2}c\right)\left(\frac{1}{2}ab+\frac{1}{6}bc+\frac{1}{3}ca\right)\geq$$
$$\geq a^{\frac{1}{6}}b^{\frac{1}{3}}c^{\frac{1}{2}}(ab)^{\frac{1}{2}}(bc)^{\frac{1}{6}}(ca)^{\frac{1}{3}}=abc,$$ which gives your inequali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3188234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the sum: $\sum_{n=1}^{\infty} \ln\left(\frac{b+n+1}{a+n+1}\right)$ Let $0<a<b$, I would like to compute the sum
$$\sum_{n=1}^{\infty} \ln\left(\frac{b+n+1}{a+n+1}\right).$$
But first I am worrying that a test convergence might lead to the divergence of this series
What do I miss here?
$$\begin{split}\sum_{n=1}^{... | HINT:
Note that if $a\ne b$, then
$$\begin{align}
\int_{1/(b+n)}^{1/(a+n)}\frac1{x+1}\,dx&=\log\left(\frac{1+1/(a+n)}{1+1/(b+n)}\right)\\\\
&=\log\left(\frac{a+n+1}{a+n}\frac{b+n}{b+n+1}\right)\\\\
&\ne \log\left(\frac{a+n+1}{b+n+1}\right)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3188733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the exact length of the curve $y = \ln(1-x^2)$ Edit: Update with the full question for context
Find the exact length of the curve $y = \ln(1-x^2), 0 \leq x \leq \frac{1}{2}$
The integral below is what I got after finding the derivative $\frac{-2}{1-x^2}$ via the chain rule.
Can someone give me a hint on how t... | Partial fraction seems to be a good tool for your quesiton.
\begin{align}I &= \int_{0}^{\frac{1}{2}} \sqrt{1+\bigg(\frac{-2x}{1-x^2}\bigg)^2}\,\mathrm dx\\
&=\int_{0}^{\frac{1}{2}} \frac{\sqrt{(1-x^2)^2+4x^2}}{1-x^2}\mathrm \,dx\\
&=\int_0^\frac12 \frac{\sqrt{1+2x^2+x^4}}{1-x^2}\mathrm\, dx \\
&=\int_0^\frac12 \frac{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3189983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Describe all integral solutions of the equation $x^2 + y^2 = 2z^2$ such that $x,y,z > 0$, gcd$(x,y,z) = 1$, and $x > y$. As the title states, the question tasks me with finding all the integral solutions of the equation under the specified constraints. I have an idea of where to start due to a somewhat similar problem ... | The trick to this one is to create three new variables
$$
a = x+y\\
b = x-y\\
c = 2z
$$
Then $x^2+y^2 = 2z^2 \implies a^2+b^2 = c^2$
Temporarily ignoring the condition that $\gcd(x,y,z) = 1$ we can now list all solutions of
$a^2+b^2 = c^2$ as
$$
a = k(n^2 - m^2)\\
b = 2kmn\\
c = k(n^2 + m^2)
$$
with $m \neq n \pmod 2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3191780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that a triangle is right angled In $\triangle ABC$ (not isosceles) $CH$, $CL$ and $CM$ are respectively height, bisector and median. Show that $\angle ACB = 90^\circ$ if and only if $\angle HCL = \angle MCL$.
I think that I have to show that $\triangle MCB$ (or $\triangle ACM$) is isosceles, but I can't figure it ... |
Let's assume $\angle C > 90 ^\circ$ as show in the picture. $CL1$ is the bisector of $\angle ACB$ , $CL2$ is the bisector of $\angle HCM$ , now we show $AL2>AL1$
let $l1=AL1,l2=AL2$
$l1=\frac{AC}{AC+BC}*(c1+c2)$,
$AC=\sqrt{h^2+c1^2},BC=\sqrt{h^2+c2^2}$,
$l1=\frac{\sqrt{h^2+c1^2}}{\sqrt{h^2+c1^2}+\sqrt{h^2+c2^2}}*(c1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3194545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
How to find the power series of $\sqrt{1+x^4}$? The complete question is to find the integral from $0$ to $1$ of $$\sqrt{1+x^4}$$
I am unsure of how to find the power series of this equation in order to do that. I haven't dealt with square root power series equations yet and any help would be appreciated. Thank you!
| $$\frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}x^n\tag{1} $$
for any $x\in[-1,1)$ is a standard result. By applying termwise integration one gets
$$ \sqrt{1-x} = \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\frac{x^n}{1-2n}\tag{2} $$
for any $x\in[-1,1]$. By replacing $x$ with $-x^4$ one gets
$$ \sqrt{1+x^4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3197148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$11x + 13 \equiv 4$ (mod 37) $11x + 13 \equiv 4$ (mod 37)
My "solution" to the problem
$11x + 13 \equiv 4$ (mod 37) $\rightarrow$
$11x + 13 = 4 + 37y $
$11x - 37y = - 9$
Euclid's algorithm.
