Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Rationalize nested radical expression $\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}$ I have a college task to rationalize this fraction.
$$\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}$$
I do not know how to simplify this fraction.
Please, explain how to remove the radical from the denominator. Thanks, for your help.
| Let use
$$\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}} \frac{\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}}{\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}} \frac{\sqrt{4-{\frac{5+\sqrt{5}}{2}}}}{\sqrt{4-{\frac{5+\sqrt{5}}{2}}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Minimize $x+2y$ subject to $x^2+y^2\le1$ and $3x+4y\le-5$. I want to minimize $x+2y$ subject to $x^2+y^2\le1$ and $3x+4y\le-5$.
I found the gradient conditions: $1+2x\lambda_1+3\lambda_2=2+2y\lambda_1+4\lambda_2=0$.
Then, by complementary slackness: $\lambda_1(x^2+y^2-1)=\lambda_2(3x+4y+5)=0$.
Wolfram-Alpha gives $(x,y,\lambda_1,\lambda_2) = (\frac1{\sqrt5},\frac2{\sqrt5},-\frac{\sqrt5}2,0)$ or $(-\frac1{\sqrt5},-\frac2{\sqrt5},\frac{\sqrt5}2,0)$. The second one gives an objective value of $-\sqrt5$, but both of these violate the second constraint.
The actual mimimum is $-\frac{11}5$ at $(x,y) = (-\frac35,-\frac45)$. Where did I go wrong?
| Your generalized Lagrange function:
$$L=x+2y+\lambda_1(1-x^2-y^2)+\lambda_2(-5-3x-4y)$$
The Kuhn-Tucker conditions:
$$\begin{cases}L_x\ge 0, xL_x=0\\
L_y\ge 0,yL_y=0\\
L_{\lambda_1}\le 0,\lambda_1\ge 0,\lambda_1L_{\lambda_1}=0\\
L_{\lambda_2}\le 0,\lambda_2\ge 0,\lambda_2L_{\lambda_2}=0\\
\end{cases} \Rightarrow
\begin{cases}1-2x\lambda_1-3\lambda_2\ge 0,xL_x=0\\
2-2y\lambda_1-4\lambda_{\color{red}2}\ge 0,yL_y=0\\
x^2+y^2\le 1,\lambda_1\ge 0,\lambda_1L_{\lambda_1}=0\\
3x+4y\le -5,\lambda_2\ge 0,\lambda_2L_{\lambda_2}=0\end{cases}$$
When $\lambda_1>0,\lambda_2>0$ we get:
$$\begin{cases}L_{\lambda_1}=0\\ L_{\lambda_2}=0\end{cases} \Rightarrow \begin{cases}x^2+y^2=1\\ 3x+4y=-5\end{cases} \Rightarrow (x,y)=\left(-\frac35,-\frac45\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3421069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I solve $13^{({11}^{7})}\equiv x\pmod{10}$
Solve $x$, where $x\in[0,9]\cap\mathbb{Z}:$
$$13^{11^7}\equiv x\pmod{10}$$
Here is my thoughts:
Since $13\equiv3\pmod{10}$
Implies $13^{11^7}\equiv 3^{11^7}\pmod{10}$
$$3^0=1,3^1=3,3^2=9,3^3=27,3^4=81\cdots$$
For mod $10$, have reminder repeat between:
(they have to repeat, but i will skip the induction here) $$R_1=\{1,3,9,7\}$$
Then solve
$$11^7+1\equiv x\pmod{|R_1|}$$
$$11^0+1=2,11^1+1=12,11^2+1=122,11^3+1=1332$$
For mod $4$, have reminder index repeat between:
$$R_2=\{2,0\}$$
In another word, the reminder repeat between second and last term in $R_1$
Then solve $$7+1\equiv x\pmod{|R_2|},\text{ that }x=0$$
$x=0$, means the reminder is the last term in $R_1$ which is $7$, that implies
$$13^{11^7}\equiv7\pmod{10}$$
Therefore $x=7$
My qeustion: is there any theorem to apply so this calculation could be simpler$?$
Any help would be appreciated.
| HINT:
Using Euler's theorem,
$\phi(10) = 4 \Rightarrow 13^4\equiv 1 \pmod{10}$
and
$11^7 \equiv (-1)^7\pmod{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3422438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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$a^2b^2+a^2c^2+b^2c^2\leq 3$ Let $a, b, c\geq 0$ s.t.$$(a^2-a+1)(b^2-b+1)(c^2-c+1)=1$$
Show that $ a^2b^2+a^2c^2+b^2c^2\leq 3$.
My idea:
I denote $a+b+c=x$, $ab+bc+ac=y$ and $abc=z$.
Then I have $$x^2+y^2+z^2-xy-xz-yz+2z-y-x=0$$
I have to show that $$y^2-2xz\leq 3$$
I tried to prove it with the sign of trinom, but it doesn't work.
| We'll prove that
$$(a^2-a+1)(b^2-b+1)\geq\frac{a^2+b^2}{2}.$$
Indeed, let $a+b=2u$ and $ab=v^2$.
Thus, we need to prove that
$$a^2b^2-ab(a+b)+a^2+b^2+ab-a-b+1\geq\frac{a^2+b^2}{2}$$ or
$$v^4-2uv^2+2u^2-2u+1\geq0$$ or
$$(v^2-u)^2+(u-1)^2\geq0,$$ which is obvious.
Thus, $$1=\prod_{cyc}(a^2-a+1)=\sqrt{\prod_{cyc}(a^2-a+1)(b^2-b+1)}\geq\sqrt{\prod_{cyc}\frac{a^2+b^2}{2}}.$$
Id est, it's enough to prove that:
$$\sqrt{\prod_{cyc}\frac{a^2+b^2}{2}}\geq\left(\sqrt[4]{\frac{a^2b^2+a^2c^2+b^2c^2}{3}}\right)^3.$$
Now, let $a^2=x$, $b^2=y$ and $c^2=z$.
Thus, we need to prove that
$$27(x+y)^2(x+z)^2(y+z)^2\geq64(xy+xz+yz)^3.$$
Now, since $$(x+y)(x+z)(y+z)\geq\frac{8}{9}(x+y+z)(xy+xz+yz)$$ it's
$$\sum_{cyc}z(x-y)^2\geq0,$$ it's enough to prove that
$$(x+y+z)^2\geq3(xy+xz+yz)$$ or
$$\sum_{cyc}(x-y)^2\geq0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3422788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the second general solution of this equation? $\frac{dx}{y(x+y)+az} = \frac{dy}{x(x+y)-az} = \frac{dz}{z(x+y)}$
By comparing the first two ratios (adding them), I found one solution. How do find the second solution?
| $$\frac{dx}{y(x+y)+az} = \frac{dy}{x(x+y)-az} = \frac{dz}{z(x+y)}$$
$\frac{dx}{y(x+y)+az} = \frac{dy}{x(x+y)-az} =\frac{dx+dy}{(x+y)^2}$
$$\frac{dx+dy}{(x+y)^2}= \frac{dz}{z(x+y)}$$
$$\frac{dx+dy}{x+y}= \frac{dz}{z}$$
$$\boxed{\frac{z}{x+y}=c_1}$$
$\frac{dx}{y(x+y)+az} = \frac{dy}{x(x+y)-az}=\frac{dx}{(y+ac_1)(x+y)} = \frac{dy}{(x-ac_1)(x+y)}$
$$\frac{dx}{y+ac_1} = \frac{dy}{x-ac_1}$$
$$(x-ac_1)dx-(y+ac_1)dy=0$$
$$\frac12(x^2-y^2)-ac_1(x+y)=c_2$$
$$\boxed{\frac12(x^2-y^2)-az=c_2}$$
The PDE is certainly
$$(y(x+y)+az)\frac{\partial z}{\partial x}+(x(x+y)-az)\frac{\partial z}{\partial y}=(x+y)z$$
So, the general solution expressed on implicit form $c_1=F(c_2)$ is :
$$\boxed{z=(x+y)\:F\left(\frac12(x^2-y^2)-az \right)}$$
$F$ is an arbitrary function, to be determined according to some boundary condition (presently not specified in the wording of the question).
AN ALTERNATIVE WAY TO SOLVE THE PDE :
Change of function
$$z=(x+y)u\quad\implies\quad \begin{cases} z_x=u+(x+y)u_x \\z_y=u+(x+y)u_y \end{cases}$$
$(y(x+y)+a(x+y)u)(u+(x+y)u_x) +(x(x+y)-a(x+y)u)(u+(x+y)u_y)=(x+y)^2u$
After simplification :
$$(y+au)\frac{\partial u}{\partial x}+(x-au)\frac{\partial u}{\partial y}=0$$
Solving with method of characteristics leads to :
$$u=F\left(\frac12(x^2-y^2)-a(x+y)u \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3424407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The limit of the ratio of polygamma functions I want to calculate this quantity:
$$\lim_{x \rightarrow \infty}\frac{\Psi_1 (x)}{\Psi_1 (x + y)}$$
where
$$\Psi_1 (x)=\frac{d^2}{dx^2}\log \Gamma (x)=\sum_{k=0}^{\infty}\frac{1}{(x+k)^2}. $$
I guess it is $1$, but I am not sure about my proof.
My proof is following:
Let $\epsilon > 0$ be given. Since $\Psi_1 (x)$ is convergent on $(0, \infty)$ and decreasing, for any $x, y >0$ there exist $K=K(\epsilon) < \infty$ such that $\sum_{k=K+1}^{\infty} \frac{1}{(x + y + k)^2} < \sum_{k=K+1}^{\infty} \frac{1}{(x + k)^2} \leq \epsilon$. Then, we have
\begin{align*}
&\frac{\Psi_1 (x)}{\Psi_1 (x + y)} \leq \Bigg( \sum_{k=0}^{K} \frac{1}{(x + k)^2} + \sum_{k=K+1}^{\infty} \frac{1}{(x + k)^2} \Bigg) \Bigg/ \Bigg( \sum_{k=0}^{K} \frac{1}{(x + y + k)^2} \Bigg)\\
%&\leq \Bigg( \sum_{k=0}^{K} \frac{1}{(x + k)^2} - \frac{1}{(x + y + k)^2} \Bigg) \Bigg/ \Bigg( \sum_{k=0}^{K} \frac{1}{(x + y + k)^2} \Bigg) + 2\epsilon\\
&\leq \Bigg( \frac{K+1}{x^2} + \sum_{k=K+1}^{\infty} \frac{1}{(x + k)^2} \Bigg) \Bigg/ \Bigg( \sum_{k=0}^{K} \frac{1}{(x + y + k)^2} \Bigg)\\
&= \Bigg( \frac{K+1}{x^2} \Bigg) \Bigg/ \Bigg( \sum_{k=0}^{K} \frac{1}{(x + y + k)^2} \Bigg)
+ \Bigg( \sum_{k=K+1}^{\infty} \frac{1}{(x + k)^2} \Bigg) \Bigg/ \Bigg( \sum_{k=0}^{K} \frac{1}{(x + y + k)^2} \Bigg)\\
& \leq \Bigg( \frac{K+1}{x^2} \Bigg) \Bigg/ \Bigg( \frac{K+1}{(x + y + K)^2} \Bigg)
+ \Bigg( \sum_{k=K+1}^{\infty} \frac{1}{(x + k)^2} \Bigg) \Bigg/ \Bigg( \sum_{k=0}^{K} \frac{1}{(x + y + k)^2} \Bigg)\\
& = \frac{(x + y + K)^2}{x^2}
+ \Bigg( \sum_{k=K+1}^{\infty} \frac{1}{(x + k)^2} \Bigg) \Bigg/ \Bigg( \sum_{k=0}^{K} \frac{1}{(x + y + k)^2} \Bigg).
\end{align*}
In the last line, the first term goes to $1$ as $x \longrightarrow \infty$ for any fixed $K$. For the second term, since we can choose arbitrarily large $K$ and $\Psi_1$ is convergent, it goes to $0$ as $K \longrightarrow \infty$ for any fixed $x>0$. Thus, we have $\frac{\Psi_1 (x)}{\Psi_1 (x + y)} \leq 1$ as $x \longrightarrow \infty$.
On the other hand, since $\Psi_1(x)$ is decreasing in $x$,
\begin{equation*}
\frac{\Psi_1 (x)}{\Psi_1 (x + y)} \geq 1 \text { for any } x, y >0.
\end{equation*}
Thus,
\begin{equation*}
\frac{\Psi_1 (x)}{\Psi_1 (x + y)} \longrightarrow 1 \text { as } x \longrightarrow \infty.
\end{equation*}
Am I correct? I think I am cheating somewhere. I can't convince myself.
| Why not to use the asmptotics
$$\Psi_1 (z)=\frac{1}{z}+\frac{1}{2 z^2}+\frac{1}{6
z^3}+O\left(\frac{1}{z^5}\right)$$ Apply it twice and continue with Taylor series to get
$$\frac{\Psi_1 (x)}{\Psi_1 (x + y)}=1+\frac{y}{x}+\frac{y}{2 x^2}+\frac{y}{12
x^3}+O\left(\frac{1}{x^4}\right)$$
Trying with $x=100$ and $y=10$, the "exact" value is $1.100500898$ while the above truncated series gives $\frac{1320601}{1200000}=1.100500833$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$(z-1)^5 = z^5$ for complex $z$ So the question asks to find $z \in \mathbb{C}$ such that $(z-1)^5 = z^5$.
I argued that:
$(\frac{z}{z-1})^5 = 1 = e^{2k \pi i}$
so $\frac{z}{z-1} = e^{\theta i}$
where $\theta = \frac{2k \pi}{5}$
so rearranging gives
$z = \frac{e^{i \theta}}{e^{i \theta} -1} = \frac{e^{i \theta}}{e^{i \phi}(e^{i \phi} - e^{-i \phi})}$ where $\phi = \frac{\theta}{2} = \frac{k \pi}{5}$
$z = \frac{e^{i \phi}}{2i \sin (\phi)} = \frac{ \cos \phi + i \sin \phi}{2i \ sin \phi} = \frac{1}{2} - \frac{i}{2} \cot \phi = \frac{1}{2} - \frac{i}{2} \cot \frac{k \pi}{5}$ for k = 1, ...4
However the answer says that it should be +cot, not -, but I can't see where I have dropped a negative.
Can anyone help me?
Many thanks
| Nothing has been dropped and everything is fine. Since
\begin{align*}
\cot\left(\frac{\pi}{5}\right)
&=\frac{\cos\left(\frac{\pi}{5}\right)}{\sin\left(\frac{\pi}{5}\right)}
=\frac{-\cos\left(\frac{4\pi}{5}\right)}{\sin\left(\frac{4\pi}{5}\right)}
=-\cot\left(\frac{4\pi}{5}\right)\\
\cot\left(\frac{2\pi}{5}\right)
&=\frac{\cos\left(\frac{2\pi}{5}\right)}{\sin\left(\frac{2\pi}{5}\right)}
=\frac{-\cos\left(\frac{3\pi}{5}\right)}{\sin\left(\frac{3\pi}{5}\right)}
=-\cot\left(\frac{3\pi}{5}\right)
\end{align*}
we have
\begin{align*}
\left\{\frac{1}{2}\color{blue}{-}\frac{i}{2}\cot\left(\frac{k\pi}{5}\right)\Bigg|1\leq k \leq 4\right\}
=\left\{\frac{1}{2}\color{blue}{+}\frac{i}{2}\cot\left(\frac{k\pi}{5}\right)\Bigg|1\leq k \leq 4\right\}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Bounding Euler's e by showing $2 \leq \left (1+ \frac{1}{n}\right)^n < 3$
(a)For $m,n \in \mathbb{N}$, $m<n$ and $k=0, ..., m$:
$$\frac{1}{m^k} \binom{m}{k} \leq \frac{1}{n^k} \binom{n}{k}$$
(b)For $n\in \mathbb{N}$ and $k=1, ..., n:$
$$\frac{1}{n^k} \binom{n}{k} \leq \frac{1}{k!} \leq \frac{1}{2^{k-1}}$$
(c)Show that for $n \in \mathbb{N}:$
$$2 \leq (1+ \frac{1}{n})^n < 3$$
Use (a) and the binomial theorem for (b). Use the following for (c): $$\sum_{j=0}^{n}q^j=\frac{1-q^{n+1}}{1-q}$$
I've already proven (a) thanks to @trancelocation. Now I got stuck on (b) and (c).
(b) I don't know how to use (a) and the binomial theorem here.
$\frac{1}{n^k} \binom{n}{k} \leq \frac{1}{k!} \leq \frac{1}{2^{k-1}} \Leftrightarrow \frac{n!}{n^k*(n-k)!* k!}\leq \frac{1}{k!} \leq \frac{1}{2^{k-1}} \Leftrightarrow \frac{n!}{n^k*(n-k)!}\leq 1 \leq \frac{k!}{2^{k-1}}$
Now I have to show: $\frac{n!}{n^k*(n-k)!}\leq 1$ and $1 \leq \frac{k!}{2^{k-1}}$
$\frac{n!}{n^k*(n-k)!} = \frac{n*(n-1)*...*(n-k+1)*[(n-k)*...*1]}{n^k*[(n-k)*...*1]}=\frac{n*...*(n-k+1)}{n^k}=\frac{n}{n}*\frac{n-1}{n}*...*\frac{n-k+1}{n} \leq 1$.
The first factor is $1$. Starting from the second factor their value gets smaller, so the entire term gets smaller then $1$.
I tried to do it like this:
for $i=0, ..., n$: $\frac{n-i}{n} \leq 1 \Leftrightarrow n-i \leq n \Leftrightarrow 0 \leq i$
For the second part:
$\frac{k!}{2^{k-1}} \geq 1 \Leftrightarrow \frac{k!}{2^k*2} \geq 1 \Leftrightarrow \frac{k!}{2^k} \geq 2\Leftrightarrow \frac{k}{2}*\frac{k-1}{2}*\frac{k-2}{2}*...*\frac{1}{2} \geq 2$
So for $\frac{k-j}{2} \geq 2$, with $j=0, ..., k-1$
$\frac{k-j}{2} \geq 2 \Leftrightarrow k-j \geq 4$
This doesn't seem to help at all.
(c)$2 \leq (1+ \frac{1}{n})^n < 3$
I didn't know how to start at c) at all. The provided formula doesn't help me.
| For the part (b) you had the right idea but you made a little mistake. Here is the correct one:
$\frac{k!}{2^{k-1}} \geq 1 \Leftrightarrow \frac{2 \cdot k!}{2^k} \geq 1 \Leftrightarrow \frac{k!}{2^k} \geq \frac12\Leftrightarrow \frac{k}{2}*\frac{k-1}{2}*\frac{k-2}{2}*...*\frac{1}{2} \geq \frac12$
or
$\frac{k}{2}*\frac{k-1}{2}*\frac{k-2}{2}*...*\frac{3}{2}*\frac{2}{2}
\geq 1$
and since all factors on the LHS are $≥1$ this is true.
For part (c), we need to solve two inequalities. First, let us rewrite
the geometric series $\sum_{j=0}^{n}q^j=\frac{1-q^{n+1}}{1-q}$ as $q^n = \frac1q (1 - (1-q)\sum_{j=0}^{n}q^j)$. Using $q = 1 + \frac1n$ and $(1 + \frac1n)^j \ge 1$ gives
$$
(1 + \frac1n)^n = \frac{1}{1 + \frac1n} (1 +\frac1n\sum_{j=0}^{n}(1 + \frac1n)^j) \\
= \frac{1}{1 + \frac1n} (1 +\frac1n+\frac1n\sum_{j=1}^{n}(1 + \frac1n)^j)
= \frac{1 +\frac1n}{1 + \frac1n} +\frac1n\sum_{j=1}^{n}\frac{(1 + \frac1n)^j}{1 + \frac1n}\\
= 1 + \frac1n\sum_{j=0}^{n-1}(1 + \frac1n)^j \ge 1 + \frac{n}{n} =2
$$
which establishes the left inequality. For the right inequality, use the binomial theorem, then (b), then the given geometric series, to write
$$
(1 + \frac1n)^n = \sum_{k=0}^{n} \binom{n}{k} \frac{1}{n^k} = 1 + \sum_{k=1}^{n} \binom{n}{k} \frac{1}{n^k}\\
\le 1 + \sum_{k=1}^{n} \frac{1}{2^{k-1}} = 1 + \sum_{k=0}^{n-1} \frac{1}{2^{k}} = 1 + \frac{1 - (\frac12)^n}{1 - \frac12} < 1 + \frac{1 }{1 - \frac12} = 3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Series expansion for $e^\frac{x^2}{2}$ $$e^{-x^2}=1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\frac{x^8}{4!}+ \cdots$$
$$e^{\frac{x^2}{2}}=1+(\frac{x^2}{2})^2+(\frac{x^4}{2\cdot2!})^2+(\frac{x^6}{2\cdot3!})^2+(\frac{x^8}{2\cdot4!})^2+\cdots$$
$=e^{\frac{x^2}{2}}=1+\dfrac{x^4}{8}+\dfrac{x^{8}}{4\cdot2!}-\dfrac{x^{12}}{8\cdot3!}+\dfrac{x^{16}}{16\cdot4!}+\cdots$
Wolfram's answer is $e^{\frac{x^2}{2}}=1+\dfrac{x^2}{2}+\dfrac{x^4}{4\cdot2!}+\dfrac{x^6}{8\cdot3!}+\dfrac{x^8}{16\cdot4!}+\cdots$
So what is wrong with my computation? Is Wolfram correct, I suspect that it cannot be wrong.
| $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...$
So you just plug in $x^2/2$ for $x$ and this gives you what Wolfram gave you.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the domain and range of $f(x)=\sin^{-1}(\sqrt{x^2+x+1})$ Find the domain and range of $f(x)=\sin^{-1}(\sqrt{x^2+x+1})$
Finding the domain:
$$x^2+x+1>=0 \text { it is always true }$$
$$-1<=\sqrt{x^2+x+1}<=1$$
$$\sqrt{x^2+x+1}>=-1 \text { and } \sqrt{x^2+x+1}<=1$$
$$\sqrt{x^2+x+1}<=1$$
$$x^2+x+1<=1$$
$$x(x+1)<=0$$
$$x\in[-1,0]$$
Finding the range
$$f(x)=\sin^{-1}\left(\sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left([-1,0]+\frac{1}{2}\right)^2+\frac{3}
{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left[\frac{-1}{2},\frac{1}{2}\right]^2+\frac{3}
{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left[0,\frac{1}{4}\right]+\frac{3}
{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left[\frac{3}{4},1\right]}\right)$$
$$f(x)=\sin^{-1}\left({\left[\frac{\sqrt{3}}{2},1\right]}\right)$$
$$f(x)\in \left[2m\pi+\dfrac{\pi}{3},2m\pi+\dfrac{2\pi}{3}\right] \text { where m is integer }$$
but actual answer is $\left[\dfrac{\pi}{3},\dfrac{\pi}{2}\right]$
| The range of $f(x)=sin^{-1} x$ is $[-\pi/2,\pi/2]$ over the domain $[-1,1]$.
So here range the range of the the given function is $[\pi/3,\pi/2].$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3426659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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if $s=\sqrt{x^2+6x+9} +\sqrt{x^2+24x+144}$ and $-10 < x < -8$ there follows that...
if $s=\sqrt{x^2+6x+9} + \sqrt{x^2+24x+144}$ and $-10 < x < -8$ there follows that $s=9$
After noticing that the first radical is $x+3$ and the second one is $x+12$ all I get is $s=2x+15$, which btw. is suggested as an alternative answer in the exam.
So, how to go about this?
| $$\sqrt{x^2+6x+9} = \pm(x+3)$$
And
$$\sqrt{x^2+24x+144} = \pm(x+12)$$
Hence $$S = \pm (x+3) \pm (x+12)$$
There are $4$ different cases :
$$S = 2x+15$$
$$S = -9$$
$$S = 9$$
And
$$S = -2x-15$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find Ratio of Integrals $I:J$ Given:
$$I=\int_{0}^{1}\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}\:dx}{12}$$
and
$$J=\int_{0}^{1}\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}\:dx}{(x+3)^8}$$
Find Value of $\frac{I}{J}$
My attempt:The Integral $I$ is easy to evaluate using Beta Function.
So i was trying to Manipulate $J$ to convert it to $I$ as follows:
We can write $J$ as:
$$J=\int_{0}^{1}\frac{x^{6}\left ( \frac{1}{x} -1\right )^\frac{7}{2}}{x^{8}\left ( 1+\frac{3}{x} \right )^8}$$
Now put $\frac{1}{x}=t$ we get:
$$J=\int_{1}^{\infty}\frac{(t-1)^{\frac{7}{2}}}{(1+3t)^8}$$
Using integration by Parts taking $u=(t-1)^{3.5}$ and $v=\frac{1}{(1+3t)^8}$ we get:
$$J=\frac{1}{6}\times \int_{1}^{\infty}\frac{(t-1)^{\frac{5}{2}}}{(1+3t)^7}$$
Repeating Parts again and again:
$$J=\frac{1}{432}\times \int_{1}^{\infty}\frac{\sqrt{t-1}}{(1+3t)^5}\:dt$$
Any way to proceed from here?
| Well, substitute $\sqrt{t-1}=z$
then,
$\dfrac{1}{2}\cdot\dfrac{1}{\sqrt{t-1}}dt=dz$
Now multiply the J by $\dfrac{\sqrt{t-1}}{\sqrt{t-1}}$
$J=\dfrac{1}{216}\cdot\int^{\infty}_{1}\dfrac{\sqrt{t-1}}{(1+3t)^{5}}\cdot\dfrac{\sqrt{t-1}}{2\sqrt{t-1}}dt$
$J=\dfrac{1}{216}\cdot\int^{\infty}_{1}\dfrac{t-1}{(1+3t)^{5}}\cdot\dfrac{1}{2\sqrt{t-1}}dt$
Now, substitute $t=z^{2}+1$ and $\dfrac{1}{2\sqrt{t-1}}dt=dz$
$J=\dfrac{1}{216}\cdot\int^{\infty}_{0}\dfrac{z^{2}}{(3z^{2}+4)}dz$
Now use Integration by parts.
Hint:
$\dfrac{d}{dx}[x-tan^{-1}(x)]=\dfrac{x^{2}}{x^{2}+1}$
| {
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Eccentric angles of points of contact of two parallel tangents in an ellipse The following statement is given in my book under the topic Tangents to an Ellipse:
The eccentric angles of the points of contact of two parallel tangents differ by $\pi$
In case of a circle, it is easy for me to visualise that two parallel tangents meet the circle at two points which are apart by $\pi$ radians as they are diametrically opposite. But in case of ellipse, as the eccentric angle is defined with respect to the auxiliary circle and not the ellipse, I am unable to understand why two parallel tangents meet the ellipse at points which differ by $\pi$.
Kindly explain the reason behind this fact.
|
Kindly explain the reason behind this fact.
The reason is that an ellipse can be obtained by stretching/shrinking a circle. The strech/shrink is a linear map (linear transformation).
