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For what $n$ do the equations $ab = c, bc = a, ca = b$ have solutions mod $n$? In $\mathbb Z/(12)$ the elements $5, 7, 11$ have the property that the product of any two of them equals the third:
$$5 \times 7 = 11$$
$$7 \times 11 = 5$$
$$11 \times 5 = 7$$
I'm interested in generalizations of this. For what integers $n$ does the set of equations $ab = c, bc = a, ca = b$ have solutions mod $n$? Is there a way to describe all such solutions?
(Also would be interested in further generalizations -- for example, is there a cycle of four elements $a, b, c, d$ so that $ab = c, bc = d, cd =a, da=b$ mod $n$ for some $n$?)
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If we write your equations $\pmod 3$ and $\pmod 4$ we get
$$\pmod 3 \quad \pmod 4\\
-1 \times 1 = -1\quad 1 \times -1 = -1\\
1 \times -1=-1 \quad -1 \times -1=1\\
-1 \times -1=1 \quad -1 \times 1=-1$$
and you can apply the Chinese Remainder Theorem to get your values. One way to get other solutions is to replace $3,4$ by any pair of coprime numbers greater than $2$. We cannot use $2$ because $-1=1 \pmod 2$. For example, if we use $5,7$ we work $\pmod {35}$ and get
$$29 \times 6=34\quad 29 \times 34=6\quad 6 \times 34=29$$
This suggests another approach. If we replace $-1$ by $a$ in the above equations, the only one that is problematic is $a \times a=1$. If we choose a modulus where $1$ has more square roots than $1, -1$ we can make it work. A semiprime will do this for us. If we replace $3$ by $15$ we have $4 \times 4=1, 11\times 11=1$ and we can get three solutions $\pmod {60}$. There are certainly many more.
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"timestamp": "2023-03-29T00:00:00",
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|
Trignometry problem: If $\sin^2\theta + 3\cos\theta = 2$ then find $\cos^3\theta + \frac{1}{\cos^3\theta}$ If $\sin^2\theta + 3\cos\theta = 2$ then find $\cos^3\theta + \frac{1}{\cos^3\theta}$
What I did:
$\sin^2\theta + 3\cos\theta = 2$
$3\cos\theta - 1 = 1 - \sin^2\theta$
$3\cos\theta - 1 = \cos^2\theta$
$\cos^3\theta + \frac{1}{\cos^3\theta} = (\cos^2\theta + 1)^3 + \frac{1}{(\cos^2\theta + 1)^3}$
Then I'll have to take LCM and it would be bigger. Can anyone suggest an easy way to solve this problem?
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From $3\cos\theta-1=\cos^2\theta$, let $u=\cos\theta$. Then you have the quadratic $u^2-3u+1=0$. Solving this quadratic gives $u_+=\frac{3+\sqrt{5}}2$ and $u_-=\frac{3-\sqrt{5}}2$. Note that $u_+>1$, and therefore it is not a valid solution to $u=\cos\theta$.
Now, you need to find $u^3+u^{-3}$. Plugging in $u_-$, you get $u^3+u^{-3}=\frac{27-27\sqrt{5}+45-5\sqrt{5}}{8}+\frac{8}{27-27\sqrt{5}+45-5\sqrt{5}}=\frac{72-32\sqrt{5}}{8}+\frac{8}{72-32\sqrt{5}}=\frac{(72-32\sqrt{5})^2}{8(72-32\sqrt{5})}+\frac{64}{8(72-32\sqrt{5})}=\frac{5184-4608\sqrt{5}+5120+64}{576-256\sqrt{5}}=\frac{10368-4608\sqrt{5}}{576-256\sqrt{5}}=\frac{64(162-72\sqrt{5})}{64(9-4\sqrt{5})}=\frac{162-72\sqrt{5})}{9-4\sqrt{5}}=\frac{(9+4\sqrt{5})(162-72\sqrt{5})}{(9+4\sqrt{5})(9-4\sqrt{5})}=\frac{1458+648\sqrt{5}-648\sqrt{5}-1440}{81-80}=\frac{18}{1}=18$
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The equation $x^2-3ax+b=0$ does not have distinct real roots, find the least possible value of $\frac{b}{a-2}$, where $a\gt2$. The following question is taken from JEE practice set.
The equation $x^2-3ax+b=0$ does not have distinct real roots, find the least possible value of $\frac{b}{a-2}$, where $a\gt2$.
My Attempt:
$$D\le0\\ \implies 9a^2-4b\le0\\ \implies9a^2-36\le4b-36\\ \implies9(a-2)(a+2)\le4(b-9)$$
Now, $a+2\gt4\implies 9(a-2)(a+2)\gt36(a-2)$
Also, $b\gt b-9\implies4b\gt4(b-9)$
Thus, $4b\gt9(a-2)+a+2)\gt36(a-2)$
Therefore, $\frac{b}{a-2}\gt9$
The answer given is $18$. How to do this?
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Another approach would be to consider that there are two cases to examine:
• $ \ x^2 - 3ax + b \ $ has a real "double zero" $ \ r \ \ , \ $ so $ \ 3a \ = \ 2r \ $ and $ \ b \ = \ r^2 \ \ ; \ $ or
• the polynomial has a "complex-conjugate pair" of zeroes $ \ \rho \ \pm \ i·\sigma \ \ , \ \ \rho \ , \ \sigma \ $ real, thus, $ \ 3a \ = \ 2 \rho \ $ and $ \ b \ = \ \rho^2 + \sigma^2 \ \ . $
For the double-zero case, the imposed condition requires $ \ a \ = \ \frac23·r \ > \ 2 \ \Rightarrow \ r \ > \ 3 \ $ in the ratio $$ \ \frac{b}{a - 2} \ = \ \frac{r^2}{\frac23·r \ - \ 2} \ = \ \frac{3·r^2}{2·(r \ - \ 3)} \ \ . $$
If we are to forgo calculus, we can look for the values of $ \ r \ $ that give "extreme" values to this ratio by solving for when it has a single value $ \ c \ \ : $
$$ \frac{3·r^2}{2·(r \ - \ 3)} \ \ = \ \ c \ \ \Rightarrow \ \ 3·r^2 \ - \ 2c·r \ + \ 6c \ \ = \ \ 0 \ \ , $$
the discriminant of which is zero for $ \ 4c^2 \ - \ 72c \ = \ 0 \ \Rightarrow \ c \ = \ 0 \ , \ 18 \ \ . \ $ Since $ \ c \ = \ 0 \ \Rightarrow \ r \ = \ 0 $ $ \Rightarrow \ a \ = \ 0 \ \ , \ $ this is excluded by the given condition, while $ \ c \ = \ 18 \ \Rightarrow \ 3·r^2 \ - \ 36·r \ + \ 108 \ \ = \ \ 0 $ $ \Rightarrow \ r \ = \ 6 \ \Rightarrow \ a \ = \ \frac23·6 \ = \ 4 \ \ , \ $ so a minimum of $ \ \mathbf{18} \ $ is acceptable for $ \ \frac{b}{a - 2} \ $ in the case of a double zero at $ \ 6 \ \ ( $ the polynomial becomes $ \ x^2 - 12x + 36 \ \ . \ ) $ [Alternatively, extremization by calculus has us set the derivative of the ratio equal to zero, or
$$ \frac{6·r^2 \ - \ 36·r}{4·(r \ - \ 3)^2} \ \ = \ \ 0 \ \ \Rightarrow \ \ r \ = \ 0 \ , \ 6 \ \ . \ $$
We can even "shift" the curve for the ratio by $ \ 3 \ $ units "to the right", transforming the ratio to $ \ u \ = \ r - 3 \ \rightarrow \ \frac{3·(u + 3)^2}{2·u} \ \ , \ $ which has odd symmetry about its vertical asymptote and the extremal value $ \ 0 \ $ for $ \ u \ = \ -3 \ , \ +3 \ \rightarrow \ r \ = \ 0 \ , \ 6 \ \ . \ ] $
With complex zeroes , the ratio is $$ \frac{b}{a - 2} \ = \ \frac{\rho^2 + \sigma^2}{\frac23 \rho \ - \ 2} \ = \ \frac{3·(\rho^2 + \sigma^2)}{2·(\rho \ - \ 3)} \ \ . $$ From this, we see that the smallest values for the ratio can be obtained for $ \ \sigma \ = \ 0 \ \ , \ $ which thereby reduces this to the real double zero case.
Hence, the minimum value 18 for the ratio is obtained for $ \ a \ = \ 4 \ \ , \ \ b \ = \ 36 \ \ . \ $ (The $ \ r \ = \ a \ = \ b \ = \ 0 \ $ result we found is a local maximum for $ \ a \ < \ 2 \ \ . \ $ )
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Find the first derivative of $y=(x^4-1)\sqrt[3]{x^2-1}$
Find the first derivative of $$y=(x^4-1)\sqrt[3]{x^2-1}$$
We can write the function as $$y=(x^4-1)\left(x^2-1\right)^\frac13$$ For the derivative we have $$y'=4x^3\left(x^2-1\right)^\frac13+\dfrac13\left(x^2-1\right)^{-\frac23}2x(x^4-1)\\=4x^3\left(x^2-1\right)^\frac13+\dfrac23x\left(x^2-1\right)^{-\frac23}(x^2-1)(x^2+1)\\=\dfrac23x\left(x^2-1\right)^\frac13\left(6x^2+x^2+1\right)\\=\dfrac23x\left(x^2-1\right)^\frac13(7x^2+1)$$ The given answer is $$y'=\dfrac{2x(7x^4-6x^2-1)}{3\sqrt[3]{\left(x^2-1\right)^2}}$$ I don't see my mistake...
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Too long for a comment
Make you life much easier using logarithmic differentiation
$$y=(x^4-1)\sqrt[3]{x^2-1} \implies \log(y)=\log(x^4-1)+\frac 13 \log(x^2-1)$$
$$\frac {y'}y=\frac {4x^3}{x^4-1}+\frac 13\frac {2x}{x^2-1}=\frac{2 x\left(7 x^2+1\right)}{3 \left(x^4-1\right)}$$
$$y'=\frac {y'}y \times y= ??? $$
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|
evaluate $\sum_{i=0}^{n-1}\sum_{j=i+1}^{n+1} {n+1\choose j}{n\choose i}$
Evaluate $\sum_{i=0}^{n-1}\sum_{j=i+1}^{n+1} {n+1\choose j}{n\choose i}$.
Below is a summary of a solution based off of a problem in the summation chapter of the book Problem Solving Through Problems by Loren Larson.
Multiply both sides of the sum by $\dfrac{1}{2^{2n+1}}.$
Consider a matching game played with two players A and B and $2n+1$ coins. Player $A$ flips $n+1$ coins and picks $n$ of them so that the number of heads flipped is maximal. Player $B$ flips $n$ coins. The player with the most heads flips wins, with ties going to B.
Then the probability $A$ wins is $\sum_{i=0}^{n-1} P(\text{ B flips i heads }) \sum_{j=i+1}^n P(\text{ A flips j heads } | \text{ B flips i heads }) = \sum_{i=0}^{n-1} P(\text{ B flips i heads }) \sum_{j=i+1}^n P(\text{ A flips j heads }) = \sum_{i=0}^{n-1} {n\choose i}\dfrac{1}{2^n}\sum_{j=i+1}^n {n+1\choose j}(\dfrac{1}2)^{n+1}$
Note that the game can be reformulated as follows: players A and B both flip $n$ coins and the one with the most heads flipped wins if there's no tie. If they both flip $n$ heads, then player B wins. If they flip less than n heads and there's a tie, then player A flips the $(n+1)$st coin and wins if it's heads and loses otherwise.
In this case, player B wins exactly two more times than player A (when both players flip n heads). So the probability player A wins is $1-P(\text{B wins}) = 1 - \dfrac{\dfrac{1}2 (2^{2n+1}+2)}{2^{2n+1}} = \dfrac{2^{2n}-1}{2^{2n+1}}.$
Hence the desired answer is $2^{2n}-1.$
I was wondering if there was a more elementary approach using basic combinatorial identities such as the hockey-stick identity, the Binomial theorem, the identity ${n\choose i} = \dfrac{n}i {n-1\choose i-1},$ etc? Basic rearranging gives that the sum in question equals $\sum_{j=1}^{n+1} {n+1\choose j} \sum_{i=0}^{j-1} {n\choose i}-1.$
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We seek to evaluate
$$\sum_{p=0}^{n-1} \sum_{q=p+1}^{n+1}
{n+1\choose q} {n\choose p}.$$
This is
$$\sum_{p=0}^{n-1} {n\choose p}
\sum_{q=0}^{n-p} {n+1\choose q}
\\ = \sum_{p=0}^{n-1} {n\choose p}
[v^{n-p}] \frac{1}{1-v} \sum_{q\ge 0} {n+1\choose q} v^q
\\ = [v^n] \frac{1}{1-v}
\sum_{p=0}^{n-1} {n\choose p} v^p (1+v)^{n+1}
\\ = -1 +
[v^n] \frac{1}{1-v} (1+v)^{n+1}
\sum_{p=0}^{n} {n\choose p} v^p
\\ = -1 + [v^n] \frac{1}{1-v} (1+v)^{2n+1}
\\ = -1 + \sum_{q=0}^n {2n+1\choose q}
\\ = -1 + \frac{1}{2} 2^{2n+1}
= -1 + 2^{2n}.$$
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|
A smarter (not bashy) way to solve this roots of unity problem? (Mandelbrot) Let $\xi = \cos \frac{2\pi}{5} + i \sin \frac{2pi}{5}$ be a complex fifth root of unity. Set $a = 20\xi^2 + 13 \xi, b = 20\xi^4 + 13\xi^2, c = 20\xi^3 + 13\xi^4, \text{and } d = 20\xi + 13\xi^3$. Find $a^3 + b^3 + c^3 + d^3$
Immediately what comes to mind is finding $(a + b + c + d)^3$ and subtracting whatever we don't need to get $a^3 + b^3 + c^3 + d^3$. However,
\begin{equation*}
(a+b+c+d)^3 = a^3+3a^2b+3a^2c+3a^2d+3ab^2+6abc+6abd+3ac^2+6acd+3ad^2+b^3+3b^2c+3b^2d+3bc^2+6bcd+3bd^2 + c^3 + 3c^2d + 3cd^2 + d^3
\end{equation*}
There is simply no good way to calculate $6abc + 6abd + 6acd + 6bcd $ without expanding everything.
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They simplified exponents when defining $\xi,$ but you could also write the equations
$$\begin{align*}a &= \xi(20\xi+13),\\b &= \xi^2(20\xi^2+13),\\c &= \xi^3(20\xi^3+13),\\d &= \xi^4(20\xi^4+13)\end{align*}$$
(actually, they swapped $c$ and $d$, but that doesn't change the problem). Consider the polynomial
$$f(x) = [x(20x+13)]^3 = 20^3x^6+3\cdot20^2\cdot13x^5+3\cdot20\cdot13^2x^4+13^3x^3.$$
A roots of unity filter* using fifth roots of unity would give the sum of the coefficients of $x^0, x^5, x^{10},$ etc., which is just $3\cdot20^2\cdot 13 = 15600.$ But $\xi$ is a fifth root of unity! So,
$$15600 = \frac{f(1)+f(\xi)+f(\xi^2)+f(\xi^3)+f(\xi^4)}{5} = \frac{33^3+a^3+b^3+c^3+d^3}{5}.$$
Rearranging gives
$$a^3+b^3+c^3+d^3=78000-33^3=\boxed{42063}.$$
*Say we have some polynomial
$$P(x) = a_0x^0+a_1x^1+a_2x^2+\cdots+a_nx^n$$
Plugging in powers of the $k$th root of unity $\omega$ gives
$$\sum_{j=0}^{k-1}P(\omega^j) = \sum_{i=0}^n a_i\sum_{j=0}^{k-1}(\omega^{j})^i.$$
Inspecting the inner summand, we find
$$\sum_{j=0}^{k-1}(\omega^i)^j = k$$
if $\omega^i = 1$ (i.e. $i$ is a multiple of $k$), otherwise
$$(\omega^i-1)\sum_{j=0}(\omega^i)^j = (\omega^{i})^k-1 = 0,$$
so it evaluates to zero.
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|
Adding 1 to each entry of continued fractions Here we denote $[a_0,...,a_n]$ as the continued fraction of some rational number. If I take $p/q=[a_0,a_1,...,a_n]$ to $p'/q'=[a_0+1,a_1+1,...,a_n+1]$, are there any nice properties I can say about $p'/q'$?
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We have $p' = p + q$ and $q' = q$. This is because, in a continued fraction, the $n$-th term is $a_n = \lfloor \frac{p}{q} \rfloor$, the greatest integer less than or equal to $\frac{p}{q}$. But we have
$$
\frac{p}{q} = a_0 \frac{1}{1} + \frac{1}{a_1 \frac{1}{1} + \frac{1}{\ddots + \frac{1}{a_n}}}
$$
so
$$
\frac{p + 1}{q + 1} = a_0 \frac{1}{1} + \frac{1}{a_1 \frac{1}{1} + \frac{1}{\ddots + \frac{1}{a_n + \frac{1}{1}}}} = \frac{p'}{q'}
$$
where $p' = p + q$ and $q' = q$.
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|
Prove that $\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln\big(\frac mn \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}\dots\big)$ page 371 in ‘Synopsis Of Elementary Results In Pure Mathematics’ contains the following result,
$$\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln \left(\frac{m}{n} \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}…\right)$$
How does one prove this step-by-step?
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First, we start with the following.
$$I(m,1)=\int_0^1\frac{x^{m-1}}{1+x}dx$$
We can substitute $u=-\ln{x}$ to get the following integral.
$$I(m,1)=\int_0^\infty\frac{e^{-mu}}{1+e^{-u}}du$$
When we apply the geometric series to $\frac{1}{1-(-e^{-u})}$, switch the order of integration and summation, then integrate, we get the following sum. We evaluate it by splitting it into even and odd terms and applying the digamma function.
$$I(m,1)=\sum_{n=0}^\infty\frac{(-1)^n}{n+m}=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n+m-1}=\sum_{n=1}^\infty\left(\frac{(-1)^{(2n-1)-1}}{(2n-1)+m-1}-\frac{(-1)^{(2n)-1}}{(2n)+m-1}\right)$$
$$=\frac{1}{2}\sum_{n=1}^\infty\left(\frac{1}{n+\frac{m-2}{2}}-\frac{1}{n+\frac{m-1}{2}}\right)=\frac{1}{2}\psi\left(\frac{m+1}{2}\right)-\frac{1}{2}\psi\left(\frac{m}{2}\right)$$
Next, we have to expand $I(m)$ with a new parameter.
$$I(m,p)=\int_0^1\frac{x^m}{1+x^p}dx$$
First, we can substitute $u=x^p$ to get the following
$$I(m,p)=\frac{1}{p}\int_0^1\frac{x^{\frac{m+1}{p}-1}}{1+x}dx=\frac{1}{2p}\psi\left(\frac{\frac{m+1}{p}+1}{2}\right)-\frac{1}{2p}\psi\left(\frac{\frac{m+1}{p}}{2}\right)$$
$$=\frac{1}{2p}\psi\left(\frac{m+p+1}{2p}\right)-\frac{1}{2p}\psi\left(\frac{m+1}{2p}\right)$$
Therefore
$$\int_0^1\frac{x^m}{1+x^p}dx=\frac{1}{2p}\psi\left(\frac{m+p+1}{2p}\right)-\frac{1}{2p}\psi\left(\frac{m+1}{2p}\right)$$
We take the definite integral with respect to m from 0 to a new parameter $n$
$$\int_0^1\frac{1}{1+x^p}\left(\int_0^n x^mdm\right)dx=\int_0^n\left(\frac{1}{2p}\psi\left(\frac{m+p+1}{2p}\right)-\frac{1}{2p}\psi\left(\frac{m+1}{2p}\right)\right)dm$$
$$\int_0^1\frac{1}{1+x^p}\frac{x^n-1}{\ln{x}}dx=\ln\left(\frac{\Gamma\left(\frac{m+p+1}{2p}\right)}{\Gamma\left(\frac{m+1}{2p}\right)}\right)\Bigg|_{m=0}^{m=n}=\ln\left(\frac{\Gamma\left(\frac{n+p+1}{2p}\right)\Gamma\left(\frac{1}{2p}\right)}{\Gamma\left(\frac{p+1}{2p}\right)\Gamma\left(\frac{n+1}{2p}\right)}\right)$$
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|
Evaluate $\oint_C \frac{z}{(z+1)(z+3)} dz$ I want to evaluate $\oint \frac{z}{(z+1)(z+3)} dz$ where $C$ is the rectangle with the edges $2\pm i$, $-2\pm i$.
My attempt:
I used partial fraction expansion to express $\frac{z}{(z+1)(z+3)} = -\frac{1}{2(z+1)} +\frac{3}{2(z+3)}$.
Therefore,
\begin{align*}
\oint_C \frac{z}{(z+1)(z+3)} dz = \oint_C \frac{3}{2(z+3)} dz - \oint_C \frac{1}{2(z+1)} dz
\end{align*}
Since $-1$ is an interior point of the rectangle, it is easy to calculate this integral with Cauchy Integral Formula
\begin{align*}
- \oint_C \frac{1}{2(z+1)} dz = -2\pi i\cdot f(-1) = -\pi i
\end{align*}
where $f(z)= 1/2$ constant.
I cannot do the same for the other integral, since $-3$ is not an interior point.
So I think I can express
\begin{align*}
C = \begin{cases}
C_1 = \{z:z=x-i, x\in[-2,2]\}\\
C_2 = \{z:z=2+iy, y\in [-1,1]\}\\
C_3 =\{z:z=x+i, x\in[2,-2]\} \\
C_4 = \{z:z=-2+iy, y\in[1,-1]\}
\end{cases}
\end{align*}
Therefore,
\begin{align*}
\frac{3}{2} \oint_C \frac{1}{(z+3)} dz = \frac{3}{2} (\log(x-i+3)|_{-2}^2+i\log(2+iy+3)|_{-1}^1 + \log(x+i+3)|_2^{-2} + i\log(-2+iy+3)|_1^{-1})
\end{align*}
But I can't really simplify this. Does anyone know how to proceed?
|
Since $-3$ is not an interior point (as you wrote), the integral is $0$, by the Cauchy integral theorem.
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|
Integrating $\int \frac{3}{(x^2 +5)^2}dx$ by parts Integrating $$\int \frac{3}{(x^2 +5)^2}dx$$
After removing the constant, it is basically integrating $\frac{1}{x^4+10x^2+25}$. I only have learnt up to integrating $\frac{1}{ax^2 + bx +c}$ with the highest power of $x$ is 2. And this cannot be broken up into partial fractions too. What happens when the power of $x$ is 4?
My thoughts are to do integration by parts (product rule)
$$\int u v' dx = uv - \int u' v dx$$
But I am unclear what do substitute $u$ and $v$ to
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$$\frac{3}{10}\int\frac{10}{x^4+10x^2+25}dx$$
Dividing numerator and denominator by $x^2$
$$\frac{3}{10}\int\frac{\frac{10}{x^2}}{x^2+10+(\frac5x)^2}dx\\=\frac{3}{10}\int\frac{\frac5{x^2}+1+\frac5{x^2}-1}{x^2+10+(\frac5x)^2}dx\\=\frac{3}{10}\int\frac{1+\frac5{x^2}}{x^2+10+(\frac5x)^2}dx-\frac3{10}\int\frac{1-\frac5{x^2}}{x^2+10+(\frac5x)^2}dx\\=I_1-I_2$$
For $I_1$, put $x-\frac5x=t$. Its a) derivative is $(1+\frac5{x^2})dx=dt$ and b) square is $x^2-10+(\frac5x)^2=t^2$
Thus, $$I_1=\frac3{10}\int\frac{dt}{t^2+20}\\=\frac{3}{20\sqrt5}\tan^{-1}\frac{t}{2\sqrt5}+c_1\\=\frac3{20\sqrt5}\tan^{-1}\frac{x^2-5}{2\sqrt5x}+c_1$$
For $I_2$, put $x+\frac5x=t$. Its a) derivative is $(1-\frac5{x^2})dx=dt$ and b) square is $x^2+10+(\frac5x)^2=t^2$
Thus, $$I_2=\frac3{10}\int\frac{dt}{t^2}\\=\frac{-3}{10t}+c_2\\=\frac{-3x}{10(x^2+5)}+c_2$$
|
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|
Why are the derivatives of $\frac{x^2+1}{x^2+x+1}$ and $\frac{-x}{x^2+x+1}$ the same?
Why are the derivatives of these functions the same?
$$\frac{x^2+1}{x^2+x+1} \qquad\qquad \frac{-x}{x^2+x+1}$$
original exercise text (See part (e).)
I have tried to answer this question and consulted the answer booklet but this did not make much sense. Would someone be able to simplify or elaborate further? Thank you :)
The answer guide said the derivates are the same as differ at most by an additive constant.
|
$$
{d\over dx} (f(x)+1)={df\over dx}+{d1\over dx}={df\over dx}
$$
so we can add $1$ to a function and the derivative of the sum is the same as the derivative of the function. So
$$
\begin{align}
\frac{-x}{x^2+x+1}+1
&=\frac{-x}{x^2+x+1}+\frac{x^2+x+1}{x^2+x+1}\\
&={x^2+x+1-x\over x^2+x+1}\\
&={x^2+1\over x^2+x+1}\\
\end{align}
$$
|
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|
Calculate $\sum_{n=2}^{\infty}\left (n^2 \ln (1-\frac{1}{n^2})+1\right)$ I am interested in evaluating
$$\sum_{n=2}^{\infty}\left (n^2 \ln\left(1-\frac{1}{n^2}\right)+1\right)$$
I am given the solution for the question is $\,\ln (\pi)-\frac{3}{2}\,.$
$$\sum_{n=2}^{\infty}\left(n^2\ln\left(\!1\!-\!\frac{1}{n^2}\!\right)+1\right)=4\ln\left(\!\frac{3}{4}\!\right)+1+9\ln\left(\!\frac{8}{9}\!\right)+1+\ldots$$
Any tricks to solve it?
|
This answer is just an elaboration on KStarGamer's comment. I myself went for the calculation idea.
Suppose the sum is converging. We have then
$$S(x):=\sum_{n=2}^{\infty}\left(n^2\ln\left(1-\frac{x^2}{n^2}\right)+x^2\right)$$
$$\implies S'(x)=\sum_{n=2}^{\infty}\left(-\frac{2x}{n^2}n^2\frac{1}{1-\frac{x^2}{n^2}}+2x\right)$$
\begin{align*}
S'(x)&=\sum_{n=2}^{\infty}\left(-2x{\frac{n^2}{n^2-x^2}}+2x\right)\\
&=2x\sum_{n=2}^{\infty}\left({\frac{n^2}{x^2-n^2}}+1\right)\\
&=2x\sum_{n=2}^{\infty}\left({\frac{n^2+x^2-n^2}{x^2-n^2}}\right)\\
&=x^2\left[2x \sum_{n=2}^{\infty}{\frac{1}{x^2-n^2}}\right]\\
&=\pi x^2\left[2\pi x \sum_{n=2}^{\infty}{\frac{1}{(\pi x)^2-(\pi n)^2}}\right]\\
&=\pi x^2\left[-\frac{2x}{\pi(x^2-1)}-\frac{1}{\pi x}+\underbrace{\frac{1}{\pi x}+2\pi x \sum_{n=1}^{\infty}{\frac{1}{(\pi x)^2-(\pi n)^2}}}_{\cot(\pi x)}{}\right]\\
\end{align*}
where $\cot(\pi x)$ is expressed using Mittag-Leffler's theorem on meromorphic functions (Look here).
We finally get that
$$S'(x)=-x+\left(\pi x^2\cot(\pi x)-\frac{2x^3}{x^2-1}\right)$$
By integrating, we get :
$$S(x)=-\frac{x^2}2+\int_0^x\left(\pi t^2\cot(\pi t)-\frac{2t^3}{t^2-1}\right)dt$$
For $x=1$, we get
$$S(1)=-\frac{1}{2}+\int_0^1\left(\pi t^2\cot(\pi t)-\frac{2t^3}{t^2-1}\right)dt=-\frac{1}{2}+\ln(\pi)-1=\ln(\pi)-\frac{3}{2}.$$
The integral $\int_0^1\left(\pi t^2\cot(\pi t)-\frac{2t^3}{t^2-1}\right)dt$ was evaluated exactly via Wolframalpha.
