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Yet another difficult integration question For $r \in (0,1)$ and $k \in \mathbb Z^+$, prove
$$ \frac{1}{\pi} \int_{0}^\pi \ln\left(1 + r \cos(u)\right) \ln \left( 1 + r \cos(3^k u)\right) du = \left(\ln\left(\frac{2(1-\sqrt{1-r^2})}{r^2}\right)\right)^2.$$
Given that for any $n \in \mathbb Z^+$
$$ - \frac{1}{\pi} \int_{0}^\pi \ln\left(1 + r \cos(nu)\right) du = \ln\left(\frac{2(1-\sqrt{1-r^2})}{r^2}\right).$$
In my mind, the strategy should be to use the fact that $\cos(u)$ and $\cos(3^ku)$ are orthogonal. Write $[0,\pi]$ as a product space where $\cos(u)$ and $\cos(3^ku)$ are constant along the (orthogonal) fibres. Then use Fubini's theorem. But I am having difficulty in putting the pieces together. Thanks for your help.
|
Lemma 1: Let $n \in \mathbb{Z}\backslash\{0\}$ and let $f$ be a function such
that $f(x+T)=f(x)$ for all $x$. Then $$
\int_0^T f(nx) \mathrm{d}x = \int_0^T f(x) \mathrm{d}x $$
${}$
Lemma 2: Let $f$ be any function such that $\int_0^{\pi} f(x)\,\mathrm{d}x $ converges, then $$\int_0^{\pi} f(\cos x)\,\mathrm{d}x=\frac{1}{2}\int_0^{2\pi} f(\cos x)\,\mathrm{d}x$$
Using these two results it follows immediately that
\begin{align*}
I_n(r)
& = - \frac{1}{\pi} \int_{0}^\pi \ln\left(1 + r \cos(nu)\right) du \\
& = - \frac{1}{2\pi} \int_{0}^{2\pi} \ln\left(1 + r \cos(nu)\right) du \\
& = - \frac{1}{\pi} \int_{0}^\pi \ln\left(1 + r \cos u\right) du \\
& = I_1(r) = I(r)
\end{align*}
The next step is to differentiate under the integral sign. I will come back to the conditions where this holds, so
\begin{align*}
I'(r)
=
- \frac{1}{2\pi} \int_{0}^{2\pi} \frac{\partial }{\partial r} \ln\left(1 + r \cos u\right)\,\mathrm{d}u
= \frac{-1}{2\pi} \int_{0}^{2\pi} \frac{\cos u}{1 + r \cos u}\,\mathrm{d}u
= \frac{1}{r} - \frac{1}{r\sqrt{1-r^2}}\end{align*}
Also note that $I(0) = 0$, so integrating the above expression from $0$ to $r$ gives
\begin{align*}
\int_0^r I(\tau) \mathrm{d}\tau & = \int_0^r \frac{1}{\tau} - \frac{1}{\tau\sqrt{1-\tau^2}} \mathrm{d}\tau \\
I(r) - I(0) & = \log \left( \frac{2}{r} \right) - \operatorname{arctanh}\left( \frac{1}{\sqrt{1-r^2}} \right)
\end{align*}
This is valid since we assume that $|r|< 1$. Hence
$$
I_n(r)
= - \frac{1}{\pi} \int_{0}^\pi \ln\left(1 + r \cos(nu)\right) du
= \log \left( \frac{2}{r} \right) - \operatorname{arctanh}\left( \frac{1}{\sqrt{1-r^2}} \right)
$$
The proof of the second lemma is trivial, and the evaluation of the trigonometric integral can be done something along the lines of
$$
\frac{1}{r}\int_0^{2\pi} \frac{r\cos x}{1 + r \cos x} \mathrm{d}x
= \frac{1}{r}\int_0^{2\pi} 1 - \frac{\mathrm{d}x}{1 + r \cos x}
= \frac{2\pi}{r} - \int_{-\infty}^\infty \frac{2\mathrm{d}u}{(1-r)u^2 + 1+r}
$$
Then use something like $u \mapsto u\frac{1+r}{\sqrt{1-r}}$ to complete the work.
The proof of the first integral is also elementary note
\begin{align}
\int_0^{T} f(kx) \mathrm{d}x
= \frac{1}{k}\int_{0}^{kT} f(x)\mathrm{d}x
= \frac{1}{k}\sum_{n=0}^{k-1} \int_{nT}^{(n+1)T} f(x)\mathrm{d}x
= \frac{1}{k}\sum_{n=0}^{k-1} \int_{0}^{T} f(u + nT)\mathrm{d}u
\end{align}
This completes the proof since, $f$ is has a period of $T$ then $f(u+nT)=f(u)$.
Eg intuitively we have $k$ integrals $\int_0^T + \int_T^{2T} + \cdots + \int_{(k-1)T}^{kT}$ equally big.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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|
Applications of the identity $ab + \left(\frac{a+b}{2} - b\right)^2 = \left(\frac{a+b}{2}\right)^2$ I am reading Euclid's elements
I found the algebraic identity
$ab + \left(\frac{a+b}{2} - b\right)^2 = \left(\frac{a+b}{2}\right)^2$
I ponder on usage of this identity for $2$ hours.
but I can't click anything.
$a^2 + b^2 = c^2$ can be used when you want to know the direction between $2$ coordinates.
Any example involving this identity?
|
This equation can be rewritten as $$ab=\frac{(a+b)^2}4-\frac{(a-b)^2}4.$$ This is the basis of tables of quarter-squares. Such a table tabulated $\left\lfloor \frac{n^2}4\right\rfloor$ for various integer values $n$. It was used as an aid to multiplying integers. To find $ab$, look up the quarter-squares of $a+b$ and $a-b$, then subtract.
If $a$ and $b$ have different parities, $a+b$ and $a-b$ are both odd, so the tabulated quarter-squares are too low by a quarter, but those quarters cancel.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Solve a system of linear equations $\newcommand{\Sp}{\phantom{0}}$There is a system of linear equations:
\begin{alignat*}{4}
&x - &&y - 2&&z = &&1, \\
2&x + 3&&y - &&z =-&&2.
\end{alignat*}
I create the matrix of the system:
$$
\left[\begin{array}{rrr|r}
1 & -1 & -2 & 1 \\
2 & 3 & -1 & -2
\end{array}\right]
$$
then with GEM,
$$
\left[\begin{array}{rrr|r}
1 & -1 & -2 & 1 \\
0 & 5 & 3 & -4
\end{array}\right]
$$
I don't know how to proceed after that?
I have found the correction of this exercise but I still don't understand the way to solve it. Can someone help me please?
|
Bringing the matrix in reduced row echelon form $R$ (the way other answers have shown) is the first step no matter what you do.
As for "reading off" the solutions off $R$, I think there are several techniques. The one I was taught, and which I find the easiest to use, is the following:
*
*remove all zero rows
In the example this leads to
$\begin{bmatrix}
1 & 0 & -\frac{7}{5} & | & \frac{1}{5} \\
0 & 1 & \frac{3}{5} & | & -\frac{4}{5}
\end{bmatrix}$, so nothing changed here, since there were no zero rows
*then insert any additional zero rows so the matrix is square and the pivots are all on the diagonal
$\begin{bmatrix}
1 & 0 & -\frac{7}{5} & | & \frac{1}{5} \\
0 & 1 & \frac{3}{5} & | & -\frac{4}{5} \\
0 & 0 & 0 & | & 0
\end{bmatrix}$
*now you can read off a special solution $\lambda$ off the right hand side
$\lambda = \begin{pmatrix}\frac{1}{5} \\ -\frac{4}{5} \\ 0\end{pmatrix}$
*replace all zero entries on the diagonal with $-1$.
$\begin{bmatrix}
1 & 0 & -\frac{7}{5} & | & \frac{1}{5} \\
0 & 1 & \frac{3}{5} & | & -\frac{4}{5} \\
0 & 0 & -1 & | & 0
\end{bmatrix}$
*the columns with $-1$ on the diagonal (the free columns) now form the basis of the null space
$N = \langle \begin{pmatrix}-\frac{7}{5} \\ \frac{3}{5} \\ -1\end{pmatrix} \rangle = \{ r\cdot \begin{pmatrix}-\frac{7}{5} \\ \frac{3}{5} \\ -1\end{pmatrix}$ }, for all $r \in \mathbb{R}$.
And then, of course, the complete solution is just $\lambda + N$, that is $L = \{ \begin{pmatrix}\frac{1}{5} \\ -\frac{4}{5} \\ 0\end{pmatrix} + r\cdot \begin{pmatrix}-\frac{7}{5} \\ \frac{3}{5} \\ -1\end{pmatrix} | r \in \mathbb{R} \}$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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|
Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$ Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$
My approach :
I used $\sin A +\sin B = 2\sin(A+B)/2\times\cos(A-B)/2 $
$\Rightarrow \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7} = 2\sin(3\pi)/7\times\cos(\pi/7)+\sin(8\pi/7) $
$= 2\sin(3\pi)/7\times\cos(\pi/7)+\sin(\pi + \pi/7) $
$= 2\sin(3\pi)/7\times\cos(\pi/7)-\sin(\pi/7) $
I am not getting any clue how to proceed further or whether it is correct or not. Please help thanks..
|
Setting $7\theta=\pi,S=\sin2\theta+\sin4\theta+\sin8\theta=(\sin2\theta+\sin8\theta)+\sin4\theta$
Using Prosthaphaeresis Formula & Double angle formula,
$S=2\sin5\theta\cos3\theta+2\sin2\theta\cos2\theta$
As $5\theta=\pi-2\theta,\sin5\theta=\sin2\theta$
and $3\theta=\pi-4\theta,\cos3\theta=-\cos4\theta$
$S=-2\sin2\theta\cos4\theta+2\sin2\theta\cos2\theta=2\sin2\theta(\cos2\theta-\cos4\theta)$
Again using Prosthaphaeresis Formula, $S=2\sin2\theta(2\sin3\theta\sin\theta)$ which is clearly $>0$
But, $\sin2\theta=\sin(\pi-2\theta)=\sin5\theta;\sin\theta=\sin6\theta;\sin3\theta=\sin4\theta$
$\implies \dfrac S4=+\sqrt{\prod_{r=1}^6\sin r\theta}\ \ \ \ (1)$
We can derive (See below) $\sin7x=7\sin x+\cdots-64\sin^7x$
If $\sin7x=0,7x=n\pi$ where $n$ is any integer, $x=\dfrac{n\pi}7,0\le n\le6$
So, $\sin\dfrac{n\pi}7,0\le n\le6$ are the roots of $7\sin x+\cdots-64\sin^7x=0$
So, $\sin\dfrac{n\pi}7,1\le n\le6$ are the roots of $7+\cdots-64\sin^6x=0\iff64\sin^6x+\cdots-7=0$
Using Vieta's formula, $\prod_{r=1}^6\sin r\theta=\dfrac7{64}$
Apply this in $(1)$
Derivation $\#1:$
Using de Moivre's Theorem, $\cos7y+i\sin7y=(\cos y+i\sin y)^7$
$=\cdots +i\left(\sin^7y -\binom72\sin^5y\cos^2y+\binom74\sin^3y\cos^4y-\binom76\sin y\cos^6y\right)$
Writing $\cos^2y=1-\sin^2y$
$\cos7y+i\sin7y=\cdots +i\left(7\sin y-64\sin^7y\right)$
Derivation $\#2:$
Using Prosthaphaeresis Formula, $\sin7x+\sin x=2\sin4x\cos3x=2(2\sin2x\cos2x)(4\cos^3x-3\cos x)=4(2\sin x\cos x)(1-2\sin^2x)(4\cos^3x-3\cos x)$
$=8(\sin x-2\sin^3x)\cos^2x(4\cos^2x-1)$
$=8(\sin x-2\sin^3x)(1-\sin^2x)\{4(1-\sin^2x)-1\}$
$=8\sin x+\cdots-64\sin^7x$
$\implies\sin7x=7\sin x+\cdots-64\sin^7x$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Best way to solve $X^3-X^2-X-1=0$ can anyone help me for this cubic equation ?
can be solved without delta method?
$X^3-X^2-X-1=0$
(answer is $\sim 1.8393$)
|
@Darksonn 's answer certainly works, but if you only want the positive real solution, there is a "cubic formula" that is reasonably useful in this case.
Theorem: If the cubic equation
$$X^3 + pX + q$$
($p, q$ real) satisfies
$$\frac{p^3}{27} + \frac{q^2}{4} \geq 0,$$
then a solution to the cubic equation is
$$X = \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{p^3}{27} + \frac{q^2}{4}}} + \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{p^3}{27} + \frac{q^2}{4}}}.$$
This is the Cardano-Tartaglia Formula.
In our case, we take @Darksonn 's $y$-substitution $\displaystyle y = x - \frac{1}{3}$ and get
$$y^3 - \frac{4}{3}y - \frac{38}{27} =0.$$
So $p = \displaystyle - \frac{4}{3}$, $q = \displaystyle - \frac{38}{27}$; one may check that in this case,
$$ \frac{p^3}{27} + \frac{q^2}{4} = \frac{- 64 \cdot 4 + 38^2}{27^2 \cdot 4} = \frac{297}{27^2}, \quad -\frac{q}{2} = \frac{19}{27},$$
so that the answer becomes
\begin{align}
y &= \sqrt[3]{\frac{19}{27} + \sqrt{\frac{297}{27^2}}} + \sqrt[3]{\frac{19}{27} - \sqrt{\frac{297}{27^2}}}\\
& = \sqrt[3]{\frac{19 + \sqrt{297}}{27}} + \sqrt[3]{\frac{19 - \sqrt{297}}{27}} \\
&= \frac{1}{3} \left(\sqrt[3]{19 + 3\sqrt{33}} + \sqrt[3]{19 - 3\sqrt{33}} \right).
\end{align}
Since $\displaystyle y = x - \frac{1}{3}$, $\displaystyle x = y + \frac{1}{3}$, and we get
$$x = \frac{1}{3} + \frac{1}{3} \left(\sqrt[3]{19 + 3\sqrt{33}} + \sqrt[3]{19 - 3\sqrt{33}} \right).$$
To see that our answer matches @Darksonn 's, note that the terms not already matching are $\sqrt[3]{19 - 3\sqrt{33}}$ on the one side and $\displaystyle \frac{4}{\sqrt[3]{19 + 3 \sqrt{33}}}$ on the other side. To see this equality, note that for all positive $a, b$,
$$a - b = \frac{a^2 - b^2}{a + b}$$
and letting $a = 19$, $b = 3 \sqrt{33}$, we get
$$19 - 3 \sqrt{33} = \frac{64}{19 + 3 \sqrt{33}}$$
Taking the cube root of both sides, we're done.
If you wish to handle the case where the discriminant
$$\frac{p^3}{27} + \frac{q^2}{4}$$ is negative, or to understand the shortcomings of the above formula, I strongly recommend looking at Lecture 4, "Equations of Degree Three and Four," in Fuchs and Tabachnikov, Mathematical Omnibus [Amer. Math. Soc., Providence, 2007]. It is a good "elementary" explanation of what is going on.
Yet other good methods are mentioned in the answers to the similar problem https://math.stackexchange.com/questions/612765/find-roots-of-the-cubic-equation-x3-x2-3-0-without-using-calculator?rq=1 .
|
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"timestamp": "2023-03-29T00:00:00",
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|
Factoring in the derivative of a rational function Given that
$$
f(x) = \frac{x}{1+x^2}
$$
I have to find
$$\frac{f(x) - f(a)}{x-a}$$
So some progressing shows that:
$$
\frac{\left(\frac{x}{1+x^2}\right) - \left(\frac{a}{1+a^2}\right)}{x-a} =
\frac{(x)(1+a^2)-(a)(1+x^2)}{(1+x^2)(1+a^2)}\cdot\frac{1}{x-a} =
\frac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)}
$$
Now, is it possible to factor $x+xa^2-a-ax^2$? I can't seem to find a way, as for simplifying the whole thing. Is there any rule I can use, and I'm unable to see?
|
$$\eqalign{x+xa^2-a-ax^2&= x-a+xa^2-ax^2 \\ & = x-a+x(a^2-ax) \\ &= x-a+x(a(a-x)) \\ &= x-a+x(-a(x-a)) \\ &=\color{blue}{x-a}-ax\color{blue}{(x-a)} \\ &=(x-a)(1-ax).\;\checkmark
}$$
Therefore you can conclude that: $$\eqalign{\require{cancel}\dfrac{f(x)-f(a)}{x-a}&=\dfrac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)} \\ &=\dfrac{\color{red}{\cancel{\color{black}{(x-a)}}}(1-ax)}{(1+x^2)(1+a^2)\color{red}{\cancel{\color{black}{(x-a)}}}} \\ &= \dfrac{1-ax}{(1+x^2)(1+a^2)}.
}\tag{$x\neq a$}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/819527",
"timestamp": "2023-03-29T00:00:00",
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|
If $f:\mathbb{R}\rightarrow \mathbb{{R}}$ and $f(x)$ satisfying $f(2x+3)+f(2x+5) = 2$ . The period of $f(x)$ If $f:\mathbb{R}\rightarrow \mathbb{{R}}$ and $f(x)$ be a function satisfying $f(2x+3)+f(2x+5) = 2$. Then period
of function $f(x)$ is.
$\bf{My\; Solution::}$ Let $(2x+3) = t\;,$ Then equation is $f(t)+f(t+2) = 2$
Now Replace $t\rightarrow (t+2)\;,$ we get $f(t+2)+f(t+4) = 2$
So from these two equation, we get $f(t+4) = f(t)$
So period of function $f(t)$ is $=4$
But answer Given as $=2$ . But in Solution it is given as $ = 2$
I did not understand where I have done Wrong.
Help me
Thanks
|
$\displaystyle f(2x+3) + f(2x+5) = 2$
Replace $x$ with $x+1$:
$\displaystyle f(2x+5) + f(2x+7) = 2$
Subtracting and rearranging, $\displaystyle f(2x+3) = f(2x+7)$
Hence $\displaystyle f(2x+3) = f((2x+3)+4)$
At this point, you can conclude that the period is $4$. If this is difficult to see, you can replace $(2x+3)$ with $t$, as the OP has done.
It might be tempting to do this:
$\displaystyle f(2(x+\frac{3}{2})) = f(2(x+\frac{7}{2}))$
and conclude that the period is $\displaystyle \frac{7}{2} - \frac{3}{2} = 2$.
But this would be a mistake. This ($2$) is the period of $f(2x)$. The period of $f(x) = 4$.
If you're still unconvinced, here's another example. If we define $f(x)=\sin 2x$, then the period of $f(x)$ is $\pi$. But if we define $f(2x)=\sin 2x$, then the period of $f(x)= \sin x$ is $2\pi$, even if the period of $\sin 2x=f(2x)$ remains $\pi$. This is what's happening in this case. The period of $f(2x)$ is $2$, but the period of $f(x)$ is actually $4$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
The 3 Integral $\int_0^\infty {x\,{\rm d}x\over \sqrt[3]{\,\left(e^{3x}-1\right)^2\,}}=\frac{\pi}{3\sqrt 3}\big(\log 3-\frac{\pi}{3\sqrt 3} \big)$ Hi I am trying evaluate this integral and obtain the closed form:$$
I:=\int_0^\infty \frac{x\,dx}{\sqrt[\large 3]{(e^{3x}-1)^2}}=\frac{\pi}{3\sqrt 3}\left(\log 3-\frac{\pi}{3\sqrt 3} \right).
$$
The integral and result has all 3's everywhere. I am not sure how to approach this on. The denominator seems to be a problem.
If $\displaystyle x=\frac{2in\pi}{3}$ we have a singularity but I am not sure how to use complex methods. We will have a branch cut because of the root function singularity.
Differentiating under the integral sign did not help either. I tried partial integration with $v=(e^{3x}-1)^{\frac{2}{3}}$ but this did not simplify since I get a power $x^n, \ (n>1)$ in the new integral.
Thanks, how can we evaluate the integral I?
|
Following @Pranav Arora's idea, if defining the following integral
$$ I(a)=\int_0^\infty\frac{\ln(1+at^3)}{1+t^3}dt, $$
the calculation will just be basic calculus without using advanced tools. In fact, $I(0)=0$ and
\begin{eqnarray}
I'(a)&=&\int_0^\infty\frac{t^3}{(1+t^3)(1+at^3)}dt\\
&=&\frac{1}{a-1}\int_0^\infty\left(\frac{1}{1+t^3}-\frac{1}{1+at^3}\right)dt\\
&=&\frac{1}{a-1}\left(\frac{2\pi}{3\sqrt3}-\frac{2\pi}{3\sqrt3\sqrt[3]a}\right)\\
&=&\frac{2\pi}{3\sqrt3}\frac{1}{\sqrt[3]a(1+\sqrt[3]a+\sqrt[3]a^2)}.
\end{eqnarray}
So
\begin{eqnarray}
I&=&\frac{2\pi}{3\sqrt3}\int_0^1\frac{1}{\sqrt[3]a(1+\sqrt[3]a+\sqrt[3]a^2)}da\\
&=&\frac{2\pi}{3\sqrt3}\int_0^1\frac{3u}{1+u+u^2}du\\
&=&\frac{\pi}{3\sqrt 3}\left(\log 3-\frac{\pi}{3\sqrt 3}\right).
\end{eqnarray}
|
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|
a problem in application of conformal mappings Obtain the complex potential
$F=A(z^
2
+\frac
{1}
{z^
2})$
for a flow in the region $r≥1,0≤θ≤π/2$. Write expressions for $V$ and $ψ$. Note how
the speed $|V|$ varies along the boundary of the region, and verify that $ψ(x, y)=0$ on
the boundary.
this is a problem in application of conformal mappings but i really can't understand the meaning of question .it has given the formula for the complex potential , so what should i obtain??
i'm making a stupidly mistake.
can someone explain it ?
probably,i can solve it by myself then.
thank you
|
This question is similar to a few other ones in MSE, among these:
How to differentiate Complex Fluid Potential
Applications of conformal mapping
From the first reference it is clear that, in general, the following formulas are valid:
$$F(z) = \phi + i\,\psi \qquad ; \qquad
\frac{dF}{dz} = u - i\,v \quad \Longrightarrow
\quad \left| \frac{dF}{dz} \right| = \sqrt{u^2+v^2} = V
$$
Where $\phi$ is the flow potential, $\psi$ is the stream function,
$(u,v)$ are the flow velocity components and $V$ is the speed.
