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If $a$ and $b$ are cathets a right triangle whose hypotenuse is $1$ determine the highest value of $2a + b$ Can some one help me out on where to go?
If $a$ and $b$ are cathets a right triangle whose hypotenuse is $1$ determine the highest value of $2a + b$ ?
|
Your problem is as follows:
\begin{align*}
&\max \,2a+b\\
\\
&\text{s.t. } a^2 + b^2=1
\end{align*}
Using substitution, let $f(b) = 2a + b = 2\sqrt{1 - b^2} +b$. Now, we wish to find the maximum of this function with respect to $b$.
Using simple derivatives, we get
$$
\frac{-2b}{\sqrt{1-b^2}} + 1 = 0\implies 2b=\sqrt{1-b^2} \implies 5b^2=1\implies b=\sqrt{1/5}\implies a = \sqrt{4/5}.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that $(a+1/a)^2+(b+1/b)^2\ge 25/2$
If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that
$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\frac{1}{b}\bigg)^2\ge \dfrac{25}{2}.$$
My work:
$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge \dfrac{25}{2}\implies a^2+\dfrac{1}{a^2}+b^2+\dfrac{1}{b^2}+4\ge \dfrac{25}{2}$$
Now, we have $a^2+\dfrac{1}{a^2}\ge 2$ and $b^2+\dfrac{1}{b^2}\ge 2$.
Here, I am stuck, I cannot use the information provided, $a+b=1$ to any use. Please help!
|
By QM-AM, $\displaystyle a^2+b^2 \geq \frac{1}{2}(a+b)^2 = \frac{1}{2}$. QM-AM again gives $\displaystyle \frac{1}{a^2}+\frac{1}{b^2} \geq \frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)^2$. By AM-HM, $\displaystyle \frac{2}{\frac{1}{a}+\frac{1}{b}} \leq \frac{a+b}{2} = \frac{1}{2}$, whence $\displaystyle\frac{1}{a}+\frac{1}{b} \geq 4$, and thus $\displaystyle \frac{1}{a^2}+\frac{1}{b^2} \geq 8$. Thus we have $$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 = a^2 + b^2 + \frac{1}{a^2}+\frac{1}{b^2} + 4 \geq \frac{1}{2} + 8 + 4 = \frac{25}{2}.$$
|
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|
Finding sum of sequence $\sum\limits_{k=2}^n \frac{1}{k^2-1}$ Here is a problem I need to solve:
$$ \sum_{k=2}^n \frac{1}{k^2-1} $$
It came with another one alreay done in the same task:
$$ \sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^n \frac{(k+1)-k}{k(k+1)} = \sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n \frac{1}{k+1} = \sum_{k=1}^n \frac{1}{k} - \sum_{k=2}^{n+1} \frac{1}{k} = 1 - \frac{1}{n+1} $$
So, the problem have to have similar solution. I can't find a proper change for the numerator. I think the numeratore could be find using factorised dominator:
$$ \sum_{k=2}^n \frac{1}{(k-1)(k+1)} $$
|
Hint
$$\frac{1}{k^2-1}=\frac 1 2\left(\frac{1}{k-1}-\frac{1}{k+1}\right)=\frac 1 2\left(\frac{1}{k-1}-\frac 1 k+\frac 1 k-\frac{1}{k+1}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/638078",
"timestamp": "2023-03-29T00:00:00",
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|
minimum value of $\cos(A-B)+\cos(B-C) +\cos(C-A)$ is $-3/2$ How to prove that the minimum value of $\cos(A-B)+\cos(B-C) +\cos(C-A)$ is $-3/2$
|
If $\displaystyle2x+2y+2z=n\pi,$
$$F=\cos2x+\cos2y+\cos2z=2\cos(x+y)\cos(x-y)+2\cos^2z-1$$
As $\displaystyle x+y=n\pi-y, \cos(x+y)=(-1)^n\cos z$
$$\implies2\cos^2z\pm2\cos z\cos(x-y)-(1+F)=0$$ which is a Quadratic Equation in $\cos z$
So, the discriminant must be $\ge0$
$$\implies (2\cos(x-y))^2\ge 4.\cdot2(-1-F)\iff 2F\ge-2-2\cos^2(x-y)\ge-3 $$
The equality occurs if $\cos^2(x-y)=1\implies \cos2(x-y)=2\cos^2(x-y)-1=1\iff 2(x-y)=2m\pi$ where $m$ is an integer
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Find all solutions of $z^5+a^5=0$ The task is as follows:
Find all solutions of $z^5+a^5=0$, where $a$ is a positive real number.
My initial attempt
(which leads nowhere)
My guess is that i'll have to find the 5 5th roots of $-z^5$:
$w_1 = |-z^5|^5(cos(\frac{\theta}{5})+isin(\frac{\theta}{5})) \\
w_1 = |z|(cos(\frac{\theta}{5})+isin(\frac{\theta}{5})$
Then, the other roots are:
$w_2 = |z|(cos(\frac{\theta+2\pi}{5})+isin(\frac{\theta+2\pi}{5}) \\
w_3 = |z|(cos(\frac{\theta+4\pi}{5})+isin(\frac{\theta+4\pi}{5}) \\
w_4 = |z|(cos(\frac{\theta+6\pi}{5})+isin(\frac{\theta+6\pi}{5}) \\
w_5 = |z|(cos(\frac{\theta+8\pi}{5})+isin(\frac{\theta+8\pi}{5})$
but i'm not sure what to do with this new information. Also, I guess $z^5$ have to be a negative real number, since added to $a^5$ its $0$. Again, not sure what to do with this either.
|
I figured it out (delayed rubber duck?)
I define $z=(r,\theta)$. Then $z^5=(r^5,5\theta)$.
Since $z^5$ have to be a real negative number,
$$ 5\theta=\pi+2\pi n \\ \theta=\frac{\pi+2\pi n}{5}$$
And since $$r^5=a^5 \\ r=a$$
Then $z=(a, \frac{\pi+2\pi n}{5})$.
Or, with a more straight forward notation:
$z=a(cos(\frac{\pi+2\pi n}{5})+i sin(\frac{\pi+2\pi n}{5}))$.
|
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|
Sum of consecutive square roots inside a square root $$\large\sqrt{1+\sqrt{1+2+\sqrt{1+2+3+\sqrt{1+2+3+4+\cdots}}}}$$
I saw this somewhere in the internet but, the website didn't provide me any further information. What is the sum of the equation above? What is it called?
|
What is the sum of the equation above? What is it called?
It's neither an equation nor a sum, but is a continued radical defined as the limit (as $n\to\infty$) of the sequence of the nested radicals
$$
u_n = \sqrt{a_1+\sqrt{a_2 + \cdots + \sqrt{a_n}}}
$$
where $a_n = \sum_{i=1}^n i = n(n+1)/2$ (the $n^{th}$ triangle number), $n = 1,2,3,...$
That this limit exists follows from a theorem proved by T. Vijayaraghavan (1927):
For any sequence of nonnegative reals $(a_n)_{n=1,2,3,...}$, the sequence of nested radicals $(u_n)_{n = 1,2,3,...}$ with
$$u_n = \sqrt{a_1+\sqrt{a_2 + \cdots + \sqrt{a_n}}}$$
converges if and only if there exists a finite upper limit
$$\overline{\lim}\ {\left(\frac{\log{a_n}}{2^n}\right)} < \infty.$$
This clearly holds in the present case, because
$$\overline{\lim}\ {\left(\frac{\log{\frac{n(n+1)}{2}}}{2^n}\right)} = 0.
$$
To bound the error of $u_n$ as an approximation of the limit, the following consequence of a theorem proved by Herschfeld (1935) can be used:
If $a_i \gt 0 \ \ (i = 1,2,3,...)$, then for all $n \ge 1$,
$$0 \le u_{n+1}-u_n \le \frac{1}{2^n}\sqrt{\frac{a_{n+1}}{a_1 a_2 \cdots a_n}}. $$
Substituting $a_i = i(i+1)/2$ and simplifying then gives
$$0 \le u_{n+1}-u_n \le \frac{1}{n!}\sqrt{\frac{n+2}{2^{n+1}}} ,
$$
whence, noting that $n! \ge \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$,
$$0 \le u_{n+1}-u_n \le A \left(\frac{B}{n}\right)^n
$$
where $A = \sqrt{\frac{3}{4\pi}} = 0.4886...$ and $B = \frac{e}{\sqrt{2}} = 1.922...$
Now for all $m > n > 1$,
$$\begin{align}
0 \le u_m - u_n & = (u_m - u_{m-1}) + (u_{m-1} - u_{m-2}) + \cdots + (u_{n+1} - u_n) \\
& \le A \left ( \left (\frac{B}{m-1}\right )^{m-1} + \left (\frac{B}{m-2}\right )^{m-2} + \cdots + \left (\frac{B}{n}\right )^{n} \right )\\
& \le A \left ( \left (\frac{B}{n}\right )^{m-1} + \left (\frac{B}{n}\right )^{m-2} + \cdots + \left (\frac{B}{n}\right )^{n} \right )\\
& \le A \left (\frac{B}{n}\right )^n \frac{1 - (\frac{B}{n})^{m-n}}{1 - \frac{B}{n}}\tag{*}
\end{align}
$$
where the geometric progression was summed in the last step (noting that $B/n < 1$ for all $n > 1$).
Finally, taking limits with $m \to \infty$ and letting $u = \lim_{m \to\infty}{u_m}$,
$$0 \le u - u_n \le A \left (\frac{B}{n}\right )^n \frac{1}{1 - \frac{B}{n}}.
$$
(Note that (*) above also gives another proof of the existence of the limit, by showing the sequence $(u_n)_{n = 1,2,3,...}$ to be Cauchy.)
The latter error bound is rather weak, but it suffices to prove, for example, that
$$u = 1.8644589581634881323520037152739437841564220698266...$$
by computing $u_n$ for sufficiently large $n$. For example, $u_n$ has at least $n$ correct digits for any $n \ge 20$, because in that case
$A \left (\frac{B}{n}\right )^n \frac{1}{1 - \frac{B}{n}} \le 0.5 \ 10^{-n}.
$
(Aside)
Here's the Sage program I used to compute the $u_n$:
def u(n):
r = 0
for k in [1..n][::-1]: r = sqrt(k*(k+1)/2 + r)
return r
for n in [10..100]: print n, u(n).n(digits=100)
|
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|
If $A,B,C$ are the angle of a triangle, then show that $\sin A+\sin B-\cos C\le \dfrac3 2$ If $A,B,C$ are the angle of a triangle, then show that $\sin A+\sin B-\cos C\le \dfrac32$
I tried substituting $C=180^\circ-(A+B)$ and got stuck. I also tried using the formula $\sin A+\sin B=2\sin \frac{A+B}{2}\cos \frac{A-B}{2}$ without any progress. Please help!
|
So, we have $\displaystyle \sin\frac{A+B}2=\sin\left(\frac\pi2-\frac C2\right)=\cos\frac C2$
Again, $\displaystyle\cos C=2\cos^2\frac C2-1$
Let $\displaystyle\sin A+\sin B-\cos C=y$
$\displaystyle\implies2\cos\frac C2\cos\frac{A-B}2-\left(2\cos^2\frac C2-1\right)=y$
$\displaystyle\implies2\cos^2\frac C2-2\cos\frac C2\cos\frac{A-B}2+y-1=0\ \ \ \ (1)$ which is a Quadratic Equation in $\cos\frac C2$ which is real as $C$ is
So, the discriminant must be $\ge0 \displaystyle\implies \left(-2\cos\frac{A-B}2\right)^2\ge 4\cdot2\cdot(y-1)$
$\displaystyle\iff y\le 1+\frac{\cos^2\dfrac{A-B}2}2\le 1+\frac12$ as $\displaystyle\cos^2\dfrac{A-B}2\le1$
The equality occurs if $\displaystyle\cos^2\dfrac{A-B}2=1\iff\cos(A-B)=2\cos^2\dfrac{A-B}2-1=1\implies A-B=2n\pi$ where $n$ is an integer
As $0<A,B<\pi, $ it needs $A-B=0\iff A=B$ and will reduces $(1)$ to $\displaystyle2\cos^2\frac C2-2\cos\frac C2+\frac32-1=0$
$\iff \cos\frac C2=\frac12\implies \frac C2=\frac\pi3 $ as $0<\frac C2<\frac\pi2$
$\displaystyle\implies C=\frac{2\pi}3$ and $\displaystyle A=B=\frac{A+B}2=\frac{\pi-C}2=\cdots$
|
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|
Solve the system using Gaussian elimination with back-substitution or Gauss-Jordan elimination Here is the system:
$$
\left\{
\begin{aligned}
x_1-3x_3&=-2 \\
3x_1+x_2-2x_3&=5 \\
2x_1+2x_2+x_3&=4
\end{aligned}
\right.
$$
This is my very first problem actually using a matrix so here is my attempt
First I setup the augmented matrix:
$$ \left[
\begin{array}{ccc|c}
1&0&-3&-2\\
3&1&-2&5 \\
2&2&1&4
\end{array}
\right]$$
Then I did $(-3)R_1+R_2->R_2$ which produced:
$$ \left[
\begin{array}{ccc|c}
1&0&-3&-2\\
0&1&7&11 \\
2&2&1&4
\end{array}
\right]$$
And then I did $(-2)R_1+R_3->R_3$:
$$ \left[
\begin{array}{ccc|c}
1&0&-3&-2\\
0&1&7&11 \\
0&2&7&8
\end{array}
\right]$$
And then $(1/2)R_3->R_3$:
$$ \left[
\begin{array}{ccc|c}
1&0&-3&-2\\
0&1&7&11 \\
0&1&\frac{7}{2}&4
\end{array}
\right]$$
And then $(-1)R_2+R_3->R_3$
$$ \left[
\begin{array}{ccc|c}
1&0&-3&-2\\
0&1&7&11 \\
0&0&\frac{-7}{2}&7
\end{array}
\right]$$
And finally $\left(\frac{-1}{7}\right)R_3->R_3$
$$ \left[
\begin{array}{ccc|c}
1&0&-3&-2\\
0&1&7&11 \\
0&0&1&-1
\end{array}
\right]$$
But this is not correct according to the solution. Can someone tell me where I went wrong? Clearly I did something really wrong here.
|
Your mistake is at row: $(-1)R_2 + R_3 \to R_3$.
|
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|
Is this polynomial solvable by radicals? The polynomial
$p(x) = x^6-9x^4-4x^3+27x^2-36x-23$.
has at least one (real, irrational) root that is expressible by radicals (can you find it?).
Are all the roots of $p$ expressible by radicals and if so, how can one find the expressions?
|
Maple says
$$
16(x^6-9x^4-4x^3+27x^2-36x-23) =
\left( i\sqrt {3}\sqrt [3]{2}-2\,\sqrt {3}-\sqrt [3]{2}-2\,x
\right) \\
\left( i\sqrt {3}\sqrt [3]{2}+2\,x+2\,\sqrt {3}+\sqrt [3]{2}
\right) \left( i\sqrt {3}\sqrt [3]{2}-2\,\sqrt {3}+\sqrt [3]{2}+2\,x
\right) \\
\left( i\sqrt {3}\sqrt [3]{2}+2\,\sqrt {3}-\sqrt [3]{2}-2\,x
\right) \left( x+\sqrt {3}-\sqrt [3]{2} \right) \left( -x+\sqrt {3}
+\sqrt [3]{2} \right)
$$
|
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|
if $A+B+C+D=\pi$, what is $\min(\cos{A}+\cos{B}+\cos{C}+\cos{D})$ Let $A,B,C,D \in R ~|~ A+B+C+D=\pi$, what is the minimum of the following function
$$f(A,B,C,D)=\cos{A}+\cos{B}+\cos{C}+\cos{D}$$
I found a post about a similar problem, butI think an even number of variables is harder than an odd number of variables, because for three variables,
$A+B+C=\pi~|~A,B,C\in R$ then it follows that $\cos{A}+\cos{B}+\cos{C}=1+4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}\to 1$, but for four variables, we can't.
My try: $$\cos{A}+\cos{B}+\cos{C}+\cos{D}=2\cos{\dfrac{A+B}{2}}\cos{\dfrac{A-B}{2}}+2\cos{\dfrac{C+D}{2}}\cos{\dfrac{C-D}{2}}=2\sin{\dfrac{C+D}{2}}\cos{\dfrac{A-B}{2}}+2\cos{\dfrac{C+D}{2}}\cos{\dfrac{C-D}{2}}$$
|
So we want to minimize $f(A,B,C) = \cos A + \cos B + \cos C + \cos(\pi-A-B-C)$. We have
\begin{align*}\partial_A f(A,B,C) &= \sin(\pi-A-B-C) - \sin A\\
\partial_B f(A,B,C) &= \sin(\pi-A-B-C) - \sin B\\
\partial_C f(A,B,C) &= \sin(\pi-A-B-C) - \sin C
\end{align*}
In a mininum, we have $f'(A,B,C) = 0$, this gives
$$ \sin A = \sin B = \sin C = \sin(\pi-A-B-C)$$
So for some $\lambda \in [-\pi, \pi]$, we must have
$$ A, B, C, \pi - A- B-C \in \{\lambda, \pi-\lambda\} + 2\pi \mathbb Z$$
Now we consider two cases (that suffices, as we may replace $\lambda$ by $\pi-\lambda$ are use the symmetry in $A, B, C$):
*
*Up to multiples of $2\pi$, $A=B=C = \lambda$. Then $$ \pi - A-B-C= \pi - 3\lambda$$ hence for some $k$, either
$$ \pi - 3 \lambda = \lambda + 2k\pi \iff \lambda = \frac 14(2k-1)\pi $$
or
$$ \pi - 3\lambda = \pi - \lambda + 2k\pi \iff 2\lambda = -2k\pi \iff \lambda = -k \pi. $$
In the first case $\cos \lambda \in \{\pm 2^{-1/2}\}$, hence
$$ f(A,B,C) = \pm 4\cdot 2^{-1/2} = \pm 2^{3/2}. $$
In the second case $\cos\lambda \in \{\pm 1\}$, hence
$$ f(A,B,C) = \pm 1 \pm 1 \pm 1 \mp 1 \in \{\pm 2\} $$
*Up to multiples of $2\pi$, $A = B = \lambda$, $C = \pi - \lambda$. Then $A+B+C = \pi + \lambda$, hence either
$$-\lambda = \lambda + 2k\pi \iff \lambda = -k\pi $$
or
$$-\lambda = \pi - \lambda + 2k\pi $$
but this is impossible for integer $\mathbb Z$. So this case gives nothing new.
Hence, the global mininum is $-2^{3/2}$, attained for example when $A= B= C = \frac 34\pi$, $D = -\frac 54\pi$.
|
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|
Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$
For $a\ge 0$ let's define $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)dx.$$ Find explicit formula for $I(a)$.
My attempt: Let
$$\begin{align*}
f_n(x)
&= \frac{\ln\left(1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2\right)-\ln\left(1-2a\cos x+a^2\right)}{\frac{1}{n}}\\
&=\frac{\ln\left(\displaystyle\frac{1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2}{1-2a\cos x+a^2}\right)}{\frac{1}{n}}\\
&=\frac{\ln\left(1+\dfrac{1}{n}\left(\displaystyle\frac{2a-2\cos x+\frac{1}{n}}{1-2a\cos x+a^2}\right)\right)}{\frac{1}{n}}.
\end{align*}$$
Now it is easy to see that $f_n(x) \to \frac{2a-2\cos x}{1-2a\cos x+a^2}$ as $n \to \infty$. $|f_n(x)|\le \frac{2a+2}{(1-a)^2}$ RHS is integrable so $\lim_{n\to\infty}\int_0^\pi f_n(x)dx = \int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2} dx=I'(a)$. But
$$\int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2}=\int_0^\pi\left(1-\frac{(1-a)^2}{1-2a\cos x+a^2}\right)dx.$$ Consider
$$\int_0^\pi\frac{dx}{1-2a\cos x+a^2}=\int_0^\infty\frac{\frac{dy}{1+t^2}}{1-2a\frac{1-t^2}{1+t^2}+a^2}=\int_0^\infty\frac{dt}{1+t^2-2a(1-t^2)+a^2(1+t^2)}=\int_0^\infty\frac{dt}{(1-a)^2+\left((1+a)t\right)^2}\stackrel{(*)}{=}\frac{1}{(1-a)^2}\int_0^\infty\frac{dt}{1+\left(\frac{1+a}{1-a}t\right)^2}=\frac{1}{(1-a)(1+a)}\int_0^\infty\frac{du}{1+u^2}=\frac{1}{(1-a)(1+a)}\frac{\pi}{2}.$$
So $$I'(a)=\frac{\pi}{2}\left(2-\frac{1-a}{1+a}\right)\Rightarrow I(a)=\frac{\pi}{2}\left(3a-2\ln\left(a+1\right)\right).$$
It looks too easy, is there any crucial lack?
$(*)$ — we have to check $a=1$ here by hand and actually consider $[0,1), (1,\infty)$ but result on these two intervals may differ only by constant - it may be important but in my opinion not crucial for this proof.
|
Old thread, but this question came up again and was marked a duplicate, so I have to post my physics lecture here.
You know the symmetries of the problem that the electric field of a line charge must be purely radial and that its magnitude has no axial or azimuthal dependence $\vec E=E(r)\hat e_r$. Thus we may apply Gauss's law over a cylinder or radius $r$ and length $\ell$ to find the magnitude of the electric field of a line charge $\lambda$ coincident with the $z$-axis:
$$\oint\vec E\cdot d^2\vec A=E(r)A_{\text{curvy}}=2\pi rE(r)=\int\frac{\rho}{\epsilon_0}d^3V=\frac{\lambda\ell}{\epsilon_0}$$
So $\vec E=\frac{\lambda}{2\pi\epsilon_0r}\hat e_r=-\vec\nabla V$. So we can get the potential $V=-\frac{\lambda}{2\pi\epsilon_0}\ln r+C$. Now that constant of integration is a bit problematic because $V$ is unbounded at both $r=0$ and $r=\infty$, but if we have a line charge $\lambda$ at $(x,y)=(b\cos\theta,b\sin\theta)$ and another charge $-\lambda$ on the $z$-axis, then the potential of the combination is
$$V(\theta)=-\frac{\lambda}{2\pi\epsilon_0}\ln\sqrt{1-2\frac br\cos\theta+\frac{b^2}{r^2}}$$
And this potential does go to $0$ as $r$ goes to $\infty$. Now we can create the potential for a cylinder centered on the $z$-axis with radius $b$ and surface charge density $\sigma$ with a line charge $-2\pi b\sigma$ running along the $z$-axis by superposition:
$$V=\int_0^{2\pi}-\frac{a\sigma}{4\pi\epsilon_0}\ln\left(1-2\frac br\cos\theta+\frac{b^2}{r^2}\right)d\theta$$
For $r>b$, applying Gauss's law over a cylinder of radius $r$ and length $\ell$ this time shows that $\vec E=\vec0$ because the cylinder contains no net charge. So this time $V=C$, a constant for $r>b$. Since
$$\lim_{r\rightarrow\infty}V(r)=0$$ by construction, so we have shown that for $0<a<1$,
$$\int_0^{2\pi}\ln(1-2a\cos\theta+a^2)d\theta=0$$
If $a>1$, then
$$\begin{align}\int_0^{2\pi}\ln(1-2a\cos\theta+a^2)d\theta&=\int_0^{2\pi}\left[2\ln a+\ln(1-2a^{-1}\cos\theta+a^{-2})\right]d\theta\\
&=4\pi\ln a+0=4\pi\ln a\end{align}$$
|
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|
Solve $2000x^6+100x^5+10x^3+x-2=0$ One of the roots of the equation $2000x^6+100x^5+10x^3+x-2=0$ is of the form $\frac{m+\sqrt{n}}r$, where $m$ is a non-zero integer and $n$ and $r$ are relatively prime integers.Then the value of $m+n+r$ is?
Tried to use the fact that another root will be $\frac{m-\sqrt{n}}r$ as coefficients are rational but there are six roots and using sum and product formulas would allow many variables in the equations.
|
We have $\displaystyle x+10x^3+100x^5=x\frac{1000x^6-1}{10x^2-1}$. (A geometric progression)
Hence $\displaystyle -2(1000x^6-1)=x \frac{1000x^6-1}{10x^2-1}$
Hence either $1000x^6-1=0$ or $x=-2(10x^2-1)$.
Therefore $20x^2+x-2=0$ for second equation.
Solving we get
$$x=\frac{-1\pm \sqrt{161}}{40}$$.
Comparing $m=-1, n=161$ and $r=40$. Hence $m+n+r=200$
|
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|
Incorrect Solution for Problem 7 of Pinter's Book of Abstract Algebra, Chapter 2? I'm just getting started with Pinter's A Book of Abstract Algebra, please be kind. The book solution for Chapter 2, Problem 7 claims that the following operator is non-associative:
$$ x * y = \frac{xy}{x+y+1}$$
The book solution:
$$ (x * y) * z = \left(\frac{xy}{x+y+1}\right) * z = \frac{xyz(x+y+1)}{x+y+z+xy+yz+xz+1}$$
$$ x * (y * z) = x * \left(\frac{yz}{y+z+1}\right) = \frac{xyz(y+z+1)}{x+y+z+xy+yz+xz+1}$$
However the $(x+y+1)$ and $(y+z+1)$ terms appear to actually cancel out:
$$ (x * y) * z = \left(\frac{xy}{x+y+1}\right) * z = \frac{\left(\frac{xy}{x+y+1}\right)z}{\frac{xy}{x+y+1} + z + 1} $$
$$ x * (y * z) = x * \left(\frac{yz}{y+z+1}\right) =\frac{x\left(\frac{yz}{y+z+1}\right)}{x+\frac{yz}{y+z+1} + 1} $$
So the denominators cancel out, the two groupings are equivalent, the operator is associative. This result is duplicated in Mathematica, but I'd like to make sure I'm not missing something. Sorry for bothering people with such a basic question.
