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Looking for other approaches to find the height $AH$ in triangle $ ABC $ where $A(1,5)$ , $B(7,3)$, $C(2,-2)$
We have a triangle with vertices $A(1,5)$ , $B(7,3)$, $C(2,-2)$. What
is the length of the height $AH$ in the triangle $ABC$ ?
$1)4\qquad\qquad2)3\sqrt2\qquad\qquad3)5\qquad\qquad4)4\sqrt2$
This is a problem from a timed exam.
Here is my approach:
The equation of the line passes through $B$ and $C$ is $y_=x-4$ so slope of $AH$ is $-1$ and its a line passes through $A(1,5)$ so the equation for this line becomes $y=-x+6$. in order to find coordinate of $H$ we should equate the formulas of those lines:
$$x-4=-x+6\quad\to\quad x=5\quad\text{and $\quad y=1\quad$ So$\quad H (5,1)$}$$
$$\text{Distance from $A$ to $H$:$\qquad$}AH=\sqrt{(5-1)^2+(1-5)^2}=4\sqrt2$$
Although I believe my answer is quick but I'm looking for other ideas to solve this problem.
|
Other than applying law of cosine or using distance formula from a point to a line that I mentioned in comment, another approach that I would consider is to recognize that $\triangle ABC$ is an isosceles triangle with,
$AC = BC = 5\sqrt2, \ AB = 2 \sqrt{10}$. If $G$ is midpoint of $AB$ then we know that $CG^2 = AC^2 - AG^2 = 50 - 10 = 40 \implies CG = 2 \sqrt{10}$
So equating area of $\triangle ABC, AH \times BC = CG \times AB$
$AH = \dfrac{2 \sqrt{10} \times 2 \sqrt{10}}{5 \sqrt2} = 4\sqrt2$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding $n>2$ such that $a_1$, $a_2$, $a_3$ are in arithmetico-geometric progression and $\left(1-x^3\right)^n=\sum_{r=0}^na_rx^r(1-x)^{3n-2r}$ This is the strangest, mind-boggling question I came across while doing binomial theorem.
Let
$$\displaystyle\left(1-x^3\right)^n=\sum_{r=0}^na_rx^r(1-x)^{3n-2r},\quad n\gt2$$
then find the value if $n$ so that $a_1,a_2,a_3$ are in A. G. P.
(original image)
Sorry but i have absolutely no clue of solving it.max i did was to expand the stuff inside the summation but it made things worse by bringing in two summations. I had aimed at comparing coefficients.
can someone help me out ? hints appreciated.
** AGP = arithmetico geometric progression
** problem encountered in my JEE coaching study material
|
Dividing both sides by $(1-x)^{3n}$ shows that the equation is equivalent to
$$
\left(1+\frac{3x}{(1-x)^2}\right)^n=\sum_{r=0}^na_r\left(\frac{x}{(1-x)^2}\right)^r\tag1
$$
Set $u=\frac{x}{(1-x)^2}$, then we have
$$
\left(1+3u\right)^n=\sum_{r=0}^na_ru^r\tag2
$$
So $a_r=3^r\binom{n}{r}$, as ZAhmed computed.
Suppose that $3\binom{n}{1},9\binom{n}{2},27\binom{n}{3}$ are in anArithmetico-Geometric Progression
$$
a_0,a_0(1+\Delta)\mathrm{R},a_0(1+2\Delta)R^2\tag3
$$
Then
$$
a_0=3n\tag4
$$
and
$$
\frac{1+2\Delta+\Delta^2}{1+2\Delta}=\frac{\binom{n}{2}^2}{\binom{n}{1}\binom{n}{3}}=\frac32\frac{n-1}{n-2}\tag5
$$
Let $\lambda=\frac32\frac{n-1}{n-2}-1$ and then
$$
\Delta=\lambda+\sqrt{\lambda(\lambda+1)}\tag6
$$
The ratio of the first two terms is
$$
(1+\Delta)\mathrm{R}=\frac{9\binom{n}{2}}{3\binom{n}{1}}=\frac32(n-1)\tag7
$$
Therefore,
$$
R=\frac{3(n-1)}{2(1+\Delta)}\tag8
$$
Examples
These can be fit to an AGP for any $n\gt2$:
For $n=3$, we get
$$
a_0=9,\Delta=2+\sqrt6,\mathrm{R}=\frac3{3+\sqrt6}\tag{Ex3}
$$
For $n=4$, we get
$$
a_0=12,\Delta=\frac{5+3\sqrt5}4,R=\frac6{3+\sqrt5}\tag{Ex4}
$$
For $n=5$, we get
$$
a_0=15,\Delta=1+\sqrt2,R=\frac6{2+\sqrt2}\tag{Ex5}
$$
For $n=6$, we get
$$
a_0=18,\Delta=\frac{7+\sqrt{105}}8,R=\frac{60}{15+\sqrt{105}}\tag{Ex6}
$$
For $n=7$, we get
$$
a_0=21,\Delta=2,R=3\tag{Ex7}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve $\frac{1}{2\sin50^\circ}+2\sin10^\circ$ I want to solve the expression $\frac{1}{2\sin50^\circ}+2\sin10^\circ$ to get a much simpler and neater result. I have tried to manipulate this expression such as using sum/difference formulas, but it didn't help (and made the expression even more messy).
Here is what I did:
\begin{align}
\frac{1}{2\sin50^\circ}+2\sin10^\circ&=\frac{1}{2\sin(60-10)^\circ}+2\sin10^\circ \\
&= \frac{1}{2\left(\frac{\sqrt{3}}{2}\cdot\cos10^\circ-\frac{1}{2}\cdot\sin10^\circ\right)}+2\sin10^\circ \\
&= \frac{1}{\sqrt{3}\cdot\cos10^\circ-\sin10^\circ}+2\sin10^\circ \\
&= \frac{\sqrt{3}\cdot\cos10^\circ+\sin10^\circ}{\left(\sqrt{3}\cdot\cos10^\circ-\sin10^\circ\right)\left(\sqrt{3}\cdot\cos10^\circ+\sin10^\circ\right)}+2\sin10^\circ \\
&= \frac{\sqrt{3}\cdot\cos10^\circ+\sin10^\circ}{3\cos^210^\circ-\sin^210^\circ}+2\sin10^\circ
\end{align}
But I don't know how to continue at this point. Multiplying in $2\sin10^\circ$ into the fraction is clearly unrealistic as it would result in trignometry of third power. Any help or hint would be appreciated. According to a calculator, the result of this expression should come to a nicely $1$, but I just want to know how to algebraically manipulate this expression to show that it is equal to $1$.
|
$$\begin{align}\frac1{2\sin50^\circ}+2\sin10^\circ&=\frac{1+4\sin50^\circ\sin10^\circ}{2\sin50^\circ}\\
&=\frac{1+2(\cos40^\circ-\cos60^\circ)}{2\cos40^\circ}\\&=1
\end{align}$$
|
{
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|
If $\dfrac{1-\sin x}{1+\sin x}=4$, what is the value of $\tan \frac {x}2$?
If $\dfrac{1-\sin x}{1+\sin x}=4$, what is the value of $\tan \frac
{x}2$?
$1)-3\qquad\qquad2)2\qquad\qquad3)-2\qquad\qquad4)3$
Here is my method:
$$1-\sin x=4+4\sin x\quad\Rightarrow\sin x=-\frac{3}5$$
We have $\quad\sin x=\dfrac{2\tan(\frac x2)}{1+\tan^2(\frac{x}2)}=-\frac35\quad$. by testing the options we can find out $\tan(\frac x2)=-3$ works (although by solving the quadratic I get $\tan(\frac x2)=-\frac13$ too. $-3$ isn't the only possible value.)
I wonder is it possible to solve the question with other (quick) approaches?
|
$$\tan \frac{x}{2} = \frac{\sin x/2}{\cos x/2} = \frac{2 \sin x/2 \cos x/2}{2 \cos^2 x/2} = \frac{\sin x}{\cos x + 1}$$
hence with $\sin x = -\frac{3}{5}, \cos x = ±\sqrt{1 - \sin^2 x} = ±\frac{4}{5}$, $\tan \frac{x}{2} = -\frac{1}{3}, -3$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Use the definition of a limit to prove the following: $\lim_{x\to4}\frac{x-2}{\sqrt{x}+2}=\frac{1}{2}$ Use the definition of a limit to prove the following:
$$\lim_{x\to4}\frac{x-2}{\sqrt{x}+2}=\frac{1}{2}$$
I'm trying to prove that:
$$\forall\varepsilon>0,\exists\delta>0; \\~\\
0<|x-4|<\delta\Longrightarrow\left|\frac{x-2}{\sqrt{x}+2}-\frac{1}{2}\right|<\varepsilon$$
I don't know what to do with $\frac{x-2}{\sqrt{x}+2}-\frac{1}{2}$
I've tried using the triangle inequality, but I get wrong results.
I know for now: $\frac{1}{\sqrt{x}+2}<\frac{1}{\sqrt{3}+2}$
|
Note that\begin{align}\frac{x-2}{\sqrt x+2}-\frac12&=\frac{2x-\sqrt x-6}{2\sqrt x+4}\\&=\frac{2(x-4)-\left(\sqrt x-2\right)}{2\sqrt x+4}\\&=\left(\sqrt x-2\right)\frac{2\sqrt x+3}{2\sqrt x+4}\\&=(x-4)\frac{2\sqrt x+3}{\left(\sqrt x+2\right)\left(2\sqrt x+4\right)}\\&\leqslant(x-4)\frac{2\sqrt x+3}8,\end{align}and therefore$$\left|\frac{x-2}{\sqrt x+2}-\frac12\right|\leqslant|x-4|\frac{2\sqrt x+3}8.$$So, if $|x-4|<1$, you have $x<5$, and therefore\begin{align}\frac{2\sqrt x+3}8&<\frac{2\sqrt5+3}8\\&<\frac{2\times3+3}8\\&=\frac98\\&<2.\end{align}Therefore, given $\varepsilon>0$, if you take $\delta=\min\left\{1,\frac\varepsilon2\right\}$, you have$$|x-4|<\delta\implies\left|\frac{x-2}{\sqrt x+2}-\frac12\right|<\varepsilon$$
|
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|
For $a,b,c>0$ and $a+b+c=6$. Prove that $\sum^{}_{cyc} \frac{ab}{\sqrt{a^2+b^2+2c^2}}\leq 3$ For $a,b,c>0$ and $a+b+c=6$. Prove that $$\sum^{}_{cyc} \frac{ab}{\sqrt{a^2+b^2+2c^2}}\leq 3$$
And this attempt $$\sum \frac{ab}{\sqrt{a^2+b^2+2c^2}}$$
$$=\sum \sqrt{\frac{a^2b^2}{\sqrt{a^2+b^2+2c^2}}}$$
$$\leq \sqrt{3}.\sqrt{\frac{(ab+ca+ca)^2}{4(a^2+b^2+c^2)}}$$
$$\leq \sqrt{3}.\sqrt{\frac{a^2+b^2+c^2}{4}}$$
|
From $(a+b+c)^2 \geqslant 3(ab+bc+ca),$ we get $ab+bc+ca \leqslant 12.$
Now, using the Cauchy-Schwarz and AM-GM inequality, we have
$$\left(\sum \frac{ab}{\sqrt{a^2+b^2+2c^2}}\right)^2 \leqslant (ab+bc+ca)\sum \frac{ab}{a^2+b^2+2c^2}$$
$$ \leqslant \frac{1}{4}(ab+bc+ca) \sum \frac{(a+b)^2}{a^2+b^2+2c^2}$$
$$ \leqslant \frac{1}{4}(ab+bc+ca) \sum \left(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\right) $$
$$ = \frac{3}{4}(ab+bc+ca) \leqslant 9.$$
Therefore
$$\sum \frac{ab}{\sqrt{a^2+b^2+2c^2}} \leqslant 3.$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Diagonalizing the matrix $A$ then finding $A^{10}$ Diagonalize the matrix
$$A= \begin{pmatrix} -3 & -14 & -10\\ 2 & 13 & 10\\ -2 & -7 & -4 \end{pmatrix}$$
Then find $A^{10}.$
We have the characteristic polynomial of $A:$
$$\left | A- \lambda I \right |= \left | \begin{pmatrix} -3 & -14 & -10\\ 2 & 13 & 10\\ -2 & -7 & -4 \end{pmatrix}- \lambda\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} \right |= \begin{vmatrix} -3- \lambda & -14 & -10\\ 2 & 13- \lambda & 10\\ -2 & -7 & -4- \lambda \end{vmatrix}=$$
$$= \left ( 6- \lambda \right )\left ( 1+ \lambda \right )\left ( 1- \lambda \right )$$
which has the roots $\lambda_{1}= 6, \lambda_{2}= -1, \lambda_{3}= 1.$ These roots are the diagonal elements as well as the eigenvalues of $A.$ These three eigenvalues correspond to the eigenvectors $v_{1}= \left ( 2, -2, 1 \right ),$
$v_{2}= \left ( 2, -1, 1 \right ), v_{3}= \left ( 1, -1, 1 \right ).$ Hence, $A= SJS^{-1}$ with $S= \begin{pmatrix} 2 & 1 & 2\\ -1 & -1 & -2\\ 1 & 1 & 1 \end{pmatrix},$ $J= \begin{pmatrix} -1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 6 \end{pmatrix}, S^{-1}= \begin{pmatrix} 1 & 1 & 0\\ -1 & 0 & -2\\ 0 & -1 & -1 \end{pmatrix}.$ I have no plans to find $A^{10}.$ I need to the helps
|
Observe that if $A=SJS^{-1}$ we also have $A^n = SJ^n S^{-1}$ for all $n \in \mathbb N$. We can see this by induction with $A^2= SJS^{-1}SJS^{-1} = SJ^2S^{-1}$ and $A^n = SJS^{-1} SJ^{n-1}S^{-1} = SJ^nS^{-1}$
Now $J^{10} = \text{diag}(1,1, 6^{10})$
and $A^{10} = SJ^{10}S^{-1}$.
|
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|
prove that $\displaystyle\int_0^1 \dfrac{1}{x^x}dx = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^3} + \dfrac{1}{4^4} + \ldots $ $\displaystyle\int_0^1 \dfrac{1}{x^x}dx = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^3} + \dfrac{1}{4^4} + \ldots $
the only idea I have is using the series expansion of $x^{-x} \approx 1 - x\log x + \dfrac{(x\log x)^2}{2!} - \ldots$. But it ends up little complicated, any idea?
|
Uniform convergence of the power series allows to interchange summation and integration:
\begin{aligned}
&\int_{0}^{1} x^{-x} \, \mathrm d x=\int_{0}^{1} e^{-x \ln x}\, \mathrm d x \Rightarrow \text { Applying Power Series of } e^{k x}=\sum_{n=0}^{\infty} \frac{(k x)^{n}}{n !} \\
&\Rightarrow \int_{0}^{1} \sum_{n=0}^{\infty} \frac{(-\ln x)^{n}}{n !} x^{n} \, \mathrm d x \Rightarrow \sum_{n=0}^{\infty} \frac{1}{n !} \int_{0}^{1} x^{n}(-\ln x)^{n} \, \mathrm d x
\end{aligned}
\begin{aligned}
&\Rightarrow\left\{u=\ln x \Leftrightarrow e^{u}=x \\ \mathrm d x=e^{u} \, \mathrm d u\right\} \Rightarrow \sum_{n=0}^{\infty} \frac{1}{n !} \int_{-\infty}^{0} e^{u(n+1)}(-u)^{n}\, \mathrm d u \\
&\Rightarrow\left\{t=-u(n+1) \Leftrightarrow u=-\frac{t}{n+1} \\ \mathrm d u=-\frac{\mathrm d t}{n+1}\right\} \Rightarrow \sum_{n=0}^{\infty} \frac{1}{n !} \int_{\infty}^{0} e^{-t}\left(\frac{t}{n+1}\right)^{n} \frac{(-1)}{n+1}\, \mathrm d t
\end{aligned}
$\Rightarrow \sum_{n=0}^{\infty} \frac{1}{n !} \int_{0}^{\infty} e^{-t} \frac{t^{n}}{(n+1)^{n+1}} \, \mathrm d t=\sum_{n=0}^{\infty} \frac{1}{n !(n+1)^{n+1}} \int_{0}^{\infty} e^{-t} t^{n} \, \mathrm d t=\sum_{n=0}^{\infty} \frac{\Gamma(n+1)}{n !(n+1)^{n+1}}=\sum_{n=1}^{\infty} \frac{1}{n^{n}}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $ without expansion It's easy to prove that if $a,b,c \neq 0 $: $$ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $$
as $\frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff \frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\iff (ab+bc+ca)(a+b+c)=abc \iff (a+b)(b+c)(c+a)=0 $
I'm curious to prove this some other ways. I tried to use function and inequality but still no progress.
|
$p(x)=(x-a)(x-b)(x-c)=x^3-ux^2+vx-w$ has roots $a,b,c$ with sum $a+b+c=u$.
$(a+b)(b+c)(c+a)=0 \iff (u-c)(u-a)(u-b)=0 \iff p(u) = 0\,$ thus:
$$
\require{cancel}
\cancel{u^3}-\cancel{u\,u^2} + v\,u-w=0 \;\;\iff\;\; \frac{v}{w}=\frac{1}{u} \;\;\iff\;\; \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}
$$
|
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|
How to check if an integer can be represented with all set bits for any given base? By all set bits, I mean all the set bits are placed consecutively together. e.g., for base 2, we have 1, 3, 7, 15, 31, 63, i.e., any integer $x$ such that $x = 2^m - 1$ for some integer $m$.
In binary, $1$ is $1$ and $3$ is $11$ and $7$ is $111$ and $15$ is $1111$ and so on.
I want to be able to do this with any base. Based on wolfram, I am inducing that the formula for checking an integer $x$ and some base $b$ and some non-negative exponent $m$ is
$$
x = \frac{1}{b - 1}\left(b^m - 1 \right)
$$
How can you intuitively derive this formula?
After I wrote out $x = b^0 + b^1 + b^2 + \ldots + b^m$, it became obvious to me how to derive this expression.
Let $f(m) = x = b^0 + b^1 + b^2 + \ldots + b^m$, then we have
$$
f(m) / b = 1/b + b^0 + b^1 + \ldots + b^{m - 1} \\
= 1/b + f(m) - b^m \\
\implies f(m) = 1 + bf(m) - b^{m + 1} \\
\therefore f(m) = \frac{b^{m + 1} - 1}{b - 1}
$$
|
Your formula for $x$ is a geometric sum of $m$ digits of all $1$s,
$$
{\overbrace{111\cdots1}^m}_b = b^0+b^1+\cdots + b^{m-1} = \frac{b^m-1}{b-1} = \frac{10_b^m-1}{10_b-1}$$
To apply the usual proof of geometric sum specifically to $x$ in base $b$,
$$\begin{array}{crcrl}
&10_b x &= &{\overbrace{111\ldots 1}^m0}_b\\
-& x &= &{\overbrace{11\ldots 11}^m}_b\\
\hline
&(10_b-1)x &= &1{\overbrace{00\ldots00}^m}_b & -1\\
&&= &10^m_b&-1\\
\hline
&x &= &\dfrac{10^m_b-1}{10_b-1}
\end{array}$$
|
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|
Throwing a coin - expected value of tails There are $10$ coins: $8$ of them are fair (equal probability for heads and tails) and for $2$ probability of heads is two times higher than for tails (so $\frac{2}{3}$ for heads). We draw one coin and throw it three times. Let $X$ denote a number of tails - I need to find a $\Bbb EX$. So:
$\Bbb P(X = k) = \Bbb P(X = k| \text{fair coin} ) \cdot \Bbb P(\text{fair coin}) + \Bbb P(X = k| \text{unfair coin} ) \cdot \Bbb P(\text{unfair coin})$
With
$\Bbb P(X = k| \text{fair coin} )={3 \choose k} (\frac{1}{2})^k(\frac{1}{2})^{3-k}$
$\Bbb P(\text{fair coin}) = \frac{8}{10} $
$\Bbb P(X = k| \text{unfair coin} )={3 \choose k} (\frac{1}{3})^k(\frac{2}{3})^{3-k} $
$\Bbb P(\text{unfair coin})= \frac{2}{10} $
So
$\Bbb P(X = k) = {3 \choose k} (\frac{1}{2})^k(\frac{1}{2})^{3-k} \cdot \frac{4}{5} + {3 \choose k} (\frac{1}{3})^k(\frac{2}{3})^{3-k}\cdot \frac{1 }{5} $
And then obviously $\Bbb EX = \sum\limits_{i=0}^3 i\cdot \Bbb P(X=i)$
Am I correct?
|
You know that the expectation of a binomial is $\mu=np$ thus you have
$$E(X)=\frac{8}{10}\times 3 \times \frac{1}{2}+\frac{2}{10}\times 3 \times \frac{1}{3}=\frac{14}{10}$$
this easy method works as you pmf is a mixture of two pmf's
$$p_X(x)=0.8p_{X_1}(x)+0.2p_{X_2}(x)$$
|
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|
How to graph and solve this equation? I'am trying to solve this equation.
\begin{equation}
x^{8}+(x+2)^{8}=2
\end{equation}
What I tried:
\begin{equation}g(x)=x^{8}+(x+2)^{8}\end{equation}
\begin{equation}
\begin{array}{l}
\text { }\\
y=x+1
\end{array}
\end{equation}
\begin{equation}
(y+1)^{8}+(y-1)^{8}=2
\end{equation}
\begin{equation}
(y+1)^{8}=y^{8}+a_{7} y^{7}+a_{6} y^{6}+\cdots+a_{1} y+1
\end{equation}
\begin{equation}
(y-1)^{8}=y^{8}-a_{7} y^{7}+a_{6} y^{6}+\cdots-a_{1} y+1
\end{equation}
\begin{equation}
\begin{array}{l}
2\left(y^{8}+a_{6} y^{6}+a_{4} y^{4}+a_{2} y^{2}+1\right)=2 \Leftrightarrow \\
y^{8}+a_{6} y^{6}+a_{4} y^{4}+a_{2} y^{2}=0 \Leftrightarrow y^{2}=0 \Leftrightarrow y=0
\end{array}
\end{equation}
y=0, x= -1
Maybe there is another way of solving this...
|
Let's try
$$ 1 + 1 = 2.$$
$$ \left\{
\begin{aligned}
x^8=1\\
(x+2)^8=1
\end{aligned}
\right.
$$
A solution is $x=-1.$
Actually, $x=-1$ is a double root:
Let $q(x) = x^8 + (x+2)^8 -2$. So we have $$q(x)=0.$$ We can factor q(x):
$$q(x) = 2 (x+1)^2 (127 + 258 x + 253 x^2 + 132 x^3 + 43 x^4 + 6 x^5 + x^6).$$
The equation $127 + 258 x + 253 x^2 + 132 x^3 + 43 x^4 + 6 x^5 + x^6=0$
has no real solutions.
|
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|
Find the value of $α^3+β^3+γ^3+δ^3$ given that $α,β,γ,δ$ are roots of $x^4+3x+1=0$ Let $α,β,γ,δ$ be the roots(real or non real) of the equation $x^4-3x+1=0$. Then find the value of $α^3+β^3+γ^3+δ^3$.
I tried this question as $S_1=0, S_2=0, S_3=3, S_4=1$ and then I used $S_1^3$ to find the the value of asked question, but i am not able to factorise it further and it seems like a dead end. Moreover it is very lengthy method so can you tell of any other more elegant way of approaching this question?
|
Using Vieta's theorem for $x^4+3x+1=0$, we get:
$$a+b+c+d=0 \Rightarrow b+c+d=-a\\
ab+ac+ad+bc+bd+cd=0 \Rightarrow bc+bd+cd=a^2\\
abc+abd+acd+bcd=-3\Rightarrow a^3=-bcd-3\\
abcd=1$$
Hence:
$$a^3+b^3+c^3+d^3=(-bcd-3)+(-acd-3)+(-abd-3)+(-abc-3)=\\
-(abc+abd+acd+bcd)-12=3-12=-9.$$
Note: $x^4+3x+1=0$ is given in the title, but $x^4-3x+1=0$ is given in the body. The same method for the second equation will produce $9$.
|
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Alternative approaches to maximize $y=x\sqrt{100-x^2}$ I could find three good approaches to find maximum of the function $y=x\sqrt{100-x^2}$. I will explain them briefly :
First: Finding $x$ satisfies $y'=0$ then plugging it in the function.
Second: Using the substitution $x=10\sin\theta$ (or $x=10\cos\theta)$ for
$\theta\in(0,\frac{\pi}2)$ to get $y=100\sin\theta\cos\theta=50\sin(2\theta)$ hence the maximum is $50$.
Third: Using AM-GM inequality: It is obvious that maximum occurs for $x>0$ So we can rewrite $y$ as $y=\sqrt{x^2(100-x^2)}$ . Now the sum of $x^2$ and $100-x^2$ is $100$ so the maximum of product happens when $x^2=100-x^2$ or $x^2=50$ Hence $y_{\text{max}}=50$.
Just for fun, can you maximize $y=x\sqrt{100-x^2}$ with other approaches?
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Let $f(x) = x\sqrt{100-x^{2}}$. Then $x\in[-10,10]$.
In order to find the maximum value, we can consider that $x\geq 0$ (otherwise $f(x) \leq 0$).
For such values of $x$, one has that
\begin{align*}
f(x) & = x\sqrt{100 - x^{2}}\\\\
& = \sqrt{100x^{2} - x^{4}}\\\\
& = \sqrt{2500 - (2500 - 100x^{2} + x^{4})}\\\\
& = \sqrt{2500 - (50-x^{2})^{2}}
\end{align*}
Since the square root function is strictly increasing, $f(x)$ attains its maximum when $x^{2} = 50$.
Hopefully this helps!
|
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How to convert $\alpha = \cos(\frac{8\pi}{11})+ i \sin(\frac{8\pi}{11})$ into $\text{n}^{th}$ roots of unity format? Context: Preparing for JEE and it is one of the practice problems.
Question:
If $\alpha = cos(\frac{8\pi}{11})+ i sin(\frac{8\pi}{11})$ then $\Re({\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5})$ is equal to
A) $\frac{1}{2}$
B) $-\frac{1}{2}$
C) $0$
D) None
Key: Option B
What I've tried:
$$\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5 = e^\frac{8\pi}{11}+e^\frac{16\pi}{11}+e^\frac{24\pi}{11}+e^\frac{32\pi}{11}+e^\frac{40\pi}{11}$$
Then I was struck here knowing that the given argument is not in form of $\frac{2k\pi}{n}$ so now I cant use $\text{n}^{th}$ root of unity
So what should I do now? Is there any way to convert in form of $\alpha = e^{i(\frac{2k\pi}{n})}$
where $k = {0, 1, ... n-1}?$
Or is there any other way to solve it?
|
Roots of unity is indeed the correct approach here, rather than geometric series.
$a^{11} = 1$ and $a \bar{a} = 1$. Thus the conjugate of $a$, $\bar{a} = \frac{1}{a} = \frac{1}{a} \cdot a^{11} = a^{10}$, which has the same real part as $a$. Now since:
$$1 + (a + a^2 + \cdots + a^5) + (a^6 + a^7 + \cdots + a^{10})$$
$$ = 1 + (a + a^2 + \cdots + a^5) + (\overline{a^5} + \overline{a^4}+ \cdots + \bar{a}) = 0,$$
it hence follows that the real part is $-1/2$.
|
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A Problem dealing with a Deck of Cards and the Hypergeometric Distribution Problem:
If $13$ cards are to be chosen at random(without replacement), determine
the probability that (a) $6$ will be picture cards, (b) none will be
picture cards.
Answer: (a)
The probability that a card drawn from a deck of cards will be a
picture card is $\dfrac{3}{13}$. It would tempting to
model this as a binomial distribution but that would be wrong because
it is done without replacement. Instead we use the hypergeometric distribution.
The general form of the hypergeometric distribution is:
$$P(X = x) = \dfrac{ {{b}\choose{x}} { {r} \choose {n-x} } }
{ {{b+r} \choose {n}} } $$
In this case, we have:
\begin{align*}
x &= 6 \\
n &= 13 \\
b &= 4(3) = 12 \\
r &= 52 - 12 = 40 \\
p &= \dfrac{12}{52} = \frac{3}{13} \\
\end{align*}
Let $p_a$ be the probability we seek.
\begin{align*}
p_a &= \dfrac{ {{12}\choose{6}} { {40} \choose {13-6} } }
{ {{12+40} \choose {13}} } \\
%
p_a &= \dfrac{ {{12}\choose{6}} { {40} \choose {7} } }
{ {{52} \choose {13}} } \\
\end{align*}
However, the book gets:
$$ \dfrac{ {{13}\choose{6}} { {39} \choose {7} } } { {{52} \choose {13}} } $$
Where did I go wrong?
|
Your answer is correct, and the book is wrong.
|
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Finding all polynomials $P(x)$ satisfying $(P(x)+P(\frac{1}{x}))^2 =P(x^2)P(\frac{1}{x^2})$
Find all polynomials $P(x)$ satisfying
$$\left(P(x)+P\left(\frac{1}{x}\right)\right)^2 =P(x^2)P\left(\frac{1}{x^2}\right)$$
If $P'(x)\neq 0$, then $P(x) = a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$
$\Rightarrow$ ($a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0+a_n\frac{1}{x^n}+a_{n-1}\frac{1}{x^{n-1}}+...+a_1\frac{1}{x}+a_0)^2 = (a_nx^{2n}+a_{n-1}x^{2n-2}+...+a_1x^2+a_0)( a_n\frac{1}{x^{2n}}+a_{n-1}\frac{1}{x^{2(n-1)}}+...+a_1\frac{1}{x^2}+a_0)$
Considering the coefficient of $x^{2n-1}$ we see : $2a_na_{n-1} = 0$ $\Rightarrow a_{n-1}= 0$
Considering the coefficient of $x^{2n-3}$ we see : $2a_na_{n-3}+2a_{n-1}a_{n-2}= 0$
$\Rightarrow a_{n-3}= 0$
Similarly, we can prove $a_{n-5}=0;a_{n-7}=0;a_{n-9}=0;...$
But at this point I have no further ideas, it is certain that all the coefficients of $P(x)$ cannot simultaneously be zero except $a_n$ (because it is easy to see that $P(x) = Cx^n$ does not satisfy the requirement)
Looking forward to getting help from everyone. Thank you very much! I sincerely apologize for my mistake. I have corrected my post, please forgive my ignorance.
|
We claim that the only solutions of the equation is $P(x) = 0$.
Let $P(x) = \sum_{k} a_k x^k$ satisfy the given equation. Write $d = \deg P $.
Lemma. $P(x)$ is an even polynomial.
Proof. The conclusion is obvious if $P(x)$ is constant, so we only consider the case where $P(x)$ is non-constant.
First, we show that $P(x)$ is either even or odd. Otherwise, there exists a smallest positive odd integer $m$ such that $a_{d-m} \neq 0$. Then by comparing the coefficients of $x^{2d-m}$ in both sides of the equation, we get $ 2a_d a_{d-m} = 0, $, and so, $a_{d-m} = 0$, a contradiction.
Next, by comparing the coefficients of $x^{2d}$ in both sides of the equation, we get $a_d^2 = a_d a_0$, and so, $a_0 = a_d$. This cannot happen if $P(x)$ is odd, and therefore the lemma is proved. $\square$
Now suppose $P(x)$ solves the equation.
