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Ring theory - cosets problem; prove specific result Let $F=\mathbb{Z}_2[x]/I$ where $f(x)=x^3+x^2+[1]\in\mathbb{Z}_2[x]$ and $I=(x^3+x^2+[1])\mathbb{Z}_2[x]$. With $X=x+I\in F$ and writing $a$ for $a+I$ with $a\in\mathbb{Z}_2$ show that the following hold:
(i) $X^3=X^2+[1]$
(ii) $X^4=X^2+X+[1]$
The way I've been taught is to say $X^3+X^2+1=0$ from the initial polynomial $f$ (not sure why we assume this to be zero). From this, $X^3=-X^2-[1]$, but its negative is $X^2+[1]$ and parity does not change under negation. Thus $X^3=X^2+[1]$
For (ii) we have $X^4=X+X^3$ (since the 'multiplication' is in fact addition) and so $X^4=X^2+x+[1]$. The idea that multiplication was intended to be addition comes from the fact that this hold immediately in this case and if we simply wrote $X^4=XX^3$ we would get the wrong answer.
How can I better understand this topic and write better solutions?
| You could use the congruence notation, in the same way as you would write $13 \equiv 3 \bmod 5$. In your case, your question amounts to show that
(i) $X^3 \equiv X^2+1 \bmod X^3 + X^2 + 1$
(ii) $X^4 \equiv X^2+X+1 \bmod X^3 + X^2 + 1$
You already observed that, in your ring of characteristic $2$, the conditions $g = h$, $g-h= 0$ and $g+h= 0$ are equivalent. In particular, (i) is equivalent to
(i') $X^3 + X^2 + 1 \equiv 0 \bmod X^3 + X^2 + 1$
which is trivially true. Similarly, (ii) is equivalent to
(ii)' $X^4 + X^2 + X + 1 \equiv 0 \bmod X^3 + X^2 + 1$
which is true since $(X^3 + X^2 +1)(X + 1) = X^4 + X^2 + X + 1$.
| {
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Showing $\frac{xy+yz+xz}{x+y+z}>5-\sqrt{4(x^2+y^2+z^2)+6}$ for $x,y,z>0$ and $xyz=1$
Let $x,y,z>0$ with $ xyz=1$ then prove that,
$$\frac{xy+yz+xz}{x+y+z}>5-\sqrt{4(x^2+y^2+z^2)+6}$$
Let $$5≤\sqrt{4(x^2+y^2+z^2)+6}\implies x^2+y^2+z^2≥\frac {25-6}{4}=\frac {19}{4}$$ then the inequality is always true.
This also shows that, if $x≥3$ or $x≥\frac {\sqrt {19}}{2}$ then the inequality is true. Since the inequality is cyclic, if $x$ or $y$ or $z≥\frac {\sqrt {19}}{2}$ then the inequality is again true.
I see that if $x,y$ is strictly increasing then $5-\sqrt{4(x^2+y^2+z^2)+6}$ is always negative.
But, I couldn't prove the general case.
| In fact, we can prove a stronger inequality
$$
\frac{xy+yz+zx}{x+y+z}+\sqrt{4(x^2+y^2+z^2)+6}\geq 1+3\sqrt{2} >5
$$
for all $x,y,z>0$ with $xyz=1$.
To prove the above inequality, we use the Cauchy-Schwarz Inequality to deduce that
$$
\left( 1+1+1+\frac{3}{2} \right)\left( 4x^2+4y^2+4z^2 +6 \right)\geq \left( 2x+2y+2z +3 \right)^2,
$$
which gives
$$
\sqrt{4(x^2+y^2+z^2)+6} \geq \frac{\sqrt{2}}{3}\left( 2x+2y+2z +3 \right).
$$
As a result, we have
\begin{equation} \begin{split}
\frac{xy+yz+zx}{x+y+z}+\sqrt{4(x^2+y^2+z^2)+6} &\geq \frac{xy+yz+zx}{x+y+z}+\frac{\sqrt{2}}{3}( 2x+2y+2z)+\sqrt 2 \\
&=\frac{xy+yz+zx}{x+y+z}+ 2\sqrt 2 \left(\frac{x+y+z}{3} \right)+\sqrt 2.
\end{split}\end{equation}
Therefore, it suffices to show
$$
\frac{xy+yz+zx}{x+y+z}+ 2\sqrt 2 \left(\frac{x+y+z}{3} \right) \geq 1 +2\sqrt{2}.
$$
To do this, we note by AM-GM inequality
\begin{align}
xy + yz +zx& \geq 3 \sqrt[3]{x^2y^2z^2} =3, \\
x+y+z &\geq 3 \sqrt[3]{xyz} =3
\end{align}
Using the above inequalities, we have
\begin{equation} \begin{split}
&\frac{xy+yz+zx}{x+y+z}+ 2\sqrt 2 \left(\frac{x+y+z}{3} \right) \\
& \quad\quad\quad =\frac{xy+yz+zx}{x+y+z}+\frac{x+y+z}{3} +(2\sqrt 2-1)\left(\frac{x+y+z}{3} \right) \\
& \quad\quad\quad \geq 2 \sqrt{\frac{xy+yz+zx}{3}}+ (2\sqrt 2-1)\left(\frac{x+y+z}{3} \right) \quad (\text{by AM-GM}) \\
& \quad\quad\quad \geq 2+ 2\sqrt 2-1 \\
& \quad\quad\quad = 1+ 2\sqrt 2.
\end{split}\end{equation}
This completes the proof.
| {
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Prove the inequality $9(a+b)(b+c)(c+a) \geq 8(a+b+c)(ab+bc+ca)$
Prove the inequality $9(a+b)(b+c)(c+a) \geq 8(a+b+c)(ab+bc+ca)$ for $a, b, c \in \mathbb{R_{>0}}$
I tried by first using AM-HM inequality on $a, b, c$ to get the following result.
$\frac{a+b+c}3 \geq \frac 3{\frac 1a+\frac1b+\frac1c}$
$\implies (a+b+c)(\frac1a+\frac1b+\frac1c) \geq 9$
$\implies (a+b+c)(ab+bc+ca) \geq 9abc$
Also I used the inequality
$(a+b)(b+c)(c+a) \geq 8abc$
But then I am not able to proceed further. Can someone please help me.
| $$9(a+b)(b+c)(c+a) - 8(a+b+c)(ab+bc+ca)$$
$$ = (a+b)(b+c)(c+a) - 8abc$$
$$\ge0$$
because $(a+b)(b+c)(c+a) = (a+b+c)(ab+bc+ca)-abc$
and $(a+b)(b+c)(c+a) \ge 8abc$ by AM-GM: $(a+b)(b+c)(c+a)=abc+\dots+bca\ge 8\sqrt[8]{a^8b^8c^8}$.
| {
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Demonstration of an imaginary ellipse I'm studying the classfication of non-degenerate conics based on the determinant of the matrix associated to it $A_Q$.
I didn't understand, in the case of a non-degenerate ellipse (det $A_{33}>0$, minor of $A_Q$, and det $A_Q \neq 0$) why we have an imaginary ellipse if $(A+C)detA_Q > 0$.
I tried looking to some examples, also from Wikipedia, $x^2+y^2+10=0$ has no real solutions. How can I demonstrate that $\{(x,y)^T \in \mathbb{R}^2 : ax^2 + bxy + cy^2 + dx + ey + f = 0\}$ is empty?
| When $D=b^2-4ac\neq 0$, you can translate the center $(\frac{2cd-be}{b^2-4ac},-\frac{bd-2ae}{b^2-4ac})$ to the origin without changing the form of the conic:
$$ax^2+bxy+cy^2+f+\frac{-ae^2+bde-cd^2}{4ac-b^2}=0\tag1$$ or
$$ax^2+bxy+cy^2+\frac{\begin{vmatrix}a&b/2&d/2\\b/2&c&e/2\\d/2&e/2&f\end{vmatrix}}{\begin{vmatrix}a&b/2\\b/2&c\end{vmatrix}}=0\tag2$$
(For $D>0,$ equation $(1)$, hence the original equation, represents a hyperbola (or a pair of lines when $f+\frac{-ae^2+bde-cd^2}{4ac-b^2}=0$) since then $ax^2+bxy+cy^2=a(x+y(b+\sqrt{D})/2a)(x+y(b-\sqrt{D})/2a);$ compare with $xy=k.$)
When $D<0,$ equation$(1)$ can represent an ellipse and $ax^2+bxy+cy^2=a(x+y(b+i\sqrt{-D})/2a)(x+y(b-i\sqrt{-D})/2a)$ which are complex lines that meet in the real point $(0,0).$ or in other words $ax^2+bxy+cy^2=0$ is the origin (or in the original conic just the center). So if in addition to $D<0$ $(a+c)>0$ then $f+\frac{-ae^2+bde-cd^2}{4ac-b^2}>0$ gives $ax^2+bxy+cy^2$ positive and there are no real points. And if $(a+c)<0$ then $f+\frac{-ae^2+bde-cd^2}{4ac-b^2}<0$ gives $ax^2+bxy+cy^2$ negative and there are no real points.
So the condition $\operatorname{trace}(\begin{pmatrix}a&b/2\\b/2&c\end{pmatrix})\cdot\begin{vmatrix}a&b/2&d/2\\b/2&c&e/2\\d/2&e/2&f\end{vmatrix}>0$ is the correct one.
| {
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$ 2 \cos ^{2021} x+\sin ^{2022} x=2 $ Someone sent me this task, it is meant to be solved quickly, nothing to think too much about.
At first I tried representing $\sin ^{2022}x$ as $(\sin ^{2}x)^{1011}$
$$
2 \cdot \frac{\left(\cos ^{1011} x\right)^{2}}{\cos (x)}+\left(1-\cos ^{2} x\right)^{1011}=2.
$$
Then using using $t=\cos x = \sqrt{1-\sin ^{2}x}$ substitution and find $x$. The expression got too complex so I gave up the idea.
After some time considering Taylor's expansion, funtion series, etc I just realized the 'only' case when it satisfies the equality it's when $x=0$.
How would you solve it. How do you solve this equation analyticaly, I mean step by step, line by line.
There is no answer proposed in the book.
| One elegant solution has been given in the comments. Here is an alternative one, using the Cauchy-Schwarz inequality. For any positive integer $n$ is
$$
\begin{align}
f(x) &= 2 \cos^{2n+1}(x)+\sin^{2n+2}(x) \\
&= \cos^{2n}(x) \cdot 2 \cos(x) + \sin^{2n}(x) \cdot \sin^2(x) \\
&\le \sqrt{\cos^{4n}(x) + \sin^{4n}(x)} \cdot \sqrt{4 \cos^2(x) + \sin^4(x)} \, .
\end{align}
$$
Now
$$
0 \le \cos^{4n}(x) + \sin^{4n}(x) \le \cos^{2}(x) + \sin^{2}(x) = 1
$$
and
$$
0 \le 4 \cos^2(x) + \sin^4(x) = 4 \cos^2(x) + (1 - \cos^2(x))^2
= (1+\cos^2(x))^2 \le 4
$$
so that $f(x) \le 2$ for all $x\in \Bbb R$.
Equality holds if and only if $\cos^2(x) = 1$ and $\cos(x) \ge 0$, that is exactly for all integer multiples of $2 \pi$.
| {
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Conjecture: Every $n \geq 20 \in \mathbb{N}$ can be written as a sum of three integers $(\geq 2)$ that are pairwise coprime This question on the sum of pairwise prime numbers piqued my interest, and I started looking at what numbers can be written as the sum of three pairwise coprime numbers (excluding $1$):
$$
\begin{align*}
10&=5+3+2\\
12&=7+3+2\\
14&=7+5+2\\
15&=7+5+3\\
16&=9+5+2\\
18&=11+4+3\\
19&=11+5+3\\
20&=11+7+2\\
21&=11+7+3\\
22&=11+7+4\\
&\ldots
\end{align*}
$$
It's fairly clear why $10$ is the smallest number possible, but I couldn't find any pattern until $20$ where it seems like all the rest follow suit.
It seems after $20$, there are so many possible combinations that every following number can be written as a sum of three integers $(\geq 2)$ which are pairwise coprime. Is this a known result, and, can it be (dis)proved?
EDIT:
Thanks for the responses all, I believe I have devised a nice case proof (inspired by Yuval Peres' case answer!)
Proof
Case 1: $n$ is even
$$
\begin{align*}
n&=6k=2+3+(6k−5)\\
n&=6k+2=4+3+(6k−7)\\
n&=6k+4=2+3+(6k−5)
\end{align*}
$$
Case 2: $n$ is odd
$$
\begin{align*}
n&=12k+1=3+(6k−7)+(6k+5) \text{ with } k≥2\\
n&=12k+3=9+(6k−5)+(6k−1) \text{ with } k≥2\\
n&=12k+5=3+(6k−5)+(6k+7) \text{ with } k≥2\\
n&=12k+7=3+(6k−1)+(6k+5)\\
n&=12k+9=3+(6k−1)+(6k+7)\\
n&=12k+11=3+(6k+1)+(6k+7) \hspace{35pt}\blacksquare
\end{align*}
$$
| Assume that $n$ is not the power of sum prime, i.e $n$ cannot be written as $n=p^{\alpha}$ for some prime $p$ and exponent $\alpha$. In this case, we can find two distinct primes $p,q$ that both divide $n$. Clearly $p+q<pq<n$ and so $n-(p+q)$ is a positive integer. This means that we can write
$$n=p+q+(n-(p+q)).$$
The first two terms, $p$ and $q$, are clearly coprime since $p$ and $q$ are prime, and hence all there is left to check is that $(n-(p+q))$ is not a multiple of $p$ or $q$. This follows from the fact that
$$n-(p+q)\equiv -q \neq 0\mod(p)$$
$$n-(p+q)\equiv -p \neq 0\mod(q)$$
since $n$ is a multiple of $p$ and $q$, and both $p$ and $q$ are prime thus are nonzero modulo each other.
In the general case you should be able to do something similar by shifting, since prime powers are 'rare enough' in a certain sense.
| {
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Solution of the recurrence relation $y_n = \frac{1}{2} + \frac{1}{2}y_{n-1}$ $y_0 = 0$ and $y_n = \frac{1}{2} + \frac{1}{2}y_{n-1}$. Solution of this reccurent equation is $y_n = 1 - \frac{1}{2^n}$, accordingly with the software. But I do not understand the minus sign since it would be $y_n = \frac{1}{2} + \frac{1}{2}\left(\frac{1}{2}+\frac{1}{2}y_{n-2}\right)$ and so forth until $y_n = \frac{1}{2} + \frac{1}{2^{n}} +x_{0}$.
Can someone try to explain it to me, I'm having a very tough time trying to understand.
| Repeat the formula $n$ times: $y_n = \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots + \dfrac{1}{2^{n-1}} + \dfrac{1}{2^n}+\dfrac{1}{2^n}y_0= 1 -\dfrac{1}{2^n}$
| {
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How many squares of each colour are in a generalized checkerboard $C$-coloured $m \times n$ rectangle?
How many squares of each colour are in a generalized checkerboard $C$-coloured $m \times n$ rectangle?
Assume an $m\times n$ rectangle has been been divided into a grid of $mn$ unit squares, and the squares have been coloured with $C$ colours in such a way that the colours in any row or column cycle, i.e. if the colours are represented by the number $1,2,\dots, C$ then
$1 \rightarrow 2,\; 2 \rightarrow 3,\; \dots, C-1 \rightarrow C,\; C \rightarrow 1$, etc.
With this setup there are actually $C$ possible coloured variants, so we choose the variant that has a black square
in the $(1,1)$ position. Here is an example for $m = 10$, $n = 13$ and $C = 4$:
It has matrix representation
$$\begin{bmatrix}
1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 \\
2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 \\
3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 \\
4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 \\
1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 \\
2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 \\
3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 \\
4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 \\
1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 \\
2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 & 1 & 2
\end{bmatrix}$$
Here we have number of squares of colour $1 = 33$,
number of squares of colour $2 = 33$,
number of squares of colour $3 = 32$,
and number of squares of colour $4 = 32$.
So what is the general formula? I'm hoping that this is a known problem and someone can point
me in the right direction (even if it is an open problem). Thank you.
| Let $a = m \bmod C$ and $b = n \bmod C$ i.e. the remainders of $m,n$ divided by $C$.
As noted in the comments above, as long as we have a square or a rectangle with at least one side divisible by $C$, all the colors will be present in that square or rectangle and their occurrences will be all equal.
We note that we can decompose the $m \times n$ rectangle into four rectangles/squares: $(m-a) \times (n-b)$, $(m-a) \times b$, $a \times (n-b)$, $a \times b$.
The total occurrence of each color in the three regions $(m-a) \times (n-b)$, $(m-a) \times b$ and $a \times (n-b)$ will be exactly $(mn-ab)/C$.
We remain with the square or rectangle $a \times b$.
Let the color index be $x$, $1 \le x \le C$.
We can note that:
*
*the upper left corner of $a \times b$ has color $1$ because $C$ divides $m-a$ and $n-b$;
*$a,b \lt C$;
*$x+C \le a+b-1 \iff x \le a+b-1-C$;
*$a+b-1-C \lt \min(a,b)$;
*$x+C \gt \max(a,b)$;
*$x+2C > a+b-1$;
*$x + [a+b-(x+C)] = a+b-C$.
Starting from the upper left corner of the region $a \times b$, we can start counting the occurrences of colors: if we look diagonally, we see that the first color $1$ appears $1$ time followed by color $2$ in two adjacent squares, followed by $3$ in three squares and so on till the minimum between $a$ and $b$ is hit. This takes place when reaching the lower left corner if $b<a$ or the upper right corner if $a<b$. After this, the occurrence of the following colors is constant and equal to $\min(a,b)$ till the maximum between $a$ and $b$ is hit. This takes place when reaching the upper right corner if $b<a$ or the lower left corner if $a<b$. After this, the occurrence of the following colors start to decrease till it reaches a minimum of $1$ at the lower right corner for color $a+b-1$ (the color index continued to increase while following half of the perimeter). Since for $\max(a,b) \le x \le a+b-1$ the function counting the occurrences in that interval is therefore $h(x) = k - x$ and $h(a+b-1) = 1$, then $k-(a+b-1)=1$ and thus $k=a+b$ and $h(x)=a+b-x$.
Up to now, we have four intervals where we know how to compute occurrences: they are $x \le \min(a,b)$, $\min(a,b) \lt x \lt \max(a,b)$, $\max(a,b) \le x \le a+b-1$ and $a+b-1 \lt x$ (this latter one is for missing colors, with occurrence equal to $0$).
We then need to examine if a color can belong to more than one interval, because given a color $x$, one of the above first three inequalities is satisfied with $x+jC$ ($j$ positive integer). For $j \ge 2$ this is not possible due to observations 2 and 6 above. We don't care about $a+b-1 \lt x+jC$ because there we already "exited" from the $a \times b$ region. We thus remain with the $x+C$ case: but due to observations 3 and 4 this can only occur in the first interval $x \le \min(a,b)$, and due to observation 4, more precisely only for $x \le a+b-1-C$. By the way, this does not take place always, but only when $a+b-1-C \ge 1$. Now, for colors $x \le a+b-1-C$ we have that the second contribute to count at $x+C$ is within the interval $\max(a,b) \le x+C \le a+b-1$ (observations 3 and 5), and therefore it contributes with a count of $a+b-(x+C)$ (see the above reasoning for that interval), which added to the contribute $x$ in the first interval gives a total of $a+b-C$ (observation 7).
Putting the above all together, we can then figure out a formula like this for the occurrence of color $x$ in $a \times b$, valid for $1 \le x \le C$:
$$g(x) =
\begin{cases}
a+b-C, & \text{if $x \le a+b-1-C$} \\
x, & \text{if $a+b-1-C \lt x \le \min(a,b)$} \\
\min(a,b), & \text{if $\min(a,b) \lt x \lt \max(a,b)$} \\
a+b-x, & \text{if $\max(a,b) \le x \le a+b-1$} \\
0, & \text{if $a+b-1 \lt x$} \\
\end{cases}
$$
And adding the first contribution above, the whole count formula, valid for $1 \le x \le C$, will be:
$$f(x) =
\begin{cases}
(mn-ab)/C + a+b-C, & \text{if $x \le a+b-1-C$} \\
(mn-ab)/C + x, & \text{if $a+b-1-C \lt x \le \min(a,b)$} \\
(mn-ab)/C + \min(a,b), & \text{if $\min(a,b) \lt x \lt \max(a,b)$} \\
(mn-ab)/C + a+b-x, & \text{if $\max(a,b) \le x \le a+b-1$} \\
(mn-ab)/C, & \text{if $a+b-1 \lt x$} \\
\end{cases}
$$
Note that the third case can be stated as $\min(a,b) \le x \le max(a,b)$ as well (the evaluation on the extremes of adjacent intervals is always the same if done for one case or the other). Note also that the function returns correctly $(mn-ab)/C=mn/C$ if $a=0$ and/or $b=0$.
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Evaluate the sum: $\sum_{n=0}^{\infty}\frac{x^{n+2}}{(n+2)\ n!}$
How to evaluate the below sum?
$$\sum_{n=0}^{\infty}\frac{x^{n+2}}{(n+2)\ n!}$$
I was trying to find the integral of $x\ e^x$ without using Integral By Parts.
Here's what I got so far:
$$\begin{aligned}e^x &= \sum_{n=0}^\infty \dfrac{x^n}{n!}\\x\ e^x &= \sum_{n=0}^\infty \dfrac{x^{n+1}}{n!} \\\int x\ e^x\ dx& = \int\sum_{n=0}^\infty \dfrac{x^{n+1}}{n!} dx\\& = \sum_{n=0}^\infty \int\dfrac{x^{n+1}}{n!} dx\\& = \sum_{n=0}^\infty\dfrac{x^{n+2}}{(n+2)\ n!} + C\end{aligned}$$
How to evaluate this series? I know the answer should be $e^x(x-1) + c$ which is the antiderivative of $x \ e^x$ and I've verified it from WolframAlpha, but not getting any idea how to evaluate the series.
I think partial fraction decomposition might work here because $\frac{1}{(n+2)n!} = \frac{1}{(n+1)!} - \frac{1}{(n+2)!}$ but not sure how to prove this and how to implement.
| You have\begin{align}\frac1{(n+1)!}-\frac1{(n+2)!}&=\frac{n+2-1}{(n+2)!}\\&=\frac{\cancel{n+1}}{(n+2)\cancel{(n+1)}n!}\require{cancel}\\&=\frac1{(n+2)n!}\end{align}
and therefore you have\begin{align}\sum_{n=0}^\infty\frac{x^{n+2}}{(n+2)n!}&=\sum_{n=0}^\infty\left(\frac1{(n+1)!}-\frac1{(n+2)!}\right)x^{n+2}\\&=\sum_{n=0}^\infty\frac{x^{n+2}}{(n+1)!}-\sum_{n=0}^\infty\frac{x^{n+2}}{(n+2)!}\\&=x\sum_{n=0}^\infty\frac{x^{n+1}}{(n+1)!}-\sum_{n=0}^\infty\frac{x^{n+2}}{(n+2)!}\\&=x(e^x-1)-(e^x-1-x)\\&=(x-1)e^x+1\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The solutions of the equation $ \sin{x} + \sin{3x} = \frac{8}{3\sqrt{3}} $ are? I tried this:
$$ \sin{x} + \sin{3x} = \frac{8}{3\sqrt{3}} $$
$$ 2\sin{2x}\cos{x} = \frac{8}{3\sqrt{3}} $$
$$ 4\sin{x}\cos{x}\cos{x} = \frac{8}{3\sqrt{3}} $$
$$ \sin{x}(1-\sin^2{x}) = \frac{2}{3\sqrt{3}} $$
Here, I tried to set $\sin x = t$
$$ t(1-t^2) = \frac{2}{3\sqrt{3}}, $$ but I don't know to resolve this.
| HINT
To begin with, notice that $\sin(3x) = 3\sin(x) - 4\sin^{3}(x)$. Hence it results that:
\begin{align*}
\sin(x) + \sin(3x) = \frac{8}{3\sqrt{3}} & \Longleftrightarrow 4\sin(x) - 4\sin^{3}(x) = \frac{8}{3\sqrt{3}}\\\\
& \Longleftrightarrow \sin(x) - \sin^{3}(x) = \frac{2\sqrt{3}}{9}
\end{align*}
By inspection, one concludes that
\begin{align*}
\sin(x) = \frac{1}{\sqrt{3}}
\end{align*}
satisfies the resulting equation.
From then on, you can factor the cubic equation to obtain a quadratic which is easy to deal with.
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Finding Laurent Series Expansion for $f(z) = \frac{z^3}{z^2-3z+2}$ for different domains So I wanted to find the Laurent series expansion for $f(z) = \frac{z^3}{z^2-3z+2}$. So I guess more specifically, my question pertains how the expansion may change for different domains: $|z| < 1; 1 < |z| < 2; |z| > 2; 0 < |z − 1| < 1; |z − 1| > 1$.
I first did partial fraction decomposition, leading to:$$f(z) = \frac{z^3}{z^2-3z+2} = z+3+ \frac{20}{z-2} - \frac{13}{z-1}$$
From here, my question is, how do I know if I take out a $\frac{1}{z}$ from the fractions or not? I'm guessing it depends on the domain, but if anyone could outline this for me given my domain choices, that would help immensely. Would I consider taking out the $\frac{1}{z}$ if $|z| >1$?
$$z+3+ \frac{20}{z-2} - \frac{13}{z-1} = z+3+ \frac{20}{z}\frac{1}{1-(\frac{2}{z})} + 13\frac{1}{1-z} $$
$$= z+3+\frac{20}{z}\sum_{n=0}^\infty (\frac{2}{z})^n +13\sum_{n=0}^\infty z^n $$
(in writing this, I'm forming my own conclusions: I'm thinking that if I wanted $|z| >1$ then I would do exactly what I did in the above line, but reversing the power series in terms of taking of a $\frac{1}{z}$ and doing it instead for the $\frac{1}{1-z}$. Is this correct (i.e, what I did above would be for the case $|z|>2$? And in general, if $|z|>n$ we would take out a $\frac{1}{z}$ for that fraction?)
And in the case the domain is $0 < |z − 1| < 1$, would I then try taking out $z-1$ from the denominator of $\frac{1}{z-2} = -\frac{1}{1-(z-1)}$ and substitute $z-1$ for $z$ in the summation notation?
As always any help would be awesome!
| If you want to find the Laurent expansion of $\frac{1}{z-b}$ in an annulus centred at $a$ with some given radii, you may rewrite the fraction as
$$
\frac{1}{z-b}=\begin{cases}
\frac{1}{a-b}\frac{1}{1-w}
&\text{when $b\ne a$ and $|z-a|<|b-a|$, where $w=\frac{z-a}{b-a}$,}\\
\frac{1}{b-a}\frac{w}{1-w}
&\text{when $b\ne a$ and $|z-a|>|b-a|$, where $w=\frac{b-a}{z-a}$,}\\
\frac{1}{z-b}
&\text{when $b=a$.}\\
\end{cases}
$$
In the first two cases, the whole point is to expand $(1-w)^{-1}$ as $1+w+w^2+\cdots$, where $w$ is a scalar multiple of either $z-a$ or $(z-a)^{-1}$ (because we want an expansion about $a$) with $|w|<1$ (so that $1+w+w^2+\cdots$ converges). In your example,
$$
\frac{1}{z-1}=\begin{cases}
-\frac{1}{1-w}
&\text{on the annulus $0<|z|<1$, where $w=z$,}\\
\frac{w}{1-w}
&\text{on the annuli $1<|z|<2$ and $|z|>2$ where $w=\frac{1}{z}$,}\\
\frac{1}{z-1}
&\text{on the annuli }0<|z-1|<1 \text{ and } |z-1|>1,\\
\end{cases}
$$
$$
\frac{1}{z-2}=\begin{cases}
-\frac12\frac{1}{1-w}
&\text{on the annuli $0<|z|<1$ and $|z|<2$, where $w=\frac{z}{2}$,}\\
\frac{1}{2}\frac{w}{1-w}
&\text{on the annulus $|z|>2$, where $w=\frac{2}{z}$,}\\
-\frac{1}{1-w}
&\text{on the annulus $0<|z-1|<1$, where $w=z-1$,}\\
\frac{w}{1-w}
&\text{on the annulus $|z-1|>1$, where $w=\frac{1}{z-1}$.}\\
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\binom{13+m}{m}-(m+1)\binom{6+m}{m}\geq m$ for $m\in \mathbb{N}\backslash \{0,1\}$ Prove the following inequality for every $m\in \mathbb{N}\backslash \{0,1\}$:
$$ \binom{13+m}{m}-(m+1)\binom{6+m}{m}\geq m.$$
By some computational arguments, the inequality seems to be true and in particular the difference between the first term $\binom{13+m}{m}$ and the second one $(m+1)\binom{6+m}{m}$ is very large, so obviously is more that $m$. Hence the use of the induction technique, in order to find a way to give a formal proof, leads to a disaster consequence.
In fact, for $m=2$ the inequality is trivial. Fix $m> 2$ and suppose the the inequality is true for $m$. We want to prove that
$$ \binom{14+m}{m+1}-(m+2)\binom{7+m}{m+1}\geq m+1.$$
Start from the first member and apply Stifiel's formula
$$ \binom{14+m}{m+1}-(m+2)\binom{7+m}{m+1}=\binom{13+m}{m+1}+\binom{13+m}{m}-(m+1+1)\Big[\binom{6+m}{m+1}+\binom{6+m}{m}\Big]=
\binom{13+m}{m}-(m+1)\binom{6+m}{m}+\binom{13+m}{m+1}-(m+1)\binom{6+m}{m+1}-\binom{6+m}{m+1}-\binom{6+m}{m}\geq...
$$
Applying the inductive hypothesis we have:
$$...\geq m+\binom{13+m}{m+1}-(m+1)\binom{6+m}{m+1}-\binom{6+m}{m+1}-\binom{6+m}{m},$$
but this is less than $m+1$. Therefore we don't get the desired conclusion.
Which can be an useful stategy to prove this inequality?
| Using the identity (well-known, easy to prove)
$$\binom{n + 1}{k} = \binom{n}{k} + \binom{n}{k - 1}, $$
we have
\begin{align*}
\binom{13 + m}{m} &= \binom{12 + m}{m} + \binom{12 + m}{m - 1}\\
&= \binom{11 + m}{m} + \binom{11 + m}{m - 1} + \binom{11 + m}{m - 1} + \binom{11 + m}{m - 2}\\
&\ge \binom{11 + m}{m} + 2\binom{11 + m}{m - 1}\\
&= \binom{10 + m}{m} + \binom{10 + m}{m - 1}
+ 2\binom{10 + m}{m - 1} + 2\binom{10 + m}{m - 2}\\
&\ge \binom{10 + m}{m} + 3\binom{10 + m}{m - 1}.
\end{align*}
After repeating the process, we have
$$\binom{13 + m}{m} \ge \binom{7 + m}{m} + 6\binom{7 + m}{m - 1}.$$
It suffices to prove that
$$\binom{7 + m}{m} + 6\binom{7 + m}{m - 1} - (m + 1)\binom{6 + m}{m} \ge m$$
or (after some simple manipulations)
$$6m(m - 1)\frac{(6 + m)!}{m!\, 8!} \ge m$$
or
$$6(m - 1)\frac{(6 + m)!}{m!\, 8!} \ge 1$$
or
$$6(m - 1)\frac{(6 + m)(5 + m)(4 + m)(3 + m)(2 + m)(1 + m)}{8!} \ge 1$$
which is true since
$$\mathrm{LHS} \ge 6(2 - 1)\frac{(6 + 2)(5 + 2)(4 + 2)(3 + 2)(2 + 2)(1 + 2)}{8!} = 3.$$
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove that $c^3 + a^3 − 2abc = 6b − 11$ Let $a, b, c$ be real numbers such that $(a + b + c)^2 = 3(ab + bc + ca + 1).$ Given,
$$a^3 + b^3 − 2abc = 6c + 2,$$
$$b^3 + c^3 − 2abc = 6a + 9,$$
Prove that $$c^3 + a^3 − 2abc = 6b − 11$$
Note that, subtracting the two equations, we get $$a^3-c^3=6(c-a)+7. $$
Moreover, we have $$a^2+b^2+c^2-ab-bc-ca=3\implies (a-b)^2+(b-c)^2+(c-a)^2=6$$
So, we get $$a^3-c^3=[(a-b)^2+(b-c)^2+(c-a)^2](c-a)+7$$ $$\implies a^3-c^3+2c^3-2abc-11=[(a-b)^2+(b-c)^2+(c-a)^2](c-a)+7+2c^3-2abc-11.$$
And $$3 a^3 - 2 a^2 b - 4 a^2 c + 2 a b^2 + 4 a c^2 - 2 b^2 c + 2 b c^2 - 3 c^3 - 7 = 0 $$
So, it's enough to show that $$[(a-b)^2+(b-c)^2+(c-a)^2](c-a)+7+2c^3-2abc=6b $$
or show that $$2 a^3 - 4 a^2 c + 4 a c^2 + 2 b^3 - 4 b^2 c + 4 b c^2 - 4 c^3=7$$
or show that $$2 (a^3+b^3) - 4 a^2 c + 4 a c^2 - 4 b^2 c + 4 b c^2 - 4 c^3=7$$
I got something ahead, but failed. Any solutions?
| Let $s=a+b+c$ and $t=ab+bc+ca$, so that $s^2=3t+3$ is the first given equation. Note that
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=s(s^2-3t)=3s.$$
So,
$$(a^3+b^3-2abc)+(b^3+c^3-2abc)+(c^3+a^3-2abc)=6s.$$
Now simply subtract the other two given equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Computing Correlation cofficient from joint pdf X and Y have the joint pdf:
$f(x,y)= \frac{2}{x} e^{-2x}$ for $0<x<\infty , 0<y<x$
I'm using the formula: $E(h(x,y))=\int_{0}^{\infty} \int_{0}^{x} h(x,y)\frac{2}{x} e^{-2x} dydx $ . And I'm simply setting $x$,$x^2$, etc. as h(x,y).
