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What is the probablity that the sum of two dice is 4 or 6?
What is the probablity that the sum of two dice is 4 or 6?
The explanation I found is as follows:
Total number of outcomes $6 \times6 = 36$
Number of outcomes where the event occurs: $1+3, 2+2, 3+1, 1+5, 2+4, 3+3, 4+2$ and $5+1$ (Total $ 8$)
The probability that the event occurs is $8/36$ or $2/9$
My doubt is why $1+3$ and $3+1$ are considered as separate outcomes. Isn't it the same as the dices are identical and we cannot really find out which dice had which number on it?
| Here is a simple way to visualize the sample space:
\begin{array}{c|llllll}
+ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
1 & 2 & 3 & \color{blue}{4} & 5 & \color{blue}{6} & 7\\
2 & 3 & \color{blue}{4} & 5 & \color{blue}{6} & 7 & 8\\
3 & \color{blue}{4} & 5 & \color{blue}{6} & 7 & 8 & 9\\
4 & 5 & \color{blue}{6} & 7 & 8 & 9 & 10\\
5 & \color{blue}{6} & 7 & 8 & 9 & 10& 11\\
6 & 7 & 8 & 9 & 10& 11 & 12\\
\end{array}
In blue we have the favourable outcomes. Hence, the probability of getting a sum of $4$ or $6$ is $8/36$, or $2/9$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3744093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
Proving $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$ for positive $a$, $b$, $c$
For $a,b,c>0$ Prove that $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$$
My attempt: By AM-GM we obtain $$\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\ge 3\sqrt[3]{\frac{a^2}{bc}}=\frac{3a}{\sqrt[3]{abc}}$$
Thus $$\sum \frac{a+c}{b}\ge \frac{2(a+b+c)}{\sqrt[3]{abc}}$$
So it suffices to show that $$6\ge \frac{2(a+b+c)}{\sqrt[3]{abc}}\Leftrightarrow 3\sqrt[3]{abc}\ge a+b+c$$
Which is clearly wrong. :"(
Thank you very much.
| Since $\prod\limits_{cyc}(a+b)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)$, by AM-GM we obtain:
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3=\frac{\sum\limits_{cyc}(a^2b+a^2c+2abc)}{abc}=$$
$$=\frac{\sum\limits_{cyc}c(a+b)^2}{abc}\geq \frac{3\sqrt[3]{abc\prod\limits_{cyc}(a+b)^2}}{abc}\geq \frac{3\sqrt[3]{abc\cdot\frac{64}{81}(a+b+c)^2(ab+ac+bc)^2}}{abc}\geq$$
$$\geq \frac{3\sqrt[3]{abc\cdot\frac{64}{81}(a+b+c)^2\cdot3abc(a+b+c)}}{abc}=\frac{4(a+b+c)}{\sqrt[3]{abc}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$f\colon\{1,2,3,4,5\}\longrightarrow\{1,2,3,4,5\}$ Find the total number of functions such that $f(f(x))=f(x)$ I tried to solve it by letting $f(x)=y$
then equation reduces to $f(y)=y$ but i didn't get it further.
Please give me hint to solve this from where I should start
| You have a very good starting point: for all $x$ in your set, $y=f(x)$ is a fixed point.
Conversely, suppose $f$ maps elements to fixed points, then it clearly satisfies $f(f(x))=f(x)$.
So the answers looks like: choose a subset of $\{1,2,3,4,5\}$ to be the fixed points of your function, and then map the remaining points arbitrarily to these fixed points. A few choices of $f$ are:
1 => 1, 2 => 1, 3 => 1, 4 => 1, 5 => 1
1 => 1, 2 => 1, 3 => 1, 4 => 1, 5 => 5
1 => 1, 2 => 1, 3 => 1, 4 => 4, 5 => 1
1 => 1, 2 => 1, 3 => 1, 4 => 4, 5 => 4
1 => 1, 2 => 1, 3 => 1, 4 => 4, 5 => 5
1 => 1, 2 => 1, 3 => 1, 4 => 5, 5 => 5
1 => 1, 2 => 1, 3 => 3, 4 => 1, 5 => 1
1 => 1, 2 => 1, 3 => 3, 4 => 1, 5 => 3
1 => 1, 2 => 1, 3 => 3, 4 => 1, 5 => 5
1 => 1, 2 => 1, 3 => 3, 4 => 3, 5 => 1
1 => 1, 2 => 1, 3 => 3, 4 => 3, 5 => 3
1 => 1, 2 => 1, 3 => 3, 4 => 3, 5 => 5
1 => 1, 2 => 1, 3 => 3, 4 => 4, 5 => 1
1 => 1, 2 => 1, 3 => 3, 4 => 4, 5 => 3
1 => 1, 2 => 1, 3 => 3, 4 => 4, 5 => 4
1 => 1, 2 => 1, 3 => 3, 4 => 4, 5 => 5
1 => 1, 2 => 1, 3 => 3, 4 => 5, 5 => 5
1 => 1, 2 => 1, 3 => 4, 4 => 4, 5 => 1
The list goes on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int\frac{dx}{(a+b\cos(x))^2},(a>b)$ Evaluate $$\int\frac{dx}{(a+b\cos(x))^2},(a>b)$$
I tried to write 1 in numerator as $p'(x)(a+b\cos(x))-p(x)(a+b \cos(x))'$,making something like quotient rule but did not get after that.
| $$I=\int \frac{dx}{(a+b\cos x)^2} \\=\int\frac{dx}{\left(a+b\cdot \frac{1-\tan^2 \frac x2}{1+\tan^2 \frac x2} \right)^2}\\=\int\frac{1+\tan^2 \frac x2}{\left(a(1+\tan^2 \frac x2) +b(1-\tan^2\frac x2)\right)^2} \sec^2\frac x2 \ dx \\ \overset{t=\tan \frac x2}=2\int\frac{1+t^2}{\left[(a-b)t^2 +a+b\right]^2} dt$$ Use partial fractions to break this down into $$ \underbrace{\frac{1}{a-b}\int \frac{dt}{(a-b)t^2 +a+b} }_{I_1} -\underbrace{\frac{2b}{a-b}\int\frac{dt}{\left((a-b)t^2 +a+b\right)^2}}_{I_2}$$ $I_1$ is easy to evaluate (use the arctan function). For $I_2$, substitute $t^2=\frac{a+b}{a-b}\tan^2y$, which results in a massive simplification: $$I_2 =\frac{1}{(a+b)\sqrt{a^2-b^2}}\int \cos^2y \ dy $$ Hopefully you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine if the integral $ \intop_{0}^{\infty}\left(\frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}\right)dx $ convergent So, I need to determine if the integral $ \intop_{0}^{\infty}\left(\frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}\right)dx $ convergent, for any $ C\in \mathbb{R} $.
I found that for C=1 the integral convergent. I think its easy to show that for $ C\leq 0 $ the integral will diverge, but I dont know how to determine for the rest of the real values.
Here's what I've done:
for any $ 0<t $
$ \intop_{0}^{t}\left(\frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}\right)dx=\intop_{0}^{t}\frac{1}{\sqrt{x^{2}+4}}dx-\intop_{0}^{t}\frac{C}{x+2}dx=\intop_{0}^{t}\frac{1}{\sqrt{x^{2}+4}}dx-C\ln\left(t+2\right)+C\ln2 $
Now, we'll calculate the integral $ \intop_{0}^{t}\frac{1}{\sqrt{x^{2}+4}}dx $
$ \intop_{0}^{t}\frac{1}{\sqrt{x^{2}+4}}dx=\intop_{0}^{t}\frac{1}{2\sqrt{\left(\frac{x}{2}\right)^{2}+1}}dx $
And let $ \frac{x}{2}=\sinh u $
so $ dx=2\cosh udu $
and we get
$ \intop_{0}^{t}\frac{1}{2\sqrt{\left(\frac{x}{2}\right)^{2}+1}}dx=\intop_{0}^{aricsinh\left(\frac{t}{2}\right)}1du=arcsinh\left(\frac{t}{2}\right)=\ln\left(\frac{t}{2}+\sqrt{1+\left(\frac{t}{2}\right)^{2}}\right) $
In order to determine if the integal converges, its enough to calculate the limit :
$ \lim_{t\to\infty}\left(\ln\left(\frac{t}{2}+\sqrt{1+\left(\frac{t}{2}\right)^{2}}\right)-c\ln\left(t+2\right)\right) $
If c=1 then the limit will be 0, and thus the integral converge, because we'll have:
$ \lim_{t\to\infty}\ln\left(\frac{\frac{t}{2}+\sqrt{1+\left(\frac{t}{2}\right)^{2}}}{t+2}\right)=\ln\left(1\right)=0 $
How can I determine for $ C\neq1 $ ?
Is there a way to calculate the limit for $ C\neq 1 $ ?
Thanks in advacnce
| Note
\begin{eqnarray}
&&\frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}\\
&=&\frac{x+2-C\sqrt{x^{2}+4}}{(x+2)\sqrt{x^{2}+4}}\\
&=&\frac{(x+2)^2-C^2(x^{2}+4)}{(x+2)\sqrt{x^{2}+4}(x+2+C\sqrt{x^{2}+4})}\\
&=&\frac{(1-C^2)x^2+4x+4(1-C^2)}{(x+2)\sqrt{x^{2}+4}(x+2+C\sqrt{x^{2}+4})}
\end{eqnarray}
If $1-C^2>0$ or $|C|<1$, then there is $N>0$ such that, for $x>N$,
$$ \frac{(1-C^2)x^2+4x+4(1-C^2)}{(x+2)\sqrt{x^{2}+4}(x+2+C\sqrt{x^{2}+4})}>0 $$
and
$$ (1-C^2)x^2+4x+4(1-C^2)=O(x^2), (x+2)\sqrt{x^{2}+4}(x+2+C\sqrt{x^{2}+4})=O(x^3)$$
which implies
$$ \frac{(1-C^2)x^2+4x+4(1-C^2)}{(x+2)\sqrt{x^{2}+4}(x+2+C\sqrt{x^{2}+4})}=O(\frac1x) $$
and hence
$$ \int_N^\infty\bigg(\frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}\bigg)dx $$
diverges. So does
$$ \int_0^\infty\bigg(\frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}\bigg)dx. $$
If $1-C^2=0$, then $C=\pm1$. For $C=-1$
$$ \frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}=\frac{1}{\sqrt{x^{2}+4}}+\frac{1}{x+2}>\frac{1}{x+2}, $$
and hence the integral diverges.
For For $C=1$
$$ \frac{1}{\sqrt{x^{2}+4}}-\frac{1}{x+2}=\frac{4x}{(x+2)\sqrt{x^{2}+4}(x+2+\sqrt{x^2+4}))}=O(\frac{1}{x^2}) $$
and hence the integral converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}$ given that $\sin\alpha-\cos\alpha=\frac12$ Given that $\sin\alpha-\cos\alpha=\frac12$. What is the value of $$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}?$$
My work:
$$\sin\alpha-\cos\alpha=\frac12$$
$$\sin\alpha\frac1{\sqrt2}-\cos\alpha\frac1{\sqrt2}=\frac1{2\sqrt2}$$
$$\sin\left(\alpha-\frac{\pi}{4}\right)=\frac1{2\sqrt2}$$
$$\alpha-\frac{\pi}{4}=\sin^{-1}\left(\frac{1}{2\sqrt{2}}\right)$$
I calculated the value of $\sin^{-1}\left(\frac{1}{2\sqrt{2}}\right)\approx 20.705^\circ$,
so I got $\alpha\approx 45^\circ+20.705^\circ=65.705^\circ$
I calculated
$$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}=\frac{1}{\sin^365.705^\circ}-\frac{1}{\cos^3 65.705^\circ}\approx -13.0373576$$
My question: Can I find the value of above trigonometric expression without using calculator? Please help me solve it by simpler method without solving for $\alpha$. Thanks
| With $s=\sin \alpha$ and $c=\cos \alpha$ we have
$$\frac 1{s^3} -\frac 1{c^3}=\left(\frac 1 s - \frac 1c\right)\left(\frac 1{s^2} + \frac 1{sc} + \frac 1{c^2}\right)$$
$$ = \frac{c-s}{sc}\left(\frac 1{sc}+\frac 1{(sc)^2}\right)$$
Now, since $s-c=\frac 12$, you have $\frac 14 = 1-2sc \Leftrightarrow sc = \frac 38$. So, you get
$$\frac 1{s^3} -\frac 1{c^3} = -\frac 12\cdot \frac 83\left(\frac 83 + \left(\frac 83\right)^2\right) = -\frac{352}{27}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{u}{(u^2+1)^2}-\dfrac{u}{(u^2+1)^3}$$
Substitute $u^2+1=t$, $2u\ du=dt$, $u\ du=dt/2$
$$\int\frac{u^3}{(u^2+1)^3}du=\int \frac{dt/2}{t^2}-\int \frac{dt/2}{t^3}$$
$$=\frac12\dfrac{-1}{t}-\frac{1}{2}\dfrac{-1}{2t^2}$$
$$=-\dfrac{1}{2t}+\dfrac{1}{4t^2}$$
$$=-\dfrac{1}{2(u^2+1)}+\dfrac{1}{4(u^2+1)^2}+c$$
My question: Can I integrate this with suitable substitution? Thank you
| This is one case where you can solve the problem without any integration.
Because of the cube in denominator, assume that
$$\int\frac{u^3}{(u^2+1)^3}du=\frac{P_n(u)}{(u^2+1)^2}$$ Differentiate both sides and remove the common denominator to get
$$u^3=\left(u^2+1\right) P_n'(u)-4 u P_n(u)$$ Comparing the degrees $n=2$; so, write $P_2(u)=a+ b u +c u^2$ to get
$$0=b+2 (c-2 a)u-3 b u^2-(2 c+1) u^3$$ Then, $b=0$, $c=-\frac 12$ and $a=-\frac 14$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 4
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Find the $x$ intercepts of $x^3+x^2-4x-4$ I am to find the $x$ intercepts of $x^3+x^2-4x-4$.
The solution in my book shows these as: $(2,0)$, $(-2,0)$, $(-1,0)$ whereas I get just $(4, 0)$, $(-1, 0)$.
My working:
$x^3+x^2-4x-4$ =
$x^2(x+1)-4(x+1)$ =
$(x^2-4)(x+1)$
Thus, intercepts when $x$ is equal to $4$ and $-1$.
Where have I gone wrong and how can I arrive at $x$ intercepts of $2, -2$ and $1$?
| Just as said in the comments, I will quickly put this into answer form.
The $x^2-4$ can be simplified to $(x+2)(x-2)$, so the roots are $2$, $-2$, and $-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $x≠y$ and $2005(x+y) = 1$; Show that $\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$ Problem:
Let $x$ and $y$ two real numbers such that $x≠0$ ; $y≠0$ ; $x≠y$ and $2005(x+y) = 1$
*
*Show that $$\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$$
*Calculate $l$:
$$l = \frac{y}{y-x} - \frac{y-x}{y} - \frac{x}{y-x} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2 $$
For the first question, I tried to work it out with algebra; I solved for x through the equation given, then multiplied it by y and I got the value of $\frac{1}{xy} = 2005\left(\frac{1}{y-2005y^2}\right) $. Then I tried proving that $\frac{1}{y-2005y^2} =\frac{1}{x} + \frac{1}{y} $ but I failed at this.
| For part 1, if $xy\not=0$, then
$$\begin{align}
1=2005(x+y)\implies{1\over xy}&=2005(x+y){1\over xy}\\
&=2005\left({x\over xy}+{y\over xy}\right)\\
&=2005\left({1\over y}+{1\over x}\right)\\
&=2005\left({1\over x}+{1\over y}\right)
\end{align}$$
For part 2, regroup the terms with denominators $y-x$, $y$, and $x$ and simplify:
$$\begin{align}
{y\over y-x}-{y-x\over y}-{x\over y-x}-{y-x\over x}+{y\over x}-{x\over y}+2
&=\left({y\over y-x}-{x\over y-x} \right)-\left({y-x\over y}+{x\over y} \right)+\left({y\over x}-{y-x\over x} \right)+2\\
&=\left(y-x\over y-x\right)-\left(y\over y\right)+\left(x\over x\right)+2\\
&=1-1+1+2\\
&=3
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Extreme points of a function at domain ends Consider the following function: $f(x) = x\sqrt{9-x^2}$
$\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad $
$f'(x) = \frac{-2x^2+9}{\sqrt{9-x^2}}$ and $D(f) = [-3,3]$ therefore the critical points of the function are $x_{c_i} = \left\{ -3, -\frac{3\sqrt2}{2}, \frac{3\sqrt2}{2}, 3 \right\}$
Apparently the points $\{-\frac{3\sqrt2}{2}, \frac{3\sqrt2}{2}\}$ are global minumum and global maximum respectively.
But what about the domain ends $\{-3, 3\}$? Are they considered to be saddle points, local minimums, or local maximums and why?
| The domain it's $[-3,-3].$
Let $0\leq x\leq 3$.
Thus, by AM-GM $$x\sqrt{9-x^2}=\sqrt{x^2(9-x^2)}\leq\frac{x^2+9-x^2}{2}=\frac{9}{2}.$$
The equality occurs for $$x^2=9-x^2$$ or
$$x=\frac{3}{\sqrt2},$$ which says that $\frac{9}{2}$ is a maximal value of $f$ on $[0,3]$.
The minimal value is $0$ and occurs for $x=0$ or $x=3$.
Let $-3\leq x\leq0$.
Here the maximal value is $0$ and occurs for $x=0$ or $x=-3$.
The minimal value we can get by the similar way:
$$f(x)=-\sqrt{x^2(9-x^2)}\geq-\frac{x^2+9-x^2}{2}=-\frac{9}{2}$$ and it occurs for $x=-\frac{3}{\sqrt2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $\int \frac{x^3}{(4x^2 + 9)^\frac{3}{2}} dx$ The integral I must solve is this:
$$\begin{equation} \int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx \tag{1}\end{equation}$$
Now I know that I'm supposed to convert the denominator into some form of $\sqrt{a^2 + x^2}$ in order to apply the substitution $x = a \ tan \ \theta$. Now I can change the form of the denominator to become easy enough to substitute as follows:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +x^2\bigg)\bigg) ^\frac{3}{2} $$
Which makes it clear that $x$ needs to be substituted as $x = \frac{3}{2} \tan \theta $, and $dx = \frac{3}{2} \sec^2 \theta $
for later use. At this point I can represent $(1)$ in terms of my substituted trignometric function. The only problem comes with the denominator where I get stuck on the power. Here is how I went about solving it:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\left(\frac{3}{2} \tan \theta\right)^2 \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\frac{9}{4} \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \frac{9}{4}\bigg (\ 1 + \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$ 9^\frac{3}{2}\bigg( \ 1 + \tan^2 \theta \bigg) ^\frac{3}{2} $$
$$ 27\ ( \sec^2 \theta ) ^\frac{3}{2} $$
Now I have no idea how to evaluate this power of $sec$. The author says that it changes into $sec^3 \theta$ but I just can't fathom how that would go about. If what I understand is correct, the power it is raised to would be added to it's own making it $ \sec^\frac{7}{2} \theta$. My question is that how exactly is my reasoning wrong here?
| So we want to solve
$${\int \frac{x^3}{\left(4x^2 + 9\right)^{\frac{3}{2}}}dx}$$
The first thing I do when I see an integral like this is ask if there are any simple substitutions I can make. Well, notice if I make the substitution ${u=4x^2 + 9}$, then...
$${\Rightarrow \int \frac{x^3}{u^{\frac{3}{2}}}\times \frac{1}{8x} dx=\int \frac{1}{8}\frac{x^2}{u^{\frac{3}{2}}} du=\frac{1}{8}\int \frac{\frac{u-9}{4}}{u^{\frac{3}{2}}} du}$$
And this equals
$${=\frac{1}{32} \int \frac{1}{\sqrt{u}} - \frac{9}{u^{\frac{3}{2}}}du}$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
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Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x}$ What I attempted thus far:
Multiply by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x} \cdot \frac{\sqrt{1 + x\sin x} + \sqrt{\cos x}}{\sqrt{1 + x\sin x} + \sqrt{\cos x}} = \lim_{x \to 0} \frac{1 + x\sin x - \cos x}{x\tan x \cdot(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$$
From here I can’t see any useful direction to go in, if I even went in an useful direction in the first place, I have no idea.
| $\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x}$
$=\lim_{x \to 0} \frac{(\sqrt{1 + x\sin x} - \sqrt{\cos x})(\sqrt{1 + x\sin x} + \sqrt{\cos x})}{x\tan x(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$
$=\lim_{x \to 0} \frac{1+x\sin x-\cos x}{x\tan x(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$
$=\lim_{x \to 0} \frac{x\sin x+2\sin^2 {x\over 2}}{x\tan x(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$
divided by $x^2$
$$=\lim_{x \to 0} \frac{\frac{\sin x}{x}+{1\over 2}\frac{(\sin {x\over 2})^2}{(x/2)^2}}{\frac{\tan x}{x}(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$$
$$=\frac{1+\frac12}{1(1+1)}$$
$$=\frac34$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why do some partial fractions have x or a variable in the numerator and others don't? Why do rational expressions like $\left(\frac{1}{(x-2)^3}\right)$ do not have x in the numerator of the partial fraction but a rational expression like $\left(\frac{1}{(x^2+2x+3)^2}\right)$ does have x in the numerator of its partial fraction?
| You can integrate $\frac{1}{(x-2)^k}$ by itself, so it is not necessary to break it into the form $\frac{A}{x-2} + \frac{Bx+C}{(x-2)^2} + \dots.$ On the other hand, $\frac{1}{(x^2+2x+3)^k}$ does not have a simple antiderivative, so we must decompose it into the form $\frac{Ax+B}{x^2+2x+3} + \frac{Cx+D}{(x^2+2x+3)^2} + \dots$
The most general explanation is that $x^2+2x+3$ has complex roots while $x-2$ has real roots. If you allow complex numbers, you can break $\frac{1}{(x^2+2x+3)^k}$ into partial fractions where the numerator is always a constant instead of a term of the form $Ax+B.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the absolute extrema of $F(x) = 2x + 5\cos(x)$
Find the absolute extrema of $F(x) = 2x + 5\cos(x)$ on the interval $[0,2\pi]$ using the extreme value theorem.
Answer should be 2 ordered pairs.
I got $\arcsin(2/5)$ for the first value of $x$, but can’t figure out the second.
Thanks in advance.
| The question says absolute extreme.
$5\cos x \le 5$ and $2x \le 4\pi$ so $2x+ 5\cos x\le 5+4\pi$ so if $2x + 5\cos x$ ever equals $5+4\pi$, which it does at $x = 2\pi$, that will be an absolute (albeit not necessarily a local) maximum. So the absolute maximum is $(2\pi, 5+4\pi)$.
We can't do the same for minimum as $2x \ge 0$ and $5\cos x \ge -5$ but when $2x=0$ we do not have $5\cos x =-5$ for $x =0$ so $2x + 5\cos x > -5$.
For a minimum we can find local minimum via $[2x+5\cos x]'= 2 -5\sin x = 0$ so $\sin x =\frac 25$. So $0<\frac 25 < 1$ there will be one such value $x_1$ in the first quadrant and a second value $x_2$ in the second quadrant. $x_1$ is $\arcsin \frac 25$ and $x_2$ is $\pi-\arcsin \frac 25$.
As $\sin x_1 =\frac 25$ then $\cos x_1 = \sqrt{1 - \frac 4{25}} = \frac {\sqrt{21}}5$. And $x_2$ being in the second quadrant has $\cos x_2 = -\frac {\sqrt {21}}5$.
So the local extrema are $(\arcsin \frac 25, 2\arcsin \frac 25 + \sqrt {21})$ and $(\pi - \arcsin \frac 25, 2\pi - 2\arcsin \frac 25-\sqrt{21})$
Now $\frac 25 < \frac 12$ so $0< \arcsin \frac 25 < \frac \pi 6$ and $5>\sqrt{21} > 4$ so $2\pi -2\arcsin \frac 25-\sqrt{21} < 2\pi - 4 < 7- 4 =3< \sqrt{21}<2\arcsin \frac 25 + \sqrt {21}< 4\pi + 5$
So $(\pi - \arcsin \frac 25, 2\pi - 2\arcsin \frac 25-\sqrt{21})$ is a local minimum and $(\arcsin \frac 25, 2\arcsin \frac 25 + \sqrt {21})$ is a local but NOT an absolute maximum.
Now $2*0 + 5\cos 0 =5> 3 >2\pi -2\arcsin \frac 25-\sqrt{21}$. So the local minimum is the absolute minimum.
So the absolute extrema at the local minimum, $(\pi - \arcsin \frac 25, 2\pi - 2\arcsin \frac 25-\sqrt{21})$ and at the end point, $(2\pi, 4\pi + 5)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $\sin x\cos3x+\cos x\sin3x=\sqrt{3}/2$ in the interval $[0,2\pi]$
Solve the following trig equation:
$$\sin(x)\cos(3x)+\cos(x)\sin(3x)=\frac{\sqrt{3}}{2}$$
in the interval $[0,2\pi]$.
Using trig identities, it is now at
$$\sin(x+3x)=\sqrt{3}/2 \tag{1}$$
Next, it has become
$$\sin(x+3x)=\sin 60^\circ \tag{2}$$
leading to
$$\sin(x+3x)-\sin60^\circ=0 \tag{3}$$
From there, it is unclear what to do.
| $\sin(\theta)=\frac{\sqrt{3}}{2}$ when $\theta=\frac{\pi}{3}$, as you pointed out, but also when $\theta=\frac{2\pi}{3}$. Not only that, because the sine function is periodic, we must also remember that adding any multiple of $2\pi$ to either of those angles gives another solution to $\sin(\theta)=\frac{\sqrt{3}}{2}$.
In general we have $\theta=\frac{\pi}{3}+2\pi\cdot k$ and $\theta=\frac{2\pi}{3}+2\pi\cdot k$, where $k$ can be any integer. Note that I am using radians because the answer needs to be in radians in the interval $[0,2\pi]$.
Since $\theta$ represents $4x$, we need to solve the two equations above, substituting $4x$ in place of $\theta$.
For the first, we have $4x=\frac{\pi}{3}+2\pi\cdot k$
Divide by $4$ to get $x=\frac{\pi}{12}+\frac{\pi}{2}\cdot k$.
