Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the sum of series $ \sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$ Let it be known that $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}
{6}.$$ Given such—find $$\sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$$
Attempt:
I have tried using the fact that $\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ and then expandi... | I'm basically just going to use the identity you noted in the comments $\frac 1n-\frac 1{n+1}=\frac 1{n(n+1)}$ repeatedly and then the closed form for $\zeta(2)$.
$$\sum_{n=1}^\infty \frac 1{n^3(n+1)^3}=\sum_{n=1}^\infty \left(\frac 1n-\frac 1{n+1}\right)^3=\sum_{n=1}^\infty \frac 1{n^3}-\frac{3}{n^2(n+1)}+\frac{3}{n(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4254127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Solving $\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 $. Simply bringing it to a common denominator does not lead me to success How can I solve this equation?
$$\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 $$
Simply bringing it to a common denominator does not lead me to success
What I tried
| \begin{equation}
\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}
=4 \\
\text{We multiply both sides by teh common denominator}\\
(1+x)(1+2x)(1+3x\quad =\quad 4\big((4+x)(4+2x)(4+3x)\big) \\
\text{we expand both sides} \\
x^3 + 11 x^2 + 6 x + 1\quad =\quad 24 x^3 + 176 x^2 + 384 x + 256\\
\text{we collect terms} \\
18 x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4254604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Difficult limit proof Let $a_1 = 1$ and define a sequence recursively by $$a_{n+1} = \sqrt{a_1+a_2+\dots+a_n}.$$ Show that $\lim_{n \to \infty} \frac{a_n}{n} = \frac{1}{2}$.
So far, I've written $a_{n+1} = \sqrt{a_n^2 + a_n}$, and have shown that if $a_n/n < 1/2$, then so is $a_{n+1}/{(n+1)}$. I'm really not sure how t... | The Stolz-Cesaro theorem states the following:
Let $a, b$ be sequences of real numbers such that $b$ is monotone increasing and $\lim\limits_{n \to \infty} b_n = \infty$. If $\lim\limits_{n \to \infty} \frac{a_{n + 1} - a_n}{b_{n + 1} - b_n} = \ell$ then $\lim\limits_{n \to \infty} \frac{a_n}{b_n} = \ell$.
It's basic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4255038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solve differential equation. $2xy'$ + $y^2 - 1 =0$ Solve differential equation.
$2xy' + y^2 - 1 =0$
My work.
$2x\dfrac{dy}{dx}+y^2-1=0$
$\dfrac{dx}{2x}=\dfrac{dy}{1-y^2}$
$\int\dfrac{dx}{2x}=\int\dfrac{dy}{1-y^2}$
getting
$\frac{1}{2}\ln|x|-\frac{1}{2}(\ln|1+y|-\ln|1-y|)=0$
but solution in the book says $y=\dfrac{cx-1}... | So $\dfrac{1}{2}\ln|\dfrac{1+y}{1-y}| = \dfrac{1}{2}\ln|x| +c \implies \ln|\dfrac{1+y}{1-y}| = \ln|x| +c_1 \implies \ln|\dfrac{1+y}{1-y}| = \ln|c_2x|$ as any number can be written as logarithm of some number. Thus, $\frac{1+y}{1-y} =cx \implies \frac{2}{1-y}-1=cx, cx+1=\frac{2}{1-y}, 1-y=\frac{2}{cx+1}, y=1-\frac{2}{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4256435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $1 - \sqrt{ \frac{2}{n-1} } \cdot \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } $ as $n \to \infty$ I stumbled upon this problem while reading about the bias of the sample standard deviation.
How to show that:
$$\bigg(1 - \sqrt{ \frac{2}{n-1} } \cdot \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \bi... | Using twice Stirling formula, we have
$$\log \left(\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma
\left(\frac{n-1}{2}\right)}\right)=\frac{1}{2} \log \left(\frac{n}{2}\right)-\frac{3}{4 n}-\frac{1}{2 n^2}-\frac{3}{8
n^3}+O\left(\frac{1}{n^4}\right)\tag 1$$
Using the binomial expansion and taking the logarithm and u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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The series representation of $\sum _{i=0}^{n-1} (-1)^{i+1} \left(\frac{i}{n}\right)^{r}$ Lets consider the:
$$\left(\frac{1}{n}\right)^r-\left(\frac{2}{n}\right)^r+\left(\frac{3}{n}\right)^r-...+\left(\frac{n-1}{n}\right)^r$$
Trying to find any formal serise representation for variable $n$. Have tried the Euler–Maclaur... | First note that
$$
\sum\limits_{j = 0}^{n - 1} {( - 1)^{j + 1} \left( {\frac{j}{n}} \right)^r } = - \frac{1}{{n^r }}\sum\limits_{j = 0}^{n - 1} {( - 1)^j j^r } .
$$
Employing the Euler–Boole summation formula with $f(x) = x^r$, $h = 0$, $m = r + 1$, we deduce
\begin{align*}
\sum\limits_{j = 0}^{n - 1} {( - 1)^j j^r ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4259826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Eliminating $\theta$ from $\cos^3\theta +a\cos\theta =b$ and $\sin^3\theta +a\sin\theta =c$
Eliminate $\theta$ from the equations.
$$\cos^3\theta +a\cos\theta =b$$
$$\sin^3\theta +a\sin\theta =c$$
Can anyone solve this question?
| So, if I understood properly your question, we have
$$
\left\{ \begin{array}{l}
\theta ,a,b,c \in R \\
\cos ^3 \theta + a\cos \theta - b = 0 \\
\sin ^3 \theta + a\sin \theta - c = 0 \\
\end{array} \right.
$$
and we are looking for which domain of the four variables $ \theta ,a,b,c$
the system of the two equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4265926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Maximum distance between mid-point of chord of ellipse Let $E$ be the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$. For any three distinct points $P,Q$ and $Q′$ on $E$, let $M(P, Q)$ be the mid-point of the line segment joining $P$ and $Q$, and $M(P, Q′)$ be the mid-point of the line segment joining $P$ and $Q′$. Then the ... | Define $X\equiv M(P,Q)$ and $Y\equiv M(P,Q')$. Using the midpoint theorem $(\triangle PXY\sim \triangle PQQ')$,
$$|XY|=\frac{1}{2}\cdot|QQ'|$$
$|QQ'|$ is maximum when $Q$ and $Q'$ lie on the major axis of the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}$ (the two farthest points on an ellipse). Hence, $Q=(\pm 4,0)$ and $Q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4269074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Showing $ 2\sqrt{\frac{x+3}{x}}+8\sqrt{\frac{x+1}{x}}-\ln\left(\frac{(x+1)^{3/2}(x+3)}{(x-1)^{5/2}}\right)\geq 10 $ for $x\geq7$
Suppose that $x\geq 7$. I would like to show that
$$ 2\sqrt{\frac{x+3}{x}} + 8\sqrt{\frac{x+1}{x}}-\ln\left(\frac{(x+1)^{3/2}(x+3)}{(x-1)^{5/2}}\right)\geq 10 $$
I rewrote the inequality as... | We need to prove that, for all $x \ge 7$,
$$2\sqrt{\frac{x+3}{x}}-2\ln \sqrt{\frac{x+3}{x}} +8\sqrt{\frac{x+1}{x}}-3\ln \sqrt{\frac{x+1}{x}} \geq 10+5\ln\sqrt{\frac{x}{x-1}} .$$
With the substitution $x = \frac{1}{y}$,
it suffices to prove that, for all $y\in [0, 1/7]$,
$$2\sqrt{1 + 3y} - \ln(1 + 3y) + 8\sqrt{1 + y} - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4270862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solve $\lim_{x\to \pi/4} \frac{\sin x - \cos x}{x-\pi/4}$ As the title suggests, we have to solve the limit:
$\lim_{x\to \frac\pi4} \frac{\sin x - \cos x}{x-\frac \pi4}$
I'm able to solve it by using L'Hospital's rule and got an answer $\sqrt2$ but the problem is that this rule is not allowed at school level.
So I trie... | Note that $\cos$ is continuous at $\pi/4$ and $\cos(\pi/4)>0$, so $\cos(x)$ is certainly nonzero near $\pi/4$. This justifies writing
\begin{align}
\frac{\sin(x)-\cos(x)}{x-\frac{\pi}{4}} &= \frac{\cos(x)\left(\frac{\sin(x)}{\cos(x)}-1\right)}{x-\frac{\pi}{4}}\\
&= \cos(x)\cdot\frac{\tan(x)-1}{x-\frac{\pi}{4}}\\
&= \co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4273783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\sum_\circ\frac{a^3+3b^3}{5a+b}\ge\frac23(a^2+b^2+c^2)$ for $a,b,c>0$
Prove the following for a, b and c being positive real numbers :-
$\sf{{ \dfrac{ a^3 + 3b^3 }{5a+b} + \dfrac{ b^3 + 3c^3}{5b+c} + \dfrac{ c^3 + 3a^3}{5c+a} \geqslant \dfrac{2}{3} (a^2 + b^2 + c^2) }} $
So here we have a question in whic... | Split our the fractions; use CS to grouped summands; use $ab+bc+ca \le a^2+b^2+c^2$:
$$\begin{align}
&\dfrac{ a^3 + 3b^3 }{5a+b} + \dfrac{ b^3 + 3c^3}{5b+c} + \dfrac{ c^3 + 3a^3}{5c+a} \\
&= \color{blue}{\dfrac{ a^4 }{5a+b}} + 3\color{fuchsia}{\frac{b^3}{5a+b}} \\
&+ \color{blue}{\dfrac{ b^3}{5b+c}} + 3\color{fuchsia}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4275449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
the value of $\lim_{x\to0}\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}$ I want to compute this limit
$$\lim_{x\to0}\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}.$$
I tried to apply Hopital rule, but I cannot compute it.
| Just to suggest another way: $\ln(1+x^2)=x^2-\frac{x^4}2+O(x^6)$ and $\sin(x^2)=x^2+O(x^6)$, so the numerator behaves like $(x^2+1)(x^2-\frac{x^4}2)-x^2\sim\frac{x^4}2$. The denominator on the other hand behaves like $x^3$.
Proof that the limit is $0$.
*
*By simple study of functions, we check that $\lvert\ln(x^2+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4277769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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A question about the maximal domain of a function So I have got the following equation:
$$h(x) = \sqrt{\frac{1}{x+1}+1}$$
I need to find the maximal domain of the function.
I have tried doing it algebraically:
As this is a square root, $\frac{1}{x+1}+1$, must be greater than $0$ and there is an asymptote at $x = -1$.
... | In what follows, I will assume that $h$ is a real-valued function of a real variable.
The expression
$$\frac{1}{x + 1}$$
is undefined when $x = -1$. Thus, we require that $x > -1$ or $x < -1$.
