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Properties of Matrix Product I am studying for my exam next week and the teacher has posted previous exams online.
I have the following question, given:
$A = \begin{bmatrix}4&-2&2\\2&4&-4\\1&1&0\end{bmatrix}$
$u = \begin{bmatrix}1\\3\\2\end{bmatrix}$
Find $A^5*u$ without any calculations...
HINT: Properties of the Matrix Product
|
Note that
$$A u = \begin{pmatrix}4 & -2 & 2\\ 2 & 4 & -4\\1 & 1 & 0\end{pmatrix}\begin{pmatrix}1\\3\\2\end{pmatrix} = \begin{pmatrix}2 \\ 6 \\ 4\end{pmatrix} = 2 \begin{pmatrix}1 \\ 3 \\ 2\end{pmatrix} = 2 u$$
Then for any $n>0$
$$A^n u = A^{n-1} (A u)= 2^n u$$
|
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|
Does $c\cdot\sum\limits_{n=k}^{\infty}a_{n}= \sum\limits_{n=k}^{\infty}c\cdot a_{n}$ provided that the series converge? I am struggling to find what is wrong about this reasoning when calculating a series that does not start at $n=0$.
For instance, let $S = \sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n$.
Then $\left(\frac{1}{2}\right)^2S=\left(\frac{1}{2}\right)^2\sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n$ (I)
And:
$$\left(\frac{1}{2}\right)^2S=\left(\frac{1}{2}\right)^2\sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n=\sum\limits_{n=2}^{\infty}\left(\left(\frac{1}{2}\right)^n\cdot\left(\frac{1}{2}\right)^2\right)=\sum\limits_{k=0}^{\infty}\left(\frac{1}{2}\right)^k=2$$
But that means (from I):
$$\left(\frac{1}{2}\right)^2S = 2 \Rightarrow \dfrac{\frac{1}{4}}{\frac{1}{4}}S=\dfrac{2}{\frac{1}{4}}\Rightarrow S = 8$$
Which is clearly wrong since:
$$\sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n=\sum\limits_{n=0}^{\infty}\left(\frac{1}{2}\right)^n-\sum\limits_{n=0}^{1}\left(\frac{1}{2}\right)^n=\sum\limits_{n=0}^{\infty}\left(\frac{1}{2}\right)^n - 1 -\frac{1}{2}=\frac{1}{2}$$
I suspect that there is something behind this step $\left(\frac{1}{2}\right)^2\sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n=\sum\limits_{n=2}^{\infty}\left(\left(\frac{1}{2}\right)^n\cdot\left(\frac{1}{2}\right)^2\right)$ that I am missing.
So my question is:
Considering that $\sum\limits_{n=k}^{\infty}a_{n}$ is well defined (converges?), does:
$$c\cdot\sum\limits_{n=k}^{\infty}a_{n}= \sum\limits_{n=k}^{\infty}c\cdot a_{n}$$
And if it does, why does the reasoning presented above is wrong?
|
Your algebra is cooky!
Note that
$$ \left( \frac{1}{2} \right)^2 \sum_{n=2}^\infty \left( \frac{1}{2} \right)^n = \sum_{n=2}^\infty \left( \frac{1}{2} \right)^{n+2} = \sum_{n=0}^\infty \left( \frac{1}{2} \right)^{n + 4}$$
|
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|
(-8)^(4/3) is equals with 16 or (-16)*(-1)^(1/3)? 1. $(-8)^{4/3}=\bigl((-8){^4\bigr)^{1/3}}=4096^{1/3}=16$.
2.
$$
\begin{align*}
(-8)^{4/3} &= (-8)^{1+1/3} \\
&= -8\times(-8)^{1/3} \\
&= -8\times (-1)^{1/3}\times 8^{1/3} \\
&= -2\times 8\times (-1)^{1/3} \\
&= -16\times (-1)^{1/3}.
\end{align*}
$$
So, which is the correct?
|
Both your solutions are correct. To see your first and second solution align note that one solution of $x = (-1)^{1/3}$ is $$x = -1$$
So one possible solution of your original problem is $$-16*(-1)^{1/3} = 16$$
This can be seen by observing that $$(-1)^3 = -1$$
In the complex plane $x= (-8)^{4/3}$ has multiple solutions of the form $-16r_i$ where $r_1,r_2,r_3$ are the complex cube roots of $-1$ which are
\begin{align*}
r_1 &= \frac{1}{2} + \frac{\sqrt{3}}{2}i \\
r_2 &= \frac{1}{2} - \frac{\sqrt{3}}{2}i \\
r_3 &= -1
\end{align*}
|
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|
Find all five solutions of the equation $z^5+z^4+z^3+z^2+z+1 = 0$ $z^5+z^4+z^3+z^2+z+1 = 0$
I can't figure this out can someone offer any suggestions?
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution.
I solved for all roots of $z^4 = -4$ but the structure for this example was more simple.
|
Hint:
Since $$z^4+z^2+1=(z^2+1)^2-z^2=(z^2+z+1)(z^2-z+1)$$
Solve $z^2+z+1=0$ and $z^2-z+1=0$ by using the quadratic formula:
$$z^2+z+1=0\implies z_{1,2}=\frac{-1\pm\sqrt{1-4}}{2}=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$$
$$z^2-z+1=0\implies z_{3,4}=\frac{1\pm\sqrt{1-4}}{2}=\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$$
|
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|
Induction proof I'm having trouble with: $1+x+x^2+x^3+...+x^n = \frac{1-x^{n+1}}{1-x}$ So I'm being asked to use induction to prove that for every $x\in\{a\ |\ a\in R, a\neq 1\}$ and for every $n\in N$
$$
1+x+x^2+x^3+...+x^n = \frac{1-x^{n+1}}{1-x}
$$
I have no trouble proving it for $n=1$ :
$$
1+x = \frac{1-x^2}{1-x}
$$
Factor the polynomial:
$$
1+x = \frac{(1-x)(1+x)}{1-x}
$$
Divide by $(1-x)$
$$
1+x=1+x
$$
And there you have it. The trouble I'm running into is with the induction step. If we assume that our claim is true for $n=k$ then
$$
1+x+x^2+x^3+...+x^k+x^{k+1} = \frac{1-x^{k+2}}{1-x}
$$
Or in other words,
$$
\frac{1-x^{k+1}}{1-x} + x^{k+1} = \frac{1-x^{k+2}}{1-x}
$$
Can someone help with this? I'm having some trouble with the factoring and the book I'm studying from isn't very clear on how they proved the last equation is true.
Thanks in advance :)
|
$$\frac{1-x^{k+1}}{1-x} +x^{k+1}= \frac{1-x^{k+1}}{1-x} + \frac{1-x}{1-x}x^{k+1} = \frac{1-x^{k+1} + (1-x)x^{k+1}}{1-x} = \dots$$
|
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|
Sum of roots of equation system It is necessary to find the number of $x+y+z$ (sum of roots of equation) in the range $\left(\frac{-5\pi}{4}; \frac{13\pi}{2} \right)$:
$$
\left\{
\begin{array}{c}
\sin x+\sin y=2\cos z\\
\sin y + \sin z= 2\cos x \\
\sin z + \sin x = 2 \cos y
\end{array}
\right.
$$
I tried different ways, trying to simplify, but even couldn't find real roots.
|
If you are interested in only one solution, choose $x=y=z$, simplifying to only one equation:
$\sin(x)=\cos(x)$
which is e.g. satisfied for $x=\frac{\pi}{4}$. Your soultion would then be $x+y+z=3x=\frac{3\pi}{4}$.
|
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|
evaluating a cosine function using sine function
If $\sin^2(\frac\pi9)+\sin^2(\frac{2\pi}9)+\sin^2(\frac{3\pi}9)+\sin^2(\frac{4\pi}9)=\frac94$, evaluate $\cos^2(\frac\pi9)+\cos^2(\frac{2\pi}9)+{}$ $\cos^2(\frac{3\pi}9)+\cos^2(\frac{4\pi}9)$.
I know the identity $\sin^2(x)+\cos^2(x)=1$
I am thinking of replacing $\sin^2(\frac{\pi}{9})$ with $1-\cos^2(\frac{\pi}{9})$ and replacing the same identity for each sin.
like this:
$$\Big(1-\cos^2\big(\frac{\pi}{9}\big)\Big) +\Big(1-\cos^2\big(\frac{2\pi}{9}\big)\Big)+\Big(1-\cos^2\big(\frac{3\pi}{9}\big)\Big) +\Big(1-\cos^2\big(\frac{4\pi}{9}\big)\Big) =\frac{9}{4}$$
and then: when moving i will get:
$$\cos^2\big(\frac{\pi}{9}\big)+\cos^2\big(\frac{2\pi}{9}\big)+\cos^2\big(\frac{3\pi}{9}\big)+\cos^2\big(\frac{4\pi}{9}\big)=1+1+1+1+\frac{9}{4}=4+\frac{9}{4}=\frac{25}{4}$$
Does this seem right?
|
No, it's not right. There is an error in the last step, you should subtract $\frac{9}{4}$ as follows
$$\cos^2\left(\frac{\pi}{9}\right)+\cos^2\left(\frac{2\pi}{9}\right)+\cos^2\left(\frac{3\pi}{9}\right)+\cos^2\left(\frac{4\pi}{9}\right)=1+1+1+1-\frac{9}{4}=\frac{16-9}{4}=\color{blue}{\frac{7}{4}}$$
|
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|
Finding Laurent Series With Square Denominator I'm lost. We very, very briefly went over Laurent series in our last class and I have no idea how to deal with them. How might one find:
$f(z)=\frac{1}{(z^2-1)^2}$
Find the Laurent series in the annulus $0 < |z-1| < 2$.
I'm not even really sure how to begin. Is there anyone who can direct me or walk me through it?
Thanks!
|
This illustrates the standard technique in pendantic detail:
$$
\begin{align}
\frac{1}{(z^{2}-1)^{2}}
& =\frac{1}{(z-1)^{2}}\frac{1}{(z+1)^{2}}\\
& =\frac{1}{(z-1)^{2}}\frac{1}{((z-1)+2)^{2}} \\
& =\frac{1}{(z-1)^{2}}\frac{1}{4(1+\frac{z-1}{2})^{2}} \\
& =-\frac{1}{(z-1)^{2}}\frac{1}{2}\frac{d}{dz}\frac{1}{1+\frac{z-1}{2}} \\
& =-\frac{1}{(z-1)^{2}}\frac{1}{2}\frac{d}{dz}\sum_{n=0}^{\infty}(-1)^{n}\left(\frac{z-1}{2}\right)^{n}, \;\;\;\; 0 < |\frac{z-1}{2}| < 1, \\
& =-\frac{1}{(z-1)^{2}}\frac{1}{2^{2}}\sum_{n=1}^{\infty}(-1)^{n}n\left(\frac{z-1}{2}\right)^{n-1} \\
& =\frac{1}{(z-1)^{2}}\frac{1}{2^{2}}\sum_{n=0}^{\infty}(-1)^{n}(n+1)\left(\frac{z-1}{2}\right)^{n} \\
& =\sum_{n=0}^{\infty}\frac{(-1)^{n}(n+1)}{2^{n+2}}(z-1)^{n-2}.
\end{align}
$$
The series converges absolutely for $0 <|z-1|< 2$.
|
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|
Derive a ϴ(1) formula for a Recurrence relation I'm given a piece wise function with sequence $a_0$ $a_1$ etc
$$a_n = \begin{cases}8 & n=0\\-7 & n=1\\25 & n=2\\7a_{(n-2)}+6a_{n-3} & otherwise\end{cases}$$
I'm asked to derive a ϴ(1) formula for $a_n$, by solving the recurrence relation.
I'm still learning about recurrence relations, so I'm wondering how to go about doing this.