In your case (assuming that $A$ is real):
$$
F(z) = A\left(z^2+\frac{1}{z^2}\right) = A\left((x+iy)^2+\frac{1}{(x+iy)^2}\right)
= A(x+iy)^2+A\frac{(x-iy)^2}{(x^2+y^2)^2} = \\
= A\left((x^2-y^2)+\frac{x^2-y^2}{(x^2+y^2)^2}\right)
+ i\,A\left(2xy-\frac{2xy}{(x^2+y^2)^2}\right) = \phi + i\,\psi
$$
On the boundaries we have $r = 1$ and so $\,x^2+y^2=1$; or $\,x=0\,$ or $\,y=0$ , hence indeed:
$$
\psi = A\left(2xy-\frac{2xy}{(x^2+y^2)^2}\right) = 0
$$
As far as the speed $V$ is concerned:
$$
V = \left|\frac{dF}{dz}\right| = A\left|2\,z - \frac{2}{z^3}\right|
$$
It's a matter of routine to calculate herefrom the speed $V$ as a function of $(x,y)$ ,
but the result may be not a quite nice formula.
Anyway, here is a contour plot of the flow $\color{red}{potential}$ together with
the $\color{green}{stream function}$ .Contour lines are at $25$ levels between the
minimum and the maximum, $A=1$ and the viewport is $(0 \le x \le 5 , 0 \le y \le 5)$ .
Augmented with a vector plot with (scaled) velocities and grey values for the speed
(lighter at greater speed, black=zero inside circle).The flow velocities can be calculated as in
this question :
$$
\phi(x,y) = A \left[ x^2-y^2+\frac{x^2-y^2}{(x^2+y^2)^2} \right] \quad \Longrightarrow
$$
$$
u = \frac{\partial \phi}{\partial x} = A \left[ 2 x + \frac{2 x}{(x^2+y^2)^2} - \frac{4(x^2-y^2)x}{(x^2+y^2)^3} \right] \\
v = \frac{\partial \phi}{\partial y} = A \left[ - 2 y - \frac{2 y}{(x^2+y^2)^2} - \frac{4(x^2-y^2)y}{(x^2+y^2)^3} \right]
$$
The speed $\,V=\sqrt{u^2+v^2}\,$ may become a fairly complicated expression in general,
but it is greatly simplified for the boundaries. Let's follow the streamline
with $\psi=0$ from bottom-right to top-left. Then we subsequently have, for the x-axis (with $y=0$ and $v=0$) :
$$
y = 0 \quad , \quad \infty > x \ge 1 \qquad \Longrightarrow \qquad V = 2\,A\frac{x^4-1}{x^3}
$$
for the quarter of a circle (by repeatedly substituting $\,x^2+y^2=1$) :
$$
r=1 \quad , \quad 0 \le \theta \le \pi/2 \qquad \Longrightarrow \qquad
V = 4\,A\,\sqrt{1-(x^2-y^2)^2} = 8\,A\,x\sqrt{1-x^2}
$$
for the y-axis (with $x=0$ and $u=0$) :
$$
x = 0 \quad , \quad 1 \le y < \infty \qquad \Longrightarrow \qquad V = 2\,A\frac{y^4-1}{y^3}
$$
Note that the speed is zero - as well as continuous - for $(x,y)=(1,0)$ and $(x,y)=(0,1)$ .
But .. it's all much easier with polar coordinates.
Flow potential $\phi$ and stream function $\psi$ :
$$
F(z) = A\left[z^2+\frac{1}{z^2}\right]
= A\left[r^2 e^{2i\theta} + \frac{1}{r^2} e^{-2i\theta} \right] \\
= A\left[\left(r^2 + \frac{1}{r^2}\right) \cos(2\theta)\right]
+ i\, A\left[\left(r^2 + \frac{1}{r^2}\right) \sin(2\theta)\right] = \phi + i\,\psi
$$
Flow velocity field $(u,v)$ :
$$
F'(z) = 2 A\left[z - \frac{1}{z^3}\right]
= 2A\left[r e^{i\theta} - \frac{1}{r^3} e^{-3i\theta} \right] \\
= 2A\left[r \cos(\theta) - \frac{1}{r^3}\cos(3\theta)\right]
- i\, 2A\left[- r \sin(\theta) - \frac{1}{r^3}\sin(3\theta)\right] = u - i\, v
$$
Speed $V$ at the boundaries:
$$
\theta = 0 \quad , \quad r > 1 \qquad \Longrightarrow \qquad
V = 2A\left[r - 1/r^3\right] \\
0 \le \theta \le \pi/2 \quad , \quad r = 1 \qquad \Longrightarrow \qquad
V = \left| 2 A \, 2\,i\, e^{- i\theta} \frac{e^{2i\theta} - e^{-2i\theta}}{2\, i}\right|
= 4A\sin(2\theta) \\
\theta = \pi/2 \quad , \quad r > 1 \qquad \Longrightarrow \qquad
V = 2A\left[r - 1/r^3\right]
$$
|
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|
How to integrate $\int \frac{1}{\sin^4x + \cos^4 x} \,dx$?
How to integrate
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx$$
I tried the following approach:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{\sin^4x + (1-\sin^2x)^2} \,dx = \int \frac{1}{\sin^4x + 1- 2\sin^2x + \sin^4x} \,dx \\
= \frac{1}{2}\int \frac{1}{\sin^4x - \sin^2x + \frac{1}{2}} \,dx = \frac{1}{2}\int \frac{1}{(\sin^2x - \frac{1}{2})^2 + \frac{1}{4}} \,dx$$
The substitution $t = \tan\frac{x}{2}$ yields 4th degree polynomials and a $\sin$ substitution would produce polynomials and expressions with square roots while Wolfram Alpha's solution doesn't look that complicated. Another approach:
$\sin^4x + \cos^4 x = (\sin^2 x + \cos^2x)(\sin^2 x + \cos^2 x) - 2\sin^2 x\cos^2 x = 1 - 2\sin^2 x\cos^2 x = (1-\sqrt2\sin x \cos x)(1+\sqrt2\sin x \cos x)$
and then I tried substituting: $t = \sin x \cos x$ and got
$$\int\frac{t\,dt}{2(1-2t^2)\sqrt{1-4t^2}}$$
Another way would maybe be to make two integrals:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{(1-\sqrt2\sin x \cos x)(1+\sqrt2\sin x \cos x)} \,dx = \\ \frac{1}{2}\int \frac{1}{1-\sqrt2\sin x \cos x} \,dx + \frac{1}{2}\int\frac{1}{1+\sqrt2\sin x \cos x} \,dx$$
... and again I tried $t = \tan\frac{x}{2}$ (4th degree polynomial) and $t=\sqrt2 \sin x \cos x$ and I get $\frac{\sqrt 2}{2} \int \frac{\,dt}{(1-t)\sqrt{1-2t^2}}$ for the first one.
Any hints?
|
$$
\begin{aligned}
\sin ^{4} x+\cos ^{4} x &=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x \\
&=1-2 \sin ^{2} x \cos ^{2} x \\
\int \frac{d x}{\sin ^{4} x+\cos ^{4} x} &=\int \frac{d x}{1-2 \sin ^{2} x \cos ^{2} x} \\
&=\int \frac{\sec ^{4} x}{\sec ^{4} x-2 \tan ^{2} x} d x \\
&\stackrel{t=\tan x}{=} \int \frac{1+t^{2}}{\left(1+t^{2}\right)^{2}-2 t^{2}} d t, \\
&=\int \frac{1+t^{2}}{t^{4}+1} d t \\
&=\int \frac{1+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}} d t \\
&=\int \frac{d\left(1-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^{2}+2} \\
&=\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+C \\
&=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\tan x-\cot x}{\sqrt{2}}\right)+C
\end{aligned}
$$
|
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|
Integrate : $\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}dx$
$$\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}\ dx$$
My approach :
Dividing the denominator by $\cos^2x$ we get $\dfrac{x^2\sec^2x }{(x -\tan x)(x\tan x +1)}$ then
$$\int \frac{x^2\sec^2x}{x^2\tan x -x\tan^2x+x-\tan x}\ dx=\int \frac{x^2(1+\tan^2x)}{x^2\tan x -x\tan^2x+x-\tan x}dx$$
But I am not getting any relation between numerator and denominator so that I will take any substitution and solve further please suggest whether it is correct and how to proceed in this. Thanks.
|
\begin{align}
I &= \int\frac{x^2dx}{(x\sin x+\cos x)(x\cos x-\sin x)} \\ &= -\int\frac{x^2dx}{(1+x^2)\left(\frac{x}{\sqrt{1+x^2}}\sin x+\frac{1}{\sqrt{1+x^2}}\cos x\right)\left(\frac{1}{\sqrt{1+x^2}}\sin x-\frac{x}{\sqrt{1+x^2}}\cos x\right)}\\
&=-\int\frac{\left(1-\frac{1}{1+x^2}\right)dx}{\cos (x-\arctan x)\sin (x-\arctan x)}\\
&=-\int\frac{d(x-\arctan x)}{\cos (x-\arctan x)\sin (x-\arctan x)}\\
&=-\int\frac{d(2(x-\arctan x))}{\sin (2(x-\arctan x))}\\
&=-\ln|\tan (x-\arctan x)|+C.
\end{align}
|
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|
Product of Gamma functions I What is the value of the product of Gamma functions
\begin{align}
\prod_{k=1}^{8} \Gamma\left( \frac{k}{8} \right)
\end{align}
and can it be shown that the product
\begin{align}
\prod_{k=1}^{16} \Gamma\left( \frac{k}{8} \right) = \frac{ 3 \Gamma(11) \ \pi}{2^{19}} \zeta(2) \zeta(4) \approx \frac{\pi^{7}}{26}
\end{align}
is a valid result?
|
For the first question, you can simply calculate:
$$\prod_{k=1}^{8}\Gamma\left(\frac{k}{8}\right) = \prod_{k=1}^{8}\frac{8}{k}\cdot\frac{k}{8}! = \frac{8^8}{8!}\cdot\prod_{k=1}^{8}\frac{k}{8}!$$
Or alternatively (even simpler):
$$\prod_{k=1}^{8}\Gamma\left(\frac{k}{8}\right) = \prod_{k=1}^{8}\frac{k-8}{8}!$$
|
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|
Help in understanding contour integration I would like help in understanding the process of contour integration. As an (hopefully straightforward) example, I have chosen the calculation of Bernoulli number $B_2$.
I should be very grateful if someone could explain how to get from (and describe the steps between) here
$$
B_n=\dfrac{n!}{2\pi i}\oint\dfrac{z}{e^z-1}\dfrac{\text{d}z}{z^{n+1}}
$$
to here
$$
\dfrac{1}{\pi i}\oint\dfrac{z}{e^z-1}\dfrac{\text{d}z}{z^{3}}=\dfrac{1}{6}
$$
|
So between $$B_n=\dfrac{n!}{2\pi i}\oint\dfrac{z}{e^z-1}\dfrac{\text{d}z}{z^{n+1}}$$ and $$\dfrac{1}{\pi i}\oint\dfrac{z}{e^z-1}\dfrac{\text{d}z}{z^{3}}$$ all they did was plug in $n=2$, so now all we need to do is find the residue at $z=0$ of $$\frac{1}{z^{2}(e^{z}-1)}=\frac{1}{z^{2}(1-1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+...)}=\frac{1}{z^{3}(1+\frac{z}{2!}+\frac{z^{2}}{3!}+...)}$$
Now, let $x=\frac{z}{2!}+\frac{z^{2}}{3!}+...$ and use the binomial expansion on $(1+\frac{z}{2!}+\frac{z^{2}}{3!}+...)^{-1}$
$$(1+\frac{z}{2!}+\frac{z^{2}}{3!}+...)^{-1}=1-x+x^{2}+...$$
We only care about the $z^{2}$ and this only has a finite number of contributions, so we'll compute that.
$$a_{2}=(-1)\cdot\frac{1}{3!}+(1)\cdot\frac{1}{2^2}=\frac{1}{12}$$
Where the first term is from x in the expansion, and the second term is from the $\frac{z}{2}$ being squared in the $x^2$ term.
So, after multiplying by $z^{-3}$, we have the residue is $\frac{1}{12}$
Using the residue theorem, we have $$\dfrac{1}{\pi i}\oint\dfrac{z}{e^z-1}\dfrac{\text{d}z}{z^{3}}=\frac{2\pi i}{\pi i}\cdot \frac{1}{12}=\frac{1}{6}$$
|
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|
What is the Laurent series of $ \exp \! \bigl( - \frac{1}{z} \bigr) $? I’m thinking that I could simply let $ x = - \dfrac{1}{z} $ in the Maclaurin series for $ e^{x} $:
$$
1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \cdots
= 1 - \frac{1}{z} + \frac{1}{2! z^{2}} - \frac{1}{3! z^{3}} + \cdots.
$$
Is that right?
|
The Taylor series for $e^x$ is
$$
e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \cdots
$$
Now as you have said, let $x = -\frac{1}{z}$. Then we have
$$
e^{-1/z} = 1 - \frac{1}{z} + \frac{1}{2z^2} - \frac{1}{3!z^3} + \cdots = \sum_{n=0}^{\infty}\frac{1}{(-z)^nn!} = \sum_{n=-\infty}^0\frac{(-z)^n}{(-n)!}
$$
where the Laurent series is
$$
\sum_{n=-\infty}^0\frac{(-z)^n}{(-n)!}
$$
since to be a Laurent series we need a principal part $\sum_{n=-\infty}^{-1}$ and an analytic part $\sum_{n=0}^{\infty}$
|
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|
Basic algebra, isolating the variable So I have the equation
$$\tan30=\frac{4.9t-\frac{10}{t}}{\frac{8.77}{t}}$$
And I want to find t, but my algebra has failed me.
This is my working so far.
$$\frac{8.77}{t}=\frac{4.9t-\frac{10}{t}}{\tan30}$$
$$\frac{8.77}{t}=\frac{4.9t}{\tan30}-\frac{10}{\tan30t}$$
$$8.77=\frac{4.9t}{\tan30t}-\frac{10}{\tan30t^2}$$
$$8.77=\frac{4.9}{\tan30}-\frac{10}{\tan30t^2}$$
$$0=\frac{4.9}{\tan30}-\frac{10}{\tan30t^2}-8.77$$
$$\frac{10}{\tan30t^2}=\frac{4.9}{\tan30}-8.77$$
Invert
$$\frac{\tan30t^2}{10}=\frac{\tan30}{4.9}-\frac{1}{8.77}$$
$$\frac{t^2}{10}=\frac{\tan30}{4.9\tan30}-\frac{1}{8.77\tan30}$$
$$t^2=\frac{10}{4.9}-\frac{10}{8.77\tan30}$$
$$t=\sqrt{0.065844}$$
$$=0.2566$$
However I know this is too long winded for the question, and the answer is wrong as well. So I am wondering 1- where I have gone wrong and 2- what is a better way of doing it. Thanks
|
Multiplying the right side of the equation by $\frac{t}{t} = 1$, with the assumption that $t \neq 0$, we have
$$\tan(30^{\circ}) = \frac{4.9t^2 - 10}{8.77} \,\,.$$
Now, we can isolate $t$ as follows
$$\begin{align} \tan(30^{\circ}) = \frac{4.9t^2 - 10}{8.77} &\implies 8.77\tan(30^{\circ}) = 4.9t^2 - 10 \\&\implies t^2 = \frac{8.77\tan(30^{\circ}) + 10}{4.9} \\&\implies t = \pm \sqrt{\frac{8.77\sqrt{3} + 10}{4.9}} \,\,. \end{align}$$
|
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|
Intersection of ellipse and hyperbola at a right angle Need to show that two functions intersect at a right angle.
Show that the ellipse
$$ \frac{x^2}{a^2} +\frac{y^2}{b^2} = 1
$$
and the hyperbola
$$
\frac{x^2}{α^2} −\frac{y^2}{β^2} = 1
$$
will intersect at a right angle if
$$α^2 ≤ a^2 \quad \text{and}\quad a^2 − b^2 = α^2 + β^2$$
Not sure how to tackle this question, graphing didn't help.
|
The gradient vector $\left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}\right)$ is normal to the curve $f(x,y)=0$.
Then
$$\left(\frac{2x}{a^2},\frac{2y}{b^2}\right)\cdot\left(\frac{2x}{\alpha^2},-\frac{2y}{\beta^2}\right)=\frac{4x^2}{a^2\alpha^2}-\frac{4y^2}{b^2\beta^2}=0.$$
We can eliminate $x^2$ and $y^2$ from the three equations by
$$\left|\begin{matrix}\dfrac1{a^2}&\dfrac1{b^2}&1\\\dfrac1{\alpha^2}&-\dfrac1{\beta^2}&1\\\dfrac1{a^2\alpha^2}&-\dfrac1{b^2\beta}&0\end{matrix}\right|=\frac{\beta^2+\alpha^2+b^2-a^2}{a^2b^2\alpha^2\beta^2}=0.$$
Hence the claim.
I have not investigated the inequality $\alpha^2<a^2$. Most probably you obtain it by expressing that $x^2,y^2>0.$ [Confirmed by the expressions provided by @mathlove.]
|
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|
Prove by induction that $(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}$ Prove by induction that $$(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}.$$
I got up to:
$n=1$ is true, and assuming $n=k$ prove for $n=k+1$.
Prove...
$$\frac{k(2k+1)(7k+1)+6(2(k+1))^2}{6} = \frac{(k+1)(2k+3)(7k+8)}{6}$$
I keep trying to expand to $6(2k+2)^2$ and factorising but I end up being short on one factor, e.g., I end up with $\frac{(k+1)(2k+3)(7k+6)}{6}$.
|
Hint :
Be careful, at the step $k+1$ the sum is
$$ (k+2)^2+\cdots+(2k+1)^2+(2k+2)^2 $$
and the sum at the step $k$ is
$$ (k+1)^2+\cdots+(2k-1)^2+(2k)^2 $$
|
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|
Show that $\int_{-\infty}^{\infty}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}=\frac{7\pi}{50}$ Show that
$$\int_{-\infty}^{\infty}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}=\frac{7\pi}{50} $$
So I figured since it's an improper integral I should change the limits
$$\lim_{m_1\to-\infty}\int_{m_1}^{0}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}+ \lim_{m_2\to\infty}\int_{0}^{m_2}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}$$
I'm however not sure how to evaluate this. Any help would be great - thanks.
|
Hint
Use partial fractions to write out the integral in the form:
$$\frac{A x+B}{x^2+1}+\frac{C x+D}{\left(x^2+1\right)^2}+\frac{E+F
x}{x^2+2 x+2}$$
You'll find that $a=-6/25, c = 2/5$, etc. Now look at at what remains, and use the fact that:
$$\int \frac{1}{x^2+1}dx=\arctan x$$
And substitute $x\rightarrow x+1$ in the last integral to simplify it.
|
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|
Factor the Expression completely$ (a+b)^2 - (a-b)^2$ I don't understand this question. The answer in the book is $4ab$, but how is that term a factor?
I was thinking along the line that this was a difference of squares example. $a^2-b^2 = (a+b)(a-b)$
My answer is $[(a+b)-(a-b)][(a+b)+(a-b)]$
What do I not understand?
|
Note that you don't even need to factor it, just expand the expression.
$$(a+b)^2-(a-b)^2$$
$$=(a^2+b^2+2ab)-(a^2+b^2-2ab)$$
$$=a^2+b^2+2ab-a^2-b^2+2ab$$
$$=4ab$$
If you have to factor it, remember the difference of squares $a^2-b^2=(a+b)(a-b)$. In this case, $a^2$ is $(a+b)^2$, and $b^2$ is $(a-b)^2$.
$$(a+b)^2-(a-b)^2$$
$$=(a+b+a-b)(a+b-(a-b))$$
$$=(2a)(a+b-a+b)$$
$$=(2a)(2b)$$
$$=4ab$$
|
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|
Use induction to show that $a_{n+1}-a_n=\biggl(-\frac{1}{2} \biggr)^n (a_1-a_0) .$
Let $a_0$ and $a_1$ be distinct real numbers. Define
$a_n=\frac{a_{n-1}+a_{n-2}}{2}$ for each positive integer $n\geq 2$. Prove that $$a_{n+1}-a_n=\biggl(-\frac{1}{2} \biggr)^n (a_1-a_0) $$
This is what I have so far:
Let $P(n)$ be the statement: For all $n\geq 2$, $n\in\mathbb{N}$, $$
a_{n+1}-a_n=\biggl(-\frac{1}{2} \biggr)^n (a_1-a_0). $$ We must show
1. $P(2)$ holds, 2. For all $k\geq 2$, $k\in\mathbb{N}$, if $P(k)$ holds, then $P(k+1)$ holds. Let $n=2$. Then $$ \begin{aligned}
a_{n+1}-a_n &=a_3-a_2\\ &=\frac{a_2+a_1}{2}-\frac{a_1+a_0}{2}\\
&=\frac{a_1+a_0+2a_1}{4}-\frac{a_1+a_0}{2}\\
&=\frac{3a_1+a_0}{4}+\frac{-2a_1-2a_0}{4}\\ &=\frac{a_1-a_0}{4}\\
&=\biggl(-\frac{1}{2}\biggr)^2(a_1-a_0). \end{aligned} $$ Hence,
$P(2)$ holds. Suppose that for all $k\geq 2$, $k\in\mathbb{N}$,
$P(k)$ holds, that is $$ \begin{aligned} a_{k+1}-a_k
&=\frac{a_k+a_{k-1}}{2}-\frac{a_{k-1}+a_{k-2}}{2}\\
&=\biggl(-\frac{1}{2} \biggr)^k (a_1-a_0). \end{aligned} $$ Then
$$ \begin{aligned} a_{(k+1)+1}-a_{k+1} &=a_{k+2}-a_{k+1}\\
&=\frac{a_{k+1}+a_k}{2}-\frac{a_k+a_{k-1}}{2}\\
&=\frac{a_k+a_{k-1}+a_{k-1}+a_{k-2}}{4}-\frac{a_{k-1}+a_{k-2}+a_{k-2}+a_{k-3}}{4}\\ &=\biggl(\frac{a_k+a_{k-1}-a_{k-2}-a_{k-3}}{2}\biggr)^2\\
&=\frac{a_{k-1}+a_{k-2}+a_{k-2}+a_{k-3}-a_{k-3}-a_{k-4}-a_{k-4}-a_{k-5}}{8}\\
&=\biggl(\frac{a_{k-1}+2a_{k-2}-2a_{k-4}-a_{k-5}}{2}\biggr)^3
\end{aligned} $$
Ok, so here it seems that I will just end up substituting forever. I'm sure that I'm missing something obvious. Any help would be greatly appreciated. Thanks.
|
Hint: Subtract $a_{n-1}$ from both sides. We get
$$a_n-a_{n-1}=-\frac{1}{2}(a_{n-1}-a_{n-2}).$$
The "new difference" is the "old difference" times $-\frac{1}{2}$.
|
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|
How to solve $\left(\dfrac{5}{3}\right)^3\left(-\dfrac{3}{5}\right)^2$ I need help in solving this problem (sorry I didn't know how to write it on here).