Since we cannot take the square root of a negative number, we require that the radicand (the term inside the square root) be n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4278979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\delta > 0$ such that $|x-2| < \delta$ implies $|x^2+2x-18| < \frac{1}{4}$ Find $\delta > 0$ such that $|x-2| < \delta$ implies $|x^2+2x-18| < \frac{1}{4}$.
I have been having trouble with this question from my Analysis class because $f(x)$ doesn't factor. Here is the first scratch work I did attempting to find ... | Let $f(x)=|x^2+2x-18|$. For all $\delta>0$, $|2-2|<\delta$ but $f(2)=10\geq\frac{1}{4}$. So no matter which value you pick for $\delta$ there is a $x$ ($2$) that wont satisfy $|x^2+2x-18|<\frac{1}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4280446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Prove that $17$ divides $3^{4n} + 4^{3n+2}$ for all natural numbers $n$ by induction. I would like to preface my very first question here by stating that this is not homework. Indeed, I am a mathematics major and computer science minor. However, I am doing extra work on my own to develop my understanding. I am unsure o... | Assume true for $n=k$. For $n=k+1$ we have (using $3^4=81$ and $4^3=64$) $81\times 3^{4k}+64\times 4^{3k+2}=17\times 3^{4k}+64\times(3^{4k}+4^{3k+2})$, which is divisible by $17$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4282561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$3 \sin x + 4 \cos y = 5$, $4 \sin y + 3 \cos x = 2$ How to find $\sin x$, $\sin y$, $\cos x$, $\cos y$, 2020 contest question https://www.cemc.uwaterloo.ca/contests/past_contests/2020/2020CSMC.pdf
https://www.cemc.uwaterloo.ca/contests/past_contests/2020/2020CSMCSolution.pdf
Question 5 from 2020 CSMC math contest:
$$3... | Hint: starting with $\sin x \cos y + \cos x \sin y=\frac{1}{6}$ and using that $\cos y=\frac{5-3\sin x}{4}$, $\sin y=\frac{2-3\cos x}{4}$ we can obtain
$$ \sin x \cdot (5-3\sin x)+\cos x \cdot (2-3\cos x)=\frac{2}{3}$$
$$ 5\sin x+2\cos x =\frac{11}{3} \implies \cos x=\frac{11}{6}-2.5\sin x $$We also have that $\cos^2 x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4282724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to use the epsilon delta definition to prove that $\lim\limits_{x\to 1} \frac{x^3-1}{x-1} = 3$
Not sure if I am doing this right, however, this is what I have:
Let $\epsilon > 0$. We need to find a $\delta > 0$ such that $0<|x-1|<\delta$ leads to the conclusion $|f(x)-3|<\epsilon$.
$$|\frac{x^3-1}{x-1} - 3| < \epsi... | We must find a $\delta > 0$ such that $|x-1|<\delta \implies \left|\frac {x^3 - 1}{x-1} - 3\right| < \epsilon$
$\left|\frac {x^3 - 1}{x-1} - 3\right|\\
\left|\frac {x^3 - 3x + 2}{x-1}\right|\\
\left|\frac {(x-1)^2(x+2)}{x-1}\right|$
$\left|\frac {(x-1)^2}{x-1}\right| = |x-1|<\delta$ when $x\ne 1$
Let $\delta \le 1$ the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4283629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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finding the argument of $1-\cos 2\theta -i\sin 2 \theta$ "Let z = $1-\cos 2\theta -i\sin 2 \theta$, $0\leq \theta \leq \pi$. Find the modulus and argument of z in terms of $\theta$ in their simplest forms."
The modulus is pretty easy, I got $2\sin \theta$ which is correct.
The problem is, when finding the argument of z... | $$z = 1-\cos 2\theta -i\sin 2 \theta$$
Modulus
$$|z| = \sqrt{(1-\cos 2 \theta)^2 + (-\sin 2 \theta)^2} = \\
= \sqrt{1 + \cos^2 2 \theta - 2 \cos 2 \theta + \sin^2 2 \theta} = \\
= \sqrt{2 - 2 \cos 2 \theta} = \\
= \sqrt{2 - 2 \cos^2 \theta + 2 \sin^2 \theta} = \\
= \sqrt{2 - 2 \cos^2 \theta + 2 (1-\cos^2 \theta)} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4284889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab}$, prove that $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$.
Problem: Let $a, b, c, x, y, z$ be real numbers
such that $abc \ne 0$, $ ~ a + b + c \ne 0$, $~ (a^2 - bc)(b^2 - ca)(c^2 - ab)\ne 0$, $xyz\ne0$, $x+y+z\ne0$, $(x-y)^2+(y-z)^2+(z-x)^2\ne0$
and $$\fr... | @Jean Marie and @orangeskid gave very nice proofs.
Their idea, in algebraic manner, leads to the following proof:
We have the following identity
$$(p^2 - qr)^2 - (q^2 - rp)(r^2 - pq)
= p(p + q + r)(p^2 + q^2 + r^2 - pq - qr - rp).$$
Using this identity, we have
$$\frac{(x^2 - yz)^2 - (y^2 - zx)(z^2 - xy)}{(a^2 - bc)^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4286648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Dealing more efficiently with fractional forms in system of equations As an example, suppose we have to solve the following system of two equations and two unknowns:
$$
\begin{cases}
-\frac{10}{x}-\frac{8}{y} &= \frac{8}{3} \\
-\frac{6}{x}+\frac{6}{y} &= -\frac{1}{3}
\end{cases}
$$
My approach and solution
I opted ... | $$
\begin{cases}
-\frac{10}{x}-\frac{8}{y} &= \frac{8}{3},\quad{|}\cdot{3}\\
-\frac{6}{x}+\frac{6}{y} &= -\frac{1}{3}.\quad{|}\cdot{(-5)}
\end{cases}
\iff
\begin{cases}
-\frac{30}{x}-\frac{24}{y} &= \frac{24}{3},\\
\frac{30}{x}-\frac{30}{y} &= \frac{5}{3}.
\end{cases}
\iff
$$
$$
\iff
\begin{cases}
-\frac{54}{y} &= \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4288341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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General form for $\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) - 1\right)$ I'm wondering if there is a general form for the following sum:
$$\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) - 1\right)$$ for $m \in \mathbb{N}$
I have obtained the following closed-forms for these speci... | $\newcommand{\D}{\mathfrak{D}}\newcommand{\d}{\,\mathrm{d}}$Contour integration naturally leads us to the situation of Random Variable’s answer.
Fix a real $s>1$ and call this series $S$. Let $\log$ be the principal logarithm, so that $z\mapsto\log\frac{zs+1}{zs-1}$ is holomorphic on $\Bbb C\setminus[-s^{-1},s^{-1}]$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 2
} |
$3$ Bracket Knockout Tournament Probability $12$ people in $3$ brackets ($A,B$, and $C$) compete against each other in a knockout tournament. The final round has $4$ contestants, the $3$ winners from brackets $A,B$ and $C$, plus $1$ of the dropouts from brackets $A$ and $B$ brought back into the game at random. What is... | *
*A random player's chance of winning four games in a row is $\left(\frac{1}{2}\right)^4 = \frac{1}{16} \neq \frac{1}{12}$.
*You haven't counted the players who win their first match and lose their second (and are awarded the final random spot).
Its hard to count all the possibilities, but here we go:
Win all game... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4299828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Understanding Heron's formula proof
I was trying to understand the proof of Heron's formula. I understood everything up to step $(d)$ except how they arrived at the expression ($a$+$b$+$c$)($b$+$c$-$a$)($a$+$b$-$c$)($a$-$b$+$c$) and the steps afterwards I still don't understand I would appreciate if someone gave me in... | Hint:
\begin{align}
h^2
&=b^2 - d^2\\
&=b^2 - \frac{(b^2+c^2-a^2)^2}{4c^2}\\
&=\frac{(2bc)^2-(b^2+c^2-a^2)^2}{4c^2}\\
&=\frac{(2bc+b^2+c^2-a^2)(2bc-b^2-c^2+a^2)}{4c^2}\\
&=\frac{\big[(b+c)^2-a^2\big]\big[-(b-c)^2+a^2\big]}{4c^2}\\
&=\frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{4c^2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4300588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $(x+1)^6=x^6$ then prove that $x=\frac{-1}2-i\cot (\frac{\theta}2)$ where $\theta=\frac{2k\pi}6$, $k=0,1,2,3,4,5$ QUESTION
If $(x+1)^6=x^6$ then prove that $x=\Large{\frac{-1}2}$$-i\cot (\large\frac\theta2)$ where $\theta=\Large\frac{2k\pi}6$, $k=0,1,2,3,4,5$
How can we prove this question using complex numbers. We... | First note that $k = 0$ does not give a solution, as $cot(0)$ is undefined. This should match with the intuition that your equation
$$(x+1)^6 = x^6 \Longrightarrow 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1 = 0$$
should only have up to 5 roots by the fundamental theorem of algebra.
Otherwise, your solution is sound. You can... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of real solutions of $\begin{array}{r} {\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]} =\frac{3 x-1}{2} \end{array}$ Solve for $x \in \mathbb{R}$ $$\begin{array}{r}
{\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]}
=\frac{3 x-1}{2}
\end{array}$$ where $[x]$ denotes greatest integer less th... | HINT: The RHS $\frac{3x-1}{2}$ of the original equation must be an integer, which implies that $3x$ has to be an odd integer [typo noted via comment below]. Your bounds for $a$ leave only a few values to check for $x$, in particular $-2<x<9$; this an $3x$ an odd integer leaves only to check each $x$ satisfying $x= \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4311152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use Induction to prove recurrence
if the above screenshot is not visible, here is the text format:
Question:
Solve the following recurrence and prove your result is correct using induction:
$a_1 = 0$
$a_n = 3(a_{n-1}) + 4^{n}$ for $n>=2$
Use induction to prove this recursive sequence.
So my approach was that, I plug ... | Here's a useful "trick".
If
$a_n
=ua_{n-1}+v^n
$
then
(here comes the trick),
dividing by $u^n$,
$\dfrac{a_n}{u^n}
=\dfrac{ua_{n-1}}{u^n}+\dfrac{v^n}{u^n}
=\dfrac{a_{n-1}}{u^{n-1}}+(v/u)^n
$.
Let $b_n = \dfrac{a_n}{u^n}$.
Then
$b_n
=b_{n-1}+r^n
$
where $r = v/u$
or
$b_n-b_{n-1}
=r^n
$.
This becomes a telescoping sum,
s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rationalize the denominator of $\frac{4}{9-3\sqrt[3]{3} + \sqrt[3]{9}}$ Rationalize the denominator of $\frac{4}{9-3\sqrt[3]{3}+\sqrt[3]{9}}$
I keep making a mess of this. I tried vewing the denominator as
$a +\sqrt[3]{9}$, where $a=9-3\sqrt[3]{3}$ and secondly as
$b -3\sqrt[3]{3}+\sqrt[3]{9}$, where $b=9$.