Now we consider what values of $k$ will make $x$ be in the interval $[0,2\pi]$
We can see that $k= 0, 1, 2, 3$, corresponds to $x=\frac{\pi}{12}, \frac{7\pi}{12}, \frac{13\pi}{12}, \frac{19\pi}{12}$
I will leave solving $4x=\frac{2\pi}{3}+2\pi\cdot k$ up to you
| {
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Question invoving using binomial identities to determine $n$ and $k$ given $\binom{n}{k-1} = 2002$ and $\binom{n}{k} = 3003$ I have been trying to do a problem in a combinatorics textbook involving using binomial identities. The problem is :
"Determine $n,k \in \mathbb{N}$ from the equalities $\binom{n}{k-1} = 2002$ and
$\binom{n}{k} = 3003$"
The chapter that the problem is associated with has a lot of identities involving binomial coefficients. Some identities included are :
*
*$\binom{n}{k} = \binom{n}{n-k} \; 0 \leq k \leq n$
*$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1} \; 1 \leq k \leq n$
*$\binom{n}{m} \binom{m}{k} = \binom{n}{k} \binom{n-k}{m-k} = \binom{n}{m-k}\binom{n-m+k}{k}$
Some other identities are also presented in examples.
I am not sure how to approach this problem. I can deduce that :
\begin{equation}
\frac{3}{2} = \frac{n-k+1}{k}
\end{equation}
given that :
\begin{equation}
\frac{3003}{2002} = \frac{1001}{1001} \frac{3}{2} = \frac{3}{2}
\end{equation}
and :
\begin{equation}
\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}
\end{equation}
But I am not sure how to use this information and the identities to solve for $n$ and $k$.
Does anyone know how to approach a problem like this? There are no examples in the text. Maybe if I see how this problem can be solved then others like in the text will be easier.
| Starting from $$\frac{n-k+1}{k} = \frac{3}{2},$$ we obtain the Diophantine equation $$5k - 2n = 2$$ where $k < n \in \mathbb Z^+$. Thus $k$ must be even, say $k = 2m$, and the above equation becomes $$5m - n = 1,$$ or $n = 5m-1$, hence $$3003 = \binom{n}{k} = \binom{5m-1}{2m} = \frac{(5m-1)!}{(2m)!(3m-1)!}.$$ Since $3003 = 3 \cdot 7 \cdot 11 \cdot 13$, we require $5m-1 \ge 13$ or $m \ge 3$. Since there are no larger prime factors, we also know $5m-1 < 17$, or $m \le 3$. Therefore, $m = 3$ is the only candidate, and $$\binom{14}{6} = \frac{14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} = \frac{(7 \cdot 13 \cdot 11 \cdot 3)(2 \cdot 12 \cdot 5 \cdot 2 \cdot 3)}{12 \cdot 5 \cdot 2 \cdot 2 \cdot 3} = 3003$$ as claimed.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$?
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$
Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\\
&=\int \dfrac{3\sec^2\theta d\theta}{81\sec^4\theta}\\
&=\dfrac{1}{27}\int \cos^2\theta d\theta\\
&=\dfrac{1}{27}\int \frac{1+\cos2\theta}{2} d\theta\\
&=\dfrac{1}{54}\left(\theta+\frac{\sin2\theta}{2}\right)+C
\end{align*}
This is where I got stuck. How can I get the answer in terms of $x$?
Can I solve it by other methods?
| Here is an alternative method. You can first use substitution to simplify the integral:
$$\int \dfrac{dx}{(x^2-4x+13)^2}$$
$$ =\int \dfrac{dx}{((x-2)^2+9)^2} = \int \dfrac{du}{(u^2+9)^2} \tag{$u=x-2, du = dx$}$$
$$=\frac{1}{81}\int \dfrac{du}{(\frac{u^2}{9}+1)^2} = \frac{1}{27} \int \dfrac{dv}{(v^2+1)^2} \tag{$u=3v, du = 3 \ dv$}$$
and then use integration by parts as shown in Rene's answer in this thread. This leads to:
$$2\int \frac{1}{(1+v^2)^2}\,dv=\frac{v}{1+v^2}+\int \frac{1}{1+v^2}\,dv$$
$$\int \frac{1}{(1+v^2)^2}\,dv=\frac{1}{2} \left(\frac{v}{1+v^2}+\arctan(v) \right)$$
and then a couple of back-substitutions leads you to the answer of the original problem.
| {
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"timestamp": "2023-03-29T00:00:00",
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$a = \log_{40}100, b = \log_{10}20$.How can I express $b$ depending only on $a$? Let $a = \log_{40}{100}, b = \log_{10}{20}$. How can I express $b$ depending only on $a$? I tried using the formula to change the base from $40$ to $10$, but couldn't get it just depending on $a$.
I used the base change formula $\log_a b = \dfrac{\log_ c b}{\log_c a}$ and got that $\log_{40} 100 = \dfrac{\log_{10} 100}{\log_{10}{40}} = \dfrac{2}{\log_{10}{20} + \log_{10}{2}}$. But then how could I express $\log_{10}2$ depending on $\log_{10}20$? I think that's what it suffices to show.
| $$a=\log_{40}{100}=2\log_{40}10=\frac{2}{\log_{10}40}=\frac{2}{1+2\log_{10}2},$$ which gives $$\log_{10}2=\frac{2-a}{2a}.$$
Id est, $$\log_{10}20=1+\log_{10}2=\frac{a+2}{2a}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Newton's evaluation of $1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots$ How might have Newton evaluated the following series?
$$\sqrt{2} \, \frac{\pi}{4} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots$$
The method of the this thread applies by setting $x=\pi/4$ in the Fourier series for $f(x) = \pi/2 - x/2$ and then subtracting the extraneous terms (which are a multiple of the Gregory-Leibniz series for $\pi/4$).
I read that this series appears in a letter from Newton to Leibniz. However, I do not have access the letter which appears in this volume.
| Looking at the text mentioned, we see something interesting on page $156$, in note $(48)$. Transcribing it (you can find a screencap of the text here):
On observing that $$1 + \frac 1 3 - \frac 1 5 - \frac 1 7 + \frac 1 9 + \text{etc.} = \int_0^1 \frac{1+x^2}{1+x^4}dx$$ by expanding the integrand as an ascending series in $x$, we may suppose that Newton obtained his result by integrating the identity $$\frac{1}{e+fz+gz^2} + \frac{1}{e-fz+gz^2} = \frac{2e + 2gz^2}{e^2 + g^2 z^4} \text{ (on putting } 2eg = f^2 \text{)}$$ in two ways. Put $z \sqrt g = x \sqrt e$ and $-x\sqrt e$, respectively, in the two fractions on the left, and integrate from $0$ to $1$ with regard to $x$. Then, on combining the terms on the left, the identity gives $$\frac 1 2 \int_{-1}^1 \frac{dx}{1 + \sqrt 2 x + x^2} = \int_0^1 \frac{1+x^2}{1+x^4}dx$$ By putting $1 + x \sqrt 2 = \tan \theta$, and integrating $\theta$ from $- \frac 1 8 \pi$ to $\frac 3 8 \pi$, the left side gives $\pi/2 \sqrt 2$, and the right is equal to the series. (Cf. Hofmann, p. $175$.) The series, when written $1 + (\frac 1 3 - \frac 1 5) - \frac 1 7 - \frac 1 9) + \text{etc.}$, leads to the next result in the letter.
Granted, I'm not sure if this is how Newton actually calculated it, and I haven't actually read this book thoroughly enough to say if it's what you're looking for (I just skimmed until I found what seemed relevant). Hopefully it's enlightening though.
| {
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"timestamp": "2023-03-29T00:00:00",
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Need help with even number problem Problem:
A random non-zero positive even integer, $E$ is picked.
We can call the bracketed section in the statement $E = \{A\cdot2^n\}$ the factored form of this number (because it is even, $n$ is at least $1$). $n$ is to be as large as possible ensuring $A$ is odd.
E.g.: $E= 80$ so $A=5$ and $n=4$ because $5\cdot2^4$ is the factored form of $80$.
Every time we get a random even number $E$, we will divide by $2$ until we find $A$. The question is, on average how much lower is $A$ than $E$, or what is the average $n$ you should expect to find in the factored form?
My approach:
$1/2$ of all numbers or every other number is even or in other words a factor of $2^1$
half of the factors of $2^1$ are factors of $2^2$
half of the factors of $2^2$ are factors of $2^3$
and so on.
I believe, adding from $n=1$ to infinity for $1/(2^n)\cdot1/(2^n)$ may represent the change from $E$ to $A$. I added a picture with actual sigma notation at the bottom.
I got 1/(2^n) because:
1/2+1/4+1/8...= 1
1/2 of all even numbers have n=1 as the largest n in the factored form
therefore 1/2 of all even numbers can only be divided once by 2 until they become odd
1/4 of all even numbers have n=2 as the largest n in the factored form.
therefore 1/4 of all even numbers can be divided once by 4 until they become odd
1/8 of all even numbers have n=3 as the largest n in the factored form.
therefore 1/8 of all even numbers can be divided once by 8 until they become odd
1/16
etc...
generally 1/2^n of all even numbers can be multiplied by 1/2^n until odd
therefore addition from 1 to infinity of (1/(2^n))^2 will represent the difference from E to A out of a whole
According to my sigma calculator, squaring each term before adding them suggesting $A=\frac{1}{3}E$. I'm convinced I messed up somewhere in trying to calculate the average difference between $E$ and $A$.
| To find the expected value of $n$, we use the following method. Assume we want to find the probability that $n=n_0$ for a fixed $n_0$. We need $2^{n_0} \mid E$ and $2^{n_0+1} \nmid E$. This is equivalent to saying $E \equiv 2^{n_0} \pmod{2^{n_0+1}}$. Thus, we have one choice for $E$ in every $2^{n_0+1}$ numbers, giving us the fact that the probability that $n=n_0$ is $\frac{1}{2^{n_0+1}}$. Now, consider summing this over all of $n$:
Thus, the expected value of $n$ is:
$$n_{\text{avg}}=\sum_{n=1}^\infty \frac{n}{2^{n+1}}=\sum_{n=1}^\infty \frac{1}{2^{n+1}}+\sum_{n=1}^\infty \frac{n-1}{2^{n+1}}=\frac{1}{2}+\frac{1}{2}\sum_{n=1}^\infty \frac{n-1}{2^{n}}$$
$$n_{\text{avg}}=\frac{1}{2}+\frac{1}{2}\sum_{n=1}^\infty \frac{n-1}{2^n}=\frac{1}{2}+\frac{1}{2}\sum_{n=2}^\infty \frac{n-1}{2^n}=\frac{1}{2}+\frac{1}{2}\sum_{n=1}^\infty \frac{n}{2^{n+1}}=\frac{n_{\text{avg}}+1}{2}$$
This would give $n_\text{avg}=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Rolling Dice Game, Probability of Ending on an Even Roll The game is described as follows. $A$ and $B$ take turns rolling a fair six sided die. Say $A$ rolls first. Then if $A$ rolls {1,2} they win. If not, then $B$ rolls. If $B$ rolls {3,4,5,6} then they win. This process repeats until $A$ or $B$ wins, and the game stops.
What is the probability that the game ends on an even turn when $A$ rolls first?
Now the book gives the answer as $\frac{4}{7}$, however, when try to calculate I end up with $\frac{2}{11}$.
Below is my work:
To calculate this probability, we decompose the event into two disjoint events, (a) the event where $A$ wins on an even roll, and (b) the event where $B$ wins on an even roll.
(a) Now, the probability $A$ wins can be calculated as follows
\begin{align*}
\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{1}{3}\biggr) + \biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{1}{3}\biggr) + \dots = \sum_{k=0}^\infty \biggr(\frac{2}{9}\biggr)^{2k+1}\frac{1}{3}\\
= \sum_{k=0}^\infty \frac{2}{27}\biggr(\frac{2}{9}\biggr)^{2k} = \sum_{k=0}^\infty \frac{2}{27}\biggr(\frac{4}{81}\biggr)^k = \frac{2}{27}\cdot \frac{1}{1- \frac{4}{81}} = \frac{6}{77}.
\end{align*}
(b) Similarly we calculate the probability $B$ wins on an even roll as
\begin{align*}
\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3}\cdot \frac{2}{3}\biggr) + \biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3}\cdot\frac{2}{3}\biggr) + \dots = \sum_{k=0}^\infty \biggr(\frac{2}{9}\biggr)^{2k+1}\frac{4}{9}\\
= \sum_{k=0}^\infty \frac{8}{81}\biggr(\frac{2}{9}\biggr)^{2k} = \sum_{k=0}^\infty \frac{8}{81}\biggr(\frac{4}{81}\biggr)^k = \frac{8}{81}\cdot \frac{1}{1- \frac{4}{81}} = \frac{8}{77}.
\end{align*}
Therefore, it follows that the probability of the game ending on an even number of rolls is
\begin{equation*}
\frac{6}{77} + \frac{8}{77} = \frac{2}{11}.
\end{equation*}
Am I missing something?
| Thanks to the comment of @JMoravitz I realized my mistake. I was interpreting turns as the rolls $A$ AND $B$, as in $\{A_1,B_1\}, \{A_2,B_2\}, \dots$. In reality the question is merely asking what the probability of $B$ winning if $A$ rolls first.
The work is as follows:
We calculate the probability of $B$ winning. Denote the probability of $B$ winning on their $i$th roll as $S_i$. Now, the probabilities of $B$ winning on her first roll, second roll, third roll, etc., are as follows:
\begin{equation*}
P(S_1) = \biggr(\frac{2}{3}\biggr)\biggr(\frac{2}{3}\biggr), \quad P(S_2) = \biggr(\frac{2}{3}\biggr)\biggr(\frac{1}{3}\biggr)\biggr(\frac{2}{3}\biggr)\biggr(\frac{2}{3}\biggr), \quad P(S_3) = \biggr(\biggr(\frac{2}{3}\biggr)\biggr(\frac{1}{3}\biggr)\biggr)^2\biggr(\frac{2}{3}\biggr)\biggr(\frac{2}{3}\biggr), \dots
\end{equation*}
It then follows that in general that $\displaystyle P(S_i) = \biggr(\frac{2}{9}\biggr)^{i-1} \biggr(\frac{4}{9}\biggr).$ Thus, it follows that the probability of $B$ winning is calculated as
\begin{equation*}
P(S) = P\biggr(\bigcup_{i=1}^\infty S_i\biggr) = \sum_{i=1}^\infty P(S_i) = \sum_{i=1}^\infty \biggr(\frac{2}{9}\biggr)^{i-1} \biggr(\frac{4}{9}\biggr) = \frac{4}{9} \sum_{i=1}^\infty \biggr(\frac{2}{9}\biggr)^{i-1} = \frac{4}{9} \cdot \frac{9}{7} = \frac{4}{7}.
\end{equation*}
| {
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"timestamp": "2023-03-29T00:00:00",
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A problem involving the roots of quartic polynomial $x^4+px^3+qx^2+rx+1$ Let $\alpha_1$, $\alpha_2$, $\alpha_3$, $\alpha_4$ be the roots of the following polynomial
$$P(x)=x^4+px^3+qx^2+rx+1$$
Show that
$$(1+{\alpha_1}^4)(1+{\alpha_2}^4)(1+{\alpha_3}^4)(1+{\alpha_4}^4)=(p^2+r^2)^2+q^4-4pq^2r.$$
I came across this problem on a facebook page as an challenge.The only way that Strike me is multiply the terms in LHS and put the corresponding values but this method would be very long as polynomial is of 4th degree.
If anybody know any other method then Please tell me. Thank you for your help!
| Rewrite $x^4+px^3+qx^2+rx+1=0$ as
$$x^2+\frac1{x^2}+q=-(px +\frac r x) $$
Square to get
$$x^4+\frac1{x^4} + (2+q^2-2pr)=(p^2-2q)x^2+ \frac{r^2-2q}{x^2}$$
Square again and rearrange to obtain the quartic equation in $x^4$
\begin{align}
f(x^4)=&x^{16}+[2(2+q^2-2pr)-(p^2-2q)^2]x^{12}\\
&+[2+(2+q^2-2pr)^2-2(p^2-2q)(r^2-2q)]x^8\\
&+[2(2+q^2-2pr)-(r^2-2q)^2]x^4+1=0
\end{align}
Then,
$$(1+\alpha_1^4)(1+\alpha_2^4)(1+\alpha_3^4)(1+\alpha_4^4)
=f(-1)=(p^2+r^2)^2+q^4-4pq^2r
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving the system $\sqrt{x} + y = 7$, $x + \sqrt{y} = 11$ I want to solve the following nonlinear system of algebraic equations. Indeed, I am curious about a step by step solution for pedagogical purposes. I am wondering if you can come up with anything. I tried but to no avail.
\begin{align*}
\sqrt{x} + y &= 7 \\
x + \sqrt{y} &= 11
\end{align*}
The answer is $x=9,\,y=4$. A geometrical investigation can give us better insights as depicted below.
$\hspace{2cm}$
| By subtracting the first equation from the second equation, we have
$$\sqrt{y}-\sqrt{x} + x-y = 4$$
and then $\sqrt{y}-\sqrt{x} + (\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) = 4$. So, we get $(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}-1) = 4$.
Put $t := \sqrt{x}-\sqrt{y}$. It follows from the above equation that $t(t+2\sqrt{y}-1) = 4$. By isolating $\sqrt{y}$, we obtain $\sqrt{y} = \frac{4+t-t^2}{2t}$. In additon, we also obtain $\sqrt{x} = t+\sqrt{y} = \frac{4+t+t^2}{2t}$.
Now, we have
\begin{align}
\sqrt{x}+y &= \frac{4+t+t^2}{2t} + \frac{(4+t-t^2)^2}{4t^2} \\
&= \frac{(8t+2t^2+2t^3) + (16+t^2+t^4 + 8t-8t^2-2t^3)}{4t^2} \\
&= \frac{16 + 16t - 5t^2 + t^4}{4t^2}.
\end{align}
Therefore, the condition $\sqrt{x}+y = 7$ yields $t^4 - 33t^2 + 16t + 16 = 0$.
We can factorize the left hand side of this equation as below:
$$
(t-1)(t^3 + t^2 - 32t - 16) = 0.
$$
There are three real roots of $t^3 + t^2 - 32t - 16$. However, in our case, $t$ must satisfy the condition $\sqrt{x} > 0$ and $\sqrt{y} > 0$. Unfortunately, none of them satisfies the condition.
Indeed, (i) $\sqrt{x} = \frac{(t+\frac{1}{2})^2+\frac{15}{4}}{4t} > 0$ requires at least $t > 0$, and (ii) under the assumption of $t > 0$, $\sqrt{y} = \frac{-(t-\frac{1}{2})^2+\frac{17}{4}}{4t} > 0$ requires at least $0 < t < 3$. However, (iii) from a sketch of the graph of $f(t) = t^3+t^2-32t-16$, it follows that $f(t) < 0$ for $0 \leq t \leq 3$.
Consequently, the only possible value of $t$ is $t = 1$, i.e. $(x,y) = (9,4)$ is the only real answer of the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
Show that $(a^3+a+1)(b^3+b+1)(c^3+c+1)\le 27$ Let $a,b,c\ge 0$ be such that $a^2+b^2+c^2=3$. Show that
$$(a^3+a+1)(b^3+b+1)(c^3+c+1)\le 27$$
I want to consider the function
$$f(x)=\ln{(x^{3/2}+x^{1/2}+1)}$$ Maybe it isn't the case $f''(x)\le 0$, so I can't use Jensen's inequality.
| Let $f,g$ be given by
$$
\left\lbrace
\begin{align*}
f(a,b,c)&=(a^3+a+1)(b^3+b+1)(c^3+c+1)\\[4pt]
g(a,b,c)&=a^2+b^2+c^2\\[4pt]
\end{align*}
\right.
$$
Using the method of Lagrange Multipliers, if a point $(a,b,c)$ satisfying the constraint $g(a,b,c)=3$ is such that $f$ has a local extremum at $(a,b,c)$, then for some $\lambda\in\mathbb{R}$ we must have
$$
\left\lbrace
\begin{align*}
(3a^2+1)(b^3+b+1)(c^3+c+1)&=2\lambda a&&(\text{eq}1)\\[4pt]
(a^3+a+1)(3b^2+1)(c^3+c+1)&=2\lambda b&&(\text{eq}2)\\[4pt]
(a^3+a+1)(b^3+b+1)(3c^2+1)&=2\lambda c&&(\text{eq}3)\\[4pt]
a^2+b^2+c^2&=3&&(\text{eq}4)\\[4pt]
\end{align*}
\right.
$$
Given $a,b,c\ge 0$, each of equations $1,2,3$ has positive $\text{LHS}$, so must also have positive $\text{RHS}$, hence $a,b,c,\lambda > 0$.
Without loss of generality, assume $a=\max(a,b,c)$.
Then from $(\text{eq}4)$, it follows that $1\le a < \sqrt{3}$.
Suppose $a\ne b$.
Dividing $(\text{eq}1)$ by $(\text{eq}2)$, we get
\begin{align*}
&
\frac
{(3a^2+1)(b^3+b+1)}
{(3b^2+1)(a^3+a+1)}
=
\frac{a}{b}
\\[4pt]
\implies\; &
a(a^3+a+1)(3b^2+1)=b(b^3+b+1)(3a^2+1)
\\[4pt]
\implies\; &
a(a^3+a+1)(3b^2+1)-b(b^3+b+1)(3a^2+1)=0
\\[4pt]
\implies\; &
(a-b)(3a^3b^2+3a^2b^3+a^3+b^3+a^2b+ab^2-3ab+a+b+1)=0
\\[4pt]
\implies\; &
3a^3b^2+3a^2b^3+a^3+b^3+a^2b+ab^2-3ab+a+b+1=0
\\[4pt]
\implies\; &
3a^3b^2+3a^2b^3+a^3+b^3+a^2b+ab^2+a+b=3ab-1
\\[4pt]
\implies\; &
(a+b)(3a^2b^2+a^2+b^2+1)=3ab-1
\\[4pt]
\implies\; &
(b+1)(4b^2+2)< 6b-1
\quad\text{[replaced $a$ by $1$ on the left and by $2$ on the right]}
\\[4pt]
\implies\; &
4b^3+4b^2-4b+3 < 0
\\[4pt]
\implies\; &
(4b^2-4b+1)+(4b^3+2) < 0
\\[4pt]
\implies\; &
(2b-1)^2+(4b^3+2) < 0
\\[4pt]
\end{align*}
contradiction.$\;$Hence $a=b$.
Analogously we get $a=c$.
Then $a=b=c$, hence from $(\text{eq}4)$, we get $a=b=c=1$.
It follows that for $a,b,c\ge 0$, the maximum value of $f(a,b,c)$ subject to the constraint $g(a,b,c)=3$ is $f(1,1,1)=27$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x$ I was generalizing the following integral:
$$\int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x \hspace{40pt} n\geq 0$$
We can start by noting that $\sin x =\dfrac{e^{ix}-e^{-ix}}{2i}$ and thus $\displaystyle \sin^{2n+1}x=\frac{(-1)^n}{2^{2n}}\sum_{r=0}^n (-1)^r \binom{2n+1}{r}\sin(2r+1)x$.
This implies $$\begin{aligned}\displaystyle \int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x &=\frac{(-1)^n}{2^{2n}}\sum_{r=0}^n (-1)^r \binom{2n+1}{r}\int_{0}^{\infty}\frac{\sin(2r+1)x}{x}\mathrm{d}x \\ &=\frac{(-1)^n\pi}{2^{2n+1}}\sum_{r=0}^n(-1)^r\binom{2n+1}{r} \\ &=\frac{(-1)^n\pi}{2^{2n+1}}\sum_{r=0}^n\left((-1)^r\binom{2n}{r}-(-1)^{r-1}\binom{2n}{r-1}\right)\end{aligned}$$
Where I have used the well known result $\displaystyle \int_{0}^{\infty} \frac{\sin(2r+1)x}{x}\mathrm{d}x=\int_{0}^{\infty}\frac{\sin x}{x}\mathrm{d}x=\frac{\pi}{2}$ and the property of binomial coefficients that $\displaystyle \binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}$
Since the above sum telescopes, we have $$\displaystyle \int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x=\frac{(-1)^n\pi}{2^{2n+1}}(-1)^n\binom{2n}{n}=\frac{\pi}{2^{2n+1}}\binom{2n}{n} ~\forall ~ n\in \mathbb{Z^{+}}$$
I would like to know other methods for evaluating this integral.
| One can use the recurrence relation of Gradshteyn and Ryzhik to give, for some $p, q, \in \mathbb{N}$
\begin{align}
\displaystyle \int_0^\infty \frac{\sin^p x}{x^q}\,\mathrm{d}x &= \frac{p}{q-1}\displaystyle \int_0^\infty \frac{\sin^{p-1} x}{x^{q-1}}\cos x\,\mathrm{d}x \qquad (p > q-1>0)\\
&= \frac{p(p-1)}{(q-1)(q-2)}\displaystyle \int_0^\infty \frac{\sin^{p-2} x}{x^{q-2}}\,\mathrm{d}x - \frac{p^2}{(q-1)(q-2)}\int_0^\infty \frac{\sin^{p} x}{x^{q-2}}\,\mathrm{d}x\qquad (p > q-1>1)
\end{align}
Now, so if $q$ and $p$ are both even or both odd, with $p \geq q$ to assure convergence of the integral, we arrive at integrals of the form
$$\int_0^\infty \frac{\sin^{2n+1}x}{x}\,dx = \frac{(2n-1)!!}{(2n)!!}\frac{\pi}{2}$$
$$\int_0^\infty \frac{\sin^{2n}x}{x^2}\,dx = \frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2}$$
Which I think satisifes your requirement of another method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Solving $x^8 - x^5 + x^2 - x + 1 > 0$ over $\mathbb{R}$ I could not find any decent approach to solve this inequality. I would appreciate any help, and input if this is even possible to solve(without a computer).
$$x^8-x^5+x^2-x+1>0$$
| Using the AM-GM inequality, we have
$$x^8 + \frac{x^2}{2} \geqslant 2\sqrt{x^8 \cdot \frac{x^2}{2}} = \sqrt{2} \cdot |x^5| \geqslant x^5,$$
and
$$1+\frac{x^2}{2} \geqslant 2\sqrt{\frac{x^2}{2}} = \sqrt 2 \cdot |x| \geqslant x$$
Therefore
$$x^8+x^2+1> x^5 + x.$$
Done.
Note. The SOS form
$$x^{8}+x^{2}+1-x^{5}-x=\left(x^{4}-\frac{x}{2}\right)^{2}+\left(\frac{x}{2}-1\right)^{2}+\frac{x^{2}}{2}>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is this a valid proof for $I(n^2) \geq \frac{5}{3}$, if $q^k n^2$ is an odd perfect number with special prime $q$? Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$, and let $I(x)=\sigma(x)/x$ be the abundancy index of $x$.
Note that both $\sigma$ and $I$ are multiplicative functions.
A number $m$ is said to be perfect if $\sigma(m)=2m$. Equivalently, $I(m)=2$.
