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what is the integral of $\int x\cos^2(\pi x)$ I have a function $f(x)=x\cos^2(\pi x)$
I'm asked to find the size of the aria that is trapped between
the function $f(x)$ and x axis between [0,1.5]
$\int_0^\frac{3}{2} x\cos^2(\pi x)$
my attempt:
I know that S=$\int_0^\frac{1}{2} x\cos^2(\pi x)$+ $\int_\frac{1}{2}^\frac{3}{2} x\cos^2(\pi x)$
I cant succeed in finding the integral of the function I tried finding the integral with (integral by parts method) and by (assignment method) with no result
I'm stuck at this question pleas help
| The answer is $\frac{9}{16}-\frac{1}{4\pi^{2}}$.
Let $f: x\mapsto x\cos^{2}(\pi x)$ and $\displaystyle I=\int_{0}^{3/2}f(x){\rm d}x$. The function $f$ is continuous on $[0;3/2]$ so $I$ there exist and is finite.
Note that $$\forall x\in \mathbb{R}:\quad \cos^{2}(\pi x)=\frac{1}{2}\left(\cos (2\pi x)+1\right).$$
Then,
\begin{eqnarray*}
\int x\cos^{2}(\pi x){\rm d}x&=&\frac{1}{2}\int x(\cos(2\pi x)+1){\rm d}x\\
&=& \frac{1}{2}\int x\cos 2\pi x {\rm d}x+\frac{1}{2}\int {\rm d}x\\
&\overset{{\rm IBP}}{=}&\frac{x\sin 2\pi x}{4\pi}-\frac{1}{4\pi}\int \sin 2\pi x{\rm d}x+\frac{x^{2}}{4}\\
&\overset{t\mapsto 2\pi x}{=}&\frac{x\sin 2\pi x}{4\pi}-\frac{1}{8\pi^{2}}\int \sin t{\rm d}t+\frac{x^{2}}{4}\\
&=&\frac{\cos t}{8\pi^{2}}+\frac{x\sin 2\pi x}{4\pi}+\frac{x^{2}}{4}+C\\
&=&\frac{2\pi x(\pi x+\sin(2\pi x))+\cos(2\pi x)}{8\pi^{2}}+C
\end{eqnarray*}
Therefore,
$$\color{red}{\boxed{\int_{0}^{3/2}x\cos^{2}(\pi x){\rm d}x=\frac{9}{16}-\frac{1}{4\pi^{2}}\approx 0.537}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the sum of $1+4k+9k^2+...+n^2k^{n-1}$ I'm having trouble using Abel's summation formula as $a_1b_1+a_2b_2+...+a_nb_n=(a_1-a_2)(b_1)+(a_2-a_3)(b_1+b_2)+...+(a_{n-1}-a_n)(b_1+b_2+...+b_{n-1})+a_n(b_1+b_2+...+b_n)$
to find the sum of $1+4k+9k^2+...+n^2k^{n-1}$.
I know $1+2k+3k^2+...+nk^{n-1}=\frac{nk^n}{k-1}-\frac{k^n-1}{(k-1)^2}$ from a previous part.
Applying Abel's formula once I get $(-3)(1)+(-5)(1+k)+...-(2n-1)(1+k+k^2+...+k^{n-2})+n^2(1+k+k^2+...+k^{n-1})$
Then I think the next step is to apply it again to get $-2(1)-2(1+1+k)-...-2(1+(1+k)+(1+k+k^2)+...+(1+k+k^2+...+k^{n-3}))+(2n-1)(1+(1+k)+...+(1+k+...+k^{n-2}))+n^2\left(\frac{k^n-1}{k-1}\right)$
$=-2[1+(1+1+k)+(1+1+k+1+k+k^2)+...+(1+(1+k)+...+(1+k+..+k^{n-3}))]+(2n-1)((n-1)+(n-2)k+...+(n-(n-1))k^{n-2})+n^2\left(\frac{k^n-1}{k-1}\right)$
$=-2[1+(1+1+k)+(1+1+k+1+k+k^2)+...+(1+(1+k)+...+(1+k+..+k^{n-3}))]+(2n-1)\left(n\left(\frac{k^{n-1}-1}{k-1}\right)-\frac{(n-1)k^{n-1}}{k-1}+\frac{k^{n-1}-1}{(k-1)^2}\right)+n^2\left(\frac{k^n-1}{k-1}\right)$
$=-2[1+(1+1+k)+(1+1+k+1+k+k^2)+...+(1+(1+k)+...+(1+k+..+k^{n-3}))]+(2n-1)\left(\frac{k^{n-1}-n}{k-1}+\frac{k^{n-1}-1}{(k-1)^2}\right)+n^2\left(\frac{k^n-1}{k-1}\right)$
$=-2[1+(1+1+k)+(1+1+k+1+k+k^2)+...+(1+(1+k)+...+(1+k+..+k^{n-3}))]+\frac{2n-1}{(k-1)^2}\left(k^n-nk+n-1\right)+n^2\left(\frac{k^n-1}{k-1}\right)$
My problem arises when trying to find $[1+(1+1+k)+(1+1+k+1+k+k^2)+...+(1+(1+k)+...+(1+k+..+k^{n-3}))]$, it feels way too contrived and the sums within a sum thing is confusing me, I feel like I'm not taking the right approach, if anyone has a faster way using this formula that would be greatly appreciated.
By the way the solution is $\frac{n^2k^n}{k-1}-\frac{(2n-1)k^n+1}{(k-1)^2}+\frac{2k^n-2}{(k-1)^3}$
| Surb’s method is much cleaner, but you can also use the method used to find the sum of geometric series to solve this.
\begin{align}
v &= \sum_{k=1}^nk^2x^{k-1} \\
v-vx &= -n^2x^n+\sum_{k=0}^{n-1}(2k+1)x^k \\
&= -n^2x^n+2\sum_{k=0}^{n-1}kx^k+\sum_{k=0}^{n-1}x^k
\end{align}
At which point solving for $v$ should be pretty easy (the same method can solve $\sum_{k=0}^{n-1}kx^k$ if you want to use it).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4336763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
prove: $15(a+b)\ge17+14\sqrt{2ab}$
Let $a,b\ge0: a^4+b^4=17$. Prove that: $$15(a+b)\ge17+14\sqrt{2ab}$$
I am looking for a nice approach. My approach is ugly by replace: $b=\sqrt[4]{17-a^4} $ and the rest is working with fuction.
I am quite sure $(1,2)$ is the only case equality so I guess we can use AM-GM in someway. Is there any better idea for this problem?
Thanks for your help!
| Due to Erik Satie's hint:
We will prove that: $$\frac{17^2}{a^4+b^4}+60(a+b)\ge 85+56\sqrt{2ab}$$
By AM-GM: $$\frac{17}{a^4+b^4}\ge 5-4\sqrt[4]{\frac{a^4+b^4}{17}}\ge 5-\frac{4}{\sqrt{17}}\sqrt{5(a^2+b^2)-4ab}$$
The inequality becomes: $$15(a+b)\ge\sqrt{17\left(5(a^2+b^2)-4ab\right)}+14\sqrt{2ab}$$
By C-S inequality, it turns out: $$17\sqrt{\frac{5(a^2+b^2)-4ab}{17}}+28\sqrt{\frac{ab}{2}}\leq\sqrt{45\left(5(a^2+b^2)-4ab+14ab\right)}=15(a+b)$$
Obviously, equality holds iff $(a,b)=(1,2); (2,1)$. The proof is done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Evalute $\sqrt[n]{\frac{20}{2^{2n+4}+2^{2n+2}}}$ Evaluate the given expression $$\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}$$ The given answer is $\dfrac{1}{4}$. My attempt:
$$\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}=\sqrt[n]{\dfrac{20}{2^{2n}\cdot2^4+2^{2n}\cdot2}}\\=\sqrt[n]{\dfrac{20}{2^{2n}\cdot18}}=\sqrt[n]{\dfrac{10}{9\cdot2^{2n}}}$$ This is as far I as I am able to reach. Thank you!
PS I don't see how one can get $\dfrac14$. For that we have to get something like $\sqrt[n]{A^n}$.
| You made a small mistake :\begin{aligned}\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}&=\sqrt[n]{\dfrac{20}{2^{2n}\cdot2^4+2^{2n}\cdot2^{\color{red}{2}}}}\\&=\sqrt[n]{\dfrac{20}{2^{2n}\cdot\color{red}{20}}}=\sqrt[n]{\dfrac{1}{2^{2n}}}=\frac{1}{4}\end{aligned}
| {
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"url": "https://math.stackexchange.com/questions/4341297",
"timestamp": "2023-03-29T00:00:00",
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Solve for $x$: $\frac{x^2-10x+15}{x^2-6x+15}=\frac{3x}{x^2-8x+15}$ Solve the equation: $\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{3x}{x^2-8x+15}$.
When $x\ne3$ and $x\ne5$ we get $$(x^2-10x+15)(x^2-8x+15)=3x(x^2-6x+15)\\(x^2-9x+15-x)(x^2-9x+15+x)-3x(x^2-6x+15)=0\\(x^2-9x+15)^2-x^2-3x(x^2-6x+15)=0.$$ I am stuck here. The Rational Root Theorem won't be useful as the equation does not have such roots.
I got $-9x$ by averaging $-10x$ and $-8x$.
I don't know if it makes sense.
| Hint: divide the numerators and denominators by $x \ne 0$ and let $z = x + \dfrac{15}{x}\,$, then:
$$
\frac{z-10}{z-6} = \frac{3}{z-8}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4345432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\sum_{k=0}^{n-1} {n-1-k\choose k}\left(\frac{1}{2}\right)^{n-1-k}+\sum_{k=0}^{n-2} {n-2-k\choose k}\left(\frac{1}{2}\right)^{n-2-k} $ How can you find
$$
\sum_{k=0}^{n-1} {n-1-k\choose k}\left(\frac{1}{2}\right)^{n-1-k}+\sum_{k=0}^{n-2} {n-2-k\choose k}\left(\frac{1}{2}\right)^{n-2-k}
$$
?
I found the value via interpritting the above formula combinatorially. (If I am correct, it is $\frac{2}{3}+\frac{1}{3}\left(-\frac{1}{2}\right)^n$)
But I want to know how to solve it by way of complex integrals or formal power series or any algebraic manipulations.
| Let $$a_m=\sum_{k=0}^{m/2} \binom{m-k}{k}\left(\frac{1}{2}\right)^{m-k}$$ so that your desired expression is $a_{n-1}+a_{n-2}$. We show that $a_m=(2+(-2)^{-m})/3$, yielding $a_{n-1}+a_{n-2}=2(2+(-2)^{-n})/3$.
Let $A(z)=\sum_{m=0}^\infty a_m z^m$ be the ordinary generating function. Then
\begin{align}
A(z) &= \sum_{m=0}^\infty \left(\sum_{k=0}^{m/2} \binom{m-k}{k}\left(\frac{1}{2}\right)^{m-k}\right) z^m \\
&= \sum_{k=0}^\infty 2^k \sum_{m=2k}^\infty \binom{m-k}{k} (z/2)^m \\
&= \sum_{k=0}^\infty 2^k (z/2)^{2k} \sum_{m=0}^\infty \binom{m+k}{k} (z/2)^m \\
&= \sum_{k=0}^\infty (z^2/2)^k \frac{1}{(1-z/2)^{k+1}} \\
&= \frac{1}{1-z/2}\sum_{k=0}^\infty \left(\frac{z^2/2}{1-z/2}\right)^k \\
&= \frac{1}{1-z/2}\sum_{k=0}^\infty \left(\frac{z^2}{2-z}\right)^k \\
&= \frac{1}{1-z/2}\cdot\frac{1}{1-z^2/(2-z)} \\
&= \frac{2}{(1-z)(2+z)} \\
&= \frac{2/3}{1-z}+\frac{2/3}{2+z} \\
&= \frac{2/3}{1-z}+\frac{1/3}{1+z/2} \\
&= \frac{2}{3}\sum_{m=0}^\infty z^m + \frac{1}{3}\sum_{m=0}^\infty (-z/2)^m \\
&= \sum_{m=0}^\infty \frac{2+(-2)^{-m}}{3} z^m,
\end{align}
so $a_m=(2+(-2)^{-m})/3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why can I square both sides? I am not used to English. I ask for your understanding in advance.
There is the equation:
$ x= 2^\frac{1}{2}$
we can square both side like this:
$ x^2= 2$
But I don't understand why that it's okay to square both sides.
What I learned is that adding, subtracting, multiplying, or dividing both sides by the same thing is okay. For example:
$ x = 1 $
$ x-1 = 1-1 $
$ x-1 = 0 $
$ x \times 2 = 1 \times 2 $
$ 2x = 2 $
like this.
But how come squaring both sides is okay too?
$ x = 2 $
$ x \times 2 = 2 \times 2 $
$ 2x = 4 $
$ 2x \times x = 4 \times x $
This does not induce it.
Can you answer this silly question?
| We start from the fact that if $x = y$, then $a\cdot x = a\cdot y$ and $x\cdot a = y\cdot a$ for any $a$.
Therefore, if $x = y$, letting $a = x$ we get $x\cdot x = x\cdot y$ and letting $a = y$ we get $x\cdot y = y\cdot y$, so $$x^2 = x\cdot x = x\cdot y = y\cdot y = y^2.$$
Thus, we have shown that given equality $x = y$ it is ok to square both sides, i.e. $x = y$ implies $x^2 = y^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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For $a, b, c$ with $a+b+c=0$ prove $\frac15\sum a^5=\frac13\sum a^3\cdot\frac12\sum a^2$ and $\frac17\sum a^7=\frac15\sum a^5\cdot\frac12\sum a^2$ Consider the following problem:
Problem. Suppose that real numbers $a$, $b$ and $c$ satisfy the condition $a+b+c=0$. Prove the following identities:
$$
\frac{a^5+b^5+c^5}{5}=\frac{a^3+b^3+c^3}{3}\cdot\frac{a^2+b^2+c^2}{2},
\\
\frac{a^7+b^7+c^7}{7}=\frac{a^5+b^5+c^5}{5}\cdot\frac{a^2+b^2+c^2}{2}.
$$
Perhaps, the shortest solution I can think of is as follows: plug $c=-a-b$ into the equation but instead of expanding everything use the following identities
$$
(a+b)^3-a^3-b^3=3ab(a+b),
\\
(a+b)^5-a^5-b^5=5ab(a+b)(a^2+ab+b^2),
\\
(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2.
$$
However, these identies are coming out of nowhere and moreover, in order to prove them one still needs to do some computations.
Question. Is it possible to solve this problem in a "smart" way (i.e. avoiding computations and preferrably elementary since it is almost a high school problem)? Any other solutions are also welcome.
Comment. It should be also noted that it is unclear (at least for me) how those identies were invented. It seems there is no nice similar identities for $\frac{a^p+b^p+c^p}{p}$ for $p$ other than 2, 3, 5, 7.
| let's have a look at the following identities called Lame-type identities (see here).
$$(x+y+z)^3 - (x^3+y^3+z^3) = 3(x+y)(x+z)(y+z)$$
$$(x+y+z)^5 - (x^5+y^5+z^5) = 5(x+y)(x+z)(y+z)(x^2+y^2+z^2+xy+xz+yz)$$
$$(x+y+z)^7 - (x^7+y^7+z^7) = 7(x+y)(x+z)(y+z)((x^2+y^2+z^2+xy+xz+yz)^2+xyz(x+y+z))$$
Algebraically manipulating the second and third identities give;
Let
$$A = x+y+z,$$
$$B = x^2+y^2+z^2,$$
$$C = x^3+y^3+z^3,$$
$$E = x^5+y^5+z^5,$$
$$G = x^7+y^7+z^7,$$
then
\begin{equation}
6E=A^5-5BA^3+5CA^2+5BC \tag{1}
\end{equation}
\begin{equation}
36G=A^7+7CA^4-21B^2A^3+28C^2A+21B^2C \tag{2}
\end{equation}
So, letting $A=0$ reduces (1) and (2) to the original problems.
Hope that solves the problems.
I have found, using computer, that an identity for the 9th powers is:
\begin{align}
72\frac{(x^9+y^9+z^9)}{(x^3+y^3+z^3)}=&27(x^2+y^2+z^2)^2 +8(x^3+y^3+z^3)^2
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Find the foot of the perpendicular through the point $A(-3;2)$ to the line $2x-y+4=0$. Find the foot of the perpendicular through the point $A(-3;2)$ to the line $2x-y+4=0$.
Let $AH\perp a:2x-y+4=0, H\in a.$
I have tried to use the fact that the vector $\vec{AH}(x_H+3;y_H-2)$ is a normal vector of the line $a$ and also $a$ passes through $H(x_H;y_H)$. Then we will have another equation for $a$ $$(x_H+3)(x-x_H)+(y_H-2)(y-y_H)=0$$
I am not familiar with the concept for slopes.
| We have $y=2x+4$. Consider the line through $A$ that is perpendicular to this line. This line hence must have slope $-\frac{1}{2}$. Thus its represented by $y=-\frac{1}{2}x+b$ for some $b$. But since it passes through $A$, $2=\frac{3}{2}+b\implies b=\frac{1}{2}$. Thus the line is represented by $y=-\frac{1}{2}x+\frac{1}{2}$. The foot of the perpendicular is simply the intersection of the two lines. Setting the $y$-coordinates, equal:
$$2x+4=\frac{1}{2}-\frac{1}{2}x\implies 4x+8=1-x\implies x=-\frac{7}{5}$$
And then $y=-\frac{14}{5}+4=\frac{6}{5}$. So the answer is $\left(-\frac{7}{5}, \frac{6}{5}\right)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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shortest distance between points avoiding eclipse What is the shortest path between two points (-3,0) and (3,0) that avoids the interior of the ellipse $\frac{x^2}{2}+\frac{y^2}{1}=1$
I should use the maximum principle for state constraints, but I have a problem with the part where the path touches ellipse on some time interval? I know it should be a straight line away from the obstacle.
| The shortest path from the point $\ (-3,0)\ $ to the point $\ (3,0)\ $ within the set $\ \left\{(x,y)\,\left|\,x^2+2y^2\ge2\right.\right\}\ $ comprises a straight line from the point $\ (-3,0)\ $ to the ellipse, and tangent to it at the point where they meet, a segment of the ellipse from that point to its reflection in the $\ y$-axis, and the straight line from the latter point to the point $\ (3,0)\ $. There are two such paths, illustrated in red and green in the diagram below. While this is intuitively obvious, and I'm sure it's true, I'm not aware of any succinct way of proving it. If you're willing to accept it as established, however, here's a way of finding the coordinates of the points where the lines from $\ (-3,0)\ $ are tangent to the ellipse.
Let $\ (a,b)\ $ be the coordinates of the point where the red line from $\ (-3,0)\ $ meets the ellipse. Then $\ b>0\ $, and the coordinates where the green line from $\ (-3,0)\ $ meets the ellipse will be $\ (a,-b)\ $, by symmetry. For the red line to be tangent to the ellipse at $\ (a,b)\ $, its slope must be the same as that of the ellipse at that point. The slope of the line is
$$
\frac{b}{a+3}\ ,
$$
and the slope of the ellipse is its derivative $\ \frac{dy}{dx}\ $ at the point in question. Differentiating the equation of the ellipse gives
$$
0=x+2y\frac{dy}{dx}=a+2b\frac{dy}{dx}\ ,\ \text{ or}\\
\frac{dy}{dx}=-\frac{a}{2b}
$$
at the point $\ (a,b)\ $. Equating this to the slope of the line gives
\begin{align}
\frac{b}{a+3}&=-\frac{a}{2b}\ ,\ \text{ or}\\
2b^2&=-a^2-3a\\
&=2-a^2\ ,
\end{align}
because the point $\ (a,b)\ $ lies on the ellipse. It follows from this last equation that $\ a=-\frac{2}{3}\ $, and then $\ b=\frac{\sqrt{7}}{3}\ $.
The equation of the line segment from the point $\ (-3,0)\ $ to the point $\ \left(-\frac{2}{3}, \frac{\sqrt{7}}{3}\right)\ $ is
$$
y=\frac{x+3}{\sqrt{7}}\ .
$$
The reflection of the point $\ \left(-\frac{2}{3}, \frac{\sqrt{7}}{3}\right)\ $ in the $\ y$-axis is $\ \left(\frac{2}{3}, \frac{\sqrt{7}}{3}\right)\ $, and the equation of the line segment from that point to the point $\ (3,0)\ $ is
$$
y=\frac{3-x}{\sqrt{7}}\ .
$$
As TonyK has pointed out in a comment, the length of the elliptical part of these paths cannot be expressed in terms of elementary functions of the coordinates of its end points and the parameters of the ellipse. Its length is given by
\begin{align}
\int_{-\frac{2}{3}}^\frac{2}{3}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx&=2\int_0^\frac{2}{3}\sqrt{1+\frac{x^2}{4y^2}}\,dx\\
&=2\int_0^\frac{2}{3}\sqrt{\frac{1-\frac{x^2}{4}}{1-\frac{x^2}{2}}}\,dx\\
&=2\int_0^\frac{2}{3}\sqrt{\frac{1-\frac{1}{2}\left(\frac{x}{\sqrt{2}}\right)^2}{1-\left(\frac{x}{\sqrt{2}}\right)^2}}\,dx\\
&=2\sqrt{2}\int_0^\frac{\sqrt{2}}{3}\sqrt{\frac{1-\frac{t^2}{2}}{1-t^2}}\,dt\\
&=2\sqrt{2}E\left(\arcsin\left(\frac{\sqrt{2}}{3}\right)\,\left|\,\frac{1}{\sqrt{2}}\right.\right)\ ,
\end{align}
where $\ E(\,\cdot\,|\,\cdot\,)\ $ is the incomplete elliptic integral of the second kind. The length of the straight line from $\ (-3,0)\ $ to $\ \left(-\frac{2}{3}, \frac{\sqrt{7}}{3}\right)\ $ is
$$
\sqrt{\left(3-\frac{2}{3}\right)^2+\frac{7}{9}}=\frac{2\sqrt{14}}{3}\ ,
$$
so the shortest distance is
$$
\frac{4\sqrt{14}}{3}+2\sqrt{2}E\left(\arcsin\left(\frac{\sqrt{2}}{3}\right)\,\left|\,\frac{1}{\sqrt{2}}\right.\right)\ ,
$$
for which WolframAlfa gives the approximation
$$
6.33877741851728513700827575564\ .
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $\sin x+\cos x=k \sin x \cos x$ for real $x$, where $k$ is a real constant. As I had solved the equation when $k=1$ in Quora and MSE by two methods, I started to investigate the equation for any real constant $k$:
$$
\sin x+\cos x=k \sin x \cos x,
$$
I first rewrite the equation as
$$
\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)=\frac{k}{2} \sin (2 x)
$$
Letting $ \displaystyle y=x-\frac{\pi}{4}$ yields
$$
\begin{array}{l}
\sqrt{2} \cos y=\frac{k}{2}\left(2 \cos ^{2} y-1\right) \\
2 k \cos ^{2} y-2 \sqrt{2} \cos y-k=0
\end{array}
$$
When $k\neq 0$, using quadratic formula gives
$$
\cos y=\frac{1 \pm \sqrt{1+k^{2}}}{\sqrt{2} k}
$$
For real $y$, we have to restrict $\displaystyle \frac{1 \pm \sqrt{1+k^{2}}}{\sqrt{2} k}$ in $[-1,1]$. Then I found that
$$
-1 \leqslant \frac{1+\sqrt{1+k^{2}}}{\sqrt{2} k} \leqslant 1 \Leftrightarrow \quad|k| \geqslant 2 \sqrt{2}
$$ and $$
-1 \leqslant \frac{1-\sqrt{1+k^{2}}}{\sqrt{2} k} \leqslant 1 \Leftrightarrow \quad k<0 \text { or } k>0
$$
Now we can conclude that
A. When $k\neq0$
$$x=n \pi-\frac{\pi}{4}$$
B. When $0\neq|k| \geqslant 2 \sqrt{2}, $
$$x=\frac{(8 n+1) \pi}{4} \pm \arccos \left(\frac{1\pm \sqrt{1+k^{2}}}{\sqrt{2} k}\right)$$
C. When $ 0 \neq|k|<2 \sqrt{2},$
$$ x= \frac{(8 n+1) \pi}{4} \pm \arccos \left(\frac{1-\sqrt{1+k^{2}}}{\sqrt{2} k}\right)
$$
where $n\in Z.$
I am looking forward to seeing other methods to solve the equation.
Furthermore, how about $$a\sin x+b\cos x+c\sin x\cos x=0?$$
| k should be necessarily $> 1$ because each part >1.
Say since there would be 4 solutions verified for a particular numerical case of given k=2
$$ 1/\sin(t) + 1/ \cos(t) = 2, $$
we get 4 approximate solutions by iteration,.. without graphing
$$(2.81901,3.45426,1.14103,-7.55065) $$
By graphing
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4358878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
If normal to curve $x=3\cos\theta-\cos^3{\theta}$ and $y=3\sin\theta-\sin^3{\theta}$ at point $P(\theta)$ passes through the origin, then $\theta=$
If normal to curve $x=3\cos\theta-\cos^3{\theta}$ and $y=3\sin\theta-\sin^3{\theta}$ at point $P(\theta)$ passes through the origin, then $\theta=$
(A) $0$
(B) $\dfrac{\pi}{4}$
(C) $\dfrac{\pi}{2}$
(D) $\dfrac{\pi}{6}$
My Approach:
I got the slope of normal as $\tan^3{\theta}.$
So equation of normal will be $$y-3\sin\theta +\sin^3{\theta}=\tan^3{\theta \cdot(x-3\cos\theta+3\cos^3{\theta})}$$
Because it passes through origin so I put $x=0,y=0$, which lead me to the equation,
$$3\sin\theta \cdot \cos{\theta} \cdot\cos{2\theta}=0$$
So value of $\theta$ will be $\theta=0, \dfrac{\pi}{2},\dfrac{\pi}{4}.$
But answer given is only $\theta = \dfrac{\pi}{4}$
My doubt:
Why option (A) and (C) are wrong.
| Your equation should be:
$$(y-3\sin{\theta}+\sin^3{\theta}) = \tan^3{\theta}(x-3\cos{\theta}+\cos^3{\theta})$$
When you rewrite $\tan^3{\theta}$ as $\frac{\sin^3{\theta}}{\cos^3{\theta}}$ and substitute $x=y=0$, you get $\tan^2{\theta}=1$ or $\sin{2\theta}=0$. Your explanation is right, $\theta=0,\frac{\pi}{2}$ work as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Taylor-expansion of numerator and denominator separately To Taylor-expand $f(x) = \frac{\ln{(x+1)}}{\sin{x}}$, can you expand the numerator and denominator separately?
So if $g(x) = \ln{(x+1)}$ and $h(x) = \sin{x}$, then:
$g'(x) = \frac{1}{x+1}$ and $g''(x) = \frac{-1}{(x+1)^2}$
$h'(x) = \cos{x}$ and $h''(x) = -\sin{x}$
So the Taylor-expansions to second degree about 0 is:
$T_{g(x)}(x) = x - \frac{1}{2}x^2$ and $T_{h(x)}(x) = x$, giving the Taylor-expansion for $f$:
$$T_{f(x)}(x) = 1 - \frac{1}{2}x$$
I don't know if this is true, however it doesn't seem to be very inaccurate if $x=0.01$...
| $$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right)$$
$$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$
$$f(x) = \frac{\log{(x+1)}}{\sin(x)}=\frac{x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right) } { x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)}$$ Now, using the long division
$$f(x)=1-\frac{x}{2}+\frac{x^2}{2}-\frac{x^3}{3}+\frac{11 x^4}{40}+O\left(x^5\right)=1-\frac{x}{2}+\frac{x^2}{2}+O\left(x^3\right)$$
Use always a few more terms when you need to divide later.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4363862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$( 1 - \sin x ) ( 1 + \csc x ) = \cos x \; , \; 0 \leq x \leq 2 \pi$ I found some solutions to this problem however I appear to be missing some solutions. Could anyone help me find the whole solution set?
$$( 1 - \sin x ) ( 1 + \csc x ) = \cos x \; , \; 0 \leq x \leq 2 \pi$$
$$1 + \csc x - \sin x - \sin x \csc x = \cos x $$
$$ 1 + \frac {1}{\sin x} - \sin x - 1 = \cos x $$
$$ \frac {1}{\sin x} - \sin x = \cos x $$
$$ \frac {1}{\sin x} = \cos x + \sin x $$
$$ 1 = \sin x (\cos x + \sin x) $$
$$ \sin^2 x + \cos^2 x = \sin x \cos x + \sin^2 x $$
$$ \cos^2 x = \sin x \cos x $$
$$ \cos x = \sin x $$
$$ \frac {\sin x}{\cos x} = 1 $$
$$\tan x = 1$$
$$ x = \frac \pi 4, \, \frac {5 \pi}{4}$$
| Using your equation $$\sin x(\sin x + \cos x) = 1$$ the additional solutions you missed were $$\sin x = 1 \rightarrow x = \dfrac {\pi}{2}, \dfrac {3 \pi}{2}$$ The other solutions are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4364147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\sqrt{x^2}=|x|$ or $\sqrt{x^2}=x$ in an indefinite integral Question:
Find the following integral: $\int{\sqrt{1+\cos(x)}}dx$
My attempt:
$$\int{\sqrt{1+\cos(x)}}dx$$
$$=\int{\sqrt{2\cos^{2}\frac{x}{2}}}dx$$
$$=\int{\sqrt{2}\ \left|\cos\frac{x}{2}\right|}dx$$
$$=\sqrt{2}\int{\left|\cos\frac{x}{2}\right|}dx\tag4$$
$$=\sqrt{2}\int{\pm\cos\frac{x}{2}}dx$$
$$=\pm\sqrt{2}\int{\cos\frac{x}{2}}dx$$
$$=\pm2\sqrt{2}\sin\frac{x}{2}+C$$
My book's attempt:
$$\int{\sqrt{1+\cos(x)}}dx$$
$$=\int{\sqrt{2\cos^{2}\frac{x}{2}}}dx$$
$$=\int{\sqrt{2}\cos\frac{x}{2}}dx$$
$$=\sqrt{2}\int{\cos\frac{x}{2}}dx$$
$$=2\sqrt{2}\sin\frac{x}{2}+C$$
Basically, my book didn't put $\pm$ sign, while I did. My book did this essentially: $\sqrt{x^2}=x$, while I did this: $\sqrt{x^2}=|x|$. Is my process more correct?
| Your solution goes further than the book's solution in terms of correctness. You have two issues. The first is that changing $\left|\cos\frac{x}{2}\right|$ to $\pm\cos\frac{x}{2}$ only creates a loss of information. Moreover, the $\pm$ sign itself depends on that value of $\cos\frac{x}{2}$, so it makes no sense to pull out the $\pm$ sign here, even if you did have it.
Integrating over absolute value signs is tricky. I'll use the definite integral to give an answer to the indefinite integral.
$$ \sqrt{2}\int{\left|\cos\frac{x}{2}\right|}dx = \sqrt{2}\int_{0}^{x}{\left|\cos\frac{u}{2}\right|}du + C $$
In the case of a definite integral, you have to split your domain into intervals on a case by case basis.
Case 1: $x>\pi$.
In this case, we have
\begin{align}
\sqrt{2}\int_{0}^{x}{\left|\cos\frac{u}{2}\right|}du + C &= \sqrt{2}\int_{0}^{\pi} \left|\cos\frac{u}{2}\right| du \\
& \quad+ \left( \sqrt{2}\int_{\pi}^{\pi+2\pi} \left|\cos\frac{u}{2}\right| du + \cdots + \sqrt{2}\int_{\pi+(n-1)2\pi}^{\pi+n2\pi} \left|\cos\frac{u}{2}\right| du \right) \\
& \quad+ \sqrt{2}\int_{\pi+n2\pi}^{x} \left|\cos\frac{u}{2}\right| du + C \qquad\qquad
\end{align}
where $n = \left\lfloor \dfrac{x-\pi}{2\pi}\right\rfloor$.
