Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find all $k \in \mathbb{R}$ such that both $\frac{12}{k+1}$ and $\frac{6}{k-1}$ are positive integers
Find all $k \in \mathbb{R}$ such that both $\frac{12}{k+1}$ and $\frac{6}{k-1}$ are positive integers
Clearly $k$ is a rational number, so I let $k=\frac{a}{b}$ where $a, b \in \mathbb{Z}$ and $b \neq 0$, this gives $\frac{12b}{a+b}$ and $\frac{6b}{a-b}$ are positive integers. I tried different things here but couldn't make progress. I want to verify that $2, 3$ and $\frac{7}{5}$ are the only possible values for $k$. How to proceed from here? Thanks in advance.
| Starting from $\frac{6}{k-1}=n \in \mathbb N$, we can generate all possible rational $k=\frac{n+6}{n}$ without regard to the second condition.
Considering the second condition, $\frac{12}{k+1}=m \in \mathbb N$, we substitute to get $m=\frac{12}{\frac{n+6}{n}+1}=\frac{12n}{2n+6}=\frac{6n}{n+3}$. We need to find positive integers $n$ that yield positive integers $m$.
Note that $\frac{6n}{n+3}$ asymptotically approaches $6$, so we only have to look at values of $n$ such that $\frac{6n}{n+3} \le 5$, which by algebraic rearrangement yields $n \le 15$. We find $m\in \mathbb N \Rightarrow n= 3,6,15$, which in turn yields $k=3,2,\frac{7}{5}$, which is exactly what you found by trial and error.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$ I am trying to evaluate
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$
My solution is as follow
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n} \Rightarrow T = {e^{\mathop {\lim }\limits_{n \to \infty } n\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}} - 1} \right)}} = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{n}} \right)}}$$
$$T = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdots + \frac{1}{{{n^2}}}} \right)}} = {e^{\left( {0 + 0 + \cdots + 0} \right)}} = {e^0} = 1$$
The solution is correct but I presume my approach $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdot + \frac{1}{n}}}{n}} \right) \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdot + \frac{1}{{{n^2}}}} \right) = 0$ is wrong.
Is there any generalized method
| Consider the partial term
$$T_n=\left(1+\frac{H_n}{n^2}\right)^n\implies \log(T_n)=n \log\left(1+\frac{H_n}{n^2}\right)$$
For large $n$
$$H_n=\log (n)+\gamma +\frac{1}{2 n}-\frac{1}{12 n^2}+\frac{1}{120
n^4}+O\left(\frac{1}{n^6}\right)$$
$$\log\left(1+\frac{H_n}{n^2}\right)=\frac{\log (n)+\gamma }{n^2}+\frac{1}{2 n^3}+O\left(\frac{1}{n^4}\right)$$
$$\log(T_n)=\frac{\log (n)+\gamma }{n}+\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
$$T_n=e^{\log(T_n)}=1+\frac{\log (n)+\gamma }{n}+\frac{(\log (n)+\gamma )^2+1}{2
n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and also how it is approached.
Moreover, this give a shortcut method for the evaluation of $T_n$. Suppose $n=10$. The exact value is
$$T_{10}=\left(\frac{259381}{252000}\right)^{10} \sim 1.334677$$ while the above truncated series gives $\sim 1.334446$ (relative error $=0.017$%).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that the sum $2^0+2+2^2+\cdots+2^{p-1}(2^p-1)$ is $2^{p-1}(2^p-1)$ Attempt: In an attempt to find the sum of the series given that $2^{p-1}$ is prime,
$$1+2+2^2+\cdots+2^{p-2}+2^{p-1}+(2^p-1)+2(2^p-1)+2^2(2^p-1)+\cdots+2^{p-1}(2^p-1)$$
A teacher grouped elements as follows,
$$(1+2^p-1)+2(1+2^p-1)+2^2(1+2^p-1)+\cdots+2^{p-2}(1+2^p-1)+2^{p-1}$$
Question: I am confused a little about the last two terms in the summation above as what I think it should be is as follows,
$$(1+2^p-1)+2(1+2^p-1)+2^2(1+2^p-1)+\cdots+2^{p-2}(1+2^p-1)+2^{p-1}(1+2^p-1)$$
Edit:
Question: I should get the sum $2^{p−1}(2^p − 1)$, but with the way I wrote it I did not get the sum wanted, but I got $2^p(2^p-1)$ for,
$$(1+2^p-1)+2(1+2^p-1)+2^2(1+2^p-1)+\cdots+2^{p-2}(1+2^p-1)+2^{p-1}(1+2^p-1)$$
Edit 2: The only solution to this is that the sum is $2^p(2^p-1)$.
| Your grouping method is correct. Your teacher seems to have made a mistake.
From your sum you get
$$(1+2^p-1)+2(1+2^p-1)+2^2(1+2^p-1)+\cdots+2^{p-2}(1+2^p-1)+2^{p-1}(1+2^p-1)$$
which is equal to
$$(2^p)+2(2^p)+2^2(2^p)+\cdots+2^{p-2}(2^p)+2^{p-1}(2^p)=$$
$$2^p(2+2^2+...+2^{p-1})$$
Now, you only have to solve
$$2+2^2+...+2^{p-1}$$
and multiply by $2^p$ and get your answer. Is that clear?
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{x\to-1}\frac{2x^2+5x+3}{2-\sqrt{2+\sqrt{3-x}}}$
What is the value of the limit:
$$\lim_{x\to-1}\frac{2x^2+5x+3}{2-\sqrt{2+\sqrt{3-x}}}$$
$1)8\quad\quad\quad\quad\quad2)12\quad\quad\quad\quad\quad3)16\quad\quad\quad\quad\quad4)24$
I evaluated the limit by L'Hopital rule:
$$\lim_{x\to-1}\dfrac{4x+5}{\dfrac1{2\sqrt{3-x}\times2\sqrt{2+\sqrt{3-x}}}}=\dfrac{1}{\dfrac{1}{16}}=16$$
But I wonder is there other approach to evaluate the limit?
| HINT
What about multiplying by the conjugate twice?
\begin{align*}
\frac{1}{2 - \sqrt{2+\sqrt{3-x}}} = \frac{2 + \sqrt{2+\sqrt{3-x}}}{2 - \sqrt{3-x}}
\end{align*}
Similarly, we have
\begin{align*}
\frac{1}{2-\sqrt{3-x}} = \frac{2 + \sqrt{3-x}}{1 + x}
\end{align*}
Finally, notice that
\begin{align*}
2x^{2} + 5x + 3 & = (2x^{2} + 2x) + (3x + 3)\\\\
& = 2x(x+1) + 3(x+1)\\\\
& = (2x+3)(x+1)
\end{align*}
Can you take it from here?
| {
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Given the number $N=3^{12}-1$. Find how many even and odd divisors of $N$ exist, which divide $N$ but don't divide $3^k-1$, for $1\le k\le 11$ Given the number $N=3^{12}-1$. Find how many even and odd divisors of $N$ exist, which divide $N$ but don't divide $3^k-1$, for $1\le k\le 11$
Initially I tried working out how many odds and evens divide $N$, without the extra restriction.
$3^{12}-1=(3-1)(3^2+3+1)(3+1)(3^2-3+1)(3^6+1)=2*13*4*7*730$
$=2^4*5*7*13*73$
Hence there are $16$ odd divisors and $64$ even divisors.
I don't know how to finish off the question with the extra condition, without exmining all $80$ possibilities individually. Could you please explain to me how to solve this question?
Also a solution to the question is the following, despite the fact that I have tried to understand it, but it is too complex for me:
If we have an integer $m$, $3^{12}\equiv 1\mod{m}$, then the smallest integer for which $d<12$ for which it holds true that $3^d\equiv 1\mod{m}$, must divide $12$, since if this is not true, then we can write $12=dp+r$, with $0<r<d$ and then we have $3^r\equiv 1\mod{m}$, which is not true, from the hypothesis that $d$ is the smallest number for which $3^d\equiv 1\mod{m}$.
Hence to be sure that $m$ does not divide $3^k-1$, for $1\le k\le 11$, we examine all values of $k$ which divide $12$, in other words $k=1,2,3,4,6, 3^1-1=2, 3^2-1=2^3,3^3-1=2*13, 3^4-1=2^4*5 and 3^6-1=2^3*7*13$.
Odd divisors of $N$ divide $N$ but do not divide $3^k-1$, for $1\le k \le 11$ is: $16-5=11$.
Even divisors of $N$ which divide $N$, but do not divide $3^k-1$ for $1\le k\le 11$, are: $64-17=47$
| Let $d$ be a divisor of $3^{12}-1$. Note $d\neq3$. Call $d$ "bad" if $d$ divides some $3^k-1$ for $k\in\{1,2,\ldots,11\}$.
If $d$ divides $3^k-1$, then mod $d$: $3^k\equiv1\implies3^{12}\equiv3^{12-k}\implies1\equiv3^{12-k}$. So if $d$ is bad, then $d$ divides some $3^k-1$ for $k\in\{1,2,\ldots,6\}$.
If $d$ divides $3^k-1$, then mod $d$: $3^k\equiv1\implies3^{2k}\equiv1$. So if $d$ is bad, then $d$ divides some $3^k-1$ for $k\in\{4,6\}$.
So you only need examine divisors of $3^4-1=2^4(5)$ and $3^6-1=2^3(7)(13)$. From the odd divisors of $3^{12}-1$, exclude the subtree of $5$ (that is, $\{5,1\}$). And exclude the subtree of $(7)(13)$ (that is, $\{91,13,7,1\}$). That is five things to exclude, and leaves $16-5=11$ such odd divisors.
From the even divisors of $3^{12}-1$, exclude the supertree of $2$ within the divisor tree for $3^4-1$ (that is, $\{2^{\{1,2,3,4\}}5^{\{0,1\}}\}$). And exclude the supertree of $2$ within the divisor tree for $3^6-1$ (that is, $\{2^{\{1,2,3\}}7^{\{0,1\}}13^{\{0,1\}}\}$). That is 17 things to exclude, and leaves $64-17=47$ such even divisors.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int x^4\sqrt{x^2-3} \, dx$ Integrate $\int x^4\sqrt{x^2-3} \, dx$
I tried substituting $\sqrt{x^2-3}=t$ than by squaring both sides and by simplifying i got $x\,dx=t\,dt$ and $x^2=t^2+3$
Now after substituting to integral i have $\int (t^2+3)^2\cdot t\frac{t\,dt}{x}$, can't get rid of $x$ :(
Don't know how to move on, need a bit help if possible.
Thanks you in advance :)
| Utilize the recursive relation
$$\int x^n\sqrt{x^2-3}\ dx=I_n=\frac{x^{n-1}}{n+2}(x^2-3)^{3/2}+\frac{3(n-1)}{n+2}I_{n-2}
$$
to evaluate the integral systematically
\begin{align}
\int x^4\sqrt{x^2-3}\ dx=
\left(\frac16 x^3 +\frac38 x \right)(x^2-3)^{3/2}+\frac98I_0
\end{align}
where
$$I_0= \int \sqrt{x^2-3}\ dx=\frac x2\sqrt{x^2-3}-\frac32\tanh^{-1}\frac{\sqrt{x^2-3}}x
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4081654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find all natural solutions that satisfy $2^ + 3^ = ^2$ It looks like an easy question but I couldn't find a way to solve it. I found (0,1,2),(3,0,3),(4,2,5) by trial and error and I'm kinda sure they are the only answers but I'm not sure how to prove it.
| First consider the cases $x=0$ and $y=0$. If $x=0$, then $3^y=z^2-1=(z-1)(z+1)$ and hence $z-1$ and $z+1$ are two powers of $3$ with distance $2$; that immediately implies $z+1=3, z-1=1$ and so $z=2, y=1$. If $y=0$, then $2^x=z^2-1=(z-1)(z+1)$ and hence $z-1$ and $z+1$ are two powers of $2$ with distance $2$; that immediately implies $z+1=4, z-1=2$ and so $z=3, x=3$. We thus have the solutions $(0,1,2)$ and $(3,0,3)$.
Now let $x,y>0$. Considering the equation mod $3$, we see that the RHS is $\equiv 0,1$ (as it is a square) and the LHS is $\equiv 2^x$. As $2^x\equiv 0\pmod{3}$ is impossible, we must have $2^x\equiv 1\pmod{3}$, which means that $x$ is even. As a consequence, $x=2x'$ for some $x'>0$. Considering the equation mod $4$, we now see that the RHS is $\equiv 0,1$ and the LHS is $\equiv 3^y$. As $3^y\equiv 0\pmod{4}$ is impossible, we must have $3^y\equiv 1\pmod{4}$, which means that $y$ is even. As a consequence, $y=2y'$ for some $y'>0$.
We now have the equation
$$2^{2x'}+3^{2y'}=z^2\Leftrightarrow \left(2^{x'}\right)^2+\left(3^{y'}\right)^2=z^2$$
and so $\left(2^{x'}, 3^{y'}, z\right)$ is a Pythagorean triple; it is even a primitive Pythagorean triple. Such triples are of the form $\left(2uv, u^2-v^2, u^2+v^2\right)$ for coprime integers $u>v$ with distinct parity. We must then have $2uv=2^{x'}$ as both of them are the only even numbers in each triple; this means that $u$ and $v$ are both powers of $2$. However, they must also be coprime and have different parity, so the only option is $u=2, v=1$. This gives $x'=2$ and $u^2-v^2=3$, which is indeed a power of $3$, so $y'=1$. Hence, $x=4$ and $y=2$, which leads to $z=5$. Therefore, we have the additional solution $(4,2,5)$.
In conclusion, the solutions $(0,1,2), (3,0,3), (4,2,5)$ are the only ones.
| {
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Problem understanding integration by parts from the book An example from my book (from unrelated subject) has the following steps:
$$ -2 \int \sqrt{9-y^2} dy = - y\sqrt{9-y^2} + 9\sin^{-1}\frac{y}{3} \ + C$$
I am trying to understand the steps in between from my books example when they use the integral forumla (second image below).
I understand that this is integration by parts and they use the integral formula on the bottom. Apparently, they set the parts to $$ u(y) = \sqrt{9-y^2} \\ v(y) = y $$
And thats how the term $y\sqrt{9-y^2}$ is calculated but I dont understand how they do with the next term in the partial integration. My attempt:
$$ - \int^3_0 \left(\sqrt{9-y^2}\right)' y \ dy = - \int^3_0 \frac{1}{2\sqrt{9-y^2}} \cdot 2y \cdot y \ dy = \dots $$
The term that popped up is $2y$ which is the result of derivative of the inner function $9-y^2$. This result does not seem to be right. So what steps do they take here?
| As mentioned by @Deepak you can use substitution and find the answer that way.
Next it should be
$$-2\int \sqrt{9-y^2}dy=-y\sqrt{9-y^2}\color{red}{-}9\sin^{-1}\big(\frac{y}{3}\big)+C$$
Here is the integration by parts method
First note that $\frac{d}{dy}\big(\sqrt{9-y^2}\big)=\color{red}{-}\frac{y}{\sqrt{9-y^2}},$ then we have
$$\int \sqrt{9-y^2}dy=\int u(y)v'(y)dy=u(y)v(y)-\int u'(y)v(y)dy$$
$$=y\sqrt{9-y^2}\color{blue}{+}\int \frac{y^2}{\sqrt{9-y^2}}dy$$
$$=y\sqrt{9-y^2}\color{blue}-\int \frac{-y^2+\color{red}{9}-\color{red}{9}}{\sqrt{9-y^2}}dy$$
$$=y\sqrt{9-y^2}-\int\frac{9-y^2}{\sqrt{9-y^2}}+9\int\frac{1}{\sqrt{9-y^2}}dy$$
$$=y\sqrt{9-y^2}-\int\sqrt{9-y^2}dy+9\sin^{-1}\big(\frac{y}{3}\big)-C$$
and rearranging gives
$$2\int{\sqrt{9-y^2}}dy=y\sqrt{9-y^2}+9\sin^{-1}\big(\frac{y}{3}\big)-C$$
or $$-2\int{\sqrt{9-y^2}}dy=-y\sqrt{9-y^2}-9\sin^{-1}\big(\frac{y}{3}\big)+C$$
| {
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Different values of $A^n$ using Cayley-Hamilton Theorem And Direct Multiplication Let's say there's a matrix A
$$
A = \begin{pmatrix}
3 & -4\\
1 & -1\\
\end{pmatrix}
$$
Now I want to find $A^n$
I tried the following two methods but get different answers. Any solution ?
By Direct Multiplication:
$$
A^2= \begin{pmatrix}
5 & -8\\
2 & -3\\
\end{pmatrix}
$$
$$
A^3= \begin{pmatrix}
7 & -12\\
3 & -5\\
\end{pmatrix}
$$
$$
A^4= \begin{pmatrix}
9 & -16\\
4 & -7\\
\end{pmatrix}
$$
Observing the pattern
$$
A^n= \begin{pmatrix}
2n+1 & -4n\\
n & 1-2n\\
\end{pmatrix}
$$
Using Cayley-Hamilton Theorem:
The characteristic equation of the matrix A is
$$
(\lambda-1)^2=0
$$
Using Cayley-Hamilton Theorem:
$$
(A-I)^2=0
$$
$$\implies A=I$$
$$\implies A^n=I$$
Now clearly
$$
I \neq \begin{pmatrix}
2n+1 & -4n\\
n & 1-2n\\
\end{pmatrix}
$$
Which is correct ??
| $(A-I)^{2}=0$ does not imply that $A=I$.
[For example, $M=\left[\begin{array}{llll}0 & 1 \\ 0 & 0 \end{array}\right]$ is a matrix whose square is $0$. Bur $M$ itself is not $0$].
C-H Theorem gives $A^{2}=2A-I$. A simple induction argument gives $A^{n}=nA-(n-1)I$ for all $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\{a_n\}$ converges.
Suppose $a_1,a_2>0$ and
$a_{n+2}=2+\dfrac{1}{a_{n+1}^2}+\dfrac{1}{a_n^2}(n\ge 1)$. Prove
$\{a_n\}$ converges.
First, we may show $\{a_n\}$ is bounded for $n\ge 3$, since $$2 \le a_{n+2}\le 2+\frac{1}{2^2}+\frac{1}{2^2}=\frac{5}{2},~~~~~~ \forall n \ge 1.$$
But how to go on?
| Note that $2<a_n<3$ for $n\ge5$. Since, for $n\ge5$,
\begin{eqnarray}
&&|a_{n+3}-a_{n+2}|\\
&=&\bigg|\frac{1}{a_{n+2}^2}-\frac{1}{a_{n}^2}\bigg|=\frac{(a_{n+2}+a_{n})|a_{n+2}-a_{n}|}{a_{n+2}^2a_{n}^2}\\
&\le&\frac{6}{2^2\cdot 2^2}|a_{n+2}-a_{n}|\le\frac12|a_{n+2}-a_{n}|\\
&\le&\cdots\le\frac{1}{2^{n-2}}|a_5-a_3|
\end{eqnarray}
we conclude that $\{a_n\}$ is Cauchy and hence $\lim_{n\to\infty}a_n=L$ exists.
| {
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$ax^2 + 2hxy + by^2 +2gx + 2fy+c = 0$ represents a point when $\Delta = 0$ and $h^2 - ab < 0$. How can I show that $ax^2 + 2hxy + by^2 +2gx + 2fy+c = 0$ represents a point when $\Delta = 0$ and $h^2 - ab < 0$.
I know I can do it by completing the square. But it turns out to be too tedious. Can anyone help me ?
| We can write $$y=\frac{hx+f\pm \sqrt{(h^2-ab)x^2+2(hf-bg)x+(f^2-ac)}}{b}~~~~~(1)$$
For a pair of lines, $y$ needs to be real for all real values of $x$. inside square root we should have a perfect square of like $(px+q)^2>0$. For a point it should be like $-(rx+s)^2$ ($y$ non real excepting one value of $x=-s/r$. For this we need to have $A<0$ and $B^2-4AC=0$ for the quadratic in the radical. The two conditions give $h^2-ab<0, (hf-bg)^2-(h^2-ab)(f^2-ac)=0=\Delta.~~~~(2)$
The examples are: $x^2+y^2=0, 3(x-1)^2+(y-x+2)^2=0, 3(x-1)^2+2(y-2)^2=0.$
Note that if $x,y$ are real then sum of perfect squares cannot be zero. The only possibility is that both squares are simultaneously zero. the points are $(0,0),(1,-1),(1,2)$ are the points represented by the three examples, respectively. Upon expansion, for them we get $a,b,c,f,g,h$ satisfying both parts of (2).
| {
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Finding variance from a beta distribution Suppose $[X\mid \theta]\sim\operatorname{Uniform}([0, \theta])$ and $B$ is is a beta distributed random variable. Additionally, $\max(X_1,…,X_n)/θ\sim \operatorname{Beta}(n+1,1)$
Show that $\hat{θ}= \frac{n+2}{n+1} \max(X_1,…,X_n)$ and find the variance of $\hat\theta$ given $\theta$.
I found: $E[B]= \frac{n+1}{n+2}$ and $V[B]= \frac{n+1}{(n+2)^2(n+3)}$
How do I find the variance of $\hat{\theta}\mid \theta?$
| I will assume you meant $B\sim\operatorname{Beta}(n+1,1).$
Recall that $\operatorname{var}(B) = \operatorname E\big(B^2\big) - \big( \operatorname E(B) \big)^2$.
\begin{align}
& \operatorname E(B^2) = \int_0^1 x^2\cdot f_B(x)\,dx \\[8pt]
= {} & \int_0^1 x^2\cdot \frac{\Gamma(n+2)}{\Gamma(n+1)\Gamma(1)} \cdot x^n\,dx \\[8pt]
= {} & (n+1) \int_0^1 x^{n+2}\,dx = \frac{n+1}{n+3}.
\end{align}
So the variance is
$$
\frac{n+1}{n+3} - \left( \frac{n+1}{n+2} \right)^2 = \frac{(n+1)(n+2)^2 - (n+1)^2}{(n+3)(n+2)^2} = \frac{n+1}{(n+3)(n+2)^2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4102584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)$ The problem
Given that $a,b>0$ and $$2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)$$
Find the value of $$\log _{a b}\left(\frac{1}{a}+\frac{1}{b}\right)$$
My attempt
We have from the given condition
$$2+\frac{\log a}{\log 2}=3+\frac{\log b}{\log 3}=\frac{\log (a+b)}{\log 6}$$
$\implies$
$$\frac{2\log 2+\log a}{\log 2}=\frac{3 \log 3+\log b}{\log 3}=\frac{\log (a+b)}{\log 6}$$
By ratio and proportion we get each ratio equal to
$$\frac{2\log 2+3\log 3+\log(ab)-\log(a+b)}{0}$$
Thus we have
$$\log\left(\frac{1}{a}+\frac{1}{b}\right)=\log(108)$$
But i am unable to find $\log(ab)$
| $$2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)=k$$
$$2+\log _{2} a=k \implies a=2^{k-2}$$
$$3+\log _{3} b=k \implies b=3^{k-3}$$
So, now, we have
$$\frac{\log \left(2^{k-2}+3^{k-3}\right)}{\log (6)}=k\tag 1$$ and we need to compute
$$\log _{a b}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{\log \left(2^{2-k}+3^{3-k}\right)}{\log \left(2^{k-2} 3^{k-3}\right)}\tag 2$$ There is no closed form for equation $(1)$. Using graphics and numerical methods, we find $k=-1.18196$ and plugging in $(2)$ the result is $-0.688556$.
This strange value of $k$ effectively gives
$$\log\left(\frac{1}{a}+\frac{1}{b}\right)=\log(108)$$
Strange problem ! Typo's may be.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4103740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding the intersection of $x= y^2 + y - 2$ and $y = -x^2 - \frac32x + 1 $ Goodmorning, I am struggling in finding the points of intersection of the following parabolas:
$$\begin{align}
x &= y^2 + y - 2 \\[4pt]
y &= -x^2 - \frac32x + 1
\end{align}$$
I know that these two can be solved either algebraically or with matrices, but at the moment the algebraic solution is the one that I'm going for.
How can I solve this system without getting lost in complex and boring calculations? (I'm asking specifically if there are some tricks to solve this more quickly end elegantly)
| $$x = y^2 + y - 2 \tag 1$$
$$y = -x^2 - \frac32x + 1 \tag 2$$
Inspection :
$x=0$ in Eq.(2) gives $y=1$ and $y=1$ in Eq.(1) gives $x=0$ . Thus a first solution is
$$(x=0\:;\:y=1)$$
$y=0$ in Eq.(1) gives $x=-2$ and $x=-2$ in Eq.(2) gives $y=0$ . Thus a second solution is
$$(x=-2\:;\:y=0)$$
Full solving :
Put $y$ from Eq.(2) into Eq.(1) :
$$x = (-x^2 - \frac32x + 1)^2 + (-x^2 - \frac32x + 1) - 2 $$
$$x^4+3x^3-\frac34 x^2-\frac{11}{2}x=0$$
Knowing the two above roots it is easy to factor :
$$x(x+2)(4x^2+4x-11)=0$$
Solving $4x^2+4x-11)=0$ leads to the two other roots. for each one Eq.(1) gives $y$.
The third solution is :
$$x=-\frac12+\sqrt{3}\quad ; \quad y=-\frac32-\frac12\sqrt{3}$$
The fourth solution is :
$$x=-\frac12-\sqrt{3}\quad ; \quad y=-\frac32+\frac12\sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4104198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$ \int \frac{x-4}{\sqrt{x^2-4x+5}}\, dx$ I'm trying to solve this irrational integral $$ \int \frac{x-4}{\sqrt{x^2-4x+5}}\, dx$$
doing the substitution
$$ x= \frac{5-t^2}{2 \cdot (2+t)}$$
according to the rule.
So the integral becomes:
$$ \int \frac{1}{2} \cdot \frac{t^2+8t+11}{t^2+4t+5}\, dt =\frac{1}{2} \int 1+\frac{4t+7}{t^2+4t+5}\, dt= \frac{1}{2}t+2 \ln (t+2)+\frac{1}{2}\frac{1}{t+2} + cost$$ with $t=-x+ \sqrt{x^2-4x+5}$
The final result according to my book is instead $\sqrt{x^2-4x+5}-2 \ln( x-2+ \sqrt{(x-2)^2+1})$
I don't understant why this difference.
Can someone show me where I'm making mistakes?
| To check your results, here's an other approach
$$\int \frac{2x-4-4}{2\sqrt{(x-2)^2+1}}dx=$$
$$\sqrt{x^2-4x+5}-2\int \frac{dx}{\sqrt{(x-2)^2+1}}$$
put
$$x-2=\sinh(t)=\frac{e^t-e^{-t}}{2}$$
the last integrale becomes
$$\int dt=t+C$$
but
$$e^{2t}-2(x-2)e^t-1=0$$
gives
$$e^t=(x-2)+\sqrt{x^2-4x+5}$$
So, the final result is
$$\sqrt{x^2-4x+5}-$$
$$2\ln\Bigl((x-2)+\sqrt{x^2-4x+5}\Bigr)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$a^4+b^4+c^4+d^4=4abcd$ , prove that $a=b=c=d$.
If $a,b,c$ and $d$ are positive real numbers satisfying the expression: $$a^4+b^4+c^4+d^4=4abcd$$ then, prove that $a=b=c=d$.
Approach:
$$a^4+b^4+c^4+d^4=4abcd$$
$$a^4-2a^2b^2+b^4+c^4-2c^2d^2+d^4=4abcd-2a^2b^2-2c^2d^2$$
$$(a^2-b^2)^2 +(c^2-d^2)^2 = -2(ab-cd)^2$$
$$(a^2-b^2)^2 +(c^2-d^2)^2+2(ab-cd)^2=0$$
so then we get
$a^2=b^2;$ $c^2=d^2;$ and $ab=cd$.
Hence:
$a=b=c=d$.
I am looking for a method using $AM-GM$ inequality. Any help would be most appreciated!
| Using AM-GM with $a^4, b^4, c^4,d^4$, we get
$$
\frac{a^4 + b^4 + c^4 + d^4}{4} \ge \sqrt[4]{a^4 b^4 c^4 d^4}
$$
This means that
$$
a^4 + b^4 + c^4 + d^4 \ge 4abcd
$$
With equality if and only if $a^4 = b^4 = c^4 = d^4 \implies a = b = c =d$
Alternatively, we can just transform your proof into an AM-GM one:
We use AM-GM on $a^4$ and $b^4$, $c^4$ and $d^4$, $a^2b^2$ and $c^2d^2$.
$$
\frac{a^4 + b^4}{2} \ge \sqrt{a^4 b^4} \\
\frac{c^4 + d^4}{2} \ge \sqrt{c^4 d^4} \\
\frac{a^2 b^2 + c^2 d^2}{2} \ge \sqrt{a^2b^2c^2d^2}
$$
Summing up the first two and then using the third inequality we get,
$$
a^4 + b^4 + c^4 + d^4 \ge 2(a^2 b^2 + c^2 d^2) \ge 2 \cdot 2abcd \ge 4abcd
$$
With equality if and only if $a^4 = b^4$, $c^4 = d^4$ and $a^2 b^2 = c^2 d^2$ which means that $a = b = c =d$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4110836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$ \int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x $ I am trying to evaluate this antiderivative $$
\int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x
$$
What i have done:
$$
\begin{split}
I
&= \int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x \\
&= \int \frac{2+\sqrt{x+1}-\sqrt{3-x}}{4+x+1+4 \sqrt{x+1}-3+x} \cdot d x\\
&= \int \frac{2+\sqrt{x+1}-\sqrt{3-x}}{2+2 x+4 \sqrt{x+1}} \cdot d x\\&=\int \frac{1}{1+x+2 \sqrt{x+1}} \cdot d x\\
&\quad +\int \frac{\sqrt{x+1}}{2(1+x)+4 \sqrt{x+1}}\cdot d x \\
&\quad -\int\frac{\sqrt{3-x}}{2(1+x)+4 \sqrt{x+1}} \cdot d x
\end{split}
$$
From this last line i can evaluate the first two antiderivatives but the last one $$\int\frac{\sqrt{3-x}}{2(1+x)+4 \sqrt{x+1}} \cdot dx$$ seems a bit hard for me two evaluate, i don't find the good substitution. I am opened to any suggestion. Thanks in advance!
| hint
$$x+1=(x-1)+2$$
$$3-x=2-(x-1)$$
put
$$x-1=2\cos(2t)$$
to get
$$x+1=4\cos^2(t)$$
and
$$3-x=4\sin^2(t)$$
the integrale becomes
$$\int\frac{-2\sin(2t)dt}{1+\sin(t)+\cos(t)}$$
which can be computed using the substitution
$$u=\tan(\frac t2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4111706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to find $\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r}$? Problem was to find $$\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r}.$$ My partial progress i tried to motivate such that upper term in binomial terms gets constant rather than variable , so i thought of changing it to form $\binom{n}{n-r}\binom{n-r}{n-2r}$. Now it's of the form $\binom{n}{m}\binom{m}{k}$. We know it is equivalent to $\binom{n}{k}\binom{n-k}{m-k}$. Doing this we get $\binom{n}{2r}\binom{2r}{r}2^{n-2r}$. Now what I need to do as next step as such still nothing the upper or lower terms r any constant values(sum to ) , all are variables still. Now by one answer by Robpratt i got by snake oil , is there any other way to solve this problem ?
| Applying
$$\binom{n}{k}\binom{n-k}{m-k}=\binom{n}{m}\binom{m}{k}$$
with $m=2r$ and $k=r$, we have
$$\binom{n}{r}\binom{n-r}{r} = \binom{n}{2r}\binom{2r}{r}.$$
Now snake oil yields
\begin{align}
\sum_{n\ge 0}\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r} z^n
&=\sum_{n\ge 0}\sum_{r\ge 0} \binom{n}{2r}\binom{2r}{r} 2^{n-2r} z^n \\
&=\sum_{r\ge 0}\binom{2r}{r} 2^{-2r} \sum_{n\ge 2r} \binom{n}{2r} (2z)^n \\
&=\sum_{r\ge 0}\binom{2r}{r} 2^{-2r} \frac{(2z)^{2r}}{(1-2z)^{2r+1}} \\
&=\frac{1}{1-2z}\sum_{r\ge 0}\binom{2r}{r} \left[\left(\frac{z}{1-2z}\right)^2\right]^r \\
&=\frac{1}{1-2z}\cdot\frac{1}{\sqrt{1-4\left(\frac{z}{1-2z}\right)^2}} \\
&=\frac{1}{\sqrt{(1-2z)^2-4z^2}} \\
&=\frac{1}{\sqrt{1-4z}} \\
&=\sum_{n\ge 0} \binom{2n}{n}z^n.