Euler proved that an odd perfect number, if one exists, must have the form
$$m = q^k n^2$$
where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Since $q$ is prime, we have
$$\frac{q+1}{q} = I(q) \leq I(q^k) = \frac{\sigma(q^k)}{q^k} = \frac{q^{k+1} - 1}{q^k (q - 1)} < \frac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1}$$
from which it follows that
$$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$
Note that we then have the lower bound
$$I(n^2) > \frac{2(q-1)}{q} \geq \frac{8}{5}$$
since $q$ is a prime satisfying $q \equiv 1 \pmod 4$.
Here is my initial question:
Can we improve the lower bound for $I(n^2)$ to
$$I(n^2) \geq \frac{5}{3}$$
using the following argument?
$$\bigg(\frac{2q}{q+1} \geq I(n^2) > \frac{5}{3}\bigg) \implies q > 5 \implies q \geq 13 \implies \bigg(I(n^2) > \frac{2(q-1)}{q} \geq \frac{24}{13} > \frac{5}{3}\bigg)$$
Thus, we have the biconditional
$$I(n^2) > \frac{5}{3} \iff q > 5.$$
Next, we have the implication
$$I(n^2) = \frac{5}{3} \implies q = 5.$$
It then suffices to prove the implication
$$q = 5 \implies I(n^2) = \frac{5}{3}$$
to finally show that
$$I(n^2) \geq \frac{5}{3},$$
since $q \geq 5$ holds.
But note that, if $q=5$, then
$$\frac{5}{3} = I(n^2) = \frac{2}{I(5^k)} = \frac{2\cdot{5^k}(5-1)}{5^{k+1}-1}$$
which implies that the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds.
Still, notice that we have
$$k=1 \implies I(q^k) = I(q) = \frac{q+1}{q} = 1 + \frac{1}{q} \leq \frac{6}{5} \implies I(n^2) = \frac{2}{I(q^k)} = \frac{2}{I(q)} \geq \frac{2\cdot{5}}{6} = \frac{5}{3},$$
which is what we set out to prove.
Here is my final question:
Would it be possible to remove the reliance of the proof on the Descartes-Frenicle-Sorli Conjecture?
| Not a complete answer, just some thoughts that recently occurred to me, which would be too long to fit in the Comments section.
Since the biconditionals
$$I(n^2) > \frac{5}{3} \iff q > 5$$
and
$$I(n^2) = \frac{5}{3} \iff \bigg(q = 5 \land k = 1\bigg)$$
hold, it remains to consider what happens to the bounds for $I(n^2)$ when $q = 5$ and $k > 1$.
Since $k > 1$ and $k \equiv 1 \pmod 4$, then $k \geq 5$. By assumption, we have $q=5$, so that we obtain
$$I(q^k) = I(5^k) \geq I(5^5) \iff I(n^2) = \frac{2}{I(q^k)} \leq \frac{2}{I(5^5)} = \frac{3125}{1953} \approx 1.6001.$$
(WolframAlpha computation for $\dfrac{2}{I(5^5)}$ is here.)
On the other hand, we have the lower bound
$$\frac{8}{5} = \frac{2\cdot(5 - 1)}{5} = \frac{2(q - 1)}{q} < I(n^2),$$
whence there is no contradiction.
It is natural then, to attempt to derive a better lower bound than
$$1.6 = \frac{8}{5} < I(n^2),$$
specifically when $q=5$ and $k>1$.
It turns out that we can do better under the case $q=5$ and $k>1$.
Cohen and Sorli ruled out $5^5$ as a possible Eulerian component $q^k$ for an odd perfect number in page 4 of their paper titled On Odd Perfect Numbers and Even 3-Perfect Numbers.
Thus, under the assumption $q=5$ and $k>1$, we have that $k \geq 9$ (since $k \equiv 1 \pmod 4$), whereupon we get
$$1.249999872 = \frac{2441406}{1953125} = \frac{5^{10} - 1}{5^9 (5 - 1)} = I(5^9) \leq I(q^k) < \frac{5}{4} = 1.25$$
$$1.6 = \frac{8}{5} < I(n^2) \leq \frac{1953125}{1220703} \approx 1.60000016384.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $(x-1)\log(1 - 2 x) -2x = \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n} \quad \text{for} \; 2|x|<1$ Prove that $$(x-1)\log(1 - 2 x) -2x = \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n} \quad \text{for} \; 2|x|<1$$
First of all, I don't really know if by proving it means finding the function Sum of $\sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n}$ and concluding that $f(x)=(x-1)\log(1 - 2 x) -2x$ or to replace $\log(1 - 2 x)$ for it's Taylor Expansion and resaulting in $ \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n}$. I've tried both ways, starting from the Taylor Expansion of $\log(1 - 2 x)=\sum_{n=1}^{\infty} -\frac{2^n x^{n}}{n}$ but I've failed both ways. Any hints on how to prove this? Thanks in advance.
Edit: What I did
$$\log(1 - 2 x)=\sum_{n=1}^{\infty} -\frac{2^n x^{n}}{n} = -2x + \sum_{n=2}^{\infty} -\frac{2^n x^{n}}{n} \\ \rightarrow (x-1)\log(1 - 2 x)=-2x^2 + x\sum_{n=2}^{\infty} -\frac{2^n x^{n}}{n} + 2x +\sum_{n=2}^{\infty} \frac{2^n x^n}{n} \\ \rightarrow (x-1)\log(1 - 2 x) -2x =-2x^2 + \sum_{n=2}^{\infty} -\frac{2^n x^{n+1}}{n} + \sum_{n=2}^{\infty} \frac{2^n x^n}{n} \\ = \sum_{n=1}^{\infty} -\frac{2^n x^{n+1}}{n} + \sum_{n=2}^{\infty} \frac{2^n x^n}{n} \\ = \sum_{n=1}^{\infty} -\frac{2^{n} x^{n+1}}{n} + \sum_{n=1}^{\infty} \frac{2^{n+1} x^{n+1}}{n+1} \\ = \sum_{n=1}^{\infty} \frac{-2^{n} x^{n+1} (n+1) + 2^{n+1} x^{n+1} n}{n(n+1)} \\ = \sum_{n=1}^{\infty} \frac{-2^{n} x^{n+1}[(n+1)-2n]}{n(n+1)} \\ = \sum_{n=1}^{\infty} \frac{-2^{n} x^{n+1}(1-n)}{n(n+1)} \\ = \sum_{n=1}^{\infty} \frac{2^{n} x^{n+1}(n-1)}{n(n+1)} \\ = 0 + \sum_{n=2}^{\infty} \frac{2^{n} x^{n+1}(n-1)}{n(n+1)} \\ = \sum_{n=2}^{\infty} \frac{2^{n} x^{n+1}(n-1)}{n(n+1)}$$
Which is what I was trying to prove.
| Let $f(x)=(x-1)\log(1-2x)-2x$, then $f$ is differentiable and
$$ f'(x)=(x-1)\frac{-2}{1-2x}+\log(1-2x)-2=\log(1-2x)+\frac{1}{1-2x}-1$$
For $|x|<\frac{1}{2}$, we have
$$ \log(1-2x)=-\sum_{n=1}^{+\infty}\frac{(2x)^n}{n} \text{ and } \frac{1}{1-2x}=\sum_{n=0}^{+\infty}(2x)^n$$
thus
$$ f'(x)=\sum_{n=1}^{+\infty}(2x)^n\frac{n-1}{n}=\sum_{n=2}^{+\infty}(2x)^n\frac{n-1}{n} $$
Now let $g(x)=\sum_{n=2}^{+\infty}\frac{2^n(n-1)x^{n+1}}{n^2+n}$, then $g$ is differentiable and
$$ g'(x)=\sum_{n=2}^{+\infty}2^n\frac{n-1}{n}x^n=f'(x) $$
Thus, since $f(0)=g(0)=0$, we have $f(x)=g(x)$ for $|x|<\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Polar form of -$2$ - $2\sqrt{2i}$ -$2$ - $2\sqrt{2i}$
So I am confused on the $\sqrt{2i}$ part, so what I did was using the rule $\sqrt{ab}$ = $\sqrt{a} \sqrt{b}$ and that gives $\sqrt{2} \sqrt{i}$
-$2$ - $2\sqrt{2i}$ = -2 -2$\sqrt{2} \sqrt{i}$ multiply by $i^\frac{9}{2}$ to get an $i$ therefore it becomes -2 -2$\sqrt{2}$i ; r = 2$\sqrt{3}$ and $\theta$ = $\tan^{-1}({\sqrt{2})}$
so the polar form is 2$\sqrt{3}$ cis($\tan^{-1}({\sqrt{2})}$)
Is this correct?
| Note
\begin{align}
-2-2\sqrt{2i}&= -2-2\sqrt2 (e^{i\frac\pi2})^{1/2}\\
&= -2-2\sqrt2e^{i\frac \pi4}\\
&= -4-2i \\
&= 2\sqrt5 e^{i\tan^{-1}\frac12}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim_{h\to 0}\frac{1}{h^2}\begin{vmatrix}\tan x&\tan(x+h)&\tan(x+2h)\\\tan(x+2h)&\tan x&\tan(x+h)\\\tan(x+h)&\tan(x+2h)&\tan x\end{vmatrix}$ Evaluate
$$
\lim_{h\to 0}\frac{\Delta}{h^2}=\lim_{h\to 0}\frac{1}{h^2}\begin{vmatrix}
\tan x&\tan(x+h)&\tan(x+2h)\\
\tan(x+2h)&\tan x&\tan(x+h)\\
\tan(x+h)&\tan(x+2h)&\tan x
\end{vmatrix}
$$
Attempt
$$
\lim_{h\to 0}\frac{\Delta}{h^2}=\begin{vmatrix}
\lim_{h\to 0}\tan x&\lim_{h\to 0}\dfrac{\tan(x+h)-\tan x}{h}&\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}\\
\lim_{h\to 0}\tan(x+2h)&\lim_{h\to 0}\dfrac{\tan x-\tan(x+2h)}{h}&\lim_{h\to 0}\dfrac{\tan(x+h)-\tan(x+2h)}{h}\\
\lim_{h\to 0}\tan(x+h)&\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}&\lim_{h\to 0}\dfrac{\tan x-\tan(x+2h)}{h}
\end{vmatrix}\\
=\begin{vmatrix}
\lim_{h\to 0}\tan x&\lim_{h\to 0}\dfrac{\tan(x+h)-\tan x}{h}&\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}\\
\lim_{h\to 0}\tan(x+2h)&-2.\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan x}{h}&-1.\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}\\
\lim_{h\to 0}\tan(x+h)&\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}&-2.\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan x}{2h}
\end{vmatrix}\\
$$
$$
\lim_{h\to 0}\dfrac{\tan(x+h)-\tan x}{h}=\frac{d}{dx}\tan x=\sec^2x\\
\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan x}{2h}=\frac{d}{dx}\tan x=\sec^2x\\
\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}=\frac{d}{dx}\tan(x+h)=\sec^2(x+h)
$$
But my reference gives the solution $9\tan x.\sec^4x$, I think by taking $\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}=\sec^2x$. Will that make a difference ?
It might be silly but could anyone clarify this confusion in my attempt ?
| $$ \Delta = \begin{vmatrix} \tan x&\tan(x+h)&\tan(x+2h)\\
\tan(x+2h)&\tan x&\tan(x+h)\\
\tan(x+h)&\tan(x+2h)&\tan x \end{vmatrix} $$
$$=\begin{vmatrix} \tan x&\tan(x+h)-\tan x&\tan(x+2h)-\tan x\\
\tan(x+2h)&\tan x-\tan(x+2h)&\tan(x+h)-\tan(x+2h)\\
\tan(x+h)&\tan(x+2h)-\tan(x+h)&\tan x-\tan(x+h) \end{vmatrix} $$
$$=\begin{vmatrix} \tan x&\dfrac{\sin h}{\cos(x+h)\cos x}&\dfrac{\sin2h}{\cos(x+2h)\cos x}\\
\tan(x+2h)&-\dfrac{\sin2h}{\cos(x+2h)\cos x}&-\dfrac{\sin h}{\cos(x+h)\cos(x+2h)}\\
\tan(x+h)&\dfrac{\sin h}{\cos(x+h)\cos(x+2h)}&-\dfrac{\sin h}{\cos x\cos(x+h)} \end{vmatrix} $$
$$=\dfrac{\sin^2h}{\cos x\cos(x+2h)\cos(x+h)}\begin{vmatrix} \sin x&\dfrac1{\cos(x+h)}&\dfrac{2\cos h}{\cos(x+2h)}\\
\sin(x+2h)&-\dfrac{2\cos h}{\cos x}&-\dfrac1{\cos(x+h)}\\
\sin(x+h)&\dfrac1{\cos(x+2h)}&-\dfrac1{\cos x} \end{vmatrix}$$
Use $\lim_{h\to0}\dfrac{\sin h}h=1$
Finally set $\lim_{h\to0} \Delta$
$$=\begin{vmatrix}\sin x&\sec x&2\sec x\\
\sin x&-\sec x&-\sec x\\
\sin x&\sec x&-\sec x\end{vmatrix}=\sin x\sec^2x\begin{vmatrix}1&1&2\\
1&-1&-1\\
1&1&-1\end{vmatrix}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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If $x^2+y^2+xy=1$ then find minimum of $x^3y+xy^3+4$
If $x,y \in \mathbb{R}$ and $x^2+y^2+xy=1$ then find the minimum value of $x^3y+xy^3+4$
My Attempt:
$x^3y+xy^3+4$
$\Rightarrow xy(x^2+y^2)+4$
$\Rightarrow xy(1-xy)+4$ (from first equation)
$\Rightarrow xy-(xy)^2+4 =f(x)$
For minimum value, $\frac{df(x)}{dx}=0$.
$\Rightarrow \frac{df(x)}{dx}=(y-2xy²) + \frac{dy}{dx}(x-2yx²)=0$
How should I proceed from here?
| By AM-GM, $x^2+y^2=|x|^2+|y|^2\ge2|xy|$ so $r^2=x^2+y^2$ has extrema $\tfrac23,\,2$ respectively achieved by$$2xy=x^2+y^2\implies1=(1+1/2)(x^2+y^2)$$and$$-2xy=x^2+y^2\implies1=(1-1/2)(x^2+y^2).$$Note the minimum of $r^2(1-r^2)$ on $[\tfrac23,\,2]$ occurs at $r^2=2$. In particular, $x=-y=1$ minimizes $r^2(1-r^2)+4$ as $2\times-1+4=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the minimum value of $\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right|$.
Given positives $a, b, c$ such that $abc = 1$, if possible, calculate the minimum value of $$\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right|$$
Without loss of generalisation, assume that $a \le b \le c$.
We have that $$\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right| \ge \frac{c^2 - ba}{b - a} + \frac{a^2 - bc}{b - c}$$
$$ = \frac{(c + a)(a^2 + b^2 + c^2 - bc - ca - ab)}{(c - b)(b - a)} \ (1)$$
Let $c' = b - a, a' = c - b \iff c = a' + b, a = b - c'$, $(1)$ becomes $$\frac{(2b - c' + a')(c'^2 + c'a' + a'^2)}{c'a'}$$
and $(b - c')b(b + a') = b^3 - (c' - a')b^2 - c'a'b = 1$
$$\iff (2b - c' + a')b^2 = b^3 + c'a'b + 1 \iff 2b - c' + a' = \frac{b^3 + c'a'b + 1}{b^2}$$
Another idea I had was that $\left|\dfrac{a^2 - bc}{b - c}\right| + \left|\dfrac{b^2 - ca}{c - a}\right| + \left|\dfrac{c^2 - ab}{a - b}\right|$
$$ = \frac{1}{2}\sum_{\text{cyc}}\left(|c - a|\left|\frac{2(b^2 - ca)}{(c - a)^2}\right|\right) = \frac{1}{2}\sum_{\text{cyc}}\left(|c - a|\left|\frac{2b^2 - c^2 - a^2}{(c - a)^2} + 1\right|\right)$$
$$ = \frac{1}{2}\left[(c - b)\left(\left|\frac{2(b^2 - a^2)}{(b - c)^2} + \frac{c^2 - b^2}{(b - c)^2} - 1\right| + \left|\frac{b^2 - a^2}{(c - a)^2} - \frac{c^2 - b^2}{(c - a)^2} + 1\right|\right)\right.$$
$$\left. + (b - a)\left(\left|\frac{b^2 - a^2}{(c - a)^2} - \frac{c^2 - b^2}{(c - a)^2} + 1\right| + \left|\frac{b^2 - a^2}{(a - b)^2} + \frac{2(c^2 - b^2)}{(a - b)^2} + 1\right|\right)\right]$$
$$ \ge \frac{1}{2}\left[(c - b)\left(\frac{2(b^2 - a^2)}{(b - c)^2} + \frac{c^2 - b^2}{(b - c)^2} + \frac{b^2 - a^2}{(c - a)^2} - \frac{c^2 - b^2}{(c - a)^2}\right)\right.$$
$$\left. + (b - a)\left(\frac{b^2 - a^2}{(c - a)^2} - \frac{c^2 - b^2}{(c - a)^2} + \frac{b^2 - a^2}{(a - b)^2} + \frac{2(c^2 - b^2)}{(a - b)^2} + 2\right)\right]$$
$$ = \frac{1}{2}\left[(c^2 - b^2)\left(\frac{1}{c - b} - \frac{1}{c - a} + \frac{2}{b - a}\right) + (b^2 - a^2)\left(\frac{2}{c - b} + \frac{1}{c - a} + \frac{1}{b - a}\right)\right] + (b - a)$$
$$ = \frac{1}{2}\left(\frac{b^2 + c^2 - 2a^2}{c - b} + \frac{2b^2 - c^2 - a^2}{c - a} + \frac{2c^2 - a^2 - b^2}{b - a}\right) + (b - a)$$
There must have been something wrong, but that's all I have for now.
| By AM-GM $$\sum_{cyc}\left|\frac{a^2-bc}{b-c}\right|=\sqrt{\left(\left|\sum\limits_{cyc}\frac{a^2-bc}{b-c}\right|\right)^2}=$$
$$=\sqrt{\left(\sum\limits_{cyc}\frac{a^2-bc}{b-c}\right)^2-2\sum_{cyc}\frac{(a^2-bc)(b^2-ac)}{(b-c)(c-a)}+2\sum_{cyc}\left|\frac{(a^2-bc)(b^2-ac)}{(b-c)(c-a)}\right|}\geq$$
$$\geq\sqrt{-2\sum_{cyc}\frac{(a^2-bc)(b^2-ac)}{(b-c)(c-a)}-2\sum_{cyc}\frac{(a^2-bc)(b^2-ac)}{(b-c)(c-a)}}=$$
$$=\sqrt{4(a+b+c)^2}=2(a+b+c)\geq6.$$
Now, prove that we got an infimum and the minimum does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
How can I integrate this function? $\frac{\sqrt{e^{2x+2y+z}}}{(1+e^x+e^y+e^{x+y+z})^2}$ I want to evaluate the following integration
$$
\int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dy \int_{-\infty}^{\infty} dz \frac{\sqrt{e^{2x+2y+z}}}{(1+e^x+e^y+e^{x+y+z})^2}.
$$
According to Mathematica12, the answer is $\pi^2$.
How can I get the answer?
| Let $x= \ln X$, $y=\ln Y$ and $z=2\ln Z$. Then
$$ I = \int_0^\infty\int_0^\infty\int_0^\infty\frac{XYZ}{(1+X+Y+XYZ^2)^2} \frac{dX}{X}\frac{dY}{Y}\frac{2\cdot dZ}{Z}.$$
This is a rational function, and thus can be easily integrated. For instance integrate first against X:
$$ I = 2\int_0^\infty\int_0^\infty \frac{1}{1+YZ^2}\frac{1}{1+Y}dY dZ. $$
Then integrate against $Z$:
$$ I = 2\int_0^\infty \frac{\pi}{2\sqrt{Y}}\frac{1}{1+Y}dY . $$
Then, let $u=\sqrt{Y}$ to obtain
$$I= \pi\int_0^\infty \frac{2u }{u}\frac{1}{1+u^2}du=\pi^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3782844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Let $\frac{\tan A}{1-\tan^2A}=\sin^220^\circ-\sin160^\circ\sin220^\circ+\sin^2320^\circ$, find $\tan6A$ Let $\dfrac{\tan A}{1-\tan^2A}=\sin^220^\circ-\sin160^\circ\sin220^\circ+\sin^2320^\circ$, find $\tan6A$
My attempt :
\begin{align*}
\dfrac{\tan2A}{2}=\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ\\
\tan2A=2(\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ)
\end{align*}
and
\begin{align*}
\tan6A&=\tan(2A-60^\circ)\tan2A\tan(2A+60^\circ)\\
&=(\dfrac{\tan2A-\sqrt{3}}{1+\sqrt{3}\tan60^\circ})(\tan2A)(\dfrac{\tan2A+\sqrt{3}}{1-\sqrt{3}\tan60^\circ})
\end{align*}
give
$$\tan6A=(\dfrac{2(\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ)-\sqrt{3}}{1+\sqrt{3}\tan60^\circ})(2(\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ))(\dfrac{2(\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ)+\sqrt{3}}{1-\sqrt{3}\tan60^\circ})$$
This method is incredibly long, there may be better way to deal with this problem.
| Let me post a different route to the simplication, as OP has begun.
So we have
$$
\sin^2 20^\circ-\sin160^\circ\sin220^\circ+\sin^2 320^\circ=
\sin^2 20^\circ(1+2\cos 20^\circ + 4\cos^2 20^\circ)
$$
I claimed for all $\theta$,
$$
\sin^2\theta(1+2\cos\theta+4\cos^2\theta)=\frac12(2+\cos\theta-\cos2\theta-\cos3\theta-\cos4\theta)
$$
Proof of claim:
\begin{align*}
LHS&=\sin^2\theta(1+2\cos\theta+4\cos^2\theta)\\
&=\frac12(1-\cos 2\theta)(3+2\cos\theta+2\cos2\theta)\\
&=\frac12(3+2\cos\theta-\cos2\theta-2\cos\theta\cos2\theta- 2\cos^22\theta)\\
&=\frac12(2+2\cos\theta-\cos2\theta-2\cos\theta\cos2\theta- \cos4\theta)\\
&=\frac12(2+\cos\theta-\cos2\theta-(4\cos^3\theta-3\cos\theta)- \cos4\theta)\\
&=\frac12(2+\cos\theta-\cos2\theta-\cos3\theta- \cos4\theta)\\
&=RHS\quad\checkmark
\end{align*}
So with $\theta=20^\circ$, this is
\begin{align*}
&\sin^2 20^\circ(1+2\cos 20^\circ + 4\cos 40^\circ)\\
&=\frac12(2+\cos 20^\circ-\cos40^\circ-\cos60^\circ- \cos80^\circ)\\
&=\frac12\left(\frac32+\cos 20^\circ-\cos40^\circ-\cos80^\circ\right)\\
&=\frac12\left(\frac32+\cos 20^\circ-2\cos60^\circ\cos20^\circ\right)\\
&=\frac34.
\end{align*}
So $\tan 2A=\frac32$ and
$$
\tan 6A=\tan 3(2A)=\frac{3\tan 2A-\tan^3 2A}{1-3\tan^2 2A}=\dots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $a^2+a b+b^2=40$ and $a^2-\sqrt{a b}+b=5$, then find $a^2+\sqrt{a b}+b$ I was given this problem to solve with elementary methods (High School level).
Knowing that
$$\begin{align}
a^2+a b+b^2 &=40 \\
a^2-\sqrt{a b}+b &=\phantom{0}5
\end{align}$$
find
$$a^2+\sqrt{a b}+b$$
I tried to look for $\sqrt{a b}$, since the requested quantity is
$$a^2+\sqrt{a b}+b=(a^2-\sqrt{a b}+b)+2\sqrt{ab}=5 + 2\sqrt{ab}$$
So I set
$$x=a^2;\;y=\sqrt{ab};\;b=z$$
and the system became
$$x+y^2+z^2=40;\;x-y-z=5$$
subtracting the two equations I got
$$z^2-z+y^2+y-35=0$$
which has one real solution when the discriminant is zero.
That is
$1-4(y^2+y-35)=0$
and then
$y=\frac{1}{2} \left(-1\pm\sqrt{142}\right)$
and finally
$$a^2+\sqrt{a b}+b=4\pm\sqrt{142}$$
I know that there are other solutions because I've found them with Wolfram Mathematica, but I couldn't find them with elementary methods.
Any help will be appreciated.
| One way to do this is to graph out these equations of lines.
Find the values of $a,b$ and plug into the desired output equation:
*
*Equation of a line:
see link here.
*Look at the possible values and
*plug answers into the final equation:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
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Find all values of $a$ for which the maximum value of $f(x)=\frac{ax-1}{x^4-x^2+1}$equals $1$. Find all values of $a$ for which the maximum value of $$f(x)=\frac{ax-1}{x^4-x^2+1}$$equals $1$.
I equated $f(x)$ equal to $1$ to obtain a polynomial $x^4-x^2-ax+2=0$. Now this must have at least one repeated root.But I could not get any further.
Is there a calculus way to do it
| $$f(x)=\frac{ax-1}{x^4-x^2+1}\implies f'(x)=\frac{a \left(-3 x^4+x^2+1\right)+4 x^3-2 x}{\left(x^4-x^2+1\right)^2}$$
So, we need
$$a \left(-3 x^4+x^2+1\right)+4 x^3-2 x=0\tag 1$$ for an extremum and $$a x-x^4+x^2-2=0\tag 2$$ to have the maximum value of $1$.
From $(2)$,we can get $a=\frac{x^4-x^2+2}{x}$. Plug it in $(1)$ to have
$$3 x^8-4 x^6+2 x^4+x^2-2=(1-x) (x+1) \left(3 x^2+2\right) \left(x^4-x^2+1\right)=0$$ So, the only real roots are $x=\pm 1$ to which correspond $a=\pm 2$.
These are the only solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
What is the minimum value of $x+y$?
Suppose $x,y$ are positive real numbers that satisfy $$xy(x+2y)=2$$ What is the minimum value of $x+y$?
My Thoughts
I’ve attempted using arithmetic-geometric mean inequality and got:
$\frac{x+y+x+2y}{3} \geq \sqrt[3]{2}$
Therefore $2(x+y)+y \geq 3\sqrt[3]{2}$, then I got trapped.
Feels like I’m in the wrong way, I need a hint.
| $xy(x+2y)=2$.