Let's consider two tangent lines on the circle $x^2+y^2=a^2$ at $(a\cos\theta,a\sin\theta)$,$(a\cos(\theta+\pi),a\sin(\theta+\pi))$. You already know that the two tangent lines are parallel.
Now, let's stretch/shrink the circle and the tangent lines. Stretching/shrinking the circle $x^2+y^2=a^2$ to obtain the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ means that you replace $y$ in $x^2+y^2=a^2$ with $\frac{a}{b}y$ to have $x^2+\left(\frac aby\right)^2=a^2$ which is nothing but $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
By this stretch/shrink, we have the followings :
*
*The circle $x^2+y^2=a^2$ is transformed to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
*The two parallel lines are transformed to two parallel lines.
*The two lines tangent to the cirlce are transformed to two lines tangent to the ellipse.
*The tangent points $(a\cos\theta,a\sin\theta)$,$(a\cos(\theta+\pi),a\sin(\theta+\pi))$ on the circle are transformed to two tangent points $(a\cos\theta,b\sin\theta)$,$(a\cos(\theta+\pi),b\sin(\theta+\pi))$ on the ellipse respectively.
From the above facts, it follows that the eccentric angles of the points of contact of two parallel tangents differ by $\pi$.
The followings are the proof for the above facts.
Let's consider the circle $x^2+y^2=a^2$ and two points $(a\cos\theta,a\sin\theta)$,$(a\cos(\theta+\pi),a\sin(\theta+\pi))$.
The equation of the tangent line at $(a\cos\theta,a\sin\theta)$ is given by
$$a\cos\theta\ x+a\sin\theta\ y=a^2\tag1$$
Similarly, the equation of the tangent line at $(a\cos(\theta+\pi),a\sin(\theta+\pi))$ is given by
$$a\cos(\theta+\pi)x+a\sin(\theta+\pi)y=a^2\tag2$$
Now, let's stretch/shrink the circle and the lines $(1)(2)$ by replacing $y$ with $\frac aby$ to have
$$(1)\to a\cos\theta\ x+a\sin\theta\cdot\frac aby=a^2\tag3$$
$$(2)\to a\cos(\theta+\pi)x+a\sin(\theta+\pi)\cdot\frac aby=a^2\tag4 $$
Here note that these lines $(3)(4)$ are parallel since the slope of each line is $\frac{-b\cos\theta}{a\sin\theta}$.
Finally, note that $(3)$ can be written as
$$\frac{a\cos\theta}{a^2}x+\frac{b\sin\theta}{b^2}y=1\tag5$$
which is nothing but the tangent line at $(a\cos\theta,b\sin\theta)$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
Similarly, $(4)$ can be written as
$$\frac{a\cos(\theta+\pi)}{a^2}x+\frac{b\sin(\theta+\pi)}{b^2}y=1\tag6$$
which is nothing but the tangent line at $(a\cos(\theta+\pi),b\sin(\theta+\pi))$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
Since $(5)(6)$ are parallel, we see that the eccentric angles of the points of contact of two parallel tangents differ by $\pi$. $\quad\square$
| {
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"source": "stackexchange",
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$2\arctan(\phi^{-n})=\arctan\frac{p}{q}$ or $\arctan\frac{p\sqrt{5}}{q}$, where $\phi$ is the Golden Ratio. Is there a pattern in the $\frac{p}{q}$s? It is very interesting to know that
$$\arctan\frac{1}{\phi} + \arctan\frac{1}{\phi^3}= \arctan 1 = \frac{\pi}{4}$$
where Golden ratio $\phi = \frac12(\sqrt5 +1)$ is in association with circle constant $\pi$.
More interesting phenomenon is evaluation of inverse tan functions of inverse of $\phi$ in its consecutive powers as follows
$$\begin{align}
2\arctan\frac{1}{\phi} &= \arctan 2 &
2\arctan\frac{1}{\phi^2} &= \arctan\frac{2\sqrt{5}}{5}\\
2\arctan\frac{1}{\phi^3} &= \arctan\frac{1}{2} &
2\arctan\frac{1}{\phi^4} &= \arctan\frac{2\sqrt{5}}{15}\\
2\arctan\frac{1}{\phi^5} &= \arctan\frac{2}{11} &
2\arctan\frac{1}{\phi^6} &= \arctan\frac{\sqrt5}{20} \\
2\arctan\frac{1}{\phi^7} &= \arctan\frac{2}{29} &
2\arctan\frac{1}{\phi^8} &= \arctan\frac{2\sqrt5}{105} \\
2\arctan\frac{1}{\phi^9} &= \arctan\frac{1}{38} &
2\arctan\frac{1}{\phi^{10}} &= \arctan\frac{2\sqrt5}{275} \\
2\arctan\frac{1}{\phi^{11}} &= \arctan\frac{2}{199}
\end{align}$$
Here are the observations
*
*Odd powers of inverse $\phi$ in double arctan functions lead to arctan of well defined fractions
*Even powers of inverse $\phi$ in double arctan functions lead to arctan of fractions involving $\sqrt5$.
My curiosity is to know, is there any pattern in these interesting series?
I will be grateful to understand more, if anyone has come across such evaluations.
| Here is the pattern:
Given $$2\arctan\frac1{\phi^n}=\arctan A_n$$
the following recursive reciprocals hold, for both odd and even $n$'s,
$$\frac1{A_{n+2}}+\frac1{A_{n-2}}=\frac3{A_{n}}$$
| {
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"question_score": "4",
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} |
$(p-1)(p+1)/24 \in \mathbb N$ for all primes $p \geq 5$ I want to show
\begin{align}
\frac{(p-1)(p+1)}{24} \in \mathbb N \quad \text{for all primes} \quad p \geq 5 \tag{1}.
\end{align}
I can show $(1)$, if the following statement is true.
Let $a,b,c,d \in \mathbb N$ and $a \geq b \cdot c \cdot d$.
\begin{align}
\text{If} \quad \frac{a}{b},\frac{a}{c},\frac{a}{d} \in \mathbb N, \quad \text{then} \quad \frac{a}{b \cdot c \cdot d} \in \mathbb N \tag{2}.
\end{align}
Given $(2)$ we show that $(1)$ is true for $a = (p-1)(p+1)$, $b = 2$, $c = 3$ and $d = 4$. Since $p$ is a prime $(p-1)$ and $(p + 1)$ are even, implying $(p-1)/2 \in \mathbb N$, $(p+1)/2 \in \mathbb N$ and thus $(p-1)(p+1)/2 \in \mathbb N$ and $(p-1)(p+1)/4 \in \mathbb N$. One of the three numbers $(p-1)$, $p$ and $(p+1)$ must be divisible by 3. Since $p$ is a prime either $(p-1)$ or $(p+1)$ is divisible by 3, implying $(p-1)(p+1)/3 \in \mathbb N$.
Question Is $(2)$ true?
| Statement (1) can also be shown by realizing that all primes $p\ge 5$ are odd numbers having no factors of $2$ or $3$, and hence have the form $6k\pm 1$.
$(p-1)(p+1)=p^2-1=(6k\pm 1)^2-1=36k^2 \pm 12k+1-1=12\cdot k \cdot (3k\pm 1)$
$k$ must be either odd or even. If $k$ is even, then the product has at least one more factor or $2$ in addition to the explicit factor of $12$ and hence is a multiple of $24$. If $k$ is odd, then $(3k\pm 1)$ is even and the product has at least one more factor or $2$ in addition to the explicit factor of $12$ and hence is a multiple of $24$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that $\sum _{cyc}\sqrt{1-\frac{\left(a+b\right)^2}{4}}\ge \sqrt{6}$ Let $a,b,c\ge 0$ such that $a^2+b^2+c^2=1$. Prove that $$\sqrt{1-\frac{\left(a+b\right)^2}{4}}+\sqrt{1-\frac{\left(b+c\right)^2}{4}}+\sqrt{1-\frac{\left(a+c\right)^2}{4}}\ge \sqrt{6}$$
My try:
$$\sqrt{1-\frac{\left(a+b\right)^2}{4}}=\sqrt{a^2+b^2+c^2-\frac{\left(a+b\right)^2}{4}}=\frac{\sqrt{3a^2-2ab+3b^2+4c^2}}{2}\ge\frac{\sqrt{2a^2+2b^2+4c^2}}{2}$$$
Then i need to prove $$\sum \sqrt{2a^2+2b^2+4c^2}\ge 2\sqrt{6}$$
But that's failed. Without using $a^2+b^2\ge 2ab$ to remove the radical i have no more idea, i tried to use holder but failed. Help me.
| By C-S
$$\sum_{cyc}\sqrt{1-\frac{(a+b)^2}{4}}=\frac{1}{2}\sum_{cyc}\sqrt{3a^2-2ab+3b^2+4c^2}=$$
$$=\frac{1}{2}\sqrt{\sum_{cyc}(10a^2-2ab+2\sqrt{(3a^2-2ab+3b^2+4c^2)(3a^2-2ac+3c^2+4b^2)}}=$$
$$=\frac{1}{2}\sqrt{\sum_{cyc}\left(10a^2-2ab+2\sqrt{((a-b)^2+2a^2+2b^2+2c^2+2c^2)((a-c)^2+2a^2+2b^2+2c^2+2b^2)}\right)}\geq$$
$$\geq\frac{1}{2}\sqrt{\sum_{cyc}(10a^2-2ab+2((a-b)(a-c)+2a^2+2b^2+2c^2+2bc))}=\frac{1}{2}\sqrt{\sum_{cyc}24a^2}=\sqrt6.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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Issue with the following limit $\bigg(2 \sqrt{1 + \frac{1}{n}}\bigg)^n$ Calculate the following limit:
$\bigg(2 \sqrt{1 + \frac{1}{n}}\bigg)^n$
When I calculate it I get to different answers.
First way (Edit: this is where I did the mistake): $$\bigg(2 * \sqrt{1 + \frac{1}{n}}\bigg)^n = \bigg({4 + \frac{4}{n} \bigg)^\frac{n}{2}} = \bigg({4 + \frac{4}{n} \bigg)^{\frac{n}{4} \cdot \frac{4}{n}\cdot\frac{n}{2}}}$$
When we do $\lim_{n \to \infty}\bigg(\bigg({4 + \frac{4}{n} \bigg)^{\frac{n}{4}\cdot \frac{4}{n}\cdot\frac{n}{2}}}\bigg)$ we get $e^2$
Now the second way:
$$\bigg(2 \cdot \sqrt{1 + \frac{1}{n}}\bigg)^n = 2^n\cdot (1 + \frac{1}{n})^{n \cdot \frac{1}{2}}$$
When we do limit out of this we get $2^\infty \cdot \sqrt{e}$ which is of course $\infty$.
Could someone point out the mistake I made?
Edit:
I just realised where my mistake lies! I mistakenly thought that $(4 + \frac{4}{n})^\frac{n}{4} = e$ which is false, actually $(1 + \frac{4}{n})^\frac{n}{4} = e$. The second way of calculating this limit is the correct one!
| Since$$\lim_{n\to\infty}4+\frac4n=4\text{ and }\lim_{n\to\infty}\frac n4\times\frac4n\times\frac n2=\infty,$$we also have$$\lim_{n\to\infty}\left(4+\frac4n\right)^{\frac n4\times\frac4n\times\frac n2}=\infty,$$instead of $e^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3437070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is it possible to swap only two elements on a 3x3 grid by swapping rows and columns? Take a 3x3 grid with a different number in each spot, like one cell of a Sudoku puzzle:
\begin{array}{|c|c|c|}
\hline
1&2&3\\
\hline
4&5&6\\
\hline
7&8&9\\
\hline
\end{array}
Suppose you can rearrange the number by swapping two rows or two columns.
$$\left.\begin{array}{|c|c|c|}\hline 1&2&3\\ \hline 4&5&6\\ \hline7 & 8 & 9\\ \hline
\end{array} \longrightarrow \begin{array}{|c|c|c|}\hline 1&2&3\\ \hline7&8&9\\ \hline4&5&6\\ \hline
\end{array} \longrightarrow \begin{array}{|c|c|c|}\hline2&1&3\\ \hline8&7&9\\ \hline5&4&6\\ \hline
\end{array}
\right.$$
Is it possible to manipulate the grid such that an arbitrary pair of numbers is swapped (with the rest returned to their original positions)?
e.g.
$$\left.\begin{array}{|c|c|c|}\hline 1&2&6\\ \hline 4&5&3\\ \hline7&8&9\\ \hline
\end{array} \qquad\text{or}\qquad \begin{array}{|c|c|c|}\hline 8&2&3\\ \hline4&5&6\\ \hline7&1&9\\ \hline
\end{array} \qquad\text{or}\qquad \begin{array}{|c|c|c|}\hline1&2&3\\ \hline4&7&6\\ \hline5&8&9\\ \hline
\end{array}
\right.$$
| Let's apply this answer from the post that I have linked in the comments.
Note that applying a row-swap followed by a column-swap has the same effect as applying that column-swap followed by the row-swap (however, column-swaps cannot generally be switched with each other, and neither can row-swaps).
With that in mind, we can suppose that all row-swaps are performed before the column-swaps. That is, we can argue that every possible rearrangement can be obtained by some series of row-swaps followed by some series of column-swaps. The net effect of the row-swaps is some permutation of the rows, and the net effect of the column-swaps is some permutation of the columns.
There are $3 \times 2 \times 1 = 6$ ways that we can permute three rows around, and likewise $6$ ways that we can permute three columns around. So all together, the total number number of rearrangements possible using row-swaps and column swaps is $6 \times 6 = 36$.
On the other hand, if we were allowed to freely arrange entries within the box, then there is a total of
$$
9 \times 8 \times \cdots \times 2 \times 1 = 362880.
$$
So, if we take a random rearrangement (for instance, switching two entries while keeping the others fixed), then the probability that the rearrangement is possible using only row and column swaps is $\frac{36}{362880} \approx 0.00992 \%$.
We can also more rigorously prove that it's impossible to switch every pair of two entries using the above calculations: the total number of switches that we would like to attain is "9 choose 2", i.e. $\frac {9 \times 8}{2} = 36$. However, there are only $36$ permutations attainable in total, and it is clear that not all of these permutations are switches of this kind.
This answer from the same post also leads to a nice argument.
Claim: If $i$ and $j$ are in the same (row/column), then they must also be in the same (row/column) after a series of row and column switches.
I will leave it to you to convince yourself that this holds. With that established, we can see what goes wrong with swaps. If we swap two elements in the same row,
$$
\pmatrix{1&2&3\\4&5&6\\7&8&9} \to \pmatrix{\color{red}3&2&\color{red}1\\4&5&6\\7&8&9}
$$
then we contradict the claim. For example, $1$ was in the same column as $4$, but is not after the switch.
A similar argument can be made if the switched elements share a row.
If we swap elements that have neither a common row nor column, then the situation is slightly different but a similar argument can be made. For instance,
$$
\pmatrix{1&2&3\\4&5&6\\7&8&9} \to \pmatrix{\color{red}8&2&3\\4&5&6\\7&\color{red}1&9}.
$$
$1$ was in the same row as $3$, but this is no longer the case after the switch.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$(z + 1/z)^2 + (z^2 + 1/z^2)^2 + (z^3 + 1/z^3)^2 + (z^4 + 1/z^4)^4 + (z^5 + 1/z^5)^2 + (z^6 + 1/z^6)^2$ if $z^2 + z + 1 = 0$
Solve $(z + \frac{1}{z})^2 + (z^2 + \frac{1}{z^2})^2 + (z^3 + \frac{1}{z^3})^2 + (z^4 + \frac{1}{z^4})^2 + (z^5 +\frac{1}{z^5})^2 + (z^6 + \frac{1}{z^6})^2$
if $z^2 + z + 1 = 0$
I tried this problem and come up with this solution:
$z^2 + z + 1 = 0$
$z^2 = -z -1$
$(z^2)^2 = (-z -1)^2$
$z^4 + (z^2 + 1) = z^2+2z+1+ (z^2 + 1)$
$z^4 + z^2 + 1 = 2(z^2 + z + 1) = 0$
and this help me to figure out that the answer is 12. However the step
$(z^2)^2 = (-z -1)^2$ isn't quite valid because it turns a quadratic equation to a quartic equation, which will lead to extreneous roots.
Therefore, I want to ask for a better solution to this. Thanks in advance.
| If lab's $z^3=1$ idea doesn't strike you immediately here is a dumb method for working sum of powers of roots.
$$t^2+t+1=0$$
Notice that if you replace $t$ by $\frac{1}{t}$, the equation doesn't change.
This means if $z$ is a root, then $\frac{1}{z}$ is also a root.
Using sum of roots, product of roots:
$s_1=z+\frac{1}{z} = -1$
$s_2=z^2+\frac{1}{z^2} = s_1^2 - 2 = -1$
Next use the following recurrence to compute remaining sum of powers
$$s_{n+1}=-s_n-s_{n-1}$$
| {
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Find $\lim_{n \to \infty} \prod_{k=1}^{n} \frac{(k+1)^2}{k(k+2)}$ I have to find the following limit:
$$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}}$$
This is what I tried:
$$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}} =
\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{k^2+2k+1}{k^2+2k}} =
\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \bigg(\dfrac{k^2+2k}{k^2+2k}} + \dfrac{1}{k^2+2k} \bigg ) = $$
$$ = \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \bigg(1 } + \dfrac{1}{k^2+2k} \bigg ) =
\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} 1 } + \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{1}{k^2+2k} }$$
Now,
$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} 1 } = 1$
and:
$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{1}{k^2+2k} } = 0$
I think the above equals $0$, since this is a product and the limit of the last term of the product is $0$, so the whole thing would be $0$, but I am not exactly sure if my intuition is right.
So that means:
$$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}} = \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} 1 } + \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{1}{k^2+2k} } = 1 + 0 = 1$$
The problem I have is that my textbook claims that the correct answer is $2$, not $1$. So I did something wrong, however, I can't spot my mistake/mistakes.
| First observe that the product is a telescopic product: $$\prod_{k = 1}^n \frac{(k+1)^2}{k(k+2)} = \frac{2^2}{3} \cdot \frac{3^2}{2\cdot 4}
\cdot \frac{4^2}{3 \cdot 5} \cdot ... \cdot \frac{n^2}{(n-1)(n+1)} \cdot \frac{(n+1)^2}{n(n+2)} = \frac{2(n+1)}{n+2}. $$
In case of need use induction to prove it.
Now it is easy:
$$ \lim_{n \to \infty} \frac{2(n+1)}{n+2} = 2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Express a sum as a definite integral I need help expressing $\lim_{n\to\infty} \sum_{i=1}^n \frac{2}{n}(1+ \frac{2i-1}{n})^\frac{1}{3} $ as a definite integral. I am usually pretty good with figuring out these problems but I don't know what to do with the $-1$ term in the numerator and I can't find similar examples online.
| We have that
$$\left(1+ \frac{2i}{n}\right)^\frac{1}{3}-\left(1+ \frac{2i-1}{n}\right)^\frac{1}{3}=\frac{\frac1n}{\left(1+ \frac{2i}{n}\right)^\frac{2}{3}+\left(1+ \frac{2i}{n}\right)^\frac{1}{3}\left(1+ \frac{2i-1}{n}\right)^\frac{1}{3}+\left(1+ \frac{2i}{n}\right)^\frac{2}{3}}\le$$
$$\le \frac1n\frac{1}{3\left(1+ \frac{2i}{n}\right)^\frac{2}{3}}$$
and
$$\lim_{n\to\infty} \left(\frac{2}{n} \cdot \frac1n \sum_{i=1}^n \frac{1}{3\left(1+ \frac{2i}{n}\right)^\frac{2}{3}}\right)=0$$
therefore
$$\lim_{n\to\infty} \sum_{i=1}^n \frac{2}{n}\left(1+ \frac{2i-1}{n}\right)^\frac{1}{3}=\lim_{n\to\infty} \frac{2}{n}\sum_{i=1}^n \left(1+ \frac{2i}{n}\right)^\frac{1}{3}=\int_0^2 (1+x)^\frac13\ \mathsf dx$$
| {
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} |
Prove that for ${a^3} = {b^2} + {c^2}$. where $a$, $b$ and $c$ are positive integers that there are an infinite number of values for $a$. Prove that for ${a^3} = {b^2} + {c^2}$. where $a$, $b$ and $c$ are positive integers that there are an infinite number of values for $a$.
| Let suppose first that exist $u,v\in\mathbb{Z}$ such that $a=u^2+v^2$ then $a^{3}=(au)^2+(av)^2.$ That is, every number $a$ that is sum of two squares is solution to the problem. Now, $5=2^{2}+1^2$ so every number of the form $5n^2$ can be written as $(2n)^2+n^2$ so there we have infinite possible values to $a.$ For example, taking $n=1$ we have $5=2^2+1^2$ so $5^3=10^2+5^2,$ for $n=2$ we have $20=4^2+2^2$ so $20^3=80^2+40^2,$ etc...
| {
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"source": "stackexchange",
"question_score": "2",
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Solve inequality $\cos(x)+2\tan(x)\le2+\sin(x)$ Solve inequality
$\cos(x)+2\tan(x)\le2+\sin(x)$
My proof:
$\cos(x)+2\tan(x)\le2+\sin(x)\\\cos^2(x)+2\sin(x)\le2\cos(x)+\sin(x)\cos(x)\\\cos(x)\left(\cos(x)-2 \right )+\sin(x)\left(2-\cos(x) \right )\le0\\\left(\cos(x)-\sin(x) \right )\left(\cos(x)-2 \right )\le0\\\sqrt{2}\left(\sin\left(x-\frac{\pi}{4} \right ) \right )\left(\cos(x)-2 \right )\le0\\-\sqrt{2}\cos\left(x+\frac{\pi}{4} \right )\left(\cos(x)-2 \right )\le0$
I stopped at this moment and I have no idea what to do now
| The inequality should be factorized as,
$$\frac{\sqrt{2}}{\cos x}\cos\left(x+\frac{\pi}{4} \right )\left(\cos x -2 \right )\le0$$
Two cases:
1) $\cos x > 0, \>\>\>\> \cos\left(x+\frac{\pi}{4} \right )\ge 0$, which leads to
$$-\frac\pi2 +2\pi k < x \le \frac\pi4+2\pi k$$
2) $\cos x < 0, \>\>\>\> \cos\left(x+\frac{\pi}{4} \right )\le 0$, which leads to
$$\frac{\pi}2 +2\pi k < x \le \frac{5\pi}4+2\pi k$$
| {
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prove two dependent sequences converge Please help with my homework question:
Define $\{a_n\}_{n=1}^{\infty}$, $\{b_n\}_{n=1}^{\infty}$ as:
$b_{n+1}=\frac{a_n+b_n}{2}, a_{n+1}=\frac{a_n^2+b_n^2}{a_n+b_n}, 0<b_1<a_1$
Prove the limits converge and that $\lim \limits_{n \to \infty}a_n = \lim \limits_{n \to \infty}b_n $.
I have tried to show that they are monotone and bounded, tried looking for a property of arithmetic mean (related to $b_n$) but I'm completely stuck as none of these yielded anything useful.
Any help would be appreciated
| You may turn it into a fixpoint problem as follows:
*
*Consider $q_n = \frac{b_n}{a_n} \Rightarrow 0<q_1 <1$ and $a_1 >0$ and you have
$$a_{n+1} = a_n\frac{1+q_n^2}{1+q_n} \mbox{ and } q_{n+1} = \frac{1}{2}\frac{(1+q_n)^2}{1+q_n^2}$$
Now, it is enough to show that $\boxed{\lim_{n\to \infty}q_n = 1}$.
To do so, consider $f(x) = \frac{1}{2}\frac{(1+x)^2}{1+x^2}$ and check the following properties:
*
*$f:[0,1] \Rightarrow [\frac{1}{2},1]$, because for $x \in [0,1]$ you have
$$f(x) = \frac{1}{2}\left(1+\frac{2x}{1+x^2} \right)\leq \frac{1}{2}\left(1+\frac{1+x^2}{1+x^2} \right)=1$$
$$f(x)=\frac{1}{2}\frac{(1+x)^2}{1+x^2}\geq \frac{1}{2}\frac{(1+x)^2}{1+x}=\frac{1}{2}(1+x)\geq \frac 12$$
*$f$ has the fixpoint $x=1$ since $f(1) = 1$.
*$f'(x) = \frac{1-x^2}{(1+x^2)^2}$ and for $x \in [\frac{1}{2},1]$ you have
$$0\leq f'(x) = \frac{1-x^2}{(1+x^2)^2} \leq 1-x^2 \leq \frac 34$$
So, $f$ is a contraction on $[\frac 12, 1]$ and you can conclude that $q_n$ converges to the fixpoint $x=1$.
Since $a_{n+1} = a_n \frac{1+q_n^2}{1+q_n} \stackrel{q_n \in [0,1]}{\leq}a_n \frac{1+q_n}{1+q_n}=a_n$, you see that $a_n$ is decreasing and bounded from below (by $0$). So, $a_n$ is convergent and hence $b_n$, too and they must have the same limit.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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$\int\frac1{(x^2+1)^2}\ dx$ by partial fraction decomposition Is there any possible way to calculate the integral of $\frac{1}{(x^2+1)^2}$ by partial fraction decomposition? I do not know the formulas for the trigonometric method.
Thank you!
| Hermite's algorithm essentially does what you want. You can read about it here. Rather than replicate that work...
Observe
$$ \frac{1}{(x^2+1)^2} = \frac{(1/2)(1-x^2)}{(x^2+1)^2} + \frac{1/2}{x^2+1} $$
and
$$ \frac{\mathrm{d}}{\mathrm{d}x} \frac{x}{x^2+1} = \frac{1-x^2}{(x^2 +1)^2} \text{.} $$
Another way to get to the second fact is to use https://math.stackexchange.com/a/68512/123905 .