*
*Assume $P(x)$ is non-constant. By the above lemma, $P(x)$ is an even polynomial. In particular, $Q(x) = P(\sqrt{x})$ is still a polynomial in $x$ and satisfies
$$ (Q(x) + Q(x^{-1}))^2 = (P(x^{1/2} + P(x^{-1/2}))^2 = P(x)P(x^{-1}) = Q(x^2)Q(x^{-2}). $$
So $Q(x)$ also solves the equation, and in particular, $\deg Q = \frac{1}{2}\deg P$ is also an even integer. Repeating this argument, we find that $\frac{1}{2^k}\deg P$ is an even integer for all $k \geq 0$, which is impossible.
*The above contradiction tells that $P(x)$ must be constant. Then by plugging $P(x) = a$ to the equation, we find that $4a^2 = a^2$, and hence $a = 0$. Therefore $P(x) = 0$.
|
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Different answers of same differentiation Question with two different methods
Question:- Find $\frac{dy}{dx}$ if $$\arccos\bigg({\frac{x^2-y^2}{x^2+y^2}}\bigg)=\arctan( a)$$
$$\arccos\bigg({\frac{x^2-y^2}{x^2+y^2}}\bigg)=\arctan( a)$$
$$\implies \frac{x^2-y^2}{x^2+y^2}=\cos(\arctan(a))$$
Taking derivative on both sides w.r.t $\space x$, we get $\frac{dy}{dx}=\frac{y}{x}$
Out of fun, I tried the substitution $y^2=x^2\cos(\theta)$ and got
$$\frac{1-\cos(\theta)}{1+\cos(\theta)}=\cos(\arctan(a))$$ for $x\ne 0$
$$\implies \tan^2{\frac{\theta}{2}}=\cos(\arctan(a))$$
Differentiate both sides w.r.t $\space \theta$, we get
$$\tan{\frac{\theta}{2}}\bigg(1+\tan^2{\frac{\theta}{2}}\bigg)=0$$
$$\implies \tan{\frac{\theta}{2}}=0$$
$$\cos(\theta)=\frac{1-\tan^2{\frac{\theta}{2}}}{1+\tan^2{\frac{\theta}{2}}}=1$$
As $y^2=x^2\cos(\theta)$,this gives $y^2=x^2$ and $\frac{dy}{dx}=\pm 1$ Which contradicts the other method, So What's wrong with the $2^{nd}$ method?
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In more detail, suppose that for some constant $c$
$$ c =\frac{x^2-y^2}{x^2+y^2}. \tag{1} $$
Note that this is undefined iff $\,x^2+y^2 = 0.\,$
Solve this for $\,y\,$ explicitly to get
$$ y = \pm \sqrt{\frac{1-c}{1+c}}x = \pm C\, x. \tag{2}$$
Take the implicit differential of equation $(1)$ to get
$$ 0 = \frac{4 x y (x\, dy - y\, dx)}{(x^2+y^2)^2}. \tag{3}$$
This implies that $\,\frac{dy}{dx}=\frac{y}{x}\,$
assuming that $\,x^2+y^2\ne 0\,$ and $\,x\ne 0.\,$
Note that $\,\frac{dy}{dx}=-1\,$ is true if $\,y=-x,\,$
but even in this case, $\, \frac{dy}{dx}=\frac{y}{x}\,$
so there is no contraction.
In other words, if $\,y = \pm C\,x,\,$ then
$\, \frac{dy}{dx} = \pm C = \frac{y}{x}. $
By the way, you can't
Differentiate both sides w.r.t $\,\theta$
because the explicit solution in equation $(2)$ implies that
$\,\theta\,$ is a constant.
|
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Is $\frac{1-\alpha}{1+\alpha}=y \Rightarrow \alpha=\frac{1-y}{y+1}$ correct even if $y=-1$? I was trying to solving this question:
If roots of the equation $a x^{2}+b x+c=0$ are $\alpha$ and $\beta$, find the equation whose roots are $\frac{1-\alpha}{1+\alpha}, \frac{1-\beta}{1+\beta}$
I was not able to solve it so I looked to the solution given in the book, it was as follows:
Let $\frac{1-\alpha}{1+\alpha}=y \Rightarrow \alpha=\frac{1-y}{1+y}$
.
Now $\alpha$ is root of the equation $a x^{2}+b x+c=0$ $\Rightarrow a \alpha^{2}+b \alpha+c=0$
$\Rightarrow \quad a\left(\frac{1-y}{1+y}\right)^{2}+b\left(\frac{1-y}{1+y}\right)+c=0$
. Hence required equation is $a(1-x)^{2}+b\left(1-x^{2}\right)+c(1+x)^{2}=0$
In the first step the auther used componendo dividend rule as follows:
$\begin{aligned} & \frac{1-\alpha}{1+\alpha}=y \\ \Rightarrow & \frac{1-\alpha+1+\alpha}{1-\alpha-(1+\alpha)}=\frac{y+1}{y-1} \\ \Rightarrow & \frac{2}{-2 \alpha}=\frac{y+1}{y-1} \\ \Rightarrow & \frac{-1}{\alpha}=\frac{y+1}{y-1} \Rightarrow \alpha=\frac{1-y}{y+1} \end{aligned}$
But I don't understand how he could use it, as we are not sure whether $y$ could be - 1 and hence $1 +y$ can be zero. So is this an error? If so then how can we solve this question and if not then why not?
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Is $\frac{1-\alpha}{1+\alpha}=y \Rightarrow \alpha=\frac{1-y}{y+1}$ correct even if $y=-1$?
The tag algebra-precalculus is given. The answer to the question is: No.
But with tag riemann-sphere, the answer to the question would be: Yes.
Calculating in the Riemann sphere, we have
$$
\frac{1-\infty}{1+\infty} = -1\qquad\text{and}\qquad \frac{1-(-1)}{(-1)+1} = \infty .
$$
|
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How to evaluate the following limit $\lim_{n\to \infty} \sum_{k=2}^n\log_{\frac{1}{3}} \left(1-\frac{2}{k(k+1)}\right)$? In order to evaluate this limit :
$$\lim_{n\to \infty} \sum_{k=2}^n\log_{\frac{1}{3}} \left(1-\frac{2}{k(k+1)}\right)$$
I have to compute the following sum :
$$ \sum_{k=2}^n\log_{\frac{1}{3}} \left(1-\frac{2}{k(k+1)}\right)$$
$$1-\frac{2}{k(k+1)}=\frac{k^2+k-2}{k^2+k}$$
So :
$$\log_{\frac{1}{3}} \left(\frac{k^2+k-2}{k^2+k}\right)=\log_{\frac{1}{3}}\left(k^2+k-2\right) -\log_{\frac{1}{3}} \left(k^2+k\right)$$
I'm really lost now, and the worse is Wolfram Alpha is giving to me this crazy value :
$$S_n=\frac{\ln (3 \Gamma(n+1) \Gamma(n+2)-\ln (\Gamma(n) \Gamma(n+3))}{\ln 3}$$
And when $$\lim_{n\to \infty} S_n =1$$
Should I really use "Gamma function" to solve this 12th grade sum ? Or is there any other method ?
|
In this answer I'm considering $b=\frac{1}{3}$.
\begin{align}
\sum_{k=2}^n \log_b\left(1-\frac{2}{k(k+1)}\right)&= \sum_{k=2}^n \log_b\left(\frac{k^2+k-2}{k(k+1)}\right)\\
&=\sum_{k=2}^n \log_b \left(\frac{k+2}{k+1}\right) +\log_b \left(\frac{k-1}{k}\right)\\
&=\sum_{k=2}^n \log_b (k+2)-\log_b(k+1) +\sum_{k=2}^n \log(k-1)-\log_b(k)\\
&=\sum_{k=2}^n \log_b (k+2)-\log_b(k+1) -\sum_{k=2}^n \log(k)-\log_b(k-1)\\
&=\log_b(n+2)-\log_b(3) -\log_b(n)+\log_b(1)\\
&=\log_b\left(\frac{n+2}{n}\right)-\frac{\ln(3)}{\ln(1/3)}\\
&=\log_b\left(\frac{n+2}{n}\right)+\frac{\ln(3)}{\ln(3)}\\
&=\log_{\frac{1}{3}} \left(1+\frac{2}{n}\right)+1\\
\lim_{n\to \infty} \sum_{k=2}^n \log_b\left(1-\frac{2}{k(k+1)}\right)&=\lim_{n\to \infty} \log_{1/3}\left(1+\frac{2}{n}\right)+1\\
&=1
\end{align}
Recall :
$$\frac{1}{n} \overset{n\to \infty}{\longrightarrow} 0 \ \ \ \ \ \ \text{;} \ \ \ \ \ \ \log_a(1)=0 \ \ \ \ \ \ \text{;}\ \ \ \ \ \ \log_b (x)=\frac{\ln x}{\ln b}\ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ln(1/a)=-\ln(a)$$
|
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With $x^2+y^2=1$ find Minimum and Maximum of $x^5+y^5$ (do not use derivative) It's easy to see that the minimum is $-1$ and maximum is $1$.
My idea is put $x=\cos(a)$, $y=\sin(a)$ and $t=x+y$, so I have $-\sqrt{2}\le t \le \sqrt{2}$ then $0 \le t^2 \le 2$
When $t=x+y$ then $(x+y)^2=1+2xy$.
Then
\begin{aligned}
x^5+y^5&=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)\\
&=x^3+y^3-x^2y^2(x+y)\\
&=(x+y)(x^2-xy+y^2)-x^2y^2(x+y)\\
&=(x+y)(1-xy-x^2y^2)\\
&=\left( {x + y} \right)\left( {1+\frac{1}{2} - {{\left( {x + y} \right)}^2} - \frac{{{{\left( {x + y} \right)}^4}}}{4} + \frac{{{{\left( {x + y} \right)}^2}}}{2} - \frac{1}{4}} \right)\\
&=t\left( {\frac{5}{4} - \frac{{{t^4}}}{4}} \right)\\
\end{aligned}
This problem is for those who have not studied derivatives so I dont have any idea for next step. Anyone can help me for the hint or other solutions? Very Thanks
|
My solution using AM -GM.
It's more complex than @Aman Kushwaha. His solution is very simple and naturally. Sorry for not using MathJax
|
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|
Find the minimum of $P(X=0)$ when $E[X]=1,E[X^2]=2,E[X^3]=5$ using the probability generating function I was given the following exercise.
Let $X$ be a random variable that takes non-negative natural number
values such that $E[X]=1,E[X^2]=2,E[X^3]=5$. Find the minimum value of
$P(X=0)$ using the taylor expansion of the probability generating function
at $z=1$.
I know the method not using the generating function stated in this question.
My attempt:
Let $G(z)=E[z^X]$ be the probability generating function of $X$. Then by definition,
$G(z) = P(X=0)+P(X=1)z+P(X=2)z^2+\cdots $
Also, since $E[\frac{X(X-1)\cdots (X-n+1)}{n!}]=\frac{G^{(n)}(1)}{n!}$ according to Wikipedia, we have
$\displaystyle G(z) = \sum_{n=0}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](z-1)^n$
Substituting $z=0$ yields
$\displaystyle P(X=0) = \sum_{n=0}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](-1)^n$
Note that for $n=0,1,2,3$, the coefficient can be calculated as follows:
$\begin{align} E[1] &= 1 \\ E[X]&= 1 \\ E[X(X-1)/2] &= E[X^2]/2 -E[X]/2 = 1-1/2 =1/2 \\ E[X(X-1)(X-2)/3!] &= E[X^3]/6 -E[X^2]/2 +E[X]/3 = 5/6-2/2+1/3 \\ &= 1/6 \end{align}$
Therefore,
$\displaystyle \begin{align} P(X=0) &= \sum_{n=0}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](-1)^n \\ &=1-1+\frac{1}{2} -\frac{1}{6} +\sum_{n=4}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](-1)^n \\ &= \frac{1}{3} + \sum_{n=4}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](-1)^n\end{align}$
According to the question linked above, $1/3$ is the minimum, so I think we need to prove that the sum is non-negative to complete the proof. However, I was unable to do so.
Am I on the right path? If not, what is the correct one? If yes, how can I finish it?
|
Conditional on a given value $X=N\geq 4$ we get the tail sum
$$\sum_{n=4}^N {N\choose n} (-1)^n=- ({N\choose 0}-{N\choose 1} +{N\choose 2}-{N\choose 3})=$$
$$-(1-N+N(N-1)/2+N(N-1)(N-2)/6)$$
$$=\frac{1}{6} (N-1)(N-2)(N-3)={N-1\choose 3}>0.$$
(We re-derived a special case of a known identity for binomial coefficients.)
Of course, conditional on $X<4$ the tail sum is $0$. Thus the tail sum is non-negative (and is zero precisely if $X$ has zero probability of being above $3$).
Observe that we get that the tail sum is the expectation of $\frac{1}{6} (X-1)(X-2)(X-3)$, and basically recover the solution to which that you linked.
|
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Converting $\intop_{0}^{a}\sqrt{\frac{a^{2}-x^{2}}{1-x^{2}}}dx$ to elliptic integral Tried using $x=a\sin\left(\theta\right)$ $$\rightarrow \intop_{0}^{\pi/2}\sqrt{\frac{a^{2}-\left(a\sin\left(\theta\right)\right)^{2}}{1-\left(a\sin\left(\theta\right)\right)^{2}}}a\cos{\left(\theta\right)d\theta}$$
$$\iff \intop_{0}^{\pi/2}\frac{a^{2}\cos^{2}\left(\theta\right)}{\sqrt{1-\left(a\sin\left(\theta\right)\right)^{2}}}{d\theta}$$
Which looks similar to the complete elliptical of first or second kind, is there a way to make the conversion? Thanks
|
You can convert this integral in terms of elliptic functions. You need to make an assumption about the parameter $a$:
If $0<a<1$ we can continue with your approach:
First, note
$$ J=\intop_{0}^{\pi/2} \frac{\sin^{2}\left(\theta\right)}{\sqrt{1-k^2\sin^2\left(\theta\right)}}{d\theta} = \frac{K(k)-E(k)}{k^2}$$
To show this recall the definition of the complete elliptic integrals of the first and second kind, respectively:
$$K(k) = \intop_{0}^{\pi/2} \frac{1}{\sqrt{1-k^2\sin^2\left(\theta\right)}}{d\theta}$$
$$E(k) = \intop_{0}^{\pi/2} \sqrt{1-k^2\sin^2\left(\theta\right)}{d\theta}$$
Hence
$$\frac{1}{k^2}\left[K(k)-E(k)\right] = \frac{1}{k^2} \intop_{0}^{\pi/2} \left[\frac{1}{\sqrt{1-k^2\sin^2\left(\theta\right)}} - \sqrt{1-k^2\sin^2\left(\theta\right)}\right]{d\theta} =\frac{1}{k^2} \intop_{0}^{\pi/2} \left[\frac{1}{\sqrt{1-k^2\sin^2\left(\theta\right)}} - \frac{1-k^2\sin^2(\theta)}{\sqrt{1-k^2\sin^2\left(\theta\right)}}\right]{d\theta} = \intop_{0}^{\pi/2} \frac{\sin^{2}\left(\theta\right)}{\sqrt{1-k^2\sin^2\left(\theta\right)}}{d\theta} $$
Then
$$ I = \intop_{0}^{\pi/2}\frac{a^{2}\cos^{2}\left(\theta\right)}{\sqrt{1-a^2\sin^2\left(\theta\right)}}{d\theta} = a^2\intop_{0}^{\pi/2}\frac{1-\sin^{2}\left(\theta\right)}{\sqrt{1-a^2\sin^2\left(\theta\right)}}{d\theta} = a^2\intop_{0}^{\pi/2}\frac{1}{\sqrt{1-a^2\sin^2\left(\theta\right)}}{d\theta} - a^2\intop_{0}^{\pi/2}\frac{\sin^{2}\left(\theta\right)}{\sqrt{1-a^2\sin^2\left(\theta\right)}}{d\theta} = a^2K(a) -K(a)+E(a) =E(a)-a'^2K(a)$$
where $a' = \sqrt{1-a^2}$ is the complementary modulus
Hence
$$\boxed{\intop_{0}^{a}\sqrt{\frac{a^{2}-x^{2}}{1-x^{2}}}dx= E(a)-a'^2K(a)}$$
Note that Wolfram use a slight different notation for Elliptic integrals.
There are some nice approximations for the complete Elliptic integrals.
$$ K(k) \approx \frac{\pi}{2} \left(\frac{16-5k^2}{16-9k^2}\right) \quad 0\leq k\leq 0.67$$
$$ E(k) \approx \frac{\pi}{2} \left(\frac{16-7k^2}{16-3k^2}\right) \quad 0\leq k \leq 0.71$$
and you can find others when $k$ is close to $1$.
You can also find series expansions for both functions.
|
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|
I don't know how to exactly compute this determinant I've tried to compute this determinant by row transformations and column transformations, but it gives me a formula that doesn't work. The determinant is:
\begin{vmatrix}
x & a & b & c & d\\
a & x & b & c & d\\
a & b & x & c & d\\
a & b & c & x & d\\
a & b & c & d & x
\end{vmatrix}
I thought I could start doing row 5 - row 4, row 4 - row 3, row 3 - row 2 and row 2 - row 1 and then you get this determinant:
\begin{vmatrix}
x & a & b & c & d\\
a-x & x-a & 0 & 0 & 0\\
0 & b-x & x-b & 0 & 0\\
0 & 0 & c-x & x-c & 0\\
0 & 0 & 0 & d-x & x-d
\end{vmatrix}
Then I made column 1 - column 2, column 2 - column 3, column 3 - column 4, column 4 - column 5 and you get:
\begin{vmatrix}
x-a & a-b & b-c & c-d & d\\
0 & x-a & 0 & 0 & 0\\
0 & 0 & x-b & 0 & 0\\
0 & 0 & 0 & x-c & 0\\
0 & 0 & 0 & 0 & x-d
\end{vmatrix}
And, as it is triangular, you can multiply the diagonal elements, so you get that the determinant is:
$(x-a)^2(x-b)(x-c)(x-d)$
But this isn't correct and I don't know what to do, could someone please help me? I'd really appreciate.
|
Another approach: Consider it as a polynomial in $x$. Note that it will vanish when $x$ is any of $a,b,c,d$, since then we see a repeated row. It will also vanish when $x = -a-b-c-d$, since then all rows sum to $0$, and so we have the eigenvalue $0$ with eigenvector $(1,1,1,1,1)$. Furthermore, we know that the leading coefficient is $1$. Therefore the determinant is
$$(x-a)(x-b)(x-c)(x-d)(x-(-a -b-c-d)).$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove $a^2 + b^2 + c^2 + ab + bc +ca \ge 6$ given $a+b+c = 3$ for $a,b,c$ non-negative real. I want to solve this problem using only the AM-GM inequality. Can someone give me the softest possible hint? Thanks.
Useless fact: from equality we can conclude $abc \le 1$.
Attempt 1:
Adding $(ab + bc + ca)$ to both sides of inequality and using the equality leaves me to prove: $ab + bc + ca \le 3$.
Final edit: I found a easy way to prove above.
$18 = 2(a+b+c)^2 = (a^2 + b^2) + (b^2 + c^2) + (c^2 + a^2) + 4ab + 4bc + 4ca \ge 6(ab + bc + ca) \implies ab + bc + ca \le 3$
(please let me know if there is a mistake in above).
Attempt 2: multiplying both sides of inequality by $2$, we get:
$(a+b)^2 + (b+c)^2 + (c+a)^2 \ge 12$. By substituting $x = a+b, y = b+c, z = c+a$ and using $x+y+z = 6$ we will need to show:
$x^2 + y^2 + z^2 \ge 12$. This doesnt seem trivial either based on am-gm.
Edit: This becomes trivial by C-S.
$(a+b).1 + (b+c).1 + (c+a).1 = 6 \Rightarrow \sqrt{((a+b)^2 + (b+c)^2 + (c+a)^2)(1 + 1 + 1)} \ge 6 \implies (a+b)^2 + (b+c)^2 + (c+a)^2 \ge 12$
Attempt 3:
$x = 1-t-u$, $y = 1+t-u$, $z = 1 + 2u$
$(1-u-t)^2 + (1-u+t)^2 + (1+2u)^2 + (1-u-t)(1-u+t) + (1+t-u)(1+2u) + (1-t-u)(1+2u)$
$ = 2(1-u)^2 + 2t^2 + (1 + 2u)^2 + (1-u)^2 - t^2 + 2(1+2u)$
expanding we get:
$ = 3(1 + u^2 -2u) + t^2 + 1 + 4u^2 + 4u + 2 + 4u = 6 + 7u^2 + t^2 + 2u\ge 6$.
Yes, this works.. (not using am-gm or any such thing).
|
A classic fact is that if $a+b+c=3$ then $a^{p}+b^{p}+c^{p}\ge 3$ for every $p\ge 1$ this can be proved by Hôlder, C-S when $p=2$ or Power Mean inequality.
Here suppose the converse $a^2+b^2+c^2+ab+bc+ac<6$, $(p=2)$ necessarily $ab+bc+ac<3$ adding this you get $a^2+b^2+c^2+2(ab+bc+ac)=(a+b+c)^2<9$ a contradiction.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate $\int\frac{x^4}{\sqrt{a^2-x^2}}dx$ using trigonometric substitution Integrate $\int\frac{x^4}{\sqrt{a^2-x^2}} \, dx$ using trigonometric substitution
Ok, so it's been a really long time since I've done a problem like this but after doing a little bit o studying, this is how far I've gotten.
$$\int\frac{x^4}{\sqrt{a^2-x^2}}\,dx$$
Focusing on the denominator and using a triangle, I found that $\sin\theta=\frac{x}{a}$, therefore $x=a\sin\theta$ and $dx=a\cos\theta \, d\theta$
Plugging this into the given problem, I got $\int\frac{x^4}{\sqrt{a^2-x^2}}\, dx$ ${}= \int\frac{a^4 \sin^4\theta}{a\cos\theta}\cdot a\cos\theta \, d \theta $= $a^4\int (\sin^4 \theta ) \, d \theta$
Since $\sin^2\theta=\frac{1-\cos2\theta}{2}$, I can say
$$a^4\int (\sin^4 \theta \,d \theta =a^4\int\frac{1-2\cos2\theta+\cos^22\theta}{4}\,)d\theta$$ Pulling out the 4 from the denominator and substituting $\cos^22\theta=1-\sin^22\theta$, I get
$$\frac{a^4}{4} \int({1-2\cos2\theta+1-\sin^22\theta})\, d\theta=\frac{a^4}{4}\int({2-2\cos2\theta-\sin^22\theta}) \, d\theta$$
Substituting $\sin^22\theta for \frac{1-\cos4\theta}{2}$ gets me
$$\frac{a^4}{4}\int(2-2\cos2\theta-\frac{1-\cos4\theta}{2})d\theta=\frac{a^4}{8}\int(4-4\cos2\theta-1+\cos4\theta) \, d\theta=\frac{a^4}{8}\int(3-4 \cos2\theta + \cos4\theta)\, d\theta$$
$$=\frac{a^4}{8}[3\theta-\frac{4\sin2\theta}{2}+\frac{\sin4\theta}{4}]+C$$
From here, there are some substitutions that we must do.
We know that $sin\theta=\frac{x}{a}$ So, $\theta=sin^{-1}(\frac{x}{a})$
Also, $sin2\theta=2sin\theta cos\theta$=$2*\frac{x}{a}*\frac{\sqrt{x^2-a^2}}{a}$ and $sin4\theta=4cos^3\theta sin\theta-4sin^3\theta cos\theta=4*({\frac{\sqrt{a^2-x^2}}{a}})^3*\frac{x}{a}-4*(\frac{x}{a})^3*\frac{\sqrt{a^2-x^2}}{a}$
Substituting this gives,
$\frac{a^4}{8}[3sin^{-1}\frac{x}{a}-\frac{8x\sqrt{a^2-x^2}}{a^2}+\frac{4x(\sqrt{a^2-x^2}^3}{a^2}-\frac{4x^3\sqrt{a^2-x^2}}{a^2}]+C$
And then finally, $\frac{3a^4}{8}sin^{-1}\frac{x}{a}-\frac{a^2x\sqrt{a^2-x^2}}{2}+\frac{a^2x(\sqrt{a^2-x^2}^3}{2}-\frac{ax^3\sqrt{a^2-x^2}}{2}]+C$
Is this correct? Please critique, comment, and help clarify. I really need to understand this problem.
|
You need to substitute $\theta=\arcsin(\frac{x}{a})$, or as you said just find $\sin(4\theta) $ and $\sin (2\theta)$ using the double angle properties and put the values in the final answer.
|
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|
Conditional inequality $2a^3+b^3≥3$
Non-negative $a$ and $b$ such that $a^5+a^5b^5=2$. How then do I prove
the following inequality $2a^3+b^3≥3$?
So, we can try using the Lagrange multiplier method:
Let $f(a, b)=2 a^{3}+b^{3}+\lambda(a^{5}+a^{5} b^{5}-2), \quad a, b \geq 0$
$$\tag1 \frac{\partial f}{\partial a}=6 a^{2}+\lambda(5 a^{4}+5 a^{4} b^{5})=0 \ldots$$
$$\tag2 \frac{\partial f}{\partial b}=3 b^{2}+\lambda(5 a^{5} b^{4})=0 \ldots$$
$$\left\{\begin{array}{c} 6+\lambda(5 a^{2}+5 a^{2} b^{5})=0 \\ 3+\lambda(5 a^{3} b^{4})=0 \end{array}\right. $$
$$ \lambda=-\frac{6}{5 a^{2}+5 a^{2} b^{5}}=-\frac{3}{5 a^{3} b^{4}} $$
Multiply by $a^3$, $\,2 a^{6} b^{4}=a^{5}+a^{5} b^{5}=2$.
$$ a^{6} b^{4}=1,\, a^{3} b^{2}=1 \Rightarrow b=\frac{1}{a^{\frac{3}{2}}}\quad a, b \geq 0 $$
$$ \begin{gathered} a^{5}+a^{5} b^{5}=2 \Rightarrow a^{5}+\frac{a^{5}}{a^{\frac{15}{2}}}=2 \Rightarrow a^{5}+\frac{1}{a^{\frac{5}{2}}}=2 \Rightarrow a^{5}-2 a^{\frac{5}{2}}+1=0 \\ \Rightarrow\left(a^{\frac{5}{2}}-1\right)^{2}=0 \Rightarrow a=1 \end{gathered} $$
And the minimum is at $(a,b)=(1,1)$.
I'm not sure I solved this inequality correctly, I would like to see a more beautiful way.
|
$b^5=\frac{2-a^5}{a^5},$ where $0<a<\sqrt[5]{2}$ and we need to prove that
$$2a^3+\left(\frac{2-a^5}{a^5}\right)^{\frac{3}{5}}\geq3$$ and since for $3-2a^3\leq0$ the inequality is obvious, we need to prove that $f(a)\geq0$ for any $0<a<\sqrt[3]{\frac{3}{2}},$ where $$f(a)=\frac{3}{5}\ln(2-a^5)-3\ln{a}-\ln(3-2a^3).$$
We see that
$$f'(a)=-\frac{3a^4}{2-a^5}-\frac{3}{a}+\frac{6a^2}{3-2a^3}=$$
$$=\frac{3(a-1)(3+3a+3a^2-a^3-a^4-a^5-a^6-a^7)}{a(2-a^5)(3-2a^3)}$$ and since
$$3+3a+3a^2-a^3-a^4-a^5-a^6-a^7>0$$ for any $0<a<\sqrt[3]{\frac{3}{2}}$, we obtain $a_{min}=1$ and we are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$(\frac{s-a_1}{n-1})^{a_1} \cdot (\frac{s-a_2}{n-1})^{a_2}\cdots (\frac{s-a_n}{n-1})^{a_n} \le (\frac{s}{n})^s$ If $a_1,a_2,\ldots,a_n$ be $n$ positive rational numbers and $s=a_1+a_2+\cdots+a_n$, prove that
$$\left(\frac{s-a_1}{n-1}\right)^{a_1} \cdot \left(\frac{s-a_2}{n-1}\right)^{a_2}\cdots \left(\frac{s-a_n}{n-1}\right)^{a_n} \le \left(\frac{s}{n}\right)^s$$
My try:
Consider $\left(\frac{s-a_1}{a_1}\right),\left(\frac{s-a_2}{a_2}\right),\ldots,\left(\frac{s-a_n}{a_n}\right)$ be $n$ numbers with associated weights $a_1,a_2,\ldots,a_n$ .
Then by Weighted AM-GM inequality
$$\frac{a_1\cdot\left(\frac{s-a_1}{a_1}\right)+a_2\cdot\left(\frac{s-a_2}{a_2}\right)+\cdots+a_n\cdot\left(\frac{s-a_n}{a_n}\right)}{a_1+a_2+\cdots+a_n}\ge \left[\left(\frac{s-a_1}{a_1}\right)^{a_1}\cdot \left(\frac{s-a_2}{a_2}\right)^{a_2}\cdots \left(\frac{s-a_n}{a_n}\right)^{a_n}\right]^{\frac{1}{a_1+a_2+\cdots+a_n}}$$
Therefore
$$\left(n-1\right)^{a_1+a_2+\cdots+a_n}\ge \left(\frac{s-a_1}{a_1}\right)^{a_1}\cdot \left(\frac{s-a_2}{a_2}\right)^{a_2}\cdots \left(\frac{s-a_n}{a_n}\right)^{a_n}$$
$$\left(\frac{s-a_1}{n-1}\right)^{a_1}\cdot \left(\frac{s-a_2}{n-1}\right)^{a_2}\cdots \left(\frac{s-a_n}{n-1}\right)^{a_n}\le a_1^{a_1}\cdot a_2^{a_2}\cdots a_n^{a_n}$$
I am stuck. please give me some hint.Thanks
|
AM-GM gives:
$\frac{\left(\frac{s-a_1}{n-1}\right)a_1 + \left(\frac{s-a_2}{n-1}\right)a_2+ \cdots +\left(\frac{s- a_n}{n-1}\right)a_n}{a_1 + a_2 +\cdots+a_n} \ge \left(\left(\frac{s-a_1}{n-1}\right)^{a_1} \cdot \left(\frac{s-a_2}{n-1}\right)^{a_2}\cdots \left(\frac{s-a_n}{n-1}\right)^{a_n}\right)^{\frac{1}{s}}$
LHS $= \frac {2\sum\limits_{1\le i\lt j\le n} a_ia_j}{(n-1)s}$
We can show AM of the numbers $\frac{s}{n} \ge \frac {2\sum\limits_{1\le i\lt j\le n} a_ia_j}{(n-1)s}$
$\implies (n-1)s^2 \ge 2n\sum\limits_{1\le i\lt j\le n} a_ia_j$
$\implies (n-1)\sum\limits_{i=1}^n a_i^2 \ge 2\sum\limits_{1\le i\lt j\le n} a_ia_j$
which is true since we can write $\binom{n}{2}$ equations of the form:
$ a_i^2 + a_j^2 \ge 2 a_i a_j$ and add them up.
|
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|
Finding $\lim \frac{(2n^{\frac 1n}-1)^n}{n^2}$. I want to find limit of $a_n= \frac{(2n^{\frac 1n}-1)^n}{n^2}$ as $n\to \infty$.