$37 = 11*3 + 4$
$11 = 4*2 + 3$
$4 = 3*1 + 1 \rightarrow GCD(37,11) = 1$
$3 = 1*3 + 0$
Write as linear equation
$1 = 4 - 1*3$
$3 ... | $\bmod 37\!:\,\ x\equiv 90\equiv 16\ $ so both are correct. Gauss's algorithm is simpler:
$\bmod 37\!:\,\ x\equiv\dfrac{-9}{11}\equiv\dfrac{-27}{33}\equiv\dfrac{10}{-4}\equiv\dfrac{5}{{-}2\,}\equiv\dfrac{-32}{-2}\equiv 16$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Evaluating limit of a function with integral involved. We need to evaluate the following limit:
$$\lim\limits_{x\to 0} \frac{1}{x} \int_0^x \sin^2\left(\frac{1}{y}\right)\,\mathrm dy.$$
I am finding hard to solve this integral as I cannot see a clear method to evaluate such integral. I have tried using substitutions, i... | I am posting an answer on the basis of hints given in comments, hope it helps if someone has the same question. Please tell if this is correct.
\begin{align*}
\lim\limits_{x \to 0}\frac{1}{x} \int_{0}^{x} \sin^2\left( \frac{1}{u}\right) \mathrm du & = \lim\limits_{x \to 0}\frac{1}{x} \int_{0}^{x} \frac{\left(1-\cos\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I prove this combinatorial identity Show that
$${2n \choose n} + 3{2n-1 \choose n} + 3^2{2n-2 \choose n} + \cdots + 3^n{n \choose n} \\ = {2n+1 \choose n+1} + 2{2n+1 \choose n+2} + 2^2{2n+1 \choose n+3} + \cdots + 2^n{2n+1 \choose 2n+1}$$
One way that I did it was to use the idea of generating functions.
For ... | Here is a combinatorial proof. Both sides of the equation answer the following question:
How many sequences are there of length $2n+1$, with entries in $\{0,1,2\}$, such that
*
*at least one of the entries is a $2$, and
*there are exactly $n$ zeroes to the left of the leftmost $2$?
LHS:
Suppose the le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3201479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 2
} |
Minimizing $\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2$ While solving a problem I came across this task, minimizing
\begin{align}
\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2.
\end{align}
One can eas... | AM-GM says $a^2+b^2\geq 2ab$ with equality when $a=b.$ Thus we cannot obtain a lower value than which gives GM.
With a bit of algebra we get that the equality really occurs, and this is iff $|\sin x|=|\cos x|,$ from where the minimum $$2.5^2+2.5^2=12.5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Describe the region $\{|z^2 - 1|<1\}$ of the complex plane Since it's complex analysis, I assume there is an easy solution without much algebra. But even in my algebra heavy solution there should be a more streamlined approach... Either approach is appreciated as an answer, or verification that my approach is the sta... | Credit to @Reinhard Meier in the comments
There is an algebra heavy answer that you can find to give an exact, graphable answer: $$ x^4 - 2x^2 + y^4 + 2y^2 + 2x^2 y^2 < 0 $$
Although we will take another approach. Rewrite the condition to be $$ |(z+1)(z-1)|<1 \implies |z+1|\cdot|z-1|<1 $$
We can see clearly 3 things... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve for $x$ and $y$, The equations are $x \cos^{3} y+3x \cos y \sin^{2} y =14 $ and $ x \sin^{3} y+3x \cos^{2} y \sin y = 13 $ Consider the system of equations
$$x \cos^{3} y+3x \cos y \sin^{2} y =14 $$
$$ x \sin^{3} y+3x \cos^{2} y \sin y = 13 $$
$1)$ the values of $x$ is /are..