Therefore, \begin{align*}
\int \frac{1}{(x^2+1)^2} \,\mathrm{d}x
&= \frac{1}{2} \int \frac{1-x^2}{(x^2+1)^2} + \frac{1}{x^2+1} \,\mathrm{d}x \\
&= \frac{1}{2}\left( \int \frac{1-x^2}{(x^2+1)^2} \,\mathrm{d}x + \int \frac{1}{x^2+1} \,\mathrm{d}x \right) \\
&= \frac{1}{2}\left( \frac{x}{x^2+1} + \tan^{-1}x + C \right) \\
&= \frac{1}{2}\left( \frac{x}{x^2+1} + \tan^{-1}x \right) + C \text{.}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the $n^{th}$ derivative of $y=\tan^{-1} \left(\frac {1+x}{1-x}\right)$ Find the $n^{th}$ derivative of $y=\tan^{-1} \left(\dfrac {1+x}{1-x}\right)$
My Attempt:
$$y=\tan^{-1} \left(\dfrac {1+x}{1-x}\right)$$
Put $x=\cos (\theta)$
$$y=\tan^{-1} \left(\dfrac {1+\cos (\theta)}{1-\cos (\theta)}\right)$$
$$y=\tan^{-1} \left(\dfrac {2\cos^2 \left(\dfrac {\theta}{2}\right)}{2\sin^2\left(\dfrac {\theta}{2}\right)}\right)$$
$$y=\tan^{-1} \left(\cot^2 \left(\dfrac {\theta}{2}\right)\right)$$
| In this question the derivatives given are incorrect starting with the third (your fourth). Also the accepted answer could be worked forward a little, so we provide another answer. First off,
$$\tan^{-1}\left(\frac{1+x}{1-x}\right)=\tan^{-1}\left(\tan\left(\tan^{-1}1+\tan^{-1}x\right)\right)=\tan^{-1}1+\tan^{-1}x+[-\pi]$$
The $[-\pi]$ happens when $x>1$. So for $x\ne1$,
$$\frac d{dx}\tan^{-1}\left(\frac{1+x}{1-x}\right)=\frac d{dx}\tan^{-1}x=\frac1{1+x^2}=\frac{-\frac12i}{x-i}+\frac{\frac12i}{x+i}$$
Then for $x\ne1$,
$$\begin{align}\frac{d^n}{dx^n}\tan^{-1}\left(\frac{1+x}{1-x}\right)&=\frac{-\frac12i(-1)^{n-1}(n-1)!}{(x-i)^n}+\frac{\frac12i(-1)^{n-1}(n-1)!}{(x+i)^n}\\
&=\frac{\frac12i(-1)^n(n-1)!\left[(x+i)^n-(x-i)^n\right]}{(x^2+1)^n}\\
&=\frac{\frac12i(-1)^n(n-1)!}{(x^2+1)^n}\sum_{k=0}^n{n\choose k}x^{n-k}\left[i^k-(-i)^k\right]\\
&=\frac{\frac12i(-1)^n(n-1)!}{(x^2+1)^n}\sum_{k=0}^n{n\choose k}x^{n-k}i^k\left[1-(-1)^k\right]\\
&=\frac{\frac12i(-1)^n(n-1)!}{(x^2+1)^n}\sum_{k=0}^{\lfloor\frac{n-1}2\rfloor}{n\choose {2k+1}}x^{n-2k-1}i(-1)^k(2)\\
&=\frac{(-1)^{n+1}(n-1)!\sum_{k=0}^{\lfloor\frac{n-1}2\rfloor}(-1)^k{n\choose {2k+1}}x^{n-2k-1}}{(x^2+1)^n}\end{align}$$
Written out,
$$\begin{array}{rl}f^{\prime}(x)&=\frac1{x^2+1}\\
f^{\prime\prime}(x)&=\frac{-(2x)}{(x^2+1)^2}\\
f^{\prime\prime\prime}(x)&=\frac{2(3x^2-1)}{(x^2+1)^3}\\
f^{(4)}(x)&=\frac{-6(4x^3-4x)}{(x^2+1)^4}\\
f^{(5)}(x)&=\frac{24(5x^4-10x^2+1)}{(x^2+1)^5}\\
f^{(6)}(x)&=\frac{-120(6x^5-20x^3+6x)}{(x^2+1)^6}\end{array}$$
| {
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"source": "stackexchange",
"question_score": "2",
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Prove trigonometric an identity $\frac{2\cos\alpha -1}{\sqrt3-2\sin\alpha}=\tan\left(\frac{\alpha}{2}+\frac{\pi}{6}\right)$
Prove trigonometric an identity
$\frac{2\cos\alpha
-1}{\sqrt3-2\sin\alpha}=\tan\left(\frac{\alpha}{2}+\frac{\pi}{6}\right)$
My proof:
$\frac{2\cos\alpha
-1}{\sqrt3-2\sin\alpha}=\frac{\cos\alpha-\frac{1}{2}}{\frac{\sqrt3}{2}-\sin\alpha}=\frac{\cos\alpha-\sin\frac{\pi}{6}}{\cos\frac{\pi}{6}-\sin\alpha}$
I have no idea what to do now
| Use that $$\tan(x+y)=\frac{2\sin(x)\cos(y)}{\cos(x)\cos(y)-\sin(x)\sin(y)}$$
| {
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Prove that the minimum values of $x^2+y^2+z^2$ is $27$ with given condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$. Question: Prove that the minimum values of $x^2+y^2+z^2$ is $27$, where $x,y,z$ are positive real variables satisfying the condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$.
From AM$\ge$ GM, we have $\left( \dfrac{x^2+y^2+z^2}{3}\right)^3\ge (xyz)^2=(xy+yz+zx)^2$. Is it possible to show thae result from this relation?
| If you wish to complete your AM-GM, note
$$1=\frac1x+\frac1y+\frac1z\geqslant \frac3{\sqrt[3]{xyz}}\implies xyz\geqslant 27$$
Using that with what you got, viz.
$$\left(\frac{x^2+y^2+z^2}3 \right)^3 \geqslant (xyz)^2\geqslant 27^2=9^3 \implies x^2+y^2+z^2\geqslant 27$$
| {
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"question_score": "7",
"answer_count": 8,
"answer_id": 3
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Limit $\lim\limits_{x\to0}{\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}}$ Without using L'Hopital's rule find:
$$\lim_{x\rightarrow 0}{\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}}$$
I found the first derivative because I planned to use Taylor series:
\begin{align}
\frac{d}{dx}{\left(\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}\right)}
&=\frac{\left(\frac{1}{x+e}-e^x\right)\left(\cos^2{x} -e^x\right)-\left(\ln(x+e)-e^x\right)\left(\sin{2x}-e^x\right)}
{(\cos^2{x} -e^x)^2}\\
&=\frac{\frac{1}{x+e}-e^x}{\cos^2{x}-e^x}-\frac{\left(\ln(x+e)-e^x\right)\left(\sin{2x}-e^x\right)}{(\cos^2{x} -e^x)^2}
\end{align}
However, it seems I haven't gone so far. Should I start from the beginning and try a different method?
Source in Croatian: 2.kolokvij, matematička analiza
| If you want the Taylor series, work separately numerator and denominator
$$\log (x+e)-e^x=\left(\frac{1}{e}-1\right) x-\frac{\left(1+e^2\right) }{2 e^2}x^2+\left(\frac{1}{3
e^3}-\frac{1}{6}\right) x^3+O\left(x^4\right)$$
$$\cos ^2(x)-e^x=-x-\frac{3 }{2}x^2-\frac{1}{6}x^3+O\left(x^4\right)$$
$$\frac{\log (x+e)-e^x } { \cos ^2(x)-e^x}=\frac{\left(\frac{1}{e}-1\right) x-\frac{\left(1+e^2\right) }{2 e^2}x^2+\left(\frac{1}{3
e^3}-\frac{1}{6}\right) x^3+O\left(x^4\right) } {-x-\frac{3 }{2}x^2-\frac{1}{6}x^3+O\left(x^4\right)}$$
Now, use the long division to get
$$\frac{\log (x+e)-e^x } { \cos ^2(x)-e^x}=\frac{e-1}{e}+\frac{1}{2} \left(-2+\frac{1}{e^2}+\frac{3}{e}\right) x+O\left(x^2\right)$$ which shows the limit and also how it is approached.
| {
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A bound for $\sqrt\frac{b+c-a}a+\sqrt\frac{c+a-b}b+\sqrt\frac{a+b-c}c$ in a triangle Assume that $ABC$ is a triangle with $a\geq b\geq c$, where the angle $A$ has a fixed value. We denote by $\Sigma$ the sum
$$\sqrt\frac{b+c-a}a+\sqrt\frac{c+a-b}b+\sqrt\frac{a+b-c}c.$$
Then the only possible values of $A$ are $\pi/3\leq A<\pi$ and we have:
(i) The smallest possible value $\Sigma$ is
$$\frac{4\sin\frac A2+\sqrt{2\left( 1-\sin\frac A2\right)}}{\sqrt{2\sin\frac A2}}.$$
(ii) If $\pi/3\leq A<\pi/2$ then the largest possible value of $\Sigma$ is
$$\frac{4\cos A+\sqrt{2\left( 1-\cos A\right)}}{\sqrt{2\cos A}}.$$
(iii) If $\pi/2\leq A<\pi$ then there is no finite upper bound for $\Sigma$.
My question is how to prove (i), (ii), and (iii).
I firstly tried to square the $LHS$, but nothing. I also tried the Radulescu-Maftei theorem, but it didn't help. Hence, I am looking forward to seeing your ideas.
| (i) With$$\frac{\pi}{3}\le A <\frac{\pi}{2}$$let $\angle B=\angle C$
In the isosceles triangle$$\sqrt{\frac{b+c-a}{a}}=\sqrt{\frac{b+c}{a}-1}=\sqrt{\csc \frac{A}{2}-1}$$Again$$\sqrt{\frac{c+a-b}{b}}=\sqrt{2\sin \frac{A}{2}}$$and also$$\sqrt{\frac{a+b-c}{c}}=\sqrt{2\sin \frac{A}{2}}$$Thus$$\Sigma=2\sqrt{2\sin \frac{A}{2}}+\sqrt{\frac{1-\sin \frac{A}{2}}{\sin \frac{A}{2}}}$$And putting over a common denominator and multiplying top and bottom by $\sqrt2$
$$\Sigma=\frac{4\sin \frac{A}{2}+\sqrt{2}\cdot\sqrt{1-\sin \frac{A}{2}}}{\sqrt2\cdot \sqrt{\sin \frac{A}{2}}}=\frac{4\sin \frac{A}{2}+\sqrt{2(1-\sin \frac{A}{2})}}{\sqrt{2\sin \frac{A}{2}}}$$the proposed expression.
(ii) In the next figure, suppose$$\frac{\pi}{3}\le B=A<\frac{\pi}{2}$$
Then$$\sqrt\frac{b+c-a}{a}=\sqrt\frac{c}{a}=\sqrt{2\cos A}$$and$$\sqrt\frac{c+a-b}{b}=\sqrt\frac{c}{b}=\sqrt{2\cos A}$$and$$\sqrt\frac{a+b-c}{c}=\sqrt{\frac{a+b}{c}-1}=\sqrt{\sec A-1}$$Hence$$\Sigma=2\sqrt{2\cos A}+\sqrt{\sec A-1}=2\sqrt{2\cos A}+\sqrt{\frac{1-\cos A}{\cos A}}$$Putting over a common denominator$$\Sigma=\frac{2\sqrt2\cdot\sqrt{\cos A}\cdot\sqrt{\cos A}+\sqrt{1-\cos A}}{\sqrt{\cos A}}$$And multiplying top and bottom by $\sqrt2$,$$\Sigma=\frac{4\cos A+\sqrt2\cdot\sqrt{1-\cos A}}{\sqrt{2\cos A}}=\frac{4\cos A+\sqrt{2(1-\cos A)}}{\sqrt{2\cos A}}$$Thus for given$$\frac{\pi}{3}\le A<\frac{\pi}{2}$$we get the other proposed expression when $B=A$.
To examine for maximum and minimum, note the limited range of positions for $B$.
Building on the first figure above, we see that for given $\angle A$, point $B$ is confined between $B'$ and $B''$, beyond which the condition that$$A\ge B\ge C$$ does not hold.
Further, since$$\triangle AB'C\sim \triangle AB''C$$then the three quantities$$\sqrt{\frac{b+c-a}{a}}, \sqrt{\frac{c+a-b}{b}}, \sqrt{\frac{a+b-c}{c}}$$ are the same, respectively, in these two limiting triangles, and hence the extreme value of $\Sigma$ is the same.
To see that it is a maximum, note that if $A=\frac{\pi}{3}$, then points $B'$ and $B''$ coincide with $B$, and since $\sin \frac{\pi}{6}=\cos \frac{\pi}{3}$, then$$\frac{4\sin \frac{A}{2}+\sqrt{2(1-\sin \frac{A}{2})}}{\sqrt{2\sin \frac{A}{2}}}=\frac{4\cos A+\sqrt{2(1-\cos A)}}{\sqrt{2\cos A}}=3$$Hence no distinction between maximum and minimum. On the other hand, if $A=\frac{\pi}{2}$, and is thus no longer acute, then$$\frac{4\sin \frac{A}{2}+\sqrt{2(1-\sin \frac{A}{2})}}{\sqrt{2\sin \frac{A}{2}}}\approx3.022$$while$$\frac{4\cos A+\sqrt{2(1-\cos A)}}{\sqrt{2\cos A}}=\frac{0+\sqrt2}{0}$$has passed all bounds.
Hence abbreviating the left sides as $f(\sin \frac{A}{2})$ and $f(\cos A)$, it is clear that, for $\frac{\pi}{3}<A<\frac{\pi}{2}$, the ratio$$\frac{f(\cos A)}{f(\sin \frac{A}{2})}$$grows beyond all bounds with increasing $A$.
$\Sigma$ is thus a maximum in (ii), when $\angle A=\angle B$ or $\angle C$, and a minimum in (i), when $\angle B=\angle C$.
(iii) Finally for$$\frac{\pi}{2}\le A<\pi$$
With $\angle A$ and length $c$ fixed, if we move point $C$ toward point $A$, then
$\sqrt{\frac{a+b-c}{c}}$ and $\sqrt{\frac{b+c-a}{a}}$ each approaches zero, while $\sqrt{\frac{a+c-b}{b}}$ exceeds all bounds. Thus $\Sigma$ is unbounded in this case.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "10",
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Show that following determinant is divisible by $\lambda^2$ and find the other factor. Show that $\begin{vmatrix}
a^2+\lambda &ab &ac \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$ is divisible by $\lambda^2$ and find the other factor.
My attempt is as follows:-
$$R_1\rightarrow R_1+R_2+R_3$$
$$\begin{vmatrix}
a(a+b+c)+\lambda &b(a+b+c)+\lambda &c(a+b+c)+\lambda \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$$
$$C_1\rightarrow C_1-\dfrac{a}{b}C_2$$
$$C_2\rightarrow C_2-\dfrac{b}{c}C_3$$
$$\begin{vmatrix}
\lambda-\dfrac{a\lambda}{b}&\lambda-\dfrac{b\lambda}{c} &c(a+b+c)+\lambda \\
-\lambda & \lambda & bc \\
0 & -\lambda & c^2+\lambda
\end{vmatrix}=0$$
Taking $\lambda^2$ common
$$\lambda^2\begin{vmatrix}
1-\dfrac{a}{b}&1-\dfrac{b}{c} &c(a+b+c)+\lambda \\
-1 & 1 & bc \\
0 & -1 & c^2+\lambda
\end{vmatrix}=0
$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b-a&c-b &c(a+b+c)+\lambda \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0
$$
$$R_1\rightarrow R_1-R_3$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b-a&2c-b &ca+bc \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0$$
$$R_1\rightarrow R_1-R_2$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
2b-a&c-b &ca \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0$$
Now expanding it
$$\dfrac{\lambda^2}{bc}\left(c(2b^2c-abc+abc)+(c^2+\lambda)(2bc-ac+bc-b^2)\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(2b^2c^2+(c^2+\lambda)(3bc-ac-b^2)\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(2b^2c^2+3bc^3-ac^3-b^2c^2+3bc\lambda-\lambda ac-\lambda b^2\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(b^2c^2+3bc^3-ac^3+3bc\lambda-\lambda ac-\lambda b^2\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(c^2(b^2+3bc-ac\right)+\lambda(3bc-ac-b^2)=0$$
So another factor seems to be $\dfrac{1}{bc}\left(c^2(b^2+3bc-ac)+\lambda\left(3bc-ac-b^2\right)\right)$
But actual answer is $a^2+b^2+c^2+\lambda$.
I tried to find my mistake, but everything seems correct. What am I missing here? Please help me in this.
| I did not go all the way, but the first (maybe the only) mistake is in the $C_1\rightarrow C_1-\dfrac{a}{b}C_2$ step. The second row will be $$ab-\frac ab(b^2+\lambda)=ab-ab-\frac ab\lambda=-\frac ab\lambda\ne-\lambda$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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Let $X\sim\mathrm{Normal}(1,4)$. If $Y=0.5^X$, find $\Bbb E[Y^2]$.
Let $X\sim\mathrm{Normal}(1,4)$. If $Y=0.5^X$, find $\Bbb E[Y^2]$.
So I have $\Bbb E[Y^2]=M^{\prime\prime}_Y(0)$
So I need to compute $M_Y(t)=M_{0.5^X}(t)=\Bbb E\left[e^{0.5^X}\right]$
$$\Bbb E\left[e^{0.5^X}\right]=\int_{-\infty}^\infty e^{0.5^x t}\cdot f_X(x)\,dx=\int_{-\infty}^\infty e^{0.5^x t}\cdot \frac{1}{2\sqrt{2\pi}}e^{-\frac{(x-1)^2}{2\cdot 4}}\,dx$$
So from here do I have
$$M_Y^{\prime\prime}(t)=\int_{-\infty}^\infty (0.5^x)^2 e^{0.5^x t}\cdot \frac{1}{2\sqrt{2\pi}}e^{-\frac{(x-1)^2}{2\cdot 4}}\,dx$$ or $$M_Y^{\prime\prime}(0)\int_{-\infty}^\infty (0.5^x)^2\cdot \frac{1}{2\sqrt{2\pi}}e^{-\frac{(x-1)^2}{2\cdot 4}}dx\,?$$
| Let $f(x)$ and $F(x)$ denote the p.d.f. and c.d.f. of $X\sim\operatorname N(1,2^2)$ respectively.
Let $g(z)$ and $G(z)$ denote the p.d.f. and c.d.f of $Z=0.5^{2X}$ respectively.
Then $$G(z)=\operatorname P(Z\le z)=\operatorname P\left(2^{2X}\ge\frac1z\right)=\operatorname P\left(2X\ge-\frac{\ln z}{\ln2}\right)=1-F\left(-\frac{\ln z}{2\ln2}\right)$$ so its derivative becomes $$g(z)=\frac1{2z\ln2}f\left(-\frac{\ln z}{2\ln 2}\right)=\frac1{2z\ln2}\cdot\frac1{2\sqrt{2\pi}}\exp\left(-\frac1{2\cdot2^2}\left(-\frac{\ln z}{2\ln2}-1\right)^2\right).$$ Hence \begin{align}\Bbb E[Z]&=\int_{\Omega_Z} zg(z)\,dz=\frac1{(4\ln2)\sqrt{2\pi}}\int_0^\infty\exp\left(-\frac18\left(\frac1{4\ln^22}\ln^2z+\frac1{\ln2}\ln z+1\right)\right)\,dy\\&=\frac{e^{-1/8}}{(4\ln2)\sqrt{2\pi}}\int_0^\infty z^{-(\ln z+4\ln2)/(32\ln^22)}\,dz\end{align} and the substitution $u=\ln z$ yields \begin{align}\Bbb E[Z]&=\frac{e^{-1/8}}{(4\ln2)\sqrt{2\pi}}\int_{-\infty}^\infty(e^u)^{1-(u+4\ln2)/(32\ln^22)}\,du\\&=\frac{e^{-1/8}e^{(8\ln2-1)^2/8}}{(4\ln2)\sqrt{2\pi}}\int_{-\infty}^\infty\exp\left(-\frac12\left(\frac{u-2(8\ln2-1)\ln2}{4\ln2}\right)^2\right)\,du\\&=e^{-1/8}e^{(8\ln2-1)^2/8}=e^{(8\ln2-2)\ln2}=2^{8\ln2-2}.\end{align} Thus $$\Bbb E[Y^2]=\Bbb E[Z]=4^{4\ln2-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3460931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Calculate the following series using telescoping I have the following series
$$\sum_{k=1}^\infty \frac{(k-1)!}{(k+N)!},\quad\text{where }N \in \mathbb{N}. $$
I have found out that the series is equal to
$$\sum_{k=1}^\infty \biggl(\frac{1}{k}\cdot \frac{1}{k+1}\cdot _{...} \cdot \frac{1}{k+N}\biggr)$$
I also know that for $N=1$ we can use partial fraction expansion and we get a telescoping sum of $1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + ... + \frac{1}{k+1}$ which leaves $1$ if $k$ is heading to $\infty$.
Does anyone have any idea how I can go about doing this ?
| The neat thing about
$p_n(x)
=\prod_{k=0}^{n-1}(x+k)
$
is that is telescopes in both
numerator and denominator.
$\begin{array}\\
p_n(x+1)-p_n(x)
&=\prod_{k=0}^{n-1}(x+1+k)-\prod_{k=0}^{n-1}(x+k)\\
&=\prod_{k=1}^{n}(x+k)-\prod_{k=0}^{n-1}(x+k)\\
&=(x+n)\prod_{k=1}^{n-1}(x+k)-x\prod_{k=1}^{n-1}(x+k)\\
&=((x+n)-x)\prod_{k=1}^{n-1}(x+k)\\
&=n\prod_{k=0}^{n-2}(x+1+k)\\
&=np_{n-1}(x+1)\\
\text{so}\\
\dfrac{p_n(x+1)-p_n(x)}{n}
&=p_{n-1}(x+1)\\
\text{or}\\
\dfrac{p_{n+1}(x)-p_{n+1}(x-1)}{n+1}
&=p_{n}(x)\\
\text{and}\\
\dfrac1{p_n(x)}-\dfrac1{p_n(x+1)}
&=\dfrac1{\prod_{k=0}^{n-1}(x+k)}-\dfrac1{\prod_{k=0}^{n-1}(x+1+k)}\\
&=\dfrac1{\prod_{k=0}^{n-1}(x+k)}-\dfrac1{\prod_{k=1}^{n}(x+k)}\\
&=\dfrac{x+n}{\prod_{k=0}^{n}(x+k)}-\dfrac{x}{\prod_{k=0}^{n}(x+k)}\\
&=\dfrac{(x+n)-x}{\prod_{k=0}^{n}(x+k)}\\
&=\dfrac{n}{p_{n+1}(x)}\\
\text{so}\\
\dfrac1{n}\left(\dfrac1{p_n(x)}-\dfrac1{p_n(x+1)}\right)
&=\dfrac{1}{p_{n+1}(x)}\\
\text{or}\\
\dfrac1{n-1}\left(\dfrac1{p_{n-1}(x)}-\dfrac1{p_{n-1}(x+1)}\right)
&=\dfrac{1}{p_{n}(x)}\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solve equation $xy^2+3y^2-x^2\frac{dy}{dx}=0$
Solve equation $xy^2+3y^2-x^2\frac{dy}{dx}=0$
$$xy^2+3y^2=\frac{dy}{dx}x^2$$
$$dx(xy^2+3y^2)-x^2 dy=0$$
$$dx\frac{x+3}{x^2}-\frac{1}{y^2}dy=0$$
$$\int \frac{x+3}{x^2}dx=\frac{1}{y^2}dy$$
$$\int \frac{x+3}{x^2}dx= (-x^{-1})(x+3)-\int\frac{1}{x^2}dx=-\frac{1}{x}(x+3)+\frac{1}{x}$$
$$=-1-\frac{3}{x}+\frac{1}{x}$$
$$=-1-\frac{2}{x}$$
$$\int \frac{1}{y^2}dy=-y^{-1}$$
Then from
$$\int \frac{x+3}{x^2}dx=\frac{1}{y^2}dy$$
$$-1-\frac{2}{x}=-y^{-1}$$
$$1+\frac{2}{x}=y^{-1}$$
$$\frac{1}{1+2/x}=y$$
| $$xy^2+3y^2-x^2\frac{dy}{dx}=0$$
This equation is not exact but it is separable:
$$(xy^2+3y^2)dx-x^2{dy}=0$$
$$\frac {(x+3)dx}{x^2}-\frac {dy} { y^2}=0$$
Integrate
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Given $a,b$, find $c$ such that $a+c, b+c, ab+c$ are perfect squares Prove that for every different parity numbers $a,b \in \Bbb N$ there exist $c \in \Bbb Z$ such that numbers $a+c, b+c, ab+c$ are perfect squares.
I tried to find separate solutions, $a=2, b=7, c=2$ also $a=6, b=13, c=3$ and etc., then all $a+c, b+c, ab+c$ are perfect squares. But have no idea how to start proof in general.
Have any ideas how to start proof? Thanks.
| Since $b$ is of a different parity than $a$, we may choose an integer $k$ such that $b=a+2k+1$. Then take $c=k^2-a$. Then we have:$$a+c=a+(k^2-a)=k^2\\b+c=(a+2k+1)+(k^2-a)=k^2+2k+1=(k+1)^2\\ab+c=a(a+2k+1)+(k^2-a)=a^2+2ka+a+k^2-a=a^2+2ka+k^2=(a+k)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
is there a real solution to this differential equation? This is the equation:
$\frac {d^4y}{dx^4}=-y$
I know the solution to this equation:
$\frac {d^2y}{dx^2}=-y$
is this:
$y(x)=a\cdot \cos(x)+b\cdot \sin(x)$
and it involves $e^{ix}$, but I wondered if there was any solution to a higher-order version of this.
| $$\frac {d^4y}{dx^4}=-y$$
The particular solutions on the form $y_p=e^{rx}$ are obtained according to the roots of $r^4=-1$, thus $r=\pm\frac{1\pm i}{\sqrt{2}}$.
$$y_p=\exp\left(\frac{\pm 1\pm i}{\sqrt{2}}x \right)=\exp\left(\pm\frac{x}{\sqrt{2}} \right)\left(\cos\left(\frac{x}{\sqrt{2}} \right)\pm i\sin\left(\frac{x}{\sqrt{2}} \right)\right)$$
So we have four independent particular solutions, wich leads to the general solution :
$y(x)=c_1\exp\left(\frac{x}{\sqrt{2}} \right)\cos\left(\frac{x}{\sqrt{2}} \right) +c_2\exp\left(-\frac{x}{\sqrt{2}} \right)\cos\left(\frac{x}{\sqrt{2}} \right) +c_3\exp\left(\frac{x}{\sqrt{2}} \right)\sin\left(\frac{x}{\sqrt{2}} \right) +c_4\exp\left(-\frac{x}{\sqrt{2}} \right)\sin\left(\frac{x}{\sqrt{2}} \right)$
Or on another equivalent form :
$y(x)=C_1\cosh\left(\frac{x}{\sqrt{2}} \right)\cos\left(\frac{x}{\sqrt{2}} \right)\:+C_2\sinh\left(\frac{x}{\sqrt{2}} \right)\cos\left(\frac{x}{\sqrt{2}} \right)\:+C_3\cosh\left(\frac{x}{\sqrt{2}} \right)\sin\left(\frac{x}{\sqrt{2}} \right)\:+C_4\sinh\left(\frac{x}{\sqrt{2}} \right)\sin\left(\frac{x}{\sqrt{2}} \right)$
All those solutions are real insofar the coefficients are real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Choosing $\alpha$ and $\beta$ so that $f(x) = \frac{\sin (x + \alpha) \sin (x + \beta)}{\cos (x + \alpha)\cos (x + \beta)}$ is independent of $x$
Consider the function:
$$f(x) = \frac{\sin (x + \alpha) \sin (x + \beta)}{\cos (x + \alpha)\cos (x + \beta)}$$
Choose the parameters $\alpha$ and $\beta$ such that $f(x)$ does not
depend on $x$.