We look at the various definite integrals separately. First, note that $\cos\frac{u}{2}$ is nonnegative for all $u\in (0, \pi)$. This means we have $\left|\cos\frac{u}{2}\right| = \cos\frac{u}{2}$ on that interval. Hence
$$ \sqrt{2}\int_{0}^{\pi} \left|\cos\frac{u}{2}\right| du = \sqrt{2}\int_{0}^{\pi} \cos\frac{u}{2} du = 2\sqrt{2}. $$
Next, using periodicity of cosine, we can deduce that
$$ \sqrt{2}\int_{\pi+(k-1)2\pi}^{\pi+k2\pi} \left|\cos\frac{u}{2}\right| du = \sqrt{2}\int_{-\pi}^{\pi} \cos\frac{u}{2} du = 4\sqrt{2} $$
for each integer $k$.
Lastly, we need to do the last integral, which is going to give us a function of $x$. Again, using properties of cosine, we can deduce that
\begin{align}
\sqrt{2}\int_{\pi+n2\pi}^{x} \left|\cos\frac{u}{2}\right| du &= \sqrt{2}\int_{-\pi}^{x - 2(n+1)\pi} \cos\frac{u}{2} du = \sqrt{2}\cdot
2\sin\left(\frac{u}{2}\right) \Big|_{-\pi}^{x-(n+1)2\pi} \\
&= 2\sqrt{2} \left[ \sin\left(\frac{x-(n+1)2\pi}{2}\right) - \sin\left(-\frac{\pi}{2}\right) \right] \\
&= 2\sqrt{2} \left[ \sin\left(\frac{x}{2}-(n+1)\pi\right) - (-1) \right] \\
&= 2\sqrt{2} \left[ \sin\left(\frac{x}{2}\right)\cos(-(n+1)\pi) + \sin(-(n+1)\pi)\cos\left(\frac{x}{2}\right) + 1 \right] \\
&= 2\sqrt{2} \left[ \sin\left(\frac{x}{2}\right)(-1)^{n+1} + 0 + 1 \right] \\
&= 2\sqrt{2} \left[ 1 + (-1)^{n+1}\sin\left(\frac{x}{2}\right) \right]
\end{align}
Going all the way back to the beginning of Case 1, and putting this all together, we have
\begin{align}
\sqrt{2}\int_{0}^{x}{\left|\cos\frac{u}{2}\right|}du + C &= 2\sqrt{2} + \underbrace{(4\sqrt{2} + \cdots + 4\sqrt{2})}_{n\text{ terms}} + 2\sqrt{2}[1 + (-1)^{n+1}\sin\left(\frac{x}{2}\right)] + C \\[1.2ex]
&= 2\sqrt{2} + n(4\sqrt{2}) + 2\sqrt{2} + 2\sqrt{2}(-1)^{n+1}\sin\left(\frac{x}{2}\right) + C \\[1.6ex]
&= (n+1)\cdot 4\sqrt{2} + 2\sqrt{2}(-1)^{n+1}\sin\left(\frac{x}{2}\right) + C
\end{align}
Thus, for $x > \pi$ we have
$$ \sqrt{2}\int{\left|\cos\frac{x}{2}\right|}dx = 4\sqrt{2}(n+1) + 2\sqrt{2}(-1)^{n+1}\sin\left(\frac{x}{2}\right) + C $$
where $n = \left\lfloor \dfrac{x-\pi}{2\pi}\right\rfloor$.
The solution might look weird, but when I graph it in desmos for $C=0$ I get a continuous function.
Cases for $0\le x\le \pi$, $-\pi\le x<0$, and $x<-\pi$ are similar. I'll leave these other cases to you :)
The reason why we have all these cases and the reason why we need to split our definite integral over various intervals is because the absolute value function is actually a piecewise function that is itself handled by cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4364920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Surface area of rotation of a circle around a tangent Self-studying integral calculus and I got this problem:
The circle $x^2+y^2=a^2$ is rotated around a line tangent to the circle. Find the area of the surface of rotation.
I made the construction:
There were a few hints given alongside this question, namely: "Set up coordinate axes and a convenient parametrization of the circle. What does the polar graph $r=2a\sin(\theta)$ look like?" I understood the first and last hints since under this new coordinate system, the circle's equation becomes:
$${x_n}^2+(y_n-a)^2=a^2$$
Which when converted to polar, gives you the last hint. However I was unable to describe this in terms of parameters, so I decided to take the upper semicircle's surface area of revolution going from $a$ to $-a$ and multiplying that by 2 to account for the lower semicircle. My integral:
$$2\int_{-a}^a 2\pi (\sqrt{a^2-x^2}+a)\sqrt{1+\frac{x^2}{a^2-x^2}} dx$$
Upon simplification:
$$4\pi\int_{-a}^a a+\frac{a^2}{\sqrt{a^2-x^2}}dx$$
Evaluation leads me to:
$$8\pi a^2 + 4\pi^2 a^2$$
However my book (Serge Lang's First Course in Calculus) gives only $4\pi^2 a^2$. Where has my logic gone wrong if I am getting an extraneous term $8\pi a^2$?
EDIT for clarity on integral setup:
I first rearranged for $y$ while taking positive square root as I want to take the upper semicircle into consideration for surface of revolution about x-axis. I'll double this to account for the lower semicircle. This gives:
$$y=\sqrt{a^2-x^2}+a$$
Using the surface of revolution formula with the derivative as
$$\frac{dy}{dx}=\frac{-x}{\sqrt{a^2-x^2}}$$
Using this into the surface of revolution integral nets me my first integral in this post (also applied $\times$2)
| After changing the coordinates, in effect you are rotating $x^2 + (y-a)^2 = a^2$ around x-axis.
The circle is $x^2 + y^2 = 2 ay$
$ \displaystyle y' = \frac{x}{a-y}$
$ \displaystyle ds = \sqrt{1 + (y')^2} ~dx = \frac{a}{|y-a|} ~ dx$
For lower half -
$y = a - \sqrt{a^2-x^2}$
So, $ \displaystyle S_1 = 2 \pi a \int_{-a}^a \frac{a - \sqrt{a^2-x^2}}{\sqrt{a^2-x^2}} ~ dx$
$ = 2 \pi a^2 (\pi - 2)$
For upper half -
$y = a + \sqrt{a^2-x^2}$
So, $ \displaystyle S_2 = 2 \pi a \int_{-a}^a \frac{a + \sqrt{a^2-x^2}}{\sqrt{a^2-x^2}} ~ dx$
$ = 2 \pi a^2 (\pi + 2)$
Adding both, $S = 4 \pi^2 a^2$
But it is easier in polar coordinates as I mentioned in comments. The circle is,
$r = 2a \sin\theta, 0 \leq \theta \leq a$
$\dfrac{dr}{d\theta} = 2a \cos\theta$
$ \displaystyle ds = \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} ~ d\theta = 2a ~ d\theta$
$y = 2a\sin^2\theta$
So the integral is,
$ \displaystyle S = 8 \pi a^2 \int_0^{\pi} \sin^2\theta ~ d\theta = 4 \pi^2 a^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Solving $\lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}$
I've been asked to solve the limit.
$$\lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}$$
Here's my approach:
$$\lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}$$
Using the identity, $\cos(x) =\sin(90^{\circ} - x)$
\begin{aligned}\implies \lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}
& = \lim_{x\to0}\frac{\sin\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}
\\& = \lim_{x\to0}\dfrac{\left(\dfrac{\pi}{2} - \dfrac{\pi}{2\cos(x)}\right)\cdot\dfrac{\sin\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}{\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}}{\sin(\sin(x^2))}
\\ & = \lim_{x\to0}\dfrac{\left(\dfrac{\pi}{2} - \dfrac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}\cdot \underbrace{\lim_{x\to0}\dfrac{\sin\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}{\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}}_{1}
\\ & = \dfrac{\lim\limits_{x\to0}\dfrac{\pi}{2}\left(\dfrac{\cos(x) - 1}{\cos(x)}\right)}{\underbrace{\lim\limits_{x\to0}\dfrac{\sin(\sin(x^2))}{\sin(x^2)}}_1\cdot\sin(x^2)}
\\ & =\dfrac{\lim\limits_{x\to0}\dfrac{\pi}{2}\left(\dfrac{\cos(x) - 1}{\cos(x)}\right)}{\underbrace{\lim\limits_{x\to0}\dfrac{\sin(x^2)}{x^2}}_1\cdot x^2}
\\ & = \color{blue}{\boxed{\lim\limits_{x\to0}\dfrac{\pi}{2x^2}\left(\dfrac{\cos(x) - 1}{\cos(x)}\right)}}
\end{aligned}
Now, I'm unable to think of anything to do with this boxed part. Can anyone check my above method and tell me what to do further with this question? Any other shorter method is also most welcomed!
| You have $\sin(x^2)=x^2+o(x^2)$ and therefore also
$$
\sin(\sin(x^2))=x^2+o(x^2)
$$
Thus you need the Taylor expansion of
$$
\cos\Bigl(\frac{\pi}{2\cos x}\Bigr)
$$
up to degree $2$. Consider $f(x)=1/\cos x$. Then $f(0)=1$ and
$$
f'(x)=\frac{\sin x}{\cos^2x},\qquad f'(0)=0
$$
and
$$
f''(x)=\frac{\cos^3x+2\sin^2x\cos x}{\cos^4x}=\frac{\cos^2x+2\sin^2x}{\cos^3x},\qquad f''(0)=1
$$
Hence
$$
\frac{\pi}{2\cos x}=\frac{\pi}{2}+\frac{\pi}{4}x^2+o(x^2)
$$
and therefore
$$
\cos\Bigl(\frac{\pi}{2\cos x}\Bigr)=
\cos\Bigl(\frac{\pi}{2}+\frac{\pi}{4}x^2+o(x^2)\Bigr)=-\sin\Bigl(\frac{\pi}{4}x^2+o(x^2)\Bigr)=-\frac{\pi}{4}x^2+o(x^2)
$$
and your limit is
$$
\lim_{x\to0}\frac{-\pi x^2/4+o(x^2)}{x^2+o(x^2)}=-\frac{\pi}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Number of integer solutions of linear equation. I have the following problem.
Assume we have an unlimited number of blocks of 1cm, 2cm and 3cm height. Ignoring the position of the blocks, how many towers of 15cm height can we build?
I know I must find the coefficient of $x^{15}$ of the function
$$
f(x)=\frac{1}{1-x}\frac{1}{1-x^2}\frac{1}{1-x^3}=(1+x+x^2)\frac{1}{1-x^2}\frac{1}{(1-x^3)^2}
$$
How do I find it?
| $\mathbf{\text{Remember :}}$ $$\sum_{k=0}^{\infty}\binom{n+k-1}{k}x^k= \bigg(\frac{1}{1-x}\bigg)^n$$
By using given formula ,
*
*$$\frac{1}{1-x}= \sum_{n=0}^{\infty}x^n=1+x+x^2+x^3+...$$ , so the coefficient of $x^n$ is $\binom{n+1-1}{n}$
*$$\frac{1}{1-x^2}= \sum_{k=0}^{\infty}(x^2)^k=1+x^2+x^4+x^6+...$$ , so the coefficient of $x^{2k}$ is $\binom{k+1-1}{k}$
*$$\frac{1}{1-x^3}= \sum_{m=0}^{\infty}(x^3)^m=1+x^3+x^6+x^9+...$$ , so the coefficient of $x^{3m}$ is $\binom{m+1-1}{m}$
where $n+2k+3m =15$ , so find the possible $n,k,m$ values that satify the equation and they are nonnegative integers such that
*
*$n=15,k=0,m=0$
*$n=13,k=1,m=0$
*$n=12,k=0,m=1$
*$n=11,k=2,m=0$
*$n=10,k=1,m=1$
*$n=9,k=3,m=0$
*$n=9,k=0,m=2$
*$n=8,k=2,m=1$
*$n=7,k=4,m=0$
*$n=7,k=1,m=2$
*$n=6,k=0,m=3$
*$n=6,k=3,m=1$
*$n=5,k=5,m=0$
*$n=5,k=2,m=2$
*$n=4,k=4,m=1$
*$n=4,k=1,m=3$
*$n=3,k=6,m=0$
*$n=3,k=3,m=2$
*$n=3,k=0,m=4$
*$n=2,k=5,m=1$
*$n=2,k=3,m=3$
*$n=1,k=7,m=0$
*$n=1,k=4,m=2$
*$n=1,k=1,m=4$
*$n=0,k=0,m=5$
*$n=0,k=3,m=3$
*$n=0,k=6,m=1$
As you see , there are $27$ possible solution that satisfy the equation .Moreover , realize that whichever integer we put in place of $n,k,m$ , the result of binomial coefficient is $1$ , so we just need to sum how many possible solution there are , so the answer is $27$.
I highly recommend you to use wolfram or other softwares to calculate generating functions. I am putting a LINK
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4368201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Using factor theorem on a multivariable function Q: Use the factor theorem to show that $a + b - c$ is a factor of $f(a, b, c) = (a+b+c)^3 - 6(a + b + c)(a^2 + b^2 + c^2) + 8(a^3 + b^3 + c^3)$ and hence factorise completely.
I know that the factor theorem states for $f(x)$ if $f(a) = 0$ then $x-a$ is a factor but when there are multiple variables how do you use factor theorem?
| Here is a compilation of my comments. Denote $c$ by $x$, write your polynomial as a cubic polynomial in $x$ whose coefficients are polynomials in $a,b$:
$$f(x)=f_0(a,b)x^3+f_1(a,b)x^2+f_2(a,b)x+f_3(a,b).$$
Plug in $x=a+b$, you get $f(a+b)=0$. So by the theorem, $x-(a+b)$ is a factor of $f(x)$: $f(x)=(x-(a+b))g(x)$. Then your original polynomial $f(c)=(c-a-b)g(c)=(a+b-c)(-g(c))$, so $a+b-c$ is a factor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4369958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\int\frac {dx}{\cos x-1}$ I was wondering if my solution to the integral: $$\int\frac{dx}{1 - \cos x}$$ is legit?
$$
\int\frac1{\cos x-1}\,\mathrm{d}x=\int-\frac12\cdot\frac1{\sin^2\left(\frac x2\right)}=\frac12\int-\csc\left(\frac12x\right)\,\mathrm{d}x=\boxed{\cot\left(\frac12x\right)+C}\\
\left(\cot\left(\frac12x\right)\right)'=-\csc^2\left(\frac12x\right)\cdot\frac12=\boxed{-\frac12\csc^2\left(\frac12x\right)}\\
\cos x-1=\cos x-\cos(0)=-2\sin\left(\frac{x+0}2\right)\sin\left(\frac{x-0}2\right)=-2\sin^2\left(\frac x2\right)
$$
orginal image
My solution is based around the fact that the derivative of $\cot x$ is $-\csc^2x$.
I basically converted $\cos x-1$ to $\cos x - \cos 0$ and from there used the $\cos a - \cos b$ trig identity.
I then googled the solution and found that there is a way to do it by multiplying by a conjugate but was still wondering if there is a flaw in my logic which I don't realize?
Thanks in advance :)
| Your logic is good, and your answer is correct. In the video, they find that $\int \frac{dx}{1-\cos x} = \csc x + \cot x + C$, and you found that $\int \frac{dx}{1-\cos x} = \cot \frac{x}{2} + C$, and you can show that these are equivalent:
$$\begin{eqnarray}\csc x + \cot x & = & \frac{1}{\sin x} + \frac{\cos x}{\sin x} \\
& = & \frac{1 + \cos x}{\sin x} \\
& = & \frac{1 + (2 \cos^2 \frac{x}{2} - 1)}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \\
& = & \frac{2 \cos^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \\
& = & \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \\
& = & \cot \frac{x}{2}
\end{eqnarray}$$
(Note that there was no guarantee that the $+C$ in both integrals would be the same, but in this case it is.)
| {
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"url": "https://math.stackexchange.com/questions/4370096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\int_0^1 |t-z|^{-1/2}\ \mathrm{d}t < c(1 + |z|)^{-1/2}$ I want to show that there is a constant $c > 0$ such that
$$
\int_0^1 |t-z|^{-1/2}\ \mathrm{d}t < c(1 + |z|)^{-1/2}
$$
for any $z \in \mathbb{C}$.
I found the assertion in a paper I'm reading and I have not succeeded in proving it. Perhaps it uses a common technique in complex analysis (my complex analysis is rusty). Power series expansions came to mind, but I still don't see how to make use of it.
| Alternative proof:
Let $z = a + b\mathrm{i}$.
We have
$$(1 + |z|)^{-1/2} = \frac{1}{\sqrt{1 + \sqrt{a^2 + b^2}}} \ge \frac{1}{\sqrt{1 + |a| + |b|}}. \tag{1}$$
Also, we have
$$|t - z| = \sqrt{(t - a)^2 + b^2} \ge \frac{|t - a| + |b|}{\sqrt2}$$
and thus
$$\int_0^1 |t - z|^{-1/2}\mathrm{d} t
\le \int_0^1 \frac{\sqrt[4]{2}}{\sqrt{|t - a| + |b|}}\mathrm{d}t =: I.$$
If $a\le 0$, we have
$$I = \int_0^1 \frac{\sqrt[4]{2}}{\sqrt{t - a + |b|}}\mathrm{d}t
= \frac{2\sqrt[4]{2}}{\sqrt{|a| + |b| + 1} + \sqrt{|a| + |b|}}
\le \frac{2\sqrt[4]{2}}{\sqrt{|a| + |b| + 1}}. \tag{2}$$
If $a \ge 1$, we have
\begin{align*}
I &= \int_0^1 \frac{\sqrt[4]{2}}{\sqrt{a - t + |b|}}\mathrm{d}t\\
&= \frac{2\sqrt[4]{2}}{\sqrt{|a| + |b|} + \sqrt{|a| + |b| - 1}}\\
&\le \frac{2\sqrt[4]{2}}{\sqrt{|a| + |b|}}\\
&\le \frac{4\sqrt[4]{2}}{\sqrt{|a| + |b| + 1}}.\tag{3}
\end{align*}
If $0 < a < 1$, we have
\begin{align*}
I &= \int_0^a \frac{\sqrt[4]{2}}{\sqrt{a - t + |b|}}\mathrm{d}t
+ \int_a^1 \frac{\sqrt[4]{2}}{\sqrt{t - a + |b|}}\mathrm{d}t\\
&= \frac{2a\sqrt[4]{2}}{\sqrt{a + |b|} + \sqrt{|b|}}
+ \frac{2(1 - a)\sqrt[4]{2}}{\sqrt{1 - a + |b|} + \sqrt{|b|}}\\
&\le \frac{2a\sqrt[4]{2}}{\sqrt{a + |b|}}
+ \frac{2(1 - a)\sqrt[4]{2}}{\sqrt{1 - a + |b|}}\\
&\le \frac{4\sqrt[4]{2}}{\sqrt{1 + |a| + |b|}}
+ \frac{4\sqrt[4]{2}}{\sqrt{1 + |a| + |b|}}\\
&= \frac{8\sqrt[4]{2}}{\sqrt{1 + |a| + |b|}}.\tag{4}
\end{align*}
Remark: Here it is easy to prove that
$\frac{2a\sqrt[4]{2}}{\sqrt{a + |b|}}
\le \frac{4\sqrt[4]{2}}{\sqrt{1 + |a| + |b|}}$
and
$\frac{2(1 - a)\sqrt[4]{2}}{\sqrt{1 - a + |b|}}
\le \frac{4\sqrt[4]{2}}{\sqrt{1 + |a| + |b|}}$.
Using (1)-(4), we have
$$I < \frac{16}{\sqrt{1 + |a| + |b|}} \le 16(1 + |z|)^{-1/2}.$$
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4373339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Showing $\sum_{k=0}^n \binom{2k}{k} \left(\frac{1}{4}\right)^k=\frac{2n+1}{4^n} \binom{2n}{n} $ I want to show that :
\begin{equation} \sum_{k=0}^n \binom{2k}{k} \left(\frac{1}{4}\right)^k=\frac{2n+1}{4^n} \binom{2n}{n} \end{equation}
I have no idea how to. I know it's the series for $\frac{1}{\sqrt{1-x}}$ at $1$ which diverges, but that doesn't give much information on the closed form of the sum. If there's a trick to solve this, I would be eager to know it.
| I think we can show this by induction. The base case $n=0$ is clear. For the induction step, we have that
$$ \sum_{k=0}^{n+1} \binom{2k}{k}\left( \frac{1}{k}\right)^k = \frac{2n+1}{4^n}\binom{2n}{n} + \frac{1}{4^{n+1}} \binom{2(n+1)}{n+1}. $$
To evaluate this sum, recall that $\binom{n}{k}k=\binom{n-1}{k-1}n$ holds for all $n, k$, and so we have
$$ \begin{align*}
\frac{2n+1}{4^n}\binom{2n}{n} &= \frac{n+1}{4^n} \binom{2n+1}{n+1} \\
&= \frac{n+1}{4^n} \binom{2n+1}{n} \\
&= \frac{2n+2}{2 \cdot 4^n} \binom{2n+1}{n} \\
&= \frac{n+1}{2 \cdot 4^n} \binom{2n+2}{n+1}.
\end{align*}$$
This way, we get that
$$ \frac{2n+1}{4^n}\binom{2n}{n} + \frac{1}{4^{n+1}} \binom{2(n+1)}{n+1} = \left( \frac{n+1}{2 \cdot 4^n} + \frac{1}{4^{n+1}} \right) \binom{2(n+1)}{n+1} = \frac{2(n+1)+2}{4^{n+1}} \binom{2(n+1)}{n+1}, $$
which shows the inductive step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
How can we evaluate the integral $\int_{0}^{\infty} \frac{d x}{1+x+x^{2}+\ldots+x^{n-1}}$, where $n\geq 3$? We are going to investigate the integral
$$
I_{n}=\int_{0}^{\infty} \frac{d x}{1+x+x^{2}+\ldots+x^{n-1}} \text {, where } n \geqslant 3.
$$
Let’s start with the simpler cases.
$$
\begin{aligned}
I_{3} &=\int_{0}^{\infty} \frac{d x}{1+x+x^{2}} \\
&=\int_{0}^{\infty} \frac{d x}{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} \\
&=\frac{2}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)\right]_{0}^{\infty} \\
&=\frac{2}{\sqrt{3}}\left(\frac{\pi}{2}-\frac{\pi}{6}\right) \\
&=\frac{2 \pi}{3 \sqrt{3}}
\end{aligned}
$$
and
$$
\begin{aligned}
I_{4} &=\int_{0}^{\infty} \frac{1}{(1+x)\left(1+x^{2}\right)} d x \\
&=\frac{1}{2} \int_{0}^{\infty}\left(\frac{1}{x+1}+\frac{1-x}{x^{2}+1}\right) d x \\
&=\frac{1}{2}\left[\ln (x+1)+\tan ^{-1} x-\frac{1}{2} \ln \left(x^{2}+1\right)\right]_{0}^{\infty} \\
&=\frac{1}{4}\left[\ln \frac{(x+1)^{2}}{x^{2}+1}\right]_{0}^{\infty}+ \left[\frac{1}{2} \tan ^{-1} x\right]_{0}^{\infty} \\
&=\frac{\pi}{4}
\end{aligned}
$$
But the integrals are difficult when $n\geq 5$.
My question: Is there any elementary method to evaluate it?
Your suggestions and solutions are warmly welcome.
| Too long for a comment
$$\int_{0}^{1} \Big(\frac{1-x}{1-x^{n}}+\frac{x^{n-3}-x^{n-2}}{1-x^{n}}\Big) dx=I_1+I_2$$
We can try to use the regularisation to get a shortcut to the answer. For example,
$$I_2=\int_{0}^{1} \frac{x^{n-3}-x^{n-2}}{1-x^{n}}dx=\lim_{\epsilon \to 0}\int_{0}^{1} (x^{n-3}-x^{n-2})(1-x^{n})^{\epsilon-1}dx$$
$$=\frac{1}{n}\lim_{\epsilon \to 0}\int_{0}^{1}(t^{-\frac{2}{n}}-t^{-\frac{1}{n}}) (1-t)^{\epsilon-1}dt$$
$$=\frac{1}{n}\lim_{\epsilon \to 0}\Gamma(\epsilon)\bigg(\frac{\Gamma(1-\frac{2}{n})}{\Gamma(1+\epsilon-\frac{2}{n})}-\frac{\Gamma(1-\frac{1}{n})}{\Gamma(1+\epsilon-\frac{1}{n})}\Bigg)$$
Given that $\Gamma(\epsilon)=\frac{1}{\epsilon}+O(1)$ and $\Gamma(1+\epsilon-\frac{1}{n})=\Gamma(1-\frac{1}{n})+\Gamma(1-\frac{1}{n})\psi(1-\frac{1}{n})\,\epsilon +O(\epsilon^2)$, where $\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}$ - digamma-function
$$I_2=\frac{1}{n}\lim_{\epsilon \to 0}\frac{1}{\epsilon}\bigg(\frac{1}{1+\psi(1-\frac{2}{n})\,\epsilon}-\frac{1}{1+\psi(1-\frac{1}{n})\,\epsilon}\bigg)=\frac{1}{n}\Big(\psi(1-\frac{1}{n})-\psi(1-\frac{2}{n})\Big)$$
$$I_1+I_2=\frac{1}{n}\bigg(\psi\Big(\frac{2}{n}\Big)-\psi\Big(\frac{1}{n}\Big)+\psi\Big(1-\frac{1}{n}\Big)-\psi\Big(1-\frac{2}{n}\Big)\bigg)$$
Using $\psi(1-x)-\psi(x)=\pi\cot\pi x$, we immediately get
$$I_1+I_2=\frac{\pi}{n}\Big(\cot\frac{\pi}{n}-\cot\frac{2\pi}{n}\Big)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Simplifying algebraic expression with fraction of polynomials I cannot seem to find a way to simplify the following expression:
$$\frac{9{a}^{4}-{a}^{2}{ b }^{ 4 } +16 { b }^{ 8 } }{ 3 { a }^{ 2 } -5a { b }^{ 2 } +4 { b }^{ 4 } }$$
I have tried factoring by $a$ and rewriting the division as multiplication with inverse.
| The numerator $9a^4 - a^2 b^4 + 16b^8$ factorises as
$(3a^2+4b^4+5ab^2)(3a^2+4b^4−5ab^2)$.
Thus, when the $3a^2+4b^4−5ab^2\not=0$,
$$\frac{9a^4 - a^2 b^4 + 16b^8}{3a^2+4b^4−5ab^2}= 3a^2+4b^4+5ab^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Limit of $\lim_{n\rightarrow∞} \sum_{k=1}^{n}\frac{n}{(n+k)^2}$ What is the limit of $\lim_{n\rightarrow∞} \sum_{k=1}^{n}\frac{n}{(n+k)^2}$? I got lower bound by plugging $k=n$ which results in limit equal to $1/4$. Similarly, the upper bound is 1, for $k=0$. I do not really know how to proceed, so I would appreciate a hint, I feel like I am missing something obvious.
| How to do it without integrals? Here is one way. But of course the method shown by Nevzat is much simpler.
Upper bound
\begin{align}
\sum_{k=1}^n\frac{n}{(n+k)^2} &< \sum_{k=1}^n\frac{n}{(n+k)(n+k-1)}
\\ &=
\sum_{k=1}^n \left[\frac{k}{n+k} - \frac{k-1}{n+k-1}\right]
\\ &=
\frac{n}{n+n} - \frac{0}{n+0}\qquad\text{(telescoping sum)}
\\ &= \frac{1}{2} .
\end{align}
Lower bound
\begin{align}
\sum_{k=1}^n\frac{n}{(n+k)^2} &> \sum_{k=1}^n\frac{n}{(n+k+1)(n+k)}
\\ &=
\sum_{k=1}^n \left[\frac{k+1}{n+k+1} - \frac{k}{n+k}\right]
\\ &=
\frac{n+1}{n+n+1} - \frac{1}{n+1}\qquad\text{(telescoping sum)}
\\ &=
\frac{n^2}{(2n+1)(n+1)} .
\end{align}
So
$$
\frac{n^2}{(2n+1)(n+1)} < \sum_{k=1}^n\frac{n}{(n+k)^2} < \frac{1}{2} ,
\\
\lim_{n\to\infty}\frac{n^2}{(2n+1)(n+1)}
\le \lim_{n\to\infty}\sum_{k=1}^n\frac{n}{(n+k)^2}
\le \lim_{n\to\infty}\frac{1}{2} ,
\\
\frac{1}{2}
\le \lim_{n\to\infty}\sum_{k=1}^n\frac{n}{(n+k)^2}
\le \frac{1}{2} .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4381287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
$(x^2+1)(y^2-1)dx+xydy=0$
Solve $(x^2+1)(y^2-1) \, dx+xy \, dy=0$
My solution :
$$\begin{align}
\frac{x^2+1}{-x} \, dx &= \frac{-y}{y^2-1} \, dy \\
\left(-x-\frac{1}{x}\right) \, dx &= -\frac{1}{2}\frac{2y}{y^2-1} \, dy \\
\int \left(-x-\frac{1}{x}\right) \, dx &= -\frac{1}{2} \int \frac{2y}{y^2-1} \, dy \\
\frac{x^2}{2}+\ln|x| &= \frac{1}{2}\ln|y^2-1| \\
y^2 &= e^{x^2+2\ln|x|}+1 \\
y &= \pm\sqrt{e^{x^2+2\ln|x|}+1}
\end{align}$$
I got these solutions using ode calculator, and I don't get why it is correct and where am I wrong?
$$y = \pm \frac{\sqrt{e^{-x^2+c_1}+x^2}}x$$
Help please.
Thanks !
| In your calculations you missed a sign and forget to add the constant when integrating.
You should have got $$\pm \sqrt{e^{-x^2-2\log|x|+c}+1}$$
Then you can transform this into the calculator answer like this:
\begin{align}
\pm \sqrt{e^{-x^2-2\log|x|+c}+1}
& = \pm \sqrt{e^{-x^2-\log(x^2)+c}+1}\\
& = \pm \sqrt{e^{-x^2+c}x^{-2}+1}\\
& = \pm \frac{\sqrt{e^{-x^2+c}+x^2}}{x}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4384778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving $5^x - 2^x = 117$ using modular arithmetic I'm trying to solve $5^x - 2^x = 117$. The solution is very easy to find by inspection, however I'm having difficulty trying to find reasoning specifically using modular arithmetic where the solution pops up naturally (i.e. not having to guess).
My work so far has been:
$5^x - 2^x\,$ is strictly increasing and positive only for $x>0$ so the original equation must only have $1$ solution. Assuming there exists an integer solution, the following holds true:
$$\begin{align*}5^x-2^x &= 117\\5^x-125&=2^x-8\\125(5^{x-3}-1)&=8(2^{x-3}-1)\end{align*}$$
This tells us $2^{x-3}-1$ has to be a multiple of $125$, so $$2^{x-3}\equiv1\mod125$$Since $\text{ord}_{125}(2)=100$, $x-3 = 100k\,$ for some $k\in\Bbb N$. Similarly, $5^{x-3}-1$ has to be a multiple of $8$, and $\text{ord}_{8}(5)=2$, so $x-3 = 2m \,$ for some $m \in\Bbb N$. The first equation already encapsulates the second one so $x=100k+3$.
This is where I'm stuck. Any help towards reaching the $x=3$ solution using modular arithmetic would be very much appreciated.
| we may also allow separate exponents for $2,5,$ Very similar to what you did, there is a certain amount of factoring to find useful primes, in this problem we need add just $41$ to the list $2,5,41.$
To begin with $$ 5^u - 125 = 2^v -8$$
As you also used, make new exponents $x,y$ so that
$$ 125 \left( 5^x -1 \right) = 8 \left( 2^y - 1 \right) $$
With the guess that we already know the largest possible $u,v:$ ASSUME that both $x,y \geq 1 $ in the above.
