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Asymptotic of integral I have this definite integral:
$$ \int_{0}^{\frac \pi 2} \left( 1 + t^2 \cot^2\theta \right )^{-\frac{n-1}{2}} d \theta , t\in (0,1)$$
where $n$ is an integer. I need to find the asymptotic as a function on $n$.
I suspect it should be $O(\frac{1}{\sqrt{n}})$ but wasn't able to complete the calculation.
Any ideas?
|
One may write:
$$
\begin{align}
I(n)=\int_{0}^{\large \frac \pi 2} \left( 1
+ t^2 \cot^2\theta \right )^{-\frac{n-1}{2}} d \theta&=\int_{0}^{\large \frac \pi 2} \left( 1 + t^2 \tan^2\theta
\right )^{-\frac{n-1}{2}} d \theta \\\\
&=\int_{0}^{\infty}
\frac{1}{1+x^2}\left( 1 + t^2 x^2 \right )^{-\frac{n-1}{2}} dx
\end{align}
$$
Upper bound ($t\leq1$):
$$
\begin{align}
I(n) &=\int_{0}^{\infty}
\frac{1}{1+x^2}\left( 1 + t^2 x^2 \right )^{-\frac{n-1}{2}} dx \\\\
&\leq\int_{0}^{\infty}
\frac{1}{1+t^2x^2}\left( 1 + t^2 x^2 \right )^{-\frac{n-1}{2}} dx \\\\
&= \int_{0}^{\infty}
\left( 1 + t^2 x^2 \right )^{-\frac{n+1}{2}} dx \\\\
&= \frac1t \int_{0}^{\infty}
\left( 1 + x^2 \right )^{-\frac{n+1}{2}} dx = \frac1t f(n)
\end{align}
$$
Lower bound(for $t\leq 1$):
$$
\begin{align}
I(n) &\geq \int_{0}^{\infty}
\frac{1}{1+x^2}\left( 1 + x^2 \right )^{-\frac{n-1}{2}} dx \\\\
&= \int_{0}^{\infty}
\left( 1 + x^2 \right )^{-\frac{n+1}{2}} dx =f(n)\\\\
\end{align}
$$
Thus:
$$ f(n) \leq I(n) \leq \frac1t f(n)$$
which is valid as a finite approximation when $t$ is strictly larger than 0.
A recursion can be obtain for $f(n)$. Laplace approximation for $f(n)$ would give that $f(n) \approx \sqrt{\pi \over 2n}$ which show that the upper bound can be tight.
|
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|
What values of x does $\sum_{n=1}^{\infty} \frac{1}{n} (\frac{3}{(-1)^n+2})^n x^n$ converge for? In a past year paper there was a question asking to find the values of x in which $\sum_{n=1}^{\infty} \frac{1}{n} (\frac{3}{(-1)^n+2})^n x^n$ converges.
I tried using the root test, and so:
$\lim_{n \rightarrow \infty} (\frac{1}{n} (\frac{3}{(-1)^n+2}x^{n})^{1/n} = \lim_{n\rightarrow \infty} \frac{1}{n^{1/n}} \frac{3}{-1+2^{1/n}} x$
And so we have a problem because $lim_{n \rightarrow \infty} 2^{1/n} = 1$ and we will get 0 in the denominator.
Help would be much appreciated!
|
First note that the coefficients of this power series is
$$a_n=\frac{1}{n}\left(\frac{3}{(-1)^n+2}\right)^n=\begin{cases}\frac{3^n}{n},&n\text{ odd}\\
\frac{1}{n},&n\text{ even}\end{cases}.$$
Then for sufficiently large $n$
$$\frac{3}{n^\frac{1}{n}}\leq \sup_{N\geq n}|a_N|^{\frac{1}{N}}\leq 3.$$
Thus the radius of convergence is
$$R=\frac{1}{\limsup_{n\to\infty}|a_n|^{\frac{1}{n}}}=\frac{1}{3}.$$
Now we consider the case $|x|=\frac{1}{3}$. It's obvious that the series diverges when $x=\frac{1}{3}$. For $x=-\frac{1}{3}$, then the odd terms of the series are $-\frac{1}{n}$ and the even terms are $\frac{1}{n3^n}$, and the series also diverges in this case.
All in all, the series converges precisely when $|x|<\frac{1}{3}$.
|
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|
How to show $\frac{2x}{2+x}<\log(x+1)$
How to show $\frac{2x}{2+x}<\log(x+1)$ for $x>0$
Without differentiating, more elementary (it looks then more complicated but OK)
$\log(x+1)=\int\limits_1^{x+1}\frac 1udu$ and $\frac{2x}{2+x}=x\frac{1}{ 1+\frac x2}$ hence
$\log(x+1)-\frac{2x}{2+x}=\int\limits_1^{x+1}\frac 1u-\frac{1}{ 1+\frac x2}du=\int\limits_1^{x+1}\frac{1+\frac x2-u}{u\left(1+\frac x2\right)}du=\int\limits_1^{\frac x2+1}\frac{1+\frac x2-u}{u\left(1+\frac x2\right)}du-\int\limits_{\frac x2+1}^{x+1}\frac{u-\left(1+\frac x2\right)}{u\left(1+\frac x2\right)}du$
The numerator of the integrand in the first integral is positive and negative in the $2$nd. But for the first one the denominatior is smaller, so the fraction is always bigger than its corresponding fraction in the second one, and so the difference is always positive is this correct, do you have an alternative proof
|
Note that
$$ \frac{4}{(x+2)^2}-\frac{1}{x+1}=-\frac{x^2}{(x+1)(x+2)^2}<0$$
and hence
$$ \frac{4}{(x+2)^2}<\frac{1}{x+1}. $$
Integrating from $0$ to $x$ ($x>0$), one has
$$ \int_0^x\frac{4}{(t+2)^2}dt<\int_0^x\frac{1}{t+1}dt $$
which gives
$$ \frac{2x}{2+x}<\ln(x+1). $$
Done.
|
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Prove that ${4\over \pi}=\prod_{k=1}^{\infty}\left(1+{1\over 4k}\right)^2\left(2k+1\over 2k+1+(-1)^{k-1}\right)^{(-1)^{k-1}}$ How do I prove this infinite product?
$${4\over \pi}=\prod_{k=1}^{\infty}\left(1+{1\over 4k}\right)^2\left(2k+1\over 2k+1+(-1)^{k-1}\right)^{(-1)^{k-1}}$$
I try:
$$\prod_{k=1}^{\infty}\left(1+{1\over 4k}\right)^2={5\over 4}\cdot{9\over 8}\cdot{13\over 12}\cdots={144\over 121}$$
$$\prod_{k=1}^{\infty}\left(2k+1\over 2k+1+(-1)^{k-1}\right)^{(-1)^{k-1}}={3\over 4}\cdot{4\over 5}\cdot{7\over 8}\cdot{8\over 9}\cdots$$
Simplified to get
$$\prod_{k=1}^{\infty}\left(2k+1\over 2k+1+(-1)^{k-1}\right)^{(-1)^{k-1}}={3\over 1}\cdot{1\over 5}\cdot{7\over 1}\cdot{1\over 9}\cdot{11\over 1}\cdot{1\over 13}\cdots$$
How do I combine theses two products to show that it is a Wallis's product? This is as far I can go.
|
Hint. Let's set
$$
P_n:=\prod_{k=1}^{n}\left(1+{1\over 4k}\right)^2\left(2k+1\over 2k+1+(-1)^{k-1}\right)^{(-1)^{k-1}}.
$$ One may observe that
$$
P_{2n+1}=\left(1+{1\over 4(2n+1)}\right)^2\left(4n+3\over 4n+4\right)P_{2n}, \quad n\ge0,
$$ since
$$
\lim_{n \to \infty}\left(1+{1\over 4(2n+1)}\right)^2\left(4n+3\over 4n+4\right)=1
$$thus, if the limits exist, we have
$$
\lim_{n \to \infty}P_{2n+1}=\lim_{n \to \infty}P_{2n}.
$$
Then one has, as $n \to \infty$,
$$
\begin{align}
P_{2n}&=\prod_{k=1}^{2n}\left(1+{1\over 4k}\right)^2\left(2k+1\over 2k+1+(-1)^{k-1}\right)^{(-1)^{k-1}}
\\\\&=\prod_{k=1}^{2n}\left(1+{1\over 4k}\right)^2\cdot\prod_{k=1}^{2n}\left(2k+1\over 2k+1+(-1)^{k-1}\right)^{(-1)^{k-1}}
\\\\&=\left(\frac{\Gamma\left(2n+\frac{5}{4}\right)}{\Gamma\left(\frac{5}{4}\right) \Gamma\left(2n+1\right)}\right)^2\cdot \prod_{p=1}^{n}\left(4p+1\over 4p\right)^{-1}\prod_{p=1}^{n}\left(4p-1\over 4p\right)
\\\\&=\left(\frac{\Gamma\left(2n+\frac{5}{4}\right)}{\Gamma\left(\frac{5}{4}\right) \Gamma\left(2n+1\right)}\right)^2\cdot\frac{\Gamma\left(\frac{5}{4}\right)\Gamma\left(n+\frac{3}{4}\right)}{\Gamma\left(\frac{3}{4}\right)\Gamma\left(n+\frac{5}{4}\right)}
\\\\& = \frac{4}\pi+\mathcal{O}\left(\frac1n \right)
\end{align}
$$ where we have used the generalized Stirling formula.
|
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Prob. 4, Chap. 3 in Baby Rudin: How to show that these are the limit superior and the limit inferior? Here's Prob. 4, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Find the upper and lower limit of the sequence $\left\{ s_n \right\}$ defined by $$s_1 = 0; \ s_{2m} = \frac{s_{2m-1}}{2}; \ s_{2m+1} = \frac{1}{2}+ s_{2m}.$$
My effort:
We note that $s_1 = 0$, $s_2 = 0$, $s_3 = \frac{1}{2}$, $s_4 = \frac{1}{4}$, $s_5 = \frac{1}{2} + \frac{1}{4}$, $s_6 = \frac{1}{4} + \frac{1}{8}$, $s_7 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8}$, $s_8 = \frac{1}{4} + \frac{1}{8} + \frac{1}{16}$, $s_9 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}$, $s_{10} = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32}$, $s_{11} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32}$, $s_{12} = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64}$, and continuing in this way we obtain
$$s_{2m} = \frac{1}{2}- \frac{1}{2^m} \ \mbox{ and } \ s_{2m+1} = 1- \frac{1}{2^m} $$ for $m = 0, 1, 2, 3, \ldots$. So $$\lim_{m \to \infty} s_{2m} = \frac{1}{2} \ \mbox{ and } \ \lim_{m \to \infty} s_{2m+1} = 1.$$
Am I right?
Using the above, how to rigorously prove that $$\lim\inf_{n \to \infty} s_n = \frac{1}{2} \ \mbox{ and } \ \lim\sup_{n\to\infty} s_n = 1?$$
|
Hint. Show inductively that
$$
s_{2n}=\frac{1}{2}-\frac{1}{2^n}\quad\text{and}\quad s_{2n+1}=1-\frac{1}{2^n}.
$$
|
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Computing $\lim_{x\to0} \frac 8 {x^8} \left[ 1 - \cos\frac{x^2} 2 - \cos\frac{x^2}4 + \cos\frac{x^2}2\cos\frac{x^2}4 \right]$ without using L'Hospital We have to find the following limit.
$$\lim_{x\to0} \frac 8 {x^8} \left[ 1 - \cos\frac{x^2} 2 - \cos\frac{x^2}4 + \cos\frac{x^2}2\cos\frac{x^2}4 \right]$$
In this I thought to use Lhopital . But using that it will become too long . Is there ny short method .
|
Beside the good hint by lab bhattacharjee, as Henry W commented, Taylor series make the problem quite simple.
Starting with $$\cos(y)=1-\frac{y^2}{2}+\frac{y^4}{24}+O\left(y^6\right)$$ and replacing $y$ successively by $\frac {x^2} 2$ and $\frac {x^2} 4$ $$\cos \left(\frac{x^2}{2}\right)=1-\frac{x^4}{8}+\frac{x^8}{384}+O\left(x^{12}\right)$$ $$\cos \left(\frac{x^2}{4}\right)=1-\frac{x^4}{32}+\frac{x^8}{6144}+O\left(x^{12}\right)$$ $$1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right)\cos \left(\frac{x^2}{4}\right)=\frac{x^8}{256}+O\left(x^{12}\right)$$
|
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Determining the number of ways a number can be written as sum of three squares I was going through Erich Friedman's "What's Special About This Number?" and there some numbers are classified based on the number of ways we can write them as sum of squares. I want to prove the following claim by Friedman:
129 is the smallest number that can be written as the sum of 3 squares
in 4 ways.
Indeed, as given in Wikipedia,
$$11^2+2^2+2^2 = 10^2+5^2+2^2 = 8^2+8^2+1^2 = 8^2+7^2+4^2 = 129$$
So what remains to prove is that this is the smallest such number.
Is it possible to write a proof for this fact using some insights
along with brute force/cases? How can we solve this problem using only
brute-force?
Also, since I know the proof of Legendre's three-square theorem. I am also curious to know:
How can we determine the number of ways we can write a non-negative
integer which satisfies Legendre's three-square theorem as sum of
three squares?
Edit1: Related discussions on MathOverflow:
*
*Is there a simple way to compute the number of ways to write a positive integer as the sum of three squares?: Note that this is not answer of my question since $r_k(n)$ counts the number of representations of $n$ by $k$ squares, allowing zeros and distinguishing signs and order.
*Efficient computation of integer representation as sum of three squares
Edit2: Related discussions on ComputerScience.SE
*
*Listing integers as the sum of three squares $m=x^2+y^2+z^2$
Edit3: Related discussions on Mathematics.SE
*
*When is a rational number a sum of three squares?
*Why can't this number be written as a sum of three squares of rationals?
*Sum of one, two, and three squares
|
For the system of equations.
$$x_1^2+x_2^2+x_3^2=x_4^2+x_5^2+x_6^2=x_7^2+x_8^2+x_9^2=x_{10}^2+x_{11}^2+x_{12}^2$$
Solutions can be parameterized.
$$x_1=a(y^2+z^2+n^2)(p^2+s^2+n^2)(k^2+t^2+n^2)$$
$$x_2=b(y^2+z^2+n^2)(p^2+s^2+n^2)(k^2+t^2+n^2)$$
$$x_3=c(y^2+z^2+n^2)(p^2+s^2+n^2)(k^2+t^2+n^2)$$
$$x_4=(ay^2-2byn+az^2-2czn-an^2)(p^2+s^2+n^2)(k^2+t^2+n^2)$$
$$x_5=(bz^2-2cyz-by^2-2ayn+bn^2)(p^2+s^2+n^2)(k^2+t^2+n^2)$$
$$x_6=(cy^2-2byz-cz^2-2azn+cn^2)(p^2+s^2+n^2)(k^2+t^2+n^2)$$
$$x_7=(ap^2-2bpn+as^2-2csn-an^2)(y^2+z^2+n^2)(k^2+t^2+n^2)$$
$$x_8=(bs^2-2cps-bp^2-2apn+bn^2)(y^2+z^2+n^2)(k^2+t^2+n^2)$$
$$x_9=(cp^2-2bps-cs^2-2asn+cn^2)(y^2+z^2+n^2)(k^2+t^2+n^2)$$
$$x_{10}=(ak^2-2bkn+at^2-2ctn-an^2)(y^2+z^2+n^2)(p^2+s^2+n^2)$$
$$x_{11}=(bt^2-2ckt-bk^2-2akn+bn^2)(y^2+z^2+n^2)(p^2+s^2+n^2)$$
$$x_{12}=(ck^2-2bkt-ct^2-2atn+cn^2)(y^2+z^2+n^2)(p^2+s^2+n^2)$$
It is interesting that such triples can be too much. The formula can be increased to any number. That is the same to write not only for 4 partitions, but for any number. The main thing that all the variables were not identical to each other.
|
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Prove that if $10 \mid (a^2+ab+b^2)$, then $1000 \mid (a^3-b^3)$.
Let $a$ and $b$ be integers. Prove that if $10 \mid (a^2+ab+b^2)$, then $1000 \mid (a^3-b^3)$.
I saw that $a^3-b^3 = (a-b)(a^2+ab+b^2)$ and $(a^2+ab+b^2) = (a+b)^2-ab$. How can we use the fact that $10 \mid (a^2+ab+b^2)$ to solve this question?
|
$10 \mid (a^2+ab+b^2)$ implies $10 \mid a^3-b^3 = (a-b)(a^2+ab+b^2)$, that is, $a^3 \equiv b^3 \bmod 10$.
Now, $x \mapsto x^3$ is a bijection mod $10$. Therefore, $a^3 \equiv b^3 \bmod 10$ implies $a \equiv b \bmod 10$.
Then $a^2+ab+b^2 \equiv 3a^2$ implies $10 \mid a$ and so $10 \mid b$.
Therefore, $1000 \mid a^3$ and $1000 \mid b^3$, hence $1000 \mid (a^3-b^3)$.
|
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Compute $\frac{p(12)+p(-8)}{10}$ where $p(x)=x^4+ax^3+bx^2+cx+d$ I have got an olympiad problem which is as follow:
Compute $\frac{p(12)+p(-8)}{10}$ where $p(x)=x^4+ax^3+bx^2+cx+d$ and $p(1)=10$, $p(2)=20$, $p(3)=30$.
I have been told that answer is $1984$.
I thought applying the values and getting a relation between $a,b,c,d$ from $p(1), p(2), p(3)$ will suffice but the problem is that I will end up with with 4 variables and three equations.
I m a newbie to polynomials and I don't think I m gonna get the answer. So, please help me in this. Thanks.
|
Let $q(x) = p(x)-10x$. Then, $q(x)$ is a monic polynomial, and $q(1) = q(2) = q(3) = 0$.
So the roots of $q(x)$ are $x = 1, 2, 3$ and $r$ for some real number $r$.
Hence, we can write $q(x) = (x-1)(x-2)(x-3)(x-r)$.
Thus, $p(x) = (x-1)(x-2)(x-3)(x-r)+10x$ for some real number $r$.
So, $p(12) = 11 \cdot 10 \cdot 9 \cdot (12-r) + 120 = 990(12-r)+120$, and $p(-8) = (-9) \cdot (-10) \cdot (-11) \cdot (-8-r) - 80 = -990(-8-r)-80$.
Therefore, $\dfrac{p(12)+p(-8)}{10} = \dfrac{990(12-r)+120-990(-8-r)-80}{10} = 1984$.
|
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Show that:$\prod_{n=1}^{\infty}\left({2n\over 2n-1}\right)^2\left(\cdots\right)={\pi\over2}\cdot{\phi\over5}\cdot\sqrt{\phi\sqrt{5}}$ $\phi$ is the golden ratio
Show that
$$\prod_{n=1}^{\infty}\left({2n\over 2n-1}\right)^2\left(10n-6\over 10n-1\right)\left(10n-4\over 10n+1\right)={\pi\over2}\cdot{\phi\over5}\cdot\sqrt{\phi\sqrt{5}}$$
I try:
$$\prod_{n=1}^{\infty}\left({2n\over 2n-1}\right)^2\left(10n-6\over 10n-1\right)\left(10n-4\over 10n+1\right)={\phi\over5}\cdot\sqrt{\phi\sqrt{5}}\prod_{n=1}^{\infty}\left({2n\over 2n-1}\cdot{2n\over 2n+1}\right)$$
$$\prod_{n=1}^{\infty}\left({2n+1\over 2n-1}\right)\left(10n-6\over 10n-1\right)\left(10n-4\over 10n+1\right)={\phi\over5}\cdot\sqrt{\phi\sqrt{5}}$$
$$\lim_{M\to \infty}(2M+1)\prod_{n=1}^{M}\left(10n-6\over 10n-1\right)\left(10n-4\over 10n+1\right)={\phi\over5}\cdot\sqrt{\phi\sqrt{5}}$$
I can't go any further. Please help!
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Rewrite this as the product of $$\frac{(2n)^2}{(2n-1)(2n+1)} \cdot \frac{(10n)^2}{(10n-1)(10n+1)} \cdot \frac{(10n/4-1)(10n/4+1)}{(10n/4)^2} \cdot \frac{(2n+1)(10n-6)}{(2n-1)(10n+4)}.$$ Using the infinite product representation of sine, you can show that $$\frac{\pi}{m \sin (\pi / m)} = \prod_{n=1}^{\infty} \frac{(mn)^2}{(mn-1)(mn+1)}$$ which takes care of the first three products. The fourth product telescopes to $4/5$. So the result is $$\frac{\pi}{2\sin(\pi/2)} \cdot \frac{\pi}{10 \sin (\pi/10)} \cdot \frac{5 \sin (2\pi/5)}{2\pi} \cdot \frac{4}{5}$$ in which you can use known values for $\sin(2\pi/5)$ and $\sin(\pi/10).$
|
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In how many $4$-digit numbers the sum of two right digits is equal to the sum of two left digits In how many $4$-digit numbers the sum of two right digits is equal to the sum of two left digits.
My attempt:We should find number of two pairs that can be digits of this number for choosing the place of digits we have $*8$ (We should notice we cant have $0$ in the beginning.But the biggest problem is finding such pairs.How can I find them?
|
This problem can be solved using Stars and Bars method.
Let the digits be $a,b,c,d$.
Then we must find integral solutions of $$a+b=c+d \Rightarrow a+b-c-d=0$$
with restrictions on $a,b,c,d\ $ i.e $1\le a\le 9$ and $0\le \{b,c,d\}\le 9$
The equation can be transformed by taking $a=x_1+1,\ b=x_2,\ c=9-x_3,\ d=9-x_4$ to:
$$\begin{align}&x_1+1+x_2+x_3-9+x_4-9=0\\\Rightarrow \ \ &x_1+x_2+x_3+x_4=17\end{align}$$
with $0\le x_1\le 8$, $0\le \{x_2,x_3,x_4\}\le 9$
The number of integral solutions of above equation can be solved using the Stars and Bars method.
Try yourself or see below for rest of the procedure:
Number of solutions for $$x_1+x_2+x_3+ x_4 = 17\\$$ where $x_i \ge 0$
for $i = 1,2,3,4$ is $\binom{17+4-1}{17}=\binom{20}{3}$ solutions.
We have to remove cases where $x_1\ge9$ and $\{x_2,x_3,x_4\}\ge10$,
let's say $x_1$ for example, is $\ge 9$, then we can find all the
number of solutions to be excluded by putting $y_1 = x_1 - 9$, we can
write $$y_1 + x_2 +x_3+x_4 = 8$$ which has
$\binom{8+4-1}{8}=\binom{11}{3}$ solutions.
Similarly for $x_i\ge 10$ for $i=2,3,4$, then we can find all the
number of solutions to be excluded by putting $y_i = x_i - 10$, we can
write, for e.g. $x=2$, $$x_1 + y_2 +x_3+x_4 = 7$$ which has
$\binom{7+4-1}{8}=\binom{10}{3}$ solutions.
Hence total number of such 4-digit numbers is:
$$\binom{20}{3}-\binom{11}{3}-3\times\binom{10}{3}=615$$
|
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|
Finding the splitting field of the polynomial $x^5+2x^4+5x^2+x+4$ over $F_{11}$. I am working on the following problem:
Find the splitting field of the polynomial $f(x)=x^5+2x^4+5x^2+x+4$ over $F_{11}$.
What I have done:
So far I found that $-2$ is the only root of $f(x)$ in $F_{11}$. I have also factored $f(x)$ as $f(x)=(x+2)(x^2+5x+1)(x^2-5x+2)$ over $F_{11}$.
Apparently the splitting field of $f(x)$ over $F_{11}$ is $F_{11^4}$. Can someone help me understand how one can conclude that.
|
The polynomial $x^2+5x+1$ has no roots in $F_{11}$: indeed
$$
x^2+5x+1=x^2-6x+9-8=(x-3)^2-8
$$
and $8$ is not a square modulo $11$.
Similarly, $x^2-5x+2=x^2+6x+9-7=(x+3)^2-7$ and $7$ is not a square modulo $11$.
Adding a root of $x^2+5x+1$ is the same as adding a square root of $2$. The elements are of the form $a+b\sqrt{2}$, and we must look whether $7$ is a square here: try and solve
$$
(a+b\sqrt{2})^2=7
$$
and you'll have your answer.
|
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|
Finding necessary and sufficient condition for $x^3+y^3+z^3+kxyz$ to be divisible by $x+y+z$ My question is that:
Find the necessary and sufficient condition for $x^3+y^3+z^3+kxyz$ to be divisible by $x+y+z$.
I know that answer is $k=-3$, as I know that $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.
But I can't prove that $k=-3$ is also a necessary condition. A mathematical proof is needed.
Thanks.
|
The condition necessary and sufficient for a polynomial $f(x)$ to be divisible by $(x-a)$ is that $f(a)=0$.
