Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Complex number inequality $|z-1| \ge \frac{2}{n-1}$
If $n \ge 3$ is an odd number and $z\in\mathbb{C}, z\neq -1$ such that $z^n=-1$, prove that
$$|z-1|\ge \frac{2}{n-1}$$
I was thinking that $-2=z^n-1=(z-1)(z^{n-1}+z^{n-2}+...+z^2+z+1)$ and because $z\neq -1$, the second factor can not be $0$:
$$|z-1|=\frac{2}{|z^{n... | If $z^n = -1$, we have
$$
z = \exp\left(\frac{(2k + 1)i \pi}{n}\right),
\qquad k \in \{0, \ldots,n - 1\}
$$
Denote for $k \in \{0, \ldots,n - 1\}$
$$
a_k := \frac{(2k + 1) \pi}{n}
$$
Then
\begin{align}
| z - 1 |^2
& = | \cos(a_k) + i \sin(a_k) - 1 |^2
= (\cos(a_k) - 1)^2 + \sin^2(a_k) \\
& = \cos^2(a_k) + \sin^2(a_k) -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Proof of $n^3/3 + n^2/2 + n/6 = [1^2 + 2^2 + ... + n^2]$? I am having a hard time following this proof. Here is how it goes.
$$
(k+1)^3 = k^3+3k^2+3k+1\\
3k^2+3k+1 = (k+1)^3-k^3\\
$$
if $ k = 1, 2, 3, ... , n-1$ we add all the 5 formulas like this
$$
3(1)^2+3(1)+1 = ((1)+1)^3-(1)^3\\
3(2)^2+3(2)+1 = ((2)+1)^3-(2)^3\\
... | $$ 3\left[1^2+2^2 + ... + (n-1)^2\right] + 3\left[1 + 2 + ... + (n-1)\right] + (n-1) = n^3-1^3 $$
$$3\left[1^2+2^2 + ... + (n-1)^2\right] + 3\left[1 + 2 + ... + (n-1)\right] + n = n^3 $$
$$3\left[1^2+2^2 + ... + (n-1)^2+n^2\right] + 3\left[1 + 2 + ... + (n-1)\right] + n = n^3+3n^2 $$
$$3\left[1^2+2^2 + ... + (n-1)^2+n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find a limit with sqrt $\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$ $$\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$$
I don't know how to rewrite or rationalize in order to find the limit.
| If you write the expression as $$x^4\left(1-\sqrt{1+6/x^2}+3/x^2\right)=\frac{1-\sqrt{1+6/x^2}+3/x^2}{1/x^4},$$ then you may be able to apply the Marquis de L'hopital's method.
After exactly two iterations, you get $$\frac92\frac{\frac{1}{\sqrt{1+6/x^2}}-1}{1/x^2},$$ where you may now take an elementary limit as $x\to+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
} |
Hard inequality problem (if $(a_2-a_1)^2 + (a_3-a_2)^2 + \ldots + (a_{2n}-a_{2n-1})^2 = 1$ ...) I have a hard problem :
If $$(a_2-a_1)^2 + (a_3-a_2)^2 + \ldots + (a_{2n}-a_{2n-1})^2 = 1$$ where $a_1,a_2...,a_{2n} \in \mathbb{R}$
What is the maximum of $$(a_{n+1}+a_{n+2}+…+a_{2n})−(a_1+a_2+…+a_n)?$$
I found that : $$1... | Following achille hui's suggestion: Let
$$s_k = a_k - a_{k-1}, \qquad k \ge 2 $$
with $\sum\limits_{k=2}^{2n} s_{k}^2 = 1$. Now the target function is
$$
f = (a_{n+1}+a_{n+2}+…+a_{2n})−(a_1+a_2+…+a_n) \\
= (a_{n+2}+…+a_{2n})−(a_1+a_2+…+a_{n-1}) +a_{n+1} - a_{n} \\
= (a_{n+2}+…+a_{2n})−(a_1+a_2+…+a_{n-1}) +s_{n+1} \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find general term of recursive sequences $ x_{n+1}=\frac{1}{2-x_n}, x_1=1/2,$ Please help to solve:
*
*$ x_{n+1}=\frac{1}{2-x_n}, x_1=1/2,$
*$x_{n+1}= \frac{2}{3-x_n}, x_1=1/2$
I know answers, but can't figure out the solution.
The first one is obvious if you calculate first 3-5 terms by hand. But how can I get th... | A recurrence of the form:
$\begin{equation*}
w_{n + 1}
= \dfrac{a w_n + b}{c w_n + d}
\end{equation*}$
with $a d \ne b c$ and $c \ne 0$ is called a Ricatti recurrence. One way to solve them is to recognize the right hand side is a Möbius transform, and those can be composed like matrix products:
$\begin{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3557190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\sum _{j=0}^n \frac{\binom{n}{j}^2}{(j+1)^2}$ When testing Mathematica, I accidentally found this equality:
$$\sum _{j=0}^n \frac{\binom{n}{j}^2}{(j+1)^2}=\frac{2^{2 n+2} \Gamma \left(\frac{1}{2} (2 n+3)\right)}{\sqrt{\pi } (n+1)^3 \Gamma (n+1)}-\frac{1}{(n+1)^2}$$
Mathematica gives this closed-form directly,... | Use Binomial identity:
$$
(1+t)^n=\sum_{j=0}^{n} {n \choose j}t^n.
\tag{1}
$$
Integration of $(1)$ from $t=0$ to $t=x$ gives
$$
\frac{(1+x)^{n+1}-1}{n+1}= \sum_{j=0}^n {n \choose j}\frac{x^{j+1}}{j+1}.\tag{2}
$$
We can change $x$ to $1/x$ in $(2)$ to get
$$
\frac{(1+1/x)^{n+1}-1}{n+1}= \sum_{j=0}^n {n \choose j}\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to factor $x^6-4x^4+2x^3+1$ by hand? I generated this polynomial after playing around with the golden ratio. I first observed that (using various properties of $\phi$), $\phi^3+\phi^{-3}=4\phi-2$. This equation has no significance at all, I just mention it because the whole problem stems from me wondering: which ot... | Your original method is tedious but it can be done.
You can show that $(x^3+Ax^2+Bx+C)(x^2+Dx+E)$ is equal to:
$$x^5+(D+A)x^4+(1+AD+B)x^3 + (AE+BD+C)x^2 + (BE+CD) + CE$$
so $A+D = 1, B+AD+1 = -3, AE+BD+C=-1, BE+CD=-1, CE=-1$.
Assuming $A,B,C,D,E$ are all integers, we either have $C=-1, E=1$ or $C=1, E=-1$.
If $C=-1, E=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is $\sqrt{-5}$ a prime in $\mathbb{Z}{[\sqrt{-5}]}$? I know it is irreducible over $\mathbb{Z}{[\sqrt{-5}]}$ but since it is not even UFD, so we can't conclude primality from irreducibility.
My guess is yes since $N(\sqrt{-5})=5$ is prime. I started with $ab=c\sqrt{-5}$ where a, b and c are in $\mathbb{Z}{[\sqrt{-5}]... | Say $5|N(a)$. Since $N(x+y\sqrt{-5}) = x^2+5y^2$, if $a=x+y\sqrt{-5}$ then $5|x$. So $a=5z+y\sqrt{-5} = \sqrt{-5}(y-z\sqrt{-5})$.
You can do this without invoking the norm, just low-tech: Note that an element $x+y\sqrt{-5}$ of $\mathbb{Z}[\sqrt{-5}]$ is a multiple of $\sqrt{-5}$ if and only if $x$ is a multiple of $5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3566189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find sum of geometric-like series with binomial coefficients using complex analysis Studying analytic number theory, I stumbled across the problem of finding the sum of the series
$\sum_{n=0}^{\infty}\binom{2n}{n}\left(\frac{1}{5}\right)^n$
A professor gave me the hint of "using basic complex analysis" but honestly, I ... | Consider the integral
$$\int_0^{2\pi}4^n\cos^{2n}(x)\:dx$$
Expanding out using Euler's formula and binomial expansion, we get that
$$\int_0^{2\pi}4^n\cos^{2n}(x)\:dx = \int_0^{2\pi}(e^{ix}+e^{-ix})^{2n}\:dx = \sum_{k=0}^{2n} {2n \choose k} \int_0^{2\pi}e^{i(2n-2k)x}\:dx$$
The integral on the right will always be $0$ un... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3566990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Simplify $(1+i^\frac{1}{2})^\frac{1}{2}$ How can I simplify $\sqrt{1+\sqrt{i}}$?
I thought about making $z^2=1+\sqrt{i}$ and then $w=z^2$
But I'm not really sure
| Using polar coordinates $\displaystyle \sqrt{1+\sqrt{i}} = \sqrt{1+\sqrt{e^{i \pi/2}}} = \sqrt{1 + e^{i \pi/4}} = \sqrt{1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}} \, $ from this we have
$\displaystyle r^2 = \left( 1 + \frac{\sqrt{2}}{2} \right)^2 + \left( \frac{\sqrt{2}}{2} \right)^2 = 1 + \sqrt{2} + \frac{1}{2} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3567232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Let $U_1, U_2, \ldots$ be a sequence of independent random variables with PDF $f(u) = 1, \;\; 0 < u < 1$. Let $U_1, U_2, \ldots$ be a sequence of independent random variables with PDF
$$f(u) = 1, \;\; 0 < u < 1$$
Find $E[N]$ when $N = \min\left\{n\mid \sum\limits_{i = 1}^n U_i > 1\right\}$.
Hint: For $x\in (0,1)$, defi... | For $x\ge 0$, let $N(x)=\min\left\{n:\sum_{i=1}^n U_i>x\right\}$ as in the hint, so that $N=N(1)$. We want to calculate $$p(x,n)=\Bbb P\big[N(x)=n\big]$$ when $n$ is a positive integer and $0\le x\le 1$.
For a positive integer $k$, let $S_k=U_1+U_2+\ldots+U_k$ with probability density $f_k$. For $0\le x\le 1$, ob... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Multinomial theorem with imposed conditions
The number of ways in which 12 identical balls can be grouped in four marked non-empty sets $P, Q, R, S$ such that $n(P) < n(Q)$ is?