Thanks to both Angelo and KStarGamer for their corrections and comments.
|
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|
Better approach to evaluate the limit $\lim_{x\to0^+}(\cot x-\frac{1}{x})(\cot x+\frac{1}{x})$ I solved it by rewriting the limit as indeterminate form $0/0$, then apply L'Hopital's rule 4 times, It was really lengthy and easy to make mistakes, If anyone got a better approach, please tell me!
$$
\begin{align}
\lim_{x\to0^+}(\cot x-\frac{1}{x})(\cot x+\frac{1}{x})&=\lim_{x\to0^+}\frac{x^2\cos^2x-\sin^2x}{x^2\sin^2 x}\\
\vdots\\
&=\lim_{x\to0^+}\frac{-16\cos 2x+\ldots}{24\cos 2x+\ldots}\\
&=-\frac{2}{3}
\end{align}
$$
|
Without using the derivatives, or Taylor or o(little) or O(big),$$\lim \limits_{x \to 0}\left( \cot^2 x-{1 \over x^2}\right) = \lim \limits_{x \to 0} \frac{x^2\cos^2 x-\sin^2 x}{x^2\sin^2 x} = $$$$\lim \limits_{x \to 0} \frac{x\cos x+\sin x}{x}\cdot\lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x^3} $$$$\cdot\lim \limits_{x \to 0}\frac{x^2}{\sin^2 x}=\lim \limits_{x \to 0} \left( \cos x +\frac{\sin x}{x}\right)\cdot\lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x^3} $$$$= 2 \cdot \lim \limits_{x \to 0}\frac{\cos x-x\sin x- \cos x}{3x^2}=-\frac 23$$
|
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|
Is there an identity for $\sum_{k=0}^{n-1}\tan^4\left({k\pi\over n}\right)$? Is there a simple relation for $\sum_{k=0}^{n-1}\tan^4\left({k\pi\over n}\right)$
like there is for $\sum_{k=0}^{n-1}\tan^2\left({k\pi\over n}\right)$?
Looking at Jolley, Summation of Series, formula 445:
$\sum_{k=0}^{n-1}\tan^2\left(\theta+{k\pi\over n}\right)=n^2\cot^2\left({n\pi\over2}+n\theta\right)+n(n-1)$
or more simply:
$\sum_{k=0}^{n-1}\tan^2\left({k\pi\over n}\right)=n(n-1)$
Does $\sum_{k=0}^{n-1}\tan^4\left({k\pi\over n}\right)$ possibly equals some integer values?
The numerical table seems to suggest so even for higher even powers of $\tan$:
$\left(
\begin{array}{ccccc}
n & \sum_{k=0}^{n-1}\tan^2\left({k\pi\over n}\right)&\sum_{k=0}^{n-1}\tan^4\left({k\pi\over n}\right)&\sum_{k=0}^{n-1}\tan^6\left({k\pi\over n}\right)&\sum_{k=0}^{n-1}\tan^8\left({k\pi\over n}\right)\\
3. & 6. & 18. & 54. & 162. \\
5. & 20. & 180. & 1700. & 16100. \\
7. & 42. & 742. & 14154. & 271558. \\
9. & 72. & 2088. & 66600. & 2.14049\times 10^6 \\
11. & 110. & 4730. & 226622. & 1.09528\times 10^7 \\
13. & 156. & 9308. & 624780. & 4.2335\times 10^7 \\
15. & 210. & 16590. & 1.48533\times 10^6 & 1.34314\times 10^8 \\
\end{array}
\right)$
I had tried to use residue methods and I particularly like falager's method in this answer using Vieta's formula but kept getting stuck.
If one could make progress with the quartic, one could possibly find for higher even powers of tan too.
|
A possible approach is to use (here $n$ is still odd!) $$\prod_{k=1}^{n-1}\left(1+x^2\tan^2\frac{k\pi}{n}\right)=\left(\frac{(1+x)^n+(1-x)^n}{2}\right)^2$$ obtained by factoring the RHS over $\mathbb{C}$, or as $P_n(1+x,x-1)/P_n(1,-1)$ where $$P_n(a,b)=\prod_{k=0}^{n-1}\left(a^2+b^2-2ab\cos\frac{2k\pi}{n}\right)=(a^n-b^n)^2$$ comes from factoring $z^n-1$ over $\mathbb{C}$. Denoting $S_{2m}:=\sum_{k=1}^{n-1}\tan^{2m}(k\pi/n)$, we get
\begin{align}
\sum_{m=1}^\infty\frac{(-1)^m}{m}S_{2m}x^{2m}&=-\sum_{k=1}^{n-1}\log\left(1+x^2\tan^2\frac{k\pi}{n}\right)
\\&=-2\log\frac{(1+x)^n+(1-x)^n}{2}
\\&=-2\log\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}x^{2k}
\\&=2\sum_{m=1}^\infty\frac{(-1)^m}{m}\left(\sum_{k=1}^{\lfloor n/2\rfloor}\binom{n}{2k}x^{2k}\right)^m.
\end{align}
Comparing the individual coefficients, we get
\begin{align}
S_2&=2\binom{n}{2}&&=n^2-n,\\
S_4&=2\binom{n}{2}^2-4\binom{n}{4}&&=\color{blue}{\frac{n^4-4n^2+3n}{3}},\\
S_6&=2\binom{n}{2}^3-6\binom{n}{2}\binom{n}{4}+6\binom{n}{6}&&=\frac{2n^6-10n^4+23n^2-15n}{15},
\end{align}
and so on; say, $S_8=(17n^8-112n^6+308n^4-528n^2+315n)/315$.
|
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|
Find all the real numbers an expression might take when $a, b, c$ are complex numbers of the same modulus.
Let $a, b, c$ be three complex numbers of the same modulus. Find all real numbers that might be equal to:
$$x = \frac{a^3+b^3+c^3}{abc}$$
It is obvious that when all three numbers are equal, we might write that:
$$x = \frac{3a^3}{a^3} = 3 \in \mathbb{R}$$
I would write, in order to obtain real values:
$$x = \overline{x}$$
So:
$$\overline{a}\overline{b}\overline{c} (a^3 + b^3 + c^3) = abc (\overline{a}^3 + \overline{b}^3 + \overline{c}^3)$$
But the calculations are quite harsh, the only thing I might use to solve them is that:
$$a\overline{a} = b\overline{b} = c\overline{c} = m^2$$
where $m$ is the complex modulus of the three numbers.
|
Let us implement Mark Bennet's idea.
Let $\frac{a^3}{abc}=e^{i\alpha}$ and $\frac{b^3}{abc}=e^{i\beta}$, where $\alpha,\beta\in[0,2\pi)$. Then $\frac{c^3}{abc}=\frac{abc}{a^3}\frac{abc}{b^3}=\frac1{e^{i\alpha}}\frac1{e^{i\beta}}=e^{i(-(\alpha+\beta))}.$
$$\begin{aligned}x &= e^{i\alpha} + e^{i\beta}+e^{i(-(\alpha+\beta))}\\
&=(\cos\alpha+\cos\beta+\cos(-(\alpha+\beta))) + i(\sin\alpha+\sin\beta+\sin(-(\alpha+\beta)))\end{aligned}$$
Since $$\begin{aligned}&\quad\sin\alpha+\sin\beta+\sin(-(\alpha+\beta))\\
&= 2\sin\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2-2\sin\frac{\alpha+\beta}2\cos\frac{\alpha+\beta}2\\
&=4\sin\frac{\alpha+\beta}2\sin\frac\alpha2\sin\frac\beta2,
\end{aligned}$$
$x$ is real $\iff$ $\sin\frac{\alpha+\beta}2=0$ or $\sin\frac\alpha2=0$ or $\sin\frac\beta2=0$ $\iff$ $\alpha+\beta=0$ or $\alpha=0$ or $\beta=0$.
*
*$\alpha=0$.
$x=1+2\cos\beta\in[-1,3]$.
Note $1+2\cos\beta$ is attainable as we can set $a=1$, $b=e^{i\frac\beta3}$, $c=1/b$.
*$\beta=0$.
$x=1+2\cos\alpha\in[-1,3]$.
*$\alpha+\beta=0$.
$x=1+2\cos\alpha\in[-1,3]$.
Hence, all possible real $x$s are $[-1,3]$.
|
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|
short weierstrass form for cubic I have some question about the derivation of the short Weierstrass form. In https://www.staff.uni-mainz.de/dfesti/EllipticCurvesNotes.pdf this note, I follow the derivation till the point $y^2 = β_0x^3 + β_1x^2 + β_2x + β_3$, but then it says that using the transformation $x' = x + β_1/3, y'=y$, we then arrive at $y^2 = x^3 + ax + b$. But my question is how it turns the coefficient $\beta_0$ into $1$? One method I come up with is to let $y^2 = β_0(x^3 + \gamma_1x^2 + \gamma_2x + \gamma_3)$, then applying $x' = x + \gamma_1/3, y'=y$, we get $y^2 = \beta_0(x^3 + ax + b)$ then change by a scalar $x' = \beta_0^{1/3}x, y'=y$, we get the form $y^2 = x^3 + ax + b$. but the problem is the final step would need to assume that $\beta_0$ has a cube root, which is not true from general rational numbers. Does anyone knows how to solve this? for a specific example, how could one turn the equation $y^2 = 5x^3+x+1$ into the short Weierstrass form?
|
You're right, there's a step missing here.
Starting from $$y^2 = \beta_0x^3 + \beta_1x^2 + \beta_2+ \beta_3$$ we can get rid of the (nonzero) $\beta_0$ as follows: send $y\mapsto \beta_0^2y$ and $x\mapsto \beta_0x$, which gives us $$\beta_0^4y^2 = \beta_0^4x^3 + \beta_0^2\beta_1x^2 + \beta_0\beta_2x+ \beta_3$$ and after dividing by $\beta_0^4$ we get $$y^2=x^3+\frac{\beta_1}{\beta_0^2}x^2+\frac{\beta_2}{\beta_0^3}x+\frac{\beta_3}{\beta_0^4}$$ so we may assume we have an equation of the form $y^2=x^3+\gamma_1x^2+\gamma_2x+\gamma_3$.
Now to get rid of $\gamma_1$, make the substitution $x=x-\frac{\gamma_1}{3}$. This sends $$x^3+\gamma_1x^2+\gamma_2x+\gamma_3 \mapsto (x-\frac{\gamma_1}{3})^3+\gamma_1(x-\frac{\gamma_1}{3})^2+\gamma_2(x-\frac{\gamma_1}{3})+\gamma_3,$$ and the right-hand side is equal to $$x^3+(\gamma_2-\frac{\gamma_1^2}{3})x+(\gamma_3-\frac{\gamma_1\gamma_2}{3}+\frac{2\gamma_1^3}{27})$$ which is the desired form.
|
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|
What is the value of $\int_{-1}^{2} 4 \, x^2 \, (2+x-x^2) \, dx$? $$\int_{-1}^{2} 4 \, x^2 \, (2+x-x^2) \, dx$$
*
*When I solved this took the constant out of the integral and then multiplied $x^2$
by the bracket and evaluated the integral to get:
$$4 \, \left[ \frac{x^4}{4} +\frac{2x^3}{3}-\frac{x^5}{5} \right]^2_{-1}=\frac{63}{20}$$
*But when I solve this integral on my calculator on (Casio 991ERX) or Wolfram I get $ \frac{63}{5}$. Why is that?
|
Let us do it step by step:
\begin{align*}
\int_{-1}^{2}4x^{2}(x + 2 - x^{2})\mathrm{d}x & = \int_{-1}^{2}(4x^{3} + 8x^{2} - 4x^{4})\mathrm{d}x\\\\
& = \left(x^{4} + \frac{8x^{3}}{3} - \frac{4x^{5}}{5}\right)\bigg\rvert_{-1}^{+2}\\\\
& = \left(16 + \frac{64}{3} - \frac{128}{5}\right) - \left(1 - \frac{8}{3} + \frac{4}{5}\right)\\\\
& = \frac{176}{15} + \frac{13}{15} = \frac{189}{15} = \frac{63}{5}
\end{align*}
Hopefully this helps!
|
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|
How is $\ln (x-2) - \frac{1}{2} \ln (x-1) = \frac{1}{2} \ln \frac{(x-2)^2}{x-1}$
How is $\ln (x-2) - \frac{1}{2} \ln (x-1) = \frac{1}{2} \ln \frac{(x-2)^2}{x-1}$
Can someone enlighten me on how is these 2 actually equals and the steps taken? the left hand side is actually the answer for $\int \frac{x}{2 (x-2)(x-1)} dx$ but I need to combine the expression to the right hand side to continue with the steps in the question I am attempting
|
$$\ln(x-2)=\ln(((x-2)^2)^{1/2})=\frac{1}{2}\ln(x-2)^2$$
Can you continue?
|
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|
Proving $\sum_{cyc}\frac{ab}{a+b+2c} \le \frac{1}{4}(a+b+c)$ for positive real $a$, $b$, $c$ Prove that $$\frac{ab}{a+b+2c} + \frac{bc}{b+c+2a} + \frac{ca}{c+a+2b} \le \frac{1}{4}(a+b+c)$$
for positive real numbers $a$, $b$, and $c$.
|
By AM-HM
$$\sum_{sym} \frac{bc}{2a+b+c} \leq \sum_{sym} \frac{bc}{4}(\frac{1}{a+b}+\frac{1}{a+c})=\frac{a+b+c}{4}.$$
Seems like you are new to inequalities. To study this topic, I recommend you this book.
|
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|
Is this general nested radical for $\pi$ true? We have,
I. Liu Hui (c. 300 AD)
$$\pi \approx 3\cdot2^{\color{red}8}\times \underbrace{\sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2+\sqrt{\color{blue}1}}}}}}}}}}}_{\color{red}{10}\text{ square roots}}$$
II. Viete (c. 1590 AD)
$$\pi = \lim_{k\to\infty} 4\cdot2^{\color{red}{k}}\times \underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{\color{blue}2}}}}}}}}}_{\color{red}{k+2}\text{ square roots}}$$
III. (yours truly)
$$\pi \approx 6\cdot2^{\color{red}8}\times \underbrace{\sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2+\sqrt{\color{blue}3}}}}}}}}}}}_{\color{red}{10}\text{ square roots}}$$
For the blue numbers, I tried $\sqrt{5}$, $\sqrt{6}$, and other $\sqrt{n}$, nothing worked, until I remembered $\sqrt{2}$ and $\sqrt{3}$ has a trigonometric context.
Question: Given,
$$\color{blue}{\beta}=4\cos^2\left(\frac{\pi}{\color{brown}{\alpha}}\right)$$
then is it true that,
$$\pi = \lim_{k\to\infty} \color{brown}{\alpha}\cdot2^{\color{red}{k}}\times \underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{\color{blue}\beta}}}}}}}}}_{\color{red}{k+2}\text{ square roots}}$$
So the first three are simply the cases $\alpha = 3,4,6$, though if the general form is true, then one can use other positive integers.
P.S. For example, note that if $\alpha = 5$, then we relate $\pi$ to our old friend the golden ratio $\phi$,
$$\pi = \lim_{k\to\infty} \color{brown}{5}\cdot2^{\color{red}{k}}\times \underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{\color{blue}{\phi^2}}}}}}}}}}_{\color{red}{k+2}\text{ square roots}}$$
|
Firstly, we define the recursive sequence
\begin{equation}
\begin{split}
A_{k+1}&=\sqrt{2+A_k}\\
A_0&=\sqrt{\beta}=2\cos\left(\frac{\pi}{\alpha}\right)\\
\end{split}
\end{equation}
and let $B_k=\sqrt{2-A_k}$. Your claim is that
$$\pi=\lim_{k\rightarrow\infty}2^k\alpha B_k$$
To prove this claim, we shall first show by induction that $A_k$ admits the simple closed form
$$A_k=2\cos\left(\frac{\pi}{2^k\alpha}\right)$$
By definition, the above equality is satisfied for $k=0$. Suppose that the above formula is true for $k$, then by the cos half-angle formula $2\cos(x/2)=\sqrt{2+2\cos(x)}$, we have that
$$A_{k+1}=\sqrt{2+A_k}=\sqrt{2+2\cos\left(\frac{\pi}{2^k\alpha}\right)}=2\cos\left(\frac{\pi}{2^{k+1}\alpha}\right)$$
and therefore, by induction $A_k=2\cos\left(\frac{\pi}{2^k\alpha}\right)$ for all $k\geq 0$. Now, applying the sin half-angle formula $2\sin(x/2)=\sqrt{2-2\cos(x)}$ gives us that
$$B_k=\sqrt{2-A_k}=\sqrt{2-2\cos\left(\frac{\pi}{2^k\alpha}\right)}=2\sin\left(\frac{\pi}{2^{k+1}\alpha}\right)$$
Finally, applying the fact that $\pi=\lim\limits_{x\rightarrow 0}\frac{\sin(\pi x)}{x}$, we have that
$$\lim_{k\rightarrow\infty}2^k\alpha B_k=\lim_{k\rightarrow\infty}2^{k+1}\alpha\sin\left(\frac{\pi}{2^{k+1}\alpha}\right)=\pi$$
as desired.
|
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|
Extracting coefficients from a generating function Recently, I found that the generating function for a sequence I am interested in is
$$s(x) = -\frac{3 \, x^{3} + x^{2} + 2 \, x}{2 \, x^{3} + x - 1}.$$
Naturally, I am now keen on extracting the $n$th coefficient of the Taylor expansion of $s(x)$ without the help of a computer. Unfortunately, the three roots of the denominator of $s(x)$, of which two are complex, are rather unfriendly creatures, so that partial fraction decomposition does not seem to be a viable method.
But what can I do then to find the desired coefficients?
Thank you!
|
We can derive the $n$-th coefficient by makeing a geometric series expansion of
\begin{align*}
\color{blue}{s(x)}&\color{blue}{=-\frac{3x^3+x^2+2x}{2x^3+x-1}}\\
&\color{blue}{=2x+3x^2+6x^3+10x^4+16x^5+28x^6+\cdots}
\end{align*}
We use the coefficient of operator $[x^n]$ to derive the coefficient of $x^n$ of a series.
We obtain
\begin{align*}
\color{blue}{[x^n]}&\color{blue}{s(x)}
=[x^n]\left(-\frac{3x^3+x^2+2x}{2x^3+x-1}\right)\\
&=\left(3[x^{n-3}]+[x^{n-2}]+2[x^{n-1}]\right)\frac{1}{1-x\left(1+2x^2\right)}\tag{1}\\
&=\left(3[x^{n-3}]+[x^{n-2}]+2[x^{n-1}]\right)\sum_{k=0}^{\infty}x^k\left(1+x^2\right)^k\tag{2}\\
&=3\sum_{k=0}^{n-3}[x^{n-3-k}]\left(1+2x^2\right)^k
+\sum_{k=0}^{n-2}[x^{n-2-k}]\left(1+2x^2\right)^k\\
&\qquad\qquad+2\sum_{k=0}^{n-1}[x^{n-1-k}]\left(1+2x^2\right)^k\tag{3}\\
&=3\sum_{k=0}^{n-3}[x^{k}]\left(1+2x^2\right)^{n-3-k}
+\sum_{k=0}^{n-2}[x^{k}]\left(1+2x^2\right)^{n-2-k}\\
&\qquad\qquad+2\sum_{k=0}^{n-1}[x^{k}]\left(1+2x^2\right)^{n-1-k}\tag{4}\\
&=3\sum_{k=0}^{\left\lfloor\frac{n-3}{2}\right\rfloor}[x^{2k}]\left(1+2x^2\right)^{n-3-2k}
+\sum_{k=0}^{\left\lfloor\frac{n-2}{2}\right\rfloor}[x^{2k}]\left(1+2x^2\right)^{n-2-2k}\\
&\qquad\qquad+2\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}[x^{2k}]\left(1+2x^2\right)^{n-1-2k}\tag{5}\\
&\color{blue}{=3\sum_{k=0}^{\left\lfloor\frac{n-3}{2}\right\rfloor}\binom{n-3-2k}{2k}2^{k}
+\sum_{k=0}^{\left\lfloor\frac{n-2}{2}\right\rfloor}\binom{n-2-2k}{2k}2^{k}}\\
&\qquad\qquad\color{blue}{+\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\binom{n-1-2k}{2k}2^{k+1}}\tag{6}\\
\end{align*}
Comment:
*
*In (1) we use the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.
*In (2) we make a geometric series expansion.
*In (3) we again apply the rule as in (1). We also restrict the upper limit since other terms do not contribute.
*In (4) we change the order of summation $k\to n-a-k, a=1,2,3$.
*In (5) we respect that only even powers contribute.
*In (6) we select the coefficients accordingly.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 − + 1$; = 1, 2, 3,... How can we prove that
$^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 − + 1$; = 1, 2, 3,...
I attempted to solve this with Mathematical Induction as follows:
Let s(n) = $x^2 - x + 1$ | $^{6n+2} − ^{6n+1} + 1$; = 1, 2, 3,..
Basic Step
Let n = 1
⇒ $x^2 - x + 1$ | $^8 − ^7 + 1$
I then proved that the remainder is 0 using polynomial long division.
$\frac{^{6n+2} − ^{6n+1} + 1}{x^2 - x + 1}$ = $x^6 - x^4 - x^3 + x + 1$ R 0
∴ s(1) is true
Assumption Step
Assume that s(m) is true
⇒ $\frac{^{6m+2} − ^{6m+1} + 1}{x^2 - x + 1}$ = Q(x) where Q(x) is a polynomial
Inductive Step
To prove that s(m+1) is true
⇒ $\frac{^{6(m+1)+2} − ^{6(m+1)+1} + 1}{x^2 - x + 1}$ = T(x) where T(x) is a polynomial(x) is a polynomial
⇒ $\frac{^{6m+8} − ^{6m+7} + 1}{x^2 - x + 1}$ = T(x)
⇒ $\frac{x^6(^{6m+2} − ^{6m+1}) + 1}{x^2 - x + 1}$ = T(x)
However, I'm unsure of how to proceed from here. I would appreciate it if anyone could help me with this. Thanks!
|
*
*Finds the roots of $x^2-x+1$; they are complex
*Pick one of them and prove that is a root of $x^{6n+2}-x^{6n+1}+1$; automatically the other one is also a root therefore the two polynomials have common roots, hence the one with lesser number of roots is a factor of the other polynomial.
|
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|
Proof Of A Gamma Function - Double Factorial Identity How does one prove the following identity?
$$\sqrt{(-1)^n\frac{\Gamma(n+1/2)}{\Gamma(1/2-n)}} = \frac{(2n-1)!!}{2^n}$$
I attempted to prove this using the definition of the double factorial, however I couldn't continue and feel like there is a better method. I would appreciate it if somebody could show me a proof. Thank you
|
We use the identity
\begin{align*}
\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin (\pi z)}\qquad\qquad z\notin \mathbb{Z}
\end{align*}
evaluated at $z=n-\frac{1}{2}$ and obtain
\begin{align*}
\Gamma\left(1-\left(n+1/2\right)\right)\Gamma\left(n+1/2\right)&=\frac{\pi}{\sin\left(\pi\,\frac{2n+1}{2}\right)}\\
\color{blue}{\Gamma(1/2-n)\Gamma(n+1/2)}&\color{blue}{=(-1)^n\,\pi}\tag{1}
\end{align*}
We obtain with (1) and application of the identities
\begin{align*}
\Gamma(z+1)&=z\Gamma(z)\qquad\qquad\qquad z\in\mathbb{C}\setminus\{0,-1,-2,\ldots\}\tag{2}\\
\Gamma(1/2)&=\sqrt{\pi}\tag{3}
\end{align*}
\begin{align*}
\color{blue}{\sqrt{(-1)\frac{\Gamma(n+1/2)}{\Gamma(1/2-n)}}}
&=\frac{1}{\sqrt{\pi}}\Gamma(n+1/2)\tag{$\to$ (1)}\\
&=\frac{1}{\sqrt{\pi}}\left(n-\frac{1}{2}\right)\Gamma(n-1/2)\tag{$\to$ (2)}\\
&=\frac{1}{\sqrt{\pi}}\left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\Gamma(n-3/2)\tag{$\to$ (2)}\\
&=\cdots\\
&=\frac{1}{\sqrt{\pi}}\left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\cdots\left(\frac{1}{2}\right)\Gamma(1/2)\tag{$\to$ (2)}\\
&=\left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\cdots\left(\frac{1}{2}\right)\tag{$\to$ (3)}\\
&=\frac{1}{2^n}(2n-1)(2n-3)\cdots 1\\
&\,\,\color{blue}{=\frac{1}{2^n}(2n-1)!!}
\end{align*}
and the claim follows.
|
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|
The parabola with the equation $y=-x^2+4x+8$ is shifted so that it passes through the points (1,1) and (3,5). Find the equation of the new parabola. Given the points $(1,1)$ and $(3,5)$, the vertex form would be:
$1=(1-h)^2+k$ for $(1,1)$ and
$5=(3-h)^2+k$ for $(3,5)$.
With a system of equations, I obtain that $h = 1$ and $k = 1$, so the new formula of the shifted parabola would be: $y=(x-1)^2+1$
What I don't understand is the answer of my exercise sheet, which says that the new equation would be $y= -x^2+6x-4$
Can someone help me understand this?
Thanks
|
The original parabola is
$ y = - x^2 + 4 x + 8 $
Shifting by $(h, k)$ gives the parabola,
$ y = - (x - h)^2 + 4 (x - h) + 8 + k $
Points $(1,1)$ and $(3,5)$ are on the new parabola, so
$ 1 = - (1 - h)^2 + 4 (1 - h) + 8 + k $
$ 5 = - (3 - h)^2 + 4 (3 - h) + 8 + k $
Substracting,
$ - 4 = - (1^2 - 3^2 - 2 h (1 - 3 )) + 4 (1 - 3) $
$ - 4 = - (-8 + 4 h ) - 8 $
Therefore,
$ -8 + 4 h = - 4 $
From which
$ 4 h = 4 $
Hence, $ h = 1$.
Now substitute this into one of the new parabola equations to find $k$.
$ 1 = - (1 - 1)^2 + 4 (1 - 1) + 8 + k $
Therefore, $ k = -7 $
Now the equation of the new parabola is
$ y = - (x - 1)^2 + 4 (x - 1) + 8 + (-7) $
Expanding, this becomes
$ y = - x^2 +6 x - 4 $
|
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|
Evaluate $\int\frac{1}{x-\sqrt{1-x^2}}dx$ The integral $I$ in question is defined as follows
$$
I \equiv \int\frac{1}{x-\sqrt{1-x^2}}dx
$$
To solve this, I tried the trig substitution $x = \sin\theta$, with $dx = \cos\theta d\theta$, and rewrote the integral as follows
$$
\int\frac{\cos\theta}{\sin\theta-\sqrt{1-\sin^2\theta}}d\theta
$$
I used to identity $1 - \sin^2\theta = \cos^2\theta$ and simplified the denominator as follows
$$
\int\frac{\cos\theta}{\sin\theta-\cos\theta}d\theta
$$
I then rewrote $\cos\theta$ as $\frac{\sin\theta + \cos\theta}{2} - \frac{\sin\theta - \cos\theta}{2}$ and rewrote the integrand as follows
$$
\int\frac{1}{2}\frac{\sin\theta+\cos\theta}{\sin\theta - \cos\theta} - \frac{1}{2}d\theta
$$
I then split the integral as follows
$$
\frac{1}{2}\int\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}d\theta - \frac{1}{2}\int1d\theta
$$
For the first integral, I substituted $\phi = \sin\theta-\cos\theta$, with $d\theta = \frac{1}{\sin\theta+\cos\theta}d\phi$
We can then rewrite our integral as
$$
\frac{1}{2}\int\frac{1}{\phi}d\phi
$$
This is trivial and after undoing the substitutions we have a result of
$$
\frac{\ln({x - \cos(\arcsin(x))})}{2}
$$
The second integral is also trivial and just evaluates to $\frac{x}{2}$
Combining everything together gives us a final simplified answer of
$$
I = \frac{\ln({x - \sqrt{1-x^2}})-x}{2} + C
$$
However, both IntegralCalculator and WolframAlpha give very different answers, so if someone could tell me where I made a mistake or another approach entirely that would be greatly appreciated.
|
You made mistake in the second integral. It should be
$$
I_2 = \frac{\theta}{2} + C_2 = \frac{\arcsin(x)}{2} + C_2
$$
|
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|
$\int_{0}^{\pi}(\sqrt{x^2+\sin^2x}+\cos x\sin x\ln(x+\sqrt{x^2+\sin^2x}))\mathrm{d}x=\frac{\pi^2}{2}$ with one-variable calculus solution Prove that $$\int_{0}^{\pi}(\sqrt{x^2+\sin^2x}+\cos x\sin x\ln(x+\sqrt{x^2+\sin^2x}))\mathrm{d}x=\frac{\pi^2}{2}$$
My textbook says that we need to note that the integral can be transformed into $\int_L (\sqrt{x^2+y^2}\mathrm{d}x+y\ln(\sqrt{x^2+y^2}+x)\mathrm{d}y)$, where $L: y=\sin x$, $0\leq x\leq\pi$. We denote $\sqrt{x^2+y^2}$ by $P(x,y)$, and $y\ln(\sqrt{x^2+y^2}+x)$ by $Q(x,y)$. And then it is not hard to compute $\int_l P\mathrm{d}x+Q\mathrm{d}y$, where $l:(0\leq x\leq\pi\wedge y=0)$, as well as $\int_{l+L} P\mathrm{d}x+Q\mathrm{d}y$ (by Green's formula), (all the directions here are omitted, but it should make no ambiguity). Hence the question is done. I think this method is hard to observe. Do we have a way merely apply theorems of one-variable calculus? I have observed that $\cos x\sin x\ln(x+\sqrt{x^2+\sin^2x})=\cos x\sin x(2\ln\sin x-\ln(x+\sqrt{x^2+\sin^2x}))$, and $\cos x\sin x\ln\sin x=(\sin x\ln\sin x)(\sin x)'$, hence its indefinite integral is computable. I do not know if this observation would help.
|
Clearly
\begin{eqnarray}
&&\int_{0}^{\pi}(\sqrt{x^2+\sin^2x}+\cos x\sin x\ln(x+\sqrt{x^2+\sin^2x}))\mathrm{d}x\\
&=&\int_{0}^{\pi}\sqrt{x^2+\sin^2x}\mathrm{d}x+\int_{0}^{\pi}\sin x\ln(x+\sqrt{x^2+\sin^2x})\mathrm{d}\sin x\\
&=&\int_L\sqrt{x^2+y^2}\mathrm{d}x+y\ln(x+\sqrt{x^2+y^2})\mathrm{d}y\\
&=&\int_{L+l}\sqrt{x^2+y^2}\mathrm{d}x+y\ln(x+\sqrt{x^2+y^2})\mathrm{d}y-\int_l\sqrt{x^2+y^2}\mathrm{d}x+y\ln(x+\sqrt{x^2+y^2})\mathrm{d}y.
