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Prob. 19, Chap. 1, in Baby Rudin: For what $\mathbf{c}$ and $r > 0$ does this equivalence hold? Here's Prob. 19 in Chap. 1 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Suppose $\mathbf{a} \in \mathbb{R}^k$, $\mathbf{b} \in \mathbb{R}^k$. Find $\mathbf{c} \in \mathbb{R}^k$ and $r > 0$ such that, for all $\mathbf{x} \in \mathbb{R}^k$, we have $$\vert \mathbf{x} - \mathbf{a} \vert = 2 \vert \mathbf{x} - \mathbf{b} \vert$$ if and only if $$\vert \mathbf{x} - \mathbf{c} \vert = r.$$
Although Rudin has given a solution, namely $3\mathbf{c} = 4 \mathbf{b} - \mathbf{a}$, $3r = 2\vert \mathbf{b} - \mathbf{a} \vert$, I'm wondering how he's obtained it.
How to attack this type of a problem?
Is this problem part of the exercises by some well-thought-out design? I mean is it going to be used later on in the book? Or, is it just to give the reader some practice?
| First a comment. What this problem really says is that the subset of $\mathbb{R}^k$ defined by the equation $|x - a| = 2|x - b|$ is a sphere. So there is really not going to be any choice in determining $c$ and $r$. The point $c$ will need to be the centre of the sphere and $r$ its radius.
I imagine the main point of this exercise is to give readers practice with dot products, particularly the relation $|x|^2 = x \cdot x$.
To prove the result, write the following succession of equivalent equations, calculations that amount to completing the square with dot products.
$$
\begin{align*}
2|x - b| &= |x - a| \\
4|x-b|^2 &= |x - a|^2 \\
4 (x - b) \cdot (x - b) &= (x - a) \cdot (x - a) \\
4 x \cdot x - 8 x \cdot b + 4 b \cdot b &= x \cdot x - 2 x \cdot a + a \cdot a \\
3 x \cdot x - 8 x \cdot b + 2 x \cdot a &= a \cdot a - 4 b \cdot b \\
x \cdot x - 2 x \cdot \left( \frac{4}{3}b - \frac{1}{3} a\right) &= \frac{1}{3}a \cdot a - \frac{4}{3}b \cdot b \\
x \cdot x - 2 x \cdot \left( \frac{4}{3}b - \frac{1}{3} a\right) +\left( \frac{4}{3}b - \frac{1}{3} a\right) \cdot \left( \frac{4}{3}b - \frac{1}{3} a\right) &= \frac{1}{3}a \cdot a - \frac{4}{3}b \cdot b + \left( \frac{4}{3}b - \frac{1}{3} a\right) \cdot \left( \frac{4}{3}b - \frac{1}{3} a\right)\\
(x - c) \cdot (x -c) &= r^2 \\
|x - c| &= r
\end{align*}
$$
where $c = \frac{4}{3}b - \frac{1}{3} a$ and $r^2 = \frac{4}{9} a \cdot a - \frac{8}{9} a \cdot b + \frac{4}{9} b \cdot b = \frac{4}{9}(a - b) \cdot (a - b)$, so $r = \frac{2}{3}|a - b|$.
Technically, the exercise is incorrect when $a = b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\int \frac{2x}{9x^2+3}dx=?$ So this one seems very easy. And it really is (I guess). But I have an issue with this one. I solved it this way:
\begin{align}
& \int \frac{2x}{9x^2+3}dx=\frac{1}{6} \int \frac{x}{3x^2+1}dx=\frac{1}{6} \left( x\frac{\arctan(\sqrt3x)}{\sqrt{3}}-\int 1\cdot\frac{\arctan(\sqrt3x)}{\sqrt3} dx\right) \\[10pt]
= {} & \frac{1}{6} \left( x\frac{\arctan(\sqrt3x)}{\sqrt{3}}-x\frac{\arctan(\sqrt3x)}{\sqrt{3}}+\frac{\arctan(\sqrt3x)}{\sqrt3}\right)=\frac{\arctan(\sqrt3x)}{\sqrt3}+C
\end{align}
But at the same time, the textbook says, that it is equal to this:
$$
\frac{1}{9}\log(3x^2+1)+C
$$
Did I do something wrong? It would be quite hard for me to believ that these to functions are the same.
Thanks for the responses!
| I'd have written $u = 9x^2+ 3$ so $du = 18x\,dx$ and $2x\,dx = \dfrac{du} 9$. Then you have
$$
\int \frac{2x}{9x^2+3} \, dx = \frac 1 9 \int \frac{du} u = \frac 1 9 \log|u|+C = \frac 1 9 \log ( 9x^2+3) + C.
$$
This is the same as
$$
\frac 1 9 \log ( 3( 3x^2+1)) + C = \frac 1 9 \left( \log(3x^2+1) + \log 3 \right) + C = \frac 1 9 \log(3x^2+1) + (\text{a different } C).
$$
Your initial integration by parts says
$$
\frac{1}{6} \int \frac{x}{3x^2+1}dx=\frac{1}{6} \left( x\frac{\arctan(\sqrt3x)}{\sqrt{3}}-\int 1\cdot\frac{\arctan(\sqrt3x)}{\sqrt3} dx\right).
$$
That's alright as far as it goes. I haven't attempted to figure out just what you're doing after that, but let's recall how to antidifferentiate the arctangent function:
\begin{align}
& \int \arctan x\,dx = \overbrace{\int u\,dx = xu - \int x\,du}^\text{integration by parts} \\[10pt]
= {} & x\arctan x - \int \frac{x\,dx}{1+x^2} = x\arctan x - \int \frac{dw/2}{w} \\[10pt]
= {} & x \arctan x - \frac 1 2 \log|w|+C \\[10pt]
= {} & x \arctan x - \frac 1 2 \log(1+x^2) + C.
\end{align}
So I suspect the first term in the result just above this line (suitably adapted with your $\sqrt3$, etc.) will cancel out the first term in your integration by parts and then the result will ultimately be the same.
Generally I wouldn't integrate by parts when a simple substitution will work, unless there were some special reason to do that.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\tan A+\tan B+\tan C=6$ and $\tan A\tan B=2 $ in $\triangle ABC$, then find the type of triangle.
In $\triangle ABC$, $\tan A+\tan B+\tan C=6 \\
\tan A\tan B=2
$
Then the triangle is
$a.)\text{Right-angled isosceles} \\
b.) \text{Acute-angled isosceles}\\
\color{green}{c.)\text{Obtuse-angled}} \\
d.)\text{equilateral} $
$\ \ \ $
$\tan A+\tan B+\tan C=6 \\
\tan A\tan B=2 \\
A+B=180-C\\
\dfrac{\tan A+\tan B}{1-\tan A\tan B}=-\tan C\\
\tan A+\tan B=\tan C\\
\tan C=3 \\
\tan A\tan B=2\ \text{and} \ \tan A+\tan B=3 \\
\implies \tan A=2, \tan B=1\ \ \text{or}\ \ \tan A=1, \tan B=2
$
Now I am stucked, I look for a short and simple way.
I have studied maths upto $12$th grade .
Note:- Calculator is not allowed.
| You have done the hard work
As $0<\tan A,\tan B,\tan C<\infty$
$0<A,B,C<\dfrac\pi2$
As $\tan A,\tan B,\tan C$ are distinct, so will be $A,B,C$
| {
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"timestamp": "2023-03-29T00:00:00",
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Area of region under 2 curves. Find the area of the region R enclosed by the line $y=2x−1$ and the parabola $y^2=4x+141$
Here is what I have done:
*
*Write each equation in terms of x. ($x=\frac{y+1}{2}$ and $x=\frac{y^2-141}{4}$)
*Find intersections: y= -11 and 13
*Write the integral (I don't quite know).
*Calculate answer.
If someone could check over my step 1 and 2 that would be awesome. Also If someone could do step 3 and 4 (would be awesome).
| I checked $1$ and $2$ and it is correct.
Step 3:
If $f(y)>g(y)$ for $x$-values between $[-11,13]$ then the integral is
$\int_{-11}^{13}{\left(f(y)-g(y)\right)}{dy}$
Since $\frac{y+1}{2}>\frac{y^2-141}{4}$ from $[-11,13]$ we have
$\int_{-11}^{13}\left(\frac{y+1}{2}-\frac{y^2-141}{4}\right)dy$
$\frac{1}{4}\int_{-11}^{13}\left({2y+2}-{y^2+141}\right)dy$
Now use the following rule that $\int{jy^k}dy=\frac{j}{k+1}y^{k+1}$
$\frac{1}{4}\int_{-11}^{13}\left(-y^2+2y+143\right)dy$
$\frac{1}{4}\left(\int_{-11}^{13}\left(-y^2\right)dy+\int_{-11}^{13}2ydy+\int_{-11}^{13}143dy\right)$
$\frac{1}{4}\left(\int_{-11}^{13}\left(-y^2\right)+\int_{-11}^{13}2y+\int_{-11}^{13}143\right)$
$\frac{1}{4}\left(|_{-11}^{13}\left(-\frac{1}{3}y^3\right)dy+|_{-11}^{13}{y^2}+|_{-11}^{13}143y\right)$
Now use $\int_{a}^{b}f(x)=F(b)-F(a)$ where $F(x)$ is the indefinite integral.
$\frac{1}{4}\left(\left(-\frac{1}{3}{(13)}^3+\frac{1}{3}{(-11)}^3\right)+\left({13}^{2}-(-11)^{2}\right)+\left(143(13)-143(-11)\right)\right)$
Now do the calculations on your calculator this ensure you did not make any mistakes.
$\frac{1}{4}(\frac{-2197-1331}{3}+48+3432)$
$\frac{1}{4}(\frac{-2197-1331+10440}{3})$
$\frac{6912}{12}=576$
You are correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate $\lim _{x\to \infty }\left(\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right)\right)$? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to \infty }\left(\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right)\right)$$
| I will assume you know that $\frac{\sin u}{u}\xrightarrow[u\to 0]{} \sin^\prime(0) = 1$.
*
*First, observe that $\frac{x+1}{4x^2+3} \xrightarrow[x\to\infty]{} 0$.
*Use this observation to rewrite
$$
\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right)
= \frac{2x^3+x}{x^2+1}\frac{x+1}{4x^2+3} \frac{\sin\left(\frac{x+1}{4x^2+3}\right)}{\frac{x+1}{4x^2+3}}
$$
*Compute the limit of the first term, now that the second is dealt with:
$$
\frac{2x^3+x}{x^2+1}\frac{x+1}{4x^2+3} = \frac{2x}{4x}\frac{2+x^{-2}}{1+x^{-2}}\frac{1+x^{-1}}{4+3x^{-2}} \xrightarrow[x\to\infty]{}\frac{1}{2}
$$
*Conclude that
$$
\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right) \xrightarrow[x\to\infty]{}\frac{1}{2} \cdot 1 = \frac{1}{2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
How to evaluate $\lim\limits _{x\to \infty }\left(\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+3}}\right)^{5x+1}$? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to \infty }\left(\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+3}}\right)^{5x+1}$$
I tried like that:
$$\lim _{x\to \infty }\left(e^{\left(5x+1\right)ln\left(\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+2}}\right)}\right)$$
Then:
$$ln\left(\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+2}}\right)=\frac{\sqrt[3]{x^4+1}-\sqrt[6]{9x^8+2}}{x+\sqrt[6]{9x^8+2}}$$
$$=(5x+1)\frac{\sqrt[3]{x^4+1}-\sqrt[6]{9x^8+2}}{x+\sqrt[6]{9x^8+2}}$$
$$=\frac{5x^2\sqrt[3]x(1-\sqrt[3]{3})}{x+x\sqrt[3]x\sqrt[3]3}$$
Now I don't understand a thing:
the result of the latter fraction is $-\infty$ so $e^{-\infty}=\color{red}0$?
I'm sure I missed a few step, help me. Thanks
| I'd set $x=1/t^3$ (with $x>0$, which is not restrictive) so the base becomes a fraction with
$$
\frac{1}{t^3}+\sqrt[3]{\frac{1}{t^{12}}+1}=
\frac{t+\sqrt[3]{1+t^{12}}}{t^4}
$$
at the numerator and
$$
\frac{1}{t^3}+\sqrt[6]{\frac{9}{t^{24}}+3}=
\frac{t+\sqrt[6]{9+3t^{24}}}{t^4}
$$
at the denominator. So, taking the logarithm, you have
$$
\lim_{t\to0^+}
\frac{5+t^3}{t^3}\log\frac{t+\sqrt[3]{1+t^{12}}}{t+\sqrt[6]{9+3t^{24}}}
=-\infty
$$
because the fraction under the logarithm has limit $1/\sqrt[3]{3}<1$.
Thus your limit is the same as $\lim_{u\to-\infty}e^u=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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induction to prove the equation $3 + 9 + 15 + ... + (6n - 3) = 3n^2$ I have a series that I need to prove with induction. So far I have 2 approaches, though I'm not sure either are correct.
$$3 + 9 + 15 + ... + (6n - 3) = 3n^2$$
1st attempt:
\begin{align*}
& = (6n - 3) + 3n^2\\
& = 3n^2 + 6n - 3\\
& = (3n^2 + 5n - 4) + (n + 1)
\end{align*}
That seems way wrong lol ^^^
2nd attempt:
\begin{align*}
f(n) & = 3 + 9 + 15 + ... + (6n - 3)\\
f(n + 1) & = 6(n + 1) - 3\\
f(n + 1) & = 6(n - 3) + 6(n + 1) - 3\\
& = ?
\end{align*}
I don't know I feel like I'm headed in the wrong direction
I guess another attempt I have would be:
\begin{align*}
f(n) & = 3n^2\\
f(n+1) & = 3(n + 1)^2\\
& = 3(n^2 + 2n + 1)\\
& = 3n^2 + 6n + 3\\
& = f(n) + (6n + 3)
\end{align*}
| Using summation notation gives
$$3 + 9 + 15 + \cdots + 6n - 3 = \sum_{k = 1}^n (6k - 3)$$
Let $P(n)$ be the statement
$$\sum_{k = 1}^{n} (6k - 3) = 3n^2$$
To prove this statement by induction, we must show that $P(1)$ holds and that whenever $P(m)$ holds for some positive integer $m$, then $P(m + 1)$ holds since we then have the chain of implications
$$P(1) \implies P(2) \implies P(3) \implies \cdots$$
Proof. Let $n = 1$. Then
$$\sum_{k = 1}^{1} (6k - 3) = 6 \cdot 1 - 3 = 6 - 3 = 3 = 3 \cdot 1 = 3 \cdot 1^2$$
so $P(1)$ holds.
Since $P(1)$ holds, we may assume there exists a positive integer $m$ such that $P(m)$ holds. Then
$$\color{green}{\sum_{k = 1}^{m} (6k - 3) = 3m^2}$$
Let $n = m + 1$. Then
\begin{align*}
\sum_{k = 1}^{m + 1} (6k - 3) & = \color{green}{\sum_{k = 1}^{m} (6k - 3)} + [6(m + 1) - 3] && \text{by definition}\\
& = \color{green}{3m^2} + 6(m + 1) - 3 && \text{by the induction hypothesis}\\
& = 3m^2 + 6m + 6 - 3\\
& = 3m^2 + 6m + 3\\
& = 3(m^2 + 2m + 1)\\
& = 3(m + 1)^2
\end{align*}
Hence, $P(m) \implies P(m + 1)$.
Since $P(1)$ holds and $P(m) \implies P(m + 1)$ for each positive integer $m$, $P(n)$ holds for all positive integers.$\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Finiteness of the sums of reciprocals of positive solutions of $\tan x = x$ and $x = \tan \sqrt x$ Let $a_n$ be the sequence of positive solutions of the equation $\tan x=x$ and $b_n$ be a sequence of positive solutions of the equation $x=\tan \sqrt x$.
Prove that $\sum \dfrac{1}{a_n}$ diverges but $\sum \dfrac{1}{b_n}$ converges.
We have $\tan {a_n}=a_n$ and $b_n=\tan \sqrt b_n$.In order to check convergence of $\sum \dfrac{1}{a_n}$ and $\sum \dfrac{1}{b_n}$ we have to check convergence of $\sum \dfrac{1}{\tan a_n}$ and $\sum \dfrac{1}{\tan \sqrt b_n}$.
I am facing problem on how to judge whether the $n^{th}$ term tends to zero or not.Any help will be appreciated.
| For each integer $n$, $\tan x$ is a bijection from $(\tfrac{2n-1}{2}\pi,\tfrac{2n+1}{2}\pi)$ to $\mathbb{R}$.
So for each positive integer $n$, there is exactly one positive solution to $x = \tan x$ in each interval $(\tfrac{2n-1}{2}\pi,\tfrac{2n+1}{2}\pi)$. Also, the only solution in $(-\tfrac{1}{2}\pi,\tfrac{1}{2}\pi)$ is $x = 0$, which isn't positive.
Hence, the $n$-th smallest positive solution to $x = \tan x$ satisfies $\tfrac{2n-1}{2}\pi < a_n < \tfrac{2n+1}{2}\pi$.
Using similar logic, you can show that the $n$-th smallest positive solution to $x = \tan \sqrt{x}$ satisfies $\left(\tfrac{2n-1}{2}\pi\right)^2 < b_n < \left(\tfrac{2n+1}{2}\pi\right)^2$.
Can you determine the convergence of $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{a_n}$ and $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{b_n}$ from this?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632696",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How to evaluate $\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2} \, \mathrm{d}\theta$ I have some trouble in how to evaluate this integral:
$$
\int_{0}^{\pi}\theta\ln\left(\tan\left(\theta \over 2\right)\right)
\,\mathrm{d}\theta
$$
I think it maybe has another form
$$
\int_{0}^{\pi}\theta\ln\left(\tan\left(\theta \over 2\right)\right)
\,\mathrm{d}\theta
=
\sum_{n=1}^{\infty}{1 \over n^{2}}
\left[\psi\left(n + {1 \over 2}\right) - \psi\left(1 \over 2\right)\right]
$$
| \begin{align}
J&=\int_0^\pi \theta\ln\left(\tan\left(\frac{\theta}{2}\right)\right)d\theta\\
&\overset{u=\tan\left(\frac{\theta}{2}\right)}=4\underbrace{\int_0^\infty \frac{\arctan u\ln u}{1+u^2}du}_{=K}\\
K&\overset{\text{IBP}}=\underbrace{\left[\arctan u\left(\int_0^u \frac{\ln t}{1+t^2}dt\right)\right]_0^\infty}_{=0}-\int_0^\infty \frac{1}{1+u^2}\left(\underbrace{\int_0^u \frac{\ln t}{1+t^2}dt}_{y(t)=\frac{t}{u}}\right)du\\
&=-\int_0^\infty \left(\int_0^1 \frac{u\ln(uy)}{(1+u^2)(1+u^2y^2)}dy\right)du\\
&=-\int_0^\infty \left(\int_0^1 \frac{u\ln u}{(1+u^2)(1+u^2y^2)}dy\right)du-\int_0^1 \left(\int_0^\infty \frac{u\ln y}{(1+u^2)(1+u^2y^2)}du\right)dy\\
&=-\int_0^\infty \left[\frac{\arctan(uy)}{1+u^2}\right]_{y=0}^{y=1}\ln udu-\frac{1}{2}\int_0^1 \left[\frac{\ln\left(\frac{1+u^2}{1+u^2y^2}\right)}{1-y^2}\right]_{u=0}^{u=\infty}\ln ydy\\
&=-K+\int_0^1 \frac{\ln^2 y}{1-y^2}dy\\
&=\frac{1}{2}\int_0^1 \frac{\ln^2 y}{1-y}dy-\frac{1}{2}\underbrace{\int_0^1 \frac{y\ln^2 y}{1-y^2}dy}_{z=y^2}\\
&=\frac{7}{16}\int_0^1 \frac{\ln^2 y}{1-y}dy\\
J&=\frac{7}{4}\int_0^1 \frac{\ln^2 y}{1-y}dy\\
J&=\frac{7}{4}\times 2\zeta(3)\\
&=\boxed{\frac{7}{2}\zeta(3)}
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Choosing a substitution to evaluate $\int \frac{x+3}{\sqrt{x+2}}dx$ Is there any other value you can assign to the substitution variable to solve this integral?
$$\int \frac{x+3}{\sqrt{x+2}}dx$$
Substituting $u = x + 2$:
$$du = dx; u +1 = x+3 ,$$
and we get this new integral that we can then split into two different ones:
$$\int \frac{u + 1}{\sqrt{u}}du = \int \frac{u}{\sqrt{u}}du + \int \frac{1}{\sqrt{u}}du .$$
We can substitute again $s = \sqrt u$ and get two immediate integrals:
$$s = \sqrt{u}; \quad ds = \frac{1}{2\sqrt{u}}du; \quad 2s^2 =u .$$ Substituting back $u$ to $s$ and $x$ to $u$ we get this result,
$$s^2 + \ln{\left | \sqrt{u} \right |} = u + \ln{\left | \sqrt{u} \right |} = x+2+\ln{\left | \sqrt{x+2} \right |},$$ which doesn't look quite to be right. What am I doing wrong? I'm pretty unsure about the second substitution, $2s^2 = u$. Is it correct?
| Let $\sqrt{x+2}=t\implies \frac{dx}{2\sqrt{x+2}}=dt$ or $dx=2t\ dt$
$$\int \frac{x+3}{\sqrt{x+2}}\ dx$$$$=\int \frac{t^2-2+3}{t}(2t\ dt)$$
$$=2\int (t^2+1)\ dt$$
$$=2\left(\frac{t^3}{3}+t\right)+C$$
$$=2\left(\frac{(x+2)^{3/2}}{3}+\sqrt{x+2}\right)+C$$
$$=\frac 23(x+5)\sqrt{x+2}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $a(x+y+z) = x(a+b+c)$ If $(a^2+b^2 +c^2)(x^2+y^2 +z^2) = (ax+by+cz)^2$
Then prove that $a(x+y+z) = x(a+b+c)$
I did expansion on both sides and got:
$a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2(abxy+bcyz+cazx) $
but can't see any way to prove $a(x+y+z) = x(a+b+c)$. How should I proceed?
| $$a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2=a^2x^2+b^2y^2+c^2z^2+2axby+2axcz+2byxz$$
$$a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2axby+2axcz+2byxz$$
$$a^2y^2-2axby+b^2x^2+a^2z^2-2axcz+c^2x^2+b^2z^2-2byxz+c^2y^2=0$$
$$(ay-bx)^2+(az-cx)^2+(bz-cy)^2=0$$
From here the answer is clear. All three terms are zero.
| {
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Find all pairs of nonzero integers $(a,b)$ such that $(a^2+b)(a+b^2)=(a-b)^3$
Find all pairs of nonzero integers $(a,b)$ such that
$(a^2+b)(a+b^2)=(a-b)^3$
My effort
Rearranging the equation I have
\begin{array}
\space (a^2+b)(a+b^2)-(a-b)^3 &=0 \\
a^2(b^2+3b)+a(-3b^2+b)+2b^3 &=0 \\
\end{array}
Letting $a=x$, we have the polynomial $Q(x)$ such that $$Q(x)=x^2(b^2+3b)+x(-3b^2+b)+2b^3 $$
So I must have that the roots $a_1,a_2$ must be such that
\begin{array}
\space a_1+a_2=-\cfrac{-3b+1}{b+3} \\
a_1\cdot a_2=\cfrac{2b^2}{b+3} \\
\end{array}
I am stuck now,should I go by brute force and verify some values or
there's still something I can do here to simplify the problem ?
| The solutions of a quadratic with integer coefficients are either both rational or both irrational. So if one of $a_1,a_2$ is an integer the other must be rational. Say $a_1\in \Bbb Z$ and $a_2=\frac nm\in\Bbb Q$ is a rational with denominator $m$. Then the denominator of $a_1+a_2$ is also $m$ and the denominator of $a_1a_2$ is a divisor of $m$. We conclude that the denominator of $a_1a_2$ divides both $b+3$ and $b+1$, hence divides $2$.
Then $2a_1a_2$ is an integer, i.e.,
$$ b+3\mid 4b^2.$$
*
*If $3\nmid b$, $b+3$ and $b$ are coprime, hence $b+3\mid 4$ and $b\in\{-7,-5,-4,-2,-1,1\}$
*If $3\mid b$, say $b=3c$, we find $c+1\mid 12c^2$. As $c+1$ and $c$ are coprime, $c+1\mid 12$ and $b\in\{-39,-21,-15,-12,-9,-6,0,3,6,9,15,33\}$
The rest of the problem is finite.
| {
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If $a \ne b$ in a ring $R$ satisfy $a^3 = b^3$ and $a^2b = b^2a$, show that $a^2 + b^2$ is not a unit. If $a \ne b$ in a ring $R$ satisfy $a^3 = b^3$ and $a^2b = b^2a$, show that $a^2 + b^2$ is not a unit.
So I am thinking that I should be able to do this by contradiction. So if I assume there is some element $z\in R$ such that $(a^2 + b^2)z=1$ then
$a^2z + b^2z=1$, but I'm not sure this helps in any way.
I also tried this just to see if it would reveal the right way to proceed, but I am still stuck.
