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Does this vector lie in the vector subspace $U$?
Given are the vectors $v_{1}=\begin{pmatrix} 1\\ 3\\ 5
\end{pmatrix}, v_{2}=\begin{pmatrix} 4\\ 5\\ 6 \end{pmatrix},
v_{3}=\begin{pmatrix} 6\\ 4\\ 2 \end{pmatrix}$ from
$\mathbb{R}^{3}$.
The vector subspace $U = \text{span}\left\{v_{1},v_{2},v_{3}\right\}$
Does the vector $\begin{pmatrix} 2\\ 4\\ 5 \end{pmatrix}$ lie in
$U$?
I'm not sure how this is done correctly but I think it's done by checking if the vector $\begin{pmatrix} 2\\ 4\\ 5 \end{pmatrix}$ is a linearly combination of $v_{1},v_{2},v_{3}$.
So I have written it all like that:
Let $a,b,c \in \mathbb{R}$.
$$a\begin{pmatrix}
1\\
3\\
5
\end{pmatrix}+b\begin{pmatrix}
4\\
5\\
6
\end{pmatrix}+c\begin{pmatrix}
6\\
4\\
2
\end{pmatrix}=\begin{pmatrix}
2\\
4\\
5
\end{pmatrix}$$
And then calculated each variable by using Gauss. In the end I had a wrong statement aka no solution and from this I conclude that the vector $\begin{pmatrix}
2\\
4\\
5
\end{pmatrix}$ doesn't lie in $U$.
I don't want write down all the calculation steps with Gauss because it would be too long, but is the way I did the correct way? Or is it done completely different?
| Equivalently, we want a+ 4b+ 6c= 2, 3a+ 5b+ 4c= 4, and 5a+ 6b+ 2c= 5. Subtract the second equation from twice the third: 7a+ 7b= 6. Subtract the first equation from three times the third: 14a+ 14b= 13. Multiplying 7a+ 7b= 6 by two, 14a+ 14b= 12. a and b cannot satisfy both 14a+ 14b= 13 and 14a+ 14b= 12. There are no numbers a and b that make the vector equation true. $\begin{pmatrix} 2 \\ 4 \\ 5\end{pmatrix}$ cannot be written as a linear combination of $v_1$, $v_2$, and $v_3$ so is not in subspace U.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to solve that logarithmic inequality? $$\log_{1\over31} (4x-5)^2 > \log_{1\over31} (5x-7)^2$$
$\begin{cases}
x\neq {7\over 5} \\
x\neq {5\over 4} \\
(4x-5)^2<(5x-7)^2
\end{cases}$
$(4x-5)^2-(5x-7)^2<0, \quad (4x-5-(5x-7))(4x-5+5x-7)<0, \quad (2-x)(x-{4\over3})<0, \quad x \in \left(-\infty; {4\over3}\right) \cup \left(2; +\infty\right)$
$x \in \left(-\infty; {5\over4}\right)\cup\left({5\over4};{4\over3}\right)\cup\left(2;+\infty\right)$
But right answer is $x \in \left({4\over3};{7\over5}\right) \cup \left({7\over5}; 2\right)$
| $\log_b$ with basis less than $1$ is decreasing. Therefore the equality is equivalent to
$$(4x-5)^2>(5x-7)^2\iff 9x^2-30x+24=3(3x^2-10x+8)>0.$$
Use the Rational root theorem to find $2$ is a root, hence by Vieta's relations, $4/3$ is the other root. The solutions are the $x$ outide of the interval of the roots:
$$(-\infty,4/3)\cup(2,+\infty)\smallsetminus\{5/4,7/5\}=(-\infty,5/4)\cup(5/4,4/3))\cup(2,+\infty).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine: $S = \frac{2^2}{2}{n \choose 1} + \frac{2^3}{3}{n \choose 2} + \frac{2^4}{4}{n \choose 3} + \cdots + \frac{2^{n+1}}{n+1}{n \choose n}$ We are given two hints: consider $(n+1)S$; and use the Binomial Theorem. But we are not to use calculus.
My consideration of $(n+1)S$ goes like this:
\begin{align*}
\sum\limits_{k=1}^{n}\frac{2^{k+1}}{k+1}{n \choose k}
&= \frac{1}{n+1}\sum\limits_{k=1}^{n}(n+1)\frac{2^{k+1}}{k+1}{n \choose k} \\
&= 2\frac{1}{n+1}\sum\limits_{k=1}^{n}2^k{n+1 \choose k+1} \\
&= 2\frac{1}{n+1}\sum\limits_{k=1}^{n}(1+1)^k{n+1 \choose k+1} \\
\end{align*}
Now I think I'm in a position to use the Binomial Theorem, giving
\begin{equation*}
2\frac{1}{n+1}\sum\limits_{k=1}^{n}\sum\limits_{i=0}^{k}{k \choose i}{n+1 \choose k+1}
\end{equation*}
I don't know if I am on the right track, but I do know that I'm stuck. Can anyone offer any advice on how to proceed?
| HINT:
Like Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$,
$$\sum_{k=1}^n\dfrac{a^{k+1}}{k+1}\binom nk=\dfrac1{n+1}\sum_{k=1}^n\binom{n+1}{k+1}a^{k+1}$$
Now $$\sum_{k=-1}^n\binom{n+1}{k+1}a^{k+1}=(a+1)^{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Five Consecutive Integers divisible by a square greater than 1 I have the task of finding 5 consecutive
integers of the
form {x, x+1, x+2, x+3, x+4, x+5}
where each number in the sequence
is divisible by a square k greater than 1..
I tried to write a simple JAVA program
to find a sequence of that form by checking
if each number in the sequence = 0 mod k
But that can't be possible, how would I go about
finding these numbers?
| Chinese remainder theorem:
$$\begin{align}x&\equiv 0\pmod{4}\\
x&\equiv -1\pmod{9}\\
x&\equiv -2\pmod{25}\\
x&\equiv -3\pmod{49}\\
x&\equiv -5\pmod{121}
\end{align}$$
That's gonna be ugly.
If you only need five, then you can ignore the last line and you get $$x\equiv 29348\pmod{4\cdot 9\cdot 24\cdot 49}.$$ Then $x,x+4$ are divisible by $4$, $x+1$ is divisible by $9$, $x+2$ is divisible by $25$, and $x+3$ is divisible by $49$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2178089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is there another transformation matrix for the bases of the image and of the preimage of this mapping? There is the following Matrix:
\begin{pmatrix}1&1&1\\ a&b&c\\ a^2&b^2&c^2\end{pmatrix}
At a point it is needed to calculate the determinant of the matrix. In the official solution it is written:
$det\begin{pmatrix}1&1&1\\ a&b&c\\ a^2&b^2&c^2\end{pmatrix} = (c-b)(c-a)(b-a)$
And I don't see how they get this. If I calculate the determinant I am always getting this:
$det\begin{pmatrix}1&1&1\\ a&b&c\\ a^2&b^2&c^2\end{pmatrix} = (bc^2-b^2c)-(ac^2-a^2c)+(ab^2-a^2b)=c(b(c-b)-a(c-a)+ab(b-a)).$
But after that point I don't know how to proceed and get the form above. Can you help me?
| Substracting the first column from the second and third, you get
$\begin{vmatrix}
1 & 1&1 \\
a&b &c \\
a^{2} & b^{2} &c^{2}
\end{vmatrix}=\begin{vmatrix}
1 & 0&0 \\
a&b-a &c-a \\
a^{2} & b^{2}-a^{2} &c^{2} -a^{2}
\end{vmatrix}=$
$=(b-a)\cdot (c-a)\begin{vmatrix}
1 & 0&0 \\
a&1 &1 \\
a^{2} & b+a &c+a
\end{vmatrix}=(b-a)\cdot (c-a)\begin{vmatrix}
1 &1 \\
b+a &c+a
\end{vmatrix}=$
$=(b-a)\cdot (c-a)\cdot (c+a-b-a)=(b-a)\cdot (c-a)\cdot(c-b).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2179649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the number $\;'c'\;$ that satisfies the conclusion of Rolle's theorem? I have a problem
What is the number $\;'c'\;$ that satisfies the conclusion of Rolle's theorem for the function $$f(x)=(x^2-1)(x-2)\;\;\text{in}\;(1,\;2]$$
I've tried$$f'(x)= 3x^2-4x-1$$
we know, to fiind 'c' $$f'(x)=0$$
So, $$\Rightarrow 3x^2-4x-1=0\\x=\dfrac {4+ \sqrt 4}{6}=1\\ x=\dfrac {4- \sqrt 4}{6}=0.33\\\therefore 1\notin(1,\;2]\\0.33\notin(1,\;2]$$
Please help. Where I've made misstake.
| Solve quadratic equation correctly
$$3x^2-4x-1=0\\\Rightarrow \dfrac{4\pm\sqrt{16-4(3)(-1)}}{2(3)}\\\Rightarrow x=\dfrac {2\pm \sqrt 7}{3}\\\Rightarrow x=\dfrac {2+ \sqrt 7}{3}\;\text{and}\; x=\dfrac {2- \sqrt 7}{3}\\x=\dfrac {2+ \sqrt 7}{3}=1.55\in (1,\;2]\\ x=\dfrac {2- \sqrt 7}{3}=-.215\notin(1,\;2]$$
Hence,
$\; c=\dfrac {2+ \sqrt 7}{3}\;$ satisfies the conclusion of Rolle's theorem for the given function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\sum_{j=0}^n \left(-\frac{1}{2}\right)^j \binom{n}{j}\binom{n+j}{j}\binom{j}{k} = 0$ when $n+k$ is odd An integral led me to a power series with these coefficients:
$$a_k = \sum_{j=k}^n \left(-\frac{1}{2}\right)^j \binom{n}{j}\binom{n+j}{j}\binom{j}{k}$$
I strongly suspect that the series should have $a_k = 0$ when $n+k$ is odd, and I've verified it for $k,n\leq 10$. I'm looking for a direct proof of this. Does anyone have a suggestion?
| Suppose we seek to verify that
$$a_{n,k} = \sum_{j=k}^n \left(-\frac{1}{2}\right)^j
{n\choose j} {n+j\choose j} {j\choose k}$$
is zero when $n+k$ is odd. We have
$${n+j\choose j} {j\choose k}
= \frac{(n+j)!}{n! k! (j-k)!}
= {n+k\choose k} {n+j\choose n+k}$$
and obtain for the sum
$${n+k\choose k} \sum_{j=k}^n \left(-\frac{1}{2}\right)^j
{n\choose j} {n+j\choose n+k}$$
Introduce
$${n+j\choose n+k} = {n+j\choose j-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{j-k+1}} (1+z)^{n+j} \; dz$$
Note that this vanishes when $j\lt k$ so we may lower $j$ to start at
zero, getting for the inner sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} z^{k-1} (1+z)^{n}
\sum_{j=0}^n \left(-\frac{1}{2}\right)^j
{n\choose j} \frac{(1+z)^j}{z^j}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} z^{k-1} (1+z)^{n}
\left(1-\frac{1+z}{2z}\right)^n
\; dz
\\ = \frac{1}{2^n} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-k+1}} (1+z)^{n}
(z-1)^n\; dz
\\ = \frac{1}{2^n} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-k+1}} (z^2-1)^n
\; dz.$$
We thus have the closed form
$$\frac{1}{2^n} {n+k\choose k} [z^{n-k}] (z^2-1)^n
\\ = \begin{cases}
\frac{1}{2^n} {n+k\choose k} (-1)^{(n+k)/2} {n\choose (n-k)/2}
\quad\text{if}\quad n-k\quad\text{is even}
\\ 0\quad\text{otherwise.}
\end{cases}.$$
We see having reached the result that we did not make use of the
differential in the integral which means the above also works using
formal power series only.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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What is the value of the $X$?
With the details, what could be the value of X?
| Let $d(x, y)$ be the distance of two vectors $x, y \in \mathbb{R}^2$. Let $p$ be the point where the two triangles touch. We aim to find $||p||$. $d((15/2, 0), p) = 15/2$ and $d((0, 10), p) = 5$. Let $p = (p_1, p_2)$. Then
\begin{align*}
\sqrt{(15/2 - p_1)^2 + p_2^2} = 15/2, \ \sqrt{p_1^2 + (10-p_2)^2} = 5
\end{align*}
The first gives $p_2 = \sqrt{15 p_1 - p_1^2}$, assuming it is positive. Inserting this into the second we get
\begin{align*}
\sqrt{p_1^2 + (10-\sqrt{15 p_1 - p_1^2})^2} = 5
\end{align*}
And solving for $p_1$ we get $p_1 = 3$, so that $p_2$ must be $6$, using the other equations.
$||p||$ must therefore be $\sqrt{ 3^2 + 6^2} = \sqrt{45} = 3 \sqrt{5}$
More detail upon request.
Edit: it has been pointed out by Andrei in the comments below that the equations that solve for the coordinates of $p$ can be geometrically interpreted as finding the point at the intersection of two circles.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Put identical coins into identical boxes In how many ways can I put 200 (identical) coins into 3 identical boxes?
I started count it, and I saw that the possibilities goes this way:
$200, 0, 0$
$199, 1, 0$
$198, 2, 0$
$198, 1, 1$
$197, 3, 0$
$197, 2, 1$
ect.
which mean I can count it this way: $2 \cdot 1 + 2 \cdot 2 + 2 \cdot 3$ ect..
I'm wondering if there is a better way to count all the possibilities.
| See OEIS sequence A001399. You are asking for the value of $a_{200}$ where $a_n$ is the number of partitions of the number $n$ into at most $3$ parts, or equivalently, the number of partitions of $n$ into parts of size at most $3$, that is, the number of ways we can write $n$ as a sum of $1$s, $2$s, and $3$s without regard to order.
Let $A_{n,i}$ be the set of all partitions of $n$ into parts of size at most $3$ with at least one part equal to $i$. By the in-and-out formula, for $n\gt0$ we have
$$a_n=|A_{n,1}\cup A_{n,2}\cup A_{n,3}|$$
$$=|A_{n,1}|+|A_{n,2}|+|A_{n,3}|-|A_{n,1}\cap A_{n,2}|-|A_{n,1}\cap A_{n,3}|-|A_{n,2}\cap A_{n,3}|+|A_{n,1}\cap A_{n,2}\cap A_{n,3}|$$
$$=a_{n-1}+a_{n-2}+a_{n-3}-a_{n-3}-a_{n-4}-a_{n-5}+a_{n-6}$$$$=a_{n-1}+a_{n-2}-a_{n-4}-a_{n-5}+a_{n-6}.$$
Now, the homogeneous linear recurrence
$$a_n=a_{n-1}+a_{n-2}-a_{n-4}-a_{n-5}+a_{n-6}$$
has the characteristic polynomial
$$t^6-t^5-t^4+t^2+t-1=(t-1)^3(t+1)(t^2+t+1)$$
with roots
$$1,\ 1,\ 1,\ -1,\ e^{2\pi i/3},\ e^{-2\pi i/3};$$
so the general solution is
$$a_n=An^2+Bn+C+D(-1)^n+E\cos\frac{2n\pi}3+F\sin\frac{2n\pi}3$$
where $A,B,C,D,E,F$ are arbitrary constants.
We use the initial values $a_0=1$ and $a_n=n$ for $n=1,2,3,4,5$ (or $a_n=0$ for $n=-1,-2,-3,-4,-5$) to evaluate the constants and get the particular solution
$$a_n=\frac{6n^2+36n+47+9(-1)^n+16\cos\frac{2n\pi}3}{72}.$$
Finally,
$$a_{200}=\frac{240000+7200+47+9-8}{72}=\boxed{3434}.$$
By the way, since
$$a_n=\frac{(n+2)(n+4)}{12}+\frac{-1+9(-1)^n+16\cos\frac{2n\pi}3}{72}$$
and since
$$\left|\frac{-1+9(-1)^n+16\cos\frac{2n\pi}3}{72}\right|\lt\frac12,$$
it follows that $a_n$ is the nearest integer to $\frac{(n+2)(n+4)}{12}$. For $n=200$ we have
$$\frac{(n+2)(n+4)}{12}=\frac{202\cdot204}{12}=3434=a_{200}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $(ab+ac+bc)\sum\limits_{cyc}\frac{1}{(a-7b)^2}\geq\frac{1}{4}$
Let $a$, $b$ and $c$ be non-negative numbers such that $\prod\limits_{cyc}(a-7b)\neq0$. Prove that:
$$(ab+ac+bc)\left(\frac{1}{(a-7b)^2}+\frac{1}{(b-7c)^2}+\frac{1}{(c-7a)^2}\right)\geq\frac{1}{4}$$
I think this inequality is very interesting
because it's similar to the known Ji Chen's inequality (Iran 1996):
$$(ab+ac+bc)\left(\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}\right)\geq\frac{9}{4}$$
Example of my trying.
BW does not help:
Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Hence, we need to prove that:
$$44064(u^2-uv+v^2)a^4+864(38u^3+17u^2v+73uv^2+38v^3)a^3-$$
$$-24(217u^4-1478u^3v-2157u^2v^2-3494uv^3+217v^4)a^2+$$
$$+4(98u^5-1631u^4v+3938u^3v^2+15698u^2v^3-1463uv^4+98v^5)a+$$
$$+uv(196u^4-2793u^3v+10490u^2v^2-2457uv^3+196v^4)\geq0,$$
which is nothing.
Thank you!
| Hints:
1)Put : $A=a$,
$AB=b$,
$AC=c$
With this following substitution you can eliminate a variable
We obtain $$(B+C+BC)(\frac{1}{(1-7B)^2}+\frac{1}{(B-7C)^2}+\frac{1}{(C-7)^2})$$
2)Try to prove this :
$(x+\alpha+\frac{1}{x+\beta}+\frac{x+\alpha}{x+\beta})(\frac{1}{(1-7(x+\alpha))^2}+\frac{1}{((x+\alpha)-7\frac{1}{x+\beta})^2}+\frac{1}{(\frac{1}{x+\beta}-7)^2})$
$\geq (x+\alpha+\frac{1}{x+\alpha}+1)(\frac{1}{(1-7(x+\alpha))^2}+\frac{1}{((x+\alpha)-7\frac{1}{x+\alpha})^2}+\frac{1}{(\frac{1}{x+\alpha}-7)^2}) $
With $0\leq \beta\leq \alpha \leq x$
For the last inequality it's easy to see that $$(x+\alpha+\frac{1}{x+\alpha}+1)(\frac{1}{(1-7(x+\alpha))^2}+\frac{1}{((x+\alpha)-7\frac{1}{x+\alpha})^2}+\frac{1}{(\frac{1}{x+\alpha}-7)^2}) $$
Is a translation of the function :
$$(x+\frac{1}{x}+1)(\frac{1}{(1-7(x))^2}+\frac{1}{((x)-7\frac{1}{x})^2}+\frac{1}{(\frac{1}{x}-7)^2}) $$
And the minimum of this last function is 0.25
| {
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"timestamp": "2023-03-29T00:00:00",
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Consider the family of functions of domain $\mathbb{R}\setminus \{a\}$, $\frac{2x^2+x-3}{x-a}$.Find a so that there are no vertical asymptotes
Consider the family of functions of domain $\mathbb{R}\setminus
\{a\}$ defined by $f(x) = \frac{2x^2+x-3}{x-a}$.
Determine $a$ so that $f$ has no vertical
asymptotes.
I know that one way to solve this is by factoring ${2x^2+x-3}$ into $(x-a)$ times something.
I'm not very good at factoring so I messed around a bit to see if I could find the solution.
My book says the solutions are x = 1 and x = -1.5. I looked at the graph of the function an realised these were the zeros of $2x^2+x-3$.
With a little help from Wolfram I found out how to factor $2x^2+x-3$ : $(x-1)(2x+3)$.
And so $a = 1$
To get the other solution I tried this:
$${2x^2+x-3} = (x-a)(y+z)$$
$x \cdot y = 2x^2 \Leftrightarrow y = 2x$
$${2x^2+x-3} = (x-a)(2x+z) = 2x^2+xz-2xa-az = 2x^2+x(z-2a) -az$$
$az = 3$
$x(z-2a) = x \Leftrightarrow z-2a = 1$
Then I put the above two conclusions into a system and calculated them and got $a = 1 \lor a = -1.5$, no problem there.
My questions:
*
*Why does $a$ take the values of the zeros of $2x^2+x-3$? Is it just a coincidence?
*Is there a simpler way of solving this?
| Note that $$f(x) = \frac{2x^2+x-3}{x-a} = \frac{(2x+3)(x-1)}{x-a} = \frac{2(x+1.5)(x-1)}{x-a}.$$ Thus if $a=-1.5$ or $a=1$, then $f(x)$ simplifies to a linear polynomial over the domain $\mathbb{R} \setminus \{a\}$, and linear polynomials do not have vertical asymptotes.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\cot(\pi/14)-4\sin(\pi/7)=\sqrt7$ Show that $\cot(\pi/14)-4\sin(\pi/7)=\sqrt7$.
This problem is from G.M. 10/2016 and I can't solve it. I tried with an isosceles triangle with angles $3\pi/7, 3\pi/7$ and $\pi/7$ and I tried to find a relation between the sides of the triangle but I couldn't find anything. I also thought to solve it with complex numbers but again I could't find anything. Any ideas?
| Let $a=\frac{\pi}{7},c=\cos a$.
This answer uses that $x=c$ is a root of
$$8x^3-4x^2-4x+1$$
The proof is written at the end of this answer.
Multiplying the both sides of
$$8c^3-4c^2-4c+1=0$$
by $2$ gives
$$16c^3-8c^2-8c+2=0,$$
i.e.
$$(16c^2-24c+9)+(16c^3-24c^2+9c)-7+7c=0,$$
i.e.
$$(1+c)(16c^2-24c+9)=7(1-c)$$
Multiplying the both sides by $1-c$ gives
$$(1-c^2)(4c-3)^2=7(1-c)^2,$$
i.e.
$$1-c^2=\frac{7(1-c)^2}{(4c-3)^2}$$
So
$$\sqrt{1-c^2}=\frac{\sqrt 7\ (1-c)}{4c-3}$$
since $4c-3\gt 4\cos\frac{\pi}{6}-3=2\sqrt 3-3\gt 0$.
So, we have
$$\frac{4c-3}{1-c}\sqrt{1-c^2}=\sqrt 7\tag1$$
By the way,
$$\begin{align}\cot\frac{a}{2}-4\sin a&=\sqrt{\frac{1+c}{1-c}}-4\sqrt{1-c^2}\\\\&=\sqrt{\frac{1-c^2}{(1-c)^2}}-4\sqrt{1-c^2}\\\\&=\frac{\sqrt{1-c^2}}{1-c}-4\sqrt{1-c^2}\\\\&=\frac{4c-3}{1-c}\sqrt{1-c^2}\tag2\end{align}$$
The claim follows from $(1)(2)$.
Finally, let us prove that $x=c$ is a root of
$$8x^3-4x^2-4x+1$$
Since $3a+4a=\pi$, we have
$$\sin(3a)=\sin(4a)$$
from which we have
$$\sin a\ (3-4\sin^2a)=2\sin(2a)\cos(2a)=4\sin a\cos a\cos(2a)$$
Dividing the both sides by $\sin a$ gives
$$3-4(1-c^2)=4c(2c^2-1),$$
i.e.
$$8c^3-4c^2-4c+1=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Using the simmetry of the function $t \rightarrow 1/t $ to change limits of an integral. I was looking at this very famous answer and unfortunately I could not get through even the first step:
\begin{align}
& 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_1^{\infty} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\
&= 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_0^{1} dt \frac{t^{1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\
\end{align}
Could someone please walk me through this step? how are we using the simmetry of $1/t$ here?
| In the integral
$$\int \limits _1 ^\infty \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \Bbb dt$$
make the change $u = \frac 1 t$. This will turn it into
$$\int \limits _1 ^0 \frac {u^{\frac 1 2}} {1 - \frac 1 {u^2}} \log \left( \frac {5 - \frac 2 u + \frac 1 {u^2}} {1 - \frac 2 u + \frac 5 {u^2}} \right) \left( - \frac 1 {u^2} \right) \Bbb d u = \int \limits _0 ^1 \frac {u^{\frac 1 2}} {1 - \frac 1 {u^2}} \log \left( \frac {5 - \frac 2 u + \frac 1 {u^2}} {1 - \frac 2 u + \frac 5 {u^2}} \right) \left( \frac 1 {u^2} \right) \Bbb d u = \\
\int \limits _0 ^1 \frac {u^{\frac 1 2}} {u^2 - 1} \log \left( \frac {5u^2 - 2u + 1} {u^2 - 2u + 5} \right) \Bbb d u = - \int \limits _0 ^1 \frac {u^{\frac 1 2}} {u^2 - 1} \log \left( \frac {u^2 - 2u + 5} {5u^2 - 2u + 1} \right) \Bbb d u = \\
\int \limits _0 ^1 \frac {u^{\frac 1 2}} {1 - u^2} \log \left( \frac {5 - 2u + u^2} {1 - 2u + 5u^2} \right) \Bbb d u .$$
Now change the letter $u$ into $t$ and you get the second integral on the second row of your question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $x^2 + y^2 + z^2 + 2xyz = 1$ then $x+ y+ z \leq \frac{3}{2}?$ True or false?
If $x \geq 0, y \geq 0, z \geq 0 $ and $x^2 + y^2 + z^2 + 2xyz = 1$ then $x+ y+ z \leq \frac{3}{2}.$
I want to know if there is a way to demonstrate this conditional inequality.
I know I can make a connection with two properties known in a triangle. I tried to find an algebraic demonstration of these problems and we did.
Thanks in advance for any suggestions.
| Let $(a,b,c)=(\lambda x,\lambda y,\lambda z)$, where $\lambda(x+y+z)=\frac32$. Then $a+b+c=\frac32$, and we have
$$a^2+b^2+c^2+2abc=\lambda^2(x^2+y^2+z^2+2\lambda xyz),$$
which is increasing in $\lambda$. Hence to show $\lambda\geq1$ it suffices to prove
$$a+b+c=\frac32\implies a^2+b^2+c^2+2abc\geq1.$$
We now use the notation for cyclic sums $\sum$ and symmetric sums $\sum_\text{sym}$. Note that $(\sum a)^2=\sum a^2+2\sum ab=\frac94$, so
\begin{align*}
\sum a^2+2\sum abc&\geq1\\
\iff\frac94\left(\sum a^2+2\sum abc\right)&\geq\sum a^2+2\sum ab\\
\iff18abc&\geq8\sum ab-5\sum a^2\\
\iff27abc&\geq\left(\sum a\right)\left(8\sum ab-5\sum a^2\right)\\
&=8\sum_\text{sym}a^2b+24abc-5\sum a^3-5\sum_\text{sym}a^2b\\
\iff5\sum a^3+3abc&\geq3\sum_\text{sym}a^2b.
\end{align*}
The last inequality follows from AM-GM ($\sum a^3\geq3abc$) and Schur's inequality ($\sum a^3+3abc\geq3\sum_\text{sym}a^2b$):
$$5\sum a^3+3abc\geq3\left(\sum a^3+3abc\right)\geq3\sum_\text{sym}a^2b,$$
and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the closed form for a recurrence relation I'm having trouble finding a closed form for a geometric recurrence relation where the term being recursively multiplied is of the form (x+a) instead of just (x).
Here's the recursive sequence:
$a_{n} = 4a_{n-1} + 5$ for $n \geq 1$ with the initial condition $a_{0} = 2$.