Let, $$f(x)=x^3+y^3+z^3+kxyz$$
For this polynomial to be divisible by $x+y+z$, it is necessary and sufficient that $f(-y-z)=0$.
However,
$$f(-y-z)=-(y+z)^3-kyz(y+z)+y^3+z^3=-(k+3)yz(y+z)=0$$
Simplifying it gives us, $k=-3$.
Thus for $x^3+y^3+z^3+kxyz$ to be divisible by $x+y+z$, it is necessary and sufficient that $k=-3$.
|
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|
If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$. I have an inequality problem which is as follow:
If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$.
I am not so good in inequalities. So, please give me some hints so that I can proceed.
Thanks.
|
The expression $ a^2 + b^2 - ab = c^2 $ is equivalent to
$$
a^2 + b^2 - ab = c^2 \longleftrightarrow a^2 - ab = c^2 - b^2 \longleftrightarrow a(a - b) = (c - b)(c + b)
$$
Now we multiply by $ b - c $ both sides (if $ b - c = 0 $ then there is nothing to prove):
$$
a(a - b)(b - c) = (c - b)(c + b)(b - c) = -(c - b)^2 (c + b)
$$
The right side is negative, because $ -(c - b)^2 \leq 0 \ $ (a square is always positive) and $ c,b \geq 0 \rightarrow c + b \geq 0 $.
Then we deduce
$$
a(a - b)(b - c) \leq 0
$$
And we know that $ a \geq 0 $, so
$$
(a - b)(b - c) \leq 0
$$
|
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|
$x + \frac1y = y + \frac1z = z + \frac1x$, then value of $xyz$ is? If $x,y,z$ are distinct positive numbers, such that $$x + \frac1y = y + \frac1z = z + \frac1x $$ then value of $xyz$ is?
$$A)\ 4\quad B)\ 3\quad C)\ 2\quad D)\ 1$$
My attempt:
1.I equaled the equation to '$k$'. Using the AM-GM inequality, I found that $k>2$ (the equality does not hold because all are distinct). However, this, I couldn't put to much use.
2. For the next attempt, I substituted the values of $x$ and $y$ in terms of $z$ and $k$ in $xyz$. What I am getting is $xyz=zk^2 - k - z$. That's the farthest I could do..Please, help.
|
Other answers have shown that the question is flawed. For fun, let's remove the condition that $x,y,z$ are positive, but keep the requirement that $x,y,z$ are distinct, and try to find all solutions $(x,y,z)$.
Trivially, $x,y,z \neq 0$. If we set $x+\dfrac{1}{y} = y+\dfrac{1}{z} = z + \dfrac{1}{x} = k$, for som real number $k$, then
\begin{align}
x &= k - \dfrac{1}{y}
\\
y &= k - \dfrac{1}{z}
\\
z &= k - \dfrac{1}{x}
\end{align}
If we substitute equation (3) into (2) and then substitute that into (1), we get $$x = k-\dfrac{1}{k-\dfrac{1}{k-\tfrac{1}{x}}}.$$
By unraveling this fraction, we eventually get the equation $$(k^2-1)(x^2-kx+1) = 0.$$
If $k \neq \pm 1$, then the solutions $x$ will satisfy $$x^2-kx+1 = 0 \iff x = k - \dfrac{1}{x} = z,$$ and we won't have distinctness.
If $k = 1$, then $z = 1-\dfrac{1}{x} = \dfrac{x-1}{x}$ and $y = 1-\dfrac{1}{z} = -\dfrac{1}{x-1}$.
So the solutions in this case are $(x,y,z) = \left(x,-\dfrac{1}{x-1},\dfrac{x-1}{x}\right)$, all of which satisfy $xyz = -1$.
If $k = -1$, then $z = -1-\dfrac{1}{x} = -\dfrac{x+1}{x}$ and $y = -1-\dfrac{1}{z} = -\dfrac{1}{x+1}$.
So the solutions in this case are $(x,y,z) = \left(x,-\dfrac{1}{x+1},-\dfrac{x+1}{x}\right)$, all of which satisfy $xyz = 1$.
Therefore, the only solutions where $x,y,z$ are distinct are all of the form $(x,y,z) = \left(x,-\dfrac{1}{x-1},\dfrac{x-1}{x}\right)$ for $x \neq 0,1$ or $(x,y,z) = \left(x,-\dfrac{1}{x+1},-\dfrac{x+1}{x}\right)$ for $x \neq 0,-1$. (It's easy to check that these satisfy the distinctness condition.) Also, the only possible values of $xyz$ are $\pm 1$.
|
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|
finding value of $ \int\frac{1}{x^3\sqrt{(1+x)^3}}dx$ finding value of $\displaystyle \int\frac{1}{x^3\sqrt{(1+x)^3}}dx$
Substituting $x=\tan^2 \theta\;, dx = 2\tan \theta \sec^2 \theta$
integral is $=\displaystyle \int \frac{2\tan \theta \sec^2 \theta }{\tan^6 \theta \cdot \sec^3 \theta}d\theta= 2\int\frac{\cos^6 \theta}{\sin^5 \theta}d\theta $
wan,t be able to proceed after, could some help me
|
Here is another way...
First substitute $$u^2=1+x$$ so that $$I=2\int\frac{udu}{(u^2-1)^3u^3}$$
Then partial fraction decomposition gives $$I=2\times\int\left(-\frac{1}{u^2}-\frac{15}{16(u+1)}-\frac{7}{16(u+1)^2}-\frac{1}{8(u+1)^3}+\frac{15}{16(u-1)}-\frac{7}{16(u-1)^2}+\frac{1}{8(u-1)^3}\right) du$$
|
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|
Prove that $A = \dfrac{f^2(1)+f^2(-1)}{2}$ is a composite number
Given $f(x) = a^{2016}x^2+bx+a^{2016}c-1$ where $a,b,c \in \mathbb{Z}$, suppose that the equation $f(x) = -2$ has two positive integer solutions. Prove that $A = \dfrac{f(1)^2+f(-1)^2}{2}$ is a composite number.
Let $g(x) = f(x)-2 = a^{2016}x^2+bx+a^{2016}c-3$, and let $-r,-s$ be the two positive integer roots of $g(x)$. Then by Vieta's Formula, $r+s = \dfrac{b}{a^{2016}}$ and $rs = \dfrac{a^{2016}c-3}{a^{2016}}$. Thus $a = \pm 1$ and so $f(x) = x^2+bx+c-1$ and so $f(1) = b+c$ and $f(-1) = c-b$. Then $$A = \dfrac{[(b+c)^2+b(b+c)+c-1]+[(c-b)^2+b(c-b)+c-1]}{2} = b^2+bc+c^2+c-1.$$ How do we prove this is composite?
|
Choosing $ a = 1$, $b = -3$, $c = 5$ gives two positive integer roots to $f(x) = 2$ ($x = 1$ and $x = 2$) and $[f^2(1)+f^2(-1)]/2 = 23$. This statement appears to be false.
If $f^2(x)$ means $f(x)^2$, then the statement fails for $a =1$, $b = -6$, $c = 11$, where the two roots are 2 and 4 and $[f(1)^2+f(-1)^2]/2 = 157$.
|
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|
Coefficient of $x^{50}$ in the expansion Find the coefficient of $x^{50}$ in the expansion of $$(1+x)^{1000}+2x(1+x)^{999}+3x^2(1+x)^{998}+\cdots+1001x^{1000}$$
|
Let $$S = (1 + x)^{1000} + 2x(1+x)^{999} +...+ 1000x^{999}(1+x)+ 1001 x^{1000}\tag1$$
This is an Arithmetic Geometric Series with $r = \frac{x}{1+x}$ and $d = 1$. Now $$\frac{x}{1+x}S = x(1 + x)^{999} + 2x^2(1 + x)^{998} +\cdots + 1000x^{1000} + \frac{1000x^{1001}}{1+x}\tag2$$
Subtracting we get,
$$(1 - \frac{x}{1+x}) S =(1+x)^{1000} + x(1+x)^{999} +\cdots + x^{1000} - \frac{1001x^{1000}}{1+x}$$
$$\Rightarrow S = (1+x)^{1001} + x(1+x)^{1000} + x^2(1+x)^{999} +...+ x^{1000}(1+x)-1001x^{1001}$$
This is a G.P. whose sum is $$S = (1+x)^{1002} - x^{1002} - 1002x^{1001}$$
So the coeff. of $x^{50}$ is $\binom{1002}{50}$. Hope it helps.
|
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|
For which $n$ this inequality is always true $a^5+1\ge a^3+a^n$ for every $a>0$? For which $n$ this inequality is always true $a^5+1\ge a^3+a^n$ for every $a>0$?
1.$1$
2.$2$
3.$3$
4.$4$
5.$5$
My attempt:It's clear that it is true for $n=2$.Because:
$(a^3-1)(a^2-1) \ge 0$
Is true because $a^3-1$ and $a^2-1$ are both negative or both positive.But there is a problem:How can we make sure it doesn't hold for any other $n$?
|
$n = 1 \implies a^5+1 - a^3-a = a^3(a^2 -1) - (a-1) = (a-1)(a^4+a^3-1)$. Observe that $a \to 1^{-} \implies a^4+a^3 - 1 \to 1 $, thus for $\epsilon = 0.01$ we can find a $\delta > 0$ such that $ -\delta < a - 1 < 0$, and $|a^4+a^3-1 - 1| < 0.01$ or $a^4+a^3-1 > 0.99$, thus $(a-1)(a^4+a^3-1) < 0$ on $(1-\delta, 1)$.
You did the case $n = 2$.
$n = 3$, $a^5+1- 2a^3 \ge 0$. For $0 < a \le 1\implies a^5 \ge a^6 \implies a^5+1 - 2a^3\ge a^6-2a^3 + 1 = (a^3-1)^2 \ge 0$, and for $\sqrt{\dfrac{6}{5}} > a > 1, f(a) = a^5+1-2a^3\implies f'(a) = a^2(5a^2-6) < 0\implies f(a) < f(1) = 0$.
$n = 5$, choose $a > 1$ it does not hold.
$n = 4$. $a^5+1 - a^3 - a^4 = a^3(a^2-1) - (a^2-1)(a^2+1) = (a^2-1)(a^3-a^2-1)$. You can choose $a$ slightly bigger than $1$, then it fails to be greater than $0$.
Thus $n = 2$ is the only value of $n$.
|
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|
Is probability that $x<2$ equal to the probability that $x^2<4$ given $1Assuming $x$ is a real number uniformly distributed over the interval $(1,3).$
so $x^2$ is also uniformly distributed over the interval $(1,9)${As for every $x=a\in (1,3) $ there exists $x^2=a^2\in (1,9)$}.
Probability that $x<2$ would be $\frac{1}{2}$ as $x$ can be in $(1,2)$ where sample set of $x$ is $(1,3)$, while probability that $x^2<4$ is $\frac{3}{8}$ as $x^2$ can be in $(1,4)$ where sample set of $x^2$ is $(1,9).$
So why is the probability that $x<2$ different from $x^2<4$ if both are identical?
|
You decided for any $x$ to match $x^2$, so you are claiming that $x<2$ is identical to $x^2<4$, but you can also match $4x-3$ in the interval $(1,9)$ for all $x$ in the interval $(1,3)$, which makes $x<2$ the same as $4x-3<5$.
That is the problem of defining probability on infinite group.
|
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|
Different results with different methods to same limit I'm considering this limit:
$$\lim\limits_{x \to 0} \frac{x \cos(x) - \sin(x)}{x \sin^2(x)}$$
If I apply de l'Hôpital's rule I get $-\frac{1}{3}$, while if I immediately simplify the $x$ like this:
$$\frac{\cos(x) - \frac{\sin(x)}{x}}{\sin^2(x)}$$
I get $-\frac{1}{2}$ as result, because I write it as (remembering that $\frac{sin(x)}{x} = 1$):
$$\frac{cos(x) - 1}{x^2}$$
which I simplify it with $-\frac{1}{2}$.
What am I getting wrong?
|
tl;dr: Your error is in the second step. You substitute $\frac{\sin x}{x}$ by its limit, $1$. But this is getting rid of the low-order terms, which matter.
One way to see it rigorously is to do a polynomial approximation near $0$, that is a Taylor expansion: when $x\to 0$,
$$
\frac{\sin x}{x} = 1-\frac{x^2}{6} + o(x^2)
$$
so that you neglect a term "behaving like" $\frac{x^2}{6}$. Does it matter? Well, again, near $0$: we have
$$
\cos x = 1-\frac{x^2}{2} +o(x^2) \tag{1}
$$
so
$$\begin{align}
\cos x - 1 &= -\frac{x^2}{2} +o(x^2) \tag{2}\\
\cos x - \frac{\sin x}{x} &= 1-\frac{x^2}{2} +o(x^2) - \left(1-\frac{x^2}{6} + o(x^2)\right)
= -\frac{x^2}{3} +o(x^2) \tag{3}
\end{align}$$
which explains the difference you get between you two methods. The low-order term you (erroneously) dismissed in the second did have a role to play.
|
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|
Polynomial not equal to a square Can we prove that for any $a, b ,c$, there exists an integer $n$ such that $\sqrt{n^3+an^2+bn+c}$ is not an integer?
I think yes, because the range of polynomial is the whole of $\mathbb{R}$. Any ideas. By the way, is this related to elliptic curves by any chance? Thanks beforehand.
|
This is a $1998$ putnam B6 problem.
An alternate approach: We write the assumed perfect square $n^3+an^2+bn+c $ in the form $(n^{3/2 }+ dn^{1/2}+f)^2$ giving us $$n^3+an^2+bn+c =n^3+2n^2d +2 (\sqrt {n})^3f+nd^2+2d\sqrt {n}f +f^2$$ Choosing $d=\frac {1}{2}a$ and $f=\pm 1$, we then, for $n $ sufficiently large, $$(n^{3/2}+\frac {1}{2}an^{1/2 }-1)^2 <n^3+an^2+bn+c <(n^{3/2}+\frac {1}{2}an^{1/2}+1)^2$$ If $n$ is a perfect square, say $n = m^2$, then the extreme left and right are
perfect squares and there is only one perfect square between them, namely $(n^{3/2}+\frac {1}{2}an^{1/2})^2$. Hence if $n=m^2$ and our cubic is a perfect square then we can get $$bm^2+c =\frac {1}{4}a^2m^2$$ For this to hold for sufficiently large $m $, we must have $c=0$ and $b=\frac {1}{4}a^2$. Thus $$n^3+an^2+bn+c =(n^{3/2}+\frac {1}{2}an^{1/2})^2 =(\sqrt {n}(n+\frac {a}{2}))^2$$ which is not a perfect square unless $n $ itself is.
|
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|
Prove that there do not exist integers that satisfy the system
Prove that there do not exist integers $x,y,z,w,t,$ that satisfy \begin{align*}(x+1)^2+y^2 &= (x+2)^2+z^2\\(x+2)^2+z^2 &= (x+3)^2+w^2\\(x+3)^2+w^2 &= (x+4)^2+t^2.\end{align*}
I thought about using a modular arithmetic argument. The given system is equivalent to \begin{align*}y^2 &= 2x+3+z^2\\z^2 &= 2x+5+w^2\\w^2 &= 2x+7+t^2.\end{align*} Thus $y^2 = 6x+15+t^2$. How can we find a contradiction from this?
|
You are asking about four $x$ values. These are consecutive, so they assume the four values $0,1,2,3 \pmod 4.$ Let us give them new names,
$$ x_1 \equiv 1 \pmod 4, \; \; x_2 \equiv 2 \pmod 4, \; \; x_3 \equiv 3 \pmod 4, \; \; x_4 \equiv 0 \pmod 4. $$
$$ n = x_1^2 + y_1^2 = x_2^2 + y_2^2 = x_3^2 + y_3^2 = x_4^2 + y_4^2 $$
First we prove that $n$ is odd. Assume $n$ is even.
$x_1^2 \equiv 1 \pmod 8.$ If $y_1$ is odd, then $n \equiv 2 \pmod 8$ is not divisible by $4.$ However, $x_4^2 \equiv 0 \pmod 8.$ With $y_4$ even, we get $n$ divisible by $4.$ The impossibility about $n$ being divisible by 4 and not divisible by 4 contradicts the assumption that $n$ was even. So, actually, $n$ is odd.
Again, $x_4^2 \equiv 0 \pmod 8.$ With the necessary odd $y_4,$ we have $y_4^2 \equiv 1 \pmod 8,$ so
$$ n \equiv 1 \pmod 8. $$
However, $x_2^2 \equiv 4 \pmod 8.$ With the necessary odd $y_2,$ we have $y_2^2 \equiv 1 \pmod 8,$ so
$$ n \equiv 5 \pmod 8. $$
The inconsistent values $1,5 \pmod 8$ contradict the assumption that there were four $ x_1 \equiv 1 \pmod 4, \; \; x_2 \equiv 2 \pmod 4, \; \; x_3 \equiv 3 \pmod 4, \; \; x_4 \equiv 0 \pmod 4. $
In turn, this contradicts the assumption that we can take the $x$ values $x+1, x+2, x+3, x+4.$
|
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|
Evaluate the Volume integral $\iiint\left(x^2+y^2+z^2\right)\mathbb dv$. Suppose $a>0$ and $S = \{(x,y,z) \in \mathbb R^3 : x^2+y^2+z^2=a^2\}$ then
MY FIRST APPROACH: $$\int\int\int\left(x^2+y^2+z^2\right)\mathbb dv=a^2\int\int\int \mathbb dv=a^2.\frac{4}{3}\pi a^3=\frac 4 3\pi a^5$$.
MY SECOND APPROACH:If i use spherical coordinate then $dV = r^2 \sin \theta dr d\theta d\phi.$ and $\left(x^2+y^2+z^2\right)=r^2$ now $$\int\int\int\left(x^2+y^2+z^2\right)\mathbb dv=\iint d\Omega \int_0^a (r^4) dr =\frac 4 5\pi a^5$$
where $\iint d\Omega = \int_0^\pi \sin(\theta)d\theta \int_0^{2\pi} d\phi = 4\pi.$
My question is that why such discrepancy in the answer$?$ and where did i commit mistake in my first approach.
(Note:2nd answer is correct but what's the problem with 1st$?$)
|
I think your mistake in first approach is that you take that as a line integral, whereas it is not: you can not put $\;x^2+y^2+z^2=a^2\;$ for the integrand, as this is a scalar function over the sphere $\;x^2+y^2+z^2=a^2\;$ , so you actually get (in spherical coordinates)
$$\iiint_S(x^2+y^2+z^2)dV=\int_0^a\int_0^{2\pi}\int_0^\pi\rho^4\sin\theta d\theta\,d\phi\,d\rho=$$
$$=\frac{a^5}5\cdot2\pi\int_0^\pi\sin\theta\,d\theta=\frac{4\pi a^5}5$$
which is basically the same as the second approach once one understands the first part...
The above is because in spherical coordinates we get $\;x^2+y^2+z^2=\rho^2\;$ , of course
|
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|
Finding the largest rectangle under the graph of $y = 1 - x^4$ using trigonometry I need to find the area of the largest rectangle possible under the curve of equation $1 - x^4$ with the base on the x-axis. The answers I've seen from other questions similar to this use calculus, but how would you solve it this using trigonometry?
|
Let $(x,1-x^4)$ is a vertex of the rectangle, where $x>0$.
Thus, by AM-GM $$S=2x(1-x^4)=2\left(x-x^5\right)=2\left(\frac{4}{5\sqrt[4]5}-\left(x^5+\frac{4}{5\sqrt[4]5}-x\right)\right)\leq$$
$$\leq\frac{8}{5\sqrt[4]5}-2\left(5\sqrt[5]{x^5\cdot\left(\frac{1}{5\sqrt[4]5}\right)^4}-x\right)=\frac{8}{5\sqrt[4]5}.$$
The equality occurs for $x=\frac{1}{\sqrt[4]5}$, which says that the answer is $\frac{8}{5\sqrt[4]5}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Verify $\cos{x}<\left(\frac{\sin{x}}{x}\right)^3$ for $0Verify by Mean Value Theory or otherwise that $\cos{x}<\left(\frac{\sin{x}}{x}\right)^3$ for $0<x<\pi/2$.
I am unable to solve the problem. Please give me a solution of the problem.
|
We need to prove that $f(x)>0$, where $f(x)=\frac{\sin{x}}{\sqrt[3]{\cos{x}}}-x$.
Indeed, by AM-GM $$f'(x)=\frac{\cos{x}\sqrt[3]{\cos{x}}+\frac{\sin^2x}{3\sqrt[3]{\cos^2x}}}{\sqrt[3]{\cos^2x}}-1=\frac{3\cos^2x+\sin^2x-3\sqrt[3]{\cos^4x}}{3\sqrt[3]{\cos^4x}}=$$
$$=\frac{1+\cos^2x+\cos^2x-3\sqrt[3]{\cos^4x}}{3\sqrt[3]{\cos^4x}}\geq\frac{3\sqrt[3]{\cos^4x}-3\sqrt[3]{\cos^4x}}{3\sqrt[3]{\cos^4x}}=0.$$
Thus, $f(x)>f(0)=0$ and we are done!
|
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|
Function is small $o$ of $x^2$ I have to solve the following exercise:
Give $a$, $b$, $c \in \mathbb R$ such that
$$\frac{1}{1-\cos x} = \frac{a}{x^2} + b +cx^2 + o(x^2)$$
for $x\to 0$.
Here's my attempt:
I know that $$\cos x = \frac{1}{2}\left( e^{ix} + e^{-ix}\right) = \frac{1}{2} \sum_{n=0}^\infty \frac{(ix)^{2n}}{(2n)!} = \frac{1}{2}\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!} = \frac{1}{2}\left(1-\frac{x^2}{2}+o(x^2)\right)$$
for $x\to 0$.
The question now is how to resolve $\frac{1}{1-\cos x}$. This much looks like an application of the geometric series here. I assume I can use it since $\cos x < 1$ for all sufficiently small $x \neq 0$. I wouldn't know how to solve this exercise otherwise. Proceeding, it follows that
$$\frac{1}{1-\cos x} = \frac{1}{1-\left(\frac{1}{2} - \frac{x^2}{4} + o(x^2)\right)} = \sum_{n=0}^\infty\left(\frac{1}{2} - \frac{1}{4}x^2+o(x^2)\right)^n = 1+ \frac{1}{2}-\frac{1}{4}x^2+o(x^2).$$
Choosing $a=0$, $b = 1,5$ and $c=-\frac{1}{4}$ establishes the case.
|
hint
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+o(x^6)$$ and after factoring out by $x^2$ ,
$$\frac{1}{1-X}=1+X+X^2+o(X^2).$$
|
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|
Approximating square roots using binomial expansion. Through the binomial expansion of $(1 - 2x)^\frac{1}{2}$, I am required to find an approximation of $\sqrt2$.
Binomial expansion
$ (1 + x)^n = 1 + \frac{n}{1}x + \frac{n(n-1)}{1*2}x^2 + ... $
Thus, the expansion of $(1 - 2x)^\frac{1}{2}$: $$
= 1 - x -\frac{1}{2}x^2 - \frac{1}{2}x^3 + ... $$
The suggested way, is to choose a value for $x$ so that $(1-2x)$ has the form $2*$'a perfect square'. This can be done by taking $x = 0.01$. Thus, $(1 - 2x)=(1-2*0.01) = 0.98 = 2*0.7^2$
And
$$
(1 - 2x)^\frac{1}{2} = 0.98^\frac{1}{2} = 0.7\sqrt2$$
Which is equal to the previously established expansion, so we can now go ahead and find $\sqrt2$. The problem I am facing, is that there was no mention of how the value of $x=0.01$ was arrived at.
Is there an easy way to determine an appropriate value for $x$?
|
We want to (manually) approximate $\sqrt{2}$ by using the first few terms of the binomial series expansion of
\begin{align*}
\sqrt{1-2x}&= \sum_{n=0}^\infty \binom{\frac{1}{2}}{n}(-2x)^n\qquad\qquad\qquad\qquad |x|<\frac{1}{2}\\
&= 1-x-\frac{1}{2}x^2-\frac{1}{2}x^3+\cdots\tag{1}
\end{align*}
Here we look for a way to determine appropriate values of $x$ using the binomial expansion.
In order to apply (1) we are looking for a number $y$ with
\begin{align*}
\sqrt{1-2x}&=\sqrt{2y^2}=y\sqrt{2}\tag{2}\\
\color{blue}{\sqrt{2}}&\color{blue}{=\frac{1}{y}\sqrt{1-2x}}
\end{align*}
We see it is convenient to choose $y$ to be a square number which can be easily factored out from the root. We obtain from (2)
\begin{align*}
1-2x&=2y^2\\
\color{blue}{x}&\color{blue}{=\frac{1}{2}-y^2}\tag{3}
\end{align*}
When looking for an appropriate $y$ which fulfills (3) there are some aspects to consider:
*
*We have to respect the radius of convergence $|x|<\frac{1}{2}$.