$a^2 + b^2=a^2 + b^2$
$(a^2 + b^2)b=(a^2 + b^2)b$
$a^2b + b^3=a^2b + b^3$
$a^2b + b^3=b^2a + a^3$
| You have, $(a-b)(a^2+b^2)=a^3+ab^2-ba^2-b^3=0$
| {
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Equation of plane perpendicular to given plane Find the equation of the plane which contains the line of intersection of the planes $x+2y+3z-4=0$ and $2x+y-z+5=0$ and which is perpendicular to the plane $5x+3y-6z+8=0$
By setting $z=0$ I found a point $(-\frac{14}{3}, \frac{13}{3}, 0)$ but not able to obtain normal vector of the plane. Would finding complete equation of line of intersection be more helpful?
| In answer to your last question, yes: because our required plane, which we shall call $P$, must contain a vector, $\mathbf{l}$ say, parallel to our line of intersection and also the vector, $\mathbf{n} := (5,3,-6)$, normal to the plane $5x+3y-6y+8=0$.
Vector $\mathbf{l}$ is perpendicular to the normals: $(1,2,3)$ and $(2,1,-1)$, which means it equals their cross product:
$$\mathbf{l} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & 3 \\
2 & 1 & -1 \\
\end{vmatrix} = (-5,7,-3).
$$
Because plane $P$ contains both $\mathbf{l}$ and $\mathbf{n}$, its normal, $\mathbf{n_p}$ say, equals their cross product:
$$\mathbf{n_p} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-5 & 7 & -3 \\
5 & 3 & -6 \\
\end{vmatrix} = (-33,-45,-50).
$$
Multiplying it by $-1$ for convenience, we have $\mathbf{n_p} = (33,45,50)$. So $P$ has equation $33x+45y+50z=c\;$ for some value $c$. Our point $\left(\dfrac{-14}{3},\dfrac{13}{3},0\right)$ lies on $P$ so we substitute it into this equation to find $c$:
$$c = 33\times \dfrac{-14}{3} + 45\times \dfrac{13}{3}= 41.$$
So our plane $P$ is:
$$33x+45y+50z=41.$$
| {
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Improper Integral $\int_0^1\frac{\arcsin^2(x^2)}{\sqrt{1-x^2}}dx$ $$I=\int_0^1\frac{\arcsin^2(x^2)}{\sqrt{1-x^2}}dx\stackrel?=\frac{5}{24}\pi^3-\frac{\pi}2\log^2 2-2\pi\chi_2\left(\frac1{\sqrt 2}\right)$$
This result seems to me digitally correct?
Can we prove that the equality is exact?
| \begin{align}\int_0^{1} \frac{\arcsin^2 x^2}{\sqrt{1-x^2}}\,dx &= \frac{1}{2}\int_0^{1} \frac{\arcsin^2 x}{\sqrt{x}\sqrt{1-x}}\,dx \tag{1}\\&= \frac{1}{2}\int_0^{\pi/2} \frac{\theta^2\cos \theta}{\sqrt{\sin \theta - \sin^2 \theta}}\,d\theta \tag{2}\\&= \frac{1}{\sqrt{2}}\int_0^{\pi/2} \frac{\left(\frac{\pi}{2} - \theta\right)^2\cos \frac{\theta}{2}}{\sqrt{1-2\sin^2 \frac{\theta}{2}}}\,d\theta \tag{3}\\&= \int_0^{\pi/2} \left(\frac{\pi}{2} - 2\arcsin \left(\frac{\sin \alpha}{\sqrt{2}}\right)\right)^2\,d\alpha \tag{4}\\&= \frac{\pi^3}{8} - 2\pi \int_0^{\pi/2} \arcsin \left(\frac{\sin \alpha}{\sqrt{2}}\right)\,d\alpha + 4\int_0^{\pi/2} \left(\arcsin \left(\frac{\sin \alpha}{\sqrt{2}}\right)\right)^2\,d\alpha \tag{5}\end{align}
where, we made $x \mapsto \sqrt{x}$ in step $(1)$. In step $(2)$ we made $\theta = \arcsin x$ and finally in $(3)$ we made the change of variable $\sin \dfrac{\theta}{2} = \dfrac{\sin \alpha}{\sqrt{2}}$.
Now we recall the famous series expansion: $\displaystyle \arcsin^2 x = \frac{1}{2}\sum\limits_{n=1}^{\infty} \dfrac{(2x)^{2n}}{n^2\binom{2n}{n}}$,
Hence, \begin{align*}\int_0^{\pi/2} \left(\arcsin \left(\frac{\sin \alpha}{\sqrt{2}}\right)\right)^2\,d\alpha &= \frac{1}{2}\sum\limits_{n=1}^{\infty} \dfrac{2^n}{n^2\binom{2n}{n}}\int_0^{\pi/2} \sin^{2n} \alpha \,d\alpha\\&= \frac{\pi}{4}\sum\limits_{n=1}^{\infty} \dfrac{1}{n^22^{n}} = \frac{\pi}{4}\operatorname{Li}_2 \left(\frac{1}{2}\right) = \frac{\pi}{8}\left(\zeta(2) - \log^2 2\right)\end{align*}
also, the infinite series expansion for $\displaystyle \arcsin x = \sum\limits_{n=0}^{\infty} \dfrac{\binom{2n}{n}x^{2n+1}}{(2n+1)4^n}$ give us,
\begin{align*}\int_0^{\pi/2} \arcsin \left(\frac{\sin \alpha}{\sqrt{2}}\right)\,d\alpha &= \frac{1}{\sqrt{2}}\sum\limits_{n=0}^{\infty} \dfrac{\binom{2n}{n}}{(2n+1)8^n}\int_0^{\pi/2} \sin^{2n+1} \alpha \,d\alpha\\&= \frac{1}{\sqrt{2}}\sum\limits_{n=0}^{\infty} \dfrac{1}{(2n+1)^2 2^n} = \chi_2 \left(\frac{1}{\sqrt{2}}\right)\end{align*}
Combining the results, $$\int_0^{1} \frac{\arcsin^2 x^2}{\sqrt{1-x^2}}\,dx = \frac{5\pi^3}{24} - \frac{\pi}{2}\log^2 2 - 2\pi \chi_2 \left(\frac{1}{\sqrt{2}}\right)$$
| {
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Proving $\tan A=\frac{1-\cos B}{\sin B} \;\implies\; \tan 2A=\tan B$
If $\tan A=\dfrac{1-\cos B}{\sin B}$, prove that $\tan 2A=\tan B$.
My effort:
Here
$$\tan A=\frac{1-\cos B}{\sin B}$$
Now
$$\begin{align}\text{L.H.S.} &=\tan 2A \\[4pt]
&=\frac{2\tan A}{1-\tan ^2A} \\[6pt]
&=\frac{(2-2\cos B)\over\sin B}{1-\frac{(1-\cos B)^2}{\sin^2 B}}
\end{align}$$
On simplification from here, I could not get the required R.H.S.
| Note that the given expression implies:
$\tan(A)=\frac{1-\cos(B)}{\sin(B)} \cdot \frac{1+\cos(B)}{1+\cos(B)}=\frac{\sin(B)}{1+\cos(B)}$
Therefore
$\tan(2A)=\frac{2\tan(A)}{1-\tan^2(A)}=\frac{2\frac{\sin(B)}{1+\cos(B)}}{1-\left(\frac{1-\cos(B)}{\sin(B)}\cdot \frac{\sin(B)}{1+\cos(B)}\right)}=\frac{2\sin(B)}{2\cos(B)}=\tan(B)$
| {
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integrate $\int dx \frac{2x+1}{(x^2-9)^\frac{5}{2}}$
$$\int dx \frac{2x+1}{(x^2-9)^\frac{5}{2}}$$
$x=\frac{3}{\sin\theta}$
$dx=\frac{3\sin\theta}{\cos^2\theta}d\theta$
$$\int d\theta \frac{\left(\frac{6}{\sin\theta}+1\right)\cdot\frac{3\sin\theta}{\cos^2\theta}}{\left(\frac{9}{\sin^2\theta}-9\right)^\frac{5}{2}}=\int d\theta \frac{\left(\frac{6}{\sin\theta}+1\right)\cdot\frac{3\sin\theta}{\cos^2\theta}}{\left(\frac{9}{\sin^2\theta}-9\right)^\frac{5}{2}}=\int d\theta \frac{\left(\frac{6}{\sin\theta}+1\right)\cdot\frac{3\sin\theta}{\cos^2\theta}}{\left(\frac{9}{\sin^2\theta}-9\right)^\frac{5}{2}}=\int d\theta \frac{\left(\frac{6}{\sin\theta}+1\right)\cdot\frac{3\sin\theta}{\cos^2\theta}}{\left(3\cos\theta\right)^{5}}$$
How should I continue?
| This is not a continuation of your method, but rather another technique.
$\int \frac{(2x+1) dx}{(x^2-9)^\frac{5}{2}}=\int \frac{2xdx}{(x^2-9)^\frac{5}{2}}+\int \frac{dx}{(x^2-9)^\frac{5}{2}}$
For the first term, let $u=x^2-9$, making $du=2x dx$, setting the integral up for a power rule.
For the second term, let $x=3\sec u$, making $du=3\sec u \tan u$. Some trigonometric manipulation will now be necessary. Can you proceed?
| {
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Find the smallest positive value taken by $a^3+b^3+c^3-3abc$
Find the smallest positive value taken by $a^3+b^3+c^3-3abc$ for positive integers $a,b,c$. Find all integers $a,b,c$ which give the smallest value.
Since it is generally hard to find the minimum of a multivariate polynomial, I tried factoring it at first. The $abc$ reminds me of AM-GM, but I wasn't sure how to relate the $a^3+b^3+c^3$ to it. Also the condition that we are working with integers may make it easier to find the minimum.
| The answer is $4$, for $a = 1, b = 1, c = 2$ (or some permutation).
We have
$$a^3 + b^3 + c^3 - 3abc = (a + b + c)[(a + b + c)^2 - 3(ab + bc + ca)].$$
This shows that any positive value taken by the expression must be at least as large as $a + b + c$, which leaves only the above cases and $a =b = c = 1$ to examine.
Edit Let $A = a^3 - 3abc + b^3 + c^3$. We will write $A$ in terms of the elementary symmetric polynomials. The leading term in lexicographic order is $a^3$, so we compute
$$A - (a + b + c)^3 = - 3a^2b - 3a^2 c - 3ab^2 - 9abc - 3ac^2 - 3b^2 c.$$
The leading term in this expression is $-3a^2b$, so we compute
$$A - (a + b + c)^3 + 3(a + b + c)(ab + ac + bc) = 0.$$
| {
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Is there an identity that says $|\sqrt {a^2+x^2} - \sqrt {a^2+y^2}| \leq |\sqrt {x^2} - \sqrt {y^2}|$? Is there an identity that says $|\sqrt {a^2+x^2} - \sqrt {a^2+y^2}| \leq |\sqrt {x^2} - \sqrt {y^2}|$?
Because of the nature of the square root function, its derivative monotonically decreases. so differences "further up" the function would be less than those lower down.
| Yes.
$$
\left|\sqrt {a^2+x^2} - \sqrt {a^2+y^2}\right| =\frac{\lvert x^2-y^2\rvert}{\left|\sqrt {a^2+x^2} + \sqrt {a^2+y^2}\right|}
= |\sqrt {x^2} - \sqrt {y^2}|\cdot \frac{|\sqrt {x^2} + \sqrt {y^2}|}{\left|\sqrt {a^2+x^2} + \sqrt {a^2+y^2}\right|}
\leq |\sqrt {x^2} - \sqrt {y^2}|
$$
| {
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Find all primes $p,q$ and even $n > 2$ such that $p^n+p^{n-1}+\cdots+1 = q^2+q+1$
Find all primes $p,q$ and even $n > 2$ such that $p^n+p^{n-1}+\cdots+1 = q^2+q+1$.
Attempt
The first thing I would do is simplify the geometric series to $\dfrac{p^{n+1}-1}{p-1} = q^2+q+1$. I was thinking from here we could use a modular arithmetic argument, perhaps using FLT. It is unclear to me though how to proceed since we are working with two different primes.
| Suppose that $n=2k$ for some integer $k>1$. Then we have $q(q+1)=p\left(\frac{p^{2k}-1}{p-1}\right)$. As $k>1$, $p\neq q$, so that $q$ divides $p^{2k}-1=\left(p^k+1\right)\left(p^k-1\right)$. That is, $q$ divides either $p^k-1$ or $p^k+1$. In any case, $q\leq p^k+1$. Therefore, $$p^{2k}+p^{2k-1}+2< p^{2k}+p^{2k-1}+\ldots+p=q^2+q\leq \left(p^k+1\right)^2+\left(p^k+1\right)=p^{2k}+3p^k+2\,.$$
Ergo, $p^{2k-1}<3p^k$, or $p^{k-1}<3$. Consequently, $k=2$ and $p=2$. Hence, the only solution $(p,q,n)$ to $p^n+p^{n-1}+\ldots+p+1=q^2+q+1$ with $p,q,n\in\mathbb{N}$ such that $p$ and $q$ are primes and that $n>2$ is even is $(p,q,n)=(2,5,4)$.
In fact, if $n$ is allowed to be an odd integer greater than $2$, then it follows that $p=2$ (otherwise $p^n+p^{n-1}+\ldots+p+1$ is even, but $q^2+q+1$ is odd). That is, we are looking for a prime $q$ such that $q^2+q=2^{n+1}-2$, or $(2q+1)^2=2^{n+3}-7$. As $n$ is odd, $n+3$ is even. That is, $$\left(2^{\frac{n+3}{2}}-2q-1\right)\left(2^{\frac{n+3}{2}}+2q+1\right)=7\,.$$ Hence, $2^{\frac{n+3}{2}}-2q-1=1$ and $2^{\frac{n+3}{2}}+2q+1=7$. Consequently, $2^{\frac{n+3}{2}}\leq 6<2^3$. Thus, $n<3$, which is a contradiction.
In conclusion, the only solution $(p,q,n)\in\mathbb{N}^3$ to $p^n+p^{n-1}+\ldots+p+1=q^2+q+1$ with $p$ and $q$ being prime is $(p,q,n)=(2,5,4)$.
| {
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Discriminant of Elliptic Curves In the study of elliptic curves, specifically in Weierstrass form, you have the equation
$E : y^2 = x^3 +ax +b$.
However I have found the discriminant comes in two different forms:
$\Delta = -16(4a^3 + 27b^2) $ or $\Delta = 4a^3 + 27b^2$
I understand how to get the second equation, but where does the $-16$ come from?
From the Wiki page: "Although the factor −16 is irrelevant to whether or not the curve is non-singular, this definition of the discriminant is useful in a more advanced study of elliptic curves."
| A cubic over $k$ in Weierstrass form (affine form) is given by $$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6.$$ The discriminant is defined by $$\Delta = -b_2^2b_8-8b_4^3-27b_6^2+9b_2b_4b_6,$$ where $ b_2=a_1^2+4a_2$, $ b_4=2a_4+a_1a_3 $, $b_6=a_3^2+4a_6$ and $b_8 = a_{1}^{2} a_{6}+4 a_{2} a_{6}-a_{1} a_{3} a_{4}+a_{2} a_{3}^{2}-a_{4}^{2}$.
Finally, a elliptic curve over $k$ is a cubic in Weiertrass form, where $\Delta \neq 0$ (i.e., is a non-singular cubic in Weiertrass form).
We can make some substitutions to simplify the equation of cubic in Weiertrass form, the first assumes $char(k)$ is not $2$. Replacing $y$ by $ \frac{1}{2} \left(y-a_1x-a_3\right)$, the result is $$y^2=4x^3+b_2x^2+2b_4x+b_6.$$
The second assumes in addition that $char(k) \neq 3$. Replace $(x,y)$ by $\left( \frac{x-3b_2}{36}, \frac{y}{108} \right) $, and the result is $$y^2=x^3-27c_4-54c_6,$$ where $c_4=b_2^2-24b_ 4$ and $c_6=-b_2^3+36b_2b_4-216b_6.$
Moreover, when $char(k)$ not is $2$ or $3$, we have $$ 1728\Delta=c_4^3 - c_6^2. $$
Now, consider the cubic $y^2=x^3+ax+b$ over $k$. If $char(k)$ not is $2$ or $3$, we have $c_4=-48a $ and $c_6=-864b$, so $$\Delta = \frac{(-48a)^3-(-864)^2}{1728} = -16(4a^3+27b^2).$$
Thus, assuming that $char(k)\neq2$ and $char(k) \neq 3$, an elliptic curve over $k$ is given by $$y^2=x^3+ax+b,$$ where $\Delta=-16(4a^3+27b^2) \neq 0$.
Note that, $\Delta=-16(4a^3+27b^2) = 0$ if, and only if, $4a^3+27b^2 = 0$, because $16=2^4 \neq 0$ in $k$ with $char(k) \neq 2$. Thus, the factor $−16$ is irrelevant in this case.
See Chapter III of the book 'Elliptic Curves' of Anthony Knapp for more information.
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For which values of $x$ does the series converge? $$\sum_{k=0}^{\infty} \frac{3^k\cdot 3^2 \cdot x^k}{2^k}$$
I have to find the values of $x$ for which the series converges.
I used the root test,the limit of the absolute value of the kth root of the series must be smaller than 1 and I ended up with $|3x/2|<1 $ which leads to
$-2/3< x <2/3 $.
and then I put the once $-2/3 $ and once $2/3 $ in the original series and tested for convergence and found that for both values the series diverges which means the series only converges in the open interval $(-2/3,2/3)$.
Am I right or did I miss something?
| The geometric series
\begin{align*}
\sum_{k=0}^{\infty}x^k=\frac{1}{1-x}\qquad\qquad |x|<1
\end{align*}
is convergent for $x\in(-1,1)$.
OPs example is a geometric series (somewhat in disguise), since
\begin{align*}
\sum_{k=0}^{\infty}\frac{3^k3^2x^k}{2^k}&=3^2\sum_{k=0}^{\infty}\left(\frac{3x}{2}\right)^k\\
&=\frac{9}{1-\frac{3}{2}x}\\
&=\frac{18}{2-3x}
\end{align*}
which is convergent for
\begin{align*}
&\left|\frac{3}{2}x\right|<1\\
\end{align*}
or equivalently for $x\in\left(-\frac{2}{3},\frac{2}{3}\right)$.
| {
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An ellipse is drawn with the major and minor axes of lengths of $10$ and $8$, Then radius of circle which touches that ellipse.
An ellipse is drawn with the major and minor axes of lengths of $10$ and $8$ respectively
Using one focus as center is drawn such that it is tangent to the ellipse.
The radius of the circle is
$\bf{My\; Try::}$ Here I have assume Center of ellipse is at origin.
So equation of ellipse is $\displaystyle \frac{x^2}{25}+\frac{y^2}{16}=1$
Using Some calculation $\displaystyle e = \frac{3}{5}$ and So focus is $\displaystyle (\pm 3,0)$
Now equation of Circle whose one center is at one focus $(3,0)$
So equation of Circle is $(x-3)^2+y^2=r^2\;,$ Where radius is $=r$
Now eliminating $y$ from these two equation, We get $\displaystyle \frac{x^2}{25}+\frac{r^2-(x-3)^2}{16} = 1$
So we get $\displaystyle 16x^2+25r^2-25(x-3)^2=400\Rightarrow 16x^2+25r^2-25x^2-225+150x=400$
So $\displaystyle -9x^2+150x+25r^2-625=0\Rightarrow 9x^2-150x-25(r^2-25)=0$
Now given Ellipse and Circle are Tangent.
So we will put $\bf{Discriminant =0}$
So $\displaystyle (150)^2-+4\cdot 25 \cdot 9 (r^2-25) =0\Rightarrow r^2=0\Rightarrow r=0$
But When I solved Using Parametric Coordinate I get $r=2$
I did not understand Why I am not getting Same answer , plz explain me
Thanks
| The circles of centers $(3,0)$ and radius $r = 5 \pm 3$ are clearly tangent to the ellipse. So why aren't the discriminant of your equation in $x$ equal to $0$ for these values of $x$? Because there are also other solutions in $x$. The problem is that the corresponding $y$'s are complex ( try solving the system obtained).
One way to find tangent circles to an ellipse is using Lagrange multipliers. You want to find level curves of the function $f(x,y) = (x-3)^2 + y^2$ that are tangent to the curve $g(x,y) = x^2/25 + y^2/16-1=0$. So get the Lagrange function $F\colon = f - t \cdot g$ and solve the system in $(x,y,t)$
$\colon \frac{\partial F}{\partial l}=0$, $l=x,y,t$. You will find in fact three circles, the two obvious ones and (surprise!) the one of radius $r=0$ ( although for the last one the points where the curves are tangent have complex coordinates).
$\bf{ Added}$: Let's see how that Lagrange system works out:
$$ 2( x-3) - t \frac{2 x}{25} =0 \\
2 y - t \frac{y}{8} = 0\\
x^2/25 + y^2/16 = 1$$
The second equation is $y(16-t) = 0$. Hence $y=0$ or $t = 16$. In the first case $y=0$ and so $x = \pm 5$. In the second case, $t=16$ and from the first equation we get $x=\frac{25}{3}$, and from the third, $y=\pm \frac{16 i}{3}$. Note that $(x-3)^2 + y^2 =( \frac{16}{3})^2 + ( \frac{16i}{3})^2=0$.
Let's also see what happens with the circle of radius $2$ with center $(3,0)$. The intersection of the circle with the ellipse is given by the system
$$(x-3)^2 + y^2 = 4 \\
x^2/25 + y^2/16 = 1$$
with solutions $(3,0)$ ( with multiplicity $2$ - but not for the equation in $x$ but as the fact that the curves are tangent there ) and two more complex solutions
$(x,y) = (35/3, \pm \frac{16 \sqrt{10} i}{3})$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Solve $3 = -x^2+4x$ by factoring I have $3 = -x^2 + 4x$ and I need to solve it by factoring. According to wolframalpha the solution is $x_1 = 1, x_2 = 3$.
\begin{align*}
3 & = -x^2 + 4x\\
x^2-4x+3 & = 0
\end{align*}
According to wolframalpha $(x-3) (x-1) = 0$ is the equation factored, which allows me to solve it, but how do I get to this step?
| Solving by factoring means reducing the polynomial to a lower degree by dividing it by factors $(x-x_i)$ for already known roots $x_i$.
The reason for this is that the lower degree polynomial might be easier to solve. Here this is not necessary, as a order $2$ polynomial is already solvable with moderate effort.
How to know roots? For this kind of task one should always try the $0, \pm 1, \pm 2$, which here gives $x_1 = 1$.
What is left is the division:
\begin{align}
& x^2 - 4x + 3 : x - 1 = x - 3 \\
-&(x^2-x) \\
= & -3x+3 = -3(x-1)
\end{align}
So we have the factored polynomial $(x+1)(x-3)$.
| {
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Proving for all integer $n \ge 2$, $\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$ Prove the following statement by mathematical induction:
For all integer $n \ge 2$, $$\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$$
My attempt: Let the given statement be $p(n)$ .
1.\begin{align*} \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} & =\frac{\sqrt 2 +1}{\sqrt 2} \\
2 &< \sqrt 2 +1 \\
\sqrt 2 &< \frac{\sqrt 2 +1}{\sqrt 2}=\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} \end{align*}
Hence, $p(2)$ is true.
2.For an arbitrary integer $k \ge 2$, suppose $p(k)$ is true.
That is, $$\sqrt k < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt k}$$
Then we must show that $p(k+1)$ is true.
We're going to show that $$\sqrt {k+1} < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt k}+\frac{1}{\sqrt {k+1}}$$
I'm stuck on this step. I can't develop it further. How can I complete this proof?
| For the last induction step
$$\sqrt{k^2 + k}+1> k +1 \implies \frac{\sqrt{k^2 + k}}{\sqrt{k+1}}+ \frac1{\sqrt{k+1}}> \sqrt{k+1}\implies \sqrt{k} +\frac1{\sqrt{k+1}}> \sqrt{k+1}.$$
Hence
$$\sum_{j=1}^{k} \frac1{\sqrt{j}} > \sqrt{k} \implies \sum_{j=1}^{k+1} \frac1{\sqrt{j}} > \sqrt{k} + \frac1{\sqrt{k+1}} > \sqrt{k+1}.$$
For a simpler proof
$$\sum_{k=1}^n \frac1{\sqrt{k}}> \frac{n}{\sqrt{n}}= \sqrt{n}.$$
| {
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Prove $\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2$ if $a+b+c=0$ Found this lovely identity the other day, and thought it was fun enough to share as a problem:
If $a+b+c=0$ then show $$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2.$$
There are, of course, brute force techniques for showing this, but I'm hoping for something elegant.
| Start off with the fact that
$$a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) = - 2(ab+bc+ca)
$$
Since $a+b+c=0$ we know that ${a,b,c}$ are the three roots of some cubic missing the $x^2$ term:
$$
x^3+kx+m = 0
$$ with $ab+bc+ca = k$. And by the way, that says that
$$a^2+b^2+c^2=-2k$$
Now start chaining upward, expressing $a^n+b^n+c^n$ in terms of $m$ and $k$. For $n=3$ we add the cubic expression with $x=a$ to that with $x=b$ and $x=c$ to get
$$
a^3+ b^3 + c^3 +k(a+b+c)+(m+m+m) = 0 \\
a^3+ b^3 + c^3 +3m = 0 \\
a^3+ b^3 + c^3 = -3m
$$
Now for $n=4$ we multiply each of the equations by $a$, $b$ and $c$ respectively before adding them:
$$
a^4+ b^4 + c^4 +k(a^2+b^2+c^2)+m(a+b+c) = 0 \\
a^4+ b^4 + c^4 -2k^2 = 0 \\
a^4+ b^4 + c^4 = 2k^2
$$
Now for $n=5$ we multiply each of the equations by $a^2$, $b^2$ and $c^2$ respectively before adding them:
$$
a^5+ b^5 + c^5 +k(a^3+b^3+c^3)+m(a^2+b^2+c^2) = 0 \\
a^5+ b^5 + c^5 -3mk -2mk = 0 \\
a^5+ b^5 + c^5 = 5mk
$$
For this problem we can afford to skip $n=6$.
$$
a^7+ b^7 + c^7 +k(a^5+b^5+c^5)+m(a^4+b^4+c^4) = 0 \\
a^7+ b^7 + c^7 +5mk^2 +2mk^2 = 0 \\
a^7+ b^7 + c^7 = -7mk^2
$$
Then your identity reads $$
\left( -\frac{3m}{3} \right) \left( -\frac{7mk^2}{7} \right) = \left( -\frac{5mk}{5} \right)^2 \\
\left( -m \right) \left( -mk^2 \right) = (mk)^2 $$
| {
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"url": "https://math.stackexchange.com/questions/1661035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Orthogonality relation as double sum of products of binomial coefficients I have stumbled upon the following sum over $x,y$ for non-negative integers $\kappa,\lambda$:
$$
\sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}\frac{\displaystyle{\kappa+\lambda \choose x+y}}{\displaystyle{2\left(\kappa+\lambda\right) \choose 2\left(x+y\right)}}=\begin{cases}
\dfrac{2^{4\kappa-1+\delta_{\kappa0}}}{\displaystyle{4\kappa \choose 2\kappa}}, & \kappa=\lambda\\
0, & \kappa\neq\lambda
\end{cases}
$$
To reiterate, this sum is zero if $\kappa\neq\lambda$, so it looks to me like some orthogonality relation. Does anyone else know how to prove this or can anyone suggest a resource where similar sums are listed?