Using Werner's formulas, I obtained:
$$f(x) = \frac{\cos(\alpha - \beta) - \cos(2x + \alpha + \beta)}{\cos(2x + \alpha + \beta) + \cos(\alpha - \beta)}$$
If $\alpha = \beta \pm \frac{\pi}{2}$, $f(x) = -1$.
Is this the only way to obtain a constant value from $f(x)$? I can't figure out if there are some alternatives, neither I am able to prove that there are not.
| $f(x) = \frac{\sin(x + \alpha)\sin(x + \beta)}{\cos(x + \alpha)\cos(x + \beta)} = \frac{\cos(\alpha-\beta) - \cos(2x + \alpha + \beta)}{\cos(\alpha-\beta) + \cos(2x + \alpha + \beta)}$.
We need $f(x; \alpha, \beta) = const$, which is equivalent to $f'(x; \alpha, \beta) = 0, \forall x$.
$f'(x; \alpha, \beta) = \frac{4\sin(2x+\alpha + \beta)\cos(\alpha - \beta)}{\left(\cos(\alpha-\beta) + \cos(2x + \alpha + \beta)\right)^2} = 0, \forall x \Leftrightarrow \cos(\alpha-\beta) = 0 \Leftrightarrow \alpha-\beta = \frac{\pi}{2} + \pi n, n = 0, \pm 1, \pm 2, \ldots$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3473892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Why is $\frac{1}{10}\left(\frac{9}{10}+\frac{8}{9}+\dots+\frac{1}{2}\right)\approx \frac{\sqrt{2}}{2}$? When doing a stochastics problem recently I noticed that
\begin{equation*}
\frac{1}{10}\sum\limits_{k=1}^9 \frac{k}{k+1}=\frac{1}{10}\left(\frac{9}{10}+\frac{8}{9}+\dots+\frac{1}{2}\right)=0.7071031746
\end{equation*}
while
\begin{equation*}
\frac{\sqrt{2}}{2}=0.7071067812\dots
\end{equation*}
These two quantities are amazingly close to each other (in fact, the discrepancy is only about $5\cdot 10^{-4}\%$) and thus, I wondered if there is any deeper reason behind this.
Is there some relationship between those two quantities that explains why they are so close to each other or is it just "pure luck"?
| A more detailed approach:
$ \frac{9}{10}+\frac{8}{9}+\frac{7}{8}+ . . . +\frac{1}{2}=(1-\frac{1}{10})+(1-\frac{1}{9})+(1-\frac{1}{8})+ . . . +(1-\frac{1}{2})=9-\big(\frac {1}{10}+\frac{1}{9}+\frac{1}{8}+ . . . + \frac {1}{2}\big )$
$s=\big(\frac {1}{10}+\frac{1}{9}+\frac{1}{8}+ . . . + \frac {1}{2}\big )
=\int ^{10}_2 \frac{dx}{x}= \big[ Ln (x)\big]^{10}_2$
⇒ $S=\frac{1}{10}\big(\frac{9}{10}+\frac{8}{9}+\frac{7}{8}+ . . . +\frac{1}{2}\big)=\frac{9-(Ln (10)- Ln (2)}{10}≈ 0.739056$
$\frac{\sqrt 2}{2}=0.707106$
$0.739056-0.707106=0.03195≈0.1-0.06805$
$0.06805≈ \gamma/10$
Where $\gamma$ is Euler constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3475524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Math theory behind the slope formula of Least Squares When studying linear regression, I see the slope formula of least squares like this: least squares. But I couldn't find the math theory behind it, i.e. how mathematicians came up with this equation, or why this equation gives out the best fitting line. Can you please explain it for me or point me to the right resource like a website or a book? Thanks.
| You can derive them by minimizing the sum of the squared errors between your "y" datavalues and the trendline predictions.
$$ SSE = \sum_{n=1}^N \Big( y_n - (ax_n+b) \Big)^2$$
The basic idea is to think of SSE as a parabolioid in the variables $a$ and $b$. The vertex of the paraboloid will tell us the values of $a$ and $b$ which minimize the SSE.
In the form written above it is not easy to discern what the vertex is. We will have to do a bit of algebra to get there.
$$ SSE = \sum_{n=1}^N \Big( y_n - (ax_n+b) \Big)^2$$
First expand the binomial
$$ = \sum_{n=1}^N \Big[ y_n^2 + (ax_n+b)^2 - 2 y_n (ax_n+b) \Big] $$
$$ = \sum_{n=1}^N \Big[ y_n^2 + a^2 x_n^2 +b^2 + 2ab
x_n - 2 a y_n x_n - 2 b y_n \Big]$$
Now distribute the sum
$$ = \sum_{n=1}^N y_n^2 + a^2 \sum_{n=1}^N x_n^2 + b^2 \sum_{n=1}^N 1 + 2ab
\sum_{n=1}^N x_n - 2 a \sum_{n=1}^N y_n x_n - 2 b \sum_{n=1}^N y_n $$
We will introduce the short hand notation,
$$ \overline{z} = \frac{1}{N} \sum_{n=1}^N z_n,$$
our expression then can be written as,
$$ SSE = N \overline{y^2} + a^2 N \overline{x^2} + b^2 N + 2ab
N\overline{x} - 2 a N \overline{xy} - 2 b N \overline{y}$$
$$ SSE = \Big( N \overline{x^2} \Big)a^2 + \Big( 2
N\overline{x} \Big) ab + \Big( N \Big) b^2 + \Big( - 2 N \overline{xy}\Big) a + \Big(- 2 N \overline{y}\Big) b + \Big(N \overline{y^2} \Big)$$
Now to find the vertex we compute the partial derivatives with respect to $a$ and $b$ setting them equal to $0$.
$$ \frac{\partial SSE}{\partial a} = \Big( 2N \overline{x^2} \Big)a + \Big( 2
N\overline{x} \Big) b + \Big( - 2 N \overline{xy}\Big) = 0 \qquad \text{(A)}$$
$$ \frac{\partial SSE}{\partial b} = \Big( 2
N\overline{x} \Big) a + \Big( 2N \Big) b + \Big(- 2 N \overline{y}\Big) = 0 \qquad \text{(B)} $$
We now need to solve this system of equations for $a$ and $b$. We can eliminate $b$ and solve for $a$ by multiplying equation B by $\overline{x}$ and subtracting it from equation A.
$$ A - B \overline{x}$$
$$ \Big( 2N \overline{x^2} \Big)a + \Big( 2
N\overline{x} \Big) b + \Big( - 2 N \overline{xy}\Big) -\Big( 2
N\overline{x}^2 \Big) a - \Big( 2N \overline{x} \Big) b - \Big(- 2 N \overline{y}\overline{x} \Big) = 0 $$
$$ \Big( 2N \overline{x^2} - 2N \overline{x}^2 \Big)a - \Big( 2 N \overline{xy} - 2N \overline{x} \overline{y} \Big) = 0 $$
$$ \Big( \overline{x^2} - \overline{x}^2 \Big)a = \Big( \overline{xy} - \overline{x} \overline{y} \Big) $$
$$ a = \frac{\overline{xy} - \overline{x} \overline{y} }{\overline{x^2} - \overline{x}^2 } $$
$$a = \frac{ \frac{1}{N} \sum x_n y_n -
\frac{1}{N^2} \sum x_n \sum y_n}{ \frac{1}{N} \sum x_n^2 - \frac{1}{N}^2 \Big(\sum x_n \Big)^2 }$$
$$ \boxed{ a = \frac{ N \sum x_n y_n -
\sum x_n \sum y_n}{ N \sum x_n^2 - \Big(\sum x_n \Big)^2 }} $$
We can eliminate $a$ from the equations by taking the difference of equation A times $\bar{x}$ and B times $\bar{x^2}$.
$$ A \overline{x} - B \overline{x}^2 $$
$$ Big( 2N \overline{x^2}\overline{x} \Big)a + \Big( 2
N\overline{x}^2 \Big) b + \Big( - 2 N \overline{xy} \overline{x} \Big) - \Big( 2
N\overline{x} \overline{x^2} \Big) a - \Big( 2N
\overline{x^2} \Big) b - \Big(- 2 N \overline{y} \overline{x^2} \Big) = 0 $$
$$ \Big( 2
N\overline{x}^2 - 2N
\overline{x^2} \Big) b - \Big(2 N \overline{xy} \overline{x} - 2 N \overline{y} \overline{x^2} \Big) = 0 $$
$$ \Big( 2
N\overline{x}^2 - 2N
\overline{x^2} \Big) b = \Big(2 N \overline{xy} \overline{x} - 2 N \overline{y} \overline{x^2} \Big) $$
$$ b = \frac{ 2 N \overline{xy} \overline{x} - 2 N \overline{y} \overline{x^2} }{2
N\overline{x}^2 - 2N
\overline{x^2}} $$
$$ b = \frac{ \overline{xy} \overline{x} - \overline{y} \overline{x^2} }{ \overline{x}^2 -
\overline{x^2}} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $2 · 1! + 5 · 2! + 10 · 3! + . . . + (n ^2 + 1)n! = n(n + 1)!$ Show that
$$2 · 1! + 5 · 2! + 10 · 3! + . . . + (n
^2 + 1)n! = n(n + 1)!$$
I got stuck after the n+1 part.
$$((n+1)^2 + 1)(n+1)! = (n+1)(n+2)!.$$
I'm not sure how to proceed from this part.
| You want to prove that
$$\sum_{k=1}^n(k^2+1)k! = n(n+1)!.$$
First case: if $n = 1$, then $(1^2+1)1! = 2 = 1 \cdot 2 = n(n+1)!$.
Then by induction, if the formula holds for $n-1$ you find:
$$\sum_{k=1}^n(k^2+1)k! = \sum_{k=1}^{n-1}(k^2+1)k! + (n^2+1)n! = (n-1)(n-1+1)! + (n^2+1)n! = (n-1)n! + (n^2+1)n! = (n+n^2)n! = n(n+1)n! = n(n+1)!$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3475801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find factor $f(x)$ conmmon to two quartic equation Let $f(x) = x^2 + bx + c$, where $b, c ∈ R$. If f(x) is a factor of both $x^4 + 6x^2 + 25$ and $3x^4 + 4x^2 + 28x + 5$, then find $f(x)$
My approach ,on dividing both quartic equation by $f(x)$ remainder is zero, but not getting the answer
| Hint :
$$\begin{align}x^4 + 6x^2 + 25 &= x^4 + 10x^2 + 25 -4x^2 \\ &= (x^2+5)^2 - (2x)^2\\&=(x^2+2x+5)(x^2-2x+5) \\ &\end{align}$$
$$\begin{align}3x^4 + 4x^2 + 28x + 5 &= (x^2-2x+5)(3x^2+6x+1)\end{align} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\left(1+\sqrt2\right)^{2011}=a+b\sqrt{2}$, for integers $a$ and $b$, then what is $\left(1-\sqrt2\right)^{2010}$ expressed using $a$ and $b$?
Being $ a $ and $ b $ integers such that $\left(1+\sqrt{2}\right)^{2011} =a+b\sqrt{2}, \left(1-\sqrt{2}\right)^{2010}$ equals:
a) $a+2b+(a-b)\sqrt{2}$
b) $a-2b+(a-b)\sqrt{2}$
c) $a+2b+(b-a)\sqrt{2}$
d) $2b-a+(b-a)\sqrt{2}$
e) $a+2b-(a+b)\sqrt{2}$
Solution:
Not going with any of the alternatives
| Let $x= \left(1-\sqrt{2}\right)^{2010}$
Then $$x(a+b\sqrt{2}) = (1-2)^{2010}\left(1+\sqrt{2}\right)=1+\sqrt{2}$$
So $$x= {1+\sqrt{2}\over a+b\sqrt{2}} = {(1+\sqrt{2})( a-b\sqrt{2})\over a^2-2b^2}$$
Now $\left(1-\sqrt{2}\right)^{2011} = a-b\sqrt{2}$ so $$a^2-2b^2 = (1-\sqrt{2})^{2011} \cdot \left(1+\sqrt{2}\right)^{2011} =-1$$
So $$x=(1+\sqrt{2})( -a+b\sqrt{2}) =...$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 2
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Definite integral of $1/(5+4\cos x)$ over $2$ periods Question:
$$\int_0^{4\pi}\frac{dx}{5+4\cos x} $$
My approach:
First I calculated the antiderivative as follows:
Using: $\cos\theta= \frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$ we have:
$\int\frac{dx}{5+4\cos x}=\int\frac{dx}{5+4\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}=\int\frac{1+\tan^2\frac{x}{2}}{5+5\tan^2\frac{x}{2}+4-4\tan^2\frac{x}{2}}dx=\int\frac{\frac{1}{\cos^2 \frac{x}{2}}}{3^2+\tan^2\frac{x}{2}}dx$
Using substitution we have:
$u=\tan\frac{x}{2}$
$du=\frac{1}{2}\frac{1}{\cos^2\frac{x}{2}}dx$
$2\int\frac{\frac{1}{2}\frac{1}{\cos^2 \frac{x}{2}}}{3^2+\tan^2\frac{x}{2}}dx=2\int\frac{du}{3^2+u^2}=\frac{2}{3}\arctan\frac{u}{3}+\mathscr{C}=\frac{2}{3}\arctan\frac{\tan\frac{x}{2}}{3}+ \mathscr{C}$
Now we can calculate the definite integral as follows:
$\int_0^{4\pi}\frac{dx}{5+4\cos x} = \frac{2}{3}\arctan\frac{\tan\frac{x}{2}}{3}\bigl|_0^{4\pi}=\frac{2}{3}(\arctan\frac{\tan\frac{4\pi}{2}}{3}-\arctan\frac{\tan\frac{0}{2}}{3})=0$
The result I get is $0$ but the correct one is $\frac{4\pi}{3}$. Can someone explain me why?
Here it shows that the correct answer is $\frac{4\pi}{3}$.
| You have everything right up to the limit taking,
$$I=\int_0^{4\pi}\frac{dx}{5+4\cos x} = \frac{2}{3}\arctan\frac{\tan\frac{x}{2}}{3}\bigl|_0^{4\pi}$$
Note that the anti derivative function on the RHS is discontinuous at $\pi$ and $3\pi$. So, the limits have to be broken into three intervals,
$$\bigl|_0^{4\pi} = \bigl|_0^{\pi}+\bigl|_\pi^{3\pi} +\bigl|_{3\pi}^{4\pi} $$
which leads to the result
$$I = \frac23 (\frac\pi2+\pi+\frac\pi2)=\frac43\pi$$
as expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Let $A,B$ be squared matrices. Given $A=I-AB$, Prove: $B^3=0$ if and only if $A=I-B+B^2$
Let $A,B$ be squared matrices.
Given $A=I-AB$,
Prove: $B^3=0 \iff A=I-B+B^2$
The question has 3 sections:
*
*Given $A=I-AB$, Prove that $A$ is invertible and that $BA=AB$.
*Prove that if $B$ is a symmetric matrix, then $A$ is symmetric
*Prove: $A=I-B+B^2$ if and only if $B^3=0$.
I proved the first two sections, and did the first direction of the third section, Thus we suppose that $A=I-B+B^2$, and I proved that $B^3=0$ as follows:
Suppose $A=I-B+B^2$, then $A=I-AB=I-B+B^2$
$$-AB=-B+B^2$$
$$\text{ we'll substitute $A$ by $I-B+B^2$}$$
$$-(I-B+B^2)B= -B+B^2$$
$$-B+B^2-B^3 =-B+B^2 \Longrightarrow B^3 =0$$
As wished.
Now let $B^3=0$. Prove: $A=I-B+B^2$.
$$AB = I-A$$
We'll multiply by $B$ on the left side:
$$AB^2 = B-AB$$
$$AB^2+AB=B$$
$$AB(B+I)=B$$
and from this point I'm stuck.
| For the second case, just do iteration of the defintion of $A$.
$A=I-AB=I-(I-AB)B=I-B+AB^2=I-B+(I-AB)B^2$
$A=I-B+B^2-AB^3$
But $B^3=O$, then you have your definition of $A$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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If two side of a Pythagorean triangle are primes then the third side has at least $4$ distinct prime factors.
Conjecture: If two side of a Pythagorean triangle are primes $> 19$ then the third side has at least $4$ distinct prime factors.
E.g. $421^2 = 420^2 + 29^2$. Here $29$ and $421$ are both primes and the third side $420$ has four distinct prime factors $2,3,5,7$.
I can show that the third side is divisible by $60$ which contributes the three prime factors $2,3$ and $5$. How do we show that there is always a fourth prime?
Update: Verified this for primes upto $2 \times 10^9$
| If $a^2+b^2=c^2$ where $a,b,c$ are relatively prime positive integers, with $b$ even, then there are integers $x,y>0$ such that $a=x^2-y^2$, $b=2xy$, and $c=x^2+y^2$. In your case, $a$ and $c$ must be primes $>19$. In particular, for $a=(x-y)(x+y)$ to be prime, we must have $x-y=1$ so we have $a=2y+1$, $b=2y(y+1)$, and $c=2y^2+2y+1$. The claim is then that if $a$ and $c$ are primes greater than $19$ then $b$ has at least $4$ distinct prime factors.
Simple mod $3$ and mod $5$ considerations show that $b$ must be divisible by $3$ and $5$. If $b$ has at most $3$ distinct prime factors, then, both $y$ and $y+1$ have no prime factors greater than $5$. It can be shown that there are only ten positive integers $y$ such that both $y$ and $y+1$ have no prime factors greater than $5$, namely $y=1,2,3,4,5,8,9,15,24,80$. Testing all of these values of $y$, none yield values of $a$ and $c$ that are both primes greater than $19$, so there are no examples where $b$ has only three distinct prime factors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How to find an ellipse equation with non-symmetrical foci values? I have tried a lot but i do not seem to find an equation.
Question is the following:
Find an equation of the specified curve, that is, of the locus of points the sum of whose distances from the points $(2,3)$ and $(4,1)$ is $8$.
I first tried to translate and rotate the axis to see the ellipse like its center lies on the origin $(0,0)$. It seemed to be tough. I know I can use the general equation with center $(h,k)$ but then there will not be any rotation though the major axis is rotated.
| Let the distances to the foci be $d_1,d_2$ and their sum be $d$. We have
$$d=d_1+d_2$$
Squaring,
$$d^2-d_1^2-d_2^2=2d_1d_2.$$
Squaring a second time,
$$d^4-2d^2(d_1^2+d_2^2)+(d_1^2+d_2^2)^2=4d_1^2d_2^2,$$
or
$$d^4-2d^2(d_1^2+d_2^2)+(d_1^2-d_2^2)^2=0.$$
Expanding the distances, we have
$$d_1^2+d_2^2=2x^2+2y^2-2(x_1+x_2)x-2(y_1+y_2)y+x_1^2+x_2^2+y_1^2+y_2^2\\
=2x^2+2y^2-2\sigma_xx-2\sigma_yy+\Sigma$$
and
$$d_1^2-d_2^2=-2(x_1-x_2)x-2(y_1-y_2)y+x_1^2-x_2^2+y_1^2-y_2^2\\
=-2\delta_xx-2\delta_yy+\Delta,$$
and
$$(d_1^2-d_2^2)^2=4\delta_x^2x^2+8\delta_x\delta_yxy+4\delta_y^2y^2-4\delta_x\Delta x-4\delta_y\Delta y+\Delta^2.$$
Finally, we get the conic
$$4(\delta_x^2-d^2)x^2+8\delta_x\delta_yxy+4(\delta_y^2-d^2)y^2-4(\delta_x\Delta+d^2\sigma_x) x-4(\delta_y\Delta+d^2\sigma_y)y\\
+\Delta^2-2d^2\Sigma+d^4=0.$$
Check:
With the usual placement of the foci, $(f,0)$ and $(-f,0)$, we have $\sigma_x=\sigma_y=\delta_y=\Delta=0$, $\delta_x=2f$ and $\Sigma=2f^2$. The equation reduces to
$$4(d^2-4f^2)x^2+4d^2y^2=d^2(d^2-4f^2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\sum_{i = 0}^{n - 1}\lfloor\sqrt{a + \frac{i}{n}}\rfloor = n\lfloor a \rfloor + \lfloor n(a - \lfloor \sqrt a \rfloor) \rfloor$.
Prove that for natural $n$ and all values $a$ such that $\lfloor a \rfloor + 1$ is a perfect square, $$\large \sum_{i = 0}^{n - 1}\left\lfloor\sqrt{a + \frac{i}{n}}\right\rfloor = n\lfloor a \rfloor + \lfloor n(a - \lfloor \sqrt a \rfloor) \rfloor$$
We have that $$\left\lfloor\sqrt{a + \frac{i}{n}}\right\rfloor \ge \lfloor \sqrt a \rfloor, i = \overline{0, n - 1} \implies \sum_{i = 0}^{n - 1}\left\lfloor\sqrt{a + \frac{i}{n}}\right\rfloor \ge n\lfloor \sqrt a \rfloor$$
It can also be seen that $$\sum_{i = 0}^{n - 1}\left\lfloor\sqrt{a + \frac{i}{n}}\right\rfloor > n\lfloor \sqrt a \rfloor \implies \left\lfloor\sqrt{a + \frac{n - 1}{n}}\right\rfloor > \lfloor \sqrt a \rfloor$$
Which means that there exists natural $k$ such that $$\sqrt a < k \le \sqrt{a + \frac{n - 1}{n}} \implies k^2 + \frac{1}{n} \le a + 1 < k^2 + 1 \implies \lfloor
a \rfloor + 1 = k^2$$
Then I'm sure what to do next... But actually, there's a solution to this problem which I have provided. It would be greatly appreciated with the provision that you have feedback about my solution.
| Let $\lfloor n(a - \lfloor \sqrt a \rfloor) \rfloor = r$ and $\lfloor a \rfloor + 1 = k^2$, we have that $$k^2 + \frac{r}{n} \le a + 1 < k^2 + \frac{r + 1}{n} \implies \sqrt{a + \frac{n - r - 1}{n}} < k \le \sqrt{a + \frac{n - r}{n}}$$
Deducing that $$\left\lfloor\sqrt{a + \frac{i}{n}}\right\rfloor = \left\{ \begin{align} k - 1 &\text{ where } j < n - r\\ k &\text{ where } n - r \le j \end{align} \right.$$
This implies that $$\sum_{i = 0}^{n - 1}\left\lfloor\sqrt{a + \frac{i}{n}}\right\rfloor = (n - r)(k - 1) + rk = n(k - 1) + r$$
Since $k - 1 = \lfloor\sqrt {a} \rfloor$, $$\sum_{i = 0}^{n - 1}\left\lfloor\sqrt{a + \frac{i}{n}}\right\rfloor = n\lfloor a \rfloor + \lfloor n(a - \lfloor \sqrt a \rfloor) \rfloor$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve it in simple ways, Find $f'(0)$ $f(x) = \dfrac{\left(x-3\right)\left(x-2\right)\left(x-1\right)x}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}$
Find $f'(0)$
By apply the Quotient Rule $\frac{f'g - g'f}{g^2} $, I can find the answer but it's too long.
Is there any other methods to simplify the steps?
| Take $\log$:
\begin{align*}
&\log f(x)\\
&=\log(x-3)+\log(x-2)+\log(x-1)+\log x-\log(x+1)-\log(x+2)-\log(x+3)\\
&\dfrac{f'(x)}{f(x)}\\
&=\dfrac{1}{x-3}+\dfrac{1}{x-2}+\dfrac{1}{x-1}+\dfrac{1}{x}-\dfrac{1}{x+1}-\dfrac{1}{x+2}-\dfrac{1}{x+3}\\
&f'(x)\\
&=f(x)\left(\dfrac{1}{x-3}+\dfrac{1}{x-2}+\dfrac{1}{x-1}+\dfrac{1}{x}-\dfrac{1}{x+1}-\dfrac{1}{x+2}-\dfrac{1}{x+3}\right)
\end{align*}
and then plugging in $x=0$. But first cancel the same factor $x$ in $f(x)\cdot\dfrac{1}{x}$ and then plugging in $x=0$. For the other factors, they are zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $a^2+b^2+c^2$ is a square when $\frac{1}{a}+\frac{1}{b} = \frac{1}{c}$ and $a,b,c\in\mathbb{Q}$
Knowing that $$\dfrac1a+ \dfrac1b=\dfrac1c$$ Prove that $a^2+b^2+c^2$ is a square, where $a,b,c\not=0$ are rational numbers.
It can probably be solved by a quick factoring trick, but I really can’t figure it out.
| Let $\dfrac1a=A$ etc.
$\implies A+B=C$
$$a^2+b^2+c^2=\dfrac{(AB)^2+(BC)^2+(CA)^2}{(ABC)^2}$$
$$(AB)^2+(BC)^2+(CA)^2$$
$$=(AB)^2+C^2(A^2+B^2)$$
$$=(AB)^2+(A+B)^2(A^2+B^2)\text{ if }C=\pm(A+B)$$
$$=(AB)^2+(A^2+B^2+2AB)(A^2+B^2)$$
$$=(AB)^2+(A^2+B^2)^2+2\cdot AB\cdot(A^2+B^2)=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof by Induction $2\cdot 7^n + 3\cdot 5^n - 5$ is multiple of $24$ Proof by induction that $2\cdot 7^n + 3\cdot 5^n - 5$ is a multiple of $24$.
I tried solving but got stuck.
Show that it is true for n=1
$$2\cdot 7^1 + 3\cdot 5^1 - 5 = 14 + 15 - 5 = 24$$
Assume it true for $n = k$
$$2\cdot 7^k + 3\cdot 5^k - 5 = 24g$$
Show it is true for $n= k + 1$
$$2\cdot 7^{k+1} + 3\cdot 5^{k+1} - 5$$
is a multiple of 24
$$(2\cdot 7)(2\cdot 7^k) + (3\cdot 5^k)(3\cdot 5) - 5$$
$$2\cdot7(24g + 5)3\cdot 5 - 5$$
I'm stuck and don't know how to proceed
| First note that $5^k-5$ is divisible by $4$ for all positive integers $k$.
Let $P(n)$ be the proposition that $2\times7^n+3\times5^n-5$ is a multiple of $24$.
$P(1)$ is the proposition that $2\times7+3\times5-5$ is a multiple of $24$ which is true
Suppose $P(k)$ is true and so $2\times7^k+3\times5^k-5$ is a multiple of $24$.