Would I first try to find the given sequence for this piece wise function, and then find a formula from that?
|
A generatingfunctionological solution is to define $A(z) = \sum_{n \ge 0} a_n z^n$, shift the recurrence by 3, multiply by $z^n$ and recognize resulting sums:
$\begin{align*}
\sum_{n \ge 0} a_{n + 3} z^n
&= 7 \sum_{n \ge 0} a_{n + 1} z^n + 6 \sum_{n \ge 0} a_n z^n \\
\frac{A(z) - a_0 - a_1 z - a_2 z^2}{z^3}
&= 7 \frac{A(z) - a_0}{z} + 6 A(z)
\end{align*}$
solve for $A(z)$ with the given values for $a_0, a_1, a_2$, as partial fractions:
$\begin{align*}
A(z)
&= \frac{8 - 7 z - 31 z^2}{1 - 7 z^2 - 6 z^3} \\
&= \frac{4}{1 + z} + \frac{3}{1 + 2 z} + \frac{1}{1 - 3 z}
\end{align*}$
As this are just geometric series:
$\begin{align*}
a_n
&= [z^n] A(z) \\
&= 4 \cdot (-1)^n + 3 \cdot (-2)^n + 3^n
\end{align*}$
|
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|
find conditional distribution $ \mathbb{P}\left(\frac{1}{4} \le Y \le \frac{3}{4} \bigm| X = \frac{1}{3}\right) $ $f(x,y) =x+y$, $0<x<1$ ,$0<y<1$
find
$ \mathbb{P}\left(\frac{1}{4} \le Y \le \frac{3}{4} \bigm| X = \frac{1}{3}\right)$
For now,
$$fx(x) = x+\frac{1}{2} $$, $$fy(y) = y+\frac{1}{2} $$
and $$(y \mid x) = \frac{x + y}{x + 1/2}$$
i get the just plug in x=$\frac{1}{3}$
$$\frac{x + y}{1/3 + 1/2}$$
this is what i got so far,
however, how can i solve the Y? should i have to use integral?
|
You are on your way. But please note that the conditional density of $Y$ given that $X=1/3$ is $\frac{1/3+y}{1/3+1/2}$, not $\frac{x+y}{1/3+1/2}$. Now integrate from $1/4$ to $3/4$.
|
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|
Find the complex number $z$ such that it satisfies: 1.$|z+\frac{1}{z}|=\frac{\sqrt{13}}{2} $, 2.$[Im (z)]^2+ [Re(z)]^2=2$ Find the complex number $z$ such that it satisfies: $$\left|z+\frac{1}{z}\right|=\frac{\sqrt{13}}{2} $$$$\Im (z)^2+ \Re(z)^2=2$$$$\frac{\pi}{2}<\arg(z)<\pi$$ then find $$z^{1991}$$
Know I was thinking that this second condition might mean whole part to the power of two, but am unsure about that, just wanted some input on how to do these types of assignments because we having done anything like this before..
|
Assuming $z=-a+bi$ with $a,b\in\mathbb{R}^+$ so $a> 0$ and $b> 0$:
*
*The Absolute value:
$$\left|(-a+bi)+\frac{1}{(-a+bi)}\right|=\left|\frac{1+(-a+bi)^2}{(-a+bi)}\right|=\frac{|1+(-a+bi)^2|}{|(-a+bi)|}=\sqrt{\frac{(1+(-a)^2-b^2)^2+(2(-a)b)^2}{(-a)^2+b^2}}=\sqrt{\frac{(1+a^2-b^2)^2+(2ab)^2}{a^2+b^2}}$$
*The real and imaginary part:
$$\Re\left(-a+bi\right)^2+\Im\left(-a+bi\right)^2=(-a)^2+b^2=a^2+b^2$$
*The argument:
$$\arg\left(-a+bi\right)=\frac{\pi}{2}+\tan^{-1}\left(\frac{a}{b}\right)$$
*
*Solving $z$:
$$
\begin{cases}
\sqrt{\frac{(1+a^2-b^2)^2+(2ab)^2}{a^2+b^2}} = \frac{\sqrt{13}}{2} \\
a^2+b^2 = 2
\end{cases}\Longleftrightarrow
\begin{cases}
a=\frac{\sqrt{\frac{11}{2}}}{2},b=\pm\frac{\sqrt{\frac{5}{2}}}{2} \\
a=-\frac{\sqrt{\frac{11}{2}}}{2},b=\pm\frac{\sqrt{\frac{5}{2}}}{2}
\end{cases}
$$
But because $\arg(z)$ has to be between $\pi$ and $\frac{\pi}{2}$ the solutions that we are looking for are the ones with a positive $b$ (because the imaginary part has to be positive)! So we can use the following solution for $a$ and $b$:
$$a=-\frac{\sqrt{\frac{11}{2}}}{2} \space , \space b=\frac{\sqrt{\frac{5}{2}}}{2}$$
So:
$$z^{1991}=\left(-\frac{\sqrt{\frac{11}{2}}}{2}+\frac{\sqrt{\frac{5}{2}}}{2}i\right)^{1991}\approx -4.66055\cdot10^{299}-8.38854\cdot 10^{298}i$$
|
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|
Proving that $\sum_{a=1}^{b} \frac{a \cdot a! \cdot \binom{b}{a}}{b^a} = b$
Prove that for all positive integers $b$ that $$\sum_{a=1}^{b} \frac{a \cdot a! \cdot \binom{b}{a}}{b^a} = b.$$
My idea is induction, but I cannot figure stuff out on the inductive step.
|
This can be reduced to a telescoping sum.
$$
\begin{align}
\sum_{a=1}^b\frac{a\cdot a!\binom{b}{a}}{b^a}
&=\sum_{a=1}^b\frac{[b-(b-a)]\frac{b!}{(b-a)!}}{b^a}\tag{1}\\
&=\sum_{a=1}^b\frac{\frac{b!}{(b-a)!}}{b^{a-1}}-\sum_{a=1}^{b-1}\frac{\frac{b!}{(b-a-1)!}}{b^a}\tag{2}\\
&=\sum_{a=1}^b\frac{\frac{b!}{(b-a)!}}{b^{a-1}}-\sum_{a=2}^b\frac{\frac{b!}{(b-a)!}}{b^{a-1}}\tag{3}\\
&=\frac{\frac{b!}{(b-1)!}}{b^{1-1}}\tag{4}\\[12pt]
&=b\tag{5}
\end{align}
$$
Explanation:
$(1)$: $a=b-(b-a)$ and $a!\binom{b}{a}=\frac{b!}{(b-a)!}$
$(2)$: left sum: $\frac{b}{b^a}=\frac1{b^{a-1}}$
$\phantom{\text{(2):}}$ right sum: $(b-a)\frac{b!}{(b-a)!}=\frac{b!}{(b-a-1)!}$, removing the $a=b$ term
$(3)$: right sum: substitute $a\mapsto a-1$
$(4)$: all terms cancel except for the $a=1$ term in the left sum
$(5)$: evaluate
|
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|
prove trig equivalence $$\sin x + \sin y = 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$
I want to use the the $e^{ix}$ identities but I'm not sure if it can be done that way, let alone how to do it.
Any tips would be appreciated.
|
From $\sin(A+B)=\sin A\cos B+\cos A\sin B$ and $\sin(A-B)=\sin A\cos B-\cos A\sin B$, get:
$$\require{cancel}\begin{align}
\sin x+\sin y&=\sin\left(\frac{x+y}{2}+\frac{x-y}{2}\right)+\sin\left(\frac{x+y}{2}-\frac{x-y}{2}\right) \\[2ex]
&=\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)+\cancel{\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)}\\&\qquad +\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)-\cancel{\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)} \\[2ex]
&=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)
\end{align}$$
|
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|
suppose p is fermat prime and p>3, show that $\left(\frac{7}{p}\right) = -1$ suppose p is fermat prime, show that $\left(\frac{7}{p}\right) = -1$
We wish to solve $\left(\frac{7}{p}\right) = -1$ for $p$. By quadratic reciprocity and the fact that $7 \equiv 3 \pmod{4}$ we know $\left(\frac{7}{p}\right) = - \left(\frac{p}{7}\right)$.
By a simple calculation, we see that $\left(\frac{7}{p}\right) = -1$ if $p \equiv 1,2,4 \pmod{7}$.
I know that fermat primes are equal to $p \equiv 1 \pmod{4}$ either.
How can i use this information to conclude that $\left(\frac{7}{p}\right) = -1$?
|
Fermat primes are of the form $p=2^{2^{k}}+1$ for some $k\in\Bbb Z_{\ge 0}$. If $k=0$, your claim is wrong; but it's true if $k\ge 1$. I'll assume $k\ge 1$.
We know $p\equiv 1\pmod{4}$, so by Quadratic Reciprocity $\left(\frac{7}{p}\right)=\left(\frac{p}{7}\right)$.
We know $2^{2^k}+1\equiv \{5,3\}\pmod{7}$, none of which is a quadratic residue, so $\left(\frac{p}{7}\right)=-1$.
|
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|
taylor expansion of $\sinh(x)$
I would like to find taylor expansion of $sh(x)$
My thoughts
indeed,
note that : $\sinh(x)=\dfrac{e^{x}-e^{-x}}{2}$ then
\begin{align}
\sinh(x)&=\frac{e^x-e^{-x}}{2} \\
&=\frac{1}{2}\left( e^x-e^{-x} \right)\\
&\underset{x\to 0}=\frac{1}{2}\left(\sum_{k=0}^n\frac{x^k}{k!}+o(x^n)-\sum_{k=0}^n\frac{(-1)^k x^{k}}{k!}-o(x^n)\right) \\
&\underset{x\to 0}=\frac{1}{2}\left(\sum_{k=0}^{n}\left( \frac{x^{k}}{k!}-\frac{(-1)^{k}x^{k}}{k!}\right)+o(x^{n})\right) \\
&\underset{x\to 0}=\frac{1}{2}\left(\sum_{k=0}^{n}\left( \frac{(1+(-1)^{k+1})x^{k}}{k!}\right)+o(x^{n})\right) \\
&\underset{x\to 0}=\frac{1}{2}\left(\sum_{k=0}^{n}\left( \frac{(1+(-1)^{k+1})x^{k}}{k!}\right)+o(x^{n})\right)\\
&\underset{x\to 0}=\begin{cases}\dfrac{1}{2}\left(\sum\limits_{k'=0}^{n}\left( \dfrac{(1+(1)^{2(k'+1)})x^{2k'+1}}{k!}\right)+o(x^{n})\right) & \text{ if } k \text{ odd }\, k=2k'+1 \text{ with } k'\in\mathbb{Z}\\\dfrac{1}{2}\left(\sum\limits_{k=0}^{n}\left( \dfrac{(1+(-1)^{2k'+1})x^{k}}{k!}\right)+o(x^{n})\right) & \text{ if } k \text{ even }\, k=2k' \text{ with } k'\in\mathbb{Z}\end{cases}\\
&\underset{x\to 0}=\begin{cases}\dfrac{1}{2}\left(\sum\limits_{k'=0}^{2n'+1}\left( \dfrac{2x^{2k'+1}}{k!}\right)+o(x^{2n'+1})\right) & \text{ if } k \text{ odd }\, k=2k'+1 \text{ with } k'\in\mathbb{Z}\\\dfrac{1}{2}\left(\sum\limits_{k=0}^{2n'}\left( \dfrac{(1-1)x^{2k'}}{2k'!}\right)+o(x^{2n'})\right) & \text{ if } k \text{ even }\, k=2k' \text{ with } k'\in\mathbb{Z}\end{cases}\\ &\underset{x\to 0}=\begin{cases}\left(\sum\limits_{k'=0}^{2n'+1}\left( \dfrac{x^{2k'+1}}{(2k'+1)!}\right)+o(x^{2n'+1})\right) & \text{ if } k \text{ odd }\, k=2k'+1 \text{ with } k'\in\mathbb{Z}\\ 0 & \text{ if } k \text{ even }\, k=2k' \text{ with } k'\in\mathbb{Z}\end{cases}
\end{align}
Update
*
*but if I want from there
$$\sinh(x)\underset{x\to 0}=\frac{1}{2}\sum_{k'=0}^{E(n/2)} \frac{(1+(-1)^{2k'+1})x^{2k'}}{(2k')!} +
\frac{1}{2}\sum_{k'=0}^{E((n-1)/2)} \frac{(1+(-1)^{2k'+2})x^{2k'+1}}{(2k'+1)!} + o(x^{n}) $$
*
*to get the desired result:
$$\sinh(x)\underset{x\to 0}=\left(\sum_{k=0}^{n}\left(\dfrac{x^{2k+1}}{(2k+1)!}\right)+o(x^{2n+1})\right)$$
indeed,
\begin{align}
sh(x)&\underset{x\to 0}=\frac{1}{2}\sum_{k'=0}^{E(n/2)} \frac{(1+(-1)^{2k'+1})x^{2k'}}{(2k')!} +
\frac{1}{2}\sum_{k'=0}^{E((n-1)/2)} \frac{(1+(-1)^{2k'+2})x^{2k'+1}}{(2k'+1)!} + o(x^{n})\\
&\underset{x\to 0}=0 +
\frac{1}{2}\sum_{k'=0}^{E((n-1)/2)} \frac{2)x^{2k'+1}}{(2k'+1)!} + o(x^{n})\\
&\underset{x\to 0}=
\sum_{k'=0}^{E((n-1)/2)} \frac{x^{2k'+1}}{(2k'+1)!} + o(x^{n})\\
&\underset{x\to 0}=\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\\
&\text{These few passing I would like to know them. }\\
sh(x)&\underset{x\to 0}=\left(\sum_{k=0}^{n}\left(\dfrac{x^{2k+1}}{(2k+1)!}\right)+o(x^{2n+1})\right)
\end{align}
*
*Is my proof correct
*I'm interested in more ways of finding taylor expansion of $\sinh(x)$.
|
How about a rather simple derivation like the one below:
$$\exp(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots $$
and
$$\exp(-x)= 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$
So when you subtract the two equations:
$$\exp(x) - \exp(-x) = 2x + 2\frac{x^3}{3!} + 2\frac{x^5}{5!}$$
Finally,
$$\frac{\exp(x) - \exp(-x)}{2} = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots = \sum_{n=0}^{\infty} \dfrac{x^{2n+1}}{(2n+1)!}$$
Odd powers remain and sine is an odd function.
|
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|
Evaluating the integral $\int (x^2-1)^{\frac{-3}{2}}dx$ I need to solve the differential equation:
$$\displaystyle f'(x) = (x^2-1)^{-\frac{3}{2}}, f(2)=\frac{3-2\sqrt 3}{3}$$
Which basically amounts to solving the integral $\int (x^2-1)^{-\frac{3}{2}} \mathrm{d} x$. I was thinkng of using $\int (x^2-1)^{-\frac{1}{2}} \mathrm{d} x = \operatorname{arcosh} x$ but I can't haven't managed to simplify it. Any ideas?
|
I thought it might be instructive to present an approach that uses integration by parts, rather than hyperbolic trigonometric substitution. To that end, we begin by writing
$$\begin{align}
\frac{1}{(x^2-1)^{3/2}}&=\frac{(1-x^2)+x^2}{(x^2-1)^{3/2}}\\\\
&=\frac{x^2}{(x^2-1)^{3/2}}-\frac{1}{\sqrt{x^2-1}}
\end{align} \tag 1$$
Using $(1)$ permits our writing
$$\int \frac{1}{(x^2-1)^{3/2}}\,dx=\int \frac{x^2}{(x^2-1)^{3/2}}\,dx-\int \frac{1}{\sqrt{x^2-1}} \,dx \tag 2 $$
Integrating by parts the first integral on the right-hand side of $(2)$ yields
$$\int \frac{x^2}{(x^2-1)^{3/2}}\,dx=-x\frac{1}{\sqrt{x^2-1}}+\int \frac{1}{\sqrt{x^2-1}}\,dx +C \tag 3$$
Finally, using $(3)$ in $(2)$ reveals
$$\int \frac{1}{(x^2-1)^{3/2}}\,dx=-x\frac{1}{\sqrt{x^2-1}}+C$$
which agrees with the result reported by @OlivierOloa from using hyperbolic trigonometric substitution!
|
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|
Find a polynomial with integral coefficients whose zeros include $\sqrt{2} + \sqrt{5}$. Find a polynomial with integral coefficients whose zeros include $\sqrt{2} + \sqrt{5}$.