To reduce the size of the numbers, we point out: as $2^y - 1 \equiv 0 \pmod 5,$ we know already that $4|x.$ Just a little more, from $2^y - 1 \equiv 0 \pmod {25},$ we find that $20|x.$
This is enough to go on: $$ 2^{20} - 1 | 2^y - 1 $$ and
$$ 2^{20} - 1 = 3 \cdot 5^2 \cdot 11 \cdot 31 \cdot 41 . $$
Thus $41 | 2^y-1.$ This forces $41 | 5^x - 1 ,$ therefore $20 | x.$
We don't need that much. It is enough to use $4|x.$ Because
$$ 5^4 - 1 = 2^4 \cdot 3 \cdot 13 $$ That is, $5^x-1$ is divisible by $16;$ we reach the needed contradiction from $$ 16 | 8 \left( 2^y - 1 \right). $$ It cannot be true unless $y=0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4386035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How do I get rid of extraneous solutions? I was dealing with this equation today
$ x - k = \sqrt{x^2 + y^2} + \sqrt{(x-a)^2 + y^2}$
Where $a$ and $k$ are constants. I wanted to get rid of the square roots, so I squared the equation. Then I realised that squaring the above equation creates a set of extraneous solutions, so, my question is, is there a way that I can get rid of the extraneous solutions in this equation ? How do I do that ?
Thank you in advance.
| You need to add condition before squaring.
$$x−k=\sqrt{x^2+y^2}+\sqrt{(x−a)^2+y^2}\Leftrightarrow$$
$$x-k-\sqrt{x^2+y^2}=\sqrt{(x−a)^2+y^2}\Leftrightarrow$$
$$\left(x-k-\sqrt{x^2+y^2}\right)^2=(x−a)^2+y^2 \land x-k-\sqrt{x^2+y^2}\geq 0\Leftrightarrow$$
$$x^2-2kx+k^2+x^2+y^2-2(x-k)\sqrt{x^2+y^2}=x^2-2ax+a^2+y^2 \land \\ x-k\geq\sqrt{x^2+y^2}\Leftrightarrow$$
$$x^2+2(a-k)x+k^2-a^2=2(x-k)\sqrt{x^2+y^2} \land x-k\geq\sqrt{x^2+y^2}\Leftrightarrow$$
$$(x^2+2(a-k)x+k^2-a^2)^2=4(x-k)^2(x^2+y^2) \land x^2+2(a-k)x+k^2-a^2 \geq 0 \land \\ x-k\geq\sqrt{x^2+y^2}$$
You can solve equation and check for inequalities. IMHO, the problem is too hard to solve it generally.
| {
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"url": "https://math.stackexchange.com/questions/4387628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $\int_{0}^\infty \frac{1}{(x^2 +b^2)^4}\ dx$ I ran into this integral in the context of Quantum Mechanics, and I don't really know how to tackle it. Here, $b$ is simply a real constant, which we can assume is positive.
$$\int\limits_0 ^\infty\frac{1}{(x^2+b^2)^4}\ dx$$
It doesn't look like I can use "traditional" methods to solve it, so I was thinking to maybe try to transform it to complex integral somehow and apply Cauchy's theorem or something, but I'm unsure if that would even work.
Any nudge in the right direction would be appreciated!
| Here's a short proof for a generalization of the problem. From this answer you know that the integral $I_{\color{blue}{n}} = \int_0^\infty\frac{\mathrm{d}x}{(x^2+1)^{\color{blue}{n}}}$ follows the recursion $I_{n+1}=\frac{2n-1}{2n}I_n$. Solving the recursion gives:
$$
I_{n} = \frac{2n-3}{2n-2}I_{n-1} = \frac{(2n-3)(2n-5)}{(2n-2)(2n-4)}I_{n-2}= \dots = \frac{(2n-3)(2n-5)\dots(5)(3)(1)}{(2n-2)(2n-4)\dots(6)(4)(2)}I_1
$$
And since $I_1 = \int_{0}^{\infty} \frac{\mathrm{d}x}{(1+x^2)^1} = \arctan(x)\Big\vert_{0}^{\infty} = \frac{\pi}{2}$, recalling the definition of the double factorial you get the closed form
$$
I_n =\int_0^\infty\frac{1}{(u^2+1)^{n}}\mathrm{d}u = \frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2}\qquad \text{for} \quad \qquad n \in \mathbb{N}
$$
But since $ \int_0^\infty\frac{1}{(u^2+1)^{n}}\mathrm{d}u = \int_0^\infty\frac{1}{\left(\frac{(ub)^2}{b^2}+\frac{b^2}{b^2}\right)^{n}}\mathrm{d}u = b^{2n}\int_0^\infty\frac{1}{\left((ub)^2+b^2\right)^{n}}\mathrm{d}u $, taking the substitution $x = ub$ gives
$$
\boxed{\int_{0}^{\infty} \frac{1}{(x^2 +b^2)^n}\mathrm{d}x=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2|b|^{2n-1}} \qquad \text{for} \quad b\neq 0, \ n \in \mathbb{N}}
$$
which for your particular case gives $\int_{0}^{\infty} \frac{1}{(x^2 +b^2)^4} = \frac{5!!}{6!!}\frac{\pi}{2b^{2(4)-1}} = \frac{5\pi}{32b^{7}} $ as expected.
| {
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"url": "https://math.stackexchange.com/questions/4391223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplification of $\frac{2\sqrt{21}-\sqrt{35}+5\sqrt{15}-16}{\sqrt7+2\sqrt5-\sqrt3}$
Simplify
$$\dfrac{2\sqrt{21}-\sqrt{35}+5\sqrt{15}-16}{\sqrt7+2\sqrt5-\sqrt3}$$
Final solution should have rational denominators.
Suppose the solution is $X$, I have tried to make up an equation for $X^2$
$$X^2 = \frac{200}{-2\sqrt{15}+2\sqrt{35}-\sqrt{21}+10}$$
My idea is that solving for $X^2$, if can be simply done, can easily give us $X$.
Please help!
| For a shortcut, let $\,a=\sqrt{3}, b=\sqrt{5}, c=\sqrt{7}\,$, then:
$$
\require{cancel}
\begin{align}
& \dfrac{2\sqrt{21}-\sqrt{35}+5\sqrt{15}-16}{\sqrt{7}+2\sqrt{5}-\sqrt{3}} = \frac{2ac-bc+5ab-16}{c+2b-a}
\\ &\quad\quad\quad\quad =\frac{c(2a-b)\color{red}{+(2b-a)(2a-b)-(2b-a)(2a-b)}+5ab-16}{c+2b-a}
\\ &\quad\quad\quad\quad = \frac{(c+2b-a)(2a-b)-\cancel{5ab}+2a^2+2b^2+\cancel{5ab}-16}{c+2b-a}
\\ &\quad\quad\quad\quad = 2a-b+\frac{\cancel{6 + 10 - 16}}{c+2b-a}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4391625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How to evaluate $\int x^n e^{x} \cos x d x$ and $\int x^n e^{x} \sin x d x?$ I am going to generalize the result in my post
$$\int xe^x\cos xdx =\frac{e^{x}}{2}[x \cos x+(x-1) \sin x]+C$$
by finding a reduction formula for
$$I_n=\int x^ne^x\cos xdx .$$
After trying for couple of hours, I found that it is hard to evaluate without its partner integral $$J_n=\int x^ne^x\sin xdx$$
Modified Version
As advised by Hans and Martin, I try to find the reduction formula (3) and (4) by complex numbers.
$$
\begin{aligned}
I_{n}+i J_{n} &=\int x^{n} e^{x}(\cos x+i \sin x) d x \\
&=\int x^{n} e^{x} \cdot e^{x i} d x \\
&=\int x^{n} e^{x(1+i)} d x \\
&=\frac{1}{1+i} \int x^{n} d\left(e^{(1+i) x}\right) \\
&=\frac{1}{1+i}\left[x^{n} e^{(1+i) x}-\int n x^{n-1} e^{(1+i) x} d x\right] \\
&=\frac{1}{1+i}\left[x^{n} e^{(1+i) x}-n\left(I_{n-1}+i J_{n-1}\right)\right] \\
&=\frac{1-i}{2}\left[x^{n} e^{x}(\cos x+i \sin x)-n\left(I_{n-1}+i J_{n-1}\right)\right]
\end{aligned}
$$
Now comparing the real and imaginary parts on both sides yields $$
\begin{array}{l}
I_{n}=\frac{1}{2}\left[x^{n} e^{x}(\cos x+\sin x)-n\left(I_{n-1}+J_{n-1}\right)\right] \\
J_{n}=\frac{1}{2}\left[x^{n} e^{x}(\sin x-\cos x)+n\left(I_{n-1}-J_{n-1}\right)\right]
\end{array}
$$
***Original Method ***
First of all, we need to evaluate the following integrals using integration by parts.
$$
I_0=\int e^{x} \cos x d x=\frac{e^{x}}{2}(\cos x+\sin x)+c_{1}
$$
and $$
J_0=\int e^{x} \sin x d x=\frac{e^{x}}{2}(\sin x-\cos x)+c_{2}
$$
Consequently, $$
\int e^{x}(\cos x +\sin x) d x=e^{x} \sin x+c_3
$$
and $$
\int e^{x}(\cos x-\sin x) d x=e^{x} \cos x+c_4
$$
We then obtain easily$$
\begin{aligned}
I_{n}+J_{n} &=\int x^{n} \cdot e^{x}(\cos x+\sin x) d x \\
&=\int x^{n} d\left(e^{x} \sin x\right) \\
&=x^{n} e^{x} \sin x-n \int x^{n-1} e^{x} \sin x d x \\
&=x^{n} e^{x} \sin x-n J_{n-1}\qquad \qquad\cdots (1)
\end{aligned}
$$
Similarly, $$
I_{n}-J_{n}=x^{n} e^{x} \cos x-n I_{n-1} \qquad\qquad \cdots(2)
$$
$(1)+(2) $ yields
$$
\boxed{I_{n}=\frac{1}{2}\left[x^{n} e^{x}(\cos x+\sin x)-n\left(I_{n-1}+J_{n-1}\right)\right]} \qquad \cdots (3)
$$
$(1)-(2) $ yields
$$\boxed{J_{n}=\frac{1}{2}\left[x^{n} e^{x}(\sin x-\cos x)+n\left(I_{n-1}-J_{n-1}\right)\right]}\qquad \cdots (4) $$
Now let’s try to find $I_2 $using $ (3) $ and $ (4).$
Using (3) yields
$$
\begin{aligned}
I_{1} &=\frac{1}{2}\left[x e^{x}(\cos x+\sin x)-\left(I_{0}+J_{0}\right)\right] \\
&=\frac{e^{x}}{2}[x \cos x+(x-1) \sin x]+C_1
\end{aligned}
$$
Using (4) yields
$$
J_{1}=\frac{e^{x}}{2}\left[x(\sin x-\cos x)+ \cos x\right]+C_2
$$
Using (3) again yields
$$
\begin{aligned}
I_{2} &=\frac{1}{2}\left[x^{2} e^{x}(\cos x+\sin x)-2\left(I_{1}+J_{1}\right)\right] \\
&=\frac{1}{2}\left[x^{2} e^{x} \cos x+x^{2} e^{x} \sin x-e^{x}(2 x \sin x-\sin x-\cos x)\right]\\&= \frac{e^{x}(x-1)}{2}[(x-1) \sin x+(x+1) \cos x]+C_3
\end{aligned}
$$
Using (4) again yields
$$
J_{2}=\frac{e^{x}(x-1)}{2}[(x+1) \sin x-(x-1) \cos x]+C_{4}
$$
My Question
Is there other alternative methods? You opinions and solutions are highly appreciated.
| Note that
\begin{align}
\int x^n e^{ax}dx =\frac{d^n}{da^n}\int e^{ax}dx
=\frac{d^n}{da^n}\left(\frac{e^{ax}}a\right)
= \sum_{k=0}^n \frac{(-1)^k n!}{(n-k)!}
\frac{x^{n-k}e^{ax}}{a^{k+1}}
\end{align}
Then
\begin{align}
\int x^n e^{x}\cos x\>dx
=&\>\Re \int x^n e^{(1+i)x}dx
=\>\Re\sum_{k=0}^n \frac{(-1)^k n!}{(n-k)!} \frac{x^{n-k}e^{(1+i)x}}{(1+i)^{k+1}}\\
=&\sum_{k=0}^n \frac{(-1)^k n! x^{n-k}e^{x}}{2^{\frac{k+1}2}(n-k)!} \cos\bigg(x - \frac{(k+1)\pi}4 \bigg)
\\
\\
\int x^n e^{x}\sin x\>dx
=&\sum_{k=0}^n \frac{(-1)^k n! x^{n-k}e^{x}}{2^{\frac{k+1}2}(n-k)!} \sin\bigg(x - \frac{(k+1)\pi}4 \bigg)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4394997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum and maximum value of $F=|a-2b|+|b-2c|+|c-2a|$
Let $a, b, c \in \Bbb R$ satisfy $a^2+b^2+c^2=21$. Find the minimum and maximum value of $$F=|a-2b|+|b-2c|+|c-2a|$$
I found $7\le F \le \sqrt{399}$ but couldn't prove it. I was thinking of the following inequality:
$$|x_1+x_2+\cdots+x_n|\le|x_1|+|x_2|+\cdots+|x_n|\le \sqrt{n \left(x_1^2+x_2^2+\cdots+x_n^2\right)}$$
but they are not really efficient. Does anyone know how to solve this problem or know where it first appeared?
| 1) For the maximum
It is easy to prove that
$$F = p a + qb + r c$$
for some $p, q, r$ (dependent on $a, b, c$) with
$p^2 + q^2 + r^2 \le 19$.
(See the remarks at the end.)
Using Cauchy-Bunyakovsky-Schwarz, we have
$$pa + qb + rc \le \sqrt{(p^2 + q^2 + r^2)(a^2 + b^2 + c^2) } \le \sqrt{19 \cdot 21} = \sqrt{399}.$$
Also, when $a = \frac{3}{19}\sqrt{399}, b = -\frac{1}{19}\sqrt{399},
c = - \frac{3}{19}\sqrt{399}$,
we have
$a^2 + b^2 + c^2 = 21$
and
$F = |a - 2b| + |b - 2c| + |c - 2a| = \sqrt{399}$.
Thus, the maximum of $F$ is $\sqrt{399}$.
Remarks:
We split into eight cases.
If $a - 2b \ge 0, b - 2c \ge 0, c - 2a \ge 0$, we have
$$F = a - 2b + b - 2c + c - 2a
= -a - b - c.$$
If $a - 2b \ge 0, b - 2c \ge 0, c - 2a < 0$, we have
$$F = a - 2b + b - 2c - c + 2a
= 3a - b - 3c.$$
If $a - 2b \ge 0, b - 2c < 0,
c - 2a \ge 0$, we have
$$F = a - 2b - b + 2c + c - 2a = -a - 3b + 3c.$$
Similarly, we deal with the remaining $5$ cases.
$\phantom{2}$
2) For the minimum
Denote $x = a - 2b, y = b - 2c, z = c - 2a$.
We have
\begin{align*}
F^2 &= (|x| + |y| + |z|)^2\\
&= x^2 + y^2 + z^2
+ 2|xy| + 2|yz| + 2|zx|\\
&\ge x^2 + y^2 + z^2 + 2|xy + yz + zx|\\
&= 105 - 4(ab + bc + ca) + |6(ab + bc + ca) - 84|.
\end{align*}
where we have used
$x^2 + y^2 + z^2 = 5(a^2 + b^2 + c^2) - 4(ab + bc + ca)$,
and $xy + yz + zx = 3(ab + bc + ca) - 2(a^2 + b^2 + c^2)
$.
If $6(ab + bc + ca) - 84 \ge 0$,
we have
\begin{align*}
F^2 &\ge 105 - 4(ab + bc + ca) + 6(ab + bc + ca) - 84\\
& = 21 + 2(ab + bc + ca)\\
&\ge 21 + 2 \cdot \frac{84}{6}\\
& = 49.
\end{align*}
If $6(ab + bc + ca) - 84 < 0$,
we have
\begin{align*}
F^2 &\ge 105 - 4(ab + bc + ca) - 6(ab + bc + ca) + 84\\
&= 189 - 10(ab + bc + ca) \\
&\ge 189 - 10\cdot \frac{84}{6} \\
&= 49.
\end{align*}
Thus, we have $F \ge 7$.
Also, when $a = 4, b = 2, c = 1$, we have
$a^2 + b^2 + c^2 = 21$ and $F = 7$.
Thus, the minimum of $F$ is $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4395233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How do we parametrize the curve determined by the intersection of $x^{2} + y^{2} + z^{2} = 1$ and $x + y + z = 1$ and find it arc length? How do we parametrize the curve determined by the intersection of $x^{2} + y^{2} + z^{2} = 1$ and $x + y + z = 1$?
Should I make a new equation $x^{2} + y^{2} + z^{2} = x + y + z$?
Or do I need to use spherical coordinates to parametrize $x^{2} + y^{2} + z^{2} = 1$ ?
Also, I want to find the arc length of this parametric equation.
| The intersection will be an ellipse in 3-space. Therefore, it is parameterisable by a single variable $t$.
First, solve the equation of the plane for $z$.
$$z = 1-x-y.$$
Then substitute into the equation of the sphere:
$$x^2+y^2+(1-x-y)^2 = 1.$$
Write this equation in the form:
$$\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2} = 1.$$
Then, the parameterisation is given by
$$x = x_0 + a\cos(t),\quad y = y_0 + b\sin(t),\quad z = 1-x_0-y_0-a\cos(t)-b\sin(t), \quad 0 \leq t \leq 2\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4397432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
$\int_{0}^{\infty} \frac{1}{x^4+x^2+1} dx$ with residue theorem Let $C$ be the integration path of a semicircle of radius $R$ denoted by $S(R)$ with a straight line in the real axis from $-R$ to $R$. Then we have that:
$$\oint_{C} \frac{1}{z^4+z^2+1}dz = \left(\int_{0}^{R} +\int_{S(R)} + \int_{-R}^{0}\right) \frac{1}{z^4+z^2+1}dz.$$
Note that:
$$\int_{0}^{R} \frac{1}{z^4+z^2+1}dz = \int_{-R}^{0}\frac{1}{z^4+z^2+1}dz$$
with the substitution $u = -z$. Now,by residue theorem we have that:
$$\oint_{C}\frac{1}{z^4+z^2+1}dz = 2i\pi \left(\frac{1}{2(1+i\sqrt{3})}+\frac{1}{2(-1+i\sqrt{3})}\right) = \frac{\sqrt{3}\pi}{2}$$
Where the terms in the LHS are $2i\pi \sum_{j=1}^2 Res(f(z_j))$ where $z_1,z_2$ are the two roots of the polynomial that are enclosed by $C$. I allready prooved that:
$$\lim_{R\to\infty} \int_{S(R)} \frac{1}{z^4+z^2+1}dz = 0$$
With all these results I could conclude that:
$$\int_{0}^{\infty} \frac{1}{x^4+x^2+1} = \sqrt{3}\frac{\pi}{4}$$
However, by a numerical simulation, the value seems to be $\pi/2\sqrt{3}$, but I have no idea where I do a wrong move.
| It looks like you've made a mistake in the process of computing the residues.
We know $f$ has poles at $\pm \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$, and assuming we're taking the semicircle in the upper half plane this means the relevant poles are $\pm \frac{1}{2} + \frac{\sqrt{3}}{2} i$.
Then we compute the residues to be $\frac{1}{\pm 3 + \sqrt{3}i}$ (which seems to disagree with what you computed in your question), so that the residue theorem gives us
$$
\oint f\ dz =
2 \pi i \left ( \frac{1}{3 + \sqrt{3}i} + \frac{1}{-3 + \sqrt{3}i} \right ) =
\frac{\sqrt{3}\pi}{3}
$$
Since, as you argue, your integral should be half of this, we get an answer that agrees with your numerical simulation.
I hope this helps ^_^
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4399156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How many instances of number $k$ in sums that add up to number $n$? Task in Combinatorics course:
$1 \leq k < n$. Prove that the number $k$ can be found $(n-k+3)2^{n-k-2}$ times in sums that add up to $n$.
Example: $n=4, k=2$.
$1+1+2$
$1+2+1$
$2+1+1$
$2+2$
These are the sums that add up to $4$ and the number $2$ can be found $5$ times.
Thus far, I have tried making sense of that formula. $n-k$ is the number or countables left if we have one $k$. $2^{n-1}$ is how many sums that add up to $n$ there are in total so in the second half of the formula it could be $2^{n-1}2^{k-1}$ that would add up to $2^{n-k-2}$. But none of that comes together. Why would there be a $+3$ in the first half.
| Fix $k$ and let $a_n$ be the number of $k$s that appear among all ($2^{n-1}$) ordered partitions of $n$. Then $a_n=0$ for $n<k$, $a_k=1$, and conditioning on the first part $i$ yields recurrence relation
$$a_n = \sum_{i=1}^n ([i=k]2^{n-i-1} + a_{n-i}) = 2^{n-k-1} + \sum_{i=1}^{n-k} a_{n-i} \quad \text{for $n > k$}.$$
Let $A(z) = \sum_{n=0}^\infty a_n z^n$ be the ordinary generating function. Then
\begin{align}
A(z)
&= \sum_{n=0}^{k-1} 0 z^n + 1z^k + \sum_{n=k+1}^\infty \left(2^{n-k-1} + \sum_{i=1}^{n-k} a_{n-i}\right) z^n \\
&= z^k + \sum_{n=k+1}^\infty 2^{n-k-1} z^n + \sum_{i=1}^\infty z^i \sum_{n=i+k}^\infty a_{n-i} z^{n-i} \\
&= z^k + \frac{z^{k+1}}{1-2z} + \sum_{i=1}^\infty z^i A(z) \\
&= z^k + \frac{z^{k+1}}{1-2z} + \frac{zA(z)}{1-z}.
\end{align}
So
\begin{align}
A(z)
&= \frac{z^k(1-z)^2}{(1-2z)^2} \\
&= z^k \left(\frac{1}{4}+\frac{1/2}{1-2z}+\frac{1/4}{(1-2z)^2}\right) \\
&= z^k \left(\frac{1}{4}+\frac{1}{2}\sum_{n=0}^\infty(2z)^n+\frac{1}{4}\sum_{n=0}^\infty\binom{n+1}{1}(2z)^n\right) \\
&= z^k \left(1+\frac{1}{2}\sum_{n=1}^\infty(2z)^n+\frac{1}{4}\sum_{n=1}^\infty(n+1)(2z)^n\right) \\
&= z^k +\sum_{n=k+1}^\infty \left(2^{n-k-1}+(n-k+1)2^{n-k-2}\right)z^n \\
&= z^k +\sum_{n=k+1}^\infty (n-k+3)2^{n-k-2}z^n,
\end{align}
which immediately implies that
$a_n = (n-k+3)2^{n-k-2}$ for $n>k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Limit of $(1+\frac{1}{n^3})^{n^2}$ I have been trying to solve the limit of $y_n = (1+\frac{1}{n^3})^{n^2}$. Through graphical analysis, I have found that $$\lim_{n \to \infty} y_n = 1$$ Which can also be intuitively be understood as $n^3 \geq n^2$. Using Bernoulli's inequality, you can easily find that $$y_n \geq (1+\frac{n^2}{n^3}) \geq 1$$ I have also found that $$y_n - y_{n+1} \geq \left( 1+\frac{1}{(n+1)^3}\right)^{n^2} - \left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^2} = \left( 1+\frac{1}{(n+1)^3}\right)^{n^2} \left(1 - \left( 1+\frac{1}{(n+1)^3} \right) ^{2n+1} \right) = \left( 1+\frac{1}{(n+1)^3}\right)^{n^2} \left( \frac{1}{(n+1)^3} \right) \left( 1 + \left( 1 + \frac{1}{(n+1)^3} \right) + \cdots + \left( 1+\frac{1}{(n+1)^3}\right)^{2n} \right) \geq 1*0*2n\geq 0 $$$$\implies yn \geq y_{n+1}$$ Thus, by using the monotone convergence theorem, we know $y_n$ converges and has a lower bound of $1$. I am however stuck at showing that $\inf{\{y_n | n \geq 1\}} = 1$, which would show that $\lim_{n \to \infty} y_n = 1$. Could I get a hint or a nudge in the right direction ?
PS: I cannot use exponential and logarithmic properties, nor l'hopital's rule, as we have not defined all these things in class
| You can use binomial expansion and force an upper bound that converges to 1.
$$
(1 + \frac1{n^3})^{n^2} = 1 + \sum_{k=1}^{n^2} {n^2 \choose k} \frac1{n^{3k}} \leq 1 + \sum_{k=1}^{n^2} \frac{n^{2k}}{k!} \frac1{n^{3k}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4402975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\lim_{n\rightarrow \infty} n^{2} \sum_{k=1}^{n}\frac{1}{(n^2+k^2)(\sqrt{n^2+k^2}+n)}$. Find $$\lim_{n\rightarrow \infty} n^{2}\sum_{k=1}^{n}\frac{1}{(n^2+k^2)(\sqrt{n^2+k^2}+n)}.$$
My approach:
\begin{align*}
I& =\lim_{n\rightarrow \infty} n^{2}\sum_{k=1}^{n}\frac{1}{(n^2+k^2)(\sqrt{n^2+k^2}+n)}=\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{(1+(\frac{k}{n})^2)(1+\sqrt{1+(\frac{k}{n})^2})}\\ & =\int_{0}^{1} \frac{1}{(1+x^2)(1+\sqrt{1+x^2})}dx \Longrightarrow
\end{align*}
\begin{align*}I & =\int_{0}^{1} \left(\frac{1}{x^2+1}-\frac{1}{\sqrt{x^2+1}}+\frac{1}{1+\sqrt{x^2+1}}\right)dx\\ & =\arctan(1)-\ln(1+\sqrt{2})+\int_{0}^{1}\frac{1}{1+\sqrt{x^2+1}}dx\end{align*} and at this point I am stuck.
Any help, please?
| For the last integral consider the indefinite integral
$$I=\int \frac{1}{\sqrt{x^2+1}+1}dx=\int \frac{\sqrt{x^2+1}-1}{x^2}dx$$
Here, you can use the hyperbolic trig substitution $x=\sinh t$ for the first summand in the numerator which yields
$$\int \frac{\sqrt{x^2+1}}{x^2}dx=\int \coth ^2 t ~ dt=t-\coth t+C$$
which after substituting back yields an antiderivative for $I$
$$I=\sinh ^{-1}x-\frac{\sqrt{x^2+1}-1}{x}+C$$
Note that this function is, as expected, regular at $x=0$, unlike the one in the 2nd equation above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to tackle the integral $\int_{0}^{1} \sqrt{-1+\sqrt{\frac{4}{x}-3}} d x$? $ \text {Let } y=\sqrt{-1+\sqrt{\frac{4}{x}-3}}\textrm{ then ,}$
$ \displaystyle \begin{aligned}I&=16 \int_{0}^{\infty} \frac{y^{2}\left(y^{2}+1\right) d y}{\left(y^{4}+2 y^{2}+4\right)^{2}}\\&=4\left[3 \underbrace{\int_{0}^{\infty} \frac{y^{2}\left(y^{2}+2\right)}{\left(y^{4}+2 y^{2}+4\right)^{2}} d y}_{J}+\underbrace{\int_{0}^{\infty} \frac{y^{2}\left(y^{2}-2\right)}{\left(y^{4}+2 y^{2}+4\right)^{2}}}_{K} d y\right] \end{aligned}\tag*{} $
Now let’s play a little trick on the integral $ J$.
$\displaystyle \begin{aligned}J &=\int_{0}^{\infty} \frac{1+\frac{2}{y^{2}}}{\left(y^{2}+\frac{4}{y^{2}}+2\right)^{2}} d y \\&=\int_{0}^{\infty} \frac{d\left(y-\frac{2}{y}\right)}{\left[\left(y-\frac{2}{y}\right)^{2}+6\right]^{2}} \\&=\int_{-\infty}^{\infty} \frac{d u}{\left(u^{2}+6\right)^{2}}\\ &\stackrel{u=\sqrt6 \tan \theta}{=}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sqrt{6} \sec ^{2} \theta d \theta}{\left(6 \sec ^{2} \theta\right)^{2}}\\&=\frac{\pi}{12 \sqrt{6}} \end{aligned} \tag*{} $
For the integral $ K$ , we first split the interval into two.
$ \displaystyle \begin{aligned}K &=\int_{0}^{\infty} \frac{1-\frac{2}{y^{2}}}{\left(y^{2}+\frac{4}{y^{2}}+2\right)^{2}} d y \\&=\int_{0}^{1} \frac{1-\frac{1}{y^{2}}}{\left(y^{2}+\frac{4}{y^{2}}+2\right)^{2}} d y+\int_{1}^{\infty} \frac{1-\frac{2}{y^{2}}}{\left(y^{2}+\frac{4}{y}+2\right)^{2}} d y \\&=\int_{0}^{1} \frac{d\left(y+\frac{2}{y}\right)}{\left[\left(y+\frac{2}{y}\right)^{2}-2\right]^{2}}+\int_{3}^{\infty} \frac{d\left(y+\frac{2}{y}\right)}{\left[\left(y+\frac{2}{y}\right)^{2}-2\right]^{2}} d y \\&=\int_{\infty}^{3} \frac{d u}{\left(u^{2}-2\right)^{2}}+\int_{3}^{\infty} \frac{d v}{\left(v^{2}-2\right)^{2}} \\&=0 \end{aligned} \tag*{} $
Now we can conclude that
$\displaystyle \boxed{I=4\left(3 \cdot \frac{\pi}{12 \sqrt{6}}\right)=\frac{\pi}{\sqrt{6}}}\tag*{} $
My Question
Is there any other substitution or method to tackle the integral?
| $$I=\int_0^1\sqrt{\sqrt{\frac{4}{x}-3}-1}dx$$
Start with integration by parts
$$u=\sqrt{\sqrt{\frac{4}{x}-3}-1}⇒du=d\left(\sqrt{\sqrt{\frac{4}{x}-3}-1}\right)$$
$$dv=dx⇒v=x$$
$$I=\int_0^1xd\left(\sqrt{\sqrt{\frac{4}{x}-3}-1}\right)$$
Substitute $u=\sqrt{\sqrt{\frac{4}{x}-3}-1}$
$$I=\int_0^\infty\frac{4du}{u^4+2u^2+4}$$
Substitute $u→\sqrt2u $
$$I=\int_0^\infty\frac{4\sqrt2dx}{4u^4+4u^2+4}=\int_{0}^\infty\frac{\sqrt2du}{u^4+u^2+1}$$
I'll let WolframAlpha take it from here
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Tiles in 2 Boxes Twenty tiles are numbered 1 through 20 and are placed into box $A$. Twenty other tiles numbered 11 through 30 are placed into box $B$. One tile is randomly drawn from each box. What is the probability that the tile from box $A$ is less than 15 and the tile from box $B$ is either even or greater than 25? Express your answer as a common fraction.
Since the two events are independent, we consider each separately. The probability of the tile from A being less than 15 is equal to $\frac{14}{20} = \frac{7}{10}$. The probability of a tile from B being even or greater than 25 is $\frac{10+2}{20} = \frac{3}{5}$. So we multiply the probabilities for the independent events, giving us probability $\frac{7}{10} \cdot \frac{3}{5} = \boxed{\frac{21}{50}}$.
| The probability that the tile from box A is less than $15$ is $\frac{14}{20} = \frac{7}{10}$, because there are $14$ tiles which are less than $15$ in box A, out of the $20$ total tiles it has.
For box B, let $E_1$ be the event of the tile being greater than $25$ and $E_2$ be the event of the tile being even. We want to calculate the proability of at least one of these happening, which can be represented by $P(E_1 \text{ or } E_2)$. By the inclusion-exclusion principle, $P(E_1 \text{ or } E_2) = P(E_1) + P(E_2) - P(E_1 \text{ and } E_2)$. If you don't know what that is, just think of the sets of situations where events $E_1$ and $E_2$ happen. Picture these sets in a Venn diagram and notice that we want the events that happen in the union of these sets. And this union can be thought of as the sum of the quantity of situations in which $E_1$ happens plus the same for $E_2$ minus the situations where they both happen (these were counted twice so we need to subtract them one time).