Based on this, I got:
$E(XY)=\frac{1}{4} , E(X)=\frac{1}{4} , E(Y)= \frac{1}{4}, E(X^2)=\frac{1}{2}, E(Y^2)=\frac{1}{6}$
$Cov(X,Y)=E(XY)-E(X)E(Y)=\frac{1}{4}, var(X)= E(X^2)-[E(X)]^2=\frac{1}{4}, var(Y)=E(Y^2)-[E(Y)]^2 = \frac{5}{48}$
$ \rho(X,Y)= \frac{Cov(X,Y)}{\sqrt{var(X)var(Y)}}= 1.5492$
But the coefficient cannot exceed 1. So where did I go wrong? Any help would be appreciated!
| Here is a useful trick. Instead of individually computing each moment, consider the generalized expression for nonnegative integers $a, b$
$$\begin{align}
\operatorname{E}[X^a Y^b] &= \int_{x=0}^\infty \int_{y=0}^x x^a y^b \frac{2}{x} e^{-2x} \, dy \, dx \\
&= \int_{x=0}^\infty 2x^{a-1} e^{-2x} \left[ \frac{y^{b+1}}{b+1} \right]_{y=0}^x \, dx \\
&= \frac{2}{b+1} \int_{x=0}^\infty x^{a-1} e^{-2x} x^{b+1} \, dx \\
&= \frac{2}{b+1} \int_{x=0}^\infty x^{a+b} e^{-2x} \, dx. \tag{1}
\end{align}$$
Now recalling the definition of the gamma function
$$\Gamma(z) = \int_{t=0}^\infty t^{z-1} e^{-t} \, dt, \tag{2}$$ this motivates the substitution $$t = 2x, \quad x = \frac{t}{2}, \quad dx = \frac{1}{2} \, dt$$ to yield $$\operatorname{E}[X^a Y^b] = \frac{2}{b+1} \int_{t=0}^\infty \frac{t^{a+b}}{2^{a+b}} e^{-t} \frac{1}{2} \, dt = \frac{\Gamma(a+b+1)}{(b+1)2^{a+b}} = \frac{(a+b)!}{(b+1)2^{a+b}}. \tag{3}$$
Now this lets us compute all desired moments rapidly; e.g.,
$$\begin{array}{c|c|c}
(a,b) & X^a Y^b & \operatorname{E}[X^a Y^b] \\
\hline
(1,0) & X & \frac{1}{2} \\
(0,1) & Y & \frac{1}{4} \\
(1,1) & XY & \frac{1}{4} \\
(2,0) & X^2 & \frac{1}{2} \\
(0,2) & Y^2 & \frac{1}{6}
\end{array} \tag{4}$$
Then $$\operatorname{Var}[X] = \frac{1}{2} - \frac{1}{2^2} = \frac{1}{4}, \\ \operatorname{Var}[Y] = \frac{1}{6} - \frac{1}{4^2} = \frac{5}{48},$$
hence $$\rho = \frac{\frac{1}{4} - \frac{1}{2} \frac{1}{4}}{\sqrt{\frac{1}{4}\frac{5}{48}}} = \sqrt{\frac{3}{5}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Solve the PDE $(xz-y)p+(yz-x)q=xy-z$ using lagrange method I have the following PDE,
$$(xz-y)p+(yz-x)q=xy-z$$where $p=z_x,\quad q=z_y$
Now having a hard time to get two solution from,
$$\frac{dx}{xz-y}=\frac{dy}{yz-x}=\frac{dz}{xy-z}$$
I can't think of any multipliers trick which can help me to get one. In fact, $dx-dy$ or $dx+dy$ also couldn't help here. I guess the $xy-z$ term is the main culprit.
Any solution or hint will be appreciated. It will be a great help if you also describe how to guess the trick (if you used so).
Thanks in advance.
update
In fact, I got two more problems where I faced the similar issue:
$$\left(x z+y^{2}\right) p+\left(y z-2 x^{2}\right) z=-\left(2 x y+z^{2}\right)\qquad (1)$$
gives
$$\frac{d x}{x z+y^2}=\frac{d y}{y z-2 x^{2}}=\frac{d z}{-\left(2 x y+z^{2}\right)}$$
$$(y+3 z)p+(z+5 x)q=x+7 y\qquad (2)$$
$$
\begin{align}
&\frac{d x}{y+3 z}=\frac{d y}{z+5 x}=\frac{d z}{x+7 y}\\
\implies&\frac{d x-3d y}{y-15x}=\frac{d y-5d z}{z-7 y}=\frac{d z-7d x}{x-21 z}
\end{align}
$$
| This is not a complete answer rather expanding the @ShengtongZhang's one as the OP commented there,
characteristic equation of that PDE,
$$
\begin{align}
&\frac{dx}{xz-y}=\frac{dy}{yz-x}=\frac{dz}{xy-z}=dt\\
&x'=xz-y,\quad y'=yz-x, \quad z'=xy-z
\end{align}
$$
Now, inspecting the $x'y-y'x$ term,
$$
\begin{align}
x'y-y'x&=(xz-y)y-(yz-x)x\\
&=xyz-y^2-xyz+x^2\\
&=x^2-y^2 \text{(free of $z$ variable)}
\end{align}
$$
Which give some idea to guess one solution as $\frac{x}{y}$,
$$
\begin{align}
\frac{d}{dt}\left(\frac{x}{y}\right)&=\frac{x'y-y'x}{y^2}\\
&=\frac{x^2-y^2}{y^2} \text{(solvable)}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4454310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Is it possible to compute $-\log\left({\sqrt{1.8\times 10^{-5}\times 0.1}}\right)$ without a calculator? The following question is part of a chemistry problem that came in the Dhaka University admission exam 2013-14.
What is $-\log\left({\sqrt{1.8\times 10^{-5}\times 0.1}}\right)$?
(a) 2.672
(b) 2.772
(c) 2.872
(d) 2.972
My attempt
$$\begin{aligned} -\log_{10}\left({\sqrt{1.8\times 10^{-5}\times 0.1}}\right) &= -\log_{10}\left({\sqrt{1.8\times 10^{-6}}}\right)\\ &\approx -\log_{10}\left({\sqrt{\left(10^{-3}\right)^2}}\right) \\ &=-\log_{10} \left(10^{-3}\right) \\ &= 3 \end{aligned}$$
The options are cruelly close to one another, so the approximation that I have done is of no use to me. How do I find the correct answer by hand quickly (as this is a competitive exam)?
| Working with whole numbers
$$1.8\times 10^{-5}\times 0.1= \frac 9 5 \times 10^{-6}$$
$$\sqrt{1.8\times 10^{-5}\times 0.1}=\frac{3}{ \sqrt{5}}\times 10^{-3}$$
$$\frac{3}{ \sqrt{5}}=\frac{3}{ \sqrt{4+1}}=\frac 3 2 \times \frac 1{\sqrt{1+\frac 14}}\sim\frac 3 2 \times \left(1-\frac 18\right)=\frac{21}{16}=1+\frac 5 {16}$$
$$\log_{e}\left(1+\frac 5 {16}\right)\sim\frac 5 {16}\implies \log_{10}\left(1+\frac 5 {16}\right)\sim \frac {5}{16 \times 2.30} \sim 0.135$$
Then $-2.865$ by hand
Edit
A good rational approximation of $\log_e(10)$ is $\frac{76}{33}$
$$\log_{10}\left(1+\frac 5 {16}\right)\sim \frac {5}{16} \times \frac{33}{76}=\frac{165}{1216}\sim\frac{165}{1215}=\frac{11}{81}\times \frac{125}{125}=\frac{1375}{10125} \sim\frac{1375}{10000}=0.1375$$
| {
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"url": "https://math.stackexchange.com/questions/4455969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
How to show that $\frac{\pi}{2} \le \sum_{n=0}^\infty \frac{1}{n^2+1} \le \frac{3\pi}{4}$ How to show that $\frac{\pi}{2} \le \sum_{n=0}^\infty \frac{1}{n^2+1} \le \frac{3\pi}{4}$ ?
My Attempt : I was using Integral Test of a Series. I got $\int_0^\infty \frac{1}{1+x^2} \le \sum_{n=0}^\infty \frac{1}{1+n^2} \le \frac{1}{1+0^2} + \int_0^\infty \frac{1}{1+x^2}$ which gives $ \frac{\pi}{2} \le \sum_{n=0}^\infty \frac{1}{1+n^2} \le 1+ \frac{\pi}{2}$.
Can anyone please help me by giving any hint ?
| This can be computed using the result of this answer:
$$
\begin{align}
\sum_{n=0}^\infty\frac1{n^2+1}
&=-\frac1{2i}\sum_{n=0}^\infty\left(\frac1{i-n}+\frac1{i+n}\right)\\
&=\frac12-\frac1{2i}\sum_{n\in\mathbb{Z}}\frac1{i+n}\\
&=\frac12-\frac1{2i}\pi\cot(\pi i)\\[3pt]
&=\frac12+\frac\pi2\coth(\pi)\\[9pt]
&\approx2.076674
\end{align}
$$
We can get bounds tighter than $\frac\pi2$ and $\frac{3\pi}4$ with telescoping series:
$$
\begin{align}
\sum_{n=0}^\infty\frac1{n^2+1}
&\ge1+\sum_{n=1}^\infty\frac1{n^2+n}\\
&=1+\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}\right)\\[6pt]
&=2
\end{align}
$$
and, although this does not give as tight an upper bound as in Zhang's answer, it only uses telescoping series
$$
\begin{align}
\sum_{n=0}^\infty\frac1{n^2+1}
&\le1+\frac12+\sum_{n=2}^\infty\frac1{n^2-\frac14}\\
&=1+\frac12+\sum_{n=2}^\infty\left(\frac1{n-\frac12}-\frac1{n+\frac12}\right)\\[3pt]
&=\frac{13}6
\end{align}
$$
Thus,
$$
2\le\sum_{n=0}^\infty\frac1{n^2+1}\le\frac{13}6
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to find the closed form of $\int_{-\infty}^{\infty} \frac{x^{2 n+1} \sin x}{\left(1+x^{2}\right)^{n+1}} d x, \textrm{ where }n=0,1,2,3,…?$ In my post, I had found the exact value of the integral
$$\displaystyle I:=\int_{-\infty}^{\infty} \frac{x^{3} \sin x}{\left(1+x^{2}\right)^{2}}dx= \frac{\pi}{2 e}\tag*{} $$
by differentiating $J(a)$w.r.t. $a$ once.
$\displaystyle J(a)=\int_{-\infty}^{\infty} \frac{x \sin x}{1+a x^{2}} dx =\frac{\pi e^{-\frac{1}{\sqrt{a}}}}{a}, \textrm{ where }a>0\tag*{} $
Then I generalize $I$ to $$
I_{n}=\int_{-\infty}^{\infty} \frac{x^{2 n+1} \sin x}{\left(1+x^{2}\right)^{n+1}} d x, \textrm{ where }n=0,1,2,3,…
$$
by differentiating $J(a)$ w.r.t. by $n$ times at $a=1$.
$$
\left. J^{(n)}(1) =\pi \frac{d^{n}}{d a^{n}}\left(a^{-1} e^{-\frac{1}{\sqrt{a}}}\right)\right|_{a=1}
$$
$$J^{(n)}(1) = \int_{-\infty}^{\infty} \frac{(-1)^{n} n ! x^{2n+1}\sin x}{\left(1+a x^{2}\right)^{n+1}} dx= (-1)^{n} n !I_n \tag*{}$$
Hence we can conclude that
$$ I_{n}=\left.\frac{(-1)^{n}}{n !} J^{(n)}(1) = \frac{(-1)^{n} \pi}{n !}\frac{d^{n}}{d a^{n}}\left(a^{-1} e^{-\frac{1}{\sqrt{a}}}\right)\right|_{a=1}
\tag*{} $$
Theoretically, we can find the exact value of $I_n$ by $J^{(n)}(1)$.
Urging to the closed form of $I_n$, I use Leibniz’s Rule to find $J^{(n)}(1)$
\begin{aligned}
\left.\frac{d^{n}}{d a^{n}}\left(a^{-1} e^{-\frac{1}{\sqrt{a}}}\right) \right|_{a=1} &= \left.\sum_{k=0}^{n}{n\choose k} \left(a^{-1}\right)^{(n-k)}\left(e^{-\frac{1}{\sqrt{a}}}\right)^{(k)}\right|_{a=1} =
\left.\sum_{k=0}^{n} {n\choose k} (-1)^{n-k}(n-k)!( e^{-\frac{1}{\sqrt{a}}} )^{(k)} \right|_{a=1}
\end{aligned}
Hence $$I_{n}=\frac{(-1)^{n}}{n !} J^{(n)}(1)=
\left. \pi\sum_{k=0}^{n} \frac{(-1)^{k}}{k!}( e^{-\frac{1}{\sqrt{a}}} )^{(k)} \right|_{a=1}$$
By Wolframalpha, we have
$$
\frac{\partial^{k} e^{-1 / \sqrt{x}}}{\partial x^{k}}=e^{-1 / \sqrt{x}} x^{-k} \sum_{j=0}^{k} \sum_{i=0}^{j} \frac{(-1)^{i}\left(-\frac{1}{\sqrt{x}}\right)^{j}\left(\frac{1}{2}(2+i-j-2 k)\right)_{(k)}}{i !(j-i) !}
$$
Putting $x=1$ yields
$$
\left.\frac{\partial^{k} e^{-\frac{1}{\sqrt{x}}}}{\partial x^{k}}\right|_{x=1}=e^{-1} \sum_{j=0}^{k} \sum_{i=0}^{j} \frac{(-1)^{i}\left(\frac{1}{2}(2+i-j-2 k)\right) _{(k)}}{i !(j-i) !}
$$
Now we can conclude that
$$I_{n} = \pi\sum_{k=0}^{n} \frac{(-1)^{k}}{k!} \left.\frac{\partial^{k} e^{-\frac{1}{\sqrt{x}}}}{\partial x^{k}}\right|_{x=1}=\frac{\pi}{e} \sum_{k=0}^{n} \sum_{j=0}^{k} \sum_{i=0}^{j} \frac{(-1)^{i+k}\left(\frac{1}{2}(2+i-j-2 k)\right)_{(k)}}{i !(j-i) !k!} $$
The closed form is rather complicated and ugly. Is there a simpler one?
| This not an answer.
Computing
$$I_{n}=\left.\frac{(-1)^{n}}{n !} J^{(n)}(1) = \pi \frac{d^{n}}{d a^{n}}\left(a^{-1} e^{-\frac{1}{\sqrt{a}}}\right)\right|_{a=1}$$
$${I_n}= \frac \pi{2^n \,e }\, b_n$$ and the $b_n$ generate the sequence
$$\{-1,2,-5,5,176,-3881,70337,-1285390,24806665,-512336809,\cdots\}$$ which is not identified by $OEIS$.
Edit
For the generation of the sequence of $b_n$, let $a=b+1$ and use Taylor expansion around $b=0$ of
$$\frac 1a e^{-\frac{1}{\sqrt{a}}}=\frac 1{1+b} e^{-\frac{1}{\sqrt{1+b}}}$$ from inside to outside
$$\frac{1}{\sqrt{1+b}}=1-\frac{b}{2}+\frac{3 b^2}{8}-\frac{5 b^3}{16}+\frac{35 b^4}{128}-\frac{63
b^5}{256}+\frac{231 b^6}{1024}-\frac{429 b^7}{2048}+O\left(b^{8}\right)$$
$$e^{-\frac{1}{\sqrt{1+b}}}=\frac{1}{e}+\frac{b}{2 e}-\frac{b^2}{4 e}+\frac{7 b^3}{48 e}-\frac{35 b^4}{384 e}+\frac{113
b^5}{1920 e}-\frac{1769 b^6}{46080 e}+\frac{16003 b^7}{645120 e}+O\left(b^8\right)$$
$$\frac{e^{-\frac{1}{\sqrt{1+b}}}}{1+b}=\frac{1}{e}-\frac{b}{2 e}+\frac{b^2}{4 e}-\frac{5 b^3}{48 e}+\frac{5 b^4}{384 e}+\frac{11
b^5}{240 e}-\frac{3881 b^6}{46080 e}+\frac{70337 b^7}{645120 e}+O\left(b^8\right)$$ Multiply by $e$ and notice that, multiplied by $2^n$ give whole numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Algebraic Manipulation involving square roots I have the following expressions:
$$E(g) := \frac{Ug^2 - 4tg}{1+g^2} \hspace{2cm} (1)$$
$$ g_0 := \frac{1}{4t} \left[ -U + \sqrt{U^2 + (4t)^2} \right] \hspace{2cm} (2) $$
By plugging (2) in (1) I should arrive at:
$$ E(g_0) = \frac{1}{2} \left( U - \sqrt{U^2 + (4t)^2} \right) \hspace{2cm} (3)$$
But I can't get there. Here is my attempt so far:
$$ 1 + g_0^2 = 1+ \frac{1}{16t^2} \left( U^2+U^2+(4t)^2 - 2U \sqrt{ U^2 + (4t)^2} \right)$$
$$ = 1+ \frac{2U^2}{16t^2} + 1 - \frac{2U}{16t^2} \sqrt{ U^2 + (4t)^2}$$
$$ = 2 + \frac{2U}{16t^2} \left( U - \sqrt{ U^2 + (4t)^2} \right)$$
$$ -4t g_0 + Ug_0^2 = U - \sqrt{ U^2 + (4t)^2} + U \left( \frac{2U^2}{16t^2} + 1- \frac{2U}{16t^2} \sqrt{ U^2 + (4t)^2} \right)$$
$$ = - \sqrt{ U^2 + (4t)^2} + U \left( \frac{2U^2}{16t^2} + 2 - \frac{2U}{16t^2} \sqrt{ U^2 + (4t)^2} \right)$$
$$ = - \sqrt{ U^2 + (4t)^2} + U \left( 2 + \frac{2U}{16t^2} \left\{ U - \sqrt{ U^2 + (4t)^2} \right\} \right)$$
So:
$$ E(g_0) = \frac{- \sqrt{ U^2 + (4t)^2} + U \left( 2 + \frac{2U}{16t^2} \left\{ U - \sqrt{ U^2 + (4t)^2} \right\} \right)}{2 + \frac{2U}{16t^2} \left( U - \sqrt{ U^2 + (4t)^2} \right)}$$
$$ = U - \frac{ \sqrt{ U^2 + (4t)^2}}{2 + \frac{2U}{16t^2} \left( U - \sqrt{ U^2 + (4t)^2} \right)}$$
I don't know how to simplify this any further.
For reference Im reading this: http://www.scholarpedia.org/article/Gutzwiller_wave_function , and working with equations (4), (5) and (6)
| It follows from $\,(2)\,$ that $\require{cancel}\,(4 t g_0+U)^2 = U^2 + 16t^2\,$, or $\,16t^2g_0^2 + 8 Utg_0 + \cancel{U^2} = \cancel{U^2} + 16t^2\,$, and assuming $\,t \ne 0\,$ it follows that $\,2tg_0^2 + Ug_0-2 t = 0 \iff \color{blue}{2tg_0^2} = -Ug_0+2t\,$. Then:
$$
\begin{align}
E(g_0) &= \frac{Ug_0^2 - 4tg_0}{1+g_0^2} \cdot \frac{2t}{2t}
\\ &= \frac{2tg_0^2U - 8t^2g_0}{2t+\color{blue}{2tg_0^2}}
\\ &= \frac{2tg_0(Ug_0- 4t)}{2t -Ug_0+2t}
\\ &= \frac{2tg_0\cancel{(Ug_0- 4t)}}{-\cancel{(Ug_0-4t)}}
\\ &= -2tg_0
\end{align}
$$
The latter is equivalent to equality $\,(3)\,$ which was to be proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4466905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculate double integral $\iint_D\frac{\sqrt{x^2+y^2 - a^2}x}{(x^2+y^2)^2}dxdy$ over unbounded region $D$ Calculate
$$I = \iint\limits_D\frac{\sqrt{x^2+y^2 - a^2}x}{(x^2+y^2)^2}\,dx\,dy$$
where the region of integration is:
$$D = \{(x,y) \in \mathbb{R}^2 \mid x+y - a\sqrt{2} \geq 0, -x+y +a\sqrt{2} \geq 0, a > 0\}$$
I tried using polar coordinates, but then, I don't know how to proceed because of the region $D$ and of the integrand:
$$I = \iint\limits_D\frac{\sqrt{r^2-a^2}\cos\theta}{r^2}\,dr\,d\theta$$
| With the variable changes $x=au$ and $y=av$
\begin{align}
&\iint_D\frac{\sqrt{x^2+y^2 - a^2}x}{(x^2+y^2)^2}\,dx\,dy\\
=&\int_0^\infty \int_{-v+\sqrt2}^{v+\sqrt2} \frac{\sqrt{u^2+v^2 - 1}\ u}{(u^2+v^2)^2}\,du\,dv\\
=& \int_0^\infty \frac12\bigg(\tan^{-1}{\sqrt{u^2+v^2 - 1}}-\frac{\sqrt{u^2+v^2 - 1}}{u^2+v^2}\bigg)\bigg|_{-v+\sqrt2}^{v+\sqrt2}dv\\
=&\ \frac12\int_0^\infty \bigg(\tan^{-1}(1+\sqrt2v) -\tan^{-1}|1-\sqrt2v|\bigg)dv\\
&-\frac12\int_0^\infty \left(\frac{1+\sqrt2v}{(v+\sqrt2)^2+v^2} -\frac{|1-\sqrt2v|}{(v-\sqrt2)^2+v^2}\right)dv\\
=& \ \frac12 \cdot \frac{\pi +2\ln2}{2\sqrt2}
-\frac12 \left(-\frac{\ln2}{\sqrt2}\right)\\
=&\ \frac1{\sqrt2}\left(\frac\pi4 +\ln2\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the line integral along a parabola
Evaluate the line integral of $f(x,y)=-y+x$ along part of the parabola $y=2(x+1)^2$ from the point $(0,2)$ to the point $(-1,0)$
I need help trying to find a good parameterization for this because what I've done just lands me in a mess.
My work so far:
Let $x=t, y=2(t+1)^2$
$$
\begin{split}
r(t) &= \left<t,2(t+1)^2\right>, \quad -1\leq t \leq 0 \\
r'(t) &= \left<1,4(t+1)\right>, \quad -1\leq t \leq 0 \\
\|r'(t)\| &= \sqrt{1+16(t+1)^2} \\
-y+x &= -2(t+1)^2+t\\
&=-2t^2-3t-2\\
\end{split}
$$
So we have
$$
\int_0^{-1}\left(-2t^2-3t-2\right)\sqrt{1+16(t+1)^2}dt
$$
This integral is really ugly. so then I tried a different method:
$$
\begin{split}
y &= 2(x+1)^2 \\
\frac{dy}{dx} &= 4x+4 \\
dS &= \sqrt{1+(4x+4)^2} dx\\
&=\sqrt{16x^2+32x+17} dx
\end{split}
$$
So we get
$$\int_0^{-1} \left(-2(x+1)^2+x\right)\sqrt{16x^2+32x+17}dx$$
Again, very ugly. Can someone please help me solve this?
| Using the substitution $4(x+1)=y$ on the latter integrand, we get:
\begin{align}\frac{1}{16}\int_{0}^{1\over4}(\frac{y^2}{2}-y-4)\sqrt{y^2+1}\,dy\end{align}
The integral$\int_{0}^{1\over4}(y+4)\sqrt{y^2+1}\,dy$ is quite straightforwardly evaluated.
For the $4\int_{0}^{1\over4}\sqrt{y^2+1}\,dy$, substituting $y=\sinh u$, we get $2(3\ln(y+\sqrt{y^2+1})-y\sqrt{1+y^2})$.
For $\int_{0}^{1\over4}y\sqrt{y^2+1}\,dy$, let $y^2=\alpha$, getting $\frac{1}{3}(y^2+1)^{3\over2}$. The integral becomes:
\begin{align}\frac{1}{32}\biggr(\int_{0}^{1\over4}y^2\sqrt{y^2+1}\,dy\biggr)-\frac{1}{16}\biggr(6\ln(y+\sqrt{y^2+1})-2y\sqrt{1+y^2}+\frac{1}{3}(y+1)^{3\over2}\biggr)\biggr|_0^{1\over4}\end{align}
To integrate the first part, using the substitution $y=\sinh u$, it becomes:
\begin{align}\frac{1}{256}\biggr(2y^{3}\sqrt{1+y^2}+\ln(y+\sqrt{y^2+1})-y\sqrt{1+y^2}\biggr)\end{align}
So the integral finally is:
\begin{align}\frac{1}{256}\biggr(2y^{3}\sqrt{1+y^2}-95\ln(y+\sqrt{y^2+1})+31y\sqrt{1+y^2}-\frac{16}{3}(y+1)^{3\over2}\biggr)\biggr|_0^{1\over4}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Calculate a triple integral where the integrand contains cubic terms Calculate
$$I = \iiint\limits_Q(x^3 + y^3 + z^3)\,dx\,dy\,dz$$
where $Q: x^2 + y^2 + z^2 - 2ax-2ay-2az + 2a^2 = 0$ is a sphere
If I try using spherical coordinates I don't go anywhere. Any hint about
how to calculate it?
| Due to symmetry
$$I = \iiint_Q(x^3 + y^3 + z^3)\,dV
=3 \iiint_Q z^3\,dV$$
Then, recenter the sphere at origin and integrate in spherical coordinates
\begin{align}
I =& \ 3 \iiint_{x^2+y^2+z^2<a^2} (z+a)^3\,dV
=\ 3 \iiint_{x^2+y^2+z^2<a^2} (3az^2 +a^3)\,dV\\
=& \ 3a \iiint_{r<a} r^2 \ r^2\sin\phi dr d\theta d\phi + 3a^3 \cdot \frac43\pi a^3=\frac{32}5\pi a^6
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\tan\frac{\alpha}{2},$ if $\sin\alpha+\cos\alpha=\frac{\sqrt7}{2}$ and $\alpha\in(0^\circ;90^\circ)$. Find $\tan\dfrac{\alpha}{2},$ if $\sin\alpha+\cos\alpha=\dfrac{\sqrt7}{2}$ and $\alpha\in(0^\circ;90^\circ)$.
So we have $$\sin\alpha+\cos\alpha=\dfrac{\sqrt7}{2}$$ Then $$\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha=\dfrac74\\2\sin\alpha\cos\alpha=\dfrac74-1=\dfrac34\\\sin2\alpha=\dfrac34\\\dfrac{2\tan\alpha}{1+\tan^2\alpha}=\dfrac34\\\tan\alpha=\dfrac{4\pm\sqrt{7}}{3},$$ Is there a more direct approach?
| Let $\beta=\alpha +45^o.$ Then $\sin \beta=\sin \alpha \cos 45^o+\cos \alpha \sin 45^o=(1/\sqrt 2)(\sin \alpha +\cos \alpha)=(\sqrt 7)/(2\sqrt 2).$ And so $\cos \beta=\pm \sqrt {1-\sin^2\beta}=\pm \sqrt {1-7/8}=\pm 1/(2\sqrt 2).$
Now $\cos \alpha =\cos (\beta -45^o)=\cos \beta \cos 45^o+\sin \beta \cos 45^o=(\cos \beta +\sin \beta)/(\sqrt 2)$ and the rest follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving $x^2-3-\operatorname{frac}(x)=0$
Problem. Solve the equation
$$ x^2 - 3 - \operatorname{frac}(x) = 0 $$
I tried to solve this question by moving $3$ to the other side:
$$x^2 - \operatorname{frac}(x) = 3$$
Since the fractional part of R.H.S. is $0$,
$$
\operatorname{frac}\left(x^2\right) = \operatorname{frac}(x)
\qquad\text{or}\qquad
\operatorname{frac}\left(x^2\right) = 1 - \operatorname{frac}(x).$$
Since there is only 1 case for $\operatorname{frac}(x^2)=\operatorname{frac}(x)$ which is $\operatorname{frac}\left(x^2\right)=\operatorname{frac}(x)=0$ and no value of $x$ satisfies the equation, we can rule this condition out.
We are left with $\operatorname{frac}\left(x^2\right)=1-\operatorname{frac}(x)$ and $x$ is negative, but I am stuck here.
Can anyone please give me an idea to proceed with this question?
| We have
$$
0 = x^2 - 3 - {\rm frac}(x) = x^2 - 3 - \left\{ x \right\}
$$
where
$$
x = \left\lfloor x \right\rfloor + \left\{ x \right\}\quad \left| {\,0 \le \left\{ x \right\} < 1} \right.
$$
So we can put first of all
$$
\eqalign{
& 3 \le x^2 = 3 + \left\{ x \right\} < 4\quad \Rightarrow \quad \sqrt 3 \le \pm x < 2\quad \Rightarrow \cr
& \Rightarrow \quad \left( {\sqrt 3 \le x < 2} \right) \cup \left( { - 2 < x \le - \sqrt 3 } \right)\quad \Rightarrow \cr
& \Rightarrow \quad \left( {1 \le \left\lfloor x \right\rfloor < 2} \right)
\cup \left( { - 3 < \left\lfloor x \right\rfloor \le - 2} \right)\quad \Rightarrow \quad \left\lfloor x \right\rfloor
= - 2,1 \cr}
$$
Then, for $\left\lfloor x \right\rfloor = 1$
$$
\eqalign{
& 0 = x^2 - 3 - \left\{ x \right\} = \left( {1 + \left\{ x \right\}} \right)^2 - \left\{ x \right\} - 3 = \cr
& = 1 + 2\left\{ x \right\} + \left\{ x \right\}^2 - \left\{ x \right\} - 3 = \cr
& = \left\{ x \right\}^2 + \left\{ x \right\} - 2\quad \Rightarrow \cr
& \Rightarrow \quad \left( {\left\{ x \right\} = - 2} \right) \cup \left( {\left\{ x \right\} = 1} \right)
\quad \Rightarrow \cr
& \Rightarrow \quad {\rm no}\;{\rm solution} \cr}
$$
while for $\left\lfloor x \right\rfloor = -2$
$$
\eqalign{
& 0 = x^2 - 3 - \left\{ x \right\} = \left( { - 2 + \left\{ x \right\}} \right)^2 - \left\{ x \right\} - 3 = \cr
& = 4 - 4\left\{ x \right\} + \left\{ x \right\}^2 - \left\{ x \right\} - 3 = \cr
& = \left\{ x \right\}^2 - 5\left\{ x \right\} + 1\quad \Rightarrow \cr
& \Rightarrow \quad \left( {\left\{ x \right\} = {{5 + \sqrt {21} } \over 2}} \right)
\cup \left( {\left\{ x \right\} = {{5 - \sqrt {21} } \over 2}} \right)\quad \Rightarrow \cr
& \Rightarrow \quad \left\{ x \right\} = {{5 - \sqrt {21} } \over 2} \cr}
$$
That means that for $x$ we get
$$
x = \left\lfloor x \right\rfloor + \left\{ x \right\}
= - 2 + {{5 - \sqrt {21} } \over 2} = {{1 - \sqrt {21} } \over 2}
$$
which is the same as the solution indicated by S. Lee
This is the graphical representation of the above
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Compute $f^{(2020)}(0)$ Problem :
Let
$$f(x)=\frac{x}{(x+1)(1-x^2)}$$
Then find $f^{(2020)}(0)$.
My Attempt :
From partial fraction decomposition,
$$f(x)=\frac{1}{4(1-x)}+\frac{1}{4(x+1)}-\frac{1}{2(x+1)^2}$$
and,
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n,\quad \frac{1}{1+x}=\sum_{n=0}^\infty (-1)^nx^n, \quad\frac{1}{(1+x)^2}=\sum_{n=1}^\infty (-1)^{n+1}nx^{n-1}=\sum_{n=0}^\infty(-1)^n(n+1)x^n$$
From these,
$$f(x)=\sum_{n=0}^\infty \frac{x^n}{4}+\sum_{n=0}^\infty \frac{(-1)^nx^n}{4}+\sum_{n=0}^\infty\frac{(-1)^{n+1}(n+1)x^n}{2} \\ =\sum_{n=0}^\infty\frac{x^n+(-1)^nx^n+(-1)^{n+1}(n+1)x^n}{4} \\ =\sum_{n=0}^\infty\frac{(1+(-1)^n+(-1)^{n+1}(n+1))x^n}{4}$$
$f^{(2020)}(0)=(2020)!\times a_{2020}$ where $f(x)=\sum a_nx^n$
In this case, $a_n = 1+(-1)^n+(-1)^{n+1}(n+1), a_{2020}=-2019$ but there is no same answer.
Where did I make a mistake?
| Since the other responses have dissected the OP's (i.e. original poster's) work, I can give an alternative approach.
$\underline{\text{Preliminary Results}}$
$\displaystyle f(u) = u^{-1} \implies f^{(2020)}(u) = [(2020)!] \times u^{-2021}.$
$\displaystyle f(u) = u^{-2} \implies f^{(2020)}(u) = [(-2) \times (-3) \times \cdots (-2021)] \times u^{-2022}$
$\displaystyle = [(2021)!] \times u^{-2022}.$
From the OP's work:
$$f(x) = \left[~\left(\frac{-1}{4}\right) \times \left(\frac{1}{x-1}\right) ~\right]
~+~ ~\left[~\left(\frac{1}{4}\right) \times \left(\frac{1}{x+1}\right) ~\right]
~+~ ~\left[~\left(\frac{-1}{2}\right) \times \left(\frac{1}{x+1}\right)^2 ~\right].$$
Let $~\displaystyle u = (x - 1), v = (x + 1) \implies
\frac{du}{dx} = 1 = \frac{dv}{dx}.$
Then
$$f(x) = \left[(-1/4)u^{-1}\right] + \left[(1/4)v^{-1}\right] + \left[(-1/2)v^{-2}\right].$$
Using the preliminary results, you have that
$$f^{(2020)}(x) = [(2020)!] ~~\times $$
$$\left\{ ~\left[(-1/4)u^{-2021}\right] ~+~ \left[(1/4)v^{-2021}\right]
~+~ \left[(-1/2)(2021)v^{-2022}\right] ~\right\}.$$
At $~x = 0,~$ you have that $~u = -1, ~v = +1.$
Therefore,
$$f^{(2020)}(x=0) = [(2020)!] ~~\times $$
$$\left\{ ~\left[(-1/4)(-1)\right] ~+~ \left[(1/4)(+1)\right]
~+~ \left[(-1/2)(2021)(+1)\right] ~\right\}$$
$$= [(2020)!] \times \left\{ ~(1/4) + (1/4) - \left[(1/2)(2021)\right] ~\right\}$$
$$= [(2020)!] \times (-1010).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Slight variation of Mertens' third theorem. Do we have an estimate of $\sum_{d \mid n\#}(-1)^{\omega(d)}\dfrac{\sigma_0(d)}{d}$? Define $f(n)$ to be:
$$
\sum_{d \mid n\#}(-1)^{\omega(d)}\dfrac{\sigma_0(d)}{d}
$$
But $\sigma_0(d) = 2^{\omega(d)}$ for any $d \mid n\#$ a primorial, so:
$$
f(n) = \prod_{p \text{ prime} \\ p \leq n} \dfrac{1}{\left(1 - \dfrac{2}{p}\right)}
$$
Mertens' third theorem has as $1$ instead of a $2$ there.
So what would be an asymptotic estimate of $f(n)$?
| We have
\begin{align*}
\prod\limits_{2<p \le n} {\left( {1 - \frac{2}{p}} \right)^{ - 1} } & = \exp \left( { - \sum\limits_{2<p \le n} {\log \left( {1 - \frac{2}{p}} \right)} } \right) \\ & = \exp \left( {\sum\limits_{2<p \le n} {\frac{2}{p}} - \sum\limits_{2<p \le n} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} } \right).
\end{align*}
Now by Mertens' second theorem
$$
\sum\limits_{2<p \le n} {\frac{2}{p}} = 2\log \log n + 2M -1+ \mathcal{O}\!\left( {\frac{1}{{\log n}}} \right),
$$
where
$$
M = \gamma + \sum\limits_p {\left[ {\frac{1}{p} + \log \left( {1 - \frac{1}{p}} \right)} \right]}
$$
is the Meissel–Mertens constant. Also
\begin{align*}
\sum\limits_{2<p \le n} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} & = \sum\limits_{p>2} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} - \sum\limits_{p > n} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]}
\\ & = \sum\limits_{p>2} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} - \sum\limits_{p > n} {\mathcal{O}\!\left( {\frac{1}{{p^2 }}} \right)} \\ &
= \sum\limits_{p>2} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} + \mathcal{O}\!\left( {\frac{1}{{n\log n}}} \right).