Below is the confirmation of the above for different $\kappa,\lambda$:
| Consider the integral
$$
I=\int_0^1 \left[\left(\sqrt{1-t^2}+it\right)^{2k}+\left(\sqrt{1-t^2}-it\right)^{2k}\right]\cdot \left[\left(\sqrt{1-t^2}+it\right)^{2\lambda}+\left(\sqrt{1-t^2}-it\right)^{2\lambda}\right]\frac{dt}{\sqrt{1-t^2}}.
$$
One can easily check that it equals
\begin{align}
I=&\ 4\sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-1\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}\int_0^1 t^{2x+2y}\left(1-t^2\right)^{k+\lambda-x-y-\frac{1}{2}}dt=\\
&2\sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-1\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}\int_0^1 t^{x+y-\frac{1}{2}}\left(1-t\right)^{k+\lambda-x-y-\frac{1}{2}}dt=\\
&2\sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-1\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}\frac{\Gamma\left(x+y+\frac{1}{2}\right)\Gamma\left(k+\lambda-x-y+\frac{1}{2}\right)}{\Gamma\left(k+\lambda+1\right)}=\\
&\frac{ 2^{1-k-\lambda}\pi}{(k+\lambda)!}\sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-1\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}(2x+2y-1)!!(2k+2\lambda-2x-2y-1)!!=\\
&\frac{ \pi (2k+2\lambda)!}{2^{2k+2\lambda-1}(k+\lambda)!^2}\sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-1\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}\frac{{\kappa+\lambda \choose x+y}}{{2\left(\kappa+\lambda\right) \choose 2\left(x+y\right)}}.
\end{align}
On the other hand, substitution $t=\sin\theta$ shows that (it is assumed that $k>0$, since the case $k=\lambda=0$ is trivial)
\begin{align}
I=4\int_0^{\frac{\pi}{2}}\cos{2k\theta}\cos{2\lambda\theta}\ d\theta=\pi\delta_{k,\lambda}.
\end{align}
So we proved that
$$
\frac{ \pi (2k+2\lambda)!}{2^{2k+2\lambda-1}(k+\lambda)!^2}\sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-1\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}\frac{{\kappa+\lambda \choose x+y}}{{2\left(\kappa+\lambda\right) \choose 2\left(x+y\right)}}=\pi\delta_{k,\lambda},\quad k\neq 0,
$$
which is equivalent to
$$
\sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}\frac{{\kappa+\lambda \choose x+y}}{{2\left(\kappa+\lambda\right) \choose 2\left(x+y\right)}}=\frac{2^{4\kappa-1}}{{4\kappa \choose 2\kappa}}\delta_{k,\lambda},\quad k\neq 0.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Make $2^8 + 2^{11} + 2^n$ a perfect square Can someone help me with this exercise? I tried to do it, but it was very hard to solve it.
Find the value of $n$ to make $2^8 + 2^{11} + 2^n$ a perfect square.
It is the same thing like $4=2^2$.
| Note the fact that $2^8+2^{11}=48^2$
This implies that we are trying to find values of $n$ where $2^n=(x-48)(x+48)$.
Thus, we must find $k,l$ where $2^k-2^l=96$(where $x+48=2^k$, $x-48=2^l$)
Note the fact that $k \ge 7$.
This implies that $2^k$ is divisible by $32$, which implies that $2^l$ is also divisible by $32$.
Also, notice that if $k \ge 9$, when $2^k-2^l \ge 256$.
This implies that $k=7$or $k=8$.
Note the fact that $k=8$ does not have an integer solution, thus $k=7$.
Thus $x=80$, and thus $n=12$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does $\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta\dots}}}}=\sqrt{1 \sqrt{1 \sqrt{1\dots}}} \implies \cos{\theta}=1$? I was solving this equation:
$$\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta\dots}}}}=1$$
I solved it like this:
The given equation can be written as:
\begin{align*}
\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta\dots}}}}&=\sqrt{1 \sqrt{1 \sqrt{1\dots}}} \\
\cos{\theta}&=1 \\
\theta&=\arccos {1}
\end{align*}
So the solution is $2n\pi, n \in \mathbb Z$.
Have I solved it the wrong way?
(The title originally contained a more general question: Does $\sqrt{a \sqrt{a \sqrt{a\dots}}}=\sqrt{b \sqrt{b \sqrt{b\dots}}} \implies a=b$? The current title is consistent with the body and the accepted answer.)
| Hint:
$$\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}}...=1.$$
Multiply by $\cos(\theta)$ and take the square root. You get
$$\sqrt{\cos(\theta)}=\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}}}...=1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How does this factoring work? $$ (z^2 - 2i ) = (z -1 -i)(z + 1 +i) $$
I see if you multiply out the right-hand side, you obtain the left-hand side, but how does one know to factor like that or this?
$$ (z^2 − 3iz − 3 + i) = (z − 1 − i)(z + 1 − 2i) $$
| If we are trying to factor $p(z) = z^2 - 2i$ we can determine the factorization by finding the zeros of the polynomial.
If $p(z) = 0$ then $z^2 = 2i$. Thus $|z| = \sqrt{2}$. Writing $z=\sqrt{2} e^{i\theta}$ we have $e^{i2\theta} = e^{i \frac{\pi}{2}}$. This yields $$\cos(2\theta) + i \sin(2\theta) = \cos(\pi/2) + i \sin(\pi/2).$$
Now if we restrict $\theta \in [-\pi,\pi]$, we find the solutions $\theta = \pi/4$ and $\theta = -\frac{3\pi}{4}$.
This means $p(\sqrt{2} e^{i \pi/4} ) = 0$ and $p(\sqrt{2} e^{-i\frac{3\pi}{4}})$, and since it is monic, it can be factored as $$(z- \sqrt{2} e^{i\pi/4}) ( z - \sqrt{2} e^{-i3\pi/4}).$$
Finally $\sqrt{2} e^{i \pi/4} = \sqrt{2} ( \cos(\pi/4) + i \sin(\pi/4) ) = 1 + i$, and $\sqrt{2} e^{i3\pi/4} = -1-i$.
| {
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If $b$ is an odd composite number and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is a prime number, what happens when $q = 2^{r + 1} - 1$? (Note: An improved version of this question has been cross-posted to MO.)
Let $\sigma(X)$ be the sum of the divisors of $X$. For example, $\sigma(2) = 1 + 2 = 3$, and $\sigma(4) = 1 + 2 + 4 = 7$.
My question is:
If $b$ is an odd composite and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is a prime number, what happens when $q = 2^{r + 1} - 1$ (with $r \geq 1$)?
Without the restriction on $q = 2^{r + 1} - 1$ being prime, I only know that $M = {2^r}{b^2}$ must be an even almost perfect number. (That is, it must satisfy $\sigma(M) = 2M - 1$.)
I guess my question can be rephrased as follows:
(1) If $b$ is an odd composite, how often is $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ a prime number?
(2) If $b$ is an odd composite, and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is a prime number, how often is $q$ a Mersenne prime?
| This is only a partial answer to my initial question.
Let $I(x) = \sigma(x)/x$ be the abundancy index of $x$.
If $b$ is an odd composite and
$$\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = \sigma(2^r) = 2^{r+1} - 1,$$
then I know that
$$b^2 - 1 = \sigma(2^r)\sigma(b^2) - {b^2}\left(2^{r+1} - 1\right)$$
so that
$$\sigma\left({2^r}{b^2}\right) = 2^{r+1}{b^2} - 1$$
whence $M = {2^r}{b^2}$ must be an even almost perfect number that is not a power of two.
If, in addition,
$$\sigma(2^r) = 2^{r+1} - 1 = q$$
is prime, then notice that we have
$$q + 1 = \sigma(q) = 2^{r+1}$$
so that we obtain
$$I(q) = \dfrac{\sigma(q)}{q} = \dfrac{2^{r+1}}{\sigma(2^r)} = \dfrac{2}{I(2^r)},$$
whence we get $I(2^r)I(q) = I({2^r}q) = 2$ (since $\gcd(2^r, q) = \gcd(2^r, \sigma(2^r)) = 1$). Consequently, this means that $N = {2^r}{q}$ is an even perfect number (since $r \geq 1$). By the Euclid-Euler characterization for even perfect numbers, $q = 2^{r+1} - 1$ must be a Mersenne prime, so that $r + 1$ must be prime.
Lastly, as an aside to Giovanni Resta's earlier answer, it is known (by work of Antalan) that $3 \nmid b$.
| {
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Integral $\int \frac{\sqrt{x}}{(x+1)^2}dx$ First, i used substitution $x=t^2$ then $dx=2tdt$ so this integral becomes $I=\int \frac{2t^2}{(1+t^2)^2}dt$ then i used partial fraction decomposition the following way:
$$\frac{2t^2}{(1+t^2)^2}= \frac{At+B}{(1+t^2)} + \frac{Ct + D}{(1+t^2)^2} \Rightarrow 2t^2=(At+B)(1+t^2)+Ct+D=At+At^3+B+Bt^2 +Ct+D$$
for this I have that $A=0, B=2, C=0, D=-2$
so now I have
$I=\int \frac{2t^2}{(1+t^2)^2}dt= \int\frac{2}{1+t^2}dt - \int\frac{2}{(1+t^2)^2}dt$
Now,
$$ \int\frac{2}{1+t^2}dt = 2\arctan t$$
and
$$\int\frac{2}{(1+t^2)^2}dt$$
using partial integration we have:
$$u=\frac{1}{(1+t^2)^2} \Rightarrow du= \frac{-4t}{1+t^2}$$
and $$dt=dv \Rightarrow t=v$$
so now we have:
$$\int\frac{2}{(1+t^2)^2}dt =\frac{t}{(1+t^2)^2} + 4\int\frac{t^2}{1+t^2}dt = \frac{t}{(1+t^2)^2} + 4\int\frac{t^2 + 1 -1}{1+t^2}dt = \frac{t}{(1+t^2)^2} + 4t -4\arctan t$$
so, the final solution should be:
$$I=2\arctan t - \frac{t}{(1+t^2)^2} - 4t +4\arctan t$$
since the original variable was $x$ we have
$$I= 6\arctan \sqrt{x} - \frac{\sqrt{x}}{(1+x)^2} - 4\sqrt{x} $$
But, the problem is that the solution to this in my workbook is different, it says that solution to this integral is $$I=\arctan \sqrt{x} - \frac{\sqrt{x}}{x+1}$$
I checked my work and I couldn't find any mistakes, so i am wondering which solution is correct?
| Integrating by parts, $$\int\dfrac{2t^2}{(1+t^2)^2}dt=t\int\dfrac{2t}{(1+t^2)^2}dt-\int\left(\dfrac{dt}{dt}\cdot\int\dfrac{2t}{(1+t^2)^2}dt\right)dt$$
$$=-\dfrac t{(1+t^2)}-\int\dfrac{dt}{1+t^2}=?$$
Alternatively, choose $\sqrt x=\arctan u\implies x=(\arctan u)^2,dx=\dfrac{\arctan u}{1+u^2}du$
| {
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How Many Ways to Arrange These Boys and Girls?
There are $7$ boys and $3$ girls. In how many ways can they be arranged in a row such that the two ends are occupied by boys and no two girls are seated together?
The answer is $6 \cdot 5 \cdot 4 \cdot 7!$ and the book gave the same explanation is the same as the one here. This was my analysis. Can you explain why it's wrong?
There are $7 \cdot 6$ ways to fill the first and the last slot. If we ignore the rules for a moment, there are $8!$ ways to fill the remaining slots.
*
*Any time we have two girls together is unacceptable. So, we can count the two girls as one person. This gives us $7!$ unacceptable arrangements before considering the different choices in the two-girl spot.
*Since there are $3 \cdot 2$ ways to fill the two-girl spots, we have $6 \cdot 7!$ unacceptable arrangements in the middle.
It follows that there are $2 \cdot 7!$ acceptable arrangements in the middle, and $7 \cdot 6 \cdot 2 \cdot 7!$ ways to fill the spots according the the rules. As you see, I am short by $6 \cdot 7!$. What am I missing?
| You have $20$ different seat-combinations for the girls:
*
*$2,4,6$
*$2,4,7$
*$2,4,8$
*$2,4,9$
*$2,5,7$
*$2,5,8$
*$2,5,9$
*$2,6,8$
*$2,6,9$
*$2,7,9$
*$3,5,7$
*$3,5,8$
*$3,5,9$
*$3,6,8$
*$3,6,9$
*$3,7,9$
*$4,6,8$
*$4,6,9$
*$4,7,9$
*$5,7,9$
Then, you can permute the girls in $3!$ different ways.
Then, you can permute the boys in $7!$ different ways.
Hence the total number of ways to arrange them is $20\cdot3!\cdot7!$.
| {
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steps to calculate the space surface area cut by a cylinder(see the picture) The space surface(in yellow) $ x^2+y^2 = 2az\ $ is cut by a cylinder(in green) $x^2+y^2=3a^2 (a>0)$
How to calculate the cut out part area $A$?
I think the part is between the two planes $z=0$ and $ z= \frac32a $, and I only have the knowledge of using a formula $ \iint\limits_D \sqrt{1 + z^2_x + z^2_y} $ to calculate the $A$, and I should use it because there is an exercise for using this formula. In this case, $ z=\frac1{2a}(x^2+y^2), D: x^2+y^2 \le 3a^2 $ , so
$$z_x = \frac1ax,\ z_y= \frac1ay$$ $$ A=\iint\limits_D \sqrt{1 + \frac{x^2}{a^2} + \frac{y^2}{a^2}} dxdy= \frac1a \iint\limits_D \sqrt{a^2+x^2+y^2} dxdy $$ using polar coordinates to do the remain steps:
$$\frac1a \int_0^{2\pi}\int_0^{\sqrt3a} \sqrt{a^2+r^2}r dr=\frac\pi a\frac23 ({a^2+r^2})^\frac32\bigg|_0^{\sqrt3a}=\frac{14} 3\pi a^2$$
| Your work is corrrect.
Note that you can find the surface also as the surface of revolution of the parabola
$$
x^2=2az
$$
around the $z-$ axis for $0\le x\le a\sqrt{3}$. This is a bit easier and gives the integral:
$$
A=2 \pi\int_0^{a\sqrt{3}}x\sqrt{1+\frac{x^2}{a^2}}dx
$$
that can be solved with the substitution $1+\frac{x^2}{a^2}=t $ and gives the same result as your.
| {
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"timestamp": "2023-03-29T00:00:00",
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The indefinite integral $\int x^{13}\sqrt{x^7+1}dx$ I used integration by parts technique to solve this problem
so $u = \sqrt{x^7+1}$ ,$du = \frac{1}{2}\frac{7x^6}{\sqrt{x^7+1}}dx$
$v =\frac{x^{14}}{14}$ $dv=x^{13}dx$
then it becomes
$\frac{x^{14}}{14}\sqrt{x^7+1}-\int\frac{x^{14}}{28}\frac{7x^6}{\sqrt{x^7+1}}dx$
and this is where i got stuck at. I tried to substitute $u =x^7$
but then the integral become $\int \frac{u^2}{28\sqrt{u+1}}du$
the final answer that I found using wolffram calculator is $\frac{2}{105}(x^7+1)^{3/2}(3x^7-2)$
| Hint: Use the substitution $u^2=x^7+1$
| {
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How to find the roots of the polynomial We consider the irreducible polynomial $g=y^4+y+1 \in \mathbb{F}_2[y]$ and let $b$ be a root of $g$.
I want to find all the roots of $g$ and also three generators of $\mathbb{F}_{16}^{\ast}$ as for the basis $\{1, b, b^2, b^3 \}$.
I have tried the following:
I applied the euclidean division of $y^4+y+1$ with $y-b$ and I got $y^4+y+1=(y-b) (y^3+by^2+b^2 y+(b^3+1))$.
Then I applied the euclidean division of $y^3+by^2+b^2 y+(b^3+1)$ with $y-(b+1)$ and I got $y^3+by^2+b^2 y+(b^3+1)=(y-(b+1))(y^2+y+(b^2+b+1))$.
Then suppose that the third root is $r_1$. Then applying the euclidean division of $y^2+y+(b^2+b+1)$ with $y-r_1$ we get $y^2+y+(b^2+b+1)=(y-r_1)(y+(1+r_1))$ when $r_1^2+r_1+b^2+b+1=0$.
How can we solve the last equation to get $r_1$? Can we use the discriminant?
| Observe that if $ b $ is a root of $ g $, then so are $ b+1, b^2, b^2 + 1 $. All of these are obviously pairwise distinct since $ g $ is an irreducible polynomial.
EDIT : $ b^2 $ is a root because we have $$ g(b^2) = (b^2)^4 + b^2 + 1 = b^8 + b^2 + 1 = (b+1)^2 + b^2 + 1 = 0 $$ From the fact that $ b $ is a root $ \implies $ $ b + 1 $ is a root, we get that $ b^2 + 1 $ is also a root.
To answer your second question, you want to find a generator for $ \mathbb{F}_{16}^{\times} \cong \mathbb{F}_2(b)^{\times} $. We know that $ b^{15} = 1 $. Further, $ b^3 \neq 1 $, $ b^5 = b^2 + b \neq 1 $, so the order of $ b $ is indeed $ 15 $ and hence it is a generator for the multiplicative group. How then, can you find all other generators?
| {
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On a remarkable system of fourth powers using $x^4+y^4+(x+y)^4=2z^4$ The problem is to find four integers $a,b,c,d$ such that,
$$a^4+b^4+(a+b)^4=2{x_1}^4\\a^4+c^4+(a+c)^4=2{x_2}^4\\a^4+d^4+(a+d)^4=2{\color{blue}{x_3}}^4\\b^4+c^4+(b+c)^4=2{x_4}^4\\b^4+d^4+(b+d)^4=2{x_5}^4\\c^4+d^4+(c+d)^4=2{x_6}^4$$
As W. Jagy pointed out, the form $x^4+y^4+(x+y)^4 = 2z^4$ appear in the context of triangles with integer sides and one $120^\circ$ angle. PM 2Ring discovered that, remarkably, the quadruple,
$$a,b,c,d = 195, 264, 325, 440$$
yields five integer $x_i$ (except $x_3$).
I found that, using an elliptic curve, it can be showed there are infinitely many non-zero integer triples with $\gcd(a,b,c)=1$ such that three $x_i$ are integers.
Q: However, are there infinitely many quadruples with $\gcd(a,b,c,d)=1$ such that at least five of the $x_i$ are integers?
| After mulling over the problem, it turned out the same elliptic curve can make five of the $x_i$ as integers. To start, note that,
$$x^4+y^4+(x+y)^4 = 2(x^2+xy+y^2)^2$$
Thus, the system is reduced to finding,
$$\color{blue}{a^2+ab+b^2 = x_1^2}\tag1$$
$$\color{blue}{a^2+ac+c^2 = x_2^2}\tag2$$
$$b^2+bc+c^2 = x_4^2\tag3$$
$$\color{brown}{b^2+bd+d^2 = x_5^2}\tag4$$
$$\color{brown}{c^2+cd+d^2 = x_6^2}\tag5$$
Assume $b,c$ as constant. Plugging them into $(1),(2)$, then a pair $\color{blue}{P_1}$ of quadratics in $\color{blue}a$ must be made a square. Since this has a rational point, it is birationally equivalent to an elliptic curve.
Plugging $b,c$ into $(4),(5)$, a pair $\color{brown}{P_2}$ of quadratics in $\color{brown}d$ must be made a square. But $P_1$ and $P_2$ have the same form. Thus, two different points in the same elliptic curve will yield the $a,d$. Clearing denominators, we then get another quadruple,
$$a,b,c,d = 232538560625,\, 670011598080,\, 824824884000,\, 749417043168$$
and an infinite more with $\gcd(a,b,c,d)=1$. (Presumably, smaller ones may exist.)
| {
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Determine all the roots of the equation given by $z^2(1-z^2)=16.$ For my third year Complex variable course, the question is
Determine all the roots of the equation given by
$$z^2(1-z^2)=16.$$
My attempt:
Let $z^2 = x$
$x(1-x) = 16$
$x-x^2 = 16$
$x^2-x-16 = 0$
$x = \frac{1 \pm \sqrt{1 -4(16)}}{2}$
$x = \frac{1 \pm \sqrt{-63}}{2}$
$x = \frac{1 \pm i\sqrt{63}}{2}$
$z^2 = \frac{1 \pm i\sqrt{63}}{2}$
Am I correct so far?
BTW the question is worth $5$ marks.
| Remember it is a fourth order equation, so the complete solution has four roots, in this example they occur in complex-conjugate pairs.
| {
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If $A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}\;,$ Then $\lfloor A \rfloor\;\;,$
If $$A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}\;,$$ Then $\lfloor A \rfloor\;\;,$ Where $\lfloor x \rfloor $ Represent floor fiunction of $x$
My Try:: Using $\bf{A.M\geq H.M\;,}$ We get
$$\frac{1980+1891+1982+....+2012}{33}>\frac{33}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}$$
So $$\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}<\frac{1980+1981+....+2012}{(33)^2}=\frac{1996}{33}\approx 60.5<61$$
Now how can i prove that the above expression $A$ is $>60$
Help me, Thanks
| The elementary estimate
$$\frac{m}{n+m} < \sum_{k = 0}^{m-1} \frac{1}{n+k} < \frac{m}{n}$$
for $m > 1$ (we have equality on the right for $m = 1$) gives
$$\frac{1}{61} = \frac{33}{2013} < \sum_{k = 1980}^{2012} \frac{1}{k} < \frac{33}{1980} = \frac{1}{60},$$
whence $\lfloor A\rfloor = 60$.
| {
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Finding Maximum Area of a Rectangle in an Ellipse
Question: A rectangle and an ellipse are both centred at $(0,0)$.
The vertices of the rectangle are concurrent with the ellipse as shown
Prove that the maximum possible area of the rectangle occurs when the x coordinate of
point $P$ is $x = \frac{a}{\sqrt{2}} $
What I have done
Let equation of ellipse be
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
Solving for y
$$ y = \sqrt{ b^2 - \frac{b^2x^2}{a^2}} $$
Let area of a rectangle be $4xy$
$$ A = 4xy $$
$$ A = 4x(\sqrt{ b^2 - \frac{b^2x^2}{a^2}}) $$
$$ A'(x) = 4(\sqrt{ b^2 - \frac{b^2x^2}{a^2}}) + 4x\left( (b^2 - \frac{b^2x^2}{a^2})^{\frac{-1}{2}} \times \frac{-2b^2x}{a^2} \right) $$
$$ A'(x) = 4\sqrt{ b^2 - \frac{b^2x^2}{a^2}} + \frac{-8x^2b^2}{\sqrt{ b^2 - \frac{b^2x^2}{a^2}}a^2} = 0 $$
$$ 4a^2\left(b^2 - \frac{b^2x^2}{a^2} \right) - 8x^2b^2 = 0 , \sqrt{ b^2 - \frac{b^2x^2}{a^2}a^2} \neq 0 $$
$$ 4a^2\left(b^2 - \frac{b^2x^2}{a^2} \right) - 8x^2b^2 = 0 $$
$$ 4a^2b^2 - 4b^2x^2 - 8x^2b^2 = 0 $$
$$ 4a^2b^2 - 12x^2b^2 = 0 $$
$$ 12x^2b^2 = 4a^2b^2 $$
$$ x^2 = \frac{a^2}{3} $$
$$ x = \frac{a}{\sqrt{3}} , x>0 $$
Where did I go wrong?
edit:The duplicate question is the same but both posts have different approaches on how to solve it so I don't think it should be marked as a duplicate..
| The ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ is a circle of radius $a$ in $(\hat x,y)$ coordinates, where $\hat x=\dfrac{a}{b}x$. This transformation multiplies areas by the constant $\dfrac{a}{b}$, so the problem is equivalent to finding the rectangle of maximum area in a circle, which is well-known to be a square.