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As far as I can tell it's associative. You've proved it twice yourself already, but my nitty-gritty arithmetic is:
\begin{align*}
(x * y) * z &= \left(\frac{xy}{x+y+1}\right) * z \\
&= \frac{\left(\frac{xy}{x+y+1}\right)z}{\frac{xy}{x+y+1} + z + 1} \\
&= \frac{\left(\frac{xyz}{x+y+1}\right)}{\frac{xy+(z+1)(x+y+1)}{x+y+1}} \\
&= \frac{xyz}{xy+(z+1)(x+y+1)} \\
&= \frac{xyz}{xy+xz+yz+x+y+z+1} \\
&= \frac{xyz}{yz+(x+1)(y+z+1)} \\
&= \frac{\left(\frac{xyz}{y+z+1}\right)}{\frac{yz+(x+1)(y+z+1)}{y+z+1}} \\
&= \frac{x\left(\frac{yz}{y+z+1}\right)}{x + \frac{yz}{y+z+1} + 1} \\
&= x*\left(\frac{yz}{y+z+1}\right) \\
&= x*(y*z).
\end{align*}
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|
Induction: show that $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + ... + \frac{1}{\sqrt{n}} < 2\sqrt{n}$ The question:
Induction: show that:
$$1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + ... + \frac{1}{\sqrt{n}} < 2\sqrt{n}$$
for $n \geq 1$
My attempt at a solution:
First we test the base case: $n=1$ this gives us:
$$1 < 2$$
Which works.
Then we do the inductive assumption that it holds true for $k=n$, this gives us:
$$1 + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n}} < 2\sqrt{n}$$
If we can prove that the following is true, we have solved the problem:
$$1 + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n+1}} < 2\sqrt{n+1}$$
we then have to show that $$2\sqrt{n} + \frac{1}{\sqrt{n+1}} \leq 2\sqrt{n+1}$$
Which can be written as: $$2\sqrt{n}\sqrt{n+1} + 1 \leq 2(n+1)$$
$$1 \leq 2(n+1) - 2\sqrt{n}\sqrt{n+1}$$
$$2\sqrt{n}\sqrt{n+1} \leq 2(n+1) - 1$$
$$2\sqrt{n}\sqrt{n+1} \leq 2n+1$$
$$(2\sqrt{n}\sqrt{n+1})^2 \leq (2n+1)^2$$
$$4n^2 + 4n \leq 4n^2 + 4n + 1$$
Can anyone please help me to continue from here or does this mean that I have proved the inductive assumption?
Thank you!
|
You have proved it; you just need to write the 6 inequalities in the reverse order, and connect them by implication $\implies$ or keep the ordering and show that each is equivalent $\iff$ with the next.
|
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How to find all polynomials P(x) such that $P(x^2-2)=P(x)^2 -2$? I am trying the fallowing exercise :
Solve $P(X^2 -2)=P(X)^2 -2$ with P a monic polynomial (non-constant)
My attempt :
Let P satisfying $P(X^2-2) = (P(X))^2-2$
Then $Q(X)=P(X^2-2) = (P(X))^2-2$
Therefore, $$Q(X^2-2) = (P(X^2-2))^2-2 = (P(X)^2-2)^2-2 = Q^2-2$$
As X is a solution, by defining the sequence:
$(P_n)_{n \geq 1}$ with $P_1 = X$ and for all $n \geq 1, P_{n+1} = P_n^2-2$
We obtain a sequence of polynomials which are solutions.
But I don't know how to prove it's the only one. If someone have an idea to prove it or an another method to solve the problem ?
Thank you in advance for your time.
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Lemma : If $P(x)^2$ is a polynomial in $x^2$, then so is either P(x) or P(x)/x.
By the lemma, there is a polynomial Q such that $P(x)=Q(x^2-2)$ or $P(x)=xQ(x^2-2)$.
Then $Q((x^2-2)^2-2)=Q(x^2-2)^2−2$ or $(x^2-2)Q((x^2-2)^2-2)=x^2Q(x^2-2)^2-2$
Substituting $x^2-2=y$ yields $Q(y^2-2)=Q(y)^2-2$ and $yQ(y^2-2)=(y+2)Q(y)^2+1$
Suppose that $yQ(y^2-2)=(y+2)Q(y)^2-2$. Setting $y=-2$ we obtain that $Q(2)=1$
Note that, if $a\neq 0$ and $Q(a)=1$ then also $aQ(a^2+1)=(a+2)-2$ and hence $Q(a^2+1)=1$.
We thus obtain an infinite sequence of points at which Q takes value 1.
Namely the sequence given by $a_{n+1}=a_{n^2}-2$.
Therefore $Q≡1$. It follows that if $Q≢1$, then $P(x)=Q(x^2-2)$.
Now we can easily list all solutions: these are the polynomials of the form $T(T(⋯(T(x))⋯))$, where $T(x)=x^2-2$.
NB: Lemma/proof,
Let $P(x)=a_nx^n+a_{n−1}x_{n−1}+⋯+a_0$, $a_n≠0$.
The coefficient at $x^{2n−1}$ (of $P(x)^2$) is $2a_na_{n−1}$, from which we get $a_{n−1}=0$. Now the coefficient at $x_{2n−3}$ equals $2a_na_{n−3}$; hence $a_{n−3}=0$, and so on. Continuing in this manner we conclude that $a_{n−2k−1}=0$ for $k=0,1,2,…,$ i.e. $P(x)=a_nx^n+a_{n−2}x^{n−2}+a_{n−4}x_{n−4}+⋯$.
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|
Solve the difference quotient
So I tried a variety of answers and they all came up incorrect. I guess I don't understand difference quotient because I feel like it's pointless and a slow way to do the problem. That being said... can't get the answer
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Evaluate $f(4)$:
$$f(4) = \frac{4+6}{4+4}=\frac{10}{8}$$
Plug in what you know into the difference quotient:
$$\frac{\frac{x+6}{x+4}-\frac{10}{8}}{x-4}$$
Simplifying the numerator:
$$\frac{x+6}{x+4}-\frac{10}{8}=\frac{8(x+6)-10(x+4)}{8(x+4)}=\frac{8-2x}{8(x+4)}=\frac{-2(x-4)}{8(x+4)}=-\frac{(x-4)}{4(x+4)}$$
$$\frac{-\frac{x-4}{4(x+4)}}{x-4}=-\frac{x-4}{4(x+4)(x-4)}=-\frac{1}{4(x+4)}$$
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Finding maximum value for a function I was working on this question to find the following function's maximum value.Let $$y=f(x)={{(\sqrt{-3+4x-x^2}+4)}}^2 + (x-5)^2$$ where $$1 \le x \le 3$$.I have to find it's maximum value.
I tried by simple method of taking $$\frac{dy}{dx}=0$$ and got the answer.But this path was pretty lenghty and tiresome.Is there any short method possible for this?
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You have simply$$y=(\sqrt{-3+4x-x^2}+4)^2 + (x-5)^2 \\
y=(\sqrt{1-(x-2)^2}+4)^2 + (x-5)^2 \\
y'=2(\sqrt{1-(x-2)^2}+4).\frac{-2(x-2)}{2\sqrt{1-(x-2)^2}} + 2(x-5)\\
y'=(\sqrt{1-(x-2)^2}+4).\frac{-2(x-2)}{\sqrt{1-(x-2)^2}} + 2(x-5)\\
y'=\frac{-2(x-2)\sqrt{1-(x-2)^2}-8x+16+2(x-5)\sqrt{1-(x-2)^2}}{\sqrt{1-(x-2)^2}}\\
y'=\frac{\sqrt{1-(x-2)^2}-8(x-2)}{\sqrt{1-(x-2)^2}}\\$$
$y'=0 \to \sqrt{1-(x-2)^2}-8(x-2)=0$ then
$$1-(x-2)^2=64(x-2)^2\\
(x-2)^2=\frac{1}{65}\\
x=2+\frac{1}{\sqrt{65}}\\
x=2-\frac{1}{\sqrt{65}}$$
You have to compare the values of $f$ at these two values and at $x=1$ and at $x=3$. I hope it helps you.
|
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|
Limit $\lim\limits_{x\to0}\frac1{\ln(x+1)}-\frac1x$ The limit is
$$\lim\limits_{x\to0}\frac1{\ln(x+1)}-\frac1x$$
The problem is I don't know if I can calculate it normally like with a change of variables or not. Keep in mind that I'm not allowed to use L'Hôpital's rule nor the $\mathcal O$-notation.
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I assume you know about the geometric series and Taylor's theorem to integrate them, but you might not. I apologize if you don't, as this will not be of much use.
We have
$$
\frac 1{\log(x+1)} - \frac 1x = \frac{x-\log(x+1)}{x \log(x+1)} = \frac{1 - \frac{\log(x+1)}{x}}{\log(x+1)}.
$$
Now
$$
\frac 1{1+x} = \frac 1{1-(-x)} = \sum_{n \ge 0} (-x)^n,
$$
hence
$$
\log(x+1) = \sum_{n \ge 0} \frac{-(-x)^{n+1}}{n+1} = \sum_{n \ge 0} \frac{(-1)^n x^{n+1}}{n+1} = x - \frac{x^2}2 + \frac{x^3}3 - \cdots.
$$
Also,
$$
\frac{\log(x+1)}{x} = \sum_{n \ge 0} \frac{(-1)^n x^n}{n+1},
$$
hence
$$
1 - \frac{\log(x+1)}x = \sum_{n \ge 1} \frac{(-1)^{n+1} x^n}{n+1} = \frac x2 - \frac{x^2}3 + \frac{x^3}4 - \cdots.
$$
Therefore,
$$
\lim_{x \to 0} \frac{1}{\log(x+1)} - \frac 1x = \lim_{x \to 0} \frac{\frac x2 - \frac{x^2}3 + \frac{x^3}4 - \cdots}{x - \frac{x^2}2 + \frac{x^3}3 - \cdots} = \lim_{x \to 0} \frac{\frac 12 - \frac{x}3 + \frac{x^2}4 - \cdots}{1 - \frac{x}2 + \frac{x^2}3 - \cdots} = \frac 12.
$$
Note : you don't need the whole series expansion, you could approximate to the second term using Taylor's theorem. But I guess that falls into your "big-O notation" category of proofs.
Hope that helps,
|
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|
Prove trig identity: $\tan(x) + \cot(x) = \sec(x) \csc(x)$ wherever defined I appreciate the help.
My attempt:
$$
\begin{align}
\tan(x) + \cot(x) &= \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} \\
&= \frac{\sin^2(x)}{\cos(x) \sin(x)}+\frac{\cos^2(x)}{\cos(x) \sin(x)} \\
&= \frac{\sin^2(x)+\cos^2(x)}{\cos(x) \sin(x)}\\
&= \frac{1}{\cos(x) \sin(x)}\\
&= \frac{1}{\frac{1}{\sec(x)}\frac{1}{\csc(x)}}\\
&=\frac{1}{\frac{1}{\sec \csc}}\\
&=\frac{1}{1}\cdot \frac{\sec(x) \csc(x)}{1}\\
&= \sec(x) \csc(x)
\end{align}
$$
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All your work is okay.
Equations last but one, last but two and last but three can be deleted and you can straight away jump to last result because you already know $ \sec,\csc $ are inverse trig functions of $\cos, \sin$.
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|
For a positive integer $n$ both $5n+1$ and $7n+1$ are perfect squares. Show that $n$ is divisible by 24. My try:
$5n + 1 = k^2$
$7n +1 = \frac{7k^2-2}5$
Just don't know how to proceed after this. Please help.
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If $x\equiv1(mod\ 3)$, we have $7n+1\equiv2(mod\ 3)$.
If $x\equiv2(mod\ 3)$, we have $5n+1\equiv2(mod\ 3)$.
But a perfect square cannot be congruent 2 mod 3, so x must be divisible by 3.
The remainders modulo 8 of the numbers are :
? for(n=0,7,print(n," ",Mod(7*n+1,8)," ",Mod(5*n+1,8)))
0 Mod(1, 8) Mod(1, 8)
1 Mod(0, 8) Mod(6, 8)
2 Mod(7, 8) Mod(3, 8)
3 Mod(6, 8) Mod(0, 8)
4 Mod(5, 8) Mod(5, 8)
5 Mod(4, 8) Mod(2, 8)
6 Mod(3, 8) Mod(7, 8)
7 Mod(2, 8) Mod(4, 8)
?
Since the only quadratic residues modulo 8 are 0,1,4 , we must have $n\equiv0$ (mod 8)
Hence, 24 must divide n.
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|
Show if $a^2+b^2 \le 2$ then $a+b \le 2$ If $a^2+b^2 \le 2$ then show that $a+b \le2$
I tried to transform the first inequality to $(a+b)^2\le 2+2ab$ then $\frac{a+b}{2} \le \sqrt{1+ab}$ and I thought about applying $AM-GM$ here but without result
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First note that if $a=0$ or $b=0$, then the question is easy. So assume that both are non-zero. Consider the value $ab$. We can show that we must have $|ab|\le1$.
Assume by contrary that we have: $|ab| > 1$. This means $$\frac{1}{|b|}<|a|$$
Expanding: $$0\le(b^2-1)^2$$
we get:$$2\le \frac{1}{b^2}+b^2,$$
Thus, using the original ineqaulity we see:
$$
2\le \frac{1}{b^2}+b^2\lt a^2+b^2\le2
$$
which is a contradiction. Hence,$|ab|\le1$, and we can continue your original inequality, by writing:
$$
a+b\le \sqrt{2+2ab}\le \sqrt{2+2}=2
$$
|
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|
Pseudo-pythagorean theorem Pythagoras' theorem is a special case of the Cosine theorem for a angle of $90°$. But also for an angle of 60° and 120°, "aesthetical" special cases derive:
$c^2=a^2+b^2\pm ab$
First question:
Are there further angle $x°$ with a rational number $x$, so that $\cos x$ is rational as well, thus creating "aesthetical" special cases?
Second question:
Does anybody know some internet sources to the equivalent of pythagorean triplets (Integers $a,b,c$, so that $c^2=a^2+b^2\pm ab$)?
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To find all integer triples $a,b,c$ for which $c^2=a^2+b^2\pm ab$, it's enough to consider only $c^2=a^2+b^2+ab$ because changing the sign of $a$ or $b$ gives a solution to the other equation and vice versa.
We'll only look for primitive solutions, i.e, satisfying $\gcd(a,b,c)=1$. Then at least one of $a$ or $b$ is odd, say it's $b$.
Note that the equation can be rewritten
$$4c^2=(2a+b)^2+3b^2.$$
Set $d=2a+b$, so $d$ is odd. We have
$$3b^2=(2c-d)(2c+d)$$
where $\gcd(2c-d,2c+d)=1$ because of the primitivity. There are two cases: $2c-d=3u^2$ and $2c+d=v^2$, or $2c-d=v^2$ and $2c+d=3u^2$. In any case we have $\gcd(u,v)=1$, $3\nmid v$ and $b=uv$, with $u$ and $v$ odd.
The first case gives $c=\frac{v^2+3u^2}4$ and $d=\frac{v^2-3u^2}2$ from which $a=\frac{v^2-2uv-3u^2}4$.
The second case yields $c=\frac{v^2+3u^2}4$ and $d=\frac{3u^2-v^2}2$ from which $a=\frac{3u^2-2uv-v^2}4$.
In both cases $b=uv$.
As said before, we chose $b$ to be odd. This means we can change the values found for $a$ and $b$. (We could be introducing doubles here in the case where $a$ and $b$ are both odd. In fact, I'm sure we do.)
To obtain all values (including the non-primitive) we simply have to multiply by an integer constant $k=\gcd(a,b,c)$.
Example
Take coprime odd integers $u$ and $v$ with $3\nmid v$. Say $u=3$ and $v=5$. Then with the first case we obtain $a=-8$, $b=15$ and $c=13$. Indeed,
$$13^2=(-8)^2+15^2+(-8)\cdot15.$$
The second case gives the same values for $b$ and $c$, but $a=-7$. Again,
$$13^2=(-7)^2+15^2+(-7)\cdot15.$$
|
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|
Help needed for Partial Fraction Decomposition The problem I need help with is this:
$$\frac{x^4} {(x-1)^3}$$
I've already put it into the form: $$ \frac{x^4} {(x-1)^3}=\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3}$$
Then, I multiplied the whole equation by the LCD ($(x-1)^3$) then got: $$x^4=A(x-1)^2+B(x-1)+C$$
I let x=1 to find C, which resulted in C being 1.
At this point I tried to replace C with 1, then expand the equation to get: $$x^4=Ax^2-2Ax+A+Bx-B+1$$
From here I don't know how to get the answer, which is $$x+3+\frac{6}{x-1}+\frac{4}{(x-1)^2}+\frac{1}{(x-1)^3}$$
Any help would be much appreciated!
|
You can't use partial fraction decomposition since the degree of the numerator is greater than the degree of the denominator. Consider using polynomial division first before applying partial fraction decomposition. Given the equation
$$\dfrac{x^4}{(x - 1)^3}$$
Expanding the denominator expression, we have
$$\dfrac{x^4}{x^3 - 3x^2 + 3x - 1}$$
By polynomial division, we have
$$\begin{aligned}
\dfrac{x^4 - x(x^3 - 3x^2 + 3x - 1)}{x^3 - 3x^2 + 3x - 1} &= x + \dfrac{x^4 - x^4 + 3x^3 - 3x^2 - x}{x^3 - 3x^2 + 3x - 1}\\
&= x + \dfrac{3x^3 - 3x^2 - x}{x^3 - 3x^2 + 3x - 1}\\
&= x + 3 + \dfrac{3x^3 - 3x^2 - x - 3(x^3 - 3x^2 + 3x - 1)}{x^3 - 3x^2 + 3x - 1}\\
&= x + 3 + \dfrac{6x^2 - 8x + 3}{x^3 - 3x^2 + 3x - 1}
\end{aligned}$$
Finally, as you said before, by partial fraction decomposition (left as exercise for you to apply partial fraction decomposition), you get:
$$x + 3 + \dfrac{6}{x - 1} + \dfrac{4}{(x - 1)^2} + \dfrac{1}{(x - 1)^3}$$
|
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|
Help with a general solution for $\int \tan^a \theta$ Working out the integrals of $\tan^6 \theta$ I saw a pattern. I would like to put it in series representation. The pattern is as follows:
$$\int \tan^a d\theta= \int\tan^{a-2}\theta(\sec^2\theta-1)\ d\theta= $$ $$ \int \tan ^{a-2}\theta\sec^2\theta\ d\theta-\int\tan^{a-2}\theta\ d\theta=$$ solving the first integral, let $u=\tan \theta, du=\sec^2\theta\ d\theta$ so $$ \int u^{a-2} du=\frac{1}{a-2+1}u^{a-2+1}=$$ $$\frac{1}{a-1}u^{a-1}= $$$$\frac{1}{a-1}\tan^{a-1}\theta-\int \tan^{a-2}\theta\ d\theta $$
then $$ \int u^{(a-2)-2} du=\frac{1}{(a-2)-1}u^{(a-2)-1}=$$ $$\frac{1}{a-3}u^{a-3}= $$$$\frac{1}{a-3}\tan^{a-3}\theta-\int \tan^{a-4}\theta\ d\theta $$ It goes on like this forever
$$\frac{1}{a-1}\tan^{a-1}\theta-\frac{1}{a-3}\tan^{a-3}\theta-\frac{1}{a-5}\tan^{a-5}\theta \text {....} \theta\ d\theta $$ So then my series should be....$$\frac{1}{a-1}\tan^{a-1}\theta-\big[\sum_{n=3}^{a-2} {\frac{1}{a-n}\tan^{a-n}\theta}\big] -\int \tan^{2}\theta\ d\theta\space \text{if a is even. OR}$$
$$\frac{1}{a-1}\tan^{a-1}\theta-\big[\sum_{n=3}^{a-2} {\frac{1}{a-n}\tan^{a-n}\theta}\big] -\int \tan\ d\theta\space \text{if a is odd}.$$
I am having problems tweaking this series so it only includes the odd values of "n"?
$addendum$
For those not reading the entire thread as discussed below. the value of n is $n=2k-1$.
The lower limit should be $k=1$ and the upper limit should be $\frac {a}{2}-1$ for even "a"'s and $\frac {a-1}{2}$ for odd. Thanks again to SemSem.
|
Simply you have
$$I_a=\int \tan^a d\theta= \int\tan^{a-2}(\sec^2-1)d\theta
\\=\frac{1}{a-1}\tan^{a-1}\theta-\int \tan^{a-2} d\theta
\\ I_a=\frac{1}{a-1}\tan^{a-1}\theta-I_{a-2}$$from this last one you can start by applying it many times to get your required formula. If $n$ is even we get
$$I_a=\frac{1}{a-1}\tan^{a-1}\theta-\frac{1}{a-3}\tan^{a-3}\theta+ \frac{1}{a-5}\tan^{a-5}\theta ....I_{2}
\\=\sum_{k\ =\ 1}^{\frac{a}{2}-1}(-1)^{k+1}\frac{1}{a-(2k-1)}\tan^{a-(2k-1)}\theta+(-1)^{\frac{a}{2}}I_2$$
If $n$ is odd we get
$$I_a=\frac{1}{a-1}\tan^{a-1}\theta-\frac{1}{a-3}\tan^{a-3}\theta+ \frac{1}{a-5}\tan^{a-5}\theta ....I_{1}
\\=\sum_{k\ =\ 1}^{\frac{a-1}{2}}(-1)^{k+1}\frac{1}{a-(2k-1)}\tan^{a-(2k-1)}\theta+(-1)^{\frac{a+1}{2}}I_2$$
|
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|
Derivative of $\sqrt{\frac{9+x}{x}}$ using first principle I am trying to find the derivative of $\sqrt{\dfrac{9+x}{x}}$ using first principle to differentiate.
For sake of simplifying the formulas, I will omit the /h part to the first principles formula, unless someone is able to show me how to integrate it with these equations.
$$\lim_{h\to 0} \frac {f(x+h)-f(x)}{h} $$
What I tried was rationalizing/simplifying first:
$$\dfrac{\sqrt{9+x}}{x} \cdot \dfrac{\sqrt{9+x}}{\sqrt{9+x}}=\dfrac{9+x}{x\sqrt{9+x}}$$
Which when subbed into original first principle formula you get:
$$\dfrac{(9+x+h)}{(x+h)\sqrt{9+x}} - \dfrac{(9+x)}{x\sqrt{9+x}}$$
I then tried to find a common denominator and this was the result:
$$\dfrac{(9x+x^2+h)x\sqrt{9+x}}{x^2+18+2x+xh}- \dfrac{(9x+9h+x^2+hx)\sqrt{9+x}}{x^2+18+2x+xh}$$
After collecting like terms I get:
$$\dfrac{\dfrac{1}{x^2+18+2x+xh} - \dfrac{9h}{x^2+18+2x+xh}}{h}$$
I know it is close because the numerator is close to what the final product looks like:
$$\dfrac{-9}{2 x^2 \sqrt{\dfrac{x+9}{x}}}$$
|
For $f(x)=\sqrt {\dfrac{9+x}{x}}$,
\begin{align}
&\dfrac{f(x+h)-f(x)}{h}
\\=&\dfrac{\sqrt{\dfrac{9+x+h}{x+h}}-\sqrt{\dfrac{9+x}{x}}}{h}
\\\\\\=&\dfrac{{\dfrac{9+x+h}{x+h}}-{\dfrac{9+x}{x}}}{h\left(\sqrt{\dfrac{9+x+h}{x+h}}+\sqrt{\dfrac{9+x}{x}}\right)}\tag{Multiplying by the conjugate}
\end{align}
The numerator simplifies as $\dfrac{9x+x^2+hx-(x^2+9h+hx+9x)}{x(x+h)}=\dfrac{-9h}{x(x+h)}$.
So, $\dfrac{f(x+h)-f(x)}{h}=\dfrac{ \dfrac{-9}{x(x+h)} }{ \left(\sqrt{\dfrac{9+x+h}{x+h}}+\sqrt{\dfrac{9+x}{x}}\right) }$.
Now as $h\to0$, $\dfrac{f(x+h)-f(x)}{h}\to\dfrac{ \dfrac{-9}{x^2} }{2\sqrt{\dfrac{9+x}{x}}}=\dfrac{-9}2 \dfrac{\sqrt{x}}{x^2\sqrt{9+x}}$ which is our sought derivative.
|
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|
If $x+\frac1x=5$ find $x^5+\frac1{x^5}$. If $x>0$ and $\,x+\dfrac{1}{x}=5,\,$ find $\,x^5+\dfrac{1}{x^5}$.
Is there any other way find it?
$$
\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=23\cdot 110.
$$
Thanks
|
Hint:
$$\left( x + \frac{1}{x}\right)^5 = x^5 + \frac{1}{x^5} + 5\left(x^3 + \frac{1}{x^3}\right)
+ 10\left(x + \frac{1}{x} \right)$$
$$x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x} \right)\left(x^2 + \frac{1}{x^2} - 1 \right)$$
$$\left( x + \frac{1}{x}\right)^2 - 2 = x^2 + \frac{1}{x^2}$$
|
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|
Proof of Divisibility of $n(n^2+20)$ by 48. This is a question from Bangladesh National Math Olympiad 2013 - Junior Category that still haunts me a lot. I want to find an answer to this question. Please prove this.
If $n$ is an even integer, prove that $48$ divides $n(n^2+20)$.
|
$48=2^4\cdot 3$, so we will work modulos $16$ and $3$.
We have $n(n^2+20)=n^3+20n$. Now in modulo 3, one notices that $n^3\equiv n\pmod 3$ for any n, so that $n^3+20n\equiv n+20n\equiv 21n\equiv 0\pmod 3$. So we're done with divisibility by 3.
We now need to show divisibility by 16. We have $n(n^2+20)\equiv n(n^2+4)\pmod {16}$. Since $n$ is even, let $n=2k$. We must show $2k(4k^2+4)\equiv 0\pmod {16}$, or $8k(k^2+1)\equiv 0\pmod {16}$. This is equivalent to $k(k^2+1)\equiv 0\pmod 2$, which is obviously true since $\gcd(k,k^2+1)=1$.
Now since $\gcd(3,16)=1$, and $3\mid n(n^3+20)$ and $16\mid n(n^2+20)$, it follows $48=3\cdot 16\mid n(n^2+20)$ and we're done. $\blacksquare$
Arkan
|
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|
Recurrence relations (Big-O notation) Say I'm given a recursive function such as:
function(n) {
if (n <= 1)
return;
int i;
for(i = 0; i < n; i++) {
function(0.8n)
}
}
how would I go about applying recurrence relations to the find the Big-O run time
(as a function of n)?
|
I solved the recurrence relation as the following: $$\text{With: }0.8 = \frac{4}{5}$$
$$
\\
T(n) = \sum_{i = 0}^{n - 1}(T(\frac{4}{5}n) + c), \text{ with } T(n) = 1 \text{ when } n \leq 1
\\\\
T(n) = nT(\frac{4}{5}n) + nc
\\\\
T(n) = (\frac{4}{5})n^2T([\frac{4}{5}]^2n) + (\frac{4}{5})n^2c + nc
\\\\
T(n) = (\frac{4}{5})^2n^3T([\frac{4}{5}]^3n) + (\frac{4}{5})^2n^3c + (\frac{4}{5})n^2c + nc
\\
...