Answer is $ \pm\sqrt 5 $
The number ... | When you render $\tan y =1/2$, it follows that
$1+\tan^2 y = \sec^2 y = 1/\cos^2 y = 5/4$
But then you have to allow both the positive and negative square roots for the cosine. One period of the cosine is $2\pi$ which is two periods of the tangent function, so both the positive and negative cosine values occur in diff... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Simplifying fractional surds I have this fractional surd:
$$\frac{5\sqrt{7}+4\sqrt{2}}{3\sqrt{7}+5\sqrt{2}}$$
I can calculate this with a calculator fairly easily obviously but what is the best tactic without one?
Thank you!
| So first remember some key rules:
*
*$\sqrt{a} \times \sqrt{a}=a$
*$\sqrt{a} \times \sqrt{b}=\sqrt{ab}$ (which you can simplify usually)
(There are more but we only really need these!)
So a good technique to use in this question is called rationalising the denominator. Which basically means take the denominator o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Different methods give different answers. Let A,B,C be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Let $A,B,C$ be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Find all possible values of $p$ such that $A,B$ and $C$ are angles of a triangle.
case 1- discriminant
We can re... | We should consider discriminants but in the way they arise directly:
Let $tx = \tan(x)$ for shorthand notation
$$t_A= t{(\pi-B-C)}$$
$$1=\frac{-(tA+tB)}{1-tA\, tB} $$
$$tA \,tB=p\quad$$
plug in and simplify to find quadratic equation roots
$$ tB^2+(1-p)tB+p=0 $$
$$ 2\,tA=-(1-p)-\sqrt{1-6p+p^2} $$
$$ 2\,tB=-(1-p)+\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 6
} |
Prove that $2^{2^{2^{\cdot^{\cdot^{2}}}}} \mod 9 = 7$
Prove that $\underbrace{2^{2^{2^{\cdot^{\cdot^{2}}}}}}_{2016 \mbox{
times}} \mod 9 = 7$
I think that it can be done by induction:
Base:
$2^{2^{2^{2}}} \equiv 2^{16} \equiv 2^8 \cdot 2^8 \equiv 2^4 \cdot 2^4 \cdot2^4 \cdot2^4 \equiv 7^2 \cdot 7^2 \equiv 4 \cdot ... | For the correct way to do modular exponent reduction see here. Using that method
$\!\!\bmod 9\!:\,\ 2^{\Large\color{#c00} 6}\equiv 1\ \ {\rm thus}\ \ 2^{\large N}\equiv\, 2^{\large N\Large \bmod\, \color{#c00}6}\ $
$\text{Applying that here}\!:\ 2^{\Large {2^{\Large 2K}}}\!\!\equiv\, 2^{\Large \color{#0a0}{2^{\Large... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3225521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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If $a \equiv 9 \pmod {12}$, find all possible values for $\gcd(a^2+21a+72,252)$ We know that
$$a^2+21a+72 \equiv 9^2 + 21 \cdot 9 + 0 \equiv 6 \pmod {12}$$
So we know that that expression, let's say $\alpha$, is such that $12 \mid \alpha - 6$. But then $12 \mid 2(\alpha - 6)=2a+12$, which means that $\alpha$'s prime fa... | Write $a=12b+9$. Then $a^2+21a+72=18 (8 b^2 + 26 b + 19)$ and so $$\gcd(a^2+21a+72,252) = 18 \gcd(8 b^2 + 26 b + 19,14)$$
Since $8 b^2 + 26 b + 19$ is always odd, we have
$$\gcd(8 b^2 + 26 b + 19,14) = \gcd(8 b^2 + 26 b + 19,7) = \gcd(b^2 -2b -2,7) = \gcd((b-1)^2 -3,7) = 1
$$
because $3$ is not a square mod $7$.
Theref... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to factorized this 4th degree polynomial? I need your help to this polynomial's factorization.
Factorize this polynomials which doesn't have roots in Q.
$ \ f(x) = x^4 +2x^3-8x^2-6x-1 $
P.S.) Are there any generalized method finding 4th degree polynomials factor?
| For all real $k$ we have:
$$x^2+2x^3-8x^2-6x-1=(x^2+x+k)^2-x^2-k^2-2kx^2-2kx-8x^2-6x-1=$$
$$=(x^2+x+k)^2-((2k+9)x^2+2(k+3)x+k^2+1).$$
Now, we'll choose a value of $k$ such that
$$(2k+9)x^2+2(k+3)x+k^2+1=(ax+b)^2.$$
Easy to see that $k=0$ is valid.
