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How do you integrate: $\int_0^r\sqrt{x-x^2}.dx$ $$\int_0^r\sqrt{x-x^2}.dx$$
I only have basic calculus and would like to know how would one go about integrating an expression of this form.
I have tried substituting say $u=x-x^2$ but I'm still left with an $x$.
The method is not in my book and I can't find a similar example anywhere.
A hint in the right direction is sufficient.
|
$$\mathcal{I}(\text{r})=\int_0^\text{r}\sqrt{x-x^2}\space\text{d}x=\int_0^\text{r}\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}\space\text{d}x=$$
Substitute $u=x-\frac{1}{2}$ and $\text{d}u=\text{d}x$.
This gives a new lower bound $u=0-\frac{1}{2}=-\frac{1}{2}$ and upper bound $u=\text{r}-\frac{1}{2}$:
$$\int_{-\frac{1}{2}}^{\text{r}-\frac{1}{2}}\sqrt{\frac{1}{4}-u^2}\space\text{d}u=$$
Substitute $u=\frac{\sin(s)}{2}$ and $\text{d}u=\frac{\cos(s)}{2}\space\text{d}s$.
Then $\sqrt{\frac{1}{4}-u^2}=\sqrt{\frac{1}{4}-\frac{\sin^2(s)}{4}}=\frac{\cos(s)}{2}$ and $s=\arcsin(2u)$.
This gives a new lower bound $s=\arcsin\left(-1\right)=-\frac{\pi}{2}$ and upper bound $s=\arcsin\left(2\text{r}-1\right)$:
$$\frac{1}{4}\int_{-\frac{\pi}{2}}^{\arcsin\left(2\text{r}-1\right)}\cos^2(s)\space\text{d}s=$$
Use $\cos^2(s)=\frac{1+\cos(2s)}{2}$:
$$\frac{1}{8}\int_{-\frac{\pi}{2}}^{\arcsin\left(2\text{r}-1\right)}\cos(2s)\space\text{d}s+\frac{1}{8}\int_{-\frac{\pi}{2}}^{\arcsin\left(2\text{r}-1\right)}1\space\text{d}s=$$
For $\int_{-\frac{\pi}{2}}^{\arcsin\left(2\text{r}-1\right)}\cos(2s)\space\text{d}s$ substitute $p=2s$ and $\text{d}p=2\space\text{d}s$.
This gives a new lower bound $s=2\cdot-\frac{\pi}{2}=-\pi$ and upper bound $s=2\arcsin\left(2\text{r}-1\right)$:
$$\frac{1}{16}\int_{-\pi}^{2\arcsin\left(2\text{r}-1\right)}\cos(p)\space\text{d}p+\frac{1}{8}\int_{-\frac{\pi}{2}}^{\arcsin\left(2\text{r}-1\right)}1\space\text{d}s=$$
Use:
*
*$$\int1\space\text{d}s=s+\text{C}$$
*$$\int\cos(p)\space\text{d}p=\sin(p)+\text{C}$$
$$\frac{1}{16}\left[\sin(p)\right]_{-\pi}^{2\arcsin\left(2\text{r}-1\right)}+\frac{1}{8}\left[s\right]_{-\frac{\pi}{2}}^{\arcsin\left(2\text{r}-1\right)}=$$
$$\frac{\sin(2\arcsin\left(2\text{r}-1\right))-\sin(-\pi)}{16}+\frac{\arcsin\left(2\text{r}-1\right)+\frac{\pi}{2}}{8}=$$
$$\frac{\sin(2\arcsin\left(2\text{r}-1\right))}{16}+\frac{\arcsin\left(2\text{r}-1\right)+\frac{\pi}{2}}{8}$$
|
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|
Range of $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$ with $A,B,C$ angles in a triangle
What is the range of $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$ if $A,B,C$ are angles in a triangle?
For $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$, I know we have to apply the AM–GM inequality. So
$$\begin{align}\frac12\left(\tan^2\frac A2+\tan^2\frac B2+\tan^2\frac C2\right)&\ge\sqrt{\tan^2\frac A2\tan^2\frac B2\tan^2\frac C2}\\
&\ge\tan\frac A2\tan\frac B2\tan\frac C2\\
\tan^2\frac A2+\tan^2\frac B2+\tan^2\frac C2&\ge2\tan\frac A2\tan\frac B2\tan\frac C2
\end{align}$$
After this, the LHS is of the required form, but the RHS is having half-angle terms. Then again, if $A+B+C=\pi$, $\tan\frac{A+B+C}2=\tan90^\circ=\infty$. Now I am stuck; how to solve further?
The given options are:
*
*$>1$
*$<1$
*$\ge1$
*$\le1$
|
HINT:
As $\tan\left(\dfrac{A+B}2\right)=\tan\left(\dfrac{\pi- C}2\right)$
$$\implies\dfrac{\tan\dfrac A2+\tan\dfrac B2}{1-\tan\dfrac A2\tan\dfrac B2}=\dfrac1{\tan\dfrac C2}$$
$$\implies\tan\dfrac A2\tan\dfrac B2+\tan\dfrac B2\tan\dfrac C2+\tan\dfrac C2\tan\dfrac A2=1$$
Now $$\left(\tan\dfrac A2-\tan\dfrac B2\right)^2+\left(\tan\dfrac B2-\tan\dfrac C2\right)^2+\left(\tan\dfrac C2-\tan\dfrac A2\right)^2\ge0$$
|
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|
What is the determinant of the matrix $A^2+B^2$
Let $A$ and $B$ be two $n\times n $ matrices such that $A\neq B$, $A^3=B^3$, and $A^2B=B^2A,$ then what is the value of $\det(A^2+B^2)$ ?
My attempt: $A^3=B^3 \Rightarrow (A-B)(A^2+AB+B^2)=0 \Rightarrow A^2+AB+B^2=0$
but I can't get any idea
|
First a couple of notes on your attempt:
$$(A-B)(A^2+AB+B^2) = A^3+A^2B+AB^2-BA^2-BAB-B^3$$
and if $A$ and $B$ do not commute, it doesn't need to be $A^3-B^3$. You have to be careful about factorization in non-commutative rings, familiar formulas do not always work.
Even it were true, as I already said in comments, $$((A-B)(A^2+AB+B^2) = 0\ \wedge\ A\neq B)\ \nRightarrow\ A^2 +AB + B^2 = 0$$
This is because there are zero divisors in matrix algebra (there are examples in the link).
Now, to the solution:
$$(A^2+B^2)A = A^3 + B^2A = B^3 +A^2B = (A^2 + B^2)B \implies (A^2 + B^2)(A-B) = 0$$
Now, if $\det(A^2 + B^2) \neq 0$, then $A^2 + B^2$ is invertible, but then $A-B = 0$ which is contradiction. Hence, $\det (A^2 + B^2) = 0$.
|
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|
Simplifying surd of degree 1/4 to 1/2 What will be the simplification of
$$\sqrt { 6 } \cdot \sqrt { 2-\sqrt { 3 } } $$
Thanks in advance!
|
$$\sqrt { 6 } \cdot \sqrt { 2-\sqrt { 3 } } =\sqrt { 3 } \cdot \sqrt { 2 } \sqrt { 2-\sqrt { 3 } } =\sqrt { 3 } \sqrt { 4-2\sqrt { 3 } } =\sqrt { 3 } \sqrt { { \left( \sqrt { 3 } -1 \right) }^{ 2 } } \\=\sqrt { 3 } \left( \sqrt { 3 } -1 \right) = 3-\sqrt { 3 } $$
|
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|
Minimum value of $a^2 + b^2$ so that the quadratic $x^2 + ax + (b+2) = 0$ has real roots
The equation $ x^2 + ax + (b+2) = 0 $ has real roots, where $a$ and $b$ are real numbers.
How would I find the minimum value of $a^2 + b^2$ ?
|
In the $(a,b)$ plane, the point $(a,b)$ has to be under the parabola $a^2-4b =8$. The point of this domain closest to origin is the vertex of the parabola, $(0,-2)$. Hence the minimum of $a^2+b^2$ is $4$.
|
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|
Is there a less trickier way to solve this differential equation? Solve the following differential equation:
$$ y^3 dy+(x + y^2)dx = 0$$
Solution:
The following solution uses the substitution $y^2=tx$.
$$y^3\frac{dy}{dx}+x+y^2=0\tag1$$
Differentiating $y^2 = tx$,
$$ 2y\frac {dy}{dx}= t + x\frac{dt}{dx}$$
Now substituting in (1),
$$tx\ (t + x\frac{dt}{dx})+2(x+tx)=0$$
$$t^2x+tx^2\frac{dt}{dx}+ 2x + 2tx=0 $$
$$2tx\ dx + t^2x\ dx + 2x\ dx+tx^2\ dt=0$$
$$x\ dx\ (t^2 + 2t + 2)=-tx^2\ dt$$
$$\int{\frac{1}{x}}dx= \int{\frac{-t}{t^2+2t+2}}dt$$
$$lnx=\int{\frac{-t-1}{t^2+2t+2}}dt + \int{\frac{1}{t^2+2t+2}dt}$$
$$lnx=\int{\frac{-t-1}{(t+1)^2+1}}dt+\int{\frac{1}{(t+1)^2+1}}dt$$
$$lnx=-\frac{1}{2}ln[(t+1)^2+1]+ \arctan(\frac{y^2}{x}+1)$$
$$lnx=-\frac{1}{2}ln[(\frac{y^2}{x}+1)^2+1)]+ \arctan(\frac{y^2}{x}+1)$$
I have the following questions:
1) How did the person who solved this problem get the idea of substituting
$y=tx^2$ ? This substitution doesn't seem to be obvious.
2) Is there any other method (without using the above substitution) to solve this problem, in a less trickier manner?
|
$y^3~dy+(x+y^2)~dx=0$
$y^3\dfrac{dy}{dx}=-x-y^2$
Let $u=y^2$ ,
Then $\dfrac{du}{dx}=2y\dfrac{dy}{dx}$
$\therefore\dfrac{y^2}{2}\dfrac{du}{dx}=-x-y^2$
$\dfrac{du}{dx}=-\dfrac{2x}{y^2}-2$
$\dfrac{du}{dx}=-\dfrac{2x}{u}-2$
Luckily this becomes a first-order homogeneous ODE.
|
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|
Find the equation of the plane that has distance $1$ from line $X: (1,0,2) + \lambda (1,0,2)$ and contains points $P=(1,1,-1)$ and $Q = (2,1,1)$ I am asked the following problem:
Find the equation of the plane that has distance $1$ from line $X: (1,0,2) + \lambda (1,0,2)$ and contains points $P=(1,1,-1)$ and $Q = (2,1,1)$
Let's call the normal vector of the plane $\vec{n}$ so that $\vec{n} = (a,b,c)$
What I have so far is:
I) Since the distance from the plane and the line is $1$, using the point $(1,0,2)$ from the line and the normal vector of the plane,
$$
d_{istance} = 1 = \frac{\vert 1 a + 0 b + 2 c + d \vert }{\sqrt{a^2+b^2+c^2}}\\
1 = \frac{\vert a+2c+d \vert }{\sqrt{a^2+b^2+c^2}}
$$
II) Since the plane must be parallel to the line, the dot product between the normal vector and $(1,0,2)$ must be zero:
$$
(a,b,c) \cdot (1,0,2) = 0\\
a+2c = 0
$$
III) Since points $P$ and $Q$ are contained on the plane,
$$
a+b-c+d=0\\
2a+b+c+d = 0
$$
I'm stuck on this part: there are more variables than equations to solve (If we combine the equations in III we get equation II). Is there something else that I'm missing?
Textbook's answer: $y-1=0$ and $6x-2y-3z-7=0$
Thank you.
|
($d=1$ can be taken arbitrarily here. If another value is chosen, the values for $a,b,c$ are scaled by a factor of $d$ and the same plane equations result.)
$$1=\frac{|a+2c+1|}{\sqrt{a^2+b^2+c^2}}\tag1$$
$$a+b-c=-1\tag2$$
$$2a+b+c=-1\tag3$$
$$a+2c=0\tag4$$
Substitute (4) into (1) twice:
$$1=\frac{|0+1|}{\sqrt{(-2c)^2+b^2+c^2}}$$
$$1=\frac1{\sqrt{5c^2+b^2}}$$
$$5c^2+b^2=1\tag5$$
Substitute (4) and (5) into (2) to leave an equation in $c$ only:
$$-2c+\sqrt{1-5c^2}-c=-1$$
$$\sqrt{1-5c^2}=3c-1$$
$$1-5c^2=9c^2-6c+1$$
$$14c^2-6c=0$$
from which we get $c=0$ or $c=\frac37$. For both cases substitute into (4) and then (2) to get two possible normals:
$$a=0,b=-1,c=0$$
$$a=-\frac67,b=\frac27,c=\frac37$$
(2) establishes that $(a,b,c)\cdot P=-1$ in both cases, so we have the plane equations
$$-y=-1,\quad-\frac67x+\frac27y+\frac37z=-1$$
which rearrange to the given answers of
$$y-1=0,\quad 6x-2y-3z-7=0$$
|
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|
Find the sum of the all possible values of $n$ such that $5\cdot 3^m+4=n^2$ $5\cdot 3^m+4=n^2$.
Find the sum of all possible values of $n$.
It is an question from prermo 2016 west Bengal exam. I try to do it using theory of congruence. But I can't proceed. I am disappointed, how do I find the sum? Can anybody can help me?
Thank you
|
Given equation can be re written as
$$5\cdot 3^m = (n-2)(n+2)$$
let $\gcd((n-2),(n+2))=d$ So $d|(n+2)-(n-2)=4$ Thus possible values for $d$ is $1,2,4$. As $d|(5\cdot 3^m), d=1$ as $5\cdot 3^m$ is odd for all values of $m$. This ensure that $5\cdot 3^m$ can only be factored as $5$ and $3^m$ , not as $5\cdot 3^k$ and $ 3^{(m-k)}$ where $1 \le k \le m$ as on that case $\gcd(5\cdot3^k,3^{(m-k)})=3^r$ where $r=min(k,m-k)$ , but that is not possible. So either $n-2=5$ and $n+2= 3^m$ or $n+2=5$ and $n-2= 3^m$. From first case we get $(m,n)=(2,7)$ and from second case we get $(m,n)=(0,3)$
|
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|
$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$ Find the values of $a$ and $b$ for which the polynomial $ax^3+8x^2+bx+6$ is divisible by $x^2-2x-3$.
|
if we divide both polynomials we get
$$xa+2\,a+8+1/4\,{\frac {27\,a+3\,b+78}{x-3}}+1/4\,{\frac {a+b-14}{x+1}}$$
and the remainder must be zero. And it must be
$$27a+3b+78=0$$ and $$a+b-14=0$$
|
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|
Lengthy partial fractions? $$\frac{1}{(x+3)(x+4)^2(x+5)^3}$$
I was told to integrate this, I see partial fractions as a way, but this absurd! Is there an easier way?
|
Splitting into partial fractions can also be done this way:
Let $u = (x+3)(x+5)$ and $v = (x+4)^2$. Then $v-u = 1$ and hence
\begin{align*}
\frac{1}{(x+3)(x+4)^2(x+5)} &= \frac{v-u}{uv} \\
&= \frac{1}{u} - \frac{1}{v} \\
&= \frac{1}{(x+3)(x+5)} - \frac{1}{(x+4)^2}\\
&= \frac{1}{2}\frac{1}{x+3} - \frac{1}{2}\frac{1}{x+5} - \frac{1}{(x+4)^2}
\end{align*}
Thus
\begin{align*}
\frac{1}{(x+3)(x+4)^2(x+5)^3} & = \frac{1}{(x+5)^2}\left(\frac{1}{2}\frac{1}{x+3} - \frac{1}{2}\frac{1}{x+5} - \frac{1}{(x+4)^2}\right)\\
&= \frac{1}{2}\frac{1}{(x+3)(x+5)^2} - \frac{1}{2}\frac{1}{(x+5)^3} - \frac{1}{(x+4)^2(x+5)^2} \\
&= \frac{1}{4}\left(\frac{(x+5)-(x+3)}{(x+3)(x+5)^2}\right)- \frac{1}{2}\frac{1}{(x+5)^3} - \frac{((x+5) - (x+4))^2}{(x+4)^2(x+5)^2} \\
&= \frac{1}{4}\left(\frac{1}{(x+3)(x+5)}-\frac{1}{(x+5)^2}\right) - \frac{1}{2}\frac{1}{(x+5)^3} \\
& \qquad \qquad - \left(\frac{1}{(x+4)^2} - \frac{2}{(x+4)(x+5)} + \frac{1}{(x+5)^2}\right)\\
&=\frac{1}{4}\left(\frac{1}{2}\left(\frac{1}{(x+3)} - \frac{1}{(x+5)}\right)-\frac{1}{(x+5)^2}\right) - \frac{1}{2}\frac{1}{(x+5)^3} \\
& \qquad \qquad - \left(\frac{1}{(x+4)^2} - \frac{2}{(x+4)} + \frac{2}{(x+5)} + \frac{1}{(x+5)^2}\right)\\
&= \frac{1/8}{x+3}+\frac{2}{x+4} - \frac{1}{(x+4)^2} - \frac{17/8}{x+5} - \frac{5/4}{(x+5)^2} - \frac{1/2}{(x+5)^3}
\end{align*}
|
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|
A different way to prove that $\int_{1+\sqrt{2}}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, dx = \frac{3 G}{4} + \frac{\pi}{16} \, \ln 2$ From the fact that $$\int_{0}^{1} \int_{0}^{1} \frac{1}{2-x^{2}-y^{2}} \, dx \, dy $$ is an integral representation of Catalan's constant ($G$), I was able to deduce that $$\int_{1+\sqrt{2}}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, dx = \int_{0}^{\pi/8} \ln(1+ \cot u) \, du = \frac{3G}{4} + \frac{\pi}{16} \, \ln 2\tag{1}.$$
What is another way to prove $(1)$ that preferably doesn't involve the dilogarithm function?
EDIT:
In response to Dr. MV's comment, the following is how I deduced $(1)$ from that integral representation of Catalan's constant.
$$ \begin{align} G&= \int_{0}^{1} \int_{0}^{1} \frac{1}{2-x^{2}-y^{2}} \, dx \, dy \\ &= \int_{0}^{1} \frac{1}{2-y^{2}} \int_{0}^{1} \frac{1}{1-\frac{x^{2}}{2-y^{2}}} \, dx \, dy \\ &= \int_{0}^{1} \frac{1}{\sqrt{2-y^{2}}} \, \text{artanh} \left(\frac{1}{\sqrt{2-y^{2}}} \right) \, dy \\ &= \int_{0}^{\pi/4} \text{artanh} \left(\frac{1}{\sqrt{2} \cos \theta} \right) \, d \theta \\ &= \frac{1}{2} \int_{0}^{\pi/4} \ln \left(\frac{\sqrt{2} \cos \theta +1}{\sqrt{2} \cos \theta -1} \right) \, d \theta \\ &= \frac{1}{2} \int_{0}^{\pi/4} \ln \left(\frac{(\sqrt{2} \cos \theta+1)^{2}}{2 \cos^{2} \theta -1} \right) \, d \theta \\ &=\int_{0}^{\pi/4} \ln (\sqrt{2} \cos \theta +1) \, d \theta - \frac{1}{2} \int_{0}^{\pi/4} \ln(\cos 2 \theta) \, d \theta \\ &= \int_{0}^{\pi/4} \ln\left(\sqrt{2} \cos \left(\frac{\pi}{4} - \phi\right)+1\right) \, d \phi - \frac{1}{4} \int_{0}^{\pi/2} \ln( \cos \tau) \, d \tau \\ &= \int_{0}^{\pi/4} \ln \left(\sin(\phi) + \cos(\phi)+1\right) \, d \phi - \frac{1}{4} \left(- \frac{\pi}{2} \, \ln 2 \right) \\ &= \int_{0}^{\pi/4} \ln (\sin \phi) \, d \phi + \int_{0}^{\pi/4} \ln \left(1+ \frac{1+ \cos \phi}{\sin \phi} \right) \, d \phi + \frac{\pi}{8} \, \ln 2 \\ &= - \frac{G}{2} - \frac{\pi}{4} \, \ln 2 + \int_{0}^{\pi/4} \ln \left(1+ \cot \frac{\phi}{2} \right) \, d \phi + \frac{\pi}{8} \, \ln 2 \\ &= - \frac{G}{2} - \frac{\pi}{8} \, \ln 2 + 2 \int_{0}^{\pi/8} \ln (1 + \cot u) \, du \end{align}$$
|
Using the identity
$$
\log(\cos(x))=-\log(2)+\sum_{k=1}^\infty(-1)^{k-1}\frac{\cos(2kx)}k\tag{1}
$$
and the evaluations
$$
\begin{align}
\sin(3k\pi/4)-\sin(k\pi/4)
&=2\color{#C00000}{\cos(k\pi/2)}\color{#00A000}{\sin(k\pi/4)}\\
&=\left\{\begin{array}{}
\color{#C00000}{0}&\text{if $k$ is odd}\\
2(-1)^{\frac{k/2+1}2}&\text{if $k/2$ is odd}\\
\color{#00A000}{0}&\text{if $k/2$ is even}
\end{array}\right.\tag{2}\\
\sin(k\pi/2)&=\left\{\begin{array}{}
(-1)^{\frac{k-1}2}&\text{if $k$ is odd}\\
0&\text{if $k$ is even}
\end{array}\right.\tag{3}
\end{align}
$$
we get
$$
\begin{align}
&\int_{1+\sqrt2}^\infty\frac{\log(1+x)}{1+x^2}\,\mathrm{d}x\tag{4a}\\
&=\int_{3\pi/8}^{\pi/2}\log(1+\tan(x))\,\mathrm{d}x\tag{4b}\\
&=\int_{3\pi/8}^{\pi/2}\left[\log(\cos(x)+\sin(x))-\log(\cos(x))\right]\,\mathrm{d}x\tag{4c}\\
&=\int_{3\pi/8}^{\pi/2}\left[\log(\sqrt2\cos(x-\pi/4))-\log(\cos(x))\right]\,\mathrm{d}x\tag{4d}\\
&=\frac\pi{16}\log(2)+\int_{\pi/8}^{\pi/4}\log(\cos(x))\,\mathrm{d}x-\int_{3\pi/8}^{\pi/2}\log(\cos(x))\,\mathrm{d}x\tag{4e}\\
&=\frac\pi{16}\log(2)+\left[\sum_{k=1}^\infty(-1)^{k-1}\frac{\sin(2kx)}{2k^2}\right]_{\pi/8}^{\pi/4}\,-\left[\sum_{k=1}^\infty(-1)^{k-1}\frac{\sin(2kx)}{2k^2}\right]_{3\pi/8}^{\pi/2}\tag{4f}\\
&=\frac\pi{16}\log(2)+\sum_{k=1}^\infty(-1)^{k-1}\frac{\sin(k\pi/2)-\sin(k\pi/4)+\sin(3k\pi/4)}{2k^2}\tag{4g}\\
&=\frac\pi{16}\log(2)+\sum_{k=0}^\infty\frac{(-1)^k}{2(2k+1)^2}+\frac{2(-1)^k}{2(4k+2)^2}\tag{4h}\\
&=\frac\pi{16}\log(2)+\frac34\mathrm{G}\tag{4i}
\end{align}
$$
Explanation:
$\text{(4b)}$: substitute $x\mapsto\tan(x)$
$\text{(4c)}$: $(1+\tan(x))\cos(x)=\cos(x)+\sin(x)$
$\text{(4d)}$: $\cos(x)+\sin(x)=\sqrt2\cos(x-\pi/4)$
$\text{(4e)}$: substitute $x\mapsto x+\pi/4$ in the left integral
$\text{(4f)}$: apply $(1)$
$\text{(4g)}$: evaluate at the limits
$\text{(4h)}$: apply $(2)$ and $(3)$
$\text{(4i)}$: apply the definition of Catalan's Constant
|
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|
CDF for Negative Binomial Distribution I am trying to show that the following statement is true.
$$
\sum_{x = r}^{X}\binom{x-1}{r-1}p^r(1-p)^{x-r} =
\sum_{x = r}^{X}\binom{X}{x}p^x(1-p)^{X-x}
$$
Where $X$ and $r$ and $p$ are constants, with $X \geq r$, and $ 0 \leq p \leq 1.$
How did I get there? Well, this is the story:
Consider a sequence of independent binomial trials, each one producing the result success or failure, with probabilities $p$, and $1-p$, respectively.
Let $x$ be the total number of trials which must be carried out in order to attain exactly $r$ successes.
Knowing that the probability mass function for this Negative Binomial Distribution is as follows,
$P(x=X)=\binom{X-1}{r-1}p^r(1-p)^{X-r}$, (for $X \geq r$),
I was trying to prove the following about the corresponding Cumulative Distribution Function.
$P(x \leq X)=\sum_{x=r}^{x=X}\binom{x-1}{r-1}p^r(1-p)^{x-r}=1-\sum_{x=0}^{x=r-1}\binom{X}{x}p^x(1-p)^{X-x}$
I started out with the following:
$\sum_{x=r}^{\infty}\binom{x-1}{r-1}p^r(1-p)^{x-r}=1$,
which can be recast as below. (Relation I)
$\sum_{x=r}^{x=X}\binom{x-1}{r-1}p^r(1-p)^{x-r}+\sum_{x=X+1}^{\infty}\binom{x-1}{r-1}p^r(1-p)^{x-r}=1$
In addition, from binomial theorem, we have:
$\left ( p+(1-p) \right )^X=\sum_{x=0}^{x=X}\binom{X}{x}p^x(1-p)^{X-x}=1$
Which can be restated in Relation II as below.
$\sum_{x=0}^{x=r-1}\binom{X}{x}p^x(1-p)^{X-x}+\sum_{x=r}^{x=X}\binom{X}{x}p^x(1-p)^{X-x}=1$
Comparing the relations I and II with the expression for the CDF, the proof boils down to verification of the following:
$\sum_{x=r}^{x=X}\binom{x-1}{r-1}p^r(1-p)^{x-r}=\sum_{x=r}^{x=X}\binom{X}{x}p^x(1-p)^{X-x}$
Any idea how to continue form this point onward?