$\displaystyle a_{n} =\frac{\left( 2n^{\frac{1}{n}} -1\right)^{n}}{n^{2}} =\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}}\right)^{n}$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\log a_{n} =n\log\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}}\right) =\frac{\log\left( 1+\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)\right)}{\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)} .\left(\frac{2n}{n^{\frac{1}{n}}} -\frac{n}{n^{\frac{2}{n}}} -n\right)\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\log\left( 1+\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)\right)}{\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)} .\left(\frac{n^{\frac{1}{n}} -1}{n^{\frac{1}{n}}}\right)^{2} .( -n)
\end{array}$
The first term on RHS has limit equal to $\displaystyle 1\ $but the second term is giving me problem.
Please help. Thanks.
|
Power Series Approach
Using the power series $e^x=1+x+O\!\left(x^2\right)$, we get
$$
\begin{align}
2n^{1/n}-1
&=2e^{\frac1n\log(n)}-1\tag{1a}\\[6pt]
&=2\left(1+\frac1n\log(n)+O\!\left(\frac{\log(n)^2}{n^2}\right)\right)-1\tag{1b}\\
&=1+\frac2n\log(n)+O\!\left(\frac{\log(n)^2}{n^2}\right)\tag{1c}
\end{align}
$$
and
$$
\begin{align}
n^{2/n}
&=e^{\frac2n\log(n)}\tag{2a}\\[6pt]
&=1+\frac2n\log(n)+O\!\left(\frac{\log(n)^2}{n^2}\right)\tag{2b}
\end{align}
$$
Therefore,
$$
\begin{align}
\frac{\left(2n^{1/n}-1\right)^n}{n^2}
&=\left(\frac{2n^{1/n}-1}{n^{2/n}}\right)^n\tag{3a}\\
&=\left(1+O\!\left(\frac{\log(n)^2}{n^2}\right)\right)^n\tag{3b}\\
&=1+O\!\left(\frac{\log(n)^2}n\right)\tag{3c}
\end{align}
$$
Explanation:
$\text{(3a)}$: $n^2=\left(n^{2/n}\right)^n$
$\text{(3b)}$: apply $(1)$, $(2)$, and $\frac{1+x+O\left(x^2\right)}{1+x+O\left(x^2\right)}=1+O\!\left(x^2\right)$
$\text{(3c)}$: Binomial Theorem
Thus,
$$
\lim_{n\to\infty}\frac{\left(2n^{1/n}-1\right)^n}{n^2}=1\tag4
$$
More Elementary Approach
For $n\ge12$,
$$
4(n-1)(n-2)-3n^2=n(n-12)+8\gt0\tag5
$$
Thus, $\frac43(n-1)(n-2)\gt n^2$. Then the Binomial Theorem says,
$$
\left(1+2n^{-2/3}\right)^n
\ge1+2n\cdot n^{-2/3}+2n(n-1)\cdot n^{-4/3}+\overbrace{\frac43n(n-1)(n-2)\cdot n^{-2}}^{\ge n}\tag6
$$
Therefore,
$$
\begin{align}
1-n^{-1/n}
&\le n^{1/n}-1\tag{7a}\\[6pt]
&\le2n^{-2/3}\tag{7b}
\end{align}
$$
Explanation:
$\text{(7a)}$: $1\le n^{1/n}$
$\text{(7b)}$: $(6)$ says that $\left(1+2n^{-2/3}\right)^n\ge n$
Finally,
$$
\begin{align}
\frac{\left(2n^{1/n}-1\right)^n}{n^2}
&=\left(\frac{2n^{1/n}-1}{n^{2/n}}\right)^n\tag{8a}\\
&=\left(1-\frac{\left(n^{1/n}-1\right)^2}{n^{2/n}}\right)^n\tag{8b}\\[3pt]
&=\left(1-\left(1-n^{-1/n}\right)^2\right)^n\tag{8c}\\[9pt]
&\ge1-n\left(1-n^{-1/n}\right)^2\tag{8d}\\[12pt]
&\ge1-4n^{-1/3}\tag{8e}
\end{align}
$$
Explanation:
$\text{(8a)}$: $n^2=\left(n^{2/n}\right)^n$
$\text{(8b)}$: expand the square
$\text{(8c)}$: $\frac{n^{1/n}-1}{n^{1/n}}=1-n^{-1/n}$
$\text{(8d)}$: Bernoulli's Inequality, applicable since $\left(1-n^{-1/n}\right)^2\in[0,1]$
$\text{(8e)}$: apply $(7)$
$\text{(8c)}$, for the upper bound, and $\text{(8e)}$, for the lower bound, yield
$$
1-4n^{-1/3}\le\frac{\left(2n^{1/n}-1\right)^n}{n^2}\le1\tag9
$$
to which we can apply the Squeeze Theorem to get $(4)$.
|
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|
Solve the integral of heat equation with only boundary condition $u\left(x,0\right)=\dfrac{1}{x^2+1}$ with $x \in R$ The most thing I want is solve this integral
$$\frac{1}{{16t\sqrt \pi }}\int\limits_{ - \infty }^{ + \infty } {\frac{{\exp \left( { - {a^2}} \right)}}{{{{\left( {a + \frac{x}{{4\sqrt t }}} \right)}^2} + \frac{1}{{16t}}}}} da$$
I have the PDE that
$
\begin{cases}
u_t=4u_{xx} & -\infty < x < + \infty\\
u\left(x,0\right)=\dfrac{1}{x^2+1} &
\end{cases}
$
When $-\infty < x < + \infty$, I use Poisson's equation, so that
$$
\begin{aligned}
u\left( {x,t} \right) = \frac{1}{{4\sqrt {\pi t} }}\int\limits_{ - \infty }^{ + \infty } {{e^{ - \dfrac{{{{\left( {\xi - x} \right)}^2}}}{{16t}}}}.\frac{1}{{{\xi ^2} + 1}}d\xi }
\end{aligned}
$$
with $\dfrac{{\xi - x}}{{4\sqrt t }} = a \Rightarrow \left\{ \begin{array}{l}
d\xi = 4\sqrt t da\\
\xi = x + 4a\sqrt t
\end{array} \right.$
I have
$$
\begin{aligned}
u\left( {x,t} \right) &= \frac{1}{{\sqrt \pi }}\int\limits_{ - \infty }^{ + \infty } {\exp \left( { - {a^2}} \right).\frac{1}{{16t{a^2} + 8xa\sqrt t + {x^2} + 1}}da} \\
& = \frac{1}{{\sqrt \pi }}\int\limits_{ - \infty }^{ + \infty } {\exp \left( { - {a^2}} \right).\frac{1}{{16t\left( {{a^2} + \frac{{xa}}{{2\sqrt t }} + \frac{{{x^2} + 1}}{{16t}}} \right)}}da} \\
& = \frac{1}{{\sqrt \pi }}\int\limits_{ - \infty }^{ + \infty } {\exp \left( { - {a^2}} \right).\frac{1}{{16t\left( {{a^2} + \frac{{xa}}{{2\sqrt t }} + \frac{{{x^2} + 1}}{{16t}}} \right)}}da} \\
& = \frac{1}{{16t\sqrt \pi }}\int\limits_{ - \infty }^{ + \infty } {\frac{{\exp \left( { - {a^2}} \right)}}{{{{\left( {a + \frac{x}{{4\sqrt t }}} \right)}^2} + \frac{1}{{16t}}}}} da
\end{aligned}
$$
I dont know how to solve this integral, anyone can help me for this integral or other solution of this PDE. Many thanks
|
We convolve the initial condition with the fundamental solution:
$$u(x,t)=\int_{\mathbb{R}}\frac{1}{1+y^2}\frac{1}{\sqrt{\pi 16t}}\exp\bigg\{-\frac{(x-y)^2}{16t}\bigg\}dy=\mathbb{E}_{Y \sim\mathcal{N}(x,8t)}\bigg[\frac{1}{1+Y^2}\bigg]$$
This has been solved on stats.stackexchange. The solution to the PDE is
$$u(x,t)=\frac{\sqrt{\frac{\pi }{2}} e^{-\frac{(x +i)^2}{16t}} \left(e^{\frac{2 i x }{8t}} \text{erfc}\left(\frac{1+i x }{\sqrt{8t} }\right)-\text{erf}\left(\frac{-1+ix}{\sqrt{8t}}\right)+1\right)}{16t }.$$
|
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|
Two sides of quadrilateral $ABCD$ intersect at $E$ Quadrilateral $ABCD$ is inscribed in a circle. The extensions of $AD$ and $BC$ intersect at $E$, such that $D$ is between $A$ and $E$. If $BC=2DE,AD:EC=7:2$ and $\cos\measuredangle AEB=\dfrac78$, find $\cos\measuredangle ABC$.
Since $\measuredangle ABC+\measuredangle ADC=180^\circ,$ angle $\measuredangle CDE$ is also equal to $\beta$. Triangles $ABE$ and $CDE$ also share $\measuredangle E$. By AA Similarity, $\triangle ABE\sim\triangle CDE$ with a ratio of $\dfrac{AB}{CD}=\dfrac{BE}{DE}=\dfrac{AE}{CE}$.
$Let DE=x, BC=2x$ and $AD=7y, EC=2y$. We have to find $$\cos\beta=\dfrac{AB^2+BE^2-AE^2}{2\cdot AB\cdot BE}$$ and we have $$\cos\varphi=\dfrac{ED^2+EC^2-CD^2}{2\cdot ED\cdot EC}=\dfrac{x^2+4y^2-CD^2}{2\cdot x\cdot 2y}=\dfrac{x^2+4y^2-CD^2}{4xy}$$ I don't see how to relate $AB$ and $CD$ (other than the similar triangles) Thank you in advance!
|
Hint:
From the similarity of $\triangle ABE$ and $\triangle CDE$ we have
$$\frac{x}{2y} = \frac{2x+2y}{x+7y}$$
$$x^2+7xy = 4xy + 4y^2 $$
$$(x+4y)(x-y) = 0$$
Can you take it from here?
|
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|
Divisibility rules with prime number Let $n \in \mathbb{Z}$ with the property that
$$
7 \mid\left(n^{3}+1\right)
$$
but $7$ does not divide $\left(n^{2}-2 n-3\right) .$ Prove that $7 \mid(4 n+1)$.
So far I got:
$n^3+1 = (n+1)\cdot(n^2-n+1)$, since $7$ is a prime, so
$7\mid(n+1)$ or $7\mid(n^2-n+1)$.
And since $7$ does not divide $\left(n^{2}-2 n-3\right) = (n-3)\cdot(n+1)$, I know that $7\mid(n^2-n+1)$.
And $7$ does not divide the difference of $(n^2-n+1)$ and $(n^2-2n-3)$ which is $n+4$.
Same since $7$ is prime, $7$ does not divide $4n+16\Rightarrow$$7$ does not divide $4n+2$.
Then I am stuck here, how could I use the information I got so far to prove $7 \mid(4 n+1)$?
Really hope someone could help/hint me with it! Thank you!
|
Two methods:
Method 1
If $n^3\equiv-1\bmod 7$ then $n\in\{3,5,6\}\bmod 7$. But $n\in\{3,6\}$ renders $n^2-2n-3=(n+1)(n-3)\equiv0$, denied by hypothesis, and the last possibility $n\equiv5$ renders $4n+1\equiv0$.
Method 2
Multiply $n^2-2n-3$ by $4n+1$:
$(n^2-2n-3)(4n+1)=4n^3-7n^2-14n-3\equiv4(n^3+1)\bmod 7$
Thus if the prime number $7$ divides $n^3+1$ it must divide (at least) one of the factors $n^2-2n-3$ or $4n+1$.
|
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|
Is there a technique in how to write certain expression in certain times of integration as in case of $\int \frac {(3x + 2)} {(5x + 1)^2}dx$
I can solve this integration $$\frac{3} {5}\int\frac{1}{5x + 1}dx + \frac{7} {5}\int\frac{1}{(5x + 1)^2}dx$$ but thing is I dont under how that online calculator wrote $$3x + 2$$ as $$\frac{3}{5}(5x + 1) + \frac{7}{5}$$
I have difficulty understanding that step. I know we have to make numerator and denominator same but is there a method to not guess that I have to multiply by 3/5 and add 7/5 but a standard way of doing some mathematical magic and getting a value that makes it easy to change $3x + 2$ to $\frac{3}{5}(5x + 1) + \frac{7}{5}$
|
It is just partial fraction decomposition. The key step is to write $3x + 2 = A(5x + 1) + B$, as you want the $5x+1$ terms to cancel so you can integrate $\frac{A}{5x+1} + \frac{B}{(5x+1)^2}$, which can be done by $u$-substitution.
Comparing coefficients will be easier than substituting $x = -\frac{2}{3}, -\frac{1}{5}$ (remember, this should be an identity that holds for all $x$). Thus $3x + 2 = 5Ax + (A + B)$, or $3 = 5A \implies A = \frac{5}{3}$ and $2 = A + B \implies B = \frac{1}{3}$. And now you can integrate.
|
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|
Solving $m^3-n^3=2mn+8$ We have to find integer solutions to the given equation, this is what i tried :-
For ease, let us denote $x=-m, \enspace n=-y$, and then we are basically considering $x,m,y,n$ as nonnegative wherever they occur below :-
We have four cases :-
CASE $1$ : $mn<0$ and $x^3+n^3=-(8+2xn)$
This case is clearly rejected as $x,n$ are both positive.
CASE $2$ : $mn<0$ and $m^3+y^3=8+2my$
For this we apply AM - GM to get,
$$9 + 2my = m^3 + y^3 + 1 \geq 3my \implies my \le 9 $$
Case bash (sadly) gives that the solutions to this case are
$$(m,y)=(0,2), \enspace (2,0), \enspace (2,2)$$
Now we are left with two cases
CASE $3$ : $mn>0$ and $m^3-n^3=8-2mn$
CASE $4$ : $mn>0$ and $x^3-y^3=2xy-8$
Now how to finish off these last $2$ cases?
Thanks in advance!
|
Observe that $x^3 - (x-1)^3 > 2x(x-1) + 8$ for all $x$ except $-2 ≤ x ≤ 3$ and that $x^3 - x^3 < 2x(x)+ 8$ for all $x$. From this, the equation lies between $n = m-1$ and $n = m$, hence the only integer solutions are when $-2 ≤ m ≤ 3$.
|
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|
square root of $1 \pmod m$ for prime $m$ and non prime $m$ While doing square root $1$ moduli i.e. $x ^2 \equiv 1 \pmod m$ I noticed the following:
If the moduli $m$ is prime it seems that the square is always $1$ and $m - 1$
But if it is not prime, it is not the case and I am not sure if there is a general formula for that.
Examples:
$1 \pmod 3$: $1$ and $2$
$1 \pmod 4$: $1$ and $3$
$1 \pmod 5$: $1$ and $4$
$1 \pmod 6$: $1$ and $5$
$1 \pmod 7$: $1$ and $6$
$1 \pmod 8$: $1$ and $3$ and $5$ and $7$
$1 \pmod 9$: $1$ and $8$
$1 \pmod {11}$: $1$ and $10$
Here we see that for prime modulo we have $1$ and $m - 1$ while for non prime it varies.
Am I right to generalize that for prime it always holds? What is the property that causes this and is there a general conclusion for non prime?
|
If $a$ and $b$ are two coprime positive integers and
$$x_1^2\equiv1\pmod{a}\qquad\text{ and }\qquad x_2^2\equiv1\pmod{b},\tag{1}$$
then by the Chinese remainder theorem there exists a unique integer $x$ with $0\leq x<ab$ such that
$$x\equiv x_1\pmod{ab}\qquad\text{ and }\qquad x\equiv x_2\pmod{ab},$$
which consequently satisfies $x^2\equiv1\pmod{a}$ and $x^2\equiv1\pmod{b}$, and hence
$$x^2\equiv1\pmod{ab}.\tag{2}$$
This shows that the solutions to $(2)$ correspond bijectively to pairs of solutions to $(1)$. So if $N(m)$ is the number of solutions to $x^2\equiv1\pmod{m}$, then $N(ab)=N(a)N(b)$. It follows that if $m$ is a positive integer which factors as $m=\prod p^{m_p}$, then
$$N(m)=\prod N(p^{m_p}).$$
So it suffices to determine the number of solutions for all prime powers.
If $m=p$ is prime and $x^2\equiv1\pmod{p}$ then $p$ divides
$$x^2-1=(x-1)(x+1),$$
and so either $p$ divides $x-1$ or $p$ divides $x+1$, or equivalently, for some integer $n$ we have
$$x=np\pm1.$$
If we require that $0\leq x<p$ then the only solutions are $x=1$ and $x=m-1$. Note that these are not distinct only for $p=2$.
Similarly, if $m=p^k$ is an odd prime power and $x^2\equiv1\pmod{p^k}$ then $p^k$ divides
$$x^2-1=(x-1)(x+1).$$
Because the two factors differ by $2$ they cannot both be divisible by $p$, and so either $p^k$ divides $x-1$ or $p^k$ divides $x+1$, or equivalently, for some integer $n$ we have
$$x=np^k\pm1.$$
If we again require that $0\leq x<p^k$ then the only solutions are $x=1$ and $x=m-1$, and so $N(p^k)=2$.
On the other hand, if $m=2^k$ and $x^2\equiv1\pmod{2^k}$ then $2^k$ divides
$$x^2-1=(x-1)(x+1).$$
Because the two factors differ by $2$ they must both be even, and they cannot both be divisibly by $4$, and so either $2^{k-1}$ divides $x-1$ or $2^{k-1}$ divides $x+1$, or equivalently, for some integer $n$ we have
$$x=n2^{k-1}\pm1.$$
If we again require that $0\leq x<2^k$ then the only solutions are
$$x=1,\qquad x=2^{k-1}-1,\qquad x=2^{k-1}+1,\qquad x=2^k-1.$$
Note that for $k=1$ and $k=2$, corresponding to $m=2$ and $m=4$, this only yields $1$ and $2$ distinct solutions, respectively. This shows that $N(2)=1$, $N(4)=2$ and $N(2^k)=4$ for $k\geq3$.
|
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|
How do I show that $\frac {|s|} {\sigma} \leq \sqrt {1 + \tan \left (\frac {\pi} {2} - \delta \right )}\ $? Show that in the region $\left \{s\ \big |\ \left |\arg \left (s \right ) \right | \leq \frac {\pi} {2} - \delta \right \}$ $$\frac {|s|} {\sigma} \leq \sqrt {1 + \tan \left (\frac {\pi} {2} - \delta \right )}$$ where $\mathfrak {R} (s) = \sigma$ and $0 \lt \delta \lt \frac {\pi} {2}.$
What I did is the following $:$
Let $\mathfrak {I} (s) = \tau.$ Then $$\begin {align*} \frac {|s|} {\sigma} & = \frac {\sqrt {\sigma^2 + \tau^2}} {\sigma} \\ & = \sqrt {1 + \left (\frac {\tau} {\sigma} \right )^2} \\ & \leq \sqrt {1 + \tan^2 \left (\frac {\pi} {2} - \delta \right )} \end{align*}$$ The last equality follows from the fact that $-\frac {\pi} {2} \lt - \frac {\pi} {2} + \delta \leq \arg \left (s \right ) \leq \frac {\pi} {2} - \delta \lt \frac {\pi} {2}$ and $\tan x$ is increasing on $\left (-\frac {\pi} {2}, \frac {\pi} {2} \right ).$ But the problem is that the last inequality what I have obtained doesn't match properly with that of the desired inequality. Did I make any mistake here? Could anybody please check it once?
Thanks for your time.
|
Your derivation of
$$
\frac {|s|} {\sigma} \leq \sqrt {1 + \tan^2 \left (\frac {\pi} {2} - \delta \right )}
$$
is correct, and equality holds if $|\arg(s)| = \pi/2 - \delta$.
If $|\arg(s)| = \pi/2 - \delta$ and $0 <\delta < \pi/4$ then
$$
\frac {|s|} {\sigma} = \sqrt {1 + \tan^2 \left (\frac {\pi} {2} - \delta \right )} > \sqrt {1 + \tan \left (\frac {\pi} {2} - \delta \right )}
$$
so that the first inequality does not hold in general.
|
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|
Calculating the Image of $A.$ Where did I err? $$A=\begin{pmatrix}
4 &-1& 1\\
8&-2&2\\
-6&1&-2\\
\end{pmatrix}$$
I have to show $p=\begin{pmatrix}
1 \\
2\\
-2 \\
\end{pmatrix} \in \mathrm{Im}A=\left\{Ax \mid x\in \mathbb{R^3} \right\}$
If I let $x=\begin{pmatrix}
0 \\
0 \\
1 \\
\end{pmatrix},$ then $Ax=\begin{pmatrix}
1 \\
2\\
-2 \\
\end{pmatrix}=p$ thus $p\in \mathrm{Im}A$.
But when I try to find what $\mathrm{Im}A$ is, I'm pazzled.
In order to find Im$A,$ I did elementary transformations for $A$, and I got $$A\to \begin{pmatrix}
1 & 0 & \frac{1}{2} \\
0&1&1\\
0&0&0\\
\end{pmatrix}.$$
And $$\begin{pmatrix}
1 & 0 & \frac{1}{2} \\
0&1&1\\
0&0&0\\
\end{pmatrix}
\begin{pmatrix}
x_1 \\
x_2\\
x_3\
\end{pmatrix}=(x_1+\frac{1}{2}x_3)\begin{pmatrix}
1\\
0\\
0\\
\end{pmatrix}
+
(x_2+x_3)
\begin{pmatrix}
0 \\
1 \\
0\\
\end{pmatrix}$$
so Im$A=$span$\left\{
\begin{pmatrix}
1\\
0\\
0\\
\end{pmatrix},
\begin{pmatrix}
0 \\
1 \\
0\\
\end{pmatrix}
\right\}$
But I cannot write $p$ as a linear combination of $\begin{pmatrix}
1\\
0\\
0\\
\end{pmatrix}$ and $
\begin{pmatrix}
0 \\
1 \\
0\\
\end{pmatrix}$
Where did I err ?
|
Your mistake is that $\mathrm{C\left(A\right) ≠ span}\left(\begin{bmatrix}1\\0\\0\end{bmatrix},\ \begin{bmatrix}0\\1\\0\end{bmatrix}\right)$.
Elimination actually changes the column space of $\mathrm A$, so $\mathrm{C\left(A\right) ≠ C\left(rref\left(A\right)\right)}$.
By reducing $\mathrm A$ you just revealed a number of linearly independent columns, but nothing more.
The actual column space of $\mathrm A$ is the span of vectors of $\mathrm A$, corresponding to pivot columns of $\mathrm{rref\left(A\right)}$.
So in your particular case $\mathrm{C\left(A\right) = span}\left(\begin{bmatrix}4\\8\\-6\end{bmatrix},\ \begin{bmatrix}-1\\-2\\1\end{bmatrix}\right)$.
Just a side note:
It's not usually written as $\mathrm {Im \left(A\right)}$, because the word "image" is referred to a transformation. Meaning, if you have your linear transformation $\mathrm {T\left(\vec x\right) = A\vec x}$, then you can speak of $\mathrm {Im \left(T\right)}$, which is the image of $\mathbb R^n$ under $\mathrm T$ (which is the range of $\mathrm T$). But $\mathrm {Im \left(T\right)}$ is equivalent to $\mathrm {C\left(A\right)}$. They're just the same thing by their definitions.
So it would be more correct to ask about $\mathrm {C\left(A\right)}$ and not $\mathrm{Im \left(A\right)}$ in your case.
|
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|
Cardinal of the set $A=\Big\{x=\frac pq\in\mathbb{Q} :|\sqrt 2-x|<\frac1{q^3}\Big\}$ Let $A=\Big\{x=\frac pq\in\mathbb{Q} :|\sqrt 2-x|<\frac1{q^3}\Big\}$
We want to prove that $A$ is finite and find its cardinal? we can prove that it's finite by using directly Roth’s theorem which is a generalized theorem of Lionville's theorem. But, I am stuck on finding the number of its elements. Any help, and thanks in advance.
|
Note that $|2 - \frac{p^2}{q^2}| \ge \frac{1}{q^2}$, and $\sqrt{2}+\frac{p}{q}< \sqrt{2} + (\sqrt{2} + \frac{1}{q^3})$ therefore
$$\frac{1}{q^3} >|\sqrt{2} - \frac{p}{q}| = \frac{|2 - \frac{p^2}{q^2}|}{\sqrt{2}+\frac{p}{q}}> \frac{\frac{1}{q^2}}{2 \sqrt{2}+\frac{1}{q^3}} $$
and so
$$2\sqrt{2}+\frac{1}{q^3}> q$$
We conclude that $q\le 2$, and so $x=1$, $x=2$, or $x=\frac{3}{2}$ ($0< \frac{3}{2} - \sqrt{2} < \frac{1}{10} < \frac{1}{8}$).
|
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|
Simplifiying $A \cos(x-\phi)$ I have to proof that we can transform this expression : $$a \cos x + b \sin x$$
to this one : $$A \cos (x-\phi)$$
Indeed, I can do the reverse path using the formula of $\cos(x-y)= \cos x \cos y + \sin x \sin y$
but I get stuck on the initial path.
Which trig formula should I use?
Thanks
|
You have
\begin{align*}
A\cos(x-\phi)&=A\cos x\cos \phi+A\sin x\sin \phi\\
&=(A\cos \phi)\cos x+(A\sin \phi)\sin x\\
&=a\cdot \cos x+ b\cdot \sin x
\end{align*}
where $a=A\cos \phi$ and $b=A\sin \phi$
Now, let us try to solve the equations
$$a=A\cos \phi\\ b=A\sin \phi$$
Squaring the equations and adding them, we get
$$a^2+b^2=A^2\cos^2\phi+A^2\sin^2\phi=A^2\left(\cos^2\phi+\sin^2\phi\right)=A^2$$
which gives
$$A=\sqrt{a^2+b^2}$$
Put this back into the first equation and see that
$$a=\sqrt{a^2+b^2}\cos \phi$$
which means
$$\phi=\cos^{-1}\frac a {\sqrt{a^2+b^2}}$$
So, we have proved that given the expression
$$a\cos x+b\sin x$$
we can write it as
$$\sqrt{a^2+b^2}\cos\left(x-\cos^{-1}\frac a {\sqrt{a^2+b^2}}\right)$$
which gives you the required form.
|
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|
Number of ternary sequences of length $n$ with same number of $1$'s and $0$'s My attempt is considering the number of $1$'s to be $k$ then for each $k$ we choose choose the inner order of the $1$'s and $0$'s which has ${2k \choose k }$ options then choosing the indices of the $2$'s which has ${ n \choose n -2k } = { n \choose 2k }$. So if we sum for each $k$ we get:
$\sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} {2k \choose k }{ n \choose 2k } = \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} \frac{(2k)!}{k!k!} \frac{n!}{(2k)!(n-2k)!} = \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} \frac{n!}{k!k!(n-2k)!} $
The problem is that this doesn't add up to a nice closed form, any idea how to get a nice closed form from here or by any other method?
|
We find using OPs approach
\begin{align*}
\color{blue}{\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{2k}{k}\binom{n}{2k}}
&=\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{n!}{k!k!(n-2k)!}=\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{n!}{k!(n-k)!}\,\frac{(n-k)!}{k!(n-2k)!}\\
&\color{blue}{=\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{n}{k}\binom{n-k}{k}}\tag{1}
\end{align*}
In the following we use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. We obtain from (1)
\begin{align*}
\color{blue}{\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{n}{k}\binom{n-k}{k}}
&=\sum_{k=0}^n\binom{n}{k}[z^k](1+z)^{n-k}\tag{2}\\
&=[z^0](1+z)^n\sum_{k=0}^n\binom{n}{k}\left(\frac{1}{z(1+z)}\right)^k\tag{3}\\
&=[z^0](1+z)^n\left(1+\frac{1}{z(1+z)}\right)^n\tag{4}\\
&\,\,\color{blue}{=[z^n]\left(1+z+z^2\right)^n}\tag{5}
\end{align*}
and observe (1) is a representation of the central trinomial coefficients (5).
Comment:
*
*In (2) we use the coefficient of operator $[z^k](1+z)^{n-k}=[z^k]\sum_{j=0}^{n-k}\binom{n-k}{j}z^j=\binom{n-k}{k}$.
*In (3) we factor out terms independent from $k$ and use the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
*In (4) we apply the binomial theorem.
*In (5) we simplify and apply the rule stated in (3) again.
The following notes from the experts show there is no closed form available for the central trinomial coefficients:
D.E. Knuth gives in Concrete Mathematics, Appendix A 7.56 the following representation of a more general expression
\begin{align*}
[z^n](a+bz+cz^2)^n=[z^n]\frac{1}{\sqrt{1-2bz+(b^2-4ac)z^2}}\tag{6}
\end{align*}
He states that according to the paper Hypergeometric Solutions of Linear Recurrences with Polynomial Coeffcients by Marko Petkovšek there exists a closed form (more precisely: a closed form solution as a finite sum of hypergeometric terms) if and only if
$$\color{blue}{abc(b^2-4ac)=0}$$
In case of central trinomial coefficients we have $a=b=c=1$. Since then the expression $abc(b^2-4ac)=-3\ne 0$
there is no such closed form in particular for the central trinomial coefficients.
Note: Some might be interested how to obtain the right-hand side expression of (6). This can be done e.g. by a clever change of variables stated as rule 5 in section 1.2 of Integral Representation and the Computation of Combinatorial Sums by G. P. Egorychev. Rule 5 adapted for this special case is:
\begin{align*}
[z^n]f^{n}(z)
&=[y^n]\left.\frac{f(z)}{f(z)-zf^{\prime}(z)}\right|_{z=g(y)}\tag{7}
\end{align*}
Here we have $f(z)=1+z+z^2$ and $g=g(y)$ is the inverse function of
\begin{align*}
\frac{z}{f(z)}=\frac{z}{1+z+z^2}=y
\end{align*}
We obtain
\begin{align*}
yz^2&+(y-1)z+y=0\\
z&=\frac{1}{2y}\left(1-y\pm\sqrt{1-2y-3y^2}\right)\tag{8}
\end{align*}
We take from (8) the root $z$ with the minus sign, since this one represents a power series.