\\
T(n) = (\frac{4}{5})^{k - 1}n^kT([\frac{4}{5}]^kn) + c\sum_{i = 1}^{k}[\frac{4}{5}]^{i - 1}n^i
\\
\text{When } [\frac{4}{5}]^kn = 1, [\frac{4}{5}]^k = [\frac{1}{n}], k = log_{[\frac{4}{5}]}([\frac{1}{n}]), k = -log_{[\frac{4}{5}]}(n)
\\
T(n) = (\frac{4}{5})^{-log_{[\frac{4}{5}]}(n) - 1}n^{-log_{[\frac{4}{5}]}(n)}T(1) + c\frac{5n[(\frac{4}{5})^{-log_{[\frac{4}{5}]}(n)}n^{-log_{[\frac{4}{5}]}(n)} - 1]}{4n - 5}
\\
\text{Therefore, } T(n) \text{ } \in \text{ } \Theta (n^{-log_{[\frac{4}{5}]}(n)})
$$
I wrote a small c program which computes the number of calls of $function()$. and I found that when n = 10, the number of calls was 93616111.
I had to break the program when I executed it with n = 100; it took forever.
Note that the formula of the series above was deduced, thanks to WolframAlpha.
|
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|
Show that the sequence $a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$ is upper bounded Let $\{a_{n}\}$ be defined with $a_{1}\in(0,1)$, and
$$a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$$
for all $n\gt 0$. Show that the sequence is upper bounded.
My idea: since
$$a_{n+1}=a_{n}\left(1+\dfrac{a_{n}}{n^2}\right)$$
then
$$\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}}-\dfrac{1}{a_{n}+n^2}$$
then
$$\dfrac{1}{a_{n}}-\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}+n^2}$$
so
$$\dfrac{1}{a_{1}}-\dfrac{1}{a_{n+1}}=\sum_{i=1}^{n}\dfrac{1}{a_{i}+i^2}$$
since
$$a_{n+1}>a_{n}\Longrightarrow \dfrac{1}{a_{i}+i^2}<\dfrac{1}{a_{1}+i^2}$$
so
$$\dfrac{1}{a_{n+1}}>\dfrac{1}{a_{1}}-\left(\dfrac{1}{1+a_{1}}+\dfrac{1}{2^2+a_{1}}+\cdots+\dfrac{1}{a_{1}+n^2}\right)$$
But the RHS might be $\lt0$ for a sufficiently large starting value; for instance, with $a_{1}=\dfrac{99}{100}$
then
$$\dfrac{1}{a_{1}}-\left(\dfrac{1}{1+a_{1}}+\dfrac{1}{2^2+a_{1}}+\cdots+\dfrac{1}{a_{1}+n^2}\right)<0,n\to\infty$$
see:
so this method won't let me bound the series and I don't know what else to do.
|
We have
$$\frac{a_{n+1}}{a_n} = 1 + \frac{a_n}{n^2}$$
Therefore, if the product
$$\prod_{n=1}^{\infty} ( 1+ \frac{a_n}{n^2})$$
is bounded, then so is $a_n$. Since $1+x \le e^x$, it is enough to show that $\sum_{n\ge 0} \frac{a_n}{n^2} < \infty$. Now, this would follow from $\frac{a_n}{n} < n^{1-\epsilon}$ for some $\epsilon>0$.
Now, for the sequence $b_n\colon = \frac{a_n}{n}$ we have
$$b_{n+1} = \frac{n b_n + b_n^2}{n+1}$$
so by induction $1 > b=b_1 > b_2 > \ldots > 0$. Now we have
$$\frac{b_{n+1}}{b_n} = \frac{1+ \frac{b_n}{n} }{1 + \frac{1}{n}}$$
Now, the partial products of the infinite product
$$\prod_{n=1}^{\infty} \frac{1 + \frac{b}{n}}{1 + \frac{1}{n}}$$
decrease like $n^{b-1}$. We are done.
$\bf{Added:}$ A similar argument shows that the sequence of holomorphic function $\phi_1(z) =z$, $\phi_{n+1}(z) = \phi_n(z) + (\frac{\phi_n(z)}{n})^2$, converges uniformly on compacts to a holomorphic function $\phi(z)$ on the unit disk $D(0, 1)$.
|
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|
How to go about solving this recurrence relation? In my discrete math class we were given the problem of finding an explicit formula for this recurrence relation and proving its correctness via induction.
$$a_n=2a_{n-1}+a_{n-2}+1, a_1 = 1, a_2=1.$$
I feel confident that I could prove the correctness of the formula through induction but I have no idea what the formula is. What kind of strategy should I use for solving recurrence relations like this?
|
Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, write your recurrence as:
$$
a_{n + 2} = 2 a_{n + 1} + a_n + 1
$$
Multiply by $z^n$, sum over $n \ge 0$ and recognize the resulting sums:
$$
\frac{A(z) - a_0 - a_1 z}{z^2} = 2 \frac{A(z) - a_0}{z} + A(z) + \frac{1}{1 - z}
$$
Solve for $A(z)$:
$$
A(z)
= \frac{3}{4} \cdot \frac{1}{1 - (1 + \sqrt{2}) z}
+ \frac{3}{4} \cdot \frac{1}{1 - (1 - \sqrt{2}) z}
- \frac{1}{2} \cdot \frac{1}{1 - z}
$$
Everything in sight are geometric series:
$$
a_n
= \frac{3}{4} \cdot (1 + \sqrt{2})^n
+ \frac{3}{4} \cdot (1 - \sqrt{2})^n
- \frac{1}{2}
$$
Proof by induction is superfluous, the derivation is the proof.
|
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|
Visual explanation of $\pi$ series definition Can you visually explain why the following is true:
$$
\frac{\pi}{4} = \sum\limits_{k=0}^\infty \frac{(-1)^k}{2k + 1} = \frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}\ldots\approx 78.5\%
$$
By visually I mean use a circle and square to help me understand rather than some calculus that I won't understand.
I have recently been very intrigued by $\pi /4$ as it seems to make geometry simpler. For example, the area of a circle is $\pi /4$ times the area of its circumscribed square, and the circle's perimeter is $\pi /4$ times the perimeter of its circumscribed square. See more here: Geometry with $\pi /4$
|
You should know that:
$$\dfrac{\pi}{4}=\tan^{-1}(1)$$
This means that:
$$\dfrac{\pi}{4}=\int_0^1 \dfrac{1}{1+x^2} \ dx$$
We will use the rule that $\dfrac{1}{1+x^2}=1-x^2+x^4-x^6+x^8\dots \ (-1\le x \le 1)$.
$$\dfrac{\pi}{4}=\int_0^1 1 -x^2+x^4-x^6+x^8\dots \ dx$$
We will solve the indefinite integral $\int 1 -x^2+x^4-x^6+x^8\dots \ dx$ first. This is basically power rule repeated an infinite number of times.
$$\int 1 -x^2+x^4-x^6+x^8\dots \ dx$$
$$= x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \dfrac{x^7}{7} + \dfrac{x^9}{9} \dots + C$$
Now we will evaluate the definite integral. We just need to use the Fundamental Theorem of Calculus to do this. The Fundamental Theorem of Calculus is this:
Suppose $G$ is an antiderivative of $f$. Then:
$$\int_a^b f(x) \ dx = G(b) - G(a)$$
To find the definite integral, we just have to plug in the value of $b$ (which is $1$) into the antiderivative (which is basically the answer to the integral) and evaluate it. Then we plug in $a$ to the antideravitave and evaluate it. Finally, we subtract the second value ($G(a)$) from the first value ($G(b)$).
$$\dfrac{\pi}{4}= \left(1 - \dfrac{1^3}{3}+\dfrac{1^5}{5}-\dfrac{1^7}{7}+\dfrac{1^9}{9}\dots + C \right) - \left(0 - \dfrac{0^3}{3}+\dfrac{0^5}{5}-\dfrac{0^7}{7}+\dfrac{0^9}{9}\dots + C \right)$$
$$\dfrac{\pi}{4}= 1 - \dfrac{1^3}{3}+\dfrac{1^5}{5}-\dfrac{1^7}{7}+\dfrac{1^9}{9}\dots + C - C$$
$$\displaystyle \boxed{\dfrac{\pi}{4}= 1 - \dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}\dots}$$
|
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|
How prove this inequality $\sum\limits_{cyc}\sqrt{\frac{\cos{A}\cos{B}}{\cos{C}}}\ge\frac{3\sqrt{2}}{2}$ let $x,y,z>0$,and such $$x^2y^2+y^2z^2+x^2z^2+2x^2y^2z^2=1$$
show that
$$x+y+z\ge\dfrac{3\sqrt{2}}{2}$$
My idea: in $\Delta ABC$,we have
$$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$
so let
$$xy=\cos{A},yz=\cos{B},zx=\cos{C},A,B,C\in(0,\dfrac{\pi}{2})$$
$$\Longleftrightarrow \sqrt{\dfrac{\cos{B}\cos{C}}{\cos{A}}}+\sqrt{\dfrac{\cos{A}\cos{C}}{\cos{B}}}+\sqrt{\dfrac{\cos{A}\cos{B}}{\cos{C}}}\ge\dfrac{3\sqrt{2}}{2}$$
But I can't.Thank you very much
|
The claim is false: If we let $x = 0$, $y=1$ and $z=1$, then we have
$x+y+z = 2 < 3 \sqrt{2} / 2$. (This conclusion holds by continuity for some $x,y,z > 0$, too.)
|
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|
Vector fields on $\Bbb S^2$ with zero construct a vector field on ${S} ^ 2$ with only one zero?
$$\mathbb{S}^2= \left\{(x_1,x_2,x_3) \in R^3: x_1^2+x_2^2+x_3^2=1\right\}$$
|
Write down the stereographic projection from the plane $R^2$ to the sphere, with projection from the north pole. Call that F.
For each $(x, y)$ in $R^2$, take
$$
DF(x, y) \begin{bmatrix} 1/(1 + x^2 + y^2) \\ 0 \end{bmatrix}
$$
That will be a vector at $F(x, y) \in S^2$. It'll be nonzero everywhere except the north pole. Fill in with the zero vector at the north pole.
Addition in response to comment: stereographic projection takes a point $P$ of the plane and joins it by a line to the north pole, $N = (0,0,1)$ of the sphere. That line meets the sphere at another point $F(P)$, called "the stereographic projection of $P$". In coordinates, it's the map
$$
(x, y) \mapsto \frac{1}{x^2 + y^2 + 1} \left( 2x, 2y, x^2 + y^2 - 1 \right).
$$
Added after further work:
Let's use $uv$ coordinates on the plane, $xyz$-coordinates on the sphere. Then projection from the plane to the sphere is
$$
P : R^2 \to S^2 : (u,v) \mapsto \frac{1}{u^2 + v^2 + 1} (2u, 2v, u^2 + v^2 -1)
$$
while projection from the sphere to the plane is
$$
H : S^2 \to R^2 : (x, y, z) \mapsto (\frac{x}{1-z}, \frac{y}{1-z}).
$$
The functions $H$ and $S$ are inverses of one another.
Letting $r$ denote $u^2 + v^2 + 1$ to keep things a little shorter, the derivative of $P$ at $(u, v)$ is
\begin{align}
DP(u, v) &= \begin{bmatrix}
\frac{2r - 4u^2}{r^2} & \frac{-4uv}{r^2} \\
\frac{-4uv}{r^2} & \frac{2r - 4v^2}{r^2}\\
\frac{2ur - 2u(u^2 + v^2 - 1)}{r^2} & \frac{2vr - 2v(u^2 + v^2 - 1)}{r^2}
\end{bmatrix}\\
&= \begin{bmatrix}
\frac{2r - 4u^2}{r^2} & \frac{-4uv}{r^2} \\
\frac{-4uv}{r^2} & \frac{2r - 4v^2}{r^2}\\
\frac{4u}{r^2} & \frac{4v}{r^2}
\end{bmatrix}
\end{align}
If we multiply this by the vector $\begin{bmatrix} 1\\0 \end{bmatrix}$, we'll get a tangent vector to $S^2$ at the point $P(u, v)$, namely
\begin{align}
V_1 &= \begin{bmatrix}
\frac{2r - 4u^2}{r^2} \\
\frac{-4uv}{r^2} \\
\frac{4u}{r^2}
\end{bmatrix}
\end{align}
which is proportional to
\begin{align}
V_2 &= \begin{bmatrix}
{2r - 4u^2} \\
{-4uv} \\
{4u}
\end{bmatrix}
\end{align}
It'd be nice, though, to have that in terms of the sphere point $(x, y, z)$ rather than its $uv$ parameters. Using the substitution provided by $H$, we can replace $u$ and $v$ by $x/(1-z)$ and $y/(1-z)$ and simplify (using $x^2 + y^2 + z^2 = 1$), and scaling up or down as needed to simplify the algebra. We eventually get a vector field $V_3(x, y, z)$ which is proportional to $V_2$, and nonzero everywhere except the north pole, namely
$$
V_3(x, y, z) = \begin{bmatrix}
x^2 + z - 1 \\
xy \\
x(z-1)
\end{bmatrix}.
$$
How do we know that $V_3$ is nonzero except at the poles? Well, the middle entry is nonzero only if $x = 0$ or $y = 0$, i.e., the union of two great circles, each of which passes through the north pole. On the circle $x = 0$, the first term is $-1 + z$, which is zero only when $z = 1$, i.e., at the north pole. On the circle $y = 0$, we have $x \ne 0$ (except at the poles), and $z \ne 1$ (except at the poles), so the third term is nonzero everywhere except the poles.
To summarize: on the union of the circles $x= 0$ and $y = 0$, ignoring the north pole, the first entry of the field is nonzero on $x = 0$, while the third is nonzero on $y = 0$, except at the south pole (at which the first entry is nonzero). In other words, at least one entry of the field is nonzero everywhere except the north pole. At the north pole, all three entries are zero, and the morse index theorem tells us the index must be -2.
Alternatively, you can compute the squared length of the vector $V_2(x, y, z)$ and find that it's $(z-1)^2$, which is evidently nonzero everywhere on the sphere except the north pole.
|
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"timestamp": "2023-03-29T00:00:00",
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Showing that $\sum_{k=0}^{\infty} a^{k} \cos(kx) = \frac{1- a \cos x}{1-2a \cos x + a^{2}}$ without using complex variables The identity $$\sum_{k=0}^{\infty} a^{k} \cos(kx) = \frac{1- a \cos x}{1-2a \cos x + a^{2}} \ , \ |a| <1$$
can be derived by using the fact that $ \displaystyle \sum_{k=0}^{\infty} a^{k} \cos(kx) = \text{Re} \sum_{k=0}^{\infty} (ae^{ix})^{k}$.
But can it be derived without using complex variables?
|
Using the identity,
$$\cos{\left(nx\right)}=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}(-1)^k\binom{n}{2k}\sin^{2k}{\left(x\right)}\cos^{n-2k}{\left(x\right)},$$
the infinite series in question may be rewritten as a double infinite series over a triangle. Changing the order of summation (if you're like me and the transformation gymnastics with multiple indices makes you dizzy, here's a very handy cheat-sheet), we're left with fairly elementary summations:
$$\begin{align}
\sum_{n=0}^{\infty}a^{n}\cos{\left(nx\right)}
&=\sum_{n=0}^{\infty}a^{n}\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}(-1)^k\binom{n}{2k}\sin^{2k}{\left(x\right)}\cos^{n-2k}{\left(x\right)}\\
&=\sum_{n=0}^{\infty}\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}(-1)^k\binom{n}{2k}a^{n}\sin^{2k}{\left(x\right)}\cos^{n-2k}{\left(x\right)}\\
&=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}(-1)^k\binom{2k+n}{2k}a^{2k+n}\sin^{2k}{\left(x\right)}\cos^{n}{\left(x\right)}\\
&=\sum_{k=0}^{\infty}(-1)^ka^{2k}\sin^{2k}{\left(x\right)}\sum_{n=0}^{\infty}\binom{2k+n}{2k}\left[a\cos{\left(x\right)}\right]^n\\
&=\sum_{k=0}^{\infty}(-1)^ka^{2k}\sin^{2k}{\left(x\right)}\frac{1}{\left(1-a\cos{\left(x\right)}\right)^{2k+1}}\\
&=\frac{1}{1-a\cos{\left(x\right)}}\sum_{k=0}^{\infty}(-1)^k\left[\frac{a\sin{\left(x\right)}}{1-a\cos{\left(x\right)}}\right]^{2k}\\
&=\frac{1}{1-a\cos{\left(x\right)}}\cdot\frac{1}{1+\left[\frac{a\sin{\left(x\right)}}{1-a\cos{\left(x\right)}}\right]^{2}}\\
&=\frac{1}{1-a\cos{\left(x\right)}}\cdot\frac{\left(1-a\cos{\left(x\right)}\right)^2}{\left(1-a\cos{\left(x\right)}\right)^2+a^2\sin^2{\left(x\right)}}\\
&=\frac{1-a\cos{\left(x\right)}}{1-2a\cos{\left(x\right)}+a^2\cos^2{\left(x\right)}+a^2\sin^2{\left(x\right)}}\\
&=\frac{1-a\cos{\left(x\right)}}{1-2a\cos{\left(x\right)}+a^2}.~~\blacksquare\\
\end{align}$$
Happy holidays!
|
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|
Sum this series $\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\ldots$ upto $n$ terms
Sum this series: $$\dfrac{1}{1+1^2+1^4}+\dfrac{2}{1+2^2+2^4}+\ldots$$ upto $n$ terms.
My approach:
$$(1-n^6)=(1-n^2)(1+n^2+n^4)\implies \dfrac{n}{1+n^2+n^4}=\dfrac{n(1-n^2)}{1-n^6}$$
So, the above series can be written as $$\sum\limits_{i=1}^n \dfrac{i(1-i^2)}{1-i^6}$$
I suppose that this can now be converted into integration which I cannot apparently. Please help. It would be better if the solution is not based upon integration but algebra.
|
HINT:
As $\displaystyle1+n^2+n^4=(1+n^2)^2-n^2=(1+n+n^2)(1-n+n^2)$
$$\frac{2n}{1+n^2+n^4}=\frac{(1+n+n^2)-(1-n+n^2)}{(1+n+n^2)(1-n+n^2)}=\cdots$$
Again if $\displaystyle f(n)=\frac1{n^2-n+1}, f(n+1)=?$
So, we are dealing with a Telescoping series
|
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|
Proof Using the Monotone Convergence Theorem for the sequence $a_{n+1} = \sqrt{4 + a_n}$ Consider the recursively defined sequence $a_0 = 1$
$a_{n+1} = \sqrt{4 + a_n}$
How to prove that the sequence converges using the Monotone Convergence Theorem, and find the limit?
|
Step 1. $\{a_n\}$ is strictly increasing, i.e., $a_{n+1}> a_n$.
This is shown inductively. For $n=1$, We have that $$a_2=\sqrt{4+a_1}=\sqrt{4+1}=\sqrt{5}>1=a_1.$$
Assume now that it is true for $n=k$, i.e., $a_{k+1}>a_{k}$. Then $a_{k+1}+4>a_{k}+4$, and hence
$$
a_{k+2}=\sqrt{4+a_{k+1}}>\sqrt{4+a_{k}}=a_{k+1},
$$
and thus it is true for $n=k+1$. Hence indeed, $\{a_n\}$ is increasing.
Step 2. $\{a_n\}$ is upper bounded.
In particular, we shall show that $a_n<3$ inductively. So, for $k=1$, we have that $a_1=1<3$.
Assume that that it is true for $n=k$ and $a_k<3$. Then $a_{k+1}=\sqrt{4+a_k}<\sqrt{4+3}=\sqrt{7}<3$. Thus $\{a_n\}$ is upper bounded by 3.
Step 1.+Step 2. imply that $\{a_n\}$ converges, as an increasing and upper bounded sequence does converge.
Step 3. Finding of the limit of $\{a_n\}$.
Assume that $a_n\to a$, then $a_n+4\to a+4$ and $a_{n+1}=\sqrt{a_n+4}\to\sqrt{a+4}$.
But $\{a_n\}$ and $\{a_{n+1}\}$ have the same limit, and hence
$$
a=\sqrt{a+4},
$$
which implies that
$$
a^2-a-4=0,
$$
and hence
$$
a=\frac{1\pm\sqrt{17}}{2}.
$$
The negative root is rejected as the terms of $\{a_n\}$ are positive.
Thus
$$
a_n\to\frac{1+\sqrt{17}}{2}.
$$
|
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|
What is the coefficient of the term $x^4 y^5$ in $(x+y+2)^{12}$? What is the coefficient of the term $x^4 y^5$ in $(x+y+2)^{12}$?
How can we calculate this expression ?
I've applied the binomial theorem formula and got $91$ terms but I am not sure if it is right or wrong.
|
You could use the binomial theorem twice. Let $[x^{k}]$ denote the coefficient of $x^k$ from the polynomial $P(x)=\sum_{j=0}^{n}a_jx^j$, i.e. $[x^{k}]P(x)=a_{k}$.
Now, $$\begin{eqnarray}[x^4y^5](x+y+2)^{12}&=&[x^4y^5](x+(y+2))^{12}\\&=&[x^4y^5]\sum_{j=0}^{12}\binom{12}{j}x^j(y+2)^{12-j}\\&=&[y^5]\binom{12}{4}(y+2)^{12-4}\\&=&\binom{12}{4}[y^5](y+2)^8\\&=&\binom{12}{4}[y^5]\sum_{j=0}^{8}\binom{8}{j}y^j2^{8-j}\\&=&\binom{12}{4}\binom{8}{5}2^{8-5}\\&=&\frac{12!}{4!8!}\frac{8!}{5!3!}2^3=\frac{12!2^3}{3!4!5!}\end{eqnarray}$$
|
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|
How do i simplify this $\cos{\theta}=\frac{1-\sin^2{\theta}}{1+\cos{\theta}}$ I was wondering if someone could help me solve this equation: $$\cos{\theta}=\frac{1-\sin^2{\theta}}{1+\cos{\theta}}$$
|
$\cos{\theta}=\dfrac{1-\sin^2{\theta}}{1+\cos{\theta}}$:
1.) Multiply through by $1 + \cos \theta$:
$\cos \theta + \cos^2 \theta = 1 - \sin^2 \theta$;
2.) Add $\sin^2 \theta$ to each side:
$\cos \theta + \cos^2 \theta + \sin^2 \theta = 1$;
3.) Use $\cos^2 \theta + \sin^2 \theta = 1$ on the left:
$\cos \theta + 1 = 1$;
4.) Subtract $1$ throughout:
$\cos \theta = 0$:
5.) Deduce from item (4) that $\theta$ must be an odd multiple of $\pi / 2$: $\theta = (2n + 1)\pi / 2$ for $n$ an integer.
And that's it!
Hope this helps! Cheerio,
and as always,
Fiat Lux!!!
|
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|
Help me prove the result of this limit: $\lim_{x\to \infty} {\log_{x^3+1} (x^2+1) \over \log_{2e^x+1} (x^2+5x)}$ The limit is
$$\lim_{x\to \infty} {\log_{x^3+1} (x^2+1) \over \log_{2e^x+1} (x^2+5x)}$$
I know that it should be equal to $\infty$ but i have yet to prove it. Please help me do so.
|
Hint 1: $\displaystyle \log_a b = \frac{\ln b}{\ln a}$. So $$\frac{\log_{x^3+1} (x^2+1)}{\log_{2e^x+1} (x^2+5x)} = \frac{\frac{\ln (x^2+1)}{\ln (x^3+1)}}{\frac{\ln (x^2+5x)}{\ln (2e^x+1)}} = \frac{\ln (x^2+1) \ln (2e^x+1)}{\ln (x^3+1) \ln (x^2+5x)}$$
Hint 2: For very large $x$, $x^2+1 \approx x^2$ and $x^3 + 1 \approx x^3$. So our equation becomes $$\frac{\ln (x^2+1) \ln (2e^x+1)}{\ln (x^3+1) \ln (x^2+5x)} \approx \frac{\ln (x^2) \ln (2e^x+1)}{\ln (x^3) \ln (x^2+5x)} \approx \frac{2\ln (x) \ln (2e^x+1)}{3\ln (x) \ln (x^2+5x)} \approx \frac{2}{3} \frac{\ln (2e^x+1)}{\ln (x^2+5x)}$$
Can you take it from here now?
|
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|
Reducing the System of linear equations \begin{align*}
x+2y-3z&=4 \\
3x-y+5z&=2 \\
4x+y+(k^2-14)z&=k+2
\end{align*}
I started doing the matrix of the system:
$$ \begin{pmatrix} 1 & 2 & -3 & 4 \\ 3 & -1 & 5 & 2 \\ 4 & 1 & k^2-14 & k+2 \end{pmatrix} $$
But I still don't know how can I reduce this, could you explain me what steps I should follow?
Which values of $k$ make the system inconsistent, determined, undetermined?
|
This is not very different from other answers. And I think that the approach using Gaussian elimination is more systematic. I wanted to show how this can be done with a little of guesswork.
If you notice that the sum of the first two equations is $4x+y+2z=6$, then you see that every solution of your system also fulfills the equations
\begin{align*}
4x+y+2z&=6 \\
4x+y+(k^2-14)z&=k+2
\end{align*}
Now if you subtract these equations you get:
$$(k^2-16)z=k-4$$
which is the same as $(k-4)(k+4)z=(k-4)$.
If $\boxed{k=4}$, then this last equation is $0=0$ and you system is equivalent to the system
\begin{align*}
x+2y-3z&=4 \\
3x-y+5z&=2
\end{align*}
If $\boxed{k=-4}$, then the this equation is $0=-8$, and the system does not have solution.
If $\boxed{k\ne\pm4}$, then the original system is equivalent to the following system of equations:
\begin{align*}
x+2y-3z&=4 \\
3x-y+5z&=2 \\
(k^2-16)z&=k-4
\end{align*}
Since $k^2-16\ne0$, the variable $z$ is uniquely determined by the last equation. Since the matrix $\begin{pmatrix}1&2\\3&-1\end{pmatrix}$ is regular, the values of $x$ and $y$ are the also uniquely determined.
|
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|
Determine the convergence of a sequence given by $a_n= \frac{a_{n-1} + a_{n-2}}2$ let $a_0$ and $a_1$ be any two real numbers, and define
$a_n= \dfrac{a_{n-1} + a_{n-2}}{2}$
Determine the convergence of a sequence.