*Since we want to calculate an approximation of $\sqrt{2}$ by hand we should take $y\in\mathbb{Q}$ with rather small numbers as numerator and denominator.
*Last but not least: We want to find a value $x$ which provides a good approximation for $\sqrt{2}$.
We will see it's not hard to find values which have these properties.
We see in (1) a good approximation is given if $x$ is close to $0$. If $x$ is close to zero we will also fulfill the convergence condition. $x$ close to zero means that in (3) we have to choose $y$ so that
$
y^2
$
is close to $\frac{1}{2}$. We have already (1) and (3) appropriately considered. Now we want to find small natural numbers $a,b$ so that
\begin{align*}
y^2=\frac{a^2}{b^2}\approx \frac{1}{2}
\end{align*}
This can be done easily. When going through small numbers of $a$ and $b$ whose squares are apart by a factor $2$ we might quickly come to $100$ and $49$. These are two small squares and we have $2\cdot 49=98$ close to $100$. That's all.
Now it's time to harvest. We choose $y^2=\frac{49}{100}$ resp. $\color{blue}{y=\frac{7}{10}}$. We obtain for $x$ from (3)
\begin{align*}
x=\frac{1}{2}-y^2=\frac{1}{2}-\frac{49}{100}=\frac{1}{100}
\end{align*}
We have now a nice value $\color{blue}{x=\frac{1}{100}}$ and we finally get from (1) the approximation:
\begin{align*}
\color{blue}{\sqrt{2}} \approx \frac{10}{7}\left(1- 10^{-2}-\frac{1}{2}\cdot 10^{-4}-\frac{1}{2}\cdot 10^{-6}\right)\color{blue}{=1.414\,213\,5}71\ldots
\end{align*}
We have $\color{blue}{\sqrt{2}=1.414\,213\,5}62\ldots$ with an approximation error $\approx 9.055\times 10^{-9}$. This result is quite impressive when considering that we have used just four terms of the binomial series.
Note: In a section about binomial series expansion in Journey through Genius by W. Dunham the author cites Newton: Extraction of roots are much shortened by this theorem, indicating how valuable this technique was for Newton.
|
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|
Proving a sequence is convergent by convergence of odd and even subsequences? I want to use this method to prove convergence of: $a_{n}=\sqrt{2-a_{n-1}}$, $a_{0}=\frac{2}{3}$.
Here is my attempt at proof:
It can be proven inductively that $0< a_{n}<2$ for all $n$.
I want to show that each of the odd and even subsequences are monotone.
$a_{2n+3}-a_{2n+1}=\sqrt{2-a_{2n+2}}-\sqrt{2-a_{2n}}=\frac{a_{2n}-a_{2n+2}}{\sqrt{2-a_{2n+2}}+\sqrt{2-a_{2n}}}=\frac{a_{2n}-\sqrt{2-\sqrt{2-a_{2n+1}}}}{\sqrt{2-a_{2n+2}}+\sqrt{2-a_{2n}}}$
I'm stuck as I'm not able to show that $(a_{2n+1})_{n\geq0}$ is decreasing.
Thank's in advance.
|
$$
\begin{align}
&\, a_{n}=\sqrt{2-a_{n-1}}\,\Rightarrow\,a_{n}\ge0 \\[2mm]
&\, a_{n}^2=2-a_{n-1}\,\Rightarrow\,a_{n-1}=2-a_{n}^2=\left(\sqrt{2}-a_{n}\right)\left(\sqrt{2}+a_{n}\right)\,\ge0 \\[2mm]
&\, \qquad\Rightarrow\,\sqrt{2}-a_{n}\ge0\,\Rightarrow\,a_{n}\le\sqrt{2} \\[8mm]
&\, a_{n}-a_{n-2}=\left(2-a_{n+1}^2\right)-\left(2-a_{n-1}^2\right)=a_{n-1}^2-a_{n+1}^2=\left(a_{n-1}-a_{n+1}\right)\left(a_{n-1}+a_{n+1}\right) \\[2mm]
&\, \qquad =\left[\left(2-a_{n}^2\right)-\left(2-a_{n+2}^2\right)\right]\left(a_{n-1}+a_{n+1}\right)=\left(a_{n+2}^2-a_{n}^2\right)\left(a_{n-1}+a_{n+1}\right) \\[2mm]
&\, \qquad =\left(a_{n+2}-a_{n}\right)\left(a_{n+2}+a_{n}\right)\left(a_{n-1}+a_{n+1}\right)\,\Rightarrow \,\text{sgn}\left(a_{n+2}-a_{n}\right)=\text{sgn}\left(a_{n}-a_{n-2}\right) \\[2mm]
&\, \qquad\Rightarrow\,\text{Either }\,\,\,\left\{\,a_{n+2}\ge a_{n}\ge a_{n-2}\,\right\}\,\,\text{ or }\,\,\left\{\,a_{n+2}\le a_{n}\le a_{n-2}\,\right\} \\[2mm]
&\, \qquad\Rightarrow\,\text{Both }\quad\left\{a_{2n-1}\right\}\,\,\text{ and }\,\,\left\{a_{2n}\right\}\,\text{ are }\,\color{red}{\text{ monotonic}} \\[2mm]
&\, \qquad\Rightarrow\begin{cases} a_0=\,\frac{2}{3}\,\,\approx0.667,\quad a_2\approx0.919\,\gt a_0 &\Rightarrow\color{red}{\left\{a_{2n}\right\}\,\text{ increasing}}\\[2mm] a_1=\frac{2}{\small\sqrt{3}}\approx1.155,\quad a_3\approx1.040\,\lt a_1 &\Rightarrow\color{red}{\left\{a_{2n-1}\right\}\,\text{ decreasing}} \end{cases} \\[2mm]
&\, \qquad\text{And because}\,\left\{\,0\le a_{n}\le\sqrt{2}\,\right\}\,\Rightarrow\,\text{Both }\,\,\left\{a_{2n-1}\right\}\,\text{ and }\,\left\{a_{2n}\right\}\,\text{ are }\,\color{red}{\text{convergent}} \\[8mm]
&\, \text{Let: }\,\,L=\lim_{n\rightarrow\infty}a_{n},\quad L_{o}=\lim_{n\rightarrow\infty}a_{2n-1},\quad L_{e}=\lim_{n\rightarrow\infty}a_{2n} \\[2mm]
&\, \qquad\Rightarrow\,L_o=2-L_e^2\,\,\,{\small\&}\,\,\,L_e=2-L_o^2 \\[2mm]
&\, \qquad\Rightarrow\,L_o-L_e=L_o^2-L_e^2=\left(L_o-L_e\right)\left(L_o+L_e\right) \\[2mm]
&\, \qquad\Rightarrow\,L_o-L_e=0 \quad\text{ or }\quad L_o+L_e=1 \\[2mm]
&\, L_o+L_e=1\,\Rightarrow\,\begin{cases} L_o^2=2-L_e=1+L_o &\Rightarrow\,L_o=\frac{1\pm\sqrt{5}}{2}\,\notin\left[0,\sqrt{2}\right]\\[2mm] L_e^2=2-L_o=1+L_e &\Rightarrow\,L_e=\frac{1\pm\sqrt{5}}{2}\,\notin\left[0,\sqrt{2}\right] \end{cases} \\[2mm]
&\, L_o-L_e=0\,\Rightarrow\,L_o=L_e=L=\sqrt{2-L} \\[2mm]
&\, \qquad\qquad\quad\,\,\Rightarrow\,L^2+L-2=0\,\Rightarrow\,L=\frac{-1\pm3}{2}\,\Rightarrow\,\color{red}{L=1} \\[8mm]
&\, \text{Generally, }\quad \color{blue}{\lim_{n\rightarrow\infty}a_{2n-1}=\lim_{n\rightarrow\infty}a_{2n}}\color{red}{=\lim_{n\rightarrow\infty}a_{n}=1\,\,\colon\,\forall\,\,a_0\in[-2,\,+2]}
\end{align}
$$
|
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|
Decomposition of this partial function I came across this
$$\int \frac{dx}{x(x^2+1)^2}$$
in "Method of partial functions" in my Calculus I book.
The thing is that, however, decomposing $\frac{1}{x(x^2+1)^2}$ would take long in this way:
$$\frac{1}{x(x^2+1)^2}= \frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$
So is there a quicker or a more practical way that I can use?
|
\begin{equation}
\int \frac{dx}{x(x^2+1)^2}
\end{equation}
\begin{equation}
\frac{1}{x(x+i)^2(x-i)^2}=\dfrac{A}{x}+\dfrac{B}{x+i}+\dfrac{C}{x-i}+\dfrac{D}{(x+i)^2}+\dfrac{E}{(x-i)^2}
\end{equation}
*
*Let $x=0$. Then $A=\dfrac{1}{(0+i)^2(0-i)^2}=1$\
*Let $x=-i$. Then $B=\dfrac{1}{-i(-i-i)^2}=-\dfrac{1}{2}$\
*Let $x=i$. Then $C=\dfrac{1}{i(i+i)^2}=-\dfrac{1}{2}$\
Therefore we have
\begin{eqnarray}
\frac{1}{x(x^2+1)^2}&=&\dfrac{1}{x}-\dfrac{1}{2}\left(\dfrac{1}{x+i}+\dfrac{1}{x-i}\right)+\dfrac{D}{(x+i)^2}+\dfrac{E}{(x-i)^2}\\
&=&\dfrac{1}{x}-\dfrac{x}{x^2+1}+\dfrac{D(x^2-2ix-1)+E(x^2+2ix-1)}{(x^2+1)^2}\\
&=&\dfrac{1}{x}-\dfrac{x}{x^2+1}+\dfrac{(D+E)x^2+2(E-D)ix-(D+E)}{(x^2+1)^2}\\
&=&\dfrac{1}{x}-\dfrac{x}{x^2+1}+\dfrac{2Eix}{(x^2+1)^2}
\end{eqnarray}
Note that since the $x^2$ term must vanish it must be the case that $D=-E$.
Thus we only have to find the value of $E$.
\begin{eqnarray}
\frac{1}{x(x^2+1)^2}&=&\dfrac{1}{x}-\dfrac{x}{x^2+1}+\dfrac{2Eix}{(x^2+1)^2}
\end{eqnarray}
This equation holds for every value of $x$ with the exception of $x=0,i,-i$. Therefore, when $x=1$ it is true that
\begin{eqnarray}
\frac{1}{4}&=&1-\dfrac{1}{2}+\dfrac{E}{2}i\\
E&=&\dfrac{i}{2}
\end{eqnarray}
Therefore
\begin{equation}
\frac{1}{x(x^2+1)^2}=\dfrac{1}{x}-\dfrac{x}{x^2+1}-\dfrac{x}{(x^2+1)^2}
\end{equation}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Differentiate by first principle,$f(x)=\frac{x^2}{\sin x}$ Differentiate by first principle,$f(x)=\frac{x^2}{\sin x}$
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
$$f'(x)=\lim_{h\to0}\frac{\frac{(x+h)^2}{\sin(x+h)}-\frac{x^2}{\sin x}}{h}$$
$$f'(x)=\lim_{h\to0}\frac{\frac{(x+h)^2\sin x-x^2\sin(x+h)}{\sin(x+h)\sin x}}{h}$$
$$f'(x)=\lim_{h\to0}\frac{(x+h)^2\sin x-x^2\sin(x+h)}{h\sin x\sin(x+h)}$$
$$f'(x)=\lim_{h\to0}\frac{x^2}{\sin x}\frac{(1+\frac{h}{x})^2\sin x-\sin(x+h)}{h\sin(x+h)}$$
I am stuck here.
|
Let's go back to here:
$$f'(x)=\lim_{h\to0}\frac{(x+h)^2\sin x-x^2\sin(x+h)}{h\sin x\sin(x+h)} =$$
$$\lim_{h\to0}\frac{x^2(\sin x - \sin(x+h)) + (2xh+h^2)\sin x}{h\sin x\sin(x+h)} =$$
$$\lim_{h\to0}\left(\frac{x^2(\sin x - \sin x\cos h - \cos x\sin h)}{h\sin x\sin(x+h)}+\frac{2x+h}{\sin(x+h)}\right) =$$
$$\lim_{h\to0}\left(\frac{hx^2}{\sin(x+h)}\cdot\frac{1-\cos h}{h^2} -\frac{x^2\cos x}{\sin x\sin(x+h)}\cdot\frac{\sin h}{h}+\frac{2x+h}{\sin(x+h)}\right) =$$
$$0 - \frac{x^2\cos x}{\sin^2x} + \frac{2x}{\sin x} =$$
$$\frac{2x\sin x - x^2\cos x}{\sin^2x}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $a_{n+2}=3a_n+2\sqrt{2a_n^2+2a_{n+1}^2}$ is an integer Given the sequence
$$a_1=a_2=1;\ a_ {n+2} = 3a_n + 2\sqrt{2a_n^2 + 2a_{n+1}^2}$$
prove that $a_n$ is an integer for all $n\in\mathbb N$.
Attempt
It is enough to show that $2a_n^2 + 2a_{n+1}^2$ is a perfect square. That means it's an even perfect square and so divisible by 4; thus $a_n^2+a_{n+1}^2=2k^2$ for some integer $k$. I think can solve this diophantine equation but I can't relate the solution to the original problem. Can anyone help?
|
$a_{n+2}\ =\ 3a_n+2\sqrt{2a_n^2+2a_{n+1}^2}$
$\implies\ \left(a_{n+2}-3a_n\right)^2\ =\ 8a_n^2+8a_{n+1}^2$
$\implies\ a_n^2-6a_{n+2}a_n-8a_{n+1}^2+a_{n+2}^2=0$
Treat this as a quadratic equation in $a_n$. The discriminant is
$\Delta_n\ =\ 36a_{n+2}^2+32a_{n+1}^2-4a_{n+2}^2 = 4\left(8a_{n+1}^2+8a_{n+2}^2\right)$
Thus
$a_n\ =\ \dfrac{6a_{n+2}-\sqrt{\Delta}}2\ (\because a_n<a_{n+2})$
$=\ 3a_{n+2}-2\sqrt{2a_{n+1}^2+2a_{n+2}^2}$
$\implies\ -a_n+3a_{n+1}+3_{a_n+2}\ =\ 3a_{n+1}+2\sqrt{2a_{n+1}^2+2a_{n+2}^2}=a_{n+3}$
Since $a_1=a_2=1$ and $a_3=7$ are integers, $a_4$ is an integer; by induction $a_n\in\mathbb Z$ for all $n\in\mathbb N$.
|
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|
Coefficient in binomial expansion for negative terms Find the exponent for $x^2$ in the expansion of:
$(\frac{x}{3} - \frac{1}{x^3})^{10}$
What trips me up is that the second term is negative. Though I find using the binomial theorem that the expansion should be a sum of terms on the form:
${ 10 \choose k}(\frac{x}{3})^k(-\frac{1}{x^3})^{10-k} $
But according to the solution, that is wrong. The form is supposed to be:
${ 10 \choose k}(\frac{x}{3})^k(\frac{1}{x^3})^{3(10-k)} $
A solution is then found for $k = 8$ which is simple. But I don't understand why my form is wrong and the solutions form is right. Why is the coefficient for the second factor $3(10-k)$ and where did the negative sign go?
The solution for the problem is supposed to be:
$3^{-8}{ 10 \choose 8 }$
|
$$\begin{align}
\left(\frac x3-\frac 1{x^3}\right)^{10}
&=\frac {x^{10}}{3^{10}}\left(1-\frac 3{x^4}\right)^{10}\\
&=\frac {x^{10}}{3^{10}}\sum_{r=0}^{10}\binom {10}r\left(- 3x^{-4}\right)^r
\end{align}$$
Putting $r=2$ gives $x^2$ term, hence coefficient of $x^2$ is
$$\frac 1{3^{10}} \binom{10}2(-3)^2=\frac {45}{3^8}=\color{red}{\frac 5{3^6}}$$
|
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|
Proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$ So I am trtying to proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$. First I tried with Cauchy–Schwarz inequality but got nowhere.
Now I am trying to find a convex function, so I can use jensen's inequality, but I can't come up with one which works.. Has anyone an idea?
|
Another way.
Let $a+b+c=3$.
Hence, we need to prove that
$$\sum\limits_{cyc}\frac{a}{9-2a}\geq\frac{3}{7}$$ or
$$\sum\limits_{cyc}\left(\frac{a}{9-2a}-\frac{1}{7}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{a-1}{9-2a}\geq0$$ or
$$\sum\limits_{cyc}\left(\frac{a-1}{9-2a}-\frac{a-1}{7}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{(a-1)^2}{9-2a}\geq0.$$
Done!
|
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|
Sum the series: $1+\frac{1+3}{2!}+\frac{1+3+3^2}{3!}+\cdots$ We have the series$$1+\frac{1+3}{2!}+\frac{1+3+3^2}{3!}+\cdots$$How can we find the sum$?$
MY TRY: $n$th term of the series i.e $T_n=\frac{3^0+3^1+3^2+\cdots+3^n}{(n+1)!}$. I don't know how to proceed further. Thank you.
|
The numerator is a geometric sum that evaluates to,
$$\frac{3^{n+1}-1}{3-1}$$
Hence what we have is,
$$\frac{1}{2} \sum_{n=0}^{\infty} \frac{(3^{n+1}-1)}{(n+1)!}$$
$$=\frac{1}{2} \sum_{n=1}^{\infty} \frac{3^n-1}{n!}$$
$$=\frac{1}{2} \left( \sum_{n=1}^{\infty} \frac{3^n}{n!}- \sum_{n=1}^{\infty} \frac{1^n}{n!} \right)$$
Recognizing the Taylor series of $e^x$ we have
$$=\frac{1}{2}((e^3-1)-(e-1))$$
|
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|
Differentiation first principles for cube Find the derivative of function $f(x) = \sqrt{x} + \dfrac{1}{x^3}$ from the first principles.
I tried to use the formula
$$
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.
$$
I try to expand for the $x^3$ first but it looks more and more complicated. The answer end up with $0$.
|
Hint: (I'm assuming you mean $\require{cancel}f(x) = \sqrt{x} + \frac1{x^3}$.) Here's the "hard" computation for a few similar functions. You can see the pattern: in the end, there is a factor of $h$ in the numerator which will cancel with the $h$ that you are going to use as the denominator in the difference quotient, and all but the first term of the remaining factor will vanish as $h\to 0$.
For $f(x) = \frac1x$:
$$\tfrac1{(x+h)}-\tfrac1{x} = \tfrac x{x(x+h)}-\tfrac{x+h}{x(x+h)}=\tfrac{\cancel{x}-(\cancel{x}+h)}{x(x+h)} = \tfrac{-h}{x(x+h)}$$
For $f(x)=\frac1{x^2}$:
$$\tfrac1{(x+h)^2}-\tfrac1{x^2} = \tfrac {x^2}{x^2(x+h)^2}-\tfrac{(x+h)^2}{x^2(x+h)^2}=\tfrac{\cancel{x^2}-(\cancel{x^2}+2hx + h^2)}{x^2(x+h)^2}=\tfrac{-2xh - h^2}{x^2(x+h)^2}=\tfrac{-h(2x + h)}{x^2(x+h)^2}$$
For $f(x) = \frac1{x^3}$ (which is what you are using):
$$\tfrac1{(x+h)^3}-\tfrac1{x^3} = \tfrac {x^3}{x^3(x+h)^3}-\tfrac{(x+h)^3}{x^3(x+h)^3}=\tfrac{\cancel{x^3}-(\cancel{x^3}+3hx^2 + 3h^2x + h^3)}{x^3(x+h)^3}=\tfrac{-3hx^2 - 3h^2x - h^3}{x^3(x+h)^3}=\tfrac{-h(3x^2 + 2hx + h^2)}{x^3(x+h)^3}$$
I presume you can handle the $\sqrt{x}$ term without difficulty.
|
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|
Prove that a sequence is increasing (check the idea)
4.5 A real sequence $\{x_n\}$ satisfies $7x_{n+1}=x_n^3+6$ for $n\ge1$. If $x_1=\frac{1}{2}$, prove that the sequence increases and find its limit. What happens if $x_1=\frac{3}{2}$ or if $x_1=\frac{5}{2}$?
In order to prove, that sequence is increasing, my idea is to take the first derivative of the function $y = (x^3 +6)/7$
Our result is $y' = 3x^2/7$ and this is positive. So our function is increasing function. Is this solution correct?
Are there other approaches that can prove the above statement without using derivatives?
Edited:
I agree, I was wrong. Here are my new arguments:
1) By induction I have proved that $0<x_n<1$.
2) $x_{n+1} - x_n = (x_n^3 + 6)/7 -x_n = x_n^3/7 - x_n + 6/7 $
But I don't know if we can conclude from this that $x_n^3/7 - x_n + 6/7 >0$
|
Given that $x_{n+1} = (x_{n}^3 + 6)/7$, when $x_1 = \frac{1}{2}$ we will find the limit in 3 steps:
*
*$\color{red}{0 < x_n < 1 \text{ for all $n$.}}$ Note that if $0 < x_n < 1$ then $0 < x_n^3 + 6 < 7$, which in turn implies that $0 < x_{n+1} < 1$. So, starting with $x_1 = \frac{1}{2}$, we will get the entire sequence satisfying the inequality $0 < x_n < 1$.
*$\color{red}{x_n \text { is increasing. }}$ To show this, it is sufficient to show that if $0 <x_n < 1$ then $x_n < x_{n+1}$. To get this, we subtract $x_n$ from both sides of $x_{n+1} = (x_{n}^3 + 6)/7$ and get:
\begin{eqnarray*}x_{n+1} - x_{n} = \displaystyle\frac{x_{n}^3 + 6 - 7 x_n}{7} = \displaystyle\frac{(x_n-1)(x_{n}^2 + x_n - 6)}{7} > 0 & \text{ when $0 < x_n < 1$} \end{eqnarray*}
*$\color{red}{x_n \text{ converges to 1}. }$ Since the sequence $(x_n)$ is increasing and bounded, it is convergent. Therefore, the limit $l \leq 1$ exists and satisfy: $l = (l^3 + 6)/7$. This gives us the limit as $l = 1$.
Similarly, when $x_1 = \frac{3}{2}$ we can again find it in 3 steps:
*
*$\color{red}{1 < x_n < 2 \text{ for all $n$.}}$ If $1 < x_n < 2$ then $1 < x_n^3 + 6 < 14$, which in turn implies that $1 < x_{n+1} < 2$. So, starting with $x_1 = \frac{3}{2}$, we will get the entire sequence satisfying the inequality $1 < x_n < 2$.
*$\color{red}{x_n \text { is decreasing. }}$ To show this, it is sufficient to show that if $1 <x_n < 2$ then $x_n > x_{n+1}$. To get this, we subtract $x_n$ from both sides of $x_{n+1} = (x_{n}^3 + 6)/7$ and get:
\begin{eqnarray*}x_{n+1} - x_{n} = \displaystyle\frac{x_{n}^3 + 6 - 7 x_n}{7} = \displaystyle\frac{(x_n-1)(x_{n}^2 + x_n - 6)}{7} < 0 & \text{ when $1 < x_n < 2$} \end{eqnarray*}
*$\color{red}{x_n \text{ converges to 1}. }$ Since the sequence $(x_n)$ is decreasing and bounded, it is convergent. Therefore, the limit $1 \leq l \leq 1.5$ exists and satisfy: $l = (l^3 + 6)/7$. This gives us the limit as $l = 1$.
Finally, when $x_{1} = \frac{5}{2}$ we can show that the sequence is unbounded above, and increasing. Therefore, limit does not exist on real line. This follows from the fact that
\begin{eqnarray*}x_{n+1} - x_{n} = \displaystyle\frac{x_{n}^3 + 6 - 7 x_n}{7} = \displaystyle\frac{(x_n-1)(x_{n}^2 + x_n - 6)}{7} >0 & \text{ when $x_n > 2$} \end{eqnarray*}
Also, when $x_n > 2$, the function $\displaystyle\frac{(x_n-1)(x_{n}^2 + x_n - 6)}{7}$ is also increasing in $x_n$. This implies that the difference between $x_{n+1}$ and $x_{n}$ is always greater than or equal to the value:
$\displaystyle\frac{(x_1-1)(x_{1}^2 + x_1 - 6)}{7} = \frac{1.5(2.75)}{7}$. Therefore, the sequence $(x_n)$ is increasing and unbounded.
|
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|
Find all values of $ac-bd$ where $a, b, c, d$ are real numbers, $a^2c^2+a^2d^2+b^2c^2+b^2d^2=2017$, and $ad+bc=44$. Find all values of $ac-bd$ where $a, b, c, d$ are real numbers, $a^2c^2+a^2d^2+b^2c^2+b^2d^2=2017$, and $ad+bc=44$.