Or, looked at another way (pun intended), this ellipse is what you see if you look at the circle of radius $a$ in the $x-y$ plane from just the right angle instead of from directly above. When you see what appears to be an inscribed rectangle in the ellipse of maximum area, what you’re looking at is an inscribed rectangle in the circle of maximum area.
| {
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Moment Generating Function I am little unsure if I am doing the following question right:
Suppose Y is a discrete random variable such that
$P(Y=0)=\frac{1}{3}$; $P(Y=1)=\frac{1}{6}$; $P(Y=2)=\frac{1}{6}$;
$P(Y=3)=\frac{1}{3}$;
a) Write down the moment generating function of $Y$.
b) Find the mean and variance of $Y$.
c) Suppose that $X_1$ and $X_2$ are independent random variables with
the same distribution as $Y$. Compute $P(X_1 + X_2 = 3)$.
My solutions so far:
a) Defintion of MGF $$M(t) = E[e^{tX}] = \sum_{x}e^{tx}p(x)$$
So
$$M_Y(s) = \frac{1}{3} + \frac{1}{6}e^{t} + \frac{1}{6}e^{2t} + \frac{1}{3}e^{3t}$$
b) Using the following defintions:
$$ M'(0) = E[X]$$
$$M''(0) = E[X^2]$$
So
$M'(t) = e^{3t} + \frac{1}{3}e^{2t} + \frac{1}{6}e^t \implies M'(0) = \frac{3}{2}$
$M''(t) = 3e^{3t} + \frac{2}{3}e^{2t} + \frac{1}{6}e^t \implies M''(0) = \frac{23}{6}$
$Var(X) = \frac{23}{6} - (\frac{3}{2})^2 = \frac{19}{12}$
I am not sure how to do part c though.
| Let $Z=X_1+X_2$. Since they are independent random variables with the same distribution as $Y$
\begin{align}
M_Z(t) &= E[e^{t(X_1+X_2)}]
\\
&= E[e^{tX_1}] \:E[e^{tX_2}]
\\
&=(\frac{1}{3} + \frac{1}{6}e^{t} + \frac{1}{6}e^{2t} + \frac{1}{3}e^{3t})^2
\\
&=\frac1{36}(4+4e^t+5e^{2t}+10e^{3t}+5e^{4t}+4e^{5t}+4e^{6t})
\end{align}
Thus $P(Z=3)$ is the coefficient of $e^{3t}$, i.e.
$$
P(Z=3)=\frac{5}{18}
$$
| {
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Prove that $f(x, y) \le 3 $ for $x \ge 0, y > 0$
Let $x \ge 0, y>0$ and
\begin{align*}
f(x,y)&=\sqrt{\dfrac{y}{y+x^2}}+4\sqrt{\dfrac{y}{(y+(x+1)^2)(y+(x+3)^2)}}\\[6pt]
&\qquad +4\sqrt{\dfrac{y}{(y+(x-1)^2)(y+(x-3)^2)}}.
\end{align*}
Prove that $f(x,y) \le 3$.
I can prove when $x\ge 2, f(x,y) < 3$, but $f(x,y)=3$ when $x=0,y=3$, so the key part is $x\le 2 $
I found when $x<2, f'_x(y \ge 1.5) <0$ but I can't prove it as it has high degree equations.
And for $x<2,y<1.5$, I have no idea how to prove $f(x,y)<3$
| No clue without an overview. With help of restrictions as found already by the OP, a contour plot / isoline chart has been produced for the function at hand:
The $y$-axis is in $\color{green}{\mbox{green}}$. Our viewport is:
xmin := -6 ; xmax := 6;
ymin := 0 ; ymax := 12;
The are 27 contour levels nivo, defined by:
for geval := 1 to 27 do
begin
nivo := geval/9.1;
The higher the level, the more black. The lower the level, the more white.
The pixels in $\color{red}{red}$ are close to the conjectured maximum $=3$ . They are defined by:
const
eps : double = 0.0002;
if (3-f < eps) and (3-f > 0) then
It is seen that the red spot is inside an area with very black lines,
meaning that the function $f(x,y)$ is increasing there, towards the red spot.
Analysis. There is a mirror symmetry in the $y$-axis, as is clear from the figure but also from some algebra.
Thus for all real $x$ and for all real $y > 0$ we have:
$$
f(x,y) = f(-x,y)
$$
It follows that:
$$
\frac{\partial f(x,y)}{\partial x} = - \frac{\partial f(-x,y)}{\partial x}
\quad \Longrightarrow \quad
\left.\frac{\partial f(x,y)}{\partial x}\right|_{x=0} = - \left.\frac{\partial f(-x,y)}{\partial x}\right|_{x=0}
\\ \Longrightarrow \quad
\left.\frac{\partial f(x,y)}{\partial x}\right|_{x=0} = 0
$$
Meaning that all function gradients near the $y$-axis are tangent to it. Since the gradient vectors are perpendicular
to the isolines, this can be observed in the figure as well. Now a sufficient condition for having a stationary point
(maximum or minimum eventually) is that the total gradient be zero. But we already have $\,\partial f / \partial x = 0\,$
at the $y$-axis. Therefore substitute $\,x=0\,$ in $\,f(x,y)\,$ and only consider function values at the $y$-axis:
$$
f(0,y)=1+8\sqrt{\frac{y}{(y+1)(y+9)}}
$$
Extremes are found for:
$$
\frac{d f(0,y)}{dy} = 0 \quad \Longrightarrow \\
\frac{1}{2} \left[ {\frac {1}{ \left( y+1 \right) \left( y+9 \right) }}-{
\frac {y}{ \left( y+1 \right) ^{2} \left( y+9 \right) }}-{\frac {y}{
\left( y+1 \right) \left( y+9 \right) ^{2}}} \right]
\sqrt{\frac{\left( y+1 \right) \left( y+9 \right)}{y}} = 0 \\
\Longrightarrow \quad (y+1)(y+9) - y(y+9) - y(y+1) = -y^2 +9 = 0 \quad \Longrightarrow \quad y = \pm 3
$$
With the restriction $\,y > 0$ . Hence the only extreme at $\,(y,f(0,y))$ is $(3,3)\,$ and it is a maximum.
Picking up the comment by nbubis, prove that $f(0,2) < 3$ and $f(0,4) < 3$:
$$
f(0,2)=1+8\sqrt{\frac{2}{(2+1)(2+9)}}=1+\sqrt{\frac{128}{33}}<1+\sqrt{\frac{128}{32}}\\
f(0,4)=1+8\sqrt{\frac{4}{(4+1)(4+9)}}=1+\sqrt{\frac{256}{65}}<1+\sqrt{\frac{256}{64}}
$$
| {
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Solving $\frac{\sin x}{4}=\frac{\sin y}{3}=\frac{\sin z}{2}$ where $x$, $y$, $z$ are angles of a triangle Can any one give me a hint to find value of $x$
where:
$$\frac{\sin x}{4}=\frac{\sin y}{3}=\frac{\sin z}{2}$$
and $x$, $y$, $z$ are angles of a triangle.
I tried to use sine law but got nothing.
| My solution involves finding the size of angle x. Then getting it's sine.
From the sine law we can tell that the triangle has sides 3, 4 and 2. And that angle x is opposite to the side with length 4.
We can employ cosine law to find the size of angle x.
$c^2 = a^2 + b^2 + 2ab\cos\theta$
$4^2 = 2^2 + 3 ^2 - 2(3*2)\cos\theta$
This evaluates to
$\cos\theta = -\frac{3}{12} = -0.25$
To get the size of the angle, feed $- 0.25$ into the inverse cos function
$\cos^{-1} (-0.25) = 104.478^\circ $
Finally, get the sine of the angle
$\sin (104.478^\circ) = 0.968 $
Therefore, $\sin x = 0.968 $
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the limit of $\lim_{x\to 0} \frac{1-\cos^n x}{x^2}$ How can I show that
$$
\lim_{x\to 0} \frac{1-\cos^n x}{x^2} = \frac{n}{2}
$$
without using Taylor series $\cos^n x = 1 - \frac{n}{2} x^2 + \cdots\,$?
| You just need to know that $\lim_{x\to 0}\frac{\sin x}{x}=1$, since:
$$\begin{eqnarray*} \frac{1-\cos^n x}{x^2} &=& \frac{1-\cos x}{x^2}\cdot \sum_{k=0}^{n-1}\cos^k(x) = \frac{2\sin^2\frac{x}{2}}{x^2}\cdot \sum_{k=0}^{n-1}\cos^k(x)\\ &=& \frac{1}{2}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2 \cdot \sum_{k=0}^{n-1}\cos^k(x)\stackrel{x\to 0}{\longrightarrow}\color{red}{\frac{n}{2}}.\end{eqnarray*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Find x for $\sqrt{(5x-1)}+\sqrt{(x-1)}=2$ Solve:
$$\sqrt{(5x-1)}+\sqrt{(x-1)}=2$$
When $x=1$, we get the following equation to equal to $2$
I've been trying to solve this problem but when I square both sides and simplify I end up with:
$$x^2+6x+2=0$$ and of course $x=1$ cannot be a solution. So im not sure what im doing wrong. Any help on this problem?
| Perhaps a slightly tidier solution:
$$\begin{align}
\sqrt{5x-1}+\sqrt{x-1}&=2 \\
\sqrt{5x-1} &= 2-\sqrt{x-1} &\text{from regrouping}\\
5x-1&=4-4\sqrt{x-1}+(x-1) & \text{after squaring}\\
4x-4&=4\sqrt{x-1} &\text{after regrouping again}\\
(x-1)^2&=x-1 &\text{divide by 4 and square again}\\
(x-1)^2-(x-1)&=0 \\
(x-1)(x)&=0 &\text{factor}
\end{align}$$
so either $x-1=0$ or $x=0$. Plugging back into the original equation we see that $x=1$ is a solution and $x=0$ is not.
| {
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Focus of the Parabola Find the Focus of $$(2x+y-1)^2=5(x-2y-3)$$.
Clearly its a Parabola whose axis is $2x+y-1=0$ and since $x-2y-3=0$ is perpendicular to $2x+y-1=0$ Tangent at the vertex is $x-2y-3=0$.Also the Vertex is $(3,-1)$, but now how to find its focus?
| More generally if we have a parabola in the form (which all parabolas can be put into) $$(l_{\rm axis})^2=e\cdot l_{\text{tangent at vertex}}$$ or $$(ax+by+c)^2=e(-bx+ay+d),$$ and use the orthonormal change of basis $x'=\frac{ax+by}{\sqrt{a^2+b^2}},y'=\frac{-bx+ay}{\sqrt{a^2+b^2}}$, we get $$(\sqrt{a^2+b^2}x'+c)^2=(\sqrt{a^2+b^2}y'+d)$$ or $$(a^2+b^2)((x'+\frac{c}{\sqrt{a^2+b^2}})^2-\frac{e}{\sqrt{a^2+b^2}}(y'+\frac{d}{\sqrt{a^2+b^2}}))=0.$$ So using that$$x''^2=4\frac{e}{4\sqrt{a^2+b^2}}y''$$ has focus $$(x'',y'')=(0,\frac{e}{4\sqrt{a^2+b^2}})$$ we nest back to $$(x',y')=(-\frac{c}{\sqrt{a^2+b^2}},\frac{e-4d}{4\sqrt{a^2+b^2}})$$ and get $$(x,y)=(-\frac{a(4d-e)+4bc}{4b^2+4a^2},-\frac{(b(e-4d)+4ac}{4b^2+4a^2}).$$
In your special case the focus is $(x,y)=(\frac54,-\frac32).$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{\sqrt{3}}{8}<\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}\,dx<\frac{\sqrt{2}}{6}$
Prove that $$\frac{\sqrt{3}}{8}<\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}\,dx<\frac{\sqrt{2}}{6}$$
My try: Using $$\displaystyle \sin x<x$$ and $$\frac{\sin x-0}{x-0}>\frac{1-0}{\frac{\pi}{2}-0}=\frac{2}{\pi}$$
So we get $$\frac{2}{\pi}<\frac{\sin x}{x}<1$$
So we get $$\frac{2}{\pi}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}1\,dx<\frac{\sin x}{x}<1\cdot \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\,$$
But this is not what I have to prove here.
| Since $\sin\left(x\right)/x
$ is a monotone decreasing function on $\left[\frac{\pi}{4},\frac{\pi}{3}\right]
$ we have $$\int_{\pi/4}^{\pi/3}\frac{\sin\left(x\right)}{x}dx\geq\frac{\sin\left(\pi/3\right)}{\pi/3}\int_{\pi/4}^{\pi/3}dx=\frac{\sqrt{3}}{8}.$$ For the upper bound, if we want more precision, we can use for example the Cauchy Schwarz inequality and get $$\int_{\pi/4}^{\pi/3}\frac{\sin\left(x\right)}{x}dx\leq\left(\int_{\pi/4}^{\pi/3}\sin^{2}\left(x\right)dx\right)^{1/2}\left(\int_{\pi/4}^{\pi/3}\frac{1}{x^{2}}dx\right)^{1/2}=\frac{1}{2}\sqrt{\frac{\pi-3\left(\sqrt{3}-2\right)}{6\pi}}<\frac{\sqrt{2}}{6}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $4k^3 + 17k^2 - 228k -1116 = 0$ The equation given to me is $$4x^4 + 16x^3 - 17x^2 - 102x -45 = 0$$
I'm asked to find it's resolvent cubic which is not so difficult to find. But the problem is that the question further asks to find the solution of resolvent cubic.
I have found resolvent cubic using Ferrari's method. The resolvent cubic came out to be $$4k^3 + 17k^2 - 228k -1116 = 0$$
The trouble begins now when I'm trying to solve this cubic it's getting very exhaustive and long calculations. But since the question asks to find the solution of the cubic.
Now can someone help me solve this cubic easily or there's no way out to this problem. I want to escape the tedious calculations while solving this cubic.
Kindly help me if you can.
| Let's use Bill Dubuque's AC method on this cubic. Start by factoring $228 = 2^3 \cdot 3 \cdot 19$ and $1116 = 2^2 \cdot 3^2 \cdot 31$. Their greatest common divisor is $2 \cdot 3 = 6$, so it is an ideal candidate to reach the form $f(6x) = a(6x)^3 + b(6x)^2 + c(6x) + d$.
To do this, transform the roots $k \mapsto 1/x$ and multiply by $-1$ to get:
$$\begin{align} (2^2 \cdot 3^2 \cdot 31)x^{3}+(2^2 \cdot 3 \cdot 19)x^{2}-17x-4 = 0 \\
\implies (2^3 \cdot 3^3 \cdot 31)x^{3}+(2^3 \cdot 3^2 \cdot 19)x^{2}-17 \cdot 6x-24 = 0 \end{align}$$
and now since this is in the form $f(6x)$, let $6x = u$:
$$31u^3 + 38u^2 - 17u - 24 = 0$$
After searching for candidates using the rational root theorem, one might stumble across $-1$ as a root. Synthetic division yields:
$$\begin{array}{c|rrr}&31&38&-17&-24\\-1&&-31&-7&24\\\hline\ &31&7&-24&0\\\end{array}$$
thus this polynomial can be factored as $(u+1)(31u^2+7u-24)$ and here $u = -1$ is again a root of the quadratic. Therefore $(u+1)^2 (31u - 24) = 0$ since the constant term and $u^3$ term must match, or that: $$k = \frac{1}{1/6}, \frac{1}{(24/31)/6} = 6, \frac{31}{4}.$$
| {
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indefinite integration $ \int \frac { x^2 dx} {x^4 + x^2 -2}$ problem : $ \int \frac { x^2 dx} {x^4 + x^2 -2}$
solution : divide numerator and denominator by $x^2$
$ \int \frac { dx} {x^2 + 1 -\frac{1}{x^2}}$
Now whats the next step $?$
Am I doing right $?$
| $$I= \int \frac { x^2dx} {x^4 + x^2 -2}$$
Divide by $x^4$.
$$I=\int \frac{\frac{1}{x^2}dx}{1+\frac{1}{x^2}-\frac{2}{x^4}}$$
Let $\frac1x=u$. Then, $du=-\frac{1}{x^2}dx$.
$$I=\int\frac{du}{2u^4-u^2-1}=\int\frac{du}{2u^4-2u^2+u^2-1}=\int\frac{du}{(2u^2+1)(u^2-1)}=\int\frac{du}{(2u^2+1)(u+1)(u-1)}$$
Now this can be solved by using partial fractions.
| {
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$C$ is a complex number.$f:C\to R$ is defined by $f(z)=|z^3-z+2|.$Find the maximum value of $f(z)$ if $|z|=1.$ $C$ is a complex number.$f:C\to R$ is defined by $f(z)=|z^3-z+2|.$Find the maximum value of $f(z)$ if $|z|=1.$
My try:
I applied $|z^3-z+2|\leq|z|^3+|-z|+|2|$,i got $f(z)\leq 4$ but book says my answer is wrong.
Correct answer is $|f(z)|$ is maximum when $z=w,$where $w$ is cube root of unity and $|f(z)|=\sqrt{13}$.
I do not know how to get correct answer.
| Let $\cos\theta=c,\sin\theta=s$ and $z=c+si$.
Then, since
$$\begin{align}z^3-z+2&=(c+si)^3-(c+si)+2\\&=c^3+3c^2si-3cs^2-s^3i-c-si+2\\&=(c^3-3cs^2-c+2)+(3c^2s-s^3-s)i\end{align}$$
we have
$$\begin{align}f(z)&=\sqrt{(c^3-3cs^2-c+2)^2+(3c^2s-s^3-s)^2}\\&=\sqrt{c^6+3 c^4 s^2-2 c^4+4 c^3+3 c^2 s^4+c^2-12 c s^2-4 c+s^6+2 s^4+s^2+4
}\\&=\sqrt{(c^6+3c^4s^2+3c^2s^4+s^6)+(s^2+c^2)-2(c^4-s^4)-12cs^2+4c^3-4c+4}\\&=\sqrt{(c^2+s^2)^3+1-2(c^2+s^2)(c^2-s^2)-12c(1-c^2)+4c^3-4c+4}\\&=\sqrt{1+1-2\cdot 1\cdot (c^2-(1-c^2))-12c(1-c^2)+4c^3-4c+4}\\&=\sqrt{16c^3-4c^2-16c+8}\end{align}$$
Here, let $g(t)=16t^3-4t^2-16t+8$. Then,
$$g'(t)=0\iff t=-\frac 12,\frac 23.$$
Hence, the maximum value of $g(t)$ for $-1\le t\le 1$ is
$$\max\{g(-1/2),g(1)\}=\max\{13,4\}=13.$$
Therefore, the maximum value of $f(z)$ is $f((-1\pm\sqrt 3\ i)/2)=\sqrt{13}$.
| {
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Identifying units in a polynomial ring
Problem Statement: Let $R$ be a domain. Identify the units in $R[x]$.
I am trying to identify the units in a domain $R$ by considering an arbitrary element $a=a_{n}x^{n}+\cdots+a_{1}x+a_{0}\in R[x]$ with an inverse $b=b_{m}x^{m}+\cdots+b_{1}x+b_{0}\in R[x]$.
If $b=a^{-1}$, then $$1=ab=ba=(a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0})(b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots+b_{1}x+b_{0})
= (a_{n}b_{m}x^{n+m}+a_{n}b_{m-1}x^{n+m-1}+\cdots+a_{n}b_{1}x^{n+1}+a_{n}b_{0}x^{n})
+ (a_{n-1}b_{m}x^{n+m-1}+a_{n-1}b_{m-1}x^{n+m-2}+\cdots+a_{n-1}b_{1}x^{n}+a_{n-1}b_{0}x^{n-1})+\cdots
+ (a_{1}b_{m}x^{m+1}+a_{1}b_{m-1}x^{m}+\cdots+a_{1}b_{1}x^{2}+a_{1}b_{0}x)
+(a_{0}b_{m}x^{m}+a_{0}b_{m-1}x^{m-1}+\cdots+a_{0}b_{1}x+a_{0}b_{0})$$
$$= a_{n}b_{m}x^{m+n}+(a_{n}b_{m-1}+a_{n-1}b_{m})x^{n+m-1} +
(a_{n}b_{m-2}+a_{n-1}b_{m-1}+a_{n-2}b_{m})x^{n+m-2}+\cdots+(a_{2}b_{0}+a_{1}b_{1}+a_{0}b_{2})x^{2}+(a_{1}b_{0}+a_{0}b_{1})x+a_0b_0$$
Thus,
$$a_{n}b_{m}=a_{n}b_{m-1}+a_{n-1}b_{m}=a_{n}b_{m-2}+a_{n-1}b_{m-1}+a_{n-2}b_{m}=\cdots=a_{2}b_{0}+a_{1}b_{1}+a_{0}b_{2}=a_{1}b_{0}+a_{0}b_{1}=0$$
Or more generally,
$$a_{i}b_{0}+a_{i-1}b_{1}+\cdots+a_{1}b_{i-1}+a_{0}b_{i}=0$$
and
$$a_{0}b_{0}=1.$$
Since $R$ is a domain, $R[x]$ is a domain, and thus, $a_{n}b_{m}=0$ implies that $a_{n}=0$ or $b_{m} =0$.
If $a_{n}=0$ and $b_{m}\neq0$, then it would recursively follow that $a_{n}=0\ \Rightarrow a_{n-1}=0$?
Similarly, If $b_{m}=0$ and $a_{n}\neq0$, then it would recursively follow that $b_{m}=0\ \Rightarrow b_{m-1}=0$?
But that would imply that either $a_{0}=0$ or $b_{0} = 0$, which is an issue, because we want $a_{0}b_{0}=1$.
I am thinking that the only units are constant polynomials? That is, unless there is some positive integer $k$ such that $x^{k}=0$. Should I be considering that case as well, or is there an easier way to approach this problem?
Any suggestions are appreciated!
| Hint: if $R$ is a domain, then $\deg{fg} = \deg{f} + \deg{g}$ for all polynomials $f, g\in R[x]$.
| {
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Number of common roots of $x^3 + 2 x^2 +2x +1 = 0$ and $x^{200} + x^{130} + 1 = 0 $ The equations $x^3 + 2 x^2 +2x +1 = 0$ and $x^{200} + x^{130} + 1 = 0 $ have
*
*exactly one common root;
*no common root;
*exactly three common roots;
*exactly two common roots.
I factored the first equation. I think the roots are $-1$, $\omega$ and $\omega^2$.
| $\gcd (x^3 + 2 x^2 +2x +1 , x^{200} + x^{130} + 1 ) = x^2+x+1$, so two roots.
Note that you don't need to know the two roots, just that the gcd has degree $2$ and is coprime with its derivative (so no repeated roots).
| {
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Unit Tangent Vector and Unit Normal Vector for a given value of t
To Find - unit tangent vector $T$ and the unit normal vector $N$ for $t=0$; I
know the following
$\mathbf r(t)=\mathbf i(t)+\frac{1}{2}\mathbf j(t^2)+\frac{1}{3}\mathbf k(t^3)$
$T=\frac{r'(t)}{|r'(t)|}$
and
$N=\frac{T'(t)}{|T'(t)|}$
now,
$\mathbf r'(t)=\mathbf i+\mathbf j(t)+\mathbf k(t^2)$
and
$\Vert\mathbf r'(t)\Vert= \sqrt{1+t^2+t^4}$
therefore,
$T(t)=\frac{\mathbf i}{\sqrt{1+t^2+t^4}}+\frac{\mathbf j(t)}{\sqrt{1+t^2+t^4}}+\frac{\mathbf k(t^2)}{\sqrt{1+t^2+t^4}}$
$T '(t)=\mathbf i\frac{-t(2t^2+1)}{({t^4+t^2+1})^\frac{3}{2}}+\mathbf j\frac{1-t^4}{(t^4+t^2+1)^\frac{3}{2}}+\mathbf k\frac{t(t^2+2)}{(t^4+t^2+1)^\frac{3}{2}}$
$\Vert T '(t)\Vert=\sqrt{\frac{(-2t^3-t)^2+(1-t^4)^2+ (t^3+2t)^2}{(t^4+t^2+1)^3}}
$
This is far as I got!
| \begin{align*}
\mathbf{v} &= \dot{s} \mathbf{T} \\
&=(1,t,t^{2}) \\
\dot{s} &= \sqrt{1+t^{2}+t^{4}} \\
\ddot{s} &= \frac{t(1+2t^{2})}{\sqrt{1+t^{2}+t^{4}}} \\
\mathbf{v}(0) &= (1,0,0) \\
\dot{s}(0) &= 1 \\
\mathbf{T}(0) &= \mathbf{i} \\
\mathbf{a} &= \ddot{s} \mathbf{T}+\kappa \dot{s}^{2}\mathbf{N} \\
&=(0,1,2t) \\
\mathbf{a}(0) &= (0,1,0) \\
\ddot{s}(0) &= 0 \\
\mathbf{N}(0) &= \mathbf{j}
\end{align*}
| {
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Which two digit number when you find the product of the digits yields a number that is half the original? Which two digit number when you find the product of the digits yields a number that is half the original?
Let x=$ab$ be the $2$-digit number. So $x=10a+b$.
Then $ab=\frac{x}{2} \implies ab=\frac{10a+b}{2} \implies 2ab=10a+b \implies b=\frac{10a}{2a-1}$. I guess $a=3$ and get $b=6$. So the answer is $36$.
But how can this be done without guessing.
| $2ab-10a-b=0$ is a simple quadratic Diophantine equation. The canonical approach in this case is the following.
You add $5$ to both sides and you obtain
$$
2ab-10a-b+5=5,\ \ \Rightarrow (2a-1)(b-5)=5
$$
so $b-5$ or $2a-1$ must be equal to $5$ (or $-5$, but it is not possible) and the other must be equal to $1$.