Alright, what I have so far. I have two cases $a_0$ > $a_1$ or $a_0$ < $a_1$
for $a_0 > a_1$
The sequence converges to $\frac{1}{3}$($a_0 - a_1$) + $a_0$
for $a_0 < a_1$
The sequence converges to $\frac{2}{3}$($a_1 - a_0$) + $a_0$
Any additional input?
|
$a_n = \dfrac{a_{n-1}+a_{n-2}}{2}$. One overkill way of finding the limit, is getting the $a_n$ in terms of $n$ via a generating function.
Write $F(x) = \sum_{i=0}^{\infty}a_i x^i$. Thus,
$$
\frac{F(x)-a_0}{x}=\sum_{i=0}^{\infty}a_{i+1}x^{i}
$$
$$
\frac{F(x)-a_0-a_1}{x^2} = \sum_{i=0}^{\infty}a_{i+2}x^{i}
$$
So,
$$
\frac{F(x)-a_0-a_1}{x^2} = \frac{1}{2}\cdot \frac{F(x)-a_0}{x} + \frac{1}{2}F(x)
$$
Solving for $F(x)$, one obtains:
$$
\frac{(x-2)a_0-2a_1}{x^2+x-2}=F(x)
$$
Using partial fractions, one then obtains:
$$
F(x) = \frac{-(a_0+2a_1)}{3(x-1)}+\frac{2(2a_0+a_1)}{3(x+2)}
$$
Now write each term as a power series, obtaining:
$$
F(x) =\frac{(a_0+2a_1)}{3}\sum_{k=0}^{\infty} x^k +\frac{(2a_0+a_1)}{3}\sum_{k=0}^{\infty} \left(-\frac{1}{2}\right)^k(x)^k
$$
Now, $a_{n+1}$ is the $(n+1)^{st}$ coefficient and therefore the coefficient of $x^n$ given above. Explicitly,
$$
a_{n+1} = \frac{(a_0+2a_1)}{3}+\frac{(2a_0+a_1)}{3}\left(\frac{-1}{2}\right)^n
$$
e.g. if $a_0=0, a_1=1$, we have:
$$
a_2=\frac{1}{2},a_3=\frac{\frac{1}{2}+1}{2}=\frac{3}{4},a_4=\frac{\frac{3}{4}+\frac{1}{2}}{2}=\frac{5}{8}
$$
Now we expect that $$a_4 = \frac{(a_0+2a_1)}{3}+\frac{(2a_0+a_1)}{3}\left(\frac{-1}{2}\right)^3 = \frac{5}{8}$$
And
$$
\frac{(a_0+2a_1)}{3}+\frac{(2a_0+a_1)}{3}\left(\frac{-1}{2}\right)^3= \\\frac{2}{3}+\frac{1}{3}\left(\frac{-1}{2}\right)^3=\frac{5}{8}
$$
Now, it is easy to see that
$$
\lim_{n\to \infty}a_{n+1} = \lim_{n\to \infty} \frac{(a_0+2a_1)}{3}+\frac{(2a_0+a_1)}{3}\left(\frac{-1}{2}\right)^n = \frac{(a_0+2a_1)}{3}
$$
|
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|
Contour integration question help I am trying to evaluate the following integral using complex analysis:
$$\int^{2\pi}_0\frac{\cos(2x)}{5-4\cos(x)}dx$$
I have written it in the following form:
$$\frac{-i}{2}\int_{|z|<1}\frac{z^4+1}{z^2(z-2)(1-2z)}dz$$
However, the residue for $z=1/2$ is $-17/6$ whereas the residue for the double pole at $z=0$ is $-5/4$ and when the sum of these residues is multiplied by $2\pi i\times\frac{-i}{2}=\pi$ the wrong result is obtained (i.e. not $\pi/6$).
What is it that I am doing wrong?
|
\begin{align}
\operatorname{Res}\limits_{z=1/2} \frac{z^4+1}{z^2(z-2)(1-2z)} &=
\lim_{z\to 1/2} (z-\frac12) \cdot \frac{z^4+1}{z^2(z-2)(1-2z)} \\
&= \lim_{z\to 1/2} \frac{z^4+1}{z^2(z-2)\cdot(-2)} = \frac{(1/2)^4+1}{(1/2)^2\cdot(-3/2)\cdot(-2)} = \frac{17}{12}
\end{align}
|
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|
If $a,b,c>0$ and $a+b+c=1$ prove inequality: $\frac a{b^2 +c} + \frac b{c^2+a} + \frac c{a^2+b} \ge \frac 94$ If $a,b,c>0$ and $a+b+c=1$ prove inequality: $$\frac a{b^2 +c} + \frac b{c^2+a} + \frac c{a^2+b} \ge \frac 94$$
|
By C-S
$$\sum_{cyc}\frac{a}{b+c^2}=\sum_{cyc}\frac{a^2}{ab+ac^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+a^2b)}.$$
Thus, it remains to prove that
$$4(a+b+c)^2\geq9(ab+ac+bc)+9(a^2b+b^2c+c^2a)$$ or
$$4(a+b+c)^3\geq9(ab+ac+bc)(a+b+c)+9(a^2b+b^2c+c^2a)$$ or
$$\sum_{cyc}(4a^3-6a^2b+3a^2c-abc)\geq0$$
and since by AM-GM $\sum\limits_{cyc}a^2c\geq3abc$, it remains to prove that
$$\sum_{cyc}(2a^3-3a^2b+a^2c)\geq0$$ or
$$\sum_{cyc}(2a^3-3a^2b+ab^2)\geq0$$ or
$$\sum_{cyc}a(a-b)(2a-b)\geq0$$ or
$$\sum_{cyc}\left(a(a-b)(2a-b)-\frac{1}{3}(a^3-b^3)\right)\geq0$$ or
$$\sum_{cyc}(a-b)^2(5a+b)\geq0.$$
Done!
|
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|
Proof this curious trigonometric identity Proof that $$\cos^2{10^\circ} + \cos^2{50^\circ} - \sin{40^\circ}\sin{80^\circ} = \frac{3}{4}$$
I notice that $10^\circ + 80^\circ = 90^\circ$, and $50^\circ +40^\circ = 90^\circ$. I tried doing some manipulation but my efforts were futile. Any hints?
|
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
$\displaystyle\cos^210^\circ+\cos^250^\circ=\cos^210^\circ-\sin^250^\circ+1=\cos60^\circ\cos40^\circ+1=\frac{\cos40^\circ}2+1$
Use Werner Formulas to find $\displaystyle2\sin40^\circ\sin80^\circ=\cos40^\circ-\cos120^\circ$
Now, $\displaystyle\cos120^\circ=\cos(180^\circ-60^\circ)=-\cos60^\circ=-\frac12$
Can you take it home from here?
Alternatively, $\displaystyle\cos^210^\circ+\cos^250^\circ-\sin40^\circ\sin80^\circ=\cos^210^\circ+\cos^250^\circ-\cos50^\circ\cos10^\circ$
$\displaystyle=\left(\cos10^\circ-\cos50^\circ\right)^2+\cos50^\circ\cos10^\circ$
Using Prosthaphaeresis Formulas, $\displaystyle\cos10^\circ-\cos50^\circ=2\sin30^\circ\sin20^\circ=\sin20^\circ$
Now use Double-Angle Formulas i.e., $\displaystyle\sin^2x=\frac{1-\cos2x}2$ for $x=20^\circ$
Using Werner Formulas, $\displaystyle2\cos50^\circ\cos10^\circ=\cos40^\circ+\cos60^\circ$
|
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|
Calculate a recursive equation in terms of theta I am struggling with the following equation for one week!
Please help me solve it.
$$T(n)=T(\frac{n}{2})+\frac{n}{logn}$$
So far, I have come to the equation $T(n)=\Sigma \frac{2^x}{x}$
|
Suppose you have $T(0)=0$ and $T(1) = 1$ and your recurrence for $n>1$ is
$$T(n) = T(\lfloor n/2 \rfloor) + \frac{n}{\lfloor\log_2 n \rfloor}.$$
By unrolling the recursion we find the exact formula for $n\ge 2:$
$$T(n) = 1 +
\sum_{k=0}^{\lfloor\log_2 n \rfloor - 1} \frac{1}{\lfloor\log_2 n \rfloor - k}
\sum_{j=k}^{\lfloor\log_2 n \rfloor} d_j 2^{j-k}$$
where the digits $d_j$ are from the binary representation of $n$ given by
$$n = \sum_{j=0}^{\lfloor\log_2 n \rfloor} d_j 2^j.$$
We now compute upper and lower bounds. For an upper bound consider a string of one digits, which gives
$$T(n)\le 1 +
\sum_{k=0}^{\lfloor\log_2 n \rfloor - 1} \frac{1}{\lfloor\log_2 n \rfloor - k}
\sum_{j=k}^{\lfloor\log_2 n \rfloor} 2^{j-k}
= 1 +
\sum_{k=0}^{\lfloor\log_2 n \rfloor - 1} \frac{1}{\lfloor\log_2 n \rfloor - k}
\left(2^{\lfloor\log_2 n \rfloor+1-k} -1\right)
\\ = 1 - H_{\lfloor\log_2 n \rfloor} +
\sum_{k=0}^{\lfloor\log_2 n \rfloor - 1}
\frac{2^{\lfloor\log_2 n \rfloor+1-k}}{\lfloor\log_2 n \rfloor - k}
= 1 - H_{\lfloor\log_2 n \rfloor} +
2\times \sum_{k=1}^{\lfloor\log_2 n \rfloor}\frac{2^k}{k}.$$
To extract the asymptotics from this we use the next-to-last form, which gives
$$1 - H_{\lfloor\log_2 n \rfloor} +
\frac{2^{\lfloor\log_2 n \rfloor+1}}{\lfloor\log_2 n \rfloor}
\sum_{k=0}^{\lfloor\log_2 n \rfloor - 1}
\frac{2^{-k}}{1 - \frac{k}{\lfloor\log_2 n \rfloor}}
\\ = 1 - H_{\lfloor\log_2 n \rfloor} +
\frac{2^{\lfloor\log_2 n \rfloor+1}}{\lfloor\log_2 n \rfloor}
\left(
\sum_{k=0}^{\lfloor\log_2 n \rfloor - 1} 2^{-k}
+ \sum_{k=0}^{\lfloor\log_2 n \rfloor - 1} \frac{k 2^{-k}}{\lfloor\log_2 n \rfloor}
+ \sum_{k=0}^{\lfloor\log_2 n \rfloor - 1} \frac{k^2 2^{-k}}{\lfloor\log_2 n \rfloor^2}
+\cdots\right)
\\= 1 - H_{\lfloor\log_2 n \rfloor} \\+
\frac{2^{\lfloor\log_2 n \rfloor+1}}{\lfloor\log_2 n \rfloor}
\left(
\sum_{k=0}^{\lfloor\log_2 n \rfloor - 1} 2^{-k}
+ \frac{1}{\lfloor\log_2 n \rfloor}\sum_{k=0}^{\lfloor\log_2 n \rfloor - 1}k 2^{-k}
+ \frac{1}{\lfloor\log_2 n \rfloor^2}\sum_{k=0}^{\lfloor\log_2 n \rfloor - 1} k^2 2^{-k}
+\cdots\right).$$
The sums all converge to numbers so we finally get
$$1 - H_{\lfloor\log_2 n \rfloor} +
\frac{2^{\lfloor\log_2 n \rfloor+1}}{\lfloor\log_2 n \rfloor}
\left(2
+ \frac{2}{\lfloor\log_2 n \rfloor}
+ \frac{6}{\lfloor\log_2 n \rfloor^2}
+ \frac{26}{\lfloor\log_2 n \rfloor^3}
+\cdots\right).$$
Continuing with the lower bound which occurs for a one digit followed by a string of zeroes we get
$$T(n)\ge 1 +
\sum_{k=0}^{\lfloor\log_2 n \rfloor - 1} \frac{1}{\lfloor\log_2 n \rfloor - k}
2^{\lfloor\log_2 n \rfloor-k}
= 1 +
\sum_{k=1}^{\lfloor\log_2 n \rfloor} \frac{2^k}{k}.$$
To extract the asymptotics we work with the next-to-last form, which gives
$$1 + \frac{2^{\lfloor\log_2 n \rfloor}}{\lfloor\log_2 n \rfloor}
\sum_{k=0}^{\lfloor\log_2 n \rfloor - 1}
\frac{2^{-k}}{1 - \frac{k}{\lfloor\log_2 n \rfloor}}.$$
The sum is the same as before so we finally have the lower bound
$$1 + \frac{2^{\lfloor\log_2 n \rfloor}}{\lfloor\log_2 n \rfloor}
\left(2
+ \frac{2}{\lfloor\log_2 n \rfloor}
+ \frac{6}{\lfloor\log_2 n \rfloor^2}
+ \frac{26}{\lfloor\log_2 n \rfloor^3}
+\cdots\right).$$
Note that these bounds are attained. Joining the upper and the lower bound we have a complexity of
$$T(n)\in\Theta\left(\frac{2^{\lfloor\log_2 n \rfloor}}{\lfloor\log_2 n \rfloor}\right)
= \Theta\left(\frac{n}{\log n}\right)$$
which to be clear about it could have been spotted by inspection, but there is no harm in doing a formal proof.
Remark. The reader may profit from studying the calculation at this MSE link which is similar but just different enough to create a learner's effect. (Unfortunately it has $j$ and $k$ in the explicit formula reversed but this does not warrant editing everything I have just written.)
Observation. When using the asymptotic expansion for fixed $n$ care must be taken because the coefficients eventually outgrow the inverse powers of the logarithm. The best value to use is the one just before the expansion starts diverging.
Addendum. The sequence of coefficients $2,2,6,26,\ldots$ in the asymptotic expansion is examined in detail at the following MSE link.
|
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|
$3^{2x} - 34(15^{x-1}) + 5^{2x} = 0$ I've never seen anything like this. Do somebody have a way to solve it?
I've tried the basci exponential functions techniques but it does not work. Even substitution does not work...
I'm really intersted in learning, this is not a homework.
|
Another way to look at this is to consider that $ \ a^{2x} = (a^x)^2 \ , $ from the properties of exponents. Suppose you had a binomial square $ \ (3^x + 5^x)^2 \ $ then: its expansion would be
$$ \ (3^x)^2 \ + \ 2 \cdot 3^x \cdot 5^x \ + \ (5^x)^2 \ = \ 3^{2x} \ + \ 2 \cdot (3 \cdot 5)^x \ + \ 5^{2x} \ = \ 3^{2x} \ + \ 2 \cdot 15^x \ + \ 5^{2x} \ \ . $$
Now, our equation is $ \ 3^{2x} - 34(15^{x-1}) + 5^{2x} = 0 \ , $
which we can arrange as
$$ 3^{2x} \ - \ 34 \ (15^x \cdot 15^{-1}) \ + \ 5^{2x} \ = \ 0 \ \ \Rightarrow \ \ 3^{2x} \ - \ \frac{34}{15} \cdot 15^x \ + \ 5^{2x} \ = \ 0 $$
$$ \Rightarrow \ \ 3^{2x} \ + \ (\frac{30}{15} - \frac{64}{15}) \cdot 15^x \ + \ 5^{2x} \ = \ 0 \ \ \Rightarrow \ \ (3^{2x} \ + \ 2 \cdot 15^x \ + \ 5^{2x}) \ = \ \frac{64}{15} \cdot 15^x $$
$$ \Rightarrow \ \ (3^x \ + \ 5^x)^2 \ = \ \frac{64}{15} \cdot \ 3^x \cdot \ 5^x \ = \ \frac{64}{15} \cdot \ [ \ (\sqrt{3})^x \ ]^2 \cdot \ [ \ (\sqrt{5})^x\ ]^2 $$
$$ \Rightarrow \ \ \left( \ \left[ \ \sqrt{ \frac{3}{5}} \ \right]^x + \ \left[ \ \sqrt{\frac{5}{3}} \ \right]^x \ \right)^2 = \ \frac{64}{15} $$
$$\Rightarrow \ \ \left( \sqrt{\frac{3}{5}} \ \right)^x + \ \left( \ \sqrt{ \frac{5}{3}} \ \right)^x \ = \ \frac{8}{\sqrt{15}} \ \ = \ \frac{3 + 5}{\sqrt{3 \ \cdot \ 5}} \ \ . $$
It's reasonably clear from this result that $ \ x = 1 \ $ is a solution by simply adding the square-root ratios. But because those ratios are reciprocals of one another, we see that $ \ x = -1 \ $ is also a solution. So we have located the two roots of this "quadratic" equation.
This approach has little to recommend itself as a general technique: it simply works out nicely because of the "middle" coefficient in the equation.
|
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|
How do I prove this trigonmetric identity? I need to prove that the following identity is true:
$$
\frac{\cos^2x-\sin^2x}{1-\tan^2x}=\cos^2x
$$
This isn't homework; just a practice exercise. But I keep getting stuck! Thanks much.
|
This is sort of the "reverse" of MPW's suggestion, using the difference of two squares:
$$ \frac{\cos^2 x \ - \ \sin^2 x}{1-\tan^2x} \ = \ \frac{(\cos x - \sin x) \ (\cos x + \sin x) }{(1 - \tan x) \ (1 + \tan x)} $$
$$ = \ \frac{(\cos x - \sin x) \ (\cos x + \sin x) }{(1 - \tan x) \ (1 + \tan x)} \ \cdot \ \frac{\cos^2 x}{\cos^2 x} $$
$$ = \ \frac{(\cos x - \sin x) \ (\cos x + \sin x) }{\cos x \cdot (1 - \tan x) \cdot \cos x \cdot (1 + \tan x)} \ \cdot \ \frac{\cos^2 x}{1} $$
$$= \ \frac{(\cos x - \sin x) \ (\cos x + \sin x) }{(\cos x - \sin x) \ (\cos x + \sin x)} \ \cdot \ \cos^2 x \ = \ \cos^2x \ \ . $$
We could also use two versions of the Pythagorean Identity to write
$$ \frac{\cos^2 x \ - \ \sin^2 x}{1 \ - \ \tan^2x} \ = \ \frac{\cos^2 x \ - \ [ 1 - \cos^2 x]}{1 - \ [\sec^2 x - 1]} \ = \ \frac{2 \cos^2 x \ - \ 1 }{2 \ - \ \sec^2 x } $$
$$ = \ \frac{2 \cos^2 x \ - \ 1 }{2 \ - \ \sec^2 x } \ \cdot \ \frac{\cos^2 x}{\cos^2 x} \ = \ \frac{2 \cos^2 x \ - \ 1 }{ ( 2 - \sec^2 x ) \ \cos^2 x } \ \cdot \ \frac{\cos^2 x}{1} $$
$$ = \ \frac{2 \cos^2 x \ - \ 1 }{2 \cos^2 x \ - \ 1 } \ \cdot \ \cos^2 x \ = \ \cos^2 x \ \ . $$
|
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|
Solve $7\sin^2(x) - 9\cos(2x) = 0$ I need to solve for x in the polynomial
$$7\sin^2(x) - 9\cos(2x) = 0$$
I have tried approaching the problem in multiple ways. I am only looking for some hints, not the actual answer. Thanks :D
|
There are two ways to solve this all leading to the same answer:
1) 7sin^2(x)−9cos(2x)=0
using $cos(2x) = $cos^2x - sin^2x$
=> $7sin^2(x)−9(cos^2x - sin^2x)=0$ {as cos(2x) = $cos^2x - sin^2x$}
=> $7sin^2(x)−9(cos^2x - sin^2x)=0$
=> $7sin^2(x)−9(1 - sin^2x - sin^2x)=0$ {as $cos^2x = 1 - sin^2x$}
=> $7sin^2(x)−9(1 - 2sin^2x)=0$
=> $7sin^2(x)−9 + 18sin^2x)=0$
=> $25sin^2x=9$
=> $sinx=(3/5)$ or $sinx= -(3/5)$
=> x = 0.643501109 rad or x = 36.8698976 degrees
2) 7sin^2(x)−9cos(2x)=0
using $sin^2(x)=(1−cos(2x))/2$
=> $7(1−cos(2x))/2−9cos(2x)=0$
=> $7−7cos(2x)−18cos(2x)=0$
=> $7=25cos(2x)$
=> $cos(2x) = 7/25$
=> $2x= arccos(7/25)$ {which is 1.28700222 rad or 73.7397953 degrees}
=> $x=$ 0.64350111 rad or x = 36.86989765 degrees
Ref : http://en.wikipedia.org/wiki/List_of_trigonometric_identities
|
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|
Values of the limit $\lim\limits_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$ Hi I have a question regarding finding the values of limit for the following question.
Let $a, b \in \mathbb R$. Find the limit
$$\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$
|
When you have these kinds of limits it's always a good idea to multiply by conjugate. In your case you have:
$$
\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)} - x\right) = \lim_{x\to+\infty}\frac{\left(\sqrt{(x+a)(x+b)} - x\right)\left(\sqrt{(x+a)(x+b)} + x\right)}{\sqrt{(x+a)(x+b)} + x} = \\ = \lim_{x\to+\infty}\frac{(x+a)(x+b) - x^2}{\sqrt{(x+a)(x+b)} + x} = \lim_{x\to+\infty}\frac{x^2+ax+bx+ab-x^2}{\sqrt{(x+a)(x+b)} + x} = \\
= \lim_{x\to+\infty}\frac{x\left(a+b+\frac{ab}{x}\right)}{x\left(\sqrt{\left(1+{a\over x}\right)\left(1+{b\over x}\right)} + 1\right)} = \lim_{x\to+\infty}\frac{a+b+{ab\over x}}{\sqrt{1+{a+b\over x} + {ab \over x^2}}+1}
$$
So as you see canceling $x$ from both nominator and denominator is going to produce the desired result.
In general consider functions in the form:
$$
\begin{cases}
f(x) = \sqrt[n]{(x + a_1)(x+a_2)\cdots (x+a_n)} - x \\
n,k \in \Bbb N \\
a_k \in \Bbb R
\end{cases}
$$
We know that:
$$
a^n – b^n = (a – b)\left(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 +
\cdots + ab^{n – 2} + b^{n – 1}\right)
$$
Denote:
$$
g(x) = \sqrt[n]{(x + a_1)(x+a_2)\cdots (x+a_n)}
$$
So based on that $f(x)$ may be rewritten as:
$$
\lim_{n\to+\infty}f(x) = \lim_{n\to+\infty}(g(x) - x) = \lim_{n\to+\infty} \frac{(g(x))^n-x^n}{\sqrt[n]{(g(x))^{n-1}} +\sqrt[n]{(g(x))^{n-2}}x + \sqrt[n]{(g(x))^{n-3}}x^2 + \cdots + \sqrt[n]{g(x)}x^{n-2} +x^{n-1}} = \\
=\lim_{n\to+\infty} \frac{x^{n-1}a_1 + x^{n-1}a_2+ \cdots + x^{n-1}a_n + \cdots}{x^{n-1}\left(\sqrt[n]{1+o\left({1\over x}\right)+\cdots} + \sqrt[n]{1+o\left({1\over x}\right) +\cdots} +\cdots + 1\right)} = \\
= \frac{a_1 + a_2 + \cdots + a_n}{n}
$$
|
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|
Can $\frac {100-100}{100-100}=2$? \begin{align*}
\frac{0}{0} &= \frac{100-100}{100-100} \\
&= \frac{10^2-10^2}{10(10-10)} \\
&= \frac{(10+10)(10-10)}{10(10-10)} \\
&= \frac{10+10}{10} \\
&= \frac{20}{10} \\
&= 2
\end{align*}
I know that $0/0$ isn't equal to $2$, the what is wrong in this proof? I wasn't satisfied after searching it on Google, any help would be appreciated.
|
I think the best way to see the folly here is to work backwards. The problem occurs when we encounter the "equality"
$$
\frac{10+10}{10}=\frac{(10+10)(10-10)}{10(10-10)}
$$
But to obtain this equality, we must multiply the rational number $\displaystyle\frac{10+10}{10}$ by the undefined expression $\displaystyle\frac{0}{0}$.
|
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|
Probability [Distinct balls in distinct boxes] 6 different balls are put in 3 different boxes, no box being empty. The probability of putting balls in the boxes in equal numbers is ? Could anyone pleases give the answer and explain it.
|
Note that the balls and boxes are both distinct.
Select 2 balls in each of 3 boxes, count the permutations of the six balls. Ways to arrange 6 balls, 2 in each box: $N(2,2,2)=\frac{6!}{2!2!2!}$ That's permutations of all six balls, divided by permutations of balls in each box (since the order in the box does not matter).
Select the 1 box to have 4 balls, permute the balls. Ways to arrange 6 balls, 1 in each of 2 boxes and 4 in the third: $N(1,1,4) = \frac{6!}{4!}\times {3 \choose 1} = \frac{6!}{4!}\times 3$
Select 1 of 3 boxes to have 3 balls, select 1 of remaining 2 to have 2 balls, permute the balls. Ways to arrange 6 balls, 1 in one box, 2 in another, 3 into the third: $N(1,2,3) = \frac{6!}{2!3!}\times{3 \choose 1}{2 \choose 1} = \frac{6!}{2!3!} \times 3 \times 2$
Given that no box can be empty, that is all the ways to fill the boxes. So we want the probability : $$\dfrac{N(2,2,2)}{N(2,2,2)+N(1,1,4)+N(1,2,3)} \\ = \dfrac{\frac{1}{2!2!2!}}{\frac{1}{2!2!2!}+\frac{3}{4!}+\frac{1}{3!2!}\cdot 3\cdot 2} \\ = \frac{4!}{4!+2!2!2!3+2!2!4!} \\ = \dfrac{1}{6}$$
|
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|
Inequality $\sum_{cyc}\frac{(a+b)^{3}}{\sqrt[3]{2(a+b)(a^{2}+b^{2})}}\ge{12}$ when $ab+bc+ca=6$ Let $a$, $b$ and $c$ be non-negative reals such that $ab+bc+ca=6$.