The answer to the above problem is the number of positive integral solutions of the expression $$P + Q + R + S = 12$$
when $P<Q$. I know that... | Let $a,b,c,d\ge0$. Then set $|P|=a+1$, $|Q|=a+b+2$, $|R|=c+1$, $|S|=d+1$ where $2a+b+c+d+5=12$. We want to count the non-negative solutions to
$$
2a+b+c+d=7\tag1
$$
Considering the coefficients of $x^n$ in
$$
\overbrace{\left(1+x^2+x^4+x^6+\dots\right)\vphantom{{x^2}^3}}^a\overbrace{\left(1+x+x^2+x^3+\dots\right)^3}^{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3571211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)+\log_{zx}\left(\frac{y^2+4}{4\sqrt{zx}}\right)+\log_{xy}\left(\frac{z^2+4}{4\sqrt{xy}}\right)=0$ Can I ask how to solve this type of equation:
$$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)+\log_{zx}\left(\frac{y^2+4}{4\sqrt{zx}}\right)+\log_{xy}\left(\frac{z^2+4}{4\... | Notice that we can use the following property $\log_a \frac{b}{c}=\log_a b-\log_a c$ to get:
$$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)=\log_{yz}\frac{x^2+4}{4}-\log_{yz}\sqrt{yz}=\log_{yz}\frac{x^2+4}{4}-\frac{1}{2}$$
The equation can, thus, be written as:
$$\log_{yz}\frac{x^2+4}{4}+\log_{zx}\frac{y^2+4}{4}+\lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3572967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding the joint distribution of two dependent random variables given conditions
Suppose $X \sim U(0,1)$ and $Y|X=x \sim U(0,1-x)$. Find the joint
distribution of $(X,Y)$ given that $X\le \frac{1}{2}$, $Y\le
\frac{1}{2}.$
My attempt:
The pdf of $X$ is $f_{X}(x) = 1$ for $x \in (0,1)$ and the pdf of $Y|X=x$ is $f_... | \begin{align}
P(Y\le y, X\le x)&= \int_{-\infty }^x P\left(Y\le y\ |\,X=t\right)f_X(t)dt\\
&=\cases{\displaystyle 0& if $\ x\le0\ $or$\ y\le0$\\
\displaystyle\int_0^x\frac{y}{1-t}dt&if $\ 0<x\le1-y,y< 1$\\
\displaystyle\int_0^{1-y} \frac{y}{1-t}dt+\int_{1-y}^xdt&if $\ 0<y<1, 1-y<x<1$\\
\displaystyle\int_0^{1-y} \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
At least one even number among $\{ \lfloor 2^{n}\sqrt{2} \rfloor, \lfloor 2^{n+1}\sqrt{2} \rfloor,..., \lfloor 2^{2n}\sqrt{2} \rfloor \}$
For any positive integer $n$, prove that the set
$$\{ \lfloor 2^{n}\sqrt{2} \rfloor, \lfloor 2^{n+1}\sqrt{2} \rfloor,..., \lfloor 2^{2n}\sqrt{2} \rfloor \}$$
contains at least one e... | For the sake of contradiction assume that each element in the set is odd. Then, for some $m \geq 1$, we have:
$$2m-1 < 2^n\sqrt{2}<2m$$
and multiplying by $2$:
$$4m-2 < 2^{n+1}\sqrt{2}<4m$$
However, since $\lfloor 2^{n+1}\sqrt{2}\rfloor$ is odd, then
$$4m-1<2^{n+1}\sqrt{2}<4m$$
Repeating the process
$$2^{n+1}m-1<2^{2n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Find the minimum value of $x^2+y^2$, where $x,y$ are nonnegative integers and $x+y=k$. Question: Let $k$ be a fixed odd positive integer. Find the minimum value of $x^2+y^2$, where $x,y$ are nonnegative integers and $x+y=k$.
My approach: After trying some examples I can conjecture that, the minimum value of $x^2+y^2$ i... | AM-GM: $x+y=k \geqslant 2 \sqrt{xy}$
${xy} \leqslant \frac {k^2} {4} $
$x^2+y^2=k^2-2xy\geqslant \frac {k^2} {2}$
If $k$ is even, then $minimum\; x^2+y^2=\frac {k^2} {2}$
If $k$ is odd, then
$minimum\; x^2+y^2=\frac {k^2+1} {2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to show that $b_n(n) \neq 0$ for $b_n(x) = -\frac{16 x}{n} b_{n-1}(x) - \frac{48 x}{n} b_{n-2}(x)$? Lets define the recursive polynomial:
$b_0(x) = 1$
$b_1(x) = -16 x$
$b_n(x) = -\frac{16 x}{n} \left( b_{n-1}(x) + 3 b_{n-2}(x) \right) = -\frac{16 x}{n} b_{n-1}(x) - \frac{48 x}{n} b_{n-2}(x)$ ; with $x \in \mathbb{R... | If we fix $n$ and define $a_k=b_k(n)$, then indeed we obtain a recursion
$$a_0=1,\quad a_1=-16n,\quad a_k=-16a_{k-1}-48a_{k-2}. $$
From here we infer that
$$\tag1 a_k=A(-12)^k+B(-4)^k$$
for suitable $A,B$.
The desired result is that this makes $a_n\ne0$.
Clearly, $a_0\ne 0$. Also, when $n=1$, we have $a_1=-16\ne 0$, so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3575346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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For how many integers $n$ is $n^6+n^4+1$ a perfect square? QUESTION
For how many integers $n$ is $n^6+n^4+1$ a perfect square?
I am completely blank on how to start. Could anyone please provide tricks on how to get a start on such questions?
Thanks for any answers!
| $n^6 + n^4 + 1 = n^6 + n*n^3 + 1=n^6 + 2\frac n2*n^3 + \frac {n^2}4 +(1-\frac {n^2}4) =(n^3 + \frac n2)^2 + (1-\frac {n^2}4)$
So if $n\ge 2$ then $n^6 + n^4 + 1 \le (n^3 + \frac n2)^2$ with equality holding only if $n = 2$.
And $(n^3 +\frac {n}2 - 1)^2 =(n^3 +\frac {n-2}2)^2 = n^6 + (n-2)n^3 + \frac {n^2-4n -4}4=n^6 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How to obtain the asymptote formula of $\int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx$? I want to obtain the formula in the form as follows:
$$ \int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx=A+\frac{B}{n}+\frac{C}{n^2}+\frac{D}{n^3}+o\left(\frac{1}{n^3}\right).$$
At least, it holds that $$\int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx = \int_{0}^{... | You properly wrote
$$I_n=\int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx =\frac{\sqrt \pi}2\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\right)}{\Gamma\left(1+\frac{1}{2n}\right)}$$ Now, take logarithms for the gamma functions, useStirling approximation of the gamma function and continue with Taylor expansions to get
$$\log \left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3581473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Simplify $(1+x^2 )^{1/2}-(1-x^2)^{1/2}$ I need to simplify the following expression in a way that introduces minimal floating point cancellation errors.
$$(1+x^2 )^{\frac{1}{2}}-(1-x^2 )^{\frac{1}{2}}$$
The errors accumulate when numbers close together are subtracted from each other. I get
$$\sqrt{2}\left[1-\left(1-x^4... | I would try this:
$$\begin{align}(1+x^2 )^{\frac{1}{2}}-(1-x^2 )^{\frac{1}{2}}&=[(1+x^2 )^{\frac{1}{2}}-(1-x^2 )^{\frac{1}{2}}]\frac{(1+x^2 )^{\frac{1}{2}}+(1-x^2 )^{\frac{1}{2}}}{(1+x^2 )^{\frac{1}{2}}+(1-x^2 )^{\frac{1}{2}}}\\&=\frac{(1+x^2 )-(1-x^2 )}{(1+x^2 )^{\frac{1}{2}}+(1-x^2 )^{\frac{1}{2}}}\\&=\frac{2x^2}{(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The integral: $\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$ The integral: $$\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$$
has been encountered today while solving a longlish problem at MSE. The question here is: How would one evaluate it?
Addendum
For an interesting use of this... | $$I=\int_{0}^{\pi/2} \frac{\sin x \cos^5x}{(1-2\sin^2 x \cos^2 x)^2} dx$$
Next using $\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx,$
we get $$I=\int_{0}^{\pi/2} \frac{\sin^5 x \cos x}{(1-2\sin^2 x \cos^2 x)^2} dx$$ Adding the two integrals, we get
So $$2I=\int_{0}^{\pi/2} \frac{\sin x \cos x(\sin^4 x+ \cos^4 x)}{(1-2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3585397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$ takes all real values for $x\in R$ are: => The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$
takes all real values for $x\in R$ are:
My question is why we need to validate end points i.e. $1,7$ (Refer the last part of my attempt)
My... | When $a=1$, $\displaystyle y=\frac{x^2+3x-4}{1+3x-4x^2}=\frac{-x-4}{1+4x}\ne-\dfrac14$ for all $x\in\mathbb{R}$.
When $a=7$, $\displaystyle y=\frac{7x^2+3x-4}{7+3x-4x^2}=\frac{7x-4}{7-4x}\ne-\dfrac74$ for all $x\in\mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3585702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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show that $\sqrt{n+1}-\sqrt{n} \rightarrow 0$ as $n \rightarrow \infty$ show that $\sqrt{n+1}-\sqrt{n} \rightarrow 0$ as $n \rightarrow \infty$
Here is the algebric proof:
We have $a_n=\sqrt{n+1}-\sqrt{n}$, and we want to show that $\lim a_n=0$.
$$\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\... | Observe that $\sqrt{n+1} > \sqrt{n}$, so, $\sqrt{n+1} + \sqrt{n} > 2\sqrt{n}$ and then
$$\frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{2\sqrt{n}} \ .$$
Now, let $\varepsilon>0$. Since $4\varepsilon^2$ is a positive real number, by the Archimedian property of $\mathbb{R}$ there exists $N \in \mathbb{N}$ such that
$$0< \fra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Particular generators of GL(2,R) I want to prove that $$ \text{GL}_{2}( \mathbb{R} ) = \left\langle \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & a \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}; \ a \in \mathbb{R} \setminus \{0\} \... | As noted by @LouisHainaut, these four matrices enable you to do the elementary row operations. Since it is possible to get to and from the identity matrix to any invertible matrix by doing elementary row operations, the result follows.
The first two allow you to multiply a row by a scalar. The last two, to add a mul... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of $\frac{a_{n+1}}{a_n}$, where $|a_{n+1}a_{n-1} - a_n^2| = 1 $ Let $(a_n)$ a non-decreasing sequence of positive real numbers such that $\lim a_n = \infty$ and $|a_{n+1}a_{n-1} - a_n^2| = 1 $. Prove that the sequence $(\frac{a_{n+1}}{a_n})$ converges.
My little ´´progress´´:
$|a_{n+1}a_{n-1} - a_n^2| = 1 $... | We separate two cases.
Case 1: for every $n \geq 0$, we have $a_{n + 1} - a_n \leq 1$.