Then using ... | To solve the general case, use the formula for the rationalization of 3 cube roots. $$\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}=\frac{\left(\sqrt[3]{a^{2}}+\sqrt[3]{b^{2}}+\sqrt[3]{c^{2}}-\sqrt[3]{ab}-\sqrt[3]{ac}-\sqrt[3]{bc}\right)\left(\left(3\sqrt[3]{abc}+a+b+c\right)^{2}-3\left(a+b+c\right)\sqrt[3]{abc}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4315306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the easier way to find the circle given three points? Given three points $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$, if $$\frac{y_2-y_1}{x_2-x_1} \neq \frac{y_3-y_2}{x_3-x_2} \neq \frac{y_1-y_3}{x_1-x_3},$$ then there will be a circle passing through them. The general form of the circle is $$x^2 + y^2 + dx + ey ... | I prefer a different approach.
You know that the center of the circle is the intersection between any two of the three perpendicular bisectors.
Therefore,
$$(x - x_1)^2 + (y - y_1)^2 = (x - x_2)^2 + (y - y_2)^2 \tag1 $$
and
$$(x - x_1)^2 + (y - y_1)^2 = (x - x_3)^2 + (y - y_3)^2. \tag2 $$
Let $f_1(x) = (x - x_1)^2 - (x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to apply the AM-GM relation in this inequality? Here is an inequality that I came across recently.
$((a+b)/2)^m$ $\le$ $(a^m + b^m)/2$, if $m<0$ or $m>1.$
Equality happens when $m = 0$ or $m = 1.$ Inequality reverses if $m$ is between 0 and 1.
I was able to prove this for $m = 2$. Here is what I did:
We know that $... | For $m>1$ (and $m\in\mathbb{Z}$), assume $0\le a\le b$ and let $c=\frac{a+b}{2}$ and $d=\frac{b-a}{2}$.
Then $a=c-d$ and $b=c+d$.
The LHS is $c^m$, and the RHS is $ \frac{(c-d)^m +(c+d)^m}{2}\ge c^m$ due to any negative terms being cancelled.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum or bounds of $\sum\frac{1}{a^n+b^n}$ I will keep it short as I have made $0$ progress on the problem (both solving it and researching) and am just interested. Is there any closed form or non-trivial bounds (so tighter than $1 + \frac{1}{a - 1}$) on the sum $\sum\frac{1}{a^n + b^n}$ where $a > 1$? Clearly it converg... | We can suppose $b<a$. I´ll assume too that the sum begins in $n=1$, so the obvious lower bound (with $b=0$) is $\sum_{n=1}^\infty\frac{1}{a^n}=\frac{1}{a-1}$. I don´t know if there is an exact way to find the sum, but one way I found to get upper/lower bounds as tight as you want uses this equality:
$$\frac{1}{a^n+b^n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4317994",
"timestamp": "2023-03-29T00:00:00",
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Decomposition of numbers: Show that there are positive integers $x, y$ and $z$, with $x$ and $y$ coprime, such that $a+b=x^2z\,$ and $\,c+d=y^2z$ The positive integers $a$, $b$, $c$ and $d$ satisfy the equation $$(ad-bc)^2=(a+b)(c+d).$$ Show that there are positive integers $x$, $y$ and $z$, with $x$ and $y$ coprime, s... | Write $a+b=vz,c+d=wz$. By the fundamental theirem of arithmatic, $v=p_1^{e_1}p_2^{e_2}..., w=q_1^{f_1}q_2^{f_2}...$. Since $v$ and $w$ are coprime, $p_i\neq q_j\forall i,j$. Finally,
$$(ad-bc)^2=(a+b)(c+d)=vwz^2$$
$$l^2:=\left(\frac{ad-bc}{z}\right)^2=vw$$
Now $l$ has its own prime factorization $r_1^{g_1}r_2^{g_2}...$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4323366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to calculate this improper integral? Calculate the improper integral
$$\displaystyle\int_0^{\infty}{\frac{1}{\theta}e^{\cos\theta}\sin(\sin\theta){d\theta}}$$
My try:
We know that for any $a\in\mathbb{C}$ the integral
$$\displaystyle\int_0^{\infty}e^{-ax^2}=\frac{1}{2}\sqrt{\frac{\pi}{a}}$$
Let $a=\cos\theta+i\sin\... | 1st Solution. Define the sine integral by
$$ \operatorname{Si}(x) = \int_{0}^{x} \frac{\sin t}{t} \, \mathrm{d}t. $$
Using integartion by parts, it can be proved that
$$ \operatorname{Si}(x) = \frac{\pi}{2} + \mathcal{O}\left(\frac{1}{x}\right) \qquad\text{as } x \to \infty. $$
Now note that $e^{\cos\theta}\sin\sin\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4323539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $(a_n)$ such that $a_1 \in (0,1)$ and $a_{n+1}=a_n+(\frac{a_n}{n})^2$. Prove that $(a_n)$ has a finite limit. Given $(a_n)$ such that $a_1 \in (0,1)$ and $a_{n+1}=a_n+(\frac{a_n}{n})^2$. Prove that $(a_n)$ has a finite limit.
Clearly $a_n$ are increasing. Also, $$\frac{1}{a_{n+1}}=\frac{1}{a_n\left(1+\frac{a_n}{n... | EDIT. I realized that this problem has been solved (even by myself) several times in the past; see the links above. A bound of the form
$$ \frac{x}{1-x} \leq \lim a_n \leq \frac{x}{1-\sqrt{x}} $$
is also proved there.
Let $x = a_1 \in (0, 1)$.
Claim 1. We have $a_n \leq nx$ for all $n \geq 1$.
Indeed, the base case ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4329102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Demonstration for the equal number of odd and unequal partitions of an integer I'm having some problems trying to resolve one exercise from The art and craft of problem solving by Paul Zeitz. What this problem asks you is to prove that $F(x)$ is equal to 1 for all $x$, where
$$F(x) = (1-x)(1+x)(1-x^3)(1+x^2)(1-x^5)(1+x... | We already know from OPs derivation
\begin{align*}
F(x)=F\left(x^2\right)=\cdots=F\left(x^{2^q}\right)\qquad\qquad q\in\mathbb{N}_0\tag{1}
\end{align*}
We set $F(x)=\sum_{n=0}^\infty a_nx^n$ and obtain from (1) by coefficient comparison
\begin{align*}
F(x)&=F\left(x^2\right)\\
\sum_{n=0}^{\infty}a_nx^n&=\sum_{n=0}^{\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4329739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Taylor expansion of $\sin \pi z$ at $z = -1$. Taylor expansion of $\sin \pi z$ at $z = -1$ is
$$\sin\pi z = -\sin(\pi(z+1)) = -\sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}(z+1)^{2n+1}$$ so that
$$\sin\pi z = \sum_{n=0}^\infty \frac{(-1)^{n+1}\pi^{2n+1}}{(2n+1)!}(z+1)^{2n+1}. \tag{$\dagger$}$$
But if I try this
\b... | Start from
$$\sin\pi z = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}z^{2n+1}$$
and
$$-1 = \cos\pi = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n}}{(2n)!}$$
Consider
$$\sin\pi z \cos\pi = \sin(\pi (z+1)) = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}z^{2n+1}\left(\sum_{n=0}^\infty \frac{(-1)^n\pi^{2n}}{(2n)!}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4329877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Let $P$ be a point inside $\triangle{ABC}$ such that $\angle{ABP}=10^\circ, \angle{BAP}=\angle{BCP} = 20^\circ$. Show $\triangle{ABC}$ is an isosceles Let $P$ be a point inside $\triangle{ABC}$ such that $\angle{ABP}=10^\circ$, $\angle{BAP}=\angle{BCP} = 20^\circ$. Show $\triangle{ABC}$ is an isosceles and find $\angle... | initial conditions: Let $P$ be a point inside $△ABC$ such that $∠ABP=10^\circ$, $∠BAP=∠BCP=20^\circ$
if $C\in R=λ_{A}∪λ_{B}∪l$ then the triangle ABC is an isosceles.
$\begin{array}{} λ_{A}(A,1) & λ_{B}(B,1) & (l) \text{ perpendicular bisector of the line segment AB.} \end{array}$
If $C$ belongs to the intersection of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$1^2 - 2^2 + 3^2 - 4^2 + \dots + 1999^2$ using closed form formula for sum of squares of first $n$ natural numbers The question is simple, the solution, not so much
Q.Find the sum of the given expression $1^2- 2^2 + 3^2 - 4^2 + \dots + 1999^2$
My idea is we know $1^2 + 2^2 + 3^2 + 4^2 + \dots + n^2 = \frac{n(n + 1)(2n ... | We can use the identity $$(x+1)^2-x^2=x+1+x$$
to tackle the sum in 2 simpler ways:
Way 1
$$ \begin{aligned}&1^2 - 2^2 + 3^2 - 4^2 + 5^2-... -1998^2+ 1999^2
\\=& 1+(2+3)+(4+5)+…+(1998+1999)
\\= &19990000\end{aligned}$$
Way 2
$$ \begin{aligned}&(1^2 - 2^2 + 3^2 - 4^2 + 5^2-6^2...-1998^2)+ 1999^2
\\=&-((1+2)+(3+4)+(5+6)+…... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integration of $\int_0^{\pi/2}\frac{d\theta}{a^2+b^2\cos^2(\theta)}$ gives results such that $\tan(\pi/2)$ comes which is "undefined". How to proceed? Integrating the below equation:
$$\int_{0}^{\pi/2}\frac{d\theta}{a^2+b^2\cos^2(\theta)}$$
gives,
$$\frac{1}{\left| a \right|\sqrt{a^2+b^2}}\left[\arctan\left(\frac{ a\... | Assume that $a>0$. You have\begin{align}\int_0^{\pi/2}\frac1{a^2+b^2\cos^2\theta}\,\mathrm d\theta&=\int_0^{\pi/2}\frac{\sec^2\theta}{a^2\sec^2\theta+b^2}\,\mathrm d\theta\\&=\frac1a\int_0^{\pi/2}\frac{a\sec^2\theta}{a^2\tan^2(\theta)+a^2+b^2}.\end{align}Now, if you do $x=a\tan\theta$ and $\mathrm dx=a\sec^2(\theta)\,\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4337434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\lim\limits_{R\to +∞}\iint_{x^2+y^2\leq R^2}\left(\frac{1+2x^2}{1+x^4+6x^2y^2+y^4}-\frac{1+y^2}{2+x^4+y^4}\right)\,\mathrm dx\mathrm dy$
Evaluate $\lim\limits_{R\to\infty} J(R)$, where
$$J(R) = \iint_{x^2 + y^2\leq R^2} \left(\frac{1+2x^2}{1+x^4+6x^2y^2 + y^4} - \frac{1+y^2}{2+x^4+y^4}\right)\,\mathrm dx\m... | $\def\d{\mathrm{d}}\def\R{\mathbb{R}}\def\vector#1#2{\begin{pmatrix}#1\\#2\end{pmatrix}}\def\abs#1{\left|#1\right|}\def\brace#1{\left\{#1\right\}}\def\paren#1{\left(#1\right)}$To prove (1), it suffices to prove\begin{gather*}
\iint\limits_{x^2 + y^2 \leqslant R^2} f(x, y) \,\d x\d y = \iint\limits_{x^2 + y^2 \leqslant ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4337990",
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$a,b>0$ then: $\frac{1}{a^2}+b^2\ge\sqrt{2(\frac{1}{a^2}+a^2)}(b-a+1)$
Let $a,b>0$. Prove that: $$\frac{1}{a^2}+b^2\ge\sqrt{2\left(\frac{1}{a^2}+a^2\right)}(b-a+1)$$
Anyone can help me get a nice solution for this tough question?