Prove that:
$$\frac{(a+b)^{3}}{\sqrt[3]{2(a+b)(a^{2}+b^{2})}}+\frac{(b+c)^{3}}{\sqrt[3]{2(b+c)(b^{2}+c^{2})}}+\frac{(c+a)^{3}}{\sqrt[3]{2(c+a)(c^{2}+a^{2})}}\ge{12}$$
|
We'll prove a stronger inequality: $\sum\limits_{cyc}\frac{(a+b)^3}{\sqrt[3]{2(a+b)(a^2+b^2)}}\geq24$.
Indeed, let $a=\sqrt2x$, $b=\sqrt2y$, $c=\sqrt2z$, $x+y+z=3u$
and $xy+xz+yz=3v^2$, where $v\geq0$.
Hence, $v=1$, $u\geq v$ and we need to prove that $\sum\limits_{cyc}\frac{(x+y)^3}{\sqrt[3]{2(x+y)(x^2+y^2)}}\geq12.$
Now, by AM-GM and Holder we obtain:
$$\sum\limits_{cyc}\frac{(x+y)^3}{\sqrt[3]{2(x+y)(x^2+y^2)}}=\sum\limits_{cyc}\frac{3(x+y)^3}{3\sqrt[3]{2(x+y)(x^2+y^2)}}\geq$$
$$\geq\sum\limits_{cyc}\frac{3(x+y)^3}{2+x+y+x^2+y^2}\geq\frac{3\left(\sum\limits_{cyc}(x+y)\right)^3}{\sum\limits_{cyc}(2+x+y+x^2+y^2)}=$$
$$=\frac{4(x+y+z)^3}{3+x+y+z+x^2+y^2+z^2}=\frac{36u^3}{1+u+3u^2-2v^2}=$$
$$=\frac{36u^3}{v^3+uv^2+3u^2v-2v^3}=\frac{36u^3}{3u^2v+uv^2-v^3}\geq12,$$
where the last inequality it's just $(u-v)(3u^2-v^2)\geq0$, which is obvious.
Done!
|
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|
Getting incorrect result on limit of square root I have a limit problem:
$\lim_{x \to \infty} \sqrt{x^2 + x} - x$
According to Wolfram|Alpha the answer is $\frac{1}{2}$
However, my calculation gives $1$. Please help me understand what I'm doing wrong. The process is:
$\lim_{x \to \infty} \sqrt{x^2 + x} - x = \lim_{x \to \infty} \sqrt{x^2(1 + \frac{1}{x})} - x = \lim_{x \to \infty} x\sqrt{1 + \frac{1}{x}} - x = \lim_{x \to \infty} x(\sqrt{1 + \frac{1}{x}} - 1)$
Let $t = \frac{1}{x}$
$\lim_{x \to 0} \frac{\sqrt{1 + t} - 1}{t} \Rightarrow (l'Hopital) \lim_{x \to 0} \frac{\frac{1}{\sqrt{1 + t}}}{1} = \lim_{x \to 0} \frac{1}{\sqrt{1+t}} = 1$
|
You've lost your 2 in the derivation!
$\frac{d}{dt}\sqrt{1+t} = \frac{1}{2}\frac{1}{\sqrt{1+t}}$
|
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|
Find Max value of this expression: $P=(x-2yz)(y-2zx)(z-2xy)$ Let $x$, $y$ and $z$ be positives and satisfying $x^2+y^2+z^2=2xyz+1$ .
Find a maximum of this expression:
$$P=(x-2yz)(y-2zx)(z-2xy).$$
|
Let $x=1$ and $y=z=\frac{1}{2}.$
Hence, $P=\frac{1}{8}$.
We'll prove that it's a maximal value.
Indeed, let $x\geq y\geq z$.
If $x\leq2yz$ then we obtain:
$$2xy\geq2xz\geq2yz\geq x\geq y\geq z,$$
which says that $P\leq0$.
Thus, we can assume that $x>2yz$.
*
*$x=1$.
In this case the condition gives $y=z$ and
$$P=(1-2z^2)z^2=\frac{1}{2}(1-2z^2)2z^2\leq\frac{1}{2}\left(\frac{1-2z^2+2z^2}{2}\right)^2=\frac{1}{8};$$
2. $x\in(0,1).$
Let $x=\cos\alpha$.
We have here $$(x-2yz)(y-2xz)(z-2xy)=(x-2yz)(yz-2x(y^2+z^2)+4x^2yz)=$$
$$=(x-2yz)(yz-2x(1+2xyz-x^2)+4x^2yz)=(x-2yz)(2x^3-2x+yz)=$$
$$=\frac{1}{2}(x-2yz)(4x^3-4x+2yz)\leq\frac{1}{2}\left(\frac{x-2yz+4x^3-4x+2yz}{2}\right)^2=$$
$$=\frac{1}{8}(4x^3-3x)^2=\frac{1}{8}\cos^23\alpha\leq\frac{1}{8}.$$
3. $x>1$.
In this case we obtain:
$$x^2+y^2+z^2-1-2xyz\geq x^2+2yz-1-2xyz=(x-1)(x-2yz+1)>0,$$
which is a contradiction and this case is just impossible.
Done!
|
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|
Recurrence relation for number sequence Let $a_n$ be the number of sequences of $n$ numbers, consisting of $0's, 1's$ and $2's$, such that a number $1$ on the $j$-th place isn't followed by a $1$ or $2$ on the $j+1$-th place for $1\leq j\leq n-1$.
I'm asked to prove that the correct recurrence relation for this sequence is given by $$a_n=2a_{n-1}+a_{n-2}, a_1=3,a_2=7$$
A shorter sequence can be extended in the following ways: let there be $x$ correct sequences with length $n$. Let's say $y$ sequences of length $n$ don't end with an $1$, that means you can extend those sequences with any of the three possibilities, giving $3y$ new sequences of length $n+1$. For the other $x-y$ sequences, ending in $1$, they can only be extended with a $0$, giving $x-y$ new sequences. The total number of new sequences now is $3y+x-y = 2y+x$. This seems going in the correct direction, I just don't see how they are linked to $a_{n-1}$ and $a_{n-2}$.
|
Got the recurrence, now solve it. You have:
$$
a_{n + 2} = 2 a_{n + 1} + a_n
$$
There is 1 zero-length sequence, and 3 one-length sequences; this checks $a_2 = 7$.
Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply the recurrence by $z^n$ and sum over $n \ge 0$. Then recognize:
\begin{align}
\sum_{n \ge 0} a_{n + 1} z^n &= \frac{A(z) - a_0}{z} \\
\sum_{n \ge 0} a_{n + 2} z^n &= \frac{A(z) - a_0 - a_1 z}{z^2}
\end{align}
to get:
$$
\frac{A(z) - 1 - 3 z}{z^2} = 2 \frac{A(z) - 1}{z} + A(z)
$$
Solving:
$$
A(z) = \frac{1 + z}{1 - 2 z - z^2}
$$
Calling $r_1 = 1 + \sqrt{2}$, $r_2 = 1 - \sqrt{2}$, we can write:
$$
A(z) = \frac{r_1 + 1}{2^{3/2} (1 - r_1 z)} - \frac{r_2 + 1}{2^{3/2} (1 - r_2 z)}
$$
This is just two geometric series:
$$
a_n = \frac{r_1 + 1}{2^{3/2}} \cdot r_1^n - \frac{r_2 + 1}{2^{3/2}} \cdot r_2^n
$$
|
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|
determine all polynomials $P(x)$ such that $(x+1)P(x-1)-(x-1)P(x)$ is a constant polynomial Determine all polynomials $P(x)$ with real coefficients such that $(x+1)P(x-1)-(x-1)P(x)$ is a constant polynomial.
*
*clearly we have to show $(x+1)P(x-1)-(x-1)P(x)=c$ for all values of $x$ ($c$ is a real constant)
*i have tried this: see if $r$ is a root of $P(x)$ then $P(r)=0$ and then $(r+1)P(r-1)=c$. Now we consider $P(x)=a(x-r)(x-c)(x-d)\cdots(x-l)$, so we have
$$(r+1)P(r-1)=(c+1)P(c-1)=(d+1)P(d-1)=\dots\\=(l+1)P(l-1)=\pm 2rcd\dots l=c$$
[because see $P(0)=P(-1)=\dfrac{c}{2}$ ]
*then i am stuck!!!!!!!
see $P(x)=v$ is a solution where $v$ is a constant.
|
Let $(x+1)P(x−1)−(x−1)P(x)=C=const$. Set $x=2$ in order to get $$P(2)=3P(1)-C.$$ Set $x=3$ to obtain $$P(3)=6P(1)-\frac{5}{2}C.$$
This motivates us to try to prove by induction that
$$P(n)=\frac{n(n+1)}{2}P(1)-\frac{(n-1)(n+2)}{4}C$$ for every integer $n\geq 2.$ Indeed, if we assume the last equality holds for some $n\geq 2,$ then for $n+1$ we just set $x=n+1$ and obtain
$$P(n+1)=\frac{1}{n}\bigg( \frac{n(n+1)(n+2)}{2}P(1)-\frac{(n-1)(n+2)^2+4}{4}C \bigg)=\frac{(n+1)(n+2)}{2}P(1)-\frac{n(n+3)}{4}C$$
as desired. But since $P$ agrees with a quadratic polynomial at infinitely many integer values, we conclude that $P$ is quadratic. If we then let $P(x)=ax^2+bx+c,$ we can plug it into the original equation and see that all solutions are of the form
$$P(x)=ax^2+ax+\frac{C}{2},$$ where $a$ is an arbitrary real number and $C$ is the constant given in the beginning.
|
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|
Calculation of a Residue Does anyone know of a good way to calculate the residue at zero of the following function? I was able to calculate it with the higher order pole formula for residues and then used Mathematica to find the limit. Really just looking for a nice trick without having to get too dirty.
$$ f(z)=\frac{1}{z^2(e^{-z}-1)} $$
Thank you for any help.
By the way, the answer is $\frac{-1}{12}$
|
$$\begin{eqnarray}
\frac{1}{z^2(e^{-z}-1)}
&=& \frac{1}{z^2\left(1-z+\frac{z^2}{2}-\frac{z^3}{6} +\ldots - 1\right)} \\
&=& -\frac{1}{z^3\left(1-\left(\frac{z}{2}-\frac{z^2}{6}+\ldots\right)\right)} \\
&=& -\frac{1}{z^3}\left(1+\left(\frac{z}{2}-\frac{z^2}{6}+\ldots\right)+\left(\frac{z}{2}-\frac{z^2}{6}+\ldots\right)^2\right) \\
&=& -\frac{1}{z^3}\left(1+\frac{z}{2}+\frac{z^2}{12}+\ldots\right) \\
&=& -\frac{1}{z^3} - \frac{1}{2z^2} - \frac{1}{12 z} + \ldots
\end{eqnarray}$$
|
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|
$ \frac{x}{4 \ \sqrt{x^2+1}} \ = \ \frac{y}{5 \ \sqrt{y^2+1}} \ = \ \frac{z}{6 \ \sqrt{z^2+1}} $ and $ \ x+y+z \ = \ xyz \ $ Consider the system of equations in real numbers $ \ x,y,z \ $ satisfying
$$ \frac{x}{4 \ \sqrt{x^2+1}} \ = \ \frac{y}{5 \ \sqrt{y^2+1}} \ = \ \frac{z}{6 \ \sqrt{z^2+1}} $$
and $ \ x+y+z \ = \ xyz \ $. Find $ \ x,y,z \ . $
I substituted $ \ x,y,z \ $ as $ \ \tan(\theta_1) \ , \ \tan(\theta_2) \ , \ \tan(\theta_3) \ $ , where
$$ \theta_1 \ + \ \theta_2 \ + \ \theta_3 \ = \ 180º \ \ . $$
|
Starting with $\frac{sint}{4} = \frac{sinu}{5} = \frac{sinw}{6}$, then use the law of sines to get: $\frac{a}{4} = \frac{b}{5} = \frac{c}{6}$. So $a = \frac{2c}{3}$, $b = \frac{5c}{6}$.
So apply the law of cosines to have: $a^2 = b^2 + c^2 - 2bccosA$. Substituting these values of $a$ and $b$ into the above equation we can solve for $cosA$, and then $tanA$, and $tanB$, and $tanC$.
|
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|
Find closed formula $f(n)$ from generating function I'm asked to find a closed formula for
$f(n)=6f(n-1)-9f(n-2)$ for $n>1$ with $f(0)=-1. f(1)=0$,
using the ordinary generating function $F(X)$.
I found $F(X)=-1/(1-3x)^2$ but from there I don't manage to get a satisfactory formula for $f(n)$. Can anyone give me a hint ?
Thanks
|
The reason you're not getting the closed form from your generating function is that the generating function is wrong. Let $F(x) = \sum_{n \ge 0} f(n) x^n$, then
$$\begin{align} F(x)
&= f(0) + f(1)x + \sum_{n \ge 2} f(n) x^n \\
&= f(0) + f(1)x + \sum_{n \ge 2} (6f(n-1) - 9f(n-2))x^n \\
&= f(0) + f(1)x + 6x\sum_{n \ge 2} f(n-1)x^{n-1} - 9x^2\sum_{n \ge 2}f(n-2)x^{n-2}\\
&= f(0) + f(1)x + 6x\sum_{n \ge 1}f(n)x^n - 9x^2\sum_{n \ge 0}f(n)x^n \\
&= f(0) + f(1)x + 6x(F(x) - f(0)) - 9x^2F(x)
\end{align}$$
so with $f(0) = -1$ and $f(1) = 0$, you get
$$F(x) = -1 + 6x(F(x) + 1) - 9x^2F(x)$$
$$(1 - 6x + 9x^2)F(x) = -1 + 6x$$
which gives
$$F(x) = \frac{-1 + 6x}{1- 6x + 9x^2} = \frac{-1+6x}{(1-3x)^2} = \frac{-1}{(1-3x)^2} + \frac{6x}{(1-3x)^2}$$
so
$$f(n) = -(n+1)3^n + 6n3^{n-1} = 3^n (-n - 1 + 2n) = 3^n(n - 1)$$
|
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|
Solving a Linear Recurrence Relation I made quick progress on this, and then of course got stumped, so here's the problem:
$$a_0 = -1, a_1 = -2, a_n = 4a_{n-1} - 3a_{n-2}$$
So, following the way I was taught to solve this type of problem, I do the following:
Convert it to $$x^n = 4x^{n-1} - 3x^{n-2}$$
Divide by $x^{n-2}$ because it is the smallest exponent, leaving me with:
$x^2 = 4x-3$
Setting equal to zero so I can factor
$x^2-4x+3 = 0$
Factored:
$(x-3)(x-1)$
Giving me roots of 3 and 1, which not being equal, allows me to continue
$$a_n = q_13^n + q_21^n$$
Using the value of $a_0$
Gets me this:
$$a_0 = q_1+q_2 = -1$$
Which is useless I think because getting either $q_1$ or $q_2$ equal to $$-1 - q_{1 or 2}$$
Gets me no where because then, plugging it in gets me:
$$a_1 = -1 - q_2 + q_21^1 = -2$$
Simplified to:
$-1 = -2$
Even I know -1 does not equal -2, I get a similar situation if I try using $a_1$ first, no clue where to go from here, any help much appreciated, thanks guys.
|
A solution using generating functions: Define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence as:
$$
a_{n + 2} = 4 a_{n + 1} - 3 a_n \qquad a_0 = -1, a_1 = -2
$$
Multiply the recurrence by $z^n$, sum over $n \ge 0$, recognize:
\begin{align}
\sum_{n \ge 0} a_{n + 1} z^n &= \frac{A(z) - a_0}{z} \\
\sum_{n \ge 0} a_{n + 2} z^n &= \frac{A(z) - a_0 - a_1 z}{z^2}
\end{align}
to get:
$$
\frac{A(z) + 1 + 2 z}{z^2}
= 4 \frac{A(z) + 1}{z} - 3 A(z)
$$
Solve for $A(z)$ and split into partial fractions:
$$
A(z) = - \frac{1 - 2z}{1 - 4 z + 3 z^2}
= - \frac{1}{2} \frac{1}{1 - z} - \frac{1}{2} \frac{1}{1 - 3 z}
$$
Those are just two geometric series:
$$
a_n = - \frac{1}{2} - \frac{3^n}{2}
= - \frac{3^n + 1}{2}
$$
|
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|
Need help to solve taylor series of $e^{\sin x}$ How to derive the taylor series of $e^{\sin x}$, up to $x^5$?
i just don't know how to get the answer
$$f(x) = 1 + x + \frac{x^2}{2} - \frac{x^4}{8} -\frac{x^5}{15}$$
really need some help. Thanks
|
$$\sin x \sim x - \frac{x^3}{6} + \frac{x^5}{120}$$
$$e^x \sim 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120}$$
so $$e^{\sin x} \sim 1 + \sin x + \frac{\sin ^2 x}{2} + \frac{\sin ^3 x}{6} + \frac{\sin ^ 4 x}{24} + \frac{\sin ^5x}{120} $$
Now substitute the expansion of $\sin x$, and you should get to the result
(remember to eliminate all those terms that have a degree higher than 5! :-) )
|
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|
What is being represented by this 2 images? image 1
image 2
It's possible that image 1 is showing some kind of methods for building polygons out of trigonometric functions ?
It's also possible that image 2 is a quadratic bezier curve ?
|
Here's how we can produce the figures in image 1. The graphs are showing the $ \ y-$ coordinate of the point being "swept along" on each curve.
The circle is simple because familiar trigonometric functions can be defined in terms of a circle. For a circle of radius $ \ a \ , $ the coordinates of the point are just $ \ (a \cos \theta \ , \ a \sin \theta) \ , $ with $ \ \theta \ $ being the angle measured (counter-clockwise) from the positive $ \ x-$ direction. [Keep in mind that, in the animations, the graphs are being drawn "right to left".]
The square of side $ \ 2a \ $ , having four straight sides lying along the lines $ \ x \ = \ \pm a \ $ and $ \ y \ = \ \pm a \ , $ requires a piecewise function with four "branches":
$$ \sigma(\theta) \ = \ \left\{ \begin{array}{cc}( \ a \ , \ a \ \tan \theta \ ) \ ,&
-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}\\( \ a \ \cot \theta \ , \ a ) \ ,&
\frac{\pi}{4} \le \theta \le \frac{3\pi}{4}\\( \ a \ , \ -a \ \tan \theta \ ) \ ,&
\frac{3\pi}{4} \le \theta \le \frac{5\pi}{4}\\( \ -a \ \cot \theta \ , \ a ) \ ,&
\frac{5\pi}{4} \le \theta \le \frac{7\pi}{4}\end{array}\right. $$
The hexagon depicted is correspondingly more complicated and I won't do all of it here. We will call the lengths of the sides $ \ a \ . $ Except for the vertical sides, the rest make 30º angles to the "horizontal", so their slopes are $ \ \pm \frac{\sqrt{3}}{3} \ . $ Since the altitude of an equilateral triangle is $ \ \frac{\sqrt{3}}{2} a \ , $ the right vertical side lies on the line $ \ x \ = \ \frac{\sqrt{3}}{2} a \ , $ so that branch of the piecewise function is
$$ ( \ \frac{\sqrt{3}}{2} a \ , \ \frac{\sqrt{3}}{2} a \ \tan \theta \ ) \ , \ -\frac{\pi}{6} \ \le \ \theta \ \le \ \frac{\pi}{6} \ \ . $$
The next side in the counter-clockwise direction has a slope of $ \ -\frac{\sqrt{3}}{3} \ $ and a $ \ y-$ intercept of $ \ a \ $ , so it lies on the line $ \ y \ = \ a \ - \ \frac{\sqrt{3}}{3} x \ . $ In polar coordinates, this is
$$ r \ = \ \frac{a}{\frac{\sqrt{3}}{3} \cos \theta \ + \ \sin \theta} \ \ . $$
The branch of the piecewise function definition covering this side is then
$$ ( \ \frac{a \ \cos \theta}{\frac{\sqrt{3}}{3} \cos \theta \ + \ \sin \theta} \ , \ \frac{a \ \sin \theta}{\frac{\sqrt{3}}{3} \cos \theta \ + \ \sin \theta} \ ) \ , \ \frac{\pi}{6} \ \le \ \theta \ \le \ \frac{\pi}{2} \ \ . $$
The branch definitions for the other four sides of the hexagon can be determined analogously.
|
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|
if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$ This is supposed to be an application of AM-GM inequality.
if $abc=1$, then the following holds true: $a^2+b^2+c^2\ge a+b+c$
First of all,
$a^2+b^2+c^2\ge 3$
by a direct application of AM-GM.Also,we have
$a^2+b^2+c^2\ge ab+bc+ca$
Next,we consider the expression
$(a+1)(b+1)(c+1)=a+b+c+ab+bc+ca+abc\le a+b+c+a^2+b^2+c^2+1$
but that hardly helps.I know that
$3(a^2+b^2+c^2)\ge (a+b+c)^2$
From the first derived inequality,we know that the left hand side of the above inequality is greater than or equal to 9.But I can't see how that can be used here.I know that we can get $a+b+c\ge 3$ using AM-GM but that does not take me a step closer to finding the solution(from what I can understand).Also,
$a^3+b^3+c^3\ge a^2b+b^2c+c^2a$
$(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]\ge a^2b+b^2c+c^2a$
A hint will be appreciated at this point.
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Use the fact that
$$a^2+a^2+a^2+b+c \geq5\sqrt[5]{a^2.a^2.a^2.b.c}=5\sqrt[5]{a^6bc}=5a$$ since $abc=1$
Similarly, we get two more results, and adding them we get:
$$3(a^2+b^2+c^2)+2(a+b+c) \geq 5(a+b+c) $$ which gives us our result.
$$a^2+b^2+c^2 \geq a+b+c$$
$$Q.E.D.$$
|
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|
2x2 Fibonacci matrix singular value decomposition $A = \left[\begin{array}[c]{rr}1 & 1\\1 & 0\end{array}\right]$
I am supposed to find all the eigenvalues and vectors for this matrix so that $Av=σu$ and then form a singular value decomposition $UΣVᵀ$ and show that $A=UΣVᵀ$.
I have attempted this so many times and just keep getting so confused with what the question really wants, since I get something very complicated, as I get when I try to solve it. However, when I look at a similar markscheme it's quite simply put, but I don't understand how.
I know that I can use $det(AᵀA)$ to get $σ^2$ or I can get σ with $det(A-λI)$. So those give me the same, which are the diagonals of $Σ$ and the rest of $Σ$ consists of zeros.
So I get
$$\Sigma = \left(
\begin{array}{cc}
\sqrt{\frac{1}{2} \left(3+\sqrt{5}\right)} & 0 \\
0 & \sqrt{\frac{1}{2} \left(3-\sqrt{5}\right)} \\
\end{array}
\right)$$
and $v_1$ and $v_2$ I calculated to be as follows
$v_1$$=$$\left[\begin{array}[c]{r}(1+√5)/2\\1\end{array}\right]$
$v_2$$=$$\left[\begin{array}[c]{r}(1-√5)/2\\1\end{array}\right]$
After that am I not supposed to convert it to the unit vector? That comes out really messy...
I just feel like I'm doing something wrong because it's not supposed to be a very complicated question, and if I continue from here it gets way ugly. I'm not sure how I am supposed to show what they're asking for.
Also, I believe $u_1=v_1$ and $u_2=-v_2$, but how are those calculated?
Another note - can this be done by keeping the eigenvectors as variables instead of substituting the values? Maybe that would not come out as complicated.
|
We are given:
$$A = \begin{bmatrix}1 & 1\\1 & 0 \end{bmatrix}$$
We have:
$$W = A^T A = \begin{bmatrix}2 & 1\\1 & 1 \end{bmatrix}$$
The characteristic polynomial and eigenvalues of W are:
$$\lambda^2 +3 \lambda -1 = 0 \implies \lambda_1 = \frac{1}{2} \left(3-\sqrt{5}\right), ~ \lambda_2 = \frac{1}{2} \left(3+\sqrt{5}\right)$$
This gives us the singular values:
$$\sigma_1= \sqrt{\frac{1}{2} \left(3-\sqrt{5}\right)}, ~ \sigma_2= \sqrt{\frac{1}{2} \left(3+\sqrt{5}\right)}$$
The eigenvectors of $W$ are:
$$v_1 = \begin{bmatrix} \frac{1}{2} \left(1-\sqrt{5}\right) \\ 1 \end{bmatrix}, ~ v_2 = \begin{bmatrix} \frac{1}{2} \left(1+\sqrt{5}\right) \\ 1 \end{bmatrix}$$
Normalizing these eigenvectors, we have:
$$v_1 = \begin{bmatrix} \frac{1+\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+1}} \\ \frac{1}{\sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+1}} \end{bmatrix}, ~ v_2 = \begin{bmatrix} \frac{1-\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+1}} \\ \frac{1}{\sqrt{\frac{1}{4} \left(1-\sqrt{5}\right)^2+1}} \end{bmatrix}$$
Now, we can write $W = U \Sigma V^T$ as:
$$U = \left(
\begin{array}{cc}
\frac{\frac{1}{2} \left(1+\sqrt{5}\right)+1}{\sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+\left(\frac{1}{2} \left(1+\sqrt{5}\right)+1\right)^2}} & \frac{\frac{1}{2} \left(1-\sqrt{5}\right)+1}{\sqrt{\frac{1}{4} \left(1-\sqrt{5}\right)^2+\left(\frac{1}{2} \left(1-\sqrt{5}\right)+1\right)^2}} \\
\frac{1+\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+\left(\frac{1}{2} \left(1+\sqrt{5}\right)+1\right)^2}} & \frac{1-\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1-\sqrt{5}\right)^2+\left(\frac{1}{2} \left(1-\sqrt{5}\right)+1\right)^2}} \\
\end{array}
\right)$$
$$\Sigma = \left(
\begin{array}{cc}
\sqrt{\frac{1}{2} \left(3+\sqrt{5}\right)} & 0 \\
0 & \sqrt{\frac{1}{2} \left(3-\sqrt{5}\right)} \\
\end{array}
\right)$$
$$V = \left(
\begin{array}{cc}
\frac{1+\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+1}} & \frac{1-\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1-\sqrt{5}\right)^2+1}} \\
\frac{1}{\sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+1}} & \frac{1}{\sqrt{\frac{1}{4} \left(1-\sqrt{5}\right)^2+1}} \\
\end{array}
\right)$$
Notes:
*
*Recall: The columns of V are called the right singular vectors. The columns of U are called the left singular vectors. To find $U$, we calculate the columns as:
$$u_1 = \dfrac{1}{\sigma_1} A v_1, ~ u_2 = \dfrac{1}{\sigma_2} A v_2$$
Care needs to be taken when one or both of the eigenvalues are zero!