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$\begin{align}\left(\dfrac{5}{3}\right)^3\left(\dfrac{-3}{5}\right)^2 & = \left(\dfrac{5^3}{3^3}\right)\cdot\left(\dfrac{(-1)^2 3^2}{5^2}\right) & \text{by commutativity of exponents} \\ ~ & = \dfrac{(-1)^2 5^1}{3^1} & \text{by associativity of exponents} \\ ~ & = \dfrac{5}{3} & \text{by }(-1)^2=1, a^1 = a \end{align}$
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Evaluating the limit of $\sqrt[2]{2+\sqrt[3]{2+\sqrt[4]{2+\cdots+\sqrt[n]{2}}}}$ when $n\to\infty$ The following nested radical $$\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$$ is known to converge to $2$.
We can consider a similar nested radical where the degree of the radicals increases:
$$\sqrt[2]{2+\sqrt[3]{2+\sqrt[4]{2+\cdots+\sqrt[n]{2}}}}$$
which converges to a constant $C=1.8695973...$. Is there a closed-form expression for this limit?
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As already mentioned by others, there is no closed form known for this expression (like the majority of infinite nested radicals).
I would consider some bounds for this number, which give quite good approximations.
First, instead of truncating the number, it's much better to replace the remainder by $1$ at any step, for example:
$$\sqrt{2+\sqrt[3]{2+\sqrt[4]{2+\cdots}}}>\sqrt{2+\sqrt[3]{2+\sqrt[4]{2+1}}}=1.8684804$$
For the upper bound we can use the approximation:
$$\sqrt{2+\sqrt[3]{2+\sqrt[4]{2+\cdots}}}<\sqrt{2+\sqrt[3]{2+\sqrt[3]{2+\cdots}}}$$
To find $\sqrt[3]{2+\sqrt[3]{2+\cdots}}$, we use the equation:
$$x^3-x-2=0$$
$$x=\frac{1}{3} \left(\sqrt[3]{27-3\sqrt{78}}+\sqrt[3]{27+3\sqrt{78}} \right)=1.5213797$$
$$\sqrt{2+\sqrt[3]{2+\sqrt[4]{2+\cdots}}}<\sqrt{2+1.5213797}=1.876534$$
For a better upper bound we find:
$$\sqrt[4]{2+\sqrt[4]{2+\cdots}}=\frac{1}{3} \left(1-2\sqrt[3]{\frac{2}{47+3 \sqrt{249}}}+\sqrt[3]{\frac{47+3 \sqrt{249}}{2}} \right)=1.3532099$$
$$\sqrt{2+\sqrt[3]{2+\sqrt[4]{2+\cdots}}}<\sqrt{2+\sqrt[3]{2+1.3532099}}=1.8699639$$
So, we get, at this stage:
$$1.8685<\sqrt{2+\sqrt[3]{2+\sqrt[4]{2+\cdots}}}<1.8700$$
We get the value with uncertainty less than $0.1 \%$ in just a couple of steps.
We won't get a closed form this way, but this is much more accurate, than just truncating the nested radical.
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solve the equation for a: (matrix) = 10
how did he expand the determinant?
I tried using the method where you take the determinent by the top left element times the det of the bottom right 4 minus the top element times the det of the bottom left 2 and bottom right 2 + the top right times the det of the bottom left 4 and I got a = 7... am I allowed to use this method to solve it? Did i just miscalculate? how did this person solve this?
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$$
\begin{vmatrix} 1 & -1 & 4 \\ -1 & a & 2 \\ 1 & -2 & 3\end{vmatrix} = 1*\begin{vmatrix} a & 2 \\ -2 & 3\end{vmatrix} - (-1)*\begin{vmatrix} -1 & 2 \\ 1 & 3\end{vmatrix} + 4*\begin{vmatrix} -1 & a \\ 1 & -2\end{vmatrix} = 10
$$
can you take it from here?
Also use:
$$
\begin{vmatrix} a & b \\ c & d\end{vmatrix} = ad-bc.
$$
note this is just to verify the OP answer.
$$
1*(3a+4) + (-3 - 2) + 4*(2-a) = 3a + 4 - 5 + 8 - 4a = -a +7 = 10
$$
this leads to a = -3
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$n \in \mathbb{N} \ 5|\ 2^{2n+1}+3^{2n+1}$ show for all $n \in \mathbb{N}$,
$$5|\ 2^{2n+1}+3^{2n+1}$$
Indeed,
we've to show that : $2^{2n+1}+3^{2n+1}=0[5] $
note that $2^{2n+1}+3^{2n+1}=2.4^n+3.9^n= $
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You can do it by induction. For $n=0$, we have
$$2^{2n+1}+3^{2n+1}=2+3=5$$
which is divisible by $5$. Now suppose that
$2^{2k+1}+3^{2k+1}$ is divisible by $5$, i.e.
$$\tag{1}2^{2k+1}+3^{2k+1}=5N$$
for some integer $N$. Now consider
$$2^{2(k+1)+1}+3^{2(k+1)+1}=4\cdot 2^{k+1}+9\cdot3^{2k+1}=
4\cdot (2^{k+1}+3^{2k+1})+5\cdot 3^{2k+1}=5(N+3^{2k+1})$$
where the last equality follows from $(1)$, which is divisible by $5$.
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How do you integrate the reciprocal of square root of cosine? I encountered this integral in physics and got stuck.
$$\int_{0}^{\Large\frac{\pi}{2}} \dfrac{d\theta}{\sqrt{\cos \theta}}.$$
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Substituting $y=\cos{\theta}$
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \dfrac{d\theta}{\sqrt{\cos \theta}} &= \int_{0}^{1} \, \frac{1}{\sqrt{y\, \left(1-y^2\right)}}\, dy \\
&= \frac{1}{2}\int_{0}^{1} \, t^{-3/4}\left(1-t\right)^{-1/2}\, dt \tag{where $t=y^2$} \\
&= \frac{1}{2}\mathrm{B}\left(\frac{1}{4}, \frac{1}{2}\right) \\
&= \frac{\sqrt\pi}{2}\frac{\Gamma\left(1/4\right)}{\Gamma\left(3/4\right)} \approx 2.62205755429212
\end{align*}
Look up Beta and Gamma functions.
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Find the last two digits of $9^{{9}^{9}}$ I have to find the last two decimal digits of the number $9^{{9}^{9}}$.
That's what I did:
$$m=100 , \phi(m)=40, a=9$$
$$(100,9)=1, \text{ so from Euler's theorem,we have :} 9^{40} \equiv 1 \pmod{100}$$
$$9^{{9}^{9}} \equiv 9^{9^{2+2+2+2+1}} \equiv 9^{9^2 \cdot 9^2 \cdot 9^2 \cdot 9^2 \cdot 9} \equiv 9^{81 \cdot 81 \cdot 81 \cdot 81 \cdot 9} \equiv 9^{(40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot 9} \equiv (9^{(40+40+1)})^{(40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot 9} \equiv 9^9 \equiv 9^4 \cdot 9^4 \cdot 9 \equiv 6561 \cdot 6561 \cdot 9 \equiv 3721 \cdot 9 \\ \equiv 21 \cdot 9 \equiv 89 \pmod{100}$$
So,the last two digits are $8 \text{ and } 9$.
$$$$But,is there also an other way two calculate the last two digits of $9^{{9}^{9}}$ or is the above the only one?
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You could also just do modular exponentiation and take note of the periods.
The powers of $9 \mod 100$ are: 1, 9, 81, 29, 61, 49, 41, 69, 21, 89, 1, 9, 81, 29, 61, 49, ... They have a period of $10$. This means that if $n \equiv 9 \mod 10$, then $9^n \equiv 89 \mod 100$. Since $9^9 = 387,420,489$, this means that $9^{9^9} \equiv 89 \mod 100$.
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Parabola and line proof Given are three non-zero numbers $a, b, c \in \mathbb{R}$. The parabola with equation $y=ax^2+bx+c$ lies above the line with equation $y=cx$.
Prove that the parabola with equation $y=cx^2-bx+a$ lies above the line with equation $y=cx-b$.
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The parabola with equation $y=ax^2+bx+c$ lies above the line with equation $y=cx$. So
$ax^2 + bx + c = cx \quad \Rightarrow \quad ax^2 + (b - c)x + c = 0 \quad \Rightarrow \quad \Delta = (b - c)^2 - 4ac$.
On the other hand, the parabola with equation $y=cx^2-bx+a$ lies above the line with equation $y=cx - b$. So
$cx^2 - bx + a = cx - b \quad \Rightarrow \quad cx^2 - (b + c)x + a + b = 0 \quad \Rightarrow$
$\Delta = (b + c)^2 - 4(a + b)c = b^2 + 2bc + c^2 - 4ac - 4bc \quad \Rightarrow \quad \Delta = (b - c)^2 - 4ac$
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Solve for x if $z$ is a complex number such that $z^2+z+1=0$ I was given a task to solve this equation for $x$:
$$\frac{x-1}{x+1}=z\frac{1+i}{1-i}$$
for a complex number $z$ such that $z^2+z+1=0$.
Solving this for $x$ is trivial but simplifying solution using the given condition is what's bothering me. Thanks ;)
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$$\frac{x-1}{x+1}=z\frac{1+i}{1-i}=z\frac{(1+i)^2}{(1+i)(1-i)}=zi$$
Applying componendo and dividend, $$x=\frac{1+zi}{1-zi}=\frac{(1+zi)^2}{(1-zi)(1+zi)}=\frac{1-z^2+2iz}{-z}=-2i+z-z^2$$ as $z^2+z+1=0\implies z^3=1$
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Find $\lim\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\ldots+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}\right)$ Find $$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\ldots+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}\right)$$
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From the Hint given by @cameron Williams, I did the following.
$$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{\sqrt{1}-\sqrt{3}}{-2}+\frac{\sqrt{3}-\sqrt{5}}{-2}+\ldots+\frac{\sqrt{2n-1}-\sqrt{2n+1}}{-2}\right)$$
Now the second and consectives gets cancelled. So,
$$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{1}{-2}-\frac{\sqrt{2n+1}}{-2}\right)=\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\sqrt{n}\left(\frac{\sqrt{2+\frac{1}{n}}}{2}\right)=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$$
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What are the members of the set $A=7^{n}+5^{n}(mod35)$ I have this set
$A=${$\ x \ \in \mathbb{N}|\ \exists \ n \ \in \mathbb{N}:$
$x \equiv 7^{n}+5^{n}$ (mod $35$) $ $, $ 35\gt x\ge 0$}
I want to know how many members has this set?
thanks in advance
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By Little Fermat's Theorem, $7^4\equiv 1\pmod5$ and $5^6\equiv1\pmod7$. In fact, $4$ is the order of $7$ modulo $5$ and $6$ is the order of $5$ modulo $7$. For any integer $k$ the number $12k$ is a multiple of $4$ and $6$. Therefore,
$$5^{n+12k}+7^{n+12k}=5^n\cdot(1+7r)+7^n\cdot(1+5r)\equiv 5^n+7^n\pmod{35}$$
We see that we get the same value for $x$ if we add $12$ (or a multiple of $12$) to both exponents. That is, $A$ has at most $12$ elements. But it is possible that the period is smaller. If it were the case, the period must be a divisor of $12$. But
$$5^0+7^0=2$$
$$5^6+7^6\equiv15+14\not\equiv2\pmod{35}$$
$$5^4+7^4\equiv-5+21\not\equiv2\pmod{35}$$
so the period (and, hence, the number of elements in $A$) is exactly $12$.
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Evaluation of $ \int \tan x\cdot \sqrt{1+\sin x}dx$ Calculation of $\displaystyle \int \tan x\cdot \sqrt{1+\sin x}dx$
$\bf{My\; Try::}$ Let $\displaystyle (1+\sin x)= t^2\;,$ Then $\displaystyle \cos xdx = 2tdt\Rightarrow dx = \frac{2t}{\sqrt{2-t^2}}dt$
So Integral is $\displaystyle = \displaystyle 2\int \frac{t^2}{\sqrt{2-t^2}} \frac{(t^2-1)}{\sqrt{2-t^2}}dt = 2\int\frac{t^4-t^2}{2-t^2}dt $
Now How Can I solve after that
Help me
Thanks
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Hint: $\frac{t^4-t^2}{2-t^2} \equiv -t^2-\frac{2}{t^2-2}-1,$ which is easy to integrate.
You should find, after a suitable hyperbolic substitution, that $\int \frac{2}{t^2-2}dt= \frac{1}{\sqrt{2}} \ln \left| \frac{\sqrt{2}-t}{\sqrt{2}+t} \right|$.
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Find the equation of a circle which is tangent to $y$-axis at a given point and cuts a chord of given length on the $x$-axis How to find the equation of the circle which touches $y$ axis at $(0,3)$ and cuts a chord of length $8$ on the $x$ axis?
It should look like this:
My approach:
Since the circle touches $y$ axis at $(0,3)$, its center has $y$-coordinate $3$. So the equation of the circle is of the form $(x-r)^2+(y-3)^2=r^2$.
How can I proceed further by using the fact that the circle passes through $(a,0)$ and $(b,0)$ with $b-a=8$? Anyway this would be very long. Is there some alternative?
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$\begin{vmatrix}
x^2+y^2&x&y&1\\
0^2+3^2&0&3&1\\
1^2+0^2&1&0&1\\
9^2+0^2&9&0&1\\
\end{vmatrix}
=
\begin{vmatrix}
x^2+y^2&x&y&1\\
9&0&3&1\\
1&1&0&1\\
81&9&0&1\\
\end{vmatrix}=
24(x^2+y^2)-240x-144y+216=
x^2+y^2-10x-6y+9=0$
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Calculation of Trigonometric Limit with Summation. If $\displaystyle f(x)=\lim_{n\rightarrow \infty}\sum_{r=0}^{n}\left\{\frac{\tan \left(\frac{x}{2^{r+1}}\right)+\tan^3\left(\frac{x}{2^{r+1}}\right)}{1-\tan^2\left(\frac{x}{2^{r+1}}\right)}\right\}.$ Then value of $f(x)$ is
$\bf{My\; Try::}$ Let $\displaystyle \left(\frac{x}{2^{r+1}}\right)=y$. So Trigonometric Expression is
$\displaystyle \frac{\tan y\cdot \left(1+\tan^2 y\right)}{\left(1-\tan^2 y\right)}=\frac{\tan y}{\cos 2y}=\frac{\sin y}{\cos 2y \cdot \cos y} = \frac{1}{2}\left\{\frac{\sin (2y-y)}{\cos 2y \cdot \cos y}\right\} = \frac{1}{2}\left\{\tan (2y)-\tan (y)\right\}$
Now How Can I solve after that
Help me
Thanks
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$\displaystyle 2f(x)=2\sum_{r=0}^{n}\left\{\frac{\tan \left(\frac{x}{2^{r+1}}\right)+\tan^3\left(\frac{x}{2^{r+1}}\right)}{1-\tan^2\left(\frac{x}{2^{r+1}}\right)}\right\}.$
$\displaystyle=\sum_{r=0}^{n}\left\{\tan \frac x{2^r}-\tan \frac x{2^{r+1}}\right\}$ which is telescopic
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Finding the remainder when a polynomial is divided by another polynomial. Find the remainder when $x^{100}$ is divided by $x^2 - 3x + 2$.
I tried solving it by first calculating the zeroes of $x^2 - 3x + 2$, which came out to be 1 and 2.
So then, using the Remainder Theorem, I put both their values, and so the remainder came out to be $1 + 2^{100}$.
But the correct answer is $(2^{100} - 1)x + (2 - 2^{100})$.
Can you please explain the exact process to reach the solution?
Thanks in advance. :)
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We write the Euclidean division:
$$x^{100}=(x^2-3x+2)Q(x)+ax+b$$
and notice that $1$ and $2$ are roots of $x^2-3x+1$ so
*
*let $x=1$ we get $1=a+b$
*let $x=2$ we get $2^{100}=2a+b$
so we find $a=2^{100}-1$ and $b=2-2^{100}$.
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Eliminate $\theta$ from the equations $\frac{\cos(\alpha-3\theta)}{\cos^3\theta}=\frac{\sin(\alpha-3\theta)}{\sin^3\theta}=m$
Eliminate $\theta$ from the equations
$$\frac{\cos(\alpha-3\theta)}{\cos^3\theta}=\frac{\sin(\alpha-3\theta)}{\sin^3\theta}=m$$
Ans: $m^2+m\cos\alpha-2=0$.
I tried using the following two identities:
$$\cos(3\theta)=4\cos^3(\theta)-3\cos(\theta)$$
$$\sin(3\theta)=3\sin(\theta)-4\sin^3(\theta)$$
but these didn't help much. I am sure that this is a simple problem but I am unable to figure out the right approach to solve it. :(
Any help is appreciated. Thanks!
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Using $\displaystyle\sin(A-B),\cos(A-B)$ and on rearrangement we have $$\cos\alpha\cos3\theta+\sin\alpha\sin3\theta-m\cos^3\theta=0\ \ \ \ (1)$$
$$\cos\alpha\sin3\theta-\sin\alpha\cos3\theta+m\sin^3\theta=0\ \ \ \ (2)$$
Solving for $\displaystyle\cos\alpha,\sin\alpha,$
$\displaystyle\dfrac{\cos\alpha}m=\cos^3\theta\cos3\theta-\sin^3\theta\sin3\theta$
$\displaystyle=\cos^3\theta(4\cos^3\theta-3\cos\theta)-\sin^3\theta(3\sin\theta-4\sin^3\theta)$
$\displaystyle=4(\cos^6\theta+\sin^6\theta)-3(\cos^4\theta+\sin^4\theta)$
$\displaystyle=4\{(\cos^2\theta+\sin^2\theta)^3-3(\cos^2\theta\sin^2\theta)(\cos^2\theta+\sin^2\theta)\}-3\{(\cos^2\theta+\sin^2\theta)^2-2\cos^2\theta\sin^2\theta\}$
$\displaystyle\implies\dfrac{\cos\alpha}m=1-6(\sin\theta\cos\theta)^2\ \ \ \ (3)$
Similarly, $\displaystyle\dfrac{\sin\alpha}m=-3(\sin\theta\cos\theta)\ \ \ \ (4)$
Compare the values of $(\sin\theta\cos\theta)^2$ from $(3),(4)$ to eliminate $\theta$ and simplify
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What is the sum of this series: $\displaystyle\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$ I tried getting it into a closed form but failed. Could someone help me out?
$$\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$$
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You should know the Taylor series for $e^x$:
$e^x = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \cdots$
Then, the Taylor series for $e^{-x}$ is:
$e^{-x} = 1 + (-x) + \dfrac{(-x)^2}{2} + \dfrac{(-x)^3}{3!} + \dfrac{(-x)^4}{4!} + \cdots$
$e^{-x} = 1 - x + \dfrac{x^2}{2} - \dfrac{x^3}{3!} + \dfrac{x^4}{4!} - \cdots$
Add these two series together, and all the odd terms cancel. So you get:
$e^x+e^{-x} = 2 + 2\dfrac{x^2}{2} + 2\dfrac{x^4}{4!} + \cdots$
$\dfrac{e^x+e^{-x}}{2} = 1 + \dfrac{x^2}{2} + \dfrac{x^4}{4!} + \cdots$
Now, plug in $x = \sqrt{a}$, and see what you get.
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Evaluation of $\int\sqrt[4]{\tan x}dx$ Evaluation of $\displaystyle \int\sqrt[4]{\tan x}dx$
$\bf{My\; Try}::$ Let $\tan x = t^4\;\;,$ Then $\sec^2 xdx = 4t^3dt$. So $\displaystyle dx = \frac{4t^3}{1+t^8}dt$
So Integral Convert into $\displaystyle 4\int\frac{t^4}{1+t^8}dt = 2\int \frac{(t^4+1)+(t^4-1)}{t^8+1}dt$
So Integral is $\displaystyle 2\int\frac{t^4+1}{t^8+1}dt+2\int\frac{t^4-1}{t^4+1}dt$
Now How can I solve after that
Help me
Thanks
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In general, with $t=\sqrt[n]{\tan x}$
\begin{align}
&\int\sqrt[n]{\tan x}\>dx=n\int \frac{t^n}{1+t^{2n}}dt\\
=&\>\frac14
\sum_{k=1}^{2n} (-1)^{k-1}\left(\sin a_k \ln(x^2-2x\cos a_k+1)-2\cos a_k \tan^{-1}\frac{\sin a_k}{x-\cos a_k}\right) +C
\end{align}
where $a_k=\frac{(2k-1)\pi}{2n}$.