Thanks.
|
Here is another variation of the theme. It is convenient to use the coefficient of operator $[z^r]$ to denote the coefficient of $z^r$ of a series. This way we can write e.g.
\begin{align*}
[z^r](1+z)^t=\binom{t}{r}
\end{align*}
We observe LHS and RHS are polynomials in $p$ with lowest degree $r$ and highest degree $X$.
We prove the polynomials
\begin{align*}
G(p)&=\sum_{j=r}^X\binom{j-1}{r-1}p^r(1-p)^{j-r}\qquad\qquad\qquad 0\leq r \leq X, 0\leq p\leq 1\\
H(p)&=\sum_{j=r}^X\binom{X}{j}p^j(1-p)^{X-j}
\end{align*}
are equal by showing equality of the coefficients
\begin{align*}
[p^t]G(p)=[p^t]H(p)\qquad\qquad\qquad\qquad\qquad\qquad& r\leq t\leq X
\end{align*}
$$ $$
We obtain
\begin{align*}
[p^t]G(p)&=[p^t]\sum_{j=r}^X\binom{j-1}{r-1}p^r(1-p)^{j-r}\\
&=\sum_{j=r}^X\binom{j-1}{r-1}[p^{t-r}]\sum_{k=0}^{j-r}\binom{j-r}{k}(-p)^k\tag{1}\\
&=\sum_{j=r}^X\binom{j-1}{r-1}\binom{j-r}{t-r}(-1)^{t-r}\tag{2}\\
&=(-1)^{t-r}\binom{t-1}{r-1}\sum_{j=t}^X\binom{j-1}{t-1}\tag{3}\\
\end{align*}
Comment:
*
*In (1) we use the linearity of the coefficient of operator and apply the rule $[z^{t-r}]A(z)=[z^t]z^rA(z)$.
*In (2) we select the coefficient of $p^{t-r}$.
*In (3) we use the binomial identity $$\binom{j-1}{r-1}\binom{j-r}{t-r}=\binom{t-1}{r-1}\binom{j-1}{t-1}$$ and we set the lower limit of the sum to $j=t$ since otherwise $\binom{j-1}{t-1}=0$.
Since
\begin{align*}
\sum_{j=t}^X&\binom{j-1}{t-1}=\sum_{j=0}^{X-t}\binom{t+j-1}{j}=\sum_{j=0}^{X-t}\binom{-t}{j}(-1)^j\\
&=\sum_{j=0}^{X-t}[z^j](1+z)^{-t}(-1)^j\\
&=[z^0](1+z)^{-t}\sum_{j=0}^{X-t}\left(-\frac{1}{z}\right)^j\\
&=[z^0](1+z)^{-t}\frac{1-\left(-\frac{1}{z}\right)^{X-t+1}}{1+\frac{1}{z}}\\
&=(-1)^{X-t}[z^{X-t}](1+z)^{-(t+1)}\\
&=(-1)^{X-t}\binom{-(t+1)}{X-t}\\
&=\binom{X}{t}
\end{align*}
we obtain from (3)
\begin{align*}
[p^t]G(p)=(-1)^{t-r}\binom{X}{t}\binom{t-1}{r-1}\qquad\qquad r\leq t\leq X
\end{align*}
And now the RHS
We obtain using the same techniques as above
\begin{align*}
[p^t]H(p)&=[p^t]\sum_{j=r}^X\binom{X}{j}p^j(1-p)^{X-j}\\
&=\sum_{j=r}^X\binom{X}{j}[p^{t-j}]\sum_{k=0}^{X-j}\binom{X-j}{k}(-p)^k\\
&=\sum_{j=r}^X\binom{X}{j}\binom{X-j}{t-j}(-1)^{t-j}\\
&=(-1)^t\binom{X}{t}\sum_{j=r}^t\binom{t}{j}(-1)^j\tag{4}
\end{align*}
Comment:
*
*In (4) we use the binomial identity
\begin{align*}
\binom{X}{j}\binom{X-j}{t-j}=\binom{X}{t}\binom{t}{j}
\end{align*}
and we also set the upper limit of the sum to $j=t$ since otherwise $\binom{t}{j}=0$.
Since
\begin{align*}
\sum_{j=r}^t&\binom{t}{j}(-1)j=\sum_{j=0}^{t-r}\binom{t}{j+r}(-1)^{j+r}\\
&=\sum_{j=0}^\infty[z^{j+r}](1+z)^t(-1)^{j+r}\tag{5}\\
&=[z^r](1+z)^t(-1)^r\sum_{j=0}^\infty\left(-\frac{1}{z}\right)^j\\
&=(-1)^r[z^r](1+z)^t\frac{1}{1+\frac{1}{z}}\\
&=(-1)^r[z^{r-1}](1+z)^{t-1}\\
&=(-1)^r\binom{t-1}{r-1}
\end{align*}
Comment:
*
*In (5) we set the upper limit to $\infty$ without changing anything since we are adding zeros only.
*In (6) we use the formula of the geometric series expansion.
we obtain from (4)
\begin{align*}
[p^t]H(p)=(-1)^{t}\binom{X}{t}\binom{t-1}{r-1}\qquad\qquad r\leq t\leq X
\end{align*}
showing the coefficients of $G(p)$ and $H(p)$ are equal. We finally conclude:
The following is valid
\begin{align*}
G(p)=H(p)=\sum_{j=r}^X(-1)^{j}\binom{X}{j}\binom{j-1}{r-1}p^j
\end{align*}
|
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|
On fifth powers $x_1^5+x_2^5+\dots = y_1^5+y_2^5+\dots$ There's a nice identity by Vandermergel. If,
$$a^3+b^3 = c^3+d^3\tag1$$
then,
$$(ac)^3+(bc)^3+(d^2)^3=(ad)^3+(bd)^3+(c^2)^3\tag2$$
Here's one by yours truly. If,
$$a^4+b^4 = c^4+d^4\tag3$$
then,
$$(a^2 + d^2)^2 - (a^2 - d^2)^2 + (2 b c)^2 = (b^2 + c^2)^2 - (b^2 - c^2)^2 + (2 a d)^2\tag{4a}$$
$$(a^2 + d^2)^4 + (a^2 - d^2)^4 + (2 b c)^4 = (b^2 + c^2)^4 + (b^2 - c^2)^4 + (2 a d)^4\tag{4b}$$
Question: Any known $5$th deg identity that will lead from $(5)$ to $(6)$ below,
$$\sum_{i=1}^m u_i^5 = \sum_{i=1}^m v_i^5\tag5$$
$$\sum_{i=1}^n x_i^5 = \sum_{i=1}^n y_i^5\tag6$$
where, like the previous two examples have $m<n$, or $(5)$ has less terms than $(6)$?
P.S. Incidentally, since $(2)$ has $x_1^3+x_2^3+x_3^3=y_1^3+y_2^3+y_3^3$ as well as $x_1x_2x_3=y_1y_2y_3$, then it obeys the high-power relation,
$$3(x_1^3+x_2^3+x_3^3)(x_1^6+x_2^6+x_3^6-y_1^6-y_2^6-y_3^6)=2(x_1^9+x_2^9+x_3^9-y_1^9-y_2^9-y_3^9)$$
though I haven't explored the properties of $(4)$ yet.
|
Vandermergel's generalizes to any power.
For any $m$, $$a^m + b^m = c^m + d^m$$ implies
$$ (ac)^m + (bc)^m + (d^2)^m = (ad)^m + (bd)^m + (c^2)^m$$
since it's really just
$$AC + BC + D^2 - AD - BD - C^2 = (A+B-C-D)(C-D)$$
Similarly,
$$A^2+AC-AD-AE-AF-B^2-BC+BD+BE+BF = (A+B+C-D-E-F)(A-B)$$
so that $$a^m + b^m + c^m = d^m + e^m + f^m$$
implies
$$ (a^2)^m+(ac)^m +(bd)^m+(be)^m+(bf)^m = (ad)^m+(ae)^m+(af)^m+(bc)^m+(b^2)^m$$
|
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|
Show that $m! (n − m)!$ divides $n!$ for all $m$, $n \in \Bbb{N}$ with $m \leq n$. I am studying Analysis by Amann and Escher by my own I am stuck at this exercise:
Show that $m! (n − m)!$ divides $n!$ for all $m$, $n\in\Bbb{N}$ with $m\leq n$.
(Hint: $(n+1)!=n!(n+1−m)+n!m$.)
Thanks in advance
|
Take any prime number $p$ and observe that within
$$1 \cdot 2 \cdot 3 \cdot \ldots \cdot k$$
it is used exactly
$$
\left\lfloor \frac{k}{p} \right\rfloor
+\left\lfloor \frac{k}{p^2} \right\rfloor
+\left\lfloor \frac{k}{p^3} \right\rfloor
+\ldots
$$ times (this series becomes zero after $\log_p k$ steps).
So to prove your theorem we need to show that
$$
\left\lfloor \frac{n}{p} \right\rfloor
+\left\lfloor \frac{n}{p^2} \right\rfloor
+\left\lfloor \frac{n}{p^3} \right\rfloor
+\ldots
\geq
\left\lfloor \frac{m}{p} \right\rfloor
+\left\lfloor \frac{m}{p^2} \right\rfloor
+\left\lfloor \frac{m}{p^3} \right\rfloor
+\ldots
+
\left\lfloor \frac{n-m}{p} \right\rfloor
+\left\lfloor \frac{n-m}{p^2} \right\rfloor
+\ldots
$$
which is easy to see when split into multiple similar parts:
$$
\left\lfloor \frac{n}{p^i} \right\rfloor
= \left\lfloor\frac{m}{p^i}+\frac{n-m}{p^i}\right\rfloor
\geq \left\lfloor \frac{m}{p^i} \right\rfloor + \left\lfloor \frac{n-m}{p^i} \right\rfloor$$
because $\lfloor a+b\rfloor \geq \lfloor a \rfloor + \lfloor b \rfloor$ for any real numbers $a$ and $b$.
I hope this helps $\ddot\smile$
|
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|
Prove $ \frac 1 2 \cdot \frac 3 4 \cdot \frac 5 6 \cdots \frac{2n-1}{2n} < \frac 1 {(2n+1)^{0.5}} $ . Prove
$$
\frac 1 2 \cdot \frac 3 4 \cdot \frac 5 6 \cdots \frac{2n-1}{2n} < \frac 1 {(2n+1)^{0.5}}
$$
Can this be done by induction using the pi function.
If no, why not.
|
Let $$A=\frac12\cdot\frac34\cdot\frac56\cdot...\frac{2n-1}{2n}$$
$$B=\frac23\cdot\frac45\cdot\frac67\cdot...\frac{2n}{2n+1}$$
Then $$A<B$$
Then $$A^2<AB=\frac1{2n+1}$$
$$A=\frac12\cdot\frac34\cdot\frac56\cdot...\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}$$
|
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|
Does $\rvert z \rvert +z= i$ have any solutions? $$z=a+i \ b$$
$$\vert z \rvert = \sqrt{a^2+b^2} \\$$
$$\sqrt{a^2+b^2}+a+i \ b=i \\$$
\begin{cases}b=1 \\
\sqrt{a^2+b^2}+a=0 \end{cases}
$$\sqrt{a^2+1}+a=0$$
$$\sqrt{a^2+1}=-a$$
$$a^2+1=a^2 \\ \\$$
The equation has no solution
Is it correct?
|
with $$z=a+bi$$ you will get the equation
$$\sqrt{a^2+b^2}+a+bi-i=0$$ this is equivalent to
$$\sqrt{a^2+b^2}+a+i(b-1)=0$$ from here we get
$$b=1$$
and $$\sqrt{a^2+b^2}+a=0$$
from the second equation we get
$$\sqrt{a^2+b^2}=-a$$
squaring gives
$$a^2+b^2=a^2$$ substracting $a^2$ gives $$b=0$$
this is a contradiction to $$b=1$$
thus the given equation does not hold
|
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|
How can we calculate this limit How can i calculate this limit :
$$\lim_{x \rightarrow 1} \frac{\sqrt[3]{x^2}-\sqrt[3]{(x-1)^2}-1}{x-1}$$
I cannot calculate this square root
Please help
|
Rewrite the expression as follows:
\begin{equation}
\lim_{x\to 1}\frac{x^{2/3}-1}{x-1} - \lim_{x\to 1} \frac{(x-1)^{2/3}}{x-1} \tag{*}
\end{equation}
The second limit is
$$\lim_{x\to 1^+} \frac{1}{(x-1)^{1/3}} = \infty,$$
$$\lim_{x\to 1^-} \frac{1}{(x-1)^{1/3}} = -\infty.$$
To simplify the first limit in (*), let $x^{1/3} -1 = t$, or $x = (t+1)^3$, then $x \to 1^+$ means $t \to 0^+$ and $x \to 1^-$ means $t \to 0^-$. Then,
$$\lim_{x\to 1^+}\frac{x^{2/3}-1}{x-1} = \lim_{x\to 1^+}\frac{(x^{1/3}-1)(x^{1/3}+1)}{x-1} = \lim_{t \to 0^+}\frac{(t+1-1)(t+1+1)}{(t+1)^3-1} = \lim_{t \to 0^+} \frac{t(t+2)}{t^3 + 3t^2 + 3t} = \lim_{t \to 0^+}\frac{t+2}{t^2 + 3t +3} = \frac{2}{3}. $$
It is easy to that
$$\lim_{x\to 1^-}\frac{x^{2/3}-1}{x-1} = \frac{2}{3}. $$
Hence,
$$\lim_\limits{x\to 1^+}\frac{\sqrt[3]{x^2} -\sqrt[3]{(x-1)^2}-1}{x-1} = \frac{2}{3} - \infty = - \infty,$$
$$\lim_\limits{x\to 1^-}\frac{\sqrt[3]{x^2} -\sqrt[3]{(x-1)^2}-1}{x-1} = \frac{2}{3} - (-\infty) = \infty.$$
|
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|
Finding the missing entry in determinant We have,
$\begin{vmatrix}
a^2+2a & 2a+1 & 1 \\
2a+1 & x & 1\\
3 & 3 & 1
\end{vmatrix}=(a-1)^3$ I am asked to prove $x=a+2$ by using properties of determinants.
I've no idea how to solve it. Though tried to add or substract one column or row with others. No combinations had worked.
|
$$\begin{vmatrix}
a^2+2a & 2a+1 & 1 \\
2a+1 & x & 1\\
3 & 3 & 1
\end{vmatrix}\underbrace{=}_{C_1-3C_3, C_2-3C_3} \begin{vmatrix}
a^2+2a-3 & 2a-2 & 1 \\
2a-2 & x-3 & 1\\
0 & 0 & 1
\end{vmatrix}=\begin{vmatrix}
a^2+2a-3 & 2a-2 \\
2a-2 & x-3\\
\end{vmatrix}$$
$$=(x-3)(a^2+2a-3)-4(a-1)^2=(a-1)^3 \iff x-3=\frac{(a-1)^3+4(a-1)^2}{a^2+2a-3}=a-1,$$ from where $x=a+2.$
|
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|
Showing that $\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6$ How does one evaluate the following limit?
$$\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)$$
The answer is $6$.
How does one justify this answer?
Edit: So it really was just combine the fraction and use L'hopital's rule twice (because function and its first derivative are of indeterminate form at $x=1$). This problem is more straightforward than it seems at first.
|
$$ =\lim_{x \to 1} \frac{23 - 23x^{11} -11 + 11x^{23}}{1 - x^{11} - x^{23} + x^{34}} = \lim_{x \to 1} \frac{-23\cdot 11 x^{10} + 11\cdot 23 x^{22}}{-11x^{10} - 23x^{22} + 34x^{33}} =$$$$= \lim_{x \to 1} \frac{-23\cdot 11 \cdot 10 x^9 + 11\cdot 23 \cdot 22x^{21}}{-11\cdot 10 \cdot x^9 - 23\cdot 22x^{21} + 34\cdot 33 x^{32}} = \frac {3036}{506} = 6 $$
where we used Hopital twice
|
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|
Find the set of $n\in\Bbb Z^+$ with $M=\{n,n+1,n+2,n+3,n+4,n+5\}$ partitionable into two sets
Find the set of all positive integers $n$ with the property that the set $M=\{n, n + 1,n + 2,n + 3,n + 4,n + 5\}$ can be partitioned into two sets such that the product of the numbers in one set equals the product of the numbers in the other set.
If $n=1$ them $M=\{1,2,3,4,5,6\}$ and there is no such partition, so $n \ge 2$.
If the prime $p|n$ then either $p|2$ or $p|3$ or $p|5$ which means either $p=2$ or $p=3$ or $p=5$.
Suppose $n=2k$. Then $M=\{2k, 2k + 1,2k + 2,2k + 3,2k + 4,2k + 5\}$. I have no idea how to proceed.
|
Hint, given
$$n \equiv 0 \pmod{n}$$
$$n+1 \equiv 1 \pmod{n}$$
$$n+2 \equiv 2 \pmod{n}$$
$$n+3 \equiv 3 \pmod{n}$$
$$n+4 \equiv 4 \pmod{n}$$
$$n+5 \equiv 5 \pmod{n}$$
if there exists such a partition, $Q\cap S=\varnothing$ and $Q\cup S=\left \{ 0,1,2,3,4,5 \right \}$
$$\prod_{q \in Q} \left ( n+q \right ) = \prod_{s \in S} \left ( n+s \right ) \Rightarrow \prod_{q \in Q} \left ( n+q \right ) \equiv \prod_{s \in S} \left ( n+s \right ) \pmod{n}$$
which is the same as
$$\prod_{q \in Q} q \equiv \prod_{s \in S} s \pmod{n}$$
And $0$ will be on one side, let's suppose the right one
$$\prod_{q \in Q} q \equiv 0 \pmod{n}$$
which has no solutions for $n > 120$. This leaves us with just a handful of cases to analyse.
|
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|
Proving that $\angle B\le60^\circ$
$A$ is the smallest angle of $\triangle ABC$. The height from $A$ over $BC$ is equal to the median from $B$. Prove that $\angle B\le 60^\circ$.
I tried extending the median to construct a parallelogram and expressed the area in terms of $BC$, the height over $BC$ and the median, but didn't succeed.
|
Note that the median partitions the triangle into two triangles with the same area. Hence if $M$ is the midpoint of $AC$ we have: $$[MBC] = \frac{[ABC]}{2} \implies \frac{BC \cdot BM \cdot \sin \angle MBC}{2} = \frac{BC \cdot h_a}{4} $$
$$\implies \sin \angle MBC = \frac 12 \implies \angle MBC = \frac{\pi}{6} $$
Similarly:
$$[MBA] = \frac{[ABC]}{2} \implies \frac{BA \cdot BM \cdot \sin \angle MBA}{2} = \frac{BA \cdot h_c}{4} $$
$$\implies \sin \angle MBC = \frac 12 \cdot \frac{h_b}{h_a} \le \frac 12 \implies \angle MBC \le \frac{\pi}{6} $$
Therefore $\angle B = \angle BMA + \angle BMC \le \frac{\pi}{3}$
|
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|
Prove by induction that $\left [ \left (3+\sqrt{5} \right )^{n} + \left (3-\sqrt{5} \right )^{n} \right ]$ is multiple of $2^{n}$ Prove by induction that :
$\left [ \left (3+\sqrt{5} \right )^{n} + \left (3-\sqrt{5} \right )^{n} \right ]$ is multiple of $ 2^{n}$
My proof is :
At $n=1$
$$\frac{\left ( 3+\sqrt{5} \right )+\left ( 3-\sqrt{5} \right )}{2} = \frac{6}{2} = 3 \in \mathbb{Z}$$
Assume $P(k)$ is true
$$\frac{\left ( 3+\sqrt{5} \right )^{k}+\left ( 3-\sqrt{5} \right )^{k}}{2^{k}} = m \in \mathbb{Z}$$
$$\Rightarrow \left ( 3+\sqrt{5} \right )^{k}+\left ( 3-\sqrt{5} \right)^{k} = 2^{k}.m \rightarrow (*)$$
At $n=k+1$
$$\left ( 3+\sqrt{5} \right )^{k+1}+\left ( 3-\sqrt{5} \right )^{k+1} = \left [ \left ( 3+\sqrt{5} \right )^{k}+\left ( 3-\sqrt{5} \right )^{k} \right ]\left [ \left ( 3+\sqrt{5} \right )+\left ( 3-\sqrt{5} \right ) \right ]-\left [ \left ( 3+\sqrt{5} \right )^{k}\left ( 3-\sqrt{5} \right )+\left ( 3-\sqrt{5} \right )^{k}\left ( 3+\sqrt{5} \right ) \right ]$$
from $(*) $:
$$\left ( 3+\sqrt{5} \right )^{k+1}+\left ( 3-\sqrt{5} \right )^{k+1} =2^{k}.m.6-\left ( 3+\sqrt{5} \right )\left ( 3-\sqrt{5} \right ) \left [ \left ( 3+\sqrt{5} \right )^{k-1}+\left ( 3-\sqrt{5} \right )^{k-1} \right ] $$
but i can't resume my proof .. i can't do any thing with $\left [ \left ( 3+\sqrt{5} \right )^{k-1}+\left ( 3-\sqrt{5} \right )^{k-1} \right ]$ .
|
You're on the right track, but you need to use a slightly extended form of induction for this problem. Note that to prove $P(k+1)$, you don't have to limit yourself to $P(k)$; you can also use $P(i)$ for $i\leq k$. In this case, the proposition $P(k-1)$ will prove helpful for dealing with the last term. (But note that this means that you're going to have to prove two base cases, rather than just one!)
To be a little more explicit, let's write $s_n=(3+\sqrt{5})^n+(3-\sqrt{5})^n$. Then the relation that you've discovered is that $s_{k+1}=6s_k-4s_{k-1}$. (Note that you have a small mistake where you write $2^k\cdot m\cdot 4$; that $4$ should be a $6$, because it's $(3+\sqrt5)+(3-\sqrt5)$. The $4$ in the second term here arises as the product $(3+\sqrt5)\cdot(3-\sqrt5)$.) Now, we know that there are $a$ and $b$ such that $s_k=a\cdot 2^k$ and $s_{k-1}=b\cdot 2^{k-1}$; what does the relation then say about $s_{k+1}$?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1949455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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|
How to determine if vector b is in the span of matrix A? Given a matrix A = \begin{bmatrix}
1 &2 &3 \\
4 &5 &6 \\
7 &8 &9
\end{bmatrix}
Determine if vector $b$ is in $span(A)$
where
$$
b = \begin{bmatrix}
1 \\
2 \\
4
\end{bmatrix}
$$
|
First we address $\mathrm{Span}(A)$,
$$\mathrm{Span}(A) = \left\{ \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ \end{array}\right] : \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ \end{array}\right] = a \left[\begin{array}{c} 1\\ 4\\ 7\\ \end{array}\right] + b \left[\begin{array}{c} 2\\ 5\\ 8\\ \end{array}\right] + c \left[\begin{array}{c} 3\\ 6\\ 9\\ \end{array}\right]. a,b,c \in \mathbb{R} \right\}.$$
From this definition we can see that asking if vector $\vec{b} \in \mathrm{Span}(A)$ is equivalent to asking if there exists a vector $\vec{x}$ such that $A\vec{x} = \vec{b}$,
because $$\vec{x} = \left[\begin{array}{c} x_1\\ x_2\\ x_3\\ \end{array}\right],$$
then $$A\vec{x} = \left[\begin{array}{ccc} 1 & 2 & 3\\ 4& 5 & 6\\ 7 & 8 & 9\\ \end{array}\right] \left[ \begin{array}{c} x_1\\ x_2\\ x_3 \end{array}\right] = x_1 \left[ \begin{array}{c} 1\\ 4\\ 7 \end{array}\right] + x_2 \left[ \begin{array}{c} 2\\ 5\\ 8 \end{array}\right] + x_3 \left[ \begin{array}{c} 3\\ 6\\ 9 \end{array}\right].$$
So if there exists $x_1, x_2, x_3$ such that the final line equals $\vec{b}$, then we know $\vec{b}$ is in $\mathrm{Span}(A)$. This means it suffices to ask if there exists a $\vec{x}$ such that $A\vec{x} = \vec{b}$.
This final equation is equivalent to the matrix equation:
$$\left[\begin{array}{ccc} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{array} \right] \vec{x} = \left[\begin{array}{c} 1\\ 2\\ 4\\ \end{array}\right],$$
which we can convert to the system:
$$\left[\begin{array}{ccc|c}
1 & 2 & 3 & 1\\
4 & 5 & 6 & 2\\
7 & 8 & 9 & 4\\
\end{array}\right].$$
As you might have learned, we solve this system by row reduction (I used technology for this step, your instructor may require row reduction by hand):
$$\mathrm{RREF}\left(\left[\begin{array}{ccc|c}
1 & 2 & 3 & 1\\
4 & 5 & 6 & 2\\
7 & 8 & 9 & 4\\
\end{array}\right]\right) = \left[\begin{array}{ccc|c}
1 & 0 & 0 & \frac{-4}{3}\\
0 & 1 & 0 & \frac{8}{3} \\
0 & 0 & 1 & -1\\
\end{array}\right]. $$
This implies the vector $\vec{x}$ we seek is given by:
$$\vec{x} = \left[\begin{array}{c} \frac{-4}{3}\\ \frac{8}{3} \\ -1\\ \end{array} \right].$$
The existence of this $\vec{x}$ alone guarantees $\vec{b} \in \mathrm{Span}(A)$, but lets check our answer by computing $A\vec{x}$.
$$A\vec{x} = \left[\begin{array}{ccc} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{array} \right] \left[\begin{array}{c} \frac{-4}{3}\\ \frac{8}{3} \\ -1\\ \end{array} \right] = \left[\begin{array}{c} 1\\ 2 \\ 4\\ \end{array} \right], $$
as desired.
Some things to think about: are there any vectors in $\mathbb{R}^3$ that are not in Span($A$)? If so can you find them, if not can you justify it? Hope this answer helps!