I noticed that $a^2c^2=(ac)^2$ and so on for all the terms in the polynomial. How does that help? Am I close to something. All help is appreciated.
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Well, $(ac-bd)^2+(ad+bc)^2=a^2c^2+a^2d^2+b^2c^2+b^2d^2$.
|
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|
Infinite sum including logarithm I would like to calculate the following sum:
$$\sum_{n=1}^{\infty}\ln \left( \frac{n^2+2n+1}{n^2+2n} \right)$$
I do know that it converges but I have gone that far:
\begin{align}
& \sum_{n=1}^{\infty}\ln \left( \frac{n^2+2n+1}{n^2+2n} \right) \Longleftrightarrow \sum_{n=1}^\infty \ln \left( \frac{(n+1)^2}{n(n+2)} \right) \\[10pt]
= {} & \ln \left( \frac{4}{3} \right)+\ln \left( \frac{9}{8} \right)+\ln \left( \frac{16}{15} \right)+\cdots+\ln \left( \frac{n}{n-1} \right)\\[10pt]
= {} & \ln \left( \frac{4}{3}\frac{9}{8}\frac{16}{15} \cdots \frac{n}{(n-1)} \right)=\ln (n)
\end{align}
which diverges as $n\to \infty.$
It looked like telescoping in the beginning but now I am confused. Where have I gone wrong?
Thanks.
|
Rewrite the general term as $\ln\dfrac{(n+1)^2}{n(n+2)}$, use the functional property of logs and you'll obtain a telescoping product for partial sums:
\begin{align}\sum_{k=1}^{n}\ln \frac{(k+1)^2}{k(k+2)}&=\ln\frac{2^2}{1\cdot 3}+\ln\frac{3^2}{2\cdot 4}+\ln\frac{4^2}{3\cdot 5}+\dotsm\dotsm\dotsm\\&\phantom{=}+\ln\frac{(n-1)^2}{(n-2)n}+\ln\frac{n^2}{(n-1)(n+1)}+\ln\frac{(n+1)^2}{n(n+2)}\\
&=\ln\frac{2^{\not2}\cdot\not 3^2\cdot4^2\dotsm\dotsm\dotsm\dotsm(n-1)^2\not n^2(n+1)^2\hspace{3em}}{1\cdot \not3\cdot\not2\cdot 4\cdot\not 3\cdot 5\dotsm(n-2)\not n(n-1)(n+1)\not n(n+2)}\\
&=\log\frac{2(n+1)}{n+2}\to \ln 2.\end{align}
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|
Surface integral on one-sheeted hyperboloid I'm working on the following exercise:
"Let $\Sigma:=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2=z^2+1,-1\leq z\leq3\}$ be a surface with orientation $\omega$.
Compute $\int\int_{\Sigma}(\nabla\times F)\cdot\omega\ dS$ knowing that $F=(-\frac{y}{3},-\frac{z}{3},-\frac{x}{3})$ and $\omega (0,1,0)=(0,1,0)$.
Verify the result you have obtained using Stokes' Theorem."
Here's what I've done:
-> $\nabla\times F=(\frac{1}{3},\frac{1}{3},\frac{1}{3})$;
-> I've divided the surface in three pieces:
(+) $0<z\leq 3$: parametrization $\varphi_{+}:=(x,y,\sqrt{x^2+y^2-1})\to \varphi_{+_x}=(1,0,\frac{x}{\sqrt{(x^2+y^2-1)}}), \varphi_{+_y}=(0,1,\frac{y}{\sqrt{(x^2+y^2-1)}})\to$
$\omega_+=\varphi_{+_x}\times\varphi_{+_y}=(-\frac{x}{\sqrt{x^2+y^2-1}},-\frac{y}{\sqrt{x^2+y^2-1}},1)$
($0$)$\ z=0$: $\omega_0=(x,y,0)$ (ok, since $\omega (0,1,0)=(0,1,0)$;
(-)$\varphi_{-}:=(x,y,-\sqrt{x^2+y^2-1})\to \varphi_{-_x}=(1,0,-\frac{x}{\sqrt{(x^2+y^2-1)}}), \varphi_{-_y}=(0,1,-\frac{y}{\sqrt{(x^2+y^2-1)}})\to$
$\omega_-=\varphi_{-_x}\times\varphi_{-_y}=(\frac{x}{\sqrt{x^2+y^2-1}},\frac{y}{\sqrt{x^2+y^2-1}},1)$
(Notice that I've not normalized the normal vector since when computing the flux of a vector field across a surface in space we have $\int\int_S F\cdot n dS=\int\int_S F\cdot (\varphi_x \times\varphi_y)dxdy$)
So $\int\int_{\Sigma_+}(\nabla\times F)\cdot\omega_+ dS=\int\int_{\Sigma_+}(\frac{1}{3},\frac{1}{3},\frac{1}{3})\cdot (-\frac{x}{\sqrt{x^2+y^2-1}},-\frac{y}{\sqrt{x^2+y^2-1}},1)dxdy=\frac{1}{3}\int\int_{\Sigma_+}(-\frac{x}{\sqrt{x^2+y^2-1}}-\frac{y}{\sqrt{x^2+y^2-1}}+1)=?$
Here is where I get stuck, since I can't figure out what the bounds for this double integral should be; can anyone give me an hint about this?
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By the divergence theorem, your integral equals
$$
\Phi = \iiint_E \underbrace{\nabla\cdot \nabla \times F}_{=0}\; dV-\iint_{S_1}\nabla \times F\; dS-\iint_{S_2}\nabla \times F\; dS,
$$
where $S_1$ and $S_2$ are the surfaces that close $\Sigma$ at $z=-1$ and $z=3$. $S_1$ can be parametrized as follows:
$$
\begin{cases}
x=x\\
y=y\quad \mbox{with }\; x^2+y^2\le 2\\
z=-1
\end{cases}
$$
And similarly for $S_2$:
$$
\begin{cases}
x=x\\
y=y\quad \mbox{with }\; x^2+y^2\le 10\\
z=3
\end{cases}
$$
It follows that
$$
\Phi = -\iint_{x^2+y^2\le 2} \pmatrix{1/3\\1/3\\1/3}\cdot \pmatrix{0\\0\\-1}\; dS -\iint_{x^2+y^2\le 10} \pmatrix{1/3\\1/3\\1/3}\cdot \pmatrix{0\\0\\1}\; dS =\frac{A(S_1)}{3}-\frac{A(S_2)}{3} = -\frac{8\pi}{3}
$$
|
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|
Find a minimum of $x^2+y^2$ under the condition $x^3+3xy+y^3=1$ As in the title, I've tried to find a maximum and mininum of $x^2+y^2$ when $x^3+3xy+y^3=1$ holds. It is not too hard to show that $x^2+y^2$ has no maximum, but I can't find a minimum.
Lagrange multiplier gives a dirty calculation so I can't handle it. Is there any elegant way to find it? Thanks for any help.
p.s. Sorry. I make a typo in the $xy$-coefficient.
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Let $x=u+v$ and $y=u-v$. Then $x^2+y^2=2(u^2+v^2)$, while the equation $x^3+3xy+y^3=1$ becomes $2u^3+6uv^2+3(u^2-v^2)=1$, or
$$2u^3+3u^2-1=-3v^2(2u-1)$$
Factoring the cubic on the left hand side, we obtain
$$(u+1)^2(2u-1)=-3v^2(2u-1)$$
So either $2u-1=0$, in which case $v$ can be anything, or else $(u+1)^2=-3v^2$, which, because of the minus sign, is possible only if $u+1=v=0$. The latter possibility leads to
$$x^2+y^2=2((-1)^2+0^2)=2$$
while the former leads to
$$x^2+y^2=2\left(\left(1\over2\right)^2+v^2\right)={1\over2}+2v^2$$
which is clearly minimized when $v=0$. So the minimum value of $x^2+y^2$ is $1\over2$, and it occurs at $(x,y)=({1\over2}+0,{1\over2}-0)=({1\over2},{1\over2})$.
|
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|
Infimum of two variable function on the closed unit disc I am trying to compute
$$\inf_{x,y} \frac{ax-by}{1+x^2+y^2}$$
subject to the constraint $x^2+y^2 \leq 1$. Here $a,b$ are any two fixed, real numbers. I am having trouble computing this using standard derivative techniques, and Wolfram alpha is unable to recognize $a,b$ as constants. I have plugged in $\pi$ and $e$ as constants into WA and gotten a result, but this obviously is not sufficient when trying to achieve a general result.
I suspect the answer is (assuming $a,b>0$, for example), that this infimum occurs at $x= \frac{-a}{\sqrt{a^2+b^2}}$ and $y= \frac{b}{\sqrt{a^2+b^2}}$, but I cannot rigorously show it, or get WA to give me a general answer.
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By using the Cauchy-Schwarz inequality, one has
$$ |f(x,y)|\le\frac{\sqrt{a^2+b^2}\sqrt{x^2+y^2}}{1+x^2+y^2}=\frac{\sqrt{a^2+b^2}}{\sqrt{x^2+y^2}+\frac{1}{\sqrt{x^2+y^2}}} $$
and "=" holds if and only if $\frac{a}{b}=\frac{x}{-y}$.
Using
$$ a^2+b^2\ge2ab$$
("=" holds if and only if $a=b$) one has
$$ |f(x,y)|\le\frac{\sqrt{a^2+b^2}\sqrt{x^2+y^2}}{1+x^2+y^2}=\frac{\sqrt{a^2+b^2}}{\sqrt{x^2+y^2}+\frac{1}{\sqrt{x^2+y^2}}}\le\frac{1}{2}\sqrt{a^2+b^2}. $$
Note that "=" holds if and only if
$$ \frac{a}{b}=\frac{x}{-y}, x^2+y^2=1.$$
Solving this, one has
$$ x=\mp\frac{a}{\sqrt{a^2+b^2}}, y=\pm\frac{b}{\sqrt{a^2+b^2}}.$$
Thus
$$ -\frac{1}{2}\sqrt{a^2+b^2}\le f(x)\le \frac{1}{2}\sqrt{a^2+b^2} $$
and hence when $x=-\frac{a}{\sqrt{a^2+b^2}}, y=\frac{b}{\sqrt{a^2+b^2}}$ the minimum of $f(x,y)$ has $-\frac{1}{2}\sqrt{a^2+b^2}$ .
|
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|
find all integer solutions for $xy+1=3(x+y)$ the original problem was to find all integer solutions for $xy+1=3(x+y)$.
$$xy+1=3(x+y)$$
$$xy+1=3x+3y$$
$$xy-3x=3y-1$$
$$x(y-3)=3y-1$$
$$x=\frac{3y-1}{y-3}$$
I was able to find all possible solutions using this method,
I was wondering if you can say that $3|xy+1$ or $x+y|xy+1$ and proceed to solve it in this fashion.
If it is possible please show me how.
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We want to find a factorization if possible, so consider $xy-3x-3y+k = k-1$ as a re-writing of the condition. Then we can take any value for $k$ to make a factorization possible, and $(x-3)(y-3) = xy-3k-3y+9$, so take $k=9$ to get
$$(x-3)(y-3) = 8$$
Then the integer ordered factor pairs of $8$ are
$$\{(-1,-8), (-2,-4), (-4,-2), (-8,-1),(1,8), (2,4), (4,2), (8,1)\}$$ and the possible solutions $$(x,y)\in\{(2,-5), (1,-1), (-1,1), (-5,2),(4,11), (5,7), (7,5), (11,4)\}$$
Looking at $(x+y)\mid(xy+1)$, it seems like the best result from that probably takes us back on the same track, ie
\begin{align}
j(x+y)&=xy+1 \\
xy-jx-jy+1 &= 0 \\
xy-jx-jy + j^2 &= j^2-1 \\
(x-j)(y-j) & = j^2-1
\end{align}
... except here of course we already know what $j$ is.
In a different problem,
$\begin{align}(x+y) &\mid(xy+1) \\
\Rightarrow(x+y) &\mid(xy+1)-(x+y) \\
\Rightarrow(x+y) &\mid(xy-x-y+1) \\
\Rightarrow (x+y) &\mid(x-1)(y-1)
\end{align}$
might be useful.
|
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|
Is it possible to integrate $\int \frac{\mathrm{d}x}{\sin^2 x+\sin x+1}$ without involving complex numbers? The integral is $$\int \frac{\mathrm{d}x}{\sin^2 x+\sin x+1}$$
Consider
$$\int \frac{\mathrm{d}x}{ax^2+bx+c}$$
There are a total of three cases, depending on the discriminant of $ax^2+bx+c$. Two of which are shown here (#1).
The third one, i.e. when $\Delta=0$, simply means evaluating $$\frac{1}{a}\int \frac{\mathrm{d}x}{\big(x+\frac{b}{2a}\big)^2} $$
For the first two cases, you can see different substitutions are used so as to prevent $i$ from appearing in the answer, hence resulting in an $\arctan$ function and a $\ln$ function respectively. The discrepancy arises when we change $+(4ac-b^2)$ to $-(b^2-4ac)$.
In general, is it possible to evaluate
$$\int \frac{\mathrm{d}x}{a\sin^2 x+b\sin x+c}$$ such that the result does not contain complex numbers when $\Delta_{\sin x}<0$?
The only approach I can think of when $\Delta_{\sin x}>0$ is by partial fraction decomposition, which differs from the method of substitution used above.
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Let $\displaystyle I = \int \frac {1}{\sin ^2x + \sin x + 1}dx$
Let $$\sin x = -\frac {(2-\sqrt {3})t + (2 +\sqrt {3})}{t + 1}\implies \cos x\ dx = \frac {2\sqrt {3}}{(t + 1)^2}dt$$
Then $$I = 2\sqrt 3 \int \frac {t + 1}{\left((6 - 3\sqrt 3)t^2 + (6 + 3\sqrt 3)\right)\sqrt {(4\sqrt {3} - 6)t^2 - (4\sqrt {3} + 6)}}\ dx$$
Now consider $$I_1 = \int \frac {t}{\left((6 - 3\sqrt 3)t^2 + (6 + 3\sqrt 3)\right)\sqrt {(4\sqrt {3} - 6)t^2 - (4\sqrt {3} + 6)}}\ dx$$
and substitute $ \ (4\sqrt {3} - 6)t^2 - (4\sqrt {3} + 6) = z^2$ Reduce into well known equation.
and Consider $$ \ I_2 = \int \frac {1}{\left((6 - 3\sqrt 3)t^2 + (6 + 3\sqrt 3)\right)\sqrt {(4\sqrt {3} - 6)t^2 - (4\sqrt {3} + 6)}}\ dx$$
and first put $\displaystyle \ t = \frac 1p$ after this substitution put $ \ (4\sqrt {3} - 6) - (4\sqrt {3} + 6)p^2 = u^2$
and we are done without using Complex number...
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|
Prove that $r = \frac{a+b-c}{2}\;\text{and}\; r_c = \frac{a+b+c}{2}$ if $\measuredangle C = 90^{\circ}$
In$\triangle ABC$, $\angle C$ is a right angle. Prove that $r = \frac{a+b-c}{2}\;\text{and}\; r_c = \frac{a+b+c}{2}.$ Legs are named in traditional way.
My Work
As $r = \frac{A}{s}$. So, here $r = \frac{ab}{a+b+c}$. Is there any way to prove this is equal to $\frac{a+b-c}{2}$?. Or else how to solve this is also the second part?
|
The first part:
$$(a+b+c)(a+b-c)=(a+b)^2-c^2 = a^2+b^2 + 2ab -c^2=2ab$$
by Pythagora's theorem.
The second part: $r_c=\frac{A}{s-c}=\frac{ab}{a+b-c}$ and it works with absolutely the same calculation above.
|
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|
How to show that $\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^3}\mathrm dx=3\pi$
Consider
$$I=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx
\qquad J=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^3}\mathrm dx$$
I want to show that $I=3\pi$ and that $I=J$.
First, we noticed that $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$
So it gives us an idea to try and factorise $x^4-x^2+1$ but cannot find any factors.
Integrate $I$ (We try some substitutions to see where it will get us to)
$x=\sqrt{u}$ then $dx={2\over \sqrt{u}}du$
$$I=16\cdot{1\over 2}\int_{0}^{\infty}{u^{3/2}\over (1-u+u^2)^4}\mathrm du$$
$u=\tan(y)$ then $du=\sec^2(y)dy$
$$I=8\int_{0}^{\pi/2}{\tan^{3/2}(y)\over (1-\tan(y)+\tan^2(y))^4}{\mathrm dy\over \cos^2(y)}$$
then simplified down to
$$I=128\int_{0}^{\pi/2}{\cos^6(y)\tan^{3/2}(y)\over (2-\sin(2y))^4}\mathrm dy$$
we further simplified down to
$$I={128\over 2^{3/2}}\int_{0}^{\pi/2}{\cos^3(y)\sin^{3/2}(2y)\over (2-\sin(2y))^4}\mathrm dy$$
Not so sure what is the next step.
|
$\begin{align}I&=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx
\qquad\\
&=\int_{0}^{\infty} \dfrac{16x^4(1+x^2)^4}{(1+x^6)^4} dx\\
&=\int_{0}^{\infty} \dfrac{16\cdot {{x}^{12}}+64\cdot {{x}^{10}}+96\cdot {{x}^{8}}+64\cdot {{x}^{6}}+16\cdot {{x}^{4}}}{(1+x^6)^4} dx\\
\end{align}$
Perform the change of variable $y=x^6$,
$\begin{align}
I&=\int_{0}^{\infty} \dfrac{\tfrac{8}{3}x^{\tfrac{7}{6}}+\tfrac{32}{3}x^{\tfrac{5}{6}}+16x^{\tfrac{1}{2}}+\tfrac{32}{3}x^{\tfrac{1}{6}}+\tfrac{8}{3}x^{-\tfrac{1}{6}}}{(1+x)^4} dx\\
&=\dfrac{8}{3}\text{B}\left(\dfrac{13}{6},\dfrac{11}{6}\right)+\dfrac{32}{3}\text{B}\left(\dfrac{13}{6},\dfrac{11}{6}\right)+16\text{B}\left(\dfrac{3}{2},\dfrac{5}{2}\right)+\dfrac{32}{3}\text{B}\left(\dfrac{7}{6},\dfrac{17}{6}\right)+\dfrac{8}{3}\text{B}\left(\dfrac{5}{6},\dfrac{19}{6}\right)\\
&=\dfrac{8}{3}\times \dfrac{35\pi}{648}+\dfrac{32}{3}\times \dfrac{35\pi}{648}+16\times \dfrac{\pi}{16}+ \dfrac{32}{3}\times\dfrac{55\pi}{648}+\dfrac{8}{3}\times \dfrac{91\pi}{648}\\
&=\boxed{3\pi}
\end{align}$
Addendum:
$B$ is the Euler beta function.
$\begin{align}
\text{B}\left(\dfrac{13}{6},\dfrac{11}{6}\right)&=\dfrac{\Gamma\left(\dfrac{13}{6}\right)\Gamma\left(\dfrac{11}{6}\right)}{\Gamma\left(4\right)}\\
&=\dfrac{1}{6} \left(\dfrac{7}{6}\Gamma\left(\dfrac{7}{6}\right)\right)\times \left(\dfrac{5}{6}\Gamma\left(\dfrac{5}{6}\right)\right)\\
&=\dfrac{1}{6} \left(\dfrac{7}{6}\times\dfrac{1}{6}\Gamma\left(\dfrac{1}{6}\right)\right)\times \left(\dfrac{5}{6}\Gamma\left(\dfrac{5}{6}\right)\right)\\
&=\dfrac{35}{1296}\Gamma\left(\dfrac{1}{6}\right)\Gamma\left(\dfrac{5}{6}\right)\\
&=\dfrac{35}{1296}\times \dfrac{\pi}{\sin\left(\pi\times \tfrac{1}{6}\right)}\\
&=\boxed{\dfrac{35\pi}{648}}
\end{align}$
|
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|
How to i solve this Exponential equation How to solve this exponential equation?
$$7 \cdot 3^{x+1} - 5^{x+2}= 3^{x+4}- 5^{x+3}$$
|
$$\\ 7 \cdot 3^{x+1} - 5^{x+2}= 3^{x+4}- 5^{x+3}\\ 125\cdot { 5 }^{ x }-25\cdot { 5 }^{ x }=81\cdot { 3 }^{ x }-21\cdot { 3 }^{ x }\\ 100\cdot { 5 }^{ x }=60\cdot { 3 }^{ x }\\ { \left( \frac { 5 }{ 3 } \right) }^{ x }=\frac { 3 }{ 5 } \\ x=-1$$
|
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|
Variance of number of heads in three flips of a coin selected from a set of fair and unfair coins We have $10$ coins, $2$ are two-tailed, $2$ are two-headed, the other $6$ are fair ones. We (randomly) pick a coin and we flip it $3$ times. Find the variance of the number of gotten heads.
My attempt:
$X$ - number of heads that we got
$\mathbb{P}\left(X=0\right)=\frac{2}{10}\cdot1\cdot1\cdot1 + \frac{6}{10}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}$ - we picked two-tailed coin or fair one
$\mathbb{P}\left(X=1\right)=\frac{6}{10}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot 3$ - we picked fair one and we have 3 possibilites: HHT, HTH, THH
$\mathbb{P}\left(X=2\right)=\frac{6}{10}\cdot{3\choose2}\cdot\left(\frac{1}{2}\right)^2\cdot\left(\frac{1}{2}\right)^1$ - again we picked fair one and we have to have 2 successes in 3 tries
$\mathbb{P}\left(X=3\right)=\frac{2}{10}\cdot1\cdot1\cdot1+\frac{6}{10}\cdot\left(\frac{1}{2}\right)^3$ - we can pick two-headed or fair one coin
And the rest is simple,
$\text{Var}X=\sum_{i=0}^3i^2\cdot\mathbb{P}\left(X=i\right)-\left(\sum_{i=0}^3i\cdot\mathbb{P}\left(X=i\right)\right)^2$
Is my solution correct?
|
That's one approach. It is good. Here's another,
We know $X\mid C \sim \mathcal{Bin}(3, C)$ and $\mathsf P(C=c)=\tfrac 15\mathbf 1_{c=0}+\tfrac 35\mathbf 1_{c=1/2}+\tfrac 15\mathbf 1_{c=1}$
So $\mathsf E(C)= \tfrac 1{2}$ and $\mathsf {Var}(C)=\tfrac 1{10}$ and $\mathsf E(C^2)=\frac 7{20}$
By the Law of Total Probability: $\mathsf E(X) \\ = \mathsf E(\mathsf E(X\mid C)) \\ = \mathsf E(3C) \\ = \frac 32$
Then by the Law of Total Variance: $\quad\mathsf {Var}(X) \\ = \mathsf{Var}(\mathsf E(X\mid C))+\mathsf E(\mathsf{Var}(X\mid C)) \\ = \mathsf {Var}(3C)+\mathsf E(3C(1-C)) \\ = \frac 9{10}+\frac 32-\frac {21}{20} \\ = \frac {27}{20}$
|
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|
Number of $(p,q)$ in $\mathbb R^2$ such that $x^4+px^2+q$ is divisible by $x^2+px+q$ I have found four through my attempt, but apparently the answer is that there are five pairs.
Mine were $(-1,0),(-1,1),(0,0),(0,1)$.
What am I missing?
|
Fun problem here's another approach using Vieta's formulas:
The sum of the roots of $x^2+px+q$ is $-p$. The sum of the roots of $x^4+px^2 + q$ is $0$. Similarly the product of the roots of both polynomials is $q$. In particular this tells us that if $x^2+px+q$ divides $x^4+px^2 + q$ the quotient would have to be $x^2-px+1$, unless $q=0$.
If $q=0$ the roots of $x^2+px+q$ are $0$ and $-p$. Zero is obiously a root of $x^4+px^2$, so we just need to plug $-p$ in and see when it's also a root. This gives us the equation $p^4+p^3 =0$. So we get two solutions $(p,q)= (0,0)$ and $(p,q)= (-1,0)$.
Suppose $q \ne 0$ now we can look at the coefficients of $x^2$ and $x$ in $(x^2+px+q)(x^2-px+1)$, and set them equal to $p$ and $0$ respectively. This gives us the equations:
$$q-p^2+1 = p$$
$$p-qp = 0$$
The second equation forces either $q=1$ or $p =0$, which if we plus into the other equation and solve gives us the ordered pairs $(p,q) = (-2,1)$, $(1,1)$, and $(0,-1)$.
|
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|
In how many ways can we roll a red die, a yellow die, and a black die, and get a sum of $9$? I know I can use generating functions. Each of the die has a generating function $x+x^2+x^3+x^4+x^5+x^6$, and so I need to find the coefficient of $x^9$ in the generating function of their sum, $(x+x^2+x^3+x^4+x^5+x^6)^3$. I am not sure how to do this, however, short of expanding it all out.