\end{align}
So $$\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r} = \binom{2n}{n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Solve the equation $\frac{x^2-10x+15}{x^2-6x+15}=\frac{4x}{x^2-12x+15}$ Solve the equation $$\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{4x}{x^2-12x+15}.$$
First we have $$x^2-6x+15\ne0$$ which is true for every $x$ ($D_1=k^2-ac=9-15<0$) and $$x^2-12x+15\ne0\Rightarrow x\ne6\pm\sqrt{21}.$$ Now $$(x^2-10x+15)(x^2-12x+15)=4x(x^2-6x+15)\\x^4-12x^3+15x^2-10x^3+120x-150x+15x^2-180x+225=\\=4x^3-24x^2+60x$$ which is an equation I can't solve. I tried to simplify the LHS by $$\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{(x^2-6x+15)-4x}{x^2-6x+15}=1-\dfrac{4x}{x^2-6x+15}$$ but this isn't helpful at all. Any help would be appreciated! :) Thank you in advance!
| let $p=x^2+15,q=x$ then we have to solve $$\frac{p-10q}{p-6q}=\frac{4q}{p-12q}$$ $$\iff (p-10q)(p-12q)-4q(p-6q)=0$$ $$\iff (p - 8 q) (p - 18 q)=0$$ provided that $p\neq 6q,p\neq 12q$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Using polar coordintaes to evaluate $\int \int {\sqrt {\frac {1-x^2-y^2} {1+x^2+y^2}} }\ dx \ dy$ I want evaluate the following integral using polar coordinates.
$$ \int \int {\sqrt {\frac {1-x^2-y^2} {1+x^2+y^2}} }\ dx \ dy $$ over the positive quadrant of the circle $$x^2+y^2=1$$
I used the substitution $x=r\cos\theta$, $y=r\sin\theta$ and reduced the integral to
$$ \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} {\sqrt {\frac {1-r^2} {1+r^2}}}r drd\theta$$
Then, I made the substitution $u=r^2$, giving me, $2rdr=du$ and got
$$\int {\sqrt{\frac{1-u}{1+u}}}du$$
I then split the integrand into partial fractions $-1 + {\frac{2}{1+u}}$
How do I now proceed? Can I reduce it to the form $\int \sqrt{x^2-a^2}\ dx$?
| Substitute $u=1+r^2$ and $\mathrm du=2r\,\mathrm dr$, so that
$$\sqrt{\frac{1-r^2}{1+r^2}}r\,\mathrm dr=\sqrt{\frac{2-(1+r^2)}{1+r^2}}r\,\mathrm dr=\frac12\sqrt{\frac{2-u}u}\,\mathrm du$$
Then take $v=\sqrt u$ (so that $v^2=u$) and $2v\,\mathrm dv=\mathrm du$:
$$\frac12\sqrt{\frac{2-u}u}\,\mathrm du=\frac12\sqrt{\frac{2-v^2}{v^2}}(2v\,\mathrm dv)=\sqrt{2-v^2}\,\mathrm dv$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4121030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Use generating functions to calculate the sum $\sum_{k=0}^n (k+1)(n-k+1)$ Use generating functions to calculate the sum $\sum_{k=0}^n (k+1)(n-k+1)$
And Prove it by induction.
Attempt:
$$A(x)= \sum_{k=0}^n (k+1)(n-k+1)$$
$$\sum_{n=0}^\infty(\sum_{k=0}^n (k+1)(n-k+1))x^n$$
The series of differences is :$\{(n+1),2(n),3(n-1),4(n-2)…,1,…\}$
The generating functions of this series is : $(1-x)A(x)$
The series of differences is :$\{(n+1),(n-1),(n-3),(n-5),…,2-n,…\}$
The generating functions of this series is : $(1-x)^2A(x)$
The series of differences is :$\{(n+1),-2,-2,-2,…,-2,…\}$
The generating functions of this series is : $(1-x)A^3(x)$
The series of differences is :$\{(n+1),-(n+3),0,0,…,0,…\}$
The generating functions of this series is : $(1-x)^4A(x)$
On the other hand, the generating functions $(n+1)+(-n-3)x$,
Therefore: $(1-x)^4A(x)=(n+1)+(-n-3)x$
$$A( x) =\frac{( n+1) +( -n-3) x}{( 1-x)^{4}} =(( n+1) +( -n-3) x) \cdot \sum _{n=0}^{\infty }\binom{n+4-1}{4-1} x^{n}$$
$$\displaystyle \sum _{k=0}^{n}( k+1)( n-k+1) =\sum _{n=0}^{\infty }( n+1) \cdot \binom{n+4-1}{4-1} x^{n} +\sum _{n=0}^{\infty }( -n-3) \cdot \binom{n+4-1}{4-1} x^{n+1}$$
Prove in with induction :
\begin{aligned}
( n+1) \cdot \binom{n+3}{3} +( -n-3)\binom{n+2}{3} & =\frac{( n+3)( n+2)( n+1)^{2} n}{3!} +\frac{( n+2)( n+1)( n-1) n( -n-3)}{3!}\\
& =\frac{( n+3)( n+2)( n+1)^{2} -( n+2)( n+1)( n+3) n}{6}\\
& =\frac{( n+3)( n+2)( n+1)}{6}\\
& =\frac{1}{6}\left( n^{3} +6n^{2} +11n+6\right)
\end{aligned}
$$\displaystyle \sum _{k=0}^{n}( k+1)( n-k+1) =\frac{1}{6}\left( n^{3} +6n^{2} +11n+6\right)$$
induction basis: $n=0$
$$\displaystyle \sum _{k=0}^{n}( 0+1)( 0-0+1) =1=\frac{1}{6}\left( 0^{3} +6\cdotp 0^{2} +11\cdotp 0+6\right)$$
For $n+1$:
\begin{aligned}
\sum _{k=0}^{n+1}( k+1)( n-k+1) & =\sum _{k=0}^{n}( k+1)( n-k+1) +( n+2)(( n+1) -( n+1) +1)\\
& =\frac{1}{6}\left( n^{3} +6n^{2} +11n+6\right) +( n+2)\\
\end{aligned}
If anyone can enlighten me, and find my mistakes I appreciate that.
Unfortunately, I did not succeed.
| Note that
\begin{align}
\sum_{n \ge 0} \sum_{k=0}^n a_k a_{n-k} x^n
&=\sum_{k \ge 0} a_k x^k \sum_{n\ge k} a_{n-k} x^{n-k} \tag1\\
&=\sum_{k \ge 0} a_k x^k \sum_{n\ge 0} a_n x^n \tag2\\
&=\left(\sum_{n \ge 0} a_n x^n\right)^2
\end{align}
Equality $(1)$ arises from interchanging the order of summation.
Equality $(2)$ arises from the change of index $n\mapsto n+k$.
Taking $a_k = k+1$ yields
\begin{align}
\sum_{n \ge 0} \sum_{k=0}^n (k+1)(n-k+1) x^n
&=\left(\sum_{n \ge 0} (n+1) x^n\right)^2 \\
&=\left(\sum_{n \ge 0} \binom{n+1}{1} x^n\right)^2 \\
&=\left(\frac{1}{(1-x)^2}\right)^2 \\
&=\frac{1}{(1-x)^4} \\
&=\sum_{n\ge 0} \binom{n+3}{3} x^n \\
\end{align}
So
$$\sum_{k=0}^n (k+1)(n-k+1) = \binom{n+3}{3}$$
for $n \ge 0$.
Alternatively, you can prove the equivalent identity
$$\sum_{k=2}^{n+2} \binom{k-1}{1}\binom{n-k+3}{1}=\binom{n+3}{3}$$
combinatorially by counting $3$-subsets of $\{1,\dots,n+3\}$ in two different ways. The RHS is clear. For the LHS, condition on the middle element $k$. You then have $k-1$ choices for the smallest element and $n-k+3$ choices for the largest element.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4124684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
An immediate alternative to a trigonometry problem (high school) I have a right triangle, $AH\perp BC$, where $\cos \beta=\sqrt 5/5$ and $\overline{AH}+\overline{CH}+\overline{HB}=7$.
I have to found the area.
My steps (or solution):
$$\mathcal A(\triangle ABC)=\frac 12\overline{AB}\cdot \overline{BC} \sin \beta$$
with $0<\beta<\pi$. Hence $\sin \beta=\sqrt{1-5/25}=2\sqrt5/5$. But $\overline{AB}=\overline{BC}\cos \beta$ and the area
$$\mathcal A(\triangle ABC)=\frac12(\overline{BC}\cos \beta)\cdot (\overline{BC}\sin \beta)=\frac12\overline{BC}^2 \tag 1$$
We know that:
$$\overline{AH}+\overline{CH}+\overline{HB}=7, \quad \text{with}\quad \overline{CH}+\overline{HB}=\overline{BC}$$
Hence
$$\overline{AH}+\overline{BC}=7 \iff \overline{BC}=7-\overline{AH}$$
and $\overline{AH}=\overline{AB}\cos\gamma=\overline{BC}\cos\gamma\cos\beta$.
Putting this last identity to the condition $\overline{BC}=7-\overline{AH}$ I will have:
$$\overline{BC}=7-\overline{BC}\cos\gamma\cos\beta \iff \overline{BC}(1+\cos\gamma\cos\beta)=7$$
After,
$$\overline{BC}=\frac{7}{1+\cos\gamma\cos\beta}$$
with $\cos \gamma=\cos(\pi/2-\cos\beta)=\sin\beta=2\sqrt5/5$.
Definitively
$$\overline{BC}^2=\frac{49}{\left(1+\frac 25\right)^2}=\frac{49}{\frac{49}{25}}$$
and
$$\mathcal{A}(\triangle ABC)=\frac15\overline{BC}^2=5 \tag 2$$
I'm very tired and often don't find immediate solutions to problems. I had thought of using Euclid's second theorem by placing
$\overline{AH}=x$ and $\overline{BC}=y$, and $x+y=7$. After I will have $x^2=\overline{BH}\cdot \overline{CH}$. But I have left this path.
Can this possible alternative be used or is there another more immediate way?
| HINT
Let $a = \overline{AH}$, $b = \overline{CH}$ and $c = \overline{HB}$. Then we have that
\begin{align*}
\begin{cases}
a = \tan(\beta)c\\\\
a = \tan(\pi/2 - \beta)b
\end{cases} \Rightarrow a(1 + \tan(\beta) + \cot(\beta)) = 7
\end{align*}
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4127013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Finding the probability density function of the $n$th largest random variable. Let $X_1,...,X_{25}$ be independent Unif $[0,1]$ random variables. Let $Y$ be the $13$th largest of the $25$ random variables. Find the probability density function of $Y$.
I already know the answer for this one, but the answer that was provided just told me to use a formula to get the answer without much explanation as to why I should use that formula. This was the answer I was given:
Formula: $g(x_\gamma)=\frac{n!}{(\gamma-1)!(n-\gamma)!}[F(x)]^{\gamma-1}\cdot f(x)\cdot[1-F(x)]^{n-\gamma}$
Then since $X_1,X_2,\ldots,X_{25}\sim Unif[0,1]$, we have
$$
f(x)=
\begin{cases}
1 & \text{if } 0\leq x\leq 1 \\
0 & \text{otherwise}
\end{cases}
$$
and
$$
F(x)=\int^x_01~dx=
\begin{cases}
0 & \text{if } x<0\\
x & \text{if } 0\leq x\leq 1\\
1 &\text{if } x>1
\end{cases}
$$
Then $Y=X_{(13)}$ so
$$f(y)=\frac{25!}{(13-1)!(25-13)!}(x)^{13-1}\cdot1\cdot(1-x)^{25-13}$$
$$={^{25}C_{13}}13(x)^{12}(1-x)^{12}$$
Thus, the p.d.f. is
$$
g(Y)=
\begin{cases}
{^{25}C_{13}}13(x)^{12}(1-x)^{12} & \text{if } 0\leq x\leq 1\\
0 & \text{otherwise}
\end{cases}
$$
Could I get an explanation of why I should use this formula or if there is another way to get the answer that is more clear? Thank you.
| First note that the $13$-largest of $25$ is also the $13$-smallest, i.e. $13$th from the bottom:
$$
\begin{array}{rcccl}
& i & & 26-i \\
\hline
\text{smallest} \rightarrow & 1 & & 25 \\ \text{2nd-smallest} \rightarrow & 2 & & 24 \\ \text{3rd-smallest} \rightarrow & 3 & & 23 \\ & 4 & & 22 \\ & \vdots & & \vdots \\
& 11 & & 15 \\ & 12 & & 14 \\ \text{13th-smallest} \rightarrow & 13 & \longleftrightarrow & 13 & \leftarrow \text{13th-largest} \\ & 14 & & 12 \\ & 15 & & 11 \\
& \vdots & & \vdots \\ & 23 & & 3 & \leftarrow \text{3rd-largest} \\ & 24 & & 2 & \leftarrow \text{2nd-largest} \\ & 25 & & 1 & \leftarrow \text{largest}
\end{array}
$$
\begin{align}
F_Y(y) = {} & \Pr(Y\le y) \\[6pt]
= {} & \Pr(\text{13 or more observations are}\le y) \\[6pt]
= {} & \Pr\big({\ge 13 \text{ successes in 25 trials with}} \\
& \qquad\quad\text{probability $y$ of success on each trial} \big) \\[6pt]
= {} & \sum_{i=13}^{25} \binom {25} i y^i (1-y)^{25-i}. \\[6pt]
f_Y(y) = {} & \frac d {dy} F_Y(y) \\[6pt]
= {} & \sum_{i=13}^{25} \binom {25} i \Big( iy^{i-1} (1-y)^{25-i} - y^i(25-i)(1-y)^{25-i-1} \Big) \\[6pt]
= {} & \left( \sum_{j=12}^{24} \binom {25}{j+1} (j+1)y^j(1-y)^{25-j-1} \right. \\
& \quad \text{(where $j=i-1$ and $i= j+1$)} \\
& \left. {} - \sum_{i=13}^{25} \binom{25} i y^i(25-i)(1-y)^{25-i-1} \right) \\[10pt]
= {} & \sum_{j=12}^{24} \binom {25}{j+1} (j+1)y^j(1-y)^{25-j-1} \\
& {} - \sum_{i=13}^{25} \binom{25} {i+1} (i+1) y^i(1-y)^{25-i-1} \\[6pt]
& \qquad \text{(since $\displaystyle \binom{25} i (25-i) = \binom{25}{i+1}(i+1) $)}
\end{align}
And now the two sums cancel each other out except the term $j=12$ and the term $i=25.$ But the term $i=25$ is $0.$ Thus we have
$$
f_Y(y) = \binom{25}{13}13 y^{12} (1-y)^{12}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4128742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $A$ and $B$ are positive-definite matrices with $A>B$, must $A^k > B^k$ for $k>0$? Assume $A$ and $B$ are positive-definite matrices. If $A>B$, can we conclude that $A^k > B^k$ for any positive scalar $k$?
Note that $A>B$ means $A-B$ is a positive-definite matrix, not an element-wise comparison. I've tried using Cayley-Hamilton theorem but could not get anywhere with it so far.
| No. In $\mathbb{R}^{2 \times 2}$, take $B=I$ and
$$ A = \left( \begin{array}{cc} 2 & -2 \\ 2 & 2 \end{array} \right) $$
$A$ is positive-definite since
$$ (\begin{array}{cc} x & y \end{array})
\left( \begin{array}{cc} 2 & -2 \\ 2 & 2 \end{array} \right)
\left( \begin{array}{c} x \\ y \end{array} \right) = 2 x^2 + 2 y^2 $$
$A-B$ is positive-definite since
$$ (\begin{array}{cc} x & y \end{array})
\left( \begin{array}{cc} 1 & -2 \\ 2 & 1 \end{array} \right)
\left( \begin{array}{c} x \\ y \end{array} \right) = x^2 + y^2 $$
But
$$ A^2-B^2 = \left( \begin{array}{cc} 0 & -8 \\ 8 & 0 \end{array} \right) - I
= \left( \begin{array}{cc} -1 & -8 \\ 8 & -1 \end{array} \right) $$
and
$$ (\begin{array}{cc} 1 & 0 \end{array})
\left( \begin{array}{cc} -1 & -8 \\ 8 & -1 \end{array} \right)
\left( \begin{array}{c} 1 \\ 0 \end{array} \right) = -1 < 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$?
For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$?
This question comes from counting the number of digits of $10^n$ in terms of the number of digits of $2^n$ and $5^n$. Number of digits of $10^n$ is $n+1$ which equals to the sum of the number of digits of $2^n$ and $5^n$. I know that the number of digits of a positive integer $x$ is $\lfloor \log_{10}(x)\rfloor + 1$.
Using programming, I've checked that this is true for $n$ less than $100$.
| We can first do the steps that are obvious: rearranging. This would give us $\left \lfloor{\log_{10}(2^n)}\right \rfloor+\left \lfloor{\log_{10}(5^n)}\right \rfloor=n-1$. Using logrithm properties, we can simplify that to $\left \lfloor{n\log_{10}(2)}\right \rfloor+\left \lfloor{n\log_{10}(5)}\right \rfloor=n-1$.
Let's first consider something easier, without the floor function. That would bring up this equation: $n\log_{10}(2)+n\log_{10}(5)=n\log_{10}(2\cdot5)=n\log_{10}(10)=n$.
Let's get back to our original problem. We can first estimate the constant terms. $\log_{10}(2)\approx0.3010$ and $\log_{10}(5)\approx1-0.3010=0.6970$. Since both of the numbers are irrational, any positive integer $n$ multiplied by that would still result in an irrational number. Thus, both of the terms in the floor functions are not integers.
The floor function gets rid of the decimal part of a number, thus, it can subtract less than $2$ with two floor functions added. ($(0.999+0.999)-(\left \lfloor{0.999}\right \rfloor+\left \lfloor{0.999}\right \rfloor)<2$). Since they are both irrational, the floor function cannot subtract $0$ either. Thus, $$n-2<\left \lfloor{\log_{10}(2^n)}\right \rfloor+\left \lfloor{\log_{10}(5^n)}\right \rfloor<n\textrm{ and }n\in\mathbb{Z}^+$$
Therefore, we get that $\left \lfloor{\log_{10}(2^n)}\right \rfloor+\left \lfloor{\log_{10}(5^n)}\right \rfloor+2=n+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find definite integral as power series I need to find the following indefinite integral as power series and then compute the definite integral
$$\int_0^{1/2}\sqrt{1+x^3}dx$$
with the accuracy of $10^{-4}$.
My attempt:
Using the binomial theorem, we can express $\sqrt{1+x^3}$ as
$$\sum_{k=0}^\infty{\frac{1}{2}\choose k}x^{3k}.$$
Thus, we can rewrite the integral:
\begin{align}
\int_0^{1/2}\sqrt{1+x^3}dx&=\int_0^{1/2}\sum_{k=0}^\infty{\frac{1}{2}\choose k}x^{3k}\\
&=\sum_{k=0}^\infty{\frac{1}{2}\choose k}\left[\frac{x^{3k+1}}{3k+1}\right]_0^{1/2}\\
&=\sum_{k=0}^\infty{\frac{1}{2}\choose k}\frac{\left(\frac{1}{2}\right)^{3k+1}}{3k+1}
\end{align}
How can I approximate the definite integral from here?
| Use the well-known Taylor series for the function $\sqrt{1+x}$ and substitute $x\to x^3$ to get
$$\sqrt{1+x^3}=\sum^{\infty}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)} x^{3n}.$$
This series converges for $\lvert x\rvert <1$. Now integrating the series yields
$$\int_0^{1/2}\sum^{\infty}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)} x^{3n}\,\mathrm{d}x=\sum^{\infty}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)}\frac{2^{-3n - 1}}{3n + 1}.$$
Theorem (Alternating series approximation). Let $S=\sum\limits_{n=0}^\infty a_n$ be an alternating series where $\vert a_{n+1} \vert < \vert a_n \vert$ and $\displaystyle\lim_{n \rightarrow \infty}a_n=0$. Let $S_n$ be the $n$th partial sum of the series. Then $\vert S-S_n \vert \leq \vert a_{n+1} \vert$.
We want to find the biggest $a_{n+1}$ which is smaller than $10^{-4}$. This is the case for $n=2$, so
$$\lvert \sum^{\infty}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)}\frac{2^{-3n - 1}}{3n + 1}-\sum^{2}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)}\frac{2^{-3n - 1}}{3n + 1}\rvert \le \frac{1}{163840}<10^{-4}.$$
So the final answer is $$\int_0^{1/2}\sqrt{1+x^3}dx\approx \sum^{2}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)}\frac{2^{-3n - 1}}{3n + 1}= \frac{3639}{7168}$$
where the approximation error is at most $10^{-4}$. See here for numerical verification.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Local maximum uniqueness of the logarithm of the Student-t distribution The negative logarithm of the Student-t distribution partial density function is
$$f(\nu,x) := -\ln\Gamma\left(\frac{\nu+1}{2}\right)
+\ln\Gamma\left(\frac{\nu}{2}\right)
+\frac{1}{2}\ln(\pi\nu)
+\frac{\nu+1}{2}\ln\left(1+\frac{x^2}\nu\right)$$
How would one prove or disprove there is only one local minimum with respect to $\nu>0$ for any given $x$?
Numerical computation seems to suggest $f(\nu,x)$ strictly decreases with $\nu\in(0,\infty)$ for $x\in[0,1.5]$, and $f(\nu,x)$ is convex in $(0,a)$ and concave in $(a,\infty)$ for some $a>0$ for $x\in (b,\infty)$ for some $b\ge 1.5$.
To facilitate the solution, I post the first and second partial derivative of $f$ as follows.
\begin{align}2\frac{\partial f}{\partial \nu}=\frac{1-x^2}{\nu+x^2}+\ln\Big(1+\frac{x^2}\nu\Big)-\int_0^\infty \frac{e^{-\frac \nu2t}}{1+e^{-\frac t2}}\,dt,
\end{align}
$$4\frac{\partial^2 f}{\partial \nu^2}=-2\frac{\nu+x^4}{\nu(\nu+x^2)^2}+\int_0^\infty \frac{te^{-\frac \nu2t}}{1+e^{-\frac t2}}\,dt$$
| Just some thoughts (I will add proofs in the future, if possible.)
For convenience, we replace $x^2$ with $y$.
We have
$$
\frac{\partial f}{\partial v}
= \frac{1 - y}{2v + 2y} + \frac12 \ln \left(1 + \frac{y}{v}\right) - \frac12 \psi\left(\frac{v + 1}{2}\right) + \frac12\psi\left(\frac{v}{2}\right)
$$
where $\psi(\cdot)$ is the digamma function defined by $\psi(u) = \frac{\mathrm{d} \ln \Gamma(u)}{\mathrm{d} u} = \frac{\Gamma'(u)}{\Gamma(u)}$.
(i) If $0 < y \le 1$, then
\begin{align*}
\frac{\partial f}{\partial v}
&\le \frac{1 - y}{2v} + \frac12 \cdot \frac{y}{v}
- \frac12 \left(\ln \frac{v + 1}{2} - \frac{2}{v + 1}\right) + \frac12\left(\ln \frac{v}{2} - \frac{1}{v}\right)\\
&= \frac{1}{v + 1} - \frac12\ln\left(1 + \frac{1}{v}\right)\\
&< 0
\end{align*}
where we have used $\ln z - \frac{1}{z} \le \psi(z) \le \ln z - \frac{1}{2z}$ for all $z > 0$, and $\ln(1 + u) \le u$ for all $u \ge 0$.
Note: The last inequality is easy to prove by taking derivative.
(ii) If $1 < y \le 1 + \sqrt{2}$, then
\begin{align*}
\frac{\partial f}{\partial v}
&\le \frac{1 - (1 + \sqrt2)}{2v + 2(1 + \sqrt2)} + \frac12 \ln \left(1 + \frac{1 + \sqrt2}{v}\right)\\
&\quad - \frac12\left(\ln \frac{v + 1}{2} - \frac{1}{v + 1} - \frac{1}{3(v + 1)^2}\right)\\
&\quad + \frac12\left(\ln \frac{v}{2} - \frac{1}{v} - \frac{1}{12(v/2 + 1/14)^2}\right)\\
&< 0
\end{align*}
where we have used $\ln u - \frac{1}{2u} - \frac{1}{12u^2} < \psi(u) < \ln u - \frac{1}{2u} - \frac{1}{12(u + 1/14)^2}$ for all $u > 0$ (Theorem 5, [1]), and $y \mapsto \frac{1 - y}{2v + 2y} + \frac12 \ln \left(1 + \frac{y}{v}\right)$ is strictly increasing on $(1, \infty)$.
Note: The last inequality is easy to prove by taking derivative.
Remark: Where does $1 + \sqrt2$ come from? We rewrite $\frac{\partial f}{\partial v}$ as
$$\frac{\partial f}{\partial v}
= \frac{v \mathrm{e}^{A}}{2(v + y)}
\left\{\left(1 + \frac{y}{v}\right)\mathrm{e}^{-A}
\ln \left[\left(1 + \frac{y}{v}\right)\mathrm{e}^{-A}\right]
+ \left(1 + \frac{1}{v}\right)\mathrm{e}^{-A}\right\}$$
where
$A = 1 + \psi\left(\frac{v + 1}{2}\right) - \psi\left(\frac{v}{2}\right)$.
From $\frac{\partial f}{\partial v} = 0$, we solve
$y = {{\mathrm e}^{{\mathrm{LambertW}}\left(
- (1 + 1/v)\mathrm{e}^{-A}
\right) +A}}\,v - v$
where $W(\cdot)$ is the Lambert W function.
Maple tells us $\lim_{v\to \infty} \left( {{\mathrm e}^{{\mathrm{LambertW}}\left(
- (1 + 1/v)\mathrm{e}^{-A}
\right) +A}}\,v - v\right) = 1 + \sqrt{2}$.
By the way, @heropup also pointed out in comment that the point $x = \sqrt{1 + \sqrt{2}}$ is a demarcation point.
(iii) If $y > 1 + \sqrt{2}$ is fixed, we claim that $f$ has exactly one global minimizer $v^\ast$ on $(0, \infty)$, furthermore, $f$ is strictly decreasing on $(0, v^\ast)$ and strictly increasing on $(v^\ast, \infty)$.
We can prove the claim if the following conjecture is true:
Conjecture 1: $Q < 0$ for all $v > 0$, where
$$Q = - 4 - v(v + 2)^2 \int_0^\infty \frac{t^2\mathrm{e}^{-vt/2}}{1 + \mathrm{e}^{-t/2}}\mathrm{d} t
+ (6v^2 + 16v + 8)\int_0^\infty \frac{t\mathrm{e}^{-vt/2}}{1 + \mathrm{e}^{-t/2}}\mathrm{d} t.$$
Numerical evidence shows its truth. However, we have not yet proved it.
References
[1] L. Gordon, “A stochastic approach to the gamma function”, Amer. Math. Monthly, 9(101), 1994, 858-865.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4139399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Let P and Q be two points and let R be the point on PQ whose distance from P is twice its distance from Q.Verify that $w= \frac{1}{3}u + \frac{2}{3}v$ Let $P$ and $Q$ be two points in space and let $R$ be the point on $PQ$ whose distance from $P$ is twice its distance from $Q$. If $U=OP$, $v=OQ$ and $W=OR$. Verify that $w= \frac{1}{3}u + \frac{2}{3}v$
I tried the following:
*
*$P=(p_1,p_2)$
*$Q=(q_1,q_2)$
*$R=(r_1,r_2)$
*Im assuming O as the origin, but im not sure.
$w= \frac{1}{3}u + \frac{2}{3}v$
$\implies 3\sqrt{r_1^2 +r_2^2}=\sqrt{p_1^2 +p_2^2} +2\sqrt{q_1^2 +q_2^2}$
Also, I know that:
$||QR||+||RP||=||QP||$ and $||QP||=3||QR||$
So, from this equality I tried to build what should be verified. However, I don't come up with a result that works for me.
| If you want to use your initial Cartesian coordinate approach:
Let $\vec{u} = OP = (u_x, u_y)$, $\vec{v} = OQ = (v_x, v_y)$, and $\vec{w} = OR = (w_x, w_y)$.
