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Find the integral of $\int \ln\left(\sqrt{x-b}+\sqrt{x-a}\right)\,dx$
Evaluate $\displaystyle\int \ln\left(\sqrt{x-b}+\sqrt{x-a}\right)\,dx$.
I am tryed to integrate it by parts by taking $du = 1$ and $v=\ln\left(\sqrt{x-b}+\sqrt{x-a}\right)$
Therefore,
$vu - \displaystyle\int v du =x\ln\left(\sqrt{x-b}+\sqrt{x-a}\right)-{\displaystyle\int}\dfrac{x\left(\frac{1}{2\sqrt{x-b}}+\frac{1}{2\sqrt{x-a}}\right)}{\sqrt{x-b}+\sqrt{x-a}}\,\mathrm{d}x$
Which further simplifies to $=x\ln\left(\sqrt{x-b}+\sqrt{x-a}\right) - 0.5{\displaystyle\int}\dfrac{x}{\sqrt{x-a}\sqrt{x-b}}\,\mathrm{d}x$
I am stuck here. I need help solving
${\displaystyle\int}\dfrac{x}{\sqrt{x-a}\sqrt{x-b}}\,\mathrm{d}x$
|
$$I=\int\frac x{\sqrt{x^2-(a+b)x+ab}}dx$$You can write $x=0.5[2x-(a+b)+(a+b)]$,$$I=\int\frac{0.5[2x-(a+b)+(a+b)]}{\sqrt{x^2-(a+b)x+ab}}dx\\=0.5\left[\int\frac{2x-(a+b)}{\sqrt{x^2-(a+b)x+ab}}dx+\int\frac{(a+b)}{\sqrt{\left[x-\frac{(a+b)}2\right]^2-(a+b)^2/4+ab}}dx\right]$$The first integral reduces to $\int\frac{du}{\sqrt u}$ where $u=x^2-(a+b)x+ab$, while the second integral reduces to one of the standard forms $\int\frac{dv}{\sqrt{v^2-k^2}}$ or $\int\frac{dv}{\sqrt{v^2+k^2}}$ where $v=\left[x-\frac{(a+b)}2\right]$ depending on the value of $a,b$.
|
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|
Conditional probability of min and max of two dice Consider the following problem, from Tijms's Understanding Probability:
Two dice are rolled. Let the random variable $X$ be the smallest of the two outcomes and let $Y$ be the largest of the two outcomes. What are the conditional mass functions $P (X = x | Y = y)$ and $P (Y = y | X = x)$?
My attempt:
$$
P(X=x, Y=y) = \begin{cases}
\frac{1}{36} & \text{if }x=y \\
0 & \text{if }x>y \\
\frac{2}{36} & \text{if }x<y \\
\end{cases}.
$$
For the individual probabilities, we have that one of the two outcomes is fixed and has to be equal to the minimum/maximum, the other dice can roll any number between the minimum and 6, or between 1 and the maximum. The order does not count, so I multiply by 2:
$$
P(X=x) = \frac16 \frac{6-x+1}{6}\cdot 2,
$$
and
$$
P(Y=y) = \frac16 \frac{y}{6}\cdot 2.
$$
Putting everything together, we have:
$$
P(X=x|Y=y) = \frac{P(X=x,Y=y)}{P(Y=y)} = \begin{cases}
\frac{1}{2y} & \text{if }x=y \\
0 & \text{if }x > y \\
\frac{1}{y} & \text{if }x < y \\
\end{cases}.
$$
and
$$
P(Y=y|X=x) = \frac{P(X=x,Y=y)}{P(X=x)} = \begin{cases}
\frac{1}{2(6-x+1)} & \text{if }x=y \\
0 & \text{if }x > y \\
\frac{1}{6-x+1} & \text{if }x<y
\end{cases}.
$$
Does it sound right?
|
Partial answer due lack of time
For $P(Y=y)$ I have something slightly different. I´ve worked with the table below:
$P(Y=y|X=x)$
$$ \begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|} \hline \text{ X/Y } & 1 &2 &3 &4 &5 &6 \\ \hline \hline \hline 1 &\frac{1}{36} &\frac{2}{36} &\frac{2}{36}& \frac{2}{36}& \frac{2}{36}& \frac{2}{36}\\ \hline 2& &\frac{1}{36} &\frac{2}{36} &\frac{2}{36} &\frac{2}{36} &\frac{2}{36} \\ \hline 3& & &\frac{1}{36} &\frac{2}{36} &\frac{2}{36} &\frac{2}{36}\\ \hline 4 & & & &\frac{1}{36} &\frac{2}{36} &\frac{2}{36} \\ \hline 5 & & & & &\frac{1}{36} &\frac{2}{36} \\ \hline 6& & & & & &\frac{1}{36} \\ \hline \end{array}$$
Now we can sum up the cells in column $y$ to obtain $P(Y=y)=\frac{2\cdot y-1}{36}$.
We can check if this is plausible by calculating the sum ( sanity check).
$\sum\limits_{y=1}^6 P(Y=y)=\sum\limits_{y=1}^6\frac{2\cdot y-1}{36}=\frac1{36}\cdot \left( 2\cdot \sum\limits_{y=1}^6 y-\sum\limits_{y=1}^61 \right)=\frac1{36}\cdot \left( 2\cdot \frac{6\cdot 7}{2}-6 \right)=1$
This is the result we have expected.
|
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|
$\{a_n\}$ be a sequence such that $ a_{n+1}^2-2a_na_{n+1}-a_n=0$, then $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in... Let $\{a_n\}$ be a sequence of positive real numbers such that
$a_1 =1,\ \ a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$.
Then the sum of the series $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in...
(A) $(1,2]$, (B) $(2,3]$, (C) $(3,4]$, (D)$(4,5]$.
Solution attempt:
Firstly, we figure out what $\frac{a_{n+1}}{a_n}$ is going to look like. We get, from the recursive formula, $\frac{a_{n+1}}{a_n}=1+\sqrt{1+\frac{1}{a_n^2}}$ (remembering the fact that $a_n>0$, the other root is rejected).
We know that, if $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}>1$, then $\lim a_n \to \infty$.
Further, $(a_{n+1}-a_n)= \sqrt{a_n(a_n+1)}>0$. (Again, the other root is rejected due to the same reason).
Hence, $(a_n)$ increases monotonically.
Therefore, the largest value of $\frac{a_{n+1}}{a_n}$ is approximately $1+\sqrt{1+\frac{1}{1}} \approx 2.15$
Now, the sum can be approximated as $\displaystyle\frac{\frac{1}{3}}{1-\frac{2.15}{3}} \approx 1.3$ (In actuality, $\mathbb{sum}< 1.3$).
So, option $(A)$ is the correct answer.
Is the procedure correct?
I have been noticing a handful of this type of questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie.
Is there any "definitive" approach that exploits the recursive formula and gives us the value, or does the approach varies from problem to problem?
|
Following your calculations and according to the ratio test
$$0<\frac{\frac{a_{n+1}}{3^{n+1}}}{\frac{a_n}{3^n}}=\frac{1}{3}\cdot \frac{a_{n+1}}{a_n}<1$$
thus
$$\sum\limits_{n=1}\frac{a_n}{3^n}< \infty$$
Now, applying the same technique from the proof on the ratio test
$$S=\frac{1}{3}+\frac{a_2}{3^2}+\frac{a_3}{3^3}+\cdots+\frac{a_n}{3^n}+\cdots=\\
\frac{1}{3}+\frac{a_2}{a_1}\cdot\frac{a_1}{3^2}+\frac{a_3}{a_2}\cdot\frac{a_2}{3^3}+\cdots+\frac{a_{n}}{a_{n-1}}\cdot\frac{a_{n-1}}{3^n}+\cdots$$
or
$$\frac{1}{3}+2\cdot\frac{1}{3^2}+2\cdot\frac{a_2}{3^3}+\cdots+2\cdot\frac{a_{n-1}}{3^n}+\cdots<
S<\\
\frac{1}{3}+2.15\cdot\frac{1}{3^2}+2.15\cdot\frac{a_2}{3^3}+\cdots+2.15\cdot\frac{a_{n-1}}{3^n}+\cdots$$
and repeating this
$$\frac{1}{3}+\frac{2}{3^2}+\frac{2^2}{3^3}+\cdots+\frac{2^{n-1}}{3^n}+\cdots<
S<\\
\frac{1}{3}+\frac{2.15}{3^2}+\frac{2.15^2}{3^3}+\cdots+\frac{2.15^{n-1}}{3^n}+\cdots$$
or
$$\frac{1}{3}\cdot\left(1+\frac{2}{3}+\frac{2^2}{3^2}+\cdots+\frac{2^{n-1}}{3^{n-1}}+\cdots\right)<
S<\\
\frac{1}{3}\cdot\left(1+\frac{2.15}{3}+\frac{2.15^2}{3^2}+\cdots+\frac{2.15^{n-1}}{3^{n-1}}+\cdots\right)$$
or
$$\color{red}{1}=\frac{\frac{1}{3}}{1-\frac{2}{3}}<\color{red}{S}<\frac{\frac{1}{3}}{1-\frac{2.15}{3}}=\frac{1}{3-2.15}<\color{red}{2}$$
This kind of squeezing technique is widely applied in analysis, functional analysis, numerical analysis. So, it makes sense to ask something similar for a master degree entrance test.
|
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|
Evaluate the limit $ \lim_\limits{n \to \infty} \frac{1^a+2^a\cdots+n^a}{n^{a+1}} $ The exact question is-
Find the real value(s) of $a (a \ne -1)$ for which the limit $$ \lim_{ n \to \infty} \frac{ 1^a+2^a\cdots+n^a}{(n+1)^{a-1}[ (na+1)+(na+2) \cdots+(na+n)]} = \frac{1}{60}$$
I simplified it a bit to get that limit. However, I got the answer from Wolfram Alpha $\frac{1}{a+1}$ without a solution and an assumption that $|a|<1$ however the answer that I got on further solving (and given) is $a=7$ and $a=-\frac{17}{2}$
How is it solved?
|
Useful link related to the first equation;
Evaluate $\lim\limits_{n\to\infty}\frac{\sum_{k=1}^n k^m}{n^{m+1}}$
Limit of the sequence $\frac{1^k+2^k+...+n^k}{n^{k+1}}$
$\lim_{{n}\to {\infty}}\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}=?$
Showing that $\lim_{n \rightarrow \infty}\frac{1^k+2^k+...+n^k}{n^{k+1}}=\frac{1}{k+1}$ using integral
Prove that limit of $\frac{1^p + 2^p + \ldots + n^p}{n^{p+1}}$ is equal to $\frac{1}{p+1}$
$$1^a+2^a+...+n^a\sim\frac{n^{a+1}}{a+1}$$ so we could write this limit :
$$ \lim_{ n \to \infty} \frac{ 1^a+2^a\cdots+n^a}{(n+1)^{a-1}[ (na+1)+(na+2) \cdots+(na+n)]} =\lim_{ n \to \infty}\frac{\frac{n^{a+1}}{a+1}}{{(n+1)^{a-1}[ n^2a+\frac{n(n+1)}{2}]}} = \lim_{ n \to \infty}\frac{2n^{a+1}}{{(a+1)n(n)^{a-1}n[2a+1]}} = \frac{2}{(a+1)(2a+1)}=\frac{1}{60}$$
So $(a+1)(2a+1)=120$ and $a=7,\frac{-17}{2}$
|
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|
Hard limit involving different order radicals $\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$ Please help me to calculate the following limit
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$$
I factored out $n$ from both radicals but it didn't work and tried to use the identity $a^2-b^2$ and $a^3-b^3$.
|
There is a way to use directly the two binomial formulas you mentioned. It only needs a little trick:
$$\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} = \sqrt[3]{n^3+3n^2}\color{blue}{-n + n}-\sqrt{n^2+2n}$$
Now, consider
\begin{eqnarray*} \sqrt[3]{n^3+3n^2}-n
& \stackrel{n=\sqrt[3]{n^3}}{=} & \frac{n^3+3n^2 - n^3}{(\sqrt[3]{n^3+3n^2})^2 + n\sqrt[3]{n^3+3n^2}+ n^2}\\& = & \frac{3}{\left(\sqrt[3]{1+\frac{3}{n}}\right)^2 + \sqrt[3]{1+\frac{3}{n}} + 1}\\
& \stackrel{n \to \infty}{\longrightarrow} & 1
\end{eqnarray*}
Similarly, you get $n - \sqrt{n^2+2n} \stackrel{n \to \infty}{\longrightarrow} -1$.
Hence, you get
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} ) = \lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-n ) + \lim_{n \to \infty} (n-\sqrt{n^2+2n} ) = 1-1 = 0$$
|
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|
Given three non-negative $a,b,c$ so that $c$ between $a$ and $b$. Prove that $2\sum\limits_{cyc}a^{3}+3abc\geqq3\sum\limits_{cyc}a^{2}b$ .
Given three non-negative $a, b, c$ so that $c$ between $a$ and $b$. Prove that $2\sum\limits_{cyc}a^{3}+ 3abc\geqq 3\sum\limits_{cyc}a^{2}b$
Inspried from: https://math.stackexchange.com/a/3264953/688846, after using Ravi-subs for
$$(\sum\limits_{cyc}a^{3}- \sum\limits_{cyc}a^{2}b)\geqq 0$$
by $a= a+ b, b= b+ c, c= c+ a$, equal to
$$(\sum\limits_{cyc}a^{3}+ \sum\limits_{cyc}a^{2}b- 2\,abc)\geqq 0$$
And again, we have
$$5(a^{3}+ b^{3}+ c^{3}- 3\,abc)= 3(\sum\limits_{cyc}a^{3}+ \sum\limits_{cyc}a^{2}b- 2\,abc)+ (\underbrace{2\sum\limits_{cyc}a^{3}+ 3abc- 3\sum\limits_{cyc}a^{2}b}_{\geqq 0(that's\,all\,we\,need\,to\,prove\,!)})\geqq 0$$
|
By Schur's inequality, the basic inequality $3(a^2+b^2+c^2)\geq (a+b+c)^2$ (which can be proved by Cauchy-Schwarz if we write $3=1^2+1^2+1^2$) and the non-negativity of $a,b$ and $c$ we have that
$$a^3+b^3+c^3+3abc\geq a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)\Rightarrow\\ 2(a^3+b^3+c^3)+3abc\geq (a+b+c)(a^2+b^2+c^2)\geq \frac{(a+b+c)^3}{3}.$$
The first line is Schur's inequality. At the second line we just added $a^3+b^3+c^3$ at both sides and did the algebra.
|
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|
Show that $2 - \sqrt{2}$ is irrational I suppose $2 - \sqrt{2} $ is rational.
so $$2- \sqrt{2} = {a/b} $$
where a,b are integers and gcd(a,b) = 1.
$$\text{Step 1. } 2 = (a/b)^2 \text{ //squared both sides }$$
$$\text{Step 2. } 2b^2 = a^2 \text{ //We see $a^2$ is even }$$
$$\text{Step 3. }2b^2 = (2k)^2$$
Step 3 since $a^2$ is even and so $a$ is even too. Also, k is an integer.
$$\text{Step 4. }b^2 = (2k)^2 \text{ //We see $b^2$ is even }$$
Step 4 since $b^2$ is even and so $b$ is even too.
We see $ a$ and $b$ are even. We see $gcd(a,b) \neq 1$
Is this proof correct ?
|
As an alternate approach, note that $2-\sqrt 2$ is a root of $x^2-4x+2$ (the conjugate root is $2+\sqrt 2$ so just compute $(x-(2-\sqrt 2))(x-(2+\sqrt 2))$.
By the Rational Root Theorem, this quadratic has no rational roots.
|
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|
Integrating double integral with spherical coordinates problem with interpret a domain Hello i have the following problem i am solving integral with spherical coordinates but i am getting wrong answer - i think i am integrating correct so i think the problem is coming from the constraints. So i have $$\iint_Dxdx$$ and i have $D = (x^2 + y^2 \le 2y, x \ge 0, y \ge \frac{1}{2})$. So i begin solving - first i complete the circle and came with $(x-0)^2 + (y-1)^2 = 1$ that is circle centered at $y = 1$ with $radius$ = 1 so i substitute $x^2 + y^2 = 2y$ with $r^2 = 2rsin\theta$ i divide both sides by r i get $r = 2sin\theta$ so i have limits from $0$ to $2sin\theta$ for r also i need to rotate $\phi$ from $0$ to $\pi$ because that is how the circle is placed. And i got $$\int_0^\pi \int_0^{2sin\theta} (rcos\theta)r \,d\varphi\,dr$$ and so after integrating $r$ i am left with $$\frac{1}{3}\int_o^\pi sin^3\theta cos\theta$$ i am using u substitution and say $u = sin\theta$ and $du = cos\theta$ so i have $$\frac{8}{3}\int_0^\pi u^3du$$ and the new limits are from $0$ to $0$ because $sin\theta$ from 0 is 0 and $sin\theta$ from $\pi$ is also $0$ that means the final answer is zero in and it should be $\frac {9}{16}$. I am not good at math so i cannot see where my error is. Thank you for any help in advance.
|
The limits are $x^2+y^2\le 2y \ ,\ x\ge0 \ , y\ge\frac{1}{2} $
You've found the upper limit i.e, $r = 2\sin\theta$ (upper limit).
For $y\ge\frac{1}{2} \ , \ r\sin\theta \ge\frac{1}{2} $
So, the lower limit of $r$ is $\frac{1}{2\sin\theta}$
Also, $x\ge0$ , $r\cos\theta\ge0$ , $\cos\theta\ge0$ So, $\theta\le\pi/2$
At $y=\frac{1}{2}$, $\sin\theta = \frac{\pi}{6}$ [as, $1.\sin(\pi/6) =1/2$]
$$I = \int^{\pi/2}_{\pi/6}\int^{2\sin\theta}_{\frac{1}{2sin\theta}}r^2\cos\theta \ dr d\theta = \frac{1}{3}\int^{\pi/2}_{\pi/6}\bigg\{8\sin^3\theta - \frac{1}{8sin^3\theta}\bigg\}\cos\theta d\theta$$
Now let $u = sin\theta$ , $du = \cos\theta d\theta$
At $\theta = \pi/6$, $u=1/2$ and at $\theta = \pi/2$, $u=1$
So, $$I = \frac{1}{3}\int^{1}_{1/2}\bigg\{8u^3- \frac{1}{8u^3}\bigg\}du = \frac{1}{3}\bigg[8\cdot\frac{1}{4}(1-\frac{1}{16}) - \frac{1}{8}\cdot\frac{-1}{2}\bigg(1 - \frac{1}{(1/2)^2}\bigg)\bigg] = \frac{1}{3}$$
$$I = \frac{1}{3}\big[ \frac{15}{8} - \frac{3}{16}\big] = \frac{1}{3}[\frac{27}{16}] = \frac{9}{16}$$
|
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Proving $\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge \:\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}$ The inequality:
$$\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge \:\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}$$
But can someone help me with a nice elegant solution. This is an olympiad question I was trying to solve, but couldn't manage an elegant solution.
|
$$\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge \:\sqrt{\left(a+b\right)^2+\left(c+d\right)^2} \\
\left( \sqrt{a^2+c^2}+\sqrt{b^2+d^2} \right)^2 \ge \: \left( \sqrt{\left(a+b\right)^2+\left(c+d\right)^2} \right)^2 \\ a^2+c^2+b^2+d^2+2\left( \sqrt{a^2+c^2}\sqrt{b^2+d^2} \right) \ge \: \left(a+b\right)^2+\left(c+d\right)^2 \\
a^2+c^2+b^2+d^2+2\left( \sqrt{(a^2+c^2)(b^2+d^2)} \right) \ge \: a^2+2ab+b^2+c^2+2cd+d^2 \\ 2\left( \sqrt{(a^2+c^2)(b^2+d^2)} \right) \ge \: 2ab+2cd \: \: \: \text{(subtract} \: a^2+b^2+c^2+d^2 \: \text{from both sides)} \\ \sqrt{(a^2+c^2)(b^2+d^2)} \ge \: ab+cd \: \: \text{(divide both sides by 2)} \\ (a^2+c^2)(b^2+d^2) \ge \: (ab+cd)^2 \\ a^2b^2+a^2d^2+c^2b^2+c^2d^2 \ge \: a^2b^2+2abcd+c^2d^2 \\ a^2d^2+c^2b^2 \ge \: 2abcd \: \: \text{(subtract } \: a^2b^2+c^2d^2 \: \text{from both sides)} \\ a^2d^2+c^2b^2 \ge \: 2abcd$$
|
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|
Find the minimum using lagrange I've got the formula $$x^3+y^3+z^3$$
With the constraint $ax+by+cz = 1$
I tried to solve this using lagrange but every possible way I try to use does not get me to the right answer
Using the lagrange I got to
$$3x^2+ka = 0 $$
$$3y^2+kb = 0 $$
$$3z^2+kc = 0 $$
=
$$3x^2bc+kabc = 0 $$
$$3y^2ac+kabc = 0 $$
$$3z^2ab+kabc = 0 $$
$$z = \sqrt{\frac{cy^2}{b}} $$
$$x = \sqrt{\frac{ay^2}{b}} $$
Fill this in the constraint and I then I cannot figure it out
Any got a clue how to solve this properly?
|
Assuming $x \ge 0, y \ge 0, z \ge 0$ to assure a bounded solution, we have from the stationary conditions
$$
x = \sqrt\frac{\lambda a}{3}\\
y = \sqrt\frac{\lambda b}{3}\\
z = \sqrt\frac{\lambda c}{3}\\
$$
and substituting into the restriction
$$
\frac{\lambda}{3} = \frac{1}{(a\sqrt a+b\sqrt b+c\sqrt c)^2}
$$
but
$$
x^3+y^3+z^3 = \left(\frac{\lambda}{3}\right)^{\frac 32}(a\sqrt a+b\sqrt b+c\sqrt c)
$$
hence
$$
\left\{\min_{x>0,y>0,z>0}(x^3+y^3+z^3)\ \ \mbox{s. t.}\ \ \ ax+by+cz=1\right\} = \frac{1}{(a\sqrt a+ b\sqrt b+c\sqrt c)^2}
$$
Now the bordered Hessian
$$
H_g = \left(
\begin{array}{cccc}
0 & -a & -b & -c \\
-a & 6 x & 0 & 0 \\
-b & 0 & 6 y & 0 \\
-c & 0 & 0 & 6 z \\
\end{array}
\right)
$$
has value
$$
\det(H_g) = -36 \left(a^2 y z+b^2 x z+c^2 x y\right)
$$
Which with $x>0,y>0,z>0$ characterizes a minimum.
|
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|
complete set of values of $a$ having modulus and linear terms
If $(9-x^2)>|x-a|$ has at least one negative real solution for $a\in\mathbb{R}.$ Then complete set of values of $a$ is
Plan
If $x>a$ Then $9-x^2>x-a\Rightarrow x^2+x-(a+9)<0$
If $x\leq a$ Then $9-x^2>a-x\Rightarrow x^2-x+a-9<0$
How do i solve these inequalityHelp me please
|
Sketch a graph!
$k=5$">
Now you can see it more clearly. You want the red curve to be above the blue one.
For $a= 3$, the two curves intersect at $(3,0)$ and $(-2,5)$. (Solve the equation $9-x^2=-x+3$) So the solution set is $x\in (-2,3)$. The solution set is $x\in (-3,2)$ for $a=-3$.
For $a \leq 3$, we need to solve the equation $9-x^2=-x+a$:
$$x^2-x+a-9=0\\
B^2-4AC=1-4(a-9)=37-4a\\
B^2-4AC\geq 0\Rightarrow a\leq \frac{37}{4}.
x=\frac{1\pm\sqrt{37-4a}}{2}.$$
So the solution set is $x\in (\frac{1-\sqrt{37-4a}}{2},\frac{1+\sqrt{37-4a}}{2})$ for $3\leq a< \frac{37}{4}$, and $\emptyset$ for $a\geq \frac{37}{4}$.
By symmetry, the solution set is $x\in (\frac{-1-\sqrt{37+4a}}{2},\frac{-1+\sqrt{37+4a}}{2})$ for $-3\geq a> -\frac{37}{4}$, and $\emptyset$ for $a\leq \frac{37}{4}$.
It remains to consider $a\in(-3,3)$. You can work out that it is $(\frac{-1-\sqrt{37+4a}}{2},\frac{1+\sqrt{37-4a}}{2}).$
|
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|
Solve $\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $ Solve $$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $$
My attempt:
Let $A = \sqrt{1+x}, B = \sqrt{1-x}$ and then by squaring the problematic equation we get:
$$(1+AB)(A^{3} + B^{3})^{2} = (AB)^{2} + 4AB + 4 $$
$$ A^{6} + B^{6} + BA^{7} + A B^{7} = -2 (AB)^{4} - 2(AB)^{3} + (AB)^{2} + 4AB + 4 $$
I also have tried using$A = (1+x), B = (1-x)$ , and some others, but none solves the problem.
I am now trying $A = (1+x)$ and $(1-x) = -(1+x) + 2 = 2 - A$, so:
$$\sqrt{1 + \sqrt{A(2-A)}}\left(\sqrt{(A)^{3}} + \sqrt{(2-A)^{3}} \right) = 2 + \sqrt{A(2-A)} $$
|
Let $p = A+B = \sqrt{1+x}+\sqrt{1-x}$. We have
$$ p^2 = 2+2\sqrt{1-x^2}$$
$$ p^3 = \sqrt{(1-x)^3}+\sqrt{(1+x)^3} + 3p\sqrt{1-x^2}$$
so
$$ 1+ \sqrt{1-x^2} = \frac12 p^2$$
$$ \sqrt{(1-x)^3}+\sqrt{(1+x)^3} = -\frac12 p^3 +3p$$
and we get
$$ \frac{p}{\sqrt{2}} (-\frac12 p^3 +3p) = 1 + \frac12 p^2$$
$$ - \frac{1}{2\sqrt{2}}p^4 +\frac{3\sqrt{2}-1}{2} p^2 - 1 = 0$$
which can be solved for $p$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
positive root of the equation $x^2+x-3-\sqrt{3}=0$ $x^2+x-3-\sqrt{3}=0$
using the quadratic formula we get
$$x=\frac{-1+\sqrt{13+4\sqrt{3}}}{2}$$ for the positive root
but the actual answer is simply $x=\sqrt3$
I am unable to perform the simplification any help would we helpful
|
$13+4\sqrt{3}=(2\sqrt{3})^2+2(2\sqrt{3})(1)+1=(2\sqrt{3}+1)^2$
So it is just $\dfrac{-1+2\sqrt{3}+1}{2}$.