I think I can use $-3= (\sqrt{2} + \sqrt{5})(\sqrt{2} - \sqrt{5})$ and a certain telescopic factorisation. The problem is that I don't know how to continue this problem. Is anyone is able to give me a hint?
|
You may observe that
$$
(\sqrt{2} + \sqrt{5})^2=7+2\sqrt{10}
$$ giving
$$\left((\sqrt{2} + \sqrt{5})^2-7\right)^2= (2\sqrt{10})^2=40.$$ Thus $\sqrt{2} + \sqrt{5}$ is a root of
$$P(X)= (X^2-7)^2-40=X^4-14X^2+9.$$
|
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|
Find : $\sqrt[6]{\frac{\sqrt{2}+(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^7}{(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^{11}}}$ in its algebraic form. Find : $$\sqrt[6]{\frac{\sqrt{2}+(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^7}{(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^{11}}}$$ in its algebraic form.
Now, I kinda think it would not be wise to try to expand this, but rather apply de Moivre formula on the complex number in the numerator and denominator, then simplify that complex number within the root, and once again apply moivres formula.
I have tried but then I get the expression:$\sqrt[6]{\frac{\sqrt{2}+\cos{\frac{21 \pi}{4}+i\sin{\frac{21 \pi }{4}}}}{\cos \frac{11\pi}{4}+i\sin\frac{11 \pi}{4}}}$ and don't know what to do with it.
|
Hint: $$\cos\left(\dfrac{21}{4}\pi\right)=\cos\left(\left(\dfrac{20}{4}+\dfrac{1}{4}\right)\pi\right),\ \,\sin\left(\dfrac{21}{4}\pi\right)=\sin\left(\left(\dfrac{20}{4}+\dfrac{1}{4}\right)\pi\right),\\ \cos\left(\dfrac{11}{4}\pi\right)=\cos\left(\left(\dfrac{10}{4}+\dfrac{1}{4}\right)\pi\right),\ \,\sin\left(\dfrac{11}{4}\pi\right)=\sin\left(\left(\dfrac{10}{4}+\dfrac{1}{4}\right)\pi\right).$$
|
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|
Solving a partial sum.... Hey can anyone help with this? This is the classic NPV equation:
$$\texttt{NPV = -CapEx} + \sum_{i=0}^n \frac{\texttt{Revenue − Costs}}{(1+\texttt{Discount})^i}$$
For my purposes all the elements are know except costs.
I need to isolate costs in this equation. Is this possible?
Thanks,
Mike
|
Your equation is
$NPV = -CapEx + \sum_{i=0}^n \frac{R − C}{(1+d)^i} \quad |+CapEX$
$NPV + CapEx = \sum_{i=0}^n \frac{R − C}{(1+d)^i} \quad $
Splitting the fraction
$NPV + CapEx = \sum_{i=0}^n \left( \frac{R }{(1+d)^i} -\frac{C }{(1+d)^i} \right) \quad $
Factoring out $\frac{1}{(1+d)^i}$
$NPV + CapEx = \sum_{i=0}^n \frac{1}{(1+d)^i}\left( R-C \right) \quad $
$NPV + CapEx = \left( R-C \right)\cdot \sum_{i=0}^n \frac{1}{(1+d)^i} \quad | \cdot (-1)$
$-NPV - CapEx = \left( -R+C \right)\cdot \sum_{i=0}^n \frac{1}{(1+d)^i} \quad $
Dividing the equation by $\sum_{i=0}^n \frac{1}{(1+d)^i}$
$\frac{-NPV - CapEx}{\sum_{i=0}^n \frac{1}{(1+d)^i}}=-R+C$
$C=R-\frac{NPV + CapEx}{\sum_{i=0}^n \frac{1}{(1+d)^i}}$
It is $\sum_{i=0}^n \frac{1}{(1+d)^i}=\sum_{i=0}^n \left( \frac{1}{1+d}\right)^ i$
This is a partial sum of a geometric series.
Therefore $\sum_{i=0}^n \left( \frac{1}{1+d}\right)^ i=\Large{\frac{1-\left( \frac{1}{1+d}\right)^{n+1}}{1-\frac{1}{1+d}}}$
In total it is $C=R-(NPV+CapEx)\cdot \Large{\frac{1- \frac{1}{1+d}}{1-\left(\frac{1}{1+d}\right)^{n+1}}}$
Solve for R-C
$R-C=(NVP+CapEx)\cdot \Large{\frac{\left( 1- \frac{1}{1+d}\right) }{\left(1-\left(\frac{1}{1+d}\right)^{n+1}\right)}}$
$R-C=(1909+1315)\cdot \Large{\frac{\left(1- \frac{1}{1+0.08}\right)}{\left(1-\left(\frac{1}{1+0.08}\right)^{24}\right)}}=\normalsize 283.5268$
|
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|
Write out the Form of the Partial Fraction Decomposition: $\int\frac{x^6}{x^2-4}$ Write out the Form of the Partial Fraction Decomposition: $\int\frac{x^6}{x^2-4}$
The book says use long division my answer was $x^3+\frac{4x^3}{x^2-4}$
The answer manual is $\frac{x^6}{x^2-4}=x^4+4x^2+16+\frac{64}{(x+2)(x-2)}$
|
Your long division is wrong and you can easily check it by doing the computation:
$$
x^3+\frac{4x^3}{x^2-4}=\frac{x^5}{x^2-4}
$$
Moreover, the degree of the remainder should be less than the degree of the denominator.
If the long division is performed correctly, you find
$$
\frac{x^6}{x^2-4}=x^4+4x^2+16+\frac{64}{x^2-4}
$$
but this is not the end of the job, because you must also split the last fraction as
$$
\frac{64}{x^2-4}=\frac{A}{x-2}+\frac{B}{x+2}
$$
that entails
\begin{cases}
A+B=0\\[6px]
2A-2B=64
\end{cases}
that means $A=16$, $B=-16$. Thus the final decomposition is
$$
\frac{x^6}{x^2-4}=x^4+4x^2+16+\frac{16}{x-2}-\frac{16}{x+2}
$$
from which the integral can be easily computed.
Here's the long division:
|
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|
How to compute $\int_0^{+\infty} \frac{dt}{1+t^4} = \frac{\pi}{2\sqrt 2}.$
How to compute
$$\int_0^{+\infty} \frac{dt}{1+t^4} = \frac{\pi}{2\sqrt 2}.$$
I'm interested in more ways of computing this integral.
There is always the straight forward method to decompose into simple elements
$\dfrac{1}{(1 + t^{4})}$ it works but it is tedious. If someone has a faster and clever method, I'm interested :)
Update :
*
*My quesion is different than this question because all the solution that state there is talk about the the straight forward method to decompose into simple elements $\dfrac{1}{(1 + t^{4})}$ as i said before i'm not intersted in that way
|
Here is another method using purely clever substitutions. Note that
$$
\int_0^\infty \frac{\mathrm{d}x}{1+x^4}
=
\int_{0}^\infty \frac{w^2}{1+w^4} \mathrm{d}w
$$
Where the last integral comes from the substitution $w \mapsto 1/x$. Addition now gives that
$$
\begin{align*}
\int_0^\infty \frac{\mathrm{d}x}{1+x^4}
& = \frac{1}{2}\int_0^\infty \frac{1+x^2}{1+x^4}\mathrm{d}x \\
& = \frac{1}{2}\int_0^\infty \frac{1+1/x^2}{x^2+1/x^2} \mathrm{d}x \\
& = \frac{1}{2}\int_0^\infty \frac{(x-1/x)'}{(x-1/x)^2+2} \mathrm{d}x \\
& = \frac{1}{2}\int_{-\infty}^\infty \frac{\mathrm{d}u}{u^2+2}
= \frac{1}{2\sqrt{2}}\int_{-\infty}^\infty \frac{\mathrm{d}y}{1+y^2}
= \frac{\pi}{2\sqrt{2}}
\end{align*}
$$
Where the substitutions $u \mapsto x - 1/x$ and $y \mapsto \sqrt{2} u$ were used.
|
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|
How do I complete a proof for an inequality by using the triangle inequality theorem? Prove that the inequality will hold for every real number, $x$
$$\left| 2+x \right| \le \left| 2x+1 \right| +\left| 1-x \right| $$
Proof: This proof is by case analysis.
Case 1:
1) Let $a=2x+1$ and let $b=1-x$ and $a,b\in\mathbb{R}$
2) Suppose $a+b \ge 0$, then $\left| a+b \right| =a+b$
3) Assume that $a\le \left| a \right|$ and $b\le \left| b \right| $, by adding these two inequalities, we get $a+b\le \left| a \right| +\left| b \right|$
4) Therefore we get that $\left| a+b \right| =a+b\le \left| a \right| +\left| b \right|$
Case 2:
1) Let $a=2x+1$ and let $b=1-x$ and $a,b\in\mathbb{R}$
2) Suppose $a+b < 0$, then $\left| a+b \right| = -a-b$
3) Assume that $-a\le \left| -a \right|$ and $-b\le \left| -b \right| $, by adding these two inequalities, we get $-a-b\le \left| a \right| +\left| b \right|$
4) Therefore we get that $\left| -a-b \right| =a+b\le \left| a \right| +\left| b \right|$
At this point, I have no idea what to do to complete my proof or to even show that my original claim is correct. I would like to be guided in the right direction. I am new to proof writing, so this is quite difficult for me to wrap my head around.
|
By the triangle inequality $|a| + |b| \ge |a + b|$
Let $a = 2x + 1$. Let $b = 1 - x.$
Then $|2x + 1| + |1 - x| \ge |2x + 1 + 1 - x| = |x + 2|$
|
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|
For $n \in \mathbb{N}$, show that $4^n + 10 \times 9^{2n-2}$ is divisible by 7 For $n \in \mathbb{N}$, show that $4^n + 10 \times 9^{2n-2}$ is divisible by 7.
I'm not sure how to do this proof so any help would be appreciated.
|
$4^n+10\cdot 9^{2n-2}=4\cdot 4^{n-1}+(7+3)(9^2)^{n-1}=4\cdot 4^{n-1}+(7+3)(81)^{n-1}$
$=4\cdot 4^{n-1}+7(81)^{n-1}+3(81)^{n-1}=4\cdot 4^{n-1}+7(81)^{n-1}+3(81^{n-1}-4^{n-1}+4^{n-1})$
$=4\cdot 4^{n-1}+7(81)^{n-1}+3(81^{n-1}-4^{n-1})+3\cdot 4^{n-1}=7\cdot 4^{n-1}+7(81)^{n-1}+3(81^{n-1}-4^{n-1})$
$=7\cdot 4^{n-1}+7(81)^{n-1}+3((77+4)^{n-1}-4^{n-1})$
$=7\cdot 4^{n-1}+7(81)^{n-1}+3(\sum_{k=0}^{n-1}{{n-1}\choose k}77^k4^{n-1-k}-4^{n-1})$
$=7\cdot 4^{n-1}+7(81)^{n-1}+3(4^{n-1}+\sum_{k=1}^{n-1}{{n-1}\choose k}77^k4^{n-1-k}-4^{n-1})$
$=7\cdot 4^{n-1}+7(81)^{n-1}+3(\sum_{k=1}^{n-1}{{n-1}\choose k}77^k4^{n-1-k})$
|
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|
Sum of the following series upto n terms:$\sum_{k=1}^n \frac {k}{(k+1)(k+2)} 2^k$ Find the sum of the following series up to n terms:
$\sum_{k=1}^n \frac {k}{(k+1)(k+2)} 2^k$
My attempt:
First,I tried to apply binomial series,but i cannot understand how to apply it.