Let's assume $\vec{w} = \frac{1}{3} \vec{u} + \frac{2}{3} \vec{v}$, i.e.
$$\left\lbrace \begin{aligned}
w_x &= \frac{1}{3} u_x + \frac{2}{3} v_x = \frac{u_x + 2 v_x}{3} \\
w_y &= \frac{1}{3} u_y + \frac{2}{3} v_y = \frac{u_y + 2 v_y}{3} \\
\end{aligned} \right.$$
The distance between $P$ and $R$ is
$$\begin{aligned}
\lVert R - P \rVert & = \lVert OR - OP \rVert = \lVert \vec{w} - \vec{u} \rVert \\
~ & = \sqrt{(w_x - u_x)^2 + (w_y - u_y)^2} \\
~ & = \sqrt{\left(\frac{u_x + 2 v_x - 3 u_x}{3}\right)^2 + \left(\frac{u_y + 2 v_y - 3 u_y}{3}\right)^2} \\
~ & = \sqrt{\left(\frac{2 v_x - 2 u_x}{3}\right)^2 + \left(\frac{2 v_y - 2 u_y}{3}\right)^2} \\
~ & = \sqrt{\frac{2^2}{3^2}\biggl( (v_x - u_x)^2 + (v_y - u_y)^2 \biggr)} \\
~ & = \frac{2}{3}\sqrt{(v_x - u_x)^2 + (v_y - u_y)^2} \\
~ & = \frac{2}{3}\sqrt{(u_x - v_x)^2 + (u_y - v_y)^2} \\
~ & = \frac{2}{3} \lVert \vec{v} - \vec{u} \rVert = \frac{2}{3} \lVert PQ \rVert \end{aligned}$$
Similarly, the distance between $Q$ and $R$ is
$$\begin{aligned}
\lVert R - Q \rVert & = \lVert OR - OQ \rVert = \lVert \vec{w} - \vec{v} \rVert \\
~ & = \sqrt{(w_x - v_x)^2 + (w_y - v_y)^2} \\
~ & = \sqrt{\left(\frac{u_x + 2 v_x - 3 v_x}{3}\right)^2 + \left(\frac{u_y + 2 v_y - 3 v_y}{3}\right)^2} \\
~ & = \sqrt{\left(\frac{u_x - v_x}{3}\right)^2 + \left(\frac{u_y - v_y}{3}\right)^2} \\
~ & = \sqrt{\frac{1}{3^2} \biggl( ( u_x - v_x )^2 + (u_y - v_y)^2 \biggr) } \\
~ & = \frac{1}{3}\sqrt{(u_x - v_x)^2 + (u_y - v_y)^2} \\
~ & = \frac{1}{3} \lVert \vec{v} - \vec{u} \rVert = \frac{1}{3} \lVert PQ \rVert \\
\end{aligned}$$
i.e. $\lVert R P \rVert = 2 \lVert R Q \rVert$.
(In case you wonder about the notation, I've used upper case letters $O$, $P$, $Q$, and $R$ for position vectors, and $\vec{u}$, $\vec{v}$, and $\vec{w}$ for displacement vectors. Furthermore, displacement vector $OP$ corresponds to the difference in their position vectors, $P - O$; thus $PQ = Q - P = (OQ) - (OP) = (Q - O) - (P - O) = Q - P$, and $O$ is actually irrelevant. However, in vector algebra, $O$ should refer to origin, unless explicitly defined otherwise.
I personally do not prefer to differentiate between position and displacement vectors – they're both vectors with the exact same rules – but the way the question is stated, I thought keeping the distinction might help understand the problem and the solution.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4142717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Showing $\tan70° = \tan20° + 2\tan50°$ Q. Prove that $\tan 70° = \tan 20° + 2\tan 50°$.
My approach:
$ LHS = \tan 70° = \dfrac1{\cot 70°} = \dfrac1{\tan 20°}$
$ \begin{align}
RHS
&= \tan 20° + 2\tan 50° \\
&= \tan 20° + 2\tan (20+30)° \\
&= \tan 20° + \dfrac{2(\tan 20° + 1/√3)}{1 - \tan 20°/√3} \\
&=\dfrac{2 + 3√3 \tan 20° - \tan^2 20°)}{√3 - \tan 20°}
\end{align}
$
Why are the two sides not equal despite being expressed in the same terms? Can someone offer some help?
Much to my surprise, my friend just expanded tan70° and cross-multiplied the terms to prove it.
| This will answer the question you asked in bold.
In fact, the two expressions you got involving $\tan 20^\circ$ are equal, not in the sense that algebraic manipulation will produce identical expressions, but rather depending upon a special fact regarding $\tan 20^\circ$. You want to show for $x=\tan 20^\circ$, $$\frac 1x = \frac{2+3\sqrt 3 x -x^2}{\sqrt 3-x}\\ \iff x^3-3\sqrt 3x^2 -3x +\sqrt 3 =0 $$
To do this, invoke the fact that $3\times 20 =60$ and the formula for $\tan 3x$: $$\sqrt 3 = \frac{3x -x^3}{1-3x^2} $$ and this rearranges to the same cubic equation above.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Finding distance from Local maximum of $f(x)=x+\sqrt{4x-x^2}$ to bisector of first quadrant
What is the distance of the local maximum of the function
$f(x)=x+\sqrt{4x-x^2}$ to bisector of first quadrant?
$1)1\qquad\qquad2)\sqrt2\qquad\qquad3)2\qquad\qquad3)2\sqrt2$
It is a question from timed exam so the fastest answers are the best.
My approach:
First I realized the function defined for $x\in[0,4]$. to find local maximum I put derivative equal to zero :
$$f'(x)=1+\frac{4-2x}{2\sqrt{4x-x^2}}=0$$
$$x-2=\sqrt{4x-x^2}$$
$$2x^2-8x+4=0\rightarrow x^2-4x+2=0\rightarrow x=2\pm\sqrt2$$
From here I noticed that the quadratic $-x^2+4x$ is symmetric along $x=2$ line so after plugging $x=2\pm\sqrt2$ in the $\sqrt{4x-x^2}$, hence $f(x)$ is maximum for the value $x=2+\sqrt2$ and $f(2+\sqrt2)=2+\sqrt2+\sqrt{2}=2+2\sqrt2$
Now we should find the distance from the point $(2+\sqrt2,2+2\sqrt2)$ from the line $y=x$. Now by recognizing that $(2+\sqrt2,2+\sqrt2)$ is a point on the line $y=x$, I guess the distance might be $1$ and because it is the least value in the choices then I conclude it is the answer.
Is it possible to solve this problem quickly with other approaches?
| This is an arguably faster way of getting to the $x$-coordinate of the local maximum. $$f(x)=2+(x-2)+\sqrt{4-(x-2)^2} $$ Let $x-2 = 2\sin t$: $$f(t)= 2+2(\sin t +\cos t)= 2+2\sqrt 2\sin\left( t+\frac{\pi}{4} \right) $$ For a local maximum, $t=\frac{\pi}{4}$ (e.g.) for which $x=2+\sqrt 2$. And then you can do the same thing from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4144247",
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"source": "stackexchange",
"question_score": "2",
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} |
Indefinite integration of $\int\frac{1}{\cos(x-1)\cos(x-2)\cos(x-3)}\,\textrm dx$
Integrate $$\int\dfrac{1}{\cos(x-1)\cos(x-2)\cos(x-3)}\,\textrm dx$$
My Attempt:
Using, $$\tan A-\tan B=\dfrac{\sin(A-B)}{\cos A\cdot \cos B}$$
The given integral can be transformed as
$$\int\dfrac{\tan(x-1)}{\cos(x-2)}\,\textrm dx - \int\dfrac{\tan(x-3)}{\cos(x-3)}\,\textrm dx$$
The right most integral can be calculated easily by writing $\tan(x-3)$ as $\frac{\sin(x-3)}{\cos(x-3)}$ and then by a substituiton $\cos(x-3)$ as $t$. But I have no clue for the left most integral. How to evaluate that?
| Little bit of trigonometry
$$\begin{align}
\frac{\tan(x-1)}{\cos(x-2)}
&= \frac{\tan(x-1)}{\cos((x-1) - 1)} \\
&= \frac{\tan(x-1)}{\cos(x-1)\cos 1 + \sin(x-1)\sin 1} \\
&= \frac{\tan(x-1)\sec(x-1)}{\cos 1 + \tan(x-1)\sin 1} \\
&= \frac{d\left(\sec(x-1)\right)}{\cos1 + \sin 1 \sqrt{\sec^2(x-1) - 1}} \\
&= \frac{du}{\cos 1 + \sin 1 \sqrt{u^2-1}}\end{align}$$
So, we are to calculate in a more general case
$$\begin{align}
\int \frac{du}{a+b\sqrt{u^2-1}} \end{align}$$
You can find a solution for this here and it gives
$$\frac{1}{2b}\ln\frac{|\sqrt{u^2-1}-1|}{|\sqrt{u^2-1}+1|} + \frac{a}{b\sqrt{a^2+b^2}}\left(\text{artanh}\frac{bu}{\sqrt{b^2+a^2}} - \text{artanh}\frac{\sqrt{b^2+a^2}\sqrt{u^2-1}}{au}\right) + C$$
Now substitute back $\sqrt{u^2 - 1} = \tan(x-1)$ and $a = \cos 1$, $b = \sin 1$ to get something prettier.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4148615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Proving $\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$ for positive reals $a$, $b$, $c$ Question $5$ of BMO1 $2008$:
For positive real numbers $\;a,\;b,\;c,\;$ prove that $$\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$$
I noticed that the right side can be grouped, but did not get further.
| Note that
$$
(a+b+c)(a+b-c)(a-b+c)(-a+b+c)=((a+b)^2-c^2)(c^2-(a-b)^2)=
\\
=(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)=(2ab)^2-(a^2+b^2-c^2)^2.
$$
Can you continue now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4150995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How can I expand on this part of this proof concerning magnetic fields? I am supposed to prove:
$$B_y=\frac{\mu_0}{4\pi}Iaz\int_{0}^{2\pi} \frac{\sin\phi}{(a^2+y^2+z^2-2ay\sin\phi)^{3/2}}d\phi=\frac{\mu_0Ia^2}{4r^3}\biggl(\frac{3yz}{r^2}\biggl)$$
and
$$B_z=\frac{\mu_0}{4\pi}Iaz\int_{0}^{2\pi} \frac{a-y\sin\phi}{(a^2+y^2+z^2-2ay\sin\phi)^{3/2}}d\phi=\frac{\mu_0Ia^2}{4r^3}\biggl(\frac{3z^2}{r^2}-1\biggl)$$
One of the steps to doing it is proving:
$$(a^2+y^2+z^2-2ay\sin\phi)^{3/2}\approx r^{-3}\biggl(1-\frac{2ay\sin\phi}{r^2}\biggl)^{-3/2}\approx r^{-3}\biggl(1+\frac{3ay\sin\phi}{r^2}\biggl)$$
using the Binomial theorem (the specific instruction reads "for this step use the binomial theorem, expand up to second term"). How can I go about this step, taking into account the following?
$$r=\sqrt{z^2+y^2}>>a$$
| First note that:
*
*$r = \sqrt{z^2+y^2} \gg a \implies z^2+y^2 \gg a^2 \implies a^2+y^2+z^2 \approx y^2+z^2$
*$r \gt y \;\land\; r \gg a \implies r^2 \gg ay \implies r^2 \gg ay \sin \phi$
Then:
$$
\require{cancel}
\begin{align}
(a^2+y^2+z^2-2ay\sin\phi)^{-3/2} &\approx (y^2+z^2-2ay\sin\phi)^{-3/2}
\\ &= \left(r^2 \left(1-\frac{2ay\sin\phi}{r^2}\right)\right)^{-3/2}
\\ &= \left(r^{2}\right)^{-3/2} \cdot \left(1- \left(-\frac{3}{\cancel{2}}\right) \cdot \frac{\cancel{2}ay\sin\phi}{r^2} + \dots\right)
\\ &\approx r^{-3}\biggl(1+\frac{3ay\sin\phi}{r^2}\biggl)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4153785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving $\frac{a}{b}+\left(\frac{3b+c}{3c+b}\right)^2+\left(\frac{2c+a}{2a+c}\right)^3\ge 3$ for positive $a$, $b$, $c$
Let $a,b,c$ be positive real number. Prove that
$$\dfrac{a}{b}+\left(\dfrac{3b+c}{3c+b}\right)^2+\left(\dfrac{2c+a}{2a+c}\right)^3\ge 3$$
It is not a symmetric inequality. I know that the equality occur when $a=b=c$ but It's hard to know how to approach it.
First, let $x=\dfrac{a}{b},y=\dfrac{b}{c},z=\dfrac{c}{a}$, then $xyz=1$ and then inequality becomes
$$x+\left(\dfrac{3y+1}{y+3}\right)^2+\left(\dfrac{2z+1}{z+2}\right)^3\ge 3 \tag1$$
My direction is use UCT method but
$$\left(\dfrac{3y+1}{y+3}\right)^2\ge y\quad\Leftrightarrow\quad y\le 1 \tag2$$ and similar to $$\left(\dfrac{2z+1}{z+2}\right)^3\ge y \tag3$$ and I don't have any information about $y,z$.
I wonder if it was the right way to solve it! Please give me some hint! Thanks!
| Using the AM-GM inequality,
$$\left(\frac{3b+c}{3c+b}\right)^2+1\ge2\left(\frac{3b+c}{3c+b}\right) $$
$$\left(\frac{2c+a}{2a+c}\right)^3+1+1\ge3\left(\frac{2c+a}{2a+c}\right)\;\;\;\;\;$$
$$\therefore\;\;\frac{a}{b}+\left(\frac{3b+c}{3c+b}\right)^2+\left(\frac{2c+a}{2a+c}\right)^3+3\ge \frac{a}{b}+2\left(\frac{3b+c}{3c+b}\right)+3\left(\frac{2c+a}{2a+c}\right)\;\;\;$$
It remains to prove:$$\frac{a}{b}+2\left(\frac{3b+c}{3c+b}\right)+3\left(\frac{2c+a}{2a+c}\right)\ge 6 \;\Longleftrightarrow \frac{a-b}{b}+4\left(\frac{b-c}{3c+b}\right)+3\left(\frac{c-a}{2a+c}\right)\ge0$$
Expanding the left hand side and completing the square leads to
$$6c(a-b)^2+2b(a-c)^2+3a(b-c)^2\ge0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Right Tangential Trapezoid with an incircle: Find the radius of the incircle and the area of the trapezoid The incircle of the trapezoid $ABCD$ that has a right angle $A$ with the center of $O$ is tangent with the point $E$ on $CD$ in a way that $CE=3$ and $ED=12$.
Find the angle of $COD$, the radius of the incircle and the area of the trapezoid.
| Suppose the $AD$ and $BC$ are the bases of the trapezoid $ABCD$:
According to the data given, $CE = 3$ and $DE = 12$
Suppose $\angle ADC = 2\alpha$. Since $\angle BAD = 90^\circ$, $\angle ADC + \angle DCB = 180^\circ = 2\alpha + \angle DCB$:
$$\therefore \, \angle DCB = 180^\circ - 2\alpha = 2(90^\circ - \alpha) \tag1$$
Since $AD$ and $DC$ are tangent to the inscribed circle, $DY = DE = 12$. Simillarly, considering tangents $CB$ and $CD$, $CE = CX = 3$.
Using the knowledge of geometry, we can prove that $\triangle DOY$ and $\triangle DOE$ are equivalent, therefore, $\angle ODY = \angle ODE = \frac{1}{2} \angle EDY = \alpha$. Simillarly, $\triangle COX$ and $\triangle COE$ are also equivalent, and therefore, $\angle OCX = \angle OCE = \frac{1}{2} \angle ECX = 90^\circ - \alpha$ (from the equation $(1)$)).
Now, consider the $\triangle COD$. We have proved that $\angle ODC = \alpha$ and $\angle OCD = 90^\circ - \alpha$.
$$\therefore \, \angle COD = 180^\circ - (\alpha + 90^\circ - \alpha) = 90^\circ \tag2$$
To find the radius $(r)$ of the incircle, we should consider right triangles, $\triangle ODE$ and $\triangle COE$. They are equal angled triangles:
$$\angle ODE = \angle COE; \quad \angle DOE = \angle ECE; \quad \angle OED = \angle OEC$$
$$\therefore \, \frac{OY}{CE} = \frac{DY}{OE} = \frac{OD}{OC} \ \Rightarrow \ \frac{r}{3} = \frac{12}{r} = \frac{OD}{OC} \tag3$$
From the equation $(3)$,
$$\frac{r}{3} = \frac{12}{r} \ \Rightarrow \ r^2 = 3 \times 12 = 36 \ \Rightarrow \ \therefore \ r = \sqrt{36} = 6$$
The area of the trapezoid $ABCD$:
$$\text{Area} = \frac{1}{2}(AD + BC)\cdot AB = \frac{1}{2}(r + 12 + r + 3)\cdot 2r = \frac{1}{2}(2r + 15)\cdot 2r = r(2r + 15) \\ = 6 \times (12 + 15) = 162$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4169310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the number of 5-digit number divisible by 6 which can be formed using $0,1,2,3,4,5$ if repetition of digits is not allowed.
Find the number of $5$-digit number divisible by $6$ which can be formed using $0,1,2,3,4,5$ if repetition of digits is not allowed.
I started by considering the following cases:
*
*Unit digit = $0$
I can fill four places with digits ${1,2,4,5}$ , so number of 5-digit numbers $= P(4,4) = 24$
*Unit digit = $2$
I can fill four places with digits ${0,1,3,4,5}$ such that (a) $0$ does not come in the first place (b) either of $1,4$ is used (c) $0,3,5$ are always used. I have to fill 4 places with $0,(1,4),3,5$. So, total number of arrangements $=P(4,4)$. Number of arrangements when $0$ comes as first digit $=P(3,3)$. Number of arrangements of $(1,4) = 2$. Therefore , number of 5-digit numbers $=2(P(4,4)-P(3,3)) = 36$
*Unit digit = $4$
Similarly, number of 5-digit numbers $= 36$
Therefore, the total number of 5-digit numbers $= 24+36+36 = 96$.
But the correct answer is $108$. Where did I make a mistake?
| Case $1$:
$1,2,3,4,5$ are chosen.
For the number to be divisible by $2$, last digit is $2$ or $4$, i.e. last digit can be selected in 2 ways and the other digits can arrange in $4!$. Thus, the number of ways =$4! \cdot 2=48$.
Case $2$:
$0,1,2,4,5$ are chosen.
a) if last digit is $0$: number of ways =$4!=24$
b) if last digit is $2$: the other digits may be filled in (from left to right): $3×3×2×1=18$ ways.
c) if last digit is $4$: the other digits may be filled in (from left to right): $3×3×2×1=18$ ways.
So, the answer $=48+24+18+18=108$ ways.
So answer is $108$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4170033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Proving $\sum_{k=1}^{n}\cos\frac{2\pi k}{n}=0$ I want to prove that the below equation can be held.
$$\sum_{ k=1 }^{ n } \cos\left(\frac{ 2 \pi k }{ n } \right) =0, \qquad n>1 $$
Firstly I tried to check the equation with small values of $n$
$$ \text{As } n=2 $$
$$ \cos\left(\frac{ 2 \pi \cdot 1 }{ 2 } \right) + \cos\left(\frac{ 2 \pi \cdot 2 }{ 2 } \right) $$
$$ = \cos\left(\pi\right) + \cos\left(2 \pi\right) $$
$$ = -1+ 1 =0 ~~ \leftarrow~~ \text{Obvious} $$
But
$$ \text{As}~~ n=3 $$
$$ \cos\left(\frac{ 2 \pi \cdot 1 }{ 3 } \right) +\cos\left(\frac{ 2 \pi \cdot 2 }{ 3 } \right) + \cos\left(\frac{ 2 \pi \cdot 3 }{ 3 } \right) $$
$$ = \cos\left(\frac{ 2 \pi }{ 3 } \right) + \cos\left(\frac{ 4 \pi }{ 3 } \right) + \cos\left( 2\pi \right) $$
$$ = \cos\left(\frac{ 2 \pi }{ 3 } \right) + \cos\left(\frac{ 4 \pi }{ 3 } \right) + 1 =?$$
What formula(s) or property(s) can be used to prove the equation?
| Your sum is the sum of the real parts of the $n$th roots of the unity. That is, they are the roots of the polynomial $z^n-1$. The sum of the roots is equal to the coefficient of $z^{n-1}$ by Vieta's formulas, hence it is zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4170548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given the distance measure for two points and the point $p = (0,2)$, which of the following points have the same distance as $p$ from the origin? Given the distance measure dist:
$$\mathrm{dist}(x,y) =
\sqrt{(x_1-y_1,x_2-y_2)\begin{pmatrix}
3 & 0\\
0 & 4
\end{pmatrix} (x_1-y_1,x_2-y_2)^{T} } $$
for two dimensional points and the point $p=(0,2)$, which of the following points have the same distance as $p$ from the origin $(0,0)$?
*
*$(4,0)$
*$\color{red}{(2,0)}$
*$(2,1)$
*$(\sqrt{6},0)$
*$(\sqrt{7},0)$
*$\left(\sqrt{\dfrac{16}{3}},0 \right)$
*$\color{red}{(0,-2)}$
The answers in red are my solutions, which I found through this work. However the correct solutions from the Prof. are $(2,1)$ , $\left(\sqrt{\frac{16}{3}}, 0\right)$, and $(0,-2)$.
I just want to know why answers $(2,1)$ and $\left(\sqrt{\frac{16}{3}}, 0\right)$ are correct, it makes no sense to me?
| For two points $x = (x_1,x_2)$ and $y =(y_1,y_2)$ we have,
$$\begin{split}
\mathrm{dist}(x,y) =&
\sqrt{(x_1-y_1,x_2-y_2)\begin{pmatrix}
3 & 0\\
0 & 4
\end{pmatrix} (x_1-y_1,x_2-y_2)^{T} } \\
=& \sqrt{(x_1-y_1,x_2-y_2)\begin{pmatrix}
3 & 0\\
0 & 4
\end{pmatrix} \begin{pmatrix}
x_1-y_1 \\
x_2-y_2
\end{pmatrix} } \\
=& \sqrt{(x_1-y_1,x_2-y_2) \begin{pmatrix}
3(x_1-y_1) \\
4(x_2-y_2)
\end{pmatrix} }\\
=& \sqrt{3(x_1-y_1)^2+4(x_2-y_2)^2}
\end{split}$$
Therefore if $y = (0,0)$ then,
$$\mathrm{dist}(x,(0,0)) = \sqrt{3x_1^2+4x_2^2}$$
Hence,
$$\boxed{\mathrm{dist}(p,(0,0)) =\sqrt{3\cdot0^2+4\cdot 2^2}=4}$$
$$\mathrm{dist}((4,0),(0,0)) =\sqrt{3\cdot4^2+4\cdot 0^2}=4\sqrt{3}$$
$$\mathrm{dist}((2,0),(0,0)) =\sqrt{3\cdot2^2+4\cdot 0^2}=2\sqrt{3}$$
$$\boxed{\mathrm{dist}((2,1),(0,0)) =\sqrt{3\cdot2^2+4\cdot 1^2}=4}$$
$$\mathrm{dist}((\sqrt{6},0),(0,0)) =\sqrt{3\cdot\sqrt{6}^2+4\cdot 0^2}=3\sqrt{2}$$
$$\mathrm{dist}((\sqrt{7},0),(0,0)) =\sqrt{3\cdot\sqrt{7}^2+4\cdot 0^2}=\sqrt{21}$$
$$\boxed{\mathrm{dist}\left(\left(\sqrt{\frac{16}{3}},0\right),(0,0)\right) =\sqrt{16+4\cdot 0^2}=4}$$
$$\boxed{\mathrm{dist}((0,-2),(0,0)) =\sqrt{3\cdot0^2+4\cdot (-2)^2}=4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4170807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $y' = (1+\frac{y-1}{2x})^2$ Solve $y' = (1+\frac{y-1}{2x})^2$
My first thought was to expand to see if I can get a linear form:
$$y' = 1 + \frac{y-1}{x} + (\frac{y-1}{2x})^2 = 1 + \frac{y-1}{x} + \frac{y^2-2y+1}{4x^2}$$
$$y' = 1 + \frac{y}{x} -\frac{1}{x} + \frac{y^2}{4x^2} + \frac{-2y}{4x^2} + \frac{1}{4x^2} \rightarrow y'+y(\frac{2}{4x^2}-\frac{1}{x}) = \frac{(2x-1)^2+y^2}{4x^2}$$
However, this is not even a bernoulli equation. How should I approach this problem?
| Hint
If you consider $$u=1+\dfrac{y-1}{2x}$$ then we get $$y'=2xu'+2u-2.$$ Thus the original equation becomes
$$2xu'+2u-2=u^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trigonometric Simplification Question I was working through a trigonometry problem, and was having some difficulty so I decided to look at the solution. Here are the steps:
$$\frac{\sin(2x+50^\circ)+\sin(150^\circ)}{\sin(2x+50^\circ)-\sin(150^\circ)}=\frac{\cos(50^\circ)-\cos(2x+50^\circ)}{\cos(50^\circ)+\cos(2x+50^\circ)}$$
$$\frac{\sin(2x+50^\circ)}{\sin150^\circ}=\frac{-\cos50^\circ}{\cos(2x+50^\circ)}$$
(Image that replaced text.)
I am not exactly how the solution got from the first step to the second one. I would just like some clarification on the intermediate step.
| This is similar to Siong's answer, but written differently using ratios:
\begin{align}
\frac{a+b}{a-b} - 1 &= \frac{c-d}{c+d} -1 \\[2ex]
\require{cancel} \frac{2b}{a-b} &= -\frac{2d}{c+d} \\[2ex]
\frac{a-b}{b} &= -\frac{c+d}{d} \\[2ex]
\frac{a}{b} - 1 &= -\frac{c}{d} -1 \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4176913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find value of $|(\overrightarrow{a} \times\overrightarrow{b}) \times\overrightarrow{c}|$
let $\overrightarrow{a} = 2 \hat {i}+ \hat {j}-2 \hat {k}$ and $\overrightarrow{b}= \hat {i} + \hat{j}$. $\:$ if $\overrightarrow{c}$ is a vector such that $\overrightarrow{a} \cdot \overrightarrow{c}+2|\overrightarrow{c}|=0$ and $|\overrightarrow{a}-\overrightarrow{c}|= \sqrt{14}$ and angle between $\overrightarrow{a} \times \overrightarrow{b}$ and $\overrightarrow{c}$ is $30 ^\circ$. Then value of $|(\overrightarrow{a} \times\overrightarrow{b}) \times\overrightarrow{c}|$ is ?
My first approach:
$|\overrightarrow{a}-\overrightarrow{c}|=\sqrt{|\overrightarrow{a}|^2+|\overrightarrow{c}|^2-2\overrightarrow{a} \cdot\overrightarrow{c}}= \sqrt{14}$
after substituting value of $\overrightarrow{a} \cdot \overrightarrow{c}=-2|\overrightarrow{c}|$ and $|\overrightarrow{a}|=3$ in above equation. I obtained $9+|\overrightarrow{c}|^2+4|\overrightarrow{c}|=14$
$\implies$ $|\overrightarrow{c}|^2+4|\overrightarrow{c}|-5=0$
$\implies$ $|\overrightarrow{c}|=1$
$\overrightarrow{a} \times \overrightarrow{b}=2 \hat{i}-2\hat{j}+ \hat{k}$
$\implies$ $|\overrightarrow{a} \times \overrightarrow{b}|=3$
So from above two result
$|(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}|= |\overrightarrow{a} \times \overrightarrow{b}||\overrightarrow{c}|sin(30^\circ)=(3)(1) \frac{1}{2}= \frac{3}{2}$.
My second Approach:
$|(\overrightarrow{a} \times\overrightarrow{b}) \times\overrightarrow{c}|=|(\overrightarrow{a} \cdot \overrightarrow{c})\overrightarrow{b}-(\overrightarrow{b} \cdot \overrightarrow{c})\overrightarrow{a}|$
To calculate $\overrightarrow{b} \cdot \overrightarrow{c}$ $\;$I took help of
$[\overrightarrow{a} \; \overrightarrow{b} \; \overrightarrow{c}]^2=\begin{vmatrix} \overrightarrow{a}\cdot \overrightarrow{a} & \overrightarrow{a}\cdot \overrightarrow{b} &\overrightarrow{a}\cdot \overrightarrow{c} \\ \overrightarrow{b}\cdot \overrightarrow{a} & \overrightarrow{b}\cdot \overrightarrow{b} &\overrightarrow{b}\cdot \overrightarrow{c} \\ \overrightarrow{c}\cdot \overrightarrow{a} & \overrightarrow{c}\cdot \overrightarrow{b} &\overrightarrow{c}\cdot \overrightarrow{c} \end{vmatrix}$
$\overrightarrow{a} \cdot \overrightarrow{a}=9$ & $\overrightarrow{b} \cdot \overrightarrow{b}=2$ $\overrightarrow{c} \cdot \overrightarrow{c}=1$ & $\overrightarrow{a} \cdot \overrightarrow{b}=3$ & $\overrightarrow{a} \cdot \overrightarrow{c}=-2$ and let $\overrightarrow{b} \cdot \overrightarrow{c}=x$
$[\overrightarrow{a} \; \overrightarrow{b} \; \overrightarrow{c}]=(\overrightarrow{a} \times \overrightarrow{b}) \cdot \overrightarrow{c} = |\overrightarrow{a} \times \overrightarrow{b}| |\overrightarrow{c}| cos(\theta)=\frac{3 \sqrt3}{2}$
here $\theta$ = angle between $\overrightarrow{a} \times \overrightarrow{b}$ and $\overrightarrow{c} $
$\frac {27}{4}=\begin{vmatrix} 9 & 3 & -2 \\ 3 & 2 & x \\ -2 & x & 1\end{vmatrix}$
After solving above determinant I obtained a equation which is $36 x^2+48x+23=0$
$\implies$ no real value of $x$ that is no real $\overrightarrow{b} \cdot \overrightarrow{c}$
Doubt: Why am I not getting answer using second method?
| The simple answer is that, the question is erroneous and $\overrightarrow{c}$ simply does not exist.
Note that from your calculations, $\overrightarrow{a}\cdot \overrightarrow{c}=-2|\overrightarrow{c}|$, and $|\overrightarrow{c}|=1$. Since $|\overrightarrow{a}|=3$, this means that if $\phi$ is the angle between these two vectors, then $\cos \phi=-\frac 23$. So, $\phi\approx 131°$.
Now let us demonstrate that this is impossible.
For this purpose, I shall shift to a different reference frame, as that'll help in calculations as well as intuition. Let us consider the frame where $\overrightarrow{a}=3\hat{i}$, with $\overrightarrow {b}$ lying on the $xz$ plane. In this frame, $\overrightarrow{a}×\overrightarrow{b}=3\hat{j}$. Since angle between $\overrightarrow {c}$ and $3\hat{j}$ would be $30°$ in this frame too, we can imagine $\overrightarrow {c}$ to form a cone of semi-vertical angle $30°$ with the $y$-axis, slant length $1$, and vertex at origin.