$P(E_1) = \frac{5}{20}$ because there are $5$ tiles greater than $25$, $P(E_2) = \frac{10}{20}$ because there are $10$ even numbers from $11$ to $30$ and $P(E_1 \text{ and } E_2) = \frac{3}{20}$ because there are only $3$ even tiles greater than $25$ in box B.
So $P(E_1 \text{ or } E_2) = \frac{5}{20} + \frac{10}{20} - \frac{3}{20} = \frac{12}{20} = \frac{3}{5}$. The problem asks for the probability of two independent events (taking tiles from one box doesn't influence the choice of tiles over the other one), therefore we just need to multiply their corresponding probabilities: $$ p = \frac{7}{10} \cdot \frac{3}{5} = \frac{21}{50}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Question about finding minima using Minkowsky's inquality I would like to find the minima of the value
$$\sqrt{(x-1)^2+(x-2)^2}+\sqrt{(x-y)^2+(x)^2}+\sqrt{(y-2)^2+(1)^2}$$
Given that $x$ and $y$ are both real numbers.
I know that I can rewrite this as
$\sqrt{(2-x)^2+(1-x)^2}+\sqrt{(x-y)^2+(x)^2}+\sqrt{(y-2)^2+(1)^2}$ or $\sqrt{(1-x)^2+(2-x)^2}+\sqrt{(x-y)^2+(x)^2}+\sqrt{(y-2)^2+(1)^2}$
But applying Minkowsky's inequality gives value $\sqrt4$ or $\sqrt{10}$, how to deal with this problem?
| You need to consider the equality condition for Minkowski's inequality.
Using Minkowski's inequality, we have
\begin{align*}
&\sqrt{(1-x)^2+(2-x)^2}+\sqrt{(x-y)^2+(x)^2}+\sqrt{(y-2)^2+(1)^2} \\
\ge\,& \sqrt{(1 - x + x - y + y - 2)^2 + (2 - x + x + 1)^2}\\
=\,& \sqrt{10}.
\end{align*}
Equality holds if $x = 5/4, y = 5/3$ (when $(1 - x, x - y, y - 2)$ and $(2 - x, x, 1)$ are proportional).
On the other hand, using Minkowski's inequality, we have
\begin{align*}
&\sqrt{(2-x)^2+(1-x)^2}+\sqrt{(x-y)^2+(x)^2}+\sqrt{(y-2)^2+(1)^2}\\
\ge\,&\sqrt{(2 - x + x - y + y - 2)^2 + (1 - x + x + 1)^2}\\
=\,& \sqrt{4}.
\end{align*}
Equality never holds
($(2 - x, x - y, y - 2)$ and $(1- x, x, 1)$ are never proportional).
| {
"language": "en",
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Prove this $4(a+b+c+d)+(a^3+b^3+c^3+d^3)\le 20$ let $a,b,c,d\in R$,and such $a^2+b^2+c^2+d^2=4$, prove or disprove
$$4(a+b+c+d)+(a^3+b^3+c^3+d^3)\le 20$$
I try use Cauchy-Schwarz inequality have
$$4(a+b+c+d)\le 4\sqrt{4(a^2+b^2+c^2+d^2)}=16$$
but
$$(a^3+b^3+c^3+d^3)^2(1+1+1+1)\ge (a^2+b^2+c^2+d^2)^3$$
so we have
$$a^3+b^3+c^3+d^3\ge 4$$
| You found $a^3+b^3+c^3+d^3\geq 4$
Now using Tchebychev inequality we have:
$$(a+b+c+d)(a^2+b^2+c^2+d^2)\leq 4(a^3+b^3+c^3+d^3)$$
Or:
$$4(a+b+c+d)\leq 4(a^3+b^3+c^3+d^3)$$
$$4(a+b+c+d)+(a^3+b^3+c^3+d^3)\leq 5(a^3+b^3+c^3+d^3)$$
But:
$$5(a^3+b^3+c^3+d^3)\geq 20$$
therefore:
$$4(a+b+c+d)+(a^3+b^3+c^3+d^3)\leq 20$$
| {
"language": "en",
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Find all solutions in set of integers for $a^2 + 5b^2 = 3c^2$ I haven't learnt Diophantine equations to solve this equation (just know Modular Arithmetic). So given the equation we are to solve for all solutions in $\Bbb Z$. So what I've done so far:
Case 1: $\Bbb Z$ $\{0\}$
When $a=b=c=0$ the solution exists (since $0^2 + 5(0^2) = 3(0^2) = 0$)
Case 2: $\Bbb Z \setminus \{0\}$
For $a^2+5b^2=3c^2$ to hold, if $3c^2$ is even (that is, when c is also even), then a and b has to be even (to make $a^2+5b^2$ even).
If $3c^2$ is odd (that is, when c is also odd), then either one of a or b has to be odd (to make $a^2+5b^2$ even).
And attempting in every other way I know I just rearranged the equations in terms of a, b and c to see what I get:
$a =\pm \sqrt{3c^2-5b^2}, b=\pm \sqrt{\frac{3c^2-a^2}5}, c= \pm \sqrt{\frac{a^2+5b^2}3}$
And from this I know that $3c^2\geq5b^2$ but I am not sure how to show this mathematically and where to go from here without using Diophantine equations. I've also looked at Find all solutions: $x^2 + 2y^2 = z^2$
and attempted as suggested in the solutions but making no progress. Any suggestions would be much appreciated.
| For all integers $n$,
$n^2$ has to be $0$, $1$ or $-1$ mod$(5)$.
Suppose $a^2 + 5b^2 = 3c^2$.
Since $a^2\equiv 3c^2$ mod$(5)$,
$c^2\not\equiv \pm1$ mod$(5)$ because $a^2\not\equiv\pm3$ mod$(5)$.
Therefore, $a\equiv c\equiv 0$ mod$(5)$. So $a$ and $c$ are both multiples of $5$.
Let $a=5d$ and $c=5e$.
We get $(5d)^2+5b^2=3(5c)^2$.
This simplifies to $5d^2+b^2=15c^2$, or equivalently $b^2=15c^2-5d^2=5(3c^2-d^2)$.
Therefore, $b$ is a multiple of $5$.
So $a$, $b$, and $c$ are all multiples of $5$.
$\\$
EDIT to help the OP:
The above proof shows that if $a,b,c$ are integers and $a^2 + 5b^2 = 3c^2$, then $\frac{a}{5},\frac{b}{5},\frac{c}{5}$ are also integers.
It is also the case that if $a^2 + 5b^2 = 3c^2$, then $(\frac{a}{5})^2 + 5(\frac{b}{5})^2 = 3(\frac{c}{5})^2$. This you can verify.
So you can apply the same logic to $\frac{a}{5},\frac{b}{5},\frac{c}{5}$. You will get that $\frac{a}{25},\frac{b}{25},\frac{c}{25}$ are integers, and $(\frac{a}{25})^2 + 5(\frac{b}{25})^2 = 3(\frac{c}{25})^2$.
You can repeat this procedure and find out that $\frac{a}{5^n},\frac{b}{5^n},\frac{c}{5^n}$ are integers for all $n\geq 1$. If one of these $a,b,c$ is a non-zero integer, we would get an immediate contradiction. Therefore, we have $a=b=c=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is one root extraneous when using complex number polynomial expansions of cos(4x) to find cos(π/8)? Using De Moivre's theorem, binomial expansion, and the Pythagorean identity, I have the following polynomial for $\cos 4\theta$:
$$\cos\left(4\theta\right) = 8\cos^4\theta -8\cos^2\theta +1 $$
I trying to find an exact value for $\cos\left(\frac{\pi}{8}\right)$.
Since $\cos\left(\frac{\pi}{2}\right)=0$, then $\cos\left(4\cdot\frac{\pi}{8}\right)=0$, so it must be true that
$$ 8\cos^4\left(\frac{\pi}{8}\right) -8\cos^2\left(\frac{\pi}{8}\right) +1 =0$$
and by the quadratic formula I have
$$ \cos \left(\frac{\pi}{8}\right) = \pm\frac{\sqrt{2\pm\sqrt{2}}}{2} $$
Since $\frac{\pi}{8}$ is in the first quadrant, we discard the negative root, so we have
$$ \cos \left(\frac{\pi}{8}\right) = \frac{\sqrt{2\pm\sqrt{2}}}{2} $$
Now, $ \cos \left(\frac{\pi}{8}\right) = \frac{\sqrt{2+\sqrt{2}}}{2} $ is true, but $ \cos \left(\frac{\pi}{8}\right) = \frac{\sqrt{2-\sqrt{2}}}{2} $ is false. (I know this from the approximate numerical value of $\cos\left(\frac{\pi}{8}\right)$.)
My question: How do we know $\frac{\sqrt{2-\sqrt{2}}}{2}$ is an extraneous root?
| Note that by setting $\theta=\frac{3\pi}{8},\frac{9\pi}{8}, \frac{11\pi}{8}$ we recover three more solutions to the very same quartic equation. Since this equation can only have 4 solutions, this is the complete set of solutions.
Now $\cos\frac{9\pi}{8}=-\cos\frac{\pi}{8}$ and $\cos\frac{11\pi}{8}=-\cos\frac{3\pi}{8}$ and therefore, of the positive roots one of them is $\cos\pi/8$ and the other is $\cos 3\pi/8.$ Since $\cos x$ ia s decreasing function in $[0,\pi]$, conclude that the desired value must be the lowest of the two positive ones, and get $\cos 3\pi/8$ for free as well.
| {
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"timestamp": "2023-03-29T00:00:00",
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Ratio of radii of two circles inscribed in a right isosceles triangle. There is a right isosceles triangle $\triangle ABC$ with the vertex $B$ facing the hypotenuse.
A circle is inscribed into the triangle with radius $r_1$, then another circle with radius $r_2$ is inscribed in the leftover space close to either $A$ or $C$ but not $B$
What is the ratio $\large{\frac{r_1}{r_2}}$ equal to?
My Attempt:
Let's call the circle with radius $r_1$, $C_1$ and the other circle with radius $r_2$, $C_2$
The smaller circle shall be closer to vertex $A$.
The Length from $A$ to $C_2$s tangents will be called $h_1$, and from these tangents to $C_1$s tangents will be called $h_2$. The length of the legs of the triangle will be called $x$.
If we draw a line from $C_1$ to $A$ we will see $r_1$ and $r_2$ are bases of similar triangles.
This means $\large{\frac{h_1}{r_2}=\frac{h_1+h_2}{r_1}}$
If we ignore $C_2$ we can see the triangle is made up of four smaller triangles and a square, since the sum of the area of these shapes will be equal to the area of the triangle: $$\large{2r_1(x-r_1)+r_1^2=\frac{x^2}{2}\\2xr_1-r_1^2=\frac{x^2}{2}\\-r_1^2+2xr_1-\frac{x^2}{2}=0}$$ From the quadratic equation:$$\large{\frac{-2x\mp\sqrt{4x^2-2x^2}}{-2}\\x\mp\frac{x}{\sqrt{2}}}$$ Since we know $r_1$ must be less than $x$
$$r_1=x-\frac{x}{\sqrt{2}}$$
$h_1+h_2$ is exactly half of the hypotenuse, this means $h_1+h_2=\frac{x}{\sqrt{2}}$
From this it follows that $$\large{\frac{h_1+h_2}{r_1}=\frac{\frac{x}{\sqrt{2}}}{x-\frac{x}{\sqrt{2}}}=\frac{1}{\sqrt{2}-1}}$$
Since $\large{\frac{h_1}{r_2}=\frac{h_1+h_2}{r_1}}$, $\large{\frac{h_1}{r_2}=\frac{1}{\sqrt{2}-1}}$ and $\large{(\sqrt{2}-1)h_1=r_2}$
Because $h_1+h_2=\frac{r_1}{\sqrt{2}-1}$$$\large{h_2=\frac{r_1-r_2}{\sqrt{2}-1}}$$
If we extend a line like so...
We can see $\large{(h_2)^2+(r_1-r_2)^2=(r_1+r_2)^2}$
| Look at your third diagram.
Follow the line from vertex A to the center of the smaller circle.
Where that first intersects the smaller circle label as point D.
Let $x$ denote the length of segment AD.
Then
\begin{equation}
\frac{r_2}{x+r_2}=\sin\left(\frac{\pi}{8}\right)\tag{1}
\end{equation}
and
\begin{equation}
\frac{r_1}{x+2r_2+r_1}=\sin\left(\frac{\pi}{8}\right)\tag{2}
\end{equation}
Solve equation (1) for $x$ in terms of $r_2$. Then substitute that for $x$ in equation (2).
The resulting equation can be solved for $r_1$ as a fixed constant times $r_2$ from which you can get the desired ratio of $r_1$ to $r_2$.
ALTERNATE APPROACH
This can also be solved using the geometric series:
When $|x|<1$ then
$$ 1+x+x^2+x^3+\cdots =\frac{1}{1-x} $$
Rather than the ratio $\dfrac{r_1}{r_2}$ let us consider its reciprocal $P=\dfrac{r_2}{r_1}<1$.
At present we have two circles, call them $C_1, C_2$. Continue that sequence of circles where $C_3$ bears the same relation to $C_2$ that $C_2$ bears to $C_1$. Keep that going so that we have a sequence of circles in the triangle approaching $A$. Let $D_1=2r_1,D_2=2r_2, D_3=2r_3, \cdots$.
The ratio of the diameters successive diameter will be the same as the ratio of their radii. So each diameter has length $P$ times the length of the diameter of the previous circle.
So the sum of all the circle diameters in the triangle is
$$ S=D_1+D_1P+D_1P^2+D_1P^3+\cdots=\frac{D_1}{1-P}=\frac{2r_1}{1-P} $$
Now we can solve for $S$ another way. It equals the sum of $r_1$ and the distance from the center of $C_1$ to $A$.
$$ S=r_1+r_1\csc\left(\frac{\pi}{8}\right) $$
So we have
\begin{eqnarray}
\frac{2r_1}{1-P}&=&r_1+r_1\csc\left(\frac{\pi}{8}\right)\\
1-P&=&\frac{2}{1+\csc\left(\frac{\pi}{8}\right)}\\
1-P&=&\frac{2\sin\left(\frac{\pi}{8}\right)}{\sin\left(\frac{\pi}{8}\right)+1}\\
P&=&\frac{1-\sin\left(\frac{\pi}{8}\right)}{1+\sin\left(\frac{\pi}{8}\right)}\\
\frac{r_1}{r_2}&=&\frac{1}{P}=\frac{1+\sin\left(\frac{\pi}{8}\right)}{1-\sin\left(\frac{\pi}{8}\right)}\end{eqnarray}
| {
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"url": "https://math.stackexchange.com/questions/4411188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculating area trapped between $g(x)=\frac{8x}{\pi}+\frac{\sqrt{2}}{2}-1$ and $f(x)=\sin(2x) $ I want to calculate the area trapped between $f(x)$ and $g(x)$ while
$$
0\leq x\leq\frac{\pi}{4}
$$
$$
g(x)=\frac{8x}{\pi}+\frac{\sqrt{2}}{2}-1, \qquad f(x)=\sin(2x)
$$
I`m given a clue to remember that:
$$
\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}
$$
What I tried:
I compared $f(x)$ and $g(x)$ and didn't find success. I tried to create new function $h(x)=g(x)-f(x)$ to find the cut points but got stuck as well.
| The given functions are:
$$ f(x) = \sin 2x, \ \ g(x) = {8 x \over \pi} + {\sqrt{2} \over 2} - 1$$
It is easy to note that
$$
f\left({\pi \over 8} \right) = \sin {\pi \over 4} = {\sqrt{2} \over 2} \ \ \mbox{and}
\ \ f\left({\pi \over 8} \right) = 1 + {\sqrt{2} \over 2} - 1
= {\sqrt{2} \over 2}.
$$
Hence, $f$ and $g$ intersect at $x = {\pi \over 8}$ in the region $\left[ 0, {\pi \over 4} \right]$.
Moreover, $f(x) \geq g(x)$ in the region $0 \leq x \leq {\pi \over 8} $.
Also, $g(x) \geq f(x)$ in the region $ {\pi \over 4} \leq x \leq {\pi \over 4} $.
Hence, the area between the curves $f$ and $g$ is found as
$$
I = I_1 + I_2
$$
where
$$
I_1 = \int\limits_{0}^{\pi \over 8} \ [f(x) - g(x)] \ dx, \ \ \
I_2 = \int\limits_{\pi \over 8}^{\pi \over 4} \ [g(x) - f(x)] \ dx.
$$
A simple integration yields
$$
I_1 = {1 \over 2} - {\sqrt{2} \over 4} - {\pi \over 16} -
\left( {\sqrt{2} \over 2} - 1 \right) {\pi \over 8} \tag{1}
$$
and
$$
I_2 = {3 \pi \over 16} +
\left( {\sqrt{2} \over 2} - 1 \right) {\pi \over 8} - {\sqrt{2} \over 4}. \tag{2}
$$
Adding $(1)$ and $(2)$, we get
$$
I = I_1 + I_2 = {1 \over 2} + {\pi \over 8} - {\sqrt{2} \over 2} = {\pi \over 8} + {1 - \sqrt{2} \over 2},
$$
which is the required area of intersection between $f$ and $g$ in the region, where $0 \leq x \leq {\pi \over 4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4414304",
"timestamp": "2023-03-29T00:00:00",
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Finding the symbolic expression for a particular matrix raised to the $n^{\text{th}}$ power Consider the following matrix, with $a$ being a real number between $0$ and $1$.
\begin{equation}
\text{M}=\left[
\begin{array}{cc}
1-\frac{a^2}{3} & \frac{1}{3} \left(2 a-a^2\right) \\
\frac{2 a^2}{3} & 1-\frac{2}{3} \left(2 a-a^2\right) \\
\end{array}
\right].
\end{equation}
For a general integer $n$, I am trying to compute the closed form expression for the entries of:
\begin{equation}
\text{FinalVector}= \text{M}^{n} \left(
\begin{array}{c}
1\\
1 \\
\end{array}
\right).
\end{equation}
It is very easy to compute what $\text{FinalVector}$ is like for a few specific values of $n$, but is there a way to find a formula for a general $n$?
Here are the values for $\text{FinalVector}$ for the first few $n$ (from Mathematica.)
$n = 1$
$$\left(
\begin{array}{c}
-\frac{2 a^2}{3}+\frac{2 a}{3}+1 \\
\frac{4 a^2}{3}-\frac{4 a}{3}+1 \\
\end{array}
\right)$$
$n = 2$
$$\left(
\begin{array}{c}
-\frac{2 a^4}{9}+\frac{10 a^3}{9}-\frac{20 a^2}{9}+\frac{4 a}{3}+1 \\
\frac{4 a^4}{9}-\frac{20 a^3}{9}+\frac{40 a^2}{9}-\frac{8 a}{3}+1 \\
\end{array}
\right)$$
$n = 3$
$$\left(
\begin{array}{c}
-\frac{2 a^6}{27}+\frac{2 a^5}{3}-\frac{22 a^4}{9}+\frac{122 a^3}{27}-\frac{14 a^2}{3}+2 a+1 \\
\frac{4 a^6}{27}-\frac{4 a^5}{3}+\frac{44 a^4}{9}-\frac{244 a^3}{27}+\frac{28 a^2}{3}-4 a+1 \\
\end{array}
\right)$$
$n = 4$
$$\left(
\begin{array}{c}
-\frac{2 a^8}{81}+\frac{26 a^7}{81}-\frac{16 a^6}{9}+\frac{440 a^5}{81}-\frac{812 a^4}{81}+\frac{308 a^3}{27}-8 a^2+\frac{8 a}{3}+1 \\
\frac{4 a^8}{81}-\frac{52 a^7}{81}+\frac{32 a^6}{9}-\frac{880 a^5}{81}+\frac{1624 a^4}{81}-\frac{616 a^3}{27}+16 a^2-\frac{16 a}{3}+1 \\
\end{array}
\right)$$
| $ M = I + B = I + \begin{bmatrix} 1 \\ -2 \end{bmatrix} \begin{bmatrix} \alpha && \beta \end{bmatrix} $
where $\alpha = -\frac{1}{3} a^2 $ and $\beta = \frac{1}{3} (2 a - a^2) $
Simple analysis shows the eigenvalues of $B$ are $0$ with corresponding eigenvector $[- \beta, \alpha]^T$, and $\alpha - 2 \beta $ with corresponding eigenvector $[1, -2]^T$
Now we want to express $[1,1]^T$ as a linear combination of these two eigenvectors.
$[1,1]^T = c_1 [-\beta, \alpha]^T + c_2 [1, -2]^T $
Solving yields,
$c_1 = \dfrac{3}{\alpha - 2 \beta} , \hspace{40pt} c_2 = \dfrac{\alpha+\beta}{\alpha - 2 \beta} $
Hence,
$M^n [1, 1]^T = c_1 (1)^n [-\beta, \alpha]^T + c_2 (\lambda_2)^n [1, -2]^T $
where $\lambda_2 = 1 + \alpha - 2 \beta \lt 1 $
Therefore,
$\displaystyle \lim_{n \to \infty} M^n [1, 1]^T = c_1 [-\beta, \alpha]^T = \dfrac{3}{\alpha-2\beta} \begin{bmatrix} -\beta \\ \alpha \end{bmatrix} $
And this simplifies to
$\displaystyle \lim_{n \to \infty} M^n [1, 1]^T = \dfrac{3}{\frac{a}{3}(a - 4) } \begin{bmatrix} - \frac{a}{3} (2 - a) \\ -\frac{1}{3} a^2 \end{bmatrix} $
And this simplifies further to,
$\displaystyle \lim_{n \to \infty} M^n [1, 1]^T = 3 \begin{bmatrix} \frac{2 -a}{4 - a} \\ \frac{a}{4 - a} \end{bmatrix} $
| {
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Prove that $(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥2(x+y+z)^2$
Let, $x,y,z>0$ such that $xyz=1$, then prove that
$$(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥2(x+y+z)^2$$
My progress:
Using the Cauchy-Schwars inequality I got
$$(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥2(xy+yz+xz)(x+y+z)≥2(x+y+z)^2 \implies xy+yz+xz≥x+y+z$$
But, this is not always true.
I also tried
$$x^2+y^2+z^2≥\frac{(x+y+z)^2}{3}$$
I got
$$(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥(xy+yz+xz)\left(\frac{(x+y+z)^2}{3}+x+y+z\right)≥2(x+y+z)^2 \implies (xy+yz+xz)(x+y+z+3)≥6(x+y+z)$$
But, again I failed.
| Let $S=x+y+z, T=xy+xz+yz$ and note that $S,T \ge 3$ by AMGM and $xyz=1$
The inequality is: $T(S^2+S-2T) \ge 2S^2$
Case 1: $S \ge T$ then $S^2(T-2) \ge S^2 \ge T^2 \ge T(2T-S)$ which rewrites to the required inequality
Case 2: $S \le T$; then we have $T(S^2+S-2T) \ge S(S^2+S-2T)$ so we need to prove only that $S^2+S-2T \ge 2S$ or that $S^2 \ge S+2T$. But (by C-S) $S^2 \le 3(S^2-2T)$ so $S^2 \ge 3T \ge 2T+S$ and we are done
| {
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"url": "https://math.stackexchange.com/questions/4421193",
"timestamp": "2023-03-29T00:00:00",
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Solving $3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8 + \sqrt{3^{16} + ...}}}}$
How to find the value of $3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8 + \sqrt{3^{16} + ...}}}}$
I tried to solve it and found a relation that if I assume the given expression to be something say $x$, then following result holds true.
$$\boxed{x = 3 + \sqrt{3x}}\hspace{4cm}\bf ...(1.)$$
Can be proved as,
$$x = 3 + \sqrt{3(3 + \sqrt{3x})}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^3x}}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^3(3 + \sqrt{3x})}}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^7x})}}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^7(3 + \sqrt{3x})})}}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8 + \sqrt{3^{15}x}}}}$$
If I continue this process repeatedly,
$$x= 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^{8} + \sqrt{3^{16} +\sqrt{...}}}}}$$
But how can I make sure that $(1.)$ is absolutely correct? Also how can I make sure that "If I continue this process repeatedly" I would get the same expression?
| If we have
$$a_{n+1}=x \sqrt{x\,a_n} \qquad \text{with}\qquad a_0=x$$
$$a_n=x^{b_n} \implies x^{b_{n+1}}=x \sqrt{x^{b_n+1}} \implies b_{n+1}=\frac 12 b_n+\frac 32$$
$$b_n=c_n+3\implies c_{n+1}=\frac 12 c_n$$
Finishing and back to $a_n$
$$a_n=x^{3-2^{1-n}}\quad \to \quad x^3$$
| {
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Geometric sequence $a,b,c,d$ and arithmetic sequence $a, \frac{b}{2},\frac{c}{4}, d-70$ The first four terms, given in order, of a geometric sequence $a,b,c,d$ and arithmetic sequence $a, \frac{b}{2},\frac{c}{4}, d-70$, find the common ratio $r$ and the values of each $a,b,c,d$.
What I have tried:
$b=ar, c=ar^2,d=ar^3$, we have find the value of $b$ by calculating the mean from the values $1$ and $3$.$$ar=a+\frac{ar^2}{4} \implies 0 = a(r^2-4r+4a) \\0 = a(r-2)(r+2)$$
To calculate for $c$ we take the second and fourth term of the sequence.
$$\frac{ar^2}{2}=\frac{ar}{2}+ar^3-70 \implies 70 = a(r^3-\frac{r^2}{2}+\frac{r}{2})$$
However, this seems a little messy because it involves complex numbers - am I still on the right track?
| $b = a r, c = a r^2, d = a r^3 $
Using a constant common difference,
$\frac{1}{2} a r - a = \frac{1}{4} a r^2 - \frac{1}{2} a r = a r^3 - 70 - \frac{1}{4} a r^2 $
Re-arranging
$ r^2 - 4 r + 4 = 0\hspace{25pt}(1) $
and
$ a r^3 - \dfrac{1}{2} a r^2 + \dfrac{1}{2} a r = 70\hspace{25pt}(2) $
From $(1)$ , $ r = 2 $
Substitute this into $(2)$:
$ a ( 8 - 2 + 1 ) = 70 $
So $a = 10 $
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the solutions to $(z+1)^7 = z^7$ have same real parts Prove that all the solutions to the equation $(z+1)^7=z^7$ have equal real parts, and find what this real part is equal to.
I found that $z=\frac{1}{e^{2\pi i/7}-1}, \frac{1}{e^{4\pi i/7}-1}, \frac{1}{e^{6\pi i/7}-1}, \frac{1}{e^{8\pi i/7}-1}, \frac{1}{e^{10\pi i/7}-1}, \frac{1}{e^{12\pi i/7}-1},$ but I'm not sure what to do next.
| Based on the proposed equation, one concludes that
\begin{align*}
(z + 1)^{7} = z^{7} & \Rightarrow |z + 1| = |z|\\\\
& \Rightarrow |z + 1|^{2} = |z|^{2}\\\\
& \Rightarrow z\overline{z} + z + \overline{z} + 1 = z\overline{z}\\\\
& \Rightarrow 2\operatorname{Re}(z) = -1\\\\
& \Rightarrow \operatorname{Re}(z) = -\frac{1}{2}
\end{align*}
and we are done.
Hopefully this helps !
| {
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Find $x$ such that $\left\vert -a x\pm \sqrt{a^2x^2-1}\right\vert\leq 1.$ Given inequality
\begin{align*}
\left\vert -a x\pm \sqrt{a^2x^2-1}\right\vert\leq 1.
\end{align*}
for
$-1\leq a\leq 1,\quad a^2x^2-1\geq 0,\quad$ $x\in\mathbb R$.
Determine the what the range of $x$?.
I try as follows.
\begin{align}
&\left\vert -a x\pm \sqrt{a^2x^2-1}\right\vert\leq 1\\
\iff& \left\vert -a x\right\vert+\left\vert\pm \sqrt{a^2x^2-1}\right\vert\leq 1 \quad (\text{triangle inequality})\\
\iff& \left\vert a\right\vert \left\vert x\right\vert+\sqrt{a^2x^2-1}\leq 1 \quad (\text{absolute value properties})\\
\iff& \left\vert x\right\vert+\sqrt{x^2-1}\leq 1 \quad (\max a =1)\\
\iff& \left\vert x\right\vert+\vert x\vert\leq 1 \quad (\sqrt{x^2-1} \leq \sqrt{x^2}=\vert x\vert)\\
\iff& \vert x\vert\leq \frac{1}{2}\\
\iff& -\frac{1}{2}\leq x\leq \frac{1}{2}.\\
\end{align}
Now take $a=0.5$ and $x=2$. Then
$$\vert-0.5\cdot 2\pm \sqrt{0} \vert= 1\leq 1.$$
Why if I take $a=0.5$ and $x=2$ then inequality satisfied? Even though I have found the range of $x$ in $-\frac{1}{2}\leq x\leq \frac{1}{2}$.
I feel in finding $x$ is correct, but the result is incorrect. Anyone know my mistake?
| The triangle inequality won't help because you want transformations that are equivalences.
Setting $y=ax$ and squaring both sides:
$$y^2 \mp 2y \sqrt{y^2-1} +y^2-1\leqslant 1$$
then rearranging gives
$$ y^2 - 1 \leqslant \pm y \sqrt{y^2-1}$$
which holds if $y^2=1$, or if $y^2>1$ we divide by the square root to get
$$\sqrt{y^2-1} \leqslant \pm y$$
*
*If the inequality is supposed to hold in either $\pm$ cases, then it follows that $y=0$ and the square root must be zero, too. However, the root is only defined if $|y|\geqslant 1$.
*If just one of the $\pm$ cases must hold, then we may use $$\sqrt{y^2-1}\leqslant |y|$$ which always holds true.
| {
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Growth of a circular ink blot grows at the rate of $2\text{ cm}^2$ per second A circular ink blot grows at the rate of $2\text{ cm}^2$ per second. Find the rate at which the radius is increasing after $2\frac{6}{{11}}$ seconds. Use $\pi = \frac{{22}}{7}$.
My solution is as follow let $A = \pi {r^2}$ where $A$ is the volume
$\frac{{dA}}{{dt}} = 2\pi r\frac{{dr}}{{dt}}$, given $\frac{{dA}}{{dt}} = 2\text{ cm}^2/\sec $ and $r = 2\frac{6}{{11}}\text{ cm} = \frac{{28}}{{11}}\text{ cm}$. Solving we get $2\text{ cm}^2/\sec = 2 \times \frac{{22}}{7} \times \frac{{28}}{{11}}\frac{{dr}}{{dt}} \Rightarrow \frac{{dr}}{{dt}} = \frac{1}{8}\text{ cm}/\sec $.
But the correct answer is $\frac{1}{4}\text{ cm}/\sec$.