I know that in general the way to solve these problems is to start by writing out all of the arithmetic for the first few values of $a_{n}$, starting with the initial condition. Here's what I have:
$a_{0} = 2$
$a_{1} = 4 (2) + 5 \equiv ((2)(2)(2) + 5)$
$a_{2} = 4(4(2)+5)+5 \equiv (2)(2)((2)(2)(2)+5)+5$
$a_{3} = 4(4(4(2)+5)+5)+5 \equiv (2)(2)((2)(2)((2)(2)(2)+5)+5)+5$
$\ldots$ etc.
So at this point it's pretty clear to me that
$a_{n} = 2^{2n + 1} + something$
My problem is figuring out how to account for all of those 5's. Especially since the first 5 is being multiplied by $2^{3}$ and all of the other 5's are being multiplied by $2^{2}$.
I guessed something like this:
$a_{n} = 2^{2n+1} + 5(4^{n-1}) + 5^{n-1}$
$\ldots$ and the results were close, but not exact.
Can anyone help me out with the correct method for solving these types of problems? Thanks very much for your time.
| Finish the distributing the operators:
$$\begin{array} {rll}
%
a_0 &= 2 \\
%
a_1 &= 4\cdot 2 + 5 \\
%
a_2 &= 4\cdot (4\cdot 2 + 5) + 5 &= 4^2\cdot 2 + 4\cdot 5 + 5 \\
%
a_3 &= 4 \cdot (4^2\cdot 2 + 4\cdot 5 + 5) + 5 &= 4^3 \cdot 2 + 4^2\cdot 5 + 4 \cdot 5 + 5 \\
%
a_4 &= 4 \cdot (4^3 \cdot 2 + 4^2\cdot 5 + 4 \cdot 5 + 5) + 5 &= 4^4 \cdot 2 + 4^3 \cdot 5 + 4^2\cdot 5 + 4 \cdot 5 + 5 \\
%
\end{array}$$
Do you see the geometric series? (Factor out the 5)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Prove that: $\sum\limits_{cyc}\frac{a}{\sqrt{a^2+3bc}}\leq\frac{9(a^2+b^2+c^2)}{2(a+b+c)^2}$ Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a}{\sqrt{a^2+3bc}}+\frac{b}{\sqrt{b^2+3ac}}+\frac{c}{\sqrt{c^2+3ab}}\leq\frac{9(a^2+b^2+c^2)}{2(a+b+c)^2}$$
I tried Cauchy-Schwarz:
$$\left(\sum\limits_{cyc}\frac{a}{\sqrt{a^2+3bc}}\right)^2\leq(a+b+c)\sum_{cyc}\frac{a}{a^2+3bc}.$$
Hence, it remains to prove that
$$\sum_{cyc}\frac{a}{a^2+3bc}\leq\frac{81(a^2+b^2+c^2)}{4(a+b+c)^5},$$
which is wrong for $c\rightarrow0^+$.
Also we can use the following C-S:
$$\left(\sum\limits_{cyc}\frac{a}{\sqrt{a^2+3bc}}\right)^2\leq(1+1+1)\sum_{cyc}\frac{a^2}{a^2+3bc}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{a^2}{a^2+3bc}\leq\frac{27(a^2+b^2+c^2)^2}{4(a+b+c)^4},$$
which is wrong again: $b=c=1$.
| Using the substitutions $(a, b, c) \to (a^2, b^2, c^2)$, the inequality becomes
$$\sum_{\mathrm{cyc}}\frac{a^2}{\sqrt{a^4+3b^2c^2}} \le \frac{9(a^4+b^4+c^4)}{2(a^2+b^2+c^2)^2}.$$
Using the Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align}
\mathrm{LHS}^2 &= \sum_{\mathrm{cyc}} \frac{a^4}{a^4 + 3b^2c^2}
+ \sum_{\mathrm{cyc}} \frac{2a^2b^2}{\sqrt{(a^4+3b^2c^2)(b^4+3c^2a^2)}}\\
&\le \sum_{\mathrm{cyc}} \frac{a^4}{a^4 + 3b^2c^2}
+ \sum_{\mathrm{cyc}} \frac{2a^2b^2}{a^2b^2 + 3abc^2}.
\end{align}
It suffices to prove that
$$\frac{81(a^4+b^4+c^4)^2}{4(a^2+b^2+c^2)^4}\ge \sum_{\mathrm{cyc}} \frac{a^4}{a^4 + 3b^2c^2}
+ \sum_{\mathrm{cyc}} \frac{2a^2b^2}{a^2b^2 + 3abc^2}.$$
After clearing the denominators, it suffices to prove that $f(a, b, c)\ge 0$ where
$f(a,b,c)$ is a homogeneous polynomial of degree $26$.
Due to symmetry and homogeneity, WLOG, assume that $1 = c \le b \le a$.
Let $b = 1 + s, \ a = 1+s + t; \ s,t \ge 0$. $f(1+s+t, 1+s, 1)$ is a polynomial in $s, t$ with non-negative coefficients.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2198935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that the gcd is 1. I need to show that for $x$ odd the $\gcd\bigg(x,\dfrac{x^2-1}{2},\dfrac{x^2+1}{2}\bigg)=1$. Im trying it doing it pairwise. I already show that $\gcd\bigg(\dfrac{x^2-1}{2},\dfrac{x^2+1}{2}\bigg)=1$, but I dont have an idea how to show that $\gcd\bigg(x,\dfrac{x^2-1}{2}\bigg)=1$ and $\gcd\bigg(x,\dfrac{x^2+1}{2}\bigg)=1$.
I will appreciate the help.
| If $p|x$ then $p|x^2$ and $p\not |x^2 \pm 1$ so $p\not|\frac{x^2 \pm 1}k$.
So $x$ and $\frac{x^2 \pm 1}2$ have no prime factors in common. So $\gcd(x, \frac{x^2 -1}2) = \gcd(x,\frac{x^2 +1}2) = 1$.
And $\gcd(\frac{x^2-1}2,\frac{x^2 + 1}2) = \gcd(\frac{x^2 -1}2, \frac{x^2 +1}2 - \frac{x^2-1}2) = \gcd(\frac{x^2-1}2, 1) = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How find this maximum of the value $\sum_{i=1}^{6}x_{i}x_{i+1}x_{i+2}x_{i+3}$? Let
$$x_{1},x_{2},x_{3},x_{5},x_{6}\ge 0$$ such that
$$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=1$$
Find the maximum of the value of
$$\sum_{i=1}^{6}x_{i}\;x_{i+1}\;x_{i+2}\;x_{i+3}$$
where
$$x_{7}=x_{1},\quad x_{8}=x_{2},\quad x_{9}=x_{3}\,.$$
| Let $x_1=a,\ x_2=b,\ x_3=c,\ x_4=d,\ x_5=e,\ x_6=f.$
Objective function is
$$Z(a,b,c,d,e,f) = abcd + bcde + cdef + defa + efab + fabc.$$
Let maximize $Z(a,b,c,d,e,f)$ using Lagrange mulptiplyers method for the function
$$F(a,b,c,d,e,f,\lambda) = abcd + bcde + cdef + defa + efab + fabc + \lambda(1-a-b-c-d-e-f).$$
Equaling to zero partial derivatives $F'_a,\ F'_b,\ F'_c,\ F'_d,\ F'_e,\ F'_f,\ $ one can obtain the system:
$$\begin{cases}
bcd+bcf+bef+def = \lambda\\
acd+acf+aef+cde = \lambda\\
abd+abf+bde+def = \lambda\\
abc+aef+bce+cef = \lambda\\
abf+adf+bcd+cdf = \lambda\\
abc+abe+ade+cde = \lambda\\
a+b+c+d+e+f = 1.
\end{cases}$$
Note that:
*
*First and third equations contains the common term $def.$
*Second and fourth - $aef.$
*Third and fifth - $abf.$
*Fourth and six - $abc.$
*Fifth and first - $bcd.$
*Sixth and second - $cde.$
So
$$\begin{cases}
bcd+bcf+bef = abd+abf+bde\qquad(1)\\
acd+acf+cde = abc+bce+cef\qquad(2)\\
abd+bde+def = adf+bcd+cdf\qquad(3)\\
aef+bce+cef = abe+ade+cde\qquad(4)\\
abf+adf+cdf = bcf+bef+def\qquad(5)\\
abc+abe+ade = acd+acf+aef\qquad(6)\\
a+b+c+d+e+f = 1.\qquad\qquad\qquad(7)
\end{cases}$$
Easy to see that:
*
*Both $Z(a,b,c,d,e,f)$ and accordingly the system $(1-7)$ WLOG allows any cyclic permutation of $(a,b,c,d,e,f)$.
*Substitution of any pair of zero unknowns equals objective function to zero if that unknowns are not neighbours in the list of unknowns.
Let us consider zero cases in detail.
"Neighbouring zeros" case $b=c=0$ (and cyclic permutations).
Equations $(1)$ and $(2)$ give $LHS=RHS=0,$ remaining equations can be simplified to the form of
$$a=e,\quad d=f,\quad a+d = \frac12,$$
with the object function
$$Z(a, 0, 0, d, a, d) = a^2\left(\frac12-a\right)^2\quad\text{for } a\in\left(0,\dfrac12\right).$$
That gives
$$Z_{max} = \frac1{256}\text{ for } a=\frac14.$$
"Single zero" case $b=0, a>0, c>0.$ (and cyclic permutations).
And condition $Z>0$ requires $d>0, e>0, f>0.$
The equations $(3,4,6)$ forms a system
$$\begin{cases}
def = adf+cdf\\
aef+cef = ade+cde\\
ade = acd+acf+aef
\end{cases}\rightarrow
\begin{cases}
e = a+c\\
(a+c)f = (a+c)d\\
de = cd+cf+ef,
\end{cases}$$
$ef=de,\ c(d+f)=0.$
This contradicts the conditions $c>0,\ d>0,\ f>0.$
Case $abcdef>0.$
This case allows reduction of $(1-6)$:
$$\begin{cases}
cd+cf+ef = ad+af+de\qquad(1a)\\
ad+af+de = ab+be+ef\qquad(2a)\\
ab+be+ef = af+bc+cf\qquad(3a)\\
af+bc+cf = ab+ad+cd\qquad(4a)\\
ab+ad+cd = bc+be+de\qquad(5a)\\
bc+be+de = cd+cf+ef\qquad(6a)\\
a+b+c+d+e+f = 1.\qquad\quad(7)
\end{cases}$$
Summations $(1a+2a),\ (2a+3a),\ (3a+4a),\ (4a+5a),\ (5a+6a),\ $ leads to the system
$$\begin{cases}
cd+cf = ab+be\qquad\qquad(1b)\\
ad+de = bc+cf\qquad\qquad(2b)\\
be+ef = ad+cd\qquad\qquad(3b)\\
af+cf = be+de\qquad\qquad(4b)\\
ab+ad = cf+ef\qquad\qquad(5b)\\
a+b+c+d+e+f = 1.\quad(7)
\end{cases}$$
Now weighted sums $(1b)\cdot d+(2b)\cdot b,$ $(2b)\cdot e+(3b)\cdot,$ $(3b)\cdot f+(4b)\cdot d,$ $(4b)\cdot a+(5b)\cdot c$ form the system
$$\begin{cases}
cd^2+cdf = b^2c+bcf\\
ade+de^2 = acd+c^2d\\
bef+ef^2 = bde+d^2e\\
a^2f+acf = cef+e^2f\\
a+b+c+d+e+f = 1
\end{cases}\rightarrow
\begin{cases}
(d-b)\cdot(b+d+f) = 0\\
(e-c)\cdot(c+e+a) = 0\\
(f-d)\cdot(f+d+b) = 0\\
(a-e)\cdot(a+e+c) = 0\\
a+b+c+d+e+f = 1
\end{cases}$$
$$f=d=b,\quad e=c=a,\quad a+b = \dfrac13.$$
The object function is
$$Z(a, b, a, b,a,b) = 6a^3\left(\dfrac13 - a\right)^3$$
for $a\in\left(0,\dfrac13\right)$.
That gives
$$Z_{max}(a) = \frac1{216}\text{ for } a=\frac16,$$
and finally
$$\boxed{\max Z(\vec x) = \dfrac 1{216}\quad\text{ for all }x_i=\dfrac16.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Surface area of a sphere with integration of disks Why it is not correct to say that the surface area of a sphere is:
$$
2 \int_{0}^{R} 2\pi r \text{ } dr
$$
In my mind we are summing up the perimeters of disks from $r=0$ to $r=R$, so by 1 integration, we would have $\frac{1}{2}$ of the surface area of the sphere.
I know that it's not correct because that will give us $2\pi R^2$ that it's different from $4\pi R^2$, but why???
Thanks!
| $$\text{Area will be}~~: 2 \int_{0}^{R} 2\pi x ~ ds$$
Where $ds$ is width of strip bounded by circles of radius $x$ and $x+dx$ situated at height $y$. Also $ds \neq dr$ it's tilted in $y$ direction too. Only horizontal projection of $ds$ is $dr$.What you have done is valid for Disk See image below.
$$(ds)^2=(dx)^2+(dy)^2$$
Therefore :
$$ds = \sqrt{1+\Bigg( \frac{dy}{dx}\Bigg)^2} \; dx$$
Now,
$$x^2 + y^2 = R^2$$
$$y = \sqrt{R^2 - x^2}$$
$$\dfrac{dy}{dx} = \dfrac{-2x}{2\sqrt{R^2 - x^2}}$$
$$\dfrac{dy}{dx} = \dfrac{-x}{\sqrt{R^2 - x^2}}$$
$$\left( \dfrac{dy}{dx} \right)^2 = \dfrac{x^2}{R^2 - x^2}$$
Thus,
$$\displaystyle A = 4\pi \int_0^R x \sqrt{1 + \dfrac{x^2}{R^2 - x^2}} \, dx$$
$$\displaystyle A = 4\pi \int_0^R x \sqrt{\dfrac{(R^2 - x^2) + x^2}{R^2 - x^2}} \, dx$$
$$\displaystyle A = 4\pi \int_0^R x \sqrt{\dfrac{R^2}{R^2 - x^2}} \, dx$$
Let
$$x = R \sin θ \implies
dx = R \cos θ dθ$$
When $x = 0, θ = 0$
When $x = R, θ = \pi/2$
Thus,
$$\displaystyle A = 4\pi \int_0^{\pi/2} R \sin \theta \sqrt{\dfrac{R^2}{R^2 - R^2 \sin^2 \theta}} \, (R \cos \theta \, d\theta)$$
$$\displaystyle A = 4\pi \int_0^{\pi/2} R^2 \sin \theta \cos \theta\sqrt{\dfrac{R^2}{R^2(1 - \sin^2 \theta)}} \, d\theta$$
$$\displaystyle A = 4\pi R^2 \int_0^{\pi/2} \sin \theta \cos \theta\sqrt{\dfrac{1}{\cos^2 \theta}} \, d\theta$$
$$\displaystyle A = 4\pi R^2 \int_0^{\pi/2} \sin \theta \cos \theta \left( \dfrac{1}{\cos \theta} \right) \, d\theta$$
$$\displaystyle A = 4\pi R^2 \int_0^{\pi/2} \sin \theta \, d\theta$$
$$A = 4\pi R^2 \bigg[-\cos \theta \bigg]_0^{\pi/2}$$
$$A = 4\pi R^2 \bigg[-\cos \frac{1}{2}\pi + \cos 0 \bigg]$$
$$A = 4\pi R^2 \bigg[ -0 + 1 \bigg]$$
$$A = 4\pi R^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 1
} |
Find sum binomial coefficients $$
\sum_{k>=1}^{\infty} {2N \choose N-k}k
$$
How to find this sum?
I know that the answer is $ \frac{1}{2}N{2N \choose N}$
But it is very interesting to know the solution :)
| Our first goal in dealing with this sum is to get rid of the $k$. The standard approach is to rewrite something like $\binom{n}{k} \cdot k$ as $\frac nk \binom{n-1}{k-1} \cdot k$, or $n \binom{n-1}{k-1}$. Here, the bottom index doesn't match the extra factor, but we can make it so with a little extra work:
\begin{align}
\binom{2N}{N-k} k &= \binom{2N}{N+k}k \\
&= \binom{2N}{N+k}(N+k) - \binom{2N}{N+k} N \\
&= \frac{2N}{N+k} \binom{2N-1}{N+k-1}(N+k) - N \binom{2N}{N+k} \\
&= 2N \binom{2N-1}{N+k-1} - N \binom{2N}{N+k}.
\end{align}
At this point, we have two sums that are both easier to deal with:
$$\sum_{k \ge 1} \binom{2N}{N-k} k = 2N \sum_{k \ge 1} \binom{2N-1}{N+k-1} - N \sum_{k \ge 1} \binom{2N}{N+k}.$$
The sum of all binomial coefficients of the form $\binom{2N-1}{i}$ is $2^{2N-1}$, and our first sum takes only those binomial coefficients of this form where $i \ge N$. These are the second half, which by symmetry is equal to the first half, so the first sum simplifies to $2^{2N-2}$.
We're in much the same position with the second sum, except that the $\binom{2N}{i}$ coefficients also have a central coefficient $\binom{2N}{N}$, which is left out here. The sum of all coefficients that aren't the central one is $2^{2N} - \binom{2N}{N}$, and this sum is half of that.
Putting these facts together, we get
\begin{align}
\sum_{k \ge 1} \binom{2N}{N-k} k
&= 2N \Bigg(2^{2N-2}\Bigg) - N \Bigg(2^{2N-1} - \frac12\binom{2N}{N}\Bigg) \\
&= N \cdot 2^{2N-1} - N \cdot 2^{2N-1} + \frac N2 \cdot \binom{2N}{N} \\
&= \frac N2 \cdot \binom{2N}{N}.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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Showing that $\sum_{k=0}^{n}(-1)^k{n\choose k}{1\over k+1}\sum_{j=0}^{k}{H_{j+1}\over j+1}={1\over (n+1)^3}$ Consider this double sums $(1)$
$$\sum_{k=0}^{n}(-1)^k{n\choose k}{1\over k+1}\sum_{j=0}^{k}{H_{j+1}\over j+1}={1\over (n+1)^3}\tag1$$
Where $H_n$ is the n-th harmonic
An attempt:
Rewrite $(1)$ as
$$\sum_{k=0}^{n}(-1)^k{n\choose k}{1\over k+1}\left(H_1+{H_2\over 2}+{H_3\over 3}+\cdots+{H_{k+1}\over k+1}\right)\tag2$$
Recall $$\sum_{k=0}^{n}(-1)^k{n\choose k}{1\over k+1}={1\over n+1}\tag3$$
Not sure how to continue
| We seek to show that
$$\sum_{k=0}^n (-1)^k {n\choose k} \frac{1}{k+1}
\sum_{j=0}^k \frac{H_{j+1}}{j+1} = \frac{1}{(1+n)^3}.$$
This is
$$\sum_{k=0}^n (-1)^k {n+1\choose k+1} \frac{k+1}{n+1} \frac{1}{k+1}
\sum_{j=0}^k \frac{H_{j+1}}{j+1} = \frac{1}{(1+n)^3}$$
or
$$\sum_{k=0}^n (-1)^k {n+1\choose k+1}
\sum_{j=0}^k \frac{H_{j+1}}{j+1} = \frac{1}{(1+n)^2}.$$
The LHS is
$$\sum_{j=0}^n \frac{H_{j+1}}{j+1} \sum_{k=j}^n (-1)^k {n+1\choose k+1}.$$
Writing
$${n+1\choose k+1} = {n+1\choose n-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+1}} (1+z)^{n+1}
\; dz$$
we get range control (vanishes for $k\gt n$) so we may write for the
inner sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^{n+1}
\sum_{k\ge j} (-1)^k z^k
\; dz
\\ = (-1)^j \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-j+1}} (1+z)^{n+1}
\sum_{k\ge 0} (-1)^k z^k
\; dz
\\ = (-1)^j \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-j+1}} (1+z)^{n+1}
\frac{1}{1+z}
\; dz
\\ = (-1)^j \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-j+1}} (1+z)^{n}
\; dz
= (-1)^j {n\choose n-j} = (-1)^j {n\choose j}.$$
We thus have to show that
$$\sum_{j=0}^n \frac{H_{j+1}}{j+1} (-1)^j {n\choose j}
= \frac{1}{(1+n)^2}$$
or alternatively
$$\sum_{j=0}^n \frac{H_{j+1}}{j+1} (-1)^j
\frac{j+1}{n+1} {n+1\choose j+1}
= \frac{1}{(1+n)^2}$$
which is
$$\sum_{j=0}^n H_{j+1} (-1)^j
{n+1\choose j+1}
= \frac{1}{1+n}$$
The LHS is
$$\sum_{j=0}^n (-1)^j
{n+1\choose j+1} \sum_{q=1}^{j+1} \frac{1}{q}
= \sum_{j=0}^n (-1)^j
{n+1\choose j+1} \sum_{q=0}^{j} \frac{1}{q+1}
\\ = \sum_{q=0}^n \frac{1}{q+1}
\sum_{j=q}^n (-1)^j {n+1\choose j+1}.$$
We re-use the computation from before to get
$$\sum_{q=0}^n \frac{1}{q+1} (-1)^q {n\choose q}
= \sum_{q=0}^n \frac{1}{q+1} (-1)^q \frac{q+1}{n+1} {n+1\choose q+1}
\\ = \frac{1}{n+1} \sum_{q=0}^n (-1)^q {n+1\choose q+1}.$$
We have reduced the claim to
$$\sum_{q=0}^n (-1)^q {n+1\choose q+1} = 1$$
which holds by inspection or by writing
$$- \sum_{q=1}^{n+1} (-1)^q {n+1\choose q}
= 1 - \sum_{q=0}^{n+1} (-1)^q {n+1\choose q} = 1 - (1-1)^{n+1} = 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2207066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Calculate the volume of the solid bounded laterally. How do i find the volume bounded below by the plane $xy$ and bounded above by $x^2+y^2+4z^2=16$ and laterally by the cylinder $x^2+y^2-4y=0$.
Since when i change to polar coordinates $x^2+y^2-4y=0$. is equal to $4sin(\theta)$.
And for the limits. $z=\frac{\sqrt{16-x^2-y^2}}{4}$ is equal to $z=\sqrt{4-\frac{r^2}{4}}$
So i think the integral for this is:
$$\int_{0}^{\pi}\int_{0}^{\sqrt{4-\frac{r^2}{4}}} 4rsin(\theta)drd\theta$$
| You have some mistake in the limits of integration. Using the symmetry of the solid around the $y-z$ plane ( see the figure), we can take for $\theta$ the values between $0$ and $\frac{\pi}{2}$ and duplicate the integral, so the limits becomes:
$$
0<\theta<\frac{\pi}{2} \qquad 0<r<4\sin \theta \qquad 0<z< \frac{1}{2}\sqrt{16-r^2}
$$
so the volume is:
$$
V= 2\int_0^{\frac{\pi}{2}}\int_0^{4\sin \theta}\int_0^{\frac{1}{2}\sqrt{16-r^2}}rdzdrd\theta=2\int_0^{\frac{\pi}{2}}\int_0^{4\sin \theta}\frac{1}{2}\sqrt{16-r^2}dr d\theta=
$$
$$
=\frac{1}{3}\int_0^{\frac{\pi}{2}}64(1-\cos^2\theta)d\theta=\frac{32}{3}\pi-\frac{128}{9}
$$
(if my calculations are correct).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$n^{n-1}-1$ is a multiple of $k$
Find the number of integers $k$ with $2 \leq k \leq 1000$ satisfying the following property:
*
*For every positive integer $n$ relatively prime to $k$, $n^{n-1}-1$ is a multiple of $k$.
Let $k = 2^{\alpha_1}3^{\alpha_2} \cdots p_n^{\alpha_n}$ be the prime decomposition of $k$. Then by the Chinese Remainder Theorem $n^{n-1} \equiv 1 \pmod{k}$ for all $n$ such that $\gcd(n,k) = 1$ if and only if\begin{align*}n^{n-1} &\equiv 1 \pmod{2^{\alpha_1}}\\n^{n-1} &\equiv 1 \pmod{3^{\alpha_2}}\\&\vdots\\n^{n-1} &\equiv 1 \pmod{p_n^{\alpha_n}}\end{align*} for all $n$ such that $\gcd(n,k) = 1$. How can we continue?
| If $\gcd(n,k)=1$ then $\gcd(n+k,k)=1$. So:
$$1\equiv (n+k)^{n+k-1}\equiv n^{n+k-1}=n^{n-1}n^k\pmod{k}$$
and hence $n^k\equiv 1\pmod{k}$ for all $n$ relatively prime to $k$.
Now if $0<n<k$ with $\gcd(n,k)=1$, then:
$$1\equiv (k-n)^{k-n-1} \equiv (-1)^{k-n-1} n^kn^{-(n-1)}n^{-2}\pmod{k}$$
But $n^k\equiv n^{-(n-1)}\equiv 1\pmod{k}$ so you have $$n^2\equiv (-1)^{k-n-1}\pmod{k}$$
That should reduce the problem greatly for case-by-case analysis. The only prime powers where every relatively prime square is $\pm 1$ are $2,4,8,3,5$. And $k$ has to be a product of these.
We know $k$ must be even, since otherwise $(2,k)=1$ and $k$ must divide $2^{2-1}-1=1$, we are most of the way.
When $k$ is even, we get that $n$ is odd, and hence $(-1)^{k-n-1}=1$ so we need $n^2\equiv 1\pmod{k}$ for all relevant $n$. But that means that $5$ can't be a factor.
This leaves us with $k=2,4,8,6,12,24$. We can quickly check each case.
Okay, more verbosely, you need to prove this lemma:
If $p^{a}\mid k$ and there is some $n_0$ such that $\gcd(n_0,p^a)=1$ and $n_0^2\not\equiv 1\pmod{p^a}$, then there is an $n$ with $\gcd(n,k)\neq 1$ such that $n^{2}\not\equiv 1\pmod{k}$.
Essentially, this is because you can write $k=p^bk'$ for $\gcd(k',p)=1$ and then solve the Chinese remainder question:
$$n\equiv n_0\pmod{p^b}\\
n\equiv 1\pmod{k'}$$
This gives you an $n$ with $n^2\not\equiv -1\pmod{p^b}$ so $n^2\not\equiv -1\pmod{k}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ I know there are various methods showing that $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$, but I want to know how to derive it from letting $t\rightarrow 0^{+}$ for the following identity:
$$\sum_{n=-\infty}^{\infty}\frac{1}{t^2+n^2}=\frac{\pi}{t}\frac{1+e^{-2\pi t}}{1-e^{-2\pi t}}$$
| Note that we have
$$\sum_{n=-\infty}^\infty \frac{1}{t^2+n^2}=\frac1{t^2}+2\sum_{n=1}^\infty\frac{1}{t^2+n^2}$$
Therefore, using $\sum_{n=-\infty}^\infty \frac{1}{t^2+n^2}=\frac\pi t \frac{1+e^{-2\pi t}}{1-e^{-2\pi t}}$, we find that
$$\sum_{n=1}^\infty\frac{1}{t^2+n^2}=\frac12\left(\frac\pi t \frac{1+e^{-2\pi t}}{1-e^{-2\pi t}}-\frac{1}{t^2}\right) \tag 1$$
The limit of the left-hand side of $(1)$ is the series of interest, $\sum_{n=1}^\infty\frac{1}{n^2}$. The limit of the term on the right-hand side is
$$\begin{align}
\frac12\lim_{t\to 0}\left(\frac\pi t \frac{1+e^{-2\pi t}}{1-e^{-2\pi t}}-\frac{1}{t^2}\right)&=\frac12\lim_{t\to 0}\left(\frac\pi t \frac{1+1-2\pi t+2\pi^2t^2+O(t^3)}{2\pi t-2\pi^2t^2-\frac43 \pi^3t^3+O(t^4)}-\frac{1}{t^2}\right)\\\\
&=\frac12\lim_{t\to 0}\left(\frac{\frac23 \pi^3t^2+O(t^3)}{2\pi t^2\left(1+O(t)\right)}\right)\\\\
&=\pi^2/6
\end{align} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Expansion of this expression Let $x$ be a real number in $\left[0,\frac{1}{2}\right].$ It is well known that
$$\frac{1}{1-x}=\sum_{n=0}^{+\infty} x^n.$$
What is the expansion or the series of the expression $(\frac{1}{1-x})^2$?