In this case, since the sequence $(a_n)_n$ tends to infinity by assumption, we have $\lim\limits_{n\rightarrow\infty}\frac{a_{n + 1} - a_n}{a_n} = 0$, and it follows that $\lim\limits_{n\rightarrow\infty}\frac{a_{n + 1}}{a_n} = 1$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculate $\mathbb{E}(X-Y\mid 2X+Y).$ if $X\sim N(0,a)$ and $Y\sim N(0,b)$
Question: Given that $X$ and $Y$ are two random variables satisfying $X\sim N(0,a)$ and $Y\sim N(0,b)$ for some $a,b>0$. Assume that $X$ and $Y$ have correlation $\rho.$
Calculate
$$\mathbb{E}(X-Y \mid 2X+Y).$$
I tried to use the fact t... | By @Kavi Rama Murthy answer (and me in other answer)
$$E(X-Y|2X+Y)=A(2X+Y)$$
Now By the Projection property ,$E(X-Y|2X+Y)$ minimized
$$E(X-Y-g(2X+Y))^2$$ conditional-expectation-as-best-predictor
I want to find $A$ by minimizing $E(X-Y-A(2X+Y))^2$
$$E(X-Y-A(2X+Y))^2=E((1-2A)X-(1+A)Y)^2$$
$$=E((1-2A)X)^2+E((1+A)Y)^2
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solutions of the equation $x=\sqrt{2 + \sqrt{2 +\sqrt 2+.....}}$ I have seen the equation $x=\sqrt{2 + \sqrt{2 +\sqrt 2+.....}}$ in many places and the answer is $x=2$ which is obtained by making the substitution for $x$ inside the radical sign i.e $x=\sqrt{2+x}$ which renders the quadratic $x^2 -x -2=0$
having the so... | Note that if that expression defines a number $x$, then $x$ satisfies $x^2-x-2$. That $x$ satisfies some other relations is not relevant. That some other numbers satisfy those other relations is even less relevant; those other numbers do not satisfy $x^2-x-2$.
As an aside; to prove that $x=2$ it is also necessary to pr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $2^{29}$ has exactly 9 distinct digits There is a problem that say: The number $2^{29}$ has exactly 9 distinct digits. Which digit is missing?
It is an Olympiad problem and I see it solved by using remainder modulo 9.
$2^{29}=536870912$ interesting!
My Question: Can we find any mathematically way to show $2^{29}... | We know that $2^{10}=1024$, so we can calculate the value of $2^{29}$ by hand as follows
$2^{30} = 1024^3 = (10^3 + 24)^3 = 10^9 + 3.24.10^6 + 3.24^2.10^3 + 24^3$
Now $24=3.8$, so
$24^2 = 9.64 = 576 \\
\Rightarrow 3.24^2 = 1,728\\
24^3 = 24.576 = 3.8.576 = 3.4,608 = 13,824 \\
\Rightarrow 1024^3 = 10^9 + 72.10^6 + 1,72... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding a coefficient of $x^{57}$ in a polynomial $(x^2+x^7+x^9)^{20}$ So the task is to find a coefficient of $x^{57}$ in a polynomial $(x^2+x^7+x^9)^{20}$
I was wondering if there is a more intelligible and less exhausting strategy in finding the coefficient, other than saying that $(x^2+x^7+x^9)^{20}=((x^2+x^7)+x^9)... | $$
\begin{align}
\left[x^{57}\right]\left(x^2+x^7+x^9\right)^{20}
&=\left[x^{57}\right]\sum_{\substack{a+b+c=20\\a,b,c\ge0}}\frac{20!}{a!\,b!\,c!}x^{2a+7b+9c}\tag1\\
&=\sum_{\substack{2a+7b+9c=57\\a+b+c=20\\a,b,c\ge0}}\frac{20!}{a!\,b!\,c!}\tag2\\
&=\sum_{\substack{5b+7c=17\\b,c\ge0}}\frac{20!}{(20-b-c)!\,b!\,c!}\tag3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Euclid's algorithm to find multiplicative inverses over a polynomial field I'm trying to find the multiplicative inverse of $\overline{x+1}$ over the field $\mathbb{F}_3[x]/(x^3 + 2x + 1)$. I know I need to use Euclid's algorithm to do so, but I keep running into some difficulties.
I let $f(x) = x^3 + 2x + 1$ and $g(x)... | You could use polynomial long division, which as discussed & shown in the linked Wikipedia article is done similarly to base $10$ long division except rather than using powers of $10$, you use powers of $x$ instead. Also, similar to what the long division method does, you can notice by adding & subtracting various ter... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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There are two straight lines passing through the point A(2,0) which make an angle of 45 deg with the tangent at $A$...
There are two straight lines passing through the point $A(2,0)$ intersecting the tangent from $A$ of the circle $x^2+y^2+4x-6y-12=0$ under $45^{\circ}$. Find the equation of the circles with radius $3... | Observe that the two straight lines intersect the circles in two points $B,C$. Now, thanks to ortogonality of straight lines, you have $B\hat A C = 90^\circ$. So the points $A,B,C$ form a part of the square inscribed in the circle: moreover the diagonal of this square is the diameter of the circle that is $10$. Thanks ... | {
"language": "en",
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Is $\sum_{n=0}^\infty \frac{(x^2-5)^n}{2^n}$ a power series? The series $\sum_{n=0}^\infty \frac{(x^2-5)^n}{2^n}$ can be expressed as the following geometric series:
$$\sum_{n=0}^\infty \left(\frac{x^2-5}{2}\right)^n.$$
This series should converge if $$\left|\frac{x^2-5}{2}\right|<1.$$
This gives the possible values ... | A power series is any series of the form
$$\sum_{n} a_{n} \, (x-b)^n$$
which leads to saying the series in question is not a power series.
The two forms
$$\sum_{n=0}^{\infty} \left(\frac{x^2 - a^2}{b}\right)^n \quad \text{and} \quad \sum_{n=0}^{\infty} \frac{1}{b^n} \, ( (x - a)^2 + 2 a \, (x-a) )^n$$
provide the sam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3605164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Cubic roots and difference of cubes in limits $\lim\limits_{n\to\infty} (\sqrt[3]{n^6-6n^4+1} - n^2)$ Find the limit:
$$\lim\limits_{n\to\infty} (\sqrt[3]{n^6-6n^4+1} - n^2)$$
I applied the identity:
$$a^3 - b^3 = (a-b)(a^2+ab+b^2)$$
by multiplying the numerator and denominator by the complementary part.
$$\lim\limits_... | A possible way is turning it into a derivative using
$$n=\frac{1}{\sqrt t}$$
and consider $t\to 0^+$.
Hence,
$$\sqrt[3]{n^6-6n^4+1} - n^2 = \frac{\sqrt[3]{1-6t+t^3}-1}{t}$$
$$\stackrel{t\to 0^+}{\longrightarrow}\left.(\sqrt[3]{1-6t+t^3})'\right|_{t=0}=\frac{-6}{3}=-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3606707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$f(x) = x + \int_{0}^{1} (xy^2 + x^2y) f(y)dy$. Find $f(x)$ $f(x) = x + \int_{0}^{1} (xy^2 + x^2y) f(y) dy$. Find $f(x)$
I've tried taking $\int_{0}^{1} (xy^2 + x^2y) f(y) dy$ to be $k(x)$ since it comes out to be a function of $x$. That transforms our equation to $f(x) = x + k(x)$.
$f(y) = y + k(y) \implies (xy^2 + x... | $$f(x) = x + \int_{0}^{1} (xy^2 + x^2y) f(y) dy$$
$$f'(x) = 1 + \int_{0}^{1} (y^2 + 2xy) f(y) dy$$
$$f''(x) = \int_{0}^{1} 2y f(y) dy=\text{constant}$$
$$f(x)=ax^2+bx+c$$
$$ax^2+bx+c=x+ \int_{0}^{1} (xy^2 + x^2y) (ay^2+by+c) dy$$
After calculus of the integral and simplification :
$$(-\frac13 a+\frac13 b+\frac12 c)x^2+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the values of $x$ in terms of $a$ in $x^2+\frac{(ax)^2} {(x+a)^2} =3a^2$ The question is as follows.
Find the values of $x$ in terms of $a$ in
$x^2+\dfrac{(ax)^2}{(x+a)^2} =3a^2 $
My solution:
Multiply both sides by $(x+a)^2$ and expand. On rearranging we get
$x^4+2ax^3-a^2x^2-6a^3x-3a^4=0$
Now dividing by $a^4$... | Rewrite and factorize the equation as
$$\begin{align}
0 & = x^2(x+a)^2+(ax)^2 -3a^2 (x+a)^2\\
& = x^4+2x^3a+2x^2a^2 -3a^2 (x+a)^2 \\
& = x^4+2x^2a(x+a)-3a^2 (x+a)^2 \\
& = (x^2-a(x+a)) (x^2+3a(x+a))\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given: $\frac{\log_{10}z}{y-x} = \frac{\log_{10}y}{x-z} = \frac{\log_{10}x}{z-y},$ prove: a) $x^x\times y^y\times z^z =1,$ b) $x\times y \times z =1.$ Given: $ \frac{\log_{10}z}{y-x} = \frac{\log_{10}y}{x-z} = \frac{\log_{10}x}{z-y} $
Prove: $ a)\ x^x\times y^y\times z^z =1 \\ b) \ x\times y \times z =1 $
Here is... | Let $k$ be equal to our fractions:
$$ \frac{\log_{10}z}{y-x} = \frac{\log_{10}y}{x-z} = \frac{\log_{10}x}{z-y}=k.$$
Thus, $$\prod_{cyc}z=\prod_{cyc}10^{k(y-x)}=10^0=1.$$
Also, $$\prod_{cyc}z^z=\prod_{cyc}10^{kz(y-x)}=10^0=1.$$
I used $$\sum_{cyc}k(y-x)=k(y-x+z-y+x-z)=0$$ and
$$\sum_{cyc}kz(y-x)=k(zy-zx+xz-xy+yx-yz)=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For what $x$ and $y$ polynomial has maximum value?
For what $x,y\in\mathbb R$ does the polynomial
$$-5x^2-2xy-2y^2+14x+10y-1$$
attain a maximum?
My attempt:
I called $\alpha$ maximum value.
$$-5x^2-2xy-2y^2+14x+10y-1\leqslant\alpha$$
$$-5x^2-2xy-2y^2+14x+10y-1-\alpha\leqslant 0$$
$$5x^2+2xy+2y^2-14x-10y+1+\alpha\... | Your idea is quite correct, but you have to complete squares in a different way:
\begin{gather}
-(5x^2+2xy+2y^2-14x-10y+1) =\\
-\left[\left(x\sqrt 5 + y\frac{1}{\sqrt 5} -\frac{7}{\sqrt 5}\right)^2 +2y^2-10y+1-\frac{1}{5}y^2-\frac{49}{5}+\frac{14}{5}y\right]=\\
-\left[\left(x\sqrt 5 + y\frac{1}{\sqrt 5} -\frac{7}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Cambridge IGCSE Additional Mathematics Challenge Q This is a challenge question from my Cambridge IGCSE Additional Maths textbook. Bear with me on the drawing. The drawing consists of a square, circle, and quarter circle. The only measurement give is that the side length of the square is $10$cm. Can someone help me fin... | Sketch. Here is an elementary way to obtain the area of the lune. Join the points of intersection of the two arcs, which gives a common chord $C$ for the two involved circles. Thus the area we seek is the difference in area of the segment of the small circle and the big circle, cut off by $C.$ Let these areas respectiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How find the maximum of the value $t$ such $x^3+y^3+z^3+t(xy^2+yz^2+zx^2)\ge (1+t)(x^2y+y^2z+z^2x)$ let $x,y,z\ge 0$, find the maximum of the $t$, such
$$x^3+y^3+z^3+t(xy^2+yz^2+zx^2)\ge (1+t)(x^2y+y^2z+z^2x)$$
maybe this take $y=x+u,z=x+t$,But I can't get the answer,because this is very ugly, Thanks
| I assumed that it means that we need to find a maximal value of $t$, for which this inequality is true for any non-negatives $x$, $y$ and $z$.
