Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Evaluate $a$ and $b$ s.t. the piecewise function is differentiable at two points Evaluate $a$ and $b$ s.t. the piecewise function is differentiable at $x=1$ and $x=2$ at the same time.
\begin{equation*}
f(x) = \left\{
\begin{array}{ll}
ax^2+bx+1 & \quad x ≥ 1 \\
-1 & \quad x < 1
\end{array}
\right.
\end{equation*}
So first, we gotta prove it's continuous at these points.
$$f(1)=a+b+1$$
$$\lim_{x \to 1^-} -1=-1$$
$$\lim_{x \to 1^+} ax^2+bx+1=a+b+1$$
$-1=a+b+1 ⟹ a+b=-2$
How do I continue this for $x=2$? And then for differentbility? Im kinda new to this and need a little help.
| In order to be differentiable at $x = 1$, then the following limit needs to exits:
$$\lim_{h\to 0}\frac{f(1+h) - f(1)}{h}.$$ For this to exist you need to consider both the left and right limits:
$$\lim_{h\to 0^{-}}\frac{f(1+h) - f(1)}{h} = \lim_{h\to 0^{-}}\frac{-1 - (a+b+1)}{h} = \lim_{h\to 0^{-}}\frac{-(a+b+2)}{h} = \lim_{h\to 0^{-}}\frac{0}{h} = 0$$ using the fact that $a + b = -2$. Now for the right:
\begin{align*}
\lim_{h\to 0^{+}}\frac{f(1+h) - f(1)}{h} &= \lim_{h\to 0^{+}}\frac{a(1+h)^2 + b(1+h) + 1 - (a+b+1)}{h}\\
&=\lim_{h\to 0^{+}}\frac{a + 2ah + ah^2+b+bh+1-a-b-1}{h}\\
&=\lim_{h\to 0^{+}}\frac{2ah + ah^2 + bh}{h}\\
&=\lim_{h\to 0^{+}}2a+ah+b\\
&=2a+b.
\end{align*}
Now, for the limit to exist, you must have $2a+b = 0$, which gives you two equations in two unknowns to solve for $a$ and $b$, and I trust you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Expected overshoot in a dice stopping problem You throw a 6-sided dice and keep track of the sum $S$ of the rolls so far. Stop throwing when $S\geq N$ for some integer threshold $N$. What is $\mathbb{E}(S)$? (I'm in particular interested in the case $N$ going to infinity)
Note: for example, if $N=20$ and you throw $6, 4, 5, 6$, giving $S=21$, so you stop.
Intuitively, since the average roll is $3.5$, we can guess that $S-X$ is "roughly uniform between $0$ and $3.5$", giving a guess of $1.75$.
Motivation: Part of the problem from here (Maths Problems, Q14).
| Here's a proof that $\ \mathbb{E}(S-N)\rightarrow\frac{5}{3}\ $. Let $\ S_r\ $ be the sum after $\ r\ $ rolls. Then the conditional expectation $e_j:= \mathbb{E}\big(S-N\,\big|\,S_r=j\,\big)\ $ is independent of $\ r\ $ as long as $\ r\le j\le 6r\ $ (it's undefined otherwise). Then
$$
\begin{align}
e_j&=\frac{1}{6}\sum_{i=1}^6e_{j+i}&&\text{ for } 0\le j\le N-1, \text{ and}\\
e_{N+i}&=i &&\text{ for } 0\le i\le 5.
\end{align}
$$
Solving the recurrence, it follows that
$$
e_j=c_0+\sum_{k=1}^5c_k\lambda_k^j\ ,
$$
where the recurrence has characteristic polynomial
\begin{align}
&x^6+x^5+x^4+x^3+x^2+x-6\\
&=(x-1)\big(x^5+2x^4+3x^3+4x^2+5x+6\big)\\
&=:(x-1)g(x)
\end{align}
and $\lambda_k$ are the roots of $g(x)$. Wolframalpha tells us that $g(x)$ has one real and two pairs of complex conjugate roots, all of which have absolute values strictly greater than $1$.
There are six initial conditions. We will combine them all to get $c_0$.
Let $\ f_j=e_j-c_0=\sum_\limits{k=1}^5c_k\lambda_k^j\ $. Then
\begin{align}
\sum_{i=0}^5(6-i)f_{j+i}&=\sum_{i=0}^5(6-i)\sum_\limits{k=1}^5c_k\lambda_k^{j+i}\\
&=\sum_\limits{k=1}^5c_k\lambda_k^j\sum_{i=0}^5\underbrace{(6-i)\lambda_k^i}_{=g(\lambda_k) = 0} = 0.\\
\end{align}
In particular, letting $j = N$,
\begin{align}
0&=\sum_{i=0}^5(6-i)f_{N+i}\\
&=\sum_{i=0}^5(6-i)\big(e_{N+i}-c_0\big)\\
&=\sum_{i=0}^5(6-i)i-21c_0\\
&=35-21c_0\ ,
\end{align}
so $\ c_0=\frac{5}{3}\ $.
The expectation we want is $\ e_0\ $, and we have
$$
e_{i+N}=i=\frac{5}{3}+\sum_{k=1}^5c_k\lambda_k^{i+N}
$$
for $\ 0\le i\le5\ $, or
$$
\pmatrix{-\frac{5}{3}\\-\frac{2}{3}\\\frac{1}{3}\\\frac{4}{3}\\\frac{7}{3}\\\frac{10}{3}}=\pmatrix{1&1&\dots&1\\
\lambda_1&\lambda_2&\dots&\lambda_5\\
\lambda_1^2&\lambda_2^2&\dots&\lambda_5^2\\
\vdots&\vdots&\ddots&\vdots\\
\lambda_1^5&\lambda_2^5&\dots&\lambda_5^5}\pmatrix{c_1\lambda_1^N\\c_2\lambda_2^N\\\vdots\\c_5\lambda_5^N}\ .
$$
Since the roots $\ \lambda_k\ $ are all distinct, the transposed Vandermonde matrix on the right of this equation is invertible. Therefore, if
$$
\pmatrix{d_1\\d_2\\\vdots\\d_5}=\pmatrix{1&1&\dots&1\\
\lambda_1&\lambda_2&\dots&\lambda_5\\
\lambda_1^2&\lambda_2^2&\dots&\lambda_5^2\\
\vdots&\vdots&\ddots&\vdots\\
\lambda_1^5&\lambda_2^5&\dots&\lambda_5^5}^{-1}\pmatrix{-\frac{5}{3}\\-\frac{2}{3}\\\frac{1}{3}\\\frac{4}{3}\\\frac{7}{3}\\\frac{10}{3}}\
$$
we have $\ c_k=d_k\lambda_k^{-N}\ $ and
$$
e_0=\frac{5}{3}+\sum_{k=1}^5d_k\lambda_k^{-N}\rightarrow\frac{5}{3}\ \ \text{as }\ N\rightarrow\infty\ .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve the integral through a contour with branch points? Consider the integral
$$ \int^{\infty}_{0} \frac{x^{\frac{1}{3}}}{1+x^2}dx $$
on the complex plane
$$ \oint_{C} \frac{z^{\frac{1}{3}}}{1+z^2}dz $$
To find the poles $ 1+z^2=0 \Rightarrow z^2=-1 \Rightarrow z= \pm \sqrt{-1} \Rightarrow z= \pm i $ , by the residue theorems $$ \oint_{C} \frac{z^{\frac{1}{3}}}{1+z^2}dz = 2 \pi i \,\ Res f(z) $$
I know that
$$ \oint_{C} \frac{z^{\frac{1}{3}}}{1+z^2}dz = \oint_{C_R} \frac{z^{\frac{1}{3}}}{1+z^2}dz +\oint_{C_2} \frac{z^{\frac{1}{3}}}{1+z^2}dz + \oint_{C_r} \frac{z^{\frac{1}{3}}}{1+z^2}dz + \oint_{C_1} \frac{z^{\frac{1}{3}}}{1+z^2}dz $$
the contour used is below:
The integral in $C_R \rightarrow 0$ because $R \rightarrow \infty$ and the integral in $C_r \rightarrow 0$ because $r \rightarrow 0$.
But how can I calculate the integrals over $C_1$ and $C_2$? How can I calculate residuals?
The teacher said the result is
$$ I(1-e^{\frac{2 \pi i}{3}}) = 2 \pi i \Big[ \frac{e^{\frac{\pi i}{6}}}{2i}\Big] (1-e^{\frac{\pi i}{3}}) \quad \quad \Rightarrow \quad \quad I= \frac{\pi}{2 \sin(\frac{\pi}{3})} = \frac{\pi}{\sqrt{3}}$$
| You may start by removing the branch point through a suitable substitution:
$$I=\int_{0}^{+\infty}\frac{x^{1/3}}{1+x^2}\,dx\stackrel{x\mapsto z^3}{=}\int_{0}^{+\infty}\frac{3x^3}{1+x^6}\,dx=\int_{0}^{1}\frac{3x^3}{1+x^6}\,dx+\int_{1}^{+\infty}\frac{3x^3}{1+x^6}\,dx$$
then realize you do not need complex analysis at all:
$$ I = 3\int_{0}^{1}\frac{x+x^3}{1+x^6}\,dx=3\int_{0}^{1}\frac{x}{1-x^2+x^4}\,dx\stackrel{x\mapsto\sqrt{u}}{=}\frac{3}{2}\int_{0}^{1}\frac{du}{1-u+u^2}=\left.\sqrt{3}\arctan\left(\frac{2u-1}{\sqrt{3}}\right)\right|_{0}^{1}$$
immediately leads to $I=\frac{\pi}{\sqrt{3}}$.
Also, all integrals of the form $\int_{0}^{+\infty}\frac{x^\alpha}{1+x^\beta}\,dx $ can be computed through the substitution $\frac{1}{1+x^\beta}\to z$, Euler's Beta function and the reflection formula for the $\Gamma$ function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Angle between Tangent of the ellipse Let $\frac{{x\cos \theta }}{2} + y\sin \theta = 1$,$\theta \in \left( {\frac{\pi }{{18}},\frac{\pi }{{15}}} \right)$ intersect the ellipse ${x^2} + 2{y^2} = 6$ at P and Q, then angle between tangents at P and Q of the ellipse is $\frac{\pi }{K}$ where K is equal to ______
My approach is as follow
$y = \frac{1}{{\sin \theta }}\left( {1 - \frac{{x\cos \theta }}{2}} \right)$
${x^2} + \frac{2}{{{{\sin }^2}\theta }}{\left( {1 - \frac{{x\cos \theta }}{2}} \right)^2} = 6$
${\sin ^2}\theta {x^2} + 2{\left( {1 - \frac{{x\cos \theta }}{2}} \right)^2} = 6{\sin ^2}\theta \Rightarrow 2{\left( {1 - \frac{{x\cos \theta }}{2}} \right)^2} = {\sin ^2}\theta \left( {6 - {x^2}} \right)$
$ \Rightarrow 2\left( {1 + \frac{{{x^2}{{\cos }^2}\theta }}{4} - x\cos \theta } \right) = \left( {1 - {{\cos }^2}\theta } \right)\left( {6 - {x^2}} \right) \Rightarrow \left( {2 + \frac{{{x^2}{{\cos }^2}\theta }}{2} - 2x\cos \theta } \right) = 6\left( {1 - {{\cos }^2}\theta } \right) - {x^2}\left( {1 - {{\cos }^2}\theta } \right)$
$ \Rightarrow {x^2}\left( {\frac{{{{\cos }^2}\theta }}{2} + \left( {1 - {{\cos }^2}\theta } \right)} \right) - 2x\cos \theta + 2 - 6\left( {1 - {{\cos }^2}\theta } \right) = 0 \Rightarrow {x^2}\left( {\frac{{{{\cos }^2}\theta + 2 - 2{{\cos }^2}\theta }}{2}} \right) - 2x\cos \theta + 2 - 6\left( {1 - {{\cos }^2}\theta } \right) = 0$
$\begin{array}{l}
\Rightarrow {x^2}\left( {\frac{{{{\cos }^2}\theta }}{2} + \left( {1 - {{\cos }^2}\theta } \right)} \right) - 2x\cos \theta + 2 - 6\left( {1 - {{\cos }^2}\theta } \right) = 0 \Rightarrow {x^2}\left( {\frac{{{{\cos }^2}\theta + 2 - 2{{\cos }^2}\theta }}{2}} \right) - 2x\cos \theta + 2 - 6\left( {1 - {{\cos }^2}\theta } \right) = 0\\
\Rightarrow {x^2}\left( {\frac{{2 - {{\cos }^2}\theta }}{2}} \right) - 2x\cos \theta + 6{\cos ^2}\theta - 4 = 0 \Rightarrow {x^2} - \frac{{2x\cos \theta }}{{\left( {\frac{{2 - {{\cos }^2}\theta }}{2}} \right)}} + \frac{{6{{\cos }^2}\theta - 4}}{{\left( {\frac{{2 - {{\cos }^2}\theta }}{2}} \right)}} = 0\end{array}$
${x_1} + {x_2} = \frac{{2\cos \theta }}{{\left( {\frac{{2 - {{\cos }^2}\theta }}{2}} \right)}};{x_1}{x_2} = \frac{{6{{\cos }^2}\theta - 4}}{{\left( {\frac{{2 - {{\cos }^2}\theta }}{2}} \right)}} \Rightarrow {\left( {{x_1} - {x_2}} \right)^2} = {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2}$
${\left( {{x_1} - {x_2}} \right)^2} = \frac{{4{{\cos }^2}\theta }}{{{{\left( {\frac{{2 - {{\cos }^2}\theta }}{2}} \right)}^2}}} - \frac{{\left( {24{{\cos }^2}\theta - 16} \right)}}{{\left( {\frac{{2 - {{\cos }^2}\theta }}{2}} \right)}}$
How will I approach from here as it is getting too complicated
| >Alternate approach<
The equation of chord of contact of tangents drawn from an external point $(h,k)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\frac{hx}{a^2}+\frac{ky}{b^2}=1$.
Let the external point where the tangents intersect be $P(h,k).$
$\Rightarrow$ Chord of contact of tangents from $P$ to the given ellipse: $$\frac{hx}{6}+\frac{ky}{3}=1$$
This line must coincide with the given line $\frac{x\cos\theta}2+y\sin\theta=1.$
$$\Rightarrow \frac{\frac h6}{\frac{\cos\theta}2}=\frac{\frac k3}{\sin\theta}=1$$
$$\Rightarrow P(h,k)\equiv(3\cos\theta,3\sin\theta)$$
Director circle of given ellipse: $$\Rightarrow x^2+y^2=6+3=9$$
Hence, $P$ represents any point on the director circle of the given ellipse.
$$\Rightarrow K=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the formal way find $a$ such that $\lim_{x\to4} \frac{ax-\sqrt{x}+6}{x-4} = \frac{3}{4}$? So I got this problem
Determine the value of $a$ that satisfies the following limit
$$\lim_{x\to4} \frac{ax-\sqrt{x}+6}{x-4} = \frac{3}{4}$$
If we substitute $x = 4$, the value of the limit will be $\frac{4a + 4}{0}$ which is undefined. So, the numerator must have the same root as the denominator. With $x - 4 = (\sqrt{x} - 2)(\sqrt{x} + 2)$, and the one which make the limit undefined is $(\sqrt{x} - 2)$. From here, I conclude that the numerator must have ($\sqrt{x} - 2$) as a root. And from here, I can do polynomial division and got that $a = -1$.
The other way is by argue that "if the limit have a value, then make the result indeterminate form: $\frac{0}{0}$. Then, we'll get $4a + 4 = 0$ and $a = -1$.
The problem is, I find my method is not good enough for essay problems.
| If $\lim_{x\to a}f(x)$ exists (in $\Bbb R$) but is non-zero, and $\lim_{x\to a}g(x)=0$, then
$$
\lim_{x\to a}\frac{f(x)}{g(x)}\quad\text{does not exist.}
$$
Therefore, if $\lim_{x\to 4}\frac{ax-\sqrt{x}+6}{x-4}$ exists, then it must be the case that $\lim_{x\to 4}ax-\sqrt{x}+6=0$. Since the function $x\mapsto ax-\sqrt{x}+6$ is elementary, it is continuous, and so we can evaluate $\lim_{x\to4}ax-\sqrt{x}+6$ by plugging in $x=4$. We find that $\lim_{x\to4}ax-\sqrt{x}+6=4a-\sqrt{4}+6=4a+4$. Hence, $a=-1$.
However, if $a=-1$, then the limit actually equals $-5/4$. To prove this, note that for $x\ge0,x\neq4$, we have
$$
\frac{-x-\sqrt{x}+6}{x-4}=-\frac{x+\sqrt{x}-6}{x-4}=-\frac{(\sqrt{x}+3)(\sqrt{x}-2)}{(\sqrt{x}+2)(\sqrt{x}-2)}=-\frac{\sqrt{x}+3}{\sqrt{x}+2} \, .
$$
Therefore, there is no value of $a$ for which the limit equals $3/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Best number to choose in a betting game Assume $3$ players are rolling a $n$-sided dice. And as the $4$-th player, you can choose an integer $1 \leq x \leq n$. Now after the rolling, the players with the highest/lowest numbers, as well as the two middle ones, team up. The team with the greater sum of the rolls pays the the team with the lower sum the difference of their sums. For example, suppose the rollings are $48, 50, 2, 10$, then the team $48, 10$ need to pay $50, 2$ an amount of $6$.
Which $x$ do you pick to maximize your expected value, and what expected value is it?
My attempt: For simplicity first consider uniform distribution on $[0,1]$. Suppose you choose $x$, there are $4$ cases: $x$ being lowest, second lowest, second greatest, and the greatest.
*
*Lowest. Probability $(1-x)^3$, $$EV = \left( \frac{2(1-x)}{4} + x + \frac{(1-x)}{4}+x \right) - \left(\frac{3(1-x)}{4}+ x + x\right) = 0$$
*Second to lowest. Probability $x(1-x)^2$, $$EV = (\frac{x}{2}+x+\frac{2(1-x)}{3} ) - (x+x+\frac{(1-x)}{3})=\frac{1}{3}-\frac{5x}{6}$$
*Second to highest. Probability $x^2(1-x)$, $$EV= (x + \frac{1-x}{2}+\frac{x}{3})-(x+\frac{2x}{3}) = \frac{1}{2} - \frac{5x}{6}$$
*$EV = 0$ by symmetry.
Thus it suffices to find $x$ that maximize
$$x(1-x)^2(\frac{1}{3}-\frac{5x}{6})+x^2(1-x)(\frac{1}{2} - \frac{5x}{6}) = \frac{x(x-1)(2x-1)}{3}$$
And thus $x \approx 0.211$ solves the maximum, with EV being $0.016$
So choosing $x$ to be approximately the rounding of $0.211n$ would give the answer and expected profit of $0.016n$. But I'm not sure of an exact solution.
The case where $n=50$ was a Jane Street Capital interview question. I am interested in the general case.
| Assume that $x$ is the value that you choose and $L$, $M$ and $U$ the lowest, middle and highest values choosen by other players. I am also assuming that when $x$ is equal to the lowest or the highest, you will team with the middle such that you are winner. Your gain is:
\begin{align}
G(x) &= \left(U + M - L\right)\mathbf 1_{U < x} + \left(U + L - M\right)\mathbf 1_{L \le x \le U} + \left(L + M - U\right)\mathbf 1_{x < L} - x\\
&= U\left(2\times\mathbf 1_{L \le x} - 1\right) + M\left(1 - 2\times \mathbf 1_{L \le x \le U}\right) + L\left(2\times\mathbf 1_{x \le U} - 1\right) - x
\end{align}
So
\begin{align}
G(x+1) - G(x) &= 2U\left(\mathbf 1_{L \le x+1} - \mathbf 1_{L \le x}\right) + 2M\left(\mathbf 1_{L \le x \le U} - \mathbf 1_{L \le x+1 \le U}\right) + 2L\left(\mathbf 1_{x+1 \le U} - \mathbf 1_{x \le U}\right) - 1\\
&= 2\left(U\mathbf 1_{L=x+1} - M\mathbf 1_{L=x+1} + M\mathbf 1_{U=x} - L\mathbf 1_{U=x}\right) - 1\\
&= 2\left((U-M)\mathbf 1_{L=x+1} + (M-L)\mathbf 1_{U=x}\right) - 1
\end{align}
Since $(L, M, U)$ and $(n-U+1, n-M+1, n-L+1)$ have the same probability law then, \begin{align}
E\left[(U-M)\mathbf 1_{L=x+1}\right] &= E\left[((n - L +1)-(n- M+1))\mathbf 1_{n-U+1=x+1}\right]\\
&= E\left[(M-L)\mathbf 1_{U=n-x}\right]
\end{align}
Let \begin{align}
F(x) &= E\left[(M-L)\mathbf 1_{U=x}\right]\\
&= \frac1{n^3}\sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^n (M(i,j,k) - L(i,j,k))\mathbf 1_{U(i,j,k)=x}\\
&= \frac1{n^3} \left(6\sum_{1\le i \le j < x} (j-i) + 3\sum_{1\le i < x}(x-i)\right)\\
&= \frac3{n^3}\left(2\sum_{1\le i \le x-1}\sum_{j=0}^{x-1} j + \sum_{1\le i \le x-1} i\right)\\
&= \frac{3}{n^3}(2x-1)\sum_{i=1}^{x-1}i\\
&= \frac{3}{2n^3}x(x-1)(2x-1).
\end{align}
\begin{align}
E\left[G(x+1)-G(x)\right] &= 2(F(n-x) + F(x)) - 1\\
&= \frac{3}{n^3}(n-1)\left(6x^2 - 6nx + n(2n-1)\right) - 1
\end{align}
You can study the sign of the function $E\left[G(x+1)-G(x)\right]$ to find the value that maximise $E[G(x)]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Range of values of x for which $a\gt b \rightarrow \frac{1}{x+a} \lt \frac{1}{x+b}$ I am working through a pure maths book as a hobby. I have come to inequalities and am struggling with this question.
Given that $a$ and $b$ are positive, state the range of values of x for which the following statement is true:$a\gt b \rightarrow \frac{1}{x+a} \lt \frac{1}{x+b}$
I have said, $\frac{1}{x+a} \lt \frac{1}{x+b} \rightarrow 1<\frac{x+a}{x+b} \rightarrow \frac{x+a}{x+b} \gt 1 \rightarrow x+a\gt x+b$
So the original statement can be expressed as
$a\gt b \rightarrow x+a\gt x+b$, which is true for all x.
But the book says the answer is: $x\lt -a$ or $x>-b$
| Multiplying the inequality by $x+a$ and conserving the < sign is only right if $x+a>0$ if $x+a<0$ you get $1>\frac{x+a}{x+b}$
multiplying inequality you always have to take in account the sign of the factor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4251892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is $\int_{-\infty}^0 \max\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx+\int_0^\infty \min\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx=\pi?$ Playing around with floor functions, I thought about using the maximum and minumum of functions to “merge” 2 asymptotic graphs which intersected each other infinitely. I then thought of the classical trigonometric Dirichlet Integral. Here is a graph of our goal integral. Please note the sign function:
$$\int_\Bbb R f(x)dx=\int_{-\infty}^0 \max\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx+ \int_0^\infty \min\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx = \int_0^\infty \min\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)+ \max\left(-\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx =\int_0^\infty \frac{\cos(x) \text{sgn}\left(\frac{\sin(x)}x - \frac{\cos(x)}x\right)}{2 x} +\frac{\sin(x) \text{sgn}\left(\frac {\cos(x)}x - \frac{\sin(x)}x\right)}{2 x} + \frac{\sin(x)}{2 x} + \frac{\cos(x)}{2 x}-\frac{\cos(x) \text{sgn}\left(\frac{\sin(x)}x + \frac{\cos(x)}x\right)}{2 x}+ \frac{\sin(x) \text{sgn}\left(-\frac {\cos(x)}x - \frac{\sin(x)}x\right)}{2 x} + \frac{\sin(x)}{2 x} - \frac{\cos(x)}{2 x} dx =\frac\pi 2+ \int_0^\infty \frac{\cos(x) \text{sgn}\left(\frac{\sin(x)}x - \frac{\cos(x)}x\right)}{2 x} +\frac{\sin(x) \text{sgn}\left(\frac {\cos(x)}x - \frac{\sin(x)}x\right)}{2 x} -\frac{\cos(x) \text{sgn}\left(\frac{\sin(x)}x + \frac{\cos(x)}x\right)}{2 x}+ \frac{\sin(x) \text{sgn}\left(-\frac {\cos(x)}x - \frac{\sin(x)}x\right)}{2 x} dx=\frac\pi 2+\frac12 \int_0^\infty\frac{|\cos(x)+\sin(x)|}{x}-\frac{|\cos(x)-\sin(x)|}{x}dx=\boxed{ \frac\pi 2+\frac1{\sqrt 2} \int_0^\infty\frac{\left|\sin\left(x+\frac\pi 4\right)\right|}{x}-\frac{\left|\sin\left(x-\frac\pi 4\right)\right|}{x}dx }=\frac{\pi}{2}+ \lim_{x\to \infty}\left(\frac12\text{Ci}(x) \left[\text {sgn}\,\sin\left(x + \frac \pi4\right) -\text{sgn}\,\cos \left(x + \frac{\pi}4\right)\right] + \frac12 \text{Si}(x) \left[\text {sgn}\,\sin\left(x + \frac{\pi}4\right) + \text{sgn}\,\cos \left(x + \frac{\pi}4\right)\right]\right)\mathop=^{???}\pi$$
Our integral uses trigonometric integrals. I need to verify this solution. Here is a graph of that final integrand. Here is a graph of the actual problem; this plot is also a visual representation of our constant:
We used the Dirichlet Integral in one step. Here is a verification of the maximum and minimum function terms. Can you please verify this solution or give another way to evaluate it? Desmos and Wolfram Alpha cannot find a numerical approximation for the problem. Please correct me and give feedback!
Similar results which should have no typos:
$$\int_\Bbb R \min \left(\frac{\sin(x)}{x},e^{-|x|}\right)dx=2$$
$$\int_\Bbb R \max\left(\frac{\sin(x)}{x},e^{-|x|}\right)dx=\pi$$
Final result using Silver Ratio $\delta_S$:
$$\int_\Bbb R f(x)dx=\int_{-\infty}^0 \max\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx+ \int_0^\infty \min\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx=\frac\pi 2+\coth^{-1}\left(\sqrt 2\right)=\frac\pi 2+\ln\left(\sqrt 2+1\right)=\frac\pi 2 +\ln(\delta_S)= 2.452169913814439644463931016619543751126745027949188…$$
| I will concentrate on this part of your proof, the calculation of this integral (the other steps of your proof are correct):
$$\int_{0}^{\infty} \frac{\Big|\sin\left(x+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(x-\frac{\pi}{4}\right)\Big|}{x} dx$$
We will make use of the Lobachevsky integral formula:
Let $f(x)$ a $\pi$-periodic function (continous or integrable over its period). with
$\displaystyle f(x+\pi) = f(x)$ and $f(\pi-x) = f(x)$, for $0\leq x <\infty$. Then
$\displaystyle \int_{0}^{\infty} \frac{\sin x}{x} f(x) dx = \int_{0}^{\frac{\pi}{2}} f(x) dx. $
Now, for your integral:
If you make the change of variable $\displaystyle x=\frac{w}{2}$
$\displaystyle I=\int_{0}^{\infty} \frac{\Big|\sin\left(x+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(x-\frac{\pi}{4}\right)\Big|}{x} dx = \int_{0}^{\infty} \frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{w} dw $
Multiplying and dividing by $\sin w$:
$$I= \int_{0}^{\infty} \frac{\sin w}{w} \frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w} dw$$
The function
$$f(w) = \frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w}$$
is $\pi$-periodic
By the Lobachevsky integral formula:
$$\displaystyle I=\int_{0}^{\infty} \frac{\sin w}{w} \frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w} dw = \int_{0}^{\frac{\pi}{2}}\frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w}dw $$
Note that for $w\in\left(0,\frac{\pi}{2}\right)$
$$\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big|= \sin\left(\frac{w}{2}+\frac{\pi}{4}\right)$$
$$\Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|= -\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)$$
Hence
$$\displaystyle\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big| = \sqrt{2}\sin\left(\frac{w}{2}\right)\displaystyle $$
Therefore
$$ I = \int_{0}^{\frac{\pi}{2}}\frac{\Big|\sin\left(\frac{w}{2}+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(\frac{w}{2}-\frac{\pi}{4}\right)\Big|}{\sin w}dw = \sqrt{2}\int_{0}^{\frac{\pi}{2}}\frac{\sin\left(\frac{w}{2}\right)}{\sin w}dw$$
Using $\displaystyle \frac{\sin\left(\frac{w}{2}\right)}{\sin w} = \frac{1}{2}\sec\left(\frac{w}{2}\right)$
we have
$$I = \sqrt{2}\int_{0}^{\frac{\pi}{2}}\frac{\sin\left(\frac{w}{2}\right)}{\sin w}dw = \frac{\sqrt{2}}{2}\int_{0}^{\frac{\pi}{2}} \sec\left(\frac{w}{2} \right)dw = \sqrt{2} \operatorname{arctanh}\left(\frac{1}{\sqrt{2}}\right)$$
We can conclude
$$\boxed{\int_{0}^{\infty} \frac{\Big|\sin\left(x+\frac{\pi}{4}\right)\Big| - \Big|\sin\left(x-\frac{\pi}{4}\right)\Big|}{x} dx = \sqrt{2} \operatorname{arctanh}\left(\frac{1}{\sqrt{2}}\right)}$$
Therefore
$$\boxed{\int_{-\infty}^0 \max\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx+ \int_0^\infty \min\left(\frac{\cos(x)}{x},\frac{\sin(x)}{x}\right)dx = \frac{\pi}{2} +\operatorname{arctanh}\left(\frac{1}{\sqrt{2}}\right)} $$
Some numerical approximations from Wolfram:
The sum is 2.44206
while
$$\frac{\pi}{2} + \operatorname{arctanh}\left(\frac{1}{\sqrt{2}}\right)\approx 2.45216..$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4256660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Find $\int{\sqrt[5]{\frac{x}{x+1}}\frac{1}{x^3}}\>dx$ How to integrate
$$\int{\sqrt[5]{\frac{x}{x+1}}\>\frac{1}{x^3}}\>dx$$
My work:
$$ \sqrt[5]{\frac{x}{x+1}}=t$$
$$t^5=\frac{x}{x+1}$$
$$5t^4dt=\frac{1}{(x+1)^2}dx$$
But here, I don't know how to get $\frac{1}{x^3}$.
| It is more desirable to utilize the substitution $t=\sqrt[5]\frac{1+x}x$, instead. As a result, $\frac1x=t^5-1$,
$\frac{dx}{x^3}= -\frac{5t^4}{x}dt= 5(t^4-t^9)dt
$
and
$$\int{\sqrt[5]{\frac{x}{x+1}}\frac{1}{x^3}}dx
=5\int(t^3-t^8)dt= \frac54t^4 - \frac59t^9+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $n=2^{10}\times 3^5$. Find number of divisors of $n^2$ which are less than $n$ but do not divide $n$
If $N=2^{10}\times 3^5$. Find number of divisors of $n^2$ which are less than $n$ but do not divide $n$
My solution:
$$n^2=2^{20}\times 3^{10}$$
Factors of $n^2=\left(20+1\right)(10+1)=21\times 11=231$
Factors which are greater than or equal to $n$ can be found as we have choices for power of 2 $\in [10,11,\dots 20]$ and that of 3 $\in[5,6,\dots10]$
We get the factors of $n^2$ which are greater than or equal to $n$ as $11\times 6=66$
Factors of $n$ which are less than $n=(10+1)(5+1)-1=65$ ($-1$ as $n$ has been counted in the above condition)
$\therefore$ factors which satisfy the condition are $231-66-65=100$
But the answer is $50$. What is the correct way to solve it and what is the error in my solution?
| $n = 248832.$
You want all numbers of the form $2^a \times 3^b$ where
*
*$a \geq 11$ or $b \geq 6$.
*$a \leq 20$ and $b \leq 10$.
*$2^a \times 3^b < n.$
Consider the following two charts:
\begin{array}{| r | r | r |}
\hline
2^a & b & \text{number of solutions} \\
\hline
2^{11} & b \leq 4 & 5 \\
\hline
2^{12} & b \leq 3 & 4 \\
\hline
2^{13} & b \leq 3 & 4 \\
\hline
2^{14} & b \leq 2 & 3 \\
\hline
2^{15} & b \leq 1 & 2 \\
\hline
2^{16} & b \leq 1 & 2 \\
\hline
2^{17} & b = 0 & 1 \\
\hline
\end{array}
\begin{array}{| r | r | r |}
\hline
3^b & a & \text{number of solutions} \\
\hline
3^{6} & a \leq 8 & 9 \\
\hline
3^{7} & a \leq 6 & 7 \\
\hline
3^{8} & a \leq 5 & 6 \\
\hline
3^{9} & a \leq 3 & 4 \\
\hline
3^{10} & a \leq 2 & 3 \\
\hline
\end{array}
$50$ solutions in the two charts combined.