\end{eqnarray}
Now you can use Green's Theorem to calculate the first term since the second term is easy to handle.
|
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|
Real Analysis fundamental theorems of Calculus contradiction? Evaluate: $\frac{d}{dx} \int_{0}^{x} x^3t^3dt.$
Solution: $\frac{d}{dx} \int_{0}^{x} x^3t^3dt = \frac{7x^6}{4}.$
Proof:
Consider the function's $F:[a,b] \rightarrow \mathbb{R}$ and $f:[a,b] \rightarrow \mathbb{R}$, define: $F(t) = \frac{1}{4}x^3t^4 + c$ and $f(t) = x^3t^3.$ Observe that by proposition 4.7 F is differentiable on (a,b) and $F'(t) = x^3t^3 = f(t)$ for all x in $(a,b).$ Moreover by the polynomial property (proposition 2.17): F is continuous on [a,b] and f is continuous on [a,b], so by theorem 3.14 we know f is bounded and by theorem 6.18 f is integrable. We satisfy the criteria in the first fundamental theorem of calculus (theorem 6.22) hence:
$\frac{d}{dx} \int_{0}^{x} x^3t^3dt = \frac{d}{dx} [F(x) - F(0)] = \frac{d}{dx} \frac{1}{4}x^7 = \frac{7}{4}x^6.$
But by the second fundamental theorem of calculus (theorem 6.29):
$\frac{d}{dx} \int_{0}^{x} x^3t^3dt = f(x) = x^6 \neq F'(x) = \frac{7}{4}x^6$. Contradiction. WHY!?
(Textbook: Advanced Calculus - Patrick M. Fitzpatrick, Chapter 6.6)
Edit
Thank you everyone for your help, I managed to figure it out from all of the advice!
|
It should be noted that the Leibniz Integral Rule deals with integrals like the one in question.
In the current example the integral takes the following form (with a slight generalization).
\begin{align}
\frac{d}{dx} \, \int_{0}^{x} (x \, t)^n \, dt &= (x \, x)^n \, \frac{d}{dx}(x) - (x \cdot 0)^n \, \frac{d}{dx}(0) + \int_{0}^{x} \left[ \frac{d}{dx}(x \, t)^n \right] \, dt \\
&= x^{2n} + n \, x^{n-1} \, \int_{0}^{x} t^{n} \, dt \\
&= x^{2 n} + n \, x^{n-1} \cdot \frac{x^{n+1}}{n+1} \\
&= \frac{2 n + 1}{n+1} \, x^{2 n}.
\end{align}
A straight forward evaluation takes the form
\begin{align}
\frac{d}{dx} \, \int_{0}^{x} (x \, t)^n \, dt &= \frac{d}{dx} \, \left( \frac{x^{2 n + 1}}{n+1} \right) = \frac{2n+1}{n+1} \, x^{2n}.
\end{align}
By using a corollary of the Fundamental Theorem of Calculus, namely,
$$ \int_{a}^{b} f(t) \, dt = F(b) - F(a), $$
where $F(x)$ is the anti-derivative of $f(x)$, then
\begin{align}
\frac{d}{dx} \, \int_{0}^{x} (x \, t)^n \, dt &= \frac{d}{dx} \, \left[ x^n \, (F(x) - F(0) ) \right] \\
&= \frac{d}{dx} \, \left[ x^n \, \left(\frac{x^{n+1}}{n+1} - \frac{0^{n+1}}{n+1}\right) \right] \\
&= \frac{d}{dx} \, \left( \frac{x^{2n+1}}{n+1}\right) \\
&= \frac{2n+1}{n+1} \, x^{2n}.
\end{align}
Each of these views yields the same result.
|
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|
Prove continuity of $f: [\frac{1}{2},\infty ) \rightarrow \mathbb{R}: x \mapsto \sqrt{2x-1}$ for $x_0>\frac{1}{2}$ with Epsilon-delta definition Prove continuity of $f: [\frac{1}{2},\infty ) \rightarrow \mathbb{R}: x \mapsto \sqrt{2x-1}$ for $x_0>\frac{1}{2}$ with Epsilon-delta definition of continuity
show:
$\forall \epsilon >0 \ \exists \delta>0 \ \forall x \in [\frac{1}{2}, \infty): ( \ |x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\epsilon \ )$
$|x-x_0|<\delta$
$|\sqrt{2x-1} - \sqrt{2x_0-1} | = |\frac{(2x-1)-(2x_0-1)}{\sqrt{2x-1} + \sqrt{2x_0-1}}| = |\frac{2(x-x_0)}{\sqrt{2x-1} + \sqrt{2x_0-1}}|=\frac{2|x-x_0|}{\sqrt{2x-1} + \sqrt{2x_0-1}}$
$\sqrt{2x-1}>0 \Rightarrow \frac{2|x-x_0|}{\sqrt{2x-1} + \sqrt{2x_0-1}}<\frac{2|x-x_0|}{\sqrt{2x_0-1}} \Rightarrow |f(x)-f(x_0)|<\frac{2\delta}{\sqrt{2x_0-1}}$
let $\delta=\frac{\epsilon \sqrt{2x_0-1}}{2} \Rightarrow |f(x)-f(x_0)|<\frac{\epsilon \sqrt{2x_0-1}}{\sqrt{2x_0-1}}=\epsilon$
Which shows
$\forall \epsilon >0 \ \exists \delta>0 \ \forall x \in [\frac{1}{2}, \infty): ( \ |x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\epsilon \ )$
and therefore the continuity of $f(x)$ for $x_0>\frac{1}{2}$ is proven.
Is this prove correct?
Edit: changed $x=\frac{1}{2}$ to $x_0>\frac{1}{2}$
|
For $x_0=\frac{1}{2}$:
$|f(x)-f(\frac{1}{2})|=|f(x)|=\left|\sqrt{2x-1}\right|<\epsilon$
$\forall x \in ( \frac{1}{2},\frac{1}{2}+\delta): \left|\sqrt{2x-1}\right| \stackrel{!}{<} \epsilon$
choose: $\delta:= \frac{\epsilon^{2}}{2} \Rightarrow \forall x \in ( \frac{1}{2},\frac{1}{2}+\delta): \left|\sqrt{2x-1}\right| < \epsilon$
And therefore the continuity of $f(x)$ in $x_0=\frac{1}{2}$ is proven
|
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|
Can we evaluate $\int_0^\infty x^k\frac{ae^{bx}}{\left(1+ae^{bx}\right)^2}e^{-\frac{x^2}{2}}dx$ This sequence of integrals come from the following expectations $$E\left(X^kf(X)\right),\quad f(x;a,b)=\frac{ae^{bx}}{\left(1+e^{bx}\right)^2},\quad a>0,~b>0,~k=0,1,2,\cdots,$$
where $X\sim N(0,1)$. So we can express them as a sequence of integrals $$\int_{-\infty}^\infty x^k\frac{ae^{bx}}{\left(1+ae^{bx}\right)^2}e^{-\frac{x^2}{2}}dx,\quad k=0,1,2,\cdots.$$
Since $f(-x;a,b)=f(x;\frac{1}{a},b)$, which have some similarities, then we only consider $$\int_0^\infty x^k\frac{ae^{bx}}{\left(1+ae^{bx}\right)^2}e^{-\frac{x^2}{2}}dx.$$ I have found that $$\left(\frac{1}{b}\cdot\frac{ae^{bx}}{1+ae^{bx}}\right)'=\frac{ae^{bx}}{\left(1+ae^{bx}\right)^2},$$ so, maybe we can use integration by parts (here we take $k\geq 1$)
\begin{align}
\int_0^\infty x^k\frac{ae^{bx}}{\left(1+ae^{bx}\right)^2}e^{-\frac{x^2}{2}}dx&=\frac{1}{b}\int_0^\infty x^ke^{-\frac{x^2}{2}}d\frac{ae^{bx}}{1+ae^{bx}}\\
&=\frac{1}{b}x^ke^{-\frac{x^2}{2}}\frac{ae^{bx}}{1+ae^{bx}}\Big\vert_0^\infty-\frac{1}{b}\int_0^\infty \frac{ae^{bx}}{1+ae^{bx}}dx^ke^{-\frac{x^2}{2}}\\
&=\frac{1}{b}\int_0^\infty x^{k+1}\frac{ae^{bx}}{1+ae^{bx}}e^{-\frac{x^2}{2}}dx-\frac{k}{b}\int_0^\infty x^{k-1}\frac{ae^{bx}}{1+ae^{bx}}e^{-\frac{x^2}{2}}dx
\end{align}
Therefore, it suffices to calculate $$\int_0^\infty x^k\frac{ae^{bx}}{1+ae^{bx}}e^{-\frac{x^2}{2}}dx,\quad k=0,1,2\cdots,$$
and I am stuck on this.
|
The farthest I could go
$$x^k\frac{ae^{bx}}{1+ae^{bx}}e^{-\frac{x^2}{2}}=\sum_{n=0}^\infty (-1)^n\, x^k\,e^{-\frac{x^2}{2}}\left(\frac{e^{-b x}}{a}\right)^n$$
Defining
$$J_{n,k}=\int_0^\infty x^k\,e^{-\frac{x^2}{2}}e^{-nb x}\,dx$$ Using Kummer confluent hypergeometric functions, they write (with $t=bn$)
$$J_{n,k}=2^{\frac{k-1}{2}} \Gamma
\left(\frac{k+1}{2}\right) \,
_1F_1\left(\frac{k+1}{2};\frac{1}{2};\frac{t^2
}{2}\right)- 2^{\frac k2} t\, \Gamma
\left(\frac{k+2}{2}\right) \,
_1F_1\left(\frac{k+2}{2};\frac{3}{2};\frac{t^2}
{2}\right)$$
They write
$$J_{n,k}=(-1)^k\,\sqrt{\frac{\pi }{2}} e^{\frac{t^2 }{2}}
\text{erfc}\left(\frac{t}{\sqrt{2}}\right)\,P_k(t)+(-1)^{k+1}\,Q_{k}(t)$$ and the first polynomials are respectively
$$\left(
\begin{array}{cc}
k & P_k(t) \\
0 & 1 \\
1 & t \\
2 & t^2+1 \\
3 & t^3+3 t \\
4 & t^4+6 t^2+3 \\
5 & t^5+10 t^3+15 t \\
6 & t^6+15 t^4+45 t^2+15 \\
7 & t^7+21 t^5+105 t^3+105 t \\
8 & t^8+28 t^6+210 t^4+420 t^2+105 \\
\end{array}
\right)$$
and
$$\left(
\begin{array}{cc}
k & Q_k(t) \\
0 & 0 \\
1 & 1 \\
2 & t \\
3 & t^2+2 \\
4 & t^3+5 t \\
5 & t^4+9 t^2+8 \\
6 & t^5+14 t^3+33 t \\
7 & t^6+20 t^4+87 t^2+48 \\
8 & t^7+27 t^5+185 t^3+279 t \\
\end{array}
\right)$$
These show interesting patterns which would be intresting to explore further.
The problem remains with the infinite summations.
|
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Prove that for every positive integer $n$ there do not exist four positive integers $a,b,c,d$ with $ad = bc$ and $n^2 < a < b < c < d < (n+1)^2$. Here is the problem Timothy Gowers was trying to solve in this YouTube video.
Prove that for every positive integer $n$ there do not exist four positive integers $a,b,c,d$ with $ad = bc$ and $n^2 < a < b < c < d < (n+1)^2$.
The problem seems to be related to the determinant of the $2 \times 2$ matrix $\begin{pmatrix}a&b\\ c&d\end{pmatrix}$. The condition $ad = bc$ is equivalent to $$\det \begin{pmatrix}a&b\\ c&d\end{pmatrix} = 0.$$
Geometrically this mean that the space spanned by the vectors $(a,b)$ and $(c,d)$ is a line. Is there a way to relate this to the given problem or is it just a coincidence?
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All variables are positive integers.
It is enough to prove the following claim.
Claim: Suppose $n^2\le a<b\le c<d\le (n+1)^2$ and $ad=bc$. Then $a=n^2$, $b=c=n(n+1)$, $d=(n+1)^2$.
Proof:
$ad=bc\implies\frac dc=\frac ba.$ Hence $$d-c=c(\frac dc-1)> a(\frac ba-1)=b-a.$$
Let $k=b-a$. Since $d-c$ and $b-a$ are integers, $d-c\ge k+1$. (This "$+\ 1$" is critical.)
$$\begin{aligned}&\quad\ \,ad-bc\\&=ad-(a+k)(d-(d-c))\\
&\ge ad-(a+k)(d-(k+1))\\
&=(k+1)a-kd+k(k+1)\\
&\ge (k+1)n^2-k(n+1)^2+k^2+k\\
&=(n-k)^2\\
&\ge0
\end{aligned}$$
Since $ad=bc$, all inequalities above must be equalities, i.e., $d-c=k+1$, $a=n^2$, $d=(n+1)^2$, and $k=n$. Hence $b=a+k=n(n+1)$ and $c=d-(k+1)=n(n+1)$.
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|
Show that $b^2+c^2-a^2\leq bc$. Let $a,b,c>0$ such that $b<\sqrt{ac}$, $c<\frac{2ab}{a+b}$. Show that $b^2+c^2-a^2\leq bc$.
I tried to construct a triangle with $a,b,c$ and to apply The cosine rule, but I am not sure that it's possible to construct it and also I have no idea how to prove that an angle it's greater than $60^{\circ}$.
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Both conditions and the hypothesis are homogenous, so we can scale the variables until $a=1$. Then the problem is equivalent to the implication:
$$b<\sqrt{c}\land c<\frac{2b}{1+b}\Rightarrow b^2+c^2\le1+bc.$$
New conditions are more convenient and can be rewritten to:
$$b^2<c<\frac{2b}{1+b}.$$
Let's devide the condition by $b$ and the hypothesis by $b^2$, the problem is equivalen to:
$$b<\frac{c}{b}<\frac{2}{1+b}\Rightarrow1+\Big(\frac{c}{b}\Big)^2\le\frac{c}{b}+\frac{1}{b^2}.$$
Let's introduce $r:=\frac{c}{b}$. The problem is equivalent to:
$$b<r<\frac{2}{1+b}\Rightarrow1-r+r^2\le\frac{1}{b^2}.$$
The condition implies:
$$b<\frac{2}{1+b}\Rightarrow b+b^2<2\Rightarrow\Big(b+\frac{1}{2}\Big)^2<\frac{9}{4}\Rightarrow b+\frac{1}{2}<\frac{3}{2}\Rightarrow b<1.$$
Returning the limitation $b>0$ together with the $\frac{2}{1+b}$ being decreasing with the supremum at $b=0$ we get:
$$0<b<1\land0<r<2.$$
The case $r\le1$ is easy:
$$1-r+r^2=\frac{3}{4}+\Big(\frac{1}{2}-r\Big)^2\le\frac{3}{4}+\max_{0<r\le1}\Big|\frac{1}{2}-r\Big|^2=\frac{3}{4}+\Big(\frac{1}{2}\Big)^2=1=\frac{1}{1}<\frac{1}{b^2}.$$
For $r>1$ we have
$$\frac{d}{dr}(1-r+r^2)=2r-1>1>0,$$
so the expression $1-r+r^2$ increases with $r$ and we have
$$1-r+r^2<1-\frac{2}{1+b}+\Big(\frac{2}{1+b}\Big)^2=\frac{3+b^2}{(1+b)^2}=\frac{1}{b^2}\frac{(3+b^2)b^2}{(1+b)^2}=$$
$$\frac{1}{b^2}\frac{b^2+2b^2+b^4}{(1+b)^2}<\frac{1}{b^2}\frac{1+2b+b^2}{(1+b)^2}=\frac{1}{b^2}.$$
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|
Probability on Number Theory Problem: Suppose that $a,b,c \in \{1,2,3,\cdots,1000\}$ are randomly selected with replacement. Find the probability that $abc+ab+2a$ is divisible by $5$.
Answer given from the worksheet: $33/125$
My answer: $\frac{641}{3125}$
Attempt: Since $abc+ab+2a = a(bc+b+2)$, either $a \equiv 0 \pmod{5}$ or $bc+b+2 \equiv 0 \pmod{5}$. The first case, which is $a \equiv 0 \pmod{5}$, has a probability of $1/5$. Now on the other case,
$$bc+b+2 \equiv 0 \pmod{5} \Rightarrow b(c+1) \equiv 3 \pmod{5}$$
happens when $b\equiv 1$ and $c \equiv 2$, $b \equiv 2$ and $c \equiv 3$, $b \equiv 3$ and $c \equiv 0$, or $b \equiv 4$ and $c \equiv 1$, total of $4$ solutions. So, this case has a probability of $4 \cdot\frac{1}{5^4} = \frac{4}{625}$. By Inclusion-exclusion principle, the final probability should be $\frac{1}{5} + \frac{4}{625} - \frac{1}{5} \cdot \frac{4}{625} = \frac{641}{3125}.$
This is apparently not the same from the given answer in the worksheet. Where did I go wrong?
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Brute forcing this in Python, I get the same answer of $41/125 = 0.328$ previously given more thoroughly by John Omielan, which differs from the worksheet's answer of $33/125$:
import itertools as it
N = 100
combos = list(it.product(range(1,N+1), repeat=3))
exprs = [a*b*c + a*b + 2*a for (a,b,c) in combos]
print(len([x for x in exprs if x%5==0]) / len(exprs))
>>> 0.328
(note: this code only selects $a, b, c \in \{ 1, 2, 3, ..., 100 \}$ for memory reasons, but this should not matter, as long as the upper limit of the set of integers is any multiple of 5.)
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Evaluation of $~\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta$ $$
I:=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta
$$
My tries
$$\begin{align}
s&:=\sin\theta\\
c&:=\cos\theta\\
I&=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta\\
&=\int_{0}^{2\pi}{c^2-s^2\over s^4+c^4}\mathrm d\theta\\
&=\int_{0}^{2\pi}{(1-s^2)-s^2\over s^4+(c^2)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+(1-s^2)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+(s^2-1)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+s^4-2s^2+1}\mathrm d\theta\\
&=\int_{0}^{2\pi}\underbrace{\color{red}{\left({1-2s^2\over 2s^4-2s^2+1}\right)}}_{\text{I got stuck here}}\mathrm d\theta\\
\end{align}$$
I need your help.
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Here is an alternate solution using residues. Starting from where you left off, we can use the complex definitions of $\sin(x)$ and $\cos(x)$ to transform the integral into
$$\int_{0}^{2\pi}\frac{4e^{2ix}\left(1+e^{4ix}\right)}{6e^{4ix}+e^{8ix}+1}dx.$$
Since the integrand is periodic on $\pi$, let $z=e^{2ix}$. The contour $\gamma$ will be the unit circle. Then
$$\oint_{\gamma} \frac{4z\left(1+z^{2}\right)}{6z^{2}+z^{4}+1}\left(\frac{dz}{2iz}\right) = -2i\oint_{\gamma}\frac{1+z^{2}}{6z^{2}+z^{4}+1}dz.$$
The set of singularities on the disc of $\gamma$ is $Z := \left\{i\sqrt{3-2\sqrt{2}}, -i\sqrt{3-2\sqrt{2}}\right\}.$ By Residue Theory, we transform the contour integral into
$$2\pi i (-2i)\sum_{z_0 \in Z}\operatorname{Res}\left(\frac{1+z^{2}}{6z^{2}+z^{4}+1}, z=z_0\right).$$
Thus,
$$2\pi i\left(-2i\right)\left(\frac{1+\left(i\sqrt{3-2\sqrt{2}}\right)^{2}}{\frac{d}{dz}\left[6z^{2}+z^{4}+1\right]_{z=i\sqrt{3-2\sqrt{2}}}}\right) + 2\pi i\left(-2i\right)\left(\frac{1+\left(-i\sqrt{3-2\sqrt{2}}\right)^{2}}{\frac{d}{dz}\left[6z^{2}+z^{4}+1\right]_{z=-i\sqrt{3-2\sqrt{2}}}}\right)$$
which equals
$$2\pi i\left(-2i\right)\left(\frac{1+\left(i\sqrt{3-2\sqrt{2}}\right)^{2}}{12\left(i\sqrt{3-2\sqrt{2}}\right)+4\left(i\sqrt{3-2\sqrt{2}}\right)^{3}}\right)+2\pi i\left(-2i\right)\left(\frac{1+\left(-i\sqrt{3-2\sqrt{2}}\right)^{2}}{12\left(-i\sqrt{3-2\sqrt{2}}\right)+4\left(-i\sqrt{3-2\sqrt{2}}\right)^{3}}\right).$$
This simplifies down to $0$.
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|
Solving the homogeneous first order ode $y' = \frac{2xy}{y^2-x^2}$ Solving the homogeneous first order ode $y' = \frac{2xy}{y^2-x^2}$
substituting $y=ux$ so that $y' = u + x\frac{du}{dx}$"
$u + x\frac{du}{dx} = \frac{2x^2u}{ux^2-x^2} = \frac{2u}{u^2-1}$
$\rightarrow x\frac{du}{dx} = \frac{2u}{u^2-1} - u = \frac{2u}{u^2-1} - \frac{u(u^2-1)}{u^2-1}$
$\rightarrow x\frac{du}{dx} = \frac{-u^3+u}{u^2-1}$
Separating variables...
$\rightarrow \frac{u^2-1}{-u^3+u}du = \frac{1}{x}dx$
Now, from here we can factor the top and bottom of the LHS as so...
$\frac{u^2-1}{-u^3+u} = \frac{(u+1)(u-1)}{-u(u+1)(u-1)} = \frac{-1}{u}$
And I'm thinking this is where I'm going wrong... Is there a reason that this simplification is not valid? I'm thinking that maybe I shouldn't simplify and that I should use partial fraction's instead?
The answer is supposed to be $3yx^2 - y^3 = k$
Edit: I continued as so:
$\frac{u^2-1}{3u-u^3} = \frac{A}{u} + \frac{B}{u-\sqrt{3}} + \frac{C}{u + \sqrt{3}}$.
I got $A = \frac{1}{3}$ by multiplying both sides by $u$ and then taking the limit as $u \rightarrow 0$. A similar method led to $B = \frac{1}{3}$ and $C = \frac{1}{3}$. Thus I arrived at:
$\frac{u^2-1}{3u-u^3} = \frac{1}{3u} + \frac{1}{3(u-\sqrt{3})} + \frac{1}{3(u+\sqrt{3})}$.
And so:
$(\frac{1}{3u} + \frac{1}{3(u-\sqrt{3})} + \frac{1}{3(u+\sqrt{3})})du = \frac{1}{x}dx$
Integrating and then exponentiating both sides:
$(u(u+\sqrt{3})(u-\sqrt{3}))^{\frac{1}{3}} = xk$
$(u(u^2-3))^{\frac{1}{3}} = xk$
$u = \frac{y}{x}$ and so:
$(\frac{y}{x}((\frac{y}{x})^2-3))^{\frac{1}{3}} = xk$
This leads to:
$(\frac{y}{x})^3 - \frac{3y}{x} = x^3k$
... Which is still not correct!!!
I haven't done ODE's quite a long time... I've been flying through Schaum's outline on the topic and I've done plenty of problems like this one... But i can't get this one correct!! help appreciated.
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$$y' = \frac{2xy}{y^2-x^2}$$
Since $y'=\dfrac 1 {x'}$:
$$\dfrac 1 {x'} = \frac{2xy}{y^2-x^2}$$
$${y^2-x^2} = {2xx'y}$$
Note that $2xx'=(x^2)'$:
$$y^2 = {(x^2)'y}+x^2$$
$$y^2 = {(x^2y)'}$$
Integrate:
$$\dfrac {y^3}3=x^2y+C$$
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Evaluating $\log_3(1+2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1))$ in the most efficient way I have come across a tricky question while studying logarithms.
$$\log_3(1+2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1))$$
While plugging it into a calculator brings a seemingly simple answer, I cannot find a way to start this without a calculator.
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Rewrite $\log_3(1+2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1))$ as $\log_3(1+(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1))$. You can use the Difference of Squares formula to find $\log_3(1+(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1))$. If you use the Difference of Squares formula 5 times, you get $\log_3(1+(3^{64}-1))$. Finally, by removing the parenthesis, you get $\log_3(3^{64}+1-1)$, which simplifies to $\log_3(3^{64})$. Now, with the logarithm power rule, this becomes $64\log_3(3)$, which is equal to $64$.
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Dodecahedron, angle between edge and face. In an effort to build a dodecahedron frame in Fusion360 I need to know some of the angles. Looking around I found out that the angle between an edge and a face on a regular dodecahedron is $121.7^\circ$ but I couldn't find the mathematical formula nor the way to calculate this angle. The formula is needed so the exact angle can be used so the simulation is precise. Can anyone help?
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This angle can be derived from the dihedral angle $\delta = \arccos (-1/\sqrt{5})$ of the dodecahedron using a vector method. (The dihedral angle $\delta$ is derived during construction of the dodecahedron - eg see this answer to How does this proof of the regular dodecahedron's existence fail?).
Consider a group of three adjacent pentagon faces as shown in Fig 1, in which pentagon $A$ is in the plane of the page, and $B$ and $C$ are tilted forwards to meet $A$ at the dihedral angle $\delta$. Each pentagon makes dihedral angle $\delta$ with the other two. The angle required is $\alpha$ between edge $e$ and the plane of pentagon $A$. If $\beta$ is the angle between the unit normal $\underline{\mathbf{u}}_A = \underline{\mathbf{k}}$ to $A$ and the unit vector $\underline{\mathbf{e}}$ along $e$, then $\alpha = \beta + 90$. Since edge $e$ is the intersection line of the planes $B$ and $C$, then if $\underline{\mathbf{u}}_B$ is a unit normal to $B$ and $\underline{\mathbf{u}}_C$ is a unit normal to $C$ then noting that the angle between $\underline{\mathbf{u}}_B$ and $\underline{\mathbf{u}}_C$ is the dihedral angle $\delta$, we have :
$$\underline{\mathbf{e}} = \frac{1}{|\underline{\mathbf{u}}_B \times \underline{\mathbf{u}}_C|} \; (\underline{\mathbf{u}}_B \times \underline{\mathbf{u}}_C) = \frac{1}{\sin \delta} \; (\underline{\mathbf{u}}_B \times \underline{\mathbf{u}}_C).
$$
Setting up an $xyz$-axes system as shown in Figs 1 and 2 we have :
\begin{eqnarray*}
\underline{\mathbf{u}}_B & = & -\sin \epsilon\; \underline{\mathbf{i}} + \cos \epsilon\; \underline{\mathbf{k}} \\
\Rightarrow \hspace{1em} \underline{\mathbf{u}}_B & = & -\sin \delta\; \underline{\mathbf{i}} - \cos \delta\; \underline{\mathbf{k}}
\end{eqnarray*}
Since pentagon $C$ is obtained from pentagon $B$ by rotating $B$ by $+72^\circ$ about the $z$-axis, we can obtain $\underline{\mathbf{u}}_C$ from $\underline{\mathbf{u}}_B$ by rotating $\underline{\mathbf{u}}_B$ by $72^\circ$ about the $z$-axis. (Note we are rotating vectors, which possess only direction and magnitude - they have no starting or ending point, in contrast to directed line segments, but can be 'placed' anywhere and will always point in the same direction). This rotation is a 2D rotation in the $xy$-plane and leaves the $\underline{\mathbf{k}}$ component unchanged, resulting in :
$$
\underline{\mathbf{u}}_C = -\sin \delta \cos 72\; \underline{\mathbf{i}} - \sin \delta \sin 72\; \underline{\mathbf{j}} - \cos \delta\; \underline{\mathbf{k}}
$$
Then :
$$
\underline{\mathbf{u}}_B \times \underline{\mathbf{u}}_C = \sin^2 \delta \sin 72\; \underline{\mathbf{k}} -\sin \delta \cos \delta\; \underline{\mathbf{j}} + \sin \delta \cos \delta \cos 72\; \underline{\mathbf{j}} - \sin \delta \cos \delta \sin 72\; \underline{\mathbf{i}}
$$
and so :
$$
\cos \beta = \underline{\mathbf{k}} \cdot \underline{\mathbf{e}} = \frac{1}{\sin \delta} \; (\underline{\mathbf{u}}_B \times \underline{\mathbf{u}}_C) \cdot \underline{\mathbf{k}} = \sin \delta \sin 72
$$
and the desired angle $\alpha$ is given by :
$$
\cos \alpha = \cos(\beta + 90) = -\sin \beta = -\sqrt{1 - \cos^2 \beta}.
$$
The 18-72-90$^\circ$ triangle (which comes from the diagonal of a regular pentagon) tells us :
$$
\sin 72 = \frac{\sqrt{4\Phi + 3}}{2\Phi}
$$
where $\Phi$ is the golden ratio $\frac{1 + \sqrt5}{2}$.