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Prove $\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\leq{3\over2}(a+b+c)$ with $a,b,c$ are nonnegative Hope someone can help on this inequality using nonanalytical method (i.e. simple elementary method leveraging basic inequalities are prefered).
Prove $\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\leq{3\over2}(a+b+c)$ with $a,b,c$ are nonnegative.
Here is what I already got.
First of all, one should notice equality holds not when $a=b=c$ as one might initially thought. Rather, the equality holds when $(a,b,c)$ is a permutation of $({1\over2},{1\over2},0)$.
Secondly and obviously, this is cyclical and homgeneous and hence we can apply the EMV theorem developed by a IMO golden medalist (reference here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1130901) and then the original inequality can be easily proved by assuming $$f(a,b,c)={\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\over {a+b+c}},$$and then prove $$f(1,1,1)\leq{3\over2},\\f(a,b,0)\leq{3\over2},\forall a,b\geq0.$$
But this kind of proof does not fit my appetite as it not only involves some additional theorem but also not very nice as the simple nice form of the question of itself.
So can people help on some elementary proof that might only taking use of the basic inequalities like AM-GM, Jensen's inequality, etc.
|
By C-S
$$\left(\sum_{cyc}\sqrt{a^2+bc}\right)^2\leq\sum_{cyc}\frac{a^2+bc}{2a^2+b^2+c^2+bc}\sum_{cyc}(2a^2+b^2+c^2+bc).$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{a^2+bc}{2a^2+b^2+c^2+bc}\sum_{cyc}(4a^2+bc)\leq\frac{9(a+b+c)^2}{4}$$ or
$$\sum_{cyc}(2a^8+2a^7b+2a^7c+9a^6b^2+9a^6c^2-23a^5b^3-23a^5c^3+20a^4b^4+11a^6bc+$$
$$+70a^5b^2c+70a^5c^2b+79a^4b^3c+79a^4c^3b+263a^4b^2c^2-195a^3b^3c^2)\geq0,$$
which is obvious because by Schur and Muirhead we obtain:
$$\sum_{cyc}(2a^8+2a^7b+2a^7c+9a^6b^2+9a^6c^2-23a^5b^3-23a^5c^3+20a^4b^4+11a^6bc+$$
$$+70a^5b^2c+70a^5c^2b+79a^4b^3c+79a^4c^3b+263a^4b^2c^2-195a^3b^3c^2)\geq$$
$$\geq\sum_{cyc}(4a^7b+4a^7c+9a^6b^2+9a^6c^2-23a^5b^3-23a^5c^3+20a^4b^4)=$$
$$=\sum_{cyc}ab(4a^6+9a^5b-23a^4b^2+20a^3b^3-23a^2b^4+9ab^5+4b^6)=$$
$$=\sum_{cyc}ab(a-b)^2(4a^4+17a^3b+7a^2b^2+17ab^3+4b^4)\geq0.$$
Done!
|
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|
Solve The Triangle I am having a tough time trying to solve this problem.
I have utilized the 30, 60, 90 triangle measures for the length of sides. However, I am stuck since the side that would be √3 has 100 as its length. How do I solve then?
This is what I have done so far:
|
To find $x$ here, you will need to look at this as 2 separate triangles, and find the base of each separate triangle, then add the bases up to get $x$.
For the triangle on the left hand side, we can use the $\tan(x)$ function to relate the $60^{\circ}$ angle with the opposite side of length 100, and our adjacent side, which we are looking for (let's call it $y$).
So, we know $\tan(x) = \dfrac{\text{opp}}{\text{adj}}$, and this gives $\tan(60^{\circ}) = \dfrac{100}{y}$. But $\tan(60^{\circ}) = \sqrt{3}$, so we have:
$\sqrt{3} = \dfrac{100}{y}$, and solving for $y$ gives $y = \dfrac{100}{\sqrt{3}}$.
Now, we need to solve for the adjacent side of the triangle on the right-hand side using the angle $30^{\circ}$. Again, we want to relate the opposite side of $30^{\circ}$ to its adjacent side (the unknown -- we can call it $z$). We will have to use $\tan(x)$ again. So we get:
$\tan(30^{\circ}) = \dfrac{\text{opp}}{\text{adj}} = \dfrac{100}{z}$. But, $\tan(30^{\circ}) = \dfrac{1}{\sqrt{3}}$, so we have the equation:
$\dfrac{1}{\sqrt{3}} = \dfrac{100}{z}$ and solving for $z$ gives:
$z = \dfrac{100}{\frac{1}{\sqrt{3}}} = 100\sqrt{3}$.
Finally, $x = y + z$, since $x$ is the length of both adjacent sides put together, so:
$x = \dfrac{100}{\sqrt{3}} + 100\sqrt{3} = \dfrac{100}{\sqrt{3}} + \dfrac{100\sqrt{3}}{1}*\dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{100 + 100*3}{\sqrt{3}} = \dfrac{400}{\sqrt{3}}$, which is your final answer.
Here is the picture to help you see what I am doing:
|
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|
Proving $\int_0^1 \frac{\mathrm{d}x}{1-\lfloor \log_2(1-x)\rfloor} = 2 \log 2 - 1$. By testing in maple I found that
$$
\int_0^1 \frac{\mathrm{d}x}{1-\lfloor \log_2(1-x)\rfloor} = 2 \log 2 - 1
$$
Does there exists a proof for this? I tried rewriting it as an series but no luck there.
$$
\frac{1}{1-\lfloor \log_2(1-x)\rfloor}
= \sum_{k=0}^\infty \lfloor \log_2(1-x) \rfloor^k
= \sum_{k=0}^\infty \Big \lfloor -\frac{1}{\log 2} \sum_{n=0}^\infty \frac{x^n}{n} \Big\rfloor^k
$$
However this expression seems to diverge
|
Start by noting that $\lfloor\log_2(1-x)\rfloor = -(n+1)$ for all $x \in (1-2^{-n},1-2^{-(n+1)})$.
Therefore, $\displaystyle\int_0^1 \dfrac{\,dx}{1-\lfloor \log_2(1-x)\rfloor} = \sum_{n = 0}^{\infty}\int_{1-2^{-n}}^{1-2^{-(n+1)}} \dfrac{\,dx}{1-\lfloor \log_2(1-x)\rfloor}$
$= \displaystyle\sum_{n = 0}^{\infty}\int_{1-2^{-n}}^{1-2^{-(n+1)}} \dfrac{\,dx}{1+(n+1)} = \sum_{n = 0}^{\infty}\dfrac{2^{-(n+1)}}{n+2}$.
To evaluate this summation, start with the geometric series $\dfrac{x}{1-x} = \displaystyle\sum_{n = 0}^{\infty}x^{n+1}$.
Integrate both sides to get $-x-\ln(1-x) = \displaystyle\sum_{n = 0}^{\infty}\dfrac{x^{n+2}}{n+2}$.
Divide both sides by $x$ to get $-1-\dfrac{\ln(1-x)}{x} = \displaystyle\sum_{n = 0}^{\infty}\dfrac{x^{n+1}}{n+2}$.
Then, plug in $x = \dfrac{1}{2}$ to get $\displaystyle\sum_{n = 0}^{\infty}\dfrac{2^{-(n+1)}}{n+2} = -1-\dfrac{\ln(1-\tfrac{1}{2})}{\tfrac{1}{2}} = 2\ln 2 - 1$.
|
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|
The closed form of $\int_0^{\pi/4}\frac{\log(1-x) \tan^2(x)}{1-x\tan^2(x)} \ dx$ What tools or ways would you propose for getting the closed form of this integral?
$$\int_0^{\pi/4}\frac{\log(1-x) \tan^2(x)}{1-x\tan^2(x)} \ dx$$
EDIT: It took a while since I made this post. I'll give a little bounty for the solver of the problem, 500 points bounty.
Supplementary question:
Calculate
$$\int_0^{\pi/4}\frac{\log(1-x)\log(x)\log(1+x) \tan^2(x)}{1-x\tan^2(x)} \ dx$$
|
Just a few notes for a series development, because a "closed" formula is very unlikely.
$$\int\limits_0^{\pi/4} \frac{\ln(1-x) \tan^2 x}{1-x\tan^2 x} dx = -\sum\limits_{n=0}^\infty \sum\limits_{k=0}^n \frac{1}{n-k+1} \int\limits_0^{\pi/4} x^{n+1} (\tan x)^{2k+2} dx$$
Or using $\int\limits_0^{\pi/4} \frac{\ln(1-x)}{x (1-x\tan^2 x)} dx$ it becomes a little more handsomely:
$$ \int\limits_0^{\pi/4} \frac{\ln(1-x) \tan^2 x}{1-x\tan^2 x} dx = \text{Li}_2\left(\frac{\pi}{4}\right) -\sum\limits_{n=0}^\infty \sum\limits_{k=0}^n \frac{1}{n-k+1} \int\limits_0^{\pi/4} x^n (\tan x)^{2k} dx$$
Then we have: $$ \int\limits_0^{\pi/4} x^n (\tan x)^{2k} dx = (-1)^k \left(\frac{\pi^{n+1}}{(n+1)4^{n+1}} + A_{n,k} - B_{n,k} - C_{n,k} \right) \enspace$$ for:
$$A_{n,k} := \left(\sin \frac{\pi n}{2}\right)\frac{n!}{2^{n+1}} \sum\limits_{~j=0 \\ j~\text{odd}}^{2k-1} c_{2k-1,j}~\eta(n-j+1)\\
B_{n,k} := \frac{1}{2^{2n+2}}\sum\limits_{~v=0 \\ v~\text{odd}}^n \left(\sin \frac{\pi v}{2}\right)\frac{n!\pi^{n-v}}{(n-v)!} \sum\limits_{~j=0 \\ j~\text{odd}}^{2k-1} 2^j c_{2k-1,j}~\eta(v-j+1)\\
C_{n,k} := \frac{1}{2^{2n+1}}\sum\limits_{~v=0 \\ v~\text{even}}^n \left(\cos \frac{\pi v}{2}\right)\frac{n!\pi^{n-v}2^v}{(n-v)!} \sum\limits_{~j=0 \\ j~\text{odd}}^{2k-1} c_{2k-1,j}~\beta(v-j+1)\\
c_{n,j} := \frac{1}{n!} \sum\limits_{v=0}^{n+1} {\binom {n+1} v} \sum\limits_{l=j}^n \begin{bmatrix}{n+1}\\{l+1}\end{bmatrix}{\binom l j}(-v)^{l-j}$$
and the Stirling numbers of the first kind $\begin{bmatrix}n\\k\end{bmatrix}$ defined by: $$\sum\limits_{k=0}^n\begin{bmatrix}n\\k\end{bmatrix}x^k:=\prod\limits_{k=0}^{n-1}(x+k),$$
so that $ c_{ \text{odd},\text{even} }=0 $ and $ c_{ \text{even},\text{odd} }=0$. Some values $\,c_{n,j}\,$ can be seen here .
The needed analytical continuation for Dirichlet eta function $\eta(s)$ and Dirichlet beta function $\beta(s)$ can be seen in my answer of the question here . With the additional information
$$ \sum\limits_{n=0}^\infty\frac{\pi^{n+1}}{(n+1) 4^{n+1}}\sum\limits_{k=0}^n \frac{(-1)^k}{n-k+1} = \\ \text{Li}_2\left(\frac{\pi}{4}\right) + \text{Li}_2 \left(\frac{1}{2} -\frac{\pi}{8} \right) + \left(\ln\left(\frac{1}{2} +\frac{\pi}{8} \right)\right)\ln\left(1-\frac{\pi}{4}\right)$$
we get:
$$ \int\limits_0^{\pi/4} \frac{\ln(1-x) \tan^2 x}{1-x\tan^2 x} dx = \\
-\text{Li}_2 \left(\frac{1}{2} -\frac{\pi}{8} \right) - \left(\ln\left(\frac{1}{2} +\frac{\pi}{8} \right)\right)\ln\left(1-\frac{\pi}{4}\right) - \sum\limits_{n=0}^\infty \sum\limits_{k=0}^n \frac{(-1)^k}{n-k+1}(A_{n,k} - B_{n,k} - C_{n,k} )$$
|
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|
Homework - Resolve the recurrence relation What's the closed formula of this recurrence relation?
$$a_n = a_{n-1}+2a_{n-2}+2^n \text{ with } a_0=1, a_1=2 $$
|
This is nonhomogeneneous difference equations. First solve the homogeneous equation
$$
a_n - a_{n-1} - 2a_{n-2} = 0 \quad (1)
$$
Let $a_n = r^n$, so that $a_{n-1} = r^{n-1}$ and $r_{n-2} = r^{n-2} = 0$. Replacing in (1), we have
$$
r^n - r^{n-1} - 2r^{n-2} = 0 \quad r^2 - r - 2 = 0 \quad \Rightarrow \ r_1 = -1, \ r_2 = 2
$$
$$
a_{nh} = C_1(-1)^n + C_22^n
$$
Particular solution: Let $a_{np} = a_n = An2^n$. The presence of n is due to the fact that $2$ is a root of the equation (1). So, $a_{n-1} = A(n-1)2^{n-1}$ and $a_{n-2} = A(n-2)2^{n-2}$. Substituting in the given equation, we have
$$
An2^n + (An - A)2^{n-1} - 2A(n-2)2^{n-2} = 2^n \quad \Rightarrow \quad A = \frac{2}{3}
$$
Thus, $a_n = a_{nh} + a_{np} = C_1(-1)^n + C_22^n + \frac{2n2^n}{3}$. But, $a_0 = a(0) = 1$ and $a_1 = a(1) = 2$. We have the system
$$
\begin{cases}
1 = a_0 = C_1 + C_2\\
2 = a_1 = -C_1 + C_2 + \frac{8}{3}
\end{cases}
\quad \Rightarrow \quad C_1 = 8/9, \quad C_2 = 1/9
$$
Thus,
$$
a_n = \frac{8(-1)^n}{9} + \frac{1}{9}2^n + \frac{2n2^n}{3}
$$
|
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|
Verify algebraically that the equation $\frac{\cos(x)}{\sec(x)\sin(x)}=\csc(x)-\sin(x)$ is an identity I am stuck when I get to this point $\frac{\cos^2(x)}{\sin(x)}$.
Am I on the right track?
Verify algebraically that the equation is an identity:
$$\frac{\cos(x)}{\sec(x)\sin(x)}=\csc(x)-\sin(x)$$
My work:
$$\frac{\cos(x)}{\frac{1}{\cos(x)}\cdot\sin(x)}=\frac{\cos(x)}{\frac{\sin(x)}{\cos(x)}}=\frac{\cos(x)}{1}\cdot\frac{\cos(x)}{\sin(x)}=\frac{\cos^2(x)}{\sin(x)}$$
|
General hints:
Agree on only using $\sin$ and $\cos$ by substituting
$$\csc(x) = \frac1{\sin(x)}, \qquad \sec(x) = \frac1{\cos(x)}$$
Then get rid of any fractions by multiplying with the denominator and finally use
$$\sin^2(x) + \cos^2(x) = 1$$
$$\begin{align*}
\frac{\cos x}{\sec x \sin x} & = \frac{\cos x}{\frac1{\cos x}\sin x} = \frac{\cos^2 x}{\sin x} \\
& = \frac{1-\sin^2 x}{\sin x} = \frac1{\sin x} - \sin x\\
& = \csc x - \sin x
\end{align*}$$
|
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|
Factor $3x^2-11xy+6y^2-xz-4yz-2z^2$ This problem is from my Math Challenge II Algebra class, and it's really confusing. How can you factor something like this? Here's the question again: Factor $3x^2-11xy+6y^2-xz-4yz-2z^2$.
|
First, split up the expression into pieces.
\begin{align*}
F(x,y,z) &= 3x^2-11xy+6y^2-xz-4yz-2z^2 \\
&= 3x^2\underbrace{-11xy-xz}_{G(x,y,z)}+\underbrace{6y^2-4yz-2z^2}_{H(y,z)}
\end{align*}
Start with the easy one : $H(y,z)$.
\begin{align*}
H(y,z) &= 6y^2-4yz-2z^2 \\
&= 6y^2-6yz+2yz-2z^2 \\
&= 6y(y-z)+2z(y-z) \\
&= 2(y-z)(3y+z)
\end{align*}
Now, the tuff one : $G(x,y,z)$. Remember, we want to find some factors like those in $H(x,y)$. So, we have to use our imagination...
\begin{align*}
G(x,y,z) &= -11xy-xz \\
&= -3x(3y+z)-2xy+2xz \\
&= -3x(3y+z)-2x(y-z) \\
\end{align*}
Finish, with all the pieces together.
\begin{align*}
F(x,y,z) &= 3x^2 + G(x,y,z) + H(y,z) \\
&= 3x^2-3x(3y+z)-2x(y-z)+2(y-z)(3y+z) \\
&= (3y+z)[2(y-z)-3x]-x[2(y-z)-3x] \\
&= [2(y-z)-3x][(3y+z)-x] \\
&= (3x-2y+2z)(x-3y-z)
\end{align*}
|
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|
Show that this expression is a perfect square? Show that this expression is a perfect square?
$(b^2 + 3a^2 )^2 - 4 ab*(2b^2 - ab - 6a^2)$
|
In fact, note that $(x + y - z)^2 = x^2 + y^2 + z^2 + 2xy - 2yz - 2xz$. Thus,
$$
(3a^2 + b^2)^2 - 8ab^3 + 4a^2b^2 + 24a^3b = 9a^4 + 10a^2b^2 + b^4 - 8ab^3 + 24a^3b
$$
$$
= 9a^4 + (16a^2b^2 - 6a^2b^2) + b^4 + 24a^3b - 8ab^3
$$
$$
= 9a^4 + 16a^2b^2 + b^4 + 24a^3b - 6a^2b^2 - 8ab^3
$$
$$
= (3a^2)^2 + (4ab)^2 + (-b^2)^2 + 2(3a^2)(4ab) + 2(3a^2)(-b^2) + 2(4ab)(-b^2)
$$
$$
= (3a^2 + 4ab - b^2)^2
$$
|
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|
How prove that $ \sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}=\sqrt[3]{\sqrt[3]2-1} $ How check that $ \sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}=\sqrt[3]{\sqrt[3]2-1} $?
|
If we abbreviate $w=\sqrt[3]2$, the left hand side is $L=\frac1{\sqrt[3]9}(1-w+w^2)$. From $(1+w)(1-w+w^2)=1+w^3=3$ we see that $L=\frac3{\sqrt[3]9(1+\sqrt[3]2)}$, hence
$$ L^3=\frac{27}{9(1+w)^3}=\frac3{1+3w+3w^2+w^3}=\frac1{1+w+w^2}$$
As above, note that $(1+w+w^2)(w-1)=w^3-1=1$, hence
$$ L^3=w-1=R^3.$$
|
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|
How to evaluate the limit $\lim_{x \to \infty} \frac{2^x+1}{2^{x+1}}$ How to evaluate the limit as it approaches infinity
$$\lim_{x \to \infty} \frac{2^x+1}{2^{x+1}}$$
|
Hint:
$\displaystyle \frac{2^x + 1}{2^{x+1}} = \frac{2^x}{2^{x+1}} + \frac{1}{2^{x+1}} = \frac{1}{2} + \frac{1}{2^{x+1}}$
|
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|
Trigonometric relation between sides and angles of a triangle $$a \cdot \sin (B-C) +b \cdot \sin(C-A) +c \cdot \sin(A-B) =0$$
where $a, b, c$ are the sides of a triangle and $A, B, C$ are the angles of a triangle.
No idea how to solve this problem.
|
$$
a \sin(B-C) + b \sin(C-A) + c \sin(A-B)\\
= a ( \sin B \cos C - \sin C \cos B) + b(\sin C \cos A - \sin A \cos C) + c (\sin A \cos B - \sin B \cos A)\\
= a \sin B \cos C - a \sin C \cos B + b\sin C \cos A - b \sin A \cos C + c \sin A \cos B - c \sin B \cos A\\
= a \sin B \cos C - b \sin A \cos C + b \sin C \cos A - c \sin B \cos A + c \sin A \cos B - a \sin C \cos B
$$
The sine rule says that
$$
\frac{\sin A }{a} = \frac{\sin B }{b} = \frac{\sin C}{c}
$$
So we have:
$$
b\sin A = a\sin B \\
c\sin B = b\sin C\\
a\sin C = c \sin A
$$
Substituting this into the final expression from above gives:
$$
a \sin(B-C) + b \sin(C-A) + c \sin(A-B)\\
= a \sin B \cos C - b \sin A \cos C + b \sin C \cos A - c \sin B \cos A + c \sin A \cos B - a \sin C \cos B \\
= a \sin B \cos C - a \sin B \cos C + b\sin C \cos A - b \sin C \cos A + c\sin A \cos B - c \sin A \cos B\\
=0
$$
|
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|
Simplify $\frac {\sqrt5}{\sqrt3+1} - \sqrt\frac{30}{8} + \frac {\sqrt {45}}{2}$ I am trying to find the value of: $$\frac {\sqrt5}{\sqrt3+1} - \sqrt\frac{30}{8} + \frac {\sqrt {45}}{2}$$
I have the key with the answer $\sqrt 5$ but am wondering how I can easily get to that answer?