Where am I commiting mistake?
| Area after ${28 \over 11} \mbox{sec} = 2 \times {28 \over 11} = \pi r^2$
Simplifying,
$$
r^2 = {56 \over 11 \pi}
$$
Thus,
$$
r = \sqrt{ {56 \over 11 \pi}}
$$
Since $A = \pi r^2$, we have
$$
{dA \over dt} = 2 \pi r {dr \over dt}
$$
Thus,
$$
2 = 2 \pi \sqrt{ {56 \over 11 \pi}} {dr \over dt}
$$
Thus,
$$
1 = \sqrt{56 \pi \over 11} \ {dr \over dt}
$$
or
$$
{dr \over dt} = \sqrt{11 \over 56 \pi}
$$
Using the approximation $\pi \approx {22 \over 7}$, we get
$$
{11 \over 56 \pi} = {11 \over 56 \times {22 \over 7}} = {11 \over 56} \times {7 \over 22} = {1 \over 16}
$$
Hence,
$$
{dr \over dt} = \sqrt{1 \over 16} = {1 \over 4}
$$
| {
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Integrate $\iint_D \frac1{(x^2+y^2)^2 }dx dy$ over a region bounded by four circles
I'm thinking of doing the substitution $x = \frac{u}{u^2 + v^2}, y = \frac{v}{u^2 + v^2},$ but I'm not sure how to exactly compute the range of values $u$ and $v$ take. Clearly, $u=\frac{x}{x^2+y^2}$ and $v = \frac{y}{x^2 + y^2}.$ The Jacobian of the result is $-\frac{1}{u^2 + v^2}$, and so the resulting integral equals $\iint_{D'} dudv,$ where $D'$ is the region of possible values for the pairs $(u,v)$. In the circle $x^2 + y^2 - 2x= 0$, for instance, we have the point $(1,0)$, which corresponds to $u=1, v=0$. Also, we have the point $(2,0)$, corresponding to $u= \frac{1}2, v = 0$. I can't seem to generalize what values u and v can take on.
| In polar coordinates, the enclosed region can be partitioned into three angular spans, with break points corresponding to the intercepts of the circles, at $\tan^{-1}\frac13, \tan^{-1}\frac23, \frac\pi4$ and $\tan^{-1}2$. Then
\begin{align}
&\iint_D \frac1{(x^2+y^2)^2 }dx dy\\
=& \int_{\tan^{-1}\frac13}^{\tan^{-1}\frac23}\int_{2\cos\theta}^{6\sin\theta}
+ \int_{\tan^{-1}\frac23}^{\frac\pi4}\int_{2\cos\theta}^{4\cos\theta}
+ \int_{\frac\pi4}^{\tan^{-1}2}\int_{2\sin\theta}^{4\cos\theta}\frac1{r^3} dr d\theta\\
=&\>\frac1{48}+ \frac1{32} +\frac1{32}=\frac1{12}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $a\gt 2 $ be a root of the equation $x^3-x-8 = 0$. Compute $\sqrt[3]{6a^2-13a} + \sqrt[3]{6a^2+13a+16}$ I came across this question in some past contest prep papers belonging to Akdeniz University. To solve it, I've tried letting the cuberoots equal some $k$ and $l$ and tried writing $k+l$ in terms of $kl$ and $k^3+l^3$ but to no avail. Any help would be appreciated.
| Using $a^3-a-8=0$ one gets:
$$(2+a)^3=a^3+6a^2+12a+8=(a^3-a-8)+6a^2+13a+16=6a^2+13a+16$$
$$(2-a)^3=-a^3+6a^2-12a+8=-(a^3-a-8)+6a^2-13a=6a^2-13a$$
So $\sqrt[3]{6a^2+13a+16}+\sqrt[3]{6a^2-13a}=2+a+2-a=4$
| {
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Proving $x^2 + y^2 + z^2 + x + y + z \geq 2(xy+yz+zx)$ for positive values with $xyz=1$
How can you prove that
$$x^2 + y^2 + z^2 + x + y + z \geq 2(xy+yz+zx)$$
given that $x,y,z > 0$ and $xyz = 1$.
We can easily prove that the equality holds when $x = y = z = 1$
I could able to prove the result when one of $x,y,z$ is $1$ considering $x,y,z$ as $x,1/x, 1$ using the inequality $x + 1/x \geq 2$ for any positive number of $x$.
But couldn't able to find a full proof.
| Let $x=a^3, y=b^3, z=c^3$ then we have that $abc=1$ and we want to prove that $a^6+b^6+c^6+(a^3+b^3+c^3)=a^6+b^6+c^6+abc(a^3+b^3+c^3)\ge 2(a^3b^3+b^3c^3+c^3a^3)$. Now by Schur's inequality for $t=4$, $a^6+b^6+c^6+abc(a^3+b^3+c^3)\ge \sum_{sym} a^5b$ and by Muirhead's inequality $\sum_{sym} a^5b \ge \sum_{sym} a^3b^3=2(a^3b^3+b^3c^3+c^3a^3)$ and thus we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Power series of $\ln\frac{k}{k-1}$ In "Concrete Mathematics", equation (6.63) says that $\ln\frac{k}{k-1}$ expands to the infinite series:
$$
\ln\frac{k}{k-1}=\frac{1}{k} + \frac{1}{2k^2}+\frac{1}{3k^3}+\cdots
$$
But I couldn't see why it can expand in such a way.
| Thanks for the quick comments, somtimes I just couldn't see such fundamental things...
$$
\begin{align}
\ln\frac{k}{k-1} &= -\ln\frac{k-1}{k} \\
&= - \ln\left({1-\frac{1}{k}}\right) \\
&= -\left(-\frac{1}{k}-\frac{1}{2k^2}-\frac{1}{3k^3} - \cdots\right) \\\\
&= \frac{1}{k} + \frac{1}{2k^2}+\frac{1}{3k^3} + \cdots
\end{align}
$$
| {
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How to find point of intersection with conic section and tangent dropped from a point not on a conic Question: We have conic section $-12x^2 + 28xy+4x-9y^2-8y=0$ and a point not on a conic $(2/5,1/5)$, how to find an intersection point with tangent dropped from $(2/5,1/5)$ to a conic section?
My ideas: to use polar points and Homogeneous coordinates
| The given equation of the conic is
$ - 12 x^2 + 28 xy - 9 y^2 + 4 x - 8 y = 0 \hspace{15pt}(1) $
And the given point is $(\dfrac{2}{5}, \dfrac{1}{5})$
Using implicit differentiation, with respect to $x$,
$ - 24 x + 28 (y + x y') - 18 y y' + 4 - 8 y' = 0 $
Therefore,
$ y' (28 x - 18 y - 8) = 24 x - 28 y - 4 $
i.e.
$ y' = \dfrac{ 24 x - 28 y - 4 }{28 x - 18 y - 8} = \dfrac{12 x - 14 y - 2}{14 x - 9 y - 4 } $
Now $y'$ is the slope of the tangent line, which is also given by
$ y' = \text{Slope} = \dfrac{ \dfrac{1}{5} - y }{\dfrac{2}{5} - x} $
Therefore, by cross multiplication,
$ ( 12 x - 14 y - 2 ) (\dfrac{2}{5} - x) = (14 x - 9 y - 4) (\dfrac{1}{5} - y) $
Multiply through by $5$
$ (12 x - 14 y - 2) (2 - 5 x) = (14 x - 9 y - 4) ( 1 - 5 y)$
Expand
$ 24 x - 60 x^2 - 28 y+ 70 x y - 4 + 10 x = 14 x - 9 y - 4 - 70 x y + 45 y^2 + 20 y$
And this reduces to
$ -60 x^2 + 140 x y - 45 y^2 + 20 x -39 y = 0 \hspace{15pt} (2)$
Now we have to solve the system of equations $(1),(2)$, and the solutions are
$(0, 0) $ and $(\dfrac{1}{3} , 0 ) $
| {
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Prove that $p$ divides $(p-1)!\sum_{k=1}^{p-1}(-1)^{k-1}x^k/k $ Let $p$ be an odd prime such that $p\mid x^3-1$ but $p\not\mid x-1$. Prove that $$p \mid (p-1)!\sum_{k=1}^{p-1}(-1)^{k-1}\frac{x^k}{k} $$
This is my work,
Lemma if $p$ is prime then $$k^{-1}\equiv (-1)^{k-1}{p \choose k}/p\pmod p$$
Now note that $$\sum_{k=1}^{p-1}(-1)^{k-1}\frac{x^k}{k}\equiv \sum_{k=1}^{p-1}x^k{p \choose k}/p\equiv\frac{1}{p}\sum_{k=1}^{p-1}x^k{p \choose k}\pmod p$$
And we know that $$(x+1)^p=\sum_{k=0}^{p}x^k{p \choose k}=1+x^p+\sum_{k=1}^{p-1}x^k{p \choose k}$$
Hence $$\sum_{k=1}^{p-1}x^k{p \choose k}\equiv 0\pmod p.$$
Also we know $(p-1)!\equiv -1\pmod p$, so $$(p-1)!\sum_{k=1}^{p-1}(-1)^{k-1}\frac{x^k}{k}\equiv -\frac{1}{p}\sum_{k=1}^{p-1}x^k{p \choose k}\pmod p$$
so we need to prove this last guy is divisible by $p$, or equivalently $$\sum_{k=1}^{p-1}x^k{p \choose k}\equiv 0\pmod{p^2}$$
| Note first that because of $p|(x^3-1)$ and $p\not | (x-1)$ we have $x^3=1$, $x\ne1$ and and $x^2+x+1=0$ modulo $p$. Besides there are only two values $a,b\in \mathbb F_p$ satisfying the equation $x^2+x+1=0$ and we have $a+b=-1$ and $ab=1$.
Further the only involved primes are those of the form $3n+1$ because if $p=3n+2$ then (Fermat's little theorem) $x^{3n+1}=x(x^3)^n=x=1$ modulo $p$. Moreover the factor $(p-1)!$ is only the factor $-1$ modulo $p$ (Wilson's theorem) therefore in
$$p \mid (p-1)!\sum_{k=1}^{p-1}(-1)^{k-1}\frac{x^k}{k}\iff(p-1)!\sum_{k=1}^{p-1}(-1)^{k-1}\frac{x^k}{k}\equiv0\pmod p$$ we can consider just $$\sum_{k=1}^{p-1}(-1)^{k-1}\frac{x^k}{k}\equiv0\pmod p$$ for the only values $a,b$ of $x$ and $p$ a prime of the form $3n+1$.
$$*****$$
So we want to prove for $x=a$ and $x=b=\dfrac1a$ and the primes $p$ with the form $p\equiv1\pmod3$ $$\sum_{k=1}^{p-1}(-1)^{k-1}\frac{x^k}{k}\equiv0\pmod p$$ We have
$$\sum_{k=1}^{p-1}(-1)^{k-1}\frac{a^k}{k}=\left(a-\frac{a^2}{2}+\frac{a^3}{3}\right)-a^3\left(\frac a4-\frac{a^2}{5}+\frac{a^3}{6}\right)+a^6\left(\frac a7-\frac{a^2}{8}+\frac{a^3}{9}\right)\cdots -a^{3(n-1)}\left(\frac {a}{p-3}-\frac{a^2}{p-2}+\frac{a^3}{p-1}\right)$$
$$\sum_{k=1}^{p-1}(-1)^{k-1}\frac{a^k}{k}=\left(\frac a1+\frac{a+1}{2}+\frac{1}{3}\right)-\left(\frac a4+\frac{a+1}{5}+\frac{1}{6}\right)+\left(\frac a7+\frac{a+1}{8}+\frac{1}{9}\right)\cdots -\left(\frac {a}{p-3}+\frac{a+1}{p-2}+\frac{1}{p-1}\right)$$
This gives for some $A,B,C,D$ in $\mathbb F_p$
$$\sum_{k=1}^{p-1}(-1)^{k-1}\frac{a^k}{k}=Aa+B$$ if $Aa+B=C\iff Aa+D=0$. But we also have $$\dfrac Aa+D=0$$ therefore $$A\left(a-\frac 1a\right)=0$$ from which $A=0$ and $D=0$ modulo $p$. We are done.
| {
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Find the minimum of $\frac{a^2+b^2}{2}-\frac{|a-b|}{2}\sqrt{1+(a+b)^2}$ When $a\neq b$ $a,b\in\mathbb{R}$
find the minimum of
$\frac{a^2+b^2}{2}-\frac{|a-b|}{2}\sqrt{1+(a+b)^2}$
I think that the answer is $-\frac{1}{4}$ but not sure.
Please tell me a solution.
| Since swapping $a$ and $b$ does not change the value of the expression, let's restrict ourselves only to solutions where $a\geq b$. This allows us to rewrite $|a-b|$ as $a-b$, which makes our expression much easier to deal with.
Let $c=\frac{a+b}{2}$ and $d=\frac{a-b}{2}$. Then $c$ can be any real number, and $d$ can be any non-negative real. Written in terms of $c$ and $d$, our expression becomes:
$c^2 + d^2 - d\sqrt{4c^2+1}$
Differentiating with respect to $d$ gives $2d - \sqrt{4c^2+1}$. This is zero when $d = \sqrt{c^2 + 1/4}$ which means this value of $d$ minimises the expression.
If you plug this value of $d$ into the expression, the $c$s cancel out and the expression reaches its minimum value of $-1/4$ as you suggested.
If you have a particular value for $a$ and you want a corressponding value of $b$ that achieves this minimum, remember that $a=c+d$ and $d = \sqrt{c^2 + 1/4}$. Solving this gives $c=\frac{a}{2}-\frac{1}{8a}$ and $d=\frac{a}{2}+\frac{1}{8a}$. We get $b=c-d =-\frac{1}{4a}$. This only works if this value of $b$ is less than or equal to $a$, which we see is only when $a>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrating the function $f(x,y)=x$ over the area inside the disc $x^2+y^2\le 4$ and outside the disc $(x-1)^2+y^2=1.$
Integrate the function $f(x,y)=x$ over the area inside the disc $x^2+y^2\le 4$ and outside the disc $(x-1)^2+y^2=1.$
My attempt:
I chose shifted polar coordinates $$\begin{cases}x=2+r\cos\varphi\\ y=r\sin\varphi.\end{cases}$$ Then $$\begin{aligned}x^2+y^2&=4\\\iff (2+r\cos\varphi)^2+(r\sin\varphi)^2&=4\\\iff r(2\cos\varphi+r)&=0\end{aligned}$$ and $$\begin{aligned}(x-1)^2+y^2&=1\\\iff (1+r\cos\varphi)^2+(r\sin\varphi)^2&=1\\\iff r(4\cos\varphi)&=0\end{aligned}$$
So, I'm integrating over the set $$\left\{(\varphi,r),\frac\pi2\le\varphi\le\frac{3\pi}2, -2\cos\varphi\le r\le -4\cos\varphi\right\}.$$
The integral becomes $$\begin{aligned}\int_{\pi/2}^{3\pi/2}\int_{-2\cos\varphi}^{-4\cos\varphi}r(2+r\cos\varphi)drd\varphi&=\int_{\pi/2}^{3\pi/2}\left(r^2+\frac{r^3}3\cos\varphi\Big|_{-2\cos\varphi}^{-4\cos\varphi}\right)d\varphi\\&=\int_{\pi/2}^{3\pi/2}12\cos^2\varphi-\frac{56}3\cos^4\varphi d\varphi\\&=\int_{\pi/2}^{3\pi/2}\frac{56}3\cos^2\varphi(1-\cos^2\varphi)d\varphi-\int_{\pi/2}^{3\pi/2}\frac{20}3\cos^2\varphi d\varphi\\&=\frac{14}3\int_{\pi/2}^{3\pi/2}(2\sin\varphi\cos\varphi)^2-\frac{10}3\int_{\pi/2}^{3\pi/2}(1+\cos(2\varphi))d\varphi\\&=\frac{14}3\int_{\pi/2}^{3\pi/2}\frac{1-\cos(4\varphi)}2d\varphi-\frac{10}3\left(\varphi+\frac{\sin(2\varphi)}2\Big|_{\pi/2}^{3\pi/2}\right)\\&=\frac{14}6\left(\varphi-\frac{\sin(4\varphi)}4\Big|_{\pi/2}^{3\pi/2}\right)-\frac{10}3\pi\\&=\frac{14}6\pi-\frac{10}3\pi\\&=\frac{-6}6\pi\\&=-\pi\end{aligned}$$
UPDATE:
I've just found my mistake, instead of dividing by $3,$ I divided by $2$ after the first integration in the $r$ variable.
| First note that $f(x) = x$ is an odd function of $x$ and as the circle $C_1: x^2 + y^2 \leq 4$ is symmetric to y-axis $(x = 0)$, its integral over the circle will be zero.
$I_1 = 0$
For integral over the circle to the right of y-axis,
In polar coordinates, $x = r \cos\theta, y = r \sin\theta$
So, $C_2: (x - 1)^2 + y^2 \leq 1 \implies r = 2 \cos\theta, - \pi/2 \leq \theta \leq \pi/2 $
The integral will be,
$ I_2 = \displaystyle \int_{-\pi/2}^{\pi/2} \int_0^{2\cos\theta} r^2 \cos\theta ~dr ~ d\theta = \pi$
As you have to find integral inside the circle $C_1$ and outside $C_2$,
$I = I_1 - I_2 = - \pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Solve $x+\sin(x) = \frac\pi2$ I was recently doing a high school geometry problem.
I believe that the intermediate steps are less than relevant, but if needed, I can post them too.
I ended up with the equation $$x+\sin(x)=\frac\pi2$$
I have no idea how to solve it. Any help would be appreciated.
| As said in comments, there is no closed form but you can have good approximations since you noticed that the solution is close to $\frac \pi 4$.
Expanding as series
$$x+\sin(x)=\left(\frac{1}{\sqrt{2}}+\frac{\pi }{4}\right)+\left(1+\frac{1}{\sqrt{2}}\right) \left(x-\frac{\pi }{4}\right)+\frac{1}{\sqrt{2}}\sum_{n=2}^\infty \frac{\sin \left(\frac{\pi n}{2}\right)+\cos \left(\frac{\pi n}{2}\right)}{n!}\left(x-\frac{\pi }{4}\right)^n$$
Truncating to some order and using series reversion,
$$x=\frac{\pi }{4}+t+\left(\frac{1}{\sqrt{2}}-\frac{1}{2}\right) t^2+\frac{8-5 \sqrt{2}}{6}
t^3+\frac{27 \sqrt{2}-37}{12}
t^4+\frac{518-363 \sqrt{2}}{60} t^5+O\left(t^{6}\right)$$ where $t=\frac{\pi -2 \sqrt{2}}{2 \left(2+\sqrt{2}\right)}$.
Using these terms only, this gives,as an approximation $x=\color{red}{0.83171119}30$ while the solution, given using Newton method, is
$x=\color{red}{0.83171119360}$
Edit
What you could also do is performing one single iteration of Halley method (instead of Newton method) and face the problem of solving for $x$ the linear equation
$$\frac \pi 2=\frac {\frac{1}{4} \left(2 \sqrt{2}+\pi \right)+\frac{\left(28+16 \sqrt{2}+\sqrt{2} \pi
\right) }{8
\left(2+\sqrt{2}\right)}\left(x-\frac{\pi }{4}\right) } {1+\frac{1}{\sqrt{2}
\left(2+\sqrt{2}\right)}\left(x-\frac{\pi }{4}\right) }$$ which gives as an estimate
$$x=\frac \pi 4 +\frac{2 \left(2+\sqrt{2}\right) \pi -8 \left(1+\sqrt{2}\right)}{4 \left(7+4
\sqrt{2}\right)-\sqrt{2} \pi }=0.831700\cdots$$
Using Householder method
$$x=\frac \pi 4 +\frac{3 \left(120+88 \sqrt{2}-16 \left(3+2 \sqrt{2}\right) \pi
+\left(1+\sqrt{2}\right) \pi ^2\right)}{-8 \left(119+83 \sqrt{2}\right)+8
\left(5+4 \sqrt{2}\right) \pi +\left(1+\sqrt{2}\right) \pi ^2}$$ which is $0.83171110\cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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A gamma summation: $\sum_{n=0}^{\infty} \frac{2}{\Gamma ( a + n) \Gamma ( a - n )} = \frac{2^{2a-2}}{\Gamma ( 2a - 1 )} + \frac{1}{\Gamma^2 (a)}$ Let $a \notin \mathbb{Z}$ and $a \neq \frac{1}{2}$. Prove that
$$\sum_{n=0}^{\infty} \frac{2}{\Gamma \left ( a + n \right ) \Gamma \left ( a - n \right )} = \frac{2^{2a-2}}{\Gamma \left ( 2a - 1 \right )} + \frac{1}{\Gamma^2 (a)}$$
Attempt
Using the fact that
\begin{align*}
\frac{1}{\Gamma\left ( a+x \right ) \Gamma \left ( \beta - x \right )} &= \frac{1}{\left ( a+x-1 \right )! \left ( \beta-x-1 \right )!} \\ &=\frac{1}{\Gamma \left ( a + \beta - 1 \right )} \frac{\left ( a + \beta-2 \right )!}{\left ( a + x -1 \right )! \left ( \beta - x -1 \right )!} \\ &=\frac{1}{\Gamma \left ( a + \beta - 1 \right )} \binom{a + \beta - 2}{a + x -1}
\end{align*}
the question really boils down to the sum
$$\mathcal{S} = \sum_{n=0}^{\infty} \binom{2a-2}{a+n-1}$$
To this end,
\begin{align*}
\sum_{n=0}^{\infty} \binom{2a-1}{a+n-1} &=\frac{1}{2\pi i} \sum_{n=0}^{\infty} \oint \limits_{|z|=1} \frac{\left ( 1+z \right )^{2a-1}}{z^{a+n}}\, \mathrm{d}z \\
&= \frac{1}{2\pi i} \oint \limits_{|z|=1} \frac{\left ( 1 + z \right )^{2a-1}}{z^a} \sum_{n=0}^{\infty} \frac{1}{z^n} \, \mathrm{d}z \\
&= \frac{1}{2\pi i} \oint \limits_{|z|=1} \frac{\left ( 1+z \right )^{2a-1}}{z^{a-1} \left ( z-1 \right )} \, \mathrm{d}z
\end{align*}
using the handy identity $\displaystyle \binom{n}{k} = \frac{1}{2\pi i } \oint \limits_{\gamma} \frac{\left ( 1+z \right )^n}{z^{k+1}} \, \mathrm{d}z$. I think I'm on the right track, but I'm having a difficult time evaluating the last contour integral. Any help?
| If you enjoy special functions
$$f(x)=\sum_{n=0}^{\infty} \frac{2\,x^n}{\Gamma \left ( a + n \right ) \Gamma \left ( a - n \right )} = \frac{2 }{\Gamma (a)^2}\,\, _2F_1(1,1-a;a;-x)$$
$$\, _2F_1(1,1-a;a;-1)=\frac{1}{2} \left(1+\sqrt{\pi }\frac{ \Gamma (a)}{\Gamma
\left(a-\frac{1}{2}\right)}\right)$$
$$f(1)=\frac{\sqrt{\pi }}{\Gamma \left(a-\frac{1}{2}\right) \Gamma (a)}+\frac{1}{\Gamma (a)^2}$$
Now, work the first term to get the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4446515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Upper bound of $\sum_{n=1}^N |1-z^n|$ where $|z| \leq 1$ How to derive an upper bound of
$$\sum_{n=1}^N |1-z^n|$$ where $z\in\mathbb{C}$ and $|z| \leq 1$?
A trivial upper bound would be $2N$ since each $|1-z^n| \leq 2$. But I am hoping for tighter bounds. I ran some numerical experiments and believe the bound should be $\frac{3}{2}N$ but don't know how to prove it.
| Here is an upper bound:
According to the link given by @Gerd in comment (Maximum of sum of finite modulus of analytic function.), the maximum of
$\sum_{n=1}^N |1 - z^n|$ is attained on $|z| = 1$.
Let $z = \mathrm{e}^{\mathrm{i}\theta}$
with $\theta \in [0, 2\pi]$. Using AM-QM, we have
\begin{align*}
\sum_{n=1}^N |1 - z^n|
&\le \sqrt{N\sum_{n=1}^N |1 - z^n|^2}\\
&= \sqrt{N\sum_{n=1}^N (2 - 2\cos n\theta)}\\
&= \sqrt{2N^2 + N \frac{\sin \frac{\theta}{2} - \sin\frac{(2N + 1)\theta}{2}}{\sin \frac{\theta}{2}}}. \tag{1}
\end{align*}
Fact 1: Let $x\in [0, \pi/2]$ and $N \ge 20$. Then
$$\frac{\sin x - \sin (2N + 1)x}{\sin x} \le \frac{N}{2}.$$
(The proof is not difficult.)
By Fact 1 and (1), we have, for all $N \ge 20$,
$$\sum_{n=1}^N |1 - z^n| \le \sqrt{5/2}\, N.$$
Some thoughts:
We have
\begin{align}
\sum_{n=1}^N |1 - z^n|
&= \sum_{n=1}^N \sqrt{2 - 2\cos n\theta }\\
&= \sum_{n=1}^N 2\left|\sin \frac{n\theta}{2}\right|\\
&= \frac{4}{\pi}N - \sum_{n=1}^N \sum_{k=1}^\infty \frac{8}{\pi(4k^2 - 1)}\cos kn\theta\\
&= \frac{4}{\pi}N - \sum_{k=1}^\infty \sum_{n=1}^N \frac{8}{\pi(4k^2 - 1)}\cos kn\theta\\
&= \frac{4}{\pi}N + \sum_{k=1}^\infty \frac{8}{\pi(4k^2 - 1)}
\frac{\sin\frac{k\theta}{2} - \sin \frac{(2N + 1)k \theta}{2}}{2\sin \frac{k\theta}{2}}\\
&= \frac{4}{\pi}N + \sum_{k=1}^\infty \frac{4}{\pi(4k^2 - 1)}
+ \sum_{k=1}^\infty \frac{4}{\pi(4k^2 - 1)}
\frac{ - \sin \frac{(2N + 1)k \theta}{2}}{\sin \frac{k\theta}{2}}\\
&= \frac{4}{\pi}N + \frac{2}{\pi} + \sum_{k=1}^\infty \frac{4}{\pi(4k^2 - 1)}
\frac{ - \sin \frac{(2N + 1)k \theta}{2}}{\sin \frac{k\theta}{2}}
\end{align}
where we have used the identity
$$\left|\sin y\right| = \frac{2}{\pi}-\sum_{k = 1}^\infty \frac{4}{\pi(4k^2-1)}\cos(2k y).$$
(Note: The RHS is the Fourier expansion of the LHS.)
We need to find bounds for
$$\sum_{k=1}^\infty \frac{4}{\pi(4k^2 - 1)}
\frac{ - \sin \frac{(2N + 1)k \theta}{2}}{\sin \frac{k\theta}{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
Find $\sqrt{\frac{1}{2}-f(1)}+\dots+ \sqrt{\frac{1}{2}-f(99)}$ Define $f(n)=\sqrt[2]{n^4+\frac{1}{4}}-n^2.$
Find $$\sqrt{\frac{1}{2}-f(1)}+\dots+ \sqrt{\frac{1}{2}-f(99)}$$
I tried to simply $f(n).$ So rationalising, we get $$\sqrt{\frac{1}{2}-f(n)}\sqrt{\frac{1}{2}+f(n)}=\sqrt{\frac{1}{2}+n^2-\sqrt{n^4+\frac{1}{4}}}\sqrt{\frac{1}{2}+n^2+\sqrt[2]{n^4+\frac{1}{4}}}=\sqrt{(\frac{1}{2}+n^2)^2-n^4-\frac{1}{4}}=\sqrt{n^2}=n$$
So $$\sqrt{\frac{1}{2}-f(n)}=\frac{n}{\sqrt{\frac{1}{2}+f(n)}}.$$
EDIT: Using the hints given below, we have to find We have to find $$\sum_{n=1}^{99} \frac{n}{\sqrt{n^2+1/4}+1/2}=\sum_{n=1}^{99} \frac{n({\sqrt{n^2+1/4}-1/2})}{(\sqrt{n^2+1/4}+1/2)(\sqrt{n^2+1/4}-1/2)}=\sum_{n=1}^{99}\frac{n({\sqrt{n^2+1/4}-1/2})}{n^2}= \frac{\sqrt{n^2+1/4}-1/2}{n}$$
Any solutions?
| Hint
Let
$$g_n=\sqrt{\frac 12-f_n}\qquad \text{with} \qquad f_n=\sqrt[2]{n^4+\frac{1}{4}}-n^2$$ and expand it as a series for "large" values of $n$
$$g_n=\frac 1{\sqrt 2}\Bigg[1-\frac{1}{8 n^2}+O\left(\frac{1}{n^4}\right)\Bigg]$$ Using generalized hamonic numbers
$$\sum_{n=1}^p g_n=\frac 1{\sqrt 2}\Bigg[p-\frac{1}{8}H_p^{(2)}+\cdots\Bigg]$$ Now, suppose that $p$ is large and use the asymptotics to obtain
$$\sum_{n=1}^p g_n=\frac 1{\sqrt 2}\Bigg[p-\frac{ \pi ^2}{48}+\frac 1{8p}+O\left(\frac{1}{p^2}\right)\Bigg]$$ which gives a relative error smaller than $0.01$% as soon as $p\geq 18$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Volume above a cone and within a sphere, using triple integrals and cylindrical polar coordinates Consider the region above the cone $z = \sqrt{x^2+y^2}$ and inside the sphere $x^2+y^2+z^2 = 16$.
Use cylindrical polar coordinates to show that the volume of region $R$ is $\frac{64\pi}{3}(2-\sqrt{2})$.
To solve this, I took the limits of $\theta$ as $0$ to $2π$.
I then took the limits of $z$ by substituting $z^2 = x^2+y^2$ (cone) into the equation of the sphere, giving me $z = 2\sqrt{2}$ as my upper bound (and $0$ as the lower bound).
I finally then took the limits of $r$ by changing the equation of the sphere into $r = \sqrt{16-z^2}$ (using $r^2 = x^2+y^2$), and $r = z$ (using $r^2 = x^2+y^2 = z^2$, the equation of the cone).
This gave me
$$\int_{0}^{2\pi}\int_{0}^{2\sqrt{2}}\int_{z}^{\sqrt{16-z^{2}}}rdrdzd\theta$$
"∫2π 0 ∫2√2 0 ∫(√16-Z^2) Z rdrdZdθ"
However, the answer I got from this integral was $2π(16√2 - 16/3(√2))$, which was incorrect.
| The volume is give by
$$V=\iiint_{E}\, {\rm d}V$$
where $E$ is the solid built by above of the cone $z=\sqrt{x^{2}+y^{2}}$ and inside of the sphere $x^{2}+y^{2}+z^{2}=4^{2}$.
Using cylindrical coordinates $(r,\theta,z)$:
$$x=r\cos\theta,\quad y=r\sin\theta,\quad z=z$$
The intersection is a projection over $(X,Y)$ is given by
$$\{x^{2}+y^{2}+z^{2}=4^{2}\}\cap\{z=+\sqrt{x^{2}+y^{2}}\}\implies x^{2}+y^{2}=(\sqrt{8})^{2}$$
Then,
\begin{align*}
V&=\int_{0}^{2\pi}\int_{0}^{\sqrt{8}}\left(\sqrt{16-r^{2}}-\sqrt{r^{2}}\right)\color{red}{r}\, {\rm d}r{\rm d}\theta,\\
&=\int_{0}^{2\pi}\frac{32}{3}(2-\sqrt{2})\, {\rm d}\theta,\\
&=\boxed{\frac{64\pi}{3}(2-\sqrt{2})}
\end{align*}
Using spherical coordinates $(r,\theta,\phi)$:
$$x=\rho\cos\theta\sin\phi,\quad y=\rho\sin\theta\sin\phi,\quad z=\rho\cos\phi$$
Then,
\begin{align*}
V&=\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{4}\color{red}{\rho^{2}\sin\phi}\,{\rm d}\rho\, {\rm d}\phi\, {\rm d}\theta,\\
&=\int_{0}^{2\pi}\frac{4^{3}}{3}\left(-\cos\frac{\pi}{4}+\cos 0\right)\, {\rm d}\theta,\\
&=\boxed{\frac{64\pi}{3}(2-\sqrt{2})}
\end{align*}
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4448471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\int _0^1f^2\left(x\right)dx-2\int _0^{\sqrt{3}-1}\:\left(x+1\right)f\left(x\right)dx\:+1=0$ Let $ f $ be an increasingly continuous function over $[0,1]$ such that:
$$\int _0^1(f\left(x\right))^2dx-2\int _0^{\sqrt{3}-1}\:\left(x+1\right)f\left(x\right)dx\:+1=0$$
Find all functions $f$ with these properties.
Attempt.:
First, the function $f$ is integrable therefore the computations with integrals work there.