The dihedral angle $\delta$ satisfies $\sin^2 \delta = 4/5$ so making use of the identity $\Phi^2 = \Phi + 1$ we now have:
\begin{eqnarray*}
\cos^2 \beta & = & \sin^2 \delta \sin^2 72 = \frac{4}{5} \cdot \frac{4\Phi + 3}{4\Phi^2} = \frac{4\Phi + 3}{5\Phi + 5} \\
\Rightarrow 1 - \cos^2 \beta & = & \frac{\Phi + 2}{5\Phi^2} \\
\Rightarrow \cos \alpha & = & -\frac{1}{\Phi \sqrt{5}} \sqrt{\Phi + 2} = -\frac{1}{\Phi \sqrt{5}} \sqrt{\frac{5 + \sqrt{5}}{2}} = -\frac{2}{1 + \sqrt{5}} \sqrt{\frac{5 + \sqrt{5}}{10}} \\
& = & -\frac{1}{2} \sqrt{\frac{(\sqrt{5} - 1)^2 (5 + \sqrt{5})}{10}} = -\frac{1}{2} \sqrt{\frac{(6 - 2\sqrt{5}) (5 + \sqrt{5})}{10}} \\
& = & -\frac{1}{2} \sqrt{\frac{20 - 4\sqrt{5}}{10}} = -\sqrt{\frac{5 - \sqrt{5}}{10}}.
\end{eqnarray*}
Thus :
$$
\alpha = \arccos \left\{ -\sqrt{\frac{5 - \sqrt{5}}{10}} \right\}.
$$
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|
Contest Math Question: simplifying logarithm expression further I am working on AoPS Vol. 2 exercises in Chapter 1 and attempting to solve the below problem:
Given that $\log_{4n} 40\sqrt{3} = \log_{3n} 45$, find $n^3$ (MA$\Theta$ 1991).
My approach is to isolate $n$ and then cube it. Observe:
\begin{align*}
\frac{\log 40\sqrt{3}}{\log 4n} = \frac{\log 45}{\log 3n} \\
\log 40\sqrt{3}\log 3n = \log 45\log 4n\\
\log 40\sqrt{3} \cdot (\log 3 + \log n) = \log 45 \cdot (\log 4 + \log n)\\
\log n \cdot (\log 40\sqrt{3} - \log 45) = \log 45\log 4 - \log 40\sqrt{3}\log 3
\end{align*}
Dividing through and putting the coefficients as powers, we have:
\begin{align*}
\log n &= \frac{\log 45^{\log 4} - \log \left[(40\sqrt{3})^{\log 3}\right]}{\log\left(\frac{40\sqrt{3}}{45}\right)}
=\frac{\log \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right) }{\log\left(\frac{40\sqrt{3}}{45}\right)} \\
&=\log \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right)^{\log\left(\frac{40\sqrt{3}}{45}\right)^{-1}}
\end{align*}
which shows that
\begin{align*}
n^3 = \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right)^{3\cdot\log\left(\frac{40\sqrt{3}}{45}\right)^{-1}}
\end{align*}
Somehow it feels like this answer may be simplified further. Are the steps shown so far correct and can the answer be expressed in a better way?
|
We could also take a "factorization" approach. Since
$$ \log_{4n} 40\sqrt{3} \ \ = \ \ \log_{3n} 45 \ \ = \ \ \alpha \ \ , $$
we can write
$$ (4n)^\alpha \ \ = \ \ 2^3·3^{1/2}·5 \ \ \ , \ \ \ (3n)^\alpha \ \ = \ \ 3^2·5 $$ $$ \Rightarrow \ \ n^\alpha \ \ = \ \ 2^{3 \ - \ 2·\alpha}·3^{1/2}·5 \ \ = \ \ (2^0)·3^{2 \ - \ \alpha}·5 \ \ . $$
The "primes-raised-to-powers" in each of these factorizations need to "match", so we have
$$ 3 \ - \ 2·\alpha \ \ = \ \ 0 \ \ \ , \ \ \ \frac12 \ \ = \ \ 2 \ - \ \alpha \ \ , $$
both of which tell us that $ \ \alpha \ = \ \frac32 \ \ . $
Thus, we have $$ n^3 \ \ = \ \ (n^\alpha)^2 \ \ = \ \ (2^{3 \ - \ 2·\alpha}·3^{1/2}·5)^2 $$ $$ = \ \ 2^{6 \ - \ 4·\alpha}·3·25 \ \ = \ \ 2^{6 \ - \ 4·[3/2]}·3·25 \ \ = \ \ 2^0·75 $$
or
$$ n^3 \ \ = \ \ 3^{4 \ - \ 2·\alpha}·5^2 \ \ = \ \ 3^{4 \ - \ 2·[3/2]}·25 \ \ = \ \ 3·25 \ \ . $$
[Since this problem has so few "moving parts", the posted answers end up being variations on the same concept.]
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|
Combination with repeated elements If I have a set of $4$ numbers, for example $(1, 2, 3, 3)$, how do I calculate the $C(4, 3)$? The order doesn't matter. So, there are three different combinations that I can get - $(1, 2, 3), (1, 3, 3)$ and $(2, 3, 3)$.
Another example, if I have the word $BABY$, how do I calculate $C(4, 2)$? The combinations that I get are - $(B, A), (B, B), (B, Y)$ and $(A, Y)$.
How do I calculate $C(n, r)$ with repeated elements? Is there a formula or a method to calculate this? Something that will work with bigger values of $n$ and $r$ as well.
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With a little algebra, we can calculate the desired numbers. We use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of a series. In this way we can write, for example
\begin{align*}
C(4,2)=\binom{4}{2}&=[x^2](1+x)^4\\
&=\color{blue}{[x^2]}\left(1+4x+\color{blue}{6}x^2+4x^3+1\right)\color{blue}{=6}
\end{align*}
If we consider the multiset $\{1,2,3,3\}$, we can choose $1$ zero or once, $2$ zero or once, and $3$ zero, once or twice. We encode this situation algebraically as
\begin{align*}
\underbrace{(1+x)}_{1}\underbrace{(1+x)}_{2}\underbrace{(1+x+x^2)}_{3}\tag{1}
\end{align*}
where the exponent $0,1$ or $2$ of $x$ denotes the number of occurrences of $1,2$ resp. $3$.
*
*The number of three-element subsets of the multiset $\{1,2,3,3\}$ is
\begin{align*}
\color{blue}{[x^3](1+x)^2(1+x+x^2)}&=\left([x^3]+[x^2]+[x^1]\right)(1+x)^2\tag{2}\\
&=0+1+2\\
&\,\,\color{blue}{=3}
\end{align*}
according to $|\{\{1,2,3\},\{1,3,3\},\{2,3,3\}\}|=\color{blue}{3}$.
In (2) we use the coefficient of operator rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. Similarly we consider now the multiset $\{A,B,B.Y\}$ and enocde the situation algebraically as
\begin{align*}
\underbrace{(1+x)}_{A}\underbrace{(1+x+x^2)}_{B}\underbrace{(1+x)}_{Y}\tag{1}
\end{align*}
*
*The number of two-element subsets of the multiset $\{A,B,B,Y\}$ is
\begin{align*}
\color{blue}{[x^2](1+x)^2(1+x+x^2)}&=\left([x^2]+[x^1]+[x^0]\right)(1+x)^2\\
&=1+2+1\\
&\,\,\color{blue}{=4}
\end{align*}
according to $|\{\{A,B\},\{A,Y\},\{B,B\},\{B,Y\}\}|=\color{blue}{4}$.
Note: See for instance formula (5.53) in Concrete Mathematics by R. L. Graham, D. Knuth and O. Patashnik for more information of the coefficient of operator $[x^k]$.
|
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|
Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$ Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$.
My effort: I tried using the fact $9^{\sqrt{2}}<9^{1.5}=27.$
Also We have $512 <729 \Rightarrow 2^9<27^2 \Rightarrow 2^{\frac{9}{2}}<27 \Rightarrow \sqrt{2}^9=2^{4.5}<27$. But both are below $27$.
|
Using Henry's approach, but with different estimates. If you know that the first three digits of $\ \sqrt{2}\ $ is $\ 1.41,\ $ then you have: $\ 1.40 < \sqrt{2} < 1.42857\ldots,\ $ i.e.,
$$\ \frac{7}{5} < \sqrt{2} < \frac{10}{7}.$$
So,
$$ \left( 9^{\sqrt{2}} \right)^7 < \left( 9^{ \frac{10}{7}} \right)^7 = 9^{10} = {81}^5 = \left( 80+1 \right)^5 $$
$$ = {80}^5 + 5\times {80}^4 + 10\times {80}^3 + 10\times {80}^2 + 5\times 80 + 1 $$
$$ < {80}^5 + 6 \times {80}^4 = 86 \times {80}^4 < 100 \times{80}^4 $$
$$ =100\times 8^4 \times {10}^4 < 4,100,000,000 = 4.1 \times 10^{9}, $$
whereas
$$ \left(\left(\sqrt{2}\right)^9\right)^7 = \left(\sqrt{2}\right)^{63} = 2^{31}\times \sqrt{2} > 1.4 \times 2^{31} = \frac{7}{2} \times 2^{31} $$
$$ = 7 \times 2^{30} = 7 \times \left( 2^{10} \right)^3 = 7 \times 1024^{3} > 7 \times 1000^3 = 7 \times 10^9. $$
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|
Making integration problems such that $\int{}f^{-1}(x)\text{ d}x$ is difficult and $\int{}f(x)\text{ d}x$ is easy. There is an amazing formula to integrate the inverse of a function:
$$\int{f^{-1}(x)\text{ d}x}=x f^{-1}(x)-F\bigg(f^{-1}(x)\bigg)+c, \text{where }F(x)=\int{f(x)\text{ d}x}$$
I know how to derive this formula. So if we have $\int{f^{-1}(x)\text{ d}x}$
Then starting with the substitution $u=f^{-1}(x)$
So $x=f(u)$ and $\text{d}x=f'(u)\text{ d}u$
Therefore, $\int{f^{-1}(x)\text{ d}x}=\int u f'(u) \text{ d}u$
Integrating by parts, $\int u f'(u) \text{ d}u=u f(u)-\int f(u) \text{ d}u=x f^{-1}(x)-F\bigg(f^{-1}(x)\bigg)+c$, where $F(x)=\int f(x) \text{ d}x$.
An example to use this is to find $\int \log(x) \text{ d}x$
We have $f^{-1}(x)=\log(x) \implies f(x)=e^x \implies F(x)=\int f(x) \text{ d}x=\int e^x \text{ d}x=e^x \implies F\bigg(f^{-1}(x)\bigg)=e^{\log(x)}=x$
So $\int \log(x) \text{ d}x=x \log(x)-x+c$
Another example is $\int \tanh^{-1}(x) \text{ d}x$
We have $f^{-1}(x)=\tanh^{-1}(x) \implies f(x)=\tanh(x) \implies F(x)=\int f(x) \text{ d}x=\int \tanh(x) \text{ d}x = \log\big|\cosh(x)\big|$
$\implies F\bigg(f^{-1}(x)\bigg)=\log\bigg|\cosh\big(\tanh^{-1}(x)\big)\bigg|$
So $\int \tanh^{-1}(x) \text{ d}x=x\tanh^{-1}(x)-\log\bigg|\cosh\big(\tanh^{-1}(x)\big)\bigg|+c$
The previous two examples are also easy without this amazing formula; the first example can be determined directly by integrating by parts (well-known classical example of integrating by parts), and the second example we can replace $\tanh^{-1}(x)$ with its equivalent expression:
$\frac{1}{2}\log\bigg(\frac{1+x}{1-x}\bigg)=\frac{1}{2}\bigg(\log(1+x)-\log(1-x)\bigg)$, then integrate by parts.
However, I am looking for integrals that cannot be done easily without the formula mentioned above (possibly can be done by a long and tedious way). For instance,
$$\int \sqrt{\frac{1}{x}-1}\text{ d}x$$
Let $f^{-1}(x)=\sqrt{\frac{1}{x}-1}$
Now using the usual way to find the inverse of a function (steps are omitted), we get $f(x)=\frac{1}{x^2+1}$
$F(x)=\int f(x) \text{ d}x= \tan^{-1}(x)$, so $F\bigg(f^{-1}(x)\bigg)=\tan^{-1}\bigg(\sqrt{\frac{1}{x}-1}\bigg)$
So $\int \sqrt{\frac{1}{x}-1}\text{ d}x=x \sqrt{\frac{1}{x}-1} - \tan^{-1}\bigg(\sqrt{\frac{1}{x}-1}\bigg)+c$
I do not want the integrand to be logarithm or inverse trigonometric function or inverse hyperbolic function, because they are easy in the first place. However, I need to make integral problems similar to the last example mentioned above.
So $\int{f^{-1}(x)\text{ d}x}=\color{red}{\text{DIFFICULT}}$ but $\int{f(x)\text{ d}x}=\color{green}{\text{EASY}}$
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Another example:
Take $$f(x) = -\frac{1}{x \sqrt{1 - x^2}} , \qquad x \in \left(0, \frac{1}{\sqrt{2}}\right)$$
(we've restricted the domain of $f$ so that it has an inverse). Then,
$$f^{-1}(x) = \frac{1}{\sqrt{2}} \frac{\sqrt{x^2 + x \sqrt{x^2 - 4}}}{x} .$$ Then, we can transform the integral
$$\int f^{-1}(x)\,dx = \frac{1}{\sqrt{2}} \int \frac{\sqrt{x^2 + x \sqrt{x^2 - 4}}}{x} \,dx$$
via the substitution $u = x^2, du = 2 x\,dx$ (dropping the resulting overall constant) to
$$\int \frac{\sqrt{u + \sqrt{u^2 - 4 u}}}{u} \,du .$$ Since $f(x) = \frac{d}{dx} \operatorname{arsech} x$, we find that
\begin{align*}\int \frac{\sqrt{u + \sqrt{u^2 - 4 u}}}{u} \,du &= 2 \sqrt{x^2 + x \sqrt{x^2 - 4}} - 2 \sqrt{2} \operatorname{arsech} \left(\frac{1}{\sqrt{2}} \frac{\sqrt{x^2 + x \sqrt{x^2 - 4}}}{x}\right) + C\\
&= 2 \sqrt{u + \sqrt{u^2 - 4 u}} - 2 \sqrt{2} \operatorname{arsech} \sqrt{\frac{u + \sqrt{u^2 - 4 u}}{2 u}} + C\end{align*}
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|
Find the value of $\frac{a+b}{10}$
If $\sin x+\cos x+\tan x+\cot x+\sec x +\csc x=7$, then assume that $\sin(2x)=a-b\sqrt7$, where $a$ and $b$ are rational numbers. Then find the value of $\frac{a+b}{10}$.
How to solve these kind of problems. I can make substitutions and convert all of them to $\sin$ and then solve for it but it'll be very lengthy. Is there any other short and nicer method.
|
Since the question asked for $\sin 2x$, after failures using other methods, I found this one:
$$\sin x + \cos x + \tan x + \cot x + \csc x+ \sec x = 7$$
$$(\sin x + \cos x) + \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x} + \frac{1}{\sin x}+\frac{1}{\cos x}=7$$
$$(\sin x + \cos x) + \frac{\sin^2x+\cos^2x}{\sin x\cos x} + \frac{\sin x + \cos x}{\sin x\cos x}=7$$
$$(\sin x + \cos x)+\frac{2(\sin x + \cos x)}{\sin 2x} = 7 - \frac{2}{\sin 2x}$$
We've converted the equation into the terms of $\sin(2x)$ (well, almost) using $\sin 2x = 2\sin x \cos x$. Notice that squaring will allow us to use this identity again, but now it's entirely in terms of $\sin(2x)$, so we can make a substitution.
$$(\sin x + \cos x)\left(1+\frac{2}{\sin 2x}\right) = 7 - \frac{2}{\sin 2x}$$
$$(\sin^2x+\cos^2x+2\sin x\cos x)\left(1+4\left(\frac{1}{\sin^22x}+\frac{1}{\sin2x}\right)\right) = 49+\frac{4}{\sin^22x}-\frac{28}{\sin 2x}$$
$$(1+\sin2x)\left(1+4\left(\frac{1}{\sin^22x}+\frac{1}{\sin2x}\right)\right) = 49+\frac{4}{\sin^22x}-\frac{28}{\sin 2x}$$
Now, you're ready to perform the substitution. Set $\sin 2x = a$
$$(1+a)\left(1+4\left({1\over a}+{1\over a^2}\right)\right) = 49 + \frac{4}{a^2}-\frac{28}{a}$$Multiply by $a^2$ to get an easier equation.
$$(1+a)(a^2+4a+4) = 49a^2+4-28a$$
$$a(a^2+4a+4) = 49a^2 - a^2-28a-4a+4-4$$
$$a(a^2+4a+4) = a(48a - 32)$$
$$a(a^2 - 44a + 36) = 0$$
One quickly discards $a=0$ as that yields the RHS of the original equation undefined for any solution of $\sin 2x = 0$.
We have
$$a^2 - 44a + 36 = 0$$
Plug this into the quadratic formula to get:
$$a = 22 \pm 8\sqrt7$$One easily sees that $22+8\sqrt7$ is out of the range of $\sin \theta$, so we have:
$$\color{blue}{\sin 2x = 22 - 8\sqrt7}$$
(At this point, I realized $a$ was already a variable in your question, I'm so sorry. The $a$ used above as a substitution is not the same $a$ which you asked for in the question.)
So, your final answer should be:
$$\color{green}{\frac{22+8}{10} = 3}$$
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|
How do we prove $x^6+3x^3+2x^2+x+1 \geq 0$ Question
How do we prove
$$x^6 + 3x^3+2x^2+x+1\geq0$$
My progress
$$x^6+3x^3+2x^2+x+1=(x+1)^2(x^4-2x^3+3x^2-x+1)$$
I appreciate your interest
|
$$x^6 + 3x^3+2x^2+x+1= (x + 1)^2 \left( {\left( {x - \tfrac{1}{2}} \right)^4 + \tfrac{1}{2}\left( {x - \tfrac{1}{2}} \right)^2 + x^2 + \tfrac{{13}}{{16}}} \right) \ge 0$$
|
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|
Primitive Polynomials over $\mathbb{Z}_3$ Which of the following polynomials are primitive over $\mathbb{Z}_3$?
*
*$x^3 + x^2 + x + 1$
*$x^3+x^2+x+2$
*$x^3+2x+1$
So I now in order to be primitive the polynomials have to be irreducible, which is only true for $x^3+x^2+x+2$ and $x^3+2x+1$.
But how do I show if they are primitive?
|
Hint
There are $4=\dfrac {\varphi (3^3-1)}3$ primitive third degree polynomials over $\Bbb F_3.$
Let $t$ be a root of $x^3+2x+1.$ Take $\alpha=t+(x^3+2x+1),$ a primitive $26$th root of unity in $\Bbb F_{27}$.
We know $\alpha $ is primitive because $t^2\neq1$ and by the Frobenius automorphism, $$t^9=(t^3)^3=(t-1)^3=t^3-1=t+1\implies t^{13}=t^{3+9+1}=t(t-1)(t+1)=t^3-t=-1.$$
Then the primitive polynomials are:
$$\begin {align}p_1=(x-\alpha)(x-\alpha ^3)(x-\alpha ^9),\, &\\p_2=(x-\alpha ^5)(x-\alpha^{15})(x-\alpha ^{19}),\,&\\p_3=(x-\alpha ^7)(x-\alpha ^{21})(x-\alpha ^{11})\,,&\\p_4=(x-\alpha ^{17})(x-\alpha ^{25})(x-\alpha ^{23})\end{align}.$$
So let's take $$p_1=(x-\bar t)(x-\bar t^3)(x-\bar t^9)=x^3-(\bar t+\bar t^3+\bar t^9)x^2+(\bar t^4+\bar t^{12}+\bar t^{10})x-(\bar t^{13})=x^3-x+1.$$
Etc.
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|
Inequality with increasing variables
If $n,k\ge2$, and $0\le a_0\le a_1\le\cdots$, prove that
\[\left(\frac{1}{k n} \sum_{l=0}^{k n-1} a_{l}\right)^{k} \geq \frac{1}{n} \sum_{i=0}^{n-1} \prod_{j=0}^{k-1} a_{n j+i}.\]
This inequality is an improvement of AM-GM inequality. For $n=3, k=2$, $0\le a_0\le a_1\le\cdots\le a_5$, the inequality is
\[\left(\frac{a_0+a_1+\cdots+a_5}{6}\right)^2\geq \frac{a_0a_3+a_1a_4+a_2a_5}{3}.\]
This is a homogeneous inequality, so WLOG, let $\displaystyle\sum_{l=0}^{kn-1}a_l=kn$. We have to prove
\[\sum_{i=0}^{n-1} \prod_{j=0}^{k-1} a_{n j+i}\le n.\label1\tag1\]
I tried to use Carlson inequality, it became
\[\left(\sum_{i=0}^{n-1} \prod_{j=0}^{k-1} a_{n j+i}\right)^n\le\prod_{i=0}^{k-1}\sum_{j=0}^{n-1}a_{ni+j}^n\overset?\le n^n.\]
However this doesn't make use of the condition $0\le a_0\le a_1\le\cdots$.
From another perspective we could use the adjustment method. I can prove that $\forall~0\le t\le k-1$, the numbers $a_{tn}$, $a_{tn+1}$ $\ldots$ $a_{(t+1)n-1}$ can take at most three different values to reach maximum.
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(With the homogenization of $\sum a_i = kn$.)
First, we consider the $ k = 2, n = 2$ case.
Suppose $ a + b + c + d = 4, a \leq b \leq c \leq d$, we want to show that $ ac + bd \leq 2$.
Intuitvely, if $a \leq d$ are fixed, we'd want to increase $b$ and decrease $c$ subject to $ b \leq c $.
Hence, we replace $(a, b, c, d)$ with $ ( a , \frac{b+c}{2}, \frac{b+c}{2}, d)$, which only increases the value.
This becomes $ a \times \frac{b+c}{2} + \frac{b+c}{2} \times d = (a+d)(b+c) / 2$, which by AM-GM, we know it has a maximum value of $2$, hence we are done.
Equality occurs when $ a + d = b+c = 2$ and $ b = c ( = 1)$. EG $(0, 1, 1, 2)$ and $(1, 1, 1, 1)$ are equality cases.
Note: The hint about the equality condition is that it encourages you to realize that $a+d = b+c = 2$, and so we'd want to force that out somehow.Otherwise, one might think of $(a+b)(c+d)$ instead, but that doesn't help us classify the equality cases, nor have I had success exploring that path.
Now we consider the $k = 3, n = 2$ case.
Suppose $ a + b + c + d + e + f = 6, a\leq b \leq c \leq d \leq e \leq f$, we want to show that $ ace + bdf \leq 2$.
If $a, d, e, f$ are fixed, then since $a \leq d, e\leq f$ so $ae \leq df$ and we'd want to increase $b$ and decrease $c$ subject to $ b \leq c$.
Similarly, if $a, b, c, f$ are fixed, since $ac \leq bf$, we'd want to increase $d$ and decrease $e$ subject to $ d \leq e$.
Thus, we replace $(a, b, c, d, e, f)$ with $(a, \frac{b+c}{2}, \frac{b+c}{2},\frac{d+e}{2}, \frac{d+e}{2},f )$, which only increase the value.
Again, the expression becomes $(a+f)(b+c)(d+e) / 4$, which by AM-GM has a maximum value of 2, and we are done.
Equality holds when $ a +f = b + c = d+e = 2, b = c ( = 1), d = e (= 1)$.
Now we consider the $k = 2, n = 3 $ case.
Suppose $ a +b + c + d + e + f = 6 , a\leq b \leq c \leq d \leq e \leq f,$ we want to show that $ ad + be + cf \leq 3$.
If $ a \leq f$ are fixed, we'd want to decrease d, increase c subject to $ c \leq d$.
Thus we replace $(a, b, c, d, e, f)$ with $(a, b, \frac{c+d}{2}, \frac{c+d}{2}, e, f)$.
*
*If $ b+e \geq c+d$, then we'd want to replace $(b, e)$ with $(\frac{c+d}{2}, b+e - \frac{c+d}{2})$. Then, we're maximizing $(c+d)/2 \times (6 - 3*(c+d)/2) $, which has a maximum of $3$ wen $ (c+d)/2 = 1$.
*Otherwise, if $b+e < c+d$, then we'd want to replace $(b, e) $ with $( b+e - \frac{c+d}{2}, \frac{c+d}{2} )$. Then, we're likewise maximizing $(c+d)/2 \times (6 - 3*(c+d)/2)$, which has a maximum of $3$ when $(c+d)/2 = 1$.
The equality cases are
*
*$(a, b, 1, 1, 1, f)$ subject to $ a \leq b \leq 1 \leq f$, $a+b+f = 3$.
*$(a, 1, 1, 1, e, f)$ subject to $ a \leq 1 \leq e \leq f$, $a+e+f = 3$.
I leave you to generalize this, in the exact same way as shown above.
Maybe think about the $ k = 3, n = 3$ case, how can we use the above ideas?
Once you've done that, please post a general solution.
|
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|
$\epsilon$–$\delta$ proof that $\lim_{x\to x_0}\frac{1}{x} = \frac{1}{x_0}$ for all $x_0\neq 0$; how to identify $\delta$? I would like to use the $\epsilon$–$\delta$ definition of the limit of a function to show that
$$\lim_{x\to x_0} \frac{1}{x} = \frac{1}{x_0}$$
But I'm having trouble identifying a $\delta>0$ for arbitrary $\epsilon>0$ and $x_0\neq 0$ so that
$$ 0<|x-x_0|<\delta \implies |\frac{1}{x} -\frac{1}{x_0}|<\epsilon$$
How can you find a $\delta$ that satisfies this condition?
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Given $\ \varepsilon > 0.$
If $\varepsilon < \frac{1}{\vert x_0 \vert },$ then $\vert \varepsilon x_0 \vert<1,\ $ and so $$ x \in \left( \underbrace{\ \frac{x_0}{1 + \varepsilon x_0}\ }_{a}, \underbrace{\ \frac{x_0}{1- \varepsilon x_0}\ }_{b} \right) \implies \frac{1}{x} \in \left( \frac{1 - \varepsilon x_0}{x_0}, \frac{1+ \varepsilon x_0}{x_0} \right) = \left( \frac{1}{x_0} - \varepsilon, \frac{1}{x_0} + \varepsilon \right). $$
Else, if $\ \varepsilon \geq \frac{1}{\vert x_0 \vert }\ \left( > \frac{1}{2}\cdot \frac{1}{\vert x_0 \vert} \right),\ $ then $$ x \in \left( \underbrace{\ \frac{1}{ \frac{1}{x_0} + \frac{1}{2} \cdot \frac{1}{\vert x_0 \vert}\ } }_{c}, \underbrace{\ \frac{1}{ \frac{1}{x_0} - \frac{1}{2} \cdot \frac{1}{\vert x_0 \vert} }\ }_{d} \right) \implies \frac{1}{x} \in \left( \frac{1}{x_0} - \frac{1}{2} \cdot \frac{1}{\vert x_0 \vert}, \frac{1}{x_0} - \frac{1}{2} \cdot \frac{1}{\vert x_0 \vert} \right)$$
$$\subset \left( \frac{1}{x_0} - \varepsilon, \frac{1}{x_0} + \varepsilon \right). $$
So let $$\delta:=
\begin{cases}
\frac{b-a}{2}&\text{if}\, \varepsilon < \frac{1}{\vert x_0 \vert }\\
\frac{d-c}{2}&\text{if}\, \varepsilon \geq \frac{1}{\vert x_0 \vert }\\
\end{cases}
$$
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Find $a+b +c$, if $\sin{x}+\sin^2{x}=1$ and $\cos^{12}{x}+a\cos^{10}{x}+b\cos^8{x}+c\cos^6{x}=1$ There is my problem :
Find $a+b +c$,
if $$\sin{x}+\sin^2{x}=1$$ and $$\cos^{12}{x}+a\cos^{10}{x}+b\cos^8{x}+c\cos^6{x}=1$$
I'm sorry, I can't solve this problem but I really want to know the solution.
I know that $\cos^2{x}=\sin{x}$, but I can't find $a+b+c$.