If so, we can say that your way (BW) helps!
Let $x=\min\{x,y,z\}$, $y=x+u$, $z=x+v$ and $u=kv$.
Thus, $u$ and $v$ are non-negatives and $$\sum_{cyc}(x^2+txy^2-(t+1)x^2y)=$$
$$=2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Identifying distributions
The pgf of a random variable $X$ has pgf
$$\frac{1}{7}(1+s+3s^2 +s^3 +s^4 )$$
What is the distribution of $X$ please?
I know that a pgf is defines as however the $z$ have just been replaced with $s$ in my example
$$
G(z)=E[z^X]=\sum_{x=0}^\infty p(x)z^x.
$$
| As in the definition, the PGF is the sum of probabilities times a variable to some power, where the probability is the probability that the power occurs. (Re-read this several times if that sentence doesn't make sense.)
So, we have a PGF of
$$
\begin{align}
\dfrac{1}{7}(1 + s + 3s^2 + s^3 + s^4) &= \dfrac{1}{7} \cdot 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3625691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Checking the argument for calculation of the limit $\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)-\frac{1}{2}} {x-\frac{\pi}{6}}$ = $\sqrt3 \over 2$
$\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)-\frac{1}{2}} {x-\frac{\pi}{6}}$ = $\sqrt3 \over 2$
In computing this limit, i have used following steps:
$\lim_{x\to \frac{\pi}{6}} \f... | You cannot partially pass to the limit. See this example:
$$
1 = \lim_{x \to 1} 1 = \lim_{x \to 1} \frac{x-1}{x-1} = \lim_{x\to1} \frac{1-1}{x-1} = \lim_{x \to 1} 0 = 0.
$$
The error lies here:
$$
\lim_{x \to 1} \frac{x-1}{x-1} = \lim_{x\to1} \frac{1-1}{x-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3627825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Inequality with 4 variables For $a,b,c,d>0$ such that $abcd\ge 1$. Prove that $$\frac{1}{3a+2b+c+6}+\frac{1}{3b+2c+d+6}+\frac{1}{3c+2d+a+6}+\frac{1}{3d+2a+b+6} \leq \frac{1}{3}$$
My attempts:
By AM-GM $$3a+2b+c+6\geq12\sqrt[12]{a^3b^2c}$$ and I have to prove that
$$\frac{1}{\sqrt[12]{a^3b^2c}}+\frac{1}{\sqrt[12]{b^3c^2... | Let $a=kx$, $b=ky$, $c=kz$ and $d=kt$, where $k>0$ and $xyzt=1$.
Thus, $$k^4xyzt\geq1,$$ which gives $k\geq1$ and we obtain:
$$\sum_{cyc}\frac{1}{3a+2b+c+6}=\frac{1}{k(3x+2y+z)+6}\leq\sum_{cyc}\frac{1}{3x+2y+z+6}.$$
Id est, it's enough to prove that:
$$\sum_{cyc}\frac{1}{3x+2y+z+6}\leq\frac{1}{3}$$ or
$$\sum_{cyc}\left... | {
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"timestamp": "2023-03-29T00:00:00",
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find $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3} \ -\sqrt{2x+2}}$ without l'hospital rule EDITED VERSION
find $$\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3}
\ -\sqrt{2x+2}}$$ without l'hospital rule.
using l'hospital rule, you'll have: $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3}
\ -\sqrt{... | Multiplying the top and the Bottom by $ \left(2x+\sqrt{x^2+3}\right)\left(\sqrt{x+3}+\sqrt{2x+2}\right) $, gives the following :
\begin{aligned} \lim_{x\to 1}{\frac{2x-\sqrt{x^2+3}}{\sqrt{x+3}-\sqrt{2x+2}}}&=\lim_{x\to 1}{\frac{-3\left(x+1\right)\left(\sqrt{2x+2}+\sqrt{x+3}\right)}{\sqrt{x^{2}+3}+2x}}\\ &=\frac{-3\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3634771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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To find the range of $f(x)=\frac {x^2-1}{x^2+3x+2}$, the "discriminant method" doesn't work—why and how can we fix it? Define $f: \mathbb R \setminus \{-1,-2\} \rightarrow \mathbb R $ by $$f(x)=\frac {x^2-1}{x^2+3x+2}$$
To find the range of $f$, we use the "discriminant method" (used in e.g. 1, 2, 3):
*
*Write $y=\f... | With $y = 1$, note your equation has a $0$ coefficient for $x^2$, so it's not a quadratic any more. Instead, you have
$$\begin{equation}\begin{aligned}
3yx + 2y + 1 & = 0 \\
3x + 3 & = 0 \\
x & = -1
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
However, $x = -1$ is one of the excluded values, which means that $y = 1... | {
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"question_score": "1",
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"answer_id": 2
} |
Evaluate $\lim_{n \to \infty}\prod_{k=0}^{n} \left(1+\frac{2}{45^{2^k}+45^{-2^k}}\right)$ Evaluate $$P=\lim_{n \to \infty}\prod_{k=0}^{n} \left(1+\frac{2}{45^{2^k}+45^{-2^k}}\right)$$
My try:
Let $a_k=45^{2^k}$ Then we have $45^{-2^k}=\frac{1}{a_k}$
So We get:
$$1+\frac{2}{a_k+\frac{1}{a_k}}=\frac{(a_k+1)^2}{a_k^2+1}$... | Note that $a_k^2=a_{k+1}$, so
$$
\prod_{k=0}^{n}\frac{(a_k+1)^2}{a_k^2+1}=\prod_{k=0}^{n}\frac{(a_k+1)^2}{a_{k+1}+1}
=\frac{a_0+1}{a_{n+1}+1}\cdot(a_0+1)(a_1+1)\ldots(a_n+1).
$$
Then,
$$
a_k+1=\frac{a_k^2-1}{a_k-1}=\frac{a_{k+1}-1}{a_k-1},~\text{so}~
$$
$$
\prod_{k=0}^{n}\frac{(a_k+1)^2}{a_k^2+1}=\frac{a_0+1}{a_{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3638344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Inequality with the $2^n$ partial sum of the 'harmonic series' I would like to know whether $p_n:=\sum\limits_{k=1}^{2^n}\frac{1}{k}\le n+1$ for all $n\in\mathbb{N}\setminus\{0\}$ holds or not.
It holds for $n=1,2,3$. Furthermore computing $p_n$ for $n$ big the graph looks similar to the graph of the log-function. On ... | Indeed, the induction step is correct. You have
$$\frac{1}{2^n +1} + \frac{1}{2^n +2}+ \ldots + \frac{1}{2^{n+1}} \leq \frac{1}{2^n +1} + \ldots + \frac{1}{2^n +1} \leq \frac{2^n}{2^n +1},$$ where in the last step we used the fact that there are $2^n$ terms in the sum. Since $\frac{2^n}{2^n +1}\leq 1,$ the proof folows... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3640714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\lim_{n \rightarrow \infty} \big( \frac{1}{n+1}-\frac{1}{n+2}+...+\frac{(-1)^{n-1}}{2n} \big)$ I need to compute
$$\lim_{n \rightarrow \infty} \big( \frac{1}{n+1}-\frac{1}{n+2}+...+\frac{(-1)^{n-1}}{2n} \big).$$
I've previously managed to show that
$$\lim_{n \rightarrow \infty} \Big( \frac{1}{n+1}+\frac{1}{n+2}+...+\f... | $\begin{array}\\
s(n)
&= \big( \frac{1}{n+1}-\frac{1}{n+2}+...+\frac{(-1)^{n-1}}{2n} \big)\\
&= \sum_{k=1}^n\frac{(-1)^{k-1}}{n+k}\\
s(2n)
&= \sum_{k=1}^{2n}\frac{(-1)^{k-1}}{2n+k}\\
&= \sum_{k=1}^{n}\left(\frac{1}{2n-1+k}-\frac{1}{2n+k}\right)\\
&= \sum_{k=1}^{n}\frac{1}{(2n-1+k)(2n+k)}\\
&\le \sum_{k=1}^{n}\frac{1}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3640926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to show that $f(x)=\sin(x)+\cos(x)$ is sinusoidal(alternative) There was this question in our trig homework; it was for plotting a graph but I found it far more interesting than that. When drawing the graph of $\sin(x)+\cos(x)$ (by hand, which I find rather pointless), I found that it looked like some sort of sine ... | Yes. Indeed all linear combinations of the form $a\cos x+b\sin x$ are sinusoids, as shown below:
Write this as $$\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}\left(a\cos x+b\sin x\right)=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x\right)=R(\sin\phi\cos x+\cos\phi\sin x)=R\sin(\phi+x),$$ wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3646388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
evaluate $\int^{\infty}_1\frac{\ln(x)}{1+x^2}dx$ I need to calculate the integral: $$\int^{\infty}_1\frac{\ln(x)}{1+x^2}dx$$
My attempt:
IBP: $u=\ln(x) \Rightarrow du=\frac{1}{x}dx, vdv=\frac{1}{1+x^2}dx \Rightarrow v=\arctan(x)$
So, the integral becomes:
$$\ln(x)\arctan(x)\Big|^{\infty}_1-\int^{\infty}_1\frac{\arctan(... | Convert the limits to $(0,1)$ to have the expansion converging,
$$\begin{align}
&\int^{\infty}_1\frac{\ln x}{1+x^2}dx
=- \int_{0}^1\frac{\ln t }{1+t^2}dt\\
& =-\ln t\arctan t\Big|_0^1+\int^{1}_0\frac{\arctan t}{t}dt \\
&=\int^{1}_0\frac1t (t-\frac{t^3}3 + \frac{t^5}5 - ...)dt
=(1-\frac{t^3}{3^2}+ \frac{t^5}{5^2}- ...)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3646585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Necessary and sufficient conditions for a right triangle
Show that if for $\triangle ABC$ the equalities $h_c^2=a_1b_1$ and $b^2=b_1c$ are true where $h_c$ is the height, $AC=b, AB=c$ and $a_1$ and $b_1$ are the the projections of $BC$ and $AC$ on $AB$, then the triangle is right angled.