Then ,I tried as follows:
$\sum_{k=1}^n \frac {k}{(k+1)(k+2)} 2^k=\sum_{k=1}^n \frac{\frac{2k+3}{2}-\frac{3}{2}}{(k+1)(k+2)}2^k=\sum_{k=1}^n\big(\frac{1}{2}-\frac{3}{2(k+1)(k+2)}\big)2^k=\sum_{k=1}^n2^{k-1}-\frac{3\cdot2^{k-1}}{(k+1)(k+2)}$
Now if we find $\sum_{k=1}^n \frac{2^{k-1}}{(k+1)(k+2)}$ then we are done.but I cannot find this.
Am I going right?If not,how should I proceed?
Thanks
|
It is a telescopic series. We have:
$$ \frac{k}{(k+1)(k+2)} = \frac{2}{k+2}-\frac{1}{k+1}$$
hence:
$$ \frac{k}{(k+1)(k+2)}2^k = \frac{2^{k+1}}{k+2}-\frac{2^k}{k+1} $$
and:
$$ \sum_{k=1}^{n}\frac{k}{(k+1)(k+2)}2^k = -1+\frac{2^{n+1}}{n+2}.$$
|
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|
Expression for recurrence relation $a_n$ using exponential generating functions $a_0 = 2$, $a_n = na_{n - 1} - n!$ for $n \geq 1$.
Let $$f(x) = \sum_{n \geq 0}a_n\frac{x^n}{n!}.$$
Multiplying each term in the relation by $\frac{x^n}{n!}$ and summing over values for which the relation is defined and we get:
$$
\begin{align*}
\sum_{n \geq 1} a_n\frac{x^n}{n!} &= \sum_{n \geq 1}na_{n - 1}\frac{x^n}{n!} - \sum_{n \geq 1}n!\frac{x^n}{n!}\\
f(x) - a_0 &= x\sum_{n \geq 1}a_{n - 1}\frac{x^{n - 1}}{(n - 1)!} - \sum_{n \geq 1}n!\frac{x^n}{n!}\\
f(x) - 2 &= x \cdot f(x) - \left(\frac{1}{1 - x} - 0!\right)\\
f(x) - 2 &= x \cdot f(x) - \frac{x}{1 - x}\\
f(x) - x \cdot f(x) &= 2 - \frac{x}{1 - x}\\
f(x)\left[1 - x\right] &= \frac{2 - 3x}{1 - x}\\
f(x) &= \frac{2 - 3x}{(1 - x)^2}
\end{align*}
$$
After this, we're supposed to find the coefficient of $\frac{x^n}{n!}$ to find a concise expression for $a_n$. However, I've hit several dead ends. Basically I don't know how to play with
$$f(x) = \frac{2 - 3x}{(1 - x)^2}$$
to find the coefficient. Do I use convolution? Partial fraction decomposition? I have no clue.
|
I'll take off where you end up. You want:
$\begin{align}
n! [x^n] \frac{2 - 3 x}{(1 - x)^2}
&= n! [x^n] (2 - 3 x) (1 - x)^{-2} \\
&= n! \left(
2 [x^n] (1- x)^{-2} - 3 [x^{n - 1}] (1 - x)^{-2}
\right) \\
&= n! \left(
2 (-1)^n \binom{-2}{n} - 3 (-1)^{n - 1} \binom{-2}{n - 1}
\right) \\
&= n! \left(
2 \binom{n + 2 - 1}{2 - 1} - 3 \binom{n - 1 + 2 - 1}{2 - 1}
\right) \\
&= n! \left(
2 \binom{n + 1}{1} - 3 \binom{n}{1}
\right) \\
&= n! \left(
2 (n + 1) - 3 n
\right) \\
&= - n! (n - 2)
\end{align}$
|
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|
Solve using the limit's definition The question is :
I am stuck here :
$$| \sqrt{x+1} - \sqrt{x} -1 | / \sqrt{x} + 1$$
i know that the numerator is negetive so i must change it in order to delete the absolute value yet i still don't know how to proceed.
Thanks in advance !
|
Let $\epsilon >0$, we are seeking for $\delta >0$, such that , for all $x\geq 0$ with $x>\delta$, we have $$ \Bigg| \frac{\sqrt{x+1}}{\sqrt{x}+1} -1\Bigg| <
\epsilon$$
Indeed, $$ \Bigg| \frac{\sqrt{x+1}}{\sqrt{x}+1} -1\Bigg| = \Bigg| \frac{\sqrt{x+1} -\sqrt{x}-1}{\sqrt{x}+1} \Bigg| $$
Note that , for all $x\geq 0$ we have $\sqrt{x+1} \leq \sqrt{x}+ 1 $ (you can see it by squaring both sides). Thus $$ \Bigg| \frac{\sqrt{x+1} -\sqrt{x}-1}{\sqrt{x}+1} \Bigg|= \frac{\sqrt{x}+1-\sqrt{x+1} }{\sqrt{x}+1} $$
But $ \sqrt{x+1} \geq \sqrt{x}$ since the square root is an increasing function, and so $-\sqrt{x+1} \leq -\sqrt{x} $ and $\frac{1}{\sqrt{x+1}} \leq \frac{1}{\sqrt{x}} $, hence
$$ \Bigg| \frac{\sqrt{x+1}}{\sqrt{x}+1} -1\Bigg| = \frac{\sqrt{x}+1-\sqrt{x+1} }{\sqrt{x}+1} \leq \frac{\sqrt{x}+1-\sqrt{x} }{\sqrt{x}+1}= \frac{1}{\sqrt{x}+1} \leq \frac{1}{\sqrt{x}} < \frac{1}{\sqrt{\delta}} $$
So if we choose $\delta $ such that $ \frac{1}{\sqrt{\delta}} < \epsilon $ , then we are done. So enough to tkae $\delta > \frac{1}{\epsilon ^2} $.
|
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|
Solving an equation by telling the value of $x^2+y^2$. I have a problem solving an equation. The equation is:
$xy+x+y=44$ and $x^2y+xy^2=448$
and we have to tell the value of $x^2+y^2$
First I tried solving this by doing the following:
$xy+x+y=44~\to~x+y=44-xy~\to~x^2+2xy+y^2=44^2-88xy+x^2y^2~\Rightarrow$
$\Rightarrow~x^2+y^2=44^2-90xy+x^2y^2$
But from here I didn't know what to do. Could you help me in solving this equation?
|
Here is a solution using vieta's formula.
Let $s=x+y$ and $p=xy$. Then $s+p=44$ and $sp=448$. Therefore $s$ and $p$ are the roots of the quadratic $z^2-44z+448=0$. Thus it follows that $s=16$ and $p=28$. Hence $$x^2+y^2=(x+y)^2-2xy=s^2-2p=256-56=\boxed{200}.$$
But note that if $s=28$ and $p=16$, then
$$x^2+y^2=(x+y)^2-2xy=784-32=\boxed{752}.$$
|
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|
How to solve this limit without using L'Hospital's Rule? $$
\lim\limits_{x\to{\infty}}\frac{\arctan \frac{3}{x}}{|\arctan \frac{2}{x}|}
$$
Can anybody help me to solve this one ?
I ve done somethig like this but im not sure if it is the correct aproach.
$$
\lim\limits_{x\to{\infty}}\frac{\arctan\frac{3}{x}}{|\arctan \frac{2}{x}|} =
\lim\limits_{x\to{\infty}}\frac{\arctan \frac{3}{x}}{\arctan \frac{2}{x}} =
\lim\limits_{x\to{\infty}} \frac{\arctan \frac{3}{x}}{\arctan \frac{2}{x}} * \frac{\frac{\frac{3}{x}}{\frac{3}{x}}}{\frac{\frac{2}{x}}{\frac{2}{x}}} = \frac{\lim\limits_{x\to{\infty}}\frac{\arctan \frac{3}{x}}{\frac{3}{x}}}{\lim\limits_{x\to{\infty}}\frac{\arctan \frac{2}{x}}{\frac{2}{x}}}*\lim\limits_{x\to{\infty}}\frac{\frac{3}{x}}{\frac{2}{x}}
$$
and then for each limit with arctan i ve substitued 3/x and 2/x by tan y and tan z
$$
\frac{\lim\limits_{y\to{0^+}}\frac{\arctan \tan y}{\tan y}}{\lim\limits_{z\to{0^+}}\frac{\arctan \tan z}{\tan z}}*\lim\limits_{x\to{\infty}}\frac{\frac{3}{x}}{\frac{2}{x}}
$$
and then for each limit which goes to 0^+ i ve done this
$$
\lim\limits_{y\to{0^+}}\frac{\arctan \tan y}{\tan y} = \lim\limits_{y\to{0^+}}\frac{y}{\tan y} = \lim\limits_{y\to{0^+}}\frac{y}{\frac{\sin y}{\cos y}} = 1
$$
so in the end
$$
\frac{1}{1}*\frac{3}{2} = \frac{3}{2}
$$
|
Let $u := \frac{1}{x}$, then $$\lim_{u\to 0}\frac{\arctan 3u}{\arctan 2u} = \frac{3}{2}$$ because $\arctan u\sim u$, when $u\to 0$. As $x\to\infty$, then $|\arctan x|\to \frac{\pi}{2}$: absolute value signs unnecessary.
|
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|
solve $x^4 + y^4 = x^3 + y^3 + 10$ for $x,y \in \mathbb{Z}$ solve $x^4 + y^4 = x^3 + y^3 + 10$ for $x,y \in \mathbb{Z}$.
I tried solving this by trying to find upper bounds for $|x|$ and $|y|$, therefor it is quite useful to write:
$x^4 + y^4 - x^3 - y^3 = .....$
where ... is in the form of a square.
I tried to write $x^4 - x^3 = x^2(x^2+x) = x^2((x+\frac{1}{2})^2 - \frac{1}{4})$, but the $'\frac{1}{4}'$ is kind of troublesome, any tips or hints on how to get a good square, so i can estimate my polynomial?
Kees
|
$x^4-x^3$, and $y^4-y^3$, are never negative and rarely less than $10$. There must be very few possibilities to check.
|
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|
Maximum value of the integral: $\int _{10}^{19} \frac{\sin x}{1+x^a}dx$
Find the minimum odd value of $a$, where$a>1,~ a \in \mathbb{N}$ such that $$\int_{10}^{19} \frac{\sin x}{1+x^a}dx<\frac{1}{9}$$
ATTEMPT:- Let $I(a)=\int _{10}^{19} \frac{\sin x}{1+x^a}dx$ then from Leibnitz's rule, $$I'(a)=-\int _{10}^{19} \frac{(\sin x)x^a\log x}{(1+x^a)^2}dx$$
Now using By Parts, $\int udv=uv-\int vdu$ with
$$\begin{aligned}
u &= \sin x\log x & &\implies & du &= \cos x\log x+\frac{\sin x}{x} \\
dv &=\frac{x^a}{(1+x^a)^2} & &\implies & v &=-\frac{1}{(1+x^a)\log a} \end{aligned} \\
\implies I'(a)=\sin x\log x\frac{1}{(1+x^a)\log a} -\int _{10}^{19} \frac{\cos x\log x+\frac{\sin x}{x}}{(1+x^a)\log a}$$
which doesn't seem solvable.
Next , since it asks for minimum odd value with respect to $a$, I put $a=3$ in the integral giving me:
$$I=\int _{10}^{19} \frac{\sin x}{1+x^3}dx$$
The indefinite integral is in terms of SinIntegral and CosIntegral functions and even the definite integral is given as visual representation of the integral by Wolfram Mathematica
The text says the answer is $3$.
How am I supposed to evaluate this integral?
|
As the question is written, if you can show $\int _{10}^{19} \frac{\sin x}{1+x^3}dx \lt \frac 19$ you are done. But $\int _{10}^{19} \frac{\sin x}{1+x^3}dx \lt \int_{10}^{19}\frac 1{1001}dx=\frac {9}{1001}\lt \frac 19$ No need to evaluate the integral.
|
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|
upper limit of $\cos (x^2)-\cos (x+1)^2$ is $2$ How to prove that upper limit of $\cos (x^2)-\cos (x+1)^2$ is $2$.
Can anyone show proof of this?
|
Let's use the sum to product formula:
$$ \cos a + \cos b = \sin \tfrac{a+b}{2}\cos \tfrac{a-b}{2} $$
Hmm is that right? I never remember the correct combinations of + and - symbols. Let's see:
$$ \frac{1}{2} \left( e^{2\pi i \, \frac{a+b}{2}} - e^{-2\pi i \, \frac{a+b}{2}}\right) \times \frac{1}{2} \left( e^{2\pi i \, \frac{a-b}{2}} + e^{-2\pi i \, \frac{a-b}{2}}\right)$$
If I am lucky we re-group the terms and it comes out correct:
$$ \frac{1}{4} \left( e^{2\pi i \, a} + e^{-2\pi i \, a}\right) +
\frac{1}{4} \left( e^{2\pi i \, b} - e^{-2\pi i \, b}\right)
$$
This is wrong... I have written $\frac{1}{2} \left(\cos a - \sin b \right)$, but you can imagine with the correct choice of signs we can get it to work.