Edit
Thanks to Calvin Lin for pointing out that $3^{11}$ was not a valid solution, since it does not divide $n^2.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help with proving the following inequality I have to prove that $0\leq \sin(x)-(x-x^3/3!+x^5/5!-x^7/7!)\leq x^9/9!$, $x>0$
With the help of Taylor's theorem I got the following inequality,
$-x^9/9!\leq \sin(x)-(x-x^3/3!+x^5/5!-x^7/7!)\leq x^9/9!$. How should I proceed further? Thanks in advance!
EDIT: Applying Taylor's theorem to the function $f(x)=\ sinx$ in $[0,x]$ with remainder after nine terms, I got,
$sinx=x-x^3/3!+x^5/5!-x^7/7!+\sin(\theta x)x^9/9!$
Now,$-1 \leq \sin(\theta x)x^9/9! \leq 1$
Putting the value of $\sin(\theta x)x^9/9!$ I got the previous inequality.
| The lower bound is obtained by applying Taylor's theorem with the remainder after seven terms. The seventh derivative of $\sin(x)$ is $-\cos(x)$. Therefore, if $ x > 0$ then we have with some $\theta \in (0, 1)$,
$$
\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!}(-\cos(\theta x))
\ge x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}
$$
since $\cos(\theta x)) \le 1$.
For the upper bound it should be
$$
\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!}(\cos(\theta' x))
\le x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}+ \frac{x^9}{9!}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4261457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Various ways to calculate $\int \sin(x) \cos(x) \, \mathrm{d}x$ Consider the integral
$$\mathcal{I} := \int \sin(x) \cos(x) \, \mathrm{d} x$$
$
\newcommand{\II}{\mathcal{I}}
\newcommand{\d}{\mathrm{d}}
$
This is one of my favorite basic integrals to think about as an instructor, because on the face of it, there are a lot of different ways to solve it, many are accessible to Calculus I students, and can give some insights into the nature of integration and to some trigonometry identities. For instance:
*
*Substitution with $u = \sin(x)$ gives $$\II = \frac{\sin^2(x)}{2} + C$$
*
*Substitution with $u = \cos(x)$ gives $$ \II = - \frac{\cos^2(x)}{2} + C$$
*
*(Noted in comments by Koro) Make the substitution
$$
u = \sec(x) \implies \d u = \sec(x) \tan(x) \, \d x= \frac{\sin(x)}{\cos^2(x)} \, \d x$$
Then
$$\begin{align*}
\II &= \int \sin(x) \cos(x) \frac{\cos^2(x)}{\sin(x)} \, \d u\\
&= \int u^{-3} \, \d u\\
&= - \frac{1}{2\sec^2(x)} + C\\
&= - \frac{\cos^2(x)}{2} + C
\end{align*}$$
A similarly motivated substitution:
$$
u = \csc(x) \implies \d u = -\cot(x) \csc(x) \, \d x = - \frac{\cos(x)}{\sin^2(x)} \, \d x
$$
yields
$$\begin{align*}
\II &= -\int \sin(x) \cos(x) \frac{\sin^2(x)}{\cos(x)} \, \d u\\
&= -\int u^{-3} \, \d u\\
&= \frac{1}{2\csc^2(x)} + C\\
&= \frac{\sin^2(x)}{2} + C
\end{align*}$$
*
*Using $\sin(2x) = 2 \sin(x) \cos(x)$ readily leads to $$\II = -\frac{\cos(2x)}{4} + C$$
*
*Integration by parts differentiating $\sin(x)$ yields $$\II = \sin^2(x) - \II$$ which will yield a previous solution on solving for $\II$.
*
*Integration by parts differentiating $\cos(x)$ yields $$ \II = -\cos^2(x)- \II$$ which, similarly, yields a previous solution once we solve for $\II$.
*
*Using the Weierstrass substitution $t = \tan(x/2)$ gives
$$
\II = \int \frac{2t}{1+t^2} \frac{1-t^2}{1+t^2} \frac{2}{1+t^2} \, \mathrm{d}t = \frac{2t^2}{(1+t^2)^2} + C=\frac{2 \tan^2(x/2)}{(1 + \tan^2(x/2))^2} + C
$$
*
*We can use the complex sine and cosine:
$$
\sin(x) = \frac{e^{ix} - e^{-ix}}{2i} \qquad
\cos(x) = \frac{e^{ix} + e^{-ix}}{2}
$$
Then
$$
\II = \int \frac{e^{2ix} - e^{-2ix}}{4i} \, \mathrm{d} x = - \frac 1 8 \left( e^{2ix} + e^{-2ix} \right) + C = -\frac 1 4 \cos(2x) + C
$$
(Warning for Novices: The $C$ constant in each expression may not be the same as in others. These answers are equivalent ones for the integral, but the solutions without the $+C$ terms are not equal expressions. These hint at certain trigonometry identities, which is why I find them interesting.)
This has given us a set of solutions to $\II$ via a few basic methods, and a few less-basic but accessible methods.
My question is, what other solutions can you come up with for $\II$?
I'm particularly interested in answers which:
*
*are obviously not functionally equivalent to those already given
*give answers other than those already expressed (perhaps a hint at other identities or concepts of note?)
*use methods that you don't see in a typical calculus class, or methods that are rarely used
*use unusual but slick and effective techniques
or any combination thereof! I have no real motivation for this but my own curiosity, but I'm curious to see what you guys can think of.
| Let's make a series out of integration by parts. Let
$$I = \int \frac{1}{2}\sin 2x\:dx = \frac{1}{2}x\sin2x-\int x\cos 2x\:dx$$
Then keep going
$$I = \frac{1}{2}x\sin 2x - \frac{1}{2}x^2\cos 2x - \int x^2\sin 2x \;dx$$
$$= \frac{1}{2}x\sin 2x - \frac{1}{2}x^2\cos 2x - \frac{1}{3}x^3\sin 2x + \int\frac{2}{3}x^3\cos 2x\:dx$$
$$= \frac{1}{2}x\sin 2x - \frac{1}{2}x^2\cos 2x - \frac{1}{3}x^3\sin 2x + \frac{1}{6}x^4\cos2x + \int \frac{1}{3}x^4\sin 2x\:dx$$
or in other words the sum of the following two series
$$I = \sin 2x\left(\frac{1}{2}x-\frac{1}{3}x^3+\frac{1}{15}x^5-\cdots\right)-\cos 2x\left(\frac{1}{2}x^2-\frac{1}{6}x^4+\frac{1}{45}x^6-\cdots\right)$$
| {
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"url": "https://math.stackexchange.com/questions/4262190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
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Prove $ \sum_{n=1}^{2q-1}\frac{n}{q}\sin\left(\frac{\pi np}{q}\right)=-\cot\left(\frac{\pi p}{q}\right)-\csc\left(\frac{\pi p}{q}\right) $ For $p$ and $q$ are positive integers, $p < q$.
How to prove this identity?
$$
\sum_{n=1}^{2q-1}\frac{n}{q}\sin\left(\frac{n\pi p}{q}\right)=-\cot\left(\frac{\pi p}{q}\right)-\csc\left(\frac{\pi p}{q}\right)=-\cot\left(\frac{\pi p}{2q}\right)
$$
This identity is comprised by two parts.
$$
\sum_{n=1}^{q-1}\frac{2n}{q}\sin\left(\frac{2n\pi p}{q}\right)=-\cot\left(\frac{\pi p}{q}\right)\\
\sum_{n=1}^{q}\frac{(2n-1)}{q}\sin\left(\frac{(2n-1)\pi p}{q}\right)=-\csc\left(\frac{\pi p}{q}\right)
$$
I found these identities during the calculations when proving the other identity in my previous question by comparison. I am curious how to prove the identities in general ways.
I also found some other identities all by comparison like:
For $q$ is odd
$$
\sum_{n=1}^{(q-1)/2}\frac{2n}{q}\sin\left ( \frac{2n\pi p}{q} \right )=\frac{(-1)^{p-1}}{2}\csc\left( \frac{\pi p}{q} \right ) \\
\sum_{n=1}^{(q-1)/2}\frac{(2n-1)}{q}\sin\left ( \frac{(2n-1)\pi p}{q} \right )=\frac{(-1)^{p-1}}{2}\cot\left( \frac{\pi p}{q} \right ) \\
\sum_{n=1}^{(q-1)/2}2\sin\left ( \frac{2n\pi p}{q} \right )=\cot\left( \frac{\pi p}{q} \right )+(-1)^{p-1}\csc\left( \frac{\pi p}{q} \right ) \\
\sum_{n=1}^{(q-1)/2}2\sin\left ( \frac{(2n-1)\pi p}{q} \right )=\csc\left( \frac{\pi p}{q} \right )+(-1)^{p-1}\cot\left( \frac{\pi p}{q} \right )
$$
For $q$ is even
$$
\sum_{n=1}^{q/2}\frac{2n}{q}\sin\left ( \frac{2n\pi p}{q} \right )=\frac{(-1)^{p-1}}{2}\cot\left( \frac{\pi p}{q} \right ) \\
\sum_{n=1}^{q/2}\frac{(2n-1)}{q}\sin\left ( \frac{(2n-1)\pi p}{q} \right )=\frac{(-1)^{p-1}}{2}\csc\left( \frac{\pi p}{q} \right ) \\
\sum_{n=1}^{q/2}2\sin\left ( \frac{2n\pi p}{q} \right )=\cot\left( \frac{\pi p}{q} \right )+(-1)^{p-1}\cot\left( \frac{\pi p}{q} \right ) \\
\sum_{n=1}^{q/2}2\sin\left ( \frac{(2n-1)\pi p}{q} \right )=\csc\left( \frac{\pi p}{q} \right )+(-1)^{p-1}\csc\left( \frac{\pi p}{q} \right )
$$
How to prove them in general ways?
| Computationally, I could transform the sum $\sum _{n=1}^{2 q-1} \frac{n \sin \left(\frac{\pi n p}{q}\right)}{q}$ to the following term:
$$
\frac{1}{4} \csc \left(\frac{p \pi }{2 q}\right) \left(-4 \cos \left(\frac{1}{2} p \pi \left(4-\frac{1}{q}\right)\right)+\frac{\csc \left(\frac{p \pi }{2 q}\right) \sin (2 p \pi )}{q}\right)
$$
Here conversions to exponential forms (which are expanded back to trigonometric forms) internally occurred. One can retrace it using Wolfram:
Trace[FullSimplify[Sum[(n/q)*Sin[Pi*n*p/q], {n, 1, 2 q - 1}]],
TraceInternal -> True]
This looks like a (possibly hard) pathway to obtain the identity given in the question. Interestingly, when setting $p=1$, we obtain a much simpler term:
$$
\sum _{n=1}^{2 q-1} \frac{n \sin \left(\frac{\pi n}{q}\right)}{q}=-\cot \left(\frac{\pi }{2 q}\right)
$$
Running $q$ from $0$ to $2\pi$, the plot of the sum $\sum _{n=1}^{2 q-1} \frac{n \sin \left(\frac{\pi n}{q}\right)}{q}$ looks as follows:
The Partial Fractions Expansion of Cotangent has been elaborated by this MSE Post or by this Proof Wiki Page.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Separate single integral into multiple integrals I can separate an integral like this, (3) parts.
$\int_{1}^{4}x^{2}dx = \int_{1}^{2}x^{2}dx + \int_{2}^{3}x^{2}dx + \int_{3}^{4}x^{2}dx$
I found out this way to separate double integral into three parts.
$\int_{1}^{4}\int_{1}^{4}x^{2}y^{2}dxdy = \int_{1}^{4}\int_{1}^{2}x^{2}y^{2}dxdy + \int_{1}^{4}\int_{2}^{3}x^{2}y^{2}dxdy + \int_{1}^{4}\int_{3}^{4}x^{2}y^{2}dxdy$
But the values of the outer integral are unchanged (1 to 4). Is there any way to split into multiple parts by changing both the inner and outer integral values?
Like, in triple integral.
$\int_{1}^{4}\int_{1}^{4}\int_{1}^{4}x^{2}y^{2}z^{2}\,dx\,dy\,dz = \int_{1}^{4}\int_{1}^{4}\int_{1}^{2}x^{2}y^{2}z^{2}\,dx\,dy\,dz + \int_{1}^{4}\int_{1}^{4}\int_{2}^{3}x^{2}y^{2}z^{2}\,dx\,dy\,dz + \int_{1}^{4}\int_{1}^{4}\int_{3}^{4}x^{2}y^{2}z^{2}\,dx\,dy\,dz$
or
$\int_{1}^{4}\int_{1}^{4}\int_{1}^{4}x^{2}y^{2}z^{2}\,dx\,dy\,dz = \int_{1}^{4}\int_{1}^{2}\int_{1}^{4}x^{2}y^{2}z^{2}\,dx\,dy\,dz + \int_{1}^{4}\int_{2}^{3}\int_{1}^{4}x^{2}y^{2}z^{2}\,dx\,dy\,dz + \int_{1}^{4}\int_{3}^{4}\int_{1}^{4}x^{2}y^{2}z^{2}\,dx\,dy\,dz$
I can only separate using any one integral. I want to separate using by changing all three integral values.
| You can only do this only one at a time, but you can do all 3. For example
\begin{align*}
\int_1^4\int_1^4\int_1^4x^2y^2z^2dxdydz
& = \int_1^2\int_1^4\int_1^4x^2y^2z^2dxdydz + \int_2^3\int_1^4\int_1^4x^2y^2z^2dxdydz + \int_3^4\int_1^4\int_1^4x^2y^2z^2dxdydz \\
& = \int_1^2\int_1^2\int_1^4x^2y^2z^2dxdydz + \int_1^2\int_2^3\int_1^4x^2y^2z^2dxdydz + \int_1^2\int_3^4\int_1^4x^2y^2z^2dxdydz \\
& + \int_2^3\int_1^2\int_1^4x^2y^2z^2dxdydz + \int_2^3\int_2^3\int_1^4x^2y^2z^2dxdydz + \int_2^3\int_3^4\int_1^4x^2y^2z^2dxdydz \\
& + \int_3^4\int_1^2\int_1^4x^2y^2z^2dxdydz + \int_3^4\int_2^3\int_1^4x^2y^2z^2dxdydz + \int_3^4\int_3^4\int_1^4x^2y^2z^2dxdydz \\
& = etc...
\end{align*}
| {
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show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$ let $a,b>0.$ Show that
$$\sqrt{\dfrac{a^b}{b}}+\sqrt{\dfrac{b^a}{a}}\ge 2\tag{1}$$
I known How to prove $a^b+b^a>1$,where $a,b>0.$ See $x^y+y^x>1$ for all $(x, y)\in \mathbb{R_+^2}$
to prove $（1）$, I want use AM-GM inequality
$$\sqrt{\dfrac{a^b}{b}}+\sqrt{\dfrac{b^a}{a}}\ge 2\left(\dfrac{a^b}{b}\cdot\dfrac{b^a}{a}\right)^{1/4}=2\left(a^{b-1}b^{a-1}\right)^{1/4}$$
But $a^{b-1}b^{a-1}$ is not always $>1$
| My try (please point out the errors if any!):
Multiplying both sides by $\sqrt{ab}$, we get that
$$\sqrt{a^{b+1}}+\sqrt{b^{a+1}}\geq 2\sqrt{ab}\tag{1}$$
$$\frac{\sqrt{a^{b+1}}+\sqrt{b^{a+1}}}{2}\geq \sqrt{ab}$$
Using AM-GM inequality,
$$\frac{\sqrt{a}+\sqrt{b}}{2}\geq \sqrt{\sqrt{ab}}$$
For $0\leq a\leq b\leq 1$, we have that $\sqrt{\sqrt{ab}}\geq\sqrt{ab}$, $\sqrt{a^{b+1}}\geq \sqrt{a}$ and $\sqrt{b^{a+1}}\geq \sqrt{b}$. Therefore,
$$\frac{\sqrt{a^{b+1}}+\sqrt{b^{a+1}}}{2}\geq\frac{\sqrt{a}+\sqrt{b}}{2}\geq \sqrt{\sqrt{ab}}\geq \sqrt{ab}$$
Thus, the original inequality holds.
For $1\leq a\leq b$, we can transform $(1)$ into $$a^{b+1}+b^{a+1}+2{a^{\frac{b+1}{2}}b^{\frac{a+1}{2}}}\ge 4ab$$
As ${a^{\frac{b+1}{2}}b^{\frac{a+1}{2}}}\geq ab$, it follows that $4ab-2{a^{\frac{b+1}{2}}b^{\frac{a+1}{2}}}\leq 2$. As $a^{b+1}+b^{a+1}\geq 2$, the original inequality holds in this case too.
QED
| {
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Numbers formed from all the digits $1113333344455678$ so that $6$ and $8$ are on dofferent sides from $7$
How many numbers formed from all the digits $1113333344455678$ are there if the digits $6$ and $8$ have to appear from the different sides of the digit $7$?
My thoughts:
I thought I should first find how many numbers are formed from all the $14$ digits $11133333444557,$ which is $$\binom{14}3\binom{11}5\binom63\binom32=\frac{14!}{3!5!3!2!},$$and, then, analyze the possibilies given a position of the digit $7,$ but then, I lose the track and I only tell for sure $6$ and $8$ can be first and last digits in every situation, so there are strictly more numbers than $2\frac{14!}{3!5!3!2!}=\frac{14!}{(3!)^25!}.$
Then, I thought, I should first consider the position $p, 1<p<16$ among $16$ overall places for $7$ and choose the positions in $p-1$ and $16-p$ ways for $6$ and $8$ and multiply the result by $2$ for each $p,$ but this didn't lead me anywhere.
How should I approach this problem?
| One of the approaches is to see that once three positions are decided for $6, 7$ and $8$, you can arrange them within, in $3! = 6$ ways but here instead of $6$, there are only $2$ ways to arrange them as $7$ must be in between $6$ and $8$.
So we first find all possible arrangements of $1113333344455678$ and then divide by $3$ to get all favorable arrangements.
So the answer is $ ~ \displaystyle \frac{1}{3} \cdot \frac{16!}{3! \cdot 5! \cdot 3! \cdot 2!}$
Also sharing a longer approach - first arrange $13$ numbers other than $6, 7$ and $8$. Then there are $14$ places between them (including the ends). We can choose three places, two places or one place out of $14$ places for $6, 7$ and $8$. That leads to,
$ \displaystyle \frac{13!}{3! \cdot 5! \cdot 3! \cdot 2!} \left[2 \cdot {14 \choose 1} + 4 \cdot {14 \choose 2} + 2 \cdot {14 \choose 3}\right]$
Note that when we choose two places, if the first place from the left has two numbers then it must have $67$ or $87$ and if the second place has two numbers then it must have $76$ or $78$. That is why we multiply second term by $4$.
| {
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Binomial expansion question of complex sine How to show that the binomial expansion of $$\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^{2n+1}$$ equals
$$\sin ^{2n+1}(x)=\sum_{k=0}^{n}(-1)^{k+n}{{2n+1}\choose{k}}\frac{\sin ((2n-2k+1))x}{2^{2n}}$$
Please help! Sorry for asking such a dumb question!
| I think the upper bound of the summation should be $2n +1$ not $n$, and it should be $2^{2n + 1}$ not $2^{2n}$. You want to use the binomial theorem
$$(x + y)^n = \sum_{k = 0}^n {n \choose k } x^k y^{n - k}$$
This tells us
$$\left( e^{ix} - e^{-ix} \right)^{2n + 1} = \sum_{k = 0}^{2n + 1} {2n + 1 \choose k} e^{i(2n + 1 - k)x} \left(-e^{-ix} \right)^k = \sum_{k = 0}^{2n + 1} { 2n + 1 \choose k} (-1)^ke^{(2n - 2k + 1)ix} \\
= \sum_{k = 0}^{2n + 1} { 2n + 1 \choose k} (-1)^k \cos{ \left((2n -2k+1)x\right) } + i \sum_{k = 0}^{2n + 1} { 2n + 1 \choose k} (-1)^k \sin{ \left((2n -2k+1)x\right) }$$
See that both sums are real. This means that the sum with the sines is the imaginary part of $\left( e^{ix} - e^{-ix} \right)^{2n + 1}$. Then because the left hand side of
$$
\sin^{2n + 1} x = \left( \frac{e^{ix} - e^{-ix}}{2i} \right)^{2n + 1} = \frac{(-1)^n}{2^{2n + 1} i} \left( e^{ix} - e^{-ix} \right)^{2n + 1}
$$
is purely real, we only want the imaginary part of $\left(e^{ix} - e^{-ix} \right)^{2n + 1}$ so that the right hand side is purely real (this also tells us that the sum with the cosines must be 0). Substituting the sum with the sines gives the answer.
$$
\sin^{2n + 1} x = \frac{1}{2^{2n + 1}}\sum_{k = 0}^{2n + 1} (-1)^{n + k} { 2n + 1 \choose k } \sin{ \left((2n -2k+1)x\right) }
$$
Edit: It seems that the upper bound being $n$ and it being $2^{2n}$ is actually correct, as the expression above simplifies because the terms in the sum are equal when $k = n + i$ and $k = n - (i - 1)$
We will split the sum as:
$$\sum_{k = 0}^{2n + 1} = \sum_{k = 0}^n + \sum_{k = n + 1}^{2n + 1}$$
We will manipulate the second sum $\sum_{k = n + 1}^{2n + 1}$ as follows by substituting $k = j + (n + 1)$ so that the sum starts at $j = 0$ and ends at $j = n$
$$
\sum_{k = n + 1}^{2n + 1} = \sum_{j = 0}^{n} (-1)^{n + j + (n + 1)} { 2n + 1 \choose j + n + 1 } \sin{ \left((2n - 2(j + n + 1) +1)x\right) } \\
= \sum_{j = 0}^{n} (-1)^{j + 1} { 2n + 1 \choose j + n + 1 } \sin{ \left(-2j - 1)x\right) }
$$
And now we substitute $j = n - k$ so the sum starts at $k = n$ and ends at $k = 0$. That is, we are changing the order of summation to be backwards so we have
$$
\sum_{j = 0}^{n} (-1)^{j + 1} { 2n + 1 \choose j + n + 1 } \sin{ \left(\left(-2j - 1\right)x\right) } = \sum_{k = 0}^n (-1)^{n - k + 1} { 2n + 1 \choose 2n - k + 1 } \sin{ \left( (-2n + 2k - 1)x\right) } \\
= \sum_{k = 0}^n (-1)^{n + k} { 2n + 1 \choose k } \sin{ \left( (2n - 2k + 1)x\right) }
$$
where in the last step we used the oddness of $\sin$ and ${ n \choose k} = { n \choose n - k}$ for any positive integers $k \le n$. This is the same as the $\sum_{k = 0}^n$ sum from before so we have
$$
\sum_{k = 0}^{2n + 1} = 2 \sum_{k = 0}^n
$$
The two in the front will cancel reduce the $2^{2n + 1}$ to $2^{2n}$ so we'll have
$$
\sin^{2n + 1} x = \frac{1}{2^{2n}}\sum_{k = 1}^{n} (-1)^{n + k} { 2n + 1 \choose k } \sin{ \left((2n -2k+1)x\right) }
$$
| {
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How many ways are there to purchase $24$ croissants which come in packages of sizes $3$ and $6$? A croissant shop has $8$ different types of croissants that come in packages of size $3$. Additionally, there are $5$ different types of gluten-free croissants available that only come in packages of size $6$. You want to buy exactly $24$ croissants and do not buy any duplicates of any package. How many different ways are there to purchase croissants?
I was thinking of counting the number of ways you could buy packages of $6$ (without duplication) from $0$ packages to $4$ packages, and filling the rest with different packages of $3$ until we reach $24$ croissants. Then, I would add these together.
$0$ packages of $6 + 8$ packages of $3$ $${8 \choose 3} $$
$1$ packages of $6 + 6$ packages of $3$ $${6\choose 1} \cdot {8\choose6}$$
$2$ packages of $6 + 4$ packages of $3$ $${6\choose 2} \cdot {8\choose4}$$
$3$ packages of $6 + 2$ packages of $3$ $${6\choose 3} \cdot {8\choose2}$$
$4$ packages of $6 + 0$ packages of $3$ $${6\choose 4} \cdot {8\choose4}$$
number of different ways = $${8 \choose 3} + {6\choose 1}{8\choose6} + {6\choose 2}{8\choose4} + {6\choose 3}{8\choose2} + {6\choose 4}{8\choose4}$$
I'm not sure if this is the correct approach or not.
| I see that @N.F.Taussig gave you what you need.Hence , i want to give you another approach that facilitate your work. We make use of generating functions. Your question says that there will not be any dublication ,so for example if we choose $9$ croissants from $8$ different types of croissants that come in packages of size $3$ , then we must select $3$ types of croissants in $8$ different types of croissants by $\binom{8}{3}$.
Now , it is the time for constructing our generating functions such that
*
*For $8$ different types of croissants that come in packages of size $3$ : $$\binom{8}{0}x^0 +\binom{8}{1}x^3 + \binom{8}{2}x^6 + \binom{8}{3}x^9 + \binom{8}{4}x^{12} +\binom{8}{5}x^{15} + \binom{8}{6}x^{18} + \binom{8}{7}x^{21} + \binom{8}{8}x^{24}$$
Then , what does that expression mean ? The answer is it represents the number of selections such that $\color{red}{x}$ means the croissants that come in packeges of size $3$ , the exponents represent the total number of croissants in packeges of size $3$ , the binomial coefficients mean the number of selection to construct the total croissant in exponentials. For example , the binomial coefficient of $\binom{8}{6}x^{18}$ means the number of all ways to construct $18$ croissants by using only the croissants in packeges of size $3$
*
*For $5$ different types of gluten-free croissants come in packages of size $6$: $$\binom{5}{0}x^0 +\binom{5}{1}x^6 + \binom{5}{2}x^{12} + \binom{5}{3}x^{18} + \binom{5}{4}x^{24} +\binom{5}{5}x^{30} $$
The explanation is similar to foregoing generating function. I guess you got the logic.
Now , it is time for reaching the conclusion. We want to know how many way there are to buy $24$ croissants. We can find it by multiplying these two expression , and find the coefficient of $x^{24}$ . It will give you the result , because , for example when we multiply them , one of the pieces that play role in constructing $24$ croissants is $\binom{8}{6}x^{18} \times \binom{5}{1}x^6$. This means that select $6$ types from $8$ different types of croissants that come in packages of size $3$ ,and select one type from $5$ different types of gluten-free croissants come in packages of size $6$ . $\binom{8}{6} \times \binom{5}{1} =140 $ is the number of making that selections.
Now , the question is "how can i find $[x^{24}]$ in the product of these two expression". Answer is that you can do it by hands if you have knowledge over generating functions , if not , just use wolfram-alpha or other softwares.
I want to point out something here , as you can realize , these two generating function form are like binomial expression such as $$\bigg(1+ x^3\bigg)^8 \text{and} \bigg(1+ x^6 \bigg)^5 \text{,respectively}.$$
Then , search $x^{24}$ in $$\bigg(1+ x^3\bigg)^8 \times \bigg(1+ x^6 \bigg)^5$$
CALCULATION FOR YOU
Result is $\color{blue}{1126}$ , it match with @N.F.Taussig's answer.
| {
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General formula for the power sum $\sum_{k=1}^{n}k^mx^k\quad \forall\; m\in \mathbb{N}$ In my last question, it turns out to be solving the formula of $\sum_{k=1}^{n}k\omega^k$. I am curious if there is a geranal formula for the power sum:
$$\sum_{k=1}^{n}k^mx^k\quad \forall\; m\in \mathbb{N}$$
I have calculated first few formula.
\begin{align}
\sum_{k=1}^{n}x^k&=\frac{x}{x-1}(x^n-1)\\
\sum_{k=1}^{n}kx^k&=\frac{x}{x-1}\left(x^n\left(n-\frac{1}{x-1}\right)+\frac{1}{x-1}\right)\\
\sum_{k=1}^{n}k^2x^k&=\frac{x}{x-1}\left(x^n\left(n^2-\frac{2n-1}{x-1}+\frac{2}{(x-1)^2}\right)-\left(\frac{1}{x-1}+\frac{2}{(x-1)^2}\right)\right)\\
\sum_{k=1}^{n}k^3x^k&=\frac{x}{x-1}\left(x^n\left(n^3-\frac{3n^2-3n+1}{x-1}+\frac{6n-6}{(x-1)^2}-\frac{6}{(x-1)^3}\right)+\left(\frac{1}{x-1}+\frac{6}{(x-1)^2}+\frac{6}{(x-1)^3}\right)\right)
\end{align}
The formula has a common form.
$$
\sum_{k=1}^{n}k^mx^k=\frac{x}{x-1}\left(x^nf_m\left(\frac{1}{x-1}\right)-\left.f_m\left(\frac{1}{x-1}\right)\right|_{n=0}\right)\\
f_m \text{ is a polynomial of } \frac{1}{x-1}
$$
It seems there are some rules for $f_m\left(\frac{1}{x-1}\right)$when $m$ increased. For example:
The constant term coefficient of $f_m\left(\frac{1}{x-1}\right)$ is $\;n^m$.
The $\frac{1}{x-1}$ term coefficient of $f_m\left(\frac{1}{x-1}\right)$ is $\;(n-1)^m-n^m$.
The $\frac{1}{(x-1)^m}$ term coefficient of $f_m\left(\frac{1}{x-1}\right)$ is $\;(-1)^mm!$.
I also find a recursive formula for the power sum.
$$\sum_{k=1}^{n}k^{m+2}x^k=\sum_{k=1}^{n}k^{m+1}x^k+n(n+1)\sum_{k=1}^{n}k^mx^k-\sum_{k=1}^{n}\left(2k\sum_{j=1}^{k}j^mx^j\right)$$
Then the formula of $(m+2)^{th}$ order can always be calculated from previous two formula of $(m+1)^{th}$ order and $m^{th}$ order.
I have calculated the formula up to $6^{th}$ order, but the calculations become extremely complex and tedious when the order increased.
The term $$\sum_{k=1}^{n}\left(2k\sum_{j=1}^{k}j^mx^j\right)$$ will be expanded to all the formula from order $1$ to order $m+1$, and each formula times a polynomial of $\frac{1}{x-1}$.
Is there a general formula like Faulhaber's formula to calculate the coefficients of the formula for this power sum in a easier way?
| Recalling the umbral calculus relation:
$$ (x \frac{d}{dx})^m g(x)= \sum_{f=0}^{\infty} { m \brace f} \: x^f \: \frac{d^f}{dx^f} g(x) $$
where ${n \brace m} $ are the Stirling Numbers of the Second Kind.
One then has:
$$ \sum_{k=1}^{n} k^m x^k = \sum_{f=0}^{\infty} { m \brace f} \: x^f \: \frac{d^f}{dx^f} \frac{1-x^{n+1}}{1-x} $$
Despite the sum going to infinity, the sum is finite for integer values $m$. For example the sum $$ \sum_{k=1}^{10} k^4 x^k = \left(\frac{1-x^{11}}{(1-x)^2}-\frac{11 x^{10}}{1-x}\right) x+7 \left(\frac{2 \left(1-x^{11}\right)}{(1-x)^3}-\frac{22 x^{10}}{(1-x)^2}-\frac{110 x^9}{1-x}\right) x^2+6 \left(\frac{6 \left(1-x^{11}\right)}{(1-x)^4}-\frac{66 x^{10}}{(1-x)^3}-\frac{330 x^9}{(1-x)^2}-\frac{990 x^8}{1-x}\right) x^3+\left(\frac{24 \left(1-x^{11}\right)}{(1-x)^5}-\frac{264 x^{10}}{(1-x)^4}-\frac{1320 x^9}{(1-x)^3}-\frac{3960 x^8}{(1-x)^2}-\frac{7920 x^7}{1-x}\right) x^4 $$
which, of course, is equivalent to $$10000 x^{10}+6561 x^9+4096 x^8+2401 x^7+1296 x^6+625 x^5+256 x^4+81 x^3+16 x^2+x $$
| {
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$S$ , $P$ are sum and product of real roots of $x^4-7x^2-5=0$ what is the value of $2P^2-3SP+2S$?
If $S$ and $P$ are sum and product of real roots of $x^4-7x^2-5=0$
what is the value of $2P^2-3SP+2S$ ?