Then $2\times7^{k+1}+3\times5^{k+1}-5=14\times 7^k+15\times5^k-5=7\times (2\times7^k+3\times5^k-5)-6\times(5^k-5)$
is divisible by $24$. Therefore $P(k+1)$ is true and we are finished.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Evaluate $\int \cos 2\theta \ln\left(\frac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)d\theta$ $$\int \cos 2\theta \ln\left(\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)d\theta$$
My attempt is as follows:-
$$\ln\left(\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)=t\tag{1}$$
$$\dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\cdot\dfrac{\left(\cos\theta-\sin\theta\right)^2-(-\sin\theta-\cos\theta)(\cos\theta+\sin\theta)}{(\cos\theta-\sin\theta)^2}=\dfrac{dt}{d\theta}$$
$$\dfrac{2}{\cos2\theta}=\dfrac{dt}{d\theta}$$
Let's calculate $\cos2\theta$ from equation $1$
$$\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}=e^t$$
$$\dfrac{1+\tan\theta}{1-\tan\theta}=e^t$$
Applying componendo and dividendo
$$\dfrac{2}{2\tan\theta}=\dfrac{e^t+1}{e^t-1}$$
$$\dfrac{e^t-1}{e^t+1}=\tan\theta$$
$$\cos2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}$$
$$\cos2\theta=\dfrac{(e^t+1)^2-(e^t-1)^2}{(e^t+1)^2+(e^t-1)^2}$$
$$\cos2\theta=\dfrac{4e^t}{2(e^{2t}+1)}$$
$$\cos2\theta=\dfrac{2e^t}{e^{2t}+1}\tag{2}$$
So integral will be
$$\dfrac{1}{2}\cdot\int \left(\dfrac{2e^t}{e^{2t}+1}\right)^2dt$$
$$\dfrac{1}{2}\cdot\int \dfrac{4e^{2t}}{(1+e^{2t})^2}$$
$$e^{2t}+1=y$$
$$2e^{2t}=\dfrac{dy}{dt}$$
$$2e^{2t}dt=dy
$$\int \dfrac{dy}{y^2}$$
$$-\dfrac{1}{y}+C$$
$$-\dfrac{1}{1+e^{2t}}+C$$
$$-\dfrac{1}{1+e^{\ln\left(\dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right)^2}}+C$$
$$-\dfrac{1}{1+\dfrac{1+\sin2\theta}{1-\sin2\theta}}+C$$
$$-\dfrac{1-\sin2\theta}{2}+C$$
$$\dfrac{\sin2\theta}{2}+C'$$
And this should be actually wrong because if we differentiate the result, it will give $\cos2\theta$, but integrand is $\cos 2\theta \ln\left(\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)$
What am I missing here, checked multiple times, but not able to get the mistake. Any directions?
| Simplify the argument of the logarithm thus:
$$\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}=\frac{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)}{(\cos\theta-\sin\theta)^2}=\frac{\cos^2\theta-\sin^2\theta}{1-2\sin\theta\cos\theta}=\frac{\cos2\theta}{1-\sin2\theta}.$$
Then it follows that $$\log\left(\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right)=\log\left(\frac{\cos2\theta}{1-\sin2\theta}\right)=\log\cos2\theta-\log(1-\sin2\theta)=\log\sqrt{1-\sin^22\theta}-\log(1-\sin2\theta)=\frac12\log[(1-\sin2\theta)(1+\sin2\theta)]-\log(1-\sin2\theta)=\frac12\log(1-\sin2\theta)+\frac12\log(1+\sin2\theta)-\log(1-\sin2\theta)=\frac12\log(1+\sin2\theta)-\frac12\log(1-\sin2\theta).$$
Thus, we may write the differential as $$\frac14\log(1+\sin2\theta)\,\mathrm d(1+\sin2\theta)+\frac14\log(1-\sin2\theta)\,\mathrm d(1-\sin2\theta),$$ which is easily done if we recall the result that the $\int\log y\,\mathrm dy=y\log y-y+C.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
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Evaluate $\int \frac{1}{\cot \frac{x}{2}\cdot\cot\frac{x}{3}\cdot\cot\frac{x}{6}}\text{d}x$ $$\int \dfrac{1}{\cot \dfrac{x}{2}\cdot\cot\dfrac{x}{3}\cdot\cot\dfrac{x}{6}}dx$$
My multiple attempts are as follows:-
Attempt $1$:
$$\int \tan\dfrac{x}{2}\cdot\tan\dfrac{x}{3}\cdot\tan\dfrac{x}{6}dx$$
$$\int \tan\dfrac{x}{2}\cdot\tan\dfrac{x}{3}\cdot\tan\left(\dfrac{x}{2}-\dfrac{x}{3}\right)dx$$
$$\int \tan\dfrac{x}{2}\cdot\tan\dfrac{x}{3}\cdot\dfrac{\tan\dfrac{x}{2}-\dfrac{x}{3}}{1+\tan\dfrac{x}{2}\cdot\tan\dfrac{x}{3}}dx$$
$$\int \dfrac{\tan^2\dfrac{x}{2}\cdot\tan\dfrac{x}{3}\cdot\cos\dfrac{x}{2}\cdot\cos\dfrac{x}{3}}{\cos\dfrac{x}{6}}-\dfrac{\tan\dfrac{x}{2}\cdot\tan^2\dfrac{x}{3}\cdot\cos\dfrac{x}{2}\cdot\cos\dfrac{x}{3}}{\cos\dfrac{x}{6}}dx$$
$$\int \dfrac{\sin^2\dfrac{x}{2}\cdot\sin\dfrac{x}{3}}{\cos\dfrac{x}{2}\cdot\cos\dfrac{x}{6}}-\dfrac{\sin\dfrac{x}{2}\cdot\sin^2\dfrac{x}{3}}{\cos\dfrac{x}{3}\cos\dfrac{x}{6}}dx$$
$$2\left(\int \dfrac{\sin^2\dfrac{x}{2}\cdot\sin\dfrac{x}{6}}{\cos\dfrac{x}{2}}-\dfrac{2\sin\dfrac{x}{2}\cdot\sin^2\dfrac{x}{6}\cos\dfrac{x}{6}}{\cos\dfrac{x}{3}}\right)dx$$
This doesn't seem to be going anywhere.
Attempt $2$:
$$\int \dfrac{\sin\dfrac{x}{2}\cdot\sin\dfrac{x}{3}\cdot\sin\dfrac{x}{6}}{\cos\dfrac{x}{2}\cdot\cos\dfrac{x}{3}\cdot\cos\dfrac{x}{6}}dx$$
$$\int \dfrac{\cos\dfrac{x}{6}\cdot\sin\dfrac{x}{6}-\cos\dfrac{5x}{6}\cdot\sin\dfrac{x}{6}}{\cos^2\dfrac{x}{6}+\cos\dfrac{x}{6}\cdot\cos\dfrac{5x}{6}}dx$$
$$\int \dfrac{\sin\dfrac{x}{3}-\sin x+\sin\dfrac{2x}{3}}{1+\cos\dfrac{x}{3}+\cos x+\cos \dfrac{2x}{3}}dx$$
As we know $\cos \alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\cdots+\cos(\alpha+(n-1)\beta)=\dfrac{\sin\dfrac{n\beta}{2}}{\sin\dfrac{\beta}{2}}\cos\left(\dfrac{\alpha+\alpha+(n-1)\beta}{2}\right)$
$$\int \dfrac{\sin\dfrac{x}{3}-\sin x+\sin\dfrac{2x}{3}}{\cos(0)+\cos\left(0+\dfrac{x}{3}\right)+\cos\left(0+\dfrac{2x}{3}\right)+\cos\left(0+\dfrac{3x}{3}\right)}dx$$
$$\int \dfrac{\sin\dfrac{x}{3}-\sin x+\sin\dfrac{2x}{3}}{\dfrac{\sin\dfrac{4x}{6}}{\sin\dfrac{x}{6}}\cdot\cos\dfrac{x}{2}}dx$$
$$\int \dfrac{\sin\dfrac{x}{3}+\sin x+\sin\dfrac{2x}{3}-2\sin x}{\dfrac{\sin\dfrac{4x}{6}}{\sin\dfrac{x}{6}}\cdot\cos\dfrac{x}{2}}dx$$
$$\int \dfrac{\dfrac{\sin\left(\dfrac{3x}{6}\right)}{\sin\dfrac{x}{6}}\cdot\sin\dfrac{2x}{3}-2\sin x}{\dfrac{\sin\dfrac{4x}{6}}{\sin\dfrac{x}{6}}\cdot\cos\dfrac{x}{2}}dx$$
$$\int \tan\dfrac{x}{2}dx-\int \dfrac{4\sin\dfrac{x}{2}\sin\dfrac{x}{6}}{\sin\dfrac{2x}{3}}dx$$
$$\int \tan\dfrac{x}{2}dx-2\int \dfrac{\cos\dfrac{x}{3}-\cos\dfrac{2x}{3}}{\sin\dfrac{2x}{3}}dx$$
$$\int \tan\dfrac{x}{2}dx-\int \mathrm{cosec}\dfrac{x}{3}dx+2\int \cot\dfrac{2x}{3}dx$$
$$2\ln\left|\sec \dfrac{x}{2}\right|-3\ln\left|\mathrm{cosec}\dfrac{x}{3}-\cot\dfrac{x}{3}\right|+3\ln\left|\sin\dfrac{2x}{3}\right|+C$$
But this got too long,any short and better approach.
| Here is the step-by-step solution:
| {
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"timestamp": "2023-03-29T00:00:00",
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Find minimum value $E(x, y) = x^2 + y^2 -6x -10y$, where $x^2 + y^2 - 2y \le 0$ I am given the expression:
$$E(x, y) = x^2 + y^2 -6x -10y$$
And I have to find the minimum value of $E(x, y)$ for $(x, y) \in D$ where:
$$D = \{ (x,y) \in \mathbb{R}^2 \hspace{0.25cm} | \hspace{0.25cm} x^2 + y^2 -2y \le 0 \}$$
Doing the following:
$$\hspace{5.8cm} x^2+y^2-2y \le 0 \hspace{5cm} |+1$$
$$x^2+(y-1)^2 \le 1$$
So I have to find the minimum value of $E(x,y)$ where $(x, y)$ is from the circle $x^2 + (y-1)^2 \le 1$.
I don't see how I should approach this.
| Note that $E(r) = r^2-34$, where $r$, given by the circle
$r^2= (x-3)^2 + (y-5)^2$, is its radius and the circle $x^2+(y-1)^2=1$ has radius 1.
So, $E(r)$ is minimum if $r$ is the smallest, which is the case when the two circles just touch, i.e. the sum of their radii is equal to the distance of their centers, $r+1=\sqrt{3^2+4^2}=5$. Thus, the minimum $E$ is $(5-1)^2-34$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3493525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int \sin^{-1}\frac{2x}{1+x^2}dx$ $\int \sin^{-1}\dfrac{2x}{1+x^2}dx$
My attempt is as follows:-
$$x=\tan\theta$$
$$dx=\sec^2\theta d\theta$$
$$\int \sin^{-1}(\sin2\theta) \cdot\sec^2\theta d\theta$$
So here should we make cases on the basis of values of $\theta$ or can we write $\sin^{-1}(\sin2\theta)$ as $2\theta$?
| I think it may be easier to apply integration by parts on the original integral (i.e. differentiate the integrand and integrate $1$) giving
\begin{align}
\int\arcsin{\left(\frac{2x}{1+x^2}\right)}\mathrm{d}x
&=x\cdot\arcsin{\left(\frac{2x}{1+x^2}\right)}-\int x\cdot\frac{2\,\text{sign}(1-x^2)}{1+x^2}\mathrm{d}x\\
&=x\cdot\arcsin{\left(\frac{2x}{1+x^2}\right)}+\text{sign}(x^2-1)\cdot\int\frac{2x}{1+x^2}\mathrm{d}x\\
&=x\cdot\arcsin{\left(\frac{2x}{1+x^2}\right)}+\text{sign}(x^2-1)\cdot\ln{(1+x^2)}+C\\
\end{align}
It turns out that if we choose $C=-\ln{(2)}\cdot\text{sign}{(x^2-1)}+C'$ then we get a continuous antiderivative namely
$$\int\arcsin{\left(\frac{2x}{1+x^2}\right)}\mathrm{d}x=x\cdot\arcsin{\left(\frac{2x}{1+x^2}\right)}+\text{sign}(x^2-1)\cdot(\ln{(1+x^2)}-\ln{(2)})+C'$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Solve a system of equations for positive ($x$, $y$, $z$) I arrived at the system of equations
$$(3+2z)x=3(1-xy)$$
$$(3+7z)y=4(1-xy)$$
$$3y-x+z(y-x)=2xy$$
while working on a geometric problem, with $x$, $y$ and $z$ representing segment lengths of a quadrilateral configuration. The system is inherently quartic. Given its non-standard appearance, I am not sure yet how to rearrange the pieces, which would lead to the solutions smoothly. A math tool I have spills out $(\frac12, \frac13, 1)$, without indicating how.
Before resorting to brute-force substitution, I would like to have a shot at anyone who may have expertise, or just happens to be conversant in such systems.
| \begin{align}
\begin{cases}
(3+2z)x&=3(1-xy)
,\\
(3+7z)y&=4(1-xy)
,\\
3y-x+z(y-x)&=2xy
.
\end{cases}
\tag{1}\label{1}
\end{align}
Rearranging the system \eqref{1} as
\begin{align}
\begin{cases}
\phantom{-}3xy+2zx &= 3-3x
,\\
\phantom{-}4xy+7yz &= 4-3y
,\\
-2xy+yz-zx &=x-3y,
\end{cases}
\tag{2}\label{2}
\end{align}
we can express $xy,yz,zx$ in terms of $x,y$:
\begin{align}
15xy &= \phantom{-}7x+36y-13
\tag{3}\label{3}
,\\
15yz &= -4x-27y+16
\tag{4}\label{4}
,\\
5zx &= -11x-18y+14
\tag{5}\label{5}
.
\end{align}
Equation \eqref{3} gives $y$ in terms of $x$:
\begin{align}
y &= \frac{7x-13}{3(5x-12)}
\tag{6}\label{6}
.
\end{align}
$z$ in terms of $x$
can be found from
equations \eqref{4} and \eqref{6}:
\begin{align}
z &= -\frac{4x^2-13x+15}{7x-13}
\tag{7}\label{7}
.
\end{align}
Equations \eqref{5}, \eqref{6} and \eqref{7} combined
result in
\begin{align}
-\frac{2(2x-1)(5x^3-45x^2+127x-117)}{(7x-13)(5x-12)}
&=0
\tag{8}\label{8}
,
\end{align}
which gives two real solutions for $x$:
\begin{align}
x_1&=\tfrac12
,\\
x_2&=
3+\tfrac1{15}\sqrt[3]{2025+15\sqrt{10545}}
+\frac 8{\sqrt[3]{2025+15\sqrt{10545}}}
\approx 4.54212431
.
\end{align}
Corresponding pairs of $y$ and $z$ are
\begin{align}
y_1&=\tfrac13
,\\
y_2&\approx 0.58492933
,\\
z_1&=1
,\\
z_2&\approx -2.047152
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Sequence of functions inequality I need some help to prove that $f_n(x)\leq 1$ , where $f_n:[0,1]\to\mathbb{R}, f_n(x)=\frac{1+nx^2}{(1+x^2)^n}, n\in\mathbb{N}$.
I can write $(1+x^2)^n=\binom{0}{n}x^n+\binom{1}{n}x^{n-1}y+\binom{2}{n}x^{n-2}y^2+...+\binom{n-2}{n}x^2y^{n-2}+\binom{n-1}{n}xy^{n-1}+\binom{n}{n}y^n=\\x^n+nx^{n-1}y+\frac{n(n-1)}{2!}x^{n-2}y^2+...+\frac{n(n-1)}{2!}x^2y^{n-2}+nxy^{n-1}+y^n$
$\\$
I unsuccessfully tried to find $1+nx^2$ in $(1+x^2)^n$ to show that $1+nx^2\leq(1+x^2)^n$.
| You did not substitute $1$ and $x^2$ for $x$ and $y$ in the binomial theorem. The correct expansion is
$$
(1+x^2)^n = \sum_{k=0}^n \binom nk 1^{n-k}(x^2)^k = \sum_{k=0}^n \binom nk x^{2k} \, .
$$
It follows (by taking only the first two terms of the sum) that
$$
(1+x^2)^n
\ge \binom n0 x^0 + \binom n1 x^2 = 1 + n x^2 \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
proof-check: $\frac{x+a}{x^2+1}$ always has $3$ inflection points I am asked to show that the function $f(x) = \frac{x+a}{x^2+1}$ always has $3$ inflection points for any $a \in \mathbb{R}$.
The second derivative of $f$ turns out to be
$$f''(x) = \frac{2(x^3 + 3ax^2 - 3x - a)}{(x^2 + 1)^3}.$$
The discriminant of that third degree polynomial is
$$108a^4 + 216a^2 + 108 > 0$$
so the polynomial always has $3$ real distinct roots.
Now, a 3rd degree polynomial with 3 distinct roots changes sign at it's roots, so the whole $f''$ must change sign at the roots of the polynomial, so the $3$ roots of the polynomial are indeed inflection points.
Is this enough?
| Also since $g(x)=x^3+3ax^2-3x-a \implies g'(x)=3x^2+ 6ax-3=0 \implies x_{1,2}=-a\pm \sqrt{a^2+1}$ has one minimum at $x=x_1$ maximum at $x=x_2$ as $g''(x_1)=6 \sqrt{1+a^2}>0$ and $g''(x_2)=-6 \sqrt{1+a^2}.$ Further, $g'(x_1)=6 \sqrt{1+a^2},$ $g(x_1)=-2(1+a^2)[\sqrt{1+a^2)}-a]<0$ and $g(x_2)=2(1+a^2)[\sqrt{1+a^2)}-a]>0.$
So as mim is negative and max is positive $g(x)=0$ has three distinct real roots.
Eventually $f(x)$ has three points of inflexion.
This is what @Simon Zhou seems to suggest
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3499979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Recognise factors for fractional exponents Hi is there a general rule to factorising these type of exponent fractions? The answer makes sense but I wouldn't have thought of it.
$\ x^{ \frac{2x}{1-2x} } - 2x^{ \frac{1}{1-2x} }$
Answer:
$\ x^{ \frac{2x}{1-2x} } (1- 2x$)
| It is a simple algebra: just factor out $x$ to any power you want. For example:
$$\ \color{red}{x^{ \frac{2x}{1-2x} }} - 2x^{ \frac{1}{1-2x} }=\ \color{red}{x^{ \frac{2x}{1-2x} }}\left(1 - 2x^{ \frac{1}{1-2x}-\frac{2x}{1-2x} }\right)=x^{\frac{2x}{1-2x}}\left(1-2x\right) \ \ \text{or}\\
\ x^{ \frac{2x}{1-2x} } - 2\color{blue}{x^{ \frac{1}{1-2x} }}=\ \color{blue}{x^{ \frac{1}{1-2x} }}\left(x^{\frac{2x}{1-2x}-\frac1{1-2x}} - 2\right)=x^{\frac{1}{1-2x}}\left(x^{-1}-2\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Problem solving homogenous second order DE, $x''-\frac{1}{t}x'+\frac{1}{t^2}x=0$ So, here is the DE:
$$x''-\frac{1}{t}x'+\frac{1}{t^2}x=0$$
I tried to solve it using a substitution of variables: $z=x' \Rightarrow x''=\frac{dx'}{t}=\frac{dz}{dt}=\frac{dz}{dx}\cdot \frac{dx}{dt}=\frac{dz}{dx}\cdot z$ Then DE can be rewritten as $$\frac{dz}{dx}z -\frac{1}{t}x'+\frac{1}{t^2}x=0$$ Then multiplying expression by $dx$ we get $$zdz-\frac{1}{t}x'dx+\frac{1}{t^2}xdx =0$$ Then taking an integrating the expression we get: $$\frac{z^2}{2}-\frac{1}{t}x+\frac{1}{2t^2}x^2=const. \; const. = k$$
$z^2=\frac{2x}{t}-\frac{x^2}{t^2}+k$ But here I already know that when I will take a square root of the right side of the expression and will try to integrate both sites, it won't give me any sensible result. So where is my mistake?
| $$x''-\frac{1}{t}x'+\frac{1}{t^2}x=0$$
$$\implies t^2x''-tx'+x=0\tag1$$
Take $~t=e^z\implies z=\ln t~$.
$~x'=\dfrac{dx}{dt}=\dfrac{dx}{dz}\cdot \dfrac{dz}{dt}\implies \dfrac{1}{t}\cdot \dfrac{dx}{dz}\implies tx'=\dfrac{dx}{dz}=Dx~,\qquad\text{where}\quad D\equiv \dfrac{d}{dz}~$.
$~x''=\dfrac{d^2x}{dt^2}=\dfrac{d}{dt}\left[\dfrac{dx}{dt}\right]=-\dfrac{1}{t^2}\cdot \dfrac{dx}{dz}+\dfrac{1}{t}\cdot \dfrac{d^2x}{dz^2}\cdot \dfrac{dz}{dt}=-\dfrac{1}{t^2}\cdot \dfrac{dx}{dz}+\dfrac{1}{t^2}\cdot \dfrac{d^2x}{dz^2}$
$ \implies t^2x''=(D^2-D)x~$.
From $(1)$ ,
$$(D^2-D)x-Dx+x=0$$
$$\implies (D^2-2D+1)x=0\tag2$$
Let $~x=e^{mz}~$ be a trial solution of $(1)$, then putting the value of $~x~$ in $(2)$, we have $$m^2-2m+1=0\implies m=1~,~~1$$
So the general solution of $(2)$ is $$x=(A+B~z)~e^z$$where $~A,~B~$are constants of integration.
And the general solution of $(1)$ is $$ x=(A+B~\ln t)~t$$where $~A,~B~$are constants of integration.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to evaluate $\int_0^{2\pi} \frac{\sin^2\theta}{1+\cos^2\theta}\,d\theta$ using residues? I have the following integral and I want to evaluate it using residues
$$I=\int_0^{2\pi} \frac{\sin^2\theta}{1+\cos^2\theta}d\theta$$
By using the transformation $\frac{1}{z}=e^{-i\theta}$, I got to show that
$$I=i\int_C\frac{z^4-2z^2+1}{z(z^4+6z^2+1)}dz$$
where $C$ is the centered unit circle.
I'm trying to use Cauchy's Residue Theorem, as we have 3 poles inside C: $z=0$, $z=i\sqrt{3-2\sqrt{2}}$ and $z=-i\sqrt{3-2\sqrt{2}}$. However, I'm really struggling to compute the residues of $z=i\sqrt{3-2\sqrt{2}}$ and $z=-i\sqrt{3-2\sqrt{2}}$ by hand.
Any tips or help to compute them?
| Note that $\sqrt{3-2\sqrt2}=\sqrt2-1$. So, since $\pm i\left(\sqrt2-1\right)$ is a simple root of $z(z^4+6z^2+1)$, you have\begin{align}\operatorname{res}_{z=\pm i\left(\sqrt2-1\right)}\frac{z^4-2z^2+1}{z(z^4+6z^2+1)}&=\operatorname{res}_{z=\pm i\left(\sqrt2-1\right)}\frac{z^4-2z^2+1}{z^5+6z^3+z}\\&=\left.\frac{z^4-2z^2+1}{5z^4+18z^2+1}\right|_{z=\pm i\left(\sqrt 2-1\right)}\\&=\frac{3-2\sqrt2}{4-3\sqrt2}.\end{align}Besides,$$\operatorname{res}_{z=0}\frac{z^4-2z^2+1}{z(z^4+6z^2+1)}=\left.\frac{z^4-2z^2+1}{5z^4+18z^2+1}\right|_{z=0}=1.$$
So, you want to compute $\int_0^{2\pi}R(\cos\theta,\sin\theta)\,\mathrm d\theta$, with $R(x,y)=\frac{y^2}{1+x^2}$. But then you define$$f(z)=\frac1zR\left(\frac{z-1/z}2,\frac{z-1/z}{2i}\right)=-\frac{z^4-2z^2+1}{z(z^4+6z^2+1)}$$and then\begin{align}\int_0^{2\pi}\frac{\sin^2\theta}{1+\cos^2\theta}\,\mathrm d\theta&=\int_0^{2\pi}R(\cos\theta,\sin\theta)\,\mathrm d\theta\\&=\frac1i\int_0^{2\pi}e^{-i\theta}R\left(\frac{e^{i\theta}+e^{-i\theta}}2,\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)ie^{i\theta}\,\mathrm d\theta\\&=\frac1i\int_C-\frac{z^4-2z^2+1}{z(z^4+6z^2+1)}\,\mathrm dz\\&=-2\pi\sum_{z_0\in\left\{0,\pm i\left(\sqrt2-1\right)\right\}}\operatorname{res}_{z=z_0}\frac{z^4-2z^2+1}{z(z^4+6z^2+1)}\\&=-2\pi\left(1+2\frac{3-2\sqrt2}{4-3\sqrt2}\right)\\&=2\pi\left(\sqrt2-1\right).\end{align}
| {
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"url": "https://math.stackexchange.com/questions/3501391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution verification: $\sup S\;\&\;\inf S,\;S:=\left\{\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}:\;a,b,c\in\langle 0,+\infty\rangle\right\}$ Find $\sup S\;\&\;\inf S$ if exist.
$$S:=\left\{\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}:\;a,b,c\in\langle0,+\infty\rangle\right\}$$
My work:
Although it is obvious due to its simmetry, I reformulated the given expression:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}=\frac{1}{1+\frac{b}{a}}+\frac{1}{1+\frac{c}{b}}+\frac{1}{1+\frac{a}{c}}<3\implies\sup S=3$$
$A-G\;\text{mean}$ for possible $\min S$ if $\;\inf S\in S:$
$$\frac{1}{1+\frac{b}{a}}+\frac{1}{1+\frac{c}{b}}+\frac{1}{1+\frac{a}{c}}\geq\frac{3}{\sqrt[3]{\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)}}$$
The equality holds $\iff\;\;a=b=c$
$$\implies \inf S=\frac{3}{2}\in S\implies\min S=\frac{3}{2}$$
Is this correct?
Note: The set was originally named $A$, but I renamed it not to mix it with $A$ for the $\text{arithmetic mean}$.
| No, your solution is not correct.
$$\sum_{cyc}\frac{a}{a+b}>\sum_{cyc}\frac{a}{a+b+c}=1.$$
Now, for $c\rightarrow0^+$ we obtain:
$$\sum_{cyc}\frac{a}{a+b}\rightarrow\frac{a}{a+b}+1,$$ which for $b\rightarrow+\infty$ is closed to $1$.
Thus, $$\inf\sum_{cyc}\frac{a}{a+b}=1.$$
Also $$\sum_{cyc}\frac{a}{a+b}<\sum_{cyc}\frac{a+c}{a+b+c}=2.$$
Can you end it now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solution verification: $\lim_{x\to \infty}\Bigg(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\Bigg)^{x\ln x}$
Find: $$\displaystyle\lim_{x\to
\infty}\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln x}$$
My attempt:
$\displaystyle\lim_{x\to\infty}\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\lim_{x\to\infty}\left(1+\frac{\ln(x^2+3x+4)-\ln(x^2+2x+3)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\\\displaystyle\lim_{x\to\infty}\left(1+\frac{\ln\left(\frac{x^2+3x+4}{x^2+2x+3}\right)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\lim_{x\to\infty}\left(1+\frac{\ln\left(1+\frac{x+1}{x^2+2x+3}\right)}{\ln(x^2+2x+3)}\right)^{x\ln x}$
What I used:
$\displaystyle\lim_{x\to\infty}\ln\left(1+\frac{x+1}{x^2+2x+3}\right)=0\;\;\&\;\;\lim_{x\to\infty}\ln(x^2+2x+3)=+\infty$
In the end, I got an indeterminate form: $\displaystyle\lim_{x\to\infty}1^{x\ln x}=1^{\infty}$
Have I made a mistake anywhere? It seems suspicious.
added: replacement $\frac{x+1}{x^2+2x+3}$ by $\frac{1}{x}$ wasn't appealing either.