What would the maximum possible angle between this cone and the $x$- axis ($3\hat{i}$, specifically) be? It is easy to see that this angle cannot exceed $120°$. If you are having trouble gaining this intuition you can observe that the component of $\overrightarrow {c}$ along the $xz$ plane traces a circle of radius $\frac 12$, and it's vertical component is fixed at $1\cdot \cos(30°)=\frac {\sqrt 3}{2}$. Hence, in general, $\overrightarrow {c}$ in this frame is given by $$\overrightarrow {c}=\frac 12 \cos \theta\hat{i}+\frac {\sqrt 3}{2} \hat {j}+\frac 12\sin\theta\hat{k}$$
Thus, doing a dot product shows that angle cannot exceed $120°$. Since $\phi>120°$, the situation is impossible.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
How to solve this integral: $\int_0^\infty{\frac{ln x}{x^2+2x+4}}\cdot dx$ I have been trying to solve this integral:- $$\int_0^\infty{\frac{\ln x}{x^2+2x+4}}\cdot dx$$ But I have been unsuccessful so far. I have tried integration by parts ( taking $1/(x^2+2x+4)$ as the second function does not work because it ends of as tan-inverse function which is more compicated to integrate.)
taking $\ln x$ as the second function was interesting, I ended up in this dead end:-
$$\begin{align} I&=\int_0^\infty{\frac{\ln x}{x^2+2x+4}}\cdot dx \\ &= \left[ {\frac{x\ln x -x}{x^2+2x+4}}\right]_0^\infty + \int_0^\infty \frac{2(x+1)(x\ln x-x)}{(x^2+2x+4)^2}\cdot dx \\&= 2\int_0^\infty \frac{(x^2+2x+4-(x+4))(\ln x-1)}{(x^2+2x+4)^2}\cdot dx \\ &= 2\int_0^\infty\frac{\ln x}{x^2+2x+4}\cdot dx - 2\int_0^\infty\frac{1}{x^2+2x+4}\cdot dx -2\int_0^\infty \frac{(x+4)(\ln x-1)}{(x^2+2x+4)^2}\cdot dx \\&= 2I - \left[\frac{1}{\sqrt{3}} \tan^{-1}{\frac{x+1}{\sqrt{3}}} \right]_0^\infty - 2\int_0^\infty \frac{(x+4)(\ln x-1)}{(x^2+2x+4)^2}\cdot dx \end{align}$$
the function in the second step limits to $0$ so I didn't mention it in the third step. Now I have quartic polynomial at the bottom and don't know how to proceed. I have also tried trigonometric substitution ($ x=\tan\theta$)
If possible please try to end this process, I am a high school student so please keep that in mind if you show another method :)
| The Path
To evaluate
$$\newcommand{\Res}{\operatorname*{Res}}
\begin{align}
\int_0^\infty{\frac{\log(x)}{x^2+2x+4}}\,\mathrm{d}x\tag1
\end{align}
$$
we will evaluate
$$
\int_\gamma{\frac{\log(x)^2}{x^2+2x+4}}\,\mathrm{d}x\tag2
$$
where
$$
\begin{align}
\gamma
&=\left[\epsilon+i\epsilon^2,R+i\epsilon^2\right]\cup\sqrt{R^2+\epsilon^4}e^{i\left[\arctan\left(\frac{\epsilon^2}R\right),2\pi-\arctan\left(\frac{\epsilon^2}R\right)\right]}\\
&\,\cup\left[R-i\epsilon^2,\epsilon-i\epsilon^2\right]\cup\sqrt{\epsilon^2+\epsilon^4}e^{i[2\pi-\arctan(\epsilon),\arctan(\epsilon)]}
\end{align}\tag3
$$
as $\epsilon\to0$ and $R\to\infty$.
The Integral Along The Pieces
The integral along both curved pieces tends to $0$.
The integral along the upper line tends to
$$
\int_0^\infty\frac{\log(x)^2}{x^2+2x+4}\,\mathrm{d}x\tag4
$$
The integral along the lower line tends to
$$
-\int_0^\infty\frac{\log(x)^2+4\pi i\log(x)-4\pi^2}{x^2+2x+4}\,\mathrm{d}x\tag5
$$
Thus, the integral in $(2)$ is equal to the sum of these four parts:
$$
\int_0^\infty\frac{4\pi^2-4\pi i\log(x)}{x^2+2x+4}\,\mathrm{d}x\tag6
$$
The imaginary part of which is the integral in question.
The Residues
Since
$$
\frac1{x^2+2x+4}=\frac1{2i\sqrt3}\left(\frac1{x+1-i\sqrt3}-\frac1{x+1+i\sqrt3}\right)\tag7
$$
we have
$$
\Res_{z=-1+i\sqrt3}\left(\frac1{z^2+2z+4}\right)=\frac1{2i\sqrt3}\tag8
$$
and
$$
\Res_{z=-1-i\sqrt3}\left(\frac1{z^2+2z+4}\right)=-\frac1{2i\sqrt3}\tag9
$$
Thus, the integral in $(6)$ is the $2\pi i$ times the sum of the residues inside $\gamma$, which is
$$
\begin{align}
&\frac\pi{\sqrt3}\log\left(-1+i\sqrt3\right)^2-\frac\pi{\sqrt3}\log\left(-1-i\sqrt3\right)^2\\
&=\frac\pi{\sqrt3}\left(\log(2)+i\frac{2\pi}3\right)^2-\frac\pi{\sqrt3}\left(\log(2)+i\frac{4\pi}3\right)^2\\
&=\frac{4\pi^3-4\pi^2i\log(2)}{3\sqrt{3}}\tag{10}
\end{align}
$$
The Results
$(10)$ and $(6)$ say that
$$
\int_0^\infty\frac1{x^2+2x+4}\,\mathrm{d}x=\frac\pi{3\sqrt3}\tag{11}
$$
$$
\int_0^\infty\frac{\log(x)}{x^2+2x+4}\,\mathrm{d}x=\frac{\pi\log(2)}{3\sqrt3}\tag{12}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
"answer_count": 3,
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} |
A series representation for $e^x$ I want to show that for $x \notin \{2,3,4,\dots\}$ $$e^x = \frac{2+x}{2-x} + \sum_{k=2}^{\infty} \frac{-x^{k+1}}{k!(k-x)(k+1-x)}$$
It is pretty clear that why $x \notin \{2,3,\dots\}$. I proved that the series converges for all other $x$ by using the ratio test. My initial thoughts were to take
$$ f(x) = \frac{2+x}{2-x} + \sum_{k=2}^{\infty} \frac{-x^{k+1}}{k!(k-x)(k+1-x)}$$ and find $f'(x)$ by doing term by term by differentiation and since we know that $f(0) = 1$ we may be able to find $f(x)$. I suspect that we can not do this without proving that term by term differentiation is valid. Neverthless I tried to do that also but got stuck here.
$$ f'(x) = \frac{4}{{\left(x - 2\right)}^{2}} - \sum_{k=2}^{\infty}\left( \frac{{\left(k + 1\right)} x^{k}}{{\left(k - x + 1\right)} {\left(k - x\right)} k!} + \frac{x^{k + 1}}{{\left(k - x + 1\right)} {\left(k - x\right)}^{2} k!} + \frac{x^{k + 1}}{{\left(k - x + 1\right)}^{2} {\left(k - x\right)} k!} \right)$$
How can I show that this term by term differentiation is valid and how can I proceed further?
Can someone suggest some other ways to do it?
|
Can someone suggest some other ways to do it?
Yes, one can prove the identity directly, without computing the derivative. We start with a partial fraction decomposition:
$$
\sum_{k=2}^{\infty} \frac{-x^{k+1}}{k!(k-x)(k+1-x)}
= \sum_{k=2}^{\infty}\frac{-x^{k+1}}{k!} \left( \frac{1}{k-x} - \frac{1}{k+1-x}\right) \, .
$$
The right-hand side is a convergent series of the form $\sum_{n=2}^\infty (a_n +b_n)$ with $a_n \to 0$, therefore it is allowed to re-group the terms to $a_2 + \sum_{n=2}^\infty (a_{n+1}+b_n)$. This gives
$$
\frac{-x^3}{2!}\frac{1}{2-x} + \sum_{k=2}^{\infty} \left( - \frac{x^{k+2}}{(k+1)!(k+1-x)}+\frac{x^{k+1}}{k!(k+1-x)} \right) \,.
$$
(The idea is to combine the terms with the same factor $(k+1-x)$ in the denominators.) And now the magic happens: the terms in the new sum simplify significantly, and we get
$$
\frac{-x^3}{2!}\frac{1}{2-x} + \sum_{k=2}^{\infty} \frac{x^{k+1}}{(k+1)!}
= \frac{-x^3}{2!}\frac{1}{2-x} + \left( e^x - 1 - x - \frac{x^2}{2} \right)
= e^x - \frac{2+x}{2-x} \, .
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
"answer_count": 1,
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} |
Prove that $(n+1)a\leq a^{n+1}+n, \forall a,n\in\mathbb{N}$. We can start from the fact that:
\begin{align*}
0\leq a^n + a^{n-1} +\ldots + a^2 + a-n,\forall a,n\in\mathbb{N}.
\end{align*}
The above is true, since if $a>1$, then $a^n>n, \forall n \in\mathbb{N}$. Also if $a = 1$, then we will have $1 + 1 + \ldots + 1 + 1-n = n-n = 0, \forall n \in \mathbb{N}$. Thus we have:
\begin{align*}
0 &\leq (a^{n} + a^{n-1} + \ldots + a^2 + a-n)(a-1)\\
0 &\leq (a^{n} + a^{n-1} + \ldots + a^2 + a)(a-1)-n(a-1)\\
0 &\leq a(a^{n-1} + a^{n-2} + \ldots + a + 1)(a-1) -na + n\\
0 &\leq a(a^{n}-1) -na + n\\
0 &\leq a^{n + 1} -a-na + n\\
0 &\leq a^{n + 1} -(n + 1) a + n\\
(n + 1)a &\leq a^{n + 1} + n\\
\end{align*}
Another form of reason is using the inequality of the arithmetic mean with the geometric mean, as follows:
\begin{align*}
\frac{a^{n + 1} + n}{n + 1} &= \frac{a^{n + 1} + 1 + 1 + \ldots + 1 + 1}{n + 1}\\
&\geq \sqrt[n + 1]{a^{n + 1}(1)(1)\ldots(1)(1)}\\
&= \sqrt[n + 1]{a^{n + 1}}\\
& = a.
\end{align*}
Finally it is solved and we have $a^{n + 1} + n \geq (n + 1)a$.
I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.
| Define $f(x)=x^{n+1}-(n+1)x+n$. Note that $f(1)=0$ and $f'(x)=(n+1)(x^n-1)$. Therefore, $f'(x)$ takes negative values in $(0,1)$ and positive values in $(1,\infty)$, as desired. So $f(x)$ takes its minimum value (which is zero) at $x=1$.
| {
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} |
How to represent Hadamard gate as product of Rx and Ry gates? As simple as question asks - how to represent Hadamard gate as product of Rx and Ry gates assuming that values for rotation of Rx and Ry gate can be different?
| We have
$$R_x(\pi)R_y(\pi/2)=\left(
\begin{array}{cc}
0 & -i \\
-i & 0 \\
\end{array}
\right)\left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{array}
\right)$$
$$=\left(
\begin{array}{cc}
-\frac{i}{\sqrt{2}} & -\frac{i}{\sqrt{2}} \\
-\frac{i}{\sqrt{2}} & \frac{i}{\sqrt{2}} \\
\end{array}
\right)=-i\left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
\end{array}
\right)=-iH$$
This suffices for the purposes of quantum computing since the overall phase of $\frac{3\pi}{2}$ (represented by $-i$) doesn't effect the final measurement output.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is $y^2+f(y)b+c$ a quadratic equation? The solution to the question:
Let $x, y, z \in R$ such that $x+y+z=6$ and $x y+y z+z x=7$. Then find the range of values of $x, y$, and $z$.
given in book is as follows:
$x, y, z \in R$
$x+y+z=6$ and $x y+y z+z x=7$
$\Rightarrow y(6-y-z)+y z+z(6-y-z)=7$
$\Rightarrow \quad-y^{2}+(6-z+z-z) y+z(6-z)-7=0$
$\Rightarrow \quad y^{2}+(z-6) y+7+z(z-6)=0$
.
Now, $y$ is real. Therefore,
$(z-6)^{2}-4[7+z(z-6)] \geq 0$...(1)
or $3 z^{2}-12 z-8 \leq 0$
or $\frac{12-\sqrt{144+96}}{6} \leq z \leq \frac{12+\sqrt{144+96}}{6}$
or $\frac{6-2 \sqrt{15}}{3} \leq z \leq \frac{6+2 \sqrt{15}}{3}$.
From symmetry, $x$ and $y$ have same range.
But I doubt if this is correct; since $y$ is function of $z$, hence the author cannot solve the question this way. Let me explain my point clearly.
Quadratic equations are defined as those which can be expressed in the form $ax^2+bx+c=0$, where $a \not=0$ and $a, b, c$ are constants. But clearly in this case $z$ is not constant, it is function of $y$, hence $(1)$ is not a quadratic equation so we cant the do the steps after $(1)$ hence the solution is wrong.
So the question is, is $y^2+f(y)b+c$ a quadratic equation?
| Just go back and recall the derivation of the quadratic formula, it never uses the fact that $a$, $b$ and $c$ are constant, you can use it any time. Just as an example, consider:
$$\begin{align*}
2x+4&= 20\\
4+2x-20&= 0\\
(1)\color{blue}{2}^2+(x)\color{blue}{2}+(-20)&= 0\\
\color{blue}{2}&= \dfrac{-x\pm\sqrt{x^2-4(1)(-20)}}{2(1)}\\
4+x&=\pm\sqrt{x^2+80}\\
16+8x+x^2&=x^2+80\\
x&= 8
\end{align*}$$
Here, the coefficient was itself a variable and the variable was a constant, but the quadratic formula, as you see works. Note the coefficient of variable squared must be non-zero (here, $1\neq 0$).
$x^2+f(x)b+c=0$ may or may not be a quadratic equation, but you can always use the quadratic formula!
Hope this helps. Ask anything if not clear :)
| {
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"url": "https://math.stackexchange.com/questions/4197443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Prove that $\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}$.
Let $a,b,c$ be positive real numbers. Prove that $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}.$$
I want to prove the inequality with elementary inequalities ( Mean inequalities, Cauchy-Schwarz etc.). I tried to use Cauchy-Schwarz, but I got $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\frac{9}{2(a^2+b^2+c^2)+ab+bc+ca}.$$
Then we have to show that $$2(a^2+b^2+c^2)+ab+bc+ca\leq (a+b+c)^2 $$
which is not true.
Multiplying the numerators of the fractions in LHS by $a^2,b^2,c^2$ also makes the problem more difficult.
So, how to solve the problem with elementary inequalities?
| Here is a proof following Momo answer.
$$\dfrac{1}{a^2+b^2+ab}+\dfrac{1}{b^2+c^2+bc}+\dfrac{1}{a^2+c^2+ac}\ge \dfrac{10.5}{a^2+b^2+c^2+2.5(ab+bc+ca)}$$ where all variables are nonnegative or positive. This is the inequality with $r=2.5$ also if the inequality in Momo answer is true for $r\ge 0$ it is also true for every $r_0\in[0,r]$ (not hard to prove), and $r=2.5$ is the maximum possible trying $a=b=1$ and $c\to 0$.
Since the inequality is quite symmetric we can multiply out denominators and with Muirhead inequality the case $r=2$ is easy. For $r=2.5$ we obtain the numerator $$7(b^5a+a^5b+c^5b+b^5c+c^5a+a^5c-(b^4a^2+a^4b^2+c^4b^2+b^4c^2+c^4a^2+a^4c^2))+N $$ where $N$ can be written as $$(a^3-c^3)(a-c)(a^2+c^2+ca-b^2)+(b^3-c^3)(b-c)(b^2+c^2+cb-a^2)+(a^3-b^3)(a-b)(a^2+b^2+ba-c^2).$$ Assuming $b\ge a\ge c$ we see that the only possible negative term is $ a^2+c^2+ca-b^2$ but $(a^2+c^2+ca-b^2)+(b^2+c^2+bc-a^2)\ge 0$ and this completes the proof.
| {
"language": "en",
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} |
Order on three real numbers given they are roots of a cubic, their sum and sum of pairwise product
Let $a,b,c$ be three real numbers which are roots of a cubic polynomial and satisfy $a+b+c=6$ and $ab+bc+ca=9.$ Suppose $a<b<c.$ Show that $$0<a<1 <b<3<c<4$$
Source: ISI Bmath Entrance 18-Jul-2021
Let the polynomial be $p(x) = (x-a)(x-b)(x-c)= x^3 - x^2 (6) + 9x - abc$, now clearly $p(a)=p(b)=p(c)$, adding the three:
$$p(a)+p(b)+p(c)=0$$
$$ (a^3 + b^3 + c^3)- 6(a^2 + b^2 +c^2) + 9 (a+b+c) - abc=0$$
Using identites we find the expression above becomes:
$$a^3 + b^3+c^3 - 6(6^2- 2 \cdot 9) + 9 \cdot 6 -abc=0$$
or
$$ a^3 + b^3 + c^3 -54 -abc=0$$
To be frank, I didn't do the manipulatons with any plan in mind, under desperation in the exam hall, I took the above to be a cubic in $a$ and applied Vieta assuming roots to be $\{a_1,a_2,a_3 \}$, where I found:
$$a_1 + a_2 + a_3 = 0$$
$$a_1 a_2 + a_2 a_3 + a_1 a_3 =0$$
$$a_1 a_2 a_3 = b^3 + c^3 - 54 - abc$$
We can write similar relation by considering it as a cubic in $b$ and $c$... but what after this..?
| Solution using the derivative:
$P(x)$ is a polynomial, hence its roots are separated by the roots of the first derivative $P'(x).$ $$P'(x)=3x^2-12x+9=3(x-1)(x-3),$$ thus $$a<1<b<3<c.$$
To complete the proof, we have to show that $a>0$ and $c<4.$
*
*Let us prove $a>0$ by contradiction.
*
*Case $a=0:$
Here, the polynomial is $P(x)=x^3-6x^2+9x=x(x-3)^2,$ which contradicts $b< c.$
*Case $a<0:$
Now, $abc<0$ because $b,c>1.$ Let us apply Descartes rule of signs. There is only one change of sign in $P(x).$ Therefore, $P(x)$ has exactly one positive root, which contradicts to $\;1<b<3<c\;$ found before.
We conclude that $a>0.$
*Let us prove that $c<4.$
From $P'(x)=3(x-1)(x-3)$ we know that $P(x)$ increases on $(-\infty,1)$ and on $(3,\infty)$ and decreases on $(1,3).$ Since $$P(4)=4-abc=P(1)>0,$$ the root $c$ must be less than $4.$
| {
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"source": "stackexchange",
"question_score": "2",
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Why and when can I just peacefully substitute into $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$ without checking range conditions? This is an example question in my book:
To solve for $x$:
$$\tan^{-1}\frac{x-1}{x+2}+\tan^{-1}\frac{x+1}{x+2}=\frac{\pi}{4}$$
and it is solved by a direct formula given by
$$\tan^{-1}x+\tan^{-1}y=\tan^{-1} \left(\frac{x+y}{1-xy}\right) $$ when $xy<1$. The book uses this formula without checking conditions many many times and it is mildly infuriating. It is alright when checking the condition is simple but in this case finding the range of $\dfrac{x^2-1}{(x+2)^2}$ is a little time consuming and has the square root of 21 in it.
Well my question is:
Why and when can I just peacefully substitute the formula without worrying about the condition as the book leisurely does so?
| We have,
$\mathsf{tan^{-1}\left(\dfrac{x-1}{x+2}\right)+tan^{-1}\left(\dfrac{x+1}{x+2}\right)=\dfrac{\pi}{4}}$
$\mathsf{\implies\,tan^{-1}\left(\dfrac{x+1}{x+2}\right)=\dfrac{\pi}{4}-tan^{-1}\left(\dfrac{x-1}{x+2}\right)}$
$\mathsf{\implies\,tan\left\{tan^{-1}\left(\dfrac{x+1}{x+2}\right)\right\}=tan\left\{\dfrac{\pi}{4}-tan^{-1}\left(\dfrac{x-1}{x+2}\right)\right\}}$
$\mathsf{\implies\,\dfrac{x+1}{x+2}=\dfrac{1-\dfrac{x-1}{x+2}}{1+\dfrac{x-1}{x+2}}}$
$\mathsf{\implies\,\dfrac{x+1}{x+2}=\dfrac{x+2-x+1}{x+2+x-1}}$
$\mathsf{\implies\,\dfrac{x+1}{x+2}=\dfrac{3}{2x+1}}$
$\mathsf{\implies\,(x+1)(2x+1)=3(x+2)}$
$\mathsf{\implies\,2x^2+2x+x+1=3x+6}$
$\mathsf{\implies\,2x^2+3x+1=3x+6}$
$\mathsf{\implies\,2x^2=5}$
$\mathsf{\implies\,x=\pm\sqrt{\dfrac{5}{2}}}$
Now, check the value(s) of x satisfying the given equation,
Those value(s) of x are not satisfying the given equation, will be neglected
There is no need to work with the conditions when solving ITF equations.
In this equation $\mathsf{x=-\sqrt{\dfrac{5}{2}}}$ does not satisfy the given equation
Thus, $\mathsf{x=\sqrt{\dfrac{5}{2}}}$ is the only real solution.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Prove that $2^n + 5^n + 56$ is divisible by $9$, where $n$ is an odd integer
Prove that $9 \mid2^n + 5^n + 56$ where n is odd
I have proved this by using division into cases based on the value of $n\bmod3$ but it seems a little bit clumsy to me and I wonder if there are other ways to prove it, probably by using modular arithmetic or induction? Below is my proof:
$\text{Case 1, }n\bmod3=0,\text{then $n=3k$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k}+5^{3k}+56
\\ & = 8^k+125^k+56
\\ & \equiv (-1)^k+(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 2, }n\bmod3=1,\text{then $n=3k+1$ for some even integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+1}+5^{3k+1}+56
\\ & = 2\cdot8^k+5\cdot125^k+56
\\ & \equiv 2\cdot(-1)^k+5\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 9\equiv0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 3, }n\bmod3=2,\text{then $n=3k+2$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+2}+5^{3k+2}+56
\\ & = 4\cdot8^k+25\cdot125^k+56
\\ & \equiv 4\cdot(-1)^k+25\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv -27\equiv0\quad&\left(\bmod9\right)
\end{align}$$
| It still involves cases, but here's a somewhat slicker proof.
By Euler's theorem, $a^{\varphi(n)}\equiv 1$ mod $n$, where $\varphi$ is Euler's totient function, which count the number of positive integers less than $n$ coprime to $n$.
As $\varphi(9)=6$, this implies that the residue of
$$2^n+5^n+56$$
mod $9$ depends only on $n$ mod $6$. If we assume $n$ is odd, then the only possible residues are $1,3,5$. As such it is sufficient to verify the claim for $n=1,3,5$.
We can further reduce the number of calculations we have to do by noting that $2$ and $5$ are multiplicative inverses mod $9$. Then, as $2^5\equiv 5$, we must have $5^5\equiv 2$, so the $n=5$ case follows from the $n=1$ case. Similarly, as $2^3\equiv -1$, we have $5^3\equiv -1$, letting us easily verify the $n=3$ case.
Though it is arguably simpler to just directly verify these three cases
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 2
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Finding $a$ such that $f(x)=a\,|x-b|$ is satisfied by $(2,1)$ and $(10,3)$
I was given the following equation: $$f(x)=a\,|x-b|$$ and the information that both $(2,1)$ and $(10,3)$ are solutions for the equation. The question asked to solve for $a$.
This is what I did: $$a=\frac{1}{2-b}\\[10pt]a=\frac{-1}{2-b}\\[10pt]a=\frac{-3}{10-b}\\[10pt]a=\frac{3}{10-b}$$ all of these seem to be possible solutions for the equation, I just need to find the one that will work with both solution sets. I solved and got multiple values for a and b. The ones that worked for both ordered pairs were the solutions $a=\frac12 \text{ and } b=4$ and $a=\frac14\text{ and } b=-2$.
I feel pretty confident about my answer (perhaps wrongly so :) ) but is there a faster, more efficient way to do this?
| Your answers are correct, but there is a more efficient way of finding them using what we know about the vertex in this context.
First, note that $f(x) = a|x-b|$ has two transformations, a shift to the right by $b$ and a vertical stretch by $a$. So, we know that the vertex has coordinates $(b,0)$. Since the vertex must be on the $x$-axis, there are only two possible cases here, each with exactly one possible solution: either the vertex is left of $x = 2$ or it's between $x = 2$ and $x = 10$.
Case 1: $b<2$
If the vertex is left of $x = 2$, we know that $(b,0)$, $(2,1)$, and $(10,3)$ are collinear with slope $a$. We start by finding the slope between $(2,1)$ and $(10,3)$:
$$ a = \frac{3-1}{10-2} = \frac{1}{4} . $$
Then, we use $a$ to find $b$:
$$ \frac{1-0}{2-b} = \frac{1}{4} \Rightarrow 2-b = 4 \Rightarrow b = -2 . $$
The first solution is $a = \frac{1}{4}, b = -2$.
Case 2: $ 2 < b < 10$
Here, we use the absolute value's vertical symmetry about the vertex - the slopes must be opposites on either side of the vertex:
$$ \frac{0-1}{b-2} = -\frac{3-0}{10-b} \Rightarrow b-10=3(b-2) \Rightarrow b = 4 . $$
Then, $a$ is the slope to the right of the vertex:
$$ a = \frac{3-0}{10-b} = \frac{3}{10-4} = \frac{1}{2} . $$
The second solution is $a = \frac{1}{2}, b = 4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the limit $\lim_{n\rightarrow\infty}\left(\frac{\prod_1^n(1 +r/n^2)}{e^{1/2}}\right)^n$ How to calculate the limit
$$\lim_{n\rightarrow\infty}\left(\frac{\prod\limits_{r=1}^n\left(1 +\frac{r}{n^2}\right)}{e^{1/2}}\right)^n?$$
I tried using the sandwich theorem by taking logarithm of the limit and using the expansion
$$ x - (x^2)/2<\ln(1+x)<x$$
but I was getting different from answer from the two sides. The correct answer is $e^{1/3}$.
| Hint:
If we take
$\begin{align}
&P(x) := \prod_{r=1}^n\left(x +\frac{r}{n^2}\right) \\[1mm]
&= x^n + \frac{\sum_{r_1=1}^n r_1}{n^2}x^{n-1} + \frac{\sum_{r_1=1}^n\sum_{r_2=1}^n r_1r_2}{n^4}x^{n-2} + \dots + \frac{\sum_{r_1=1}^n\sum_{r_2=1}^n\dots\sum_{r_n=1}^n r_1r_2\cdot...\cdot r_n}{n^{2n}} \\[1mm]
&= x^n + \frac{\sum_{r_1=1}^n r_1}{n^2}x^{n-1} + \frac{\left(\sum_{r_1=1}^n r_1\right)^2}{n^4}x^{n-2} + \dots + \frac{\left(\sum_{r_1=1}^n r_1\right)^n}{n^{2n}} \\[1mm]
&= x^{n} + \frac{S}{n^2}x^{n-1} + \frac{S^2}{n^4}x^{n-2} + \dots + \frac{S^n}{n^{2n}} \tag{$\star$} \end{align}$
where $$S = \sum_{r_1=1}^n r_1 = \frac{n(n+1)}{2}$$
Then the numerator of the ratio of the limit
$$\lim_{n\rightarrow\infty}\left(\frac{\prod_{r=1}^n(1 +r/n^2)}{e^{1/2}}\right)^n$$
will be
$$\begin{align}
P(1) &= 1 + \frac{S}{n^2} + \frac{S^2}{n^4} + \dots + \frac{S^n}{n^{2n}} \\[1mm]
&= \frac{1 - \frac{S^{n+1}}{n^{2(n+1)}}}{1 - \frac{S}{n^2}} \\[1mm]
&= \frac{n\left(-n^{-n-1}\left(n+1\right)^{n+1}+2^{n+1}\right)}{2^n\left(n-1\right)}\\[1mm]
&= \frac{n}{n-1}\frac{\left(2^{n+1} - \left(1+\frac{1}{n}\right)^{n+1}\right)}{2^n} \\[1mm]
&\sim 2 - \frac{1}{2^n}\left(1+\frac{1}{n}\right)^{n}\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Extrema of $f(x,y)=(1-x^2-y^2)\cdot xy$ subject to $x^2+y^2\leq 1$ I want to determine the extrema of $f(x,y)=(1-x^2-y^2)\cdot xy=xy-x^3y-xy^3$ subject to $x^2+y^2\leq 1$.
We use Lagrange Multipliers to check the critical points on the circle $x^2+y^2=1$.
To check the critical points inside the circle, $x^2+y^2<1$, we calculate the critical points of $f(x,y)$.
Let's start with the Lagrange-Multipliers.
We have $f(x,y)=(1-x^2-y^2)\cdot xy=xy-x^3y-xy^3$ und $g(x,y)=x^2+y^2-1=0$.
We consider
\begin{equation*}L(x,y,\lambda )=f(x,y)+\lambda g(x,y)=xy-x^3y-xy^3+\lambda \left (x^2+y^2-1\right )\end{equation*}
We calculate the partial derivatives of $L$.
\begin{align*}&\frac{\partial{L}}{\partial{x}}=y-3x^2y-y^3+2\lambda x \\ &\frac{\partial{L}}{\partial{y}}=x-x^3-3xy^2+2\lambda y \\ &\frac{\partial{L}}{\partial{\lambda }}= x^2+y^2-1 \end{align*}
To calculate the extrema we set each equation equal to zero and solve the system:
\begin{align*}&y-3x^2y-y^3+2\lambda x=0 \ \ \ \ \ \ \ \ (1) \\ & x-x^3-3xy^2+2\lambda y =0 \ \ \ \ \ \ \ \ (2) \\ & x^2+y^2-1=0 \ \ \ \ \ \ \ \ (3)\end{align*}
We get the critical points \begin{equation*}P_1\left (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right ) \ \ , \ \ \ P_2\left (-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right )\ \ , \ \ \ P_3\left (-1, 0\right )\ \ , \ \ \ P_4\left (1, 0\right ) \ \ , \ \ \ P_5\left (0, -1\right ) \ \ , \ \ \ \ P_6\left (0, 1\right )\end{equation*}
Now we check inside the circle.
We set the partial derivatives of $f$ equal to zero and solve the system:
\begin{align*}&f_x=0 \Rightarrow y-3x^2y-y^3=0 \ \ \ \ \ \ \ \ (1)\\ &f_y=0 \Rightarrow x-x^3-3xy^2=0\ \ \ \ \ \ \ \ (2)\end{align*}
Then we get the critical points \begin{align*}&Q_1\left (0, 0\right ) \ \ , \ \ \ Q_2\left (\frac{1}{2}, \frac{1}{2}\right )\ \ , \ \ \ Q_3\left (-\frac{1}{2}, -\frac{1}{2}\right )\ \ , \ \ \ Q_4\left (0, -1\right )\ \ , \ \ \ Q_5\left (0, 1\right )\ \ , \ \ \ Q_6\left (-2, -1\right ) \ \ , \ \ \ Q_7\left (0, -1\right )\ \ , \ \\ & \ Q_8\left (-\frac{3}{2}, -\frac{1}{2}\right ) \ \ , \ \ \ Q_9\left (\frac{1}{2}, -\frac{1}{2}\right )\ \ , \ \ \ Q_{10}\left (2, 1\right )\ \ , \ \ \ Q_{11}\left (\frac{3}{2}, \frac{1}{2}\right ) \ \ , \ \ \ Q_{12}\left (-\frac{1}{2}, \frac{1}{2}\right )\end{align*}
Then we have to find the maximal and minimum value of $f$ at all these critical points.