\end{align*}
Taking exponentials and simplifying
$$
\prod\limits_{2<p \le n} {\left( {1 - \frac{2}{p}} \right)^{ - 1} } = C(\log n)^2 +\mathcal{O}(\log n),
$$
where
$$
C = \frac{{e^{2\gamma } }}{4}\prod\limits_{p > 2} {\left( {1 + \frac{1}{{p(p - 2)}}} \right)} = \frac{{e^{2\gamma } }}{4\Pi_2}=1.201303559967362\ldots,
$$
with $\Pi_2$ being the twin prime constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4472059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding the minimum value of $(x_1-x_2)^2 +\left(\frac{9}{x_1} - \sqrt{1-x_2^2}\right)^2$ by using trigonometry
Find the minimum value of the given function by using trigonometry
$$(x_1-x_2)^2 + \left(\dfrac{9}{x_1} - \sqrt{1-x_2^2}\right)^2$$
I know the distance formula method but is there any other suitable method to solve this? I’m thinking along the lines of trigonometry or AM-GM inequality.
| Expand as usual, and using Cauchy-Schwarz inequality: $f(x_1,x_2) = x_1^2 - 2x_1x_2 + \dfrac{81}{x_1^2} - \dfrac{18\sqrt{1-x_2^2}}{x_1}+1 = 1+x_1^2+\dfrac{81}{x_1^2} - 2\left(x_1x_2+\dfrac{9\sqrt{1-x_2^2}}{x_1}\right)\ge 1+x_1^2+\dfrac{81}{x_1^2} - 2\sqrt{x_1^2+\dfrac{81}{x_1^2}}\sqrt{x_2^2+(1-x_2^2)} = 1 + y^2- 2y = g(y)$, with $y = \sqrt{x_1^2+\dfrac{81}{x_1^2}}\ge 3\sqrt{2}$ by AM-GM inequality. Thus the problem reduces to finding the min value of $g(y) = y^2-2y+1, y \ge 3\sqrt{2}$. We have again: $g(y) = (y-1)^2 \ge (3\sqrt{2}-1)^2= 19-6\sqrt{2}$, and thus $f(x_1,x_2)_{\text{min}} = 19-6\sqrt{2}$ and this min value achieved when $x_1 = \pm 3, \dfrac{x_1}{x_2}=\dfrac{\dfrac{9}{x_1}}{\sqrt{1-x_2^2}}\implies x_1 = \pm 3, x_2 = \sqrt{1-x_2^2}\implies x_2^2=1-x_2^2\implies x_2 = \pm \dfrac{1}{\sqrt{2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4472906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Answer:
$a+b = 7, ab = 2$
$$\begin{align}
(a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt]
a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \\
&= (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)
\end{align}$$
now,
$$\begin{align}
a^4 + b^4 &= (a+b)^4 - (4a^3b + 6a^2b^2 + 4ab^3) \\
&= (a+b)^4 - (4ab(a^2 + b^2) + 6(ab)^2) \\
&= (a+b)^4 - (4ab((a + b)^2 - 2ab) + 6(ab)^2) \\
&= 7^4 - (4(2)(7^2 - 2(2)) + 6(2)^2) \\
&= 2017
\end{align}$$
so
$$\begin{align}
&\phantom{=}\; (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)\\
&= 7^6 - (6\cdot2\cdot(2017) + 15(2)^2 (7^2 - 2(2)) + 20(2)^3) \\
&= 90585
\end{align}$$
correct?
| A better way to handle this is to observe the factorization
$$a^6 + b^6 = (a^2 + b^2)(a^4 - a^2 b^2 + b^4),$$ from which we observe
$$a^2 + b^2 = (a+b)^2 - 2ab,$$ and
$$a^4 - a^2 b^2 + b^4 = (a^2 + b^2)^2 - 3a^2 b^2.$$
Since we know that $$x^2 - 7x + 2 = (x - a)(x - b) = x^2 - (a+b)x + ab,$$ it follows that $a+b = 7$ and $ab = 2$. So we just put everything together:
$$a^2 + b^2 = 7^2 - 2(2) = 49 - 4 = 45,$$
$$a^4 - a^2 b^2 + b^4 = (45)^2 - 3(2)^2 = 2013,$$
therefore
$$a^6 + b^6 = 45(2013) = 90585.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 3
} |
Evaluate $\sum _{n=0}^{\infty } \frac{16^n}{(2n+1)^3 \binom{2 n}{n}^2}$ In this post give the result of $$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=8\pi\text{G}-14 \zeta (3)$$
Then i wonder if i change $n^3$ to $(2n+1)^3$ so what is the result. I check with CAS give the result: $$\sum _{n=0}^{\infty } \frac{16^n}{(2n+1)^3 \binom{2 n}{n}^2}=-\pi\text{G}+\frac{7}{2} \zeta (3)\tag1$$.
My attemp is: $\frac{2}{(2n+1)^3}=\int_{0}^{1}x^{2n}\log^{2}{x}dx$, then (1) become: $\sum _{n=0}^{\infty } \frac{16^n}{(2n+1)^3 \binom{2 n}{n}^2}=\frac{1}{2}\sum _{n=0}^{\infty } \frac{16^n}{\binom{2 n}{n}^2}\int_{0}^{1}x^{2n}\log^{2}{x}dx=\int_{0}^{1}\log^{2}{x}dx\sum _{n=0}^{\infty } x^{2n}\frac{16^n}{\binom{2 n}{n}^2}$. But the sum is seem harder. I don't have much experience with central binomial coefficient. Can you give some hints? Thank you.
| Other than @Frank W's elegant solution, here is a different approach.
It is well-known that:
$\displaystyle \tag*{}\frac {\arcsin x}{\sqrt{1-x^2}}=\sum\limits_{n\ge 0}\frac {4^nx^{2n+1}}{(2n+1)\binom {2n}n}$
Dividing by $x$ and doing $\int_0^x \,dx$ on both sides gives:
$\displaystyle \tag{1}\int_0^x\frac {\arcsin x}{x\sqrt{1-x^2}}dx=\sum\limits_{n\ge 0}\frac {4^nx^{2n+1}}{(2n+1)^2\binom {2n}n}$
Now, we use the fact that
$$ \int_0^{\pi/2} \sin^{2n+1}(\theta) \, d\theta = \frac {4^n}{(2n+1)\binom {2n}n} $$
This tells us that if we let $x = \sin (\theta)$ and taking $\int_0^{\pi /2} \, d\theta$ both sides of $(1)$ gives our sum. Thus,
$$ \sum\limits_{n\ge 0}\frac {16^n}{(2n+1)^3\binom {2n}n} = \underbrace{\int_0^{\pi /2} \int_0^\theta \frac{x}{\sin x} \, dx \, d\theta}_{I}$$
Integrating by parts with $u=x$ and $v' = \log(\tan\frac x2)$, we get
$$I = \int_0^{\pi /2} \theta \log \tan \frac \theta 2 \, d\theta -\int_0^{\pi/2}\int_0^\theta \log \tan \frac x2 \, dx \, d\theta $$
Now, with the help of Fourier series, we have $\forall \ x \in (0, \pi)$
$$\begin{align} \tag{2} \log \sin \frac x2 &= -\log 2 - \sum_{n \ge 1} \frac{\cos nx}{n} \\ \tag{3}\log \cos \frac x2 &= -\log 2 + \sum_{n \ge 1} \frac{(-1)^{n-1}\cos nx}{n}\end{align}$$
Doing $(2)-(3)$ gives the series for $\log \tan \frac x2$. Now,plugging in the value, we get
$$ \begin{align} I &= \frac \pi 2\sum_{n\ge 1} \frac{(-1)^n \sin (n \pi/2)}{n^2} - \underbrace{\sum_{n\ge 1} \frac{(-1)^n}{n^3}}_{\dfrac{-3\zeta(3)}{4}} - \frac \pi 2 \sum_{n \ge 1} \frac{\sin (n \pi /2)}{n^2} + \underbrace{\sum_{n \ge 1} \frac{1}{n^3}}_{\displaystyle \zeta(3)} + \underbrace{2\sum_{n \ge 1} \frac{1}{(2n+1)^3} + 2}_{\dfrac{7 \zeta(3)}{4}} \\\\ &= \frac{7 \zeta(3)}{2} - \frac \pi 2 \sum_{n \ge 1} \frac{\sin(n\pi /2)((-1)^n-1)}{n^2} \\ &= \frac{7 \zeta(3)}{2} - \pi \sum_{n \ge 1} \frac{(-1)^n}{(2n+1)^2} \\ &= {\color{Red}{\frac{7 \zeta(3)}{2} - \pi G}}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Given $0\leq a\leq b\leq c$ and $b > 0$, prove that: $2\sqrt{\frac{a}{c}}\leq \frac{b}{c} + \frac{a}{b}\leq 1 + \frac{a}{c}$ This is an exercise that got me stuck. It's from the book "Functions of Several Real Variables 1.1 - exercise 10" by Martin Moskowitz and Fotios Paliogiannis.
Exercise:
Given $0\leq a\leq b\leq c$ and $b > 0$, prove that:
$2\sqrt{\frac{a}{c}}\leq \frac{b}{c} + \frac{a}{b}\leq 1 + \frac{a}{c}$
What I did so far:
$2\sqrt{\frac{a}{c}}\leq \frac{b}{c} + \frac{a}{b}\leq 1 + \frac{a}{c}$
$4\frac{a}{c}\leq \frac{b^2}{c^2} +2\frac{a}{c} + \frac{a^2}{b^2}\leq 1 + 2\frac{a}{c}+\frac{a^2}{c^2}$ [power of 2]
$2\frac{a}{c}\leq \frac{b^2}{c^2} + \frac{a^2}{b^2}\leq 1 +\frac{a^2}{c^2}$ [minus $2\frac{a}{c}$]
$2acb^2 \leq b^4+a^2c^2 \leq c^2b^2+a^2b^2$ [multiply by $c^2b^2$]
$2acb^2 -2a^2c^2\leq b^4\leq c^2b^2+a^2b^2 - a^2c^2$
And now I have to prove the inequalities for $b^4$. However, I can not advance (actually I have no way of knowing if this is the right path since I have very little experience writing proofs).
| Let $a=x^2, b=y^2, c=z^2.$ For fixed $b,c$ let $f(x)=(b/c)+(a/b)-2\sqrt {a/c}\,=(y^2/z^2)+(x^2/y^2)-2(x/z)$ and let $g(a)=1+(a/c)-(b/c)-(a/b).$
Show that $f'(x)>0\iff x>y^2/z.$ So $\min f(x)=f(y^2/z)=0.$
Show that $g'(a)\le 0.$ So $\min \{g(a):a\le b\}=g(b)=0.$
The substitution $a=x^2, b=y^2, c=z^2$ is not really necessary. It just makes the calculations easier for the left inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4475725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Laurent series on different areas? Looking for the Laurent series of
$f(z)=\frac{1}{z(z-1)(3i-z)}=\frac{i}{3 z} + \frac{\frac{1}{10} - \frac{i}{30}}{z - 3 i} - \frac{\frac{1}{10} + \frac{3i}{10}}{z - 1}$
So we look for $z_0=0,z_1=1,z_2=3i$ on
$A=\{z \in \Bbb C: |z| <1\}$
$B=\{z \in \Bbb C: 1< |z| <3\}$
$C=\{z \in \Bbb C: 3< |z| < \infty\}$
On $A$:
$\frac{i}{3 z}=\frac{i}{3 (z-1+1)}=\frac{i}{3 }\sum_{n=0}^\infty \frac{i}{3 }(z-1)^n(-1)^n $
$\frac{\frac{1}{10} - \frac{i}{30}}{z - 3 i} =\frac{1}{z}\frac{\frac{1}{10} - \frac{i}{30}}{1 - 3 i/z}=\frac{1}{z }\sum_{n=0}^\infty (\frac{1}{10} - \frac{i}{30})(-\frac{3i}{z })^n=\sum_{n=0}^\infty (\frac{1}{10} - \frac{i}{30})((-\frac{3i}{z })^{n+1}\sum_{n=1}^\infty (\frac{1}{10} - \frac{i}{30})((-\frac{3i}{z })^n$
$\frac{\frac{1+3i}{10}}{z - 1}=-\frac{1+3i}{10}\sum_{n=0}^\infty z^n $
Is this correct and how is this different to B and C ? How would I begin with in B ?
| Hint: for $|z|<1$ we have $$\frac{1}{1-z} = \sum_{n=0}^\infty z^n$$ but for $|z|>1$ we have $$\frac{1}{1-z} =-\frac{1}{z} \frac{1}{1-\frac{1}{z}} = -\sum_{n=1}^\infty\frac{1}{z^n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What will be the value of $ 2\cos^{2}\theta - 1 $ , if $ \cos^{4}\theta - \sin^{4}\theta = \frac{2}{3}$ The question was What will be the value of $ 2\cos^{2}\theta - 1 $, if $ \cos^{4}\theta - \sin^{4}\theta = \frac{2}{3}$
Here is my working:
$\cos^{4}\theta - \sin^{4}\theta = \frac{2}{3} $
$(\cos^{2}\theta)^{2}-(\sin^{2}\theta)^{2} = \frac{2}{3}$
$(\cos^{2}\theta-\sin^{2}\theta)\cdot(\cos^{2}\theta+\sin^{2}\theta) = \frac{2}{3}$
$\cos2\theta = \frac{2}{3}$
But I am stuck at this step so can anyone help me ?
| I finally got the answer
$\cos2\theta = 2\cos^{2}\theta - 1$
$2\cos^{2}\theta - 1 = \frac{2}{3} $
This is our required answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrating $\int { \arctan{t} \over \sqrt{t} } $ Integrate: $$\int { \arctan{t} \over \sqrt{t} }\,\mathrm dt$$
My attempt:
I first substituted $y = \sqrt{t}$
then $$\int { \arctan{t} \over \sqrt{t} }\,\mathrm dt = 2\int { \arctan{y^2}}\,\mathrm dy$$
then I integrated by parts:
$u = \arctan{y^2}$, $v' = 1$
$$ = 2\left(y\arctan{y^2} - \int { y^2 \over 1+y^4}\,\mathrm dy\right)$$
but I am stuck, I cannot think of a way to solve this other integral.
I believe there is a more straightforward approach than the one I am using.
Thanks in advance.
| Elaborating on the comment, the key is to note that the denominator can indeed be further factored
$$\begin{align*}y^4+1 & =y^4+2y^2+1-2y^2\\ & =\left(y^2+1\right)^2-2y^2\\ & =\left(y^2+y\sqrt2+1\right)\left(y^2-y\sqrt2+1\right)\end{align*}$$
Therefore, we can use standard partial fraction decomposition to arrive at
$$\frac {y^2}{y^4+1}=\frac {Ay+B}{y^2+y\sqrt2+1}+\frac {Cy+D}{y^2-y\sqrt2+1}=\frac 1{2\sqrt{2}}\left(\frac {y}{y^2-y\sqrt{2}+1}-\frac {y}{y^2+y\sqrt{2}+1}\right)$$
The algebra is a bit messy for the partial fraction decomposition. One cool method you can use to find the coefficients is, starting off with$$\frac {y^2}{(y^2-y\sqrt{2}+1)(y^2+y\sqrt2+1)}=\frac {Ay+B}{y^2+y\sqrt2+1}+\frac {Cy+D}{y^2-y\sqrt2+1}$$Equate one of the factors to zero, say $$y^2-y\sqrt{2}+1=0\qquad\implies\qquad y^2=y\sqrt{2}-1$$Now, covering up the $y^2-y\sqrt{2}+1$ factor in the denominator on the left-hand side, take the limit as $y^2$ tends towards $y\sqrt{2}-1$; the resulting expression will be your $Cy+D$ term. To put it in equation-form$$Cy+D=\lim\limits_{y^2\to y\sqrt{2}-1}\frac {y^2}{y^2+y\sqrt{2}+1}=\frac {y\sqrt2-1}{2y\sqrt2}=\frac {y}{2\sqrt{2}}$$To get to the last line, I expanded the fraction into its constituents and used the equation$$\frac 1y=\sqrt2-y$$Directly comparing coefficients, it's clear that we must have$$C=\frac 1{2\sqrt2}\qquad\qquad D=0$$Repeat the same procedure with $y^2\to -y\sqrt2-1$ to arrive at$$A=-\frac 1{2\sqrt2}\qquad\qquad B=0$$
The integral becomes
$$\int\frac {y^2}{y^4+1}\,\mathrm dy=\frac 1{2\sqrt{2}}\int\frac {y}{y^2-y\sqrt{2}+1}\,\mathrm dy-\frac 1{2\sqrt2}\int\frac {y}{y^2+y\sqrt{2}+1}\,\mathrm dy$$
The resulting integral can be evaluated by first observing that $(y^2\pm y\sqrt{2}+1)'=2y\pm\sqrt{2}$. Thus, adding and subtracting $1/\sqrt2$ from the numerator then gives
$$\begin{align*}\int\frac {y}{y^2\pm y\sqrt{2}+1}\,\mathrm dy & =\frac 12\int\frac {2y\pm\sqrt{2}\mp\sqrt{2}}{y^2\pm y\sqrt{2}+1}\,\mathrm dy\\ & =\frac 12\int\frac {2y\pm\sqrt{2}}{y^2\pm y\sqrt{2}+1}\,\mathrm dy\mp\frac 1{\sqrt{2}}\int\frac {\mathrm dy}{\left(y\pm\frac 1{\sqrt2}\right)^2+\frac 12}\\ & =\frac 12\log(y^2\pm y\sqrt{2}+1)\mp\arctan\left(y\sqrt{2}\pm 1\right)\end{align*}$$
Putting everything together, then we see that
$$\begin{align*}\int\frac {y^2}{y^4+1}\,\mathrm dy=\frac 1{4\sqrt2}\log\left(\frac {y^2-y\sqrt{2}+1}{y^2+y\sqrt{2}+1}\right) & +\frac 1{2\sqrt{2}}\arctan\left(y\sqrt{2}-1\right)\\ & +\frac 1{2\sqrt{2}}\arctan\left(y\sqrt{2}+1\right)+C\end{align*}$$
Thus, our integral becomes
$$\begin{align*}\int\arctan y^2\,\mathrm dy=y\arctan y^2-\frac 1{2\sqrt2}\log\left(\frac {y^2-y\sqrt2+1}{y^2+y\sqrt2+1}\right) & -\frac 1{\sqrt2}\arctan\left(y\sqrt2-1\right)\\ & -\frac 1{\sqrt2}\arctan\left(y\sqrt2+1\right)+C\end{align*}$$
Confirmed by Wolfram Alpha
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda} \leq \frac{3}{\lambda +1}$
If $a+b+c=3$ ,$ a,b,c>0$ and $\lambda \geq 1$,
prove that : $$\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda} \leq \frac{3}{\lambda +1}.$$
my attempt:
using CBC twice and AM_AG inequality
$\frac{3}{3}\Big(\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda}\Big)^2\leq 3 \Big(\frac{a}{a^2+\lambda}\Big)^2+\Big(\frac{b}{b^2+\lambda}\Big)^2+\Big(\frac{c}{c^2+\lambda}\Big)^2\leq 3\Bigg( \bigg(\frac{a}{\frac{(a+\sqrt{\lambda})^2}{2}}\bigg)^2+ \bigg(\frac{b}{\frac{(b+\sqrt{\lambda})^2}{2}}\bigg)^2+\bigg(\frac{c}{\frac{(c+\sqrt{\lambda})^2}{2}}\bigg)^2\Bigg) =12\bigg(\frac{a^2}{(a+\sqrt{\lambda })^4}+\frac{b^2}{(b+\sqrt{\lambda})^4}+\frac{c^2}{(c+\sqrt{\lambda })^4}\bigg)\leq12 \bigg(\frac{a^2}{\sqrt{\lambda}^4}+\frac{b^2}{\sqrt{\lambda}^4}+\frac{c^2}{\sqrt{\lambda}^4}\bigg)=\frac{48(a^2+b^2+c^2)}{4\lambda ^2 }\leq \frac{48\cdot9}{4\lambda^2} \leq \frac{9}{(\lambda +1)^2}$
beacuse $f(\lambda )=\frac{9}{(\lambda+1)^2}-\frac{48\cdot9}{4\lambda^2} \geq 0 $ for $\lambda \geq 1$
so finally
$\frac{3}{3}\Big(\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda}\Big)^2\leq\frac{9}{(\lambda +1)^2}$$\Leftrightarrow$$\Big(\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda}\Big)\leq\frac{3}{(\lambda +1)}$
I have just one question:
-does my attempt is true?
| A comment on the tangent line method. Why does it work?
The function $f$ is not concave everywhere. What's going on?
The function $f_{\lambda}=f\colon x\mapsto \frac{x}{x^2 + \lambda}$
has derivatives
$$f'(x) = \frac{\lambda - x}{(\lambda + x^2)^2} \\
f^{''}(x) = \frac{2x(x^2 -3 \lambda) }{(x^2 + \lambda)^3}$$
Therefore, $f$ is strictly increasing on the interval $[0, \sqrt{\lambda}]$ and strictly concave on the interval $[0, \sqrt{3 \lambda}]$. We conclude that the function
$$f_1(x) = \begin{cases} f(x) &\textrm{ for }& x \in [0, \sqrt{\lambda}] \\
f(\sqrt{\lambda}) &\textrm{ for }& x \in [\sqrt{\lambda}, \infty)
\end{cases}$$
is concave and $\ge f(x)$. Now, the tangent line at $x=1$ for the graph $f_1$ is the tangent line for the graph of $f$. Therefore, the tangent line is above the graph of $f_1$, and therefore, above the graph of $f$.
$\bf{Added:}$ With all the preparations above, we can state:
if $a_1$, $\ldots$, $a_n$ are positive, $t_i \ge 0$, $\sum_{i=1}^n t_i = 1$, and $a\colon = \sum_{i=1}^n t_i a_i \le \sqrt{\lambda}$ then
$$\sum_{i=1}^n t_i f_{\lambda}(a_i) \le f_{\lambda}(a)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Closed form of $\sum _{k\ge 1} \frac{(-1)^{\binom{k}{p}}}{k}$, an alternating harmonic series with the signs determined by a binomial coeffcient In a comment to Evaluating $\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$ for $n \in \mathbb{N}$ I proposed to study this alternating harmonic sum
$$s(p) = \sum _{k\ge 1} \frac{(-1)^{\binom{k}{p}}}{k}\tag{1}$$
where the sign of the terms is determined by a binomial coefficient.
This turned out to be interesting, and the results can be formulated in three statements:
Statement 1: the sum is convergent only if $p$ is power of $2$: $p = 2^q, q=0,1,2,...$
Statement 2: the closed expressions of $s$ for the first few values of $q$ are
$$s(1) = -\log (2)\simeq -0.693147$$
$$s(2) = \frac{\pi }{4}-\frac{\log (2)}{2}\simeq 0.438825$$
$$s(4)= \frac{\pi}{8} \left(1+2 \sqrt{2}\right) -\frac{\log (2)}{4}\simeq 1.33013 $$
$$s(8) = \frac{\pi}{16} \left(4 \sqrt{2+\sqrt{2}}+2 \sqrt{2}+1\right) -\frac{\log (2)}{8}\simeq 2.11629$$
The next step was hard, and I did not expect to find a simple expression. But here it is
$$\begin{align}s(16) = \frac{\pi}{16} \left(4 \left(\sqrt{2-\sqrt{2+\sqrt{2}}}+\sqrt{\left(2+\sqrt{2}\right) \left(2-\sqrt{2+\sqrt{2}}\right)}\right)+\sqrt{2}+2 \sqrt{2+\sqrt{2}}+2 \sqrt{2-\sqrt{2+\sqrt{2}}}+2 \sqrt{\sqrt{2+\sqrt{2}}+2}+\frac{1}{2}\right) -\frac{\log (2)}{16}\simeq 2.85436\end{align}$$
Statement 3: The general structure seems to be
$$s(p=2^q) = \frac{\pi}{p}a(p)-\frac{1}{p} \log(2) \tag{2}$$
where $a(p)$ is an algebraic number composed of the number $2$.
Questions
(1) are these results known in the literature?
(2) can you derive/prove the three statements?
(3) can you give the expression $a(p)$ to extend the given range?
| Here, I will examine Statement 2 and 3.
Step 1. To determine the parity of $\binom{k}{2^q}$, we consider its generating function in the ring $\mathbb{F}_2[\![x]\!]$ of formal power series with coefficients in $\mathbb{F}_{2}$:
\begin{align*}
\sum_{k=0}^{\infty} \binom{k}{2^q} x^k
&= \frac{x^{2^q}}{(1-x)^{2^q + 1}}
= \frac{x^{2^q}}{(1-x)(1 - x^{2^q})}
= \frac{1 - x^{2^q}}{1 - x} \cdot \frac{x^{2^q}}{(1 - x^{2^q})^2} \\
&= (1 + x + \cdots + x^{2^q-1})(x^{2^q} + x^{3\cdot2^q} + x^{5\cdot 2^q} + \cdots).
\end{align*}
(In the second step, we utilized the Freshman's dream.) By expanding this product, we find that
$$ \binom{k}{2^q} \equiv \begin{cases}
0, & \lfloor k/2^q \rfloor = 0, 2, 4, \ldots; \\
1, & \lfloor k/2^q \rfloor = 1, 3, 5, \ldots; \\
\end{cases} \pmod{2} $$
Using this, the series $s(2^q)$ can be recast as
\begin{align*}
s(2^q)
&= \int_{0}^{1} \frac{1}{x} \left[ \sum_{k=0}^{\infty} (-1)^{\binom{k}{2^q}} x^k - 1 \right] \, \mathrm{d}x \\
&= \int_{0}^{1} \frac{1}{x} \left[ \frac{1 - x^{2^q}}{(1 - x)(1 + x^{2^q})} - 1 \right] \, \mathrm{d}x \\
&= \int_{0}^{1} \frac{1 - 2x^{2^q - 1} + x^{2^q}}{(1-x)(1 + x^{2^q})} \, \mathrm{d}x.
\end{align*}
Step 2. Now write
$$ P(x) = 1 - 2x^{2^q - 1} + x^{2^q}
\qquad\text{and}\qquad
Q(x) = (1-x)(1 + x^{2^q})$$
for the numerator and denominator of the integrand, respectively. Also, let
$$ Z = \{\omega \in \mathbb{C} : 1 + \omega^{2^q} = 0 \} $$
be the zero set of $1 + x^{2^q}$. Since $P(1) = 0$, $\omega \in Z$ are the only poles of $P(x)/Q(x)$, all of which being simple. So,
\begin{align*}
\frac{P(x)}{Q(x)}
= \sum_{\omega \in Z} \frac{P(\omega)}{Q'(\omega)} \frac{1}{x - \omega}
= \sum_{\omega \in Z} \frac{-2^{1-q}}{1-\omega} \frac{1}{x - \omega}.
\end{align*}
Integrating both sides from $0$ to $1$,
\begin{align*}
s(2^q)
&= - 2^{1-q} \sum_{\omega \in Z} \frac{1}{1-\omega} [\log(1 - \omega) - \log(-\omega)],
\end{align*}
where $\log(\cdot)$ is the principal branch of the complex logarithm.
Step 3. From this point on, we will assume that $q \geq 1$ so that $2^q$ is an even number. Then the map $\omega \mapsto -\omega$ is a permutation of the set $Z$, and so, we may permute the sum to rewrite
\begin{align*}
s(2^q)
&= - 2^{1-q} \sum_{\omega \in Z} \frac{1}{1+\omega} [\log(1 + \omega) - \log \omega]
\end{align*}
Also, note that
$$ \frac{1}{1+\omega} = \frac{1 - i\tan(\frac{1}{2}\arg\omega)}{2}
\qquad\text{and}\qquad
\operatorname{Im} \left[ \log(1 + \omega) - \log \omega \right] = - \frac{\arg\omega}{2}. $$
Since $s(2^q)$ is a real number and each $\log \omega$ is pure imaginary, the sum reduces to
\begin{align*}
s(2^q)
&= \frac{1}{2^q} \sum_{\omega \in Z} \left[ \frac{\arg\omega}{2} \tan\left(\frac{\arg\omega}{2}\right) - \operatorname{Re}[\log(1 + \omega)] \right] \\
&= \frac{1}{2^q} \sum_{\omega \in Z} \frac{\arg\omega}{2} \tan\left(\frac{\arg\omega}{2}\right) - \frac{1}{2^q}\operatorname{Re}\left[\log \prod_{\omega \in Z} (1 + \omega)\right] \\
&= \bbox[color:navy;padding:3px;border:1px dotted navy;]{\frac{\pi}{4^q} \sum_{k=1}^{2^{q-1}} (2k-1) \tan\left(\frac{(2k-1)\pi}{2^{q+1}}\right) - \frac{\log 2}{2^q}.}
\end{align*}
Step 4. Finally, we note that the first sum is algebraic. Since $\tan(k\theta)$ is a rational function of $\tan\theta$, it suffices to verify that $\tan(\pi/2^{q+1})$ is algebraic for all $q$. Indeed, if we assume that $0 \leq \theta \leq \frac{\pi}{4}$ and if $\alpha = 2\cos(2\theta)$, then
$$ \tan \theta = \sqrt{\frac{2-\alpha}{2+\alpha}}
\qquad\text{and}\qquad \tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{2-\sqrt{2+\alpha}}{2+\sqrt{2+\alpha}}}. $$
Using this, we get
\begin{align*}
q = 1
&\quad\implies\quad \tan\left(\frac{\pi}{2^2}\right) = 1 = \sqrt{\frac{2-0}{2+0}}, \\[0.25em]
q = 2
&\quad\implies\quad \tan\left(\frac{\pi}{2^3}\right) = \sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}, \\[0.25em]
q = 3
&\quad\implies\quad \tan\left(\frac{\pi}{2^4}\right) = \sqrt{\frac{2-\sqrt{2 + \smash[b]{\sqrt{2}} }}{2+\sqrt{2 + \smash[b]{\sqrt{2}} }}}, \\[0.25em]
&\qquad \vdots
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Trying to solve $|2x-15| = -x^2 - 5x -8$ My first instinct was to take the positive and negative of the right hand side, resulting in
$2x-15 = -x^2 - 5x - 8$, and $2x-15 = x^2 + 5x + 8$, which results in the first giving me two real answers using the quadratic equation, and the second being two imaginary solutions. The problem is that, when graphed, these 2 graphs do not intersect at all, so there should be no real solutions. As for the given complex solutions, neither were considered by wolfram alpha.
Knowing that there are no real solutions, I'm confused as to how I might get complex solutions through means not already attempted and explained above.
| For complex numbers, the expression $ \ |2x - 15| \ $ is interpreted as the modulus of the number(s) in brackets. If we use $ \ z \ = \ a + bi \ \ , \ \ a \ , \ b \ $ real, the given equation becomes
$$ \sqrt{(2a - 15)^2 \ + \ (2b)^2 } \ \ = \ \ -(a + bi)^2 \ - \ 5(a + bi) \ - \ 8 $$
$$ \Rightarrow \ \ \sqrt{4a^2 - 60a + 225 + 4b^2 } \ \ = \ \ -(a^2 - b^2 + 5a + 8) \ - \ (2ab + 5b)i \ \ . $$
Since the modulus is a real number, the imaginary part of the expression on the right side of this equation must be zero. So we have $ \ 2ab + 5b \ = \ 0 \ \Rightarrow \ a \ = \ -\frac52 \ \ \text{or} \ \ b \ = \ 0 \ \ . $
Since it is easy to show that the "downward-opening" parabola $ \ y \ = \ -x^2 - 5x - 8 \ = \ -\left(x + \frac52 \right)^2 - \frac74 \ \ $ has its vertex "below" the $ \ x-$axis, there can be no purely real solutions to the original equation (as you also found from a graph).
Inserting $ \ a \ = \ -\frac52 \ \ $ into the equation instead leads to
$$ \ \sqrt{25 + 150 + 225 + 4b^2 } \ \ = \ \ -(\frac{25}{4} - b^2 - \frac{25}{2} + 8) \ \ \Rightarrow \ \ 2·\sqrt{100 + b^2 } \ \ = \ \ b^2 - \frac{7}{4} $$
$$ \Rightarrow \ \ b^4 \ - \ \frac{15}{2}·b^2 \ - \ \frac{6351}{16} \ \ = \ \ 0 \ \ , $$
a biquadratic equation with two real solutions for $ \ b \ $ . The two complex roots of the original equation are thus $ \ z \ = \ -\frac52 \ \pm \ i·\frac{ \sqrt{15 \ + \ 4 \sqrt{411}}}{2} \ \ . $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Better upper bound for $ \sum_{k=2}^{\infty} \frac{1}{2^{k-1}} \sum_{n=1}^{\infty} (n! \: \text{mod} \: k) $ Since the sum $\sum_{n=1}^{\infty} (n! \: \text{mod} \: k)$ will be zero beyond $k-1$, the series could be interpreted as an finite sum of length $k-1$. Also, the max value of $n! \: \text{mod} \: k$ is naturally $k-1$.
Taken together one has a max value of:
$$ \sum_{k=2}^{\infty} \frac{(k-1)^2}{2^{k-1}} = 6 $$
Mathematica gives the value of
$$ \sum_{k=2}^{\infty} \frac{1}{2^{k-1}} \sum_{n=1}^{\infty} (n! \: \text{mod} \: k) \approx 3.005674093 $$ so my upper bound is clearly quite crude. What would be a better upper bound?
Edit: To be clear my upper bound is $\frac{x (x+1)}{(1-x)^3}$ for $ \sum_{k=2}^{\infty} x^{k-1} \sum_{n=1}^{\infty} (n! \: \text{mod} \: k)$.
| Calculate more terms we could reach more accurate bound.
Since $\sum_{k=n}^{\infty}\frac{(k-1)^2}{2^{k-1}}=
\frac{(n-1)^2}{2^{n-4}}-\frac{2n^2-6n+3}{2^{n-3}}+\frac{(n-2)^2}{2^{n-2}}$
So we have the sum is between $\sum_{k=2}^{h-1}\frac1{2^{k-1}}\sum_{n=1}^{\infty}n!\pmod k$ and $\left(\sum_{k=2}^{h-1}\frac1{2^{k-1}}\sum_{n=1}^{\infty}n!\pmod k\right) +\frac{(h-1)^2}{2^{h-4}}-\frac{2h^2-6h+3}{2^{h-3}}+\frac{(h-2)^2}{2^{h-2}}$.