This SVD can also be written as:
$$\sigma_1~u_1~v_1^T + \sigma_2~u_2~v_2^T$$
*
*Recall we need to form $V^T$ when writing this out.
|
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|
let a,b,c be the sides of a triangle and $a+b+c=2$ then prove that $ a^2 + b^2 + c^2 +2abc < 2$ let a,b,c be the sides of a triangle and $a+b+c=2$ then prove that
$ a^2 + b^2 + c^2 +2abc < 2$
I used the fact that $ a+b-c>0 $ and multiplied the 3 cyclic equations but coudn't reach the final equation.
|
Let's homogenize the equation by multiplying with $ a+ b + c = 2$.
We need to show that
$$ 2 ( a + b + c) (a^2 + b^2 + c^2) + 8abc < (a+b+c)^3.$$
Expanding this, it is equivalent to showing that
$$ (a+b-c) ( b+c-a) ( c+a-b) > 0. $$
Hence we are done.
|
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|
Area of part of parametric function I need to get area of function: $x= 2\sqrt{2}\cos ^3 t$ and $y= 4\sqrt{2}{\sin ^3 t}$, but only the part when $x\geq1$. How can I do that? I know that area of full function would be
$$S= \int_a^b y(t)x'(t) \, dt $$
$a \leq t \leq b$
|
UPDATE. The area from $x=1$ to $x(0)=a=2\sqrt{2}$ enclosed by the parametric curve
\begin{equation*}
\left( x(t),y(t)\right) =\left( 2\sqrt{2}\cos ^{3}t,4\sqrt{2}\sin
^{3}t\right)
\end{equation*}
is
\begin{equation*}
S=2\int_{1}^{a}y\,dx.
\end{equation*}
Since $x^{\prime }(t)=-6\sqrt{2}\cos ^{2}t\sin t$ and
\begin{equation*}
x(t)=1\Rightarrow 2\sqrt{2}\cos ^{3}t=1\Leftrightarrow t_{1}=\frac{\pi }{4}
,t_{2}=-t_{1},
\end{equation*}
the same area, using your formula, can be expressed as
\begin{eqnarray*}
S &=&2\int_{\pi /4}^{0}\left( 4\sqrt{2}\sin ^{3}t\right) \left( -6\sqrt{2}%
\cos ^{2}t\sin t\right) \,dt \\
&=&96\int_{0}^{\pi /4}\cos ^{2}t\sin ^{4}t\,dt \\
&=&96\int_{0}^{\pi /4}\left( 1-\sin ^{2}t\right) \sin ^{4}t\,dt \\
&=&96\int_{0}^{\pi /4}\sin ^{4}t\,dt-96\int_{0}^{\pi /4}\sin ^{6}t\,dt \\
&=&\cdots \\
&=&-2+\frac{3}{2}\pi \\
&\approx &2.7124.
\end{eqnarray*}
Warning. This answer is no longer updated. The question was changed to the parametric curve $$(x= 2\sqrt{2}\cos ^3 t,y= 4\sqrt{2}{\sin ^3 t})$$
The area from $x=1$ to $x(0)=a=2\sqrt{2}$ enclosed by the parametric curve
$$\left( x(t),y(t)\right) =\left( 2\sqrt{2}\cos t^{3},4\sqrt{2}\sin
t^{3}\right) $$
is given by
\begin{equation*}
S=2\int_{1}^{a}y\,dx.
\end{equation*}
Since $x^{\prime }(t)=-6\sqrt{2}t^{2}\sin t^{3}$ and
\begin{equation*}
x(t)=1\Rightarrow 2\sqrt{2}\cos t^{3}=1\Leftrightarrow t_{1}=\arccos ^{1/3}
\frac{\sqrt{2}}{4},t_{2}=-t_{1},
\end{equation*}
the same area, using your formula, can be expressed as
\begin{eqnarray*}
S &=&2\int_{t_{1}}^{0}\left( 4\sqrt{2}\sin t^{3}\right) \left( -6\sqrt{2}
t^{2}\sin t^{3}\right) \,dt \\
&=&96\int_{0}^{t_{1}}t^{2}\sin ^{2}t^{3}\,dt=96\left[ \frac{1}{6}\left(
t^{3}-\cos t^{3}\sin t^{3}\right) \right] _{0}^{t_{1}} \\
&=&16\left. \left( t^{3}-\cos t^{3}\sin t^{3}\right) \right\vert
_{0}^{\arccos ^{1/3}\frac{\sqrt{2}}{4}} \\
&=&16\left( \arccos \frac{\sqrt{2}}{4}-\frac{\sqrt{2}}{4}\sin \left( \arccos
\frac{\sqrt{2}}{4}\right) \right) \\
&=&-2\sqrt{7}+16\arccos \frac{\sqrt{2}}{4} \\
&\approx &14.059.
\end{eqnarray*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral $ \int \frac{dx}{\cos^3 x+2\sin(2x)-5\cos x}$ $$
I\equiv \int \frac{dx}{\cos^3 x+2\sin(2x)-5\cos x}.
$$
This integral does have a closed form. I am not sure where to start. We can factorize the denominator as
$$
\cos^3 x+2\sin(2x)-5\cos x=(\cos^2 x+4\sin x-5)\cos x,
$$
but I am not sure where to go from there. Thank you
|
The present integrand is a rational function (i.e. a ratio of two polynomials) of $\cos x$ and $\sin x$, which is considered a common question. So I take it partially from the answer by Arturo Magidin to the question Evaluating $\int P(\sin x, \cos x) \text{d}x$ by Aryabhata, which is an entry of the List of Generalizations of Common Questions.
Weierstrass Substitution
A method that always works is Weierstrass substitution, which will turn such an integral into an integral of rational functions, which in turn can always be solved, at least in principle, by the method of partial fractions. This works even for rational functions of sine and cosine, as well as functions that involve the other trigonometric functions.
Weierstrass substitution replaces sines and cosines (and by extension, tangents, cotangents, secants, and cosecants) by rational functions of a new variable. The identities begin by the trigonometric substitution $t = \tan\frac{x}{2}$, with $-\pi\lt x\lt \pi$, which yields
$$\begin{align*}
\sin x &= \frac{2t}{1+t^2}\\
\cos x &= \frac{1-t^2}{1+t^2}\\
dx &= \frac{2\,dt}{1+t^2}.
\end{align*}$$
Now we have
$$I =\int \frac{dx}{\cos ^{3}x+4\sin x\cos x-5\cos x}.$$
Using the substitution above we obtain:
\begin{eqnarray*}
I
&=&\int \frac{2}{\left( \frac{1-t^{2}}{1+t^{2}}\right) \left( \left( \frac{
1-t^{2}}{1+t^{2}}\right) ^{2}+4\frac{2t}{1+t^{2}}-5\right) \left(
1+t^{2}\right) }\,dt \\[2ex]
&=&\frac{1}{2}\int \frac{\left( 1+t^{2}\right) ^{2}}{\left( -1+t^{2}\right)
\left( t^{2}-t+1\right) ^{2}}dt,
\end{eqnarray*}
which can be integrated by the method of partial fractions:
\begin{eqnarray*}
I&=&\frac{1}{2}\int \left( \frac{2}{-1+t}-\frac{2}{9\left( 1+t\right) }-\frac{
1}{9}\frac{-5+16t}{t^{2}-t+1}-\frac{1}{3}\frac{-2+t}{\left( t^{2}-t+1\right)
^{2}}\right) dt \\[2ex]
&=&\ln (\left\vert -1+t\right\vert -\frac{1}{9}\ln \left( \left\vert
9+9t\right\vert \right) -\frac{4}{9}\ln \left( \left\vert
t^{2}-t+1\right\vert \right) +\frac{1}{6}\frac{t}{t^{2}-t+1}+C \\[2ex]
&=&\ln \left( \left\vert -1+\tan \frac{x}{2}\right\vert \right) -\frac{1}{9}
\ln \left( \left\vert 9+9\tan \frac{x}{2}\right\vert \right) -\frac{4}{9}\ln
\left( \left\vert \tan ^{2}\frac{x}{2}-\tan \frac{x}{2}+1\right\vert \right)
+\frac{1}{6}\frac{\tan \frac{x}{2}}{\tan ^{2}\frac{x}{2}-\tan \frac{x}{2}+1}
+C.
\end{eqnarray*}
PS. I've checked that the derivative of this last integral is
$$
\frac{1}{4}\frac{3\tan ^{2}\frac{x}{2}+\tan ^{6}\frac{x}{2}+3\tan ^{4}\frac{x}{2}+1}{\left( -1+\tan \frac{x}{2}\right) \left( 1+\tan \frac{x}{2}\right) \left( \tan ^{2}\frac{
x}{2}-\tan \frac{x}{2}+1\right) ^{2}},
$$
which is equal to
$$\frac{1}{\cos ^{3}x+4\sin x\cos x-5\cos x}.$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Help with this combinatorial proof $\sum\limits_{k=1}^nk^2(k-1){n\choose k}^2 = n^2(n-1){2n-3\choose n-2}$ considering $n\ge2$
$\displaystyle\sum\limits_{k=1}^nk^2(k-1){n\choose k}^2 = n^2(n-1)
{2n-3\choose n-2}$ considering $n\ge2$
Can somebody help with this combinatorial proof?
I'm struggling a lot.
Thanks.
EDIT: Ok. I could figure it out, if we had $\displaystyle\sum\limits_{k=1}^nk^2{n\choose k}^2 = n^2
{2n-2\choose n-1}$.
The problem is, i don't understand what to do with that $(k-1)$ and how it leads to ${2n-3\choose n-2}$.
I know $k{n\choose k} = n{n-1\choose k-1}$
Choosing a team of $k$ elements from $n$ and from that $k$ elements, pick a captain is the same as choose a captain first, and then, complete the team, choosing $k-1$ elements from $n-1$
But, what about $k(k-1){n\choose k}$ ?
|
Hint: Note that because choosing $k$ elements from a set of $n$ is the same as choosing the complement of the $k$ elements, we have
$$
\binom{n}{k}=\binom{n}{n-k}\tag{1}
$$
and since choosing a team of $k$ people and then a leader from those chosen is the same as choosing a leader and then choosing the remaining $k-1$ from the remaining $n-1$, we get
$$
k\binom{n}{k}=n\binom{n-1}{k-1}\tag{2}
$$
and
$$
k^2(k-1)=k(k-1)(k-2)+2k(k-1)\tag{3}
$$
Then consider Vandermonde's Identity.
Full Solution:
$$
\hspace{-5mm}\begin{align}
&\sum_{k=1}^nk^2(k-1)\binom{n}{k}^2\\
&=\sum_{k=1}^nk(k-1)(k-2)\binom{n}{k}\binom{n}{n-k}+2\sum_{k=1}^nk(k-1)\binom{n}{k}\binom{n}{n-k}\tag{4}\\
&=n(n-1)(n-2)\sum_{k=1}^n\binom{n-3}{k-3}\binom{n}{n-k}+2n(n-1)\sum_{k=1}^n\binom{n-2}{k-2}\binom{n}{n-k}\tag{5}\\
&=n(n-1)(n-2)\binom{2n-3}{n-3}+2n(n-1)\binom{2n-2}{n-2}\tag{6}\\[4pt]
&=(n-1)(n-2)^2\binom{2n-3}{n-2}+4(n-1)^2\binom{2n-2}{n}\tag{7}\\[4pt]
&=n^2(n-1)\binom{2n-3}{n-2}\tag{8}
\end{align}
$$
Explanation:
$(4)$: apply $(1)$ and $(3)$
$(5)$: apply $(2)$ several times
$(6)$: Vandermonde Identity
$(7)$: $\binom{2n-3}{n-3}\stackrel{(1)}=\binom{2n-3}{n}\stackrel{(2)}=\frac{2n-3}{n}\binom{2n-4}{n-1}\stackrel{(1)}=\frac{2n-3}{n}\binom{2n-4}{n-3}\stackrel{(2)}=\frac{n-2}{n}\binom{2n-3}{n-2}$
$(7)$: $\binom{2n-2}{n-2}\stackrel{(1)}=\binom{2n-2}{n}\stackrel{(2)}=\frac{2n-2}{n}\binom{2n-3}{n-1}\stackrel{(1)}=\frac{2n-2}{n}\binom{2n-3}{n-2}$
|
{
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"url": "https://math.stackexchange.com/questions/748815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Proof that $ p $ | ${ p^n \choose k } $ for any prime $p$ and $ k < p^n$ I know how to prove the fact that $ p$ | ${ p \choose k } $ (when writing it as a fraction, $p$ cannot be divided by any of the $1\times2\times...\times k$ or $1\times2\times...\times(p-k)$ because $p$ is prime).
When I try to apply the same rationale for $ p$ | ${ p^n \choose k } $ I get stumped because $p^n$ is in fact divisible by one of the $k! = 1\times2\times...\times k$, where $0 < k < p^n$.
If we take the example of $2^3$ and $k = 7$, we can easily see that $k! = 1 \times 2 \times ... \times 6 \times 7$, and $2$ clearly divides $2^3$.
How else can I approach this?
|
The highest prime power $p$ dividing $n!$ is expressed as $$\sum_{x=1}^\infty \lfloor \frac{n}{p^x} \rfloor.$$
Since $\binom{p^n}{k}=\frac{(p^n)!}{k!(p^n-k)!},$ we compare the highest prime power $p$ which divide numerator and denominator.
$$v_p((p^n)!)=\sum_{x=1}^n \lfloor \frac{p^n}{p^x} \rfloor,$$
$$v_p(k!(p^n-k)!)=v_p(k!)+v_p((p^n-k)!)=\sum_{x=0}^n \left( \lfloor \frac{k}{p^x} \rfloor + \lfloor \frac{p^n-k}{p^x} \rfloor \right).$$
(We've replaced $\infty$ bound with $n$ since if $x>n,$ the terms evaluate to $0$).
Notice that $\lfloor a+b \rfloor \ge \lfloor a \rfloor + \lfloor b \rfloor,$ and the that if $a+b$ is an integer while $a$ and $b$ are not, then $\lfloor a+b \rfloor = \lfloor a \rfloor + \lfloor b \rfloor - 1.$ We can conclude with the first fact that $$\sum_{x=1}^n \lfloor \frac{p^n}{p^x} \rfloor \ge \sum_{x=0}^n \left( \lfloor \frac{k}{p^x} \rfloor + \lfloor \frac{p^n-k}{p^x} \rfloor \right).$$
Because
*
*$0<k<p^n,$
*$\frac{p^n}{p^x} \in \mathbb{Z}$ for integers $1 \le x \le n,$
*$\frac{k}{p^x}+\frac{p^n-k}{p^x} \in \mathbb{Z}$ for same bounds,
*When $x=n,$ neither $\frac{k}{p^x}$ nor $\frac{p^n-k}{p^x}$ are integral $\implies \lfloor \frac{p^n}{p^n} \rfloor = 1 > \lfloor \frac{k}{p^n} \rfloor + \lfloor \frac{p^n-k}{p^n} \rfloor = 0.$
Finally, we can conclude $$\sum_{x=1}^n \lfloor \frac{p^n}{p^x} \rfloor > \sum_{x=0}^n \left( \lfloor \frac{k}{p^x} \rfloor + \lfloor \frac{p^n-k}{p^x} \rfloor \right).$$ This implies the divisibility.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find centre,vertics,foci,focal radii,letus rectum... when exists of a general quadratic equation in x and y Is there a generalized way( a particular conic section of any shape,for instance an ellipse without determining its major/minor axis) to find the centre,vertices,foci,focal radii,letus rectum,eccentricity,etc when exists of any parabola,ellipse,hyperbola?(dealing separately)
$ax^2+by^2+cx+dy+exy+c=0$ ($a,b,c$ not all zero)
|
One can use a list of equations to determine the property you require. Note, however, that in many cases it is easier to use derived value(s) as opposed to using equations that rely solely on the coefficients of the equation in general quadratic form. Below are examples of equations one can use.
Properties of an ellipse from equation for conic sections in general quadratic form
Given the equation for conic sections in general quadratic form:
$ a x^2 + b x y + c y^2 + c x + e y + f = 0 $
The equation represents an ellipse if:
$ b^2 - 4 a c < 0 $
or similarly,
$ 4 a c - b^2 > 0 $
The coefficient normalizing factor is given by:
$ q = 64 {{f (4 a c - b^2) - a e^2 + b d e - c d^2} \over {(4ac - b^2)^2}} $
The distance between center and focal point (either of the two) is given by:
$ s = {1 \over 4} \sqrt { |q| \sqrt { b^2 + (a - c)^2 }} $
The semi-major axis length is given by:
$ r_\max = {1 \over 8} \sqrt { 2 |q| {\sqrt{b^2 + (a - c)^2} - 2 q (a + c) }} $
The semi-minor axis length is given by:
$ r_\min = \sqrt {{r_\max}^2 - s^2} $
The latus rectum is given by:
$ l = 2 {{ {r_\min}^2 } \over {r_\max}} $
The eccentricity is given by:
$ g = {{s} \over {r_\max}} $
The distance between center and closest directix point (either of the two) is given by:
$ h = {{{r_\max}^2} \over {s}} $
The center of the ellipse is given by:
$ x_\Delta = { b e - 2 c d \over 4 a c - b^2} $
$ y_\Delta = { b d - 2 a e \over 4 a c - b^2} $
The top-most point on the ellipse is given by:
$ y_T = y_\Delta + {\sqrt {(2 b d - 4 a e)^2 + 4(4 a c - b^2)(d^2 - 4 a f)} \over {2(4 a c - b^2)}} $
$ x_T = {{-b y_T - d} \over {2 a}} $
The bottom-most point on the ellipse is given by:
$ y_B = y_\Delta - {\sqrt {(2 b d - 4 a e)^2 + 4(4 a c - b^2)(d^2 - 4 a f)} \over {2(4 a c - b^2)}} $
$ x_B = {{-b y_B - d} \over {2 a}} $
The left-most point on the ellipse is given by:
$ x_L = x_\Delta - {\sqrt {(2 b e - 4 c d)^2 + 4(4 a c - b^2)(e^2 - 4 c f)} \over {2(4 a c - b^2)}} $
$ y_L = {{-b x_L - e} \over {2 c}} $
The right-most point on the ellipse is given by:
$ x_R = x_\Delta + {\sqrt {(2 b e - 4 c d)^2 + 4(4 a c - b^2)(e^2 - 4 c f)} \over {2(4 a c - b^2)}} $
$ y_R = {{-b x_R - e} \over {2 c}} $
The angle between x-axis and major axis is given by:
if $ (q a - q c = 0) $ and $ (q b = 0) $ then $ \theta = 0 $
if $ (q a - q c = 0) $ and $ (q b > 0) $ then $ \theta = {1 \over 4} \pi $
if $ (q a - q c = 0) $ and $ (q b < 0) $ then $ \theta = {3 \over 4} \pi $
if $ (q a - q c > 0) $ and $ (q b >= 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} $
if $ (q a - q c > 0) $ and $ (q b < 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} + {\pi} $
if $ (q a - q c < 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} + {1 \over 2}{\pi} $
The focal points are given by:
$ F_{1,x} = x_\Delta - s \ cos (\theta) $
$ F_{1,y} = y_\Delta - s \ sin (\theta)) $
$ F_{2,x} = x_\Delta + s \ cos (\theta) $
$ F_{2,y} = y_\Delta + s \ sin (\theta)) $
(I tried to enter the equations without error. If you find an error, please post a comment)
|
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|
Evaluate $\int x \sqrt{1 - x^4} \,\mathrm{d}x$ I have the following question
$$\int x \sqrt{1 - x^4} \,\mathrm{d}x$$
I know we have to use trig. substitution for this and therefore, I did the following by letting $x = \sin \theta$ and $dx = \cos \theta \,\mathrm{d}\theta$
\begin{align}
&\int x \sqrt{1-x^4} \,\mathrm{d}x \\
&=\int \sin \theta \cos \theta\sqrt{1 - (\sin \theta)^4} \, \mathrm{d}\theta \\
&=\int \sin \theta \cos \theta \sqrt{(1-\sin^2 \theta)(1+\sin^2\theta)} \,\mathrm{d}\theta \\
&=\int \sin \theta \cos \theta \sqrt{(\cos^2 \theta)(1+\sin^2\theta)} \,\mathrm{d}\theta \\
\end{align}
Now, I'm confused. How do I proceed?
Thanks!
EDIT: Taking from the answer, I have a (nearly) full solution below for future users.
Instead of letting $x = \sin \theta$. We'll let $x^2 = \sin \theta$ and this will greatly simplify everything. Since, $\mathrm{d}x = \frac{\cos \theta}{2x} \,\mathrm{d\theta}$
\begin{align}
&\int x \sqrt{1-x^4} \quad \mathrm{d}x \\
&=\frac{1}{2}\int \cos \theta\sqrt{1 - (\sin x)^2} \quad \mathrm{d}\theta \\
&= \frac{1}{2} \int \cos \theta \cos \theta \quad \mathrm{d}\theta \\
&= \frac{1}{4} \int 1 + \cos 2\theta \quad \mathrm{d}\theta \\
&= \frac{1}{4} \left(\theta + \frac{\sin2\theta}{2} \right) \\
&= \frac{\theta}{4} + \frac{\sin \theta \cos \theta}{4} \\
\end{align}
After this, you only have to put $\sin \theta$ back in terms of $x$ and you're done!
|
HINT :
Let $u^2=1-x^4\;\Rightarrow\; x^2=\sqrt{1-u^2}\;\Rightarrow\; x\ dx=-\dfrac{u\ du}{2\sqrt{1-u^2}}$, then rewrite
$$
\begin{align}
\int x\sqrt{1-x^4}\ dx&=-\frac12\int\dfrac{u^2}{\sqrt{1-u^2}}\ du\\
&=\frac12\int\dfrac{1-u^2-1}{\sqrt{1-u^2}}\ du\\
&=\frac12\left(\int\dfrac{1-u^2}{\sqrt{1-u^2}}\ du-\int\dfrac{1}{\sqrt{1-u^2}}\ du\right).
\end{align}
$$
The left part integral can be solved by using IBP and the right part integral can be solved by using trigonometry substitution.
ADDENDUM :
I have just found that the fastest way is letting $u=x^2$.
|
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|
Geometric intuition: Seeing the regions in double integrals Context: solving double integrals.
I had the formula $$x^2+y^2=1-x-y$$ yet I could not see what shape it had. This is even more true with 3D pictures like $$2x^2+2y^2 \le 1+z^2.$$ Is there a summary somewhere of shapes to learn, so that I can get this.
|
Note
\begin{align}
x^2+y^2&=1-x-y \\
\Leftrightarrow x^2+y^2+x+y & =1 \\
\Leftrightarrow x^2+x+\frac{1}{4}+y^2+y+\frac{1}{4}&=1+\frac{1}{4}+\frac{1}{4} \\
\Leftrightarrow \left(x+\frac{1}{2}\right)^2+ \left(y+\frac{1}{2}\right)^2 & =\frac{3}{2}
\end{align}
Last equation describes a circular cilinder with axis parallel to $z$ axis and radius $\frac{\sqrt{6}}{2}$.
|
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|
Minimal polynomial: is $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$? I was wondering about the minimal polynomial of real number
$$u=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$$
over field $\mathbb{Q}$.
As you can see here, I worked out that $u$ is a root of monic rational polynomial $x^3+3x-4$. This is not irreducible:
$$x^3+3x-4=\left(x-1\right)\left(x^2+x+4\right)$$
and the second, quadratic, factor has complex roots $\frac{-1\pm i\sqrt{15}}{2}$.
Can I claim that the minimal polynomial of $u$ over $\mathbb{Q}$ is $x-1$?
In other words: does this prove that $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$?
In case it does, how can we demonstrate $u=1$ working on the roots in a way different from how I did in the link?
Edit:
$\sqrt[3]{2-\sqrt{5}}$ is meant to be the real cube root of $2-\sqrt{5}$.
|
A calculation shows that $(1+\sqrt{5})^3=16+8\sqrt{5}$, so the real cube root of $2+\sqrt{5}$ is $\frac{1}{2}(1+\sqrt{5})$.
Similarly, or by using conjugation, the real cube root of $2-\sqrt{5}$ is $\frac{1}{2}(1-\sqrt{5})$.
Add.
|
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|
Range of f(x) = $\frac{\sqrt3\,\sin x}{2 + \cos x}$ Can you give any idea about the range of the following function?
$$f(x) = \frac{\sqrt{3}\,\sin x}{2 + \cos x}$$
|
Yes:
*
*$-1 \leq \sin x \leq 1 \implies -\sqrt{3} \leq \sqrt{3}\sin x \leq \sqrt{3}$
*$-1 \leq \cos x \leq 1 \implies 1 \leq 2+\cos x \leq 3$
Hence: $$-\sqrt{3} \leq \frac{\sqrt{3}\sin x}{2+\cos x} \leq \sqrt{3}$$
Of course, you can do better than this using $f'(x)$ in order to calculate min/max points of $f(x)$:
*
*$f'(x)=\frac{\sqrt{3}(1+2\cos x)}{(2+\cos x)^2}$
*$f'(x)=0 \implies \cos x=-\frac{1}{2} \implies x=\frac{2}{3}\pi+2\pi k$
And then using $f''(x)$ in order to be sure that they are indeed min/max (and not curve) points:
*
*$f''(x)=\frac{4\sqrt{3}(\sin\frac{x}{2})^2\sin x}{(2+\cos x)^3}$
*$f''(x)=0 \implies x=\pi k \implies x \neq \frac{2}{3}\pi+2\pi k$
Finally, you can calculate the values of $f(x)$ at these points:
*
*$f(x)=\frac{\sqrt{3}\sin x}{2+\cos x}$
*$x=\frac{2}{3}\pi+2\pi k \implies -1 \leq f(x) \leq 1$
|
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|
Prove this identity: $ \tan(2x)-\sec(2x) =\tan(x-\pi/4)$ I've been having a time with this problem. I tried to start with the left side but I hit a dead end quick... I then tried the right side and had a little more luck but I've hit a block. I first used the tan difference identity then converted everything into sins and cosines. Then I used a difference identity on sin and cosine and then converted into exact values. $\dfrac{\pi}{4}$ for sine and cosine is $\dfrac{\sqrt{2}}{2}$. All the radicals cancelled out leaving me with $\dfrac{\sin x - \cos x }{\cos x + \sin x}$... I can't think anything to do with this... where did I go wrong?