We may also start from the original equation.
\begin{align*}
x^2+x-3-\sqrt{3}&=0\\
x^2-3+x-\sqrt{3}&=0\\
(x-\sqrt{3})(x+\sqrt{3})+x-\sqrt{3}&=0\\
(x-\sqrt{3})(x+\sqrt{3}+1)&=0
\end{align*}
|
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|
Find the amplitude of a complex number
If
$$z = (1-\cos{\theta}) + i \sin{\theta},$$
then the amplitude of $z$ is:
$$a) \frac{\pi}{4}-\frac{\theta}{2}$$ $$b)
\frac{\pi}{2}-\frac{\theta}{2}$$ $$c) \frac{\pi}{2}-\theta$$ $$d)
\frac{\pi}{2}-\frac{\theta}{4}$$ where 0<$\theta$ <$\pi$.
I tried :
$x= 1-\cos{\theta}, y= \sin{\theta}$ then
$\tan{\phi}=\frac{\sin{\theta}}{1-\cos{\theta}}$.
However I could not solve the last equation.
|
Your $z$ can be nicely written as
$$z=1-e^{-i \theta}$$
Now,
\begin{align}
|z| &= |1-e^{-i \theta}| \\
&=\bigg |e^{-i \frac{\theta}{2}}\left(e^{i \frac{i\theta}{2}} -e^{-i \frac{\theta}{2}} \right)\bigg | \\
&=\bigg |e^{-i \frac{i\theta}{2}} \left( \cos \left( \frac{\theta}{2}\right) + i \sin \left( \frac{\theta}{2}\right)-\cos \left( \frac{-\theta}{2}\right) - i \sin \left( \frac{-\theta}{2}\right) \right)\bigg |\\
&= \bigg |2i \sin \left( \frac{\theta}{2}\right)e^{i \frac{i\theta}{2}}\bigg|\\
&= |2i|. \bigg |\sin \left( \frac{\theta}{2}\right)\bigg|.\bigg|e^{i \frac{-\theta}{2}}\bigg| \\
&= 2 \bigg |\sin \left(\frac{\theta}{2}\right)\bigg |
\end{align}
Whence
\begin{align}
\angle z &= \angle (2i)+\angle\left( \sin \left(\frac{\theta}{2}\right) \right)+\angle (e^{-i \frac{\theta}{2}}) \\
&= \frac{\pi}{2}+0-\frac{\theta}{2}
\end{align}
as required.
|
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|
Find $a,b,c,d$ such that $2^a + 2^b + 2^c = 4^d$ Let $a,b,c,d$ be whole numbers that satisfy
$$2^a + 2^b + 2^c = 4^d$$
What values of $(a,b,c,d)$ would make this equation true?
Here is my work so far.
Without loss of generality, assume $a\ge b\ge c$. Then one trivial solution by inspection is $(1,0,0,1)$. Playing around, I also found a solution at $(3,2,2,2)$. Then I checked $a=5,b=4,c=4$ and found that it also worked.
It seems that there is a family of solutions at $(2n-1,2n-2,2n-2,n)$. I can prove this easily:
\begin{align}
LHS&=2^{2n-1}+2^{2n-2}+2^{2n-2}\\
&=2^{2n-1}+2^{2n-1}\\
&=2^{2n}\\
&=4^n\\
&=RHS
\end{align}
Is this the only solution? If it is, how do I go about proving it?
|
$$2^a+2^b+2^c=2^c(2^{a-c}+2^{b-c}+1)$$
Since the second factor must be even, we see that $a>b=c$, and then it becomes
$$2^c(2^{a-c}+2)=2^{c+1}(2^{a-c-1}+1)$$
And now we see that $a=c+1$.
Now,
$$2^a+2^b+2^c=4\cdot2^c$$
so $c$ is even.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $ \lim_{k \rightarrow \infty } \frac{ \sum_{x = 0}^{k} {x^{n + 1}}} {(k + 1)k^{n + 1}}$ How to compute the limit
$$ \lim_{k \rightarrow \infty } \frac{ \sum_{x = 0}^{k} {x^{n + 1}}} {(k + 1)k^{n + 1}}$$
Does exist an explicit formula for finding the numerator?
|
By the Stolz-Cesro Theorem , one has
\begin{eqnarray}
&&\lim_{k \rightarrow \infty } \frac{ \sum_{x = 0}^{k} {x^{n + 1}}} {(k + 1)k^{n + 1}}\\
&=&\lim_{k \rightarrow \infty } \frac{{(k+1)^{n + 1}}} {(k + 2)(k+1)^{n + 1}-(k + 1)k^{n + 1}}\\
&=&\lim_{k \rightarrow \infty } \frac{1} {(k + 2)-(k + 1)\bigg(\frac{k}{k+1}\bigg)^{n + 1}}\\
&=&\lim_{k \rightarrow \infty } \frac{1} {(k + 2)-(k + 1)\bigg(1-\frac{1}{k+1}\bigg)^{n + 1}}\\
&=&\lim_{k \rightarrow \infty } \frac{1} {(k + 2)-(k + 1)\bigg(1-(n+1)\frac{1}{k+1}+O(\frac1{(k+1)^2})\bigg)}\\
&=&\frac 1{n+2}.
\end{eqnarray}
|
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|
Find Maxima and Minima of $f( \theta) = a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta$ Show that, whatever the value of $\theta$, the expression
$$a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta\ $$
Lies between
$$\dfrac{a+c}{2} \pm \dfrac 12\sqrt{ b^2 + (a-c)^2} $$
My try:
The given expression can be reduced as sum of sine functions as:
$$(a-c) \sin^2 \theta + \dfrac b2 \sin 2 \theta + c \tag{*} $$
Now, there is one way to take everything as a function of $\theta$ and get the expression in the form of $ a \sin \theta + b \cos \theta = c$ and dividing it by $ \sqrt{ a^2 + c^2} $ both sides, but square in sine function is a big problem, also both have different arguments.
Other way, I can think of is taking $ \tan \dfrac \theta 2 = t$ and getting sine and cosine function as $ \sin \theta = \dfrac{ 2t}{1+t^2} $ while cosine function as $ \dfrac{ 1-t^2} {1+t^2}$ solving. So getting $(*)$ as a function of $t$, and simplifying we get,
$$ f(t) = \dfrac{2 Rt + 2 R t^3 + R_0 t - R_0 t^3}{1+t^4 + 2t^2} + c\tag{1}$$
For $R_0 = 2b, R = (a-c)$ , but this is where the problem kicks in!, The Range of given fraction seems $ (-\infty,+ \infty)$ and is not bounded!
So what's the problem here? Can it be solved?
Thanks :)
Edit : I'd like to thank @kaviramamurthy for pointing out that as $t \rightarrow \pm \infty, f(t) \rightarrow c$. That's a mistake here.
|
By C-S we obtain:
$$a\sin^2\theta+b\sin\theta\cos\theta+c\cos^2\theta=a\cdot\frac{1-\cos2\theta}{2}+b\cdot\frac{\sin2\theta}{2}+c\cdot\frac{1+\cos2\theta}{2}=$$
$$=\frac{1}{2}\left(a+c+b\sin2\theta-(a-c)\cos2\theta\right)\leq\frac{1}{2}\left(a+c+\sqrt{(b^2+(-a+c)^2)\left(\sin^22\theta+\cos^22\theta\right)}\right)=$$
$$=\frac{1}{2}\left(a+c+\sqrt{b^2+(a-c)^2}\right)$$ and
by C-S again we obtain:
$$a\sin^2\theta+b\sin\theta\cos\theta+c\cos^2\theta=\frac{1}{2}\left(a+c+b\sin2\theta-(a-c)\cos2\theta\right)\geq$$
$$\geq \frac{1}{2}\left(a+c-\sqrt{(b^2+(-a+c)^2)\left(\sin^22\theta+\cos^22\theta\right)}\right)=$$
$$=\frac{1}{2}\left(a+c-\sqrt{b^2+(a-c)^2}\right).$$
The equality in the both cases occurs for $$(\sin2\theta,\cos2\theta)||(b,-a+c),$$
which says that we got a minimal and the maximal value of the expression.
Now, since $f$ is a continuous function, we obtain that the range of $f$ it's:
$$\left[\frac{1}{2}\left(a+c-\sqrt{b^2+(a-c)^2}\right),\frac{1}{2}\left(a+c+\sqrt{b^2+(a-c)^2}\right)\right]$$
|
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|
Determining the coefficients of $(1 + x + x^2 +\cdots+x^n)^{n-1}$ Suppose we have the following polynomials:
$$f_1(x)=(1 + x + x^2)$$
$$f_2(x)=(1 + x + x^2 + x^3)^2$$
$$f_3(x)=(1 + x + x^2 + x^3 + x^4)^3$$
$$f_4(x)=(1 + x + x^2 + x^3 + x^4 + x^5)^4$$
$$\vdots$$
$$f_{n-1}(x)=(1 + x + x^2 + x^3 +x^4+ x^5+\cdots+x^n)^{n-1}$$
upon expanding them we get:
$$f_1(x)=1 + x + x^2$$
$$f_2(x)=1 + 2 x + 3 x^2 + 4 x^3 + 3 x^4 + 2 x^5 + x^6$$
$$f_3(x)=1 + 3 x + 6 x^2 + 10 x^3 + 15 x^4 + 18 x^5 + 19 x^6 + 18 x^7 +
15 x^8 + 10 x^9 + 6 x^{10} + 3 x^{11} + x^{12}$$
$$f_4(x)=1 + 4 x + 10 x^2 + 20 x^3 + 35 x^4 + 56 x^5 + 80 x^6 + 104 x^7 +
125 x^8 + 140 x^9 + 146 x^{10} + 140 x^{11} + 125 x^{12} + 104 x^{13} +
80 x^{14} + 56 x^{15} + 35 x^{16} + 20 x^{17} + 10 x^{18} + 4 x^{19} + x^{20}$$
$$\vdots$$
$$f_{n-1}(x)=1 + ?x + ?x^2 + ?x^3 +?x^4+ ?x^5+\cdots+?x^{n(n-1)}$$
I'm wondering how to determine the coefficients for the n-th order? I can observe that the coefficients are symmetric.
|
We use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of a series.
We obtain for $0\leq k\leq n(n-1)$:
\begin{align*}
\color{blue}{[x^{k}]}&\color{blue}{\left(1+x+x^2+\cdots+x^n\right)^{n-1}}\\
&=[x^k]\left(\frac{1-x^{n+1}}{1-x}\right)^{n-1}\tag{1}\\
&=[x^k]\sum_{j=0}^\infty\binom{-(n-1)}{j}(-x)^j\sum_{l=0}^{n-1}\binom{n-1}{l}\left(-x^{n+1}\right)^l\tag{2}\\
&=[x^k]\sum_{j=0}^\infty\binom{n-j-2}{j}x^j\sum_{l=0}^{n-1}\binom{n-1}{l}(-1)^lx^{(n+1)l}\tag{3}\\
&=\sum_{j=0}^{k}\binom{n-j-2}{j}[x^{k-j}]\sum_{l=0}^{n-1}\binom{n-1}{l}(-1)^lx^{(n+1)l}\tag{4}\\
&=\sum_{j=0}^{k}\binom{n-k+j-2}{k-j}[x^{j}]\sum_{l=0}^{n-1}\binom{n-1}{l}(-1)^lx^{(n+1)l}\tag{5}\\
&=\sum_{j=0}^{\left\lfloor\frac{k}{n+1}\right\rfloor}\binom{n-k+(n+1)j-2}{k-(n+1)j}[x^{(n+1)j}]\sum_{l=0}^{n-1}\binom{n-1}{l}(-1)^lx^{(n+1)l}\tag{6}\\
&\,\,\color{blue}{=\sum_{j=0}^{\left\lfloor\frac{k}{n+1}\right\rfloor}\binom{(n+2)j-k-2}{k-(n+1)j}\binom{n-1}{j}(-1)^j}\tag{7}\\
\end{align*}
Comment:
*
*In (1) we use the finite geometric series formula.
*In (2) we use the binomial series expansion and apply the binomial theorem.
*In (3) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (4) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$ and set the upper limit of the outer sum to $k$ since indices $j>k$ do not contribute.
*In (5) we reverse the order of summation of the outer sum: $j\to k-j$.
*In (6) we consider only $(n+1)$-multiples of $j$ since other values do not occur as exponent of $x$ in the inner sum.
*In (7) we finally select the coefficients of $x^{(n+1)j}$ by taking $l=j$.
|
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|
Prove that $\sum_{cyc}\frac{1}{x^2 + 1} \ge \frac{2}{3}\bigl(\sum_{cyc}\frac{x}{\sqrt{x^2 + 1}}\bigr)^3$ where $x, y, z > 0$ and $xy + yz + zx = 1$.
$x$, $y$ and $z$ are positives such that $xy + yz + zx = 1$. Prove that $$\large \frac{1}{x^2 + 1} + \frac{1}{y^2 + 1} + \frac{1}{z^2 + 1} \ge \frac{2}{3}\left(\frac{x}{\sqrt{x^2 + 1}} + \frac{y}{\sqrt{y^2 + 1}} + \frac{z}{\sqrt{z^2 + 1}}\right)^3$$
We have that $$\frac{x}{\sqrt{x^2 + 1}} = \frac{x}{\sqrt{x^2 + xy + yz + zx}} = \frac{x}{\sqrt{(x + y)(z + x)}} \le \frac{x}{x + \sqrt{yz}} = 1 - \frac{\sqrt{yz}}{x + \sqrt{yz}}$$
$$ \implies \sum_{cyc}\frac{x}{\sqrt{x^2 + 1}} \le 3 - \sum_{cyc}\frac{\sqrt{yz}}{x + \sqrt{yz}} = 3 - \sqrt{xyz}\sum_{cyc}\frac{1}{x\sqrt x + \sqrt{xyz}} \le 3\left(1 - \frac{3\sqrt{xyz}}{x\sqrt x + y\sqrt y + z\sqrt z + 3\sqrt{xyz}}\right) = \frac{3(x\sqrt x + y\sqrt y + z\sqrt z)}{x\sqrt x + y\sqrt y + z\sqrt z + 3\sqrt{xyz}}$$
Furthermore, $$\frac{1}{x^2 + 1} = \frac{1}{x^2 + xy + yz + zx} = \frac{1}{(x + y)(z + x)} \ge \frac{4}{(z + 2x + y)^2}$$
$$\ge \frac{8}{4x^2 + (y + z)^2} = \frac{2}{x^2}\left[1 - \frac{(y + z)^2}{4x^2 + (y + z)^2}\right] = \frac{2}{x^2}\left[1 - \frac{(y + z)^2}{4(x^2 + yz)}\right]$$
$$\implies \sum_{cyc}\frac{1}{x^2 + 1} \ge \sum_{cyc}\frac{2}{x^2}\left[1 - \frac{(y + z)^2}{4(x^2 + yz)}\right]$$
$$\ge \frac{18}{x^2 + y^2 + z^2} - \frac{1}{2}\sum_{cyc}x^2\left[\frac{1}{y^2(y^2 + zx)} + \frac{1}{z^2(z^2 + xy)}\right]$$
And I'm stuck.
|
We need to prove that:
$$\sum_{cyc}\frac{1}{x^2+1}\geq\frac{2}{3}\left(\sum_{cyc}\frac{x}{\sqrt{x^2+1}}\right)^3$$ or
$$\sum_{cyc}xy\sum_{cyc}\frac{1}{(x+y)(x+z)}\geq\frac{2}{3}\left(\sum_{cyc}\frac{x}{\sqrt{(x+y)(x+z)}}\right)^3.$$
Now, by C-S
$$\sum_{cyc}xy\sum_{cyc}\frac{1}{(x+y)(x+z)}=\frac{\sum\limits_{cyc}x(y+z)\sum\limits_{cyc}x}{\prod\limits_{cyc}(x+y)}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}x\sqrt{y+z}\right)^2}{\prod\limits_{cyc}(x+y)}=\left(\sum_{cyc}\frac{x}{\sqrt{(x+y)(x+z)}}\right)^2.$$
Id est, it's enough to prove that
$$\sum_{cyc}\frac{x}{\sqrt{(x+y)(x+z)}}\leq\frac{3}{2},$$
which is true by AM-GM:
$$\sum_{cyc}\frac{x}{\sqrt{(x+y)(x+z)}}=\sum_{cyc}\left(\sqrt{\frac{x}{x+y}}\cdot\sqrt{\frac{x}{x+z}}\right)\leq$$
$$\leq\frac{1}{2}\sum_{cyc}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{x}{x+y}+\frac{y}{y+x}\right)=\frac{3}{2}.$$
Done!
|
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|
Prove that $\sum_{k=0}^{n}{\frac{1}{k!}}\leq 3$ $\forall n\in \mathbb{N}$ Prove that $\sum_{k=0}^{n}{\frac{1}{k!}}\leq 3$ $\forall n\in \mathbb{N}$
I proved it by induction:
"$n=1$" $ \quad \sum_{k=0}^{1}{\frac{1}{k!}}\leq 3 \quad \checkmark$
"$n \implies n+1$": $$\quad \sum_{k=0}^{n+1}{\frac{1}{k!}}\leq 3 \\ \left(\sum_{k=0}^{n}{\frac{1}{k!}}\right)+\frac{1}{(n+1)!}\leq 3 \\ $$$$3+\overbrace{\frac{1}{(n+1)!}}^{\geq 0}\leq 3 \\$$ That's of course true.
|
$$\sum_{k=0}^{\infty}\frac{1}{k!} = 1 + 1 + \frac{1}{2!}+\frac{1}{3!}+ \cdots$$ $$ < 1+1+\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} +\frac{1}{4\cdot 5}+ \cdots$$ $$ = 1 + 1 +\left (1-\frac{1}{2}\right) + \left (\frac{1}{2}-\frac{1}{3}\right) + \left (\frac{1}{3}-\frac{1}{4}\right)+\cdots$$ $$ = 1+1 +1 =3.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)} \ge xy + yz + zx$ given that where $x, y, z > 0$ and $xyz = \frac{1}{2}$.
$x$, $y$ and $z$ are positives such that $xyz = \dfrac{1}{2}$. Prove that $$ \frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)} \ge xy + yz + zx$$
Before you complain, this problem is adapted from a recent competition. I have put my solution down below, there might be more practical and correct answers. In that case, please post them.
|
Let $x=\frac{1}{a},$ $y=\frac{1}{b}$ and $z=\frac{1}{c}.$
Thus, $abc=2$ and by C-S we obtain:
$$\sum_{cyc}\frac{xy}{z^2(x+y)}=\sum_{cyc}\frac{c^2}{a+b}\geq\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}=xy+xz+yz.$$
|
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|
Find all ordered pairs $(x,y)$ that satisfy both $\frac{3x-4y}{xy} = -8$ and $\frac{2x+7y}{xy} = 43$ The following is listed under the "multiple variable" category of my Algebra I homework.
Find all ordered pairs $(x,y)$ that satisfy both $\frac{3x-4y}{xy} = -8$ and $\frac{2x+7y}{xy} = 43$
I can't wrap my head around what to do. Thanks in advance for your help.
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$$\frac{3x-4y}{xy} = \frac{3}{y} - \frac{4}{x} = -8$$
$$\frac{2x+7y}{xy} = \frac{2}{y}+\frac{7}{x} = 43$$
Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$.
Then,
$3v - 4u = -8$ and $2v + 7u = 43$
Solve these linear equations and replace $x$ and $y$.
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|
How to find the sum of 2 discrete random variables Let X and Y be independent random discrete variables and
$X = \begin{pmatrix}-1&0&1\\\frac13&\frac13&\frac13\end{pmatrix}$ and $Y = \begin{pmatrix}0&1\\\frac13&\frac23\end{pmatrix}$
Then what is $X + Y$? As a discrete random variable.. I understood that I have to add the first row just like in matrix addition but what happens when they don't have the same size and how to add the probabilities? Also what is the difference between independent variables and incompatible variables?
The missing entries were:
$X + Y = \begin{pmatrix}-1&0&1&2\\\frac19&\frac39&\frac39&\frac29\end{pmatrix}$
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Hint: Can you enumerate all the cases? For example, if $X+Y=1$ it means either $X=0$, $Y=1$ (probability $\frac{1}{3}\cdot\frac{2}{3}=\frac{2}{9}$) or $X=1$, $Y=0$ (probability $\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}$) $\Longrightarrow P\left(X+Y=1\right)=\frac{2}{9}+\frac{1}{9}=\frac{1}{3}$.
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|
Solve system of congruences $k^2 + l^2 \equiv 0 \mod 17 \wedge k^3+l^3 \equiv 0 \mod 17$
Solve system of congruences
$$k^2 + l^2 \equiv 0 \mod 17 \wedge k^3+l^3 \equiv 0 \mod 17$$
Is there any faster way to solve that congurence than looking at table and finding such pairs of $i$ that in both cases it gives $0$? When we have this table it is very fast job but making it is... time-consuming
\begin{array}{|c|c|c|c|}
\hline
i^2 \mod 17 & i^3 \mod 17 \\ \hline
0 & 0 \\ \hline
1 & 1 \\ \hline
4 & 8 \\ \hline
9 & 10 \\ \hline
16 & 13 \\ \hline
8 & 6 \\ \hline
2 & 12 \\ \hline
15 & 3 \\ \hline
13 & 2 \\ \hline
13 & 15 \\ \hline
15 & 14 \\ \hline
2 & 5 \\ \hline
8 & 11 \\ \hline
16 & 4 \\ \hline
9 & 7 \\ \hline
4 & 9 \\ \hline
1 & 16 \\ \hline
0 & 0 \\ \hline
1 & 1 \\ \hline
4 & 8 \\ \hline
\end{array}
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From $k^3 + l^3 \equiv 0 \pmod{17}$, we have
$$k^3 \equiv (-l)^3 \pmod {17}.$$
But, as you can see from your table, cubes are distinct modulo $17$, which effectively lets us take cube roots: thus, $k \equiv -l \pmod {17}$. Therefore
$$k^2 + (-k)^2 \equiv 0 \pmod {17},$$
so $2k^2 \equiv 0 \pmod {17}$. Divide by $2$ and conclude that $k \equiv l \equiv 0 \pmod{17}$.
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Find K such that the line is tangent to $y=-\frac{3}{4}x+4$ The function is $f(x)= \frac{k}{x}$ and the line is $y =-\frac{3}{4}x+3$
So the derivative of $f(x)$ should equal $-\frac{3}{4}$
If I take $k$ to be a constant I can remove it from the fraction like so
$$f'(x)=k \frac{d}{dx} \frac{1}{x}$$
to get $-\frac{k}{x^2}$ but then I try to solve for $k$
$$-\frac{k}{x^2}=-\frac{3}{4}$$
and this is where I get stuck. So should I not even remove $k$ from the differentiation or have I gone in a completely wrong direction?
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If the tangent occurs at $x = t$, we have $\displaystyle\frac{k}{t^2} = \frac{3}{4}$, as you've already found, but we also must have $\displaystyle\frac{k}{t} = -\frac{3}{4}t + 3$.
So we obtain $\displaystyle k = -\frac{3t^2}{4} + 3t$, and subbing into the first equation gives $\displaystyle -\frac{3}{4} + \frac{3}{t} = \frac{3}{4}$, from which we obtain $\displaystyle \frac{1}{t} = \frac{1}{2}$ and finally $t=2$. So $\displaystyle \frac{k}{t^2} = \frac{3}{4}$ now becomes $\displaystyle\frac{k}{4} = \frac{3}{4}$, and so $k=3$.
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|
Trig identities analogous to $\tan(\pi/5)+4\sin(\pi/5)=\sqrt{5+2\sqrt{5}}$ The following trig identities have shown up in various questions on MSE:
$$-\tan\frac{\pi}{5}+4\sin\frac{2\pi}{5}=\tan\frac{\pi}{5}+4\sin\frac{\pi}{5}=\sqrt{5+2\sqrt{5}}$$
$$-\tan\frac{2\pi}{7}+4\sin\frac{3\pi}{7}=-\tan\frac{\pi}{7}+4\sin\frac{2\pi}{7}=\sqrt{7}$$
$$\tan\frac{\pi}{11}+4\sin\frac{3\pi}{11}=\tan\frac{3\pi}{11}+4\sin\frac{2\pi}{11}=\sqrt{11}$$
$$\tan\frac{6\pi}{13}-4\sin\frac{5\pi}{13}=\tan\frac{2\pi}{13}+4\sin\frac{6\pi}{13}=\sqrt{13+2\sqrt{13}}$$
Does anyone know of any analogous identities for larger primes? I have not been able to find anything similar for $p=17$ or $p=19$.
(I am not asking for proofs of the above equations.)
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There is also the following: $$\tan\frac{\pi}{19}+4\sin\frac{7\pi}{19}+4\sin\frac{8\pi}{19}-4\sin\frac{6\pi}{19}=\sqrt{19},$$
$$-\tan\frac{3\pi}{19}+4\sin\frac{\pi}{19}+4\sin\frac{2\pi}{19}+4\sin\frac{5\pi}{19}=\sqrt{19},$$
$$\tan\frac{5\pi}{19}+4\sin\frac{2\pi}{19}-4\sin\frac{3\pi}{19}-4\sin\frac{8\pi}{19}=\sqrt{19},$$
$$\tan\frac{7\pi}{19}+4\sin\frac{\pi}{19}-4\sin\frac{4\pi}{19}+4\sin\frac{8\pi}{19}=\sqrt{19}.$$
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|
Evaluate $\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}$ How to prove that
$$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}=\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\ln2+\frac{\pi^4}{32}-\frac1{256}\psi^{(3)}\left(\frac14\right)$$
where $H_n=1+\frac1{2}+\frac1{3}+...+\frac1{n}$ is the $n$th harmonic number.
This sum was proposed by Cornel and I solved it using integration but can we solve it using series manipulation?
The integral representation of the sum is $\ \displaystyle\frac12\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx$ in case it is needed.
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\begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}&=\sum_{n=1}^\infty(-1)^{n-1} H_n\int_0^1\frac12x^{2n}\ln^2 x\ dx\\
&=-\frac12\int_0^1\ln^2x\sum_{n=1}^\infty(-x^2)H_n\\
&=\frac12\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx\tag{1}
\end{align}
\begin{align}
I&=\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx\\
&=\int_0^\infty\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx-\underbrace{\int_1^\infty\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx}_{\large x\mapsto1/x}\\
&=\underbrace{\int_0^\infty\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx}_{\large x^2\mapsto x}-I+2\int_0^1\frac{\ln^3x}{1+x^2}\ dx\\
2I=&\frac18\int_0^\infty\frac{\ln^2x\ln(1+x)}{\sqrt{x}(1+x)}\ dx+2(-6\beta(4))\\
I&=\frac1{16}\lim_{a\ \mapsto1/2\\b\ \mapsto1/2}\frac{-\partial^3}{\partial a^2\partial b}\text{B}(a,b)-6\beta(4)\\
&=\frac1{16}(14\pi\zeta(3)+2\pi^3\ln2)-6*\frac1{768}\left(\psi^{(3)}\left(\frac14\right)-8\pi^4\right)\\
&=\frac{7\pi}{8}\zeta(3)+\frac{\pi^3}{8}\ln2-\frac1{128}\left(\psi^{(3)}\left(\frac14\right)-8\pi^4\right)\tag{2}
\end{align}
Plugging $(2)$ in $(1)$ we get
$$\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}=\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\ln2+\frac{\pi^4}{32}-\frac1{256}\psi^{(3)}\left(\frac14\right)$$
Notes:
$\displaystyle\beta(s)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}\ $ is the Dirichlet beta function and the value of $\beta(4)$ can be found here.