I realize this is a very basic question on radicals and I know that square roots can be separated into parts, for instance $\sqrt {45} = \sqrt 9 \cdot \sqrt 5$ and I can see how that would be useful here in canceling out radicals but I have not yet been able to reach a solution using this method.
|
$$\begin{align}
\frac{\sqrt{5}}{\sqrt{3}+1} - \sqrt{\frac{30}{8}} + \frac{\sqrt{45}}{2} &= \frac{\sqrt{5}(\sqrt{3}-1)}{(\sqrt{3}+1)((\sqrt{3}-1))} - \sqrt{\frac{15}{4}} + \frac{\sqrt{9·5}}{2}\\
&= \frac{\sqrt{5}(\sqrt{3}-1)}{2} - \frac{\sqrt{15}}{2} + \frac{3\sqrt{5}}{2}\\
&= \frac{\sqrt{15}-\sqrt{5} -\sqrt{15} + 3\sqrt{5}}{2}\\
&= \sqrt{5}
\end{align}$$
|
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|
Why does this infinite series equal one? $\sum_{k=1}^\infty \binom{2k}{k} \frac{1}{4^k(k+1)}=1$ Why does
$$\sum_{k=1}^\infty \binom{2k}{k} \frac{1}{4^k(k+1)}=1$$
Is there an intuitive method by which to derive this equality?
|
Your series is a telescoping one, since:
$$\begin{eqnarray*}\frac{1}{4^{k+1}}\binom{2k+2}{k+1}-\frac{1}{4^k}\binom{2k}{k}&=&\frac{1}{4^{k+1}}\binom{2k}{k}\left(\frac{(2k+2)(2k+1)}{(k+1)^2}-4\right)\\&=&-\frac{1}{2(k+1)4^{k}}\binom{2k}{k},\end{eqnarray*}$$
hence:
$$\sum_{k=1}^{+\infty}\binom{2k}{k}\frac{1}{4^k(k+1)}=2\sum_{k=1}^{+\infty}\left(\frac{1}{4^k}\binom{2k}{k}-\frac{1}{4^{k+1}}\binom{2k+2}{k+1}\right)=\frac{2}{4}\binom{2}{1}=1.$$
No need of generating functions or integrals.
|
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|
How are these two integrals related? How to express the integral $$\int_{-2}^{2} (x-3) \sqrt{4-x^2} \ dx $$
in terms of the integral $$ \int_{-1}^{1} \sqrt{1-x^2} \ dx?$$ I know that the latter integral is equal to $\pi / 2$.
We can't use substitution. We can only use the following two results:
*
*$$ \int_{a}^{b} f(x) \ dx = \int_{a+c}^{b+c} f(x-c) \ dx.$$
And
*
*$$ \int_{a}^{b} f(x) \ dx = \frac{1}{k} \int_{ka}^{kb} f(\frac{x}{k}) \ dx$$ for any $k \neq 0$.
|
Using $k=\frac{1}{2}$ in the second rule, we get
$$\begin{align}
\int\limits_{-2}^{2} (x-3) \sqrt{4-x^2} \ dx &=2\int\limits_{-1}^{1} (2x-3) \sqrt{4-4x^2} \ dx
\\&=2\int\limits_{-1}^{1} 2(2x-3) \sqrt{1-x^2} \ dx
\\&=4\int\limits_{-1}^{1} (2x-3) \sqrt{1-x^2} \ dx
\\&=8\int\limits_{-1}^{1}x\sqrt{1-x^2} \ dx-12\int\limits_{-1}^{1}\sqrt{1-x^2}\ dx
\\&=-12\int\limits_{-1}^{1}\sqrt{1-x^2}\ dx
\end{align}
$$
where the last equality occurs because $x\sqrt{1-x^2}$ is an odd function and we are integrating from $-1$ to $1$, so $\int_{-1}^{1}x\sqrt{1-x^2} \ dx=0$.
|
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|
Vectors with given angle and magnitude Give an example of vectors $\mathbf{v}$ and $\mathbf{w}$ such that the angle between $\mathbf{v}$ and $\mathbf{w}$ is $\frac{2\pi}{3}$ and $\|\mathbf{v} \text{ x } \mathbf{w}\|=\sqrt{3}$.
Should I just use something random for $\mathbf{v}$ and then solve for $\mathbf{w}$?
Thank you.
|
Let $v = (1,0,0)$ and $w = (a,b,0)$.
$$
v\cdot w = \mid\mid v \mid\mid.\mid\mid w\mid\mid\cos \dfrac{2\pi}{3} \quad \Rightarrow \quad -2a = \sqrt{a^2 + b^2} \ \Rightarrow \ 3a^2 = b^2 \quad (1)
$$
But,
$$
\sqrt{3} = \mid\mid v \mid\mid.\mid\mid w\mid \mid \sin \dfrac{2\pi}{3} \quad \Rightarrow \quad \sqrt{3} = \sqrt{a^2 + b^2}.\dfrac{\sqrt{3}}{2} \quad \Rightarrow \quad a^2 + b^2 = 4 \quad (2)
$$
From (1) and (2), $4a^2 = 4 \ \Rightarrow \ a=\pm 1$ and $b = \pm \sqrt{3}$. Thus,
$$
v = (1,0,0) \quad \text{and} \quad w = (-1,\sqrt{3},0)
$$
So that the angle between $v$ and $w$ is $2\pi/3$, we must have $a = -1$
|
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|
Prove that $\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$
Given $a$, $b$ and $c$ are positive real numbers. Prove that:$$\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$$
Additional info: We can't use induction. We should mostly use Cauchy inequality. Other inequalities can be used rarely.
Things I have done so far: The inequality look is similar to Nesbitt's inequality.
We could re-write it as: $$\sum \limits_{cyc}\frac {a}{(b+c)^2}(2(a+b+c)) \geq \frac{9}{2}$$
Re-write it again:$$\sum \limits_{cyc}\frac {a}{(b+c)^2}\sum \limits_{cyc}(b+c) \geq \frac{9}{2}$$
Cauchy appears:
$$\sum \limits_{cyc}\frac {a}{(b+c)^2}\sum \limits_{cyc}(b+c) \geq \left(\sum \limits_{cyc}\sqrt\frac{a}{b+c}\right)^2$$
So, if I prove $\left(\sum \limits_{cyc}\sqrt\frac{a}{b+c}\right)^2 \geq \frac {9}{2}$ then problem is solved.
Re-write in semi expanded form:$$2\left(\sum \limits_{cyc}\frac{a}{b+c}+2\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)}\right) \geq 9$$
We know that $\sum \limits_{cyc}\frac{a}{b+c} \geq \frac {3}{2}$.So$$4\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)} \geq 6$$
So the problem simplifies to proving this $$\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)} \geq \frac{3}{2}$$
And I'm stuck here.
|
Let $x=b+c,y=c+a,z=a+b$, then $a=\frac{y+z-x}{2},\dots$. Your inequality becomes
$$(x+y+z)\left(\frac{y+z-x}{x^2}+\cdots\right)\geq 9.$$
Write $y+z-x=(x+y+z)-2x,\dots$ we need to show
$$(x+y+z)^2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right) -2(x+y+z)\left(\frac1x+\frac1y+\frac1z\right)\geq 9.$$
Use $3\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{x^2}\right)\geq \left(\frac1x+\frac1y+\frac1z\right)^2$, we only need to show
$$\frac13(x+y+z)^2\left(\frac1x+\frac1y+\frac1z\right)^2-2(x+y+z)\left(\frac1x+\frac1y+\frac1z\right)\geq 9.$$
The last inequality is correct because $(x+y+z)\left(\frac1x+\frac1y+\frac1z\right)\geq 9$.
|
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|
How does $x^3 - \sin^3 x$ become $x^3 + \frac{1}{4}\sin{3x}-\frac{3}{4}\sin x$? I was going through answers on this question and came across this answer and I was wondering how the user arrived at the first line where they state:
$$f(x) \equiv x^3 - \sin^3 x = x^3 + {1 \over 4} \,\sin {3x} - {3 \over 4}\,\sin x$$
How does $x^3 - \sin^3 x$ become $x^3 + \frac{1}{4}\sin{3x}-\frac{3}{4}\sin x$?
Are they using some simple identity or is there some other observation happening?
Thanks!
|
Hint:
$$\sin^3(x) = \left( \frac{e^{ix}-e^{-ix}}{2i} \right)^3 = \dots\,. $$
|
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|
A certain “harmonic” sum Is there a simple, elementary proof of the fact that:
$$\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)=0$$
I have thought of a very simple notation for "harmonic" sums like these: just write down the numerators. So, for example:
$[\overline{1}]=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots=\infty\;$ is the harmonic series
$[\overline{1,-1}]=\frac{1}{1}+\frac{-1}{2}+\frac{1}{3}+\dots=\ln2\;$ is well known
$[\overline{1,1,-2}]=\frac{1}{1}+\frac{1}{2}+\frac{-2}{3}+\dots=\ln3\;$ is slightly less well known (I think)
$[\overline{1,0,-1,0}]=\frac{1}{1}+\frac{0}{2}+\frac{-1}{3}+\dots=\frac{\pi}{4}\;$ is the Gregory-Leibniz series for $\pi$
What I claim is that $[\overline{1,-1,-2,-1,1,2}]$ is equal to $0$. I wonder if there are any simple proofs of this (i.e. definitely without using calculus, preferably without appealing to complex numbers/taylor series/etc.)
P.S. I know a method that doesn't use any integrals or derivatives, but requires knowledge of the taylor series for $\ln(x)$ and the Euler formula for $e^{ix}$.
The reason I believe that there should be an elementary proof is that the sum, $0$, is a very simple number.
|
We may rewrite your series in the following manner:
\begin{align}
&\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)\\
&=\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{1}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{-1}{6n+6}\right)\\
&\hspace{1cm}-\sum_{n=0}^\infty\left(\frac{3}{6n+3}-\frac{3}{6n+6}\right)\\
\end{align}
But these summations are both the alternating series $\sum_{n=0}^\infty \dfrac{(-1)^n}{n+1}$. Therefore they cancel and the summation is equal to zero.
|
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|
How to prove $(1-\frac1{36})^{25}\lt\frac12$? How to prove the inequality?
$(1-\frac1{36})^{25}\lt\frac12$
I'm in trouble.
Thank you very much for your help
|
We have that $\displaystyle\left(1-\frac{1}{36}\right)^{25}=\sum_{r=0}^{25} \binom{25}{r}(-1)^{r}\frac{1}{36^r}=1-\frac{25}{36}+\frac{25\cdot24}{2\cdot36^2}-\frac{25\cdot24\cdot23}{6\cdot36^3}+\frac{25\cdot24\cdot23\cdot22}{24\cdot36^4}-\cdots$.
Since the sum is alternating and has terms that are decreasing in absolute value, its value is less than
$\displaystyle1-\frac{25}{36}+\frac{25\cdot24}{2\cdot36^2}-\frac{25\cdot24\cdot23}{6\cdot36^3}+\frac{25\cdot24\cdot23\cdot22}{24\cdot36^4}=\frac{29}{54}-\frac{25\cdot23\cdot122}{36^4}<\frac{1}{2}$ since
$27(25)(23)(122)>27(24)(23)(120)=9(12^{2})(3)(2)(23)(10)=36^{2}(1380)>36^2(1296)=36^{4}$
$\implies\displaystyle\frac{25\cdot23\cdot122}{36^{4}}>\frac{1}{27}.$
|
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|
Limit of a definite integral We need to calculate $$\lim_{x \to 0}\int_{\sin x}^{x}\frac{dt}{t^3(1+\sqrt{t})}$$
Integral itself doesn't seem to be the problem here. When making a substitution $\sqrt{t}=u$, we get $$\lim_{x \to 0}2\int_{\sqrt{\sin x}}^{\sqrt{x}}\frac{du}{u^5(1+u)}=2\lim_{x \to 0}\int_{\sqrt{\sin x}}^{\sqrt{x}}\frac{du}{u^5+u^6}$$
Then by partial fractions, which I did manually and chcecked with WolframAlpha afterwards, it becomes $$\begin {align}
2\lim_{x \to 0}\int_{\sqrt{\sin x}}^{\sqrt{x}}\left(\frac{1}{u^5}-\frac{1}{u^4}+\frac{1}{u^3}-\frac{1}{u^2}+\frac{1}{u}-\frac{1}{1+u}\right) du =\\
\lim_{x \to 0}\int_{\sqrt{\sin x}}^{\sqrt{x}}2\left(\log{u}-\log{(1+u)}+\frac{1}{u}-\frac{1}{2u^2}+\frac{1}{3u^3}-\frac{1}{4u^4}\right) du =\\
\lim_{x \to 0}\int_{\sin x}^{x}\left(\log{t}-2\log{(1+\sqrt{t})}+\frac{2}{\sqrt{t}}-\frac{1}{t}+\frac{2}{3t^{3/2}}-\frac{1}{2t^2}\right) dt
\\\end{align}$$
Fianlly we obtain the following limit:
$$\lim_{x \to 0}\left(\log {x}-\log {\sin x}+2\log {(1+\sqrt{x})}-2\log {(1+\sqrt{\sin x})}+\frac{2}{\sqrt{x}}-\frac{2}{\sqrt{\sin x}}-\frac{1}{x}+\frac{1}{\sin x}+\frac{2}{3x^{3/2}}-\frac{2}{3\sin^{3/2} x}-\frac{1}{2x^2}+\frac{1}{\sin^2 x}\right)$$
Here's where I stuck. It gets messy when I try to calculate $\frac{2}{3x^{3/2}}-\frac{2}{3\sin^{3/2} x}$ and $\frac{2}{\sqrt{x}}-\frac{2}{\sqrt{\sin x}}$. The rest is rather doable - de l'Hospital's rule is useful with $-\frac{1}{x}+\frac{1}{\sin x}$ which is $0$ in limit, so as logarithm expressions (obviously) and Taylor expansion helps with $-\frac{1}{2x^2}+\frac{1}{\sin^2 x}$ which, in turn, equals $1/6$ when $x$ approaches $0$.
Did I make any mistakes? I hope not, but even so I'm not certain what to do with this horrible limit. I'd be glad if anyone could point out what to do.
|
Your proof is nice but there's more simple:
Let consider this simplified integral
$$\int_{\sin x}^x\frac{dt}{t^3}=-\frac12t^{-2}\Bigg|_{\sin x}^x=-\frac12\left(\frac1{x^2}-\frac1{\sin^2x}\right)\sim_0-\frac12\frac{(x-\frac{x^3}{6})^2-x^2}{x^4}\sim_0\frac16$$
and now we prove that the two integrals have the same limit by:
\begin{align}
0\le\int_{\sin x}^x\left(\frac{1}{t^{3}}-\frac{1}{t^{3}(1+\sqrt t)}\right)dt=\int_{\sin x}^x\frac{\sqrt t}{t^3(1+\sqrt t)}dt &\le\int_{\sin x}^x t^{-5/2}dt = -\frac25t^{-3/2}\Bigg|_{\sin x}^x \\
&=-\frac25\left(\frac1{x^{3/2}}-\frac1{\sin^{3/2}x}\right)\sim_0\frac25\frac{x^{7/2}}{4x^3}\xrightarrow{x\to0}0
\end{align}
|
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|
Maximum and minimum of $z=\frac{1+x-y}{\sqrt{1+x^2+y^2}}$ Find the maximum and minimum of the function
$$z=\frac{1+x-y}{\sqrt{1+x^2+y^2}}$$
I have calculated
$\frac{\partial z}{\partial x}=\frac{1+y^2+xy-x}{(1+x^2+y^2)^{\frac{3}{2}}}$
$\frac{\partial z}{\partial y}=\frac{-1-x^2-xy-y}{(1+x^2+y^2)^{\frac{3}{2}}}$
I have problem to solve the system of equations:
$$1+y^2+xy-x=0$$
$$1+x^2+xy+y=0$$
|
Another solution for solving your two equations $$1+y^2+xy-x=0$$ $$1+x^2+xy+y=0$$ From the first, you can extract $$x=-\frac{y^2+1}{y-1}$$ and put it in the second, which after vsimplification reduces to $$\frac{2 \left(y^3+1\right)}{(y-1)^2}=0$$ So, $y=-1$ and back to the definition of $x$, $x=1$.
|
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|
Chinese Remainder Theorem, redundant information I want to solve the following system of congruences:
$ x \equiv 1 \mod 2 $
$ x \equiv 2 \mod 3 $
$ x \equiv 3 \mod 4 $
$ x \equiv 4 \mod 5 $
$ x \equiv 5 \mod 6 $
$ x \equiv 0 \mod 7 $
I know, but do not understand why, that the first two congruences are redundant. Why is this the case? I see that the modulo of the congruences are not pairwise relatively prime, but why does this cause a redundancy or contradiction? Further, why is it that in the solution to this system, we discard the first two congruences and not
$ x \equiv 3 \mod 4 $
$ x \equiv 5 \mod 6 $
being that $ gcd(3,6) = 3 $ and $gcd(2,4) = 2$ ?
EDIT:
How is the modulo of the unique solution effected if I instead consider the system of congruences without the redundancy i.e. does $M = 4 * 5 * 6 * 7$ or does it remain $M= 2*3*4*5*6*7$?
|
$x \equiv 1 \mod 2$
$x \equiv 2 \mod 3$
$x \equiv 3 \mod 4 \implies x \equiv 1 \pmod 2$
$x \equiv 4 \mod 5$
$x \equiv 5 \mod 6 \iff
\left.\begin{cases}
x \equiv 1 \mod 2 \\
x \equiv 2 \mod 3
\end{cases} \right\}$
By the CRT.
$x \equiv 0 \mod 7$
So first we replace $x \equiv 5 \mod 6$ with
$\left.\begin{cases}
x \equiv 1 \mod 2 \\
x \equiv 2 \mod 3
\end{cases} \right\}$
$x \equiv 1 \mod 2$
$x \equiv 2 \mod 3$
$x \equiv 3 \mod 4 \implies x \equiv 1 \pmod 2$
$x \equiv 4 \mod 5$
$x \equiv 1 \mod 2$
$x \equiv 2 \mod 3$
$x \equiv 0 \mod 7$
Now, because $x \equiv 1 \pmod 2$ is redundant, we remove all instanced of it and we remove all but one instance of $x \equiv 2 \mod 3$.
$x \equiv 2 \mod 3$
$x \equiv 3 \mod 4$
$x \equiv 4 \mod 5$
$x \equiv 0 \mod 7$
In this case, we can cheat a little if we change the first three congruences to equivalent congruences.
$
\left.
\begin{array}{l}
x \equiv -1 \mod 3 \\
x \equiv -1 \mod 4 \\
x \equiv -1 \mod 5
\end{array}
\right\} \ \iff x \equiv -1 \mod 60$
(Again, by the CRT.)
$x \equiv 0 \mod 7$
So we now have
$x \equiv -1 \mod 60$
$x \equiv 0 \mod 7$
To solve this, we start with $x \equiv 0 \mod 7$, which implies that
$x = 7n$ for some integer $n$. Substitute that into $x \equiv -1 \mod 60$ and you get
$7n \equiv -1 \mod 60$
So we need to find $\dfrac 17 \pmod{60}$ The most fundamental way to do this is to inspect numbers congruent to $1 \pmod{60}$ until we find one that is a multiple of $7$. At worst, we will have to examint $7$ such numbers.
$1, 61, 121, 181, 241, \color{red}{301}$
Since $7 \times 43 = 301$, then $\dfrac 17 \equiv 43 \pmod{60}$. So we conclude that $n \equiv -43 \equiv 17 \mod{60}$.
Then $x = 7n = 119 \mod{420}$.
|
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|
If $a_1a_2\cdots a_n=1$, then the sum $\sum_k a_k\prod_{j\le k} (1+a_j)^{-1}$ is bounded below by $1-2^{-n}$ I am having trouble with an inequality. Let $a_1,a_2,\ldots, a_n$ be positive real numbers whose product is $1$. Show that the sum
$$ \frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+\frac{a_3}{(1+a_1)(1+a_2)(1+a_3)}+\cdots+\frac{a_n}{(1+a_1)(1+a_2)\cdots(1+a_n)}$$
is greater than or equal to $$ \frac{2^n-1}{2^n}$$
If someone could help approaching this, that would be great. I don't even know where to start.
|
Note that for that every positive integer $i$ we have
\begin{eqnarray}
\frac{a_i}{(1+a_1)(1+a_2) \cdots (1+a_i)} & = & \frac{1 + a_i}{(1+a_1)(1+a_2) \cdots (1+a_i)} - \frac{1}{(1+a_1)(1+a_2) \cdots (1+a_i)} \nonumber \\
& = & \frac{1}{(1+a_1) \cdots (1+a_{i-1})} - \frac{1}{(1+a_1) \cdots (1+a_i)}. \nonumber
\end{eqnarray}
Let $b_i = (1+a_1)(1+a_2) \cdots (1+a_i)$, with $b_0= 0$. Then by telescopy
$$\sum\limits_{i=1}^n \left( \frac{1}{b_{i-1}} - \frac{1}{b_i} \right) = 1 - \frac{1}{b_n}.$$
Since $1+x\geq 2\sqrt{x}$ for all $x\ge 0$, we have
$$b_n = (1+a_1)(1+a_2) \cdots (1+a_n) \geq (2 \sqrt{a_1})(2 \sqrt{a_2}) \cdots (2 \sqrt{a_n}) = 2^n,$$
with equality precisely if $a_i=1$ for all $i$. It follows that
$$1 - \frac{1}{b_n} \geq 1 - \frac{1}{2^n} = \frac{2^n-1}{2^n}.$$
|
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|
Trigonometric simplification for limits: $\lim_{x\to 0} \frac{1-\cos^3 x}{x\sin2x}$ Have to evaluate this limit, but trigonometry part is :(
$$\lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}.$$
Had written the denominator as $2x\sin x\cos x$, no idea what to do next. Please help...
|
By Taylor series $$\cos x \sim_0 1 - \frac{x^2}{2}, \,\sin 2x \sim_0 2x\, \text{ and }\,\Big(1-\frac{x^2}{2}\Big)^3 \sim_0 1 - \frac{3x^2}{2}$$ so $$\dfrac{1-\cos^3 x}{x\sin2x} \sim_0 \frac{1 - \Big(1-\frac{x^2}{2}\Big)^3}{2x^2} \sim_0 \frac{ \frac{3x^2}{2} }{2x^2} = \frac{3}{4}.$$
|
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|
Evaluate the $\displaystyle\lim_{x\to 0}\left(\frac{1}{x\sqrt{1+x}}-\frac 1x\right)$ Evaluate the limit as $x$ approaches $0$ : $$\frac{1}{x\sqrt{1+x}}-\frac 1x$$
I have so far $$\frac{x -x\sqrt{1+x}}{x(x\sqrt{1+x})}$$
Do I multiply the numerator and denominator by the conjugate of $x - x\sqrt{1+x}$? Or do I multiply the top and bottom by $x\sqrt{1+x}$?
|
You can get
$$\frac{1}{\color{red}{x}\sqrt{1+x}}-\frac{1}{\color{red}{x}}=\frac{1}{x\sqrt{1+x}}-\frac{\sqrt{1+x}}{x\sqrt{1+x}}=\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}.$$
Then, multiply it by $$\frac{1+\sqrt{1+x}}{1+\sqrt{1+x}}.$$
|
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|
Max of $3$-Variable Function I'm trying the find the maximum of the function
$$f(a,b,c)=\frac{a+b+c-\sqrt{a^2+b^2+c^2}}{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}$$
for all nonnegative real numbers $a, b, c$ with $ab + bc + ca > 0$.