Then I did the following:
$$\int _0^1(f\left(x\right))^2dx-2\int _0^{\sqrt{3}-1}\:\left(x+1\right)f\left(x\right)dx\:+1=$$
$$\int _0^{\sqrt{3}-1}(f\left(x\right))^2dx+\int _{\sqrt{3}-1}^1(f\left(x\right))^2dx-2\int _0^{\sqrt{3}-1}\:\left(x+1\right)f\left(x\right)dx\:+1=$$
$$\int _0^{\sqrt{3}-1}\left(f\left(x\right)-\left(x+1\right)\right)^2dx+\int _{\sqrt{3}-1}^1(f\left(x\right))^2dx-\int _0^{\sqrt{3}-1}\left(x+1\right)^2dx+1=0$$
Since
$$-\int _0^{\sqrt{3}-1}\left(x+1\right)^2dx=-\frac{6\sqrt{3}-10}{3}-3+\sqrt{3}$$
Then
$$\int _0^{\sqrt{3}-1}\left(f\left(x\right)dx-\left(x+1\right)\right)^2dx+\int _{\sqrt{3}-1}^1(f\left(x\right))^2dx-\frac{6\sqrt{3}-10}{3}-2+\sqrt{3}=0$$
$$\int _0^{\sqrt{3}-1}\left(f\left(x\right)dx-\left(x+1\right)\right)^2dx+\int _{\sqrt{3}-1}^1(f\left(x\right))^2dx+\frac{-3\sqrt{3}+4}{3}=0$$
How should I continue from there? I got stuck there. I am not sure this is going to lead me somewhere. Maybe I should have started differently.
Has somebody an idea what the functions look like?
| Substituting $f(x)=1+g(x)$, your equation becomes
$$\int_0^{\sqrt{3}-1} {\rm d}x \, g(x)\left[g(x)-2x\right] + \int_{\sqrt{3}-1}^1 {\rm d}x \, g(x) \left[g(x)+2\right]=0 \tag{1}$$
where $g(x)$ is continuously increasing on $(0,1)$.
Evidently this equation is fulfilled if $g=0$. However, it would be not increasing. We show that the LHS is always positive if $g\neq 0$. To this end, we want to bound it from below:
$$\text{LHS}(1) > -2g(\sqrt{3}-1) \int_0^{\sqrt{3}-1} {\rm d}x \,x + 2g(\sqrt{3}-1)\int_{\sqrt{3}-1}^1 {\rm d}x \\
=g(\sqrt{3}-1) \left( -(\sqrt{3}-1)^2+2(2-\sqrt{3}) \right) = 0 \, .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4452396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Proving by Induction a Combinatorial Binomial I'm currently stuck trying to prove that for all items in the sequence $\binom{n+1}{2}$, for all $n\geq 1$, that $\binom{n+1}{2}=\sum \limits _{i=0}^ni$.
My first assumption is to solve for my base case, which would be $1$, so I calculated that $\binom{n}{k}=\frac{n!}{k!(n-k)!}$, therefore $\binom{n+1}{2}=\frac{(n+1)!}{(2)!(n+1-2)!}=\frac{(n+1)!}{2(n+1-2)!}$.
When $n=1$, $\frac{(n+1)!}{2(n+1-2)!}=\frac{(1+1)!}{2(1+1-2)!}=\frac{2!}{2(2-2)!}= \frac{2}{2\cdot 0!}=\frac{2}{2\cdot 1}=1$, and $\sum \limits _{i=0}^ni$ when $n=1$ is $1$, so the base case is true.
So for the inductive hypothesis I state that $n=k$, and for the inductive step I then try to prove that since it's true for $k$, how it should be true for $k+1$.
First I substitute $n$ in the formula for $k+1$:
$\binom{n+1}{2}=\frac{(n+1)!}{2(n+1-2)!}=\binom{k+1}{2}=\frac{(k+1)!}{2(k+1-2)!}$, therefore $\binom{(k+1)+1}{2}=\frac{(k+1+1)!}{2(k+1+1-2)!}=\frac{(k+2)!}{2(k+2-2)!}=\frac{(k+2)!}{2k!}$.
And for the other side of the equality, $\sum \limits _{i=0}^ni=\sum \limits _{i=0}^ki$, therefore $\sum \limits _{i=0}^{k+1}i$.
From this, $\frac{(k+2)!}{2k!}$ should equal $\sum \limits _{i=0}^{k+1}i$.
It's from this point I'm stumped on what to do next on how to actually prove this statement. Are there certain properties I'm supposed to employ to substitute some of the terms, am I just supposed to plug in the next value for $n+1$ as $k+1$? Because if I do that then:
$\frac{(2+2)!}{2(2!)}=\sum \limits _{i=0}^2i\rightarrow \frac{4!}{4}=2\rightarrow 6\neq 2$.
| $$\binom{k+2}{2} = \frac{(k+2)!}{2k!} = \frac{(k+2)(k+1)}{2} = \frac{(k+1)k}{2} + (k+1) = \binom{k+1}{2} + (k+1).$$
By induction you know $\binom{k+1}{2} = \sum_{i=1}^k i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4452954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to evaluate the sum of $\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1}$ I hava an infinite sum
$$\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1}$$
I factored the denominator
$$\sum_{n=0}^{\infty}\frac{1}{\left(3n+1\right)\left(n+1\right)}$$
Then I separated the fraction
$$\frac{1}{2}\sum_{n=0}^{\infty}\frac{3}{\left(3n+1\right)}-\frac{1}{\left(n+1\right)}$$
Then I set 1 (the numerator) to be equal to x to some power which I don't know if I can do
$$\frac{3}{2}\sum_{n=0}^{\infty}\frac{x^{3n+1}}{3n+1}-\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}$$
Then I set the integral which would satisfy the previous terms
$$\frac{3}{2}\sum_{n=0}^{\infty}\int_{0}^{1}x^{3n}dx$$
and
$$-\frac{1}{2}\sum_{n=0}^{\infty}\int_{0}^{1}x^{n}dx$$
Then I changed the order of summation and integration and I got
$$\frac{3}{2}\int_{0}^{1}\frac{1}{1-x^{3}}dx$$ and $$-\frac{1}{2}\int_{0}^{1}\frac{1}{1-x}dx$$
The first integral can be factored to
$$\frac{3}{2}\int_{0}^{1}\frac{1}{\left(1-x\right)\left(1+x+x^{2}\right)}dx$$
Then separated
$$\frac{1}{2}\int_{0}^{1}\frac{1}{1-x}+\frac{x+2}{1+x+x^{2}}dx$$
The first one will cancel out with
$$-\frac{1}{2}\int_{0}^{1}\frac{1}{1-x}dx$$
And I'm left with
$$\frac{1}{2}\int_{0}^{1}\frac{x+2}{1+x+x^{2}}dx$$
which is
$$\frac{\sqrt{3}\pi}{12}+\frac{\ln\left(3\right)}{4}$$
But the correct answer is
$$\frac{\sqrt{3}\pi}{12}+\frac{3\ln\left(3\right)}{4}$$
So I would like to ask if this approach is invalid or if I'm just missing something.
| $$ \begin{align}\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1} &= \frac{1}{2} \sum_{n=0}^{\infty} \left(\frac{3}{3n+1}-\frac{1}{n+1} \right) \\ &= \frac{1}{2} \sum_{n=0}^{\infty} \left(3\int_{0}^{1}x^{3n} \, \mathrm dx- \color{red}{3\int_{0}^{1} x^{3n+2} \, \mathrm dx}\right) \\ &= \frac{3}{2} \sum_{n=0}^{\infty} \int_{0}^{1} \left(x^{3n}-x^{3n+2} \right) \, \mathrm dx \\ &= \frac{3}{2} \sum_{n=0}^{\infty} \int_{0}^{1} x^{3n}\left(1-x^{2}\right) \, \mathrm dx \end{align}$$
Since $x^{3n}(1-x^{2})$ is nonnegative for all $n$ and all $x$, Tonelli's theorem allows us to interchange the order of integration and summation.
Therefore, we have $$ \begin{align} \sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1} &= \frac{3}{2} \int_{0}^{1} (1-x^{2}) \sum_{n=0}^{\infty} x^{3n}\, \mathrm dx \\ &= \frac{3}{2} \int_{0}^{1} \frac{1-x^{2}}{1-x^{3}} \, \mathrm dx \\ &= \frac{3}{2} \int_{0}^{1} \frac{x+1}{x^{2}+x+1} \, \mathrm dx \\ &= \frac{3}{4} \int_{0}^{1} \left( \frac{2x+1}{x^{2}+x+1} +\frac{1}{x^{2}+x+1} \right)\, \mathrm dx \\ &= \frac{3}{4} \, \ln(3) + \frac{3}{4} \int_{0}^{1} \frac{\mathrm dx}{\left(x+\frac{1}{2} \right)^{2} + \frac{3}{4}} \, \mathrm dx \\ &= \frac{3}{4} \, \ln(3) + \frac{3}{2\sqrt{3}} \, \arctan \left(\frac{2}{\sqrt{3}} \left( x+\frac{1}{2}\right) \right) \Bigg|_{0}^{1} \\ & = \frac{3}{4} \, \ln(3) + \frac{3}{2 \sqrt{3}} \left(\frac{\pi}{3}- \frac{\pi}{6} \right) \\ &=\frac{3}{4} \, \ln(3) + \frac{\sqrt{3} \pi}{12}. \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Show that if $0 < \frac{1}{j},\frac{1}{k} \leq \frac{1}{N} \leq 1$, then $|\frac{1}{j}-\frac{1}{k}| \leq \frac{1}{N}$. ($j,k,N \in \mathbb{N}^+$)) I am working through Tao's analysis I book and I am trying to prove that the sequence $a_n = \frac{1}{n}$ is a Cauchy sequence (Proposition 5.1.11). I understand the proof, but I am having trouble showing why it is exactly that
$0 < \frac{1}{j},\frac{1}{k} \leq \frac{1}{N} \leq 1 \rightarrow |\frac{1}{j}-\frac{1}{k}| \leq \frac{1}{N}$.
I have tried manipulaing single inequalitis on their own and combining them in order to reach the final statement, but with no success. I am always left with an extra $\frac{1}{j},\frac{1}{k}$ on the side of the $\frac{1}{N}$ which doesn't allow me to conclude the reasoning. I suspect I am neglecting some simpler inequalities which would allow me to compare $|\frac{1}{j}-\frac{1}{k}|$ with $\frac{1}{N}$, but I don't see it yet. Any clue is more than welcome.
| We have $0 < \frac{1}{j} \leq \frac{1}{N}$ and $0 < \frac{1}{k} \leq \frac{1}{N}$.
Then we can subtract $\frac{1}{j}$ from the second inequality to get:
$$-\frac{1}{j} < \frac{1}{k} - \frac{1}{j} \leq \frac{1}{N} - \frac{1}{j}$$ but we know $\frac{1}{j} > 0$ so $\frac{1}{N} - \frac{1}{j} \leq \frac{1}{N}$ and we know $\frac{1}{j} \leq \frac{1}{N}$ so $-\frac{1}{N} \leq -\frac{1}{j}$ hence
$$-\frac{1}{N} \leq -\frac{1}{j} < \frac{1}{k} - \frac{1}{j} \leq \frac{1}{N} - \frac{1}{j} \leq \frac{1}{N}$$
But we can simplify this to get:
$$-\frac{1}{N} \leq \frac{1}{k} - \frac{1}{j} \leq \frac{1}{N} \implies |\frac{1}{k} - \frac{1}{j}| \leq \frac{1}{N}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4454706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$ I am asked to prove that $$\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac{\pi^2}{8}.$$
However, I am asked to prove it using the fact that
$$\frac{\pi}{2}\tan\left(\frac{\pi}{2}z\right)=\sum_{m \text{ odd}}\left(\frac{1}{m-z}-\frac{1}{m+z}\right),$$
where $z\in \mathbb{C}$, which is something I proved in a previous exercise.
My first thought was using the fact that
$$\frac{1}{m-z}-\frac{1}{m-z}=\frac{2z}{m^2-z^2}$$
and therefore
$$\sum_{m \text{ odd}}\left(\frac{1}{m-z}-\frac{1}{m+z}\right)=\sum_{m \text{ odd}}\frac{2z}{m^2-z^2}=\sum_{n=0}^\infty \frac{2z}{(2n+1)^2-z^2}.$$
This last series is similar to the one I am aiming at, but I don't know how to transform it into the one that I want. Can someone help me?
| Following the comment of @RaymondManzoni, we obtain continuing OP's approach
\begin{align*}
\frac{\pi}{4z}\tan\left(\frac{\pi}{2}z\right)=\sum_{n=0}^\infty\frac{1}{(2n+1)^2-z^2}\tag{1}
\end{align*}
We can evaluate the right-hand side of (1) at $z=0$. Since $\tan\left(\frac{\pi}{2}z\right)$ has a series expansion for $|z|<1$ we obtain by taking the limit as $z\to 0$ and the tangent series expansion:
\begin{align*}
\color{blue}{\sum_{n=0}^\infty\frac{1}{(2n+1)^2}}&
=\lim_{z\to 0}\frac{\pi}{4z}\tan\left(\frac{\pi}{2}z\right)\\
&=\lim_{z\to 0}\frac{\pi}{4z}\left(\frac{\pi}{2}z+O(z^3)\right)\\
&\,\,\color{blue}{=\frac{\pi^2}{8}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4455694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
For what value of $b$ does the system $3x-2y+z=b,5x-8y+9z=3,2x+y-3z=-1$ have infinite solutions?
For what value of $b$ does the system $3x-2y+z=b,5x-8y+9z=3,2x+y-3z=-1$ have infinite solutions?
My Attempt:
Let $A$ be the coefficient matrix. And $B$ be the column matrix with elements $b,3,-1$. Also, $det(A)=0$
In Matrix method, for infinite solutions, $(\text{adj}(A))B=0$. My $(\text{adj}(A))B$ is coming a column matrix with entries $15b-25,33b-55,21b-35$. For this to be zero, $b=5/3$.
But the answer given is $1/3$.
If I use Crammer's method, calculating $D_x$ (i.e. in $A$, replacing coefficient of $x$ with constants), and making it zero, I do get $b=1/3$.
Did I make some mistake in the matrix method?
For $\text{adj}(A)$, my first row has elements $15,-5,-10$, second row $33,-11,-22$, and third $21,-7,-14$.
| We have the adjoint matrix (this agrees with your result)
$$\begin{pmatrix}
15 & -5 & -10 \\ 33 & -11 & -22 \\ 21 & -7 & -14 \\
\end{pmatrix}$$
We then form
$$\begin{pmatrix}
15 & -5 & -10 \\ 33 & -11 & -22 \\ 21 & -7 & -14
\end{pmatrix}.\begin{pmatrix}
b \\ 3 \\ -1
\end{pmatrix} = \begin{pmatrix}
15 b-5\\33 b-11\\21 b-7 \end{pmatrix}$$
It appears you had a simple sign error in $b$ with the last value, $1$ instead of $-1$, when doing the multiplications.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
An Application of Cauchy Schwarz - AM-GM in Discrete Probability Measures Suppose $p_m \geq 0$ and $\sum_{m \in \mathbf{Z}} p_m =1 .$ That is $p$ is a probability measure on integers. Then how can I show (is it true) that
$$ \sum_{m \in \mathbf{Z}} (p_m + p_{m+1} + p_{m+2})^n \leq 2^n \sum_{m \in \mathbf{Z}} (p_m + p_{m+1})^n - (2^n-1) \sum_{m \in \mathbf{Z}} p_m ^n $$
for any $n \in \mathbf{N}.$
If I am on not too far, PLEASE do NOT answer, yet provide me a hint (I do not want to ruin the joy). Yet I have been stuck for a bit, thank you!
| I tried to use cauchy schwarz, and I believe some form of such an inequality should work. As a simple case, consider $n=2:$ After changing indices, we have
\begin{align}
\textbf{Left} = \sum_{m \in \mathbf{Z}} (p_m + p_{m+1} + p_{m+2})^2 =3 \sum_{m \in \mathbf{Z}} p_m ^2 + 4 \sum_{m \in \mathbf{Z}} p_m p_{m+1} + 2 \sum_{m \in \mathbf{Z}} p_m p_{m+2}
\end{align}
and
\begin{align}
\textbf{Right} = 2^2 \sum_{m \in \mathbf{Z}} (p_m + p_{m+1})^2 - (2^2-1) \sum_{m \in \mathbf{Z}} p_m ^2 &= 4 \Big[ 2 \sum_{m \in \mathbf{Z}} p_m ^2 + 2 \sum_{m \in \mathbf{Z}} p_m p_{m+1} \Big] - 3 \sum_{m \in \mathbf{Z}} p_m ^2 \\
&= 5 \sum_{m \in \mathbf{Z}} p_m ^2 + 8 \sum_{m \in \mathbf{Z}} p_m p_{m+1} .
\end{align}
By cauchy schwarz,
\begin{align}
\sum_{m \in \mathbf{Z}} p_m p_{m+2} \leq \Big( \sum_{m \in \mathbf{Z}} p_m ^2 \Big) ^{\frac{1}{2}} \Big( \sum_{m \in \mathbf{Z}} p_{m+2} ^2 \Big) ^{\frac{1}{2}} = \sum_{m \in \mathbf{Z}} p_m ^2
\end{align}
which suffices to show $\textbf{Left} \leq \textbf{Right}.$ As another way, by AM-GM we have
\begin{align}
\sum_{m \in \mathbf{Z}} p_m ^2 = \frac{1}{2} \sum_{m \in \mathbf{Z}} p_m ^2 + \frac{1}{2} \sum_{m \in \mathbf{Z}} p_{m+2} ^2 = \sum_{m \in \mathbf{Z}} \frac{p_m ^2 + p_{m+2} ^2 }{2} \geq \sum_{m \in \mathbf{Z}} \sqrt{p_m ^2 p_{m+2} ^2 } = \sum_{m \in \mathbf{Z}} p_m p_{m+2} .
\end{align}
For $n=3 :$
After changing indices, we have
\begin{align}
\textbf{Left} = \sum_{m \in \mathbf{Z}} (p_m + p_{m+1} + p_{m+2})^3 &=3 \sum_{m \in \mathbf{Z}} p_m ^3 + 6 \sum_{m \in \mathbf{Z}} p_m^2 p_{m+1} + 3 \sum_{m \in \mathbf{Z}} p_m^2 p_{m+2} \\
&+ 6 \sum_{m \in \mathbf{Z}} p_m p_{m+1} ^2 + 3 \sum_{m \in \mathbf{Z}} p_m p_{m+2} ^2 + 6 \sum_{m \in \mathbf{Z}} p_m p_{m+1} p_{m+2}
\end{align}
and
\begin{align}
\textbf{Right} = 2^3 \sum_{m \in \mathbf{Z}} (p_m + p_{m+1})^3 - (2^3-1) \sum_{m \in \mathbf{Z}} p_m ^3 &= 8 \Big[ 2 \sum_{m \in \mathbf{Z}} p_m ^3 + 3 \sum_{m \in \mathbf{Z}} p_m ^2 p_{m+1} + 3 \sum_{m \in \mathbf{Z}} p_m p_{m+1} ^2 \Big] - 7 \sum_{m \in \mathbf{Z}} p_m ^3 \\
&= 9 \sum_{m \in \mathbf{Z}} p_m ^3 + 24 \sum_{m \in \mathbf{Z}} p_m ^2 p_{m+1} + 24 \sum_{m \in \mathbf{Z}} p_m p_{m+1} ^2 .
\end{align}
One can perform cancelations and similarly show by cauchy schwarz or AM-GM
\begin{align}
\sum_{m \in \mathbf{Z}} p_m ^2 p_{m+2} \leq \sum_{m \in \mathbf{Z}} p_m ^3
\end{align}
and
\begin{align}
\sum_{m \in \mathbf{Z}} p_m p_{m+2} ^2 \leq \sum_{m \in \mathbf{Z}} p_m ^3
\end{align}
and using AM-GM
\begin{align}
\sum_{m \in \mathbf{Z}} p_m ^2 p_{m+1} + \sum_{m \in \mathbf{Z}} p_m p_{m+1} ^2 = \sum_{m \in \mathbf{Z}} \frac{p_m ^2 p_{m+1} + p_{m+1} p_{m+2}^2 }{2} \geq \sum_{m \in \mathbf{Z}} p_m p_{m+1} p_{m+2} .
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4460936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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When $a^2 + b^2 + c^2 + d^2 = 1$ and the expression $(a - b)(b - c)(c - d)(d - a)$ reaches its minimum value, determine the value of product $abcd$.
Consider real number $a, b, c, d$ such that $a^2 + b^2 + c^2 + d^2 = 1$. When the expression $(a - b)(b - c)(c - d)(d - a)$ reaches its minimum value, determine the value of product $abcd$.
[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)
By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]
So I took the mock exam which was organised by our school today... It didn't go well. ｡ﾟ( ﾟஇ‸இﾟ)ﾟ｡ That's why this question is here.
Here's what I'd attempted in the official amount of time that was given, (so no external thoughts, I still have more subjects that need to be taken care of, sorry~)
(I lied, there are some afterthoughts sprinkled in here.)
First of all, $$\begin{aligned} (a - b)(b - c)(c - d)(d - a) &= [(ac + bd) - (ab + cd)][(ac + bd) - (bc + da)]\\ &= (ac + bd)^2 - (ab + bc + cd + da)(ac + bd)\\ &+ (ab + cd)(bc + da) \end{aligned}$$
(From this point on, it's all what I've thought of after the official time period has ended.)
Let $ac + bd = x$. The minimum value of the quadratic function $$x^2 - (ab + bc + cd + da)x + (ab + cd)(bc + da)$$ is $\dfrac{-\Delta}{4} = \dfrac{4(ab + cd)(bc + da) - (ab + bc + cd + da)^2}{4} = -\left[\dfrac{(a - c)(b - d)}{2}\right]^2$.
The equal sign occurs when $x = \dfrac{ab + bc + cd + da}{2} \iff 2(ac + bd) = ab + bc + cd + da \ (1)$. Now, we just need to find the maximum value of $\left[\dfrac{(a - c)(b - d)}{2}\right]^2$.
(∩ᗒ.ᗕ)⊃━☆ﾟ.❉*, $\left[\dfrac{(a - c)(b - d)}{2}\right]^2 \le \dfrac{(a - c)^2 + (b - d)^2}{8} \le \dfrac{a^2 + b^2 + c^2 + d^2}{4} = \dfrac{1}{4}$.
The equal sign occurs when $a = b = -c = -d \ (2)$. Combining $(1)$ and $(2)$, we have that $$\left\{ \begin{aligned} 2(ac + bd) &= ab + bc + cd + da\\ a = b &= -c = -d \end{aligned} \right. \implies a = b = c = d = 0$$
This feels wrong, and of course, it is. Let's elaborate on $(1)$ further, in that, $$\begin{aligned} 2(ac + bd) = ab + bc + cd + da &\iff 2[(ac + bd) - (bc + da)] = ab - bc + cd - da\\ &\iff 2(a - b)(c - d) = (a - c)(b - d) \end{aligned}$$
Now, does this mean anything? It is with deep regret that I have to inform you that, I don't freaking know.
Hmmm~ this doesn't seem to work. Let's try something different, for example, assuming that $a \ge b \ge c \ge d$, we have that $$\begin{aligned} (a - b)(b - c)(c - d)(d - a) &\ge \dfrac{[(a - b) + (b - c) + (c - d)]^3(d - a)}{27} = \dfrac{-(a - d)^4}{27} \end{aligned}$$
The equal sign occurs when $a - b = b - c = c - d \iff \left\{ \begin{aligned} b = \dfrac{d + 2a}{3}\\ c = \dfrac{2d + a}{3} \end{aligned} \right. \ (3)$.
(From this point on, it's all what I've thought of after the official time period has ended, the sequel.)
Plugging the above results into $a^2 + b^2 + c^2 + d^2 = 1$, we have that $7d^2 + 4da + 7a^2 = \dfrac{9}{2}$, and since $(7d^2 + 4da + 7a^2) - \dfrac{5(d - a)^2}{2} = \dfrac{9(d + a)^2}{2} \ge 0, \forall d, a \in \mathbb R$, it means that $$\dfrac{5(d - a)^2}{2} \le \dfrac{9}{2} \iff (d - a)^2 \le \dfrac{9}{5} \iff\dfrac{-(a - d)^4}{27} \ge -\dfrac{3}{25}$$
The equal sign occurs when $d = -a \ (4)$.
Combining $(3)$ and $(4)$, we have that $$\left[ \begin{aligned} (a; b; c; d) &= \left(\dfrac{3\sqrt{5}}{10}; \dfrac{\sqrt{5}}{10}; -\dfrac{\sqrt{5}}{10}; -\dfrac{3\sqrt{5}}{10}\right)\\ (a; b; c; d) &= \left(-\dfrac{3\sqrt{5}}{10}; -\dfrac{\sqrt{5}}{10}; \dfrac{\sqrt{5}}{10}; \dfrac{3\sqrt{5}}{10}\right) \end{aligned} \right. \implies abcd = \dfrac{9}{400}$$
According to WolframAlpha, the minimum value of the above expression is $-\dfrac{1}{8}$, happening when $$(a; b; c; d) = \left(-\dfrac{\sqrt 3 + 1}{4}; \dfrac{\sqrt 3 + 1}{4}; \dfrac{\sqrt 3 - 1}{4}; -\dfrac{\sqrt 3 - 1}{4}\right) \implies abcd = \dfrac{1}{64}$$
So my friend was right after all...（；¬＿¬) How am I supposed to do this in 90 minutes?
All anger aside, that's all for now. This took way more time to write down than it needed to. Anyhow, have a wonderful tomorrow, everyone~
By the way, the choices were $-\dfrac{1}{64}$; $-\dfrac{1}{8}$; $\dfrac{1}{8}$ and $\dfrac{1}{64}$.
| My second solution:
Let $x = (a + b)/2, y = (a - b)/2, z = (c + d)/2, w = (c - d)/2$ (correspondingly, $a = x + y, b = x - y, c = z + w, d = z - w$).
We have $x^2 + y^2 + z^2 + w^2 = (a^2 + b^2 + c^2 + d^2)/2 = 1/2$.
We have
$$f(a,b,c,d) : =(a-b)(b-c)(c-d)(d-a) = 4yw(w+y)^2-4yw(x-z)^2
$$
and
$$g(a,b,c,d) := abcd = -w^2x^2 + w^2y^2 + x^2z^2 - y^2z^2.$$
If $yw \le 0$, we have
\begin{align*}
f &\ge 4yw(w + y)^2\\
& = - 4 \cdot |yw| \cdot (w^2 + y^2 - 2|yw|)\\
&\ge -2\cdot \frac{(2|yw| + w^2 + y^2 - 2|yw|)^2}{4}\\
& = -\frac{(w^2 + y^2)^2}{2}\\
&\ge - \frac{(w^2 + y^2 + x^2 + z^2)^2}{2}\\
&= - \frac18
\end{align*}
with equality if and only if $x = z = 0$
and $y^2 + w^2 = 1/2$ and $2|yw| = w^2 + y^2 - 2|yw|$ and $yw \le 0$ which results in $x = z = 0, |yw| = 1/8$.
Then we have $g = 1/64$.
If $yw > 0$, we have
\begin{align*}
f &= 4yw (w^2 + y^2 + 2yw - x^2 - z^2 +2zx) \\
&\ge 4yw (w^2 + y^2 + 2yw - x^2 - z^2 - (x^2 + z^2))\\
&= 4yw (w^2 + y^2 + 2yw - (1 - 2y^2 - 2w^2))\\
&= 4yw (3w^2 + 3y^2 + 2yw - 1)\\
&\ge 4yw (6yw + 2yw - 1)\\
&= 32(yw - 1/16)^2 - 1/8\\
&\ge -1/8
\end{align*}
with equality if and only if $yw = 1/16$
and $y = w$ and $x + z = 0$ and $x^2 + y^2 + z^2 + w^2 = 1/2$ which results in $y^2 = w^2 = 1/16, x^2 = z^2 = 3/16$. Then we have $g = 1/64$.
As a result, $f_{\min} = -1/8$ and at minimum $abcd = 1/64$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Maximization Of Multivariable function let $F(a,b,c,d,e,f) = abcdef$ and our constraint is $a^2+b^2+c^2+d^2+e^2+f^2=6$ ,$a+b+c+d+e+f=0$ then lagrange
we get
$abcdef = 2a^2k +ma=2b^2.k+m.b =.....=2f^2.k+mf$ but this is quite complicated to do is there any elegant way like trig sub or something
for $a,b,c,d,e,f$ real numbers
| The following satisfy the constraints:
$$a=b=c=d=\frac{1}{\sqrt2}$$
$$e=f=-\frac{2}{\sqrt2}$$
$$abcdef=\frac12$$
In fact, this will turn out to be the maximum, but for now, we just need to note that the maximum is greater than zero.
$a+b+c+d+e+f=0$ tells us that some are positive and some are negative. In fact, the number of positive values must be even because $abcdef>0$.
We assume that there are 4 positive values (a,b,c,d) and two negative (e,f). The proof for 2 positive and 4 negative is much the same.
Define
$$E=-e$$
$$F=-f$$
$$a+b+c+d=E+F=S$$
$$E^2+F^2=Q$$
$$a^2+b^2+c^2+d^2=6-Q$$
Now apply the inequality of geometric, arithmetic and quadratic means:
$$(abcd)^{\frac14} \leqslant \frac{S}{4} \leqslant \left(\frac{6-Q}{4}\right)^{\frac12} \tag1$$
$$(EF)^{\frac12} \leqslant \frac{S}{2} \leqslant \left(\frac{Q}{2}\right)^{\frac12} \tag2$$
From the second parts of (1) and (2):
$$S^2 \leqslant 24-4Q$$
$$S^2 \leqslant 2Q$$
and combining these gives
$$S\leqslant 2 \sqrt 2 \tag3 $$
From the first parts of (1) and (2):
$$abcd \leqslant \frac{S^4}{256}$$
$$ef=EF \leqslant \frac{S^2}{4}$$
Multiply:
$$abcdef \leqslant \frac{S^6}{1024} \leqslant \frac{( 2 \sqrt 2 )^6}{1024}=\frac12 $$
As required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Value of $x^6 - 2\sqrt{3}x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt{3}$ given $x = \frac{1}{2-\sqrt{3}}$? It is given that $x = \frac{1}{2-\sqrt{3}}$. Find the value of $x^6 - 2\sqrt{3}x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt{3}$.
Well I tried rationalising and I came to know that $x = 2 + \sqrt{3}$. And I know that directly putting the values wont help either but I am not able to factorize the polynomial or manipulate it to help me.
I would be grateful if anybody could help me.
| As OP noted, $\,x = 2 + \sqrt{3}\,$, and also $\,\dfrac{1}{x} = 2 - \sqrt{3}\,$. Then $\,\color{blue}{x+ \dfrac{1}{x} = 4}\,$, $\,\color{green}{x - \dfrac{1}{x} = 2\sqrt{3}}\,$, and:
$$
\require{cancel}
\begin{align}
& \;\;x^6 - \color{green}{2\sqrt{3}}\, x^5 - x^4 + x^3 - \color{blue}{4}\,x^2 + 2x - \sqrt{3}
\\ = &\; \cancel{x^6} - \left(\color{green}{\cancel{x} - \bcancel{\frac{1}{x}}}\right)x^5 - \bcancel{x^4} + \xcancel{x^3} - \left(\color{blue}{\xcancel{x}+\frac{1}{x}}\right)x^2 + 2x - \sqrt{3}
\\ = &\;\; x - \sqrt{3}
\\ = &\; \;2
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Calculate $\int x(x+1)^{1/2} dx$. The integral in question:
$$I=\int x(x+1)^{1/2} dx$$
I've done integration by parts to get to
$I=x\frac{2}{3}(x+1)^{3/2} - \int\frac{2}{3}(x+1)^{3/2} dx$.