Attempt
I used substitute $t=\sin(x)$, and number $1=t^2+t$ put on the left. Then I divided by $t$ as long as I can, then I got polynomial with degree $3$, but I can't conclude what is $a+b+c$.
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$\sin^2 x+\sin x-1=0$ gives $\sin x = \frac{-1\pm\sqrt{5}}{2}$, so that
$$\cos^2 x = 1-\sin^2 x = 1-\frac{3\pm\sqrt{5}}{2} = \frac{-1\pm\sqrt{5}}{2}.$$
Then you want to solve
$$(\cos^2 x)^6+a(\cos^2 x)^5 + b(\cos^2 x)^4 + c(\cos^2 x)^3 = 1.$$
After some algebra, this reduces to (for the positive sign on $\sqrt{5}$)
\begin{gather*}
-\frac{11 }{2}a+\frac{7 }{2}b-2 c+9=1 \\
4 - \frac{5}{2}a + \frac{3}{2} b - c = 0
\end{gather*}
which has the solution $a=b$, $c=4-b$. So the best you can do is $b+c=4$.
Taking the negative sign on $\sqrt{5}$ produces the same result.
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|
Where've I gone wrong on this Maclaurin series expansion I keep getting the same answer, and I know it's wrong, but I'm not seeing where I've gone wrong.
I'm supposed to find the first 4 terms of $e^{-x} \cos(x)$ using Taylor series expansion and multiplying the terms.
The answer should be:
$$1-x+\frac{1}{3}x^3-\frac{1}{6}x^4$$
But I keep getting:
$$1-x+\frac{1}{3}x^3-\frac{5}{24}x^4$$
The expansion for $e^{-x}$ and $\cos(x)$ are:
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\rightarrow e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...$$
$$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-...$$
When I multiply and combine terms I get:
$$1\left ( 1-\frac{x^2}{2!}+\frac{x^4}{4!} \right )-x\left ( 1-\frac{x^2}{2!} \right )+\frac{x^2}{2!}\left ( 1-\frac{x^2}{2!} \right )-\frac{x^3}{3!}\left ( 1 \right )$$
$$\rightarrow \left ( 1-\frac{x^2}{2} +\frac{x^4}{24}\right )+\left ( -x+\frac{x^3}{2} \right )+\left ( \frac{x^2}{2}-\frac{x^4}{4} \right )+\left ( -\frac{x^3}{6} \right )$$
$$\rightarrow 1-x+\left ( -\frac{1}{2}+\frac{1}{2} \right )x^2+\left ( \frac{1}{2}-\frac{1}{6} \right )x^3+\left ( \frac{1}{24}-\frac{1}{4} \right )x^4$$
$$\rightarrow 1-x+\left ( 0 \right )x^2+\left ( \frac{2}{6} \right )x^3+\left ( -\frac{5}{24} \right )x^4$$
$$\rightarrow 1-x+\frac{1}{3}x^3-\frac{5}{24}x^4$$
|
You are missing the term $\frac{x^4}{4!} \cdot 1$ from $(\cos x)(e^{-x})$. Everything else is correct.
|
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|
Find the sum of $\sum_{k=1}^n k(k-1) {{n} \choose {k-1}}$
Find the sum of $$\sum_{k=1}^n k(k-1) {{n} \choose {k-1}}$$
The solution in the book is a lot different than what I tried
what they did in the book is say $k-1=t$ then they expanded from this point $\sum_{t=0}^{n-1} (t+1)(n) {{n-1} \choose {t-1}}$ and the final answer is $n[(n-1)(2^{n-2}-1)+2^n-2]$ but I did not understand their way as it complicated to me
I thought about using derivatives but I got stuck and I am not sure if it really works I just remember using derivatives for some cases in class and I tried it
$(x+b)^n=$ ${n \choose 0} x^0 b^{n-0}+ {n \choose 1} x^1 b^{n-1} +...+ {n \choose n} x^n b^{n-n} $
derivative with respect to $x$ is
$n(x+b)^n-1=$ $0+ {n \choose 1} b^{n-1} +...+ {n \choose n} nx^{n-1} b^{n-n} $
then derivative again
$n(n-1)(x+b)^n-2=$ $0+0+ {n \choose 2} 2b^{n-2} +...+ {n \choose n} n(n-1)x^{n-2} b^{n-n} $
but I could not continue from here
is it ok to use derivatives ? is there another way other than the book(just expanding)?
thanks for any tips and help!
|
Since $(1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k$, differentiate twice to get
$$n(n-1)(1+x)^{n-2} = \sum_{k=2}^n k(k-1)\binom{n}{k}x^{k-2}$$
Substituting $n+1$ for $n$ we also get
$$(n+1)n(1+x)^{n-1} = \sum_{k=2}^{n+1} k(k-1)\binom{n+1}{k}x^{k-2}.$$
Substitute $x=1$ in each of these, giving
\begin{align*}
n(n-1)2^{n-2} &= \sum_{k=2}^n k(k-1)\binom{n}{k} \\
n(n+1)2^{n-1} &= \sum_{k=2}^{n+1} k(k-1)\binom{n+1}{k}.
\end{align*}
Subtract the first of these from the second, giving
\begin{align*}
\tag{*}2^{n-2}n(2(n+1)-(n-1)) &= \sum_{k=2}^{n+1}k(k-1)\left(\binom{n+1}{k}-\binom{n}{k}\right)\\
&= \sum_{k=2}^{n+1}k(k-1)\binom{n}{k-1}.
\end{align*}
Finally, this gives
$$\sum_{k=1}^{n}k(k-1)\binom{n}{k-1} = n2^{n-2}(n+3) - n(n+1)$$
since the summand of (*) is zero for $k=1$ and is $n(n+1)$ for $k=n+1$. The right-hand side indeed simplifies to
$$n((n-1)(2^{n-2}-1)+2^n-2).$$
|
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|
How do I change the order of integration $\int_{\pi/2}^{5\pi/2} \int_{\sin x}^{1} f(x,y)dydx$ How do I change the order of integration in $$\int_{\pi/2}^{5\pi/2} \int_{\sin x}^{1} f(x,y)dydx\;?$$
$y=\sin x$; $y=1$, $x=\pi/2$; $x= 5\pi/2$.
I can guess from here that $y$ is from $-1$ to $1$.
Then $x=\sin^{-1}(y)$ and $\sin^{-1}(-1) = -\pi/2$ and $\sin^{-1}(1)=\pi/2$.
What will be the limits of $x$?
|
You need to find the equation of the form $x=a+b \cdot \text{sin}^{-1}(y)$ passing through $\left( \frac{3\pi}{2},-1\right)$ and $\left( \frac{\pi}{2},1\right).$ So $\frac{3\pi}{2}=a + b \cdot \text{sin}^{-1}(-1)$ and $\frac{\pi}{2}=a + b \cdot \text{sin}^{-1}(1)$. Now you have a system of 2 linear equations in $a$ and $b$. Solve for $a$ and $b$.
$$ $$
Also find the equation of the form $x=a+b \cdot \text{sin}^{-1}(y)$ passing through $\left( \frac{3\pi}{2},-1\right)$ and $\left( \frac{5\pi}{2},1\right).$
|
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|
Asymptotic behavior of the function $f(x) = x \int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy.$ In one of my analysis course, we considered the function
$$f(x) = x \int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy.$$
Then my teacher told us that $f$ had the following behavior
$$
f(x)\sim\begin{cases}
\frac{-1}{4}x\log x & \text{as } x \to 0^+,\\
\frac{2}{x} & \text{as } x \to \infty.
\end{cases}$$
For the case $x \to \infty$ I understand as in this case $e^{-\frac{1}{4}y^2}$ is negligeable and then
$$f \sim x \left[\frac{-1}{2y^2}\right]^\infty_x = \frac{1}{2x}.$$
However I do not have any intuition of the form of
$$\int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy,$$
which should behave as a $\log$. I thought maybe we could expand $e^{-\frac{1}{4}y^2}$ to the first order so that
$$\frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy = \frac{1}{4}\frac{1}{y} + O(y)$$
and by integrating we indeed get a $\log$. However I don't get how exactly we should get this behavior of $f$. Is there someone more skilful than me who could answer to this question ?
|
First note that $$ x \int_1^{\infty} \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy=o(x\log{x})$$
as $x\to 0^{+}$. Indeed, the integral $\int_1^{\infty} \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy$ is convergent and $\lim_{x\to 0^{+}}\frac{x}{x\log{x}}=0$.
Second, using l'Hopital's rule one can compute
$$\lim_{x\to 0^{+}}
\frac{x \int_x^1 \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy}{x\log{x}}=
-\lim_{x\to 0^{+}}
\frac{1 - e^{-\frac{1}{4}x^2}}{x^2}=-\lim_{x\to 0^{+}}
\frac{\frac{1}{2}x e^{-\frac{1}{4}x^2}}{2x}=-\frac{1}{4}.$$
This shows that $f(x)\sim -\frac{1}{4} x\log{x}$
as $x\to 0^{+}$.
|
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|
Power series of inverse function Let, $f$ be a bijective function on set of Real numbers. Let, $f(x) =\sum_{n=1}^{\infty}a_{n}x^n$ such that $a_{1}=2,a_{2}=4$ let, $f^{-1}(x) =\sum_{n=1}^{\infty} b_nx^n$
Then find value of $b_1$.
My approach:
we know,
$$\frac{1}{1-2x}=1+2x+4x^2+8x^3+\ldots $$
$$\frac{2x}{1-2x} = \sum_{n=1}^{\infty}(2x)^n $$
Hence,
$$f(x) = \frac{2x}{1-2x}$$
Let, $f(x)=y$
then, $$2x =y(1-2x)$$
$$2x= \frac{y}{1+y}$$
$$x= \frac{y}{2(1+y)}$$
Now,
$$\frac{1}{1+y}= 1-y+y^2-y^3+\ldots$$
$$\frac{y}{2(1+y)}= \frac{y}{2} -\frac{y^2}{2}+\ldots $$
Hence, $b_1=\frac{1}{2}$.
Am I correct?
|
You cannot assume anything about $f$ other than the facts that $a_1=2$ and $a_2=4$.
If $f^{-1}(x)=\sum_{n=1}^\infty b_nx^n$, then $f^{-1}\bigl(f(x)\bigr)=x$ means that$$b_1(a_1x+a_2x^2+a_3x^3+\cdots)+b_2(a_1x+a_2x^2+a_3x^3+\cdots)^2+\cdots=x.\label{a}\tag1$$But the coefficient of $x$ on the LHS of \eqref{a} is $b_1a_1$. So, $b_1a_1=1$. But $a_1=2$. Therefore, $b_1=\frac12$.
|
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|
Computing $\int_0^{\infty} x^n\left\lfloor\frac{1}{x}\right\rfloor d x=\frac{\zeta(n+1)}{n+1}$, where $n\in \mathbb N$. In the post, it was found that
$$
\int_0^{\infty} x\left\lfloor\frac{1}{x}\right\rfloor d x=\frac{\pi^2}{12}
$$
I want to generalise the integral as
$$
\int_0^{\infty} x^n\left\lfloor\frac{1}{x}\right\rfloor d x=\frac{\zeta(n+1)}{n+1},\; \textrm{ where } n\in \mathbb N.
$$
Now, let’s see whether it can be proved in a similar manner.
Noting that for any $\displaystyle x\in \Big(\frac{1}{k+1}, \frac{1}{k}\Big]$ , $$
\frac{1}{k+1}<x \leq \frac{1}{k} \Leftrightarrow k+1>\frac{1}{x} \geqslant k \Leftrightarrow\left\lfloor\frac{1}{x}\right\rfloor=k
$$
and
for any $\displaystyle x\in \big(1, \infty\big]$, we have
$\left\lfloor\frac{1}{x}\right\rfloor=0$.
$$
\begin{aligned}
\int_0^{\infty} x^n\left\lfloor\frac{1}{x}\right\rfloor d x
& =\int_0^1 x^n\left\lfloor\frac{1}{x}\right\rfloor d x \\
& =\sum_{k=1}^{\infty} \int_{\frac{1}{k+1}}^{\frac{1}{k}} k x^n d x \\
& =\sum_{k=1}^{\infty}\left[\frac{k x^{n+1}}{n+1}\right]_{\frac{1}{k+1}}^{\frac{1}{k}} \\
& =\frac{1}{n+1} \sum_{k=1}^{\infty}\left[\frac{1}{k^n}-\frac{(k+1)-1}{(k+1)^{n+1}}\right] \\
&= \frac{1}{n+1} \sum_{k=1}^{\infty}\left[\left(\frac{1}{k^n}-\frac{1}{(k+1)^n}\right)+\frac{1}{(k+1)^{n+1}}\right]\\&= \frac{1}{n+1}\left(1+\sum_{k=1}^{\infty} \frac{1}{(k+1)^{n+1}}\right)\\&= \frac{\zeta(n+1)}{n+1}
\end{aligned}
$$
Comments and alternative methods are highly appreciated.
|
Here is an alternative method. Similarly to the comment by reuns in the linked post, we can do the following:
\begin{align*}
\int_0^\infty x^n\left\lfloor\frac 1x\right\rfloor dx
&=\int_0^\infty y^{-n}\lfloor y\rfloor\frac{dy}{y^2}\\
&=\int_0^\infty y^{-n-2}\sum_{\substack{m\in\mathbb N\\ 1\leq m\leq y}}1 dy\\
&=\sum_{m=1}^\infty \int_m^\infty y^{-n-2}dy\\
&=\sum_{m=1}^\infty \frac{m^{-n-1}}{n+1}=\frac{\zeta(n+1)}{n+1}.
\end{align*}
|
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|
Foci of ellipse My question is given a ellipse of the equation : $$\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$$ where $a>b$ then how we can find the coordinates of the foci. I want to find those coordinates without the presuming that the foci exists because most proofs I found online assume the properties of foci to be true and then take some extreme case to find foci's coordinates.
So to rephrase my question: Given a closed curve of the equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$$ prove that there exists two points inside the curve such that if we take any point on the boundary of the curve and join it to those two points then the sum of those lengths will give a fixed constant based on $a$ and $b$ (Assume $a>b$).
Here's my attempt which gave me nothing:
Let's take a point $P$ on the curve as $\left(x,\ b\sqrt{1-\frac{x^2}{a^2}}\right)$
Let those two points be $(-f,0)$ and $(f,0)$ then the sum of lengths from point $P$ becomes
$$S = \sqrt{(f+x)^2+b^2\left(1-\frac{x^2}{a^2}\right)}+\sqrt{(f-x)^2+b^2\left(1-\frac{x^2}{a^2}\right)}$$
Differentiating this wrt to $f$ and equating to $0$ to find the stationary case
$$\dfrac{f+x}{\sqrt{\left(f+x\right)^2+b^2\left(1-\frac{x^2}{a^2}\right)}}+\dfrac{f-x}{\sqrt{\left(f-x\right)^2+b^2\left(1-\frac{x^2}{a^2}\right)}}=0$$
Squaring and simplifying
$$4bfx\left(1-\frac{x^2}{a^2}\right)=0 \implies f =0$$
which is obviously wrong...
Note that in my attempt I too assumed two points, that the foci will be symmetrical and on the major axis, if we can even take these assumptions out that would be amazing. It's just that with these assumptions I was able to at least start somewhere.
P.S.
$\textbf{thanks to the comments by @Blue}$
differentiating $S$ wrt $x$ and equating to zero
$$\dfrac{2\left(x+f\right)-\frac{2b^2x}{a^2}}{2\sqrt{\left(x+f\right)^2+b^2\cdot\left(1-\frac{x^2}{a^2}\right)}}+\dfrac{-\frac{2b^2x}{a^2}-2\left(f-x\right)}{2\sqrt{b^2\cdot\left(1-\frac{x^2}{a^2}\right)+\left(f-x\right)^2}}=0$$
Squaring and simplifying
$$ a^4b^2fx(f^2-(a^2-b^2))=0$$
So,
$$f= \sqrt{a^2-b^2}$$
|
The sum of lengths of $(-f, 0)$ and $(f, 0)$ from point P is
$$S = \sqrt{(f+x)^2+b^2\left(1-\frac{x^2}{a^2}\right)}+\sqrt{(f-x)^2+b^2\left(1-\frac{x^2}{a^2}\right)}$$
Your approach was correct upto this step. However, you need to realize what happens when you differentiate S with respect to f.
The condition $\frac{{\partial}S}{{\partial}f} = 0$ considers the points $(-f, 0)$ and $(f, 0)$ to be variable, and finds the minimum value of the sum of distances $S$. Intuitively, the sum of distances is minimum when both the points coincide with the center of the ellipse. Hence, you get the value $f = 0$.
What you really need to do to find the focal points is to find a value of $f$ such that the expression for $S$ is independent of the parameter $x$.
With a little bit of algebraic manipulation you get
$$f = a\sqrt{1-\dfrac{b^2}{a^2}}$$
And this is how you get the coordinates of the focal points.
If you want to solve the problem without assuming that the foci are symmetrically placed on the major axis, you can assume arbitrary coordinates $(x_1, y_1)$ and $(x_2, y_2)$ for them, derive a general expression for $S$, and then solve for either $\frac{{\partial}S}{{\partial}h} = 0$ or $\frac{{\partial}S}{{\partial}k} = 0$, where $(h, k)$ are the coordinates of a point P on the ellipse.
|
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How many methods are there to evaluate $\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^{2n}}$? Background
As I had found the integral
$$I=\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^2} d x =\frac{\pi}{4}, $$
by using $x\mapsto \frac{1}{x}$ yields
$\displaystyle I=\int_0^{\infty} \frac{\frac{1}{x^2}}{\left(\frac{1}{x}+x\right)^2} d x\tag*{} $
Averaging them gives the exact value of the integral
$\displaystyle \begin{aligned}I & =\frac{1}{2} \int_0^{\infty} \frac{1+\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2} d x \\& =\frac{1}{2} \int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+4} d x \\& =\frac{1}{4}\left[\tan ^{-1}\left(\frac{x-\frac{1}{x}}{2}\right)\right]_0^{\infty} \\& =\frac{1}{4}\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right] \\& =\frac{\pi}{4}\end{aligned}\tag*{} $
I guess that we can similarly evaluate the general integral
$$
I_n=\int_0^{\infty} \frac{d x}{\left(x+\frac{1}{x}\right)^{2 n}}
$$
by mapping $x\mapsto \frac{1}{x}$ and then averaging.
$$
I_n=\frac{1}{2} \int_0^{\infty} \frac{1+\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^{2 n}} d x=\frac{1}{2} \int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left[\left(x-\frac{1}{x}\right)^2+4\right]^n}
$$
Letting $x-\frac{1}{x}=\tan \theta$ yields
$$
\begin{aligned}
I_n & =\frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2 \sec ^2 \theta d \theta}{4^n \sec ^{2 n} \theta} =\frac{1}{4^n} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos ^{2 n-2} \theta d \theta= \boxed{\frac{\pi(2 n-3) ! !}{4^n(2 n-2) ! !}}
\end{aligned}
$$
where the last answer comes from the Wallis cosine formula.
My question: How many methods are there to evaluate $\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^{2n}}$?
|
In terms of the Gaussian hypergeometric function
$$\int \frac{dx}{\left(x+\frac{1}{x}\right)^{2n}}=\frac{x^{1-2 n}}{2 n+1}\, _2F_1\left(2 n,\frac{2n+1}{2};\frac{2n+3}{2};-x^2\right)$$ which can also write
$$\int \frac{dx}{\left(x+\frac{1}{x}\right)^{2n}}=(-1)^{n+1}\frac{i}{2} B_{-x^2}\left(n+\frac{1}{2},1-2 n\right)$$
$$\int_0^{\infty} \frac{dx}{\left(x+\frac{1}{x}\right)^{2n}}=\sqrt{\pi }\,\,\frac{ \Gamma
\left(n-\frac{1}{2}\right)}{ 4^{n}\,\Gamma (n)}$$
|
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|
Proving $\sin\frac{\pi}{13}+\sin\frac{3\pi}{13}+\sin\frac{4\pi}{13}=\frac12\sqrt{\frac{13+3\sqrt{13}}2}$
Prove that $$\sin\left(\frac{\pi}{13}\right)+\sin\left(\frac{3\pi}{13}\right)+\sin\left(\frac{4\pi}{13}\right)=\frac{1}{2}\sqrt{\frac{13+3\sqrt{13}}{2}}$$
My Attempt
Let $$x = \frac{1}{2}\sqrt{\frac{13+3\sqrt{13}}{2}} \implies 16x^4-52x^2+13=0$$
And through some donkey work we can calculate the chebyshev polynomial for $\sin\left(\frac{\pi}{13}\right),\sin\left(\frac{3\pi}{13}\right),\sin\left(\frac{4\pi}{13}\right)$ which will all be the same as $\sin(n\pi)=0,\text{ for all } n \in \mathbb{I} $, so
$$P(x) = 4096x^{12}-13312x^{10}+16640x^8-9984x^6+2912x^4-364x^2+13$$
where $x = \sin\left(\frac{2i\pi}{13}\right), \text{ from } 1 \le i \le 12 \text{ where } i \in \mathbb{I}$, are the roots of $P(x)$.
Now I am not getting how to connect these two into a possible solution and even it is possible (probably is), its still a pretty donkey method as you need to find the $13^{th}$ chebyshev polynomial, so if possible maybe give some another method of approach to this question.
|
with $ \; w = \cos \frac{2 \pi}{13} + i \sin \frac{2 \pi}{13} \; \; \; $ in mind, let
$$ x = -i \left( w - w^{25} + w^3 - w^{23} + w^9 - w^{17} \right) $$
we may calculate polynomials in $x.$ We may then apply the relation $w^{26 } = 1 $ repeatedly and express the outcome as sums of $w^{25}, w^{24}, ..., w^2, w,1 $
I get
$$ x^4 - 13 x^2 + 13 = w^{24} + w^{22} + \cdots w^4 + w^2 + 1 $$
with no odd exponents and all the coefficients $1$
Multiply by the (nonzero) $w^2 - 1 $ to get $ x^4 - 13 x^2 + 13 $
The original ideas were due to Gauss, see Cox chapter
Many examples were worked out by Reuschle;
This comes from page 529: Reuschle 1875
|
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|
Proof that $n^3+2n$ is divisible by $3$ I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!
Problem:
For any natural number $n , n^3 + 2n$ is divisible by $3.$
This makes sense
Proof:
Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$
$2 \times 0 = 0.$ So it is divisible by $3.$
Induction: Assume that for an arbitrary natural number $n$,
$n^3+ 2n$ is divisible by $3.$
Induction Hypothesis: To prove this for $n+1,$ first try to express $( n + 1 )^3 + 2( n + 1 )$ in terms of $n^3 + 2n$ and use
the induction hypothesis. Got it
$$( n + 1 )^3+ 2( n + 1 ) = ( n^3 + 3n^2+ 3n + 1 ) + ( 2n + 2 ) \{\text{Just some simplifying}\}$$
$$ = ( n^3 + 2n ) + ( 3n^2+ 3n + 3 ) \{\text{simplifying
and regrouping}\}$$
$$ = ( n^3 + 2n ) + 3( n^2 + n + 1 ) \{\text{factored out
the 3}\}$$
which is divisible by $3$, because $(n^3 + 2n )$ is divisible by $3$
by the induction hypothesis. What?
Can someone explain that last part? I don't see how you can claim $(n^3+ 2n ) + 3( n^2 + n + 1 )$ is divisible by $3.$
|
Presumably you're only looking for a way to understand the induction problem, but you can note that $n^3+2n = n^3 - n + 3n = (n-1)(n)(n+1) + 3n$. Since any three consecutive integers has a multiple of three, we're adding two multiples of three and so get another multiple of 3.
|
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|
How can I understand and prove the "sum and difference formulas" in trigonometry? The "sum and difference" formulas often come in handy, but it's not immediately obvious that they would be true.
\begin{align}
\sin(\alpha \pm \beta) &= \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\
\cos(\alpha \pm \beta) &= \cos \alpha \cos \beta \mp \sin \alpha \sin \beta
\end{align}
So what I want to know is,
*
*How can I prove that these formulas are correct?
*More importantly, how can I understand these formulas intuitively?
Ideally, I'm looking for answers that make no reference to Calculus, or to Euler's formula, although such answers are still encouraged, for completeness.
|
Consider a unit circle with $O$ as the centre. Let $P_{1}$, $P_2$ and $P_{3}$ be points on the circle making angles $A$, $B$ and $A−B$, respectively, with the positive direction of the X-axis.
We know that if two chords subtend equal angle at the centre, then the chords are equal and chords $P_{3}P_{0}$ and $P_1P_2$ subtend equal angles at $O$. Therefore, $\overline{P_{3}P_{0}}=\overline{P_1P_2}$. By distance formula, the distance between the points $P_{0}(1,0)$ and $P_{3}(\cos(A−B),\sin(A−B))$ is $$P_{3}P_{0}=\sqrt{\left(\cos \left( A-B \right) - 1\right) ^{2} + \left(\sin \left( A-B \right) - 0\right) ^{2}}$$ Similarly, the distance between the points $P_{1}(\cos A,\sin A)$ and $P_{2}(\cos B,\sin B)$ is $$P_1P_2= \sqrt{\left( \cos B-\cos A\right) ^{2}+\left( \sin B - \sin A \right) ^{2}}$$
On squaring both sides, we get
$$
\begin{array}{ll}
\Rightarrow & \left\{\cos (A-B)-\left.1\right|^{2}+\sin ^{2}(A-B)=(\cos B-\cos A)^{2}+(\sin B-\sin A)^{2}\right. \\
\Rightarrow & \cos ^{2}(A-B)-2 \cos (A-B)+1+\sin ^{2}(A-B)=\cos ^{2} B+\cos ^{2} A-2 \cos A \cos B \\
& +\sin ^{2} B+\sin ^{2} A-2 \sin A \sin B \\
\Rightarrow & 2-2 \cos (A-B)=2-2 \cos A \cos B-2 \sin A \sin B \\
\Rightarrow & \cos (A-B)=\cos A \cos B+\sin A \sin B \\
\text { Hence, } & \cos (A-B)=\cos A \cos B+\sin A \sin B
\end{array}
$$
|
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|
The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?
|
I have another method as well. From skimming the previous solutions, I don't think it is a duplicate of any of them
In Complex analysis, we learn that $\sin(\pi z) = \pi z\Pi_{n=1}^{\infty}\Big(1 - \frac{z^2}{n^2}\Big)$ which is an entire function with simple zer0s at the integers. We can differentiate term wise by uniform convergence. So by logarithmic differentiation we obtain a series for $\pi\cot(\pi z)$.
$$
\frac{d}{dz}\ln(\sin(\pi z)) = \pi\cot(\pi z) = \frac{1}{z} - 2z\sum_{n=1}^{\infty}\frac{1}{n^2 - z^2}
$$
Therefore,
$$
-\sum_{n=1}^{\infty}\frac{1}{n^2 - z^2} = \frac{\pi\cot(\pi z) - \frac{1}{z}}{2z}
$$
We can expand $\pi\cot(\pi z)$ as
$$
\pi\cot(\pi z) = \frac{1}{z} - \frac{\pi^2}{3}z - \frac{\pi^4}{45}z^3 - \cdots
$$
Thus,
\begin{align}
\frac{\pi\cot(\pi z) - \frac{1}{z}}{2z} &= \frac{- \frac{\pi^2}{3}z - \frac{\pi^4}{45}z^3-\cdots}{2z}\\
-\sum_{n=1}^{\infty}\frac{1}{n^2 - z^2}&= -\frac{\pi^2}{6} - \frac{\pi^4}{90}z^2 - \cdots\\
-\lim_{z\to 0}\sum_{n=1}^{\infty}\frac{1}{n^2 - z^2}&= \lim_{z\to 0}\Big(-\frac{\pi^2}{6} - \frac{\pi^4}{90}z^2 - \cdots\Big)\\
-\sum_{n=1}^{\infty}\frac{1}{n^2}&= -\frac{\pi^2}{6}\\
\sum_{n=1}^{\infty}\frac{1}{n^2}&= \frac{\pi^2}{6}
\end{align}
|
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|
Proving an identity involving terms in arithmetic progression. If $a_1,\ldots,a_n$ are in arithmetic progression and $a_i\gt 0$ for all $i$, then how to prove the following two identities:
$ (1)\large \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$
$(2) \large\frac{1}{a_1 \cdot a_n} + \frac{1}{a_2 \cdot a_{n-1}} + \frac{1}{a_3 \cdot a_{n-2}}+ \cdots + \frac{1}{a_n \cdot a_1} = \frac{2}{a_1 + a_n} \biggl( \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} \biggr)$
|
This question is now old enough for some more complete answers.