For example when h=10, we have the result is between $\frac{93}{32} \approx2.91$ and $\frac{423}{128}\approx3.30$; when h=20, the result is between $\frac{393939}{131072}\approx3.0055$ and $\frac{98535}{32768}\approx3.0070$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4484476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Derivative of $\tan^{-1} \sqrt{\frac{1-\cos x}{1-\sin x}}$ Is there any way to differentiate $\tan^{-1} \sqrt{\frac{1-\cos x}{1-\sin x}}$ without any messy calculations? Here are my thoughts:
We write $\cos x$ and $\sin x$ in terms of $\tan{\frac{x}{2}}$. Then our function becomes $\tan^{-1} \frac{\sqrt{2}\tan \frac{x}{2}}{1-\tan \frac{x}{2}}$. But I am afraid this doesn't seem to work since we can't extrapolate any triginometric formula of $\tan$ from here.
| $\newcommand{\d}{\mathrm{d}}$Using some algebra and trigonometric identities, we can save ourselves a lot of work: $$\frac{1-\cos x}{1-\sin x}=\frac{(1-\cos x)(1+\sin x)}{\cos^2 x}=(\sec x-1)(\tan x+\sec x)$$
Using the chain rule: $$\frac{\d}{\d x}\arctan\left(\sqrt{\frac{1-\cos x}{1+\sin x}}\right)=\frac{1}{2}\cdot\frac{1}{\sqrt{(\sec x-1)(\tan x+\sec x)}}\cdot\frac{1}{1+(\sec x-1)(\tan x+\sec x)}\\\times\frac{\d}{\d x}[(\sec x-1)(\tan x+\sec x)]$$
This last derivative comes as, using the product rule: $$\tan x\sec x(\tan x+\sec x)+(\sec x-1)(\sec^2x+\tan x\sec x)=\\\sec^3 x+2\sec^2x\tan x+\sec x\tan x-\sec^2x-\tan x\sec x$$
Given $a^3+2a^2b+b^2a-a^2-ab$, and that we are interested in the quantities $(b-1)$ and $(a+b)$, with $a=\sec x$ and $b=\tan x$, we can factorise strategically: $$\begin{align}a^3+2a^2b+b^2a-a^2-ab&=a\cdot(a^2+2ab+b^2-a-b)\\&=a\cdot((a+b)^2-(a+b))\\&=a(a+b)\cdot(a+b-1)\end{align}$$
We can then bring it all together, rather simply, cancelling through $\frac{a+b}{\sqrt{a+b}}$: $$\begin{align}\frac{\d}{\d x}\arctan\left(\sqrt{\frac{1-\cos x}{1+\sin x}}\right)&=\frac{1}{2}\sqrt{\frac{\tan x+\sec x}{\sec x-1}}\cdot\frac{\sec x(\sec x+\tan x-1)}{1+(\sec x-1)(\tan x+\sec x)}\\&=\frac{\sec x(\tan x+\sec x-1)}{2(1+(\sec x-1)(\tan x+\sec x))}\cdot\sqrt{1+\cot^2x(\sec x+1)(\tan x+1)}\end{align}$$
If another user sees a way to simplify the last two expressions, please let me know! I think the problem was intended to have this neat factorisation. Be advised that the derivative of this function is highly discontinuous, more so than my expressions reveal. They should be understood as equal only when everything is well-defined and every stage of the working is too. Occasionally there are absolute value and sign errors too. This expression is valid on $(-\frac{\pi}{2},\frac{\pi}{2})\setminus\{0\}$ and on other regions too, periodically.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Compute $\int_0^1 \frac{\sqrt{t(1-t)}}{a+(t-b)^2} \ dt$ for $a,b>0$ I would like to compute the integral
$$\int_0^1 \frac{\sqrt{t(1-t)}}{a+(t-b)^2} \ dt$$
where $a$ and $b>0$ are positive parameters.
Wolfram Alpha is able to provide an answer for the indefinite integral, but is struggeling with computing the definite one
https://www.wolframalpha.com/input?i=integrate+sqrt%28t*%281-t%29%29%2F%28a%2B%28t-b%29%5E2%29+dt+
| Hint Here's a sketch of one method: The substitution $$t = \frac{u^2}{(1 + u^2)} , \qquad dt = \frac{2 u \,du}{(1 + u^2)^2}$$ transforms the original integral in $t$ to the improper integral
$$2 \int_0^\infty \frac{u^2 \,du}{(1 + u^2) (A u^4 + B u^2 + C)},$$
where
$$A := a + (b - 1)^2, \quad B := 2 [a + b(b - 1)], \quad C := a + b^2.$$
If we write $q(v) := A v^2 + B v + C$, so that $A u^4 + B u^2 + C = q(u^2)$, the discriminant of $q$ is $\Delta(q) = -4a < 0$, hence all the roots of the quartic are complex, and we can factor
$$A u^4 + B u^2 + C = (D u^2 - E u + F) (D u^2 + E u + F) ,$$ so that $$A = D^2, \qquad B = 2 DF - E^2, \qquad C = F^2 ,$$ where $D u^2 \pm E u + F$ are irreducible, equivalently, $E^2 - 4 D F < 0$.
At this point we can take either of the following approaches:
*
*Use the Method of Partial Fractions explicitly to find that our integrand is
$$\frac{D (D - F) u + EF}{2 E \Lambda} \left(\frac{1}{D u^2 - E u + F} - \frac{1}{D u^2 + E u + F}\right) - \frac{1}{\Lambda} \cdot \frac{1}{1 + u^2},$$ where $\Lambda := (D - F)^2 + E^2$, after which we can integrate each term as usual.
*Use the fact that the integrand of our new integral is even to write it as
$$\int_{-\infty}^\infty \frac{u^2 \,du}{(1 + u^2) (A u^4 + B u^2 + C)} ,$$ and then apply a standard Residue Theorem argument.
In several special cases the computation simplifies significantly:
*
*When $b \in \{0, 1\}$: $$\pi\left(\sqrt{\frac{1}{2}\left(1 + \sqrt{1 + \frac{1}{a}}\right)} - 1\right)$$
*When $b = \frac{1}{2}$, $$\pi \left(\sqrt{1 + \frac{1}{4 a}} - 1\right)$$
*In the limiting case $a = 0$, $$\pi \left(\frac{2 b - 1}{2 \sqrt{b(b - 1)}} - 1\right) ,$$ provided that $b \not\in [0, 1]$.
*As $a \to \infty$, the integral decays as $$\frac{\pi}{8 a} - \frac{(16 b^2 - 16 b + 5) \pi}{128 a^2} + O\left(\frac{1}{a^3}\right) .$$
*As $b \to \infty$, the integral decays as $$\frac{\pi}{8 b^2} + \frac{\pi}{8 b^3} + O\left(\frac{1}{b^4}\right) .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Inverse Triple Laplace Transform of $\frac{-1}{s^2_{1} + s^2_{2} + s^2_{3}}$ I want to find the inverse triple Laplace transform of $L^{-1}_{x_{3}} L^{-1}_{x_{2}} L^{-1}_{x_{1}} \left[ \frac{-1}{s^2_{1} + s^2_{2} + s^2_{3}} \right]$. I did
\begin{align*}
L^{-1}_{x_{3}} L^{-1}_{x_{2}} L^{-1}_{x_{1}} \left[ \frac{-1}{s^2_{1} + s^2_{2} + s^2_{3}} \right] &= L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[L^{-1}_{x_{1}} \left[ \frac{-1}{s^2_{1} + s^2_{2} + s^2_{3}} \right] \right] \right]
\\
&= (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[\frac{1}{a} L^{-1}_{x_{1}} \left[ \frac{a}{s^2_{1} + a^2} \right] \right] \right], \ \ a^2 = s^2_{2} + s^2_{3}
\\
&= (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[\frac{ \sin \left( x_{1} \sqrt{ \left( s^2_{2} + s^2_{3}\right)} \right) }{\sqrt{ \left( s^2_{2} + s^2_{3}\right)}} \right] \right]
\\
&= (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[ \frac{ \displaystyle\sum_{k = 0}^{\infty} \frac{(-1)^k \left(x_{1} \sqrt{s^2_{2} + s^2_{3}} \right)^{2k+1}}{(2k+1)!} }{\sqrt{ \left( s^2_{2} + s^2_{3}\right)}} \right] \right]
\\
&\approx (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[ \frac{ x_{1} \sqrt{s^2_{2} + s^2_{3}} - \frac{1}{6} \left(x_{1} \sqrt{s^2_{2} + s^2_{3}} \right)^3 }{\sqrt{ \left( s^2_{2} + s^2_{3}\right)}} \right] \right]
\\
&= (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[ x_{1} - \frac{1}{6} x_{1}^3 \left( s^2_{2} + s^2_{3} \right) \right] \right]
\\
&= (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[ \left( x_{1} - \frac{1}{6} x_{1}^3 s^2_{3} \right) - \frac{1}{6} x_{1}^3 s^2_{3} \right] \right]
\\
&= (-1) L^{-1}_{x_{3}} \left[ \left( x_{1} - \frac{1}{6} x_{1}^3 s^2_{3} \right) \delta(x_{2}) - \frac{1}{6} x_{1}^3 \delta^{"}(x_{2}) \right]
\\
&= (-1) \left( \left( x_{1} \delta(x_{3}) - \frac{1}{6} x_{1}^3 \delta^{"}(x_{3}) \right) \delta(x_{2}) - \frac{1}{6} x_{1}^3 \delta^{"}(x_{2}) \delta(x_{3}) \right)
\end{align*}
I am wondering if this solution is correct or not? I would appreciate your help.
| I tried another way to solve the problem by using the Taylor series of three variables and I am wondering whether it is correct :
\begin{align*}
L^{-1}_{x_{3}} L^{-1}_{x_{2}} L^{-1}_{x_{1}} \left[ \frac{-1}{s^2_{1} + s^2_{2} + s^2_{3}} \right] &= L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[L^{-1}_{x_{1}} \left[ \frac{-1}{s^2_{1} + s^2_{2} + s^2_{3}} \right] \right] \right]
\\
&= L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[L^{-1}_{x_{1}} \left[ \frac{-1}{a^2 + b^2 + c^2} + \frac{2}{a^2 + b^2 + c^2} \left[ a(s_{1} - a) + b(s_{2} - b) + c(s_{3} - c) \right] + \ldots \right] \right] \right]
\\
& \approx L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[L^{-1}_{x_{1}} \left[ \frac{-1}{a^2 + b^2 + c^2} + \frac{2}{a^2 + b^2 + c^2} \left[ a(s_{1} - a) + b(s_{2} - b) + c(s_{3} - c) \right] \right] \right] \right] \\
&\approx \frac{-1}{a^2 + b^2 + c^2} \delta(x_{1}) \delta(x_{2}) \delta(x_{3}) \\
& \ \ \ + \frac{2}{a^2 + b^2 + c^2} \Big[ a \left(\delta^{\prime}(x_{1}) \delta(x_{2}) \delta(x_{3}) - a \delta(x_{1}) \delta(x_{2}) \delta(x_{3}) \right) \\
& \ \ \ + b \left(\delta(x_{1}) \delta^{\prime}(x_{2}) \delta(x_{3}) - b \delta(x_{1}) \delta(x_{2}) \delta(x_{3}) \right) \\
& \ \ \ + c \left(\delta(x_{1}) \delta(x_{2}) \delta^{\prime}(x_{3}) - c \delta(x_{1}) \delta(x_{2}) \delta(x_{3}) \right) \Big],
\end{align*}
for arbitrary $a, b, c \in [0, \infty)$.
| {
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"url": "https://math.stackexchange.com/questions/4488585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The number of binary strings of length $n$ without the sequence (1;0;1) in it. First i know that a similat question had been answered here: The number of binary strings of length $n$ with no three consecutive ones But i am really a numb on this subject so i want to be sure to understand it correctly.
But my question is a little bit different as i am looking for the number of binary word of length $n$ without a different sequence, more precisally without the sequence $(1;0;1)$ in it.
Let writte $s_{n}$ all the correct sequence of length $n$.
So $s_{n}=s_{n-1}0 + s_{n-1, valid}1$ that means the total number of valid sequences of length $n$ equal the total number of valid sequences length $n-1$ that end with a zero + the total number of valid sequences length $n-1$ that end with a 1.
No matter what is the valid sequence of length $n-1$ if we add a zero at the end our sequence is still valid. The difficulty is concerning: $s_{n-1, valid}1$. Indeed if $s_{n-1}$ is a valid sequence of length $n-1$ that end with $10$ if we add a $1$ at the end to it we get the forbidden sequence.
So let focus on the total number of valid sequence of length $n-1$ that end with a $1$
$s_{n-1, valid}1 = s_{n-2}11 + s_{n-2, valid}01$
Now we encounter the same difficulty as describe above for $s_{n-2, valid}01$.
But similarly we can writte:$s_{n-2,valid}01=s_{n-3}001$
That means that the only valid sequences of length $n-2$ that after we add 01 to them are still valid sequences of numbers are the one who finish with a zero.
So we can conclude that:
$s_{n}=s_{n-1}0+s_{n-1, valid}1=s_{n-1}0+s_{n-2}11+s_{n-2, valid}01=s_{n-1}0+s_{n-2}11+s_{n-3}001$
So: $s_{n}=s_{n-1}+s_{n-2}+s_{n-3}$
With as initial condition: $s_1=2, s_2=4$ and $s_3=7=2^3-1$
Am i correct?
| Hint: First I'd like to point at the somewhat problematic part of OPs conclusion. Let's have a look at
\begin{align*}
s_{n}&=s_{n-1}0+s_{n-1, \text{valid}}1\\
&=s_{n-1}0+\color{blue}{s_{n-2}11}+s_{n-2, \text{valid}}01\\
&=s_{n-1}0+s_{n-2}11+s_{n-3}001\\
\\
&s_{n}\stackrel{?}{=}s_{n-1}+s_{n-2}+s_{n-3}
\end{align*}
The pitfall here is the blue marked part $\color{blue}{s_{n-2}11}$. We are not allowed to take any valid sequence of length $n-2$ followed by $11$, since there are valid sequences ending in $10$ which combined with $11$ give the invalid sequence $\color{blue}{101}1$.
Note: Besides the nicely given answers here you might also find this answer helpful which essentially follows your approach (and which makes this post regrettably a duplicate).
Generating function approach
Nevertheless I'd like to add another approach based upon generating functions, from which the recurrence relation can be easily derived. It is the so-called Goulden-Jackson Cluster Method.
We consider binary words and the set $B=\{101\}$ of bad words, which are not allowed to be part of the words we are looking for. We derive a generating function $A(z)$ with the coefficient of $z^n$ being the number of wanted words of length $n$ and derive from it the currence relation.
According to the referred paper (p.7) the generating function $A(z)$ is
\begin{align*}
A(z)=\frac{1}{1-dz-\text{weight}(\mathcal{C})}\tag{1}
\end{align*}
with $d=2$, the size of the alphabet and $\mathcal{C}$ the weight-numerator of bad words with
\begin{align*}
\text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[101])
\end{align*}
We calculate according to the paper
\begin{align*}
\text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[101])&=-z^3-z^2\text{weight}(\mathcal{C}[101])\tag{2}\\
\end{align*}
where $z^3$ marks the length $3$ of the bad word $101$ and the term $z^2$ respects overlapping of bad words as in $101\color{blue}{01}$.
It follows from (1) and (2):
\begin{align*}
\color{blue}{A(z)}&=\frac{1}{1-dz-\text{weight}(\mathcal{C})}\\
&=\frac{1}{1-2z+\frac{z^3}{1+z^2}}\\
&\,\,\color{blue}{=\frac{1+z^2}{1-2z+z^2-z^3}}\tag{3}\\
\end{align*}
We recall if a generating function has a representation as rational function of the form
\begin{align*}
A(z)=\sum_{n=0}^\infty a_n z^n=\frac{P(z)}{Q(z)}
\end{align*}
with $P(z), Q(z)$ polynomials, $\deg Q=q>\deg P$ and
\begin{align*}
Q(z)=1+\alpha_1 z+\alpha_2 z^2+\cdots + \alpha_q z^q
\end{align*} then the coefficients $a_n$ follow the recurrence relation
\begin{align*}
a_{n+q}+\alpha_1 a_{n+q-1}+\alpha_2 a_{n+q-2}+\cdots +\alpha_q a_{n}=0\qquad\qquad n\geq 0
\end{align*}
See for instance theorem 4.1.1 in Enumerative Combinatorics, Vol. I by R. P. Stanley.
Thanks to this theorem we can derive the recurrence relation from (3) as
\begin{align*}
a_{n+3}-2a_{n+2}+a_{n+1}-a_{n}=0\qquad\qquad n\geq 0
\end{align*}
resp. by shifting the indices we get
\begin{align*}
\color{blue}{a_n}&\color{blue}{=2a_{n-1}-a_{n-2}+a_{n-3}\qquad\qquad n\geq 3}\tag{4}\\
\color{blue}{a_0}&\color{blue}{=1,a_1=2,a_2=4,a_3=7}
\end{align*}
in accordance with already given answers. The initial conditions of the recurrence relation follow easily.
Two recurrence relations: Besides (4) another recurrence relation
\begin{align*}
\color{blue}{s_n=s_{n-1}+s_{n-2}+s_{n-4}\qquad\qquad n\geq 4}\tag{5}
\end{align*}
is stated by @Mateo. In terms of generating functions we observe that expanding in (3) $A(z)$ with $1+z$ results in
\begin{align*}
A(z)&=\frac{1+z^2}{1-2z+z^2-z^3}=\\
&=\frac{\left(1+z^2\right)(1+z)}{\left(1-2z+z^2-z^3\right)(1+z)}\\
&=\frac{1+z+z^2+z^3}{\color{blue}{1-z-z^2-z^4}}\tag{6}
\end{align*}
and the denominator in (6) provides us with the recurrence relation (5).
| {
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"answer_count": 3,
"answer_id": 2
} |
$\epsilon-N$ for $\lim\limits_{n \to \infty} \sqrt{n^{2} +3n-3} -n = \frac{3}{2}$ First, I tried to use the triangle inequality only once to find an N:
$$
\left | \sqrt{n^2+3n-3}-n-\frac{3}{2} \right | \leqslant \left | \sqrt{n^2+3n-3}-n \right | + \left | \frac{3}{2} \right | = \epsilon
$$
$$
N=\left \lfloor \frac{(\epsilon -\frac{3}{2})^{2}+3}{6-2\epsilon } \right \rfloor +1
$$
I choose epsilon to be 0.01, and N is 1, which is incorrect.
Then I manipulated the inequality again by using the triangle inequality one more time:
$$
\left | \sqrt{n^2+3n-3}-n-\frac{3}{2} \right | \leqslant \left | \sqrt{n^2+3n-3}-n \right | + \left | \frac{3}{2} \right |
$$
$$
\leqslant \left | \sqrt{n^2+3n-3} \right | +n+ \frac{3}{2} =\epsilon
$$
$$
N=\left \lfloor \frac{(\epsilon -\frac{3}{2})^{2}+3}{2\epsilon } \right \rfloor +1
$$
and this time, when epsilon is 0.01, N is 2624, which is correct
I would like to know why the first approach is wrong and the second one is right, Thank you.
| You have
$$\sqrt{n^2+3n-3}-n = \frac{ (\sqrt{n^2+3n-3}-n)(\sqrt{n^2+3n-3}+n)}{\sqrt{n^2+3n-3}+n}$$
$$= \frac{3n-3}{\sqrt{n^2+3n-3}+n}.$$
Divide top an bottom by $n$ to get
$$\frac{3-\frac{3}{n}}{\sqrt{1+\frac{3}{n} - \frac{3}{n^2}}+1} $$
| {
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} |
Profit maximization with Markov Chains
Problem : An opera singer is due to perform a long series of concerts. Having a bad temper, they are liable to pull out each night with probability $1/2$ . Once this has happened they will not sing again until the promoter convinces them of the promoter’s high regard. This the promoter does by sending flowers every day until the singer returns. Flowers costing $x$ thousand pounds, $0 ≤ x ≤ 1$,
bring about a reconciliation with probability $\sqrt{x}$ .The promoter stands to
make $£750$ from each successful concert. How much should they spend on flowers?
I'd like to have solution verification or alternative approaches .
(Problem's from exercise 1.10.4 of Markov Chains by J.R. Norris)
Interpretation : I suppose the singer pulls out independent of the past given that they performed last night. Also I suppose $x$ is "how much they[the promoter] should spend on flowers", which is time-homogeneous. I suppose promoter wants to maximize their expected profit.
Plan : I'll use a 2 state MC to track whether the singer performs or not . Then I'll find the long-run proportion of time that singer performs : $v(x)$ . Finally $x$ will be the maximizer of $f(x) = 750v(x)-x(1-v(x))$ , I think $f(x)$ should approximate the expected profit .
If the MC turns out to be irreducible , then by Theorem 1.10.2 , $v(x)$ will almost surely be the inverse of expected return time of the state that the singer performs .
Theorem 1.10.2 Let $P$ be irreducible and let $\lambda$ be any distribution . If $(X_n)_{n\ge 0}$ is Markov$(\lambda,P)$ then
$$
\mathbb{P}\left( \frac{V_i(n)}{n} \to \frac{1}{m_i} \text{ as } n \to \infty \right) = 1
$$
where $V_i(n) = \sum_{k=0}^{n-1} 1_{\{X_k = i\}}$ and $m_i$ is expected return time to state $i$ .
Attempt :
Let
$(X_n)_{n\ge 0}$ be a Markov chain such that the initial distribution is uniform and
$
X_n =
\left\{\begin{array}{cc}
1 & \text{if singer performs } \\
0 & \text{else }
\end{array}\right.
$ . So transition probabilities are
$$
\left\{\begin{array}{cc}
p_{10} = 1/2 , & p_{11} = 1/2 \\
p_{01} = \sqrt{x} , & p_{00} = 1 - \sqrt{x}
\end{array}\right.
$$ with $x\in (0,1] $ . $(X_n)_{n\ge 0}$ is irreducible on state space $\{0,1\}$ . The expected return time to $1$ is $m_1 = 1 + \frac{1}{2}\frac{1}{\sqrt{x}} $ , so
$v(x) = 1/m_1 $ .
When $x = 0 $ , $\{0\}$ becomes the absorption state , the expected profit is
$\frac{1}{2} 750 \sum_{x=1}^{\infty} x\left(\frac{1}{2}\right)^x = \frac{1}{2} 750(2) = 750 $ .
Numerically I found the maximum of $f(x) , x\in (0,1]$ is around $500$ at $x=1$ .
So should I conclude that he should spend $x=0$ GBP on her ?
| Suppose the opera singer has $n$ performance days. Let $X_i = \begin{cases} 1 & \text{if performance occurs on day $i$}\\ 0 & \text{else} \end{cases}$ for $1 \leq i \leq n$.
Then the profit is $M = \left (\frac{3}{4} + x \right )\left (X_1 + \dots + X_n\right ) - nx$ (in thousands of pounds).
We have transition matrix
$$P = \begin{bmatrix}
1 - \sqrt{x} & \sqrt{x}\\
\frac{1}{2} & \frac{1}{2}
\end{bmatrix} = \begin{bmatrix}
1 & -2\sqrt{x}\\
1 & 1
\end{bmatrix} \begin{bmatrix}
1 & 0\\
0 & \frac{1}{2}-\sqrt{x}
\end{bmatrix} \begin{bmatrix}
\frac{1}{1 + 2\sqrt{x}} & \frac{2\sqrt{x}}{1 + 2\sqrt{x}}\\
-\frac{1}{1 + 2\sqrt{x}} & \frac{1}{1 + 2\sqrt{x}}
\end{bmatrix},$$
so
\begin{align*}
X_1 &= \begin{bmatrix}
\frac{1}{2}\\
\frac{1}{2}
\end{bmatrix}\\
\implies X_i = X_1^TP^{i-1} &= \begin{bmatrix}
\frac{1}{2}\\
\frac{1}{2}
\end{bmatrix}^T \begin{bmatrix}
1 & -2\sqrt{x}\\
1 & 1
\end{bmatrix} \begin{bmatrix}
1 & 0\\
0 & \left (\frac{1}{2}-\sqrt{x}\right )^{i-1}
\end{bmatrix} \begin{bmatrix}
\frac{1}{1 + 2\sqrt{x}} & \frac{2\sqrt{x}}{1 + 2\sqrt{x}}\\
-\frac{1}{1 + 2\sqrt{x}} & \frac{1}{1 + 2\sqrt{x}}
\end{bmatrix}\\
&= \begin{bmatrix}
\frac{1 - \left (\frac{1}{2} - \sqrt{x} \right )^i}{1 + 2\sqrt{x}}\\
\frac{2\sqrt{x} + \left (\frac{1}{2} - \sqrt{x} \right )^i}{1 + 2\sqrt{x}}
\end{bmatrix}
\end{align*}
Thus,
\begin{align*}
\mathbb{E}[X_i] &= \frac{2\sqrt{x} + \left (\frac{1}{2} - \sqrt{x} \right )^i}{1 + 2\sqrt{x}}\\
\implies E_n(x) = \mathbb{E}[M] &= \left ( \frac{3}{4} + x \right ) \left (\sum_{i = 1}^n \frac{2\sqrt{x} + \left (\frac{1}{2} - \sqrt{x} \right )^i}{1 + 2\sqrt{x}}\right ) - nx\\
&= \left ( \frac{3}{4} + x \right ) \left (\frac{(4nx + 1) + 2(n - 1)\sqrt{x} - 2\left (\frac{1}{2} - \sqrt{x}\right )^{n+1}}{\left (1 + 2\sqrt{x}\right )^2}\right ) - nx\\
&= \left ( \frac{3}{4} + x \right ) \left (\frac{1 - 2\sqrt{x} - 2\left (\frac{1}{2} - \sqrt{x}\right )^{n+1}}{\left (1 + 2\sqrt{x}\right )^2}\right ) + \frac{n\left (3\sqrt{x} - 2x \right )}{2 + 4\sqrt{x}}
\end{align*}
Thus as $n \to \infty$, the total profit is infinite as long as $3\sqrt{x} > 2x$, i.e. $x > 0$.
On the other hand, if we wanted to maximise the eventual daily profit then we go back to our distribution for $X_i$ and take $i \to \infty$ to get $X = \begin{bmatrix}
\frac{1}{1 + 2\sqrt{x}}\\
\frac{2\sqrt{x}}{1 + 2\sqrt{x}}
\end{bmatrix}$, so the expected profit made is $\left (\frac{3}{4} + x\right )\mathbb{E}[X] - x = \frac{3\sqrt{x} - 2x}{2 + 4\sqrt{x}}$, which is maximised when $x = \frac{1}{4}$.
Therefore, to maximise daily profit, the club should spend $250$ pounds.
| {
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"answer_count": 3,
"answer_id": 2
} |
$\frac{dx^5}{dx^2},$ the derivative of $x^5$ with respect to $x^2$
Let $f(x)=x^5.$ Find its derivative with respect to $x^2$ , i.e., find $\frac{dx^5}{dx^2}.$
We know that $\frac{dx^5}{dx}=5x^4.$
But what should I do when it is needed to take derivative with respect to $x^2 ,x^3$ or other weird things different from classical derivatives?
My approaches:
$1-)$ Say $x^2=a$ , then $x^5 =a^{5/2}$. Hence , find $$\frac{da^{5/2}}{da}=\frac{5}{2}a^{3/2}=\frac{5}{2}x^{3}$$
$2-)$ Find $$\lim_{h\to 0}\frac{f(x^2+h)-f(x^2)}{h}=\lim_{h\to 0}\frac{(x^2+h)^5-(x^2)^5}{h}=5x^8$$
My approaches conflict. Why do they give different results?
Addendum: Extra examples will be appreciated, for example, $dx^7 /dx^3,\;dx^6/dx^5.$
| In your second approach you should substitute as follows:
(note:in my solution $f(x) \neq x^5$)
$\frac{dx^5}{dx^2}=\frac{d((x^2)^{\frac{5}{2}})}{dx^2}$
So $f(x^2)=(x^2)^{\frac{5}{2}}$ which implies $f(x)=x^{\frac{5}{2}} $
\begin{align*} \lim_{h\to 0}\frac{f(x^2+h)-f(x^2)}{h}&=\lim_{h\to 0}\frac{(x^2+h)^{\frac{5}{2}}-(x^2)^{\frac{5}{2}}}{h}\\&=\lim_{h\to 0}\frac{(x^2+h)^{\frac{5}{2}}-(x^2)^{\frac{5}{2}}}{h} \cdot\frac{(x^2+h)^{\frac{5}{2}}+(x^2)^{\frac{5}{2}}}{(x^2+h)^{\frac{5}{2}}+(x^2)^{\frac{5}{2}}}\\&=\lim_{h\to 0}\frac{1}{(x^2+h)^{\frac{5}{2}}+(x^2)^{\frac{5}{2}}}\cdot\lim_{h\to 0}\frac{(x^2+h)^5-(x^2)^5}{h}\\&=\frac{1}{2x^5}\cdot\lim_{h\to 0}\frac{(x^2+h)^5-(x^2)^5}{h}\\&=\frac{1}{2x^5}\cdot5x^8\\&=\frac{5}{2}x^3
\end{align*}
| {
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Probability of not playing an opponent in a contest with $2^n$ players, from Ross Introduction to Probability In a certain contest, the players are of equal skill
and the probability is $\frac{1}{2}$ that a specified one of
the two contestants will be the victor. In a group
of $2^n$ players, the players are paired off against
each other at random. The $2^{n-1}$ winners are again
paired off randomly, and so on, until a single winner
remains. Consider two specified contestants, $A$
and $B$, and define the events $A_i$, $i\leq n$,$E$ by
$A_i$ : $A$ plays in exactly $i$ contests:
$E$: $A$ and $B$ never play each other.
We have to calculate the probability of $P(A_i)$ and $P(E)$.
Though an elegantly simple approach has been stated here
https://math.stackexchange.com/a/2481789/496972 but I wanted to know the flaw in my reasoning
My attempt
$P(A_i)$ is pretty simple. After every elimination round $2^{n-i}$ contestants remain $i=\{0,1,2..n\}$. By symmetry every contestant has an equal chance to make to the $i^{th}$ round. Hence,
$$
P(A_i)= \dfrac{2^{n-i}}{2^n}= \left(\frac{1}{2} \right)^i
$$
If $A$ is eliminated after $i^{th}$ round, he has played against $i+1$ players. Then the probability of not playing $B$ is same as choosing $i+1$ players out of $2^n- 2$ players. Then,
$$
\begin {align*}
P(E)&= \sum_{i=0}^{n-1} P(E|A_i)P(A_i)\tag{because $A_i$'s are disjoint}\\
&= \sum_{i=0}^{n-1} \dfrac{\binom{2^n-2}{i+1} }{\binom{2^n-1}{i+1}}\times \left(\frac{1}{2} \right)^i\\
&=\sum_{i=0}^{n-1} \dfrac{2^n-2-i}{2^n-1} \times \left(\frac{1}{2} \right)^i\\
&=\dfrac{2^n-2}{2^n-1}\sum_{i=0}^{n-1}\left (\dfrac{1}{2} \right)^i-\frac{1}{2(2^n-1)}\sum_{i=1}^{n-1}i \left(\frac{1}{2} \right)^{i-1}
\end{align*}
$$
When I simplify this expression, it is not remotely close to $1- \dfrac{1}{2^{n-1}}$ which is the answer. Where am I going wrong?
| Sorry that I still stick to counting $i$ from $1$ to $n$. This sounds more natural to me.
I think that the calculation is incorrect because
First, in your formula, $\sum_{i=1}^n P(A_i)=\sum_{i=1}^n \frac {1}{2^i}=1-\frac{1}{2^n} \neq 1 $
It means that something is wrong. Actually $P(A_n)$ should be $\frac {1}{2^{n-1}}$ (See Leander Tilsted Kristensen's comment).
Second, the formula for $P(E|A_i)$ should be $$P(E|A_i)=\frac { 2^n-1 \choose i}{2^n-1 \choose i}=\frac {2^n-1-i}{2^n-1}$$
Accordingly
\begin{align}
P(E) &= \sum_{i=1}^nP(E|A_i)P(A_i) \\
&= \sum_{i=1}^{n-1}P(E|A_i)P(A_i)+P(E|A_n)P(A_n) \\ &= \sum_{i=1}^{n-1} \frac {2^n-1-i}{2^n-1}\times \frac{1}{2^i}+ \frac{2^n-1-n}{2^n-1}\times\frac{1}{2^{n-1}} \\
&= \sum_{i=1}^{n-1}\frac{1}{2^i}-\frac{1}{2^n-1}\sum_{i=1}^{n-1}\frac{i}{2^i}+ \frac{2^n-1-n}{2^n-1}\times\frac{1}{2^{n-1}} \\
&= 1- \frac{1}{2^{n-1}}
\end{align}
Note:
We can prove that $$-\frac{1}{2^n-1}\sum_{i=1}^{n-1}\frac{i}{2^i}+ \frac{2^n-1-n}{2^n-1}\times\frac{1}{2^{n-1}}=0$$
or equivalently
$$\sum_{i=1}^{n-1}\frac{i}{2^i}=\frac {2^n-1-n}{2^{n-1}}$$
as follows:
Let $$S=\sum_{i=1}^{n-1}\frac{i}{2^i}$$ we have
$$S = \frac{1}{2}+2\left(\frac{1}{2} \right)^2+3\left(\frac{1}{2} \right)^3+ \dots +(n-1)\left(\frac{1}{2} \right)^{n-1} $$
$$\frac{S}{2}= \left(\frac{1}{2} \right)^2+2\left(\frac{1}{2} \right)^3+ \dots +(n-2)\left(\frac{1}{2} \right)^{n-1}+ (n-1)\left(\frac{1}{2} \right)^{n} $$
Hence
$$S-\frac{S}{2}=\frac{1}{2}+\left(\frac{1}{2} \right)^2+ \dots +\left(\frac{1}{2} \right)^{n-1} -
(n-1)\left(\frac{1}{2} \right)^n$$
$$\frac{S}{2}=1-\left(\frac{1}{2} \right)^{n-1}-(n-1)\left(\frac{1}{2} \right)^n$$
$$S=\frac {2^n-1-n}{2^{n-1}}$$
| {
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"answer_count": 2,
"answer_id": 0
} |
What is the number of different sums of the items of a set of consecutive natural numbers? I have been playing around with sets of consecutive natural numbers like say $S=\{1,2,3,...,10\}$ and I have come up with this problem for which however I have not yet found an answer and I don't know if there is one. My problem is the following: Considering a set $S=\{1,2,3,...,n\}$ where $n$ is an arbitrary natural number, can we derive a formula that will indicate the number of different sums that will be produced by summing the items of the set $S$ in all different combinations? For example for the set $S=\{1,2,3,4,5\}$ we get:
$$1+2=3\\
1+2+3=6\\
1+2+3+4=10\\1+3+4+5=13\\1+2+4+5=12\\1+2+3+4+5=15\\
1+2+3+5=11\\1+4+5=10\\1+3+5=9\\1+3+4=8\\1+2+5=8\\1+2+4=7\\1+3=4\\1+4=5\\1+5=6\\
2+3=5\\2+4=6\\2+5=7\\
2+3+4=9\\2+3+5=10\\2+4+5=11\\
2+3+4+5=14\\
3+4=7\\3+5=8\\
3+4+5=12\\
4+5=9\\$$
So we have $26$ different summations and $13$ different sums(I hope I didn't make any mistake)
| Here is an answer with the same content as Thomas Andrews' answer, but presented a bit more heuristically.
Since the sum of the first $n$ positive integers is given by $\frac{n(n+1)}{2}$, you can obviously make that sum with the $n$ integers. By removing one integer addend at a time, you can make a number which is $1$ smaller, and $2$ smaller, etc. all the way up to $n$ smaller, which will provide sums equal to all of the integers down to $\frac{n(n+1)}{2}-n$
But $\frac{n(n+1)}{2}-n=\frac{n(n+1)}{2}-\frac{2n}{2}=\frac{n(n+1)-2n}{2}=\frac{n(n+1-2)}{2}=\frac{(n-1)n}{2}$
You should recognize the final expression as the form of the sum of the first $n-1$ positive integers, which in any event is what you have when you remove $n$ as the addend from the set of the first $n$ integers.