We obtain from (5), (7) and (8)
\begin{align*}
\color{blue}{[z^n]}&\color{blue}{\left(1+z+z^2\right)^n}\\
&=[y^n]\left.\frac{1+z+z^2}{1+z+z^2-z(1+2z)}\right|_{z=\frac{1}{2y}\left(1-y-\sqrt{1-2y-3y^2}\right)}\\
&=[y^n]\left.\frac{1+z+z^2}{1-z^2}\right|_{z=\frac{1}{2y}\left(1-y-\sqrt{1-2y-3y^2}\right)}\\
&\,\,\color{blue}{=[y^n]\frac{1}{\sqrt{1-2y-3y^2}}}
\end{align*}
corresponding to the right-hand side of (6).
|
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"url": "https://math.stackexchange.com/questions/4237197",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
Evaluate $\int_0^\infty \frac{e^{-x}}{x}\ln\big(\frac{1}{x}\big) \sin(x)dx$ I'm having trouble with this integral
Evaluate $$\int_0^\infty \frac{e^{-x}}{x}\ln\left(\frac{1}{x}\right) \sin(x)dx$$
$$I=\int_0^\infty \frac{e^{-x}}{x}\ln\left(\frac{1}{x}\right) \sin(x)dx=\Im\left[\int_0^\infty \frac{e^{-x+ix}}{x}\ln\left(\frac{1}{x}\right) dx\right]$$
How would one proceed from here ? $u$ substitute $u=\ln(x)$? How do you solve this integral? Thank you for your time.
|
Using the identities
\begin{align*}
\frac{1}{x} &= \int_0^\infty \mathrm{e}^{-x t}\, \mathrm{d} t, \\
\ln x &= \int_0^\infty \frac{\mathrm{e}^{-s} - \mathrm{e}^{-xs}}{s}\,\mathrm{d} s,
\end{align*}
we have
\begin{align*}
I &= \int_0^\infty \mathrm{e}^{-x}\sin x
\left(\int_0^\infty \mathrm{e}^{-x t}\, \mathrm{d} t\right)
\left(\int_0^\infty \frac{\mathrm{e}^{-xs} - \mathrm{e}^{-s}}{s}\,\mathrm{d} s\right)\mathrm{d}x\\[8pt]
&= \int_0^\infty \frac{1}{s}
\left[\int_0^\infty \left(\int_0^\infty \mathrm{e}^{-x(1 + s + t)}\sin x \, \mathrm{d} x - \int_0^\infty \mathrm{e}^{-x(1 + t) - s}\sin x\, \mathrm{d} x\right)\mathrm{d} t\right]\mathrm{d} s\\[8pt]
&= \int_0^\infty \frac{1}{s}
\left[\int_0^\infty \left(\frac{1}{(1 + s + t)^2 + 1} - \mathrm{e}^{-s}\, \frac{1}{(1 + t)^2 + 1}\right)\mathrm{d} t\right]\mathrm{d} s \tag{1}\\[8pt]
&= \int_0^\infty \frac{1}{s}
\left[\frac{\pi}{2} - \arctan(1 + s) - \mathrm{e}^{-s}\,\frac{\pi}{4}\right]\mathrm{d} s\\[8pt]
&= \lim_{c\to 0^{+}}
\int_c^\infty \frac{1}{s}
\left[\frac{\pi}{2} - \arctan(1 + s) - \mathrm{e}^{-s}\,\frac{\pi}{4}\right]\mathrm{d} s\\[8pt]
&= \lim_{c\to 0^{+}}
\left[ \Big(\frac{\pi}{4}\mathrm{e}^{-c} - \frac{\pi}{2} + \arctan(1 + c)\Big)\ln c - \frac{\pi}{4}\int_c^\infty \mathrm{e}^{-s}\ln s \, \mathrm{d} s\right. \\[6pt]
&\qquad\qquad\left. + \int_c^\infty \frac{\ln s}{s^2 + 2s + 2}\,\mathrm{d} s\right] \tag{2}\\[12pt]
&= \frac{\pi}{4}\gamma + \frac{\pi}{8}\ln 2. \tag{3}
\end{align*}
Explanations:
(1): $\int \mathrm{e}^{-bx}\sin x \, \mathrm{d} x = -\frac{b\sin x + \cos x}{b^2 + 1}\mathrm{e}^{-bx} + C$;
(2): the IBP (Integration By Parts);
(3): $\lim_{c\to 0^{+}} (\frac{\pi}{4}\mathrm{e}^{-c} - \frac{\pi}{2} + \arctan(1 + c))\ln c = 0$ (use L'Hopital rule),
and $\int_0^\infty \mathrm{e}^{-s}\ln s \, \mathrm{d} s = - \gamma$
where $\gamma$ is the Euler-Mascheroni constant,
and $\int_0^\infty \frac{\ln s}{s^2 + 2s + 2}\,\mathrm{d} s = \frac{\pi}{8}\ln 2$ (the proof is given at the end).
A very nice proof of $\int_0^\infty \frac{\ln s}{s^2 + 2s + 2}\,\mathrm{d} s = \frac{\pi}{8}\ln 2$ by @Kelenner:
Let $J = \int_0^\infty \frac{\ln s}{s^2 + 2s + 2}\,\mathrm{d} s$.
With the substitution $s = \frac{2}{u}$, we have
$$J = \int_0^\infty \frac{\ln 2 - \ln u}{u^2 + 2u + 2}\,\mathrm{d} u
= \int_0^\infty \frac{\ln 2 }{u^2 + 2u + 2}\,\mathrm{d} u - J$$
which results in
$J = \frac{1}{2} \int_0^\infty \frac{\ln 2 }{u^2 + 2u + 2}\,\mathrm{d} u = \frac{\ln 2}{2}\arctan(1 + u)\vert_0^\infty = \frac{\pi}{8}\ln 2$. We are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding $x^8+y^8+z^8$, given $x+y+z=0$, $\;xy +xz+yz=-1$, $\;xyz=-1$
The system says
$$x+y+z=0$$
$$xy +xz+yz=-1$$
$$xyz=-1$$
Find
$$x^8+y^8+z^8$$
With the first equation I squared and I found that $$x^2+y^2+z^2 =2$$
trying with
$$(x + y + z)^3 =
x^3 + y^3 + z^3 + 3 x^2 y + 3 x y^2 + 3 x^2 z++ 3 y^2 z + 3 x z^2 +
3 y z^2 + 6 x y z$$
taking advantage of the fact that there is an $xyz=-1$ in the equation, but I'm not getting anywhere, someone less myopic than me.how to solve it?
Thanks
Edit : Will there be any university way to solve this problem , they posed it to a high school friend and told him it was just manipulations of remarkable products.
His answers I understand to some extent but I don't think my friend understands all of it.
|
The following is an elementary, entirely self-contained solution which does not assume knowledge of Vieta's relations, Newton's identities, or anything other than simple algebra. (This is not the best or recommended way to solve it - see the other answers and comments for that - but it answers OP's edit asking for a more basic solution.)
*
*From $x+y+z=0$ it follows that:
$$y+z = -x \tag{1}$$
*
*From $xy +xz+yz=-1$ and $(1)$ it follows that:
$$-1=x(y+z)+yz=-x^2 + yz \;\;\implies\;\; yz = x^2-1 \tag{2}$$
*From $xyz=-1$ and $(2)$ it follows that:
$$-1=xyz=x(x^2-1)=x^3-x \;\;\implies\;\; x^3=x-1 \tag{3}$$
Multiplying $(3)$ by $x$ gives:
$$x^4=x^2-x \tag{4}$$
Squaring $(4)$ and using that $x^3=x-1$ per $(3)\,$:
$$
\begin{align}
x^8 &= x^4-2x^3+x^2
\\ &=x \cdot x^3-2x^3+x^2
\\ &=x(x-1)-2(x-1)+x^2
\\ &=2x^2-3x+2 \tag{5}
\end{align}
$$
Repeating the steps $(1)\dots(5)$ for the other variables, it follows by symmetry that:
$$
\begin{align}
y^8 = 2y^2-3y+2 \tag{6}
\\ z^8 = 2z^2-3z+2 \tag{7}
\end{align}
$$
Adding $(5)+(6)+(7)\,$ and using that $x+y+z=0$:
$$
\require{cancel}
\begin{align}
x^8+y^8+z^8 &= 2(x^2+y^2+z^2) - \cancel{3(x+y+z)} + 6
\\ &= 2\left(\bcancel{(x+y+z)^2}-2(xy+yz+zx)\right)+6
\\ &= -4 \cdot (-1) + 6
\\ &= 10
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve $\sin(2\theta)$ questions Given that: $\sin\theta=\displaystyle{}\frac{12}{13}$ and $0<\theta<\displaystyle{}\frac{\pi}{2}$ the value of $\sin(2\theta)$ is:
I figured out a way to solve it, though I'm not sure if it is the best solution.
Here we will combine two different trigonemtric identities. First:
$\begin{align}
\sin(2\theta) & = 2\sin\theta\cos\theta \\
& = 2\cdot\frac{12}{13}\cdot\cos\theta \\
& = \frac{24}{13}\cdot\cos\theta
\end{align}$
Also:
$\begin{align}
1 & = \cos^2\theta+\sin^2\theta \\
1 & = \cos^2\theta+\Bigg(\frac{12}{13}\Bigg)^2 \\
1 & = \cos^2\theta+\frac{144}{169} \\
1-\frac{144}{169} & = \cos^2\theta \\
\frac{25}{169} & = \cos^2\theta \\
\sqrt\frac{25}{169} & = \sqrt{\cos^2\theta} \\
\frac{5}{13} & = \cos\theta \\
\end{align}$
Then we insert this into the previous equation:
$\begin{align}
\sin(2\theta) & = \frac{24}{13}\cdot\cos\theta \\
& = \frac{24}{13}\cdot\frac{5}{13} \\
& = \frac{120}{169}
\end{align}$
And I believe this is the correct answer. I'm just not sure if this was a super round about way of solving it or if there is something better.
|
The line
$$\cos^2\theta = \frac{25}{169}$$
simplifies to
$$|\cos\theta| = \frac{5}{13}$$
since $\sqrt{x^2} = |x|$. Since $\cos\theta > 0$ if $0 < \theta < \dfrac{\pi}{2}$, $|\cos\theta| = \cos\theta$ in this interval, which allows you to conclude that
$$\cos\theta = \frac{5}{13}$$
The rest of your work is correct.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate $\int^{\pi}_{0}\left\{\frac{\tan^2\left(\frac{x}{2}\right)}{\sin^2(x)\cdot\,\cos^2(x)}\right\}^{\frac{1}{9}}\,dx$ $\displaystyle\int^{\pi}_{0}\left\{\dfrac{\tan^2\left(\dfrac{x}{2}\right)}{\sin^2(x)\cdot\,\cos^2(x)}\right\}^{\frac{1}{9}}\,dx$
$\sf{\color{blue}{My\,\,approach\,}:}$
$=\displaystyle\int^{\pi}_{0}\left\{\dfrac{\tan^2\left(\dfrac{x}{2}\right)}{\dfrac{4\,\tan^2(\frac{x}{2})}{\sec^4(\frac{x}{2})}\cdot\,\dfrac{(1-\tan^2(\frac{x}{2}))^2}{\sec^4(\frac{x}{2})}}\right\}^{\frac{1}{9}}\,dx$
$=\displaystyle\int^{\pi}_{0}\left\{\dfrac{\sec^8\left(\dfrac{x}{2}\right)}{4\cdot\,\left(1-\tan^2\left(\dfrac{x}{2}\right)\right)^2}\right\}^{\frac{1}{9}}\,dx$
$=\displaystyle\int^{\pi}_{0}\dfrac{\sec^{\frac{8}{9}}\left(\dfrac{x}{2}\right)}{2^{\frac{2}{9}}\cdot\,\left(1-\tan^2\left(\dfrac{x}{2}\right)\right)^{\frac{2}{9}}}\,dx$
$=\displaystyle\int^{\pi}_{0}\dfrac{\sec^{\frac{-10}{9}}\left(\dfrac{x}{2}\right)\cdot\dfrac{1}{2}\sec^2\left(\dfrac{x}{2}\right)dx}{2^{\frac{-7}{9}}\cdot\,\left(1-\tan^2\left(\dfrac{x}{2}\right)\right)^{\frac{2}{9}}}$
If I substitute $\color{orange}{\tan\left(\dfrac{x}{2}\right)=t,}$ a weird expression occurs..
Help me to figure out this
|
$I = \dfrac{{0,028 \cdot {{2,496}^3}}}{{12}} + 2 \cdot \left( {\dfrac{{0,62 \cdot {{0,052}3}}}{{12}} + 0,62 \cdot 0,052 \cdot {{1,274}2}} \right)$
|
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|
Maximum value of $(ab+1)(bc+1)(cd+1)(da+1)$: mistake in solution? Find maximum value of $(ab+1)(bc+1)(cd+1)(da+1)$ if $abcd=1$ and $\frac{1}{2}≤a,b,c,d≤2$
$$ab+cd≥2(abcd)^{1/2}=2$$
$$da+bc≥2(abcd)^{1/2}=2$$
$$(ab+1)(bc+1)(cd+1)(da+1)≤(\frac{ab+1+bc+1+cd+1+da+1}{4})^4$$
$$\implies ≤(\frac{2+2+4}{4})^4=16$$
However, $(2,2,1/2,1/2)$ gives us $25$, which is clearly greater than $16$. Where is the mistake in my solution? (I already have a different solution that gives the right answer of 25 but if someone has a method similar to my approach which does produce the correct answer, then please share)
|
The direction of your second inequlity sign in the followilng should be reversed
$$(ab+1)(bc+1)(cd+1)(da+1)≤(\frac{ab+1+bc+1+cd+1+da+1}{4})^4$$
$$\implies ≤(\frac{2+2+4}{4})^4=16$$
Thus your implication is not valid.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove $\int_{0}^{\pi/2}\frac{\sqrt{\cos \theta}}{1 + \cos^2\theta}d\theta = \frac{\pi}{4}$ Prove that
$$
\int_{0}^{\pi/2}\frac{\sqrt{\cos \theta}}{1 + \cos^2\theta}d\theta = \frac{\pi}{4}
$$
My attempt :I tried to use the beta function, but I couldn't.
|
In search of a simple solution and after many attempts, I found this solution. Let $\cos \theta = \tan \alpha$ and $d\theta = -\dfrac{\sec^2 \alpha \ d\alpha}{\sqrt{1 - \tan \alpha}}$. Thus,
$$
I = \int_{0}^{\pi/2}\dfrac{\sqrt{\cos \theta}}{1 + \cos^2 \theta}d\theta = \int_{\pi/4}^{0}\dfrac{\sqrt{\tan \alpha}(-\sec^2 \alpha)d\alpha}{(1 + \tan^2 \alpha)\sqrt{1 - \tan^2 \alpha}} = \int_{0}^{\pi/4}\sqrt{\dfrac{\tan \alpha}{1 - \tan^2 \alpha}}d\alpha \quad \Rightarrow
$$
$$
I = \dfrac{1}{\sqrt{2}}\int_{0}^{\pi/4}\sqrt{\dfrac{\sin(2\alpha)}{\cos(2\alpha)}}d\alpha = \dfrac{1}{2\sqrt{2}}\int_{0}^{\pi/2}\sin^{1/2}\alpha \cos^{-1/2}\alpha d\alpha \quad \Rightarrow
$$
$$
I = \dfrac{1}{4\sqrt{2}}B(1/4,3/4)= \dfrac{1}{4\sqrt{2}}\cdot \dfrac{\pi}{\sin(\pi/4)} = \dfrac{\pi}{4}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the minimum value of: $\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$ Let $a,b,c>0$, find the minimum value of: $$\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$$
I have tried:
$\bullet$ The minimum value is $\dfrac{3}{4}$ occur when $a=b=c$
$\bullet \dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}= \dfrac{a^2}{3a^2+ab}+\dfrac{b^2}{3b^2+bc}+\dfrac{c^2}{3c^2+ca}\ge\dfrac{(a+b+c)^2}{3(a^2+b^2+c^2)+ab+bc+ca}=\dfrac{(a+b+c)^2}{3(a+b+c)^2-5(ab+bc+ca)}$
$\bullet$ $ab+bc+ca\le\dfrac{(a+b+c)^2}{3}$ so we have $\dfrac{(a+b+c)^2}{3(a+b+c)^2-5(ab+bc+ca)}\le \dfrac{3}{4}$ but we need to prove $\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c} \ge \dfrac{3}{4}$
Can you give me some hint ?
|
The value $\frac 3 4$ is a maximum, indeed we have that by $x=\frac b a$, $y=\frac cb$, $z=\frac a c$ with $xyz=1$
$$\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c} = \dfrac{1}{x+3}+\dfrac{1}{y+3}+\dfrac{1}{z+3} \le \frac 3 4$$
and
$$\frac34-\left(\dfrac{1}{x+3}+\dfrac{1}{y+3}+\dfrac{1}{z+3}\right) =\frac{3 x y z+ 5 y z+ 5 x z+ 5 x y+ 3 x + 3 y + 3 z-27}{4 (3 + x) (3 + y) (3 + z)}\ge 0$$
since by AM-GM
$$3 x y z+5 y z+ 5 x z+ 5 x y+ 3 x + 3 y + 3 z \ge 27\sqrt[27]{(xyz)^{18}}=27$$
with equality for $x=y=z=1$.
As noticed the expression has not a minimum, indeed by
*
*$b=ka \implies x=k$
*$c=kb \implies y=k$
*$z=\frac1 {xy}=\frac1{k^2}$
we obtain
$$\dfrac{1}{x+3}+\dfrac{1}{y+3}+\dfrac{1}{z+3}=\dfrac{1}{k+3}+\dfrac{1}{k+3}+\dfrac{k^2}{1+3k^2} \to \frac13$$
as $k \to \infty$ and
$$\dfrac{1}{x+3}+\dfrac{1}{y+3}+\dfrac{1}{z+3}-\frac13=\frac{53 + 9 (x+y+z)}{3 (x + 3) (y + 3) (z + 3)}>0$$
which shows that the value $\frac13$ is an infimum for the expression that is
$$\dfrac{1}{x+3}+\dfrac{1}{y+3}+\dfrac{1}{z+3}>\frac13$$
|
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|
Number of different value of given expression for $x,y,z \in \{1,2,3,4,5\}$ Let $x,y,z \in \{1,2,3,4,5\}$, then find number of different values of
$$\dfrac{x^3}{(x-z)(x-y)}+\dfrac{y^3}{(y-x)(y-z)}+\dfrac{z^3}{(z-x)(z-y)}$$
Could someone please share the approach to start this question. I tried to factorize the expression after taking L.C.M. but the whole gets very messy. How should I start?
|
First, by observing the form of the denominator, we can immediately pose the condition that $x \neq y \neq z$ for the fraction to be well-defined.
Next, because all $3$ fractions are identical, permutation of a set of $(x,y,z)$ will all result in the same value (i.e. $(x,y,z) = (1,2,3)$ will give the same value as $(x,y,z) = (2,3,1)$ and so on).
With this, we narrow down the $125$ possibilities of $(x,y,z)$ to ${5 \choose 3} = 10$ possibilities; namely
$$
(1,2,3)\quad (1,2,4)\quad (1,2,5)\quad (1,3,4)\quad (1,3,5) \\
(1,4,5)\quad (2,3,4)\quad (2,3,5)\quad (2,4,5)\quad (3,4,5)
$$
(Optional) We can simplify the equation as follows:
\begin{align}
\frac{x^3}{(x-y)(x-z)} + \frac{y^3}{(y-x)(y-z)} + \frac{z^3}{(z-x) (z-y)}
&= \frac{x^3(y-z) - y^3(x-z) + z^3(x-y)}{(x-y)(x-z)(y-z)} \\
&= \frac{x^3(y-z) + yz(y^2 - z^2) - x(y^3 - z^3)}{(x-y)(x-z)(y-z)} \\
&= \frac{x^3(y-z) + yz(y-z)(y+z) - x(y-z)(y^2+yz+z^2)}{(x-y)(x-z)(y-z)} \\
&= \frac{x^3 + yz(y+z) - x(y^2+yz+z^2)}{(x-y)(x-z)} \\
&= \frac{x(x^2-y^2)-yz(x-y)-z^2(x-y)}{(x-y)(x-z)} \\
&= \frac{x(x+y)-yz-z^2}{x-z} \\
&= x+y+z
\end{align}
From here, just manually substitute these $10$ combinations and you will find that the possible values are $\{6,7,8,9,10,11,12\}$.
|
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|
How to prove $\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$? I was recently searching for interesting looking integrals. In my search, I came upon the following result:
$$ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$$
and I wanted to try and prove it.
Inspired by this answer by Jack D'Aurizio, I took the Weierstrass product for $\cosh(x)$ to obtain
$$
\cosh\left(\frac{\pi x}{2} \right) = \prod_{n \ge 1}\left(1 + \frac{x^2}{(2n-1)^2} \right)
$$
And by logarithmically differentiating twice we get
$$
\frac{\pi^2}{4}\text{sech}^2\left(\frac{\pi x}{2} \right) = \sum_{n \ge 1} \frac{4(2n-1)^2}{\left(x^2 + (2n-1)^2\right)^2} - \frac{2}{x^2 + (2n-1)^2}
$$
Which means we get
\begin{align*}
\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx & =\frac{4}{\pi^2}\sum_{n\ge 1} \int_{0}^{\infty} \frac{(1-x^2)}{(1+x^2)^2}\left( \frac{4(2n-1)^2}{\left(x^2 + (2n-1)^2\right)^2} - \frac{2}{x^2 + (2n-1)^2}\right)\, dx
\end{align*}
However, after this, I couldn't figure out how to evaluate the resulting integral.
Does anyone know how I could continue this method? Or alternatively, does anyone know another way in which the result can be proven? Thank you very much!!
Edit:
Per jimjim's request, I'll add that I found this integral on the Wikipedia article for $\zeta(3)$. I believe the reference is to this text where the following formula is given
$$
(s-1) \zeta(s) = 2\pi \int_{\mathbb{R}}\frac{\left(\frac{1}{2} + xi \right)^{1-s}}{\left(e^{\pi x} +e^{-\pi x} \right)^2}\, dx
$$
which for the case of $s=3$ reduces to the surprisingly concise
$$
\int_{\mathbb{R}}\frac{\text{sech}^2(\pi x)}{(1+2xi)^2} \, dx = \frac{\zeta(3)}{\pi}
$$
And I presume that one can modify the previous equation to get to the original integral from the question, but it is not apparent to me how this may be done.
Edit 2:
Random Variable has kindly posted in the comments how to go from $\int_{\mathbb{R}}\frac{\text{sech}^2(\pi x)}{(1+2xi)^2} \, dx$ to $ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx$. Thank you very much!
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A somewhat simpler approach using contour integration is to first notice that $$ \begin{align} \int_{-\infty}^{\infty} \frac{\operatorname{sech}^{2}\left(\frac{\pi x}{2}\right)}{\left(x+i \right)^{2}} \, \mathrm dx &= \int_{-\infty}^{\infty} \frac{x^{2}-1}{(1+x^2)^{2}} \operatorname{sech}^{2} \left(\frac{\pi x}{2} \right) \, \mathrm dx - i\int_{-\infty}^{\infty} \frac{2x}{(1+x^2)^2} \operatorname{sech}^{2} \left(\frac{\pi x}{2} \right) \, \mathrm dx \\ &= \int_{-\infty}^{\infty} \frac{x^{2}-1}{(1+x^2)^{2}} \operatorname{sech}^{2} \left(\frac{\pi x}{2} \right) \, \mathrm dx.\end{align} $$
Integrating by parts, we get $$\int_{-\infty}^{\infty} \frac{\operatorname{sech}^{2} \left(\frac{\pi x}{2} \right)}{(x+i)^2} \, \mathrm dx = \frac{4}{\pi}\int_{-\infty}^{\infty} \frac{\tanh \left(\frac{\pi x}{2} \right)}{(x+i)^3} \, \mathrm dx. $$
Now let's integrate the function $$ f(z) =\frac{\tanh \left(\frac{\pi z}{2} \right)}{(z+i)^3} $$ around a square contour with vertices at $ \pm N$, $\pm N + 2iN,$ where $N$ is some positive integer.
Since $\tanh \left(\frac{\pi z}{2} \right) $ is $2$- periodic in the imaginary direction and tends to $\pm 1$ as $\Re(z) \to \pm \infty$, the magnitude of $\tanh \left(\frac{\pi x}{2} \right) $ remains bounded on the contour as $N \to \infty$ through the positive integers since the contour stays away from the poles on the imaginary axis.
We can therefore use the estimation lemma to argue that the integral vanishes on the vertical sides of the contour and the upper side of the contour as $N \to \infty$.
Inside the contour there are simple poles at $z= i(2n+1), n \in \mathbb{N}_{\ge 0}$.
Therefore, we have $$\begin{align} \int_{-\infty}^{\infty} \frac{\tanh \left(\frac{\pi x}{2} \right)}{(x+i)^3} \, \mathrm dx &= 2 \pi i \sum_{n=0}^{\infty}\operatorname{Res} [f(z), i(2n+1)] \\ &= 2 \pi i \sum_{n=0}^{\infty} \lim_{z \to i(2n+1)} \frac{\sinh \left(\frac{\pi z}{2} \right)}{\frac{\mathrm d}{\mathrm dz}\left((z+i)^3 \cosh \left(\frac{\pi z}{2} \right)\right)} \\ &= -2 \pi i \sum_{n=0}^{\infty} \frac{1}{4 \pi i (n+1)^3} \\ &= -\frac{\zeta(3)}{2}, \end{align}$$
from which it follows that $$\begin{align} \int_{0}^{\infty} \frac{1-x^2}{(1+x^2)^{2}} \operatorname{sech}^{2} \left(\frac{\pi x}{2} \right) \, \mathrm dx &= -\frac{1}{2} \int_{-\infty}^{\infty} \frac{x^2-1}{(1+x^2)^{2}} \operatorname{sech}^{2} \left(\frac{\pi x}{2} \right) \, \mathrm dx \\ &= -\frac{2}{\pi} \int_{-\infty}^{\infty} \frac{\tanh \left(\frac{\pi x}{2} \right)}{(x+i)^3} \, \mathrm dx \\ &=- \frac{2}{\pi} \left(-\frac{\zeta(3)}{2} \right) \\ &= \frac{\zeta(3)}{\pi}. \end{align}$$
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problem with power series $\sum \frac{x^n}{n+3}$ I need to evaluate the following power series, and I don’t really know how to do it, this is the series $$\sum \frac{x^n}{n+3}$$
This is how I tackled this problem, but the final solution looks very dumb.
So we know that $$\frac{1}{1-x}=\sum x^n$$
But we also know that we can get that $n+3$if we integrate $x^n+2$
So what we want is this series $$\sum x^{n+2}$$
Now, this series is equal to the original times $x^2$, so we can do this$$\sum x^{n+2} = x^2 \frac{1}{1-x}$$ Now we take the integral of both sides and we get something like this $$\sum \frac{x^{n+3}}{n+3} = (-\frac{x^2}{2}-1-\ln|x-1|)$$ But know we see that the left-hand side is $$x^3 \sum \frac{x^n}{n+3}$$ so we can multiply both sides by $\frac{1}{x^3}$ and we get the following $$\sum \frac{x^n}{n+3}=-\frac{1}{2x}-\frac{1}{x^3}-\frac{\ln|x-1|}{x^3}$$Can someone tell me if this answer is acceptable or if I have completely messed up?
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Note that
$$-\int\frac{x^2}{x-1}dx=-\int\frac{x^2-1+1}{x-1}dx=-\int\frac{(x+1)(x-1)+1}{x-1}dx$$
$$=-\int(x+1)dx-\int\frac{1}{x-1}dx=-\frac{x^2}{2}-\color{red}{x}-\ln|x-1|+C$$
So
$$\sum_{n=0}^{\infty}\frac{x^{n+3}}{n+3}=-\frac{x^2}{2}-\color{red}{x}-\ln|x-1|$$
and we have
$$\sum_{n=0}^{\infty}\frac{x^n}{n+3}=-\frac{1}{2x}-\frac{1}{x^2}-\frac{1}{x^3}\ln|x-1|$$
and as mentioned by Gary in the comments, note that $\ln|x-1|$ can be replaced by $\ln(1-x)$ since the series converges absolutely for $|x|<1$
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What are relations of $S_4$, if generators are $a=(12)$, $b=(1234)$? I think, relations could be $a^2=e$, $b^4=e$, $ababab=e$, $b^2 a b^2 a b = b a b^2 a$.
By two first relations and $b^3 = ababa$ (it is result of our relations), we can present any element that is generated by $a$ and $b$, as a sequence, where there are not more than one consecutive $a$ and not more than two consecutive $b$. By the last relation we can put all squares of $b$ to the beginning of sequence. We can remove 3 or more consecutive $ab$ by third relations. And also we have $ab^2ab^2ab^2ab^2=e$ as result of relations, so we can remove 4 or more of consectutive $ab^2$ from the beginning of sequence.
But, I'm not sure that we need exactly 4 relations, that there are not 3 relations that would be sufficient.
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As Derek Holt mentions in the comments you have $S_4 \cong \langle x,y | x^2 = y^4 = (xy)^3 = 1 \rangle$, so for the generators $a = (1\ 2)$ and $b = (1\ 2\ 3\ 4)$ three relations $a^2 = b^2 = (ab)^3 = 1$ suffice.
Let $G = \langle x,y | x^2 = y^4 = (xy)^3 = 1 \rangle$. In $S_4$ you have a normal subgroup $V \cong C_2 \times C_2$ of order $4$, generated by $b^2 = (1\ 3)(2\ 4)$ and $ab^2a^{-1} = (1\ 4)(2\ 3)$.
So consider in $G$ the subgroup $N = \langle z, w \rangle$ where $z = y^2$ and $w = xy^2x^{-1}$. We would expect that $N$ is a normal subgroup and $N \cong C_2 \times C_2$.
For normality, it is clear that $xzx^{-1} = w$, $xwx^{-1} = z$, and $yzy^{-1} = z$. It remains to check that $ywy^{-1} \in N$. For this, using $(yx)^3 = 1$ and $x^2 = 1$ we find that $yxy = xy^{-1}x$ (as suggested in a comment). Then \begin{align*}
ywy^{-1} &= yxy^2xy^3 \\ &= (yxy)(yxy)y^2 \\ &= (xy^{-1}x)(xy^{-1}x)y^2 \\ &= xy^2xy^2 \\ &= wz.
\end{align*}
Certainly $N$ has $C_2 \times C_2$ as a quotient. Check that $(zw)^2 = 1$ so in fact $N \cong C_2 \times C_2$: \begin{align*} (zw)^2 &= y^2xy^2xy^2xy^2x \\ &= y(yxy)(yxy)(yxy)yx \\ &= y(xy^{-1}x)(xy^{-1}x)(xy^{-1}x)yx \\ &= y(xy^{-3}x)yx \\ &= yxyxyx = (yx)^3 = 1.\end{align*}
Then $G/N$ is generated by $\overline{x}$, $\overline{y}$ with $(\overline{x})^2 = (\overline{y})^2 = (\overline{x}\overline{y})^3 = 1$, so $G/N$ is a quotient of $S_3$.
Conclude $G \cong S_4$.
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Find all functions $f: \mathbb{R} \to\mathbb{R} $ such that if $a+f(b)+f^2(c) = 0$ then $f(a)^3 + bf(b)^2 + c^2f(c) = 3abc$
Find all functions $ f : \mathbb R \to \mathbb R $ such that for all $ a , b , c \in \mathbb R $, if
$$ a + f ( b ) + f ^ 2 ( c ) = 0 \tag {Eq1} \label {eqn1} $$
then
$$ f ( a ) ^ 3 + b f ( b ) ^ 2 + c ^ 2 f ( c ) = 3 a b c \text . \tag {Eq2} \label {eqn2} $$
I've been trying to solve this "implication-based" functional equation. My first thought was to use substitution to \eqref{eqn2} from \eqref{eqn1}.
$$ f ( a ) = - f ^ 2 ( c ) - a $$
$$ f ( a ) ^ 2 = f ^ 2 ( c ) ^ 2 + 2 a f ^ 2 ( c ) + a ^ 2 $$
therefore after some manipulation:
$$ f ( a ) ^ 3 + b f ^ 2 ( c ) ^ 2 + 2 a b f ^ 2 ( c ) + c ^ 2 f ( c ) = a b ( 3 c - a ) $$
Then, I tried plugging in some values, for instance the simplest case $a=b=c=1$
$$f(1)^3 + f^2(1)^2 + 2f^2(1) + f(1) = 2$$
I have not been able to deduce any useful information from this, nor from similar substitutions with 1.
I'd be grateful for your help with this problem.
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First of all, note that the problem can be stated in the following equivalent form.
Find all functions $ f : \mathbb R \to \mathbb R $ such that
$$ f \left( - f ( x ) - f ^ 2 ( y ) \right) ^ 3 + 3 x y \left( f ( x ) + f ^ 2 ( y ) \right) + x f ( x ) ^ 2 + y ^ 2 f ( y ) = 0 \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $. Here, $ f ^ 2 ( y ) $ means $ f \bigl( f ( y ) \bigr) $, and $ f ( x ) ^ 2 $ means $ f ( x ) \cdot f ( x ) $.