Is everything correct? I am not really sure about the part inside the circle, if all critical points are correct.
| Let $x=r\cos t$ and $y=r\cos t$, where $0\leq r\leq1$ and $t\in[0,2\pi)$.
Thus, by AM-GM $$(1-x^2-y^2)xy=(1-r^2)r^2\sin t\cos t=\frac{1}{2}(1-r^2)r^2\sin2t\leq$$
$$\leq\frac{1}{2}(1-r^2)r^2\leq\frac{1}{2}\left(\frac{1-r^2+r^2}{2}\right)^2=\frac{1}{8}.$$
The equality occurs for $1-r^2=r^2,$ which gives $r=\frac{1}{\sqrt2}$ and $\sin2t=1,$ which says that we got a maximal value.
By the similar way we obtain that $-\frac{1}{8}$ is a minimal value.
Another way.
By AM-GM twice we obtain:
$$(1-x^2-y^2)xy\leq(1-x^2-y^2)\cdot\frac{1}{2}(x^2+y^2)\leq\frac{1}{2}\left(\frac{1-x^2-y^2+x^2+y^2}{2}\right)^2=\frac{1}{8}.$$
The equality occurs for $x=y$ and $1-x^2-y^2=x^2+y^2,$ which says that we got a maximal value.
Also, $$(1-x^2-y^2)xy\geq(1-x^2-y^2)\cdot\left(-\frac{1}{2}(x^2+y^2)\right)\geq$$
$$\geq-\frac{1}{2}\left(\frac{1-x^2-y^2+x^2+y^2}{2}\right)^2=-\frac{1}{8}.$$
The equality occurs for $x=-y$ and $1-x^2-y^2=x^2+y^2,$ which says that we got a minimal value.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$a^a\cdot{b^b}\ge \bigl(\frac{a+b}{2}\bigl)^{a+b}\ge{a^b}\cdot{b^a}$ If $a$ and $b$ are positive rational numbers, prove that
$$a^a\cdot b^b\ge \left(\frac{a+b}{2}\right)^{a+b} \ge a^b \cdot{b^a}$$
My try:
consider $\frac{a}{b}$ and $\frac{b}{a}$ be two positive numbers with associated weights $b$ and $a$.
Then $\displaystyle\frac{b\cdot\frac{a}{b}+a\cdot\frac{b}{a}}{a+b}\ge \biggl[\left(\frac{a}{b}\right)^b\cdot \left(\frac{b}{a} \right)^a\biggl]^\frac{1}{a+b}$ implies $a^a\cdot{b^b}\ge a^b\cdot{b^a}$
Please help me to solve this problem. Thanks
| The second inequality.
We need to prove that:
$$\ln\frac{a+b}{2}\geq\frac{b}{a+b}\ln{a}+\frac{a}{a+b}\ln{b}.$$
Now, since $\frac{b}{a+b}+\frac{a}{a+b}=1$ and $\ln$ is a concave function, by Jensen and AM-GM we obtain:
$$\frac{b}{a+b}\ln{a}+\frac{a}{a+b}\ln{b}\leq\ln\left(\frac{ba}{a+b}+\frac{ab}{a+b}\right)\leq\ln\frac{a+b}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $2x^3-5ix^2+3x+4i=0$, need $(21x_1-1)^{-2}+(21x_2-1)^{-2}+(21x_3-1)^{-2}$ I have this polynomial from Problemas selectos (Lumbreras editors):
$$2x^3-5ix^2+3x+4i=0$$
I make $x=it$. then:
$$2(it)^3-5i(it)^2+3it+4i=0$$
$$i(-2it^3+5it^2+3it+4)=0$$
$$2t^3-5t^2-3t-4=0$$
$$8t^3-20t^2-12t-16=0$$
$$(2t)^3-3(2t)^2(5/3)+(50/3)t-125/27-(14/3)t-307/27=0$$
$$(2t-5/3)^3-(86/3)t-307/27=0$$
If $2t-5/3=z$, then:
$$z^3-(43/3)z-952/27=0$$
Where $p=-\frac{43}{3}$ and $q=-\frac{952}{27}$
When $A=-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}$ and $B=-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}$
We have $z_1=\sqrt[3]{A}+\sqrt[3]{B}$, $z_2=-\frac{\sqrt[3]{A}+\sqrt[3]{B}}{2}+i\sqrt{3}\frac{\sqrt[3]{A}-\sqrt[3]{B}}{2}$, $z_3=-\frac{\sqrt[3]{A}+\sqrt[3]{B}}{2}-i\sqrt{3}\frac{\sqrt[3]{A}-\sqrt[3]{B}}{2}$
There for $x_n=\frac{i}{2}(z_n+\frac{5}{3})$,
| Starting from @mjw's answer, you face a cubic equation with only one real root in $y$ since $\Delta=-5447$.
Using the hyperbolic method for this root, we then have
$$y_1=\frac{5}{6}+\frac{\sqrt{43}}{3} \cosh \left(\frac{1}{3} \cosh
^{-1}\left(\frac{476}{43 \sqrt{43}}\right)\right)=3.17174\cdots$$
Now, deflate the cubic and get the quadratic equation
$y^2+Ay+B=0$ where
$$A=y_1-\frac{5}{2}\qquad \text{and} \qquad B=\frac 2 {y_1}$$
Numerically, they are $y_{2,3}=-0.335868\pm 0.719557\, i$ as already given by @mjw.
| {
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"timestamp": "2023-03-29T00:00:00",
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The Diophantine equation $x^5-2y^2=1$ I'm trying to solve the Diophantine equation $x^5-2y^2=1$.
Here's my progress so far. We can write the Diophantine equation as
$$\frac{x-1}{2}\cdot(x^4+x^3+x^2+x+1)=y^2.$$
If $x\not\equiv1\pmod{5}$, then $\gcd(\frac{x-1}{2},x^4+x^3+x^2+x+1)=1$, so both $\frac{x-1}{2}$ and $x^4+x^3+x^2+x+1$ must be perfect squares (note: $x^4+x^3+x^2+x+1>0$).
In particular, $4(x^4+x^3+x^2+x+1)$ is a perfect square.
Comparison with $(2x^2+x)^2$ and $(2x^2+x+1)^2$ forces $-1\leq x\leq3$.
This results in the solutions $(3,\pm11)$.
If $x\equiv1\pmod{5}$, then we can write the Diophantine equation as
$$\frac{x-1}{10}\cdot\frac{x^4+x^3+x^2+x+1}{5}=\left(\frac{y}{5}\right)^2,$$
where $\gcd(\frac{x-1}{10},\frac{x^4+x^3+x^2+x+1}{5})=1$, so both $\frac{x-1}{10}$ and $\frac{x^4+x^3+x^2+x+1}{5}$ must be perfect squares.
Thus,
$$x=10a^2+1,$$
$$x^4+x^3+x^2+x+1=5b^2.$$
Unfortunately, this is where I get stuck.
I can substitute the first equation into the second, giving
$$10000a^8+5000a^6+1000a^4+100a^2+5=5b^2,$$
$$2000a^8+1000a^6+200a^4+20a^2+1=b^2,$$
$$2000a^8+1000a^6+200a^4+20a^2=(b-1)(b+1),$$
$$5a^2(100a^6+50a^4+10a^2+1)=\frac{b-1}{2}\cdot\frac{b+1}{2},$$
but this doesn't seem to be making progress, even with modular arithmetic considerations.
| Also not a super elementary argument, but the only non-trivial result I use here is that $K=\Bbb Q(\sqrt{-2})$ has class number $1$, i.e. $\Bbb Z[\sqrt{-2}]$ is a PID.
Let $x^5-2y^2=1$ with $x,y\in\Bbb Z$. Note that $x$ is odd. We can factor the equation as $$x^5=(1-\sqrt{-2}y)(1+\sqrt{-2}y).$$ Let $d$ be a gcd of $(1-\sqrt{-2}y),(1+\sqrt{-2}y)$ in $\Bbb Z[\sqrt{-2}]$. Note that $d\mid 2$ and $d\mid x^5$. As $x^5$ is odd this implies that $d\mid 1$, i.e. the elements are coprime, hence (as $\Bbb Z[\sqrt{-2}]$ is a UFD) there is a unit $\varepsilon\in \Bbb Z[\sqrt{-2}]^\times=\{-1,1\}$ and $z=a+b\sqrt{-2}\in\Bbb Z[\sqrt{-2}]$ such that $1+\sqrt{-2}y=\varepsilon z^5$. As $(-1)^5=-1$ we may assume that $\varepsilon=1$. Then expanding the fifth power and comparing coefficients we get:
\begin{align*}
1&=a^5-20a^3b^2+20ab^4\\
y&=5a^4b-20a^2b^3+4b^5
\end{align*}
The first equation implies $a=\pm1$ and for $a=-1$ we don't get any solutions and for $a=1$ we get $b^4-b^2=0$, i.e. $b=0$ or $b=\pm1$. These correspond to $y=0$ and $y=\pm11$. Hence the only possible solutions for the original equation are $(1,0),(3,-11),(3,11)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding foot of perpendicular of centre of ellipse on its variable tangent
Find the locus of the foot of perpendicular from the centre of the ellipse $${x^2\over a^2} +{y^2\over b^2} = 1$$
on the chord joining the points whose eccentric angles differ by $π/2.$
Ref:Locus of foot of perpendicular of origin from tangent of ellipse, related
The points may be given as $(a\cos \alpha, b\sin \alpha)$ and $(-a\sin \alpha , b \cos \alpha )$, then their midpoint is given as: $\left( \frac{a}{2} (\cos \alpha - \sin \alpha) , \frac{b}{2} (\cos \alpha + \sin \alpha) \right)=(\kappa, \beta)$, the equation of chord given midpoint of ellipse is:
$$ \frac{\kappa X}{a^2} + \frac{\beta Y}{b^2} = \frac{\kappa^2}{a^2} + \frac{\beta^2}{b^2} $$
Let $u= \frac{\cos \alpha - \sin \alpha}{2}$ and $ v=\frac{\cos \alpha + \sin \alpha}{2}$, then our equation becomes:
$$\frac{ uX}{a} + \frac{ vY}{b} = u^2 + v^2 $$
Foot of perpendicular from origin for above line is given as:
$$ \frac{ax}{u}= \frac{by}{v}= \frac{(ab)^2 (u^2 +v^2)}{b^2u^2 +a^2v^2}$$
$ax= u(\frac{(ab)^2 (u^2 +v^2)}{b^2u^2 +a^2v^2})$
and $by =v (\frac{(ab)^2 (u^2 +v^2)}{b^2u^2 +a^2v^2})$, following the last line of this answer, I squared and added:
$$(ax)^2 + (by)^2 = (ab)^4 \left[ \frac{u^2 +v^2}{b^2 u^2 +a^2v^2} \right]^2= \frac{(ab)^4}{4} \left[ \frac{1}{b^2 u^2 + a^2 v^2}\right]^2$$
How do I write bracketed term in purely $(x,y)$?
| Here's one method to simplify things. First note that :
$$u^2+v^2=1/2$$
Now, consider the equation you found:
$$\frac{ax}{u}= \frac{by}{v}= \frac{a^2b^2 (u^2 +v^2)}{b^2u^2 +a^2v^2}$$
Then:
$$axv=byu\Rightarrow a^2v^2=b^2u^2y^2/x^2\Rightarrow b^2u^2+a^2v^2=b^2u^2\left(\frac{x^2+y^2}{x^2}\right)$$
Plugging this back into one of the equations gives:
$$\frac{ax}{u}=\frac{a^2b^2x^2}{2b^2u^2(x^2+y^2)}\Rightarrow \boxed{u=\frac{ax}{2(x^2+y^2)}}$$
Similarly you will also find
$$\boxed{v=\frac{by}{2(x^2+y^2)}}$$
Squaring and adding them, along with the first equation at the extreme top gives:
$$\frac12=u^2+v^2=\frac{a^2x^2}{4(x^2+y^2)^2}+\frac{b^2y^2}{4(x^2+y^2)^2}$$
$$\Rightarrow \boxed{2(x^2+y^2)^2=a^2x^2+b^2y^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Geometry Problem from Olympiad book Given the base and the vertical angle of a triangle show that its area is greatest when it's isoceles
I am stuck on how to proceed and which theorem to use here
Thanks in advance
| Denote $S = \dfrac{1}{2}ab\sin C$. Thus: $S \le \dfrac{1}{8}(a+b)^2\sin C$, with equality occur at $a = b$ which shows $\triangle ABC$ isosceles. To show that $S$ reaches a maximum value, you need to show: $(a+b)^2$ also reaches a maximum in terms of $c$. To this end, use the law of cosines: $c^2 = a^2+b^2 - 2ab\cos C= (a+b)^2 - 2ab(1+\cos C)\ge (a+b)^2 - \dfrac{1+\cos C}{2}(a+b)^2= \dfrac{1-\cos C}{2}(a+b)^2= \sin^2(\frac{C}{2})(a+b)^2\implies (a+b)^2 \le \dfrac{c^2}{\sin^2(\frac{C}{2})}\implies S \le \dfrac{1}{8}\cdot\dfrac{c^2}{\sin^2(\frac{C}{2})}\cdot \sin C= \dfrac{c^2}{4\tan(\frac{C}{2})}$. This is the maximum value of the area $S$ of $\triangle ABC$, and this value occurs when $a = b$ by AM-GM inequality. So $\triangle ABC$ is isosceles when the max occurs.
| {
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To prove: $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \cot^{-1}3$ To prove: $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \cot^{-1}3$
My Attempt:
First Method: we know that $\cot^{-1}x = \tan^{-1}\frac{1}{x}$ for $x>0$ and $\tan^{-1}x+\tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}, xy<1$
Now $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18$ = $\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{18}$ = $(\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{8})+\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7}\frac{1}{8}}+\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{\frac{15}{56}}{\frac{55}{56}}+\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{3}{11} +\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{\frac{3}{11}+\frac{1}{18}}{1-\frac{3}{11}\frac{1}{18}}$ = $\tan^{-1}\frac{\frac{65}{198}}{\frac{195}{198}}$ = $\tan^{-1}\frac{1}{3}$ = $\cot^{-1}3$
Second Method: we know that $\tan^{-1}x+\cot^{-1}x = \frac{π}{2}$ for $x \in \Bbb R$. So $\cot^{-1}x = \frac{π}{2} - \tan^{-1}x$.
Now $$\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \frac{π}{2} - \tan^{-1}7+ \frac{π}{2} - \tan^{-1}8+ \frac{π}{2} - \tan^{-1}18 \implies \\ \frac{3π}{2} - (\tan^{-1}7+\tan^{-1}8+\tan^{-1}18) = \frac{3π}{2} - \tan^{-1}\frac{7+8+18-7×8×18}{1-7×8-8×18-7×18} \\ = \frac{3π}{2} - \tan^{-1}\frac{-975}{-325} = \frac{3π}{2} - \tan^{-1}3 = π + (\frac{π}{2} - \tan^{-1}3) = π + \cot^{-1}3 .$$ Also we know that $\tan x$ and $\cot x$ are periodic function with period $π$. Please help me in Second Method. Is $π + \cot^{-1}3 = \cot^{-1}3$ ?. If yes, then elaborate it.
| The statement is equivalent to
$$\DeclareMathOperator{\arccot}{arccot}
\arccot7+\arccot8=\arccot3-\arccot18
$$
The right-hand side is positive because the arccotangent is decreasing, so it is in the interval $(0,\pi/2)$ and we have to ensure that the same holds for the left-hand side, but this is easy, because $7>1$ and $8>1$, hence
$$
\arccot7+\arccot8<2\arccot1=\frac{\pi}{2}
$$
The two sides are equal if and only if their cotangents are.
Since
$$
\cot(\alpha+\beta)=\frac{\cot\alpha\cot\beta-1}{\cot\alpha+\cot\beta}
\qquad
\cot(\alpha-\beta)=\frac{\cot\alpha\cot\beta+1}{\cot\beta-\cot\alpha}
$$
this amounts to proving that
$$
\frac{7\cdot8-1}{7+8}=\frac{3\cdot18+1}{18-3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Roots of polynomials with repeated roots Let $a$ and $b$ be real numbers. Consider the cubic equation $$x^3+2bx^2-ax^2-b^2=0$$
(i) Show that if $x=1$ is a solution of the cubic then $$ -1+\sqrt{2}\leq b\leq1+\sqrt{2} $$
(ii) Show that there is no value of $b$ for which $x=1$ is a repeated root of the cubic
I need some assistance on (ii). In order to show that there is no value of $b$ for which $x=1$ is a repeated root of the cubic. I attempted to factorise the cubic. If $x=1$ is a solution, then the cubic may be expressed as follows with no remainder. $x^3+2bx^2-a^2x-b^2 \equiv (x-1)(Ax^2+Bx+C)$. Comparing coefficients, $A = 1, C = b^2, B = a^2+b^2$ $\therefore (x-1)(x^2+(a^2+b^2)x+b^2)$. Finally, if we sub $x=1$ into the resulting quadratic, we get $1+a^2+b^2+b^2 > 0$ for all real values of $a,b$. Therefore, $x=1$ cannot be a repeated root.
I have been told that $x^2+(a^2+b^2)x+b^2$ is not the quadratic equation for this and in fact it's $x^2+(2b+1)x+b^2$
$(x-1)(x^2+(2b+1)x+b^2) = x^3+2bx^2+(b^2-2b-1)x-b^2$. The coefficient of the $x$ term is correct, but the $b^2-2b-1 \not= -a^2$ therefore this cannot be right aswell.
My solution:
$(x-1)(x^2+(a^2+b^2)x+b^2) = x^3+(a^2+b^2-1)x^2-a^2x-b^2$
So my solution seems to get the correct coefficient for $x$ term, but the incorrect $a^2+b^2-1$ coefficient for the $x^2$ term.
What is going on here? Has math been broken?
| If $x=1$ is a root of $P(x) = x^3 + 2bx^2 - a^2x - b^2$, then
\begin{align}
P(1) &= 0 \\
1+2b-a^2-b^2 &= 0 \\
a^2 &= 1 + 2b - b^2
\end{align}
So
\begin{align}
P(x) &= x^3 + 2bx^2 + (b^2-2b-1)x - b^2 \\
&= (x-1)(x^2+(2b+1)x + b^2)
\end{align}
If $x=1$ is a double root of $P(x)$, then
\begin{align}
(x^2+(2b+1)x + b^2)|_{x=1} &= 0 \\
1 + (2b+1) + b^2 &= 0 \\
(b+1)^2 + 1 &= 0
\end{align}
Which is impossible.
OR, if you do synthetic division by $x-1$ on $P(x)$, you get
\begin{array}{rrrrrrr}
\phantom{1 }&|1 & 2b & -a^2 & - b^2 \\
1 &|0 & 1 & 2b+1 & -a^2+2b+1\\
\hline
\phantom{1 }& 1 & 2b+1 & -a^2+2b+1 & -a^2-b^2+2b+1\\
\end{array}
Which means $P(x) = (x-1)(x^2 +(2b+1)x +(-a^2+2b+1)) + (-a^2-b^2+2b+1)$
And so $P(1) = 0$ implies $-a^2-b^2+2b+1=0$
and so on
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is it possible to change $x^3\prod_{n=1}^{\infty}\left(1-\frac{x^4} {2\pi^2n^2}-\frac{x^6}{2\pi^3n^3}\right)$ to equal $x\sin{(x^2)}$? Consider a function $$f(x)=x^3\prod_{n=1}^{\infty}\left(1-\frac{x^4} {2\pi^2n^2}-\frac{x^6}{2\pi^3n^3}\right)$$
This function has its zeros when $x=0$ or $x^2=k\pi$.
This function is very similar to another function which is $$g(x)=x\sin{(x^2)}$$. They have the same zeros and just that their increasing speed is different.
Is it possible to make some changes such that they are equal? One seems to be increasing and one is decreasing. Also, they are the functions that appears to have the exponential function $\frac 1 {x^3}$.
| Note :
Consider the Weierstass factorization of $\sin z$ :
$$\sin z = z \cdot \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{\pi^2 n^2}\right).$$
let $z = x^2$ then
$$\sin(x^2 ) = x^2 \cdot \prod_{n=1}^{\infty} \left( 1 - \frac{x^4}{\pi^2 n^2} \right)$$
and
$$x \sin(x^2 ) = x^3 \cdot \prod_{n=1}^{\infty} \left( 1 - \frac{x^4}{\pi^2 n^2} \right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Solve in $\mathbf{N}$ the equation $9x^2+p=y^2$ In these days, I have been trying to solve this problem:
Let $p \in \mathbf{N}$ a positive large integer ($> 10^9$). Find all $x, y \in\mathbf{N}$ such that:
$$9x^2+p=y^2$$
The first approach that I have tried is the following. We know that:
$$(t+n)^2-t^2=t^2+2\cdot t \cdot n +n^2-t^2=2\cdot t \cdot n +n^2$$
Also, we can build every possible square $w_n$ greater than $p$ in this way:
$$w_n=\left\lfloor\sqrt{p}\right\rfloor^2+2\cdot \left\lfloor\sqrt{p}\right\rfloor \cdot n +n^2 = \left(n+\left\lfloor\sqrt{p}\right\rfloor^2\right)^2$$
For example, let $p=5$. Follows that: $w_1=\left(1+\left\lfloor\sqrt{5}\right\rfloor^2\right)^2=9$, $w_2=\left(2+\left\lfloor\sqrt{5}\right\rfloor^2\right)^2=16$ and so on.
Now, using this idea and applying to the first equation:
$$9x^2=\left(n+\left\lfloor\sqrt{p}\right\rfloor^2\right)^2-p$$
I am not allowed to apply Pell's equation because $9=3^2$ and calcultaing $\Delta$ in $x$ doesn't help anymore.
Another approach is based on Pell's equation. I thought to express $9x^2=8x^2+x^2$ and then:
$$8x^2+x^2+p=y^2\leftrightarrow 8x^2+p=y^2-x^2\leftrightarrow (y^2-x^2)-8x^2=p \leftrightarrow u^2-8x^2=p$$
But then, in order to generate all the solutions, I have to guess the first one (or one of them) that is pretty complicated for big $p$.
So, how can we do that? Are there any other solutions?
Thanks.
| If we let $\quad A^2+B^2=C^2
\qquad A=3x\quad B=\sqrt{p}\quad C=y\qquad$
we can use Euclid's formula for Pythagorean triples to find infinite solutions based on the value of $(x)$. We start with the formula
$$ \qquad A=m^2-k^2\qquad B=2mk \qquad C=m^2+k^2\qquad$$
and solve the $A$-function for $(k).\quad$
Then we test a defined range of $m$-values to see which yield integers.
\begin{equation}
A=m^2-k^2\implies k=\sqrt{m^2-A}\\
\text{for}\qquad \lfloor\sqrt{A+1}\rfloor \le m \le \frac{A+1}{2}
\end{equation}
The lower limit ensures $k\in\mathbb{N}$ and the upper limit ensures $m> k.$
There are Pythagorean triples for all $A=2n+1$ but A=$3x$, so we are limited to odd muliples of $3:$
$$ x\in\big\{1,3,5,7,\cdots\big\}\implies
A\in\big\{3,9,15,21,\cdots\big\}.\quad $$
Here, we will use $x=5\implies A=15$
$$A=15\implies \lfloor\sqrt{15+1}\rfloor=4\le m \le \frac{15+1}{2} =8\\
\land\quad m\in\{4,8\}\implies k \in\{1,7\} $$
$$F(4,1)=(15,8,17)\implies (x,p,y)=(5,8^2,17)\\
F(8,7)=(15,112,113)\implies (x,p,y)=(5,112^2,113) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Finding the square root of a radical expression I have this (not necessarily optimum) method for finding the value of sin $15^{\circ}$.
(Meant to be an application of Angle Bisector Theorem).
In a equilateral triangle with side length 1, draw the altitude to get two $30^{\circ}-60^{\circ}-90^{\circ}$ triangles
Then, in one of the $30^{\circ}-60^{\circ}-90^{\circ}$ triangles, draw the angle bisector again to get a $15^{\circ}-90^{\circ}-75^{\circ}$ triangle.
Apply the Angle bisector theorem to find that the side opposite $15$ degrees is $\dfrac{(2\sqrt{3}-3)}{2}$.
Combine with height $\dfrac {\sqrt{3}}{2}$ to get the hypotenuse squared as $6 \sqrt{3}$, which then has as its square root $\dfrac {(3-\sqrt{3})}{\sqrt{2}}$.
Apply sin = opp/hyp to get the usual formula for $\sin 15^{\circ}$.
I got $\sqrt{6-3\sqrt {3}}$ as $\dfrac{(3-\sqrt{3})}{\sqrt{2}}$ through trial and error.
Is there a systematic way of finding it?
| Presumably you mean
$\sqrt{6-\color{blue}{3}\sqrt3}=(3-\sqrt3)/\sqrt2.$
Suppose we are given two positive rational numbers $a,b$ and we seek $x,y$ such that
$\sqrt{a-\sqrt{b}}=\sqrt{x}-\sqrt{y}$
Then we will have the conjugate relation
$\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y}$
And upon multiplication
$\color{blue}{\sqrt{a^2-b}=x-y}$
We can also square each of the conjugate equations and add those squared results. The radicals cancel and we end with
$\color{blue}{a=x+y}$
Now, if $a^2-b$ is a perfect square, then the left sides of both blue equations are rational numbers and we can simply solve these equations as a linear system for $x$ and $y$. For the problem at hand we have $a=6,b=27,\therefore a^2-b=9=3^2$ and this process will work. We get $x=9/2,y=3/2$ which matches the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A simple general expression for a definite integral Let
$$I_k = \int_0^{\frac{\pi}{2}} \sin x \cos^{2k} x \sqrt{1 + \sin x} \, dx,$$
where $k = 0,1,2,\ldots$. I wish to find a simple general expression for $I_k$ in terms of $k$. Simple here is the operative word.
Making a tangent half-angle substitution of $t = \tan \frac{x}{2}$ produces
$$I_k = 4\int_0^1 \frac{t(1 + t)^{2k + 1} (1 - t)^{2k}}{(1 + t^2)^{2k + \frac{5}{2}}} \, dt.$$
Trying to guess its general form one notices:
\begin{align*}
k = 0 : \quad I_0 &= 4\int_0^1 \frac{t(1 + t)}{(1 + t^2)^{\frac{5}{2}}} \, dt = 4 \left [\frac{(t - 1)(t^2 + t + 1)}{3(t^2 + 1)^{\frac{3}{2}}} \right ]_0^1 = \frac{4}{3}\\
k = 1 : \quad I_1 &= 4\int_0^1 \frac{t(1 + t)^3(1 - t)^2}{(1 + t^2)^{\frac{9}{2}}} \, dt = 4 \left [\frac{(t - 1)^3(11t^4 + 33t^3 + 52t^2 + 33t + 11)}{105(t^2 + 1)^{\frac{7}{2}}} \right ]_0^1 = \frac{44}{105}\\
k = 2 : \quad I_2 &= 4 \int_0^1 \frac{t(1 + t)^5(1 - t)^4}{(1 + t^2)^{\frac{13}{2}}} \, dt = 4\left [\frac{(1 - t)^5(211t^6 + 1055t^5 + 2593 t^4 + 3370t^2 + 1055t + 211)}{3465(t^2 + 1)^{\frac{11}{2}}} \right ]_0^1 = \frac{844}{3465}
\end{align*}
So it appears
$$I_k = 4 \left [\frac{(t - 1)^{2k + 1}p_k(t)}{a_k(t^2 + 1)^{2k + \frac{3}{2}}} \right ]_0^1,$$
where $a_k$ is a positive integer and $p_k(t)$ is a symmetric polynomial of degree $(2k + 2)$. But what is $a_k$ and $p_k(t)$?
Alternatively, a closed-form expression in terms of the hypergeometric function can be found. It is
$$I_k = \frac{2}{3\sqrt{2}} {}_2 F_1 \left (\frac{3}{2},-2k;\frac{5}{2};1 \right ) + \frac{1}{2^{2k + \frac{3}{2}}(2k+1)} {}_2F_1 \left (2k + 1,2k + \frac{5}{2};2k + 2; \frac{1}{2} \right ),$$
but this is hardly simple since ultimately $I_k$ is just a positive rational number. Perhaps this expression can be simplified in some way.
So my question is, can a simple expression for $I_k$ be found?
| I think that we could simplify the problem letting
$$\sqrt{1+\sin(x)}=t \implies x=-\sin ^{-1}\left(1-t^2\right)\implies dx=\frac{2 t}{\sqrt{1-\left(1-t^2\right)^2}}\,dt$$ This makes
$$I_k=2\int_1^{\sqrt 2}\left(t^2-1\right) t^{2 k+1} \left(2-t^2\right)^{k-\frac{1}{2}}\,dt$$
$$\color{blue}{I_k=\frac{\sqrt{\pi }\, \Gamma (2 k+1)-4^k \Gamma \left(2 k+\frac{3}{2}\right) \left((2
k-1) B_{\frac{1}{2}}\left(k+1,k-\frac{1}{2}\right)-4 k
B_{\frac{1}{2}}\left(k+1,k+\frac{1}{2}\right)\right)}{\sqrt{2}\, \Gamma \left(2
k+\frac{5}{2}\right)}}$$ which is valid $\forall k\geq 0$.
For integer $k$, this generates the sequence
$$\left\{\frac{4}{3},\frac{44}{105},\frac{844}{3465},\frac{7676}{45045},\frac{271372}
{2078505},\frac{785244}{7436429},\frac{40395868}{456326325},\frac{343225244}{450
8102925},\frac{3309796628}{49589132175},\cdots\right\}$$
These sequences are not found in $OEIS$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
$\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$ How to find $\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$? Here is my attempt.