A quick look at SOS math gives:
$$ \boxed{ \cos a + \cos b = 2 \,\cos \frac{a+b}{2} \, \cos \frac{a-b}{2} } $$
even pros get it wrong :-)
Let's try to solve your problem now... Let $a = (x+1)^2$ and $b = x^2$. Then:
$$ \cos (x+1)^2 - \cos x^2 = 2 \,\sin \left(x^2 + x + \frac{1}{2} \right) \, \sin \left( x + \frac{1}{2} \right) \leq 2$$
We know that $\boxed{\cos x \leq 1}$ always. So our number is always less than $2$. In fact, the $\limsup$ is $2$. This boils down to solving for integers:
\begin{eqnarray*}
x + \frac{1}{2}&\approx& \frac{\pi}{2} + 2\pi n_1 \hspace{0.5in}\text{ with }n_1 \in \mathbb{Z}\\
x^2 + x + \frac{1}{2}&\approx& \frac{\pi}{2} + 2\pi n_2 \hspace{0.5in}\text{ with }n_2 \in \mathbb{Z}
\end{eqnarray*}
It may be surprising at first, how these questions about trigonometry turn into questions of diophantine approximation. This system of equation is easy to solve let's substitute one equation into the other
$$ x^2 + \left(x + \frac{1}{2}\right)\approx
x^2 + \left(\frac{\pi}{2} + 2\pi n_1 \right)\approx
\frac{\pi}{2} + 2\pi n_2 $$
This leads to $x \approx \sqrt{2\pi (n_2 - n_1)}= \sqrt{2\pi n_3}$ and yet $x \approx \left(\frac{\pi}{2} - \frac{1}{2} \right) + 2\pi n_1$ where we are looking mod $2\pi$.
Remark I If I choose units correctly we can look mod $1$ and just igonore all the numbers before the decimal point looking only at the "cents" $\pi = \color{red}{3}.14159 \mod 1 \equiv \color{green}{0}.14159$
Remark II Proving that $\sqrt{2\pi n}$ is equidistributed mod $1$ is clear since the numbers are getting closer and closer together
$$ \sqrt{n+1} - \sqrt{ n} \approx \frac{1}{\sqrt{n} + \sqrt{n+1}} \approx \frac{1}{2\sqrt{n}} $$
Using a computer we can plot this function and see that we have a chance at getting the numbers to be close:
Further analysis of the numbers appearing after the decimal point in $\sqrt{n}$ is extremely difficult. Here, two Harvard professors duking it out over the gap distribution [1].
|
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|
Is the numerator of $\sum_{k=0}^n \frac{(-1)^k}{2k+1}\binom{n}{k}$ a power of $2$? I stumbled on something numerically, and was just starting to work on it, but it seemed fun enough to share.
Let $$f(n)=\sum_{k=0}^{n} \frac{(-1)^{k}}{2k+1}\binom{n}{k}$$
It appears, from the first few values, that $f(n)$ always has numerator equal to a power of $2$. Is this true? If so, why?
The first values:
$$\frac{1}{1},
\frac{2}{3},
\frac{8}{15},
\frac{16}{35},
\frac{128}{315},
\frac{256}{693},
\frac{1024}{3003},
\frac{2048}{6435},
\frac{32768}{109395},\\
\frac{65536}{230945},
\frac{262144}{969969},
\frac{524288}{2028117},
\frac{4194304}{16900975}
$$
Alternative ways you can see this value: $$f(n)=\int_0^{1}(1-x^2)^n\,dx=\int_{0}^{\pi/2}\cos^{2n+1} t\,dt=\frac{1}{2^{2n}}\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}\binom{2n+1}{n-k}$$
To summarize some of the comments, the sequence above appears to match:
$$\begin{align}
f(n)&=\frac{(2n)!!}{(2n+1)!!} \\
&= \frac{2}{3}\cdot \frac{4}{5}\cdot \frac{6}{7}\cdots \frac{2n}{2n+1}
\end{align}$$
Thus, if this is correct (and the answer below proves it is,) we have $$f(n)=f(n-1)\cdot \frac{2n}{2n+1}=f(n-1)\left(1-\frac{1}{2n+1}\right).$$
There might be a proof of this recursion using integration by parts for one of the integral forms above.
|
Posting my own answer to my question after a few months because:
*
*This answer is directly related to the source of the question.
*This answer shows the numerator is a power of 2 without finding the explicit closed formula for the value.
This question originally came from taking this question and asking myself the obvious generalization:
Find $g(x)$, a polynomial of degree $2n+1$, such that $g(x)+1$ is divisible by $(x-1)^{n+1}$ and $g(x)-1$ is divisible by $(x+1)^{n+1}$.
One answer is to note that $g'(x)$ must be divisible by $(1-x)^n$ and $(1+x)^n$, hence must be a multiple of $(1-x^2)^{n}$. Find the anti-derivative $G(x)$ of $(1-x^2)^n$ with $G(0)=0$, and then your answer is $g(x)=\frac{1}{G(1)}G(x)$.
Turns out, $G(1)$ is the value in my question above. Indeed, that was the origin of my question. I started computing these values and saw the pattern.
But one new answer to that original question gives a clear reason why there is a power of $2$ in the numerator of $G(1)$. I've adjusted that argument to handle the more general case:
Note that $$\begin{align}2^{2n+1} &= \left((1+x)+(1-x)\right)^{2n+1}\\
&=\sum_{k=0}^{n}\binom{2n+1}{k}\left((1+x)^k(1-x)^{2n+1-k} +(1+x)^{2n+1-k}(1-x)^{k}\right)\\
&=\sum_{j=0}^{n}\binom{2n+1}{n-j}\left((1+x)^{n-j}(1-x)^{n+1+j} + (1+x)^{n+1+j}(1-x)^{n-j}\right)
\end{align}$$
Dividing by $2^{2n}$ and rearranging, we get:
$$1-\frac{1}{2^{2n}}\sum_{j=0}^{n} \binom{2n+1}{n-j}(1+x)^{n+1+j}(1-x)^{n-j} = -1 + \frac{1}{2^{2n}}\sum_{j=0}^{n} \binom{2n+1}{n-j}(1+x)^{n-j}(1-x)^{n+1+j}$$
So if $g(x)=1-\frac{1}{2^{2n}}\sum_{j=0}^{n} \binom{2n+1}{n-j}(1+x)^{n+1+j}(1-x)^{n-j}$, then $g(x)-1$ is divisible by $(1+x)^{n+1}$ and $g(x)+1$ is divisible by $(x-1)^{n+1}$.
Now, the lead coefficient of $g(x)$ is $\frac{(-1)^n}{(2n+1)G(1)}$ in the original solution for $g(x)$, and it is $$\frac{1}{2^{2n}}\sum_{j=0}^{n}\binom{2n+1}{n-j}(-1)^{n-j+1}=\frac{M}{2^{2n}}$$ for an integer $M$ in this new solution.
So we only need to know that there is a unique solution for $g(x)$. That's relatively easy to do: If $g_1,g_2$ are solutions, then $g_1(x)-g_2(x)$ must be divisibly by both $(x-1)^{n+1}$ and $(x+1)^{n+1}$, and hence must be zero since it is of degree at most $2n+1.$
So this shows that $G(1)$ must have a numerator equal to a power of $2$.
Indeed, we get the general formula for $f(n)$ if we prove:
$$M=\sum_{j=0}^n (-1)^{j+1}\binom{2n+1}{j} = (-1)^n\binom{2n}{n}$$
Then we get that $f(n)=\frac{2^{2n}}{(2n+1)\binom{2n}{n}}$.
|
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|
Find the limit $\lim_{x \to 0} (2^x + \sin (3x)) ^{\cot(3x)}$ Please help, I have already tried every thing I can, but nothing works. I have no I idea what to do.
$$\lim_{x \to 0} \; (2^x + \sin (3x)) ^{\cot(3x)}$$
|
While evaluating limit of expressions of type $\{f(x)\}^{g(x)}$ it is best to take logarithms. Let $L$ be the desired limit so that
\begin{align}
\log L &= \log\left(\lim_{x \to 0}(2^{x} + \sin 3x)^{\cot 3x}\right)\notag\\
&= \lim_{x \to 0}\log(2^{x} + \sin 3x)^{\cot 3x}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0}\cot 3x\log(2^{x} + \sin 3x)\notag\\
&= \lim_{x \to 0}\frac{\log(2^{x} + \sin 3x)}{\tan 3x}\notag\\
&= \lim_{x \to 0}\frac{\log(2^{x} + \sin 3x)}{3x}\cdot\frac{3x}{\tan 3x}\notag\\
&= \frac{1}{3}\lim_{x \to 0}\frac{\log(1 + 2^{x} - 1 + \sin 3x)}{x}\notag\\
&= \frac{1}{3}\lim_{x \to 0}\frac{\log(1 + 2^{x} - 1 + \sin 3x)}{2^{x} - 1 + \sin 3x}\cdot\frac{2^{x} - 1 + \sin 3x}{x}\notag\\
&= \frac{1}{3}\lim_{x \to 0}1\cdot\left(\frac{2^{x} - 1}{x} + 3\cdot\frac{\sin 3x}{3x}\right)\notag\\
&= \frac{1}{3}(\log 2 + 3) = 1 + \frac{\log 2}{3}\notag
\end{align}
and hence $L = 2^{1/3}e$.
|
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How to evaluate $\sin^{-1} (\sqrt{2} \sin \theta) + \sin^{-1} (\sqrt{\cos2 \theta})$ $$\sin^{-1} (\sqrt{2} \sin \theta) + \sin^{-1} (\sqrt{\cos2 \theta})$$
to evaluate the above equation, I used the formula: $$\sin^{-1} (x) + \sin^{-1} (y) = \sin^{-1}[x(1-y^2) + y(1-x^2)]$$
therefore I have got,
$$\sin^{-1} (\sqrt{2} \sin \theta \sin2 \theta + \cos \theta )$$
The result is supposed to be 1.
How can I do this?
|
Let $\sin^{-1}(\sqrt2\sin\theta)=x\implies\sqrt2\sin\theta=\sin x$ and $-\dfrac\pi2\le x\le\dfrac\pi2$
Now $\sqrt{\cos2\theta}=\sqrt{1-\sin^2x}=|\cos x|$
As $\cos x\ge0,|\cos x|=+\cos x\implies\sin^{-1}\sqrt{\cos2\theta}=\sin^{-1}(\cos x)=\dfrac\pi2-\cos^{-1}(\cos x)$
Now $\cos^{-1}(\cos x)=\begin{cases} x &\mbox{if } 0\le x\le\dfrac\pi2\iff\sin\theta\ge0\\
-x & \mbox{ if } -\dfrac\pi2\le x<0\end{cases} $
Can you take it from here?
|
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Let z be a complex number such that $z^2 +z + 1/z^2 + 1/z + 1=0 $ If n is a natural number then find the value of $ z^{2012n} + z^{1006n} + 1/z^{2012n} +1/z^{1006n} $ is equal to.
I tried rewriting it as $ t^2+t-1=0 $ where $ t=z+1/z $ and then find roots but I don't know how to use it to get required value.
|
HINT:
As $z\ne0,$ multiply throughout by $z^2$
$$z^4+z^3+z^2+z+1=0\implies z^5-1=(z-1)(z^4+z^3+z^2+z+1)=0\implies z=e^{2\pi m i/5}$$ where $m\equiv1,2,3,4\pmod5$
|
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"timestamp": "2023-03-29T00:00:00",
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|
A new approach to find value of $x^2+\frac{1}{x^2}$ When I was teaching in a college class, I write this question on board.
If we now $\bf{x+\dfrac{1}{x}=4}$ show the value of $\bf{x^2+\dfrac{1}{x^2}=14}$
Some student asks me for a multi idea to show or prove that.