(I'm not just trying to get the answer but also the method behind it. Thanks a lot.)
|
Robert Frost's approach is probably the simplest, but here's a way to do it with generating functions.
We could just multiply it out. But we can save a bit of work by noticing that $(x + x^2 + x^3 + x^4 + x^5 + x^6)^3 = \left(\frac{x(1 - x^6)}{1 - x}\right)^3 = \frac{x^3 - 3x^9 + 3x^{15} - x^{21}}{(1 - x)^3}$.
We can drop the $x^{15}$ and $x^{21}$ terms because the powers are larger than 9, so we're left with finding the coefficient of $x^9$ in $(x^3 - 3x^9)(1-x)^{-3}$.
That's $\left[x^6\right](1 - x)^{-3} - 3 \left[x^0\right](1 - x)^{-3} = \left(\binom{3}{6}\right) - 3 \left(\binom{3}{0}\right) = 25$.
|
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|
Asymptotics for $k^c +k - 1 \choose k$ when $c\geq 1$ I am looking for the asymptotics of $n +k - 1 \choose k$ when $n = k^c$ for integer constant $c\geq 1$.
From this we get:
$$\frac{(n+k-1)!}{k! (n-1)!} \approx \frac{\sqrt{2 \pi(n+k-1)}\left(\frac{n+k-1}{e}\right)^{n+k-1}}{\sqrt{2 \pi k}\left(\frac{k}{e}\right)^k \sqrt{2\pi(n-1)}\left(\frac{n-1}{e}\right)^{n-1}} = \frac{\sqrt{(n+k-1)}\left(n+k-1\right)^{n+k-1}}{\sqrt{ k}\left(k\right)^k \sqrt{2\pi(n-1)}\left(n-1\right)^{n-1}}$$
We also have that,
$$
\frac{(n+k-1)!}{k! (n-1)!} = \frac{(k^c+k-1)!}{k! (k^c-1)!}
$$
I am not sure where to go from here.
Blindly copying from the wiki and assuming that "$k$ is much smaller than $n$" is satisfied, you get:
$$
{k^c +k - 1 \choose k} \approx \frac{(\frac{k^c+k-1}{k} - \frac{1}{2})^k e^k}{\sqrt{2 \pi k}}
$$
Does this mean that $${k^c +k - 1 \choose k} \sim \frac{k^{ck-1}e^k}{\sqrt{2 \pi k}}\;?$$
|
Note that $$\frac{k!}{(k^c+k-1)^k}\binom{k^c+k-1}{k}=\prod_{i=1}^k\frac{k^c+i-1}{k^c+k-1}=\prod_{i=1}^{k-1}\left(1-\frac{i}{k^c+k-1}\right)$$ Since $c>1$, the last product is asymptotically equivalent to $$\prod_{i=1}^{k-1}\left(1-\frac{i}{k^c}\right)\approx\prod_{i=1}^{k-1}\exp\left(-\frac{i}{k^c}\right)\approx\exp\left(-\frac{k^2}{2k^c}\right)$$ Hence, if $c>2$, $$\binom{k^c+k-1}{k}=\frac{(k^c+k-1)^k}{k!}\left(1+O\left(\frac1{k^{c-2}}\right)\right)$$ while $$\binom{k^2+k-1}{k}\sim\frac{(k^2+k-1)^k}{k!}e^{-1/2}\sim\frac{k^{2k}}{k!}e^{+1/2}$$ and, for every $1<c<2$, $$\binom{k^c+k-1}{k}\sim\frac{(k^c+k-1)^k}{k!}\exp\left(-\frac{k^{2-c}}2\right)\sim\frac{k^{ck}}{k!}\exp\left(+\frac{k^{2-c}}2\right)$$
|
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|
$2 \times 2$ Examples
Find examples of $2 \times 2$ matrices with $a_{12} = \frac{1}{2}$ for which (a) $A^2 = I$, (b) $A^{-1}=A^T$, and (c) $A^2 = A$.
What I thought was a trivial problem is turning out to be rather irritating. At the moment I am dealing with part (a). To solve the problem, I narrowed my search space of examples by taking $A$ to be invertible with unit determinant. If $A$ is invertible, then $A^2 = I$ implies $A = A^{-1}$, further narrowing the search space.
Taking $A = \begin{bmatrix} a & 1/2 \\ c & d \\ \end{bmatrix}$, $A = A^{-1}$ would imply $a=d$ and $c = - \frac{1}{2}$. From this I got $a = \frac{\sqrt{3}}{2}$. However, the calculations never add up. I have done these calculations just short of a trillion times, and even had wolfram do some of the computations, but $A^2$ never equals $I$. What I am doing wrong? Is there some error in my logic?
|
Let's start with what you have:
$$A = \begin{bmatrix} a & \frac{1}{2} \\ -\frac{1}{2} & d \end{bmatrix}$$
So then
$$A^2 = \begin{bmatrix} a & \frac{1}{2} \\ -\frac{1}{2} & d \end{bmatrix}\cdot \begin{bmatrix} a & \frac{1}{2} \\ -\frac{1}{2} & d \end{bmatrix} = \begin{bmatrix} a^2-\frac{1}{4} & \frac{1}{2}(a+d) \\ -\frac{1}{2}(a+d) & -\frac{1}{4}+d^2 \end{bmatrix}=I$$
So $a^2 - \frac{1}{4}=1$, or $a = \pm\frac{\sqrt{5}}{2}$. And as Jason points out in the comments, $d = \mp\frac{\sqrt{5}}{2}$.
|
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|
Solve differential equation $(x^2+xy)y'=x\sqrt{x^2-y^2}+xy+y^2$ Solve and find a particular solution that satisfies $y(1)=1$.
What is the type of this differential equation?
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$$(x^2+xy)y'=x\sqrt{x^2-y^2}+xy+y^2$$
$$y'=\dfrac{x\sqrt{x^2-y^2}+xy+y^2}{(x^2+xy)}$$
is an homogeneous differential equation. Let $u=\dfrac{y}{x}$ so
$$u'x+u=\dfrac{\sqrt{1-u^2}+u+u^2}{(u+1)}$$
or
$$\dfrac{u+1}{\sqrt{1-u^2}}du=\dfrac{dx}{x}$$
after integration (let $u=\sin t$) we have
$$t-\cos t=\ln Cx$$
with $y(1)=1$ we get $C=\exp(\dfrac{\pi}{2})$.
|
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|
Evaluate $ \int \frac{1}{\sin{x}\cos^{3}x}dx $ $$ \int \frac{1}{\sin{x}\cos^{3}x}dx $$
$$\Rightarrow \int \frac{1}{\sin{x}\cos^{3}x}{\cos{x}\over \cos{x}}dx$$
$$\Rightarrow \int \frac{\sec^{4}{x}}{\tan{x}}dx$$
$$\Rightarrow \int \frac{\sec^{2}(1+\tan^{2}x)}{\tan{x}}dx$$
$$Substitution \tan{x}=t \Rightarrow \sec^{2}xdx=dt$$
$$\Rightarrow \int \frac{1+t^{2}}{t}dt$$
$$\Rightarrow \int \frac{1}{t}dt+\int tdt $$
$$\Rightarrow \log|t|+\frac{t^{2}}{2}+C$$
$$\Rightarrow \log|\tan{x}|+\frac{1}{2}\tan^{2}x+C$$
Any other elegant way to do this?
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Another way, I am not sure it is an elegant way but is different than yours
$$\int \frac { 1 }{ \sin { x } \cos ^{ 3 } x } dx=\int { \frac { \sin ^{ 2 }{ x+\cos ^{ 2 }{ x } } }{ \sin { x } \cos ^{ 3 } x } dx= } \int { \frac { \sin { x } }{ \cos ^{ 3 }{ x } } } dx+\int { \frac { 1 }{ \sin { x\cos { x } } } dx } =\\ =-\int { \frac { d\cos { x } }{ \cos ^{ 3 }{ x } } +\int { \frac { \sin ^{ 2 }{ x+\cos ^{ 2 }{ x } } }{ \sin { x } \cos { x } } } } dx=\frac { 1 }{ 2\cos ^{ 2 }{ x } } +\int { \frac { \sin { x } }{ \cos { x } } dx+\int { \frac { \cos { x } }{ \sin { x } } dx } = } \\ =\frac { 1 }{ 2\cos ^{ 2 }{ x } } -\ln { \left| \cos { x } \right| +\ln { \left| \sin { x } \right| } +C= } \frac { 1 }{ 2\cos ^{ 2 }{ x } } +\ln { \left| \tan { x } \right| } +C$$
|
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|
How to integrate ${dx}/{\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{R}\right)}}$
If
$\displaystyle\frac{dx}{dt}=-\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{R}\right)}$
and $x(0)=R$, find $t$ when $x=0$.
I have no idea how to integrate $\displaystyle\frac{dx}{\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{R}\right)}}$. Any suggestions?
|
$$\displaystyle\frac{dx}{dt}=-\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{R}\right)}$$
$$ \int \frac{1}{\sqrt{\frac{2K}{m}\left( \frac{1}{x}-\frac{1}{R}\right)}}dx= \int - dt $$
$$ \frac{1}{\sqrt{\frac{2K}{m}}} \int \frac{1}{\sqrt{\frac{1}{x}-\frac{1}{R}}} dx=\int - dt$$
$$ \frac{1}{\sqrt{\frac{2K}{m}}} \int \frac{1}{\frac{1}{\sqrt{R}}\sqrt{\frac{R}{x}-1}} dx=\int - dt$$
$$ \frac{1}{\frac{1}{\sqrt{R}} \sqrt{\frac{2K}{m}}} \int \frac{1}{\sqrt{\frac{R}{x}-1}}=\int-dt$$
$$ u=\frac{R}{x}-1 \implies du=-\frac{R}{x^2}dx \implies x^2=\frac{R^2}{(u+1)^2}$$
$$ \frac{-R}{\frac{1}{\sqrt{R}} \sqrt{\frac{2K}{m}}} \int \frac{1}{\sqrt{u}(u+1)^2} du=\int-dt$$
Are you able to continue?
A hint would be to use the substiution $t=\sqrt{u}$ and then you will get a nice looking integral :)
|
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|
Find the sum of the infinite series: $ 1 + \frac{1+2}{2!} + \frac{1+2+2^2}{3!} +\frac{1+2+2^2+2^3}{4!}+...$
Find the sum of the infinite series
$$ 1 + \frac{1+2}{2!} + \frac{1+2+2^2}{3!} +\frac{1+2+2^2+2^3}{4!}+... ....$$
What I have done let
$$ S = \underbrace{\frac{1}{1!}}_{\text{1st Term}} + \underbrace{\frac{1+2}{2!}}_{\text{2nd Term}} + \underbrace{\frac{1+2+2^2}{3!}}_{\text{3rd Term}} + \underbrace{\frac{1+2+2^2+2^3}{4!}}_{\text{4th Term}} +... ....$$
I can see the denominator can be written as such but I'm not sure how to manipulate the numerator?
$$ S = \sum^{\infty}_{n=1} \frac{\text{?}}{n!} $$
|
The numerator is simply geometric progression,
$$1 + 2 + 2^2 + \cdots + 2^{n-1} = 2^n - 1$$
Therefore,
$$\sum_{n=1}^{\infty} \frac{2^n - 1}{n!} = \sum_{n = 1}^{\infty}\frac{2^n}{n!} - \sum_{n = 1}^{\infty}\frac{1}{n!} = (e^2 - 1) - (e - 1) = e^2 -e$$
|
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|
Limit of function with natural logarithm I need help solving this problem. I tried L'hospital and rearranging but nothing worked.
$$
\lim_\limits{x→∞} x^2\left(\ln\left(1 + \frac{1}{x}\right)- \frac{1}{x+1}\right)
$$
|
We have
$$x(\frac{\ln (1+\frac{1}{x})}{\frac{1}{x}})-\frac{x^2}{1+x}$$
Let $x=\frac{1}{y}$. We have,
$$=\frac{1}{y}\frac{ \ln (1+y)}{y}-\frac{1}{y^2+y}$$
This is,
$$=\frac{1}{y} \frac{ \ln (1+y)}{y}-\frac{1}{y}+\frac{1}{y+1}$$
$$=\frac{1}{y}(\frac{\ln(1+y)}{y}-1)+\frac{1}{y+1}$$
$$=\frac{\ln(1+y)-y}{y^2}+\frac{1}{y+1}$$
The first limit, as $y \to 0^+$ can be computed by LH used twice.
$$ \to -\frac{1}{2}+1=\frac{1}{2}$$
|
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|
Find the minimum real number $\lambda$ so that the following relation holds ($x>y$): $\lambda\frac{x^3-y^3}{(x-y)^3}\geq \frac{x^2-y^2}{(x-y)^2}$
ّFind the minimum real number $\lambda$ so that the following relation holds for arbitrary real numbers $x,y$($x>y$): $$\lambda\frac{x^3-y^3}{(x-y)^3}\geq \frac{x^2-y^2}{(x-y)^2}$$
I simplified the relation into the following: $(\lambda-1)x^2+\lambda xy+(\lambda +1)y^2\geq0$ , but can't go further...
|
For $\lambda$, any value above $0$ is possible.
$\lambda\dfrac{x^3-y^3}{(x-y)^3}\geq \dfrac{x^2-y^2}{(x-y)^2}$
$\lambda\dfrac{x^3-y^3}{(x-y)}\geq (x^2-y^2)$
$\lambda (x^2+xy+y^2) \geq x^2-y^2$
$\lambda \geq \dfrac{x^2-y^2}{x^2+xy+y^2} \cdots(1)$
$(x-y)^2 \gt 0 \implies x^2+y^2 \gt 4xy$
Also $(x^2-y^2) \gt 0$
Since numerator and denominator of equation $(1)$ is postive, $\lambda$ is positive.
As $x$ becomes close to $y$, $\lambda$ tends to $0$
|
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|
Find the number of integer solutions of $x_1 + x_2 + x_3 = 16$, with $x_i \geq 0$, $x_1$ odd, $x_2$ even, $x_3$ prime. Find the number of integer solutions to $x_1+x_2+x_3=16$, with $x_i \geq 0$, $x_1$ odd, $x_2$ even , $x_3 $ prime.
My attempt:
Is the same as the number of ways to choose $16$ objects from $3$ distinct objects, i.e., $${16+3-1\choose 3-1}={18\choose 2}$$
|
We need this:
$x_1+x_2+x_3=16$ and $x_1=2k_1+1,~x_2=2k_2,~k_1,k_2\ge0$ and $x_3\in\{2,3,5,7,11,13\}$.
Since $x_1+x_2$ is odd then $x_3$ should be odd so $x_3\in\{3,5,7,11,13\}$.
Let's find the answer for all $x_3$.
Note that: $x_1+x_2=n$, has $\left(\binom{n}{2}\right)$, since $n$ is odd then $x_1$ is odd and $x_2$ is even or $x_1$ is even and $x_2$ is odd, then when $x_1$ is even and $x_2$ is odd we will change them, but whit this we are counting 2 times the number of solutions, then there are $\displaystyle\frac{\left(\binom{n}{2}\right)}{2}=\frac{\binom{n+2-1}{2-1}}{2}=\frac{\binom{n+1}{1}}{2}=\frac{\frac{(n+1)!}{1!n!}}{2}=\frac{n+1}{2}$ ways to choose $x_1$ and $x_2$ such that $x_1+x_2=n$ and $x_1$ is odd and $x_2$ is even.
1-) $x_3=3\Rightarrow x_1+x_2=13\Rightarrow$ ways: $\frac{14}{2}=7$.
2-) $x_3=5\Rightarrow x_1+x_2=11\Rightarrow$ ways: $\frac{12}{2}=6$.
3-) $x_3=7\Rightarrow x_1+x_2=9\Rightarrow$ ways: $\frac{10}{2}=5$.
4-) $x_3=11\Rightarrow x_1+x_2=5\Rightarrow$ ways: $\frac{6}{2}=3$.
5-) $x_3=13\Rightarrow x_1+x_2=3\Rightarrow$ ways: $\frac{4}{2}=2$.
Then there are $7+6+5+3+2=23$ integer solutions to $x_1+x_2+x_3=16$, with $x_i\ge0,~x_1$ odd, $x_2$ even and $x_3$ prime.
|
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|
Proof in Arithmetic Progression My maths teacher at school asked a question which I am finding difficult to crack down.
We are given that $a^2 , b^2$ and $c^2$ are in AP. We need to prove that $\frac{a}{b+c} , \frac{b}{a+c}$ and $\frac{c}{a+b}$ are in AP.
This is what I tried.
Let the common difference of the AP be d . So,
$b^2 - a^2 = d \implies b-a = \frac{d}{a+b} ........(1)$
Similarly
$c^2 - b^2 = d \implies c-b = \frac{d}{b+c}. ........(2)$
Also,
$a^2 - c^2 = -2d \implies a-c = \frac{-2d}{a+c} .........(3) $
Now adding the three equations,
$$0 = \frac{d}{a+b} + \frac{d}{b+c} - \frac{2d}{a+c}
\implies \frac{2d}{a+c} = \frac{d}{a+b} + \frac{d}{b+c}$$
$$\implies \frac{2}{a+c} = \frac{1}{a+b} + \frac{1}{b+c}$$
So, $\frac{1}{a+b}, \frac{1}{a+c}$ and $\frac{1}{b+c}$ are in AP. How should I go further? Or if I am going wrong anywhere, please tell.
|
$$\frac{2b}{a+c}=\frac{a}{b+c}+\frac{c}{a+b}\Leftrightarrow \\
2b[b^2+b(a+c)+ac]=a[a^2+a(b+c)+bc]+c[c^2+c(a+b)+ab]\Leftrightarrow\\
2b^3+2b^2(a+c)=a^3+a^2(b+c)+c^3+c^2(a+b)\Leftrightarrow\\
2b^2(a+b+c)=a^2(a+b+c)+c^2(a+b+c)\Leftrightarrow \\
(a+b+c)(2b^2-a^2-c^2)=0$$
Can you finish?
|
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|
Find the remainder of $\frac{3^{11}-1}{2}$ divided by $9$ I have the following question:
Find the remainder of $$\frac{3^{11}-1}{2}$$ divided by $9$
I tried to reformat the question:
$$\frac{3^{11} -1 } {2} \times \frac{1}{9} = \frac{3^{11}-1}{18}$$
Since $3^2 = 9$
$$\frac{3^2(3^9) -1}{3^2 \times2}$$
I don't know where to go next. Anyway, this is one of my many attempts to solve this question, and most of them ends with a complicated solution. I don't want to use modular arithmetic for this question. A hint or anything will help me.
|
$$\frac{3^{11}-1}{2}=\frac{3-1}{2}(3^{10}+3^{9}+3^{8}+3^{7}+3^{6}+3^{5}+3^{4}+3^{3}+3^{2}+3+1)\equiv4$$
|
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|
Sum with irrational powers and binomial coefficients What is the value of:
$$\sum_{k=1}^{n} {\left(\left(-1\right)^{n-k}{n+1 \choose k+1}\left(\frac{\sqrt5}{1+\sqrt5}\left(3+\sqrt5\right)^k-\frac{\sqrt5}{1-\sqrt5}\left(3-\sqrt5\right)^k-\frac{5}{2}\right)\right)}?$$
I am stuck because of the binomial coefficient there, because without it the sum would just be a bunch of geometric series.
|
$\sum_{k=1}^{n} {\left(\left(-1\right)^{n-k}{n+1 \choose k+1}\left(\frac{\sqrt5}{1+\sqrt5}\left(3+\sqrt5\right)^k-\frac{\sqrt5}{1-\sqrt5}\left(3-\sqrt5\right)^k-\frac{5}{2}\right)\right)}?
$
Here's a start.
I'm feeling too tired right now
to do more.
$\begin{array}\\
\sum_{k=1}^{n} {n+1 \choose k+1} x^k
&=\sum_{k=2}^{n+1} {n+1 \choose k} x^{k-1}\\
&=\frac1{x}\sum_{k=2}^{n+1} {n+1 \choose k} x^{k}\\
&=\frac1{x}\left(-1-(n+1)x+\sum_{k=0}^{n+1} {n+1 \choose k} x^{k}\right)\\
&=\frac1{x}\left((1+x)^{n+1}-1-(n+1)x\right)\\
\end{array}
$
Note that
$(3-\sqrt{5})(3+\sqrt{5})
=4
$.
If $x = -(3+\sqrt{5})$,
$\begin{array}\\
\sum_{k=1}^{n} {n+1 \choose k+1}(-1)^k (3+\sqrt{5})^k
&=\frac1{-(3+\sqrt{5})}\left((1-(3+\sqrt{5}))^{n+1}-1+(n+1)(3+\sqrt{5})\right)\\
&=\frac{-(3-\sqrt{5})}{4}\left((-1)^{n+1}(2+\sqrt{5})^{n+1}-1+(n+1)(3+\sqrt{5})\right)\\
\end{array}
$
That's all.
|
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|
How many passwords of length 8 with digits non-decreasing sequence and start with 5? Passwords have length of 8 characters.
The usable characters are the 10 digits 0,1,2,3,4,5,6,7,8,9.
How many passwords with digits only have non-decreasing sequence of digits and start with 5?
The sequence of eight digits $D_1, D_2, D_3, D_4, D_5, D_6, D_7, D_8$ is non-decreasing if $D_1 \leq D_2 \leq D_3 \leq D_4 \leq D_5 \leq D_6 \leq D_7 \leq D_8$
How can we solve this using combinations in a way that takes into account the possibilities of using any number of digits 5, 6, 7, 8 and 9.
|
One way to choose a password is to just choose where the strict increases in the password are, and how big they are. For example, the password
(5, 5, 5, 7, 7, 7, 8, 8)
has strict increases after positions 3 and 6, the first being a jump of size 2, and the second of size 1. Also, any sequence of positions to increase at, and jump sizes at those sites of total height at most 4 -- since we can't use digits bigger than 9 -- corresponds to a password.
One way to count these guys is to split them up first by what the last digit in the password is, i.e. how much total 'jumping' we did, and then by partitions of that number.
For example, say the last digit is 8. Then we jumped by 3 total: the three partitions of 3 are
$3 = 1 + 1 + 1 = 2 + 1,$
and the jumps could go in any of 7 possible positions (after any of the digits in the password except the last). Thus, in this case, there are
$\binom{7}{1} + \binom{7}{3} + 2 \binom{7}{2} = 84$
possible passwords.
Just do this for the other possible total jump heights 1, 2 and 3: the counts come out to
(Total jump = 0) One password only: all 5's.
(Total jump = 1, only partition of 1 is 1) $\binom{7}{1} = 7$.
(Total jump = 2, partitions of 2 are 2 and 1 + 1) $\binom{7}{1} + \binom{7}{2} = 28$.
(Total jump = 4, partitions of 4 are 4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1) $\binom{7}{1} + 2 \binom{7}{2} + \binom{7}{2} + \binom{7}{1}\binom{6}{2} + \binom{7}{4} = 210.$
Thus there are a total of $1 + 7 + 28 + 84 + 210 = 330$ such passwords.
|
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|
Prove that $\sum\limits_{cyc}\frac{a^2-bd}{b+2c+d}\geq0$
Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\frac{a^2-bd}{b+2c+d}+\frac{b^2-ca}{c+2d+a}+\frac{c^2-db}{d+2a+b}+\frac{d^2-ac}{a+2b+c}\geq0$$
This inequality is a similar to the following inequality of three variables.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^2-bc}{b+c+2a}+\frac{b^2-ca}{c+a+2b}+\frac{c^2-ab}{a+b+2c}\geq0,$$
which we can prove by the following reasoning.
$$\sum\limits_{cyc}\frac{a^2-bc}{b+c+2a}=\frac{1}{2}\sum\limits_{cyc}\frac{(a-b)(a+c)-(c-a)(a+b)}{b+c+2a}=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)\left(\frac{a+c}{b+c+2a}-\frac{b+c}{c+a+2b}\right)=\frac{1}{2}\sum_{cyc}\frac{(a-b)^2}{(b+c+2a)(c+a+2b)}\geq0,$$
but this idea does not help for the starting inequality.