My approach works for 2 cases:
Case 1: $b-a+1>0$ then squaring both side, we get equival... | Since OP posted their own version, I'd post my solution.
WTS $ b^2 - \sqrt{2 ( \frac{1}{a^2 } + a^2 )} \cdot b + [\frac{1}{a^2} + \sqrt{2 ( \frac{1}{a^2 } + a^2 )} \cdot (a-1) ] \geq 0$.
We will prove this by viewing it as a quadratic in $b$, and showing that the discriminant is $ \leq 0$ for $ a > 0$, hence the value... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proving $\frac{1}{abc}+\frac{12}{a^2b+b^2c+c^2a}\ge5$
Let $a,b,c>0$, $a+b+c=3$. Prove that$$\frac1{abc}+\frac{12}{a^2b+b^2c+c^2a}\ge5$$
My approach using a well-known result:$$a^2b+b^2c+c^2a+abc\le\frac4{27}(a+b+c)^3$$
We need to prove that $\frac1{abc}+\frac{12}{4-abc}\ge5$ but this inequality does not hold for all ... | Remark
$$\displaystyle\sum f(a,b,c)=f(a,b,c)+f(b,c,a)+f(c,a,b)$$
means cyclic sum.
Homogenize it, then we get
denote
$$
\begin{aligned}
f\left( a,b,c \right) :=&\sum{a^5b}+3\sum{a^4b^2}+3\sum{a^3b^3}+\sum{a^2b^4}\\
&+15\sum{a^4bc}-92\sum{a^3b^2c}+42\sum{a^2b^3c}+81a^2b^2c^2
\end{aligned}
$$
we gonna prove $f(a,b,c)\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4341189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding perfect two terms Egyptian fraction If I have two unity fractions, like $\frac{1}{12} + \frac{1}{180}$, for instance. These two fractions can be re-writen as $\frac{1}{15} + \frac{1}{45}$ or even $\frac{1}{18} + \frac{1}{30}$, which satisfies the Egyptian Fraction concept of maximising the value of the smallest... | For a given rational number $\dfrac{c}{d}$, we wish to find all pairs of positive integers $(a,b)$ such that $$\dfrac{c}{d} = \dfrac{1}{a}+\dfrac{1}{b}.$$
We can manipulate the equation as follows: $$cab = da+db$$ $$cab - da-db = 0$$ $$c^2ab-cda-cdb = 0$$ $$c^2ab-cda-cdb+d^2 = d^2$$ $$(ca-d)(cb-d) = d^2.$$
Hence, $ca-d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4341333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How to solve system of linear equations that came up in looking at a probability problem, any tricks? The following system of $k$ equations came up in a probability problem I was looking at:
$$(n+1)x_1 - x_2 = 1, \quad x_1 + nx_2 - x_3 = 1 , \quad x_1 + nx_3 - x_4 = 1, \quad x_1 + nx_4 - x_5 = 1, \quad \ldots \quad x_1... | "Smart" Brute force isn't that bad. We have the following pattern:
$S_1: (n+1) x_1 - x_2 = 1$
$S_2 + nS_1: (n^2+n+1) x_1 - x_3 = 1+n$
$S_3 + nS_2 + n^2 S_1: (n^3+n^2+n+1) x_1 - x_4 = 1+n+n^2$.
$\vdots$
$S_{k-1}+nS_{k-2} + \ldots + n^{k-2}S_1: \frac{n^k-1}{n-1} x_1 - x_k = \frac{n^{k-1} - 1 } { n-1}$
$S_k: x_1 + nx_k =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is the series $\sum_{n=1}^\infty\frac{1}{n}\left(\sum_{k=1}^n\frac{1}{k}\left(\frac{1}{2}\right)^{n-k}\right)$ convergent? $$
\sum_{n = 1}^\infty\dfrac{1}{n}\left(\sum_{k = 1}^n\dfrac{1}{k}\left(\dfrac{1}{2}\right)^{n - k}\right)
$$
Does the series converge?
I calculate it using Matlab, and it seems that the sum conver... | \begin{aligned}\sum_{k=1}^{n}{\frac{2^{k-n}}{k}}&=\frac{1}{2^{n-1}}\sum_{k=1}^{n}{\int_{0}^{1}{\left(2x\right)^{k-1}\,\mathrm{d}x}}\\ &=\frac{1}{2^{n-1}}\int_{0}^{1}{\frac{1-2^{n}x^{n}}{1-2x}\,\mathrm{d}x}\\ &=\frac{1}{2^{n}}\int_{-\frac{1}{2}}^{\frac{1}{2}}{\frac{1-\left(1-2x\right)^{n}}{x}\,\mathrm{d}x}\end{aligned}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$\lim_{n\to+\infty} {n^{(4/3)}} \Big( \sqrt[3]{n^2 +1} + \sqrt[3]{3- n^2} \Big)$ $\lim_{n\to+\infty} {n^{(4/3)}} \Big( \sqrt[3]{n^2 +1} + \sqrt[3]{3- n^2} \Big)$
I used an extension of the appropriate 1 (A ^ 2 - AB + B ^ 2) / (A ^ 2 - AB + B ^ 2). I got here
$\lim_{n\to+\infty} {n^{(4/3)}} \Big(4/ (({n^2 +1} )^{(2/3)}... | WLOG $\dfrac1{n^2}=h$ to find $\sqrt[3]{3-n^2}=\sqrt[3]{\dfrac{3h-1}h}=\dfrac{\sqrt[3]{3h-1}}{h^{1/3}}=-\dfrac{\sqrt[3]{1-3h}}{h^{1/3}}$
So, the limit in question reduces to
$$\lim_{h\to0^+}\dfrac{\sqrt[3]{1+h}-\sqrt[3]{1-3h}}h=\lim_{h\to0^+}\dfrac{\sqrt[3]{1+h}-1}h-\lim_{h\to0^+}\dfrac{\sqrt[3]{1-3h}-1}h$$
Now for s... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A 3d line intersecting 2 other 3d lines The equation of the line parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and intersecting the lines $9x + y + z + 4 = 0 = 5x + y + 3z$ & $x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3$
My solution is as follow
$\frac{{x - a}}{2} = \frac{{y - b}}{3} = \frac{{z - c}}{4}$ represent ... | Your way of squeezing more than one equation in a line is confusing me and I will do it the usuual way, one line for an equation.
A linear equation represent a plane. A line is represented as the intersection of two planes, that is a systeme of two equations. The coefficients of $x,y,$ and $z$ of a linear equation are ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Hopf fibration from $SO(3)$ Lie algebra generators? One can use the Pauli matrices $\sigma_i$ to generate $Cl_3(\mathbb{R})$ and taking commutators of these matrices gives the $SU(2)$ Lie algebra $\mathfrak{su}(2)=\biggl(\begin{matrix}
ia&-z\\
z&-ia\\
\end{matrix}\biggr)$
However, one can also generate $Cl_3(\mathbb{R}... | The $3\times 3$ Lie algebra matrices, you have quoted, would act on the quaternion expressed as a $3\times 3$ matrix, i.e. the SO(3) group.
https://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation
Rather than on a quaternion represented as a 4-column, a $2\times2$ complex matrix [SU(2)], or a $4\times4$ real matr... | {
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"source": "stackexchange",
"question_score": "3",
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Let $x,y,z$ be real numbers in the interval $[-1,2]$ such that $x+y+z=0$. Prove that $\sum_{cyc}\sqrt{\frac{(2-x)(2-y)}{(2+x)(2+y)}}\ge 3$ Let $x,y,z$ be real numbers in the interval $[-1,2]$ such that $x+y+z=0$. Prove that
$ \sqrt{\frac{(2-x)(2-y)}{(2+x)(2+y)}}+\sqrt{\frac{(2-y)(2-z)}{(2+y)(2+z)}}+\sqrt{\frac{(2-z)(2-... | As you suggested, let $p=a+b+c$, $q=ab+bc+ca$, $r=abc$. We use contradiction: if $q<3$, then $f(p,q,r)=3+a^2+b^2+c^2-a^2b^2-b^2c^2-c^2a^2-3a^2b^2c^2=3+p^2-2q-q^2+2pr-3r^2>0$ and thus leads to a contradiction. If $p,q$ is fixed, we must have $f$ takes its minimum when $r$ is at its extreme. Consider $F(X)=x^3-p^2X+qX$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $x$ in the figure
Find $x$ in the figure.
(Answer: $20^\circ$)
My progress:
Let $P$ such that $PDHC$ is cyclic.
$\angle ACH = 180^\circ-20^\circ-90^\circ = 70^\circ\implies AHC = 100^\circ$
$\therefore ∠PDC=∠AHC=100^\circ$ and $∠PDA=180^\circ-60^\circ-100^\circ=20^\circ$
$\therefore ∠BDH+x=60^\circ$
If I find ... | Leveraging colleague Vasily's perpendicular idea:
Let $P$ be a point on $AH$ such that the quadrilateral $PDHC$ is cyclic.
$\angle ACH = 180^\circ-20^\circ-90^\circ = 70^\circ\implies AHC = 100^\circ.$
$\therefore ∠PDC=∠AHC=100^\circ$ and $∠PDA=180^\circ-60^\circ-100^\circ=20^\circ.$
Trace $DE \perp AH ~ (E \in AC).$... | {
"language": "en",
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"source": "stackexchange",
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Finding integer solutions to $(x-y)^2+(y-z)^2+(z-x)^2=2022$ This was from round A of the Awesomemath Summer Program application. The deadline was yesterday and we can discuss the problems now.
Find all integer triples $(x,y,z)$ which satisfy
$$(x-y)^2+(y-z)^2+(z-x)^2=2022$$
I tried a couple of things. First I noticed... | I am going from here: instead of dividing two, I am going to multiple two.
$$4044=2((x-y)^2+(y-z)^2+(z-x)^2)=(2x-y-z)^2+3(y-z)^2$$
WLOG $x\ge y\ge z$ Therefore set $2x-y-z=a$ and $y-z=b$ we are finding the solution of $a^2+3b^2=4044$. After trying, we have three solutions: $(63,5),(39,29),(24,34)$. $(63,5)$ yields $x=3... | {
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determinant of $a_{ij}=2$ where if $i=j$ and $1$ if $|i-j|=1$, otherwise zero I want to find the general formula for $n\times n$ matrix $A$ whose entries $a_{ij}$ is $2$ if $i=j$ and $1$ for $|i-j|=1$ and $0$ otherwise.