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1949704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Calculate $\lim_{x\to 0^+}\frac{\frac{4}{\pi}\arctan(\frac{\arctan x}{x})-1}{x}$ without Taylor's theorem or L'Hospital rule Calculate this limit without using taylor or hopital
$$\lim_{x\rightarrow 0^+}\frac{\frac{4}{\pi}\arctan(\frac{\arctan x}{x})-1}{x}$$
I have no idea to start the problem please help
|
We can proceed as follows
\begin{align}
L &= \lim_{x \to 0^{+}}\dfrac{\dfrac{4}{\pi}\arctan\left(\dfrac{\arctan x}{x}\right) - 1}{x}\notag\\
&= \lim_{x \to 0^{+}}\frac{4}{\pi}\cdot\dfrac{\arctan\left(\dfrac{\arctan x}{x}\right) - \arctan 1}{x}\notag\\
&= \frac{4}{\pi}\lim_{x \to 0^{+}}\frac{1}{x}\arctan\left(\frac{\arctan x - x}{\arctan x + x}\right)\tag{1}\\
&= \frac{4}{\pi}\lim_{x \to 0^{+}}\frac{1}{x}\cdot\dfrac{\arctan x - x}{\arctan x + x}\cdot\dfrac{\arctan\left(\dfrac{\arctan x - x}{\arctan x + x}\right)}{\dfrac{\arctan x - x}{\arctan x + x}}\tag{2}\\
&= \frac{4}{\pi}\lim_{x \to 0^{+}}\frac{1}{x}\cdot\dfrac{\arctan x - x}{\arctan x + x}\tag{3}\\
&= \frac{4}{\pi}\lim_{x \to 0^{+}}\dfrac{\arctan x - x}{x^{2}}\cdot\frac{x}{\arctan x + x}\notag\\
&= \frac{4}{\pi}\lim_{x \to 0^{+}}\dfrac{\arctan x - x}{x^{2}}\cdot\dfrac{1}{\dfrac{\arctan x}{x} + 1}\notag\\
&= \frac{4}{\pi}\cdot 0 \cdot\frac{1}{1 + 1}\notag\\
&= 0\notag
\end{align}
We have made use of the standard limit $$\lim_{x \to 0}\frac{\arctan x}{x} = 1$$ and also note that from this answer we have $$\lim_{x \to 0^{+}}\frac{\arctan x - x}{x^{2}} = 0$$ and hence $$\lim_{x \to 0^{+}}\frac{\arctan x - x}{\arctan x + x} = \lim_{x \to 0^{+}}\frac{\arctan x - x}{x^{2}}\cdot x\cdot\dfrac{1}{\dfrac{\arctan x}{x} + 1} = 0$$ and therefore the steps from $(1)$ to $(2)$ to $(3)$ are justified.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Challenging Integral [indefinite] Integrate: $$\int \frac{x^2+n(n-1)}{(x\sin x+n\cos x )^2}dx$$
I've been beating my head around this problem for quite some time now, but I've got nowhere. I'd request the person writing the solution to please explain his thought process because I would like to learn how to appraoch such Integrals in the future.
|
Using $\displaystyle (x\cdot \sin x+n\cdot \cos x) = \sqrt{x^2+n^2}\left\{\frac{x}{\sqrt{x^2+n^2}}\cdot \sin x+\frac{n}{\sqrt{x^2+n^2}}\cdot \cos x\right\}$
$$\displaystyle = \sqrt{x^2+n^2}\cdot \cos\left(x-\phi\right)\;,$$ where $\displaystyle \sin \phi = \frac{x}{\sqrt{x^2+n^2}}$ and $\displaystyle \cos \phi = \frac{n}{\sqrt{x^2+n^2}}$ and $\displaystyle \tan \phi = \frac{x}{n}\Rightarrow \phi = \tan^{-1}\left(\frac{x}{n}\right)$
So Integral is $$\displaystyle = \int \sec^2(x-\phi)\cdot \left(\frac{x^2+n(n-1)}{x^2+n^2}\right)dx$$
Now Let $$\displaystyle (x-\phi) = y\Rightarrow \left(x-\tan^{-1}\left(\frac{x}{n}\right)\right)=y$$. Then $$\displaystyle \left(\frac{x^2+n(n-1)}{x^2+n^2}\right)dx = dy$$
So Integral is $$\displaystyle \int \sec^2(y)dy = \tan y +\mathbb{C} = \tan\left(x-\tan^{-1}\left(\frac{x}{n}\right)\right)+\mathbb{C}$$
So $$\displaystyle \int \frac{x^2+n(n-1)}{(x\cdot \sin x+n\cdot \cos x)^2}dx = \left(\frac{n\cdot \tan x-x}{n+x\cdot \tan x}\right)+\mathcal{C}= \frac{n\sin x-x\cos x}{n\cos x+x\sin x}+\mathcal{C}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1952415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluation of the value of a function at a point in a Non-linear initial value problem Given $y''y^3=1,\quad y(0)=1,\quad y'(0)=2$. I have to show that $y(1)=\sqrt{10}$. I tried to substitute $y^2=z$ but did not succeed.
|
You have
$$y^{\prime} y^{\prime \prime} y^3= y^{\prime} \iff y^{\prime} y^{\prime \prime}= \frac{y^{\prime}}{y^3} \text{ (for } y^{\prime} \neq 0)$$ hence
$$\frac{1}{2}\left( y^{\prime} \right)^2 - \frac{1}{2}\left( y^{\prime}_0 \right)^2 = \frac{1}{2} \frac{1}{y_0^2} - \frac{1}{2} \frac{1}{y^2}$$ or
$$\left( y^{\prime} \right)^2-4=1-\frac{1}{y^2} \iff y^\prime = \sqrt{5-\frac{1}{y^2}}$$ as $y^\prime(0) = 2 >0$ which leads to
$$\frac{1}{5} y \sqrt{5 -\frac{1}{y^2}} - \frac{1}{5} y_0 \sqrt{5 -\frac{1}{y^2_0}} = x-x_0$$ hence
$$\frac{1}{5} \sqrt{5 y(1)^2 -1} - \frac{2}{5} = 1$$ and by solving this algebraic equation
$$y(1) = \sqrt{10}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1957645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Fourier series for $|x|$ not in complex form $\mathbf{f(x)=|x|=\frac{1}{2}a_0+ \sum_{n=1}^{\infty}(a_n \cos{nx}+b_n \sin{nx})}$
for $-\pi <x< \pi$
Since $|x|$ is even $\rightarrow b_n=0$
f(x) =
\begin{cases}
-x, & {-\pi<x<0} \\
x, & {0<x<\pi}
\end{cases}
$a_0=\frac{1}{2\pi}[\int_{-\pi}^0 f(x)$ $dx$ + $\int^{\pi}_0 f(x)$ $dx$]
$a_0=\frac{1}{2\pi}[\int_{-\pi}^0 -x$ $dx$ + $\int^{\pi}_0 x$ $dx$]
$=\frac{1}{\pi}\int ^{\pi}_0 x$ $dx$
$\frac{1}{\pi}(\frac{\pi^2}{2})=\frac{\pi}{2}$
Similarly,
$a_n=\frac{1}{\pi}[\int_{-\pi}^0 -x \cos nx$ $dx$ + $\int^{\pi}_0 x \cos nx$ $dx$]
$=-\frac{4}{\pi}[\frac{1}{n^2}+\frac{(-1)^n}{n^2}]$
$\therefore |x|=\frac{\pi}{2}-\frac{4}{\pi}\sum^{\infty}_0[\frac{1+(-1)^n}{n^2} \cos nx]$
But the answer is
$\therefore |x|=\frac{\pi}{2}-\frac{4}{\pi}\sum^{\infty}_0[\frac{\cos (2n+1)x}{n^2}]$
In order for $1+(-1)^n \cos nx$ to become $\cos (2n+1)x$ , How do I reach here, unless there is an error somewhere.
|
Note that
$$1+(-1)^n=\begin{cases}
2 & \text{if $n$ is even, i.e. } n=2k \\
0 & \text{if $n$ is odd, i.e. }n=2k+1
\end{cases}$$
And since $\frac{\cos nx}{n}$ is indeterminate for $n=0$:
$$\sum_{n=\color{red}{1}}^\infty\frac{1+(-1)^n}{n^2}\cos nx=\sum_{k=\color{red}{1}}^\infty\frac{2}{(2k)^2}\cos(2k x)+0=\frac{1}{2}\sum_{n=\color{red}{1}}^\infty\frac{\cos 2nx}{n^2}$$
So it looks like there is an error somewhere. Because
$$\int_{-\pi}^{\pi}|x|\cos nx\,dx=\frac{2}{n^2}(\color{red}{-1}+(-1)^n)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1958930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find a closed-form expression for $\binom n1+3\binom n3+5\binom n5+\cdots ,$ where $n > 1$. Find a closed-form expression for
$$\binom n1+3\binom n3+5\binom n5+\cdots ,$$
where $n > 1$. You may find the identity $k\binom{n}{k} = n\binom{n-1}{k-1}$ helpful.
I got $2^{n-2}$ but it was wrong! I don't know where I miscalculated...
|
Consider
$$(1+x)^n=1+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+\binom{n}{4}x^4+\cdots$$
Differentiating with respect to $x$,
$$n(1+x)^{n-1}=\binom{n}{1}+2x\binom{n}{2}+3x^2\binom{n}{3}+\cdots$$
Put $x=1$,
$$n2^{n-1}=\binom{n}{1}+2\binom{n}{2}+3\binom{n}{3}+\cdots\tag1$$
Put $x=-1$,
$$0=\binom{n}{1}-2\binom{n}{2}+3\binom{n}{3}-+\cdots\tag2$$
Taking $\frac{(1)+(2)}{2}$,
$$\binom{n}{1}+3\binom{n}{3}+5\binom{n}{5}+\cdots=n2^{n-2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1959387",
"timestamp": "2023-03-29T00:00:00",
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|
Sum of the series $\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$
The sum of the series $$\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$$
$\bf{My\; Try::}$ We can write it as $\displaystyle \binom{n}{0} = $ Coefficient of $x^0$ in $(1+x)^n$
Similarly $\displaystyle \binom{n-1}{1} = $ Coefficient of $x^1$ in $(1+x)^{n-1}$
Similarly $\displaystyle \binom{n-2}{2} = $ Coefficient of $x^2$ in $(1+x)^{n-2}$
Now, how can I solve it after that, Help Required, Thanks
|
$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{{n \choose 0} - {n - 1 \choose 1} + {n - 2 \choose 2} -
{n-3 \choose 3} + \cdots:\ ?}$.
\begin{align}
\sum_{k = 0}^{n}\pars{-1}^{k}{n - k \choose k} & =
\sum_{k = 0}^{\infty}\pars{-1}^{k}{n - k \choose n - 2k} =
\sum_{k = 0}^{\infty}\pars{-1}^{k}
\oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{n - k} \over z^{n - 2k + 1}}
\,{\dd z \over 2\pi\ic}
\\[5mm] & =
\oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{n} \over z^{n + 1}}
\sum_{k = 0}^{\infty}\pars{-\,{z^{2} \over 1 + z}}^{k}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
\oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{n} \over z^{n + 1}}
{1 \over 1 + z^{2}/\pars{1 + z}}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
\oint_{\verts{z}\ =\ 1^{-}}
{\pars{1 + z}^{n + 1} \over z^{n + 1}\pars{z^{2} + z + 1}}\,{\dd z \over 2\pi\ic}
\,\,\,\stackrel{z\ \mapsto\ 1/z}{=}\,\,\,
\oint_{\verts{z}\ =\ 1^{\color{#f00}{+}}}
{\pars{1 + z}^{n + 1} \over z^{2} + z + 1}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
{\pars{1 + r}^{n + 1} \over 2r + 1} +
{\pars{1 + \bar{r}}^{\, n + 1} \over 2\bar{r} + 1}\quad
\mbox{where}\quad
r \equiv -\,{1 \over 2} + {\root{3} \over 2}\,\ic =
\exp\pars{{2\pi \over 3}\,\ic}
\end{align}
Then $\ds{\pars{~\mbox{note that}\ 1 + r = \exp\pars{{\pi \over 3}\,\ic}~}}$,
\begin{align}
\sum_{k = 0}^{n}\pars{-1}^{k}{n - k \choose k} & =
2\,\Re\pars{\bracks{1 + r}^{n + 1} \over 2r + 1} =
2\,\Re\pars{\expo{\bracks{n + 1}\pi\ic/3} \over \root{3}\ic}
\\[5mm] & =
\bbox[10px,border:1px groove navy]{{2\root{3} \over 3}
\,\sin\pars{\bracks{n + 1}\pi \over 3}} =
\bbox[10px,border:1px groove navy]{%
{\root{3} \over 3}\,\sin\pars{n\pi \over 3} + \cos\pars{n\pi \over 3}}
\end{align}
This result generates the sequence $\ds{\pars{~\mbox{starting with}\ n = 0~}}$:
$$
\underbrace{1, 1, 0, -1, -1, 0}_{},\
\underbrace{1, 1, 0, -1, -1, 0}_{},\ \underbrace{1, 1, 0, -1, -1, 0}\ldots
$$
because
$\ds{\sin\pars{\bracks{n + 1}\pi \over 3} =
\sin\pars{\bracks{n + \color{#f00}{6} + 1}\pi \over 3} =
\sin\pars{\bracks{n + 7}\pi \over 3}}$.
|
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"url": "https://math.stackexchange.com/questions/1960944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Prove $\csc(x)=\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{x+k\pi}$
Prove $$\csc(x)=\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{x+k\pi}$$
Hardy uses this fact without proof in a monograph on different ways to evaluate $\int_0^{\infty}\frac{\sin(x)}{x} dx$.
|
Presumably the principal value of the two-sided infinite sum is what was intended in the question.
We'll solve this just using Euler's product formula for the sine function:
\begin{align}
\sin x=x\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2 \pi^2}\right).
\end{align}
Compute the logarithmic derivative:
\begin{align}
\cot x &= \frac1{x}+\sum_{n=1}^\infty \frac{-2x/n^2\pi^2}{1-\frac{x^2}{n^2 \pi^2}}
\\&=\frac1{x}+\sum_{n=1}^\infty \frac{2x}{x^2-n^2\pi^2}
\\&=\frac1{x}+\sum_{n=1}^\infty\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big),
\end{align}
so
$$ \cot x-\frac1{x}=\sum_{n=1}^\infty\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big).$$
It follows that
\begin{align}
\cot\big(\frac{x}{2}\big)-\frac2{x}&=\sum_{n=1}^\infty\big(\frac{1}{x/2+n\pi}+\frac{1}{x/2-n\pi}\big)
\\&=2\sum_{n=1}^\infty\big(\frac{1}{x+2n\pi}+\frac{1}{x-2n\pi}\big)
\\&=2\sum_{\substack{n\ge 1\\n\text{ even}}}\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big).
\end{align}
Subtracting now, we obtain
\begin{align}
\cot\big(\frac{x}{2}\big)-\cot(x)-\frac1{x}&=2\cdot\!\!\!\sum_{\substack{n\ge 1\\n\text{ even}}}\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big)-\sum_{n=1}^\infty\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big)
\\&=\sum_{\substack{n\ge 1\\n\text{ even}}}\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big)-\sum_{\substack{n\ge 1\\n\text{ odd}}}\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big)
\\&=\sum_{n=1}^\infty (-1)^n \big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big).
\end{align}
But
\begin{align}
\cot\big(\frac{x}{2}\big)-\cot(x)&=\frac{\cos(x/2)}{\sin(x/2)}-\frac{\cos x}{\sin x}
\\&=\frac{\cos(x/2)}{\sin(x/2)}\cdot\frac{2\sin(x/2)\cos(x/2)}{\sin x}-\frac{\cos x}{\sin x}
\\&=\frac{2\cos^2(x/2)-\cos x}{\sin x}
\\&=\frac1{\sin x}
\\&= \csc x.
\end{align}
So we've shown that
\begin{align}
\csc x &= \frac1{x}+\sum_{n=1}^\infty (-1)^n \big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big),
\end{align}
which is the principal value of $$\sum_{n=-\infty}^\infty (-1)^n \frac{1}{x+n\pi},$$
as desired.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\sum_{n=1}^\infty \frac{1}{(2n+1)^4}$ using Fourier Series I am given $f(x) = x$ for $0 \le x \le 2$. The question wants me to evaluate $\sum_{n=1}^\infty \frac{1}{(2n+1)^4}$ by first evaluating the Fourier sine of $f(x)$ by extending it outside the interval.
I get $f(x) = 2 + \sum_{n=1}^\infty \frac{-4}{\pi}\sin(\frac{n\pi x}{2})$. And I know that I probably have to make use of Parseval's theorem. Yet, I am not sure how to proceed as I don't know how to get an expression to the power of $4$ from the Fourier sine expression of $f(x)$.
Thanks!
|
Using your function, $f(x)=x$ on $[0,2]$ then extended by periodicity (that is, the period is $T=2$), you get Fourier coefficients
$$a_0=\frac1T\int_0^T f(x)\mathrm{d}x=1$$
And for all $n>0$
$$a_n=\frac2T\int_0^T f(x)\cos \frac{2n\pi x}{T}\mathrm{d}x=\int_0^2 x\cos (n\pi x)\mathrm{d}x=\left[x\frac{\sin(n\pi x)}{n\pi}\right]_0^2-\int_0^2 \frac{\sin(n\pi x)}{n\pi}\mathrm{d}x=0$$
$$b_n=\frac2T\int_0^T f(x)\sin \frac{2n\pi x}{T}\mathrm{d}x=\int_0^2 x\sin (n\pi x)\mathrm{d}x=\left[-x\frac{\cos(n\pi x)}{n\pi}\right]_0^2+\int_0^2 \frac{\cos(n\pi x)}{n\pi}\mathrm{d}x\\=-\frac{2}{n\pi}$$
Thus, since the function $f$ is piecewise continuous, the Fourier series converges to $\hat f$, and on $]0,2[$, where it's continuous, you have
$$x=1-\frac2\pi\sum_{n=1}^\infty \frac{\sin(n\pi x)}n$$
You can integrate the Fourier series, and you get
$$\frac{x^2}2-x+C=\frac2{\pi^2}\sum_{n=1}^\infty \frac{\cos(n\pi x)}{n^2}$$
The mean of the RHS is zero on $[0,2]$, and to get zero mean on the LHS, you need $C=\frac13$.
You may also write
$$\frac{x^2}2-x=-\frac13+\frac2{\pi^2}\sum_{n=1}^\infty \frac{\cos(n\pi x)}{n^2}$$
Then, Parseval's theorem yields
$$\frac12\int_0^2 \left(\frac{x^2}2-x\right)^2\mathrm{d}x=\frac19+\frac{2}{\pi^4}\sum_{n=1}^\infty \frac1{n^4}$$
Then
$$\frac12\int_0^2 \left(\frac{x^2}2-x\right)^2\mathrm{d}x=\frac12\left[\frac{x^5}{10}+\frac{x^3}{3}-\frac{x^4}{4}\right]_0^2=\frac{2}{15}$$
And $\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{1}{45}$, hence
$$\sum_{n=1}^\infty \frac1{n^4}=\frac{\pi^4}{90}$$
Now,
$$\sum_{n=1}^\infty \frac1{(2n)^4}=\frac{1}{16}\sum_{n=1}^\infty \frac1{n^4}=\frac{\pi^4}{90\times16}$$
And finally
$$\sum_{n=0}^\infty \frac1{(2n+1)^4}=\sum_{n=1}^\infty \frac1{n^4}-\sum_{n=1}^\infty \frac1{(2n)^4}=\frac{\pi^4}{90}\left(1-\frac1{16}\right)=\frac{\pi^4}{96}$$
But!
Your sum starts from $n=1$, so your sum is
$$\sum_{n=1}^\infty \frac1{(2n+1)^4}=\frac{\pi^4}{96}-1$$
|
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|
Probability and Conditional Probability Could someone please provide a worked solution to this question?
A box contains 5 amber, 7 blue and 9 green balls. Six of the balls are removed from the box at random and without replacement.
Find the probability that
(a) three out of the six balls are blue;
(b) four of the balls are blue and two are green;
(c) the second ball to be selected is amber, given that the final ball to be selected is blue.
I was wondering if there is a combinatorial argument for these types of problems, because multiplying fractions here seems a little time consuming and messy.
My answers:
(a) $\frac{7}{21}\times\frac{6}{20}\times\frac{5}{19}\times\frac{14}{18}\times\frac{13}{17}\times\frac{12}{16}\times\frac{6!}{(3!)^2}$
(b) $\frac{7}{21}\times\frac{6}{20}\times\frac{5}{19}\times\frac{4}{18}\times\frac{9}{17}\times\frac{8}{16}\times\frac{6!}{(4!)(2!)}$
Any help very much appreciated, as always.
|
(A) The probability that when selecting $6$ from $21$ balls, you select $3$ from $7$ blue and $3$ from $14$ non-blue is: $$\dfrac{\binom 73\binom {14}3}{\binom {21}6} \tag{$= \dfrac{\frac{7\cdot 6\cdot 5}{3\cdot 2\cdot 1}\cdot \frac{14\cdot 13\cdot 12}{3\cdot 2\cdot 1}}{\frac{21\cdot 20\cdot 19\cdot 18\cdot 17\cdot 16}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}}$}$$
Which matches your answer, in more compact form.
The count of Blue balls so selected follows a Hypergeometric Distribution.
(B) The probability that when selecting $6$ from $21$ balls, you select $4$ from $7$ blue, $2$ from $9$ green (, and $0$ from $5$ amber) is: $$\dfrac{\binom 74\binom {9}2\color{silver}{\binom 5 0}}{\binom {21}6}$$
Which again matches your answer.
(C) The probability that when given that $1$ of the blue balls will be reserved for the sixth draw, the second ball selected will be one from the $5$ of the $20$ unreserved balls that are amber, is just:
$$\mathsf P(C_2=a\mid C_{6}=b) = \frac{5}{20}$$
No need to overcomplicate it.
|
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|
If $abc=1$ so $\sum\limits_{cyc}\frac{a}{\sqrt{a+b^2}}\geq\frac{3}{\sqrt2}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{a}{\sqrt{a+b^2}}+\frac{b}{\sqrt{b+c^2}}+\frac{c}{\sqrt{c+a^2}}\geq\frac{3}{\sqrt2}$$
After substitution $a=\frac{y}{x}$... I tried C-S, but without success.
|
Using Hölder's inequality, we have in general:
$$
\left( \frac{a}{\sqrt{X}}+\frac{b}{\sqrt{Y}}+\frac{c}{\sqrt{Z}}\right)^2(aX+bY+cZ) \geq (a+b+c)^3
$$
Substitute $\sqrt{a+b^2}, \sqrt{b+c^2}, \sqrt{c+a^2}$ for $X, Y, Z$:
$$
\left( \frac{a}{\sqrt{a+b^2}}+\frac{b}{\sqrt{b+c^2}}+\frac{c}{\sqrt{c+a^2}} \right)^2 \geq \frac{(a+b+c)^3}{a^2+b^2+c^2+ab^2+bc^2+ca^2}
$$
Only we have to do is show (the right side) $\geq \frac{9}{2}$.
$$
f(a,b,c) \equiv 2(a+b+c)^3-9(a^2+b^2+c^2+ab^2+bc^2+ca^2) \\
= 12+2(a^3+b^3+c^3)-9(a^2+b^2+c^2)+6(a^2b+b^2c+c^2a)-3(ab^2+bc^2+ca^2)
$$
Since this is a symmetric polynomial, the min should be at $a=b=c(=1)$. Though this is not the strict proof.
Here, I tried Muirhead's inequality, for example,
$$
a^2b+b^2c+c^2a \geq ab^2+bc^2+ca^2
$$
but without success.
Instead, by partial differential, we can say that at $a=b=c=1$ we have the global minimum $0$ (for $a,b,c>0$. but the same thing if $a \geq b \geq c \space and \space b,c<0$). The inequality is true.