In the general we need to solve the following equation.
$$(k+3)^2-(2k+9)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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How to get sum of $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+...+\frac{1}{(1+x^2)^n}$ using mathematical induction Prehistory: I'm reading book. Because of exercises, reading process is going very slowly. Anyway, I want honestly complete all exercises.
Theme in the book is mathematical induction. There were examples, where w... | For any $n$ in your calculations(which are right but missed the right trick) use the following:
first set
$${{S}_{n}}=q+{{q}^{2}}+\ldots +{{q}^{n}}$$
then
$$q{{S}_{n}}={{q}^{2}}+{{q}^{3}}+\ldots +{{q}^{n+1}}$$
al last
$$\begin{align}
& {{S}_{n}}-q{{S}_{n}}=q-{{q}^{n+1}} \\
& {{S}_{n}}=\frac{q\left( 1-{{q}^{n}} \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Given three positve numbers $a,b,c$. Prove that $\sum\limits_{cyc}\frac{a}{\sqrt{b(a+b)}}\geqq \sum\limits_{cyc}\frac{a}{\sqrt{b(c+a)}}$ .
Given three positive numbers $a, b, c$, prove that $$\sum\limits_{cyc}\frac{a}{\sqrt{b(a+ b)}}\geqq \sum\limits_{cyc}\frac{a}{\sqrt{b(c+ a)}}.$$
I tried on Holder inequality and ht... | The inequality is true. The Buffalo Way works.
Here is a hint. Currently, I do not give the Buffalo Way part for people to attempt.
$\phantom{2}$
With the substitutions $(a, b, c) \to (a^2, b^2, c^2)$, the inequality becomes
$$\sum_{\mathrm{cyc}} \frac{a^2}{b\sqrt{a^2+b^2}} \ge \sum_{\mathrm{cyc}} \frac{a^2}{b\sqrt{c^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Hard limit involving different order radicals $\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$ Please help me to calculate the following limit
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$$
I factored out $n$ from both radicals but it didn't work and tried to use the identity $a^2-b^2$ and $a^3-b^... | First let's get rid of the square root:
$$\lim_{n \to \infty} (\sqrt[3]{n^3-3n^2}-\sqrt{n^2+2n})=\lim_{n \to \infty} \frac{\sqrt[3]{(n^3+3n^2)^2}-(n^2+2n)}{\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n}}\\= -\lim_{n \to \infty} \frac{n^2+2n-\sqrt[3]{(n^3+3n^2)^2}}{\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n}}.$$
Now let's get rid of the cub... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
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Even stronger than Sophomore's dream Sophomore's dream states that:
$$
\int_0^1x^{-x}dx=\sum_{n=1}^\infty n^{-n}
$$
and
$$
\int_0^1x^{x}dx=-\sum_{n=1}^\infty(-n)^{-n}
$$
A friend of mine noticed that numerically:
$$
\int_0^1\int_0^1(xy)^{xy}dxdy\approx 0.7834... \approx \int_0^1x^{x}dx
$$
Are these two integrals reall... | Yes they are equal! The trick is to use the binomial theorem to separate the integrals:
\begin{align}
\int_0^1\int_0^1(xy)^{xy}dxdy &= \int_0^1\int_0^1\exp(xy\log{xy})dxdy \\
&= \sum_{n=0}^\infty\frac{1}{n!}\int_0^1\int_0^1(xy\log{xy})^ndxdy \\
&= \sum_{n=0}^\infty\frac{1}{n!}\int_0^1\int_0^1(xy)^n(\log{x}+\log{y})^ndx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Can we show a sum of symmetrical cosine values is zero by using roots of unity? Can we show that
$$\cos\frac{\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{5\pi}{7}+\cos\frac{6\pi}{7}=0$$
by considering the seventh roots of unity? If so how could we do it?