I tried in vain to prove that $\max_{a,b,c}f(a,b,c)=1-\frac{\sqrt{3}}{3}$
|
WLOG,let $c=$Min{$a,b,c$},
$f(a,b,c) \le \dfrac{a+b-\sqrt{a^2+b^2}}{\sqrt{ab}}=\dfrac{2ab}{\sqrt{ab}(a+b+\sqrt{a^2+b^2})} \le \dfrac{2ab}{\sqrt{ab}(2\sqrt{ab}+\sqrt{2ab})}=2-\sqrt{2}$
$\dfrac{a+b+c-\sqrt{a^2+b^2+c^2}}{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}} \le \dfrac{a+b-\sqrt{a^2+b^2}}{\sqrt{ab}} \iff \sqrt{ab}(a+b+c-\sqrt{a^2+b^2+c^2})\le (\sqrt{ab}+\sqrt{bc}+\sqrt{ca})(a+b-\sqrt{a^2+b^2}) \iff \sqrt{abc}((a^2+b^2)\sqrt{a}+(a^2+b^2)\sqrt{b}+c(a\sqrt{a}+b\sqrt{b}+a\sqrt{b}+b\sqrt{a})-2ab\sqrt{c}-\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})) \ge 0$
"=" will be hold only when $c=0$
so the max of $f(a,b,c)$ is $2-\sqrt{2}$ when $a=b,c=0$ or cycle.
the $1-\dfrac{\sqrt{3}}{3}$ is a paddle point.
|
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|
A cotangent series related to the zeta function $$\sin x = x\prod_{n=1}^\infty \left[1-\frac{x^2}{n^2\pi^2}\right]$$
If you apply $\log$ to both sides and derivate:
$$\cot x = \frac{1}{x} - \sum_{n=1}^\infty \left[\frac{2x}{n^2\pi^2} \frac{1}{1-\frac{x^2}{n^2\pi^2}}\right]$$
I have to make the expansion:
$$\frac{1}{1-\frac{x^2}{n^2\pi^2}} = 1 + {\left(\frac{x^2}{n^2\pi^2}\right)} + \left(\frac{x^2}{n^2\pi^2}\right)^2 + \left(\frac{x^2}{n^2\pi^2}\right)^3 + \cdots \tag{1}$$
But for this, I'm gonna expand the infinite sum:
$$\cot x = \frac{1}{x} - \left(\frac{2x}{1^2\pi^2} \frac{1}{1-\frac{x^2}{1^2\pi^2}} + \frac{2x}{2^2\pi^2} \frac{1}{1-\frac{x^2}{2^2\pi^2}} + \frac{2x}{3^2\pi^2} \frac{1}{1-\frac{x^2}{3^2\pi^2}}+\cdots\right)$$
So I can understand the boundaries of this expansion (cause the series $\frac{1}{1-x}=1 + x + x^2 + x^3 + \cdots$ as long as $|x|<1$.
So, am I right in saying that I can do the expansion $(1)$ as long as $|x|<\pi$? Because for $x=\pi$ we have: $$\frac{1}{1-\frac{\pi^2}{1^2\pi^2}} = \frac{1}{1-1}$$
And if $|x|<\pi$ the other terms like:
$$\frac{1}{1-\frac{\pi^2}{2^2\pi^2}}$$
Can be expanded by $(1)$ too.
So:
$$\cot x = \frac{1}{x} - \sum_{n=1}^\infty \left[\frac{2x}{n^2\pi^2} \left(1 + {\left(\frac{x^2}{n^2\pi^2}\right)} + \left(\frac{x^2}{n^2\pi^2}\right)^2 + \left(\frac{x^2}{n^2\pi^2}\right)^3 + \cdots \right)\right]\tag{$|x|<\pi$}$$
$$x \cot x = 1 - 2\sum_{n=1}^\infty \left[\frac{x^2}{n^2\pi^2} + \frac{x^4}{n^4\pi^4} + \frac{x^6}{n^6\pi^6} + \frac{x^8}{n^8\pi^8} + \cdots\right]$$
And finally the famous:
$$x \cot x = 1 - 2\sum_{n=1}^{\infty} \left[\zeta(2n)\frac{x^{2n}}{\pi^{2n}}\right]\tag{$|x|<\pi$}$$
Am I right with the boundaries for $x$?
|
Actualy (see [1] pg. 75)
$$
\cot(z)=\frac{1}{z}-\sum^{\infty}_{n=1}\frac{(-1)^{n-1}2^{2n}B_{2n}}{(2n)!}z^{2n-1}\textrm{, }|z|<\pi
$$
and ([1] pg.807)
$$
\zeta(2n)=\frac{(2\pi)^{2n}(-1)^{n-1}}{2(2n)!}B_{2n}\textrm{, }n\in\textbf{N}
$$
where $B_{2n}$ are the Bernoulli numbers i.e.
$$
\frac{x}{e^x-1}=\sum^{\infty}_{n=0}\frac{B_n}{n!}x^n\textrm{, }|x|<2\pi
$$
Your expansion is very right. But unfortunately this is quite known. In [1] you can find more of these results. For example
$$
\log\left(\frac{\sin(z)}{z}\right)=-\sum^{\infty}_{n=1}\frac{\zeta(2n)}{n}\frac{z^{2n}}{\pi^{2n}}
$$
Actualy for an analytic function in $(-1,1)$ which is also absolutly convergent at 1 we have
$$
\sum^{\infty}_{k=1}\left(f\left(\frac{x}{2\pi i k}\right)+f\left(\frac{-x}{2\pi i k}\right)-2f(0)\right)=2\sum^{\infty}_{k=1}\frac{f^{(2k)}(0)}{(2k)!}\frac{B_{2k}}{(2k)!}x^{2k}\textrm{, }|x|<2\pi
$$
[1]: Milton Abramowitz and Irene A. Stegun. "Handbook of Mathematical functions". Dover publications, inc., New York. (1972)
|
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|
Convergence of series with root Given
$$\sum_{n=1}^\infty {((-1)^n + \alpha^3) (\sqrt{n+1} - \sqrt{n})}$$
find all values of $\alpha$ such that the series converges.
My try:
By multiplying the series with the expression
$$\frac {\sqrt{n+1} + \sqrt n} {\sqrt{n+1} + \sqrt n}$$
we get
$$\sum_{n=1}^\infty {((-1)^n + \alpha^3) (\sqrt{n+1} - \sqrt{n}) \left(
\frac {\sqrt{n+1} + \sqrt n} {\sqrt{n+1} + \sqrt n}
\right)}$$
$$=
\sum_{n=1}^\infty {
((-1)^n + \alpha^3) \left(\frac 1{\sqrt{n+1} + \sqrt{n}} \right)
}
$$
Let $$a_n = ((-1)^n + \alpha^3)~,~b_n = \frac 1{\sqrt{n+1} + \sqrt{n}} $$
then, it is easy to show that $b_n$ is monotonously decreasing and
$$\lim_{n \to \infty} b_n = 0$$
I know that by Dirichlet rule, if i can find $\alpha$ such that $\sum_{n=1}^\infty a_n$ is bound, than the series will converge.
1.How do i find such an alpha? (if this is the right way...)
2.The Dirichlet rule is not iff. can i use it in this case?
|
You may write
$$
\begin{align}
\sum_{n=1}^N {((-1)^n + \alpha^3) (\sqrt{n+1} - \sqrt{n})} & =\sum_{n=1}^N {\frac{(-1)^n}{\sqrt{n+1} + \sqrt{n}}+ \alpha^3 \sum_{n=1}^N (\sqrt{n+1} - \sqrt{n})} \\&= \sum_{n=1}^N {\frac{(-1)^n}{\sqrt{n+1} + \sqrt{n}}+ \alpha^3 (\sqrt{N+1}-1)}
\end{align}
$$
and, as $N \rightarrow \infty$, it is easy to conclude.
|
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|
How prove $n|2^{\frac{n(n-1)}{2}}\cdot (2-1)(2^2-1)(2^3-1)\cdots (2^{n-1}-1)$ Question:
Today, when I solve other problem, I found this follow interesting result
$$n\mid\left(2^{\frac{n(n-1)}{2}}\cdot (2-1)(2^2-1)(2^3-1)\cdots (2^{n-1}-1)\right),n\ge 1$$
It is clear $n=1$ is true.
$n=2$
$$2\mid2^{\frac{n(n-1)}{2}}\cdot (2-1)(2^2-1)(2^3-1)\cdots (2^{n-1}-1)=2$$
when $n=3$.then
$$3\mid2^{\frac{n(n-1)}{2}}\cdot (2-1)(2^2-1)(2^3-1)\cdots (2^{n-1}-1)=8\cdot 3=24$$
when
$n=4$,then
$$4\mid64\cdot 3\cdot 7$$
$$\cdots,\cdots,$$
then How prove this general?
I know this Euler theorem
$$a^{n-1}-1\equiv 1\pmod n,(a,n)=1$$
How prove it?
|
Let $n=2^st$ where $t$ is odd. If $t>1$ then
$$t\mid 2^{\phi(t)}-1\ ,$$
and $\phi(t)$ is one of the numbers $1,2,\ldots,n-1$, so
$$t\mid (2-1)(2^2-1)(2^3-1)\cdots (2^{n-1}-1)\ ;$$
obviously this is also true when $t=1$. Also, $s<n\le\frac{1}{2}n(n-1)$, so
$$2^s\mid 2^{n(n-1)/2}\ ,$$
and hence
$$n=2^st\mid 2^{n(n-1)/2}(2-1)(2^2-1)(2^3-1)\cdots (2^{n-1}-1)$$
as claimed.
|
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|
For any base-$n$ number system, what is the average length of a number $\leq100$? By this I mean the amount of symbols (bits / digits / etc. ) for any number from $0$ to $100$. I don't know whether this can be answered, I'm just asking out of interest - not homework or anything.
|
Any integer between $n^{d-1}$ and $n^d-1$ in base-$n$ requires exactly $d$ digits. Also $0$ needs $1$ digit.
For instance, in base-$2$:
$0 = 0_2$ through $1 = 1_2$ require $1$ digit each
$2 = 10_2$ through $3 = 11_2$ require $2$ digits each
$4 = 100_2$ through $7 = 111_2$ require $3$ digits each
$8 = 1000_2$ through $15 = 1111_2$ require $4$ digits each
$16 = 10000_2$ through $31 = 11111_2$ require $5$ digits each
$32 = 100000_2$ through $63 = 111111_2$ require $6$ digits each
$64 = 1000000_2$ through $100 = 1100100_2$ require $7$ digits each
Writing all $101$ integers between $0$ and $100$ inclusive requires $2 \cdot 1 + 2 \cdot 2 + 4 \cdot 3 + 8 \cdot 4 + 16 \cdot 5 + 32 \cdot 6 + 37 \cdot 7 = 581$ digits, so the average length of an integer between $0$ and $100$ in base-$2$ is $\frac{581}{101} \approx 5.75$ digits.
In general, writing an integer $k$ in base $d$ requires $\lfloor \log_n k \rfloor + 1$ digits, and so, the average length of an integer between $0$ and $100$ in base-$n$ is $1 + \dfrac{1}{101}\displaystyle\sum_{k = 1}^{100}\lfloor \log_n k \rfloor$.
|
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|
Derivatives of trig polynomials do not increase degree? Let $c = \cos x$ and $s = \sin x$,
and consider a trigonometric polynomial $p(x)$
in $c$ and $s$.
The degree of $p(x)$ is the maximum of
$n+m$ in terms $c^n s^m$.
Is it the case that repeated derivatives of $p(x)$,
expressed again in terms of $c$ and $s$,
never increase the degree?
For example, $p(x)=c^4 s^2$ has degree $6$, and
here are its first $5$ derivatives.
$$
c^4 s^2 \\
d^1 = 2 c^5 s-4 c^3 s^3 \\
d^2 = 2 c^6-22 c^4 s^2+12 c^2 s^4 \\
d^3 = -56 c^5 s+136 c^3 s^3-24 c s^5 \\
d^4 = -56 c^6+688 c^4 s^2-528 c^2 s^4+24 s^6 \\
d^5 = 1712 c^5 s-4864 c^3 s^3+1200 c s^5
$$
Because $d^3$ and $d^5$ has the same
$c^5 s + c^3 s^3+ c s^5$ structure, we are in a loop,
establishing that all derivatives of $p(x)$ have degree $6$.
|
To show this is true for any polynomial of degree $n+m$, it's sufficient to show that this is the case for an arbitrary term $c^ns^m$ - since every term must be of this form for some $n,m$ - and to check that adding such terms together won't affect the derivative.
Note that $$\frac{d}{dx}(c^n s^m)=mc^{n+1}s^{m-1}-nc^{n-1}s^{m+1}$$
This gives another degree $n+m$ polynomial.
You do have to be careful that your original polynomial cannot be reduced further. For instance, $$c^2s + s^3=s(s^2 + c^2)$$ could be seen as degree $3$, but since $s^2 + c^2 = 1$, it is actually degree $1$.
However, noticing that $$c^ns^m=\begin{cases}c^n(1-c^2)^{m/2}&\text{$m$ even}\\
c^ns(1-c^2)^{\frac{m-1}2}&m\text{ odd}\end{cases}$$if we initially write the polynomial in terms of $c^{n+m-1}s$ and $c^{n+m}$ we can be sure of avoiding this problem.
|
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|
Finding multivariable limit I would like to find the following limit $$\lim_{(x,y,z)\to(0,0,0)}\frac{x^3yz+xy^3z+xyz^3}{x^4+y^4+z^4}.$$
It looks like it would be zero since if we put $M=\max\{x,y,z\}$ and $m=\min\{x,y,z\}$, then $$\Big|\frac{x^3yz+xy^3z+xyz^3}{x^4+y^4+z^4}\Big| \leq \Big|\frac{M^5}{m^4}\Big|$$ so the exponent in the denominator is bigger than the exponent in the numerator.
But how do I actually calculate this limit?
Thank you.
|
Let $r = \sqrt{x^2+y^2+z^2}$. Then, using the generalized mean inequality, we have:
$\sqrt[4]{\dfrac{x^4+y^4+z^4}{3}} \ge \sqrt[2]{\dfrac{x^2+y^2+z^2}{3}} \ge \sqrt[3]{|xyz|}$.
Therefore, $x^4+y^4+z^4 \ge 3\left(\dfrac{x^2+y^2+z^2}{3}\right)^2 = \dfrac{r^4}{3}$ and $|xyz| \le \left(\dfrac{x^2+y^2+z^2}{3}\right)^{3/2} = \dfrac{r^3}{3\sqrt{3}}$.
Hence, $\left|\dfrac{x^3yz+xy^3z+xyz^3}{x^4+y^4+z^4}\right| = \dfrac{|xyz| \cdot |x^2+y^2+z^2|}{|x^4+y^4+z^4|} \le \dfrac{\frac{r^3}{3\sqrt{3}} \cdot r^2}{\frac{r^4}{3}} = \dfrac{r}{9\sqrt{3}} = \dfrac{\sqrt{x^2+y^2+z^2}}{9\sqrt{3}}$.
Now, apply the squeeze theorem.
If you don't like this approach, you can try using spherical coordinates. Let $x = \rho \sin\phi \cos\theta$, $y = \rho\sin\phi\sin\theta$, and $z = \rho\cos\phi$. After making those substitutions, the function will simplify to $\rho \cdot \left(\text{some trig expression with} \ \phi \ \text{and} \ \theta \ \text{that I hope is bounded}\right)$, which will approach $0$ as $\rho \to 0$.
|
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|
How to prove this inequality without using Muirhead's inequality? I ran into a following problem in The Cauchy-Schwarz Master Class:
Let $x, y, z \geq 0$ and $xyz = 1$.
Prove $x^2 + y^2 + z^2 \leq x^3 + y^3 + z^3$.
The problem is contained in the chapter about symmetric polynomials and Muirhead's inequality.
The proof based on Muirhead's inequality is pretty quick:
We multiply the left hand side with $\sqrt[3]{xyz} = 1$ and prove
$$x^{\frac{7}{3}}y^{\frac{1}{3}}z^{\frac{1}{3}} + x^{\frac{1}{3}}y^{\frac{7}{3}}z^{\frac{1}{3}} + x^{\frac{1}{3}}y^{\frac{1}{3}}z^{\frac{7}{3}} \leq x^3 + y^3 + z^3$$
with Muirhead ( $(3, 0, 0)$ majorizes $(\frac{7}{3}, \frac{1}{3}, \frac{1}{3})$).
I'm curious if there's a way to prove this without machinery of Muirhead's inequality and majorization. Also, this approach readily generalizes to proving
$$x^n + y^n + z^n \leq x^{n+1} + y^{n+1} + z^{n+1}$$
for non-negative $x, y, z$ such that $xyz = 1$.
Is there a way to prove this generalization without Muirhead?
|
From Chebyshev's sum inequality we have
\begin{align*}
\frac{x^{n+1}+y^{n+1}+z^{n+1}}{3}\geq \frac{x^n+y^n+z^n}{3}\cdot\frac{x+y+z}{3}.
\end{align*}
By AM-GM we have $\frac{x+y+z}{3}\geq\sqrt[3]{xyz}=1$ and that proves the desired inequality.
|
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|
Integral $\int\frac{dx}{x\sqrt{x^2+4}}$ While solving this:
$$
\begin{align}
\int\frac{dx}{x\sqrt{x^2+4}}
&=\int\frac{t(-1/t^2)dt}{\sqrt{(1/t)^2+4}}\tag{t=1/x}\\
&=\int\frac{(-1/t)dt}{(1/t)\sqrt{1+4t^2}}\\
&=-\frac12\int\frac{dt}{\sqrt{t^2+1/4}}\\
&=-\frac12\ln(t+\sqrt{t^2+1/4})+C\\
&=-\frac12\ln(1/x+\sqrt{1/x^2+1/4})+C\\
&=-\frac12\ln\left(\frac{2+\sqrt{4+x^2}}{2x}\right)+C\\
&=\color{green}{\frac12(\ln(x)-\ln(2+\sqrt{4+x^2}))}+C\tag{$\ln(2x)=\ln(x)+\ln2$}\\
\end{align}$$
I got this, but answer given in textbook is $\displaystyle\color{blue}{ \ln[(x+1/2)+\sqrt{x^2+x+1}]}+C$
I see no relation between them, is my answer wrong? Along with answering this question please shed some light on this related issue which can happen with other integrals and etc.
|
You can see that your answer is correct and that the given answer is wrong because
$$\left(\frac 12\left(\ln (x)-\ln\left(2+\sqrt{4+x^2}\right)\right)\right)'=\frac{1}{x\sqrt{x^2+4}}$$
$$\left(\ln\left[(x+1/2)+\sqrt{x^2+x+1}\right]\right)'=\frac{1}{\sqrt{x^2+x+1}}$$
|
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|
$\lim_{N\to \infty} \sum_{n=N}^{2N} c_n = 0$ $\Rightarrow \sum_{n=1}^{\infty} c_n$ converges? If $\lim_{N\to \infty} \sum_{n=N}^{2N} c_n = 0$ do we have that $\sum_{n=1}^{\infty} c_n$ converges? At first this did not seem true($\sum_{n=N}^{2N} (-1)^n$ is $0$ when N is odd), but I've failed to find a proper counter-example. I've tried thinking in terms of partial sums to prove that is it true but have had no luck. Can someone help?
|
Let
\begin{align*}
c_1 = c_2 &= \frac{1}{2}\\
& \\
c_3 = c_4 = c_5 = c_6 &= \frac{1}{4}\times\frac{1}{2} = \frac{1}{8}\\
& \\
c_7 = c_8 = c_9 = c_{10} = c_{11} = c_{12} = c_{13} = c_{14} &= \frac{1}{8}\times\frac{1}{3} = \frac{1}{24}
\end{align*}
and so on. More precisely, let $c_n = \dfrac{1}{2^kk}$ where $k = \lfloor\log_2(n+1)\rfloor$.
Then if $S_m = \displaystyle\sum_{n=1}^ma_n$, we have
$$S_{2^{p+1}-2} = \sum_{n=1}^{2^{p+1}-2} c_n = \sum_{k=1}^{p}\sum_{n=2^k-1}^{2^{k+1}-2}c_n = \sum_{k=1}^{p}\sum_{n=2^k-1}^{2^{k+1}-2}\frac{1}{2^kk} = \sum_{k=1}^{p}2^k\frac{1}{2^kk} = \sum_{k=1}^p\frac{1}{k}.$$
As the harmonic series diverges, $S_{2^{p+1}-2}$ diverges. Therefore $\displaystyle\sum_{n=1}^{\infty}c_n$ diverges.
|
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|
Geometry problem involving infinite number of circles What is the sum of the areas of the grey circles? I have not made any progress so far.
|
Consider the picture below.