You can reductively repeat this process to generate all smaller integer sums until you reach the set $\{1,2\}$, which can give you the sum $3$, but there is no way (using sums of two or more integers) that you can generate the sum $2$ or $1$.
Hence, the set of the first $n$ positive integers can generate $\frac{n(n+1)}{2}-2$ distinct sums.
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Getting the area between $y$-axis, $f(x)$, $f(2-x)$ when both function is given by their subtraction?
*
*For a polynomial $f(x)$, let function $g(x) = f(x) - f(2-x)$.
*$g'(x) = 24x^2 - 48x + 50$
What is the area between $y=f(x)$, $y=f(2-x)$, and $y$-axis?
My approach:
*
*$g(1) = f(1) - f(1) = 0$.
*From $g'(x)$, $g(x) = 8x^3 - 24x^2 + 50x + C$.
*From 1 and 2, $C = -34$.
*Since $f(x)$ is a cubic function, let $f(x) = ax^3 + bx^2 + cx + d$ and compare coefficients of $g$ and $f(x) - f(2-x)$.
*$a = 4$, $2b + c = 1$.
And I'm stuck. I can't see how I can get more info about $f(x)$ using these conditions and get areas out of it.
| From where you left off,
$$ g(x) \ \ = \ \ f(x) \ - \ f(2 - x) $$ $$ = \ \ 2a·x^3 \ - \ 6a·x^2 \ + \ (12a + 4b + 2c)·x \ - \ (8a + 4b + 2c) \ \ , $$
which does match up with your integration of $ \ g'(x) \ $ with $ \ a \ = \ 4 \ \ , \ \ 2b + c \ = \ 1 \ \ : $
$$ g(x) \ \ = \ \ 2·4·x^3 \ - \ 6·4·x^2 \ + \ (12·4 + 2·1)·x \ - \ (8·4 + 2·1) $$ $$ = \ \ 8·x^3 \ - \ 24·x^2 \ + \ 50·x \ - \ 34 \ \ . $$
We would need to express $ \ f(x) \ $ as, say, $ \ 4x^3 + bx^2 + (1 - 2b)·x + d \ \ , \ $ giving us
$$ \ f(2 - x) \ = \ -4x^3 \ + \ (b + 24)·x^2 \ - \ (2b + 49)·x \ + \ (d + 34) \ \ \ . $$
This tells us that the $ \ y-$intercept of $ \ f(2 - x) \ $ is above that of $ \ f(x) \ \ . \ $ Since $ \ f(2 - x) \ $ is the vertical reflection of $ \ f(x) \ $ about the $ \ y-$axis, "shifted to the right" by 2 units, it is also the vertical reflection of $ \ f(x) \ $ about the line $ \ x \ = \ 1 \ \ . \ $ So the two function curves intersect at $ \ x \ = \ 1 \ \ $ [the geometrical meaning of your result $ \ g(1) \ = \ 0 \ ] \ $ with $ \ f(2 - x) \ $ "above" $ \ f(x) \ \ . $ [We know that there are no other intersections in the interval $ \ ( 0 \ , \ 1 ) \ $ because $ \ g(0) \ = \ -34 \ $ and $ \ g'(x) \ = \ 24·(x - 1)^2 + 26 \ > \ 0 \ \ . \ ] $
For the problem then, we wish to integrate $$ \ \int_0^1 \ [ \ f(2 - x) \ - \ f(x) \ ] \ \ dx \ \ = \ \ \int_0^1 \ -g(x) \ \ dx \ \ . $$
[The value of $ \ d \ $ would be irrelevant in any case for the area between the curves and we find that we also don't need to know $ \ b \ \ . \ $ We do obtain a positive value for this area between the curves, as we should.]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Quadratic formula $x = \frac{- (b +\sqrt{b^2- 4ac})}{ \pm2a}$ In the proof of the quadratic formula
$$x = \frac{- b +\sqrt{b^2- 4ac}}{2a}$$
shouldn't there be $\pm 2a$ instead of $2a$, since both can be the square root of $4a^2$?
| Another way to approach it which does not rely on the study of the sign of $a$:
\begin{align*}
ax^{2} + bx + c = 0 & \Longleftrightarrow 4a^{2}x^{2} + 4abx + 4ac = 0\\\\
& \Longleftrightarrow (4a^{2}x^{2} + 4abx + b^{2}) = b^{2} - 4ac\\\\
& \Longleftrightarrow (2ax + b)^{2} = b^{2} - 4ac\\\\
& \Longleftrightarrow 2ax + b = \pm\sqrt{b^{2} - 4ac}\\\\
& \Longleftrightarrow x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}
\end{align*}
Hopefully this helps!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Issues with integrating $\int x^2 \sqrt{x^3 +1}$ via integration by parts I want to integrate
$$\int x^2 \sqrt{x^3 +1}~dx$$
I tried it with integration by parts (because we have a product here), but an online calculator did it with integration by substition.
Would this still be correct?
$$\frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \int \frac{1}{3} x^3 \frac{1}{2} (x^3+1)^{-\frac{1}{2}} \cdot 3~dx \\
= \frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \int x^3 \frac{1}{2} (x^3+1)^{-\frac{1}{2}} ~dx\\
= \frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \frac{1}{4} x^4\cdot 2(x^3 +1)^{-\frac{1}{2}} \\
= \frac{1}{3}x^3 \sqrt{x^3+1} - \frac{1}{2} x^4 \frac{1}{\sqrt{x^3+1}}$$
I think this is wrong because when $x=1$ I get a different result than when I insert $x=1$ into
Can someone tell me where I went wrong and why we rather use integration by substition instead of integration by parts here?
| The first line of your calculation is missing a factor of $x^2$, which arises when using the chain rule to compute $d(\sqrt{x^3 + 1})$.
Instead, applying integration by parts with $$u = \sqrt{x^3 + 1}, \qquad dv = x^2$$ gives
$$\left(\frac{1}{3} x^3\right) \left(\sqrt{x^3 + 1}\right) - \int \left(\frac{1}{3} x^3\right) \left( \frac{3 x^2 \,dx}{2 \sqrt{x^3 + 1}} \right)
= \frac{\sqrt{x^3 + 1}}{3 x^3} - \frac{1}{2} \int \frac{x^5 \,dx}{\sqrt{x^3 + 1}} .$$
At this point the remaining integral is arguably worse than the one we started with, but we can still evaluate it using the same substitution the online calculator used, $$w = x^3 + 1, \qquad dw = 3 x^2;$$
we have
$$\int \frac{x^5 \,dx}{\sqrt{x^3 + 1}} = \int \frac{[(x^3 + 1) - 1] \cdot x^2 \,dx}{\sqrt{x^3 + 1}} = \frac{1}{3} \int \frac{(w - 1) \,dw}{\sqrt{w}},$$ and the lattermost integrand is a sum of power functions. But it's easier just to apply that substitution to the original integral instead:
$$\int x^2 \sqrt{x^3 + 1} \,dx = \frac{1}{3} \int \sqrt{w} \,dw .$$
| {
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"source": "stackexchange",
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If a, b, c are positive real numbers such that $a^2+ b^2+ c^2 = 1$ $( \frac{1}{a} +\frac{1}{b} +\frac{1}{c}) +a +b +c \geq 4\sqrt{3}$
If a, b, c are positive real numbers such that $a^2+ b^2+ c^2 = 1$
Show that:
$$\frac{1}{a} +\frac{1}{b} +\frac{1}{c}+a +b +c \geq 4\sqrt{3}.$$
My attempt:
First , I used Holder's : $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{27}{3(a+b+c)}$$
similarly:
$$ a+b+c \geq \frac{(a+b+c)^3}{3(a^2+b^2+c^2)} = \frac{(a+b+c)^3}{3}$$
Which gives us:
$$LHS \geq \frac{27}{3(a+b+c)}+ \frac{(a+b+c)^3}{3}$$
By AM-GM:
$$LHS \geq 2\sqrt{3} \times (a+b+c)$$
So all we need to do is prove that $$2\sqrt{3} \times (a+b+c)\geq 4\sqrt{3}$$
And this means we need to prove $$a+b+c \geq 2 $$
Which I can't.
I am not looking for a solution , only hints that would guide me through solving it .
Thanks in advance for any communicated help !
| By your argument, we already have
$$\frac1a+\frac1b+\frac1c+a+b+c\geq\frac9{a+b+c}+\frac{(a+b+c)^3}3=\frac9u+\frac{u^3}3\equiv f(u)$$
where $u=a+b+c>0$. Try to show that $f'(x)=0$ only has one root $x^*=\sqrt3$ when $x>0$ and $f''(x^*)>0$, therefore $f(u)\geq\min_{x>0}f(x)=f(x^*)=4\sqrt3$.
| {
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"source": "stackexchange",
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Evaluating the limit of $\lim_{n \to \infty} \frac{4n^2 +9}{2n^2+2n+3}$ with definition I have troubles with "estimating".
We want to evaluate the limit with the definition. For all $\epsilon >0$ we have to find an $N$, such that $|a_n-a|<\epsilon$ holds for $n>N$.
The limit in question is:
$$\lim_{n \to \infty} \frac{4n^2 +9}{2n^2+2n+3}$$
We know the limit is $2$.
My approach:
Let $\epsilon > 0$.
$$\begin{align*}
|a_n-2|
&= \left|\frac{4n^2+9}{2n^2+2n+3} -2\right| \\
&= \left| \frac{4n^2+9 - 2(2n^2+2n+3)}{2n^2+2n+3} \right| \\
&= \left| \frac{4n^2+9 - 4n^2 - 4n -6}{2n^2+2n+3} \right| \\
&= \left| \frac{-4n+3}{2n^2+2n+3} \right| \\
&\leq \left| \frac{-4n}{2n^2+2n+3}\right| \\
&\leq \frac{4n}{2n^2} \\
&= \frac{2}{n} < \epsilon
\end{align*}$$
If we choose $N = \frac{2}{\epsilon}$ it holds that $\forall \epsilon > 0: |a_n -a| < \epsilon$, if $n > N$.
My professors approach:
$$\begin{align*}
\left| \frac{-4n+3}{2n^2+2n+3} \right|
&\leq \left|\frac{3-4n}{2n^2} \right| \\
&\leq \left| \frac{3}{2n^2} \right| + \left|\frac{4n}{2n^2}\right| \\
&= \frac{3}{2} \cdot \frac{1}{n^2} + 2 \cdot \frac{1}{n}
\end{align*}$$
We note that:
$$\begin{align*}
&\frac{3}{2} \cdot \frac{1}{n^2} < \frac{\epsilon}{2} \Leftrightarrow \frac{3}{n^2} < \epsilon \iff n > \sqrt{\frac{1}{\epsilon}} \\
&2 \cdot \frac{1}{n} < \frac{\epsilon}{2} \Leftrightarrow 4 \cdot \frac{1}{n} < 2 \iff \frac{4}{\epsilon} < n
\end{align*}$$
Then:
$$\begin{align*}
&\text{Let } N > \max\bigg\{\sqrt{\frac{1}{\epsilon}}, \frac{4}{\epsilon} \bigg\} \\
&\implies\forall n > N \text{ it holds that } |a_n-a| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon
\end{align*}$$
I have the following questions:
$1.$ What is wrong with my estimation?
$2.$ How does my prof get to $\frac{3}{2} \cdot \frac{1}{n^2} < \frac{\epsilon}{2}$
$3.$ What exactly does $\max\bigg\{\sqrt{\frac{1}{\epsilon}}, \frac{4}{\epsilon} \bigg\}$ tell us?
| In your approach you need to find an upper bound for the numerator and a lower bound for the denominator.
Your calculation was ok but I would present it as below to make it a little bit more explicit:
*
*$4n-3<4n$
*$2n^2+2n+3>2n^2$
In fact your way of proceeding is totally fine, you can even skip the epsilon part and just conclude by $\frac 2n\to 0$.
In analysis except in rare exceptions when you need a really tight bound, don't bother with heterogeneous things like $\frac 32n^2+2\frac 1n$.
Just reduce it to a rough bound, i.e. $\frac 32<2$ and $\frac 1{n^2}<\frac 1n$ so that in the end you get $<\frac 4n$, and only then proceed with the epsilon if necessary.
I really don't get why professors go on with square roots of epsilons and so on...
So my advice, is just continue doing it the way you did !
Remark: in the same vein, when you have an epsilon-delta proof in $0$ and not infinity, you can also use this fact.
$x\to x_0$ then use $u=x-x_0\to 0$ in particular $u^n<\cdots<u^3<u^2<u<1$.
So you can bound any polynomial in $u$ by the sum of coefficients, i.e.
$|7u^3+2u^2-5u|<(7+2+5)|u|=14|u|\to 0$
Rather than bothering with $\epsilon, \sqrt{\epsilon}, \sqrt[3]{\epsilon}$...
| {
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How to show that two matrices are similar without using anything about eigenvalues or eigenvectors Problem: Show that $A$ and $B$ are similar without using eigenvalues or eigenvectors:
$$
A =
\left[ {\begin{array}{ccc}
1 & 1 & 1 \\
0 & 2 & 2 \\
0 & 0 & 3 \\
\end{array} } \right]
$$
$$
B =
\left[ {\begin{array}{ccc}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3 \\
\end{array} } \right]
$$
I see that they are row-equivalent. But how to determine the transition matrix $S$ such that $S^{-1}AS = B$? I understand diagonalizability (which they are) but I am instructed not to use that concept.
I also know that I could set up a system of equations to solve for the entries of $S$ like $AS = SB$, but that strikes me as crude.
|
I also know that I could set up a system of equations to solve for the entries of $S$ like $AS=SB$, but that strikes me as crude.
Crude as it may be, it's certainly doable.
$$S = \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & j\end{bmatrix}$$
$$AS = \begin{bmatrix}a + d + g & b + e + h & c + f + j \\ 2 d + 2 g & 2 e + 2 h & 2 f + 2 j \\ 3 g & 3 h & 3 j\end{bmatrix}$$
$$SB = \begin{bmatrix}a & 2b & 3c \\ d & 2e & 3f \\ g & 2h & 3j\end{bmatrix}$$
From the bottom row, we have $3g = g$ and $3h = 2h$, which means $g = 0$ and $h = 0$.
$$\begin{bmatrix}a + d & b + e & c + f + j \\ 2 d & 2 e & 2 f + 2 j \\ 0 & 0 & 3 j\end{bmatrix} = \begin{bmatrix}a & 2b & 3c \\ d & 2e & 3f \\ 0 & 0 & 3j\end{bmatrix}$$
From the middle row, we have $2d = d \implies d = 0$, and $2f + 2j = 3f \implies f = 2j$.
$$\begin{bmatrix}a & b + e & c + 3j \\ 0 & 2 e & 6j \\ 0 & 0 & 3 j\end{bmatrix} = \begin{bmatrix}a & 2b & 3c \\ 0 & 2e & 6j \\ 0 & 0 & 3j\end{bmatrix}$$
And from the top row, we have $b + e = 2b \implies e = b$ and $c + 3j = 3c \implies 3j = 2c \implies j = \frac{2}{3}c$.
$$\begin{bmatrix}a & 2b & 3c \\ 0 & 2 e & 4c \\ 0 & 0 & 3 j\end{bmatrix} = \begin{bmatrix}a & 2b & 3c \\ 0 & 2e & 4c \\ 0 & 0 & 2c\end{bmatrix}$$
Putting this together, we have:
$$S = \begin{bmatrix}a & b & c \\ 0 & b & \frac{4}{3}c \\ 0 & 0 & \frac{2}{3}c\end{bmatrix}$$
Where $a$, $b$, and $c$ can be anything (except 0, if you want $S^{-1}$ to exist). I'll pick $a = 1$, $b = 1$ and $c = 3$.
$$S = \begin{bmatrix}1 & 1 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 2\end{bmatrix}$$
| {
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"source": "stackexchange",
"question_score": "2",
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What to conclude from $x y \le N$ and $x\le 1$? For real $x, y, N$, what can we conclude for $y$ from two inequalities:
$$x y \le N\qquad {\rm and} \qquad x\le 1?$$
| Let $x \ne 0$. Then you have six cases following from $x y \le N$:
*
*For $0 < x \le 1$ and $N \ge 0$ you have $0 \le y \le \frac{N}{x}$ .
*For $0 < x \le 1$ and $N < 0$ you have $ y \le \frac{N}{x} < N < 0 $ .
*For $-1 \le x <0$ and $N \ge 0$ you have $0 \ge y \ge \frac{N}{x}$ .
*For $-1 \le x <0$ and $N < 0$ you have $0 \le -N \le \frac{N}{x} \le y$ .
*For $x < -1 $ and $N \ge 0$ you have $0 \ge y \ge\frac{N}{x} > -N$ .
*For $x < -1 $ and $N < 0$ you have $ y \ge \frac{N}{x} \ge 0$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Interesting integral $\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{2}}$ Latest Edit
Inspired by @J.G., I find a formula in general,
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{n}} &=2 \int_{0}^{\pi} \frac{d x}{(3-\cos x)^{n}} \\
&=\left.\frac{2(-1)^{n}}{(n-1) !} \frac{\partial^{n}}{\partial a^{n}}\left(\int_{0}^{\pi} \frac{d x}{a-\cos x} \right)\right|_{a=3} \\
&=\left.\frac{2(-1)^{n} \pi}{(n-1) !} \frac{\partial^{n}}{\partial a^{n}}\left(\frac{1}{\sqrt{a^{2}-1}}\right)\right|_{a=3}
\end{aligned}
$$
Multiplying both the numerator and denominator $\sec^4x$ yields
$\displaystyle I=\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{2}}=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{4} x}{\left(\sec ^{2} x+\tan ^{2} x\right)^{2}} d x \tag*{} $
Letting $ t=\tan x$ gives
$\displaystyle \begin{aligned}I&= \int_{0}^{\infty} \frac{1+t^{2}}{\left(1+2 t^{2}\right)^{2}} d t\\&=\int_{0}^{\infty} \frac{1+\frac{1}{t^{2}}}{\left(2 t+\frac{1}{t}\right)^{2}} d t\\&=\int_{0}^{\infty} \frac{\frac{3}{4}\left(2+\frac{1}{t^{2}}\right)-\frac{1}{4}\left(2-\frac{1}{t^{2}}\right)}{\left(2 t+\frac{1}{t}\right)^{2}} d t\\&=\frac{3}{4} \int_{0}^{\infty} \frac{d\left(2 t-\frac{1}{t}\right)}{\left(2 t-\frac{1}{t}\right)^{2}+8}-\frac{1}{4} \int_{0}^{\infty} \frac{d\left(2 t+\frac{1}{t}\right)}{\left(2 t+\frac{1}{t}\right)^{2}}\\&=\left[\frac{3}{8 \sqrt{3}} \tan ^{-1}\left(\frac{2 t-\frac{1}{t}}{2 \sqrt{2}}\right)+\frac{1}{4\left(2 t+\frac{1}{t}\right)}\right]_{0}^{\infty}\\&=\frac{3 \pi}{8 \sqrt{2}}\end{aligned}\tag*{} $
Is there any method other than tangent half-angle substitution?
| \begin{align}
\int_{0}^{\frac{\pi}{2}} \frac{dx}{(1+\sin ^{2} x)^{2}}
= &\int_{0}^{\frac{\pi}{2}} \frac1{4\tan^3x} \ d\left(\frac{\tan^2x}{1+\csc^2x}\right)\\
\overset{ibp}=&\ \frac34 \int_{0}^{\frac{\pi}{2}} \frac{d(\tan x)}{1+2\tan^2x}
=\frac{3\pi}{8\sqrt2}
\end{align}
| {
"language": "en",
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"source": "stackexchange",
"question_score": "8",
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Find the function $f(x)$ when $f(f(x))=1-x$, for $x\in [0,1]$
The function f(x) is continuous and $f(f(x))=1-x$, for $x\in [0,1]$ then,
(A) $f(\frac{1}{8})+f(\frac{7}{8})=3$
(B) $f(\frac{2}{3})+f(\frac{1}{3})=2$
(C) $f(\frac{5}{6})+f(\frac{1}{6})=1$
(D) None of These
My approach is as follow $f\left( {f\left( x \right)} \right) = 1 - x \Rightarrow f'\left( {f\left( x \right)} \right) \times f'\left( x \right) = - 1$
$f'\left( {f\left( 0 \right)} \right) \times f'\left( 0 \right) = - 1\& f'\left( {f\left( 1 \right)} \right) \times f'\left( 1 \right) = - 1$
$f\left( {f\left( 0 \right)} \right) = 1;f\left( {f\left( 1 \right)} \right) = 0$
Not able to procced from here
| From $f(f(x)) =1-x$ follows
$$1-f(x)=f(f(f(x)))=f(1-x) .$$
Hence,
$$f(x) +f(1-x) =1$$
for all $x\in[0, 1]$.
| {
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"source": "stackexchange",
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How to find sum of this series using GP formula??? Given G.P:- $1 + 2^{1/n} + 2^{2/n} + . . . + 2^{(N-1)/N}$
My approach:-
We know, $S_{n} = \frac{a(r^n - 1)}{r - 1}$
Hence, here as we have n-1 terms, the formula will be:-
$S_{n-1} = \frac{a(r^{n-1} - 1)}{r - 1}$
We can see, $a = 1$, $r = 2^{1/n}$
Hence, $S_{n-1} = \frac{2^{(n-1)/n} - 1}{2^{1/n} - 1}$
Which is not same as the answer which is:- $\frac{1}{2^{1/n} - 1}$
Am I missing something?
Any help is appreciated.
Thank you!
| Answer edited.
The sum of the first $N$ terms of a geometric series with first term $a$ and common ratio $r$ is: $a + ar + \ldots + ar^{N-1} = \frac{a(1-r^N)}{1-r}.$
To see that this finite series has $N$ terms and not $N-1$ terms, write it as:
$$a + ar + \ldots + ar^{N-1} = ar^0 + ar^1 + \ldots + ar^{N-1} = ar^{-1}\underbrace{ \left( r^1 + r^2 + \ldots + r^N \right)}_{N \text{ terms}}\ . $$
Assuming you meant your sequence to be: $1 + 2^{1/N} + 2^{2/N} + . . . + 2^{(N-1)/N},$ then your sequence is the above with $a=1$ and $r = 2^{1/N}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\int_0^\pi x\frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx$
How am I supposed to solve the following definite integral?
$$
\mathcal{I} = \int_0^\pi x \cdot \frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx
$$
This definite integral is solved if the minus sign is replaced by a plus sign, and it yields $\pi^2$.
$$
\mathcal{I} = \int_0^\pi x \cdot \frac{\sin{\frac{x}{2}} + \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx \text{
— (I)} \\
\implies \mathcal{I} = \int_0^\pi (\pi - x) \cdot \frac{\cos{\frac{x}{2} + \sin{\frac{x}{2}}}}{\sqrt{\sin{x}}} dx \text{
— (II)}
$$
On (I) + (II), we have,
$$
\mathcal{I} = \frac{\pi}{2}\int_0^\pi \frac{\sin{\frac{x}{2}} + \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx = \frac{\pi}{2} \int_0^\pi \frac{\sin{\frac{x}{2}}+\cos{\frac{x}{2}}}{\sqrt{1 - (\sin{\frac{x}{2}-\cos{\frac{x}{2}}})^2}} dx
$$
On substitution,
$$
\sin{\frac{x}{2}} - \cos{\frac{x}{2}} = u \implies \left(\sin{\frac{x}{2}} + \cos{\frac{x} {2}}\right) dx = 2 \cdot du
$$
The upper and lower limits changes to 1 and -1. Now, we have
$$
\mathcal{I} = \frac{\pi}{2} \int_{-1}^1 \frac{2 \cdot du}{\sqrt{1 - u^2}} du = \pi \cdot \left[\arcsin{u}\right]_{-1}^1 = \pi^2
$$
But...
*
*The sign was not supposed to be changed. We get $(2x - \pi)$ instead of $\pi$ in the nominator when adding both integrals. It complicates the problem.
*Using integral-calculator.com or a scientific calculator is helpless.
*The answer to the original problem should be $2\pi \cdot \ln{2}$ (approx 4.35.)
| Here is a solution using real analysis. First, denote the two integrals as
\begin{align*}
J & \equiv\int\limits_0^{\pi}\frac {x\left(\sin\frac x2\color{red}-\cos\frac x2\right)}{\sqrt{\sin x}}\,\mathrm dx\\K & \equiv\int\limits_0^{\pi}\frac {x\left(\sin\frac x2\color{red}+\cos\frac x2\right)}{\sqrt{\sin x}}\,\mathrm dx
\end{align*}
And from your post, recall that $K=\pi^2$. Adding the two integrals together removes the $\cos\frac x2$ factor inside the integrand, leaving only
\begin{align*}
J+K & =2\int\limits_0^{\pi}\frac {x\sin\frac x2}{\sqrt{\sin x}}\,\mathrm dx\\ & =\sqrt 2\int\limits_0^{\pi}x\sqrt{\tan\frac x2}\,\mathrm dx\\ & =4\sqrt{2}\int\limits_0^{+\infty}\frac {\sqrt t}{1+t^2}\arctan t\,\mathrm dt
\end{align*}
Where a double angle identity was utilized in the second equation and the half-angle tangent substitution in the third equation. The last integral has been evaluated here before using Complex Analysis, Feynman's Trick, etc. Here is an alternative approach using double integrals. First, enforce the substitution $x=\sqrt t$ so that the integral becomes
$$\int\limits_0^{+\infty}\frac {\sqrt t}{1+t^2}\arctan t\,\mathrm dt=2\int\limits_0^{+\infty}\frac {x^2}{1+x^4}\arctan x^2\,\mathrm dx$$
Next, use the identity
$$\arctan x^2=\int\limits_0^1\frac {x^2}{1+x^4 y^2}\,\mathrm dy$$
Swapping the order of integration and using partial fraction decomposition, then we get
\begin{align*}
2\int\limits_0^{+\infty}\,\int\limits_0^1\frac {x^4}{(1+x^4)(1+x^4y^2)}\,\mathrm dy\,\mathrm dx & =2\int\limits_0^1\frac 1{y^2-1}\int\limits_0^{+\infty}\frac 1{1+x^4}-\frac 1{1+x^4y^2}\,\mathrm dx\,\mathrm dy\\ & =\frac {\pi}{\sqrt 2}\int\limits_0^1\frac 1{y^2-1}\left(1-\frac 1{\sqrt y}\right)\,\mathrm dy\\ & =\frac {\pi^2}{4\sqrt 2}+\frac {\pi\log 4}{4\sqrt 2}
\end{align*}
To recap, we have the equation
$$J+\pi^2=4\sqrt{2}\int\limits_0^{+\infty}\frac {\sqrt t}{1+t^2}\arctan t\,\mathrm dt=\pi^2+\pi\log 4$$
Subtracting a $\pi^2$ from both sides, then
$$\int\limits_0^{\pi}\frac {x\left(\sin\frac x2-\cos\frac x2\right)}{\sqrt{\sin x}}\,\mathrm dx\color{blue}{=\pi\log 4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 1
} |
Computing $\int_{0}^{\pi} \ln (\sin x+2) d x$ and $\int_{0}^{\pi} \ln (2-\sin x) d x$ I first encountered this integral
$$
I=\int_{0}^{\pi} \ln (\sin x+2) d x
$$
several months ago without any idea and had tried many methods such as integration by parts, substitution and Fourier series etc. but all are in vain. Today, I tried the tangent substitution and succeeded. Now I want to share with you and seek any other alternatives.
For simplicity, I convert, by symmetry, the integral into
$$
I=2 \int_{0}^{\frac{\pi}{2}} \ln (\sin 2 x+2) d x
$$
and substitute $t=\tan x$, and get
$$
\begin{aligned}
I &=2 \int_{0}^{\frac{\pi}{2}} \ln (\sin 2 x+2) d x \\
&=2 \int_{0}^{\infty} \ln \left(\frac{2 t}{1+t^{2}}+2\right) \frac{d t}{1+t^{2}} \\
&=2 \int_{0}^{\infty} \frac{\ln 2+\ln \left(t^{2}+t+1\right)-\ln \left(1+t^{2}\right)}{1+t^{2}} d t \\
&=\pi \ln 2+2 \int_{0}^{\infty} \frac{\ln \left(t^{2}+t+1\right)}{1+t^{2}}-2 \int_{0}^{\infty} \frac{\ln \left(1+t^{2}\right)}{1+t^{2}} d t \\
&=\pi \ln 2+2 \underbrace{ \left[\frac{\pi}{3} \ln (2+\sqrt{3})+\frac{4}{3} G\right]}_{(*)} -2 \underbrace{\pi \ln 2}_{(**)} \\
&=-\pi \ln 2+\frac{2 \pi}{3} \ln (2+\sqrt{3})+\frac{8}{3} G
\end{aligned}
$$
Note:
(*): post 1, (**):post 2
For the second integral, we use the similar technique and arrive at
\begin{aligned}
J&=2 \int_{0}^{\infty} \frac{\ln 2+\ln \left(1-t+t^{2}\right)-\ln \left(1+t^{2}\right)}{1+t^{2}} d t \\
&=2 \int_{0}^{\infty} \frac{\ln 2}{1+t^{2}} d t+2 \int_{0}^{\infty} \frac{\ln \left(1-t+t^{2}\right)}{1+t^{2}} d t-2 \int \frac{\ln \left(1+t^{2}\right)}{1+t^{2}} d t \\
&=\pi \ln 2+2 \underbrace{\left(\frac{2 \pi}{3} \ln (2+\sqrt{3})-\frac{4}{3} G\right)}_{(***)}-2 \pi \ln 2 \\
&=-\pi \ln 2+\frac{4 \pi}{3} \ln (2+\sqrt{3})-\frac{8}{3} G
\end{aligned}
Note:(***) post 3
Eager to know whether there are any alternatives!
Comments and alternative solutions are highly appreciated.
| Let $ I_{\pm}=\int_0^{\pi/2}\ln(2\csc2x\pm 1)dx$ and evaluate
\begin{align}
I_+ +I_- =& \int_0^{\pi/2}\ln(4\csc^2 2x-1)dx
=\int_0^{\pi/2}\ln(\tan^4 x+\sec^2x)dx\\
=& \int_0^{\pi/2}\int_0^1 \frac{2y\sec^2 x}{\tan^4 x+y^2(1+\tan^2x)}dy\ dx\\
=&\int_0^1 \frac\pi{\sqrt{(y+1)^2-1}}dy=\pi\ln(2+\sqrt3)\\
\\
I_+ -I_- =& \int_0^{\pi/2}\ln\frac{2\csc 2x+1}{2\csc 2x-1} dx
=\int_0^{\pi/2}\ln\frac{1+\frac12\sin 2x}{1-\frac12\sin 2x} dx\\
=& \int_0^{\pi/2} \int_{0}^{\pi/12}
\frac {4\cos 2y \sin 2x}{1-\sin^2 2x \sin^2 2y} dy \ dx
=\int_{0}^{\pi/12} \frac{8y }{\sin 2y }\ dy\\
\overset{ibp}=& \ -\frac{\pi}3\ln\tan\frac\pi{12}- 4 \int_{0}^{\pi/12} \ln (\tan y) \ dy
=- \frac{\pi}3\ln(2+\sqrt3)+\frac{8}3G
\end{align}
Then
\begin{align}
\int_{0}^{\pi} \ln (2\pm \sin x) d x
= & \int_{0}^{\pi} \ln (\sin x) d x+ 2I_\pm \\
= & -\pi \ln 2+\frac{(3\mp 1) \pi}{3} \ln (2+\sqrt{3})\pm \frac{8}{3} G
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How to compute derivative of $\sin(x^3)$ by definition? I am trying to proof that derivative of $\sin(x^3) = 3x^2\cos(x^3)$ by definition.
But I don't know an identity for $\sin(x^3)$ for getting $\cos(x^3)$.
Even I try to find a quantity similar to $\frac{\sin(x)}{x}$.
| $$\frac{d\sin x^3}{dx}=\lim_{h\rightarrow 0} \frac{\sin(x+h)^3-\sin x^3}{h}$$
$(x+h)^3=x^3(1+h/x)^3$ Let us use $(1+z)^k=1+kz+\frac{k(k-1)}{2} z^2+O(z^3)$ to get
$$\frac{d\sin x^3}{dx}=\lim_{h\rightarrow 0}\frac{\sin[x^3(1+h/x)^3]-\sin x^3}{h}$$
$$=\lim_{h\rightarrow 0}\frac{\sin[x^3+3hx^2+O(h^2)]-\sin x^3}{h}$$
$$=\lim_{h\rightarrow 0} \frac{\sin x^3\cos(3hx^2)+\cos x^3 \sin(3hx^2)-\sin x^3 }{h}$$
$$=\lim_{h\to 0} \frac{\sin x^3[(\cos 3hx^2)-1]}{h}+\cos x^3 \lim_{h\to 0}\frac {\sin 3hx^2}{h}$$
Use $\cos z=1-z^2/2+O(z^4), \sin z=z+O(z^3)$, then we get
$$\frac{d\sin x^3}{dx}=\sin x^3\lim_{h\to 0} \frac{9h^2x^4}{h}+\cos x^3 \lim_{h\to 0}\frac{3hx^2}{h}=0+3x^2 \cos x^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is wrong with my solution to find the size of an angle of a triangle? The sides of a triangle $a$, $b$, and $c$ has the following lengths.
$$a=x^2+x+1$$
$$b=2x+1$$
$$c=x^2-1$$
Since $a,b,c>0$, $x>1$.
$$a-b=x(x-1)>0\ (\because x>1)$$
$$a-c=x+2>0$$
Therefore, $a$ is the longest side, and its opposite angle $A$ is the largest angle. Let $B$ and $C$ be the opposite angle of $b$ and $c$, respectively.
$$a:b:c=x^2+x+1:2x+1:x^2-1=A:B:C$$
$$A=(x^2+x+1)k$$
$$B=(2x+1)k$$
$$C=(x^2-1)k$$
$$A+B+C=(2x^2+3x+1)k=\pi$$
$$k=\frac{\pi}{2x^2+3x+1}$$
$$A=\frac{(x^2+x+1)\pi}{2x^2+3x+1}$$
The correct value of A is $\frac{2}{3}\pi$, but the given solution doesn't produce the right answer. I got the correct value by an alternative method, but I can't figure out what's wrong with the solution above.
| Well, the angles A,B,C are not "proportional" as you wrote.
By the cosine law, you have
$$
a^2=b^2+c^2-2bc \cos(A).
$$
Hence
$$
\cos(A)=\frac{b^2+c^2-a^2}{2bc}=\frac{(2x+1)^2+(x^2-1)^2-(x^2+x+1)^2}{2(2x+1)(x^2-1)}=-\frac{1}{2}.
$$
Therefore $A=\frac{2}{3}\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4515335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solving $\cos(2\theta)=\sin\left(\frac{\theta}{2}\right)$ Consider an acute angle $\theta$, which has the property that
$$\cos(2\theta)=\sin\left(\frac{\theta}{2}\right).$$
I am trying to find the value of $\theta$. Using the identity
$$\cos(2\theta)=\sin\left(\frac{\pi}{2}-2\theta\right),$$
I can write the first equation as
$$\sin\left(\frac{\pi}{2}-2\theta\right)=\sin\left(\frac{\theta}{2}\right).$$
Equating the argument of the $\sin$ functions, we see that
\begin{align}
\frac{\pi}{2}-2\theta&=\frac{\theta}{2} \\
\frac{5}{2}\theta&=\frac{\pi}{2} \\
\theta&=\frac{\pi}{5}.