$1)59-7\sqrt{69}\qquad\qquad2)7+\sqrt{69}\qquad\qquad3)50\qquad\qquad4)59+7\sqrt{69}$
I solved this problem as follow:
Since $f(x)=x^4-7x^2-5$ is an even function, sum of the roots is equal to zero Hence we are looking for the value of $2P^2$.
$$x^2=\dfrac{7\pm\sqrt{69}}{2}\quad\text{Since $x^2\ge0$, $\quad x^2=\dfrac{7+\sqrt{69}}{2}$ is acceptable}$$
$$x_1=+\sqrt{\dfrac{7+\sqrt{69}}{2}}\quad\text{,}\quad x_2=-\sqrt{\dfrac{7+\sqrt{69}}{2}}$$
$$2P^2=2\times \left(\dfrac{7+\sqrt{69}}{2}\right)^2=\dfrac{49+69+14\sqrt{69}}{2}=59+7\sqrt{69}$$
Is my answer right? Also I'm very eager to see if it is possible to solve this with other approaches.
| Observe that $f$ is a function of the quadratic in $z = x^2$: $$f(x) = z^2 - 7z - 5.$$ This has roots $$x^2 = z = \frac{7 \pm \sqrt{(-7)^2 - 4(-5)}}{2} = \frac{7 \pm \sqrt{69}}{2}.$$ Since the negative root for $z$ yields non-real roots for $x$, we can discard these, and find that the two real roots of $f$ are $$x \in \pm \sqrt{\frac{7 + \sqrt{69}}{2}}.$$ Their sum $S$ is indeed zero, and their product is simply $$P = - \frac{7 + \sqrt{69}}{2}.$$ Thus $$2P^2 = 2 \left(\frac{7 + \sqrt{69}}{2}\right)^2 = \frac{1}{2} (49 + 14\sqrt{69} + 69) = 59 + 7 \sqrt{69}.$$ This is nearly identical to your solution.
| {
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Show by induction that for all $n \in \mathbb{N}$, $3^n \geq 1+2^n$. I started with the base case of P(1), which does work.
$3^1 = 3 \geq 1+2^1 = 3$
I then assumed P(k), which is
$3^k \geq 1+2^k$.
To prove P(k+1), which is $3^{k+1} \geq 1+2^{k+1}$, I have algebraically manipulated the following:
$3^{k+1} = 3 \cdot 3^k > 3^k \geq 1+2^k$
$2 \cdot 3^k \geq 2 + 2^{k+1}$
$2 \cdot 3^k - 1 \geq 1 + 2^{k+1}$.
At this juncture, I would obviously want to prove that $3^{k+1} > 2 \cdot 3^k - 1$. This is where I have run into trouble. If anyone could either help prove that, or show an alternate way of going about the problem, that would be fantastic.
| If $3^k\geqslant1+2^k$, then\begin{align}3^{k+1}&=3\times3^k\\&\geqslant3\times(1+2^k)\\&=3+3\times2^k\\&\geqslant1+2\times2^k\\&=1+2^{k+1}.\end{align}
| {
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Is $\sum_{a=0}^m\sum_{b=0}^n\cos(abx)$ always positive? Fix integers $m,n\geq0$.
Do we have the inequality $\displaystyle\sum_{a=0}^m\sum_{b=0}^n\cos(abx)>0$ for all $x\in\mathbb{R}$?
We can also write this function as
\begin{align*}
\sum_{a=0}^m\sum_{b=0}^n\cos(abx)&=m+n+1+\sum_{a=1}^m\sum_{b=1}^n\cos(abx)\\
&=m+n+1+\sum_{a=1}^m\frac{1}{2}\left(\frac{\sin((n+1/2)ax)}{\sin(ax/2)}-1\right)\\
&=\frac{m}{2}+n+1+\frac{1}{2}\sum_{a=1}^mD_n(ax),
\end{align*}
where
$$D_n(x)=\frac{\sin((n+1/2)x)}{\sin(x/2)}$$
is the Dirichlet kernel (up to a factor of $2\pi$, depending on your convention).
Using this formula, it is easy to check the conjecture for small values of $m$ and $n$ (desmos link).
| No! Consider the case $m = n$ and $x = \frac{8π}{4n + 3}$. We have
\begin{split}
\sum_{a=0}^n \sum_{b=0}^n \cos(abx)
&= \frac{3n}{2} + 1 + \frac12 \sum_{a=1}^n \frac{\sin\left(\left(n + \frac12\right)ax\right)}{\sin\left(\frac12ax\right)} \\
&= \frac{3n}{2} + 1 + \frac12 \sum_{a=1}^n \frac{\sin\left(2πa - \frac{2πa}{4n + 3}\right)}{\sin\left(\frac{4πa}{4n + 3}\right)} \\
&= \frac{3n}{2} + 1 + \frac12 \sum_{a=1}^n \frac{\sin\left(-\frac{2πa}{4n + 3}\right)}{\sin\left(\frac{4πa}{4n + 3}\right)} \\
&= \frac{3n}{2} + 1 - \frac14 \sum_{a=1}^n \sec\left(\frac{2πa}{4n + 3}\right) \\
&< \frac{3n}{2} + 1 - \frac14 \int_0^n \sec\left(\frac{2πa}{4n + 3}\right)\,da \\
&= \frac{3n}{2} + 1 + \frac{4n + 3}{8π} \ln \tan \frac{3π}{4(4n + 3)} \\
&\sim \frac{3n}{2} + 1 + \frac{4n + 3}{8π} \ln \frac{3π}{4(4n + 3)} \\
&→ -∞ \quad \text{as $n → ∞$}.
\end{split}
We can confirm this by plotting the exact sum at $m = n$ and $x = \frac{8π}{4n + 3}$ (blue); it first becomes negative at $n = 3286$. By optimizing $x$ near $\frac{8π}{4n + 3}$ to minimize the sum (orange), we find a slightly earlier negative value at $m = n = 3161$ and $x = 0.001987239$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4279731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
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Find an maxima such that $\sqrt a+\sqrt b$, when $27a^2+b^2=27$ (In high-school level!) How to get a maximum value of $\sqrt a+\sqrt b$, when $27a^2+b^2=27$?
I know that using Lagrange multiplier is easiest way in college-level math, but I believe there is a elementary and elegant way to prove that in high-school level.
I made $a=\cos\theta$ and $b=3\sqrt3\sin\theta$, but in that case, the process of finding the maximum value becomes too messy.
I need your help.
| Here's an entirely elementary solution, although it may not be very satisfying. We have, using the power-mean inequality,
\begin{align*}
\frac{\sqrt a + \sqrt b}4
&=\frac{\sqrt a + \sqrt{\frac b9} + \sqrt{\frac b9} + \sqrt{\frac b9}}4\\
&\leq \left(\frac{(\sqrt a)^4+\left(\sqrt{\frac b9}\right)^4+\left(\sqrt{\frac b9}\right)^4+\left(\sqrt{\frac b9}\right)^4}{4}\right)^{1/4}\\
&=\left(\frac{a^2+\frac{b^2}{81}+\frac{b^2}{81}+\frac{b^2}{81}}{4}\right)^{1/4}\\
&=\left(\frac{\frac{27a^2+b^2}{27}}{4}\right)^{1/4}=\frac1{4^{1/4}}=\frac1{\sqrt 2}.
\end{align*}
As a result, $\sqrt a+\sqrt b\leq 2\sqrt 2$. Equality is reached at $a=1/2$ and $b=9/2$, as can be traced through the application of the power-mean inequality.
The power-mean inequality is the statement that
$$\left(\frac{\sum_{i=1}^n a_i^p}{n}\right)^{1/p}$$
is an increasing function in $p$. This can be proven for the necessary exponents ($1$ and $4$) in four variables by first showing it for two variables:
$$\frac{x+y}{2}\leq \left(\frac{x^4+y^4}2\right)^{1/4}$$
since
$$\frac{x^4+y^4}2-\left(\frac{x+y}2\right)^4=\frac{(x+y)^4(7x^2+10xy+7y^2)}{16},$$
and then using this to show
$$\frac{w+x+y+z}{4}\leq \frac{\left(\frac{w^4+x^4}2\right)^{1/4}+\left(\frac{y^4+z^4}2\right)^{1/4}}2\leq \left(\frac{w^4+x^4+y^4+z^4}{4}\right)^{1/4}.$$
(These techniques are similar to the proof of the AM-GM inequality.)
It's probably easier to simply use differential calculus: letting $x=\sqrt a$, we wish to maximize
$$f(x) = x+(27-27x^4)^{1/4}.$$
We compute
$$f'(x)=1-\frac{27x^3}{(27-27x^4)^{3/4}}; f''(x)=-\frac{3^{7/4}x^2}{(1-x^4)^{7/4}}<0,$$
so the function has a unique maximum where
$$1=\frac{27x^3}{(27-27x^4)^{3/4}} \implies x=\frac{\sqrt 2}2.$$
| {
"language": "en",
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"source": "stackexchange",
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What is the probability of strictly increasing digits in a randomly generated 4 figure number? Suppose I make a $4$ digit number with each digit chosen at random from $\{0,1,...,9\}$, with repetition allowed. How do I find the probability that the digits in the number are in strictly increasing order?
I don't quite see how to proceed (unless naively writing out all possible combinations)?
| The first digit cannot be greater than 6, because with 7, 8, or 9 as the first digit it is not possible to create a 4 digit number with strictly increasing digits.
Let's denote the possibility of the first digit as A, then as the first digit is restricted with the set {0, 1, ..., 6} we have $$P(A)=\frac{7}{10}$$
The second digit is restricted with the set {1, 2, 3, 4, 5, 6, 7} by the same requirement of strictly increasing digits. Furthermore, the probability of the second digit is conditioned by the first digit. Denoting the probability of the second digit as B and the value of the first digit as x we have this
1st digit
0
1
2
3
4
5
6
P(B|x)
$\frac{7}{10}$
$\frac{6}{10}$
$\frac{5}{10}$
$\frac{4}{10}$
$\frac{3}{10}$
$\frac{2}{10}$
$\frac{1}{10}$
Further, the third digit is also restricted with the relative set {2, 3, 4, 5, 6, 7, 8} and its probability is conditioned by the second digit. Let's denote the probability of the third digit as C and the value of the second digit as y.
2nd digit
1
2
3
4
5
6
7
P(C|y)
$\frac{7}{10}$
$\frac{6}{10}$
$\frac{5}{10}$
$\frac{4}{10}$
$\frac{3}{10}$
$\frac{2}{10}$
$\frac{1}{10}$
The fourth digit is restricted with the set {3, 4, 5, 6, 7, 8, 9} and its probability is conditioned by the third digit. Let's denote the probability of the fourth digit as D and the value of the third digit as z.
3d digit
2
3
4
5
6
7
8
P(D|z)
$\frac{7}{10}$
$\frac{6}{10}$
$\frac{5}{10}$
$\frac{4}{10}$
$\frac{3}{10}$
$\frac{2}{10}$
$\frac{1}{10}$
So actually the probability of the number with strictly increasing order is the sum of all the possible products of the 4 digits probabilities like $\frac{7}{10}*\frac{7}{10}*\frac{7}{10}*\frac{7}{10}$.
Eventually we can define the final formula as $$\sum\frac{7*a*b*c}{10^4}$$ for each a, b, and c $\in \{1, 2, 3, 4, 5, 6, 7\}$ where $a\geq b \geq c$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Eliminating $\theta$ from $x=\cos\theta(2-\cos 2\theta)$ and $y=\sin\theta(2-\sin 2\theta)$
If $$x=2\cos\theta-\cos\theta\cos 2\theta$$
$$y=2\sin\theta-\sin\theta\sin 2\theta$$
find a relation between $x$ and $y$ (not involving $\theta$).
Another trig elimination that has me stumped. One approach is to express as homogeneous functions,
$$x=3\sin^2\theta \cos\theta+\cos^3\theta$$
$$y=2\sin^3\theta+2\sin\theta\cos^2\theta-2\sin^2\theta\cos \theta$$
and by forming linear combinations of $x,y$ find some third power trig polynomials eg$(\cos \theta+\sin\theta)^3$. But I am not finding any success with this problem.
Note that this problem is from Hobson, Treatise on Plane Trigonometry 2 ed pg.97 #47. Slightly altered in the form I ask it. And may contain a typo, see discussion.
| Let $u=\cos\theta$ so $x=3u-2u^3$ and $y=-2u\pm2\sqrt{1-u^2}+2u^3$.
Let $z=x+y$ so $4(1-u^2)=(y+2u-2u^3)^2=(z-u)^2$ giving $$5u^2-2zu+z^2-4=0\implies5u=z\pm2\sqrt{5-z^2}.$$ Apply the quadratic equality repeatedly to get \begin{align}x&=3u-2u\cdot5^{-1}(2zu-z^2+4)=(3+2\cdot5^{-1}z^2-8\cdot5^{-1})u-4\cdot5^{-1}zu^2\\&=(3+2\cdot5^{-1}z^2-8\cdot5^{-1})u-4\cdot5^{-2}z(2zu-z^2+4)\\&=(3+2\cdot5^{-1}z^2-8\cdot5^{-1}-8\cdot5^{-2}z^2)5^{-1}(z\pm2\sqrt{5-z^2})+4\cdot5^{-2}(z^2-4).\end{align} Simplification leads to $$125x=22(x+y)^3-45(x+y)\pm2(35+2(x+y)^2)\sqrt{5-(x+y)^2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Is there a closed form for this sequence $a_n = m, \ \binom{m}{2}\le n < \binom{m+1}{2}$? I was playing around with the sequence:
$$
2,2, 3,3,3,4,4,4,4,\dots
$$
If I denote the first element of said sequence by $a_1$ I realized the sequence can be written as
$$
a_n = m, \quad \binom{m}{2}\le n < \binom{m+1}{2}
$$
for $m \ge 2$. Although the above does work, I would like to find a closed for of a sequence that gives $a_n$ in terms of common functions like the floor function. Does anyone know how I could do this?
I managed to re-arrange the condition $ \binom{m}{2}\le n < \binom{m+1}{2}$ into
$$
0\le \frac{n}{m} + \frac{1-m}{2} < 1
$$
with hopes of using something like the floor function on it, but I couldn't seem to make it work.
| We have
$$
\begin{array}{l}
\left( \begin{array}{c}
m \\ 2 \\ \end{array} \right) =
\frac{{m\left( {m - 1} \right)}}{2} \le n < \left( \begin{array}{c}
m + 1 \\ 2 \\ \end{array} \right) = \frac{{m\left( {m + 1} \right)}}{2}\quad \Rightarrow \\
\Rightarrow \quad m^2 - m - 2n \le 0 < m^2 + m - 2n\quad \Rightarrow \\
\Rightarrow \quad \frac{{\sqrt {1 + 8n} - 1}}{2} = \frac{{\sqrt {1 + 8n} + 1}}{2} - 1 <
m \le \frac{{\sqrt {1 + 8n} + 1}}{2} \\
\end{array}
$$
and since
$$
x - 1 < \left\lfloor x \right\rfloor \le x
$$
then
$$
a_n = \left\lfloor {\frac{{\sqrt {1 + 8n} + 1}}{2}} \right\rfloor
$$
where $a_0=1, a_1=2, a_2=2, \cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4286449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Integral of $n$-th Bernoulli polynomial of the second kind We have
\begin{align}
\int x(x−1)(x−2)...(x−n)\,dx=(n+1)!\cdot\psi_{n+2}(x),
\end{align}
where, $\psi_n(x)$ is the $n$-th Bernoulli polynomial of the second kind. We have
\begin{align}
I &= \int_0^{n} \underbrace{x}_{\color{red}{=\,u}}\cdot \underbrace{x(x-1)\cdots(x-n)}_{\color{red}{=\,v'}}\,d u \\
&= x\cdot (n+1)!\,\psi_{n+2}(x)\,\Big|_0^{n} - (n+1)! \int_0^{n} \psi_{n+2}(u)\,d u \\
&= (n+1)!\left[n\cdot\psi_{n+2}(n) - \int_0^n \psi_{n+2}(u)\,d u\right] \\
&= -(n+1)!\left[n\cdot|G_{n+2}| + \int_0^n \psi_{n+2}(u)\,d u\right]
\end{align}
What is $\displaystyle\int_0^n \psi_{n+2}(u)\,d u$ ?
| It was so simple:
\begin{align}
(n+2)!\cdot\psi_{n+3}(x) &= \int x(x-1)(x-2)\cdots(x-n)(x-(n+1))\ dx \\
&=\underbrace{\int x \cdot x(x-1)(x-2)\cdots(x-n) \ dx}_{\color{red}{=\,I}} - (n+1) \underbrace{\int x(x-1)(x-2)\cdots(x-n) \ dx}_{\color{red}{=\,(n+1)!\,\cdot\,\psi_{n+2}(x)}} \\
&= I - (n+1)\cdot(n+1)!\,\cdot\,\psi_{n+2}(x)
\end{align}
Thus applies:
\begin{align}
I &= (n+1)\cdot(n+1)!\,\cdot\,\psi_{n+2}(x) + (n+2)! \cdot \psi_{n+3}(x) \color{white}{\frac{1}{2}} \\
&= (n+1)! \cdot \left[(n+1)\cdot \psi_{n+2}(x) + (n+2)\cdot \psi_{n+3}(x)\right] \color{white}{\frac{1}{2}}
\end{align}
Now the only question is what is $\psi_{n+3}(n)$....
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Eliminate ${\theta, \varphi}$ from $\tan\theta+\tan\varphi=a$, $\sec\theta+\sec\varphi=b$, $\csc\theta+\csc\varphi=c$ Here is the problem:
Eliminate $\theta, \varphi$ from the equations
$$\tan\theta+\tan\varphi=a$$
$$\sec\theta+\sec\varphi=b$$
$$\csc\theta+\csc\varphi=c$$
What I have so far.
$$\tan \theta=a-\tan\varphi$$
$$\sec \theta=b-\sec\varphi$$
$$\csc \theta=c-\csc\varphi$$
And using
$$\tan \theta \csc\theta =\sec\theta$$ gives us
$$c\tan\varphi+a\csc\varphi-2\sec \varphi=ac-b$$
And also
$$\tan^2\theta+1=\sec^2\theta$$
gives
$$2a\tan\varphi-2b\sec\varphi=a^2-b^2$$
Thus we have two linear relations. But I cannot think of any more properties to use. Things like $\sin^2+\cos^2=1$ give a big mess. So what are the end steps ?
Update: The method used in this question: How to eliminate $\theta$ & $\phi$ from above equations
will work and solves the problem.
Following that method gives me
$$[(b^2-a^2)^2+4a^2]a^2c=4a^2[c(a^2+b^2)-2ab]$$
What a fantastic relation !
| Disclosure: This almost seems to be the answer that @Rene Schipperus found (originally by @John Bentin on this post), but has some variations. So I am posting it. Although some lines are exactly as the previous answer, I will include them here for the sake of completeness.
$$\tan\theta+\tan\varphi=a$$
$$\sec\theta+\sec\varphi=b$$
$$\csc\theta+\csc\varphi=c$$
Changing the shape of these equations, $$\sin\theta\cos\varphi+\sin\varphi\cos\theta=a\cos\theta\cos\varphi\tag1$$
$$\cos\theta+\cos\varphi=b\cos\theta\cos\varphi\tag2$$
$$\sin\theta+\sin\varphi=c\sin\theta\sin\varphi\tag3$$
Further moulding,
$$2\sin\left(\frac{\theta+\varphi}2\right)\cos\left(\frac{\theta+\varphi}2\right)=a\cos\theta\cos\varphi\tag4$$
$$2\cos\left(\frac{\theta+\varphi}2\right)\cos\left(\frac{\theta-\varphi}2\right)=b\cos\theta\cos\varphi\tag5$$
$$2\sin\left(\frac{\theta+\varphi}2\right)\cos\left(\frac{\theta-\varphi}2\right)=c\sin\theta\sin\varphi\tag6$$
Dividing (4) by (5),
$$\frac{\sin\left(\frac{\theta+\varphi}2\right)}{\cos\left(\frac{\theta-\varphi}2\right)}=\frac ab\tag7$$
Multiply (6) by (7) and 1/(7) respectively,
$$2\sin^2\left(\frac{\theta+\varphi}2\right)=\frac{ac}b\sin\theta\sin\varphi$$
$$2\cos^2\left(\frac{\theta-\varphi}2\right)=\frac{bc}a\sin\theta\sin\varphi$$
From the last two we get,
$$1-\cos(\theta+\varphi)=\frac{ac}b\sin\theta\sin\varphi\tag8$$
$$1+\cos(\theta-\varphi)=\frac{bc}a\sin\theta\sin\varphi\tag9$$
(9) - (8)
$$\cos(\theta+\varphi)+\cos(\theta-\varphi)=\left(\frac{bc}a-\frac{ac}b\right)\sin\theta\sin\varphi$$
$$2\cos\theta\cos\varphi=\frac{c(b^2-a^2)}{ab}\sin\theta\sin\varphi$$ $$\tan\theta\tan\varphi=\frac{2ab}{c(b^2-a^2)}=\lambda\quad\text{ (say)}$$
Ok, leave it here and let's move back to first few equations.
Dividing (6) by (5),
$$\tan\left(\frac{\theta+\varphi}2\right)=\frac cb\tan\theta\tan\varphi=\frac cb\lambda$$
Now using, $$\tan(\theta+\varphi)=\frac{2\tan\left(\frac{\theta+\varphi}2\right)}{1-\tan^2\left(\frac{\theta+\varphi}2\right)}$$
$$\frac{\tan\theta+\tan\varphi}{1-\tan\theta\tan\varphi}=\frac{2\tan\left(\frac{\theta+\varphi}2\right)}{1-\tan^2\left(\frac{\theta+\varphi}2\right)}$$
$$\frac a{1-\lambda}=\frac{2\dfrac cb\lambda}{1-\left(\dfrac cb\lambda\right)^2}$$
Further simplifications lead to, $$c[(b^2-a^2)^2-4a^2]=4[c(b^2-a^2)-2ab]$$ and finally, $$(b^2-a^2)^2c=4b^2c-8ab$$ which is indeed the answer obtained using the other method.
An alternate form is $$\boldsymbol{(2b-a^2+b^2)(2b+a^2-b^2)c=8ab}$$ which seems better!
| {
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"url": "https://math.stackexchange.com/questions/4288989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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$\frac{\cos(xM)-\cos(xM+x)}{1-\cos(x)}=\frac{x\sin(M+\frac{1}{2})}{\sin(\frac{x}{2})}$ I am trying to prove the identity
$$
\frac{\cos(xM)-\cos(xM+x)}{1-\cos(x)}=\frac{x\sin(M+\frac{1}{2})}{\sin(\frac{x}{2})}
$$
I start with$$
\frac{\cos(xM)-\cos(xM+x)}{1-\cos(x)}=\frac{2\sin\left(Mx+\frac{x}{2}\right)\sin\left(\frac{x}{2}\right)}{1-\cos(x)}
$$
$$
\Longrightarrow\frac{2\sin\left(Mx+\frac{x}{2}\right)\sin\left(\frac{x}{2}\right)}{\cos^2(\frac{x}{2})+\sin^2(\frac{x}{2})-(\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2}))}
$$
$$
\Longrightarrow
\frac{\sin\left(Mx+\frac{x}{2}\right)\sin\left(\frac{x}{2}\right)}{\sin^2(\frac{x}{2})}
$$
$$
\Longrightarrow
\frac{\sin\left(Mx+\frac{x}{2}\right)}{\sin(\frac{x}{2})}
$$
How do I proceed from here?
Context:
This problem arose when I solve for the Discrete-time Fourier Transform (DTFT) of the function $f[n]=1$ for $-M≤ n≤M$
Let $F_1(x)$ be the DTFT of $f_1[n]=1$ where $n=0,1\cdots M$ and $F_2(x)$ be thr DTFT of $f_1[n]=1$ where $n=-1,-2\cdots -M$
$$
F(x)=F_1(x)+F_2(x)
$$
$$
F(x)=\frac{1-e^{-ix(M+1)}}{1-e^{-ix}}+\frac{1-e^{-ix(M)}}{1-e^{ix}}e^{ix}
$$
$$
F(x)=\frac{e^{ixM}+e^{-ixM}-e^{ix(M+1)}-e^{-ix(M+1)}}{2-e^{-ix}-e^{ix}}
$$
leading to the equation above:
$$
\frac{\cos(xM)-\cos(xM+x)}{1-\cos(x)}
$$
which, according to the book, should yield $$
\frac{x\sin(M+\frac{1}{2})}{\sin(\frac{x}{2})}
$$
| The correct answer is $$\frac{\sin\left(Mx+\frac{x}{2}\right)}{\sin(\frac{x}{2})}$$
This is a bit embarrassing but I misunderstood the book on its written formula as seen below
$$
X(\omega ) = A\frac{{\sin (M + \frac{1}{2})\omega }}{{\sin (\omega /2)}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4291861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Reduce error when $x$ is much smaller than 1 I have the following function $$\frac{1}{1+2x}-\frac{1-x}{1+x} $$
How to find equivalent way to compute it but when $x$ is much smaller than 1? I assume the problem here is with $1+x$ since it probably would be equal to 1. I don't know if multiplying by $(1-x)$ would be helpful as it would be
$$ \frac{1-x}{1+x-2x^2}-\frac{(1-x)^2}{1-x^2} $$ so there's still term $1+x$.
| If the therms in $x^2$ can be neglegted:
$$\frac{1}{1+2x}\approx\frac{1-4x^2}{1+2x}=1-2x$$
$$\frac{1-x}{1+x}\approx\frac{(1-x)(1-x^2)}{1+x}=(1-x)^2\rightarrow$$
$$f(x)\approx1-2x-1+2x-x^2=-x^2$$
If therm in $x^3$ can be neglegted:
$$\frac{1}{1+2x}\approx\frac{1-8x^3}{1+2x}=1-2x+4x^2$$
$$\frac{1}{1+x}\approx\frac{1-x^3}{1+x}=1-x+x^2$$
And so on...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4292618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Extraneous solution when solving $x^2+x+1=0$ by getting $x^2=1/x$ Let's assume that we have$$x^2+x+1=0.\tag1$$
Substituting $x=0$, we get $1=0$, so $0$ is not a root for the quadratic equation and thus, $x\neq0$. Therefore, there exists $\frac{1}{x}$, which we'll multiply by both sides of $(1)$, giving us $$x+1+\frac{1}{x}=0.$$
We will, then, move $\frac{1}{x}$ to the other side and get $$x+1=-\frac{1}{x}.$$ If we add $x^2$ to both sides and note that $x^2+x+1=0$, we will have $x^2-\frac{1}{x}=0$. The real root of this equation is $x=1$, which is not a root of $(1)$.
I was wondering at which step did I do something that was incorrect and resulted in this supposed root.
| Hint:
$x^2+x+1=0$
is true only for some value of $x$ not $\forall x$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\int_0^1\frac{\text{Li}_2(-x^2)}{\sqrt{1-x^2}}\,dx=\pi\int_0^1\frac{\ln\left(\frac{2}{1+\sqrt{1+x}}\right)}{x}\,dx$ I managed here to prove $$\int_0^1\frac{\text{Li}_2(-x^2)}{\sqrt{1-x^2}}\,dx=\pi\int_0^1\frac{\ln\left(\frac{2}{1+\sqrt{1+x}}\right)}{x}\,dx$$
but what I did was converting the LHS integral to a series then converting the series to the RHS integral. Is it possible to relate the two integrals without going through the series?
| First consider the following definition:
$$\operatorname{Li}_2\left(x\right)=-\int _0^x\frac{\ln \left(1-t\right)}{t}\:dt.$$
This means that:
$$\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:dx=-\int _0^1\frac{1}{\sqrt{1-x^2}}\underbrace{\int _0^{-x^2}\frac{\ln \left(1-t\right)}{t}\:dt}_{y=-\frac{t}{x^2}}\:dx$$
$$=-\int _0^1\frac{1}{y}\int _0^1\frac{\ln \left(1+yx^2\right)}{\sqrt{1-x^2}}\:dx\:dy.$$
Now if we use differentiation under the integral sign for that integral we obtain:
$$=-\int _0^1\frac{1}{y}\left(\pi \ln \left(\frac{1+\sqrt{1+y}}{2}\right)\right)\:dy.$$
Therefore:
$$\boxed{\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:dx=\pi \int _0^1\frac{\ln \left(\frac{2}{1+\sqrt{1+y}}\right)}{y}\:dy.}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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Give a complete non-redundant list of elements $x\in\mathbb{C}$ such that $x^{10}+x^5+1=0$ Since, $x^{10}+x^5+1 = (x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1)=0$.
I know that $x^2+x+1=0$ has root $x=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$ and $\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$.
How am I supposed to find the roots in $\mathbb{C}$ for $x^8-x^7+x^5-x^4+x^3-x+1$ ?
| Since $x^{10}+x^5+1=\frac{x^{15}-1}{x^5-1}$, root of $x^{10}+x^5+1=0$ must be root of $x^{15}-1=0$, but not root of $x^5-1=0$. $x^5-1=0$ and $x^{10}+x^5+1=0$ has no common root.
So, the zero set of $x^{10}+x^5+1$ is...
$$\{e^{\frac{2ni\pi}{15}}\mid 3\not\mid n\in\mathbb Z \}.$$
Because the zero set of $x^{15}-1=0$ is $\{e^{\frac{2ni\pi}{15}}\mid n\in\mathbb Z \}$, and the zero set of $x^5-1=0$ is $\{e^{\frac{2ni\pi}{5}}\mid n\in\mathbb Z \}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4302816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Combinatoric question - arrange balls in cells [Probability] We want to arrange $10$ (different) balls in $4$ cells, what is the chance that $5$ will be entered to cell number $1$, and the rest of them will enter an even cell ($2$ or $4$)?
My attempt:
Note that we have a symmetric space, in which we have $4^{10}$ options to arrange the balls.
Let us choose $5$ balls to enter cell $1$ - $\binom{10}{5}$.
Then we have to arrange the next $5$ balls in an even cell, so each one of them has $2$ options, to enter cell number $2$ or cell number $4$, which is: $\binom{5}{5} \cdot 2^5$.
So in total we have: $$\frac{\binom{10}{5} + \binom{5}{5} \cdot 2^5}{4^{10}}=0.027 ~\% $$
Is that correct?
Thanks!
| We might view this problem as an example of the Multinomial Distribution. There are $10$ independent trials, and in each trial either (1) a ball lands in bin $1$, with probability $1/4$; (2) a ball lands in bin $2$ or bin $4$, with probability $2/4$; or (3) a ball lands in bin $3$, with probability $1/4$. Then the probability of the outcome $(5,5,0)$ is
$$\binom{10}{5 \quad 5 \quad 0} \left( \frac{1}{4} \right)^5 \left( \frac{2}{4} \right)^5 = \frac{10!}{5! 5! 0!} \left( \frac{1}{4} \right)^5 \left( \frac{2}{4} \right)^5 = 0.00769043$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Question about divergence of $\vec{F} = \frac{1}{r^2} \hat{r}$ I am looking at the divergence of this famous expression:
$\vec{F} = \frac{1}{r^2} \hat{r}$
i saw this calculation which looks reasonable ...
$\hat{r} = (x,y,z)/\sqrt{x^2 + y^2 + z^2}$
$$
F(x,y,z) = \frac{1}{(x^2 + y^2 + z^2)^{3/2}} (x,y,z) \\
\frac{\partial}{\partial x} F_x = \frac{\partial}{\partial x} \frac{x}{(x^2 + y^2 + z^2)^{3/2}} = \frac{-2x^2 + y^2 + z^2}{(x^2 + y^2 + z^2)^{5/2}} \\
\frac{\partial}{\partial y} F_y = \frac{\partial}{\partial y} \frac{y}{(x^2 + y^2 + z^2)^{3/2}} = \frac{x^2 -2y^2 + z^2}{(x^2 + y^2 + z^2)^{5/2}} \\
\frac{\partial}{\partial z} F_z = \frac{\partial}{\partial z} \frac{z}{(x^2 + y^2 + z^2)^{3/2}} = \frac{x^2 + y^2 -2z^2}{(x^2 + y^2 + z^2)^{5/2}} \\
$$
Putting together :
$$
\nabla\cdot F = \frac{\partial}{\partial x} F + \frac{\partial}{\partial y} F + \frac{\partial}{\partial z} F = \frac{0}{(x^2 + y^2 + z^2)^{5/2}} = 0
$$
However I am confused about one thing.
*
*Suppose vector field $\vec{F}$ has a value $F_1$ at (1,1,1).
*Now if x changes by dx, that is x2 = 1+dx, y and z remaining unchanged,
*Then the change in x component vector F will be 0 since partial derivative of $\vec{F_x}$ w.r.t to x evaluates to 0. Similarly the change in y and z component will also be 0, since their partial derivatives also evaluate to 0 at (1,1,1).
This seems confusing to me. How can there be no change in the vector when x changes by dx, at (1,1,1) ?