We can make the following. By Holder
$$\sum_{cyc}\frac{a^2}{b+2c+d}=\sum_{cyc}\frac{a^3}{ab+2ac+ad}\geq\frac{(a+b+c+d)^3}{4\sum\limits_{cyc}(ab+2ac+ad)}=\frac{(a+b+c+d)^3}{8\sum\limits_{cyc}(ab+ac)}$$
and
$$\sum_{cyc}\frac{bd}{b+2c+d}=2(a+b+c+d)\left(\frac{bd}{(b+2c+d)(d+2a+b)}+\frac{ac}{(a+2b+c)(c+2d+a)}\right).$$
Thus, it remains to prove that
$$\frac{(a+b+c+d)^2}{16\sum\limits_{cyc}(ab+ac)}\geq\frac{bd}{(b+2c+d)(d+2a+b)}+\frac{ac}{(a+2b+c)(c+2d+a)}$$
and I don't see, what is the rest.
|
Update
By chance, I saw a nice proof as follows.
\begin{align}
\sum_{\mathrm{cyc}} \frac{a^2-bd}{b+2c+d}
&\ge \sum_{\mathrm{cyc}} \frac{a^2-\frac{(b+d)^2}{4}}{b+2c+d}\\
&= \sum_{\mathrm{cyc}} \Big(\frac{a^2-\frac{(b+d)^2}{4}}{b+2c+d} + \frac{b+d-2c}{4}\Big)\\
&= \sum_{\mathrm{cyc}} \frac{a^2-c^2}{b+2c+d}\\
&= \Big(\frac{a^2-c^2}{b+2c+d} + \frac{c^2-a^2}{d+2a+b}\Big) + \Big(\frac{b^2-d^2}{c+2d+a} + \frac{d^2-b^2}{a+2b+c}\Big)\\
&= \frac{2(a+c)(a-c)^2}{(b+2c+d)(d+2a+b)} + \frac{2(b+d)(b-d)^2}{(c+2d+a)(a+2b+c)}\\
&\ge 0.
\end{align}
Previously written
Donald Splutterwit gave a nice SOS (Sum of Squares) solution.
Actually, the Buffalo Way works.
After clearing the denominators, we need to prove that $f(a,b,c, d)\ge 0$ where $f(a,b,c,d)$ is
a homogeneous polynomial of degree four.
WLOG, assume that $d = \min(a,b,c,d)$. Let $c = d+s, \ b=d+t, \ a = d+r; \ s, t, r\ge 0$.
We have $f(d+r, d+t, d+s, d) = Ad^2 + Bd + C$ where
\begin{align}
A &= 24 r^2-48 r s+24 s^2+24 t^2, \\
B &= 16 r^3-16 r^2 s+8 r^2 t-16 r s^2-16 r s t+8 r t^2+16 s^3+8 s^2 t+8 s t^2+16 t^3, \\
C &= 2 r^4+2 r^3 s+3 r^3 t-8 r^2 s^2-3 r^2 s t-r^2 t^2+2 r s^3-3 r s^2 t \\
&\quad +4 r s t^2+3 r t^3+2 s^4+3 s^3 t-s^2 t^2+3 s t^3+2 t^4.
\end{align}
It suffices to prove that $A, B, C\ge 0$. Clearly $A\ge 0$. It is easy to prove that $B\ge 0$ using discriminant. The proof of $C\ge 0$ may be a little harder. Omitted here. However I also verified it by Mathematica Resolve.
Maybe someone can find a nice proof of $B, C\ge 0$.
|
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|
How prove $ax(a+x)+by(b+y)+cz(c+z)\ge 3(abc+xyz)$ let $a,b,c,x,y,z\ge 0$ and such
$$a+b+c=x+y+z$$
show that
$$ax(a+x)+by(b+y)+cz(c+z)\ge 3(abc+xyz)$$but it does not help for a proof of the starting inequality (at least I don't see, how it helps).
I tried also BW, but we get there something, which impossible to kill during a competition.
|
It's enough to prove that $x^2a+y^2b+z^2c+xyz\geq4abc$.
Let $x^2a+y^2b+z^2c+xyz<4abc$, $x=kp$, $y=kq$ and $z=kr$, where $k>0$ and
$p^2a+r^2b+q^2c+pqr=4abc$.
Hence, $p^2a+r^2b+q^2c+pqr=4abc>x^2a+y^2b+z^2c+xyz=k^2(p^2a+r^2b+q^2c+kpqr)$,
which says that $0<k<1$ and $a+b+c=x+y+z=k(p+q+r)<p+q+r$,
which is contradiction because we'll prove now that $a+b+c\geq p+q+r$,
where $a$, $b$, $c$, $p$, $q$ and $r$ are positives such that $p^2a+r^2b+q^2c+pqr=4abc$.
Indeed, we'll rewrite the last condition in the following form.
$$\frac{p^2}{4bc}+\frac{q^2}{4ac}+\frac{r^2}{4ab}+\frac{pqr}{4abc}=1.$$
Let $\cos\alpha=\frac{p}{2\sqrt{bc}}$, $\cos\beta=\frac{q}{2\sqrt{ac}}$ and $\cos\gamma=\frac{r}{2\sqrt{ab}}$.
Hence, $$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=1,$$
which says that $\alpha+\beta+\gamma=180^{\circ}$ and we need to prove that
$$a+b+c\geq2\sqrt{bc}\cos\alpha+2\sqrt{ac}\cos\beta+2\sqrt{ab}\cos\gamma.$$
Let $\Delta ABC$ is a triangle such that $\measuredangle A=\alpha$, $\measuredangle B=\beta$ and $\measuredangle C=\gamma$,
$\vec{u}\uparrow\uparrow \vec{CB}$, $|\vec{u}|=\sqrt{a}$, $\vec{v}\uparrow\uparrow \vec{BA}$, $|\vec{v}|=\sqrt{b}$, $\vec{w}\uparrow\uparrow \vec{AC}$, $|\vec{w}|=\sqrt{c}$ and since
$$(\vec{u}+\vec{v}+\vec{w})^2\geq0,$$
we are done!
|
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|
Use Newtons iterative process to show the following For the function
$$ f: = \cot(\sqrt x) + \frac{1}{\sqrt{x}}$$
with the initial approximate
$$x_0 = \pi^2\left(n-\frac{1}{2}\right)^2$$
Show that after one iteration
$$ x_{n+1} = \pi^2\left(n-\frac{1}{2}\right)^2\left(1+ \frac{1}{1+ \pi^2\left(n-\frac{1}{2}\right)^2}\right)^2$$
So we have
$$ f' = \frac{-x\csc^2(\sqrt{x})-1}{2x^{3/2}}$$
Then
$$ x_{n+1} = \pi^2\left(n-\frac{1}{2}\right)^2 - \frac{\cot(\pi\left(n-\frac{1}{2}\right))+\frac{1}{\pi\left(n-\frac{1}{2}\right)}}{\frac{-\pi^2\left(n-\frac{1}{2}\right)^2\csc^2\left(\pi\left(n-\frac{1}{2}\right)\right)-1}{2\pi^3\left(n-\frac{1}{2}\right)^3}}$$
I think that
$$ \cot\left(\pi\left(n-\frac{1}{2}\right)\right) = 0, \forall n \in \mathbb{N}$$
and
$$ \csc^2\left(\pi\left(n-\frac{1}{2}\right)\right) = 1, \forall n \in \mathbb{N}$$
So,
$$ x_{n+1} = \pi^2\left(n-\frac{1}{2}\right)^2 - \frac{\frac{1}{\pi\left(n-\frac{1}{2}\right)}}{\frac{-\pi^2\left(n-\frac{1}{2}\right)^2-1}{2\pi^3\left(n-\frac{1}{2}\right)^3}}$$
$$ = \pi^2\left(n-\frac{1}{2}\right)^2+ \frac{2\pi^2\left(n-\frac{1}{2}\right)^2}{\pi^2\left(n-\frac{1}{2}\right)^2-1}$$
$$ = \pi^2\left(n-\frac{1}{2}\right)^2\left(1 + \frac{2}{1+ \pi^2\left(n-\frac{1}{2}\right)^2}\right)$$
Which is not the correct answer! Can anyone spot a flaw?
|
From your resulto of $x_1$, I followed the calculations and get the same result and a usefull thing should be this.
If we use that, calling $\Delta := \pi(n-1/2)$
$$\left(1 + \frac{1}{1+\Delta^2}\right)^2 = 1 + \frac{2}{1 + \Delta^2} + \frac{1}{(1 + \Delta^2)^2}$$
From your result we have that
$$x_1 = \Delta^2\left(1 + \frac{1}{1+\Delta^2}\right)^2 - \frac{\Delta^2}{(1 + \Delta^2)^2} = R - C$$
Where $R$ is the answer and $C$ is a positive real number.
|
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How to solve $x^2 +\left(\frac{x}{x-1}\right)^2 =8$? I tried to solve this question but it turns into a 4th degree equation and I could only get one solution for this equation,i.e., 2.
It is to be evaluated for solutions. Thanks.
|
$$x^2 +\frac{x^2}{(x-1)^2} = 8$$
$$x^2 +\frac{x^2}{(x^2-2x + 1)} = 8$$
$$x^2 = (8 - x^2)(x^2-2x + 1)$$
$$x^2 = (8 - x^2)x^2-(8 - x^2)2x + (8 - x^2)$$
$$x^2 = 8x^2 - x^4 -16x + 2x^3 - x^2 + 8$$
$$0 = - x^4 -16x + 2x^3 + 6x^2 + 8 $$
$$(x-2)^2(x^2+2x-2)= 0$$
$$x= 2$$
$$x= \sqrt{3}-1 $$
$$x= -1 - \sqrt{3}$$
|
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|
$a\sqrt{\sin x}+b\sqrt{\cos x}≤(a^{4/3}+b^{4/3})^{3/4}$ prove that :
$$a,b>0\\,0<x<\pi/2$$
$$a\sqrt{\sin x}+b\sqrt{\cos x}≤(a^{4/3}+b^{4/3})^{3/4}$$
my try :
$$a\sqrt{\sin x}+b\sqrt{\cos x}=a(\sqrt{\sin x}+\frac{a}{b}\sqrt{\cos x})$$
$$\frac{a}{b}=\tan y$$
$$a\sqrt{\sin x}+b\sqrt{\cos x}=a(\sqrt{\sin x}+\tan y\sqrt{\cos x})$$
$$\frac{\sin y}{\cos y}=\tan y$$
$$a\sqrt{\sin x}+b\sqrt{\cos x}=a(\sqrt{\sin x}+\frac{\sin y}{\cos y}\sqrt{\cos x})$$
now?
|
Using AM-GM inequality,
$a^{4/3}+b^{4/3} \ge 2\sqrt{a^{4/3}b^{4/3}}$
$a^{4/3}+b^{4/3} \ge 2ab^{2/3}$
$\left(a^{4/3}+b^{4/3}\right)^{3/4} \ge 2^{3/4}{ab^{1/2}} \cdots(1)$
Using AM-GM inequality,
$a\sqrt{\sin x}+b\sqrt{\cos x} \ge 2\sqrt{ab\sqrt{\sin x \cos x}}$
$a\sqrt{\sin x}+b\sqrt{\cos x} \ge 2(ab)^{1/2}(\sin x\cos x)^{1/4} \cdots(2)$
Divide the equations
$\dfrac{a\sqrt{\sin x}+b\sqrt{\cos x}}{\left(a^{4/3}+b^{4/3}\right)^{3/4}} \ge 2^{1/4}(\sin x\cos x)^{1/4}$
$\dfrac{a\sqrt{\sin x}+b\sqrt{\cos x}}{\left(a^{4/3}+b^{4/3}\right)^{3/4}} \ge 2^{1/4}\left[\dfrac{\sin(2x)}{2}\right]^{1/4}$
$\dfrac{a\sqrt{\sin x}+b\sqrt{\cos
x}}{\left(a^{4/3}+b^{4/3}\right)^{3/4}} \ge [\sin(2x)]^{1/4} \cdots(3)$
Note that
$0 \le \sin(2x) \le 1$
or
$0 \le [\sin(2x)]^{1/4} \le 1$
Therefore, numerator of equation (3) must be less than or equal to denominator
i.e., $a\sqrt{\sin x}+b\sqrt{\cos x} \le
\left(a^{4/3}+b^{4/3}\right)^{3/4}$
|
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Polynomial system If there are 3 numbers $x,y,z$ satisfying
$f=x+y+z=3$ , $g=x^2+y^2+z^2=5$ , $h=x^3+y^3+z^3=7$ then prove that they also satisfy
$x^4+y^4+z^4=9$ but not $x^5+y^5+z^5=11$
I dont know how to tackle this to be honest, i have started trying to write $x^4+y^4+z^4-9$ as $r_1\times f + r_2\times g+ r_3\times h$ where $r_1,r_2,r_3$ in $C[x,y,z]$ but then i realised that i cant possibly do that by hand.This was an exercise on my first computational algebra course so we havent really learnt anything much yet. I would like a hint if possible.
|
Systematically work order by order
\begin{eqnarray*}
(x+y+z)^2= x^2+y^2+z^2+2(xy+yz+zx)
\end{eqnarray*}
right firstly we do not need to write everything out; We shall use the following short hand
\begin{eqnarray*}
(\sum x)^2= \sum x^2+2\sum xy
\end{eqnarray*}
so $\sum xy=2$
\begin{eqnarray*}
(\sum x)(\sum x^2)= \sum x^3+\sum x^2y
\end{eqnarray*}
so $\sum x^2y=8$
\begin{eqnarray*}
(\sum x)(\sum xy)= \sum x^2y+3xyz
\end{eqnarray*}
so $xyz=-2/3$.
\begin{eqnarray*}
(\sum xy)^2= \sum x^2y^2+2xyz(\sum x)
\end{eqnarray*}
so $\sum x^2y^2=8$
\begin{eqnarray*}
(\sum x^2)^2= \sum x^4+2(\sum x^2y^2)
\end{eqnarray*}
so $\sum x^4=9$ ... the first part is shown.
\begin{eqnarray*}
(\sum xy)(\sum x^3)= \sum x^4y+xyz(\sum x^2)
\end{eqnarray*}
so $\sum x^4y=52/3$
\begin{eqnarray*}
(\sum x) (\sum x^4)= \sum x^5+(\sum x^4y)
\end{eqnarray*}
so $\sum x^5=29/3$ ... the second part is shown.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $1+\left(\frac{1+\frac12}{2}\right)^2+\left(\frac{1+\frac12+\frac13}{3}\right)^2+\left(\frac{1+\frac12+\frac13+\frac14}{4}\right)^2+...$ Evaluate:
$$S_n=1+\left(\frac{1+\frac12}{2}\right)^2+\left(\frac{1+\frac12+\frac13}{3}\right)^2+\left(\frac{1+\frac12+\frac13+\frac14}{4}\right)^2+...$$
a_n are the individual terms to be summed.
My Try :
\begin{align}
&a_1=1\\
&a_2=\left(\frac{3}{4}\right)^2=\frac{9}{16}\\
&a_3=\left(\frac{11}{18}\right)^2\\
&a_4=\left(\frac{25}{48}\right)^2
\end{align}
now :?
|
By setting $H_n = \sum_{k=1}^{n}\frac{1}{k}$ we have to compute $\sum_{n\geq 1}\left(\frac{H_n}{n}\right)^2$. We may notice that
$$ \sum_{n=1}^{N}\frac{H_n}{n}=\sum_{1\leq m\leq n\leq N}\frac{1}{mn}=\frac{H_N^2+H_N^{(2)}}{2}\tag{1}$$
and for the same reason
$$ \sum_{n=1}^{N}\frac{H_n^{(2)}}{n^2} = \frac{1}{2}\left[\left(\sum_{n=1}^{N}\frac{1}{n^2}\right)^2+\sum_{n=1}^{N}\frac{1}{n^4}\right]\stackrel{N\to +\infty}{\longrightarrow}\frac{\zeta(2)^2+\zeta(4)}{2}=\frac{7\pi^4}{360} \tag{2}$$
Since $-\log(1-x)=\sum_{n\geq 1}\frac{1}{n}\,x^n$, by multiplying both sides by $\frac{1}{1-x}$ and applying termwise integration we have
$$ \sum_{n\geq 1}\frac{H_{n}}{n}\,x^{n} = \text{Li}_2(x)+\frac{1}{2}\log^2(1-x) \tag{3}$$
hence by $(1)$ it follows that:
$$ \sum_{N\geq 1}\frac{H_N^2+H_N^{(2)}}{2}x^{N} = \frac{\text{Li}_2(x)}{1-x}+\frac{1}{2}\cdot\frac{\log^2(1-x)}{1-x}\tag{4} $$
and by multiplying both sides of $(4)$ by $-\frac{2\log x}{x}$ and performing termwise integration over $(0,1)$:
$$ \sum_{N\geq 1}\frac{H_N^2+H_N^{(2)}}{N^2} = -\int_{0}^{1}\left[\frac{2\text{Li}_2(x)\log(x)}{x(1-x)}+\frac{\log^2(1-x)\log(x)}{x(1-x)}\right]\,dx.\tag{5} $$
The integral $-\int_{0}^{1}\frac{\log^2(1-x)\log(x)}{x(1-x)}\,dx$ can be computed by differentiating Euler's beta function, and it equals $\frac{\pi^4}{36}$. Since $\int\frac{\log(x)}{x(1-x)}\,dx=\frac{1}{2}\log^2(x)+\text{Li}_2(1-x)$ and $\frac{d}{dx}\text{Li}_2(x)=-\frac{\log(1-x)}{x}$, by integration by parts the whole problem boils down to computing:
$$ I = \int_{0}^{1}\frac{\text{Li}_2(x)\log(x)}{1-x}\,dx \tag{6}$$
but we have already done that in $(2)$, since $\frac{\text{Li}_2(x)}{1-x}=\sum_{n\geq 1}H_n^{(2)}x^n.$ Collecting pieces,
$$ \sum_{n\geq 1}\left(\frac{H_n}{n}\right)^2 = \color{red}{\frac{17\pi^4}{360}}.$$
|
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|
All possible ways scores $(1,2,3)$ add up to $n$ Could someone please explain mathematical explanation behind this?
You can win three kinds of basketball points, 1 point, 2 points, and 3 points. Given a total score $n$, print out all the combination to compose $n$.
Examples:
For n = 1, the program should print following:
1
For n = 2, the program should print following:
1 1
2
For n = 3, the program should print following:
1 1 1
1 2
2 1
3
For n = 4, the program should print following:
1 1 1 1
1 1 2
1 2 1
1 3
2 1 1
2 2
3 1
Algorithm:
*
*At first position we can have three numbers 1 or 2 or 3. First put 1 at first position and recursively call for n-1.
*Then put 2 at first position and recursively call for n-2.
*Then put 3 at first position and recursively call for n-3.
*If n becomes 0 then we have formed a combination that compose n, so print the current combination.
I've solved in using JS as per below. But I don't quite understand the mathematical reasoning behind it.
https://jsfiddle.net/d2Lft7d1/
|
It is a Fibonacci like sequence.
Let $\,A_{\small n}\,$ be the total number of $\,(1,\,2,\,3)\,$ combinations that compose $\,n\,$, Then:
$$ A_{\small 1}=1,\,A_{\small 2}=2,\,A_{\small 3}=4,\quad\color{red}{A_{\small n}=A_{\small n-1}+A_{\small n-2}+A_{\small n-3}} \\[4mm] \Rightarrow\quad \left\{A_{\small n}\right\}=\left\{1,\,2,\,4,\,7,\,13,\,24,\,44,\,\cdots\right\} $$
And the idea behind that for $\,n\gt3\,$, you will have the ability to add a Most Significant Digit (MSD) equals $\,1,\,2,\text{ or } \,3\,$. This should left you with $\,n-1,\,n-2,\text{ and } \,n-3\,$ respectively. For Example:
$$ \begin{align}
n &=5 \\[2mm]
\text{MSD} &=\color{red}{1} \quad\Rightarrow\text{ The comination of }\,(n-1=4)= \begin{cases} \color{red}{1}\,1\,1\,1\,1 \\ \color{red}{1}\,1\,1\,2 \\ \color{red}{1}\,1\,2\,1 \\ \color{red}{1}\,1\,3 \\ \color{red}{1}\,2\,1\,1 \\ \color{red}{1}\,2\,2 \\ \color{red}{1}\,3\,1 \end{cases} \\[2mm]
\text{MSD} &=\color{blue}{2} \quad\Rightarrow\text{ The comination of }\,(n-2=3)= \begin{cases} \color{blue}{2}\,1\,1\,1 \\ \color{blue}{2}\,1\,2 \\ \color{blue}{2}\,2\,1 \\ \color{blue}{2}\,3 \end{cases} \\[2mm]
\text{MSD} &=\color{Green}{3} \quad\Rightarrow\text{ The comination of }\,(n-3=2)= \begin{cases} \color{Green}{3}\,1\,1 \\ \color{Green}{3}\,2 \end{cases} \\[2mm]
A_{\small5} &= \color{red}{A_{\small4}}+\color{blue}{A_{\small3}}+\color{green}{A_{\small2}} = \color{red}{7}+\color{blue}{4}+\color{green}{2} = 13
\end{align} $$
For other similar combination $\,\left({\small\text{e.g }}\,(1,2)\,,(1,2,4)\,,\cdots\right)\,$, we start by computing the first required terms, then we apply the concept of Fibonacci sequence and Most Significant Digit (MSD).
$\underline{\bf(1,2)}$:
$$ \begin{align}
(n=1) &\rightarrow \begin{cases} \color{red}{1} \end{cases} \qquad\Rightarrow\, A_{\small 1}=1 \\[2mm]
(n=2) &\rightarrow \begin{cases} \color{blue}{1}\,\color{red}{1} \\ \color{blue}{2} \end{cases} \quad\Rightarrow\, A_{\small 2}=2 \\[2mm]
A_{\small n} &= A_{\small n-1}+A_{\small n-2} = \left\{1,\,2,\,3,\,5,\,8,\,13,\,21,\,\cdots\right\}
\end{align} $$
$\underline{\bf(1,2,4)}$:
$$ \begin{align}
(n=1) &\rightarrow \begin{cases} \color{red}{1} \end{cases} \qquad\qquad\Rightarrow\, A_{\small 1}=1 \\[2mm]
(n=2) &\rightarrow \begin{cases} \color{blue}{1}\,\color{red}{1} \\ \color{blue}{2} \end{cases} \quad\qquad\Rightarrow\, A_{\small 2}=2 \\[2mm]
(n=3) &\rightarrow \begin{cases} \color{green}{1}\,\color{blue}{1}\,\color{red}{1} \\ \color{green}{1}\,\color{blue}{2} \\ \color{green}{2}\,\color{red}{1} \end{cases} \qquad\Rightarrow\, A_{\small 3}=3 \\[2mm]
(n=4) &\rightarrow \begin{cases} 1\,\color{green}{1}\,\color{blue}{1}\,\color{red}{1} \\ 1\,\color{green}{1}\,\color{blue}{2} \\ 1\,\color{green}{2}\,\color{red}{1} \\ 2\,\color{blue}{1}\,\color{red}{1} \\ 2\,\color{blue}{2} \\ 4 \end{cases} \quad\Rightarrow\, A_{\small 4}=6 \\[2mm]
A_{\small n} &= A_{\small n-1}+A_{\small n-2}+A_{\small n-4} = \left\{1,\,2,\,3,\,6,\,10,\,18,\,31,\,\cdots\right\}
\end{align} $$
|
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|
How to simplify $\frac{75}{8}\times\sqrt{\frac{a^3}{9}-\frac{a^3}{25}}$ How would you go about simplifying the expression $\frac{75}{8}\times\sqrt{\frac{a^3}{9}-\frac{a^3}{25}}$?
|
$$\frac{75}{8}\sqrt{\frac{a^3}{9}-\frac{a^3}{25}}=\frac{75}{8}\sqrt{\frac{25a^3-9a^3}{225}}=\frac{75}{8}\frac{\sqrt{16}\sqrt{a^3}}{\sqrt{225}}=\frac{75}{8}\frac{4}{15}\sqrt{a^3}=2.5\sqrt{a^3}$$
|
{
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"url": "https://math.stackexchange.com/questions/2166160",
"timestamp": "2023-03-29T00:00:00",
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|
Limit of a infinite series $$\frac{2}{2} + \frac{2\cdot 5}{2\cdot 9} + \frac{2\cdot 5\cdot 10}{2\cdot 9\cdot 28} + \cdots + \frac{2\cdot 5\cdot 10 \cdots (n^2+1)}{2\cdot 9\cdot 28\cdots (n^3+1)}\tag1$$
For this series $(1)$, how would one go about applying the comparison test to check for convergence or divergence?
|
The general term of the series, $a_n$ is given by
$$a_n=\prod_{k=1}^n\frac{(k^2+1)}{(k^3+1)}$$
Then, we see that
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2+1}{(n+1)^3+1}\to 0\implies \text{the series converges}$$
|
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|
Finding a basis for a set of matrices in a vector space. I am trying to find a basis for the following vector space:
V = {2x2 matrices A | $\bigl( \begin{smallmatrix} 1 & 2 \\ 2 & 3 \end{smallmatrix} \bigr)$A = A$\bigl( \begin{smallmatrix} 1 & 2 \\ 2 & 3 \end{smallmatrix} \bigr)$}
So far, I have augmented $\bigl( \begin{smallmatrix} 1 & 2 \\ 2 & 3 \end{smallmatrix} \bigr)$ with the identity matrix to find that A = $\bigl( \begin{smallmatrix} -3 & 2 \\ 2 & -1 \end{smallmatrix} \bigr)$. Is the identity matrix the only basis for this question?
|
No, it is not. For example, also $\;A\;$ and all its powers commute with A, and also for example
$$\begin{pmatrix}-1&1\\1&0\end{pmatrix}\;\ldots$$
To solve this, write
$$B:=\begin{pmatrix}a&b\\c&d\end{pmatrix}\;,\;\;\text{so that}\;\;AB=BA\iff \begin{pmatrix}1&2\\2&3\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&2\\2&3\end{pmatrix}\iff$$
$$\iff\begin{cases}a+2c=a+2b\\b+2d=2a+3b\\2a+3c=c+2d\\2b+3d=2c+3d\end{cases}$$
Now solve this system....for example, from the first eq. we get $\;b=c\;$ already. Play around a little with this. The set of all the matrices commuting with $\;A\;$ is a subspace of dimension $\;2\;$ , and thus you'll need two linearly independent such matrices to have a basis.
|
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|
Number of ways in which 3 people can throw a normal die to have a total score of 11
Number of ways in which 3 people can throw a normal die to have a total score of 11
My approach:
The answer can be obtained by finding the coefficient of $x^{11}$ in the expansion of $(x+x^2+x^3+x^4+x^5+x^6)^3$.