Many thanks.
| Or do by brute force multiplication and gathering terms:
$$\frac{1}{(1-x)^2}=$$
$$
(1+x+x^2+x^3+x^4+\ldots)(1+x+x^2+x^3+x^4+\ldots)\\
=(1+x+x^2+x^3+x^4+\ldots)\\
+x+x^2+x^3+x^4+\ldots\\
\quad +x^2+x^3+x^4+\ldots\\
\quad \quad +x^3+x^4+\ldots\\
=1+2x+3x^2+4x^3+\ldots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2210103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
What is wrong with this way of solving trig equations? Let's suppose I have to find the values of $\theta$ and $\alpha$ that satisfy these equations:
*
*$\cos^3 \theta$ = $\cos \theta $
*$3\tan^3 \alpha = \tan \alpha$
on the interval $[0; 2 \pi]$.
If I try to solve, for instance, the first equation like this:
$$\cos^3 \theta = \cos \theta $$
$$\cos^2 \theta * \cos \theta = \cos \theta$$
$$\cos^2 \theta = \cos \theta \div \cos \theta $$
$$\cos^2 = 1$$
$$\cos \theta = \pm \sqrt {1}$$
$$\cos \theta = \pm 1$$
I end up getting only:
$\theta = 0 ; \pi ; 2 \pi$
But I know that $\pi /2$ and $3\pi /2$ would also make the equation true since $\cos^3 (\pi /2) = cos (\pi /2) = 0$ and $\cos^3 ( 3 \pi /2) = cos (3 \pi /2) = 0$.
The same problem arises when I try to solve the second equation:
$$3\tan^3 \alpha = \tan \alpha$$
$$3\tan^2 \alpha * \tan \alpha = \tan \alpha$$
$$3\tan^2 \alpha = \tan \alpha \div \tan \alpha$$
$$3\tan^2 \alpha =1$$
$$\tan^2 \alpha =1/3$$
$$\tan \alpha =\pm \sqrt{1/3}$$
$$\tan \alpha =\pm 1/\sqrt{3}$$
The values of $\alpha$ that make the tangent equal to $\pm 1/\sqrt{3}$ between 0 and $2 \pi$ are only these:
$ \alpha = \pi / 6 ; 5 \pi /6 ; 7 \pi / 6 ; 11 \pi /6$
However, I should also find $0$ and $\pi$ because $3\tan^3 (0) = \tan (0) = 3 tan^3 (\pi) = tan ( \pi ) = 0 $.
Something similar happens when I'm looking for the local minima and maxima of this function:
$$ f(x) = \sin^2(x) + \cos(x)$$
on the interval $[0; 2 \pi]$
$f'(x) = 2 \sin(x) \cos(x) - \sin(x)$
$ 0 = 2 \sin(x) \cos(x) - \sin(x)$
$ \sin(x) = 2 \sin(x) \cos(x)$
$ \sin(x) / \sin(x) = 2 \cos(x)$
$1 = 2 \cos(x)$
$1/2 = \cos(x)$
$x = \pi / 3 ; 5 \pi /3$
And again, plugging $x = 0$ or $x= \pi$ or $x = 2 \pi$ also make the derivative 0.
I've noticed that in all of these 3 cases I have the same trigonometric function on both sides of the equation and I'm dividing both sides by that function. This is making one side of the equation equal to 1 at some point. What is wrong or incomplete with this method? Why am I missing some results when I do this?
| The equation $x^3=x$ has three solutions: you can write it as
$$
x^3-x=0
$$
so
$$
x(x-1)(x+1)=0
$$
and the roots are $0$, $1$ and $-1$.
You cannot “divide by $x$”, which is the mistake you make when you “divide by $\cos\theta$”.
Thus your equation becomes
$$
\cos\theta=0
\quad\text{or}\quad
\cos\theta=1
\quad\text{or}\quad
\cos\theta=-1
$$
The second equation is similar:
$$
\tan\alpha=0
\quad\text{or}\quad
\tan\alpha=\frac{1}{\sqrt{3}}
\quad\text{or}\quad
\tan\alpha=-\frac{1}{\sqrt{3}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2210945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that for every natural number $n$ and for every real numbers Prove that for every natural number $n$ and for every real numbers $x\neq \dfrac {k\pi}{2^t }$ $(t=0,1,....n;)$ (where $k$ is any integer)
$$\dfrac {1}{\sin 2x}+\dfrac {1}{\sin 4x}+.....+\dfrac {1}{\sin 2^n x}=\cot x-\cot 2^nx$$
| Since
\begin{align}
\frac{1}{\sin 2x}=\frac{2{\cos}^2x-\cos2x}{2\sin x\cos x}=\cot x-\cot 2x,
\end{align}
then, if we regard $x$ as $2x$
\begin{align}
\frac{1}{\sin 4x}=\frac{2{\cos}^22x-\cos 4x}{2\sin 2x \cos 2x}=\cot 2x-\cot 4x.
\end{align}
Similarly, repeat the above step until $n$,
we have
\begin{align}
LHS=\cot x-\cot 2x+\cot 2x-\cot 4x+...+\cot 2^{n-1}x-\cot 2^nx=\cot x-\cot 2^nx=RHS.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $\sum_{n=1}^\infty \frac{n^2}{(n+1)!}=e-1$ Show that:
$$\sum^\infty_{n=1} \frac{n^2}{(n+1)!}=e-1$$
First I will re-define the sum:
$$\sum^\infty_{n=1} \frac{n^2}{(n+1)!} = \sum^\infty_{n=1} \frac{n^2-1+1}{(n+1)!} - \sum^\infty_{n=1}\frac{n-1}{n!} + \sum^\infty_{n=1} \frac{1}{(nm)!}$$
Bow I will define e:
$$e^2 = 1+ \frac{2}{1!} + \frac{x^2}{2!} + ... + \infty$$
$$e' = 1 + \frac{1}{1!} + \frac{1}{2!} + ... + \infty$$
$$(e'-2) = \sum^\infty_{n=1} \frac{1}{(n+1)!}$$
Now I need help.
| A differential approach follows with by use of $\delta = x D = x \frac{d}{dx}$ and integration. Since $\delta x^n = n x^n$ then
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2 \, x^n}{n!} &= \delta^{2} \, e^{x} = x(x+1) \, e^{x}
\end{align}
now, by integration,
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2 \, t^{n+1}}{(n+1)!} &= \int_{0}^{t} x(x+1) \, e^{x} \, dx = (1-t +t^2) \, e^{t} -1.
\end{align}
This can also be seen in the form
$$\sum_{n=1}^{\infty} \frac{n^2 \, t^{n}}{(n+1)!} = \frac{1 + t^3}{t(1+t)} \, e^{t} - \frac{1}{t}.$$
Setting $t=1$ leads to the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Calculate the rotation area when $y = \cos(\frac{\pi x}{6})$, $\left| x \right| \leq 3$ rotates around the x-axis. Question
Calculate the rotation area when $y = \cos(\frac{\pi x}{6})$, $\left| x \right| \leq 3$ rotates around the x-axis.
Attempt at solution
$$ A = 2\pi \int_{-3}^{3} f(x) \sqrt{1+ f'(x)^2} dx$$
Which in our case can be written as:
$$ A = 2\pi \int_{-3}^{3} \cos \left(\frac{\pi x}{6}\right) \sqrt{ 1 + \frac{1}{36} \pi^2 \sin^2\left(\frac{\pi x}{6}\right)} dx$$
By substitution $u = \sin \left(\frac{\pi x}{6}\right)$ and $du = \cos \left(\frac{\pi x}{6}\right)dx$, we get:
$$ A = 2\pi \int_{-1}^{1} \sqrt{ 1 + \frac{1}{36} \pi^2 u^2 }dx$$
Here I am stuck...
| $$ A = 2\pi \int_{-3}^{3} \cos \left(\frac{\pi x}{6}\right) \sqrt{ 1 + \frac{1}{36} \pi^2 \sin^2\left(\frac{\pi x}{6}\right)} dx$$
By substitution $u = \sin \left(\dfrac{\pi x}{6}\right)$ and $du = \dfrac{\pi }{6}\cos \left(\dfrac{\pi x}{6}\right)dx$, we get:
$$ A = 2\pi\dfrac{6}{\pi} \int_{-1}^{1} \sqrt{ 1 + \frac{1}{36} \pi^2 u^2 }dx$$
let $\dfrac{1}{6}\pi u=\sinh t\,$:
$$A=\frac{72}{\pi } \int_{\sinh ^{-1}\left(-\frac{\pi }{6}\right)}^{\sinh ^{-1}\left(\frac{\pi }{6}\right)} \cosh ^2t \, dt$$
and with $2\cosh ^2t=1+\cosh2t$ we arrive this answer
$$ A = 2 \sqrt{\pi ^2+36}+\frac{72}{\pi } \sinh ^{-1}\left(\frac{\pi }{6}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
One particular equation involving floor function I've got this equation to solve
$ \left\lfloor{\frac {x + 1}{x+7}}\right\rfloor = \left\lfloor{\frac {x }{x+3}}\right\rfloor $
It's not very hard to solve it if using properties of rational functions but I wonder if it can be solved using only arithmetics. Anyone got idea?
| If $x \ge 0$ then $0 < \frac {x+1}{x+7} < 1$ and $0 \le \frac x{x +3} < 1$ so $[\frac{x+1}{x+1}] = [\frac x{x+3}] = 0$. So all $x \ge 0$ are solutions.
If $x < 0$ is a matter of solving $[\frac {y - 1}{y-7}] = [\frac y{y- 3}]$ for $y = - x > 0$.
If $y = 1$ then $[\frac {y - 1}{y-7}] = 0 \ne [\frac y{y- 3}] = [- 1/2] = -1$.
Otherwise $\frac {y-1}{y-7} = 1 + \frac 6{y-7}$ and $\frac y{y-3} = 1 + \frac 3{y-3} $
so $[\frac {y - 1}{y-7}]=[\frac y{y- 3}]\iff [\frac 6{y-7}] =[\frac 3{y-3}] $
$[\frac 6{y-7}] = k \iff k \le \frac 6{y-7}< k + 1 \iff \frac 6{k}+7\ge y >\frac6{k+1} +7 $
Likewise $[\frac 3{y-3}] = k \iff \frac 3k + 3 \ge y > \frac 3{k+1} +3$
So $y >0 $ $[\frac 3{y-3}]= [\frac 6{y-7}] \iff \min(\frac 6k + 7,\frac 3k + 3) \ge y > \max (\frac 6{k+1} + 7, \frac 3{k+1} + 3)$.
If $k = 0$ then this is true if $y > 13$ and so $x < -13 \implies [\frac {x+1}{x+7} ]=[\frac {x}{x+3}] = 1$
If $k \ge 1$ then this is true if $\frac 3k + 3 \ge y > \frac 6{k+1} + 7 \ge \frac 6{2k} + 7 = \frac 3k +7$ which is impossible.
If $k = -1$ then to be true, $0 \ge y $ which is impossible.
If $k = -2$ then $\frac 32 \ge y > 1$ allows $[\frac{x+1}{x+7}] =[\frac{x}{x+3}] = -1$.
if $k = -3$ then $\frac 3{k} + 3 \ge y > \frac 6{k + 1} + 7 > \frac 6{2k} + 7 = \frac 3k + 7$ which is impossible.
So solution is $x < -13; -\frac 32 \le x < -1$; $x \ge 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proving $\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$.
Prove the following identity:
$$\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$$
How can I express $\cos(4\theta) $ in other terms?
| $$\cos (4\theta)=2\cos^2 (2\theta)-1=2(1-2\sin^2 \theta)^2-1=8\sin^4 \theta-8\sin^2 \theta+1$$
$$4\cos 2\theta = 4(1-2\sin^2 \theta)=4-8\sin^2 \theta$$
$$\cos 4\theta-4\cos 2\theta= (8\sin^4 \theta-8\sin^2 \theta+1)-(4-8\sin^2 \theta)=8\sin^4 \theta-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2218085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Need direction for solving integral
Evaluate: $$\int \frac{\sin^2x}{\cos^2x +4}dx.$$
I tried to this things:
First
$$\tan\left(\frac{x}{2}\right) = t$$
$$dx = \frac{2~dt}{1+t^2}$$
$$\sin x= \frac{2t}{1+t^2}$$
$$\cos x= \frac{1-t^2}{1+t^2}$$
Tried also but this is same thing?
$$\tan\left(\frac{x}{2}\right) = t$$
There I don't have to place $2$ left to integral ( same thing )
Second
I tried trigonometric transformations like
$$\sin^2x \equiv 1-\cos^2x$$
$$\int \frac{1-\cos^2x}{\cos^2x +4}dx$$
$$-1\cdot \int \frac{\cos^2x-1}{\cos^2x +4}~dx$$
$$-1\cdot \int \frac{\cos^2x+4-5}{\cos^2x +4}~dx$$
But I get stucked in this part
$$5\int \frac{1}{\cos^2x +4}dx.$$
And mix something from first step
Third
I tried to divide everything with $\sin^2x$, but didn't succeed to solve + mix something from first step
Can anyone give me direction how to solve this?
sorry, for late update....
My goal is to complete this task without trigonometry function sec
| HINT:
$$\frac{x^2}{5-x^2}=-\frac{-x^2+5-5}{5-x^2}=-1+\frac{5}{5-x^2}=-1+\frac{A}{\sqrt{5}-x}+\frac{B}{\sqrt{5}+x}\\A(\sqrt{5}+x)+B(\sqrt{5}-x)=5$$
Now when you solve for $A,B$ instead of $x$ plug in $\sin(x)$ [$\cos^2x+4=5-\sin^2 x$]
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\sqrt{x(3x+y)} + \sqrt{y(3y+z)} +\sqrt{z(3z+x)} \le 2(x+y+z)$ Prove $\sqrt{x(3x+y)} + \sqrt{y(3y+z)} +\sqrt{z(3z+x)} \le 2(x+y+z)$ using triangle inequality
To answer the comment: I'm pretty stumped on the question, i know i can substitute some variables in to make it simpler but from there im clueless, thats why im here, this is my last resort haha
| By AM-GM
$$\sum_{cyc}\sqrt{x(3x+y)}=\sqrt{\sum_{cyc}(3x^2+xy+2\sqrt{xy(3x+y)(3y+z)}}=$$
$$=\sqrt{\sum_{cyc}\left(3x^2+xy+2\sqrt{x(3y+z)\cdot y(3x+y)}\right)}\leq$$
$$\leq\sqrt{\sum_{cyc}(3x^2+xy+x(3y+z)+y(3x+y))}=2(x+y+z).$$
Done!
The proof by triangle inequality (Minkowski).
$$2(x+y+z)=\frac{1}{4}\sum_{cyc}(4x+3x+y)=\frac{1}{4}\sum_{cyc}\left(\sqrt{(4x)^2+0^2}+\sqrt{(3x+y)^2+0^2}\right)\geq$$
$$\geq\frac{1}{4}\sum_{cyc}\sqrt{(4x+3x+y)^2+(0+0)^2}=\frac{1}{4}\sum_{cyc}\sqrt{49x^2+14xy+y^2}=$$
$$=\frac{1}{4}\sum_{cyc}\sqrt{48x^2+16xy+(x-y)^2}\geq\frac{1}{4}\sum_{cyc}\sqrt{48x^2+16xy}=\sum_{cyc}\sqrt{x(3x+y)}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $2^p+p^2$ is prime for $p=3$ only I do know that all prime numbers larger than $3$ can be expressed as $3k + 1$ and or $3k + 2$. Plugging those in I still see no solution.
EDIT: $p$ can only be a prime number.
| So you know that all primes larger than 3 are of the form $3k + 1$ or $3k + 2$.
Do you know which of those forms $2^p$ is? Obviously $2 = 3 \times 0 + 2$. Then $4 = 3 \times 1 + 1$, $8 = 3 \times 2 + 2$, $16 = 3 \times 5 + 1$, $32 = 3 \times 10 + 2$, etc. In short, $2^n \equiv 2 \pmod 3$ if $n$ is odd, so if $p$ is a prime number greater than 2, it's odd, and consequently $2^p \equiv 2 \pmod 3$. Let's say $2^p = 3m + 2$ for later reference.
Now let's review what happens with the squares of primes. Suppose $p = 3k + 2$. Then $p^2 = (3k + 2)^2 = 9k^2 + 12k + 4 \equiv 1 \pmod 3$.
And so $2^p + p^2 = (3m + 2) + (9k^2 + 12k + 4) = 9k^2 + 12k + 3m + 6$, and, just in case it's not obvious enough: $$\frac{9k^2 + 12k + 3m + 6}{3} = 3k^2 + 4k + m + 2.$$
| {
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"source": "stackexchange",
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How to evaluate this limit??
Evaluate: $$\ f(x)= \lim_{n\rightarrow \infty}\left( \dfrac{n^n(x+n)\left( x+\dfrac{n}{2}\right)\left( x+\dfrac{n}{3}\right)... \left( x+\dfrac{n}{n}\right)}{n!(x^2+n^2)\left( x^2+\dfrac{n^2}{4}\right)\left( x^2+\dfrac{n^2}{9}\right)...\left( x^2+\dfrac{n^2}{n^2}\right)}\right)^{\dfrac{x}{n}}$$
$x\in R^+$
Find the coordinates of the maxima of $f(x)$.
| We solve it using Riemann sum. You can write it as $$\left(\frac {(\frac {x}{n}+1)\cdots (\frac {nx}{n}+1)}{((\frac{x}{n})^2+1)\cdots ((\frac {nx}{n})^2+1)}\right)^{x/n}$$
Now take logs and use the fact of Riemann sum. Put $\dfrac {1}{n}=\mathrm dt,\dfrac {r}{n}=t$ in numerator and denominator where $r$ varies from $1-n $ thus the limit changes to integration. The equation becomes $\displaystyle \ln f (x))=x\int _0 ^1\ln \left(\frac {xt+1}{x^2t^2+1}\right)\mathrm dt $
Note we are integrating wrt $t $ so treat $x$ as constant. Put $xt=p $ thus $x\, \mathrm dt=\mathrm dp $ . Now limits change from $0$ to $x$ so we have $\displaystyle \ln (f (x))=\int _0 ^x \ln \left(\frac {1+p}{1+p^2}\right)\, \mathrm dp. $ Now use log property to separate the integral and then by parts . So we get $\displaystyle \ln (f (x))=(x+1)(\ln (x+1)-1)-\int _0 ^1 \frac {2p^2+2-2 }{1+p^2}\, \mathrm dp $ thus the result is $\ln (f (x))=(x+1)(\ln (x+1)-1)+1-x\ln (1+x^2)+2x-2\arctan (x)$ .
Now you can see that $f'(x)=\ln (1+x)-\ln (1+x^2) $ thus maxima is achived either at $0$ or $1$ . Using second derivative test we see the maxima is at $1$ thus putting $x=1$ we have $f (1)=e^{\ln (2)+1-\frac {\pi}{2}}=2.e.e^{-pi/2}=2ei^i\approx 1.12$ thus coordinates are $(1,1.13)$.
| {
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Evaluate $\int_0^{2\pi}\frac{1}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{ for }\ A,B <<1$ I need to evaluate the definite integral $$\int_0^{2\pi}\frac{1}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{ for various}\ A,B \text{; with}\ A,B<<1.$$
Wolfram Alpha provides the following indefinite general solution:-
$$\int \frac{1}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta = -(2/K) \tanh^{-1} ( \frac{A-(B-1)\tan(\theta/2)}{K}) $$
where $K = \sqrt{A^2 + B^2 -1}$.
But I am having trouble checking it for the simple case when $A=B=0$ when I would expect the answer to be given by:-
$$\int_0^{2\pi}\frac{1}{1 + 0 + 0}\, \mathrm{d}\theta = 2\pi.$$
I have approached the Wolfram Alpha solution thus:-
$$ -(2/K) \tanh^{-1} ( \frac{\tan(2\pi/2)}{K}) +(2/K) \tanh^{-1} ( \frac{\tan(0/2)}{K}) $$
$$ -(2/K) \tanh^{-1} ( \frac{\tan(\pi)}{K}) +(2/K) \tanh^{-1} ( \frac{\tan(0)}{K}) $$
$$ -(2/K) \tanh^{-1} ( \frac{0}{K}) +(2/K) \tanh^{-1} ( \frac{0}{K}) $$
which gives the result of zero.
I presume this error comes from trying to integrate across the range $0, 2\pi$ where the $\tan$ function has singularities at $\pi/2$ and $3\pi/2$.
However when I try and break the integration into the three continuous ranges $0,\pi/2$ and $\pi/2,3\pi/2$ and $3\pi/2,2\pi$ I am still getting a result of zero thus:-
$$ -(2/K) \tanh^{-1} ( \frac{\tan(2\pi/2)}{K}) +(2/K) \tanh^{-1} ( \frac{\tan(3\pi/4)}{K}) + $$
$$ -(2/K) \tanh^{-1} ( \frac{\tan(3\pi/4)}{K}) +(2/K) \tanh^{-1} ( \frac{\tan(\pi/4)}{K}) +$$
$$ -(2/K) \tanh^{-1} ( \frac{\tan(\pi/4)}{K}) +(2/K) \tanh^{-1} ( \frac{\tan(0)}{K}) $$
leading to
$$ -(2/K) \tanh^{-1} ( \frac{0}{K}) +(2/K) \tanh^{-1} ( \frac{-1}{K}) + $$
$$ -(2/K) \tanh^{-1} ( \frac{-1}{K}) +(2/K) \tanh^{-1} ( \frac{1}{K}) +$$
$$ -(2/K) \tanh^{-1} ( \frac{1}{K}) +(2/K) \tanh^{-1} ( \frac{0}{K}) $$
which gives the same result of zero.
I would be grateful if somebody could tell me where I am going wrong here?
EDIT 1: I have accepted the solution provided kindly by Dr. MV.
I have posted a related question which seeks to understand where my original evaluation of the definite integrand goes wrong.
EDIT 2: In the related question comments from user mickep pointed out that the wrong partitions had been used in the original evaluation. Using the correct partitions ($0...\pi$) and ($\pi...2\pi$) leads to the correct answer, for $A=B=0$, of $2\pi$ (as described in my self-answer to that same question).
EDIT 3: It was pointed out by user mickep in comments to the related question that the Wolfram Alpha solution
$$ -\left(\frac{2}{K1}\right) {\arctan}h \left( \frac{A-(B-1)\tan(\theta/2)}{K1}\right) $$
where $K1= \sqrt{A^2 + B^2 -1}$.
is not as friendly as an alternative solution (reported by user mickep) which is:
$$ +\left(\frac{2}{K2}\right) \arctan \left( \frac{A+(1-B)\tan(\theta/2)}{K2}\right) $$
where $K2 = \sqrt{1 - A^2 - B^2}$.
| Note that the integral of interest fails to converge if $\sqrt{A^2+B^2}\ge 1$. So, we restrict $A$ and $B$ such that $\sqrt{A^2+B^2}< 1$.
Then, we can write
$$\begin{align}
\int_0^{2\pi}\frac{1}{1+A\sin(\theta)+B\cos(\theta)}\,d\theta&=\int_0^{2\pi}\frac{1}{1+\sqrt{A^2+B^2} \cos(\theta-\arctan(A/B)}\,d\theta\\\\
&=\int_{-\arctan(A/B)}^{2\pi-\arctan(A/B)}\frac{1}{1+\sqrt{A^2+B^2} \cos(\theta)}\,d\theta\\\\
&=2\int_{0}^{\pi}\frac{1}{1+\sqrt{A^2+B^2} \cos(\theta)}\,d\theta\tag1
\end{align}$$
where we exploited both the $2\pi$-periodicity and the evenness of the cosine function.
Next, we enforce the Weierstrass Substitution, $ t=\tan(\theta/2)$, in $(1)$ to obtain
$$\begin{align}
\int_0^{2\pi}\frac{1}{1+A\sin(\theta)+B\cos(\theta)}\,d\theta&=4\int_0^\infty \frac{1}{(1+\sqrt{A^2+B^2})+(1-\sqrt{A^2+B^2})t^2}\,dt\\\\
&=\frac{4}{1-\sqrt{A^2+B^2}}\int_0^\infty \frac{1}{\frac{1+\sqrt{A^2+B^2}}{1-\sqrt{A^2+B^2}}+t^2}\,dt \tag 2\\\\
&=\frac{4}{1-\sqrt{A^2+B^2}} \left.\left(\frac{\arctan\left(\frac{\sqrt{1-\sqrt{A^2+B^2}}}{\sqrt{1+\sqrt{A^2+B^2}}}t\right)}{\sqrt{\frac{1+\sqrt{A^2+B^2}}{1-\sqrt{A^2+B^2}}}}\right)\right|_{0}^{\infty} \tag 3\\\\
&=\frac{2\pi}{\sqrt{1-A^2-B^2}} \tag 4
\end{align}$$
Note that we could have written $(2)$ as
$$\begin{align}
\frac{4}{1-\sqrt{A^2+B^2}}\int_0^\infty \frac{1}{\frac{1+\sqrt{A^2+B^2}}{1-\sqrt{A^2+B^2}}+t^2}\,dt&=\frac{4}{1-\sqrt{A^2+B^2}}\int_0^\infty \frac{1}{t^2-\frac{\sqrt{A^2+B^2}+1}{\sqrt{A^2+B^2}-1}}\,dt\\\\
&=\frac{4}{\sqrt{A^2+B^2}-1}\left.\left( \frac{\text{arctanh}\left(\sqrt{\frac{\sqrt{A^2+B^2}-1}{\sqrt{A^2+B^2}+1}}t\right)}{\sqrt{\frac{\sqrt{A^2+B^2}+1}{\sqrt{A^2+B^2}-1}}}\right)\right|_{0}^\infty\\\\
&=\frac{4}{\sqrt{A^2+B^2}-1}\,\left(\frac{i\pi/2}{\sqrt{\frac{\sqrt{A^2+B^2}+1}{\sqrt{A^2+B^2}-1}}}\right)\\\\
&=\frac{2\pi}{\sqrt{1-A^2-B^2}}
\end{align}$$
as expected!
| {
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Proving three lines are concurrent
In the triangle $ABC$ the segment $AD$ is altitude. $M$ is the midpoint of $BC$ and $N$ is the reflection of $M$ in $D$. The circumcircle of $ANM$ meets $AB$ at $P$ and $AC$ at $Q$. Show that $AN, BQ, CP$ are concurrent.