What am I doing wrong ?
| $
\def\l{\lambda}
\def\n{\nabla}
\def\L#1{\l^{-#1}}
\def\qiq{\quad\implies\quad}
\def\tr{\,{\rm Trace}}
\def\c#1{\color{red}{#1}}
$Differentiate the length $\l$ of the position vector $r$
$$\eqalign{
\l^2 &= r\cdot r \\
2\l\:d\l &= 2r\cdot dr \\
d\l &= \L1r\cdot dr \\
}$$
Then calculate the gradient of $F$
$$\eqalign{
F &= \L3r \\
dF &= \L3dr - r(3\L4d\l) \\
&= \L3I\cdot dr - r(3\L5\,r\cdot dr) \\
&= (\L3I - 3\L5rr)\cdot dr \\
\n F &= \L3I - 3\L5rr \\
}$$
Finally, calculate the divergence
$$\eqalign{
\n\cdot F &= \tr(\n F) \\
&= \L3\tr(I) - 3\L5\tr(rr) \\
&= \L3(n) - 3\L5(\l^2) \\
&= (n-3)\L3 \\
}$$
Therefore in three dimensions $(n=3)$ the divergence is equal to zero, but in any other dimension it is non-zero.
Note however, that the Taylor series at $r=r_1$ uses the gradient of $F\,$ not its divergence
$$\eqalign{
F(r_2) - F(r_1) &= (r_2 - r_1)\cdot\c{\n F(r_1)}
\;+\; {\cal O}\left(\|r_2-r_1\|^2\right) \\
dF &\approx dr\cdot\c{(\L3I - 3\L5r_1r_1)} \\
dF &\approx \L3dr - 3\L5(dr\cdot r_1)\,r_1 \\
}$$
So the change in the function value is $dF$ which (as you suspected) is not equal to zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Calculate $f^{(n)}(0)$ where $f(x) = e^{x^2}$ The problem is stated as:
Calculate $f^{(n)}(0)$ where $f(x) = e^{x^2}$ and $n \in \mathbb{N}$
Attempt:
We know that $e^{x} = \sum_{k=0}^{n} \frac{1}{k!}x^k +\frac{x^{n+1}e^{\theta x}}{(n+1)!}$ for some $\theta = \theta (x) \in (0,1)$
Hence, $e^{x^2}=\sum_{k=0}^{n} \frac{1}{k!}x^{2k} +\frac{x^{2n+1}e^{\theta x^2}}{(n+1)!}$
Comparing terms with the general Maclaurin polynomial, we get that:
$\frac{1}{k!}x^{2k} = \frac{f^{(n)}(0)}{n!}x^{n}$
First, we have to equate the exponent of the $x$ - terms, so $2k = n \iff k = n/2$
Therefore, we have that:
$\Rightarrow f^{n}(0) = \frac{n!}{(n/2)!}$
However, since we had natural numbers in our expression, I wonder what would happen if $n$ was odd in this case. How could I generalize this even further?
Thanks!
| You can make the problem more general and compute the $n^{th}$ derivative at any value of $n$.
They all write
$$f^{(n)}(x)=e^{x^2} P_n(x)$$ Computing the first ones (using the chain rule makes it easy)
$$f^{(n+1)}(x)=e^{x^2}\left(2x P_n(x)+P'_n(x) \right)$$
So, you have
$$\left(
\begin{array}{cc}
n & P_n(x) \\
1 & 2 x \\
2 & 4 x^2+2 \\
3 & 8 x^3+12 x \\
4 & 16 x^4+48 x^2+12 \\
5 & 32 x^5+160 x^3+120 x \\
6 & 64 x^6+480 x^4+720 x^2+120 \\
7 & 128 x^7+1344 x^5+3360 x^3+1680 x \\
8 & 256 x^8+3584 x^6+13440 x^4+13440 x^2+1680 \\
9 & 512 x^9+9216 x^7+48384 x^5+80640 x^3+30240 x \\
10 & 1024 x^{10}+23040 x^8+161280 x^6+403200 x^4+302400 x^2+30240
\end{array}
\right)$$ where you can notice interesting patterns.
In fact
$$\color{red}{P_n=i^{-n} H_n(i x)\implies f^{(n)}(x)=i^{-n} H_n(i x)e^{x^2}}$$
Now, if you make $x=0$, as expected since $e^{x^2}$ is a even function, $f^{(2n+1)}(0)=0$ and for $f^{(2n)}(0)$ generates the sequence
$$\{2,12,120,1680,30240,665280,\cdots\}$$ which is your result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How do we solve $x^2 + \{x\}^2 = 33$ without computer? This is a problem taken from a group on Facebook. I wonder how to solve this without numerical process.
$x^2 + \{x\}^2 = 33\tag{1}$
My unfinished attempt:
$$\begin{align}
x^2 + \{x\}^2 &= 33\\
x^2 + \left(x - \lfloor x \rfloor\right)^2 &= 33\\
x^2 + x^2 - 2x \lfloor x \rfloor + \lfloor x \rfloor^2 &= 33\\
2x\left(x - \lfloor x \rfloor\right) + \lfloor x \rfloor^2 &= 33\\
2x\{x\} + \lfloor x \rfloor^2 = 33
\end{align}$$
I'm stuck at there. Also we know that the fractional part of $x$ is bounded i.e.:
$$0\leq\{x\}<1$$
From that, I can predict where the two solutions are placed at:
$$\begin{align}
0&\leq \{x\} < 1\\
0 &\leq \{x\}^2 < 1\\
x^2 &\leq x^2 + \{x\}^2 < x^2 + 1\\
x^2 &\leq 33 < x^2 + 1 \tag{$x^2 + \{x\}^2 = 33$}\\
S &= -\sqrt{33}\leq x \leq \sqrt{33} \quad \bigcap \quad x< -\sqrt{32} \lor x > \sqrt{32}\\
\therefore S &= -\sqrt{33}\leq x < -4\sqrt2 \quad \bigcup \quad 4\sqrt{2} < x \leq \sqrt{33}\\
S &\approx -5.74456 \leq x < -5.65685 \quad \bigcup \quad 5.65685 < x \leq 5.74456
\end{align}$$
I don't know if that's even going to help. Anyway, here are the two solutions (to the original problem) that's given by Wolfram Alpha:
$$\begin{align}
x_1 &= \frac12 \left(3\sqrt{17} - 1\right) \approx -5.815\\
x_2 &= \frac12 \left(-1 - 3\sqrt{113}\right) \approx 5.685
\end{align}$$
You see $x_1$ isn't at the interval. I'm confused.
| ADDED EDIT BELOW.
First note roughly we know $5\le x \le 6$ or $-6\le x \le -5$.
So we have $\{x\}=x-5$ when $x>0$ and $\{x\}=x-(-6)=x+6$ when $x<0$, using the fact that $\{x\} = x - \lfloor x \rfloor$, where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$.
Now you solve each case separately.
When $x>0$, solve $x^2 +(x-5)^2=33$, noting we want the positive solution to this. With quadratic formula we get $\frac{5}{2} + \frac{\sqrt{41}}{2}$
And when $x<0$ solve $x^2 + (x+6)^2=33$, noting we want the negative solution between -6 and -5 to this. We get $-3 - \sqrt{15/2}$.
This matches what Wolfram give: https://www.wolframalpha.com/input/?i=x%5E2+%2B+%28x-floor%28x%29%29%5E2+%3D+33%2C+solve+for+x
(Which didn't match what you quoted above.)
Edit.
As it turns out $\{x\} = frac(x)$ is not universally agreed when $x$ is negative. According to Wikipedia: "However, in case of negative numbers, there are various conflicting ways to extend the fractional part function to them" https://en.wikipedia.org/wiki/Fractional_part.
So one might also consider the alternate definition, when $x<0$, take $\{x\} = x - \lceil x \rceil$ instead, and resolve it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4308317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Finding Binomial Expansion as x approaches infinity of a function I am trying to find the series that is valid for
$(x^3 + x^2 + 1)^{\frac{1}{3}} - (x^2 + x)^{\frac{1}{2}}$
as x approaches infinity.
The question says this should be possible using the binomial expansion but I can't work out how to do that.
| $$(x^3 + x^2 + 1)^{\frac{1}{3}} - (x^2 + x)^{\frac{1}{2}}=x\left(1+\frac{x^2+1}{x^3}\right)^{\frac{1}{3}}-x \left(1+\frac{1}{x}\right)^{\frac{1}{2}}$$
Now use
$$(1+\epsilon )^{\frac{1}{n}}=\sum_{n=0}^\infty \binom{\frac{1}{n}}{n}\epsilon^n$$
Fot the first term, make $n=\frac 13$ and $\epsilon= \frac{x^2+1}{x^3}$; for the second , make $n=\frac 12$ and $\epsilon= \frac{1}{x}$.
Expand a little the powers of $\frac{x^2+1}{x^3}$ and you will have
$$(x^3 + x^2 + 1)^{\frac{1}{3}} - (x^2 + x)^{\frac{1}{2}}=-\frac{1}{6}+\frac{1}{72 x}+\frac{431}{1296 x^2}-\frac{6977}{31104
x^3}+O\left(\frac{1}{x^4}\right)$$
Try for $x=1000$; the truncated series will give
$$-\frac{5183557662977}{31104000000000} =\color{red}{-0.166652445440}361$$ while
$$\sqrt[3]{1001000001}-10 \sqrt{10010}=\color{red}{-0.166652445440174}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4308841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the area of this closed curve $\alpha(t) = (\sin(2t), 2\sin t)$ on the plane? The closed curve $\alpha(t) = (\sin(2t), 2\sin t)$ with $0 \leq t \leq \pi$.
Find the area of the $A$ enclosed by the $\alpha$ on the $\mathbb{R}$.
I took the formula $\int_{0}^{\pi} \frac{1}{2}{f(\theta)}^2 d\theta$ for the $r = f(\theta)$.
In my thought, $\alpha(t) =(x(t), y(t)) = (\sin(2t), 2\sin t)$. So the $r^2(t) = x^2(t) + y^2(t) = \sin^2(2t) + 4\sin^2t$
My answer is $\int_{0}^{\pi} \frac{1}{2}{f(t)}^2 dt = \int_{0}^{\pi} \frac{1}{2} r^2 dt = \int_{0}^{\pi} \frac{1}{2} (\sin^2(2t) + 4\sin^2t) dt = \frac{5\pi}{4}$
But the answer was not my thing. Actually the answer was $\frac{8}{3}$. So my question is What the point did I have a mistake? Why is the formula $\int_{0}^{\pi} \frac{1}{2}{f(\theta)}^2 d\theta$ not hold for this case?
| In polar coordinates, $x = r \cos\theta, y = r\sin\theta$ where $\theta$ is the angle with the positive x coordinate axis. That is not the case here and hence the formula you write cannot be applied as is. You can check it yourself. For the $r$ you obtained, does $r \cos\theta ~$ give you $x$?
$y = 2\sin t$
$x = \sin2t \implies x^2 = 4 \sin^2 t (1 - \sin^2 t)$
$ ~ \displaystyle x^2 = y^2 - \frac{y^4}{4} ~ $ is the equation of the curve in cartesian coordinates.
That leads to $\displaystyle - \sqrt{y^2 - \frac{y^4}{4}} \leq x \leq \sqrt{y^2 - \frac{y^4}{4}}$
So the integral to find area can be written as,
$ 2 \displaystyle \int_0^2 \sqrt{y^2 - \frac{y^4}{4}} ~ dy$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4309485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Calculating $\lim_{x \to 0}\frac{x \tan2x-2x\tan x}{(1-\cos2x)^2}$ I'm a bit confused regarding this question. I've been trying to solve it and have gotten to the same answer ($3/8$) thrice now. I have no idea where I'm going wrong and would really appreciate some help figuring it out. Here's my solution:
$$\lim_{x \to 0}\frac{x\tan2x-2x\tan x}{(1-\cos2x)^2}$$
Applying L'Hospital's rule,
$$\lim_{x \to 0}\frac{\tan2x+x(\sec^22x)(2)-2\tan x-2x\sec^2x}{2(1-\cos2x)(\sin2x)(2)}$$
Dividing numerator and denominator by $2x$,
$$\begin{align*}
\lim_{x \to 0}&\frac{(\frac{\tan2x}{2x})+(\frac{2x(\sec^22x)}{2x})-(\frac{2\tan x}{2x})-(\frac{2x\sec^2x}{2x})}{4(1-\cos2x)\frac{(\sin2x)}{2x}}\\
&=\lim_{x \to 0}\frac{1+\sec^22x-1-\sec^2x}{4(1-\cos2x)}\\
&=\lim_{x \to 0}\frac{\sec^22x-\sec^2x}{4(1-\cos2x)}
\end{align*}$$
Applying L'Hôpital's rule (again),
$$\lim_{x \to 0}\frac{(2)(\sec2x)(\sec2x)(\tan2x)(2)-(2)(\sec x)(\sec x\tan x)}{4(\sin2x)(2)}$$
$$=\lim_{x \to 0}\frac{(2)(\sec^22x)(\tan2x)-(\sec^2x)(\tan x)}{4(\sin2x)}$$
Dividing numerator and denominator by $2x$,
$$\begin{align*}
&=\lim_{x \to 0}\frac{(2)(\sec^22x)(\frac{\tan2x}{2x})-(\sec^2x)(\frac{\tan x}{2x})}{4(\frac{\sin2x}{2x})}\\
&=\lim_{x \to 0}\frac{(2)(\sec^22x)(\frac{\tan2x}{2x})-(\sec^2x)(\frac{\tan x}{x})(\frac{1}{2})}{4(\frac{\sin2x}{2x})}\\
&=\lim_{x \to 0}\frac{(2)(\sec^22x)(1)-(\sec^2x)(1)(\frac{1}{2})}{4(1)}\\
&=\frac{(2)(\sec^20)(1)-(\sec^20)(1)(\frac{1}{2})}{4(1)}\\
&=\frac{(2)(1)(1)-(1)(1)(\frac{1}{2})}{4(1)}\\
&=\frac{2-\frac{1}{2}}{4}\\
&=\frac{\frac{3}{2}}{4}\\
&=\frac{3}{8}.
\end{align*}$$
| I am now going to find the limit without L’Hospital Rule.
$$\begin{array}{l}
\displaystyle \quad \lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}}\\
=\displaystyle \lim _{x \rightarrow 0} \frac{\frac{x \sin 2 x}{\cos 2 x}-\frac{2 x \sin x}{\cos x}}{(1-\cos 2 x)^{2}}\\
=\displaystyle \lim _{x \rightarrow 0} \frac{x \sin 2 x \cos x-2 x \sin x \cos 2 x}{\cos 2 x \cos x(1-\cos 2 x)^{2}}\\
=\displaystyle \lim _{x \rightarrow 0} \frac{x\left(\sin 2 x \cos x-\sin x \cos 2 x\right)-x \sin x \cos 2 x}{\tan 2 x \cos x(1-\cos 2 x)^{2}}\\
=\displaystyle \lim _{x \rightarrow 0} \frac{x \sin x(1-\cos 2 x)}{\cos 2 x \cos x(1-\cos 2 x)^{2}}\\
=\displaystyle \left(\lim _{x \rightarrow 0} \frac{x \sin x}{1-\cos 2 x}\right)\left(\lim _{x \rightarrow 0} \frac{1}{\cos 2 x \cos x}\right)\\
=\displaystyle \lim _{x \rightarrow 0} \frac{x \sin x}{2 \sin ^{2} x}\\
=\displaystyle \frac{1}{2} \lim _{x \rightarrow 0} \frac{x}{\sin x}\\
=\displaystyle \frac{1}{2}
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Evaluating $\sum_{cyc} \frac{a^4}{(a-b)(a-c)}$, where $a=-\sqrt3+\sqrt5+\sqrt7$ , $b=\sqrt3-\sqrt5+\sqrt7$, $c=\sqrt3+\sqrt5-\sqrt7$
Let $a=-\sqrt{3}+\sqrt{5}+\sqrt{7}$ , $b=\sqrt{3}-\sqrt{5}+\sqrt{7}$, $c=\sqrt{3}+\sqrt{5}-\sqrt{7}$. Evaluate:
$$\sum_{cyc} \frac{a^4}{(a-b)(a-c)}$$
What I have tried so far is writing the denominators in terms of $(a+b+c)$ but that didn't help too much. I have also tried establishing a lower bound on the sum, but that won't work either.
Any help is appreciated!
| HINT:
Check first that the algebraic expression in $a$, $b$, $c$ simplifies to
$$ a^2 + b^2 + c^2 + a b + a c + b c$$
(there are many ways to get that, including just calculations)
Now reduce it to
$$\frac{1}{2}((a+b)^2 + (a+c)^2 + (b+c)^2)$$
Now it should be simple. Note that the expression in $a$, $b$, $c$ is a Schur function in $a$, $b$, $c$, being $s_{(2,0,0)}(a,b,c)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4319845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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What is $ \lim_{x \rightarrow 0}\left(\frac{1}{\ln\cos (x)}+\frac{2}{\sin ^{2}(x)}\right) $? The answer of the following limit:
$$
\lim_{x \rightarrow 0}\left(\frac{1}{\ln \cos (x)}+\frac{2}{\sin ^{2}(x)}\right)
$$
is 1 by Wolfram Alpha.
But I tried to find it and I got $2/3$ :
My approach :
$1)$
$
\ln(\cos x)=\ln\left(1-\frac{x^{2}}{2}+o\left(x^{3}\right)\right)=-\frac{x^{2}}{2}+o\left(x^{3}\right)
$
$2)$
$
\sin ^{2}(x)=\left(x-\frac{x^{3}}{3!}+o\left(x^{3}\right)\right)^{2}=x^{2}-\frac{x^{4}}{3}+o\left(x^{4}\right)
$
$3)$
$\begin{aligned} \frac{1}{-\frac{x^{2}}{2}+o\left(x^{3}\right)}+\frac{2}{x^{2}-\frac{x^{4}}{3}+o\left(x^{4}\right)}=\frac{-x^{2}+x^{2}-\frac{x^{4}}{3}+o\left(x^{3}\right)}{-\frac{x^{4}}{2}+o\left(x^{5}\right)}=\frac{-\frac{1}{3}+o\left(x^{3}\right)}{-\frac{1}{2}+o\left(x^{5}\right)} \end{aligned}$
$4)$
$\lim _{x \rightarrow 0} \frac{-\frac{1}{3}+o\left(x^{3}\right)}{-\frac{1}{2}+o\left(x^{5}\right)}=\frac{-\frac{1}{3}}{-\frac{1}{2}}=\frac{2}{3}$
So where is the mistake in my approach?
Note: $o$ denotes the little-o notation
Edit : I've understood where's my mistake is, but another question popped up reading the answers which is : does $o(1/x)$ tends to zero as x tends to zero?
| If we put $$y=\sin^2(x)$$
the limit becomes
$$\lim_{y\to 0}2\frac{y+\ln(1-y)}{y\ln(1-y)}$$
$$=\lim_{y\to 0}-2\frac{y+\ln(1-y)}{y^2}$$
$$=\lim_{y\to 0}-2\frac{y-y-\frac{y^2}{2}(1+o(1))}{y^2}$$
$$=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $x\ y\ dx=(y^3+x^2y+x^2)\ dy$ differential equation. Solve $x\ y\ dx=(y^3+x^2y+x^2)\ dy$ differential equation.
What I tried so far.
$M(x,y)=x\ y$ $N(x,y)=y^3+x^2y+x^2$
$\frac{\partial M}{\partial y}=x$ $\frac{\partial N}{\partial x}=2x+2xy$
They are not equal so we need integrating factor.
$$\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x} }{N}=\frac{2xy-x}{y^3+x^2y+x^2}$$
so $$\mu(x)=e^{-{\huge \int}{\frac{(2xy-x)dx}{y^3+x^2+x^2y}}}$$
from here I can't continue.
EDIT.
I am getting $$\mu=|(1+y)\ x^2+y^3|^\frac{1-2y}{2+2y}$$
| $$xydx=(y^3+x^2y+x^2)dy$$
$$x(ydx-xdy)=y(y^2+x^2)dy$$
$$(ydx-xdy)=\dfrac yx(y^2+x^2)dy$$
Divide by $y^2$:
$$d \left (\dfrac {x}y \right)=\dfrac y x \left(1+\dfrac {x^2}{y^2} \right)dy$$
This is separable:
$$\dfrac {du}{1+u}=2dy$$
With $u=\left (\dfrac xy\right)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4326716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the ordinary differential equation.
Solve the equation: $$x(2y+1)dx=y(x^2-3x+2)dy$$
To solve this, I first move $y(x^2-3x+2)dy$ to the LHS, then divide both sides by $(2 y + 1)(x^2 - 3 x + 2)$, the equation becomes:
$$\dfrac{x}{x^2-3x+2}dx-\dfrac{y}{2y+1}dy=0$$
After that, I integrate both sides: $$\int\dfrac{x}{x^2-3x+2}dx-\int\dfrac{y}{2y+1}dy=C$$
Which gives: $$2\ln|x-2|-\ln|x-1|+\dfrac{\ln|2y+1|-2y-1}{4}=C$$
Did I do this correctly?
| You want to get all the y stuff to one side, and all the x stuff to the other. These are called "separable equations" for this reason.
$x(2y+1)dx=y(x^2-3x+2)dy$ $\Rightarrow$
$\frac{x}{x^2-3x+2}dx=\frac{y}{2y-1}dy$ $\Rightarrow$
$\int\frac{x}{x^2-3x+2}dx=\int\frac{y}{2y-1}dy$
Now we integrate.
The left side looks like partial fraction decomposition and the right side looks like a u-sub with $u=2y-1$.
| {
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"url": "https://math.stackexchange.com/questions/4327237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $x$, $y$, $z$ be three positive reals such that $x+y+z=\sqrt{10+\sqrt{19}}$ and $\frac1x+\frac1y+\frac1z=\sqrt{10-\sqrt{19}}$ ...
Let $x,y,$ and $z$ be three positive real numbers such that
$x+y+z=\sqrt{10+\sqrt{19}}$ and
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\sqrt{10-\sqrt{19}}$. If
$a=x^2+y^2+z^2$ and $b=a+\frac{9}{a}$, find $b$.
Solution:
Just take the product and you get $\sum_{cyc}\frac{x}{y}+\frac{y}{x}=6$ after simplifying. Note that by AM-GM on LHS we have $\sum_{cyc}\frac{x}{y}+\frac{y}{x}\ge 3(2)=6$, and because equality occurs in AM-GM iff all the variables are equal we find that $\frac{x}{y}=\frac{y}{x}\Leftrightarrow x=y$, and cyclically. Hence $x=y=z$ so from the first equation we get $x=\frac{\sqrt{10+\sqrt{19}}}{3}$. Plugging into $a$ gives $a=3x^2=\frac{10+\sqrt{19}}{3}$, and plugging into $b$ yields $b=a+\frac{9}{a}=\frac{10+\sqrt{19}}{3}+\frac{27}{10+\sqrt{19}}=\frac{20}{3}$.
Question: Why is $x = y = z $ in this solution?
| Let's write $\sum_{cyc} \frac{x}{y}+\frac{y}{x}$ explicitly as $\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z}$ and denote this by $A$. Then the arithmetic mean of these $6$ numbers is $\frac{A}{6}$ and the geometric mean of these $6$ numbers is $\sqrt[6]{1}=1$. By AM-GM we have $\frac{A}{6}\geq 1$ with equality if and only if all $6$ numbers are equal, that is, $A\geq 6$ with equality if and only if $\frac{x}{y}=\frac{y}{x}=\frac{y}{z}=\ldots$. Since it was shown before that indeed $A=6$ then all $6$ numbers must be equal and in particular $\frac{x}{y}=\frac{y}{x}$ therefore $x=y$, and similarly for $y=z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing $f(kx,ky)\geq f(x,y)$ We have a function $$f(x,y)={y\choose x}$$ for $1<x<y$. We are trying to show that for a positive integer $k$, $f(kx,ky)\geq f(x,y)$.
So what I did so far was to write
$$\frac{ky(ky-1)\cdots(ky-kx+1)(ky-kx)!}{(ky-kx)!(kx)!} \geq \frac{y(y-1)\cdots(y-x+1)(y-x)!}{(y-x)!x!}$$
So we get
$$\frac{ky(ky-1)\cdots(ky-kx+1)}{(kx)!} \geq \frac{y(y-1)\cdots(y-x+1)}{x!}$$
Then I got stuck because every term on the numerator of the left hand side is greater than or equal to that of its right hand side but same goes for the denominator!
Could you anyone please give me a hint on what to do next? Is this even the right direction?
| Introduce a dummy index $0 \leq j \leq x-1$ to express the right hand side as
$$\begin{align}
\binom{y}x &=\frac{y (y-1)(y-2) \cdots (y-x+2)(y-x+1)}{ x(x-1)(x-2)\cdots 2\cdot1} \\
&= \prod_{j=0}^{x-1} \frac{y-j}{x-j} \\
&=\frac{y-0}{x-0}\frac{y-1}{x-1}\frac{y-2}{x-2}\cdots\frac{y-j}{x-j}\cdots\frac{y-(x-2)}{x-(x-2)} \cdot \frac{y-(x-1)}{x-(x-1)}
\end{align}$$
Similarly examine the left hand side, but only pay attention to terms at every $k$ multiples.
$$\begin{align}
\binom{ky}{kx} &=\frac{ \mathbf{ky} (ky-1)(ky-2) \cdots \mathbf{(ky-k)} \cdots \mathbf{(ky-2k)} \cdots (ky-kx+2)(ky-kx+1)}{ \mathbf{kx}(kx-1)(kx-2) \cdots \mathbf{(kx-k)} \cdots \mathbf{(kx-2k)} \cdots 2\cdot1} \\
&=\mathbf{\frac{ky \! - \! 0}{kx \! - \! 0}} \cdot \frac{ky \! - \! 1}{kx \! - \! 1} \cdot \frac{ky \! - \! 2}{kx \! - \! 2} \cdots \frac{ky \! - \! k \! + \! 1}{kx \! - \! k \! + \! 1} \cdot \mathbf{\frac{ky \! - \! k}{kx \! - \! k}} \cdot \frac{ky \! - \! k \! - \! 1}{kx \! - \! k \! - \! 1} \cdots \\
& \quad \cdots \frac{ky \! - \! 2k \! + \! 1}{kx \! - \! 2k \! + \! 1} \cdot \mathbf{\frac{ky \! - \! 2k}{kx \! - \! 2k}} \cdot \frac{ky \! - \! 2k \! - \! 1}{kx \! - \! 2k \! - \! 1} \cdots \frac{ky \! - \! jk \! + \! 1}{kx \! - \! jk \! + \! 1} \cdot \mathbf{\frac{ky \! - \! jk}{kx \! - \! jk}} \cdot \frac{ky \! - \! jk \! - \! 1}{kx \! - \! jk \! - \! 1} \cdots \\
&\qquad\qquad \small\text{(note that $j$ runs from $0$ to $x-1$)} \vphantom{\frac12}\\
& \quad\cdots \frac{ky-(x-1)k+1}{kx-(x-1)k+1} \cdot \mathbf{\frac{ky-(x-1)k}{kx-(x-1)k}} \cdot \frac{ky-(x-1)k-1}{kx-(x-1)k-1} \cdots \\
& \quad \cdots \frac{ky- kx + 2 }{kx - (kx-2)} \cdot \frac{ky- kx + 1}{kx - (kx-1)}
\end{align}$$
At each $j$, the term highlighted in boldface matches that on the right hand side (the left "covers" the right).
$$\frac{ky}{kx}=\frac{y}x~,\quad \frac{ky-k}{kx-k}=\frac{y-1}{x-1}~,\quad \frac{ky-2k}{kx-2k}=\frac{y-2}{x-2}~\ldots \\
\ldots,~\frac{ky -jk}{kx-jk}=\frac{y-j}{x-j}~,\ldots,~\frac{ky -(x-1)k}{kx-(x-1)k}=\frac{y-(x-1)}{x-(x-1)} $$
At the same time, since $y > x$, all the other (non-bold) terms on the left hand side are strictly greater than one: $(ky - \Delta)/(kx - \Delta) > 1$. Consequently, we have
$$\begin{align}
& \mathbf{\frac{ky-0}{kx-0}} \cdot \frac{ky-1}{kx-1} \cdot \frac{ky-2}{kx-2} \cdots \frac{ky-k+1}{kx-k+1} & &> \mathbf{\frac{ky-0}{kx-0}} = \frac{y}x \\
&\mathbf{\frac{ky-k}{kx-k}} \cdot \frac{ky-k-1}{kx-k-1} \cdots \frac{ky-2k+1}{kx-2k+1} & & > \mathbf{\frac{ky-k}{kx-k}} = \frac{y-1}{x-1} \\
& \mathbf{\frac{ky-2k}{kx-2k}} \cdot \frac{ky-2k-1}{kx-2k-1} \cdots & &> \mathbf{\frac{ky-2k}{kx-2k}} = \frac{y-2}{x-2} \\
&\qquad \vdots & &\qquad \vdots \\
&\mathbf{\frac{ky-jk}{kx-jk}} \cdot \frac{ky-jk-1}{kx-jk-1} \cdots & &> \mathbf{\frac{ky-jk}{kx-jk}} = \frac{y-j}{x-j} \\
&\qquad \vdots & &\qquad \vdots \\
&\mathbf{\frac{ky-(x-1)k}{kx-(x-1)k}} \cdot \frac{ky-(x-1)k-1}{kx-(x-1)k-1} \cdots & &> \mathbf{\frac{ky-(x-1)k}{kx-(x-1)k}} = \frac{y-x+1}{x-(x-1)}
\end{align}$$
As a product of all these fractions, the left hand side $\displaystyle\binom{ky}{kx}$ is strictly larger than the right hand side $\displaystyle\binom{y}x$, where equality holds only when $k=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $~ {a,b,c \ge 0 ~ , ~ a+b+c=1 } ~ $ then prove $~ {\sum\limits_{cyc} \sqrt{a+b^2 } \geqslant 2 } $
Given $~ a,b,c \ge 0 ~ , ~ a+b+c=1 ~ $, prove that
$$\sum\limits_{\mathrm{cyc}} \sqrt{a+b^2 } := \sqrt{a + b^2} + \sqrt{b + c^2} + \sqrt{c + a^2}\ge 2.$$
Here's what I've tried : $\sum\limits_{cyc} \sqrt{a+b^2 } = \sum\limits_{cyc} \sqrt{\sqrt a^2 +b^2 }=\sum\limits_{cyc}\sqrt{\frac{1}{2}(\sqrt a^2 +b^2 )(1^2 +1^2 ) } \ge \\ \ge \sum\limits_{cyc} \sqrt{\frac{1}{2} (\sqrt{a} +b )^2 }=\frac{1}{\sqrt 2} (1+ \sqrt a +\sqrt b + \sqrt c ). $
Any ideas how to proceed and am I on the right track ?
| COMMENT.- A method to solve these inequalities graphically. Leaving fixed $a$, we have to study the relative position of the line $L: x + y = 1-a$ with respect to the curve
$$\Gamma: \sqrt{a+x^2}+\sqrt{x+y^2}+\sqrt{y+a^2}=2$$
where $0\lt a, x, y \lt1$. which is decreasing in the first quadrant because $\dfrac{dy}{ dx}\lt 0$.
$L$ is tangent to $\Gamma$ at the point $(\frac13,\frac13)$ and it is the only point where it touches the curve ($L$ cannot cut the curve in the first quadrant because if it does then the inequality is false in a certain interval $(x_1, x_2)$ , in other words the line $L$ should be always in top of $\Gamma$ , at most it can be tangent to $\Gamma$). The calculations can be hard in some cases. Anyway this is what happen for all $a$ and it is easily verified graphically.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove: $\frac{\frac{1}{a}+b}{\sqrt{\frac{1}{a}+a}}+\frac{\frac{1}{b}+c}{\sqrt{\frac{1}{b}+b}}+\frac{\frac{1}{c}+a}{\sqrt{\frac{1}{c}+c}}\ge3\sqrt{2}$
Let $a,b,c>0$. Prove that: $$\frac{\frac{1}{a}+b}{\sqrt{\frac{1}{a}+a}}+\frac{\frac{1}{b}+c}{\sqrt{\frac{1}{b}+b}}+\frac{\frac{1}{c}+a}{\sqrt{\frac{1}{c}+c}}\ge3\sqrt{2}$$
Anyone help me a hint to solve above problem?
I tried by AM-GM without success: $$LHS\ge3\sqrt[3]{\frac{\left(\frac{1}{a}+b\right)\left(\frac{1}{b}+c\right)\left(\frac{1}{c}+a\right)}{\sqrt{\left(\frac{1}{a}+a\right)\left(\frac{1}{b}+b\right)\left(\frac{1}{c}+c\right)}}}$$
The rest is proving: $$\left(\frac{1}{a}+b\right)\left(\frac{1}{b}+c\right)\left(\frac{1}{c}+a\right)\ge2\sqrt{2\left(\frac{1}{a}+a\right)\left(\frac{1}{b}+b\right)\left(\frac{1}{c}+c\right)}$$ which is not true by a=60,b=0.02 and c=0.9
Thank you for your help!
| Using Radon's inequality we need to show :
$$\frac{\left(\left(\frac{1}{a}+b\right)^{\frac{2}{3}}+\left(\frac{1}{b}+c\right)^{\frac{2}{3}}+\left(\frac{1}{c}+a\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}}{\sqrt{a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}}-3\sqrt{2}\geq 0$$
Can you end now ?