I want to demonstrate that to... | First Solution:
It's known that a triangle is right angled if and only if the three edges satisfies the Pythagorean theorem (this statement follows from the Low of Cosines). So we have only to verify the Pythagorean theorem for the triangle.
Named $a=BC$ we have:
\begin{gather}
a^2+b^2 = h_c^2 + a_1^2 + b_1 c = a_1b_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3648486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Theorem needed to prove a summation I know that the following relation holds:
$$\sum_{x=1}^y\frac{x(5x+6)}{45}=\frac{y(y+1)(10y+23)}{270}$$
But what theorem should I use to prove that relation?
| It is known that $\sum_{x=1}^y x=\frac{y(y+1)}{2}$ and $\sum_{x=1}^y x^2=\frac{y(y+1)(2y+1)}{6}$. Refer to here for example.
Your required sum will become:
$$\sum_{x=1}^y \frac{x^2}{9} + \frac{2}{15}x$$
$$=\frac{1}{9}\cdot\frac{y(y+1)(2y+1)}{6}+\frac{2}{15}\cdot\frac{y(y+1)}{2}$$
$$=y(y+1)\left[\frac{2y+1}{54}+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3655022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why does this Olympiad inequality proving technique (Isolated Fudging) work? A few years ago, I was in a math olympiad training camp and they taught us a technique to prove inequalities. I just came across it again recently. However, I am not able to understand why it works. So, here is how it goes. Suppose, you want t... | It not always works.
More exactly, we not always can find this trick during a competition without computer.
For example, there is the following estimation (Ji Chen):
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$.
Prove that:
$$\frac{1}{2+a^2+b^2}\leq\frac{3(6a^2+b^2+c^2+2ab+2ac+4bc)}{32(a^2+b^2+c^2+ab+a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3657950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
show this inequality $\sum_{cyc}\frac{1}{5-2xy}\le 1$ let $x,y,z\ge 0$ and such $x^2+y^2+z^2=3$ show that
$$\sum_{cyc}\dfrac{1}{5-2xy}\le 1$$
try:
$$\sum_{cyc}\dfrac{2xy}{5-2xy}\le 2$$
and
$$\sum_{cyc}\dfrac{2xy}{5-2xy}\le\sum_{cyc}\dfrac{(x+y)^2}{\frac{5}{3}z^2+\frac{2}{3}x^2+\frac{2}{3}y^2+(x-y)^2}\le\sum\dfrac{3(x+y... | We need to prove
$$ \sum \dfrac{xy}{5-2xy} \leqslant 1,$$
or
$$ \sum \dfrac{3xy}{5(x^2+y^2+z^2)-6xy} \leqslant 1,$$
Indeed, because
$$[13(x^2+y^2)+10z^2-18z(x+y)+36xy][5(x^2+y^2+z^2)-6xy]-108xy (x^2+y^2+z^2)$$
$$=\left[63(x^2+y^2)+\frac{109z^2}{2}+112xy-72z(x+y)\right](x-y)^2+\frac{(2x+2y-5z)^2(x+y-2z)^2}{2}$$
$$\geqsl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3658374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Catalan triangle created by $[t^{n+1}](\frac{1-\sqrt{1-4t}}{2})^{k+1}=\frac{k+1}{n+1}{2n-k \choose n-k}$? I tried to get the catalan triangle from its Riordan array. I failed at $d_{n,k}=[t^{n+1}](\frac{1-\sqrt{1-4t}}{2})^{k+1}=\frac{k+1}{n+1}{2n-k \choose n-k}$. How do I get there?
| Working with the generating function $C(z)$ of the Catalan numbers
the claim is equivalent to
$$[z^{n-k}] C(z)^{k+1} = \frac{k+1}{n+1}
{2n-k\choose n-k}.$$
The generating function
$$C(z) = \frac{1-\sqrt{1-4z}}{2z}$$
has the functional equation
$$C(z) = 1 + z C(z)^2.$$
Note that $[z^0] C(z)^{k+1} = 1$ which matches the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3660670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplifying $\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}$
The expression $\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}$, where $0<x<1$, is equal to $x$ or $\sqrt{(1+x^2)}$ or $\frac1{\sqrt{(1+x^2)}}$ or $\frac x{\sqrt{(1+x^2)}}$? (one of these 4 is correct).
My attempt: $$0<x<1\implies 0<\arctan x<\frac{\pi}{4}\im... | Your function at $x=0$ is $0$. Thus it can only be $x$ or $x/\sqrt{1+x^2}$. As $x\to 1$ it tends to $1$. Hence, it must be $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Conditional Probability: Papoulis/Pillai Ex 2-13 Question: A box contains white and black balls. When two balls are drawn without replacement, suppose the probability that both are white is 1/3. (A) Find the smallest number of balls in the box.
Solution from the text:
Let a and b denote the number of white and black ba... | $$a^2 + b^2 + 2ab \lt 3a^2 \\
2a^2 - b^2 - 2ab \gt 0 \\
a \gt \frac{2b \ \pm \sqrt{4b^2 + 8b^2}}{4} \\
a \gt \frac{2b \ \pm 2b\sqrt{3}}{4} \\
\bbox[yellow]{a \gt \frac{( \sqrt{3} + 1)b}2} \\
3(a-1)^2 \lt (a-1)^2 + b^2 + 2b(a-1) \\
2(a-1)^2 - b^2 - 2b(a-1) \lt 0 \\
a-1 \lt \frac{2b \ \pm \sqrt{4b^2 +8b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Given the Rodrigues' formula for Legendre's polynomials, show that it satisfies the ODE. The Rodrigues Formula for Legendre's Polynomials is $P_{l}(x)=\frac{1}{2^{l}l!}\frac{d^{l}}{dx^{l}}(x^2-1)^l$.
I wrote $P_{l}(x)=\frac{1}{2^{l}l!}\frac{d^l}{dx^l}\sum_{k=0}^l(-1)^{k-l}\frac{l!}{k!(l-k)!}x^{2k}=\sum_{k=0}^l\frac{(-... | This is another approach that does not explicitly use integration. Use $ D $ to stand for $ d/dx$.
First, apply Leibniz rule for the $n+2$ derivative of a product,
$$
\begin{align}
D^{n+2}(x^2-1)^{n+1} &= (x^2-1) D^{n+2}(x^2-1)^n \\
& \quad + \left( n+2 \atop 1 \right) 2x D^{n+1}(x^2-1)^n \\
&\quad\quad + 2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3667895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Quadratic form on $\mathbb{C}^3$ Let
$$\Phi(x)= \frac{1}{4}x_1^2+\frac{3}{4}i(\bar{x_1}x_2-\bar{x_2}x_1)-\frac{1}{2\sqrt{2}}(\bar{x_1}x_3+\bar{x_3}x_1)+\frac{1}{4}x_2^2-\frac{1}{2\sqrt{2}}i(\bar{x_2}x_3+\bar{x_3}x_2)-\frac{1}{2}x_3^2$$
Show that $\Phi$ is hermitian and find its signature.
So in order to solve this I ... | Your matrix is $\ast$-congruent to a diagonal real matrix.
We need not worry about the eigenvalues, those may or may not be nice.
$$
R =
\left(
\begin{array}{rrr}
1& -\frac{3i}{2} & - \sqrt 2 \\
0& 1 & - i \sqrt 2 \\
0& 0 & 1 \\
\end{array}
\right)
$$
$$
R^\ast A R =
\left(
\begin{array}{rrr}
\frac{1}{4} &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to derive the condition? (I am not sure it is correct.) Given a polynomial with real coefficients $\alpha x^2 + \beta x + a^2 + b^2 + c^2 - ab- bc - ca $ has imaginary roots, how do we prove $ 2(\alpha - \beta) + ((a - b)^2 + (b - c)^2 + (c - a)^2) > 0 $?
Given all coefficients are real, roots are $ z $ and $\overl... | You can try with $2f(-1)>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3673235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute Taylor series $\frac{1}{x^2+4x+3}$ at $x = 2$ I was trying to solve a textbook exercise stated in the following:
Use completing the square and the geometric series to get the Taylor expansion about $x=2$ of $
\frac{1}{x^2+4x+3}$
My early attempt was
$\frac{1}{x^2+4x+3} = -1 (\frac{1}{1-(x+2)^2})$, even though... | Using partial fraction decomposition we get that
$$\frac{1}{x^2+4x+3} = \frac{1}{(x+3)(x+1)} = \frac{1}{2}\left(\frac{1}{x+1} - \frac{1}{x+3}\right) = \frac{1}{2}\left(\frac{1}{3}\frac{1}{1+\frac{x-2}{3}} - \frac{1}{5}\frac{1}{1+\frac{x-2}{5}} \right)$$
where we can now use geometric series to get
$$\frac{1}{2}\sum_{n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3675629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int _0^{\infty }\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx$ with real methods I started like this:
$$\int _0^{\infty }\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx\:
=\int _0^1\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx+\overset {x=\frac{1}{x}}{\int _1^{\infty }\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx}$$
$$=2\int _0^1... | Continue to reduce the result in the post as follows
\begin{align}
& \text{Ti}_2\left(\frac{\sqrt{1-\frac{2}{\sqrt{5}}}-1}{\sqrt{1-\frac{2}{\sqrt{5}}}+1}\right)+\text{Ti}_2\left(\frac{\sqrt{1+\frac{2}{\sqrt{5}}}-1}{\sqrt{1+\frac{2}{\sqrt{5}}}+1}\right)\\
=&\ \text{Ti}_2(\tan\frac\pi{20}) - \text{Ti}_2(\tan\frac{3\pi}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3675897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Compute $\int \frac{\cos ^2x+\cos (\sin x)}{\sin x \sin (\sin x)+1} \, dx$ From software, I get
$$\int \frac{\cos ^2 x+\cos (\sin x)}{\sin x \sin (\sin x)+1} \, dx=-2 \tan ^{-1}\left(\cos \left(\frac{x}{2}-\frac{\sin x}{2}\right) \csc \left(\frac{x}{2}+\frac{\sin x}{2}\right)\right)$$
I tried $t=\sin x,x=\arcsin t$ an... | Rewrite the denominator
\begin{align}
\sin x \sin (\sin x)+1 = &\frac12[\cos(x-\sin x) - \cos(x+\sin x)]+1
\\ = &\frac12[1-\cos(x+\sin x)] + \frac12[1+\cos(x-\sin x)]
\\ = & \sin^2t_+ + \cos^2t_-
\end{align}
where $t_\pm=\frac x2 \pm \frac{\sin x}2$. Also rewrite the numeritor
\begin{align}
\cos ^2x+\cos (\sin x)= &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3677185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\sum_{n,k} \binom{n}{k}^{-1} $
Evaluate $$\sum_{n,k} \frac{1}{\binom{n}{k}}, $$ where the summation ranges over all positive integers $n,k$ with $1<k<n-1$.