It's straightforward to verify that the constant zero function, the identity function and the additive inversion function all satisfy the required condition. We prove that these are the only solutions.
Letting $ k = f ( 0 ) $ and putting $ x = y = 0 $ in \eqref{0}, we get $ f \bigl( - k - f ( k ) \bigr) = 0 $. Then, plugging $ x = - k - f ( k ) $ and $ y = 0 $ in \eqref{0} we can see that $ f \bigl( - f ( k ) \bigr) = 0 $. Now, set $ x = - f ( k ) $ in \eqref{0} and consider the equation once with $ y = - k - f ( k ) $ and another time with $ y = - f ( k ) $. Comparing the two results, you can conclude $ k ^ 2 f ( k ) = 0 $, which together with $ f \bigl( - f ( k ) \bigr) = 0 $ implies $ k = 0 $.
Putting $ y = 0 $ in \eqref{0} gives
$$ f \bigl( - f ( x ) \bigr) ^ 3 = - x f ( x ) ^ 2 \tag 1 \label 1 $$
for all $ x \in \mathbb R $, while plugging $ x = 0 $ in \eqref{0} yields
$$ f \left( - f ^ 2 ( y ) \right) ^ 3 = - y ^ 2 f ( y ) \tag 2 \label 2 $$
for all $ y \in \mathbb R $. Letting $ x = f ( y ) $ in \eqref{1}, we get $ f \left( - f ^ 2 ( y ) \right) ^ 3 = - f ( y ) f ^ 2 ( y ) ^ 2 $, which together with \eqref{2} shows that for any $ y \in \mathbb R \setminus \{ 0 \} $ with $ f ( y ) \ne 0 $ we have $ f ^ 2 ( y ) ^ 2 = y ^ 2 $. But also note that if $ y , f ( y ) \ne 0 $ and $ f ^ 2 ( y ) = - y $, then \eqref{2} gives $ f ( y ) \in \{ - y , y \} $; $ f ( y ) = y $ cannot happen since it implies $ f ^ 2 ( y ) = y $ (which contradicts $ f ^ 2 ( y ) = - y $ as $ y \ne 0 $), and $ f ( y ) = - y $ is not possible because it implies $ f ( - y ) = f ^ 2 ( y ) = - y $, which then substituting $ - y $ for $ y $ in \eqref{2} implies $ f ( y ) = y $ (contradicting $ f ( y ) = - y $ as $ y \ne 0 $). Therefore, the case $ f ^ 2 ( y ) = - y $ is ruled out, and we must have
$$ f ( y ) \ne 0 \implies f ^ 2 ( y ) = y \tag 3 \label 3 $$
for all $ y \in \mathbb R \setminus \{ 0 \} $. Now, assume that there exists $ y _ 0 \in \mathbb R \setminus \{ 0 \} $ with $ f ( y _ 0 ) \ne 0 $. Putting $ y = y _ 0 $ in \eqref{2} and using \eqref{3} we have $ f ( - y _ 0 ) ^ 3 = - y _ 0 ^ 2 f ( y _ 0 ) $. Using this and \eqref{3} and setting $ y = y _ 0 $ in \eqref{0}, we can see that if $ f ( x ) = 0 $ for some $ x \in \mathbb R $, then we must have $ x = 0 $. Hence, using \eqref{3} we can conclude $ f ^ 2 ( y ) = y $ for all $ y \in \mathbb R $. Consequently, substituting $ f ( x ) $ for $ x $ in \eqref{0}, we get
$$ f ( - x - y ) ^ 3 + 3 y ( x + y ) f ( x ) + x ^ 2 f ( x ) + y ^ 2 f ( y ) = 0 \tag 4 \label 4 $$
for all $ x , y \in \mathbb R $. Interchanging $ x $ and $ y $ in \eqref{4} and comparing with \eqref{4} itself, it's straightforward to see that
$$ y f ( x ) = x f ( y ) $$
for all $ x , y \in \mathbb R $. In particular, this gives $ f ( x ) = f ( 1 ) x $ for all $ x \in \mathbb R $, and as $ f ^ 2 ( 1 ) = 1 $, we have $ f ( 1 ) \in \{ - 1 , 1 \} $. Therefore, we've proven that if $ f $ differs from the constant zero function, we must either have $ f ( x ) = x $ for all $ x \in \mathbb R $ or $ f ( x ) = - x $ for all $ x \in \mathbb R $, which proves what was claimed.
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Evaluating $(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})...(2^{64} + 3^{64})$ I'm trying to evaluate the following expression$$(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})...(2^{64} + 3^{64})$$
I'm not really used to these types of problems, so I first tried using logarithms but I'm not sure what to do from there. See:
Let $P = (2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})...(2^{64} + 3^{64})$.
Then we have:
$$\log_2{P} = \log_2((2^1 + 3^1)...(2^{64} + 3^{64}))$$
$$\log_2{P} = \log_2(2^1 + 3^1) + \log_2(2^2 + 3^2) + ... + \log_2(2^{64} + 3^{64})$$
$$P = 2^{2^1 + 3^1} + 2^{2^2 + 3^2} +... + 2^{2^{64} + 3^{64}}$$
$$P = 2^{2^1}2^{3^1} + 2^{2^2}2^{3^2} +... + 2^{2^{64}}2^{3^{64}}$$
Would factoring $2^2$ be of any use here?
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Sayan Dutta shows a clever way. Another way is to note that when you multiply the parentheses, each term will be of the form $2^n 3^{127-n}$:
$$
(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})...(2^{64} + 3^{64})
= \prod_{k=0}^{6} (2^{2^k}+3^{2^k})
\\
= \sum_{n=0}^{127} 2^n 3^{127-n}
= 3^{127} \sum_{n=0}^{127} (2/3)^n
= 3^{127} \frac{1-(2/3)^{128}}{1-2/3}
= 3^{128} \frac{1-(2/3)^{128}}{3-2}
= 3^{128} - 2^{128}.
$$
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Find the minimum value of $a^8+b^8+c^8+2(a-1)(b-1)(c-1)$ Let $a,b,c$ be the lengths of the three sides of the triangle, $a+b+c=3$. Find the minimum value of $$a^8+b^8+c^8+2(a-1)(b-1)(c-1)$$
My attempts:
$\bullet$ The minimum value is $3$, equality holds iff $a=b=c=1$ so by AM-GM, we have: $$a^8+b^8+c^8\ge 8(a+b+c)-21=3$$
$\bullet$ We need to prove $$3+2(a-1)(b-1)(c-1)\ge3$$ or $$abc-(ab+bc+ca)+a+b+c-1\ge0$$
$\bullet$ Note that $ab+bc+ca\le \dfrac{(a+b+c)^2}{3}=3 $ so we need to prove: $$abc-3+3-1\ge0$$ or $$abc\ge1$$
But I have no idea from here, please help me
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Let $a=b=c=1$.
Thus, we obtain a value $3$.
We'll prove that it's a minimal value, for which it's enough to prove that $$a^8+b^8+c^8+2(a-1)(b-1)(c-1)\geq3.$$
We'll prove that this inequality is true even for any reals $a$, $b$ and $c$ such that $a+b+c=3$.
Indeed, by AM-GM $$\sum_{cyc}a^8\geq\sum_{cyc}(2a^4-1)$$ and it's enough to prove that:
$$2(a^4+b^4+c^4)-3+2(abc-ab-ac-bc+3-1)\geq3$$ or
$$a^4+b^4+c^4+abc-ab-ac-bc-1\geq0$$ or
$$81(a^4+b^4+c^4)+27(a+b+c)abc-9(a+b+c)^2(ab+ac+bc)-(a+b+c)^4\geq0$$ or
$$\sum_{cyc}(80a^4-13a^3b-13a^3c-24a^2b^2-30a^2bc)\geq0$$ or
$$13\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+67\sum_{cyc}(a^4-a^2b^2)+43\sum_{cyc}(a^2b^2-a^2bc)\geq0$$ or
$$13\sum_{cyc}(a-b)^2(a+b-c)^2+67\sum_{cyc}(a^2-b^2)^2+43\sum_{cyc}c^2(a-b)^2\geq0$$ and we are done!
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Range of $\frac{x-y}{1+x^2+y^2}=f(x,y)$ I have a function $\frac{x-y}{1+x^2+y^2}=f(x,y)$. And, I want to find the range of it. I analyzed this function by plotting it on a graph and found interesting things. Like if the level curve is $0=f(x,y)$, then I get $y=x$ which is a linear function. But if the level curve is something not 0, then the level curve becomes a circle. And for big values of level curves, the circle disappears. Is there something I can use to find the range of this function?
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The denominator of the expression for the function $ \ z \ = \ f(x,y) \ = \ \frac{x-y}{1+x^2+y^2} \ $ has "four-fold" symmetry about the origin, which is "broken" by the "diagonal" (anti-)symmetry in the numerator. So this function has the symmetry property $ \ f(x,-y) \ = \ -f(-x,y) \ \ $ and it is reasonable to expect that the extremal points of the function will be described by $ \ (\pm x \ , \ \mp x) \ \ . $
If we take "slices" $ \ x + y \ = \ c \ $ through the function surface, we obtain "cross-sectional" curves
$$ z \ \ = \ \ \frac{x \ - \ (c-x)}{1 \ + \ x^2 \ + \ (c-x)^2} \ \ = \ \ \frac{2x \ - \ c}{ 2x^2 \ - \ 2cx \ + \ (1 + c^2)} \ \ . $$
It is somewhat unappealing to differentiate this as it stands. By completing-the-square in the denominator, we find that we can write
$$ z \ \ = \ \ \frac{2x \ - \ c}{ 2·\left(x^2 \ - \ cx \ + \ \frac{c^2}{4} \right) \ \ + \ (1 + c^2) \ - \ \frac{c^2}{2}} \ \ = \ \ \frac{2·\left(x \ - \ \frac{c}{2} \right)}{ 2·\left(x \ - \ \frac{c}{2} \right)^2 \ \ + \ \left(1 + \frac{c^2}{2} \right) } $$ $$ \rightarrow \ \ \frac{2·u}{ 2·u^2 \ \ + \ \left(1 + \frac{c^2}{2} \right) } \ \ . $$
It is clear that the denominator of the expression is (still) never equal to zero, so we can establish that the derivative of this transformed function is zero for $ \ 4·u^2 \ + \ 2·\left(1 + \frac{c^2}{2} \right) \ - \ 2u·4u \ = \ 0 $ $ \Rightarrow \ u^2 \ = \ \frac12·\left(1 + \frac{c^2}{2} \right) \ \ . $ This locates the extremal points on the cross-sectional curve, with the extremal values being
$$ z \ \ = \ \frac{\pm \ \sqrt2 \ · \ \sqrt{1 + \frac{c^2}{2}}}{ \left(1 + \frac{c^2}{2} \right) \ + \ \left(1 + \frac{c^2}{2} \right) } \ \ = \ \ \pm \frac{ \sqrt2}{2} \ · \ \frac{1}{\sqrt{1 + \frac{c^2}{2}}} \ \ = \ \ \frac{\pm \ 1}{\sqrt{2 \ + \ c^2 }} \ \ . $$
We see from this result that $ \ \lim_{|c| \ \rightarrow \ \pm \infty} \ \ z \ = \ 0 \ \ , $ as you observed. The extremal values are greatest in absolute-value for $ \ c = 0 \ \ , $ confirming that the extremal points occur on the line $ \ x + y \ = \ 0 \ \ . $ The absolute maxima and minima for the function are thus $ \ \pm \frac{ 1}{\sqrt2} \ \ . $
[We did not need to know the locations of the extrema, but we use our relations at $ \ c = 0 \ $ to find $ \ u \ = \ x - \frac{0}{2} \ = \ x \ = \ \pm \frac{ 1}{\sqrt2} \ $ and $ \ y = \ 0 - x \ = \ \mp \frac{ 1}{\sqrt2} \ \ . \ ] $
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How to find numbers like 174 We know Ramanujan number:
$1729 = 1^3+12^3 = 9^3+10^3$
The smallest number expressible as the sum of cubes of two positive integers in two different ways.
We also know how to find other Ramanujan numbers:
$n^3 +(12n)^3 = (9n)^3 + (10n)^3$
In the same way, I want to find $174$
$$\begin{align*} 174
&=1^2+2^2+5^2+12^2\\
&=1^2+3^2+8^2+10^2\\
&=1^2+4^2+6^2+11^2\\
&=2^2+5^2+8^2+9^2\\
&=3^2+4^2+7^2+10^2\\
&=5^2+6^2+7^2+8^2\\
\end{align*}$$
Thus $174$ is the smallest number which can be expressed as the sum of squares of $4$-different integers in $6$ different ways.
How can I find such numbers and don't want to miss any number?
Any algorithm/logical idea is welcome.
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There's a lot you could throw at it actually. Some of these include :
$$y=mx+b\land p\mid m, p\in \mathbb{P} \implies y^p\equiv b^p\pmod {pm}$$
$$y=a^2+b^2+c^2+d^2\implies a\leq b\leq c\leq d$$
The fact that distinct $(a,b,c,d)$ for each sum is only possible above $576=24^2$ because there will be $24$ squares involved. Any with duplicates are based on numbers representable as a sum of $3$ or fewer squares.
If $(a,b,c,d)$ have a pair that form two lower legs of a Pythagorean triple then the current number plus a square will always have at least $1$ representation as $4$ squares
If $(a,b,c,d)$ works then $x^2(a,b,c,d)$ is a representation for a multiplier that's square. Therefore, we can reduce the search to square-free numbers somewhat, because if the radical of $\frac{n}{p}$ works, then it's sufficient to show $np$ has at least as many as required , we just need to rule out it having too many.
Another avenue is looking at the product of two numbers that are a sum of two squares. It always has a representation with four squares.
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How to find $a$ and $b$ given this limit if $f(x)=ax+b$? Question
Suppose $f(x)=ax+b$ and $\lim\limits_{x \to 25} \dfrac{\sqrt{x}-5}{f(x)}=\dfrac{1}{20}$.
Find $a$ and $b$.
Attempt
I figured out that
$\lim\limits_{x \to 25} \sqrt{x}-5=0\space$ but $0$ divided by any denominator will give $0.\quad$ So I do not know how to find $a$ and $b$ given the limit is $\dfrac{1}{20}$.
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Following the hints in the comments, $ax+b$ must go to $0$ when $x$ goes to 25. So $$\lim_{x\to 25}ax+b=25a+b=0$$
That way the limt has the form of $0/0$:$$\lim_{x\to 25}\frac{\sqrt x-5}{f(x)}\to\frac 00$$
Since $25$ is a root of $ax+b$, we can write $$f(x)=a(x-25)=a(\sqrt x -5)\sqrt x+5)$$
It is equivalent of saying $b=-25a$.
To find $a$, we calculate the limit:
$$\lim_{x\to 25}\frac{\sqrt x-5}{f(x)}=\lim_{x\to 25}\frac{\sqrt x-5}{a(\sqrt x-5)(\sqrt x+5)}=\frac1{a(\sqrt{25}+5)}=\frac1{10a}=\frac1{20}$$
So $a=2$ and $b=-50$
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|
Binomial Expansion related problem If ${}^n{C_0} - {}^n{C_1} + {}^n{C_2} - {}^n{C_3} + .. + {\left( { - 1} \right)^r}{}^n{C_r} = 28$. Then find the value of $n$.
My approach is as follow ${}^n{C_0} - {}^n{C_1} + {}^n{C_2} - {}^n{C_3} + .. + {\left( { - 1} \right)^r}{}^n{C_r} = 28 = {}^8{C_2} = {}^8{C_6}$
${\left( {1 - x} \right)^n} = {}^n{C_0} - {}^n{C_1}x + {}^n{C_2}{x^2} - {}^n{C_3}{x^3} + .. + {\left( { - 1} \right)^n}{}^n{C_n}.{x^n}$
Putting x=1
${}^n{C_0} - {}^n{C_1} + {}^n{C_2} - {}^n{C_3} + .. + {\left( { - 1} \right)^n}{}^n{C_n} = 0$
$ \Rightarrow 28 + {\left( { - 1} \right)^{r + 1}}{}^n{C_{r + 1}} + .. + {\left( { - 1} \right)^n}{}^n{C_n} = 0$
How do I approach from here
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Pascal's identity is sufficient :
$$ {N-1 \choose R-1} + {N-1 \choose R} = {N \choose R}$$
Since $${n \choose 0} = 1 = {n-1 \choose 0}$$
given expression is
$${n-1 \choose 0} - {n \choose 1} + {n \choose 2} - {n \choose 3} + \ldots + (-1)^r {n \choose r}$$
Operating on first two terms every step,
$$= \color{red}{- {n-1 \choose 1}} + {n \choose 2} - {n \choose 3} + \ldots + (-1)^r {n \choose r}$$
$$= \color{blue}{ n-1 \choose 2} - {n \choose 3} + \ldots + (-1)^r {n \choose r}$$
$$ \cdots $$
$$= \color{brown}{(-1)^r{n-1 \choose r}}$$
Thus as you have identified, $n-1 = 8$ and $r=2$ or $6$.
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Does the (Stirling number of the second kind) equality ${2n\brace 2} = 2^{2n-1}-1$ hold? I filled in from the definition of a Stirling number of the second kind that the following holds.
$${2n\brace 2} = \frac{1}{2} \sum_{i=0}^{2} (-1)^i \binom{2}{i} (2-i)^{2n}$$
And I've visually confirmed in Desmos that the following equality 'appears' to hold.
$${2n\brace 2} = 2^{2n-1}-1$$
How do I prove that this equality does in fact hold?
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I guess the most straight-forward way is to expand the summation:
\begin{align}
{2n\brace 2} = &\frac{1}{2} \sum_{i=0}^{2} (-1)^i \binom{2}{i} (2-i)^{2n} = \\
& \frac{1}{2} \Bigg( \Big[(-1)^0 \binom{2}{0} (2-0)^{2n}\Big] + \Big[(-1)^1 \binom{2}{1} (2-1)^{2n}\Big] + \Big[(-1)^2 \binom{2}{2} (2-2)^{2n}\Big] \Bigg) = \\
&\frac{1}{2} \Bigg( \Big[1 \cdot 1 \cdot 2^{2n}\Big] + \Big[(-1) \cdot 2 \cdot 1^{2n}\Big] + \Big[1 \cdot 1 \cdot 0^{2n}\Big] \Bigg) = \\
&\frac{1}{2} \Bigg( 2^{2n} - 2 + 0 \Bigg) = \\
& \frac{2}{2} \Bigg( 2^{2n-1} - 1 \Bigg) = 2^{2n-1} - 1
\end{align}
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How to solve for the zeros of $3x^4 - 7x^3 - 12x^2 + 12x + 11$? I’ve tried to see if a rational factor would work from $q/p$, but that hasn’t worked.
I tried grouping and then factoring but that does not work either.
I do not know any further methods that I could use since it doesn’t seem like I could use the quadratic formula or completing the square.
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Hints: You can convert the equation such that factorization becomes easier:
$3x^4-7x^3-12x^2+12x+11=0$
rewrite as:
$x^4-\frac 73 x^3-\frac 4x^2+4x+\frac {11}3=0$
Let y=kx :
$y^4-\frac 73 \cdot \frac{y^3}{k^3}-4\cdot \frac{y^2}{k^2}+4\cdot\frac yk +\frac{11}3=0$
let $k=3$:
$y^4-7y^3-36y^2+108y+297=0$
Now you can factor and tranform equation to:
$y^2(y^2-7y-18)-18y^2+108y+297=0$
$y^2(y-9)(y+2)-9[2(y-9)(y+3)+21]=0$
$(y-9)[y^2(y+2)-18(y+3)]=1\times 9\times 21$
Now equate factors on both sides; for example:
$y-9=1\Rightarrow y=10\rightarrow x=\frac{10}3=3.3$
$y^2(y+2)-18(y+3)= 9\times 21$
apply similar method to find more solutions.
Wolfram gives $x=-1.345, -0.69, 1.28, 3.08$.
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Proving central limit theorem in a specific case( fourier analysis course) I am stuck with the following problem: Show that for any given $R > 0$
$$\lim_{N \to \infty}\sup_{\left|\xi \right| \le R} \left |\left(\frac{1}{6}\sum_{k=1}^{6}e^{2\pi i\xi \frac{k-\frac{7}{2} }{\sqrt{N}} }\right)^N -e^{\frac{-35\pi^2\xi^2}{6}}\right|=0.$$I have tried to taylor expand $ \eta \mapsto e^{i\eta} $ and then noticed that all of the terms that had the imaginary part raised to an odd power cancelled(as the terms $k-\frac{7}{2} $ when summing from 1 to 6 has a negative counterpart) but i did not end up with anything good in the end. Any ideas? Thanks in advance.
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I will show what a possible start could be and let you fill in the technical details.
Using $z + z^* = 2 \mathrm{Re}(z)$ we have
$$
\left(\frac{1}{6} \sum_{k = 1}^{6}{e^{2 \pi i \xi \frac{k - \frac{7}{2}}{\sqrt{N}}}} \right)^N
= \left(\frac{1}{3} \sum_{k \in \{1, 3, 5\}} \frac{e^{\pi i \xi \frac{k}{\sqrt{N}}} + e^{-\pi i \xi \frac{k}{\sqrt{N}}}}{2} \right)^N
= \left(\frac{1}{3} \sum_{k \in \{1,3,5\} } \cos\left(\pi \xi \frac{k}{\sqrt{N}}\right) \right)^N.
$$
Remembering the (first two summands of the) power series representation of cosine, we get
$$
\cos\left(\pi \xi \frac{k}{\sqrt{N}}\right) \approx 1 - \frac{k^2}{2}\frac{\pi^2 \xi^2 }{N}.
$$
Therefore
$$
\left(\frac{1}{6} \sum_{k = 1}^{6}{e^{2 \pi i \xi \frac{k - \frac{7}{2}}{\sqrt{N}}}} \right)^N
\approx \left(1 - \frac{1}{3}\left(\frac{1^2}{2} + \frac{3^2}{2} + \frac{5^2}{2} \right)\frac{\pi^2 \xi^2 }{N} \right)^N
= \left(1 - \frac{35 \pi^2 \xi^2 }{6 N}\right)^N
\longrightarrow e^{-\frac{35 \pi^2 \xi^2}{6}}.
$$
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What is the proof for variance of triangular distribution? In Wikipedia, the formula for the variance of the triangular distribution is given here. However, I don't know how to find it. I have tried a brute force method but the formula is quite complicated (polynomial of degree 5 in a, b, c) and I can't simplify it (I tried manually and with Xcas).
The following part is edited thanks to @Imaosome remark:
I came to this question with the following problem:
Say $X$ and $Y$ are independent random variables with uniform distribution between $0$ and $1$.
I want to study $Z = X-Y$. With convolution, I find that the distribution is triangular, centered in $0$ with extremities $-1$ and $1$ (the proof is also available in this pdf here). With Wikipedia notations, it gives $a=-1, b=1, c=0$. And in this case, we sum $2$ independent variables therefore the variance shoud be
$Var(X)+Var(-Y) = Var(X)+Var(Y)=2 Var(X)$. The variance of $X$ is $1/12$ (see for instance formula here). Therefore the variance of $Z$ is $Var(Z) = 2 * 1/12 = 1/6$.
If I come back to Wikipedia formula, I find:
$$
\frac{a^2+b^2+c^2-ab-ac-bc}{18}=\frac{1+1+0+1-0-0}{18}=1/6.
$$
This shows, at least in that particular case that the formula is correct. But I would like to try and prove Wikipedia result. Once again, I know that it should be possible to prove it by integration but I did not succeed and I hope somebody has a simple way to get this formula.
Here are the details. By definition, we want to compute:
\begin{align*}
\sigma^2 &= \int_a^c \frac{2(x-a)}{(b-a)(c-a)} \left( x - \frac{a+b+c}{3} \right)^2 dx + \int_c^b \frac{2(b-x)}{(b-a)(b-c)} \left( x - \frac{a+b+c}{3} \right)^2 dx \\
&= \frac{2}{(b-a)(c-a)} \left[ \frac{1}{3} (x-a) \left( x - \frac{a+b+c}{3} \right)^3 - \frac{1}{12} \left( x - \frac{a+b+c}{3} \right)^4 \right]_a^c \\
& ~~~~~~ \frac{2}{(b-a)(b-c)} \left[ \frac{1}{3} (b-x) \left( x - \frac{a+b+c}{3} \right)^3 - \frac{1}{12} \left( x - \frac{a+b+c}{3} \right)^4 \right]_c^b
\end{align*}
From there, one can see that terms in $\left( x - \frac{a+b+c}{3}\right)$ cancel out leaving us with:
\begin{align*}
\sigma^2 &= \frac{2}{12(b-a)(c-a)} \left( - \left( \frac{2c-a-b}{3} \right)^4 + \left( \frac{2a-b-c}{3} \right)^4 \right) \\
& ~~~~~~ \frac{2}{12(b-a)(b-c)} \left( - \left( \frac{2b-a-c}{3} \right)^4 + \left( \frac{2c-a-b}{3} \right)^4 \right) \\
&= \frac{((c-a)-(b-c))^5 +(b-c)((b-a)+(c-a))^4 - (c-a)((b-a)+(b-c))^4}{ 2 \times 3^5 (b-a)(b-c)(c-a) }
\end{align*}
And from this point, I am stuck. After all, maybe the last line is not helping much. I don't know.
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It's a little easier to use the identity $\ \sigma^2=\mathbb{E}\big(Z^2\big)-\mathbb{E}(Z\,)^2\ $:
\begin{align}
\mathbb{E}\big(Z^2\big)&=\frac{2}{(b-a)(c-a)}\int_a^cx^2(x-a)\,dx\\
&\hspace{2em}+\frac{2}{(b-a)(b-c)}\int_c^bx^2(b-x)\,dx\\
&=\Bigg(\frac{2}{(b-a)(c-a)}\Bigg)\Bigg(\frac{c^4-a^4}{4}-\frac{a\big(c^3-a^3\big)}{3}\Bigg)\\
&\hspace{2em}+\Bigg(\frac{2}{(b-a)(b-c)}\Bigg)\Bigg(\frac{b\big(b^3-c^3\big)}{3}-\frac{b^4-c^4}{4}\Bigg)\\
&=\Bigg(\frac{1}{(b-a)(c-a)}\Bigg)\Bigg(\frac{3\big(c^4-ac^3\big)+a^4-ac^3}{6}\Bigg)\\
&\hspace{2em}+\Bigg(\frac{1}{(b-a)(b-c)}\Bigg)\Bigg(\frac{3\big(c^4-bc^3\big)+b^4-bc^3}{6}\Bigg)\\
&=\Bigg(\frac{1}{(b-a)(c-a)}\Bigg)\Bigg(\frac{3c^3(c-a)+a(a-c)\big(a^2
+ac+c ^2\big)}{6}\Bigg)\\
&\hspace{2em}+\Bigg(\frac{1}{(b-a)(b-c)}\Bigg)\Bigg(\frac{3c^3(c-b)+b(b-c)\big(b^2+bc+c^2\big)}{6}\Bigg)\\
&=\frac{3c^3-a^3-a^2c-ac^2}{6(b-a)}+\frac{b^3+b^2c+bc^2-3c^3}{6(b-a)}\\
&=\frac{a^2+b^2+c^2+ab+ac+bc}{6}\\
\mathbb{E}(Z\,)^2&=\Big(\frac{a+b+c}{3}\Big)^2\\
&=\frac{a^2+b^2+c^2+2ab+2ac+2bc}{9}\\
\sigma^2&=\mathbb{E}\big(Z^2\big)-\mathbb{E}(Z\,)^2\\
&=\frac{a^2+b^2+c^2-ab-ac-bc}{18}
\end{align}
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Let $x$,$y$ be rationals such that $\frac{x^2+x+\sqrt{2}}{y^2+y+\sqrt{2}}$ is also rational, prove that either =, or +=−1. I followed one convincing proof in this related post but I don't understand the assumption and conclusion.
The proof goes as follows:
Let $r\in\mathbb{Q}$ such that
$\frac{x²+x+\sqrt{2}}{y²+y+\sqrt{2}}=r$ and suppose that $\frac{x}{y}$ is a irreducible fraction. Then
$$\frac{x²+x+\sqrt{2}}{y²+y+\sqrt{2}}=r\Leftrightarrow
> x²+x+\sqrt{2}=(y²+y+\sqrt{2})r\Leftrightarrow
> x²+x-y²r-yr=\sqrt{2}(r-1).$$ If $r\neq 1$ we have
$\sqrt{2}=\frac{x²+x-y²r-yr}{r-1}\in\mathbb{Q}$, but this is absurd.
So we get $r=1$ and $$x²+x+\sqrt{2}=y²+y+\sqrt{2}\Rightarrow
> y(y+1)=x(x+1).$$ If $x=-1$, we have $y(y+1)=0$ and so $y=0=x+1$ or
$y=x=-1$. If $x\neq -1$ we have $$x=y(\frac{y+1}{x+1}).$$ If $y=-1$,
it is analogous. We can suppose that $y\neq -1$. If $y=0$, $x=0$.
Suppose $-1\neq y\neq 0$. We have $$\frac{x}{y}=\frac{y+1}{x+1}.
$$Since $\frac{x}{y}$ is irreducible, follows the last equality that $x=y$.
I have two questions:
*
*Is it valid to suppose that $\frac{x}{y}$ is an irreducible fraction? Wouldn't that ignore other values?
*How assuming $\frac{x}{y}$ is irreductible follows that $x$=$y$ for $\frac{x}{y}=\frac{y+1}{x+1}$?
I am particularly interested in the first question as it conflicts with my developing rigour in Mathematics.
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No, it's not allowed to even consider $x$ and $y$ integers.
When you arrive at
$$
x(x+1)=y(y+1)
$$
you're almost done, because this can be rewritten as
$$
x^2-y^2+x-y=0
$$
hence
$$
(x-y)(x+y+1)=0
$$
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Factoring a third degree polynomial with a given root My professor gave us the following polynomial:
$f(x) = 3x^3-4x^2-x+2$
Given is that $x = 1$ is a root of this function. We are asked to find the other ones.
He then told us that, given $x=1$ is a root, we now know that we can factorize this polynomial into $(x-1)$, and a second factor starting with $3x^2...$.
I'm familiar with the process of factoring polynomials, and of finding the roots by setting each factor to zero. But what are you actually doing when you factor a polynomial? In my mind it's just a formulaic way to find zeros of such functions. That's why I was a little confused when we are asked to do the reverse; finding the factor of a given root. More generally, if you know a root of a polynomial, for example $x=7$, can you then always conclude that one factor is $(x-7)$?
How should I proceed finding the other factor(s)?
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Yes, that is correct. If $\alpha$ is the root of the polynomial $P(x)$, then $x-\alpha \mid P(x)$.
Also, we have several methods to factorise your polynomial:
*
*Polynomial long division
*Synthetic division
Finally, I can suggest the following general method:
$$\begin{align}&3x^3-4x^2-x+2=(x-1)(3x^2+ax+b)\\
\implies &3x^3-4x^2-x+2=3x^3+x^2(a-3)-x(a-b)-b\\
\implies &\begin{cases}a-3=-4\\ a-b=1\\ b=-2\end{cases}\\
\implies &(a,b)=(-1,-2)
\end{align}$$
Then applying the quadratic formula we get,
$$3x^2-x-2=(3x+2)(x-1)$$
Therefore, we have
$$3x^3-4x^2-x+2=(3x+2)(x-1)^2$$
Small supplement:
Since $f'(x)=9x^2-8x-1$ and $f'(1)=0$, then we see that $x_1=1$ and $x_2=1$ are repeated root of the polynomial $f(x)=3x^3-4x^2-x+2$.