I hope it's OK as a hint.
|
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|
Integration involving rational function and exponentials I hope I can find closed form solution for two following definite integrals. Unfortunately I don't have Mathematica and I can't find similar integrals in Tables. Can any one help me please?
\begin{equation}
\int_0^\infty \frac{e^{-x}}{x^{1/2}+a x^{3/2}} dx
\end{equation}
\begin{equation}
\int_0^\infty \frac{xe^{-x}}{x^{1/2}(1+bx)^2} dx
\end{equation}
|
The second integral is calculated in the same fashion:
\begin{eqnarray}
&&\int\limits_0^\infty \frac{x e^{-x}}{\sqrt{x} (1+b x)^2} dx \underbrace{=}_{z=\sqrt{x}}\\
&&\int\limits_0^\infty \frac{z^2}{(1+b z^2)^2} 2 e^{-z^2} dz=\\
&&\int\limits_0^\infty \left(\frac{1}{b} \frac{1}{1+b z^2} - \frac{1}{b} \frac{1}{1+b z^2)^2} \right) 2 e^{-z^2} dz=\\
&&\frac{2}{b^{3/2}} \left( \int\limits_0^\infty \frac{e^{-\frac{z^2}{b}}}{1+z^2}dz - \int\limits_0^\infty \frac{e^{-\frac{z^2}{b}}}{(1+z^2)^2}dz\right)=\\
&&\frac{2}{b^{3/2}} \left( \frac{\pi}{2} \exp(\frac{1}{b}) erfc(\frac{1}{\sqrt{b}}) - \int\limits_0^\infty \frac{e^{-\frac{z^2}{b}}}{(1+z^2)^2}dz\right)=\\
&&\frac{2}{b^{3/2}} \left( \frac{\pi}{2} \exp(\frac{1}{b}) erfc(\frac{1}{\sqrt{b}}) - \int\limits_0^\infty \frac{1}{b} \left( \frac{z}{1+z^2} + \arctan(z)\right) z e^{-\frac{z^2}{b}} dz\right)=\\
&& \frac{\pi}{b^{3/2}}\exp(\frac{1}{b})erfc(\frac{1}{\sqrt{b}}) - \frac{2}{b^{5/2}} \int\limits_0^\infty \left( 1-\frac{1}{1+z^2} + z \arctan(z) \right) e^{-\frac{z^2}{b}} dz=\\
&&\frac{\pi}{b^{3/2}}\exp(\frac{1}{b})erfc(\frac{1}{\sqrt{b}}) - \frac{2}{b^{5/2}}\left( \sqrt{b} \frac{\sqrt{\pi}}{2} - \frac{\pi}{2} \exp(\frac{1}{b}) erfc(\frac{1}{\sqrt{b}}) + \int\limits_0^\infty z e^{-\frac{z^2}{b}} \arctan(z) dz \right)=\\
&&\frac{\pi}{b^{3/2}}\exp(\frac{1}{b})erfc(\frac{1}{\sqrt{b}}) - \frac{2}{b^{5/2}}\left( \sqrt{b} \frac{\sqrt{\pi}}{2} - \frac{\pi}{2} \exp(\frac{1}{b}) erfc(\frac{1}{\sqrt{b}}) + \frac{b}{2} \int\limits_0^\infty e^{-\frac{z^2}{b}} \frac{1}{1+z^2} dz \right)=\\
&&\frac{\pi}{b^{3/2}}\exp(\frac{1}{b})erfc(\frac{1}{\sqrt{b}}) - \frac{2}{b^{5/2}}\left( \sqrt{b} \frac{\sqrt{\pi}}{2} - \frac{\pi}{2} \exp(\frac{1}{b}) erfc(\frac{1}{\sqrt{b}}) + \frac{b}{2} \frac{\pi}{2} \exp(\frac{1}{b}) erfc(\frac{1}{\sqrt{b}}) \right)=\\
&& -\frac{\sqrt{\pi}}{b^2} + \frac{\pi}{2 b^{3/2}} \exp(\frac{1}{b}) erfc(\frac{1}{\sqrt{b}}) \left( 1+ \frac{2}{b} \right)
\end{eqnarray}
|
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|
How do I solve differential equation $\frac{d}{dt}x(t)=(2x(t)+8)(t^4+2t^2-t)$? How do I solve differential equation $\frac{dx}{dt}=(2x+8)(t^4+2t^2-t)$?
$\frac{dx}{2x(t)+8}=dt(t^4+2t^2-t)$
$\frac{1}{2}\ln(2x+8)=\frac{t^5}{5}+2\frac{t^3}{3}-\frac{t^2}{2}$
$e^{\ln(2x+8)}=e^{2(\frac{t^5}{5}+2\frac{t^3}{3}-\frac{t^2}{2})}$
$2x+8=e^{2(\frac{t^5}{5}+2\frac{t^3}{3}-\frac{t^2}{2})}$
$2x=e^{2(\frac{t^5}{5}+2\frac{t^3}{3}-\frac{t^2}{2})}-8$
$x=\frac{e}{2}^{2(\frac{t^5}{5}+2\frac{t^3}{3}-\frac{t^2}{2})}-4$
$x(0)=3$
Is the above calculation correct?
I can put that values, so I get:
$\frac{1}{2}\ln(14)=-C$
Where do I write value of the constant C in the solution $x(t)$?
|
$\frac{1}{2}\ln(2x+8)=\frac{t^5}{5}+2\frac{t^3}{3}-\frac{t^2}{2}+C$
I advocate waiting until the end to solve for $C.$ There is a lot of algebra to follow, and it is easier if there is one less thing to think about.
For example, $C$ is an arbitrary constant. Multiply by $2, 2C$ is equally arbitrary, so just leave it as $C.$
$\ln(2x+8)=2\frac{t^5}{5}+4\frac{t^3}{3}-{t^2}+C$
$e^{\ln(2x+8)} = e^{(2\frac{t^5}{5}+4\frac{t^3}{3}-{t^2}+C)}\\
2x+8 = e^{(2\frac{t^5}{5}+4\frac{t^3}{3}-{t^2})}e^C\\
2x+8 = Ce^{(2\frac{t^5}{5}+4\frac{t^3}{3}-{t^2})}$
Same game, $e^C$ and $C$ are equally arbitrary constants.
$x+4 = Ce^{(2\frac{t^5}{5}+4\frac{t^3}{3}-{t^2})}\\
x = Ce^{(2\frac{t^5}{5}+4\frac{t^3}{3}-{t^2})}-4$
Now, lets tackle $C$ in the initial condition.
$x(0) = C - 4 = 3$
$x = 7e^{(2\frac{t^5}{5}+4\frac{t^3}{3}-{t^2})}-4$
|
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|
When does $\int_a^\infty g(x)/f(x)\,dx$ converge?
When does $\int_a^\infty g(x)/f(x) \,dx$ converge?
Where $g$ and $f$ are polynomials.
I am not so sure about this pattern. Consider,
$\int_1^\infty \frac{1}{x^3 + x^2 + x + 1} \, dx$, this integral converges, but $\int_1^\infty \frac{4x^2}{x^3 + x^2 + x + 1} \,dx$ does not converge.
How can I tell when it converges? And what is the pattern?
|
In the integral $\displaystyle \int_1^\infty \frac{4x^2}{x^3 + x^2 + x + 1} \,dx$ the leading terms in the numerator and denominator are $4x^2$ and $x^3$, so as $x\to\infty$ it goes to $0$ at a rate comparable to that of $4x^2/x^3 = 4/x$, so the integral diverges.
$$
\frac{4x^2}{x^3+x^2+x+1} = \frac 4 {x + 1 + \dfrac 1 x + \dfrac 1 {x^2}} \ge \frac 1 x \text{ if } x\ge 1, \text{ and } \int_1^\infty \frac{dx} x = \infty.
$$
By contrast $\dfrac 1 {x^3+x^2+x+1} \le \dfrac 1 {x^2}$ and $\displaystyle\int_1^\infty \dfrac{dx}{x^2}<\infty.$
|
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|
Investigate the convergence of $\sum_{n=1}^{\infty} u_{n}(x)$, where $u_{n}(x)=(-1)^n \frac{x^2+n}{n^2}$ Let $u_{n}(x)=(-1)^n \dfrac{x^2+n}{n^2}$. Show that the series $\sum_{n=1}^{\infty} u_{n}(x)$ converges uniformly on any interval $[a;b]$ but does not converge absolutely at any point $x\in R$.
For the absolute convergence, we have $\lim_{n\to\infty} \dfrac{\frac{x^2+n}{n^2}}{\frac{1}{n}}=1$ and the series $\sum_{n=1}^{\infty} \dfrac{1}{n}$ diverges, thus the given series does not converge absolutely at any point $x\in R$.
For the uniform convergence, I have no idea. Can anyone help me please?
|
(Rough Sketch of Proof): Fix an interval $[a, b]$, then there exists $N_1$ such that for all $n>N_1$ we have
\begin{align}
\frac{x^2+n}{n^2} \leq \frac{b^2+a^2+n}{n^2}
\end{align}
which tends to $0$ as $n\rightarrow \infty$. Hence $|u_n(x)|$ tends to $0$ uniformly for all $x \in [a, b]$. Thus, by the alternating series theorem, we have that
\begin{align}
\sum^\infty_{k=0}(-1)^k\frac{x^2+k}{k^2}
\end{align}
converges for all $x \in [a, b]$. Moreover, we have
\begin{align}
\left|\sum^n_{k=m} (-1)^k \frac{x^2+k}{k^2} \right|\leq \sum^n_{k=m}\frac{x^2}{k^2} + \left|\sum^n_{k=m}(-1)^k\frac{1}{k} \right| \leq \sum^n_{k=m}\frac{a^2+b^2}{k^2} + \left|\sum^n_{k=m}(-1)^k\frac{1}{k} \right|
\end{align}
for all $x \in [a, b]$.
Here's the proof. Fix $\epsilon>0$. Since
\begin{align}
\sum^\infty_{k=1}(-1)^k\frac{1}{k} \ \ \text{ and } \ \ \sum^\infty_{k=1}\frac{a^2+b^2}{k^2}
\end{align}
are convergent then there exists $N$ such that for all $m, n>N$, we have
\begin{align}
\left|\sum^n_{k=m} (-1)^k \frac{1}{k} \right|<\epsilon \ \ \text{ and } \ \ \left|\sum^n_{k=m} \frac{a^2+b^2}{k^2} \right|<\epsilon
\end{align}
which means
\begin{align}
\left|\sum^n_{k=m} (-1)^k \frac{x^2+k}{k^2} \right|<2\epsilon
\end{align}
i.e. the partial sums are uniformly Cauchy. Hence the series converges uniformly on $[a, b]$.
|
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|
Find the determinant of $I + A$ What is the determinant of an $n \times n$ matrix $B=A+I$, where $I$ is the $n\times n$ identity matrix and $A$ is the $n\times n$ matrix
$$
A=\begin{pmatrix}
a_1 & a_2 & \dots & a_n \\
a_1 & a_2 & \dots & a_n \\
\vdots & \vdots & \ddots & \vdots \\
a_1 & a_2 & \dots & a_n \\
\end{pmatrix}
$$
|
First look at what happens when $n=2$, $n=3$, renaming our matrix $B_n(a_1,\cdots,a_n)$ one gets
$$\begin{align}\det{B_2(a_1,a_2)}&=1+a_1+a_2\\\det{B_3(a_1,a_2,a_3)}&=1+a_1+a_2+a_3\end{align}$$
Assume that $\det{B_{n-1}(a_1,\cdots,a_{n-1})}=1+a_1+\cdots a_{n-1}$ and consider
$$\begin{vmatrix}1+a_1 & a_2 & \cdots & a_n\\
a_1 & 1+a_2 & \cdots & a_n\\
\vdots & \vdots & \ddots & \vdots\\
a_1 & a_2 & \cdots & 1+a_n
\end{vmatrix}$$
Substract the second row from the first to get
$$\begin{vmatrix} 1 & -1 & 0 & \cdots & 0\\
a_1 & 1+a_2 & a_3 &\cdots & a_n\\
a_1 & a_2 & 1+a_3 & \cdots & a_n\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
a_1 & a_2 & a_3 &\cdots & 1+a_n
\end{vmatrix}$$
Then add the first column to the second and get
$$\begin{vmatrix} 1 & 0 & 0 & \cdots & 0\\
a_1 & 1+a_1+a_2 & a_3 &\cdots & a_n\\
a_1 & a_1+a_2 & 1+a_3 & \cdots & a_n\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
a_1 & a_1+a_2 & a_3 &\cdots & 1+a_n
\end{vmatrix}$$
Developing along the first row one gets
$$\det{B_n(a_1,\cdots,a_n)}=\det{B_{n-1}(a_1+a_2,\cdots ,a_n)}$$
And this by the induction assumption is $$1+a_1+a_2+\cdots+a_n$$
|
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|
Solving a complex equation with $z=x \pm iy$ I'm trying to solve part of the question below:
The solution is below:
I managed to get to the line $x-y=\pm 2$ , $y=\pm x$ but I can't understand how you can deduce $y=-x$ from these equations. I tried substituting them into each other but got nothing.
|
The equation (b) becomes
$$
x^2-y^2+2xyi=(x^2+y^2-4)i
$$
translates into the system
$$
\begin{cases}
x^2-y^2=0 \\[6px]
2xy=x^2+y^2-4
\end{cases}
$$
and both equations should be satisfied. From the second one you get $(x-y)^2=4$, so $x-y\ne0$. The first equation tells us
$$
(x-y)(x+y)=0
$$
so we get $x+y=0$. Now we have two linear systems
$$
\begin{cases}
x+y=0 \\[6px]
x-y=2
\end{cases}
\qquad\text{or}\qquad
\begin{cases}
x+y=0 \\[6px]
x-y=-2
\end{cases}
$$
that you can easily solve.
For (a) we get instead
$$
\begin{cases}
x^2-y^2=x^2+y^2-4\\[6px]
2xy=0
\end{cases}
$$
The first equation is $y^2=2$; the second equation now gives $x=0$.
A different approach. We have, for equation (a), $z^2=z\bar{z}-4$, so also $\bar{z}^2=\bar{z}z-4$. In particular, $z^2=\bar{z}^2$, so either
$z=\bar{z}$ or $z=-\bar{z}$. In the first case we obtain $z^2=z^2-4$, which has no solution, in the second case $z^2=-z^2-4$, so $z^2=-2$.
|
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|
Show that if $x + \frac1x = 1$, then $x^5 + \frac1{x^5} = 1$. Suppose $x + \frac{1}{x} = 1$.
Without first working out what $x$ is, show that
$x^5 + \frac{1}{x^5} = 1$ as well.
|
Notice that $x+x^{-1}=1$ implies that
$$1 = (x+x^{-1})^3 = (x^3+x^{-3}) + 3(x+x^{-1}),$$
and thus
$$x^3+x^{-3} = -2.$$
Therefore,
$$1 = (x+x^{-1})^5 = (x^5+x^{-5}) + 5(x^3+x^{-3}) + 10(x+x^{-1}),$$
giving
$$x^5+x^{-5} = 1.$$
|
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|
Prove that if $2b^2+1 = 3^b$ where $b$ is a positive integer, then $b \leq 2$
Prove that if $2b^2+1 = 3^b$ where $b$ is a positive integer, then $b \leq 2$.
Is there some way we can transform the equation in order to get the inequality? We have $2b^2 = 3^b-1$.
|
Subtract $b^3$ from both sides.
$2b^2 + 1 -b^3 = 3^b-b^3$
$\Rightarrow b^2(2-b) = 3^b - (b^3+1)$
Now think about the expression
$3^b - (b^3+1)$.
Both $3^b$ and $b^3+1$ are monotonically increasing. Being an exponential function, $3^b$ grows faster than $b^3+1$ for positive b. These two facts ensure that if at any non-negative point $3^b$ dominates $b^3+1$, it will continue to dominate $b^3+1$ at subsequent points. And as a matter of fact, $3^b=b^3+1$ for $b=0$
Thus, for $b\geqslant0$ we have $3^b-(b^3+1)\geqslant0$
$\Rightarrow b^2(2-b)\geqslant0$
$\Rightarrow 2 \geqslant b$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How would you find the trigonometric roots of a cubic?
Question: How would you find the roots of the cubic$$x^3+x^2-10x-8=0\tag{1}$$
I'm not too sure where to begin. I'm thinking of somehow, implementing $\cos 3\theta=4\cos^3\theta-3\cos\theta$.
I've tried substituting $x$ with $t+t^{-1}$, but didn't get anywhere, and using Vieta's trigonometric solution formula, I got
$$x_1=\frac {2\sqrt{31}}3\cdot\cos\left(\frac {\arccos \frac {2}{\sqrt{31}}}3\right)-\frac 13\\x_2=\frac {2\sqrt{31}}3\cdot\cos\left(\frac {2\pi+\arccos\frac {2}{\sqrt{31}}}3\right)-\frac 13\\x_3=\frac {2\sqrt{31}}{3}\cdot\cos\left(\frac {4\pi+\arccos\frac 2{\sqrt{31}}}3\right)-\frac 13$$
But that's not the form I want. I'm looking for a form of $2\left(\cos\frac {\text{something}}{31}+\cos\frac {\text{something}}{31}+\cos\frac {\text{something}}{31}\right)$.
I've spent so much time, that I'm practically burnt out. -.-
|
1) Reduce the standard way your equation $x^3+x^2-10x-8=0$ to the form $$x^3+ax+b=0\qquad(*)$$
2) Since you have the identity $$4\cos^3\theta-3\cos \theta-\cos 3\theta=0\qquad(**)$$ make $x=u\cos\theta$ in $(*)$ so you get
$$u^3\cos^3\theta+au\cos\theta+b=0\iff4\cos^3\theta+\frac{4au}{u^3}\cos\theta+\frac{4b}{u^3}\qquad(***)$$
3) In order to refer $(***)$ to $(**)$ you need to take $$-3=\frac{4au}{u^3}\iff u=2\sqrt{\frac{-2a}{3}}$$ and $$\frac{4b}{u^3}=-\frac{3b}{2a}\sqrt{\frac{-3}{a}}$$ which gives $$\cos3\theta=\frac{3b}{2a}\sqrt{\frac{-3}{a}}$$
Hence your roots are given by $$x_k=2\sqrt{\frac{-a}{3}}\cos\frac{\theta_k}{3}$$ where $$\theta_k=\arccos\left(\frac{3b}{2a}\sqrt{\frac{-3}{a}}-\frac{2k\pi}{3}\right);\space k=0,1,2$$
Note.-You can use this method because the three roots of your equation are real.
|
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|
How to solve $\sin(x) + 2\sqrt{2}\cos x =3$ How to solve $\sin(x) + 2\sqrt{2}\cos x=3$ ?
What is general method for doing these kind of questions?
Thanks
|
Your problem is of the following form
$$a\sin x+b\cos x = c$$
where $a = 1$, $b = 2\sqrt{2}$ and $c = 3$.
Let $R = \sqrt{a^2 + b^2}$. We can define $$A=\dfrac{a}{R} =\cos\theta$$ and $$B=\dfrac{b}{R} =\sin\theta$$
Therefore
$$\begin{align*}
a\sin x+b\cos x&=R(A\sin x+B\cos x)=R(\cos\theta\sin x+\sin\theta\cos x)=R\sin(x+\theta)\;.
\end{align*}$$
Hence $$\sin(x+\theta)=\frac{c}{R}\;$$
or
$$x=(\sin^{-1}\frac{c}{R})- \theta = (\sin^{-1}\frac{c}{R})- (\sin^{-1}\frac{b}{R}) = (\sin^{-1}\frac{3}{3})- (\sin^{-1}\frac{2\sqrt{2}}{3}) = 0.3398$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve this limit problem? $\lim_{n \to \infty} \left ( \sqrt{n + \sqrt{n}} - \sqrt{n - \sqrt{n}} \right )$ I am stuck with the following question from my homework:
$\lim_{n \to \infty} \left ( \sqrt{n + \sqrt{n}} - \sqrt{n - \sqrt{n}} \right )$
Using wolfram alpha gives me 1 for the solution. However, I would like to know how you come up with this result. I hope someone can explain this to me.
|
$$\lim _{ n\to \infty } \left( \sqrt { n+\sqrt { n } } -\sqrt { n-\sqrt { n } } \right) =\lim _{ n\to \infty } \frac { \left( \sqrt { n+\sqrt { n } } -\sqrt { n-\sqrt { n } } \right) \left( \sqrt { n+\sqrt { n } } +\sqrt { n-\sqrt { n } } \right) }{ \left( \sqrt { n+\sqrt { n } } +\sqrt { n-\sqrt { n } } \right) }\\\\ =\lim _{ n\to \infty } \frac { 2\sqrt { n } }{ \left( \sqrt { n+\sqrt { n } } +\sqrt { n-\sqrt { n } } \right) } =\lim _{ n\to \infty } \frac { 2\sqrt { n } }{ \sqrt { n } \left( \sqrt { 1+\frac { 1 }{ \sqrt { n } } } +\sqrt { 1-\frac { 1 }{ \sqrt { n } } } \right) } =\\=\lim _{ n\to \infty } \frac { 2 }{ \left( \sqrt { 1+\frac { 1 }{ \sqrt { n } } } +\sqrt { 1-\frac { 1 }{ \sqrt { n } } } \right) } =1$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the determinants below using the fact: $\det \left[\begin{smallmatrix}a&b&c\\d&e&f\\g&h&i\end{smallmatrix}\right]=7$ a. \begin{bmatrix}g&h&i\\2d&2e&2f\\3a&3b&3c\end{bmatrix}
b. \begin{bmatrix}a&b&c\\d-2a&e-2b&f-2c\\5g&5h&5i\end{bmatrix}
Hello, I am not sure how to go about answering this question. I don't need and exact answer, but I just need to know how to get started on answering part A and B of this question.
|
Lets try solve a similar determinant:
$$\det\begin{pmatrix}2d & 2e & 2f \\ a-3g & b - 3h & c - 3i \\4g & 4h & 4i\end{pmatrix} $$
We have that:
\begin{align*}
\det\begin{pmatrix}2d & 2e & 2f \\ a-3g & b - 3h & c - 3i \\4g & 4h & 4i\end{pmatrix} & =
2\det\begin{pmatrix}d & e & f \\ a-3g & b - 3h & c - 3i \\4g & 4h & 4i\end{pmatrix} \\
& \text{Multiply the first row by a constant} \\
& = 8\det\begin{pmatrix}d & e & f \\ a-3g & b - 3h & c - 3i \\g & h & i\end{pmatrix} \\
& \text{Multiply the third row by a constant} \\
& = 8\det\begin{pmatrix}d & e & f \\ a & b & c \\g & h & i\end{pmatrix} \\
&\text{Add three times the third row to the second row} \\
& = -8\det\begin{pmatrix} a & b & c \\d & e & f \\g & h & i\end{pmatrix} \\
&\text{Switch the first and second rows} \\
& = -8\times (7) = -56 \\
\text{Substitute in the determinant we know}
\end{align*}
Hopefully you see how you can solve your problems from this.
|
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|
Mathemathic induction proof I need to prove that $$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{n(n+1)}=\frac{n}{n+1}$$
Here is what I tried:
\begin{align}
& \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{n(n+1)}+\frac{1}{n+1(n+2)} \\[10pt]
= {} & \frac{n}{n+1}+\frac{1}{n+1(n+2)} = \frac{(n+1)(n+2)+(n+1)}{n^2(n+2)}
\end{align}
I am stuck from here, I feel there must be someway to simplify the last part to get to
$$\frac{n+1}{n+2}$$
|
$$\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}=\frac{n(n+2)+1}{(n+1)(n+2)}=\frac{(n+1)^2}{(n+1)(n+2)}$$
|
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|
roots of the equation $(abc^2)x^2+3a^2 cx+b^2cx-6a^2-ab+2b^2=0$ are rational $a,b,c$ are non zero , unequal rational numbers then prove that roots of the equation $$(abc^2)x^2+3a^2 cx+b^2cx-6a^2-ab+2b^2=0$$ are rational
quadratic eqn. in standard form $(abc^2)x^2+(3a^2c+b^2c)x-(6a^2+ab-2b^2) = 0$
$\displaystyle D=(3a^2c+b^2c)^2+4(abc^2)(6a^2+ab-2b^2) = 9a^4c^2+b^4c^2+6a^2b^2c^2+24a^3bc^2+4a^2b^2c^2-8ab^3c^2$
could some help me with this
|
$$B=3a^2c+b^2c=c(3a^2+b^2)$$
$$-C=6a^2+ab-2b^2=6a^2+4ab-3ab-2b^2=2a(3a+2b)-b(3a+2b)=(3a+2b)(2a-b)$$
$$A=abc^2$$
Break $$C\cdot A=(3a+2b)ac\cdot bc(2a-b)$$ as $$(3a+2b)ac-bc(2a-b)=B$$
|
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|
What are the conditions on $a, b, c$ so that $x^3+ax^2+bx+c$ is bijective? I would like to find the conditions on $a$, $b$, $c$ so that function $$f(x)=x^3+ax^2+bx+c$$ is bijective.
I thought about resolving the equation
$$x^3+ax^2+bx+c=y$$
but I didn't succeed. And our math teacher told us that we cannot prove that a function is bijective by proving that this function is strictly increasing or decreasing.
Thanks for your help!
Marie
|
Surjectivity is clear, because a third degree equation always has at least a real root.
Suppose $f(x)=f(y)$, with $x\ne y$; then
$$
(x^3-y^3)+a(x^2-y^2)+b(x-y)=0
$$
that becomes
$$
(x-y)(x^2+xy+y^2+a(x+y)+b)=0
$$
and so $x^2+xy+y^2+a(x+y)+b=0$.
Set $s=x+y$ and $p=xy$: then $s^2-4p>0$. We have $s^2+as+b-p=0$, so
$$
s^2-4s^2-4as-4b>0
$$
or
$$
3s^2+4as+4b<0
$$
This is only possible if the discriminant of this polynomial in $s$ is positive: indeed, if the discriminant is $\ge0$, the inequality $3s^2+4as+4b\ge0$ holds for every $s$, contrary to the assumption that for the particular $s=x+y$ the $<0$ inequality holds. Thus we obtain
$$
a^2-3b>0
$$
Can you finish up?
Conversely, suppose $a^2-3b\le0$. Then, for every $s$, we have $3s^2+4as+4b\ge0$, so $s^2-4(s^2+as+b)\le0$. If $f(x)=f(y)$, and we set $s=x+y$, $p=xy$, we have either $x=y$ or $s^2-4p\le0$, but this implies $x=y$ again.
|
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|
How to integrate $\int_{0}^{\infty} \frac{x \arctan(x) \ dx}{(1+x^2)^2}$ I am asked to solve the following integral:
$$\int_{0}^{\infty} \frac{x \arctan(x) \ dx}{(1+x^2)^2}$$
I've made a couple of guesses but none of them took me anywhere. One of my ideas was to do, at first, an u-substitution:
$$
u = 1+x^2\\
du = 2x dx\\
\frac{du}{2} = x dx\\
\\
x = 0 \Rightarrow u = 1\\
x = a \Rightarrow u = 1+a^2
$$
but then there's a $\arctan(x)$ left there untouched. Does anyone have an idea I can pursue?
Source: James Stewart's Calculus book.
Thank you.
|
Let us integrate by parts and put
$u'=\frac{x}{(1+x^2)^2}$
and
$v=arctan(x).$
thus
the integral becomes
$$I=[-\frac{1}{2(1+x^2)}arctan(x)]_0^\infty$$
$$+\frac{1}{2}\int_0^\infty\frac{dx}{(1+x^2)^2}$$
$$=\frac{1}{2}\int_0^\infty\frac{1+x^2-x^2}{(1+x^2)^2}dx$$
$$=\frac{\pi}{4}+\frac{1}{4}\int_0^\infty x \frac{-2x}{(1+x^2)^2}dx$$
$$=\frac{\pi}{4}+\frac{1}{4}([\frac{x}{1+x^2}]_0^\infty-\int_0^\infty\frac{1}{1+x^2}dx)$$
$$\color{red}{=\frac{\pi}{4}-\frac{\pi}{8}=\frac{\pi}{8}}$$
|
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|
What is the expected determinant of a symmetric $2\times2$ matrix, whose three elements are distinct and draw from$ [-n,n]$
The symmetric matrix has three elements:
\begin{pmatrix} A&B\\ B&C \end{pmatrix}
$A,B$ and $C$ are integers draw randomly from $\{-n,-n+1,\dots,n-1,n\}$, $n\ge2$ and they are distinct with each other. What is the expected determinant of this matrix?