Thouhgts:
We are trying to evaluate $$\sum_{n=4}^{\infty} \sum_{k=2}^{n-2} \binom{n}{k}^{-1}$$
We may try to find a closed form of the inner summation ... | Let $\ell = n -k$, we have
$$\begin{align} \sum_{n=4}^\infty\sum_{k=2}^{n-2} \binom{n}{k}^{-1}
&= \sum_{k=2}^\infty\sum_{n=k+2}^\infty \binom{n}{k}^{-1}\\
&= \sum_{k=2}^\infty\sum_{\ell=2}^\infty \binom{k+\ell}{k}^{-1}
= \sum_{k=2}^\infty\sum_{\ell=2}^\infty \frac{k!\ell!}{(k+\ell)!}\\
&= \sum_{k=2}^\infty\sum_{\ell=2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3677292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
$\sum(-1)^{k}k^{2}C_{n}^{k}$ I'm asked to compute $\sum^{n}_{k=1}(-1)^{k}k^{2}C_{n}^{k}$ for $n\geqslant 1$. I tried to use generating functions to solve the problem: $A(x):=\sum^{n}_{k=1}k^{2}C_{n}^{k}x^{k}$. So, I need to compute $A(-1)$.
For $n=1$ $A(-1)=-1$, for $n=2$ $A(-1)=2$, for $n=2$ $A(-1)=2$, for $n=3,4,5,6... | First note that
\begin{align}
\sum_{k=1}^\infty k^2 x^k
&=x^2\sum_{k=1}^\infty k(k-1) x^{k-2} + x\sum_{k=1}^\infty k x^{k-1}\\
&=x^2\frac{d^2}{dx^2}\sum_{k=0}^\infty x^k + x\frac{d}{dx} \sum_{k=0}^\infty x^k\\
&=x^2\frac{d^2}{dx^2}\frac{1}{1-x} + x\frac{d}{dx} \frac{1}{1-x}\\
&=x^2\frac{2}{(1-x)^3}+x\frac{1}{(1-x)^2}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3679399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Compute $\int_0^1 \frac{\text{Li}_2(-x^2)\log (x^2+1)}{x^2+1} \, dx$ How can we evaluate: $$\int_0^1 \frac{\text{Li}_2\left(-x^2\right) \log \left(x^2+1\right)}{x^2+1} \, dx$$
Any help will be appreciated.
| Using double integration we have:
$$\scriptsize I=\frac{\pi ^2 C}{12}+2 C \log ^2(2)-16 \Im(\text{Li}_4(1+i))-\frac{21 \pi \zeta (3)}{8}+\frac{1}{6} \pi \log ^3(2)+\frac{5}{24} \pi ^3 \log (2)+\frac{11 \psi ^{(3)}\left(\frac{1}{4}\right)}{768}-\frac{11 \psi ^{(3)}\left(\frac{3}{4}\right)}{768}$$
And a corollary (than... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum value of $x+2y$ given $\frac{1}{x + 2} + \frac{1}{y + 2} = \frac{1}{3}.$
Let $x$ and $y$ be positive real numbers such that
$$\frac{1}{x + 2} + \frac{1}{y + 2} = \frac{1}{3}.$$Find the minimum value of $x + 2y.$
I think I will need to use the Cauchy-Schwarz Inequality here, but I don't know how I ... | Hint: $y = \left(\dfrac{1}{3} - \dfrac{1}{x+2}\right)^{-1}-2= \dfrac{3x+6}{x-1}-2= \dfrac{x+8}{x-1}\implies x+2y=x+\dfrac{2x+16}{x-1}= \dfrac{x^2+x+16}{x-1}=f(x)$.From this point, you simply set $f'(x) = 0$ and solve for critical points and take it from there. It should be standard calculus problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find vertex coordinates of a square given their distances $p,\>s,\>q,\>r$ to an inner point
The figure represents a square, with P a point inside the square. The four segments are drawn from P to the four vertices of the square and they are named p, q, r, s.
If the bottom-left vertex is at origin and the sides of the... |
Let $P(x,y)$ and $a$ the side length of the square. Then,
$$x^2+y^2=p^2, \>\>\>
(a-x)^2+y^2=s^2,\>\>\> x^2+(a-y)^2=r^2
$$
The 2nd and 3rd equations leads to
$x= \frac{a^2+p^2-s^2 }{2a}$ and $y= \frac{a^2+p^2-r^2}{2a}$.
Substitute them into the 1st equation to get
$$ a^4 -(s^2+r^2)a^2 +\frac12[(s^2-p^2)^2+ (r^2-p^2)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving $C^{12}_{x} + C^{12}_{x+1} = C^{13}_{2x}$. $C^{12}_{x} + C^{12}_{x+1} = C^{13}_{2x}$
I did find by brute force the solutions $n=1$ and $n=4$, through the inequalities $2x \le 13, x \ge 0 \implies x \in \{0,1,2,3,4,5,6\}$
But is there a more analytical way to solve this?
Here is my attempt:
$C^{12}_{x} + C^{12}... | Using the property
$$C_r^n+C_{r-1}^n=C_r^{n+1}$$
On the LHS, with $n=12,r=x+1$ you have
$$C^{13}_{x+1}=C^{13}_{2x}$$
You now have either $x+1=2x$ or $2x+(x+1)=13$.
From here, it's fairly easy to get $x=1,4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculating a crazy trigonometric integral What is the fastest and best way to do this crAZy integral?
$$ \int\frac{1-\tan^4\theta d\theta}{\tan^{\frac{9}{6}}\theta(\sec^2 \theta+\tan\theta)^{\frac{1}{2}}+\tan^{\frac{10}{6}}\theta (\sec^2 \theta+ \tan\theta)^{\frac{1}{3}}}$$
I tried substituting:
$ \tan x = t$
but tha... | $$ \int\frac{(1-\tan^4\theta) d\theta}{\tan^{\frac{9}{6}}\theta(\sec^2 \theta+\tan\theta)^{\frac{1}{2}}+\tan^{\frac{10}{6}}\theta (\sec^2 \theta+ \tan\theta)^{\frac{1}{3}}}$$
$$ \rightarrow \int \frac{ \sec^2 \theta (1- \tan^2 \theta) d\theta}{ \tan^{\frac{3}{2}} \theta \left[\sec^2 \theta + \tan \theta \right]^{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Help to evaluate the integral $\iint_D\frac{y}{\sqrt{x^2+y^2}}dxdy$ I'm solving a problem about integrals in curves, and I got this integral:
$$\int_1^2\int_1^2\frac{y}{\sqrt{x^2+y^2}}dxdy.$$
I have been struggling to solve it. I'm sure i have to do some variable change to polar coordinates (to simplify the denominator... | Use change of order of integration $$\int_1^2\int_1^2\frac{y}{\sqrt{x^2+y^2}}dxdy$$
$$=\int_1^2\int_1^2\frac{y}{\sqrt{x^2+y^2}}dydx$$
$$=\int_1^2\left(\frac12\int_1^2\frac{d(x^2+y^2)}{\sqrt{x^2+y^2}}\right)dx$$
$$=\int_1^2\left(\sqrt{x^2+y^2}\right)_1^2dx$$
$$=\int_1^2\left(\sqrt{x^2+4}-\sqrt{x^2+1}\right)dx$$
$$=\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3696069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the asymptotic of an integral I came across the following exercise on asymptotic behavior of integrals:
$$I(a) = \int_0^\infty\frac{\cos x}{x^a} \, dx, \text{ where } a\to0^+.$$
I have tried integrating by parts or replacing $\cos x$ with the first summands of its Taylor series, but I end up with something equa... | Integration by parts is a good idea (as it usually is when one factor of the integrand is oscillating): when $0<a<1$,
$$
I(a) = \int_0^\infty \frac{\cos x}{x^a} \,dx = \frac{\sin x}{x^a}\bigg|_0^\infty - \int_0^\infty \frac{-a\sin x}{x^{1+a}} \,dx = a \int_0^\infty \frac{\sin x}{x^{1+a}} \,dx.
$$
This is enough to prov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3697492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Measure of an angle "subtended by each pentagon" in a truncated icosahedron A soccer ball is a truncated icosahedron; it consists of 12 black regular pentagons and 20 white regular hexagons; the edge lengths of the pentagons and hexagons are congruent. What is the measure of the solid angle (in ste-radian) subtended by... | The wikipedia about solid angle says that
The solid angle of a right $n$-gonal pyramid, where the pyramid base is a regular $n$-sided polygon of circumradius $r$, with a pyramid height $h$ is
$$2\pi -2n\arctan\bigg(\frac{\tan(\frac{\pi}{n})}{\sqrt{1+r^2/h^2}}\bigg)$$
Note here that our $r$ is the circumradius of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3697649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
identity for sum of binomial coefficients I am trying to prove that $$\sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor}\binom{n}{2k+1}(-1)^k = \sqrt{2^n}\sin{}n\pi/4$$
This follows from the subtraction of $\sum_{k=0}^{...}\binom{n}{4k+1}$ and $\sum_{k=1}^{...}\binom{n}{4k-1}$, which according to wikipedia can be evaluated using... | Because $$\frac{1+(-1)^k}{2}=\begin{cases}1&\text{if $k$ is even}\\0&\text{if $k$ is odd}\end{cases}$$
we have $$\sum_{k\ge 0} a_{2k} = \sum_{k\ge 0} a_k \frac{1+(-1)^k}{2}.$$
Now take $a_k=\binom{n}{k+1}i^k$, where $i=\sqrt{-1}$, to obtain
\begin{align}
\sum_{k\ge 0} \binom{n}{2k+1}(-1)^k &= \sum_{k\ge 0} \binom{n}{2k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3698077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove or disprove that the ellipse of largest area (centered at origin) inscribed in $y=\pm e^{-x^2}$ has the equation $x^2+y^2=\frac12(1+\log2)$. I can show that $x^2+y^2=\frac12(1+\log2)$ is the equation of the circle of largest area inscribed in $y=\pm e^{-x^2}$:
The minimum distance $r$ (which will be the radius o... | $$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \implies
\frac{dy}{dx}=-\frac{b^2}{a^2}\frac xy=-\frac{b}{a^2}\frac x{\sqrt{1-\frac{x^2}{a^2}}}.
$$
Hence we have the following system to find the tangent point:
$$\begin{cases}
e^{-x^2}=b\sqrt{1-\frac{x^2}{a^2}}\\
2xe^{-x^2}=\frac{b}{a^2}\frac x{\sqrt{1-\frac{x^2}{a^2}}}
\end{case... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find infimum of the sets of number $x + \frac{1}{x} $ Let $A = \{ z = x + \frac{1}{x} : x > 0 \} $ and $B = \{z = 2^x + 2^{1/x} : x > 0 \} $
I want to find $\inf A $ and $\inf B $.
Proof.