Follow up
Let $x_i>0$, $1\leq i\leq 3=n$, such that $x_1=x_{4}$ then we have :
$$\sum_{i=1}^{n}\frac{\frac{1}{x_i}+x_{i+1}}{\sqrt{\frac{1}{x_i}+x_i}}\geq 3\sqrt{2}\tag{I}$$
At first glance it seems that Am-Gm is too weak so as above I use Radon's inequality (see vivid edit) we need to show :
$$\frac{\left(\sum_{i=1}^{n}\left(\frac{1}{x_i}+x_{i+1}\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}}{\sqrt{\sum_{i=1}^{n}\left(\frac{1}{x_i}+x_i\right)}}\geq 3\sqrt{2}$$
Then I used Minkowski's inequality with $p=\frac{2}{3}$ we need to show :
$$\frac{\left(\sum_{i=1}^{n}\left(\frac{1}{x_i}\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}+\left(\sum_{i=1}^{n}\left(x_{i}\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}}{\sqrt{\sum_{i=1}^{n}\left(\frac{1}{x_i}+x_i\right)}}\geq 3\sqrt{2}$$
Edit :
We can apply Jensen's inequality and we need to show :
$$\frac{2^{-\frac{1}{2}}\left(\sum_{i=1}^{n}\left(\frac{1}{x_{i}}\right)^{\frac{2}{3}}+\sum_{i=1}^{n}\left(x_{i}\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}}{\sqrt{\sum_{i=1}^{n}\left(\frac{1}{x_{i}}+x_{i}\right)}}-3\sqrt{2}\geq 0$$
Idea to conclude for the case $n=3$:
the function $x>0$ and $a,b\geq 4$:
$$f(x)=\left(x^{\frac{2}{3}}+\frac{1}{x^{\frac{2}{3}}}+a\right)^{\frac{3}{2}}-\left(x+\frac{1}{x}+b\right)^{\frac{1}{2}}\cdot6$$
We have :
$$f(x)\geq f(1)$$
Using this lemma a judicious number of times the conclusion follow for the case $n=3$.
| {
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"url": "https://math.stackexchange.com/questions/4335241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Eliminate $\theta$ from $\sin3\theta=a\cos\theta$ and $\cos3\theta=b\sin\theta$
Eliminate $\theta$ from $$\sin3\theta=a\cos\theta$$ $$\cos3\theta=b\sin\theta$$
I came to this point while trying to solve this problem: Eliminating $\theta$ from trigonometric system (Remark: Symbols differ from the original question)
We can find $$b+a=\frac{2\cos2\theta}{\sin2\theta}$$ $$b-a=\frac{2\cos4\theta}{\sin2\theta}$$
Though at the moment I can't imagine how to arrange these to get a proper relation between $a$ and $b$. We can replace $\sin$ and $\cos$ with $\tan$ and then proceed, but it is cumbersome.
I suspect that there should be a clever way to end this, giving a beautiful answer. For reference, like in this question, a specific method gives the desired answer.$\leftarrow\small \text{(not strictly relevant)}$
So, what is the happy ending of this problem?
| I used WolframAlpha to get the following relation between $a$, $b$
$$4 - 3 a^2 - 2 a b + a^3 b - 3 b^2 + 2 a^2 b^2 + a b^3=0$$
which is equivalent to
$$(a-b)^2 = \frac{ ((a+b)^2 -4)^2}{(a+b)^2 + 4}$$
or
$$v^2 = \frac{ (u^2 -1)^2}{u^2 + 1}$$
where $u=\frac{a+b}{2}$ and $v=\frac{a-b}{2}$.
The contour curve is the union of two graphs
$$v = \pm \frac{ u^2 -1}{\sqrt{u^2 + 1}}$$
$\bf{Added:}$ With trigonometry, one checks that for
$$a = \frac{\sin 3 \theta}{\cos \theta}\\
b= \frac{\cos 3 \theta}{\sin \theta}$$
we have
$$\frac{a+b}{2} = \cot 2 \theta \\
\frac{a b -1}{2} = \cos 4 \theta$$
and from here we get a relation between $a$, $b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
find cokernel of a matrix
Let $$T=\begin{pmatrix} 1&2&3 \\ 5&4&2\\1&2&3\\2&1&4
\end{pmatrix}.$$ Find the cokernel of $T$.
I have tried to find Smith Normal Form of this matrix $S=\begin{pmatrix} 1&0&0 \\
0&1&0\\0&0&27\\0&0&0\end{pmatrix}$
so the cokernel should be $\mathbb Z/1\oplus\mathbb Z/1\oplus\mathbb Z/27\oplus\mathbb Z$ but we can wrote it as $\mathbb Z/27\oplus\mathbb Z$ is it right?
| All of your work is correct, well done.
Regarding the Smith normal form: we have $UTV = S$, with the $S$ that you wrote and
$$
U = \left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
2 & 0 & 0 & -1 \\
-35 & -1 & 0 & 20 \\
-1 & 0 & 1 & 0 \\
\end{array}
\right), \quad
V = \left(
\begin{array}{ccc}
1 & -4 & -5 \\
0 & -1 & -2 \\
0 & 2 & 3 \\
\end{array}
\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4338152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Show that $\int_{0}^{n} \left (1-\frac{x}{n} \right ) ^n \ln(x) dx = \frac{n}{n+1} \left (\ln(n) - 1 - 1/2 -...- 1/{(n+1)} \right )$ The problem is stated as:
Show that $\int_{0}^{n} \left (1-\frac{x}{n} \right ) ^n \ln(x) dx = \frac{n}{n+1} \left (\ln(n) - 1 - 1/2 -...- 1/{(n+1)} \right )$
My attempt
First of all, we make the substitution $1-\frac{x}{n} = t$, we then have that the integral can be rewritten as:
$\int_{1}^{0} -n t^n \ln(n(1-t)) dt = \int_{0}^{1} n t^n \ln(n(1-t)) dt$
Using logarithmic laws, we can split the integral into two seperate ones as follows:
$\int_{0}^{1} n t^n \ln(n(1-t)) dt = \int_{0}^{1} n t^n \ln(n) dt + \int_{0}^{1} n t^n \ln(1-t) dt$
We calculate each integral from the sum above:
$ I_1 := \int_{0}^{1} n t^n \ln(n) dt = \frac{n}{n+1}\ln(n)$
$ I_2 := \int_{0}^{1} n t^n \ln(1-t) dt = -n\int_{0}^{1} t^n \sum_{k=1}^{\infty}\frac{t^k}{k} dt$
Since the radius of convergence of $\sum_{k=1}^{\infty}\frac{t^k}{k}$ is 1, and we are integrating from $0$ to $1$, we can interchange the order of limit operations. Meaning, we can calculate the integral first.
$ I_2 = -n\sum_{k=1}^{\infty}\int_{0}^{1}\frac{t^{(n+k)}}{k} dt = -\sum_{k=1}^{\infty} \frac{n}{k(n+k+1)} = \frac{-n}{n+1} \sum_{k=1}^{\infty} \frac{n+1}{k(n+k+1)}$
Using partial fraction decomposition, we have that $I_2$ can be written as:
$\frac{-n}{n+1}\sum_{k=1}^{\infty} \frac{n+1}{k(n+k+1)} = \frac{-n}{n+1} \sum_{k=1}^{\infty} \frac{1}{k} + \frac{n}{n+1}\sum_{k=1}^{\infty} \frac{1}{n+k+1}$
Putting it all together we get:
$I_1 + I_2 = \frac{n}{n+1} \left ( \ln(n) - \sum_{k=1}^{\infty} \frac{1}{k} + \sum_{k=1}^{\infty} \frac{1}{n+k+1} \right )$
Which is indeed close the the result sought, however, I don't really know what to do with the last sums, and why I did wrong in choosing $\infty$ as an upper limit in the summation. I see that the sum of $1/k's$ diverge, but how can I avoid this?
Thank you for any help that could help me complete the last step of this problem.
| Hint: $\sum_{k=1}^{\infty} \frac{1}{k}$ and $ \sum_{k=1}^{\infty} \frac{1}{n+k+1} $ are both infinity so you cannot write these sums separately. Instead, you should write $\lim_{N \to \infty} [-\sum_{k=1}^{N} \frac{1}{k} +\sum_{k=1}^{N} \frac{1}{n+k+1}]$.
Now $[-\sum_{k=1}^{N} \frac{1}{k} +\sum_{k=1}^{N} \frac{1}{n+k+1}]$ simplifies to $\frac 1 {N+1}+\frac 1 {N+2}+..+\frac 1 {n+N+1}-(1+\frac 1 2+\frac 1 3+...+\frac 1 {n+1})$ (for $N>n+1$). Note that $\frac 1 {N+1}+\frac 1 {N+2}+..+\frac 1 {n+N+1} \to 0$ as $N \to \infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the volume of the solid bounded by $z^2=xy$; $x+y=a$; $x+y=b$ $(0Find the volume of the solid bounded by $z^2=xy$; $x+y=a$; $x+y=b$ $(0<a<b)$ by applying variable substitution. $xy>0$ and $x,y$ can't be negative at the same time so $x>0;y>0.$ Now, $a - x < y < b - x$ and we get that $a < x < b$. So the integral to compute is $\int_{a}^{b}\int_{a-x}^{b-x}{\sqrt{xy}\,dxdy}$. Now this integral, as requested by the problem, should be solved by applying variable substitution. But the answer in the textbook is $\dfrac{\pi(b^3-a^3)}{12}$. I don't get where this $\pi$ comes from. The only way I know to obtain such a result is by changing to polar coordinate system, but I don't see any way to obtain relations between $a,b$, $a-x$, $b-x$ and the new coordinate system especially because in none of these equations are $x$ or $y$ squared. Am I doing some mistake or how does this $\pi$ appear in the answer? Thanks in advance.
| The quadric $z^2 = x y $ corresponds to the quadratic form
$ r^T Q r = 0 $
where $r = [x, y, z]^T$ and
$Q = \begin{bmatrix} 0 && -1/2 && 0 \\ -1/2 && 0 && 0 \\ 0 && 0 && 1 \end{bmatrix} $
Diagonalizing $Q$ it becomes $Q = R D R^T $ where
$ D = \begin{bmatrix} -0.5 && 0 && 0 \\ 0 && 0.5 && 0 \\ 0 && 0 && 1 \end{bmatrix} $
$ R = \begin{bmatrix} \dfrac{1}{\sqrt{2}} && - \dfrac{1}{\sqrt{2}} && 0 \\ \dfrac{1}{\sqrt{2}} && \dfrac{1}{\sqrt{2}} && 0 \\ 0 && 0 && 1 \end{bmatrix} $
Define the change of variable as follows:
$ w = R^T r $ i.e. $r = R w $. So $x + y = [1, 1, 0] r = [1, 1, 0] R w = [\sqrt{2}, 0, 0 ] w $
The equation of the quadric is
$ 0.5 w_1^2 = 0.5 w_2^2 + w_3^2 $
which simplifies to
$ w_1^2 = w_2^2 + 2 w_3^2 $
Now the equation of the boundary $r^T Q r = 0 $ is a cone whose axis is along the the first column of $R$, so in the changed variable it is a cone whose axis is the $w_1$ axis. The limits of $w_1$ are from $\dfrac{a}{\sqrt{2}} $ to $\dfrac{b}{\sqrt{2}} $. If we take the volume between $w_1 = 0$ and $w_1=c$, then the volume is just one third of the area of base (in the $w_2, w_3$ plane) times the height $c$; the area is $ \pi \left( \dfrac{c^2}{\sqrt{2}} \right) $ so that volume is
$ \dfrac{1}{3 \sqrt{2}} \pi c ^ 3 $
Now we want to take the difference of volumes between $c = \dfrac{b}{\sqrt{2}}$ and $c = \dfrac{a}{\sqrt{2}} $ , and this is
$ V = \dfrac{\pi}{3\sqrt{2}} \left( \dfrac{b^3 }{2 \sqrt{2}} - \dfrac{a^2}{2 \sqrt{2}} \right) = \dfrac{ \pi (b^3 - a^3) }{12}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4343289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many ways to deal with the integral $\int \frac{d x}{1-\sin x \cos x}$? Multiplying both numerator and denominator of the integrand by $\sec^2 x$ yields
\begin{aligned}
& \int \frac{d x}{1-\sin x \cos x} \\
=& \int \frac{\sec ^{2} x}{\sec ^{2} x-\tan x} d x \\
=& \int \frac{d(\tan x)}{\tan ^{2} x-\tan x+1} \\
=& 2 \int \frac{d(2 \tan x-1)}{(2 \tan x-1)^{2}+(\sqrt{3})^{2}} \\
=& \frac{2}{\sqrt{3}} \arctan\left(\frac{2 \tan x-1}{\sqrt{3}}\right)+C .
\end{aligned}
Is there any simpler solution? Let me know if you have any, thank you for your attention.
Wish you enjoy the solution!
| another method:
$\sin(x) = \frac{tan(x)}{sec(x)}$ and $\cos(x) = \frac{1}{sec(x)}$
now you have to do u-substitution and complete the square.
| {
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Solving $D=\begin{vmatrix}a&\omega b&\omega^2c\\\omega^2b&c&\omega a\\\omega c&\omega^2a&b\end{vmatrix}$
If $D=\begin{vmatrix}a&\omega b&\omega^2c\\\omega^2b&c&\omega a\\\omega c&\omega^2a&b\end{vmatrix}$ and $D'=\begin{vmatrix}a& b&c\\b&c& a\\\ c&a&b\end{vmatrix}$, where $\omega$ is the non real cube root of unity, then which of the following does not hold:
*
*$1)\ D=0$ if $a+b+c=0$ and $a,b,c$ all distinct.
*$2)\ D'=0$ if $a=b=c$ and $a+b+c\ne0$
*$3)\ D=-D'$
*$4)\ D=D'$
My Attempt:
I am getting $D'=-\frac12(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]$. So, $2)$ holds.
For $D$, I tried the row operation $R_1\to R_1+R_2+R_3$ but couldn't conclude.
I also tried multiplying each row with $\omega$ twice but in vain.
I tried writing the value of $\omega$ in $i$ but couldn't finish again.
Any help?
| Throughout this solution, I shall be making use of the fact that $\omega^3=1$ as $\omega$ is the complex cube root of unity.
Now, consider the determinant
$$D_1=\begin{vmatrix}
1 & 0 & 0\\
0 & \omega & 0\\
0 & 0 & \omega^2
\end{vmatrix} = 1\cdot \omega \cdot \omega^2=1$$
Multiplying the matrix of $D'$ with the matrix of determinant $D_1$ we get,
$$\begin{bmatrix}
a & b\omega & c\omega^2\\
b & c\omega & a\omega^2\\
c & a\omega & b\omega^2
\end{bmatrix}$$
Taking the determinant of that matrix and using the property that $det(A\cdot B)=det(A)\cdot det(B)$ we obtain
$$D'=\begin{vmatrix}
a & b\omega & c\omega^2\\
b & c\omega & a\omega^2\\
c & a\omega & b\omega^2
\end{vmatrix}$$
I shall be multiplying with $\omega^2$ on both sides to get -
$$\omega^2\cdot D'=\omega^2\cdot\begin{vmatrix}
a & b\omega & c\omega^2\\
b & c\omega & a\omega^2\\
c & a\omega & b\omega^2
\end{vmatrix}$$
Now from the property of the determinant that any scalar factor may be multiplied across any one row or column of the determinant, I am multiplying the outside factor with the second row ($R_2$) on the RHS -
$$\omega^2\cdot D'=\begin{vmatrix}
a & b\omega & c\omega^2\\
b\omega^2 & c & a\omega\\
c & a\omega & b\omega^2
\end{vmatrix}$$
Here the property of $\omega^3=1$ was used twice. Now multiplying with $\omega$ on both sides, we get -
$$1\cdot D'=\omega\cdot\begin{vmatrix}
a & b\omega & c\omega^2\\
b\omega^2 & c & a\omega\\
c & a\omega & b\omega^2
\end{vmatrix}$$
Again I shall be multiplying the scalar factor outside with the third row of the determinant ($R_3$) -
$$ D'=\begin{vmatrix}
a & b\omega & c\omega^2\\
b\omega^2 & c & a\omega\\
c\omega & a\omega^2 & b
\end{vmatrix}$$
Again the property $\omega^3=1$ was used. If you look carefully, the determinant on the RHS is $D$. Therefore, we have -
$$\boxed{D=D'}$$
Therefore, the correct answer to the question must be (3).
This question could have been done through direct row expansion of the determinant without too much effort and you could have observed that $D=D'$.
| {
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"question_score": "2",
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} |
Solve $x^5=\frac{133x-78}{133-78x}$ Solve $$x^5=\dfrac{133x-78}{133-78x}$$ We have $D:133-78x\ne0,x\ne\dfrac{133}{78}$. In $x\in D$ the given equation is equivalent to $$x^5(133-78x)=133x-78\\78x^6-133x^5+133x-78=0$$ As $x=0$ is not a solution, we can divide both sides by $x^3\ne0$ to get $$78x^3-133x^2+\dfrac{133}{x^2}-\dfrac{78}{x^3}=0\\78\left(x^3-\dfrac{1}{x^3}\right)-133\left(x^2-\dfrac{1}{x^2}\right)=0$$ I am not able to come up with a subtle substitution. It would have been easier if we had $x^2\color{red}{+}\dfrac{1}{x^2}.$
$x=1$ is an "obvious" solution. What else? Thank you!
| Another obvious solution is $x=-1$. So we can write the equation as, $$(x^2-1)(78x^4+ax^3+bx^2+cx+78)=0$$ and comparing coefficients, we can easily find $a=c=-133$, $b=78$.
Now divide this (second polynomial) by $x^2$ and substitute $t=x+\frac1x$ to get a quadratic in $t$. It's easy to solve then.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Integral of $\int \frac{\sqrt{x^2-1}}{x}dx$
Evaluate the integral of $$\int \frac{\sqrt{x^2-1}}{x}dx$$
Attempt: I've tried taking $x=\sec y$,
$$ \int \frac{\sqrt{\sec^2y-1}}{\sec y}dx $$
How to proceed further?
| The answer is $$\int \frac{\sqrt{x^{2}-1}}{x}{\rm d}x= \sqrt{x^{2}-1}-\arctan\left( \sqrt{x^{2}-1}\right)+C.$$
In the integral perform the change of variable $x=\sec u$,
\begin{align*}
\int\frac{\sqrt{x^{2}-1}}{x}{\rm d}x&= \int \frac{\sqrt{\sec^{2}u-1}}{\sec u}\tan u\sec u{\rm d}u\\
&=\int \tan u \tan u{\rm d}u\\
&=\int \tan^{2} u{\rm d}u\\
&=\int \sec^{2}u -1 {\rm d}u\\
&=\tan(u)-u+C
\end{align*}
Therefore,
\begin{align*} \int \frac{\sqrt{x^{2}-1}}{x}{\rm d}x&=\tan(u)-u+C\\ &= \tan\left( \sec^{-1}(x)\right)-\sec^{-1}(x)+C\\ &=\sqrt{x^{2}-1}-\sec^{-1}(x)+C\\ &=\boxed{\sqrt{x^{2}-1}-\arctan\left( \sqrt{x^{2}-1}\right)+C} \end{align*}
| {
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"source": "stackexchange",
"question_score": "1",
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How to solve $\lfloor x \rfloor + \lfloor \frac{1}{x} \rfloor = 1$? I am stuck with this equation. All I could do is this:
$\lfloor x \rfloor$ = $\lfloor n + m \rfloor$ such that $n \in N$ and $m<1$.
We get:
$\lfloor x \rfloor + \lfloor \frac{1}{x} \rfloor = 1$
$\lfloor n + m \rfloor + \lfloor \frac{1}{n+m} \rfloor = 1$
$n + \lfloor m \rfloor + \lfloor \frac{1}{n+m} \rfloor = 1$
$n + 0 + \lfloor \frac{1}{n+m} \rfloor = 1$
$n + \lfloor \frac{1}{n+m} \rfloor = 1$
From here on I have no idea what to do!
Edit: It is easy to see that any value $1<x<2$ satisfies the equation, but can I find all the solutions?
| We can rule out $x<0$. If $x\geqslant 2$, then $1-\lfloor x \rfloor \leqslant -1$, which is problematic. If $x\leqslant 1/2$, then $\lfloor \frac{1}{x} \rfloor = 1-\lfloor x \rfloor$ is again impossible. For any $x\in (1/2, 2)\setminus \{1\}$ we have
$$ \left\lfloor x \right\rfloor + \left\lfloor\frac{1}{x} \right\rfloor =1. $$
| {
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"answer_count": 4,
"answer_id": 3
} |
ind the largest natural number $n$ for which $50\lfloor x\rfloor-\lfloor x\lfloor x\rfloor \rfloor=100n-27\lceil x\rceil$has a real solution for $x$. Find the largest natural number $n$ for which$$50\lfloor x\rfloor-\lfloor x\lfloor x\rfloor \rfloor=100n-27\lceil x\rceil$$has a real solution for $x$.
I tried taking $x=a+r, 0\le r<1.$
We get $$50a-\lfloor (a+r)a\rfloor= 100n-27(a+1)$$
$$\implies 50a-a^2-\lfloor ra\rfloor= 100n-27(a+1)$$
$$\implies 100n+a^2+\lfloor ra\rfloor-27=77a$$
This is a quadratic in $a,$ so we get $$a^2-77a-27+100n+\lfloor ra\rfloor=0\implies 77^2-4\cdot (27+100n+\cdot\lfloor ra\rfloor )\text{ is square } $$
Hence $$ 4\cdot (27+100n+\lfloor ra\rfloor) \le 77^2$$
So $4\cdot (27+100n+\lfloor ra\rfloor) \le 77^2\implies n\le 58.$
I can't progress after this.
| This answer assumes that your $a$ is an integer.
We get $$50a-\lfloor (a+r)a\rfloor= 100n-27(a+1)$$
Note that $\lceil x\rceil=a+1$ holds only when $x$ is not an integer.
If $x$ is an integer, then since $\lceil x\rceil=a$, one has $$50a-a^2=100n-27a\implies a^2-77a+100n=0$$ Considering the discriminant, $D=(-77)^2-400n\geqslant 0\implies n\leqslant 14$.
If $x$ is not an integer, as you did, since $\lceil x\rceil=a+1$, we have
$$100n+a^2+\lfloor ra\rfloor-27=77a$$
This is a quadratic in $a$
No, it isn't since it has $\lfloor ra\rfloor$.
*
*If $a=0$, then $n=\frac{27}{100}\not\in\mathbb N$.
*If $a\lt 0$, then one has
$$-a^2+77a-100n+27=\lfloor ra\rfloor\geqslant a\implies a^2-76a+100n-27\leqslant 0$$
Here, it is necessary that $(-76)^2-4(100n-27)\geqslant 0$ which implies $n\leqslant 14$.
*If $a\gt 0$, then one has
$$-a^2+77a-100n+27=\lfloor ra\rfloor\geqslant 0\implies a^2-77a+100n-27\leqslant 0$$
Here, it is necessary that $(-77)^2-4(100n-27)\geqslant 0$ which implies $n\leqslant 15$. For $n=15$, one has $36\leqslant a\leqslant 41$. For $a=36$, one has $3=\lfloor 36r\rfloor$, so the equation has a solution $x=36+\frac{1}{12}$.
Therefore, the answer is $\color{red}{n=15}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove derivative of an implicit function The question:
$\sqrt{1+x^2} + \sqrt{1+y^2} = a(x-y)$ prove that $\dfrac{dy}{dx} = \sqrt{\dfrac{1+y^2}{1+x^2}}$
I tried doing it the normal way and got
$\dfrac{dy}{dx} = \dfrac{(a\sqrt{1+x^2}-x)(\sqrt{1+y^2})}{(a\sqrt{1+y^2} + y)(\sqrt{1+x^2})}$
I also tried doing a trigonometry sub to get a relation of $a$ into a independent constant
$x=\tan(\alpha)$
$y=\tan(\beta)$
And got
$$a = \dfrac{\cos(\alpha) + \cos(\beta)}{\sin(\alpha - \beta)}$$ I don't know if this one can be further manipulated.
Is there something I missed? or is there another way of solving it?
| Note that $$a=\frac{\sqrt{1+x^{2}}+\sqrt{1+y^{2}}}{x-y}.$$
so that
$$
\begin{align*}
\frac{a\sqrt{1+x^{2}}-x}{a\sqrt{1+y^2}+y} &=\frac{\frac{(1+x^{2})+\sqrt{1+x^{2}}\sqrt{1+y^{2}}-x^2+xy}{x-y}}{\frac{(1+y^{2})+\sqrt{1+x^{2}}\sqrt{1+y^2}+xy-y^{2}}{x-y}}\\
&=\frac{1+\sqrt{1+x^{2}}\sqrt{1+y^{2}}+xy}{1+\sqrt{1+x^{2}}\sqrt{1+y^{2}}+xy}\\
&=1.
\end{align*}$$
Therefore from your own calculations
$$\frac{dy}{dx}=\frac{(a\sqrt{1+x^{2}}-x)(\sqrt{1+y^{2}})}{(a\sqrt{1+y^{2}}+y)(\sqrt{1+x^{2}})}=\frac{\sqrt{1+y^{2}}}{\sqrt{1+x^{2}}}.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Distributing (not necessarily all) $6$ identical white and $8$ identical black balls to $3$ children
We have $6$ identical white balls and $8$ identical black balls. In how many ways can we distribute the balls to $3$ children so that each gets at least one? We don't necessarily have to distribute all $14$ balls we have.
My thoughts:
There are already similar questions and a generalized answer to one.
In my task, I thought of separating the problem into $2$ cases: when we distribute all $16$ balls to $3$ children and when we don't. For the latter, I thought I could bring an 'auxiliary' child that should also get at least one ball. So, let $A_i=\{\text{ distributions where }i^{\mathrm{th}}\text{ child gets no ball }\}.$ In the first case we're looking for $$\begin{aligned}\left|\bigcap_{i=1}^3 A_i^c\right|&=\left|\left(\bigcup_{i=1}^3A_i\right)^c\right|\\&=\binom{3+6-1}6\binom{3+8-1}8-\left|\bigcup_{i=1}^3 A_i\right| \quad (1)\end{aligned}$$
We can now compute $(1)$ analyzing when $i,1\le i\le 3$ children get no balls, that is $3-i$ of them do, which can be done in $\binom3k\binom{3-i+6-1}6\binom{3-i+8-1}8=\binom{3}i\binom{8-i}6\binom{10-i}6$ ways. Therefore $(1)$ equals $$\begin{aligned}\sum\limits_{i=0}^3(-1)^i \binom3i\binom{3-i+6-1}6\binom{3-i+8-1}8&=\sum\limits_{i=0}^3(-1)^i\binom{3}i\binom{8-i}6\binom{10-i}8\\&=28\cdot45-3\cdot7\cdot9+3\\&=1074\end{aligned}$$
Now, for the second case, analogously, $$\begin{aligned}\left|\bigcap_{i=1}^4A_i^c\right|&=\sum_{i=0}^4(-1)^i\binom4i\binom{9-i}6\binom{11-i}8\\&=84\cdot165-4\cdot28\cdot 45+6\cdot7\cdot 9-4\\&=9194\end{aligned}$$
so my answer is $\#=10268$
Could somebody check this solution?
| This is not as nice as your solution, but I get the same answer.
So each child has $3$ choices: receive only white balls, receive only black balls, or receive both. Represent these conditions as $-1,1,0$ respectively. If we have $x$ $-1$'s $y$ $1$'s and $z$ $0$'s, then we will first subtract give the $x+z$ children $1$ white ball and the $y+z$ children $1$ black ball. We are left with $6-x-z$ white balls and $8-y-z$ black balls to distribute accordingly.
The remaining $6-x-z$ white balls must go to the $x+z$ children who will receive white balls (or to no one). We can define an auxilary child who will receive a nonnegative amount of unused balls. We can then see that there are $\binom{6}{x+z}$ ways to distribute these by stars and bars. Similarly there are $\binom{8}{y+z}$ ways to distribute the black balls.
Because there must be a total of $3$ $-1$'s, $1$'s, and $0$'z we should sum over all $x+y+z=3$. However, there are $\binom{3}{x,y,z}$ ways to arrange these conditions (i.e. assign children which choice they get), so our answer is
$$\sum_{\substack{x+y+z=3\\x,y,z\geq 0}} \binom{3}{x,y,z}\binom{6}{x+z}\binom{8}{y+z}$$
We can change the bounds into
$$=\sum_{z=0}^3 \sum_{\substack{x+y=3-z\\x,y\geq 0}} \binom{3}{x,y,z}\binom{6}{x+z}\binom{8}{y+z}$$
$$=\sum_{z=0}^3 \sum_{y=0}^{3-z} \binom{3}{3-y-z,y,z}\binom{6}{3-y}\binom{8}{y+z}$$
$$=\sum_{z=0}^3 \binom{3}{z}\sum_{y=0}^{3-z} \binom{3-z}{y}\binom{6}{3-y}\binom{8}{y+z}$$
I'm not sure if this has any nicer form, but possibly exponential generating functions could be helpful. However, one can evaluate with a calculator that this sum is $10268$
| {
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Find minimum $|z|$ satisfying $|z + 1/z |= 2$. When I tried this using normal complex inequalities like $|z_{1} - z_{2}| \ge ||z_{1}| - |z_{2}||$. $\sqrt 2 - 1$ came up but the real answer seems to be $(3 - 2\sqrt 2)^{1/2}$. Some online answers on other sites support my answer as well, but I am confused which ones correct. If the latter is correct please explain how.
| $|z^2+1| = 2|z|$. Let $z = a+bi \implies |z^2+1| = |(a+bi)^2+1|=|a^2+2abi-b^2+1|=\sqrt{(a^2-b^2+1)^2+4a^2b^2}=2\sqrt{a^2+b^2}\implies (a^2-b^2+1)^2+4a^2b^2=4a^2+4b^2\implies a^4+b^4+1-2a^2b^2+2a^2-2b^2+4a^2b^2=4a^2+4b^2\implies a^4+b^4+1+2a^2b^2-2a^2-2b^2=4b^2\implies (a^2+b^2-1)^2=4b^2\le 4(a^2+b^2)$. Let $c = a^2+b^2=|z|^2\implies (c-1)^2 \le 4c \implies c^2-2c+1 - 4c \le 0\implies c^2-6c+1\le 0\implies (c-3)^2\le 8\implies |c-3| \le 2\sqrt{2}\implies -2\sqrt{2} \le c - 3\le 2\sqrt{2}\implies c \ge 3 - 2\sqrt{2}$. Thus $|z|_{\text{min}}=\sqrt{c_{\text{min}}}=\sqrt{3-2\sqrt{2}}$. This is achieved when $a = 0, b^2-1 = \pm 2b\implies a = 0, (b\pm1)^2=2\implies a = 0, |b\pm 1| = \sqrt{2}\implies a = 0, b\pm 1=\pm \sqrt{2}\implies a = 0, b = \mp1 \pm \sqrt{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Confusion about the use of principal roots and their use in expressions I'm struggling to derive the quadratic equation using the principal root $(\sqrt{x^2} = |x|).$
Taking $ax^2 + bx + c = 0$ and solving for $x$, I got $$\left(x+\frac{b}{2a}\right)^2 = \frac{b^2-4ac}{4a^2}.$$
Taking the principal root both sides gave me $$\left|x+\frac{b}{2a}\right| = \frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}\\x+\frac{b}{2a} = \frac{±\sqrt{b^2-4ac}}{2|a|}.$$
And there's where I struggle. While for $a \geq 0, |a|=a,$ and it goes just well.
But that's not true if $a<0$, I can´t get it right with $a<0.$
Edit : Changed a typing error into the expression of the discriminant
|
$$x+\frac{b}{2a} = \frac{±\sqrt{b^2-4ac}}{2|a|}.\tag1$$
While for $a \geq 0, |a|=a,$ and it goes just well.
But that's not true if $a<0$, I can´t get it right with $a<0.$
When $a<0,\;|a|=-a\;$ (for example, $|-7|=-(-7)$).
In this case, equation $(1)$ becomes \begin{align}x+\frac{b}{2a} &= \frac{±\sqrt{b^2-4ac}}{-2a}\\x&=\frac{b\pm\sqrt{b^2-4ac}}{-2a}\\&=\frac{-b\mp\sqrt{b^2-4ac}}{2a}\\&=\frac{-b\pm\sqrt{b^2-4ac}}{2a},\end{align} as required.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $1/ac + 1/bc \ge 16 $ if $a, b, c>0$ and $a+b+c=1$ (Seen on Quora at
https://www.quora.com/How-can-I-prove-1-ac-1-bc-16-if-a-b-c-0-and-satisfy-a-b-c-1)
Show that
$1/ac + 1/bc \ge 16
$ if
$a, b, c>0$ and
$a+b+c=1$.