Followings are my explicit trials;
For $n=3$ case I have
\begin{align}
\begin{vmatrix}
2 & 1 & 0 \\... | You're on the right track. Call the matrix you've defined in dimension $n$ $A_n$. Using expansion by minors on the first row, you get
$$\det(A_n)=2\det(A_{n-1})-\det(B_{n-1}),$$
where $B_{n-1}$ is $A_n$ but with the first row and second column removed. Now, we wish to determine the determinant of $B_n$. The only nonzer... | {
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"source": "stackexchange",
"question_score": "3",
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Integrate using integration by parts $\int e^{-\frac{x}{2}}\frac{\sqrt{1-\sin{x}}}{1+\cos{x}}dx$ Integrate using integration by parts $\int e^{-\frac{x}{2}}\frac{\sqrt{1-\sin{x}}}{1+\cos{x}}dx$
My Attempt
I tried taking $u=e^{-\frac{x}{2}}$ to evaluate using the general formula $\int udv=uv-\int vdu$, but I ended up wi... | The answer is not-that correct: we assume $-\pi/2\le x\le\pi/2$. Look at this, we first consider the half-angle transform
$$\frac{\sqrt{1-\sin{x}}}{1+\cos{x}}=\frac{\cos \frac x2-\sin \frac x2}{2\cos^2 \frac x2}$$
Notice that $$\frac{{\rm d}}{{\rm d}x}\frac{1}{\cos \frac x2}=\frac{\sin \frac x2}{2\cos^2 \frac x2}$$
So ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Describe the locus of $w$ if $w=\frac{1-z}{1+z}$ and $z=1+iy$ (i.e $z$ is a complex number that moves along the line $x=1$) So I'm trying to solve the following problem: If $z=x+iy$, express $w=\frac{1-z}{1+z}$ in the form $a+bi$ and hence find the equation of the locus of $w$ if $z$ moves along the line $x=1$.
My atte... | A more systematic approach to get the result would be as follows:
First rearrange the relationship $w=\frac{1-z}{1+z}$ to make $z$ the subject, then apply the condition that the real part of $z$ is $1$.
Therefore, writing $w=u+iv$, we have $$z=\frac{1-w}{1+w}=\frac{1-u-iv}{1+u+iv}\cdot\frac{1+u-iv}{1+u-iv}$$
The real p... | {
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"source": "stackexchange",
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Prove that $\sum \frac{a^3}{a^2+b^2}\le \frac12 \sum \frac{b^2}{a}$
Let $a,b,c>0$. Prove that
$$ \frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\le \frac12 \left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right).\tag{1}$$
A idea is to cancel the denominators, but in this case Muirhead don't work because th... | (Just AM-GM is sufficient.)
From OP's work / River Li's solution, the stated inequality is equivalent to $\sum a - \frac{a^3}{a^2+b^2} \geq \sum a - \frac{1}{2} \sum \frac{b^2}{a}$, which is:
$$ \sum \frac{ab^2 } { a^2 + b^2 } \geq \sum a - \frac{1}{2} \sum \frac{ b^2}{a}$$
Let $ X = \sum \frac{ab^2 } { a^2 + b^2 } , Y... | {
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How to solve in integers the equation $2x^2+x+2x^2y-y^2+y=1$? Any hints? I could've reduced it to Pell's equation and solved it, if there wasn't $2x^2y$ part.
This is a part of a bigger problem, and that's all I have left to solve the main one.
| \begin{align*}
2x^2 &+x+2x^2y-y^2+y=1
\\
\\
2(y+1)x^2 &+(1)x-(y^2-y+1)=0\\
x &= \dfrac{\sqrt{8 y^3 + 9} - 1}{4 (y + 1)}
\\ \\
y^2& - (2x^2+1)y -(2x^2+x-1)=0
\\
y &= \dfrac{2 x^2 + 1+ \sqrt{4 x^4 + 12 x^2 + 4 x - 3}}{2}
\end{align*}
Solving for $\space x\space$ suggests that
$\space y\ne-1,\space$
but solving for $\spa... | {
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"source": "stackexchange",
"question_score": "1",
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Six people (half are female, half are male) for seven chairs. Problem:
Suppose there are $7$ chairs in a row. There are $6$ people that are going to randomly
sit in the chairs. There are $3$ females and $3$ males. What is the probability that
the first and last chairs have females sitting in them?
Answer:
Let $p$ be th... | Here is how I would think about it. There are seven of them - three male, three female and an empty chair (say, a ghost). In other words, for any given chair, there are seven equally likely possibilities.
So, $ \displaystyle P = {3 \choose 2} / {7 \choose 2} = \frac 17$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$A+B+C+D=\pi$, and $0\leq A,B,C,D \leq \frac{\pi}{2}$. Prove that $\sin^4(A)+\sin^4(B)+\sin^4(C)+\sin^4(D)\leq 2$ I tried to simplify it from some ways.
(1).$\sin^4(A)+\sin^4(B)+\sin^4(C)+\sin^4(D)$
$=\left(\frac{1-\cos(2A)}{2}\right)^2+\left(\frac{1-\cos(2B)}{2}\right)^2+\left(\frac{1-\cos(2C)}{2}\right)^2+\left(\frac... | Remark i): The desired inequality is still true without the condition $A, B, C, D \le \pi/2$.
Remark ii): According to my proof, for $A + B + C + D = \pi$, we have the following identity:
\begin{align*}
&2 - \sin^4 A - \sin^4 B - \sin^4 C - \sin^4 D \\
=\, & \sin^2 (A - B) \cos^2 (A + B)
+ \frac12(\cos (A + B) + \co... | {
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To find the number of ways to put $14$ identical balls into $4$ bins with the condition that no bin can hold more than $7$ balls. To find the number of ways to put $14$ identical balls into $4$ bins with the condition that no bin can hold more than $7$ balls.
I have tried the following:
The total no of ways to distrib... | Your required sum is the coefficient of $x^{14}$ in
$$(1+x+\dots+x^7)^4$$
or
$$[x^{14}]:\left(\frac{1-x^8}{1-x}\right)^4$$
We can expand $(1-x^8)^4$ into $1-4x^8$, and the other term is $\sum_{i=0}^\infty \binom{i+3}{3}x^i$.
We want the values only when $i=6$ or $i=14$, which gives your equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4376723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Integer solutions of $ x^4+34x^2 y^2+y^4=z^2$ Now I'm trying to solve one problem, and already twice an expression of the form $a x^4+b x^2 y^2+a y^4 \,\,(x > y > 1)$ appears under the square root.
First was: $9 x^4 - 14 x^2 y^2 + 9 y^4$
Second: $x^4 + 34 x^2 y^2 + y^4$
I suppose that there will be other equations of t... | At first, it looks like the equation can be rewritten as a Pythagorean triple
$\space A^2+B^2=C^2\quad$ here shown by Euclid's formula:
$\quad A=(m^2-k^2) \quad B=2mk\quad C=(m^2+k)^2\space $ and a triple may be found given any legitimate value of $A,\space B, \space \text{or}\space C.$
$\text{Here we have}\quad(x^2-y^... | {
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Using Lagrange Mult. to find maximum of $2x + y^2$ with constraints: $x^2 + y^2 + z^2 = 2$ and $x + y^2 + z = 0$ I was trying to maximize the function $2x + y^2$ with the following constraints: $x^2 + y^2 + z^2 = 2$ and $x + y^2 + z = 0$ using Lagrange multipliers. I already did all the work of proving that the maximu... | To simplify things, solve for $z$ from the second constraint, and plug it into the first constraint, as follows
From $x + y^2 + z = 0 $ you get $ z = - x - y^2 $
Plug this into $ x^2 + y^2 + z^2 = 2$ you get $ x^2 + y^2 +(x + y^2)^2 = 2 $
which simplifies to $ 2 x^2 + y^2 (1 + 2 x) + y^4 = 2 $
The Lagrange function is... | {
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Why $\sin(x) = x - \frac{x^3}{6} + o(x^5)$ is false? I'm wondering why the following use of little o is incorrect.
$$
\sin(x) = x - \frac{x^3}{6} + o(x^5)
$$
I know that te definition of little o is the following:
$$
f(x) = o(g(x)) \quad \text{when } x \to x_0 \qquad \text{if } f(x)=g(x) \, w(x) \quad \text{with } \lim... | Your problem is that the sentence $\sin(x) = x - \frac{x^3}{6} + o(x^5)$ is NOT saying that $f(x) = o(x^5)$ where $f(x) = x - \frac{x^3}{6}$. It is saying that $f(x) = o(x^5)$ where $f(x) = \sin(x) - x + \frac{x^3}{6}$. Now use the Taylor expansion for sine.
| {
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"source": "stackexchange",
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Functional analysis problem. Find the smallest value of $f(x) = \frac{x^2}{x-9}$ on the interval $(9, +\infty)$.
We should basically find the biggest $a$ such that $\frac{x^2}{x-9} \geq a$. We can multiply both sides by $x-9$ since it's positive and than we get $x^2-ax+9a \geq 0.$ I don't know how to proceed.
| Here is a purely algebraic way using the inequality between arithmetic and geometric mean (AM-GM):
For $x>9$ we have
\begin{eqnarray*} \frac{x^2}{x-9}
& = & \frac{x^2-81+81}{x-9} \\
& = & x+9 + \frac{81}{x-9} \\
& = & x-9 + \frac{81}{x-9} + 18 \\
& \stackrel{AM-GM}{\geq} & 2\sqrt{81} + 18\\
& = & 36
\end{eqnarray*}
Equ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the area of the triangle here?
We are given that the angle of $BAD$ is $2\alpha$ and the angle of $DAC$ is $\alpha$. $|AC| = 10$, $|BD| = 6$, $|DC| = 5$ units. Find the area of the triangle $ABC$.
The answer would be $33$. We need to show that if we drop an altitude from $A$ to $BC$ at point $E$, $|AE| = 6, |E... | Let area of $\triangle BAD = A_1$ and area of $\triangle CAD = A_2$. Let $AD = m$ and $AB = x$.
As they share a common height, the ratio $\frac{A_1}{A_2} = \frac 65$
Also, using an alternative formula for triangular area, $\frac{A_1}{A_2} = \frac{\frac 12xm\sin 2\alpha}{\frac 12(m)(10)\sin \alpha} = \frac{x\cos\alpha}{... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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$y'=2-\frac{3}{x}y+\frac{2}{x^2}y^2$ (Riccati) $y'=2-\frac{3}{x}y+\frac{2}{x^2}y^2$ (Riccati)
(a) Find the solutions.