For number 1:
$$\sum_{k=1}^{n-1} \frac{1}{ \sqrt{a_k}+ \sqrt{a_{k+1}}} =
\sum_{k=1}^{n-1} \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$$
where $d$ is the common difference,
$$ = \frac{1}{d} \left( \sqrt{a_n} - \sqrt{a_1} \right)
= \frac{a_n - a_1}{d(\sqrt{a_n} + \sqrt{a_1})}$$
$$= \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}.$$
And for number 2:
$$ S = 2 \sum_{k=1}^n \frac{1}{a_k} =
\left( \frac{1}{a_1} + \frac{1}{a_n} \right) +
\left( \frac{1}{a_2} + \frac{1}{a_{n-1}} \right) + \cdots +
\left( \frac{1}{a_n} + \frac{1}{a_1} \right)$$
$$ = \sum_{k=1}^n \frac{a_{n-k+1}+ a_k}{a_k a_{n-k+1}}.$$
Now $ a_{n-k+1}+ a_k = 2a_1 + (n-1)d = a_1 + a_n$ and so
$$ S = (a_1+a_n) \sum_{k=1}^n \frac{1}{a_k a_{n-k+1}}$$
from which the result follows.
|
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|
Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $ How to find the value of
$$\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ})$$
manually ?
|
Let $a\cos\theta+b\cot\theta=c$
$\implies a\cos\theta=c-b\cot\theta=\frac{c\sin\theta-b\cos\theta}{\sin\theta}$
$\implies a\cos\theta\sin\theta=c\sin\theta-b\cos\theta$
Putting $c=r\cos\alpha,b=r\sin\alpha$ where $r>0$
Squaring & adding we get $r^2=c^2+b^2\implies r=+\sqrt{b^2+c^2}$ and $\frac{\sin\alpha}b=\frac{\cos\alpha}c=\frac1r=\frac1{\sqrt{b^2+c^2}}$
$\implies c\sin\theta-b\cos\theta=\sqrt{b^2+c^2}\sin(\theta-\alpha)$ and $\cos\theta\sin\theta=\frac{\sin2\theta}2$
$$a\sin2\theta=2\sqrt{b^2+c^2}\sin(\theta-\alpha)$$
Now, the solution of $P\sin x= Q\sin A $ is in general intractable unless $P=0$ or $Q=0$ or $P=\pm Q\ne0$
Here
$1:$ if $a=0,$ the problem can be solved easily.
$2:$ if $b^2+c^2=0\implies b=c=0$ ( as $b,c$ are real), the problem can be solved easily.
$3:$ So, for non-trivial cases, either $a=2\sqrt{b^2+c^2}$ or $a=-2\sqrt{b^2+c^2}$
$$\begin{array}{|c|c|c|}
\hline \text{ Case } & a=2\sqrt{b^2+c^2} & a=-2\sqrt{b^2+c^2} \\
\hline \text{ Comparison} & \sin2\theta=\sin(\theta-\alpha)
& \sin2\theta=-\sin(\theta-\alpha)=\sin(\alpha-\theta),\text{ as }\sin(-x)=-\sin x \\
\hline \text{General Solution} & 2\theta=n180^\circ+(-1)^n(\theta-\alpha)\text{ where }n\text{ is any integer } & 2\theta=n180^\circ+(-1)^n(\alpha-\theta)\text{ where }n\text{ is any integer } \\
\hline n=2m & \alpha=m360^\circ-\theta\equiv-\theta\pmod{360^\circ} & \alpha=3\theta-m360^\circ\equiv3\theta \\
\hline n=2m+1& \alpha=3\theta-(2m+1)180^\circ\equiv 3\theta-180^\circ & \alpha=(2m+1)180^\circ-\theta\equiv180^\circ-\theta \\
\hline
\end{array} $$
Here $a=-4,b=\sqrt3,\theta=20^\circ, $
So, $a=-2\sqrt{b^2+c^2}$ as $\sqrt{b^2+c^2} > 0$
$\implies\sqrt{b^2+c^2}=-\frac a2=2, \alpha=3\theta=60^\circ$ or $\alpha=180^\circ-\theta=160^\circ$
So, $\sin\alpha=\frac b{\sqrt{b^2+c^2}}=\frac{\sqrt3}2\implies \alpha=60^\circ$
So, $c=\cos\alpha\cdot \sqrt{b^2+c^2}=\frac12\cdot 2=1$
|
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|
Trigonometry Expression Is $(\sin \phi)^2$ is equal to $\sin^2\phi$?
Can any one tell what is the ans for the below expression
$\sin^260$ + $\cos^260$ + $\tan^245$ + $\sec^260$ - $\csc^260$
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($sin \phi$)^2 is equal to $sin^2\phi$.
$sin^260$ + $cos^260$ + $tan^245$ + $sec^260$ - $cosec^260$
The trick to that is to use a few trigonometric identities.
$\sin^2\theta + \cos^2\theta = 1$
$\tan 45 = 1$
The value of $\cos 60$ is $\frac{1}{2}$, so $\sec^260$ will evaluate to 4.
The value of $\sin 60$ is $\sqrt{\frac{3}{4}}$, so $\csc^260$ will evaluate to $\frac{4}{3}$.
Adding it all up you have $1 + 1 + 4 - \frac{4}{3} = \frac{14}{3}$
|
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|
Completing the square problem The problem i am supposed to solve for $x$ by completing the square:
$3x^2+9x+5 = 0$
step 1. move constant to right: $3x^2+9x\quad\quad+5 = 0$
step 2. divide by $3$: $x^2+3\quad\quad+\frac{5}{3}$
step 3. $(\frac{1}{2}b)^2$: $(\frac{1}{2}\cdot 3)^2 = \frac{9}{4}$
step 4. add and subtract result from step 3 into equation: $(x^2+3+\frac{9}{4}) -\frac{9}{4} + \frac{5}{3} = 0$
step 5. simplify: $(x+\frac{3}{2})^2 -\frac{7}{12} = 0$
step 6. subtract constant on both sides: $(x+\frac{3}{2})^2 = \frac{7}{12}$
step 7. solve for $x$: $x = -\frac{3}{2} + \sqrt{\frac{7}{12}}$ or $-\frac{3}{2} - \sqrt{\frac{7}{12}}$
Wolfram seems to be telling me my answer is wrong.
the correct answer listed is: $x = \frac{-9-\sqrt{21}}{6}$ or $x = \frac{\sqrt{21}-9}{6}$
If anyone could let me know where my error is I would greatly appreciate it!
|
As Theo Buehler said in a comment, your answer and technique are correct, just yielding a different form of the same answer (rationalize the denominator in your answer, then combine the fractions by finding a common denominator).
As an aside, while I'm sure that you're applying the technique as you were taught (those steps are fairly common in first-year algebra courses), I prefer a slightly different process for completing the square. The process, described below, is a bit more compatible with uses of completing the square that show up in later courses. Using a quadratic expression (not the whole equation):
*
*$3x^2+9x+5$
*Factor out the leading coefficient from the variable terms: $3(x^2+3x)+5$
*Determine the desired perfect-square constant term: $(\frac{1}{2}\cdot 3)^2=\frac{9}{4}$
*Add and subtract the desired constant term inside the parentheses: $3(x^2+3x+\frac{9}{4}-\frac{9}{4})+5$
*Factor the perfect square and distribute the leading coefficient: $3(x+\frac{3}{2})^2-3\cdot\frac{9}{4}+5$
*Combine the constant terms: $3(x+\frac{3}{2})^2-\frac{7}{4}$
|
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|
How can I prove that $n^7 - n$ is divisible by $42$ for any integer $n$? I can see that this works for any integer $n$, but I can't figure out why this works, or why the number $42$ has this property.
|
Problems like this appear frequently here. There are a couple of standard approaches. One is to use Fermat's little theorem, which says that if $p$ is a prime number, then $n^p-n$ is divisible by $p$ for all $n$.
Since $42=2\times 3\times 7$, what we need to do is to check that 2, 3, and 7 divide $n^7-n$, no matter what $n$ is.
That 7 does is direct from Fermat's little theorem.
The theorem also ensures that 3 divides $n^3-n$. Now: $$n^7-n=n(n^6-1)=n((n^2)^3-1)=n(n^2-1)(n^4+n^2+1)=(n^3-n)(n^4+n^2+1),$$ so 3 indeed divides $n^7-n$.
Finally, $n$ and $n^7$ always have the same parity, so $n^7-n$ is even.
We can actually argue without Fermat's little theorem in this case. An approach that only requires patience is as follows: The idea is to factor the polynomial $x^7-x$ and then analyze the result when $x=n$ is an integer. (This is a trick that Bill Dubuque suggests sometimes in his solutions.)
We have: $x^7-x=x(x^6-1)=x(x^3+1)(x^3-1)=x(x+1)(x^2-x+1)(x-1)(x^2+x+1)$. When $x=n$, we have
$$ n^7-n=(n-1)n(n+1)(n^2-n+1)(n^2+n+1). $$
Now we analyze this prime by prime, as before. Note that one of $n$ and $n-1$ is always even, so the product is even. Also, of 3 consecutive numbers, such as $n-1,n,n+1$, one is always divisible by 3, so it only remains to verify divisibility by 7.
We may assume that $n=7k+b$ where $b=\pm2$ or $\pm3$, since otherwise $(n-1)n(n+1)$ is a multiple of 7. In that case, $n^2\equiv 4$ or $2\pmod 7$, and one of $n^2+n$, $n^2-n$ is $\equiv 6\pmod 7$, so $(n^2-n+1)(n^2+n+1)$ is a multiple of 7.
The disadvantage of this approach over the previous one, of course, is the need to analyze different cases. Fermat's little theorem allows us to analyze all cases simultaneously, which typically (as here) results in a much faster approach.
If you are comfortable with the method of induction, this gives us a way of verifying divisibility by 7 which is not without some elegance (divisibility by 2 and 3 is probably best approached as before). Note that $(-n)^7-(-n)=-(n^7-n)$, so we may as well assume that $n\ge 0$. If $n=0$ it is obvious that 7 divides $n^7-n$.
Suppose then that $7|n^7-n$, and argue that $7|(n+1)^7-(n+1)$. For this, actually expand $(n+1)^7$ using the binomial theorem: $$ (n+1)^7=n^7+7n^6+{7\choose 2}n^5+{7\choose 3}n^4+\dots+1.$$ The point is that $${7\choose k}=\frac{7!}{k!(7-k)!}$$ is obviously divisible by 7 as long as $k\ne0,7$, so (modulo 7) we have that $(n+1)^7-(n+1)\equiv (n^7+1)-(n+1)=(n^7-n)$. Now we invoke the induction hypothesis, that precisely says that the latter is divisible by 7, and we are done.
Of course, exactly the same inductive argument gives us a proof of Fermat's little theorem: $p|n^p-n$ for any $p$ prime and any integer $n$.
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|
How to reduce congruence power modulo prime? If I have a congruence equation, says
$$x^{15} - x^{10} + 4x - 3 \equiv 0 \pmod{7}$$
Then can I use Fermat's little theorem like this:
$$(x^{6})^2 \cdot x^3 - x^6 \cdot x^4 + 4x - 3 \equiv 0 \pmod{7}$$
$$ x^3 - x^4 + 4x - 3 \equiv 0 \pmod{7}$$
Update
Should it be
$$x^{14}x - x^7x^3 - 4x - 3 \equiv 0 \pmod{7}$$
$$x^2x - x.x^3 - 4x - 3 \equiv 0 \pmod{7}$$
$$x^3 - x^4 - 4x - 3 \equiv 0 \pmod{7}$$
?
Thanks,
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Yes, if you're looking for solutions of the equation mod $7$ then, since $\rm\:x=0\:$ is not a solution, you can in fact deduce that $\rm\:x^6 = 1\:$. If you couldn't exclude $\rm\:x=0\:$ then you'd instead need $\rm\:x^7 = x\:.$
|
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Name of Formula $x^3+y^3=z^3+1$ I encountered the formula
$$x^3+y^3=z^3+1$$
with the condition, that
$$x<y<z$$
and wonder, whether it has got a specific name or whether it can be easily transformed into another well-known (family of) formula(s).
|
$$X^3+ Y^3+ Z^3=1$$ is the formula which is known as harder factor and yours is a distorted and conditional form of harder factor
If $X+Y+Z=0$ then $X^3+ Y^3+ Z^3=1$.
In your question $X$ is less than $Y$ and $Y$ is less than $Z$ means the minimum possible difference between $X$ and $Y$, $Y$ and $Z$ is $1$.
At the same time the minimum possible difference between $X$ and $Z$ will be $2$.
So there will in all the cases except $X=-2$, $Y=-1$, $Z=3$ where $X+Y+Z$ is not equal to zero then it must be that $X^3+ Y^3+ Z^3$ is not equal to $1$. So $X^3+ Y^3+ Z^3$ must be greater/less than $1$. As it is given that $Z>Y>X$ then $X^3+ Y^3$ must be unequal to $Z^3$. It means
$X^3+ Y^3$ may be equal to $Z^3+1$.
In this way $X^3+ Y^3+ Z^3=1$ is related to the question asked by the poster
|
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Solve recursion $a_{n}=ba_{n-1}+cd^{n-1}$ Let $b,c,d\in\mathbb{R}$ be constants with $b\neq d$. Let
$$\begin{eqnarray}
a_{n} &=& ba_{n-1}+cd^{n-1}
\end{eqnarray}$$
be a sequence for $n \geq 1$ with $a_{0}=0$. I want to find a closed formula for this recursion. (I only know the german term geschlossene Formel and translated it that way I felt it could be right. So if I got that wrong, please correct me)
First I wrote down some of the chains and I got
$$\begin{eqnarray}
a_{n} &=& ba_{n-1}+cd^{n-1}\\
&=& b\left(ba_{n-2}+cd^{n-2}\right)+cd^{n-1}\\
&=& b\left(b\left(ba_{n-3}+cd^{n-3}\right)+cd^{n-2}\right)+cd^{n-1}\\
&=& b\left(b\left(b\left(ba_{n-4}+cd^{n-4}\right)+cd^{n-3}\right)+cd^{n-2}\right)+cd^{n-1}\\
&=& \dots\\
&=& \sum_{k=0}^{n}b^{k}cd^{n-k-1}\\
&=& \sum_{k=0}^{n}b^{k}cd^{n-\left(k+1\right)}
\end{eqnarray}$$
So I catched the structure in a serie. Now I am asking myself how to proceed. I took the liberty to have a little peek at what WolframAlpha wood say to this serie. I hoped for inspiration and I got
$$\sum_{k=0}^{n-1}b^{k} c d^{n-(k+1)} = (c (b^n-d^n))/(b-d)$$
How did this came to be? And more important: Is my approach useful?
Thank you in advance for any advice!
Edit: My final Solution (recalculated)
$$\begin{eqnarray}
a_{n} &=& ba_{n-1}+cd^{n-1}\\
&=& b\left(ba_{n-2}+cd^{n-2}\right)+cd^{n-1}\\
&=& b\left(b\left(ba_{n-3}+cd^{n-3}\right)+cd^{n-2}\right)+cd^{n-1}\\
&=& b\left(b\left(b\left(ba_{n-4}+cd^{n-4}\right)+cd^{n-3}\right)+cd^{n-2}\right)+cd^{n-1}\\
&=& b^{4}a_{n-4}+b^{3}cd^{n-4}+b^{2}cd^{n-3}+bcd^{n-2}+cd^{n-1}\\
&=& b^{5}a_{n-5}+b^{4}cd^{n-5}+b^{3}cd^{n-4}+b^{2}cd^{n-3}+bcd^{n-2}+cd^{n-1}\\
&=& b^{n}a_{0}+b^{n-1}c+\dots+b^{4}cd^{n-5}+b^{3}cd^{n-4}+b^{2}cd^{n-3}+cbd^{n-2}+cd^{n-1}\\
&=& \dots\\
&=& 0+b^{n-1}c+\dots+b^{4}cd^{n-5}+b^{3}cd^{n-4}+b^{2}cd^{n-3}+cbd^{n-2}+cd^{n-1}\\
&=& \sum_{k=0}^{n-1}b^{k}cd^{n-1-k}\\
&=& cd^{n-1}\sum_{k=0}^{n-1}b^{k}d^{-k}\\
&=& cd^{n-1}\sum_{k=0}^{n-1}\left(\frac{b}{d}\right)^{k}\\
&=& cd^{n-1}\frac{1-\left(\frac{b}{d}\right)^{n}}{1-\left(\frac{b}{d}\right)}\\
&=& cd^{n-1}\frac{1-\frac{b^{n}}{d^{n}}}{1-\frac{b}{d}}\\
&=& cd^{n-1}\frac{\frac{d^{n}-b^{n}}{d^{n}}}{\frac{d-b}{d}}\\
&=& cd^{n-1}\frac{d^{n}-b^{n}}{d^{n}}\cdot\frac{d}{d-b}\\
&=& \frac{c\left(d^{n}-b^{n}\right)}{d-b}
\end{eqnarray}$$
|
If you know how to solve linear recurences, this would simplify your computations:
\begin{eqnarray}
a_{n} &=& ba_{n-1}+cd^{n-1}
\end{eqnarray}
\begin{eqnarray}
da_{n-1} &=& dba_{n-2}+cd^{n-1}
\end{eqnarray}
and subtract....
|
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|
How to factor quadratic $ax^2+bx+c$? How do I shorten this? How do I have to think?
$$ x^2 + x - 2$$
The answer is
$$(x+2)(x-1)$$
I don't know how to get to the answer systematically. Could someone explain?
Does anyone have a link to a site that teaches basic stuff like this? My book does not explain anything and I have no teacher; this is self-studies.
Please help me out; thanks!
|
As Ross pointed out, and as was previously discussed, we know that
$$
(x+a)(x+b) = x(x+b) + a(x+b) = x^2 + bx + ax + ab = x^2 + (a+b)x + ab.
$$
Therefore, to factor a quadratic expression $x^2 + cx + d$, all one has to do is find two numbers that multiply to $d$ and add to $c$. Let $m$ and $n$ be those two numbers that add to $c$ and multiply $d$. Then:
$$
x^2 + cx + d = (x+m)(x+n).
$$
Note that it doesn't matter which order we pick the two numbers since $a+b = b+a$ and $ab = ba$.
In your case, to factor $x^2 + x - 2 = x^2 + 1x - 2$, we need to find two numbers that add to $1$ and multiply to $-2$. Just by picking and trying different numbers, we find the two numbers are $2$ and $-1$.
A good way to go about finding the numbers is to determine all the factors of the constant term, which for us is $-2$. Since $-2 = 2 \cdot (-1) = (-2) \cdot 1$, we only have to check whether $2 + (-1)$ or $(-2) + 1$ equals one. Since it is the first pair of numbers with this property, these are the two numbers we were looking for. Therefore,
$$
x^2 + x - 2 = (x+2)(x+(-1)) = (x+2)(x-1).
$$
The reason this works: Suppose that we want to factor $x^2 + cx + d$, and we have found the numbers $m$ and $n$ such that $m + n = c$ and $mn = d$. Then:
$$
\begin{align}
x^2 + cx + d &= x^2 + (m+n)x + mn
\\
&= x^2 + mx + nx + mn
\\
&= x(x+m) + n(x+m)
\\
&= (x+n)(x+m)
\end{align}
$$
|
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|
Calculating arc length of a curve, stuck on dy/dx part (algebra mostly) The equation is:
$$x=\frac{1}{8}y^4 + \frac{1}{4}y^{-2},\qquad 1\leq y\leq 2.$$
I have the formula. I'm not sure how to write it out but this is what it says:
Length is equal to the integral (with $b$ and $a$ for limits) of the square root of $1+(dy/dx)^2 dx$
So I first took the derivative of the equation and got $(1/2)y^3 - (1/2)y^{-3}$.
Now when I plug that back into the formula, I have to square it and I got $(1/4)y^6 - (1/4)y^{-6}$. I factored out a $1/4$ and turned the $y^{-6}$ into $1/y^6$. Just a mess at this point, that doesn't seem right. Can someone help me out with that? Haha I know it's ridiculous but my algebra skills are lacking.
|
You squared incorrectly. The square of $a-b$ is not $a^2-b^2$. (Also, you should be using $\frac{dx}{dy}$, not $\frac{dy}{dx}$, because here your independent variable is $y$ and your dependent variable is $x$; your integral will be with respect to $y$, not with respect to $x$).
The square of $\displaystyle\frac{1}{2}y^3 - \frac{1}{2}y^{-3}$ is not just the difference of the squares (which is what you wrote). Rather, it is equal to:
$$\left(\frac{1}{2}y^3 - \frac{1}{2}y^{-3}\right)^2 = \frac{1}{4}y^6 - 2\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)y^3y^{-3} + \frac{1}{4}y^{-6} = \frac{1}{4}y^6+\frac{1}{4}y^{-6} - \frac{1}{2}.$$
When you add $1$, you get
$$\frac{1}{4}y^6 + \frac{1}{4}y^{-6} + \frac{1}{2} = \left(\frac{1}{2}y^3 + \frac{1}{2}y^{-3}\right)^2.$$
Remember: $(a+b)^2 = a^2 + 2ab+b^2$, and $(a-b)^2 = a^2-2ab+b^2$. The square of the sum is not the sum of the squares.
|
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|
Complex number to polar form I need to take a raincheck with this problem. I want to make sure I haven't messed up some fundamental idea.
Convert the complex number $$-\dfrac{1}{2} + \dfrac{\sqrt3}{2}i$$ to polar form.
I took the modulus as below,
$$\lvert-\dfrac{1}{2} + \dfrac{\sqrt3}{2}i \rvert = \sqrt{(\dfrac{-1}{2})^2 + (\dfrac{\sqrt 3}{2})^2} = 1$$
And the argument as below,
$$arg(-\dfrac{1}{2} + \dfrac{\sqrt3}{2}i) = \tan^{-1} (\dfrac{\sqrt 3}{2} \times \dfrac{-2}{1}) = -60 = -\dfrac{\pi}{3}$$
Hence the complex number in polar form is,
$$-\cos \dfrac{\pi}{3} + i \sin \dfrac{\pi}{3}$$
But, The required answer is $$\cos \dfrac{2\pi}{3} + i \sin \dfrac{2\pi}{3}$$
I thought of converting the -60 to positive, as 360 - 60 = 300, ie:- $$\dfrac{5\pi}{3}$$. I have a feeling I am missing something important. Can you guys tell me where I am going wrong? Thanks for all your help!
|
The problem is you forgot to add $\pi$ in calculating the argument. See the wiki page which shows all the formulas for computing the argument depending on $x$ and $y$.
Taking $z=x+yi$ to be your complex number, here $x$ is negative and $y$ is positive, so
$$
\text{arg}(-\dfrac{1}{2} + \dfrac{\sqrt3}{2}i) = \tan^{-1} (\dfrac{\sqrt 3}{2} \times \dfrac{-2}{1})+\pi = -\dfrac{\pi}{3}+\pi=\frac{2\pi}{3}.
$$
|
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|
What are the fields with 4 elements? What are the fields with 4 elements?
|
Let $K$ be your field.
The additive group of $K$ is an abelian group with four elements. The order of $1$ in this group divides $4$, so it is either $2$ or $4$. Were it $4$, we would have $1+1\neq0$ and $(1+1)\cdot(1+1)=0$, which is absurd in a field. It follows that $1+1=0$ in $K$. But then for all $x\in K$ we have $x+x=x\cdot(1+1)=0$, and we see that all elements have order $2$. In particular, $-1=1$.
Let $a$ be an element in $K$ which is neither $0$ nor $1$. Then $a+1$ is neither $a$ nor $1$ and if we had $a+1=0$, then $a=-1=1$ which again is a not true. We conclude that the four elements of $K$ are $0$, $1$, $a$ and $a+1$.
You should check that this knowledge complete determines the addition in $K$.
We have to determine the multiplication now. Since $0$ and $1$ are what they are, we need only see what $a\cdot a$, $a\cdot(a+1)$ and $(a+1)\cdot(a+1)$ are:
*
*We can't have $a^2=a$, for then $a(a-1)=0$ and we are supposing that $a\not\in\{0,1\}$; similarly, $a^2\neq0$. If $a^2=1$, then $(a-1)^2=a^2-1=0$ , which is also impossible. We must have thus $a^2=1+a$.
*Next, using this, $a\cdot(a+1)=a^2+a=1+a+a=1$.
*Finally, using that $1+1=0$, $(a+1)\cdot(a+1)=a^2+1=a$.
Multiplication is completely determined.
Now we have to check that with this operations we do have a field... You should have no trouble with that :)
|
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|
Generating sequences using the linear congruential generator I came across the linear congruential generator on Wikipedia:
http://en.wikipedia.org/wiki/Linear_congruential_generator
I gather that for a particular choice of the modulus, multiplier and increment, the generator would generate a unique sequence. However, is there any way to determine the values of modulus, multiplier and increment that I need to create a particular finite sequence?
For instance, if I chose the modulus as 8, the multiplier as 1, and the increment as 5, I obtain the sequence 5, 2, 7, 4, 1, 6, 3, 0, assuming a seed value of 0. Now if I wanted the sequence 2, 11, 5, 9, 6, 22, how do I determine what values of the parameters to choose?
|
Not every finite sequence can be obtained by a linear congruence generator. In fact, the one you request cannot.
Note that we must have $m\geq 23$ in order to "get" $22$ as an answer. We have that you are requiring:
$$\begin{align*}
2a + c &\equiv 11 \pmod{m}\\
11a+c &\equiv 5\pmod{m}\\
5a+c &\equiv 9\pmod{m}\\
9a+c &\equiv 6\pmod{m}\\
6a+c&\equiv 22\pmod{m}.
\end{align*}$$
Subtracting the first congruence from the second, we get $9a \equiv -6\pmod{m}$; subtracting this from the fourth congruence, we obtain $c\equiv 12\pmod{m}$. So now we know the value of $c$, which gives
$$\begin{align*}
2a &\equiv -1\pmod{m}\\
11a &\equiv -7\pmod{m}\\
5a &\equiv -3\pmod{m}\\
9a &\equiv -6\pmod{m}\\
6a &\equiv 10\pmod{m}.
\end{align*}$$
Multiply the first congruence by $3$ to get $6a\equiv -3\pmod{m}$. Since we also need $6a\equiv 10\pmod{m}$, we must have $10\equiv -3\pmod{m}$, or $13\equiv 0\pmod{m}$. But that means that $m=1$ or $m=13$, which contradicts our requirement that $m\geq 23$.
There are other contradictions in this system: multiply the third congruence by $2$ to get $10a\equiv -6\pmod{m}$, and subtracting from the second congruence we get $a\equiv 3\pmod{m}$. But subtracting $9a\equiv -6\pmod{m}$ from $10a\equiv -6\pmod{m}$ would give $a\equiv 0\pmod{m}$, so we would need $3\equiv 0\pmod{m}$, again a problem.
Or note that $2a\equiv -1\pmod{m}$ tells you that $\gcd(2,m)=1$, so that $6a\equiv 10\pmod{m}$ is equivalent to $3a\equiv 5\pmod{m}$, and multiplying by $3$ gives $9a\equiv 15\pmod{m}$; comparing with $9a\equiv -6\pmod{m}$ gives $21\equiv 0\pmod{m}$, so $m$ would have to divide $21$. Etc.
|
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Inequality: $(x + y + z)^3 \geq 27 xyz$ Edit: $a,b,c$ and $x,y,z$ are positive, real numbers.
Since $(a-b)^2 \geq 0~$, $a^2 + b^2 - 2ab\geq0~$ and $a^2 + b^2 \geq 2ab~$. Similarly, $a^2 + c^2 \geq 2ac~$ and $b^2 + c^2 \geq 2bc~$.
Adding these inequalities together, $2(a^2 +b^2 + c^2) \geq 2(ab + ac +bc)~$ and accordingly, $a^2 +b^2 + c^2 \geq ab + ac +bc~$
Multiplying both sides by $(a + b + c)$:
$(a^2 +b^2 + c^2)(a+b +c) \geq (ab + ac +bc)(a + b + c)~~$ and simplifying this, $ a^3 + b^3 + c^3 + \Sigma a^2 b \geq 3abc + \Sigma a^2 b $
Therefore, it follows that $a^3 + b^3 + c^3 \geq 3abc~$, and letting $a^3 = x~$, $b^3 = y~$, $c^3 = z~$: $x + y + z\geq 3\sqrt[3]{xyz}$
Cubing both sides, $(x + y + z)^3 \geq 27 xyz~$ which was to be proven.
I was wondering if there are alternative approaches to solve this problem (possibly using higher-level maths), and is my proof entirely correct?
|
I was wondering if there are alternative approaches to solve this problem
Yes, there is.
and is my proof entirely correct?
I think so. I couldn't find anything wrong with it.
Now let's start with another proof.