Let z=x+y,
$(z-y)y(z+y)=2$
$(z^2-y^2)y=2 $
$z^2-y^2=2/y $
$z^2=2/y+y^2$
$z=(2/y+y^2)^{0.5}$
$\frac {d}{dx} (2/y-y^2)^{0.5}=\frac{y^3-1}{y^2((y^3+2)/y)^{0.5}} $which equals $0$ at $y=1$
$z^2=2/1+1=3$, $z=3^{0.5}$, $x=3^{0.5}-1$,
$x+y=3^{0.5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Compute the Surface area of $S=\left\{(x,y,z)\in\mathbb{R}^3:x^2+y^2=z^2,1\le z\le2\right\}$ $\def\hl#1#2{\bbox[#1,1px]{#2}}
\def\box#1#2#3#4#5{\color{#2}{\bbox[0px, border: 2px solid #2]{\hl{#3}{\color{white}{\color{#3}{\boxed{\underline{\large\color{#1}{\text{#4}}}\\\color{#1}{#5}\\}}}}}}}
\def\verts#1{\left\vert#1\right\vert}
\def\Verts#1{\left\Vert#1\right\Vert}
\def\R{\mathbb{R}}$
$\box{black}{black}{}
{Question}
{\text{Compute the Surface area of $S=\left\{(x,y,z)\in\R^3:x^2+y^2=z^2,1\le z\le2\right\}$}}$
My Attempts
Since $z>0$, consider a injective fuction $G:T\subseteq\R^2\to S$ such that $G(x,y)=(x,y,\sqrt{x^2+y^2})$ have
\begin{align}
G_x=&\left(1,0,\frac{x}{\sqrt{x^2+y^2}}\right)\hspace{5ex}
G_y=\left(0,1,\frac{y}{\sqrt{x^2+y^2}}\right)\\
G_x\times G_y=&\left(-\frac{x}{\sqrt{x^2+y^2}},-\frac{y}{\sqrt{x^2+y^2}},1\right)\\
\Verts{G_x\times G_y}=&\sqrt{2}
\end{align}
Therefore
\begin{align}
\text{area}(S)=&\iint_T\Verts{G_x\times G_y}dA\\
=&\int_0^{2\pi}\int_1^2\sqrt{2}\cdot r~drd\theta\\
=&3\sqrt{2}~\pi
\end{align}
Is my solution correct?
| I checked it directly in cylindrical coordinates as parametric surface and obtain
$$\begin{cases}
E=\cos^2 \phi + \sin^2 \phi +1 = 2 \\
G=r^2 \sin^2 \phi + r^2 \cos^2 \phi +0 = r^2 \\
F = r\cos \phi (-\sin \phi) + r\sin \phi \cos \phi +0 =0
\end{cases}
$$
So $\sqrt{EG-F} = r\sqrt{2}$ so your integral is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3786939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that $\sum\frac{(-1)^{n+1}} {{n}^r} \sum\frac{(-1)^{n+1}} {{n}^s} $ by Abel's rule forms a series that doesn't converge when r+s=1. It is a similar problem to that in Show that the series $\frac{1} {\sqrt{1}} -\frac{1} {\sqrt{2}} +\frac{1} {\sqrt{3}} +\dots$ converges, and its square (formed by Abel's rule) doesn't.. It may provide some hint to the latter.
Show that $\frac{1} {{1}^r} -\frac{1} {{2}^r} +\frac{1} {{3}^r} +\dots$ and $\frac{1} {{1}^s} -\frac{1} {{2}^s} +\frac{1} {{3}^s} +\dots$, where 0 < r < 1, being multiplied by Abel's rule, forms a series (say $\sum \nu_n$) that doesn't converge, when r+s=1.
Abel's rule: given $\sum a_n, \sum b_n$, $\sum_{n=0} ^\infty c_n=\sum_{n=0} ^\infty [\sum_{i=0} ^n a_{n-i}b_i]$ is the infinite series gotten from multiplication of two series.
Initial steps are similar to those in the post,
$(\frac{1} {{1}^r} -\frac{1} {{2}^r} +\frac{1} {{3}^r} +\dots)(\frac{1} {{1}^s} -\frac{1} {{2}^s} +\frac{1} {{3}^s} +\dots)\\
=\frac{1} {{1}^r}\frac{1} {{1}^s}+\dots
+[(-\frac{1} {{1}^r} \frac{1} {{(2k)}^s}+\frac{1} {{1}^r} \frac{1} {{(2k+1)}^s}
-\frac{1} {{2}^r} \frac{1} {{(2k-1)}^s}+\frac{1} {{2}^r} \frac{1} {{(2k)}^s}+\dots
-\frac{1} {{k}^r}\frac{1} {{(k+1)}^s}+\frac{1} {{k}^r}\frac{1} {{(k+2)}^s}
-\frac{1} {{(k+1)}^r}\frac{1} {{k}^s}+\frac{1} {{(k+2)}^r}\frac{1} {{k}^s}
\dots-\frac{1} {{(2k)}^r}\frac{1} {{1}^s}+ \frac{1} {{(2k+1)}^r}\frac{1} {{1}^s})
+\frac{1} {{(k+1)}^{r+s}}]+\dots,$
where
$\sum_{m=1}^{2k}|(-\frac{1} {m^r} \frac{1} {(2k+1-m)^s}+\frac{1} {m^r} \frac{1} {(2k+2-m)^s})|
=\sum_{m=1}^{2k}\frac{1} {m^r} \frac{1} {(2k+1-m)^s}(1-\frac{1} {(1+\frac{1}{2k+1-m})^s})\\
=\sum_{m=1}^{2k}\frac{1} {m^r} \frac{1} {(2k+1-m)^s}(s\frac{1}{2k+1-m}+O(\frac{1}{(2k+1-m)^2}))
=\sum_{m=1}^{2k}\frac{1} {m^r} \frac{s} {(2k+1-m)^{s+1}},$
for $1-(1+x)^{-s}=-\frac{(-s)}{1!}x-\frac{(-s)(-s-1)}{2!}x^2+\dots.$ We can not easily use $\frac{1}{\sqrt{ab}}>\frac{1}{a+b}$ here, instead we use Taylor expansion. It seems the above sum approximates $\sum_{m=1}^{2k}\frac{1} {k^r} \frac{1} {(k)^{s+1}}\approx \frac{k}{k^{r+s+1}}=\frac{1}{k},$ and so the series (say $\sum \psi_n$) it forms diverges as well.
But here we are gonna show that the serives diverges more than $\sum \frac{1}{k+1}$, which we can't yet show in the above post.
Let s go near 1-0 (i.e. r+s-0), then $\sum \psi_n$ goes near
$\sum_{m=1}^{2k}\frac{1} {m^0} \frac{r+s} {(2k+1-m)^{r+s+1}}
=\sum_{m=1}^{2k}\frac{1} {(2k+1-m)^{2}}=\frac{1}{(2k)^2}+\frac{1}{(2k-1)^2}+\dots+\frac{1}{1^2}>\frac{2}{k+1}$ (A note for myself: for calculation of this sum and the sum in the above post, see Formula for $\frac{1}{(n)^2}+\frac{1}{(n-1)^2}+\dots+\frac{1}{1^2}$.. According to results there, the left side tends to $\frac{\pi^2}{6}$ which is obviously larger than the right side that tends to 0. So $\sum \nu_n\approx \sum_{k=0}^\infty \frac{\pi^2}{6}$, it doesn't oscillates between two values but oscillates to infinity.)
when k $\geq$ 3 (i.e. $\frac{2}{k+1}\leq \frac{1}{2^2}$). Therefore when s is near 0, $|\sum \nu_n|>\sum\frac{2}{k+1}-\sum\frac{1}{k+1}$, which diverges.
My question is how to, in general, prove that it the series $\sum \nu_n$ diverges?
| So here we have, for example,
if r<s,
$
|-\frac{1} {{1}^r} \frac{1} {{(2k)}^s}
-\frac{1} {{2}^r} \frac{1} {{(2k-1)}^s}+\dots
-\frac{1} {{k}^r}\frac{1} {{(k+1)}^s}
-\frac{1} {{(k+1)}^r}\frac{1} {{k}^s}
\dots-\frac{1} {{(2k)}^r}\frac{1} {{1}^s}|>
|-\frac{1} {{1}^s} \frac{1} {{(2k)}^s}
-\frac{1} {{2}^s} \frac{1} {{(2k-1)}^s}+\dots
-\frac{1} {{k}^s}\frac{1} {{(k+1)}^s}
-\frac{1} {{(k+1)}^s}\frac{1} {{k}^s}
\dots-\frac{1} {{(2k)}^s}\frac{1} {{1}^s}|>
\sum (\frac{2}{2k+1})^{2s}>\sum (\frac{2}{2k+1})^1
,$
which doesn't coverge to 0. Similar for 2k+1.
For a general solution:
According to https://math.stackexchange.com/a/3787267/577710, (well, it seems to require some modifications, and so does what follows.)
$(n-1)^{1-\max(r,s)} \leq \sum_{m=1}^{n-1}\frac{1}{m^r}\frac{1}{(n-m)^s}
$, and so $s<(2k)^{1-s}s<s\cdot(2k+1-1)^{1-\max(r,s)} \leq \sum_{m=1}^{2k}\frac{1} {m^r} \frac{s} {(2k+1-m)^{s+1}}$, and so the series (at index n being odd) approximates $\sum_{k=0}^\infty\frac{1}{k+1}-\sum_{k=0}^\infty s$ which diverges to infinity, which fits our discussion for s=1.
When r=s=1/2, we see $\sqrt{2k}s<s\cdot(2k+1-1)^{1-\max(r,s)} \leq \sum_{m=1}^{2k}\frac{1} {m^r} \frac{s} {(2k+1-m)^{s+1}}$. And so the series diverges (oscillates actually) even more, not (as I thought in https://math.stackexchange.com/a/3787073/577710) oscillating between two values.
Correction:
$(n-1)(\frac{2}{n})^{2\max(r,s)} \leq \sum_{m=1}^{n-1}\frac{1}{m^r}\frac{1}{(n-m)^s}
$, so
$\frac{2s}{{(k)}^{2s+2}}\approx s\cdot (2k)(k+1/2)^{-2\max(r,s+1)} \leq \sum_{m=1}^{2k}\frac{1} {m^r} \frac{s} {(2k+1-m)^{s+1}}$.
(In this calculation we omit the fact that r and s+1 are not symmetrical; as a result we have not 1+2...+m+...2k which equals to O(k^2) and so the lower bound is very small.)
Therefore it's clear (but not from the above inequality) that when s is near 0, the series diverges as described above, but it is still unclear that when s=r=1/2, the series diverges to infinity or oscillates between two values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove $\int_0^1 \frac{dx}{(x-2) \sqrt[5]{x^2{(1-x)}^3}} = -\frac{2^{11/10} \pi}{\sqrt{5+\sqrt{5}}}$ $$
\mbox{Prove}\quad
\int_{0}^{1}{\mathrm{d}x \over
\left(\,{x - 2}\,\right)\,
\sqrt[\Large 5]{\,x^{2}\,\left(\,{1 - x}\,\right)^{3}\,}\,}
=
-\,{2^{11/10}\,\pi \over \,\sqrt{\,{5 + \,\sqrt{\,{5}\,}}\,}\,}
$$
*
*Being honest I havent got a clue where to start. I dont think any obvious substitutions will help ($x \to 1-x, \frac{1}{x}, \sqrt{x},$ more).
*The indefinite integral involves hypergeometric function so some miracle substitution has to work with the bounds I suspect.
*Maybe gamma function is involved some how ??.
If anyone has an idea and can provide help I would appreciate it.
| $$
\begin{aligned}
\text { Let } x&=\sin ^2 \theta, \quad \textrm{ then } d x=2 \sin \theta \cos \theta d \theta, \textrm{ and }\\
I&=\int_0^{\frac{\pi}{2}} \frac{2 \sin \theta \cos \theta d \theta}{\left(\sin ^2 \theta-2\right) \sqrt[5]{\sin ^4 \theta \cos ^6 \theta}}\\
&=2 \int_0^{\frac{\pi}{2}} \frac{\tan ^{\frac{1}{5}} \theta d \theta}{\cos ^2 \theta\left(\tan ^2 \theta-2 \sec ^2 \theta\right)}\\
&=-2 \int_0^{\infty} \frac{t^{\frac{1}{5}}}{t^2+2} d t ,\quad \textrm{ where }t=\tan \theta\\
&=-2\left(\frac{2^{\frac{1}{10}} \pi}{\sqrt{5+\sqrt{5}}}\right) \quad \textrm{(via beta function)} \\
&=-\frac{2^{\frac{11}{10}} \pi}{\sqrt{5+\sqrt{5}}}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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} |
Evaluate $\frac{1+3}{3}+\frac{1+3+5}{3^2}+\frac{1+3+5+7}{3^3}+\cdots$ It can be rewritten as
$$S = \frac{2^2}{3}+\frac{3^2}{3^2}+\frac{4^2}{3^3}+\cdots$$
When $k$ approaches infinity, the term $\frac{(k+1)^2}{3^k}$ approaches zero. But, i wonder if it can be used to determine the value of $S$. Any idea?
Note: By using a programming language, i found that the value of $S$ is $3.5$
| Let $\dfrac{(k+1)^2}{3^k}=f(k)-f(k-1)$ where $f(n)=\dfrac{an^2+bn+c}{3^n}$
$\implies\dfrac{(k+1)^2}{3^k}=\dfrac{(ak^2+kb+c)-3(a(k-1)^2+b(k-1)+c)}{3^k}$
$\implies k^2+2k+1=-2ak^2+k(b+6a-3b)+c-3(a-b+c)$
Compare the coefficients of $k^2$ to find $$-2a=1$$
and by the coefficients of $k,$ $$6a-2b=2\iff b=?$$
and by the constants, $$1=3b-3a-2c\iff c=?$$
Now use Telescoping series
Of course we need to establish $$\lim_{n\to\infty}f(n)=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Check the convergence of the series : Which test do we apply? Check the convergence of the following series:
*
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{n^2+1}{n^4+n}}$
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{n^{n^2}}{(n+1)^{n^2}}}$
For the first series do we use the comparison test? But with which sequence do we compare $\frac{n^2+1}{n^4+n}$ ? Do we maybe do the following? $$\frac{n^2+1}{n^4+n}<\frac{n^2+n^2}{n^4+n}<\frac{n^2+n^2}{n^4}=\frac{2n^2}{n^4}=\frac{2}{n^2}$$
As for the second series I thought that we could use the ratio test but I am not sure if that works because we get $$\frac{a_{n+1}}{a_n}=\frac{\frac{(n+1)^{(n+1)^2}}{(n+2)^{(n+1)^2}}}{\frac{n^{n^2}}{(n+1)^{n^2}}}=\frac{(n+1)^{(n+1)^2}(n+1)^{n^2}}{(n+2)^{(n+1)^2}n^{n^2}}=\frac{(n+1)^{(n+1)^2+n^2}}{(n+2)^{(n+1)^2}n^{n^2}}$$ So is it better to apply an other test here?
| For the second, it is better to use the root test.
$$\lim_{\infty}a_n^\frac 1n=\lim_{\infty}(\frac{n}{n+1})^n$$
think of
$$\lim_{\infty}(\frac{n+1}{n})^n=e$$
| {
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"url": "https://math.stackexchange.com/questions/3788940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Double integration with Indicator function The integral of interest is:
$$Q = \int_{\mathbb{R}^2}\int_{\mathbb{R}^2} I\left(\frac{1}{2}\frac{x_2^2 - x_1^2 + y_2^2 - y_1^2}{x_2-x_1} \in [0,1]\right) \nonumber \\
\times I\left(2\arcsin\left(\frac{\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}}{2\sqrt{y_1^2 + \left(x_1 - \frac{1}{2}\frac{x_2^2 - x_1^2 + y_2^2 - y_1^2}{x_2-x_1}\right)^2}}\right) > \theta\right) \\ \times \exp\left(-C\left(y_1^2 + \left(x_1 - \frac{1}{2}\frac{x_2^2 - x_1^2 + y_2^2 - y_1^2}{x_2-x_1}\right)^2\right)\right)\mathrm{d}Z_1\mathrm{d}Z_2,$$
where $C > 0$, $Z_1 = (x_1, y_1)$, $Z_2 = (x_2,y_2)$, $I(\cdot)$ is an indicator function, and $\theta \in (0, \pi]$.
| Start with the stretch-rotation transformations $$\begin{pmatrix} z_1 \\ z_2\end{pmatrix} = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2\end{pmatrix} \qquad \text{and} \qquad \begin{pmatrix} z_3 \\ z_4\end{pmatrix} = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2\end{pmatrix}$$
so that ${\rm d}x_1 {\rm d}x_2 {\rm d}y_1 {\rm d}y_2=\frac{1}{4} {\rm d}z_1 {\rm d}z_2 {\rm d}z_3 {\rm d}z_4$ and the integral becomes $$4Q = \int_{\mathbb{R}^4} I\left(\frac{1}{2}\left(z_1+\frac{z_3z_4}{z_2}\right) \in [0,1]\right) \times I\left(2\arcsin\left(\sqrt{\frac{{z_2^2 + z_4^2}}{{z_2^2+z_4^2 + z_3^2\left(1 + \frac{z_4^2}{z_2^2}\right)}}}\right) > \theta\right) \times \exp\left[-\frac{C}{4}\left( z_2^2 + z_4^2 + z_3^2 \left(1+\frac{z_4^2}{z_2^2} \right) \right)\right] \, \mathrm{d}z_1\mathrm{d}z_2\mathrm{d}z_3\mathrm{d}z_4$$ which eliminates the first indicator function by carrying out the integral over $z_1$ and gives a factor of 2. Furthermore the second indicator function proposes the transformation to polar coordinates i.e. $(z_2,z_4)=(r\cos\phi,r\sin\phi)$ and we have $$2Q=\int_{\mathbb{R}}{\rm d}z_3\int_0^\infty r\,{\rm d}r \int_0^{2\pi}{\rm d}\phi \, I\left(2\arcsin\left(\frac{r}{\sqrt{r^2+\frac{z_3^2}{\cos^2\phi}}}\right)>\theta\right) \exp\left[ -\frac{C}{4} \left( r^2 + \frac{z_3^2}{\cos^2\phi} \right) \right] .$$ Next we carry out the $z_3$ integration which is 2 times the integral from $0$ to $\infty$. For $z_3=0$ the argument of the arcsine is $1$ and so $\pi>\theta$ is definitely true. The other bound follows from solving $$\frac{r}{\sqrt{r^2+\frac{z_3^2}{\cos^2\phi}}}=\sin\frac{\theta}{2} \qquad \Longrightarrow \qquad z_3=r|\cos \phi| \cot \frac{\theta}{2} \, .$$ Hence
$$Q= \sqrt{\frac{\pi}{C}} \int_0^\infty r \, {\rm d}r \int_0^{2\pi} {\rm d}\phi \, |\cos\phi| \, \exp\left[-\frac{Cr^2}{4}\right] \, {\rm erf}\left( \frac{r\sqrt{C} \, \cot \frac{\theta}{2}}{2} \right) \\ = 4 \sqrt{\frac{\pi}{C}} \int_0^\infty {\rm d}r \, r \, \exp\left[-\frac{Cr^2}{4}\right] \, {\rm erf}\left( \frac{r\sqrt{C} \, \cot \frac{\theta}{2}}{2} \right) \\
= \frac{8\cot \frac{\theta}{2}}{C} \int_0^\infty {\rm d}r \, \exp\left[-\frac{Cr^2}{4\sin^2 \frac{\theta}{2}} \right] \\
=\frac{8 \sqrt{\pi} \cos \frac{\theta}{2}}{C^{3/2}} \, .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3789355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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How to prove this equality of the determinant of matrix?
Prove that
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}
=(a^2-bc)(b^2-ca)(c^2-ab)\end{equation*}
My attempt:
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}
=\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
a(b-a) & b(c-b) & c(a-c) \\
(b+a)(b-a) & (c+b)(c-b) & (a+c)(a-c)
\end{pmatrix}
\end{equation*}
But I think my direction is incorrect. Can anyone give me some hints or the solution of this question?
| Try this. Multiply column $1$ by $c$, column $2$ by $a$ and column $3$ by $b$. Then
\begin{aligned}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}
&=\frac{1}{abc}\det\begin{pmatrix}
a^2c & b^2a & c^2b\\
abc & abc & abc\\
b^2c& c^2a & a^2b
\end{pmatrix}\\
&=\det\begin{pmatrix}
a^2c & b^2a & c^2b\\
1& 1 & 1\\
b^2c& c^2a& a^2b
\end{pmatrix}\\
&=\det\begin{pmatrix}
a^2c-c^2b & b^2a-c^2b & c^2b\\
0& 0 & 1\\
b^2c-a^2b& c^2a-a^2b & a^2b
\end{pmatrix}\\
&=(a^2-bc)(c^2-ab)\det\begin{pmatrix}
c & -b & c^2b\\
0& 0 & 1\\
-b& a & a^2b
\end{pmatrix}\\
&=(a^2-bc)(c^2-ab)(b^2-ac).
\end{aligned}
(https://i.stack.imgur.com/D6GF9.jpg)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove that this number is divisible by 7 Without using induction, how can it be proved that 7 divides $3^{2n+1}+2^{n+2}$ for each $n\in\mathbb{N}$?
I tried to expand it using $\frac{x^{n+1}-1}{x-1}=1+x+..+x^n$ but I had no success.
It would be great if more than one proof is provided.
| $3^{2n + 1} + 2^{n+2} = 3\cdot 3^{2n} + 2^2\cdot 2^n = 3\cdot(9)^n + 4\cdot s2^n\equiv 3\cdot(2)^n + 4\times 2^n = 7\cdot 2^n\equiv 0\pmod 7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3795659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove/Disprove: $A - \lfloor A/B \rfloor - \lceil A/B \rceil \leq \lfloor A/B \rfloor \times (B+1)$ for $A \geq B$ For $A \geq B$, both are strictly positive integers, is the following true?
$$A - \lfloor A/B \rfloor - \lceil A/B \rceil \leq \lfloor A/B \rfloor \times (B+1)$$
I tried the technique used in proving a very similar question:
Prove/Disprove: $A - \lfloor A/B \rfloor - \lceil A/B \rceil \leq (\lfloor A/B \rfloor + 1) \times B$ for $A \geq B$
But it seems like it didn't work in proving this. I also tried empirically generating random A and B's, but also can't find a counterexample.
| $\newcommand{f}[1]{\left\lfloor #1 \right\rfloor}$
Let $B = 100$ and $A = 199$. Then:
\begin{align*}
LHS &= 199 - 1 - 2 = 196 \\
RHS &= 1(100 + 1) = 101
\end{align*}
So the inequality is false.
EDIT: In response to OP's comment, suppose we restrict further that $\f{A/B} \geq N$ for some $N \in \Bbb{Z}^+$. Let $B = 3N + 3$, and let $A = (N + 1)(3N + 3) - 1$. Clearly $A \geq B$ and $\f{A/B} = N$.
\begin{align*}
LHS &= (N + 1)(3N + 3) - 1 - N - (N + 1) \\
&= (N + 1)(3N + 1) \\
\end{align*}
\begin{align*}
RHS &= N(3N + 4) \\
&= N(3N + 1) + 3N \\
&= (N + 1)(3N + 1) - (3N + 1) + 3N \\
&= (N + 1)(3N + 1) - 1 < LHS
\end{align*}
So the inequality will still fail.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Two inequalities with parameters $a,b,c>0$ such that $ca+ab+bc+abc\leq 4$
Let $a,b,c>0$ be such that $bc+ca+ab+abc\leq 4$. Prove the following inequalities:
(a) $8(a^2+b^2+c^2)\geq 3(b+c)(c+a)(a+b)$, and
(b) $\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{a^2b}+\dfrac{2}{b^2c}+\dfrac{2}{c^2a}\geq 9$.
Prove also that the unique equality case for both inequalities is given by $a=b=c=1$.
Below are some probably useful or relevant results.
*
*https://artofproblemsolving.com/community/c6h1241430p6342224
*https://artofproblemsolving.com/community/c6h284290p1535893
*https://artofproblemsolving.com/community/c6h608971p3619202
*https://artofproblemsolving.com/community/c6h1804479p11995588
*If $ab+bc+ca+abc=4$, then $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3\leq a+b+c$
Techniques used in solving the inequalities in these links may prove useful in proving our inequalities.
Attempt. In the simplest case, $a=b=c=:t$, we have $t^3+3t^2-4\leq 0$, whence $0<t\leq 1$. Therefore, the inequalities (a) and (b) become
$$24t^2\geq 24t^3$$
and
$$\frac{3}{t^2}+\frac{6}{t^3}\geq 9\,,$$
which are obviously true. How to prove these inequalities in general?
| This is an approach using Lagrange multipliers.
For the first part, we can write the problem as \begin{align}\min&\quad8(a^2+b^2+c^2)-3(a+b)(a+c)(b+c)\\\text{s.t.}&\quad ab+ac+bc+abc=4-\epsilon\\&\quad a,b,c>0\quad\land\quad0\le\epsilon<4.\end{align} Then we have $\mathcal L=f-\lambda g$ where $f(a,b,c)=8(a^2+b^2+c^2)-3(a+b)(a+c)(b+c)$ and $g(a,b,c)=ab+ac+bc+abc-(4-\epsilon)$. The partial derivatives are \begin{align}\mathcal L_a&=16a-3(b+c)(2a+b+c)-\lambda(b+c+bc)\\\mathcal L_b&=16b-3(a+c)(2b+a+c)-\lambda(a+c+ac)\\\mathcal L_c&=16c-3(a+b)(2c+a+b)-\lambda(a+b+ab)\\\mathcal L_\lambda&=ab+ac+bc+abc-(4-\epsilon).\end{align} Next, we have \begin{align}\mathcal L_a-\mathcal L_b&=0\implies16+3(a+b)+\lambda(1+c)=0,a=b\\\mathcal L_a-\mathcal L_c&=0\implies16+3(a+c)+\lambda(1+b)=0,a=c\\\mathcal L_b-\mathcal L_c&=0\implies16+3(b+c)+\lambda(1+a)=0,b=c,\end{align} so without loss of generality we have $a=b$. Letting $c=ka$ yields $f(a,b,c)=8(2+k^2)a^2-6(1+k)^2a^3$ and $g(a,b,c)=(1+2k)a^2+ka^3-(4-\epsilon)$. Since $f\to0^+$ as $\epsilon\to4^-$ we aim to find $k,\epsilon$ such that $f\le0$.