All I have tried to do is angle chasing and hopefully get the sum of the three angles at the point to equal 180 degrees, but I'm stuck. I have also got that AMN is isosceles with AM=AN.
| One can also solve it without trigonometry:
First, show that $AP \cdot AB = QA \cdot CA$. This can be done by producing $AD$ until it intersects the circle at $L$ for the second time. Then $\angle \, LPA = \angle \, BDA = 90^{\circ}$ so triangles $LPA$ and $BDA$ are similar and this latter fact implies the ratio $\frac{AB}{AL} = \frac{AD}{AP}$ which is equivalent to the identity $AP \cdot AB = AD \cdot AL\,$. Analogously, $AD \cdot AL = QA \cdot CA \,$
Second, by the intersecting secant theorem,
$$BN \cdot BM = PB \cdot AB \,\,\, \text{ and } \,\,\, NC \cdot MC = CA \cdot CQ$$ which means that
$$BN = \frac{PB \cdot AB}{BM} \,\,\, \text{ and } \,\,\, NC = \frac{CA \cdot CQ}{MC} = \frac{CA \cdot CQ}{BM}$$ because $BM = MC$.
Third, calculate the ratio
$$\frac{BN}{NC}\cdot\frac{CQ}{QA}\cdot\frac{AP}{PB} = \frac{PB \cdot AB}{BM}\cdot \frac{BM}{CA \cdot CQ} \cdot \frac{CQ}{QA}\cdot\frac{AP}{PB} = \frac{PB \cdot AB}{CA \cdot CQ} \cdot \frac{CQ}{QA}\cdot\frac{AP}{PB} = \frac{AP \cdot AB}{QA \cdot CA} $$ However, by $AP \cdot AB = QA \cdot CA$
$$\frac{BN}{NC}\cdot\frac{CQ}{QA}\cdot\frac{AP}{PB} = \frac{AP \cdot AB}{QA \cdot CA} = 1$$ which, by Ceva's theorem, is possible if and only if the lines $AN, \, BQ$ and $CP$ are concurrent.
| {
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$\int ^{\infty}_0 \frac{\ln x}{x^2 + 2x+ 4}dx$ We have to find the integration of
$$\int ^{\infty}_0 \frac{\ln x}{x^2 + 2x+ 4}dx$$
In this I tried to do substitution of $x=e^t$
After that got stuck .
| May be this will be a more elementary solution.
Let us start with a the substitution $x=2y$. Then
\begin{gather*}
I = \int_{0}^{\infty}\dfrac{\ln(x)}{x^2+2x+4}\, dx = \int_{0}^{\infty}2\dfrac{\ln(2y)}{4y^2+4y+4}\, dy =\\[2ex] \dfrac{1}{2}\int_{0}^{\infty}\dfrac{\ln(2)}{y^2+y+1}\, dy + \dfrac{1}{2}\int_{0}^{\infty}\dfrac{\ln y}{y^2+y+1}\, dy=
\dfrac{\ln(2)}{2}I_1+ \dfrac{1}{2}I_2\tag{1}
\end{gather*}
where
\begin{equation*}
I_1 = \int_{0}^{\infty}\dfrac{1}{y^2+y+1}\, dy =\dfrac{4}{3}\int_{0}^{\infty}\dfrac{1}{\left(\frac{2y+1}{\sqrt{3}}\right)^{2}+1}\, dy = \dfrac{2}{\sqrt{3}}\left[\arctan\left(\dfrac{2y+1}{\sqrt{3}}\right)\right]_{0}^{\infty} = \dfrac{2\pi}{3\sqrt{3}}\tag{2}
\end{equation*}
and
\begin{equation*}
I_2 = \int_{0}^{\infty}\dfrac{\ln(y)}{y^2+y+1}\, dy =\left[y=\dfrac{1}{z}\right] = \int_{0}^{\infty}\dfrac{-\ln(z)}{z^2+z+1}\, dz = -I_2. \tag{3}
\end{equation*}
But (3) implies that $I_2=0$. This and (2) substituted in (1) give us $I = \dfrac{\pi\ln(2)}{3\sqrt{3}}$.
| {
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Find the probability that $m^2-n^2$ is divisible by $4$ Two integers are $m$ , $n$ chosen at random with replacement from the set of integers $1,2,3..9$ . Find the probability that $m^2-n^2$ is divisible by $4$ .
my solution : Two integers can be chosen (with replacement ) in $9\times 9=81$ ways .
$m^2-n^2 $ is divisible iff either both $m,n$ are odd or both are even .
There are $4\times 4=16 $ ways way of obtaining both even integers and $5\times 5=25 $ ways of obtaining both odd integer.
So total number of favorable outcome $=16+25=41$ .
Hence required probability = $\frac{41}{81}$ .
Now i would like to slightly change the question :
Two integers are $m$ , $n$ chosen at random without replacement from the set of integers $1,2,3..9$ . Find the probability that $m^2-n^2$ is divisible by $4$
my solution when order consideration is relevant :
Total number of obtaining two integers $= 2\times 9\times 8=2\times 72$
Numbers ways of obtaining two even numbers $=2\times 4\times 3 $ and number of ways obtaining odd numbers $=2\times 5\times 4$ .
So total number of favorable cases= $2(4\times 3+5 \times 4)=2\times 32 $ .
Hence required probability $=\frac{2 \times 32 }{2 \times 72 }=\frac{32}{72}$
my solution when order is considered irrelevant
Total numbers of cases $=9\times 8=72$
Numbers of ways obtaining both even and both odd integers is $4\times3=12$ and $5\times4=20$ respectably .
So number of favorable cases$=12+20=32$ .
Hence probability =$\frac{32}{72}$
Is my understanding in above problems correct ?
Please point out if i calculated anything wrong?
Thank you
| We will use the following three things:
1)
$(x^2-y^2) = (x-y)(x+y)$
2) Odd + Odd = Even
Odd - Odd = Even
Even + Even = Even
Even - Even = Even
Odd + Even = Even + Odd = Odd
Odd - Even = Even - Odd = Odd
3)
Even $\cdot$ Even = Multiple of $4$
Odd $\cdot$ Odd = Odd $\not = $ Multiple of $4$
Hence if $x,y $ both Odd or both Even we have $(x^2-y^2) = (x-y)(x+y) =$ Even $\cdot$ Even = Multiple of $4$
And if one is Even and the other is Odd we have $(x^2-y^2) = (x-y)(x+y) =$ Odd $\cdot$ Odd $\not =$ Multiple of $4$
Hence for $(x^2-y^2) $ to be divisible by 4 we need them to be "same" , with respect to divisibility by two. Think of this as there being $9$ cards. $4$ Red and $5$ Black. What is the probability that if, with replacement, we pick the same colour twice? Clearly this is $\frac{5}{9}\cdot \frac{5}{9} + \frac{4}{9}\cdot\frac{4}{9} = \frac{41}{81} \approx 0.5 $
| {
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Find the value of $\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}$ Find the value of $\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}$
My Attempt:
$$\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}$$
$$\sin \dfrac {180}{10} - \sin \dfrac {3\times 180}{10}$$
$$\sin 18^\circ - \sin 54^\circ$$
Now,
Let $A=18^\circ$.
$$5A=90^\circ$$
$$2A+3A=90^\circ$$
$$3A=90^\circ - 2A$$
Taking 'sin' on the both sides,
$$\sin 3A=\sin (90^\circ - 2A)$$
$$3\sin A-4\sin^3 A=\cos 2A$$
What should I do further?
| $\cos 2A=1-2\sin^2 A$, then substitute $x=\sin A$.
| {
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Find the constant term in the expansion of $(x^2+1)(x+\frac{1}{x})^{10}$ I can't solve this problem. How to solve it?
The Problem is
"Find the constant term in the expansion of $
\left({{x}^{2}\mathrm{{+}}{1}}\right){\left({{x}\mathrm{{+}}\frac{1}{x}}\right)}^{\mathrm{10}}
$"
| Let $c_k$ be the coefficient of $\left(x+\dfrac{1}{x}\right)^{10}$ before $x^k$. Then
$$(x^2+1)\left(x+\dfrac{1}{x}\right)^{10} = (x^2+1)\sum\limits_{-\infty}^{+\infty} c_kx^k = \sum\limits_{-\infty}^{+\infty} c_kx^{k+2} + \sum\limits_{-\infty}^{+\infty} c_kx^{k} = \sum\limits_{-\infty}^{+\infty} (c_{k-2} + c_k)x^{k}$$
Since we are looking for the coefficient before $x^0$, the answer to your question is $c_{-2} + c_{0}$.
According to binomial formula, we have:
$$\left(x+\dfrac{1}{x}\right)^{10} = \sum\limits_{k=0}^{10} \binom{10}{k}x^{k}\left(\dfrac{1}{x}\right)^{10-k} = \sum\limits_{k=0}^{10} \binom{10}{k}x^{2k - 10}$$
Thus,
$$c_{-2} = \{2k-10=-2 \Rightarrow k = 4\} = \binom{10}{4}$$
$$c_0 = \{2k-10=0 \Rightarrow k = 5\} = \binom{10}{5}$$
So, the answer is $$c_{-2} + c_0 = \binom{10}{5} + \binom{10}{4} = \binom{11}{5} = 462$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove inequality $\sum\frac{{{a^3} - {b^3}}}{{{{\left( {a - b} \right)}^3}}}\ge \frac{9}{4}$ Given $a,b,c$ are positive number. Prove that $$\frac{a^3-b^3}{\left(a-b\right)^3}+\frac{b^3-c^3}{\left(b-c\right)^3}+\frac{c^3-a^3}{\left(c-a\right)^3}\ge \frac{9}{4}$$
$$\Leftrightarrow \sum \frac{3(a+b)^2+(a-b)^2}{(a-b)^2} \ge 9$$
$$\Leftrightarrow \frac{(a+b)^2}{(a-b)^2}+\frac{(b+c)^2}{(b-c)^2}+ \frac{(c+a)^2}{(c-a)^2}\ge 2$$
Which $$\frac{a+b}{a-b}.\frac{b+c}{b-c}+\frac{b+c}{b-c}.\frac{c+a}{c-a}+\frac{c+a}{c-a}.\frac{a+b}{a-b} =-1$$
Use $x^2+y^2+z^2 \ge -2(xy+yz+zx)$ then inequality right
I don't know why use $x^2+y^2+z^2 \ge -2(xy+yz+zx)$ then inequality right?
P/s: Sorry i knowed, let $x=\frac{a+b}{a-b}$
We have $(x+1)(y+1)(z+1)=(x-1)(y-1)(z-1) => xy+yz+zx=-1 $
| I think it means that the variables are reals.
By your identity we obtain $$\left(\sum_{cyc}\frac{a+b}{a-b}\right)^2=\sum_{cyc}\frac{(a+b)^2}{(a-b)^2}-2\geq0$$
Thus, $$\sum_{cyc}\frac{(a+b)^2}{(a-b)^2}\geq2$$ or
$$\sum_{cyc}\left(\frac{(a+b)^2}{(a-b)^2}-1\right)\geq-1$$ or
$$\sum_{cyc}\frac{4ab}{(a-b)^2}\geq-1$$ or
$$\sum_{cyc}\frac{3ab}{(a-b)^2}\geq-\frac{3}{4}$$ or
$$\sum_{cyc}\left(\frac{3ab}{(a-b)^2}+1\right)\geq3-\frac{3}{4}$$ or
$$\sum_{cyc}\frac{a^2+ab+b^2}{(a-b)^2}\geq\frac{9}{4}$$ or
$$\sum_{cyc}\frac{a^3+b^3}{(a-b)^3}\geq\frac{9}{4}.$$
Done!
| {
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Triangle - length of the sides - proof a, b and c are the lengths of the sides of a triangle. Prove that
$$a^2+b^2 \ge \frac{1}{2}c^2$$
Let $\gamma$ be the angle between sides a and b. then:
$$a^2 + b^2 - 2ab\cos(\gamma) = c^2$$
Hence we need to prove that
$$a^2+b^2 \ge \frac{1}{2}c^2$$
$$2a^2+2b^2 \ge a^2+b^2 -2ab\cos(\gamma) $$
$$a^2+b^2+2ab\cos(\gamma) \ge0$$
Knowing that $\cos(\gamma) \ge-1$, we get $a^2+b^2+2ab\cos(\gamma) \ge a^2 +b^2 -2ab =(a-b)^2$
I have carried this problem so far but now I am stuck. How should I proceed?
| The squared length of the median through $C$ is exactly $\frac{1}{4}\left(2a^2+2b^2-c^2\right)$ and that gives $a^2+b^2\geq \frac{1}{2}c^2$ as a straightforward consequence.
| {
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What the set $I$ of invertible elements in $\mathbb{Z}_{9}$? $5.2 = 1$
The set of invertible elements in $\mathbb{Z}_{9}$ is $I = \{1,2,5\}$.
Q1: Is it correct?
I was also asked to show that it is not isomorphic to $S_{3}$
How can I show that it is not isomorphic? Thank you.
| $$\begin{array}{c|c}
\times&0&1&2&3&4&5&6&7&8\\\hline
0&0&0&0&0&0&0&0&0&0\\\hline
1&0&\color{red}1&2&3&4&5&6&7&8\\\hline
2&0&2&4&6&8&\color{red}1&3&5&7\\\hline
3&0&3&6&0&3&6&0&3&6\\\hline
4&0&4&8&3&7&2&6&\color{red}1&5\\\hline
5&0&5&\color{red}1&6&2&7&3&8&4\\\hline
6&0&6&3&0&6&3&0&6&3\\\hline
7&0&7&5&3&\color{red}1&8&6&4&2\\\hline
8&0&8&7&6&5&4&3&2&\color{red}1
\end{array}$$
Therefore, the invertible elements of $\Bbb Z_9$ are $\{1,2,4,5,7,8\}$.
Indeed, $S_3$ has $6$ elements while $\Bbb Z_9$ has $9$ elements, so they are not isomorphic to each other.
In general, for $\Bbb Z_n$, $x$ is invertible iff $\gcd(x,n)=1$.
By Bézout's Lemma, $\gcd(x,n)=1 \iff \exists a,b:ax+bn=1 \iff \exists a:ax=1$.
| {
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Help calculating a math limit Can anyone help with this limit?
\begin{equation*}
\lim_{x \rightarrow 4}
\frac{16\sqrt{x-\sqrt{x}}-3\sqrt{2}x-4\sqrt{2}}{16(x-4)^2}
\end{equation*}
I've tried a variable change of \begin{equation*} y=\sqrt{x} \end{equation*} but this didn't help.
| The variable change is a good idea to begin with, but I'd use $2y=\sqrt{x}$
$$
\lim_{y\to 1}
\frac{16\sqrt{4y^2-2y}-12\sqrt{2}\,y^2-4\sqrt{2}}{16(4y^2-4)^2}=
\frac{4\sqrt{2}}{256}
\lim_{y\to 1}
\frac{4\sqrt{2y^2-y}-(3y^2+1)}{(y^2-1)^2}
$$
You can do this with a Taylor expansion or by “rationalizing”: note that
$$
16(2y^2-1)-(3y^2+1)^2=-9y^4+26y^2-17=-(y^2-1)(9y^2-17)
$$
| {
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Finding $\int x \sqrt{1- x^2 \over 1 + x^2}dx$.
$$J =\int x \sqrt{1- x^2 \over 1 + x^2}dx$$
Substituting $u = \sqrt{1 + x^2}$
$$J = \int \sqrt{2 - u^2} du = \sqrt{2}\int \sqrt{1 - \left({u \over \sqrt{2}}\right)^2} du$$
Now substituting $\sin z = u/\sqrt{2}$
$$J = 2\int \cos^2 z dz = z + (\sin 2z)/2 + C = \arcsin\left(\sqrt{1+ x^2}\over 2\right) + {\sqrt{1-x^4}\over 2} + C $$.
The give answer is $\displaystyle \arcsin(x^2) + {\sqrt{1-x^4}\over 2} + C$.
What went wrong in my attempt ?
| Put $t=x^2$
and
$$\frac {1-t}{1+t}=u^2$$
or
$$t=\frac{1-u^2}{1+u^2} $$
$$=\frac {2}{1+u^2}-1$$
and
$$dt=\frac{-4u}{(1+u^2)^2} $$
| {
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Decide if the series $\sum\frac{4^{n+1}}{3^{n}-2}$ converges or diverges and, if it converges, find its sum
Decide if the series $$\sum_{n=1}^\infty\frac{4^{n+1}}{3^{n}-2}$$ converges or diverges and, if it converges, find its sum.
Is this how you would show divergence attempt:
For $n \in [1,\infty), a_n = \frac{4^{n+1}}{3^n -2} \geq 0$
For $n \in [1,\infty), a_n = \frac{4^{n+1}}{3^n-2} \geq \frac{4^{n+1}}{3^n} = b_n$
Since $\sum_{n=1}^{\infty} \frac{4^{n+1}}{3^n}$ is a geometric series with $r = \frac{4}{3} > 1$. Therefore it diverges by the geometric series test and by the comparison test $\sum a_n$ diverges too.
| Assume that the sequence converges for some $n$. Then:
$$\lim_{n \to \infty} \left| \ \frac{4^{n+2}}{3^{n+1}-2} \cdot \frac{3^n-2}{4^{n+1}}\ \right| <1$$
$$\Rightarrow \lim_{n \to \infty} \left| \ \frac{4^{n+2}}{3^{n+1}} \cdot \frac{3^n}{4^{n+1}} \ \right| < 1$$
$$\Rightarrow \lim_{n \to \infty} \left| \ 4 \cdot \left(\frac{4}{3}\right)^{n+1} \cdot \left(\frac{3}{4}\right)^{n+1} \cdot \frac{1}{3} \ \right| < 1$$
$$\Rightarrow\left| \ \frac{4}{3} \ \right| < 1 \ \text{which is obviously false}$$
Therefore, the sum never converges for any $n$.
| {
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Find $\int \sqrt{a- x \over x} dx$.
$$ \int \sqrt{a- x \over x} dx$$
Substituting $u = \sqrt{a - x}$,
$$\int \sqrt{a- x \over x} dx = -2\int {u^2 \over \sqrt{a - u^2}} du$$
Now, for $\sqrt{a}\sin t = u$,
$$-2\int {a\sin^2 t \over \sqrt{a} \cos t} \ \ dt \ \ \sqrt{a }\cos t = -a\int(1 - \cos 2t) dt = a(t - \cos t \sin t) + C$$
Substituting for $\displaystyle t = \arcsin\left({u\over \sqrt{a}}\right) =\arcsin\left({\sqrt{a - x}\over \sqrt{a}}\right) $ I get the answer as
$$\bbox[7px,Border:2px solid black]{a\left(\sqrt{x\over a} \sqrt{a -x \over a}- \arcsin\sqrt{a -x \over a}\right) + C}$$,
But the given answer is $\displaystyle {a\left(\sqrt{x\over a} \sqrt{a -x \over a}+ \arcsin\sqrt{x \over a}\right) + C}$
I am sure that $\displaystyle \arcsin\sqrt{x \over a} \ne - \arcsin\sqrt{a -x \over a}$.
Where did I got wrong ?
| You didn't go wrong - actually, we have that
$$\arcsin\sqrt{x \over a} = \frac{\pi}{2} - \arcsin\sqrt{a -x \over a}$$
as $$\left(\sqrt{x \over a} \right)^2 + \left(\sqrt{a -x \over a} \right)^2=1.$$
| {
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Determine the number of permutations of $ \ \{1,2,3,4,5,6,7,8,9,10\} \ $ that have exactly 3 numbers in their natural position Determine the number of permutations of $ \ \{1,2,3,4,5,6,7,8,9,10\} \ $ that have exactly 3 numbers in their natural position. $$ $$ Is it $ \ \ \begin{pmatrix} 10 \\ 3 \end{pmatrix} \times 7 ! \ $ ?
| You need choose $3$ elements that stay on their positions in $\binom{10}{3} = 120$ and multiply it by the number $!7$ of permutations of special kind called derangement on remaining $7$ elements. There are many ways to compute number $!k$ (see link), particularly it is known to be equal to $\frac{k!}{e}$ rounded to the nearest integer. Therefore $\frac{5040}{e} \approx 1854 =\,!7$ and the desired number of permutations on $10!$ elements keeping exactly $3$ of them in place is $\binom{10}{3} \cdot\,!7 = 120 \cdot 1854 = 222\,480$.
Another way to compute number of such permutation is inclusion-exclusion principle. There are $\binom{10}{3}\cdot 7!$ ways to select $3$ elements to stay in place and permute all remaining. But for each of $\binom{10}{4}$ ways to select $4$ elements each permutation that keeps them in place was computed $4$ times, but should not be computed at all. Continuing these thoughts we get that the right number of desired permutations is
$$\binom{10}{3}\cdot 7! - \binom{10}{4} \cdot 6! \cdot \binom{4}{3} + \binom{10}{5} \cdot 5! \cdot \binom{5}{3} - \binom{10}{6} \cdot 4! \cdot \binom{6}{3}\\
+ \binom{10}{7} \cdot 3! \cdot \binom{7}{3} - \binom{10}{8} \cdot 2! \cdot \binom{8}{3} + \binom{10}{9} \cdot 1! \cdot \binom{9}{3} - \binom{10}{10} \cdot 0! \cdot \binom{10}{3}\\
= 604800 - 604800 + 302400 - 100800 + 25200 - 5040 + 840 - 120 = 222480.$$
| {
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How do I compute the taylor series for $(1+x+x^2)^\frac{1}{x}$ I tried rewriting $(1+x+x^2)^\frac{1}{x}$ as $e^{\frac{1}{x}\ln(1+x+x^2)}$ and then computing the taylor series of $\frac{1}{x}$ and $\ln(1+x+x^2)$ but I'm still not getting the correct answer..
| Note that $$f(x) = (1 + x + x^{2})^{1/x} = \exp\left(\frac{\log(1 + x + x^{2})}{x}\right) = \exp\left(\frac{\log(1 - x^{3}) - \log(1 - x)}{x}\right)$$ so that we can get the series $$\frac{\log(1 - x^{3}) - \log(1 - x)}{x} = \sum_{n = 1}^{\infty}\frac{x^{n - 1} - x^{3n - 1}}{n} = 1 + \frac{x}{2} - \frac{2x^{2}}{3} + \cdots$$ and we can exponentiate this to get $$(1 + x + x^{2})^{1/x} = e\cdot\exp(x/2 - 2x^{2}/3 + \cdots) = e\left(1 + \frac{x}{2} - \frac{13}{24}x^{2} + \cdots\right)$$ Note that the calculation of coefficients in general is complicated with no specific formula.
| {
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What is $x$ in the formula? What is $x$ (assume it's integer) in $512p+ 1 = x^3$, where $p$ is a prime number.
Attempt:
$$a^3 - b^3 = (a-b) (a^2 + ab+b^2) \Longrightarrow 512p = x^3 -1 = (x-1)(x^2+x+1).$$
Here I got stuck. Am I suppose to plug in $p$ and try it one by one?
What about $16p + 1 = x^3$??
| You are on the right track. Since $512p+ 1 = x^3$, we know that $x^3$ is odd, so that $x$ must be odd.
When you have factored it into $512p = x^3 -1 = (x-1)(x^2+x+1)$, since $x$ is odd, $x^2+x+1$ is odd, so $x-1$ must be some multiple of 512. However, if $x-1$ is not exactly 512, $p$ won't be a prime. So we have $x-1 = 512$, hence $x=513$. Substitute this solution back, you will find that $p$ is not a prime, so this equation has no solution when $p$ is prime.
| {
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Point On a Plane Closest to a Point
*
*Let $\mathcal{P}$ be the plane containing the points $(-3,4,-2)$, $(1,4,0)$, and $(3,2,-1)$.
Find the point in this plane that is closest to $(0,3,-1)$.
*Let $\mathcal{P}$ be the plane containing the points $(-3,4,-2)$, $(1,4,0)$, and $(3,2,-1)$.
Let $\ell$ be the line containing the points $(2, 4, -3)$ and $(-1, -1, -9)$.
Find the intersection of the plane $\mathcal{P}$ and the line $\ell$.
I don't know where to start. I don't know any formulas for calculating a closest point.
| Problem 1.
The equation of the plane containing the points $(−3,4,−2)$, $(1,4,0)$ and $(3,2,−1)$ is given by
\begin{align*}
\begin{vmatrix}
x & y & z & 1 \\
-3 & 4 & -2 & 1 \\
1 & 4 & 0 & 1 \\
3 & 2 & -1 & 1
\end{vmatrix}
&= 0
\end{align*}
or, $x + 2y - 2z - 9 = 0$
The vector normal to this plane is given by $(1, 2, -2)$
The point in this plane that is closest to $(0,3,−1)$ must be the foot of the perpendicular from this point to the above plane.
The equation of the perpendicular is
$\frac{x - 0}{1} = \frac{y-3}{2} = \frac{z+1}{-2} = k$
or, $x = k, y = 2k + 3, z = -2k-1$
Putting this in the equation of the plane,
$k + 2(2k+3) + 2(2k+1) - 9 = 0$
or, $k = \frac{1}{9}$
Hence the point is $(\frac{1}{9}, \frac{29}{9}, -\frac{11}{9})$
Problem 2.
The equation of the line containing the points $(2,4,−3)$ and $(−1,−1,−9)$ is given by
$\frac{x - 2}{-1-2} = \frac{y-4}{-1-4} = \frac{z+3}{-9+3} = t$
or, $x = -3t+2, y = -5t+4, z = -6t-3$
Putting this in the equation of the plane,
$-3t+2 + 2(-5t+4) + 2(6t+3) - 9 = 0$
or, $t = 7$
So the point of intersection is $(-19, -31, -45)$
[Please check the calculations]
| {
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Why can we see if a binary number is divisible by 3 when we look at the $1$'s position Why is a binary number $z=b_n\dots b_1b_0$ divisible by $3$ iff the amount of $1$'s on the even position $b_{2n}$ minus the amount of $1$'s on odd positions $b_{2n+1}$is divisible by $3$?
So meaning $11000_b$ $(=24)$ is divisible by 3 because we have one $1$ on even and one $1$ on odd Position and $1-1=0$ is divisible by $3$.
$101_b$ $(=5)$is not divisible. Because $2-0=2$ is not divisible by $3$. What is the mathematical explanation?
| Even powers of $2$ are congruent to $1$ modulo $3$. Odd powers of $2$ are congruent to $2$ modulo $3$.* So suppose that in the binary representation of some number $k$, there are $m$ $1$'s in the even positions, and $n$ $1$'s in the odd positions. Then,
\begin{align}
k & \equiv m+2n \mod 3 \\
& \equiv m+2n-3n \mod 3 \\
& \equiv m-n \mod 3
\end{align}
Therefore, in order for $k$ to be congruent to $0$ modulo $3$ (that is, divisible by $3$), $m$ and $n$ must be congruent modulo $3$. In other words, $m$ and $n$ must differ by a multiple of $3$.
*There are a few ways to see this. One of the simplest is induction: Observe that $2^0 = 1$ and $2^1 = 2$, then
\begin{align}
2^{r+2} & = 4 \times 2^r \\
& = 2^r + 3 \times 2^r \\
& \equiv 2^r \mod 3
\end{align}
| {
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factorization of polynomials in $ \mathbb F_7$ I am not sure in finding a factorization of $f(X) = 6 x^6 +1 \in \mathbb F_7$.