General term $T=\frac{3}{a!b!c!d!e!f!} x^{a+2b+3c+4d+5e+6f}$
where $a+b+c+d+e+f=3$ and $a+2b+3c+4d+5e+6f=11$
But I can't figure out how to solve further.
|
Here's a pen-and-paper approach.
First, let's compute the number of ways to throw $2$ through $12$ with two dice.
\begin{eqnarray}
N(2)&= 1 &: \{1+1\}\\
N(3)&= 2 &: \{1+2,2+1\}\\
N(4)&= 3 &: \{1+3,2+2,3+1\}\\
N(5)&= 4 &: \{1+4,2+3,3+2,4+1\}\\
N(6)&= 5 &: \{1+5,2+4,3+3,4+2,5+1\}\\
N(7)&= 6 &: \{1+6,2+5,3+4,4+3,5+2,6+1\}\\
N(8)&= 5 &: \{2+6,3+5,4+4,5+3,6+2\}\\
N(9)&= 4 &: \{3+6,4+5,5+5,6+4\}\\
N(10)&= 3 &: \{4+6,5+5,6+4\}\\
N(11)&= 2 &: \{5+6,6+5\}\\
N(12)&= 1 &: \{6+6\}\\
\end{eqnarray}
Then for three dice, these are all the possibilities, where $N(11-d)$ is the case where the first person rolls $d$.
\begin{eqnarray}
N(11-1) &+& N(11-2) + N(11-3) + N(11-4) + N(11-5) + N(11-6)\\
&=& N(10) + N(9) + N(8) + N(7) + N(6) + N(5)\\
&=& 3+4+5+6+5+4\\
&=& 27
\end{eqnarray}
|
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|
Prove that $\frac{(2a+b+c)^2}{2a^2+(b+c)^2}+\frac{(2b+c+a)^2}{2b^2+(c+a)^2}+\frac{(2c+a+b)^2}{2c^2+(a+b)^2} \le 8$
Prove that $$\frac{(2a+b+c)^2}{2a^2+(b+c)^2}+\frac{(2b+c+a)^2}{2b^2+(c+a)^2}+\frac{(2c+a+b)}{2c^2+(a+b)^2} \le 8$$.
MY ATTEMPT:I want to make a relation between $a,b,c$. By trial I found that if we put $a=b=c=1$ then the above inequality holds(equality also holds). So by trial I assume that $a+b+c=3$. After that the three functions become of the form of the function below:
$f(x)=\frac{(x+3)^2}{2x^2+(3-x)^2}$.
I calculate the function and found that :
$f(x) \le ⅓(4x+4)$.
Am I do right . Anybody has other ideas.
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For non-negatives $a$, $b$ and $c$ let $a+b+c=3$.
Hence,
$$8-\sum_{cyc}\frac{(2a+b+c)^2}{2a^2+(b+c)^2}=\sum_{cyc}\left(\frac{8}{3}-\frac{(a+3)^2}{2a^2+(3-a)^2}\right)=$$
$$=\frac{1}{3}\sum_{cyc}\frac{(a-1)(7a-15)}{a^2-2a+3}=\frac{1}{3}\left(\sum_{cyc}\frac{(a-1)(7a-15)}{a^2-2a+3}+4(a-1)\right)=$$
$$=\sum_{cyc}\frac{(a-1)^2(4a+3)}{3(a^2-2a+3)}\geq0$$
Also there is the following.
It's enough to prove that
$$\frac{(2a+b+c)^2}{2a^2+(b+c)^2}\leq\frac{4(4a+b+c)}{3(a+b+c)},$$
which is $(2a-b-c)^2(5a+b+c)\geq0$.
|
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|
Solve an integral $\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$ Solve an integral $$\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$$
Using the third substitution of Euler,
$$\sqrt{x^2-1}=(x-1)t\Rightarrow x=\frac{1+t^2}{1-t^2},dx=\frac{4t}{(1-t^2)^2}$$
we get an integral $$\int\frac{1-t^2}{1+t^4}dt=\frac{1}{2}\int\frac{1-\sqrt 2 t}{t^2-\sqrt 2 t+1}dt+\frac{1}{2}\int\frac{1+\sqrt 2t}{t^2+\sqrt 2t+1}dt$$
Using substitutions $$u=t^2-\sqrt 2t+1$$ and $$v=t^2+\sqrt 2t+1$$
we get $$\int\frac{1-t^2}{1+t^4}dt=-\frac{\sqrt 2}{4}\ln|u|+\frac{\sqrt 2}{4}\ln|v|=-\frac{\sqrt 2}{4}\ln|t^2-\sqrt 2t+1|+\frac{\sqrt 2}{4}\ln|t^2+\sqrt 2t+1|$$
From $$x=\frac{1+t^2}{1-t^2}\Rightarrow t=\sqrt{\frac{x-1}{x+1}}\Rightarrow$$
$$\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx=$$$$-\frac{\sqrt 2}{4}\ln\left|\frac{x-1}{x+1}-\sqrt 2\sqrt{\frac{x-1}{x+1}}+1\right|+\frac{\sqrt 2}{4}\ln\left|\frac{x-1}{x+1}+\sqrt 2\sqrt{\frac{x-1}{x+1}}+1\right|+c$$
Is there another, quicker method to solve this integral, rather than Euler substitution.
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$$ \begin{align}\int \frac{1}{(1+x^2)\sqrt{x^2 - 1}} \, dx &= \int \dfrac{\sinh u}{(\cosh^2 u + 1)\sqrt{\cosh^2 u - 1}} \, du && x = \cosh u \\ &= \int \dfrac{\sinh u}{(\cosh^2 u + 1)\sinh u} \, du \\ &= \int \dfrac{1}{\cosh^2 u + 1}\, du \\ && \end{align}$$
After which things become simple in terms of hyperbolic tsngent of u and it's inverse. Unsubstitute after that.
|
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|
Sum of series $\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$ Find the radius of convergence and the sum of power series
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$$
Radius of convergence is $R=1$, and the interval of convergence is $-1<x<1$.
I am having trouble in finding the sum.
Here is what I have tried.
$$\sum_{n=0}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}=\sum_{n=1}^{+\infty}(n+1)x^{2n+1}\int_0^1t^{2n}dt=\int_0^1\left(\sum_{n=1}^{+\infty}(n+1)x^{2n+1}t^{2n}\right)dt$$
$$=x\int_0^1\left(\sum_{n=1}^{+\infty}(n+1)(xt)^{2n}\right)dt$$
$$\sum_{n=1}^{+\infty}(n+1)(xt)^{2n}=\sum_{n=1}^{+\infty}n(xt)^{2n}+\sum_{n=1}^{+\infty}(xt)^{2n}$$
$$\sum_{n=1}^{+\infty}(xt)^{2n}=\frac{(xt)^2}{1-(xt)^2}$$
How to find the sum of $$\sum_{n=1}^{+\infty}n(xt)^{2n}?$$
EDIT:
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}=\left(\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}\right)'$$
$$\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}=\frac{1}{2}x^2\sum_{n=1}^{+\infty}\frac{x^{2n}}{2n+1}$$
$$=\frac{1}{2}x^2\sum_{n=1}^{+\infty}x^{2n}\int_0^1t^{2n}dt=\frac{1}{2}x^2\int_0^1\left(\sum_{n=1}^{+\infty}(xt)^{2n}\right)dt$$
$$=\frac{1}{2}x^2\int_0^1\frac{(xt)^2}{1-(xt)^2}=\frac{1}{2}x^4\int_0^1\frac{t^2}{1-(xt)^2}dt$$
$$=\frac{1}{2}x^4\cdot\frac{1}{x^3}(-x-\frac{1}{2}\ln|x-1|+\frac{1}{2}\ln|x+1|)\Rightarrow$$
$$\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}=\frac{1}{2}x\left(-x-\frac{1}{2}\ln|x-1|+\frac{1}{2}\ln|x+1|\right)\Rightarrow$$
$$\left(\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}\right)'=-x-\frac{1}{4}\left(\ln|x-1|+\frac{x}{x-1}\right)+\frac{1}{4}\left(\ln|x+1|+\frac{x}{x+1}\right)\Rightarrow$$
Finally,
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}=-x-\frac{1}{4}\left(\ln|x-1|+\frac{x}{x-1}\right)+\frac{1}{4}\left(\ln|x+1|+\frac{x}{x+1}\right)$$
Question: Is this correct?
|
$$S=\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$$
For the converge use Ratio Test.
$$2S=\sum_{n=1}^{+\infty}\frac{2n+1+1}{2n+1}x^{2n+1}=\sum_{n=1}^{+\infty}x^{2n+1}+\sum_{n=1}^{+\infty}\frac{x^{2n+1}}{2n+1}$$
The first sum is clearly an Infinite Geometric Series.
Now for the second part $$\ln(1+x)-\ln(1-x)=2\sum_{n=1}^{+\infty}\frac{x^{2n+1}}{2n+1}$$
|
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|
Integration on $ \int \sec^3 (2x) dx$ The answer is this but the coefficient is $\frac{1}{4}$. Why?
My steps:
$$\int \sec^3(2x)dx$$
Let $u = 2x$, then $\frac{1}{2}du = dx$
$$\frac{1}{2} \int \sec^3(u) du $$
Using int my parts: Let $a = \sec(u)$, then $da = \sec(u)\tan(u)du$. let $dv = \sec^2 (u) du$, then $v = \tan(u)$
$$\frac{1}{2} \int \sec^3(u) du = \frac{1}{2}\big(\sec(u)\tan(u) - \int \sec(u)\tan^2(u)du\big)$$
$$= \frac{1}{2}\big(\tan(u)\sec(u) - \int \sec^3 (u)du + \int \sec(u)du\big)$$
$$\frac{1}{2} \int \sec^3(u)du + \frac{1}{2} \int \sec^3(u)du = \frac{1}{2}\big(\tan(2x)\sec(2x) + \ln |\sec(2x)+\tan(2x)|\big) + C$$
$$\int \sec^3(2x)dx = \frac{1}{2}\big(\tan(2x)\sec(2x) + \ln |\sec(2x)+\tan(2x)|\big) + C$$
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What you derived is that $\int \sec^3 (u)~du=\text{blah}$. Then note that $\int \sec^3(2x)~dx=\frac12\int \sec^3(u)~du$
|
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|
Prove, $\gcd((a^b-1)/(a-1), (a^c-1)/(a-1))=1$ Problem : Prove (or disprove) that, $\gcd((a^b-1)/(a-1), (a^c-1)/(a-1))=1$ (greatest common divisor),
when $a, b, c $ are prime numbers and $a, b, c \geq 3, b \neq c$. $(a^b-1), (a^c-1)$ are factors of $(a^{bc}-1)$.
Click here to see the related question .
|
assume $c > b$
Also gcd $(a, 1 + a + ... + a^b) = 1$
Let $x = (a^c - 1)/(a-1) = 1 + a + a^2 + ... + a^{c-1}$
Let $y = (a^b - 1)/(a-1) = 1 + a + a^2 + ... + a^{b-1}$
$gcd(x,y) = gcd(x-y, y) = gcd((a^b + ... + a^{c-1}), y) = gcd ((a^b(1 + a + ... + a^{c-b-1}), y) = gcd((1 + a + ... + a^{c-b-1}), y)$
as we keep doing, we get $gcd(x,y) = gcd((1 + a + ... + a^{k-1}), (1 + a + ... + a^{b-1}))$ with $k < b$
eventually ending up with $gcd(1 + a + ... + a^m, a^n) = 1$ for some integer $m$ and $n$.
Therefore there is no common factor between $x$ and $y$.
|
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|
How to find limit of $\lim_{x\to 1} \frac{f(x) - f(1)}{x-1}$ if $f(x)=-\sqrt{25-x^2}$ I have a question , if then find $$\lim_{x\to 1} \frac{f(x) - f(1)}{x-1}$$
I got $f(1)=-\sqrt {24}$. Should I get limit = $\frac{\sqrt{24} - \sqrt{25-x^2}}{x-1}$ but answer is $\frac{1}{\sqrt{24}}$ . Should I use $f'(x)$? If I use $f'(x)$ then I got $-\frac{1}{\sqrt{24}}$ .I should not get minus in solution.
Please help, Thanks in advance.
|
First, let us evaluate $f(1)=f(x)=-\sqrt{25-1^2}=-2\sqrt{6}$
So we wish to evaluate,
$$\lim _{x \to 3}\frac{-\sqrt{25-x^2}+2\sqrt6}{x-1}=\frac{-\sqrt{25-3^2}+2\sqrt6}{3-1}=\frac{-4+2\sqrt6}{2}= -2 + \sqrt{6}$$
$EDIT$
$$\lim _{x \to 1}\frac{-\sqrt{25-x^2}+2\sqrt6}{x-1}$$
Rationalizing
$$\lim _{x \to 1}\frac{(-\sqrt{25-x^2}+2\sqrt6)(\sqrt{25-x^2}+2\sqrt6)}{(x-1){(\sqrt{25-x^2}+2\sqrt6)}}=\frac{24-25+x^2}{(x-1){(\sqrt{25-x^2}+2\sqrt6)}}=\frac{(x+1)(x-1)}{(\sqrt{25-x^2}+2\sqrt6)(x-1)}=\frac{1+1}{(\sqrt{25-1^2}+2\sqrt6)}=
\frac{1}{2\sqrt6}$$
|
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|
Differential Equation involving System of Equations Find a solution to the boundary value problem \begin{align}y''+ 4y &= 0 \\ y\left(\frac{\pi}{8}\right) &=0\\
y\left(\frac{\pi}{6}\right) &= 1\end{align}
if the general solution to the differential equation is $y(x) = C_1 \sin(2x) + C_2 \cos (2x)$.
I was able to compute the following equations:
\begin{align}C_1 \left(\frac 12\right)\sqrt2 + C_2 \left(\frac 12\right)\sqrt2 &= 0\\
C_1 \left(\frac 12\right)\sqrt3 + C_2 \left(\frac 12\right) &= 1\end{align}
However I am unable to solve the system of equations. The books says the answer is $\frac{2}{\sqrt3 -1}$for $C_1$ and $-C_2$. I am not sure how to go about manipulating the equations to get on variable.
|
\begin{align*}
\frac{\sqrt 2 C_1}{2} + \frac{\sqrt 2 C_1}{2} &= 0 \\
C_1 + C_2 &= 0 \\
C_2 &= -C_1 \\
\frac{\sqrt{3}C_1}{2} + \frac{C_2}{2} &= 1 \\
\sqrt{3}C_1 + C_2 &= 2 \\
\sqrt{3}C_1 - C_1 &= 2 \\
(\sqrt{3} - 1)C_1 &= 2 \\
C_1 &= \frac{2}{(\sqrt{3} - 1)} \\
C_2 &= -\frac{2}{(\sqrt{3} - 1)} \\
\end{align*}
Which is what the book got.
|
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|
Summation of divergent series $1-2^p+3^p-4^p+\ldots$ I am trying to obtain that, for $p=1,2,3\ldots$ ,
$$
s_p\equiv\sum_{n=1}^\infty (-1)^{n-1} n^p = \frac{2^{p+1}-1}{p+1}B_{p+1},
$$
where $s_0=1/2$ and $B_m$ are the Bernoulli numbers. The (divergent) series is regularized by means of the Euler summation
$$
\sum_{n=1}^\infty (-1)^{n-1} n^p = \lim_{t\to1^-}\sum_{n=1}^\infty (-1)^{n-1} n^pt^n.
$$
My try: For $|t|<1$,
$$
\sum_{n=1}^\infty (-1)^{n-1} n^pt^n=\left(t\frac{d}{dt}\right)^p \sum_{n=1}^\infty (-1)^{n-1} t^n=\left(t\frac{d}{dt}\right)^p\frac{t}{1+t};
$$
hence, letting $t=e^z$,
$$
s_p=\frac{d^p}{dz^p} \frac{e^z}{1+e^z}\Bigg|_{z=0}.
$$
This is equivalent to formally defining the generating function
$$
G(z) = \sum_{p=0}^\infty s_p \frac{z^p}{p!} = \sum_{n=1}^\infty (-1)^{n-1}\sum_{p=0}^\infty \frac{(nz)^p}{p!}=\sum_{n=1}^\infty (-1)^{n-1} e^{nz} = \frac{e^z}{1+e^z}.
$$
Using the definition of the Euler numbers
$$
\frac{2}{e^z+e^{-z}}=\sum_{m=0}^\infty E_m \frac{z^m}{m!},
$$
we can recast $G(z)$ as follows
$$\begin{aligned}
G(z)=&\ \frac{2}{e^{z/2}+e^{-z/2}}\frac{e^{z/2}}{2}=\sum_{n,m=0}^\infty \frac{1}{2^{n+m+1}}\frac{E_m}{m!n!}z^{n+m}\\
=&\ \sum_{p=0}^\infty \frac{1}{2^{p+1}}\sum_{m=0}^p \binom{p}{m} E_m \frac{z^p}{p!}.
\end{aligned}$$
Hence,
$$
s_p=\frac{1}{2^{p+1}}\sum_{m=0}^p\binom{p}{m}E_m,
$$
which is a bit different at first sight.
As far as I know, the Bernoulli numbers should be
$$
\frac{z}{e^{z}-1}=\sum_{m=0}^\infty B_m\frac{z^m}{m!}
$$
up to sign conventions for $B_1$, so I am unsure on how to proceed.
|
We need to reshuffle the generating function as follows:
$$\begin{aligned}
\frac{e^z}{e^z+1}=&\ \frac{1}{1+e^{-z}}=\frac{(e^{-z}-1)^2}{(e^{-z}+1)(e^{-z}-1)^2}=\frac{e^{-2z}-2e^{-z}+1}{(e^{-z}-1)(e^{-2z}-1)}\\
=&\ \frac{1}{e^{-z}-1}-\frac{2}{e^{-2z}-1}=\frac{1}{z}\left(\frac{-2z}{e^{-2z}-1}-\frac{-z}{e^{-z}-1}\right)\\
=&\ \frac{1}{z}\sum_{m=0}^\infty \frac{B_m}{m!}\left[(-2z)^m-(-z)^m\right]\\
=&\ \sum_{p=0}^\infty (-1)^{p+1}\frac{2^{p+1}-1}{p+1} B_{p+1} \frac{z^p}{p!}.
\end{aligned}$$
Hence, for $p=0,1,2\ldots$
$$
s_p=(-1)^{p+1}\frac{2^{p+1}-1}{p+1} B_{p+1},
$$
which also proves the identity
$$
\sum_{m=0}^p \binom{p}{m}E_m =(-1)^{p+1}\frac{4^{p+1}-2^{p+1}}{p+1}B_{p+1}.
$$
For $p=1,2,\ldots$ the factor of $(-1)^{p+1}$ is inessential because odd Bernoulli numbers are zero except $B_1$, so that we may write
$$
s_p=\frac{2^{p+1}-1}{p+1} B_{p+1}.
$$
|
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|
Show that $I=J={\ln(7+4\sqrt{3})\over 4\sqrt3}?$ Consider the integrals $(1)$ and $(2)$, how does on show that
(1): $I=J$
(2): and $I=J={\ln(7+4\sqrt{3})\over 4\sqrt3}?$
$$\int_{0}^{\pi/2}{\tan x\over \sqrt{(1+\tan^2 x)(1+\tan^6 x)}}\mathrm dx=I\tag1$$
$$\int_{0}^{\pi/2}{\tan^3 x\over \sqrt{(1+\tan^2 x)(1+\tan^6 x)}}\mathrm dx=J\tag2$$
An attempt:
$u=\tan x$ $\implies du=\sec^2 x dx$
$(1)$ becomes
$$\int_{0}^{\infty}{u\over\sqrt{(1+u^2)(1+u^6)}}\cdot{\mathrm du\over 1+u^2}\tag3$$
$v=1+u^2$ then
$${1\over 2}\int_{1}^{\infty}{\mathrm dv\over \sqrt{v[1+(v-1)^3]}}\tag4$$
$u=\tan^3 x$ $\implies du=3\tan^2 x\sec^2 x dx$
$(2)$ becomes
$${1\over 3}\int_{0}^{\infty}{u^{1/3}\over\sqrt{(1+u^2)(1+u^6)}}\cdot{\mathrm du\over 1+u^{2/3}}\tag5$$
We are not sure how to continue from here...
|
To prove that $I = J$ , note that
$$I-J = \int^{\pi/2}_0 \frac{\tan x-\tan^3 x}{\sqrt{(1+\tan^2 x)(1+\tan^6 x)}}\mathrm dx = \int^{\pi/4}_{-\pi/4} \frac{\tan (\pi/4-x)-\tan^3 (\pi/4-x)}{\sqrt{(1+\tan^2 (\pi/4-x))(1+\tan^6 (\pi/4-x))}}\mathrm dx$$
We need to prove the follwing function is odd
$$f(x) = \frac{\tan (\pi/4-x)-\tan^3 (\pi/4-x)}{\sqrt{(1+\tan^2 (\pi/4-x))(1+\tan^6 (\pi/4-x))}}$$
Now use that
$$\tan\left(\frac{\pi}{4} -x\right) = \frac{\cos x - \sin x}{\cos x + \sin x}$$
After expanding the denominator and nominator we realize
$$\tan (\pi/4-x)-\tan^3 (\pi/4-x)=\frac{-\cos(x) + \cos(3 x) + \sin(x) + \sin(3 x)}{(\cos x + \sin x)^3}$$
$$(1+\tan^2 (\pi/4-x))(1+\tan^6 (\pi/4-x))=\frac{10 - 6 \cos(4 x)}{(\cos x + \sin x )^8}$$
Hence
$$f(x) = \frac{(-\cos(x) + \cos(3 x) + \sin(x) + \sin(3 x))(\sin x+ \cos x)}{\sqrt{10 - 6 \cos(4 x)}}$$
Interestingly this simplifies to
$$f(x) = \frac{\sin(4x)}{\sqrt{10 - 6 \cos(4 x)}}$$
Hence $f(x)$ is odd which imlies
$$I-J = \int^{\pi/4}_{-\pi/4} f(x) \,dx = 0$$
Now using lab bhattacharjee result and $I=J$
$$4I=\int_0^{\infty}\dfrac{4du}{(3+2u-1)\sqrt{3+(2u-1)^2}} = {\log(7+4\sqrt{3})\over \sqrt3}$$
which implies
$$J=I= {\log(7+4\sqrt{3})\over 4\sqrt3}$$
|
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|
Finding a basis for a column space Let A be the matrix: $$\begin{pmatrix} 1&2&3&2&1&0\\2&4&5&3&3&1\\1&2&2&1&2&1 \end{pmatrix}$$.
Show that {$\bigl( \begin{smallmatrix} 1 \\ 4\\3\end{smallmatrix} \bigr)$, $\bigl( \begin{smallmatrix} 3\\4\\1 \end{smallmatrix} \bigr)$} is a basis for the column space of A. Find a "nice basis for the column space of A.