Here is my answer.
I would be interested in any others.
$1/ac + 1/bc \ge 16\\
\iff b+a \ge 16abc\\
\iff 1-c \ge 16abc\\
\iff 1 \ge c(1+16ab)\\
$
If $a=b$,
$1=2a+c$
so
$a=b
=(1-c)/2
$
so
$ab=(1-c)^2/4
$
so
$\begin{array}\\
c(1+16ab)-1
&=c(1+4(1-c)^2)-1\\
&=4c^3-8c^2+5c-1\\
&=(2c-1)^2(c-1)\\
&\le 0
\qquad\text{for } 0 \le c \le 1\\
\end{array}
$
So this is true when $a=b$.
If we can show that
$\dfrac1{a}+\dfrac1{b}
\ge 2\dfrac1{(a+b)/2}$,
we are done since
$2\dfrac1{(a+b)/2}
\ge 16c
$.
This is
$\dfrac1{a}+\dfrac1{b}
\ge 4\dfrac1{a+b}
$.
$\begin{array}\\
\dfrac1{a}+\dfrac1{b}- 4\dfrac1{a+b}
&=\dfrac{b(a+b)+a(a+b)-4ab}{ab(a+b)}\\
&=\dfrac{(a+b)^2-4ab}{ab(a+b)}\\
&=\dfrac{(a-b)^2}{ab(a+b)}\\
&\ge 0\\
\end{array}
$
and we are done.
| Another way (basically, we reduce the given inequality to one variable inequality):
$$
\frac{1}{ac}+\frac{1}{bc}=\frac{a+b}{abc}\ge\frac{4(a+b)}{(a+b)^2c}=\frac{4}{c(1-c)}\ge16.
$$
Comment. As @DanielWainfleet pointed out in the comments, this approach shows that we get equality only if $a=b$ (the first inequality) and $c=1-c$ (the second one), i.e. when $(a,b,c)=(1/4,1/4,1/2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4358492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove: $\frac{1}{a^2+bc}+\frac{1}{b^2+ca}+\frac{1}{c^2+ab}+\frac{a^3+b^3+c^3+9abc}{(a+b)(b+c)(c+a)}\ge4$
Let $a,b,c\ge0: (ab)^2+(bc)^2+(ca)^2=1.$ Prove that: $$\frac{1}{a^2+bc}+\frac{1}{b^2+ca}+\frac{1}{c^2+ab}+\frac{a^3+b^3+c^3+9abc}{(a+b)(b+c)(c+a)}\ge4$$
My approach using well- known inequality:
For all $a,b,c\ge0$ then: $$\frac{1}{a^2+bc}+\frac{1}{b^2+ca}+\frac{1}{c^2+ab}\ge\frac{3}{ab+bc+ca}$$ It implies that we need to prove: $$\sqrt{a^2b^2+b^2c^2+c^2a^2}\frac{3}{ab+bc+ca}+\frac{a^3+b^3+c^3+9abc}{(a+b)(b+c)(c+a)}\ge4$$ But this one is not true for $(a,b,c)=(0,9;0,5;\dfrac{11}{28})$
I hope we can find a good way to solve the problem. Thanks!
| Now, I have a partly solution due to my teacher's hint.Using the following one:
$$\frac{a}{a^{2}+bc}\ge\frac{3ab+3ca-a^{2}-bc}{(a+b)(b+c)(c+a)}$$$$\iff (a^2+bc -ab-ac)^2 \geq 0$$
$$\implies \frac{a^2}{a^{2}+bc}\ge\frac{3a^2b+3ca^2-a^{3}-abc}{(a+b)(b+c)(c+a)}$$ Hence $$ \frac{a^2}{a^{2}+bc}+ \frac{b^2}{b^{2}+ca}+ \frac{c^2}{c^{2}+ab}+\frac{a^3+b^3+c^3+9abc}{(a+b)(b+c)(c+a)}\ge3$$
Now, we need to prove that: $$\frac{1-a^2}{a^2+bc}+\frac{1-b^2}{b^2+ca}+\frac{1-c^2}{c^2+ab}\ge1 (*)$$
But how to prove this nice one and using the original condition?
Edit: (*) is equivalent to: $$\sum_{cyc}{\frac{a^2(b^2+c^2)\left(b^2+c^2-a^2-2bc\right)^2}{(a^2+bc)\left(a^2+b^2+c^2+(b^2+c^2-a^2)bc+2a^2(b^2+c^2)\right)}}\ge0$$
I see a similar problem on AOPS which help very well. Also, I hope to see more approach for (*).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4358760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 $$a\sin x+b\cos x+c\sin x\cos x=0?$$
| Let $a=\cos x$ and $b=\sin x$, then
$$
\left\{\begin{array}{l}
a+b=k a b \\
a^{2}+b^{2}=1
\end{array}\right.
$$
Denoting the sum of $a$ and $b$ by $h$ yields
$$a+b=h \textrm{ and }a b=\frac{h}{k}$$
Now we can construct a quadratic equation (*) with roots $a$ and $b$
$$
k x^{2}-h k x+h=0 \tag*{(*)}
$$
Since $a$ and $b$ are roots of (*), therefore $$
\left\{\begin{array}{l}
k a^{2}-h k a+h=0 \quad \cdots(1) \\
k b^{2}-h k b+h=0 \quad \cdots(2)
\end{array}\right.
$$
(1) + (2) yields
$$\begin{array}{l}
k\left(a^{2}+b^{2}\right)-h k(a+b)+2 h=0 \\
\qquad k h^{2}-2 h-k=0
\end{array}
$$
When $k\neq 0$, using quadratic formula gives
$$
\begin{aligned}
h
&=\frac{1 \pm \sqrt{1+k^{2}}}{k}
\end{aligned}
$$
By $k a b=h$, $$
\begin{aligned}
\frac{k \sin 2 x}{2} &=\frac{1 \pm \sqrt{1+k^{2}}}{k} \\
\sin 2 x &=\frac{2 \pm 2 \sqrt{1+k^{2}}}{k^{2}}
\end{aligned}
$$
Noting that
$$-1 \leqslant \frac{2+2 \sqrt{1+k^{2}}}{k^{2}} \leqslant 1 \Leftrightarrow |k| \geqslant 2 \sqrt{2}$$
$$
-1 \leqslant \frac{2-2 \sqrt{1+k^{2}}}{k^{2}} \leqslant 1 \Leftrightarrow k\neq 0,
$$
we can conclude that
A. If $k=0,$
$$\displaystyle x=n \pi-\frac{\pi}{4}$$
B. If $0\neq|k| \geqslant 2 \sqrt{2}$,
$$
x=\frac{1}{2}\left[n \pi+(-1)^{n} \sin ^{-1}\left(\frac{2 \pm 2 \sqrt{1+k^{2}}}{k^{2}}\right)\right]
$$
C. If $0\neq |k| <2 \sqrt{2}$,
$$
x=\frac{1}{2}\left[n \pi+(-1)^{n} \sin ^{-1}\left(\frac{2-2 \sqrt{1+k^{2}}}{k^{2}}\right)\right],
$$
where $n \in \mathbb{Z}$.
| {
"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": 0
} |
About the diophantine equation $x^3+y^3=z^8$ After solving the exercise
*
*The equation $x^n+y^n=z^{n+1}$ has infinite solutions in the positive integers.
I realized that this can be easily generalized to
*
*If $m\equiv 1 \pmod n$, the equation $x^n+y^n = z^m$ has infinite solutions in the positive integers.
Then, for example, $x^3+y^3 = z^7$ has infinite solutions. Then I wondered about the equation $$x^3+y^3=z^8$$
After a numeric search and some simplifications I got two family of solutions:
$$(x,y,z) = (32 n^8,32 n^8,4 n^3), \text{ with } n\ge 1$$
$$(x,y,z) = (18 n^8,9n^8,3n^3), \text{ with } n\ge 1$$
It's clear that from any "primitive" solution $(a,b,c)$ one can get other solutions $(a n^8, b n^8, c n^3)$. The numeric search for $x, y \le 10000$ gives only the two primitive solutions $(32,32,4)$ and $(18,9,3)$
I'm wondering: Are all the solutions of $x^3+y^3=z^8$ covered by these two families?
| More solutions can be found. For example $x=3\cdot 28^5$, $y=28^5$, $z=28^2$ when
$$x^3+y^3=(28^5)^3(3^3+1^3)=28^{16}=z^8.$$
The ones you found can be explained as follows:
*
*Starting with $1^3+1^3=2$ we see that $(2^a)^3+(2^a)^3=2\cdot2^{3a}=2^{3a+1}$, and it is all about making $3a+1$ divisible by eight, so $a=5$ yields $x=y=32, z=4$.
*Starting with $2^3+1^3=9=3^2$ we similarly see that $(2\cdot3^a)^3+(3^a)^3=3^{3a+2}$, and we need to choose $a$ so that $3a+2$ is divisible by eight. This time $a=2$ gives $x=18, y=9, z=3$.
The same method gives other families. You can start with a random "seed" choice for $x=a$ and $y=b$, and then use $a^3+b^3$ as a "scale parameter" (like $2$, $3$ and $28$ in the three families above) and find solutions with a given $[x:y]=[a:b]$ ratio
$$x=a(a^3+b^3)^5,\qquad y=b(a^3+b^3)^5,\qquad z=(a^3+b^3)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4359593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do we evaluate the integral $\int_{0}^{\frac{\pi}{4}} \ln (\sin x) d x?$ As I have found the integral
$$\int_{0}^{\frac{\pi}{2}} \ln (\sin x) d x=-\frac{\pi}{2}\ln 2 ,$$
I started to investigate a similar integral with a different upper limit $\frac{\pi}{4}$ by similar methods with failure. Once I found the Fourier Series of sine on $[0, \pi]$, I can find it easily.
By Fourier Series
$$\ln (\sin x)=-\ln 2-\sum_{n=1}^{\infty} \frac{\cos (2 n x)}{n},$$
we integral from $0$ to $\frac{\pi}{4} $,
\begin{aligned}
\int_{0}^{\frac{\pi}{4}} \ln (\sin x) d x&=-\int_{0}^{\frac{\pi}{4}} \ln 2 d x-\sum_{n=1}^{\infty} \int_{0}^{\frac{\pi}{4}} \frac{\cos (2 n x)}{n} d x \\
&=-\frac{\pi}{4} \ln 2-\sum_{n=1}^{\infty}\left[\frac{\sin (2 n x)}{2 n^{2}}\right]_{0}^{\frac{\pi}{4}} \\
&=-\frac{\pi}{4} \ln 2-\frac{1}{2} \sum_{n=1}^{\infty}\left(\frac{\sin \frac{n \pi}{2}}{n^{2}}\right)\\
&=-\frac{\pi}{4} \ln 2-\frac{1}{2} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{2}} \\
&=-\frac{\pi}{4} \ln 2-\frac{1}{2} G,
\end{aligned}
where $G$ is the Catalan's constant.
Is there any method other than using Fourier Series?
| Let $I_0=\int_0^\frac{\pi}{2}\ln(\sin x) dx$.
Making the substitution $t=\frac{\pi}{2}-x\,,\,\, I_0=\int_0^\frac{\pi}{2}\ln(\cos x) dx$
$$2I_0=\int_0^\frac{\pi}{2}\ln(\sin x) dx+\int_0^\frac{\pi}{2}\ln(\cos x) dx=\int_0^\frac{\pi}{2}\ln(\cos x\sin x) dx$$
$$=\int_0^\frac{\pi}{2}\ln\frac{1}{2}\,dx\,+\int_0^\frac{\pi}{2}\ln(\sin 2x) dx=-\frac{\pi}{2}\ln 2+I_0\,\,\Rightarrow\,\, I_0=-\frac{\pi}{2}\ln 2$$
Exactly the same approach can be applied to the integrals $I_1=\int_0^\frac{\pi}{4}\ln(\sin x) dx\,$ and $\,I_2=\int_0^\frac{\pi}{4}\ln(\cos x) dx$
$$I_1+I_2=\int_0^\frac{\pi}{4}\ln\frac{1}{2}\,dx\,+\int_0^\frac{\pi}{4}\ln(\sin 2x) dx=-\frac{\pi}{4}\ln 2-\frac{\pi}{4}\ln 2=-\frac{\pi}{2}\ln 2$$
$$I_1-I_2=\int_0^\frac{\pi}{4}\ln(\tan x) dx$$
Making the substitution $x=\operatorname{arctan}t$
$$I_1-I_2=\int_0^1\frac{\ln t}{1+t^2}dt=-G$$
From two equations we get
$$I_1=\int_0^\frac{\pi}{4}\ln(\sin x) dx=-\frac{\pi}{4}\ln 2-\frac{G}{2}$$
$$I_2=\int_0^\frac{\pi}{4}\ln(\cos x) dx=-\frac{\pi}{4}\ln 2+\frac{G}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How do I find the lengths of the sides of a triangle that is inscribed in the unit circle that has the greatest perimeter? I know how to find the sides of the triangle when seeking to maximize the area of the triangle, since the following relationship exists $Area= \frac{abc}{4R}$ where $a, b,c$ are the sides and $R$ the radius.
But I don't know what to occupy to maximize the perimeter.
| Recall that the Law of Sines tells you that
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R,$$
where $R$ is the circumradius. Then the perimeter is
$$2R(\sin A + \sin B + \sin C).$$
Note that
$$\sin A + \sin B = 2\cos(\tfrac{A - B}{2})\sin(\tfrac{A + B}{2}) \le 2\sin(\tfrac{A + B}{2})$$
with equality only when $A = B$. Thus, the perimeter is maximized when $A = B$. Using similar logic, you can show that $B = C$ as well when the perimeter is maximized. So, the perimeter is maximized when $A = B = C = 60^\circ$ with a value of
$$2R(\sin 60^\circ + \sin 60^\circ + \sin 60^\circ) = 3\sqrt3 R.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\alpha,\beta,\gamma$ are the roots of $x^3+x^2-x+1=0$, find the value of $\prod\left(\frac1{\alpha^3}+\frac1{\beta^3}-\frac1{\gamma^3}\right)$
If $\alpha,\beta,\gamma$ are the roots of the equation $x^3+x^2-x+1=0$, find the value of $\prod\left(\frac1{\alpha^3}+\frac1{\beta^3}-\frac1{\gamma^3}\right)$
My Attempt:
$\alpha+\beta+\gamma=-1$
$\alpha\beta+\beta\gamma+\gamma\alpha=-1$
$\alpha\beta\gamma=-1$
$\prod\left(\frac1{\alpha^3}+\frac1{\beta^3}-\frac1{\gamma^3}\right)=\prod\left(\frac{(\beta\gamma)^3+(\alpha\gamma)^3-(\alpha\beta)^3}{(\alpha\beta\gamma)^3}\right)=-\prod\left((\beta\gamma)^3+(\alpha\gamma)^3-(\alpha\beta)^3\right)$
Now, one option is to expand the product and put the values where required, but is there a shorter, better approach to do this?
| 1. Let $P(x) = x^3 + x^2 - x + 1$. Since all the zeros of $P(x)$ are non-zero, $\gcd(P(x), x^3) = 1$. Then the Bézout's identity tells that we can find polynomials $A(x)$ and $B(x)$ satisfying
$$ A(x) P(x) + B(x) x^3 = 1. $$
Although there is a systematic way of determining $A(x)$ and $B(x)$, called the extended GCD algorithm, it is not hard to see that $A(x) = x+1$ and $B(x) = -x-2$ from the computation
$$ (x+1)(x^2 - x + 1) = x^3 + 1 \qquad\implies\qquad (x+1)P(x) = x^4 + 2x^3 + 1. $$
The upshot of this computation is that, if $x = x_0$ is any zero of $P(x) = 0$, then
$$ B(x_0) x_0^3 = 1 \qquad\text{and hence}\qquad \frac{1}{x_0^3} = B(x_0) = -x_0-2. $$
2. Plugging this to OP's product, we get
$$ \prod_{\text{cyc}} \left( \frac{1}{\alpha^3} + \frac{1}{\beta^3} - \frac{1}{\gamma^3} \right)
= \prod_{\text{cyc}} \left( -\alpha -\beta + \gamma - 2 \right)
= \prod_{\text{cyc}} \left(2\gamma - 1\right)
= (-2)^3 P\left(\frac{1}{2}\right)
= -7. $$
| {
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"url": "https://math.stackexchange.com/questions/4365119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\sum^\infty_{n=0}\frac{\sum^n_{k=0}\frac{(-1)^k2^k}{k+1}}{2^n}$
Determine
$$\sum^\infty_{n=0}\frac{\sum^n_{k=0}\frac{(-1)^k2^k}{k+1}}{2^n}\text{.}$$
For large even $n$,
\begin{aligned}
0&\leq1+\left(-\frac{2}{2}+\frac{4}{3}\right)+\cdots+\left(-\frac{2^{n-1}}{n}+\frac{2^n}{n+1}\right)\\
&=\left(1-\frac{2}{2}\right)+\cdots+\left(\frac{2^{n-2}}{n-1}-\frac{2^{n-1}}{n}\right)+\frac{2^n}{n+1}\leq\frac{2^n}{n+1},
\end{aligned}
so that $\frac{\sum^n_{k=0}\frac{(-1)^k2^k}{k+1}}{2^n}\rightarrow0$ as $n\rightarrow0$. Similarly, for large odd $n$, $\frac{\sum^n_{k=0}\frac{(-1)^k2^k}{k+1}}{2^n}\rightarrow0$ and $\frac{\sum^n_{k=0}\frac{(-1)^k2^k}{k+1}}{2^n}<0$. Thus the series converges.
Since
$$\sum^\infty_{k=0}\sum^\infty_{n=k}\frac{1}{(k+1)2^{n-k}}=\sum^\infty_{k=0}\frac{2}{k+1}$$
diverges, we cannot simply say $\sum^\infty_{n=0}\frac{\sum^n_{k=0}\frac{(-1)^k2^k}{k+1}}{2^n}=\sum^\infty_{k=0}\frac{(-1)^k2}{k+1}$.
How to find the sum?
| Because the series $\displaystyle{\sum \frac{(-1)^n}{n+1}}$ is convergent and the series $\displaystyle{\sum \frac{1}{2^n}}$ is absolutely convergent, then the Cauchy product of these two series is well-defined and one has
\begin{align*}
\sum^\infty_{n=0}\frac{\sum^n_{k=0}\frac{(-1)^k2^k}{k+1}}{2^n} & =\sum^\infty_{n=0}\sum^n_{k=0}\frac{(-1)^k}{(k+1)2^{n-k}}\\
&=\left( \sum_{n=0}^{+\infty} \frac{(-1)^n}{n+1}\right)\left(\sum_{n=0}^{+\infty} \frac{1}{2^n} \right)
\end{align*}
which gives directly that
$$\boxed{\sum^\infty_{n=0}\frac{\sum^n_{k=0}\frac{(-1)^k2^k}{k+1}}{2^n} = 2 \log(2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The diagonals of a rhombus, given area and tangent The area of the rhombus $ABCD$ is $24$ $cm^2$, if $\tan\measuredangle ABC=\dfrac{24}{7}$, find the diagonals $AC$ and $BD$.
I think we can say that $\measuredangle ABC$ is an acute angle. Is that true? Then $$\begin{cases}\tan\beta=\dfrac{24}{7}\\\sin^2\beta+\cos^2\beta=1\end{cases}$$ gives $\cos\beta=\dfrac{7}{25},\sin\beta=\dfrac{24}{25}.$ The area of $ABCD$ is $$S_{ABCD}=a^2\sin\beta=24\\a^2\cdot\dfrac{24}{25}=24\\a=5>0.$$ Now the Cosine Rule in triangle $ABC$ gives $$AC^2=2\cdot5^2-2\cdot5^2\dfrac{7}{25}=36,AC=6$$ The relationship $d_1d_2=48$ (from the area with the formula $S_{ABCD}=\frac{d_1d_2}{2}$) is very "clear". Can we come up with something else with the diagonals to make the solution better?
| Let $\angle ABC=\theta$. Then $\angle ABD=\frac{\theta}{2}$. Suppose $AC\cap BD=P$ and $AC=2y$, $BD=2x$. In triangle $\triangle ABP$, we have $\tan\frac{\theta}{2}=\frac{y}{x}$. We also know that $(2x)(2y)=48\implies xy=12$. Using the tangent condition:
$$\frac{24}{7}=\frac{2(y/x)}{1-(y/x)^2}$$
Letting $y/x=z$, we have that $z=\frac{3}{4}$, hence $y=\frac{3}{4}x$, implying the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4369214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trigonometric equation $\tan x \tan 2x = \cot 2x \cot 3x$
Find $\cos 8x$ if:
$$\tan x \tan 2x = \cot 2x \cot 3x$$
We can verify quickly that $\tan 2x \to \infty$ and $\tan 3x \to \infty$ are not solutions of the trig equation, so the equation may be rewritten as:
$$\tan x \tan^2 2x \tan 3x = 1$$
Using expansion formulas, we may see that:
$$\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$$
And also:
$$\tan 3x = \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x}$$
By long calculations, I have obtained that:
$$\tan x \in \{-1 - \sqrt{2}, -1 + \sqrt{2}, 1 - \sqrt{2}, 1 + \sqrt{2}\}$$
However, I am not able to take those calculations further. I have learnt about Chebyshev polynomials, but using WolframAlpha, the expression is really hairy and not realy easy to cope with.
Is there a smarter trick to solve the question? If not, how should I simplify my calculations?
|
$\tan x \tan^2 2x \tan 3x = 1$
Using the product-to-sum identity $\,\tan \theta \tan \varphi ={\frac {\cos(\theta -\varphi )-\cos(\theta +\varphi )}{\cos(\theta -\varphi )+\cos(\theta +\varphi )}}\,$:
$$1 = \tan^2 2x \cdot\frac{\cos 2x - \cos 4 x}{\cos 2 x + \cos 4 x}= \frac{1 - \cos^2 2x}{\cos^2 2x}\cdot\frac{\cos 2x - 2 \cos^2 2x+1}{\cos 2x+2\cos^2 2x - 1} \tag{1}$$
With $\,t = \cos 2x\,$:
$$
\begin{align}
t^2(2t^2+t-1)=(1-t^2)(-2t^2+t+1) \;\;&\iff\;\; 2 t^3 + 2 t^2 - t - 1 = 0
\\ &\iff\;\; (t+1)(2t^2-1) = 0
\\ &\iff\;\; t \in \left\{-1, \pm \frac{\sqrt{2}}{2} \right\}
\end{align}
$$
The root $\,t=-1\,$ must be excluded because the denominator in $\,(1)\,$ vanishes when $\,\cos 2x=-1\,$, which leaves $\,\cos 2x = \pm \frac{\sqrt{2}}{2}\,$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\displaystyle\int_{0}^{\frac{1}{\sqrt{3}}}\frac{\arctan\left(\frac{1}{\sqrt{1-2x^2}}\right)}{1+x^2} \mathrm{dx}=\frac{13\pi^2}{288}$ I need to prove that,
$\displaystyle \tag*{} \int_{0}^{\frac{1}{\sqrt{3}}}\frac{\arctan\left(\frac{1}{\sqrt{1-2x^2}}\right)}{1+x^2} \mathrm{dx}=\frac{13\pi^2}{288}$
Here is what I tried:
I tried to solve the integral by Integrating by parts.
We have:
$\displaystyle \tag{1} \int f(x) g'(x) \mathrm{ dx} = f(x)g(x) - \int f'(x) g(x) \mathrm{ dx}$
I noticed ,
$\displaystyle \tag*{} \dfrac{1}{1+x^2} = \arctan '(x)$
$\displaystyle \tag*{} g'(x) = \dfrac{1}{1+x^2} \Leftrightarrow g(x) = \arctan(x)$
Now, I defined $f(x)$
$\displaystyle \tag*{} f(x) = \arctan \left( \dfrac{1}{\sqrt{1-2x^2}}\right )$
and
$\displaystyle \tag*{} f'(x) = \arctan'\left(\dfrac{1}{\sqrt{1-2x^2}} \right ) = \text{arccot}'(\sqrt{1-2x^2}) = \dfrac{x}{\sqrt{1-2x^2}(1-x^2)}$
Now, using $(1)$ and substituting the values of $f(x)$ and $g(x)$, My indefinite integral becomes:
$\displaystyle \tag*{} \arctan(x) \arctan \left (\dfrac{1}{\sqrt{1-2x^2}} \right ) - \int \dfrac{x \arctan(x)}{\sqrt{1-2x^2}(1-x^2)} \mathrm{ dx}$
Now, we want to evaluate:
$\displaystyle \tag*{} \int \dfrac{x \arctan(x)}{\sqrt{1-2x^2}(1-x^2)} \mathrm{ dx}$
Now this is integral I am having trouble solving. Any hints or different methods would be greatly appreciated. Thank you.
| Integrate as follows
\begin{align}
&\int_{0}^{\frac{1}{\sqrt{3}}}\frac{\tan^{-1}\frac{1}{\sqrt{1-2x^2}}}{1+x^2} {dx}\\
\overset{ibp}=&\ \frac{\pi^2}{18}-\int_{0}^{\frac{1}{\sqrt{3}}}\frac{x\tan^{-1} x}{(1-x^2)\sqrt{1-2x^2}}dx
\>\>\>\>\>\>\>\sqrt2 x=\cos t\\
=& \ \frac{\pi^2}{18} -\int_{\cot^{-1}\sqrt2}^{\pi/2}
\frac{\tan^{-1}(\cos t/\sqrt2)\cos t}{2-\cos^2 t}dt\\
=& \ \frac{\pi^2}{18} -\int_{\cot^{-1}\sqrt2}^{\pi/2}\int_0^1
\frac{\sqrt2\cos^2t}{(2-\cos^2 t)(2+y^2\cos^2t)}dy\ dt\\
=& \ \frac{\pi^2}{18} -\int_0^1\frac1{1+y^2}\bigg(\frac\pi4-\frac\pi{2\sqrt{2+y^2}}+\frac{\cot^{-1} \sqrt{2+y^2}}{\sqrt{2+y^2}}\bigg)dy\\
=& \ \frac{11\pi^2}{144}- \int_0^1 \frac{\cot^{-1} \sqrt{2+y^2}}{\sqrt{2+y^2}}dy
=\frac{11\pi^2}{144}-\frac{\pi^2}{32}= \frac{13\pi^2}{288}
\end{align}
where
\begin{align}
I=&\int_0^1 \frac{\cot^{-1} \sqrt{2+y^2}}{\sqrt{2+y^2}}dy
=\int_0^1 \int_0^1 \frac1{(1+y^2)(y^2+x^2+2)}dx \ dy\\
=& \int_0^1 \int_0^1 \frac1{1+x^2}\bigg(\frac1{1+y^2}-\frac1{y^2+x^2+2}\bigg)dy \ dx
=\frac{\pi^2}{16}-I=\frac{\pi^2}{32}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
How to integrate this $\int\frac{\sin x}{3-2\sin x}dx$? How can I evaluate following integral $$\int\frac{\sin x}{3-2\sin x}dx$$
My first attempt:
Substitute $3-2\sin x=t$ or $\sin x=\frac{3-t}{2}$, $$\cos xdx=-\frac{1}{2}dt, \ dx=-\frac{dt}{\sqrt{6t-t^2-7}}$$$$\int\frac{\sin x}{3-2\sin x}dx=\int\frac{\frac{3-t}{2}}{t}\cdot -\frac{dt}{2\sqrt{6t-t^2-7}}$$$$=-\int\frac{(3-t)dt}{t\sqrt{6t-t^2-7}}$$
Here I got stuck by using this substitution
My second attempt:
I used half angle formula: $\sin x=\frac{2\tan \frac{x}{2}}{1+\tan^{2}\frac{x}{2}}$ which led me
$$\int\frac{\sin x}{3-2\sin x}dx=\int\frac{\frac{2\tan \frac{x}{2}}{1+\tan^{2}\frac{x}{2}}}{3-2\frac{2\tan \frac{x}{2}}{1+\tan^{2}\frac{x}{2}}}$$$$=\int\frac{2\tan \frac{x}{2}}{3+3\tan^2 \frac{x}{2}-4\tan\frac{x}{2}}$$
Denominator made perfect square $$=\frac{2}{3}\int\frac{2\tan \frac{x}{2}}{\left(\tan \frac{x}{2}-\frac23\right)^2-\frac13}$$
I got stuck here, don't know how to proceed.
Please help me solve this integration. Thanks in advance
| Use partial fractions then half angle formula as follows
$$\int\frac{\sin x}{3-2\sin x}dx=\frac12\int\frac{3-(3-2\sin x)}{3-2\sin x}dx$$
$$=\frac12\int\left(\frac{3}{3-2\sin x}-1\right)dx$$
$$=\frac32\int\frac{dx}{3-2\sin x}-\frac12\int 1dx$$
$$=\frac32\int\frac{dx}{3-2\cdot\frac{2\tan\frac x2}{1+\tan^2\frac x2}}-\frac12x+C$$
$$=\frac12\int\frac{\sec^2\frac x2dx}{\left(\tan\frac x2-\frac23\right)^2-\frac13}-\frac12x+C$$
$$=\int\frac{d\left(\tan\frac x2-\frac23\right)}{\left(\tan\frac x2-\frac23\right)^2-\frac13}-\frac12x+C$$
$$=\frac{1}{2\cdot \frac{1}{\sqrt3}}\ln\left|\frac{\tan\frac x2-\frac23-\frac{1}{\sqrt3}}{\tan\frac x2-\frac23+\frac{1}{\sqrt3}}\right|-\frac12x+C$$
$$=\frac{\sqrt3}{2}\ln\left|\frac{3\tan\frac x2-2-\sqrt3}{3\tan\frac x2-2+\sqrt3}\right|-\frac12x+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Rectangle in a circle of radius a that maximizes x^n+ y^n
Consider a rectangle with sides $2x$ and $2y$ inscribed in a given fixed circle $$x^2 + y^2 = a^2,$$ and let $n$ be a positive number. We wish to find the rectangle that maximizes the quantity $$z = x^n + y^n.$$ If $n = 2$, it is clear that $z$ has the constant value $a^2$ for all rectangles.
*
*If $n < 2$, show that the square maximizes $z$; and
*if $n > 2$, show that $z$ is maximized by a degenerate rectangle in which $x$ or $y$ is $0$.
My approach.
$$x^2 + y^2 = a^2$$
$$\therefore\, y = (a^2 - x^2)^{1\over 2}$$
\begin{align}
z &= x^n + y^n \\
{d\over dx}z & = nx^{n-1} - nx(a^2-x^2)^{n-2\over 2}
\end{align}
Equating with $0$ gives:
$$nx^{n-1} = nx(a^2-x^2)^{n-2\over 2}$$
$$x^{n-2} = (a^2 - x^2)^{n-2\over 2}$$
Squaring on both sides gives:
$$(x^2)^{n-2} = (a^2 - x^2)^{n-2}$$
Taking $\ln$ on both sides gives
$$(n-2) \ln(x^2) = (n-2)\ln(a^2 - x^2)$$33
Equating $x^2 = a^2 - x^2$ , we get $x = {a\over \sqrt{2}}$ and $y = {a\over\sqrt{2}}$. In particular $x=y$ implies that the optimal rectange is a square.
My computation appears to be independent of $n$, so I am puzzled about the statement in the question for $n>2$ and $n<2$.
| For $n < 2$ your expression for $\frac{dz}{dx}$ is not defined where $y = 0$, so one must check those points separately. Also, between the two equations following, "equating it with $0$...", division by $0$ assumes that $x \neq 0$.
Here's an alternative approach: This problem is a typical candidate for the method of Lagrange multipliers, and applying that method has the advantage that it leaves $x$ and $y$ on equal footing and in particular avoid the complications introduced by solving (partially) for $y$ in terms of $x$.
Our objective function is $$f(x, y) := x^n + y^n ,$$ and our constraint is $g(x, y) = 0$, where
$$g(x, y) := x^2 + y^2 - a^2 .$$ (Herein I'll assume $a > 0$, so that the circle is nondegenerate.)
By symmetry we may as well assume that any (constrained, local) extremum $(c, d)$ lies in the closed first quadrant, i.e., are both nonnegative.
In fact, if $c = 0$, then the constraint implies $d = a$, and vice versa, so if a (constrained, local) extremum occurs where $c = 0$ or $d = 0$, its value is $f(a, 0) = f(0, a) = a^n$, and we may as well restrict the rest of our search to the interior of the first quadrant, i.e., the case $c, d > 0$.
The method of Lagrange multipliers yields that any (constrained, local) extremum $(c, d)$ satisfies $$(\nabla f)(c, d) = \lambda (\nabla g)(c, d) ,$$
that is, the system
\begin{align}
n c^{n - 1} &= 2 c \lambda \\
n d^{n - 1} &= 2 d \lambda .