I've used mathematica and for the integral $\int \frac{2}{3}(x+1)^{3/2} dx$ when I calculate it by hand, I get $\frac{4}{15}(x+1)^{5/2}$ but apparently it should be just $\frac{2}{3}(x+1)^{5/2}$. I think something is wrong with my arithmetic. I thought $\int x^ndx = x^{n+1}/(n+1)$ which, in my case, I thought it would be $n+1=\frac{5}{2}$, $\frac{2}{3}\colon\frac{5}{2} = 4/15$. Therefore $I=\frac{4}{15}(x+1)^{5/2}$.
| I use $$\begin{align} x\sqrt{1+x}&=(1+x)\sqrt{1+x}-\sqrt{1+x}\\&=(1+x)^{3/2}-(1+x)^{1/2}.\end{align} $$ With the right side easily integrated to give (ignoring the constant:)
$$\begin{align} \frac25(1+x)^{5/2}-\frac23(1+x)^{3/2}&=\left(\frac25(1+x)-\frac23\right)(1+x)^{3/2}\\&=\left(\frac25x-\frac4{15}\right)(1+x)^{3/2}\end{align} $$
Your answer is: $$\begin{align} \frac23x(1+x)^{3/2}-\frac4{15}(1+x)^{5/2}&=\left(\frac23x-\frac4{15}(1+x)\right)(1+x)^{3/2}\\&=\left(\frac25x-\frac4{15}\right)(1+x)^{3/2} \end{align} $$
Same answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to find $\int_{0}^{\pi} \ln (b \cos x+c)$ without using Feynman’s integration technique? I shall find the integral by Feynman’s Technique Integration on a particular integral
$\displaystyle I(a)=\int_{0}^{\pi} \ln (a \cos x+1) d x,\tag*{} $
where $-1\leq a \leq 1.$
$\displaystyle \begin{aligned}I^{\prime}(a) &=\int_{0}^{\pi} \frac{\cos x}{a \cos x+1} d x, \\&=\frac{1}{a} \int_{0}^{\pi} \frac{(a \cos x+1)-1}{a \cos x+1} d x \\&=\frac{\pi}{a}-\frac{1}{a} \int_{0}^{\pi} \frac{d x}{a \cos x+1} \\&\stackrel{t=\tan \frac{x}{2}}{=} \frac{\pi}{a}-\frac{1}{a} \int_{0}^{\infty} \frac{1}{1+\frac{a\left(1-t^{2}\right)}{1+t^{2}}} \cdot \frac{2 d t}{1+t^{2}} \\&=\frac{\pi}{a}-\frac{2}{a} \int_{0}^{\infty} \frac{d t}{(1-a) t^{2}+(1+a)} \\&=\frac{\pi}{a}-\frac{2}{a \sqrt{1-a^{2}}} \tan^{-1}\left[\frac{\sqrt{1-a} t}{\sqrt{1+a}}\right]_{0}^{\infty} \\&=\frac{\pi}{a}-\frac{\pi}{a \sqrt{1-a^{2}}}\end{aligned}\tag*{} $
Integrating both sides w.r.t. $a$ yields
\begin{aligned}\int I^{\prime}(a) d a &=\pi\int\left(\frac{1}{a}-\frac{1}{a \sqrt{1-a^{2}}}\right) da \\& \stackrel{a=\sin \theta}{=} \pi\int\left(\frac{1}{\sin \theta}-\frac{1}{\sin \theta \cos \theta}\right) \cos \theta d \theta \\&=\pi\int \frac{\cos \theta-1}{\sin \theta} d \theta\\&I(a) =\pi \int \frac{-\sin ^{2} \theta}{\sin \theta(\cos \theta+1)} d \theta\\&=\pi \ln (1+\cos \theta) +C\end{aligned}
Putting $a=0$ gives $C=-\pi\ln 2$ and hence
$$
\boxed{\int_{0}^{\pi} \ln (a \cos x+1) d x =\pi \ln \left[1+\cos \left(\sin ^{-1} a\right)\right]= \pi \ln \left(\frac{1+\sqrt{1-a^{2}}}{2} \right)}
$$
I now want to generalize it to
$$I(b,c)=\displaystyle \int_{0}^{\pi} \ln (b \cos x+c),\tag*{} $$
where $c\neq 0$ and $-1\leq \frac{b}{c} \leq 1.$
$$
\begin{aligned}
I(b,c)&=\int_{0}^{\pi} \ln (b \cos x+c) \\
&=\int_{0}^{\pi} \ln \left[c\left(\frac{b \cos x}{c}+1\right)\right] \\
&=\pi \ln c+\int_{0}^{\pi} \ln \left(\frac{b}{c} \cos x+1\right) d x \\
&=\pi \ln c+I\left(\frac{b}{c}\right)
\end{aligned}
$$
Putting $a=\frac{b}{c}$ yields
$$\boxed{\int_{0}^{\pi} \ln (b \cos x+c) = \pi\left[\ln c+\ln \left(1+\sqrt{1-\frac{b^{2}}{c^{2}}}\right)\right] = \pi \ln \left(\frac{c+\sqrt{c^{2}-b^{2}}}{2}\right)}
$$
For example,
$$
\int_{0}^{\pi} \ln (\cos x+1)=\pi \ln \left(\frac{1}{2}\right)=-\pi \ln 2;
$$
$$
\int_{0}^{\pi} \ln (\sqrt{3} \cos x+2) d x=\pi\ln \frac{3}{2}
$$
Is there any method other than Feynman’s integration technique?
| Another option as opposed to using an infinite sequence and piecewise integration is to use a Weierstrass substitution and then a double integral.
Take $x\mapsto\tan\left(\tfrac x2\right)$ such that $\cos x\mapsto(1-x^2)/(1+x^2)$. The integral then becomes
$$\begin{align*}I & \equiv\int\limits_0^\pi\mathrm dx\,\log(1+a\cos x)\\ & =2\int\limits_0^{+\infty}\frac {\mathrm dx}{1+x^2}\log\left(1+a\frac {1-x^2}{1+x^2}\right)\\ & =2\int\limits_0^{+\infty}\frac {\mathrm dx}{1+x^2}\left[\log(1+a)+\log\left(1+\frac {1-a}{1+a}x^2\right)-\log(1+x^2)\right]\end{align*}$$
The first integral is trivial
$$\log(1+a)\int\limits_0^{+\infty}\frac {\mathrm dx}{1+x^2}=\frac \pi2\log(1+a)$$
The last two integrals can be evaluated using the general case
$$J\equiv\int\limits_0^{+\infty}\mathrm dx\,\frac {\log(1+z^2 x^2)}{1+x^2}$$
With the middle integral being the special case $z^2=(1-a)/(1+a)$ and the last integral being the case $z=1$. To evaluate $J$, we make use of a double integral, namely
$$\log(1+z^2)=\int\limits_0^1\mathrm dy\,\frac {2yz^2}{1+y^2z^2}$$
Substituting the identity above into $J$ and switching the order of integration gives
$$\begin{align*}J & =\int\limits_0^1\mathrm dy\,\int\limits_0^{+\infty}\mathrm dx\,\frac {2x^2yz^2}{(1+x^2)(1+x^2y^2z^2)}\\ & =\int\limits_0^1\mathrm dy\,\frac {2yz^2}{y^2z^2-1}\int\limits_0^{+\infty}\mathrm dx\,\left(\frac 1{1+x^2}-\frac 1{1+x^2y^2z^2}\right)\\ & =\pi z^2\int\limits_0^1\mathrm dy\,\frac y{y^2z^2-1}\left(1-\frac 1{yz}\right)\\ & =\pi z\int\limits_0^1\frac {\mathrm dy}{1+yz}\\ & =\pi\log(1+z)\end{align*}$$
Hence
$$\int\limits_0^{+\infty}\frac {\mathrm dx}{1+x^2}\log\left(1+\frac {1-a}{1+a}x^2\right)=\pi\log\left(1+\sqrt{\frac {1-a}{1+a}}\right)$$
And
$$\int\limits_0^{+\infty}\frac {\mathrm dx}{1+x^2}\log(1+x^2)=\pi\log 2$$
Putting everything together, then
$$\int\limits_0^\pi\mathrm dx\,\log(1+a\cos x)\color{blue}{=\pi\log\left(\frac {1+a}4\right)+2\pi\log\left(1+\sqrt{\frac {1-a}{1+a}}\right)}$$
Let $a=1$ and we get the integral stated in the question
$$\int\limits_0^\pi\mathrm dx\,\log(1+\cos x)=\pi\log\left(\frac 12\right)=-\pi\log 2$$
The more general case considering $\log(a+b\cos x)$ can be evaluated by transforming the integrand such that the constant term becomes one. Factoring out an $a$ gives
$$K\equiv\int\limits_0^\pi\mathrm dx\,\log(a+b\cos x)=\pi\log a+\int\limits_0^\pi\mathrm dx\,\log\left(1+\frac ba\cos x\right)$$
Using the formula we derived above and after some simplification, then
$$\int\limits_0^\pi\mathrm dx\,\log(a+b\cos x)\color{blue}{=\pi\log\left(\frac {a+b}4\right)+2\pi\log\left(1+\sqrt{\frac {a-b}{a+b}}\right)}$$
Substituting $a=2$ and $b=\sqrt3$ gives
$$\int\limits_0^\pi\mathrm dx\,\log(2+\sqrt3\cos x)=\pi\log\left(\frac {2+\sqrt3}4\right)+2\pi\log\left(1+\sqrt{\frac {2-\sqrt3}{2+\sqrt3}}\right)=\pi\log\left(\frac 32\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4470820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Expected number of ball tosses to have at least 5 balls in 4 out of 5 bins (Skyrim application) I have a bit of an interesting probability question that has an application to Skyrim and the number of quests you need to complete to get an achievement for the Thieves Guild. I can generalize the problem in terms of balls and bins.
Say you have an infinite number of balls available, and there are 5 bins, we can label them bins 1-5 (the bins are distinct). When you toss a ball, it is equally likely to fall into each bin (1/5 chance). What is the expected number of tosses so bins 1-4 have at least 5 balls in them? Each bin can hold an infinite number of balls, and we don't care about the balls falling into bin 5 (meaning it can't necessarily be the first 4 bins to have 5 balls).
I know that if I only cared about 1 bin reaching 5 balls, the expected value would be 5/p where p is the probability (1/5), but I can't continue this logic once one of the bins has 5 balls since the other bins may already have balls in them (the "misses" from trying to fill the first bin) so I have to use some other reasoning.
I wrote some code that I think simulates the rules above and I am getting around 29.7, which is lower than I would expect (the absolute minimum tosses is 20) so I would like to confirm or disprove this result as well as know how to generate a mathematical formula and calculate this without code.
Link to the code:https://github.com/nodnarb22/Skyrim-Thieves-Guild-Radiant-Quest-Simulator/blob/main/thievesguild
Any help or input would be much appreciated!
| Let's start with considering that
$$
\begin{array}{l}
\left( {x_1 + x_2 + x_3 + x_4 + x_5 } \right)^n = \cdots
+ x_{k_{\,1} } x_{k_{\,2} } \cdots x_{k_{\,n} }
+ \cdots \quad \left| {\;k_j \in \left\{ {1,2, \cdots ,5} \right\}} \right.\quad = \\
= \cdots + x_{\,j_{\,1} } ^{r_{\,1} } x_{\,j_{\,2} } ^{r_{\,2} } \cdots x_{\,j_{\,n} } ^{r_{\,n} }
+ \cdots \quad \left| \begin{array}{l}
\;j_i \in \left\{ {1, \ldots ,5} \right\} \\
\;\sum\limits_i {r_i } = n \\ \end{array} \right.\quad = \\
= \sum\limits_{\left\{ {\begin{array}{*{20}c}
{0\, \le \,k_{\,j} \,\left( { \le \,n} \right)} \\
{k_{\,1} + k_{\,2} + \, \cdots + k_{\,5} \, = \,n} \\
\end{array}} \right.\;} {\left( \begin{array}{c}
n \\ k_{\,1} ,\,k_{\,2} ,\, \cdots ,\,k_{\,5} \\
\end{array} \right)x_{\,1} ^{k_{\,1} } x_{\,2} ^{k_{\,2} } \cdots x_{\,5} ^{k_{\,5} } } \\
\end{array}
$$
is enumerating all possible sequences of $n$ tosses ending with $k_j$ balls in box $j$, and
$$
\begin{array}{l}
\left( {1 + 1 + 1 + 1 + 1} \right)^n = 5^n = \\
= \sum\limits_{\left\{ {\begin{array}{*{20}c}
{0\, \le \,k_{\,j} \,\left( { \le \,n} \right)} \\
{k_{\,1} + k_{\,2} + \, \cdots + k_{\,5} \, = \,n} \\
\end{array}} \right.\;} {\left( \begin{array}{c}
n \\ k_{\,1} ,\,k_{\,2} ,\, \cdots ,\,k_{\,5} \\ \end{array} \right)} \\
\end{array}
$$
Now let's consider the configuration of boxes having respectively $\ge 5, \ge 5,\ge 5,\ge 5, \le 4 $ balls:
last box has a different content, it is distinguishable and we have $5$
ways to choose it out of the five.
So the number of sequences that have such a configuration after $n$ tosses is
$$
\begin{array}{l}
N(n) = 5\sum\limits_{\left\{ {\begin{array}{*{20}c}
{5\, \le \,k_{\,1,2,3,4} \,\left( { \le \,n} \right)} \\
{\,0 \le k_{\,5} \le 4} \\
{k_{\,1} + k_{\,2} + \, \cdots + k_{\,5} \, = \,n} \\
\end{array}} \right.\;} {\left( \begin{array}{c}
n \\ k_{\,1} ,\,k_{\,2} ,\, \cdots ,\,k_{\,5} \\
\end{array} \right)} = \\
= 5\sum\limits_{\left\{ {\begin{array}{*{20}c}
{0\, \le \,j_{\,1,2,3,4} \,\left( { \le \,n - 5} \right)} \\
{\,0 \le k\left( { \le 4} \right)} \\
{j_{\,1} + j_{\,2} + \,j_{\,3} + j_{\,4} \, = \,n - 24 + k} \\
\end{array}} \right.\;} {\left( \begin{array}{c}
n \\ 5 + j_{\,1} ,\,5 + j_{\,2} ,\,5 + j_{\,3} ,5 + j_{\,4} ,\,4 - k \\
\end{array} \right)} = \\ = \quad \ldots \\
\end{array}
$$
there are many ways to rewrite the multinomial in terms of binomials etc. and I will omit them.
Clearly
$$
\begin{array}{l}
N(n) = 0\quad \left| {0 \le n \le 19} \right. \\
N(20) = 5\frac{{20!}}{{\left( {5!} \right)^4 0!}} \\
\quad \vdots \\
\end{array}
$$
But to answer to your question, the above is not much of interest.
We need in fact to find the number of sequences that becomes "successful" at the n-th toss.
The $n-1$ -sequences which can become successful just at the following step $n$ are only
of these two types
$$
\begin{array}{l}
\left\{ { \ge 5,\; \ge 5,\; \ge 5,\; = 4,\; = 4} \right\}, \\
\left\{ { \ge 5,\; \ge 5,\; \ge 5,\; = 4,\; < 4} \right\} \\
\end{array}
$$
and since they can be permuted, we have respectively
$$
\left( \begin{array}{c} 5 \\ 2 \\ \end{array} \right),\;
2\left( \begin{array}{c} 5 \\ 2 \\ \end{array} \right)
$$
ways to arrange them, and thereafter
*
*two ways to place the $n$th ball for the first,
*one way the second.
Therefore
$$
\begin{array}{l}
N_{first} (n) = 2\left( \begin{array}{c}
5 \\ 2 \\ \end{array} \right)\left( {\sum\limits_{\left\{ {\begin{array}{*{20}c}
{5\, \le \,k_{\,1,2,3} \,\left( { \le \,n - 9} \right)} \\
{k_{\,1} + k_{\,2} + \, \cdots + k_{\,5} \, = \,n} \\
\end{array}} \right.\;} {\left( \begin{array}{c}
n - 1 \\ k_{\,1} ,\,k_{\,2} ,k_{\,3} ,4,\,4 \\
\end{array} \right)} + \sum\limits_{\left\{ {\begin{array}{*{20}c}
{5\, \le \,k_{\,1,2,3} \,\left( { \le \,n - 5 - j} \right)} \\
{0 \le j \le 3} \\
{k_{\,1} + k_{\,2} + k_{\,3} \, + j\, = \,n - 5} \\
\end{array}} \right.\;} {\left( \begin{array}{c}
n - 1 \\ k_{\,1} ,\,k_{\,2} ,k_{\,3} ,4,\,j \\
\end{array} \right)} } \right) = \\
= \quad \cdots \\
\end{array}
$$
and for the probability
$$
P_{first} (n) = \frac{{N_{first} (n)}}{{5^n }}
$$
and then the expected $n$ follows obviously.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4472027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Answer:
$a+b = 7, ab = 2$
$$\begin{align}
(a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt]
a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \\
&= (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)
\end{align}$$
now,
$$\begin{align}
a^4 + b^4 &= (a+b)^4 - (4a^3b + 6a^2b^2 + 4ab^3) \\
&= (a+b)^4 - (4ab(a^2 + b^2) + 6(ab)^2) \\
&= (a+b)^4 - (4ab((a + b)^2 - 2ab) + 6(ab)^2) \\
&= 7^4 - (4(2)(7^2 - 2(2)) + 6(2)^2) \\
&= 2017
\end{align}$$
so
$$\begin{align}
&\phantom{=}\; (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)\\
&= 7^6 - (6\cdot2\cdot(2017) + 15(2)^2 (7^2 - 2(2)) + 20(2)^3) \\
&= 90585
\end{align}$$
correct?
| For either root of the equation:
*
*$x^2 = 7x - 2$
*$x^4 = (7x - 2)^2 = 49 x^2 - 28 x + 4 = 49(7x-2) - 28 x + 4 = 315 x - 94$
*$x^6 = (7x - 2)(315 x - 94) = 2205 x^2 - 1288 x + 188 = 14147 x - 4222$
It follows that $\,a^6+b^6 = 14147 (a+b) - 2 \cdot 4222 = 14147 \cdot 7 - 8444 = 90585\,$.
[ EDIT ] $\;$ The underlying idea in the above is based on the Euclidean division:
$$
x^6 = (x^4 + 7 x^3 + 47 x^2 + 315 x + 2111) \cdot(x^2 - 7 x + 2) + (14147 x - 4222)
$$
Then substituting $x = a,b$ the product is zero, which leaves only the remainder to evaluate, and adding the two equalities gives the same result as in the last line of the answer.
The shortcut here was to derive the remainder directly, without doing the full long division and calculating the quotient, which is not needed in this case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 0
} |
How to evaluate $\int_{0}^{\infty} \frac{\ln \left(a x^{2}+c\right)}{1+x^{2}} d x$ for positive constants $a$ and $c$? First of all, we deal with the simple case.
$$
\begin{aligned}
\int_{0}^{\infty} \frac{\ln \left(ax^{2}+1\right)}{1+x^{2}} d x \stackrel{x=\tan \theta}{=} & \int_{0}^{\frac{\pi}{2}} \ln \left(1+a \tan ^{2} \theta\right) d \theta \\
=& \int_{0}^{\frac{\pi}{2}} \ln \left(\frac{\cos ^{2} \theta+a \sin ^{2} \theta}{\cos ^{2} \theta}\right) d \theta \\
=& \int_{0}^{\frac{\pi}{2}} \ln \left(\cos ^{2} \theta+a \sin ^{2} \theta\right) d \theta-2 \int_{0}^{\frac{\pi}{2}} \ln (\cos \theta) d \theta \\
\end{aligned}
$$
Using the result in my post and $2 \int_{0}^{\frac{\pi}{2}} \ln (\cos \theta) d \theta =-\pi \ln 2$ yields
$$
\begin{aligned}
\int_{0}^{\infty} \frac{\ln \left(1+ax^{2}\right)}{1+x^{2}} d x &=\pi \ln \left(\frac{1+\sqrt{a}}{2}\right)+\pi \ln 2 =\pi \ln (1+\sqrt{a})
\end{aligned} \tag*{(*)}
$$
\begin{aligned}
& \int_{0}^{\infty} \frac{\ln \left(a x^{2}+c\right)}{1+x^{2}}d x \\
=& \int_{0}^{\infty} \frac{\left.\ln \left[c(\frac{a}{c} x^{2}+1\right)\right]}{1+x^{2}} d x \\
=& \int_{0}^{\infty} \frac{\ln c}{1+x^{2}} d x+\int_{0}^{\infty} \frac{\ln \left(\frac{a}{c} x^{2}+1\right)}{1+x^{2}}
\end{aligned}
Replacing $a$ by $\frac{a}{c}$ in $(*)$ yields
$$ \boxed{\int_{0}^{\infty} \frac{\ln \left(a x^{2}+c\right)}{1+x^{2}}d x =\pi \ln (\sqrt{a}+\sqrt{c})}$$
My question: Can we go further to evaluate a more general integral
$$\int_{0}^{\infty} \frac{\ln \left(ax^{2}+bx+1\right)}{1+x^{2}} d x?$$
| Alternatively, we can tackle the problem using contour integration on the upper semi-circle.
$$I=\int_{0}^{\infty} \frac{\ln \left(a x^{2}+c\right)}{1+x^{2}} d x= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\ln \left(a x^{2}+c\right)}{1+x^{2}} d x =\operatorname{Re} \left(\int_{-\infty}^{\infty} \frac{\ln (\sqrt{a} x+i \sqrt{c})}{1+x^{2}} d x \right)$$
Now let’s take the contour integration on the last integral along anti-clockwise direction of the path
$$\gamma=\gamma_{1} \cup \gamma_{2} \textrm{ where } \gamma_{1}(t)=t+i 0(-R \leq t \leq R) \textrm{ and } \gamma_{2}(t)=R e^{i t} (0<t<\pi) ,$$
$$
\begin{aligned}
\int_{-\infty}^{\infty} \frac{\ln (\sqrt{a} x+i \sqrt{c})}{1+x^{2}} d x &=\int_{-\infty}^{\infty} \frac{\ln (\sqrt{a} x+i \sqrt{c})}{1+x^{2}} d x \\
&=\oint_{\gamma} \frac{\ln (\sqrt{a} z+i \sqrt{c})}{1+z} d z \\
&=2 \pi i \operatorname{Res}\left(\frac{\ln (\sqrt{a} z+i \sqrt{c})}{1+z^{2}}, z=i\right)\\
&=2 \pi i \cdot \frac{\ln (\sqrt{a} + \sqrt{c})i}{2 i}\\
&=\pi \ln (\sqrt{a}+\sqrt{c})+\frac{\pi}{2}i
\end{aligned}
$$
Hence we conclude that
$$
\boxed{I=\pi \ln (\sqrt{a}+\sqrt{c})}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Finding triple sum of the dependant variables method checking Find:
A) $\sum_{i \geq j \leq k} \frac{1}{3^i 4^j 5^k}$ $i,j,k$ vary from $[0,\infty)$
B) $\sum_{i<j<k} \frac{1}{3^i 3^j 3^k}$ $i,j,k$ vary from $[0,\infty)$
My method as both are dependent summation , but second one is symmetric with the variables (function wise $\frac{1}{3^x}$ only ) so i thought maybe solving B gives hint for A) :for B) i can use the method as we do for two variable case , we know the product of those terms if they(variables) were all independent would be having these order of subscripts in the terms : $i=j=k , i<j>k,i<j<k,i<j=k,i>j>k,i>j=k,i>j<k ,i=j>k, i=j<k$ we can argue that product of the terms of forms $i<j<k$ would be same as $i>j>k$ as symmetric . Similarily for $i>j =k$ and $i=j<k$ and $i=j>k$ , $i<j=k$ So we get total sum for independent one to be equal to these individual order terms : $2(i=j<k) + 2(i=j>k) + (i=j=k) + 2(i<j<k) + (i<j>k) + (i>j<k)$ , we wanted the fourth sum in the above for B part , but for that i need to evaluate others how would i do so ? Or there is a different approach to one can think of too ?
| Both triple sums can be treated in the same way by transforming the index regions to obtain a geometric series in the form $\sum_{n=0}^{\infty}q^n=\frac{1}{1-q}$ valid for $|q|<1$.
Case A: We obtain
\begin{align*}
\color{blue}{\sum_{0\leq j\leq i,k<\infty}\frac{1}{3^i 4^j 5^k}}
&=\sum_{j=0}^\infty\frac{1}{4^j}\left(\sum_{i=j}^\infty \frac{1}{3^i}\right)\left(\sum_{k=j}^\infty \frac{1}{5^k}\right)\tag{1}\\
&=\sum_{j=0}^\infty\frac{1}{4^j}\left(\sum_{i=0}^\infty \frac{1}{3^{i+j}}\right)\left(\sum_{k=0}^\infty \frac{1}{5^{k+j} }\right)\tag{2}\\
&=\sum_{j=0}^\infty\frac{1}{4^j3^j5^j}\left(\sum_{i=0}^\infty \frac{1}{3^{i}}\right)\left(\sum_{k=0}^\infty \frac{1}{5^{k} }\right)\tag{3}\\
&=\frac{1}{1-\frac{1}{60}}\,\frac{1}{1-\frac{1}{3}}\,\frac{1}{1-\frac{1}{5}}\tag{4}\\
&=\frac{60}{59}\cdot\frac{3}{2}\cdot\frac{5}{4}\\
&\,\,\color{blue}{=\frac{225}{118}\doteq1.906}
\end{align*}
Comment:
*
*In (1) we write the triple sum somewhat more conveniently by rearranging terms as preparation for the next steps.
*In (2) shift the index of the inner sums to start with index $i=0$ and $k=0$.
*In (3) we collect factors with like exponents.
*In (4) we apply the geometric series expansion
Case B: We proceed similarly as in (A) and obtain
\begin{align*}
\color{blue}{\sum_{0\leq i<j<k<\infty}\frac{1}{3^i 3^j 3^k}}
&=\sum_{i=0}^\infty\frac{1}{3^i}\sum_{j=i+1}^\infty\frac{1}{3^j}\sum_{k=j+1}^\infty\frac{1}{3^k}\\
&=\sum_{i=0}^\infty\frac{1}{3^i}\sum_{j=i+1}^\infty\frac{1}{3^j}\sum_{k=0}^\infty\frac{1}{3^{k+j+1}}\\
&=\frac{1}{3}\sum_{i=0}^\infty\frac{1}{3^i}\sum_{j=i+1}^\infty\frac{1}{9^j}\sum_{k=0}^\infty\frac{1}{3^{k}}\\
&=\frac{1}{3}\sum_{i=0}^\infty\frac{1}{3^i}\sum_{j=0}^\infty\frac{1}{9^{j+i+1}}\sum_{k=0}^\infty\frac{1}{3^{k}}\\
&=\frac{1}{27}\sum_{i=0}^\infty\frac{1}{27^i}\sum_{j=0}^\infty\frac{1}{9^{j}}\sum_{k=0}^\infty\frac{1}{3^{k}}\\
&=\frac{1}{27}\,\frac{1}{1-\frac{1}{27}}\,\frac{1}{1-\frac{1}{9}}\,\frac{1}{1-\frac{1}{3}}\\
&=\frac{1}{27}\cdot\frac{27}{26}\cdot\frac{9}{8}\cdot\frac{3}{2}\\
&\,\,\color{blue}{=\frac{27}{416}\doteq 0.064}
\end{align*}
| {
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Evaluate the indefinite integral $\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}$ without trigonometric substitution. In order to find
$$
\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}
$$
we set $t=\arctan x$. Then $x=\tan t$ and $dt=\frac{dx}{x^2+1}$, so
$$
\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}=\int\frac{dt}{\sqrt{\frac{1}{\cos^2t}}}=\int\cos tdt\\
=\sin t+C=\sin(\arctan x)+C
$$
Now, since
$$
\sin(\arctan x)=\sqrt{\frac{\tan^2(\arctan x)}{\tan^2(\arctan x)+1}}
$$
the answer is $\frac{x}{\sqrt{x^2+1}}+C$.
My Question: Is there another way to find this integral without using trigonometry?
| You can use hyperbolic trigonometry. Set $x=\sinh t$, $x^2+1=\cosh^2 t$, $dx=\cosh t dt$, and you get:
$$\begin{array}{rcl}\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}&=&\int\frac{\cosh t\,dt}{\cosh^3 t}\\&=&\int\frac{dt}{\cosh^2 t}\\&=&\tanh t+C\\&=&\frac{\sinh t}{\cosh t}+C\\&=&\frac{x}{\sqrt{x^2+1}}+C\end{array}$$
| {
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} |
$F$ such that $F(A_{0}B_{0}+A_{1}B_{1})=F(A_{0})+F(B_{0})+F(A_{1})+F(B_{1})$? We know that the only way of turning a multiplication into addition is by a logarithmic function; so if $$C=A_{0}B_{0}$$ then we know that: $$\log(C)=\log(A_{0})+\log(B_{0})$$ but can we extend this idea? .
In the case of:
$$C=A_{0}B_{0}+A_{1}B_{1}$$
can we prove or disprove the existing of a function F(x) such as: $$F(C)=F(A_{0})+F(B_{0})+F(A_{1})+F(B_{1})$$
| No such function exists, except the identically zero function. With the notation $\,f_k = F(k)\,$:
$$
\begin{align}
2 = 1 \cdot 1 + 1 \cdot 1 \quad\implies\quad f_2 &= 4 f_1
\\ 3 = 2 \cdot 1 + 1 \cdot 1 \quad\implies\quad f_3 &= 3 f_1 + f_2 = 7 f_1
\\ 4 = 2 \cdot 1 + 2 \cdot 1 \quad\implies\quad f_4 &= 2f_1 + 2 f_2 = 10 f_1
\\ 7 = 3 \cdot 2 + 1 \cdot 1 \quad\implies\quad f_7 &= 2 f_1 + f_2 + f_3\color{red}{ = 13 f_1}
\\ 7 = 4 \cdot 1 + 3 \cdot 1 \quad\implies\quad f_7 &= 2 f_1 + f_3 + f_4\color{red}{ = 19 f_1}
\end{align}
$$
Since $\,F(7)\,$ can only take one value, it follows that $\,13f_1=19f_1\,$ $\,\implies f_1 = F(1) = 0\,$, then $\,F(k)=0\,$ for $\,k \in \mathbb N\,$ by induction, and therefore for all $\,k \ge 2\,$:
$$
\require{cancel}
k\cdot x = (k-1) \cdot x + 1 \cdot x \;\;\implies\;\; F(k \cdot x) = \cancel{F(k-1)} + F(x) + \cancel{F(1)} + F(x) = 2 F(x)
$$
But then:
$$
\begin{align}
F(2x) &= F( 2 \cdot x) = 2 F(x)
\\ F(4x) &= F(4 \cdot x) = 2 F(x)
\\ F(4x) &= F(2 \cdot 2x) = 2 F(2x) = 4 F(x)
\end{align}
$$
The last two equalities imply $\,2 F(x) = 4 F(x)\,$, so in the end $\,F(x) = 0\,$.
| {
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Geometrical series extra term confusion Given
$$
P = \frac{A}{1+x} + \frac{A}{(1+x)^2} + \dots + \frac{A}{(1+x)^{N-1}} + \frac{A}{(1+x)^{N}},
$$
how can I arrive at the textbook's expression
$$
P = A\left( \frac{1 - \frac{1}{(1+x)^N}}{x}\right)
$$
Attempt at solution
This seems like a divergent geometrical series, I know that the partial sum
$$
S_n = \frac{1-r^N}{1 - r},
$$
where $r$ is the common ratio $1 / (1+x)$, but when I write down $S_n$, I find an extra $1+x$ factor
$$
P = A\frac{S_n}{1+x}.
$$
It led me to think that maybe because in the definition of $S_n$ for an infinite geometrical series we start at $x^0$, i.e., $a + ax + ax^2 + \dots$, and in this problem we start at $ax + ax^2 + \dots$, then there is some term missing around, but I can't seem to make it work.