Elementary calculus reveals that $f(a;k)$ increases monotonically in the interval $[0,k^*]$ where $k^*=8(2+k^2)/(9(1+k)^2)$, from $0$ to $f(k^*;k)>0$. For $a>k^*$, the function $f(a;k)$ decreases monotonically to $-\infty$, where it meets the axis at $a=3k^*/2$. Notice that positive root of $g$ is largest when $4-\epsilon$ is greatest; that is, $\epsilon=0$. At this value, it suffices to notice that $$g\left(\frac{3k^*}2;k\right)=(1+2k)\left(\frac{4(2+k^2)}{3(1+k)^2}\right)^2+k\left(\frac{4(2+k^2)}{3(1+k)^2}\right)^3-4=(k-1)^2\cdot\frac{P(k)}{Q(k)}$$ where $P,Q$ are polynomials with positive coefficients. As $k>0$ the only solution to $g=0$ is $k=1$, from which it follows that $a=b=c=1$. $\square$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $f(x)=\sin^{-1} (\frac{2x}{1+x^2})+\tan^{-1} (\frac{2x}{1-x^2})$, then find $f(-10)$ Let $x=\tan y$, then
$$
\begin{align*}\sin^{-1} (\sin 2y )+\tan^{-1} \tan 2y
&=4y\\
&=4\tan^{-1} (-10)\\\end{align*}$$
Given answer is $0$
What’s wrong here?
| Hint:
$$\sin^{-1}\dfrac{2(-10)}{1+(-10)^2}=\sin^{-1}\left(-\dfrac{20}{101}\right)$$
$$\tan^{-1}\dfrac{2(-10)}{1-(-10)^2}=\tan^{-1}\dfrac{20}{99}=u(\text{say})$$
$\implies\dfrac\pi2>u>0$
$\sec u=+\sqrt{1+\left(\dfrac{20}{99}\right)^2}=\dfrac{101}{99}$
$\implies\sin u=\dfrac{\tan u}{\sec u}=?$
$u=\sin^{-1}?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
solve for $x$, $(\sqrt{a+ \sqrt{a^2-1}})^x+(\sqrt{a- \sqrt{a^2-1}})^x=2a$ Find the value of $x$ when $$(\sqrt{a+ \sqrt{a^2-1}})^x+(\sqrt{a- \sqrt{a^2-1}})^x=2a.$$See, by hit and trial method it is clear that $x=2$ is a solution. But I failed to solve this explicitly to get the solutions.
My Attempt: \begin{align*} &(\sqrt{a+ \sqrt{a^2-1}})^x+(\sqrt{a- \sqrt{a^2-1}})^x=2a \\ \implies \> & (a+ \sqrt{a^2-1})^{x/2}+(a- \sqrt{a^2-1})^{x/2}=2a\\ \implies \>& (a+ \sqrt{a^2-1})^x+(a- \sqrt{a^2-1})^x+2(a+ \sqrt{a^2-1})^{x/2}(a- \sqrt{a^2-1})^{x/2} = 4a^2.\end{align*} I have no idea how to proceed after this. Also I tried by multiplying conjugate up and down, but I failed. Please help me to solve this.
Thanks in advance.
| Hint : Let $y=a+\sqrt{a^2-1}$. The equation you have to solve is equivalent to
$$y^{x/2} + \frac{1}{y^{x/2}} = y +\frac{1}{y}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the second derivative of the absolute function $\left|\frac{x+1}{x+2}\right|$? I calculated the derivative of $\left|\frac{x+1}{x+2}\right|$ in the same way that I would do with $ \frac{x+1}{x+2}$ in order to study the function.
But when I verified on wolfram, I noticed it is all wrong. Wolfram uses the chain rule as you can see here.
I don't get it. The only rule I've been taught as far as absolute function derivatives are concerned, is $|x|' = \frac{x}{|x|}$. Does a similar rule apply for $f(x)$? And why does wolfram uses chain rule?
Edit
I calculated the derivatives as there is no absolute and then, at the result, I applied the absolute.
My answers are $|(\frac{x+1}{x+2})|' = |(\frac{x+1}{x+2})'| = \frac{1}{\left(x+2\right)^2}$ and $|(\frac{x+1}{x+2})|'' = |(\frac{x+1}{x+2})''| = \frac{2}{\left(x+2\right)^3}$
Wolfram's answer is
$\left(\left|\frac{x+1}{x+2}\right|\right)'\:=\frac{\left|x+2\right|\left(x+1\right)}{\left|x+1\right|\left(x+2\right)^3}$
| This is how I deal with absolute functions:
$$
\begin{align}
\left|\frac{x+1}{x+2}\right|&=\sqrt{\left(\frac{x+1}{x+2}\right)^{2}}\\
\\
\frac{d}{dx} \left|\frac{x+1}{x+2}\right|&=\frac{d}{dx} \sqrt{\left(\frac{x+1}{x+2}\right)^{2}}\\
&=\frac{1}{2 \sqrt{\left(\frac{x+1}{x+2}\right)^{2}}}\cdot 2 \left(\frac{x+1}{x+2}\right)\cdot\frac{1}{\left(x+2\right)^{2}}\\
&=\frac{x+1}{\left(x+2\right)^{3}\cdot\left|\frac{x+1}{x+2}\right|}
\end{align}
$$
Notice the chain rule when I differentiate the square root
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Evaluate $ \lim_{ n \to \infty }\int_{0}^{ \infty } \frac{1}{(1+x^3)(1+x^n)} dx$ Evaluate $$\displaystyle \lim_{ n \to \infty }\displaystyle \int_{0}^{ \infty } \frac{1}{(1+x^3)(1+x^n)} dx$$
I tried some substitution like $t=x^3+1$ and $t=x^n+1$ but did not work, and also division $$\lim_{ n \to \infty }\displaystyle \int_{0}^{ \infty } \frac{1}{(1+x^3)(1+x^3)(x^{n-3}+1-x^{n-1})} dx$$ but got stuck when I tried partial fraction decomposition.
| $$I=\lim_{n \to \infty}\int_{0}^{\infty} \frac{1}{(1+x^3)(1+x^n)}dx=\int_{0}^{1} \frac{1}{1+x^3} dx+\int_1^{\infty} 0~ dx$$
As $\displaystyle \lim_{n \to \infty} x^n =0, 0<x<1$; $\displaystyle\lim_{n \to \infty} x^n= \infty ,x>1.$
$$I=\int_{0}^{1}\left( \frac{1}{3(1+x)}-\frac{2x-1-3}{6(1-x+x^2)}\right) dx$$ $$I=\left[\frac{1}{3} \ln (1+x) -\frac{1}{6} \ln(x^2-x+1)+\frac{1}{\sqrt{3}}\tan^{-1}(2x-1)/\sqrt{3})\right]_{0}^{1}$$
$$I=\frac{1}{3}\ln 2+\frac{\pi}{3\sqrt{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all $x\in \mathbb{R}$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$
Find all $x\in \mathbb{R}$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$$
Letting $a=2^x$ and $b=3^x$ we get $$\frac{a^3+b^3}{a^2b+ab^2} = \frac{7}{6}$$
from the numerator we have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=7$$
since $7$ is a prime we can say that $$a+b=1, a^2-ab+b^2=7.$$
It follows that $$a=1-b$$
from where $$(1-b)^2-(1-b)b+b²=7$$
this quadratic has solutions $b=1, b=0.$
what I now did was consider cases. Firstly $a=b=0$ which has no solutions for $x$.
case $a=b=1$ has the solution $x=0$.
However the actual solutions for this were $x=1, x=-1$ which I don't see how they came up with. What is wrong with my approach?
| $$\frac{a^3 + b^3}{a^2b + ab^2} = \frac{7}{6} \implies 7 a^2b + 7ab^2 = 6 a^3 + 6b^3 \implies 7ab(a + b) = 6(a + b)(a^2 -a b + b^2)$$ $$ \implies (a + b)(6a^2 -13ab + 6b^2) =0 $$
Then either $(a + b) = 0$ or $(6a^2 -13ab + 6b^2) = (2a -3b)(3a - 2b) = 0 $
Hence, we have that
$$ a = - b \implies 2^x = - 3^x \implies \left(\frac{2}{3}\right)^x = -1 $$ which is not possible. Hence,
$$ 2a = 3b \implies 2^{x+ 1} = 3^{x +1 } \implies \left(\frac{2}{3}\right)^{x + 1} = \left(\frac{2}{3} \right)^0 \implies x = -1$$
Or, $$ 3a = 2b \implies 2^{x- 1} = 3^{x -1 } \implies \left(\frac{2}{3}\right)^{x + 1} = \left(\frac{2}{3} \right)^0 \implies x = 1 $$
Thus, $x = 1, -1$ are the only soluions.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\frac{1 - x^{n+1} }{n+1} \lt \frac{1-x^n}{n}$ given $n$ is a positive integer and $0 < x \lt 1$. Problem Statement: If $n$ is a positive integer and $0 < x \lt 1$, show that
$$ \frac{1 - x^{n+1} }{n+1} \lt \frac{1-x^n}{n}.$$
My Solution:
$$
\frac{ 1- x^{n+1} }{n+1} \lt \frac{1-x^n}{n} ~~~~\text{is true} \\
\text{if}~~~~ \frac{n}{1-x^n} \lt \frac{n+1}{1- x^{n+1} }$$
If we see the LHS we find that it is of the form of sum of a geometric series with the first term $n$ and the common ration $x^n$ (which is less than 1) similarly the RHS represents the sum of a geometric series with the first term as $n+1$ and common ratio $x^{n+1}$ (which is less than $x^n$, that is less than 1)
Now, my point is that the series represented by the LHS have the first term lesser than the first term of the series represented by the RHS, and the series represented by LHS decreases fastly in comparison to the series represented by RHS (because the common ratio $x^{n+1} \lt x^n$), hence the sum of series represented by the RHS is greater than the sum of the series represented by the LHS.
Is my solution and reasoning correct?
| We have that
$$\frac{1 - x^{n+1} }{n+1} \lt \frac{1-x^n}{n} \iff n-nx^{n+1} \lt n-nx^n+1-x^n$$
$$\iff nx^n(1-x) \lt 1-x^n \iff nx^n<\overbrace{1+x+x^2+\ldots+x^{n-1}}^{\color{red}{\text{n terms}\,> \,x^n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solving $\cos(2x)\cos\left(x - \frac{\pi}{6}\right) = \sin(2x)\sin\left(\frac{\pi}{6} - x\right)$ for $x\in(0,\pi/2)$
Solve this equation for $x\in (0 , \frac{\pi}{2})$
$$\cos(2x)\cos\left(x - \frac{\pi}{6}\right) = \sin(2x)\sin\left(\frac{\pi}{6} - x\right)$$
I gave a try using the $\cos(a-b)$ and $\sin(a-b)$ formulas, but it seems the problem complicated a little bit more.
Is any other elegant solution for this?
| The equation is $\cos (2x) \cos (x-\frac {\pi} 6)+\sin (2x)\sin (x-\frac {\pi} 6)=0$. This is same as $\cos (2x-(x-\frac {\pi} 6))=0$ or $\cos (x+\frac {\pi} 6)=0$. So $x+\frac {\pi} 6=\frac {(2n+1)\pi} 2$ for some integer $n$. For $x \in (0,\frac {\pi} 2)$ we must have $n=0$ so $x =\frac {\pi} 3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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Finding remainder of $123^{456}$ divided by 88 using Chinese Remainder Theorem I tried using Chinese remainder theorem but I kept getting 19 instead of 9.
Here are my steps
$$
\begin{split}
M &= 88 = 8 \times 11 \\
x_1 &= 123^{456}\equiv 2^{456} \equiv 2^{6} \equiv 64 \equiv 9 \pmod{11} \\
y_1 &= 9^{-1} \equiv 9^9 \equiv (-2)^9 \equiv -512 \equiv -6 \equiv 5 \pmod{11}\\
x_2 &= 123^{456} \equiv 123^0 \equiv 1 \pmod{8}\\
y_2 &= 1^{-1} \equiv 1 \pmod{8} \\
123^{456}
&\equiv \sum_{i=1}^2 x_i\times\frac{M}{m_i} \times y_i
\equiv 9\times\frac{88}{11}\times5 + 1\times\frac{88}{8} \times1 \equiv 371
\equiv 19 \pmod{88}
\end{split}
$$
| Your calculations look correct except for the last line which I don't understand.
One you get $x_1$ and $x_2$, you could simply write
$x=123^{456}=9+11k$ (from $x_1$)
so reducing mod $8$ yields
$x \equiv 1+3k \pmod{8} \equiv 1 \pmod{8}$ (from $x_2$)
therefore
$3k\equiv0 \pmod{8}$ and since $\gcd(3,8)=1$, $3$ is invertible mod $8$, so $k \equiv 0 \pmod{8}$
hence $x=9+11(0+8k')=9+88k' \equiv 9 \pmod{88}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate integral $\int \frac{x^2}{(15+6x-9x^2)^{\frac{3}{2}}} \, dx$ Evaluate the integral
$$\int \frac{x^2}{(15+6x-9x^2)^{\frac{3}{2}}} \, dx$$
Edit:
Here is Quanto's brilliant method of doing this problem with the details filled out:
SOLUTION:
First, lets complete the square $-9x^2+6x+15$:
$-9x^2+6x+15$
$=-9(x^2-\frac{2}{3}x)+15$
$=-9(x^2-\frac{2}{3}x+(\frac{2}{6})^2)+15-(-9(\frac{2}{6})^2)$
$=-9(x-\frac{1}{3})^2+16$...
$=-(1-3x)^2+16$
Now we are going to do a few extra steps in completing the square then usual, so this will be easier to integrate.
$=(-(\frac{1-3x}{4})^2+1)16$
Thus our original integral becomes:
$\int \frac{x^2}{(-(\frac{1-3x}{4})^2+1)16)^{\frac{3}{2}}}dx$
Now set $\sin(t) = \frac{1-3x}{4}$. Thus $x=\frac{1-4\sin(t)}{3}$ and $dx = \frac{-4\cos(t)}{3} \, du$
Note the strategic choice of defining $\sin(t) = \frac{1-3x}{4}$, it has a perfect place to plug in in the denominator. Let's make these substitutions
$=\int \frac{(\frac{1-4\sin(t)}{3})^2}{((-\sin^2(t)+1)16)^{\frac{3}{2}}}(\frac{-4\cos(u)}{3} \, du)$
$=\frac{-1}{432}\int \frac{(1-4\sin(t))^2}{(-\sin^2(t)+1)^{\frac{3}{2}}}(\cos(u) \, du)$
$=\frac{-1}{432}\int \frac{(1-4\sin(t))^2}{\cos^2(t)} \, du)$
$=\frac{-1}{432}\int \sec^2(t)+16\tan^2(t)-\frac{8\sin(t)}{\cos^2(t)}$
Now, Applying the identity $\tan^2(x)=\sec^2(x)-1$ we arrive at:
$=\frac{-1}{432}\int 17\sec^2(t)-16-\frac{8\sin(t)}{\cos^2(t)} \, dt$
Let's evaluate each of these separately:
$\int 17\sec^2(t)\,dt = 17\tan(t)$
$\int 16 \, dt = 16t$
$\int \frac{8\sin(t)}{\cos^2(t)}$ can be solved with u-substitution after setting $u=\cos(t)$. to get:
$\int \frac{8\sin(t)}{\cos^2(t)}=\frac{8}{\cos(t)}$
Thus our answer is:
$=\frac{1}{432}(16t-17\tan(t)+\frac{8}{\cos(t)})+C$
| Let $\sin t = \frac{1-3x}4$. Then,
\begin{align}
\int \frac{x^2}{(15+6x-9x^2)^{\frac{3}{2}}}dx
&=-\frac1{432}\int \frac{(1-4\sin t)^2}{\cos^2t}dt\\
&=\frac1{432}\int\left( 16-17\sec^2 t +\frac{8\sin t}{\cos^2t}\right)dt
\\
&= \frac1{432}\left( 16t -17\tan t+\frac8{\cos t}\right) +C
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving that inequality holds under condition.
Let $a$ and $b$ be positive numbers. Prove that inequality $$\frac{ax+by}{2} \leqslant \sqrt{\frac{ax^2+by^2}{2}}$$
holds for all real $x$ and $y$ only and only if $a+b \leqslant2$
Problem needs to be done using "basic" algebraic methods.
I tried expanding this into form $$2ax^2+2by^2-a^2x^2-b^2y^2-2abxy \geqslant 0$$
and then take oustide parenthesis $2-a-b$. Inequalities between means did not help either.
Can you give me some clues?
| Suppose $\frac{ax+by}2 \le \sqrt{\frac{ax^2+by^2}{2}}$ is true, let $x=y=1$,
$$\frac{a+b}2\le \sqrt{\frac{a+b}2}$$
$$\frac{(a+b)^2}{4}\le \frac{a+b}2$$
Hence we must have $a+b \le 2$.
Suppose we have $a+b \le 2$, we want to investigate when does
$$(2a-a^2)x^2+(2b-b^2)y^2-2abxy \ge 0, \forall x, y$$
View it as a quadratic equation in $x$, since the coefficient $2a-a^2$ is positive, this is equivalent to the discriminant being non-positive.
$$4a^2b^2y^2 -4(2a-a^2)(2b-b^2)y^2 \le 0, \forall y$$
Equivalently,
$$ab - (2-a)(2-b) \le 0$$
$$-4+2a+2b \le 0$$
$$a+b \le 2$$
which is true as that is our assumption. That is $a+b \le 2 \implies \frac{ax+by}2 \le \sqrt{\frac{ax^2+by^2}{2}}$.
Conclusion: $a+b \le 2 \iff \frac{ax+by}2 \le \sqrt{\frac{ax^2+by^2}{2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Solving $\frac{dP}{dt} = P(a - b\sqrt P)$ using separation of variables I need to solve the following differential equation using separation of variables but I am unsure on how to do this
$$
\frac{dP}{dt} = P(a - b\sqrt P)
$$
Where $p(0) = 4$, $a = 0.302$ and $b =0.002$
I tried substituting $P = u^2$ as a change of variables to help me solve the equation.
| $$\begin{aligned}\frac{dP}{dt}& = P(a - b\sqrt P)\\
(\text{Replacing }P(t)=(u(t)^2))\implies 2u\frac{du}{dt} &=u^2(a-bu) \\ (u(t)\equiv_t 0,\text{ i.e.} u(t)=0 \ \forall t, \text{ or}\cdots ) \implies2\dfrac{du}{dt} & = u(a-bu) \\ \implies \int dt = \int \dfrac{2du}{u(a-bu)}&=\dfrac{2b}{a}\int \dfrac{a}{bu(a-bu)}du\\ = \dfrac{2}{a}\int\left(\dfrac1{\frac{a}{b}-u}+\dfrac1{u}\right)du &=\dfrac{2}{a}\ln\left(\dfrac{u}{\frac{a}{b}-u}\right) +c\\ \implies t &=\dfrac2a\ln\left(\dfrac{bu}{a-bu}\right)+C\end{aligned}$$
Thus we have
$$\begin{aligned} \exp\left(\dfrac{a(t-C)}2\right)&=\dfrac{bu}{a-bu} \\ \implies \exp\left(\dfrac{a(C-t)}2\right)&=\dfrac{a-bu}{bu}=\dfrac{a}{bu}-1 \\ \implies 1+\exp\left(\dfrac{a(C-t)}2\right) &= \dfrac{a}{bu(t)} \\ \implies \sqrt{P(t)}= u(t)&=\dfrac{a}{b\left(1+\exp\left(\dfrac{a(C-t)}2\right)\right)} \\ \implies P(t) & = \dfrac{a^2}{b^2\left(1+\exp\left(\dfrac{a(C-t)}2\right)\right)^2} \end{aligned}$$
where $C$ is the constant of integration, which can be found by substituting the values of $a,b$ and $t=0, P(0)=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simple String of 6-digits This is the problem:
A string of 6 digits, each taken from the set $[0, 1, 2]$ is to be formed. The string should not contain any of the substrings $012$, $120$, and $201$. How many such 6-digit strings can be formed?
I really have trouble in using PIE in here. I'm badly needing some clear explanation in the PIE part because it's not what I usually deal with.
I have also a question: 'Does avoiding $000$, $111$, and $222$ instead will still gain the same result?'.
Thanks for all the help in advance :)
| You want to avoid 12 properties:
\begin{matrix}
000xxx & x000xx & xx000x & xxx000 \\
111xxx & x111xx & xx111x & xxx111 \\
222xxx & x222xx & xx222x & xxx222 \\
\end{matrix}
The inclusion-exclusion formula is $$\sum_{k=0}^{12} (-1)^k N_k,$$
where $N_k$ is an overcount of the number of $6$-strings that satisfy $k$ properties. Clearly, $N_k=0$ for $k>4$.
We have
\begin{align}
N_0 &= \binom{12}{0} 3^6 = 729 \\
N_1 &= \binom{12}{1} 3^{6-3} = 324 \\
N_2 &= 2 \binom{3}{2} 3^{6-6} + 3 \left(3 \cdot 3^{6-4} + 2 \cdot 3^{6-5} + 1 \cdot 3^{6-6}\right) = 108 \\
N_3 &= 3 \left(2 \cdot 3^{6-5} + 2 \cdot 3^{6-6}\right) = 24 \\
N_4 &= 3 \left(1 \cdot 3^{6-6}\right) \\
\end{align}
Putting it all together, we obtain
$$729 - 324 + 108 - 24 + 3 = 492$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proving $2\left(b^2+c^2\right)-a^2\leqslant 12$ with some condition. Problem. Let $a,b,c\in\mathbb{R}$ such that $a+b+c=6,$ $a^2+b^2+c^2\in\left[12,\frac{68}3\right]$ and $a\geq b\geq c.$ Prove
$$2\left(b^2+c^2\right)-a^2\leqslant 12.$$
When do we have equality?
I can only prove it for $a,b,c \geqslant 0.$
From $a+b+c=6$ we have $a=6-b-c \geqslant 0.$ We need to prove the inequality when $$\Big[6-b-c\geqslant 0, \left( 6-b-c \right) ^{2}+{b}^{2}+{c}^{2}-12\geqslant 0,\\{\frac {68}{3}}-
\left( 6-b-c \right) ^{2}-{b}^{2}-{c}^{2}\geqslant 0,6-c-2\,b\geqslant 0,b-c\geqslant 0\Big]$$
By computer we have$:$ $$12+ \left( 6-b-c \right) ^{2}-2\,{b}^{2}-2\,{c}^{2}$$
$$= \left[ \left( 6-b-c \right) ^{2}+{b}^{2}+{c}^{2}-12 \right] \Big[\left(
b-c \right) \left( {\dfrac {5}{32}}\,{c}^{2}+\dfrac{1}{16}\,bc \right) +\left(
6-c-2\,b \right) \left( {\frac {3}{64}}\,{b}^{2}+\dfrac{1}{4}b+{\frac {5}{32
}}\,bc \right) \Big]
$$
$$+{\frac {3}{32}} \left[ {\frac {68}{3}}- \left( 6-b-c \right) ^{2}-{b
}^{2}-{c}^{2} \right] \left( b-c \right) {c}^{2}+\dfrac{1}{2} \left( 6-b-c
\right) \left( 6-c-2\,b \right) c$$
$$+\dfrac{1}{6}\, \left( 6-c-2\,b \right) ^{2}
{b}^{2}+ \left( \left( 6-b-c \right) ^{2}+{b}^{2}+{c}^{2}-12 \right) ^{2}
\left( \dfrac{1}{12}+{\frac {3}{256}}\,{c}^{2}+\dfrac{1}{32}c+{\frac {3}{128}}\,bc
\right) +\dfrac{1}{6} \left( 6-c-2\,b \right) \left( b-c \right) bc+\dfrac{1}{6}\,
\left( b-c \right) ^{2}{c}^{2}$$
$$+ \left[ \left( 6-b-c \right) ^{2}+{b}^{2}+{c}^{2}-12 \right] \left[
{\frac {68}{3}}- \left( 6-b-c \right) ^{2}-{b}^{2}-{c}^{2} \right]
\left( {\frac {3}{256}}\,{c}^{2}+{\frac {3}{128}}\,bc \right) $$
$$+ \left[ {\frac {68}{3}}- \left( 6-b-c \right) ^{2}-{b}^{2}-{c}^{2}
\right] \left( 6-c-2\,b \right) \left( {\frac {3}{64}}\,{b}^{2}+{
\frac {3}{32}}\,bc \right)\geqslant 0 $$
For the text of the above decomposition, please see in my file: Click here.
| Let $s = b - c \ge 0$. From $a = 6 - b - c \ge b$, we have $c \le 2 - \frac{2s}{3}$.
We have $a^2 + b^2 + c^2 = (6-b-c)^2 + b^2 + c^2 = 6(c - 2 + s/2)^2 + s^2/2 + 12$.
From $a^2 + b^2 + c^2 \in [12, 68/3]$, we have
$$6(c - 2 + s/2)^2 + s^2/2 + 12 \le \frac{68}{3}$$
and thus
$$0 \le s \le \frac{8}{\sqrt{3}}, \quad 2 - s/2 - \frac{1}{6}\sqrt{64-3s^2}
\le c \le 2 - s/2 + \frac{1}{6}\sqrt{64-3s^2}.$$
Clearly, $2 - \frac{2s}{3} \le 2 - s/2 + \frac{1}{6}\sqrt{64-3s^2}$
for $0 \le s \le \frac{8}{\sqrt{3}}$.
From $2 - s/2 - \frac{1}{6}\sqrt{64-3s^2} \le 2 - \frac{2s}{3}$,
we have $0 \le s \le 4$.
Then, we have $2(b^2+c^2) - a^2 = s^2+24c+12s-36
\le s^2 + 24(2 - \frac{2s}{3}) + 12s - 36 = 12 - s(4-s) \le 12$.
We are done.
| {
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"url": "https://math.stackexchange.com/questions/3813662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Binomial identity involving negative values I want to check this identity that appears in Feller's first book on probability:
$${2n\choose n}=(-1)^n4^n{-1/2\choose n}$$
So I apply this definition
$${x\choose r}=\frac{x(x-1)(x-2)\ldots(x-r+1)}{r!}$$
obtaining
$$\frac{1/2(1/2-1)(1/2-2)\ldots(1/2-n+1)}{n!}=\frac{(-1)(-1)^{n-1}(1/2)(1/2+1)(1/2+2)\ldots(1/2+n-1)}{n!}\\=\frac{(-1)^n\left(\frac{1}{2}\right)^n(1+2)(1+4)(1+6)\ldots(1+2n-2)}{n!}$$
But I can't see how to advance from here. I also tried adding the terms (1/2+s) and then extracting the 2 in the denominator but the numerator is a factorial of odd terms and I'm stucked again. Thanks.
| You got as far as $${-1/2\choose n}=\frac{(-1)^n\left(\frac{1}{2}\right)^n(1+0)(1+2)(1+4)(1+6)\ldots(1+2n-2)}{n!}$$ though you still need to multiply by $(-1)^n 4^n$. Writing this as
$$(-1)^n4^n{-1/2\choose n}=4^n\frac{\left(\frac{1}{2}\right)^n \times 1 \times 3 \times 5 \times 7 \ldots\times (2n-1)}{ n!} \\
= 2^n \frac{ 1 \times 3 \times 5 \times 7 \ldots(2n-1)}{ n!} \times \frac{n!}{n!} \\
= \frac{ 1 \times 3 \times 5 \times 7 \ldots(2n-1)}{ n! } \times \frac{2 \times 4 \times 6 \times \cdots \times 2n }{n!} \\ = \frac{(2n)!}{n!\,n!} \\= {2n \choose n}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Let $a$ and $b$ be positive integers such that $(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = 49 + 20 \sqrt[3]{6}.$ Find $a + b.$ Let $a$ and $b$ be positive integers such that $$(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = 49 + 20 \sqrt[3]{6}.$$ Find $a + b.$
We let $\sqrt[3]a, \sqrt[3]b$ be $x,y,$ respectively. We expand LHS to get $$x^2+y^2+2xy-2x-2y+1=49+20\sqrt[3]6.$$ Now, I'm stuck. Help? Please share a clear solution :).
| Okay. Hint: consider rational and irrational parts separately.