\end{align}
Dividing the first equation by the second equation and rearranging yields $c^{n - 2} = d^{n - 2}$, so either (a) $n = 2$ (which corresponds to the tautology that there is a maximum at every point on the unit circle), or (b) the only critical point is $c = d = \frac{a}{\sqrt{2}}$. In the latter case computing gives $f\left(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}\right) = a^n (\sqrt{2})^{2 - n} ,$ and whether this quantity is greater than $f(a, 0) = f(0, a) = a^n$ manifestly depends just on the sign of $2 - n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculating $\int_{0}^\infty \frac{\cos(2x)}{(x^2 + 1)(x^2 + 4)} dx$ using residue calculus I want to calculate
$$\int_{0}^\infty \frac{\cos(2x)}{(x^2 + 1)(x^2 + 4)} dx$$
using residue calculus
My solution
First observation is that:
$$\int_\mathbb R \frac{e^{2ix}}{(x^2 + 1)(x^2 + 4)} dx = \int_\mathbb R \frac{\cos(2x)}{(x^2 + 1)(x^2 + 4)} dx + i \int_\mathbb R \frac{\sin(2x)}{(x^2 + 1)(x^2 + 4)}dx$$
whereas $\int_\mathbb R \frac{\sin(2x)}{(x^2 + 1)(x^2 + 4)}dx = 0$ since this function is odd, as multiplication of odd function with even function.
If we define now $$f(z) = \frac{e^{2iz}}{(z^2 + 1)(z^2 + 4)} = \frac{e^{2iz}}{(z^2 - i^2)(z^2 - 4i^2)} = \frac{e^{2iz}}{(z-i)(z+i)(z-2i)(z+2i)}$$
We will see that we have four poles in points $i, -i, 2i, -2i$ with rank $1$.
Now let's define our contour for $r > 2$:
$$\gamma_1: [-r ,r] \ni t \rightarrow t \in \mathbb C$$
$$\gamma_2: [0, \pi] \ni t \rightarrow Re^{it}$$
Our final contour is $\gamma = \gamma_1 + \gamma_2$
Firstly let's consider our integral on $\gamma_2$:
$$|\int_ {\gamma_2} f(z) dz| = |\int_0 ^ \pi \frac{e^{2ire^{it}}}{((re^{it})^2 + 1)((re^{it})^2 + 4)}ire^{it} dt| \le \int_0^\pi|\frac{|e^{-2\sin(t)}|}{((re^{it})^2 + 1)((re^{it})^2 + 4)}r dt$$
Now we know that:
$$\frac{1}{|(re^{it})^2 + 1|} \le \frac{1}{r^2 - 1}$$
$$\frac{1}{|(re^{it})^2 + 4|} \le \frac{1}{r^2 - 4}$$
$$\int_0^\pi e^{-A \sin t}dt < \frac{\pi}{A}$$
so
$$\int_0^\pi|\frac{|e^{-2\sin(t)}|}{((re^{it})^2 + 1)((re^{it})^2 + 4)}r dt < \frac{r\pi}{2} \cdot \frac{1}{(r^2 - 1)(r^2 + 4)} \rightarrow 0 $$
So when going with $r$ to infinity we'll have our integral only on $\gamma_1$. To calculate so, we'll just calculate residuals in $i$ and $2i$:
$$\textrm{res}_if(z) = \lim_{z \rightarrow i} = \lim_{z \rightarrow i}\frac{e^{2iz}}{(z + i)(z - 2i)(z + 2i)} = \frac{e^{-2}}{6}$$
$$\textrm{res}_{2i}f(z) = \lim_{z \rightarrow 2i} = \lim_{z \rightarrow 2i}\frac{e^{2iz}}{(z - i)(z - i)(z + 2i)} = \frac{e^{-4}}{12}$$
So finally we have that our integral equals to:
$$\pi(\frac{e^{-2}}{6} + \frac{e^{-4}}{12})$$
which unfortunately is different from what wolfamalpha suggests. Could you please tell me where did I make the mistake?
| Those residues are wrong. You have\begin{align}\operatorname{res}_{z=i}\frac{e^{2iz}}{(z^2+1)(z^2+4)}&=\lim_{z\to i}\frac{e^{2iz}}{(z+i)(z^2+4)}\\&=\frac{e^{-2}}{6i}\end{align}and, by the same argument,$$\operatorname{res}_{z=2i}\frac{e^{2iz}}{(z^2+1)(z^2+4)}=-\frac{e^{-4}}{12i}.$$Therefore\begin{align}\int_{-\infty}^\infty\frac{\cos(2x)}{(x^2+1)(x^2+4)}\,\mathrm dx&=\int_{-\infty}^\infty\frac{e^{2ix}}{(x^2+1)(x^2+4)}\,\mathrm dx\\&=2\pi i\left(\frac{e^{-2}}{6i}-\frac{e^{-4}}{12i}\right)\\&=2\pi\left(\frac{e^{-2}}6-\frac{e^{-4}}{12}\right).\end{align}So,$$\int_0^\infty\frac{\cos(2x)}{(x^2+1)(x^2+4)}\,\mathrm dx=\pi\left(\frac{e^{-2}}6-\frac{e^{-4}}{12}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4386019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $AT \cdot XY = AX \cdot BY$ In figure below $T,X,Y$ are tangency points and the big circles are tangent to each other. And $TB$ is parallel to that common tangent below.
So, I made more calculations than I should. Let $R$ be the radius of the blue $\Omega$ circle and $r$ be the radius of the green circle.
I managed, through heavy calculation, to come to:
$R = r + \frac{AB^2}{16r}$
I guess this is a good start, there should be an easy way to prove it.
By defining $TA = t, AX = a, XY = 2k, BY = b$
I also came to the following relation:
$(4t^2 + 4t(a+b+2k) + (a+b+2k)^2)\frac1{\sqrt{8t^2 + 8t(a+b+2k) + (a+b+2k)^2}} k = k^2 + (a+k)(b+k)$
which is nice, has the same variables that the question asks, can anyone solve this one?
My calculations:
$M =$ midpoint of $\overline{AB}$
$C = \Omega \cap$ the common tangent, $CA=CB$
$D$ is such that $TMCD$ is a rectangle and $E$ is the tangency point between $\Omega$ and green circle. $TE = x \implies EC = \frac{R}{r} x$ (homothety centered at $E$)
$TC^2 = TD^2 + DC^2 = (2r)^2 + (2\sqrt{Rr})^2 = (x\frac{R+r}r)^2$
so $x = \frac{2r}{R+r} \sqrt{r(r+R)}$ RIGHT?
$x \cdot TC = TA \cdot TB \implies 4r^2 = TA \cdot TB$ RIGHT?
but $TB = 2\sqrt{Rr} + \frac{AB}2$
So now:
$TA = \frac{\sqrt{16^2-AB^2}-AB}2$ oops, I think I see my problem.... I tried to isolate $R$ in $TA \cdot (2\sqrt{Rr} + \frac {AB}2) = 4r^2$ and I changed a signal.... welp if someone wants to continue from here.
| Besides $R$ and $r$, let $h$ be the distance from the center of the large circle to $AB$ and let $AB = 2l,$ for simplicity. Since the two lines are parallel $R+h = 2r$ and you can find $AT$ by Pythagora:
$$(AT+l)^2 + (R-r)^2 = (R+r)^2\implies AT = 2\sqrt{Rr} - l.$$
If you let $AX = a$ and $BY = b$, then it suffices to show that:
$$\dfrac{AX\cdot BY}{XY} = \dfrac{ab}{2l-a-b} = AT = 2\sqrt{Rr} - l.$$
First step is to find an equation relating $a$ and $b.$ Let $x$ be the radius of the small circle corresponding to $a$, then another Pythagora's theorem says:
$$(l-a)^2 + (h+x)^2 = (R-x)^2\implies x = \dfrac{R^2-h^2 - (l-a)^2}{2(R+h)} = \dfrac{a(2l-a)}{4r}.$$
This is because:
$$R+h = 2r\quad \text{ and }\quad R^2- h ^2 = l^2.$$
Similarly, if $y$ is the radius of the small circle corresponding to $b$, then $y = \dfrac{b(2l-b)}{4r}.$ Finally, if you do another Pythagora's theorem on the two small circles being tangent to each other:
$$XY^2 = (x+y)^2 - (x-y)^2 = 4xy\implies (2l-a-b)^2 = \dfrac{ab(2l-a)(2l-b)}{4r^2}.$$
You can manipulate this easily into an equivalent form:
$$\dfrac{ab}{2l-a-b} = \dfrac{4r^2}{2l + \dfrac{ab}{2l-a-b}}\implies \dfrac{ab}{2l-a-b} = -l + \sqrt{l^2+4r^2} = -l + \sqrt{R^2 - h^2 + 4r^2} = $$
$$ = -l + \sqrt{R^2-(R-2r)^2+4r^2} = -l+2\sqrt{Rr},$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4388128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve \begin{cases} x'=2x+4y+\cos t\\ y'=-x-2y+\sin t \end{cases} Solve
\begin{cases}
x'=2x+4y+\cos t\\
y'=-x-2y+\sin t
\end{cases}
My solution:
First I solve :
\begin{cases}
x'=2x+4y\\
y'=-x-2y
\end{cases}
$$\begin{pmatrix} {x'}\\ {y'}
\end{pmatrix}=\begin{pmatrix} {2} & 4\\ {-1} & -2
\end{pmatrix} \begin{pmatrix} {x}\\ {y}
\end{pmatrix}$$
The eigenvalue is $0$.
The eigevector is \begin{pmatrix} {-2}\\ {1}
\end{pmatrix}
The generalized eigenvector is \begin{pmatrix} {0}\\ -{1\over 2}
\end{pmatrix}
Then, the solution is:
$y=c_1\begin{pmatrix} {0}\\ {-1\over 2}
\end{pmatrix} + c_2 \bigg [t\begin{pmatrix} {-2}\\ {1}
\end{pmatrix}+\begin{pmatrix} {0}\\ {-1\over 2}
\end{pmatrix}\bigg]$
Using variation of parameters method:
$$\begin{pmatrix} {0} & -2t\\ {-1\over 2} & t- {1\over 2}
\end{pmatrix} \begin{pmatrix} {c'_1}\\ {c'_2}
\end{pmatrix}=\begin{pmatrix} {\cos t}\\ {\sin t}
\end{pmatrix}$$
\begin{cases}
-2tc'_2=\cos t\\
-\frac{1}{2} c'_1+ (t- {1\over 2})c'_2=\sin t
\end{cases}
$c'_2=\frac{cos t}{ -2t} \implies c_2=\int \frac{\cos t}{-2t}$
I don't know how to solve it
Where am I wrong?
Thanks !
| $$\begin{cases}
x'=2x+4y+\cos t\\
y'=-x-2y+\sin t
\end{cases}$$
It's easier to add both equations:
$$x'+2y'=\cos t +2 \sin t$$
$$x+2y=\sin t -2 \cos t +C_1$$
$$
\begin{cases}
x(t)&=-2y+\sin t -2 \cos t +C_1 \\
x'(t)&=-2y'+\cos t +2 \sin t
\end{cases}
$$
$$x'-2x=4y + \cos t$$
$$ y'=2\cos t -C_1$$
$$\implies y(t)=2\sin t -C_1t+C_2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4389523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many method to evaluate the integral $\int_{0}^{1} \frac{\ln ^{n}(1-x)}{x} d x , \textrm{ where }n\in N?$ $$
\begin{aligned}
\int_{0}^{1} \frac{\ln (1-x)}{x} d x &\stackrel{x \rightarrow 1-x}{=} \int_{0}^{1} \frac{\ln x}{1-x} d x \\
&=\sum_{k=0}^{\infty} \int_{0}^{1} x^{k} \ln x d x \\
&=\sum_{k=0}^{\infty} \int_{0}^{1} \ln x d\left(\frac{x^{k+1}}{k+1}\right) \\
& \stackrel{I B P}{=} \sum_{k=0}^{\infty}\left(\left[\frac{x^{k+1} \ln x}{k+1}\right]_{0}^{1}-\int_{0}^1{\frac{x^{k}}{k+1}} d x\right) \\
&=-\sum_{k=0}^{\infty} \frac{1}{(k+1)^{2}} \\
&=-\zeta(2)
\end{aligned}
$$
Similarly,
$$
\begin{aligned}
\int_{0}^{1} \frac{\ln ^{2}(1-x)}{x} d x
\stackrel{x\mapsto 1-x}{=} & \int_{0}^{1} \frac{\ln ^{2} x}{1-x} d x \\
=& \sum_{k=0}^{\infty} \int_{0}^{1} x^{k} \ln ^{2} x d x \\
\stackrel{IBP}{=} & \sum_{k=0}^{\infty} \frac{1}{k+1}\left(\left[x^{k+1} \ln ^{2} x\right]_{0}^{1}-\int_{0}^{1} 2 x^{k} \ln x d x\right) \\
\stackrel{IBP}{=}&-2 \sum_{k=0}^{\infty} \frac{1}{k+1}\left(\left[\frac{x^{k+1}\ln x}{k+1}\right]_{0}^{1}-\int_{0}^{1} \frac{x^{k}}{k+1} d x\right) \\
=& 2 \sum_{k=0}^{\infty} \frac{1}{(k+1)^{3}} \\
=& 2 \zeta(3)
\end{aligned}
$$
Replacing the power of $\ln x$ by $n$ and performing integration by parts by $n$ times yields $$
\begin{aligned}\int_{0}^{1} \frac{\ln ^{n}(1-x)}{x} d x &\stackrel{x\mapsto 1-x}{=} \int_{0}^{1} \frac{\ln ^{n} x}{1-x} d x\\ &\qquad\qquad \vdots \\&= (-1)(-2)(-3) \cdots(-n) \sum_{k=0}^{\infty} \frac{1}{(k+1)^{n+1}}\\&= (-1)^{n} n ! \zeta (n+1)\end{aligned}
$$
My Question
Is there an alternative method to evaluate the integral?
| I would like to switch the integration into differentiation process by defining an integral partner $$I(a):=\int_{0}^{1} \frac{\ln(1-x)(1-x)^{a}}{x} d x.$$
Then
$$ \begin{aligned}I(a)& \stackrel{x \mapsto 1-x}{=} \int_{0}^{1} \frac{x^{a}\ln x}{1-x} d x\\&=\sum_{k=0}^{\infty} \int_{0}^{1} x^{k+a} \ln x d x
\\& \stackrel{IBP}{=} -\sum_{k=0}^{\infty} \frac{1}{(k+a+1)^2}
\end{aligned}$$
Differentiating both sides w.r.t. $a$ by $n-1$ times yields $$
\begin{aligned} \int_{0}^{1} \frac{\ln (1-x)}{x} \frac{\partial^{n-1}}{\partial a^{n-1}}(1-x)^{a} d x&=-(-2)(-3)\cdots(-n) \sum_{k=0}^{\infty} \frac{1}{(k+a+1)^{n+1}}\\\int_{0}^{1} \frac{1}{x}(1-x)^{a} \ln ^{n}(1-x) dx&= (-1)^nn!\sum_{k=0}^{\infty} \frac{1}{(k+a+1)^{n+1}}
\end{aligned}
$$
Now we can conclude that
\begin{aligned}
\int_{0}^{1} \frac{\ln ^{n}(1-x)}{x} d x &=I^{(n)}(0)\\ &=(-1)^{n} n ! \sum_{k=0}^{\infty} \frac{1}{(k+1)^{n+1}} \\
&=(-1)^{n} n !\zeta(n+1)
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4392101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
What is $\frac{\binom{n}{1}}{1} + \frac{\binom{n}{2}}{2} + \frac{\binom{n}{3}}{3} + \cdots + \frac{\binom{n}{n}}{n}$ Good Day
Today, I learnt that $$\frac{\binom{n}{0}}{1} + \frac{\binom{n}{1}}{2} + \frac{\binom{n}{2}}{3} + \cdots + \frac{\binom{n}{n}}{n + 1} = \frac{2 ^ {n + 1} - 1}{n + 1}$$
I changed it a little and tried to find out $$\frac{\binom{n}{1}}{1} + \frac{\binom{n}{2}}{2} + \frac{\binom{n}{3}}{3} + \cdots + \frac{\binom{n}{n}}{n}$$
Going with a similar approach, I figured out the expression is $$n \cdot {(\frac{\binom{n - 1}{0}}{1} + \frac{\binom{n - 1}{1}}{4}} + \frac{\binom{n - 1}{2}}{9} + \cdots \frac{\binom{n - 1}{n - 1}}{n ^ 2})$$
Similarly, then
*
*I integrated $$(1 + x) ^ {n - 1} = \binom{n - 1}{0} + \binom{n - 1}{1}x + \cdots + \binom{n - 1}{n - 1}x ^ {n - 1}$$ from $0$ to $x$.
*Divide both sides by $x$.
*Integrate from $0$ to $x$ again.
*Substitute $x = 1$
But, in the final step, I got stuck on $$\int_{0}^{x}{\frac{(1 + x) ^ {n} - 1}{xn}}dx$$ Putting in WolframAlpha gives a very non-elementary answer.
There are two scenarios.
*
*I am wrong in these calculations.
*This expression does not have a simple value.
I'd appreciate if somebody could help me figure this out. Sorry if I am making a silly mistake, I don't know calculus very well.
Thanks.
| As you thought, the answer to your calculation does not have a simple value. What you have donne looks correct to me
| {
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What's the measure of angle $PCB$ in the figure below? For reference: In the interior of a triangle ABC, a point $P$ is marked in such a way that: $PC=BC$ and the measure of the angle $PAB$ is equal to the measure of the angle $PAC$ which is $17°$. calculate the measure of angle $PCB$, if the measure of angle $B=107^o$ (Answer:$26^o$)
My progress
$\triangle ABC: \angle C = 180-107-34 = 124+\theta\\
\angle CBP=\angle CPB=90^o - \frac{\theta}{2}\\
\triangle APC: \angle APC = 124^o+\theta\\
\triangle ABP: \angle BPA = 146-\frac{\theta}{2} $
...?
| Applying Trigonometric form of Ceva's theorem,
$ \displaystyle \sin \angle PAC \cdot \sin \angle PCB \cdot \sin \angle PBA$
$$= \sin \angle ACP \cdot \sin \angle CBP \cdot \sin \angle BAP $$
i.e. $~ \displaystyle \sin 17^\circ \cdot \sin \theta \cdot \sin \left(17^\circ + \frac {\theta}{2}\right)$
$$= \sin (39^\circ - \theta) \cdot \sin \left(90^\circ - \frac {\theta}{2}\right) \cdot \sin 17^\circ$$
$~ \displaystyle 2 \sin \frac {\theta}{2} \cdot \sin \left(17^\circ + \frac {\theta}{2}\right) = \sin (39^\circ - \theta)$
$\cos 17^\circ - \cos (17^\circ + \theta) = \cos (51^\circ + \theta)$
$\displaystyle \cos 17^\circ = \cos ((34^\circ + \theta) - 17^\circ) + \cos ((34^\circ + \theta) + 17^\circ)$
$\displaystyle \cos 17^\circ = 2 \cos (34^\circ + \theta) \cos 17^\circ$
$\cos (34^\circ + \theta) = \frac 12 = \cos 60^\circ$
$ \therefore \theta = 26^\circ$
| {
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Find all integers $x,y$ such that $x^5-x^3-x^2+1=y^2$ Find all integers $x,y$ such that $x^5-x^3-x^2+1=y^2$
I think we need to factor this out and I've managed to factor it to $(x-1)(x+1)^2(x^2-x+1)=y^2$, but I'm not sure what to do here. Have I done something wrong? Am I on the right path? Please help, thanks in advance.
| There is a much simpler way to finish the above answer.
We first show that there is no integer $x$ satisfying $x \ge 2$ that is in a solution. Indeed, let us assume that $x$ is an integer satisfying $x \ge 2$ that is in an solution $(x,y)$ to this equation. Then as noted, $x^2+x+1$ is a square for such an $x$. [To elaborate, factoring the LHS of the equation $x^5-x^3-x^2+1 = y^2$ yields
$$(x-1)^2(x+1)(x^2+x+1) = y^2.$$
So for $(x,y)$ to be an integral solution, the LHS to be an integral square, and thus $(x+1)(x^2+x+1)$ has to be an integral square. However, as $x^2+x+1 = x(x+1)+1$, it follows that $x^2+x+1$ and $x+1$ are positive [in particular, non-zero] and relatively prime to each other for any integral $x \ge 2$ i.e., they share no integral factors. So, the only way the product $(x+1)(x^2+x+1)$ can be an integral square is if each of $x+1$, $x^2+x+1$ is itself an integral square. Thus, for $(x,y)$ to be an integral solution, each of $x+1$, $x^2+x+1$ has to be an integral square, and thus in particular indeed, $x^2+x+1$ has to be an integral square.] However, it is impossible for $x^2+x+1$ to be an integral square as
$$(x+1)^2 =x+2x+1>x^2+x+1>x^2.$$
Then one can check that there is no negative integer $x$ satisfying $|x|\ge 2$ that can be a solution, as the LHS of the equation $x^5-x^3-x^2+1 = y^2$ is negative for such $x$.
Thus if $x$ is in a solution, it follows that $|x| \le 1$ and thus $x \in \{-1,0,1\}$. However, note that $x=-1,y=0$, $x=0,y \in \{-1,1\}$, and $x=1,y=0$ are solutions. These from the above paragraph are in fact all the solutions.
| {
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binomial identity: $\sum_{k=x+y}^{\infty}\binom{k-1}{y-1}\binom{k-y}{x}u^k = \binom{x+y-1}{y-1}\left(\frac{u}{1-u}\right)^{x+y}$? I met a problem which gave me the left part, and I can compute left part and get right part by Mathematica. However, I don't know how to prove:
$$\sum_{k=x+y}^{\infty}\binom{k-1}{y-1}\binom{k-y}{x}u^k = \binom{x+y-1}{y-1}\left(\frac{u}{1-u}\right)^{x+y}$$
with $x, y \in \mathbb{Z}, x \ge 0, y \ge 1, 0 \le u < 1$.
My Questions:
*
*How to prove above binomial identity?
*Is there simple argument behind it? Since it's quite simple, maybe we can construct two equivalent counting processes.
| As explained in this thread, the taylor polynomial of
$$\left(\frac{1}{1-u}\right)^{x+y} = \sum_{k=0}^\infty {k+x+y-1 \choose x+y-1 } u^k$$
So
\begin{align}\binom{x+y-1}{y-1}\left(\frac{u}{1-u}\right)^{x+y}
&= \sum_{k=0}^\infty \binom{x+y-1}{y-1}{k+x+y-1 \choose x+y-1 } u^{k+x+y}
\\
&= \sum_{k=0}^\infty \frac{(k+x+y-1)!}{x!(y-1)!k!} u^{k+x+y}
\\
&= \sum_{k=x+y}^\infty \frac{(k-1)!}{x!(y-1)!(k-x-y)!} u^{k}
\\
&= \sum_{k=x+y}^\infty \binom{k-1}{y-1}{k-y \choose x }u^{k}
\end{align}
| {
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Triangle-geometry problem Here is the question:
$\cos(A-B) = \frac{7}{8}$, $\cos(C) = ?$
By the Law of Cosines, I get:
$AB^2 = 41-40\cos(C)$
I also tried to expand $\cos(A-B)$ by the compound angle formula, getting:
*
*$\cos(A)\cos(B) + \sin(A)\sin(B)$
Which by the Law of Sines becomes:
*
*$\cos(A)\cos(B) + \frac{5}{4} \sin(B)^2 = \frac{7}{8}$
That's where I have been able to get so far. One thing though that has been bothering me is whether $AB =3$. I am tempted to go down that way because of the Pythagorean triple $3^2 + 4^2 = 5^2$. However, they have not specified that $\angle{A} = \frac{\pi}{2}$, so I am worried about wrongly assuming it. Any assistance would be greatly appreciated.
| You can write the angles $A$ and $B$ as follows
$ A = \left( \dfrac{A + B}{2} \right) + \left( \dfrac{A - B}{2} \right) $
and
$ B = \left( \dfrac{A + B}{2} \right) - \left( \dfrac{A - B}{2} \right ) $
The angle $\theta = \dfrac{A - B}{2}$ is known, it is
$\theta = \dfrac{1}{2} \cos^{-1} (\frac{7}{8} ) $
From the law of sines we have
$ \dfrac{\sin A}{\sin B} = \frac{5}{4} $
Hence, since $C = \pi - (A+B) $ then $\phi = \dfrac{A + B}{2} = \dfrac{\pi - C}{2} $
$ \dfrac{\sin( \phi + \theta ) }{\sin (\phi - \theta) } = \dfrac{5}{4} $
The only unknown here is $\phi$.
Cross multiplying and expanding
$ 4 \sin (\phi + \theta ) = 5 \sin(\phi - \theta) $
$ 4 ( \sin(\phi) \cos(\theta) + \cos(\phi) \sin(\theta) ) = 5 (\sin(\phi) \cos(\theta) - \cos(\phi) \sin(\theta) )$
Collecting terms
$ \sin(\phi) \cos(\theta) = 9 \cos(\phi) \sin(\theta) $
Hence,
$ \tan(\phi) = 9 \tan(\theta) $
Now
$\theta = \frac{1}{2} \cos^{-1} \frac{7}{8} $
Therefore,
$\cos(\theta) = \sqrt{ \dfrac{1 + \frac{7}{8} }{2} } = \dfrac{\sqrt{15}}{4} $
and
$\tan(\theta) = \sqrt{ \frac{16}{15} - 1 } = \dfrac{1}{\sqrt{15}} $
Hence
$\tan(\phi) = \dfrac{9}{\sqrt{15}} $
from which
$\sec(\phi) = \sqrt{\frac{32}{5}} $
and
$\cos(\phi) = \sqrt{\frac{5}{32}} $
Thus
$\cos(2 \phi) = 2 \left( \frac{5}{32} \right) - 1 = - \dfrac{11}{16} $
But $ 2 \phi = \pi - C $, therefore, $C = \pi - 2 \phi $
From which $\cos(C) = - \cos(2 \phi) = \boxed{\dfrac{11}{16}}$
| {
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Quick way to find solution $(\sqrt3,3)$ to the system $x^2+y^2-4y=0$ and $y=(1-\frac{\sqrt3}{3})x^2+x$? I would like to propose an intersection from an exam exercise for a Mathematical Analysis assignment at the university. This is a small part that involves calculating a double integral.
I have the following system to solve:
\begin{cases} x^2+y^2-4y=0 \\ y=(1-\frac{\sqrt3}{3})x^2+x \end{cases}
I know the points of intersection are $(0,0)$ $(\sqrt3,3)$.
I would like to be able to quickly find the point $(\sqrt3,3)$ without using Cardano's equations. Is there something I'm missing?
I have tried various approaches but I always find myself with laborious calculations that waste a lot of time. Is there any geometric approach or some parabola or circumference property to help find that point without going through a third degree polynomial? Unfortunately, the resolution of the resulting third degree polynomial is not trivial.
| $y=(1-\frac{\sqrt3}{3})x^2+x$
$x^2+y^2-4y=0$
$x^2+y^2-4y+4=x^2+(y-2)^2=4$
$x^2+(y-2)^2=x^2+((1-\frac{\sqrt3}{3})x^2+x-2)^2=4$
$x^2-3+((1-\frac{\sqrt3}{3})x^2+x-2)^2-1=0$
$(x-\sqrt3)(x+\sqrt3)+((1-\frac{\sqrt3}{3})x^2+x-2-1)((1-\frac{\sqrt3}{3})x^2+x-2+1)=0$
$(x-\sqrt3)(x+\sqrt3)+\frac{1}{9}((3-\sqrt3)x^2+3x-9)((3-\sqrt3)x^2+3x-3)=0$
$(x-\sqrt3)(x+\sqrt3)+\frac{1}{9}((x-\sqrt3)((3-\sqrt3)x+3\sqrt3))((3-\sqrt3)x^2+3x-3)=0$
$(x-\sqrt3)[(x+\sqrt3)+\frac{1}{9}((3-\sqrt3)x+3\sqrt3))((3-\sqrt3)x^2+3x-3)]=0$
$x-\sqrt3=0$
$x=\sqrt3$
| {
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Finding the partial fractions decomposition of $\frac{9}{(1+2x)(2-x)^2} $ So this is basically my textbook work for my class, where we are practicing algebra with partial fractions.
I understand the basics of decomposition, but I do not understand how to do it when then the denominator is a power of $x^2$?
e.g. this question - $$\frac{9}{(1+2x)(1-x)^2} $$
I understand that it will turn into- $$\frac{9}{(1+2x)(2-x)^2} = \frac {A}{1+2x} + \frac {B}{(1-x)}+ \frac {C}{(1-x)^2}$$ and then it will become
$$\frac{9}{(1+2x)(1-x)^2} =\frac{A(1-x)^2 +B(1+2x)(1-x)+C(1+2x)}{(1+2x)(1-x)^2}$$
but what do you do once you are at this step? The example on the textbook isn't very clear, so if anyone could tell me what I do after doing this, and why that is the case, I would be very thankful.
| You have $$\frac{9}{(1+2x)(2-x)^2} =\frac{A(2-x)^2 +B(1+2x)(2-x)+C(1+2x)}{(1+2x)(2-x)^2}$$
By equaling coefficients of the terms $x^0$, $x^1$ and $x^2$, you have the following equations to solve :
$$
\begin{cases}
4A+2B+C=9 \\
-4A+3B+2C=0\\
A-2B=0
\end{cases}$$
Can you continue from here ?
| {
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Showing $\sin^s\theta+\cos^s\theta\le1-K_s\cos^2\theta\sin^2\theta$, for $s>2$ and $K_s=\min\{\frac14s(s-2),2\}$ I was reading through an article (X. Lu, B. Wennberg, Solutions with increasing energy for the spatially homogeneous Boltzmann equation (2002)) and this elementary inequality is used in a proof:
For $s>2$ and $\theta\in [0,\pi/2]$,
$$\sin^s(\theta) + \cos^s(\theta) \le 1- K_s \cos^2(\theta)\sin^2(\theta)$$
where $K_s=\min\left\{\frac{1}{4}s(s-2), 2 \right\}$.
I'm trying to show it but I'm not succeeding.
I tried to use another another elementary inequality found in the same proof: for $s>2$ and $a,b \ge 0$
$$a^s+b^s \le (a+b)^s \le a^s + b^s + 2^s(a^{s-1}b+ab^{s-1}),$$
but I didn't succeed as well. Do you have any suggestions? Thank you in advance
| We split into two cases:
Case 1: $s \ge 4$
We have $K_s = 2$.
We have
$$\sin^s\theta + \cos^s\theta + K_s \cos^2\theta \sin^2\theta
\le \sin^4\theta + \cos^4\theta + 2 \cos^2\theta \sin^2\theta
= 1.$$
$\phantom{2}$
Case 2: $2 < s < 4$
We have $K_s = \frac14 s(s - 2)$.
Using Bernoulli inequality
$(1 + x)^r \le 1 + rx$ for all $x \ge -1, \, 0 < r \le 1$, we have
$$(\sin^2\theta)^{s/2 - 1} = (1 - \cos^2 \theta)^{s/2 - 1}\le 1 - \cos^2 \theta\cdot (s/2 - 1)$$
and
$$(\cos^2\theta )^{s/2 - 1} = (1 - \sin^2 \theta)^{s/2 - 1} \le 1 - \sin^2\theta \cdot (s/2 - 1).$$
Thus, we have
\begin{align*}
&\sin^s\theta + \cos^s\theta + K_s \cos^2\theta \sin^2\theta \\
=\,& \sin^2\theta\, (\sin^2\theta)^{s/2 - 1}
+ \cos^2 \theta\, (\cos^2\theta )^{s/2 - 1} + \frac14 s(s - 2)\cos^2\theta \sin^2\theta\\
\le\,& \sin^2\theta \left[ 1 -\cos^2\theta\cdot (s/2 - 1)\right] + \cos^2\theta \left[1 - \sin^2\theta\cdot (s/2 - 1)\right] + \frac14 s(s - 2)\cos^2\theta \sin^2\theta\\
=\,& 1 - \frac{(s - 2)(4 - s)}{4}\cos^2\theta \sin^2\theta\\
\le\,& 1.
\end{align*}
We are done.
| {
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Prove that $\frac{a-c}{k}=\frac{d+b}{n}$, where $N=a^2+b^2=c^2+d^2$ and $k=\gcd(a-c,d-b)$, $n=\gcd(a+c, d+b)$ Let $N$ be odd and $N = a^2 + b^2 = c^2 + d^2$, where $a, b, c, d \in \mathbb{N}$ and WLOG let $a, c$ be odd, $b, d$ be even, $a > c$, and $b < d$. Prove that $\frac{a-c}{k}=\frac{d+b}{n}$.
I first began the proof my showing that $k$ and $n$ are positive integers.