\end{align}
While this yields the correct answer, I do not know how we can justify equating the arguments of the $\sin$ functions.
| Note that the solution to $$\sin(x) = \sin(y)$$ is $$x = 2k\pi + y \text{ and } x = 2k\pi + \pi - y$$ So in this case we have $$\frac\pi 2 - 2\theta = 2k\pi + \frac\theta 2 \text{ and } \frac\pi 2 - 2\theta = 2k\pi + \pi - \frac\theta 2$$The solutions are $$\frac{5\theta}{2} = \frac \pi 2 - 2k\pi \implies \theta = \frac \pi 5 - \frac{4k\pi}{5}, \ \ k \in \mathbb{Z} \\ \frac{3\theta}{2} = -\frac \pi 2 - 2k\pi \implies \theta = -\frac \pi 3 - \frac{4k\pi}{3}, \ \ k \in \mathbb{Z}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
If the matrix A is given, then find the value of $A^{2022}$ If it is given that matrix $ A = \begin{bmatrix}
\dfrac{5}{2} & \dfrac{3}{2}\\
-\dfrac{3}{2} & -\dfrac{1}{2}
\end{bmatrix}$ then find the value of $A^{2022}$.
Here is my try on it : $$\\$$
If we look at the pattern while multliplying A with itself it comes out to be something like this,
$$A^2 = \begin{bmatrix}
4 & 3\\
-3 & -2
\end{bmatrix}$$
$$ A^3 = \begin{bmatrix}
\dfrac{11}{2} & \dfrac{9}{2}\\
-\dfrac{9}{2} & -\dfrac{7}{2}
\end{bmatrix}$$
$$A^4 = \begin{bmatrix}
7 & 6\\
-6 & -5
\end{bmatrix}$$
so $$ A^n = \begin{bmatrix}
\dfrac{3n}{2}+1 & \dfrac{3n}{2}\\
-\dfrac{3n}{2} & -\dfrac{3n}{2}+1
\end{bmatrix} \tag{1}\label{eq1}$$
thus after substituting $n=2022$ we get $ A^{2022} = \begin{bmatrix}
3034 & 3033\\
-3033 & -3032
\end{bmatrix}$
$$\\$$
Is there is any other way to find the value of $A^{2022}$ without finding out the pattern and then general equation \eqref{eq1} like I did above?
| Another option is to see that $A = I + B$, where
$B = \begin{pmatrix}
\frac 3 2 & \frac 3 2 \\
-\frac 3 2 & -\frac 3 2
\end{pmatrix}$
and $B^2 = 0$.
So $A^n = I + nB$.
But that's a post-hoc explanation: it is not obvious to remark that when looking at the matrix - although such a simplification can be expected, in an exercise like this one, especially if it comes from a math competition where speed matters.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Can't understand why cosh(1/x) changes the result for this limit I have this limit:
$$\lim_{x \to -\infty} x^2\left({(x^2+1)\cosh{\frac 1x}\over x^2}-1\right)$$
If I solve this limit by hand, I get 1
$\require{cancel}$
$$\lim_{x \to -\infty} x^2\left({(x^2+1)\cosh{\cancelto{0}{\frac 1x}}\over x^2}-1\right)=$$
$$\lim_{x \to -\infty} x^2\left({(x^2+1)\cancelto{1}{\cosh{0}}\over x^2}-1\right)=$$
$$\lim_{x \to -\infty} \cancel{x^2}\left({\cancel{x^2}+1\cancel{-x^2}\over \cancel{x^2}}\right)=1$$
However, both Wolfram Alpha and the exercise say the correct result is ${\frac 32}$
$$\lim_{x \to -\infty} x^2\left({(x^2+1)\cosh{\frac 1x}\over x^2}-1\right) = \frac 32$$
However, if I remove $\cosh{\frac 1x}$, which for ${x \to -\infty}$ tends to $1$, suddenly Wolfram Alpha says the result is 1:
$$\lim_{x \to -\infty} x^2\left({(x^2+1)\over x^2}-1\right) = 1$$
I'm at a complete loss. Wolfram Alpha uses L'Hopital's rule to solve the first limit, so it doesn't help me undersand the discrepancy
| Note that the Taylor series of $\cosh(z)$ at $z=0$ is $1+\frac{z^2}{2}+o(z^2)$. Hence, as $x\to -\infty$, we find
$$\begin{align}
x^2\left({(x^2+1)\cosh\left({\frac 1x}\right)\over x^2}-1\right)
&=x^2\left({(x^2+1)(1+\frac{1}{2x^2}+o(1/x^2))\over x^2}-1\right)
\\
&=(x^2+1)\left(1+\frac{1}{2x^2}+o(1/x^2)\right)-x^2\\
&=1+\frac{1}{2}+o(1)\to \frac{3}{2}.\end{align}$$
In your work you implicitly used a less precise Taylor series, namely $1+o(1)$, which yields
$$x^2\left({(x^2+1)(1+o(1))\over x^2}-1\right)=(x^2+1)(1+o(1))-x^2=1+x^2\cdot o(1)+o(1)$$
and we can't conclude that the limit is $1$ because $x^2\cdot o(1)$ is an indeterminate form $+\infty\cdot 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Calculate the integral of $\frac{1}{x^2 +x + \sqrt x}$ How to correctly calculate the integral:
$$\int_0^\infty \frac{1}{x^2 +x + \sqrt x}dx$$
Edit: I tried to figure out if the limit exists:
Step 1: break the integral to two parts: from 0 to 1, from 1 to infinity.
Step 2: use limit comparison test for both of the integral: the first integral compared at 1 to 1/sqrt(x) and the second is compared at infinity to 1/x^2.
Step 3: conclude that both converge, hence the original integral also converges.
Step 4: (this is the one im trying to figure out, how to actually calculate it, because the limit exists).
| $$\int \frac{dx}{x^2 +x + \sqrt x}=2\int \frac{dy}{y^3 +y + 1}$$
Write
$$\frac{1}{y^3 +y + 1}=\frac 1{(y-a)(y-b)(y-c)}$$ Use partial fraction decomposition
$$\frac{1}{y^3 +y + 1}=\frac{1}{(a-b) (a-c) (y-a)}+\frac{1}{(b-a) (b-c) (y-b)}+\frac{1}{(c-a) (c-b) (y-c)}$$ Then three logarithms to be recombined before using the bounds.
Look how nice is the real root of the cubic
$$a=-\frac{2}{\sqrt{3}}\sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{3 \sqrt{3}}{2}\right)\right)$$ Then from Vieta
$$b=-\frac{a}{2}-i\frac{\sqrt{a^3+4}}{2 \sqrt{|a|}}\quad \text{and} \quad c=-\frac{a}{2}+i\frac{ \sqrt{a^3+4}}{2 \sqrt{|a|}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Calculate $\frac{4-5\sin\alpha}{2+3\cos\alpha}$
Calculate $$\dfrac{4-5\sin\alpha}{2+3\cos\alpha}$$ if $\cot\dfrac{\alpha}{2}=-\dfrac32$.
My first approach was to somehow write the given expression only in terms of the given $\cot\frac{\alpha}{2}$ and just put in the value $\left(-\dfrac{3}{2}\right)$. Now I don't think that's possible because we have constants (4 and 2). My try, though: $$\dfrac{4-5\sin\alpha}{2+3\cos\alpha}=\dfrac{4-5.2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2+3\left(\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}\right)}=\dfrac{4-10\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2+3\cos^2\frac{\alpha}{2}-3\sin^2\frac{\alpha}{2}}$$ My second idea was to find the value of the trig functions of $\alpha$. I don't know if this is the most straight-forward approach, but $$\cot\alpha=\dfrac{\cot^2\frac{\alpha}{2}-1}{2\cot\frac{\alpha}{2}}=\dfrac{\frac94-1}{-2.\frac32}=-\dfrac{5}{12}.$$ Am I now supposed just to find the values of $\sin\alpha$ and $\cos\alpha$? Nothing more elegant? We would have $\dfrac{\cos\alpha}{\sin\alpha}=-\dfrac{5}{12}\Rightarrow\cos\alpha=-\dfrac{5}{12}\sin\alpha$ and putting into $\sin^2\alpha+\cos^2\alpha=1$ we'd get $\cos\alpha=\pm\dfrac{12}{13}$.
| $$\dfrac{4-5\sin\alpha}{2+3\cos\alpha}=\dfrac{4-5.2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2+3\left(1-2\sin^2\frac{\alpha}2\right)}=\dfrac{4-10\sin\frac{\alpha}2\cos\frac{\alpha}2}{5-6\sin^2\frac{\alpha}2}$$Now divide numerator and denominator by $\sin^2\frac{\alpha}2$,
$$\dfrac{\dfrac{4}{\sin^2\frac{\alpha}2}-10\cot\frac{\alpha}2}{\dfrac5{\sin^2\frac{\alpha}2}-6}=\dfrac{4(1+\cot^2\frac{\alpha}2)-10\cot\frac{\alpha}2}{5(1+\cot^2\frac{\alpha}2)-6}=\frac{112}{41}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What is wrong in this solution of $\sec x + \csc x = 2 \sqrt{2}$
Find number of solutions in the interval $[0,2\pi]$ of the equation - $$\csc x + \sec x = 2 \sqrt{2}.$$
$⇒ \dfrac{1}{\sin x}+ \dfrac{1}{\cos x }= 2 \sqrt2$
$⇒ \dfrac{\sin x +\cos x}{\sin x \cos x} = 2 \sqrt 2 ⇒ (\sin x+\cos x)^2=8\sin^2x\cos^2x⇒1+2\sin x \cos x=8\sin^2x \cos^2x$
$⇒\sin x \cos x=-1/4 ,\sin x \cos x=1/2⇒2\sin x \cos x=-1/2 , 2\sin x\cos x=1$
$⇒\sin2x =-1/2 , \sin2x=1.$ But if we check manually , $\sin2x=-1/2$ is not a solution.
$⇒\sin 2x =1⇒2x=(-1)^{n}\dfrac{\pi}{2}+n\pi⇒x=(-1)^{n}\dfrac{\pi}{4}+\dfrac{n\pi}{2}⇒$ For $[0,2\pi], \boxed{x=\dfrac{\pi}{4},\dfrac{5 \pi}{4}}$ are solutions.
But if we check the graph neither are the solutions . But $\pi /4 $ is a solution but $5\pi /4 $ is not a solution if we plug values manually.
Can someone point out the mistake? Thanks
$\text{References:}$ Link for above graph : https://www.desmos.com/calculator/eaakqvd9y5
| As noticed in the comments, the issue is that by squaring we can add solutions, as for the following trivial example
$$x=-1 \implies x^2 =1 \implies x=1 \lor x=-1$$
In this case to avoid this issue we can proceed as follows (since $\sin x\cos x\neq 0$):
$$\dfrac{1}{\sin x}+ \dfrac{1}{\cos x }= 2 \sqrt2 \iff \cos x + \sin x = 2 \sqrt2 \sin x \cos x$$
with
$$2 \sin x \cos x =(\cos x + \sin x)^2-1$$
and thus
$$\sqrt 2(\cos x + \sin x)^2-(\cos x + \sin x)-\sqrt 2=0$$
from wich we obtain
$$\cos x + \sin x=\sqrt 2 \:\lor \: \cos x + \sin x=-\frac{\sqrt 2}2$$
then we can use that $\cos x + \sin x= \sqrt 2 \sin \left(x+\frac \pi 4\right)$ to obtain the result.
As an alternative from here
$$\cos x + \sin x = 2 \sqrt2 \sin x \cos x$$
we have that
$$\sqrt 2 \sin \left(x+\frac \pi 4\right)= \sqrt 2 \sin (2x)$$
$$ \sin \left(x+\frac \pi 4\right)= \sin (2x)$$
which leads to
$$x+\frac \pi 4 = 2x + 2k\pi \: \lor \: x+\frac \pi 4 = \pi -2x + 2k\pi$$
that is $x=\frac \pi 4 +\frac23k\pi$.
Here is a graph with a visualization for the solutions
| {
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"url": "https://math.stackexchange.com/questions/4526494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving a function is continuous using rigorous definition of a limit Im trying to prove that the function $$\begin{cases}f(x,y)=\dfrac{(2x^2y^4-3xy^5+x^6)}{(x^2+y^2)^2}, & (x,y)≠0\\ 0, & (x,y)=0\end{cases}$$ is continuous at point (0,0) using the rigorous defintion of a limit.
Attempting to find the upper limit of the function:
$$|f(x)-f(x_0)|= \left|\frac{(2x^2y^4-3xy^5+x^6)}{(x^2+y^2)^2}-0\right|$$
I see the denominator is always positive so this is equal to
$\dfrac{|2x^2y^4-3xy^5+x^6|}{(x^2+y^2)^2}$.
Using the triangle inequality i know that this is equal or less than
$\dfrac{|(2x^2y^4)-(3xy^5)|+|x^6|}{(x^2+y^2)^2}$.
From here I would like to continue finding expressions which are equal or greater than this, which allow me to cancel some terms against $((x^2+y^2)^2)$.
Im thinking i can write
$$x^6 = (x^2)^3 ≤ (x^2+y^2)^3 $$
for instance, but i am unsure of how to "handle" $|(2x^2y^4)-(3xy^5)$|.
Could someone give me any pointers?
| Use the fact that $|xy| \le \frac{x^2 + y^2}{2}$. You can then get,
$$3|xy^5| = 3|xy|y^4 \le \frac{3}{2}(x + y)^3$$
and
$$2x^2y^4 = 2|xy|^2y^2 \le \frac{1}{2}(x + y)^3.$$
You can then proceed as you were.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
How do you compute $\int_{0}^{\pi}\sqrt{\sin^{-1}(\cos(x))\tan^{-1}(\cot(x))}dx$ I tried using the identity of $\frac{\pi}{2}-x$ and doing substitutions like $u=\cos(x)$, but I keep ending up with $0$.
Is there a concept I'm missing?
| $$\int_{0}^{\pi}\sqrt{\sin^{-1}(\cos(x))\tan^{-1}(\cot(x))}dx$$
$$= \int_{0}^{\pi} \sqrt{\left(\frac{\pi}{2} - x\right)^2} dx$$
$$= \int_{0}^{\pi} \left|\frac{\pi}{2} - x\right| dx$$
$$= \int_{0}^{\pi/2} \left(\frac{\pi}{2} - x\right) ~dx + \int_{\pi/2}^\pi \left(x-\frac{\pi}{2}\right) ~dx$$
$$= \left[\frac{\pi}{2}x - \frac{1}{2}x^2\right]_0^{\pi/2} + \left[\frac{1}{2}x^2 - \frac{\pi}{2}x \right]_{\pi/2}^{\pi}$$
$$=\left(\frac{\pi^2}{4} - \frac{\pi^2}{8}\right) + \left(\frac{\pi^2}{2} - \frac{\pi^2}{2}\right) - \left( \frac{\pi^2}{8} - \frac{\pi^2}{4} \right)$$
$$=\frac{\pi^2}{4}$$
If you're getting 0, you may have forgotten the absolute value in going from $\sqrt{\left(\frac{\pi}{2} - x\right)^2}$ to $\left|\frac{\pi}{2} - x\right|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4532858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Density of hitting time for a two-sided barrior for Brownian motion with drift For the following Brownian motion with drift $X_t = X_0 + \mu t + \sigma B_t$ where $\mu \in \mathbb{R}$, $ \sigma > 0$ and $X_0 \in \mathbb{R}$ which is a solution to the stochastic differential equation:
\begin{align*}
dX_t = \mu dt + \sigma dB_t \ .
\end{align*}
I am interested in finding the probability density funtion of the following stopping time:
\begin{align*}
\tau = \inf[t: X_t \notin (a,b) ] \ .
\end{align*}
Where we for the initial position have $X_0 \in (a,b)$. The corresponding forward Kolmologov equation to the above SDE is:
\begin{align*}
\frac{\partial u}{\partial t} = \frac{\sigma^2}{2} \frac{\partial^2u}{\partial x^2} - \mu \frac{\partial u}{\partial x} \ .
\end{align*}
From the answers in the link Density of first hitting time of Brownian motion with drift it should be possible to find the density of $\tau$ as a function of $t$ by solving the forward Kolmologov equation with the initial condition $u(x,t=0) = \delta(x-x_0)$ and with boundary condition $u(a,t)=u(b,t) = 0$ for all $t>0$. I have already tried the same strategy as mentioned in the link, but it doesn't seem to work. Do anyone have a suggesten to another strategy to solve this problem?
| I have now found all mistakes, so here is the final answer.
We use separation of variables to find a solution we define $u(x,t) = f(x)g(t)$ then we get the two equations:
\begin{align*}
\begin{cases}
D\partial_x^2 f - \mu \partial_x f + \lambda f = 0 \\
\partial_t g = - \lambda g
\end{cases}
\end{align*}
where we have introduced the constant $-\lambda$ and where $D=σ^2/2$. We start by solving the time dependent equation, this equation has the general solution:
\begin{align*}
g(t) = g(t_0) \exp\big(-\lambda(t-t_0) \big) \ .
\end{align*}
For the spatial dependent solution $f$, we put $f(x) = \exp(rx)$ then we get:
\begin{align*}
D r^2 \exp(rx) - \mu r \exp(rx) + \lambda \exp(rx) = 0 \ .
\end{align*}
We divide by $\exp(rx)$ and solve the corresponding second order polynomial:
\begin{align*}
r_1 &= \frac{\mu + \sqrt{\mu^2 - 4 D \lambda}}{2 D} \ , \\
r_2 &= \frac{\mu - \sqrt{\mu^2 - 4 D \lambda}}{2 D} \ .
\end{align*}
Now if $\mu^2 - 4 D \lambda>0$ and $r_1 \neq r_2$ is not equal then we have the solution:
\begin{align*}
f(x) = A \exp(r_1 x) + B \exp(r_2 x) \ .
\end{align*}
If $\mu^2 - 4 D \lambda=0$ then $r_1 = r_2$ and we then have the solution:
\begin{align*}
f(x) = A \exp(r x) + B x\exp(r x) \ .
\end{align*}
If $\mu^2 - 4 D \lambda<0$ then the roots are complex, thus we have $r_1 = \frac{\mu + i \sqrt{4 D \lambda - \mu^2}}{2D}$, $r_2 = \frac{\mu - i \sqrt{4 D \lambda - \mu^2}}{2D}$ and $\beta = \frac{\sqrt{4 D \lambda - \mu^2}}{2D}$. The solution is:
\begin{align*}
f(x) = \exp\left(\frac{\mu x}{2D}\right) \left(A \sin(\beta x) + B \cos(\beta x) \right) \ .
\end{align*}
Using the boundary condition only the last expression for $f$ will give us a non-trivial solution. Thus we have:
\begin{align*}
f(a) &= \exp\left(\frac{\mu a}{2D}\right) \left(A \sin(\beta a) + B \cos(\beta a) \right) = 0 \ \Rightarrow \ A = - B \frac{\cos(\beta a)}{ \sin(\beta a)} \ .
\end{align*}
Using the other boundary point we get:
\begin{align*}
f(b) &= \exp\left(\frac{\mu b}{2D}\right) \left( - B \frac{\cos(\beta a)}{ \sin(\beta a)} \sin(\beta b) + B \cos(\beta b) \right) \ .
\end{align*}
Form this we conclude that:
\begin{align*}
B \left ( \cos(\beta b) - \frac{\cos(\beta a)}{ \sin(\beta a)} \sin(\beta b) \right) = 0 \ .
\end{align*}
If $B$ is zero we get a trivial solution, thus $B$ is not zero and then we conclude that $\cos(\beta b) - \frac{\cos(\beta a)}{ \sin(\beta a)} \sin(\beta b) = 0$. By multiplication with $-\sin(\beta a)$ and using the following formula:
\begin{align*}
\sin(\beta b)\cos(\beta a) - \cos(\beta b) \sin(\beta a) = \sin(\beta ( b - a) ) \ ,
\end{align*}
we see that:
\begin{align*}
\sin(\beta ( b - a) ) = 0 \ .
\end{align*}
This is true for $\beta ( b - a) = n \pi$ where $n \in \mathbb{Z}$. By the formula $\beta = \frac{\sqrt{4 D \lambda - \mu^2}}{2D}$ we see that there is not just one possible constant $\lambda$, but one for every $n \in \mathbb{Z}$:
\begin{align*}
\lambda_n = D \left( \left( \frac{n \pi}{b-a} \right)^2 + \left( \frac{\mu}{2D} \right)^2 \right) \ , \ \ n \in \mathbb{Z} \ .
\end{align*}
Where we have introduced a subscript $n$ to the possible $\lambda$ constants to separate them. Now by the linearity of the equation we get a Fourier series as a solution:
\begin{align*}
u(x,t) = \exp\left(\frac{\mu x}{2D}\right) \sum_{n \in \mathbb{Z}} \left(A_n \sin\left(\frac{n \pi x}{b-a} \right) + B_n \cos\left(\frac{n \pi x}{b-a} \right) \right) \exp\big(-\lambda_n(t-t_0) \big) \ .
\end{align*}
Where $g(t_0)$ has been left out since it can be absorbed into the constants $A_n$ and $B_n$. Now because of the odd and even properties of sine and cosine we see that:
\begin{align*}
A_n \sin\left(\frac{n \pi x}{b-a} \right) + A_{-n} \sin\left(\frac{-n \pi x}{b-a} \right) &= (A_n - A_{-n}) \sin\left(\frac{n \pi x}{b-a} \right) \ , \ n \in \mathbb{N} \ , \\
B_n \cos\left(\frac{n \pi x}{b-a} \right) + B_{-n} \cos\left(\frac{-n \pi x}{b-a} \right) &= (B_n + B_{-n}) \cos\left(\frac{n \pi x}{b-a} \right) \ , \ n \in \mathbb{N} \ .
\end{align*}
Because of these formulas and the squared $n^2$ in the expression of $\lambda_n$ we can limit our sum to $n \in \mathbb{N}_0$ thus:
\begin{align*}
u(x,t) = \exp\left(\frac{\mu x}{2D}\right) \sum_{n \in \mathbb{N}_0} \left(A_n \sin\left(\frac{n \pi x}{b-a} \right) + B_n \cos\left(\frac{n \pi x}{b-a} \right) \right) \exp\big(-\lambda_n(t-t_0) \big) \ .
\end{align*}
One can show the following for sine and cosine:
\begin{align*}
\int_{a}^b \cos\left(\frac{n \pi x}{b-a} \right) \sin\left(\frac{m \pi x}{b-a} \right) dx &= 0 \ , \ \forall n,m \ . \\
\int_a^b \cos\left(\frac{n \pi x}{b-a} \right) \cos\left(\frac{n \pi x}{b-a} \right) dx &= 0 \ , \ n \neq m \ . \\
\int_a^b \sin\left(\frac{n \pi x}{b-a} \right) \sin\left(\frac{n \pi x}{b-a} \right) dx &= 0 \ , \ n \neq m \ . \\
\end{align*}
For the two last expressions, in the case $m=n$, we have:
\begin{align*}
\int_a^b \cos^2\left(\frac{n \pi x}{b-a} \right) dx &= \frac{b-a}{2} \ ,
\\
\int_a^b \sin^2\left(\frac{n \pi x}{b-a} \right) dx &= \frac{b-a}{2} \ .
\end{align*}
From these formulas one can deduce that:
\begin{align*}
B_n = \frac{2}{b-a} \exp\big(\lambda_n(t-t_0) \big) \int_a^b u(x,t) \cos\left(\frac{n \pi x}{b-a} \right) \exp\left(\frac{-\mu x}{2D}\right) dx \ , \\
A_n = \frac{2}{b-a} \exp\big(\lambda_n(t-t_0) \big) \int_a^b u(x,t) \sin\left(\frac{n \pi x}{b-a} \right) \exp\left(\frac{-\mu x}{2D}\right) dx \ .
\end{align*}
Since this has to be true for all $t \geq t_0$ we can put $t=t_0$ then we get:
\begin{align*}
B_n = \frac{2}{b-a} \int_a^b \delta(x-x_0) \cos\left(\frac{n \pi x}{b-a} \right) \exp\left(\frac{-\mu x}{2D}\right) dx = \frac{2}{b-a} \cos\left(\frac{n \pi x_0}{b-a} \right) \exp\left(\frac{-\mu x_0}{2D}\right) \ , \\
A_n = \frac{2}{b-a} \int_a^b \delta(x-x_0) \sin\left(\frac{n \pi x}{b-a} \right) \exp\left(\frac{-\mu x}{2D}\right) dx = \frac{2}{b-a} \sin\left(\frac{n \pi x_0}{b-a} \right) \exp\left(\frac{-\mu x_0}{2D}\right) \ .
\end{align*}
Inserting this back into our solution we get:
\begin{align*}
u(x,t) = \frac{2}{b-a} \exp\left(\frac{\mu(x-x_0)}{2D}\right) \sum_{n \in \mathbb{N}_0} \cos\left(\frac{n \pi (x-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big)
\end{align*}
We now define the survival probability as:
\begin{align*}
S(t) = \int_a^b u(x,t) dx \ .
\end{align*}
We see that:
\begin{align*}
S(t) &= \frac{2}{b-a} \sum_{n \in \mathbb{N}_0} \exp\big(-\lambda_n(t-t_0) \big) \int_a^b \exp\left( \frac{\mu (x-x_0)}{2D} \right) \cos\left( \frac{n \pi (x-x_0)}{b-a} \right) dx \ .
\end{align*}
Calculating the integral we have:
\begin{align*}
\int_a^b \exp\left( \frac{\mu (x-x_0)}{2D} \right) &\cos\left( \frac{n \pi (x-x_0)}{b-a} \right) dx = \\
\frac{D}{\lambda_n} \Bigg( \frac{\mu}{2D} & \left( \exp\left(\frac{\mu(b-x_0)}{2D} \right) \cos\left( \frac{n \pi (b-x_0)}{b-a} \right) - \exp\left(\frac{\mu(a-x_0)}{2D} \right) \cos\left( \frac{n \pi (a-x_0)}{b-a} \right) \right) \\
+ \frac{n \pi}{b-a} & \left( \exp\left(\frac{\mu(b-x_0)}{2D} \right) \sin\left( \frac{n \pi (b-x_0)}{b-a} \right) - \exp\left(\frac{\mu(a-x_0)}{2D} \right) \sin\left( \frac{n \pi (a-x_0)}{b-a} \right) \right) \Bigg)\ .
\end{align*}
By inserting this in the expression for $S(t)$ we get:
\begin{align*}
S(t) &= \frac{2}{b-a} \exp\left(\frac{\mu(b-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} \frac{D}{\lambda_n} \frac{\mu}{2D} \cos\left( \frac{n \pi (b-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \\
&- \frac{2}{b-a} \exp\left(\frac{\mu(a-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} \frac{D}{\lambda_n} \frac{\mu}{2D} \cos\left( \frac{n \pi (a-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \\
&+ \frac{2}{b-a} \exp\left(\frac{\mu(b-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} \frac{D}{\lambda_n} \frac{n \pi}{b-a} \sin\left( \frac{n \pi (b-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \\
&- \frac{2}{b-a} \exp\left(\frac{\mu(a-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} \frac{D}{\lambda_n} \frac{n \pi}{b-a} \sin\left( \frac{n \pi (a-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \ .
\end{align*}
Now the first hitting time density $f_{\tau}(t)$ can be found by the expression:
\begin{align*}
f_{\tau}(t) = - \frac{dS}{dt} \ .
\end{align*}
$f_\tau$ is also known as the first passage time density. we see that:
\begin{align*}
f_{\tau}(t) &= \frac{2}{b-a} \exp\left(\frac{\mu(b-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} \frac{\mu}{2} \cos\left( \frac{n \pi (b-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \\
&- \frac{2}{b-a} \exp\left(\frac{\mu(a-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} \frac{\mu}{2} \cos\left( \frac{n \pi (a-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \\
&+ \frac{2}{b-a} \exp\left(\frac{\mu(b-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} D \frac{n \pi}{b-a} \sin\left( \frac{n \pi (b-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \\
&- \frac{2}{b-a} \exp\left(\frac{\mu(a-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} D \frac{n \pi}{b-a} \sin\left( \frac{n \pi (a-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \ .
\end{align*}
For convenience we define:
\begin{align*}
v(x,t) = \frac{2}{b-a} \exp\left(\frac{\mu(x-x_0)}{2D} \right) \sum_{n \in \mathbb{N}_0} \frac{n \pi}{b-a} \sin\left( \frac{n \pi (x-x_0)}{b-a} \right) \exp\big(-\lambda_n(t-t_0) \big) \ .
\end{align*}
Then we have:
\begin{align*}
f_{\tau}(t) &= \frac{\mu}{2} \Big( u(b,t) - u(a,t) \Big) + D \Big( v(b,t) - v(a,t) \Big) \ .
\end{align*}
By our boundary condition $u(b,t) = u(a,t) = 0$ thus:
\begin{align*}
f_{\tau}(t) &= D \Big( v(b,t) - v(a,t) \Big) \ .
\end{align*}
The following plots are for $a=-1$, $b=1$, $x_0=0$ and $t_0 = 0$. In the plots we have $u:=\mu$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4533948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $x=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2015^2}$, show that $\frac{201}{403}
If $$x=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2015^2}$$
show that
$$\frac{201}{403}<x<\frac{2014}{2015}$$
So, I manage to do the RH inequality using that $$x<\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2014\cdot2015}=1-\frac{1}{2015}=\frac{2014}{2015}$$ Unfortunately I can't find a way to do the LH inequality. I saw that
$$\frac{1}{2015}+\frac{1}{2015}+\cdots+\frac{1}{2015}$$ for 1005 times is equal to $\frac{201}{403}$ but I don't see why
$$x> \frac{1}{2015}+\frac{1}{2015}+\cdots+\frac{1}{2015}$$ thx!
| We have
$$\frac{1}{k}-\frac{1}{k+1} < \frac{1}{k^2} < \frac{1}{k-1} - \frac{1}{k}$$
and so (telescoping sum)
$$\frac{1}{2} - \frac{1}{2016} < \sum_{k=2}^{2015} \frac{1}{k^2} < \frac{1}{1} - \frac{1}{2015}$$
that is
$$\frac{1007}{2016}< \sum_{k=2}^{2015} \frac{1}{k^2} < \frac{2014}{2015}$$
Now, LHS $= \frac{2014}{4032}= \frac{201.4}{403.2}> \frac{201}{403}$ (mediant)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4539222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Finding all $x \in \mathbb{R}$ with $|x+1| + |x-1| < 4$ We want to find all $x \in \mathbb{R}$ with
$$|x+1| + |x-1| < 4$$
I tried the following, but I'm uncertain if this is correct:
Critical points: $x= -3$ and $x = 1$.
Case $1$:
$$x \in (-\infty, -3) \\
|x+1| = -(x+1) = -x-1 \\
|x-1| = -(x-1) = -x+1 \\
|x+1| + |x-1| < 4\\
\iff -x-1-x+1 < 4\\
\iff -2x < 4 \\
\iff x < -2 \\ \Rightarrow \text{ holds } \forall x \in (-\infty,-2[$$
Case $2$:
$$x \in [-3, 1) \\
|x+1| + |x-1| < 4\\
\iff -x-1-x+1 < 4\\
\iff -2x < 4 \\
\iff x < -2 \\ \Rightarrow \text{ holds } \forall x \in [-3,-2[$$
Case $3$:
$$x \in (1, \infty) \\
|x+1| = x+1 \\
|x-1| = x-1 \\
x+1+x-1 < 4\\
\iff 2x < 4 \\
\iff x < 2 \\ \Rightarrow \text{ holds } \forall x \in [1,2)$$
The solution would then be $x \in (-\infty,-2) \cup [1,2)$.
Is this correct/wrong? Is this the common approach for questions like these?
| Consider a cord of length $4$ with ends attached to the points $-1$ and $1.$ When the cord is maximaly extended to the right or left it reaches the points $2$ and $-2,$ respectively. Therefore the answer is $(-2,2).$
Similarly let the cord of length $2L$ be attached to the points $a<b$ such that $2L>b-a.$ Then the solution of
$$|x-a|+|x-b|<2L$$
is $(-L+{a+b\over 2}, L+{a+b\over 2}).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $\int_{0}^{\infty} \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2-1}\:dx$ Find $$I=\int_{0}^{\infty} \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2-1}dx$$
I have a very small query. I used the substitution $x+\frac{1}{x}=v$, then the limits of integration are both infinity. Can we say the integral is zero?
Instead what i did is:
after using that substitution we get:
$$I=\frac{1}{2} \lim _{x \rightarrow \infty} \ln \left|\frac{x^2-x+1}{x^2+x+1}\right|-\frac{1}{2} \lim _{x \rightarrow 0} \ln \left|\frac{x^2-x+1}{x^2+x+1}\right|=0$$
Is the second one better?
| If $I$ exists, then $I = I_1 + I_2$ where $$I_1 = \int_{x=0}^1 f(x) \, dx + \int_{x=1}^\infty f(x) \, dx, \quad f(x) = \frac{1-x^{-2}}{(x+x^{-1})^2-1}.$$
Since the antiderivative of $f$ is
$$F(x) = \int f(x) \, dx = \int \frac{dv}{v^2-1} = \frac{1}{2} \int \frac{1}{v-1} - \frac{1}{v+1} \, du = \frac{1}{2} \log \left|\frac{x-1+x^{-1}}{x+1-x^{-1}}\right| + C,$$ we have
$$I_1 = F(1) - \lim_{x \to 0^+} F(x) = -\frac{\log 3}{2} - 0 = -\frac{\log 3}{2},$$ and
$$I_2 = \lim_{x \to \infty} F(x) - F(1) = 0 + \frac{\log 3}{2}.$$
Therefore, $I = I_1 + I_2 = 0$.
Alternatively, observe that the substitution $$x = 1/u, \quad dx = -u^{-2} \, du$$ gives
$$\begin{align}I
&= \int_{x=0}^\infty f(x) \, dx \\
&= \int_{u=\infty}^0 -\frac{f(1/u)}{u^2} \, du \\
&= \int_{u=0}^\infty \frac{1-u^2}{u^2 ((u^{-1} + u)^2 - 1)} \, du \\
&= \int_{u=0}^\infty \frac{-(1-u^{-2})}{(u + u^{-1})^2 - 1} \, du \\
&= -I. \end{align}$$ Therefore $I = 0$.
This same idea can also be used to show $I_1 = -I_2$, rather than computing their values explicitly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the value of given integral The given integral is :$$\displaystyle\int_0^1 \dfrac{\sin{\theta}(\cos^2{\theta} -\cos^2{\frac{\pi}{5}})(\cos^2{\theta} -\cos^2{\frac{2\pi}{5}})}{\sin{5\theta}} \, d\theta$$
I tried solving it with some trigonometric identities but comes out it does not work here.