I have no clue how to do this, this seems multiple random variables, I don't know how to attack this problem systematically.
Edit
Computer run give the result of $n =[2,10]$:
-2.49301 -4.66934 -7.52694 -11.0413 -15.18877 -19.97767 -25.64429 -31.6045 -38.66689
Therefore we have a correct answer which is $$\displaystyle-\frac{(n+1)(2n+1)}6$$ However, an more easier to understand answer is still very welcomed.
Code:
import random
N = 100000
for n in range(2,11):
total = 0
for i in range(0,N):
a,b,c=random.sample(range(-n,n+1),3)
total += a*c-b*b
print(total/N,end = ' ')
|
The problem is in the notation. When you write $B$ you actually mean $B$ conditional on $B\neq A=a$, which is not uniformly distributed on $[-n,n]$ but on $[-n,n]\backslash\{a\}$. Similarly, conditional on the values of $A=a,B=b$ the random variable $C$ is uniformly distributed on $[-n,n]\backslash\{a,b\}$. So, actually when you write $B$ you abuse notation for $B\mid A=a$ and when you write $C$ you abuse notation for $C\mid A=a, B=b$, plus the condition that $B\neq A\neq C$. All that said, you can solve it by conditioning, as is your intuition:
\begin{align}\mathbb E[AC]-\mathbb E[B^2]&=\sum_{a=-n}^n\left(\mathbb E\left[AC\mid A=a\right]-\mathbb E\left[B^2\mid A=a\right]\right)P(A=a)\\[0.3cm]&=\frac{1}{2n+1}\sum_{a=-n}^n\left(a\cdot \mathbb E\left[C\mid A=a\right]-\mathbb E\left[B^2\mid A=a\right]\right)\end{align} with
\begin{align}\mathbb E[B^2\mid A=a]&=\frac1{2n}\sum_{b=-n,b\neq a}^nb^2=\frac{1}{2n}\left(\sum_{b=-n}^nb^2-a^2\right)=\frac{1}{2n}\left(2\sum_{b=-n}^nb^2-a^2\right)\\[0.3cm]&=\frac{1}{2n}\left(2\frac{n(n+1)(2n+1)}{6}-a^2\right)=\frac{(n+1)(2n+1)}{6}-\frac{a^2}{2n}\end{align} and (now we must condition also on $B$):
\begin{align}\mathbb E[C\mid A=a]&=\sum_{b=-n,b\neq a}^n\mathbb E[C\mid A=a,B=b]P(B=b)\\[0.3cm]&=\frac{1}{2n}\sum_{b=-n,b\neq a}^n\left(\sum_{c=-n,c\neq a,b}^n c\cdot\frac{1}{2n-1}\right)=\frac{1}{2n(2n-1)}\sum_{b=-n,b\neq a}^n\left(\sum_{c=-n}^nc-b-a\right)\\[0.3cm]&=\frac{1}{2n(2n-1)}\sum_{b=-n,b\neq a}^n(0-b-a)=-\frac{1}{2n(2n-1)}\left(\sum_{b=-n}^n(b+a)-(a+a)\right)\\[0.3cm]&=-\frac{1}{2n(2n-1)}\left(0+(2n+1)a-2a\right)=-\frac{a}{2n}\end{align} And now, we substitute in the first equation to get
\begin{align}\mathbb E[AC]-\mathbb E[B^2]&=\sum_{a=-n}^n\left(\mathbb E\left[AC\mid A=a\right]-\mathbb E\left[B^2\mid A=a\right]\right)P(A=a)\\[0.3cm]&=\frac{1}{2n+1}\sum_{a=-n}^na\cdot \mathbb E\left[C\mid A=a\right]-\mathbb E\left[B^2\mid A=a\right]\\[0.3cm]&=\frac{1}{2n+1}\sum_{a=-n}^n\left(a\left(-\frac{a}{2n}\right)-\frac{(n+1)(2n+1)}{6}+\frac{a^2}{2n}\right)\\[0.3cm]&=\frac{1}{2n+1}\sum_{a=-n}^n\left(\frac{a^2}{2n}-\frac{a^2}{2n}-\frac{(n+1)(2n+1)}{6}\right)\\[0.3cm]&=\frac{1}{2n+1}\cdot(2n+1)\cdot\frac{(n+1)(2n+1)}{6}=-\frac{(n+1)(2n+1)}{6}\end{align}
|
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|
Can somebody prove this infinite series? Transformation of the Leibniz formula for $\pi$ results in the infinite series:
$$
\frac 1 {1 \times 3} + \frac 1 {5 \times 7} + \frac 1 {9 \times 11} + \frac 1 {13 \times 15} +\cdots = \frac \pi 8
$$
If you recombine the numbers in the denominator you get e.g. the following series:
$$
\frac 1 {1 \times 5} + \frac 1 {3 \times 7} + \frac 1 {9 \times 13} + \frac 1 {11 \times 15} + \cdots
$$
which seems to approach $\dfrac{\pi / 8}{\sqrt 2}$.
Can somebody prove this?
|
The same type of method as used to prove that $\pi/4 = 1 - 1/3 + 1/5 - 1/7 + \ldots$ can be used here, as soon as you write $\frac1{(2n + 1)(2n + 5)} = \frac1{4}(\frac1{2n + 1} - \frac1{2n + 5})$.
So your series is
$$\frac1{4}\left(1 + \frac1{3} - \frac1{5} - \frac1{7} + \ldots\right)$$
where the expression inside parentheses is $f(1)$ where $f(x) = x + x^3/3 - x^5/5 - x^{7}/7 + \ldots$. We calculate
$$\begin{array}{ccl}
f'(x) & = & 1 + x^2 - x^4 - x^6 + \ldots \\
& = & (1 + x^2)(1 - x^4 + x^8 - x^{12} + \ldots) \\
& = & \frac{1 + x^2}{1 + x^4}
\end{array}
$$
and so, since $f(0) = 0$, we have
$$f(1) = \int_0^1 \frac{1 + x^2}{1 + x^4} dx$$
Mathematica can give you the answer to this pretty quickly, but if you want to do this by hand, you can. We can factor $1 + x^4 = (1 - \sqrt{2} x + x^2)(1 + \sqrt{2} x + x^2)$; use a partial fraction decomposition to obtain
$$\frac{1 + x^2}{1 + x^4} = \frac{1/2}{1 - \sqrt{2} x + x^2} + \frac{1/2}{1 + \sqrt{2} x + x^2}$$
which after a little work integrates to
$$\frac{\sqrt{2}}{2} \left(\arctan(\frac{x - \sqrt{2}/2}{\sqrt{2}/2}) + \arctan(\frac{x + \sqrt{2}/2}{\sqrt{2}/2})\right)$$
and now plug in $1$ and $0$ and subtract, etc. to obtain
$$f(1) = \frac{\sqrt{2}}{2}\left(\arctan(\sqrt{2} - 1) + \arctan(\sqrt{2} + 1)\right)$$
The expressions inside the arctangents are reciprocals of each other, so the arctangent angles are complementary: they sum to $\pi/2$. Hence $f(1) = \sqrt{2}\pi/4$, and your answer is $f(1)/4 = \sqrt{2}\pi/16$, as advertised.
|
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|
Prove that $\lim ( \sqrt{n^2+n}-n) = \frac{1}{2}$ Here's what I have so far:
Given $\epsilon > 0$, we want to find N such that $\sqrt{n^2+n}-n < \epsilon$ for all $n>N$. And so:
$( \sqrt{n^2+n}-n-\frac{1}{2}) \cdot \frac{\sqrt{n^2+n}-(n+\frac{1}{2})}{\sqrt{n^2+n}-(n+\frac{1}{2})}$
$= \frac{(n^2+n)-(n+\frac{1}{2})}{\sqrt{n^2+n} + (n+\frac{1}{2})}$
And I'm not sure how to go on from here. Help would be appreciated.
|
For $n>0$ we have
(1). $ \sqrt {n^2+n}-n=\frac {n}{\sqrt {n^2+n}+n}.$
(2). $ n=\sqrt {n^2}<\sqrt {n^2+n}<\sqrt {n^2+n+\frac {1}{4}}=n+\frac {1}{2}.$
(3). Therefore $ \frac {1}{2}-\frac {1}{8n+2}=\frac {n}{( n+\frac {1}{2}) +n}<\frac {n}{\sqrt {n^2+n} +n}<\frac {n}{n+n}=\frac {1}{2}.$
|
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|
Summation of Central Binomial Coefficients divided by even powers of $2$ Whilst working out this problem the following summation emerged:
$$\sum_{m=0}^n\frac 1{2^{2m}}\binom {2m}m$$
The is equivalent to
$$\begin{align}
\sum_{m=0}^n \frac {(2m-1)!!}{2m!!}&=\frac 12+\frac {1\cdot3}{2\cdot 4}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}+\cdots +\frac{1\cdot 3\cdot 5\cdot \cdots \cdot(2n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot 2n}\\
&=\frac 12\left(1+\frac 34\left(1+\frac 56\left(1+\cdots \left(1+\frac {2n-1}{2n}\right)\right)\right)\right)
\end{align}$$
and terms are the same as coefficients in the expansion of $(1-x)^{-1/2}$.
Once the solution
$$ \frac {n+1}{2^{2n+1}}\binom {2n+2}{n+1}$$
is known, the telescoping sum can be easily derived, i.e.
$$\frac 1{2^{2m}}\binom {2m}m=\frac {m+1}{2^{2(m+1)-1}}\binom {2(m+1)}{m+1}-\frac m{2^{2m-1}}\binom {2m}m$$
However, without knowing this a priori, how would we have approached this problem?
|
Using an Extension of Pascal's Rule
$$
\begin{align}
\sum_{m=0}^n\frac1{2^{2m}}\binom{2m}{m}
&=\sum_{m=0}^n\frac{(2m-1)!!}{(2m)!!}\tag{1a}\\
&=\sum_{m=0}^n\binom{m-\frac12}{m}\tag{1b}\\
&=\sum_{m=0}^n\left[\binom{m+\frac12}{m}-\binom{m-1+\frac12}{m-1}\right]\tag{1c}\\
&=\binom{n+\frac12}{n}\tag{1d}\\[6pt]
&=\frac{(2n+1)!!}{(2n)!!}\tag{1e}\\[6pt]
&=\frac{2n+1}{2^{2n}}\binom{2n}{n}\tag{1f}
\end{align}
$$
Explanation:
$\text{(1a)}$: $\frac{\color{#C00}{(2m)!}}{\color{#C00}{2^mm!}\,\color{#090}{2^mm!}}=\frac{\color{#C00}{(2m-1)!!}}{\color{#090}{(2m)!!}}$
$\text{(1b)}$: divide numerator and denominator by $2^m$
$\text{(1c)}$: apply $(4)$ with $\alpha=\frac12$
$\text{(1d)}$: telescoping sum
$\text{(1e)}$: multiply numerator and denominator by $2^n$
$\text{(1f)}$: $\frac{(2n+1)!!}{(2n)!!}=\frac{(2n+1)\color{#C00}{(2n-1)!!}}{\color{#090}{(2n)!!}}=\frac{(2n+1)\color{#C00}{(2n)!}}{\color{#C00}{2^nn!}\,\color{#090}{2^nn!}}$
Extension of Pascal's Rule
Newton's Generalized Binomial Theorem says
$$
\begin{align}
\sum_{m=0}^\infty\binom{m+\alpha}{m}x^m
&=\sum_{m=0}^\infty(-1)^m\binom{-1-\alpha}{m}x^m\tag{2a}\\
&=(1-x)^{-1-\alpha}\tag{2b}
\end{align}
$$
Explanation:
$\text{2a}$: convert to negative binomial coefficient
$\text{2b}$: Binomial Theorem
Thus,
$$
\begin{align}
\sum_{m=0}^\infty\binom{m-1+\alpha}{m}x^m
&=(1-x)^{-\alpha}\tag{3a}\\[6pt]
&=(1-x)(1-x)^{-1-\alpha}\tag{3b}\\[9pt]
&=(1-x)\sum_{m=0}^\infty\binom{m+\alpha}{m}x^m\tag{3c}\\
&=\sum_{m=0}^\infty\binom{m+\alpha}{m}\left(x^m-x^{m+1}\right)\tag{3d}\\[3pt]
&=\sum_{m=0}^\infty\left[\binom{m+\alpha}{m}-\binom{m-1+\alpha}{m-1}\right]x^m\tag{3e}
\end{align}
$$
Explanation:
$\text{(3a)}$: apply $\text{(2b)}$
$\text{(3b)}$: factor out $(1-x)$
$\text{(3c)}$: apply $\text{(2b)}$
$\text{(3d)}$: distribute $(1-x)$
$\text{(3e)}$: substitute $m\mapsto m-1$ in the subtrahend
Thus, for arbitrary $\alpha\in\mathbb{R}$, we can extend Pascal's Rule to
$$
\binom{m-1+\alpha}{m}+\binom{m-1+\alpha}{m-1}=\binom{m+\alpha}{m}\tag4
$$
|
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|
Point on surface closest to a plane using Lagrange multipliers Find the point on $z=1-2x^2-y^2$ closest to $2x+3y+z=12$ using Lagrange multipliers.
I recognize $z+2x^2+y^2=1$ as my constraint but am unable to recognize the distance squared I am trying to minimize in terms of 3 variables. May someone help please.
|
A simpler approach for computing the distance between these objects: the given plane is orthogonal to the vector $(2,3,1)^T$, so the point(s) on the surface of minimal distance from the plane are the ones for which the tangent plane of the surface at such point(s) is orthogonal su $(2,3,1)^T$. If for some $k$
$$\left\{\begin{array}{rcl} 2x^2+y^2+z &=& 1 \\ 2x+3y+z &=& k \end{array}\right. $$
has exactly one solution, such a solution is a point of minimal distance. By eliminating $z$, we are looking for the values of $k$ such that
$$ 2x^2-2x+y^2-3y = 1-k $$
has exactly one solution. By completing the squares, it is trivial that the only point of minimal distance occurs at $(x,y)={\left(\frac{1}{2},\frac{3}{2}\right)}$, from which $(x,y,z)=\color{red}{\left(\frac{1}{2},\frac{3}{2},-\frac{7}{4}\right)}$ and
$$ d(\text{plane},\text{surface}) = \frac{\left|2\cdot\frac{1}{2}+3\cdot\frac{3}{2}-\frac{7}{4}-12\right|}{\sqrt{3^2+2^2+1^2}}=\color{red}{\frac{33}{4\sqrt{14}}}.$$
|
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|
Prove: $\log_{2}{3} < \log_{3}{6}$ How should I prove that $\log_{2}{3} < \log_{3}{6}$?
I tried something like this: $2^{\log_{2}{3}}< 2^{\log_{3}{6}}$, $3<6^{\log_{3}{2}}$, $\log_{6}{3}<\log_{6}{6^{\log_{3}{2}}}=\log_{3}{2}$, $\frac{1}{\log_{3}{6}}< \log_{3}{2}$. $\frac{1}{1+\log_{3}{2}}<\log_{3}{2}$, but still nothing.
|
Hint
You can use the fact that
$$\log_a(b)=\frac{\ln(b)}{\ln(a)}$$
so you can rewrite your inequation
$$\frac{\ln(3)}{\ln(2)}<\frac{\ln(6)}{\ln(3)}$$
if and only if
$$\ln^2(3)<\ln(2\times 3)\ln 2=\ln(2)(\ln(2)+\ln(3)).$$
|
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|
How to do SVD of the differentiation operator of $P_{2}$ Find the singular values of the differentiation operator $D \colon P_{2} \rightarrow P_{2}$ where $P_{2}$ is the vector space of real polynomials of degree less than or equal to $2$, and the inner product on $P_{2}$ is given by
$$ \left<f,g \right> :=\int_{-1}^{1}f(t)g(t) \,dt. $$
Using the brute-force method, I cannot even find the adjoint of this map.
I have just came up with the idea that the only eigenvalue of the differential operator is $0$, as the conjugation of $0$, $0$ is the only eigenvalue of the $D^*$. But I cannot say anything about the composition $D^{*}D$...
So could someone provide some ideas of solving this? Thanks so much.
|
You can start with finding an orthonormal basis for $P_2$ by applying Gram-Schmidt to the standard basis $(1,x,x^2)$. Using the fact that the integral of an odd function on $[-1,1]$ vanishes, we get:
$$ e_1 = \frac{1}{\|1\|} = \frac{1}{\sqrt{2}}, \\
e_2 = \frac{x - \left< x, e_1 \right> e_1}{x - \left< x, e_1 \right> e_1} = \frac{x}{\|x\|} = \sqrt{\frac{3}{2}} x, \\
e_3 = \frac{x^2 - \left< x^2, e_2 \right> e_2 - \left< x^2, e_1 \right> e_1}{\|x^2 - \left< x^2, e_2 \right> e_2 - \left< x^2, e_1 \right> e_1\|} = \frac{x^2 - \frac{1}{2} \int_{-1}^1 x^2 \, dx}{\| x^2 - \frac{1}{2} \int_{-1}^1 x^2 \, dx \|} = \frac{x^2 - \frac{1}{3}}{\| x^2 - \frac{1}{3} \|} = \sqrt{\frac{45}{8}} \left( x^2 - \frac{1}{3} \right).$$
Then, we calculate:
$$
D(e_1) = 0,\\
D(e_2) = \sqrt{\frac{3}{2}} = \sqrt{3} e_1, \\
D(e_3) = \sqrt{\frac{45}{2}}x = \sqrt{15} e_2.
$$
From here, since the $e_i$ are orthonormal, we already see that the singular values of $D$ are $\sqrt{15}, \sqrt{3}, 0$. Alternatively, we can see that with respect to $\mathcal{B} = (e_1, e_2, e_3)$, the operator $D$ is represented by the matrix
$$ [D]_{\mathcal{B}} = \begin{pmatrix} 0 & \sqrt{3} & 0 \\
0 & 0 & \sqrt{15} \\
0 & 0 & 0 \end{pmatrix}. $$
and $D^{*}$ is represented by the transpose
$$ [D^{*}]_{\mathcal{B}} = [D]_{\mathcal{B}}^t = \begin{pmatrix} 0 & 0 & 0 \\
\sqrt{3} & 0 & 0 \\
0 & \sqrt{15} & 0 \end{pmatrix}. $$
Hence,
$$ [D^{*}D]_{\mathcal{B}} = [D^{*}]_{\mathcal{B}} [D]_{\mathcal{B}} = \begin{pmatrix} 0 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 15 \end{pmatrix} $$
which also shows that the eigenvaleus of $\sqrt{D^{*}D}$ which are the singular values of $D$ are $\sqrt{15}, \sqrt{3}, 0$.
|
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|
Show that $f(x)=\pi-2\arctan(\sqrt{x-1})$
Given, for every $x>1$,
$$f(x)=4\arctan\frac{1}{\sqrt{x-1}+\sqrt{x}}$$
Show that $f(x)=\pi-2\arctan(\sqrt{x-1})$
I have tried to use the fact that $\arctan(x)+\arctan(1/x)=\frac{\pi}{2
}$
So I obtain: $f(x)=4(\frac{\pi}{2}-\arctan(\sqrt{x-1}+\sqrt{x})$
I am stuck here !
|
Observe that
$$\frac1{\sqrt{x-1}+\sqrt x}=\sqrt x-\sqrt{x-1}$$
and thus
$$\left(4\arctan(\sqrt x-\sqrt{x-1})\right)'=4\left(\frac1{2\sqrt x}-\frac1{2\sqrt{x-1}}\right)\cdot\frac1{1+(\sqrt x-\sqrt{x-1})^2}=$$
$$=2\left(\frac1{\sqrt x}-\frac1{\sqrt{x-1}}\right)\frac1{2\sqrt x(\sqrt x-\sqrt{x-1})}=\frac{\sqrt{x-1}-\sqrt x}{x\sqrt{x-1}(\sqrt x-\sqrt{x-1})}=$$
$$=\color{red}{-\frac1{x\sqrt{x-1}}}$$
And on the other hand:
$$\left(\pi-2\arctan\sqrt{x-1}\right)'=-\frac1{\sqrt{x-1}}\frac1{1+x-1}=\color{red}{-\frac1{x\sqrt{x-1}}}$$
Thus, both forms of the $\;f\;$ have the same derivative and thus they differ only by a constant, say $\;K\;$ :
$$\pi-2\arctan\sqrt{x-1}=4\arctan(\sqrt x-\sqrt{x-1})+K$$
and observe that in the above form both sides are well defined for $\;x=1\;$, so substituting $\;x=1\;$ :
$$\pi=\pi-2\arctan0=4\arctan(1)+K=4\frac\pi4+K=\pi+K\implies K=0$$
and we get the wanted equality
|
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|
Prove $(1+a+b)(1+b+c)(1+c+a)\ge 9(ab+bc+ca)$ How one can prove the following.
Let $a$, $b$ and $c$ be non-negative real numbers. Then the inequality holds:
$(1+a+b)(1+b+c)(1+c+a)\ge9(ab+bc+ca).$
WLOG one can assume that $0\le a\le b\le c$.
It is not difficult to prove the statement when $0\le a\le b\le c\le 1$ or $1\le a\le b\le c$.
|
$a=0, b=0, c=2 \ \implies \ 1 \cdot 3 \cdot 3 \ge 3 \cdot 2^2$, not true.
|
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|
Calculating limit of $\lim_{x\to\infty}\dfrac{\sqrt{x+1}-2\sqrt{x+2}+\sqrt{x}}{\sqrt{x+2}-2\sqrt{x}+\sqrt{x-4}}$ As the title says we want to calculate:
$$\lim_{x\to\infty}\dfrac{\sqrt{x+1}-2\sqrt{x+2}+\sqrt{x}}{\sqrt{x+2}-2\sqrt{x}+\sqrt{x-4}}$$
By multiplying nominator and denominator in their conjugates
$=\lim_{x\to\infty}\dfrac{(\sqrt{x+2}+2\sqrt{x}+\sqrt{x-4})(x+1+x+2\sqrt{x(x+1)}-4(x+2))}{(\sqrt{x+1}+2\sqrt{x+2}+\sqrt{x})(x+2+x-4+2\sqrt{(x+2)(x-4)})-4x)}$
$=\lim_{x\to\infty}\dfrac{(\sqrt{x+2}+2\sqrt{x}+\sqrt{x-4})(-2x-7+2\sqrt{x^2+x})}{(\sqrt{x+1}+2\sqrt{x+2}+\sqrt{x})(-2x-2+2\sqrt{x^2-2x-8})}$
I think now we can take $$2x\approx2\sqrt{x^2+x}\approx2\sqrt{x^2-2x-8}\\[2ex]
\sqrt{x}\approx\sqrt{x+1}\approx\sqrt{x+2}\approx\sqrt{x-4}$$ as $x$ goes to infinity. Hence the limit of above fraction would be $\dfrac{7}{2}$, but wolframalpha gives me $\dfrac{3}{2}$ as the limit of the above fraction.
What am I doing wrong?
|
You need one more conjugate-multiplication.
You already have
$$\dfrac{(\sqrt{x+2}+2\sqrt{x}+\sqrt{x-4})(-2x-7+2\sqrt{x^2+x})}{(\sqrt{x+1}+2\sqrt{x+2}+\sqrt{x})(-2x-2+2\sqrt{x^2-2x-8})}$$
First, note that
$$\dfrac{\sqrt{x+2}+2\sqrt{x}+\sqrt{x-4}}{\sqrt{x+1}+2\sqrt{x+2}+\sqrt{x}}=\frac{\sqrt{1+\frac 2x}+2+\sqrt{1-\frac 4x}}{\sqrt{1+\frac 1x}+2\sqrt{1+\frac 2x}+1}\to 1\ (x\to \infty)$$
Now multiplying
$$\dfrac{-2x-7+2\sqrt{x^2+x}}{-2x-2+2\sqrt{x^2-2x-8}}$$
by
$$\frac{-2x-7-2\sqrt{x^2+x}}{-2x-7-2\sqrt{x^2+x}}\cdot\frac{-2x-2-2\sqrt{x^2-2x-8}}{-2x-2-2\sqrt{x^2-2x-8}}\ (=1)$$
gives
$$\frac{(-2x-2-2\sqrt{x^2-2x-8})((-2x-7)^2-4(x^2+x)}{(-2x-7-2\sqrt{x^2+x})((-2x-2)^2-4(x^2-2x-8))}=\frac{(-2x-2-2\sqrt{x^2-2x-8})(24x+49)}{(-2x-7-2\sqrt{x^2+x})(16x+36)}=\frac{(-2-\frac 2x-2\sqrt{1-\frac 2x-\frac{8}{x^2}})(24+\frac{49}{x})}{(-2-\frac 7x-2\sqrt{1+\frac 1x})(16+\frac{36}{x})}\to \frac 32\ (x\to\infty)$$
|
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|
Show that exactly half of $1^{\frac{p-1}{2}}, 2^{\frac{p-1}{2}}, \dots, (p-1)^{\frac{p-1}{2}}$ are congruent to 1 modulo $p$
Let $p$ be an odd prime number. Look at the numbers in the set
\begin{align*}
S \in \{1^{\frac{p-1}{2}}, 2^{\frac{p-1}{2}}, \dots, (p-1)^{\frac{p-1}{2}}\}
\end{align*}
Show that exactly half of these numbers are congruent to 1 modulo $p$.