Let $k= \gcd(a -c, d-b)$ and $n= \gcd(a + c, d + b)$. If $a$ and $c$ are both odd, then $a\pm c$ is even. Similarly, if $d$ and $b$ are both even, then $b\pm d$ is also even. Thus, $k$ and $n$ are also both even.
Now I wish to show that $\frac{a-c}{k}=\frac{d+b}{n}$.
I first tried to use the fact that $N = a^2 + b^2 = c^2 + d^2$.
$$
\begin{align*}
a^2 + b^2 &= c^2 + d^2\\
a^2 -c^2 &= d^2-b^2\\
(a+c)(a-c)&=(d+b)(d-b)\\
\end{align*}
$$
This seems like I'm getting close but I can't quite see how to finish it. Any hints are appreciated.
| It seems to me that those sums and differences are there to muddy the waters.
Let's write $A=a+c$, $C=A-c$, $D=d+b$ and $B=d-b$. As you observed, the given equation $a^2+b^2=c^2+d^2$ is equivalent to
$$AC=DB.\qquad(*)$$
And the claim is that $(*)$ implies
$$\frac C{\gcd(B,C)}=\frac{D}{\gcd(A,D)}.\qquad(\dagger)$$
It is simple to do this one prime factor at a time. So let $p$ be a prime number, and assume that in the prime factorizations we have $p^n$ appearing in $A$, $p^m$ appearing in $C$, $p^r$ in $D$ and $p^s$ in $B$. By uniqueness of factorization $(*)$ implies that
$$n+m=r+s.\qquad(**)$$
The claim $(\dagger)$ is then translated to
$$
m-\min\{m,s\}=r-\min\{r,n\}.\qquad(\dagger\dagger)
$$
This is now easy to see. From $(**)$ it follows that $m-s=r-n$. Hence $\min\{m,s\}=s$ if and only if $\min\{r,n\}=m$ in which case the claim is exactly $m-s=r-n$. In the complementary case $\min\{m,s\}=m$ and $\min\{r,n\}=r$ the claim $(\dagger\dagger)$ reads $0=0$.
| {
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If $\log_3(A)$ is rational, can $\log_3(A+1)$ be rational? (Where $A$ is a positive real number.) Question:
If $\log_3(A)$ is rational, can $\log_3(A+1)$ be rational? (Where $A$ is a positive real number.)
My Working:
Starting by assuming
$$\log_3(A) = x \ {\rm{and}} \ \log_3(A+1) = y$$
it gives $$3^x = A \ {\rm {and}} \ 3^y = A + 1$$
Which further implies that,
$$3^y = 3^x + 1$$
and
$$3^y - 3^x = 1$$
$$3^{x}(3^{y-x}- 1) = 1$$
$$3^x = 1\ {\mathrm {and} }\ 3^{y-x} = 2 $$
$$x = 0\ {\mathrm {and} }\ 3^{y} = 2 $$
Which further gives us:
$$y = \frac{\log(2)}{\log(3)}$$
and hence $y$ is irrational.
Is my work correct?
| More generally, let's determine when $\log_3(A)$ and $\log_3(A+m)$ are both rational for any positive integer $m$. Set $x = \log_3(A)$ and $y=\log_3(A+m)$, so that $3^y = 3^x + m$. Suppose that $x=\frac ab$ and $y=\frac cd$ are rational (in lowest terms).
*
*Suppose first that $b>d$. Then $3^c = (3^x+m)^d$, and so $3^x$ is a root of the polynomial $(T+m)^d - 3^c$ which has integer coefficients. But this has smaller degree than the minimal polynomial of $3^x = \sqrt[b]{3^a}$, namely $T^b - 3^a$, which is a contradiction.
*Similarly, suppose that $d>b$. Then $3^a = (3^y-m)^b$, and so $3^y$ is a root of $(T-m)^b + 3^a$; but this has smaller degree than the minimal polynomial of $3^y = \sqrt[d]{3^c}$, which is $T^d-3^c$.
*Finally, suppose that $b=d$. Then $3^x$ again has minimal polynomial $T^b - 3^a$ but also is a root of the polynomial $(T+m)^b - 3^c$. Since these are monic of the same degree, we must have $T^b - 3^a = (T+m)^b - 3^c$. This is impossible if $b>1$ (look at the coefficient of $T^k$ for any $0<k<b$), and therefore it's necessary that $b=d=1$, so that $x$ and $y$ are actually integers.
In conclusion, we can only have $\log_3(A)$ and $\log_3(A+m)$ simultaneously rational if both $A$ and $A+m$ are integer powers of $3$. In particular, this doesn't happen when $m=1$.
| {
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Computing $\int_{0}^{1} \frac{\sin ^{-1} (x) \ln (1+x)}{x^{2}} d x$ Applying integration by parts splits the integral into 3 integrals,
$\displaystyle \begin{aligned}I&=\int_{0}^{1} \frac{\sin ^{-1} x \ln (1+x)}{x^{2}} d x\\&=-\int_{0}^{1} \sin ^{-1} x \ln (1+x) d\left(\frac{1}{x}\right) \\&=-\left[\frac{\sin ^{-1} x \ln (1+x)}{x}\right]_{0}^{1}+\underbrace{\int_{0}^{1} \frac{\ln (1+x)}{x \sqrt{1-x^{2}}}}_{K} +\underbrace{\int_{0}^{1}\frac{\sin ^{-1} x}{x}}_{L} d x-\underbrace{\int_{0}^{1} \frac{\sin ^{-1} x}{1+x}}_{M} d x \end{aligned} \tag*{} $
Letting $x= \cos \theta$ for $K$ and $\sin^{-1}x \mapsto x$ for $L$ and $M$, yields
$\displaystyle I=-\frac{\pi}{2} \ln 2 +\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{\ln (1+\cos \theta)}{\cos \theta} d \theta}_{K}+\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{x\cos x }{\sin x} d x}_{L}-\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{x\cos x }{1+\sin x} d x }_{M}\tag*{} $
For the integral $ K,$putting $ a=1$ in my post yields
$\displaystyle \boxed{K=\frac{\pi^{2}}{8}}\tag*{} $
For the integral $ L,$ integration by parts yields
$\displaystyle \begin{aligned}L &=\int_{0}^{\frac{\pi}{2}} x d \ln (\sin x) \\&=[x \ln (\sin x)]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} \ln (\sin x) d x \\&=\boxed{\frac{\pi}{2} \ln 2}\end{aligned}\tag*{} $
For the integral $ M,$ integration by parts yields
$\displaystyle \begin{aligned}M &=\int_{0}^{\frac{\pi}{2}} x d \ln (1+\sin x)\\&=[x \ln (1+\sin x)]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} \ln (1+\sin x) d x \\&=\frac{\pi}{2} \ln 2-\underbrace{\int_0^{\frac{\pi}{2} }\ln (1+\sin x) d x}_{N}\end{aligned}\tag*{} $
For the integral $ N,$ using my post in the second last step yields
$\displaystyle \begin{aligned}N \stackrel{x\mapsto\frac{\pi}{2}-x}{=} &\int_{0}^{\frac{\pi}{2}} \ln (1+\cos x) d x \\=&\int_{0}^{\frac{\pi}{2}} \ln \left(2 \cos ^{2} \frac{x}{2}\right) d x \\=&\frac{\pi}{2} \ln 2+2 \int_{0}^{\frac{\pi}{2}} \ln \left(\cos \frac{x}{2}\right) d x \\=&\frac{\pi}{2} \ln 2+4 \int_{0}^{\frac{\pi}{4}} \ln (\cos x) d x \\=&\frac{\pi}{2} \ln 2+4\left(\frac{1}{4}(2 G-\pi \ln 2)\right) \\=&\boxed{-\frac{\pi}{2} \ln 2+2 G}\end{aligned}\tag*{} $
where $ G$ is the Catalan’s Constant.
Putting them together yields
$\displaystyle \boxed{I=-\pi \ln 2+\frac{\pi^{2}}{8}+2G} \tag*{} $
Question: Is there any shorter solution?
| Perhaps not 100% satisfactory, as I've performed each step via Mathematica rather than a step-by-step derivation, but the following could be considered a more simple solution. Note that $$\ln(1+x) = \sum_{n=1}^\infty(-1)^{n+1}\frac{x^n}{n}$$ and $$\int_0^1\sin^{-1}x\,x^n dx = \frac{\sqrt{\pi}}{n+1}\left( \frac{\sqrt\pi}{2} - \frac{\Gamma\big(1+\frac n2\big)}{(n+1)\Gamma\big(\frac{n+1}{2}\big)} \right),$$ where, for $n=-1$, the above is understood to be $\frac{\pi\ln2}{2}$. Performing the infinite sum yields the answer given in the original question.
| {
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Assume $f$ is analytic on $\left|z-1\right|<1$,$f'(z)=1/z$ and $f(1)=0$, Prove that on this region $f(z)=\text{Log}(z)$. Assume $f$ is analytic on $\left|z-1\right|<1$,$f'(z)=1/z$ and $f(1)=0$, Prove that on this region $f(z)=\text{Log}(z)$.
Since $f$ is analytic on $\left|z-1\right|<1$, so for every $z=x+iy$ in this region we have that $u_x(x,y)=v_y(x,y)$ and $u_y(x,y)=-v_x(x,y)$, on the other hand $f'(z)=1/z$ implies that $u_x(x,y)=\frac{x}{x^2+y^2}$ and $v_x(x,y)=\frac{-y}{x^2+y^2}$, So $u(x,y)=1/2 \ln(x^2+y^2)+c_1(y)$, Moreover $$\frac{y}{x^{2}+y^{2}}+c_1'\left(y\right)=\frac{y}{x^{2}+y^{2}}$$
From which we conclude that $c_1(y)=c$
$$v\left(x,y\right)=\arctan\left(\frac{y}{x}\right)+c_{2}\left(y\right)$$
$$\frac{1}{\sqrt{x^{2}+y^{2}}}+c'_{2}\left(y\right)=\frac{x}{x^{2}+y^{2}}$$
Is this the correct way to go?
| $u_x=\dfrac{x}{x^2+y^2}$ implies $u(x,y)=\dfrac{1}{2}\ln(x^2+y^2)+c(y)$
whereas $u_y=\dfrac{y}{x^2+y^2}$ implies $u(x,y)=\dfrac{1}{2}\ln(x^2+y^2)+l(x)$
which forces $c(y)$ and $l(x)$ to be equal to the same constant function.
Also,
$v_y=\dfrac{x}{x^2+y^2}$ implies $v(x,y)=\arctan\left(\dfrac{y}{x}\right)+g(x)$
whereas $v_x=\dfrac{-y}{x^2+y^2}$ implies $v(x,y)=\arctan\left(\dfrac{y}{x}\right)+h(y)$, which you can verify by partial-differentiating $\arctan\left(\dfrac{y}{x}\right)$ with respect to $x$.
| {
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A circle of radius $1$ is tangent to $y=x^2$ at two points. Find the area bounded by the circle and the parabola. (Alternate solutions?)
This is from Rambo's Math subject GRE book.
One solution to this problem is to note that the equation of the circle is $x^2+(y-k)^2=1$. By taking the derivative of this and solving for $y'$, and then setting this equal to the derivative of the parabola, $y'=2x$, we can solve for $x$ and $k$ and get all the information we need to set up the necessary integeral:
$2\int_0^{\frac{\sqrt{3}}{2}}\frac{5}{4}-\sqrt{1-x^2}-x^2dx=\frac{3\sqrt{3}}{4}-\frac{\pi}{3}$
Where the expression involving a square root can is evaluated with trig substitution and power reduction identities.
The author of the text mentions that this problem also be solved by using some trigonometry and the fact that:
$2\int_0^{\frac{\sqrt{3}}{2}}x^2dx=\frac{\sqrt{3}}{2}$
However I can't figure out what he means by this. Can anyone show me how to solve this problem using less calculus and more trigonometry? Also, if anyone has any other alternative methods, I'd greatly appreciate it. Thanks!
|
Equation of the circle is $~x^2 + (y-k)^2 = 1$
Equation of the parabola is $~y = x^2$
If they are tangent to each other,
$y + (y-k)^2 = 1 \implies y^2 - (2k-1) y + (k^2-1) = 0~$ should have a double root.
So we must have, $(2k-1)^2 - 4 (k^2-1) = 0 \implies k = \dfrac 54$
At point of tangency, $ \displaystyle y = \frac{2k-1}{2} = \frac 34, x = \pm \sqrt y = \pm \frac{\sqrt{3}}{2}$
$ \displaystyle \cos \alpha = \frac{\sqrt3}{2} \implies \alpha = \frac {\pi}{3}$
We can just focus on the desired area to the right of y-axis and then using symmetry, we can multiply the result by $2$.
Half of the desired area $~ = [OABC] - [BCE] - [OAB]$ (note we use the curved path $BE$ and $OB$)
$ \displaystyle [OABC] = \frac 12 \left(\frac 54 + \frac 34\right) \cdot \frac {\sqrt3}{2} = \frac{\sqrt3}{2}$
$ \displaystyle [BCE] = \frac {\pi}{6}$
$ \displaystyle [OAB] = \int_0^{\sqrt3/2} x^2 ~ dx = \frac{\sqrt3}{8} $
So, the desired area $ ~ = \displaystyle 2 \cdot \left(\frac{\sqrt3}{2} - \frac{\sqrt3}{8} - \frac{\pi}{6}\right) = \frac{3 \sqrt3}{4} - \frac{\pi}{3}$
| {
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Show $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{1999}-\frac{1}{2000} =\frac{1}{1001}+\frac{1}{1002}+\ldots+\frac{1}{1999}+\frac{1}{2000}$ I am trying to show that
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{1999}-\frac{1}{2000}
=\frac{1}{1001}+\frac{1}{1002}+\ldots+\frac{1}{1999}+\frac{1}{2000}.
$$
It seems there is some sort of generalizable pattern here, so I will verify it for smaller numbers:
$$
\begin{align*}
\text{Say, }n=4 \hspace{35pt} 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}&=\frac{1}{3}+\frac{1}{4}\\
\frac{12}{12}-\frac{6}{12}+\frac{4}{12}-\frac{3}{12}&=\frac{4}{12}+\frac{3}{12}\\
\frac{7}{12}&=\frac{7}{12}
\end{align*}
$$
So, my guess on the general formula is
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{n-1}-\frac{1}{n}=\frac{1}{n/2+1}+\ldots+\frac{1}{n-1}+\frac{1}{n}.
$$
This really seems like I am getting somewhere, but how can I finish off the proof? Is induction viable?
| above, both are excellent answers. This is another way to look at it.
$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{1999}-\frac{1}{2000} =\frac{1}{1001}+\frac{1}{1002}+\ldots+\frac{1}{1999}+\frac{1}{2000}$
After cancelling out positive matching terms on both sides and moving negative matching terms to RHS, we get:
$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{999}-\frac{1}{1000} =\frac{1}{501}+\frac{1}{502}+\ldots+\frac{1}{999}+\frac{1}{1000}$
Continuing the descent, we will end up with:
$1-\frac{1}{2}+\frac{1}{3}=\frac{1}{2}+\frac{1}{3}$
which is true.
| {
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Distributing 10 swords among 3 champions. Right, so the protocol here seems to be stars and bars. Now let's say the blacksmith is to make sure that each champion gets two swords. We may fix two for each champion and count the combinations for the remaining $10-(3\times2)=4$ swords.
$$
\binom{4+3-1}{3-1} = 15
$$
But my first take to this "two for each" requirement yielded the use of the theorem one on the linked Wikipedia article. I thought if we distribute 5 swords to 3 champions making sure each gets at least one and then distribute again 5 swords giving each at least one then we have distributed 10 swords making sure each champion gets two.
$$
2\times\binom{5-1}{3-1} = 12 \neq 15
$$
I am inclined to believe that 15 is the correct count but can't see why the second approach is counting under. Could someone tell what cases the second approach misses and why?
| There are two problems:
*
*If you (a) distribute 5 swords to 3 champions making sure each gets at least one and (b) do that again, the total amount of outcomes is not $\binom{5-1}{3-1} + \binom{5-1}{3-1} = 6+6 = 12$ but $\binom{5-1}{3-1} \cdot \binom{5-1}{3-1} = 6 \cdot 6 = 36$. So we should be looking for an overcount, not an undercount.
*The reason this is bigger than $15$ is because certain outcomes are counted multiple times. If a champion gets $2$ swords in the first round and $1$ sword in the second round, that should be the same outcome as getting $1$ sword in the first round and $2$ swords in the second round, but $6 \cdot 6$ counts these separately.
We can confirm that $15$ is correct by brute force: the options are
\begin{array}{ccc}
6+2+2 & 2+6+2 & 2+2+6 \\
5+3+2 & 3+5+2 & 3+2+5 \\
5+2+3 & 2+5+3 & 2+3+5 \\
4+3+3 & 3+4+3 & 3+3+4 \\
4+4+2 & 4+2+4 & 2+4+4
\end{array}
| {
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Evaluating a limit of a Fourier series I have a Fourier series representation of a solution to the heat equation, given by
$\displaystyle u(t,x) = \\ \displaystyle\sin \omega t + \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\omega\left\lbrack \frac{\omega\sin \omega t + \left( n - \frac{1}{2} \right)^{2}\pi^{2}\left( \cos \omega t - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right)}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} \right\rbrack\cos\left\lbrack \left( n - \frac{1}{2} \right)\pi x \right\rbrack}.$
How would one go about computing
$\displaystyle \lim_{\omega\rightarrow\infty}{u(t,0)}$
explicitly?
| A quicker method to obtain the same result is as follows. Firstly, merge the $\sin\omega t$ term with the rest of the Fourier series. This is achieved by noting that the coefficient of $\sin\omega t$ is unity, and so the Fourier series of unity in terms of the spatial variable $x$ may be employed:
\begin{align*}
\sin\omega t &= \sum_{n=1}^{\infty}{\frac{4\left( -1 \right)^{n+1}}{(2n-1)\pi}\sin\omega t\cos\left[\left( n - \frac{1}{2}\right)\pi x\right]} \\
&= - \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\sin\omega t\cos\left\lbrack \left( n - \frac{1}{2} \right)\pi x \right\rbrack}.
\end{align*}
Thus
\begin{align*}
u(t,x) &= \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\omega\left\lbrack \frac{\omega\sin\omega t + \left( n - \frac{1}{2} \right)^{2}\pi^{2}\left( \cos\omega t - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right)}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} - \frac{\sin\omega t}{\omega} \right\rbrack\cos\left\lbrack \left( n - \frac{1}{2} \right)\pi x \right\rbrack}.
\end{align*}
Hence,
\begin{align*}
u(t,0) &= \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\left\lbrack \frac{\omega^{2}\sin\omega t + \left( n - \frac{1}{2} \right)^{2}\pi^{2}\omega\left( \cos\omega t - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right)}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} - \sin\omega t \right\rbrack} \\
&= \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\frac{\omega^{2}\sin\omega t + \left( n - \frac{1}{2} \right)^{2}\pi^{2}\omega\left( \cos\omega t - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right) - \left( \omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4} \right)\sin\omega t}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}}} \\
&= \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\frac{\left( n - \frac{1}{2} \right)^{2}\pi^{2}\omega\left( \cos\omega t - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right) - \left( n - \frac{1}{2} \right)^{4}\pi^{4}\sin\omega t}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}}}.
\end{align*}
To determine the limit as $\omega \rightarrow \infty$ of $u(t,0)$, consider each term separately:
\begin{align*}
\text{Limit 1:} && \lim_{\omega \rightarrow \infty}\left\lbrack \frac{\left( n - \frac{1}{2} \right)^{2}\pi^{2}\omega\left( \cos\omega t \right)}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} \right\rbrack, \\
\text{Limit 2:} && \lim_{\omega \rightarrow \infty}\left\lbrack \frac{\left( n - \frac{1}{2} \right)^{2}\pi^{2}\omega\left( - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right)}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} \right\rbrack, \\
\text{Limit 3:} && \lim_{\omega \rightarrow \infty}\left\lbrack \frac{- \left( n - \frac{1}{2} \right)^{4}\pi^{4}\sin\omega t}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} \right\rbrack,
\end{align*}
which all tend to zero, giving
\begin{equation*}
\lim_{\omega\rightarrow\infty}{u(t,0)} = 0
\end{equation*}
as required.
| {
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Differential equation $y'' - \frac{2y}{x^2} = 3\ln(-x)$ Hello there I am trying to solve
$$y'' - \frac{2y}{x^2} = \ln(-x)$$
I get this differential equation from book - filippov differential equations link on book (problem #606).
And same way I have solve (in this book)
$$y = \frac{C_1}{x} + \left[ C_2 - \frac{1}{3}\ln(-x) + \frac{1}{2}\ln^2(-x) \right] x^2$$
But I can not get this solve.
My solve:
$$y'' - \frac{2y}{x^2} = \ln(-x)$$
$$x^2y'' - 2y = x^2 \ln(-x)$$
$$k(k-1) - 2 = 0$$
$$k_1 = -1, k_2= 2, y_1=x^{-1}, y_2=x^2$$
$$y_0 = C_1y_1 + C_2y_2 = \frac{C_1}{x} + C_2x^2$$
$$W = \begin{vmatrix}
y_1 &y_2 \\
y'_1 &y'_2
\end{vmatrix} = \begin{vmatrix}
x^{-1} &x^2 \\
-x^{-2} &2x
\end{vmatrix} = 2+1 = 3$$
$$W_1 = \begin{vmatrix}
0 & y_2 \\
g(x) & *
\end{vmatrix} = \begin{vmatrix}
0 & x^2 \\
\ln(-x) & 2x
\end{vmatrix} = -x^2 \ln(-x)$$
$$W_2 = \begin{vmatrix}
y_1 & 0 \\
* &g(x)
\end{vmatrix} = \begin{vmatrix}
x^{-1} &0 \\
-x^{-2} & \ln(-x)
\end{vmatrix} = \frac{\ln(-x)}{x}$$
$$y_h = u_1y_1 + u_2y_2$$
$$u_1 = \int{\frac{W_1}{W}} = \int{\frac{-x^2 \ln(-x)}{3}} = -\frac{1}{9}x^3(3\ln(-x) - 1)$$
$$u_2 = \int{\frac{W_2}{W}} = \int{\frac{3\ln(-x)}{x}} = \frac{3}{2}\ln^2(-x)$$
$$y_h = -\frac{1}{9}x^3(3\ln(-x) - 1) x^{-1} + \frac{3}{2}\ln^2(-x) x^2 = -\frac{1}{3}x^2 \ln(-x) - \frac{x^2}{9} + \frac{3}{2}\ln^2(-x) x^2$$
$$y = y_0 + y_h$$
$$y = \frac{C_1}{x} + C_2x^2 - \frac{1}{3}x^2 \ln(-x) - \frac{x^2}{9} + \frac{3}{2}\ln^2(-x) x^2$$
General solution:
$$y = \frac{C_1}{x} + \left[ C_2 - \frac{1}{3} \ln(-x) - \frac{1}{9} + \frac{3}{2}\ln^2(-x) \right] x^2$$
Please help me where I'm wrong?
May be in book have wrong?
| HINT
You can also approach it as follows:
\begin{align*}
y'' - \frac{2y}{x^{2}} = \ln(-x) & \Longleftrightarrow x^{2}y'' - 2y = x^{2}\ln(-x)\\\\
& \Longleftrightarrow (x^{2}y'' + 2xy') - (2xy' + 2y) = x^{2}\ln(-x)\\\\
& \Longleftrightarrow (x^{2}y')' - (2xy)' = x^{2}\ln(-x)
\end{align*}
Can you take it from here?
| {
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Integral $\int_{-\infty}^\infty \frac{\tan^{-1}(x) - \tan^{-1}(x+a)}{(x-i)(x+a-i)} dx$ For $a\in\mathbb R$, I want to compute the integral
$$I = \int_{-\infty}^\infty \frac{\tan^{-1}(x) - \tan^{-1}(x+a)}{(x-i)(x+a-i)} dx.$$
Previously, I asked a question on a similar integral:
Integral $ \int_{-\infty}^\infty \frac{1}{x^2+1} \left( \tan^{-1} x + \tan^{-1}(a-x) \right) dx$
In that question, several ideas, such as (1) taking derivative with respect to $a$ or (2) contour integration, are suggested. However, both method seems to be not working for this integral.
This integral is motivated from physics calculation.
How can I evaluate $I$?
| Note that
\begin{align}
I(a)=&\int_{-\infty}^\infty \frac{\tan^{-1}x - \tan^{-1}\overset{x+a\to x}{(x+a)}}{(x-i)(x+a-i)} dx
=\int_{-\infty}^\infty \frac{\tan^{-1}x}{x-i}\Big(\frac{1}{x+a-i}-\frac{1}{x-a-i}\Big) dx
\end{align}
and it is straightforward to verify that $I=I^*$. Thus, it suffices to perform the integration on the real part, which can be expressed as
\begin{align}
J(a)=aI(a)
=& \int_{-\infty}^\infty \tan^{-1}x\Big(\frac{2x}{x^2+1}-\frac{x-a}{(x-a)^2+1} -\frac{x+a}{(x+a)^2+1}\Big) dx\\
=& \int_{-\infty}^\infty \frac{2x\tan^{-1}- x\tan^{-1}(x-a)-x\tan^{-1}(x+a)} {x^2+1}dx
\end{align}
$$J’(a)=\int_{-\infty}^\infty \frac{x} {(x^2+1)((x-a)^2+1)}-\frac{x} {(x^2+1)((x+a)^2+1)}\> dx=\frac{2\pi a}{a^2+4}
$$
Then
$$I(a)=\frac1a \int_0^a J’(s)ds= \frac1a \int_0^a \frac{2\pi s}{s^2+4}ds= \frac{\pi}{a}\ln\Big(1+\frac{a^2}{4}\Big)
$$
| {
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Simplify $\sqrt[3]{9\sqrt3-11\sqrt2}$ Simplify $$\sqrt[3]{9\sqrt3-11\sqrt2}$$
How can we actually simplify this radical?
| Starting with the observation that $\,9^2 \cdot 3 - 11^2 \cdot 2 = 1\,$, let $\,a = \sqrt[3]{9\sqrt3-11\sqrt2}\,$, $\,b = \sqrt[3]{9\sqrt3+11\sqrt2}\,$. Then $\require{cancel}\,a^3+b^3=9\sqrt3-\cancel{11\sqrt2}+9\sqrt3+\cancel{11\sqrt2}=18\sqrt{3}\,$ and $\,ab = \sqrt[3]{9^2 \cdot 3 - 11^2 \cdot 2} = 1\,$.
It follows that $\,(a+b)^3 = a^3+b^3+3ab(a+b)=18 \sqrt{3} + 3(a+b)\,$, so $\,t=a+b\,$ satisfies the equation $\,t^3 - 3 t - 18 \sqrt{3}=0\,$. The substitution $\,t = \sqrt{3}\,t'\,$ gives $\,3 \sqrt{3}\, t'^{\,3} - 3 \sqrt{3}\, t' - 18 \sqrt{3} = 0\,$, or $\,t'^{\,3}- t' - 6 = 0 \iff (t' - 2) (t'^{\,2} + 2 t' + 3) = 0\,$ with the only real root $\,t' = 2 \iff t = 2 \sqrt{3}\,$.
Then $\,a+b=2\sqrt{3}\,$, $\,ab=1\,$, so $\,a,b\,$ are the roots of $\,x^2 - 2 \sqrt{3} x + 1 = 0 \iff x = \sqrt{3} \pm \sqrt{2}\,$ with $\,a\,$ being the smaller root i.e. $\,a = \sqrt{3} - \sqrt{2}\,$, $\,b = \sqrt{3} + \sqrt{2}\,$.
More generally, consider the case of $\,a,b = \sqrt[3]{m \sqrt{p} \pm n\sqrt{q}}\,$ with $\,m^2 \cdot p - n^2 \cdot q = r^3\,$. Then, similarly to above, $\,a^3+b^3 = 2m\sqrt{p}\,$ and $\,ab = r\,$, so $\,(a+b)^3 = 2m \sqrt{p} + 3r(a+b)\,$, or $\,t^3 - 3r\,t - 2m \sqrt{p} = 0\,$ where $\,t=a+b\,$, or $\,p\, t'^{\,3} - 3r\, t' - 2m = 0\,$ where $\,t' = \frac{1}{\sqrt{p}}\,t\,$.
If the latter cubic has an eligible rational root $\,t' = 2s\,$, then $\,a+b=t=\sqrt{p}\,t'=2s\,\sqrt{p}\,$ and $\,ab = r\,$, so $\,a,b\,$ are the roots of $\,x^2 - 2s\sqrt{p}\, x + r = 0\,$ $\,\iff a,b = s\,\sqrt{p} \pm \sqrt{s^2 p - r}\,$.
Therefore a sufficient condition for $\,a,b = \sqrt[3]{m \sqrt{p} \pm n\sqrt{q}}\,$ to denest is for $\,m^2 \cdot p - n^2 \cdot q\,$ to be the cube of a rational $\,r\,$, and for the cubic $\,p\, t'^{\,3} - 3r\, t' - 2m\,$ to have an appropriate rational root, and in that case $\,a,b = \frac{1}{2}\left(t'\,\sqrt{p} \pm \sqrt{t'^{\,2} p-4r}\right)\,$.
| {
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An inequality $16(ab + ac + ad + bc + bd + cd) \le 5(a + b + c + d) + 16(abc + abd + acd + bcd)$ Let $a, b, c, d \ge 0$ be nonnegative real numbers such that $a + b + c + d \le 1$.
Show that $16(ab + ac + ad + bc + bd + cd) \le 5(a + b + c + d) + 16(abc + abd + acd + bcd)$.
If I set the power means as $A := (a + b + c + d)/4$ and $Q := \sqrt{(a^2 + b^2 + c^2 + d^2)/4}$ and $C := \sqrt[3]{(a^3 + b^3 + c^3 + d^3)/4}$, then
$$RHS - LHS = \frac{4}{3}(128A^3 - 96A^2 - 96QA + 15A + 16C^3 + 24Q^2),$$
where $0 \le A \le 1/4$. The power mean inquality states that $A \le Q \le C$. What can I do with the $-96QA$ term?
| Remark: Inspired by @Macavity's answer
Let $x = a + b, \ y = c + d$.
We have
\begin{align*}
\mathrm{RHS} - \mathrm{LHS} &=
5(x + y) + 16(ab y + cd x) - 16(ab + cd + xy)\\
&= 5(x + y) - 16 xy - 16(1 - y)ab - 16(1 - x)cd\\
&\ge 5(x + y) - 16 xy - 16(1 - y)\cdot \frac{x^2}{4} - 16(1 - x)\cdot \frac{y^2}{4} \tag{1}\\
&= 5(x + y) - 16xy - 4x^2 - 4y^2 + 4x^2y + 4xy^2\\
&= 5(x + y) - 8xy - 4(x + y)^2 + 4xy(x + y)\\
&= 5(x + y) - 4xy (2 - x - y) - 4(x + y)^2\\
&\ge 5(x + y) - (x + y)^2(2 - x - y) - 4(x + y)^2 \tag{2}\\
&= (x + y)(1 - x - y)(5 - x - y)\\
&\ge 0
\end{align*}
where we have used $ab \le (a + b)^2/4 = x^2/4$
and $cd \le (c + d)^2/4 = y^2/4$ in (1),
and $4xy \le (x + y)^2$ in (2).
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4438928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find number of positive integer solutions to the equation $a+b+c=15$ where a,b are even and c is odd This is equivalent to:
Coefficient of $x^{15}$ in $(1+x^2+x^4+\cdots+x^{14})^2(x+x^3+\cdots+x^{15})$
Or
Coefficient of $x^{14}$ in $(1+x^2+x^4+\cdots+x^{14})^3$
Is this approach correct?
| You say "positive" rather than "non-negative", so it is
*
*the coefficient of $x^{15}$ in $(x^2+x^4+\cdots+x^{14})^2(x+x^3+\cdots+x^{15})$
which is
*
*the coefficient of $x^{15}$ in $(x^2+x^4+\cdots+x^{12})^2(x+x^3+\cdots+x^{11})$
*the coefficient of $x^{16}$ in $(x^2+x^4+\cdots+x^{12})^3$
*the coefficient of $x^{10}$ in $(1+x^2+x^4+\cdots+x^{10})^3$
*the coefficient of $x^{5}$ in $(1+x+x^2+\cdots+x^{5})^3$
which is ${7 \choose 2}=21$
That is short enough to list as a check:
2 + 2 + 11
2 + 4 + 9
2 + 6 + 7
2 + 8 + 5
2 + 10 + 3
2 + 12 + 1
4 + 2 + 9
4 + 4 + 7
4 + 6 + 5
4 + 8 + 3
4 + 10 + 1
6 + 2 + 7
6 + 4 + 5
6 + 6 + 3
6 + 8 + 1
8 + 2 + 5
8 + 4 + 3
8 + 6 + 1
10 + 2 + 3
10 + 4 + 1
12 + 2 + 1
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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.