This implies,
$$(x-1)^2\mid 3x^3-4x^2-x+2$$
Then, Vieta's formulas tells us,
$$\begin{align}1\times 1\times x_3=-\frac 23\implies x_3=-\frac 23\end{align}$$
Therefore, we conclude that
$$\begin{align}&3x^3-4x^2-x+2=3(x-1)^2\left(x+\frac 23\right)\\
\iff &3x^3-4x^2-x+2=(3x+2)(x-1)^2.\end{align}$$
|
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How do you prove by induction that $\frac{1}{2} + \frac{2}{2^2} + \ldots + \frac{n}{2^n} = 2 - \frac{n+2}{2^n}$? For $n=1$ this is true because $\frac{1}{2^{1}}=2-\frac{1+2}{2^{1}}=\frac{1}{2}$. Further, it is a little more complicated, can we now assume that this is true up to the number $n-1$? Then do the induction step from $n-1$ to $n$.
So what I've tried:
$$\frac{1}{2}+\frac{1}{2^{2}}+\ldots+\frac{n-1}{2^{n-1}}+\frac{n}{2^{n}}=$$
$$=2-\frac{n+1}{2^{n-1}}+\frac{n}{2^{n}}=$$
|
\begin{align*}
\frac{1}{2} + \frac{1}{2^2} + ... + \frac{n-1}{2^{n-1}}+\frac{n}{2^n}
&=2-\frac{n+1}{2^{n-1}} + \frac{n}{2^n}\\
&= 2+\frac{-2n-2 + n}{2^{n}}\\
&= 2- \frac{n+2}{2^n}
\end{align*}
|
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Factoring $4x^2-2xy-4x+3y-3$. Why isn't it working? I want to factorise the following expression.
$$4x^2-2xy-4x+3y-3$$
Here are the ways I tried
$$4x^2-2xy-4x+3y-3=\left(2x-\frac y2\right)^2+3y-3-\frac{y^2}{4}-4x=\left(2x-\frac y2\right)^2+\frac{12y-12-y^2}{4}-4x=\left(2x-\frac y2\right)^2-\frac 14(y^2-12y+12)-4x$$
Now I need to factor the quadratic $y^2-12y+12$.
So, I calculated discriminant
$$D=12^2-4\times 12=96\implies \sqrt D=4\sqrt 6.$$
This means that the multipliers of quadratic are not rational. So I don't know how to proceed anymore.
|
When $2x=3$, then the expression becomes $$4x^2-2xy-4x+3y-3=9 - 3y -6 + 3y - 3 = 0$$
Hence the expression can be factored by $(2x-3)$, and you can see easily that
$$4x^2-2xy-4x+3y-3 = (2x-3)(2x-y+1)$$
|
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Prove the inequality $\displaystyle\sum_{s y m} x^{4} y^{2} z \geqslant 2 \sum_{s y m} x^{3} y^{2} z^{2}$ Prove for positive $x,y,z$ the inequality $x^{4} y^{2} z+x^{4} z^{2} y+y^{4} x^{2} z+y^{4} z^{2} x+z^{4} y^{2} x+z^{4} x^{2} y \geqslant 2(x^{3} y^{2} z^{2}+x^{2} y^{3} z^{2}+x^{2} y^{2} z^{3})$.
I found the following proof, it was without explanation:
\begin{gathered}
x^{3}(y+z)+y^{3}(x+x)+z^{3}(x+y) \geq 2(x^{2} y z+y^{2} z x+z^{2} x y) \\
2 x^{3}(y+z)+y^{3} z+z^{3} x \geq 6 x^{2} y z \\
2 y^{3}(z+x)+z^{3} x+x^{3} z \geq 6 y^{2} z x \\
2 z^{3}(x+y)+x^{3} y+y^{3} x \geq 6 z^{2} x y
\end{gathered}
I don't understand what starts happening from the 2nd line, can you explain and provide an alternative method, if possible?
|
The last term of the left side of the second line should be $z^3y$ and not $z^3x$. Its a typo. Also, they used the AM-GM inequality:
$2x^3(y+z) + y^3z + z^3y = x^3y+x^3y+x^3z+x^3z+y^3z+z^3y \ge 6\sqrt[6]{x^3y\cdot x^3y\cdot x^3z\cdot x^3z\cdot y^3z\cdot z^3y}=6x^2yz$
|
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What's the measure of the side of the rombus in the figure below? For rereence:In a rhombus the sum of the measures of its diagonals is $70$ cm
and the radius of the inscribed circle is $12$ cm. Calculate the measure of the rhombus side,
My progress:
$CO +BO = \frac{70}{2} = 35$
$\frac{1}{h^2}=\frac{1}{a^2}+\frac{1}{c^2}\\
\frac{1}{144} = \frac{1}{(35-CO)^2}+\frac{1}{CO^2}(CO = x)\implies\\
\frac{1}{144} = \frac{2x^2-70x+1225}{x^2(x-35)^2}$
therefore (by Wolfran) $CO = 15 \implies OB = 20$
or
$CO = 20 \implies OB = 15\\
\therefore BC^2 = 15^2+20^2 \implies \boxed{BC = 25}$
Would there be a way to get a simpler equation?
|
$\mathsf{
AO(x)+OB(y) = \frac{70}{2}=35\\
(x+y)^2 = 35^2\implies x^2+y^2+2xy = 1225\\
Por~ métrica: x^2+y^2 = AB(a)^2\\
a.h = x.y\implies 12a = xy\\
\therefore : a^2+24a-1225 = 0\\
\therefore \boxed{\color{red}a =AB = 25} }$
|
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|
Mistake computing $\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(x)}-\frac{a}{e^{2x}} \right)\frac{dx}{x}$ I am looking to evaluate the integral
$$\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(x)}-\frac{a}{e^{2x}} \right)\frac{dx}{x}=\ln\left(\frac{\pi}{\Gamma^2\left(\frac{1+a}{2b}\right)\cos\left(\frac{a\pi}{2}\right)}\right)$$
To this end I considered
$$I(w)=\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(bx)}-\frac{ad}{e^{cx}} \right)\frac{e^{-wx}}{x}\,dx \tag{1}$$
Note that as $w \to \infty$ the integrand vanishes. And as $w =0$ we recover the desired integral.
Differentiating $(1)$ w.r. to $w$ we obtain
$$
\begin{aligned}
I^\prime(w)&=-\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(bx)}-\frac{ad}{e^{cx}} \right)e^{-wx}\,dx\\
&=ad\int_0^\infty e^{-(c+w)x}\,dx- \int_0^\infty \frac{\sinh(ax)}{\sinh(bx)}e^{-wx}\,dx\\
&=\frac{ad}{c+w}-\int_0^\infty \frac{e^{ax}-e^{-ax}}{e^{bx}-e^{-bx}}e^{-wx}\,dx\\
&=\frac{ad}{c+w}-\int_0^\infty \frac{e^{-(w-a)x}-e^{-(w+a)x}}{e^{bx}-e^{-bx}}\,dx\\
&=\frac{ad}{c+w}-\int_0^\infty \frac{e^{-bx}}{e^{-bx}}\cdot\frac{e^{-(w-a)x}-e^{-(w+a)x}}{e^{bx}-e^{-bx}}\,dx\\
&=\frac{ad}{c+w}-\int_0^\infty \frac{e^{-(w-a+b)x}-e^{-(w+a+b)x}}{1-e^{-2bx}}\,dx\\
&=\frac{ad}{c+w}-\frac{1}{2b}\int_0^\infty \frac{e^{-\frac{(w-a+b)}{2b}x}-e^{-\frac{(w+a+b)}{2b}x}}{1-e^{-x}}\,dx \qquad (2bx \to x)\\
&=\frac{ad}{c+w}-\frac{1}{2b}\int_0^1 \frac{x^{\frac{(w-a+b)}{2b}-1}-x^{\frac{(w+a+b)}{2b}-1}}{1-x}\,dx \qquad (e^{-x} \to x)\\
&=\frac{ad}{c+w}-\frac{1}{2b}\left(\psi\left(\frac{w+a+b}{2b}\right)-\psi\left(\frac{w-a+b}{2b}\right)\right)\\
I(w)&=ad\int\frac{1}{c+w}\,dw-\frac{1}{2b}\left(\int\psi\left(\frac{w+a+b}{2b}\right)\,dw-\int\psi\left(\frac{w-a+b}{2b}\right)\,dw\right)\\
&=ad\ln(c+w)-\left(\ln\left(\Gamma\left(\frac{w+a+b}{2b}\right)\right)\,-\ln\left(\Gamma\left(\frac{w-a+b}{2b}\right)\right)\right)\\
&=ad\ln(c+w)+\ln\left(\frac{\Gamma\left(\frac{w-a+b}{2b}\right)}{\Gamma\left(\frac{w+a+b}{2b}\right)}\right)\\
\end{aligned}
$$
Now,our integral is equal to
$$I=-\int_0^\infty I^\prime(w)\,dw=I(0)$$
Letting $w=0$
$$\begin{aligned}
\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(bx)}-\frac{ad}{e^{cx}} \right)\frac{dx}{x}&=\ln\left(\frac{c^{ad}\Gamma\left(\frac12-\frac{a}{2b}\right)}{\Gamma\left(\frac12+\frac{a}{2b}\right)}\right)\\
&=\ln\left(\frac{c^{ad}\Gamma\left(\frac12-\frac{a}{2b}\right)\Gamma\left(\frac12+\frac{a}{2b}\right)}{\Gamma\left(\frac12+\frac{a}{2b}\right)\Gamma\left(\frac12+\frac{a}{2b}\right)}\right)\\
&=\ln\left(\frac{c^{ad}\pi}{\Gamma^2\left(\frac12+\frac{a}{2b}\right)\cos\left(\frac{a\pi}{2b}\right)}\right) \qquad \blacksquare\\
\end{aligned}$$
setting $b=1$, $c=2$ and $d=1$ I obtained
$$\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(x)}-\frac{a}{e^{2x}} \right)\frac{dx}{x}=\ln\left(\frac{2^{a}\pi}{\Gamma^2\left(\frac{1+a}{2b}\right)\cos\left(\frac{a\pi}{2}\right)}\right)$$
Which has an extra term leading to an incorrect answer. Can someone please point out where I am mistaking?
|
With a simple renaming of variables, I solved the OP's question using Ramanujan's generalization of Frullani's integral. The OP's method should work as well.
Evaluate $\int_0^{\infty } \Bigl( 2qe^{-x}-\frac{\sinh (q x)}{\sinh \left(\frac{x}{2}\right)} \Bigr) \frac{dx}x$
|
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|
integral with disks We are looking for the volume of the space between $y=\frac{-1}{2}x+\frac{3}{2}$ and $y=x$ as rotated about the x-axis. I used the disk/washer method ($\pi r^2$) and broke it up into 3 pieces. I notice that there is a bit of overlap under the x-axis, so I have excluded it in my computation. Please see the attached picture for a more detailed view. image
$\pi (\int_{-2}^0 [(-1/2)x+(3/2)]^2)dx+\pi(\int_0^1[(-1/2)x+(3/2)]^2-[x]^2)dx+\pi\int_1^3(x^2-[(-1/2)x+(3/2)]^2-[x]^2)dx=$
$\frac{-\pi}{4}\int_0^{-2}(x^2-6x+9)dx+\frac{\pi}{4}\int_0^1(x^2-6x+9)dx+\frac{\pi}{4}\int_1^3(3x^2+6x-x)dx=$
$\frac{\pi}{4}[(-\frac{x^3}{3}+3x^2-9x)_0^{-2}+(-x^3-3x^2+9x)_0^1+(x^3+3x^2-9x)_1^3]=$
$\frac{\pi}{4}[8/3+12+18-1+3+9+27+27-27-1-3+9]=$
$\frac{\pi}{4}[8/3+73]=$
$\frac{\pi}{4}(227/3)=$
$59.42846...$
|
You split up the integral into three parts and these parts should be
$$ \pi \int_{-2}^{0}\left(\frac{-x}{2}+\frac 3 2\right)^2dx$$
$$\pi \int_{0}^1\left(\frac{-x}{2}+\frac 3 2\right)^2 - x^2dx$$
$$\pi \int_{1}^3 x^2 - \left(\frac{-x}{2}+\frac 3 2\right)^2dx$$
Notice that your third term has an extra $-x^2$ inside the integral which shouldn't be there, I guess that's a typo.
The way you calculate the first term is quite odd (why switch the bounds of the integral ?) but looks mostly fine.
The main problem is the computation of the second term since you drop the $-x^2$ inside the integral instead of what you wrote you should have
\begin{align}
\pi \int_{0}^1\left(\frac{-x}{2}+\frac 3 2\right)^2 - x^2dx &= \frac \pi 4 \int_{0}^1(-x + 3)^2 - 4x^2dx \\
&=\frac \pi 4 \int_{0}^1x^2 - 6x + 9 - 4x^2dx\\
&= \frac \pi 4 \int_{0}^1 -3x^2 - 6x + 9dx
\end{align}
For the third term you shouldn't obtain $3x^2+6x-x$ inside the integral rather you should have $3x^2 + 6x - 9$.
|
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|
Generating Function of Riordan numbers I would like to find generating function of $f(n)$, where $f(n)$ is defined as following:
$$f(n)=\sum_k^n \binom{n}{k}(-1)^{n-k}C_k\text{.}$$
With $C_k=\frac{1}{k+1}\binom{2k}{k}$($C_k$ is the $k^{th}$ Catalan's number).
Thanks in advantage for Your help
Good Evening
|
We have for the sum
$$\sum_{k=0}^n {n\choose k} (-1)^{n-k} C_k
= \sum_{k=0}^n {n\choose k} (-1)^k C_{n-k}
\\ = [z^n] \frac{1-\sqrt{1-4z}}{2z}
\sum_{k=0}^n {n\choose k} (-1)^k z^k
= [z^n] \frac{1-\sqrt{1-4z}}{2z}
(1-z)^n
\\ = \; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{n+1}} (1-z)^n
\frac{1-\sqrt{1-4z}}{2z}.$$
Now put $z/(1-z) = w$ so that $z=w/(1+w)$ and $dz = 1/(1+w)^2 \; dw$
to find
$$\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{n+1}} (1+w)
\frac{1-\sqrt{1-4w/(1+w)}}{2w/(1+w)}
\frac{1}{(1+w)^2}
\\ = \; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{n+1}}
\frac{1+w-\sqrt{(1+w)^2-4w(1+w)}}{2w(1+w)}
\\ = \; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{n+1}}
\frac{1+w-\sqrt{1-2w-3w^2}}{2w(1+w)}.$$
It follows that the desired OGF is
$$\frac{1+w-\sqrt{1-2w-3w^2}}{2w(1+w)}.$$
|
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|
Show that if $0 \leq a \leq b$ then $f_m(a) \leq f_n(b)$ for $m \leq n.$
Let $n \in \mathbb N$ and let $f_n$ be the function $t \mapsto t \left (t + \frac {1} {n} \right )^{-1}$ on $\mathbb R_{+}.$ Let $a$ and $b$ be two positive elements in a $C^{\ast}$-algebra $A$ such that $a \leq b$ and let $m \leq n.$ Then $f_m (a) \leq f_n (b),$ where $f_m(a)$ and $f_n(b)$ are the images of $f_m$ and $f_n$ under the continuous functional calculus for $a$ and $b$ respectively.
My Attempt $:$ Here $$f_m(a) = a \left (a + \frac {1} {m} \right )^{-1} = \left (a + \frac {1} {m} \right )^{-1} a\ \ \text{and}\ \ f_n (b) = b \left (b + \frac {1} {n} \right )^{-1} = \left (b + \frac {1} {n} \right )^{-1} b.$$
From here I couldn't quite able to show the required inequality.
Since $m \leq n$ we have $0 \leq a + \frac {1} {n} \leq a + \frac {1} {m}.$ Hence $0 \leq \left (a + \frac {1} {m} \right )^{-1} \leq \left (a + \frac {1} {n} \right )^{-1}.$ So $$\begin{align*} a \left (a + \frac {1} {m} \right )^{-1} & = a^{\frac {1} {2}} \left (a + \frac {1} {m} \right )^{-1} a^{\frac {1} {2}} \\ & \leq a^{\frac {1} {2}} \left (a + \frac {1} {n} \right )^{-1} a^{\frac {1} {2}} \\ & = a \left (a + \frac {1} {n} \right )^{-1} \\ & = \left (a + \frac {1} {n} \right )^{-\frac {1} {2}} a \left (a + \frac {1} {n} \right )^{-\frac {1} {2}} \\ & \leq \left (a + \frac {1} {n} \right )^{-\frac {1} {2}} b \left (a + \frac {1} {n} \right )^{-\frac {1} {2}} \\ & = b \left (a + \frac {1} {n} \right )^{-1} \\ & = b^{\frac {1} {2}} \left (a + \frac {1} {n} \right )^{-1} b^{\frac {1} {2}} \end{align*}$$ At this stage I got stuck. I know that $0 \leq a + \frac {1} {n} \leq b + \frac {1} {n}.$ So we have $0 \leq \left (b + \frac {1} {n} \right )^{-1} \leq \left (a + \frac {1} {n} \right )^{-1},$ which is not what I wanted to have. Could anyone give me some suggestion regarding this?
Thanks for your time.
EDIT $:$ I think the last two equalities doesn't hold in general unless $a$ and $b$ commute.
|
As you showed, what you want is to prove that $f_n(a)\leq f_n(b)$ if $a\leq b$. The terminology for this is that $f_n$ is operator monotone.
For notational simplicity we may take $f(t)=t(t+c)^{-1}$, with $c>0$. As a motivation, how do we show that $f$ is (number) monotone? We want to show that
$$
s<t\implies\frac s{s+c}<\frac t{t+c}.
$$
This is trivial if one notices that
$$
\frac t{t+c}=\frac1{1+\frac ct}.
$$
the big advantage being that there is a single $t$ on the right-hand-side.
To see that $f$ is operator monotone, assume first that $a$ is invertible; then so is $b$ and we have
$$
f(a)=a(a+c)^{-1}=(1+ca^{-1})^{-1}.
$$
Then
$$
a\leq b\implies b^{-1}\leq a^{-1}\implies 1+cb^{-1}\leq 1+ca^{-1}\implies (1+ca^{-1})^{-1}\leq (1+cb^{-1})^{-1}.
$$
When $a$ is not invertible, we can apply the above to $a+\varepsilon 1\leq b+\varepsilon 1$. So $$f(a)=\lim_{\varepsilon\to0}f(a+\varepsilon 1)\leq \lim_{\varepsilon\to0} f(b+\varepsilon 1)=f(b),
$$
where the equalities are given by the continuity of $f$ at $t=0$. Explicitly, if $\Gamma$ is the Gelfand transform for $a$, let $g_\varepsilon(t)=f(t+\varepsilon)-f(t)$. Then $g_\varepsilon\to0$ uniformly on $\sigma(a)$, and so
$$
f(a+\varepsilon)-f(a)=\Gamma(g_\varepsilon)\to0.
$$
Edit: the key technical point in the argument is the fact that $0\leq a\leq b$ with $a$ invertible, implies that $b^{-1}\leq a^{-1}$. The usual proof comes from recognizing that, for $x$ positive, $x\leq 1$ if and only if $\sigma(x)\subset[0,1]$, together with the fact that $\{0\}\cup\sigma(xy)=\{0\}\cup\sigma(yx)$.
So, if $a\leq b$, then multiplying left and right by $b^{-1/2}$ we get $b^{-1/2}ab^{-1/2}\leq 1$. This is the same as $\sigma(b^{-1/2}ab^{-1/2})\subset[0,1]$. Now
$$
\sigma(b^{-1/2}ab^{-1/2})=\sigma(b^{-1/2}a^{1/2}a^{1/2}b^{-1/2})
=\sigma(a^{1/2}b^{-1/2}b^{-1/2}a^{1/2})=\sigma(a^{1/2}b^{-1}a^{1/2}).
$$
Thus $a^{1/2}b^{-1}a^{1/2}\leq1$. Multiplying left and right by $a^{-1/2}$ we obtain $b^{-1}\leq a^{-1}$.
|
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|
$f(n) = f(2n)$, and $f(2n + 1) = f(n) + 1$, find expression of such $f$ The question is
a. $f(n) = f(2n)$
b. $f(2n + 1) = f(n) + 1$,
with $f(1)=1$, find expression of such $f$ that defined on positive integers
Got an initial idea about the pattern but find the difficulty to find the exact form. It's related to exponent of 2 (obviously). the number of times it can be expressed as addition of 2 exponents
for example, for $13 = 2^3+2^2+2^0, f(13) = 3$
and $16=2^4, f(16)=1; 15 = 2^3+2^2+2^1+2^0, f(15)=4$
in other words, it's the sum of all digits in its binary form, right? if so what's the formula of it.
during such time as I wrote my thoughts down, it seems I come up with a formula
$$a-\sum_{i=1}^\infty \lfloor a/2^i\rfloor.$$
which seems to be the answer to the question
|
Expanding on my comment, let the binary representation of $a$ be $\,\overline{a_na_{n-1}\dots a_1a_0}\,$ with $\,a_k \in \{0,1\}\,$, then $\,a = \sum_{k \ge 0} 2^k a_k\,$ and $\,a_k = \left\lfloor \frac{a}{2^k}\right\rfloor \bmod 2\,$, so the sum of binary digits can be written as $\,\sum_{k \ge 0} \left(\left\lfloor \frac{a}{2^k}\right\rfloor \bmod 2\right)\,$.
Using the identity $\,n \bmod 2 = n - 2 \left\lfloor \frac{n}{2} \right\rfloor\,$ leads to the form in OP's question:
$$
\require{cancel}
\begin{align}
\sum_{k \ge 0} \left(\left\lfloor \frac{a}{2^k}\right\rfloor \bmod 2\right) &= \sum_{k \ge 0} \left( \left\lfloor \frac{a}{2^k}\right\rfloor - 2 \left\lfloor \frac{a}{2^{k+1}}\right\rfloor \right)
\\ &= \sum_{k \ge 0} \left( \left\lfloor \frac{a}{2^k}\right\rfloor - \left\lfloor \frac{a}{2^{k+1}}\right\rfloor \right) - \sum_{k \ge 0} \left\lfloor \frac{a}{2^{k+1}}\right\rfloor
\\ &= \left(\left\lfloor \frac{a}{2^0}\right\rfloor - \cancel{\left\lfloor \frac{a}{2^1}\right\rfloor} + \cancel{\left\lfloor \frac{a}{2^1}\right\rfloor} - \bcancel{\left\lfloor \frac{a}{2^2}\right\rfloor} + \bcancel{\left\lfloor \frac{a}{2^2}\right\rfloor} - \dots\right) - \sum_{k \ge 1} \left\lfloor \frac{a}{2^{k}}\right\rfloor
\\ &= a - \sum_{k \ge 1} \left\lfloor \frac{a}{2^{k}}\right\rfloor
\end{align}
$$
|
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Solve the equation $\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$ Solve the equation $$\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$$
We have $DM:3x+2>0,x>-\dfrac23, x\in DM=\left(-\dfrac23;+\infty\right)$, so we can multiply the whole equation by $\sqrt{3x+2}\ne0$. Then we will have $$3x+2+x^2=2x\sqrt{3x+2}\\x^2+3x+2-2x\sqrt{3x+2}=0\\(x+1)(x+2)-2x\sqrt{3x+2}=0$$ What next? Thank you!
|
Rewrite the equation as $$\left(\sqrt[4]{3x + 2} - \frac x{\sqrt[4]{3x+2}}\right)^2 = 0.$$ It follows that $\sqrt{3x+2} = x$ and then $x^2 - 3x - 2 = 0$. Note that we require $x \geq 0$.
|
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For positive real numbers $a,b,c$ such that $a+b+c=1$. Prove that...
For positive real numbers $a,b,c$ such that $a+b+c=1$. Prove that $$\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$$
Here's what I've done so far:
$\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
$=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
$=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}+3\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
$=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
Not sure where to go from here, any help's appreciated. I think the $AM-GM$ inequality should be used here in some way.
|
Hint:
$$(\frac{a}{b}+\frac{c}{b})+(\frac{a}{c}+\frac{b}{c})+(\frac{b}{a}+\frac{c}{a})+6$$
$$ = \frac{1-b}{b}+\frac{1-c}{c}+\frac{1-a}{a}+2+2+2$$
Now check if $$\frac{1-x}{x}+2\geq 2\sqrt2 \sqrt\frac{1-x}{x}$$ is true?
|
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|
Lagrange inversion theorem and Legendre polynomials generating function I have been trying to solve the following problem (taken from Spiegel's Complex variables, problem 6.105).
By considering the equation $z=a+w\frac{(z^2-1)}{2}$, show that
$$\frac{1}{\sqrt{1-2aw+w^2}}=1+\sum_{n=1}^{\infty} \frac{w^n}{2^n n!}\frac{d^{n}}{da^{n}}[(a^2-1)^n]$$
I managed to come up with two tries.
Attempt 1:
Starting with the equation $z=a+w\frac{(z^2-1)}{2}$ we can write $w=2\frac{z-a}{z^2-1}$, so we have:
$$\frac{1}{\sqrt{1-2aw+w^2}}=\frac{1}{\sqrt{1-2a(2\frac{z-a}{z^2-1})+(2\frac{z-a}{z^2-1})^2}}=\underbrace{\dots}_\text{some algebra later}=\pm \frac{z^2-1}{z^2-2az+1}=\widetilde{F}(z)$$
We want to apply this $\widetilde{F}(z)$ function to the Lagrange's inversion formula (6.11):
Being $F(z)$ the root of the equation $F(z)=a+wf(z)$, then:
$$F(z)=F(a)+\sum_{n=1}^{\infty} \frac{w^n}{n!}\frac{d^{n-1}}{da^{n-1}}[F'(a)f^n(a)]$$
Then we want to make $\widetilde{F}(a)=1$, so we choose the negative branch of $\widetilde{F}(z)$, so this way
$$\widetilde{F}(a)=\left.-\frac{z^2-1}{z^2-2az+1}\right|_{z=a}=-\frac{a^2-1}{a^2-2a^2+1}=\frac{1-a^2}{1-a^2}=1$$
So, let $F(z)=\frac{1-z^2}{z^2-2az+1}$, then we have $F(a)=1$, and:
$$F'(a)=\frac{d}{dz}F(z)|_{z=a}=\frac{d}{dz}(\frac{1-z^2}{z^2-2az+1})|_{z=a}=\dots=\frac{2a}{a^2-1}$$
while $f(z)=\frac{z^2-1}{2}$. So putting this in Lagrange's formula, we come up with:
$$F(z)=1+\sum_{n=1}^{\infty} \frac{w^n}{n!}\frac{d^{n-1}}{da^{n-1}}[\frac{2a}{a^2-1}(\frac{a^2-1}{2})^n]$$
$$F(z)=1+\sum_{n=1}^{\infty} \frac{w^n}{2^n n!}\frac{d^{n-1}}{da^{n-1}}[\frac{2a}{a^2-1}(a^2-1)^n]$$
$$F(z)=1+\sum_{n=1}^{\infty} \frac{w^n}{2^n n!}\frac{d^{n-1}}{da^{n-1}}[2a(a^2-1)^{n-1}]$$
$$F(z)=1+\sum_{n=1}^{\infty} \frac{w^n}{2^n n!}\frac{d^{n-1}}{da^{n-1}}[\frac{1}{n}\frac{d}{da}(a^2-1)^{n}]$$
$$F(z)=1+\sum_{n=1}^{\infty} \frac{w^n}{2^n n! \color{red}{n}}\frac{d^{n}}{da^{n}}[(a^2-1)^{n}]$$
Therefore:
$$\frac{1}{\sqrt{1-2aw+w^2}}=1+\sum_{n=1}^{\infty} \frac{w^n}{2^n n! \color{red}{n}}\frac{d^{n}}{da^{n}}[(a^2-1)^{n}]$$
which doesn't match with what we wanted to show.
Attempt 2:
After searching for more similar problems, I found Lunt's book (A collection of problems on complex analysis), specifically problem 985, that states:
Let $z=z(w)$ be a single-valued function, defined for sufficiently small $|w|$ by the equation $z-a-wf(z)=0$, the function $f(z)$ analytic at the point $z=a$ and $f(a)\neq 0$.Prove that for every function $\Phi(z)$ analytic at the point $z =a$, for sufficiently small $|w|$ there holds the expansion:
$$\frac{\Phi(z)}{1-wf'(z)}=\Phi(a)+\sum_{n=1}^{\infty} \frac{w^n}{n!}\frac{d^{n}}{da^{n}}[\Phi(a)f^n(a)]$$
Proven this, taking $f(z)=\frac{z^2-1}{2}$ and $z-a-wf(z)=0$ as in attempt 1, and $\Phi(z)=1$ we have:
$$\frac{1}{1-wf'(z)}=\frac{1}{1-(2\frac{z-a}{z^2-1})(z)}=\dots=\frac{1-z^2}{z^2-2az+1}=F(z)$$
and then, the result is immediate:
$$\frac{\Phi(z)}{1-wf'(z)}=\Phi(a)+\sum_{n=1}^{\infty} \frac{w^n}{n!}\frac{d^{n}}{da^{n}}[\Phi(a)f^n(a)]$$
$$F(z)=1+\sum_{n=1}^{\infty} \frac{w^n}{n!}\frac{d^{n}}{da^{n}}[1(\frac{a^2-1}{2})^n]$$
$$\therefore \frac{1}{\sqrt{1-2aw+w^2}}=1+\sum_{n=1}^{\infty} \frac{w^n}{2^n n!}\frac{d^{n}}{da^{n}}[(a^2-1)^n]$$
The Question
I think the 2nd attempt answers the problem but for me it doesn't 'feel correct' in some sense (don't know why). I believe that it should be possible to get the correct answer from attempt 1 but I cannot figure out how.
Actually, Lunt's problem 986 states that starting from problem 985 we can prove that
$$\Phi(z)=\Phi(a)+\sum_{n=1}^{\infty} \frac{w^n}{n!}\frac{d^{n-1}}{da^{n-1}}[\Phi'(a)f^n(a)]$$
if we apply 985 result to the function $\Phi(z)[1-wf'(z)]$. But I have not been able to solve it yet (only assuming that $f'(a)=0$ but it doesn't seems right to do without proving). So maybe this one is the real question here. Any insight, suggestion or further references on this kind of problems will be appreciated.
I haven't found more of this topic related to complex analysis (is it obsolete for some reason?), only in combinatorics and it doesn't seem to be related (?!) to what I'm looking for.