$\displaystyle\text{B}(a,b)=\int_0^\infty\frac{x^{a-1}}{(1+x)^{a+b}}\ dx$ is beta function.
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|
How to get the value of the root? I have this statement:
If $\frac{\sqrt{7}}{\sqrt{5}+\sqrt{3}} \approx \frac{2}{3}$, Which of
the following values are the closest to $\sqrt{21}$ ?
A) 68/15 B) 14/3 C) 19/4 D) 55/12 E) 9/2
My development was:
$\frac{\sqrt{7}}{\sqrt{5}+\sqrt{3}} = \frac{\sqrt{35}-\sqrt{21}}{2} \approx \frac{2}{3}$
My idea was to treat the sign $\approx$ as a sign $=$ and thus eliminate roots and clear $\sqrt21$, with which I have obtained $55/12$ but I do not know if this is correct.
$\frac{\sqrt{35}-\sqrt{21}}{2} = \frac{2}{3}$
$35 = (\frac{4}{3} +\sqrt{21})^2$
$\sqrt{21} = \frac{110}{9} * \frac{3}{8} = 55/12$
So my doubt is: Can I treat a $\approx$ sign as a $=$ sign to work like an normal equation?
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Rearrange:
$$\frac{\sqrt{7}}{\sqrt{5}+\sqrt{3}} \approx \frac{2}{3} \iff \frac{7}{\sqrt{35}+\sqrt{21}} \approx \frac{2}{3} \Rightarrow \sqrt{21}\approx \frac{21}{2}-\sqrt{35}$$
Use linear approximation to find $f(35)=\sqrt{35}$:
$$f(x_0+\Delta x)\approx f(x_0)+f'(x_0)\cdot \Delta x \ ;\\
f(36-1)\approx f(36)+\frac1{2\sqrt{36}}\cdot (-1)=6-\frac1{12}=\frac{71}{12}$$
So:
$$\frac{21}{2}-\frac{71}{12}=\frac{126-71}{12}=\frac{55}{12}.$$
|
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|
If $x$ and $y$ are integers such that $5 \mid x^2 - 2xy - y$ and $5 \mid xy - 2y^2 - x$, prove that $5 \mid 2x^2 + y^2 + 2x + y$.
Given that $x$ and $y$ are integers satisfying $5 \mid x^2 - 2xy - y$ and $5 \mid xy - 2y^2 - x$, prove that $5 \mid 2x^2 + y^2 + 2x + y$.
I have provided a (dumbfounding) solution down below if you want to check out. There should be simpler solutions, I believe so.
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If $5\mid x^2-2xy-y$ and $5\mid xy-2y^2-x$ then $5$ also divides
$$(3y-1)(x^2-2xy-y)-(3x+2)(xy-2y^2-x)=2x^2+y^2+2x+y.$$
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Arithmetic sequences. Writing the general term differently So I'm dealing with this problem that has to do with arithmetic sequences and I was wondering if I could write the general term of this kind of sequences (finite) like $$a_{k-1}=a_1+(k-\lfloor \frac{n}2 \rfloor)d$$ where $k$ can take values from $2$ to $n+1$
$\lfloor \rfloor$ is the floor function and d is the common difference
For example, if we take the sequence $1,4,7,10,13$.
In this case $n=5$ and $d=a_n-a_{n-1}=3$
From the formula above for $k=2$ $$a_1=1+3(2-\lfloor \frac{5}2 \rfloor)=1$$
$k=3$ $$a_2=1+3(3-\lfloor \frac{5}2 \rfloor)=4$$
$k=4$ $$a_3=1+3(4-\lfloor \frac{5}2 \rfloor)=7$$
$k=5$ $$a_4=1+3(5-\lfloor \frac{5}2 \rfloor)=10$$
$k=6$ $$a_5=1+3(6-\lfloor \frac{5}2 \rfloor)=13$$
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If you really want to, you can, but $n$ isn't a free parameter. If you want the formula $$ a_{k-1} = a_1 + (k-\lfloor \frac{n}{2}\rfloor)d $$
to hold for $k=2$, you get a condition
$$ a_1 = a_1 + (2-\lfloor \frac{n}{2}\rfloor)d$$
that means that for $d\neq 0$ you'll need
$$\lfloor \frac{n}{2}\rfloor = 2$$
So, for natural $n$ it's only correct for $n=4$ and $n=5$ (like in your example). The formula simplifies then to the usual
$$ a_{k-1} = a_1 + (k-2)d $$
$$ a_k = a_1 + (k-1) d $$
What may be of more use to you are the formulae
$$ a_{k} = a_{\lfloor \frac{n}{2}\rfloor} + (k-\lfloor \frac{n}{2}\rfloor)d $$
or
$$ a_{k-1} = a_{\lfloor \frac{n}{2}\rfloor-1} + (k-\lfloor \frac{n}{2}\rfloor)d $$
that are also variations on the general formula, but allow you to use an arbitrary $n$.
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What am I doing wrong solving $\sqrt{x^2+1}- 2x+1>0$? First, I started looking at where $\sqrt{x^2+1}$ is defined: $\sqrt{x^2+1}>0$ is defined everywhere. Next I
$\sqrt{x^2+1}- 2x+1>0$
$\sqrt{x^2+1}> 2x-1$
$0> 3x^2-4x$
$0> x(3x-4)$ and I solve this for $x\in (0,\frac{4}{3}]$.
I know that this solution is wrong, because I went and drew this graph. The correct result is $x\in (-\infty,\frac{4}{3}]$.
I'm totally confused on how to solve irrational inequalities now, because the official solving in my textbook looks like this:
$\sqrt{x^2+1}> 2x-1$. This inequality is fulfilled , if the right side is negative, therefore $x<\frac{1}{2}$. If $x\geq\frac{1}{2}$, the right side is positive or equals to $0$ and we get $0> 3x^2-4x$ which is true for $x\in (0,\frac{4}{3}]$. Now with previous condition $x\geq\frac{1}{2}$ we get the solution $[\frac{1}{2},\frac{4}{3})$.The complete solution set is $x\in (-\infty,\frac{4}{3}]$.
I never solved this type of inequality in this way, because it's messy- Why would I look at the conditions $x<\frac{1}{2}$ and $x\geq\frac{1}{2}$ when I can immediately tell where irrational part is defined and where not?
In the end, this textbook solution process only confused me. Could anyone please explain it why the correct solution is $(-\infty,\frac{4}{3}]$ or more concretely: Where did I miss the part of the solution $(-\infty,0]$?
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If $A,B\in \Bbb R$,$$\sqrt A>B\iff \begin{cases}A\ge 0\\ B<0\end{cases}\lor\begin{cases}A\ge 0\\ B\ge 0\\ A>B^2\end{cases}$$ In this instance $A=x^2+1$, $B=2x-1$ $$\begin{cases}x^2+1\ge 0\\ 2x-1<0\end{cases}\lor\begin{cases}x^2+1\ge 0\\ 2x-1\ge 0\\ x^2+1>4x^2-4x+1\end{cases}\iff x<\frac12\lor\begin{cases}x\ge \frac12\\ x(3x-4)<0\end{cases}\iff\\ \iff x<\frac12\lor \begin{cases}x\ge\frac12\\ 0<x<\frac43\end{cases}\iff x<\frac12\lor \frac12\le x<\frac43\iff x<\frac43$$
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Proving inequality for induction proof: $\frac1{(n+1)^2} + \frac1{n+1} < \frac1n$ In an induction proof for $\sum_{k = 1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$ (for $n \geq 1$), I was required to prove the inequality $\frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}$.
This is my attempt:
$$ \begin{align} \frac{1}{(n+1)^2} + \frac{1}{n+1} &= \frac{1+(n+1)}{(n+1)^2} \\
&=\frac{n+2}{(n+1)^2} \\
&=\frac{n+2}{n^2 + 2n + 1} \\
&<\frac{n+2}{n^2 + 2n} (\text{since } n \geq 1)\\
&=\frac{n+2}{n(n+2)}\\
&=\frac{1}{n} \end{align} $$
I am just wondering if there is a simpler way of doing this.
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Multiply through by $n(n+1)^2$ (a positive number since $n \ge 1$ ) to get $$n+n(n+1) < (n+1)^2$$
$$n < (n+1)^2 - n(n+1)$$
$$n < (n+1)(n+1-n)$$
$$n < n+1$$
Which is true of course for all $n$; in particular $n\ge 1$
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sum of series $\frac{1}{1\cdot 3}+\frac{1}{4\cdot 5}+\frac{1}{7\cdot 7}+\frac{1}{10\cdot 9}+\cdots $
The sum of series
$$\frac{1}{1\cdot 3}+\frac{1}{4\cdot 5}+\frac{1}{7\cdot 7}+\frac{1}{10\cdot 9}+\cdots $$
My attempt $$\displaystyle \sum^{n}_{k=1}\frac{1}{(3k-2)(2k+1)}=\frac{1}{7}\sum^{n}_{k=1}\bigg[\frac{3}{3k-2}-\frac{2}{2k+1}\bigg]$$
$$=\frac{1}{7}\sum^{n}_{k=1}\int^{1}_{0}\bigg(3x^{3k-3}-2x^{2k}\bigg)dx$$
$$=\frac{1}{7}\int^{1}_{0}\bigg(\sum^{n}_{k=1}3x^{3k-3}-2x^{2k}\bigg)dx$$
$$=\frac{1}{7}\int^{1}_{0}\bigg[\frac{3(1-x^{3n})}{1-x^3}-\frac{2(1-x^{2n+2})}{1-x^2}\bigg]dx$$
How do i solve it Help me please
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(CONSIDERING INFINITE SUM)
We have $$I_n=\frac 37\int_0^1 \frac {1-x^{3n}}{1-x^3} dx-\frac 27\int_0^1 \left(\frac {1-x^{2n+2}}{1-x^2} +\frac {x^2-1}{1-x^2}\right) dx$$
Let the first integral be denoted by $I_{1,n}$ and the second be denoted by $I_{2,n}$
Hence for the sum we need $$\lim_{n\to\infty} \frac 37 I_{1,n}-\frac 27 I_{2,n}$$
Now note that using the substitution $x^3=t$ in $I_{1,n}$ we get $$I_{1,n}=\frac 13 \int_0^1 \left(\frac {1-t^{n-\frac 23}}{1-t} -\frac {1-t^{-\frac 23}}{1-t}\right)dt$$
Now using the Integral relation if Digamma function I.e. $$\psi(z+1)+\gamma=\int_0^1 \frac {1-x^z}{1-x}dx$$
We get $$I_{1,n}=\frac 13\left(\psi\left(n+\frac 13\right)-\psi\left(\frac 13\right)\right)$$
Similarly using the substitution $x^2=u$ and the same integral relation in $I_{2,n}$ we get $$I_{2,n}= \frac 12\left(\psi\left(n+\frac 32\right)-\left[\psi\left(\frac 12\right)+\frac {1}{1/2}\right]\right)$$
But using that $$\psi(z+1)=\psi(z)+\frac 1z$$ we have $$I_{2,n}= \frac 12\left(\psi\left(n+\frac 32\right)-\psi\left(\frac 32\right)\right)$$
Now taking the limit we see that $$\lim_{n\to\infty} I_n=\frac 17\left[\lim_{n\to\infty} \left(\psi\left(n-\frac 23\right)-\psi\left(n+\frac 32\right)\right)\right]+\frac {\psi\left(\frac 32\right)-\psi\left(\frac 13\right)}{7}$$
Now for large $n$; $\psi(n)\sim \ln n$ . Using this the limit inside square brackets turns $0$
Thus the sum is equal to $$\frac {\psi\left(\frac 32\right)-\psi\left(\frac 13\right)}{7}$$
Which can be simplified using known values of Digamma function.
Edit: You can use from Wikipedia that $$\psi\left(\frac 13\right)=-\frac {\pi}{2\sqrt 3}-\frac {3\ln 3}{2}-\gamma$$
And $$\psi\left(\frac 32\right)=2-2\ln 2-\gamma$$
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unknown polynomial divided by $x^2(x-1)$, find the remainder. I took an exam today and there's a problem stuck in my head; I still can't figure out yet.
Here's the question (just the concept as I can't remember precisely).
An unknown polynomial divided by $(x-1)^2$ leaves the remainder of $x + 3$ (not sure about the number) and when this polynomial is divided by $x^2 $, it leaves $2x + 4$ (again, not sure about the number). From the given conditions, if this polynomial is divided by $(x-1)x^2$, what would be the remainder?
The solution as far as I figured out is this:
first, from the division of $(x-1)^2$, I got that $f(1) = 3$
in the same way from division of $x^2$, I got $f(0) = 4.$
I can write the polynomial as follows:
$f(x) = (x-1)(x)(x) g(x) + ax^2 +bx +c$
$ax^2 + bx + c$ is the remainder. And to find $a,b,c$, I can use the conditions above, so I got $c = 4$ by substituting $x = 0,$ and I got $a+b+4 = 3$ by substituting $x = 1.$
This leaves $a + b = -1,$ and I can't figure out how to continue; please help.
Edit : I made a mistake $f(1)$ should be equal to $4$ and $a+b+c = 4$
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$f(x)\equiv{x+3}\pmod{(x-1)^2}\wedge f(x)\equiv{2x+4}\pmod{x^2} \overset{CRT}{\implies} f(x)\equiv{3x^3 - 5x^2 + 2x + 4}\pmod{(x-1)^2x^2}$
? chinese(Mod(x+3,(x-1)^2),Mod(2*x+4,x^2))
%1 = Mod(3*x^3 - 5*x^2 + 2*x + 4, x^4 - 2*x^3 + x^2)
Then $f(x)\equiv{-2x^2 + 2x + 4}\pmod{(x-1)x^2}$
? Mod(3*x^3 - 5*x^2 + 2*x + 4,(x-1)*x^2)
%2 = Mod(-2*x^2 + 2*x + 4, x^3 - x^2)
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|
Fresnel integral $\int\limits_0^\infty\sin(x^2) dx$ calculation I'm trying to calculate the improper Fresnel integral $\int\limits_0^\infty\sin(x^2)dx$ calculation.
It uses several substitutions. There's one substitution that is not clear for me.
I could not understand how to get the right side from the left one. What subtitution is done here?
$$\int\limits_0^\infty\frac{v^2}{1+v^4} dv = \frac{1}{2}\int\limits_0^\infty\frac{1+u^2}{1+u^4} du.$$
Fresnel integral calculation:
In the beginning put $x^2=t$ and then: $$\int\limits_0^\infty\sin(x^2) dx = \frac{1}{2}\int\limits_0^\infty\frac{\sin t}{\sqrt{t}}dt$$
Then changing variable in Euler-Poisson integral we have: $$\frac{2}{\sqrt\pi}\int_0^\infty e^{-tu^2}du =\frac{1}{\sqrt{t} }$$
The next step is to put this integral instead of $\frac{1}{\sqrt{t}}$.
$$\int\limits_0^\infty\sin(x^2)dx = \frac{1}{\sqrt\pi}\int\limits_0^\infty\sin(t)\int_0^\infty\ e^{-tu^2}dudt = \frac{1}{\sqrt\pi}\int\limits_0^\infty\int\limits_0^\infty \sin (t) e^{-tu^2}dtdu$$
And the inner integral $\int\limits_0^\infty \sin (t) e^{-tu^2}dt$ is equal to $\frac{1}{1+u^4}$.
The next calculation: $$\int\limits_0^\infty \frac{du}{1+u^4} = \int\limits_0^\infty \frac{v^2dv}{1+v^4} = \frac{1}{2}\int\limits_0^\infty\frac{1+u^2}{1+u^4} du = \frac{1}{2} \int\limits_0^\infty\frac{d(u-\frac{1}{u})}{u^2+\frac{1}{u^2}} $$
$$= \frac{1}{2} \int\limits_{-\infty}^{\infty}\frac{ds}{2+s^2}=\frac{1}{\sqrt2}\arctan\frac{s}{\sqrt2} \Big|_{-\infty}^\infty = \frac{\pi}{2\sqrt2} $$
In this calculation the Dirichle's test is needed to check the integral $\int_0^\infty\frac{\sin t}{\sqrt{t}}dt$ convergence. It's needed also to substantiate the reversing the order of integration ($dudt = dtdu$). All these integrals exist in a Lebesgue sense, and Tonelli theorem justifies reversing the order of integration.
The final result is $$\frac{1}{\sqrt\pi}\frac{\pi}{2\sqrt2}=\frac{1}{2}\sqrt\frac{\pi}{2}$$
|
Well, if one puts $v=\frac{1}{u}$ then:
$$I=\int_0^\infty\frac{v^2}{1+v^4} dv =\int_0^\infty\frac{1}{1+u^4} du$$
So summing up the two integrals from above gives:
$$2I=\int_0^\infty\frac{1+u^2}{1+u^4} du$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3273031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Given three non-negative numbers $a, b, c$ so that $a+ b+ c= 3,\,a^{2}+ b^{2}+ c^{2}= 5$. Prove $a^{3}b+ b^{3}c+ c^{3}a\leqq 8$ . Problem. Given three non-negative numbers $a, b, c$ so that $a+ b+ c= 3,\,a^{2}+ b^{2}+ c^{2}= 5$. Prove:
$$a^{3}b+ b^{3}c+ c^{3}a\leqq 8$$
My solution in M&Y : (and I'm looking forward to seeing a nicer one(s), thanks for your interests !)
Because of the invariant of this circular permutation in $a, b, c$, suppose that $a\equiv \max\{\!a,\,b,\,c\!\}> 1$ :
$$\therefore\,ab+ bc+ ca= \frac{(a+ b+ c)^{2}- (a^{2}+ b^{2}+ c^{2})}{2}= \frac{3^{2}- 5}{2}= 2$$
Again, we have $5- a^{2}= b^{2}+ c^{2}\leqq (b+ c)^{2}= (3- a)^{2}$, therefore $a\geqq 2$ or $a\leqq 1\,(\!impossible\,\,!\!)$ .
$$\begin{align}
\therefore\,b,\,c\leqq 1\,\therefore\,a^{3}b+ b^{3}c+ c^{3}a & \leqq a^{3}b+ bc+ ca= ab(a^{2}- 1)+ 2\\
& \leqq \frac{1}{4}(a^{2}+ 4\,b^{2})(a^{2}- 1)+ 2\\
& \leqq \frac{1}{4}(5+ 3\,b^{2})(4- b^{2})+ 2\\
& = 8- \frac{1}{4}(1- b^{2})(4- 3\,b^{2})\leqq 8
\end{align}$$
The equality condition $\{\!a= 2,\,b= 1,\,c= 0\!\}\bigcup\{\!a= 1,\,b= 0,\,c= 2\!\}\bigcup\{\!a= 0,\,b= 2,\,c= 1\!\}$/q.e.d
|
Let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.
Thus, since $x+y+z=3$ and $xy+xz+yz=2,$ we obtain:
$$x+z=3-y,$$
$$xz=2-y(x+z)=2-y(3-y)=y^2-3y+2,$$
which gives $0\leq y\leq1$ because $y\geq2$ is impossible.
Thus, by Rearrangement we obtain:
$$a^3b+b^3c+c^3a=a^2\cdot ab+b^2\cdot bc+c^2\cdot ca\leq x^2\cdot xy+y^2\cdot xz+z^2\cdot yz=$$
$$=y(x^3+z^3+xyz)=y\left((3-y)^3-3(y^2-3y+2)(3-y)+y(y^2-3y+2)\right)=$$
$$=y(3y^3-12y^2+8y+y).$$
Id est, it's enough to prove that
$$3y^4-12y^3+8y^2+9y-8\leq0$$ or
$$3y^4-3y^3-9y^3+9y^2-y^2+y+8y-8\leq0$$ or
$$(y-1)(3y^3-9y^2-y+8)\leq0$$ or
$$(1-y)(8(1-y)+y(3y^2-9y+7))\geq0,$$ which is true because $9^2-4\cdot3\cdot7<0.$
Done!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $ \lim_{n\to\infty}[\frac{1}{\sqrt n}+\frac{1}{\sqrt {n+1}}+\frac{1}{\sqrt {n+2}}.......\frac{1}{\sqrt {2n}}] = \infty$ Show that $$ \lim_{n\to\infty}[\frac{1}{\sqrt n}+\frac{1}{\sqrt {n+1}}+\frac{1}{\sqrt {n+2}}.......\frac{1}{\sqrt {2n}}] = \infty$$
LHS : $ \lim_{n\to\infty}\frac{1}{n}[\frac{n}{\sqrt n}+\frac{n}{\sqrt {n+1}}+\frac{n}{\sqrt {n+2}}.......\frac{n}{\sqrt {2n}}] = \infty$
Let $a_n=\frac{n}{\sqrt{2n}}$ then $a_n = \frac{1}{\sqrt 2}\sqrt{n}$
hence $\lim_{n\to\infty}a_n = \infty$
I continue the problem therefore using Cauchy's first theorem on limits.
But the way I took $a_n$ is it correct?
Does it not make $a_1 = \frac{1}{\sqrt{2}}$
If my way of taking $a_n$ is wrong, please suggest me the right way. Thank you
|
Here is an
elementary proof that
$\dfrac{\sqrt{2}}{3n}
\lt \dfrac{s_n}{\sqrt{n}}-2(\sqrt{2}-1)
\lt \dfrac{2\sqrt{2}}{n}
$.
$\begin{array}\\
\sqrt{m+1}-\sqrt{m}
&=(\sqrt{m+1}-\sqrt{m})\dfrac{\sqrt{m+1}+\sqrt{m}}{\sqrt{m+1}+\sqrt{m}}\\
&=\dfrac{1}{\sqrt{m+1}+\sqrt{m}}\\
&>\dfrac{1}{2\sqrt{m+1}}\\
\text{and}\\
\sqrt{m+1}-\sqrt{m}
&<\dfrac{1}{2\sqrt{m}}\\
\end{array}
$
so
$2(\sqrt{m+1}-\sqrt{m})
<\dfrac{1}{\sqrt{m}}
< 2(\sqrt{m}-\sqrt{m-1})
$.
Therefore
$\begin{array}\\
s_n
&=\sum_{k=0}^n \dfrac1{\sqrt{n+k}}\\
&\gt\sum_{k=0}^n 2(\sqrt{n+k+1}-\sqrt{n+k})\\
&=2(\sqrt{2n+1}-\sqrt{n})\\
&=2\sqrt{n}(\sqrt{2+1/n}-1)\\
&=2\sqrt{n}(\sqrt{2}\sqrt{1+1/(2n)}-1)\\
&\gt2\sqrt{n}(\sqrt{2}(1+1/(6n))-1)
\quad\text{since }\sqrt{1+x} > 1+x/3
\text{ for } 0 < x < 3\\
&\gt2\sqrt{n}(\sqrt{2}-1)+\dfrac{\sqrt{2}}{3\sqrt{n}}\\
\text{and}\\
s_n
&=\sum_{k=0}^n \dfrac1{\sqrt{n+k}}\\
&\lt\sum_{k=0}^n 2(\sqrt{n+k}-\sqrt{n+k-1})\\
&=2(\sqrt{2n}-\sqrt{n-1})\\
&=2\sqrt{n}(\sqrt{2}-\sqrt{1-1/n})\\
&\lt2\sqrt{n}(\sqrt{2}-(1-1/n))
\quad\text{since }\sqrt{1-x} > 1-x
\text{ for } 0 < x < 1\\
&=2\sqrt{n}(\sqrt{2}-1)+\dfrac{2\sqrt{2}}{\sqrt{n}}\\
\end{array}
$
Therefore
$\dfrac{\sqrt{2}}{3n}
\lt \dfrac{s_n}{\sqrt{n}}-2(\sqrt{2}-1)
\lt \dfrac{2\sqrt{2}}{n}
$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
AM-GM inequality in $3$ variables I am trying to prove AM-GM inequality in $3$ variables.
Prove that $a^3 + b^3 + c^3 \ge 3abc$ for all $a,b,c \in \mathbb{R}^+$.
Could you please verify if my attempt contains logical gaps/errors?
My attempt:
Lemma: $a^2 + b^2 \ge 2ab$ for all $a,b \in \mathbb{R}^+$.
It follows from our lemma that $a^3 + b^3 \ge 2 \sqrt{a^3b^3} = 2ab\sqrt{ab}$ and $c^3 +abc \ge 2\sqrt{c^4ab} = 2c^2 \sqrt{ab}$, and that $ab\sqrt{ab} + c^2 \sqrt{ab} \ge 2\sqrt{ab\sqrt{ab}c^2 \sqrt{ab}} = 2abc$.
As a result, $(a^3 + b^3) + (c^3 +abc) \ge 2ab\sqrt{ab} + 2c^2 \sqrt{ab} \ge 4abc$ and so $a^3 +b^3 +c^3 \ge 3abc$. This completes the proof.
|
Hint: Better is to use that $$a^3+b^3+c^3-3abc= \left( a+b+c \right) \left( {a}^{2}-ab-ac+{b}^{2}-bc+{c}^{2}
\right)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3275555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
General solution of $\tan (2x)\tan (x)=1$ For the question, $\tan (2x) \tan x=1$, I divided it by $\tan x$, and got the solution as $\frac{(2n+1)\pi}{6}$.
$\tan 2x= \cot x= \tan\left(\frac{\pi}{2}-x\right)$. So, $2x=n\pi+ \frac{\pi}{2}-x$. So, $3x= \frac{(2n+1)\pi}{2}$
But the book solved using the formula of $\tan (2x)$, and got the solution as $\frac{(6n \pm 1)\pi}{6}$. I can see that my solution has odd multiples of $\pi/2$, which should be discarded, but I thought of it only after checking the solution. Also, in a way, it suggests that we can't solve these questions in different ways because that way we might get extra solutions. So, how to ensure which method to follow?
|
Your way is better, I think:
The domain gives $\cos2x\neq0$,$\cos{x}\neq0$ and $\sin{x}\neq0$ and by your idea we obtain:
$$\tan2x=\cot{x}$$ or
$$\frac{\sin2x\sin{x}-\cos2x\cos{x}}{\cos2x\sin{x}}=0$$ or
$$\cos3x=0$$ or
$$x=\frac{\pi}{6}+\frac{\pi}{3}k,$$ where $k\in\mathbb Z$.
Now, delete from these roots, roots of $\sin{x}=0$ (which is impossible), $\cos2x=0$ (which is impossible again) and $\cos{x}=0.$
We can get an answer by the following way.