Can anyone point out what I am missing ?
| $ P = \frac{A}{1+x} + \frac{A}{(1+x)^2} + \dots + \frac{A}{(1+x)^{N-1}} + \frac{A}{(1+x)^{N}}$
$=\frac{A}{1+x}(1 + \frac{A}{1+x} + \frac{A}{(1+x)^2} + \dots + \frac{A}{(1+x)^{N-2}} + \frac{A}{(1+x)^{N-1}})$
$=\frac{A}{1+x}(\frac{1-\frac{1}{(1+x)^N}}{1-\frac{1}{1+x}})$
$= \frac{A}{1+x}(\frac{1-\frac{1}{(1+x)^N}}{\frac{x}{1+x}})$
$=A(\frac{1-\frac{1}{(1+x)^N}}{x})$
| {
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Show that $(j,k)\mapsto 3k + j + (k+j)^2$ and $g$ are bijections from $\mathbb{Z}_{\ge 0}^2$ to $2\mathbb{Z}_{\ge 0}$.
Show that $(j,k)\mapsto 3k + j + (k+j)^2$ and $(j,k)\mapsto 3j + k + (k+j)^2$ are two bijections from $\mathbb{Z}_{\ge 0}^2$ to $2\mathbb{Z}_{\ge 0}$.
Clearly both maps are well defined (consider congruences modulo $2$).
Let $g :\mathbb{Z}_{\ge 0}^2\to 2\mathbb{Z}_{\ge 0}$ be the described map. Note that $g$ is bijective if and only if $f :\mathbb{Z}_{\ge 0}^2 \to 2\mathbb{Z}_{\ge 0}, (j,k)\mapsto 3j + k + (k+j)^2$ is bijective. Indeed, if $n$ is a nonnegative even integer and $g^{-1}(n) = (j,k)$, then $f((k,j)) = n$. And if $f((j,k)) = f((s,t))$ then $g((k,j)) = g((t,s))\Rightarrow (k,j) = (t,s)$. The other direction is symmetric. So it suffices to show that $f$ is bijective.
Trying a few small values gives $(0,0)\mapsto 0, (0,1)\mapsto 2,(1,0)\mapsto 4, (0,2)\mapsto 6, (1,1) \mapsto 8, (2,0)\mapsto 10.$ For fixed $x,$ there are $x+1$ pairs $(j,k)$ of nonnegative integers summing to $x$. For a given nonnegative even integer $y = 2n - 2, n \ge 1 1,$ find $x\ge -1$ so that $\frac{(x+1)(x+2)}2 < n \leq \frac{(x+2)(x+3)}2$. We claim that $f((i, x+1 - i)) = y,$ where $ i = n - \frac{(x+1)(x+2)}2 - 1 \ge 0$. Observe that $i \leq x+1$ so $(i, x+1-i)$ is indeed a pair of nonnegative integers. Also,
$\begin{align}f((i, x+1 - i)) &= 3i + (x+1-i)+(x+1)^2\\
&= 2i + (x+1)(x+2)\\
&= 2n -2.\end{align}$
To find x, first set $z = x+1$. Multiplying both sides of the inequality by $8$ gives $4z^2 + 4z < 8n \leq 4(z+1)(z+2) \Rightarrow (2z + 1)^2 < 8n + 1 \leq (2z + 3)^2\Rightarrow 2z + 1 < \sqrt{8n + 1} \leq 2z + 3\Rightarrow z < \frac{\sqrt{8n+1}-1}2 \leq z + 1\Rightarrow z = \lceil\frac{\sqrt{8n+1}-3}2\rceil\Rightarrow x = \lceil\frac{\sqrt{8n+1}-5}2\rceil.$
Now we claim that all $(j,k)$ with $j+k\leq x$ get mapped to values that are at most $n$ and that if $n > \frac{(x+1)(x+2)}2,$ it appears in the list of pairs $(j,k)$ with $j+k = x+1$. In fact, if $n = \frac{(x+1)(x+2)}2$, then $n = f(x,0)$. Otherwise, if $n > \frac{(x+1)(x+2)}2$, then $n = f(i, x+1 - i)$, where $i = n - \frac{(x+1)(x+2)}2$.
From the above, we define the inverse of $y = 2n-2,n\ge 1$ to be $(i, x+1-i)$, where $i,x$ are given in terms of $n$ above.
Is there a simpler method?
| Let $$f(j,k)=\frac{3k+j+(k+j)^2}2$$
Then $f(j-1,k+1)=f(j,k)+1$ and $f(k+1,0)=f(0,k)+1$, hence the assertion follows by induction.
For $f(0,0)=0$.
Assume $n=f(j,k)>0$.
If $j>0$, then $n+1=f(j-1,k+1)$.
Otherwise, $n=f(0,k)$, hence $n+1=f(k+1,0)$.
To prove uniqueness, note that if $j+k<j'+k'$, then
\begin{align}
f(j,k)&<f(0,j+k)\\
&<f(j+k+1,0)\\
&\leq f(j'+k',0)\\
&<f(j',k')
\end{align}
On the other hand, if $j+k=j'+k'$ and $j'<j$, then
\begin{align}
f(j,k)&<f(j',k+j-j')\\
&=f(j',k')
\end{align}
| {
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Why do you have to check all the 'sections' of an inequality when solving for x? For instance, the inequality
$x^2+3x+2>0$ factors into $(x+2)(x+1)>0$
If this were just an equation (i.e ...=0) you would know the solutions are x={-2, -1}.
But, because it's an inequality, you have to check the intervals. For instance, by drawing a number line and evaluating the inequality expression at each of the intervals and marking whether it satisfies the condition or not. How come the usual method doesn't work?
If you use the inequality symbol is an equals sign and do it the usual way you would get $x>-1$ and $x>-2$ which is not right.
| You can alternatively tackle the problem as follows:
\begin{align*}
(x + 2)(x + 1) > 0 & \Longleftrightarrow ((x + 1) + 1)(x + 1) > 0\\\\
& \Longleftrightarrow (x + 1)^{2} + (x + 1) > 0 \\\\
& \Longleftrightarrow (x + 1)^{2} + (x + 1) + \frac{1}{4} > \frac{1}{4}\\\\
& \Longleftrightarrow \left[(x + 1) + \frac{1}{2}\right]^{2} > \frac{1}{4}\\\\
& \Longleftrightarrow \left(x + \frac{3}{2}\right)^{2} > \frac{1}{4}\\\\
& \Longleftrightarrow \left|x + \frac{3}{2}\right| > \frac{1}{2}\\\\
& \Longleftrightarrow \left(x < -2\right)\vee(x > -1)
\end{align*}
Hopefully this helps!
| {
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"source": "stackexchange",
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Show that no non-zero integers satisfy this pair of equations (from Baltic Way 2021)
Show that no non-zero integers $a, b, x, y$ satisfy:
\begin{cases} ax-by=16. \\ ay+bx=1. \end{cases}
From Baltic-Way 2021.
\begin{align}
&(a+bi)(x+yi)=(ax-by)+i(ay+bx)=16+i. \\
&|(a+bi)(x+yi)|=|16+i|. \\
&\sqrt{a^2+b^2}\sqrt{x^2+y^2}=\sqrt{257}. \\
&(a^2+b^2)(x^2+y^2)=257. \\
&\therefore a^2+b^2=257, x^2+y^2=1 \text{ or } a^2+b^2=1, x^2+y^2=257.(\because 257: \text{ prime.)} \\
\Rightarrow & \text{No solution.}
\end{align}
Is my solution right?
| Your solution is indeed valid, and as FShrike writes in the comments, quite elegant. I'm putting this confirmation in a community wiki answer to mark this question as answered. (I think this is how that works!)
| {
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Why do prime factors of odd term Lucas numbers only end in 1 or 9? I'm working on a problem, which I've eventually reduced to the following question: show that every odd term Lucas number has a prime factor that ends with either 1 or 9. Here the Lucas sequence is defined by the recursion: $$a_0=2, a_1=1, a_{n+1}=a_n+a_{n-1}.$$
I looked up a list of prime factorizations of Lucas numbers on the following page:
https://r-knott.surrey.ac.uk/Fibonacci/lucas200.html
Surprisingly, I find all the odd term Lucas numbers to have only prime factors that end with either 1 or 9, with the exception that $a_{6k+3}$ is also divisible by 4. It's easy to show using modulo arithmetic that all $a_{6k+3}$ are divisible by 4 and no higher powers of 2, while all other odd term Lucas numbers are odd. If we count out those factors of 4, all the other prime factors of the odd term Lucas numbers seem to end only with 1 or 9. This is a curious observation, but I don't know how one can prove this. Can anyone help please!
| I made another Conway Topograph diagram for $x^2 - 5 y^2$ The $x,y$ pairs are written vertically in green. The values of $x^2 - 5 y^2$ are in pink.
Right, see displayed $11^2 - 5 \cdot 5^2 = -4, \; \; $ $29^2 - 5 \cdot 13^2 = -4, \; \; $ $199^2 - 5 \cdot 89^2 = -4, \; \; $ $521^2 - 5 \cdot 233^2 = -4, \; \; $
Also, $38^2 - 5 \cdot 17^2 = -1, \; \; $ and doubling this gives
$76^2 - 5 \cdot 34^2 = -4, \; \; $
If $x$ is one of your alternate Lucas numbers, there is a Fibonacci number $y$ such that
$$ x^2 - 5 y^2 = -4 $$
From $x^2 = 5 y^2 - 4, $ we see $x^2 = 5 y^2 - 2^2 = - (2^2 - 5 y^2 )$ and
$$ x | 2^2 - 5 y^2 $$
Now if there is any prime $q$ with $q \equiv \pm 2 \pmod 5,$ and if $q | x,$ then
$$ q | 2^2 - 5 y^2 $$
However, for any such prime, Legendre symbol $(20|q) = -1.$ This forces $q | y $ along with $q | 2.$
Thus, it is allowed for $x$ to be even (divisible by 2) and divisible by primes $p \equiv 1,4 \pmod 5,$ but not by $q \equiv \pm 2 \pmod 5$ when $q > 2$
On this site, you may have seen this fact: for a prime $r \equiv 3 \pmod 4,$ if $r | u^2 + v^2,$ then both $r|u$ and $r|v$ This is a general thing about the Legendre symbol.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral $\int_0^1 \frac{(1+t^2)}{1-t^2+t^4}\ln(t)dt$ $$I=\int_0^1 \frac{(1+t^2)}{1-t^2+t^4}\ln(t)dt=-\frac{4}{3}G $$ G is Catalan's constant.
$$I=\int_0^1 \frac{(1+t^2)^2}{(1+t^2)(1-t^2+t^4)}\ln(t)dt=\int_0^1 \frac{(1+t^2)^2}{(1+t^6)}\ln(t)dt$$
Do series expansion:
$$I=\sum_{n=0}^\infty (-1)^n \int_0^1(1+2t^2+t^4)t^{6n}\ln(t)dt$$
Integrate term by term:
$$I=-\sum_{n=0}^\infty (-1)^n\left( \frac{1}{(6n+1)^2}+\frac{2}{(6n+3)^2}+\frac{1}{(6n+5)^2} \right)$$
Re-organize them:
$$I=-\sum_{n=0}^\infty (-1)^n\left( \frac{1}{(6n+1)^2}-\frac{1}{(6n+3)^2}+\frac{1}{(6n+5)^2}\right)-\sum_{n=0}^\infty (-1)^n \frac{3}{(6n+3)^2} $$
The first part is Catalan's constant:
$$I=-G-\sum_{n=0}^\infty (-1)^n \frac{3}{(6n+3)^2}=-G-\frac{3}{9}G=-\frac{4}{3}G$$
Done.
| Alternatively, with
$ \frac{1+t^2}{1-t^2+t^4}= \frac{1}{1+t^2} +\frac{3t^2}{1+t^6}$
\begin{align}
&\int_0^1 \frac{(1+t^2)\ln t}{1-t^2+t^4}dt
=\int_0^1 \underset{=-G}{\frac{\ln t}{1+t^2}}dt
+ \int_0^1 \underset{=-\frac13G}{\frac{3t^2\ln t}{1+t^6} \ \overset{t^3\to t}{dt}}=-\frac43G
\end{align}
| {
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$\log_{3}\frac{16^{x-3}+14}{4^{x-3}+2}=243$ , find the value of $x$. $\log_{3}\frac{16^{x-3}+14}{4^{x-3}+2}=243=3^5$
$\Rightarrow \frac{16^{x-3}+14}{4^{x-3}+2}= {3^3}^5=3^{243}$
I am not able to proceed from here. Please help !!!
The options given for this problem are :-
*
*$x$ is a rational number.
*$x$ is a natural number.
*$x$ is an even natural number.
*$x$ is a rational number less than $0$.
Thanks in advance !!!
| Let $x$ starts from $-\infty$ and jumps only on integers waits at $0$ and again starts journey towards $\infty$
Let,
$$f(x) = \frac {16^{x-3} + 14}{4^{x -3} + 2}$$
$$\implies \log_3 \frac {16^{x-3} + 14}{4^{x -3} + 2} = L(x) = \log_3(f(x))$$
*
*$x\in (-\infty, 0)$
$$\lim_{x\to-\infty}f(x) = 7 \text{ so also the value of }L(x\to-\infty) \to \text{some constant number}$$
Let me interpret this as the derivative of $L(x)$ should tend to $0$ as the value of $x\to-\infty$ or $\color{blue}{\text{Should we say this as the value of $L(x)$ is not changing as values of $x$ does?}} $
*
*$x=0$
$f(0) \approx 7$ which is far enough to not to expect $L(0) = 243$
*$x\in (0, \infty)$
Lets write a function: $\lambda(x)$ = $\text{last digit function giving you the last digit of $x$}$
$$\begin{align}
\\ \lambda(3^2)=9
\\ \lambda(3^3)=7
\\ \lambda(3^4)=1
\\ \lambda(3^5)=3
\\
\\
\\ \implies \lambda(3^{243}) = 7
\end{align}$$
similarly
$$\lambda(L(i)) = \frac {\lambda(16^{i-3})+14}{\lambda(4^{i-3})+ 2} = \frac {\text{something ending with zero}}{\text{something ending with 6 or 8}} \ne \text{something ending with 7}$$
$x \notin \mathbb{Z}$
| {
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Find all pairs $(m,n)$ such that the quotient $q$ and reminder $r$ of $\frac{m^2+n^2}{m+n}$ satisfies $q^2+r=17$ Let $m,n\in \mathbb{N}$. Let $q,r$ be the quotient and remainder of $\frac{m^2+n^2}{m+n}$. If $q^2+r=17$, find all pairs of $(m,n)$.
My first thought was trying polynomial division, but I did not see how it could have helped. I also thought at factoring $m^2+n^2$, but to no success. How should I proceed?
| Clearly, $1\leq q \leq 4$. We can consider the 4 possible values of $q$ in turn.
If $q=1$, then $r=16$, and we have $m^2+n^2 = 1(m+n)+16$, or $(2m-1)^2 + (2n-1)^2 = 66$, which has no solutions $m, n \in \mathbb{N}$.
If $q=2$, then $r=13$, and we have $m^2+n^2 = 2(m+n)+13$, or $(m-1)^2 + (n-1)^2 = 15$, which has no solutions $m, n \in \mathbb{N}$.
If $q=3$, then $r=8$, and we have $m^2+n^2 = 3(m+n)+8$, or $(2m-3)^2 + (2n-3)^2 = 50$, which has no solutions $m, n \in \mathbb{N}$ satisfying $m+n>r$.
If $q=4$, then $r=1$, and we have $m^2+n^2 = 4(m+n) + 1$, or $(m-2)^2+(n-2)^2 = 9$, which has solutions $(m,n) \in \{(5,2), (2,5)\}$.
Thus, the full set of solutions is $(m,n) \in \{(5,2), (2,5)\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrating $\displaystyle\int_2^3 \dfrac{\text{d}x}{2x(1-2x)}$ in W. Mathematica giving wrong result. I calculated by hand the following integral
$$\int_2^3 \dfrac{\text{d}x}{2x(1-2x)}$$
What I did it to use partial fractions:
$$\dfrac{1}{x(1-2x)} = \dfrac{1}{x} + \dfrac{-2}{1-2x}$$
Whence
$$\dfrac{1}{2}\int_2^3 \left(\dfrac{1}{x} + \dfrac{-2}{1-2x}\right) = \dfrac{1}{2}\left(\ln|x|\big|_2^3 + \ln|1-2x|\big|_2^3\right) = \dfrac{1}{2}\ln\left(\dfrac{5}{2}\right)$$
Now, checking with W. Mathematica (NOT W. Alpha) it does return the value of $\ln\dfrac{3}{\sqrt{10}}$
How is it possible? Even with some manipulation, the max I can get is $\ln\dfrac{\sqrt{10}}{2}$... Thank you!
| The partial fraction that you got is wrong.
$$\frac{1}{x(1-2x)} \ne \frac{1}{x} + \frac{-2}{1-2x}$$
Since,
$$\frac{1}{x} + \frac{-2}{1-2x} = \frac{(1-2x)-2x}{x(1-2x)} = \frac{1-4x}{x(1-2x)}$$
| {
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"url": "https://math.stackexchange.com/questions/4487311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\frac{dx^5}{dx^2},$ the derivative of $x^5$ with respect to $x^2$
Let $f(x)=x^5.$ Find its derivative with respect to $x^2$ , i.e., find $\frac{dx^5}{dx^2}.$
We know that $\frac{dx^5}{dx}=5x^4.$
But what should I do when it is needed to take derivative with respect to $x^2 ,x^3$ or other weird things different from classical derivatives?
My approaches:
$1-)$ Say $x^2=a$ , then $x^5 =a^{5/2}$. Hence , find $$\frac{da^{5/2}}{da}=\frac{5}{2}a^{3/2}=\frac{5}{2}x^{3}$$
$2-)$ Find $$\lim_{h\to 0}\frac{f(x^2+h)-f(x^2)}{h}=\lim_{h\to 0}\frac{(x^2+h)^5-(x^2)^5}{h}=5x^8$$
My approaches conflict. Why do they give different results?
Addendum: Extra examples will be appreciated, for example, $dx^7 /dx^3,\;dx^6/dx^5.$
| Here is how a Physicist would do it
$$\frac{dx^5}{dx^2}\frac{dx^2}{dx}=\frac{dx^5}{dx}=5x^4$$
Since $\frac{dx^2}{dx}=2x$, we get
$$\frac{dx^5}{dx^2}=\frac{5x^3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Compute $\int \frac{\sin(\pi x)^2}{\sin(\pi x/2)^2-a^2} \ dx$ Let $a \in (0,1)$, I would like to integrate
$$\int_0^{x_{\downarrow}} \frac{\sin(\pi x)^2}{\sin(\pi x/2)^2-a^2} \ dx +\int_{x_{\uparrow}}^1 \frac{\sin(\pi x)^2}{\sin(\pi x/2)^2-a^2} \ dx. $$
Now $\sin(\pi x/2)^2$ is a monotonically increasing function from $0$ to $1$, therefore there exists a unique $x^*$ such that $\sin(\pi x^*/2)^2 = a^2.$ Now $x^* \in (x_{\downarrow},x_{\uparrow})$ such that both of the above integrals are well-defined.
Please let me know if you have any questions.
Here $x_{\downarrow}$ and $x_{\uparrow}$ are two values of which one is below $x^*.$
| Note that
\begin{align}
\frac{\frac14\sin^2\pi x}{\sin^2\frac{\pi x}2-a^2}
=-a^2 + \cos^2\frac{\pi x}2 + \frac{a^2(1-a^2)}{\sin^2\frac{\pi x}2-a^2}
\end{align}
Integrate respectively to obtain
\begin{align}
&\int_ 0^{x_{\downarrow}} \frac{\sin^2\pi x}{\sin^2\frac{\pi x}2-a^2}dx\\
=& \ 2(1-2a^2) x_{\downarrow} +\frac2\pi\sin\pi x_{\downarrow}
-\frac{8a\sqrt{1-a^2}}\pi\tanh^{-1}\frac{\sqrt{1-a^2} \tan\frac{\pi x_{\downarrow}}2}a
\\\\
&\int^1_{x_{\uparrow}} \frac{\sin^2\pi x}{\sin^2\frac{\pi x}2-a^2}dx\\
=& \ 2(1-2a^2)(1- x_{\uparrow})-\frac2\pi\sin\pi x_{\uparrow}
+\frac{8a\sqrt{1-a^2}}\pi\coth^{-1}\frac{\sqrt{1-a^2} \tan\frac{\pi x_{\uparrow}}2}a
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate the determinant of $4 \times 4$ matrix with a variable on the diagonal? I need to calculate the determinant of the following $4 \times 4$ matrix:
\begin{bmatrix}x&1&1&1\\1&x&1&1\\1&1&x&1\\1&1&1&x\end{bmatrix}
I heard there is a way by separating the matrix into blocks, but I couldn't succeed in doing that.
| Useful trick - if the sum of each row of the matrix is the same (in this case - $x+3$), then you can simplify the determinant via the following elementary operations:
*
*Add all columns to the first column.
*Subtract the first row from all rows.
In your case:
$$\begin{vmatrix}x&1&1&1\\1&x&1&1\\1&1&x&1\\1&1&1&x\end{vmatrix}
=\begin{vmatrix}x+3&1&1&1\\x+3&x&1&1\\x+3&1&x&1\\x+3&1&1&x\end{vmatrix}
=\begin{vmatrix}x+3&1&1&1\\0&x-1&0&0\\0&0&x-1&0\\0&0&0&x-1\end{vmatrix}$$
Your matrix is now triangular, and the determinant is the product of diagonal elements.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4497212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How does $\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}$ become $\sqrt{2(2+\sqrt{2})}$? I'd like to know how can one simplify the following expression
$$\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}$$
into
$$\sqrt{2(2+\sqrt{2})}.$$
Wolfram alpha suggests it as an alternative form, and numerically it's easy to verify, but I can't find the right algebra to show they are indeed equivalent.
Note I ran into this problem, trying to do: $2\cos(\pi/8)+2\sin(\pi/8)$, where
$$2\cos(\pi/8)=\sqrt{2+\sqrt{2}},$$
$$2\sin(\pi/8)=\sqrt{2-\sqrt{2}}.$$
| Rather than focus on the specifics of this one question, let's try to figure out the general case that this is an instance of.
We are given an expression of the form $$\sqrt{a + b} + \sqrt{a - b}$$
and we would like to somehow write this as the square root of a single quantity. So, let's give it a name:
$$X = \sqrt{a + b} + \sqrt{a - b}$$
Square both sides, carefully using the identity $(u + v)^2 = u^2 + 2uv + v^2$. We get
$$X^2 = (a + b) + 2\sqrt{a+b}\sqrt{a-b} + (a-b)$$
which simplifies (slightly) as
$$X^2 = 2a + 2\sqrt{a+b}\sqrt{a-b}$$
Now let's focus on the product of those two square roots. We know that
$$\begin{aligned} \sqrt{a+b}\sqrt{a-b} &= \sqrt{(a+b)(a-b)} \\ &= \sqrt{a^2 - b^2} \end{aligned}$$
which allows us to write our main result as
$$X^2 = 2a + 2\sqrt{a^2 - b^2}$$
and finally we obtain
$$X = \sqrt{2\left( a + \sqrt{a^2 - b^2} \right) }$$
So we have shown the identity
$$\sqrt{a+b} + \sqrt{a-b} = \sqrt{2\left( a + \sqrt{a^2 - b^2} \right) }$$
The question in the OP is a specific instance of this, in which $a = 2$ and $b = \sqrt{2}$; in this case, the identity just discovered reads
$$\sqrt{2 + \sqrt{2}} + \sqrt{2 - \sqrt{2}} = \sqrt{2\left(2 + \sqrt{4-2} \right)}$$
which is exactly what we were asked to show.
But there are other interesting applications of the general identity. For example, let $a = 5$ and $b = 3$. Then our identity reads
$$\sqrt{5+3} + \sqrt{5-3} = \sqrt{2\left( 5 + \sqrt{5^2 - 3^2} \right) }$$
or equivalently
$$\sqrt{8} + \sqrt{2} = \sqrt{2\left( 5 + \sqrt{25-9} \right)}$$
which can, of course, be simplified further as
$$\sqrt{8} + \sqrt{2} = \sqrt{2\left( 5 + 4 \right)}$$
which is easily seen to be true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that all the roots of $p(z)= z^5 – z^3 + 1$ are in the ring $\frac{1}{2}\lt \vert z \vert \lt \frac{3}{2}$. I am going to try check that $\vert z^5 – z^3\vert \lt 1$ for $\vert z \vert \le \frac{1}{2}.$
$1=|-1|=|z^5+z^3|=|z|^3\cdot |z^2+1|\le |z^2+1|\le |z|^2+|1|=|z|^2+1 \le 1.$
Now I'm stuck. What do I do next?
Thank's for any help.
| Using the triangle inequality, we get:
$|z^5-z^3|=|z^3||z^2-1|\le\dfrac 18|z^2-1|\le\dfrac 18(|z^2|+1)\le\dfrac 18\left(\dfrac 14+1\right)=\dfrac 18\cdot \dfrac 54=\dfrac 5{32}\lt1$ for $|z|\le\dfrac 12$
Now let $g(z)=z^5$. Since $|z^3-1|\le|z^3|+1\le\dfrac {27}8+1=\dfrac {35}8\lt|z^5|=\left(\dfrac 32\right)^5$, on $|z|=\dfrac 32$, by Rouche we get that there are $5$ zeros in the given annulus.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sqrt 2 + \sqrt 3 + \sqrt 5$ is irrational. Note that my attempted method below is distinct from the solutions in this question.
I also know this is generally true for $\sqrt p + \sqrt q + \sqrt r$ where $p,q,r$ are prime, but I am asking if my particular method works.
Prove that $\sqrt 2 + \sqrt 3 + \sqrt 5$ is irrational.
Assume that $\sqrt 2 + \sqrt 3 + \sqrt 5 = \frac pq$ for some $p,q \in \mathbb Z$, in lowest terms, with $q \neq 0$. Then at most one of $p$ and $q$ can be even. Then
\begin{align*}
\sqrt 2 + \sqrt 3 + \sqrt 5 &= \frac pq\\
\\
(\sqrt 2 + \sqrt 3 + \sqrt 5)^2 &= \left(\frac pq \right)^2\\
\\
2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})&=\frac{p^2}{q^2}\\
\\
2q^2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})&=p^2
\end{align*}
which shows that $p^2$ is even, so $p$ is even and let $p=2k$ for some $k\in \mathbb Z$. Substituting this we get
\begin{align*}
2q^2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})&=(2k)^2\\
\\
q^2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})&=2k^2\\
\\
q^2 &= 2 \left(\frac{k^2}{5+\sqrt 6 + \sqrt{10} + \sqrt{15}}\right),
\end{align*}
which shows as well that $q$ is even, a contradiction. Hence $\sqrt 2 + \sqrt 3 + \sqrt 5$ is irrational.
| No. What if $5+\sqrt{6}+\sqrt{10}+\sqrt{15} = \frac{17}{2}$? (I don't think it is, but you certainly haven't shown that it isn't.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the limit of $\mathop {\lim }\limits_{x \to \infty } \left( {{x^p}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right)$ Value of p such that $\mathop {\lim }\limits_{x \to \infty } \left( {{x^p}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right)$ is some finite | non-zero number.
My approach is as follow
$\mathop {\lim }\limits_{x \to \infty } \left( {{x^p}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right) \Rightarrow \mathop {\lim }\limits_{x \to \infty } \left( {{x^{\frac{{3p}}{3}}}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right)$
$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt[3]{{{x^{3p}}}}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right) \Rightarrow \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt[3]{{{x^{3p + 1}} + {x^{3p}}}} + \sqrt[3]{{{x^{3p + 1}} - {x^{3p}}}} - 2\sqrt[3]{{{x^{3p + 1}}}}} \right)$
How do we proceed
| $\lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x+1} + \sqrt[3]{x-1} -2\sqrt[3]{x})} = \lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x}\sqrt[3]{1+1/x} + \sqrt[3]{x}\sqrt[3]{1-1/x} -2\sqrt[3]{x})}$
$\lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x+1} + \sqrt[3]{x-1} -2\sqrt[3]{x})} = \lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x}(1+(1/3x)-(1/9x^2) + \cdots) + \sqrt[3]{x}(1-(1/3x)-(1/9x^2) + \cdots) -2\sqrt[3]{x})}$
$\lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x+1} + \sqrt[3]{x-1} -2\sqrt[3]{x})} = \lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x}(-(1/9x^2) + \cdots) + \sqrt[3]{x}(-(1/9x^2) + \cdots) )}$
$\lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x+1} + \sqrt[3]{x-1} -2\sqrt[3]{x})} = \lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x}(-2(1/9x^2) + \cdots)) }$
$\lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x+1} + \sqrt[3]{x-1} -2\sqrt[3]{x})} = \lim \limits_{x \to \infty } {x^{P+(1/3)-2}(((-2/9) + \cdots)) }$
The Higher Negative Powers will tend to $0$ in the limit.
The Only Power term we have is $x^{P+(1/3)-2}$ which must be $x^0$ hence ${P+(1/3)-2} = {0}$
${P+(1/3)-2} = {5/3}$
P less than that will give limit $0$
P more than that will give limit $\infty$
When $P=5/3$ : The Limit is $-2/9$
| {
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"url": "https://math.stackexchange.com/questions/4510808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Bound for a sum of poisson probabilities I'd like to understand a bound that has appeared in a paper I'm reading. It has the following expression:
$$\begin{align*}[...] &\leq \sum_{m \geq n }4^mP[Poisson(kt)\geq m]
\\&=e^{-kt}\sum_{m \geq n}4^m \sum_{j \geq m}\frac{(kt)^j}{j!}
\\&\leq e^{-kt}\sum_{m\geq n}e^{kt}\frac{(4kt)^m}{m!}\\
&{\color{red}\leq} e^{4kt}\frac{(4kt)^n}{\left(\frac{n}{2}\right)^{\frac{n}{2}}}
\\&{\color{blue}\leq} e^{-\frac{1}{8}n \log n}\end{align*}$$
if $n \geq (16kt)^8$.
Now, I do understand everything except the two colored inequalities.
For the blue one, I believe the idea is the following: Since $n\geq(16kt)^8\implies 4kt\leq \frac{n^{\frac{1}{8}}}{4}$ and so
$$e^{4kt}\frac{(4kt)^n}{\left(\frac{n}{2}\right)^{\frac{n}{2}}}\leq e^{\frac{n^{1/8}}{4}}\frac{\left(\frac{n^{1/8}}{4}\right)^n}{\left(\frac{n}{2}\right)^{n/2}} = e^{\frac{n^{1/8}}{4}} n^{n/8}n^{-n/2}4^{-n}2^{n/2} = e^{\frac{n^{1/8}}{4}}n^{-\frac{3}{8}n}2^{-\frac{3}{2}n}$$
$$=\exp\left(\frac{n^{1/8}}{4}-\frac{3}{8}n \log n -\frac{3}{2}n \log 2\right)$$
Now, for $n$ big enough $\frac{n^{\frac{1}{8}}}{4}-\frac{3}{2}n \log 2<0$ and so
$\exp\left(\frac{n^{1/8}}{4}-\frac{3}{8}n \log n -\frac{3}{2}n \log 2\right)\leq \exp\left(-\frac{3}{8}n \log n\right)\leq \exp \left(-\frac{1}{8}n \log n\right)$
Now, the red one I have absolutely on idea on how to proceed.
| For the red inequality, we have
\begin{equation}
\sum_{m=n}^\infty\frac{(4kt)^m}{m!}=\sum_{m=0}^\infty \frac{(4kt)^{n+m}}{(m+n)!}
\end{equation}
Since $n!\geq(\frac{n}{2})^{\frac{n}{2}}$ and $(m+n)!\geq m!n!$, it follows that the last sum is less than or equal to
\begin{equation}
\frac{(4kt)^n}{(\frac{n}{2})^{\frac{n}{2}}}\sum_{m=0}^\infty\frac{(4kt)^m}{m!}=\frac{(4kt)^n}{(\frac{n}{2})^{\frac{n}{2}}}e^{4kt}.
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.