Solution:
Consider $(1+x\sqrt[3]{a}+y\sqrt[3]{b})^2$ for some rational $x,\,y$.
If $a$ or $b$ are other than $6$ and $6^2$ we would have $\sqrt[3]{a}$ and $\sqrt[3]{b}$ in the irrational part (as $(1+\color{blue}a+\color{red}b)^2=1+a^2+b^2+\color{blue}{2a}+\color{red}{2b}+2ab$), but we don't. So we search in the form
$$(1+x\sqrt[3]{6}+y\sqrt[3]{6^2})^2=\\
12xy+1+(x^2+2y)6^{2/3}+\sqrt[3]{6}(2x+6y^2)=\\
49+20\sqrt[3]{6}$$
$$\begin{cases}
12xy+1=49\\
2x+6y^2=20\\
x^2+2y=0
\end{cases}$$
$$\begin{cases}
x=-2\\y=-2\end{cases}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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proof that $ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} = \frac{3}{2} - \frac{1}{n} $
Proof that
$$ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} = \frac{3}{2} - \frac{1}{n} $$
by induction.
Proof
Base case: Statement clearly holds for $n = 1$. Now assume that statement holds for some $n = k$ and lets show that it implies $n = k + 1$ holds. The proof:
$$ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} + \frac{1}{n(n+1)} = \frac{3}{2} - \frac{1}{n} + \frac{1}{n(n+1)} = \frac{3}{2} - \frac{1}{n} + \frac{1}{n} - \frac{1}{n + 1} $$
$$ = \frac{3}{2} - \frac{1}{n+1} $$
Now the problem is I can't find the error. The statement doesn't clearly work for $ n = 2 $. However, the assumption seems to be correct since if I assume it's true for some $n = k$ and it is true for $ n = 1$? It shouldn't be possible to show that $p(n) \implies p(n+1)$ when $p(n)$ is true and $p(n+1)$ is false. This means that $p(n)$ has to be false in this case since when $p(n)$ is false then $p(n) \implies p(n+1)$ is tautology. The problem is I don't really see how? Isn't the whole point of induction to show that $p(n)$ is true for some specific $n = k$ (not all $n$) and then show $p(n+1)$ by assuming $p(n)$. Now when $p(n)$ is false you can show anything since it's tautology but how can you be sure $p(n)$ is true if you don't show it for all $n$? And wouldn't that defeat the purpose of induction (if you have already shown it's true for all $n$)?.
| The base of induction cannot be $n=1$ because then $1/(n(n-1))$ is not defined. For this sum you do not need induction. The sum is equal to $$(1-1/2)+(1/2-1/3)...(1/(n-1)-1/n)=1-1/n.$$
| {
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Comparing the expansions of $\sin ^3 (x)$ and $\cos ^3 (x)$ By using $(\cos x+i\sin x)^3=\cos 3x+ i \sin 3x$,
We have:
$\sin ^3x= \frac{3}{4} \sin x- \frac{1}{4} \sin 3x$
$\cos ^3 x=\frac{3}{4}\cos x + \frac{1}{4}\cos 3x$
If we compare the coefficients of the equations we see coefficient of $\sin x$ and $\cos x$ are both $\frac{3}{4}$ and coefficient of $\sin 3x $ and $\cos 3x$ are $-\frac{1}{4}$ and $\frac{1}{4}$ respectively. the coefficients are very similar to each other !
Is there any approach to make me understand by intuition that why the coefficients are so similar to each other in these two equations?
| Simply do the substitution $x \mapsto x + \frac \pi2$: Then $\sin x$ becomes $\sin (x + \frac\pi2)$, which is $\cos x$. And $\sin 3x$ becomes $\sin (3x + \frac{3\pi}2)$, which is $-\cos 3x$.
Note that similar substitutions do not exist for $\sin 2x$ and $\cos 2x$, so there's nothing similar for $\sin 2x$. That's where I got the false impression that you can't express $\sin 3x$ as a polynomial of $\sin x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate below expression of modular arithmetic? How to evaluate (10^18)%(10^9 + 7) using modular arithmetic? How to proceed and what will be the steps?
| (10^9 + 7) * (10^9 - k) = 10^18 + 7*10^9 - k*10^9 - 7k
k=7: (10^9 + 7) * (10^9 - 7) = 10^18 - 49
10^18 = (10^9 + 7) * (10^9 - 7) + 49
10^18 % (10^9 + 7) = 49
Thought process:
The form of this problem is a^2 % (a+k).
Recall (a+k)*(a-k) = a^2 - k^2
So a^2 % (a+k) = k^2 % (a+k)
Now because k is so small, k^2 % (a+k) = k^2.
| {
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How to deduce this factorization of $x^5+x+1$ by looking at $\int\frac{3x^4+2x^3-2x+1}{x^5+x+1}dx$? The question is:
$$\int\frac{3x^4+2x^3-2x+1}{x^5+x+1}dx$$
I tried a lot but couldn't solve it so I looked at the solution which is:
$$x^5+x+1=(x^2+x+1)(x^3-x^2+1)$$ and we can write $$3x^4+2x^3-2x+1=(x^3-x^2+1)+(3x^2-2x)(x^2+x+1)$$
which effectively reduces the integral to very simple ones.
My question is how they deduced the factorization of the denominator. After looking at the solution I think that if we put $x=1,x=\omega$ and $x=\omega^2$ we can deduce this but this was not immediately obvious to me. Is there some sort of hint you can get by looking at the integrand or is it simply a matter of less experience?
Any help would be appreciated.
| Hmm. I'd look at it and say "There's no rational root" because $x = \pm 1$ doesn't work. So there's some irrational root, $\alpha$, and two complex-conjugate pairs.
So there's no nice obvious linear factor I can write down.
Then I'd say "Maybe there's a quadratic factor." I can assume it's monic, so I'm looking to write
$$
x^5 + x + 1 = (x^2 + ax + b) (x^3 + px^2 + qx + r).
$$
from which I can expand to get
$$
x^5 + x + 1 = x^5+(p + a)x^4 + (q + ap + b) x^3 + (ra + bq)x + br
$$
if I've done the algebra right. Equating coefficients I see that
\begin{align}
0 &= a + p\\
0 &= q + ap + b\\
1 &= ra + bq\\
1 &= br
\end{align}
so $ p = -a$, and $r = \frac1b$,and these equations become
\begin{align}
0 &= a + -a\\
0 &= q - a^2 + b\\
1 &= \frac1b a + bq\\
1 &= b(1/b)
\end{align}
which simplify down to
\begin{align}
q &= a^2 - b\\
b &= a + b^2q\\
\end{align}
or
\begin{align}
q &= a^2 - b\\
0 &= b^2 q - b + a
\end{align}
That last equation is a quadratic in $b$ or $q = 0$. The first choice yields
$$
b = \frac{1 \pm \sqrt{1-4aq }}{2q}
$$
The second choice yields $b = a, q = 0$, from which we find that $r = 1, p = -1$, which is a nice solution, so we can stop looking at the first case. (Yay!)
| {
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Finding a volume using the method of cylindrical shells which is generated by a parabola This problem is from the 7th edition of the book "Calculus and Analytic Geometry" by George Thomas
and Ross Finney. It is problem number 3 of section 5.4
Problem:
Find the volume generated when the region bounded by the given curves and lines is revolved about the
x-axis. (Note: $x = 0$ is the y-axis and $y = 0$ is the x-axis.)
\begin{align*}
y &= 3x - x^2 \\
y &= x
\end{align*}
Answer:
I am going to use the method of cylindrical shells to compute the volume. Let $V$ be the volume we are asked to find.
The general form for $V$ when we are revolving about the x-axis is:
$$ V = \int_a^b 2 \pi x f(x) \,\, dx $$
where $2 \pi x$ represents the circumference of the region. Since we are going around the x-axis $a = 0$. To find $b$ be we set up the following equation:
\begin{align*}
3x - x^2 &= x \\
2x - x^2 &= 0 \\
x^2 - 2x &= 0 \\
x(x-2) &= 0 \\
x = 0 \,&\text{ or } \, x = 2 \\
V &= \int_0^2 2 \pi x ( 3x - x^2 - x) \,\, dx \\
V &= 2 \pi \int_0^2 x ( 2x - x^2 ) \,\, dx \\
\int_0^2 x ( 2x - x^2 ) \,\, dx &= \int_0^2 2x^2 - x^3 \,\, dx \\
\int_0^2 x ( 2x - x^2 ) \,\, dx &= \frac{2x^3}{3} - \frac{x^4}{4} \Big|_0^2 \\
\int_0^2 x ( 2x - x^2 ) \,\, dx &= \frac{2(8)}{3} - \frac{16}{4} = \frac{16}{3} - 4 \\
\int_0^2 x ( 2x - x^2 ) \,\, dx &= \frac{4}{3} \\
%
V &= 2 \pi \left( \frac{4}{3} \right) \\
V &= \frac{8 \pi}{3}
\end{align*}
The book's answer is: $\frac{56\pi}{15}$
Where did I go wrong?
| You said that you would use shells but you have actually used washers. The volume should be $$V=\int_0^2 2\pi y f(x) dx$$ Here $f(x)=(3x-x^2)-x$ is the length of the rotated vertical element and $y=\frac12 ((3x-x^2)+x)$ is the distance of the centroid (= mid-point) of the rotated element from the $x$ axis.
If you were to use shells then the volume would be $$V=\int_0^{9/4} 2\pi y g(y) dy$$ This is more difficult. The upper limit of integration is the peak of the parabola. $y$ is the distance of the rotated horizontal element from the $x$ axis. $g(y)$ is the length of the rotated horizontal element. It is a compound function. For $y=0..2$ it spans between the left side of the parabola and the straight line : $g(y)=x_0-x_1$ where $x_0=y$ and $x_1$ is the smaller solution of $x^2-3x+y=0$. For $y=2..9/4$ it spans between the left and right sides of the parabola : $g(y)=x_2-x_1$ where $x_2$ is the larger solution of $x^2-3x+y=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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contour integration with dogbone, branch cut Compute the following integral
$$\frac1{2\pi\mathrm{i}}\int_{|z|=2}\frac{\sqrt{z^2-1}}{z-3}\,\mathrm{d}z.$$
Taking a branch of $\sqrt{z^2-1}$, satisfying $\sqrt{z^2-1}>0$ for $z>0$, I tried this problem with a 'dogbone' contour and I get,
$$\int_C\frac{\sqrt{z^2-1}}{z-3}\,\mathrm{d}z=-2\int_{-1}^1 \frac{\sqrt{x^2-1}}{x-3}\,\mathrm{d}x,$$
considering the integrations at the branch points are tends to zero as $\epsilon$ goes to zero.
After that, I got stuck because I cannot use the Cauchy integral theorem because the singularity is outside the domain. Please give an idea for this kind of problem. I feel I am wrong, and I want to know the right figure for the contour.
| For $R>3$, Cauchy's Integral Theorem guarantees that
$$\begin{align}
\oint_{\text{Dogbone}}\frac{\sqrt{z^2-1}}{z-3}\,dz&=\oint_{|z|=2}\frac{\sqrt{z^2-1}}{z-3}\,dz\\\\
&=\oint_{|z|=R}\frac{\sqrt{z^2-1}}{z-3}\,dz-2\pi i \text{Res}\left(\frac{\sqrt{z^2-1}}{z-3},z=3\right)\\\\
&=-2\pi i \text{Res}\left(\frac{\sqrt{z^2-1}}{z-3},z=\infty\right)-2\pi i \text{Res}\left(\frac{\sqrt{z^2-1}}{z-3},z=3\right)
\end{align}$$
where the integral around the dog bone contour is taken counter clockwise.
The Residue at Infinity of $f(z)=\frac{\sqrt{z^2-1}}{z-3}$ is equal to the residue at $z=0$ of $-\frac1{z^2}f\left(\frac1z\right)=\frac{\sqrt{1-z^2}}{z^2(3z-1)}$. Therefore, we have
$$\begin{align}
\text{Res}\left(\frac{\sqrt{z^2-1}}{z-3},z=\infty\right)&=\text{Res}\left(-\frac1{z^2}\frac{\sqrt{1/z^2-1}}{1/z-3},z=0\right)\\\\
&=\lim_{z\to 0}\frac{d}{dz}\left(\frac{\sqrt{1-z^2}}{3z-1} \right)\\\\
&=-3
\end{align}$$
and the reside at $3$ is $2\sqrt 2$.
Hence, we find that
$$\oint_{\text{Dogbone}}\frac{\sqrt{z^2-1}}{z-3}\,dz=2\pi i (3-2\sqrt 2)$$
where we have tacitly selected the branch of the square root on which $\sqrt{z^2-1}$ is of positive sign when $z\in \mathbb{R}$, $z>1$.
| {
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Find the roots of $x^3 - 6x = 4$ This exercise is from the book Complex Analysis by Joseph Bak, and it says:
"Find the three roots of $x^{3}-6x=4$ by finding the three real-valued possibilities for $\sqrt[3]{2+2i}+\sqrt[3]{2-2i}$".
I know that these numbers were found by Cardan's method, but I don't understand why they give these numbers, because I found three real roots by common methods.
Pd: the three real roots are $-2$, $1-\sqrt{3}$ and $1+\sqrt{3}$.
| I find exponential form useful for a case known to have 3 real roots. One can get to a real expression with $\cos\left(\frac{\theta}{3}+\frac{2\pi}{3}k\right)$ and keep everything real from that point forward:
$$\begin{align*}\sqrt[3]{2+2i}+\sqrt[3]{2-2i} &= \sqrt[3]{2\sqrt{2}\left(\dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}}\right)}+\sqrt[3]{2\sqrt{2}\left(\dfrac{1}{\sqrt{2}}-i\dfrac{1}{\sqrt{2}}\right)}\\
\\
&= \left[2\sqrt{2}e^{i\left(\frac{\pi}{4}+2\pi k\right)}\right]^{\frac{1}{3}}+\left[2\sqrt{2}e^{-i\left(\frac{\pi}{4}+2\pi k\right)}\right]^{\frac{1}{3}} \quad k \in \{0,1,2\}\\
\\
&= \sqrt{2}\left[e^{i\left(\frac{\pi}{12}+\frac{2\pi}{3}k\right)}+e^{-i\left(\frac{\pi}{12}+\frac{2\pi}{3}k\right)}\right]\quad k \in \{0,1,2\} \\
\\
&= 2\sqrt{2}\cos\left(\frac{\pi}{12}+\frac{2\pi}{3}k\right) \quad k \in \{0,1,2\} \\
\\
&= 2\sqrt{2}\left[\cos\left(\frac{\pi}{12}\right)\cos\left(\frac{2\pi}{3}k\right) - \sin\left(\frac{\pi}{12}\right)\sin\left(\frac{2\pi}{3}k\right)\right] \quad k \in \{0,1,2\}\\
\\
&= \left(1+\sqrt{3}\right)\cos\left(\frac{2\pi}{3}k\right) -\left(\sqrt{3}-1\right) \sin\left(\frac{2\pi}{3}k\right) \quad k \in \{0,1,2\}\\
\\
&= \left\{1+\sqrt{3},-2 ,1-\sqrt{3}\right\}\\
\end{align*}$$
| {
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for which values of $n$, $1+n+n^{2}+n^{3}+n^{4}$ is a perfect square? I saw a question my number theory notes:
for which values of $n$ , $1+n+n^{2}+n^{3}+n^{4}$ is a perfect square where $"n"$ is non negative integer?
It was solved my professor using inequalities,but i think that his method is too long.Because of that, i did not write it here.
I tried to solve it using modular arithmetic but i could not do it.
I am looking for hints or solutions to solve it using modular artihmatic.
Note= I tried to use $(mod4)$ because for all $x^{2} \equiv \pm 1 (mod4)$ but i stuck in it.
| Following from the comments.
We want to find integers $(x,y)$ such that $x^4+x^3+x^2+x+1=y^2.$ Clearly if $x^4+x^3+x^2+x+1$ is a square then so is $f(x)=4x^4+4x^3+4x^2+4x+4.$
Note that $f(x)> (2x^2+x)^2$ $\forall x\in\mathbb R$ and also $f(x)=4x^4+4x^3+4x^2+4x+4=(2x^2+x+1)^2-(x+1)(x-3).$ Since the last term is non-positive in the interval $[-1,3]$ it follows that
$$(2x^2+x)^2<f(x)<(2x^2+x+1)^2$$
for all $x\notin[-1,3].$ So $f(x)$ cannot itself be a square for $x\notin[-1,3]$ since it is bounded by two consecutive squares.
Thus we only need to check for integers in $[-1,3]$ and we find the solutions $(x,y)=(3,11),(0,1)\space\space\text{and}\space\space(-1,1).$
| {
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Predetermine the middle term of an quadratic equation when there are two possibilities
Factorise completely:
*
*$2x^2+13x+15$
*$3x^2-5x+2$
*
*As i figured, there could be two ways to break the middle term. Either, $15x-2x$ or, $10x+3x$.
If I continue with $15x-2x$:
$2x^2+13x+15$
$2x^2+15x-2x+15$
$2x(x-1)+15(x+1)$
But If I go down with $10x+3x$:
$2x^2+13x+15$
$2x^2+10x+3x+15$
$2x(x+5)+3(x+5)$
I know the former is incorrect and the later one is correct. But I would really like to know how to predetermine this problem and avoid continuing with the former one.
NB: No. 2 has the exact same approach.
| There is a way to factorize a general quadratic trinomial
$$Ax^2 + Bx + C = (Dx + E)(Fx + G),$$
that is, assuming that the trinomial can be factored in the first place. (This method will also let you determine whether the trinomial cannot be factored at all.)
Set
$$Ax^2 + Bx + C = 0$$
and solve for $x$ using the Quadratic Formula:
$$x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}.$$
Note that, in order for this method to work, then the discriminant
$$d = B^2 - 4AC$$
needs to be a perfect square.
Now, let us apply this method to your first quadratic trinomial:
$$2x^2 + 13x + 15$$
Set $2x^2 + 13x + 15 = 0$, then solve for $x$ using the Quadratic Formula, where $A_1 = 2, B_1 = 13, C_1 = 15$:
$$x = \frac{-{B_1} \pm \sqrt{{B_1}^2 - 4{A_1}{C_1}}}{2{A_1}} = \frac{-13 \pm \sqrt{169 - 4\cdot{2}\cdot{15}}}{4} = \frac{-13 \pm \sqrt{49}}{4}.$$
Note that the discriminant
$$d_1 = {B_1}^2 - 4{A_1}{C_1} = 49$$
is a perfect square. Hence, this means that the first quadratic trinomial is indeed factorable.
We obtain the $x$-values
$$x = \frac{-13 \pm 7}{4} = -5, -\frac{3}{2}$$
which means that we have the factors
$$(x + 5)(2x + 3) = 2x^2 + 13x + 15.$$
I leave it as an exercise for you to apply the Quadratic Formula Method for factoring your second quadratic trinomial:
$$3x^2 - 5x + 2$$
| {
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Simplifying the following mathematical expression using a computer? I have this following beastly expression typed up very nicely in LaTeX formatting, as you can see. What is the easiest way that I can get a computer to simplify this expression for me? I have zero programming experience. I installed sagemath but it seems pretty complicated.
$W_{(1,1)}(t,v)=\frac{-t^{-2k}v^k}{3}(\frac{v^{\frac{3}{2}}-v^{\frac{-3}{2}}}{t^{\frac{3}{2}}-t^{\frac{-3}{2}}})(\frac{v^{\frac{1}{2}}-v^{\frac{-1}{2}}}{t^{\frac{1}{2}}-t^{\frac{-1}{2}}})+\frac{t^{-2k}v^k}{4}(\frac{v-v^{-1}}{t-t^{-1}})^2+\frac{t^{-2k}v^k}{12}(\frac{v^{\frac{1}{2}}-v^{\frac{-1}{2}}}{t^{\frac{1}{2}}-t^{\frac{-1}{2}}})-\frac{t^{-k}v^k}{4}(\frac{v^2-v^{-2}}{t^2-t^{-2}}) + \frac{t^{-k}v^k}{8}(\frac{v-v^{-1}}{t-t^{-1}})^2+\frac{t^{-k}v^k}{4}(\frac{v-v^{-1}}{t-t^{-1}})(\frac{v^{\frac{1}{2}}-v^{\frac{-1}{2}}}{t^{\frac{1}{2}}-t^{\frac{-1}{2}}})^2-\frac{t^{-k}v^k}{8}(\frac{v^{\frac{1}{2}}-v^{\frac{-1}{2}}}{t^{\frac{1}{2}}-t^{\frac{-1}{2}}})^4+\frac{-v^kt^{k}}{4}(\frac{v^2-v^{-2}}{t^2-t^{-2}})+\frac{v^kt^{k}}{3}(\frac{v^{\frac{3}{2}}-v^{\frac{-3}{2}}}{t^{\frac{3}{2}}-t^{\frac{-3}{2}}})(\frac{v^{\frac{1}{2}}-v^{\frac{-1}{2}}}{t^{\frac{1}{2}}-t^{\frac{-1}{2}}})+\frac{v^kt^{k}}{8}(\frac{v-v^{-1}}{t-t^{-1}})^2-\frac{v^kt^{k}}{4}(\frac{v-v^{-1}}{t-t^{-1}})(\frac{v^{\frac{1}{2}}-v^{\frac{-1}{2}}}{t^{\frac{1}{2}}-t^{\frac{-1}{2}}})^2+\frac{v^kt^{k}}{24}(\frac{v^{\frac{1}{2}}-v^{\frac{-1}{2}}}{t^{\frac{1}{2}}-t^{\frac{-1}{2}}})^4$
| Wolfram|Alpha supports Latex. Unfortunately, the engine didn't seem to understand your query, perhaps because of its length. If possible, I would suggest splitting the expression up into chunks, and entering them one by one into Wolfram|Alpha.
| {
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Real roots of $x^7+5x^5+x^3−3x^2+3x−7=0$? The number of real solutions of the equation,
$$x^7+5x^5+x^3−3x^2+3x−7=0$$
is
$$(A) 5 \quad (B) 7 \quad (C) 3 \quad (D) 1.$$
Using Descartes rule we may have maximum no. of positive real roots is $3$ and negative real root is $0.$ So there can be either $3$ real roots or $1$ real root but how to conclude what will be the no. of real roots exactly. Can you please help me?
| Descare's rukles say that there will be at most 3 positive roots and no negative root. Since $f(1)=0$, so $(x-1)$ is factor of $f(x)=x*7+5x^5+x^3-3x^2+3x-7$
Note that $f(x)=(x-1)(x^6+x^5+6x^3+7x^2+4x+7)=(x-1)g(x)$, It can be seen that
the number of sign changes in $g(x)$ are none, so $g(x)$ will have no positive roots. Finally, $f(x)=0$ will have exactly one real root, namely $x=1$.
| {
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If $\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$ ; $\frac{1}{a} + \frac{1}{c} + \frac{1}{a+y}$ ; $\frac{1}{a} + \frac{1}{x} + \frac{1}{y} = 0$.
If $a \neq 0$ , $b \neq 0$ , $c \neq 0$ and if :- $\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$ ; $\frac{1}{a} + \frac{1}{c} + \frac{1}{a+y}=0$ ; $\frac{1}{a} + \frac{1}{x} + \frac{1}{y} = 0$ , find $(a+b+c)$ .
What I Tried :- No information is given about $x$ and $y$ . So I thought of putting $x = y = 1$ , and this silly thing came out in the end .
Now, as $x = y = 1$ , I have $a = \frac{1}{-2}$ from the $3$rd equation .
So from the $1$st equation I get :- $$\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$$
$$ \rightarrow -2 + \frac{1}{b} + 2 = 0$$
$$ \rightarrow \frac{1}{b} = 0$$
This definitely looks absurd (also it's given that $b \neq 0$), so I guess putting $x = y = 1$ was a big mistake .
I don't have any other cool ideas for now as I see that doing it algebraically is going to include a lot of simplification and stuffs, and since there are $5$ variable there must be some shortcut of this .
Can anyone help?
| From the first and the second equality we obtain:
$$x=-\frac{a(a+2b)}{a+b}$$ and $$y=-\frac{a(a+2c)}{a+c}.$$
Thus, $$\frac{1}{a}-\frac{a+b}{a(a+2b)}-\frac{a+c}{a(a+2c)}=0$$ or
$$(a+2b)(a+2c)=(a+b)(a+2c)+(a+c)(a+2b)$$ or
$$a(a+b+c)=0$$ or $$a+b+c=0.$$
| {
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Where did I go wrong in my proof that for all $n \in \mathbb{Z}^+$, $\sqrt{2} < a_n$ with $(a_n)$ being a particular recursive sequence? I am in Introduction to Abstract Math, and I have taken Calculus 1, Linear Algebra, and Discrete Math. I got stuck with an apparent contradiction in my proof and wanted to know where I messed up:
Consider the sequence $(a_n)$ defined recursively by:
$$a_1 = 3, \:\:\:\:\:\:\: a_{n+1} = \frac{a_n}{2}+\frac{1}{a_n} \:\:\:\:\:\:\: (n \geq 1).$$
Prove that $\sqrt{2}$ is a lower bound for $(a_n)$.
This can be rewritten as: Prove for all $n \in \mathbb{Z}^+$, $\sqrt{2} < a_n$, which we can prove with strong induction.
Base Step
Check for $n=1$.
$$a_1 = 3 > \sqrt{2}$$
Check for $n=2$.
$$a_2 = \frac{a_1}{2} + \frac{1}{a_1} = \frac{3}{2} + \frac{1}{3} = \frac{11}{6} > \sqrt{2}$$.
Therefore, the statement holds true for $n=1, n=2$.