I tried these : $$ $$1 :$$\left(x+\frac{1}{x}\right)^2=4^2\\x^2+\frac{1}{x^2}+2x\times \frac{1}{x}=16\\x^2+\frac{1}{x^2}=16-2 $$
2:solving quadratic equation ,and putting one of roots$$x+\frac{1}{x}=4\\\frac{x^2+1}{x}=4\\x^2+1=4x\\x^2-4x+1=0\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-4)\pm \sqrt{(-4)^2-4(1)(+1)}}{2}=\\x=\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}\\x=2+\sqrt3 \to x^2=4+4\sqrt3+3=7+4\sqrt3\\x^2+\frac{1}{x^2}=7+4\sqrt3+\frac{1}{7+4\sqrt3}=\\7+4\sqrt3+\frac{1}{7+4\sqrt3}\cdot\frac{7-4\sqrt3}{7-4\sqrt3}=\\7+4\sqrt3+\frac{7-4\sqrt3}{49-16\cdot 3}=\\7+4\sqrt3+\frac{7-4\sqrt3}{1}=14$$
3: visual approach .assume side length of a square is $x+\frac{1}{x}=4$
now I am looking for new idea to proof.Any hint will be appreciated.(more visual proof - geometrical - trigonometrical - using complex numbers ...)
|
Some obfuscation using linear algebra:
Write $x + \frac{1}{x} = a$ and let
$$ p(\lambda) = \left( \lambda - x \right) \left( \lambda - \frac{1}{x} \right) = \lambda^2 - a\lambda + 1$$ be a polynomial whose roots are $x$ and $\frac{1}{x}$ and consider the companion matrix
$$ A = \left( \begin{matrix} 0 & -1 \\ 1 & a \end{matrix} \right). $$
The eigenvalues of $A$ are $x$ and $\frac{1}{x}$ so $\mathrm{tr}(A) = x + \frac{1}{x} = a$. The eigenvalues of $A^2$ are $x^2$ and $\frac{1}{x^2}$ so
$$ x^2 + \frac{1}{x^2} = \mathrm{tr}(A^2) = \mathrm{tr} \left( \begin{matrix} -1 & -a \\ a & a^2 - 1 \end{matrix} \right) = a^2 - 2. $$
|
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|
Convert integral to a series I have to find an infinitite series expansion for the integral:
$$\int \frac{x}{8+x^3} \, dx$$
First, I started by determining the Taylor series of the integrand
$$\frac{x}{8+x^3}=\frac{x}{8} \cdot \frac{1}{1-(-(x/2)^3)} = \frac{x}{8} \cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i}$$
Then, I integrate
$$\int \frac{x}{8+x^3} \, dx = -\frac{1}{8} \int x \cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i} \, dx$$
But, I'm not sure how to continue.
Thank you for your help.
|
$$\frac{x}{8} \cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i}=-\frac{1}{4} (-\frac{x}{2})\cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i} \,=-\frac{1}{4}\sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i+1} \,$$
$$\int -\frac{1}{4}\sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i+1} \,dx=\frac{1}{2}\sum_{i=0}^{\infty}\frac{1}{3i+2} \left(-\frac{x}{2}\right)^{3i+2} \,$$
|
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"url": "https://math.stackexchange.com/questions/1541514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Mathematics Olympiad Question $a+b+c=7$, ... Given $a+b+c=7$ and $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} = 0.7$, need to find $\frac{c}{a+b} + \frac{a}{b+c} + \frac{b}{a+c}$.
I have noted that these two differ by a factor of $10$. So I divided the first equation by $10$ and equated the two. But that did not lead me anywhere. I have also tried to multiply one by the other, but the result obtained is identical to just multiplying the second expression by $7$.
|
The trick it's just to add and subtract $1$ from each fraction and then to factorize:
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a+b+c}{b+c}-1+\frac{a+b+c}{c+a}-1+\frac{a+b+c}{a+b}-1=(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})-3=7\cdot 0.7 -3=1.9$$
|
{
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"url": "https://math.stackexchange.com/questions/1542891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Finding the Maximum value.
Maximize $xy^2$ on the ellipse $b^2x^2 +a^2y^2= a^2b^2$
The steps I tried to solve:
$$\nabla f = (y^2,2yx)\lambda\qquad g = (2xb^2,2y^2a^2)\lambda$$
$$y^2= 2xb^2\lambda$$
$$2yx= 2y^2a^2\lambda$$
$$
\left.
\begin{array}{l}
\text{}&y^2= 2xb^2\lambda\\
\text{}&
\end{array}
\right\}
*a^2y
$$
$$
\left.
\begin{array}{l}
\text{}&2yx= 2y^2a^2\lambda\\
\text{}&
\end{array}
\right\}
*b^2x
$$
$$y^3a^2= 2yxa^2b^2 \lambda$$
$$2yx^2b^2 = 2y^2a^2b^2x\lambda$$
$\color{maroon}{\mathbf{Equalize}}$
$$2a^2b^2xy = 2a^2b^2xy^2$$
$$y=1$$
My main problem is which equations does one set each equal to. If anyone knows which one equals the other this will lead me to the right place in finding the solution.
|
First, transform your equation:
$$b^2x^2+a^2y^2=a^2b^2$$
$$a^2y^2=a^2b^2-b^2x^2$$
$$y^2=b^2-\frac{b^2}{a^2}x^2$$
$$xy^2=b^2x-\frac{b^2}{a^2}x^3$$
Next, differentiate and solve to find extrema:
$$b^2-3\frac{b^2}{a^2}x^2=0$$
$$b^2=3\frac{b^2}{a^2}x^2$$
$$a^2=3x^2$$
$$x=\pm\sqrt{\frac{1}{3}}a$$
Finally, evaluate $xy^2$ at $x=+\sqrt{\frac{1}{3}}a$:
$$xy^2=b^2x-\frac{b^2}{a^2}x^3=b^2\frac{a}{\sqrt{3}}-\frac{b^2}{a^2}\frac{a^3}{3\sqrt{3}}=\frac{ab^2}{\sqrt{3}}-\frac{ab^2}{3\sqrt{3}}=\frac{2}{3\sqrt{3}}ab^2$$
That's your maximum. The minimum is at $x=-\sqrt{\frac{1}{3}}a$:
$$xy^2=b^2x-\frac{b^2}{a^2}x^3=-b^2\frac{a}{\sqrt{3}}+\frac{b^2}{a^2}\frac{a^3}{3\sqrt{3}}=-\frac{ab^2}{\sqrt{3}}+\frac{ab^2}{3\sqrt{3}}=-\frac{2}{3\sqrt{3}}ab^2$$
Wolfram Alpha seems to agree.
P.S. Let's also solve the original equation for $y$ at $x=\pm\sqrt{\frac{1}{3}}a$:
$$b^2x^2+a^2y^2=a^2b^2$$
$$b^2\frac{a^2}{3}+a^2y^2=a^2b^2$$
$$\frac{b^2}{3}+y^2=b^2$$
$$y=\pm\sqrt{\frac{2}{3}}b$$
|
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|
Evaluating $\iiint_\Omega z\,\mathrm dx\mathrm dy\mathrm dz$, wrong book solution?
Evaluate $$\iiint_\Omega z\,\mathrm dx\mathrm dy\mathrm dz,$$ where
$$\Omega = \{(x, y, z) \in \mathbb R^3 \mid y \geq 0,\ z \geq 0,\ x^2 + y^2 \leq 4,\ (y - 1)^2 + z^2 \leq 1\}.$$
From the second and last conditions we get a semi-cylinder parallel to the $x$ axis, tangent to it and with radius $1$. The other two conditions give another cylinder, this time parallel to the $z$ axis. Therefore I choose to integrate as follows:
$$\iint_K\int_0^{\sqrt{1 - (y - 1)^2}} z\,\mathrm dz\,\mathrm dx\mathrm dy,$$
where $K$ is the projection on the $xy$ plane of the intersection of the two cylinders. That is, the red region in this graph:
The integral becomes
$$\begin{align}
\iint_K\left[\frac12 z^2\right]_0^{\sqrt{2y - y^2}}\,\mathrm dx\mathrm dy &= \frac12\int_0^1\int_{-\sqrt{4 - y^2}}^{\sqrt{4 - y^2}}(2y - y^2)\mathrm dx\mathrm dy =\\
&= \int_0^1(2y - y^2)\sqrt{4 - y^2}\mathrm dy =\\
&= \frac{16}3 - 2\sqrt3 + \frac{\sqrt3}4 - \frac\pi3 =\\
&= \frac{16}3 - \frac{7\sqrt3}4 - \frac\pi3
\end{align}$$
However, the book's solution is just
$$\frac{16}3 - \pi$$
Is there an error in my solution or in the book?
|
I don't think you interpreted the integral correctly. These are the limits of integration I got.
\begin{align}
-2\leq &x\leq 2\\
0\leq &y\leq \sqrt{ 4-x^2}\\
0\leq &z\leq \sqrt{2y-y^2}
\end{align}
Then the integral is
\begin{gather}
\int_{-2}^2 \int_0^{\sqrt{ 4-x^2}}\int_0^{\sqrt{2y-y^2}}z\,dz\,dy\,dx\\
\int_{-2}^2 \int_0^{\sqrt{ 4-x^2}}\left.\frac{z^2}{2}\right|_0^{\sqrt{2y-y^2}}\,dy\,dx\\
\int_{-2}^2 \int_0^{\sqrt{ 4-x^2}}\frac{2y-y^2}{2}\,dy\,dx\\
\int_{-2}^2 \left.\frac{y^2}{2}-\frac{y^3}{6}\right|_0^{\sqrt{ 4-x^2}}\,dx\\
\int_{-2}^2 \frac{4-x^2}{2}-\frac{\left(4-x^2\right)^\frac{3}{2}}{6}\,dx\\
\int_{-2}^2 \left(4-x^2\right)\frac{3-\sqrt{4-x^2}}{6}\,dx\\
\frac{16}{3}-\pi
\end{gather}
I evaluated the last integral using wolfram alpha. Nevertheless, I believe your error stems from your interpretation of the volume. The cylinders intersect so that you can find the boundaries of $z$ and $y$ directly from the inequalities already given to you. $K$ in your solution should be the entire semicircle above the $x$ axis. If you change your $y$ integration limits to be from $0$ to $2$ you get the correct answer.
|
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|
To find the greatest and least distances of a point on the ellipse $x^2+4y^2=4$ from the straight line $x+y=4$ How to use the method of Lagrange multipliers to find the greatest and least distances of a point on the ellipse $x^2+4y^2=4$ from the straight line $x+y=4$ ?
|
$$x+y=k$$
$$x^2+4y^2=(k-y)^2+4y^2=5y^2-2ky+k^2=4$$
$$5y^2-2ky+k^2-4=0$$
$$D=k^2-5(k^2-4)=20-4k^2\geq0$$
$$k^2\leq5$$
$$-\sqrt5\leq k\leq\sqrt5$$
$$x+y=4 \quad vs.\quad x+y=\sqrt5$$
$$\text{Distance between above two lines }=\frac{4-\sqrt5}{\sqrt2}=2\sqrt2-\frac{\sqrt10}{2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Maclaurin polynomial of tan(x) The method used to find the Maclaurin polynomial of sin(x), cos(x), and $e^x$ requires finding several derivatives of the function. However, you can only take a couple derivatives of tan(x) before it becomes unbearable to calculate.
Is there a relatively easy way to find the Maclaurin polynomial of tan(x)?
I considered using tan(x)=sin(x)/cos(x) somehow, but I couldn't figure out how.
|
Long division of series.
$$ \matrix{ & x + \frac{x^3}{3} + \frac{2 x^5}{15} + \dots
\cr 1 - \frac{x^2}{2} + \frac{x^4}{24} + \ldots & ) \overline{x - \frac{x^3}{6} + \frac{x^5}{120} + \dots}\cr
& x - \frac{x^3}{2} + \frac{x^5}{24} + \dots\cr & --------\cr
&
\frac{x^3}{3} - \frac{x^5}{30} + \dots\cr
& \frac{x^3}{3} - \frac{x^5}{6} + \dots\cr
& ------\cr
&\frac{2 x^5}{15} + \dots \cr
&\frac{2 x^5}{15} + \dots \cr
& ----}$$
|
{
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"url": "https://math.stackexchange.com/questions/1546539",
"timestamp": "2023-03-29T00:00:00",
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|
How to calculate a determinant of a 2x2 symmetry block matrix? I'd like to calculate the determinant of the matrix:
$$
\begin{pmatrix}
-A & B^\star \\
-B & A^\star \\
\end{pmatrix}
$$
$A$, $B$ are $L\times L$ complex matrix.
I know that if $A$ and $B$ are real matrix, the determinant can be easily calculated:
$$
\begin{pmatrix}
-A & B \\
-B & A \\
\end{pmatrix}
=
\begin{pmatrix}
A & B \\
B & A \\
\end{pmatrix}
.
\begin{pmatrix}
-1 & 0 \\
0 & 1 \\
\end{pmatrix}
$$
and
$$
\det
\begin{pmatrix}
A & B \\
B & A \\
\end{pmatrix}
=\det(A-B)\det(A+B)
$$
Is there a similar formula when $A$ and $B$ are complex matrix? Thank you for your help.
|
Hint.