I define two polynomials
\begin{align*}
&f(x) = x^{\frac{p-1}{2}} - 1 \\
&g(x) = x^{\frac{p-1}{2}} + 1
\end{align*}
According to Lagranges theorem, the congruences
\begin{align*}
&f(x) \equiv 0 \pmod{p} \\
&g(x) \equiv 0 \pmod{p}
\end{align*}
or
\begin{align*}
&x^{\frac{p-1}{2}} \equiv 1 \pmod{p} \\
&x^{\frac{p-1}{2}} \equiv -1 \pmod{p}
\end{align*}
will have maximum $\frac{p-1}{2}$ slutions each. Thus, we can say that maximum half of the numbers in $S$ will be congruent to either 1 or -1 modulo $p$. How can I show that exactly half of the numbers are congruent to 1 or -1 modulo $p$?
|
Recall that each of $1, \dots, p-1$ is a root of $X^{p-1}-1$.
Now as you alluded to $X^{p-1}-1 = (X^{(p-1)/2} -1)(X^{(p-1)/2} +1)$.
This means that for each $a=1, \dots, p-1$:
$$(a^{(p-1)/2} -1)(a^{(p-1)/2} +1)=0$$
thus at least one of the two $(a^{(p-1)/2} -1)$ and $(a^{(p-1)/2} +1)$ is zero, and of course not both can be equal to $0$.
Thus for each $a=1, \dots, p-1$ exactly one of $(a^{(p-1)/2} -1)$ and $(a^{(p-1)/2} +1)$ is zero.
As you observed correctly for each of $(X^{(p-1)/2} -1)=0$ and $(X^{(p-1)/2} -1)=0$ there can be at most $(p-1)/2$ solutions.
Yet together this means that $(X^{(p-1)/2} -1)=0$ and $(X^{(p-1)/2} -1)=0$ both have exactly $(p-1)/2$ solutions.
You know each of $p-1$ is a solution to one of the two, and each can have at most $(p-1)/2$ solution, whence the only way is both have exactly $(p-1)/2$ solutions.
|
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|
simplify $\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$ simplify $\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$.
1.$90^{\frac{3}{2}}$
2.$106\sqrt{41}$
3.$4\sqrt{41}$
4.$504$
5.$508$
My attempt:I do like this but I didn't get any of those five.
$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}={\sqrt{45+4\sqrt{41}}}^3-\sqrt{45-4\sqrt{41}}^3$
$=(\sqrt{45+4\sqrt{41}}-\sqrt{45-4\sqrt{41}})(45+4\sqrt{41}+45-4\sqrt{41}+(\sqrt{45+4\sqrt{41}})(\sqrt{45-4\sqrt{41}})$
Now I do the nested radicals formula and I get $254\sqrt{41}$ which is none of those where did I mistaked?
|
Hint:
$$45\pm4\sqrt{41}=(2\pm\sqrt{41})^2.$$
|
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|
Evaluate the integral $\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$ $$\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$$
I have tried to form a square above i also tried to get the x below under the root
but got nothing
|
The standard way is to make the substitution
$y=x+\sqrt{x^2+2x+2}$
so that $x=\frac{y^2-2}{2(y+1)}$
$\sqrt{x^2+2x+2}=\frac{y^2+2y+2}{2(y+1)}$
$dx=\frac{y^2+2y+2}{2(y+1)^2}dy$
which changes your integral into an integral of rational function, solvable by partial fraction expansion:
$\int\frac{(y^2+2y+2)^2}{2(y^2-2)(y+1)^2}dy$
It is messy to solve it, but it is definitely doable with the right ammount of determination.
The partial fraction expansion of $\frac{(y^2+2y+2)^2}{2(y^2-2)(y+1)^2}$ is:
$\frac{4}{y^2-2}+\frac{1}{y+1}-\frac{1}{2(y+1)^2}+\frac{1}{2}$
which you integrate term by term.
There might be a easier trick to solve it, but i don't see it.
EDIT: after more thinking i think that one can make the following substitution
$x+1=\tan\theta$ with $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
$\sqrt{x^2+2x+2}=\sqrt{(\tan\theta)^2+1}=\sqrt\frac{1}{\cos^2\theta}=\frac{1}{\cos\theta}$
as $\cos\theta>0$ when $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
Then $dx=\frac{1}{\cos^2\theta} d\theta$
So finally you integrate:
$\int \frac{1}{\cos^2\theta(\sin\theta-\cos\theta)}d\theta$, with $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
From here one might go to the tangent of half angle, but again I see no easy formula.
|
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|
Limit square roots of polynomials I am trying to find $\lim \limits_{n \to \infty} {\sqrt{n^3+1}-n\sqrt{n} \over \sqrt{n^2+1}-n}$. I rewrite the fraction as
$${(\sqrt{n^3+1}-n\sqrt{n})(\sqrt{n^3+1}+n\sqrt{n}) \over (\sqrt{n^2+1}-n)(\sqrt{n^3+1}+n\sqrt{n})} = {1 \over (\sqrt{n^2+1}-n)(\sqrt{n^3+1}+n\sqrt{n})}$$
I notice $$\sqrt{n^2+1}-n > 0$$
so the denominator is growing to infinity while the whole limit is $0$. In the material I'm covering, the properties of polynomials haven't been discussed (but the properties of the square root have been). Is there a more basic way to conclude about the denominator going to infinity?
|
Observe that $${\sqrt{n^3+1}-\sqrt{n^3} \over \sqrt{n^2+1}-n}=\dfrac{n^{3/2}}{n}\left({\sqrt{1+\dfrac{1}{n^3
}}-1 \over \sqrt{1+\dfrac{1}{n^2}}-1}\right)$$ and by multiplying conjugates of both top and bottom yields $${\sqrt{1+\dfrac{1}{n^3
}}-1 \over \sqrt{1+\dfrac{1}{n^2}}-1}=\dfrac{\dfrac1{n^3}}{\dfrac1{n^2}}\left({\sqrt{1+\dfrac{1}{n^2
}}+1 \over \sqrt{1+\dfrac{1}{n^3}}+1}\right).$$ Combinig these results we have $${\sqrt{n^3+1}-\sqrt{n^3} \over \sqrt{n^2+1}-n}=\dfrac{1}{n^{1/2}}\left({\sqrt{1+\dfrac{1}{n^2
}}+1 \over \sqrt{1+\dfrac{1}{n^3}}+1}\right)\to 0.$$
|
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|
Find the limit of $q(0) = \lim\limits_{x\to 0}\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}$ Find the limit of $q(0) = \lim\limits_{x\to 0}\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}$
For this I think I should use De l'Hopital's rule but it takes a lot time and I can't get to answer.
Can we use the De l'Hopital's rule twice?or three times?If yes what is the limit of that?and how can we find the limit of polynomial multiplys?
|
The limit above gives: $$\lim\limits_{x\to 0}\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}=\lim\limits_{x\to 0}\frac{x^4+10x^3+35x^2+50x}{x^2+5x}$$
Now, it's enough to do the classical polynomial division, so we have: $$\lim\limits_{x\to 0}\frac{x^4+10x^3+35x^2+50x}{x^2+5x}=\lim\limits_{x\to 0}x^2+5x+10=10\hspace{75pt} (*1)$$
You can also factorize $x^4+10x^3+35x^2+50x$, whose two solutions are $x_1=0$ and $x_2=-5$, so we have $x^4+10x^3+35x^2+50x=x(x+5)(x^2+5x+10)$. By simplifying, we got the limit in $(*1)$.
I hope I told you what you were looking for.
|
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|
Solve the Diophantine equation $a+2b=2ab$, where $(a,b)$ are positive integers. Let $a$ be a positive integer. Show that $\gcd(a, a-1) = 1.$
Let $d$ be the greatest common divisor of $a$ and $a-1.$ I.e. $\gcd(a, a-1) = d$
Therefore, $d$ must divide $a-(a-1),$ following the rule that if any number, when some number d divides two numbers, then d must also divide the difference.
In other words, $d\mid (a-(a-1)).$
If we simplify, d must divide $a-a+1= 1$.
I have shown that the $\gcd(a, a-1) = 1,$ since d = 1. QED.
b) Use the result of part a) to solve the Diophantine equation $a+2b=2ab,\,$ where $a, b \in \mathbb Z$.
Let $2b = 2k,$ where $k$ is an integer and therefore $2b$ is even.
Case 1: $a$ is even
If $a$ is even, then $a = 2k,$ where k is an integer.
$2k + 2k = 2(2k)(1)$
$4k = 4k$
Both sides equal each other...
This is where I stopped. I realized I wasn't using the previous proof to solve this equation. This is where I need help.
|
Note we can rewrite your equation as
$$a + 2b = 2ab \iff a = 2b (a - 1) $$
Both sides of the equation are integers so it follows that $a $ divides $2b(a -1) $. But because of your lemma we know $a $ and $a-1$ are coprime so $a \mid 2b \rightarrow 2b = ak $ for some integer $k $. Substituting we get that
$$a = 2b (a - 1) \iff a = ak(a - 1) \iff 1 = \frac {ak(a-1)}{a} \iff 1 = k(a-1)$$
Because both $a $ and $k $ are positive integers it follows that $a - 1 = 1 \rightarrow a = 2$ and $k = 1 \rightarrow b = 1$.
We divided by $a $ in our calculations so $a \not= 0$. A second of thought shows $a = b = 0$ would also be a possible solution.
Another way of going about it, without the lemma $gcd(a, a-1) = 1$:
$2b $ is always an even number and so is $2ab $ so it follows, from the equality $a + 2b = 2ab $ that $a $ is even.
If $a $ is even, then $a = 2k $ for some integer k. Substituting in first equality we get $2k + 2b = 2 (2k)b \iff 2 (k + b) = 4kb \iff k + b = 2kb $
Now we divide both sides by $b $ and $k $ separately to get
$$\frac{k + b}{b} = 2k \iff \frac{k}{b} + 1 = 2k$$
$$ \frac{k + b}{k} = 2b \iff \frac{b}{k} + 1 = 2b$$
From the left sides of the right equalities you get that both $\frac{b}{k}$ and $\frac{k}{b}$ must be integers so you conclude $b = k $.
Plugging in you get that
$$ k + b = 2kb \iff 2k = 2k^2 \iff k = k^2 \iff k = 0, 1$$
If you pick $k=0$ then $b = a = 0$. If you pick $k=1$ then $b = 1$ and $a = 2k = 2$.
|
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|
Minimizing $|a+bw+cw^2|$ where $a,b,c\in\mathbb Z$ and $w = \zeta_3$.
If $a,b,c\in\mathbb Z$ are not all equal and $w$ is a cube root of unity $(w\neq 1$), then the minimum value of:
$$|a+bw+cw^2|$$
is what?
I'm pretty stuck with the above problem. Could someone help me out?
|
Using $|z|^2 = z\bar{z}$
and Using $\displaystyle \bar{\omega} = \omega^2$ and $\bar{\omega^2} = \omega$ and $\omega^3 = 1$ and $1+\omega+\omega^2 = 0$
So $$|a+b\omega+c\omega^2|^2 = (a+b\omega+c\omega^2)\cdot \bar{(a+b\omega+c\bar{\omega^2})} = (a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$$
So we get $$|a+b\omega+c\omega^2|^2=a^2+b^2+c^2+ab(\omega+\omega^2)+bc(\omega+\omega^2)+ca(\omega+\omega^2)$$
So $$|a+b\omega+c\omega^2|^2=a^2+b^2+c^2-ab-bc-ca = \frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$$
Now Given $a,b,c$ not all are equal integers. So for $\min$ of right side expression
Let $a=n\;,b=n\;,c=n+1\;,$ Where $n\in \mathbb{Z}$
So $$\min|a+b\omega+c\omega^2|^2 = \frac{1}{2}\left[1^2+1^2\right] = 1$$
So $$\min |a+b\omega+c\omega^2| = 1$$
|
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|
Prove $27195^8 - 10887^8 + 10152^8\equiv 0\pmod{26460}$ So far I've tried:
1) $26460 = 2^2 * 3^3 * 5 * 7^2$
2) $10152 = 2^3 * 3^3 * 47$
3) $27195 - 10887 = 16308 = 2^2 * 3^3 * 151$ (I know $a^8 - b^8\equiv0\pmod{a - b}$)
Therefore I conclude that $27195^8 - 10887^8 + 10152^8$ is divisible by $2^2 * 3^3$, as well is 1).
But what about $5 * 7^2$ part?
|
Let $a=27195, b=10887, c=10152$ and consider this:
$$
\matrix{
m &a \bmod m &b \bmod m &c \bmod m \\
2^2 &3 &3 &0 \\
3^3 &6 &6 &0 \\
5\hphantom{^1} &0 &2 &2 \\
7^2 &0 &9 &9 \\
}
$$
Note how for each $m$, we have $b \equiv a$ or $b \equiv c \bmod m$, and the other is $0 \bmod m$.
Therefore, the terms in $27195^8 - 10887^8 + 10152^8$ reduce to $0 \bmod m$.
|
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|
Induction divisibility question Q. Prove by induction that $2^{3n-1} + 5(3^n)$ is divisible by $11$ for any even number $n$, where $n$ is an element of natural numbers.
What is have so far:
(base case): $p(2) = 77$, $77/11 = 7$. so base case holds
$p(k) = 2^{3k-1} +5(3^k) $
$p(k+2) = 2^{3k+5} + 5(3^{k+2}) $
$p(k+2) = 2^{3k+1}2^4 + 5(3^{k})(3^2) $
$p(k+2) = 2^{3k+1+(1-1)}2^4 + 5(3^{k})3^2 $
$p(k+2) = 2^{3k-1}2^6 +5(3^{k})3^2$
I am new to induction and I don't know how to continue.
A point in the right direction would be greatly appreciated, thank you.
|
The next step I would take would be to write this in terms of $p(k)$ somehow:
$$p(k+2)=(2^{3k-1}+5(3^k))3^2+(2^6-3^2)2^{3k-1},$$
and see what the remainder is.
|
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|
What is the mistake? $$1=1$$
$$\Rightarrow\frac{-1}{1}=\frac{1}{-1}$$
$$\Rightarrow \sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}$$
$$\Rightarrow\frac{i}{1}=\frac{1}{i}$$
$$\Rightarrow\frac{i}{2}=\frac{1}{2i}$$
$$\Rightarrow\frac{i}{2}+\frac{3}{2i} = \frac{1}{2i} +\frac{3}{2i}$$
$$\Rightarrow i(\frac{i}{2}+\frac{3}{2i} ) = i(\frac{1}{2i} +\frac{3}{2i})$$
$$\Rightarrow\frac{-1}{2}=\frac{1}{2}$$
$$\Rightarrow1=2$$
What is wrong in this?
|
$$\sqrt{\frac{a}{b}} \neq \frac{\sqrt{a}}{\sqrt{b}} $$ in general unless both $a$ and $b$ are positive.
|
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|
Throwing a dice: specify possible combinations of $(i, j, k) \in$ $\Omega = \{1, 2, 3, 4, 5, 6\}^3$ with $i - j + k = l$
Let $(\Omega, P)$ be a laplacian probability space with $\Omega = \{1, 2, 3, 4, 5, 6\}^3$, and let $X: \Omega \rightarrow \Bbb Z$, $(i, j, k) \rightarrow i - j + k$ be a random variable. Specify $P_x$ with $P_x(F) := P(\{X \in F\})$ for $F \in \mathscr P(\Omega)$.
I already calculated the following things:
There are $6^3 = 216$ possible combinations of the $(i, j, k) \in \Omega$, and the possible results range from $-4$ to $11$. In order to specify $P_x$, I have to find the number of all possible $(i, j, k) \in \Omega$ such that $i - j + k = l$ for $l \in \{-4, ..., 11\}$. I have written it down for the numbers $11, 10, 9, 8$ and $7$. I calculated that there is one combination of $(i, j, k)$ such that $i - j + k = 11$, three combinations for $10$, six combinations for $9$, ten combinations for $8$ and fifteen combinations for $7$.
I guess that there is a much more easy way to calculate all of these possible combinations, but I wasn't able to see a pattern yet. Plus, what would be a good way to define $P_x$? I thought about something like splitting it up in the sets $\{-4, ..., -1\}$, $\{0, ..., 6\}$ and $\{7, ..., 11\}$.
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Generating functions work nicely here.
Let $p=\frac{1}{6}\left(x+x^2+x^3+x^4+x^5+x^6\right)$.
Let $q=\frac{1}{6}\left(x^{-1}+x^{-2}+x^{-3}+x^{-4}+x^{-5}+x^{-6}\right)=\frac{p}{x^7}$.
Then the distribution you seek is given by $p^2q = \frac{p^3}{x^7}$:
\begin{align*}
p^2q =&\frac{1}{216}\left(x^{11}
+ 3 x^{10}
+ 6 x^{9}
+ 10 x^{8}
+ 15 x^{7}
+ 21 x^{6}
+ 25 x^5 \\
+ 27 x^4
+ 27 x^3
+ 25 x^2
+ 21 x^1
+ 15 x^0
+ 10 x^{-1}
+ 6 x^{-2}
+ 3 x^{-3}
+ x^{-4}\right)
\end{align*}
Thus, for example, $P_9=\frac{6}{216}$ and $P_5=\frac{25}{216}$.
|
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|
Prove that $a+b\ge6(2^{\frac{1}{3}}+4^{\frac{1}{3}})$ if equation $2x^3+ax^2+bx+4=0$ has $3$ real roots
Peoblem Staement:-
If equation $2x^3+ax^2+bx+4=0$ has $3$ real roots ($a,b\gt0$), then prove that $a+b\ge6(2^{\frac{1}{3}}+4^{\frac{1}{3}})$.
Attempt at a solution:-
Let the roots of the equation $2x^3+ax^2+bx+4=0$ be $\alpha_1, \alpha_2$ and $\alpha_3$.
We have,
$$\sum_{i=1}^3{\alpha_i}=-\dfrac{a}{2}\\
\sum_{1\le i\lt j\le3}{\alpha_i\alpha_j}=\dfrac{b}{2}\\
\alpha_1\alpha_2\alpha_3=-\dfrac{4}{2}=-2$$
Now from AM-GM inequality we get,
$$\dfrac{\displaystyle\sum_{i=1}^3{\alpha_i}}{3}\ge\sqrt[3]{\prod_{i=1}^3{\alpha_i}}\implies -\dfrac{a}{6}\ge\sqrt[3]{-2}\implies a\ge6\cdot{2}^{\frac{1}{3}}$$
Equality occurring iff $\alpha_1=\alpha_2=\alpha_3$
Similarly form the AM-GM inequality, we get
$$\dfrac{\displaystyle\sum_{1\le i\lt j\le3}{\alpha_i\alpha_j}}{3}\ge \sqrt[3]{\prod_{i=1}^3{\alpha_i}^2}\implies \dfrac{b}{6}\ge 4^{\frac{1}{3}}\implies b\ge 6\cdot{4}^{\frac{1}{3}}$$
Equality only occurring when $\alpha_1\alpha_2=\alpha_2\alpha_3=\alpha_3\alpha_1\implies\alpha_1=\alpha_2=\alpha_3$
Hence, $a+b\ge 6\cdot(2^{\frac{1}{3}}+4^{\frac{1}{3}})$
The solution looks satisfying but the only problem is that AM-GM inequality can be used only for non-negative numbers whereas $\sum_{i=1}^3{\alpha_i}$ is clearly a negative number as $a\gt0$. I was also not so sure about adding the two inequalities to get at the inequality the question asks to prove.
I am also not so sure when I can do this, i.e. add the inequalities. In this case is it valid due to the equality occurring at the same values in both the inequalities.
And, finally if you have a better solution please do post it.
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Since $a, b > 0$, the polynomial is strictly positive for any nonnegative $x$. But this means that all $\alpha_i$ are negative. Now for example
$$2^{1/3} = (|\alpha_1| |\alpha_2| |\alpha_3|)^{1/3} \le \frac{1}{3}(|\alpha_1| + |\alpha_2| + |\alpha_3|) = -\frac{1}{3}(\alpha_1 + \alpha_2 + \alpha_3) = \frac{a}{2}.$$
Note that you also made a mistake when handling the inequalities, since $-\frac{a}{6} \ge -2^{1/3}$ is equivalent to $a \le 6 \cdot 2^{1/3}$ (multiplying an inequality with a negative number changes the direction of the inequality).
Adding both inequalities is also correct, since $a \ge b$ and $c \ge d$ implies $a + c \ge b + d$.
|
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|
Mac-Laurin-series for $\psi(\frac{2x+1}{2x})-\psi(\frac{x+1}{2x})$? Consider $$f(x)=\psi(\frac{2x+1}{2x})-\psi(\frac{x+1}{2x})$$
$\psi(x)$ is the digamma-function. This function occurs in the calculation of the definite integral
$$\int_0^1 \ln(x^n+1)dx=\ln(2)-\frac{f(n)}{2}$$ for $n>0$. Wolfram alpha gives a series expansion of $f(x)$ for $x\rightarrow\infty$ , but not for $x\rightarrow 0$ and I could not even calculate $$\lim_{x\rightarrow 0} f'(x)$$, which should be $1$ due to numerical calculation.
Does the Mac-Laurin-series for $f(x)$ exist ? If yes, how can I find the series ? And, finally, which convergent radius does it have ?
|
Let's take a look at the chapter about polygamma functions in Abramowitz and Stegun.
The multiplication formula (6.4.8) shows:
$$\begin{align*}
\psi(2x) &= \ln(2) + \frac{1}{2}(\psi(x) + \psi(x+1/2)) \\
\iff \psi(x + 1/2) &= 2\psi(2x) - \ln(4) - \psi(x)
\end{align*}$$
The recurrence formula (6.4.6) shows:
$$\psi(1 + x) = \psi(x) + \frac{1}{x}.$$
Combining both we get the following:
$$\begin{align*}
f(x) &= \psi(1 + 1/(2x)) - \psi(1/2 + 1/(2x)) = \psi(1/(2x)) + 2x - 2\psi(2/(2x)) + \ln(4) + \psi(1/(2x)) \\
&= 2\big\{\psi(1/(2x)) - \psi(1/x)\big\} + 2x + \ln(4)
\end{align*}$$
Now we can plug in the asymptotic formula (6.3.18):
$$\begin{align*}
\psi(1/(2x)) - \psi(1/x) &= -\ln(2x) - x - \sum \limits_{k = 1}^n \frac{B_{2k}}{2k} 2^{2k} x^{2k} + \ln(x) + x/2 + \sum \limits_{k = 1}^n \frac{B_{2k}}{2k} x^{2k} + O(x^{2n + 1}) \\
&= -\ln(2) - x/2 - \sum \limits_{k = 1}^n\frac{B_{2k}}{2k}(4^k - 1) x^{2k} + O(x^{2n + 1})
\end{align*}$$
All together:
$$f(x) = x - \sum \limits_{k = 1}^n \frac{B_{2k}}{k} (4^k - 1)x^{2k} + O(x^{2k + 1}).$$
Edit: Since $|B_{2n}| \sim 4 \sqrt{\pi n} \left(\frac{n}{\pi e}\right)^{2n}$ the corresponding series has a convergence radius of $0$.
|
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|
factoring $(x^6 - y^6)$: what is going on here? I apologize if this question already exists, but it was quite difficult to word.
In my math class today, we learned how to factor a difference of two perfect cubes. One of our practice questions was:
factor $x^6 - 64$
almost everyone other than me (including my teacher, as he was rushing) ended up with:
$(x^2 - 4)(x^4 + 4x^2 + 16)$
I pointed out two things:
1:
$(x^6 - 64)$ was also a difference of two perfect squares, and we could factor it into this:
$(x^3 + 8)(x^3 - 8)$
Which could then be factored further into:
$(x + 2)(x^2 - 2x + 4)(x - 2)(x^2 + 2x + 4)$
and 2:
the $(x^2 - 4)$ was also a difference of two perfect squares, and we could factor further:
$(x + 2)(x - 2)(x^4 + 4x^2 + 16)$
Now for my question:
How are $(x + 2)(x^2 - 2x + 4)(x - 2)(x^2 + 2x + 4)$ and $(x + 2)(x - 2)(x^4 + 4x^2 + 16)$ equivalent?
The teacher didn't seem to know off the top of his head, and I can't figure it out after trying for a half an hour. I've heard that you can factor the sum of two perfect squares with imaginary numbers, so maybe i can do something there to help explain this?
edit: fixed instances of "- 16" to "+ 16", and "4x" to "4x^2" (thanks for pointing that out)
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"How are $(x+2)(x^2−2x+4)(x−2)(x^2+2x+4)$ and $(x+2)(x−2)(x^4+4x^2+16)$ equivalent?"
Because $(x^2−2x+4)(x^2+2x+4)=x^4+4x^2+16$
===
Nother way of looking at it.
$a^3-b^3=(a-b)(a^2+ab+b^2) $
$a^2-b^2=(a-b)(a+b) $
So therefore $a^6-b^6=(a^2-b^2)(a^4+a^2b^2+b^4) =(a-b)(a+b) (a^4+a^2b^2+b^4)$
Yet, $a^6-b^6=(a^3-b^3)(a^3+b^3)=(a-b)(a^2+ab+b^2)(a^3+b^3) $
So it must be that $ (a+b) (a^4+a^2b^2+b^4)= (a^2+ab+b^2)(a^3+b^3) $ somehow.
So $(a+b)|(a^2+ab+b^2) $ or $(a+b)|(a^3+b^3) $.
$a^2+ab+b^2=a (a+b)+b^2$ which doesn't seem to work.
$a^3+b^3=a^2 (a+b)-a^2b+b^3=a^2(a+b)-ab (a+b)+ab^2+b^3=a^2 (a+b)-ab (a+b)+b^2 (a+b)-b^3+b^3=(a+b)(a^2-ab+b^2) $
So now we must have $ (a+b) (a^4+a^2b^2+b^4)= (a+b)(a^2+ab+b^2)(a^2-ab+b^2)$ and so it must be that:
$a^4+a^2b^2+b^4=(a^2+ab+b^2)(a^2-ab+b^2)$
And if we try to multiply it out, we see that it is so.
I don't blame the students for not seeing this, but the teacher ought to have recognize that it had to happen.
|
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|
Estimates for solution of Laplace equation The function
$\Phi(x):= \left\{\begin{array}{lr}
- \frac{1}{2\pi} log(|x|), & \text{for } n=2\\
\frac{1}{n*(n-2)\alpha(n)} \frac{1}{|x|^{n-2}}, & \text{for } n \ge 3
\end{array}\right\}
$ defines for $x \in \mathbb{R}^n, x \not=0$ is the fundamental solution of Laplace´s equation.