Inductive Step
Assume that $a_n > \sqrt{2}$ for $1 \leq n \leq k$ with $k \in \mathbb{Z}^+$. Prove $a_{k+1} > \sqrt{2}$.
$$a_k = \frac{a_{k-1}}{2} + \frac{1}{a_{k-1}} > \sqrt{2}$$
$$a_{k+1} = \frac{a_k}{2} + \frac{1}{a_k}$$
$$a_{k+1} = \frac{\frac{a_{k-1}}{2} + \frac{1}{a_{k-1}}}{2} + \frac{1}{\frac{a_{k-1}}{2} + \frac{1}{a_{k-1}}}$$
$$a_{k+1} = \frac{2a_{k-1}}{a_{k-1}^2 + 2}+\frac{a_{k-1}}{4}+\frac{1}{2a_{k-1}}$$
$$a_{k+1} = \frac{2a_{k-1}}{a_{k-1}^2 + 2}+\frac{1}{2}(\frac{a_{k-1}}{2}+\frac{1}{a_{k-1}})$$
$$a_{k+1} = \frac{2a_{k-1}}{a_{k-1}^2 + 2}+\frac{1}{2}(a_k)$$
$$a_{k+1} = (\frac{2a_{k-1}}{a_{k-1}^2 + 2} \times \frac{1}{a_k})a_k+\frac{1}{2}(a_k)$$
$$a_{k+1} = a_k (\frac{2a_{k-1}}{a_{k-1}^2 + 2} \times (\frac{2}{a_{k-1}}+a_{k-1}) +\frac{1}{2})$$
Here's where I got stuck. I plugged it into Wolfram Alpha to do the algebra for that step and see if there was equivalent forms, and I saw that what was in the parenthesis was equal to $\frac{5}{2}$, which gives
$$a_{k+1} = \frac{5a_{k}}{2}$$
which doesn't make sense as the sequence would be increasing way too fast.
Where did I mess up in my proof?
| You didn't really "mess up". You just got stuck because this approach leads nowhere. However, there is certainly something wrong in what you typed in Wolfram Alpha.
You could do this, assuming $a_n>0$:
$$a_{n+1}-\sqrt{2}=\frac{a_n}{2}+\frac{1}{a_n}-\sqrt{2}\\=\frac{1}{2}\left(a_n+\frac{2}{a_n}-2\sqrt{2}\right)=\frac{1}{2}\left(\sqrt{a_n}-\sqrt{\frac{2}{a_n}}\right)^2\ge0$$
And there is equality only if $a_n=\sqrt{2}$. Given that $a_1=3>\sqrt{2}$, you get that $a_n>\sqrt{2}$ for all $n\ge1$.
| {
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An urn has $4$ red, $6$ white and $3$ blue marbles, what is the probability of drawing with replacement $2$ red, $2$ white and $2$ blues? An urn has $4$ red, $6$ white and $3$ blue marbles, what is the probability of drawing with replacement $2$ red, $2$ white and $2$ blues? The order for marbles been picked up does not matter.
For example, I know the possible combination of selecting $6$ marbles from the total set should be ${13+6-1 \choose 6}$ because the selection comes with replacement. But I am not sure the numerator term. I have thought it is possible combinations from each kind of marbles and then multiply them all together: ${4+2-1 \choose 2}{6+2-1 \choose 2}{3+2-1 \choose 2}$.
| The selections you are making are not equally likely to occur. For instance, there are $3^6 = 729$ ways for all six marbles to be blue since there are three ways to select a blue marble on each draw. However, there are
$$\binom{6}{2}4^2\binom{4}{3}6^3\binom{1}{1}3^1 = 1,036,800$$
ways to select two red, three white, and two blue marbles since we must choose two of the six positions for the two red marbles, one of the four red marbles for each of those positions, three of the remaining four positions for the three red marbles, one of the six red marbles for each of those positions, and choose one of the three blue marbles for the remaining position.
There are $4 + 6 + 3 = 13$ choices for each of the $6$ selections, so there are $13^6$ possible sequences of marbles. Of these,
$$\binom{6}{2}4^2\binom{4}{2}6^2\binom{2}{2}3^2$$
contain exactly two marbles of each color. To see this, observe that we must choose which two of the six positions will be filled with red marbles, one of the four red marbles for each of those two positions, two of the remaining four positions for the white marbles, one of the six white marbles for each of those two positions, both of the remaining two positions for the blue marbles, and one of the three blue marbles for each of those positions. Hence, the desired probability is
$$\frac{\dbinom{6}{2}4^2\dbinom{4}{2}6^2\dbinom{2}{2}3^2}{13^6}$$
| {
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How to solve this absolute value inequality? I am new to absolute value inequalities.I was looking trough a book and I found this inequality,
I know a little bit about absolute value inequalities.
The inequality is given below:
$$
\left| \frac{n+1}{2n+3} - \frac{1}{2} \right| > \frac{1}{12}, \qquad n \in \mathbb{Z}
$$
| For case 1:
$ \frac{n+1}{2n+3} - \frac{1}{2} > \frac{1}{12} $
$ \frac{n+1}{2n+3} - \frac{2n+3}{2(2n+3)} > \frac{1}{12} $
$ \frac{-1}{2(2n+3)} > \frac{1}{12} $
$ \frac{-6}{2n+3} > 1 $
$ \Rightarrow 2n+3 < 0 $ and $ 2n+3 > -6 $
$ \frac{-9}{2} > n < \frac{-3}{2} $
| {
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$\frac{a}{a^{2}+b^{2}+2}+\frac{b}{b^{2}+c^{2}+2}+\frac{c}{c^{2}+d^{2}+2}+\frac{d}{d^{2}+a^{2}+2}\le 1$ Let be a,b,c,d non negative real numbers. Prove that :
$$\frac{a}{a^{2}+b^{2}+2}+\frac{b}{b^{2}+c^{2}+2}+\frac{c}{c^{2}+d^{2}+2}+\frac{d}{d^{2}+a^{2}+2}\le 1$$
I tried many attempts but still can't find the result. Attempt 1 :
$$\sum_{cyc}\frac{a}{a^2+b^2+2}\leq\frac{1}{2}\sum_{cyc}\frac{a}{ab+1}\leq\frac{1}{4}\sum_{cyc}\sqrt{\frac{a}{b}}$$
but the final term is superior to 1
Attempt 2 :
$$\sum_{cyc}\frac{a}{a^2+1+b^2+1}\leq\frac{1}{2}\sum_{cyc}\frac{a}{a+b}=2-\frac{1}{2}\sum_{cyc}\frac{b}{a+b}=2-\frac{1}{2}\sum_{cyc}\frac{1}{1+\frac{a}{b}}.$$
So I need to prove that: $$\sum_{cyc}\frac{1}{1+\frac{a}{b}}\geq2$$
so taking $x=\frac{a}{b}$$y=\frac{b}{c}$ $z=\frac{c}{d}$ $t=\frac{d}{a}$ with $xyzt=1$
The inequality I need to prove is
$$\sum_{cyc}\frac{1}{1+x}\geq2$$
I tried using Jensen inequality to $f(x)=\frac{1}{1+x}$ but it doesn't work.
| By AM-GM twice we obtain:
$$\sum_{cyc}\frac{a}{a^2+b^2+2}=\sum_{cyc}\frac{a}{a^2+1+b^2+1}\leq\frac{1}{2}\sum_{cyc}\frac{a}{\sqrt{(a^2+1)(b^2+1)}}=$$
$$=\frac{1}{2}\sum_{cyc}\sqrt{\frac{a^2}{a^2+1}\cdot\frac{1}{b^2+1}}\leq\frac{1}{4}\sum_{cyc}\left(\frac{a^2}{a^2+1}+\frac{1}{b^2+1}\right)=1.$$
| {
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Show $\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx=\frac{5\pi^3}{64}+\frac\pi{16}\ln^22$ Tried to evaluate the integral
$$I=\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx$$
and managed to show that
\begin{align}
I &= \int_0^1 \frac{\ln^2x}{(x+1)^2+1} \, dx + \int_0^1 \frac{\ln^2x}{(x+1)^2+x^2} \, dx\\
&= \int_0^1 \frac{\ln^2x}{(x+1+i)(x+1-i)} \, dx
+ \int_0^1 \frac{\ln^2x}{(x+1+ix )(x+1-ix )} \, dx\\
&= -2\operatorname{Im}\operatorname{Li}_3\left(-\frac{1+i}2\right)
-2\operatorname{Im} \operatorname{Li}_3(-1-i)
\end{align}
which is equal to $ \frac{5\pi^3}{64}+\frac\pi{16}\ln^22$.
It is perhaps unnecessary, though, to resort to evaluation in complex space. I would like to work out an elementary derivation of this integral result.
| We can use the integral of
\begin{equation}
f(z)=\frac{\ln^3z}{(z+1)^2+2}
\end{equation}
along the classical keehole contour: above the positive real axis , along the large circle, back from below along the positive axis and avoiding the origin by a small circle. Except the horizontal parts, the other contributions can be shown to vanish. Above the axis, $z=x+i0^+$ and $\ln^3(z)=\ln^3(x)$, while below, $z=x+i0^-$ and $\ln^3(z)=(\ln(x)+2i\pi)^3$. The integral can then be written as
\begin{align}
J&=\int_0^\infty\frac{\ln^3x}{(x+1)^2+1}dx-\int_0^\infty\frac{(\ln(x)+2i\pi)^3}{(x+1)^2+1}dx
\end{align}
Considering the imaginary part of the integral,
\begin{align}
\Im J&=-3 \left( 2\pi \right)\int_0^\infty\frac{\ln^2x}{(x+1)^2+1}dx+\left( 2\pi \right)^3\int_0^\infty\frac{1}{(x+1)^2+1}dx\\
\Im J&=-6\pi I+2\pi^4
\end{align}
The poles of the function are $z_{\pm}=-1\pm i$, or $z_+=\sqrt{2}e^{i3\pi/4},z_-=\sqrt{2}e^{i5\pi/4}$. Corresponding residues $R_\pm=\ln^3(z_\pm)/\left( 2(1+z_\pm) \right)$, we deduce
\begin{equation}
J=2i\pi\left[\frac{\left( \ln\sqrt{2} +3i\pi/4\right)^3}{2i}-\frac{\left( \ln\sqrt{2} +5i\pi/4\right)^3}{2i}\right]
\end{equation}
Then,
$$\Im J=\frac{49\pi^4}{32}-\frac{3\pi^2}{8}$$
Finally,
\begin{equation}
I=\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx = \frac{5\pi^3}{64}+\frac\pi{16}\ln^22
\end{equation}
| {
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Given $|b-c|\leq a\leq b+c$ show that $\frac{a}{1+a}\leq\frac{b}{1+b}+\frac{c}{1+c}$ I am trying to show the following:
Given that $|b-c| \leq a \leq b+c$ show that:
$$\frac{a}{1+a} \leq \frac{b}{1+b} + \frac{c}{1+c}$$
So far I have done the following:
$a/(1+a) \leq b/(1+b) + c/(1+c) \iff$
$1-1/(1+a) \leq 2 - 1/(1+b) -1/(1+c) \iff$
$1/(1+b) -1/(1+c) \leq 1 + 1/(1+a) \iff$
$ (2+b+c)/(1+b)(1+c) \leq (2+a)/(1+a) \iff$
$ (2+b+c)(1+a) \leq (2+a)(1+b)(1+c) \iff$
$ a \leq bc(2+a)$
I am not really sure how to proceed from here- any help or hints would be much appreciated.
| I will assume that $a,b,c > 0$ because without such an assumption we cannot even have $1+b$ and $1+c$ in the denominator.
$f(x)=\frac x {1+x}$ is an increasing function of $x$ on $[0,\infty)$. Hence $\frac a {1+a} \leq \frac {b+c} {1+b+c} =\frac b {1+b+c}+\frac c {1+b+c} \leq \frac b {1+b} +\frac c {1+c} $.
| {
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Find the area between $r=1$ and $r=3\cos\theta$
Find the area between $r=1$ and $r=3\cos\theta$.
I squared both sides to get $r^2 = 1$, then did $r^2(\cos^2 \theta + \sin^2 \theta) = (r \cos \theta)^2 + (r \sin \theta)^2$$ = x^2+y^2 = 1$ to get $x^2+y^2=1$.
For $r = 3 \cos \theta$, I multiplied by $r$ on both sides to get $r^2 = 3r \cos \theta$, then substituted $x = r \cos \theta$ to get $x^2+y^2 =3x$. However, I don't know if it is easier to do it this way.
If not, how can I find this area?
| Geometric solution:
Rewrite $x^2+y^2=3x$ as $x^2-3x+(\frac{3}{2})^2 + y^2 = (\frac{3}{2})^2 \Rightarrow (x-\frac{3}{2})^2 + y^2 = \frac{9}{4}$. Then solving for the intersections:
$$(x-\frac{3}{2})^2 + y^2 = \frac{9}{4} \tag{1}$$
$$x^2+y^2 = 1 \tag{2}$$
$(1) - (2)$ gives $-3x + \frac{9}{4} = \frac{5}{4} \Rightarrow x = \frac{1}{3}$. Substituting back into equation $(2)$, we get that $y = ±\frac{2 \sqrt2}{3}$.
The area of $\Delta AFC$ can be found in multiple ways, including the shoelace formula. Using this gives the area as $\frac{1}{\sqrt2}$.
Now the central angle of sector $ACF$ is just $\tan^{-1} \frac{2 \sqrt{2}/3}{3/2 - 1/3} = \tan^{-1} \frac{4 \sqrt2}{7}$ radians. Therefore its area is $\frac{1}{2} \cdot (\frac{3}{2})^2 \tan^{-1} \frac{4 \sqrt2}{7} = \frac{9}{8} \tan^{-1} \frac{4 \sqrt2}{7}$, so area $a$ is just $\frac{9}{8} \tan^{-1} \frac{4 \sqrt2}{7} - \frac{1}{\sqrt2}$.
Finally, the central angle of sector $AFB$ is $\tan^{-1} \frac{2 \sqrt{2}/3}{1/3}
= \tan^{-1} 2 \sqrt{2}$, so the area of that is $\frac{1}{2} \tan^{-1} 2 \sqrt{2}$.
Therefore circular area $AFB$ is $\frac{9}{8} \tan^{-1} \frac{4 \sqrt2}{7}- \frac{1}{\sqrt2} + \frac{1}{2} \tan^{-1} 2 \sqrt{2}$, and by symmetry, the total area is just twice that, or:
$$\frac{9}{4} \tan^{-1} \frac{4 \sqrt2}{7} -\sqrt{2} + \tan^{-1} 2 \sqrt{2} \approx 1.346.$$
| {
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By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$ By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x$
$\cos^6x + \sin^6x = -3\cos^2x\sin^2x$
$\cos^6x + \sin^6x = (-3/2)(2\cos^2x\sin^2x)$
$\cos^6x + \sin^6x = (-3/2)(\sin^22x)$
How can I get it into $ 1 - (3/4)\sin^2(2x)$?
| Following along your approach,
After expansion,
$$(\cos^2x + \sin^2x)^3 = \cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$$
$$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x$$
$$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x = 1$$
Hence,
$$\cos^6x + \sin^6x = 1 - 3\cos^2x\sin^2x$$
$$ = 1 - \frac{3}{4}(4\sin^2x\cos^2x)$$
$$ = 1 - \frac{3}{4}\sin^22x$$Hence Proved. There just were $2$ errors in your solution, that I have corrected here.
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Find $\sqrt a + \sqrt b + \sqrt c$ only in terms of $p$ .
If :- $$a^2x^3 + b^2y^3 + c^2z^3 = p^5$$
$$ax^2 = by^2 = cz^2$$
$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{p}$$ Find $\sqrt a + \sqrt b + \sqrt c$ in terms of $p$ .
What I Tried :- How do you use the information here? I assumed that :-
$$ax^2 = by^2 = cz^2 = k$$
And then I can write the $1$st equation as :- $$k(ax + by + cz) = p^5$$
But I guess that just bring another extra variable into the scene so it does not help .
How can I use the $3$rd equation? I can write it as $\frac{xyz}{xy + yz + zx} = p$ . How do you use it?
| Now, for positive variables $a=\frac{k}{x^2},$ $b=\frac{k}{y^2}$, $c=\frac{k}{z^2}$ and $$a^2x^3+b^2z^3+c^2z^3=k^2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=p^5,$$ which gives $$k^2=p^6$$ or
$$k=p^3.$$
Id est, $$\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{k}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\sqrt{p}.$$
| {
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Is the representation of any prime of the form $6n+1$ as $a^2+3b^2$ essentially unique? Two well-known results in number theory are:
Fermat's $4n+1$ theorem: Every prime of the form $4n+1$ can be represented as $a^2+b^2 (a,b \in\mathbb{N})$.
Euler's $6n+1$ theorem: Every prime of the form $6n+1$ can be represented as $a^2+3b^2 (a,b \in\mathbb{N})$.
Looking at the Mathworld entries on these theorems here and here, I notice that representation of primes of the form $4n+1$ is stated to be unique (up to order), but that there is no mention of uniqueness in respect of representation of primes of the form $6n+1$. Uniqueness does however seem to hold at least for small primes of this form.
Question: Is the representation of any prime of the form $6n+1$ as $a^2+3b^2$ essentially unique?
If this is the case then a reference to a proof would be appreciated.
| If you require $a, b \in \mathbb{N}$ then the representation is literally unique, and we can argue as Daniel Fischer does in the comments: the ring of Eisenstein integers $\mathbb{Z}[\omega]$, where $\omega = \frac{-1 + \sqrt{-3}}{2}$ is a primitive third root of unity, is a unique factorization domain, and the primes congruent to $1 \bmod 6$ admit a further factorization
$$p = (x - y \omega)(x - y \omega^2) = x^2 + xy + y^2$$
where $x - y \omega$ is a prime in $\mathbb{Z}[\omega]$ and $x - y \omega^2$ is its conjugate. (This can be proven by inspecting the quotient $\mathbb{Z}[\omega]/p \cong \mathbb{F}_p[\omega]/(\omega^2 + \omega + 1)$.) We can write $x - y \omega = x + \frac{y}{2} - \frac{y}{2} \sqrt{-3}$ which gives
$$p = \left( x + \frac{y}{2} \right)^2 + 3\left( \frac{y}{2} \right)^2$$
and this representation consists of integers as long as $y$ is even. Now, $x - y \omega$ in the factorization above is unique up to multiplication by units, and the units of $\mathbb{Z}[\omega]$ are $\pm 1, \pm \omega, \pm \omega^2$. Multiplying by $\pm 1$ doesn't affect the values of $|x|$ and $|y|$ so now we inspect only the results of multiplying by $\omega$ and $\omega^2$. This gives
$$(x - y \omega) \omega = x \omega - y \omega^2 = x \omega + y (\omega + 1) = y + (x + y) \omega$$
$$(x - y \omega) \omega^2 = x \omega^2 - y = - x (\omega + 1) - y = (- x - y) - x \omega.$$
This means that if $p = a^2 + 3b^2 = (a + \sqrt{-3} b)(a - \sqrt{-3} b)$ then $b$ can only take on one of the six values $\pm \frac{y}{2}, \pm \frac{x+y}{2}, \pm \frac{x}{2}$, and some casework gives that depending on the parities of $x$ and $y$, exactly one of $x, y, x + y$ is even (the case where $x$ and $y$ are both even can't occur because then $p$ would be divisible by $4$). So $b$ is (the absolute value of) this unique even value divided by $2$, which uniquely determines $a$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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find the largest integer $m$ such that $2^m$ divides $3^{2n+2}-8n-9$
find the largest integer $m$ such that $2^m$ divides $\space 3^{2n+2}-8n-9$ when $n$ is a natural number.
If the answer was known it will be easy induction.
I started out like this :
$\space 3^{2n+2}-8n-9=9(3^{2n}-1)-8n=9\underbrace{(3^n-1)(3^n+1)}-8n$
Now we have $\frac{3^n-1}{3-1}$ is some integer (sum of GP),or
$ 2|\space 3^n-1$
also we have $3^n+1$ is even ,or
$2|3^n+1....(3)$
From this we conclude $4|(3^n-1)(3^n+1) ...(1)$
Let n be even then $3^n-1=3^{2m}-1=(3^m-1)(3^m+1)$,
by $(1)$ :
$4|(3^m+1)(3^m-1)$ meaning $4|3^n-1...........(2)$
combining $(2),(3)$ we have $8|3^{2n+2}-8n-9$
Similarly i was able to work out the same when $n=2m+1$ by noting that $3^n+1=3^{2m+1}+1$ is divisible by $4$.
I got the largest integer as $3$.
But i am wrong as the MCQ did not have the option $m=3$
how do i proceed.
Note: I have not learnt about fermat's little theorem
Also i am looking for Hints rather than complete solutions .use of >! may help
| In such problems, it's common to check for some small values to see if there's a pattern early on. Let's to do that here:
$$\begin{align}
n=1&: 3^4 - 8- 9 = 64 = 2^6 \\
n=2&: 3^6 - 16 - 9 = 704 = 64\cdot 11 = 2^6 \cdot11 \\
n=3&: 3^8 - 24 - 9 = 6528 = 128\cdot 51 = 2^7 \cdot51 \\
n=4&: 3^{10} - 32 - 9 = 59008 = 128\cdot 461 = 2^7 \cdot461
\end{align}$$
So far we are seeing that $2^6 = 64$ does the job. Since you said that you could do induction if you knew the answer, I will let you work it out and add the type with spoilers below:
If $a_n = 3^{2n+2} - 8n - 9$ then $a_{n+1} = 9a_n + 64(n+1)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Advice on integrating $\int \frac{x^2 +n(n-1)}{(x \sin x +n \cos x)^2} dx$ Our integral is:
$$\int \frac{x^2 +n(n-1)}{(x \sin x + n\cos x)^2} dx$$
I considered that this can be turned into some sort of quotient differential:
$$ d \frac{u}{v} = \frac{ v du - u dv}{v^2}$$
Now, comparing this with our integral:
$$ v = x \sin x + \cos x$$
And,
$$ dv = \sin x + x \cos x - n \sin x = (1-n) \sin x + x \cos x$$
Now the problem is I can't figure out $u$ / make the numerator of the form $ vdu - u dv$... what do I do next?
| Instead of looking for a function $u$ such that the integrand is the derivative of the quotient of two functions, I will present a different approach.
\begin{align*}
\int \frac{x^2 +n(n-1)}{(x \sin x + n\cos x)^2} \; \mathrm{d}x &= \int \frac{x^2 +n(n-1)}{\left(\sqrt{x^2+n^2} \cos \left(x-\arctan{\left(\frac{x}{n}\right)}\right)\right)^2} \; \mathrm{d}x \tag{1}\\
&=\underbrace{\int \frac{x^2 +n(n-1)}{\left(x^2+n^2\right) \cos^2 \left(x-\arctan{\left(\frac{x}{n}\right)}\right)} \; \mathrm{d}x}_{t=x-\arctan{\left(\frac{x}{n}\right)}}\\
&= \int \frac{x^2 +n(n-1)}{\left(x^2+n^2\right) \cos^2 \left(t\right)} \; \left(\frac{x^2+n^2}{x^2+n(n-1)}\; \mathrm{d}t\right) \\
&= \int \sec^2{t} \; \mathrm{d}t \\
&= \tan{t}+C \\
&= \tan{\left(x-\arctan{\left(\frac{x}{n}\right)}\right)}+\mathrm{C} \\
&= \frac{n \sin{x}-x\cos{x}}{n\cos{x}+x\sin{x}}+\mathrm{C} \\
\end{align*}
$(1)$: $\mathrm{A}\sin{x}+\mathrm{B}\cos{x}=\sqrt{\mathrm{A}^2+\mathrm{B}^2}\cos{\left(x-\arctan{\left(\frac{\mathrm{A}}{\mathrm{B}}\right)}\right)}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
prove that $\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$
prove that $$\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$$ for positives $a,b,c$
Attempt: By C-S; $$\left(\sum_{cyc}\frac{a}{b^2+c^2} \right) \left(\sum_{cyc} a(b^2+c^2) \right)\ge {(a+b+c)}^2$$ .
or as inequality is homogenous we take $a+b+c=1$.
or we have to prove (i am skipping the steps as it is just algebra) :
$$ 5(ab+bc+ca-abc)\ge 4(1+ab+bc+ca)(ab+bc+ca-3abc)$$
But i am not able to prove this by expanding.
How do i proceed?
Other methods are welcome!
| Proof by full expanding.
After your work we need to prove that:
$$5\left(\left(\sum_{cyc}a\right)^3\sum_{cyc}ab-abc\left(\sum_{cyc}a\right)^2\right)\geq4\left(\left(\sum_{cyc}a\right)^2+\sum_{cyc}ab\right)\left(\sum_{cyc}a\sum_{cyc}ab-3abc\right)$$ or
$$5\sum_{cyc}(a^2+2ab)\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)\geq4\sum_{cyc}(a^2+3ab)\sum_{cyc}(a^2b+a^2c)$$ or
$$\sum_{cyc}(a^2+2ab)\sum_{cyc}(a^2b+a^2c)+10abc\sum_{cyc}(a^2+2ab)\geq4\sum_{cyc}ab\sum_{cyc}(a^2b+a^2c)$$ or
$$\sum_{cyc}(a^2-2ab)\sum_{cyc}(a^2b+a^2c)+10abc\sum_{cyc}(a^2+2ab)\geq0$$ or
$$\sum_{cyc}(a^4b+a^4c+a^3b^2+a^3c^2+2a^2b^2c-2a^3b^2-2a^3c^2-4a^3bc-4a^2b^2c+$$
$$+10a^3bc+10a^2b^2c)\geq0$$ or
$$\sum_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2+6a^3bc+8a^2b^2c)\geq0,$$ which is true by Muirhed.
Of course, I could write a last line only, but I showed, how we can get it: three minutes of work!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\int_0^\infty \frac{x^a-1}{x^2-1}\, dx=\frac{\pi}{2}\tan\frac{a\pi}{2}$ for $0\lt a\lt 1$ How could I prove
$$\int_0^\infty \frac{x^a-1}{x^2-1}\, dx=\frac{\pi}{2}\tan\frac{a\pi}{2}$$
where $0\lt a\lt 1$?