However solution given starts with these equations mentioned down below and later they substituted $z= \cos{\theta}+ i\sin(\theta)$
$$\begin{array}{r}
z^{10}-1=\left(z^2-1\right)\left[z^2-2 \cos \left(\frac{\pi}{5}\right) z+1\right]\left[z^2-2\left(\cos \frac{2 \pi}{5}\right) z+1\right] \\
\quad \times\left(z^2-2 \cos \frac{4 \pi}{5} z+1\right)\left(z^2-2 \cos \frac{6 \pi}{5} z+1\right)
\end{array}$$
$$\begin{aligned}
z^5-\frac{1}{z^5}=\left(z-\frac{1}{z}\right) &\left(z-2 \cos \frac{\pi}{5}+\frac{1}{z}\right)\left(z-2 \cos \frac{2 \pi}{5}+\frac{1}{z}\right) \\
& \times\left(z-2 \cos \frac{4 \pi}{5}+\frac{1}{z}\right)\left(z-2 \cos \frac{6 \pi}{5}+\frac{1}{z}\right)
\end{aligned}$$
Can anyone help me understand how I can derive these equations and use them here in this integral?
Thank you for your help.
| Note that
$$\begin{aligned}
z^5-\frac{1}{z^5}
=&\ \left(z-\frac{1}{z}\right)\left(z-2 \cos \frac{\pi}{5}+\frac{1}{z}\right)\left(z-2 \cos \frac{2 \pi}{5}+\frac{1}{z}\right) \\
& \>\>\> \times\left(z+2 \cos \frac{2 \pi}{5}+\frac{1}{z}\right)\left(z+2 \cos \frac{\pi}{5}+\frac{1}{z}\right)\\
\end{aligned}$$
Then, set $z=e^{i\theta}$ to get
$$\begin{aligned}
\sin5\theta=&\ \sin\theta\left(2\cos\theta -2 \cos \frac{\pi}{5}\right)\left(2\cos\theta -2 \cos \frac{2 \pi}{5}\right) \\
& \>\>\>\times\left(2\cos\theta+2 \cos \frac{\pi}{5}\right)\left(2\cos\theta+2 \cos \frac{2 \pi}{5}\right)\\
=&\ 16\sin{\theta}\left(\cos^2{\theta} -\cos^2{\frac{\pi}{5}}\right)\left(\cos^2{\theta} -\cos^2{\frac{2\pi}{5}}\right)
\end{aligned}$$
Thus
$$\int_0^1 \dfrac{\sin{\theta}(\cos^2{\theta} -\cos^2{\frac{\pi}{5}})(\cos^2{\theta} -\cos^2{\frac{2\pi}{5}})}{\sin{5\theta}} \, d\theta
= \int_0^1 \frac1{16}d\theta = \frac1{16}$$
| {
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"url": "https://math.stackexchange.com/questions/4548238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a_n$ be a sequence of positive real numbers satisfying $a_n$ < $\frac{1}{2}a_{n-1}$ for all $n \ge 2$. Prove that it converges to zero. Let $a_n$ be a sequence of positive real numbers satisfying $a_n$ < $\frac{1}{2}a_{n-1}$ for all $n \ge 2$. Use an appropriate theorem and the fact that $\lim_{n\to\infty} (\frac{1}{2})^n = 0$ to prove that $\lim_{n\to\infty} a_n = 0$.
My attempt: $a_n$ is a decreasing sequence because $a_n < \frac{1}{2}a_{n-1}$, for all $n \ge 2$. And we can continue: $\frac{1}{2}a_{n+1} < \frac{1}{2}a_{n}$; $\frac{1}{2}a_{n+1} < \frac{1}{2}a_{n} <\frac{1}{2}a_{n-1}$. Then, $(\frac{1}{2})^na_{n+1} < (\frac{1}{2})^na_{n} <(\frac{1}{2})^na_{n-1}$ is also true. I see that the squeeze theorem is applicable but I am not sure how to proceed from here. Please advise.
| Hint: $0 < a_n < a_1\cdot \left(\dfrac{1}{2}\right)^{n-1}, n \ge 2$ .
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the smallest possible value for $x^2-3x+2y^2+4y+2$ Find the smallest possible value for $x^2-3x+2y^2+4y+2$.
I know this: $x^2-3x+2y^2+4y+2=(x-\frac{3}{2})^2+2(y+1)^2-\frac{9}{4}$ so the smallest value is $-\frac{9}{4}$.
Now we also know this: $x^2-3x+2y^2+4y+2=(x-1)^2-1-x+2(y+1)^2=(*)(x-1)^2+2(y+1)^2-(1+x)$
My question is: why we cannot say that the smallest value possible is $-(1+x)$ when $(x-1)^2=0$ (i.e. when $x=1$) and $(y+1)^2=0$, and because $x=1$ so the smallest value is $-2$?
| If you (think you) have completed the square, but your left-over term isn't a constant, then you haven't completed the square.
The fact that $x=1$ minimises the expression $(x-1)^2$ has nothing to do with minimising $g_1(x) = (x-1)^2 + 3x\ $ or $\ g_2(x) =(x-1)^2 - 26x^3 + 45,\ $ or $\ g_3(x) =(x-1)^2 + f(x),\ $ unless $f(x)$ is a constant.
If $f(x)$ is a constant, for example, $\ g_3(x) = (x-1)^2 + 24,\ $ then $x=1$ minimises $\ g_3(x)\ $; the reason being that $\ \min \{ u^2: u\in\mathbb{R} \} = 0.\quad g_3(x)$ attains this minimum when $x=1.$
This reasoning goes out the window for $g_1(x), $ because $g_1(x) = (x+1/2)^2 + 3/4,$ and so the minimum of $\ g_1(x)\ $ is $\ 3/4\ $ occurs when $x=-1/2.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the integers $n\geq 2$ that satisfy $0 \leq 1 +\frac{n}{n-1}\cos\big(\pi\cdot\frac{n}{n+1}\big)$ I want to find the integers $n\geq 2$ that satisfy the inequality
$$
x_1x_2+x_2x_3+\cdots+x_{n-1}x_n\leq\frac{n-1}{n}\Big(x_1^2+x_2^2+\cdots+x_n^2\Big)
$$
for all $x_i>0$.
I see that if $n=2$, the geometric mean and quadratic mean inequality
$$
\sqrt{x_1x_2}\leq\sqrt{\frac{x_1^2+x_2^2}{2}}
$$
solve this question, but if $n>2$, the problem is difficult. I see inequality that solve it with a positive semidefinite quadratic form. But this lead to the former inequality, I see $n=2$ satisfies that inequality but, for example, $n=3$, $n=5$ not. I suspect that $n=2$ is the only solution but I don't see how to prove it.
| Suppose the inequality is true for some $n\in\mathbb N,\ n> 1$. For $n=2$, we know the inequality is true. We show that for $n\ge 3$, there are counterexamples to the inequality.
To that end, first let $n=3$ and $x_1=2,\ x_2=3,\ x_3=2$. Then if the inequality is satisfied, we have $$x_1x_2+x_2x_3\le \frac{3-1}{3}(x_1^2+x_2^2+x_3^2)\Rightarrow 3(6+6)\le 2(4+9+4)\Rightarrow 36\le 34$$
which is a contradiction. Thus the inequality does not work for $n=3$.
Now, say it holds for some $n>3$. Then let $x_1=1/2,\ x_j=1$ for $2\le j\le n$. Then again, if the inequality is satisfied, we must have
\begin{align*}
\sum_{j=1}^{n-1} x_{j}x_{j+1}\le \frac{n-1}{n}\sum_{j=1}^n x_j^2&\Rightarrow \frac12+(n-2)\le \frac{n-1}{n}\left(\frac14+(n-1)\right)=\frac{n-1}{n}\left(n-\frac34\right)\\
&\Rightarrow n\left(n-\frac32\right)\le (n-1)\left(n-\frac34\right)\\
&\Rightarrow n^2-\frac{3n}{2}\le n^2-\frac{7n}{4}+\frac34\\
&\Rightarrow \frac{n}{4}\le \frac34\\
&\Rightarrow n\le 3
\end{align*}
which is again a contradiction.
Thus the inequality only holds for $n=2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying $\frac{a^4}{(a-b)(a-c)} + \frac{b^4}{(b-a)(b-c)} + \frac{c^4}{(c-a)(c-b)}$ using Lagrange’s polynomial I have a question of symplifying this expression
$$
A=\frac{a^4}{(a-b)(a-c)} + \frac{b^4}{(b-a)(b-c)} + \frac{c^4}{(c-a)(c-b)} \tag{1}
$$
where $a$, $b$ and $c$ are distinct nonzero real numbers. There is a quite similar problem, which is simplifying
$$
B=\frac{a^2}{(a-b)(a-c)} + \frac{b^2}{(b-a)(b-c)} + \frac{c^2}{(c-a)(c-b)}. \tag{2}
$$
In this problem, they use Lagrange’s interpolation polynomial for $P(x) =x^2$ at nodes $a$, $b$ and $c$. Then, they compare the coefficient of $x^2$ to obtain that $B=1$.
My effort is to use the similar method with $P(x)=x^4$, but I am struggling with the fact that there are only three nodes here. So the polynomial I need to choose must be a second degree polynomial but it is hard for me to find such one.
Can someone have some idea about using different polynomial to use the Lagrange’s interpolation polynomial? Thank you very much!
| For your problem,
I threw it at Wolfram Alpha and got
$a^2 + a b + a c + b^2 + b c + c^2
$.
This can be rewritten as
$(a+b+c)^2-(ab+ac+bc)$.
With 3 instead of 4 this is
$a+b+c$,
and with 5 this is
$a^3 + a^2 b + a^2 c + a b^2 + a b c + a c^2 + b^3 + b^2 c + b c^2 + c^3
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find $\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}$ without using derivatives This limit is proposed to be solved without using the L'Hopital's rule or Taylor series:
$$
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}},
$$
where $a>0$, $b>0$ are some constants.
I know how to calculate this limit using the L'Hopital's rule:
$$
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}=
e^{\lim\limits_{x\to 0} \ln\left(\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}\right)};
$$
$$
\lim\limits_{x\to 0} \ln\left(\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}\right)=
\lim\limits_{x\to 0} \frac{\ln\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)}{x}$$
$$
=
\lim\limits_{x\to 0} \frac{2}{a^{\sin x}+b^{\sin x}}\cdot\frac12\cdot
\left( a^{\sin x}\cos x \ln a+b^{\sin x}\cos x \ln b \right)=
\frac12\left( \ln a+ \ln b \right)
$$
$$
\Rightarrow
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}=
e^{\frac12\left( \ln a+ \ln b \right)}=\sqrt{ab}.
$$
I'm allowed to use the limits $\lim_{x\to0}\frac{\sin x}{x}=1$,
$\lim_{x\to0}\frac{a^x-1}{x}=\ln a$,
$\lim_{x\to0}\frac{\log_a(1+x)}{x}=\log_a e$ and $\lim_{x\to0} (1+x)^{1/x}=e$.
| This post has multiple answers that explain why, if $\displaystyle\lim_{x\to a} f(x)=1$ and $\displaystyle\lim_{x\to a}g(x)=\infty$ then $$\lim_{x\to a} (f(x))^{g(x)}=e^{\lim\limits_{x\to a} (f(x)-1)g(x)}$$ Using this formula, calling the limit as L, we have $$L=e^{\lim\limits_{x\to 0}\left(\frac{a^{\sin x}+b^{\sin x}-2}{2x}\right)}$$ so $$\ln L=\lim_{x\to 0}\left(\dfrac{a^{\sin x}-1}{2x}+ \dfrac{b^{\sin x}-1}{2x}\right)$$$$\ln L=\lim_{x\to 0}\left(\dfrac{a^{\sin x}-1}{2\sin x}\cdot\frac{\sin x}{x}+ \dfrac{b^{\sin x}-1}{2\sin x} \frac{\sin x}{x}\right)$$$$=\frac{\ln a}{2}+\frac{\ln b}{2}=\frac{\ln ab}{2}=\ln\sqrt {ab}$$ whence $L=\sqrt{ab}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that there are no integer solutions to: $x^4+6x^2+1=8y^4$
Prove that there are no integer solutions to:
$$x^4+6x^2+1=8y^4$$
where $x>1$.
My attempts:
Let $x^2=u, y^2=v$
$$u^2+6u+(1-8v^2)=0$$
$$\Delta=36-4(1-8v^2)=w_0^2$$
$$32v^2+32=(4w_1)^2$$
$$2v^2+2=w_1^2=(2w_2)^2$$
$$v^2+1=2w_2^2$$
$$v=w_2=1\implies x=y=1$$
I don't know how can I proceed.
| First, observe that $8y^4$ is even, so $x^4+6x^2+1$ is even, so $x^4+6x^2$ is odd. Therefore, $x$ is odd. Rewrite $x$ as $2n+1$, where n is an integer and $n\geq0$. So the original equation can be rewritten as $(2n+1)^4+6(2n+1)^2+1=8y^4$. By simplifying it we see that $2n^4+4n^3+6n^2+4n+1=y^4$, which can be rewritten as $(n+1)^4+n^4=y^4$, where $n$ and $y$ are integers. It's obvious that when $n=0$ and $y=1$ the equation holds, but since $x>0$ this cannot be the case. So $n\geq1$. But by Fermat's Last Theorem, such $n$ and $y$ does not exist.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum\limits_{\mathrm{cyc}}\sqrt{xy^2(y+z)} \le {1 \over 2} \sqrt{(x + y + z)^4 - (x^2 + y^2 + z^2)^2}$
Prove the following inequality for all positive real numbers $x$, $y$, and $z$:
$$\sqrt{xy^2(y+z)} + \sqrt{yz^2(x+z)} + \sqrt{zx^2(x+y)} \le {1 \over 2} \sqrt{(x + y + z)^4 - (x^2 + y^2 + z^2)^2}$$
My attempt. I tried to prove it myself but I didn't know how to. I tried to transform this inequality but made it even more complicated. Is there an easy way to prove it? Help would be very appreciated.
| Let's raise both sides of the inequality to the second power. Indeed, we just need to show:
$$2(\sqrt {xy^2(y+z)}\sqrt {yz^2(x+z)}+\sqrt {yz^2(x+z)}\sqrt {zx^2(x+y)}+\sqrt {xy^2(y+z)}\sqrt {zx^2(x+y)})\le x^3y+xz^3+y^3z+x^2y^2+x^2z^2+y^2z^2+2x^2yz+2xy^2z+2xyz^2.$$
Now, notice that: $$2\sqrt {xy^2(y+z)}\sqrt {yz^2(x+z)}\le yz[x(y+z)+y(x+z)]=2xy^2z+xyz^2+y^2z^2.$$
Hence, it suffices to prove that:
$$x^2yz+xy^2z+xyz^2\le x^3y+xz^3+y^3z; $$
and, this is almost obvious because;
$$(x^2yz+xy^2z+xyz^2)^2\le (x^3y+xz^3+y^3z)(xyz^2+xy^2z+x^2yz).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4564490",
"timestamp": "2023-03-29T00:00:00",
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Find slope of the tangent line of $4\sqrt x + 2e^\frac {3x-12}{x+2}$ at $ x_0$ Find the slope and the equation of the tangent line to the
graph $y = f(x)$ at $x_0=4$, $$4\sqrt x + 2e^\frac {3x-12}{x+2} $$
$$\lim_{h\to 0}\tfrac{4\sqrt {4+h} + 2e^\frac {12+3h-12}{4+h+2} - 10}{h} = \lim_{h\to 0}\frac{4\sqrt {4+h} + 2e^\frac {3h}{6+h} - 10}{h} $$ I am stuck at this part
| From where you left by students's L'hospital
$$f'(4)=\lim_{h\rightarrow 0}\frac{\frac{2}{\sqrt{4+h}}+2e^{\frac{3h}{6+h}}\frac{3(6+h)-3h}{(6+h)^2}}{1}=\frac{2}{\sqrt{4}}+2.e^0.\frac{3.6}{6^2}=1+1=2$$
and the tangent line at $(4,10)$ is
$$\frac{y-10}{x-4}=2\implies y=2x+2.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve for $n: \cos\left(\frac\pi4(n^2+2n)\right)=\sin\left(\frac\pi4(n^2+n+1)\right), n\in \mathbb Z$
Solve for $n: \cos\left(\frac\pi4(n^2+2n)\right)=\sin\left(\frac\pi4(n^2+n+1)\right), n\in \mathbb Z$
My Attempt:
$$\cos\left(\frac\pi4(n^2+2n)\right)=\cos\left(\frac\pi2-\frac\pi4(n^2+n+1)\right)\\\implies\frac\pi4(n^2+2n)=2p\pi\pm\left(\frac\pi2-\frac\pi4(n^2+n+1)\right), p\in \mathbb Z$$
With minus sign,
$$\frac\pi4(n^2+2n)=2p\pi-\left(\frac\pi2-\frac\pi4(n^2+n+1)\right)\\\implies n^2+2n=8p-(2-n^2-n-1)\\\implies n=8p-1$$
With plus sign,
$$\frac\pi4(n^2+2n)=2p\pi+\left(\frac\pi2-\frac\pi4(n^2+n+1)\right)\\\implies n^2+2n=8p+(2-n^2-n-1)\\\implies 2n^2+3n=8p+1$$
How to conclude from this?
The answer given is $8p-1$ or $8p-3$.
| $$2n^2+3n-5=(2n+5)(n-1)=8p+1-5=4(2p-1)$$
As $2n-5$ is odd, $4$ must divide $n-1=4r$(say)
$\implies(2(4r+1)-5)r=2p-1\implies r$ must be odd $2t-1$(say)
$\implies n=4r+1=4(2t-1)+1=8t-3$
$$\implies p=\dfrac{2n^2+3n-1}8=\dfrac{2(8t-3)^2+3(8t-3)-1}8=16t^2-9t-1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find highest common factor of two polynomials Find the highest common factor of the following expressions:
$6-8a-32a^2-18a^3$, $20-35a-95a^2-40a^3$
My attempt:
The polynomial on the left is divisible by $2$, the polynomial on the right is not divisble by $2$. Conversely the polynomial on the right is divisble by $5$ and not the one on the left. Hence simplify the polynomials, then toss out $2$ and $5$ knowing they cannot be in the final answer since we are finding the highest common factor.
\begin{array}{r|ll} 3-4a-16a^2-9a^3 & 4-7a-19a^2-8a^3 \end{array}
The lowest common multiple of $-9a^3$ and $-8a^3$ is $-72a^3$.
\begin{array}{|r|ll} 24-32a-128a^2-72a^3 & 36-63a-171a^2-72a^3 \\ 36-63a-171a^2-72a^3 \end{array}
Subtract the left hand side.
\begin{array}{r|ll} 43a^2+31a-12 & 36-63a-171a^2-72a^3 \end{array}
Now usually the next step is to multiply the polynomial on the left by $a$ and subtract on the right hand side, but the coefficients are a big problem here. $43$ is a prime number. I actually went ahead and found the lowest common multiple of $72$ and $43$ and kept calculating, but naturally the coefficients morphed into four places and I couldn't find the highest common factor. I don't believe the problem would require that kind of tedious calculation (and I still got it wrong anyway), so I feel I must've made a mistake somewhere. Any help would be appreciated thanks.
The answer is $1+a$.
| It looks like you're kind of drifting towards applying the Euclidean algorithm. In this case, though, the easiest way is probably to (1) prove that $1+a$ is a factor of both polynomials, (2) find the quotient of each polynomial divided by $1+a$, (3) find the GCD of the quotients.
If you do that here, then first you get that $9a^3+16a^2+4a-3 = (a+1)(9a^2+7a-1)$ and $8a^3+19a^2+7a-4 = (a+1)(8a^2+11a-4)$. Then we note that $(9a^2+7a-1)-(8a^2+11a-4) = a^2-4a+3 = (a-1)(a-3)$. Then it's pretty easy to spot that $a-1$ and $a-3$ are not factors of either polynomial, so only $a+1$ is the GCD.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find tangent plane of level surface of $f(x, y, z)= \cos(x^2+2y+3z)$ at $(\frac{\pi}{2}, \pi, \pi)$ I was wondering if someone could verify my solution to this problem. As stated in the title, I was requested to find the equation for the tangent plane of the level surface of $f(x, y, z)= \cos(x^2+2y+3z)$ at $(\frac{\pi}{2}, \pi, \pi)$. Here is what I did.
$I$. Firstly I computed the partial derivatives of $f$, which make up the gradient $\triangledown f$ = $<f_x, f_y, f_z>$.
*
*$f_x(x, y, z) = -2x\sin(x^2+2y+3z)$
*$f_y(x, y, z) = -2\sin(x^2+2y+3z)$
*$f_z(x, y, z) = -3\sin(x^2+2y+3z)$
$II$. Using the fact that $\triangledown f(\frac{\pi}{2}, \pi, \pi) \perp \vec{r}'(t_0)$ where $\vec{r}(t)$ is any curve that touches $P=(\frac{\pi}{2}, \pi, \pi)$ when $t=t_0$, we know the plane tangent to the level surface of $f$ will have a normal vector $\vec{n}=\triangledown f(\frac{\pi}{2}, \pi, \pi)$. Therefore the plane is given by
$$-\pi\sin(\frac{\pi^2}{4} +5\pi)(x-\frac{\pi}{2})-2\sin(\frac{\pi^2}{4} +5\pi)(y-\pi) -3\sin(\frac{\pi^2}{4} +5\pi)(z-\pi) = 0$$
Let $u=\sin(\frac{\pi}{4}+5\pi)$. Then we have
$$\begin{align}
-\pi u(x-\frac{\pi}{2})-2u(y-\pi)-3u(z-\pi)&=0 \\ \pi ux+2uy+3uz&=u(\frac{\pi^2+10\pi}{2})
\end{align}$$
or rather
$$ \pi \sin(\frac{\pi^2}{4} +5\pi)x+2\sin(\frac{\pi^2}{4} +5\pi)y+3\sin(\frac{\pi^2}{4} +5\pi)z=\sin(\frac{\pi^2}{4} +5\pi)(\frac{\pi^2+10\pi}{2})$$
I wanted to know if my reasoning and my results are correct. I'm very new to multivariate calculus and am still trying to develop a grip of the basics. Thank you in advance.
| A bit of background:
*
*If $f: \mathbf{R}^{n}\to \mathbf{R}$ is differentiable function, $x_{0}\in {\rm Dom}(f)$ and $\nabla f(x_{0})\not=0$, then gradient vector $\nabla f(x_{0})$ is orthogonal to the level set $\{x\in \mathbf{R}^{n}: f(x)=c\}$ of passing through $x_{0}$. Then the tangent plane to surface $S$ defined by $f(x)=x_{0}$ at $x_{0}$ is just $\nabla f(x_{0})\cdot \vec{xx_{0}}=0$ for all $x\in \mathbf{R}^{n}$.
The level surface for $f$ is given by $f(x,y,z)=\cos(x^{2}+2y+3z)=k$ for $k\in {\rm Im}(f)$. Then,
*
*Indeed, $$\color{blue}{\nabla f\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{bmatrix}-2x\sin(x^{2}+2y+3z)\\-2\sin(x^{2}+2y+3z)\\-3\sin(x^{2}+2y+3z)\end{bmatrix}},$$
as you said.
*Then,
$$\color{blue}{\nabla f\begin{pmatrix} \pi/2\\ \pi\\\pi\end{pmatrix}=\begin{bmatrix}\pi\sin(\pi^{2}/4)\\2\sin(\pi^{2}/4)\\3\sin(\pi^{2}/4)\end{bmatrix}}$$
*Finally, the tangent plane of level surface $f$ at $(\pi/2,\pi,\pi)$ is given by
$$\begin{bmatrix}\pi\sin(\pi^{2}/4)\\2\sin(\pi^{2}/4)\\3\sin(\pi^{2}/4)\end{bmatrix}\cdot \begin{bmatrix}x-\frac{\pi}{2}\\ y-\pi\\z-\pi\end{bmatrix}=0,$$
i.e.,
$$\color{blue}{\pi\sin\left(\frac{\pi^{2}}{4}\right)\left(x-\frac{\pi}{2}\right)+2\sin\left(\frac{\pi^{2}}{4}\right)(y-\pi)+3\sin\left(\frac{\pi^{2}}{4}\right)(z-\pi)=0}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to Evaluate the Integral? $\int_{0}^{1}\frac{\ln\left( \frac{x+1}{2x^2} \right)}{\sqrt{x^2+2x}}dx=\frac{\pi^2}{2}$ I am trying to find a closed form for
$$
\int_{0}^{1}\ln\left(\frac{x + 1}{2x^{2}}\right)
{{\rm d}x \over \,\sqrt{\,{x^{2} + 2x}\,}\,}.
$$
I have done trig substitution and it results in
$$
\int_{0}^{1}\ln\left(\frac{x + 1}{2x^{2}}\right)
{{\rm d}x \over \,\sqrt{\,{x^{2} + 2x}\,}\,} =
\int_{0}^{\pi/3}\sec\left(\theta\right)
\ln\left(\frac{\sec\left(\theta\right)}
{2\left[\sec\left(\theta\right) - 1\right]^{\,2}} \right){\rm d}\theta
$$
which doesn't help.
By part integration with
$\displaystyle u = \ln\left(\frac{x + 1}{2x^{2}} \right)$, $\displaystyle\,\,{\rm d}v=\frac{\displaystyle\,\,{\rm d}x}{\,\sqrt{\,{x^{2} + 2x}\,}\,}$ also makes it more complicated.
I appreciate any help on this problem.
| Hyperbolic substitution is very convenient for this integral. We can substitute $x+1=\cosh t$
$$I=\int_0^1 \ln{\left(\frac{x+1}{2x^2}\right)}\frac{dx}{\sqrt{x^2+2x}}=\int_0^{\ln{(2+\sqrt 3)}}\ln\left(\frac{\cosh t}{2(\cosh t-1)^2}\right)dt$$$$=\int_0^{\ln{(2+\sqrt 3)}}\ln\left(\frac{e^{-t}(1+e^{-2t})}{(1-e^{-t})^4}\right)dt$$
When we split the integral we get
$$I=-\frac{\ln^2(2+\sqrt 3)}{2}+\int_0^{\ln{(2+\sqrt 3)}}\ln (1+e^{-2t})dt-4\int_0^{\ln{(2+\sqrt 3)}}\ln(1-e^{-t})dt$$
For the first integral we can apply the substitution $u=-e^{-2t}$, and for the second integral we can use $v=e^{-t}$
This will allow us to use the dilogarithm.
$$I=-\frac{\ln^2(2+\sqrt 3)}{2}+\frac{1}{2}\int_{-1}^{-7+4\sqrt3}\frac{-\ln (1-u)}{u}du-4\int_1^{2-\sqrt 3}\frac{-\ln(1-v)}{v}dv$$
Using $\int \frac{-\ln(1-x)}{x}dx=\operatorname{Li}_2(x)+C$ we get
$$I=\frac{17\pi^2}{24}-\frac{\ln^2(2+\sqrt3)}{2}+\frac{1}{2}\operatorname{Li}_2(-7+4\sqrt3)-4\operatorname{Li}_2(2-\sqrt3)$$
ATTEMPT 2
Refer to C. Leibovici's answer, apply partial fractions and integration by parts
$$I=8\int_0^{\frac{1}{\sqrt3}}\frac{\operatorname{artanh} x}{x(1-x^4)}dx=8\int_0^{\frac{1}{\sqrt3}}\frac{\operatorname{artanh}x}{x}dx+2\int_0^{\frac{1}{\sqrt3}}\frac{4x^3\operatorname{artanh}x}{1-x^4}dx$$$$=8X_2\left(\frac{1}{\sqrt3}\right)+\ln(2-\sqrt3)\ln\left(\frac{8}{9}\right)+2\int_0^\frac{1}{\sqrt3} \frac{\ln(1-x^4)}{1-x^2}dx$$
$X_2(t)$ is the Legendre chi function
Using $x^4-1=(x+1)(x-1)(x+i)(x-i)$ in conjunction with logarithm properties and partial fractions, we can decompose the integral into 8 integrals of the form $\int \frac{\ln(x+a)}{x+b}dx$. 2 of them, $\int \frac{\ln(1+x)}{1+x}dx$ and $\int \frac{\ln(1+x)}{1+x}dx$ can be evaluated without dilogarithms. The other 6 each give us 2 dilogarithmic terms. This gives us 12 dilogarithms to simplify in total, and it might work
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "35",
"answer_count": 5,
"answer_id": 0
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how to do this using properties of definite integrals? $$\int_1^2 ln_2(x^3+1)dx$$
which is the same as
$\int_1^2 ln_2((3)-(x^3+1))dx$
which gives $\int_1^2 ln_2(2-x^3)\ dx$
which does not help.
Or integrate this by parts, but that's really tedious.
| Use $x^3 + 1 = (x+1)(1 - x + x^2)$.
$$\int\limits_{x=1}^2 \log_2 (x^3 + 1)\ dx = \int\limits_{x=1}^2 \log_2 (x + 1)\ dx +\int\limits_{x=1}^2 \log_2 (1 - x + x^2)\ dx$$
Then use (from integral tables): $\int \log_2 (x+1)\ dx = (x+1) \log (x+1)-x$
and
$\int \log_2 (1 - x + x^2)\ dx = \left(x-\frac{1}{2}\right) \log \left(x^2-x+1\right)-2 x+\sqrt{3} \tan ^{-1}\left(\frac{2
x-1}{\sqrt{3}}\right)$
sum, to find the full integral is:
$$\frac{1}{6} \left(\sqrt{3} \pi +27 \log_2 (3)-6 (3+\log_2 (4))\right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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For what $x$ the expression of the equation has the same remainder after division. For what $x$ does the expression $(50x+7)(3x+11)$ have the same remainder as $(26x − 86)x$ when divided by $111$?
$(50x+7)(3x+11) \equiv y \mod 111$
$(26x − 86)x \equiv y \mod 111$
$(50x+7)(3x+11) - (26x − 86)x \equiv 0 \mod 111 $
$124x^2+657x+77 \equiv 0 \mod 111$
$111 = 3 \times 37$
There is a system
*
*$124x^2+657x+77 \equiv 0 \mod 3$
*$124x^2+657x+77 \equiv 0 \mod 37$
First:
$(11x+12)^2\equiv 1 \mod 3$
And $x\equiv -1 \mod 3$
But I don't know how to solve the second equation and what to do next.
How to combine these two equations and get at what $x$ these expressions have the same remainder from division.
| First you have $x= 3y\pm 1$ for some integer $y$.
You can reduce the coeficents mod $37$ so \begin{align}-20x^2+30x+40 &=_{37} -5(4x^2-6x-8)\\
&=_{37} -5(4x^2+68x-8) \\&=_{37}-5 (4x^2+68x +289)+5\\
&=_{37}-5 (2x+17)^2+5\\
\end{align}
So $$(2x-20)^2 =_{37} 1 \implies 2x-20 =_{37}\pm1 $$
So you have 4 possibilites.
*
*First $x=3y+1$ and $2x-20 =_{37} 1$ and thus $ 6y =_{37} 19$ so $$y =_{37}-6\cdot 19 =_{37}-3$$ and finaly $y= 37t-3$ for some integer $t$ which gives $$x=111t-8$$ in first case.
| {
"language": "en",
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Integral Involving Harmonic Numbers: $\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx$ (Motivation) In an attempt to answer this question, I got stuck on evaluating a certain integral. I have made up the following conjecture:
$$\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx = \frac{\pi^{2}}{4}+\frac{3}{2}\ln\left(2\right)\ln\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right).$$
(Attempt) Let $H_n$ denote the n-th harmonic number $\displaystyle H_{n}=\sum_{k=1}^{n}\frac{1}{k}$. I will warn this process gets ugly, but it is the best I have so far. Expanding the integrand as a series, we get:
$$
\eqalign{
\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx &= \sum_{n=0}^{\infty}\int_{\sqrt{3}}^{\infty}\frac{4\ln\left(x\right)-H_{n}}{x^{4n+2}}dx+\sum_{n=0}^{\infty}\int_{\sqrt{3}}^{\infty}\frac{4\ln\left(x\right)-H_{n}}{x^{4n+4}}dx, \cr
}
$$
which simplifies down to
$$\sum_{n=0}^{\infty}\left(\frac{H_{n}\sqrt{3}^{-4n-1}}{-4n-1}-\frac{4\ln\left(\sqrt{3}\right)\sqrt{3}^{-4n-1}}{-4n-1}+\frac{4\sqrt{3}^{-4n-1}}{\left(-4n-1\right)^{2}}\right)+\sum_{n=0}^{\infty}\left(\frac{H_{n}\sqrt{3}^{-4n-3}}{-4n-3}-\frac{4\ln\left(\sqrt{3}\right)\sqrt{3}^{-4n-3}}{-4n-3}+\frac{4\sqrt{3}^{-4n-3}}{\left(-4n-3\right)^{2}}\right).$$
Since both series converge, we can split up the first series like
$$-\sum_{n=0}^{\infty}\frac{H_{n}\sqrt{3}^{-4n-1}}{4n+1}+4\ln\left(\sqrt{3}\right)\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-1}}{4n+1}+4\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-1}}{\left(4n+1\right)^{2}}$$
and the second series like
$$-\sum_{n=0}^{\infty}\frac{H_{n}\sqrt{3}^{-4n-3}}{4n+3}+4\ln\left(\sqrt{3}\right)\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-3}}{4n+3}+4\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-3}}{\left(4n+3\right)^{2}}.$$
Next, I found that
$$\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-1}}{4n+1}=\frac{\pi}{12}+\frac{1}{2}\operatorname{arctanh}\left(\frac{1}{\sqrt{3}}\right)$$
and
$$\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-3}}{4n+3}=\frac{1}{2}\operatorname{arctanh}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{12}.$$
However, after trying for a while, I am out of ideas for evaluating the other sums.
I realize I skipped a lot of steps, but that is because I don't want this question to be too long. So for your convenience, I put all of these into Desmos, so I believe my process is correct so far based on numerical approximations.
(Question) Does anyone have an idea of how to evaluate the integral in question, or how to evaluate the sums I am stuck on? Any hints and ideas are appreciated.
(Miscellaneous) Here are some other ideas I have:
$$
\eqalign{
\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx &= \int_{\operatorname{arcsec}\left(\sqrt{3}\right)}^{\frac{\pi}{2}}\frac{\ln\left(\left(\sec x\right)^{4}-1\right)}{\sec^{2}x-1}\sec\left(x\right)\tan\left(x\right)dx \cr
\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx &= \int_{0}^{\infty}\frac{\ln\left(\left(x+\sqrt{3}\right)^{4}-1\right)}{\left(x+\sqrt{3}\right)^{2}-1}dx.