How i can observe that we hve the following esimates:$$|D\Phi(x)| \le \frac{C}{|x|^{n-1}},\quad\quad |D^2 \Phi(x)| \le \frac{C}{|x|^n} (x \not=0)$$
for some constant C>0.
I already calculate: $(D \Phi(x))_i= \left\{\begin{array}{lr}
- \frac{1}{2\pi} \frac{1}{|x|}\frac{x_i}{|x|}, & \text{for } n=2\\
\frac{1}{n*(n-2)\alpha(n)} \frac{|x|^{-n+3}}{-n+3}\frac{x_i}{|x|}, & \text{for } n \ge 3
\end{array}\right\} $ because $ \frac{\partial |x|}{\partial x_i}= \frac{x_i}{|x|}$ for $x \not=0$.
For n=2: $|D \Phi(x)|= \frac{1}{2\pi} \frac{|x|}{|x|^2}= \frac{1}{2\pi} \frac{1}{|x|} \le \frac{C}{|x|^{2-1}}$
But for n=3: $|D \Phi(x)|= \frac{1}{n(n-2) \alpha(n)*(3-n)} \frac{|x|}{|x|^{n-2}}=\frac{1}{n(n-2) \alpha(n)*(3-n)} \frac{1}{|x|^{n-3}}$
|
You did the calculation wrong for $n\geq 3$. We have
$$\nabla |x|^{2-n} = (2-n)|x|^{2-n-1}\frac{x}{|x|} = (2-n)\frac{x}{|x|^{n}}$$
|
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|
Is it true that $\frac1n<\sum\limits_{j=n}^{m}\frac{1}{j^2}<\frac1n+\frac1{m^2}$ for $n\geq2$ and $m=\lceil 2n^2-\frac{2}{3}n \rceil$?
Let $n\geq 2$ and $m=\lceil 2n^2-\frac{2}{3}n \rceil$. Is it true
that
$$
\sum_{j=n}^{m}\frac{1}{j^2} > \frac{1}{n} > \sum_{j=n}^{m-1}\frac{1}{j^2}\ ?
$$
I have checked this for $n\leq 40$.
My thoughts : The classical comparison with an integral
yields $$\sum\limits_{j=n}^{m}\frac{1}{j^2}
\geq \frac{1}{n} -\frac{1}{m+1}$$ which is close but not enough.
|
Using the Euler-Maclaurin Sum Formula, we get that
$$
\begin{align}
f(n)
&=\sum_{k=n}^\infty\frac1{k^2}\\
&=\frac1n+\frac1{2n^2}+\frac1{6n^3}-\frac1{30n^5}+\frac1{42n^7}-\frac1{30n^9}+O\left(\frac1{n^{11}}\right)
\end{align}
$$
Then
$$
\begin{align}
\sum_{k=n}^{\left\lceil2n^2-\frac23n\right\rceil}\frac1{k^2}
&=f(n)-f\!\left(\left\lceil2n^2-\tfrac23n\right\rceil+1\right)\\
&\ge f(n)-f\!\left(2n^2-\tfrac23n+1\right)\\[6pt]
&=\frac1n+\frac5{72n^4}+\frac{17}{540n^5}+O\!\left(\frac1{n^6}\right)
\end{align}
$$
and
$$
\begin{align}
\sum_{k=n}^{\left\lceil2n^2-\frac23n\right\rceil-1}\frac1{k^2}
&=f(n)-f\left(\left\lceil2n^2-\tfrac23n\right\rceil\right)\\
&\le f(n)-f\left(2n^2-\tfrac23n+\tfrac23\right)\\[6pt]
&=\frac1n-\frac1{72n^4}-\frac{13}{540n^5}+O\!\left(\frac1{n^6}\right)
\end{align}
$$
|
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|
Finding the equation of an ellipse using eccentricity and directrix with focus at (0,0) The ellipse $\varepsilon$ has eccentricty $\frac{1}{2}$, focus $(0,0)$ and the line $x=-1$ as the corresponding directrix. Find the equation of $\varepsilon$. Find the other focus and directrix of $\varepsilon$.
I'm confused by this due to fact the focus is at $(0,0)$. As far as I was aware the center of the ellipse should be at $(0,0)$ so that the foci are at $(c,0)$ and $(-c,0)$, where $c^2=a^2-b^2$. The directrix corresponding to the focus will then be given by the equation $x=\frac{a^2}{c}$, and the eccentricity of the ellipse is $\frac{c}{a}$. Is there another set of equations I can use to determine the equation of the ellipse?
|
From your information:
\begin{align*}
\frac{\sqrt{x^2+y^2}}{x+1} &= \varepsilon \\
x^2+y^2 &= \varepsilon^2(x+1)^2 \\
(1-\varepsilon^2)x^2-2\varepsilon^2 x+y^2 &= \varepsilon^2 \\
\left[
(1-\varepsilon^2)x^2-2\varepsilon^2 x+
\frac{\varepsilon^4}{1-\varepsilon^2}
\right]+
y^2 &= \varepsilon^2+\frac{\varepsilon^4}{1-\varepsilon^2} \\
\frac{(1-\varepsilon^2)^2}{\varepsilon^2}
\left(
x-\frac{\varepsilon^2}{1-\varepsilon^2}
\right)^2+\frac{1-\varepsilon^2}{\varepsilon^2} y^2 &= 1 \\
a &= \frac{\varepsilon}{1-\varepsilon^2} \\
b &= \frac{\varepsilon}{\sqrt{1-\varepsilon^2}} \\
c &= \frac{\varepsilon^2}{1-\varepsilon^2}
\end{align*}
The other focus is
$$(2c,0)=
\left( \frac{2\varepsilon^2}{1-\varepsilon^2},0 \right)$$
The other directrix is
$$x=2c+1$$
$$x=\frac{1+\varepsilon^2}{1-\varepsilon^2}$$
|
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|
Gauss-Jordan Elimination Trying to solve this using Gauss-Jordan Elimination.
$x_1 + x_2 - x_3 = -4$
$-x_1 - 2x_2 + x_3 = 3+t$
$2x_1 + x_2 + (s-3)x_3 = st-9$
$2x_1 + (s-3)x_3 = st+2t-10$
I came across:
$t=0$
And
$s\neq 1$
Need to find the values of s and t that the equations will have: no solution, 1 solution, unlimited solutions.
$\left[\begin{array}{ccc|c}1&1&-1&-4\\-1&-2&1&3+t\\2&1&s-3&st-9\\2&0&s-3&st+2t-10\end{array}\right]$
$\left[\begin{array}{ccc|c}1&1&-1&-4\\0&1&0&1-t\\0&0&s-1&t(s-1)\\0&1&0&1-2t\end{array}\right]$
$\left[\begin{array}{ccc|c}1&1&-1&-4\\0&1&0&1-t\\0&0&1&t\\0&0&0&-t\end{array}\right]$
$\left[\begin{array}{ccc|c}1&0&0&2t-5\\0&1&0&1-t\\0&0&1&t\\0&0&0&-t\end{array}\right]$
|
From your final system, $0=-t$, hence if $t \neq 0$, there is no solution.
If $t=0$, we can drop the very last equation as it is just $0=0$.
The $3 \times 3 $ matrix on the LHS is non-singular, hence the system has a unique solution.
|
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|
Let $a,b,c$ be the roots of $x^3 - x - 1= 0$ find $a^5 + b^5 + c^5$ Consider a cubic polynomial $x^3 - x - 1 = 0$ I want the sum of the fifth powers of the roots $\sum a^5$. I know that
\begin{eqnarray*}
a + b + c &=& 0 \\
ab + bc + ca &=& 1 \\
-abc &=& 1
\end{eqnarray*}
but I have no way of combining this information into the answer.
|
$a^3=a+1$ implies that $a^5=a^3+a^2=a+1+a^2$.
$a^5+b^5+c^5=3+a+b+c+a^2+b^2+c^2=3+(a+b+c)^2-2(ab+ac+bc)=5$.
|
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|
Prove that $\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{8^{n}}= \sqrt{2}$ using $f(x) = \frac{1}{\sqrt{1-4x}}$ Prove that $\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{8^{n}}= \sqrt{2}$ using $f(x) = \frac{1}{\sqrt{1-4x}}$
For the past 2 days I've tried to prove this but with no results.
I've done some research but with no results.
Any ideas how can I finally solve this?
|
Us the extended binomial theorem to show that:
$$(1-x)^{-1/2} = 1 +\frac{1}{1!}\frac{1}{2}x+\frac{1}{2!}\frac{1}{2}\cdot\frac{3}{2}x^2+\frac{1}{3!}\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}x^3\dots$$
And then prove (by induction or just directly) that:
$$\frac{1}{n!}\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}...\frac{2n-1}{2} = \frac{1}{4^n}\binom{2n}{n}$$
This means that $$\sum_{n=0}^{\infty} \frac{x^n}{4^n}\binom{2n}{n} = \frac{1}{\sqrt{1-x}}$$
Then substitute $4x$ for $x$ and you get a power series for $\frac{1}{\sqrt{1-4x}}$ and then use $x=1/8$.
|
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|
Cubics with one real root
$(1) \quad f(x) = x^3- 3ax + b$
$(4) \quad f(x) = x^3 - 3hkx - (h^3 + k^3)$
By comparing the coefficients in equations $(1)$ and $(4)$, obtain two equations that relate $h$ and $k$ to $a$ and $b$.
One of these is solved to give $k$ in terms of $a$ and $h$.
Substitute this into the other equation and set $t = h^3$ to deduce a quadratic equation satisfied by $t$.
Choose $h^3$ to be one root of this quadratic and then show that the other root must be $k^3$.
By summing the cube roots of the roots of the quadratic one finds $x = h + k$.
If equation $(1)$ has exactly one real root, show that $h^3$ and $k^3$ are distinct real numbers.
I have worked out the quadratic to be $t^2 + bt + a^3$.
Not sure how to carry this forward.
|
Obviously,
$$a=hk,\\b=-h^3-k^3.$$
Then
$$k=\frac ah,\\b=-h^3-\frac{a^3}{h^3}=-t-\frac{a^3}t$$
or
$$t^2+bt+a^3=0.$$
From the Vieta formulas, the sum of the roots of this quadratic equation is $-b$, and from the definition,
$$t_1+t_2=-b=h^3+k^3.$$ So if $t_1=h^3$, then $t_2=k^3$.
Now $x=h+k$ is a root of the cubic because
$$(h+k)^3-3a(h+k)+b=h^3+3hk(h+k)+k^3-3a(h+k)+b=0$$ from the defintions of $h,k$.
The cubic equation has a single real root when it has no extrema, i.e. the derivative doesn't cancel:
$$3x^2-3a$$ is never zero for $a<0$, so that $h^2=k^2=a$ is not possible.
A personal note.
I find this exposition rather disconcerting. If prefer this one:
To solve $x^3-3ax+b=0$, decompose in two terms, $x=h+k$. Then substituting, the equation becomes
$$h^3+3hk(h+k)+k^3-3a(h+k)+b=0.$$
If we set $a=hk$, only
$$h^3+k^3+b=0$$ remains.
Now in the form
$$h^3+k^3=-b,\\h^3k^3=a^3$$
we get a sum/product problem which is easily solved by means of a bicubic equation:
$$h^6+h^3k^3=-bh^3,\\
(h^3)^2+b(h^3)+a^3=0.$$
|
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|
Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 - \frac{1}{8} x^4 + O \left( x^5 \right) $$
I notice powers of two in the denominator, but I'm not sure of the pattern (and to calculate the next term to confirm would entail another tedious product rule).
Any ideas? Thanks!
|
The pattern is
$$
\frac{1}{x^2+2x+2}=\frac12+\sum_{k\in \Bbb{Z}^+ | k-3\not\in \Bbb{Z}} (-1)^\left\lfloor \frac34 k\right\rfloor2^{-\left\lfloor \frac{k}2+1\right\rfloor}
$$
where $\lfloor s \rfloor$ means the greatest integer not exceeding $s$, and the sum on $k$ skips numbers of the form $4n+3$.
Thus
$$
\frac12 - \frac{x}2 +\frac{x^2}{4} - \frac{x^4}{8} + \frac{x^5}{8} - \frac{x^6}{16} + \frac{x^8}{32} -\frac{x^9}{32} + \frac{x^{10}}{64} \cdots
$$
|
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|
Arithmetic proof of absolute value function of complex numbers I am looking for the arithmetic proof that:
$ |z| = \sqrt{(x^2 + y^2)} $ where $ z = x + i y $
Previously I assumed squaring a function then square rooting it would be analogous to the absolute value function (modulus) but it seems not to be the case in the complex domain. Consider the following simple counter example:
Let $ z = \cos(x) + i \times \sin (x) $
$|z| = |\cos(x) + i \times \sin(x)| = \sqrt{((\cos(x) + i \times \sin(x))^2} $
$ = \sqrt{2 \times i \times \cos(x) \times \sin(x) + \cos^2(x) +i^2 \times \sin^2(x) }$
$ = \sqrt{2i \times \cos(x)\sin(x) + \cos^2(x) - \sin^2(x) } \neq \cos^2(x) + \sin^2(x)$
Why is that?
|
First,
the absolute value
of a complex number
is a real number.
The standard definition is
$|z|^2
=z\, \bar{z}
$.
From this,
if $z = x+iy$,
$|z|^2
=z\, \bar{z}
=(x+iy)(x-iy)
=x^2+y^2
$.
|
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|
What will be the 49th derivative? Let be the function $f=(x^3+3x) \cdot sin(x)$
$f'(x)=\left(3x^2+3\right)\sin\left(x\right)+\left(x^3+3x\right)\cos\left(x\right)$
$f''(x)=\left(3x-x^3\right)\sin\left(x\right)+\left(6x^2+6\right)\cos\left(x\right)$
$f'''(x)=\left(-9x^2-3\right)\sin\left(x\right)+\left(15x-x^3\right)\cos\left(x\right)$
$f''''(x)=\left(x^3-33x\right)\sin\left(x\right)+\left(12-12x^2\right)\cos\left(x\right)$
I still can't the exact pattern in the results, I would appreciate any help.
|
Going off A.Riesen's comment, the General Leibniz Rule states that
$$(uv)^{(n)} = \sum_{k=0}^n {n \choose k} u^{(k)} v^{(n-k)} %seriously, did nobody notice that I had written down the formula incorrectly the first time?$$
where $u,v$ are functions of $x$ and $f^{(n)}$ is the $n^{th}$ derivative of $f$ with respect to $x$. Plugging in $u = x^3 + 3x$, $v = \sin x$ and $n = 49$, we get
$$((x^3+3x)(\sin x))^{(49)} = \sum_{k=0}^{49} {n \choose k} (x^3+3x)^{(k)}(\sin x)^{(49-k)}$$
Fortunately, since $(x^3+3x)^{(n)} = 0$ for all $n \ge 4$, we can reduce this to
$$\sum_{k=0}^{3} {n \choose k} (x^3+3x)^{(k)}(\sin x)^{(49-k)}$$
$$= {49 \choose 0} (x^3+3x)^{(0)} (\sin x)^{(49)} + {49 \choose 1} (x^3+3x)^{(1)} (\sin x)^{(48)} + {49 \choose 2} (x^3+3x)^{(2)} (\sin x)^{(47)} + {49 \choose 3} (x^3+3x)^{(3)} (\sin x)^{(46)}$$
Using the derivatives of sines and cosines, this comes down to
$$= {49 \choose 0} (x^3+3x) \cos x + {49 \choose 1} (3x^2+3) \sin x + {49 \choose 2} (6x) (-\cos x) + {49 \choose 3} 6 (-\sin x)$$
I'll leave the number crunching and simplification for you.
|
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|
Find $\int_{|z|=3} {1 \over P(z)}dz$ where $P(z)=2z^5-16z^4-17z^3-11z^2+20z-18$
Evaluate
$$\int_{|z|=3} {1 \over P(z)}dz$$
where $P(z)=2z^5-16z^4-17z^3-11z^2+20z-18$.
I proved that $P(z)\neq 0$ for all $z$ outside of the ball of radius 3, except for z=9, so all the poles of ${1 \over P(z)}$ except $z=9$ are inside the ball.
$$P(z)=(z-9)(2z^4+2z^3+z^2-2z+2)=(z-9)(z-a_1)(z-\bar a_1)(z-a_2)(z-\bar a_2).$$
|
Let
$$
P(z)=2z^5-16z^4-17z^3-11z^2+20z-18
$$
Since
$$
\begin{align}
&(-218840155-162129667x-67092610x^2+13150850x^3)P(z)\\
&-(3927717-68602966x-46831817x^2-17626794x^3+2630170x^4)P'(z)\\
&=3860568450
\end{align}
$$
$\gcd(P,P')=1$. Therefore, $P$ has no repeated roots.
$P(9)=0$ and $\frac{P(x)}{x-9}=2x^4+2x^3+x^2-2x+2$ implies that the remaining roots have absolute value no greater than $\frac{2\cdot\frac12+1\cdot\frac14+2\cdot\frac18+2\cdot\frac1{16}}{2}\cdot2=\frac{13}8$. Therefore, all the roots are simple and contained in $|z|\lt3$ except for the root at $z=9$. The sum of the residues of $\frac1{P(z)}$ is $0$ so the sum of the residues inside $|z|=3$ is the negative of the residue at $z=9$, which is
$$
-\frac1{P'(9)}=-\frac1{14645}
$$
Therefore,
$$
\int_{|z|=3}\frac{\mathrm{d}z}{2z^5-16z^4-17z^3-11z^2+20z-18}=-\frac{2\pi i}{14645}
$$
|
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|
Given $a^2+b^2=2$ prove $a+b\le2$
*
*Given $a^2+b^2=2$ prove $a+b\le2$
*Given $a+b=2$ prove $a^4+b^4\ge2$
I was trying to prove these using the fact that we know $a^2+b^2\ge2ab$ but not sure where to start.
|
Using AM-GM inequality:
$\displaystyle\frac{a^2+b^2}{2} \geq \sqrt {a^2b^2}$
$\displaystyle{a^2+b^2}\geq {2ab}$
We know :
$\displaystyle\ a^2+b^2=2$
Equating the above equations, we get
$\displaystyle\ 2ab \leq 2$
Adding two equations above we get:
$\displaystyle\ a^2+b^2+2ab \leq 4$
Taking Square Roots both the sides:
$\displaystyle\ -2\leq a+b\leq 2$
Therefore,
$\displaystyle\ a+b\leq 2$
$\displaystyle\frac{a^4+b^4}{2}\geq \sqrt {a^4b^4}$
$\displaystyle{a^4+b^4}\geq 2a^2b^2$
$\displaystyle{a^4+b^4}\geq 2(ab)^2$
We can also write it as:
$\displaystyle{a^4+b^4}\geq 2$
Because
$\ 2\leq 2(ab)^2$
|
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|
Suppose we have two positive integers $a$ and $b$ which satisfy the condition $a^3 − 2b^3 = 2$. Suppose we have two positive integers $a$ and $b$ which satisfy the condition
$a^3 − 2b^3 = 2$.
It then follows that the greatest common divisor of $a$ and $b$ must be either $1$ or $2$. True or false?
I tried solving by supposing $a$ and $b$ are composite numbers sharing $\beta$ as a common factor.
Letting $a = \alpha \beta$ and $b = \gamma \beta$, I factorised $\beta^3 (\alpha^3 + 2 \gamma^3) = 2$, $$\beta^3 = \frac{2}{\alpha^3 + 2 \gamma^3},$$ which is lesser than a whole number therefore both $a$ and $b$ must be coprime to each other, i.e $\gcd(a, b) = 1$ only.
|
Let be $d=gcd(a,b)$ so $a=d.a_1$ and $b=d.b_1$, with $gcd(a_1,b_1)=1$ and backing to the equation we have:
$$a^3-2b^3=2 \Rightarrow d^3(a_1^3-2b_1^3)=2 \Rightarrow d^3|2 \Rightarrow d=1.$$
|
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|
Convergence of $\sum_{n=2}^{\infty}{1 \over \sqrt{n}}\ln{n+1 \over n-1}$ I am checking for convergence of
$$\sum_{n=2}^{\infty}{1 \over \sqrt{n}}\ln{n+1 \over n-1}$$
First, we notice
$${n+1 \over n-1}= {1+{2 \over n-1}}$$
Then, we use $\ln(1+x) \le x$ and get
$$\ln\left({1 + {2 \over n-1}}\right) \le {2 \over n-1}$$
$${1 \over \sqrt{n}}\ln{n+1 \over n-1} \le {1 \over \sqrt{n}}{2 \over n-1}$$
We now use $n-1 > {n \over 2}$
$${1 \over \sqrt{n}}\ln{n+1 \over n-1} \le {1 \over \sqrt{n}}{2 \over n-1} < {1 \over \sqrt{n}}{2 \over {n \over 2}} = 4{1 \over n\sqrt{n}}$$
$\sum_{n=2}^{\infty}4{1 \over n\sqrt{n}}$ converges and so does $\sum_{n=2}^{\infty}{1 \over \sqrt{n}}\ln{n+1 \over n-1}$. Is this correct? Alternatively, is there a quicker way?
|
If we use the well-known equivalence,
$$\ln(1+X)\sim X\;\;(X\to 0),$$
we get that
$$\ln(1+\frac{2}{n-1})\sim\frac{2}{n-1}\sim\frac{2}{n}\;\;(n\to +\infty)$$
$$\implies \frac{1}{\sqrt{n}}\ln(\frac{n+1}{n-1})\sim\frac{2}{n^\frac{3}{2}}\;(n\to+\infty)$$
the general terms are positive and equivalent, the series are both convergent since $\frac{3}{2}>1$.
|
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|
Error computing determinant of a $4\times 4$ matrix I am trying to compute the determinant of
$$C = \begin{bmatrix}-2 & 3 & 2 & 1 \\ 0 & 3 & 1 & -3 \\ -2 & -2 & -1 & 2 \\ 0 & 2 & 3 & 1 \end{bmatrix}$$
I first did the row operation $R_3 \leftarrow R_1-R_3$ so it doesn't change the determinant. So you want the determinant of:
$$ \begin{bmatrix}-2 & 3 & 2 & 1 \\ 0 & 3 & 1 & -3 \\ 0 & 5 & 3 & -1 \\ 0 & 2 & 3 & 1 \end{bmatrix}$$
and this simplifies to the $-2$ times the determinant of:
$$C = \begin{bmatrix} 3 & 1 & -3 \\ 5 & 3 & -1 \\ 2 & 3 & 1 \end{bmatrix}$$
And this $3\times 3$ matrix has determinant of $-16$ so the determinant of $C$ is $(-2)\cdot (-16) = 32$. However, the book says that the answer is $-32$ and not $32$.
Any hints where the negative comes from?
|
Recall that the determinant does not change if you add to a row a linear combination of the others.
You should replace $R_3$ with $R_3-R_1$. Then you obtain
$$\begin{bmatrix}-2 & 3 & 2 & 1 \\ 0 & 3 & 1 & -3 \\ 0 & -5 & -3 & 1 \\ 0 & 2 & 3 & 1 \end{bmatrix}$$
and the determinant of
$$\begin{bmatrix} 3 & 1 & -3 \\ -5 & -3 & 1 \\ 2 & 3 & 1 \end{bmatrix}$$
is just the opposite of $-16$ (your previous computation).
|
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|
Evaluate the sum $\sum ^{\infty}_{n=1} \frac{1}{n^2+a^4}$ Evaluate the sum $\sum ^{\infty}_{n=1} \frac{1}{n^2+a^4}$
I don't to know how to processed this problem. Can any one help with problem please. thanks
|
Complex analysis gives a very clear explanation of the identity
$$ \sum_{n=1}^{\infty} \frac{1}{n^2+z^2} = \frac{1}{z}\bigg( \frac{\pi}{2}\coth(\pi z) - \frac{1}{2z} \bigg) $$
where $z = a^2$. This equality follows from the observation that both sides have exactly the same poles and decay to $0$ as $\Re(z) \to \pm \infty$.
If you want to avoid complex-analysis method, there are some alternatives. Let me briefly discuss the idea. Let $b = a^2$ and assume that $b > 0$. Using the following Fourier transform,
$$ \int_{-\infty}^{\infty} \cos(nx) e^{-b|x|} \, dx = \frac{2b}{n^2 + b^2}, $$
we can perform the following heuristic computation:
$$\sum_{n=1}^{\infty} \frac{1}{n^2 + b^2}
= \frac{1}{2b} \sum_{n=1}^{\infty} \int_{-\infty}^{\infty} \cos(nx)e^{-b|x|} \, dx
\color{red}{\stackrel{!?}{=}} \frac{1}{2b} \int_{-\infty}^{\infty} \bigg( \sum_{n=1}^{\infty} \cos(nx) \bigg) e^{-b|x|} \, dx. $$
Now using the (distributional) Poisson summation formula
$$ 2\pi \sum_{n=-\infty}^{\infty} \delta(x - 2\pi n)
= \sum_{n=-\infty}^{\infty} e^{inx}
= 1 + 2 \sum_{n=1}^{\infty} \cos(nx), $$
we find that
\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{n^2 + b^2}
&\color{red}{\stackrel{!?}{=}} \frac{1}{2b} \int_{-\infty}^{\infty} \bigg( \pi \sum_{n=-\infty}^{\infty} \delta(x - 2\pi n) - \frac{1}{2} \bigg) e^{-b|x|} \, dx \\
&\color{red}{\stackrel{!?}{=}} \frac{1}{2b} \bigg( \pi \sum_{n=-\infty}^{\infty} e^{-2\pi b |n|} - \frac{1}{b} \bigg) \\
&= \frac{1}{2b}\bigg( \pi\coth(\pi b) - \frac{1}{b} \bigg).
\end{align*}
Up to this point, all the computations are only heuristic and desperately require some rigorous justification. Fortunately not much extra work is required. All this nonsense can be salvaged by adopting an appropriate regularization.