(b) $y(x_0)=y_0$, prove two cases:
$$0<y_0<x_0 \implies \text{solution's domain is} [x_0,\infty) $$
$$0<x_0<y_0 \implies \text{solution's domain is} [x_0,x_0+\alpha) , \alpha \in \mathbb {R}.$$
I will b... | An alternative approach parametrizes solutions as $y=2\frac{u}{u'}$ to get
$$
y''=2-2\frac{u''u}{u'^2}=2-\frac6{x}\frac{u}{u'}+\frac{8}{x^2}\frac{u^2}{u'^2}
$$
This simplifies to
$$
0=u''-\frac{3}{x}u'+\frac{4}{x^2}u
$$
This now is an Euler-Cauchy equation, with correspondingly simple solutions.
| {
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Show that it is impossible to find $a,b \in \mathbb{Q}$ such that $ \frac{\sqrt{5}}{10}=a+b \frac{\sqrt{3}}{7} $ Show that it is impossible to find $a,b \in \mathbb{Q}$ such that
$$
\frac{\sqrt{5}}{10}=a+b \frac{\sqrt{3}}{7}
$$
We have $\frac{1}{20}=a^{2}+\frac{2 \sqrt{3}}{7} a b+b^{2} \frac{3}{49}=\left(a^{2}+b^{2} \f... |
Show that it is impossible to find $a,b \in \mathbb{Q}$ such that
$$
\frac{\sqrt{5}}{10}=a+b \frac{\sqrt{3}}{7}
$$
Alternative approach.
Lemma 1
$\sqrt{15}$ is not a rational number.
$\underline{\text{Proof}}$
Suppose that $~\displaystyle \sqrt{15} = \frac{p}{q} ~:$
$~p,q~$ relatively prime integers, and $~q > 0.$ ... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Determinant identity for symmetric 4x4 matrix with zero diagonal Does anybody know a reference or attribution for this identity?
$$
\det
\begin{pmatrix}
0 & {a_{12}}^2 & {a_{13}}^2 & {a_{14}}^2 \\
{a_{12}}^2 & 0 & {a_{23}}^2 & {a_{24}}^2 \\
{a_{13}}^2 & {a_{23}}^2 & 0 & {a_{34}}... | The dehomogenized identity with $\,a_{j4}=1\,$ is Heron's formula written in terms of a Cayley–Menger determinant.
$$
\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}
\;=\; \frac{1}{4}\sqrt{\,-
\begin{vmatrix}
0 & a^2 & b^2 & 1 \\
a^2 & 0 & c^2 & 1 \\
b^2 & c^2 & 0 & 1 \\
1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4391570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is the average of primes up to $n$ smaller than $\frac{n}{2}$, if $n > 19$? For $n = 19$, the average $\frac{(2 + 3 + 5 + 7 + 11 + 13 + 17 + 19)}{ 8} = \frac{77}{8}$ is greater than $\frac{19}{2}$ but for $n > 19$, it seems that the average is smaller than $n/2$.
I confirmed it holds for all $n < 10^7$ but have not bee... | The answer is yes for sufficiently large $n$. Sketch of proof:
*
*Using summation by parts (as in either one of these two existing answers), the average in question is
$$
\frac1{\pi(n)}\sum_{p\le n} p = \frac1{\pi(n)} \biggl( n\pi(n) - \sum_{k<n} \pi(k) \biggr) = n - \frac1{\pi(n)} \sum_{k<n} \pi(k).
$$
*Instead of ... | {
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Reflection across an Ellipsoid A reflective ellipsoid is given by
$$ \dfrac{x^2}{16} + \dfrac{y^2}{9} + \dfrac{(z - 2)^2}{4} = 1 $$
A light source, emitting rays in all directions, is placed at $A=(10,4,3)$. Find the point $C=(x,y,z)$ on the surface of the ellipsoid that minimizes the sum of distances $AC + CB$, where... | Take a generic point on the ellipsoid, $(a,b,c)$ (first condition on $a,b,c$). To mimimize your sum, you need the ray of light going from $A$ to $B$ bouncing at the ellipsoid to be reflected at $C$. The normal to the ellipsoid at $(a,b,c)$ is the gradient $(a/8, 2b/9, (c-2)/2)$. You need two additional conditions: the ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Finding the volume with triple integrals I want to find the volume of a function described by:
$$ G= \{(x,y,z)|\sqrt{x^2+y^2} \le z \le 1, (x-1)^2+y^2 \le 1\}$$
This question can be best solved in cylindrical coordinates. So if I follow that process, I get the following limits:
$$ r \le z \le 1$$
$$ 0 \le r \le 2\cos(\... | Your limits are not completely correct. Note that it should be
$$0 \le r \le 2\cos(\theta)\quad\textbf{and}\quad 0 \le r \leq 1$$
that is the upper limit is $1$ for $|\theta| \le \frac{\pi}{3}$ and it is $2\cos(\theta)$ for $\frac{\pi}{3}\leq|\theta| \le \frac{\pi}{2}$.
Therefore, by symmetry,
$$\begin{align}
V&=2\int^... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Minimize $xy$ over $x^2+y^2+z^2=7$ and $xy+xz+yz=4$.
Let $x,$ $y,$ and $z$ be real numbers such that $x^2 + y^2 + z^2 = 7$ and $xy + xz + yz = 4.$ Find the smallest possible value of $xy.$
I used Cauchy to get $$(x^2+y^2+z^2)(1^2+1^2+1^2)\geq (x+y+z)^2$$ and $$(x^2+y^2+z^2)(y^2+z^2+x^2)\geq(xy+xz+yz)^2,$$ but this d... | $x^2+y^2+z^2 = 7$ and $xy +xz + yz = 4$
implies
$x+y+z = \pm\sqrt{15}$
Now using the second equality that is given in the question, we get
$xy =4 -(x+y)z = 4 - (\pm\sqrt{15} - z)z$
*
*Case 1: When $x+y+z=\sqrt{15}$, minimising the right hand side with respect to $z$ gives $z = \frac{\sqrt{15}}{2}$. So, minimum valu... | {
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"source": "stackexchange",
"question_score": "1",
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How can I show that $\underset{\left(x,y\right)\rightarrow\left(0,0\right)}{\lim}{\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}}$ exists, using limit def? I am trying to solve an exercise to show that this function $$f(x,y)=\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}$$ has a limit as $(x,y)$ approaches $(0,0)$:
$$\underset{\left(x,y\right... | You have\begin{align}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}&=\frac{(x^2+y^2)\left(\sqrt{x^2+y^2+1}+1\right)}{\left(\sqrt{x^2+y^2+1}-1\right)\left(\sqrt{x^2+y^2+1}+1\right)}\\&=\sqrt{x^2+y^2+1}+1\end{align}and therefore$$\left|\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}-2\right|=\left|\sqrt{x^2+y^2+1}-1\right|.$$But$$\require{cancel... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I determine the value of $d$ from the tangent graph of the form $y = a\times \tan (bx + c) + d$?
I understand that $d$ is the vertical shift, and I have already worked out the values of $a$, $b$ and $c$ to be the following (please tell me if I am wrong):
$\pmb{a}\quad$ Given point is $(-1, -7)$ so I substituted... | The graph of a transformed $\tan$ function is
$ f(x) = A \tan( B (x - C) ) + D $
The first thing is that $ B = \dfrac{\pi}{T} $ where $T $ is the period.
Here the period is $ 4 $ , so $B = \dfrac{\pi}{4} $
$C $ is the horizontal shift to the right. Here the graph is shifted to the left by $1$, therefore, C = -1 $
Now ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If a convex quadrilateral is inscribed and circumscribed and has perpendicular diagonals, then one diagonal divides it into congruent right triangles
A convex quadrilateral $ABCD$ is inscribed and circumscribed. If the diagonals $AC$ and $BD$ are perpendicular, show that one of them divides the quadrilateral into two ... | Let the diagonals intersect at $O$ and set: $a=AO$, $b=BO$, $c=CO$, $d=DO$.
The sums of opposite sides must be the same, giving:
$$
\strut\sqrt{a^2+b^2}+\sqrt{c^2+d^2}=
\sqrt{a^2+d^2}+\sqrt{b^2+c^2}.
$$
Squaring and simplifying we get:
$$
(a^2+b^2)(c^2+d^2)=
(a^2+d^2)(b^2+c^2),
$$
which reduces to
$$
(a^2-c^2)(b^2-d^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding max{f(x)} without derivative Consider the function $f(x)=\frac{\sqrt{x^3+x}}{x^2+x+1}$
the question is about: to find max{f} without using derivative.
I can find max with derivation and it is not hard to find. it is $f'(x)=0 \to x=1 $ so $max\{f\}=\frac{\sqrt 2}{3}$
but I am looking for an idea to do as the que... | Maybe a bit late answer but i thought it might be worth mentioning it.
Here is a direct elementary calculation of the maximum.