But $0<b<10$ so $b-5$ can't be $5$, it must be $1$, hence $b=6$ and thus $2a-1=5$, from which $a=3$.
| {
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If the range of the function $f(x)=\frac{x^2+ax+b}{x^2+2x+3}$ is $[-5,4],a,b\in N$,then find the value of $a^2+b^2.$ If the range of the function $f(x)=\frac{x^2+ax+b}{x^2+2x+3}$ is $[-5,4],a,b\in N$,then find the value of $a^2+b^2.$
Let $y=\frac{x^2+ax+b}{x^2+2x+3}$
$$x^2y+2xy+3y=x^2+ax+b$$
$$x^2(y-1)+x(2y-a)+3y-b=0$$
As $x$ is real,so the discriminant of the above quadratic equation has to be greater than or equal to zero.
$$(2y-a)^2-4(y-1)(3y-b)\geq0$$
$$-8y^2+y(-4a+4b+12)+a^2-4b\geq0$$
As above quadratic inequality is greater than or equal to zero,so its discriminant is less than or equal to zero.
$$(-4a+4b+12)^2+32(a^2-4b)\leq0$$
I am stuck here,i cant find $a^2+b^2$.
| $$-5 \le \frac{x^2+ax+b}{x^2+2x+3} \le 4 $$
As the denominator is positive, this is equivalent to
$$-5x^2-10x-15 \le x^2+ax+b \le 4x^2+8x + 12$$
which can be considered as two quadratic inequalities,
$$6x^2+(a+10)x+(b+15) \ge 0, \quad 3x^2+(8-a)x+(12-b) \ge 0$$
Note that equality must happen as well for some $x$. For this to be true for all reals, the discriminants must be then zero, and $-15 \le b \le 12$ to ensure the quadratics are both above the x axis. So
$$(a+10)^2 = 24(b+15), \quad (8-a)^2 = 12(12-b)$$
Solving, $a=14$ and $b = 9$ giving $a^2+b^2 = 277$.
| {
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Deriving the formula for the height of a trapezoid The bases of a trapezoid have lengths $a$ and $b$, and its legs have lengths $c$ and $d$. A formula for the height is
\begin{equation*}
h = \frac{
\sqrt{(-a + b + c + d)(a - b + c + d)(a - b - c + d)(a - b + c - d)}
}
{2\vert a - b \vert} .
\end{equation*}
The formula is reminiscent of Heron's Formula. I would like to see a derivation of it.
| Your formula "is reminiscent of Heron's Formula" because it is based on Heron's Formula.
One formula for the area of a triangle is
$$A=\frac 12bh$$
and Heron's formula gives
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
where $s$ is the semiperimeter given by
$$s=\frac{a+b+c}2$$
Here is a diagram for the derivation for the height of your trapezoid, assuming that $a>b$ (i.e. $a$ is the larger base and $b$ is the smaller one).
Note that I constructed a line segment (in green) parallel to side $c$ through the end of side $b$ that is not on side $c$. This line segment also has length $c$, of course, and it makes a triangle with sides $a-b,\ c,\ d$ that has the same height $h$ (dotted) as the trapezoid.
Using those sides of the triangle rather than $a,\ b,\ c$ gives us the equations
$$A=\frac 12(a-b)h$$
and
$$A=\sqrt{s(s-[a-b])(s-c)(s-d)}$$
where
$$s=\frac{(a-b)+c+d}{2}$$
Solving for $h$ in $A=\frac 12(a-b)h$, substitutions, and simplifications give us
$$\begin{align}
h &= \frac{2}{a-b}A \\[2ex]
&= \frac{2}{a-b}\sqrt{s(s-[a-b])(s-c)(s-c)} \\[2ex]
&= \frac{2}{a-b}\sqrt{\frac{(a-b)+c+d}{2}\left(\frac{(a-b)+c+d}{2}-[a-b]\right)\left(\frac{(a-b)+c+d}{2}-c\right)\left(\frac{(a-b)+c+d}{2}-d\right)} \\[2ex]
&= \frac{2}{a-b}\sqrt{\frac{a-b+c+d}{2}\left(\frac{-a+b+c+d}{2}\right)\left(\frac{a-b-c+d}{2}\right)\left(\frac{a-b+c-d}{2}\right)} \\[2ex]
&= \frac{1}{2(a-b)}\sqrt{(a-b+c+d)(-a+b+c+d)(a-b-c+d)(a-b+c-d)} \\[2ex]
&= \frac{\sqrt{(-a+b+c+d)(a-b+c+d)(a-b-c+d)(a-b+c-d)}}{2|a-b|} \\[2ex]
\end{align}$$
which is your formula.
It is easily seen that if we assume $a<b$ we end up with the same result, thanks to the absolute value in the denominator. If $a=b$ this formula fails, but we then get a parallelogram whose height is not uniquely determined, so no formula is possible for $a=b$.
I tested this formula in Geogebra, and it checks.
| {
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How to express $f(n\alpha)$ in terms of $f(\alpha)$
Original question: Let $f:\mathbb{R}\to\mathbb{R}$ be a function defined by $f(x)=\dfrac{a^x-a^{-x}}{2}$, where $a>0$ and $a\ne 1$, and $\alpha$ be a real number such that $f(\alpha)=1$. Find $f(2\alpha)$.$^1$
A few years ago, I was a high school student and solved it. Now I am reading the book again, because I has begun teaching my cousin since last week. When I revisited the question, suddenly I wanted to find $f(2\alpha)$, $f(3\alpha)$, $f(4\alpha),\;\dots$
\begin{align}
f(2\alpha)&=\frac{a^{2\alpha}-a^{-2\alpha}}{2}\\
&=\frac{(a^{\alpha}-a^{-\alpha})(a^{\alpha}+a^{-\alpha})}{2}\\
&=f(\alpha)\sqrt{a^{2\alpha}+2+a^{-2\alpha}}\\
&=f(\alpha)\sqrt{(a^{\alpha}-a^{-\alpha})^2+4}\\
&=f(\alpha)\sqrt{4(f(\alpha))^2+4}\\
f(3\alpha)&=f(\alpha)(4(f(\alpha))^2+3)\;(\text{calculations skipped})\\
f(4\alpha)&=f(\alpha)(4(f(\alpha))^2+2)\sqrt{4(f(\alpha))^2+4}
\end{align}
My question: Can we express $f(n\alpha)$ in terms of $f(\alpha)$? (Here $f(\alpha)$ isn't necessarily $1$.)
Attempt: It is known that $x^n - y^n = (x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+y^{n-1})$ for $n\in \mathbb{N}$, so
$$f(n\alpha)=\frac{a^{n\alpha}-a^{-n\alpha}}{2}=\frac{(a^{\alpha}-a^{-\alpha})(a^{(n-1)\alpha}+a^{(n-3)\alpha}+\cdots+a^{(3-n)\alpha}+a^{(1-n)\alpha})}{2}.$$
However, $(a^{(n-1)\alpha}+a^{(n-3)\alpha}+\cdots+a^{(3-n)\alpha}+a^{(1-n)\alpha})$ term is annoying me. I think it will be $2(f((n-1)\alpha)+f((n-3)\alpha)+\cdots+?)$, but I have no idea how to do next.
Partial solution is also appreciated.
$^1$ It was translated from Korean to English by me. Reference: Sunwook Hwang and 12 other authors (2010).『수학Ⅰ 익힘책』. Seoul: (주)좋은책신사고. page 62.
| From $f(\alpha)=\frac{a^\alpha - a^{-\alpha}}{2}$, we can induce the quadratic equation for $a^{\alpha}$:
$$
a^{2\alpha}-2f(\alpha)a^{\alpha}-1=0.
$$
By the quadratic formula, we get
$$
a^{\alpha}=f(\alpha)\pm \sqrt{(f(\alpha))^2+1}
$$
and using $a^{\alpha}>0$ we eliminate one possibility. Thus
$$
a^{n\alpha}=(a^{\alpha})^n =\left(\sqrt{f(\alpha)^2+1} + f(\alpha)\right)^n
$$
and
\begin{align}
a^{-n\alpha}&=\frac{1}{\left(\sqrt{f(\alpha)^2+1} + f(\alpha)\right)^n}\\
&=\frac{\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^n}{\left(\sqrt{f(\alpha)^2+1} + f(\alpha)\right)^n\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^n}\\
&=\frac{\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^n}{(f(\alpha)^2+1-f(\alpha)^2)^n}\\
&=\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^n.
\end{align}
Therefore
$$
f(n\alpha)=\frac{1}{2}\left(\left(\sqrt{f(\alpha)^2+1} + f(\alpha)\right)^n -\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^n\right)
$$
Checking the formula for $n=2$:
\begin{align}
f(2\alpha)&=\frac{1}{2}\left(\left(\sqrt{f(\alpha)^2+1} + f(\alpha)\right)^2 -\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^2\right)\\
&=\frac{1}{2}(f(\alpha)^2+1+2f(\alpha)\sqrt{f(\alpha)^2+1} + f(\alpha)^2 -f(\alpha)^2-1+2f(\alpha)\sqrt{f(\alpha)^2+1}-f(\alpha)^2)\\
&=\frac{1}{2}\cdot 4f(\alpha)\sqrt{f(\alpha)^2+1}\\
&=2f(\alpha)\sqrt{f(\alpha)^2+1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
Determine all integers $x $ and $ y$ such that $|2^x − 3^y| =1$ I am having trouble solving this problem:
Determine all integers x and y such that $|2^x − 3^y| =1$
I would think that the only solutions to it is $x = y = 1$.
How can I show that there is no other solutions?
If there are other solutions, how can I find all the solutions?
| Hint: If $2^x - 3^y = 1$, then $2^x = 1 + 3^y$. Take both sides modulo $16$:
$$2^x \equiv 1 + 3^y \pmod{16}.$$
If $x\ge 4$, then the left-hand side is $0$. The sequence of $3^y \pmod{16}$ is $3,9,27=11,81=1,3,9,11,\ldots$, so the right-hand side is $1+3=4$, $1+9=10$, $1+11=12$, or $1+1=2$, none of which is equivalent to $0\pmod{16}$. Hence, there are no solutions when $x\ge 4$.
Now, what happens if $2^x - 3^y = -1$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
Evaluate $\int_0^{2}\sqrt{4x + 1} \text{d}x$ Evaluate $\displaystyle\int_0^{2}\sqrt{4x + 1}~\text{d}x$
This becomes:
$\displaystyle\int_0^{2}(4x + 1)^\frac{1}2~\text{d}x$
I am not sure where to go from here, I suspect it might use the chain rule or reverse chain rule.
| $$\int_{0}^{2}\sqrt{4x+1}\space\text{d}x=$$
Substitute $u=4x+1$ and $\text{d}u=4\space\text{d}x$.
This gives a new lower bound $u=4\cdot0+1=1$ and upper bound $u=4\cdot2+1=9$:
$$\frac{1}{4}\int_{1}^{9}\sqrt{u}\space\text{d}u=\frac{1}{4}\int_{1}^{9}u^{\frac{1}{2}}\space\text{d}u=$$
Use $\int y^{n}\space\text{d}y=\frac{y^{1+n}}{1+n}+\text{C}$
$$\frac{1}{4}\left[\frac{u^{1+\frac{1}{2}}}{1+\frac{1}{2}}\right]_{1}^{9}=\frac{1}{4}\left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{9}=\frac{1}{4}\cdot\frac{2}{3}\left[u^{\frac{3}{2}}\right]_{1}^{9}=\frac{1}{6}\left[u^{\frac{3}{2}}\right]_{1}^{9}=\frac{1}{6}\left(9^{\frac{3}{2}}-1^{\frac{3}{2}}\right)=\frac{1}{6}(26)=\frac{13}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the general solution to differential equation $x(x+1)^2(y'-\sqrt x)=(3x^2+4x+1)y$ Equation can be transformed to linear differential equation:
$$LHS=x^3y'+2x^2y'+xy'-x^{7/2}-2x^{5/2}-x^{3/2}$$
$$RHS=3x^2y+4xy+y$$
$$\Rightarrow y'(x^3+2x^2+x)+y(-3x^2-4x-1)=x^{7/2}+2x^{5/2}+x^{3/2}$$
After dividing by $(x^3+2x^2+x),x\neq 0\land x\neq -1$
$$\Rightarrow y'+\frac{-3x^2-4x-1}{x^3+2x^2+x}y=\frac{x^{7/2}+2x^{5/2}+x^{3/2}}{x^3+2x^2+x}$$
The general solution is
$$y=e^{-\int \frac{-3x^2-4x-1}{x^3+2x^2+x}\mathrm dx}\left(c+\int \frac{x^{7/2}+2x^{5/2}+x^{3/2}}{x^3+2x^2+x}{e^{\int \frac{-3x^2-4x-1}{x^3+2x^2+x}\mathrm dx}}\mathrm dx\right)$$
Integral $\int \frac{-3x^2-4x-1}{x^3+2x^2+x}\mathrm dx$ can be found using partial fractions, $$\int \frac{-3x^2-4x-1}{x^3+2x^2+x}\mathrm dx=\ln(|x|(x+1)^2)+c$$
How to evaluate integral
$$\int \frac{x^{7/2}+2x^{5/2}+x^{3/2}}{x^3+2x^2+x}{e^{\int \frac{-3x^2-4x-1}{x^3+2x^2+x}\mathrm dx}}\mathrm dx=\int \frac{x^{7/2}+2x^{5/2}+x^{3/2}}{x^3+2x^2+x}e^{\ln(|x|(x+1)^2)}\mathrm dx?$$
Is there an easier method than transforming to linear equation?
| Starting from the equation in the title, we have
$$x(x+1)^2y'-(3x^2+4x+1)y=x(x+1)^2\sqrt x$$
It just so happens we have
$$[x(x+1)^2]'=3x^2+4x+1$$
This suggests use of the quotient rule.
$$\dfrac{x(x+1)^2y'-(3x^2+4x+1)y}{x^2(x+1)^4}=\left[\dfrac{y}{x(x+1)^2}\right]'=\dfrac{\sqrt x}{x(x+1)^2}$$
One messy integral left and we should have the solution. Next step would likely be partial fractions, although substitution may be possible.
$$x=\tan^2\theta,dx=2\tan\theta\sec^2\theta d\theta$$
$$\int\dfrac{\sqrt xdx}{x(x+1)^2}=\int\dfrac{2\tan^2\theta\sec^2\theta d\theta}{\tan^2\theta\sec^4\theta}=\int\cos^2\theta d\theta=\int(\frac12+\frac12\cos2\theta)d\theta=$$
$$\frac12\theta+\frac14\sin2\theta$$
That just leaves a messy backsubstitution, quite possibly using the power reduction formula for tangent.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Through the point $A(4,5)$ a line is drawn. Through the point $A(4,5)$ a line is drawn inclined at $45°$ with the $+ve$ X - axis. It meets $x+y=6$ at the point $B$. Find the equation of $AB$.
My solution..
Equation of $AB$
$$(y-y_1)=m(x-x_1)$$
$$(y-5)=1(x-4)$$
$$x-y+1=0$$.
But the answer in my book is $3x-y=7$.
Can anyone tell me where I made mistake?.
| The equation of the first line is: $y - 5 = \tan(45^{\circ})(x - 4)= 1(x-4) = x-4\Rightarrow y = 5+x-4 = x+1$. Thus the intersection is found by: $x + 1 = 6- x \Rightarrow x = \dfrac{5}{2}\Rightarrow y = 6 - \dfrac{5}{2} = \dfrac{7}{2}\Rightarrow B = (\frac{5}{2}, \frac{7}{2})$. Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1716427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$a$,$b$ and $c$ are roots of the equation $x^3-x^2-x-1=0$ The roots of the equation $x^3-x^2-x-1=0$ are $a$,$b$ and $c$.
if $n \gt 21 $ and $n \in \mathbb{N}$ The find the possible values of $$E=\frac{a^n-b^n}{a-b}+\frac{b^n-c^n}{b-c}+\frac{c^n-a^n}{c-a}$$ in $[0 \: 2]$ are?
Since $x^3=x^2+x+1$ from the graphs of $x^3$ and $x^2+x+1$ its clear that they meet at one point. So number of real roots of $x^3-x^2-x-1=0$ is one which is positive and remaining two are complex conjugates. So let the roots be $a$ and $b=re^{i\theta}$ and $c=re^{-i\theta}$ But product of the roots is $$ar^2=1$$ hence $$r=\frac{1}{\sqrt{a}}$$
Now assuming $b=z$ and $c=\bar{z}$ then
$$E=\frac{a^n-z^n}{a-z}+\frac{z^n-\bar{z}^n}{z-\bar{z}}+\frac{\bar{z}^n-a^n}{\bar{z}-a}$$ so
$$E=2 \Re\left(\frac{a^n-z^n}{a-z}\right)+r^{n-1}\frac{\sin(n\theta)}{\sin\theta}$$
I am not able to proceed from here
| Isn't this a trick question?
Let $$A_{n}=\frac{a^n-b^n}{a-b}$$
Then note $$a^{n+3}=a^{n+2}+a^{n+1}+a^{n}$$$$b^{n+3}=b^{n+2}+b^{n+1}+b^{n}$$
Thus subtract the two, and divide by $a-b$.
This gives us $$A_{n+3}=A_{n+2}+A_{n+1}+A_{n}$$
In a similar fashion, if $$B_{n}=\frac{b^n-c^n}{b-c}$$$$C_{n}=\frac{c^n-a^n}{c-a}$$
Then $$B_{n+3}=B_{n+2}+B_{n+1}+B_{n}$$$$C_{n+3}=C_{n+2}+C_{n+1}+C_{n}$$
Thus if $$E_n=\frac{a^n-b^n}{a-b}+\frac{b^n-c^n}{b-c}+\frac{c^n-a^n}{c-a}$$
Then $$E_{n+3}=E_{n+2}+E_{n+1}+E_{n}$$ since $E_{n}=A_{n}+B_{n}+C_{n}$.
Now, $E_0=0, E_1=3, E_2=2$ and use our above recurrence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1718713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How many base $10$ numbers are there with $n$ digits and an even number of zeros?
How many base $10$ numbers are there with
$n$ digits and an even number of zeros?
Solution:
Lets call this number $a_n$.
This is the number of $n-1$ digits that have an even number of zeros
times $9$ possibilities for the $n$th digit
+ number of $n-1$ digits
that have an odd number of zeros and a zero for the $n$th digit.
$a_n = 9a_{n-1} + (10^{n-1} - a_{n-1})$
$a_n = 8a_{n-1} + 10^{n-1}$
We define
$a_0 = 1$
$a_1 = 9$
The generating function is
$G(x) = 1 + 9x + 82x^2 + 756x^3 + \cdots $
$G(x) = \sum_{n=0}^{\infty}a_n x^n$
$$\begin{align}
G(x) - 1 & = \sum_{n=1}^{\infty} ([8a_{n-1} + 10^{n-1}] x^n)\\
& = \sum_{n=1}^{\infty} 8a_{n-1}x^n +
\sum_{n=1}^{\infty}10^{n-1}x^n\\
& = 8x\sum_{n=1}^{\infty} a_{n-1}x^{n-1} +
x\sum_{n=1}^{\infty}10^{n-1}x^{n-1}\\
& = 8x\sum_{n=0}^{\infty} a_{n}x^{n} +
x\sum_{n=0}^{\infty}10^{n}x^{n}\\
& = 8x G(x) +
x\left(\frac{1}{1-10x}\right)
&
\end{align}
$$
$(1-8x)G(x) = x\left(\frac{1}{1-10x}\right) + 1$
$G(x) = \frac{1-9x}{(1-8x)(1-10x)}$
$G(x) = \frac{1/2}{1-8x} + \frac{1/2}{1-10x}$
$\therefore$ $a_n=\frac{1}{2}(8^n+10^n)$
Is this solution/method valid?
Note that the way i have set up the solution, and defined $a_0$ and $a_1$, there are supposed to be $82$ numbers in $a_2$. I am including the $0$ numbers of zeros, i.e. there are $9 \times 9 = 81$ numbers with $0$ zeros and $1$ number $00$.
| Here's another solution: clearly the number of $n$-digit numbers with $k$ zeroes is $f(n,k) = {n \choose k} 9^k$. Then we have
$$f(n,0) + f(n,1) + \cdots + f(n,n) = \sum_{k=0}^n {n \choose k} 9^k$$
and by the binomial theorem this is $(9+1)^n = 10^n$. On the other hand,
$$f(n,0) - f(n,1) + \cdots + (-1)^n f(n,n) = \sum_{k=0}^n {n \choose k} 9^k (-1)^k$$
and this is, again by the binomial theorem, $(9-1)^n = 8^n$. Adding the two equations together, we get
$$2 f(n,0) + 2 f(n,2) + \cdots + 2 f(n, n) = 8^n + 10^n$$
if $n$ is even, and
$$2 f(n,0) + 2 f(n,2) + \cdots + 2 f(n, n-1) = 8^n + 10^n$$
if $n$ is odd. Dividing through by 2 gives the result.
To be fair, this solution is not the first one that springs to mind unless you know the answer in advance.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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proof $ \binom{n+k+1}{k+1} \times \left( \frac{1}{\binom{n+k}{k}}- \frac{1}{\binom{n+k+1}{k}} \right)=\frac{k}{k+1} $ I would appreciate if somebody could help me with the following problem
Q: How to proof (by combinatorial proof)
$$ \binom{n+k+1}{k+1} \times \left( \frac{1}{\binom{n+k}{k}}
- \frac{1}{\binom{n+k+1}{k}} \right)=\frac{k}{k+1} $$
M1: I study (not combinatorial proof)
$$\binom{n+k+1}{k+1} \times \left( \frac{1}{\binom{n+k}{k}}
- \frac{1}{\binom{n+k+1}{k}} \right)=\frac{(n+k+1_)!}{(k+1)!n!}\times \left(\frac{k!n!}{(n+k)!}-\frac{k!(n+1)!}{(n+k+1)!} \right)
=\frac{n+k+1}{k+1}-\frac{n+1}{k+1}=\frac{k}{k+1}$$
I wonder combinatorial proof!!!
| It’s a bit involved, but here is a combinatorial argument.
As usual, for $m\in\Bbb Z^+$ let $[m]=\{1,\ldots,m\}$. Let $\mathscr{A}$ be the family of $k$-element subsets of $[n+k+1]$, $\mathscr{B}$ the family of $(k+1)$-element subsets of $[n+k+1]$, $\mathscr{C}$ the family of $(k+1)$-element subsets of $[n+k+1]$ that contain the integer $n+k+1$, and $\mathscr{D}=\mathscr{B}\setminus\mathscr{C}$, the family of $(k+1)$-element subsets of $[n+k+1]$ that do not contain the integer $n+k+1$.
Each $A\in\mathscr{A}$ can be expanded to a member of $\mathscr{B}$ in $n+1$ different ways, but each member of $\mathscr{B}$ contains $\binom{k+1}k=k+1$ different members of $\mathscr{A}$, so
$$\binom{n+k+1}{k+1}=|\mathscr{B}|=\frac{n+1}{k+1}|\mathscr{A}|=\frac{n+1}{k+1}\binom{n+k+1}k\;,$$
and hence
$$\frac{\binom{n+k+1}{k+1}}{\binom{n+k+1}k}=\frac{n+1}{k+1}\;.$$
Now consider a set $C\in\mathscr{C}$. From $C$ we can form $n$ different members of $\mathscr{D}$ by replacing the integer $n+k+1$ with one of the $n$ elements of $[n+k]\setminus C$. However, each of the resulting members of $\mathscr{D}$ can be produced from any of $\binom{k+1}k=k+1$ members of $\mathscr{C}$, one for each of its $k$-element subsets, so
$$|\mathscr{D}|=\frac{n}{k+1}|\mathscr{C}|\;,$$
and
$$\binom{n+k+1}{k+1}=|\mathscr{B}|=|\mathscr{C}|+|\mathscr{D}|=|\mathscr{C}|\left(1+\frac{n}{k+1}\right)=\binom{n+k}k\left(1+\frac{n}{k+1}\right)\;,$$
and hence
$$\frac{\binom{n+k+1}{k+1}}{\binom{n+k}k}=1+\frac{n}{k+1}=\frac{n+k+1}{k+1}\;.$$
Thus,
$$\begin{align*}
\binom{n+k+1}{k+1}\left(\frac1{\binom{n+k}k}-\frac1{\binom{n+k+1}k}\right)&=\frac{\binom{n+k+1}{k+1}}{\binom{n+k}k}-\frac{\binom{n+k+1}{k+1}}{\binom{n+k+1}k}\\\\
&=\frac{n+k+1}{k+1}-\frac{n+1}{k+1}\\\\
&=\frac{k}{k+1}\;.
\end{align*}$$
| {
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"url": "https://math.stackexchange.com/questions/1720097",
"timestamp": "2023-03-29T00:00:00",
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Solving the following trigonometric equation: $\sin x + \cos x = \frac{1}{3} $ I have to solve the following equation:
$$\sin x + \cos x = \dfrac{1}{3} $$
I use the following substitution:
$$\sin^2 x + \cos^2 x = 1 \longrightarrow \sin x = \sqrt{1-\cos^2 x}$$
And by operating, I obtain:
$$ \sqrt{(1-\cos^2 x)} = \dfrac{1}{3}-\cos x$$
$$ 1 - \cos^2 x = \dfrac{1}{9} + \cos^2 x - \dfrac{2}{3}\cos x$$
$$ -2\cos^2 x + 2/3\cos x +\dfrac{8}{9}=0$$
$$ \boxed{\cos^2 x -\dfrac{1}{3}\cos x -\dfrac{4}{9} = 0}$$
Can I just substitute $\cos x$ by $z$ and solve as if it was a simple second degree equation and then obtain $x$ by taking the inverse cosine? I have tried to do this but I cannot get the right result. If I do this, I obtain the following results:
$$ z_1 = -0.520517 \longrightarrow x_1 = 121.4º\\
z_2= 0.8538509 \longrightarrow x_2 = 31.37º$$
I obtain $x$ from $z$ by taking the inverse cosine.