Put $f(x)=\frac{x}{\log x}(x^{\frac{1}{x}}-1)$ for $x>1$. Then
\begin{aligned}
\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1) &= \lim\limits_{x\rightarrow\infty} f(x) \\
&= \lim\limits_{x\rightarrow\infty}\frac{x^{\frac{1}{x}}-1}{\frac{\log x}{x}} \\
&= \lim\limits_{x\rightarrow\infty} \frac{x^{\frac{1}{x}-2}(1-\log x)}{\frac{x^2+\log x}{x^4}} \\
&= \lim x^{\frac{1}{x}}\lim x^2\lim\frac{1-\log x}{x^2+\log x} \\
&= 1 \cdot (+\infty)\cdot (0)\mbox{.}
\end{aligned}
This method does not work.
| As an alternative
$$\frac{n}{\log n}(n^{\frac{1}{n}}-1)=\frac{n^{\frac{1}{n}}-1}{\log \left(n^{\frac{1}{n}}\right)}$$
with $x+1=n^{\frac{1}{n}} \to 1$ that is as $x \to 0$
$$\lim_{x\to 0}\frac{x}{\log (1+x)}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Inductively prove $1 + \frac12 + \frac14 + \cdots + \frac{1}{2^n} = 2 - \frac{1}{2^n}$. Inductively prove that the formula holds for all $n\in\Bbb{N}$:
$$1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}=2-\frac{1}{2^n}.$$
What I have so far:
base: n = 1: $$1+\frac{1}{2}=2-\frac{1}{2}=1.5$$
inductionstep: n = k: $$1+\frac{1}{2}+\cdots+\frac{1}{2^k}=2-\frac{1}{2^k}$$
inductionhypothesis: n=k+1: $$1+\frac{1}{2}+\cdots+\frac{1}{2^k+1}=1+\frac{1}{2}+\cdots+\frac{1}{2^k}+\frac{1}{2^(k+1)}=(2-\frac{1}{2})+\frac{1}{2^k+1}$$
This is where I am stuck and not sure what to do next.
| The formula
$$
1+\dots+\frac{1}{2^n}=2-\frac{1}{2^n}
$$
is true for $n=0$ (in which case the LHS of the equation actually only has one term in it). Therefore, it is better to choose $n=0$ as your base case. Now suppose it is true when $n=k$, that is
$$
1+\dots+\frac{1}{2^k}=2-\frac{1}{2^k} \, .
$$
Then, for $n=k+1$, we have
\begin{align}
1+\dots+\frac{1}{2^{k+1}}&=\left(1+\dots+\frac{1}{2^k}\right)+\frac{1}{2^{k+1}} \\[5pt]
&= 2-\frac{1}{2^k}+\frac{1}{2^{k+1}} \\[5pt]
&= 2-\frac{2}{2^{k+1}}+\frac{1}{2^{k+1}} \\[5pt]
&= 2-\frac{1}{2^{k+1}} \, .
\end{align}
Hence, if the statement is true for $n=k$, it is true for $n=k+1$. Since it is true for $n=0$, by the principle of mathematical induction it must be true for all nonnegative integers $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the quadratic equation from given relatioship the quadratic equation whose roots are a and b where $a^2 +b^2=5$ and
$3(a^5+b^5)=11(a^3+b^3)$
What I Tried
$a^2 +b^2=5$ $(a+b)^2-2ab=5$ $(\text{sum of roots})^2 -2(\text{products of roots})=5$
$3(a^5+b^5)=11(a^3+b^3)$ $a^3(3a^2-11)=b^3(11-3b^2)$
| There is no single answer here. It turns out there are three quadratics with real coefficients and two more with complex coefficients, the roots of which satisfy the given conditions.
The quadratic equation with roots $a$ and $b$ is $x^2-sx+p$, where $s=a+b$ and $p=ab$. Some careful algebra tells us
$$a^2+b^2=s^2-2p,\quad a^3+b^2=s^3-3sp,\quad\text{and}\quad a^5+b^5=s^5-5s^3p+5sp^2$$
If $3(a^5+b^5)=11(a^3+b^3)$, then
$$3s^5-11s^3-(15s^3-33s)p+15sp^2=0$$
Now if $a^2+b^2=5$, then $2p=s^2-5$, and the equation above (multiplied by $4$) becomes
$$12s^5-44s^3-2(15s^3-33s)(s^2-5)+15s(s^2-5)^2=0$$
which simplifies to
$$0=3s^5-22s^3+45s=s(s^2-9)(3s^2+5)$$
The real roots here are $s=0$, $3$, and $-3$, and these give values $p=-5/2$, $2$, and $2$, respectively, for quadratics
$$x^2-5/2,\quad x^2-3x+2,\quad\text{and}\quad x^2+3x+2$$
The complex roots are $s=\pm\sqrt{5/3}i$, both of which give $p=-10/3$, for quadratics
$$x^2-\sqrt{5/3}ix-10/3\quad\text{and}\quad x^2+\sqrt{5/3}ix-10/3$$
Remark: Any conditions of the form $a^2+b^2=A$, $B(a^5+b^5)=C(a^3+b^3)$, will lead to a quintic in $s$ with $s$ as one factor and a quadratic in $s^2$ as another. Presumably the problem at hand was designed so that $\{a,b\}=\{1,2\}$ and $\{-1,-2\}$ are among the roots that satisfy the given conditions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the maximum value of $\frac{2\sin\theta\cos\theta}{(1+\cos\theta)(1+\sin\theta)}$? Let $\theta\in\left(0,\frac{\pi}{2}\right)$, then find the maximum value of $\dfrac{2\sin\theta\cos\theta}{(1+\cos\theta)(1+\sin\theta)}$
I found the derivative, which is equal to $\dfrac{(\cos\theta−\sin\theta)(\sin\theta+\cos\theta+1)^2}{(1+\sin\theta+\cos\theta+\sin\theta\cos\theta)^2}$
But there were complicated calculations involved. Is there a simpler way to find the maximum of this function?
| Let $$f(\theta)=\frac{2\sin\theta\cos\theta}{(\sin\theta+1)(\cos\theta+1)}=\frac{2\sin\theta\cos\theta}{1+\sin\theta+\cos\theta+\sin\theta\cos\theta}$$
Taking $\sin2\theta=t,\ t\in(0,1],$ $$f(t)=\frac t{1+\sqrt{1+t}+\frac t2}=\frac{2t}{2+2\sqrt{1+t}+t}=\frac2{\left(\frac 1t+\sqrt{\frac 1t+1}\right)^2}$$
In order to maximize the given function in the question one needs to find the minimum value of the denominator of the expression after the manipulations done in the image. The minimum value of denominator will be obtained at t=1 and it's easy to see that. Putting t=1 gives the maximum value of function to be $6-4\sqrt2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 2
} |
How was the closed form of this alternating sum of squares calculated? I am reading through this answer at socratic.org.
The question is to find the closed form of the sum
$$1^{2}-2^{2}+3^{2}-4^{2}+5^{2}-6^{2}+\ldots.$$
I understand that, if the terms were added, the sum would be
$$
\sum_{n=1}^{N} n^{2}=1^{2}+2^{2}+\ldots+N^{2}.
$$
The person goes on to say, if the series were not alternating, the sum would be
$$
S=\frac{N(N+1)}{2}
$$
But is that correct? I thought the sum of the first $N$ squares would be
$$
\frac{N(N+1)(2N+1)}{6}.
$$
Lastly, I understand moving the $-1$ constant out of the summation as such
$$
=-\sum_{n=1}^{N}(-1)^{n} n^{2}
$$
But I am completely missing how the final closed form was calculated
$$
S_{N}=-\frac{(-1)^{N} N(N+1)}{2}
$$
| It is indeed not true that $\sum_{n=1}^Nn^2=\frac{N(N+1)}{2}$ for each positive integer $N$. This is easily verified by plugging in any value $N\geq2$. Of course it is true that $\sum_{n=1}^Nn=\frac{N(N+1)}{2}$. And you are indeed correct when you say that
$$\sum_{n=1}^Nn^2=\frac{N(N+1)(2N+1)}{6}.$$
Do note that if $N$ is even, say $N=2M$, then
\begin{eqnarray*}
\sum_{n=1}^N(-1)^{n+1}n^2
&=&\sum_{n=1}^M\big((-1)^{2n}(2n-1)^2+(-1)^{2n+1}(2n)^2\big).
\end{eqnarray*}
Of course $(-1)^{2n}=1$ for all $n$, and similarly $(-1)^{2n+1}=-1$. It follows that for all $n$ we have
\begin{eqnarray*}
(-1)^{2n}(2n-1)^2+(-1)^{2n+1}(2n)^2
&=&(2n-1)^2-(2n)^2\\
&=&(4n^2-4n+1)-4n^2\\
&=&1-4n.
\end{eqnarray*}
This shows that
$$\sum_{n=1}^N(-1)^{n+1}n^2=\sum_{n=1}^M(1-4n).$$
From here we can use the fact that $\sum_{n=1}^Mn=\frac{M(M+1)}{2}$ to find that
\begin{eqnarray*}
\sum_{n=1}^M(1-4n)&=&M-4\sum_{n=1}^Mn\\
&=&M-4\cdot\frac{M(M+1)}{2}\\
&=&-2M^2-M\\
&=&-\frac{N(N+1)}{2}.
\end{eqnarray*}
That proves the case where $N$ is even. If $N$ is odd, say $N=2M+1$, then
\begin{eqnarray*}
\sum_{n=1}^N(-1)^{n+1}n^2
&=&\left(\sum_{n=1}^{2M}(-1)^{n+1}n^2\right)+N^2\\
&=&-\frac{(N-1)N}{2}+N^2\\
&=&\frac{N(N+1)}{2}.
\end{eqnarray*}
This shows that although the reasoning is incorrect, the conclusion does indeed hold; that
$$\sum_{n=1}^N(-1)^{n+1}n^2=-(-1)^N\frac{N(N+1)}{2}.$$
Alternatively, you could note that
$$1^2-2^2+3^2-4^2+5^2-\ldots=(1^2+2^2+3^2+4^2+5^2+\ldots)-2(2^2+4^2+8^2+10^2+\ldots).$$
So from your observation that
$$\sum_{n=1}^Nn^2=\frac{N(N+1)(2N+1)}{6},$$
you could conclude that if $N$ is even, say $N=2M$, then
\begin{eqnarray*}
\sum_{n=1}^N(-1)^{n+1}n^2
&=&\left(\sum_{n=1}^{2M}n^2\right)-2\left(\sum_{n=1}^M(2n)^2\right)\\
&=&\frac{2M(2M+1)(4M+1)}{6}-8\frac{M(M+1)(2M+1)}{6}\\
&=&\frac{-12M^2-6M}{6}\\
&=&-\frac{N(N+1)}{2}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Simple but interesting problem about the binomial coefficient from Olympiad
"Let's define $a_n=\sum\limits_{k=0}^{\lfloor n/2 \rfloor} {n-k \choose k}\left(-\frac{1}{4}\right)^k$. Evaluate $a_{1997}$."
This problem is from the final round of an old South Korean Mathematical Olympiad (1997 KMO).
I think this problem is very simple, but requires some combinatoric ideas, and also is very interesting.
But as a lot of time has passed by, I cannot find any solutions or guidelines about it.
I tried to divide the explicit form of $(x+y)^{2k}$ with $x^k$, but it doesn't work well.
Would you help me?
| Snake oil:
\begin{align}
\sum_{n=0}^\infty a_n z^n &= \sum_{n=0}^\infty \left(\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n-k}{k}\left(-\frac{1}{4}\right)^k \right) z^n \\
&= \sum_{k=0}^\infty \left(-\frac{1}{4}\right)^k\sum_{n=2k}^\infty \binom{n-k}{k} z^n \\
&= \sum_{k=0}^\infty \left(-\frac{1}{4}\right)^k z^{2k}\sum_{n=0}^\infty \binom{n+k}{k} z^n \\
&= \sum_{k=0}^\infty \left(-\frac{z^2}{4}\right)^k\frac{1}{(1-z)^{k+1}} \\
&= \frac{1}{1-z}\sum_{k=0}^\infty \left(-\frac{z^2}{4(1-z)}\right)^k \\
&= \frac{1}{1-z}\cdot\frac{1}{1+\frac{z^2}{4(1-z)}} \\
&= \frac{1}{(1-z/2)^2} \\
&= \sum_{n=0}^\infty \binom{n+1}{1} (z/2)^n \\
&= \sum_{n=0}^\infty \frac{n+1}{2^n} z^n
\end{align}
So $a_n= (n+1)/2^n$ for $n \ge 0$.
Note that this approach derives the formula without the need to guess the pattern.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
The equation $x^k = -1$ over finite fields Let $k$ be a positive integer and let $F_q$ be a finite field with $q$ elements, and consider the equation $X^k + 1 = 0$. When does this equation has a solution in $F_q$? For example if $k =2$ and $q$ is a prime congruent to $3 \pmod{4}$ the equation has no solution in $F_q$. Does the same hold for other values of $k$ too?
| If $q$ is even then $x = 1$ is a solution, so assume $q$ is odd. Write $k = 2^a b$ where $b$ is odd. We can solve $x^k=-1$ iff we can solve $y^{2^a} = -1$: if $x^k = -1$ then $(x^b)^{2^a} = -1$, and conversely if $y^{2^a} = -1$ then $y^k = (-1)^b = -1$. Now $y^{2^a} = -1$ if and only if $y$ has order $2^{a+1}$, and because $F_q^\times$ is cyclic of order $q-1$ such an element exists if and only if $2^{a+1} \mid q-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to show $f\left(\frac{1}{2}\right) \le \frac{M}{4} + \int_0^1 f(x)\, dx$ Let $f$ be a differentiable function such that $f'(x)$ is continuous and $|f'(x)| \le M$ for all $x \in [0,1]$. Show that
\begin{align}
f\left(\frac{1}{2}\right) \le \frac{M}{4} + \int_0^1 f(x)\, dx
\end{align}
I have no idea how to begin proving this. I think Fundamental Theorem of Calculus will be used somewhere in the proof, but I can't see how that can be related to $f'(x)$. Any help is appreciated; thanks in advance!
| Let $f$ be as in the hypothesis. We know that $|f'|\leq M$. Therefore, it follows that for $x,s\in[0,1]$ with $s>x$
$$
-(s-x)M \leq \int_{x}^{s}f'(t) dx \leq (s-x)M.
$$
Indeed, by the Fundamental Theorem of Calculus, and by letting $s=\frac{1}{2}$, we find that
$$
f\left(\frac{1}{2}\right) - f(x) \leq \left(\frac{1}{2}-x\right)M. \tag{1}
$$
Integrating again over $[0,1/2]$ with respect to $x$, we find that
$$
\int_{0}^{1/2}f\left(\frac{1}{2}\right)dx \leq M\int_{0}^{1/2} \left(\frac{1}{2}-x\right) dx + \int_{0}^{1/2} f(x) dx
$$
Simplifying, we end up with
$$
\frac{1}{2}f\left(\frac{1}{2}\right) \leq M \left(\frac{x}{2}-\frac{x^2}{2}\right)_{x=0}^{x=1/2} + \int_{0}^{1/2} f(x) dx
$$
and in particular, that
$$
\frac{1}{2}f\left(\frac{1}{2}\right) \leq \frac{M}{8} + \int_{0}^{1/2} f(x) dx.
$$
Similarly, now instead integrating (1) over $[1/2,1]$, we have
$$
\frac{1}{2}f\left(\frac{1}{2}\right) \leq \frac{M}{8} + \int_{1/2}^{1} f(x) dx.
$$
Adding these two together gives you the desired result. In particular, we find that
$$\begin{align}
f\left(\frac{1}{2}\right) &= \frac{1}{2}f\left(\frac{1}{2}\right) + \frac{1}{2} f\left(\frac{1}{2}\right) \\
&\leq \frac{M}{8} + \frac{M}{8} + \int_{0}^{1/2} f(x) dx + \int_{1/2}^{1} f(x) dx \\
&= \frac{M}{4} + \int_{0}^{1} f(x) dx.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4240031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Difficult limit: $ \lim_{n \rightarrow \infty} n^{3/2} \int _0^1 \frac{x^2}{(x^2+1)^n}dx $ I found this question online
Find the limit:$$\lim_{n \rightarrow \infty} n^{3/2} \int _0^1 \frac{x^2}{(x^2+1)^n}dx $$
I've been told that I need to use the gamma function by converting $n^{3/2}=n\sqrt{n}$ and doing $u$ sub $u=x\sqrt{n}\rightarrow du=\sqrt{n} dx$
$$\lim_{n \rightarrow \infty} n^{3/2} \int _0^1 \frac{x^2}{(x^2+1)^n}dx =
\lim_{n \rightarrow \infty} \int _0^{\sqrt{n}} \frac{nx^2\sqrt{n}}{(x^2+1)^n}dx =
\lim_{n \rightarrow \infty} \int _0^{\sqrt{n}} \frac{u^2}{(\frac{u^2}{n}+1)^n}dx
$$
$$=\lim_{n \rightarrow \infty} \int _0^{\sqrt{n}} \frac{u^2}{(\frac{u^2}{n}+1)^n}dx
$$
How does one proceed from here to get the gamma function?
| Using the inequalities $x^2-\frac12 x^4\le \log(1+x^2)\le x^2$, we find that
$$\int_0^1 x^2e^{-nx^2}e^{-nx^4/2}\,dx\le \int_0^1 \frac{x^2}{(x^2+1)^n}\,dx\le \int_0^1 x^2e^{-nx^2}\,dx\tag1$$
Then, enforcing the substitution $x\mapsto x/\sqrt n$ in $(1)$ reveals
$$\frac1{n^{3/2}}\int_0^{\sqrt n} x^2e^{-x^2}e^{-x^4/2n}\,dx\le \int_0^1 \frac{x^2}{(x^2+1)^n}\,dx\le \frac1{n^{3/2}}\int_0^{\sqrt n} x^2e^{-x^2}\,dx\tag2$$
Finally, multiplying $(2)$ by $n^{3/2}$, letting $n\to \infty$, and applying the squeeze theorem yields the coveted limit
$$\lim_{n\to\infty}n^{3/2} \int_0^1 \frac{x^2}{(x^2+1)^n}\,dx=\frac{\sqrt \pi}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4242116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the square root of a polynomial with radical I have two positive integers, $a$ and $b$, such that
$$a^2=28b\sqrt{8b^2+1}+80b^2+5, \tag{$\star$}$$
and I’d like to find $a$ in terms of $b$ (including $\sqrt{8b^2+1}$, if necessary/appropriate), with a range like $$6b + \sqrt{8b^2+1} ≤ a ≤ 2b^2+ 3\sqrt{8b^2+1}$$ being an acceptable compromise.
I recently learned how to use Laurent series, evaluated at infinity, to approximate the square root of a polynomial, but can't figure out exactly how to apply that technique here (Wolfram gives somewhat unhelpful results) — or even figure out if it’s applicable [because of the radical].
Obviously, $b^2$ is a square triangular number (because $8b^2+1$ is a square), but I haven’t figured out how to use that detail to help me.
What’s the best approximation to the square root of ($\star$)?
| Suppose that $a$ and $b$ are integers such that
$$a^2=28b\sqrt{8b^2+1}+80b^2+5.$$
Then in particular $\sqrt{8b^2+1}$ is an integer, and so $8b^2+1=c^2$ for some integer $c$. Then
$$a^2=28bc+10c^2-5,$$
and a bit of rearranging shows that
$$(10c^2-(a^2+5))^2=(28bc)^2=28^2b^2c^2=98(8b^2)c^2=98(c^2-1)c^2.$$
Expanding the left hand side and rearranging, we see that $c^2$ is a root of the quadratic
$$2X^2-(20a^2+2)X+(a^2+5)^2=0.$$
As $c^2$ is an integer the discriminant $\Delta$ of this quadratic is a perfect square, where
$$\Delta=(20a^2+2)^2-4\cdot2\cdot(a^2+5)^2=14^2(2a^4-1).$$
It follows that also $2a^4-1$ is a perfect square, say $d^2$, and so
$$d^2-2a^4=-1.$$
This shows that $(X,Y)=(d,a^2)$ is a solution to the Pell equation $X^2-2Y^2=-1$, and hence that $a^2$ is a Pell number that is a perfect square. That $a=13$ is the only such number is shown in this article.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Number of roots of the equation $ax^2+ bx + c = 0$ in $(1,2)$
Let $a, b, c \in R, a \ne 0$ such that $a$ and $4a + 3b + 2c$ have the same sign. Then the
number of roots of the equation $ax^2+ bx + c = 0$ that lie(s) in $(1,2)$ is(are)?
I began by writing that $4a^2+3ab+2ac>0$. I tried finding the sign of $f(1)\cdot f(2)$ to use this condition and Bolzano's theorem. Unfortunately, I couldn't find a way to proceed.
| One can say that if the number of the roots lying in $(1,2)$ is $2$, then the axis of symmetry of the parabola $y=ax^2+bx+c$ lies in $(1,2)$.
Therefore, one can say that if the axis of symmetry of the parabola $y=ax^2+bx+c$ does not lie in $(1,2)$, then the number of the roots lying in $(1,2)$ is not $2$.
One has
$$0\lt \frac{2a(4a+3b+2c)}{4a^2}\leqslant \frac{8a^2+6ab+b^2}{4a^2}=\bigg(1-\bigg(-\frac{b}{2a}\bigg)\bigg)\bigg(2-\bigg(-\frac{b}{2a}\bigg)\bigg)$$
i.e.
$$\bigg(1-\bigg(-\frac{b}{2a}\bigg)\bigg)\bigg(2-\bigg(-\frac{b}{2a}\bigg)\bigg)\gt 0$$
which means that the axis of symmetry of the parabola $y=ax^2+bx+c$ does not lie in $(1,2)$.
Therefore, one can say that the number of the roots lying in $(1,2)$ is not $2$.
If $a=1,b=1$ and $c=-1$ satisfying $a(4a+3b+2c)\gt 0$, then the number of roots lying in $(1,2)$ is $0$.
If $a=1,b=1$ and $c=-3$ satisfying $a(4a+3b+2c)\gt 0$, then the number of roots lying in $(1,2)$ is $1$.
Added :
If $a(4a+3b+2c)\gt 0$ and $b^2-4ac=0$, then the number of the roots lying in $(1,2)$ is $0$.
Under $a(4a+3b+2c)\gt 0$ and $b^2-4ac\gt 0$, the number of the roots lying in $(1,2)$ is
$$\begin{cases}0&\text{if $\ (a+b+c)(4a+2b+c)\geqslant 0$}
\\\\ 1&\text{otherwise}\end{cases}$$
This can be seen by noting that the axis of symmetry of the parabola $y=ax^2+bx+c$ does not lie in $(1,2)$ and that $y=ax^2+bx+c$ is monotone on $(1,2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Prove that $\int_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$ Prove that $$\int\limits_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$$
On simplifying by parts we get:
$$\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\frac{1}{2(n+2)}+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$$
Thus if we prove that$$\max\left(\displaystyle\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx\right)<\frac{1}{2(n+1)(n+2)}$$
We will be able to prove the above inequality
| $$\int_0^1 \frac{x^{n+1}}{x+1}\le \frac{1}{2}\int_0^1\frac{x^{n+1}+x^{n}}{x+1}=\frac{1}{2(n+1)}$$ here we used when $x\in [0,1] $ $x^{n+1}\le x^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Points of discontinuity of $\left[\frac{f(x)}3\right], f(x)=\lim_{n\to\infty}\ln(\sqrt{e^{\cos x}\sqrt{e^{3\cos x}...\sqrt{e^{(2n+1)\cos x}}}})$
Let $f(x)=\lim_{n\to\infty}\ln\left(\sqrt{e^{\cos x}\sqrt{e^{3\cos x}\sqrt{e^{5\cos x}...\sqrt{e^{(2n+1)\cos x}}}}}\right)$ and if $g(x)=\left[\frac{f(x)}3\right]$, then the number of points in $[0,2\pi]$ where $g(x)$ is discontinuous, is/are (where $[*]$ represents greatest integer function)...
$\cos x$ varies from $-1$ to $1$.
If $\cos x=-1$ then as $n\to\infty$, $e^{(2n+1)\cos x}=0$
If $\cos x=0$ then as $n\to\infty$, $e^{(2n+1)\cos x}=1$
If $\cos x=\infty$ then as $n\to\infty$, $e^{(2n+1)\cos x}=\infty$
Also, since $\cos x$ is changing these values in continuous manner, can we use that to comment about the continuity of $f(x)$?
But before that, I would need to solve the limit. Not able to do so. Can you give any pointers? Thanks.
EDIT: $f(x)=\lim_{n\to\infty}\ln\left(e^{\left(\frac12+\frac3{2^2}+\frac5{2^3}+...+\frac{(2n+1)}{2^{n+1}}\right)\cos x}\right)$
Let $$S=\frac12+\frac3{2^2}+\frac5{2^3}+...+\frac{(2n+1)}{2^{n+1}}\\\frac S2=\frac1{2^2}+\frac3{2^3}+\frac5{2^4}+...+\frac{(2n+1)}{2^{n+2}}$$
Subtracting, $$\frac S2=\frac12+\frac12+...\text{(n+1) times}-\frac{2n+1}{2^{n+2}}=\frac {n+1}2-\frac{2n+1}{2^{n+2}}\\\implies S=n+1-\frac{2n+1}{2^{n+1}}$$
I think my $S$ is coming out to be $\infty$. Can you point out the mistake? Thanks.
EDIT$2$: After reading the comments, I reworked and found $S=3$.
Thus, $f(x)=3\cos x\implies\left[\frac{f(x)}3\right]=[\cos x]$
Thus, there are $4$ points of discontinuity $\{0,\frac\pi2,\frac{3\pi}2,2\pi\}$.
| Hint:
Let $r = e^{\cos(x)}$. Then verify by induction or otherwise that
$$ \sqrt{r \sqrt{r^3 \sqrt{r^5 \ldots \sqrt{r ^{2n+1} }}} } = r^{\frac12+ \frac{3}{2^2} + \ldots+ \frac{2n+1}{2^{n+1}} }$$
Now evaluate $S(n) = \sum_{k=0}^{n}\frac{2k+1}{2^{k+1}}$ using this and geometric series.
Then evaluate $f(x) = \lim_{n\to \infty} \ln( e^{S(n)\cos(x)}) = \lim_{n\to \infty} S(n)\cos (x)$ then proceed to find the answer.
Update
I think there is a problem in calculating $S(n)$. To evaluate it
use the formula given here, then we see that
$$ \sum_{k=0}^{n}\frac{k}{2^k} = \frac{2^{n + 1} - n - 2}{2^{n}}$$
So $$S(n) = \frac{{3 \cdot 2^{n + 1} - 2 \, n - 5}} {2^{n + 1}}$$
Now can you find the $\lim_{n\to \infty}S(n)$?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to evaluate$J(k) = \int_{0}^{1} \frac{\ln^2x\ln\left ( \frac{1-x}{1+x} \right ) }{(x-1)^2-k^2(x+1)^2}\text{d}x$ I am trying evaluating this
$$J(k) = \int_{0}^{1} \frac{\ln^2x\ln\left ( \frac{1-x}{1+x} \right ) }{(x-1)^2-k^2(x+1)^2}\ \text{d}x.$$
For $k=1$, there has
$$J(1)=\frac{\pi^4}{96}.$$
Maybe $J(k)$ doesn't have an explicit closed-form.
An integral relation:
$$\int_{0}^{\infty} \frac{\arctan^3x}{1+k^2x^2} \text{d}x
=-\frac{3}{2}J(k)+\frac{\pi^2}{8k}
\left ( \operatorname{Li}_2\left ( \frac{1}{k} \right )
-\operatorname{Li}_2\left ( -\frac{1}{k} \right ) \right )
+\frac{\ln k}{8k} \ln^3\left ( \frac{k-1}{k+1} \right ),\qquad (k>1).$$
With some calculations, we followed that
$$\int_{0}^{\infty} \frac{\arctan^3x}{1+k^2x^2} \text{d}x
=\frac{\pi^4}{64k}
+\frac{3}{4k}\left ( \operatorname{Li}_4\left ( \frac{k-1}{k+1} \right )
-\operatorname{Li}_4\left (- \frac{k-1}{k+1} \right )
\right ) +\frac{3\pi^2}{8k} \operatorname{Li}_2\left (- \frac{k-1}{k+1} \right ),\qquad(k>1).$$
Then the final result of $J(k)$ is
$${\color{Green}{J(k)
=\frac{\pi^2}{12k}
\left ( \operatorname{Li}_2\left ( \frac{1}{k} \right )
-\operatorname{Li}_2\left ( -\frac{1}{k} \right ) \right )
+\frac{\ln k}{12k} \ln^3\left ( \frac{k-1}{k+1} \right )
-\frac{\pi^4}{96k} -\frac{1}{2k}\left ( \operatorname{Li}_4\left ( \frac{k-1}{k+1} \right )
-\operatorname{Li}_4\left (- \frac{k-1}{k+1} \right )
\right ) -\frac{\pi^2}{4k} \operatorname{Li}_2\left (- \frac{k-1}{k+1} \right ),\qquad (k>1).}}$$
| Since you have a relation with another integral, here are 2 Maclaurin series representations for a general anti derivative which you can then use to solve for J(x). Here is a Demo Graph and Integral Demo:
$$\int \frac{\arctan^3(x)}{k^2x^2+1}dx=\int \sum_{n=0}^\infty\frac{\left(\frac{d^n}{dx^n} \frac{\arctan^3(x)}{k^2x^2+1}\right)_{x=0}\ x^n}{n!}= \sum_{n=4,6,8,…}^\infty\frac{\left(\frac{d^{n-1}}{dx^{n-1}} \frac{\arctan^3(x)}{k^2x^2+1}\right)_{x=0}\ x^n}{n!}=C+\frac{x^4}{4}-\frac{(k^2+1)x^6}{6}+\frac{(15k^4+15k^2+14)x^8}{120}-\frac{(471k^6+4712k^4+ 441k^2+409)x^{10}}{4725}+\frac{211(4725k^8+4725k^6+4410k^4+4090k^2+3807)x^{12}}{11975040} $$
Here is a simpler series expansion related to the main inverse tangent expansion. The integration includes the Lerch Zeta function:
$$ \int\frac{\arctan^{3}(x)}{k^2 x^2+1}dx= \int \sum_{n=0}^\infty\frac{\left(\frac{d^n}{dx^n} \arctan^3(x)\right)_{x=0}\ }{n!} \int \frac{x^n}{k^2x^2+1}dx= \sum_{n=1}^\infty\frac{\left(\frac{d^{n-1}}{dx^{n-1}} \arctan^3(x)\right)_{x=0}\ }{(n-1)!}x^{n}\frac{1}{2}Φ\left(-k^2 x^2,1, \frac{1}{2}\right) $$
The nth derivative has a simple expansion.
From this you can possibly use it to solve for J(x). I have more to add.
| {
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Is there a unique partial fraction decomposition for every rational polynomial Consider this
$$({2x-3})/({x^2-1})(2x+3)$$
Here can i do decomposition as
$$Ax+B/(x^2-1)+C/(2x+3)$$
Instead of $$A/(x-1)+B/(x+1)+C/(2x+3)$$
And if not then why?
| Note tha in your case there are three unknowns in RHS. If you simplify yowill get
$A(x+1)(2x+3)+B(x-1)(2x+3)+C(x^2-1)=0x^2+2x-3$. Simplify this and equate the cpefficients of the same powers of $x$ on both sides, you get three eqns. to find $A,B,C$.