Would:$$\lim_{x\to\infty}\Big(\Big(1+\frac{1}{x}\Big)^x\Big)^{\ln x}=x=\infty$$
be wrong?
//a few days after users had provided hints and answered the question,we discussed this with our assistant and he suggested a table formula that can also be applied (essentialy the last step in methods provided in the answers I recieved).:$$\lim_{x\to c}f(x)=1\;\&\;\lim_{x\to c}g(x)=\pm\infty$$then$$\lim_{x\to c}f(x)^{g(x)}=e^{\lim_{x\to c}(f(x)-1)g(x)}//$$
| Use $\lim_{t\to 1}\frac{\ln(t)}{t-1}=1$ twice: first for $t=\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}$ and then for $t=\frac{x^2+3x+4}{x^2+2x+3}$. The rest is simplification.
$$\begin{align}
\lim_{x\to\infty}\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln x} &=
\exp\left(\lim_{x\to\infty}\ln\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)x\ln x\right) \\
&= \exp\left(\lim_{x\to\infty}\frac{\ln\frac{x^2+3x+4}{x^2+2x+3}}{\ln(x^2+2x+3)}x\ln x\right) \\
&=\exp\left(\lim_{x\to\infty}\frac{(x+1)x}{x^2+2x+3}\frac{\ln x}{\ln(x^2+2x+3)}\right)
\end{align} $$
We have
$$\frac{(x+1)x}{x^2+2x+3}=\frac{x^2+x}{x^2+2x+3}\to 1 $$
For the other fraction, use that $\ln(x^2+2x+3)=\ln(x^2(1+2/x+3/x^2))=2\ln x+\ln(1+2/x+3/x^2)$
$$\frac{\ln x}{\ln(x^2+2x+3)}=\frac{\ln x}{2\ln x+\ln(1+2/x+3/x^2)}=\frac{1}{2+\frac{\ln(1+2/x+3/x^2)}{\ln x}}\to \frac 12 $$
So the limit is $e^{\frac 12 \cdot 1}=\sqrt{e}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How should I solve combination addition like this?
$${2\choose 2} {10\choose 3}+{3 \choose 2}{9 \choose 3}+{4 \choose 2}{8 \choose 3}+{5 \choose 2}{7 \choose 3}+{6 \choose 2}{6 \choose 3}+{7 \choose 2}{5 \choose 3}+{8 \choose 2}{4 \choose 3}+{9 \choose 2}{3 \choose 3}={13 \choose 6}$$
I am supposed to get the answer ${13\choose 6}$. I wonder if there is any formula for combination addition like this or are there some tricks to do this?
| Count 6-subsets of $\{1,\dots,13\}$ by conditioning on the third smallest element $k$, which must be at least 3 and at most 10. For example, if $k=5$, then there are $\binom{4}{2}$ ways to choose two smaller elements from $\{1,\dots,4\}$ and $\binom{8}{3}$ ways to choose three larger elements from $\{6,\dots,13\}$. This combinatorial proof shows that $$\sum_{k=3}^{10} \binom{k-1}{2}\binom{13-k}{3}=\binom{13}{6}.$$
More generally, Identity 137 in Proofs That Really Count is:
$$\sum_{j=r}^{n+r-k} \binom{j-1}{r-1}\binom{n-j}{k-r}=\binom{n}{k},$$
and the same combinatorial proof is given.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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how to parameterize diophant triples on the elliptic curve? How to parameterize Diophantine triples on the elliptic curve?
Example: In the article of the Andrej Dujella ,
The parameterized state of the diophantine triples $(a, b, c)$ on the
$$E : y^2 = x^3 + (3t^4 − 21t^2 + 3)x^2+(3t^8 +12t^6 +18t^4 +12t^2 +3)x+(t^2+1)^6$$
is
$$a = \dfrac{18t(t−1)(t+1)}{(t^2 − 6t + 1)(t^2 + 6t + 1)}\\
b=\dfrac{(t−1)(t^2 +6t+1)^2}{6t(t+1)(t^2 −6t+1)}\\ c=\cdots$$
How are these parameters?
| The idea in the paper
A. Dujella, M. Kazalicki, M. Mikic and M. Szikszai, There are infinitely many rational Diophantine sextuples, Int. Math. Res. Not. IMRN 2017 (2) (2017), 490-508.
is to construct rational Diophantine triples $a,b,c$ (i.e. $ab+1$, $ac+1$, $bc+1$) such that
on the induced elliptic curve $y^2=(x+ab)(x+ac)(x+bc)$ the point with $x$-coordinate equal to $1$ has order $3$.
The condition can be written in the form
$$ \sigma_2=(\sigma_1^2 \sigma_3^2-12\sigma_3^2-6\sigma_1 \sigma_3-3)/(4+4\sigma_3^2), $$
where $\sigma_1=a+b+c$, $\sigma_2=ab+ac+bc$, $\sigma_3=abc$.
Inserting this in the condition that $(ab+1)(ac+1)(bc+1)$ is a perfect square,
we get $1+\sigma_3^2$ is a square, i.e. $\sigma_3=\frac{t^2-1}{2t}$.
The polynomial $X^3-\sigma_1X^2+\sigma_2X-\sigma_3$ should have rational roots,
so its discriminant has to be a perfect square.
From this condition we got the quartic in $\sigma_1$
$$ (t^6-2t^4+t^2)\sigma_1^4+(23t^3-23t^5+t^7-t)\sigma_1^3+(-45t^2-45t^6+126t^4)\sigma_1^2
+(162t^5-162t^3-54t^7+54t)\sigma_1-27-54t^4+108t^2+108t^6-27t^8=w^2, $$
which is equivalent to the elliptic curve written in the question.
The curve in question has an obvious point $R$ with $x$-coordinate equal to $0$.
We compute the point $2R$, which has $x$-coordinate $-3/4(t^2-6t+1)(t^2+6t+1)$.
Transfering it back to the quartic, we get
$$ \sigma_1 = \frac{t^8+130t^6-390t^4+130t^2+1}{3(t-1)(t+1)(t^2-6t+1)(t^2+6t+1)t}. $$
If we insert obtained values of $\sigma_1,\sigma_2,\sigma_3$ in $X^3-\sigma_1X^2+\sigma_2X-\sigma_3=0$,
we get three rational solutions for $X$, which are exactly $a,b,c$ written in the question.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that $\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}} = 3$ without using Cardano's formula. (Hint, what is $(3\pm \sqrt{13})^3$ Prove that $\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}} = 3$ without using Cardano's formula. (Hint, what is $(3\pm \sqrt{13})^3$
I have that $$(3 + \sqrt{13})^3 = 144 + 40 \sqrt{13} $$ and $$(3 - \sqrt{13})^3 = 144 - 40 \sqrt{13} $$
A cursory look into Bombelli's method led me to the following system of equations:
$$\sqrt[3]{18 + \sqrt{325}} = a + b^{1/2}$$
$$\sqrt[3]{18 - \sqrt{325}} = a - b^{1/2}$$
I am unsure how to solve this system of equations without making a mess of the radicals...I know however that the given cube roots on the LHS of the above system are solutions to the cubic $x^3 + 3x = 36 $
| Hint, a link to a nice resource and to complement Michael's answer. There is this famous statement
If $a+b+c=0$ then $a^3+b^3+c^3=3abc$
source, Mathematical Olympiad Treasures, by Titu Andreescu and Bogdan Enescu, first chapter is available in preview, the proof is there.
As a result
$$x=\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}} \Rightarrow \\
x-\sqrt[3]{18 + \sqrt{325}} - \sqrt[3]{18 - \sqrt{325}} =0 \Rightarrow \\
x^3- (18 + \sqrt{325}) - (18 - \sqrt{325}) = 3x\sqrt[3]{(18 + \sqrt{325})(18 - \sqrt{325})} \Rightarrow \\
x^3-36=-3x$$
which has $3$ as the only real solution.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solution verification: $\lim_{x\to 0}\frac{e^{2\sin x}\cosh x-1}{\sqrt[3]{x(\cosh x-\cos x)}}$
Find (without L'Hospital):$$\lim_{x\to 0}\frac{e^{2\sin x}\cosh x-1}{\sqrt[3]{x(\cosh x-\cos
x)}}$$
My attempt:
$$\lim_{x\to 0}\cosh x=1$$
$$\lim_{x\to 0}e^{2\sin x}\cosh x-1=\lim_{x\to 0}\frac{e^{2\sin x}-1}{2\sin x}\cdot2\sin x=2\lim_{x\to 0}\sin x$$
$$\lim_{x\to 0}(\cosh x-\cos x)=\lim_{x\to 0}(1-\cos x)=\lim_{x\to 0}\frac{1-\cos x}{x^2}\cdot x^2=\frac{1}{2}\lim_{x\to 0}x^2$$
$$\lim_{x\to 0}\sqrt[3]{x(\cosh x-\cos x)}=\lim_{x\to 0}\sqrt[3]{x\cdot\frac{1}{2}x^2}=\frac{1}{\sqrt[3]{2}}\lim_{x\to 0}x$$
$$2\sqrt[3]{2}\lim_{x\to 0}\frac{\sin x}{x}=2\sqrt[3]{2}$$
Is this legitimate?
| As $\cosh(x)=\cos(ix)$
using Prosthaphaeresis Formula
$$\lim_{x\to0}\dfrac{e^{2\sin x}\cosh x-1}{\sqrt[3]{x(\cos h x-\cos x)}} =\lim_{x\to0}\dfrac{e^{2\sin x}(e^x+e^{-x})-2}{2\sqrt[3]{2x\sin\dfrac{x(1+i)}2\sin\dfrac{x(1-i)}2}}$$
$$=\lim_{x\to0} \dfrac{x\left(\dfrac{e^{2\sin x+x}-1}x+\dfrac{e^{2\sin x-x}-1}x\right)}{2x\sqrt[3]{2\cdot\dfrac{\sin\dfrac{x(1+i)}2}{\dfrac{x(1+i)}2}\cdot\dfrac{\sin\dfrac{x(1-i)}2}{\dfrac{x(1-i)}2}}\cdot\dfrac{(1+i)(1-i)}4}$$
$$=\dfrac{2+1+2-1}{2}$$
| {
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"url": "https://math.stackexchange.com/questions/3509650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Find a basis of a subspace spanned by matrices $G = \langle \begin{bmatrix}1&1\\1&1\\ \end{bmatrix}, \begin{bmatrix}1&1\\1&0\\ \end{bmatrix}, \begin{bmatrix}2&-3\\1&1\\ \end{bmatrix},\begin{bmatrix}4&-1\\3&2\\ \end{bmatrix}$
First I write all the matrices as vectors:
$M_1 = \begin{bmatrix}1\\1\\1\\1 \end{bmatrix}$
$M_2 = \begin{bmatrix}1\\1\\1\\0 \end{bmatrix}$
$M_3 = \begin{bmatrix}2\\-3\\1\\1 \end{bmatrix}$
$M_4 = \begin{bmatrix}4\\-1\\3\\2 \end{bmatrix}$
Then I found the rref:
$$\begin{bmatrix}1 & 1 & 1 & 1\\1 & 1 & 1 & 0\\2 & -3 & 1 & 1\\4 & -1 & 3 & 2 \end{bmatrix} \rightarrow (...) \rightarrow \begin{bmatrix}1 & 0 & \frac{4}{5} & 0\\0 & 1 & \frac{1}{5} & 0\\0 & 0 & 0 & 1\\0 & 0 & 0 & 0 \end{bmatrix}$$
The basis I get is
$$(\begin{bmatrix}1 &0\\4/5&0 \end{bmatrix},\begin{bmatrix}0 &1\\\frac{1}{5}&0 \end{bmatrix}),\begin{bmatrix}0 &0\\0&1 \end{bmatrix}$$
Is this correct? It doesn't even come close to the solution in my book which is
$$(\begin{bmatrix}1 &1\\1&1 \end{bmatrix},\begin{bmatrix}1 &1\\1&0 \end{bmatrix}),\begin{bmatrix}2 &-3\\1&1 \end{bmatrix}$$
| Your solution isn’t wrong per se. There’s an infinite number of them. It looks like you were meant to pick out a subset of the original four matrices that form a basis, which you can do by assembling those same flattened vectors into as columns instead of rows. After row-reduction, you can pick out the original vectors that correspond to pivot columns in the reduced matrix.
| {
"language": "en",
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"answer_count": 2,
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} |
How to find the sum of the number series written recursively? I have the following number series, written in next way:
$x_1 = 2$
$x_{n+1} = x_n^2 - x_n + 1$, $n \geq 1$
and it needs to find sum of:
$$\sum_{n=1}^{\infty} \frac{1}{x_n}$$
At the moment, I haven't had much of progress, namely:
I know, that in the end of solving this task I'll have next type of situation:
$(x_1+x_2) - (x_2 +x_3) - (x_3 + x_4) ... (x_{n-1} + x_n) $
so elements mutually cancel out inside the sum and we get desired result
But every action requires mathematical proof unless it is obviously true
I've had also next observation:
$x_n$ is growing, thus $\frac{1}{x_n}$ decreases, power of decreasing $\gt 1$ thus the number series is convergent, but yet again, I don't have strictly math proof of this fact, but it requires
I'll be very grateful for any help
| Duplicate of this question, found on approach0.xyz.
But this question doesn't have any answer, so I'll just share the important hint given in the comments by @Kelenner:
$$
\frac{1}{x_n -1} - \frac{1}{x_{n + 1} -1}
= \frac{1}{x_n}
$$
As you detailed in the comments, you have used the hint to show the sum is equal to $\frac{1}{x_1 - 1}$.
For the sake of completeness, I'll write it down here.
First, we have
\begin{align}
\frac{1}{x_n -1} - \frac{1}{x_{n + 1} -1}
& = \frac{1}{x_n -1} - \frac{1}{x_n^2 - x_n + 1 -1} \\
& = \frac{x_n}{x_n^2 -x_n} - \frac{1}{x_n^2 - x_n} \\
& = \frac{x_n - 1}{x_n^2 -x_n}
= \frac{1}{x_n}.
\end{align}
Therefore, we obtain by a telescoping argument
\begin{align}
\sum_{n = 1}^{\infty} \frac{1}{x_{n}}
& = \sum_{n = 1}^{\infty} \frac{1}{x_n -1} - \frac{1}{x_{n + 1} -1}
= \lim_{m \to \infty} \left(\sum_{n = 1}^{m} \frac{1}{x_n - 1} - \frac{1}{x_{n + 1} - 1}\right) \\
& = \lim_{m \to \infty}\left( \frac{1}{x_1 - 1} - \frac{1}{x_{m + 1} - 1}\right)
= \frac{1}{x_1 - 1} + \lim_{n \to \infty} \frac{-1}{x_{n + 1} -1}
= \frac{1}{x_1 - 1}
\end{align}
As we have $x_n \to \infty$.
This can be seen by showing $x_n \ge n$ by induction.
For $n = 1$ its clear, so we begin at $n = 2$.
Base step: $x_2 = 3 \ge 2$.
Induction hypothesis: $x_n \ge n$ holds for some $n \in \mathbb{N}_{\ge 2}$.
Induction step: $n \to n + 1$.
Since $x_n \ge 2$ for all $n$ we have
$$
x_{n + 1}
= x_n(x_n - 1) + 1
\overset{\text{IH}}{\ge} (n)(n - 1) + 1
= n^2 - n + 1 \ge n + 1.
$$
The last inequality rearranges to $n^2 \ge 2n \iff n \ge 2$. So the statement is true for $n \ge 2$ but we know its also true for $n = 1$, so we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3512803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Trouble finding solution via implicit differentiation I'm currently using the book University Calculus: Alternate Edition (Hass et al., 2008) to study calculus. I was studying implicit differentiation and ran into an exercise problem that I was having trouble solving. The question is:
Use implicit differentiation to find $dy/dx$ and $d^2y/dx^2$ for:
$$x^{2/3} + y^{2/3} = 1$$
I was able to find $y'$ fairly simply as:
$$y' = -\left( \dfrac{y}{x} \right)^{1/3}$$
but I'm having trouble finding $y''$.
My approach is as follows:
$$
\begin{align}
y'' & = \frac{d}{dx}y' \\
& = \frac{d}{dx} \left( \frac{-y^{1/3}}{x^{1/3}} \right) \\
& = \frac{(-y^{1/3})'x^{1/3} - (-y^{1/3})(x^{1/3})'}{(x^{1/3})^2} \\
& = \frac{-\frac{1}{3}y^{-2/3} y' x^{1/3} + y^{1/3} \frac{1}{3}x^{-2/3}}{x^{2/3}} \\
& = \dfrac{\dfrac{-y^{-2/3} y' x^{1/3}}{3} + \dfrac{y^{1/3}}{3}}{x^{4/3}} \\
& = \dfrac{\left( -y^{-2/3} \times \dfrac{-y^{1/3}}{\phantom{-}x^{1/3}} \times x^{1/3}\right) + y^{1/3}}{3x^{4/3}} \\
& = \dfrac{y^{1/3} + y^{-1/3}}{3x^{4/3}}
\end{align}
$$
However, the textbook solution states that the correct answer for $y''$ is:
$$y'' = \dfrac{x^{2/3}y^{-1/3} - y^{1/3}}{3x^{4/3}}$$
and I'm having trouble reverse engineering where I went wrong. Any tips or pointers are appreciated. Thanks in advance.
| Using the Quotient Rule in finding higher derivatives from implicit differentiation can create opportunities for errors to "creep into" the calculations. Starting from $ \ x^{2/3} + y^{2/3} \ = \ 1 \ \ , $ you would have produced $ \ \frac23·x^{-1/3} + \frac23·y^{-1/3}·y' \ = \ 0 \ \ $ to obtain $ \ y' \ = \ -\frac{y^{1/3}}{x^{1/3}} \ \ . $ We will want to hold that aside for what follows.
We can return to the equation from which we found the first derivative and differentiate it implicitly one more:
$$ \ \frac{d}{dx} \ \left[ \ \frac23·x^{-1/3} \ + \ \frac23·y^{-1/3}·y' \ \right] \ \ = \ \ \frac{d}{dx} \ [ \ 0 \ ] \ $$
$$ \Rightarrow \ \ \frac23·\left(-\frac13 \right)·x^{-4/3} \ + \ \underbrace{\left[ \ \frac23· \left(-\frac13 \right)·y^{-4/3}·y' \ \right] · y' \ + \ \frac23·y^{-1/3}·y''}_{\text{by Product Rule}} $$
$$ = \ \ -\frac29 ·x^{-4/3} \ - \ \ \frac29· y^{-4/3}·(y')^2 \ \ + \ \frac23·y^{-1/3}·y'' \ \ = \ \ 0 $$
$$ \Rightarrow \ \ y'' \ \ = \ \ \frac{x^{-4/3} \ + \ \ y^{-4/3}·(y')^2}{3·y^{-1/3}} \ · \ \frac{x^{4/3}y^{4/3}}{x^{4/3}y^{4/3}} \ \ = \ \ \large{\frac{y^{4/3} \ + \ \ x^{ 4/3}·\left(-\frac{y^{1/3}}{x^{1/3}} \right)^2}{3·x^{4/3}·y }} $$
$$ = \ \ \frac{y^{4/3} \ + \ \ x^{ 2/3}·y^{ 2/3}} {3·x^{4/3}·y } \ \ = \ \ \frac{y^{1/3} \ + \ \ x^{ 2/3}·y^{-1/3}} {3·x^{4/3} } \ \ . $$
So by this alternative calculation, I agree with John Omielan and believe that the textbook solution either contains a "typo" or the solver lost track of a "minus-sign". (They could also have made use of the original curve equation -- that of an astroid -- to write
$$ \ y'' \ \ = \ \ \frac{y^{1/3} \ + \ \ x^{ 2/3}·y^{-1/3}} {3·x^{4/3} } \ · \ \frac{y^{1/3}}{y^{1/3}} \ \ = \ \ \frac{y^{2/3} \ + \ \ x^{ 2/3}} {3·x^{4/3}· y^{1/3} } \ \ = \ \ \frac{1} {3·x^{4/3}· y^{1/3} } \ \ . \ ) $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum value of $(\tan C – \sin A)^2 + (\cot C – \cos B)^2$ for the following given data
Let $A, B, C$ be real numbers such that
(i) $(\sin A, \cos B)$ lies on a unit circle centered at origin.
(ii) $\tan C$ and $\cot C$ are defined.
Find the minimum value of $(\tan C – \sin A)^2 + (\cot C – \cos B)^2$.
My multiple attempts are as follows:-
Attempt $1$:
$$\sin^2A+\cos^2B=1$$
$$\tan^2C+\sin^2A-2\sin A\tan C+\cot^2C+\cos^2 B-2\cot C\cos B$$
$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin A\sin C}{\cos C}+\dfrac{\cos C\cos B}{\sin C}\right)$$
$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin A\sin^2 C+\cos^2 C\cos B}{\sin C\cos C}\right)$$
$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin^2C(\sin A-\cos B)+\cos B}{\sin C\cos C}\right)\tag{1}$$
Now from here how to proceed further.
Attempt $2$:
$$\sin^2A+\cos^2B=1$$
$$\sin^2A=\sin^2B$$
$$A=n\pi\pm B$$
Considering only the principal range, $A=B$, $A=-B$, $A=n\pi-B$, $A=n\pi+B$
Case $1$: $A=B,A=-B$
Put $B=A$ or $B=-A$ in equation $(1)$
$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin A\sin^2 C+\cos^2 C\cos A}{\sin C\cos C}\right)$$
$$(\tan^2C+\cot^2C)+1-2\sqrt{\sin^4C+\cos^4C}\cdot\dfrac{\sin(A+\alpha)}{\sin C\cos C}$$
$$(\tan^2C+\cot^2C)+1-2\sqrt{\tan^2C+\cot^2C}\cdot \sin(A+\alpha)$$
So minimum value will be $3-2\sqrt{2}$
Case $1$: $A=n\pi-B,A=n\pi+B$
Put $B=n\pi-A$ or $B=A-n\pi$
$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin A\sin^2 C-\cos^2 C\cos A}{\sin C\cos C}\right)$$
$$(\tan^2C+\cot^2C)+1-2\sqrt{\sin^4C+\cos^4C}\cdot\dfrac{\sin(A-\alpha)}{\sin C\cos C}$$
$$(\tan^2C+\cot^2C)+1-2\sqrt{\tan^2C+\cot^2C}\cdot\sin(A-\alpha)$$
So minimum value will be $3-2\sqrt{2}$
Any other way to solve this question?
| Let $P = (\tan C, \cot C)$ lie on the curve $xy = 1$, $Q = (\sin A, \cos B)$ lie on the unit circle $x^2 + y^2 = 1$ and $O = (0, 0)$ be the origin.
It's easy to prove by AM-GM inequality that $PO \geqslant \sqrt2$, so $(\tan C - \sin A)^2 + (\cot C - \cos B)^2 = |PQ|^2 \geqslant (|PO| - |QO|)^2 \geqslant 3 - 2\sqrt 2$, where the equality is reached iff $P = (1, 1)$ and $Q = \left(\frac{\sqrt2}{2}, \frac{\sqrt2}{2}\right)$ or their images under O-reflection.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Probabilities of the second highest roll in a dice pool. I've been trying to figure out the probabilities of rolling a specific number in a pool of 4 20-sided dice, assuming I discard the highest and two lowest rolls. My instinct was to compound the probabilities of two dice being lower or equal to a target n, one being higher or equal, and one being the specific roll I want, like so $P(n) = \frac{(n^2)(21-n)}{20^4}$.