On the left hand side we have the original picture which we have put in the complex plane, and on the right hand side is its image under the Möbius transformation $f(z) = \frac{z}{2-z}$. The inverse transformation is $g(z) = \frac{2z}{z+1}$. In the new coordinates the green circles have radius $\frac{1}{2}$ and center at $C_k = \frac{1}{2} + i(k + \frac{1}{2})$, $k=0,1,2,\dots$.
We have $\text{Jac}\, g(z) = |g'(z)|^2 = \frac{4}{|z+1|^4}$. Therefore the area of the disks in the original picture is
$$\sum_{k=0}^\infty \int_{B_k} \frac{4}{|(x+iy)+1|^4} \, dx \, dy,$$
where $B_k = \{z : |z - C_k| \le \frac{1}{2}\}$ are the discs in the second picture. Calculating the integral inside the sum turned out to be tedious, however, so I opted for another route. EDIT: achille hui showed how to calculate this integral in an answer to my question at Integral related to a geometry problem. This yields a shorter way to get the answer.
The points where a green circle touches the red line or the blue line are at $i(k + \frac{1}{2})$ and $1 + i(k + \frac{1}{2})$ respectively. Therefore in the original picture they are at
$$A = g(i(k+\frac{1}{2})) = \frac{2(k+\frac{1}{2})^2}{1 + (k + \frac{1}{2})^2} + i \frac{2(k+\frac{1}{2})}{1 + (k + \frac{1}{2})^2}$$
and
$$B = g(1 + i(k + \frac{1}{2})) = \frac{4 + 2(k+\frac{1}{2})^2}{4 + (k + \frac{1}{2})^2} + i \frac{2(k+\frac{1}{2})}{4 + (k+\frac{1}{2})^2}.$$
Now to find the center of the green circle we calculate the intersection of the lines $1 + t(A-1)$ and $\frac{3}{2} + s(B - \frac{3}{2})$. Real and imaginary part give us two linear equations for $s$ and $t$, and we end up with the solution
$$s = \frac{4k^2 + 4k + 17}{4k^2 + 4k + 9}, \quad t = \frac{4k^2 + 4k + 5}{4k^2 + 4k + 9}.$$
Thus the center is at
$$1 + t(A-1) = \frac{8k^2 + 8k + 6}{4k^2 + 4k + 9} + i \frac{8k + 4}{4k^2 + 4k + 9}.$$
The radius is then
$$|A - 1 - t(A-1)| = |A-1| |t-1| = \frac{4}{4k^2 + 4k + 9}.$$
Hence the answer to the problem is
$$\sum_{k=0}^\infty \pi \frac{16}{(4k^2 + 4k + 9)^2},$$
for which Wolfram Alpha gives the closed form
$$\frac{1}{16} \pi^2 (\sqrt{2} \tanh(\sqrt{2}\pi) - 2 \pi \text{sech}^2(\sqrt{2}\pi)) \approx 0.8699725\dots$$
|
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|
Test the convergence of a series To test the convergence of a series:
$$
\sum\left[\sqrt[3]{n^3+1}-n\right]
$$
My attempt:
Take out $n$ in common: $\displaystyle\sum\left[n\left(\sqrt[3]{1+\frac{1}{n^3}}-1\right)\right]$.
So this should be divergent.
But, the given answer says its convergent.
|
$$ \sqrt[3]{n^3+1}-n = \left({\sqrt[3]{n^3+1}-n)}\right) \frac{ \left({ (n^3 + 1)^{\frac 2 3} + (n^3 + 1)^{\frac 1 3}n + n^{2 } }\right)}{\left({ (n^3 + 1)^{\frac 2 3} + (n^3 + 1)^{\frac 1 3}n + n^{2 } }\right)} = \frac {1}{\left({ (n^3 + 1)^{\frac 2 3} + (n^3 + 1)^{\frac 1 3}n + n^{2 } }\right)} \le \frac{1}{(n^3 + 1)^{\frac 2 3} } \le \frac{1}{(n^3)^{\frac 2 3}} = \frac 1 {n^2}$$
And we know that $\sum \frac {1}{n^2}$ converges and by the Comparison Test our series converges too.
And a series will not diverge just because $n$ is a factor. What about $$ \sum n \cdot \frac{1}{n!} \;\;\; ?? $$
|
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|
What is the maximum value of $ \sin x \sin {2x}$ What is the maximum value of $$ \sin x \sin {2x}$$
I have done my work here
$$f (x)=\sin x \sin 2x =\frac{\cos x - \cos3x}2 $$
$$f'(x)= \frac{- \sin x+3 \sin 3x}2 =4\sin x (2-3\sin^2 x)=0$$
$$x=0,\pi; \sin x= \sqrt{\frac {2}{3}}$$
$$f (0)=f (\pi)=0$$
$$f \left(\arcsin \sqrt{\frac{2}{3}}\right) =\frac{4}{3 \sqrt{3}}$$
If my work is not much right then please rectify it
|
We have $$f(x) = \sin (x)\sin (2x) = 2{\sin ^2}(x)\cos (x) = 2(\cos (x) - {\cos ^3}(x))$$for a maximum we should have $f'(x) = 0$, so that $$ - \sin (x) + 3\sin (x){\cos ^2}(x) = 0$$so finally$$\left\{ \begin{array}{l}\sin (x) = 0 \to f(x) = 0\\\cos (x) = \frac{1}{{\sqrt 3 }} \to f(x) = \frac{4}{{3\sqrt 3 }}\\\cos (x) = - \frac{1}{{\sqrt 3 }} \to f(x) = - \frac{4}{{3\sqrt 3 }}\end{array} \right.$$which clearly shows the maximum.
|
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|
For how many integral value of $x\le{100}$ is $3^x-x^2$ divisible by $5$? For how many integral value of $x\le{100}$ is $3^x-x^2$ divisible by $5$?
I compared $3^x$ and $x^2$ in $\mod {5}$ i found some cycles but didn't get anything
|
HINT : For $n\in\mathbb N$,
$n$ can be divided by $5$ $\iff$ The right-most digit of $n$ is either $0$ or $5$.
You'll find some patterns in the followings from $x=1$ to $x=20$.
The right-most digit of $3^x$ : $3,9,7,1,3,9,7,1,3,9,7,1,3,9,7,1,3,9,7,1.$
The right-most digit of $x^2$ : $1,4,9,6,5,6,9,4,1,0,1,4,9,6,5,6,9,4,1,0.$
|
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How find this $\frac{1}{x-y}+\frac{1}{y-z}+\frac{1}{x-z}$ minimum of the value let $x,y,z\in R$,and such $x>y>z$,and such
$$(x-y)(y-z)(x-z)=16$$
find this follow minimum of the value
$$I=\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{x-z}$$
My idea: since
$$\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{x-z}=\dfrac{x-z}{(x-y)(y-z)}+\dfrac{1}{x-z}$$
so
$$I=\dfrac{(x-z)^2}{16}+\dfrac{1}{x-z}=\dfrac{(x-z)^2}{16}+\dfrac{1}{2(x-z)}+\dfrac{1}{2(x-z)}\ge\dfrac{3}{4}$$
if and only if $(x-z)=2$,so $(x-y)(y-z)=8$
But we know $$(x-z)^2=[[(x-y)+(y-z)]^2\ge 4(x-y)(y-z)$$
so this is wrong,
Now I let $x-z=t$ it is clear $t\ge 4$,so
$$\dfrac{t^2}{16}+\frac{1}{t}=f(t)\Longrightarrow f'(t)\ge 0,t\ge 4$$
so
$$f(t)\ge f(4)=\dfrac{5}{4}$$
My Question: I fell my methods is ugly,I think this problem have other simple methods.Thank you
|
Let $\displaystyle a=x-y,b=y-z, c=x-z\implies abc=16$ and $\displaystyle a+b-c=0\iff c=a+b$
$$\frac1a+\frac1b+\frac1c=\frac{ab+bc+ca}{abc}=\dfrac{\dfrac{16}c+c(a+b)}{16}=\dfrac{\dfrac{16}c+c(c)}{16}$$
Now use Second Derivative Test
|
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|
Primitive of $\frac{3x^4-1}{(x^4+x+1)^2}$ How to find primitive of:
$$\frac{3x^4-1}{(x^4+x+1)^2}$$
I am having a faint idea of a type which may or maynot be in the primitve, i.e.:
$$\frac{p(x)}{x^4+x+1}$$
The problem is I am not getting an idea of a substitution to solve this problem.
I might show my work but it is totally useless, atleast in this case.
For reference:
|
$$ \int \frac{3x^4-1}{(x^4+x+1)^2} = \int \frac{3x^4+4x^3-4x^3-1}{(x^4+x+1)^2}$$
$$ \int \frac{3x^4+4x^3-4x^3-1}{(x^4+x+1)^2} = \int \frac{3x^4+4x^3}{(x^4+x+1)^2}- \int \frac{4x^3+1}{(x^4+x+1)^2}$$
Consider,
$$ \int \frac{3x^4+4x^3}{(x^4+x+1)^2} = \int \frac{4x^3(x+1)-x^4}{(x^4+x+1)^2} = \int \frac{(4x^3+1)(x+1)+(-1)(x^4+x+1)}{(x^4+x+1)^2}= -\frac{x+1}{x^4+x+1}$$
Hint: Can you see quotient rule here?
And , obviously
$$ \int \frac{4x^3+1}{(x^4+x+1)^2} = -\frac{1}{x^4+x+1} $$
Hint: Use substitution $t=x^4+x+1$
So concluding,
$$ \int \frac{3x^4-1}{(x^4+x+1)^2} = -\frac{x}{x^4+x+1} \Box$$
|
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|
Evaluating the sum $1\cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 + \dots + n\cdot 10^n$ How can I calculate
$$1\cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 + 4\cdot 10^4+\dots + n\cdot 10^n$$
as a expression, with a proof so I could actually understand it if possible?
|
Put
$$ S_n = \sum_{k=1}^{n} k\cdot 10^k.$$
Then:
$$ 9 S_n = (10-1)S_n = \sum_{k=1}^{n}k\cdot 10^{k+1}-\sum_{k=1}^{n}k\cdot 10^k=\sum_{k=2}^{n+1}(k-1)\cdot 10^k-\sum_{k=1}^{n}k\cdot 10^k$$
hence:
$$ 9 S_n = n\cdot 10^n -\sum_{k=1}^n 10^k = n\cdot 10^n-\frac{10^{n+1}-10}{9},$$
so:
$$ S_n = \color{red}{\frac{10}{81}\left(1+(9n-1)\cdot 10^n\right)}.$$
|
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|
How prove this inequality $(1+\frac{1}{16})^{16}<\frac{8}{3}$
show that
$$(1+\dfrac{1}{16})^{16}<\dfrac{8}{3}$$
it's well know that
$$(1+\dfrac{1}{n})^n<e$$
so
$$(1+\dfrac{1}{16})^{16}<e$$
But I found this $e=2.718>\dfrac{8}{3}=2.6666\cdots$
so how to prove this inequality by hand?
Thank you everyone solve it,I want see don't use $e=2.718$,because a most middle stundent don't know this value.
before I have use this well know
$$(1+\dfrac{1}{2n+1})(1+\dfrac{1}{n})^n<e$$
so
$$(1+\dfrac{1}{16})^{16}<e\cdot\dfrac{33}{34}\approx 2.638<\dfrac{8}{3}$$ to solve this, But Now we don't use $e=2.718$.
to prove this inequality by hand
|
$$ (1+\dfrac{1}{16})^{16}
= \sum \frac{1}{16^k}\binom{16}{k}
= \sum \frac{1}{ 16^k}\frac{16!}{(16-k)!} \frac{1}{k!} < 1 + 1 + \frac{1}{2} + \frac{1}{6} + \dots <\dfrac{8}{3} + \frac{2}{4!}$$
How do we know the remaining terms are small enough? Let's try an inequality. In our case, $n=4$.
$$ \sum_{k=n}^\infty \frac{1}{k!}
= \frac{1}{n!}\sum_{k=0}^\infty \frac{1}{n^k} = \frac{1}{n!}\frac{1}{1-\frac{1}{n}}< \frac{2}{n!}$$
If we are more careful we can actually prevent the overshooting that @StevenStadnicki points out.
$$1 + 1 + \frac{1}{2}\frac{15}{16} + \frac{1}{6}\frac{15\times 14}{16\times 16 } + \frac{1}{12}= 2.688 > \frac{8}{3} = 2.66\overline{6}$$
If you extend out to the 4th term the result is 2.654 which is bigger than 2.6379 which is the exact answer up to 4 digits.
|
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|
When is a sum of products of positive powers of 2 and 3 divisible by $2^b-3^n$? Here we have a really tough exercise.
Find all natural solution:
$$\frac{\sum\limits_{k=1}^n 2^{a_k} 3^{n-k}}{c}+3^n=2^{b} ,\quad
b\geq a_n; \quad a_k, b, c ,n\in \mathbb N $$
Any ideas, hints?
|
$$2^a3^{n-1}+2^{2a}3^{n-2}+\cdots+2^{an}3^0=c(2^b-3^n)$$
$$3^n\left(\left(\frac{2^a}3\right)^1+\left(\frac{2^a}3\right)^2+\left(\frac{2^a}3\right)^3+\cdots+\left(\frac{2^a}3\right)^n\right)=c(2^b-3^n)$$
$$3^n\left(\frac{\frac{2^a}3((2^a/3)^n-1)}{2^a/3-1}\right)=c(2^b-3^n)$$
$$2^a\left(\frac{2^a-3^n}{2^a-3}\right)=c(2^b-3^n)$$
$$2^a(2^a-3^n)=c(2^a-3)(2^b-3^n)$$
I hope you can solve now.
|
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|
Dividing by $\sqrt n$ Why is the following equality true?
I know I should divide by $\sqrt n$ but how is it done exactly to get the RHS?
$$ \frac{\sqrt n}{\sqrt{n + \sqrt{n + \sqrt n}}} = \frac{1}{\sqrt{1 + \sqrt{\frac{1}{n} + \sqrt{\frac{1}{n^3}}}}}$$
|
Slightly modifying the answer from @evinda:
$\begin{align}
\frac{\frac{\sqrt{n}}{ \sqrt{n}}}{\frac{\sqrt{n+\sqrt{n+\sqrt{n}}}}{\sqrt{n}}}&=\frac{1}{\sqrt{\frac{n+\sqrt{n+\sqrt{n}}}{n}}} \\
&=\frac{1}{\sqrt{1+ \frac{\sqrt{n+\sqrt{n}}}{n}}} \\
&=\frac{1}{\sqrt{1+ \sqrt{\frac{n+\sqrt{n}}{n^2}}}} \\
&=\frac{1}{\sqrt{1+ \sqrt{\frac1n+\frac{\sqrt{n}}{n^2}}}} \\
&=\frac{1}{\sqrt{1+ \sqrt{\frac1n+\sqrt{\frac{n}{n^4}}}}} \\
&=\frac{1}{\sqrt{1+ \sqrt{\frac1n+\sqrt{\frac{1}{n^3}}}}}
\end{align}$
|
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|
Short form of few series Is there a short form for summation of following series?
$$\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\alpha^{2n}((2y-1)^{2k+1}+1)}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}$$
$$\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}(\cos^{-1}(2y-1)-\pi)}{2^{4n+3}n!(n+1)!}$$
$$\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\alpha^{2n+1}(2y-1)(1-(2y-1)^2)^{k+\frac{1}{2}}}{2^{4n-2k+3}n!(n+1)!(2k+1)!}$$
Even sum of any combination of above terms could help. This is result of some integral. I guess they should contain Bessel and Struve functions.
In the case of $y=0.5$ it seems the sum of above terms to be
$$
-\frac{\pi}{2}\left[I_3(\frac{\alpha}{2}) +\frac{3}{\frac{\alpha}{2}}I_2(\frac{\alpha}{2})- I_{1}(\frac{\alpha}{2})-\frac{1}{8}\left[L_{-3}(\frac{\alpha}{2}) - L_{-1}(\frac{\alpha}{2}) - L_{1}(\frac{\alpha}{2})+L_{3}(\frac{\alpha}{2}) -\frac{2\alpha^{-2}}{\pi} -\frac{8}{3\pi}+\frac{2\alpha^2}{15\pi}\right]\right]
$$
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For $\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\alpha^{2n}(2y-1)^{2k+1}}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}$ ,
$\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\alpha^{2n}(2y-1)^{2k+1}}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}$
$=\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(-1)^kn!\alpha^{2n}(2y-1)^{2k+1}}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}$
$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^k(n+k)!\alpha^{2n+2k}(2y-1)^{2k+1}}{2^{2n+2k+1}(2n+2k)!k!n!(2k+1)}$
$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^k\alpha^{2n+2k}(2y-1)^{2k+1}\sqrt\pi}{2^{4n+4k+1}\Gamma\left(n+k+\dfrac{1}{2}\right)n!k!\left(k+\dfrac{1}{2}\right)}$ (according to https://en.wikipedia.org/wiki/Multiplication_theorem#Gamma_function-Legendre_function)
For $\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\alpha^{2n+1}(2y-1)(1-(2y-1)^2)^{k+\frac{1}{2}}}{2^{4n-2k+3}n!(n+1)!(2k+1)!}$ ,
$\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\alpha^{2n+1}(2y-1)(1-(2y-1)^2)^{k+\frac{1}{2}}}{2^{4n-2k+3}n!(n+1)!(2k+1)!}$
$=\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(k!)^2\alpha^{2n+1}(2y-1)(1-(2y-1)^2)^{k+\frac{1}{2}}}{2^{4n-2k+3}n!(n+1)!(2k+1)!}$
$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(k!)^2\alpha^{2n+2k+1}(2y-1)(1-(2y-1)^2)^{k+\frac{1}{2}}}{2^{4n+2k+3}(n+k)!(n+k+1)!(2k+1)!}$
$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(k!)^2\alpha^{2n+2k+1}(2y-1)(1-(2y-1)^2)^{k+\frac{1}{2}}\sqrt\pi}{16^{n+k+1}(n+k)!(n+k+1)!\Gamma(k+1)\Gamma\left(k+\dfrac{3}{2}\right)}$ (according to https://en.wikipedia.org/wiki/Multiplication_theorem#Gamma_function-Legendre_function)
which both relate to Kampé de Fériet function function
|
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|
Squeeze Theorem: Finding the limit of a trig function I'm stuck on finding the limit of a complex fraction/trig function. Could someone please assist, or point out where I'm going wrong?
Determine
$$\lim\limits_{x \to 0} \frac{(x+1)\cos(\ln(x^2))}{\sqrt{(x^2+2)}}$$
For all $x$: $$-1 \le \cos(\ln(x^2)) \le 1$$
Multiply by $(x+1)$:
$$-(x+1) \le (x+1)\cos(\ln(x^2)) \le (x+1)$$
Divide by $\sqrt{(x^2+2)}$:
$$\frac{-(x+1)}{\sqrt{(x^2+2)}} \le \frac{(x+1)\cos(\ln(x^2))}{\sqrt{(x^2+2)}} \le \frac{(x+1)}{\sqrt{(x^2+2)}}$$
Now to find the limits:
$$\lim\limits_{x \to 0} \frac{-(x+1)}{\sqrt{(x^2+2)}} = \frac{-1}{\sqrt{(2)}}$$
$$\lim\limits_{x \to 0} \frac{(x+1)}{\sqrt{(x^2+2)}} = \frac{1}{\sqrt{(2)}}$$
This is where I get stuck. My limits are not equal, so I cannot solve using the Squeeze Theorem. I must've skipped a step or used the wrong approach, but I'm not sure where.
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$$f(x)=\frac{(x+1)\cos(\ln(x^2))}{\sqrt{x^2+2}}$$
Take two sequences tending to $0$:
$$\begin{array}{}
a_n=\exp\left(\tfrac{\pi}{4}(1-2n)\right),&\lim\limits_{n\to\infty} a_n=0\\
b_n=\exp(-n\pi),&\lim\limits_{n\to\infty} b_n=0
\end{array}$$
We have
$$f(a_n)=\frac{(a_n+1)\cos\left(\ln\left(\exp\left(\tfrac{\pi}{4}(1-2n)\right)^2\right)\right)}{\sqrt{a_n^2+2}}
=\frac{(a_n+1)\cos\left(\tfrac{\pi}{2}(1-2n)\right)}{\sqrt{a_n^2+2}}=0$$
$$f(b_n)=\frac{(x+1)\cos(\ln(\exp(-n\pi)^2))}{\sqrt{x^2+2}}
=\frac{(x+1)\cos(-2n\pi)}{\sqrt{x^2+2}}
=\frac{b_n+1}{\sqrt{b_n^2+2}}$$
Calculating the limits of $f(a_n)$ and $f(b_n)$ we get:
$$\lim_{n\to\infty}f(a_n)=0$$
$$\lim_{n\to\infty}f(b_n)=\tfrac{\sqrt2}{2}$$
Thus the limit
$$\lim_{x\to0}f(x)$$
doesn't exist.
|
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Is sum of square of primes a square of prime? I would like to know if it has been proved that :
*
*There are no $a$, $b$ and $c$, all prime numbers, such that $a^2 + b^2 = c^2$
*There are no $a$, $b$, $c$ and $d$, all prime numbers, such that $a^2 + b^2 + c^2 = d^2$
*There are no $a$, $b$, $c$, $d$ and $e$, all prime numbers, such that $a^2 + b^2 + c^2 + d^2 = e^2$
Thanks for your answers.
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It turns out that the (full) solution is longer than a comment. So here it goes.
Suppose that for $a,b,c,d$ primes,
$$a^2+b^2+c^2=d^2$$
Obviously $d=3$ is not a solution. That implies
$$a^2+b^2+c^2\equiv 1\pmod{3}.$$
So exact two of $a,b,c$ are equal to $3$. We are left to solve
$$d^2-a^2=(d-a)(d+a)=18.$$
But $d-a$ and $d+a$ are of the same parity and $18$ cannot be factorized into such a product.