Sorry for the long post ☺.
|
Starting with $$z = a + \frac{1}{2} w (z^2 -1) \tag{1}$$
and applying Lagrange's expansion, we have
$$z = a + \sum_{n=1}^{\infty} \frac{w^n}{2^n n!} \frac{d^{n-1}}{da^{n-1}} (a^2-1)^n \tag{2}$$
On the other hand, $(1)$ is also a quadratic equation which we can solve for $z$, with result
$$z = \frac{1 \pm \sqrt{1-2aw+w^2}}{w}$$
We choose the negative sign in order that $\lim_{w \to 0} z = a$, so
$$z = \frac{1 - \sqrt{1-2aw+w^2}}{w}$$
Differentiating with respect to $a$,
$$\frac{dz}{da} = \frac{1}{\sqrt{1-2aw+w^2}} \tag{3}$$
Also differentiating $(2)$, we have
$$\frac{dz}{da} = 1 + \sum_{n=1}^{\infty} \frac{w^n}{2^n n!} \frac{d^{n}}{da^{n}} (a^2-1)^n \tag{4}$$
Equating $(3)$ and $(4)$,
$$\frac{1}{\sqrt{1-2aw+w^2}} = 1 + \sum_{n=1}^{\infty} \frac{w^n}{2^n n!} \frac{d^{n}}{da^{n}} (a^2-1)^n$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Probability a random spherical triangle has area $> \pi$ From Michigan State University's Herzog contest:
Problem 6, 1981
Three points are taken at random on a unit sphere. What is the probability that the area of the spherical triangle exceeds the area of a great circle?
I assume we always take the unique proper spherical triangle, with sides and angles not greater than $\pi$. I assume "the area of a great circle" is the plane area $\pi$, since a proper spherical triangle never has area larger than the spherical area $2 \pi$ on each side of a great circle. I assume the random distribution is uniform, such that the probability each point is within a given measurable subset of the sphere is proportional to the spherical area of that subset.
First, given triangle vertices $A,B,C$ on the unit sphere, we can rotate and/or mirror the coordinate system so that $C = (0,0,1)$, $A$ is on the $x \geq 0, y=0$ half-plane, and $B$ is on the $y \geq 0$ hemisphere. Label the side lengths opposite $A$, $B$, and $C$ as $a$, $b$, and $c$ respectively. Label the (dihedral) angles at $A$, $B$, and $C$ as $\alpha$, $\beta$, and $\gamma$ respectively. Ignore a degenerate triangle case with any of these variables exactly equal to $0$ or $\pi$; this happens with probability zero. Then all six variables are in the interval $\{a,b,c,\alpha,\beta,\gamma\} \subset (0,\pi)$. We can parameterize the probability distribution as:
$$ \begin{align*}
P(a < a_0) &= \frac{1-\cos a_0}{2} \\
P(b < b_0) &= \frac{1-\cos b_0}{2} \\
P(\gamma < \gamma_0) &= \frac{\gamma_0}{\pi}
\end{align*}$$
Variables $a$, $b$, and $\gamma$ are independent, so the overall probability density function is
$$ dP = \frac{1}{4\pi} \sin a \sin b\, da\, db\, d\gamma $$
The spherical area $\sigma$ of the triangle is
$$ \sigma = \alpha + \beta + \gamma - \pi $$
The outcome $\sigma > \pi$ is only possible if $\gamma > \frac{\pi}{2}$, since the spherical triangle covers a subset of the region between the planes containing $\{O,A,C\}$ and $\{O,A,B\}$ with area, $2 \gamma$. Similarly $\sigma > \pi$ implies $\alpha > \frac{\pi}{2}$ and $\beta > \frac{\pi}{2}$.
We also have Napier's analogy
$$ \tan \frac{\alpha+\beta}{2} = \frac{\cos \frac{a-b}{2}}{\cos \frac{a+b}{2}} \cot \frac{\gamma}{2} $$
Since $\{a,b\} \subset (0, \pi)$, $|a-b|<\pi$ and $\cos \frac{a-b}{2} > 0$. If $\sigma > \pi$ then $\{\alpha, \beta, \gamma\} \subset (\frac{\pi}{2}, \pi)$ which implies $\tan \frac{\alpha+\beta}{2} < 0$ and $\cot \frac{\gamma}{2} > 0$. Therefore $\cos \frac{a+b}{2} < 0$, and $\frac{a+b}{2} > \frac{\pi}{2}$.
So given $\{\alpha, \beta, \gamma, \frac{a+b}{2}\} \subset (\frac{\pi}{2}, \pi)$, all these inequalities are equivalent:
$$ \begin{align*}
\sigma &> \pi \\
\alpha + \beta + \gamma &> 2\pi \\
\pi - \frac{\gamma}{2} &< \frac{\alpha+\beta}{2} \\
\tan \left(\pi - \frac{\gamma}{2}\right) &< \tan \frac{\alpha+\beta}{2} \\
-\tan \frac{\gamma}{2} &< \frac{\cos \frac{a-b}{2}}{\cos \frac{a+b}{2}} \cot \frac{\gamma}{2} \\
\tan^2 \frac{\gamma}{2} &> - \cos \frac{a-b}{2}\, \sec \frac{a+b}{2} \\
\gamma &> 2\tan^{-1} \sqrt{- \cos \frac{a-b}{2}\, \sec \frac{a+b}{2}} = \gamma_L(a,b)
\end{align*}$$
where the function $\gamma_L(a,b)$ is defined where $0 \leq a \leq \pi$, $0 \leq b \leq \pi$, and $a+b>\pi$.
The probability in question is
\begin{align}P(\sigma > \pi) &= \int_0^\pi \int_{\pi-a}^\pi \int_{\gamma_L(a,b)}^\pi \frac{1}{4\pi} \sin a \sin b\, d\gamma\, db\, da\\&= \frac{1}{4\pi} \int_0^\pi \int_{\pi-a}^\pi [\pi - \gamma_L(a,b)] \sin a \sin b\, db\, da\end{align}
I've found we can also write
$$ \gamma_L(a,b) = \pi - \cos^{-1}\left(\cot \frac{a}{2} \cot \frac{b}{2}\right) $$
Either way, integration by parts can get rid of the inverse trig function in the integrand, leaving a square root, but I haven't been able to make much progress beyond that in solving the definite integral.
Wolfram Alpha gives the numeric answer as apparently $\frac{1}{6}$ (1) (2). But it seems it can't solve it symbolically either.
Given this was in a limited-time competition, maybe there's some other simpler way to go about this. Some sort of symmetry grouping related points or triangles? How can we prove the exact probability?
|
The substitution $u=\cot(a/2)\cot(b/2)$ yields $$I(a)=\int_{\pi-a}^\pi\arccos\left(\cot\frac a2\cot\frac b2\right)\sin b\,db=\int_0^1\frac{4u\tan^2(a/2)\arccos u}{(1+u^2\tan^2(a/2))^2}\,du$$ so that $$P(\sigma>\pi)=\frac1{4\pi}\int_0^\pi I(a)\sin a\,da=\frac4\pi\int_0^{\pi/2}J(a)\sin^2a\tan a\,da$$ where $\displaystyle J(a)=\int_0^1\frac{u\arccos u}{(1+u^2\tan^2a)^2}\,du$. Integration by parts yields \begin{align}J(a)&=-\frac{\arccos u}{2(1+u^2\tan^2a)\tan^2a}\bigg\vert_0^1-\frac1{2\tan^2a}\int_0^1\frac{du}{(1+u^2\tan^2a)\sqrt{1-u^2}}\\&=\frac\pi{4\tan^2a}-\frac1{2\tan^2a}\cdot\frac\pi{2\sqrt{\tan^2a+1}}\\&=\frac{\pi(1-\cos a)}{4\tan^2a}\end{align} where the last integral is evaluated by taking $t=u/\sqrt{1-u^2}$. Therefore, \begin{align}P(\sigma>\pi)=\int_0^{\pi/2}(1-\cos a)\sin a\cos a\,da=\frac16.\end{align} Not sure if there is a quicker geometrical answer.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the quadratic equation with real coefficients and solutions $x_1,x_2$ if you know that $\Delta =b^2-4ac=-36$ and $x_1+x_2=6m$ What I've done so far is:
We know that $\Delta \lt 0$, which means that $x_1$ and $x_2$ are two conjugate complex numbers with the form: $x_1=r+n\cdot i \space\text{ and }\space x_2=r-n\cdot i.$
We now need to find the sum $S=x_1+x_2$ and the product $P=x_1\cdot x_2$ to form an equation $x^2-Sx+P=0$ which has the solutions $x_1$ and $x_2$.
\begin{align*}
sum = x_1+x_2&=6m \tag{1}\\
\iff(r+n\cdot i)+(r-n\cdot i)=2r&=6m \tag{2}\\
\implies r&=3m \tag{3}
\end{align*}
The $product = x_1\cdot x_2=(r+n\cdot i)(r-n\cdot i)=r^2+n^2 = (\cdots).\quad $
From here I don't know what relation to find between $\Delta$ and $x_1\cdot x_2.$
I tried: $x_2=\dfrac{-b\pm\sqrt{-36}}{2a}
=\dfrac{-b}{2a}\pm\dfrac{6i}{2a}
=3m\pm\dfrac{3i}{a}\space
\text{ so }\space x_1\cdot x_2
=\dfrac{9m^2+9}{a^2} (\cdots)$
|
\begin{align*}
b^2-4ac=-36\implies 4ac\ge36\implies ac&\ge9\tag{1}\\
\\
x_1+x_2=
\bigg(\dfrac{-b+\sqrt{-36}}{2a}\bigg)+
\bigg(\dfrac{-b-\sqrt{-36}}{2a}\bigg)=
\dfrac{-b}{a}&=6m\tag{2}\\
\\
x_1\cdot x_2=
\bigg(\dfrac{-b+\sqrt{-36}}{2a}\bigg)
\bigg(\dfrac{-b-\sqrt{-36}}{2a}\bigg)
= \dfrac{b^2+36}{4 a^2}& \tag{3}
\end{align*}
From $(1)$ we could say $\space x^2+36=0\space$ but
$(2)\longrightarrow\dfrac{-0}{a}=6m\implies m=0.$
The first "candidates" for $\space (a, b)\space$ that will yield $\space -36\space$ are $a=1, b=2$
\begin{align*}
\dfrac{-b}{a}=\dfrac{-2}{1}
\implies a=1, b=2, 4ac=40\\
\implies 2^2-4(1)(10)=-36\\
\implies c=\dfrac{2^2+36}{4 (1^2)}=10\\
\\
\implies
x^2+2x+10=0
\\ \implies
x_1= -1+3i\quad x_2=-2-3i\\ \\
x_1\cdot x_2=( -1+3i)(-1-3i)=10 \\
\dfrac{-2}{1}=6m\implies m=\dfrac{-1}{3}
\end{align*}
These last equations are in no particular order but I think we have the quadratic equation we seek, i.e.
$\quad x^2+2x+10=0$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Number of solution of ${\left( {\sin x - 1} \right)^3} + {\left( {\cos x - 1} \right)^3} + {\sin ^3}x = {\left( {2\sin x + \cos x - 2} \right)^3}$ Number of solution of the equation ${\left( {\sin x - 1} \right)^3} + {\left( {\cos x - 1} \right)^3} + {\sin ^3}x = {\left( {2\sin x + \cos x - 2} \right)^3}$ in the interval $[0,2\pi]$ is equal to_____
My approach is as follow
$a = \sin x - 1;b = \cos x - 1;c = \sin x$
${a^3} + {b^3} + {c^3} = {\left( {a + b + c} \right)^3}$
$ \Rightarrow {a^3} + {b^3} + {c^3} = {\left( {a + b + c} \right)^3}$
${\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 6abc + 3ab\left( {a + b} \right) + 3bc\left( {b + c} \right) + 3ac\left( {a + c} \right)$
${a^3} + {b^3} + {c^3} = {\left( {a + b + c} \right)^3} \Rightarrow 6abc + 3ab\left( {a + b} \right) + 3bc\left( {b + c} \right) + 3ac\left( {a + c} \right) = 0$
$ \Rightarrow 2abc + ab\left( {a + b} \right) + bc\left( {b + c} \right) + ac\left( {a + c} \right) = 0$
How do we approach from here
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Having $a = \sin x - 1$ and $c = \sin x$ is redundant, it hides potential cancellation and simplification, and leaves you trying to factor a polynomial in three variables instead of just two. Instead, everywhere you have a $c$, replace it with $a+1$.
\begin{align*}
a^3 + b^3 + (a+1)^3 &= (a+b+a+1)^3 \\
a^3 + b^3 + (a+1)^3 &= (2a+b+1)^3 \\
1 + 3a + 3a^2 + 2a^3 + b^3 &= 8 a^3+12 a^2 b+12 a^2+6 a b^2+12 a b+6 a+b^3+3 b^2+3 b+1 \end{align*}\begin{align*}
-6 a^3-12 a^2 b-9 a^2-6 a b^2-12 a b-3 a-3 b^2-3 b &= 0 \\
-3(2 a^3+4 a^2 b+3 a^2+2 a b^2+4 a b+a+b^2+b) &= 0 \\
-3(2 a^3 + 2 a^2 b + 2 a^2 b + 2 a b^2 + 3 a^2 + 3 a b + a b + b^2 + a+b) &= 0 \\
-3(a+b)(2 a^2 + 2 a b + 3 a + b + 1) &= 0 \\
-3(a+b)(2 a^2 + 2 a b + 2 a + a + b + 1) &= 0 \\
-3(a+b)(a+b+1)(2 a + 1) &= 0
\end{align*}
Note that the factor $a+b+1 = \sin x + \cos x - 1$ leads to three solutions in $[0,2\pi]$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the result of $\sqrt{-\frac{1}{4}+\sqrt{-\frac{1}{4}+\sqrt{-\frac{1}{4}+\sqrt{-\frac{1}{4}+...}}}}$ Given the golden ratio is equal to $\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}$ you get $f(x) = \sqrt{1+f(x)}$.
Solving for $f(x)$:
$f(x)^2 = x + f(x)$
$f(x)^2 - f(x) = x$
$f(x)^2 - f(x) + \frac{1}{4} = x + \frac{1}{4}$
$(f(x) - \frac{1}{2})^2 = x + \frac{1}{4}$
$f(x) - \frac{1}{2} = \sqrt{x+\frac{1}{4}}$
$f(x) = \frac{1}{2} + \sqrt{x+\frac{1}{4}}$
Checking values for $f(x)$:
x
f(x)
1
$\phi$
2
2
6
3
12
4
...
...
x
$\frac{1}{2}+\sqrt{x+\frac{1}{4}}$
I've checked all these between using the equation and the nested square roots and they all check out, but I have a hard time believing the nested radical $\sqrt{-\frac{1}{4}+\sqrt{-\frac{1}{4}+\sqrt{-\frac{1}{4}+\sqrt{-\frac{1}{4}+...}}}}$ would equal $\frac{1}{2}$
|
The trick is to see if there exists a sequence that would work. The trouble is that many times, recursive sequences can be sensitive to their starting seeds (exhibit chaotic behavior), thus these sequences can not be said to have well defined limits, only regularized values useful for specific applications. In this case though, there is an insensitivity to starting seed provided $x_0$ that $x_0\geq \frac{1}{2}$. Consider the sequence
$$x_{n+1} = \sqrt{x_n - \frac{1}{4}} \hspace{20 pt} x_0 \geq \frac{1}{2}$$
Heuristically, one can see from this graph of the "discrete derivative" $f(x_n)=x_{n+1}-x_n$ that $x=\frac{1}{2}$ is an unstable fixed point:
Points less than $\frac{1}{2}$ move away from $\frac{1}{2}$ whereas points greater than $\frac{1}{2}$ move closer toward it (that is the nature of this discrete derivative, it is nonpositive everywhere). All we have to prove now is that in iterating this dynamical system, our sequence never jumps to the left of our limit point. In this case,
$$x_n \geq \frac{1}{2} \implies x_n-\frac{1}{4} \geq \frac{1}{4} \implies x_{n+1} = \sqrt{x-\frac{1}{4}} \geq \frac{1}{2}$$
So we are in the clear and we have proven sufficient insensitivity to starting seed to say that this one-sided limit of the function you found does exist. Interestingly enough though, since the discrete derivative has a zero of multiplicity greater than $1$, the convergence of the sequence will be very slow ($\sim \frac{1}{n}$) as opposed to exponential convergence for a root with multiplicity $1$
|
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|
Find the sum $ \sum\limits_{a,b=0}^ \infty \begin{vmatrix} x^{a+b} & y^{a+b} & z^{a+b}\\ x^b & y^b & z^b \\ 1 & 1 & 1 \end{vmatrix} u^a v^b. $ Find a close determinant expression for the sum $$ \sum_{a,b=0}^ \infty \begin{vmatrix}
x^{a+b} & y^{a+b} & z^{a+b}\\
x^b & y^b & z^b \\
1 & 1 & 1
\end{vmatrix} u^a v^b.
$$
Is is not too hard expand the determinant and get that
\begin{gather*}
\sum_{a,b=0}^ \infty \begin{vmatrix}
x^{a+b} & y^{a+b} & z^{a+b}\\
x^b & y^b & z^b \\
1 & 1 & 1
\end{vmatrix} u^a v^b=\\=\sum_{a,b=0}^ \infty ({x}^{a+b}{y}^{b}+{y}^{a+b}{z}^{b}-{x}^{a+b}{z}^{b}-{
x}^{b}{y}^{a+b}+{x}^{b}{z}^{a+b}-{z}^{a+b}{y}^{b}) u^a v^b=
\\=
{\frac {1}{
\left( xu-1 \right) \left( vxy-1 \right) }}+{\frac {1}{ \left( yu-1
\right) \left( vyz-1 \right) }}-{\frac {1}{ \left( xu-1 \right)
\left( vxz-1 \right) }}-\\-{\frac {1}{ \left( yu-1 \right) \left( vxy-1
\right) }}+{\frac {1}{ \left( zu-1 \right) \left( vxz-1 \right) }}-{\frac {1}{ \left( zu-1 \right) \left( vyz-1 \right) }}=\\=
{\frac {uv \left( y-z \right) \left( x-z \right) \left( x-y
\right) \left( uvxyz-1\right) }{ \left( zu-1 \right) \left( vyz-1
\right) \left( xu-1 \right) \left( vxy-1 \right) \left( yu-1
\right) \left( vxz-1 \right) }}
\end{gather*}
Question. Is it possible to go back to determinant and express the result
$$
{\frac {uv \left( y-z \right) \left( x-z \right) \left( x-y
\right) \left( uvxyz-1\right) }{ \left( zu-1 \right) \left( vyz-1
\right) \left( xu-1 \right) \left( vxy-1 \right) \left( yu-1
\right) \left( vxz-1 \right) }}
$$
again as determinant ( or sum ( product) of determinants)?
Ant help?
|
$$
\sum_{a,b=0}^ \infty \begin{vmatrix}
x^{a+b} & y^{a+b} & z^{a+b}\\
x^b & y^b & z^b \\
1 & 1 & 1
\end{vmatrix} u^a v^b = \\
\sum_{a,b=0}^ \infty \begin{vmatrix}
u^ax^{a+b} & u^ay^{a+b} & u^az^{a+b}\\
x^b & y^b & z^b \\
1 & 1 & 1
\end{vmatrix} v^b =\\
\sum_{b=0}^ \infty \begin{vmatrix}
\frac{1}{1-xu}x^{b} & \frac{1}{1-yu}y^{b} & \frac{1}{1-zu}z^{b}\\
x^b & y^b & z^b \\
1 & 1 & 1
\end{vmatrix} v^b =\\
\sum_{b=0}^ \infty \begin{vmatrix}
\frac{1}{1-xu} & \frac{1}{1-yu} & \frac{1}{1-zu}\\
1 & 1 & 1 \\
x^{-b} & y^{-b} & z^{-b}
\end{vmatrix} (vxyz)^b =\\
\sum_{b=0}^ \infty \begin{vmatrix}
\frac{1}{1-xu} & \frac{1}{1-yu} & \frac{1}{1-zu}\\
1 & 1 & 1 \\
(yzv)^{b} & (xzv)^{b} & (xyv)^{b}
\end{vmatrix} =\\
\begin{vmatrix}
\frac{1}{1-xu} & \frac{1}{1-yu} & \frac{1}{1-zu}\\
1 & 1 & 1 \\
\frac{1}{1-yzv}& \frac{1}{1-xzv} & \frac{1}{1-xyv}
\end{vmatrix}
$$
|
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|
Strange inequality in an NT problem If $a\neq b$ are positive integers and $a^2+ab+b^2 | ab(a+b)$ then show that $|a-b|>\sqrt[3]{ab}$.
WLOG $a>b$. $a^2+ab+b^2 | ab(a+b)-a(a^2+ab+b^2)=-a^3$, so we have $a^2+ab+b^2$ divides $a^3$ as well as $b^3$. From here I tried lots of things, but I don't think they are worth of mentioning. If you have any idea ,do not hesitate to suggest.
|
Notice that $$a^2+ab+b^2\mid ab(a+b)\iff a^2+ab+b^2\mid a^3\iff a^2+ab+b^2\mid b^3$$ $$\iff a^2+ab+b^2\mid \gcd(a,b)^3$$
Thus, let $d:=\gcd(a,b)$, and let $x:=\frac{a}d, y:=\frac{b}d$. It follows that $$a^2+ab+b^2\mid \gcd(a,b)^3\iff x^2+xy+y^2\mid d\implies d>xy$$
You are left to show that $\lvert a-b\rvert>\sqrt[3]{ab}\iff \lvert dx-dy\rvert^3=\lvert a-b\rvert^3> ab = d^2\cdot xy\iff d\lvert x-y\rvert > xy$. But this is trivial, since $d>xy$ and $\lvert x-y\rvert\geqslant 1.$
|
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|
Power of Stochastic Matrix Given the stochastic matrix
$$Q = \begin{pmatrix}0&2/3&1/3\\1/3&0&2/3\\2/3&1/3&0\\\end{pmatrix}\in \mathbb{R}^{3\times3}$$
I wish to compute $Q_{1,1}^n$ (the entry in the first row and first column) for $n\in \mathbb{N}$.
I did so by diagonalizing $Q=PDP^{-1}\implies Q^n=PD^nP^{-1}$. The computation of the (complex) eigenvectors however takes time.
Is there a more elegant way to find $Q_{1,1}^n$?
|
Following the comment by @kimchi lover, since the matrix is stochastic and circulant, the eigenvalues are $1, \lambda = \frac{2}{3} e^{2 \pi i/3} + \frac{1}{3} e^{4 \pi i/3} = \frac{1}{2}\left(-1 + \frac{i}{\sqrt{3}} \right)$, and $\overline{\lambda} = \frac{1}{2}\left(-1 - \frac{i}{\sqrt{3}} \right)$. Thus $D^n = \text{diag} \, (1, \lambda^n, \bar{\lambda}^n)$.
The matrix $P$ of eigenvectors is a Fourier matrix, with entries $P_{k\ell} = \frac{1}{3} e^{2(k-1)(\ell-1) \pi i/3}$, and $P^{-1} = P$.
Note that $\lambda^6 = - \frac{1}{27}$, thus $\lambda^n = \left( \frac{-1}{27} \right)^k \lambda^\ell$ if $n = 6k + \ell$. So you only have to find $\lambda^2, \dots, \lambda^5$.
Also note that $P_{1\ell} = P_{\ell 1} = \frac{1}{\sqrt{3}}$ for all $\ell$.
Then
$$
Q^n_{1,1} = P_{11}^2 + \lambda^n P_{12}^2 + \bar{\lambda}^n P_{13}^2
= \frac{1}{3} \left(1 + 2 \, \Re \, \lambda^n \right)
$$
The answer can then be worked out explicitly. Since $|\lambda| < 1$, this also shows explicitly that $Q_{1,1} \to \frac{1}{3}$ as $n \to \infty$.
|
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"url": "https://math.stackexchange.com/questions/4330115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Proving $1/(a+b) + 1/(b+c) + 1/(c+a) > 3/(a+b+c)$ for positive $a, b, c\,$? I have to prove that:
$$ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} > \frac{3}{a+b+c},$$
where $a, b , c$ are positive real numbers.
I am thinking about using arithmetical and geometrical averages:
$$A_{3} = \frac{a_{1}+a_{2}+a_{3}}{3},$$
$$G_{3} = \sqrt[3]{a_{1}×a_{2}×a_{3}},$$
$$A_{3}\ge G_{3}.$$
However, I am not sure how to do this.
I have tried to substract one side from the other:
$$ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} - \frac{3}{a+b+c} > 0.$$
I have also tried to reverse both sides of inequality.
I would appreciate if you could just give me a hint.
|
By AM-HM inequality
$$\frac13 \left(\frac1{a+b}+\frac1{b+c}+\frac1{c+a}\right)\ge
\frac{3}{(a+b)+(b+c)+(c+a)}.$$
Hence $\displaystyle \frac1{a+b}+\frac1{b+c}+\frac1{c+a}\ge \frac{9}2\frac1{a+b+c}$. Because $\dfrac92>3$ the problem is solved.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating an improper integral using the Residue Theorem I have:
\begin{equation}
\int_{-\infty}^{\infty}\frac{x^2}{(x^2+1)(x^2+9)}dx
\end{equation}
and I want to solve it using a complex closed contour on C. I do the following:
\begin{equation}
\int_{-\infty}^{\infty}\frac{z^2}{(z^2+1)(z^2+9)}dz
\end{equation}
which clearly has 4 poles, $\pm i,\pm3i $. The greatest radius of the clontour is thus $\rho=3$, therefore I get the following integral form:
\begin{equation}
\int_{\gamma_3}\frac{z^2}{(z^2+1)(z^2+9)}dz=2\pi i Res(f;i,+3i)
\end{equation}
which gives for i:
\begin{equation}
Res(i)=\lim_{z\longrightarrow i}(z-i)\frac{z^2}{(z-i)(z+i)(z+3i)(z-3i)}dz=i/16
\end{equation}
and for 3i:
\begin{equation}
Res(3i)=\lim_{z\longrightarrow 3i}(z-3i)\frac{z^2}{(z-i)(z+i)(z+3i)(z-3i)}dz=-i/48
\end{equation}
Plugging into the formula given above, I get:
\begin{equation}
\int_{\gamma_3}\frac{z^2}{(z^2+1)(z^2+9)}dz=2\pi i (i/16-i/48)=-\frac{\pi}{12}
\end{equation}
But this is not correct, as the integral is negative. Where is the error?
Thanks!
|
Also, using Calculus
$$
\frac{x^2}{(x^2+1)(x^2+9)}=\frac{1}{8}\left(\frac{9}{x^2+9}-\frac{1}{x^2+1}\right)
$$
and
$$
\int_{-\infty}^\infty\frac{dx}{x^2+1}=\tan^{-1}(x)\Big|_{x=-\infty}^{x=\infty}=\pi, \qquad
\int_{-\infty}^\infty\frac{dx}{x^2+9}=\frac{1}{3}\tan^{-1}(x/3)\Big|_{x=-\infty}^{x=\infty}=\frac{\pi}{3},
$$
and hence
$$
\int_{-\infty}^\infty\frac{x^2}{(x^2+1)(x^2+9)}=\frac{\pi}{4}.
$$
|
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|
Given two distinct intersecting circles, length of that chord of larger circle which is bisected by the smaller circle is equal to? Two circles whose centres lie on the x axis, whose radii are $\sqrt2cm$ and $1cm$ and whose centres are 2 cm apart intersect at a point A.The chord AC of the larger circle cuts the smaller circle at a point B and is bisected by that point. What is the length of chord AC?
.
My attempt :
.
.
As shown in the diagram above I started by assuming the centre of smaller circle S1 (Of radius 1) to be origin and the centre of the larger circle S2 (Of radius $\sqrt2$ ) to be $(2,0)$ as the centres are separated by 2 units.
I then solved :
S1 : $ x^2 + y^2 = 1 $ ; and
S2 : $ (x-2)^2 + y^2 = 2 $ to obtain $A(\frac{3}4 , \frac{\sqrt7}4)$
I noted the following equations:
*
*Since B is given to be mid point of chord AC:
XB = $\frac{X_A + X_C}2$ ; $Y_B = \frac{Y_A +Y_C}2$
*Since C lies on $S_2$ :
$ (X_C-2)^2 + Y_C^2 = 2 $
*The distance of B from origin is 1 unit :
$(X_B-0)^2 + (Y_B-0)^2 = 1$
$(\frac{X_A + X_C}2 - 0)^2$ +$(\frac{Y_A +Y_C}2 - 0)^2 = 1$
$(\frac{\frac{3}4 + X_C}2 - 0)^2$ +$(\frac{\frac{\sqrt7}4 +Y_C}2 - 0)^2 = 1$
This equation and equation generated in point 2 together are two equations in two variables and i should be able to solve them to get the co-ordinate of C. This however is proving to be cumbersome.
.
Is there a better way to avoid this approach.
|
Yes there is a different path using trigonometry:
Let, with your coordinate system:
$$B=(\cos \alpha, \sin \alpha) \ \text{and} \ C=(2+\sqrt{2}\cos \beta, \sqrt{2}\sin \beta)$$
We just have to express that B is the midpoint of [AC] by writing that $$2B=A+C \ \iff \ \begin{cases}2\cos \alpha&=& \dfrac34+2+\sqrt{2}\cos\beta & (1a)\\2 \sin \alpha&=&\dfrac{\sqrt{7}}{4}+\sqrt{2}\sin \beta & (1b)\end{cases}$$
This gives you 2 equations in the 2 unknowns $\alpha$ and $\beta$.
Squaring and adding (1a) and (1b) gives :
$$4=(\dfrac{11}{4}+\sqrt{2}\cos\beta)^2+(\dfrac{\sqrt{7}}{4}+\sqrt{2}\sin \beta)^2$$
Expanding and using once more $\cos^2a +\sin^2 a=1$:
$$4=(\dfrac{11}{4})^2+2(\dfrac{11}{4})\sqrt{2}\cos\beta+(\dfrac{7}{16})^2+2\dfrac{\sqrt{7}}{4}\sqrt{2}\sin \beta+2$$
which is the very classical equation $A \cos a + B\sin a=C$
|
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"timestamp": "2023-03-29T00:00:00",
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|
relation between roots and coefficient in a cubic polynomial If $\alpha,\beta,\gamma$ are roots of the cubic equation
$$2x^{3}+3x^2-x-1=0$$ then I want to find the equation whose roots are $\frac{\alpha}{\beta+\gamma}, \frac{\beta}{\gamma+\alpha}, \frac{\gamma}{\alpha+\beta}$.
I have $\alpha+\beta+\gamma$= - $\frac{3}{2}$.
Therefore roots become $\frac{\alpha}{-\frac{3}{2} -\alpha}$,$\frac{\beta}{-\frac{3}{2} -\beta}$, $\frac{\gamma}{-\frac{3}{2} -\gamma}$.
To find the equation we have to find their sums and products, but it looks like complicated.
Is their any easy trick ?
|
While this is not a general method for zeroes of a cubic polynomial, this particular polynomial is amenable to familiar methods. We can apply the Rational Zeroes Theorem to find that one zero is $ \ \alpha \ = \ -\frac12 \ \ , $ so we may factor $ \ 2x^3 + 3x^2 - x - 1 \ = \ \left(x + \frac12 \right)·2·(x^2 + x - 1) \ = \ 0 \ \ . $ This polynomial also has the helpful property that it has symmetry about $ \ x \ = \ -\frac12 \ \ : \ 2·\left(u - \frac12 \right)^3 + 3·\left(u - \frac12 \right)^2 - \left(u - \frac12 \right) - 1 $ $ = \ 2u·\left(u^2 - \frac54 \right) \ \ . $ The other zeroes of the original polynomial are then $ \ \beta \ , \ \gamma \ = \ \alpha \ \pm \ \Delta \ = \ -\frac12 \pm \frac{\sqrt5}{2} \ \ . \ $ (We will have need of $ \ \Delta^2 \ = \ \frac54 \ $ shortly. These two zeroes are in fact $ \ \frac{1}{\phi} \ $ and $ \ -\phi \ \ , $ with $ \ \phi \ $ being the "Golden Ratio", but we won't be making use of that in this discussion.)