Any answer is appreciated.
|
You're almost there. Dividing the numerator and denominator by $\cos x$ gives $$\frac{\sin x - \cos x}{\cos x + \sin x} = \frac{\tan x - 1}{1 + \tan x},$$ and then observe that $\tan \frac{\pi}{4} = 1$. Then recall the tangent addition identity $$\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}.$$ For what suitable choices of $\alpha$, $\beta$, does the identity match the expression you have?
Okay, so since there's some confusion about the direction in which you're going, and since you already basically have done the proof and just need to see it put all together, here it is: $$\begin{align*} \tan(x - \tfrac{\pi}{4}) &= \frac{\tan x - \tan \frac{\pi}{4}}{1 + \tan x \tan \frac{\pi}{4}} \\ &= \frac{\tan x - 1}{1 + \tan x} \\ &= \frac{\sin x - \cos x}{\cos x + \sin x} \\ &= \frac{-(\sin x - \cos x)^2}{\cos^2 x - \sin^2 x} \\ &= \frac{- \sin^2 x + 2 \sin x \cos x - \cos^2 x}{\cos 2x} \\ &= \frac{-1 + \sin 2x}{\cos 2x} \\ &= \tan 2x - \sec 2x. \end{align*}$$ Going in the other direction, we might write $$\begin{align*} \tan 2x - \sec 2x &= \frac{\sin 2x}{\cos 2x} - \frac{1}{\cos 2x} \\ &= \frac{2 \sin x \cos x - 1}{\cos^2 x - \sin^2 x} \\ &= \frac{2 \sin x \cos x - \sin^2 x - \cos^2 x}{(\cos x + \sin x)(\cos x - \sin x)} \\ &= \frac{-(\cos x - \sin x)^2}{(\cos x + \sin x)(\cos x - \sin x)} \\ &= \frac{\sin x - \cos x}{\cos x + \sin x} \\ &= \frac{\tan x - 1}{1 + \tan x} \\ &= \frac{\tan x - \tan \frac{\pi}{4}}{1 + \tan x \tan \frac{\pi}{4}} \\ &= \tan(x - \tfrac{\pi}{4}).\end{align*}$$
|
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|
Calculating absolute value and argument of a complex number I want to calculate the absolute value and argument of the complex number $a = \left(\sqrt{3} - i\right)^{-2}$.
In order to calculate these two values I tried to reform the number into the form $z = x + y \cdot i$:
$$a =\left(\sqrt{3} - i\right)^{-2} = \frac{1}{\left(\sqrt{3}-i\right)^2} = \frac{1}{2 +\left(-2\sqrt{3}\right)}$$
$$ = -\frac{2}{8} - i \frac{-2\sqrt{3}i}{4-12} = -\frac{1}{4} - i \frac{\sqrt{3}i}{4}$$
I now want to calculate the polar form of a:
$$r = |z| = \sqrt{x^2 + y^2} = \sqrt{\frac{1}{8} + \left( -1 -2i \left( \frac{\sqrt{3}i}{4}\right) + \left(\frac{\sqrt{3}i}{4}\right)^2 \right)}$$
$$ = \sqrt{\frac{1}{8} + \left( -1 -2i \left( \frac{\sqrt{3}i}{4}\right) + \frac{3 + 2 \sqrt{3}i - 1}{16} \right)} =
\sqrt{\frac{1}{8} + \left( -1 -i \left( \frac{\sqrt{3}i}{2}\right) + \frac{3 + 2 \sqrt{3}i - 1}{16} \right)} $$
$$=\sqrt{\frac{1}{8} + -1 \frac{\sqrt{3}-1}{2} + \frac{2 + 2 \sqrt{3}i}{16}}$$
but finally am stuck here. Can you please help me to go on? Is there a better way to get to the absolute value and argument?
|
Rewrite:
\begin{align}
(\sqrt{3}-i)^{-2}&=\frac{1}{(\sqrt{3}-i)^{2}}\\
&=\frac{1}{2-2\sqrt{3}\ i}\\
&=\frac{1}{2-2\sqrt{3}\ i}\cdot\frac{2+2\sqrt{3}\ i}{2+2\sqrt{3}\ i}\\
&=\frac{2+2\sqrt{3}\ i}{2^2-(2\sqrt{3}\ i)^2}\\
&=\frac{2+2\sqrt{3}\ i}{16}\\
&=\frac18+\frac18\sqrt{3}\ i
\end{align}
It should be easy from this to obtain its absolute value and argument. The rest, I leave it to you. Good luck! :)
It looks like you didn't know to evaluate the absolute value and argument of $|z|$. The absolute value can be obtained by using this formula
$$
|z|=\sqrt{x^2+y^2}
$$
and the argument of $z$ is
$$
\arg(z)=\arctan\left(\frac yx\right).
$$
In this case, we have $x=\Re(z)=\frac18$ and $y=\Im(z)=\frac18\sqrt{3}$. You may refer to this link to learn more about complex number. I hope this helps.
|
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|
Three-Digit numbers divisbile by 3 How many three digit numbers are divisible by 3 and have an additional property that the sum of of their digits is 4 times the middle digit?
My approach: let the number be $abc$ so $$abc \equiv 0\pmod{3}$$ and $$a + b+ c= 4b$$
I'm stuck now. Any help?
|
First, $b$ should be dividable by $3$, since $a+c=3b$, we have $0<3b<18$. So $b$ can be $3$ and $6$.
For $b=3$, we have $a+c=9$, there are 9 choices;
for $b=6$, we have $a+c=18$, then the only choice is $a=c=9$.
|
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|
Find the limit $\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8} }{x}$ I am trying to evaluate the following limit
$$\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8} }{x}$$
If we use the binomial expansion of the numerator term, the answer is $\frac{1}{4}$. The same answer is obtained if we apply L'Hospital's rule.
Is it possible to solve this using something even more basic ? i.e. without recourse to binomial expansion or L'Hospital's rule or definition of derivative ? The reason I ask this is because this problem (from a textbook) comes up before the introduction of L'Hospital's rule.
user37238 wanted to know how the binomial expansion can be used.Basically, we have
$(1+x)^n= 1 + nx + \frac{n (n-1)}{2} + \frac{n (n-1)(n-2)}{2!} + \ldots$
Similarly,
$(1-x)^n= 1 - nx + \frac{n (n-1)}{2} - \frac{n (n-1)(n-2)}{2!} + \ldots$
Using these expansions with $n=\frac{1}{8}$ in the given problem gives the answer.
|
For example, use
\begin{align}
&\;\;\;\frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8}}{x}=\\&=\frac{1}{x}\cdot\frac{(1+x)^\frac{1}{4}-(1-x)^\frac{1}{4}}{(1+x)^\frac{1}{8}+(1-x)^\frac{1}{8}}=\\
&=\frac{1}{x}\cdot\frac{(1+x)^\frac{1}{2}-(1-x)^\frac{1}{2}}{\left[(1+x)^\frac{1}{8}+(1-x)^\frac{1}{8}\right]\left[(1+x)^\frac{1}{4}+(1-x)^\frac{1}{4}\right]}=\\
&=\frac{1}{x}\cdot\frac{(1+x)-(1-x)}{\left[(1+x)^\frac{1}{8}+(1-x)^\frac{1}{8}\right]\left[(1+x)^\frac{1}{4}+(1-x)^\frac{1}{4}\right]\left[(1+x)^\frac{1}{2}+(1-x)^\frac{1}{2}\right]}=\\
&=\frac{2}{\left[(1+x)^\frac{1}{8}+(1-x)^\frac{1}{8}\right]\left[(1+x)^\frac{1}{4}+(1-x)^\frac{1}{4}\right]\left[(1+x)^\frac{1}{2}+(1-x)^\frac{1}{2}\right]}.
\end{align}
Now to compute the limit, it suffices to set $x=0$ in the last expression.
|
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|
Finding an integer $c$ for which $a+c^2 \equiv 0 \pmod {2b-2c}$, $a$ & $b$ constant Looking for a solution for such a challenge, I have a decision problem that is solved if there is a positive integer $c$, which for given integer constants $a$, $b$ satisfies the equation
$a+c^2 \equiv 0 \pmod {2b-2c}$
or simply to say $a+c^2$ is divisible by $2b - 2c$.
Additional facts, maybe helpful
*
*$0<c<b$
*$a<2b$
Not sure how to approach this, beside brute force, looking for an elegant solution.
|
There are many choices for $c$, depending on the given integers $a$ and $b$.
Given
$a+c^2 \equiv 0 \pmod {2b-2c}$, subtract $b^2$ from both sides;$a+c^2-b^2 \equiv -b^2\pmod {2b-2c}$
multiplying through by $2$;
$2a+(2c-2b)(c+b) \equiv -2b^2 \pmod {2b-2c}$
or
$2(a+b^2) \equiv 0 \pmod {2b-2c}$. Or $(a+b^2) \equiv 0 \pmod {b-c}$
From here, input the given values of $a$ and $b$, then equate any factor of $a+b^2$ (that is less than $b$) to $b-c$
|
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|
Show the limit of the following is $\dfrac{1}{12}$
Show that $$\lim_{n\to \infty}n^2 \log \left(\dfrac{r_{n+1}}{r_{n}} \right)=\dfrac{1}{12}$$ where $r_{n}$ is defined as;
$$r_{n}=\dfrac{\sqrt{n}}{n!} \left(\dfrac{n}{e} \right)^n$$.
Now I simplified $\dfrac{r_{n+1}}{r_{n}}$ to $$\dfrac{r_{n+1}}{r_{n}}= \left(\dfrac{n+1}{n} \right)^{n+0.5} \dfrac{1}{e}$$
And so simplified $n^{2} \log \left(\dfrac{r_{n+1}}{r_{n}} \right)$ to $$n^{2}\left((n+0.5) \log \left(\dfrac{n+1}{n} \right)-1 \right).$$
Now I checked this on wolfram and as $n \rightarrow \infty$ this does converge t $\dfrac{1}{12}$ but I am not able to show it.
I have tried L'Hopital i.e $$\lim_{n\to \infty} \dfrac{n \log \left(\dfrac{n+1}{n} \right) +0.5 \log \left(\dfrac{n+1}{n} \right)-1}{\dfrac{1}{n^2}}$$
Now as $n \rightarrow \infty$ $\log \left(\dfrac{n+1}{n} \right) \rightarrow 0$ but you end up with a $\dfrac{-1}{0}$ type limit.
Any help with this would be much appreciated.
EDIT: I have realised I have made a mistake and that $$\lim_{n\to \infty} \dfrac{n \log \left(\dfrac{n+1}{n} \right) +0.5 \log \left(\dfrac{n+1}{n} \right)-1}{\dfrac{1}{n^2}}$$ gives you a $\dfrac{0}{0}$ type limit, will delve into the algebra and get back.
|
Here is the remaining part -
$$\lim_{n\to \infty} n^{2}((n + 0.5)\log(1+\frac{1}{n})-1)\\\\\,\\
\lim_{n\to \infty} n^{2}((\frac{n+0.5}{n}-\frac{n+0.5}{2n^2}+\frac{n+0.5}{3n^3} - ...) -1)\\\\\,\\
\lim_{n\to \infty} n^{2}(\frac{0.5}{n}-\frac{n+0.5}{2n^2}+\frac{n+0.5}{3n^3} - ...)\\\\\,\\
\lim_{n\to \infty} (0.5n-0.5n - \frac{1}{4} + \frac{1}{3} + \frac{0.5}{3n} - ...)\\\\\,\\
1/3 - 1/4 = 1/12
$$
|
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|
How should I simplify this $\tan^{-1}$ expression? Integral question I have to integrate this function
$$I = \int_0^4\frac{20x-5x^2}{x^2+9} \mathrm{d}x$$
obtaining
$$20\ln(5/3) + 15\tan^{-1}(4/3) -20.$$
However, my calculator, even after somehow simplifying it a bit, gives this:
$$\frac{40\ln(5/3) -30\tan^{-1}(3/4) +15\pi -40}{2}$$
As you can see there is something wrong with arctan integration, can anybody help or know how to simplify this with some identity?
I am asking this because it is one of the questions in the practise exam, and in the exam i will have to use this calculator, no others allowed, so if a question like this comes up, i will get stuck..
|
By first simplifying the fraction, we get that $$\frac{40\ln(5/3) -30\tan^{-1}(3/4) +15\pi -40}{2} = 20\ln(5/3) -15\tan^{-1}(3/4) +\frac{15}{2}\pi -20$$
Subtracting $20\ln(5/3) + 15\tan^{-1}(4/3) -20$ from it, we get $$(20\ln(5/3) -15\tan^{-1}(3/4) +\frac{15}{2}\pi -20) - (20\ln(5/3) + 15\tan^{-1}(4/3) -20)$$
$$ = \frac{15}{2}\pi - 15\tan^{-1}(4/3) -15\tan^{-1}(3/4) = 15 \left(\frac{\pi}{2} - \tan^{-1} (4/3) - \tan^{-1} (3/4)\right)$$
Here we bring in the definition of tangent. Given a right triangle $ABC$ with $C$ being the right angle, $$\tan \angle ABC = \frac{AC}{BC}$$ From this, we get that $$\angle ABC = \tan^{-1} \frac{AC}{BC}$$
Now, we take a $3-4-5$ triangle and note that the two inverse tangents are the two angles that are not $\displaystyle \frac{\pi}{2}$. Because the angles of a triangle add up to $\pi$, we note that they sum to $\displaystyle \frac{\pi}{2}$. Therefore, $$15 \left(\frac{\pi}{2} - \tan^{-1} (4/3) - \tan^{-1} (3/4)\right) = 15 \left(\frac{\pi}{2} - \frac{\pi}{2}\right) = 0$$
Q.E.D.
|
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|
Prove $n^2(n^4-1)$ is divisible by 60 using Mathematical Induction. Base step:
p(2)=4 * 15= 60
Inductive Hypothesis:
Assuming p(k) = $k^2(k^4-1)$ = 60q
Induction:
p(k+1)= $(k+1)^2[(k+1)^4-1]$
= $(k+1)^2[(k+1)^2 + 1][(k+1)^2 - 1]$
= $(k+1)^2(k^2+2k+2)(k^2+2k+1- 1)$
= $(k+1)^2(k^2+2k+2) k (k+2)$
= $k(k+1)(k+2)(k^2+2k+2)(k+1)$
Now $k(k+1)(k+2)$ is the product of 3 consecutive natural numbers. Their product HAS to be a multiple of 2 and 3 and hence also a multiple of 6. So all that I have to do is prove that $(k^2+2k+2)(k+1)$ is a multiple of 10. But I can't seem to do so. Also through my induction, I have not at all made use of my hypothesis! How do I go about solving this question through the Principle of Mathematical Induction?
|
Does it have to be by induction?
Write $n^2(n^4-1)= n^2 (n-1) (n+1) (n^2+1) = n(n-1) n (n+1) (n^2+1)$.
$2$ divides $n(n-1)$ and $n(n+1)$ and so $4$ divides $n^2(n^4-1)$.
$3$ divides $(n-1)n(n+1)$ and so $3$ divides $n^2(n^4-1)$.
Since $60=4\cdot3\cdot5$, it remains to prove that $5$ divides $n^2(n^4-1)=n(n^5-n)$. But this follows from Fermat's little theorem and also by induction.
|
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|
Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$ Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$.
I have no clue how to proceed and tried to prove that the whole equation becomes $0$ when $\sin\frac{\pi}{14}$ is placed in place of $x$ but couldn't do anything further. I think the symbols might be different but can be the same. If it is correct, please help me to solve this; if the equation is wrong, then please modify it and solve it.
|
$$\sin\frac\pi{14}=\cos\left(\frac\pi2-\frac\pi{14}\right)=\cos\frac{3\pi}7$$
Let $\displaystyle z^7=-1=\cos\pi+i\sin\pi$
Using this, $\displaystyle z=\cos\frac{(2n+1)\pi}7+i\sin\frac{(2n+1)\pi}7$ where $n\equiv-3,-2,-1,0,1,2,3\pmod7$
So, the equation whose roots are $\displaystyle z=\cos\frac{(2n+1)\pi}7+i\sin\frac{(2n+1)\pi}7$ where $n\equiv-3,-2,-1,0,1,2\pmod7$ is $$\frac{z^7+1}{z+1}=0\iff z^6-z^5+z^4-z^3+z^2-z+1=0$$
Like this divide either sides by $z^3\ne0$ to get $$z^3+\frac1{z^3}-\left(z^2+\frac1{z^2}\right)+z+\frac1z-1=0$$
$$\iff\left(z+\frac1z\right)^3-3\left(z+\frac1z\right)-\left[\left(z+\frac1z\right)^2-2\right]+z+\frac1z-1=0$$
$$\iff \left(z+\frac1z\right)^3-\left(z+\frac1z\right)^2-2\left(z+\frac1z\right)+1=0 $$
Observe that $\displaystyle n\equiv-2,1\pmod7\implies z+\frac1z=2\cos\frac{3\pi}7$
Similarly for $\displaystyle n\equiv-1,0\pmod7; \equiv-3,2\pmod7$
So, $\displaystyle2\cos\frac{3\pi}7$ is a root of $$u^3-u^2-2u+1=0 $$
$\displaystyle\implies\cos\frac{3\pi}7$ is a root of $?$
|
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|
Closed form of $ \int_0^{\pi/2}\ln\big[1-\cos^2 x(\sin^2\alpha-\sin^2\beta \sin^2 x)\big]dx$ Hello I am trying to solve an incredible integral given by
$$
\int_0^{\pi/2}\ln\big[1-\cos^2 x(\sin^2\alpha-\sin^2\beta \sin^2 x)\big]dx=\pi \ln\bigg[\frac{1}{2}\left(\cos^2\alpha +\sqrt{\cos^4 \alpha +\cos^2\frac{\beta}{2} \sin^2 \frac{\beta}{2}}\right)\bigg],\qquad \alpha > \beta >0.
$$
I defined
$$
I\equiv
\int_0^{\pi/2}\ln\big[1-\cos^2 x(\sin^2\alpha-\sin^2\beta \sin^2 x)\big]dx
$$
and using $\cos^2 x=1-\sin^2 x$ but obtained a more complicated expression. Usually it is easier to work with a closed form of Log Sine so I was trying this. I am not really sure how else to approach this at all.
The result looks like very nice and similar to something we all know :)
I am looking for real or complex methods to solve this problem. I am not sure of what substitutions to make but maybe we could work in hyperbolic space
|
Here is a sketchy calculation. It matches numerically for all values of $a,b$ that I tried.
Abbreviate $a = \cos^2 \alpha$ and $b = \sin^2 \beta$. Then $0 \leq a,b \leq 1$.
We calculate
$$
I(a,b)= \int _{0} ^{\pi/2} \ln(1-\cos^2 x(1-a-b \sin^2 x)) \,dx
$$
We have
$$
\partial_aI(a,b)= \int _{0} ^{\pi/2} \frac {\cos^2 x} {1-\cos^2 x(1-a-b \sin^2 x)} \,dx \\
=
\int _{0} ^{\pi/2} \frac {1} {\tan^2 x+a+b \sin^2 x} \,dx \\
=\int _{0} ^{\infty} \frac {1} {(1+t^2)(t^2+a+b \frac {t^2} {1+t^2})} \,dt \\
$$
(by substituting $\tan x = t$ and using some trig identities)
$$
= \frac{1}{2} \int _{-\infty} ^{\infty} \frac {1} {t^4+(a+b+1)t^2+a} \,dt \\
= \frac{1}{2} \frac {\pi} {\sqrt{a(2 \sqrt{a}+a+b+1)}}
$$
(for example using residues; here it is crucial that $(a+b+1)^2 - 4a^2 \geq 0 $, which is the case because $0 \leq a,b \leq 1$)
By integrating with respect to $a$ (for example by setting $u = \sqrt{a}$), we hence obtain
$$
I(a,b) = \pi \ln{ \left(1 + \sqrt{a} + \sqrt{1 + 2 \sqrt{a} + a + b} \right) } + C(b)
$$
Here $C(b)$ is an integration constant depending on $b$ only. To fix it, we substitute $a = 1$ in the integral. We obtain, using some trig identities,
$$
I(1,b) = \int _{0} ^{\pi/2} \ln(1+b \sin^2 x \cos^2 x) \,dx \\
= \frac{1}{2} \int _{0} ^{\pi} \ln\left(1+ \frac{1}{4} b \sin^2 x\right) \,dx \\
= \int _{0} ^{\pi/2} \ln\left(1+ \frac{1}{4} b \sin^2 x\right) \,dx \\
= \pi \ln \left(\frac{1+\sqrt{1+b/4}}{2} \right) \\
$$
Here I used a previously calculated integral. On the other hand, we have $I(1,b) = C(b) + \pi \ln(2(1+\sqrt{1+b/4}))$. Comparing yields the value of $C(b)$. Plugging this back in, we get:
$$
I(a,b)=\pi \ln\left( \frac{1+ \sqrt{a} + \sqrt{2\sqrt{a} + a+b + 1}}{4} \right)
$$
|
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|
Expectation with square root I don't know how to calculate the expectation when there is some square root in the expression. My problem is this: we have three real random variables $X,Y,Z$, independent and with standard normal distribution $N(0,1)$ and we want to calculate
$$E\left(\frac{1}{2} \left(X + Z + \sqrt{X^2 + 4 Y^2 - 2 XZ + Z^2}\right)\right).$$
How can this be done?
Thanks.
|
I am assuming that $X$, $Y$, and $Z$ are mutually independent, standard Gaussian.
The expression of your expectation can be simplified.
Let us denote your expectation as $E$.
First, notice that $X$ and $Y$ are zero-mean, so
$$
E = \frac{1}{2}\mathsf{E}\left[\sqrt{X^2 + 4Y^2 - 2XZ + Z^2}\right]
$$
Grouping terms, we have
$$
\begin{align}
E
&= \frac{1}{2}\mathsf{E}\left[\sqrt{4Y^2 + (X-Z)^2}\right] \\
&= \frac{1}{2}\mathsf{E}\left[\sqrt{4Y^2 + (X+Z)^2}\right].
\end{align}
$$
The last equality holds because $Z$ and $-Z$ have the same distribution, due to the fact that the distribution of $Z$ is symmetric around $0$. In addition, since $X$ and $Z$ are independent standard Gaussian, their sum is zero-mean Gaussian with variance $2$. That is, the sum $X+Z$ has the same distribution as $\sqrt{2}X$. Hence,
$$
\begin{align}
E
&= \frac{1}{2}\mathsf{E}\left[\sqrt{4Y^2 + (\sqrt{2}X)^2}\right] \\
&= \frac{\sqrt{2}}{2}\mathsf{E}\left[\sqrt{2Y^2 + X^2}\right]
\end{align}
$$
@bobbym has provided a closed-form expression of this expectation computed with Mathematica in terms of an elliptic integral (see below). The way to derive it is as follows:
$$
\begin{align}
E &= \frac{\sqrt{2}}{2}\int_0^\infty\int_0^\infty\sqrt{2y^2 + x^2} \cdot \frac{1}{\sqrt{2\pi}}e^{-x^2/2} \cdot \frac{1}{\sqrt{2\pi}}e^{-y^2/2} \ \mathrm{d}x\,\mathrm{d}y \\
&= \frac{\sqrt{2}}{4\pi}\int_0^\infty\int_{-\pi}^{\pi}r^2\sqrt{1+\sin^2\theta} e^{-r^2/2} \ \mathrm{d}r\,\mathrm{d}\theta.
\end{align}
$$
This is obtained by means of the substitution
$$
\begin{align}
x &= r\cos(\theta) \\
y &= r\sin(\theta)
\end{align}
$$
Then you use
$$
\int_0^\infty r^2 e^{-r^2/2} \mathrm{d}r = \sqrt{2\pi}
$$
and the symmetry of the integral over $\theta$ to show
$$
\begin{align}
E
&= \frac{1}{\sqrt{\pi}}\int_{0}^{\pi}\sqrt{1+\sin^2\theta} \,\mathrm{d}\theta \\
&= \frac{\operatorname{E}(-1)}{\sqrt{\pi}}
\end{align}
$$
consistently with what @bobbym computed.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show $ \frac {a^3}{a^2+b^2+c^2-bc}+\frac {b^3}{a^2+b^2+c^2-ca} +\frac {c^3}{a^2+b^2+c^2-ab} \le \frac {(a+b+c)^2 (a^4+b^4+c^4)}{6abc(ab+bc+ca)} $ If $a,b,c$ are positive real numbers then how do we prove that $ \dfrac {a^3}{a^2+b^2+c^2-bc}+\dfrac {b^3}{a^2+b^2+c^2-ca} +\dfrac {c^3}{a^2+b^2+c^2-ab} \le \dfrac {(a+b+c)^2 (a^4+b^4+c^4)}{6abc(ab+bc+ca)} $
Please help. Thanks
|
$(a+b+c)^2\geq3(ab+ac+bc)$.
Thus, it remains to prove that
$$\sum_{cyc}\frac{a^3}{a^2+b^2+c^2-bc}\leq\frac{a^4+b^4+c^4}{2abc}$$ or
$$\sum_{cyc}\left(\frac{a^3}{2bc}-\frac{a^3}{a^2+b^2+c^2-bc}\right)\geq0$$ or
$$\sum_{cyc}\frac{a^3(a^2+b^2+c^2-3bc)}{bc(a^2+b^2+c^2-bc)}\geq0$$ or
$$\sum_{cyc}\frac{a^4(a^2+b^2+c^2-3bc)}{a^2+b^2+c^2-bc}\geq0.$$
But $$\left(\frac{a^4}{a^2+b^2+c^2-bc},\frac{b^4}{a^2+b^2+c^2-ac},\frac{c^4}{a^2+b^2+c^2-ab}\right)$$ and
$$(a^2+b^2+c^2-3bc,a^2+b^2+c^2-3ac,a^2+b^2+c^2-3ab)$$
are the same ordered.