So far, I have row reduced A to $$\begin{pmatrix} 1&2&0&-1&4&3\\0&0&1&1&-1&-1\\0&0&0&0&0&0 \end{pmatrix}$$ where the pivots occur in column 1 and column 3, so {(1,2,1),(3,5,2)} should be a "nice" column space? I do not see where {$\bigl( \begin{smallmatrix} 1 \\ 4\\3\end{smallmatrix} \bigr)$, $\bigl( \begin{smallmatrix} 3\\4\\1 \end{smallmatrix} \bigr)$} come from though.
|
Building on the insights of @puhsu, you know you have 2 vectors that span the image, columns 1 and 3. Can you construct the target vectors in this basis?
$$
\left[ \begin{array}{r}
1 \\ 4 \\3
\end{array} \right]
=
\alpha
\left[ \begin{array}{r}
1 \\ 2 \\ 1
\end{array} \right]
+
\beta
\left[ \begin{array}{r}
3 \\ 5 \\ 2
\end{array} \right], \quad
\alpha = 7, \ \beta = -2.
$$
$$
\left[ \begin{array}{r}
3 \\ 4 \\ 1
\end{array} \right]
=
\alpha
\left[ \begin{array}{r}
1 \\ 2 \\ 1
\end{array} \right]
+
\beta
\left[ \begin{array}{r}
3 \\ 5 \\ 2
\end{array} \right], \quad
\alpha = -3, \ \beta = 2.
$$
|
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|
Maximum value of expression: $\frac{ab+bc+cd}{a^2+b^2+c^2+d^2}$ What is the maximum value of
$$\frac{ab+bc+cd}{a^2+b^2+c^2+d^2}$$
where $a,b,c$, and $d$ are real numbers?
|
The main goal of my approach to the solution was to bring out an equality of the form $ab+bc+cd \le k(a^2+b^2+c^2+d^2)$, and thus, having our maximum value of this expression to be $k$
An initial attempt at the AM-GM inequality yields:
$\frac{1}{2}a^2+b^2+c^2+\frac{1}{2}d^2 \ge ab+bc+cd$
This result, as it turns out, isn't very useful.
Next I try splitting the expression uniformly:
$a^2+b^2+c^2+d^2 = a^2+mb^2+(1-m)b^2+mc^2+(1-m)c^2+d^2$
It is important to note that $m$ is a real number within $(0,1)$, so that both $mb^2,(1-m)b^2$ are positive, as the AM-GM inequality holds good only for positive real numbers.
Now, I individually apply the AM-GM inequality:
$a^2+mb^2 \ge 2\sqrt{m}ab$
$(1-m)b^2+(1-m)c^2 \ge 2(1-m)ab$
$mc^2+d^2 \ge 2\sqrt{m}cd$
Now, I take a value $m$ such that $\sqrt{m} = (1-m)$
We have: $m^2-3m+1 = 0$
$$m = \frac{3 \pm \sqrt{5}}{2}$$
But since, $m < 1$, we have: $m = \frac{3-\sqrt{5}}{2}$
Thus, $2\sqrt{m}=2(1-m)=\sqrt{5}-1$
Thus,we have:
$$a^2+b^2+c^2+d^2 \ge (\sqrt{5}-1)(ab+bc+cd)$$
So, to conclude:
$$\frac{ab+bc+cd}{a^2+b^2+c^2+d^2} \le \frac{1}{\sqrt{5}-1} = \frac{\sqrt{5}+1}{4}$$
The maximum value is $\frac{\sqrt{5}+1}{4}$
|
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|
Factoring a quartic polynomial over the integers with roots that are not integers The quartic polynomial
$$
1728(x - 3) - x^2(12^2 - x^2)
$$
factors "nicely" as
$$
(x^2 - 12x + 72) (x^2 + 12x - 72) = (x^2 - 12x + 72)(x - 6\sqrt{3} + 6)(x + 6\sqrt{3} + 6) \, .
$$
(Note that $1728 = 3(24^2)$.) How is this factorization obtained?
|
$$
\begin{align}
1728(x - 3) - x^2(12^2 - x^2) & = x^4 - 144 x^2 + 1728 x - 5184 \\
& = x^4 - 12^2 x^2 + 12 \cdot 12^2 x - 36 \cdot 12^2 \\
& = x^4 - 12^2(x^2 - 2 \cdot 6 \,x + 6^2) = \\
& = x^4 - 12^2(x-6)^2 = \\
& = \big(x^2 - 12(x-6) \big)\big(x^2 + 12(x-6)\big) = \cdots
\end{align}
$$
|
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|
How can I prove this trigonometric equation with squares of sines? Here is the equation:
$$\sin^2(a+b)+\sin^2(a-b)=1-\cos(2a)\cos(2b)$$
Following from comment help,
$${\left(\sin a \cos b + \cos a \sin b\right)}^2 + {\left(\sin a \cos b - \cos a \sin b\right)}^2$$
$$=\sin^2 a \cos^2b + \cos^2 a \sin^2 b + \sin^2 a \cos^2 b + \cos^2 a \sin^2 b$$
I am stuck here, how do I proceed from here?
Edit: from answers I understand how to prove,but how to prove from where I am stuck?
|
Well, let's the start the manipulation at the left hand side. Using the identity
$$\sin^2\theta=\frac{1-\cos 2\theta}{2}$$
we get
$$
\begin{align}
\sin^2(a+b)+\sin^2(a-b)&=\frac{1-\cos(2a+2b)}{2}+\frac{1-\cos(2a-2b)}{2}\\
&=1-\frac{1}{2}\bigg[\cos(2a+2b)+\cos(2a-2b)\bigg]\\
&=1-\frac{1}{2}\bigg[(\cos 2a\cos 2b-\sin 2a\sin 2b)\\
&\qquad\qquad\qquad +(\cos 2a\cos 2b+\sin 2a\sin 2b)\bigg]\\
&=1-\frac{1}{2}\bigg[2\cos 2a\cos 2b\bigg]\\
&=1-\cos 2a\cos 2b.
\end{align}
$$
|
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|
Prove that if $x \equiv 5 \pmod{10}$, then $y \equiv 0 \pmod{7}$
Let $x,y$ be positive integers satisfying $2x^2-y^2 = 1$. Prove that if $x \equiv 5 \pmod{10}$, then $y \equiv 0 \pmod{7}$.
I wasn't sure how to use the fact that $x,y$ are positive integers satisfying $2x^2-y^2 = 1$. We could use the theory of Pell's equations to find the solutions, but that would get complicated. Is there a simpler way?
|
The solutions of this Pell equation are
$$ \pmatrix{x_n\cr y_n\cr} = \pmatrix{ 3 & 2\cr 4 & 3\cr}^n \pmatrix{1\cr 1\cr} $$
for nonnegative integers $n$. If $M = \pmatrix{3 & 2\cr 4 & 3\cr}$, we have
$M^6 \equiv I \mod 70$. Thus $(x_n, y_n) \mod 70$ (and therefore mod $10$ and mod $7$) is periodic with period $6$. We find that $x_n \equiv 5 \mod 10$ for $n = 1$ and $4$, and therefore whenever $n \equiv 1$ or $4 \mod 6$. For those $n$ we find $y_n \equiv 0 \mod 7$.
|
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|
Apostol Calculus Vol 2 Exercise 8.17 Q No. 3 Evaluate the directional derivative of $f$ for the points and directions specified
$f(x,y,z)=x^2+y^2-z^2$ at $(3,4,5)$ along the curve of intersection of the two surfaces $2x^2+2y^2-z^2=25$ and $x^2+y^2=z^2.$
The answer is supposed to be $0.$ I am not getting the right answer maybe because I don't know how to get the curve of intersection or I don't really understand the directional derivative enough. This is what I did though:
\begin{align}
2x^2+2y^2-z^2&=25\\
\implies 2(x^2+y^2)-z^2&=25\\
\implies z^2&=25
\end{align}
Thus the curve of intersection is $x^2+y^2=25.$ which is a circle centred at orgin with radius $5.$ Hence if $(x,y)$ is any point of the circle, then the unit direction towards that point is $\frac{1}{5}(x,y).$ Also $\nabla f=(2x,2y,-2z)$ and hence we have $\nabla f(3,4,5)=(6,8,-10)$.
Thus the directional derivative at the point $(3,4,5)$ is $\nabla f(3,4,5)\cdot \big(\frac{3}{5},\frac{4}{5}\big)=\frac{58}{5}.$(a wrong answer)
|
I think your unit direction vector should be : $\left(
\begin{array}{c}
-\frac{4}{5}\\
\frac{3}{5}\\
0\\
\end{array}
\right)$
(maybe I'll include picture later).
And that the answer should be : $0$ accordingly : $\left(0=\left(
\begin{array}{c}
-\frac{4}{5}\\
\frac{3}{5}\\
0\\
\end{array}
\right) \cdot \left(
\begin{array}{c}
6\\
8\\
-10\\
\end{array}
\right)\right)$.
Here a picture of the tangent line in $(3,4)$ :
|
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|
Find the roots of $f(x)=x^2+x+1$ modulo 7, and modulo 13, and modulo 91 Find the roots of $f(x)=x^2+x+1$ modulo $7$, and modulo $13$, and modulo $91$
I think for mod $7$ and $13$, it can be done by trial and error. But how about mod $91$?
|
For any odd modulus,
$$x^2+x+1\equiv0\iff4x^2+4x+4\equiv0\iff(2x+1)^2\equiv-3$$
For the prime modulus $7$,
$$(2x+1)^2\equiv-3\equiv4\implies2x+1\equiv\pm2\implies4(2x+1)\equiv\pm8\implies x+4\equiv\pm1$$
which implies $x\equiv2$ or $x\equiv4$ mod $7$.
For the prime modulus $13$,
$$(2x+1)^2\equiv-3\equiv36\implies2x+1\equiv\pm6\implies7(2x+1)\equiv\pm42\equiv\pm3\implies x+7\equiv\pm3$$
which implies $x\equiv3$ or $x\equiv9$ mod $13$.
As for $91=7\cdot13$, The Chinese Remainder Theorem tells us there are four solutions to $x^2+x+1\equiv0$ mod $91$, and provides a procedure for finding them. However, for this problem we're in luck: It's easy to see that $x=9$ satisfies $x\equiv2$ mod $7$ and $x\equiv9$ mod $13$, while $x=16$ satisfies $x\equiv2$ mod $7$ and $x\equiv3$ mod $13$, so that $x=9$ and $16$ are two solutions. This implies the other two solutions correspond to
$$2x+1\equiv-(2\cdot9+1)\equiv-19\quad\text{and}\quad2x+1\equiv-(2\cdot16+1)\equiv-33$$
or $2x\equiv-20$ and $2x\equiv-34$, i.e., $x\equiv-10\equiv81$ and $x\equiv-17\equiv74$ mod $91$. So the four solutions mod $91$ are $x=9$, $16$, $74$, and $81$.
|
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|
Computing $ \int_0^1 \frac{1 + 3x +5x^3}{\sqrt{x}}\ dx$ $$ \int_0^1 \frac{1 + 3x +5x^3}{\sqrt{x}}\ dx$$
My idea for this was to break each numerator into its own fraction as follows
$$ \int_0^1 \left(\frac{1}{\sqrt{x}} + \frac{3x}{\sqrt{x}} + \frac{5x^3}{\sqrt{x}}\right)dx$$
$$ \int_0^1 (x^{-1/2} + 3x^{1/2} +5x^{5/2})\ dx $$
$$ \int_0^1 2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2} $$
Not really sure where to go from there. Should I sub 1 in for the x values and let that be the answer?
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Your work is correct, but for the last step write:
$$
\left[ 2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2}\right]_0^1
$$
because you have just done the integration as antiderivative: $F(x)=2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2}$ and you have only to evaluate the primitive at the two limits of integration, so that your definite integrale is done by $F(1)-F(0)$.
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|
Polynomial division. If an f(x) is divided by x²+4, has 2x+3 of remainder.
If an f(x) is divided by x²+6, has 6x-1 of remainder.
If an f(x) is divided by (x²+4)(x²+6), has S(x) of remainder, then find S(4)!
I have this.
I wrote it in polynomial formula.
$f(x)=(x²+4).R(x)+(2x+3)$
And yes, i have x²=-4.
But, i had to express the 2x+3 function into the (x²) one. So, basically if we have p(x)=2x+3, then p(x²)=2x²+3. and i can put -4 to the x, and i got :
f(-4)=2(-4)+3 = -5 ( and the same thing happened with x²+6 ). Is that true so far?
Sorry for bad english.
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Approach using complex numbers.
Notice that $x^2+4=0$ when $x=\pm 2i$ and $x^2+6=0$ when $x=\pm i\sqrt{6}$ since we have that $$f(x)=(x^2+4)R(x)+2x+3$$
From this we have that $$f(2i)=4i+3\\f(-2i)=-4i+3$$
Also we have that
$$f(x)=(x^2+4)Q(x)+6x-1\\f(i\sqrt{6})=6i\sqrt{6}-1\\f(-i\sqrt{6})=-6i\sqrt{6}-1$$
From this we have that
$$f(x)=(x^2+4)(x^2+6)T(x)+ax^3+bx^2+cx+d\\f(2i)=-8ai-4b+2ci+d\\f(-2i)=8ai-4b-2ci+d\\f(i\sqrt{6})=-6a\sqrt{6}i-6b+ci\sqrt{6}+d\\f(-i\sqrt{6})=6a\sqrt{6}i-6b-ci\sqrt{6}+d$$
Now $$f(2i)+f(-2i)=2d-8b=6\\ f(i\sqrt{6})+f(-i\sqrt{6})=2d-12b=-2$$ from this we have that $b=2$ and $d=11$. We also have that
$$f(2i)-f(-2i)=4ci-16ai=8i\\c-4a=2\\f(i\sqrt{6})-f(-i\sqrt{6})=2ci\sqrt{6}-12ai\sqrt{6}=12\sqrt{6}i\\c-6a=6$$
Solving this we get $a=-2$ and $c=-6$
This implies that $S(x)=-2x^3+2x^2-6x+11$ and $S(4)=-128+32-24+11=-109$
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Integration of double integral Integration of double integral of.
\begin{equation}
\int_{-\frac{1}{4}}^{0}{\int_{\frac{1}{2}-\sqrt{x+\frac{1}{4}}}^{\frac{1}{2}+\sqrt{x+\frac{1}{4}}}} {e^{y^2}dy dx} + \int_{0}^{2}{\int_{-1+\sqrt{x+1}}^{1/2+\sqrt{x+\frac{1}{4}}}} {e^{y^2}dy dx} + \int_{2}^{8}{\int_{-1+\sqrt{x+1}}^{2}} {e^{y^2}dy dx}
\end{equation}
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If we reverse the order of integration the integrands can be transformed into $ye^{y^2}\,dy$ and $y^2e^{y^2}\,dy$.
Here is the region of integration from desmos.com/calculator
First divide the region into five sections having unique left and right bounding functions of $y$
Giving the following region and subregions
Then change the order of integration over the five sub-regions to obtain
\begin{eqnarray}
& &\int_0^1\int_{y^2-y}^{0}e^{y^2}dxdy\\&+&\int_{0}^{\sqrt{3}-1}\int_{0}^{y^2+2y}e^{y^2}dxdy\\
&+&2(2-\sqrt{3})\\
&+&\int_{1}^{2}\int_{y^2-y}^2e^{y^2}dxdy\\
&+&\int_{\sqrt{3}-1}^{2}\int_2^{y^2+2y}e^{y^2}dxdy
\end{eqnarray}
You should double check the limits but I believe this is correct.
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How to write the form of partial fraction decomposition of the function. How to write the form of partial fraction decomposition of the function.
$$I= \dfrac{2\,x^3+24\,x^2+20\,x+10}{x^2+10\,x+25}$$
I used long division I get$$Quotient= 2\,x+4\\Remainder =-70\,x - 90\\$$
But don't know how to solve further steps. Thanks in advance.
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Dividing the function you get$\text{Quotient}= 2\,x+4\\\text{Remainder} =-70\,x - 90\\\text{Divisor}= x^2+10\,x+25\tag*{}$So,$2\,x^3+24\,x^2 +20\,x+10=(2x+4)(x^2+10x+25)+(-70x-90)\tag*{}$Rewrite the function,$I = \dfrac{(x^2+10\,x+25)(2\,x+4)+(-70\,x-90)}{x^2+10\,x+25}\\=(2\,x+4) -\dfrac{70\,x+90}{(x+5)^2}\tag*{}$Now decompose $\dfrac{70\,x+90}{(x+5)^2}$ by using partial fraction.$\Rightarrow\dfrac{70\,x+90}{(x+5)^2} = \dfrac{A}{(x+5)} +\dfrac{B}{(x+5)^2}...(1)\\=70\,x+90 = A\,x+5\,A+B\tag*{}$comparing coefficients of both side of equation$Ax=70x,5\,A+B=90\\A=70,B=-260\tag*{}$Therefore,$ I=(2\,x+4) -\dfrac{70}{x+5}+ \dfrac{260}{(x+5)^2}\tag*{}$
here is the similar problem with complete solution:
partial fraction decomposition
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|
Polar form of a complex with square root
Find the polar form of the roots of the polinomial
$p(z)=iz^2-2z+1+2i$
I found the roots $z_1=-i-\sqrt{i-3}$ and $z_2=-i+\sqrt{i-3}$ but I don't know how to deal with the complex numbers inside the square root in order to isolate both real and imaginary parts.
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To compute the square root of a complex number: $(x+iy)^2=-3+i$, expand the l.h.s. and identify the real and imaginary parts:
$$x^2-y^2=-3, \quad xy=\frac12.$$
You simplify the computation observing the square of the modulus of $x+iy$ is the modulus of $-3+i$:
$$x^2+y^2=\sqrt{10}.$$
So we have a linear system in $x^2$ and $y^2$:
$$\begin{cases}x^2+y^2=\sqrt{10}\\x^2-y^2=-3\end{cases}\iff\begin{cases}x^2=\dfrac{\sqrt{10}+3}2\\ y^2=\dfrac{\sqrt{10}-3}2\end{cases}$$
Observe $xy>0$, so $x$ and $y$ have the same sign, and ultimately
$$x+iy=\pm\frac12\biggl(\sqrt{2\sqrt{10}+6}+i\sqrt{2\sqrt{10}-6}\biggr).$$
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Probability Interview Question - Brain teaser
Player A has a thirty-sided and Player B has a twenty-sided die. They both roll the dies and whoever gets the higher roll wins. If they roll the same amount Player B wins. What is the probability that Player B win?
So I have
$$1 - \left(\frac{1}{3}+\frac{1}{2}\cdot\frac{2}{3}-\frac{1}{30}\right) = \frac{11}{30}.$$
There is $100\%$ chance of winning when A roll from 20 - 30 but for the rest $2/3$ there is only $50\%$ chance of winning, in addition, there is $1/30$ of chance that player A could roll the same number as B
The answer for this question should be $0.35$, I am not sure where I did wrong.
I just figured maybe I should do this instead
$1 - (\frac{1}{3}+(\frac{1}{2}-\frac{1}{30})\cdot\frac{2}{3}) = \frac{16}{45}$ is this right ?
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I've reasoned like this, as usual starting from the simplest case:
A has a 3-sided dice, B a 2-sided one.
$P(\mathbf{B} wins)$ = $P(\mathbf{B}_{dice}=1)$ * $P(\mathbf{A}_{dice}=1)$ +
$P(\mathbf{B}_{dice}=2)$ * $P(\mathbf{A}_{dice}<=2)$ =
= $\frac{1}{2}$ * $\frac{1}{3}$ + $\frac{1}{2}$ * $\frac{2}{3}$ = $\frac{1}{2}$
Generalizing ($n$ being the number of sides for B):
$P(\mathbf{B} wins)$ = $\frac{1}{n} * \sum_{j=1}^{n}(\frac{j}{\frac{3}{2}*n})$
= $\frac{2}{3n^2} * \frac{n*(n+1)}{2}$ = $\frac{1}{3n} * {(n+1)}$ = $\frac{1}{3} + \frac{1}{3n}$
So, with $n = 20$, $P(\mathbf{B} wins)$ = $\frac{1}{3} + \frac{1}{60}$ = $\frac{21}{60}$
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|
Who came up with the identity $a^3+b^3+c^3-3abc=(a+b+c)\left[a^2+b^2+c^2-ab-bc-ca\right]$ Though we can prove this it is not something that comes up intutively.
Our ancestors must have been interested in factorising $a^3+b^3+c^3$ but why find it for $a^3+b^3+c^3-3abc$ ?
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I don't know the history of this identity, but here is a derivation based on the properties of determinants.
$$\begin{vmatrix}
a &c &b\\
b &a &c\\
c &b &a
\end{vmatrix} =
\begin{vmatrix}
a+b+c &a+b+c &a+b+c\\
b &a &c\\
c &b &a
\end{vmatrix} = (a+b+c) \begin{vmatrix}
1 &1 & 1\\
b &a &c\\
c &b &a
\end{vmatrix} $$
Expanding the determinant on the left, we have
$$a^3+b^3+c^3-3abc$$
and expanding the determinant on the right we have
$$(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$
so these two expressions are equal.
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|
If $a$ and $b$ are odd perfect squares, then $a + b$ is not a perfect square.
Prove: If $a$ and $b$ are odd perfect squares, then $a + b$ is not a perfect square.
Proof by Contradiction:
If $a$ and $b$ are odd perfect squares then $a = (2k+1)^2$ and $b = (2r + 1)^2$.
Assume $a + b$ is a perfect square.
\begin{align}a + b &= (2k+1)^2 + (2r + 1)^2\\
&= 4(k^2 + r^2) + 4(k + r) + 2\\
&= 4(k + r)^2 + 4(k + r) -2kr + 2\\
&= z^2 + 2z -2kr + 2 \hspace{0.5cm} \text{where } z = 2(k+r)\\
&= (z+1)^2 -2kr + 1\\
&= (z+1)^2 - q \hspace{0.5cm} \text{where } q = 2kr + 1\end{align}
Hence, $a+b$ is in inexpressible as a perfect square when $a, b$ are odd.
Is this a valid proof? I'm not sure how to approach it any other way.
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In you second line you have shown that a+b mod 4 = 2. Is this possible if a+b is a perfect square ?
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Prove that sum is an integer. Prove that: $\sqrt{2-\frac{1}{1^2+\sqrt{1^4+\frac{1}{4}}}}+\sqrt{2-\frac{1}{2^2+\sqrt{2^4+\frac{1}{4}}}}+\sqrt{2-\frac{1}{3^2+\sqrt{3^4+\frac{1}{4}}}}+\cdots+\sqrt{2-\frac{1}{119^2+\sqrt{119^4+\frac{1}{4}}}}$ is an integer.
I noticed that $n^4+\frac{1}{4}$ is rather interesting, because it would be written as $(n^2+n+\frac{1}{2})(n^2-n+\frac{1}{2})$, but I do not know how to connect this to the rest of the problem. The square roots seem to further complicate the problem. Furthermore, I don't see what's special with 119, but upon using a program, I noticed that the only such numbers (such that the sum is an integer) below 500 are 3, 20, and 119.
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Since
$$\frac{1}{k^2+\sqrt{k^4+\frac 14}}=\frac{k^2-\sqrt{k^4+\frac 14}}{(k^2)^2-(k^4+\frac 14)}=-4k^2+2\sqrt{4k^4+1}$$
we have
$$\sum_{k=1}^{n}\sqrt{2-\frac{1}{k^2+\sqrt{k^4+\frac 14}}}=\sum_{k=1}^{n}\sqrt{2+4k^2-2\sqrt{4k^4+1}}\tag1$$
Here, since we have that
$$4k^4+1=4k^4+4k^2+1-4k^2=(2k^2+1)^2-(2k)^2=(2k^2+2k+1)(2k^2-2k+1)$$
and that
$$(2k^2+2k+1)+(2k^2-2k+1)=2+4k^2$$
we have
$$\begin{align}(1)&=\sum_{k=1}^{n}\sqrt{\left(\sqrt{2k^2+2k+1}-\sqrt{2k^2-2k+1}\right)^2}\\\\&=\sum_{k=1}^{n}\left(\sqrt{2k^2+2k+1}-\sqrt{2k^2-2k+1}\right)\\\\&=\sum_{k=1}^{n}\left(\sqrt{2k^2+2k+1}-\sqrt{2(k-1)^2+2(k-1)+1}\right)\\\\&=\sqrt{2n^2+2n+1}-\sqrt{1}\end{align}$$
This is an integer for $n=119$ since $2\cdot 119^2+2\cdot 119+1=169^2$.
Added :
$(1)$ is an integer if and only if there is an integer $m$ such that
$$2n^2+2n+1=m^2\iff (2n+1)^2-2m^2=-1$$
This is a Pell's equation, so we have
$$n=-\frac 12+\left(\frac{1-\sqrt 2}{4}\right)\cdot (3-2\sqrt 2)^k+\left(\frac{1+\sqrt 2}{4}\right)\cdot (3+2\sqrt 2)^k$$
where $k=1,2,3,\cdots$.
These satisfy $a_n=6a_{n-1}-a_{n-2}+2$ with $a_1=3,a_2=20$ :
$$a_3=119,\quad a_4=696,\quad a_5=4059,\quad a_6=23660,\quad \cdots$$
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.