The process of partial fractions is interesting and requires some rules and it yields the unknown. $F(x)=\frac{P_m(x)}{P_n(x)}, m<n$. $P_n(s)$ should contain linear factors , yheir powers or quadratic factors. For a repeated linear factor, we write $$\frac{A}{(x-a)}+\frac{B}{(x-a)^2}+\frac{C}{(x-a)^3}$$ for $$\frac{1}{(x-a)^{3}}$$
For $\frac{1}{ax^2+bx+c}$, we write $\frac{Ex+G}{ax^2+bx+c}.$
For $\frac{1}{(ax^2+bx+)^2}$, we write $\frac{Ex+G}{ax^2+bx+c}+\frac{Ux+V}{(ax^2+bx+c)^2}.$
For a good example: $$\frac{1}{(x-1)(x-2)^2(x^2+1)^2(x^2+3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{(x-2)^2}+\frac{Dx+E}{x^2+3}+\frac{Gx+H}{x^2+1}+\frac{Jx+K}{(x^2+1)^2}.$$
Thre are 9 unknowns and then there will be 9 equations. Some time one may also go for $x^2+1=(x+i)(x-i).$ Normally, partial fractions have linear or quadratic factors.
| {
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Replacing 8 red balls with blue balls You have a bag with 8 red balls in it.
Every turn you randomly select a ball from the bag. If it is red you replace it with a blue ball, and if it is blue you replace it with a red ball.
What is the expected number of turns you need to make before all 8 balls in the bag are blue?
If the answer is $\frac{a}{b}$, where a and b are coprime positive integers, what is $a+b$?
My Approach:
$E_n$ is the expected number of moves given that you have $n$ blue balls
$E(7) = \frac{1}{8}(1) + \frac{7}{8}[1 + E(6)]$
$E(6) = \frac{2}{8}[1 + E(7)] + \frac{6}{8}[1 + E(5)]$
$E(5) = \frac{3}{8}[1 + E(6)] + \frac{5}{8}[1 + E(4)]$
$E(4) = \frac{4}{8}[1 + E(5)] + \frac{4}{8}[1 + E(3)]$
$E(3) = \frac{5}{8}[1 + E(4)] + \frac{3}{8}[1 + E(2)]$
$E(2) = \frac{6}{8}[1 + E(3)] + \frac{2}{8}[1 + E(1)]$
$E(1) = \frac{7}{8}[1 + E(2)] + \frac{1}{8}[1 + E(0)]$
$E(0) = \frac{8}{8}[1 + E(1)] $
But in this way it doesn't seem to go well.
| I am not finding anything wrong in the equations.
Putting it in a Jordan- Gauss calculator, I get an answer of $\dfrac{a}{b}= \dfrac{32768}{105}$
Thus $a+b= 32873$
The calculator is here
I strongly advise you to check the computations.
| {
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how to find the smallest parameter for trigonometric equation? Find the smallest value of parameter $\alpha$ such that equation $${\mathrm{sin}}^{2}x \times \mathrm{cos}2x + \alpha ({\mathrm{cos}}^{4}x - {\mathrm{sin}}^{4}x) = -10(2\alpha + 1{)}^{2}$$ has at least one real solution.
simplifying the expression I got $${\mathrm2{sin}}^{4}x - \mathrm{sin}^{2}x(1-2a) - (a+40a^2 + 40a + 10) = 0 $$
And solved $$ (1-2a)^2 + 8(a + 40a^2 + 40a + 10) = 0 $$
At the end I have a = - 1/2. There is only 1 answer?
| To find solutions to the trigonometric equation that are real, we solve the equation with respect to $sin(x)^{2}$; we get two roots:
$sin(x)^{2}=-\frac{9|2\alpha+1|}{4}-\frac{2\alpha-1}{4}$,
$sin(x)^{2}=+\frac{9|2\alpha+1|}{4}-\frac{2\alpha-1}{4}$.
We impose that
$0≤sin(x)^{2}≤1$.
We then write the system:
$0≤-\frac{9|2\alpha+1|}{4}-\frac{2\alpha-1}{4}≤1$,
$0≤+\frac{9|2\alpha+1|}{4}-\frac{2\alpha-1}{4}≤1$.
The solution is:
$-\frac{3}{5}≤\alpha≤-\frac{2}{5}$.
| {
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Given the triangle ABC. Let BC = a, AC = b, AB = c. Find the minimum value of the following expression Given the triangle ABC. Let BC = a, AC = b, AB = c. Find the minimum value of the following expression:
a) $$P=\frac{4a}{b+c-a} + \frac{9b}{c+a-b} + \frac{16c}{a+b-c}$$
b) $$P=\frac{a^3}{2a+bc} + \frac{b^3}{2b+ac} + \frac{c^3}{2c+ab}, a+b+c=2$$
Only use AM-GM and Cauchy-Schwarz inequalities.
The problem here is that a, b, c are the 3 sides of the triangle so it has the triangle inequality so I can't find the equal condition.
Edit: My attempt:
a) I tried multiplying each fraction with the numerator variable, and then use Cauchy-Schwarz Engel's form and then it will become $$P\geq\frac{(2a+3b+4c)^2}{2ab+2bc+2ac-a^2-b^2-c^2}$$Let the RHS be Q. After that I use AM-GM with $2ab,2bc,2ac$ and then it will become $$Q\geq\frac{(2a+3b+4c)^2}{a^2+b^2+c^2}$$And then I'm stuck.
b) I tried dividing each fraction with the numerator, and then use Cauchy-Schwarz and get $$P\geq\frac{4}{6+\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}}$$And I'm stuck too.
| For the second problem, notice that $ 2a+bc = (a+b+c)a+bc = (a+b)(a+c)$. This allows us to normalize the inequality and (mostly) forget the condition. WTS
$$ \sum \frac{ a^3(b+c) } { (a+b)(b+c)(c+a) } \geq \frac{1}{2} $$
Approach 1: It is almost immediately obvious that by AM-GM / Muirhead (since there are no $a^4$ terms on the RHS, so $(3,1,0)$ majorizes everything),
$$ \sum a^3 (b+c) \geq \frac{1}{4} ( a+b)(b+c)(c+a)(a+b+c) $$
Hence the result follows.
Approach 2 via CS: WTS $$ \sum \frac{ a^3}{(a+b)(a+c) } \geq \frac{a+b+c } { 4 } $$
CS gives us
$$\sum \frac{a^4}{ a(a+b)(a+c) } \geq \frac{ (a^2+b^2+c^2)^2 } { \sum a(a+b)(b+c) } $$
Hence, it RTS (after cross multiplying and simplifying)
$$ \sum_{sym} 3/2 a^4 + 3 a^2b^2 \geq \sum_{sym} 2a^3b + 5/2 a^2bc$$
This is true because $(4, 0, 0 ) + (2, 2, 0 ) \geq 2 (3, 1, 0), \frac{1}{2}(4, 0, 0 )\geq \frac{1}{2} (2, 1, 1), (2, 2, 0 ) \geq (2, 1, 1)$
| {
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From a group of three biologists, two physicists and one mathematician, a committee of two people is to be randomly selected. From a group of three biologists, two physicists and one mathematician, a committee of two people is to be randomly selected. Denote by $X$ the random variable representing the number of biologists and by $Y$ the random variable representing the number of physicists on the committee. Calculate: $f_{X, Y}$, $f_{X}$, $f_{Y}$.
Attempt
$$f_{X}=\frac{\binom{3}{1} \binom{2}{1}}{\binom{5}{2}}+\frac{\binom{3}{1} \cdot 1}{\binom{5}{2}}$$
$$f_{Y}=\frac{\binom{2}{1} \binom{3}{1}}{\binom{5}{2}}+\frac{\binom{2}{1} \cdot 1}{\binom{5}{2}}$$
Then, $$f_{X, Y}=f_{X}+f_{Y}$$
Am I understanding the problem?
| No that is not correct. You are supposed to find probability mass function $ \small f_X(x), f_Y(y), f_{XY} (x, y)$. Below is a tabular representation of what you need to come up with,
$ \small \begin{array}{|c|c|c|}
\hline
& Y=0 & Y=1 & Y=2 & f_X\\
\hline
X=0 & & \\
\hline
X=1 & & \\
\hline
X=2 & & \\
\hline
f_Y & & & & - \\
\hline
\end{array}$
We have $3$ biologists, $2$ physicists and $1$ mathematician. We can first find pmf $ \small f_{XY}(x, y)$. For example,
$\small f_{XY}(X = 0, Y = 0) = 0 ~ $ as a two member committee cannot be formed with just mathematician.
$ \displaystyle \small f_{XY}(X = 1, Y = 0) = {3 \choose 1} {1 \choose 1} / {6 \choose 2} ~ $ as we choose one of the biologists and the other member is mathematician.
$ \displaystyle \small f_{XY}(X = 2, Y = 0) = {3 \choose 2} / {6 \choose 2}$
Similarly find
$\small f_{XY}(X = 0, Y = 1), f_{XY}(X = 1, Y = 1), f_{XY}(X = 0, Y = 2)$
Then,
$ \small f_Y(Y = 0) = f_{XY}(X = 0, Y = 0) + f_{XY}(X = 1, Y = 0) + f_{XY}(X = 2, Y = 0)$
You can similarly find $ \small f_Y(Y = 1), f_Y(Y = 2), f_X(X = 0), f_X(X = 1)$ and $ \small f_X(X = 2)$
| {
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How to solve $ \sin(x) - \cos(x) -3\sin(2x) +3\cos(2x) +\sin(3x) -\cos(3x) =0$ Hello friends please help me to solve this equation.
$ \sin(x) - \cos(x) -3\sin(2x) +3\cos(2x) +\sin(3x) -\cos(3x) =0.$
Here is what I have done so far
$ \Rightarrow \sqrt2(\frac{1}{\sqrt2}\sin x - \frac{1}{\sqrt2}\cos x ) - 3\sqrt2(\frac{1}{\sqrt2}\sin( 2x) - \frac{1}{\sqrt2}\cos (2x) )+ \sqrt2(\frac{1}{\sqrt2}\sin x - \frac{1}{\sqrt2}\cos x )=0$
$\Rightarrow 2(\frac{1}{\sqrt2}\sin x - \frac{1}{\sqrt2}\cos x ) - 3(\frac{1}{\sqrt2}\sin(2x) - \frac{1}{\sqrt2}\cos (2x) )=0$
$\Rightarrow 3\cos(\frac{\pi}{4} + 2x) -2\cos(\frac{\pi}{4}+x) =0$
Please help me to continue this method and I'd there is an easier way please tell me it too.
| We use $$\sin A+\sin B=2\sin \left( \frac {A+B}{2}\right) \cos \left(\frac {A-B}{2}\right)$$ and $$\cos A+\cos B=2\cos \left( \frac {A+B}{2} \right) \cos \left( \frac {A-B}{2} \right)$$
Hence, $$(\sin x+\sin 3x)-3\sin 2x=2\sin 2x \cos x-3\sin 2x=\sin 2x(2\cos x-3)$$
Moreover, $$(\cos x+\cos 3x)-3\cos 2x=2\cos 2x \cos x-3\cos 2x=\cos 2x(2\cos x-3)$$
Thus, the left hand side of your equation is-
$$\sin 2x(2\cos x-3)-\cos 2x(2\cos x-3)=(\sin2x-\cos 2x)(2\cos x-3)$$
This has been equated to $0$, hence solutions can be found by solving $\cos x=\frac 32$ or $\tan 2x=1$. Note that $\cos x=\frac 32$ has no real solutions.
| {
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Solving $\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 $. Simply bringing it to a common denominator does not lead me to success How can I solve this equation?
$$\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 $$
Simply bringing it to a common denominator does not lead me to success
What I tried
| We have that for $(4+x)(4+2x)(4+3x)\neq0 $
$$\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 \iff 18 x^3 + 165 x^2 + 378 x + 255=0$$
$$\iff 6 x^3 + 35 x^2 + 126 x + 85=0$$
and by rational root theorem we can find that $x=-\frac 5 3$ is a root and then we obtain
$$6 x^3 + 35 x^2 + 126 x + 85=(3 x + 5) (2 x^2 + 15 x + 17)$$
| {
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Prove that there are don't exist integers $a,b,c$ such that for every integer $x$ the number $A=(x+a)(x+b)(x+c)-x^3-1$ is divisible by $9$. Problem: Prove that there are don't exist integers $a,b,c$ such that for every integer $x$ the number $$A=(x+a)(x+b)(x+c)-x^3-1$$ is divisible by $9$.
We can consider $0\le a,b,c<9$ as we only care $\pmod 9.$
So $(x+a)(x+b)(x+c)\equiv 1,2,0\mod 9.$
Note that $$\frac{\prod_{x=0}^8 (x+a)}{(a+2)(a+5)(a+8)}\cdot \frac{\prod_{x=0}^8(x+b)}{(b+2)(b+5)(b+8)}\cdot \frac{\prod_{x=0}^8(x+c)}{(c+2)(c+5)(c+8)}= [1\cdot 2\cdot 4\cdot 5\cdot 7\cdot 8]^3\equiv -1^3\equiv -1 $$
I am stuck. Solutions are appreciated.
| The $0 \leq a,b,c<9$ step is fine, after all if $a,b,c$ satisfy the equation then so do $a \%9,b\%9$ and $c\%9$. (Where $\%9$ denotes the remainder upon division by $9$).
I don't think, however, that the step that realizes that $x^3+1$ can only be $0,1,2$ and reflects this on the RHS of the modulo, is a good step (even if it is correct), because this step loses the individuality of the term $x^3+1$. Note that the equations given by $$
(x+a)(x+b)(x+c) \equiv x^3+1 \pmod{9} \\
(x+a)(x+b)(x+c) \equiv 0,1,2 \pmod{9}
$$
are different equations, and it's likely that the second is significantly weaker than the first one, weak enough to actually admit solutions. I don't think the fact that the $x^3+1$ term depends on $x$ can be compromised, which is what that particular step was doing.
If we went with $(x+a)(x+b)(x+c) \equiv x^3+1 \pmod{9}$ for all $x$ (i.e. $x=0,1,...,8$) , then this is likelier to be helpful, particularly when you substitute values of $x$ and check for inconsistencies. Note that the above holds for all $x$ if and only if it holds for $x=0,1,2,...,8$.
The way we use the term $x^3+1$ , for example, is to find when it's zero. For what values of $x$ is $x^3+1 \equiv 0 \pmod{9}$? You can check that $x=2,5,8$ are the solutions to this equation.
Therefore, $(2+a)(2+b)(2+c) \equiv 0 \pmod{9}$. In particular, either :
*
*At least one of the terms is a multiple of $9$, or :
*At least two of the terms are multiples of $3$.
This translates to : either one of $a,b,c = 7$ OR at least two of $a,b,c$ are among the values $1,4,7$.
Repeating the same arguments with $2$ replaced by $5,8$ , which are the other numbers that yield $x^3+1= 0$ would give us that (putting the $x=2$ conclusion in here as well) all of these conclusions are true :
*
*Either one of $a,b,c = 1$ OR at least two of $a,b,c$ are among the values $1,4,7$.
*Either one of $a,b,c = 4$ OR at least two of $a,b,c$ are among the values $1,4,7$.
*Either one of $a,b,c = 7$ OR at least two of $a,b,c$ are among the values $1,4,7$.
What can we conclude about $a,b,c$ from here? I claim that we can conclude that at least two of $a,b,c$ are among the values $1,4,7$. I'll hide the answer below if you can't deduce this from the given statements.
If this were NOT the case, then the first part (coming before the OR) of each of the above statements must be true, but in that case we know that at least one of $a,b,c$ is $1$ and $4$ and $7$, so $a,b,c$ must be $1,4,7$ in some order, a contradiction because then the second part is satisfied! Thus, it is definitively true that at least two of $a,b,c$ are $1,4,7$ in some order.
We then substitute $x=0$ in the original equation to get that $abc \equiv 1 \pmod{9}$. Now, the symmetry of $a,b,c$ means that it's enough to show that each of the cases
*
*$a = 1,b=4$
*$a=1,b=7$
*$a=4,b=7$
*$a=1,b=1$
*$a=4,b=4$
*$a=7,b=7$
are impossible. But there is , in fact, something happening here which is very interesting.
You see, IF $a=1,b=4$, then $abc \equiv 1 \pmod{9}$ forces $c=7$, since this is the only value of $c$ which can satisfy this equation! Similarly, if $a=1,b=7$, then $c=4$ is forced, and similarly the third case.
For the fourth case, we are led to similar conclusions : indeed, each of $1,4,7$ satisfy $x^3 \equiv 1 \pmod{9}$, therefore IF any of the last three cases held, we must have $a=b=c = 1,4,7$.
Therefore, we conclude that if we consider the equations for just $x=0,2,5,8$, $a=1,b=4,c=7$ (or a symmetric change-up) and $a=b=c=1,4,7$ are the only possibilities for solutions to the initial equation . We've just used four values of $x$.
So if we can prove that the equation
$$
(x+1)(x+4)(x+7) \equiv x^3+1 \pmod{9}
$$
doesn't hold for some other $x$ which isn't $0,2,5,8$, then we are done with the first case. What better $x$ than $x=1$ to choose, where the RHS is $2$ and the LHS is $2 \times 5 \times 8 = 80$, and of course $80-2$ is not a multiple of $9$.
For the other cases, you can check that $(x+a)^3 = x^3+1 \pmod{9}$ is contradicted by $x=1$, for each of $a=1,4,7$ as well. In particular, the equation cannot hold for any value of $a,b,c$, for just the values $x=0,1,2,5,8$ alone. This proves the result.
| {
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Making $6$ digits numbers by the digits $2,3,3,5,5,5,5$
How many $6$ digits number we can generate by the digits $2,3,3,5,5,5,5$ ?
To solve this problem I considered three cases because we have $7$ digits and looking for six digit number.
*
*Numbers without $2\Rightarrow\quad\dfrac{6!}{4!2!}=15$
*Numbers without one $3\Rightarrow\quad\dfrac{6!}{4!}=30$
*Numbers without one $5\Rightarrow\quad\dfrac{6!}{3!2!}=60$
So the answer is $15+30+60=105$.
But The book I'm reading from used another method:
Number of $6$ digits permutation from these $7$digits=Number of $7$ digits permutation from these $7$digits= $\dfrac{7!}{2!4!}=105$.
But I don't completely why permutation of $6$ digits is equal to permutation of all the digits.
| Once you have arranged the first six digits, there is only one choice for the seventh digit. Thus, the number of arranging all seven digits also gives the total number of ways of arranging the first six of those seven digits.
| {
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How to find the derivative with respect to $x^2$? I would like to find the derivative of this function with respect to x^2.
$-0.5\ln({2\pi x^{2}}) - \frac{1}{2 x^{2}}$
The answer I got is:
$-(0.5)(\frac{1}{2\pi x^{2}})(2\pi)(2) - (0.5)(-1)(\frac{1}{x^{4}})(2)$
$=-\frac{1}{x^{2}} + \frac{1}{x^{4}}$
Is it correct?
| If in some problems you are having difficulties then you can use chain rule.
y = $-0.5\ln({2\pi x^{2}}) - \frac{1}{2 x^{2}}$ = $-0.5ln(2{\pi}) - ln(x) - \frac{x^{-2}}{2} $
To find: $\frac{dy}{dx^2}$
$\frac{dy}{dx}\frac{dx}{dx^2}$ = $\frac{dy}{dx}\left({\frac{dx^2}{dx}}\right)^{-1}$ = $(2x)^{-1}\frac{dy}{dx}$
$\frac{dy}{dx^2}$ = $\frac{1}{2x}$$(-\frac{1}{x}+x^{-3})$
Or Simply:
$-0.5\frac{1}{2\pi{x^2}}2{\pi}$ - $\frac{1}{2}(-\frac{1}{x^4})$ = $-\frac{0.5}{x^2}$ +$\frac{1}{2x^4}$
| {
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Prove that $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge ab + bc + ca$ For all positive $a,b,c $ satisfying $a+b+c = 3$,Prove:
$$
\sum_{cyc} \sqrt[3]{a} \ge \sum_{cyc} ab
$$
This is a hard problem and I tried it myself, but it's really hard without using advanced techniques(e.g. EV theorem or HCF theorem).
Is there any easy way without computer and HCF to this problem?
(Updated) Sorry for that I posted this in a hurry. I will show you more in this updated part.
First of all, HCF kills it.(If you don't know it, click here) Consider
$$
2\sum_{cyc} ab = (\sum_{cyc} a)^2 - \sum_{cyc} a^2 = 9 - \sum_{cyc} a^2
$$
Thus, we only need to prove
$$
\sum_{cyc} (a^2 + 2 \sqrt[3]{a}) \ge 9
$$
or
$$
\sum_{cyc} f(a) \ge 9
$$
where $f(x)=x^2 + 2\sqrt[3]{x}$. And because $f''(x)=2-\frac{4}{9\sqrt[3]{x^5}}$, Thus we know $f(x)$ is convex on $\left [ \frac{\sqrt[5]{8}}{3\sqrt[5]{3}}, 3\right)$. So by HCF theorem we only need to prove when $b=a$,$c=3-2a$ or
$$
6a^2 - 12a + 9 + 4\sqrt[3]{a} +2 \sqrt[3]{3-2a} \ge 9
$$
The rest is easy with derivative.
Second, a friend of mine told me that using tangent line can solve it. But I didn't see how that's relevant.
Third, I found this problem on the Internet, but I'm sure I've solved it or saw the solution to it before.
| I think, TL method does not help here.
Another way.
Let $a=x^3$, $b=y^3$ and $c=z^3$.
Thus, $x^3+y^3+z^3=3$ and we need to prove that:
$$x+y+z\geq x^3y^3+x^3z^3+y^3z^3$$ or
$$(x+y+z)^3(x^3+y^3+z^3)^5\geq243(x^3y^3+x^3z^3+y^3z^3)^3.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, we need to prove that $f(w^3)\geq0,$ where
$$f(w^3)=u^3(9u^3-9uv^2+w^3)^5-(9v^6-9uv^2w^3+w^6)^3.$$
But, $$f'(w^3)=5u^3(9u^3-9uv^2+w^3)^4+2(9uv^2-2w^3)(9v^6-9uv^2w^3+w^6)^2\geq0,$$ which says that $f$ increases and it's enough to prove our inequality for the minimal value of $w^3$.
Now, $x$, $y$ and $z$ are positive roots of the equation
$$(t-x)(t-y)(t-z)=0$$ or
$$t^3-3ut^2+3v^2t-w^3=0,$$ which says that a graph of $f(t)=t^3-3ut^2+3v^2t$ and the line $g(t)=w^3$ have three common points.
But $$f'(t)=3t^2-6ut+3v^2=3(t-(u+\sqrt{u^2-v^2}))(t-(u-\sqrt{u^2-v^2})),$$
which gives $t_{min}=u+\sqrt{u^2-v^2}$ and since $f(0)=0$, we obtain two cases:
*
*$f\left(t_{min}\right)>0$.
In this case $w^3$ gets a minimal value, when the line $g(t)=w^3$ is a tangent line to the graph of $f$, which says that in this case it's enough to prove our inequality for equality case of two variables.
We'll prove this for the inequality $$(x+y+z)^3(x^3+y^3+z^3)^5\geq243(x^3y^3+x^3z^3+y^3z^3)^3.$$
It's enough to assume $y=z=1$ and we need to prove that:
$$(x+2)^3(x^3+2)^5\geq243(2x^3+1)^3$$ or $h(x)\geq0,$ where
$$h(x)=3\ln(x+2)+5\ln(x^3+2)-3\ln(2x^3+1)-5\ln3.$$
But $$h'(x)=\frac{(x-1)(3x^5+7x^4+7x^3+6x^2-x-1)}{(x+2)(x^3+2)(2x^3+1)},$$
which gives $x_{min}=1$, $x_{max}=0.375...$ and since $h(1)=0$ it's enough to prove that $h(0)\geq0$ or $256\geq243,$ which is true.
*$w^3\rightarrow0^+$.
Let $z\rightarrow0^+$ and $y=1$.
Thus, we need to prove that $$(x+1)^3(x^3+1)^5\geq243x^9,$$ which is true by AM-GM because $256>243.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4261358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the relationship between the length of the circumradius and the inradius in $ \triangle ABC $? For reference: In a right angle triangle ABC, an interior bisector BD is traced, where I is or incenter, $\measuredangle B = 90 ^ o$ and $3BI = 4ID$. Find the relationship between the circumraio and inraio lenght of $\triangle ABC$. (Answer:3)
My progress:
I made the drawing
Inradius = r
Circumradius = R
$r=\frac{a+c-b}{2}=\frac{ab}{a+b+c}\\
R = \frac{b}{2}\\
\frac{R}{r} = \frac{b}{a+c-b}$
| T.Poncelet:$BA+BC = AC+2r\\
R = \frac{AC}{2}\\
\frac{BA}{BI}=\frac{DA}{DI}\\
\frac{BC}{BI}=\frac{DC}{DI}\\
\therefore BA+BC = AC+2r\\
\frac{DA \cdot BI}{DI}+\frac{DC \cdot BI}{DI} = 2R+2r\\
\frac{4}{3}\cdot (DA+DC) = 2R+2r\\
\frac{4}{3}\cdot (AC) = 2R+2r\\
\frac{4}{3}\cdot (2R) = 2R+2r\\
\boxed{R =3r}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show $\frac{n^2}{2^{\sqrt{\log(n)}}} \geq \frac{n}{2}. $ So we know that $\frac{n^2}{2} \geq \frac{n}{2}$, but I'm stuck proving that $\frac{n^2}{2} \geq \frac{n^2}{2^{\sqrt{\log n}}}\geq \frac{n}{2}$. Am I missing something?
| Notice that the expression $\frac{x^2}{2^\sqrt{\ln(x)}}$ is well-defined if and only if $x\geq1$. Division by $x$ then shows that your inequality is equivalent to
$$\frac{x}{2^\sqrt{\ln(x)}}\geq\frac{1}{2}$$
We can prove this inequality with calculus:
Consider the function $f:(1,\infty)\to\mathbb{R}$ defined by
$$f(x)=\frac{x}{2^\sqrt{\ln(x)}}$$
This function is differentiable, so any maxima or minima in its domain will have derivative $0$.
Using the quotient rule and the chain rule, we find that for every $x>1$, the derivative of $f$ at $x$ is
\begin{align}
f'(x) &= \frac{2^\sqrt{\ln(x)}-x\cdot2^\sqrt{\ln(x)}\ln(2)\cdot\frac{1}{2\sqrt{\ln(x)}}\cdot\frac{1}{x}}{\left(2^\sqrt{\ln(x)}\right)^2}\\
&= \frac{2^\sqrt{\ln(x)}\left(1-\frac{\ln(2)}{2\sqrt{\ln(x)}}\right)}{\left(2^\sqrt{\ln(x)}\right)^2}\\
&= \frac{1-\frac{\ln(2)}{2\sqrt{\ln(x)}}}{2^\sqrt{\ln(x)}}
\end{align}
We now solve the equation $f'(x)=0$ to find $f$'s critical points, if any exist.
\begin{align}
f'(x)=0 &\iff 1-\frac{\ln(2)}{2\sqrt{\ln(x)}}=0\\
&\iff \frac{\ln(2)}{2\sqrt{\ln(x)}}=1\\
&\iff \frac{\ln(2)}{2}=\sqrt{\ln(x)}\\
&\iff \ln(x)=\frac{\ln^2(2)}{4}\\
&\iff x=\exp\left(\frac{\ln^2(2)}{4}\right)
\end{align}
Thus, $e^{\frac{\ln^2(2)}{4}}$ is the only critical point of $f$. This actually corresponds to a minimum of $f$, which we can prove by showing that $f'(x)$ is negative for $1<x<e^{\frac{\ln^2(2)}{4}}$ and positive for $x>e^{\frac{\ln^2(2)}{4}}$.
\begin{align}
1<x<e^{\frac{\ln^2(2)}{4}} &\implies 0<\ln(x)<\frac{\ln^2(2)}{4}\\
&\implies 0<\sqrt{\ln(x)}<\frac{\ln(2)}{2}\\
&\implies \frac{\ln(2)}{2\sqrt{\ln(x)}}>1\\
&\implies 1-\frac{\ln(2)}{2\sqrt{\ln(x)}}<0\\
&\implies f'(x)=\frac{1-\frac{\ln(2)}{2\sqrt{\ln(x)}}}{2^\sqrt{\ln(x)}}<0\text{, since }2^\sqrt{\ln(x)}>0
\end{align}
This proves that $f'(x)$ is negative for $1<x<e^{\frac{\ln^2(2)}{4}}$.
\begin{align}
x>e^{\frac{\ln^2(2)}{4}} &\implies \ln(x)>\frac{\ln^2(2)}{4}>0\\
&\implies \sqrt{\ln(x)}>\frac{\ln(2)}{2}\\
&\implies \frac{\ln(2)}{2\sqrt{\ln(x)}}<1\\
&\implies 1-\frac{\ln(2)}{2\sqrt{\ln(x)}}>0\\
&\implies f'(x)=\frac{1-\frac{\ln(2)}{2\sqrt{\ln(x)}}}{2^\sqrt{\ln(x)}}>0\text{, since }2^\sqrt{\ln(x)}>0
\end{align}
This proves that $f'(x)$ is positive for $x>e^{\frac{\ln^2(2)}{4}}$. We infer that $f$ has an absolute minimum at $e^{\frac{\ln^2(2)}{4}}$, so for every $x>1$,
\begin{align}
f(x) &\geq f\left(e^{\frac{\ln^2(2)}{4}}\right)\\
&= \frac{e^{\frac{\ln^2(2)}{4}}}{2^\sqrt{\ln\left(e^{\frac{\ln^2(2)}{4}}\right)}}\\
&= \frac{e^{\frac{\ln(2)}{4}\cdot\ln(2)}}{2^\sqrt{\frac{\ln^2(2)}{4}}}\\
&= \frac{2^\frac{\ln(2)}{4}}{2^\frac{\ln(2)}{2}}\\
&= 2^{\frac{\ln(2)}{4}-\frac{\ln(2)}{2}}\\
&= 2^{-\frac{\ln(2)}{4}}
\end{align}
All that remains to be shown is that $2^{-\frac{\ln(2)}{4}}>\frac{1}{2}$:
\begin{align}
2^{-\frac{\ln(2)}{4}}>\frac{1}{2} &\iff e^{-\frac{\ln(2)}{4}\cdot\ln(2)}>\frac{1}{2}\\
&\iff -\frac{\ln(2)}{4}\cdot\ln(2)>\ln\left(\frac{1}{2}\right)\\
&\iff -\frac{\ln(2)}{4}\cdot\ln(2)>-\ln(2)\\
&\iff \frac{\ln(2)}{4}<1\\
&\iff \ln(2)<4
\end{align}
The last of these is true because $\ln(2)<1$, so we have that $2^{-\frac{\ln(2)}{4}}>\frac{1}{2}$. Thus, for every $x>1$,
$$f(x)>\frac{1}{2}$$
or
$$\frac{x}{2^\sqrt{\ln(x)}}>\frac{1}{2}$$
| {
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"url": "https://math.stackexchange.com/questions/4264808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int \frac{1}{a+b\sec(x)}dx$ I tried to find a closed-form expression for the integral
$$\int\frac{1}{a+b\sec(x)}\>dx$$ and, afterwards, set $a=0$, $b=1$ to recover the result for the simplified integral
$$\int\cos(x)\>dx=\sin(x)+C$$
The above integral was inspired by Américo Tavares’s rather illuminating solution to Ways to evaluate $\sec(t)$,
Here's the progress I've made which makes use of the integral
$$(\star)\int\frac{dx}{m^2+n^2x^2}=\frac{1}{mn}\arctan\left(\frac{nx}{m}\right)\quad m,n>0$$
Using $\cos(x)=2\cos^2(x/2)-1$, we get the following
$$I=\int\frac{dx}{a+b\sec(x)}=\int\frac{\cos(x)}{a\cos(x)+b}dx=\int\frac{2\cos^2(x/2)-1}{2a\cos^2(x/2)+b-a}dx\\=2\underbrace{\int\frac{dx}{2a+(b-a)\sec^2(x/2)}}_\color\red{(1)}-\underbrace{\int\frac{\sec^2(x/2)}{2a+(b-a)\sec^2(x/2)}dx}_\color\red{(2)}$$
$\color\red{(2)}$ becomes
$$\int\frac{\sec^2(x/2)}{2a+(b-a)\sec^2(x/2)}dx=\int\frac{\sec^2(x/2)}{a+b+(b-a)\tan^2(x/2)}dx\\=\frac{2}{\sqrt{b^2-a^2}}\arctan\left(\sqrt{\frac{b-a}{a+b}}\tan(x/2)\right)$$
from the substitution $u=\tan(x/2)$ and then applying $(\star)$.