Also I have observed that
$$\cos\f... | Pointing at the link I left in the comments
$$1 + \cos\theta + \cos2\theta +... + \cos n\theta = \frac{1}{2} + \frac{\sin[(n+\frac{1}{2})\theta]}{2\sin(\frac{\theta}{2})}$$
Then for $\forall n\in\mathbb{N}, n>0$
$$\cos\frac{\pi}{n+1}+ \cos\frac{2\pi}{n+1} +... + \cos \frac{n\pi}{n+1} = \frac{\sin\left[(n+\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Determine this limit $\lim_{n\to\infty} \frac{\ln\left(\frac{3\pi}{4} + 2n\right)-\ln\left(\frac{\pi}{4}+2n\right)}{\ln(2n+2)-\ln(2n)}.$ how can I determine the following limit? $$\lim_{n\to\infty} \frac{\ln\left(\frac{3\pi}{4} + 2n\right)-\ln\left(\frac{\pi}{4}+2n\right)}{\ln(2n+2)-\ln(2n)}.$$
This question stems from... | $$a_n= \frac{\ln\left(\frac{3\pi}{4} + 2n\right)-\ln\left(\frac{\pi}{4}+2n\right)}{\ln(2n+2)-\ln(2n)}=\frac{\log \left(1+\frac{\pi }{4 n+\frac{\pi }{2}}\right) } {\log \left(1+\frac 1n \right)}$$ Using Taylor expansions
$$a_n=\frac{\frac{\pi }{4 n}-\frac{\pi ^2}{16 n^2}+O\left(\frac{1}{n^3}\right) } {\frac{1}{n}-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Solve $\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $ Solve $$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $$
My attempt:
Let $A = \sqrt{1+x}, B = \sqrt{1-x}$ and then by squaring the problematic equation we get:
$$(... | Substituting $$a=\sqrt{1+x},b=\sqrt{1-x}$$ and using that $$a^2+b^2=2$$ and $$a^3+b^3=(a+b)(a^2+b^2-ab)$$ we get$$\sqrt{1+ab}(a+b)(2-ab)=2+ab$$
And we get by squaring $$(1+ab)(2+2ab)(2-ab)^2=(2+ab)^2$$
and with $$u=ab$$ we get
$$2(1+u)^3(2-u)^2=(2+u)^2$$
The solutions are $$\left\{\left\{x\to
-\sqrt{-\frac{7}{4}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 2
} |
Proof that $ \frac{3\pi}{8}< \int_{0}^{\pi/2} \cos{\sin{x}} dx < \frac{49\pi}{128}$ Proof that
$$
\frac{3\pi}{8} <
\int_{0}^{\pi/2} \cos\left(\sin\left(x\right)\right)\,\mathrm{d}x <
\frac{49\pi}{128}
$$
Can somebody give me some instruction how to deal with inequality like that? My current idea is:
I see $\frac{3\pi}{... | Hint for one part, using this inequality and this one
$$\cos{x} \geq 1 - \frac{x^2}{2}$$
we have
$$\int\limits_{0}^{\frac{\pi}{2}}\cos{\sin{x}} dx > \int\limits_{0}^{\frac{\pi}{2}} \left(1-\frac{\sin^2{x}}{2}\right)dx=\frac{3 \pi}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Finding the angle between two tangents of a curve I just got back from my math test. In the test I encountered the following question:
A curve is defined by the equation $$x^3+y^3=3xy$$ Find the angle between the tangents at points (-1,1) and (1,2).
After differentiating implicitly, we get:
$\frac{dy}{dx} = \frac{y-x... | Slope formula: $m = \frac{y-x^2}{y^2-x}$
At $P_1 \equiv( -1,1)$, $m_1 = \frac{1-0}{1+1} = 0$
At $P_2\equiv(1,2)$, $m_2 = \frac{2-1}{4-1}=\frac{1}{3}$
Acute angle between the 2 tangents,
$\phi = \tan^{-1}\big\vert\frac{m_1-m_2}{1+m_1m_2}\big\vert = \tan^{-1}\big\vert\frac{0-\frac{1}{3}}{1+0} \big\vert = \tan^{-1}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$ How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$
| Writing $\cos^2\theta=c,\sin^2\theta=s\implies c+s=1$
$\cos^32\theta + 3\cos2\theta$
$=(c-s)^3+3(c-s)$
$=c^3-s^3-3cs(c-s)+3(c-s)$
$=c^3-s^3+3(c-s)(1-cs)$
$=c^3-s^3+3(c-s)((c+s)^2-cs)$
$=c^3-s^3+3(c^3-s^3)=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the slope of a normal line to a curve given curve equation and x coordinate What are the steps to solving this problem? (Thank you in advance)
Find the slope of the normal line to the curve (x^2)-(xy)+(y^2) = 7 at the point where x=1. There are two possible answers - accept the SMALLER answer.