However, I noticed that the sum of the probabilities for each numbers was only ~10% instead of a flat 1. My first instinct was that this was the probability of a specific dice being the second highest, so not all outcomes were accounted for, but factoring in permutations brought me to ~242%. Still not close. Is this a case where the sum of the individual probabilities doesn't need to equal 1, or is my math wrong and where?.
| The probability for the highest die to show $k$ and all other dice to show at most $n\lt k$ is $4\cdot\frac{n^3}{20^4}$, so the probability for the highest die to show $k$ and the next highest die to show $n\lt k$ is $4\cdot\frac{n^3-(n-1)^3}{20^4}$. So the probability for the second-highest die to be lower than the highest and have the value $n$ is
$$
4\sum_{k=n+1}^{20}\frac{n^3-(n-1)^3}{20^4}=4(20-n)\frac{n^3-(n-1)^3}{20^4}\;,
$$
whereas the probability for the second-highest die to be equal to the highest and have the value $n$ is
$$\frac{n^4-(n-1)^4-4(n-1)^3}{20^4}\;.$$
The probability you seek is the sum of those two probabilities,
$$
\frac{4(20-n)\left(n^3-(n-1)^3\right)+n^4-(n-1)^4-4(n-1)^3}{20^4}=\frac{83-252n+258n^2-12n^3}{20^4}\;.
$$
Here's a table of the numerators for $n=1,\ldots,20$:
\begin{array}{r|r}
n&\\\hline
1&77\\
2&515\\
3&1325\\
4&2435\\
5&3773\\
6&5267\\
7&6845\\
8&8435\\
9&9965\\
10&11363\\
11&12557\\
12&13475\\
13&14045\\
14&14195\\
15&13853\\
16&12947\\
17&11405\\
18&9155\\
19&6125\\
20&2243
\end{array}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Deriving the BBP identify for $\pi$ I was given a problem to learn how to use Mathematica. I should derive the identity from the paper 1 known as the BBP formula for $\pi$. But I can't figure it out why
$$\begin{equation} \sum_{k=0}^{\infty} \frac{1}{16^k}(\frac{4}{8i+1} - \frac{2}{8i+4}-\frac{1}{8i+5}-\frac{1}{8i+6}) = \int_0^{\frac{1}{\sqrt{2}}} \frac{4\sqrt{2}-8x^3-4\sqrt{2}x^4-8x^5}{1-x^8} {dx} \end{equation}$$
holds while using
$$\begin{equation}
\int_0^{\frac{1}{\sqrt{2}}} \frac{x^{k-1}}{1-x^8} {dx} = \int_0^{\frac{1}{\sqrt{2}}} \sum_{i=0}^{\infty} x^{k-1}x^{8i} {dx} = \frac{1}{\sqrt{2}^k} \sum_{i=0}^{\infty} \frac{1}{16i(8i+k)}
\end{equation}$$
this. I don't see how I can figure this out neither in Mathematica nor by hand.
| Begin by writing
$$
\begin{align}
\int_0^{\frac{1}{\sqrt{2}}} &\frac{4\sqrt{2}-8x^3-4\sqrt{2}x^4-8x^5}{1-x^8} dx =
\\&
4 \sqrt{2}\int_0^{\frac{1}{\sqrt{2}}} \frac{x^0}{1-x^8} dx
-
8\int_0^{\frac{1}{\sqrt{2}}} \frac{x^3}{1-x^8} dx
\\ & \qquad -
4 \sqrt{2}\int_0^{\frac{1}{\sqrt{2}}} \frac{x^4}{1-x^8} dx
-
8\int_0^{\frac{1}{\sqrt{2}}} \frac{x^5}{1-x^8} dx
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Equality in an inequality problem I have an inequality problem :
Let $a,b,c$ are real numbers such that $a, b, c \geq 0$ and $ab + bc + ca = 1$. Prove that $\sum \dfrac{3ab+1}{a+b} \geq 4$
Solution : $$\sum \dfrac{3ab+1}{a+b} = \sum \dfrac{4ab+bc+ca}{a+b} = \sum \dfrac{4ab}{a+b} + \sum a \geq \dfrac{4(ab+bc+ca)^{2}}{ab(a+b)+bc(b+c)+ca(c+a)} + \sum a = \dfrac{4}{(a+b+c)(ab+bc+ca)-abc} + \sum a \geq \dfrac{4}{\sum a} + \sum a \geq 4$$
Manual testing shows that equality holds on when (a, b, c) = (1; 1; 0). However if a number equals 0, WLOG it is $a$, the Cauchy-Schwarz inequality holds on when $\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}$, making $b = 0$ or $c = 0$ too. Hence, $ab + bc + ca = 0$, a contracdition. Any explanation ?
| You used the following C-S:
$$\sum_{cyc}ab(a+b)\sum_{cyc}\frac{ab}{a+b}\geq(ab+ac+bc)^2.$$
The equality occurs, when $$\left(\sqrt{ab(a+b)},\sqrt{bc(b+c)},\sqrt{ca(c+a)}\right)||\left(\sqrt{\frac{ab}{a+b}},\sqrt{\frac{bc}{b+c}},\sqrt{\frac{ca}{c+a}}\right),$$
which for $(a,b,c)\rightarrow(1,1,0)$ gives
$$(\sqrt2,0,0)||\left(\frac{1}{\sqrt2},0,0\right),$$ which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3518629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Definite Integral of $(a^2-x^2)^\frac{3}{2}$
Prove the following:
$$\frac{4c}3\int\limits_0^a\left(a^2 - x^2\right)^\frac32\,\mathrm dx = \frac{\pi a^4c}4.$$
Taking $x = a\sin\theta$, how will the limit change?
| $$\dfrac{d(x(a^2-x^2)^n)}{dx}=(a^2-x^2)^n+nx(a^2-x^2)^{n-1}(-2x)$$
$$=(a^2-x^2)^n+2n(a^2-x^2-a^2)(a^2-x^2)^{n-1}$$
$$=(1+2n)(a^2-x^2)^n-2na^2(a^2-x^2)^{n-1}$$
Integrate both sides with respect to $x$ and writing $\displaystyle I_m=\int(a^2-x^2)^mdx$
$$(1+2n)I_n-2na^2I_{n-1}=x(a^2-x^2)^n$$
Now if $\displaystyle J_m=\int_0^a(a^2-x^2)^mdx$
$$(1+2n)J_n-2na^2J_{n-1}=0$$
Set $n=3/2,1/2$
$$J_{-1/2}=\int_0^a\dfrac{dx}{\sqrt{a^2-x^2}}=\dfrac1{|a|}\arcsin\dfrac xa\big|_0^a=\dfrac\pi{2|a|}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3519727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\frac{a+b}{\csc x}=\frac{a-b}{\cot x}$, show that $\csc x \cot x=\frac{a^2-b^2}{4ab}$
If $$\frac{a+b}{\csc x}=\frac{a-b}{\cot x}$$
show that
$$\csc x \cot x=\frac{a^2-b^2}{4ab}$$
(original problem image)
I tried getting rid of the denominator and then expanding the given equation, but couldn't get to the answer.
| Suppose $\sin x \neq 0$. Then we have that $\csc x = \dfrac{1}{\sin x}$ and $\dfrac{\cos x}{\sin x} = \cot x$
This implies in $\cos x = \dfrac{a-b}{a+b}$.
Therefore, $\csc x \cdot \cot x = \dfrac {\cos x}{\sin^2 x} = \dfrac {\cos x}{1-\cos^2 x}.$
Notice that $1- \cos^2 x = 1 - \left(\dfrac{a-b}{a+b}\right)^2 = \dfrac{4ab}{(a+b)^2}$.
Then, $\dfrac {\cos x}{1-\cos^2 x}=\dfrac{(a-b)(a+b)}{4ab}=\dfrac{a^2-b^2}{4ab}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there alternative factoring of a quintic equation? In a paper here
the author appears to be able to factor a Bring-Jerrard quintic making
$$P=2mn(m^2-n^2)(m^2+n^2)=2m^5n-2mn^5\\ \implies n^5-m^4n+\frac{P}{2m}=0 \rightarrow x^5+px+q=0$$
become
$$(x^3+bx^2+cx+d)(x^2+ex+f)=0$$
but I haven't been able to follow how he got there. If I could, I would have what I need to find the one or more valid values of $n$ in the equation:
$$n^5-m^4n+\frac{P}{2m}=0$$
given that I will know the values of $P$ and $m$.
Can anyone help me figure out how the 'factored' equation would look in terms of $p,q$?
| The usual way to solve that would be to expand
$(x^3 +bx^2 +cx+ d)(x^2 + ex + f)=x^5 + (e + b)x^4 + (eb+c+f)x^3+(bf + d + ce)x^2 + (ed+cf)x + fd $
Want this to equal to $x^5 + px +q$ for all $x$ which means the two ploynomails are equal and thats only true if the coefficients for the same powers are equal so
$$\begin{array}{cccCC}
e+b &=&0 \Rightarrow &e&=&-b \\
eb+c+f &=& 0\Rightarrow &c+f&=&b^2\\
bf+d+ce&=&0 \Rightarrow &d+b(f-c)&=&0 \\
ed+cf &=&p\Rightarrow &-bd+cf&=&p\\
fd &=& q\Rightarrow &f &= &\frac{q}{d}
\end{array}$$
$$\begin{array}{ccc}
cd + q &=& db^2 \\
d^2 + bq-bdc&=&0 \\
-bd^2+cq&=& dp\end{array}$$
Lets eliminate $c$
$$\begin{array}{ccc}
cbdq + q^2b & = & qdb^3 \\
-bdcq + qd^2+bq^2 & = & 0 \\
cqbd - b^2d^3 & = & d^2bp \\
\end{array}$$
Then
$$\begin{array}{ccc}
q^2b+qd^2+bq^2 &=& qdb^3\\ qd^2+bq^2-b^2d^3&=&d^2bp \end{array}$$
We need to solve the two equations
$$\begin{array}{ccc}
2qb + d^2 &=& db^3 \\
qd^2+bq^2-b^2d^3&=&d^2bp \end{array}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Minimal polynomial of $\sqrt{3+2\sqrt{3}}$ Let $a=\sqrt{3+2\sqrt{3}}$. Then
\begin{align*}
&a=\sqrt{3+2\sqrt{3}}\\
&\implies a^2=3+2\sqrt{3}\\
&\implies a^2-3 = 2\sqrt{3}\\
&\implies (a^2-3)^2 = 4\cdot 3=12\\
&\implies (a^2-3)^2-12=0.
\end{align*}
So, $a$ a root of
$$(x^2-3)^2-12=x^4-6x^2-3.$$
By Eisenstein's criterion, $x^4-6x^2-3$ is irreducible over $\mathbb Q$. So, $a$ is algebraic of degree $4$.
However, I've seen that the degree of $\mathbb Q(\sqrt{3+2\sqrt{3}})$ over $\mathbb Q$ is $2$.
What is wrong with my attempt?
| You are unquestionably right. A much more advanced viewpoint:
The ring $R=\Bbb Z[\sqrt3\,]$ is the integer-ring of the quadratic number field $K=\Bbb Q(\sqrt3\,)$. It’s “well known” (and easy to prove) that $R$ is a Principal Ideal Domain (class number is $1$).
We have the factorization $z=3+2\sqrt3=\sqrt3(2+\sqrt3\,)$, in which the two factors are respectively, an indecomposable element, and a generator of the (free part of the) unit group. In particular, neither is a square, and they are independent modulo squares. Therefore, the square root $\sqrt z$ generates a quadratic extension of $K$. That does it.
| {
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"url": "https://math.stackexchange.com/questions/3523674",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Prove that the equation:$x^{2}+x=3+\ln(x+2) $ has only one real solution at $[0,\infty ) $ My attempt:
$x^{2}+x-(3+\ln(x+2))=0$
$x_{1,2}=\dfrac{-1\pm \sqrt{1+12+4\ln(x+2)}}{2} \Rightarrow 13+4\ln(x+2)>0 $
$ \Rightarrow 4\ln(x+2)>-13 $
$\ln(x+2)>-\frac{13}{4} $
$x+2>e^{-\frac{13}{4}} $
$x>e^{-\frac{13}{4}}+2 $
What did I do wrong?
Edit: Thank you all for the help:)
| Consider that you look for the zero of function
$$f(x)=x^2+x-\log (x+2)-3$$ Its derivatives are
$$f'(x)=2 x+1-\frac{1}{x+2} \qquad \text{and} \qquad f''(x)=\frac{1}{(x+2)^2}+2 \quad >0 \forall x$$
The first derivative cancels at negative values. So the function is invreasing and only one root.
By inspection, $f(1)<0$ and $f(2)>0$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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} |
How to calculate the volume given by $(x^2+y^2+z^2)^2 = x^2+y^2$ I am working on the following exercise:
Calculate the volume of the body bounded by the following surface:
$$(x^2+y^2+z^2)^2 = x^2+y^2$$
I would solve this with a triple integral, but I do not see how I can derive the boundaries from $(x^2+y^2+z^2)^2 = x^2+y^2$. Could you help me?
| HINT
Consider the transformation $x = \rho\cos(\theta)$, $y = \rho\sin(\theta)$ and $z = z$. Then you have
\begin{align*}
(x^{2} + y^{2} + z^{2})^{2} = x^{2} + y^{2} & \Longleftrightarrow z^{2} = \pm\sqrt{x^{2} + y^{2}} - x^{2} - y^{2}\\
& \Longleftrightarrow z^{2} = \pm\rho - \rho^{2} \Longleftrightarrow z^{2} = \rho - \rho^{2}
\end{align*}
since $z^{2} \geq 0$.
| {
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Understanding the steps in factoring $x^2+5xy+4y^2+5x+23y-6$ I understand the first two steps in factoring this expression, but the third step is unclear to me:
\begin{align}x^2+5xy+4y^2+5x+23y-6 &= x^2+(5y+5)x+4y^2+23y-6 \\
&= x^2+(5y+5)x+(4y-1)(y+6) \\
&= (x+(4y-1))(x+(y+6)) \\
&= (x+4y-1)(x+y+6).\end{align}
Why does the $5y+5$ disappear?
How can I learn to see such patterns and come up with this myself? I see that $5y +5 = (4y -1) + (y+6)$. Does this observation help me in this case? Usually when factoring by grouping the goal is to get one factor multiple times and then turn addition into multiplication.
In the same way, I don't understand the second step in here:
\begin{align}(x^2-5x)^2-18x^2+90x-144 &= (x^2-5x)^2-18(x^2-5x)-144 \\
&= (x^2-5x+6)(x^2-5x-24) \\
&= (x-2)(x-3)(x+3)(x-8) \\
&= (x-2)(x-3)(x-8)(x+3).\end{align}
I have a feeling that the same thing is happening and I don't understand why I can do that. I see that $6\times 24 = 144$, and that these constants have to be at the end of the factored terms, but how does this 'enhancing' of the content of the parenthesis work? Also, the first term was squared and then the square disappears.
| Begin with $$(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2,\tag1$$ where LHS says that $r_1,r_2$ are the roots of a quadratic polynomial, and RHS gives the coefficients of the polynomial in terms of the roots (see also Vieta's formulae).
In your equation, the coefficient multiplying $x$ is the SUM (or better, opposite of the sum) and the last coefficient is the PRODUCT of the roots.
You already noticed that $$5y +5 = (4y -1) + (y+6),$$
or $$-(5y +5) = -(4y -1) +(- (y+6)).$$
Therefore, in $(1)$ is $r_1=-(4y -1)$ and $r_2=- (y+6).$
| {
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"timestamp": "2023-03-29T00:00:00",
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Vector Triple Product from Lagrange’s Identity How can one obtain (5.12) from (5.11)?
The relation
$$\tag{5.11}
\left( \mathbf{a} \times \mathbf{b}\right) \cdot \left( \mathbf{c} \times \mathbf{d}\right)
= \left( \mathbf{a} \cdot \mathbf{c}\right) \left( \mathbf{b} \cdot \mathbf{d}\right) -
\left( \mathbf{a} \cdot \mathbf{d}\right) \left( \mathbf{b} \cdot \mathbf{c}\right)
$$
is called the Identity of Langrange. From ($5.11$) follows
$$\tag{5.12}
\left(\mathbf{a} \times \mathbf{b} \right) \times \mathbf{c} = \left(\mathbf{a}\cdot \mathbf{c} \right)\mathbf{b} - \left( \mathbf{b}\cdot \mathbf{c} \right)\mathbf{a}
$$
Can someone give a hint?
| A key observation is that if $\mathbf{A}$ and $\mathbf{B}$ are two vectors such that $\mathbf{A}\cdot \mathbf{C} = \mathbf{B}\cdot \mathbf{C}$ for all vectors $\mathbf{C}$, then $\mathbf{A} = \mathbf{B}$. This fact will be used here.
If $\mathbf{d}$ is an arbitrary vector, then $$[(\mathbf{a}\times \mathbf{b})\times \mathbf{c}]\cdot \mathbf{d} = (\mathbf{a}\times \mathbf{b})\cdot (\mathbf{c}\times \mathbf{d})$$ Use (5.11) to obtain the equivalent expression $$[(\mathbf{a}\cdot \mathbf{c})\,\mathbf{b} - (\mathbf{b}\cdot \mathbf{c})\,\mathbf{a}]\cdot \mathbf{d}$$ Since $\mathbf{d}$ was arbitrary, we deduce (5.12).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3529079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\frac{az+b}{cz+d}=\frac{ez+f}{gz+h}$, then, $a = z_0 e, b = z_0 f, c = z_0 g, d = z_0 h$ for some $z_0 \in \mathbb{C}-\{0\}$. How to prove the following proposition elegantly?
My proof is very bad.
Let $a,b,c,d,e,f,g,h \in \mathbb{C}$ and $ad-bc\ne 0, eh-fg\ne 0$.
If
$$\frac{az+b}{cz+d}=\frac{ez+f}{gz+h}$$
for all $z \in \mathbb{C} \cup \{\infty\}$,
then, $a = z_0 e, b = z_0 f, c = z_0 g, d = z_0 h$ for some $z_0 \in \mathbb{C}-\{0\}$.
Proof:
Assume that $a=0\ne e$.
Then,
$$0\ne \frac{b}{c\frac{-f}{e}+d}=\frac{e \frac{-f}{e}+f}{g \frac{-f}{e} +h}=0.$$
This is a contradiction.
So, if $a=0$ then, $e =0$.
Similarly, if $e=0$ then, $a=0$.
*
*Suppose $a=e=0$.
Then, $b, c, f, g \notin \{0\}$.
Assume $d =0\ne h$.
Then, $\infty = \frac{b}{c 0}=\frac{f}{g 0 + h} \in \mathbb{C}$.
This is a contradiction.
So, if $d=0$, then $h=0$.
Similarly, if $h=0$, then $d=0$.
a. Suppose $h=d=0$.
Then, $\frac{b}{c} = \frac{b}{c 1} = \frac{f}{g 1} = \frac{f}{g}$.
So, in this case, $a =0 = \frac{b}{f} 0=\frac{b}{f} e, b = \frac{b}{f} f, c = \frac{b}{f} g, d=0 = \frac{b}{f} 0=\frac{b}{f} h$.
b. Suppose $h, d \notin \{0\}$.
Then $\infty=\frac{b}{c \frac{-d}{c}+d}=\frac{f}{g \frac{-d}{c}+h}$.
So, $g \frac{-d}{c}+h=0, -g d + c h = 0, \frac{c}{g} = \frac{d}{h} =: z_0.$
$\frac{b}{c z + d} = \frac{b}{z_0 g z + z_0 h} = \frac{\frac{b}{z_0}}{g z + h} = \frac{f}{g z+h}$ for all $z \in \mathbb{C} \cup \{\infty\}$.
So, $\frac{b}{z_0} = f, b = z_0 f$.
So, in this case, $a = 0 = z_0 0 = z_0 e,b = z_0 f, c = z_0 g, d = z_0 h$.
*Suppose $a, e \notin \{0\}$.
Assume $c=0 \ne g$.
Then $$\mathbb{C} \ni \frac{a \frac{-h}{g} + b}{d} = \frac{e \frac{-h}{g} + f}{g \frac{-h}{g} + h} = \infty.$$
This is a contradiction.
So, if $c=0$, then $g=0$.
Similarly, if $g = 0$, then $c=0$.
a. Suppose $c=g=0$.
Assume $b=0 \ne f$.
Then, $0 = \frac{a 0}{d} = \frac{e 0 + f}{h} = \frac{f}{h} \ne 0$.
This is a contradiction.
So, if $b=0$, then $f=0$.
Similarly, if $f=0$, then $b=0$.
*
*Suppose $b=f=0$.
Then, $\frac{a}{d} = \frac{a 1}{d} = \frac{e 1}{h} = \frac{e}{h}$.
So, $\frac{a}{e}=\frac{d}{h} =: z_0$.
So, in this case, $a = z_0 e, b = 0 = z_0 0 = z_0 f, c = 0 = z_0 0 = z_0 g, d = z_0 h$.
*Suppose $b, f \notin \{0\}$.
$$0=\frac{a \frac{-b}{a}+b}{d} = \frac{e \frac{-b}{a}+f}{h}, -eb+af=0, \frac{a}{e}=\frac{b}{f}=:z_0.$$
$$\frac{a z +b}{d} = \frac{z_0 e z + z_0 f}{d} = \frac{e z + f}{\frac{d}{z_0}} = \frac{e z + f}{h}$$
for all $z \in \mathbb{C} \cup \{\infty\}$.
So, $\frac{d}{z_0}=h$.
So, in this case, $a = z_0 e, b=z_0 f, c =0=z_0 0 = z_0 g, d = z_0 h$.
$\cdots$
| Let $$P =
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}\,\,\,\text{ and }\,\,\,Q =
\begin{pmatrix}
e & f \\
g & h
\end{pmatrix}
$$
The first two conditions mean that $P$ and $Q$ are both invertible.
Now, let $$u_z=\begin{pmatrix}z\\1
\end{pmatrix}$$
Then the third condition means that there exists $\lambda_z\in\mathbb C-\{0\}$ such that
$$Pu_z=\lambda_z Qu_z$$
In other words $Q^{-1}Pu_z=\lambda_z u_z$ for all $z\in\mathbb C\cup \{\infty\}$.
It's easy to verify that it implies that any vector of $\mathbb C^2$ is an eigenvector of $Q^{-1}P$ and therefore there must exist $z_0\in\mathbb C-\{0\}$ such that $$Q^{-1}P=z_0 I$$
where $I$ is the identity matrix. This yields the desired result.
| {
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"timestamp": "2023-03-29T00:00:00",
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Generalized version of calculating expected minimum value rolled until $5$ is obtained from a fair die The following is a modified interview question.
Given an $n$-sided fair die where $n\geq 1.$
You roll a die until you get a $m$ where $1\leq m\leq n$.
Calculate the expected value of the minimum rolled.
The original interview question takes $n=6$ (standard fair die) and $m=5$.
I mange to solve the problem and I reproduce my attempt below.
The expected minimum value rolled is $\frac{137}{60}$ because if $X$ is the minimum value rolled up to and including $5,$ then
$$P(X=x) = \frac{1}{x(x+1)} \quad \text{for }x=1,2,3,4 \quad \text{and} \quad P(X=5) = \frac{1}{5}.$$
So,
$$E(X) = \sum_{x=1}^5 xP(X=x) = \frac{137}{60}.$$
I am trying to solve the generalized version of the problem.
By the same spirit, let $Y$ be the minimum value rolled up to and including $m.$ Then
$$P(Y=y) = \frac{(y-1)!}{(y+1)!} = \frac{1}{y(y+1)} \quad \text{for }y =1,2,...,m-1 \quad \text{and}\quad P(Y=m) = \frac{1}{m}.$$
Therefore,
$$E(Y) = \sum_{y=1}^m y P(Y=y) = 1 + \sum_{y=1}^{m-1} \frac{1}{y+1} = 1 + \sum_{y=2}^{m} \frac{1}{y}.$$
| Your work looks correct.
A slightly different approach: for $1 \le y \le m$,
$$P(Y \ge y) = P(\text{all rolls are $\ge y$}) =\sum_{k=0}^\infty \left(\frac{n-y}{n}\right)^k \frac{1}{n} = \frac{1}{y}.$$
Then since $Y$ is nonnegative,
$$E[Y] = \sum_{y=1}^m P(Y \ge y) = \sum_{y=1}^m \frac{1}{y}.$$
| {
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Prove: $9\mid a^6+a^{6^2}+\ldots+ a^{6^{27}},\;a\in\mathbb N$
Prove $$9\mid a^6+a^{6^2}+\ldots +a^{6^{27}},a\in\mathbb N$$
My attempt:
\begin{aligned}&a \equiv 2\pmod{9}\implies a^{6} \equiv 2^{6} \equiv 1\pmod{9}\\&\begin{array}{l}{a \equiv 3\pmod{9}\implies a^{6} \equiv 3^{6} \equiv 0\pmod{9}} \\{a \equiv 4\pmod{9}\implies a^{6} \equiv\left(2^{6}\right)^{2} \equiv 1\pmod{9}}\\a\equiv 5\equiv -4\pmod{9}\implies a^6\equiv(-4)^6=4^6\equiv 1\pmod{9}\end{array}\end{aligned}
$$a^{6^i}=\left(a^6\right)^{6^{i-1}}\equiv [0,1]\pmod{9}\implies a^{6^i}=9k\;\lor\;a^{6^i}=9k+1,\;k\in\mathbb N_0$$
$$a^6+a^{6^2}+\ldots +a^{6^{27}}=\sum_{i=1}^{27}\left(a^6\right)^{6^{i-1}}=\sum_{i=1}^{27}(9k_i+[0,1])=\underbrace{\sum_{i=1}^{27}9k_i+[0,27]}_{\text{multiple of}\; 9}$$
//edited mistake [27,54]//
Is this correct and is there a more efficient method?
| It is correct, but a bit long.
Another method will use Euler's theorem: $\;\varphi(9)=6$ so, if $a$ is not divisible by $3$, we have $a^6\equiv 1\mod 9$, and by an easy induction, $\;a^{6^k}=\bigl(a^6 \bigr)^{6^{k-1}}\equiv 1$. Therefore
$$a^6+a^{6^2}+\ldots+ a^{6^{27}}\equiv\underbrace{1+1+\ldots+1}_{27 \:1\text{s}}\equiv 0\mod 9 $$
for this case.
On the other hand, if $a$ is divisible by $3$, $a^2\equiv 0\bmod 9$, hence $a^6\equiv 0$ too.
| {
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How to find the minimum of $abcd+\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}$ when $ab+bc+cd+da+ac+bd=6$ If $a,b,c,d\geq0$ are such that $ab+bc+cd+da+ac+bd=6$, then what is the minimum value of $$f(a,b,c,d)=abcd+\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}$$ ?
My attempts: I think the minimum $5$ is obtained at $a=b=c=d=1$. I tried the Hölder inequality: $$f(a,b,c,d)\geq abcd + (1+\sqrt{a b c d})^2,$$ but the last term is $1<5$ for $d=0$.
Also, direct AM-GM on the two terms of $f$ did not work either. What to do?
Remark: This problem is from AoPS. There, it is also asked what happens for $ab+bc+cd+da+ac+bd=7$, which seems even harder.
| Using the known identity $(u^2 + v^2)(x^2 + y^2) = (ux - vy)^2 + (uy + vx)^2$,
we have
$$(a^2 + 1)(b^2 + 1) = (ab - 1)^2 + (a + b)^2 \tag{1}$$
and
$$(c^2+1)(d^2+1) = (cd - 1)^2 + (c + d)^2. \tag{2}$$
From (1) and (2), using Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align*}
&(a^2 + 1)(b^2 + 1)(c^2+1)(d^2+1)\\
=\,& [(ab - 1)^2 + (a + b)^2]\,[(cd - 1)^2 + (c + d)^2]\\
\ge\,& [(ab - 1)(cd - 1) - (a + b)(c + d)]^2\\
=\,& (abcd - ab - cd + 1 - ac - ad - bc - bd)^2
\end{align*}
which results in
$$\sqrt{(a^2 + 1)(b^2 + 1)(c^2+1)(d^2+1)}
\ge ab + bc + cd + da + ac + bd - 1 - abcd.$$
Thus, we have
$$abcd + \sqrt{(a^2 + 1)(b^2 + 1)(c^2+1)(d^2+1)}
\ge ab + bc + cd + da + ac + bd - 1 = 5.$$
Also, when $a = b = c = d = 1$,
we have $abcd + \sqrt{(a^2 + 1)(b^2 + 1)(c^2+1)(d^2+1)} = 5$.
Thus, the minimum of $abcd + \sqrt{(a^2 + 1)(b^2 + 1)(c^2+1)(d^2+1)}$ is $5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute the volume of the region $x^4 + y^2 + z^6 = 1.$
Compute the volume of the region $x^4 + y^2 + z^6 = 1.$
I can let $f(y) = 1-y^2 = x^4 + z^6.$ Then $z= \sqrt[6]{f(y)-x^4}.$ I suppose I can use symmetry to calculate the volume of one portion of the integral, but I'm not sure how to come up with the integral.
| Technique due to Dirichlet. In the positive octant, volume is
$$ \frac{\Gamma \left( 1 + \frac{1}{2} \right)\Gamma \left( 1 + \frac{1}{4} \right)\Gamma \left( 1 + \frac{1}{6} \right)}{\Gamma \left(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{6} \right)} $$
and the full volume is $8$ times that. If you can read the scans below, a crucial step is that
$$ \frac{1}{p} \; \Gamma \left( \frac{1}{p} \right) = \Gamma \left( 1+ \frac{1}{p} \right) $$
For your problem, the volume in the positive octant is about $0.770182765,$ so the volume of the whole solid is about
$$ 6.1614621 $$
From gp-pari:
? gamma( 3/2)* gamma( 5/4) * gamma( 7/6) / gamma(23/12)
%6 = 0.7701827651926944255287920651
?
? 8 * gamma( 3/2)* gamma( 5/4) * gamma( 7/6) / gamma(23/12)
%7 = 6.161462121541555404230336521
?