From $\cos3x=0$ on the interval $(0,2\pi)$ we got the following numbers:
$$\left\{\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{7\pi}{6},\frac{3\pi}{2},\frac{11\pi}{6}\right\}.$$
$\frac{\pi}{2}$ and $\frac{3\pi}{2}$ we need to delete and since
$$\frac{11\pi}{6}=\pi+\frac{5\pi}{6}$$ and
$$\frac{7\pi}{6}=\pi+\frac{\pi}{6},$$ we got the following series:
$$x=\frac{\pi}{6}+\pi k$$ and
$$x=\frac{5\pi}{6}+\pi k$$ or $$\left\{\frac{(\pm1+6)\pi}{6}|k\in\mathbb Z\right\}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3276826",
"timestamp": "2023-03-29T00:00:00",
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|
Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$ Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$
I tried to induct on n:
For $n = 0$ we have $3+5 = 8$ and $8 \equiv 1 \pmod{7^{n+1}}$.
Suppose it is true for $n = k$:
$$3^{7^k}+5^{7^k}\equiv 1 \pmod{7^{k+1}}$$
so $3^{7^k}+5^{7^k}=7^{k+1}*q_1+1$
For $n = k+1$:
$$3^{7^{k+1}}+5^{7^{k+1}}\equiv r \pmod{7^{k+2}}$$
so $3^{7^{k+1}}+5^{7^{k+1}}=7^{k+2}*q_2+r$
Now if we subtract them we get that:
$$3^{7^{k+1}}+5^{7^{k+1}}-3^{7^k}-5^{7^k}=7^{k+1}(7q_2-q1)+r-1$$
From Euler's theorem we know that $a^{\phi(7^{k+1})}\equiv 1 \pmod{7^{k+1}}$
and $\phi(7^{k+1}) = 6\cdot7^{k}$ so $3^{6\cdot7^{k}}\equiv 1 \pmod{7^{k+1}}$ and $5^{6 \cdot 7^{k}}\equiv 1 \pmod{7^{k+1}}$.
So $3^{7^{k+1}}+5^{7^{k+1}}-3^{7^k}-5^{7^k}\equiv3^{7^k}\cdot3^{6\cdot{7^{k}}}+5^{7^k}\cdot5^{6\cdot{7^{k}}}-3^{7^k}-5^{7^k}\equiv 0 \pmod{7^{k+1}}$
Finally I get that $r\equiv1\pmod{7^{k+1}}$.
Here I got stuck. I don't know how to show that $r=1$.
|
Lemma: Suppose $n \geq 0$. Then ${7^n\choose k}7^k$ is divisible by $7^{n+1}$ for all $1 \leq k \leq 7^n$.
Proof: Legendre's formula says that the largest power of $7$ dividing $m!$ is $\sum\limits_{i=0}^\infty\left\lfloor\frac{m}{7^i}\right\rfloor$ (this is really a finite sum). Now use the formula ${m\choose k} = \frac{m!}{k!(m-k)!}$ to conclude that the largest power of $7$ dividing ${m\choose k}$ is $\sum\limits_{i=0}^\infty\left(\left\lfloor\frac{m}{7^i}\right\rfloor - \left\lfloor\frac{k}{7^i}\right\rfloor - \left\lfloor\frac{m - k}{7^i}\right\rfloor\right)$.
When $m = 7^n$, this is $\sum\limits_{i=0}^n\left(\left\lfloor\frac{7^n}{7^i}\right\rfloor - \left\lfloor\frac{k}{7^i}\right\rfloor - \left\lfloor\frac{7^n - k}{7^i}\right\rfloor\right) = \sum\limits_{i=0}^n\left(7^{n-i} - \left\lfloor\frac{k}{7^i}\right\rfloor - \left\lfloor7^{n-i}-\frac{k}{7^i}\right\rfloor\right) = \sum\limits_{i=0}^n\left(-\left\lfloor\frac{k}{7^i}\right\rfloor-\left\lfloor-\frac{k}{7^i}\right\rfloor\right)$
This equals $\sum\limits_{i=v_7(k)+1}^{n} 1 = n - v_7(k)$, where $v_7(k)$ is the largest power of $7$ dividing $k$. I used the result that $-\lfloor x\rfloor - \lfloor -x\rfloor = 0$ if $x$ is an integer and $1$ otherwise (in our case, $\frac{k}{7^i}$ is an integer iff $i \leq v_7(k)$).
So the largest power of $7$ dividing ${7^n\choose k}7^k$ is $n + k - v_7(k) \geq n + k - \log_7(k) \geq n + 1$ when $k \geq 1$.
Proof of main result
Now use the binomial theorem and the preceding lemma to find $5^{7^n} \equiv (7-2)^{7^n} \equiv \sum\limits_{k=0}^{7^n}{7^n\choose k}7^k(-2)^{7^n-k} \equiv (-2)^{7^n} \equiv -2^{7^n} \pmod{7^{n+1}}$.
Similarly, $3^{7^n} \equiv (7 - 4)^{7^n} \equiv \sum\limits_{k=0}^{7^n}{7^n\choose k}7^k(-4)^{7^n-k} \equiv (-4)^{7^n} = -2^{2\cdot 7^n} \pmod {7^{n+1}}$.
Then $3^{7^n} + 5^{7^n} \equiv -2^{2\cdot 7^n} - 2^{7^n}\pmod{7^{n+1}}$.
Now let $x = 2^{7^n}$. Since $2^3 = 8 = 7 + 1$, we have $x^3 \equiv (7+1)^{7^n} \equiv 1 \pmod{7^{n+1}}$, by another use of the binomial theorem and the lemma.
From here it follows that $(x-1)(x^2+x+1) \equiv 0 \pmod{7^{n+1}}$.
Note that $3^{7^n} + 5^{7^n} \equiv -x^2-x \pmod {7^{n+1}}$.
So if $x \not\equiv 1 \pmod 7$ (we do not use $7^{n+1}$ here, this needs to be prime) then it follows that $x^2 + x + 1 \equiv 0 \pmod{7^{n+1}}$, which means $3^{7^n} + 5^{7^n} \equiv 1 \pmod{7^{n+1}}$.
We check that $2^{7^n} \not\equiv 1 \pmod{7}$. But the order of $2$ modulo $7$ is $3$, and $7^n$ is not a multiple of $3$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplify $\frac{a+b-\sqrt{a^2-b^2}}{\sqrt{a+b}}$ I have to simplify the following expression:
$A =\frac{a+b-\sqrt{a^2-b^2}}{\sqrt{a+b}}$
Answer: $\sqrt{a+b}-\sqrt{a-b}$
I am trying to find the constraints of $a$ and $b$. I think that $a^2-b^2 \ge 0$ and $a+b \ge 0$. How can I simplify them? (the inequalities)
|
1) You have $\sqrt{a + b}$ in the denominator before simplification, so you can not have $\sqrt{a+ b} = 0$. So $a+b \ne 0$ and $a \ne -b$.
2) You have $\sqrt{a+b}$ before simplification so $a+b \ge 0$ and by 1) you have $a+b \ne 0$ so $a+ b > 0$ or $a > -b$.
3) You have $\sqrt{a^2 - b^2}$ before simplification so $a^2 -b^2 \ge 0$.
$a^2 - b^2 = (a+b)(a-b)$ and by 2) we have $a+b > 0$ so $a-b \ge 0$ so $a \ge b$.
We have $a \ge b$ and $a > -b$ so $a \ge |b|$.
If $b < 0$ then $-b > 0$ and $a > -b > 0$. If $b=0$ then $a > -b = 0$. If $b> 0$ then $a \ge |b| =b > 0$ so we have:
$a > 0$ and $a \ge |b|$ and if $b < 0$ then $a > |b|$.
...
Alternatively $a^2 -b^2 \ge 0$ so $a^2 \ge b^2$ so $|a| \ge |b|$.
but $a + b > 0$ so $a > -b$ and .... same results of above.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}dx$ using $\sinh x$
How to solve definite Integral
$$\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}dx=\frac{\ln(2+\sqrt{3})}{\sqrt{3}}$$
with hyperbolic function $\sinh(x)$?
Below is using $\tan x$ to do it. I thought it is same with this because $\sinh^2(x)+1=\cosh^2(x)$ is similar with $\tan^2(x)+1=\sec^2(x)$ ,but It isn't...
Substituting $$x=\tan t, dt=\sec^2{t} \ dx ,(0\leq t\leq\pi/2)$$
So, we get
$$
\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}dx =\int_{0}^{\frac{\pi}{2}}\frac{\sec{t}}{1+\tan{t}+\tan^2{t}}dt \\=\int_{0}^{\frac{\pi}{2}}\frac{\cos{t}}{1+\sin{t}\cos{t}}dt\\=I
$$
Note that $\cos{\left(\frac{\pi}{2}-t\right)}=\sin{t}$, hence substituting $t=\frac{\pi}{2}-\theta$, we have
$\cos{\left(\frac{\pi}{2}-t\right)}=\sin{t}$, hence substituting $t=\frac{\pi}{2}-\theta$, we have
$$
\int_{0}^{\frac{\pi}{2}}\frac{\cos{t}}{1+\sin{t}\cos{t}}dt=\int_{0}^{\frac{\pi}{2}}\frac{\sin{\theta}}{1+\sin{\theta}\cos{\theta}}d\theta=I
$$
Therefore, by substitution $ \sin{t}-\cos{t}=\xi, (\sin{t}+\cos{t})dt=d\xi $
$$
2I=\int_{0}^{\frac{\pi}{2}}\frac{\cos{t}+\sin{t}}{1+\sin{t}\cos{t}}dt\\=\int_{-1}^{1}\frac{1}{1+\frac{1-\xi^2}{2}}d\xi \\=\int_{-1}^{1}\frac{1}{3-\xi^2}d\xi \\=\frac{1}{\sqrt3}\int_{-1}^{1}\frac{1}{\sqrt3-\xi}+\frac{1}{\sqrt3+\xi}d\xi \\=\frac{1}{ \sqrt3 }\left(\ln(\sqrt3-\xi)-\ln(\sqrt3+\xi) \bigg\rvert_{-1}^{1}\right)\\=\frac{2\ln(2+\sqrt{3})}{\sqrt3}
$$
Because of the result, given integral is
$$I=\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}dx=\frac{\ln(2+\sqrt{3})}{\sqrt{3}}$$
And,what is actually different with this and using $\sinh(x)$?
|
Hint: Substituting $$\sqrt{x^2+1}=x+t$$ we get $$x=\frac{1-t^2}{2t}$$ so $$dx=-\frac{t^2+1}{2 t^2}dt$$
and $$\sqrt{x^2+1}=\frac{1+t^2}{2t}$$
|
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|
Prove Pascal's theorem by homogeneous coordinates I was trying to prove Pascal's theorem by using homogeneous coordinates with the following configurations (interactive graph at Desmos):
*
*A,B,C,D,E,F (homogeneous coordinates) are on a conic.
*G,H,I are the intersection points of pairs of lines (AB,DE), (CD,FA), (EF,BC) respectively.
Because G is the intersection of lines AB, DE, we can write (by using vector quadruple prodcut):
$$
\begin{array}{rcl}
G &= & (A \times B) \times (D \times E) \\
&= & (A \cdot (D \times E))B - (B \cdot (D \times E))A \\
&= & \begin{vmatrix} A & D & E \end{vmatrix} B
- \begin{vmatrix} B & D & E \end{vmatrix} A \\
&= & bB - aA
\end{array}
$$
where $b=\begin{vmatrix} A & D & E \end{vmatrix}$, $a=\begin{vmatrix} B & D & E \end{vmatrix}$
Note: $\begin{vmatrix} A & D & E \end{vmatrix}$ means the determinant determined by the three points A,D,E.
Similarly, we have:
$$
\begin{array}{rcl}
H &= & (C \times D) \times (F \times A) \\
&= & (C \cdot (F \times A))D - (D \cdot (F \times A))C \\
&= & \begin{vmatrix} C & F & A \end{vmatrix} D
- \begin{vmatrix} D & F & A \end{vmatrix} C \\
&= & dD - cC
\end{array}
$$
$$
\begin{array}{rcl}
I &= & (E \times F) \times (B \times C) \\
&= & (E \cdot (B \times C))F - (F \cdot (B \times C))E \\
&= & \begin{vmatrix} E & B & C \end{vmatrix} F
- \begin{vmatrix} F & B & C \end{vmatrix} E \\
&= & fF - eE
\end{array}
$$
where:
$$
d=\begin{vmatrix} C & F & A \end{vmatrix} \\
c=\begin{vmatrix} D & F & A \end{vmatrix} \\
f=\begin{vmatrix} E & B & C \end{vmatrix} \\
e=\begin{vmatrix} F & B & C \end{vmatrix}
$$
If we were to prove that G, H, I are collinear, it would be suffice to prove that $\begin{vmatrix} G & H & I \end{vmatrix}=0$, so I went for it.
$$
\require{cancel}
\begin{array}{ccl}
\begin{vmatrix} G \\ H \\ I \end{vmatrix}
& = & \begin{vmatrix} bB - aA \\ dD - cC \\ fF - eE \end{vmatrix} \\
& = &
bdf \begin{vmatrix} B \\ D \\ F \end{vmatrix}
-bde \begin{vmatrix} B \\ D \\ E \end{vmatrix}
-bcf \begin{vmatrix} B \\ C \\ F \end{vmatrix}
+bce \begin{vmatrix} B \\ C \\ E \end{vmatrix} \\
& &
-adf \begin{vmatrix} A \\ D \\ F \end{vmatrix}
+ade \begin{vmatrix} A \\ D \\ E \end{vmatrix}
+acf \begin{vmatrix} A \\ C \\ F \end{vmatrix}
-ace \begin{vmatrix} A \\ C \\ E \end{vmatrix} \\
& = &
bdf \begin{vmatrix} B \\ D \\ F \end{vmatrix}
\color{red}{\cancel{-bdea}}
\color{blue}{\cancel{-bcfe}} \color{blue}{\cancel{+bcef}}
\color{olive}{\cancel{-adfc}}
\color{red}{\cancel{+adeb}} \color{olive}{\cancel{+acfd}}
-ace \begin{vmatrix} A \\ C \\ E \end{vmatrix} \\
& = &
bdf \begin{vmatrix} B \\ D \\ F \end{vmatrix}
-ace \begin{vmatrix} A \\ C \\ E \end{vmatrix}
\end{array}
$$
At this point, if we wanted to prove the theorem of Pappus, it would be obvious:
Since A,C,E and B,D,F are collinear respectively, it is obvious that:
$$
\begin{vmatrix} A & C & E \end{vmatrix}=0 \\
\begin{vmatrix} B & D & F \end{vmatrix}=0
$$
and hence we can conclude immediately that $\begin{vmatrix} G & H & I \end{vmatrix}=0$
But for the case of Pascal's theorem, how can we prove that:
$$
bdf \begin{vmatrix} B \\ D \\ F \end{vmatrix}
-ace \begin{vmatrix} A \\ C \\ E \end{vmatrix} = 0
$$
i.e.
$$
\begin{vmatrix} A \\ D \\ E \end{vmatrix}
\begin{vmatrix} C \\ F \\ A \end{vmatrix}
\begin{vmatrix} E \\ B \\ C \end{vmatrix}
\begin{vmatrix} B \\ D \\ F \end{vmatrix}
=
\begin{vmatrix} B \\ D \\ E \end{vmatrix}
\begin{vmatrix} D \\ F \\ A \end{vmatrix}
\begin{vmatrix} F \\ B \\ C \end{vmatrix}
\begin{vmatrix} A \\ C \\ E \end{vmatrix}
$$
where A,B,C,D,E,F are points on a conic?
|
Define $$\phi: X \longmapsto
\begin{vmatrix} X \\ D\\E\end{vmatrix}
\begin{vmatrix} C \\ F\\X\end{vmatrix}
\begin{vmatrix} E \\ B \\C\end{vmatrix}
\begin{vmatrix} B \\ D \\F\end{vmatrix}-\begin{vmatrix} B \\ D \\E\end{vmatrix}
\begin{vmatrix} D \\ F \\X\end{vmatrix}
\begin{vmatrix} F \\ B \\C\end{vmatrix}
\begin{vmatrix} X \\ C \\E\end{vmatrix}.$$
We can assume $\phi$ is a nonzero quadratic polynomial (else we are done).
It is easy to check that $\phi$ vanishes at $X=C,D,E,F$, and not much harder to see that $\phi$ vanishes at $B$.
Assume that $B,C,D,E,F$ are pairwise distinct. Then the vanishing set of $\phi$ is a conic going through $B,C,D,E,F$: now, there is only one projective conic (up to a scalar factor) going through any five given pairwise distinct points. Thus $\phi$ must vanish on the conic containing $A,B,C,D,E,F$, hence $\phi(A)=0$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3281512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Other methods for a limit I know that I can compute the limit
$$
\lim_{x\to1}\frac{Nx^{N+1}-(N+1)x^N+1}{(x-1)^2}=\frac{N(N+1)}{2}
$$
using L'Hospital's rule (not one but two times) but I am looking for other ways. Are there any of them?
p.s.: the limit follows from a shortcut used in order to find the value of $$
\sum_{k=1}^Nkx^k\Big|_{x=1}.
$$
|
$$Nx^{N+1}-(N+1)x^N+1 = Nx^N(x-1)-(x^N-1)$$
$$ = Nx^N(x-1)-(x-1)(x^{N-1}+...x^2+x+1)$$
$$ = (x-1)\Big(\underbrace{Nx^N-(x^{N-1}+...x^2+x+1)}_{p(x)}\Big)$$
Now $$p(x) = (x^N-x^{N-1})+...+\color{red}{(x^N-x^2)}+\color{blue}{(x^N-x)}+\color{green}{(x^N-1)}$$
$$= x^{N-1}(x-1)+...+\color{red}{ x^2(x^{N-2}-1)}+\color{blue}{ x(x^{N-1}-1)}+\color{green}{(x^N-1)}$$
$$=(x-1)\Big(x^{N-1}+...+\color{red}{ x^2(x^{N-3}+...+x^2+x+1)}+\color{blue}{x(x^{N-2}+...+x^2+x+1)}+\color{green}{(x^{N-1}+...+x^2+x+1)}\Big)$$
So $$...=\lim_{x\to1}\Big(x^{N-1}+...+x^2(x^{N-3}+...+x^2+x+1)+x(x^{N-2}+...+x^2+x+1)+(x^{N-1}+...+x^2+x+1)\Big)$$
$$ 1+2+...+(N-2)+(N-1)+N=...$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3281658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Show That the Series $\frac{1}{n}+ \frac{1}{n+1} +\dots + \frac{1}{2n}$ Converges to a Number Between $\frac{1}{2}$ and $1$ $a_n = \frac{1}{n}+ \frac{1}{n+1} +\dots + \frac{1}{2n}$ converges to a number between $\frac{1}{2}$ and $1$.
I know that it converges to Ln(2) but we still haven't learned about integrals in my class, so calculating its limits kills the purpose of the exercise.
I've been trying for a while, I tried with this theorem that states that if $a_n \leq b_n$ for all n greater or equal to some N, then $\lim_{n\to\infty} a_n \leq \lim_{n\to\infty} b_n$
So my aproach was to find two sequences such that their limits are $\frac{1}{2}$ and $1$, I used these $\frac{n}{2n+1}$ and $\frac{n}{n+1}$, the limit of the first one is $\frac{1}{2}$ and the limit of the second one is $1$, and we have that
$$\frac{n}{2n+1} \leq \frac{1}{n}+ \frac{1}{n+1} +\dots + \frac{1}{2n} \leq \frac{n}{n+1}$$ for all $n \geq 6$, I've been trying to prove that with induction but I have been unable to do it. Is there any other way to prove that the limit is between $\frac{1}{2}$ and $1$?, or if someone could give me a hint with the induction proof, I would be very grateful.
|
Hint 1:
$$a_{n+1}-a_n=\left( \frac{1}{n+1}+ \frac{1}{n+1} +\dots + \frac{1}{2n+2} \right)-\left(\frac{1}{n}+ \frac{1}{n+1} +\dots + \frac{1}{2n}\right)\\ = \frac{1}{2n+1}+\frac{1}{2n+2}- \frac{1}{n}<\frac{1}{2n}+\frac{1}{2n}- \frac{1}{n}=0$$
Hint 2:
$$a_n =\frac{1}{n}+ \frac{1}{n+1} +\dots + \frac{1}{2n} \geq \frac{1}{2n}+ \frac{1}{2n} +\dots + \frac{1}{2n}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3283605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Study the sign of the quadratic matrix. Let $Q$ be the quadratic form associated with the matrix :
$$
\begin{pmatrix}
2 & 1 & 1 \\
1 & k & 0 \\
1 & 0 & 1 \\
\end{pmatrix}
$$
I reduced the matrix to the row echelon form and obtain :
$$
\begin{pmatrix}
1 & 0 & 1 \\
0 & 1 & -1 \\
0 & 0 & k-1 \\
\end{pmatrix}
$$
So for $k=1$, $\operatorname{rank}(A)=2$.
I inserted $k=1$ in the matrix and solved for eigenvalues which are : $\lambda = 0$ with multiplicity $2$ and
$\lambda = 1$ , so the signature is $(1,0)$ then the sign is positive semi definite.
Is that enough to study the sign if the quadratic matrix in this case ?
|
Careful, eigenvalues of row echelon form are in general not equal to the eigenvalues of the original matrix. In this case the eigenvalues are not easy to calculate so we can directly consider the form $Q$ itself.
We have
$$Q(x,y,z) = \left\langle \begin{pmatrix}
2 & 1 & 1 \\
1 & k & 0 \\
1 & 0 & 1 \\
\end{pmatrix}\begin{pmatrix}
x \\
y \\
z \\
\end{pmatrix},\begin{pmatrix}
x \\
y \\
z \\
\end{pmatrix}\right\rangle = (x+y)^2+(x+z)^2+(k-1)y^2$$
*
*Assume $k > 1$.
Let $(x,y,z) \ne 0$. If $y \ne 0$, clearly $Q(x,y,z) \ge (k-1)y^2 > 0$.
If $y = 0$ then $Q(x,y,z) = x^2+(x+z)^2$ so if $x \ne 0$ clearly $Q(x,y,z) \ge x^2 > 0$, and if $x = 0$ then $Q(x,y,z) = z^2 > 0$ because it must be $z \ne 0$.
Therefore $Q(x,y,z) > 0$ so we conclude that $Q$ is positive definite.
*Assume $k = 1$. Then $$Q(x,y,z) = (x+y)^2+(x+z)^2 \ge 0$$
but $Q(1,-1,-1) = 0$ so $Q$ is positive semidefinite.
*Assume $k < 1$. Then $Q(1,0,0) = 2 > 0$ and $Q(1,-1,-1) = k-1 < 0$ so $Q$ is indefinite.
|
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"timestamp": "2023-03-29T00:00:00",
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|
On the Sum of Some of the Factors of $2310$
Given 3 natural numbers a,b,c such that $a\times b\times c=2310$. Find the sum of $\sum a+b+c$
My attempt:
Now, $2310=2\times 3\times 5\times 7\times 11=a\times b\times c$
So now since $a,b,c$ are symmetric , their sum should be the same , and I should get , $\sum a+b+c= 3 \sum a$
But computing this is becoming tedious. As when $a$ is $1$ ,then $b,c$ will have $2^5 $combinations. When $a = 2$ or $3$ or $5$ or $7$ or $11$ ,then $b,c$ will have $2^4$ combination. So, the sum. Will become $$3 \sum a = 3 \{2^5 + (2+3+5+7+11) 2 ^4 + (2×3 +2×5... +3×5 +....) 2^3 + (2×3×5 +... ) 2^2 ..... +( 2×3×5×7×11)2^0 \} $$
What next ?? , Because this sum really starts to become too big. I'm sure there must be a shorter way as it is an exam type problem supposed to be solved in a limited time. Or is there any better approach ?? Please help :)
|
Silly overkill method:
For any (positive) natural number $N$, set
$$d_N=\sum_{abc=N}(a+b+c)\text{.}$$
Consider the Dirichlet series
$$f(s)=\sum_{N}\frac{d_N}{N^s}\text{.}$$
Then
$$\begin{split}f(s)&=\sum_{a,b,c}\frac{a+b+c}{a^sb^sc^s}\\
&=3\zeta(s-1)\zeta(s)^2\end{split}$$
so that $f(s)$ admits the Euler product
$$\begin{align}
\frac{f(s)}{3}&=\prod_{p\text{ prime}}f_p(p^{-s}) \\
f_p(T)&=\frac{1}{(1-pT)(1-T)^2}=1+(p+2)T+\ldots
\end{align}$$
Now suppose that $N$ has no repeated prime factors. Then reading off the coefficient of $N^{-s}$ gives
$$d_N=3\prod_{p|N}(p+2)\text{.}$$
|
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"url": "https://math.stackexchange.com/questions/3291122",
"timestamp": "2023-03-29T00:00:00",
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|
Rational points on $2x^2+2y^2=1$ and integral solutions to $2X^2+2Y^2=Z^2$
(a) Find a solution to the diophantine equation $2X^2+2Y^2=Z^2$; hence find a solution for rational numbers of the form $2x^2+2y^2=1$.
We have
$2X^2+2Y^2 \equiv Z^2\pmod{2} \implies (X,Y,Z) =c(1,1,2), ~c \in\mathbb{Z}$ is a solution. And $(1/2, 1/2, 1)$ is a possible solution to the rational equation.
(b) Find all rational solutions of the equation $2x^2+2y^2=1$; hence find all integer solutions to $2X^2+2Y^2=Z^2$
I'm not sure about this.
I know we can solve $y = m(x-1/2)+\frac{1}{2}$ and $2x^2+2y^2=1$ to get
$\displaystyle x= \frac{m^2-2m-1}{2(m^2+1)}, y= \frac{1-2m-m^2}{2(m^2+1)}$
(1) I don't know if $m$ is rational. So I'm not sure if I've found any rational solutions.
(2) If I force $\displaystyle m=\frac{p}{q}$ be rational I get
$\displaystyle x = \frac{(p^2-2pq-q^2)}{ 2(p^2 + q^2)} $, $\displaystyle y = \frac{-(p^2 + 2 p q - q^2)}{ 2(p^2 + q^2)}$.
Then solution to the integer equation $2X^2+2Y^2=Z^2$ is:
$\left(x, y, z \right) = \left(p^2-2pq-q^2, -(p^2 + 2 p q - q^2), 2(p^2 + q^2) \right)$.