I got $f(X) =6\cdot (x-1)\cdot(x+1)\cdot(x-2)\cdot(x+2)\cdot(x-3)\cdot(x+3) $
Is that correct?
| In $\Bbb{F}_7$ one has on one hand
$$\begin{align}
6x^6+1&=6(x^6-1)\\
&=6(x^2-1)(x^4+x^2+1)\\
&=6(x-1)(x+1)(x^4+x^2+1)
\end{align}$$
On the other hand
$$(x-2)(x+2)(x-3)(x+3)=(x^2-4)(x^2-2)=x^4-6x+8=x^4+x^2+1$$
So the factorisation is correct.
| {
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Angle bisector in a right angled triangle In a right angled triangle, the legs adjacent to the right angle are equal to $a$ and $b$. Prove that the length of the bisector (of the right angle) is equal to $$\frac{a\cdot b\cdot \sqrt{2}}{a+b}.$$
While approaching this question, I was very puzzled as to how I would end up with this expression.
Additionally, I couldn't figure out where the $\sqrt{2}$ would come from, other than the sine or cosine of $45$ degrees (from the bisector).
| In a General Triangle
Using the Law of Sines, we get that side $c$ is broken up as follows by the angle bisector:
Using the Law of Cosines we get
$$
\left(\frac{ac}{a+b}\right)^2+d^2-\frac{2acd}{a+b}\cos(D)=a^2\tag{1}
$$
and
$$
\left(\frac{bc}{a+b}\right)^2+d^2+\frac{2bcd}{a+b}\cos(D)=b^2\tag{2}
$$
Multiply $(1)$ by $b$ and $(2)$ by $a$ and add to get
$$
(a+b)ab\left(\frac{c}{a+b}\right)^2+(a+b)d^2=(a+b)ab\tag{3}
$$
Solving $(3)$ for $d^2$ yields
$$
\bbox[5px,border:2px solid #C0A000]{d^2=ab\frac{(a+b)^2-c^2}{(a+b)^2}}\tag{4}
$$
In a Right Triangle
Since, in a right triangle, $a^2+b^2=c^2$, $(4)$ becomes
$$
\bbox[5px,border:2px solid #C0A000]{d^2=\frac{2a^2b^2}{(a+b)^2}}\tag{5}
$$
| {
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Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some books, I came to this:
$$2x^2+2y^2+2z^2 \ge 2xy + 2xz + 2yz$$ $$2xy+2yz+2xz = 1-(x^2+y^2+z^2) $$ $$2x^2+2y^2+2z^2 \ge 1 - x^2 -y^2 - z^2 $$ $$x^2+y^2+z^2 \ge \frac{1}{3}$$
But this method is very unintuitive to me and I don't think this is the best way to solve this. Any remarks and hints will be most appreciated.
| It is sufficed prove that: $3(x^2 +y^2 +z^2) \geq (x+y+z)^2$
$\Leftrightarrow \frac{1}{2}(x+y-2z)^2 +\frac{3}{2}(x-y)^2 \geq 0$
Which is obvious.
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Every complex number has 2 square roots - Rudin I am using Baby Rudin for self study and as many of you are aware, it is one thing to follow Rudin's calculations to verify that his proof is correct. It is quite another to work backwards to see how he came up with his proof in the first place. I find that unless I do the latter, the former does absolutely nothing for me. Chapter 1 exercise 10 is as follows:
Suppose $z = a + bi$, $w = u + vi$ and
$a = \left(\frac{|w| + u}{2}\right)^{1/2}$, $b = \left(\frac{|w| - u}{2}\right)^{1/2}$. Prove that $z^2 = w$ if $v \geq 0$ and $\overline{z}^2 = w$ if $v \leq 0$. Conclude that every complex number(with one exception) has two complex square roots.
It is pretty easy to verify this calculation:
If $v \geq 0$ then
$(a+bi)^2 = (-a - bi)^2= a^2 - b^2 + 2abi = \frac{|w| + u}{2} + \frac{|w| - u}{2} + 2i\left(\frac{|w| + u}{2}\right)^{1/2}\left(\frac{|w| - u}{2}\right)^{1/2} = u + 2i \sqrt{\frac{|w|^2 - u^2}{4}} = u + 2i\sqrt{\frac{u^2 + v^2 - u^2}{4}} = u + 2i \sqrt{\frac{v^2}{4}} = u + 2i\frac{|v|}{2} = u + |v|i = u + vi$
On the other hand if $v \leq 0$ then it is similar to show that
$(a - bi)^2 = (-a + bi)^2 = u - |v|i$
This completes the proof. Now for the backwards derivation attempt:
We want $(a + bi)^2 = u + vi \implies a^2 - b^2 + 2abi = u + vi \implies a^2 - b^2 = u$ and $2ab = v$. So essentially we have a system of equations to solve for $a$ and $b$. I have been stuck on this for a couple hours. Maybe I am missing something obvious, but any help would be greatly appreciated.
| I figured it out:
$a^2 - b^2 = u$ and $2ab = v$
$b = \frac{v}{2a}$
$a^2 - \frac{v^2}{4a^2} - u = 0$
$4a^4 -4ua^2 - v^2 = 0$
By the quadratic formula:
$a^2 = \frac{4u \pm \sqrt{16^2 - 4(4)(-v^2)}}{8} = \frac{4u \pm \sqrt{16u^2 + 16v^2}}{8} = \frac{4u+4\sqrt{u^2 + v^2}}{8} = \frac{u \pm |w|}{2}$
So that $a = \sqrt{\frac{u + |w|}{2}}$ since we are only interested in real values of $a$. Then
$a^2 - b^2 = u \implies b^2 = a^2 - u = \frac{|w| + u}{2} - u = \frac{|w| - u}{2}$
so that $b = \sqrt{\frac{|w| - u}{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Limit of product of mixed root indexes I am unable to recover the limit of this function.
$$
\lim_{x \to +\infty} \left (\frac{1}{\sqrt{x}} \times \sqrt[3]{x+1} \right )
$$
I have tried many ways of solving it. These are the two simplest:
*
*Variable substitution
Using $y = \sqrt[3]{x+1} $, with which I got a common factor $\sqrt{y-1}$ and got
$$
\lim_{x \to +\infty} \left (\frac{1}{\sqrt{y^3-1}} \times y \right) =
\lim_{x \to +\infty} \left (\frac{\sqrt{y^2}}{\sqrt{y^3-1}} \right) =
\lim_{x \to +\infty} \left (\frac{\sqrt{y^2}}{\sqrt{y^3(1-\frac{1}{y^3})}} \right) =
\lim_{x \to +\infty} \left (\frac{1}{\sqrt{y \left (1-\frac{1}{y^3} \right)}} \right) = \frac{1}{+\infty \times 1} = 0
$$
*
*Using the common factor $\sqrt{x}$
$\sqrt[3]{x+1} = \sqrt[6]{(x+1)^2} = \sqrt[6]{x^2+2x+1} = \sqrt[6]{x^3 (\frac{1}{x}+\frac{2}{x^2}+\frac{1}{x^3})}=\sqrt{x} \left (\sqrt[6]{\frac{1}{x}+\frac{2}{x^2}+\frac{1}{x^3}} \right )$
and therefore I got
$$
\lim_{x \to +\infty} \left (\frac{\sqrt{x} \left (\sqrt[6]{\frac{1}{x}+\frac{2}{x^2}+\frac{1}{x^3}} \right )}{\sqrt{x}} \right ) =
\lim_{x \to +\infty} \left (\sqrt[6]{\frac{1}{x}+\frac{2}{x^2}+\frac{1}{x^3}} \right ) = \sqrt[6]{\frac{1}{+\infty}+\frac{1}{+\infty}+\frac{1}{+\infty}} = 0
$$
However, I found that the limit is positive when I plotted a function with this expression. I have plotted it here, you can look at it yourselves. This means my calculations are wrong. What am I doing wrong and how do I solve it correcly?
Thank you for the help.
| Multiplying and dividing by $x^{1/3}$ we get $\frac{1}{\sqrt{x}}(x+1)^{1/3}=\frac{x^{1/3}}{\sqrt{x}}(\frac{x+1}{x})^{1/3}$. The factor $(\frac{x+1}{x})^{1/3}$ tends to $1$ as $x \to \infty$. The factor $\frac{x^{1/3}}{\sqrt{x}}$ equals $x^{-1/6}$ that tends to $0$ as $x \to \infty$. So the limit you're looking for is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Four points, sine, Taylor's series I am creating a program that finds the cubic equation $y = ax^3+bx^2+cx+d$ through four points; I have already found how to calculate the coefficients, but I am also attempting to use a Taylor’s series to approximate a sine equation $A\sin(Bx+C)+D$ through roughly the same points based on the cubic function. Is there anyway to find $A, B, C$, or $D$ using solely the points’ coordinates and/or $a,b,c$, or $d$?
| Note that the Taylor expansion you need is
$$
\sin x = \sum_{k=1}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}
= x - x^3/6 + \Theta\left(x^5\right)
$$
Hence
$$
A\sin(Bx+C) + D
\approx D + A\left[ (Bx+C) - (Bx+C)^3/6 \right]
$$
and now expand the brackets and do arithmetic. The next term in the series would be $x^5$, so the coefficients of $x^2$ is zero and you are only looking upto cubic term.
UPDATE
Expanding the right-hand side you get 4 equations
$$
\left\{\begin{matrix}
6a = AB^3\\
2b = AB^2C \\
c = B\left(A-3C^2\right)\\
d = D + AC\left(1 - \frac{C^2}{6}\right)
\end{matrix}\right\|
$$
You need to solve these for $A,B,C,D$ in terms of $a,b,c,d$. Here is one approach.
If $a=b=c=0$ then set $A=B=C=0$ and $D=d$.
If one of $a,b,c$ is not zero, we can assuming $B \ne 0$. Then first equation implies
$$
A = \frac{6a}{B^3}
$$
and second equation implies
$$
C = \frac{2b}{AB^2} = \frac{2b}{B^2 \frac{6a}{B^3}} = \frac{Bb}{3a}.
$$
The last equation implies
$$
D = d - AC\left(1 - \frac{C^2}{6}\right)
$$
so everything boils down to knowing the value of $B$. From the 3rd equation, we have
$$
\frac{c}{B} = A - 3C^2 = \frac{6a}{B^3} - \frac{B^2 b^2}{3a^2}.
$$
Take the left and right expressions (exclude the middle one) and cross multiply by $3a^2B^3$ to get
$$
3a^2cB^2 = 18a^3 - B^5b^2
$$
and so we must solve
$$
B^5b^2 + 3a^2cB^2 = 18a^3,
$$
which does not have an analytic solution. However, given the values of $a,b,c$ you can use numerical root-finding (like Newton's method, for example) to find the value(s) of $B$ for which this equation is true (and it is guaranteed that $B$ will have at least one real value).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
why$2 (-1 + 2^{1 + n})$ is the answer to the recurrence relation $a_{n}=2a_{n-1}+2$? $a_{0}=2$
$a_{1}=2(2)+2$
$a_{2}=2(2(2)+2)+2$
$a_{3}=2(2(2(2)+2)+2)+2$
$a_{4}=2(2(2(2(2)+2)+2)+2)+2$
$a_{5}=2(2(2(2(2(2)+2)+2)+2)+2)+2$
To simplifiy
$a_{6}=2^{6}+2^{5}...2^{1}$
so my answer is
$a_{n}=2^{n+1}+2^{n}+...2^{1}$
The correct answer is $2 (-1 + 2^{1 + n})$
How do I make this transition?
| If you divide both sides by $2^n$, you will get :
$$\forall k\ge1,\,\frac{a_k}{2^k}-\frac{a_{k-1}}{2^{k-1}}=\frac1{2^{k-1}}$$
Now, cancellation occurs when summing up ($k$ from $1$ to $n$) :
$$\frac{a_n}{2^n}-a_0=\sum_{k=1}^n\frac1{2^{k-1}}=\frac{1-\frac1{2^n}}{1-\frac12}=2-\frac1{2^{n-1}}$$
that is :
$$a_n=2^n\left(4-\frac1{2^{n-1}}\right)=2^{n+2}-2$$
and finally :
$$a_n=2\left(2^{n+1}-1\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
The inequality $\frac{3\sqrt{3}}{2}\frac{27(A+B)(B+C)(C+A)}{8(A+B+C)^3}\ge \sin{A}+\sin{B}+\sin{C}$ In $\triangle ABC$, show that
$$\dfrac{3\sqrt{3}}{2}\dfrac{27(A+B)(B+C)(C+A)}{8(A+B+C)^3}\ge \sin{A}+\sin{B}+\sin{C}$$
My attempt: Since $A+B+C=\pi$, it suffices to show that
$$\dfrac{3\sqrt{3}}{2}\cdot \dfrac{27(\pi-A)(\pi-B)(\pi-C)}{8\pi^3}\ge\sin{A}+\sin{B}+\sin{C}$$
then it seem to hard to prove it (because Jensen's inequality can't solve it, and the tangent line also can't solve it)
| Not a full solution just yet, but we can go like this:
$$\sin A + \sin B + \sin C = 4 \sin\frac{A+B}2\sin\frac{B+C}2\sin\frac{C+A}2.$$
So if we let $x= \frac{B+C}2,y=\frac{C+A}2,z=\frac{A+B}2$, we need to show that
$$xyz\ge \lambda \sin x\sin y\sin z$$
for $x,y,z\in(0,\pi/2)$, and $ x+y+z=\pi$ and some constant $\lambda$ that make it an equality when $x=y=z$.
Now using Wolfram Alpha, we see that (this is where the proof is not complete, but I'm too lazy)
$$f(x) = \ln \frac{\sin x}{x}=\ln (\sin x) -\ln x$$
has negative second derivative in $(0,\pi/2)$. The inequality follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2251326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2}=\frac{c^2}{2}$ without L'Hospital's rule How to show that without using L'Hospital's rule
\begin{align}
\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2}=\frac{c^2}{2}
\end{align}
I was able to show the upper bound by using the bound $\cosh(x) \le e^{x^2/2}$
\begin{align}
\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2} \le \lim_{x \to 0} \frac{\left(x^2-xc\right)^2}{2x^2}=\frac{c^2}{2}
\end{align}
My question: How finish this argument.
| We can proceed as follows
\begin{align}
L&=\lim_{x\to 0}\frac{\log\cosh(x^{2}-cx)}{x^{2}}\notag\\
&=\lim_{x\to 0}\frac{\log\cosh(x^{2}-cx)}{\cosh(x^{2}-cx)-1}\cdot\frac{\cosh(x^{2}-cx)-1}{x^{2}}\notag\\
&=\lim_{x\to 0}\frac{2\sinh^{2}((x^{2}-cx)/2)}{x^{2}}\notag\\
&=\lim_{x\to 0}2\cdot\frac{\sinh^{2}((x^{2}-cx)/2)}{(x^{2}-cx)^{2}/4}\cdot\frac{x^{2}(x-c)^{2}}{4x^{2}}\notag\\
&=2\cdot 1\cdot\frac{c^{2}}{4}\notag\\
&=\frac{c^{2}}{2}\notag
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Show that there are no positive integer solutions to $x^4-y^4=z^4$ Also, Show that the only solution in relatively prime positive integers to $x^4+y^4=2z^4$ is $x=y=z=1$.
Here I tried a similar contradiction proof to the proof for $x^4+y^4=z^4$, but this doesn't work the same way.
| We show that if there exist $x,y,z \in \mathbb N$ with $x^2=y^4+z^4$ then there exist $x',y',z'\in \mathbb N$ with $x'^2=y'^4+z'^4$ and $x'<x.$ This implies that, if $S=\{x\in \mathbb N:\;\exists y,z \in \mathbb N\;(x^2=y^4+z^4)\}$ is not empty, then $S$ is a non-empty subset of $\mathbb N$ with no least member, which is not possible. So we will conclude that $S$ is empty.
Assume $x^2=y^4+z^4.$
(i). If $\gcd(y,z)>1,$ let $p$ be a prime divisor of $\gcd (y,z).$ Then $p^4 $ divides $x^2.$ Since $p$ is prime this implies that $p^2$ divides $x.$ So let $x'=x/p^2,\;y'=y/p,\;z'=z/p.$
(ii). If $\gcd (y,z)=1$ then $(y^2,z^2,x)$ is a primitive Pythagorean triplet. So there exist $m,n\in \mathbb N,$ not both odd, with $$\{y^2,z^2\}=\{m^2-n^2,2mn\} \;\land \;x=m^2+n^2$$and with $\gcd(m,n)=1.$... Now WLOG, $$y^2=m^2-n^2 \;\land \;z^2=2mn.$$
Note that $y$ is odd. Else $2$ divides both $y^2$ and $z^2=2mn,$ implying that $2$ divides both $y$ and $z,$ contrary to $\gcd (y,z)=1.$
Now $y^2+n^2=m^2$ and we must have $\gcd (y,n)=1.$ Because if a prime $p$ divides both $y$ and $n$ then $p^2$ divides $m^2=y^2+n^2,$ implying (since $p$ is prime) that $p$ divides $m ,$ implying $\gcd(m,n)\geq p>1,$ contrary to $\gcd (m,n)=1.$
Therefore $(y,n,m)$ is a primitive Pythagorean triplet. So there exist $a,b \in \mathbb N,$ not both odd, with $$\{y,n\}=\{a^2-b^2,2ab\} \;\land\; m=a^2+b^2$$ and with $\gcd(a,b)=1.$ Now since $y$ is odd we have $$y=a^2-b^2 \;\land \;n=2ab.$$
Applying this to $z,$ we have $$z^2=2mn=2(a^2+b^2)(2ab)=4(a^2+b^2)ab.$$ So $z/2\in \mathbb N,$ and $(a^2+b^2)ab=(z/2)^2$ is the square of a member of $\mathbb N.$ Now any prime that divides any two of $(a^2+b^2,a,b)$ will divide the third one. But $\gcd (a,b)=1.$ So $(a^2+b^2,a,b)$ are pair-wise co-prime.
From the uniqueness of prime decompositions on $\mathbb N$ we have: For any pair-wise co-prine $(U,V,W)$ in $\mathbb N$, if $UVW$ is the square of a member of $\mathbb N$ then each of $U,V,W$ is the square of a member of $\mathbb N.$ Applying this with $U=a^2+b^2,\; V=a,\;W=b$ we infer that there exist $x',y',z' \in \mathbb N$ with $$a^2+b^2=x'^2\;\land\; a=y'^2\;\land\; b=z'^2.$$ Therefore $$x'^2=a^2+b^2=y'^4+z'^4.$$ And we have $x'=\sqrt {a^2+b^2}\;=\sqrt m<2mn=z^2<\sqrt {z^4+y^4}=\sqrt {x^2}=x.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
The line $y=mx+c$ is a tangent to $x^2+y^2=a^2$ if: The line $y=mx+c$ is a tangent to $x^2+y^2=a^2$ if:
$1$. $c=a\sqrt {1+m^2}$
$2$. $c=\pm a\sqrt {1+m^2}$
$3$. $c^2=\pm a\sqrt {1+m^2}$
$4$. $\textrm {None}$
My Attempt:
The tangent to circle $x^2+y^2=a^2$ at point $(x_1,y_1)$ is given by :
$$xx_1+yy_1=a^2$$
Now, what should I do next?
| A possible approach:
1) Calculate distance from (0,0) to line $y = mx + c$.
The perpendicular line passing through the origin is
$y = - (1/m)x$.
Point of intersection: $- (1/m)x = mx + c$.
$- x = m^2x + mc$
$x ( m^2 + 1) + mc = 0$;
$x = (-mc)/(m^2 +1)$;
Corresponding $y$:
$y = - (1/m) (-mc) /(m^2 + 1)$;
$y = c/(m^2 +1)$.
Distance (squared) to the origin:
$x^2 + y^2 = [(mc)^2 + c^2]/(m^2 +1)^2$ =
= $c^2(m^2 +1)/(m^2 + 1)^2 $=
$c^2/(m^2 + 1)$.
2) For this point to be a point of tangency:
Distance (squared) = radius (squared) =
$a^2$.
Putting together:
$c^2/(m^2 +1) = a^2$.
Answer 2).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2254073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find a non-zero integer matrix $X$ such that $XA=0$ where $X,A,0$ are all $4 \times 4$ Let $A$ be the following $4 \times 4$ matrix.
\begin{bmatrix}1&2&1&3\\1&3&2&4\\2&5&3&7\\1&4&1&5\end{bmatrix}
How can we find a non-zero integer matrix $C$ such that $CA = 0_{4 \times 4}$
Note that $0$ is a $4 \times 4$ matrix.
| One option is use the projector onto the $\color{red}{null}$ space
$$
\color{red}{\mathbf{P}_{\mathcal{N}\left(\mathcal{A^{*}}\right)}} = \mathbf{I}_{4} - \mathbf{A}\mathbf{A}^{+}
$$
The target matrix and pseudoinverse are
$$
\mathbf{A} =
\left(
\begin{array}{cccc}
1 & 2 & 1 & 3 \\
1 & 3 & 2 & 4 \\
2 & 5 & 3 & 7 \\
1 & 4 & 1 & 5 \\
\end{array}
\right),
\qquad
\mathbf{A}^{+}
=
\frac{1}{18}
\left(
\begin{array}{rrrr}
26 & -19 & 7 & -9 \\
-16 & 11 & -5 & 9 \\
-12 & 15 & 3 & -9 \\
10 & -8 & 2 & 0 \\
\end{array}
\right)
$$
The projector matrix is
$$
\color{red}{\mathbf{P}_{\mathcal{N}\left(\mathcal{A}\right)}} = \mathbf{I}_{4} - \mathbf{A}^{+}\mathbf{A} =
\frac{1}{3}
\left(
\begin{array}{rrrr}
1 & 1 & -1 & 0 \\
1 & 1 & -1 & 0 \\
-1 & -1 & 1 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)
$$
Verify:
$$
\color{red}{\mathbf{P}_{\mathcal{N}\left(\mathcal{A}\right)}} \mathbf{A}=
\left(
\begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)
$$
If you look carefully at the solution matrix, you will appreciate the insight of @gt6989b.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
$\dfrac{1}{2\pi i}\int_{C}\dfrac{e^{z}}{z^{3}-1}dz = \sum_{n=0}^{\infty}\dfrac{1}{(3n+2)!}$ a) Use residue at infinity to solve the problem:
If $C$ is a circle $C(0,2)$ traversed in the counter clockwise direction, then $$\dfrac{1}{2\pi i}\int_{C}\dfrac{e^{z}}{z^{3}-1}dz = \sum_{n=0}^{\infty}\dfrac{1}{(3n+2)!}$$
b) By evaluating $$\dfrac{1}{2\pi i}\int_{C}\dfrac{e^{z}}{z^{3}-1}dz$$
using Cauchy's integral formula, show that $$\sum_{n=0}^{\infty}\dfrac{1}{(3n+2)!} = \dfrac{1}{3}(e-\dfrac{2}{\sqrt{e}}\cos(\dfrac{\pi}{3}-\dfrac{\sqrt{3}}{2}))$$
Hint: One may use the identity $\prod_{j=1,j\neq k}^{m}(e^{2\pi ij/m}-e^{2\pi ik/m}) = \dfrac{m}{e^{2\pi ik/m}}$
a) Note that the function $f(z) = \dfrac{e^{z}}{z^{3}-1}$ and we see that there will be 3 singularity, namely $z =1, z = e^{2\pi i/3}\text{ and } z = e^{4\pi i/3}$. All of which are contained in the $C(0,2)$. Well since we are given the hint to solve the first part with residue at infinity, we will follow it.
$$\dfrac{1}{2\pi i}\int_{C}f(z)dz = \text{Res}\left[\dfrac{1}{z^{2}}f\left(\dfrac{1}{z}\right),0\right] = \text{Res}\left[\dfrac{1}{z^{2}}\dfrac{e^{1/z}}{\dfrac{1}{z^{3}}-1},0\right] = \text{Res}\left[\dfrac{ze^{1/z}}{1-z^{3}},0\right]$$
We want to find the reside of $\dfrac{ze^{1/z}}{1-z^{3}}$ at $0$. Hence we expand this equation about $0$ in Laurent Series form, $$ze^{1/z}\dfrac{1}{1-z^{3}} = z\sum_{k=0}^{\infty}\dfrac{1}{k!z^{k}}\sum_{m=0}^{\infty}z^{3m} = \sum_{m=0}^{\infty}\left(\sum_{k=0}^{\infty}\dfrac{1}{k!}z^{3n-k+1}\right)$$
We want to find the coefficient of $z^{-1}$ and it happens when $3n-k+1 = -1 \Rightarrow k = 3m+2$. Hence the residue is $$\sum_{m=0}^{\infty}\dfrac{1}{(3m+2)!}$$
First question is part a), why can't i get the answer if i interchange the summation????? By right it should be the same answer.
b) We easily find out that the three singularities for the function is $z=1,z=e^{2\pi i/3},z=e^{4\pi i/3}$
And hence we are solving for $$\dfrac{1}{2\pi i}\int_{C}\dfrac{e^{z}}{(z-1)(z-e^{2\pi i/3})(z-e^{4\pi i/3})}dz$$
Since we are required to use CIF to solve, we construct 3 circles $C_1,C_2,C_3$ of radius $\epsilon_1,\epsilon_2,\epsilon_3$ and centered at $z=1,z=e^{2\pi i/3},z=e^{4\pi i/3}$.
Then from here, i can only use brute force (REALLY TOOK ME VERY LONG) to solve it, anyone knows how to solve it with the hint?
| We first note that for a simple pole we have
$$\mathrm{Res}(f/g,z_0) = \frac{f(z_0)}{g'(z_0)}$$
The sum of the residues
$$\sum \mathrm{Res}(f,z) =\frac{e}{3}+\dfrac{e^{e^{2\pi i/3}}}{3e^{4\pi i/3}}+\dfrac{e^{e^{4\pi i/3}}}{3e^{8\pi i/3}}$$
Note that
$$\dfrac{e^{e^{2\pi i/3}}}{e^{4\pi i/3}}+\dfrac{e^{e^{4\pi i/3}}}{e^{8\pi i/3}} = e^{e^{2\pi i/3}+2\pi i /3}+ e^{e^{-2\pi i/3}-2\pi i /3} = 2\mathrm{Re}\left(e^{e^{2\pi i/3}+2\pi i /3}\right)$$
Note by expnading and using trignometric properites we have
$$e^{e^{2\pi i/3}+2\pi i /3} = i\frac{\cos\left(\frac{\sqrt{3}}{2} + \frac{\pi}{6}\right)}{\sqrt{e}} - \frac{\sin\left(\frac{\sqrt{3}}{2} + \frac{\pi}{6}\right)}{\sqrt{e}}$$
Hence
$$\sum \mathrm{Res}(f,z)=\frac{e}{3}- \frac{2\sin\left(\frac{\sqrt{3}}{2} + \frac{\pi}{6}\right)}{3\sqrt{e}}=\frac{e}{3}- \frac{2\cos\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right)}{3\sqrt{e}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Hermitian matrices in an arbitrary inner product I was wondering if anyone can give me any hints for how to prove the following proposition:
Proposition: Let $V = F^n$ and let $A\in M_{n\times n}(F)$. Prove that $\langle A^* x,y \rangle = \langle x, A y \rangle$.
The problem that I am having is how to prove this for an arbitry inner product on $F^n$, an arbitrary field. Some textbooks take this as an axiom or the definition of the adjoint of $A$, but the book that I got this from does not (Linear Algebra by Friedberg).
| This is true only for $V$ with the standard inner product.