I only thought of splitting it like this:
$$\int_0^\infty \frac{x^a-1}{x^2-1}\, dx=\int_0^\infty \frac{x^a}{x^2-1}\, dx-\int_0^\infty \frac{dx}{x^2-1},$$
but there's a new unremovable singularity. Partial fraction decomposition would be helpful for natural $a$ but I can't think of anything else here...
| Your integral is$$\int_0^1\frac{1-x^a}{1-x^2}dx+\int_0^1\frac{y^{-a}-1}{y^{-2}-1}\frac{dy}{y^2}=\int_0^1\frac{x^{-a}-x^a}{1-x^2}dx.$$For integer $n\ge0$,$$\int_0^1(x^{2n-a}-x^{2n+a})dx=\frac{1}{2n-a+1}-\frac{1}{2n+a+1}=\frac{2a}{(2n+1)^2-a^2}.$$Now finish with a result proven here, with $z=a/2$:$$\sum_{k\ge0}\frac{4a}{(2k+1)^2-a^2}=\pi\tan\frac{a\pi}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Definite integral evaluation of $\frac{\sin^2 x}{2^x + 1}$
Compute
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{2^x + 1} \ dx.$$
So looking at the limits I first checked whether the function is even or odd by substituting $x = -x$ but this gave me:
$$\frac{\sin^2 x}{2^x + 1} \cdot 2^x$$
so the function is clearly neither odd or even and I don't know what to infer from this. I can't think of a good substitution either.
Any help would be appreciated!
| Let $$\;I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{\sin^2 x}{2^x+1} dx\;.$$
By substituting $\;x\to -x\;$ we get that $$I=-\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \dfrac{\sin^2 x}{2^{-x}+1} dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{2^x\sin^2 x}{2^x+1} dx=\\=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\sin^2 x -\dfrac{\sin^2 x}{2^x+1}\right)dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2 x\ dx-I\;.$$
Since
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2 x\ dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{1-\cos(2x)}{2}dx=\\=\frac{x}{2}-\dfrac{\sin(2x)}{4}\bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\cfrac{\pi}{2}\;,$$
we get that $\;I=\cfrac{\pi}{2}-I\;,\;$ hence
$I=\cfrac{\pi}{4}\;.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that $\sum_{k=0}^{n} \binom{n}{k} ka^k = an(a+1)^{n-1}$ Problem:
Show that $\sum_{k=0}^{N} \binom{N}{k} ka^k = aN(a+1)^{N-1}$
Attempt:
I tried using induction but got stuck.
$N=0$ implies $\sum_{k=0}^{0} \binom{0}{k} ka^k = 0$
$N=1$ implies $\sum_{k=0}^{1} \binom{1}{k} ka^k = a = a(1)(a+1)^{1-1}$
$N=2$ implies $\sum_{k=0}^{2} \binom{2}{k} ka^k = 2a(a+1)^{2-1}$
Assume that for $N=n$, $\sum_{k=0}^{n} \binom{n}{k} ka^k = na(a+1)^{n-1}$. Then, for $N=n+1$, we find
$$\sum_{k=0}^{n+1} \binom{n+1}{k} ka^k = \sum_{k=0}^{n+1} \left(\binom{n}{k-1}+\binom{n}{k}\right) ka^k$$
Splitting up the sum on the right gives
$$\sum_{k=0}^{n+1} \binom{n}{k-1}ka^k + \sum_{k=0}^{n} \binom{n}{k}ka^k + \binom{n}{n+1}ka^k$$
and since $\binom{n}{n+1}=0$, we have
$$\sum_{k=0}^{n+1} \binom{n+1}{k} ka^k = \sum_{k=0}^{n+1} \binom{n}{k-1}ka^k + an(a+1)^{n-1}$$
and this is where I'm stuck. If we look at it from the other direction, we have
$$(n+1)a(a+1)^{n+1-1} = (n+1)a(a+1)^n = na(a+1)^n + a(a+1)$$
But I'm not sure where to go from here.
| Just like Thomas Andrews suggested, we can use the fact that
$$
k\binom{n}{k}=n\binom{n-1}{k-1}.
$$
We have
\begin{align}
\sum_{k=0}^n\binom{n}{k}ka^k&=\sum_{k=0}^nn\binom{n-1}{k-1}a^k\\
&=n\sum_{k=0}^n\binom{n-1}{k-1}a^k \ \ \ \ \ \ (\text{setting}\ \ell=k-1),\\
&=n\sum_{\ell=0}^{n-1}\binom{n-1}{\ell}a^{\ell +1}\\
&=na\sum_{\ell=0}^{n-1}\binom{n-1}{\ell}a^{\ell}\ \ \ \ \ \ (\text{by the Binomial Theorem}),\\
&=na(a+1)^{n-1}.
\end{align}
| {
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"source": "stackexchange",
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Find the limits between which $a$ must lie in order that $\frac{ax^2-7x+5}{5x^2-7x+a}$ can take all real values Find the limits between which $a$ must lie in order that $$\frac{ax^2-7x+5}{5x^2-7x+a}$$ may be capable of taking all real values for all possible real values that $x$ may take.
My Attempt:
Let $$y=\frac{ax^2-7x+5}{5x^2-7x+a}$$
$(5y-a)x^2-7(y-1)x+ay-5=0$
Since $x\in R$, we should have Discriminant $\geq 0$
$49(y-1)^2-4(5y-a)(ay-5)\geq 0$
which simplifies to
$(49-20a)y^2+2(2a^2+1)y+(49-20a)\geq 0$
For above to hold true for all real $y$ we must have $49-20a>0$
and Discriminant $\leq 0$
$4(2a^2+1)^2-4(49-20a)^2\leq 0$
which simplifies to
$(a-5)^2(a+12)(a-2)\leq 0$
Thus $-12\leq a\leq 2$
But for $a=-12$ and $a=2$ the Numerator and Denominator of $y$ develop common factors due to which $y$ assumes the form $\frac{Ax+B}{Cx+D}$ which obviously can't take all real values.
So $-12<a<2$
Can we not simply find value(s) of $a$ such that Numerator and Denominator have common factors and somehow justify that $-12<a<2$. I want to avoid the Discriminant part.
| If $$y=\frac{f(x)}{g(x)}$$ where $f(x)$ and $g(x)$ are both quadratics.
If $y$ is of the form $$y=\frac{(x-a)(x-c)}{(x-b)(x-d)}$$ where $a<b<c<d$ then from the plot of graph it is clearly seen that $-\infty<y<+\infty$ when $x\in (b,d)$
It appears reasonable enough to say that if one root of numerator(denominator) lies between the roots of denominator(numerator) then $y$ can take all real values.
Let $x_{1}$ and $x_{2}$ be the roots of $5x^2-7x+a=0$
i.e. $5x_{1}^2-7x_{1}+a=0$ and
$5x_{2}^2-7x_{2}+a=0$
Also, $x_{1}+x_{2}=\frac{7}{5}$ ; $x_{1}x_{2}=\frac{a}{5}$
Let $f(x)=ax^2-7x+5$
$$f(x_{1})=ax_{1}^2-7x_{1}+5=a\left(\frac{7x_{1}-a}{5}\right)-7x_{1}+5=\frac{(a-5)(7x_{1}-a-5)}{5}$$
Similarly, $$f(x_{2})=ax_{2}^2-7x_{2}+5=\frac{(a-5)(7x_{2}-a-5)}{5}$$
If exactly one root of numerator i.e. $ax^2-7x+5=0$ lies between $x_{1}$ and $x_{2}$ then
$$f(x_{1})f(x_{2})<0$$
$$\frac{(a-5)^2(7x_{1}-a-5)(7x_{2}-a-5)}{25}<0$$
$$49x_{1}x_{2}-7(a+5)(x_{1}+x_{2})+(a+5)^2<0$$
$$49(\frac{a}{5})-7(a+5)(\frac{7}{5})+(a+5)^2<0$$
$$(a+12)(a-2)<0$$
$$-12<a<2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove Segner's Recurrence Relation $C_{n+1} = \sum\limits_{i=0}^n C_i C_{n-i}$ on Catalan Numbers $C_n = \frac{1}{n+1} \binom{2n}{n}$
Prove Segner's Recurrence Relation $C_{n+1} = \sum\limits_{i=0}^n C_i C_{n-i}$ on Catalan Numbers $C_n = \frac{1}{n+1} \binom{2n}{n}$
Plugging in the Catalan Equation, we want to prove:
\begin{align*}
\frac{1}{n+2} \binom{2(n+1)}{n+1} &= \sum\limits_{i=0}^n \frac{1}{i+1} \binom{2i}{i} \frac{1}{n-i+1} \binom{2(n-i)}{n-i} \\
\end{align*}
Expanding the binomial coefficients to factorials (again this is a formula we want to prove, not a result we have demonstrated):
\begin{align*}
\frac{1}{n+2} \frac{(2n+2)!}{(n+1)!(n+1)!} &= \sum\limits_{i=0}^n \frac{1}{i+1} \frac{(2i)!}{i!i!} \frac{1}{n-i+1} \frac{(2n-2i)!}{(n-i)!(n-i)!} \\
\end{align*}
How would one go from here?
| As somebody suggested the method of generating functions is good.
We have:
\begin{eqnarray}
g(x)&:=&\sum\limits_{i=0}^\infty \binom{2 i}{i} \frac{x^i}{i+1}\\
&=& \frac{1}{x} \int_0^x \sum\limits_{i=0}^\infty \binom{2 i}{i} \xi^i d\xi\\
&=& \frac{1}{x} \int_0^x \frac{d \xi}{\sqrt{1-4 \xi}} \\
&=& - \frac{-1+\sqrt{1-4 x}}{2 x}
\end{eqnarray}
Now we square and then invert. We have:
\begin{eqnarray}
g(x)^2 &=& -\frac{1}{2} \sum\limits_{n=2}^\infty \binom{1/2}{n} (-4)^n x^{n-2}\\
&=& -\frac{1}{2} \sum\limits_{n=0}^\infty \binom{1/2}{n+2} (-4)^{n+2} x^n
\end{eqnarray}
Now we just need to simplify the coefficient. We have:
\begin{eqnarray}
&&-\frac{1}{2} \binom{1/2}{n+2} (-4)^{n+2} = \\
&&-\frac{1}{2} \frac{\prod\limits_{j=0}^{n+1} (1/2-j)}{(n+2)!} (-4)^{n+2} = \\
&&-\frac{1}{2} \frac{1}{2^{n+2}} (-1)^{n+1} \frac{(2n+2)!}{(2n )!! (n+2)!} (-4)^{n+2} = \\
&& \frac{(2n+2)!}{(n+1)!(n+2)!}
\end{eqnarray}
As expected.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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System of equations (Problem $50$ from $101$ algebra by Titu)
Let $a$ and $b$ be given real numbers.
Solve the system of equations
$$\begin{aligned} \frac{x-y \sqrt{x^{2}-y^{2}}}{\sqrt{1-x^{2}+y^{2}}} &=a \\ \frac{y-x \sqrt{x^{2}-y^{2}}}{\sqrt{1-x^{2}+y^{2}}} &=b \end{aligned}$$
for real $x$ and $y$.
Solution -
Let $u=x+y$ and $v=x-y .$ Then
$$
0<x^{2}-y^{2}=u v<1, x=\frac{u+v}{2}, \text { and } y=\frac{u-v}{2}
$$
Adding the two equations and subtracting the two equations in the original system yields the new system
$$
\begin{aligned}
u-u \sqrt{u v} &=(a+b) \sqrt{1-u v} \\
v+v \sqrt{u v} &=(a-b) \sqrt{1-u v}
\end{aligned}
$$
Multiplying the above two equations yields
$$
u v(1-u v)=\left(a^{2}-b^{2}\right)(1-u v)
$$
hence $u v=a^{2}-b^{2} .$ It follows that
$$
u=\frac{(a+b) \sqrt{1-a^{2}+b^{2}}}{1-\sqrt{a^{2}-b^{2}}} \text { and } v=\frac{(a-b) \sqrt{1-a^{2}+b^{2}}}{1+\sqrt{a^{2}-b^{2}}}
$$
I did not get how they found values of $u$ and $v$ from
$u v=a^{2}-b^{2} .$ I mean obviously we can substitute the value in one of the equations and we will get some quadratic and we can find solution from there but the quadratic that I am getting is very large to handle, so is there some obvious step that directly lead solutions from $u v=a^{2}-b^{2}?$
Thank you
| It is $a^2-b^2=x^2-y^2$ hence $$\dfrac{\frac{u+v}{2}-\frac{u-v}{2}\sqrt{a^2-b^2}}{\sqrt{1-a^2+b^2}}=a$$ and $$\dfrac{\frac{u-v}{2}-\frac{u+v}{2}\sqrt{a^2-b^2}}{\sqrt{1-a^2+b^2}}=b$$
If you add them you get $u$ and if you subtract them you get $v$.
| {
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Solving the equation $X^2 + (3 +i)X + 1 +i =0$ I have to find the roots of the equation
$$X^2 + (3 +i)X + 1 +i =0.$$
The first step is to find the discrimant which is $4 + 2i$. Then, I assume that the square root of the discriminant is in the form of $a+bi$, so we have to solve the below system
$$a^2-b^2=4\qquad\text{ and }\qquad ab=1.$$
How can I solve the above system ? Please do not provide another solution (e.g using trigonometric form)
| You're work so far is good, and indeed you want to find $a,b\in\Bbb{R}$ such that $(a+bi)^2=4+2i$, i.e.
$$a^2-b^2=4\quad\text{ and }\qquad 2ab=2.$$
The latter shows you that $a$ and $b$ are nonzero, and that $b=a^{-1}$. Plugging this into the former yields
$$a^2-a^{-2}=4,$$
and multiplying everything by $a^2$ and rearranging shows that
$$a^4-4a^2-1=0.$$
This is a quadratic in $a^2$, so by the quadratic formula $a^2=2\pm\sqrt{5}$. Because $a$ is real
$$a=\pm\sqrt{2+\sqrt{5}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3873340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Find an asymptotic equivalent of the sequence $(\int_{-\infty}^{+\infty} \frac{1}{\cosh^n(x)} dx)_n$.
Find an asymptotic equivalent of the sequence $(\int_{-\infty}^{+\infty} \frac{1}{\cosh^n(x)} dx)_n$.
I found the result using a trick (using $e^x = \tan(\theta/2)$ as mentioned here) but I wanted to know if there isn't another less astute approach.
| There is a different approach: You can derive a recursive relation for $a_n:=\int_{\mathbb{R}} \cosh^{-n}(x) dx$ using the identity $\cosh^2-\sinh^2=1$ and integration by parts. In fact, for $n\geq 3$, we have
$a_n= \frac{n-2}{n-1}a_{n-2}$.
Indeed,
$a_n=\int_{\mathbb{R}} \frac{1}{\cosh^{n}(x)} dx= \int_{\mathbb{R}} \frac{\cosh^2(x)-\sinh^2(x)}{\cosh^{n}(x)} dx= \int_{\mathbb{R}} \frac{1}{\cosh^{n-2}(x)}dx -\int_{\mathbb{R}}\frac{\sinh^2(x)}{\cosh^{n}(x)}dx=a_{n-2}-\int_{\mathbb{R}}\frac{\sinh^2(x)}{\cosh^{n}(x)}dx.$
Now, integration by parts shows that the remaining integral is equal to $\frac{1}{n-1}a_{n-2}$. Indeed, recalling the identites $\cosh'=\sinh$ and $\sinh'=\cosh$, we have
$\int_{\mathbb{R}}\frac{\sinh^2(x)}{\cosh^{n}(x)}dx = \int_{\mathbb{R}}\frac{\sinh(x)}{\cosh^{n}(x)} \cdot \sinh(x) dx= \left[-\frac{1}{n-1}\cosh^{-(n-1)}(x)\cdot \sinh(x)\right]_{-\infty}^{\infty}$
$+ \int_{\mathbb{R}} \frac{1}{n-1}\cosh^{-(n-1)}(x) \cdot \cosh(x) dx
= 0 + \frac{1}{n-1}a_{n-2}=\frac{1}{n-1}a_{n-2}$.
So,
$a_n=a_{n-2}-\int_{\mathbb{R}}\frac{\sinh^2(x)}{\cosh^{n}(x)}dx = a_{n-2}-\frac{1}{n-1}a_{n-2}=\frac{n-2}{n-1}a_{n-2}$.
By induction, it follows that
$a_{n}=\frac{n-2}{n-1}\frac{n-4}{n-3}\dots \frac{2}{3}a_2$ for $n$ even.
and
$a_{n}=\frac{n-2}{n-1}\frac{n-4}{n-3}\dots \frac{1}{2}a_1$ for $n$ odd.
Therefore, it remains to compute $a_1$ and $a_2$. Now, as $\cosh(x)=\frac{1}{2}(e^x+e^{-x})$, we have
$a_1=2\int_{\mathbb{R}} \frac{1}{e^x+e^{-x}} dx
= 2\int_{\mathbb{R}} \frac{e^x}{(e^x)^2+1} dx
=2 \int_{\mathbb{R}} \frac{1}{y^2+1} dy = 2 [\arctan(y)]_{-\infty}^{\infty}=2\pi.$
Similarly, one can show $a_2=2$. Hence,
$a_{n}=\frac{n-2}{n-1}\frac{n-4}{n-3}\dots \frac{2}{3}\cdot 2$ for $n$ even.
and
$a_{n}=\frac{n-2}{n-1}\frac{n-4}{n-3}\dots \frac{1}{2}2 \pi$ for $n$ odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3873828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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cyclic rational inequalities $\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$ when $a+b+c=1$ I've been practicing for high school olympiads and I see a lot of problems set up like this:
let $a,b,c>0$ and $a+b+c=1$. Show that
$$\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$$
Any problem that involves cyclic inequalities like these always stump me. I know I'm supposed to use Cauchy-Schwarz or AM-GM at some point, but I can never get to a place where this might be useful. My first instinct is to get common denominators and hope stuff simplifies, but I can never get farther than that. For example, in this problem I did the following:
$$\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}$$
$$=\frac{(a^2+3)(b^2+3)+(b^2+3)(c^2+3)+(c^2+3)(a^2+3)}{(a^2+3)(b^2+3)(c^2+3)}$$
$$=\frac{a^2b^2+b^2c^2+c^2a^2+6(a^2+b^2+c^2)+27}{(a^2+3)(b^2+3)(c^2+3)}$$
but this is where I get stuck. I've tried using Cauchy-Schwarz on parts of this fraction to simplify it, but I can never get anything to work. How could you prove this inequality, and what are the important things to look out for in problems of this nature
| Another way.
Since $$\left(\frac{1}{x^2+3}\right)''=\frac{6(x^2-1)}{x^2+3)^2}<0$$ for $0<x<1$, by Jensen for the concave function $f(x)=\frac{1}{x^2+3}$ we obtain:
$$\sum_{cyc}(\frac{1}{a^2+3}\leq\frac{3}{\left(\frac{\sum\limits_{cyc}a}{3}\right)^2+3}=\frac{3}{\left(\frac{1}{3}\right)^2+3}=\frac{27}{28}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3878235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find $\sum_{k=1}^n P(z_k)$ where P is a polynomial and $z_k$ the $k$th root of $z^{13}=1$ Question: Given a complex number z such that $z^{13}=1$, find the sum of all possible values of $z+z^3+z^4+z^9+z^{10}+z^{12}$.
I know we have to use roots of unity and try to manipulate the polynomial we want to evaluate, but can't find a pattern.
| If $z^{13} = 1$ and $z \not= 1$, then $z = e^{2\pi i \frac{k}{13}}$ where $(k, 13) = 1$.
Since $z^{12} = z^{-1}$, $z^{10} = z^{-3}$ and $z^{9} = z^{-4}$ we have
$z+z^3 + z^4 + z^9 + z^{10}+z^{12} = 2(\cos(2 \pi \frac{k}{13}) + \cos(2 \pi \frac{3k}{13}) + \cos(2 \pi\frac{4k}{13}))$ where $z = e^{2\pi i \frac{k}{13}}$.
The total is $2\sum_{k=1}^{12} \cos(2 \pi \frac{k}{13}) + \cos(2 \pi \frac{3k}{13}) + \cos(2 \pi\frac{4k}{13}) = -6$.
If $z = 1$ is included, the total is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3878905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.
Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.
What I Tried: Here is a picture :-
I know the centroid divides each of the medians in the ratio $2:1$ . So $AD = 3\sqrt{3}$ , $BE = 3\sqrt{2}$ , $CF = 3$ .
From this site :- https://mathworld.wolfram.com/TriangleMedian.html, I find that the area of the triangle will be :- $$\frac{4}{3}\sqrt{s_m(s_m - m_1)(s_m - m_2)(s_m - m_3)}$$
Where $m_1,m_2,m_3$ are the medians of the triangle and $s_m = \frac{m_1 + m_2 + m_3}{2}$ .
After putting the respective values for the medians I get that $[\Delta ABC]$ is :-
$$\frac{4}{3}\sqrt{\Bigg(\frac{3(\sqrt{3} + \sqrt{2} + 1)}{2}\Bigg)\Bigg(\frac{3(\sqrt{2} + 1 - \sqrt{3})}{2}\Bigg)\Bigg(\frac{3(\sqrt{3} + 1 - \sqrt{2})}{2}\Bigg)\Bigg(\frac{3(\sqrt{3} + \sqrt{2} - 1)}{2}\Bigg)}$$
$$\rightarrow \frac{4}{3}\sqrt{\frac{81(\sqrt{3} + \sqrt{2} + 1)(\sqrt{2} + 1 - \sqrt{3})(\sqrt{3} + 1 - \sqrt{2})(\sqrt{3} + \sqrt{2} - 1)}{16}}$$
I am almost to the answer (assuming I made no mistake), but I think this simplification is getting complicated. How do I proceed next?
Can anyone help me?
| From where you left,
$A = \displaystyle 3 \sqrt{{(\sqrt{3} + \sqrt{2} + 1)(\sqrt{2} + 1 - \sqrt{3})(\sqrt{3} + 1 - \sqrt{2})(\sqrt{3} + \sqrt{2} - 1)}}$
Take the first two terms, it is of the form $(a-b)(a+b)$ so we have,
$(\sqrt{3} + \sqrt{2} + 1)(\sqrt{2} + 1 - \sqrt{3}) = 2\sqrt2$
Next two terms can be taken as (a+b-c)(a-b+c)
$((\sqrt{3} + 1) - \sqrt{2})((\sqrt{3} -1) + \sqrt{2}) = 3 - 1 - 2 + \sqrt 2 (\sqrt3 + 1) - \sqrt2 (\sqrt3 - 1) = 2 \sqrt2$
So $A = 6 \sqrt2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3882933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Three vertices of a triangle lie on $y^2=x^3$
If ${P(a,b),Q(c,d),R(m,n)}$ are the centroid, orthocenter, circumcenter respectively of a scalene triangle with vertices on the curve $y^2=x^3$, then find the value of $$\frac{a}{b}+\frac{c}{d}+\frac{m}{n}$$
I know that the centroid, orthocenter and circumcenter lie on the straight line and centroid divides orthocenter and circumcenter in ratio $2:1$. Thus
$$a=\frac{2m+c}{3},b=\frac{2n+d}{3}$$ As you can see I am not able to use $y^2=x^3$.
I think there is some trick involved!
| $\begin{array}{} G=(a,\sqrt{a^3}) & H=(c,\sqrt{c^3}) & O=(m,-\sqrt{m^3}) \end{array}$
$\begin{array}{} \frac{x_{H}-x_{G}}{x_{G}-x_{O}}=2 & \frac{c-a}{a-m}=2 & a=\frac{2m+c}{3} \end{array}$
$\begin{array}{} \frac{y_{H}-y_{G}}{y_{G}-y_{O}}=2 & \frac{\sqrt{c^3}-\sqrt{a^3}}{\sqrt{a^3}+\sqrt{m^3}}=2 \end{array}$
$\frac{\sqrt{c^3}-\sqrt{(\frac{2m+c}{3})^3}}{\sqrt{(\frac{2m+c}{3})^3}+\sqrt{m^3}}=2$
One solution that checks the system is:
$m=\frac{1}{2}\left( 2+\sqrt{6+\sqrt[3]{9}}-\sqrt{12-\sqrt[3]{9}}+\frac{30}{\sqrt{6+\sqrt[3]{9}}} \right) ·c$
$\begin{array}{} m=α·c & α≈0.158873884116 \end{array}$
numerically checking the value of the expression k
$k=\frac{a}{b}+\frac{c}{d}+\frac{m}{n}$
$k=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{c}}-\frac{1}{\sqrt{m}}$
$\begin{array}{} m=α·c & a=\frac{1+2α}{3} \end{array}$
$k=\left(\frac{\sqrt{3}}{\sqrt{1+2α}}+1-\frac{1}{\sqrt{α}} \right)·\frac{1}{\sqrt{c}} ≈\frac{-2·10^{-15}}{\sqrt{c}}$
Indicating that the value of k is zero
Proof:
$k=\left(\frac{\sqrt{3}}{\sqrt{1+2β}}+1-\frac{1}{\sqrt{β}} \right)·\frac{1}{\sqrt{c}} =0$
$\begin{array}{} \text{Solving} & \frac{\sqrt{3}}{\sqrt{1+2β}}+1-\frac{1}{\sqrt{β}}=0 & β=0.158873884146 \end{array}$
β is equal to α which is the solution of the initial system.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3888000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the recurrence relation: $na_n = (n-4)a_{n-1} + 12n H_n$ I want to solve
$$ na_n = (n-4)a_{n-1} + 12n H_n,\quad n\geq 5,\quad a_0=a_1=a_2=a_3=a_4=0. $$
Does anyone have an idea, what could be substituted for $a_n$ to get an expression, which one could just sum up? We should use
$$ \sum_{k=0}^n \binom{k}{m} H_k = \binom{n+1}{m+1} H_{n+1} - \frac{1}{m+1} \binom{n+1}{m+1} $$
to simplify the result.
| Let $f\left( x \right) = \sum\limits_{n = 0}^\infty {a_n x^n } $ be the generating function of the solution of the recurrence.
$$
\sum\limits_{n = 0}^\infty {na_n x^n } = \sum\limits_{n = 0}^\infty {\left( {n + 1} \right)a_n x^{n + 1} - 4\sum\limits_{n = 0}^\infty {a_{n - 1} x^n } } +
12\sum\limits_{n = 5}^\infty {na_n x^n } H_n
$$
which gives the following DE
$$
xf'\left( x \right) = x^2 f'\left( x \right) - 3xf\left( x \right)-\frac{2 x \left(50 x^5-67 x^4+2 x^3+3 x^2+6 x+6 \log (1-x)\right)}{(x-1)^2}$$
Whose solution is
$$f(x)=\frac{2}{75 (x-1)^2} \left(-7553 x^5+3750 x^5 \log (1-x)+\\+24040 x^4-18750 x^4 \log (1-x)-29405 x^3+37500 x^3 \log (1-x)+\\+16830 x^2-37500 x^2 \log (1-x)-3840 x+18750 x \log (1-x)-3840 \log (1-x)\right)$$
Coefficients of the MacLaurin expansion of $f(x)$ are
$$0,0,0,0,0,\frac{137}{5},\frac{578}{15},\frac{1667}{35},\frac{395}{7},\frac{41137}{630},\frac{13007}{175},\frac{964849}{11550},\ldots$$
There is no closed form for $a_n$, though.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
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