The correct result should be around 329º which corresponds to 4.165 rad. My question is if what I am doing is wrong because I have tried multiple times and I obtain the same result (or in the worst case, I have done the same mistake multiple times).
| You can approach this problem with the substitution
$$
\begin{cases}
X=\cos x\\[4px]
Y=\sin x
\end{cases}
$$
that transforms the equation into
$$
\begin{cases}
X+Y=\dfrac{1}{3} \\[6px]
X^2+Y^2=1
\end{cases}
$$
Rewriting the second equation as $(X+Y)^2-2XY=1$, we can substitute and get
$$
\begin{cases}
X+Y=\dfrac{1}{3} \\[6px]
XY=-\dfrac{4}{9}
\end{cases}
$$
that leads to the solving equation
$$
z^2-\frac{1}{3}z-\frac{4}{9}=0
$$
(finding two numbers knowing their sum and product). After rewriting it as $9z^2-3z-4=0$, we find the roots
$$
\frac{1-\sqrt{17}}{6},\qquad \frac{1+\sqrt{17}}{6}
$$
The solutions to the original problem are therefore
$$
\begin{cases}
\cos x=\dfrac{1-\sqrt{17}}{6},\\[6px]
\sin x=\dfrac{1+\sqrt{17}}{6}
\end{cases}
\qquad
\begin{cases}
\cos x=\dfrac{1+\sqrt{17}}{6},\\[6px]
\sin x=\dfrac{1-\sqrt{17}}{6}
\end{cases}
$$
We can express the solutions in terms of the arctangent by noting that in the first case the (principal) angle is in the interval $(\pi/2,\pi)$ and its tangent is
$$
\frac{1+\sqrt{17}}{1-\sqrt{17}}=-\frac{9+\sqrt{17}}{8}
$$
so the corresponding solution is
$$
\pi-\arctan\frac{9+\sqrt{17}}{8}+2k\pi
$$
In the second case the (principal) angle is in the interval $(-\pi/2,0)$ and its tangent is
$$
\frac{1-\sqrt{17}}{1+\sqrt{17}}=\frac{\sqrt{17}-9}{8}
$$
so the corresponding solution is
$$
\arctan\frac{\sqrt{17}-9}{8}+2k\pi
$$
There is another procedure that doesn't introduce extraneous solutions: remember the relations
$$
\cos x=\frac{1-t^2}{1+t^2},\sin x=\frac{2t}{1+t^2}
$$
where $t=\tan\dfrac{x}{2}$.
This is possible because $x=\pi$ is not a solution of the equation. Then you get
$$
\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}=\frac{1}{3}
$$
that simplifies into
$$
2t^2-3t-1=0
$$
so you get
$$
\tan\frac{x}{2}=\frac{3+\sqrt{17}}{4}
\qquad\text{or}\qquad
\tan\frac{x}{2}=\frac{3-\sqrt{17}}{4}
$$
and so
$$
x=2\arctan\frac{3+\sqrt{17}}{4}+2k\pi
\qquad\text{or}\qquad
x=2\arctan\frac{3-\sqrt{17}}{4}+2k\pi
$$
In degrees, the first solution corresponds to $\approx121.367^\circ$ and the second one to $\approx-58.633^\circ$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Find last two digits of $33^{100} $ Find last two digits of $33^{100}$.
My try:
So I have to compute $33^{100}\mod 100$
Now by Euler's Function $a^{\phi(n)}\equiv 1\pmod{n}$
So we have $33^{40}\equiv 1 \pmod{100}$
Again by Carmichael Function : $33^{20}\equiv 1 \pmod{100}$
Since $100=2\cdot40+20$ so we have $33^{100}=1\pmod{100}$
So last two digits are $01$
Is it right?
| Method$\#1:$
$$33^2=(30+3)^2\equiv3^2+2\cdot3\cdot30\equiv190-1\pmod{100}$$
$$\implies33^{100}\equiv(190-1)^{50}$$
$$(190-1)^{50}=(1-190)^{50}\equiv1-\binom{50}1\cdot190\pmod{100}\equiv?$$
Method$\#2:$
Alternatively, Carmichael Function $\lambda(100)=\cdots=20\implies33^{20}\equiv1\pmod{100}$
As $100\equiv0\pmod{20},33^{100}\equiv33^0\pmod{100}$
Method$\#3:$
As $33\equiv1\pmod4$
Again, $33\equiv8\pmod{25}\equiv2^3,2^{10}\equiv-1\pmod{25}$
$\implies33^{10}\equiv(2^3)^{10}\equiv(2^{10})^3\equiv(-1)^3\equiv-1$
$\implies33^{20}\equiv1\pmod{25}$
$\implies$ord$_{100}33=$lcm$(1,20)=20$
| {
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"url": "https://math.stackexchange.com/questions/1724246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find the exponential generating function for the number of ways to distribute $r$ distinct objects into five different boxes Find the exponential generating function for the number of ways to distribute $r$ distinct objects into five different boxes when $b_1<b_2\le 4$, where $b_1,b_2$ are the numbers of objects in boxes $1$ and $2$, respectively.
I understand that boxes $3, 4$, and $5$ will all just have the fundamental exponential generating function and that their combined generating function will be $e^{3x}$. What I don't understand is the first two boxes with the equality parameter given.
Can someone explain how to do this?
|
Three boxes have no restrictions with respect to the objects which results in $e^{3x}$. The restriction $b_1<b_2\leq 4$ of the other objects together with the objects of the three boxes is encoded as
\begin{align*}
e^{3x}&\left(\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\right)+x\left(\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\right)\right.\\
&\quad+\left.\frac{x^2}{2!}\left(\frac{x^3}{3!}+\frac{x^4}{4!}\right)
+\frac{x^3}{3!}\left(\frac{x^4}{4!}\right)\right)\\
&=e^{3x}\left(\frac{x}{1!}+\frac{x^2}{2!}+\frac{4x^3}{3!}+\frac{5x^4}{4!}+
\frac{15x^5}{5!}+\frac{15x^6}{6!}+\frac{35x^7}{7!}\right)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Greatest $n<1000$ such that $\left \lfloor{\sqrt{n}}\right \rfloor-2 \mid n-4$ and $\left \lfloor{\sqrt{n}}\right \rfloor+2 \mid n+4$? My first attempt was incorrect, and it is supposed to be a middle school problem.
So, if $n=k^2$
Then $k-2 \mid n-4$ and $k+2 \mid n+4$, so $n-4 \mid (n-4)(n+4)$. I then assumed the answer would be a perfect square (not a great assumption) to get $n=961=31^2$. I then reasoned that $n$ would be around this value, so I checked these numbers and I think it was $958$. Is there a better way to do this, rather than just checking values less than $1000$?
| Let $k=[\sqrt n]$. Then we know that
$$
k^2\le n<k^2+2k+1.
$$
So $n-4$ is in the range $[k^2-4,k^2+2k-4]$. Of the numbers in that range
only $k^2-4$, $k^2+k-6$ and $k^2+2k-8$ are divisible by $k-2$, so $n$ has to be one of $k^2,k^2+k-2, k^2+2k-4$.
We also see that $n+4$ is in the range $[k^2+4,k^2+2k+4]$. Of those numbers only
$k^2+k-2$ and $k^2+2k$ are divisible by $k+2$. Therefore $n$ also needs to be one of $k^2+k-6$, $k^2+2k-4$.
Looks like $n=k^2+2k-4=(k+1)^2-5$ is the only remaining alternative.
Therefore the answer is ____
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $a = \cos x - b\sin x$ where $a$ and $b$ are real numbers? I found this equation when solving a physics problem related to finding an angle when entering a river, that has a known current, and trying to get to a specific point on the other side. I'm not sure how to solve this equation explicitly for the angle $x$.
| Another way to solve this to use $x=y+z$ and expand the trig into
$$ a = \cos(y) \left( \cos(z) - b \sin(z) \right) - \sin(y) \left( b \cos(z) + \sin(z) \right) $$
Solve $\cos(z) - b \sin(z)=0$ with $z =\tan^{-1}(\frac{1}{b})$ and simplify the above into $$ a = - \sin(y) \left( b \cos(z) + \sin(z) \right) = - \sqrt{1+b^2} \sin(y)$$
This has a solution of $y = -\sin^{-1} \left(\frac{a}{\sqrt{1+b^2}}\right)$.
Altogether the solution is
$$\boxed{ x = \tan^{-1} \left( \frac{1}{b} \right) - \sin^{-1} \left( \frac{a}{\sqrt{1+b^2}} \right) }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Determinant of $N \times\ N$ matrix So the question asks:
For $n \geq 2$, compute the determinant of the following matrix:
$$
B =
\begin{bmatrix}
-X & 1 & 0 & \cdots & 0 & 0 \\
0 & -X & 1 & \ddots & \vdots & \vdots \\
\vdots & \ddots & \ddots & \ddots & 0 & \vdots \\
\vdots & & \ddots & \ddots & 1 & 0 \\
0 & \cdots & \cdots & 0 & -X & 1 \\
a_0 & a_1 & \cdots & \cdots & a_{n-2} & (a_{n-1} - X)
\end{bmatrix}
$$
Looking at the $2 \times 2$ and $3 \times 3$ forms of this matrix:
$\det \begin{bmatrix} -X & 0 \\ 0 & (a_1-X) \end{bmatrix} = -X(a_1-X) - 0 = X^2 - a_1X $
by expansion along the first row:
$\det \begin{bmatrix} -X & 1 & 0 \\ 0 & -X & 0 \\ 0 & 0 & (a_2-X) \end{bmatrix} = (-X) \times\det \begin{bmatrix} -X & 0 \\ 0 & a_2-X \end{bmatrix} - 1 \det\begin{bmatrix} 0 & 0 \\ 0 & a_2-X \end{bmatrix}$
$= (-X)[(-X)(a_2-X) -0] - 0 = X^3 - a_2X^2 $
So it looks like:
$\det \begin{bmatrix}
-X & 1 & 0 & \cdots & 0 & 0 \\
0 & -X & 1 & \ddots & \vdots & \vdots \\
\vdots & \ddots & \ddots & \ddots & 0 & \vdots \\
\vdots & & \ddots & \ddots & 1 & 0 \\
0 & \cdots & \cdots & 0 & -X & 1 \\
a_0 & a_1 & \cdots & \cdots & a_{n-2} & (a_{n-1} - X)
\end{bmatrix}
= X^{n} - a_{n-1}X^{n-1} - a_{n-2}X^{n-2} ... - a_1X$
Does this look right? Is "prove by induction" valid to use here?
| To add a final touch somewhat as a synthesis of the (thorough) answers of @user5713492 and @akech: the global result is that the companion matrix of polynomial p(X) is diagonalized with a Vandermonde matrix V(r_1,r_2,\cdots r_n) where the $r_k$ are the roots of p(X)" see https://en.wikipedia.org/wiki/Vandermonde_matrix
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Number of solutions of $x_1+2\cdot x_2+2\cdot x_3 = n$ I have to find number of solutions of $x_1+2\cdot x_2+2\cdot x_3 = n$. I guess it would be $[x^n](1+x+x^2 \dots)(1 + x^2 + x^4 \dots)^2$, but how to compute it? I know only that $\frac{1}{1-x} = 1+x+x^2 \dots$.
| If $n$ is odd, then $x_1$ must be odd, and if $n$ is even, then $x_1$ must be even. Thus there is an integer $y$ such that $x_1=2y+1$ for former and $x_1=2y$ for latter. Thus
$$
y+x_2+x_3=\frac{n-1}{2}\text{ or }y+x_2+x_3=\frac{n}{2}.
$$
Therefore, the answer is
\begin{align}
\binom{\frac{n+3}{2}}{\frac{n-1}{2}} &= \binom{\frac{n+3}{2}}{2}\\
&=\frac{n+3}{2}\cdot\frac{n+1}{2}\cdot\frac{1}{2}
\end{align}
for odd $n$ and
\begin{align}
\binom{\frac{n+4}{2}}{\frac{n}{2}}&=\binom{\frac{n+4}{2}}{2}\\
&=\frac{n+4}{2}\cdot\frac{n+2}{2}\cdot\frac{1}{2}
\end{align}
for even $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
What can we do to solve the following equation with $6$ variables with some information provided?
Q) There are unique integers $a_2, a_3, a_4, a_5, a_6, a_7$ such that $$\frac{a_2}{2!}+\frac{a_3}{3!}+\frac{a_4}{4!}+\frac{a_5}{5!}+\frac{a_6}{6!}+\frac{a_7}{7!}=\frac 57$$,where $0\le a_i < i$. Then the value of $a_2+a_3+a_4+a_5+a_6+a_7$(1)$8$(2)$9$(3)$10$(4)$11$
Answer: $(2)$
My workout
all these are unique integers hence all are having different values.
now multiplying the big one equation by $7!$
$$7\left\{6!\frac{a_2}{2!}+6!\frac{a_3}{3!}+6!\frac{a_4}{4!}+6!\frac{a_5}{5!}+6!\frac{a_6}{6!}\right\}+a_7= 6!.5$$
which is like the remainder theorem which implies, $6!.5=3600$, when divided by 7 leaves remainder $a_7$ so dividing 3600 by 7 gives 2 hence $a_7=2$ . Now from inequality equation we can say that
$a_2\in\{0,1\}$
$a_3\in\{0,1,2\}$
$a_4\in\{0,1,2,3\}$
$a_5\in\{0,1,2,3,4\}$
$a_6\in\{0,1,2,3,4,5\}$
$a_7\in\{0,1,2,3,4,5,6\}$
so from all the data above i can conclude that $a_2=0, a_3=1, a_4=3, a_5=4, a_6=5, a_7=2$ hence summing them up $0+1+3+4+5+2=15$ which is none of the above option
| 5xum has already pointed out your error.
Multiplying the both sides by $6!$ gives
$$\text{(integer)}+\frac{a_7}{7}=\frac{5\cdot 6!}{7}\quad\Rightarrow\quad a_7\equiv 5\cdot 6!\pmod 7\quad\Rightarrow\quad a_7=2$$
Multiplying the both sides by $7!/6$ gives
$$\text{(integer)}+\frac{7a_6+a_7}{6}=5!\times 5\quad\Rightarrow\quad 7a_6+a_7\equiv 0\pmod 6\quad\Rightarrow \quad a_6=4$$
Multiplying the both sides by $7!/5$ gives
$$\small\text{(integer)}+\frac{42a_5+7a_6+a_7}{5}=6\times 4!\times 5\quad\Rightarrow\quad 42a_5+7a_6+a_7\equiv 0\pmod 5\quad\Rightarrow\quad a_5=0$$
I think you can continue from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all $x,y$ so that $\dfrac{x+y+2}{xy-1}$ is an integer. I am trying to find the integers $x,y$ so that
$\dfrac{x+y+2}{xy-1}$ is an integer.
What I have done:
I suppose there exists $t$ such that $$t=\dfrac{x+y+2}{xy-1}$$ where $xy\neq 1$ then consider the following scenarios:
$$x=y$$ $$x>y>0$$ $$x>0>y$$ ,etc.
This approach helps me find the solutions but it is very long. Any simpler method?
| If $x=y=-1$, then $x+y+2=xy-1=0$. Therefore, some of the solutions below need to be prefaced with "except for $x=y=-1$".
If $x=y$, then $\frac{x+y+2}{xy-1}=\frac{2}{x-1}$, which says that
$$
x=y\in\{0,2,3\}\quad\text{are the only solutions with $x=y$}\tag{1}
$$
Furthermore,
$$
x+y=-2\quad\text{is a solution}\tag{2}
$$
We also have that
$$
xy=0\quad\text{is a solution}\tag{3}
$$
If $x=-1$ or $y=-1$, $\frac{x+y+2}{xy-1}=-1$. Therefore,
$$
(x+1)(y+1)=0\quad\text{is a solution}\tag{4}
$$
If $y=1$, then the solutions to $\frac{x+y+2}{xy-1}=\frac{x+3}{x-1}\in\mathbb{Z}$ must have $-3\le x\le5$. Checking gives
$$
\{(-3,1),(-1,1),(0,1),(2,1),(3,1),(5,1)\}\tag{5}
$$
If $x=1$, then the solutions to $\frac{x+y+2}{xy-1}=\frac{y+3}{y-1}\in\mathbb{Z}$ must have $-3\le y\le5$. Checking gives$$
\{(1,-3),(1,-1),(1,0),(1,2),(1,3),(1,5)\}\tag{6}
$$
Since
$$
|xy-1|-|x+y+2|\ge(|x|-1)(|y|-1)-4\tag{7}
$$
if $x+y\ne-2$ and $(|x|-1)(|y|-1)\ge5$, then $(x,y)$ cannot be a solution. Since all the solutions with $|x|\le1$ or $|y|\le1$ are covered above, we only need to check $2\le|x|,|y|\le5$. Checking these $64$ cases, the only new solutions not counted above are
$$
\{(2,5),(5,2)\}\tag{8}
$$
Therefore, all solutions are given by $(1)$-$(6)$ and $(8)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
What is the simplified average rate of change What is the simplified average rate of change between $x = 3$ and $x = 3 + h$ for the function $f(x) = -x^2 + 5$?
I know you start off with $\frac{f(x+h)-f(x)}{h}$ but after substitution I got $-15+h+x^2$. What is the correct solution?
| $$
\begin{eqnarray}
\frac{f(x+h)-f(x)}{h} &=& \frac{(-(x+h)^2 + 5) - (-x^2 + 5)}{h} \\
&=& \frac{(-x^2 - 2xh - h^2 + 5) - (-x^2 + 5)}{h} \\
&=& \frac{-2xh - h^2}{h} \\
&=& -2x - h
\end{eqnarray}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find a maximum of: $x^{2016} \cdot y+y^{2016} \cdot z+z^{2016} \cdot x $ $x,y,z \ge 0 $ , $ x+y+z =1$
Find a maximum of:
$$x^{2016} \cdot y+y^{2016} \cdot z+z^{2016} \cdot x $$
and when it is reached.
my attempt:
1) $$x^{2016} \cdot y+y^{2016} \cdot z+z^{2016} \cdot x \le $$
$$\le \left( \frac{2016x+y}{2017}\right)^{2017}+...$$
2) $z=1-x-y$
$$f(x;y)=x^{2016} \cdot y+y^{2016} \cdot (1-x-y)+(1-x-y)^{2016} \cdot x $$
$$\frac{df}{dx}=2016x^{2015}\cdot y-y^{2016}+(1-x-y)^{2016} -2016(1-x-y)^{2015}x $$
$$\frac{df}{dy}=...$$
| *
*$x\geq y\geq z$
Hence, $x^{2016}y+y^{2016}z+z^{2016}x\leq(x+z)^{2016}y=2016^{2016}\left(\frac{x+z}{2016}\right)^{2016}y\leq$
$\leq2016^{2016}\left(\frac{2016\cdot\frac{x+z}{2016}+y}{2017}\right)^{2017}=\frac{2016^{2016}}{2017^{2017}}$.
*$x\geq z\geq y$.
In this case $x^{2016}y+y^{2016}z+z^{2016}x\leq x^{2016}z+y^{2016}x+z^{2016}y\leq(x+y)^{2016}z=$
$=2016^{2016}\left(\frac{x+y}{2016}\right)^{2016}z\leq2016^{2016}\left(\frac{2016\cdot\frac{x+y}{2016}+z}{2017}\right)^{2017}=\frac{2016^{2016}}{2017^{2017}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1744968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to find a formula for a periodic sequence? I would like to find the formula for a periodic sequence such as 4, 1, 1/4, 1/4, 1, 4... with a period of 6
|
Here is a variation which utilizes the special structure of the sequence.
The sequence can be written as
\begin{align*}
(a_n)_{n\geq 0}&=\left(4,1,\frac{1}{4},\frac{1}{4},1,4,4,1,\frac{1}{4},\ldots\right)\\
&=\left(4^1,4^0,4^{-1},4^{-1},4^0,4^1,4^1,4^0,4^{-1},\ldots\right)
\end{align*}
Since
\begin{align*}
\left(1-\left(n \text{ mod} (3)\right)\right)_{n\geq 0}=(1,0,-1,1,0,-1,1,0,-1,\ldots)
\end{align*}
and
\begin{align*}
\left((-1)^{\left\lfloor\frac{n}{3}\right\rfloor}\right)_{n\geq 0}=(1,1,1,-1,-1,-1,1,1,1,\ldots)
\end{align*}
we obtain
\begin{align*}
\left((-1)^{\left\lfloor\frac{n}{3}\right\rfloor}\left(1-\left(n \text{ mod} (3)\right)\right)\right)_{n\geq 0}
=(1,0,-1,-1,0,1,1,0,-1,\ldots)
\end{align*}
We conclude the sequence $(a_n)_{n\geq 0}$ can be written as
\begin{align*}
(a_n)_{n\geq 0}=\left(4^{(-1)^{\left\lfloor\frac{n}{3}\right\rfloor}\left(1-\left(n \text{ mod} (3)\right)\right)}\right)_{n\geq 0}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Induction proof of the identity $\cos x+\cos(2x)+\cdots+\cos (nx) = \frac{\sin(\frac{nx}{2})\cos\frac{(n+1)x}{2}}{\sin(\frac{x}{2})}$
Prove that:$$\cos x+\cos(2x)+\cdots+\cos (nx)=\frac{\sin(\frac{nx}{2})\cos\frac{(n+1)x}{2}}{\sin(\frac{x}{2})}.\ (1)$$
My attempt:$$\sin\left(\frac{x}{2}\right)\sum_{k=1}^{n}\cos{(kx)}$$$$=\sum_{k=1}^{n}\sin\left(\frac{x}{2}\right)\cos{(kx)} $$ (Applying $\sin x\cos y =\frac{1}{2}\left[\sin(x+y)+\sin(x-y)\right]$)
$$=\sum_{k=1}^{n}\frac{1}{2}\sin\left(\frac{x}{2}-kx\right)+\frac{1}{2}\sin\left(\frac{x}{2}+kx\right).$$Multiplying $(1)$ by $\sin\left(\frac{x}{2}\right)$ ,gives: $$\sum_{k=1}^{n}\sin\left(\frac{x}{2}\right)\cos(kx)=\sin\left(\frac{nx}{2}\right)\cos{\frac{(n+1)x}{2}}$$ $$\Leftrightarrow\sum_{k=1}^{n}\left[\sin\left(\frac{x}{2}-kx\right)+\sin\left(\frac{x}{2}+kx\right)\right]=\sin\left(-\frac{x}{2}\right)+\sin\left(\frac{x}{2}+nx\right).$$Then, by induction on $n$ and using $(\sin x\sin y)$ and $(\sin x +\sin y)$ formulas, I end in here: $$\sum_{k=1}^{n+1}\sin\left(\frac{x}{2}\right)\cos(kx)=\left[2\sin\left(\frac{nx}{2}\right)\cos\frac{(n+1)x}{2}\right]+\left[2\sin\left(\frac{x}{2}\right)\cos(n+1)x\right].$$ Any help would be appreciated, thanks!
| You want to show that
$$
\frac{\sin\frac{nx}{2}\cos\frac{(n+1)x}{2}}{\sin\frac{x}{2}}
+\cos((n+1)x)=
\frac{\sin\frac{(n+1)x}{2}\cos\frac{(n+2)x}{2}}{\sin\frac{x}{2}}
$$
which is the same as
$$
\sin\frac{nx}{2}\cos\frac{(n+1)x}{2}+\sin\frac{x}{2}\cos((n+1)x)=
\sin\frac{(n+1)x}{2}\cos\frac{(n+2)x}{2}
$$
Set $s=nx/2$ and $t=x/2$; the relation becomes, after multiplying by $2$ for later applying the product to sum formulas,
$$
2\sin s\cos(s+t)+2\sin t\cos(2s+2t)=2\sin(s+t)\cos(s+2t)
$$
Recall
$$
2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta)
$$
so the left-hand side becomes
$$
\sin(2s+t)+\sin(-t)+\sin(2s+3t)+\sin(-2s-t)=
\sin(2s+3t)-\sin t
$$
The right-hand side becomes
$$
\sin(2s+3t)+\sin(-t)=\sin(2s+3t)-\sin t
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluation of $\int \frac{\sqrt{\sin ^4x+\cos ^4x}}{\sin ^3x \cos x }dx$ Evaluate the following integral:
$$\int \frac{\sqrt{\sin ^4x+\cos ^4x}}{\sin ^3x \cos x }dx$$ where $x \in \big(0,\frac{\pi}{2} \big)$
Could some give me hint as how to approach this question?
I tried to use the fact that $\sin ^4x+\cos ^4x=1-\frac{\sin^22x}{2}$ but it didn't help. How should I proceed?
| Dividing by $\cos^2x$ on the top and bottom gives $\displaystyle\int\frac{\sqrt{\tan^4x+1}}{\frac{\sin^3x}{\cos x}}dx=\int\frac{\sqrt{\tan^4x+1}}{\tan^3x}\sec^2x \;dx$.
Now let $u=\tan x$ to get $\displaystyle\int\frac{\sqrt{u^4+1}}{u^3}du,\;$ and then let $t=u^2$ to get $\displaystyle\frac{1}{2}\int\frac{\sqrt{t^2+1}}{t^2}dt$.