For the third equations, note that $n^2\equiv 1\pmod{4}$ if $n$ is odd. Thus if
$$a^2+b^2+c^2+d^2=e^2,$$
then exactly $3$ of $a,b,c,d$ are $2$. We are left to solve
$$(e-d)(e+d)=12=6\cdot 2.$$
It follows that $e+d=6$, and $e-d=2$. That means $e=4$, not a prime.
|
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|
Infinite Series $\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdots$ How do I find the sum of the following infinite series:
$$\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdots$$
The series definitely seems to be convergent.
|
Your sum is equal to:
\begin{align}
\sum_{i=1}^{\infty} \left ( \frac{1}{3i-1}+\frac{1}{3i+1}-\frac{2}{3i}\right ) &=\sum_{i=1}^{\infty} \frac{(3i+1) \cdot 3i+3i \cdot (3i-1)-2(3i-1) \cdot (3i+1)}{(3i-1) \cdot 3i \cdot (3i+1)} \\
&=\sum_{i=1}^{\infty}\frac{9i^2+3i+9i^2-3i-2(9i^2-1)}{(3i-1) \cdot 3i \cdot (3i+1)} \\
&=\sum_{i=1}^{\infty}\frac{18i^2-18i^2+2}{(3i-1) \cdot 3i \cdot (3i+1)} \\
&=\sum_{i=1}^{\infty}\frac{2}{(3i-1) \cdot 3i \cdot (3i+1)} \\
&= 2 \sum_{i=1}^{\infty} \frac{1}{(3i-1) \cdot 3i \cdot (3i+1)}
\end{align}
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|
Prove $a^3+b^3+c^3\ge a^2+b^2+c^2$ if $ab+bc+ca\le 3abc$
if $a,b,c$ are positive real numbers and $ab+bc+ca\le 3abc$ Prove:$$a^3+b^3+c^3\ge a^2+b^2+c^2$$
Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.
Things I have done so far: We are given $ab+bc+ca\le 3abc$ So$$2(ab+bc+ca)\le 6abc$$
so I can rewrite question as$$a^3+b^3+c^3+6abc \ge (a+b+c)^2$$
Also We can say$$9a^2b^2c^2 \ge (ab+bc+ca)^2 \ge 3abc(a+b+c)$$
My problem is is finding a way to use this facts and change the form of inequality to apply AM-GM.
|
First
$$3abc\ge ab+bc+ca\ge 3\sqrt[3]{a^2b^2c^2}.$$
So $abc\ge 1$.
It's then follows that $$a+b+c\ge 3\sqrt[3]{abc}\ge 3.$$
Now
$$3(a^3+b^3+c^3)\ge (a^2+b^2+c^2)(a+b+c)$$
by Rearrangement Inequality or by
$$(a+b+c)(a^3+b^3+c^3)\ge (a^2+b^2+c^2)^2\ge (a^2+b^2+c^2)\frac{(a+b+c)^2}{3}.$$
|
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|
Prove $(x+y)(y+z)(z+x)\ge\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$
If $x$, $y$ and $z$ are positive numbers,prove:
$$(x+y)(y+z)(z+x)\ge\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}.$$
Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.
Things I have done so far: We know that $$(x+y)(y+z)(z+x)\ge8xyz$$$$\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}\ge8xyz$$
So $(x+y)(y+z)(z+x)\ge\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$ is stronger inequality than $(x+y)(y+z)(z+x)\ge8xyz$. My Problem in Proving this inequality is that $\frac{1}{3}$ and $\sqrt[3]{}$ makes me think applying $3$ term AM-GM.but there is $8$ terms on Left hand side.Also I can't think of way to make $8$ on right hand side by $3$ term AM-GM.
By writing LHS in expanded form I was able to reach this$$LHS+xyz \ge 3\sum_{cyc}x\sqrt[3]{x^2y^2z^2}$$
|
Well I think i figured it out.by following what I have written in my post:$$LHS\ge 3(\sum_{cyc}x)\sqrt[3]{x^2y^2z^2} - xyz \ge \frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$$
For Proving $3(\sum_{cyc}x)\sqrt[3]{x^2y^2z^2} - xyz \ge \frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$ :$$\sqrt[3]{x^2y^2z^2}(3(\sum_{cyc}x) - \sqrt[3]{xyz})\ge\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$$ $$3(\sum_{cyc}x) - \sqrt[3]{xyz} \ge \frac{8}{3}(x+y+z)$$ $$\frac{x+y+z}{3}\ge\sqrt[3]{xyz}$$
Which is true by AM-GM.
|
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|
$2^{nd}$ order ODE $4x(1-x)y''-y=0$ with $y'(0)=1$ at $x=0$ it has two singular point $x_0 =1$ and $x_0=1$
I rearrange the equation to
$$y'' - \frac{1}{4x(1-x)}y =0$$
and by $v(x)=\frac{(x-x_0)^2}{4x(1-x)}$ are analytic at $x_0$ for both $0$ and $1$;
that is, $x_0=1,0$ are regular singular points,
so that, I can find a solution by series... (Here, what kind of method should I apply..?)
I just tried the Method of Frobenius:
where the indicial equation is $$r(r-1)-1=0$$ and $r_1=\frac{1+\sqrt5}{2}$and $r_2=\frac{1-\sqrt5}{2}$ where $r_1 - r_2 =\sqrt5$ is not ingeter.
I require to use $$y_1 = x^\frac{1+\sqrt5}{2} \sum_{n=0}^{\infty}a_n x^n$$
$$y_2 = x^\frac{1-\sqrt5}{2} \sum_{n=0}^{\infty}b_n x^n$$
it seems not sensible to solve this with $x^\frac{1-\sqrt5}{2}$ trem,
should I keep working with Method of Frobenius...?... or in any other way..?
|
$$a_n=\frac{(4*2*1+1)(4*3*2+1)...(4(n-1)(n-2)+1)}{4^{n-1}n!(n-1)!}=\frac{4^{n-2}(2*1+\frac{1}{4})(3*2+\frac{1}{4})...((n-1)(n-2)+\frac{1}{4})}{4^{n-1}n!(n-1)!}=\frac{(2*1+\frac{1}{4})(3*2+\frac{1}{4})...((n-1)(n-2)+\frac{1}{4})}{4*n!(n-1)!}$$
Thanks, I just tried it again..for this part
$$I=(2*1+\frac{1}{4})(3*2+\frac{1}{4})...((n-1)(n-2)+\frac{1}{4})$$
$$I=\prod_{n=3}^{+\infty}[(n-1)(n-2)+\frac{1}{4}]$$$$=\prod_{n=3}^{+\infty}[n^2 - 3n +\frac{9}{4}]$$$$=\prod_{n=3}^{+\infty}[(n-\frac{3}{2})^2]$$
$$I=\Gamma^2(n-\frac{3}{2}+1)=\Gamma^2(n-\frac{1}{2})$$
substitute this back, there is $$a_n=\frac{\Gamma^2(n-\frac{1}{2})}{4*\Gamma(n+1)\Gamma(n)}$$
where has a difference of $\pi$ and $4$ of denominator..
b.tw..the computer also gives $\pi$... how does this comes?
|
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|
Two overlapping squares $ABCD$ is a square. $BEFG$ is another square drawn with the common vertex $B$ such that $E,\ F$ fall inside the square $ABCD$. Then prove that $DF^2=2\cdot AE^2$.
|
$(NB)^2+(EN)^2=(EB)^2=y^2$$(NB)^2=y^2-(EN)^2$$(x-AN)^2=y^2-(y\sin\theta)^2=(y\cos\theta)^2$$x-AN=\pm y\cos\theta$$AN=x-y\cos\theta(\because y,\cos\theta \text{ both positive})$$(AE)^2=(AN)^2+(EN)^2$$(AE)^2=(x-y\cos\theta)^2+(y\sin\theta)^2$$(AE)^2=x^2-2xy\cos\theta+y^2\cos^2\theta+y^2\sin^2\theta=x^2-2xy\cos\theta+y^2$
$\sin(90-\theta)=\frac{EM}{y}, \therefore EM=y \cos\theta \therefore MN=y(\cos\theta+\sin\theta)$
$\therefore DP=x-y(\cos\theta+\sin\theta)$
$\cos(90-\theta)=\frac{MF}{y} \therefore MF=y\sin\theta \text{ and }PM=x-NB=x-y\cos\theta$
$PF=x-y(\cos\theta-\sin\theta)$
$(DF)^2=(DP)^2+(PF)^2$
Substitute and simplify to get $2(x^2+y^2-2xy\cos\theta)$
Hence $(DF)^2=2(AE)^2$
|
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|
Question about initial value problem This seems like it should be something simple.
While solving some HW problems I ran across this:
$dy/dx = ({1-2x})/y$
which I separated and integrated:
$\int{Ydy} = \int{1-2x}dx$
$y^2/2= x-x^2 + c$
$y^2=2(x-x^2+c)$
$y=\sqrt{2(x-x^2+c)}$
$y=\sqrt{2x-2x^2+2c}$
I was given $y(1)=-2$ and asked to solve. So i did this:
$-2=\sqrt{2(1)-2(1^2)+2c}$
$-2=\sqrt{2-2+2c}$
$-2=\sqrt{2c}$
So, I assumed c must be some complex number? However, the final solution given by the book pops a negative sign into the equation and just has $c=2$, so the solution looks like this:
$y=-\sqrt{2x-2x^2+2c}$
where $c=2$
$y=-\sqrt{2x-2x^2+4}$
which is obviously a true statement. I'm just confused where the negative sign on the right side of the equation came from. I'm sure its some silly algebra thing I keep overlooking.
|
$$\frac {y^2}{2} = -x^2+x+c \iff y=-\sqrt2 \sqrt{-x^2+x+c}$$ or $$y=\sqrt2 \sqrt{-x^2+x+c}$$
The first one when solved for the initial condition gives $c=2$ while the second one has no solution, thus can't satisfy the initial condition, ergo no solution exists. So you're left with only $$y=-\sqrt2 \sqrt{-x^2+x+2}$$
|
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|
Why does $\lim_{x\rightarrow\infty} x-x^{\frac{1}{x}^{\frac{1}{x}}}-\log^2x=0?$ Why does $$\lim_{x\rightarrow\infty} x-x^{\frac{1}{x}^{\frac{1}{x}}}-\log^2x=0?$$
Moreover, why is $$x-x^{\frac{1}{x}^{\frac{1}{x}}}\approx\log^2 x?$$
|
By a Taylor expansion,
$$
\left( \frac{1}{x} \right)^{1/x} = \exp\left( -\frac{1}{x}\ln x \right) = 1-\frac{\ln x}{x} + O\left(\frac{\ln^2 x}{x^2}\right) .
$$
Thus the second term equals
$$
(1+\delta)\exp\left( \ln x \left( 1 - \frac{\ln x}{x} \right) \right) = x \left( 1-\frac{\ln^2 x}{x} + O\left(\frac{\ln^4 x}{x^2}\right) \right)(1+\delta)
$$
with $\delta = O(\ln^3 x/x^2)$, so this gives the claim.
|
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|
How do I prove that any unit fraction can be represented as the sum of two other distinct unit fractions? A number of the form $\frac{1}{n}$, where $n$ is an integer greater than $1$, is called a unit fraction.
Noting that
$\frac{1}{2} = \frac{1}{3} + \frac{1}{6}$
and
$\frac{1}{3} = \frac{1}{4} + \frac{1}{12}$,
find a general result of the form
$\frac{1}{n} = \frac{1}{a} + \frac{1}{b}$
and hence prove that any unit fraction can be expressed as the sum of two other distinct unit
fractions.
|
$$\frac{1}{N+1}+\frac{1}{N(N+1)}=\frac{N}{N(N+1)}+\frac{1}{N(N+1)}=\frac{N+1}{N(N+1)}=\frac{1}{N}.$$
Here's a question for further investigation: is the above decomposition of a unit fraction into a pair of distinct unit fractions unique? After all, there is more than one way to split a unit fraction into a triplet of unit fractions.
$$\frac12=\frac14+\frac16+\frac{1}{12},$$
but also
$$\frac12=\frac13+\frac18+\frac{1}{24}.$$
|
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|
How to show $\lim\limits_{x \to \infty}[x(\sqrt {x^2+a} - \sqrt {x^2+b})]=\frac{a-b}{2}$ I need to prove the result without using L'Hopitals rule
$$\lim\limits_{x \to \infty}[x(\sqrt {x^2+a} - \sqrt {x^2+b})]=\frac{a-b}{2}$$
but this seems quite miraculous to me and I'm not quite sure what to do as everything I do seems to make it worse. Any help would be appreciated.
|
HINT
Multiply by a special kind of $1$ :
$$\lim\limits_{x \to \infty}\left[x(\sqrt {x^2+a} - \sqrt {x^2+b})\right]=\lim\limits_{x \to \infty}\left[ \dfrac{x(\sqrt {x^2+a} - \sqrt {x^2+b})(\sqrt {x^2+a} + \sqrt {x^2+b})}{\sqrt {x^2+a} + \sqrt {x^2+b}}\right] $$
and use difference of squares
|
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|
How to integrate $\sqrt{1+(2/3)x}$? How would you solve the following (step by step please!):
$$\int^6_5\sqrt{1+\frac23x}\ dx$$
I started with $u=1+\frac23x$, $du=\frac23\,dx$, now what?
|
You have $$\int^6_5\sqrt{1+\frac23x}\ dx$$
Now let $u = 1 + \frac{2}{3}x$, and $du = \frac{2}{3}\ dx$, so that means $dx = \frac{3}{2}du$, so make the subsitution:
$$\int^{6 = x}_{5=x}\sqrt{u}\frac{3}{2}\ du = \frac{3}{2} \int^{6 = x}_{5 = x}u^{\frac{1}{2}}\ du$$
Now we need to find the bounds for $u$, and then we will use the power rule for antiderivatives:
Since $ u = 1 + \frac{2}{3}x$, when $x = 6$, $u =5$, and when $x = 5$, $u =\frac{13}{3}$, so those are our new bounds. So the integral is: $$\frac{3}{2} \int^{5}_{\frac{13}{3}}u^{\frac{1}{2}} du = \frac{3}{2}(\frac{2}{3}(5^\frac{3}{2}) - \frac{2}{3}(\frac{13}{3})^\frac{3}{2}) = 5^\frac{3}{2} - (\frac{13}{3})^\frac{3}{2} = \sqrt{125} - \sqrt{(\frac{13}{3})^3}$$$$ = 5\sqrt{5} - \frac{13\sqrt{\frac{13}{3}}}{3}$$
Alternately, instead of finding the new bounds for $u$, you can just put the antiderivative back in terms of $x$ and solve from there.
|
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|
Integral of $x^2 \cos(a x)\; \mathrm{d}x$ I am trying to solve the following problem:
$\int x^2 \cos(a x)\; \mathrm{d}x$
I thought this would be simple and I am pretty sure this is the answer:
$I =\frac{x^2\sin(ax)}{a}+\frac{2x\cos(ax)}{a^2}+\frac{2\sin(ax)}{a^3}$
I have been told that this isn't the correct answer?
I just applied integration by parts twice. Is my answer essentially correct other than typographical errors(sign change) etc, or have I seriously diverged from the correct calculation?
|
$$\int x^2\cos(ax)dx=|u=x^2\Rightarrow du=2xdx, dv=\cos(ax)dx\Rightarrow v=\frac{1}{a}\sin(ax)|=$$
$$=\frac{x^2}{a}\sin(ax)+\frac{2}{a}\int x\sin(ax)dx=\frac{x^2}{a}\sin(ax)+\frac{2}{a}I_1$$
$$I_1=\int x\sin(ax)dx=|u=x\Rightarrow du=dx, dv=\sin(ax)dx\Rightarrow v=\frac{1}{a}\cos(ax)|=$$
$$=\frac{x}{a}\cos(ax)+\frac{1}{a}\int\cos(ax)dx=\frac{x}{a}\cos(ax)-\frac{1}{a}\cdot\frac{1}{a}\sin(ax)dx=\frac{x}{a}\cos(ax)-\frac{1}{a^2}\sin(ax)dx$$
$$\int x^2\cos(ax)dx=\frac{x^2}{a}\sin(ax)+\frac{2}{a}I_1=\frac{x^2}{a}\sin(ax)+\frac{2}{a}\left(\frac{x}{a}\cos(ax)-\frac{1}{a^2}\sin(ax)dx\right)=$$
$$=\frac{x^2}{a}\sin(ax)+\frac{2x}{a^2}\cos(ax)-\frac{2}{a^3}\sin(ax)$$
i.e.
$$\int x^2\cos(ax)dx=\frac{x^2}{a}\sin(ax)+\frac{2x}{a^2}\cos(ax)-\frac{2}{a^3}\sin(ax)$$
|
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How does one simplify the expression $\sqrt[3]{2 \sqrt{2}}\left(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\right)$ I don't know how to solve this problem. What I know is $\sqrt[3]{2 \sqrt{2}}=\sqrt{2}$. But I don't know how to continue.
|
$$\sqrt[3]{2 \sqrt{2}}=\sqrt2$$
$$\sqrt{2-\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt 2},\sqrt{2+\sqrt{3}}=\frac{\sqrt{3}+1}{\sqrt 2}$$
$$\sqrt[3]{2 \sqrt{2}}\left(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\right)=\sqrt[3]{2^{3/2}}\left(\frac{\sqrt{3}-1}{\sqrt 2}+\frac{\sqrt{3}+1}{\sqrt 2}\right)=$$
$$=\sqrt 2\frac{\sqrt3-1+\sqrt3+1}{\sqrt 2}=2\sqrt3$$
|
{
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|
Find $ \int \frac{dx}{x\sqrt{1-x^4}}$ Find $\displaystyle \int \dfrac{dx}{x\sqrt{1-x^4}}$
I cannot figure out how start this problem, can anyone explain
|
$$
\begin{aligned}
\int\frac{\mathrm{d}x}{x\sqrt{1-x^4}}&=\int\frac{\mathrm{d}x}{x^3\sqrt{1/x^4 - 1}}\\
&=\int\frac{1}{\sqrt{(1/x^2)^2 - 1}}\frac{\mathrm{d}x}{x^3}\\
&=\int\frac{1/x^2+\sqrt{(1/x^2)^2 - 1}}{\sqrt{(1/x^2)^2 - 1}}\,\frac{1}{1/x^2 + \sqrt{(1/x^2)^2 - 1}}\frac{\mathrm{d}x}{x^3}\\
&=-\frac{1}{2}\int \frac{1}{1/x^2 + \sqrt{(1/x^2)^2 - 1}}\left(1 + \frac{1/x^2}{\sqrt{(1/x^2)^2 - 1}}\right)\frac{-2}{x^3}\,\mathrm{d}x
\end{aligned}
$$
Set $t = 1/x^2 + \sqrt{(1/x^2)^2 - 1}$, so $\mathrm{d}t=\left(1 + \frac{1/x^2}{\sqrt{(1/x^2)^2 - 1}}\right)\frac{-2}{x^3}\,\mathrm{d}x$. Then
$$ \int\frac{\mathrm{d}x}{x\sqrt{1-x^4}}=-\frac{1}{2}\log\left(\frac{1+\sqrt{1-x^4}}{x^2}\right) + C. $$
|
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|
Finding x using the pythagoras theorem $$x^2 = (x+1)^2 + (x-7)^2$$
can someone please find $x$? Also this is a quadratic equation problem solving question.
|
As in user$169852$'s answer
$$x^2 = (x+1)^2 + (x-7)^2 \quad
\implies x^2 - 12 x + 50 = 0$$
and the quadratic formula shows that $x$ is complex.
$$x=\dfrac{12\pm\sqrt{144-4(1)50}}{2(1)}
=
\dfrac{12\pm2i\sqrt{14}}{2}
= 6\pm i\sqrt{14}
$$
however by the same logic
$$ x^2 = (x+1)^2 + (x+7)^2\quad \implies
x=-8\pm \sqrt{14}$$
and
$$x^2 = (x+1)^2 - (x - 7)^\quad \implies x\in\{4,12\}$$
I suspect that the latter is what you are looking for where
$A= x-7=5\qquad B=x=12\qquad C=x+1=13
$
$$\implies (A,B,C)=(5,12,13)$$
|
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|
$\int\frac{dx}{x-3y}$ when $y(x-y)^2=x$?
If y is a function of x such that $y(x-y)^2=x$
Statement-I: $$\int\frac{dx}{x-3y}=\frac12\log[(x-y)^2-1]$$
Because
Statement-II: $$\int\frac{dx}{x-3y}=\log(x-3y)+c$$
Question: Is Statement-I true? Is Statement-II true? Is Statement-II a correct explanation for Statement-I?
I can say that II is false because y is not a constant but indeeed a function, I don't know answers to other two questions.I am thinking that probably we have to eleiminate y from the given functional equation or do a clever substitution?
|
Differentiating
$$
x=y(x-y)^2\tag{1}
$$
we get
$$
\begin{align}
1&=y'(x-y)^2+2y(x-y)(1-y')\\
&=y'\left[(x-y)^2-2y(x-y)\right]+2y(x-y)\\
&=y'(x-3y)(x-y)+2y(x-y)\\
1-2y(x-y)&=y'(x-3y)(x-y)\\
(x-y)^2-1&=(1-y')(x-3y)(x-y)\tag{2}
\end{align}
$$
Differentiating Statement-I, we get
$$
\frac{(x-y)(1-y')}{(x-y)^2-1}=\frac1{x-3y}\tag{3}
$$
which matches $(2)$. Therefore, with the exception of a constant of integration, Statement-I is correct. That is, dividing $(2)$ by $(x-3y)\left[(x-y)^2-1\right]$ and integrating, we get
$$
\begin{align}
\int\frac{\mathrm{d}x}{x-3y}
&=\int\frac{(x-y)\,\mathrm{d}(x-y)}{(x-y)^2-1}\\
&=\int\frac{\frac12\,\mathrm{d}[(x-y)^2-1]}{(x-y)^2-1}\\[4pt]
&=\frac12\log\left[(x-y)^2-1\right]+C\tag{4}
\end{align}
$$
As you surmised, Statement-II implies that $y'=0$, which is false.
|
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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.