Does this sound right? Can I just assume that $m$ is rational? Why can one rational solution generate all rational solutions (if it does)? Could I have gotten the general solution for the integer version at the beginning with just modular arithmetic?
|
You have found one rational solution $(x,y)=(\frac12,\frac12)$ to
$$
2|x+iy|^2=1
$$
This you can now use to transform any other solution as in
$$
2=|1-i|^2\implies |(1-i)(x+iy)|^2=1,
$$
that is, $(x+y, y-x)$ is a rational point on the unit circle. Now these are all parametrized by Pythagorean triples $a+ib=(p+iq)^2$, $c=|p+iq|^2$. All together,
$$
x=\frac{p^2-q^2-2pq}{2(p^2+q^2)},~~~y=\frac{p^2-q^2+2pq}{2(p^2+q^2)}.
$$
All reflections and rotations can be obtained via permutation and sign-flips in the pair $(p,q)\in\Bbb Z^2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Logarithmic integral $\int_0^{1}\frac{\ln x\ln (1-x^{2})}{1+x^{2}}\mathrm dx$
Evaluate the integral in a closed-form :
$$I=\int_0^{1}\frac{\ln x\ln (1-x^{2})}{1+x^{2}}\mathrm dx$$
My attempt :
After put $x=\tan y$ we obtain:
$$I=\int_0^{\frac{π}{4}}\ln (\tan x)\ln (1-\tan^{2} x)dx$$
$$1-\tan^{2} x=(1+\tan x)(1-\tan x)$$
$$\ln (1+\tan x)=\ln(\sin x+\cos x)-\ln(\cos x)=\ln (\cos (\frac{π}{4}-x))-\ln (\cos x)$$
Also:
$$\ln (1-\tan x)=\ln(\cos x-\sin x)-\ln(\cos x)=\ln (\cos (\frac{π}{4}+x))-\ln (\cos x)$$
So:
$$\ln (\tan x)\ln (1-\tan^{2} x)$$
$$=(\ln (\sin x)-\ln(\cos x)(\ln(\cos (\frac{π}{4}-x))-\ln (\cos x))(\ln(\cos (\frac{π}{4}+x))-\ln(\cos x))$$
Now I have many integrals. How can I evaluate them? Let me know if anyone has other ideas.
|
Presented below is a self-contained evaluation. Let $J=\int_0^1 \frac{\ln^2(1+x)}{1+x^2}dx$
\begin{align}
&\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx
\overset{x\to\frac{1-x}{1+x}}= J
-2 \int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx-2G\ln2+\frac{\pi^3}{16}\\
&\int_0^\infty \frac{\ln^2(1+x)}{1+x^2}dx
\overset{(0,1)+\overset{x\to 1/x}{(1,\infty)}} = 2J -2 \int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx+\frac{\pi^3}{16}
\end{align}
Eliminate $J$ to get
$$\int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx
=\frac12\int_0^\infty \frac{\ln^2(1+x)}{1+x^2}dx-K-2G\ln2+\frac{\pi^3}{32}
\tag1$$
where $K=\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx =2\text{Im}\>\text{Li}_3\left(\frac{1+i}2\right)
$. Also
$$\int_0^1 \frac{\ln x\ln(1-x)}{1+x^2}dx
=\frac12 K-\frac12\int_0^1\frac{\ln\frac x{1-x}}{1+x^2}dx +\frac12\int_0^1\frac{\ln^2x}{1+x^2}dx\tag2
$$
Then, (1) + (2)
$$
\int_0^1 \frac{\ln x\ln(1-x^2)}{1+x^2}dx
=-2\text{Im}\>\text{Li}_3\left(\frac{1+i}2\right)-2G\ln2+\frac{\pi^3}{16} +\frac12 A \tag3$$
where
\begin{align}
A =& \int_0^1 \underset{t=1-x}{\frac{\ln^2(1-x)}{1+x^2}}dx- \int_0^1\underset{t=\frac x{1-x}}{\frac{\ln\frac x{1-x}}{1+x^2}}dx +\underset{t=1+x}{
\int_0^\infty \frac{\ln^2(1+x)}{1+x^2}}dx \\
=& \int_0^\infty {\frac{4t \ln^2 t}{4+t^4}}dt
\overset{t^2=2u}= \frac14 \int_0^\infty \frac{\ln^2(2u)}{1+u^2}du
= \frac{\pi}{8}\ln^22+\frac{\pi^3}{32}
\end{align}
Plug into (3) to obtain
$$
\int_0^1 \frac{\ln x\ln(1-x^2)}{1+x^2}dx =-2\text{Im}\>\text{Li}_3\left(\frac{1+i}2\right) -2G\ln2
+\frac{\pi}{16}\ln^22+\frac{5\pi^3}{64}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$ I'm trying to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$. I am getting x = 4 but the book gets x = 81. What am I doing wrong?
\begin{align*}
\log_{3}(x) & = \log_{\frac{1}{3}}(x) + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{\log_{3}(\frac{1}{3})} + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{-\log_{3}(3)} + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{-1} + 8\\
\log_{3}(x) & = -\log_{3}(x) + 8\\
2\log_{3}(x) & = 8\\
\log_{3}(x) & = 4\\
x & = 4
\end{align*}
What am I doing wrong?
|
The mistake:
It should be $$2\log_3x=8$$ or
$$\log_3x=4$$ or
$$\log_3x=\log_3{3^4},$$
which gives $x=81$.
Actually, in the first step you can use the following property.
$$\log_{a^{\beta}}x=\frac{1}{\beta}\log_ax,$$ where $a>0$, $a\neq1$, $x>0$ and $\beta\neq0$.
Since $\frac{1}{3}=3^{-1},$ we obtain $$\log_3x=-\log_3x+8$$ immediately.
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the remainder of polynomial $f(x) g(x)$ divided by $x^{2} + x - 6$? $F(x),g(x)$ be polynomials. $f(x)$ divided by $x^{2}-4$ has remainder $ax+a$. $g(x)$ divided by $x^{2}-9$ has remainder $ax+a-5$. It is known that the remainder
$$ Remainder \: f(x) \: divided \: by \: x+2 = Remainder \: g(x) \: divided \: by \: x-3 $$
and $f(-3)=g(2)=-2$.
What is the remainder of $f(x)g(x)$ divided by $x^{2}+x-6$?
Attempt:
By the first known information we get $f(2) = 3a, f(-2) = -a, g(3) = 4a-5, g(-3) = -2a-5$
Then using the 2nd known information we get
$$ -a = 4a - 5 \implies a = 1 $$
So we have $f(x)$ is of the form
$$ f(x) = (x^{2} - 4) F(x) + (x+1) $$
where $F(x)$ is a polynomial. Also
$$ g(x) = (x^{2}-9)G(x) + (x-4) $$
Now we have to find the form of
$$ f(x) g(x) = (x+3)(x-2) H(x) + R(x) $$
with $f(2)g(2) = -6 =R(2)$, and $ f(-3) g(-3) = 14 = R(-3)$
|
Since you're dividing by a quadratic polynomial, the remainder has to be a linear polynomial, because its degree must be strictly less than that of the divisor. So you can write your remainder as $R(x)=px+q$. Knowing that $R(2)=-6$ and $R(-3)=14$, you can set up and solve a system of two equations with two unknowns $p$ and $q$.
|
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|
Minimizing distance between an ellipse and a point Problem
An ellipse has the formula:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
What is the shortest distance from the ellipse to the point $P = (a,0)?$
Attempted solution:
I have previously solved the same problem with P = (1,0) and P = (2,0). For P = (1,0), the minimum where x = 1.8 and y = 1.6 and a distance of $\frac{4\sqrt{5}}{5}$.
For P = (2,0), the local minimum was larger than the distance between P and the ellipse at y = 0., which was equal to 1. So there is some threshold for $a$ for which the local minimum is not truly the smallest distance.
Basic strategy:
1) Solve the ellipse formula for y and put that into the distance formula for the distance between the ellipse and the point (Pythagorean theorem).
2) Take the derivative.
3) Set the derivative to 0 and solve for x.
4) Use the ellipse formula to solve for y.
5) Put in x and y in the distance formula and get the distance.
However, at some threshold the distance between the point and the ellipse where y = 0 is even shorter. So the shortest distance will have one of two cases, one between 0 and just below the threshold and one for the threshold value and above.
I figured that I could apply the same method I did for the previous cases of P = (1,0) and P = (2,0) and see if I can find some way to find this threshold.
Use the ellipse formula and solve for y:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1 \Rightarrow y = \sqrt{4\Big(1-\frac{x^2}{9}\Big)}$$
Write down distance formula
$$D = \sqrt{(x-a)^2 + y^2}$$
Put in y from the ellipse formula
$$D = \sqrt{(x-a)^2 + \Big(\sqrt{4\Big(1-\frac{x^2}{9}\Big)}\Big)^2}$$
Simplify:
$$D = \sqrt{\frac{5}{9} \cdot x^2 - 2ax +4}$$
Take the derivative:
$$D' = \frac{\frac{10}{9}x-2a}{2\sqrt{\frac{5}{9}x^2 - 2ax +4}}$$
Set the derivative to zero and solve for x:
$$\frac{10}{18}x -a = 0 \Rightarrow x = 1.8a$$
Use ellipse function and solve for y:
$$y = \sqrt{4\Big(1-\frac{(1.8a)^2}{9}\Big)} = a\sqrt{0.76}$$
Putting in x and y to the distance formula
$$D = \sqrt{(1.8a-a)^2 + (a\sqrt{0.76})^2} = \sqrt{0.64a^2 + 0.76a^2} = \sqrt{1.4a^2} = a\sqrt{1.4}$$
...but I feel like I am not getting anywhere. This is not the correct distance and the threshold is nowhere to be seen.
Expected answer is:
$$\sqrt{4-\frac{4}{5}a^2}$$
for $0 < a < \frac{5}{3}$ and
$$|a-3|$$
if $a \geq \frac{5}{3}$
|
The key is to remember $-3\leq x\leq 3.$
It is much easier, without even using calculus, to minimize $D^2$ than the minimize $D.$
We have that the distance $D$ satisfying:
$$\begin{align}D^2&=(x-a)^2+y^2\\
&=x^2-2ax+a^2+4\left(1-\frac{x^2}{9}\right)\\
&=\frac{5}{9}x^2-2ax+a^2+4\\
&=\frac{5}{9}\left(x-\frac{9}{5}a\right)^2-\frac{9}{5}a^2+a^2+4\tag{1}
\\
&=\frac{5}{9}\left(x-\frac{9}{5}a\right)^2+4-\frac{4a^2}{5}
\end{align}$$
the line (1) by completing the square.
We want the minimum value of $D^2$ for $x\in[-3,3].$
Now, the right side is is minimized when $x=\frac{9}{5}a,$ but $\frac{9}{5}a\in[-3,3]$ if and only if $|a|\leq \frac{5}{3}.$
If $|a|>\frac{5}{3}$ then minimum is when $x=\pm 3$, and $y=0$ so the minimum $D=|3-|a||.$
Otherwise, if $|a|\leq \frac{5}{3},$ then $x=\frac{9}{5}a$ gives the minimum $D^2=4-\frac{4a^2}{5}$ or $$D=2\sqrt{1-\frac{a^2}{5}}$$
Another approach
If $(x,y)$ is a local minimum-maximum distance point to $(a,0)$ then the tangent angle the curve must be perpendicular to $(x-a,y).$
But the tangent at $(x,y)$ is $(x',y')=(9y,-4x).$ So:
$$0=(x',y')\cdot (x-a,y)=9y(x-a)-4xy = y(5x-9a)$$
So either $y=0,$ then $x=\pm 3.$ Or $5x-9a=0.$ But that is only possible if $\frac{9a}{5}\in[-3,3].$
The maximum $D$ is given when $x=\pm 3$ with the opposite sign of $a.$
The rest is the same. If $\frac{9}{5}a\in[-3,3],$ you get $x=\frac{9}{5}a$ yields the minimum, otherwise $x=\operatorname{sgn}(a)\cdot 3$ yields the minimum.
More generally, if we use the curve $x^2+Cy^2=B^2$ for any $B,C$, $B>0,$ then the potential points nearest to $(a,0)$ when $y=0,x=\pm B$ and when $(C-1)x-Ca=0.$
(In your case, $B=2,C=\frac{9}{4}.)$
If $C=1,$ you have a circle, and if $a\neq 0,$ you get $(\pm B,0)$ as your potential points. If $C=1,a=0,$ then $(a,0)$ is the center of the circle and the distance is constant for all $(x,y)$ on the circle.
If $C=0$ then your "curve" is $x\pm B,$ and we get $(C-1)x-Ca=0$ when $x=0,$ which is not in the valid domain for $x.$
If $C\neq 0,1,$ you get the additional (optional) case $x=\frac{C}{C-1}a.$
If $0<C<1,$ the case $x=\frac{C}{C-1}a$ is the opposite sign of $a,$ so $(x,y)$ is further from $(a,0)$ than $(-x,y),$ so that cases in not a minimum. [This case is one of an ellipse oblong in the $y$ direction, so this value is a local maximum.
If $C<0,$ then we include the optional case when $|a|\geq B\frac{C-1}{C}.$
If $C>1$, you include the optional case when $|a|\leq B\frac{C-1}{C}.$
For example, if $B=1,C=-1,$ then the closest point to $(a,0)$ in $x^2-y^2=1$ has distance:
$$D=\begin{cases}||a|-1|&|a|\leq 2\\ \frac{1}{2}\sqrt{2a^2-4}&|a|>2\end{cases}$$
When $|a|>2,$ you have $x=\frac{a}{2},y=\pm\frac{1}{2}\sqrt{a^2-4}.$
|
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|
How to determine the coefficient of $x^{10}$ in the expansion $(1+x+x^2+x^3+.....+x^{10})^4$ I have a question
Find the coefficient of $x^{10}$ in the expansion $(1+x+x^2+x^3+.....+x^{10})^4$
There ARE questions like this on stack exchange already I know, but I'm not able to formulate a pattern or know how to apply that thing here... I've tried making combinations of $1$'s and $x^{10}$'s, $x$'s and $x^9$'s etc but I am unable to solve it. Please help.
PS. How to do it using combinations exclusively.
|
The standard geometric series tells us that
$$1 +x + x^2 + x^3 + \ldots +x^{10}= \frac{1-x^{11}}{1-x}$$
So taking this to the fourth power we get
$$\left( \frac{1-x^{11}}{1-x} \right)^4 = (1-x^{11})^4 (1-x)^{-4}\tag{1}$$
This would not seem to really help much, except that for negative powers of $n$ there
is a generalised binomial formula:
$$(1-x)^{-4} = \sum_{k=0}^\infty \binom{k+3}{k}x^k$$
The standard binomial formula gives us for the positive power $4$ that:
$$(1-x^{11})^{4} = \sum_{k=0}^{4} \binom{4}{k}(-1)^k x^{11k}$$
So $(1)$ becomes
$$\left(1 -\binom{4}{1}x^{11} + \binom{4}{2}x^{22} - \binom{4}{3}x^{33} + \binom{4}{4}x^{44} \right)\left(\binom{3}{0} +\binom{4}{1}x + \binom{5}{2}x^2 + \ldots\right)\tag{2}$$
And we still want the coefficient of $x^{10}$. But terms with $x^{11}$ or more we cannot use from the left hand side when multiplying out, so we only get the term where we use $1$ from the left hand sum and $\binom{13}{10}$ (the coefficient of $x^{10}$) from the right hand sum.
So the answer is $\binom{13}{10} = 286$. Thanks to the (generalised) binomial formula and geometric series.
|
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|
Find all pairs $(k, n)$ of positive integers such that $k! = (2^n − 1)(2^n − 2)(2^n − 4) · · · (2^n − 2^{n−1})$ Find all pairs $(k, n)$ of positive integers such that
$$k! = (2^n − 1)(2^n − 2)(2^n − 4) · · · (2^n − 2^{n−1})$$
I tried to solve this problem but only found one solution $(1,1)$. Please help me to solve this problem.
|
Find the exponent of the largest power of 2 that divides both sides. In RHS it is $0 + 1 + \ldots + (n-1) = \frac{(n-1)n}{2}$. In LHS it can be found with Legendre's formula, which gives $\sum_{i=1}^{\infty} \lfloor \frac{k}{2^i} \rfloor$. Since $\sum_{i=1}^{\infty} \lfloor \frac{k}{2^i} \rfloor < \sum_{i=1}^{\infty} \frac{k}{2^i} = k$ (the inequality is strict because at least one term in the bracket is not an integer), we have $k > \frac{(n-1)n}{2}$, or $k \ge \frac{(n-1)n}{2}+1$.
Since $RHS < 2^{n^2}$, and factorial grows faster than any exponential function, for large values of $n$, LHS will be larger than RHS. We need to find out an upper bound for $n$.
When $n \ge 6$, $$k! \ge (\frac{(n-1)n}{2}+1)! \ge 7! \cdot 8^{\frac{(n-1)n}{2}-6} > 2^{12} \cdot 2^{\frac{3}{2}n^2 - \frac{3}{2}n - 18} = 2^{n^2} \cdot 2^{\frac{1}{2}n^2 - \frac{3}{2}n - 6}$$
And $\frac{1}{2}n^2 - \frac{3}{2}n - 6 > 0$, so LHS > RHS. Therefore there are no solutions with $n \ge 6$. Manually checking the remaining cases gives the only solutions $(1, 1)$ and $(3, 2)$.
|
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|
Given $x-\sqrt {\dfrac {8}{x}}=9$, what is $x-\sqrt{8x}$ equal to?
Given $x-\sqrt {\dfrac {8}{x}}=9$, what is $x-\sqrt{8x}$ equal to?
My attempt:
We have
\begin{align}
x-\sqrt {\dfrac {8}{x}}=9
\implies -\sqrt {\dfrac {8}{x}}=9-x
\implies \dfrac {8}{x}=(9-x)^2
\end{align}
How can I proceed?
|
$x-\dfrac {\sqrt {8x}}{x}=9,\Rightarrow x^{2}-\sqrt {8x}=9x$
$\begin{aligned}x^{2}-9x=\sqrt {8x}\\ x^{2}-8x=\sqrt {8x}+x\\ \left( x-\sqrt {8x}\right) \left( x+\sqrt {8x}\right) =\sqrt {8x}+x\\ x -\sqrt {8x}=1\end{aligned}$
|
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|
Proving non-trivial integer solutions don't exist for $z^2=10x^2-5y^2$ Given $z^2=10x^2-5y^2$.
I am wanting to show this equation has integer solutions only $0=x=y=z$.
I attempted by considering modulo $2$, however this didn't give me a contradiction. Any help or hints would be appreciated.
Solution:
Consider modulo $5$. Then we have that $z^2 \equiv 0 \mod5$. This is impossible as:
$0^2=0 \mod5$
$1^2=1 \mod5$
$2^2=4 \mod5$
$3^2=4 \mod5$
$4^2=1 \mod5$
|
Outline of proof:
*
*If there is a non-trivial solution, there is a solution with $x$ odd.
*There are no solutions when $x$ odd.
We'll look entirely modulo $8.$
First note that if $x,y,z$ are solutions with $x=0$, then $y=0,z=0.$
So if there is any solution $(x,y,z)\neq (0,0,0),$ there must be one with $x>0.$ Then find the least $x>0$ such that there are integers $y,z$ such that $z^2=10x^2-5y^2.$
If $x$ is even, then $10x^2\equiv 0\pmod{8}.$ We then have $z^2+5y^2\equiv 0\pmod 8.$
But $$z^2\equiv 0,1,\text{ or } 4\pmod 8\\\text{and}\\5y^2\equiv 0,5,\text{ or } 4\pmod{8}.\tag{1}$$ So to get $z^2+5y^2\equiv 0\pmod 8,$ we must have $y,z$ both even. Then we get another solution $(x',y',z')=(x/2,y/2,z/2),$ contradicting that $(x,y,z)$ is a solution with minimal $x>0.$
So $x$ must be odd.
When $x$ is odd, we have $10x^2\equiv 2\pmod{8}.$
By (1) we get that $$z^2+5y^2\equiv 0,1,4,5,\text{ or }6\pmod{8}$$
So it is not possible for $z^2+5y^2\equiv 2\pmod{8}.$ But that means there is no $(x,y,z)\neq 0$ such that $z^2=10x^2-5y^2.$
More generally, if $a\equiv 1\pmod{8},$ $b\equiv 5\pmod 8$ and $c\equiv 2,3,7\pmod 8$ then there are no integer solutions $(x,y,z)$ with $x\neq 0$ to the equation:
$$az^2+by^2=cx^2.$$
|
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|
shortest distance of $x^2 + y^2 = 4$ and $3x + 4y = 12$ $x^2 + y^2 = 4$ and $3x + 4y = 12$
line from origin with gradient $\frac{4}{3}$ intersect circle at $(\frac{6}{5} , \frac{8}{5}) $.
intersect $3x + 4y = 12$ at $(\frac{36}{25}, \frac{36}{25})$.
so distance is $\sqrt{{(\frac{36}{25}- \frac{6}{5})}^2 + {(\frac{36}{25}- \frac{8}{5})}^2} = \frac{\sqrt{52}}{25}$
but my answer is wrong. Where am i wrong?
|
Your idea is correct.
You need to redo your calculation of intersects.
You have the point $(\frac {6}{5}, \frac {6}{5})$ on the line $y=\frac {4}{3} x$ which does not make sense.
|
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|
Variance of a Fair Coin Consider Vamshi decides to toss a fair coin repeatedly until he gets a tail. He makes atmost $4$ tosses.The value of variance $T$ is ($T$ denoted number of tosses) ______
I tried like this Standard Deviation on marks of students
$x-----1 ----- 2 ------3-----4$
$P(x)----\frac{1}{2}-----\frac{1}{4}------\frac{1}{8}----\frac{1}{16}$
Mean is $\frac{1}{4}\left ( \frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}} +\frac{1}{2^{4}}\right )=\frac{15}{64}$
So, variance will be,
$\frac{1}{4}\left [ \left ( \frac{15}{64}-\frac{1}{2} \right )^{2}+\left ( \frac{15}{64}-\frac{1}{4} \right )^{2}+\left ( \frac{15}{64}-\frac{1}{8} \right )^{2} +\left ( \frac{15}{64}-\frac{1}{16} \right )^{2}\right ]=$$\frac{460}{16384}$
But answer given like this
$E\left ( X^{2} \right )=1^{2}\times \frac{1}{2}+2^{2}\times \frac{1}{4}+3^{2}\times \frac{1}{8}+4^{2}\times \frac{1}{16}$
$E\left ( X \right )=1\times \frac{1}{2}+2\times \frac{1}{4}+3\times \frac{1}{8}+4\times \frac{1}{16}$
$V\left ( X \right )=E\left ( X^{2} \right )-\left ( E\left ( X \right ) \right )^{2}$$=\frac{252}{256}$
Why my approach is giving incorrect result??
|
$$P(X=4)=1-P(X\le3)=\frac18\ne\frac1{16}$$ It is clear that one of the probabilities must be wrong as you currently don't have $\sum_{\forall x} P(X=x)=1$.
|
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|
Finding maximum of a given function
Show that $f(x)=\sin x(1+\cos x)$ attains its maximum at $x= \pi/3$.
I differentiated the function $f$ and got $f'(x)=\cos x(1+2\cos x)$. After equating with $0$, I got $x=\pi/2$ and $x =\pi/3 + n\pi$ with $n\neq 0$. So I did not get $x= \pi/3$ even as an extreme value.
|
Probably your differentiation is not correct,
$$f(x)=\sin x(1+\cos x)$$
$$f'(x)=\sin x\frac{d}{dx}(1+\cos x)+(1+\cos x)\frac{d}{dx}\sin x$$
$$=\sin x(-\sin x)+(1+\cos x)\cos x$$
$$=-\sin^2 x+\cos x+\cos^2 x$$
$$=2\cos ^2x+\cos x-1$$
$$=(2\cos x-1)(\cos x+1)$$
$$\implies f''(x)=-4\sin x\cos x-\sin-1$$
for maximum value, setting $f'(x)=0$
$$(2\cos x-1)(\cos x+1)=0$$
$$2\cos x-1=0\ \ \ \text{Or} \ \ \ \cos x+1=0$$
$$\cos x=\frac12\ \ \ \text{Or} \ \ \ \cos x=-1$$
$$\cos x=\cos \frac{\pi}{3}\ \ \ \text{Or} \ \ \ \cos x=\cos \pi$$
$$x=2n\pi\pm\frac{\pi}{3}\ \ \ \text{Or} \ \ \ \ x=2n\pi\pm\pi$$
$$x=\ldots, -\frac{5\pi}{3}, -\frac{\pi}{3}, \frac{\pi}{3}, \frac{5\pi}{3}, \ldots\ \ \ \text{Or} \ \ \ \ x=\ldots, -3\pi,-\pi, \pi, 3\pi, \ldots$$
Now, setting $x=\pi/3$ in $f''(x)$, we get
$$f\left(\frac{\pi}{3}\right)=-4\sin\frac{\pi}{3}-\sin\frac{\pi}{3}-1$$
$$\implies f\left(\frac{\pi}{3}\right)=-2\sqrt3-\frac{\sqrt3}{2}-1<0$$
hence the function $f(x)=\sin x(1+\cos x)$ is maximum at $x=\frac{\pi}{3}$
|
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|
A lucky proof for the Basel problem. I'll modify this part since I want the proof to be here.
$$\sum_{n=1}^\infty \frac{1}{n^2}=\frac43\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=-\frac43\sum_{n=0}^\infty \int_0^1 x^{2n}\ln x dx=\frac43\int_0^1 \frac{\ln x}{x^2-1}dx$$
$$\int_0^1 \frac{\ln x}{x^2-1}dx\overset{x\rightarrow \frac{1}{x}}=\int_1^\infty \frac{\ln x}{x^2-1}dx\Rightarrow \sum_{n=1}^\infty \frac{1}{n^2}=\frac23 \int_0^\infty \frac{\ln x}{(x+1)(x-1)}dx$$
$$=\frac23 I(1,-1)=\frac23 \frac{\ln^2 (1)-\ln^2(-1)}{2(1-(-1))}=\frac23 \frac{\pi^2}{4}=\frac{\pi^2}{6}$$
Where we considered the following integral:
$$I(a,b)=\int_0^\infty \frac{\ln x}{(x+a)(x+b)}dx\overset{x\rightarrow \frac{ab}{x}}=\int_0^\infty \frac{\ln\left(\frac{ab}{x}\right)}{(x+a)(x+b)}dx$$
Summing up the two integrals from above gives:
$$2I(a,b)=\ln(ab)\int_0^\infty \frac{1}{(x+a)(x+b)}dx=\frac{\ln(ab)}{a-b}\ln\left(\frac{x+b}{x+a}\right)\bigg|_0^\infty $$
$$\Rightarrow I(a,b)=\frac{\ln(ab)}{2}\frac{\ln\left(\frac{a}{b}\right)}{a-b}=\frac{\ln^2 a-\ln^2 b}{2(a-b)}$$
From here we know that:
$$\int_0^\infty \frac{\ln x}{(x+a)(x-1)}dx=\frac{\ln^2 a+\pi^2}{2(a+1)} $$
Also by plugging $b=-1$ in $I(a,b)$ we get:
$$I(a,-1)=\int_0^\infty \frac{\ln x}{(x+a)(x-1)}dx=\frac{\ln^2a -\ln^2 (-1)}{2(a-(-1))}=\frac{\ln^2 a+\pi^2 }{2(a+1)}$$
We already know that this is true from the linked post, but let's ignore it, since the linked post uses the Basel problem to prove the result.