Where I started:
y' ... |
Slope of the normal at a point $P =m= -\frac{dx}{dy}|_P = -\frac{1}{\frac{dy}{dx}}\bigg|_P = -\frac{1}{y'|_P}$
As $x^2 - xy + y^2 =7$, by differentiating w.r.t $x$,
$2x - xy' - y + 2yy' = 0$
$y' = \frac{2x-y}{x-2y}$
and $$-\frac{1}{y'} = -\frac{x-2y}{2x-y}$$
At $x=1$, $1 -y+y^2 = 7 \implies y^2 -y -6 = 0\implies ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the roots of equation based on some geometry hints Plots of the equations $y = 8 - x^2$ and $|y|=\sqrt{8+x}$ are symmetric w.r.t. the line $y=-x$. We have to solve the equation $$8-x^2=\sqrt{8+x}$$
| $y_1=8-x^2$ and $|y_2|=\sqrt{8+x}$ are symmetric w.r.t. the line $y=-x$ implies the tangent lines at the intersection points are reflected over $y=-x$.
Refer to the graph:
$\hspace{3cm}$
The line $y_1=ax+b$ is reflected over $y=-x$ to the line $y_2=\frac1ax+\frac ba$. It implies $a\cdot \frac1a=1$.
Reference: $y_1=8-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Help with $-\int_0^1 \ln(1+x)\ln(1-x)dx$ I have been attempting to evaluate this integral and by using wolfram alpha I know that the value is$$I=-\int_0^1 \ln(1+x)\ln(1-x)dx=\frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2$$
My Attempt:
I start off by parametizing the integral as $$I(a)=\int_0^1 -\ln(1+x)\ln(1-ax)dx$$
where $I=I(1)$... | \begin{align}J&=\int_0^1 \ln(1+x)\ln(1-x)\,dx\\
&=\Big[\left(\left(1+x\right)\ln(1+x)-x-2\ln 2+1\right)\ln(1-x)\Big]_0^1+\int_0^1\frac{\left(1+x\right)\ln(1+x)-x-2\ln 2+1}{1-x}\,dx\\
&=\int_0^1\frac{\left(1+x\right)\ln(1+x)-x-2\ln 2+1}{1-x}\,dx\\
\end{align}
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3246020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
Inverse Laplace Transform of $\frac{1}{\lambda\cosh(\sqrt{as}+\sqrt{bs})+(1-\lambda)\cosh(\sqrt{as}-\sqrt{bs})}$. I am trying to find the inverse Laplace transform of some function of the form:
$$
\mathrm{F}\left(s\right) =
\frac{1}{\lambda\cosh\left(\,\sqrt{\, as\,}\, + \,\sqrt{\, bs\, }\right) +
\left(1 - \lambda\rig... | Concerning the zeros for
$$
\lambda\cosh(\sqrt{as}+\sqrt{bs})+(1-\lambda)\cosh(\sqrt{as}-\sqrt{bs}) = 0
$$
making $s = x + i y$ and taking the real and the imaginary parts
$$
R=\lambda \cos \left(\left(\sqrt{a}+\sqrt{b}\right) \sqrt[4]{x^2+y^2} \sin \left(\frac{1}{2} \arg (x+i
y)\right)\right) \cosh \left(\left(\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3248835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Rationalizing $\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}$ Problem
Rationalize $\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}$
So, I'm training for the Mexican Math Olimpiad. This is one of the algebra problems from my weekly training.
Before this problem, there was other very similar, after proving it, there's an ... | Maple rationalizes it as
$$ -{\frac { \left( {x}^{2/3}+2\,\sqrt [3]{x}\sqrt [3]{y}-\sqrt [3]{z}
\sqrt [3]{x}+{y}^{2/3}-\sqrt [3]{z}\sqrt [3]{y}+{z}^{2/3} \right)
\left( 3\,{x}^{5/3}\sqrt [3]{y}-6\,{x}^{2/3}{y}^{4/3}+3\,{x}^{2/3}
\sqrt [3]{y}z-6\,{x}^{4/3}{y}^{2/3}+3\,\sqrt [3]{x}{y}^{5/3}+3\,\sqrt
[3]{x}{y}^{2/3}z-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"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.