Let
$$
\{a,b,c,x,y,z \in \mathbb R_{\geq0} : a=x^{\frac{1}{3}},b=y^{\frac{1}{3}},c=z^{\frac{1}{3}}\}
$$
We know that
$$
\begin{equation}
\begin{aligned}
&a^2+b^2 \geq 2ab \\
\implies &(a+b)(a^2+b^2-ab) \geq ab(a+b) \\
\end{aligned}
\end{equation}
$$
$$
a^3+b^3 \geq ab(a+b)
\label{eq1}
\tag{1}
$$
Utilizing this result across $a$, $b$ and $c$ and adding them, we get
$$
2(a^3+b^3+c^3) \geq a^2(b+c)+b^2(c+a)+c^2(a+b)
\label{eq2}
\tag{2}
$$
Now,
$$
a+b \geq 2\sqrt{ab}
$$
Utilizing this result across $a$, $b$ and $c$ and multiplying them, we get
$$
(a+b)(b+c)(c+a) \geq 8abc
$$
On further simplification, we get
$$
a^2(b+c)+b^2(c+a)+c^2(a+b) \geq 6abc
\label{eq3}
\tag{3}
$$
From $\eqref{eq2}$ and $\eqref{eq3}$, it's clear that
$$
2(a^3+b^3+c^3) \geq a^2(b+c)+b^2(c+a)+c^2(a+b) \geq 6abc
$$
Therefore,
$$
\begin{aligned}
&2(a^3+b^3+c^3) \geq 6abc \\
\implies &a^3+b^3+c^3 \geq 3abc \\
\implies &x+y+z \geq 3(xyz)^{\frac{1}{3}}
\end{aligned}
$$
Cubing on both sides, we get
$$
(x+y+z)^3 \geq 27xyz
$$
|
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|
Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$ Let $a,b,c$ be positive, not equal. Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$.
I know the proof by subtracting LHS by RHS and then doing some arrangement.
But isn't there any inequality which can be used in $(1+1+1)(a^3+b^3+c^3) >(a+b+c)(a^2+b^2+c^2)$, or an easy proof?
|
As is often the case, we can use Cauchy-Schwarz.
Since the $L^3$ norm is larger then the $L^4$ norm, we have that $$\left(a^{3}+b^{3}+c^{3}\right)^{\frac{1}{3}}\geq\left(a^{4}+b^{4}+c^{4}\right)^{\frac{1}{4}}$$ with equality if and only if $a=b=c.$ Using Cauchy Schwarz, we have that $$\sqrt{3\left(a^{4}+b^{4}+c^{4}\right)}\geq\left(a^{2}+b^{2}+c^{2}\right)$$ and $$\sqrt{3\left(a^{2}+b^{2}+c^{2}\right)}\geq\left(a+b+c\right).$$ Multiplying these two inequalities, we get $$3\left(a^{4}+b^{4}+c^{4}\right)^{\frac{3}{4}}\geq\left(a+b+c\right)\left(a^{2}+b^{2}+c^{2}\right),$$ which implies the result.
|
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|
Trigonometric equality: $\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \frac{A}{2}$ Can you guys give me a hint on how to proceed with proving this trigonometric equality? I have a feeling I need to use the half angle identity for $\tan \frac{\theta}{2}$. The stuff I have tried so far(multiplying numerator and denominator by $1 + \sin A - \cos A$) has lead to a dead end.
Prove that,
$$
\dfrac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \dfrac{A}{2}
$$
|
First write $A=2b$. Then make appropriate use of $\tan b=\sin b/\cos b$, $\sin2b=2\sin b\cos b$, and $\cos2b=2\cos^2b-1=1-2\sin^2b$.
|
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How to solve these inequalities? How to solve these inequalities?
*
*If $a,b,c,d \gt 1$, prove that $8(abcd + 1) \gt
(a+1)(b+1)(c+1)(d+1)$.
*Prove that $ \cfrac{(a+b)xy}{ay+bx} \lt \cfrac{ax+by}{a+b}$
*Find the greatest value of $x^3y^5z^7$ when $2x^2+2y^2+2x^2=15$
Any hints/solution are welcome.
|
Solution:
*
*Since $a,b,c,d>1$, then the following inequalities are true based on Rearrangement inequalities: if $x>1$ and $y>1$ then $(x-1)(y-1) > 0$, ie $xy+1 >x+y$.
$$
\begin{aligned}
abcd + 1 &> abc + d
\\
abcd + 1 &> abd + c
\\
abcd + 1 &> acd + b
\\
abcd + 1 &> bcd + a
\\
abcd + 1 &> ab + cd
\\
abcd + 1 &> ad + bc
\\
abcd + 1 &> ac + bd
\\
abcd + 1 &= abcd + 1
\end{aligned}
$$
Adding them all up you get $8(abcd + 1)> (a+1)(b+1)(c+1)(d+1)$.
*Assuming $a,b,x,y>0$, and $x\neq y$, using Jensen's inequality
$$
\varphi(\frac{a_1 t_1+ a_2 t_2}{a_1 + a_2}) \leq \frac{a_1 \varphi(t_1)+a_2 \varphi(t_2)}{a_1 + a_2}
$$
where $\varphi$ is a convex function. Here take $\displaystyle \varphi(t) = \frac{1}{t}$, $\displaystyle t_1=\frac{1}{x}, t_2=\frac{1}{y}$, $a_1=axy, a_2=bxy$ , apply the inequality you have:
$$
\frac{(a+b)xy}{ay+bx} = \varphi\left(\frac{axy\cdot \frac{1}{x} + bxy\cdot \frac{1}{y}}{axy + bxy}\right)<\frac{axy \cdot \varphi(\frac{1}{x})+bxy\cdot \varphi(\frac{1}{x})}{axy + bxy}= \frac{ax+by}{a+b}.
$$
*Let $r^2 = 15/2$, then use spherical coordinates, or calculus. Or using AM-GM inequality, write
$$
\begin{aligned}
x^3 y^5 z^7 &= \frac{1}{3^{5/2}\cdot 5^{3/2}\cdot (15/7)^{7/2}} \cdot \left((5x^2)^{1/5} \cdot (3y^2)^{1/3}\cdot (\frac{15}{7} z^2)^{7/15}\right)^{15/2}
\\
&\leq \frac{1}{3^{5/2}\cdot 5^{3/2}\cdot (15/7)^{7/2}} \cdot \left(\frac{1}{5}\cdot 5x^2 + \frac{1}{3}\cdot 3y^2+ \frac{7}{15}\cdot \frac{15}{7} z^2\right)^{15/2}
\\
&= \frac{1}{3^{5/2}\cdot 5^{3/2}\cdot (15/7)^{7/2}}\cdot (\frac{15}{2})^{\frac{15}{2}}
\end{aligned}
$$
the maximum is obtained at $5x^2 = 3y^2 = \frac{15}{7} z^2$, ie, $x = \sqrt{3/2}, y= \sqrt{5/2}, z= \sqrt{7/2}$.
|
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|
Difficulty in integrating I tried to integrate this by parts but it didn't work out. Any simple means of doing it.$$\int\sin^{-1}\biggl(\frac{2x+2}{\sqrt{4x^{2}+8x+13}}\biggr) \ dx$$
|
Put $2x+2 = 3 \tan(\theta)$ and see what happens.
$\textbf{Added.}$ First observe that $$4x^{2}+8x+13= (2x+2)^{2} + 3^{2}.$$ So I hope you are aware of the fact that $\text{if you have an integral of the form}$, $1+x^{2}$, then one generally substitutes, $x= \tan(\theta)$. That's the case here as well. By doing that we get,
\begin{align*}
\int \sin^{-1}\biggl(\frac{2x+2}{\sqrt{4x^{2}+8x+13}}\biggr) \ dx &= \int\sin^{-1}\biggl(\frac{3\cdot \tan\theta}{3 \cdot \sec\theta}\biggr) \cdot 3 \sec^{2}\theta \ d\theta \\ &= 3\cdot\int \sin^{-1}(\sin\theta) \cdot \sec^{2}\theta \ d\theta \\ &= 3 \cdot \int \theta \cdot\sec^{2}\theta \ d \theta
\end{align*}
Use integration by parts to evaluate the last integral. Put $u = \theta$ and $dv = \sec^{2}(\theta) \ d\theta$. So the answer for the last part should be $$\int \theta \cdot \sec^{2}\theta \ d \theta = \theta\cdot\tan\theta + \ln(\cos\theta) + C$$
|
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|
Show that $PSL(3,4)$ has no element of order $15$.
$PSL(3,4)$ has no element of order $15$. Thus it is no isomorphic to $A_8$.
Here, $PSL(3,4)$ denotes the $3 \times 3$ projective special linear group on the field with $4$ elements.
As listing all the elements takes too much work, is there any better way to prove there is no element of order $15$ in $PSL(3,4)$?
Thank you very much.
|
Here is a proof without use of Maschke's theorem.
D.J.S. Robinson, A Course in the Theory of Groups, 2d edition, exerc. 3.2.6, p. 79 (before stating Maschke's theorem, which only happens in chapter 8), asks for a proof that $PSL(3, 4)$ has no element of order 15. The author gives the following hint : "Suppose there is such an element : consider the possible rational canonical forms of a preimage in $SL(3, 4)$."
I found a proof, but it doesn't use a rational canonical form and I don't see how a rational canonical form can help.
Here is my proof.
The center of $SL(3, 4)$ is formed by the scalar matrices in $SL(3, 4)$. (In this particular case, the scalar matrices in $SL(3, 4)$ are exactly the scalar matrices in $GL(3, 4)$, but this fact plays no role.) Thus the statement amounts to say that if $M$ is a matrix in $SL(3, 4)$ such that $M^{15}$ is scalar, then either $M^{3}$ or $M^{5}$ is scalar.
Let $F$ denote the field with 4 elements. In this field, $-1 = 1$ and $2 = 0$. Let $a$ and $b$ denote the two elements of $F$ other than $0$ and $1$.
Then $a^{2} + a + 1 = 0$, $b^{2} + b + 1 = 0$, $a^{3} = 1$, $b^{3} = 1$, $a + b = 1$, $a = b^{2}$, $b = a^{2}$, $ab = 1$.
Let $X$ be a variable. We have
(1) $X^{3} - 1 = (X - 1) (X - a) (X - b)$.
Replacing $X$ by $X^{5} $, we get
(2) $X^{15} - 1 = (X^{5} - 1) (X^{5} - a) (X^{5} - b)$.
On the other hand,
(3) $X^{5} - 1 = (X - 1) (X^{2} + aX + 1) (X^{2} + bX + 1)$.
Replacing $X$ by $X/a^{2}$ in (3) gives
(4) $X^{5} - a = (X - b) (X^{2} + X + a) (X^{2} + aX + a)$.
Similarly,
(5) $X^{5} - b = (X - a) (X^{2} + X + b) (X^{2} + bX + b)$.
Bringing (3), (4) and (5) in (2), we get
(6) $X^{15} - 1 = (X - 1) (X^{2} + aX + 1) (X^{2} + bX + 1) (X - b) (X^{2} + X + a) (X^{2} + aX + a) (X - a) (X^{2} + X + b) (X^{2} + bX + b)$.
None of the four elements of $F$ is a root of the first polynomial of degree $2$ in the right member, thus this polynomial is irreducible over $F$. Since the hypotheses are symmetric in $a$ and $b$, the second polynomial of degree $2$ in the right member is also irreducible over $F$. In view of the manner in which the other polynomials of degree $2$ were obtained from the first two ones,
(7) all factors in the right member of (6) are irreducible polynomials over $F$.
Let $M$ be a matrix in $SL(3, 4)$ such that $M^{15}$ is scalar. We have to prove that either $M^{3}$ or $M^{5}$ is scalar.
Asume first that $M^{15} = 1$. Then $M$ annihilates the polynomial $X^{15} - 1$, thus, in view of (7), every irreducible factor of its characteristic polynomial is equal to one of the nine factors in the right member of (6). (Classical property of the characteristic polynomial.) Thus, since the characteristic polynomial of $M$ is of degree 3, it is equal either to $ (X - 1) (X - a) (X - b) $ or to the product of one of the three polynomials $X - 1$, $X - a$, $X - b$ by one of the six polynomials of degree 2 appearing in the scond member of (6). In the first case, the characteristic polynomial is equal to $X^{3} - 1$ (see (1) ), so $M^{3}$ is equal to 1 and is thus scalar. In the second case, note that the hypothesis $M \in SL(3, 4)$ implies $det(M) = 1$, thus the independent coefficient of the characterisic polynomial is $(-1)^{3} = 1$ in $F$. We deduce that the six possibilities for the characterisic polynomial are $(X-1) (X^{2} + aX + 1) $, $(X-1) (X^{2} + bX + 1) $, $(X-a) (X^{2} + X + b) $, $(X-a) (X^{2} + bX + b) $, $(X-b) (X^{2} + X + a) $, $(X-b) (X^{2} + aX + a) $. Thus, from (3), (4) and (5), the characterisic polynomial divides one of the three polynomials $X^{5} - 1$, $X^{5} - a$, $X^{5} - b$, thus $M^{5}$ is scalar.
Our thesis is thus right if $M^{15} = 1$. There remains to prove it in the case where, for example, $M^{15} = a$.
This can be written $(M^{3}/a^{2})^{5} - 1 = 0$, thus, from (3),
(8) $M$ annihilates the polynomial $(X^{3} - a^{2} ) (X^{6} + X^{3} + a) (X^{6} + aX^{3} + a)$.
The only nonzero cube in $F$ is 1, thus the polynomial $X^{3} - a^{2}$ has no root in $F$. Since it is of degree $3$, it is irreducible over $F$.
Let us prove that the polynomial $X^{6} + X^{3} + a$ is irreducible over $F$.
Let $\theta$ denote a root of $X^{6} + X^{3} + a$. It suffices to prove that $F(\theta)$ is an extension of degree $6$ of $F$. Now, $F(\theta)$ is obtained by adjoining a cubic root of $\lambda$ to the field $F(\lambda)$, where $\lambda$ is a root of $X^{2} + X + a$. The norm of $\lambda$ in $F$ is $a$ and is thus not a cube in $F$, thus $\lambda$ is not a cube in $F(\lambda)$. Thus, since $X^{3} - \lambda$ is a polynomial of degree 3, it is irreducible over $F(\lambda)$, so $F(\theta)$ has degree 3 on $F(\lambda)$. On the other hand, we saw that the polynomial $X^{2} + X + a$ is irreducible over $F$, thus $F(\lambda)$ has degree $2$ over $F$. So, as announced, $F(\theta)$ is an extension of degree $6$ of $F$. As we saw, it proves that the polynomial $X^{6} + X^{3} + a$ is irreducible over $F$.
Similarly, the polynomial $X^{6} + aX^{3} + a$ is irreducible over $F$.
Thus (8) implies that every irreducible factor of the characteristic polynomial of $M$ is equal to one of the polynomials $X^{3} - a^{2}$, $X^{6} + X^{3} + a$, $X^{6} + aX^{3} + a$. But the characteristic polynomial of $M$ has degree $3$, thus it must be $X^{3} - a^{2}$, so $M^{3} = a^{2}$ and $M^{3}$ is scalar, which completes the proof.
Can the proof be simplified by use of a rational canonical form ? Thanks in advance for the answers.
|
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|
What is the maximum value of this trigonometric expression What is the maximum value of the expression
$1/(\sin^2 \theta + 3\sin\theta \cos\theta+ 5\cos^2 \theta$).
I tried reducing the expression to $1/(1 + 3\sin\theta$ $\cos\theta + 4\cos^2 \theta)$.
How do I proceed from here?
|
Another method is to rewrite $\sin^2 \theta + 3 \sin \theta \cos \theta + 5 \cos^2\theta$ as
$$\frac{1 + (\sin\theta + 3\cos\theta)^2}{2}.$$
The minimum value of this expression is $\frac12$, so the maximum value of the original expression is 2.
|
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|
Find the square root of a matrix Let $A$ be the matrix
$$A = \left(\begin{array}{cc}
41 & 12\\
12 & 34
\end{array}\right).$$
I want to decompose it into the form of $B^2$.
I tried diagonalization , but can not move one step further.
Any thought on this? Thanks a lot!
ONE STEP FURTHER:
How to find a upper triangular $U$ such that $A = U^T U$?
|
This is an expansion of Arturo's comment.
The matrix has eigenvalues $50,25$, and eigenvectors $(4,3),(-3,4)$, so it eigendecomposes to $$A=\begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix} \begin{pmatrix}50 & 0 \\ 0 & 25\end{pmatrix} \begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix}^{-1}.$$
This is of the form $A=Q\Lambda Q^{-1}$. If this is $B^2$, then there will be a $B$ of the form $Q\Lambda^{1/2} Q^{-1}$ (square this to check this is formally true). A square root of a diagonal matrix is just the square roots of the diagonal entries, so we have
$$B=\begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix} \begin{pmatrix}\sqrt{50} & 0 \\ 0 & \sqrt{25}\end{pmatrix} \begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix}^{-1}$$
$$=\frac{1}{5}\begin{pmatrix}9+16\sqrt{2} & -12+12\sqrt{2} \\ -12+12\sqrt{2} & 16+9\sqrt{2}\end{pmatrix}.$$
Here we used $\sqrt{50}=5\sqrt{2},\sqrt{25}=5$, and a quick formula for the inverse of a $2\times 2$ matrix:
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}.$$
Keep in mind that matrix square roots are not unique (even up to sign), but this particular method is guaranteed to produce one example of a real matrix square root whenever $A$ has all positive eigenvalues.
Finding an upper triangular $U$ such that $A=U^TU$ is even more straightforward:
$$A=\begin{pmatrix} a&0 \\ b&c \end{pmatrix} \cdot \begin{pmatrix} a&b \\ 0&c \end{pmatrix} $$
This is $a^2=41$ hence $a=\sqrt{41}$, $ab=12$ hence $b=\frac{12}{41}\sqrt{41}$, and $b^2+c^2=34$ hence $c=25\sqrt{\frac{2}{41}}$.
In other words,
$$U=\sqrt{41}\begin{pmatrix}1&\frac{12}{41}\\0&\frac{25}{41}\sqrt{2}\end{pmatrix}. $$
|
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|
If $a+\frac1b=b+\frac1c=c+\frac1a$ for distinct $a$, $b$, $c$, how to find the value of $abc$?
If $a, b, c$ be distinct reals such that $$a + \frac1b = b + \frac1c = c + \frac1a$$ how do I find the value of $abc$?
The answer says $1$, but I am not sure how to derive it.
|
I'm not sure how different this is, but here is my version
$a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}
\quad$ (Note this implies $abc \ne 0$)
$a^2bc + ac = ab^2c + ab = abc^2 + bc$
$a(abc) + ac = b(abc) + ab = c(abc) + bc$
$\quad a(abc) + ac = b(abc) + ab \implies (a-b)abc=a(b-c)$
$$ a=b=c \ne 0 \; \text{ or } \; abc = \dfrac{a(b-c)}{a-b}$$
$\quad a(abc) + ac = c(abc) + bc \implies (a-c)abc = c(b-a)$
$$a=c=b \ne 0 \; \text{ or } \; abc = \dfrac{c(b-a)}{a-c}$$
$\quad b(abc) + ab = c(abc) + bc \implies (b-c)abc=b(c-a)$
$$b=c=a \ne 0 \; \text{ or } \; abc = \dfrac{b(c-a)}{b-c}$$
So, one solution is $\; a=b=c \ne 0$.
But if $a,b,$ and $c$ are distinct and non zero, then
$(abc)^3 =
\dfrac{a(b-c)}{a-b} \cdot \dfrac{c(b-a)}{a-c} \cdot \dfrac{b(c-a)}{b-c}$
$(abc)^3 = abc$
$(abc)^2 = 1$
$abc = \pm 1$
$|abc| = 1$
|
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|
System of $\sqrt{7x+y}+\sqrt{x+y}=6$ and $\sqrt{x+y}-y+x=2$ $$\begin{align*}\sqrt{7x+y}+\sqrt{x+y}=6\\\sqrt{x+y}-y+x=2\end{align*}$$
I have tried various things squaring, summing but nothing really helped, got some weird intermediate results which are probably useless such as:
$$(y-x)(y+x+4)+4-x-y=0$$
or
$$x_{1,2}=\frac{2y-3\pm\sqrt{8y-7}}{2}$$
Could you please give me some hints?
|
This is the way I would solve the given system in my school days. It is very similar to the hint provided by Ross Millikan. It turns out that one of the equations becomes linear. I omit some intermediate steps. The system
$$\left\{
\begin{array}{c}
\sqrt{7x+y}+\sqrt{x+y}=6 \\
\sqrt{x+y}-y+x=2
\end{array}
\right. \tag{1}$$
is equivalent to
$$\left\{
\begin{array}{c}
\sqrt{7x+y}=4+x-y \\
\sqrt{x+y}=2-x+y.
\end{array}
\right. $$
By squaring both sides, we get
$$\left\{
\begin{array}{c}
7x+y=\left( 4+x-y\right) ^{2} \\
x+y=\left( 2-x+y\right) ^{2}
\end{array}
\right. \tag{2}$$
or
$$\left\{
\begin{array}{c}
y=\left( 4+x-y\right) ^{2}-7x \\
\left( 4+x-y\right) ^{2}-7x=\left( 2-x+y\right) ^{2}-x,
\end{array}
\right. $$
which is equivalent to
$$\left\{
\begin{array}{c}
y=\left( 4+x-y\right) ^{2}-7x \\
-12y+6x+12=0
\end{array}
\right. $$
and to
$$\left\{
\begin{array}{c}
1+\frac{1}{2}x=9-4x+\frac{1}{4}x^{2} \\
y=1+\frac{1}{2}x.
\end{array}
\right. \tag{3}$$
This system has two pairs of solutions $\left( x,y\right) =(2,2)$ and $
\left( x,y\right) =(16,9)$, but only $\left( x,y\right) =(2,2)$ is a
solution of $(1)$.
|
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|
Simplifying a simple fraction with exponent I am trying to simplify this fraction :
$
\dfrac{(3^2)(5^4)}{15^3}
$
The answer is : $ \dfrac{5}{3} $
I am trying to do the following: $ \dfrac{3^2}{15^3} \cdot \dfrac{5^4}{15^3} $ so ... $ \dfrac{1^{-3}}{3} \cdot \dfrac{1^1}{3} $
But that's not giving me the right answer, what am I doing wrong, please ?
Thanks !
|
For starters, $\displaystyle\frac{3^2}{15^3} \cdot \frac{5^4}{15^3} = \frac{(3^2)(5^4)}{(15^3)(15^3)} = \frac{(3^2)(5^4)}{15^6}$, not $\displaystyle\frac{(3^2)(5^4)}{15^3}$, so your first step isn’t right. Next, $\displaystyle\frac{3^2}{15^3}$ isn’t $\displaystyle\frac{1^{-3}}{3}$, and $\displaystyle\frac{5^4}{15^3}$ isn’t $\displaystyle\frac{1^1}{3}$; these mistakes show that you have some fundamental misconception, but I’m not sure exactly what it is.
Go back to the beginning and write out the fraction without any exponents: $$\begin{align*}
\frac{(3^2)(5^4)}{15^3} &= \frac{3\cdot 3\cdot 5\cdot 5\cdot 5\cdot 5}{15\cdot 15\cdot 15}\\&=\frac{3\cdot 3\cdot 5\cdot 5\cdot 5\cdot 5}{(3\cdot 5)(3\cdot 5)(3\cdot 5)}\\&=\frac{3\cdot 3\cdot 5\cdot 5\cdot 5\cdot 5}{3\cdot 3\cdot 3\cdot 5\cdot 5\cdot 5}\\&=\frac33\cdot\frac33\cdot\frac53\cdot\frac55\cdot\frac55\cdot\frac55\\&=1\cdot 1\cdot\frac53\cdot 1\cdot 1\cdot1\\&=\frac53.
\end{align*}$$ When you cancel, you’re really just getting rid of factors that are equal to $1$.
I suspect that you’re actually supposed to be learning to manipulate exponents at this point, but that manipulation is just a shortcut for what I did above. You use the law that $(ab)^n = a^nb^n$ to rewrite the denominator, $15^3$, as $(3\cdot 5)^3 = 3^3 \cdot 5^3$, making your fraction $\displaystyle\frac{3^2 \cdot 5^4}{3^3 \cdot 5^3}$; if you write that out in full, you have the fraction on the third line of the displayed expressions above. Then you split it, correctly this time, as $\displaystyle\frac{3^2}{3^3}\cdot\frac{5^4}{5^3}$. Now, finally, you can use the rule that $\displaystyle\frac{a^n}{a^m}=a^{n-m}$ twice to get $3^{2-3} \cdot 5^{4-3} = 3^{-1} \cdot 5^1 = \displaystyle\frac53$.
|
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|
Difficult Integral: $\int\frac{x^n}{\sqrt{1+x^2}}dx$ How to calculate this difficult integral: $\int\frac{x^2}{\sqrt{1+x^2}}dx$?
The answer is $\frac{x}{2}\sqrt{x^2\pm{a^2}}\mp\frac{a^2}{2}\log(x+\sqrt{x^2\pm{a^2}})$.
And how about $\int\frac{x^3}{\sqrt{1+x^2}}dx$?
|
Since $\frac{d}{dt}\sqrt{1+t^2} = \frac{t}{\sqrt{1+t^2}}$, we can integrate by parts to get
$$
\int \frac{t^2}{\sqrt{1+t^2}}\mathrm dt = \int t\cdot \frac{t}{\sqrt{1+t^2}}\mathrm dt
= t\sqrt{1+t^2} - \int \sqrt{1+t^2}\mathrm dt.
$$
Cheating a little bit by looking at a table of integrals, we get that since
$$
\frac{d}{dt} \left [ t\sqrt{1+t^2} + \ln(t + \sqrt{1+t^2}) \right ]
= t\frac{t}{\sqrt{1+t^2}} + \sqrt{1+t^2} + \frac{1}{t + \sqrt{1+t^2}}
\left [ 1 + \frac{t}{\sqrt{1+t^2}} \right ]
$$
which simplifies to $2\sqrt{1+t^2}$, the integral on the right above is
$\frac{1}{2}[t\sqrt{1+t^2} + \ln(t + \sqrt{1+t^2})]$ and thus we have
$$
\int \frac{t^2}{\sqrt{1+t^2}}\mathrm dt
= \frac{1}{2}\left [t\sqrt{1+t^2} - \ln(t + \sqrt{1+t^2})\right ]
$$
which matches the answer given by Charles Bao if we set $X=x$ and $a=1$ in his original post.
|
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|
Solving Radical Equations $x-7= \sqrt{x-5}$ This the Pre-Calculus Problem:
$x-7= \sqrt{x-5}$
So far I did it like this and I'm not understanding If I did it wrong.
$(x-7)^2=\sqrt{x-5}^2$ - The Square root would cancel, leaving:
$(x-7)^2=x-5$ Then I F.O.I.L'ed the problem.
$(x-7)(x-7)=x-5$
$x^2-7x-7x+14=x-5$
$x^2-14x+14=x-5$
$x^2-14x-x+14=x-x-5$
$x^2-15x+14=-5$
$x^2-15x+14+5=-5+5$
$x^2-15x+19=0$
$(x-1)(x-19)=0$
Now this is where I'm stuck because when I tried to see if I got the right numbers in the parentheses I got this....
$x^2-19x-1x+19=0$
$x^2-20x+19=0$
As you may see I'm doing something bad because I don't get $x^2-15x+19$
Could anyone please help me and tell me what I'm doing wrong?
|
$x^2-14x+49=x-5$
$x^2-15x+54=0$
$x^2-9x-6x+54=0$
$x(x-9)-6(x-9)=0$
$(x-6)(x-9)=0$
$x=6$, or $x=9$
|
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|
A Particular Two-Variable System in a Group Suppose $a$ and $b$ are elements of a group $G$.
If $a^{-1}b^{2}a=b^{3}$ and $b^{-1}a^{2}b=a^{3}$, prove $a=e=b$.
I've been trying to prove but still inconclusive. Please prove to me.
Thanks very much for proof.
|
In this answer I give credit to Jyrki Lahtonen for the answer he posted.
There are holes in his post, so I sensed the need for a step by step answer (firstly to convince myself, but also other people in doubt), and so here it is.
$\bbox[5px,border:2px solid]{\begin{array}{cc}a^3=b^{-1}a^2b&(\alpha)\\b^3=a^{-1}b^2a &(\beta)\end{array}}$
$b^{6}=b^3b^3=(a^{-1}b^2a)(a^{-1}b^2a)=a^{-1}b^4a\quad(\alpha)$
$b^{12}=b^6b^6=(a^{-1}b^4a)(a^{-1}b^4a)=a^{-1}b^8a\quad(\alpha)$
$b^{18}=b^{12}b^6=(a^{-1}b^8a)(a^{-1}b^4a)=a^{-1}b^{12}a\quad(\alpha)$
$b^{27}=b^{18}b^6b^3=(a^{-1}b^{12}a)(a^{-1}b^4a)(a^{-1}b^2a)=a^{-1}b^{18}a\quad(\alpha)$
$b^{27}=a^{-1}b^{18}a=a^{-2}b^{12}a^2=a^{-3}b^8a^3=\ (\beta)\ =(b^{-1}a^2b)^{-1}b^8(b^{-1}a^2b)=(b^{-1}a^{-2}b)b^8(b^{-1}b^{-1}a^2b)=b^{-1}a^{-2}b^8a^2b=b^{-1}(a^{-1}(a^{-1}b^8a)a)b=b^{-1}(b^{18})b=b^{18}$
$b^{27}=b^{18}\Rightarrow b^9=1$
$b^9=b^3b^3b^3=(a^{-1}b^2a)(a^{-1}b^2a)(a^{-1}b^2a)=(a^{-1}b^6a)=1\Rightarrow b^6=1\quad (\beta)$
$b^3=b^9b^{-6}=1=a^{-1}b^2a\Rightarrow b^2=1$
$b=b^3b^{-2}=1$ and then $a^3=b^{-1}a^2b=a^2\Rightarrow a=1\quad (\alpha)$.