Clearly, by AM-GM inequality one has $x + \dfrac{1}{x} \geq 2 $ and $2^x + 2^{1/x} \geq 2 \sqrt{2^{x+1/x} } \geq 2 \sqrt{2^2} = 4 $.... | Looks good to me. In fact, you prove not only that $\inf A=2$, but also that $\min A=2$. Similarly, you have proved that $\min B=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Strategy to calculate $ \frac{d}{dx} \left(\frac{x^2-6x-9}{2x^2(x+3)^2}\right) $. I am asked to calculate the following: $$ \frac{d}{dx} \left(\frac{x^2-6x-9}{2x^2(x+3)^2}\right). $$
I simplify this a little bit, by moving the constant multiplicator out of the derivative:
$$ \left(\frac{1}{2}\right) \frac{d}{dx} \left(... | HINT
To begin with, notice that
\begin{align*}
x^{2} - 6x - 9 = 2x^{2} - (x^{2} + 6x + 9) = 2x^{2} - (x+3)^{2}
\end{align*}
Thus it results that
\begin{align*}
\frac{x^{2} - 6x - 9}{2x^{2}(x+3)^{2}} = \frac{2x^{2} - (x+3)^{2}}{2x^{2}(x+3)^{2}} = \frac{1}{(x+3)^{2}} - \frac{1}{2x^{2}}
\end{align*}
In the general case, p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3707227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the distance between the points of two tangents along a circle I have the following problem: there is a circle with the $R = 5$ and center of the circle located on coordinate $(0, 0)$. I have two points $A(6, 8)$ and $B(-4, -6)$. From points, tangents to the circle were drawn. It is better illustrated as:
Let us ... |
\begin{align}
|OE|=|OF|=
R&=5
,\quad
|OA|=10
,\quad
|OB|=2\sqrt{13}
,\quad
|AB|=2\sqrt{74}
,\\
\triangle AOE:\quad
|AE|&=5\sqrt3
,\\
\triangle BFO:\quad
|BF|&=3\sqrt3
.
\end{align}
\begin{align}
\angle EOF&=\angle AOB-\angle AOE-\angle FOB
,
\end{align}
\begin{align}
\angle AOB&=\arccos\frac{|OA|^2+|OB|^2-|AB|^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3712422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
To find $A$ given that $2I + A +A^2 = B$ where $B$ is given. How to find a matrix $A$ such that the following holds:
$$2I + A +A^2 = B,$$ where the matrix $B$ is given. I tried with char poly of $B$ but not getting any idea.
Note that it is also given that $B$ is invertible.
P.S. $B = \begin{pmatrix}-2&-7&-4\\ \:12&22&... | $$A=SJS^{-1},\\ 2I+A+A^2=S(2I+J+J^2)S^{-1}=B\\
2I+J+J^2=S^{-1}BS$$
So incorporating (yeah, stealing) the idea of the Mostafa Ayaz's answer we get $$\left(J+\frac{1}{2}I\right)^2=S^{-1}BS-\frac{7}{4}I$$
Letting, for a second, that $S$ is the same for $$B=S\begin{pmatrix}2&0&0\\0&4&1\\0&0&4\end{pmatrix}S^{-1}$$
we feed t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
$\sqrt{a+b} (\sqrt{3a-b}+\sqrt{3b-a})\leq4\sqrt{ab}$ I was training for upcoming Olympiads, working on inequalities, and the following inequality came up:
$$\sqrt{a+b} (\sqrt{3a-b}+\sqrt{3b-a})\leq4\sqrt{ab}$$ with the obvious delimitations $3b\geq a;\: 3a\geq b.$
I've been pondering the question for quite a while, and... | Presumably we also have the restriction $a,b\ge 0$.
With that assumption we can proceed as follows . . .
If $a+b=0$, then $a=b=0$, and for that case, the inequality clearly holds.
So assume $a+b > 0$.
Since the inequality is homogeneous, the truth of the inequality remains the same if $a,b$ are scaled by an arbitra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Simplification of ${0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots}$ Simplify
$$0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots,$$ where $n \ge 2.$
I think we can write this as the summation $\displaystyle\sum_{i=0}^{n} 2i\binom{n}{2i},$ which simplifies to $\boxed... | Sure. Notice that
$$\sum _{i = 0}^{n}2i\binom{n}{2i}=\sum _{i = 0}^{n}\binom{2i}{1}\binom{n}{2i}=\sum _{i = 1}^{n}\binom{n}{1}\binom{n-1}{2i-1}=n\sum _{i=1}^n\binom{n-1}{2i-1}=n\cdot 2^{n-2}.$$
Notice that the last step is because you are adding half of the binomials, and the odd half equals the even half by the binomi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Find the domain and range of $f(x) = \frac{x+2}{x^2+2x+1}$: The domain is: $\forall x \in \mathbb{R}\smallsetminus\{-1\}$
The range is: first we find the inverse of $f$:
$$x=\frac{y+2}{y^2+2y+1} $$
$$x\cdot(y+1)^2-1=y+2$$
$$x\cdot(y+1)^2-y=3 $$
$$y\left(\frac{(y+1)^2}{y}-\frac{1}{x}\right)=\frac{3}{x} $$
I can't find t... | Alternate way to find the range :
$$f(x) = y =\frac{x+2}{x^2+2x+1}$$
$$yx^2+(2y-1)x+(y-2)=0 $$
Now this quadratic has real roots (since real points exist belonging to the function for all values of x (we can remove the case of -1 later)) . So applying the condition for real roots : ($b^2-4ac \geq 0$)
$$(2y-1)^2-4(y)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Name of the rule allowing the exchanging $\sin$ and $\cos$ in integrals with limits $0$ and $\pi/2$? As in $0$ to $\frac{\pi}{2}$ limits the area under curve of $\sin \theta$ and $\cos \theta$ are same, so in integration if the limits are from $0$ to $\frac{\pi}{2}$ we can replace $\sin \theta$ with $\cos \theta$ and v... | This is not a specific rule. It is the property of definite integral: $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$ i.e. substitute $x=a+b-x$ everywhere in the integrand as follows
$$\int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^3x-\cos x}{\cos^3x-\sin x} dx$$
$$=\int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^3\left(\frac{\pi}{2}-x\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Finding the number of solutions to $\sin^2x+2\cos^2x+3\sin x\cos x=0$ with $0\leq x<2\pi$
For $0 \leq x<2 \pi$, find the number of solutions of the equation
$$
\sin^2 x+2 \cos^2 x+3 \sin x \cos x=0
$$
I have dealed the problem like this
$\sin ^{2} x+\cos ^{2} x+\cos ^{2} x+3 \sin x \cos x=0$
LET, $\sin x=t ;\quad \si... | Other idea to solve $a\sin^2 x+b \cos^2x+ c \sin x \cos x =d $ is divide by $\cos^2x$ or $\sin^2x$ to turn to quadratic like equation of $\tan x$ function
$$\sin ^{2} x+2 \cos ^{2} x+3 \sin x \cos x=0 \div \sin ^{2} x \to \\ 1+2 \cot^2 x+3\cot x=0\\ and \div \cos^2x \to \tan^2x+2+3\tan x=0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $a+b+c=0, ab+bc+ca=1$ and $abc=1,$ then find the value of $\frac ab+\frac bc+\frac ca.$ Clearly $a, b, c$ are the roots of the cubic equation: $x^3+x-1=0\tag{1}.$ We have to find:
\begin{align}
\frac ab+\frac bc+\frac ca&=\frac{a^2c+b^2a+c^2b}{abc}\\\\
&=a^2c+b^2a+c^2b\\\\
&=p,\text{ say}.
\end{align} ($p$ is not a ... | $p$ and $q$ are not symmetric to the roots of the polynomial, but they are both conjugates
They form the quadratic $t^2+3t+10=0$, as it happens the root of the quadratic are the values of both $p$ and $q$, both of them satisfy the quadratic, which in turn satisfies symmetric function of the polynomial
I can therefore w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int_{0}^{\frac{\pi}{2}} \sqrt{\cos^4{(x)} + \sin^2{(2x)}}\, dx.$
How to evaluate this integral:
$$\int_{0}^{\frac{\pi}{2}} \sqrt{\cos^4{(x)} + \sin^2{(2x)}}\, dx.$$
I can take the $\cos(x)$ out of the square root by expanding $\sin(2x)$ so it becomes $$\int_{0}^{\frac{\pi}{2}} \cos(x) \sqrt{\cos^2(x) +4\si... | You can proceed as
$$\int_{0}^{\frac{\pi}{2}} \cos x\sqrt{\cos^2{(x)} + 4\sin^2{(x)}}\quad dx$$
$$=\int_{0}^{\frac{\pi}{2}} \cos x\sqrt{1-\sin^2{(x)} + 4\sin^2{(x)}}\quad dx$$
$$=\int_{0}^{\frac{\pi}{2}} \cos x\sqrt{1+ 3\sin^2{(x)}}\quad dx$$
Let $\sin x=\frac{t}{\sqrt 3}\implies \cos x\ dx=\frac{1}{\sqrt 3}dt$
$$=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Conditions for positive roots for cubic equation $$
\begin{array}{l}\text { The minimum value of ab if roots of the equation } x^{3}-a x^{2}+b x-2=0 \\ \text { are positive, is }\end{array}
$$
$$
\begin{array}{l}\text { Let } f(x)=x^{3}-a x^{2}+b x-2 \\ \therefore f^{\prime }(x)=3 x^{2}-2 a x+b \\ x_{-}=\frac{2 a-\sqr... | Starting with a generic, factored cubic equation
$$\begin{align*}f(x) &= (x-r_0)(x-r_1)(x-r_2) \\
\\
&= x^2 -(r_0 + r_1 +r_2)x^2 + (r_0r_1 + r_0r_2+r_1r_2)x - r_0r_1r_2 \\
\end{align*}$$
In your specific case, you have
$$\begin{align*}a &= (r_0 + r_1 +r_2) \\
b&= (r_0r_1 + r_0r_2+r_1r_2) \\
2 &= r_0r_1r_2 \\
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3720309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Classifying Critical Point in 3D Question: $f(x, y, z) = px^2 +q(y^2 + z^2) +rxy + syz$ where $p,q,r,s \in \mathbb{R}$ has a critical point at $(0, 0, 0)$. Classify this critical point. You can assume the product of $p$ and $q$ is positive. Also, $r$ and $s$ cannot be both equal to zero (either one is zero and the othe... | We have $f(x, y, z) = \frac{1}{2} u^\mathsf{T} H u$ where $u = [x, y, z]^\mathsf{T}$.
Results:
If $p, q > 0$,
then $(0, 0, 0)$ is a local minimizer iff $2q - \frac{r^2}{2p} - \frac{1}{2q}s^2 \ge 0$.
If $p, q > 0$ and $2q - \frac{r^2}{2p} - \frac{1}{2q}s^2 < 0$, then $(0, 0, 0)$ is a saddle point.