Counter-example:
Assume $V=\mathbb{R}^2$ and we define the inner product in the following way:
$$\langle x | y\rangle = x^T \begin{pmatrix}1 & \frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2} & 1 \end{pmatrix} y$$
Lets take for simplicity $x = \begin{pmatrix} 1\\0 \end{pmatrix} , y = \begin{pmatrix} 0\\1 \end{pmatrix} $
Lets take $A$ to be $\begin{pmatrix}1 & 2\\3 & 4 \end{pmatrix} $
Now some calculations:
$$\langle x | Ay \rangle = x^T\begin{pmatrix}1 & \frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2} & 1 \end{pmatrix} \begin{pmatrix}1 & 2\\3 & 4 \end{pmatrix} y = x^T\begin{pmatrix}1+\frac{3\sqrt{3}}{2} & 2(1+\sqrt{3})\\\frac{1}{2}(6+\sqrt{3}) & 4+\sqrt{3} \end{pmatrix} y = 2(1+\sqrt{3}) $$
$$\langle A^*x | y \rangle = \left (\begin{pmatrix}1 & 3\\2 & 4 \end{pmatrix}
x\right )^T\begin{pmatrix}1 & \frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2} & 1 \end{pmatrix} y = \begin{pmatrix} 1& 2 \end{pmatrix}\begin{pmatrix}1 & \frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2} & 1 \end{pmatrix} \begin{pmatrix} 0\\ 1 \end{pmatrix} = 2 + \frac{\sqrt{3}}{2} $$
$$ \langle x | Ay \rangle \neq \langle A^*x | y \rangle $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 2,
"answer_id": 0
} |
Floors Complicated Proof Problem:
Prove that $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$ for all real $x$ and $y$.
I have my proof down below. Its a little complicated though.
This is not a duplicate of the same question on another page on math.stackexchange.com because I provide my answer.
| Let $n=\lfloor x \rfloor$ and $m=\lfloor y \rfloor$. Consider $4$ cases:
Case 1: $x \in [n,n+\dfrac{1}{2})$ and $y \in [m,m+\dfrac{1}{2})$
Case 2: $x \in [n+\dfrac{1}{2},n+1)$ and $y \in [m,m+\dfrac{1}{2})$
Case 3: $x \in [n,n+\dfrac{1}{2})$ and $y \in [m+\dfrac{1}{2},m+1)$
Case 4: $x \in [n+\dfrac{1}{2},n+1)$ and $y \in [m+\dfrac{1}{2},m+1)$
For Case 1, we have $$\lfloor 2x \rfloor = \lfloor [2n,2n+1) \rfloor=2n$$ because $n$ is an integer.
Similarly, $$\lfloor 2y \rfloor = \lfloor [2m,2m+1) \rfloor=2m$$ because $m$ is an integer.
We also have $$\lfloor x+y \rfloor = \lfloor [n+m,n+m+1) \rfloor=n+m.$$
Now we can plug this into the original inequality:
$$\lfloor 2x \rfloor + \lfloor 2y \rfloor=2n+2m$$
and $$\lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor=2n+2m$$
This gives $$\lfloor 2x \rfloor + \lfloor 2y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$$
For Case 2, we have $$\lfloor 2x \rfloor = \lfloor [2n+1,2n+2) \rfloor=2n+1$$ because $n$ is an integer.
As in Case 1, $$\lfloor 2y \rfloor = \lfloor [2m,2m+1) \rfloor=2m$$ because $m$ is an integer.
We also have $$\lfloor x+y \rfloor = \lfloor [n+m+\dfrac{1}{2},n+m+1\dfrac{1}{2}) \rfloor$$
This implies that $$\lfloor x+y \rfloor = n+m\;or\;n+m+1$$
Now we can plug this into the original inequality:
$$\lfloor 2x \rfloor + \lfloor 2y \rfloor=2n+1+2m$$
and $$\lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor=2n+2m\;or\;2n+2m+1$$
This gives $$\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$$
Case 3 is similar to Case 2 by switching the $n$'s and $m$'s.
For Case 4, we have $$\lfloor 2x \rfloor = \lfloor [2n+1,2n+2) \rfloor=2n+1$$ because $n$ is an integer.
Similarly, $$\lfloor 2y \rfloor = \lfloor [2m+1,2m+2) \rfloor=2m+1$$ because $m$ is an integer.
We also have $$\lfloor x+y \rfloor = \lfloor [n+m+1,n+m+2) \rfloor=n+m+1.$$
Now we can plug this into the original inequality:
$$\lfloor 2x \rfloor + \lfloor 2y \rfloor=2n+2m+2$$
and $$\lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor=2n+2m+1$$
This gives $$\lfloor 2x \rfloor + \lfloor 2y \rfloor > \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor.$$
Therefore, the original inequality is true!
| {
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Solve the Differential equation $x^3 \frac{dy}{dx}=y^3+y^2\sqrt{x^2+y^2}$ Solve the Differential equation $$x^3 \frac{dy}{dx}=y^3+y^2\sqrt{x^2+y^2}$$
i reduced the equation as
$$x^3\frac{dy}{dx}=y^3\left(1+\sqrt{1+\left(\frac{x}{y}\right)^2}\right)$$ $\implies$
$$\frac{x^3}{y^3}\frac{dy}{dx}=\left(1+\sqrt{1+\left(\frac{x}{y}\right)^2}\right) \tag{1}$$ Next put $$\frac{x}{y}=v$$ we get
$$ x=vy$$ then
$$ \frac{dx}{dy}=v+y\frac{dv}{dy}$$ Then $(1)$ becomes
$$\frac{v^3}{v+y\frac{dv}{dy}}=1+\sqrt{1+v^2}$$ Reciprocating we get
$$\frac{v+y\frac{dv}{dy}}{v^3}=\frac{1}{\sqrt{1+v^2}+1}$$ Rationalizing RHS we get
$$\frac{v+y\frac{dv}{dy}}{v^3}=\frac{\sqrt{1+v^2}-1}{v^2}$$ Rearranging we get
$$\frac{dv}{v \times \left(\sqrt{1+v^2}-2\right)}=\frac{dy}{y}$$
EDIT: i am posting here the clue given by paul using Substitution $v=\tan z$:
$$\int\frac{dv}{v \times \left(\sqrt{1+v^2}-2\right)}=\int\frac{\sec^2 z\: dz}{\tan z(\sec z-2)}=\int\frac{dz}{\sin z(1-2\cos z)}$$ So
$$\int\frac{dz}{\sin z(1-2\cos z)}=\int \frac{\sin z\: dz}{\sin^2 z(1-2\cos z)}$$
Put $\cos z=t$ and use partial fractions
| Well, let $\text{y}\left(x\right)=x\cdot\text{r}\left(x\right)$, which gives $\text{y}'\left(x\right)=x\cdot\text{r}'\left(x\right)+\text{r}\left(x\right)$:
$$x^3\cdot\text{y}'\left(x\right)=\text{y}\left(x\right)^3+\text{y}\left(x\right)^2\cdot\sqrt{x^2+\text{y}\left(x\right)^2}\space\Longleftrightarrow\space$$
$$x^3\cdot\left(x\cdot\text{r}'\left(x\right)+\text{r}\left(x\right)\right)=\left(x\cdot\text{r}\left(x\right)\right)^3+\left(x\cdot\text{r}\left(x\right)\right)^2\cdot\sqrt{x^2+\left(x\cdot\text{r}\left(x\right)\right)^2}\tag1$$
Solve for $\text{r}'\left(x\right)$:
$$\text{r}'\left(x\right)=\frac{\text{r}\left(x\right)\cdot\left(\text{r}\left(x\right)^2-1+\text{r}\left(x\right)\cdot\sqrt{1+\text{r}\left(x\right)^2}\right)}{x}\tag2$$
Devide both sides by the numerator of the RHS and integrate both sides:
$$\int\frac{\text{r}'\left(x\right)}{\text{r}\left(x\right)\cdot\left(\text{r}\left(x\right)^2-1+\text{r}\left(x\right)\cdot\sqrt{1+\text{r}\left(x\right)^2}\right)}\space\text{d}x=\int\frac{1}{x}\space\text{d}x\tag3$$
For the integrals:
*
*Substitute $\text{u}=\text{r}\left(x\right)$:
$$\int\frac{\text{r}'\left(x\right)}{\text{r}\left(x\right)\cdot\left(\text{r}\left(x\right)^2-1+\text{r}\left(x\right)\cdot\sqrt{1+\text{r}\left(x\right)^2}\right)}\space\text{d}x=$$
$$\int\frac{1}{\text{u}\cdot\left(\text{u}^2-1+\text{u}\cdot\sqrt{1+\text{u}^2}\right)}\space\text{d}\text{u}=$$
$$\frac{\ln\left|\left(\text{u}+\sqrt{1+\text{u}^2}\right)^2\right|}{3}-\ln\left|\left(\text{u}+\sqrt{1+\text{u}^2}\right)^2-1\right|+\frac{2\ln\left|\left(\text{u}+\sqrt{1+\text{u}^2}\right)^2-3\right|}{3}+\text{C}_1\tag4$$
*$$\int\frac{1}{x}\space\text{d}x=\ln\left|x\right|+\text{C}_2\tag5$$
So, we get:
$$\frac{\ln\left|\left(\text{u}+\sqrt{1+\text{u}^2}\right)^2\right|}{3}-\ln\left|\left(\text{u}+\sqrt{1+\text{u}^2}\right)^2-1\right|+\frac{2\ln\left|\left(\text{u}+\sqrt{1+\text{u}^2}\right)^2-3\right|}{3}=\ln\left|x\right|+\text{C}\tag6$$.
Now, use:
$$\text{u}=\frac{\text{y}\left(x\right)}{x}\tag7$$
| {
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"answer_count": 2,
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Given $A^2$, find all $A$ that satisfy Find the set of all matrices $A\in M_{n\times n}(\mathbb R)$ satisfying $A^2=\begin{pmatrix} 0 &1& 0 \\ 0& 0& 0 \\ 0& 0& 0 \end{pmatrix}$
I have been given this problem. Do I just need to take $A=\begin{pmatrix} a &b& c \\ d& e& f \\ g& h& i \end{pmatrix}$ and solve for the variables? or is there a better way
| Let $$A=\begin{pmatrix} a &b& c \\ d& e& f \\ g& h& i \end{pmatrix}$$ as OP suggests. Now clearly,
$$A^4=0$$ and thus the minimal polynomial of $A$ divides $x^4$. However the minimal polynomial has degree $\leq 3$ and therefore the minimal polynomial divides $x^3$. Thus we have
$$A^3=0$$ and so
$$\begin{pmatrix} a &b& c \\ d& e& f \\ g& h& i \end{pmatrix}
\begin{pmatrix} 0 &1& 0 \\ 0& 0& 0 \\ 0& 0& 0 \end{pmatrix}=0$$
And $$\begin{pmatrix} 0 &1& 0 \\ 0& 0& 0\\0& 0& 0 \end{pmatrix}
\begin{pmatrix} a &b& c \\ d& e& f \\ g& h& i \end{pmatrix}
=0$$. Which imply that
$$A=\begin{pmatrix} 0 &b& c \\ 0& 0& 0 \\ 0& h& i \end{pmatrix}$$
The original equation now becomes much easier to solve we get
$$A=\begin{pmatrix} 0 &b& c \\ 0& 0& 0 \\ 0& c^{-1}& 0 \end{pmatrix}
$$
| {
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"source": "stackexchange",
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Prove that limit = 0. So the question is as in the title.
$$\lim_{(x,y,z)\to (0,0,0)}\frac{\sin(xyz)}{\sqrt{(x^2+y^2+z^2)}}$$
I decided to prove it from the definition.
so $L = 0$
$$\left\lvert\frac{\sin(xyz)}{\sqrt{(x^2+y^2+z^2)}}-0 \right \rvert \leq \frac {1}{\sqrt{(x^2+y^2+z^2)}} $$
and then trnasform it to this
$${\frac{\sqrt{(x^2+y^2+z^2)}}{x^2+y^2+z^2}}$$
and now i don't know what to do.
| Use $|\sin t|\le|t|$. Then
$$\left|\frac{\sin(xyz)}{\sqrt{x^2+y^2+z^2}}\right|
\le\left|\frac{xyz}{\sqrt{x^2+y^2+z^2}}\right|.$$
If we let $r=\sqrt{x^2+y^2+z^2}$ then $|xyz|\le r^3$ etc.
| {
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Interesting integral: $\int_0^1{\frac{nx^{n-1}}{x+1}}dx$ Find the value of $$\int_0^1{\frac{nx^{n-1}}{x+1}}dx.$$
I had no luck while integrating it. I also tried differentiating w.r.t n but still couldn't reach anywhere. Need help.
| We can write the integral as
$$
F(n) = \int_0^{\;1} {\frac{{nx^{n - 1} }}{{1 + x}}dx} = \int_0^{\;1} {\frac{1}{{1 + x}}dx^n } = \int_0^{\;1} {\left( {1 - \frac{x}{{1 + x}}} \right)dx^n } = 1 - \int_0^{\;1} {\frac{x}{{1 + x}}dx^n }
$$
$F(n+1)$ will be
$$
F(n + 1) = \int_0^{\;1} {\frac{{\left( {n + 1} \right)x^n }}{{1 + x}}dx} = \int_0^{\;1} {\frac{x}{{1 + x}}dx^n } + \int_0^1 {\frac{{x^n }}{{1 + x}}dx} = 1 - F(n) + \frac{1}{{n + 1}}F(n + 1)
$$
i.e.:
$$
\begin{array}{l}
\left( {1 - \frac{1}{{n + 1}}} \right)F(n + 1) = 1 - F(n) \\
\frac{{F(n + 1)}}{{n + 1}} = \frac{1}{n} - \frac{{F(n)}}{n} \\
\left( { - 1} \right)^{\,n} \frac{{F(n + 1)}}{{n + 1}} = \frac{{\left( { - 1} \right)^{\,n} }}{n} + \left( { - 1} \right)^{\,n - 1} \frac{{F(n)}}{n} \\
\end{array}
$$
So we have
$$
\left\{ \begin{array}{l}
G(n) = \left( { - 1} \right)^{\,n} \frac{{F(n + 1)}}{{n + 1}} \\
G(0) = F(1) = \ln 2 \\
G(n + 1) - G(n) = \frac{{\left( { - 1} \right)^{\,n + 1} }}{{n + 1}} \\
\end{array} \right.
$$
which leads to
$$ \bbox[lightyellow] {
\begin{array}{l}
G(n) = \ln (2) + \sum\limits_{k = 0}^{n - 1} {\frac{{\left( { - 1} \right)^{\,k + 1} }}{{k + 1}}} = \left( { - 1} \right)^{\,n} \Phi \left( { - 1,1,n + 1} \right) \\
F(n) = n\;\Phi \left( { - 1,1,n} \right) \\
\end{array}
}$$
where $\Phi$ denotes the Lerch Transcendent.
In fact, by the integral representation of $\Phi$
$$
\Phi \left( {z,s,a} \right)\mathop \equiv \limits^{def} \frac{1}
{{\Gamma (s)}}\int_0^\infty {\frac{{t^{\,s - 1} e^{\, - a\,t} }}
{{1 - z\,e^{\, - \,t} }}dt}
$$
and
$$
\Phi \left( { - 1,1,n} \right) = \int_{t\, = \,0}^\infty {\frac{{e^{\, - n\,t} }}
{{1 + \,e^{\, - \,t} }}dt} \quad \xrightarrow{{e^{\, - \,t} = x}}\quad - \int_{x = 1}^0 {\frac{{x^{\,n - 1} }}
{{1 + \,x}}dt}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Prove that $x^p+y^p \equiv (x+y)^p \pmod{x^2+xy+y^2}$
Let $p > 3$ be a prime, and let $x,y$ be integers such that $\gcd(x,y) = 1$. Prove that $$x^p+y^p \equiv (x+y)^p \pmod{x^2+xy+y^2}.$$
I thought about expanding $(x+y)^p$ using the binomial theorem, but I didn't see how that would help because the modulus is an algebraic expression. How can we prove the statement?
| Not true, you can use polynomial modular arithmetic, i will give an example for what i am talking about :
when $p=3$ we have $(x+y)^3=x^3+3x^2 y+3x y^2+y^3 $, and we want to find the remainder module $x^2+x y+y^2$ , the leading expression in the bigger polynomial is $x^3$ which means we need to multiply the smaller polynomial by $x$ and subtract from each, we arrive at $ 2 x^2 y+2 x y^2+y^3$ and the leading $x$ term is $2x^2 y$ in this new polynomial, so we need to multiply by $2y$ and subtract, we arrive at $-y^3$ which is the smallest term we can reach.
Now for the expression $(x+y)^p = ? \mod (x^2+x y+y^2)$ we have $6$ cases :
$p=6n$ => $y^{6n}$
$p=6n+1$ => $xy^{6n}+y^{6n+1}$
$p=6n+2$ => $xy^{6n+1}$
$p=6n+3$ => $-y^{6n+3}$
$p=6n+4$ => $-xy^{6n+3}-y^{6n+4}$
$p=6n+5$ => $-xy^{6n+4}$
And for the expression $x^p+y^p=? \mod (x^2+xy+y^2)$ we have $3$ cases :
$p=3n$ => $2y^{3n}$
$p=3n+1$ => $xy^{3n}+y^{3n+1}$
$p=3n+2$ => $-xy^{3n+1}$.
And because $5,5+6,5+2*6,5+3*6,5+4*6$ are all primes, one can conclude mistakenly that it works just for primes, but actually it also works for $p=5+6*5=35$, also $p=77$.
In General : it works for any $p=\pm 1 \mod 6$.
For Instance : take $x=2,y=3,p=35$ , so $2^35+3^35 = 5^35 \mod 19$, checking it on Wolfram Alpha, they both give remainder $4$.
| {
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"source": "stackexchange",
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Determining the first return distribution of 2-states Markov chain
A Markov chain has the transition matrix $$
P= \begin{pmatrix}
1-a & a \\
b & 1-b \\
\end{pmatrix}
$$
Find the first return distribution $f_{00}^n=\Pr\{X_1\neq 0,...,X_{n-1} \neq0,X_n=0\mid X_0=0\}$.
My attempt:
For $n=0$, $f_{00}^0$=$0$.
For n=1, $f_{00}^1$=$1=a$.
For n=2, $f_{00}^2$=Pr{$X_1\neq 0,X_2=0 |X_0=0$}=$(1-a)^2$+ab
For n=3, $f_{00}^3$=$(1-a)^3$+2ab(1-a)+ab(1-b)
And my idea was to make a generalization knowing the first results but I couldn't find the pattern because from n=2 to n=3 the expressions changes a lot and I think I'll change more for n=4.
Is there another way to find the first return distribution $f_{00}^n$, without using software?
| Draw the state transition diagram and find out the possible transitions.
The following is the state transition diagram:
Note that $f_{00}^{(1)}=p_{00}^{(1)}=1-a$.
\begin{eqnarray*}
f_{00}^{(1)}&=&p_{00}^{(1)}=1-a\\
f_{00}^{(2)}&=&p_{01}^{(1)}p_{10}^{(1)}=a\cdot b\\
f_{00}^{(3)}&=&p_{01}^{(1)}p_{11}^{(1)}p_{10}^{(1)}=a\cdot (1-b)\cdot b\\
f_{00}^{(4)}&=&p_{01}^{(1)}\underbrace{p_{11}^{(1)}p_{11}^{(1)}}p_{10}^{(1)}=a\cdot (1-b)^2\cdot b\\
f_{00}^{(4)}&=&p_{01}^{(1)}\left(p_{11}^{(1)}\right)^2p_{10}^{(1)}=a\cdot (1-b)^2\cdot b\\
\vdots&=&\vdots\\
f_{00}^{(n)}&=&p_{01}^{(1)}\left(p_{11}^{(1)}\right)^{n-2}p_{10}^{(1)}=a\cdot (1-b)^{n-2}\cdot b,\qquad \text{ for } n\geq 2.\\
\end{eqnarray*}
However, if the state space is slightly large, then the following recursive formula will be useful:
\begin{equation*}
F^{(n+1)}=P\left( F^{(n)}-F_{d}^{(n)}\right)
\end{equation*}
where $P$ is the transition probability matrix of a Markov chain, $F^{(n)}=\left(f_{ij}^{(n)}\right)$ and $F_{d}^{(n)}$ denotes a diagonal matrix containing the diagonal elements of $F^{(d)}$. \
Consider a Markov chain whose TPM is
\begin{equation*}
P=\left( \begin{array}{cc}
1-a & a\\
b & 1-b
\end{array}
\right)=F
\end{equation*}
Then,
\begin{equation*}
F_{d}^{(1)}=\left( \begin{array}{cc}
1-a & 0\\
0 & 1-b
\end{array}
\right)
\end{equation*}
\begin{equation*}
F^{(2)}=P\left(F^{(1)}-F_{d}^{(1)}\right)=\left( \begin{array}{cc}
1-a & a\\
b & 1-b
\end{array}
\right)\cdot \left( \begin{array}{cc}
0 & a\\
b & 0
\end{array}
\right)=\left( \begin{array}{cc}
ab & a(1-a)\\
b(1-b) & ab
\end{array}
\right)
\end{equation*}
\begin{equation*}
F^{(3)}=P\left(F^{(2)}-F_{d}^{(2)}\right)=\left( \begin{array}{cc}
1-a & a\\
b & 1-b
\end{array}
\right)\cdot \left( \begin{array}{cc}
0 & a(1-a)\\
b(1-b) & 0
\end{array}
\right)=\left( \begin{array}{cc}
ab(1-b) & a(1-a)^2\\
b(1-b)^2 & ab(1-a)
\end{array}
\right)
\end{equation*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Suppose that $\cos a=-\frac{8}{17}$ for $\pi < a < 3\pi/2$ find $\sin 2a$ Suppose that $\cos A= -\frac{8}{17}$ for $\pi < A < 3\pi/2$ find $\sin 2A$. how is this question solved and in what steps ?
| Tools
$$
\sin 2\theta = 2\cos\theta \sin \theta
\tag{1}
$$
$$
\sin \theta = \pm \sqrt{ 1 - \cos^{2} \theta }
$$
Data
Let the angle
$$
\theta = \arccos \left( -\frac{8}{17} \right)
$$
then the sine is
$$
\sin \theta = -\frac{15}{17}
$$
Solution
$$
\sin 2\theta = 2 \cos\theta \sin \theta =
2
\left( -\frac{8}{17}\right)
\left( -\frac{15}{17}\right)
=
\frac{240}{289}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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$A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ $A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ and $\dfrac {3\sin A}{\sin B}=\dfrac {2\cos B}{\cos A}$. Then $A+2B$ is equal to:
$1$. $\dfrac {\pi}{4}$
$2$. $\dfrac {\pi}{3}$
$3$. $\dfrac {\pi}{6}$
$4$. $\dfrac {\pi}{2}$.
My Attempt
$$\dfrac {3\sin A}{\sin B}=\dfrac {2\cos B}{\cos A}$$
$$3\sin A.\cos A= 2\cos B.\sin B$$
$$\dfrac {3}{2} \sin 2A=2\sin 2B$$.
How do I proceed further?
| HINT: we have:
$$\frac{3\sin\left(\text{A}\right)}{\sin\left(\text{B}\right)}=\frac{2\cos\left(\text{B}\right)}{\cos\left(\text{A}\right)}\space\Longleftrightarrow\space3\sin\left(\text{A}\right)\cos\left(\text{A}\right)=2\cos\left(\text{B}\right)\sin\left(\text{B}\right)\space\Longleftrightarrow\space$$
$$\frac{3\sin\left(2\text{A}\right)}{2}=\sin\left(2\text{B}\right)\tag1$$
Now, we also have:
$$3\cos^2\left(\text{A}\right)+2\cos^2\left(\text{B}\right)=1+3\cdot\frac{1+\cos\left(2\text{A}\right)}{2}+\cos\left(2\text{B}\right)=4\tag2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 4,
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} |
How to prove this property involving the floor function? How does one prove that $$\lfloor x \rfloor = \lfloor x/2 \rfloor + \lfloor (x+1)/2 \rfloor $$
where $\lfloor x \rfloor$ represents the greatest integer less than or equal to $x$?
I tried to do this by using inequalities but end up with $\lfloor x \rfloor + 1/2$ on the lhs. Can someone help?
| I here assume that $x$ is a positive real. We then have that $x = n + r$, where $n \in \mathbb{N}$ and $r$ is a real number with $r \in [0;1)$. We have that $\lfloor x \rfloor = n$.
Either $n$ or $n+1$ is even. If $n$ is even, then $\lfloor x/2 \rfloor = \frac{n}{2}$ and $\lfloor (x+1)/2 \rfloor = \lfloor (n+1)/2 \rfloor = \lfloor \frac{n}{2} + \frac{1}{2} \rfloor = \frac{n}{2}$. Therefore $\lfloor x/2 \rfloor + \lfloor (x+1)/2 \rfloor = \frac{n}{2} + \frac{n}{2} = n$.
The case where $n$ is odd is similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Domain of $f(x)=\sqrt{\lfloor x\rfloor-1+x^2}$ I drew the number line and tested with different values, getting the correct domain $(-\infty,-\sqrt3)\cup[1,\infty)$. However, how do I solve this faster by manipulating the function?
| Let $g(x) = x^2 + \lfloor{x}\rfloor - 1$.
Then the domain of $f$ is $\{x \in \mathbb{R} \mid g(x) \ge 0\}$.
Note that for all $x \in \mathbb{R}$, we have $x-1 < \lfloor{x}\rfloor \le x$, hence for all $x \in \mathbb{R}$,
\begin{align*}
g(x) &= x^2 + \lfloor{x}\rfloor - 1\\[4pt]
&\ge x^2 + (x-1) - 1\\[4pt]
&=x^2 + x - 2\\[4pt]
&=(x+2)(x-1)\\[4pt]
\end{align*}
Since $(x+2)(x-1) \ge 0$ if $x \le -2\;\,$or$\;x \ge 1$, it follows that
$$(-\infty,-2] \cup [1,\infty)$$
is a subset of the domain of $f$.
It remains to find which parts of the interval $(-2,1)$ are in the domain of $f$.
Analyze it piecewise . . .
If $-2 < x < -1$, then
\begin{align*}
g(x) &= x^2 + \lfloor{x}\rfloor - 1
\qquad\qquad\qquad\qquad\qquad\;\\[4pt]
&= x^2 + (-2) - 1\\[4pt]
&=x^2 - 3\\[4pt]
\end{align*}
hence the interval $(-2,-\sqrt{3}]$ is in the domain of $f$, but not the rest of the interval $(-2,-1)$.