Let $x\geq 0$:
We only have to find the maximum of
$$\frac{x^3+x}{(x^2+x+1)^2} = \frac{x(x^2+1)}{(x^2+x+1)^2}$$
Simple rearrangements give
\begin{eqnarray*} \frac{x(x^2+1)}{(x^2+x+1)^2}
& = & \... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Solving $\frac{\ 3x}{x^2+x+1}+\frac{2x}{x^2-x+1}=3$ with other approaches
$$\frac{\ 3x}{x^2+x+1}+\frac{2x}{x^2-x+1}=3$$
$$x=?$$
I solved this problem as follow:
$x=0$ is not a root, we can divide numerator and denominator of each fraction by $x$:
$$\frac{3}{t+1}+\frac{2}{t-1}=3\quad\quad\text{where $t=x+\frac1x$}$$
... | I don't know if you consider this elegant:
\begin{align}
0&=3-\frac{3x}{x^2+x+1}-\frac{2x}{x^2-x+1}\\
&=\left(1-\frac{3x}{x^2+x+1}\right)+2\left(1-\frac{x}{x^2-x+1}\right)\\
&=\frac{(x-1)^2}{x^2+x+1}+2\frac{(x-1)^2}{x^2-x+1}\\
&=(x-1)^2\left(\frac{1}{x^2+x+1}+\frac{2}{x^2-x+1}\right)
\end{align}
Since $x^2+x+1$ and $x^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Is there any closed form for $\int_{0}^{1} \frac{\ln ^nx}{\sqrt{1-x^{2}}}\,d x$? In my post, I found that $$\int_{0}^{1} \frac{\ln x}{\sqrt{1-x^{2}}} d x = -\frac{\pi}{2} \ln 2 .$$
Then I try to generalize the result to the integral by the same technique. $$
J_{n}:=\int_{0}^{1} \frac{\ln ^nx}{\sqrt{1-x^{2}}} d x
$$
Let... | Under the substitution $x^2 \to x$ we get that
$$I_{n,m} =\int_{0}^{1} \frac{\ln^n(x)}{\left(1-x^2\right)^m}\mathrm{d}x = \frac{1}{2^{n+1}}\int_{0}^{1}\ln^n(x) x^{-\frac{1}{2}}(1-x)^{-m}\mathrm{d}x $$ Recalling the definition of the Beta function we see that
\begin{align*}
B\left(\frac{1}{2}+t, 1-m\right) =& \int_0^{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4403512",
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"source": "stackexchange",
"question_score": "3",
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Prove that the sequence $t_n$ defined by $\dfrac{1}{n}(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}})$ converges determine if the following sequence converges
$t_n=\dfrac{1}{n}(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}})$
my solution:
$\text{ ... | You can calculate the limit in an elementary way using
$$\sqrt{k+1}-\sqrt k = \frac 1{\sqrt{k+1}+\sqrt k} > \frac 1{2\sqrt{k+1}}$$
Hence,
\begin{eqnarray*} \frac 1n\sum_{k=1}^n \frac 1{\sqrt k}
& = & \frac 1n\left(1+ \sum_{k=1}^{n-1} \frac 1{\sqrt{k+1}}\right) \\
& < & \frac 1n\left(1+ 2\sum_{k=1}^{n-1} (\sqrt{k+1}-\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Why is $\frac{1}{1+(\frac{-\sin x}{1+\cos x})^2} \equiv \frac{(1+\cos x)^2}{\sin^2x+1+2\cos x+\cos^2x}$? Currently I have a problem to understand why $$\frac{1}{1+(\frac{-\sin x}{1+\cos x})^2} \equiv \frac{(1+\cos x)^2}{\sin^2x+1+2\cos x+\cos^2x}\,?$$
My calculations get me:
$$\frac{1}{1+(\frac{-\sin x}{1+\cos x})^2}\e... | If you multiply the numerator and the denominator by $(1+\cos x)^2$, you get:
$$\frac{1}{1+(\frac{-\sin x}{1+\cos x})^2}=\frac{(1+\cos x)^2}{(1+\cos x)^2}\times\frac{1}{1+(\frac{-\sin x}{1+\cos x})^2}=\frac{(1+\cos x)^2}{(1+\cos x)^2\left(1+(\frac{-\sin x}{1+\cos x})^2\right)}\\=\frac{(1+\cos x)^2}{(1+\cos x)^2+\sin^2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding the equation of the Parabola The parabola $y = x^2 + bx + c$ has the following properties:
*
*The point on the parabola closest to $(12,3)$ is the intersection with the $y$ axis of the parabola.
*The parabola passes through $(-5,0).$
How can I find $(b, c)$?
Here is my attempt:
The point $(0, c)$ is the int... | $ y = x^2 + bx + c $
$ y' = 2 x + b $
At $(0,c)$ we have $y' = b $, and thus
$ b \bigg( \dfrac{c - 3}{0 - 12} \bigg) = -1 $
which simplifies to $ b (c - 3) = 12 $
We also have $(-5, 0)$ on the parabola, then $0 = 25 - 5 b + c $
Solving the above two equations yields two solutions
Case I: $b = -0.4, c = -27 $
Case II:... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Why does $\lim_{n \to \infty}(1+\frac{1}{4^n})^{2^n} $ not equal $\sqrt{e}$? I tried to evaluate $\lim_{n \to \infty}(1+\frac{1}{4^n})^{2^n}$ with
$$\lim_{n \to \infty}(1+\frac{1}{4^n})^{2^n} = \lim_{n \to \infty}(1+\frac{1}{4^n})^{(4^\frac{1}{2})^n} = \lim_{n \to \infty}(1+\frac{1}{4^n})^{4^\frac{n}{2}} = \lim_{n \to ... | Note that in general $\sqrt{A^x} = A^{\frac{x}{2}} \not = A^{\sqrt{x}}$ [as in general $\frac{x}{2} \not = \sqrt{x}$]. Try $A=2$ and $x=9$ to see for yourself.
In particular here: $$\left(1+\frac{1}{4^n}\right)^{2^n} = \left(1+\frac{1}{4^n}\right)^{\sqrt{4^n}} \not = \sqrt{\left(1+\frac{1}{4^n}\right)^{4^n}}.$$
Instead... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4410441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find an orthonormal basis and the signature of the quadratic form
Consider the quadratic form given by the matrix below (in the canonical basis)
\begin{pmatrix}
1 & 1 & -1\\
1 & 1 & 3\\
-1 & 3 & 1
\end{pmatrix}
Find an orthonormal basis of it and find its signature.
First I calculated the eigenvalues, which are $4, ... | This business with eigenvalues and eigenvectors is not how you diagonalize a quadratic form. It will give a correct result if done correctly, but is way too long and computationally painful. Simply use a "complete the square" method. In the canonical coordinates, your quadratic form is
$$x^2 + y^2+z^2+2xy-2xz+6yz.$$
No... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Integral $\int( 3x^2 +5x + 1 )\sqrt{2x^2 + 2x + 1}dx$ I tried to integrate
$$\int( 3x^2 +5x + 1 )\sqrt{2x^2 + 2x + 1}dx$$
by multipling by $\sqrt{2x^2 + 2x + 1}$ in the numerator and the denominator to break it into $5$ fractions
The answer is possible but it's too long
Another better solution
| A systematic approach after $t=2x+1$
\begin{align}
I=&\int (3x^2 +5x + 1 )\sqrt{2x^2 + 2x + 1}\>dx\\
= &\frac1{8\sqrt2}\int (4t+3)(t^2+1)^{3/2}-6 (t^2+1)^{1/2}\>dt\tag1
\end{align}
Integrate below by parts
\begin{align}
&\int (t^2+1)^{3/2}dt=\frac14t(t^2+1)^{3/2}+\frac34\int(t^2+1)^{1/2}dt\\
&\int (t^2+1)^{1/2}dt=\fra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Derivative of the nth order and evaluation in $x=0$ I want to solve the following exercise; be $f(x)=\frac{1}{x}e^x-\frac{1}{x^2}e^x+\frac{1}{x^2}$ with $x\neq 0$ we have that the function has derivatives in all orders for $x=0$
I want to determine $f^{(n)}(0)$
It seems that the above is equivalent to the expression $\... | We know $f$ admits a Taylor expansion at $x=0$:
\begin{align*}
f(x)=\sum_{k=0}^\infty\frac{f^{k}(0)}{k!}x^k
\end{align*}
Denoting with $[x^n]$ the coefficient of $x^n$ of a series $f(x)$ we want to calculate
\begin{align*}
f^{(n)}(0)=n![x^n]f(x)
\end{align*}
Recalling the series expansion
\begin{align*}
e^x=\sum_{j=0}... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find area of $x^2+axy+y^2=1$, $|a|\leq1$ I was wondering how to find the area of
$$x^2+axy+y^2=1,\>\>\>\>\>|a|\leq1$$
I have solved
$$\rho^{2}(\theta)=\frac{1}{1+\frac{a}{2}\sin 2\theta}$$
However, integrating this equation using trigonometric substitution is very cumbersome.
| The equation of the ellipse can be written compactly using the position vector $r=[x,y]^T$ as
$ r^T Q r = 1 $
where $Q = \begin{bmatrix} 1 && \frac{1}{2} a \\ \frac{1}{2} a && 1 \end{bmatrix} $
To find the area you need the product of the semi-minor and semi-major axes.
If you diagonalize $Q$ and write it as $ Q = R D ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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why not $G \cong \frac{\mathbb{Z}}{3\mathbb{Z}} \times \frac{\mathbb{Z}}{3\mathbb {Z}} \times \frac{\mathbb{Z}}{3\mathbb{Z}} ?$
Determine the Galois group of $(x^2-2) (x^2-3)(x^2-5)$. Determine all the subfields of the splitting field of this polynomial.
My attempt: I found the answer here
$G= Gal(K/\mathbb{... | $\sigma_2$ doesn't have "three choices" it has 8:
$$
\begin{cases}
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\end{cases}\\
\begin{cases}
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\end{cases}\\
\begin{cases}
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Range of $\sqrt{a^2 + (1 - b)^2} + \sqrt{b^2 + (1 - c)^2} + \sqrt{c^2 + (1 - d)^2} + \sqrt{d^2 + (1 - a)^2}$ on $0 \le a,$ $b,$ $c,$ $d \le 1.$ Let $0 \le a,$ $b,$ $c,$ $d \le 1.$ Find the possible values of the expression
$$\sqrt{a^2 + (1 - b)^2} + \sqrt{b^2 + (1 - c)^2} + \sqrt{c^2 + (1 - d)^2} + \sqrt{d^2 + (1 - a)^... | Denote the expression by $P(a, b, c, d)$.
First, using Minkowski inequality, we have
\begin{align*}
P &\ge \sqrt{(a + b + c + d)^2 + (1 - b + 1 - c + 1 - d + 1 - a)^2}\\
&= \sqrt{x^2 + (4 - x)^2}\\
&= \sqrt{2(x - 2)^2 + 8}\\
&\ge 2\sqrt 2
\end{align*}
where $x = a + b + c + d$.
Also, $P(1/2, 1/2, 1/2, 1/2) = 2\sqrt... | {
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"source": "stackexchange",
"question_score": "4",
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Find $k\in\mathbb{R}$ given $w = k+i$ and $z=-4+5ki$ and $\arg(w+z)$ I am working on problem 2B-11 from the book Core Pure Mathematics, Edexcel 1/AS. The question is:
The complex numbers $w$ and $z$ are given by $w = k + i$ and $z = -4 + 5ki$ where $k$ is a real constant. Given at $\arg(w + z) = \frac{2\pi}{3}$, find t... |
$w = k + i$ and $z = -4 + 5ki$
Let $w + z = a + bi$.
$a = k - 4 \\b =i + 5ki$
Typo: $b=1+5k.$
The angle of the triangle formed by the complex number $w + z$ is $$\pi -\frac{2\pi}{3} = \frac{\pi}{3}.$$
"The triangle formed by $w + z$" is ambiguous.
In any case, your critical error is here, where you assume that $a$ ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Maclaurin series of $\ln \left( 1+\frac{\ln(1+x)}{1+x} \right)$ So the first thing I done was
$$\begin{align}\ln(1+x)&=x-\frac{1}{2}x^2+\frac{1}{3}x^3+o(x^3)\\&=(1+x)\left(x-\frac{3}{2}x^2+\frac{11}{6}x^3\right)+o(x^3)\end{align}$$
I've never seen this done but I'm pretty sure I can do this.