$\bbox[5px,border:2px solid]{a=b=1}$
|
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|
If $a^3 =a$ for all $a$ in a ring $R$, then $R$ is commutative. Let $R$ be a ring, where $a^{3} = a$ for all $a\in R$. Prove that $R$ must be a commutative ring.
|
To begin with
$$ 2x=(2x)^3 =8 x^3=8x \ . $$
Therefore $6x=0 \ \ \forall x$.
Also
$$ (x+y)=(x+y)^3=x^3+x^2 y + xyx +y x^2 + x y^2 +yxy+ y^2 x + y^3 $$
and
$$ (x-y)=(x-y)^3=x^3-x^2 y - xyx -y x^2 + x y^2 +yxy+ y^2 x -y^3
$$
Subtracting we get
$$ 2(x^2 y +xyx+yx^2)=0 $$
Multiply the last relation by $x$ on the left and right to get
$$ 2(xy+x^2yx+xyx^2)=0 \qquad 2(x^2yx+xyx^2+yx)=0 \ . $$
Subtracting the last two relations we have
$$ 2(xy-yx)=0 \ . $$
We then show that $3( x+x^2)=0 \ \ \forall x$. You get this from
$$ x+x^2=(x+x^2)^3=x^3+3 x^4+3 x^5+x^6=4(x+x^2) \ . $$
In particular
$$ 3 (x+y +(x+y)^2) =3( x+x^2+ y+ y^2+ xy+yx)=0 \, $$
we end-up with $3(xy+yx)=0$. But since $6xy=0$, we have $3(xy-yx)=0$.
Then subtract $2(xy-yx)=0$ to get $xy-yx=0$.
|
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|
The $3 = 2$ trick on Google+ I found out this on Google+ yesterday and I was thinking about what's the trick. Can you tell?
How can you prove $3=2$?
This seems to be an anomaly or whatever you call in mathematics. Or maybe I'm just plain dense.
See this illustration:
$$ -6 = -6 $$
$$ 9-15 = 4-10 $$
Adding $\frac{25}{4}$ to both sides:
$$ 9-15+ \frac{25}{4} = 4-10+ \frac{25}{4} $$
Changing the order
$$ 9+\frac{25}{4}-15 = 4+\frac{25}{4}-10 $$
This is just like $a^2 + b^2 - 2a b = (a-b)^2$. Here $a_1 = 3, b_1=\frac{5}{2}$ for L.H.S, and $a_2 =2, b_2=\frac{5}{2}$ for R.H.S. So it can be expressed as follows:
$$ \left(3-\frac{5}{2} \right) \left(3-\frac{5}{2} \right) =
\left(2-\frac{5}{2} \right) \left( 2-\frac{5}{2} \right) $$
Taking positive square root on both sides:
$$ 3 - \frac{5}{2} = 2 - \frac{5}{2} $$
$$ 3 = 2 .$$
I think it's something near the root.
|
Back when I was in academia, I taught the "how to prove stuff" course, and one of the first problems that I'd give (which, I admit, I borrowed from my graduate adviser) was along the same vein, namely: criticize the "proof" of the following "theorem" or rethink your life!
"Theorem": You have all the money you need.
"Proof:" Let $M$ denote the amount of money you have and $N$ denote the amount of money you need. Let $A=\frac{M+N}{2}$ be the average of $M$ and $N$. Then, we have:
$2A=M+N$
$2A(M-N)=(M+N)(M-N)=M^2 - N^2$
$M^2-2AM = N^2-2AN$
$M^2-2AM + A^2 = N^2-2AN + A^2$
$(M-A)^2 = (N-A)^2$
And taking the square root of both sides, we have $M-A=N-A$, and hence $M=N$. $\blacksquare$
|
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|
Find $\int^{\frac{\pi}{4}}_{0} \tan^3{x}$ given $2\tan^3x = \frac{d}{dx}\left(\tan^2x+2\ln \cos x\right)$ Find $\int\nolimits^{\frac{\pi}{4}}_{0} ( \tan^3{x} ) \space dx$ given $2\tan^3x = \frac{d}{dx}( \tan^2x+2\ln \cos x )$
$$\int\nolimits^{\frac{\pi}{4}}_{0} \tan^3{x} \space dx = \frac{1}{2}\left[\tan^2{x} + 2 \ln{\cos{x}}\right]^{\frac{\pi}{4}}_0$$
I could go on but $\cos{\frac{\pi}{4}}$ is a decimal. Answer is $\frac{1}{2}(1 - \ln{2})$. How do I simplify down to that
|
You should know that $\cos\frac\pi4=\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. So $$\ln \cos\frac\pi4=\ln\frac{1}{\sqrt{2}} = -\ln\sqrt{2} = -\ln\left(2^{1/2}\right)= -\frac12\ln 2.$$
|
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|
Next number in series What are the basic/advanced strategies used to find the next number in series.
I know the simple ones such as addition, multiplication etc. But recently I came into a question that goes on something like 812, 819, 823, 835, 834, 851(Don't try to solve this, I changed some numbers and there is no sequence). This is to illustrate that the sequence goes up and goes down at 835 then goes back up. What strategies can you use to solve it?
These questions are for an aptitude test so you have about 60-90 secs to solve each question.
P.S. I can't really post the question that was asked as I might be disqualified from the job process if I post it.
|
About your series:
Let
$$
\begin{eqnarray*}
\begin{split}
P(x)&:= 812\frac{(x-2)(x-3)x-4)(x-5)(x-6)}{(1-2)(1-3)(1-4)(1-5)(1-6)}+819\frac{(x-1)(x-3)(x-4)(x-5)(x-6)}{(2-1)(2-3)(2-4)(2-5)(2-6)}\\
&+823\frac{(x-1)(x-2)(x-4)(x-5)(x-6)}{(3-1)(3-2)(3-4)(3-5)(3-6)} +835\frac{(x-1)(x-2)(x-3)(x-5)(x-6)}{(4-1)(4-2)(4-3)(4-5)(4-6)}\\
&+834\frac{(x-1)(x-2)(x-3)(x-4)(x-6)}{(5-1)(5-2)(5-3)(5-4)(5-6)}+851\frac{(x-1)(x-2)(x-3)(x-4)(x-5)}{(6-1)(6-2)(6-3)(6-4)(6-5)}
\end{split}
\end{eqnarray*}
$$
Then $P(x)$ is a polynomial which fits your data. This method is called Lagrange interpolation method.
BUT if $f(x)$ is any function defined on the integers, then
$$Q(x)=P(x)+(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)f(x)$$
also fits your data.
Most of these problems are actually asking for the simplest formula which fits that data, but really is there such a mathematical thing as "simplest". Lets say that our sequence is $2,3,5$ ( I picked only 3 to keep things simple). Which one would you say that it is the simplest description?
*
*$2^{n-1}+1$
*$\frac{n^2-n+4}{2}$
*the nth prime..
|
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|
If $n$ is an odd pseudoprime , then $2^n-1$ is also odd pseudoprime I have some problems understanding the following proof:
Definition: A composite number $n \in \mathbb{N}$ is called pseudo prime if $n \mid 2^{n-1}-1$ holds.
Theorem: If n is a odd pseudo prime number, then $2^n-1$ is also an odd pseudo prime number, too.
Proof: Let n be an odd pseudo prime number. Then we get $2^{n-1}-1=kn$ with $n \in \mathbb{N}$. It follows $2^{2^n-2}=2^{2kn}$ and further $2^{2^n-2}-1=(2^n)^{2k}-1$. We get $2^n-1 \mid 2^{2^n-2}-1$, but also $2^n-1 \mid 2^{2^n-1}-2$. Let $d \mid n$ with $1<d<n$, then it follows $2^d-1\mid 2^n-1$ and $1<2^d-1<2^n-1$. So $2^n-1$ is also an pseudo prime number.
Now my questions: What is the reason, that the author made this step: $2^{2^n-2}=2^{2kn}$, just to get rid of the "k"? The next step is also not clear: $2^{2^n-2}-1=(2^n)^{2k}-1$.
If anyone could explain the steps to me, I would be thankful. Greetings
|
This solution is essentially the same as the one quoted by OP, with some detail added and some removed. The most useful change is the introduction of $N$, which allows us to have one less level of exponentiation.
We have $2^{n-1}\equiv 1 \pmod{n}$. Let $N=2^n-1$. We first show that $N$ is composite. Since $n$ is composite, $n=ab$ for some $a>1$, $b>1$. Then $2^a-1$ is a non-trivial divisor of $2^n-1$.
Now we need to show that $2^{N-1} \equiv 1 \pmod{N}$. Since $n$ divides $2^{n-1}-1$, set $2^{n-1}-1=nk$. Multiply by $2$. We get $2^n-2=2nk$. But $2^n-2=N-1$, so $N-1=2nk$.
Thus $2^{N-1}=(2^n)^{2k}=(1+N)^{2k}\equiv 1 \pmod{N}$, which is the desired result.
Comment: One reason we let $N=2^n-1$ is that then the messy $2^{2^n-2}$ of the post becomes the understandable $2^{N-1}$, which is the object we need to compute modulo $N$ to show that $2^n-1$, that is, $N$, is a pseudo-prime to the base $2$.
|
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|
Finding $\lim\limits_{n \to \infty}{\frac{1^1+2^2+3^3+\cdots+n^n}{n^n}}$ $$\lim_{n \to \infty}{\frac{1^1+2^2+3^3+\cdots+n^n}{n^n}}.$$
With a first look this must give $1$ as a result but have a problem to explain it.
How can I do it?
Edit
I noticed that it is $\frac{\infty}{\infty}$.
$$\lim_{n \to \infty}{n^{n}\frac{(\frac{1^1}{n^{n}}+\frac{2^2}{n^{n}}+\frac{3^3}{n^{n}}+\cdots+1)}{n^n}}= \lim_{n \to \infty}{\frac{1^1}{n^{n}}+\frac{2^2}{n^{n}}+\frac{3^3}{n^{n}}+\cdots+1}=1$$
Is this correct?
|
We can write $(1^1+2^2+\cdots+n^n)/n^n$ as $a_n + b_n + 1$, where
$$
a_n = \frac{1^1+2^2+\cdots+(n-2)^{n-2}}{n^n} \text{ and } b_n = \frac{(n-1)^{n-1}}{n^n}.
$$
Both $a_n$ and $b_n$ are positive, and also
$$
a_n < \frac{(n-2)(n-2)^{n-2}}{n^n} < b_n < \frac{n^{n-1}}{n^n} = \frac1n.
$$
The squeeze theorem should allow you to prove that your answer of 1 is correct.
|
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|
How can I prove the inequality $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z}$? For $x > 0$, $y > 0$, $z > 0$, prove:
$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z} .$$
I can see that this is true, I also checked it with a few numbers. But I guess that is not enough to prove it. So how can I prove it?
(You don't need to show the whole proof, I think just a hint will be enough. I simply don't know how to start.)
|
$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z}$$
$\leftrightarrow (x+y+z)(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \geq 9$.
Using Cauchy Inequality, we have $x+y+z\geq 3\sqrt[3]{xyz}$ and $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\geq 3\sqrt[3]{\frac{1}{xyz}}$
$$\Rightarrow (x+y+z)\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \geq 3\sqrt[3]{xyz}\cdot 3\sqrt[3]{\frac{1}{xyz}} = 9$$.
Equality occurs when $x=y=z$.
|
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|
Finding the slope of the tangent line to $\frac{8}{\sqrt{4+3x}}$ at $(4,2)$ In order to find the slope of the tangent line at the point $(4,2)$ belong to the function $\frac{8}{\sqrt{4+3x}}$, I choose the derivative at a given point formula.
$\begin{align*}
\lim_{x \to 4} \frac{f(x)-f(4)}{x-4} &=
\lim_{x \mapsto 4} \frac{1}{x-4} \cdot \left (\frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{4+3 \cdot 4}} \right )
\\ \\ & = \lim_{x \to 4} \frac{1}{x-4} \cdot \left ( \frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{16}} \right ) \\ \\ & = \lim_{ x \to 4} \frac{1}{x-4} \cdot \left ( \frac{8}{\sqrt{4+3x}}-2\right)
\end{align*}$
But now I can't figure it out, how to end this limit.
I know that the derivative formula for this function is $-\frac{12}{(4+3x)\sqrt{4+3x}}$.
Thanks for the help.
|
The definition of the derivative, $\lim\limits_{x\to 4} \dfrac{f(x)-f(4)}{x-4}$ will always give you the indeterminate form $0/0$ if you plug in the number that $x$ is approaching. I.e. you get $\dfrac{f(4)-f(4)}{4-4}$.
So when you see
$$\lim_{ x \to 4} \frac{1}{x-4} \cdot \left ( \frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{3+3\cdot4}}\right)$$
what you want is to find a factor of $x-4$ in the numerator that will cancel the $x-4$ in the denominator. To do that, you want to write that difference of two fractions as just one fraction. For that you use a common denominator:
$$
\frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{3+3\cdot4}} = \frac{8}{\sqrt{4+3x}} - 2 = \frac{8}{\sqrt{4+3x}} - \frac{2\sqrt{4+3x}}{\sqrt{4+3x}} = \frac{8-2\sqrt{4+3x}}{\sqrt{4+3x}}
$$
When you plug $4$ into this, then as expected, you get $0$. Now rationalize the numerator:
$$
\frac{8-2\sqrt{4+3x}}{\sqrt{4+3x}} = \frac{8-2\sqrt{4+3x}}{\sqrt{4+3x}} \cdot \frac{8+2\sqrt{4+3x}}{8+2\sqrt{4+3x}} = \frac{64-4(4+3x)}{\sqrt{4+3x}(8+2\sqrt{4+3x})}
$$
$$
= \frac{-12(x-4)}{\sqrt{4+3x}(8+2\sqrt{4+3x})}
$$
When you multiply this by $\dfrac{1}{x-4}$, you get a cancellation, and then you can find the limit just by substituting $4$ for $x$.
|
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|
Proofs from the BOOK: Bertrand's postulate: $\binom{2m+1}{m}\leq 2^{2m}$ I have a very hard proof from "Proofs from the BOOK". It's the section about Bertrand's postulate, page 8:
It's about the part, where the author says:
$$\binom{2m+1}{m}\leq 2^{2m}$$
because $\binom{2m+1}{m}=\binom{2m+1}{m+1}$ are the same in $\sum \limits_{k=0}^{2m+1} \binom{2m+1}{k}=2^{2m+1}$
I see, why they are the same, but I don't see the reason to say $\binom{2m+1}{m}\leq 2^{2m}$. Any help would be fine :)
|
Since $2m+1$ is odd, there will be an even number of coefficients (since you have everything from $\binom{2m+1}{0}$ to $\binom{2m+1}{2m+1}$). Because the coefficients are symmetric, you can write:
$\sum \limits_{k=0}^{m} \binom{2m+1}{k}=\sum \limits_{k=m+1}^{2m+1} \binom{2m+1}{k}$
But $\sum \limits_{k=0}^{2m+1} \binom{2m+1}{k}=\sum \limits_{k=0}^{m} \binom{2m+1}{k} + \sum \limits_{k=m+1}^{2m+1} \binom{2m+1}{k}=2^{2m+1}$
$2 \times \sum \limits_{k=0}^{m} \binom{2m+1}{k} = 2^{2m+1}$
$\sum \limits_{k=0}^{m} \binom{2m+1}{k} = 2^{2m}$
Since $\binom{2m+1}{m}$ is part of the sum $\sum \limits_{k=0}^{m} \binom{2m+1}{k}$, it follows that $\binom{2m+1}{m}\leq 2^{2m}$, equality when $m=0$
|
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|
Using the Taylor expansion for ${(1+x)}^{-1/2}$, evaluate $\sum_{n=0}^\infty \binom{2n}{n} a^n$ Using the Taylor expansion for $${(1+x)}^{-1/2}$$ we have $${(1+x)}^{-1/2}= \sum_{n=0}^\infty \binom{-1/2}{n} (x^n)$$
for $|x|<1$.
But if $|a| <1$, how can we use the above fact to find
$$\sum_{n=0}^\infty \binom{2n}{n} a^n?$$
Thanks! Help much appreciated.
|
Write out what $\binom{-1/2}{n}$ means; i.e.,
$$
\begin{align}
\binom{-1/2}{n} &= \frac{(-1/2)(-3/2) \cdots ((-2n+1)/2)}{n!} = \frac{(-1)^n}{2^n} \frac{(1)(3) \cdots (2n-1)}{n!} \\
&= \frac{(-1)^n}{2^n} \frac{(2n)!}{2(4) \cdots (2n)n!} = \left(\frac{-1}{4}\right)^n \frac{(2n)!}{n!n!} \\
&= \left(\frac{-1}{4}\right)^n \binom{2n}{n}.
\end{align}
$$
This should be enough for you to be able to find $\sum_{n=0}^{\infty} \binom{2n}{n} a^n$.
Are you sure you mean $\sqrt{1+x}$, though? That would give $\sqrt{1+x} = \sum_{n=0}^{\infty} \binom{1/2}{n} x^n$. Then, following through the same argument as above you would obtain $\binom{1/2}{n} = \frac{-1}{2n-1} \left(\frac{-1}{4}\right)^n \binom{2n}{n}$, which would be a bit more difficult to deal with because of the $2n-1$ in the denominator.
|
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|
how to find this generating function this is the power series:
$$\sum_{i=0}^\infty n(n-1)^2 (n-2) z^n.$$
how can I find a generating function from it? I could use the third derivative but the $n-1$ is squared so I don't know what to do..
|
We can use
$$
\sum_{n=0}^\infty\binom{n}{k}z^n=\frac{z^k}{(1-z)^{k+1}}\tag{1}
$$
which follows from differentiating $\sum\limits_{n=0}^\infty z^n=\frac{1}{1-z}$ repeatedly $k$ times and multiplying by $\dfrac{z^k}{k!}$, and
$$
n(n-1)^2 (n-2)=n(n-1)(n-2)(n-3)+2n(n-1)(n-2)\tag{2}
$$
which is an example of the fact that any polynomial in $n$ can be written as a sum of $\binom{n}{k}$, we get
$$
\begin{align}
f(z)
&=\sum_{n=0}^\infty n(n-1)^2 (n-2) z^n\\
&=\sum_{n=0}^\infty n(n-1)(n-2)(n-3)z^n + 2n(n-1)(n-2)z^n\\
&=\sum_{n=0}^\infty 24\binom{n}{4}z^n + 12\binom{n}{3}z^n\\
&=\frac{24z^4}{(1-z)^5}+\frac{12z^3}{(1-z)^4}\\
&=\frac{12z^3(1+z)}{(1-z)^5}\tag{3}
\end{align}
$$
|
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|
Puzzle: numerical pattern recognition IF
7 - 3 = 10124
6 + 3 = 3279
5 – 2 = 763
11 + 2 = 92613
Then,
15 - 3 =?
Any ideas ? I dont know how I am supposed to go about solving puzzles like this one ? is there any strategy ? any algorithm ?
|
This is all I can come up with:
1: Do the operation, the answer to that will be placed on the far right:
$
\begin{align}
7-3 &= 4 \\
6 + 3 &=9 \\
5-2 &= 3 \\
11 + 2 &= 13
\end{align}$
2: Take the solution from above, multiply by the second digit in the original expression. Place it to the left of the answer we got in 1.:
$
\begin{align}
4\cdot 3 &= 12 \\
9\cdot 3&=27 \\
3\cdot 2&= 6 \\
13 \cdot 2 &= 26
\end{align}$
So, we have
$
\begin{align}
124 \\
279 \\
63 \\
2613
\end{align}$
3: If the original operation was addition, subtract, and vice versa. Stick this number to the left of the concatenated solutions to 1 and 2:
$
\begin{align}
7+3 &= 10 \\
6 - 3 &=3 \\
5+2 &= 7 \\
11 - 2 &= 9
\end{align}$
Thus, we have
$
\begin{align}
10124 \\
3279 \\
763 \\
92613
\end{align}$
|
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|
Integral Inequality Question I am currently working on the problem below and I am in need of help.
Consider the definite integral $\int_{1}^{2}\frac{1}{t}dt$.
(a)By the dividing the interval $1\leq t\leq 2$ into $n$ equal parts and choosing appropriate sample points, show that
$$\sum_{j=1}^{n}\frac{1}{n+j}< \int_{1}^{2}\frac{1}{t}dt< \sum_{j=0}^{n-1}\frac{1}{n+j}$$
(b)How large should $n$ be to approximate $\int_{1}^{2}\frac{1}{t}dt$ with an error of at most $5(10^{-6}) $ using one of the sums in part (a)?
Hint: What is the difference between the underestimate and over estimate?
|
For part (a), we divide the interval $[1,2]$ into $n$ intervals which only interests at boundary points, i.e. $[1+\frac{j-1}{n},1+\frac{j}{n}]$ where $1\leq j\leq n$. Note that $\frac{1}{t}$ is an decreasing function, which implies that for $1\leq j\leq n$
$$\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{1+\frac{j}{n}}dt<\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{t}dt<\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{1+\frac{j-1}{n}}dt$$
where
$$\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{1+\frac{j}{n}}dt=\frac{1}{1+\frac{j}{n}}\cdot\Big[(1+\frac{j}{n})-(1+\frac{j-1}{n})\Big]=\frac{1}{1+\frac{j}{n}}\cdot\frac{1}{n}=\frac{1}{n+j}$$
and
$$\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{1+\frac{j-1}{n}}dt=\frac{1}{1+\frac{j-1}{n}}\cdot\Big[(1+\frac{j}{n})-(1+\frac{j-1}{n})\Big]=\frac{1}{1+\frac{j-1}{n}}\cdot\frac{1}{n}=\frac{1}{n+j-1}.$$
That is
$$\frac{1}{n+j}<\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{t}dt<\frac{1}{n+j-1}\mbox{ for }1\leq j\leq n.$$
Now summming up the above inequalities give the result:
$$\sum_{j=1}^n\frac{1}{n+j}<\sum_{j=1}^n\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{t}dt=\int_1^2\frac{1}{t}dt<\sum_{j=1}^n\frac{1}{n+j-1}=\sum_{j=0}^{n-1}\frac{1}{n+j}.$$
For part (b), it follows from part (a) that
$$0<\int_1^2\frac{1}{t}dt-\sum_{j=1}^n\frac{1}{n+j}<\sum_{j=0}^{n-1}\frac{1}{n+j}-\sum_{j=1}^n\frac{1}{n+j}=\frac{1}{n}-\frac{1}{2n}=\frac{1}{2n}.$$
To approximate $\int_1^2\frac{1}{t}dt$ with an error of at most $5(10^{−6})$ using the sum $\sum_{j=1}^n\frac{1}{n+j}$ in part (a), we must have
$$\frac{1}{2n}\leq 5(10^{−6}),$$
that is $n\geq 10^5$.
|
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|
Expected value and Variance of $Y=\frac{1}{a} X-b$ where $X \sim N(\mu, \sigma^2)$ I absolutely know I am not doing this right. :[
Could I get some input or point back in the right direction?
My work done so far is shown below.
Let $X$ be a normal random variable with parameters $N(\mu, \sigma^2)$. Please find the Expected value and Variance of random variable $Y=\frac{1}{a} X-b$, where $a$ and $b$ are constant values.
My work.
$$
\begin{align*}
E(Y) &= aE(x)-b = \sum_x \Big(\frac{1}{a} x - b \Big) p_x(x) = \frac{1}{a} \sum_x x p_x(x) - b
\\ &= \frac{1}{a} \sum_x x p_x(x) - b \sum_x p_x(x) = a E(x) - b \cdot 1.
\end{align*}
$$
If $a = 0$, then $E(x-b) = E(x)$ and if $b = 0$, then $E(ax)= \frac{1}{a}E(x)$.
$$
\begin{align*}
\mu &= E(X) = \int_{-\infty}^{\infty} x f(x) dx = \int_{0}^{1} x \Big(\frac{1}{a} X - b \Big) dx = \frac{1}{a} \int_0^1 X^2 - xb ~dx
\\ & = \frac{1}{a} \Big( \frac{x^3}{3} - \frac{bx^2}{2} \Big) \text{ from } 1 \text{ to } 0
\\ & = \frac{1}{a} \Big( \frac{1}{3} - \frac{b}{2} \Big) .
\end{align*}
$$
$$
\text{RV } Y = (X - E(x)^2).
$$
$$
\sigma^2 = \operatorname{Var}(x) = E[(x) - E[x])^2]
$$
$$
\operatorname{Var} \Big( \frac{1}{a} X - b \Big) = a^2 \operatorname{Var}(x).
$$
$$
\int_{\Box}^{\Box}\Big( \frac{1}{a} X - b \Big) - \Big( \frac{1}{a} X - b \Big)^2 \ldots
$$
|
$$
\operatorname{var}\left( \frac 1 a X - b\right) = \operatorname{var}\left( \frac 1 a X \right) = \frac 1 {a^2} \operatorname{var}(X).
$$
$$
\operatorname E\left( \frac 1 a X-b\right) = \frac 1 a \operatorname E(X) - b.
$$
|
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|
Last coupon collected in the coupon collectors problem Consider the classical coupon collectors problem. Given a particular coupon $i$ we can ask for the probability that $i$ is the last coupon collected. We asked this question on cstheory and got a wonderful answer from James Martin introducing us to the idea of Poissonization in coupon collecting:
$Pr(L=i) = \int_{0}^{\infty} n_i e^{n_i t} \prod_{j\neq i} (1-e^{n_j t}) dt$
where $L$ is the last coupon collected and $n_i$ is the number of coupons of type $i$, with $\sum_i n_i = k$.
For example:
$\int_0^\infty a e^{-a t} \left(1-e^{-b t}\right) \left(1-e^{-c t}\right) \left(1-e^{-d t}\right) \left(1-e^{-e t}\right) \left(1-e^{-f t}\right) \, dt$,
gives the following in Mathematica:
$a \left[ \frac{1}{a}-\frac{1}{a+b}-\frac{1}{a+c}+\frac{1}{a+b+c}-\frac{1}{a+d}+\frac{1}{a+b+d}+\frac{1}{a+c+d}-\frac{1}{a+b+c+d}- \right.$
$\left. \frac{1}{a+e}+\frac{1}{a+b+e}+\frac{1}{a+c+e}-\frac{1}{a+b+c+e}+\frac{1}{a+d+e}-\frac{1}{a+b+d+e}-\frac{1}{a+c+d+e}+ \right.$
$\left. \frac{1}{a+b+c+d+e}-\frac{1}{a+f}+\frac{1}{a+b+f}+\frac{1}{a+c+f}-\frac{1}{a+b+c+f}+\frac{1}{a+d+f}-\frac{1}{a+b+d+f}- \right.$
$\left. \frac{1}{a+c+d+f}+\frac{1}{a+b+c+d+f}+\frac{1}{a+e+f}-\frac{1}{a+b+e+f}-\frac{1}{a+c+e+f}+\frac{1}{a+b+c+e+f}- \right.$
$\left. \frac{1}{a+d+e+f}+\frac{1}{a+b+d+e+f}+\frac{1}{a+c+d+e+f}-\frac{1}{a+b+c+d+e+f} \right]$.
The trouble is that this expression takes exponential time in the number of coupons to evaluate and has an exponential number of terms. We are looking for a way to bound this asymptotically in $k$, but any further references or information would be greatly appreciated.
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I have no idea if this bound will be of any use to you or not (or even how it compares to the inclusion-exclusion bound already posted), but you can get a quick(er) upper bound as follows. Without loss of generality let's assume the coupon we care about is coupon $k$ and that $n_1 \leq n_2 \leq \dots \leq n_{k-1}$. We can view the event that coupon $k$ is last as the intersection of $k-1$ different events.
$E_1$: The first coupon selected is not of type $k$.
$E_2$: The first coupon selected which does not match the first type of coupon drawn is not of type $k$.
...
$E_{n-1}$ the first coupon drawn does not match any of the first $n-2$ types of coupons drawn is not of type $k$.
The probability of $E_1$ is just $(1-\frac{n_k}{n})$. $P(E_2)$ is more complicated, but we know an expression for the probability of $E_2$ given that the first coupon drawn is of type $j$: $(1-\frac{n_k}{n-n_j})$, which is largest for $j=1$. It follows by averaging over all $j$ that
$$P(E_2 | E_1) \leq 1-\frac{n_k}{n-n_1}$$
In general, we have
$$P(E_j| E_1, E_2, \dots, E_{j-1}) \leq 1-\frac{n_k}{n-n_1-n_2-\dots-n_{j-1}}$$
And you can multiply this bound over all $1 \leq j \leq n-1$ to get a bound on the probability that $k$ is last:
$$P(k \textrm{ last }) \leq \prod_{i=1}^{j-1} \left(1-\frac{n_k}{n-S_{k-1}}\right)$$
where $S_k=n_1+\dots+n_k$.
This bound is exact when all of the $n_j$ are equal, but in general it gets worse the further apart the $n_j$ are.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/87120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
}
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