Can someone prove rigorously that we are allowed to plug in $b=-1$ in order to get the correct result?
|
Here is an 'elementary' proof of its validity. Note that
$$ \frac{c+1}{(x+c)(x-1)}= \frac{c-1}{(x+c)(x+1)} +\frac{2}{x^2-1} $$
Then
$$ \int_0^\infty \frac{\ln x}{(x+c)(x-1)}dx = \frac{c-1}{c+1}I_1 + \frac{2}{c+1}I_2 \tag{1}$$
where
$$ I_1 =\int_0^\infty \frac{\ln x }{(x+c)(x+1)}
\overset{t=c/x}{dx}=\int_0^\infty \frac{\ln c }{(t+c)(t+1)}dt- I_1 = \frac{\ln^2 c}{2(c-1)}\tag{2}$$
and
$I_2 =\int_0^\infty \frac{\ln x}{x^2-1}dx$. Define
$J(\alpha)=\int_0^\infty \frac{\ln (1-\alpha^2 + \alpha^2 x^2)}{2(x^2-1)}dx $
\begin{align}
J'(\alpha)=\int_0^\infty \frac{\alpha}{1-\alpha^2 + \alpha^2 x^2} dx = \frac{\pi/2}{\sqrt{1-\alpha^2}}
\end{align}
Then
$$ I_2 = J(1) = \int_0^1 J'(\alpha) d\alpha = \frac{\pi}{2}\int_0^1 \frac{1}{\sqrt{1-\alpha^2}} d\alpha = \frac{\pi^2}{4} \tag{3}$$
Substitute (2) and (3) into (1), we obtain the sought-after result
$$ \int_0^\infty \frac{\ln x}{(x+c)(x-1)}dx = \frac{\pi^2 + \ln^2 c}{2(c+1)}$$
|
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|
Order of elements in $\Bbb Z_{24}$ What elements of $\Bbb Z_{24}$ are order $2$? Order $3$? Order $4$? Order $6$?
Order $2: 12$
Order $3: 8,16$
Order $4: 6,18$
Order $6: 4$
I had listed element $20$ as order $6$, but erased it late at night. A week later, I can't for the life of me remember why I didn't include it.
Am I correct or incorrect (and why) that element $20$ is order $6$?
|
You can brute force calculate:
\begin{align*}
20 \cdot 1 \equiv 20 \mod 24 \\
20 \cdot 2 \equiv 16 \mod 24 \\
20 \cdot 3 \equiv 12 \mod 24 \\
20 \cdot 4 \equiv 8 \mod 24 \\
20 \cdot 5 \equiv 4 \mod 24 \\
20 \cdot 6 \equiv 0 \mod 24 \\
\end{align*}
So the first (positive) multiple of $20$ that produces the group identity $0$ is $6$. Therefore that is the order of $20$.
For bigger calculations in the future, you can instead remember this formula:
The order of any element $k \in \Bbb Z_n$ of the cyclic group with $n \in \Bbb N$ elements is
$$
\boxed{\text{ord}(k) = \frac{n}{\gcd(k, n)}}
$$ In this case, $n = 24$ and $k = 20$ and the greatest common divisor of $20$ and $24$ is $\gcd(20, 24) = 4$. So $\text{ord}(k) = \frac{24}{4} = 6$ as expected.
|
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|
what is the value of $9^{ \frac{a}{b} } + 16^{ \frac{b}{a} }$ if $3^{a} = 4^{b}$ I have tried to solve this expression but I'm stuck:
$$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} }$$
and since $3^{a} = 4^{b}$:
$$3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} } = 4^{ \frac{2b}{b} } + 3^{ \frac{2a}{a} } = 4^{ b } + 3^{ a }$$
Firstly is this right? Secondly how to complete?
|
I think the most straightforward approach is computing $a$ in terms of $b$.
$3^a=4^b$ so $a = \log_34^b=b\log_34$
$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 9^{ \log_34}+16^{\log_43}=3^{ 2\log_34}+4^{2\log_43}=16+9=25$
|
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|
Splitting field in math software. Can anyone help me to find a method for calculating the splitting field for a polynomial over a function field?
I think this feature is not currently supported..
|
You can find the splitting field with the Isprime and PrimaryDecomposition function
P<x,y,z> := PolynomialRing(Rationals(),3);
I := ideal<P| x^2+y^3+y, z^3+z*(x^3+y)+1>;
IsPrime(I); // true
R := quo<P|I>; x:= R.1; y := R.2; z := R.3; Q<w> := PolynomialRing(R,1);
f := w^3+w*(x^3+y)+1; g:= f div (w-z); g
P<x,y,z,w> := PolynomialRing(Rationals(),4);
I := ideal<P|x^2+y^3+y, z^3+z*(x^3+y)+1,w^2 + z*w + y^9*z + 3*y^7*z + 3*y^5*z + y^3*z + y^2*z + 2*y*z^3 + 2*y + z^5 + 2*z^2>;
IsPrime(I); // true
R := quo<P|I>; x:= R.1; y := R.2; z := R.3; w := R.4; Q<u> :=
PolynomialRing(R,1);
f := u^2 + z*u + y^9*z + 3*y^7*z + 3*y^5*z + y^3*z + y^2*z + 2*y*z^3 + 2*y + z^5 +
2*z^2; f div (u-w) // = u + z + w
thus $$T^3+T(x^3+y)+1 = (T-z)(T-w)(T-(u+z+w)) \\\in \Bbb{Q}[x,y,z,w]/J [T]$$
where $$J=(x^2+y^3+y , \qquad \qquad \qquad \qquad \qquad \qquad \\ \qquad \qquad z^3+z(x^3+y)+1, \qquad \qquad \qquad\\ \qquad \qquad w^2 + z w + y^9 z + 3 y^7 z + 3 y^5 z + y^3 z + y^2 z + 2 y z^3 + 2 y + z^5 + 2 z^2) $$
|
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|
Solutions of bivariate cubic Diophantine equation The problem is what are the integer solutions to
$$7x^2 - 40xy + 7y^2 = (|(x - y)| + 2)^3$$
I only got $x - y = 4 (\mod 13)$. Also, how to solve it?
|
Let $(x,y)$ be an integral solution to the equation. Both sides of the equation are invariant under the transformation $(x,y)\ \mapsto\ (-x,-y)$, so without loss of generality $x\geq y$ and hence
\begin{eqnarray*}
7x^2 - 40xy + 7y^2&=& (|(x - y)| + 2)^3=(x-y+2)^3.
\end{eqnarray*}
Expanding the right hand side and bit of rearranging shows that this is equivalent to
$$(x-y)^3-14(x-y)^2+12(x-y)=-8-13x^2-13y^2,\tag{1}$$
where the right hand side is strictly negative. The left hand side is a cubic in $(x-y)$ which factors as
$$(x-y)^3-14(x-y)^2+12(x-y)=\big(x-y\big)\big((x-y)-7-\sqrt{37}\big)\big((x-y)-7+\sqrt{37}\big),$$
so the left hand side of $(1)$ is positive if $x-y\geq7+\sqrt{37}$. So we see that $0\leq x-y\leq 13$.
You have already shown that $x-y \equiv 4\pmod{13}$, so it follows that $y=x-4$. Plugging this back into $(1)$ leaves you with a cubic polynomial in $x$, which you can solve by standard methods, e.g. the rational root theorem.
|
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|
Tips on solving $a^2 (a + 2 b) (a + b^2 + b)^{\frac{3}{2}}>b^3 (a^2 + a + b)^{\frac{3}{2}}$ $$a^2 (a + 2 b) (a + b^2 + b)^{\frac{3}{2}}>b^3 (a^2 + a + b)^{\frac{3}{2}}$$
This is true with $a>b>0$, according to Wolfram Alpha, but I am not able to prove this.
I to simplify using the fact that $a^2(a+2b)=a^3+2ba^2>3b^3$, and then proving a stronger result:
$$3b^3(a + b^2 + b)^{\frac{3}{2}}>b^3 (a^2 + a + b)^{\frac{3}{2}}$$
$$\iff 3(a + b^2 + b)>(a^2 + a + b)$$
$$\iff 3a + 3b^2 + 3b>a^2 + a + b$$
But this is not true, e.g. $a=100, b=1$.
|
We prove the stronger result that comes from replacing $a+2b$ on the left by $a$
Note that for $0 \lt b \lt a, \frac 1b \gt \frac 1a$ and $\frac a{b^2} \gt \frac b{a^2}$ Then
$$1+\frac 1b+\frac a{b^2} \gt 1+\frac 1a+\frac b{a^2}\\
\sqrt{1+\frac 1b+\frac a{b^2}} \gt \sqrt{1+\frac 1a+\frac b{a^2}}\\
ab\sqrt{1+\frac 1b+\frac a{b^2}} \gt ab \sqrt{1+\frac 1a+\frac b{a^2}}\\
a\sqrt{a+b^2+b} \gt b\sqrt{a^2+a+b}\\
a^3(a+b^2+b)^{3/2} \gt b^3(a^2+a+b)^{3/2}\\
a^2(a+2b)(a+b^2+b)^{3/2}\gt b^3(a^2+a+b)^{3/2}$$
|
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|
How to find value of $k$? I have a question. How can I find value of $k$?
The given below
$$x + 4$$ is a factor of $$x^4+kx^3-4x^2$$
What is value of $k$?
Thanks for the solutions.
|
Two ways.
1) Hard but straightforward way:
$x^4 + kx^3 -4x^2 = (x+4)(x^3 + ax^2 + bx + c)$ so
$x^4 + kx^3 - 4x^2 = x^4 + (a+4)x^3 + (b+4a)x^2 + (c + 4b)x + 4c$
So $4c = 0; c+4b=0; b+4a=-4; a+4 =k$
So $c = 0$ and $b = 0$ and $a =-1$ and $-1 +4 = 3 = k$.
$x^4 + 3x^3 -4x^2 = (x+4)(x^3 - x^2)$
2) Very easy way
If $x + 4$ is a factor then $-4$ is a root.
So $(-4)^4 + k(-4)^3 -4(-4)^2 = 0$
$4^4 - 4^3k - 4^3 = 0$
Solve for $k$. $4^3k = 4^4 - 4^3$
So $k =\frac {4^4 -4^3}{4^3} =4-1 =3$.
3) Division.
$\frac {x^4 + kx^3 - 4x^2}{x+4} =$
$\frac {x^4 + 4x^3}{x+4} + \frac {(k-4)x^3 -4x^2}{x+2} =$
$x^3 + \frac {(k-4)x^3 + 4(k-4)x^2}{x+4} + \frac {-4(k-4)x^2 -4x^2}{x+2} =$
$x^3 + (k-4)x^2 + \frac {(12-4k)x^2}{x+4} =$
$x^3 + (k-4)x^2 + \frac {(12-4k)x^2+ (48- 16k)x}{x+4}+ \frac {(16k-48)x}{x+4} =$
$x^3 + (k-4)x^2 + (12-4k)x + \frac {(16k-48)x + (64k - 4*48)}{x+4}-\frac {64k-4*48}{x+4} =$
$x^3 + (k-4)x^2 + (12-4k)x + (16k-48)-\frac {64k-4*48}{x+4} =$
But $x+4$ has to divide evenly so $64k - 4*48 = 0$. Solve for $k$. $64k = 4*48= 4*16*3 = 64*3$ so $k =3$.
(and we get $(x^3 -x^2)(x+4) = x^4 + 3x^3 - 4x^2$ )
|
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|
Max of $(a-x^2)(b-y^2)(c-z^2)$ when $x+y+z=a+b+c=1$, $x,y,z,a,b,c \geq 0$ What's the maximum value of
$$(a-x^2)(b-y^2)(c-z^2)$$
given $x+y+z=a+b+c=1$, $x,y,z,a,b,c \geq 0$
The tricky part, as you could see in one of the attempted answer, is how to handle the case when two of the 3 factors are negative
|
Partial solution
It is enough to consider the case where the product is positive.
1) If there are two negatives and one positive amongst $a-x^2, b-y^2, c-z^2$: I'm working on this.
2) If all $a-x^2, b-y^2, c-z^2 \geq 0$, then by applying AM-GM, we have
$$(a-x^2)(b-y^2)(c-z^2) \leq \Big( \frac{a+b+c-(x^2+y^2+z^2)}{3} \Big)^3 = \Big( \frac{1 -(x^2+y^2+z^2)}{3} \Big)^3$$
And $x^2+y^2+z^2 \geq \frac{(x+y+z)^2}{3} = \frac{1}{3}$. Thus,
$$(a-x^2)(b-y^2)(c-z^2) \leq \frac{8}{729}$$
Equality happens when: $a-x^2 = b-y^2 = c-z^2 \geq 0$ and $x=y=z$, which means $x=y=z=1/3$ and $a=b=c=1/3$.
|
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|
Compute 100th derivative A friend suggested me a rather tricky problem, namely find the $100^{th}$ derivative of
$$
f(x)=\frac{x^2+1}{x^3-x}.
$$
I have computed the zeroth derivative
$$
\frac{x^2+1}{x^3-x}
$$
and the first derivative
$$
\frac{2x(x^3-x)-(3x^2-1)(x^2+1)}{(x^3-x)^2}=\frac{1-x^4-4x^2}{(x^3-x)^2}
$$
but I don't see any obvious structure.
|
Write your function as
$$
\frac{x^2 + 1}{x^3 - x} = \frac{1}{x-1} + \frac{1}{x+1} - \frac{1}{x}
$$
|
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|
Find $\lim_{n\rightarrow \infty}\int_0^\infty \frac{x^{n-2}\cos(n\pi x)}{1+x^n} dx$ I feel it goes to zero I separate the integral $\int_0^1 \frac{x^{n-2}\cos(n\pi x)}{1+x^n}$ the sequence goes to zero and it bounded by $0.5$ so by bounded convergence theorem the limit is zero. Now for the $\int_1^\infty \frac{x^{n-2}\cos(n\pi x)}{1+x^n} $ first $\frac{x^{n-2}}{1+x^n}\leq \frac{1}{x^2}$ which integrable. Another thought as $n$ goes large enough the $\int_1^\infty\cos(n\pi x)$ will be zero and the other part will be almost constant. Any help with full details.
Trying Partial derivative as @EuklidAlexandria suggested
$$\int_1^\infty \frac{x^{n-2} \cos(n\pi x) }{1+x^n} dx = 0 -\int_1^\infty \underbrace{\frac{x^{n-3}(n-2-2x^n)\sin(n\pi x)}{n\pi(1+x^n)^2}}_{f_n}$$
then the sequence
$$\left| f_n \right| \leq \left|\frac{x^{n-3}(n-2x^n)}{n\pi x^{2n}}\right|\leq \frac{n+2x^n}{n x^{n+3}}\leq \frac{1}{x^n} + \frac{1}{x^3}\leq \frac{2}{x^3}$$ since we are concern about limit we can consider $n\geq 3$
Now by Lebesgue Dominated convergence theorem we have $f_n \rightarrow 0$ p.w and $|f_n| \leq \frac{2}{x^3}$ which is integrable on $[1,\infty)$ Hence
$$\lim_{n\rightarrow \infty} \int_{1}^{\infty} \frac{x^{n-2}\cos(n\pi x)}{1+x^n} = \lim_{n\rightarrow \infty}- \int_1^\infty \frac{x^{n-3}(n-2-2x^n)\sin(n\pi x)}{n\pi(1+x^n)^2} =\int 0 = 0$$
Is that correct?
|
Hint: If $x \geq 1$ then
$$ \frac{x^{n-2}}{1+x^n} \leq \frac{1}{x^2}$$
and if $0\leq x \leq 1$ then
$$ \frac{x^{n-2}}{1+x^n} \leq x^{n-2} \leq 1 $$
|
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|
Proof by induction - stuck on simple question! Question:
(Part 1) Show that the inequality
$$
\frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \ldots + \frac{1}{(n-1)\cdot n} < \frac{1}{2}
$$
works for all natural numbers $n > 2$.
(Part 2) Deduce that for all natural numbers $n$, the following inequality holds:
$$
\frac{1}{1^2} + \frac{1}{2^2} + \ldots + \frac{1}{n^2} < \frac{7}{4}.
$$
This is a problem I found when rummaging around through old proof by induction questions and has been quite an issue for me. You see, I struggled with the first part, I proved it works for $n=3$ and then wrote out the result for $n=k$ and $n=k+1$ but I couldn't progress from there.
What I did:
*
*When $n = 3$:
$$
\frac{1}{2\cdot 3} = \frac{1}{6}
$$
and since $1/6 < 1/2$, the proposition is true for $n=3$.
*Assuming the truth of the proposition when $n = k$ i.e.
$$
\frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \ldots + \frac{1}{(k-1)\cdot k} < \frac{1}{2}
$$
and considering $n = k+1$, prove that
$$
\frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \ldots + \frac{1}{(k+1)\cdot k} < \frac{1}{2}
$$
From there I couldn't seem to find a way to link the assumption into the $n=k+1$ inequality in order to prove that that is also $< 1/2$.
Can't solve the first part in order to link to the second part in order to fulfill the "deduce" part.
Thanks in advance and sorry if it's just a stupid error I've made.
|
$\require{cancel}$No need for induction here. Just note that\begin{align}\frac1{2\times3}+\frac1{3\times3}+\cdots+\frac1{(n-1)n}&=\frac12-\cancel{\frac13}+\cancel{\frac13}+\cdots-\cancel{\frac1{n-1}}+\cancel{\frac1{n-1}}-\frac1n\\&=\frac12-\frac1n\\&<\frac12.\end{align}And then\begin{align}\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\cdots+\frac1{n^2}&<1+\frac14+\frac1{2\times3}+\cdots+\frac1{n(n-1)}\\&<1+\frac14+\frac12\\&=\frac74.\end{align}
|
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|
Separate equations of the pair of lines $x^2+2xy\sec\theta+y^2=0$
Find the separate equations of the lines $x^2+2xy\sec\theta+y^2=0$
Attempt 1
$$
x^2+2xy\sec\theta+y^2=0\\
\frac{x^2}{y^2}+2\frac{x}{y}\sec\theta+1=0\\
\frac{x}{y}=\frac{-2\sec\theta\pm\sqrt{4\sec^2\theta-4}}{2}=-\sec\theta\pm\tan\theta=\frac{-1\pm\sin\theta}{\cos\theta}\\
x\cos\theta=-y(1\pm\sin\theta)\\
x\cos\theta+y(1\pm\sin\theta)=0
$$
Attempt 2
$$
x^2+2xy\sec\theta+y^2=0\\
\frac{y^2}{x^2}+2\frac{y}{x}\sec\theta+1=0\\
\frac{y}{x}=\frac{-2\sec\theta\pm\sqrt{4\sec^2\theta-4}}{2}=-\sec\theta\pm\tan\theta=\frac{-1\pm\sin\theta}{\cos\theta}\\
y\cos\theta=-x(1\pm\sin\theta)\\
y\cos\theta+x(1\pm\sin\theta)=0
$$
Why do I seems to get different solutions for the lines in attempts 1 and 2 ?
|
\begin{align}
\dfrac{-1 \pm \sin \theta}{\cos \theta}
&= \dfrac{-1 \pm \sin \theta}{\cos \theta} \cdot
\dfrac{-1 \mp \sin \theta}{-1 \mp \sin \theta} \\
&= \dfrac{1-\sin^2 \theta}{\cos \theta (-1 \mp \sin \theta)} \\
&= \dfrac{\cos^2 \theta}{\cos \theta (-1 \mp \sin \theta)} \\
&= \dfrac{\cos \theta}{-1 \mp \sin \theta} \\
\end{align}
|
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|
Find $y$ in $\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$ I would like to solve this equation for $y$. Any tips? It seems like you cant really do it analytically?
$$\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$$
|
Set $y-3=z,$ then square both sides and simplify to get $$z^2-3z-8=-\sqrt{4+(z-3)^2}\sqrt{16+z^2}.$$ Squaring again and simplifying the right hand side gives $$(z^2-3z-8)^2=(z^2-6z+13)(16+z^2).$$ Now let $16+z^2=w,$ then expand and simplify to get $$z^2+16z+64=5w.$$ Substituting back for $w,$ simplifying and factoring gives the quadratic $$(z-2)^2=0.$$ Thus, $z=y-3=2,$ so that if there is a solution, then it has to be $y=5.$ Substituting this in the original equation assures us that this is indeed the only solution.
|
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|
The sequence $\left\{\frac{n^2}{9^n}\right\}_{n=1}^{\infty}$
*
*Show that the sequence $\left\{\frac{n^2}{9^n}\right\}_{n=1}^{\infty}$ is monotone decreasing and bounded below.
Let $a_n = \frac{n^2}{9^n}$. For $n\geq 1$ we have
\begin{equation*}
a_n-a_{n+1} = \frac{n^2}{9^n}-\frac{(n+1)^2}{9^{n+1}}.
\end{equation*}
To show this sequence is decreasing we want to show that this quantity is positive. Now we know that $n^2 > 0$, and $(n+1)^2 > 0$ for $n\geq 1$ so $a_n-a_{n+1} > 0$, i.e. $a_n > a_{n+1}$. Since it is decreasing it must be monotone decreasing as required.
A sequence is bounded below if there is a number $m$ such that $m\leq a_n$ for all $n$. Since all the terms in this sequence are all positive, the sequence is bounded below by $0$, i.e. $a_n\geq 0$ as required.
*Use the result in (i) to prove that $\lim_{n\to\infty} \frac{n^2}{9^n} = 0$.
Since this sequence is monotone decreasing and bounded below, it converges by the monotone convergence theorem. Lets re-write this a little. We can do this by writing
\begin{equation*}
\frac{n^2}{9^n} = \frac{n^2}{\left(e^{\ln{(9)}}\right)^n} = \frac{n^2}{e^{n\ln{(9)}}}
\end{equation*}
Now we have a limit of the form $\frac{\infty}{\infty}$ so we can apply L'Hopital's rule which gives
\begin{equation*}
\begin{split}
\lim_{n\to\infty} \frac{n^2}{9^n} &= \lim_{n\to\infty} \frac{n^2}{e^{n\ln{(9)}}} \\
&= \lim_{n\to\infty} \frac{2n}{\ln{(9)}e^{n\ln{(9)}}} \\
&= \lim_{n\to\infty} \frac{2}{\left(\ln{(9)}\right)^2e^{n\ln{(9)}}} \\
&= 0
\end{split}
\end{equation*}
as required.
Is this good? Thanks!
|
*
*First note that $a_n > 0$ for all $n.$ Therefore $$a_{n+1} \leq a_n \iff \frac{a_{n+1}}{a_n} \leq 1$$ and in our case, we have $$\frac{a_{n+1}}{a_n} = \frac{1}{9}\left(1 + \frac{1}{n}\right)^2 \leq \frac{1}{9}\left(1 + \frac{1}{1}\right)^2 = \frac{4}{9} < 1$$
*We know that the limit exists, so write $L = \displaystyle\lim_{n\to\infty} a_n$ and suppose $L > 0.$ Then there will be some $N$ such that $n \geq N$ implies $$a_n - L < L \\ \implies \\ a_n < 2L \\ \implies \\ a_{n+1} = a_n \cdot \left(\frac{a_{n+1}}{a_n}\right) < (2L) \cdot \left(\frac{4}{9}\right) < L$$
which is a contradiction since a monotone decreasing sequence cannot ever be less than its limit.
|
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|
Prove inequality for a sum to $2^{n+1}$ Problem:I realize I made a mistake on one of the last questions I posted.Here is the problem:
Prove for all $n \in \mathbb{N}$
$\sum\limits_{k=1}^{2^{n+1}} \frac{1}{k} > \frac{n}{2}$
Attempt:
I'm having trouble recognizing the pattern for the general term of the inequality, can someone help me?
trying cases
for $n=1$ then $\sum\limits_{k=1}^{2^{1+1}} \frac{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}$
for $n=2$ then $\sum\limits_{k=1}^{2^{2+1}}\frac
{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+ \frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}$
Can someone tell me if this is the correct pattern to use in the proof?Would this be a proof by induction or a direct proof?
Direct proof attempt
$\sum\limits_{k=1}^{2^{n+1}} \frac{1}{k}>\frac{n2^{n}}{2^{n+1}}=\frac{n}{2}$
|
$\int_k^{k+1} \frac 1 t dt <\frac 1 k$. Summing over $k$ we see that the given sum exceeds $\int_1^{2^{n+1}+1} \frac 1 t dt=\ln (2^{n+1}+1) >ln (2^{n+1})=(n+1)ln 2>nln 2 >\frac n 2$ since $e <4$.
|
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|
Verify the following limit using epsilon-delta definition: $ \lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^2}=0$ Show that $$ \lim\limits_{(x,y)\to(0,0)}\dfrac{x^2y^2}{x^2+y^2}=0$$
My try:
We know that, $$ x^2\leq x^2+y^2 \implies x^2y^2\leq (x^2+y^2)y^2 \implies x^2y^2\leq (x^2+y^2)^2$$
Then, $$\dfrac{x^2y^2}{x^2+y^2}\leq x^2+y^2 $$
So we chose $\delta=\sqrt{\epsilon}$
|
Or alternatively, by AM-GM we get
$${x^2+y^2}\geq 2|xy|$$ so
$$\frac{x^2y^2}{x^2+y^2}\le \frac{x^2y^2}{2|xy|}=\frac{1}{2}|xy|$$ and this tends to zero if $x,y$ tend to zero.
|
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|
In simplifying the formula that I've derived for finding the square root of a complex number to the standard formula. So by easy means, I derived
$\sqrt{a+ib} = \sqrt{\frac{a(a+1)+b^2}{2}}+i(\frac{b}{\sqrt{2}\sqrt{a(a+1)+b^2}})$
But then I checked for the actual formula it is this;
$\sqrt{a+ib} = \sqrt{\frac{\sqrt{a^2 + b^2}+a}{2}}± i(\sqrt{\frac{\sqrt{a^2 + b^2}-a}{2}})$
So how do you simplify that to this?