For tackling $\color\red{(1)}$, the same u-substitution and applying partial fractions converts the integral into
$$\int\frac{dx}{2a+(b-a)\sec^2(x/2)}=2\int\frac{du}{(a+b+(b-a)u^2)(1+u^2)}\\=2\left[\frac{a-b}{2a}\int\frac{du}{(a+b+(b-a)u^2)}+\frac{1}{2a}\int\frac{du}{1+u^2}\right]$$
Thus
$$\int\frac{dx}{2a+(b-a)\sec^2(x/2)}=\frac{a-b}{a\sqrt{b^2-a^2}}\arctan\left(\sqrt{\frac{b-a}{a+b}}u\right)+\frac{1}{a}\arctan(u)\\=\frac{a-b}{a\sqrt{b^2-a^2}}\arctan\left(\sqrt{\frac{b-a}{a+b}}\tan(x/2)\right)+\frac{1}{a}\arctan(\tan(x/2)))$$
However, the problem is that setting $a=0$ afterwards results in the denominator becoming zero. Also, I couldn't think another approach to this.
Any help?
| To derive the result valid for $a\to 0$, integrate as follows
\begin{align}
\int \frac{1}{a+b\sec x}dx
=& \frac1a \int \left(1- \frac{b}{b+a\cos x}\right)dx\\
=& \frac1a \bigg(x-2b\int \frac{d(\tan\frac x2)}{(b+a)+(b-a)\tan^2\frac x2}\bigg)\\
=& \frac1a \bigg( x- \frac{2b}{\sqrt{b^2-a^2}}\tan^{-1}\frac{\tan\frac x2}{\sqrt{\frac{b+a}{b-a}} \>}\bigg) +C\\
\end{align}
Note that
$$\lim_{a\to 0 } \frac{2b}{\sqrt{b^2-a^2}}\tan^{-1}\frac{\tan\frac x2}{\sqrt{\frac{b+a}{b-a}} }
= x - \frac ab \sin x+O(a^2)
$$
Thus
$$ \lim_{a\to 0}\> \int \frac{1}{a+b\sec x}dx=\frac1b \sin x+C
$$
as expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4266408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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solve the equation $1+2z+2z^2+\ldots +2z^{n-1}+z^n=0$ im trying to solve the equation $$(E)\quad 1+2z+2z^2+\ldots +2z^{n-1}+z^n=0$$
attempt : because $1$ isnt a solution we have $$\begin{aligned}
1+2 z+2 z^{2}+\cdots+2 z^{n-1}+z^{n} &=2\left(z^{0}+z+z^{2}+\cdots+z^{n-1}\right)-1+z^{n} \\
&=2 \frac{1-z^{n}}{1-z}-1+z^{n} \\
&=\frac{2-2 z^{n}+z^{n}-z^{n+1}}{1-z}-1 \\
&=\frac{2-z^{n}-z^{n+1}}{1-z}-1
\end{aligned}$$ then (E) becomes $$z^{n+1}+z^{n}-z-1=0$$ but i dont know how to get any further.
| $$\begin{align}
1 + 2z + 2z^2 + \cdots + 2z^{n-1} + z^n
&= (1 + z + \cdots + z^{n-1}) + (z + z^2 + \cdots + z^n) \\
&= (1+z)(1 + z + z^2 + \cdots + z^{n-1}) \\
&= \frac{(z+1)(z^n - 1)}{z - 1}.
\end{align}$$
Thus the roots are $$z \in \{ e^{2\pi i k/n} : k \in \{1, 2, \ldots, n-1\} \} \cup \{ -1 \}.$$ When $n$ is even, $-1$ is a double root; when $n$ is odd, it is a single root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4269596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Isn't my book doing this math about differentiation wrongly? Problem:
Differentiate with respect to $x$: $\ln(e^x(\frac{x-1}{x+1})^{\frac{3}{2}})$
My attempt:
Let,
$$y=\ln(e^x(\frac{x-1}{x+1})^{\frac{3}{2}})$$
Both $e^x$ and $\frac{x-1}{x+1}$ are positive: $e^x$ can never be negative, and you can take the square root of only positive numbers, so $\frac{x-1}{x+1}$ is positive as well. So, we can apply logarithm properties:
$$=\ln e^x+\ln(\frac{x-1}{x+1})^{\frac{3}{2}}$$
$$=x+\frac{3}{2}\ln(\frac{x-1}{x+1})$$
Now,
$$\frac{dy}{dx}=1+\frac{3}{2}.\frac{x+1}{x-1}.\frac{d}{dx}(\frac{x-1}{x+1})$$
$$=1+\frac{3}{2}.\frac{x+1}{x-1}.\frac{x+1-x+1}{(x+1)^2}$$
$$=1+\frac{3}{2}.\frac{x+1}{x-1}.\frac{2}{(x+1)^2}$$
We can cancel $(x+1)$ and $(x+1)$ in the numerator and denominator because the graph doesn't change.
$$=1+\frac{3}{x^2-1}$$
$$=\frac{x^2+2}{x^2-1}$$
My book's attempt:
Let,
$$y=\ln(e^x(\frac{x-1}{x+1})^{\frac{3}{2}})$$
$$=\ln e^x+\ln(\frac{x-1}{x+1})^{\frac{3}{2}}$$
$$=x+\frac{3}{2}\ln(\frac{x-1}{x+1})$$
$$=x+\frac{3}{2}(\ln(x-1)-\ln(x+1))\tag{1}$$
$$...$$
$$\frac{dy}{dx}=\frac{x^2+2}{x^2-1}$$
Question:
*
*In my book's attempt, is line $(1)$ valid? I think my book's assumption in $(1)$ that $(x-1)$ and $(x+1)$ must be positive is unfounded. I deduced that $\frac{x-1}{x+1}$ is positive; $\frac{x-1}{x+1}$ can be positive even when both $(x-1)$ and $(x+1)$ is negative. So, is line $(1)$ of my book valid?
| Giving a restriction of the domain is an issue of this problem, because that quantity is defined only for $x<-1\lor x>1$. In my opinion, if this were part of a "gauntlet of exercises" in some textbook, then an indication of the domain should be specified, if the purpose of the exercise lies elsewhere: at the very least, to avoid wasting time.
For instance, I think that such a thing would be very important in the chapters about antiderivatives.
You're right: based on the information you've given, $\ln\frac{x-1}{x+1}=\ln(x-1)-\ln(x+1)$ is unwarranted, because the natural way to address the problem would be for all $x$ such that $x<-1\lor x>1$. The book could have solved the issue by saying that $\ln\frac{x-1}{x+1}=\ln\lvert x-1\rvert-\ln\lvert x+1\rvert$ (which is true in the natural domain) and then using $\frac{d}{dt}\ln\lvert t\rvert=\frac1t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4274506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Prove that: $\frac{1}{ka+b}+\frac{1}{kb+c}+\frac{1}{kc+a}\ge\frac{6}{(k+1)\sqrt[3]{(a+b)(b+c)(c+a)}}$ Problem: Let $a,b,c>0; k\ge1$. Prove that: $$\frac{1}{ka+b}+\frac{1}{kb+c}+\frac{1}{kc+a}\ge\frac{6}{(k+1)\sqrt[3]{(a+b)(b+c)(c+a)}}$$
My attempt: Using C-S inequality: $$\frac{1}{ka+b}+\frac{1}{kb+c}+\frac{1}{kc+a}\ge\frac{9}{(k+1)(a+b+c)}$$
It is desired to prove: $$\sqrt[3]{(a+b)(b+c)(c+a)}\ge\frac{2}{3}(a+b+c)$$ Which is false.
The problem has nice form but for me, it is quite hard to get nice proof. Please help me, thanks!
| I just get a part by AM-GM. Maybe stronger tools can help to make up the rest $1<k<2$
For $k=1$: the inequality holds obviously by AM-GM
For $k\ge2$: Using C-S inequality: $$k(a+b+c)\sum{\frac{1}{ka+b}}=3+(k-1)\sum{\frac{b}{ka+b}}+k\sum{\frac{c}{ka+b}}\ge\frac{6k}{k+1}+\frac{k(a+b+c)^2}{(k+1)(ab+bc+ca)}$$Also by AM-GM inequality and well- known result: $9(a+b)(b+c)(c+a)\ge8(ab+bc+ca)(a+b+c)$, we obtain:
$$2.\frac{3k}{k+1}+\frac{k(a+b+c)^2}{(k+1)(ab+bc+ca)}\ge3\sqrt[3]{\frac{9k^3}{(k+1)^3}.\frac{(a+b+c)^2}{ab+bc+ca}}\ge\frac{6k(a+b+c)}{(k+1)\sqrt[3]{(a+b)(b+c)(c+a)}}$$Hence, the proof is cleared!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is this function of partitions one-to-one? Suppose we have a set of integers $H=\{1,2, ...n\}$. Let $A$ a set of partitions of H into $n/2$ pairs $\{\{x_1,y_1\},\{x_2,y_2\}, ...,\{x_{n/2},y_{n/2}\}\}$ and
function $f:A \rightarrow Z^n$ where $f(\{\{x_1,y_1\},\{x_2,y_2\}, ...,\{x_{n/2},y_{n/2}\}\})=\cup \{x_i*y_i\}$.
For a set of integers $\{1,2, ...8\}$ and the pairing $\{\{3,6\},\{1,7\},\{2,4\},\{5,8\}\}$
$f(\{\{3,6\},\{1,7\},\{2,4\},\{5,8\}\})=\{18,7,8,40\}$
Is this function injective?
| No. For $n$ large enough, $f$ will not be injective.
It's easier to consider the case of addition, where we have
$$1 + (-1) = 0, \qquad 2 + (-3) = -1, \qquad 3 + (-2) = 1 \\
1 + (-2) = -1, \qquad 2 + (-1) = 1, \qquad 3 + (-3) = 0$$
That is, partition $\{-3,-2,-1, 1, 2, 3\}$ into three parts of size $2$ in at least two ways that give the same set of sums.
This answers your question by applying the map $x \mapsto 2^{x+3}$ to find
$$2^{1+3}2^{(-1)+3} = 2^{0+6}, \qquad 2^{2+3}2^{(-3)+3} = 2^{-1+6}, \qquad 2^{3+3}2^{(-2)+3} = 2^{1+6} \\
2^{1+3}2^{(-2)+3} = 2^{-1+6}, \qquad 2^{2+3}2^{(-1)+3} = 2^{1+6}, \qquad 2^{3+3}2^{(-3)+3} = 2^{0+6}$$
so we find that $f$ is not injective for $n \geq 2^6 =64$. [it likely fails to be injective for smaller values as well]
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab}$, prove that $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$.
Problem: Let $a, b, c, x, y, z$ be real numbers
such that $abc \ne 0$, $ ~ a + b + c \ne 0$, $~ (a^2 - bc)(b^2 - ca)(c^2 - ab)\ne 0$, $xyz\ne0$, $x+y+z\ne0$, $(x-y)^2+(y-z)^2+(z-x)^2\ne0$
and $$\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab} \ne 0.$$
Prove that $$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}.$$
The converse can be proved easily.
My Attempt:
$
\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab}\\
\Rightarrow\frac{x^2-yz-y^2+zx}{a^2-bc-b^2+ca}=\frac{y^2-zx-z^2+xy}{b^2-ca-c^2+ab}=\frac{z^2-xy-x^2+yz}{c^2-ab-a^2+bc}\ [By\ Addendo]\\\Rightarrow\frac{x-y}{a-b}\cdot\frac{x+y+z}{a+b+c}=\frac{y-z}{b-c}\cdot\frac{x+y+z}{a+b+c}=\frac{z-x}{c-a}\cdot\frac{x+y+z}{a+b+c}\\\Rightarrow\frac{x-y}{a-b}=\frac{y-z}{b-c}=\frac{z-x}{c-a}\ \left[Considering\ (x+y+z)\ and\ (a+b+c)\ \neq0\right]
$
Known Exceptional Counter-examples:-
Macavity: $a=1, b=2, c=3, x=y=z$
River Li: $a = 1, b=2, c = -3, x = 7, y = 7, z = -14$
The conditions in the question have been edited to exclude exceptional counter-examples.
| Here is a solution that necessitates to have had a previous introduction
*
*to conic curves (maybe it is the case) and to
*barycentric coordinates (less likely, unless you have prepared some Mathematics Olympiads) ; here or here are a nice introduction to barycentric coordinates (b.c. in short). One of the most elementary (and important) property is that b.c. are defined up to a multiplicative factor.
Indeed, for a fixed triangle of reference, your issue amounts to assert the bijectivity (one-to-one onto) of the following transformation,
$$M\underbrace{(x, \ y, \ z)}_{\text{b.c.}} \ \ \xrightarrow{(H)} \ \ M'\underbrace{(x^2-yz,\ \ y^2-zx, \ \ z^2-xy)}_{\text{b.c.}} \tag{1}$$
known as Hirst transform (see here).
This transformation can be defined geometrically: $M'$ is the intersection point of line $GM$ (where $G$ is the center of mass of the triangle) and polar line of $M$ with respect to the Steiner interior ellipse. See figure below and proof in the Appendix at the bottom.
Fig. 1: Steiner ellipse (S), tangent to the triangle's sides in their midpoints. $G$ is the center of mass. Being given point $M$, its Hirst transform is $M'=(P) \cap GM$, where $(P)$ is the polar line of $M$ with respect to $(S)$ (in green). Please note that in this case of figure where $M$ is outside the ellipse, (P) has been obtained by connecting the tangency points of the tangents to (S) issued from $M$.
Knowing $M'$, it is easy to retrieve its unique preimage $M$, establishing the bijectivity of Hirst transform $(H)$, even more its involutivity $(H)^{-1}=H$. This involutivity can be checked, of course, in an algebraic way.
Remarks :
*
*Strictly speaking, we have to exclude the case where $M$ has identical barycentric coordinates $(x,y,z)$ with $x=y=z$, which means that $M$ is situated at the center of mass $G$ ; indeed, in this case, line $GM$ is undefined. It is a confirmation of the exceptional "counterexample" $x=y=z=1$ found by @Macavity.
*See the very interesting analysis in the other answer by @orangeskid.
*Hirst transform is mentioned but in a handful of texts such as this one in connection with the study of certain cubic curves.
*The "nonclassical" life of Mr Hirst is described here.
Appendix: Proof of the geometrical interpretation:
The internal Steiner ellipse has equation:
$$x^2+y^2+z^2-2(xy+yz+zx)=0$$
which is equivalent to:
$$2(xy+yz+zx)=\frac12\tag{2}$$
(see here p. 119). Therefore, the polar of point $(x,y,z)$, is given by the bilinear expression associated with quadratic form in the LHS of (2) has equation:
$$(x'y+y'x)+(y'z+z'y)+(z'x+x'z)=\frac12 \iff $$
$$x'(y+z)+y'(z+x)+z'(x+y)=\frac12\tag{3}$$
Besides, the equation of line $GM$ is:
$$\begin{vmatrix}1&x&x'\\1&y&y'\\1&z&z'\end{vmatrix}=0 \iff x'(z-y)+y'(x-z)+z'(y-x)=0\tag{4}$$
The intersection of straight lines given by (3) and (4) is obtained by solving the system (3)+(4), giving, up to a multiplicative constant (as always with b.c.) :
$$\begin{pmatrix}x'\\y'\\z'\end{pmatrix}=\begin{pmatrix}-x^2+yz\\-y^2+zx\\-z^2+xy\end{pmatrix}$$
as awaited.
| {
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"timestamp": "2023-03-29T00:00:00",
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Eliminate $\theta$ from $4x=5\cos\theta -\cos 5\theta$ and $4y=5\sin\theta -\sin 5\theta$ Eliminate $\theta$ from the equations.
$$4x=5\cos\theta -\cos 5\theta$$
$$4y=5\sin\theta -\sin 5\theta$$
Alternative forms are
$$x=5\cos^3\theta-4\cos^5\theta$$
$$y=5\sin^3\theta-4\sin^5\theta$$
and $$x=\cos^3\theta(3-2\cos2\theta)$$
$$y=\sin^3\theta(3+2\cos2\theta)$$
| Let $c=\cos2\theta$.
$$4x=5\cos\theta -\cos 5\theta\tag1$$
$$4y=5\sin\theta -\sin 5\theta\tag2$$
Squaring and adding ,$$16(x^2+y^2)=25+1-10\cos4\theta$$ $$8(x^2+y^2)=13-5(2c^2-1)$$ $$5c^2=9-4(x^2+y^2)=k$$
Squaring and subtracting, $$16(x^2-y^2)=25\cos2\theta+\cos10\theta-10\cos6\theta$$ $$16(x^2-y^2)=25c+(16c^5-20c^3+5c)-10(4c^3-3c)$$ $$4(x^2-y^2)=c(4c^4-15c^2+15)$$
Substituting $k$,
$$100(x^2-y^2)=c(4k^2-75k+375)$$
Further simplifying leads to, $$25(x^2-y^2)=c(16\lambda^2+3\lambda+6)$$ where $\lambda=(x^2+y^2)$.
Squaring again and replacing all values finally we get, $$3125(x^2-y^2)^2=[9-4(x^2+y^2)][16(x^2+y^2)^2+3(x^2+y^2)+6]^2$$
And this is that epicycloid, as being described in @orangeskid's answer!
| {
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General form for $\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) - 1\right)$ I'm wondering if there is a general form for the following sum:
$$\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) - 1\right)$$ for $m \in \mathbb{N}$
I have obtained the following closed-forms for these special cases:
Where $G$ is Catalan's constant and $\text{Cl}_2$ is the Clausen function of order 2.
$$\sum_{n=1}^{\infty}(-1)^n \left(2n \, \text{arccoth} \, (2n) - 1\right) = \frac{1}{2} - \frac{2G}{\pi}$$
$$\sum_{n=1}^{\infty} (-1)^{n}\left( 3n \, \text{arccoth} \, (3n)-1\right) = \frac{1}{2} - \frac{5}{2\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{3}\right) + \frac{1}{4} \ln (3)$$
$$\sum_{n=1}^{\infty} (-1)^n \left(4n \, \text{arccoth} \, (4n) - 1\right) = \frac{1}{2}+ \frac{G}{\pi} - \frac{4}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{4}\right) - \frac{1}{4} \ln \left(3- 2\sqrt{2}\right)$$
etc.
Given that $G = \text{Cl}_2 \left(\frac{\pi}{2}\right)$, I am curious to know if the general sum is expressible in terms of the Clausen function. These above sums were determined by using the Mittag-Leffler expansion of $\csc (z)$, i.e $\csc(z) = \frac{1}{z} + 2z \sum_{n=1}^{\infty} (-1)^n \frac{1}{z^2 - \left(\pi n\right)^2}$ and substituting it into the integral $\int_{0}^{\pi/m} x \csc (x) \, dx$
If one uses the following other method, we can determine the odd and even terms of the sums
$$\sum_{n=1}^{\infty} \left( 4n \, \text{arccoth} \, (4n)-1\right) = \frac{1}{2} - \frac{G}{\pi}- \frac{1}{4} \ln (2)$$
$$\sum_{n=1}^{\infty} \left( (4n-2) \, \text{arccoth} \, (4n-2) - 1\right) = \frac{G}{\pi} - \frac{1}{4} \ln (2)$$
$$\sum_{n=1}^{\infty} \left(6n \, \text{arccoth} \, (6n) - 1\right) = \frac{1}{2} - \frac{3}{2\pi} \, \text{Cl}_2 \left( \frac{\pi}{3}\right)$$
$$\sum_{n=1}^{\infty} \left( (6n-3) \, \text{arccoth} \, (6n-3) - 1\right) = \frac{1}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{3}\right) - \frac{1}{4} \ln(3)$$
$$\sum_{n=1}^{\infty} \left( 8n \, \text{arccoth} \, (8n) - 1\right) = \frac{1}{2} - \frac{2}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{4}\right) - \frac{1}{4} \ln (2-\sqrt{2})$$
$$\sum_{n=1}^{\infty} \left( (8n-4) \, \text{arccoth} \, (8n-4) - 1\right) = \frac{2}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{4}\right) - \frac{G}{\pi} - \frac{1}{4} \ln(2+\sqrt{2})$$
EDIT
I am now interested in
$$\sum_{n=1}^{\infty} \left(m n \, \text{arccoth} \, (m n) - 1\right)$$
for $|m|>1$ as asked here.
Here is how I originally proved it for the odd terms:
For the first odd term sum, begin with the known result $$2G = \int_{0}^{\frac{\pi}{2}} x \csc (x) \, dx$$
then integrate by parts to get:
$$2G = \int_{0}^{\frac{\pi}{2}} \ln \left(\cot (x) + \csc (x) \right) \, dx = \int_{0}^{\frac{\pi}{2}} \ln (\cos (x) + 1) \, dx - \int_{0}^{\frac{\pi}{2}} \ln (\sin (x)) \,dx$$
Then use the well-known result $\int_{0}^{\frac{\pi}{2}} \ln (\sin(x)) \,dx = - \frac{\pi}{2} \ln (2)$ and use the identity $\cos (x) + 1 = 2 \cos^2\left( \frac{x}{2}\right)$ and make the substitution $\frac{x}{2} = u$.
$$\implies 2G - \pi \ln (2) = 4 \int_{0}^{\frac{\pi}{4}} \ln (\cos (u)) \, du$$
Now use the Weierstrass product for $\cos (z)$, namely $\cos(z) = \prod_{n=1}^{\infty} \left(1-\frac{4z^2}{\pi^2 (2n-1)^2}\right)$ to obtain:
$$2G - \pi \ln (2) = 4 \sum_{n=1}^{\infty} \int_{0}^{\frac{\pi}{4}} \ln \left( 1-\frac{4u^2}{\pi^2 (2n-1)^2}\right) \, du$$
After integrating, obtain $\pi \sum_{n=1}^{\infty} \ln \left(1-\frac{1}{4(1-2n)^2}\right) = -\frac{\pi}{2} \ln (2)$ and the result quickly follows. The other odd sums are the same idea. The even sums just comes from combining the two results from the alternating sum and the odd term sum.
| I didn't want to rain on mathstackuser12's parade, but now that the bounty has been awarded, here's a simpler approach using the theory of the Barnes G function. (Mathstackuser12 seems to have reinvented a lot of it.) As that answer notes, use the log representation of arccoth. Sum in pairs up to $2N,$ where $N$ will be driven to $\infty.$ The sum over $(-1)^n \cdot (-1)$ will then be zero (it is included so the summand makes sense.) Then
$$S:=\sum_{n=1}^\infty (-1)^n\big(m \, n \, \text{arccoth}(m \, n) -1 \big) =
\lim_{N \to \infty} \frac{m}{2}\log\Big( \prod_{n=1}^{2N} \Big(\frac{1+1/(m \, n)}{1-1/(m\,n)} \Big)^{(-1)^n n } \Big)$$
This is done because Adamchik has proven, (Proposition 5 of reference below)
$$ \lim_{N \to \infty} \prod_{n=1}^{2N} \Big(1+2x/n \Big)^{(-1)^{n+1} n }=\frac{e^{-x}\Gamma(x+1/2)}{\Gamma(1/2)} \Big(\frac{G(x+1/2)}{G(x+1)G(1/2)}\Big)^2$$
where $G$ is the Barnes G function.
Apply this formula to numerator and denominator of right hand side of top formula with $x=1/(2m)$ and $x=-1/(2m):$
$$S=\frac{m}{2} \log\Bigg( \Big( \frac{ G(1/2-1/(2m))G(1+1/(2m))}{G(1/2+1/(2m))G(1-1/(2m))} \Big)^2 \frac{\Gamma(1/2-1/(2m))}{\Gamma(1/2+1/(2m))} \exp{(1/m)} \Bigg)$$
Now use 2 functional equations of the Barnes G
$$\frac{G(1-x)}{G(1+x)} =\left(\frac{\sin{\pi x}}{\pi}\right)^x \exp{\left(\frac{\text{Cl}_2(2 \pi x)}{2\pi}\right) }\, , \quad G(1+x)=G(x)\Gamma(x)$$
where Cl$_2(x)$ is the Clausen function mentioned in the statement of the problem, and $0<x<1.$
The second equation is needed to to shift the ratio G(1/2-x)/G(1/2+x), which is the first factor in right hand side of the penultimate equation, so that it looks like the first functional equation. Finally, some algebra and a gamma function identity puts the equation in the form
$$S=\frac{1}{2} + \frac{1}{2}\log\big({\cot(\frac{\pi}{2m})}\big) + \frac{m}{2\pi}\big(\text{Cl}_2(\pi/m- \pi)-\text{Cl}_2(\pi/m) \big) $$
Besides an economy of getting to the answer, the answer is stated in terms of Clausen functions, which is what the OP found in his examples. Note also that there is no need for $m$ being an integer. (Numerical checks have verified it.)
Ref.: 'Multiple Gamma Function and Its Application to Computation of Series,' V.S. Adamchik, arXiv:math.CA/0308074v1 7 Aug 2003.
| {
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"timestamp": "2023-03-29T00:00:00",
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Expected Value of |H-T| A coin is flipped $n$ times, with $H$ heads and $T$ tails. What is the expected value of $|H-T|$?
Here's where I'm at so far:
There are $2^n$ total different flip sequences. Each is equally likely. Thus the expected value is the sum of all the $|H-T|$ values for each flip divided by $2^n$. We can sort these flip sequences by the value of $H$. Given $n$ flips, the number of sequences with $H$ heads is ${n\choose H} = \frac{n!}{(n-H)!H!}$. The sum therefore is $\sum\limits_{i=0}^{n} \frac{n!}{(n-H)!H!} |H-T|.$
This did not really get me anywhere, so I started to evaluate it numerically. I wrote up some python code to calculate it for $n$, and came up with the following sequence (starting with $1$ flip):
$1, 1, \frac32, \frac32, \frac{15}{8}, \frac{15}{8}, \frac{35}{16},\frac{35}{16}\dots$. According to Wolfram Alpha, the general term for this sequence is $a_{n+2} = a_n + \frac{a_{n+1}}{n+1}$, which holds for the first 10000 terms calculated by my program, so I am fairly confident that this is in fact the general formula. I just need to prove it.
Any help would be much appreciated. This seems like a fairly basic statistical property since it seems so related to standard deviation so I am surprised that my scourings of the internet have not gleaned results.
| Let $a_n = \operatorname{E}[|H-T|] = \operatorname{E}[|2H - n|]$ for $H \sim \operatorname{Binomial}(n,1/2)$ and $T = n - H$. Then we first observe from your sequence that $a_{2m} = a_{2m-1}$ for each positive integer $m$. So let's aim to prove this property first. Note
$$\begin{align}
a_{2m} &= \frac{1}{2^{2m}} \sum_{h=0}^{2m} | 2h - 2m | \binom{2m}{h} \\
&= \frac{1}{2^{2m-1}} \sum_{h=0}^{2m} |h - m| \left(\binom{2m-1}{h-1} + \binom{2m-1}{h}\right) \\
&= \frac{1}{2^{2m-1}} \left( \sum_{h=1}^{2m} |h - m| \binom{2m-1}{h-1} + \sum_{h=0}^{2m-1} |h-m| \binom{2m-1}{h} \right) \\
&= \frac{1}{2^{2m-1}} \sum_{h=0}^{2m-1} \bigl(|h - (m-1)| + |h - m| \bigr)\binom{2m-1}{h} \\
&= \frac{1}{2^{2m-1}} \sum_{h=0}^{2m-1} |2h - (2m-1)| \binom{2m-1}{h} \\
&= a_{2m-1},
\end{align}$$
where the penultimate step holds because
$$|h-(m-1)| + |h-m| = \begin{cases} 2m-1 - 2h, & 0 \le h \le m-1 \\
2h-2m+1, & m \le h \le 2m-1.\end{cases}$$ The only case where equality is violated is if $m-1 < h < m$, but this is not possible since $h, m$ are integers.
So it suffices to consider the case where $n$ is, say, even. Then we have
$$\begin{align}
a_{2m} &= \frac{1}{2^{2m-1}} \sum_{h=0}^{2m} |h-m| \binom{2m}{h} \\
&= \frac{1}{2^{2m-2}} \sum_{h=0}^m (m-h) \binom{2m}{h} \\
&= \frac{1}{2^{2m-2}} \left( m \sum_{h=0}^m \binom{2m}{h} - \sum_{h=0}^m h \binom{2m}{h}\right) \\
&= \frac{1}{2^{2m-2}} \left( \frac{m}{2}\left( \binom{2m}{m} + \sum_{h=0}^{2m} \binom{2m}{h} \right) - \sum_{h=1}^m \frac{(2m)!}{(h-1)!(2m-h)!} \right) \\
&= \frac{1}{2^{2m-2}} \left( \frac{m}{2} \binom{2m}{m} + 2^{2m-1} m - 2m \sum_{h=1}^m \binom{2m-1}{h-1}\right) \\
&= \frac{m}{2^{2m-1}} \left( \binom{2m}{m} + 2^{2m} - 4 \sum_{h=0}^{m-1} \binom{2m-1}{h} \right) \\
&= \frac{m}{2^{2m-1}} \left( \binom{2m}{m} + 2^{2m} - 2\sum_{h=0}^{2m-1} \binom{2m-1}{h} \right) \\
&= \frac{m}{2^{2m-1}} \left( \binom{2m}{m} + 2^{2m} - 2(2^{2m-1}) \right) \\
&= \frac{m}{2^{2m-1}} \binom{2m}{m}.
\end{align}$$
This was a quick and dirty proof--I don't like the computation for $a_{2m}$ because I strongly suspect there is a much more simple and elegant way to obtain this result, but this was the first thing that came to mind.
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