}
$$
Maybe I could construct a keyhole contour for the last integral?
| I will answer my own question by proving
$$\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx = \frac{\pi^{2}}{4}+\frac{3}{2}\ln\left(2\right)\ln\left(2+\sqrt{3}\right).$$
Let the integral in question equal $I$. Letting $x \to \dfrac{1-x}{1+x}$, we rewrite the integral as
$$I = \int_{\sqrt{3}-2}^{-1}\frac{\ln\left(\left(\frac{1-x}{1+x}\right)^{4}-1\right)}{\left(\frac{1-x}{1+x}\right)^{2}-1}\left(\frac{-2}{\left(1+x\right)^{2}}\right)dx.$$
Doing some simplifications, we get
$$I = \frac{3\ln\left(2\right)}{2}\int_{\sqrt{3}-2}^{-1}\frac{1}{x}dx+\frac{1}{2}\int_{\sqrt{3}-2}^{-1}\frac{\ln\left(-x\right)}{x}dx+\frac{1}{2}\int_{\sqrt{3}-2}^{-1}\frac{\ln\left(1+x^{2}\right)}{x}dx-\frac{1}{2}\int_{\sqrt{3}-2}^{-1}\frac{\ln\left(\left(x+1\right)^{4}\right)}{x}dx.$$
Trivially, we can solve the first two integrals. For the last two, we can use the dilogarithm definition. Thus,
$$I = \frac{3\ln\left(2\right)}{2}\left(-\ln\left(2-\sqrt{3}\right)\right)+\frac{1}{2}\left(-\frac{1}{2}\ln^{2}\left(2-\sqrt{3}\right)\right)+\frac{1}{2}\left(\frac{1}{24}\left(12\operatorname{Li}_2 \left(4\sqrt{3}-7\right)+\pi^{2}\right)\right)-\frac{1}{2}\left(4\left(\operatorname{Li}_2 \left(2-\sqrt{3}\right)-\frac{\pi^{2}}{6}\right)\right)$$
which simplifies down to
$$I = \ln\left(\frac{1}{2-\sqrt{3}}\right)\left(\frac{1}{4}\ln\left(2-\sqrt{3}\right)+\frac{3}{2}\ln\left(2\right)\right)+\frac{17\pi^{2}}{48}+\frac{1}{4}\operatorname{Li}_2 \left(-\left(2-\sqrt{3}\right)^{2}\right)-2\operatorname{Li}_2 \left(2-\sqrt{3}\right)$$
From this answer, we can prove that
$$\frac{1}{4}\operatorname{Li}_2 \left(-\left(2-\sqrt{3}\right)^{2}\right)-2\operatorname{Li}_2 \left(2-\sqrt{3}\right) = \frac{\ln^{2}\left(2-\sqrt{3}\right)}{4}-\frac{5\pi^{2}}{48}.$$
Combining the results, we conclude that the integral $I$ is
$$\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx = \frac{\pi^{2}}{4}+\frac{3}{2}\ln\left(2\right)\ln\left(2+\sqrt{3}\right).$$
| {
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"answer_count": 3,
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limit of a geometric-type series I came across this series while studying probability theory.
Let $$T(x)=\sum_{n=1}^\infty x^n \left(\frac{1}{n}-\frac{1}{n+1}\right),$$
where $T(x)$ converges for some range of $x$.
My question is how to prove that $T(x)=1+\frac{1-x}{x}\log(1-x)$ for all $|x|<1$.
The following is my attempt.
It is tempting to differentiate $T$ w.r.t. $x$, which I cannot justify because $\sum_{n=1}^\infty x^{n-1} \frac{1}{n+1}$ is not uniformly convergent in $(-1,1)$:
$$T'(x)=\sum_{n=1}^\infty nx^{n-1} \left(\frac{1}{n}-\frac{1}{n+1}\right)=\sum_{n=1}^\infty x^{n-1} \frac{1}{n+1}.$$
It is also tempting to do the followings:
$$x^2 T'(x)=\sum_{n=1}^\infty x^{n+1} \frac{1}{n+1}:=P(x)$$
and differentiate $P(x)$ to get:
$$P'(x)=\sum_{n=1}^\infty x^n=\frac{x}{1-x}.$$
So, $P(x)=\int P'(x)dx=-x-\log(|x-1|)+C$, where $C$ is a constant, cf. here.
Then we have $T'(x)=-\frac{1}{x}-\frac{\log(|x-1|)}{x^2}+\frac{C}{x^2}$. But I don't know how to integrate the second term to recover $T(x)$.
Another crucial problem with my attempt is that, every step is formal and we lost track of the radius of convergence of $x$. Kind of being stuck here.
Thanks for any help.
| How about using the known series for on(1-x) and elementary series rearrangements?
$\ln(1-x) = -\sum_{n=1}^\infty\frac{x^n}{n}$ for $-1<x\le 1$
so
\begin{align*}
T(x)
&=
\sum_{n=1}^\infty x^n\left(\frac{1}{n} - \frac{1}{n+1}\right)
\\&=
\sum_{n=1}^\infty \frac{x^n}{n} - \sum_{n=1}^\infty\frac{x^n}{n+1}
\\&=
\sum_{n=1}^\infty \frac{x^n}{n} - \frac{1}{x}\sum_{n=2}^\infty\frac{x^n}{n}
\\&=
\sum_{n=1}^\infty \frac{x^n}{n} - \frac{1}{x}\left(-x+\sum_{n=1}^\infty\frac{x^n}{n}\right)
\\&=
-\ln(1-x) + 1 + \frac{1}{x}\ln(1-x)
\\&=
1 + \frac{1-x}{x}\ln(1-x)
\end{align*}
All steps are valid for $-1<x<1$, except possibly x=0 (but either think of it a removable singularity or check the original equation at x=0).
| {
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Evaluate $\lim\limits_{n \to \infty} \sqrt[n^2]{1! + 2! + 3! + \dots + n!}$ $$\lim_{n \to \infty} \sqrt[n^2]{1! + 2! + 3! + \dots + n!}$$
I can try at least to evaluate it from bottom:
$$1 < \sqrt[n^2]{n^2} < \sqrt[n^2]{n!} < \sqrt[n^2]{1! + 2! + \dots + n!}$$
Well, it doesn't say anything. Ok, gonna try evaluation from top.
$$\sqrt[n^2]{1! + 2! + \cdots + n!} < \sqrt[n^2]{n \times n!}$$
More than that,
$$\lim_{n \to \infty} \frac{n!}{1! + 2! + \cdots + n!} = 1$$
but this equation is always less than one. So instead of writing $\sqrt[n^2]{n \times n!}$ I can try fixed coefficient $1 < a < n$. It will work for $a = 2$, $a =1.5$, but I'm not sure though about $a = (1 + \frac{1}{n})$
$$\sqrt[n^2]{1! + 2! + \cdots + n!} < \sqrt[n^2]{a \times n!}$$
Just to get broad idea how it looks I'll assume that $a = 1$ and I'll try to use Stirling's approximation
$$\sqrt[n^2]{1 \times n!} \approx \left((2\pi n)^{\frac{1}{2}}\left(\frac{n}{e}\right)^n \right)^\frac{1}{n^2} = (2 \pi n)^{\frac{1}{2n^2}} \times \left(\frac{n}{e}\right)^{\frac{1}{n}}$$
$\lim_{n \to \infty} \sqrt[n]{n} = 1$, $2n^2 > 2 \pi n$, $\frac{n}{e} < n$ (for big numbers) mean, that
$$\lim_{n \to \infty}(2 \pi n)^{\frac{1}{2n^2}} \times (\frac{n}{e})^{\frac{1}{n}} \approx 1$$
So I can guess, that
$$\lim_{n \to \infty} \sqrt[n^2]{1! + 2! + 3! + \cdots + n!} = 1$$
But I sincerely doubt that what I wrote here is barely a solution. Would you mind helping me? Hope I didn't make huge mistakes
| Notice that :
$$1 \leq \sqrt[n^2]{1! + \cdots + n!} \leq \sqrt[n^2]{n! + \cdots + n!} = \sqrt[n^2]{n \, n!} \leq \sqrt[n^2]{n n^n} = \sqrt[n^2]{n^{n + 1}} \to 1$$
because :
$$\ln \sqrt[n^2]{n^{n + 1}} = \dfrac{1}{n^2} \ln n^{n + 1} = \dfrac{n + 1}{n} \, \dfrac{\ln n}{n} \to 1 \times 0 = 0$$
| {
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How to attempt power series questions ? I'm confused with this topic
Use the power series representation of $\frac{1}{1-x}$ to obtain the power series representation of $\frac{2x}{4-3x}$ and the corresponding values of $x$.
Firstly, I tried to make $\frac{2x}{4-3x}$ in the form of $\frac{1}{1-x}$
$\frac{x}{2} \frac{1}{(1-3x/4)} = \frac{x}{2} \sum_{n=0}^{\infty} (\frac{3x}{4})^n $ for $-1 < 3x/4 < 1$
I know that the power series representation of $x=a$ is $\sum_{n=0}^{\infty} C_n (x-a)^n $ where a is the centre. Am I supposed to determine what is $a$?
What am I supposed to do next? I do not understand the question and what formula am I supposed to be following. I just learnt about power series.
The answer is $\sum_{n=0}^{\infty} (\frac{1}{2}) (\frac{3}{4})^n x^{n+1}$ for $-4/3 < x < 4/3$
What is being done above?
| Maybe it is not necessary to note that:
Another way is Maclaurin series. $f(x)=\frac{2x}{4-3x}=-\frac{2}{3}+\frac{8}{3}\frac{1}{4-3x}$. With this form we can easily compute derivatives: $n$-th derivative is $f^{(n)}(x)=\frac{8}{3}3^{n}n!(4-3x)^{-n-1}$ and $f^{(n)}(0)=\frac{1}{2}\frac{3^{n-1}}{4^{n-1}}n!$ for $n>0$.
Noting that $f(0)=0$, $f(x)=\sum_{n=1}^{\infty}\frac{1}{2}\frac{3^{n-1}}{4^{n-1}}{n!}\frac{x^n}{n!}=\sum_{n=0}^{\infty}\frac{1}{2}\frac{3^{n}}{4^{n}}x^{n+1}=\frac{x}{2}\sum_{n=0}^{\infty}\frac{3^{n}}{4^{n}}x^{n}.$
For the interval of convergence, we apply the ratio test which gives $|\frac{3}{4}x|<1$ and hence $-\frac{4}{3}<x<\frac{4}{3}.$
Sometimes at boundary points, here they are $x=\pm\frac{4}{3}$, the series may be convergent. But it is not the case for this series.
| {
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Is there any better way of finding the required value If $$z=\cos\theta+i\sin\theta$$ find the value of $$\frac{1+z}{1-z}$$
The solution that I have is this
$$z=\cos\theta+i\sin\theta \implies$$
$$\frac{1+z}{1-z}=\frac{1+(\cos\theta+i\sin\theta)}{1-(\cos\theta+i\sin\theta)}=\frac{(1+\cos\theta)+i\sin\theta}{(1-\cos\theta)+i\sin\theta}$$
$$=\frac{2\cos^2\frac{\theta}{2}+i\:\:2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}-i\:\:2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}=\frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}\cdot \frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{\sin\frac{\theta}{2}-i\cos\frac{\theta}{2}}$$
$$=\cot\frac{\theta}{2}\cdot\frac{\color{red}{i}}{\color{red}{i}}×\frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{\sin\frac{\theta}{2}-i\cos\frac{\theta}{2}}$$
$$=\color{red}{i}\cot\frac{\theta}{2}\cdot\frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{\color{red}{i}\sin\frac{\theta}{2}-\color{red}{i}i\cos\frac{\theta}{2}}$$
$$=\color{red}{i}\cot\frac{\theta}{2}\cdot\frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}+\color{red}{i}\sin\frac{\theta}{2}}$$
$$=\color{red}{i}\cot\frac{\theta}{2}$$
Now how on earth one will imagine the steps written in red. I am looking for an easy and a logical answer to this question. The solution that I have is impractical as you all can see. I tried it doing by $e^{i\theta}$ but no good happen.
Any help is greatly appreciated.
| Observe that $$\dfrac{\cos t+i\sin t}{\cos t-i\sin t}=\cos2t+i\sin2t$$
If $\dfrac z1=\cos2t+i\sin2t=\dfrac{\cos t+i\sin t}{\cos t-i\sin t}$
Using Componendo and Dividendo
$$\dfrac{1+z}{1-z}=\cdots=?$$
| {
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"question_score": "4",
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"answer_id": 1
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Verify that $\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})\le\frac{1}{8}$ in a general triangle $\triangle ABC$ So, this problem is inspired by a contest preparation problem I saw back in Japan, and it is as follows:
In a general triangle $\triangle ABC$, show that $\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})\le\frac{1}{8}$
Now, while I still haven't figured out a geometric interpretation of this inequality, here is my attempt to prove this:
Recall that:
$$\sin^2(\frac{A}{2})=\frac{(1-\cos(A))}{2}$$
$$\sin^2(\frac{A}{2})=\frac{1}{2}(1-\frac{b^2+c^2-a^2}{2bc})$$
$$\sin^2(\frac{A}{2})=\frac{1}{2}(\frac{a^2-b^2-c^2+2bc}{2bc})$$
$$\sin^2(\frac{A}{2})=\frac{a^2-(b-c)^2}{4bc}$$
Now, obviously $\frac{a^2-(b-c)^2}{4bc} \le \frac{a^2}{4bc}$, therefore:
$$\sin^2(\frac{A}{2}) \le \frac{a^2}{4bc}$$
$$\sin(\frac{A}{2}) \le \frac{a}{2\sqrt{bc}}$$
This can be done for $\sin(\frac{A}{2}), \sin(\frac{B}{2})$ and $\sin(\frac{C}{2})$.
$$\sin(\frac{A}{2}) \le \frac{a}{2\sqrt{bc}}$$
$$\sin(\frac{B}{2}) \le \frac{b}{2\sqrt{ac}}$$
$$\sin(\frac{C}{2}) \le \frac{c}{2\sqrt{ab}}$$
Therefore:
$$\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2}) \le (\frac{a}{2\sqrt{bc}})(\frac{b}{2\sqrt{ac}})(\frac{c}{2\sqrt{ab}})$$
$$\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2}) \le \frac{abc}{8abc}$$
$$\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2}) \le \frac{1}{8}$$
However I'm not sure if this is correct or if it is, I don't think this brute force approach is good. Are there any better options to prove this inequality? Please share your answers!
| Being angles of the triangles, the sines are all positive so we can use AM-GM inequallity:
$$
\left(\frac{\sin\left(\frac{A}{2}\right)+\sin\left(\frac{B}{2}\right)+\sin\left(\frac{C}{2}\right)}{3}\right)^{3}
\geq
\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)
$$
The sine function is convex in the range of $[0,\frac{\pi}{2}]$ so we can use the convex inequality:
$$
\sin\left(\frac{\frac{A}{2}+\frac{B}{2}+\frac{C}{2}}{3}\right)
\geq
\frac{\sin\left(\frac{A}{2}\right)+\sin\left(\frac{B}{2}\right)+\sin\left(\frac{C}{2}\right)}{3}
$$
Then simply combine them and use $A+B+C=\pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4583104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Right triangle $\triangle ABC$, $D$ lies on $AB$, inscribed in a circle of radius $9$. Find the measure of $CD$ I saw this mathematical puzzle on an Instagram post today. As title suggests, we have a right angled triangle inscribed in a circle with radius $D$ and some angles. The goal is to find the length of $CD$. I'll share my approach as an answer below. I'm not quite sure if my answer is correct, so please feel free to point out any faults in my approach and/or post your own approaches too!
| Law of Tangents, let $t = \tan α$
$\displaystyle \frac{x+6}{x-6} =
\frac{\tan \frac{(3α-90°)+(90°-α)}{2}}{\tan \frac{(3α-90°)-(90°-α)}{2}}
= \frac{\tan α}{\tan (2α-90°)} = -(\tan α)(\tan 2α) = \frac{-2t^2}{1-t^2}$
$\displaystyle \frac{x+12}{x-12} =
\frac{\tan \frac{(180°-3α)+(α)}{2}}{\tan \frac{(180°-3α)-(α)}{2}}
= \frac{\tan (90°-α)}{\tan (90°-2α)} = \frac{\tan 2α}{\tan α} = \frac{2}{1-t^2}$
$→ \displaystyle \frac{x+6}{x-6} + \frac{x+12}{x-12} = 2$
$→ x^2-6×12 = (x-6)(x-12) $
$\displaystyle → x = \frac{6×12}{(6+12)÷2} = \frac{72}{9} = 8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
given N and M count the number of solutions to some diophantine equations, prove that $N>M.$
Let $N$ be the number of integer solutions to the equation $x^3 - y^3 = z^4-w^4$ with the property that $0\leq x,y,z,w\leq 2022^{2022}$. Let $M$ be the number of integer solutions to the equation $x^3-y^3 = z^4-w^4+1$ with $0\leq x,y,z,w\leq 2022^{2022}$. Prove that $N>M.$
This is problem 8 from this problem set
Let $a_i$ denote the number of solutions to the equation $s^3 + t^4 = i$ where $0\leq s,t\leq 2022^{2022}$. The maximum value of $i$ for which $a_i > 0$ is $A := 2022^{6066} + 2022^{8088}.$ Let $f(x,y) = x^3 + y^4$ for real numbers x and y. $N$ counts the number of integer solutions to the equation $x^3 + w^4 = y^3 + z^4$ where $0\leq x,y,z,w\leq 2022^{2022}.$ For each $0\leq i\leq A,$ there are $a_i$ pairs $(x,y)$ with $f(x,y)=i$. For each pair $(x,w)$, there are $a_{x^3 + w^4}$ pairs $(y,z)$ so that $f(y,z) = f(x,w)$. Hence $N$ must equal $\sum_{i=0}^A a_i^2$ (so $a_i^2$ is the number of solutions $x,y,z,w$ to the equation with $f(x,w)=f(y,z)=i$). On the other hand, for a pair $(y,z)$ with $f(y,z) = i$, there are $a_{i+1} $ pairs $(x,w)$ with $f(x,w)=i+1$. Thus, $M$ equals $\sum_{i=0}^A a_i a_{i+1}$. Hence to show that $N > M,$ it suffices to show that $\sum_{i=0}^A a_i(a_i-a_{i+1}) > 0$. But I'm not sure how to show the latter inequality. It is clearly not always true that $a_i \ge a_{i+1},$ as this does not hold for $i = 0.$ $a_2 = 1, a_i = 0$ for $3\leq i\leq 7, a_8 = 1, a_9 = 1, a_i = 0$ for $10\leq i\leq 15, a_{16} = 1, a_{17} = 1, a_i = 0$ for $18\leq i\leq 23, a_{24} = 1, a_{25} = 0, a_{26}=0,a_{27}=1, a_{28} = 1.$ There's no noticeable pattern.
| Continuing from what you've provided, first note that $a_0 = 1$ and, as you stated about how $A$ is defined, we have $a_{A+1}=0$. Next,
similar to what's done in the related AoPS thread Number of solutions of two systems, we get
$$\begin{equation}\begin{aligned}
2\sum_{i=0}^{A}a_{i}(a_{i}-a_{i+1}) & = \color{blue}{\sum_{i=0}^{A}a_{i}^{2}} + \sum_{i=0}^{A}a_{i}^{2} - 2\sum_{i=0}^{A}a_{i}a_{i+1} \\
& = \color{blue}{a_{0}^{2} - a_{A+1}^2 + \sum_{i=1}^{A+1}a_{i}^{2}} + \sum_{i=0}^{A}a_{i}^{2} - \sum_{i=0}^{A}2a_{i}a_{i+1} \\
& = \color{blue}{a_{0}^{2} + (\sum_{i=0}^{A}a_{i+1}^{2}} - \sum_{i=0}^{A}2a_{i}a_{i+1} + \sum_{i=0}^{A}a_{i}^{2}) \\
& = a_{0}^{2} + \sum_{i=0}^{A}(a_{i+1}^{2} - 2a_{i}a_{i+1} + a_{i}^{2}) \\
& = a_{0}^{2} + \sum_{i=0}^{A}(a_{i+1} - a_{i})^{2} \\
& \gt 0
\end{aligned}\end{equation}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How many ways to place plus and minus signs in front of numbers from 0 to 12 so that the sum is divisible by 5? This question is from an old NIMO contest.
Find the number of ways a series of + and − signs can be inserted between the numbers 0, 1, 2, · · ·, 12 such that the value of the resulting expression is divisible by 5.
A computer program seems to indicate that the answer is 816.
Approach: Reduce the problem to mod 5. There are three copies of 0,1,2 and two copies of 3,4. Now suppose there are exactly $x_i$ plus signs for copy $i$. We see that for the sum to be divisible by 5, we must have $\sum ix_i =$ 4 mod 5. Now I was planning to crunch it out or use generating functions, but it looks complicated.
Suggestions?
| It does matter what we put in front of the $0,5,10$. Of the $2^3\pm$ combinations you can do for those the result will have the same divisibility as any other. So It suffices to solve for $1,2,3,4,6,7,8,9,11,12$ and multiply the result by $8$.
Consider $A_1 = \pm 1 \pm 2\pm 3\pm 4$ and $A_2=\pm 6\pm 7\pm 8 \pm 9$ and $B = \pm 11\pm 12$. $B$ can never be divisible by $5$ but of the four values it can have it can be by divisible by $1,2,3$ or $4$ meaning $A_1 + A_2$ must be $\equiv -B\pmod 5$.
Going through the 16 possible values of $A_1$ (and equivalently $A_2$) There are $4$ that are congruent to $0$ $(\pm(1+4)\pm(2+3)$; $3$ that are congruent to $1$ (basically $-7+3$); $3$ that are congruent to $2$ ($6-4=1+2+3-4; 8-2=1-2+3+4;9-1=-1+2+3+4$; $3$ that are congruent to $3\equiv -2$ (symmetry; and $3$ that are congruent to $4$.
Okay: If $B\equiv 1 \pmod 5$ (that is $-11+12$) we must have $A_1+A_2 \equiv 4$ so we must have $A_1,A_2\equiv (0,4),(1,3), (2,2)$ there are $4\times 3$ ways to do $(0,4)$ and $3\times 3$ ways to do $(1,3)$ or $(2,2)$. so there are $12 + 9+9=30$ ways to do that.
If $B \equiv 2 \pmod 5$ (that is $-11-12$) we must have $A_1+A_2\equiv 3$ so we must have $A_1,A_2 \equiv (0,3), (1,2),(4,4)$ and there are $4\times 3=12$ and $3\times 3$ and $3\times 3$ respectively to do that. So there are $30$ ways.
By symmetry the some is true for $B\equiv 3,4\pmod 5$.
So there should by $8\times 4\times 30 = 960$ ways.
I don't know if I or you have an error (probably I do) but that is how I solved it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How to solve $\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n$ Question:
$$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=?$$
My work:
$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=\lim_{n\to\infty}\left(\dfrac{n^2(1+5/n+3/n^2}{n^2(1+1/n+2/n^2)}\right)^n=\lim_{n\to\infty}\left(\dfrac{1+5/n+3/n^2}{1+1/n+2/n^2}\right)^n$
$\log L=n\log\left(\dfrac{1+5/n+3/n^2}{1+1/n+2/n^2}\right)$
Is this equal to 0? Then the answer would be $e^0=1$.
The answer was given as $e^4$ and I have no idea how to get to that.
| Another way
$$\dfrac{n^2+5n+3}{n^2+n+2}=\dfrac{n^2+5n+3}{n^2\left(1+\dfrac{1}{n}+\dfrac{2}{n^2}\right)}\sim \left(1+\dfrac{5}{n}+\dfrac{3}{n^2}\right)\left(1-\dfrac{1}{n}-\dfrac{2}{n^2}\right)\sim 1+\dfrac{4}{n}$$
$$\lim_{n\to\infty}\left( 1+\dfrac{4}{n}\right)^n=\lim_{t\to\infty}\left( 1+\dfrac{4}{4t}\right)^{4t}=\lim_{t\to\infty}\left( 1+\dfrac{1}{t}\right)^{4t}=e^4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Fair Value of a Dice Game 1 die. Up to 3 rolls. Your winnings are equal to the value of a single roll. You can stop after roll 1 or roll 2. If you proceed to the next roll (i.e the 2nd or 3rd), you forfeit the previous value (no memory). Therefore, if you roll 3 times, the value of the 3rd roll is what you win.
What is the fair price of this game?
Strategy
Since the expected value of a single roll is 3.5, I will stop on any roll if I get a 4, 5, or 6.
If Fair Price = Expected Value
And Expected Value = P(i_th roll) x E(i_th roll) i=1 to 3
Then Fair Value = P(1 roll) E(1st roll) + P(2 rolls) E(2nd roll) + P(3 rolls) * E(3rd roll)
E(i_th roll) = 3.5 for any roll
P(only 1 roll) = 1/2 :: must roll 4|5|6
P(only 2 rolls) = 1/2 * 1/2 :: must roll 1|2|3 then 4|5|6
P(3 rolls) = 1/2 * 1/2 :: must roll 1|2|3 then 1|2|3
$\frac{1}{2} * 3.5 + \frac{1}{4} * 3.5 + \frac{1}{4} * 3.5 = $
$\frac{1}{2}*\frac{7}{2} + \frac{1}{4}*\frac{7}{2} + \frac{1}{4}*\frac{7}{2} = $
$\frac{7}{4} + \frac{7}{8} + \frac{7}{8} = $
$\frac{14}{8} + \frac{7}{8} + \frac{7}{8} = $
$\frac{28}{8} = 3.50$
Whn trying to solve this before, I think I made a mistake and end up with something like $4\frac{5}{8} = 4.625$
Should 3.5 be the answer, or should the 1st and 2nd rolls be "wieghted" by what values you must roll to move on (1,2 or 3) instead of the general 3.5 expected value for a single roll?
| There is $\frac{1}{2}$ chance that we get $4, 5$ or $6$ in the first throw and $\frac{1}{2}$ chance that we get $1, 2$ or $3$.
Thus for exactly one throw , the expected value is $$\frac{1}{2} \times \frac{4+5+6}{3}=\frac{1}{2} \times 5$$
For exactly two throws , we must get $ \{1, 2 \text { }\mathrm{or} \text { }3 \}$ in the first throw and $ \{4, 5 \text { }\mathrm{or} \text { }6 \}$ in the $2$nd throw.
The expected value for exactly $2$ throws is: $$\frac{1}{2} \times 5$$
For exactly 3 throws , we must get $ \{1, 2 \text { }\mathrm{or} \text { }3 \}$ in the first $2$ throws and accept whatever we get in the $3$rd throws.
So the expected value for exactly $3$ throws is $$\frac{1}{4} \times 3.5$$
Thus the final expected value is
$$\frac{1}{2} \times 5 + \frac{1}{4} \times 5 + \frac{1}{4} \times 3.5 = 4.625$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4589195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Can we evaluate $\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x$ without expanding $\ln(1-x)$? For $|x|<1,$ we have
$$
\begin{aligned}
& \frac{1}{1-x}=\sum_{k=0}^{\infty} x^k \quad \Rightarrow \quad \ln (1-x)=-\sum_{k=0}^{\infty} \frac{x^{k+1}}{k+1}
\end{aligned}
$$
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x)}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k dx \\
& =-\sum_{k=0}^{\infty} \frac{1}{(k+1)^2} \\
& =- \zeta(2) \\
& =-\frac{\pi^2}{6}
\end{aligned}
$$
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln xdx \\
& =\sum_{k=0}^{\infty} \frac{1}{k+1}\cdot\frac{1}{(k+1)^2} \\
& =\zeta(3) \\
\end{aligned}
$$
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^2 x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln ^2 xdx \\
& =-\sum_{k=0}^{\infty} \frac{1}{k+1} \cdot \frac{2}{(k+1)^3} \\
& =-2 \zeta(4) \\
& =-\frac{\pi^4}{45}
\end{aligned}
$$
In a similar way, I dare guess that
$$\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x =(-1)^{n+1}\Gamma(n)\zeta(n+2),$$
where $n$ is a non-negative real number.
Proof:
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln ^n xdx \\
\end{aligned}
$$
Letting $y=-(k+1)\ln x $ transforms the last integral into a Gamma function as
$$
\begin{aligned}
\int_0^1 x^k \ln ^n x d x & =\int_{\infty}^0 e^{-\frac{k}{k+1}}\left(-\frac{y}{k+1}\right)^n\left(-\frac{1}{k+1} e^{-\frac{y}{k+1}} d y\right) \\
& =\frac{(-1)^n}{(k+1)^{n+1}} \int_0^{\infty} e^{-y} y^n d y \\
& =\frac{(-1)^n \Gamma(n+1)}{(k+1)^{n+1}}
\end{aligned}
$$
Now we can conclude that
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x & =(-1)^{n+1} \Gamma(n+1) \sum_{k=0}^{\infty} \frac{1}{(k+1)^{n+2}} \\
& =(-1)^{n+1} \Gamma(n+1)\zeta(n+2)
\end{aligned}
$$
Can we evaluate $\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x$ without expanding $\ln (1-x)$?
Your comments and alternative methods are highly appreciated?
| Without expanding using series at all (taking for granted any properties of special functions whose derivations require series manipulation):
$$\begin{align*}
I &= \int_0^1 \frac{\log(1-x) \log^n(x)}x \, dx \\[1ex]
&= \int_0^1 \frac{\log(x) \log^n(1-x)}{1-x} \, dx \tag{1} \\[1ex]
&= (-1)^n \int_0^\infty x^n \log\left(1-e^{-x}\right) \, dx \tag{2} \\[1ex]
&= (-1)^{n+1} n \int_0^\infty x^{n-1} \operatorname{Li}_2(e^{-x}) \, dx \tag{3} \\[1ex]
&= (-1)^{n+1} n (n-1) \int_0^\infty x^{n-2} \operatorname{Li}_3(e^{-x}) \, dx = \cdots \tag{4} \\
&\;\vdots \\
&= (-1)^{n+1} n! \int_0^\infty \operatorname{Li}_{n+1}(e^{-x}) \, dx \\[1ex]
&= (-1)^{n+1} n! \operatorname{Li}_{n+2}(1) \\[1ex]
&= \boxed{(-1)^{n+1} \Gamma(n+1) \zeta(n+2)} \tag{5}
\end{align*}$$
*
*$(1)$ : substitute $x\mapsto1-x$
*$(2)$ : substitute $x\mapsto 1-e^{-x}$
*$(3)$ : integrate by parts, recalling $\displaystyle \frac d{dx}\operatorname{Li}_2(x) = -\frac{\log(1-x)}x$ where $\operatorname{Li}_2$ is the dilogarithm
*$(4)$ : integrate by parts ad nauseam, using the recurrence $\displaystyle\frac d{dx}\operatorname{Li}_n(x)=\frac{\operatorname{Li}_{n-1}(x)}x$
*$(5)$ : $n!=\Gamma(n+1)$ and $\operatorname{Li}_n(1)=\zeta(n)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Does $(x^2+xy+y^2)$ divide $(x+y)^n-x^n-y^n$ if and only if n has no prime factors less than 5 I noticed that for $n$ with prime factors greater than 5, that $xy(x+y)(x^2+xy+y^2)$ always seems to divide $(x+y)^n-x^n-y^n$. The $xy(x+y)$ factors seem fairly obvious, but I can't figure out where the $x^2+xy+y^2$ term comes from. Strangely, if $n$ is congruent to 6 mod 1, then $(x^2+xy+y^2)^2$ seems to divide $(x+y)^n-x^n-y^n$.
Is this true for all $n$ as specified? Is the converse true, i.e. if $x^2+xy+y^2$ divides $(x+y)^n-x^n-y^n$ then $n$ has no prime factors less than 5, and if $(x^2+xy+y^2)^2$ divides $(x+y)^n-x^n-y^n$, then n is congruent to 1 mod 6?
| Here is an alternative direct approach, an induction with step six, and the "small cases" of $n$ among $0,1,2,3,4,5$ are easily checked. (We already have a quick, good, compact answer by Aphelli - i just wanted to see if an arguably more elementary proof avoiding complex numbers and derivatives can be worked out. The answer was started long time ago, at some firefox update i was asked if i really want to leave the page...)
We de-homogenize the situation, and replace at all places $y$ by $1$.
Alternatively, divide by $y^n$ and use $X=x/y$.
(It is enough to clear the question(s) for this special value of $y$.)
Let $F$ be the polynomial $X^2 + X + 1$. We will work in the rings $R_1=\Bbb Q[X]/(F)$ of polynomials taken modulo $F$, and in $R_2=\Bbb Q[X]/(F^2)$ of polynomials taken modulo $F^2$.
Notation: $P(n) = (X+1)^n - X^n - 1$.
Let us show the following:
$P(n)$ is divisible by $F$, iff $n$ is $\pm1$ modulo six.
$P(n)$ is divisible by $F^2$, iff $n$ is $1$ modulo six.
For small values of $n$ the divisibility (with $F$, and in the positive case with $F^2$) is easily checked:
$$
\begin{aligned}
(X+1)^3 &\equiv -1 \equiv X^3 && &&\text{ modulo }F\\[2mm]
P(0) &=1+1-1&&\ne0 &&\text{ modulo }F\\
P(1) &=0&&=\color{blue}{0} &&\text{ modulo }F\\
P(2) &=2x&&\ne0 &&\text{ modulo }F\\
P(3) &=3x^2+3x&&\equiv -3\ne0 &&\text{ modulo }F\\
P(4) &=4x^3+6x^2+4x&&\equiv 2x^2\ne0 &&\text{ modulo }F\\
P(5) &=5(x^4 +2x^3+2x^2+x)&&=\color{blue}{ 0} &&\text{ modulo }F\\
\end{aligned}
$$
The induction with step six is based on
$$
\begin{aligned}
&P(n+6) - P(n)\cdot(X+1)^6
\\
&\qquad=
(X+1)^{n+6} - X^{n+6} -1\\
&\qquad\ -(X+1)^{n+6} + X^n(X+1)^6 +(X+1)^6\\
&\qquad
=
X^n\underbrace{\Big(\ (X+1)^3+X^3\ \Big)}_{0\text{ modulo }F}\Big(\ (X+1)^3-X^3\ \Big)
+\underbrace{\Big(\ (X+1)^3+1\ \Big)}_{0\text{ modulo }F}\Big(\ (X+1)^3-1\ \Big)
\ .
\end{aligned}
$$
So $P(n+6)$ is zero modulo $F$ iff $P(n)$ is so. This elucidates the divisibility modulo $F$, this happens for $1,5$ plus a multiple of six, as claimed. If this happens, we want to test divisibility modulo $F^2$. So we extract the factor $F$ from getting
$$
\begin{aligned}
&\frac 1F\Big(P(n+6) - P(n)\cdot(X+1)^6\Big)
\\
&\qquad
=
X^n\cdot\frac 1F\Big(\ (X+1)^3+X^3\ \Big)\cdot\Big(\ (X+1)^3-X^3\ \Big)
+\frac1F\Big(\ (X+1)^3+1\ \Big)\cdot\Big(\ (X+1)^3-1\ \Big)
\\
&\qquad
=
X^n\cdot\Big(\ 2X+1\ \Big)\cdot\Big(\ (X+1)^3-X^3\ \Big)
+\Big(\ X+2\ \Big)\cdot\Big(\ (X+1)^3-1\ \Big)
\\
&\qquad\qquad\text{ and working now modulo $F$}
\\
&\qquad
\equiv
-X^n\cdot2X(X-1) + 2X^2(X-1)
\\
&\qquad
=-2X(X-1)(X^n - X)
\ .
\end{aligned}
$$
We can now conclude with the information modulo $F^2$.
*
*If $n=6k+1$ then $(X^n-X)=X(X^{6k}-1)=X(X^6-1)(\dots)$, which is divisible by $(X^3-1)$, and thus also by $F$. Inductively, $P(6k+1)=P(n)$ is divisible by $F^2$.
*If $n=6k+5$ then $(X^n-X)=X(X^{6k+4}-1)=X(X^{6k+4}-X^4+X^4-1)=X^4(X^6-1)(\dots) + X(X^4-1)\equiv X(X-1)$ modulo $F$, and inductively $P(6k+5)=P(n)$ is modulo $F^2$ equal to $-nF$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.