Btw here is the derivation;
Let $\sqrt{a+ib} = x+iy$
$a+ib=x^2 - y^2 + i2xy$
We know, $(x^2-y^2)^2 + (2xy)^2 = x^2 + y^2$
or, $x^2 + y^2 = a^2 + b^2$
Hence we get $x = \sqrt{\frac{a+a^2+b^2}{2}}$ and
$y=\frac{b}{\sqrt{2}\sqrt{a^2+a+b^2}}$
Therefore;
$\sqrt{a+ib} =
> \sqrt{\frac{a(a+1)+b^2}{2}}+i(\frac{b}{\sqrt{2}\sqrt{a(a+1)+b^2}})$
|
One of the approache is use Euler form if Z is a complex numbers such that $ Z\, = \,r.e^{ix}$ then
Square root of Z will be $Z_{1}= \, \sqrt{r} . e^{iy} $
Where angle y will be x/2 and (x + 2π)/2.
|
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|
Evaluating $\frac {\cos 81^{\circ}}{\sin3^{\circ}\sin57^{\circ}\sin63^{\circ}} $
Find $$\frac {\cos 81^{\circ}}{\sin3^{\circ}\sin57^{\circ}\sin63^{\circ}} $$
This expression seemed quite easy to solve, but now the $63$ and $57$ in the equation are posing me with difficulties. I tried multiplying by $\cos9$ but I was unable to solve it further.
Would someone please help me to solve this question?
Thanks in advance.
|
Let $\alpha = 60^\circ$ and $\theta = 3^\circ$. The numerator is
$$\begin{align}
\cos(81^\circ) = \sin(9^\circ) &= \sin(3\theta)\\
&= \sin\theta\cos(2\theta) + \cos\theta\sin(2\theta)\\
&= \sin\theta(\cos(2\theta) + 2\cos^2\theta)\\
&= \sin\theta(2\cos(2\theta) + 1)\end{align}$$
while the denominator equals to
$$\begin{align}
\sin 3^\circ\sin 57^\circ \sin 63^\circ
&= \sin\theta\sin(\alpha-\theta)\sin(\alpha+\theta)\\
&= \frac12\sin\theta(\cos(2\theta) - \cos(2\alpha))\\
&= \frac14\sin\theta(2\cos(2\theta) + 1)\end{align}$$
This means the expression at hand equals to $4$.
|
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|
If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$. If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$.
I did a little bit of manipulation and got $4(x+1)(x-2) + (y+1)^2=0$. I then got $x=2$ and $y=-1$ which means the maximum must be $4$, but the answer key says it's $16$. How come?
|
This can be solved as a tangency problem.
Given the non degenerated conic $C(x,y) = 0$ determine the value for $\lambda$ such that $a x + b y = \lambda$ is tangent to $C(x,y)=0$ so we proceed as follows:
first we substitute $y = \frac 1b(\lambda-a x)$ into $C(x,y) = 0$ giving the relationship
$$
C\left(x, \frac 1b(\lambda-a x)\right)=0
$$
second, we solve for $x$ giving the condition
$$
x^* = \frac{1}{169} \left(102\pm12 \sqrt{368-\lambda^2-7 \lambda}+5 \lambda\right)
$$
but at tangency $x^*$ is unique so follows the condition
$$
368-\lambda^2-7 \lambda = 0
$$
which solved gives
$$
\lambda^* = \{-23, 16\}
$$
hence the answer is $\lambda^* =16$
|
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|
Solve System of equation using elimination? \begin{align}
I:&& ~~ x+\frac12y &= 6
\\[.5em]
II:&& ~~ \frac32x + \frac{3}{2}y &= {17 \over 2}
\end{align}
when $x$ was multiplied by $(-3/2)$ in first equation the $x$ will be canceled and the resulting $y = -2/3$ and $x = 19/3$.
But when fractions were simplified first the resulting equation is
$$-8x-4y= -48$$
$$9x+4y = 51$$
here $y$ get canceled and results will be $x=3 , y=6$.
why these two attempts give two different results..I only need answer in elimination technique.
|
Multiplying the second equation by $\frac{2}{3}$ we get
$$x+\frac{1}{2}y=6,$$
and
$$x+y=\frac{17}{3}.$$
Now multiplying the first equation by $-1$ and adding to the second
$$\frac{1}{2}y=-6+\frac{17}{3}$$
Can you finish?
|
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|
Calculate in how many different ways can rotated square
We put a square $3 \text{ x } 3$ from $9$ square tiles in two types: or which can be freely rotated (we allow also symmetries). Calculate in how many different ways you can do that if we identify such systems that one goes to the other with some rotation or symmetry of big square.
My solution:
I am going to use polya counting theory https://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem
$$\begin{array}{|c|c|c|c||c|c|}
\hline
&G \in G & \text{How many}& \text{Id x }\\ \hline
a) &\text{id} & 1 & x_{1}^{9} \\ \hline
b) &\text{90 degree rotation} & 2 & x_1 \cdot x_4^{2} \\ \hline
c) &\text{180 degree rotation } & 1 & x_1 \cdot x_2^{4}\\ \hline
d) &\text{ diagonal symmetry} & 2 & x_{1}^{3} \cdot x_2^{3}\\ \hline
e) &\text{centres symmetry } & 2 & x_1^3 \cdot x_2^{3}\\ \hline
\end{array}$$
We must include our tiles:
a) Here freely, so we have $6$ types: $6^9$
b)
For each $4$-element cycle we choose whether the first will be:
The rest for him is clearly designated:
The middle one can't be adjusted to fit all $4$ rotations, so: $0$
c) Middle: and the rest of the cycles as in b): $2\cdot 6^4$
d) In the middle in $6$ ways, the rest as in b): $6^3 \cdot 6^3=6^6$
e) $6^6$ as in d)
Solution: $$\frac{1}{8} \cdot (2\cdot 6^6 + 2\cdot 6^6 +2\cdot 6^4+0+6^9)$$
Is my solution correct?
|
Let the small square with 1 shaded triangle be called A and the one with 2 shaded triangles be called B. We shall apply Burnside's Lemma to the group of symmetries of the large square acting on the $6^9$ patterns.
The identity fixes all $6^9$ patterns.
The rotations of 90 and 270 cannot fix the centre square and so fix no patterns.
The rotation of 180 fixes the centre square if and only if it is the square B. We can then have 2 choices for this square. The other 8 squares are in pairs where we have a free choice for one of each pair and therefore the total number of patterns is $2\cdot6^4$.
The two reflections in a diagonal cannot fix a corner square and so fix no patterns.
Finally consider a reflection in a horizontal line. (The result for the reflection in a vertical line will be similar.) The central row of squares must each consist of a square B or a square A in just two of its orientations. The remaining six squares are in pairs where we have a free choice for one of each pair and therefore the total number of patterns is $4^3.6^3$.
The solution is $\frac{1}{8} \cdot (6^9+2\cdot 6^4 + 2\cdot 4^3\cdot6^3)$.
|
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|
Locus of midpoint of the line cutting the other two lines Given two lines $L_1: y = 2x-2$ and $L_2: y = -2x-2$. The variable line $L$ passes through the point $R(0,1)$ and meets $L_1$ and $L_2$ at $A$ and $B$, respectively. The slope of variable line is $m$. Find the equation of the locus of the midpoint $M$ of $AB$.
So far I've got to the point that coordinates of $A$ and $B$ are:
$\left(x_A, y_A\right) = \left( \dfrac{3}{2-m}, \dfrac{2+2m}{2-m} \right) $
$\left(x_B, y_B\right) = \left( - \dfrac{3}{2+m}, \dfrac{2-2m}{2+m} \right) $
I can also calculate the coordinates of the midpoint, but don't know how to eliminate $m$. Just give me a hint how to proceed from here. I should get the equation $4x^2-y^2-y+2=0$ as final solution.
|
Let $(h,k)$ be the midpoint $M$ of $AB$.
\begin{align*}
h&=\frac{x_A+x_B}{2}=\frac{3m}{4-m^2}
\end{align*}
But $R(0,1)$ also lies on the line $L$, so $m=\frac{k-1}{h-0}=\frac{k-1}{h}$. Thus
$$h=\frac{3m}{4-m^2}=\frac{3\left(\frac{k-1}{h}\right)}{\left(4-\left(\frac{k-1}{h}\right)^2\right)}.$$
Simplify this and you get
$$\color{blue}{4h^2-k^2-k+2=0}.$$
|
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|
How to determine algebraically whether an equation has an infinite solutions or not? I was learning for the first time about partial fraction decomposition. Whoever explains it, emphasises that the fraction should be proper in order to be able to decompose the fraction. I was curious about knowing what happens if I try to decompose an improper fraction, So I tried to do one:
$\frac{x^2 - 4}{(x + 5)(x - 3)}$
I got the equation:
$\frac{(Ax + B)(x - 3) + (Cx + D)(x + 5)}{(x + 5)(x - 3)} = \frac{x^2 - 4}{(x + 5)(x - 3)}$. I have 4 unknowns: A, B, C and D.
$\therefore (Ax + B)(x - 3) + (Cx + D)(x + 5) = x^2 - 4$
After expanding and regrouping the coefficients:
$(A + C) x^2 + (-3A + B + 5C + D)x + (-3B + 5D) = x^2 - 4$
Here the coefficient of the term $x^2$ is 1 therefore:
$(A+C) = 1$
similarly:
$(-3A + B + 5C + D) = 0$
$(-3B + 5D) = -4$
I still have to get one more equation to be able to solve this system so I substituted 1 for x and I got this equation:
$-2A - 2B + 6C + 6D = -3$
After getting four equations I used this site to solve the system of equations. Unfortinetly I got no soultion. Tried another site and also the same result.
I've tried to use different values for x and got another equaitons like:
for x = 2 : $-2A - B + 24C + 7D$
for x = -1 : $4A - 4B - 4C + 4D$
for x = -2 : $10A - 5B - 6C + 3D$
But also that didn't work. Always the system of equations have an infinite solutions.
After tring to figure out why this is happening, I've managed to prove logically that this equation:
$(Ax + B)(x - 3) + (Cx + D)(x + 5) = x^2 - 4$
has an infinite solutions and my approach was as follows:
After doing polynomial long division and decomposing the fraction using the traditional way, the result should be:
$\frac{5}{8(x - 3)} - \frac{21}{8(x - 5)} + 1$
Now I can add the last term (the one) to the first term and get the follows:
$\frac{8x-19}{8(x-3)} - \frac{21}{8(x+5)}$
From that solution I can see that $A = 0, B = \frac{-21}{8}, C = 1, D = \frac{-19}{8}$. After all these are just the coefficients of the terms. and this solution worked fine.
Alternatively I can add the one to the second term instead and get:
$\frac{5}{8(x-3)} + \frac{8x + 19}{8(x+5)}$
Now $A = 1, B = \frac{19}{8}, C = 0, D = \frac{5}{8}$
Generally, after adding the one to any of the terms, I can add any number to one of the terms and add its negative to the other term and the equation will remain the same, But the value of the 4 constants (A, B, C, and D) will change. And from that I got convinced that there are an infinite solutions to this equation.
But Algebraically? I'm not able to prove that it has an infinite solutions algebraically. And my questions is how to prove algebraically that this equation has an infinite solutions? Or generally how to know whether the equation has just one solution or an infinite?
|
Here's a straightforward way to see that this particular equation ($(Ax+B)(x−3)+(Cx+D)(x+5)=^2−4$) has infinitely many solutions once you've found one: suppose we have some solution $\langle A, B, C, D\rangle$. We know that $(x+5)(x-3)+(3-x)(x+5)=0$ (since after all this is just saying that $(x+5)(x-3)-(x-3)(x+5)=0$!), so if we choose $A'=A+1, B'=B+5, C'=C-1, D'=D+3$ then we have $(A'x+B')(x-3)+(C'x+D')(x+5)$ $=(Ax+x+B+5)(x-3)+(Cx-x+D+3)(x+5)$ $=(Ax+B)(x-3)+(x+5)(x-3)+(Cx+D)(x+5)+(3-x)(x+5)$ $=(Ax+B)(x-3)+(Cx+D)(x+5)$, and this means that $\langle A', B', C', D'\rangle$ is also a solution. In terms of the partial fraction decomposition, this amounts to adding $\frac{x+5}{x+5}$ and subtracting $\frac{x-3}{x-3}$.
This is an example of a general phenomenon; since we have three (linear) equations in four unknowns (the three equations come from equating terms of the quadratics in $x$), the system is said to be underdetermined, and a version of the procedure shown above will prove that if an underdetermined system of linear equations has any solutions, then it has infinitely many.
|
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|
A double integral for $\frac{\pi}{2} \ln 2$.
Show that
\begin{eqnarray*}
I=\int_0^1 \int_0^1 \frac{dx \,dy}{\sqrt{1-x^2y^2}} = \frac{\pi}{2} \ln 2.
\end{eqnarray*}
My try ... from this question here we have
\begin{eqnarray*}
\int_0^1 \frac{ \sin^{-1}(x)}{x} \,dx = \frac{\pi}{2} \ln 2 .
\end{eqnarray*}
And from this question here we have
\begin{eqnarray*}
\int_0^1 \ln \left( \frac{1+ax}{1-ax} \right) \frac{dx}{x\sqrt{1-x^2}}=\pi\sin^{-1} a,\qquad |a|\leq 1.
\end{eqnarray*}
It is easy to show
\begin{eqnarray*}
\int_0^1 \frac{dy}{1+yz}=\frac{1}{z} \ln(1+z).
\end{eqnarray*}
So we have (with a little tad of algebra)
\begin{eqnarray*}
\frac{\pi}{2} \ln 2 &=& \int_0^1 \frac{ \sin^{-1}(x)}{x}\, dx \\
&=& \frac{1}{\pi} \int_0^1 \int_0^1 \ln \left( \frac{1+xt}{1-xt} \right) \frac{dt}{xt\sqrt{1-t^2}} x\, dx \\
&=& \frac{2}{\pi} \int_0^1 \int_0^1 \int_0^1 \frac{dx \,dy\, dt}{(1-x^2y^2t^2)\sqrt{1-t^2} } \\
\end{eqnarray*}
This suggests we should consider the integral (sub $t=\sin(\theta)$)
\begin{eqnarray*}
\frac{2}{\pi} \int_0^1 \frac{ dt}{(1-x^2y^2t^2)\sqrt{1-t^2} } \\
= \int_0^{\pi/2} \frac{d \theta }{1- x^2 y^2 \sin^2(\theta)}.
\end{eqnarray*}
Now it is well known (Geometrically expand, integrate term by term & sum the familiar plum) that
\begin{eqnarray*}
\frac{2}{\pi} \int_0^{\pi/2} \frac{d \theta }{1- \alpha \sin^2(\theta)}=\frac{2}{\pi} \frac{1}{\sqrt{1-\alpha}}
\end{eqnarray*}
and using this we have
\begin{eqnarray*}
\frac{\pi}{2} \ln 2 =\int_0^1 \int_0^1 \frac{dx\, dy}{\sqrt{1-x^2y^2}} .
\end{eqnarray*}
The above double integral reminds of
\begin{eqnarray*}
\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}=\int_0^1 \int_0^1 \frac{dx\, dy}{1-x^2y^2} = \frac{\pi^2}{8}
\end{eqnarray*}
which can be evaluated using the substitution $x= \frac{\sin u}{\cos v}$, $y= \frac{\sin v}{\cos u}$.
My solution above used some pretty heavy machinery to establish the result. So my question is: is there an easier method ?
|
$$\int_0^1 \int_0^1 \frac{dxdy}{\sqrt{1-x^2y^2}}\overset{xy=t}=\int_0^1 \frac{1}{y}\int_0^y \frac{1}{\sqrt{1-t^2}}dtdy=\int_0^1 \frac{\arcsin y}{y}dy
$$
$$\overset{IBP}=-\int_0^1 \frac{
\ln y}{\sqrt{1-y^2}}dy\overset{y=\sin x}=-\int_0^\frac{\pi}{2}\ln(
\sin x)dx=\frac{\pi}{2}\ln 2$$
See here for the above integral.
|
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|
Find minimum of function $\frac{\left| x-12\right| }{5}+\frac{\sqrt{x^2+25}}{3}$
Find minimum of function $f(x) = \frac{\left| x-12\right|
}{5}+\frac{\sqrt{x^2+25}}{3}$
I tried compute min using by definition of abs function.
I consider two cases:
*
*when $x > 12$ we have:
$f_{1}(x) = \frac{x-12}{5}+\frac{\sqrt{x^2+25}}{3}$
using standard method $f_{1}'(x)=0$ for $x_{0}=-15/4$ but $x_{0}$ is not in $ [12, \infty]$
*when $x \leq 12$ we have $f_{2}(x) = \frac{-x+12}{5}+\frac{\sqrt{x^2+25}}{3}$ similarly $f_{2}'(x) = 0$, and $x = 15/4$
So using observation from this method I computed minimum of
$f(15/4) = 56/15$.
Does it be a correct way?
|
We can use C-S and the triangle inequality:
$$\frac{|x-12|}{5}+\frac{\sqrt{x^2+25}}{3}=\frac{|x-12|}{5}+\frac{\sqrt{(3^2+4^2)(x^2+5^2)}}{15}\geq$$
$$\geq \frac{|x-12|}{5}+\frac{|3x+20|}{15}=\left|\frac{12}{5}-\frac{x}{5}\right|+\left|\frac{4}{3}+\frac{x}{5}\right|\geq $$
$$\geq \left|\frac{12}{5}-\frac{x}{5}+\frac{4}{3}+\frac{x}{5}\right|=\frac{56}{15}.$$
The equality occurs for $(3,4)||(x,5)$ and $\left(\frac{12}{5}-\frac{x}{5}\right)\left(\frac{4}{3}+\frac{x}{5}\right)\geq0,$ which gives $x=\frac{15}{4},$
which says that we got a minimal value.
|
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|
Finding the length of the arc for function $8y = x^4 +2x^{-2}$
Find the length of the arc of curve $8y = x^4 +2x^{-2}$ from $x=1$ to $x=2$
I first isolated for $y$ and derived:
$$ f(x) = {1 \over 8 }x^{4} + {1 \over 4}x^{-2} $$
$$f'(x) = {1 \over 2} x^3 - {1 \over 2}x^{-3}$$
Then found the arc length:
$$L = \int ^2 _1 \sqrt{1+[f'(x)]^2} dx
\\ = \int ^2 _1 \sqrt{1+({1\over 2}x^3 -{1 \over 2}x^{-3})^2} dx
\\ = \int ^2 _1 \sqrt{{1 \over 4} x^6 + {1\over 4}x^{-6} +{1 \over 2}} dx$$
Let $u = {1 \over 4} x^6 + {1\over 4}x^{-6} +{1 \over 2}$
$du = {{3 \over 2} x^5 - {3\over 2}x^{-7} dx}$
However, I cannot seem to integrate this
How can I integrate this equation?
|
Hint
$$u = {1 \over 4} x^6 + {1\over 4}x^{-6} +{1 \over 2}= \left({1 \over 2} x^3 + {1\over 2}x^{-3} \right)^2$$
|
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|
Difficult integral involving trigonometric and hypertrigonometric functions This is definitely the most difficult integral that I've ever seen.
Of course, I'm not able to solve this.
Could you help me?
$$\int { \sin { x\cos { x } \cosh { \left( \ln { \sqrt { \frac { 1 }{ 1-\sin { x } } } +\tanh ^{ -1 }{ \left( \sin x \right) +\tanh ^{ -1 }{ \left( \cos { x } \right) } } } \right) } } dx } $$
|
Using the fact that $\cosh t = \frac{1}{\sqrt{1-\tanh^2 t}}$ and $\sinh t = \frac{\tanh t}{\sqrt{1-\tanh^2 t}} $ $$\cosh(a+b+c) = \cosh a \cosh b \cosh c + \sinh a \sinh b \cosh c + \sinh a \cosh b \sinh c + \cosh a \sinh b \sinh c$$
the integrand simplifies to
$$\int dx \left[\cosh\left(\log\left(\frac{1}{\sqrt{1-\sin x}}\right)\right)\left(1+\sin x \cos x\right) + \sinh\left(\log\left(\frac{1}{\sqrt{1-\sin x}}\right)\right)\left(\sin x + \cos x \right) \right]$$
Next we'll substitute $x = 2z+\frac{\pi}{2}$:
$$\int 2dz \left[\cosh\left(\log\left(\frac{1}{\sqrt{1-\cos 2z}}\right)\right)\left(1-\sin 2z \cos 2z\right) + \sinh\left(\log\left(\frac{1}{\sqrt{1-\cos 2z}}\right)\right)\left(\cos 2z - \sin 2z \right) \right]$$
$$ = \frac{1}{\sqrt2}\int dz (2\sin z + \csc z )(1-\sin 2z \cos 2z)-(2\sin z - \csc z )(\cos 2z - \sin 2z)$$
$$ = \frac{1}{\sqrt2}\int dz\left[4\sin^2 z \cos z(1-\cos 2z) - 2\cos z(1+2\cos 2z)+2\sin z(1-\cos 2z)+ \csc z(1+\cos 2z) \right]$$
$$ =\frac{1}{\sqrt{2}} \int(8\sin^4 z + 4 \sin^ z - 4)\cos z - (4\cos^2 z - 2)\sin z + 2 \csc z dz$$
$$= \sqrt{2}\left(\frac{4}{5}\sin^5 z + \frac{2}{3}\sin^3 z - 2 \sin z + \frac{2}{3}\cos^3 z - 2\cos z - \log|\csc z + \cot z|\right)$$
Therefore our final answer is
$$\sqrt{2}\left(\frac{4}{5}\sin^5 \left(\frac{x}{2}-\frac{\pi}{4}\right) + \frac{2}{3}\sin^3 \left(\frac{x}{2}-\frac{\pi}{4}\right) - 2 \sin \left(\frac{x}{2}-\frac{\pi}{4}\right) + \frac{2}{3}\cos^3 \left(\frac{x}{2}-\frac{\pi}{4}\right) - 2\cos \left(\frac{x}{2}-\frac{\pi}{4}\right) - \log{\left|\csc \left(\frac{x}{2}-\frac{\pi}{4}\right) + \cot \left(\frac{x}{2}-\frac{\pi}{4}\right)\right|}\right) + C$$
Edit: We can simplify this a bit further with some trig shenanigans. Applying the angle subtraction formulas, we get:
$$\frac{1}{5}\left(\sin \left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right)^5 + \frac{1}{3}\left(\sin \left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right)^3 - 2 \left(\sin \left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right) + \frac{1}{3}\left(\sin \left(\frac{x}{2}\right)+\cos\left(\frac{x}{2}\right)\right)^3 - 2\left(\sin \left(\frac{x}{2}\right)+\cos\left(\frac{x}{2}\right)\right) - \sqrt{2}\log{\left|\frac{\sqrt{2}+\cos\left(\frac{x}{2}\right)+\sin\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)}\right|} + C$$
$$=\frac{1}{5}\left(\sin \left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right)^5 + \frac{2}{3}\sin \left(\frac{x}{2}\right)\left(\sin^2 \left(\frac{x}{2}\right)+3\cos^2\left(\frac{x}{2}\right)\right) - 4 \sin \left(\frac{x}{2}\right) - \sqrt{2}\log{\left|\frac{\sqrt{2}+\cos\left(\frac{x}{2}\right)+\sin\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)}\right|} + C$$
$$=\frac{1}{5}\left(\sin \left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right)^5 + \frac{2}{3}\sin \left(\frac{x}{2}\right)\cos x - \frac{8}{3} \sin \left(\frac{x}{2}\right) - \frac{1}{\sqrt{2}}\log{\left(\frac{3+2\sqrt{2}(\cos\left(\frac{x}{2}\right)+\sin\left(\frac{x}{2}\right))+\sin x}{1-\sin x}\right)} + C$$
And I think I will stop there.
|
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|
$\sum_{i=0}^n (-1)^i\frac{C_i}{i+2}$ $C_i$ being the coefficient of $x^i$ in $(1+x)^n$ Evaluate: $\displaystyle\sum_{i=0}^n (-1)^i\frac{C_i}{i+2}$ where $C_i$ is the coefficient of $x^i$ in $(1+x)^n$
Attempt: We consider $f(x)=x(1-x)^n$
$\displaystyle\int_0^1x(1-x)^n dx= \int_0^1x^{(2-1)}(1-x)^{(n+1)-1} dx=B(2,n+1)=\frac{\Gamma(2)\Gamma(n+1)}{\Gamma(n+3)}= \frac{ n!}{(n+2)!}= \frac{1}{(n+1)(n+2)}=\frac{C_0}{2} - \frac{C_1}{3}+ \frac{C_2}{4}+...+ \frac{(-1)^nC_n}{n+2} $
Is this all?
|
Hint:
Another way without calculus
$$\dfrac{\binom nr}{r+2}=\dfrac{(r+2-1)\binom nr}{(r+1)(r+2)}=\dfrac1{n+1}\binom{n+1}{r+1}-\dfrac1{(n+2)(n+1)}\binom{n+2}{r+2}$$
Now set $m=n+1,n+2$ in $$(1-1)^m=\sum_{r=0}^m(-1)^r\binom mr$$
|
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|
Find the reflection of the point $(4,-13)$ in the line $5x+y+6=0$
Find The image(or reflection) of the point $(4,-13)$ in the line $5x+y+6=0$
Method 1
$$
y+13=\frac{1}{5}(x-4)\implies x-5y-69=0\quad\&\quad 5x+y+6=0\implies (3/2,-27/2)\\
(3/2,-27/2)=(\frac{x+4}{2},\frac{y-13}{2})\implies(x,y)=(-1,-14)
$$
Method 2
$m=\tan\theta=-5$
Ref$(\theta)$=$\begin{bmatrix}
\cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta)
\end{bmatrix}$
$$
\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-25}{1+25}=\frac{-24}{26}=\frac{-12}{13}\\
\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{-10}{26}=\frac{-5}{13}\\
Ref(\theta)\begin{bmatrix}4\\-13\end{bmatrix}=\begin{bmatrix}
\cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta)
\end{bmatrix}\begin{bmatrix}4\\-13\end{bmatrix}=\begin{bmatrix}
\dfrac{-12}{13} & \dfrac{-5}{13} \\ \dfrac{-5}{13} & \dfrac{12}{13}
\end{bmatrix}\begin{bmatrix}4\\-13\end{bmatrix}\\
=\frac{1}{13}\begin{bmatrix}-48+65\\-20-156\end{bmatrix}=\frac{1}{13}\begin{bmatrix}17\\-176\end{bmatrix}
$$
Why am I not getting the required solution in Method two using matrix method ?