Thus, by Chebyshov we obtain
$$\sum_{cyc}\frac{a^4(a^2+b^2+c^2-3bc)}{a^2+b^2+c^2-bc}\geq\frac{1}{3}\sum_{cyc}\frac{a^4}{a^2+b^2+c^2-bc}\sum_{cyc}(a^2+b^2+c^2-3bc)=$$
$$=\sum_{cyc}\frac{a^4}{a^2+b^2+c^2-bc}\sum_{cyc}(a^2-ab)\geq0.$$
Done!
|
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|
Show $a^2+b^2+c^2=1$ $\implies$ $ \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 $ If $a$, $b$ and $c$ are real numbers such that $a^2+b^2+c^2=1$ , then how to prove that
$ \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 $
( don't apply Schur's inequality )? Thanks.
|
By C-S
$$\sum_{cyc}\frac{b^2c^2}{1+a^2}=\sum_{cyc}\frac{b^2c^2}{a^2+b^2+a^2+c^2}\leq$$
$$\leq\sum_{cyc}\frac{b^2c^2}{4}\left(\frac{1}{a^2+b^2}+\frac{1}{a^2+c^2}\right)=\frac{1}{4}\sum_{cyc}\left(\frac{b^2c^2}{a^2+b^2}+\frac{b^2c^2}{a^2+c^2}\right)=$$
$$=\frac{1}{4}\sum_{cyc}\left(\frac{b^2c^2}{a^2+b^2}+\frac{a^2c^2}{a^2+b^2}\right)=\frac{1}{4}.$$
Done!
|
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|
Show that an integer of the form $8k + 7$ cannot be written as the sum of three squares. I have figured out a (long, and tedious) way to do it. But I was wondering if there is some sort of direct correlation or another path that I completely missed.
My attempt at the program was as follows:
A number of the form, $8k + 7 = 7 (mod 8)$. That is, we are looking for integers a, b, c such that $a^2 + b^2 + c^2 = 7 (mod 8)$.
LONG and TEDIOUS way:
$$(8k)^2 = 0 (mod 8)$$
$$(8k+1)^2 = 1 (mod 8)$$
$$(8k+2)^2 = 4 (mod 8)$$
$$(8k+3)^2 = 1 (mod 8)$$
$$(8k+4)^2 = 0 (mod 8)$$
$$(8k+5)^2 = 1 (mod 8)$$
$$(8k+6)^2 = 4 (mod 8)$$
$$(8k+7)^2 = 1 (mod 8)$$
That is, using three of these modulo there is no way to arrive at
$$a^2 + b^2 + c^2 = 7 (mod 8)$$
|
A variation:
Let $n = 8k + 7 = x^2 + y^2 +z^2$.
Then $n \equiv 3 \pmod 4$. Every square is congruent to $0$ or $1$ mod $4$, so $x^2, y^2, z^2 \equiv 1 \pmod 4$ for these numbers to add up to $ 3 \bmod 4$. Then $x, y, z$ are odd.
One checks immediately that $(\pm 3)^2 \equiv 1 \pmod 8, (\pm 1)^2 \equiv 1 \pmod 8$, which accounts for all odd numbers, so $n = x^2 + y^2 +z^2 \equiv 3 \pmod 8$. But this contradicts $n \equiv 7 \pmod 8$.
This is only a speedup up if you think it's faster to reduce mod $4$ to learn that you don't have to calculate $(\pm 2)^2 \bmod 8$ and $4^2 \bmod 4$.
|
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|
How to expand out negative powers for complex numbers I have the following expansions but I don't know how my teacher gets them. Apparently there is a formula for it (though the guy who told me didn't know it well), but I cannot find it in my notes.
For example, let $z$ be a complex number. Then,
$$\frac{1}{z^2+1}=\frac{i/2}{z+i}-\frac{i/2}{z-i}$$
It makes total sense to me in this case, but I get completely confused in this second case:
$$\frac{1}{(z^2+1)^2}=\frac{i/4}{z+i}-\frac{1/4}{(z+i)^2}-\frac{i/4}{z-i}-\frac{1/4}{(z-i)^2}$$.
What is/are the formulas that I need to compute this? Thanks!
|
Start from the original equation
$$
\frac{1}{z^2+1} = \frac{i/2}{z+i} - \frac{i/2}{z-i}
$$
And just square it.
$$
\frac{1}{(z^2+1)^2} = \left(\frac{i/2}{z+i}\right)^2 + \left(\frac{i/2}{z-i}\right)^2 - 2\frac{i/2}{z+i}\frac{i/2}{z-i}
= -\frac{1/4}{(z+i)^2} - \frac{1/4}{(z-i)^2} + \frac{1/2}{z^2+1}
$$
Reuse the first equation to get:
$$
\frac{1}{(z^2+1)^2} = -\frac{1/4}{(z+i)^2} - \frac{1/4}{(z-i)^2} + \frac{i/4}{z+i} - \frac{i/4}{z-i}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove that the following function is convex? I want to prove convexity of the following function:
$$f(x) = log_x \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)$$
for any fixed $a, b \in (0, 1)$ and:
*
*$x\in(0,1)$
*$x\in(1, \infty)$
I'm trying to solve it for a very long time, tried to investigate the sign of second derivative, but it leads to very ugly equation which I am unable to handle. Any hints are appreciated. Thanks!
Edit:
$f''(x)$ equals to:
$$-\frac{{\left(\frac{x^{x - 1} {\left(p^{y} - 1\right)} x}{p - 1} + \frac{{\left(p^{x} - 1\right)}
x^{y - 1} y}{p - 1} - \frac{{\left(p^{x} - 1\right)} {\left(p^{y} - 1\right)}}{{\left(p -
1\right)}^{2}}\right)}^{2}}{{\left(\frac{{\left(x^{x} - 1\right)} {\left(p^{y} - 1\right)}}{p - 1} +
1\right)}^{2} \log\left(x\right)} +\\
+ \frac{\frac{x^{x - 2} {\left(p^{y} - 1\right)} {\left(x - 1\right)} x}{p - 1} + \frac{2 \, p^{x -
1} x^{y - 1} x y}{p - 1} + \frac{{\left(p^{x} - 1\right)} p^{y - 2} {\left(y - 1\right)} y}{p - 1} -
\frac{2 \, x^{x - 1} {\left(p^{y} - 1\right)} x}{{\left(p - 1\right)}^{2}} - \frac{2 \, {\left(p^{x}
- 1\right)} x^{y - 1} y}{{\left(p - 1\right)}^{2}} + \frac{2 \, {\left(p^{x} - 1\right)}
{\left(x^{y} - 1\right)}}{{\left(p - 1\right)}^{3}}}{{\left(\frac{{\left(p^{x} - 1\right)}
{\left(x^{y} - 1\right)}}{p - 1} + 1\right)} \log\left(p\right)} -\\
- \frac{2 \, {\left(\frac{x^{x - 1} {\left(p^{y} - 1\right)} x}{p - 1} + \frac{{\left(p^{x} -
1\right)} x^{y - 1} y}{p - 1} - \frac{{\left(p^{x} - 1\right)} {\left(p^{y} - 1\right)}}{{\left(p
- 1\right)}^{2}}\right)}}{{\left(\frac{{\left(x^{x} - 1\right)} {\left(p^{y} - 1\right)}}{p - 1} +
1\right)} x \log\left(p\right)^{2}} + \frac{\log\left(\frac{{\left(p^{x} - 1\right)} {\left(p^{y} - 1\right)}}{p - 1} + 1\right)}{p^{2}
\log\left(x\right)^{2}} + \frac{2 \, \log\left(\frac{{\left(p^{x} - 1\right)} {\left(p^{y} - 1\right)}}{p - 1} + 1\right)}{p^{2} \log\left(p\right)^{3}}
$$
Some elements of that equation are positive, some of them are negative - I do not see any way of how to prove that $f''(x) > 0$.
I have also tried to find counter example by testing the Jensen's inequality on randomly generated data, but everything seems OK - the function seems to be convex, but I am unable to prove it.
|
You can write
$$
f(x) = \log_x \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)
= \frac{\ln \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)}{\ln x},
$$
and prove that $f''(x) > 0$ where needed.
|
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|
Let f be a continuous function defined on [-2009,2009] such that f(x) is irrational for each $x \in [-2009,2009]$ ... Problem : Let f be a continuous function defined on [-2009,2009] such that f(x) is irrational for each $x \in [-2009,2009]$ and $f(0) =2+\sqrt{3}+\sqrt{5}$ Prove that the equation $f(2009)x^2 +2f(0)x +f(2009)=0$ has only rational roots.
Solution :
Let us assume the function f(x) be $2+(\sqrt{3}+\sqrt{5})(x+1)$ which satisfies $f(0) =2+\sqrt{3}+\sqrt{5}$
now $f(2009)x^2 =(2010((\sqrt{3}+\sqrt{5})x^2$ .....(1)
Also $2f(0)x =2(2+\sqrt{3}+\sqrt{5})x$ ...(2)
f$(2009)=(2010)(\sqrt{3}+\sqrt{5})$ ......(3)
Now solving for $f(2009)x^2 +2f(0)x +f(2009)=0$ by putting the values of $f(2009)x^2 , 2f(0)x ; f(2009)$ from 1,2 & 3 I am getting roots which are irrational please suggest some other method for this ... thanks....
|
A continuous function that takes only irrational values on an interval must be constant. Hence $f(2009)=f(0)\ne0$ and the quadratic is equivalent to $x^2+2x+1=0$ (with double root $x=-1$.
|
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|
Prove that $\lim_{n\to\infty} (\sqrt{n^2+n}-n) = \frac{1}{2}$ Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?
$\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| < \epsilon$
$\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| < \epsilon$
$\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} < \epsilon$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} > \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \sqrt{n} > \frac{1-2 \epsilon}{2}$
$\Rightarrow n > \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
|
Since this post became active after a long period of time and there are many linked questions, here is a generalization for such kinds of limits. Consider the set of functions in the following form:
$$
f(x) = \sqrt[n]{(x + a_1)(x+a_2)\cdots (x+a_n)} - x
$$
where
$$
n,k \in \Bbb N \\
a_k \in \Bbb R
$$
Consider the following expansion:
$$
a^n – b^n = (a – b)\left(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 +
\cdots + ab^{n – 2} + b^{n – 1}\right) \tag1
$$
Define a function in the form:
$$
g(x) = \sqrt[n]{(x + a_1)(x+a_2)\cdots (x+a_n)}
$$
Thus $f(x)$ may be rewritten as:
$$
f(x) = g(x) - x
$$
Now using $(1)$ we may rewrite $f(x)$ as:
$$
f(x) = \frac{(g(x))^n-x^n}{\sqrt[n]{(g(x))^{n-1}} +\sqrt[n]{(g(x))^{n-2}}x + \sqrt[n]{(g(x))^{n-3}}x^2 + \cdots + \sqrt[n]{g(x)}x^{n-2} +x^{n-1}} \tag2
$$
Now taking the limit of $(2)$ one may obtain:
$$
\begin{align*}
&\lim_{x\to\infty}\frac{x^{n-1}a_1 + x^{n-1}a_2+ \cdots + x^{n-1}a_n }{x^{n-1}\left(\sqrt[n]{1+o\left({1\over x}\right)+\cdots} + \sqrt[n]{1+o\left({1\over x}\right) +\cdots} +\cdots + 1\right)} \\
= &\lim_{x\to\infty}\frac{a_1 + a_2+ \cdots + a_n}{\sqrt[n]{1+o\left({1\over x}\right)+\cdots} + \sqrt[n]{1+o\left({1\over x}\right) +\cdots} +\cdots + 1} \\
= &\frac{a_1 + a_2 + \cdots + a_n}{n} \tag3
\end{align*}
$$
Or summarizing:
$$
\boxed{\lim_{x\to\infty}f(x) = \frac{a_1 + a_2 + \cdots + a_n}{n}}
$$
Lets now use that result in your limit:
$$
\sqrt{n^2 + n} - n = \sqrt{n(n+1)} - n = \sqrt{(n+0)(n+1)} - n
$$
Which gives $a_1 = 0$ and $a_2 = 1$ and the highest power is $2$ meaning $n = 2$ in $(3)$. Thus we get:
$$
\boxed{\lim_{n\to\infty}x_n = \frac{a_1 + a_2}{2} = \frac{0 + 1}{2} = {1\over 2}}
$$
As desired.
|
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|
Solve $X_n=\lfloor \sqrt{X_n} \rfloor+X_{n-1}$ How do I solve $X_n=\lfloor \sqrt{X_n} \rfloor+X_{n-1}$? The initial terms are $1,2,3,5,7,10,13,17,21,26,31$. A search on oeis.org/ gave $\lfloor n/2 \rfloor\cdot\lceil n/2 \rceil$ + 1 which should be proven by induction. Is there a different approach?
|
The difference equation
\begin{align}
x_{n} = \lfloor \sqrt{x_{n}} \rfloor + x_{n-1},
\end{align}
where $x_{0}=1 $ and $x_{1}=2$ can, upon writing out several terms, ie $x_{2}$, $x_{3}$, $x_{4}$, and so on can the given set of values stated. It is without much difficulty to show that the difference equation for $x_{n}$ can also be seen as
\begin{align}
a_{n} = a_{n-2} + n
\end{align}
where $a_{o}=1$ and $a_{2}=2$. The set generated by $a_{n}$ is $a_{n} \in \{ 1,2,3,5,7,10,13,17,21,26,31, \cdots \}$. Now the even and odd values can be considered and yields
\begin{align}
a_{2n} = n^{2} +1
\end{align}
for $n\geq 0$ for the even case and
\begin{align}
a_{2n+1} = n^{2} + n+1 = n(n+1) + 1
\end{align}
for the odd case. From the two cases it can readily be sen that the general result is given by
\begin{align}
a_{n} = \left\lfloor \frac{n}{2} \right\rfloor \cdot \left\lceil \frac{n}{2} \right\rceil + 1.
\end{align}
|
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|
How to prove Fibonacci sequence with matrices? How do you prove that:
$$
\begin{pmatrix}
1 & 1\\
1 & 0
\end{pmatrix}^n
=
\begin{pmatrix}
F_{n+1} & F_n\\
F_{n} & F_{n-1}
\end{pmatrix}$$
|
$$\begin{align}
F(n+1) &= 1\,F(n) + 1\,F(n-1)\\
F(n) &= 1\,F(n) + 0\,F(n-1)\\
\\
\begin{bmatrix} F(n+1) \\ F(n) \end{bmatrix} &= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} F(n) \\ F(n - 1) \end{bmatrix} \\
\begin{bmatrix} F(n+1) \\ F(n) \end{bmatrix} &= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n \begin{bmatrix} F(1) \\ F(0) \end{bmatrix} \\
\\
\text{as well as} \\
\begin{bmatrix} F(n) \\ F(n-1) \end{bmatrix} &= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n \begin{bmatrix} F(0) \\ F(-1) \end{bmatrix} \\
\\
\text{from which it follows}\\
\begin{bmatrix} F(n+1) & F(n) \\ F(n) & F(n-1) \end{bmatrix} &= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n \begin{bmatrix} F(1) & F(0) \\ F(0) & F(-1) \end{bmatrix} \\
\\
\text{and choosing} \\
F(1) &= 1 \\
F(0) &= 0 \\
F(-1) &= 1
\end{align}$$
|
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|
Computing the sum $\sum_{n=2}^{2011}\sqrt{1+\frac{1}{n^2}+\frac{1}{(n-1)^2}}$ Find the value of $$\sqrt{1 + \frac{1}{1^2} + \frac{1}{2^2}} + \sqrt{1 + \frac{1}{2^2} + \frac{1}{3^2}} + \sqrt{1 + \frac{1}{3^2} + \frac{1}{4^2}} +...+ \sqrt{1 + \frac{1}{2010^2} + \frac{1}{2011^2}}$$
I have not been able to simplify the terms. Although there is a clear pattern, it does not result in any easy solution. This was a MCQ type question and the options were really simple terms, so I am assuming that this can be simplified really neatly. Please help.
|
Let us first note that
$$n^2+(n-1)^2+n^2(n-1)^2=(n^2-n+1)^2,$$
and therefore the sum can be rewritten as
$$\sum_{n=2}^{2011}\sqrt{1+\frac{1}{n^2}+\frac{1}{(n-1)^2}}=\sum_{n=2}^{2011}\frac{n^2-n+1}{n(n-1)}=\sum_{n=2}^{2011}\left(1+\frac{1}{n-1}-\frac1n\right)=2011-\frac{1}{2011}.$$
The sum of $1$'s at the last step gives $2010$. The other two terms produce a telescoping series so that only the first (equal to $1$) and last (equal to $-\frac{1}{2011}$) terms survive.
|
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|
Understanding substitution to compute primitive To compute the primitive
$$\int x^2 \sqrt[3]{x^3+3}\ dx$$
I am trying this:
$t = x^3+3$
$dt = 3x^2dx$
but
$\int x^2 \sqrt[3]{x^3+3}\ dx=\ ?$
How to continue from here?
|
Take the differential form ($dx$) out from underneath the squareroot sign. It reads $\int x^2\cdot (x^3+3)^{1/3} dx = 1/3 \int (3x^2 dx)\cdot (x^3+3)^{1/3} = 1/3\cdot (x^3+3)^{4/3} / (4/3) + c = \frac{(x^3+3)^{4/3}}{4} + c$.
|
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|
Integral $\int_0^1 \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}}dx=-\left(\frac{\pi}{2c}\right)^2\sec ^2 \frac{\pi}{2c}$ Hi I am trying to prove this result $$
I:=\int_0^1 \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}}dx=-\left(\frac{\pi}{2c}\right)^2\sec ^2 \frac{\pi}{2c},\quad c>1.
$$
Thanks. Since $x\in[0,1]
$ we can write$$
I=\sum_{n=0}^\infty \int_0^1 \log x (1+x^2) x^{c-2+2cn}dx.
$$
since $\sum_{n=0}^\infty x^n= (1-x)^{-1} , |x| < 1.$ Simplifying
$$
I=\sum_n \int_0^1 \log x\, x^{c-2+2n} dx+\sum_n \int_0^1 \log x \, x^{c+2cn}\, dx.
$$
Now we can write
$$
I=-\frac{1}{4}\psi_1 \left(\frac{c+1}{2}\right)-\frac{1}{4} \psi_1\left( \frac{3+c}{2}\right)
$$
where I summed the results of the integrals using general result of $\int_0^1 \log x \, x^n \, dx=-\frac{1}{(n+1)^2}$. Howeever this is not the result $\sec^2...$. Thanks
The function $\psi$ is the polygamma function which is defined in general by
$$
\psi_m(z)=\frac{d^{m+1}}{dz^{m+1}} \log \Gamma(z)
$$
and $\Gamma(z)=(z-1)!$, for this case $m=1$.
|
Using the identity
\begin{align}
\psi(1-x) - \psi(x) = \pi \cot(\pi x)
\end{align}
then the derivative with respect to $x$ yields
\begin{align}
\psi_{1}(1-x) + \psi_{1}(x) = \pi^{2} \csc^{2}(\pi x).
\end{align}
Now the integral
\begin{align}
I &= \int_{0}^{1} \frac{ x^{c-2} (1+x^{2}) \ln(x) }{ 1 - x^{2c} } \, dx
\end{align}
is evaluate as follows.
\begin{align}
I &= \sum_{n=0}^{\infty} \, \int_{0}^{1} (1+x^{2}) \, x^{2cn+c-2} \ln(x) \, dx \\
&= - \sum_{n=0}^{\infty} \left[ \frac{1}{(2cn+c-1)^{2}} + \frac{1}{(2cn+c+1)^{2}} \right] \\
&= - \frac{1}{4 c^{2}} \left[ \psi_{1}\left(\frac{c-1}{2c}\right) + \psi_{1}\left( \frac{c+1}{2c}\right) \right] \\
&= - \left(\frac{\pi}{2c}\right)^{2} \csc^{2}\left(\frac{\pi}{2} + \frac{\pi}{2c}\right) \\
&= - \left(\frac{\pi}{2c}\right)^{2} \sec^{2}\left(\frac{\pi}{2c}\right).
\end{align}
It can now be stated that
\begin{align}
\int_{0}^{1} \frac{ x^{c-2} (1+x^{2}) \ln(x) }{ 1 - x^{2c} } \, dx = - \left(\frac{\pi}{2c}\right)^{2} \sec^{2}\left(\frac{\pi}{2c}\right).
\end{align}
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"language": "en",
"url": "https://math.stackexchange.com/questions/791870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
$(X, \circ)$ is a group, define $a\circ b$ Let $X$ be $\{1,2,3\}$ and $\circ$ a binary operation over $X$ such that $(X,\circ)$ forms a group, and $3\circ 3 = 3$.
Define $a\circ b$ for all $ a,b \in X$.
I think that the identity element is 3. But I am not sure how to proceed.
|
Knowing only $3\circ 3=3$, we start with the table: $$\begin{array}{c|ccc}\circ & 1 & 2 & 3 \\ \hline 1 & & & \\ 2 & & & \\ 3 & & & 3\end{array}$$
By definition, a group is closed, associative, has a unique identity element, and every element has an unique inverse.
Closed: $\forall a,b\in X \implies a\circ b \in X$. Thus the only image of the operator are within the group, $X$.
Associative: $(a\circ b)\circ c = a\circ (b\circ c)$
Identity: $\forall a\in X: a\circ e = e\circ a = a$. NB: since $3\circ 3 = 3$ then $3$ is the identity element.
Inverse: $\forall a\in X, \exists a^{-1}\in X: a\circ a^{-1} = a^{-1}\circ a = 3$. The inverse is unique for each element.
Thus we can fill in some more of the graph from the existence of the identity:
$$\begin{array}{c|ccc}\circ & 1 & 2 & 3 \\ \hline 1 & & & 1 \\ 2 & & & 2 \\ 3 & 1 & 2 & 3\end{array}$$
Since the identity is unique $1\circ 1\neq 1$, $2\circ 2 \neq 2$, $1\circ 2\neq 1\neq 2\circ 1$, $1\circ 2\neq 1\neq 2\circ 1$. This lets us fill in two more squares.
$$\begin{array}{c|ccc}\circ & 1 & 2 & 3 \\ \hline 1 & & 3 & 1 \\ 2 &3 & & 2 \\ 3 & 1 & 2 & 3\end{array}$$
Noting that these give us the inverse for $1$ and $2$, and since the inverse is unique for each element, then this lets us fill in the final squares:
$$\begin{array}{c|ccc}\circ & 1 & 2 & 3 \\ \hline 1 & 2 & 3 & 1 \\ 2 &3 & 1 & 2 \\ 3 & 1 & 2 & 3\end{array}$$
All that remains to do is check that this table is associative.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/792394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
no. of ways of forming a garland using $6$ red roses and $4$ white roses. The no. of ways that the garland can be made out of $6$ red and $4$ white roses so that all white roses not come together.
$\bf{My\; Solution::}$ First we will form a garlad using $6$ red roses, This can be done in $\displaystyle (6-1)!$ ways.
Now we will arrange $4$ white roses in $6$ gap produced by $6$ red roses, This can be done in $\displaystyle \binom{6}{4}\times 1$
because all roses are white.
So Total no. of ways is $ = \displaystyle 5! \times \binom{6}{4}$
But my answer is wrong.
please explain me in detail where my answer is wrong
Thanks
|
This problem can be solved with the Polya Enumeration Theorem
(PET). We assume that garlands that can be transformed into each other
by rotations or reflections are considered the same.
Furthermore we do not treat the problem where the forbidden garlands are
precisely those that contain four white roses all in one block,
as that is a trivial variation on the ordinary bracelet / necklace problem.
We will solve the slightly more general problem of having $n$ red roses
where $n\ge 4.$
First place the four white roses on the garland, leaving some space
between them. These spaces are where the red roses go but there must
be at least one red rose in every one of the four spaces. The dihedral
group $D_4$ acts on these slots and therefore the answer is given by
$$[z^n] Z(D_4)\left(\frac{z}{1-z}\right).$$
We compute $Z(D_4)$ next. The rotations contribute
$$a_1^4+a_2^2+2a_4$$
and the reflections
$$2a_1^2 a_2 + 2 a_2^2$$
for a total result of
$$Z(D_4) = \frac{1}{8}
\left(a_1^4 + 3 a_2^2 + 2 a_1^2 a_2 + 2a_4\right).$$
The substituted cycle index becomes
$$Z(D_4)\left(\frac{z}{1-z}\right)
\\= \frac{1}{8}
\left(
\left(\frac{z}{1-z}\right)^4
+ 3 \left(\frac{z^2}{1-z^2}\right)^2
+ 2 \left(\frac{z}{1-z}\right)^2\left(\frac{z^2}{1-z^2}\right)
+ 2 \left(\frac{z^4}{1-z^4}\right)\right).$$
Introduce the predicate
$$q_m(n) = \begin{cases}
1 & \quad\text{if}\quad m|n \\
0 & \quad\text{otherwise.}
\end{cases}.$$
Extracting coefficients from the inner terms we get for the first term
$${n-4 +3\choose 3} = {n-1\choose 3}.$$
For the second term we get
$$q_2(n)\left(\frac{n}{2}-1\right).$$
For the third term we obtain
(use e.g. the partial fraction decomposition)
$$\frac{1}{4} n^2 - n + \frac{7}{8} + \frac{1}{8} (-1)^n.$$
Finally the last term contributes $$q_4(n).$$
Collecting all contributions we get
$$\frac{1}{48} n^3 - \frac{1}{16} n^2 - \frac{1}{48} n + \frac{3}{32}
+ \frac{1}{32} (-1)^n
+ \frac{3}{8} q_2(n)\left(\frac{n}{2}-1\right)
+ \frac{1}{4} q_4(n).$$
This yields the following sequence (starting at $n=1$):
$$0, 0, 0, 1, 1, 3, 4, 8, 10, 16, 20, 29, 35, 47, 56, 72, 84,
104, 120, 145,\ldots$$
which is OEIS A005232.
In particular with six red beads there are $3$ garlands, which are
(first distribute four red roses into the spaces as that is the minimum,
leaving two red roses):
one, the two red roses go into the same slot,
two, the two red roses go into adjacent slots and
three, the two red roses go into opposite slots.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/792670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
To solve $2\cdot(5^y)-7^x=1$ How do we find all non-negative integers $x,y$ such that $2\cdot(5^y)-7^x=1$ ?
Does there exist any solution with $x>2, y>2$ ?
|
If $y \geq 3$, then taking $\pmod{125}$, $7^x \equiv -1 \pmod{125}$ so $x \equiv 10 \pmod{20}$. Write $x=10a$ where $a$ is odd. Then $2(5^y)=7^x+1 \equiv 7^{10a}+1 \equiv (-1)^a+1 \equiv 0 \pmod{7^{10}+1}$. However $281 \mid 7^{10}+1$, so we get a contradiction.
Thus there are no solutions with $y \geq 3$. We may check $y=0, 1, 2$ directly.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/792850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.