If $-1 \le x < 0$, then
\begin{align*}
g(x) &= x^2 + \lfloor{x}\rfloor - 1\\[4pt]
&= x^2 + (-1) - 1\\[4pt]
&=x^2 - 2\\[4pt]
&<0\qquad\text{[since $x^2-2<0\;\;$if$\;\,-1\le x < 0$]}\\[4pt]
\end{align*}
If $0 \le x < 1$, then
\begin{align*}
g(x) &= x^2 + \lfloor{x}\rfloor - 1\\[4pt]
&= x^2 + (0) - 1\\[4pt]
&=x^2 - 1\\[4pt]
&<0\qquad\text{[since $x^2-1 < 0\;\;$if$\;\,0 \le x < 1$]}
\;\;\;\\[4pt]
\end{align*}
Thus, the domain of $f$ is
\begin{align*}
&(-\infty,-2] \cup (-2,-\sqrt3] \cup [1,\infty)\\[8pt]
&=(-\infty,-\sqrt3] \cup [1,\infty)\\[4pt]
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the value of $A$ for the inequality if $x,y,z$ are positive reals and
$$S=\frac{x}{(x+y)(x+z)}+\frac{y}{(y+x)(y+z)}+\frac{z}{(z+x)(z+y)} \le A$$ find $A$
i have taken L.C.M getting
$$S=\frac{2(xy+yz+zx)}{(x+y)(y+z)(z+x)}$$ dividing by $xyz$ we get
$$S=\frac{2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}{xyz\left((\frac{1}{x}+\frac{1}{y})(\frac{1}{y}+\frac{1}{z})(\frac{1}{z}+\frac{1}{x})\right)}$$
Letting $a=\frac{1}{x}$, $b=\frac{1}{y}$ and $c=\frac{1}{z}$ we get
$$S=\frac{2abc(a+b+c)}{(a+b)(b+c)(c+a)}$$ $\implies$
$$S=\frac{abc(a+b+b+c+c+a)}{(a+b)(b+c)(c+a)}$$ $\implies$
$$S=abc \times \left(\frac{1}{(a+b)(b+c)}+\frac{1}{(b+c)(c+a)}+\frac{1}{((a+b)(c+a)}\right)$$
Now by AM-GM inequality
$$(a+b) \ge 2 \sqrt{ab}$$ and $(b+c) \ge 2 \sqrt{bc}$ so
$$(a+b)(b+c) \ge 4c\sqrt{ab}$$ so
$$S \le \frac{abc}{4c\sqrt{ab}}+\frac{abc}{4b\sqrt{ac}}+\frac{abc}{4a\sqrt{bc}}$$
so
$$S \le \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{4} \le \frac{a+b+c}{4}$$ by cauchy inequality
but how can we get $A$ from here
| For $x=y=z\rightarrow0^+$ we get $\sum\limits_{cyc}\frac{x}{(x+y)(x+z)}\rightarrow+\infty$,
which says that the needed value of $A$ does not exist.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove Ramanujan's formula for nested cubic roots $\sqrt[3]{{a}+b\sqrt[3]{r}}$ Ramanujan found that
$$\begin{align*} & \sqrt[3]{(m^2+mn+n^2)\sqrt[3]{(m-n)(m+2n)(2m+n)}+3mn^2+n^3-m^3}\\ =&\sqrt[3]{\tfrac {(m-n)(m+2n)^2}9}-\sqrt[3]{\tfrac {(2m+n)(m-n)^2}9}+\sqrt[3]{\tfrac {(m+2n)(2m+n)^2}9}
\end{align*}$$
for arbitrary $m$ and $n$. The problem is that I am not sure how to prove it.
Question: Is there a way to prove it?
Similarly to the Ramanujan's formula for $$\sqrt{\sqrt[3]{a}+\sqrt[3]{b}}$$
I tried starting with a polynomial and deriving above formula through some clever manipulation. Unfortunately, none of that worked. If I were to do it that way, the terms in the polynomial would have to be a square root.
The book provides its way of proof by cubing both sides brute-force and slugging out the difficult algebra. Is there another way to prove it?
| Denote
$$c_1=\sqrt[3]{\tfrac {(m-n)(m+2n)^2}9},\>\>\>
c_2=-\sqrt[3]{\tfrac {(2m+n)(m-n)^2}9},\>\>\>
c_3=\sqrt[3]{\tfrac {(m+2n)(2m+n)^2}9}
$$
and it is straightforward to verify that
\begin{align}
c_1c_2c_3 &=- \frac{1}{9}(m-n)(2m+n)(m+2n)\\
c_1^3+c_2^3+c_3^3 &= \frac{1}{3}(m^3+6m^2n+3mn^2-n^3)\\
c_1^3c_2^3+c_2^3c_3^3+c_3^3c_1^3 &=
\frac{1}{3} c_1c_2c_3(m^3-3m^2n-6mn^2-n^3)\\
\end{align}
Next, let
$$A=c_1+c_2+c_3, \>\>\> B=c_1c_2+c_2c_3+c_3c_1,\>\>\>C= c_1c_2c_3$$
and evaluate
\begin{align}
A^3 &= 3AB+ c_1^3+c_2^3+c_3^3-3c_1c_2c_3\\
&=3AB+ (m^3+3m^2n-n^3)\tag1 \\
B^3 &= 3c_1c_2c_3 AB + c_1^3c_2^3+c_2^3c_3^3+c_3^3c_1^3-3(c_1c_2c_3)^2\\
&=3(AB)C +(m^3-3mn^2-n^3)C\tag2
\end{align}
and their product
$$
A^3B^3 = 9(AB)^2C-27(AB)C^2+27C^3
-\frac13C(m^2+mn+n^2)^3\tag3
$$
where the followings are recognized in arriving at (3)
\begin{align}
& (m^3+3m^2n-n^3)+(m^3-3mn^2-n^3)=-9C \\
& (m^3+3m^2n-n^3)(m^3-3mn^2-n^3)=27C^3-\frac13(m^2+mn+n^2)^3\\
\end{align}
Rearrange (3)
$$(AB-3C)^3=-\frac13C(m^2+mn+n^2)^3$$
and substitute $AB$ via (1) to obtain the equation for $A$
$$[A^3-(3mn^2+n^3-m^3)]^3 = -9C(m^2+mn+n^2)^3$$
which leads to the Ramanujan formula
$$A= \sqrt[3]{(m^2+mn+n^2)\sqrt[3]{(m-n)(m+2n)(2m+n)}+3mn^2+n^3-m^3}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaluate $ \prod_{n=1}^{80}n^{k-n} \pmod{83}$
Evaluate $\displaystyle \prod_{n=1}^{80}n^{80-n} \pmod{83}$.
Attempt:
The product is \begin{align*}\prod_{n=1}^{80}n^{80-n} &= 1^{79} \cdot 2^{78} \cdot 3^{77} \cdots 79^1 \cdot 80^0\\&\equiv -2^{78} \cdot 3^{77} \cdot 4^{77} \cdots 41^{77}\\&\equiv -2 \cdot (1 \cdot 2 \cdots 41)^{77} \\&\equiv -2(1 \cdot 2 \cdots 41)^{77}\pmod{83}\end{align*} Note that $$1^2 \cdot 2^2 \cdot 3^2 \cdots \left(\dfrac{p-1}{2}\right)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod{p}$$ for a prime $p$. Thus for $83$ we find $$(1 \cdot 2 \cdot 3 \cdots 41)^2 \equiv (-1)^{\frac{83+1}{2}} \equiv (-1)^{42} \equiv 1 \pmod{83}.$$ But if $x^2 \equiv 1 \pmod{p}$, then $x \equiv \pm 1 \pmod{p}$. How do we eliminate the case that $1 \cdot 2 \cdots 41 \equiv -1 \pmod{83}$?
| For $\prod_{n=1}^{82} n^{80-n} \mod 83$, we could pair each $n$ with $83-n$:
$$ n^{80-n} (83-n)^{n-3} \equiv n^{80-n} (-n)^{n-3} \equiv (-1)^{n-3} n^{77}$$
so that
$$ \prod_{n=1}^{82} n^{80-n} \equiv \prod_{n=1}^{41} (-1)^{n-3} n^{77} \mod 83$$
$\prod_{n=1}^{41} (-1)^{n-3}$ has $20$ factors of $-1$ (one for each even $n$) so it is $1$. $\prod_{n=1}^{41} n \equiv 1 \mod 83$, so $\prod_{n=1}^{41} n^{77} \equiv 1 \mod 83$ as well. Thus
$$ \prod_{n=1}^{82} n^{80-n} \equiv 1 \mod 83$$
But you're missing the cases $n=81 \equiv -2$ and $n = 82 \equiv -1$, so multiply by
$81^{81-80} \equiv 81 \mod 83$ and $82^{82-80} \equiv (-1)^2 \equiv 1 \mod 83$
resulting in
$$ \prod_{n=1}^{80} n^{80-n} \equiv 81 \mod 83$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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To prove $\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \cdots \frac{(2n-1)}{2n} \le \frac{1}{\sqrt{3n+1}}$ To prove $$P=\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \cdots \frac{(2n-1)}{2n}\le \frac{1}{\sqrt{3n+1}}$$
i have written $P$ as
$$P=\frac{(2n)!}{2^{2n}(n!)^2}=\frac{(2n)!}{4^{n}(n!)^2}=\frac{\binom{2n}{n}}{4^n}$$
Now
$$P=\frac{\binom{2n}{n}}{(1+3)^n} \lt \frac{\binom{2n}{n}}{1+3n}$$ since
$$(1+3)^n=1+3n+\binom{n}{2}3^2+\cdots$$
Any help here..
| We need to prove that $$\binom{2n}{n}\leq\frac{4^n}{\sqrt{3n+1}}.$$
Indeed, by induction for $n=1$ we get an equality.
Now, $$\binom{2n+2}{n+1}=\binom{2n}{n}\cdot\frac{(2n+1)(2n+2)}{(n+1)^2}\leq\frac{4^n}{\sqrt{3n+1}}\cdot\frac{2(2n+1)}{n+1}.$$
Thus, it remains to prove that
$$\frac{4^n}{\sqrt{3n+1}}\cdot\frac{2(2n+1)}{n+1}\leq\frac{4^{n+1}}{\sqrt{3n+4}},$$
which is
$$(2n+2)^2(3n+1)\geq(2n+1)^2(3n+4)$$ or $$n\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$. If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$.
$$\sec \theta + \tan \theta = x$$
$$\dfrac {1}{\cos \theta }+\dfrac {\sin \theta }{\cos \theta }=x$$
$$\dfrac {1+\sin \theta }{\sqrt {1-\sin^2 \theta }}=x$$
$$1+\sin \theta =x\sqrt {1-\sin^2 \theta }$$
$$1+2\sin \theta + \sin^2 \theta = x^2-x^2 \sin^2 \theta $$
$$x^2 \sin^2 \theta + \sin^2 \theta + 2\sin \theta = x^2-1$$
$$\sin^2 \theta (x^2+1) + 2\sin \theta =x^2-1$$
| From where you are:
You obtained a quadratic function in $\sin(\theta)$. Perform the substitution $u =\sin(\theta)$.
We obtain the quadratic (in $u$):
$$(x^2+1)u^2 + 2u - x^2 +1 = 0$$
$$\Rightarrow u_{1,2} = \frac{- 2 \pm \sqrt{4 - 4(x^2+1)(1-x^2)}}{2(x^2+1)}$$
$$ = \frac{- 2 \pm \sqrt{4 + 4(x^4 -1)}}{2(x^2+1)}$$
$$ = \frac{- 2 \pm \sqrt{4x^4}}{2(x^2+1)}$$
$$ = \frac{- 2 \pm 2x^2}{2(x^2+1)}$$
$$ = \frac{- 1 \pm x^2}{x^2+1}$$
Therefore,
$$\sin(\theta)_{1,2} = \frac{- 1 \pm x^2}{x^2+1}$$
One of those solutions will not work out.
This happened because you squared the equation multiple times and we know that $a = b$ is not equivalent with $a^2 = b^2$, so you should fill in both solutions in the original expression and see which one works and which one doesn't.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding a point on an ellipsoid Find a point on the ellipsoid $x^2+4y^2+z^2=9$ where the tangent plane is perpendicular to the line with parametric equations \begin{align}x&=2+2t\\y&=1+2t\\z&=3-t\end{align}
The answer to this question is: $$\left(\frac{6\sqrt{13}}{13}, \frac{6\sqrt{13}}{13}, -\frac{3\sqrt{13}}{13}\right)\\
\left(-\frac{6\sqrt{13}}{13}, -\frac{6\sqrt{13}}{13}, \frac{3\sqrt{13}}{13}\right)$$
I know the normal is $(x, 4y, z)$. What step comes after this?
Thanks in advance.
| The normal to the plane must be parallel to the given line, hence
$$(x,4y,z)=\lambda(2,2,-1).$$
Combining with the equation of the ellipsoid,
$$x^2+4y^2+z^2=9,$$
we get
$$2^2\lambda^2+4\frac1{2^2}\lambda^2+(-1)^2\lambda^2=6\lambda^2=9,$$ and solve for $\lambda$.
$$(x,y,z)=\pm\sqrt{\frac32}\left(2,\frac12,-1\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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let $x ,y ,z \in \mathbb{N} ,x \leq y\leq z$ and : $\frac{100}{336}=\frac{1}{x}+\frac{1}{xy}+\frac{1}{xyz}$then : $x+y+z =?$ let $x ,y ,z \in \mathbb{N} ,x \leq y\leq z$
and :
$$\frac{100}{336}=\frac{1}{x}+\frac{1}{xy}+\frac{1}{xyz}$$
then :
$$x+y+z =?$$
my try :
$$\frac{100}{336}=\frac{1}{x}+\frac{1}{xy}+\frac{1}{xyz}=\frac{xy+z+1}{xyz}=\frac{25}{84}$$
now ?
| It should be $x=4$, $y=6$, $z=7$ and the answer to the original question is 17.
You can use the method described here for obtaining the Engel expansion of a number.
The Engel expansion for $x$ is written $x=\frac{1}{a_1}+\frac{1}{a_1a_2}+\frac{1}{a_1a_2a_3}+\cdots$
As described in the wiki, the non-decreasing values $a_1, a_2, \ldots$ can be obtained using
$$u_1=x
\qquad a_k=\left \lceil \frac{1}{u_k} \right \rceil
\qquad u_{k+1}=u_ka_k-1
$$
and stopping when we reach $u_k = 0$.
Starting with $x=\frac{25}{84}$ leads to the required solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Probability of a quarter final combination There are three Spanish teams, two German, one Italian, one French, and one English. What is the probability that Germany plays itself? What is the probability that Spain plays itself? What is the intersection of this probability?
The first question I think is $1/28$ because ${8 \choose 2}$ = 28 and there is only one combination with Germany vs Germany so 1/28. Spain I think is 3/28 because ${3 \choose 2}$= 3 (three teams to choose from and only two teams needed). I would have thought that the intersection is where Spain play Spain and Germany play Germany. Apparently it's 3/35 but I don't know why. I tried using the $P(A \bigcup B)=P(A)+P(B)-P(A\bigcap B)$ but I don't know the union area.
| We can list the teams in some order, say alphabetically. There are seven ways we can match a team with the first team on the list. This leaves six teams. There are five ways to match a team with the first team remaining on the list. This leaves four teams. There are three ways to match a team with the first team remaining on the list. The final two teams must play each other. Hence, the number of possible quarterfinal pairings is
$$7 \cdot 5 \cdot 3 \cdot 1$$
What is the probability that the two German teams play each other in the quarterfinal?
There is one way to match the German teams. The remaining six teams can be matched in $5 \cdot 3 \cdot 1$ ways. Hence, the probability that the German teams play each other is
$$\frac{5 \cdot 3 \cdot 1}{7 \cdot 5 \cdot 3 \cdot 1} = \frac{1}{7}$$
which makes sense since given a German team, only one of the seven possible opposing teams is German.
What is the probability that two Spanish teams play each other in the quarterfinals?
As you observed, there are $\binom{3}{2} = 3$ ways to choose a pair of Spanish teams to play each other. This leaves six teams, which can be matched in $5 \cdot 3 \cdot 1$ ways. Hence, the number of quarterfinal schedules in which two of the Spanish teams play each other is
$$\binom{3}{2} \cdot 5 \cdot 3 \cdot 1 = 3 \cdot 5 \cdot 3 \cdot 1$$
so the probability that two Spanish teams play each other in the quarterfinals is
$$\frac{3 \cdot 5 \cdot 3 \cdot 1}{7 \cdot 5 \cdot 3 \cdot 1} = \frac{3}{7}$$
What is the probability that the two German teams play each other and two Spanish teams play each other in the quarterfinals?
There is one way to match the German teams and $\binom{3}{2} = 3$ ways to match two of the three Spanish teams. This leaves four teams. They can be matched in $3 \cdot 1$ ways. Hence, the number of favorable schedules is
$$\binom{3}{2} \cdot 3 \cdot 1 = 3 \cdot 3 \cdot 1$$
Hence, the desired probability is
$$\frac{3 \cdot 3 \cdot 1}{7 \cdot 5 \cdot 3 \cdot 1} = \frac{3}{7 \cdot 5} = \frac{3}{35}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inverse element of
Given the extension $ \dfrac{\mathbb Q[x]}{\langle x^3-5\rangle}: \mathbb Q$ and $\alpha =1+x^2+\langle x^3-5\rangle \in \dfrac{\mathbb Q[x]}{\langle x^3-5 \rangle}$, find $\alpha^{-1}$.
Can somebody give a hint? or an answer? please?
| A general element of this ring looks like $a + bx + cx^2$. Suppose this were the inverse of $1+x^2$. Then:
\begin{align*}
1 &= (1+x^2)(a+bx+cx^2)\\
&= a + bx + (a+c)x^2 + bx^3 +cx^4
\end{align*}
In this ring, $x^3 = 5$ so:
\begin{align*}
1 &= a + bx + (a+c)x^2 +5b + 5cx \\
&=(a+5b) + (b+5c)x + (a+c)x^2.
\end{align*}
This tells us that $$a + 5b =1\\ b+5c = 0 \\ a + c = 0$$
I trust you can take it from here.
I can't resist adding an alternative approach. If you don't have a background in linear algebra, or this seems arcane, feel free to disregard it.
Let's consider your ring as a vector space over $\mathbb{Q}$. Then it is three dimensional, and an easy basis would be $1,x,x^2$. Define the linear transformation on this vector space that multiplies a polyomial by $x$. Now let us find the matrix (call it $A$) of this transformation with respect to the basis above:
$$A = \begin{pmatrix} 0 & 0 & 5 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}_.$$
The cool thing is that if you want to find the inverse of $x^2+1$, you can instead find the inverse of the matrix $A^2 + I$ and then express it as a linear combination of $I, A, A^2$! In this case we have that:
\begin{align*}
(A^2 + I)^{-1} &= \frac{1}{26}\begin{pmatrix} 1 & -5 & 25 \\ 5 & 1 & -5 \\ -1 & 5 & 1 \end{pmatrix} \\
&= \frac{1}{26} I + \frac{5}{26} A - A^2.
\end{align*}
This tells us the the inverse of $x^2+1$ is $$\frac{1}{26} + \frac{5}{26}x - \frac{1}{26}x^2.$$
| {
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Prove that $n^5+n^4+1$ is composite for $n>1.$
Prove that $f(n)=n^5+n^4+1$ is composite for $n>1, n\in\mathbb{N}$.
This problem appeared on a local mathematics competition, however it looks like there is no simple method to solve it. I tried multiplying it by $n+1$ or $n-1$ and then tried factorizing it, but it was too tough for me.
Any help will be appreciated.
| Building upon OP's idea:
I tried multiplying it by $n+1$ or $n-1$ and then tried factorising it
$$
\begin{align}
(n-1)(n^5+n^4+1) = \color{red}{n^6} \color{blue}{- n^4 + n} \color{red}{- 1} &= \color{red}{(n^3-1)(n^3+1)} \color{blue}{-n(n^3-1)} \\
&= (n^3-1)(n^3-n+1) \\
&= (n-1)(n^2+n+1)(n^3-n+1)
\end{align}
$$
| {
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How can we prove the equality of the following determinants? Assume that $A$ is an $n\times n$ real matrix whose entries are all $1$. Then how can we show the following determinant equality for any $x$?
$\det(A-xI)$=\begin{vmatrix}
1 -x & 1 & 1 & \cdots & 1 \\
1 & 1 -x & 1 & \cdots & 1 \\
1 & 1 & 1 -x & \cdots & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & 1 & 1 & \cdots & 1 -x
\end{vmatrix}=
\begin{vmatrix}
n-x & n-x & n-x & \cdots & n -x \\
1 & 1 -x & 1 & \cdots & 1 \\
1 & 1 & 1 -x & \cdots & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & 1 & 1 & \cdots & 1 -x
\end{vmatrix}
Thanks.
| Recall that if an $n\times n$ matrix $B$ is obtained from an $n\times n$ matrix $A$ by adding a scalar multiple of one row of $A$ to another row, then $\det(B)=\det(A)$.
In your problem the second matrix is obtained from the first by adding rows $2$ through $n$ to row $1$. So the determinants are the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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PDF of $\frac{X}{1+X^2}$ in terms of the PDF of $X$ Given a r.v. $X$ with a know probability density $g$, I would like to find the probability density of $\frac{X}{1+X^2}$. Below I write my calculations:
denote
$f(X) := \frac{X}{1+X^2}$
Now $f$ is not monotone so we can't apply the known theorem, we proceed by finding the solutions to $\frac{X}{1+X^2} < c$ for the solutions to exist we need $-1/2<c < 1/2$ and the solutions to the associated quadratic equation will be $x_{1-2} = 1 \pm\sqrt{1-4c^2}/ (2c)$.
I obtain that
$$P(f(X) < c) = P( \{X <1 -\sqrt{1-4c^2}/ (2c)\} \cup \{ X > 1 +\sqrt{1-4c^2}/ (2c) \}) = 1-P(X < 1 +\sqrt{1-4c^2}/ (2c)) + P(X <1 -\sqrt{1-4c^2}/ (2c))$$
So the resulting density of $\frac{X}{1+X^2}$ will be $$1+ \frac{ 4c/ \sqrt{1-4c^2} + 2(1+\sqrt{1-4c^2})}{4c^2}g(1 +\sqrt{1-4c^2}/ (2c)) + \frac{ 4c/ \sqrt{1-4c^2} - 2(1-\sqrt{1-4c^2})}{4c^2}g(1 -\sqrt{1-4c^2}/ (2c)$$
with the condition that $-1/2<c<1/2$.
Is this the correct approach?
| Using the CDF
Let
$$
G(x)=\int_{-\infty}^xg(t)\,\mathrm{d}t\tag{1}
$$
be the CDF of $X$ and $Y=\frac{X}{X^2+1}$. Then the CDF of $Y$ is
$$
\begin{align}
F(\lambda)
&=P(y\lt\lambda)\\
&=P\left(\frac{x}{x^2+1}\lt\lambda\right)\\
&=\left\{\begin{array}{}
P\left(x\lt\frac{2\lambda}{1+\sqrt{1-4\lambda^2}}\right)-P\left(x\lt\frac{1+\sqrt{1-4\lambda^2}}{2\lambda}\right)+1&\text{if $0\le\lambda\lt\frac12$}\\
P\left(x\lt\frac{2\lambda}{1+\sqrt{1-4\lambda^2}}\right)-P\left(x\lt\frac{1+\sqrt{1-4\lambda^2}}{2\lambda}\right)&\text{if $-\frac12\le\lambda\lt0$}
\end{array}\right.\\
&=\left\{\begin{array}{}
G\left(\frac{2\lambda}{1+\sqrt{1-4\lambda^2}}\right)-G\left(\frac{1+\sqrt{1-4\lambda^2}}{2\lambda}\right)+1&\text{if $0\lt\lambda\le\frac12$}\\
G(0)&\text{if $\lambda=0$}\\
G\left(\frac{2\lambda}{1+\sqrt{1-4\lambda^2}}\right)-G\left(\frac{1+\sqrt{1-4\lambda^2}}{2\lambda}\right)&\text{if $-\frac12\le\lambda\lt0$}
\end{array}\right.\tag{2}
\end{align}
$$
Then
$$
\begin{align}
f(\lambda)
&=F'(\lambda)\\[9pt]
&=\tfrac2{1-4\lambda^2+\sqrt{1-4\lambda^2}}g\left(\tfrac{2\lambda}{1+\sqrt{1-4\lambda^2}}\right)+\tfrac{1+\sqrt{1-4\lambda^2}}{2\lambda^2\sqrt{1-4\lambda^2}}g\left(\tfrac{1+\sqrt{1-4\lambda^2}}{2\lambda}\right)\tag{3}
\end{align}
$$
Using Only the PDF
Suppose that $Y=g(X)$ and $f_X(x)$ is the PDF for $X$. Consider the case of an infinitesimal interval, $\mathrm{d}x$, in $X$ and the corresponding infinitesimal interval, $\mathrm{d}y$, in $Y$:
When $g'(x)\gt0$, The probability of an event occurring in $\mathrm{d}x$ is $f_X(x)\,\mathrm{d}x=\frac{f_X(x)}{g'(x)}\,\mathrm{d}y$.
When $g'(x)\lt0$, The probability of an event occurring in $\mathrm{d}x$ is $f_X(x)\,\mathrm{d}x=-\frac{f_X(x)}{g'(x)}\,\mathrm{d}y$.
Thus, the probability of an event occurring in $\mathrm{d}x$ is $f_X(x)\,\mathrm{d}x=\frac{f_X(x)}{|g'(x)|}\,\mathrm{d}y$.
The probability of and event occurring in $\mathrm{d}y$ is the sum of the probabilities over all the points $x_k$ where $g(x_k)=y$. That is,
$$
f_Y(y)\,\mathrm{d}y=\sum_{g(x_k)=y}\frac{f_X(x_k)}{|g'(x_k)|}\,\mathrm{d}y\tag{4}
$$
Thus, the PDF of $Y$ is
$$
f_Y(y)=\sum_{g(x_k)=y}\frac{f_X(x_k)}{|g'(x_k)|}\tag{5}
$$
Which is the formula cited in Did's answer.
| {
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"url": "https://math.stackexchange.com/questions/2288244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solving a rational inequality with an unknown using cases So the question is,
For what values of x is $\dfrac{1}{1 + x}> -1 $ ?
Now, one way to do it is write the inequality relative to zero, and then find the behavior of the graph relative to the zeroes of the numerator and denominator:
$\dfrac{1}{1 + x}> -1 \implies \dfrac{1}{1+x} + 1> 0 \implies \dfrac{1}{1+x} + \dfrac{1+x}{1+x} > 0 $
$\implies \dfrac{2+x}{1+x} > 0$, and we simply find what happens when $x < -2, -2 < x < -1$, and $x > -1$.
What I don't understand is why I can't answer this by multiplying through by $1 +x$ if I consider both a) $1 +x >0$ and b) $1+x < 0$.
Here's the attempt:
Case a):
$1 + x > 0 \implies x > -1$, so
$\dfrac{1}{1+x} > -1$
$\implies (1+x) \dfrac{1}{1+x} > -1 (1+x) $ (we can multiply both sides by 1+x because we assume 1+x is positive)
$\implies x > -2 $.
Hence, the inequality is true when $x > -2$ and $x > -1$--so $x > -1$.
Case b): $1+x < 0 \implies x < -1$, so
$\dfrac{1}{1+x} > -1$
$\implies (1+x) \dfrac{1}{1+x} < -1 (1+x) $ (we assume $1+x$ is negative, so we reverse the sign)
$\implies 1 < 1 + x$
$\implies x > 0$.
Hence, the inequality is true when $x < -1$ and $x > 0$--this is never true.
So the real answer is $x > -1$ or $x <-2$, and I think the reason my attempt at multiplying through by an unknown doesn't work is that, in my assumption $x\in (-1, \infty)$ clearly includes both negative and positive values, and that leads the answer to be wrong.