Indeed, using the dominated convergence theorem, we may safely write
\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{n^2 + b^2}
&= \lim_{r\uparrow 1} \sum_{n=1}^{\infty} \frac{r^n}{n^2 + b^2} \\
&= \frac{1}{2b} \lim_{r\uparrow 1} \sum_{n=1}^{\infty} r^n \int_{-\infty}^{\infty} \cos(nx)e^{-b|x|} \, dx \\
&= \frac{1}{2b} \lim_{r\uparrow 1} \int_{-\infty}^{\infty} \bigg( \sum_{n=1}^{\infty} r^n \cos(nx) \bigg) e^{-b|x|} \, dx
\end{align*}
Then from the formula of the 2-dimensional Poisson kernel
$$ \frac{1 - r^2}{1 - 2r\cos x + r^2} = 1 + 2 \sum_{n=1}^{\infty} r^n \cos(nx), \qquad |r| < 1$$
we have
\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{n^2 + b^2}
&= \frac{1}{2b} \lim_{r\uparrow 1} \int_{-\infty}^{\infty} \bigg( \frac{1}{2} \frac{1 - r^2}{1 - 2r\cos x + r^2} - \frac{1}{2} \bigg) e^{-b|x|} \, dx \\
&= \frac{1}{2b} \bigg( \frac{1}{2} \lim_{r\uparrow 1} \int_{-\pi}^{\pi} \frac{1 - r^2}{1 - 2r\cos x + r^2} \bigg( \sum_{n=-\infty}^{\infty} e^{-b|x+2\pi n|} \bigg) \, dx - \frac{1}{b} \bigg).
\end{align*}
Notice that $x \mapsto \sum_{n=-\infty}^{\infty} e^{-b|x+2\pi n|}$ defines a continuous function. Thus by appealing to the approximation-to-the-identity property of the Poisson kernel, we obtain the same answer as in the heuristics.
|
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|
Show a sequence is increasing
How can you show that :
$$D_n=\sum_{k=1}^{n } \frac{1}{k}-\int_{1}^{n+1} \frac{1}{x} \ dx $$ is increasing and bounded (and hence convergent). I'm having trouble.
|
Assuming you meant
$$D_n=\sum_{k=1}^n \frac{1}{k}-\int_1^{n+1}\frac{1}{x}\ dx$$
we have
\begin{align*}
D_{n+1}-D_n &= \left(\sum_{k=1}^{n+1} \frac{1}{k}-\int_1^{n+2}\frac{1}{x}\ dx\right)-\left(\sum_{k=1}^n \frac{1}{k}-\int_1^{n+1}\frac{1}{x}\ dx\right)\\
&= \frac{1}{n+1}-\int_1^{n+2}\frac{1}{x} \ dx+\int_1^{n+1}\frac{1}{x}\ dx\\
&=\frac{1}{n+1}-\int_{n+1}^{n+2} \frac{1}{x} \ dx.
\end{align*}
Now on $[n+1,n+2]$ we have $x \geq n+1$ so $\frac{1}{x} \leq \frac{1}{n+1}$. Hence $\int_{n+1}^{n+2} \frac{1}{x} \ dx \leq \int_{n+1}^{n+2} \frac{1}{n+1} \ dx = \frac{1}{n+1}$. In fact the inequality is strict because we only have $\frac{1}{x}=\frac{1}{n+1}$ when $x=n+1$. This shows $D_{n+1}-D_n>0$ so $D_n$ is increasing.
To see that $D_n$ is bounded, note that $\sum_{k=1}^n \frac{1}{k}$ is a left Riemann sum for $\int_1^{n+1} \frac{1}{x} \ dx$. Hence $D_n$ is the difference between an integral and left Riemann sum, so $D_n$ is bounded by $|R_n-L_n|$ where $R_n$ is a right Riemann sum and $L_n$ is a left Riemann sum for $\int_1^{n+1} \frac{1}{x} \ dx$. One particular choice of left and right Riemann sums gives
$$D_n \leq \left|\sum_{k=2}^{n+1} \frac{1}{k} - \sum_{k=1}^{n}\frac{1}{k}\right|=\left|\frac{1}{n+1}-1\right|\leq \frac{1}{n+1}+1 \leq 2.$$
|
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|
How to show $\left(1+\frac{1}{3}x+\frac{1}{24}x^2+\frac{1}{300}x^3+...\right)^2 = 1 +\frac{2}{3}x+\frac{7}{36}x^2+\frac{1}{30}x^3+...$? I tried to find the right handside of the equation by manipulating the series but I failed at getting the right handside of it.
$$\left(1+\frac{1}{3}x+\frac{1}{24}x^2+\frac{1}{300}x^3+...\right)^2 = 1 +\frac{2}{3}x+\frac{7}{36}x^2+\frac{1}{30}x^3+...$$
Closed form of the left handside in the parantheses would be
$$\sum_{n=0}^{\infty}\frac{2}{n!(n+2)!}x^n$$
Any hint would be appreciated.
|
So it doesn't explicitly use generating functions as I supposed, but it is pretty close.
You have (for ease of visualization):
$$\left(\sum_{n=0}^{\infty}\frac{2}{n!(n+2)!}x^n\right)^2=\left(\sum_{n=0}^{\infty}\frac{2}{n!(n+2)!}x^n\right)\left(\sum_{n=0}^{\infty}\frac{2}{n!(n+2)!}x^n\right)=\sum_{n=0}^{\infty}\left[\frac{2}{n!(n+2)!}x^n\left(\sum_{j=0}^{\infty}\frac{2}{j!(j+2)!}x^j\right)\right].$$
I think it is fairly obvious that each power of $x$ will appear, since all coefficients are positive and such. But we will soon see that this is the case anyhow.
To begin writing out the result explicitly, we need to consider each power of $x$. To get the constant term (ie, the $x^0$ term), the only option is $n=j=0$ and so we just have $\frac{2}{0!(0+2)!}\cdot\frac{2}{0!(0+2)!}=1$, so our first term in the result is $1$.
Next, we need the $x^1$ term, and there are two ways to do this. We get $x^1$ when $n=0, j=1$ and when $n=1, j=0$. So adding the two results gives $\frac{2}{0!(0+2)!}\cdot\frac{2}{1!(1+2)!}+\frac{2}{1!(1+2)!}\cdot\frac{2}{0!(0+2)!}=\frac{2}{3},$ so the next term in our result is $\frac{2}{3}x$.
I'll do one more. For the $x^2$ term, we have three options: $n=j=1$, or $n=2, j=0$, or $n=0, j=2$. I hope by now it is clear that if $j\neq n$, we can just double the result, so our coefficient is: $\frac{2}{1!(1+2)!}\cdot\frac{2}{1!(1+2)!}+(2)\cdot\frac{2}{2!(2+2)!}=\frac{3}{37}$.
So, to find the coefficient for a general $x^k$, you need to find all the ways $n+j=k$ in non-negative integers $n,j$, and then add all such results up. It gets worse as you go. There are ${k+1 \choose k-1}$ ways to do this for a given $k>0$. If $k=10$, we have $55$ ways to consider!
But now you can give the result clearly to whatever accuracy you'd like. And with a little effort, you could find a closed form for the result, too.
|
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|
Prove that $\sum_{n=1}^{+\infty}\frac{(-1)^n n^3}{4^n}=-0.0064$ (Motivation) As homework, we have been asked to prove that the following series converges: $$\sum_{n=1}^{+\infty}\frac{(-1)^n n^3}{4^n}$$
I did it in two ways:
*
*Using the alternating series test (Leibniz criterion), proving that $\frac{n^3}{4^n}$ is decreasing and also $\lim_\limits{n\to +\infty}{\frac{n^3}{4^n}}=0$.
*Using the n-th root test (Cauchy's criterion) and absolute convergence, proving that $\lim_\limits{n\to +\infty}{\frac{(\sqrt[n]n)^3}{4}}=\frac{1}{4}<1$.
However, Wolfram Alpha states another interesting result: That this series sums exactly to $-0.0064$. I would like to see how that result is obtained, so a proof for it.
|
You can get this by clever sum manipulations.
Start from
$$
\sum \frac1{(-4)^n}=-\frac14+\frac1{16}-\frac1{64} + \cdots = -\frac15
$$
Then look at
$$
\sum \frac{n}{(-4)^n}=\begin{array}{lllll}
-\frac14&+\frac1{16}&-\frac1{64} &+ \cdots \\
&+\frac1{16}&-\frac1{64}&- \cdots \\
&&-\frac1{64}&+ \cdots
\end{array}\\
=\sum \frac1{(-4)^n} - \frac14\sum \frac1{(-4)^n} + \frac1{16}\sum \frac1{(-4)^n} \cdots = -\frac45\cdot \frac{1}{5} = -\frac{4}{25}
$$
Next do the same sort of vertical breakup, but with $\sum \frac{n}{(-4)^n}$ in each row:
$$
\begin{array}{lllll}
-\frac14&+\frac2{16}&-\frac3{64} &+\frac4{256} &-\cdots \\
&+\frac1{16}&-\frac2{64}&+\frac3{256} &-\cdots \\
&&-\frac1{64}&+\frac2{256} &- \cdots \\
&&&+\frac1{256} &-\cdots
\end{array} = \sum\frac{n(n+1)}{2}\frac1{(-4)^n} \\
=\sum \frac{n}{(-4)^n} - \frac14\sum \frac{n}{(-4)^n} + \frac1{16}\sum \frac{n}{(-4)^n} \cdots = -\frac45\cdot \frac{4}{25} = -\frac{16}{125}
$$
And the next step of the same trick, using rows of $\sum\frac{n(n+1)}{2}\frac1{(-4)^n}$, will give
$$\sum\frac{n(n+1)(n+2)}{6}\frac1{(-4)^n} = -\frac45\cdot\frac{16}{125}=-\frac{64}{625}
$$
Finally, write
$$
n^3 = 6 \frac{n(n+1)(n+2)}{6} -6 \frac{n(n+1)}{2} + n
$$
and you get the answer
$$
-6\cdot\frac{64}{625} +6\cdot\frac{16}{125} -\frac{4}{25} =-0.0064
$$
|
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|
Need help evaluating $ \int_{0}^{2\pi} \frac{d\theta}{5 + 4\sin{\theta}} $. From Churchill and Brown's Complex Analysis I am trying to conclude:
$$ \int_{0}^{2\pi} \frac{d\theta}{5 + 4\sin{\theta}} = \frac{2\pi}{3}$$
I have tried substituting in:
$$\sin{\theta} = \frac{z + z^{-1}}{2i}$$
$$d\theta = \frac{dz}{zi}$$
To get (note $C$ is $|z| = 1$):
$$ \int_{C} \frac{\frac{dz}{zi}}{5 + 4\frac{z + z^{-1}}{2i}} $$
$$ \int_{C} \frac{dz}{5 - 2i(z + z^{-1})zi} $$
$$ \int_{C} \frac{dz}{5 + 2(z + z^{-1})z} $$
$$ \int_{C} \frac{dz}{5 + 2(z^2 + 1)} $$
$$ \int_{C} \frac{dz}{7 + 2z^2} $$
$$ \frac12 \int_{C} \frac{dz}{z^2 + \frac72}$$
But then I concluded the singularities of $\pm i\frac72$ were outside of the circle so the integral became $0$ instead of $\frac{2\pi}{3}$. Where did I go wrong?
|
Note that
$$\left(5+4\left(\frac{z-z^{-1}}{2i}\right)\right)iz=2z^2+i5z-2$$
|
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|
Find general sequence equation linear algebra Suppose that the sequence $x_0, x_1, x_2,\dots$ is defined by $x_0 = 7$, $x_1 = 2$, and $x_{k+2} = −x_{k+1}+2x_k$ for $k\geq0$. Find a general formula for $x_k$. Be sure to include parentheses where necessary, e.g. to distinguish $1/(2k)$ from $1/2k$.
I have no idea how to do this question! Someone please help.
|
Just to show an alternative way, through generating function.
Allow me to change notation from $x$ to $a$ to avoid confusion in the following.
Starting from your recurrence
$$
\left\{ \begin{gathered}
a_{\,0} = 7 \hfill \\
a_{\,1} = 2 \hfill \\
a_{\,k + 2} = - a_{\,k + 1} + 2a_{\,k} \hfill \\
\end{gathered} \right.
$$
rewrite it so as to incorporate the initial conditions:
$$
\left\{ \begin{gathered}
a_{\,k < 0} = 0 \hfill \\
a_{\,k} = - a_{\,k - 1} + 2a_{\,k - 2} + 9\left[ {k = 1} \right] + 7\left[ {k = 0} \right] \hfill \\
\end{gathered} \right.
$$
where $[P]$ indicates the Iverson bracket
$$
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
1 & {P = TRUE} \\
0 & {P = FALSE} \\
\end{array} } \right.
$$
Then multiply by $z^k$ and sum up
$$
\begin{gathered}
\sum\limits_{0\, \leqslant \,k} {a_{\,k} z^{\,k} } = - \sum\limits_{0\, \leqslant \,k} {a_{\,k - 1} z^{\,k} } + 2\sum\limits_{0\, \leqslant \,k} {a_{\,k - 2} z^{\,k} } + 9\sum\limits_{0\, \leqslant \,k} {\left[ {k = 1} \right]z^{\,k} } + 7\sum\limits_{0\, \leqslant \,k} {\left[ {k = 0} \right]z^{\,k} } = \hfill \\
= - z\sum\limits_{0\, \leqslant \,k} {a_{\,k - 1} z^{\,k - 1} } + 2z^2 \sum\limits_{0\, \leqslant \,k} {a_{\,k - 2} z^{\,k - 1} } + 9z + 7 \hfill \\
\end{gathered}
$$
So we get:
$$
\sum\limits_{0\, \leqslant \,k} {a_{\,k} z^{\,k} } = F(z) = - zF(z) + 2z^2 F(z) + 9z + 7
$$
$$
\begin{gathered}
F(z) = \frac{{9z + 7}}
{{1 + z - 2z^2 }} = - \frac{{9z + 7}}
{{\left( {2z + 1} \right)\left( {z - 1} \right)}} = \frac{5}
{{3\left( {2z + 1} \right)}} + \frac{{16}}
{{3\left( {1 - z} \right)}} = \hfill \\
= \frac{5}
{3}\sum\limits_{0\, \leqslant \,k} {\left( { - 2} \right)^{\,k} z^{\,k} } + \frac{{16}}
{3}\sum\limits_{0\, \leqslant \,k} {z^{\,k} } \hfill \\
\end{gathered}
$$
$$
a_{\,k} = \frac{{5\left( { - 2} \right)^{\,k} + 16}}
{3}
$$
|
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|
How can I find the $n$'th derivative of $f(x)=x^2 \ln(x+1)$
How can I find the $n$'th derivative of $f(x)=x^2 \ln(x+1)$
I have tried this
$$f(x)=x^2\ln(x+1)$$
$$ f'(x) = 2x\ln(x+1) + \frac{x^2}{x+1}$$
$$ f''(x)=2\ln(x+1) + \frac{4x}{x+1} - \frac{x^2}{(x+1)^2}$$
However I do not see any pattern :(
|
Let $u=x^2$ and $v=\ln (1+x)$ then,
$$u'=2x,u''=2,u'''=0$$
And
$v^{(n)}=\frac{(-1)^{n-1}(n-1)!}{(1+x)^n}$ for $n \geq 1$.
Let's compute $(uv)^{(3)}=\sum_{k=0}^{3} {3 \choose k} u^{(k)}v^{(n-k)}={3 \choose 0} x^2\frac{(-1)^{n-1}(n-1)!}{(1+x)^n}+{3 \choose 1}2x\frac{(-1)^{n-2}(n-2)!}{(1+x)^{n-1}}+{3 \choose 2}2\frac{(-1)^{n-3}(n-3)!}{(1+x)^{n-2}}+{3 \choose 3}0$
With $n=3$ by the general product rule
Now it's easy to see that for $n \geq 3$ we have
$$(uv)^{(n)}={n \choose 0} x^2\frac{(-1)^{n-1}(n-1)!}{(1+x)^n}+{n \choose 1}2x\frac{(-1)^{n-2}(n-2)!}{(1+x)^{n-1}}+{n \choose 2}2\frac{(-1)^{n-3}(n-3)!}{(1+x)^{n-2}}$$
As if the third derivative of $u$ is $0$ ,and the fourth, the fifth, the sixth, etc.
This simplifies to,
$$x^2\frac{(-1)^{n-1}(n-1)!}{(1+x)^n}+2x\frac{(-1)^{n-2}n!}{(n-1)(1+x)^{n-1}}+\frac{(-1)^{n-3}n!}{(n-2)(1+x)^{n-2}}$$
|
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|
How do I find all prime solutions $p, q, r$ of the equation $\displaystyle p(p+1)+q(q+1) = r(r+1)$?
Find primes $p, q, r$ of the equation $$p(p+1)+q(q+1)
= r(r+1)$$
I know that it has only one solution namely $p = q = 2,r = 3$. But i can't show that.
Thank you for any help
|
May this lead to a simple proof for your problem according to your unic example of solution . There is only one
solution, namely $p = q = 2,r = 3$. To see that, we shall find all solutions
of the equation $p(p+1)+q(q+1) = n(n+1)$ where $p$ and $q$ are primes and
$n$ is a positive integer. Our equation yields
$p(p+1) = n(n+1)-q(q+1) = (n-q)(n+q+1)$,
and we must have $n > q$. Since $p$ is a prime, we have either $p|n-q$ or
$p|n+q+1$. If $p|n-q$, then we have $p\leq n-q$, which implies $p(p+1)
\leq (n-q)(n-q+1)$, and therefore $n+q+1 \leq n-q+1$, which is impossible.
Thus we have $p|n+q+ 1$, which means that for some positive integer $k$
,$n+q+1 = kp$, which implies $p+1 = k(n-q)\tag1$.
If we had $k = 1$, then $n+q+ 1 = p$ and $p+ 1 = n-q$, which gives $p-q
= n+ 1$ and $p+q = n- 1$, which is impossible, because $(p+q)-(p-q)=2q>0$ and $(n-1)-(n+1)=-2<0$. Thus, $k > 1$. From $(1)$ we
easily obtain:
$$\begin{align}
2q &= (n+q)-(n-q) \\
&= kp-1-(n-q) \\
&= k[k(n-q)-1]-1-(n-q) \\
&= (k+1)[(k-1)(n-q)-1].
\end{align}$$
Since $k \geq 2$, we have $k+1 \geq 3$. The last equality, whose left-hand side has
positive integer divisors $1, 2, q$, and $2q$ only, implies that either $k+ 1 = q$
or $k+1 = 2q$. If $k+1 = q$, then $(k-1)(n-q) = 3$, hence $(q-2)(n-q) = 3$.
This leads to either $q-2 = 1$, $n-q = 3$, that is $q = 3, n = 6, k = q-1 = 2$,
and, in view of $(1)$, $p = 5$, or else, $q-2 = 3$, $n-q = 1$, which gives $q = 5,
n = 6, k = 4$, and in view of $(1)$, $p = 3$.
On the other hand, if $k+1 = 2q$, then $(k-1)(n-q) = 2$, hence
$2(q-1)(n-q) = 2$. This leads to $q-1 = 1$ and $n-q = 1$, or $q = 2, n = 3$,
and, in view of $(1)$, $p = 2$. Thus, for positive integer $ n$, we have the
following solutions in primes $p$ and $q$:
*
*$(p = q = 2, n = 3; 2)$,
*$ (p = 5,
q = 3, n = 6)$, and
*$(p = 3, q = 5, n = 6)$.
Only in the first solution all
three numbers are primes.
Note: If we denote by $\displaystyle t_n = \frac{n(n+1)}{2}$ the nth triangular number,
then the equation $t_p+t_q = t_r$
has only one solution in prime numbers, namely $p = q = 2, r = 3$.
|
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|
Solution of Logarithmic Inequalities: $\log_{0.5}(\log_{5}(x^2-4))>\log_{0.5}1$
If $\log_{0.5}(\log_5 (x^2-4)) >\log_{0.5}1$ then x lies in the interval:
(a) $(-3,-\sqrt{5})\cup(\sqrt{5}, 3)$
(a) $(-3,-\sqrt{5})\cup(\sqrt{5}, 3\sqrt{5})$
(c) $(\sqrt{5}, 3\sqrt{5})$
(d) $\phi $
I have solved quite a few logarithmic inequalities and I am familiar with the procedure. I was just not able to get the answer that matches with the given options in the question.
My Working:
$\log_{0.5}(\log_{5}(x^2-4))>\log_{0.5}1 $
$\log_{5}(x^2-4)<1$
$\log_{5}(x^2-4)<\log_5(5)$
$x^2-4<5$
$x^2<9$
$\therefore -3<x<3$
Now the second equation comes from the fact that
$x^2-4>0$
$\therefore x<-2, x>2$
The intervals formed are $(-3,-2)\cup(2,3)$
|
From the inequality we get
$x^2-4>1$
Or$x^2>5$
Which provide one condition.
|
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|
Is $g(u)= \frac{E [ \frac{1}{\sqrt{X}} e^{-\frac{a^2u^2}{2X}} ] }{E [ \frac{1}{\sqrt{X}} e^{-\frac{u^2}{2X}} ]}$ decreasing in $u$ Let $X$ be a positive random variable, let us define a function
\begin{align}
g(u,a)= \frac{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{a^2u^2}{2X}} \right] }{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{u^2}{2X}} \right]}.
\end{align}
Question: Can we show that the above integral is monotonically decreasing in $u$ ( for $u>0$ ) for all $a > 1$.
Note that $X$ here represents the variance of standard normal. That is we consider the variance to be a random variable.
I can show that $g(u,a)$ is bounded by $1$ and continuous but can not establish that it is decreasing. Also, note that the function $g(u,a)$ is symmetric around $u=0$.
What I tried:
I was able to show that for $p,q\ge 1$ and $\frac{1}{q}+\frac{1}{p}=1$ and $a^2 \ge \frac{1}{p}$ we have
\begin{align}
g(u,a) \le \left( g( \beta \cdot u, a ) \right)^{\frac{1}{q}},
\end{align}
where $\beta=\sqrt{\frac{q(a^2-\frac{1}{p})}{a^2}}$.
Proof:
By using Holder's inequality
\begin{align}
E \left[ \frac{1}{\sqrt{X}} e^{-\frac{a^2u^2}{2X}} \right] &=E \left[ \frac{1}{\sqrt{X}} e^{-\frac{ (a^2-\frac{1}{p})u^2}{2X}} e^{-\frac{ \frac{1}{p}u^2}{2X}} \right] \\
&\le E^\frac{1}{q} \left[ \frac{1}{\sqrt{X}} e^{-\frac{ q(a^2-\frac{1}{p})u^2}{2X}} \right] E^\frac{1}{p} \left[ \frac{1}{\sqrt{X}} e^{-\frac{ u^2}{2X}} \right].
\end{align}
Therefore,
\begin{align}
g(u,a) \le \left( \frac{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{q(a^2-\frac{1}{p})u^2}{2X}} \right] }{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{u^2}{2X}} \right]} \right) ^\frac{1}{q}
&= \left( g( \beta \cdot u, a ) \right)^{\frac{1}{q}},
\end{align}
and
\begin{align}
g(u,a) \le \left( g( \beta \cdot u, a ) \right)^{\frac{1}{q}},
\end{align}
where $\beta=\sqrt{\frac{q(a^2-\frac{1}{p})}{a^2}}$.
Thank you. Looking forward to seeing your approaches.
|
This is false in general.
Change the variables as in my comment: $1/(2X) \to X$, $a^2 \to a$, $u^2 \to u$. Then the problem is to show that
$$
f(u) = \frac{\mathsf{E}[\sqrt{X}e^{-auX}]}{\mathsf{E}[\sqrt{X}e^{-uX}]}.
$$
decreases.
Set $a=1.1$ and let $X=1$ or $100$ with probability $1/2$. Then
$$
f(u) = \frac{e^{-1.1u} +10e^{-110u}}{e^{-u} +10e^{-100u}}.
$$
However, it is not decreasing for small values of $u$: Wolfram Alpha
|
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}
|
what is the value of $ x$? If $\log_{2}3^4\cdot\log_{3}4^5\cdot\log_{4}5^6\cdot....\log_{63} {64}^{65}=x!$, what is the value of $ x$?
I've tried
$$\log_2 3^4\cdot\log_3 4^5\cdot\log_4 5^6\cdot.... \log_{63} 64^{65}$$
$$=\dfrac{4\log 3}{\log 2}\cdot\dfrac{5\log 4}{\log 3}\cdot\dfrac{6\log 5}{\log 4}\cdots\dfrac{65\log 64}{\log 63}$$
don't know how to solve futher steps, please help.
Thanks
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\begin{align}\dfrac{4\log 3}{\log 2}\cdot\dfrac{5\log 4}{\log 3}\cdot\dfrac{6\log 5}{\log 4}\cdots\dfrac{65\log 64}{\log 63}&=\frac{65!}{3!}\frac{\log 64}{\log 2}
\\&=\frac{65!}{3!}\frac{\log 2^6}{\log 2} \\
&=\frac{65!}{3!}\frac{\log 2}{\log 2}.6
\\&=65!
\end{align}
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2058168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Strange limit problem to be solved without Hospital's Rule...? Having trouble solving this limit problem without L'Hôpital's Rule...
$$\lim_{x \to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
Tried multiplying the function by the conjugate/inverse-conjugate, of both the numerator and denominator... but no avail.... any ideas?
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$$\lim_{x \to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\lim_{x \to 2} \frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{(\sqrt{3-x}-1)(\sqrt{3-x}+1)}=\lim_{x \to 2} \frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}$$
$$=\lim_{x \to 2} \frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)(\sqrt{6-x}+2)}{(2-x)(\sqrt{6-x}+2)}=\lim_{x \to 2} \frac{(\sqrt{3-x}+1)(2-x)}{(2-x)(\sqrt{6-x}+2)}=$$
$$\lim_{x \to 2} \frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}=\frac{1}{2}$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2060328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
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.