Next let $t=\tan\theta\;$ to get
$\displaystyle\frac{1}{2}\int\frac{\sec\theta}{\tan^2\theta}\;\sec^2\theta\;d\theta=\frac{1}{2}\int\frac{\sec\theta}{\tan^2\theta}\;(\tan^2\theta+1)\;d\theta=\frac{1}{2}\int\big(\sec\theta+\cot\theta\csc\theta\big)d\theta$
$\displaystyle=\frac{1}{2}\big[\ln\big|\sec\theta+\tan\theta\big|-\csc\theta\big]+C=\frac{1}{2}\left[\ln\left(\sqrt{1+u^4}+u^2\right)-\frac{\sqrt{1+u^4}}{u^2}\right]+C$
$\displaystyle=\frac{1}{2}\left[\ln\left(\sqrt{1+\tan^4x}+\tan^2x\right)-\frac{\sqrt{1+\tan^4x}}{\tan^2x}\right]+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can you prove this by strong induction? The sequence $b_1,b_2,...$ is defined recursively as:\begin{align} b_1&=0;\\ b_2&=1;\\ b_n&=2b_{n-1}-2b_{n-2}-1 \ \text{for} \ n\geq3. \end{align} Prove that this means: $$\forall n\geq1: b_n=(\sqrt{2})^n \sin{\left(\frac{1}{4}\pi n \right)}-1$$
Edit:
I have tried to prove this by strong induction and have verified than $P(1)$ and $P(2)$ is true, where $P(n)$ is the statement $b_n=(\sqrt{2})^n \sin{\left(\frac{1}{4}\pi n \right)}-1$.
edit: I have assumed that $P(k-2)$ and $P(k-1)$ is true for $k\in\mathbb{N}$. Then I have managed to simplify to: $$b_n=(\sqrt{2})^n \left[\sqrt{2}\sin{\left(\frac{1}{4}\pi (n-1) \right)}- \sin{\left(\frac{1}{4}\pi (n-2) \right)}\right]-5$$
I can't simplify any further.
| Since you have verified the base case, we proceed to the inductive case. Fix $n >2$ and suppose that for all $k <n$, $P(k)$ is true. We want to show that $P(n)$ is true.
$$
\begin{aligned}
b_{n}&=2b_{n-1}-2b_{n-2}-1\\
&=2\left((\sqrt{2})^{n-1}\sin\left(\frac{1}{4}\pi (n-1)\right)-1\right)-2\left((\sqrt{2})^{n-2}\sin\left(\frac{1}{4}\pi (n-2)\right)-1\right)-1\\
&=(\sqrt{2})^{n}\left(\sqrt{2}\sin\left(\frac{\pi (n-1)}{4}\right)-\sin\left(\frac{\pi (n-2)}{4}\right)\right) - 1
\end{aligned}
$$
Now, using $\sin(A+B)=\sin A\cos B+\cos A\sin B$, we get
$$
\begin{aligned}
&\sqrt{2}\sin\left(\frac{\pi (n-1)}{4}\right)-\sin\left(\frac{\pi (n-2)}{4}\right)=\\
&\sqrt{2}\left(\sin\left(\frac{\pi n}{4}\right)\cos\left(\frac{-\pi}{4}\right)-\cos\left(\frac{\pi n}{4}\right)\sin\left(\frac{-\pi}{4}\right)\right)-\sin\left(\frac{\pi n}{4}\right)\cos\left(\frac{-\pi}{2}\right)+\cos\left(\frac{\pi n}{4}\right)\sin\left(\frac{-\pi}{2}\right)\\
&=\sqrt{2}\left(\sin\left(\frac{\pi n}{4}\right)\frac{1}{\sqrt{2}}-\cos\left(\frac{\pi n}{4}\right)(\frac{-1}{\sqrt{2}})\right)-\cos\left(\frac{\pi n}{4}\right)\\
&=\sin\left(\frac{\pi n}{4}\right)+\cos\left(\frac{\pi n}{4}\right)-\cos\left(\frac{\pi n}{4}\right)\\
&=\sin\left(\frac{\pi n}{4}\right)
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Find $a, b, c$ values from function $y=ax^2-bx+c$ and minimum value $D$ The problem reads like this:
The quadratic function which takes the value $41$ at $x = -2$ and $20$ at $x = 5$, is: $y = Ax^2-Bx+C$
The minimum value for this function is: $D$
I order it like this:
$$
\left\{
\begin{array}{1}
A(-2)^2-B(-2)+C=41\\
A(5)^2-B(5)+C=20
\end{array}
\right.
$$
$$
\left\{
\begin{array}{1}
4A+2B+C=41\\
25A-5B+C=20
\end{array}
\right.
$$
And try to solve it as a system of equations. But I'm stuck there. In fact, I'm not even sure if I'm approaching it the correct way.
EDIT
I got the answers for each variable, I just don't know the procedure to get them.
A=3; B=12; C=5; D=-7
| The question as given in the post does not make sense as the value can be as small as you want, according to this Desmos graph.
However, if we add in the restriction that $A, B, C$ are all positive integers, then we can solve the problem. Following on from egreg's solution, we need $A>0, B = 3A+3 ≥ 0$, which gives $A > -1$, and $C = 35-10A > 0$ which gives $A < 3.5$. Combining these two inequalities gives $-1 < A < 3.5$.
As $A$ increases, the value of $C$ decreases, which means the minimum value of the function decreases. Therefore, the maximum value of $A$ is $3$, which means $B=3(3)+3=12$, and $C = 35-10(3)=5$. The vertex thus has $x$-coordinate $\frac{-b}{2a} = \frac{3(3)+3}{2(3)} = 2$, so the minimum value is equal to $ax^2+bx+c = 3(2)^2- 12 \cdot (2) + 5 = -7$, exactly as described in your question.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is the following solution correct? Question: $ \sqrt{x^2 + 1} + \frac{8}{\sqrt{x^2 + 1}} = \sqrt{x^2 + 9}$
My solution: $(x^2 + 1) + 8 = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) - \sqrt{x^2 + 9} \sqrt{x^2 + 1} = 0$
$=> \sqrt{x^2 + 9} (\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$
So, either $\sqrt{x^2 + 9} = 0$ or $(\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$
From the first expression, I get $x = \pm 3 i$ and from the second expression, I get nothing.
Now, notice how in the 2nd step, I could've divided both the sides by $\sqrt{x^2 + 9}$, but I didn't because I learned here that we must never do that and that we should always factor: Why one should never divide by an expression that contains a variable.
So, my question is: is the solution above correct? Would it have been any harm had I divided both the sides by $\sqrt{x^2 + 9}$?
| Yes it is correct. It would do harm dividing out $\sqrt{x^2 + 9}.$ Perhaps an easier example of why this is so... Consider
\begin{equation}
x^2 = x.
\end{equation}
Obviously the two roots are 0,1, but do you see what happens when you divide by $x?$ You reduce the order of the polynomial, hence throwing away a root.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Any hint to solve given integral $\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}$? Show that for $ab>0$ $$\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}={{2\pi}\over ab}$$
I'm not sure how to go about this. Any solutions or hints are greatly appreciated.
| Let $z=e^{i\theta}$, then
\begin{align}
\int_0^{2\pi}\frac{d\theta}{a^2\cos^2\theta+b^2\sin^2\theta}&=\int_C \frac{1}{a^2\left(\frac{z+\frac{1}{z}}{2}\right)^2+b^2\left(\frac{z-\frac{1}{z}}{2i}\right)^2}\frac{dz}{iz}\\
&=\int_C \frac{-4iz}{(a^2-b^2)z^4+2(a^2+b^2)z^2 +(a^2-b^2)}dz,
\end{align}
where $C:|z|=1$. If $a=b$, then the integral becomes
$$
\int_C \frac{-i}{a^2 z}dz = 2\pi i \operatorname{Res}\left(-\frac{i}{a^2 z};0\right)=\frac{2\pi}{a^2}
$$
If $a\ne b$, Solve $(a^2-b^2)z^4+2(a^2+b^2)z^2+(a^2-b^2)=0$, then
$$
z^2=-\frac{a+b}{a-b}\text{ or }z^2=-\frac{a-b}{a+b}
$$
by quadratic formula. Since $\left|\frac{a+b}{a-b}\right|>1$ and $\left|\frac{a-b}{a+b}\right|<1$, there exist two simple poles inside $C$. Assuming $a>b>0$,
\begin{align}
\int_C \frac{-4iz}{(a^2-b^2)z^4+2(a^2+b^2)z^2 +(a^2-b^2)}dz &= 2\pi i\left(\operatorname{Res}\left(f;\sqrt{\frac{a-b}{a+b}}i\right)+\operatorname{Res}\left(f;-\sqrt{\frac{a-b}{a+b}}i\right)\right)
\end{align}
Compute residues:
\begin{align}
\operatorname{Res}\left(f;\sqrt{\frac{a-b}{a+b}}i\right) &= \frac{4\sqrt{\frac{a-b}{a+b}}}{a^2-b^2}\frac{1}{2\sqrt{\frac{a-b}{a+b}}i\left(-\frac{a-b}{a+b} + \frac{a+b}{a-b}\right)}\\
&=\frac{2}{4iab}=\frac{1}{2abi}
\end{align}
\begin{align}
\operatorname{Res}\left(f;-\sqrt{\frac{a-b}{a+b}}i\right) &=\frac{-4\sqrt{\frac{a-b}{a+b}}}{a^2-b^2}\frac{1}{-2\sqrt{\frac{a-b}{a+b}}i\left(-\frac{a-b}{a+b} + \frac{a+b}{a-b}\right)}\\
&=\frac{1}{2abi}
\end{align}
$$
\therefore \int_C \frac{-4iz}{(a^2-b^2)z^4+2(a^2+b^2)z^2 +(a^2-b^2)}dz = 2\pi i \left(\frac{1}{2abi}+\frac{1}{2abi}\right)=\frac{2\pi}{ab}
$$
We can get same conclusion when $b>a>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrating $\int \frac{\sqrt{x^2-x+1}}{x^2}dx$ Evaluate $$I=\int\frac{\sqrt{x^2-x+1}}{x^2}dx$$ I first Rationalized the numerator and got as
$$I=\int\frac{(x^2-x+1)dx}{x^2\sqrt{x^2-x+1}}$$ and splitting we get
$$I=\int\frac{dx}{\sqrt{x^2-x+1}}+\int\frac{\frac{1}{x^2}-\frac{1}{x}}{\sqrt{x^2-x+1}}dx$$ i.e.,
$$I=\int\frac{dx}{\sqrt{x^2-x+1}}+\int\frac{\frac{1}{x^3}-\frac{1}{x^2}}{\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}}dx$$
First Integral can be evaluated using standard integral. But second one i am not able to do since numerator is not differential of expression inside square root in denominator.
| I will give you a hint:
For second integral take $-1/x^2$ common becase if you take $t= 1/x$ then $dt=-1/x^2dx $ Now substitute t in place of $1/x$ so it will be converted into simple integral.
so next step becomes
$$I=\int\frac{dx}{\sqrt{x^2-x+1}}+\int\frac{\frac{1}{x^3}-\frac{1}{x^2}}{\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}}dx$$
$$I=\int\frac{dx}{\sqrt{x^2-x+1}}+\int-1/x^2\frac{-1/x + 1}{\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}}dx$$
$$I=\int\frac{dx}{\sqrt{x^2-x+1}}+\int\frac{1-t}{\sqrt{1-t+t^2}}dt$$
Now I hope you can solve further easily by method of making into perfecting square and then using the standard formula.Solve the integration and place value of $t=1/x$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Diophantine equations $x^n-y^n=2016$
Solve equation
$$x^n-y^n=2016,$$
where $x,y,n \in \mathbb N$
My work so far:
If $n=1$, then $y=k, x=k+2016, k\in \mathbb N$
If $n=2$, then $2016=2^5\cdot 3^2 \cdot 7$
$x-y=1; x+y=2016$
$x-y=2; x+y=1008$
...
If $n=3$, then $x^3-y^3=(x-y)(x^2+xy+y^2)=2016$
$n\ge4 $. I need help here.
| You found the solutions for $n=1$ and your method for $n=2$ works (with Hagen von Eitzen's caveat) to give $(x,y)=(45,3),(46,10),$ $(50,22),(54,30),(65,47),$ $(71,55),(79,65),(90,78),$ $(130,122),(171,165),$ $(254,250),(505,503)$.
For $n=3$ we need $(x-y)(x^2+xy+y^2)=2016$. Since $x^2+xy+y^2>x^2-2xy+y^2>(x-y)^2$ and $12^3<2016<13^3$ we need $(x-y)\le12$ and $x^2+xy+y^2\ge168$. That gives the only possibilities for $(x-y,x^2+xy+y^2)$ as $(12,168),(9,224),(8,252),$ $(7,288),(6,336),$ $(4,504),(3,672),(2,1008)$. But if $x=y\bmod 3$, then $x^2+xy+y^2=0\bmod 3$, which rules out the first two. Some similar arguments and some brute force establish there are no solutions for $n=3$.
For any solution for $n=4$ must correspond to a solution $(x^2,y^2)$ for $n=2$ and there are no such solutions in the list. Similarly we can rule out $n=6,8,9,10$.
For $n=5$ we have $x^5-y^5=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$, so $x^4+x^3y+x^2y^2+xy^3+y^4$ must be a factor of $2016$. Since $7^4>2016$, there are only a small number of candidates for $(x,y)$ that do not make $x^4+\dots$ too big, and it is easy to check that there are no solutions for $n=5$. Similarly for $n=7$.
For $n>10$ the difference between two successive positive $n$th powers is at least $2^{11}-1^{11}=2047$, so there are no solutions for $n>10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If one angle of a triangle be $60^\circ,$ the area $10\sqrt3$ sq cm.,and the perimeter $20$ cm,find the lengths of the sides. If one angle of a triangle be $60^\circ$, the area $10\sqrt3\ \mbox{cm}^2$, and the perimeter $20\ \mbox{cm}$, find the lengths of the sides.
Let $\angle A=60^\circ$ and $\frac{1}{2}bc\sin A=10\sqrt3$
$$bc=40,\quad a+b+c=20$$
$$2R\sin A+2R\sin B+2R\sin C=20$$
I am stuck here.
| You have $(b+c)^2=(20-a)^2$ and so $b^2+c^2-bc=a^2-40a+400-120$. But by the cosine formula that is $a^2$, so you have $40a=280$ and hence $a=7$, so $b+c=13,bc=40$. Hence $b,c=5,8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots =\ln 2$? while doing the Integration problem using Limit of a sum approach i have a doubt how
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots =\ln2$$
by infinite geometric series we have
$$1-x+x^2-x^3+x^4-x^5+\cdots =\frac{1}{1+x}$$ for $|x| \lt 1$ Integrting both sides we get
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots $$ for $|x| \lt 1$
Since this series is valid only for $|x| \lt 1$ how can we substitute $x=1$ and conclude $$\ln2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots $$
| Abel's Theorem shows that you can substitute $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\sin\frac{\pi}{3}+\frac{1}{2}\sin\frac{2\pi}{3}+\frac{1}{3}\sin\frac{3\pi}{3}+\cdots$ Find $$\sin\frac{\pi}{3}+\frac{1}{2}\sin\frac{2\pi}{3}+\frac{1}{3}\sin\frac{3\pi}{3}+\cdots$$
The general term is $\frac{1}{r}\sin\frac{r\pi}{3}$
Let $z=e^{i\frac{\pi}{3}}$
Then, $$\frac{1}{r}z^r=\frac{1}{r}e^{i\frac{r\pi}{3}}$$
I have to find the imaginary part of $$P=\sum_{r=1}^\infty \frac{1}{r}z^r$$
Let $$S=1+z+z^2+\cdots$$
Hence, $$P=\int_0^z Sdz=z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots$$
which is the required sum.
$$S=\frac{1}{1-z}$$
$$P=\int_0^z \frac{1}{1-z}dz$$
$$P=\ln\left(\frac{1}{1-z}\right)=\ln\left(\frac{1}{1-e^{i\frac{\pi}{3}}}\right)=i\frac{\pi}{3}$$
Hence, the imaginary part of $P$ is $\frac{\pi}{3}$
Is this method correct? Is there any method that does not require complex numbers?
| The analysis in the OP works. Here, we address the question regarding the necessity of using complex analysis.
To that end, we first note that using standard trigonometric identities, we can evaluate the sum
$$\begin{align}\sum_{n=1}^N \cos(nx)&=\frac{\sin(x)}{\sin(x)}\sum_{n=1}^N \cos(nx)\\\\
&=\csc(x)\sum_{n=1}^N\left(\frac{\sin((n+1)x)-\sin((n-1)x)}{2}\right)\\\\
&=\csc(x)\left(\frac{\sin((N+1)x)+\sin(Nx)-\sin(x)}{2}\right)\\\\
&=\csc(x)\left(\sin\left(\frac{(2N+1)x}{2}\right)\cos\left(\frac{x}{2}\right)\right)-\frac12\\\\
&=\frac{\sin\left(\frac{(2N+1)x}{2}\right)}{2\sin(x/2)}-\frac12
\end{align}$$
Then, we can write the series of interest as
$$\begin{align}
\sum_{n=1}^\infty \frac{\sin(nx)}{n}&=\lim_{N\to \infty}\int_0^x \sum_{n=1}^N \cos(nx') \,dx'\\\\
&=\lim_{N\to \infty}\int_0^x \frac{\sin\left(\frac{(2N+1)x'}{2}\right)}{2\sin(x'/2)} \,dx'-\frac12 x \tag 1\\\\
\end{align}$$
Finally, evaluating the limit in $(1)$ reveals for $|x|<2\pi$
$$\begin{align}
\lim_{N\to \infty}\int_0^x \frac{\sin\left(\frac{(2N+1)x'}{2}\right)}{2\sin(x'/2)} \,dx'&=\lim_{N\to \infty}\int_0^{(N+1/2)x} \frac{\sin(x')}{(2N+1)\sin\left(\frac{x'}{2N+1}\right)} \,dx' \\\\
&=\text{sgn}(x)\int_0^\infty \frac{\sin(x)}{x}\,dx \\\\
&=\frac{\pi}{2}\text{sgn}(x)\tag 2
\end{align}$$
Putting $(1)$ and $(2)$ together yields for $0<x<2\pi$
$$\sum_{n=1}^\infty \frac{\sin(nx)}{n}=\frac12(\pi-x)$$
Therefore, for $x=\pi/3$, we find that
$$\sum_{n=1}^\infty \frac{\sin(n\pi/3)}{n}=\frac{\pi}{3}$$
as expected!
| {
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"url": "https://math.stackexchange.com/questions/1767590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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System of equations that can be solved by inequalities: $(x^3+y^3)(y^3+z^3)(z^3+x^3)=8$ and $\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=\frac32$
S367. Solve in positive real numbers the system of equations:
\begin{gather*}
(x^3+y^3)(y^3+z^3)(z^3+x^3)=8,\\
\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=\frac32.
\end{gather*}
Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam
From https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2016-02/mr_2_2016_problems.pdf
I think I am smelling inequalities here. In the first equation I used Holder's inequality to show, $xyz \le 1$ , But in the second equation I used Titu's Lemma to get $x+y+z \le 3$ .But I think there would an equality case in one of the two equations. Can anyone help? The original source is Facebook
| Note that
$$
\begin{align}\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}&=\left(\frac{x^2}{x+y}+y-x\right)+\left(\frac{y^2}{y+z}+z-y\right)+\left(\frac{z^2}{z+x}+x-z\right)
\\
&=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=\frac32\,.
\end{align}$$
Therefore,
$$\frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{z+x}=3\,.$$
By the AM-GM Inequality, we see that
$$\left(\frac{x^2+y^2}{x+y}\right)\left(\frac{y^2+z^2}{y+z}\right)\left(\frac{z^2+x^2}{z+x}\right)\leq 1\tag{*}\,.$$
We claim that, for $a,b>0$,
$$\frac{a^3+b^3}{2}\leq \left(\frac{a^2+b^2}{a+b}\right)^3\,,\tag{#}$$
and the inequality becomes an equality if and only if $a=b$. To justify the claim, let $p(t)\in\mathbb{R}[t]$ be the polynomial
$$p(t):=2\,\left(t^2+1\right)^3-(t+1)^3(t^3+1)\,.$$
Since $p(1)=16-16=0$, we know that $(t-1)$ is a factor of $p(t)$. Now,
$$p'(t)=12\,t(t^2+1)^2-3\,(t+1)^2(t^3+1)-3\,t^2(t+1)^3\,,$$
which satisfies $p'(1)=48-24-24=0$ again. That is, $(t-1)^2$ is a factor of $p(t)$. We proceed further:
$$p''(t)=48\,t^2(t^2+1)+12\,(t^2+1)^2-6\,(t+1)(t^3+1)-18\,t^2(t+1)^2-6\,t(t+1)^3\,.$$
We have again that $p''(1)=96+48-24-72-48=0$, and so $(t-1)^3$ is a factor of $p(t)$. Now,
$$p'''(t)=144\,t(t^2+1)+96\,t^3-6\,(t^3+1)-54\,t^2(t+1)-54\,t(t+1)^2-6\,(t+1)^3\,,$$
so $p'''(1)=288+96-12-108-216-48=0$, whence $(t-1)^4$ is a factor of $p(t)$. Because $p$ is a monic polynomial of degree $6$, we must have
$$p(t)=(t-1)^4\,(t^2+\alpha t+\beta)$$
for some $\alpha,\beta\in\mathbb{R}$. With $p(0)=1$, we get $\beta=1$. As $p(-1)=16$, we conclude that $1-\alpha+\beta=1$, so that $\alpha=1$, as well. Consequently,
$$p(t)=(t-1)^4\,(t^2+t+1)\,,$$
which is a nonnegative polynomial (i.e., $p(\mathbb{R})\subseteq \mathbb{R}_{\geq 0}$), and the only real root of $p(t)$ is $t=1$.
Now, (#) is equivalent to
$$2\,\left(a^2+b^2\right)^3-(a+b)^3\left(a^3+b^3\right)=b^6\,p\left(\frac{a}{b}\right)=(a-b)^4\,\left(a^2+ab+b^2\right)\geq 0\,,$$
which is an equality iff $a=b$. Hence, the claim is established, but then we conclude that
$$\left(\frac{x^3+y^3}{2}\right)\left(\frac{y^3+z^3}{2}\right)\left(\frac{z^3+x^3}{2}\right)\leq 1\,,$$
using (#) in (*). Therefore,
$$(x^3+y^3)(y^3+z^3)(z^3+x^3)\leq 8\,.$$
However, the problem statement demands that the inequality above is an equality. That is, $x=y=z$ must hold. Ergo, the only positive real solution to this system of equations is
$$(x,y,z)=(1,1,1)\,.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Problem in solving a question of vector space. The question is :
Let, $V$ be the subspace of all real $n \times n$ matrices such that the entries in every row add up to zero and the entries in every column also add up to zero. What is the dimension of $V$?
Now,I find that for $n=2$ the given condition yields a symmetric matrix of order $2$ which generates a subspace of dimension $3$ and for $n=3$ the matrix is not symmetric and the dimension of the corresponding subspace becomes $6$.But I fail to generalise this concept to get my required result. Please help me.Thank you in advance.
| Actually when $n=2$, you should get 1. Suppose we decide to make the top entry $a$. Then the top right and lower left entries must be $-a$. And once we see that, the bottom right entry must be $a$. So $$V = \textrm{span}\left[\begin{pmatrix}
1 & -1 \\
-1 & 1
\end{pmatrix} \right].$$
If $n=3$, you can fill in the the top-left $2\times 2$ sub-matrix without worry.
$$\begin{pmatrix}
a_1 & a_2 & - \\
a_3 & a_4 & - \\
- & - & -
\end{pmatrix}_.$$
Now the conditions tell us that the rest of the entries must be as follows
$$\begin{pmatrix}
a_1 & a_2 & -(a_1+a_2) \\
a_3 & a_4 & -(a_3+a_4) \\
-(a_1+a_3) & -(a_2+a_4) & a_1 + a_2 + a_3 + a_4
\end{pmatrix}_.$$
So we really only had 4 choices to make. That is, $dim(V) = 4$.
A basis for $V$ when $n=3$ is $$ $$\begin{pmatrix}
1 & 0 & -1 \\
0 & 0 & 0\\
-1 & 0 & 1
\end{pmatrix} \begin{pmatrix}
0 & 1 & -1 \\
0 & 0 & 0 \\
0 & -1 & 1
\end{pmatrix} \begin{pmatrix}
0 & 0 & 0 \\
1 & 0 & -1 \\
-1 & 0 & 1
\end{pmatrix} \begin{pmatrix}
0 & 0 & 0 \\
0 & 1 & -1 \\
0 & -1 & 1
\end{pmatrix}
The general pattern is $dim(V) = (n-1)^2$.
You can see this in a similar fashion to the $3\times 3$ case. I.e. Once you fill in the upper-left $(n-1)\times (n-1)$ submatrix, the rest of the matrix is uniquely determined.
| {
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"answer_count": 2,
"answer_id": 0
} |
Algebra, finding the elements of the field and solving irreducible polynomials I'm trying to do this problem from a practice final but there are no solutions.
I honestly am pretty stumped.
My thought was since it has 7 elements, then the degree of the polynomial must be one and we must be working with the field $F_7$
1a) I have no idea what that means, would the squares and cubes just be $0^2, 1^2, 2^2, 3^2 ... , 7^2$ and $0^3, ... , 7^3$?
1b) My thought is, the field F is 7, so L must be a field with a polynomial of degree 2, but can it just be any polynomial? Don't know how to solve the polynomials $x^2-3$ and $x^2+x-1$
| The squares in $\mathbb{F}_7$ are indeed
$$
0^2=0,\quad
1^2=1,\quad
2^2=4,\quad
3^2=2,\quad
4^2=2,\quad
5^2=4,\quad
6^2=1,
$$
so you get $\{0,1,2,4\}$.