If $p, q < 0$, then $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3720699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Find all possible values $m$ such that $f(x)=x^2-5mx+10m-4$ has 2 roots and one of them is twice as the other Consider the polynomial $f(x)=x^2-5mx+10m-4$, find $m$ such that there exist a number $a$ that satisfies $f(a)=f(2a)=0$. This was my attempt:
$f(a)=f(2a)$
$a^2-5ma+10m-4=4a^2-10ma+10m-4$
$a^2-5ma=4a^2-10ma$
... | Let the roots be $\alpha, 2\alpha$.
By Vieta's formulas,
Sum of roots $=3\alpha = 5m$, product of roots $=2\alpha^2 = 10m-4$.
Hence, $2(\frac{5m}3)^2 = 10m-4 \\ \implies 25m^2 - 45m +18 = 0$, we get (by simple factorisation) $m=0.6$ or $1.2$.
Test by plugging these back into the original expression and getting $f(x) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Evaluating matrix equation A $2x2$ matrix $M$ satisfies the conditions $$M\begin{bmatrix}
-8 \\
1
\end{bmatrix} = \begin{bmatrix}
3 \\
8
\end{bmatrix}$$
and
$$M\begin{bmatrix}
1 \\
5
\end{bmatrix} = \begin{bmatrix}
-8 \\
7
\end{bmatrix}... | If you find two numbers $x,y\mathbb{R}$ such that $x\begin{bmatrix}-8\\1\end{bmatrix}+y\begin{bmatrix}1\\5\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}$, then, make a left multiply by $M$:
$$xM\begin{bmatrix}-8\\1\end{bmatrix}+yM\begin{bmatrix}1\\5\end{bmatrix}=M\begin{bmatrix}1\\1\end{bmatrix}$$
finally,
$$ ~~~~~~~~... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Show that the size of the Turan graph $T_r(n)$ is at least $(1 - \frac{1}{r}) \binom{n}{2}$. A Turan graph $T_r(n)$ is defined as the complete $r$-partite graph of order $n$ such that the number of vertices in each of the $r$ classes is either $\lfloor \frac{n}{r}\rfloor$ or $\lceil \frac{n}{r} \rceil$. For fixed $n$ a... | Let $e$ be the number of edges and $a$ the average degree of a vertex. We want to show that
$$e\ge\left(1-\frac1r\right)\binom n2.$$
Since $e=\frac{na}2$ and $\binom n2=\frac{n(n-1)}2$, that's the same as showing
$$\frac{na}2\ge\left(1-\frac1r\right)\frac{n(n-1)}2,$$
that is, we have to show that
$$a\ge\left(1-\frac1r\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find a matrix $B$ such that $B^2=A$.. Let $A$ be a $2 × 2$ matrix and $I$ be the identity matrix. Assume that the null spaces of $A − 4I$
and $A − I$ respectively are spanned by
$\begin{bmatrix}3\\2\end{bmatrix}$
and $\begin{bmatrix}1\\1\end{bmatrix}$
respectively. Find a matrix $B$ such that
$B^2 = A$.
How to approach... | Observe that $A \dbinom{3}{2} = 4 \dbinom{3}{2}$ and that $A \dbinom{1}{1} = \dbinom{1}{1}$. So, if we define $P := \bigg( \begin{matrix} 3&1\\2&1 \end{matrix} \bigg)$ then
$$P^{-1}AP \dbinom{1}{0} = \dbinom{4}{0} \textrm{ and } P^{-1}AP \dbinom{0}{1} = \dbinom{0}{1},$$
that is, $P^{-1}AP = \bigg( \begin{matrix} 4&0\\0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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What is the solution of this summation? $$S(x) = \frac{x^4}{3(0)!} + \frac{x^5}{4(1)!}+\frac{x^6}{5(2)!}+.....$$
If the first term was $$x^3$$ and the next terms were $$x^{3+i}$$ then differentiating it would have given $$x^2.e^x$$ and then it was possible to integrate it. But how to solve this one?
| I'm assuming you've made a typo and you actually have
$$S(x) = \frac{x^4}{3(0)!} + \frac{x^5}{4(1)!}+\frac{x^{{\color{red}6}}}{5(2)!}+ \cdots.$$
Consider $P(x) = S(x)/x$. We have
\begin{align}
P(x) &= \frac{x^3}{3(0)!} + \frac{x^4}{4(1)!}+ \dfrac{x^5}{5(2)!} + \cdots\\
&=\int_0^x\left(\dfrac{t^2}{0!} + \dfrac{t^3}{1!} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Counting the number of strings with at least $2$ numbers Let $k$ and $n \ge 3$ be two natural numbers. How many strings in $\{1,...,n\}^k$ contain at least one occurrence of $1$ and $2$, or at least one occurrence of $2$ and $3$ or at least one occurrence of $1$ and $3$?
I tried to break it down and first count the num... | Use Inclusion/Exclusion.
Let $A$ be the subset of strings containing at least one $1$.
Let $B$ be the subset of strings containing at least one $2$.
Let $C$ be the subset of strings containing at least one $3$.
Then:
$$|(A\cap B)\cup (A\cap C)\cup (B\cap C)| = |A\cap B|+|A\cap C|+|B\cap C| - |(A\cap B)\cap(A\cap C)|-|(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What is $P(\min\{X, Y\} = 1)$? If $y=1,2,3$ and $x=0,1,2$ where $P(X=x, Y=y) = k(2x+3y)$
I need to find $P(\min\{X, Y\} = 1)$.
I thought I need to use that the CMF of the minimum is $1-P(X)$, and maybe to find k by doing derivative on the equation and to sum it up to 1?
would love any direction on this.
| Given that $P(X=x, Y=y) = k(2x+3y)$
$\sum_{(x,y)}P(X=x,Y=y) = k(0+3) +k(2 + 3) + k(4 +3) + k(0+6) + k(2+6) + k(4+6) + k(0+9) + k(2+9) + k(4+9) = 72k = 1$
Hence $k = \frac{1}{72}$
Now, $P(\min(X,Y)=1) = P(X=1,Y=2,3) + P(X=2, Y=1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Gradient of squared norm I am looking for the gradient of the function
$$
f(x)= \dfrac{1}{2} \Vert A^Tx \Vert^2 - b^Tx \, .
$$
Well so far I came up with
$$
\nabla f(x) = AA^Tx-b
$$
because
$$
\begin{aligned}
f(x) &= \dfrac{1}{2} \Vert A^Tx \Vert^2 - b^Tx \\
&= \dfrac{1}{2} (A^Tx)^T(A^Tx) - b^Tx \\
&= \dfrac{... | We can compute in the following way as well, using the identity $\|x-y\|^2 = \|x\|^2 - 2\langle x,y\rangle +\|y\|^2$:
$$\begin{align}f(x+h) - f(x) &= \frac{1}{2}\|A^T(x + h)\|^2 - \frac{1}{2}\|A^Tx\|^2 -b^T(x+h) + b^Tx\\
&= \langle A^Tx,A^T h\rangle - b^T h\\
&= \langle AA^Tx - b,h\rangle\end{align}$$
so the gradient $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
solving diophantine problem $a^3+b^3=2019(1+ab)$ for coprime $a$ and $b$ We're interested in solving
$$\begin{cases} a^3+b^3=2019(1+ab) \\ \gcd(a,b)=1 \end{cases}$$
I'm stuck with the deduction
*
*Say why $a^3 \equiv -b^3 \pmod{2019}$ . (done)
*Using Fermat, prove that $a^{672} \equiv b^{672} \pmod{2019}$ . (done)
... | Here's a method that will work, though you'll need a computer like Wolfram Alpha to do the computations... unless you really want to do it by hand, which is also doable.
The idea is to use the factorization $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx).$$ Luckily, $$2019 = 3\cdot 673,$$ where the factor of $3$ is ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Determining the closed form of $\sum\limits_{r = 2}^n \binom{n}{r} \binom{r}{2}$ Here is a sum in combinatorics, $\sum\limits_{r = 2}^n \binom{n}{r} \binom{r}{2}$ where $n>2$, does this have a closed form?
| $$\sum_{r=2}^n \binom{n}{r} \binom{r}{2}
=\sum_{r=2}^n \binom{n}{2} \binom{n-2}{r-2}
=\binom{n}{2} \sum_{r=2}^n \binom{n-2}{r-2}
=\binom{n}{2} 2^{n-2}
=n(n-1) 2^{n-3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Maximize $\boxed{\mathbf{x}+\mathbf{y}}$ subject to the condition that $2 x^{2}+3 y^{2} \leq 1$ Maximize $\mathbf{x}+\mathbf{y}$ subject to the condition that $2 x^{2}+3 y^{2} \leq 1$
My approach
$\frac{x^{2}}{1 / 2}+\frac{y^{2}}{1 / 3} \leq 1$
Let $z=x+y$
$\mathrm{Now}, 4 \mathrm{x}+6 \mathrm{y} \frac{d y}{d x}=0 \Ri... | Use Schwarz inequality
$$(x+y)^2\le \left((\frac{1}{\sqrt{2}})^2+(\frac{1}{\sqrt{3}})^2\right)\left( (\sqrt{2}x)^2+(\sqrt{3}y)^2\right)\le 5/6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Symmetrical point with respect to a plane I have the point $A = (10,0,20)$, and I want the coordinates of its symmetrical point $B$ with respect to the plane $\pi = \begin{cases} x=2+3\alpha+\beta\\y=\alpha\\z=\alpha + 2\beta \end{cases}$, how can I do that? Look at the image:
image
These are the question's alternative... | An equation of the plane is
\begin{equation}
x = 2 + 3 y + (z-y)/2 \Longleftrightarrow 2 x - 5 y -z -4 = 0
\end{equation}
Let $A = (x_0,y_0,z_0) = (10, 0, 20)$ and $B = (x_1, y_1, z_1)= A + t (2, -5, -1)$ because the vector $(2, -5, -1)$ is perpendicular to the plane. We have $2 x_0-5 y_0 -z_0 - 4 = -4$, hence we need ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove that $\sum_{k=0}^n{(-1)^k{4n-2k\choose 2n}{2n\choose k}}=2^{2n}$? $$\sum_{k=0}^n{\left( -1 \right) ^k\left( \begin{array}{c} 4n-2k\\ 2n\\\end{array} \right) \left( \begin{array}{c} 2n\\ k\\\end{array} \right)}=2^{2n}$$
I know the correctness of this formula, but how can I prove it?
Thanks for your help.
|
Prove that$$\sum_{k=0}^n{\left( -1 \right) ^k\left( \begin{array}{c} 4n-2k\\ 2n\\\end{array} \right) \left( \begin{array}{c} 2n\\ k\\\end{array} \right)}=2^{2n}$$
Consider the following binomial expansion
\begin{align*}
(x^2-1)^{2n}&=\sum_{k=0}^{2n}(-1)^kx^{4n-2k}{2n\choose k}
\end{align*}
Put $x=1$ after differentia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"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.