But I'm not 100 percent certain it's impossible, and would like some help!
| Shortly $${ \frac { 1 }{ 1+x } }>-1\\ \frac { 1 }{ 1+x } +1>0\\ \frac { x+2 }{ x+1 } >0\\ \frac { \left( x+1 \right) \left( x+2 \right) }{ { \left( x+1 \right) }^{ 2 } } >0\\ \left( x+1 \right) \left( x+2 \right) >0\\ x\in \left( -\infty ;-2 \right) \cup \left( -1;+\infty \right) \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2290816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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The distance from the incenter to an acute vertex of a right triangle I'm seeking an alternative proof of this result:
Given $\triangle ABC$ with right angle at $A$. Point $I$ is the intersection of the three angle lines. (That is, $I$ is the incenter of $\triangle ABC$.) Prove that
$$|CI|^2=\frac12\left(\left(\;|BC|-|AB|\;\right)^2+|AC|^2\right)$$
My Proof. Draw $ID \perp AB$, $IE\perp BC$, and $IF\perp CE$.
We have $|ID|=|IE|=|IF|=x$. Since $\triangle ADI$ is right isosceles triangle, we also have that $|AD|=|ID|=x$. Respectively, we have: $$|ID|=|IF|=|IE|=|AD|=|AF|=x$$
$\triangle BDI=\triangle BEI \Rightarrow |BD|=|BE|=y$. And $|CE|=|CF|=z$
We have:
$$|CI|^2=|CE|^2+|IE|^2=x^2+z^2 \tag{1}$$
And
$$\begin{align}\frac12\left(\left(|BC|-|AB|\right)^2+|AC|^2\right) &=\frac12\left(\left(\;\left(y+z\right)-\left(x+y\right)\;\right)^2+\left(x+z\right)^2\right) \\[4pt]
&=\frac12\left(\left(x-z\right)^2+\left(x+z\right)^2\right) \\[4pt]
&=\frac22\left(x^2+z^2\right) \\[4pt]
&=x^2+z^2
\tag{2}\end{align}$$
From $(1);(2)$ we are done. $\square$
| Let $AB = c, AC = b, BC = a$, and let $IT\perp AC $ at $I$. Then, $CI$ can be written as $$\begin{split}CI^2 &= CT^2 + IT^2 \\&= \left(\frac{a+b-c}{2}\right)^2+\left(\frac{b+c-a}{2}\right)^2\\&= \frac{2a^2+2b^2+2c^2-4ac}{4}\\&=\frac{(a-c)^2+b^2}{2} \\&=\frac{(BC-AB)^2+AC^2}{2}\end{split}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
From $a_{n+1}= \frac{3a_n}{(2n+2)(2n+3)}$ to $a_n$, Case 2
Find and prove by induction an explicit formula for $a_n$ if $a_1=1$ and, for $n \geq 1$,
$$P_n: a_{n+1}= \frac{3a_n}{(2n+2)(2n+3)}$$
Checking the pattern:
$$a_1=1 $$
$$a_2= \frac{3}{4 \cdot 5}$$
$$a_3= \frac{3^2} { 4 \cdot 5 \cdot 6 \cdot 7}$$
$$a_4= \frac{3^3} { 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 }$$
$$a_n = \frac{3^{n-1}}{ \frac {(2n+1)!}{3!} }$$
$$a_n = \frac {3! \cdot 3^{n-1}} {(2n+1)!}$$
Proof by induction:
$$a_1 = \frac {3! \cdot 3^{0}} {3!} =1$$
$$a_{n+1} = \frac {3! \cdot 3^{n}} {(2n+3)!}$$
Very grateful for the feedback given before. I am new to this. I am a bit stock at the end, what is the most efficient way to reach back to $a_n$?
| You assume the expression to be true for $n=m$ and then prove it is true for $n=m+1$.
Expression:
\begin{gather}
a_n = \frac {3! \cdot 3^{n-1}} {(2n+1)!}\\
\\
\text{For } n = m \text{ assume true}\\
\\
a_{m+1}= \frac{3a_m}{(2m+2)(2m+3)} \\
a_{m+1}= \frac{3}{(2m+2)(2m+3)}\cdot \frac {3! \cdot 3^{m-1}} {(2m+1)!}\\
a_{m+1}= \frac {3! \cdot 3^{m}} {(2m+3)!} = \frac {3! \cdot 3^{(m+1)-1}} {(2(m+1)+1)!}
\end{gather}
And proved by the principle of MI.
Edit: To elaborate a bit on this, what you've done is that you've shown for a base case, that the expression is true. Then you've picked a arbitrary index (in this case $m$) and shown it's true for $m+1$. This way you've proved that if $1$ is true, then so is $2$ and if $2$ is true so is $3$ and so on, because $m$ is arbitrary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prob. 24, Chap. 5 in Baby Rudin: For $\alpha>1$, let $f(x) = (x+\alpha/x)/2$, $g(x) = (\alpha+x)/(1+x)$ have $\sqrt{\alpha}$ as their only fixed point Here is Prob. 24, Chap. 5 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
The process described in part (c) of Exercise 22 can of course also be applied to functions that map $(0, \infty)$ to $(0, \infty)$.
Fix some $\alpha > 1$, and put $$ f(x) = \frac{1}{2} \left( x + \frac{\alpha}{x} \right), \qquad g(x) = \frac{\alpha+x}{1+x}. $$
Both $f$ and $g$ have $\sqrt{\alpha}$ as their only fixed point in $(0, \infty)$. Try to explain, on the basis of properties of $f$ and $g$, why the convergence in Exercise 16, Chap. 3, is so much more rapid than it is in Exercise 17. (Compare $f^\prime$ and $g^\prime$, draw the zig-zags suggested in Exercise 22.)
Do the same when $0 < \alpha < 1$.
Here are the links to my posts here at Math SE on Prob. 22, Chap. 5, Prob. 16, Chap. 3, and Prob. 17, Chap. 3, in Baby Rudin, 3rd edition:
Prob. 22, Chap. 5 in Baby Rudin: Fixed Points of Real Functions
Prob. 16, Chap. 3 in Baby Rudin: $x_{n+1} = (x_n + \alpha/x_n)/2$, with $x_1 > \sqrt{\alpha}$, $\alpha > 0$
Prob. 17, Chap. 3 in Baby Rudin: For $\alpha > 1$, how to obtain these inequalities from this recurrence relation?
My Attempt:
We note that, for $0 < x < \infty$,
$$ g(x) = 1 + \frac{\alpha-1}{x+1},$$
and so
$$f^\prime(x) = \frac{1}{2} \left( 1 - \frac{\alpha}{x^2} \right), \qquad g^\prime(x) = - \frac{\alpha-1}{(x+1)^2},$$
and for $\alpha > 1$, we see that, if $x > \sqrt{\alpha}$, then $x^2 > \alpha$, and so
$$0 < f^\prime(x) < \frac{1}{2}. $$
which implies (by part (c) of Prob. 22, Chap. 5, in Baby Rudin, 3rd edition) that the sequence $x_0 > \sqrt{\alpha}$, $x_{n+1} = f \left( x_n \right)$, for $n = 0, 1, 2, 3, \ldots$, does converge to the only fixed point of $f$, which is $\sqrt{\alpha}$. In fact, for all $n$, we have
$$ \left| x_{n+1} - x_n \right| \leq \frac{1}{2} \left| x_n - x_{n-1} \right| \leq \cdots \leq \frac{1}{2^n} \left| x_1 - x_0 \right|, $$
and so for any $m < n$, we have
$$
\begin{align}
\left| x_n - x_m \right| &\leq \left| x_n - x_{n-1} \right| + \cdots+ \left| x_{m+1} - x_m \right| \\
&\leq \left( \frac{1}{2^{n-1}} + \cdots + \frac{1}{2^m} \right) \left| x_1 - x_0 \right| \\
&= \frac{1}{2^m} \left( 1 + \cdots + \frac{1}{2^{n-m-1}} \right) \left| x_1 - x_0 \right| \\
&= \frac{1}{2^m} \frac{ 1 - \frac{1}{2^{n-m}} }{ 1 - \frac{1}{2} } \left| x_1 - x_0 \right| \\
&= \left( \frac{1}{2^{m-1}} - \frac{1}{2^{n-1}} \right) \left| x_1 - x_0 \right|,
\end{align}
$$
and upon letting $n \to \infty$, while keeping $m$ fixed, we obtain
$$ \left| x_m - x \right| \leq \frac{1}{2^{m-1}} \left| x_1 - x_0 \right| = \frac{1}{2^{m-1}} \left| \frac{1}{2} \left(x_0 + \frac{\alpha}{x_0} \right) - x_0 \right| = \frac{1}{2^m } \left( x_0 - \frac{\alpha}{x_0} \right), $$
for $m = 1, 2, 3, \ldots$, which gives the rate of convergence of this recursive algorithm.
And, the similar situation occurs for $0 < \alpha < 1$ as well, provided we choose $x_0 > \sqrt{\alpha}$.
Is my analysis correct? Or, have I erred anywhere or missed something of substance?
And, what about $g$? How to analyze it? How to show what Rudin has demanded to be shown?
| For $f:$
Suppose $x_0>0$ and $a>0$.Let $x_{n+1}=(x_n+a/x_n)/2.$ Assume $x_0\ne \sqrt a \;$( because if $x_0=\sqrt a$ then $x_n=\sqrt a$ for all $n,$ which is a trivial case). Then we have:
(I). $x_1^2>a$ so $x_1>\sqrt a.$
(II). If $x_n^2>a$ then $x_n>x_{n+1}>\sqrt a.$ So by (I), $n\geq 1\implies x_n>x_{n+1}>\sqrt a.$
(III). If $n\geq 1$ and $x_n=k\sqrt a$ with $k\geq 2$ then $x_{n+1}=\sqrt a\;(k+1/k)/2<\sqrt a\;(k+1)/2$, so for some $m>n$ we will have $1<x_m/\sqrt a\;<2.$
(IV). If $1<x_m/\sqrt a<2$ then let $x_n=(1+d_n)\sqrt a$ for each $n$. Then $0<d_m<1.$ We have $$2ad_{m+1}<2ad_{m+1}+ad_{m+1}^2=x_{m+1}^2-a=$$ $$=\frac {1}{4}\left(x_m+\frac {a}{x_m}\right)^2-a=\frac {1}{4}\left(x_m-\frac {a}{x_m}\right)^2=$$ $$=\frac{(x_m^2-a)^2}{4x_m^2}=\frac {(2ad_m+ad_m^2)^2}{4a(1+d_m)^2}<\frac{(2ad_m+2ad_m^2)^2}{4a(1+d_m)^2}=ad_m^2,$$ $$\text {so we have }\quad 0<d_{m+1}<d_m^2<1.$$ By induction on $n,$ for all $n\geq m$ we have $0<d_{n+1}<d_n^2<1.$
Another way to do this: With $x_1>\sqrt a\;,$ let $x_n/\sqrt a\;=\coth T_n$ for $n\geq 1.$ Then when $n\geq 1$ we have $T_{n+1}=2T_n$, so $T_n=2^{n-1}T_1.$
| {
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"url": "https://math.stackexchange.com/questions/2292135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Compute $gcd\left(1714, 1814\right)$ using Euclidean Algorithm So I know the answer for this is $2$, but based on my own work, I can't get to that solution. I haven't done a gcd before where $b>a$. I thought I could just flip the numbers and use the same method but that didn't seem to work. Here's what I have so far, what I am doing wrong?
$$\begin{align}
\mathrm{gcd}(1714, 1814) &= \mathrm{gcd}(1814, 1714) \\
\mathrm{gcd}(1814, 1714) &= (1714, 100)\\
&= (100, 14)\\
&= (14, 9)\\
&= (9, 5)\\
&= (5, 4)\\
&= (4, 1)\\
&= (1, 0)\\
&= 1\\
\end{align}
$$
I basically tried using the Euclidian algorithm method where you keep doing long division into each number to get the remainder and continue with that process.
| written as a continued fraction:
$$ \frac{ 1814 }{ 1714 } = 1 + \frac{ 100 }{ 1714 } $$
$$ \frac{ 1714 }{ 100 } = 17 + \frac{ 14 }{ 100 } $$
$$ \frac{ 100 }{ 14 } = 7 + \frac{ 2 }{ 14 } $$
$$ \frac{ 14 }{ 2 } = 7 + \frac{ 0 }{ 2 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccccc}
& & 1 & & 17 & & 7 & & 7 & \\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 1 }{ 1 } & & \frac{ 18 }{ 17 } & & \frac{ 127 }{ 120 } & & \frac{ 907 }{ 857 }
\end{array}
$$
$$ $$
$$
\begin{array}{ccc}
\frac{ 1 }{ 0 } & \mbox{digit} & 1 \\
\frac{ 1 }{ 1 } & \mbox{digit} & 17 \\
\frac{ 18 }{ 17 } & \mbox{digit} & 7 \\
\frac{ 127 }{ 120 } & \mbox{digit} & 7 \\
\frac{ 907 }{ 857 } & \mbox{digit} & 0 \\
\end{array}
$$
$$ 907 \cdot 120 - 857 \cdot 127 = 1 $$
$$ \gcd( 1814, 1714 ) = 2 $$
$$ 1814 \cdot 120 - 1714 \cdot 127 = 2 $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $x^2+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}} $
Find the real roots of the equation $$x^2+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}} $$ $$(0<a<\frac{1}{4})$$
My attempt,
$(x^2+2ax+\frac{1}{16}+a)^2=a^2+x-\frac{1}{16}$
I did an expansion which becomes $x^4+4ax^3(4a^2+2a+\frac{1}{8})x^2+(4a^2+\frac{1}{4}a-1)x+\frac{1}{8}a+\frac{1}{16}+\frac{1}{256}=0$
$$\left [ x^4+(2a-1)x^3+\frac{1}{16}x^2 \right ]+\left [ (2a-1)x^3+(4a^2-1)x^2+(\frac{a}{8}+\frac{1}{16})x \right ] +\left [ (2a+\frac{17}{16})x^2+(4a^2+\frac{1}{8}a-\frac{17}{16})x+(\frac{1}{8}a+\frac{1}{16}+\frac{1}{256}) \right ]=0$$
$$\left [ x^2+(2a-1)x+\frac{1}{16} \right ]\left [ x^2(2a+1)x+(2a+\frac{17}{16}) \right ]=0$$
Since $0<a<\frac{1}{4}$,
$x^2+(2a-1)x+\frac{1}{16}=0$
$x=\frac{1-2a}{2}\pm \sqrt{(\frac{1-2a}{2})^2-\frac{1}{16}}$
My solution is really messy and I'm unsure my answer.
My question: is my answer correct? And is there another way to solve this algebra question? Thanks in advance.
| Hint
$$x^2+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}}\\
(x+a)^2-a^2+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}}\\
(x+a)^2=a^2-\frac{1}{16}-a+\sqrt{a^2+x-\frac{1}{16}}\\
(x+a)^2+(x+a)=a^2+x-\frac{1}{16}+\sqrt{a^2+x-\frac{1}{16}}$$
Now call:
$$y=\sqrt{a^2+x-\frac{1}{16}}\to y^2=a^2+x-\frac{1}{16}$$
and then
$$(x+a)^2+(x+a)=y^2+y\\
(x+a)^2-y^2+(x+a)-y=0\\
(x+a-y)(x+a+y+1)=0$$
Can you finish?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Can we prove the divergence of harmonic series like this? Let $S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\cdots$.
Or $S=(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots)+(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots)$.
$\implies S=(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots)+\frac{1}{2}({1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots)$.
$\implies S=(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots)+\frac{1}{2}(S)$.
$\implies \frac{S}2=1+(\frac{1}{3}+[\frac{1}4-\frac{1}4]+\frac{1}{5}+[\frac{1}6-\frac{1}6]+\frac{1}{7}+[\frac{1}8-\frac{1}8]+\cdots)$.
$\implies \frac{S}2=1+(\frac{1}{3}-\frac{1}4+\frac{1}{5}-\frac{1}6+\frac{1}{7}-\frac{1}8+\cdots)+\frac{1}2(\frac{1}2+\frac{1}{3}+\frac{1}{4}+\cdots)$.
$\implies \frac{S}2=1+(\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\cdots)+\frac{1}2(S-1)$.
$\implies 0=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\cdots$.
Which we know is not correct.
Also, we know that we do face this type of discrepancies when we are dealing with divergent series(as is the case when we put $x=1$ in the infinite geometric series of $\frac{1}{1-x}$) in the usual manner of dealing with convergent series.
Therefore, $S$ must be a divergent series.
Can we conclude something more/else from the last equation?
EDIT
Can we conclude that since the series is divergent and the partial sum of the terms is always increasing so, the only way it can be divergent is it($S$) has to be $\infty$?
| Yes, we can prove the divergence of the harmonic series like that, but one must at the beginning state something along the lines of "suppose the harmonic series is convergent".
Once that assumption is introduced, it is a valid proof by contradiction. Since all terms in the harmonic series are positive, if it were convergent, it would be absolutely convergent, hence also all of its subseries would be absolutely convergent, and all manipulations you made are valid for absolutely convergent series.
Since at the end you reach a manifestly false assertion by means that are valid if the assumption is true, it correctly follows that the assumption must have been false, i.e. that the harmonic series diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to show that $\int_{-\alpha}^{\alpha}\arccos\left({x\over \alpha}\right)\ln(\alpha+x)\mathrm dx=\alpha \pi \ln\left({\alpha \over 2}\right)?$ Given that:
Where $\alpha >0$
$$\int_{-\alpha}^{\alpha}\arccos\left({x\over \alpha}\right)\ln(\alpha+x)\mathrm dx=\alpha \pi \ln\left({\alpha \over 2}\right)\tag1$$
Setting $y=\arccos(x/\alpha)\implies \cos(y)={x\over \alpha}$
$dx=-\alpha \sin(y)dy$
$$\alpha\int_{0}^{\pi}y\sin(y)\ln(\alpha+\alpha\cos y)\mathrm dy\tag2$$
$$\alpha \ln\alpha \int_{0}^{\pi}y\sin y\mathrm dy+\alpha\int_{0}^{\pi}y\sin y\ln(1+\cos y)\mathrm dy\tag3$$
$$\alpha \pi \ln\alpha +\alpha\int_{0}^{\pi}y\sin y\ln(1+\cos y)\mathrm dy\tag4$$
| $\displaystyle J=\int_{-\alpha}^{\alpha} \arccos\left(\dfrac{x}{\alpha}\right)\ln\left(\alpha+x\right)dx$
Perform the change of variable $y=\dfrac{x}{\alpha}$,
$\begin{align} J&=\alpha\int_{-1}^{1} \arccos\left(x\right)\ln\left(\alpha+\alpha x\right)dx\\
\end{align}$
Perform the change of variable $y=\arccos x$,
$\begin{align} J&=\alpha\int_{0}^{\pi} x\sin x\ln\left(\alpha+\alpha \cos x\right)dx\\
&=\alpha\int_{0}^{\pi} x\sin x\ln\left(1+ \cos x\right)dx+\alpha\ln \alpha\int_{0}^{\pi} x\sin x\,dx\\
&=\alpha\int_{0}^{\pi} x\sin x\ln\left(1+ \cos x\right)dx+\alpha\ln\alpha\Big[-x\cos x\Big]_{0}^{\pi}+\alpha\ln\alpha\int_{0}^{\pi}\cos x\,dx\\
&=\alpha\int_{0}^{\pi} x\sin x\ln\left(1+ \cos x\right)dx+\pi\alpha\ln\alpha+\alpha\ln\alpha\Big[\sin x\Big]_{0}^{\pi}\\
&=\alpha\int_{0}^{\pi} x\sin x\ln\left(1+ \cos x\right)dx+\pi\alpha\ln\alpha
\end{align}$
$\begin{align}
K&=\int_{0}^{\pi} x\sin x\ln\left(1+ \cos x\right)dx\\
&=\int_{0}^{\tfrac{\pi}{2}} x\sin x\ln\left(1+ \cos x\right)dx+\int_{\tfrac{\pi}{2}}^{\pi} x\sin x\ln\left(1+ \cos x\right)dx
\end{align}$
In the latter integral perform the change of variable $y=\pi-x$,
$\begin{align}K&=\int_{0}^{\tfrac{\pi}{2}} x\sin x\ln\left(1+ \cos x\right)dx+\int_{0}^{\tfrac{\pi}{2}}\left(\pi-x\right)\sin x\ln\left(1-\cos x\right)dx\\
&=\int_{0}^{\tfrac{\pi}{2}} x\sin x\ln\left(\dfrac{1+ \cos x}{1-\cos x}\right)\,dx+\pi\int_{0}^{\tfrac{\pi}{2}}\sin x\ln\left(1-\cos x\right)dx\\
&=\pi\int_{0}^{\tfrac{\pi}{2}}\sin x\ln\left(1-\cos x\right)dx-2\int_{0}^{\tfrac{\pi}{2}} x\sin x\ln\left(\tan\left(\dfrac{x}{2}\right)\right)\,dx\\
\end{align}$
In the first integral perform the change of variable $y=1-\cos x$,
$\begin{align}K&=\pi\int_{0}^1 \ln x dx-2\int_{0}^{\tfrac{\pi}{2}} x\sin x\ln\left(\tan\left(\dfrac{x}{2}\right)\right)\,dx\\
&=\pi\Big[x\ln x\Big]_0^1-\pi\int_0^1 1\,dx-2\int_{0}^{\tfrac{\pi}{2}} x\sin x\ln\left(\tan\left(\dfrac{x}{2}\right)\right)\,dx\\
&=-\pi-2\int_{0}^{\tfrac{\pi}{2}} x\sin x\ln\left(\tan\left(\dfrac{x}{2}\right)\right)\,dx
\end{align}$
In the latter integral perform the change of variable $y=\dfrac{x}{2}$,
$\begin{align}K&=-\pi-8\int_{0}^{\tfrac{\pi}{4}} x\sin(2x)\ln\left(\tan x\right)\,dx\\
&=-\pi+4\Big[x\cos(2x)\ln\left(\tan x\right)\Big]_{0}^{\tfrac{\pi}{4}}-4\int_{0}^{\tfrac{\pi}{4}} \cos(2x)\left(\ln\left(\tan x\right)+\dfrac{x(1+\tan^2 x)}{\tan x}\right)\,dx\\
&=-\pi-4\int_{0}^{\tfrac{\pi}{4}}\left(\dfrac{1-\tan^2 x}{1+\tan^2 x}\right)\left(\ln\left(\tan x\right)+\dfrac{x(1+\tan^2 x)}{\tan x}\right)\,dx\\
&=-\pi-4\int_{0}^{\tfrac{\pi}{4}}\left(\dfrac{1-\tan^2 x}{1+\tan^2 x}\right)\left(\ln\left(\tan x\right)\right)\,dx-4\int_{0}^{\tfrac{\pi}{4}}\dfrac{x}{\tan x}\,dx+4\int_{0}^{\tfrac{\pi}{4}}x\tan x\,dx\\
&=-\pi-4\int_{0}^{\tfrac{\pi}{4}}\ln\left(\tan x\right)dx+8\int_{0}^{\tfrac{\pi}{4}}\dfrac{\tan^2 x\ln\left(\tan x\right)}{1+\tan^2 x}dx-4\int_{0}^{\tfrac{\pi}{4}}\dfrac{x}{\tan x}dx+4\int_{0}^{\tfrac{\pi}{4}}x\tan x dx\\
&=-\pi-4\int_{0}^{\tfrac{\pi}{4}}\ln\left(\tan x\right)dx+\\
&4\left(\left[\left(x-\dfrac{\tan x}{1+\tan^2 x}\right)\ln(\tan x)\right]_{0}^{\tfrac{\pi}{4}}-\int_{0}^{\tfrac{\pi}{4}}\left(x-\dfrac{\tan x}{1+\tan^2 x}\right)\left(\dfrac{1+\tan^2}{\tan x}\right)dx\right)-\\
&4\int_{0}^{\tfrac{\pi}{4}}\dfrac{x}{\tan x}dx+4\int_{0}^{\tfrac{\pi}{4}}x\tan x dx\\
&=4G-8\int_{0}^{\tfrac{\pi}{4}}\dfrac{x}{\tan x}dx\\
&=4G-8\left(\Big[x\ln(\sin x)\Big]_{0}^{\tfrac{\pi}{4}}-\int_{0}^{\tfrac{\pi}{4}}\ln(\sin x)\,dx\right)\\
&=4G+\pi\ln 2+8\int_{0}^{\tfrac{\pi}{4}}\ln(\sin x)\,dx\\
&=4G+\pi\ln 2+4\left(\int_{0}^{\tfrac{\pi}{4}}\ln(\tan x)\,dx+\int_{0}^{\tfrac{\pi}{4}}\ln(\sin x\cos x)\,dx\right)\\
&=\pi\ln 2+4\int_{0}^{\tfrac{\pi}{4}}\ln(\sin x\cos x)\,dx\\
&=\pi\ln 2+4\int_{0}^{\tfrac{\pi}{4}}\ln\left(\dfrac{\sin(2x)}{2}\right)\,dx\\
&=\pi\ln 2-\pi\ln 2+4\int_{0}^{\tfrac{\pi}{4}}\ln\left(\sin(2x)\right)\,dx\\
&=4\int_{0}^{\tfrac{\pi}{4}}\ln\left(\sin(2x)\right)\,dx
\end{align}$
In the latter integral perform the change of variable $y=2x$,
$\begin{align}K&=2\int_{0}^{\tfrac{\pi}{2}}\ln\left(\sin x\right)\,dx\\
&=-\pi\ln 2
\end{align}$
Therefore,
$\boxed{\displaystyle J=\pi\alpha\ln\left(\dfrac{\alpha}{2}\right)}$
Nota Bene:
I have used the following results:
$\displaystyle \int_{0}^{\tfrac{\pi}{2}}\ln\left(\sin x\right)\,dx=-\dfrac{\pi}{2}\ln 2$
$\displaystyle G=-\int_{0}^{\tfrac{\pi}{4}}\ln\left(\tan x\right)\,dx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Limit of function calculations I must solve limit of next function:
$$\lim_{x\to \infty}\frac{2x^3+x-2}{3x^3-x^2-x+1}$$
Does my calculations are proper? If not where is my mistake?
$$=\lim_{x\to \infty}\frac{x^3\left(2+\frac{1}{x^2}-\frac{2}{x^3}\right)}{x^3\left(3-\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}\right)} \\
\ =\lim_{x\to \infty}\frac{x^3\left(2+0-0\right)}{x^3\left(3-0-0+0\right)} \\
\ =\frac{2}{3}$$
| Put $y=1/x $ and compute
$$\lim_ {y\to 0^+}\frac { 2+y^2-2y^3 }{3-y -y^2 +y^3}=2/3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
General Solution for a PDE I would like to know if someone could provide me the solutions for the next PDE:
\begin{align*}
f_{x}(x,y) - f_{y}(x,y) = \frac{y-x}{x^{2}+y^{2}}f(x,y)
\end{align*}
I empirically found the following particular solution given by $f(x,y) = \displaystyle\frac{x+y}{\sqrt{x^{2}+y^{2}}}$. Any contribution is appreciated. Thank you in advance.
| One can use method of characteristics $\gamma = \left(x(t),y(t)\right)$:
$$f_x + f_y = \dfrac{y-x}{x^2+y^2}\,f
$$
$$
x' = y'=1, \qquad \frac{df}{dt} = \dfrac{y-x}{x^2+y^2}\,f
$$
Solving system of ODEs above get
$$
x = t+a,\quad y = t+b = x + c, \implies c = y-x \quad - \quad \operatorname{const}\\
f' = \frac{f}{2t^2 + 2(a+b)t + a^2 + b^2 }
\\
\ln f = \int \frac{dt}{2t^2 + 2(a+b)t + a^2 + b^2 }
\implies f = \exp \int \frac{dt}{2t^2 + 2(a+b)t + a^2 + b^2 }
$$
Hope you can proceed from here with computing the integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
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