The cubes are
$$
0^3=0,\quad
1^3=1,\quad
2^3=1,\quad
3^3=6,\quad
4^3=1,\quad
5^3=6,\quad
6^3=6,
$$
so you get $\{0,1,6\}$.
In order to find a field with $7^2$ elements, you just need to find an irreducible degree two polynomial in $\mathbb{F}_7[x]$. For instance $x^2-3$.
The field $\mathbb{L}=\mathbb{F}_7(\sqrt{3})$, where we add a root of $x^2-3$ is, up to isomorphism, the unique field with $49$ elements. In particular, also $x^2+x-1$ must have roots in it. A root must be of the form $a+b\sqrt{3}$:
$$
(a+b\sqrt{3})^2+(a+b\sqrt{3})-1=0
$$
becomes
$$
a^2+3b^2+a-1=0,\qquad 2ab+b=0
$$
so either $b=0$ or $a=-1/2=3$. Since $x^2+x-1=0$ has no roots in $\mathbb{F}_7$, the condition $b=0$ must be discarded and we get
$$
3b^2=-a^2-a+1=-9-3+1=3
$$
so $b=\pm1$. Thus the roots are $3+\sqrt{3}$ and $3-\sqrt{3}$.
You get a field with $343$ elements by adding the roots of a cubic irreducible polynomial, for instance $x^3-2$ (it has no roots, so it's irreducible because it has degree $3$).
There is no root of $x^2-3$ in this field, because this field has no subfield with $49$ elements: a field with $7^m$ elements has a subfield with $7^n$ elements if and only if $n\mid m$.
| {
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"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
} |
Solve for unknown matrix Let $A = \begin{bmatrix}
2 & 3 \\
4 & 5
\end{bmatrix}$
and let $B =
\begin{bmatrix}
3 & 4 \\
5 & 6
\end{bmatrix}$
Solve $A X = B$ for a matrix $X$
My guess is that i:
let $X =
\begin{bmatrix}
x \\
y
\end{bmatrix}$
Then solve it using a linear equation but i'm not too sure. Any help would be appreciated.
| Let
$$X=\left[\begin{matrix} a &b\\ c &d\end{matrix}\right].$$
Expanding the matrix product, you need to solve
$$\begin{cases}2a+3c=3\\4a+5c=5\\2b+3d=4\\4b+5d=6\end{cases}$$
Hint:
$$\begin{cases}2a+3c=3\\4a+5c=5\end{cases}$$ $$\begin{cases}2b+3d=4\\4b+5d=6\end{cases}$$
The real, efficient method is by solving the system (by Gaussian elimination) for all right-hand sides simultaneously.
$$\begin{cases}\begin{align}2x+3y=3|4\\4x+5y=5|6\end{align}\end{cases}$$
$$\begin{cases}2x+3y=\ \ \ 3|\ \ \ \ \ 4\\0x-1y=-1|-2\end{cases}$$
$$\begin{cases}2x+0y=\ \ \ 0|-2\\0x-1y=-1|-2\end{cases}$$
$$\begin{cases}x=0|-1\\y=1|\ \ \ \ \ 2\end{cases}$$
| {
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"answer_count": 3,
"answer_id": 0
} |
Finding all real roots of the equation $(x+1) \sqrt{x+2} + (x+6)\sqrt{x+7} = x^2+7x+12$
Find all real roots of the equation
$$(x+1) \sqrt{x+2} + (x+6)\sqrt{x+7} = x^2+7x+12$$
I tried squaring the equation, but the degree of the equation became too high and unmanageable. I also tried substitutions, but it didn't work out correctly. This question was in my weekly class worksheet as were this and this question which I previously asked.
Any help will be appreciated.
Thanks.
| Let $\sqrt{x+2}=a$ and $\sqrt{x+7}= b$.
So, $x + 1 = a^2 -1$
$x+6 = b^2 -1$
$ x^2 + 7x + 12= (x+3)(x+4)= (b^2-4)(a^2+2)$
Substituting and solving the system of equations.
| {
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} |
Find triples $(a,b,c)$ of positive integers such that... Find the triples $(a,b,c)$ of positive integers that satisfy $$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=3. $$
I found this on a local question paper, and I am unable to solve it.
Any help will be appreciated.
| We have $(1+\frac{1}{3})(1+\frac{1}{2})^2=3$, so that is one solution. It also shows that at least one of $a,b,c$ must be $<3$. wlog we may take $a\le b\le c$. So $a=1$ or $2$.
Suppose $a=1$. Then $(1+\frac{1}{b})(1+\frac{1}{c})=\frac{3}{2}$. Since $(1+\frac{1}{5})^2<\frac{3}{2}$ we must have $b<5$. Obviously we need $b>2$. We find $b=3$ gives the solution $(a,b,c)=(1,3,8)$ and $b=4$ gives the solution $(1,4,5)$.
Suppose $a=2$. Then we have $(1+\frac{1}{b})(1+\frac{1}{c})=2$. Since $(1+\frac{1}{3})^2<2$ we must have $b<3$. That gives the solution already noted of $(2,2,3)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Hurwitz Zeta in terms of Bernoulli polynomials. @Raymond Manzoni showed nicely in
this post how the Riemann zeta function is related to the Bernoulli numbers using the Euler-Maclaurin sum. The result is :
\begin{eqnarray}
\zeta(1-k) = -\frac{B_k}{k}.
\end{eqnarray}
(see his equation (4) )
Following a similar process but this time, instead of the Riemann Zeta function we use the Hurwitz zeta function $\zeta(s)=\sum_{n=0}^{\infty} 1/(n+x)^s$ , $\mathrm{Re}(x)>0$, we should find the extension of the equation above:
\begin{eqnarray}
\zeta(1-k,x) = -\frac{B_k(x)}{k}.
\end{eqnarray}
I am having difficulties with this. Any help is appreciated.
Thanks.
| I found an answer:
We will need the following easy to prove (I can expand here if necessary)
relations:
\begin{eqnarray*}
B_{k+1}(x) = \sum_{i=0}^{k+1} B_i
\binom{k+1}{i}
x^{k+1-i} \quad \quad (1),
\end{eqnarray*}
and
\begin{eqnarray*}
\sum_{i=0}^{n-1} (i+m)^{k-1} =
\frac{B_{k}(n+m) - B_{k}(m)}{k} \quad \quad (2)
\end{eqnarray*}
Let us start:
\begin{eqnarray*}
\sum_{k=0}^{\infty} \frac{1}{(k+x)^s} =
\sum_{k=0}^{N-1} \frac{1}{(k+x)^s} +
\sum_{k=N}^{\infty} \frac{1}{(k+x)^s}
\end{eqnarray*}
and, in the second sum, use the Euler-Maclaurin series with
$h=1, a=N, n= \infty$, $f(k)=1/(k+x)^s$. Then
\begin{eqnarray*}
\sum_{i=0}^{n-1} f(a + i h) h = \int_a^b f(t) dt +
\left . \sum_{k=1}^{m} \frac{B_k
h^k}{k!} f^{(k-1)} (t) \right |_a^b + R_m \\
\sum_{i=0}^{\infty} \frac 1{(N + i+x)^s} =
\int_{N}^{\infty} \frac {dt}{(t+x)^s} + \left .
\sum_{k=1}^{m} \frac{B_k}{k!} (-1)^{k-1}\frac{\Gamma(s+k-1)}{\Gamma(s) (t+x)^{s+k-1}}
\right |_{N}^{\infty}+ R_m\\
\sum_{k={N}}^{\infty} \frac 1{(k+x)^s} = \frac {1}{(s-1)(N+x)^{s-1}} -
\sum_{k=1}^{m} \frac{B_k}{k!} (-1)^{k-1}\frac{\Gamma(s+k-1)}{\Gamma(s) \,
(N+x)^{s+k-1}}+ R_m
\end{eqnarray*}
with
\begin{eqnarray*}
R_m = \int_N^{\infty} B_m \left \{ \frac{t-a}{h} \right \}
\frac{\Gamma(s+m)}{ \Gamma(s)} \frac{1}{ (t+x)^{s+m+1}}.
\end{eqnarray*}
Then
\begin{eqnarray*}
\zeta(s,x)= \sum_{k=0}^{N-1} \frac 1{(k+x)^s} +
\frac {1}{(s-1)\,(N+x)^{s-1}} -
\sum_{k=1}^{m} \frac{B_k}{k!} \frac{\Gamma(s+k-1)}{\Gamma(s) (N+x)^{s+k-1}}+ R_m,
\end{eqnarray*}
and taking the limit as $m \to \infty$ (since the
sum converges and $R_m \to 0$, this is easy to show for this function), we find
\begin{eqnarray*}
\zeta(s,x) = \sum_{k=0}^{N-1} \frac{1}{(k+x)^s} +
\frac{1}{ (s-1) (N+x)^{s-1}} + \sum_{k=1}^{\infty} \frac{B_k }{k!}
\frac{\Gamma(s+k-1)}{\Gamma(s)}.
\end{eqnarray*}
We want to choose $s=1-j$, for $j$ a positive integer. First,
we consider the
Pochhammer Symbold
\begin{eqnarray*}
(s)_{k-1} = \frac{ \Gamma(s + k -1)}{\Gamma(s)}.
\end{eqnarray*}
We prove the reflection formula for the Pochhammer symbol,
\begin{eqnarray*}
(-t)_n = (-1)^n (t -n + 1)_n
\end{eqnarray*}
This is,
\begin{eqnarray*}
(-t)_n &=& \frac{ \Gamma(-t+n)}{\Gamma(-t)} \\
&=& \frac{ (-t+n-1)(-t+n-2) \cdots (-t) \cancel{\Gamma(-t)}}{\cancel{\Gamma(-t)}} \\
&=& (-1)^n (t-n+1)(t-n+2) \cdots t \\
&=& (-1)^n t (t-1) \cdots (t-n+1) \\
&=& (-1)^n \frac{\Gamma(t+1)}{\Gamma(t-n+1)} \\
&=& (-1)^n (t-n+1)_n
\end{eqnarray*}
Then we find that for $s=1-j$
\begin{eqnarray*}
\frac{\Gamma(s+k-1)}{k! \; \Gamma(s)} = \frac{(1-j)_{k-1}}{k!} =
(-1)^{k-1} \frac{(j-1-k+1 +1)_{k-1}}{k!}
= (-1)^{k-1} \frac{ (j-k+1)_{k-1}}{k!}
= (-1)^{k-1} \frac{ (j-1)!}{k! \, \Gamma(j-k)}
\end{eqnarray*}
and we can write the formula for $\zeta(1-j,x)$ as
\begin{eqnarray*}
\zeta(1-j,x) &=&
\sum_{k=0}^{N-1} (k+x)^{j-1}
- \frac{ (N+x)^j}{j}
+ \frac{1}{j} \sum_{k=1}^j (-1)^{k-1} B_k \;
(N+x)^{j-k} \binom{j}{k} .
\\
&=&
\sum_{k=0}^{N-1} (k+x)^{j-1}
- \frac{1}{j} \sum_{k=0}^j (-1)^{k} B_k \;
(N+x)^{j-k} \binom{j}{k} .
\end{eqnarray*}
where we introduced the independent term into the second sum. We will change
the alternating sign $(-1)^{k-1}$ for a $-1$ sign, since for $k > 1$, odd $B_{k}=0$, and the
first two terms (indices $k=0,1$) have the right sign.
We recognize, from equation (1), into the equation above that
\begin{eqnarray*}
\zeta(1-j,x) =
\sum_{k=0}^{N-1} (k+x)^{j-1}
- \frac{1}{j} \, B_{j}(N+x).
\end{eqnarray*}
where we removed the alternating sign $(-1)^{k-1}$ since for $i>1$ odd $B_k=0$.
We now use equation (2) to write
\begin{eqnarray*}
\zeta(1-j,x) =
\frac{B_{j}(N+x) - B_{j}(x)}{j}
- \frac{1}{j} \, B_{j}(N+x)
= -\frac{B_j(x)}{j}.
\end{eqnarray*}
This is what we wanted to show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the Exponential of a Matrix that is not Diagonalizable Consider the $3 \times 3$ matrix
$$A =
\begin{pmatrix}
1 & 1 & 2 \\
0 & 1 & -4 \\
0 & 0 & 1
\end{pmatrix}.$$
I am trying to find $e^{At}$.
The only tool I have to find the exponential of a matrix is to diagonalize it. $A$'s eigenvalue is 1. Therefore, $A$ is not diagonalizable.
How does one find the exponential of a non-diagonalizable matrix?
My attempt:
Write
$\begin{pmatrix}
1 & 1 & 2 \\
0 & 1 & -4 \\
0 & 0 & 1
\end{pmatrix} = M + N$,
with $M = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}$ and $N = \begin{pmatrix}
0 & 1 & 2 \\
0 & 0 & -4 \\
0 & 0 & 0
\end{pmatrix}$.
We have $N^3 = 0$, and therefore $\forall x > 3$, $N^x = 0$. Thus:
$$\begin{aligned}
e^{At}
&= e^{(M+N)t} = e^{Mt} e^{Nt} \\
&= \begin{pmatrix}
e^t & 0 & 0 \\
0 & e^t & 0 \\
0 & 0 & e^t
\end{pmatrix} \left(I + \begin{pmatrix}
0 & t & 2t \\
0 & 0 & -4t \\
0 & 0 & 0
\end{pmatrix}+\begin{pmatrix}
0 & 0 & -2t^2 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}\right) \\
&= e^t \begin{pmatrix}
1 & t & 2t \\
0 & 1 & -4t \\
0 & 0 & 1
\end{pmatrix} \\
&= \begin{pmatrix}
e^t & te^t & 2t(1-t)e^t \\
0 & e^t & -4te^t \\
0 & 0 & e^t
\end{pmatrix}.
\end{aligned}$$
Is that the right answer?
| Hint:
Write your matrix $A$ as $I+N$ where $I$ is the identity matrix and $N$ is a nilpotent matrix. Then use the definition of $e^{At}$ as a power series, noting that $N^k=0$ for some $k$.
| {
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"question_score": "13",
"answer_count": 7,
"answer_id": 0
} |
Question on Indefinite Integration: $\int\frac{2x^{12}+5x^9}{\left(x^5+x^3+1\right)^3}\,\mathrm{d}x$ Give me some hints to start with this problem: $${\displaystyle\int}\dfrac{2x^{12}+5x^9}{\left(x^5+x^3+1\right)^3}\,\mathrm{d}x$$
| Because of the cube in denominator, you can say that $$\int \frac{2x^{12}+5x^9}{\left(x^5+x^3+1\right)^3}\,dx=\frac{P_n(x)}{(x^5+x^3+1)^2}$$ where $P_n(x)$ is a polynomial fo degree $n$.
Now differentiate both sides to get $$\frac{2x^{12}+5x^9}{\left(x^5+x^3+1\right)^3}=\frac{\left(x^5+x^3+1\right) P_n'(x)-2 x^2 \left(5 x^2+3\right)
P_n(x)}{\left(x^5+x^3+1\right)^3}$$ So, the right hand side is of degree $n+4$; then, since the left hand side is of degree $12$, this implies $n=8$.
So, set $P_8(x)=\sum_{i=0}^8 a_i x^i$; replacing and comparing the terms of same power then gives the $a_i$'s.
Edit
This works but it is just junk when compared to the so beautiful solution given by André Nicolas.
| {
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"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Integral of the function $\frac{\cos ^2 x}{1+\tan x}$ Evaluate $$\int \frac{\cos ^2 x}{1+\tan x}dx$$ I tried converting in double angle and making the derivative of the denominator in the numerator. But, it didn't work out. Some help please. Thanks.
| Another possibility is a phase shift in the argument.
$$\begin{align}\int\frac{\cos^2x}{1+\tan x}dx&=\int\frac{\cos^3x}{\cos x+\sin x}dx=\int\frac{\cos^3x}{\sqrt2\cos\left(x-\frac{\pi}4\right)}dx\\
&=\frac1{\sqrt2}\int\frac{\cos^3\left(u+\frac{\pi}4\right)}{\cos u}du\\
&=\frac1{\sqrt2}\int\frac{\left(\frac1{\sqrt2}\cos u-\frac1{\sqrt2}\sin u\right)^3}{\cos u}du\\
&=\frac14\int\left[\cos^3 u-3\sin u\cos u+3\sin^2u-\frac{\sin u}{\cos u}+\sin u\cos u\right]du\\&=\frac14\int\left[2-\cos2u-\sin2u-\frac{\sin u}{\cos u}\right]du\\
&=\frac14\left[2u-\frac12\sin2u+\frac12\cos2u+\ln|\cos u|\right]+C_1\\
&=\frac14\left[2x-\frac{\pi}2-\frac12\sin\left(2x-\frac{\pi}2\right)+\frac12\cos\left(2x-\frac{\pi}2\right)+\ln\left|\cos\left(x-\frac{\pi}4\right)\right|\right]+C_1\\
&=\frac14\left[2x-\frac{\pi}2+\frac12\cos2x+\frac12\sin2x+\ln\left|\frac1{\sqrt2}\cos x+\frac1{\sqrt2}\sin x\right|\right]+C_1\\
&=\frac14\left[2x+\cos^2x+\sin x\cos x+\ln\left|\cos x+\sin x\right|\right]+C\end{align}$$
Differentiation confirms this result.
| {
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"answer_count": 3,
"answer_id": 1
} |
If $3^x +3^y +3^z=9^{13}$.Find value of $x+y+z$ Problem: If $3^x +3^y +3^z=9^{13}$.Find value of $x+y+z$.
Solution: $3^x +3^y +3^z=9^{13}$
$3^x +3^y +3^z=3^{26}$
I am unable to continue from here.
Any assistance is appreciated.
Edited
$9^{13} =3^{26}$
$=3^{25} (3)$
$=3^{25} (1+1+1)$
$=3^{25} + 3^{25} + 3^{25}$
So $x+y+z =75$
| You can do this if $x,y,z$ are integers, otherwise there are infinitely many solutions.
First suppose that $x \geq y \geq z$, this is possible because of symmetry.
Suppose that $x \geq 26$, then $3^x\geq3^{26}$ and hence $3^x+3^y+3^z>3^{26}$. Hence $x \leq 25$.
If $z \leq 24$, then $3^x+3^y+3^z \leq 3^{25}+3^{25}+3^{24}<3^{26}$ hence $z\geq 25$. Hence $x=y=z=25$ and $x+y+z=75$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Eigenvalues of matrices of order $n$ How to find eigenvalues of following matrices of order $n$?
$$A=\begin{bmatrix}
1 & 1 & 1 & 1 & \cdots & 1 \\
1 & 0 & 1 & 1 & \cdots & 1 \\
1 & 1 & 0 & 1 & \cdots & 1 \\
1 & 1 & 1 & 0 & \cdots & 1 \\
\vdots & \vdots & \vdots & \vdots &\ddots & \vdots \\
1 & 1 & 1 & 1 & \cdots & 0 \\
\end{bmatrix}_n$$
$$B=\begin{bmatrix}
-1 & 1 & 1 & 1 & \cdots & 1 \\
1 & 0 & 1 & 1 & \cdots & 1 \\
1 & 1 & 0 & 1 & \cdots & 1 \\
1 & 1 & 1 & 0 & \cdots & 1 \\
\vdots & \vdots & \vdots & \vdots &\ddots & \vdots \\
1 & 1 & 1 & 1 & \cdots & 0 \\
\end{bmatrix}_n$$
I had apply matrix calculator to find eigenvalues and I found that exactly $n-2$ eigenvalues of both matrices are $-1$,but I was not able to find rest of the eigenvalues.
| Using row reduction,
$$
\det A=\begin{vmatrix}
1 & 1 & 1 & 1 & \cdots & 1 \\
1 & 0 & 1 & 1 & \cdots & 1 \\
1 & 1 & 0 & 1 & \cdots & 1 \\
1 & 1 & 1 & 0 & \cdots & 1 \\
\vdots & \vdots & \vdots & \vdots &\ddots & \vdots \\
1 & 1 & 1 & 1 & \cdots & 0 \\
\end{vmatrix}_n
=\begin{vmatrix}
1 & 1 & 1 & 1 & \cdots & 1 \\
0 & -1 & 0 & 0 & \cdots & 0 \\
0 & 0 & -1 & 0 & \cdots & 0 \\
0 & 0 & 0 & -1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots &\ddots & \vdots \\
0 & 0 & 0 & 0 & \cdots & -1 \\
\end{vmatrix}_n=(-1)^{n-1}
$$
One can see explicitly that there are $n-2$ linearly independent eigenvectors for the eigenvalue $-1$. Indeed if we solve the system $(A+I)x=0$, we easily get
$$
x_1=0; \ \ x_2+x_3+\cdots+x_n=0
$$
(so the dimension of the solution space, the eigenvectors of $-1$, is $n-2$). We know from $\det A=(-1)^{ n-1}$ that the product of the eigenvalues is $(-1)^{n-1}$. So, as $A $ is real and symmetric (thus selfadjoint) the geometric multiplicities agree with the algebraic ones; we then have
$$
(-1)^{n-1}=\lambda_1\lambda_2\,(-1)^{n-2}.
$$
It follows that $\lambda_1\lambda_2=-1$.
We also know that the trace of $A$ is $1$; so
$$
1=\lambda_1+\lambda_2+(n-2)(-1)=\lambda_1\lambda_2+2-n.
$$
Thus
$$
\lambda_1+\lambda_2=n-1,\ \ \lambda_1\lambda_2=-1.
$$
This is a quadratic, and one finds that
$$
\lambda_1=\frac{n-1+\sqrt{n^2-2n+5}}2,
\ \ \ \ \
\lambda_2=\frac{n-1-\sqrt{n^2-2n+5}}2
$$
The case for $B$ is similar. We get the same $n-2$ dimensional subspace of eigenvectors of $-1$. For the other two eigenvalues, working with the trace and the determinant we get
$$
\det B=(-1)^{n-1}\,(2n-3)
$$
Thus
$$
\lambda_1\lambda_2\,(-1)^{n-2}=(-1)^{n-1}\,(2n-3),\ \ \lambda_1+\lambda_2=-1.
$$
This reduces to
$$
\lambda_1\lambda_2=3-2n,\ \ \lambda_1+\lambda_2=-1.
$$
One then obtains
$$
\lambda_1=-\frac{1+\sqrt{8n-11}}2,\ \ \ \ \ \lambda_2=-\frac{1-\sqrt{8n-11}}2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1780437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the range of $y = \sqrt{x} + \sqrt{3 -x}$ I have the function $y = \sqrt{x} + \sqrt{3 -x}$. The range in wolfram is $y \in\mathbb R: \sqrt{3} \leq y \leq \sqrt{6}$
(solution after correction of @mathlove)
$\sqrt{x} + \sqrt{3 -x} = y$
$$
\begin{cases}
x \geq 0\\
x \leq 3
\end{cases}
$$
then
$(\sqrt{x} + \sqrt{3 -x})^2 = y^2$
$x+3-x+2\sqrt{x(3-x)}=y^2$
$2\sqrt{x(3-x)}=y^2-3$
irrational equation, therefore:
$$
\begin{cases}
y^2-3 \geq 0\\
4(3x-x^2)=(y^2-3)^2
\end{cases}
$$
The $y^2 \geq 3$ is verified when $y \leq -\sqrt{3}$ or $y \geq \sqrt{3}$
regarding the second element of the system $4(3x-x^2)=(y^2-3)^2$
$4(3x-x^2)=(y^2-3)^2$
$12x-4x^2=y^4+9-6y^2$
$12x-4x^2-y^4-9+6y^2 = 0$
$12x-4x^2-y^4-9+6y^2 = 0$
$4x^2-12x+y^4-6y^2+9 = 0$
the quadratic equation is verified when the discriminant is $\geq 0$, then
$b^2 - 4ac = (-12)^2-16(y^4-6y^2+9) \geq 0$
$144-16y^4+96y^2-144 \geq 0$
$16y^4-96y^2 \leq 0$
change $t=y^2$ and $t^2=y^4$
$16t^2-96t \leq 0$
the inequality is verified when $t_1 \leq t \leq t_2$ because the discriminant in t is $\geq 0$
$t(16t^2-96) = 0$
then
$t_1 = 0$ and $t_2 = 6$ but $t=y^2$ and
$\sqrt(0) \leq y \leq \sqrt{6}$
$-\sqrt{6} \leq y \leq \sqrt{6}$
Finally, the solutions in $y$ in the system are:
$y\leq -\sqrt{3}$ or $y\geq \sqrt{3}$
and
$-\sqrt{6} \leq y \leq \sqrt{6}$
the solutions in $y$ (=range) of the system is:
$\sqrt{3} \leq y \leq \sqrt{6}$ because the function in the domain is satisfy only for $0 \leq x \leq 3$
| Note that $y$ is positive, and $$y^2=3+2\sqrt{x(3-x)}$$ It's clear from this that a minimum value for $y$ is $\sqrt{3}$.
Meanwhile, the largest that $x(3-x)$ can get is $\frac32\left(3-\frac32\right)$, since the expression $x(3-x)$ is quadratic with negative leading coefficient, and $x=\frac32$ is the axis of symmetry. So the maximum that $y^2$ can be is $3+2\sqrt{\frac32\left(3-\frac32\right)}=6$. Again since $y$ is positive, $y$ can maximally be $\sqrt{6}$.
Lastly since the function is continuous, the range is $\left[\sqrt{3},\sqrt{6}\right]$, with $\sqrt{3}$ achieved at $x=0$ and $x=3$, and $\sqrt{6}$ achieved at $x=\frac{3}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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