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality with x,y,z fractions $\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x}\ge 2$ If $x,y,z>0$, show:
$$\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x}\ge 2$$
I expand and to prove
$$x^3 - 2 x^2 y + x^2 z + x y^2 - x y z + y z^2\ge 0$$
I don't know how to do this.
| By C-S and AM-GM $$\frac{x}{y}+\frac{y}{x+z}+\frac{z}{x}=\frac{x^2}{xy}+\frac{y^2}{xy+yz}+\frac{z^2}{xz}\geq\frac{(x+y+z)^2}{xy+xy+yz+xz}=$$
$$=\frac{x^2+y^2+z^2+2xy+2xz+2yz}{2xy+xz+yz}\geq \frac{4xy+2xz+2yz}{2xy+xz+yz}=2.$$
Also, we can end your work.
Indeed, we need to prove that:
$$yz^2+(x^2-xy)z+x(x-y)^2\geq0,$$ which is true for $x\geq y$.
But for $x<y$ it's enough to prove that:
$$x^2(x-y)^2-4xy(x-y)^2\leq0$$ or
$$x(x-y)^2(x-4y)\leq0,$$ which is obvious.
| {
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Integral of $\int{\frac{2x-3}{(3x^2-2x+4)\sqrt{x^2-3x+1}}}dx$ I need to solve the following integral:
$$\int{\frac{2x-3}{(3x^2-2x+4)\sqrt{x^2-3x+1}}}dx$$
First I tried factorizing the denominators but it wasn't helpful. Then I thought about completing the square in $3x^2-2x+4$ but nothing came from there. I don't know if there might be any useful trigonometric substitution I could use or maybe some other substitution. I also tried with $t = \sqrt{x^2-3x+1}$ and $t = \frac{1}{x^2-2x+4}$ but also got me nowhere. Thanks for reading and your help.
| Using the substitution $u^2 = x^2-3x+1$ we get that
$$2du = \frac{2x-3}{\sqrt{x^2-3x+1}}dx$$
which is exactly $2$ out of the $3$ terms that we need to kill for our integral. Then using quadratic equation to solve for $x$ we get that
$$x = \frac{3}{2}\pm\sqrt{u^2+\frac{5}{4}}$$
$$x^2 = u^2 + \frac{7}{3} \pm 3\sqrt{u^2+\frac{5}{4}}$$
so our integral becomes the slightly nicer looking
$$\int \frac{2\:du}{3u^2+8\pm 7\sqrt{u^2+\frac{5}{4}}}$$
and at this point so far trig and hyperbolic substitutions haven't been working nicely. Wolfram is not giving a straight answer, either, so who knows if an elementary real answer exists.
| {
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How many of the integers between $1$ and $200$ are odd numbers or divisible by $3$ or divisible by $5$? How many of the integers between $1$ and $200$ are odd numbers or divisible by $3$ or divisible by $5$?
\begin{align*}
A_1 & = \left\lfloor{\frac{200}{3}} \right\rfloor = 66 && \text{(divisible by $3$)}\\
A_2 & = \left\lfloor{\frac{200}{5}} \right\rfloor = 40 && \text{(divisible by $5$)}\\
A_3 & = \left\lfloor{\frac{200}{2}} \right\rfloor = 100 && \text{(odd)}\\
| A_1 \cap A_2 | & = \left\lfloor{\frac{200}{3 \cdot 5}}\right\rfloor = 13\\
| A_1 \cap A_3 | & = \left\lfloor{\frac{200}{3 \cdot 2}}\right\rfloor = 33\\
| A_2 \cap A_3 | & = \left\lfloor{\frac{200}{5 \cdot 2}} \right\rfloor= 20\\
| A_1 \cap A_2 \cap A_3 | & = \left\lfloor{\frac{200}{5 \cdot 2 \cdot 3}}\right\rfloor = 6
\end{align*}
Therefore, by the principle exclusion inclusion theorem
$= 66 + 40 + 100- (13 + 33 + 20) + 6 = 146$
Is this logically right?
| There is a simplier way to count the number.
First let separate the even and odd numbers. There are $100$ odd numbers between $1$ and $200$. All of them are part of the answer. We will continue only with even number. We will use the inclusion-exclusion principle, as you did.
How many even number between $2$ and $200$ are divisible by $3$? Every third number is divisible by $3$ ( divisible by $6$, actually).
$$\left\lfloor\frac{200}6\right\rfloor=33$$
How many even number between $2$ and $200$ are divisible by $5$? Every fifth number is divisible by $5$ ( divisible by $10$, actually).
$$\left\lfloor\frac{200}{10}\right\rfloor=20$$
We counted twice the even numbers that where divisible by $15$. How many even number between $2$ and $200$ are divisible by $15$? Every fifteenth number is divisible by $15$ ( divisible by $30$, actually).
$$\left\lfloor\frac{200}{30}\right\rfloor=6$$
Final answer
$$100+33+20-6=147$$
My answer is one more than yours. You have a mistake in your last intersection. Your formula
$$|A_1\cap A_2\cap A_3|=\left\lfloor\frac{200}{2\cdot3\cdot5}\right\rfloor$$
counts how many are divisible by $30$, but we need how many have a remainder of $15$ when divided by $30$. And there is one more. It could calculated like this.
$$|A_1\cap A_2\cap A_3|=\left\lfloor\frac{200-15}{2\cdot3\cdot5}\right\rfloor+1$$
Substract $15$ to every numbers and forget the negative ones. We now have $185$ numbers. How many are divisible by $30$? We add one to count the number $15$ (which became $0$ and wasn't accounted for).
| {
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Find values of $x$, such as $\log_3 \sqrt{x+3}−\log_3(9−x^2) < 0$ The Function is $$f(x) = \log_3\sqrt{(x+3)}−\log_3(9−x^2)$$
and I need to figure out arguments for which $$ f(x) < 0 $$
So I calculated the domain of function which is $ D: (-3;3)$
However I am still unable to solve $f(x) < 0$
I simplified $ \log_3\sqrt{(x+3)}−\log_3(9−x^2) < 0$ to $\log_3\frac{\sqrt{(x+3)}}{(9-x^2)} < 0$
which gets me to $$ \frac{\sqrt{(x+3)}}{9-x^2} < 1$$
But the solutions of the above equation are complex numbers and I definitely should get them as my result. So, what I am doing wrong here?
Would apprecite every answer
| First of all, we need $9-x^2>0\iff-3<x<3$
As $\log_ax$ is increasing function for $a>1$
We need $$\sqrt{x+3}<9-x^2$$
Let $\sqrt{x+3}=y>0\implies x=y^2-3$
$$\implies0<9-(y^2-3)^2-y=-y+6y^2-y^4=-y(1-6y+y^3)$$
As $y>0,$ we need $$y^3-6y+1<0$$
| {
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Prove by induction that for n > 8, ${n \choose k} < 2^{n-2}$ for each $k \in \mathbb{Z}$ This problem also comes with a second-half that then asks again to prove by induction that for every $n > 7$, $\displaystyle {n \choose k} < (n-3)!$ for each integer $k$.
I have gotten to what I think is the base case for the first half, but I am very lost. Any help?
| You know that the maximum value of ${n \choose k}$ is given as follows:
If n is even, $max {n\choose k} = \Large\frac {n!}{\frac {n}{2}!\frac {n}{2}!}$, or $k=\frac{n}{2}$ and $n-k=\frac{n}{2}$
If n is odd, $max {n\choose k} = \Large\frac {n!}{\frac {n-1}{2}!\frac {n+1}{2}!}$, or $k=\frac{n-1}{2}$ and $n-k=\frac{n+1}{2}$.
So in order to prove that ${n \choose k}<2^{n-2}$, it is sufficient to prove the inequality for the max values for both odd and even cases.
First we prove that n=9 holds.
${9 \choose 4}=\frac {9!}{4!5!}= 126 < 128=2^{9-2} $. So n=9 holds.
Let $n=p$ hold for some value of $p>9$
Case 1:
p is even. $\therefore\Large\frac{p!}{\frac{p}{2}!\frac{p}{2}!}<2^{p-2}$
Now $\Large\frac{p+1}{\frac{p}{2}+1}<2$.
Thus we can multiply the 2 given inequalities to get
$\Large\frac{p!*(p+1)}{\frac{p}{2}!\frac{p}{2}!(\frac{p}{2}+1)}<2^{p-2}*2$
Or $\Large\frac{(p+1)!}{\frac{p}{2}!(\frac{p}{2}+1)!}<2^{p-1}$. Thus it holds when p is even.
Case 2:
p is odd. $\therefore\Large\frac{p!}{\frac{p-1}{2}!\frac{p+1}{2}!}<2^{p-2}$
now $\frac{p+1}{\frac{p-1}{2}+1}=2$.
We can still multiply the inequality and the equality to get
$\Large\frac{p!*(p+1)}{\frac{p-1}{2}!\frac{p+1}{2}!(\frac{p-1}{2}+1)}<2^{p-2}*2$
Or $\large\frac{(p+1)!}{\frac{p+1}{2}!(\frac{p+1}{2})!}<2^{p-1}$. Thus it holds when p is odd also.
$\Rightarrow$ The given inequality is true for n>8
| {
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maximum value of $ac+bd$
If $a,b,c,d\in \mathbb{R}$ and $a^2+b^2\leq 2$ and $c^2+d^2\leq 4.$ Then maximum value of $ac+bd$ is
what i try
$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$
$\underbrace{(ac+bd)^2}_{\max}=\underbrace{(a^2+b^2)(c^2+d^2)}_{\max}-\underbrace{(ad-bc)^2}_{\min}$
$(ac+bd)^2\leq 8\Rightarrow (ac+bd)\leq 2\sqrt{2}$
but answer is $3$
How do i get right answer help me please
| One way of obtaining the maximum value is by doing maximisation with constraints. First observe that $\nabla_{a,b,c,d} (ac+bd)\neq \overrightarrow{0} \text{ for all } a,b,c,d\neq 0$. As it's a continuous function and the domain we're interested in is compact (it's the intersection of two perpendicular cylinders), the maximal value has to be reached somewhere in the domain. But for it to be reached in the interior of the domain, the gradient would have to become 0 at that position, which we observed not to happen. So we have to check the boundary by applying Lagrange multipliers. On the boundary both constraints $a^2+b^2=2$ and $c^2+d^2=4$ are fulfilled. So we add them with Lagrange multipliers and calculate the gradient: $\nabla_{a,b,c,d,\lambda_1,\lambda_2}(ab+cd+\lambda_1(a^2+b^2-2)+\lambda_2(c^2+d^2-4))=\begin{pmatrix}c-\lambda_1\cdot 2a\\d-\lambda_1\cdot2b\\ a - \lambda_2\cdot 2c\\ b - \lambda_2\cdot 2d\\a^2+b^2-2\\c^2+d^2-4\end{pmatrix}$. So now we want this to be zero. The first two lines give $\lambda_1=\frac{c}{2a}=\frac{d}{2b}$, the second two lines give $\lambda_2=\frac{a}{2c}=\frac{b}{2d}$, the third two lines represent our constraints. We're not interested in the exact values of the multipliers, so we just take the second equalities, which are equivalent for both. Solving this equality for d yields $d=\frac{bc}{a}(*)$. This we can insert into the constraint and obtain $c^2+d^2=c^2(1+\frac{b^2}{a^2})=4$. This means $c^2(a^2+b^2)=c^2 \cdot 2=4a^2$ and $c^2=2a^2$. With our equality from above this also means $d^2=2b^2$. We cannot obtain more, but this is already enough, as $(ac+bd)^2=(ac)^2+2acbd+(bd)^2=$
$=(ac)^2+(ad)^2+(bc)^2+(bd)^2=a^2(c^2+d^2)+b^2(c^2+d^2)=4(a^2+b^2)=8$ using the condition $ad=bc$, equivalent to $(*)$ from above.
So the maximum value is the positive choice leading to $(ac+bd)=\sqrt{8}=2\sqrt{2}$.
| {
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Find all triples $(x,y,z)\in \Bbb{R}$ that satisfy the following conditions: The question is to find all real numbers solutions to the system of equations:
*
*$y=\Large\frac{4x^2}{4x^2+1}$,
*$z=\Large\frac{4y^2}{4y^2+1}$,
*$x=\Large\frac{4z^2}{4z^2+1}$,
This seems simple enough, so I tried substituting the values of x, y and z into the different equations but I only ended up with a huge degree 8 equation which clearly doesn't seem like the right approach. I really have no idea on how to go about solving this if substitution is not the answer.
Any help would be greatly appreciated :)
| We can note that $4a^2+1 \ge 4a \ \ \forall a \in R$. Also, since $\frac{4a^2}{4a^2+1} \ge 0$, then we know that $x,y,z \ge 0$
Therefore $y=\frac{4x^2}{4x^2+1} \le \frac{4x^2}{4x}=x$ for nonezero values of $x$. Similarly $z \le y$ and $x \le y$. Therefore $x \le y \le z \le x \implies x=y=z$.
Now solving equation $a=\frac{4a^2}{4a^2+1} \implies 4a^2+1=4a \implies a = \frac{1}{2} \implies x=y=z=\frac{1}{2}$.
Finally, we assumed that numbers are non-zero, so we should include solution $(0,0,0)$
| {
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Prove if $5 \nmid n$, then $n^2 = 5k \pm 1$ I am having difficulty finishing this proof. At first, the proof is easy enough. Here's what I have thus far:
Because $5 \nmid n$, we know $\exists q \in \mathbb{Z}$ such that $$n = 5q + r$$ where $0 < r < 4$. Note $r \neq 0$ because if $r = 0$, then $5 \mid n$. Also note that $n^2 = 25q^2 + 10qr + r^2$. Then we have four cases: when $r=1$, $r=2$, $r = 3$, and $r = 4$. This is where I run into difficulty. In each of these cases, we can prove that either $n^2 = 5k + 1$ or $n^2 = 5k - 1$ for some integer $k$, but I cannot see how to prove both for each case. Any ideas?
As a side note on how I went to prove each case, I simply plugged $r$ into the formula $n^2 = 25q^2 + 10qr + r^2$. This results in $n^2 = 25q^2 + 10q + 1$.
Continuing, we get $n^2 = 5(5q^2 + 2q) + 1$, and because $5q^2 + 2q$ is still an integer, this is of the form $n^2 = 5k + 1$ for some integer $k$. But I cannot find how to make the 1 a negative to prove both cases.
| Just do it.
$n^2 = 25q^2 + 10qr + r^2 = 5(5q^2 + 2qr) + r^2$.
ANd $r^2 = 1,4=5-1, 9 =10-1$ or $16=15+1$.
So if $n = 5q + 1$ then $n^2 = 5(5q^2 + 2q) + 1$
If $n=5q+2$ then $n^2 = 5(5q^2 + 4q)+4 = 5(5q^2+4q + 1) -1$
If $n = 5q+3$ then $n^2= 5(5q^2 + 6q) + 9 = 5(5q^2 + 6q + 2) -1$
And if $n = 5q+4$ then $n^2 = 5(q^2 + 8q) + 16=5(5q^2 + 8q + 3) + 1$.
That's it.
| {
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Roots of the equation $(x^2+3x+4)^2+3(x^2+3x+4)+4=x$ The equation $(x^2+3x+4)^2+3(x^2+3x+4)+4=x$ has
(A) all its solution real but not all positive
(B) only two of its solution real
(C) two of its solution positive and two negative
(D) none of its solution real
My approach is as follow $(x^2+3x+4)^2+3(x^2+3x+4)+4-x=y$
$f(x)=y=x^4+6x^3+20x^2+32x+32$
$f(-x)=y=x^4-6x^3+20x^2-32x+32$
Using Descartes rule no positive roots but the possible ways we can have either 4,2,0 negative roots.
$f'(x)=4x^3+18x^2+40x+32$
$f''(x)=12x^2+36x+40$ which is imaginary
From here I am not able to approach.
| $$(x^2+3x+4)^2+3(x^2+3x+4)+4=x$$
$$(x^2+3x+4)^2+4(x^2+3x+4)+4=x+x^2+3x+4$$
$$(x^2+3x+4+2)^2=(x+2)^2$$
$$(x^2+3x+6)^2-(x+2)^2=0$$
then use the fact
$$a^2-b^2=(a-b)(a+b)$$
| {
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How do I get the inverse of this equation? The equation is the following:
$$y = \frac{x}{x-2}$$
What I did is:
$$y = \frac{x}{x-2}$$
$$y \times (x-2) = x$$
$$\frac{x-2}{x} = \frac{1}{y}$$
$$\frac{x}{x}-\frac{2}{x} = \frac{1}{y}$$
$$1-\frac{2}{x} = \frac{1}{y}$$
$$1-\frac{x}{2} = y$$
I've got a lot of doubts about that last step. Besides that, the result according to the textbook is the following:
$$ f^{−1}(x)=-\frac{2x}{x−1} $$
| Sorry, no, you can't take recirocals of the terms the way you have.
That is
$\frac {1}{a+b} \ne \frac 1a + \frac 1b$
Anyway
$y = \frac{x}{x-2}\\
y(x-2) = x\\
yx - 2y = x\\
yx - x = 2y\\
x(y-1) = 2y\\
x = \frac {2y}{y-1}$
$f^{-1}(y) = \frac {2y}{y-1}$
| {
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Let a,b and c be the roots of the equation $x^3-9x^2+11x-1=0$ and $s=\sqrt a +\sqrt b +\sqrt c$
Then Find the value of $s^4-18s^2-8s$
$$s=\sqrt a + \sqrt b +\sqrt c$$
$$s^2=a+b+c+2(\sqrt {ab} +\sqrt {bc} +\sqrt {ac})$$
I can’t seem to find a way around this obstacle. Squaring it again gives another $\sqrt {ab}+\sqrt {bc} +\sqrt {ac}$, and is seemingly never ending.
This drove me to the conclusion that the value of s cannot actually be found, and the question must be manipulated to get the required from. I don’t know how to do that though.
| How about evaluating the square of $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}$ independently?
$$(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^2=ab+bc+ca+2\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c}) = 11+2s$$
so
$$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\sqrt{11+2s}$$
From your work, we then get:
$$s^2=9+2\sqrt{11+2s}\Rightarrow s^2-9=2\sqrt{11+2s}\Rightarrow s^4-18s^2+81=44+8s$$
In conclusion $s^4-18s^2-8s=-37$.
| {
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$z^2 = -1$ Why what i do is wrong? I want to calculate $z^2 = -1$ i think i should get to $z = i$.
$-1 = -1+0i = 1(\cos(2\pi k)+i\sin(2\pi k)), k \in Z$
$z^2 = r^2(\cos(2 \theta) + i\sin(2\theta))$
So i get:
$$r^2(\cos(2 \theta) + i\sin(2\theta)) = 1(\cos(2\pi k)+i\sin(2\pi k))$$
Namely:
$r = 1, 2 \theta = 2 \pi k \Rightarrow \theta = \pi k$
So for $k = 0$ i get $z_0 = 1$ for $k=1: z_1 = -1$
What i do wrong?
| $$(a+b\,i)^2=a^2-b^2+2ab\,i=-1\iff a^2-b^2=-1,2ab=0.$$
The only way is by $a=0,b^2=1$, hence $\pm i$.
Your mistake was in $2k\pi$ instead of $(2k+1)\pi$.
| {
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"answer_count": 4,
"answer_id": 3
} |
Stirling numbers of first kind in a power series We need to prove that $\cfrac{(-\ln{(1-x)})^k}{k!}=\displaystyle\sum_{n=k}^\infty \begin{bmatrix} n\\ k \end{bmatrix}\cfrac{x^n}{n!}$. For this, I've done induction in $k$. The case $k=1$ is trivial. When I do the inductive step, I consider the next:
\begin{equation}
\begin{split}
\cfrac{(-\ln{(1-x)})^{k+1}}{(k+1)!}&=\left(\cfrac{(-\ln{(1-x)})^k}{k!}\right)\left(\cfrac{-\ln{(1-x)}}{k+1}\right)\\
&=\frac{1}{k+1}\left(\sum_{n=k}^\infty\begin{bmatrix} n\\ k \end{bmatrix}\frac{x^n}{n!}\right)\left(\sum_{n=1}^{\infty}\frac{x^n}{n}\right)\\
&=\frac{1}{k+1}\sum_{n=k+1}^\infty x^n\sum_{m=k}^{n-1} \begin{bmatrix} m\\ k \end{bmatrix}\frac{1}{(n-m)m!}.
\end{split}
\end{equation}
So, now my problem is showing that $\cfrac{n!}{k+1}\displaystyle\sum_{m=k}^{n-1}\begin{bmatrix} m\\ k \end{bmatrix}\frac{1}{(n-m)m!}=\begin{bmatrix} n\\ k+1 \end{bmatrix}$. On other hand, with the recurrence relation we can deduce the next:
\begin{equation}
\begin{split}
\begin{bmatrix} n\\ k+1 \end{bmatrix}&=\begin{bmatrix} n-1\\ k \end{bmatrix}+(n-1)\begin{bmatrix} n-1\\ k+1 \end{bmatrix}\\
&=\begin{bmatrix} n-1\\ k \end{bmatrix}+(n-1)\left(\begin{bmatrix} n-2\\ k \end{bmatrix}+(n-2)\begin{bmatrix} n-2\\ k+1 \end{bmatrix}\right)\\
&=\begin{bmatrix} n-1\\ k \end{bmatrix}+(n-1)\begin{bmatrix} n-2\\ k \end{bmatrix}+(n-1)(n-2)\begin{bmatrix} n-2\\ k+1 \end{bmatrix}\\
&\ \ \vdots\\
&=\frac{(n-1)!}{(n-1)!}\begin{bmatrix} n-1\\ k \end{bmatrix}+\frac{(n-1)!}{(n-2)!}\begin{bmatrix} n-2\\ k \end{bmatrix}+\frac{(n-1)!}{(n-3)!}\begin{bmatrix} n-3\\ k \end{bmatrix}+\ldots + \frac{(n-1)!}{k!}\begin{bmatrix} k\\ k \end{bmatrix}\\
&=(n-1)!\sum_{m=k}^{n-1}\frac{1}{m!}\begin{bmatrix} m\\ k \end{bmatrix}.
\end{split}
\end{equation}
| We seek to prove that
$$\frac{1}{k!} \left(\log\frac{1}{1-z}\right)^k
= \sum_{n\ge k} {n\brack k} \frac{z^n}{n!}$$
by induction. We get for the base case
$$\log\frac{1}{1-z} = \sum_{n\ge 1} \frac{z^n}{n}$$
which holds by inspection. We also have
$$n! [z^n] \frac{1}{(k+1)!} \left(\log\frac{1}{1-z}\right)^{k+1}
\\ = (n-1)! [z^{n-1}]
\left(\frac{1}{(k+1)!} \left(\log\frac{1}{1-z}\right)^{k+1}\right)'
\\ = (n-1)! [z^{n-1}]
\left(\frac{1}{k!} \left(\log\frac{1}{1-z}\right)^{k}\right)
\left(\log\frac{1}{1-z}\right)'
\\ = (n-1)! [z^{n-1}]
\left(\frac{1}{k!} \left(\log\frac{1}{1-z}\right)^{k}\right)
\frac{1}{1-z}.$$
We now use the induction hypothesis and the fact that we have
a convolution of two EGFs to get for the coefficient being extracted
$$\sum_{q=k}^{n-1} {n-1\choose q} {q\brack k} (n-1-q)!$$
We may now conclude by combinatorics, re-writing the sum as follows:
$$\sum_{q=k}^{n-1} {n-1\choose n-1-q} (n-1-q)! {q\brack k}$$
Here we are counting partitions of $[n]$ into $k+1$ cycles by classifying
according to the cycle where $n$ resides. We choose $n-1-q$ companions on
that cycle where each choice generates $(n-q)!/(n-q)$ possible cycles.
The remaining $q$ elements are partitioned into $k$ cycles. In this way
we have counted all ${n\brack k+1}$ partitions exactly once, and we have
the claim.
We may also continue algebraically using the OGF of the Stirling
numbers of the first kind:
$$\sum_{q=k}^{n-1} {n-1\choose n-1-q} (n-1-q)!
[w^k] {w+q-1\choose q} q!
\\ = (n-1)! [w^k] \sum_{q=k}^{n-1} {w+q-1\choose q}.$$
Now
$$\sum_{q\ge 0} {w+q-1\choose q} z^q = \frac{1}{(1-z)^w}$$
so we get
$$(n-1)! [w^k] [z^{n-1}] \frac{1}{(1-z)^{w+1}}
- (n-1)! [w^k] [z^{k-1}] \frac{1}{(1-z)^{w+1}}
\\ = (n-1)! [w^k] {w+n-1\choose n-1}
- (n-1)! [w^k] {w+k-1\choose k-1}
\\ = (n-1)! [w^k] \frac{n}{w} {w+n-1\choose n}
= n! [w^{k+1}] {w+n-1\choose n} = {n\brack k+1}.$$
Remark. In the induction step we have used the fact that when we
multiply two exponential generating functions of the sequences
$\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!}
\sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0}
\left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
An indeterminate limit form of infinity/infinity I am trying to solve the limit:
$$\lim_{x\to\infty}x^\frac{5}{3}\left(\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-x^\frac{1}{3}\right)$$
I was trying to find a way to bring it into a fraction form to apply L'Hospital's rule, and I tried using
$$a-b=\frac{a^3-b^3}{a^2+ab+b^2}$$
But it made it even more complex and after applying L'Hospital's rule I got stuck with all the terms. Is there a smarter way to evaluate it?
| You have that
$$
\begin{gathered}
\mathop {\lim }\limits_{x \to + \infty } x^{\frac{5}
{3}} \left[ {\left( {x + \sin \left( {\frac{1}
{x}} \right)} \right)^{\frac{1}
{3}} - x^{\frac{1}
{3}} } \right] = \hfill \\
\hfill \\
\mathop {\lim }\limits_{x \to + \infty } x^{\frac{5}
{3}} \left[ {x^{\frac{1}
{3}} \left( {1 + \frac{1}
{x}\sin \left( {\frac{1}
{x}} \right)} \right)^{\frac{1}
{3}} - 1} \right] = \hfill \\
\hfill \\
= \mathop {\lim }\limits_{x \to + \infty } x^2 \left[ {\frac{1}
{{3x}}\sin \left( {\frac{1}
{x}} \right)} \right] = \hfill \\
\hfill \\
= \frac{1}
{3}\mathop {\lim }\limits_{x \to + \infty } x\left[ {\sin \left( {\frac{1}
{x}} \right)} \right] = \frac{1}
{3}\mathop {\lim }\limits_{x \to + \infty } \left[ {\frac{{\sin \left( {\frac{1}
{x}} \right)}}
{{\frac{1}
{x}}}} \right] = \frac{1}
{3} \hfill \\
\end{gathered}
$$
where we used the fact that
$$
\left( {1 + f\left( x \right)} \right)^\alpha - 1 \sim \alpha f(x)\,\,\,\,\,\,\left( {x \to x_0 } \right)
$$
provided
$$
\mathop {\lim }\limits_{x \to x_0 } f(x) = 0
$$
and the fact that
$$
\mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}
{t} = 1
$$
with $t=1/x$ as $x \to +\infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
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.