Now I want to divide:
$$(1+... | Let $u=1+x$ and $w=u+\log(u$). Rewrite
$$
f(x)=\log(1+\frac{\log(1+x)}{1+x})=\log(w) -\log(u)
$$
We seek the coefficients in the Taylor series expansion
$$
f(x) = \sum_{n=0}^{\infty}f^{\{n\}}(0)\frac{x^n}{n!}
$$
where the coefficient $f^{\{n\}}(0)$ is the nth derivative of the function evaluated at $x=0$. We find
$$
f... | {
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"source": "stackexchange",
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"answer_count": 3,
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If $f(x)=\frac {x^ 2 -2x +4}{ x^ 2 +2x+4}$ for $x \in \mathbb{R}$, prove that the range of $f(x)$ is $[1/3, 3]$ One method to solve it would by putting $y = f(x)$ then multiplying the denominator with $y$ hence making a quadratic equation in x then we can just use the inequalities for $x$ being real to prove it.
For an... | Function range definition : The set value of the dependent variable for which a function is defined.
Rewrite the equation given above as :
$$\frac{x^2-2x+4}{x^2+2x+4}=y$$
Multiply both sides by $x^2+2x+4$ :
$$\frac{x^2-2x+4}{x^2+2x+4}\left(x^2+2x+4\right)=y\left(x^2+2x+4\right)$$
Simplify :
$$x^2-2x+4=y\left(x^2+2x+4\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is $3 \times 2 = 3+3$ or $2+2+2$? $3\times 2$ is $3+3$ or $2+2+2$. We know both are correct as multiplication is commutative for whole numbers. But which one is mathematically accurate?
| In every ring and even in every semiring, $3 \times 2$ is by definition $(1+1+1)\times (1+1)$ . Since multiplication distributes over addition, you get on the one hand
$$
(1+1+1) \times (1+1) = (1+1+1) + (1+1+1) = 3 +3
$$
and on the other hand
$$
(1+1+1) \times (1+1) = (1+1) + (1+1) + (1+1) = 2 + 2 + 2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4432047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
A basis that makes a matrix triangular. Find a basis for $\mathbb C^3$ so that the following matrix is in triangular form:
\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{pmatrix}
What are the eigenvalues?
I found the eigenvalues to be $\lambda_1 = 1, \lambda_2 = \frac{1}{2}(-1 - i \sqrt{3} ), \lambda_3 = \fra... | My calculations showed the following:
(Kindly try yourself and verify these results)
Eigenvalue: $\lambda_1 = 1$
Eigenvector for $\lambda_1 = 1$:
We solve the equation
$$
(A - \lambda_1 I) \mathbf{x} = \mathbf{0}
$$
The RREF of $A - \lambda_1 I$ is obtained as
$$
R_1 = \left[ \begin{array}{ccc}
1 & 0 & -1 \\
0 & 1 & -1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4432935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Equation of bisector between two straight lines given in symmetrical form Please help in solving the attached question.
I know in 2D, it is solved as
$\dfrac{ax+by+c}{\sqrt{a^2+b^2}}=\pm \dfrac{px+qy+s}{\sqrt{p^2+q^2}}$
Not sure if we use the same formula for 3D? And if yes, what’s the value of the constant c and s her... | In $3D$ the equation of the two lines is given as follows
$ \dfrac{ x - x_1}{a} = \dfrac{y - y_1}{b} = \dfrac{z - z_1}{c} $
for the first line, and
$ \dfrac{ x - x_1}{d} = \dfrac{y - y_1}{e} = \dfrac{z - z_1}{f} $
and these two symmetrical form equations correspond to the parametric equations
$P_1(t) = (x_1, y_1, z_1 )... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4434879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
An inequality $2(ab+ac+ad+bc+bd+cd)\le (a+b+c+d) +2(abc+abd+acd+bcd)$
Let $a, b, c, d \ge 0$ such that $a+b, a+c, a+d, b+c, b+d, c+d \le 1$. Show that
$$2(ab+ac+ad+bc+bd+cd)\le (a+b+c+d) +2(abc+abd+acd+bcd).$$
I am trying to maximise
$$f(a,b,c,d)=(a+b+c+d) /(1-(1-a)(1-b)(1-c)(1-d) +abcd) $$ over the set of nonnegativ... | The form of the constraints on the variables suggest a transform based around their differences from each other and from $1/2$. Let $a = (1-x+y)/2$, $b = (1 - x - y)/2$, $c = (1-z+w)/2$, $d = (1-z-w)/2$. The positivity constraints are $x + |y| \le 1$ and $z + |w| \le 1$. The sum constraints are $x,z\in[0,1]$ and $|y|+|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4435584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is there a mistake in solving this limit? I want to solve this:
\begin{equation}
L=\lim_{x\rightarrow 0} \frac{\sum\limits_{m=1}^{M}a_{m}\exp\left\{\frac{bm^2}{ (m^2+2x^2)x^2}\right\}{(m^2+2x^2)^{-3/2}} } {\sum\limits_{m=1}^{M} c_{m}\exp\left\{\frac{bm^2}{ (m^2+2x^2)x^2}\right\}{(m^2+2x^2)^{-3/2}}}
\end{equation}
wh... | Since @justt gave the answer, just try.
Take an example : $a_m=m$, $c_m=m^2$, $b=1$, $M=4$.
Without any simplification at all, the limit is
$$\frac{144+36 e^{3/2}+16 e^{16/9}+9 e^{15/8}}{144+72 e^{3/2}+48 e^{16/9}+36
e^{15/8}}=0.465476$$
Ignoring the term $2x^4$, the limit becomes $\frac{41}{60}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How do I get to this formula for the area of a triangle I am new here and new student to geometry.
In my geometry skript there is a task:
Show that the formula for the area $F$ of a triangle with sidelengths $a,b,c $ is given by
$$F^2 = - \frac{1}{16} \det \begin{pmatrix} 0 & c^2 & b^2 & 1 \\ c^2 & 0 & a^2 & 1 \\ b^2 &... | Start with
\begin{pmatrix} 0 & c^2 & b^2 & 1 \\ c^2 & 0 & a^2 & 1 \\ b^2 & a^2 & 0 & 1 \\ 1& 1& 1& 0 \end{pmatrix}
Perform the column transformations $C_2 \to C_2 - C_1$ and $C_3 \to C_3- C_1$ which don't change the determinant.
\begin{pmatrix} 0 & c^2 & b^2 & 1 \\ c^2 & -c^2 & a^2-c^2 & 1 \\ b^2 & a^2-b^2 & -b^2 & 1 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Prove that, for any positive real numbers a, b, and c, $a^3b^2 + a^2b^3 + a^3c^2 + a^2c^3 + b^3c^2 + b^2c^3 \geq \frac{a^5 + b^5 + c^5}{5}$. For this question I assume you would start with a rewritten form so that $a^3b^2 + a^2b^3 + a^3c^2 + a^2c^3 + b^3c^2 + b^2c^3$ = $(a^3 + b^3 + c^3)(a^2 + b^2 + c^2) - (a^5 + b^5 +... | $(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$ .for easy simplification note that($(a+b)^5=\sum_{k=0}^{5}{5 \choose k}a^{5-k}b^k$)
$(a+c)^5=a^5+5a^4c+10a^3c^2+10a^2c^3+5ac^4+c^5$
$(c+b)^5=c^5+5c^4b+10c^3b^2+10c^2b^3+5cb^4+b^5$
then $(a+b)^5+(a+c)^5+(c+b)^5=2a^5+2b^5+2c^5+5ab^4+5ac^4+5cb^4+10a^2b^3+10a^2c^3+10c^2b^3+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4437820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Proving a combinatorial identity related to convolution of central binomial coefficients I'm trying to calculate the coefficient values for $f(x) = \sum_{k=0}^n {n \choose k}^2 (1+x)^{2n-2k}(1-x)^{2k}$. TL;DR I don't know where to begin in order to prove:
$$\sum_{k=0}^r (-4)^k{n \choose k}{2n-2k \choose n}{n-2k \choose... | Suppose we seek to evaluate the coefficients
$$[x^q] f_n(x) = [x^q] \sum_{k=0}^n {n\choose k}^2
(1+x)^{2n-2k} (1-x)^{2k}.$$
Recall the generating function of the Legendre polynomials
$$\sum_{n\ge 0} P_n(y) t^n = \frac{1}{\sqrt{1-2yt+t^2}}$$
and the known fact
$$P_n(y) = \left[\frac{y-1}{2}\right]^n
\sum_{k=0}^n {n\choo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4438506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that $p^k \mid \mid (x-y)$ iff $p^k \mid \mid (x^6-y^6)$. Definition. Let $n>1$ be an integer and $p$ be a prime. We say that $p^k$ fully divides $n$ and write $p^k \mid \mid n$ if $k$ is the greatest positive integer such that $p^k \mid n$.
Let $p>6$ be a prime. Let $x$ and $y$ be two distinct integers such that ... | Your attempt so far looks good (as far as I can tell). What you really need to show is that $p\nmid (x^5+\dots+y^5)$ because by now you have only shown that $p^k \mid (x^6-y^6)$. Let's focus on this point, then the reverse direction gets easy. We have $p|(x-y)$ thus $x-y \equiv 0 \mod p$. From this we conclude $x\equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find non-zeroes $a$ and $b$ such that $\lim[a^n(u_n - 1)] = b$ where $u_1 = 0$ and $4u_{n + 1} = u_n + \sqrt{6u_n + 3}, \forall n \in \mathbb Z^+$.
Consider sequence $(u_n)$, defined as $\left\{ \begin{aligned} u_1 &= 0\\ u_{n + 1} &= \dfrac{u_n + \sqrt{6u_n + 3}}{4}, \forall n \in \mathbb Z^+ \end{aligned} \right.$. ... | Well, there may be something "lost in translation", indeed: if all those questions were multiple choice, which were the choices for $a+b$?
I'm asking because $u_n$ is converging to $1$ quite rapidly, and since
$$a=\lim_{n\to\infty}\frac{u_n-1}{u_{n+1}-1},$$ computing that ratio for a few $n$ (up to $n=10$, say) with a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Is there any closed form for $\frac{d^{2 n}( \cot z)}{d z^{2 n}}\big|_{z=\frac{\pi}{4}}$? Latest Edit
Thanks to Mr Ali Shadhar who gave a beautiful closed form of the derivative which finish the problem as
$$\boxed{S_n = \frac { \pi ^ { 2 n + 1 } } { 4 ^ { n + 1 } ( 2 n ) ! } | E _ { 2 n } | }, $$
where $E_{2n}$ is an ... | Using $\csc(z)=\cot(z/2)-\cot(z)$, we have
$$\lim_{z\to \frac{\pi}{2}}\frac{d^{2n}}{dz^{2n}}\csc(z)=\underbrace{\lim_{z\to \frac{\pi}{2}}\frac{d^{2n}}{dz^{2n}}\cot(z/2)}_{z=2x\to\, dz=2 dx}-\underbrace{\lim_{z\to \frac{\pi}{2}}\frac{d^{2n}}{dz^{2n}}\cot(z)}_{0}$$
$$=2^{-2n}\lim_{x\to \frac{\pi}{4}}\frac{d^{2n}}{dx^{2n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"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.