If $\mathbf {A^*}$ is invertible you can use the general result:
$$
\det
\begin{bmatrix}
\mathbf A&\mathbf B\\
\mathbf C&\mathbf D
\end{bmatrix}= \det (\mathbf D) \det(\mathbf A-\mathbf B\mathbf D^{-1}\mathbf C)
$$
that is a consequence of the identity:
$$
\begin{bmatrix}
\mathbf A&\mathbf B\\
\mathbf C&\mathbf D
\end{bmatrix}
\begin{bmatrix}
\mathbf I&\mathbf 0\\
\mathbf{-D^{-1}}\mathbf C&\mathbf D
\end{bmatrix}=
\begin{bmatrix}
\mathbf A -\mathbf B\mathbf {D^{-1}}\mathbf C&\mathbf B\\
\mathbf 0&\mathbf D
\end{bmatrix}=
\begin{bmatrix}
\mathbf I&\mathbf B\\
\mathbf 0&\mathbf D
\end{bmatrix}
\begin{bmatrix}
\mathbf A -\mathbf B\mathbf {D^{-1}}\mathbf C&\mathbf 0\\
\mathbf 0&\mathbf D
\end{bmatrix}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\sum_{n=1}^{50}\arctan\left(\frac{2n}{n^4-n^2+1}\right)$ Find the value of
$$\sum_{n=1}^{50}\arctan\left(\frac{2n}{n^4-n^2+1}\right)$$
$$\frac{2n}{n^4-n^2+1}=\frac{2n}{1-n^2(1-n^2)}$$
I am not able to split it into sum or difference of two $\arctan$s.Please help me.
|
Notice that we can write
$$ \frac{2n}{n^4 - n^2 + 1} = \frac{(n^2+n) - (n^2-n)}{1 + (n^2+n)(n^2-n)}. $$
In view of the addition formula for the tangent, we find that
$$ \arctan \bigg( \frac{x-y}{1+xy} \bigg) = \arctan x - \arctan y $$
for any $x > y > 0$. Thus we have
$$ \arctan\bigg( \frac{2n}{n^4 - n^2 + 1} \bigg) = \arctan(n^2+n) - \arctan(n^2-n). $$
Then you can easily compute the sum by telescoping.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Infinite product equality $\prod_{n=1}^{\infty} \left(1-x^n+x^{2n}\right) = \prod_{n=1}^{\infty} \frac1{1+x^{2n-1}+x^{4n-2}}$ Prove the following equation ($|x|<1$)
$$\prod_{n=1}^{\infty} \left(1-x^n+x^{2n}\right) = \prod_{n=1}^{\infty} \frac1{1+x^{2n-1}+x^{4n-2}}$$
I made this question and I have the following answer but I think it may be incomplete.
If anyone can point out a flaw in my proof or give a better proof then it would be appreciated.
My solution:
$$\begin{align}
&f(x)=\prod_{n=1}^{\infty} \left(1-x^n+x^{2n}\right)\\
&N(p,q)=\{n\in\mathbb N \mid n\ne (2m-1)2^{k-1};m,k\in\mathbb N,2m-1\leq p,k\leq q\}\\
&f(x)(1+x+x^2)=(1+x+x^2)(1-x+x^2)\prod_{n\in N(1,1)}^{\infty} \left(1-x^n+x^{2n}\right)\\
&=(1+x^2+x^4)\prod_{n\in N(1,1)}^{\infty} \left(1-x^n+x^{2n}\right)=(1+x^2+x^4)(1-x^2+x^4)\prod_{n\in N(1,2)}^{\infty} \left(1-x^n+x^{2n}\right)\\
&=(1+x^4+x^8)\prod_{n\in N(1,2)}^{\infty} \left(1-x^n+x^{2n}\right)=(1+x^4+x^8)(1-x^4+x^8)\prod_{n\in N(1,3)}^{\infty} \left(1-x^n+x^{2n}\right)\\
&=(1+x^8+x^{16})\prod_{n\in N(1,3)}^{\infty} \left(1-x^n+x^{2n}\right)=\cdots\\
&=\lim_{k\to\infty}(1+x^{2^k}+x^{2^{k+1}})\prod_{n\in N(1,k)}^{} \left(1-x^n+x^{2n}\right)=\prod_{n\in N(1,\infty)}^{} \left(1-x^n+x^{2n}\right)\\
&\text{Similarly,}\\
&f(x)(1+x+x^2)(1+x^3+x^6)=\prod_{n\in N(3,\infty)}^{} \left(1-x^n+x^{2n}\right)\\
&f(x)(1+x+x^2)(1+x^3+x^6)(1+x^5+x^{10})=\prod_{n\in N(5,\infty)}^{} \left(1-x^n+x^{2n}\right)\\
&\cdots\\
&f(x)\prod_{m=1}^{\infty} \left(1+x^{2m-1}+x^{2(2m-1)}\right)=\lim_{p\to\infty} \prod_{n\in N(p,\infty)}^{} \left(1-x^n+x^{2n}\right)=1
\end{align}$$
(*) Is it obvious that $\{(2m-1)\cdot2^{k-1}\mid m,k\in \mathbb N\}$ is equivalent to $\mathbb N$, or should I also prove it?
Thanks.
|
Here is a different/more streamlined/possibly less formal approach. To prove that $$\prod_{n=1}^{\infty}(1-x^n+x^{2n})(1+x^{2n-1}+x^{4n-2})=1$$
Separate the first type of factors into factors coming from even $n$ and odd $n$:
$$\begin{align}
&\prod_{n=1}^{\infty}(1-x^{2n}+x^{4n})(1-x^{2n-1}+x^{4n-2})(1+x^{2n-1}+x^{4n-2})\\
&=\prod_{n=1}^{\infty}(1-x^{2n}+x^{4n})(1+x^{4n-2}+x^{8n-4})
\end{align}$$
Note this is the same as the product from the first line, with $x$ replaced by $x^2$. You can repeat inductively so that the product has higher and higher powers of $x$. If it is convergent at all, it must converge to $1$.
|
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|
Uniform Random Variable: Correlation and Independence Let X be a uniform random variable defined on the interval $(0,1)$. If $Y = 6X^2−6X+1$, compute the correlation of X and Y . Are X and Y independent? Are X and Y uncorrelated?
So my work is.
$F(X) = \int{1} dx = x$
$E(X) = \frac{a+b}{2} = \frac{1}{2}$
$F(Y) = \int 6X^2-6X+1 dx = 2x^3-3x^2+x$
So is $E(XY) = \int\int f(x)f(y)dxdy$?
|
That's too much work, you would rather do
$$E[XY] = E[X(6X^2-6X+1)].$$
Then use the properties of expectation.
For the covariance, I would proceed as follows:
\begin{align*}
\text{Cov}(X,Y) &= \text{Cov}(X,6X^2-6X+1)\\
&=6\text{Cov}(X,X^2)-6\text{Cov}(X,X)+\text{Cov}(X,1)\\
&=6\left[E[X^3]-E[X]E[X^2]\right]-6\text{Var}(X)+0\\
&=6\left[\int_0^1x\cdot x^3\,dx-\frac{1}{2}\left(\frac{1}{12}+\frac{1}{4}\right)\right]-\frac{1}{2}\\
&=6\left[\frac{1}{5}\cdot 1 -\frac{1}{2}\cdot \frac{4}{12}\right] -\frac{1}{2}\\
&=-\frac{3}{10}.
\end{align*}
|
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|
Finding the angles of $\triangle ABC$ In a $\triangle ABC$, from vertex $C$, the median to $AB$, the angle bisector of $\angle BCA$ and the perpendicular to $AB$ divides angle $\angle BCA$ into four equal parts.
The task is to compute angles in $\triangle ABC$.
Thanks for any help.
|
Applying sine law to triangle ACD, we have $\dfrac {s}{t} = \dfrac {\sin (90 – 3x)}{\sin x} = \dfrac {\cos 3x}{\sin x}$
Applying sine law to triangle BCD, we have $\dfrac {s}{t} = \dfrac {\sin (90 – x)}{\sin 3x} = \dfrac {\cos x}{\sin 3x}$
Then $\sin 3x \cos3x – \sin x \cos x = 0$
By Wolframalpha, we have x = $\dfrac {\pi}{2} - \dfrac {3 \pi}{8} radian = 22.5^0$ (an answer that matches @Lucian 's)
[Note: the trig equation can also be solved as $2 \sin 3x \cos3x = 2\sin x \cos x$. Then, $\sin 6x = \sin 2x$. This means $x = 0$ or $6x = \pi - 2x$; and this further yields $x = \dfrac {\pi}{8}$.]
Result follows.
|
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|
how do I verify that this converges uniformly to $f(x)$? I had to find the fourier series for $f(x)=|x|, -\pi \le x \le \pi$
I got the fourier series as
$$f(x)=\frac{\pi}{2}-\sum_{n=0}^{\infty}\frac{4}{\pi(2n+1)^2}cos((2n+1)x)$$
but now I have to verify that this converges uniformly to $f(x)$ and I have to evaluate the series when $x=0$. I do not know how to show convergence but for evaluating the series at $x=0$ all I have to do is find the first few terms of the sum and substitute $0$ for $x$?
|
You may find p. 4 here useful.
Anyway,
$$|\frac{\pi}{2}-\sum_{n=0}^{\infty}\frac{4}{\pi(2n+1)^2}cos((2n+1)x)|$$
$$\le \frac{\pi}{2}+|\sum_{n=0}^{\infty}\frac{4}{\pi(2n+1)^2}cos((2n+1)x)|$$
$$\le \frac{\pi}{2}+\sum_{n=0}^{\infty}|\frac{4}{\pi(2n+1)^2}cos((2n+1)x)|$$
$$\le \frac{\pi}{2}+\sum_{n=0}^{\infty}\frac{4}{\pi(2n+1)^2}|cos((2n+1)x)|$$
$$\le \frac{\pi}{2}+\sum_{n=0}^{\infty}\frac{4}{\pi(2n+1)^2}$$
$$\le \frac{\pi}{2}+\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}$$
Now
$$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2} < \infty$$
Hence
$$\frac{\pi}{2}-\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2} < \infty$$
$\therefore$, by the Weierstrass M-test,
$$\frac{\pi}{2}-\sum_{n=0}^{\infty}\frac{4}{\pi(2n+1)^2}cos((2n+1)x)$$
converges uniformly QED
|
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|
If three distinct integers are chosen at random, show that there will exist two among them, say $a$ and $b$, such that $30 | (a^3b-ab^3)$ Problem:
If three distinct integers are chosen at random, show that there will exist two among them, say $a$ and $b$, such that $30 | (a^3b-ab^3)$
My work:
$a^3b-ab^3=ab(a+b) (a-b)$ and if $30 | (a^3b-ab^3)$, then each of $2,3,5$ divides $(a^3b-ab^3)$ since $2\cdot3\cdot5=30$ and $(2,3,5)=1$
Case for 2:
If 2 divides either $a$ or $b$, then all good.
If not, then $a$ and $b$ are both odd and so $a+b$=sum of odds = even is divisible by 2. Done.
Case for 3:
If 3 divides a or b, good.
Else, let:
*
*$a=3k+1$ and $b=3k+1$. Then, $3|(a-b)$
*$a=3k+1$ and $b=3k+2$. Then, $3|(a+b)$
*$a=3k+2$ and $b=3k+2$. Then, $3|(a-b)$
*$a=3k+2$ and $b=3k+1$. Then, $3|(a+b)$
So far, so good.
Case for 5:
Completely stuck here. Five possible remainders ($0,1,2,3,4$) seem to be impossible to do. Any hints?
Questions:
*
*Need hints for last case.
*Why do we need to pick three integers when we only need to check with two?
|
You can remove "distinct" to make it slightly stronger (but it's a trivial case). Notice $30=2\cdot 3\cdot 5$ and $2,3,5$ are primes. Your problem is: Among any three integers $a,b,c$, exist at least two $x,y\in\{a,b,c\}$ such that $2,3,5\mid xy(x+y)(x-y)$.
By Pigeonhole principle, either $abc\equiv 0\pmod{5}$ or $a\equiv \pm b\pmod{5}$ or $a\equiv \pm c\pmod{5}$ (because $a,b,c$ are three integers that belong to $0,1,2,-2,-1$ mod $5$), so exist $x,y\in\{a,b,c\}$ such that $5\mid xy(x+y)(x-y)$.
But then by Pigeonhole principle either $xy\equiv 0\pmod{3}$ or $x\equiv \pm y\pmod{3}$. Similarly, either $xy\equiv 0\pmod{2}$ or $x\equiv \pm y\pmod{2}$.
One generalization can be: Among any $k\ge 3$ integers $a_1,a_2,\ldots, a_k$ exist $x,y\in\{a_1,a_2,\ldots,a_k\}$ such that $6h\mid xy(x+y)(x-y)$, where $h\ge 5$ is any integer such that $\gcd(h,6)=1, \frac{h-1}{2}<k$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
The coefficient of $x^3$ in $(1+x)^3 \cdot (2+x^2)^{10}$ Find the coefficient of $x^3$ in the expansion $(1+x)^3 \cdot (2+x^2)^{10}$.
I did the first part, which is expanding the second equation at $x^3$ and I got: $\binom {10} 3 \cdot 2^7 \cdot (x^2)^3 = 15360 (x^2)^3$, but I can't figure out what to do from here.
|
It's also convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in an algebraic expression.