We seek a cubic polynomial with the transformed zeroes $$ \alpha' \ = \ \frac{\alpha}{\beta \ + \ \gamma} \ = \ \frac{-1/2}{-1} \ = \ \frac12 \ \ \ , \ \ \ \beta' \ = \ \frac{\beta}{\alpha \ + \ \gamma} \ = \ \frac{\alpha \ + \ \Delta}{2\alpha \ - \ \Delta} \ = \ \frac{1 \ + \ 2\Delta}{2 \ - \ 2\Delta} \ \ , $$
$$\gamma' \ = \ \frac{\gamma}{\alpha \ + \ \beta} \ = \ \frac{\alpha \ - \ \Delta}{2\alpha \ + \ \Delta} \ = \ \frac{1 \ - \ 2\Delta}{2 \ + \ 2\Delta} \ \ . $$
From these, we compute the coefficients of a new monic polynomial from
$$ c' \ \ = \ \ -(\alpha' · \beta' · \gamma') \ \ = \ \ -\frac{(1 \ + \ 2\Delta)·(1 \ - \ 2\Delta)}{2 ·(2 \ - \ 2\Delta)·(2 \ + \ 2\Delta)} \ \ = \ \ \frac{4\Delta^2 \ - \ 1}{8 · (1 \ - \ \Delta^2)} $$ $$ = \ \ \frac{5 \ - \ 1}{-2} \ \ = \ \ -2 \ \ ; $$
$$ a' \ \ = \ \ -(\alpha' \ + \ \beta' \ + \ \gamma') \ \ = \ \ - \left[ \ \frac12 \ + \ \frac{1 \ + \ 2\Delta}{2 \ - \ 2\Delta} \ + \ \frac{1 \ - \ 2\Delta}{2 \ + \ 2\Delta} \ \right] \ \ = \ \ \frac32· \left(\frac{\Delta^2 \ + \ 1}{ \Delta^2 \ - \ 1} \right) $$ $$ = \ \ \frac32· \left(\frac{5 \ + \ 4}{ 5 \ - \ 4} \right) \ \ = \ \ \frac{27}{2} \ \ ; $$
$$ \frac{b'}{c'} \ \ = \ \ \frac{\alpha' · \beta' \ + \ \alpha' · \gamma' \ + \ \beta' · \gamma'}{- \ (\alpha' · \beta' · \gamma')} \ \ = \ \ - \left(\frac{1}{\alpha'} \ + \frac{1}{\beta'} \ + \ \frac{1}{\gamma'} \right) $$
$$ = \ \ - \left[ \ 2 \ + \ \frac{2 \ - \ 2\Delta}{1 \ + \ 2\Delta} \ + \ \frac{2 \ + \ 2\Delta}{1 \ - \ 2\Delta} \ \right] \ \ = \ \ \frac{6}{ 4\Delta^2 \ - \ 1} \ \ = \ \ \frac{6}{ 5 \ - \ 1} \ \ = \ \ \frac32 $$
$$ \Rightarrow \ \ b' \ \ = \ \ \frac32 \ · \ (-2) \ \ = \ \ -3 \ \ . $$
A cubic polynomial having the transformed zeroes which has integer coefficients is then $ \ 2x^3 \ + \ 27x^2 \ - \ 6x \ - \ 4 \ \ . $
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $\frac{1}{2^{2n+1}}\sum_{i=0}^n \left[\binom{n}{i} \cdot \sum_{j=i+1}^{n+1} \binom{n+1}{j} \right]= \frac{1}{2}$ I have a hard time showing that that
$$ \frac{1}{2^{2n+1}}\sum_{i=0}^n \left[\binom{n}{i} \cdot \sum_{j=i+1}^{n+1} \binom{n+1}{j} \right]= \frac{1}{2}$$
Namely, I try to show hat
$$\sum_{i=0}^n \left[\binom{n}{i} \cdot \sum_{j=i+1}^{n+1} \binom{n+1}{j}\right] = 2^{2n} $$
Any help would be appreciated. Thank you all.
Online demo: https://www.desmos.com/calculator/cjo7zggjlf
|
We have
$$
\begin{split}
S=\sum_{0\le i<j\le n+1}\binom{n}{i}\binom{n+1}{j}&=\sum_{0\le i\le n}\sum_{0\le j\le n+1}\binom{n}{i}\binom{n+1}{j}-\sum_{0\le j\le i\le n+1}\binom{n}{i}\binom{n+1}{j}\\
&=2^n\cdot 2^{n+1}-\sum_{0\le j\le i\le n+1}\binom{n}{n-i}\binom{n+1}{n-j+1}\\
&=2^{2n+1}-\sum_{0\le i'<j'\le n+1}\binom{n}{i'}\binom{n+1}{j'}\\
&=2^{2n+1}-S,
\end{split}
$$
where $i'=n-i$ and $j'=n-j+1$. Therefore, $2S=2^{2n+1}$, i.e. $S=2^{2n}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Number of zeros of $f(x)= \frac{1}{2} E\left[ \tanh \left( \frac{x+Z}{2} \right) \right]-\tanh(x)+\frac{x}{2}$ where $Z$ is standard normal Consider the following function:
\begin{align}
f(x)= \frac{1}{2} E\left[ \tanh \left( \frac{x+Z}{2} \right) \right]-\tanh(x)+\frac{x}{2},
\end{align}
where $Z$ is standard normal.
Question: How to show that this function has only three zeros? Note, that we are not interested in the locations just the number of zeros.
By using that $\tanh(x)$ is an odd function, it is not difficult to show that $f(0)=0$.
However, I am not sure how to show the existence of the other two zeros.
I know that two more zeros exist from the numerical simulation (see the attached figure).
Edit: The current answer shows that there are at least 3 zeros. Now we need to show that there can be no more than 3 zeros.
Edit 2 Idea for a proof. Consider only positive $x$. I think if we can show the following:
*
*$f(x)>0$ for all $x>x_1$,
*$f(x)$ is convex for $x \in (0,x_2)$, and
*$x_2>x_1$.
Then this will imply that the function is convex in the regime while it changes a sign. Therefore, it can only have at most one sign change.
|
Not a complete proof, but rather a sketch. First, observe that as $x\to +\infty$, than $\tanh (x) \to 1$, so $f(x)$ asymptotically asymptotically equivalent to $\frac{1}{2} E [1] - 1 + \frac{x}{2} = \frac{x-1}{2}$, so for some large enough positive value of $x$ one has $f(x) > 0$. After that, let's look at the graph of the function and notice that $f'(0)$ must be negative, which is not hard to prove: As
$$
\frac{1}{2} E\left[ \tanh \left( \frac{x+Z}{2} \right) \right] =
\frac{1}{2} \int_\mathbb{R} \tanh \left( \frac{x+z}{2} \right) \frac{1}{\sqrt{2\pi}} e^{-z^2/2}dz
$$
So
$$
f'(x) = \frac{1}{2} \int_\mathbb{R}
\frac{\partial}{\partial x}\left[
\tanh \left( \frac{x+z}{2} \right)\right] \frac{1}{\sqrt{2\pi}} e^{-z^2/2}dz +
\frac{d}{dx} \left[-\tanh (x) + \frac{x}{2}\right] = \\ =
\frac{1}{4} \int_\mathbb{R} \frac{1}{\cosh^2 \left( \frac{x+z}{2} \right)} \frac{1}{\sqrt{2\pi}} e^{-z^2/2}dz - \frac{1}{\cosh^2 (x)} + \frac{1}{2} \\
f'(0) = \frac{1}{4} \int_\mathbb{R} \frac{1}{\cosh^2 \left( \frac{z}{2} \right)} \frac{1}{\sqrt{2\pi}} e^{-z^2/2}dz - 1 + \frac{1}{2} \leq
\frac{1}{4} \int_\mathbb{R} \frac{1}{\sqrt{2\pi}} e^{-z^2/2}dz - \frac{1}{2} =
\frac{1}{4} - \frac{1}{2} = -\frac{1}{4}
$$
where the inequality is obtained from $\frac{1}{\cosh^2 \left( \frac{z}{2} \right)} \leq 1 \;\forall z \in \mathbb{R}$. So, it means that $f(x)$ is decreasing in some neighborhood of zero.
So, there exists some $x^+ > 0$ s.t. $f(x^*) > 0$ and some $x^- > 0$ s.t. $f(x^-) < f(0) = 0$. So, by intermediate value theorem (it's also an exercise to check the continuity of $f(x)$) there exists some $x^* \in [x^-, x^+]$ s.t. $f(x^*) = 0$. As $f(x)$ is an odd function, it means that $-x^*$ is also a root.
EDIT: as $\frac{1}{2} - \frac{1}{\cosh^2(x)} \leq f'(x) \leq \frac{3}{4} - \frac{1}{\cosh^2(x)}$ and $f'(0) = 0$, I suppose that with similar methods one can prove that $f$ also has a stationary point somewhere on positive axis, which can help in proving that there are no other roots.
|
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|
Solve system of 2 equations with 3 unknowns We are given a triangle $ABC$ with sides $a, b, c$ respectively and for which the following relationships hold:
$a^2+bc\sqrt 3 = b^2+c^2$,
$c^2+ba = a^2+b^2$
We want to prove that angle $B$ is right.
I am trying to express sides $b$ and $c$ in relation to $a$ and then prove that they satisfy the Pythagorean theorem.
By combining the two equations, I am getting:
$b = \frac {c\sqrt 3+1}{2}$
Then I am plugging this expression into the second equation, in order to eliminate $b$:
$4a^2-2a(c\sqrt 3+1)-c^2+2c \sqrt 3 +1=0$
and I must now solve for $a$ in relation to $c$ but I am getting a complex expression, which, by no means, satisfies Pythagorean.
I input the two initial equations in Wolfram and it gives as a solution (apart from the ones which are rejected):
$b=2a$ and $c=a\sqrt 3$ which clearly satisfy Pythagorean, because $b^2 = 4a^2 = c^2+a^2$.
Any ideas?
Thank you!!
|
Comparing the first equation with cosine rule:
$$a^2=b^2+c^2-2bc\cos A,$$
$$a^2=b^2+c^2-\sqrt3bc,$$
$$\implies\cos A=\frac{\sqrt3}2\implies \measuredangle A=30^\circ.$$
Now using the two given equations we get, $2b=\sqrt3c+a$.
Applying sin rule, $$2\sin B=\sqrt3\sin C+\sin A$$ $$2\sin B=\sqrt3\sin(150^\circ-B)+\sin30^\circ$$
Continue this to solve for $\measuredangle B$ and you will get the answer.
Alternate solution: As I've mentioned previously, it should be $2b=\sqrt3c+a$, not $2b=\sqrt3c+\color{red}1$.
So if you substitute the value of $b$ in the second relation in terms of $a$ and $c$, you will get $c^2=3a^2$. Hence, the problem is solved.
|
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|
Struggling to compute a power series for a complex value function I am struggling to compute the power series expansion of $$f(z) = \frac{1}{2z+5}$$ about $z=0$, where $f$ is a complex function. I tried comparing it to the geometric series as follows,$$ f(z) = \frac{1}{2z+5} = \frac{1}{1-\omega} = \sum_{n=0}^\infty\omega^n $$ which gives us $$ 2z+ 5 = 1 - \omega \implies \omega = -2z - 4 \implies f(z) = \sum_{n-0}^\infty(-2z + 4)^n = \sum_{n=0}^\infty(-2)^n(z+2)^n$$ which is clearly not even an expansion about $z=0$. This is the wrong answer as well according to my textbook which stated that the answer was $$\sum_{n=0}^\infty\frac{2^n}{5^{n+1}}(-1)^nz^n$$
could someone please explain where I have gone wrong.
|
$$\begin{array}{lcl}
\dfrac{1}{2 z + 5} & = & \dfrac{1}{5} \dfrac{1}{1 + \dfrac{2 z}{5}} \\[3mm]
& = & \displaystyle \dfrac{1}{5} \sum_{n = 0}^{+\infty} (-1)^n \left(\dfrac{2 z}{5}\right)^n \\[3mm]
& = & \displaystyle \sum_{n = 0}^{+\infty} \dfrac{2^n}{5^{n + 1}} (-1)^n z^n \\[3mm]
\end{array}$$
with the condition :
$$\left|\dfrac{2 z}{5}\right| < 1$$
|
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|
Find $a$ such that ${x_1}^2+{x_2}^2$ takes the minimal value where $x_1, x_2$ are solutions to $x^2-ax+(a-1)=0$ DO NOT USE CALCULUS My thinking:
Let $x_1 = \frac{a+\sqrt{a^2-4a+4}}{2}$ and $x_2 = \frac{a-\sqrt{a^2-4a+4}}{2}$
By the AGM (Arithmetic-Geometric Mean Inequality):
We have
$x_1\cdot x_2\le \left(\frac{x_1\cdot \:x_2}{2}\right)^2$
$=\:x_1\cdot \:x_2\le \:\frac{\left(x_1\cdot \:\:x_2\right)^2}{4}\:$
$=\:\:x_1\cdot \:\:x_2\le \:\:\frac{{x_1}^2+2x_1x_2+{x_2}^2}{4}$
$=4\left(x_1\cdot x_2\right)\:\le {x_1}^2+2x_1x_2+{x_2}^2$
$=4\left(x_1\cdot \:x_2\right)\:-2x_1x_2\le \:{x_1}^2+{x_2}^2$
Substituting in the values for $x_1$ and $x_2$ we get:
$4\left(\frac{a+\sqrt{a^2-4a+4}}{2}\cdot \:\frac{a-\sqrt{a^2-4a+4}}{2}\right)\:-2\left(\frac{a+\sqrt{a^2-4a+4}}{2}\right)\left(\frac{a-\sqrt{a^2-4a+4}}{2}\right)\le \:\left(\frac{a+\sqrt{a^2-4a+4}}{2}\right)^2+\left(\frac{a-\sqrt{a^2-4a+4}}{2}\right)^2$
$= 4\left(a-1\right)\:-\left(2a-2\right)\le \:a^2-2a+2$
$ = 0\le a^2-4a+4$
It seems as if I walked in circles through this process, can anyone help? Thanks in advance!
|
Another way to go... From $x^2 - a x + (a-1)$, we know that any root satisfies
$$ x^2 = ax - (a-1) \text{,} $$
so \begin{align*}
x_1^2 + x_2^2 &= (ax_1 - (a-1)) + (ax_2 - (a-1)) \\
&= a (x_1 + x_2) - 2a + 2 \text{.}
\end{align*}
From \begin{align*}
x^2 -ax +(a-1) &= (x-x_1)(x-x_2) \\
&= x^2 -(x_1 + x_2)x + x_1x_2 \text{,} \\
\end{align*}
we have $x_1 + x_2 = a$, so \begin{align*}
x_1^2 + x_2^2 &= a(a) - 2a + 2 \\
&= a^2 - 2a + 1 - 1 + 2 \\
&= (a-1)^2 + 1 \text{,}
\end{align*}
which is a nonnegative term plus $1$, so is minimized when the nonnegative term is zero, that is, when $a = 1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
The solution set of equation $\sin^{-1}{\sqrt{1-x^2}}\;+\cos^{-1}x=\cot^{-1}{\dfrac{\sqrt{1-x^2}}{x}}-\sin^{-1}x$
Find the solution set of equation $\sin^{-1}{\sqrt{1-x^2}}\;+\cos^{-1}x=\cot^{-1}{\dfrac{\sqrt{1-x^2}}{x}}-\sin^{-1}x$
My Approach:
$\sin^{-1}{\sqrt{1-x^2}}\;+\cos^{-1}x \; +\sin^{-1}x=\cot^{-1}{\dfrac{\sqrt{1-x^2}}{x}}$
$\implies$$\sin^{-1}{\sqrt{1-x^2}}\;+ \frac{\pi}{2}=\cot^{-1}{\dfrac{\sqrt{1-x^2}}{x}}$
$\implies$$\sin^{-1}{\sqrt{1-x^2}}\;+ \frac{\pi}{2}-\cot^{-1}{\dfrac{\sqrt{1-x^2}}{x}}=0$
$\implies$$\sin^{-1}{\sqrt{1-x^2}}\;+\tan^{-1}{\dfrac{\sqrt{1-x^2}}{x}}=0$
$\implies$$2\sin^{-1}{\sqrt{1-x^2}}=0$
I am obtaining two values of $x$ and those value are $x=1,{-1}.$
But given answer is $(-1,0)\cup \{1\}$.
What am I doing wrong?
|
The part where you went wrong is:
$\implies$$\sin^{-1}{\sqrt{1-x^2}}\;+\tan^{-1}{\dfrac{\sqrt{1-x^2}}{x}}=0$
$\implies$$2\sin^{-1}{\sqrt{1-x^2}}=0$
Because, $\sin^{-1}{\sqrt{1-x^2}}=$$\tan^{-1}{\dfrac{\sqrt{1-x^2}}{|x|}}$
.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4362461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to solve the Diophantine equation $x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0$? How can one solve the following Diophantine equation in $x, y \in \mathbb{Z}$?
$$x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0$$
|
$x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0\iff(x^2+x)^2-4(x^2+x)=(y+1)^2+5$
Putting $X=x^2+x$ we get $(X-2)^2=(y+1)^2+9$ whose only solutions are clearly
$$(X-2,y+1)=(\pm3,0),(\pm5,\pm4)$$ in both cases, since $X=x^2+x$, we have the equations for the unknown $x$
$$x^2+x-5=0\text{ and } x^2+x+1=0\\x^2+x-7=0\text{ and } x^2+x+3=0$$ these four equations have not integral roots, consequently the diophantine equation $x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0$ has no solution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4362637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Determine the greatest of the numbers $\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$ Determine the greatest of the numbers $$\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$$ The least common multiple of $2,3,4,5$ and $6$ is $LCM(2,3,4,5,6)=60$, so $$\sqrt2=\sqrt[60]{2^{30}}\\\sqrt[3]3=\sqrt[60]{3^{20}}\\\sqrt[4]4=\sqrt[60]{4^{15}}=\sqrt[60]{2^{30}}\\\sqrt[5]{5}=\sqrt[60]{5^{12}}\\\sqrt[6]{6}=\sqrt[60]{6^{10}}=\sqrt[60]{2^{10}\cdot3^{10}}$$ Now how do we compare $2^{30},3^{20},4^{15},5^{12}$ and $6^{10}$? I can't come up with another approach.
|
It is also possible to work with logarithms. We may write for each of these numbers
$$ \log_2 \sqrt2 \ \ = \ \ \log_2 2^{1/2} \ \ = \ \ \frac12 \ \ , \ \ \log_3 \sqrt[3]3 \ \ = \ \ \frac13 \ \ , \ldots , \ \ \log_6 \sqrt[6]6 \ \ = \ \ \frac{1}{6} \ \ . $$
It will be more helpful to our purpose to express these in terms of a common logarithmic base, say, $ \ \log_2 x \ \ : $
$$ \log_2 \sqrt2 \ \ = \ \ \frac12 \ \ , \ \ \log_2 \sqrt[3]3 \ \ = \ \ \frac{\log_3 \sqrt[3]3}{\log_3 \ 2} \ \ = \ \ \frac{1}{3 \ \log_3 \ 2} \ \ , $$ $$ \log_2 \sqrt[4]4 \ \ = \ \ \frac{1}{4 \ \log_4 \ 2} \ \ , \ \ \log_2 \sqrt[5]5 \ \ = \ \ \frac{1}{5 \ \log_5 \ 2} \ \ , \ \ \log_2 \sqrt[6]6 \ \ = \ \ \frac{1}{6 \ \log_6 \ 2} \ \ . $$
We have one evident equality, $ \ \large{\frac{1}{4 \ \log_4 \ 2} \ = \ \frac{1}{4 \ \log_4 \ [4^{1/2}]} \ = \ \frac{1}{4 \ · \ (1/2)} \ = \ \frac{1}{2} } \ \ . $
For the rest, we need to examine the consequences of some inequalities. (All of the denominators in the ratios we've found are positive, so there are no undesired complications.)
$$ 2^3 \ < \ 3^2 \ \ \Rightarrow \ \ 2 \ < \ 3^{2/3} \ \ \Rightarrow \ \ \log_3 2 \ < \ \frac23 \ \ \Rightarrow \ \ \frac12 \ < \ \frac{1}{3 \ \log_3 2} \ \ ; $$
$$ 2^5 \ > \ 5^2 \ \ \Rightarrow \ \ 2 \ > \ 5^{2/5} \ \ \Rightarrow \ \ \log_5 2 \ > \ \frac25 \ \ \Rightarrow \ \ \frac12 \ > \ \frac{1}{5 \ \log_5 2} \ \ ; $$
$$ 5^6 \ = \ 15625 \ > \ 6^5 \ = \ 7776 \ \ \Rightarrow \ \ 5 \ > \ 6^{5/6} \ \ \Rightarrow \ \ \log_6 5 \ = \ \frac{\log_6 2}{\log_5 2} \ > \ \frac56 $$ $$ \Rightarrow \ \ \frac{1}{5 \ \log_5 2} \ > \ \frac{1}{6 \ \log_6 2} \ \ . $$
Since $ \ \log_2 x \ $ is an increasing function of $ \ x \ $ on its domain $ \ x \ > \ 0 \ \ , \ $ the chain of inequalities
$$ \log_2 \sqrt[3]3 \ \ > \ \ \log_2 \sqrt2 \ = \ \log_2 \sqrt[4]4 \ \ > \ \ \log_2 \sqrt[5]5 \ \ > \ \ \log_2 \sqrt[6]6 $$
implies
$$ \sqrt[3]3 \ \ > \ \ \sqrt2 \ = \ \sqrt[4]4 \ \ > \ \ \sqrt[5]5 \ \ > \ \ \sqrt[6]6 \ \ . $$
|
{
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"url": "https://math.stackexchange.com/questions/4363451",
"timestamp": "2023-03-29T00:00:00",
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|
Solving the system $\tan x + \tan y = 1$ and $\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$ How can I solve this system of trigonometric equations:
$$\tan x + \tan y = 1$$
$$\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$$
I tried to write tangent as $\sin/\cos$ and then multiply the first equation with the second one but it is not that brought me in a right way.
Do you have any idea how to solve this one?
|
Substitute $u=\cos x, \sqrt{1-u^2}=\sin x, v=\cos y, \sqrt{1-v^2}=\sin y$.
The second equation:
$$ u\sqrt{1-v^2} = \frac{1}{\sqrt2} $$
$$\Rightarrow u=\frac{1}{\sqrt{2-2v^2}}$$
The first equation:
$$ \frac{\sqrt{1-u^2}}{u} + \frac{\sqrt{1-v^2}}{v} = 1 $$
$$ \Rightarrow \sqrt{1-2v^2} + \frac{\sqrt{1-v^2}}{v} = 1 $$
$$\Rightarrow 4v^8-4v^6+9v^4-6v^2+1=0$$
$$\Rightarrow (2v^2-1)(2v^6-v^4+4v^2-1)=0$$
If $2v^2-1=0$,
$$(x,y)= (2\pi m, 2\pi n + \frac{\pi}{4}), (2\pi m \pm \pi, 2\pi n - \frac{3\pi}{4})$$
$2v^6-v^4+4v^2-1=0$ has two solutions (same absolute value, different signs), and they are not neat. Though we can approximate $v \approx \pm \frac{1}{2}$ since when $v = \pm \frac{1}{2}$, $2v^6-v^4+4v^2-1 = \frac{1}{32}$.
|
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|
Solve ODE $y'(y'+y)=x(x+1)$ Solve ODE $$y'(y'+y)=x(x+1)$$
I tried to remove $y'^2$ term by differentiate it wrt x and then replace value in hope that it will turn out some exact form but got stuck after
$$2y'y''-yy'+y''=x-x^2+1$$
How i proceed further or my method is wrong ?
Edit:
Exact problem
If $y'-x\neq0$ is a solution of the differential equation $y'(y'+y)=x(x+1)$ then y(x) is given by
*
*$1-x-e^x$
*$1-x-e^{-x}$
*$1+x+e^x$
*$1+x+e^{-x}$
|
The equation can be rewritten as $(y')^2+yy'-x^2-x=0.$ One can solve $y'$ in terms of $y$ and $x.$ Notice that the above equation is equivalent to $(2y')^2+2y(2y')-4x^2-4x=0=(2y')^2+2y(2y')+y^2-(y^2+4x^2+4x)=(2y'+y)^2-(y^2+4x^2+4x),$ hence $(2y'+y)^2=y^2+4x^2+4x,$ implying $y^2\geq-(4x^2+4x).$ Notice that $4x^2+4x=4x^2+4x+1-1=(2x+1)^2-1,$ so $y^2\geq1-(2x+1)^2.$ This allows us to say that $2y'+y=\pm\sqrt{y^2+(2x+1)^2-1}.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
In a triangle ABC, if certain areas are equal then P is its centroid Let $P$ be a point in the interior of $\triangle ABC$. Extend $AP$, $BP$, and $CP$ to meet $BC$, $AC$, and $AB$ at
$D$, $E$, and $F$, respectively. If $\triangle APF$, $\triangle BPD$, and $\triangle CPE$, have equal areas, prove that $P$ is the centroid of
$\triangle ABC$.
I am trying to do with Ceva's theorem: $\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1$ and also with the result $\frac{AP}{PD}=\frac{AF}{FB}+\frac{AE}{EC}$ but having some difficulties. Please give any hint.
|
I will use barycentric coordinates, an excellent reference is
Barycentric Coordinates for the Impatient, Max Schindler, Evan Chen
and it is by chance compact, and easily accesible/readable. We will use the formula in Theorem 10 in it.
Let $P$ be $P=(x,y,z)$ in barycentric coordinates w.r.t. $\Delta ABC$, $x,y,z >0$, $x+y+z=1$.
We may and do assume that the area of the given triangle is normed to $[ABC]=1$.
Then the area $[AFP]$ of $\Delta AFP$ is computed using
$$
\begin{aligned}
A &= (1,0,0)\ ,\\
P &= (x,y,z)\ , \\
F &= [x:y:0]=\left(\frac x{x+y},\frac y{x+y},0\right)\ ,\text{ so}\\
[AFP]
&=
\begin{vmatrix}
1 & 0 & 0\\
\frac x{x+y} &\frac y{x+y} & 0\\
x & y & z
\end{vmatrix}
=
\frac 1{x+y}
\begin{vmatrix}
1 & 0 & 0\\
x & y & 0\\
x & y & z
\end{vmatrix}
=
\frac {yz}{x+y}
=
\frac {xyz}{x(x+y)}
\ .
\end{aligned}
$$
The other two areas are obtained by cyclic permutation of the letters $x,y,z$ (taken in this order).
Then the given equality of areas becomes:
$$
\frac {xyz}{x(x+y)} =
\frac {xyz}{y(y+z)} =
\frac {xyz}{z(z+x)} \ .
$$
So $x(x+y)=y(y+z)=z(z+x)$. Let $s$ be the common value of this expression, and let us obtain some further relation, here with all details...
$$
\begin{aligned}
s &= x(x+y) = x(1-z) = x - xz\ ,\\
s &= y(y+z) = y(1-x) = y - yx\ ,\\
s &= z(z+x) = z(1-y) = z - zy\ ,\\[2mm]
3s &= 1 - (xy+yz+zx)\ ,\\
s &= ys + zs + xs = y(x-xz)+z(y-yx) +x(z-zy)\\
&= (xy+yz+zx) -3xyz\ ,\\[2mm]
s(1-3s) &= s(yz+zx+xy) =\sum yz(x-xz)= 3xyz - xyz = 2xyz\ ,\\
3s(1-3s) &= 6xyz = 2(xy+yz+zx) - 2s = 2(1-3s)-2s\ ,
\end{aligned}
$$
and we obtain an equation of degree two in $s$, which is
$$
0 = (2-3s)(1-3s) - 2s =2-11s + 9s^2=(1-s)(2-9s)\ .
$$
Since $s=x(x+y)<1$, we only accept $s=2/9$. The simple elementary polynomials in $x,y,z$ are
$$
\begin{aligned}
e_1 &=x+y+z=1\ ,\\
e_2 &=xy+yz+zx=1-3s=\frac 13\ ,\\
e_3 &=xyz=\frac 12s(1-3s)=\frac 1{27}\ ,
\end{aligned}
$$
so $x,y,z$ are (Vieta) the roots of the polynomial in $T$
$$
T^3 - T^2+\frac 13T -\frac 1{27}=\left(T-\frac 13\right)^3\ ,
$$
so the point $P$ is $P=(x,y,z)=\left(\frac 13,\frac 13,\frac 13\right)=[1:1:1]$,
the centroid.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inequality $2a^2 + 8b^2 + \frac{1}{2ab} \ge 4$ I saw the following question:
Prove that the least value of
$$2a^2+8b^2+\frac{1}{2ab}$$
is 4, where $a,b$ are positive numbers.
I believe that this can be shown using the criterion of the saddle and critical points that is taught usually in Calculus 3.
Can this be solved using another easier method?
|
Differently from the proposed solutions, you can also proceed as follows:
\begin{align*}
2a^{2} + 8b^{2} + \frac{1}{2ab} & = 2(a^{2} + 4b^{2}) + \frac{1}{2ab}\\\\
& = 2(a^{2} - 4ab + 4b^{2}) + 8ab + \frac{1}{2ab}\\\\
& = 2(a - 2b)^{2} + \frac{1}{2ab} + 8ab\\\\
& \geq \frac{1}{2ab} + 8ab
\end{align*}
Finally, according to the AM-GM inequality, one arrives at
\begin{align*}
2a^{2} + 8b^{2} + \frac{1}{2ab} \geq \frac{1}{2ab} + 8ab \geq 2\sqrt{\frac{8ab}{2ab}} = 4
\end{align*}
and we are done.
Hopefully this helps !
|
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|
Triple integration $\iiint_{x^2+y^2+z^2 \leq 2z} x^2y^2 dxdydz$ $$\iiint_{x^2+y^2+z^2 \leq 2z} x^2y^2 dxdydz$$
I want to solve it with cylindrical coordinates:
So we get:
$$\iiint_{r^2+z^2 \leq 2z} r^5\sin^2(\phi)\cos^2(\phi) dxdydz$$
I see that the bounds are actually a circle with radius of $\sqrt{2z}$.
How can I use this fact to solve the above integral?
|
It's better to use spherical coordinates
The region is $ x^2 + y^2 + z^2 \le 2 z $, i.e., $x^2 + y^2 + (z - 1)^2 \le 1 $
So we can define the spherical coordinates $r, \theta, \phi $ such that
$ x = r \sin \theta \cos \phi $
$ y = r \sin \theta \sin \phi $
$ z = 1 + r \cos \theta $
And the integral becomes
$ I = \displaystyle \int_{r = 0 }^ 1 \int_{\theta = 0 }^ \pi \int_{\phi = 0}^{2 \pi}
r^6 \sin^5 \theta \cos^2 \phi \sin^2 \phi \text{d}\phi \text{d}\theta \text{d} r $
Integrating with respect $r $ and with respect to $\phi$ is straight forward, and the integral reduces to
$ I = \dfrac{\pi}{28} \displaystyle \int_{\theta = 0}^\pi sin^5 \theta \text{d}\theta $
The integral with respect to $\theta$ is solved using the substitution $ u = \cos \theta$, then
$ \displaystyle \int \sin^5 \theta \text{d} \theta = -\int (1 - u^2)^2 du =-( u -\dfrac{2}{3} u^3 +\dfrac{u^5}{5}) $
Hence,
$I = \dfrac{4 \pi}{105}$
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.