Thanx @ganeshie8 for the remarks, so in that case how do I find the operator for reflection of a point over the line not passing through the origin ?
|
The simplest (and shortest) way is to do some affine geometry:
Find first the projection of the point $A(4,-13)$ onto the line $5x+y+6=0$. As a directing vector for this projection is $\vec n (5,1)$, you have a parametric equation of the line of projection:
$$\overrightarrow{OM}=A+t\mkern 1.5mu\vec n, $$it suffices to find $t$ so the point $M$ satisfies the equation $5x+y+6=0$. Then, the reflection of $A$ is the point $$A'=A+2t\mkern 1.5mu \vec n.$$
|
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|
Proof: if $x^5-4x^4+3x^3-x^2+3x-4≥0$ then $x≥0$ I have proved this question by finding the solution of the polynomial using 17 steps Newthon Rhapson and found out that x ≥ 3,0735. I am wondering is there any more simple way to solve this?
|
You have that
$$
\begin{gathered}
p(x) = x^5 - 4x^4 + 3x^3 - x^2 + 3x - 4 = \hfill \\
\hfill \\
= x^3 \left( {x^2 - 4x + 3} \right) - \left( {x^2 - 3x + 4} \right) \hfill \\
\end{gathered}
$$
Now:
*
*$$-\left( {x^2 - 3x + 4} \right)<0 \;\;\forall x\;\; \in \mathbb R$$
*$$
{x^2 - 4x + 3}>0\;\; \forall x\;\;<0$$
*$$x^3<0 \forall x\;\;<0$$
Therefore, for every x<0 you have that p(x)<0. Hence, if p(x)=0 it must be x>0
In order to prove that if p(x)=0 then x>3 we have that
*
*If $$ 0< x <1 $$ then $$ f(x) > 0 $$
*If $$ 1< x <3 $$ then $$ f(x) < 0 $$
*If $$x=0, x=1,x=3$$ then $$f(x)=0$$
*f has a local maximum at
$$
x_0 = \frac{1}
{5}\left( {8 - \sqrt {19} } \right)
$$
and
$$
f(x_0 ) = \frac{{2\left( {2147\sqrt {19} - 8986} \right)}}
{{3125}} < \frac{1}
{2}
$$
*$$g(x)=-(x^2-3x+4) \leq -\frac{7}{4} \forall x \in \mathbb R$$
Therefore if
$$
0 \leqslant x \leqslant 3
$$
it is
$$
p(x) = f(x) + g\left( x \right) \leqslant \frac{1}
{2} - \frac{7}
{4} < 0
$$
It follows that if p(x)=0 then x>3
|
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|
A family of irreducible polynomials Does someone has an idea how to prove that the polynomials
\begin{align*} P_a=(a+2)(a+1)X^{a+4}-2(a+4)(a+1)X^{a+3}+(a+4)(a+3)X^{a+2}-2(a+4)X+2(a+1)
\end{align*}
are all of the form $(X-1)^4Q_a$ with $Q_a$ irreducible (over $\mathbb{Q}$)? Sage tell me it's true until 100 but I can't prove it. In fact it's enough for me that it works for an infinity of such polynomial and even that the biggest degree of all its factor tends to infinity. I tried the classical Eisenstein and reduction modulo $p$ but witout success. Thanks for all your ideas!
|
$$P_a=(a+2)(a+1)x^{a+4}-2(a+4)(a+1)x^{a+3}+(a+4)(a+3)x^{a+2}-2(a+4)x+2(a+1)$$
Offset $a$ down by one:
$$P'_a=(a+1)ax^{a+3}-2(a+3)ax^{a+2}+(a+3)(a+2)x^{a+1}-2(a+3)x+2a$$
Subst $z = x-1$
$$P'_a=(a+1)a(z+1)^{a+3}-2(a+3)a(z+1)^{a+2}+(a+3)(a+2)(z+1)^{a+1}-2(a+3)(z+1)+2a \\
= \sum_i \left[ (a+1)a\binom{a+3}{i} - 2(a+3)a\binom{a+2}{i} + (a+3)(a+2)\binom{a+1}{i}\right]z^i - 2(a+3)z - 6 \\
= \sum_{i\ge 4} \left[ (a+1)a\binom{a+3}{i} - 2(a+3)a\binom{a+2}{i} + (a+3)(a+2)\binom{a+1}{i}\right]z^i \\
%= \sum_{i\ge 4} \left[ (a+1)a \frac{(a+3)!}{(a+3-i)!i!} - 2(a+3)a \frac{(a+2)!}{(a+2-i)!i!} + (a+3)(a+2)\frac{(a+1)!}{(a+1-i)!i!} \right]z^i \\
%= \sum_{i\ge 4} \frac{(a+3)(a+2)(a+1)a - 2(a+3)(a+2)a(a+3-i) + (a+3)(a+2)(a+3-i)(a+2-i)}{(a+3-i)(a+2-i)} \binom{a+1}{i} z^i \\
%= \sum_{i\ge 4} \frac{(a+3)(a+2)(i-2)(i-3)}{(a+3-i)(a+2-i)} \binom{a+1}{i} z^i \\
= \sum_{i\ge 4} (i-2)(i-3) \binom{a+3}{i} z^i \\
$$
So
$$\frac{P'_a}{z^4} = \sum_{j=0}^{a-1} (j+2)(j+1) \binom{a+3}{j+4} z^{j}$$
In fact it's enough for me that it works for an infinity of such polynomial
When $a+3$ is a prime, Eisenstein's criterion works.
|
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|
Finding an integrating factor for $(3t + \frac{6}{y})dt + (\frac{t^2}{y} + \frac{3y}{t})dy = 0$ I want to solve the differential equation
$$
\begin{equation}
(3t + \frac{6}{y})dt + (\frac{t^2}{y} + \frac{3y}{t})dy = 0\
\end{equation}
$$
using an integrating factor of the form $t^a y^b$ in order to make it exact, where
$$
M(t, y) = 3t + \frac{6}{y},\\
N(t, y) = \frac{t^2}{y} + \frac{3y}{t}.
$$
So we want
$$
\begin{align}
\frac{\partial}{\partial y} (t^ay^bM) &= \frac{\partial}{\partial t} (t^ay^bN)\\
b t^a y^{b-1} M+ t^ay^b \frac{\partial}{\partial y}M &= at^{a-1}y^bN + t^a y^b \frac{\partial}{\partial t}N.
\end{align}
$$
By susbtituting
$$
\begin{align}
\frac{\partial}{\partial y}M &= -\frac{6}{y^2} \\
\frac{\partial}{\partial t}N &= \frac{2t}{y} - \frac{3y}{t^2},
\end{align}
$$
I arrived at
$$
3b + 6(b-1)t^{-1}y^{-1} = (a+2) + 3(a-1)t^{-3}y^{2}.
$$
But how do I find the values of $a$ and $b$ from here?
(Also, if my calculations could be verified, I'd really appreciate it!)
Edit #1: Apparently, on the last line, I had arrived at $(a+1)$, when in fact it should have been $(a+2)$.
|
Here is a different approach to this problem. In the first place, notice that
\begin{align*}
3t + \frac{6}{y} + \left(\frac{t^{2}}{y} + \frac{3y}{t}\right)y^{\prime} = 0
\end{align*}
As suggested by Moo, multiply it by $ty$ in order to obtain
\begin{align*}
3t^{2}y + 6t + (t^{3} + 3y^{2})y^{\prime} = 0
\end{align*}
Then we can rearrange it as
\begin{align*}
(3t^{2}y + t^{3}y^{\prime}) + 3y^{2}y^{\prime} + 6t = 0 \Longleftrightarrow (t^{3}y)^{\prime} + (y^{3})^{\prime} + 6t = 0
\end{align*}
After integration, it becomes
\begin{align*}
t^{3}y + y^{3} + 3t^{2} + k =0
\end{align*}
which is a cubic in the dependent variable $y$.
|
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|
We have the equation $2x^2-\sqrt{3}x-1=0$ and have to find $|x_1-x_2|$
We have the following quadratic equation:
$2x^2-\sqrt{3}x-1=0$ with roots $x_1$ and $x_2$.
I have to find $x_1^2+x_2^2$ and $|x_1-x_2|$.
First we have: $x_1+x_2=\dfrac{\sqrt{3}}{2}$ and $x_1x_2=-\dfrac{1}{2}$
So $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=\dfrac{7}{4}$
Can someone help me with the second one?
I forgot to tell that solving the equation is not an option in my case.
|
Well, you know that, by the Quadratic Formula, $\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, so the difference between the two roots is $\frac{1}{a}\sqrt{b^2-4ac}=\frac{1}{a}\sqrt{3+4(2)(1)}=\frac{1}{2}\sqrt{11}$
|
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|
Prove a trigonometric identity: $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$ when $A+B+C=\pi$ There is a trigonometric identity:
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C\equiv 1\text{ when }A+B+C=\pi$$
It is easy to prove it in an algebraic way, just like that:
$\quad\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C\\=\cos^2A+\cos^2B+\cos^2\left(\pi-A-B\right)+2\cos A\cos B\cos \left(\pi-A-B\right)\\=\cos^2A+\cos^2B+\cos^2\left(A+B\right)-2\cos A\cos B\cos \left(A+B\right)\\=\cos^2A+\cos^2B+\left(\cos A\cos B-\sin A\sin B\right)^2-2\cos A\cos B\left(\cos A\cos B-\sin A\sin B\right)\\=\cos^2A+\cos^2B+\cos^2A\cos^2B+\sin^2A\sin^2B-2\sin A\cos A\sin B\cos B-2\cos^2A\cos^2B+2\sin A\cos A\sin B\cos B\\=\cos^2A+\cos^2B-\cos^2A\cos^2B+\left(1-\cos^2A\right)\left(1-\cos^2B\right)\\=\cos^2A+\cos^2B-\cos^2A\cos^2B+1-\cos^2A-\cos^2B+\cos^2A\cos^2B\\=1$
Then, I want to find a geometric way to prove this identity, as $A+B+C=\pi$ and it makes me think of the angle sum of triangle. However, it is quite hard to prove it in a geometric way. Therefore, I hope there is someone who can help. Thank you!
|
Since the accent in the OP is put on a purely geometric solution, i can not even consider the chance to write $\cos^2 =1-\sin^2$, and rephrase the wanted equality, thus having a trigonometric function which is better suited to geometrical interpretations.
So this answer has two steps, first we reformulate the given identity in a mot-a-mot geometric manner, the geometric framework is introduced, some strictly geometrically transposed equivalent relations are listed, then we give a proof:
In the triangle $\Delta ABC$ let $AA'$, $BB'$, $CC'$ be the heights, $A'\in BC$, $B'\in CA$, $C'\in AB$, intersecting in $H$, the orthocenter. We assume that the diameter $2R$ of the circumcircle is normed to be the unit. Then we have the following situation for the lengths of some segments in the picture:
$$
\begin{aligned}
AH &=\cos A\ , \qquad & HA'&=\cos B\cos C\ ,\\
BH &=\cos B\ , \qquad & HB'&=\cos A\cos C\ ,\\
CH &=\cos C\ , \qquad & HC'&=\cos A\cos B\ .
\end{aligned}
$$
Proof: We have:
$$
\sin \hat B =\sin \widehat{C'HA}
=\frac{C'A}{AH}
=\frac{AC\;\cos A}{AH}
=\frac{2R\sin B\; \cos A}{AH}
=\frac{\sin B\; \cos A}{AH}
\ ,
$$
which implies $AH=\cos A$, and the similar relations. Then we express twice the area of $\Delta HBC$ as
$$ HA'\cdot BC =2[HBC]=HB\cdot HC\cdot \sin\widehat{BHC}\ ,$$
thus getting $HA'=\cos B\cos C$.
We are in position to give a geometric mask to the given equality:
We use the above notations in $\Delta ABC$. We denote by $a,b,c$ the lenghts of the sides. Let $M_A, M_B,M_C$ be the mid points of the sides $BC$, $CA$, respectively $AB$. Let $G=AM_A\cap BM_B\cap CM_C$ be the intersection of the medians, the centroid. Let $A^*, B^*, C^*$ be the mid points of $HA$, $HB$, $HC$. Let $N$ be the center of the Euler circle $(N)$ passing through the nine points $A',B',C'$; $M_A,M_B,N_C$; $A^*, B^*,C^*$. It is the mid point of $OH$, and $M_AA^*$, $M_BB^*$, $M_CC^*$ are diameters in $(N)$, having the lenght $R=OA=OB=OC$. (For $OM_AA^*A$ is a parallelogram.)
Then we have the following relations:
$$
\begin{aligned}
HA^2+HB^2+HC^2 + 2 HA\cdot HA' &= 4R^2\ ,\\
HA^*{}^2+HB^*{}^2+HC^*{}^2 + HA^*\cdot HA' &= R^2\ ,\\
4OM_A^2+4OM_B^2+4OM_C^2 &= 3R^2+OH^2\ ,\\
9R^2 &= a^2 +b^2 + c^2 +OH^2\\
9R^2 &= a^2 +b^2 + c^2 +9OG^2\ .
\end{aligned}
$$
Proof: The relations above are equivalent:
*
*$AH^2=4A^*H^2=4OM_A^2$, and $2 HA\cdot HA'=4 HA^*\cdot HA'$ is the power of $H$ in the circle $(N)$, so it can be rewritten using its radius $NA^*=\frac 12 R$ and the distance to its center, $NH=\frac 12 OH$ as
$2 HA\cdot HA'=4 HA^*\cdot HA'=R^2-OH^2$.
*From the triangle $OBM_A$,
$4OM_A^2+BC^2 =4(OM_A^2+BM_A^2)=4OB^2=4R^2$.
*Note that $G$ cuts the median $AM_A$ in the proportion $AG:GM_A=2:1$, so it projects on $BC$ in the same proportion. This also holds for the colinear points $H,G,O$, so $HG:GO=2:1$, so $HO=3GO$.
*The last relation, $OG^2 = R^2-\frac 13(a^2+b^2+c^2)$, is a standard formula. We have in general the formula for an arbitrary point $P$:
$$PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+3GP^2\ .$$
We apply it for $P=O$, getting $3R^2=3OG^3+\sum AG^2=3OG^3+\frac 49\sum AM_A^2=3OG^3+\frac 49\sum \left(\frac 12b^2+\frac 12 c^2-\frac 14 a^2\right)=3OG^3+\frac 49\sum \frac 34a^2=3OG^3+\frac 13\sum a^2\ .$
$\square$
|
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|
Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$. Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$.
Attempt:
Let $f(x) = x^{81} + x^{49} + x^{25} + x^{9} + x$, then the remainder of $f(x)$ divided by $(x-1)$ is $f(1)$, divided by $(x+1)$ is $f(-1)$.
Now since $x^{3}-x = x(x^{2}-1)=x(x-1)(x+1)$ is there any relation between the remainder of $f(x)$ divided by $x^{3}-x$ and remainder when $f(x)$ divided by $(x-1)$ and $(x+1)$?
|
The remainder is a quadratic polynomial $g(x)$ that is divisible by $x$ (since $f(x)$ is divisble by $x$ and has remainders $f(1)$ and $f(-1)$ from division on $(x-1)$ and $(x+1)$ respectively.
One may argue that:
$$
g(x)=f(1)\frac{x+1}{2}+f(-1)\frac{x-1}{-2} + \gamma(x^2-1)
$$
From the fact $g(0) = 0$, one can find the coefficient $\gamma=\frac{f(1)-f(-1)}2$.
|
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|
complex series summation up to infinity
Sum of series
$$S(x)=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots , |x|<1$$
What I tried.
Let $$A=\ln\bigg[(1+x)(1+x^2)(1+x^3)(1+x^4)\cdots \cdots \bigg]$$
$$A = \ln\bigg[(1+x+x^2+2x^3+2x^4+3x^5+\cdots+ )\bigg]$$
How do is solve it help me please
|
Regarding the relationship with the $A$ term
$$
S_n(x) = \sum_{k=1}^n\frac{g_k'(x)}{g_k(x)} = \frac{d}{dx}\sum_{k=1}^n\ln g_k(x) = \frac{d}{dx}\ln\prod_{k=1}^ng_k(x) = \frac{d}{dx}A
$$
here $g_k(x) = 1+x^k$. Also $\prod_{k=1}^n g_k(x) = \frac 12(-1;x)_{n+1}$ (see) and
$$
S_n(x) = \frac{d}{dx}\ln\left(\frac 12(-1;x)_{n+1}\right)
$$
|
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|
Calculate the maximum value of $\sum_{cyc}\frac{bc}{(b + c)^3(a^2 + 1)} $ where $a, b, c \in \mathbb R^+$ satisfying $abc = 1$.
Calculate the maximum value of $$\large \frac{bc}{(b + c)^3(a^2 + 1)} + \frac{ca}{(c + a)^3(b^2 + 1)} + \frac{ab}{(a + b)^3(c^2 + 1)}$$ where $a, b, c$ are positives satisfying $abc = 1$.
We have that $$\sum_{cyc}\frac{bc}{(b + c)^3(a^2 + 1)} \le \frac{1}{2} \cdot \sum_{cyc}\frac{1}{(b + c)(a^2 + 1)} \le \sum_{cyc}\frac{1}{(b + c)(a + 1)^2}$$
$$ = \sum_{cyc}\frac{1}{a(ab + ca + b + c)(bc + 1)} \le \dfrac{1}{2} \cdot \sum_{cyc}\frac{1}{a\sqrt{bc}(ab + ca + 2\sqrt{bc})}$$
$$ = \frac{1}{2} \cdot \sum_{cyc}\frac{1}{a\sqrt a(b + c) + 2}$$
That's all I got, not because I can't go for more, but since I went overboard with this, there's no use trying to push for more.
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The inequality $$\sum_{cyc}\frac{x}{x^4+1}\leq\frac{3}{2}$$ for positives $x$, $y$ and $z$ we can prove also by the following way.
We'll prove that
$$\frac{x}{x^4+1}\leq\frac{3(x^2+1)}{4(x^4+x^2+1)}$$ is true for all positive $x$.
Indeed, let $x^2+1=2ux.$
Thus, by AM-GM $u\geq1$ and it's enough to prove that
$$3\cdot2u\cdot(4u^2-2)\geq4(4u^2-1)$$ or
$$6u^3-4u^2-3u+1\geq0$$ or $$(u-1)(6u^2+2u-1)\geq0,$$ which is obvious.
Id est, it's enough to prove that
$$\sum_{cyc}\frac{x^2+1}{x^4+x^2+1}\leq2$$ or
$$\sum_{cyc}\frac{x+1}{x^2+x+1}\leq2$$ or
$$\sum_{cyc}\left(\frac{x+1}{x^2+x+1}-1\right)\leq-1$$ or
$$\sum_{cyc}\frac{x^2}{x^2+x+1}\geq1$$ or
$$\sum_{cyc}(x^2y^2-x)\geq0$$ or
$$\sum_{cyc}(x^2y^2-x^2yz)\geq0$$ or
$$\sum_{cyc}z^2(x-y)^2\geq0$$ and we are done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3356213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Is the series $\sum_{n=1}^{\infty} \frac{4+3^n}{2^n}$ convergent or divergent $\sum_{n=1}^{\infty} \frac{4+3^n}{2^n}$
$\begin{align}
\sum_{n=1}^{\infty} \frac{4+3^n}{2^n} &= \sum_{n=1}^{\infty} \frac{4}{2^n} + \sum_{n=1}^{\infty} \frac{3^n}{2^n} \\
&= \sum_{n=1}^{\infty} \frac{4}{2 \cdot 2^{n-1}} + \sum_{n=1}^{\infty} \bigg(\frac{3}{2}\bigg)^n \\
&= 2\sum_{n=1}^{\infty}\frac{1}{2^{n-1}} + \sum_{n=1}^{\infty}\frac{3}{2} \bigg(\frac{3}{2}\bigg)^{n-1}
\end{align}$
Now I observe that the two terms are both geometric series and although the first one converges because $\frac{1}{2} < 1$, the second one doesn't because $\frac{3}{2} > 1$. Then can I assume that the series diverges and that is it?
|
Your answer is correct!
You can also argue that $
\frac{4+3^{n}}{2^{n}} \underset{n \rightarrow \infty}{\longrightarrow} \infty
$ that is the necessary condition for a series to converge doesn't hold in our case that is- the limit of a sequence defining a convergent series is $0$. Or the limit of a general term in the series is $0$. Or as peter kindly commented- the sequence of the summands to converge to $0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3362764",
"timestamp": "2023-03-29T00:00:00",
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|
Definite integral of rational expression involving quadratics I was given the following exercise on my Calculus class:
$$
\int_2^4 \frac{x^2+4x+24}{x^2-4x+8}dx
$$
I studied from a book (author, Stewart) various methods to solve integrals of the form $\int \frac{P(x)}{Q(x)}dx$; a basic idea common to all of this methods is that $P(x)$ should have a smaller degree than $Q(x)$, in order for us to make a useful factorization. Since this is not the case on the integral I was given, I first computed the division of $P(x)=x^2+4x+24$ over $Q(x)=x^2-4x+8$, which took me to the fact that
$P(x)= Q(x) + (8x+32)$. So the integral can be rewritten in the following manner:
$$
\int_2^4 \frac{(x^2-4x+8)+(8x+32)}{x^2-4x+8}dx=\int_2^4 dx+\int_2^4 \frac{8x+32}{x^2-4x+8}dx
$$
Putting aside the first term of the resulting expression, which is a very simple integral, it all comes down to solving the second term. Gladly I have managed to translate the first integral into a new one in which the numerator is of smaller degree than the denominator. But how do I go about solving $\int_2^4 \frac{8x+32}{x^2-4x+8}dx$.
Well, according to Stewart, once we have managed to solve our little degree problem, we have to factorize the denominator, and from there rewrite the whole expression. But the quadratic on the denominator does not seem to have a complete factorization, and so I am clueless on what to do. This is the only method I was taught to solve this kind of integrals. Am I making a mistake somewhere, or how would you go about solving this problem? Thanks in advance.
|
It appears you have written your integral in the wrong manner, as $8+32\neq 24$.
\begin{align}
\int_2^4 \frac{x^2+4x+24}{x^2-4x+8}\ dx&=\int_2^4 \frac{(x^2-4x+8)+(8x+16)}{x^2-4x+8}\ dx\\
&=\int_2^4 dx + 8\int_2^4\frac{x+2}{x^2-4x+8}\ dx\\
&=x\big]_2^4+8\int_2^4 \frac{x+2}{(x-2)^2+4}\ dx\\
&=2+8\int_0^2 \frac{u+4}{u^2+4}\ du&\textrm{(let } u=x-2)\\
&=2+8\int_0^2\frac{u}{u^2+4}\ du+32\int_0^2 \frac{1}{u^2+4}\ du\\
&=2+4\int_4^8\frac1{v}\ dv+16\arctan(\frac{u}2)\big]_0^2&\textrm{(let } v=u^2+4)\\
&=2+4\ln v\big]_4^8+16\arctan(\frac{u}2)\big]_0^2\\
&=2+4\ln2+16(\frac{\pi}4)\\
&=2+4\ln2+4\pi
\end{align}
A few steps may have been unnecessary, but my thought process uses really simple substitutions which may fall in line with the thinking of certain students.
|
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|
Find the exact value of trigonometric expression: $ \arctan\frac{2\cos \frac{\pi}{30} \cos\frac{\pi}{15}}{1+2\cos \frac{\pi}{30} \sin\frac{\pi}{15}}$? How can I simplify this trigonometric expression?
$$ \arctan\frac{2\cos \frac{\pi}{30} \cos\frac{\pi}{15}}{1+2\cos \frac{\pi}{30} \sin\frac{\pi}{15}}$$
I used
$$\sin \frac{\pi}{15}=2 \sin \frac{\pi}{30} \cos \frac{\pi}{30}$$ and $$\cos\frac{\pi}{15}=2\cos^2\frac{\pi}{30}-1$$
But these give me more complicated expressions.
|
$\begin{align}
T &= \large \left(\frac{2\cos 6° \cos 12°}{1\;+\;2\cos 6° \sin 12°}\right)
\left({2\sin 6° \over 2\sin 6°}\right)\cr
&= \large \frac{\sin 24°}{2\sin 6°+\;2\sin^2 12°}\cr\cr
{1\over T}&= \large \frac{2\sin(30°-24°)+\;(1-\cos 24°)}{\sin 24°}\cr
&=\large {(\cos 24° - \sqrt3\sin24°) + (1-\cos 24°) \over \sin 24°}\cr
&=\large{1\over\sin(60°-36°)} - \sqrt3 \cr
&=\large{4\over 2\sqrt3 \cos36° - 2\sin 36°}-\sqrt3 \cr
\end{align}$
From wikipedia on Golden triangle,
$\quad2\cos(36°)=\phi,\quad 2\sin(36°)=\sqrt{4-\phi^2}$
$\begin{align}
\large{1\over T}&=\left({4\over \sqrt3\phi-\sqrt{4-\phi^2}}\right)
\left({\sqrt3\phi+\sqrt{4-\phi^2} \over \sqrt3\phi+\sqrt{4-\phi^2}} \right)
-\sqrt3 \cr
&= {4(\sqrt3\phi + \sqrt{4-\phi^2}) \over 3\phi^2-(4-\phi^2)} -\sqrt3 \cr
&= {\sqrt3\phi + \sqrt{4-\phi^2} \over \phi^2-1}-\sqrt3 \cr
&= {\sqrt3\phi + \sqrt{4-\phi^2} \over \phi} -\sqrt3 \cr
&= {\sqrt{4-\phi^2} \over \phi} = \tan(36°)
\end{align}$
$$T = \tan(90°-36°) = \tan(54°)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Numerical bases and prime number theory Let $N = 3^x$. $5^y$. $7^z$. Find $N$ such that $5N$ and $27N$ have $8$ and $18$ dividers, respectively, more than $N$.
I did what the statement asks and it was here:
dividers: $(x + 1) (y + 2) (z + 1) = (x + 1) (y + 1) (z + 1) + 8; (x + 4) (y + 1) (z + 1) = (x + 1) (y + 1) (z + 1) + 18$
System
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If $N=3^x5^y7^z$ then $N$ has $(x+1)(y+1)(z+1)$ divisors.
$5N = 3^x5^{y+1}7^z$ has $(x+1)(y+2)(z+1)$ divisors
And $27N = 3^{x+4}5y7^z$ has $(x+5)(y+1)(z+1)$ divisors.
It may be easier to replace $x+1 =j; y+1=k; z+1=m$
So
So $(x+1)(y+2)(z+1)-(x+1)(y+1)(z+1)= 8$. So $j(k+1)m - jkm=jkm +jm - jkm = jm=8$.
And $(x+4)(y+1)(z+1)- (x+1)(y+1)(z+1)= = 18$.
So $(j+3)km - jkm = 3km =18$ or $km=6$
So $(x+1)(z+1) = jm= 8$ and $(y+1)(z+1)= km=6$.
So $z+1=m$ is a common divisor of $8,6$ so $m = 1,2$.
So we have either $z+1=1$ and $z=0$ and $y+1=6$ and $y=5$ and $x+1=8$ and $x=7$ and $N = 3^7*5^5$ which has $48$ divosors while $5N=3^7*5^6$ has $56$ divisors and $27N=3^{10}5^5$ has $66$ divisors.
(But presumably $z > 0$ otherwise they wouldn't have specified that $7$ was a factor. So I suppose the text isn't counting this as an answer although it fits all the requirements)
Or
$z+1=2$ and $z=1$ and $y+1=3$ and $y=2$ and $x+1=4$ and $x =3$ and $N=3^3*5^2*7=4725$ which has $24$ divisors while $5N = 3^3*5^3*7$ has $32$ divisors and $27N= 3^6*5^2*7$ has $42$ divisors.
|
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"timestamp": "2023-03-29T00:00:00",
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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.