We obtain
\begin{align*}
[x^3](1+x)^3(2+x^2)^{10}&=[x^3]\left(\sum_{k=0}^{10}\binom{10}{k}x^{2k}2^{10-k}\right)(1+x)^3\\
&=[x^3]\left(\binom{10}{0}2^{10}x^0+\binom{10}{1}2^9x^2\right)(1+x)^3\tag{1}\\
&=\left(2^{10}[x^3]+10\cdot2^9[x^1]\right)\sum_{k=0}^{3}\binom{3}{x}x^k\tag{2}\\
&=\left(2^{10}\binom{3}{3}+10\cdot2^9\binom{3}{1}\right)\\
&=2^{10}+30\cdot 2^9\\
&=16384
\end{align*}
Comment:
*
*In (1) we ignore all summands in the first sum with powers of $x$ greater than 3
*In (2) we use the rule $[x^{k+l}]f(x)=[x^k]x^{-l}f(x), k,l\geq 0$ and the linearity of the coefficient of operator.
|
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|
Solve for $x$, correct to two significant figures, the equation: $4^{x}-2^{x+1}-3=0$ Solve for $x$, correct to two significant figures, the equation:
$$4^{x}-2^{x+1}-3=0$$
My answer: $x\log4=\log3+(x+1)\log2 \Rightarrow 0.602x-0.301x=0.477+0.301 \Rightarrow x = 2.6$ (Conflicting with book answer)
Answer in book: $x=1.6$
|
$f(x) = 4^x - 2^{x+1} - 3 = (2^x)^2 - 2^{x+1} - 3 = (2^x)^2 - 2(2^x) - 3$
Let $y = 2^x \Rightarrow f(x) = y^2 - 2y - 3$
$f(x) = 0 \Rightarrow y^2 - 2y - 3 = 0 \Rightarrow (y-3)(y+1) = 0$
The roots for this equation are $y = 3$ or $y = -1$. Since $y = 2^x$, the solutions in terms of $x$ are $2^x = 3$ or $2^x = -1$ Therefore, the solutions are either $\log _2 \left( 3 \right) $ or $\log _2 \left( -1 \right) $. The latter is undefined, so your answer is $\log _2 \left( 3 \right) $ which is equal to $1.6$ to 2 significant figures. Therefore, $x = 1.6$ and the answer in the book is correct.
|
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|
Sum of an arithmetic progression Hi I cannot for the life of me remember how to use the arithmetic progression formula can someone help?
I just need to find it for this sequence:
$$5n^2 + (5n^2 - 1) + (5n^2 - 2) + \ldots + (5n^2 - n)$$
|
We have
\begin{align*}5n^2 + (5n^2 - 1) + (5n^2 - 2) + \ldots + (5n^2 - n)&= \overbrace{5n^2+5n^2+\ldots+5n^2}^{(n+1) \text{ times}}-(1+2+\ldots+n)\\
&= 5n^2(n+1)-(1+2+\ldots+n)\end{align*}
Now, if $S=1+2+\ldots+n$, then
\begin{align*}2S &= (1+2+\ldots+n)+(1+2+\ldots+n)=(n+1)+(n-1+2)+\ldots+\big(n-(n-1)+n\big)\\ &=n(n+1)\end{align*}
So that $S=n(n+1)/2$ and thus
$$ 5n^2 + (5n^2 - 1) + (5n^2 - 2) + \ldots + (5n^2 - n)= 5n^2(n+1)-\frac{n(n+1)}{2}=\frac{n (n+1) (10 n-1)}{2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Let z = 1 + i. Find the real and imaginary parts of z^19 Let z = 1 + i. Find the real and imaginary parts of $z^{19}$
|
$$\left(1+i\right)^{19}=\left(|1+i|e^{\arg(1+i)i}\right)^{19}=$$
$$\left(\sqrt{2}e^{\frac{\pi i}{4}}\right)^{19}=\sqrt{2^{19}}e^{\frac{19\pi i}{4}}=\sqrt{524288}e^{\frac{19\pi i}{4}}=512\sqrt{2}e^{\frac{19\pi i}{4}}$$
So:
$$\Re\left(512\sqrt{2}e^{\frac{19\pi i}{4}}\right)=512\sqrt{2}\cos\left(\frac{19\pi}{4}\right)=512\sqrt{2}\cdot -\frac{1}{\sqrt{2}}=-512$$
$$\Im\left(512\sqrt{2}e^{\frac{19\pi i}{4}}\right)=512\sqrt{2}\sin\left(\frac{19\pi}{4}\right)=512\sqrt{2}\cdot\frac{1}{\sqrt{2}}=512$$
|
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|
How does this determinant calculation work? Given that $a_0, a_1,...,a_{n-1} \in \mathbb{C}$ I am trying to understand how the following calculation for the determinant of the following matrix follows:
$$
\text{det}
\begin{bmatrix}
x & 0 & 0 & ... & 0 & a_0 \\
-1 & x & 0 & ... & 0 & a_1 \\
0 & -1 & x & ... & 0 & a_2 \\
. \\
. \\
. \\
0 & 0 & 0 & ... & -1 & x + a_{n-1} \\
\end{bmatrix}
\\ =
(x) \text{ det}
\begin{bmatrix}
x & 0 & 0 & ... & 0 & a_1 \\
-1 & x & 0 & ... & 0 & a_2 \\
0 & -1 & x & ... & 0 & a_3 \\
. \\
. \\
. \\
0 & 0 & 0 & ... & -1 & x + a_{n-1} \\
\end{bmatrix}
+
\text{det}
\begin{bmatrix}
0 & 0 & ... & 0 & a_0 \\
-1 & x & ... & 0 & a_2 \\
. \\
. \\
. \\
0 & 0 & ... & -1 & x + a_{n-1} \\
\end{bmatrix}
\\ = x(x^{n-1} + a_{n-1}x^{n-2}+...+a_1) + (-1)^{n-1}\text{det}
\begin{bmatrix}
-1 & x & ... & 0 & a_2 \\
. \\
. \\
. \\
0 & 0 & ... & -1 & x + a_{n-1} \\
0 & 0 & ... & 0 & a_0 \\
\end{bmatrix}
$$
I do not understand:
(1) how the determinant can be broken up into the sum of the determinants of the 2 smaller matrices and (2) how are the determinants of the 2 smaller matrices what they are?
|
If $A=(a_{i,j}) \in M_n(\mathbb{C})$, $$\det(A) = \sum_{k=1}^{n} (-1)^ka_{1,k} \det(\Delta_{1,k}) $$
Where $\Delta_{1,k}$ is A minus the column and the line of $a_{1,k}$
Here the sum only has 2 terms because others are equals to $0$.
|
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|
Roots of: $2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$ This is maybe a stupid question, but I want to find the roots of:
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$
What that I did:
$$\underbrace{2(x+2)(x-1)(x-1)(x-1)}_{A}-\underbrace{3(x-1)(x-1)(x+2)(x+2)}_{B}=0$$
So the roots are when $A$ and $B$ are both zeros when $x=1$ and $x=-2$
My questions:
$1)$ Is there an easy way to see that $x=-8$ is a root too?
$2)$ The degree of this polynomial is $4$, so I should have $4$ roots, and here I have only $3$
|
Well
$2(x+2)(x-1)^{3}-3(x-1)^{2}(x+2)^{2}=0 \Rightarrow (x-1)^{2}(x+2)[2(x-1)-3(x+2)]=0 \Rightarrow (x-1)^{2}(x+2)(-x-8)=0 $
From here it should be clear why $x=-8$ is also a root.
This is called factoring.
|
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|
Algebraic way to see why only $n=3$ is a valid coefficient I'm a bit of a sucker for brute force calculations. Say I want to calculate a coefficient with Fourier theory, in my case
\begin{align*}
a_n = \int_0^1 \sin (3\pi x) \cos (n\pi x) dx.
\end{align*}
Clearly, only $a_3$ is nonzero. But if I calculate this integral explicitly, I get
\begin{align*}
a_n = \frac{3(-1)^{n+1} - 3}{\pi(n^2-9)}.
\end{align*}
I'm not immediately seeing that only $n=3$ will yield a nonzero result here... So how can I evaluate the integral so that only $a_3$ remains?
EDIT: Some extra notes on my calculation. We have
\begin{align*}
I = \int \sin(3\pi x)\cos(n\pi x) dx = \int u \frac{dv}{dx} dx,
\end{align*}
where
\begin{align*}
u = \sin(3\pi x) \Rightarrow \frac{du}{dx} = 3\pi \cos(3\pi x),\\
\frac{dv}{dx} = \cos(n\pi x) \Rightarrow v = \frac1{n\pi}\sin(n\pi x),
\end{align*}
such that
\begin{align*}
I &= \sin(3\pi x) \cdot \frac1{n\pi}\sin(n\pi x) - \int 3\pi \cos(3\pi x) \cdot \frac1{n\pi}\sin(n\pi x) dx \\
&= \frac1{n\pi}\sin(3\pi x)\sin(n\pi x) - \frac 3n\int \cos(3\pi x)\sin(n\pi x) dx \\
&= \frac1{n\pi}\sin(3\pi x)\sin(n\pi x) - \frac 3n J.
\end{align*}
Now repeat the calculation with $J$,
\begin{align*}
J = \int \cos(3\pi x)\sin(n\pi x) dx
\end{align*}
where
\begin{align*}
u = \cos(3\pi x) \Rightarrow \frac{du}{dx} = -3\pi \sin(3\pi x),\\
\frac{dv}{dx} = \sin(n\pi x) \Rightarrow v = -\frac1{n\pi}\cos(n\pi x),
\end{align*}
such that
\begin{align*}
J &= \cos(3\pi x) \cdot \left(-\frac1{n\pi}\cos(n\pi x)\right) - \int 3\pi \sin(3\pi x) \cdot \frac1{n\pi}\cos(n\pi x) dx \\
& = -\frac1{n\pi}\cos(3\pi x) \cos(n\pi x) - \frac3n \int \sin(3\pi x) \cos(n\pi x) dx \\
& = -\frac1{n\pi}\cos(3\pi x) \cos(n\pi x) - \frac3n I
\end{align*}
Plugging in $J$ in the expression for $I$, we get
\begin{align*}
I &= \frac1{n\pi}\sin(3\pi x)\sin(n\pi x) - \frac 3n J \\
&= \frac1{n\pi}\sin(3\pi x)\sin(n\pi x) - \frac 3n \left[-\frac1{n\pi}\cos(3\pi x) \cos(n\pi x) - \frac3n I\right] \\
&= \frac1{n\pi}\sin(3\pi x)\sin(n\pi x) + \frac3{n^2\pi}\cos(3\pi x) \cos(n\pi x) + \frac9{n^2} I \\
I &= \frac{\frac1{n\pi}\sin(3\pi x)\sin(n\pi x) + \frac3{n^2\pi}\cos(3\pi x) \cos(n\pi x)}{1-\frac9{n^2}} \\
&= \frac1\pi \frac{n\sin(3\pi x)\sin(n\pi x) + 3\cos(3\pi x) \cos(n\pi x)}{n^2-9}.
\end{align*}
On the interval $0 < x < 1$, we then get
\begin{align*}
I \Big|_{0}^1 &= \frac1\pi \frac{n\sin(3\pi)\sin(n\pi) + 3\cos(3\pi) \cos(n\pi)}{n^2-9} - \frac1\pi \frac{n\sin(0)\sin(0) + 3\cos(0) \cos(0)}{n^2-9}\\
&= \frac1\pi \frac{3\cos(3\pi) \cos(n\pi)}{n^2-9} - \frac1\pi \frac{3}{n^2-9} \\
&= \frac1\pi \frac{-3(-1)^n - 3}{n^2-9},
\end{align*}
where in the last step I used $\cos(n\pi) = (-1)^n$.
|
Considering $$a_n = \int_0^1 \sin (3\pi x) \cos (n\pi x)\, dx$$ we can make the calculations faster using $$\sin(a) \cos(b)=\frac 12\big(\sin(a+b)+\sin(a-b)\big)$$ So $$2a_n=\int_0^1 \sin\big((3+n)\pi x\big)\, dx+\int_0^1 \sin\big((3-n)\pi x\big)\, dx$$ which already shows that, for $n=3$, the second integral disappears.
So $$a_3=\frac{1+\cos (3\pi )}{2 \pi (n+3)}=0$$ For the other cases $(n\neq 3)$, after simplifications, $$a_n=-\frac{3 (1+\cos (\pi n))}{\pi \left(n^2-9\right)}$$ which implies $$a_{2n+1}=0 \quad \quad \quad a_{2n}=-\frac{6}{\pi \left(4 n^2-9\right)}$$
|
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|
Sum of n terms of the series $\frac{1}{1 \cdot 3}+\frac{2}{1 \cdot 3 \cdot5}+\frac{3}{1 \cdot 3 \cdot 5 \cdot 7}+\cdots$ I need to find the sum of n terms of the series
$$\frac{1}{1\cdot3}+\frac{2}{1\cdot 3\cdot 5}+\frac{3}{1\cdot 3\cdot 5\cdot 7}+\cdots$$
And I've no idea how to move on. It doesn't look like an arithmetic progression or a geometric progression. As far as I can tell it's not telescoping. What do I do?
|
It is telescoping. Consider that:
$$ \frac{1}{1\cdot 3} = \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right),\quad \frac{2}{1\cdot 3\cdot 5} = \frac{1}{2}\left(\frac{1}{1\cdot 3}-\frac{1}{3\cdot 5}\right), $$
$$ \frac{3}{1\cdot 3\cdot 5\cdot 7}=\frac{1}{2}\left(\frac{1}{1\cdot 3\cdot 5}-\frac{1}{3\cdot 5\cdot 7}\right),\quad \ldots$$
so:
$$ \sum_{k=1}^{n}\frac{k}{(2k+1)!!} = \frac{1}{2}\left(1-\frac{1}{(2n+1)!!}\right). $$
As usual, $(2k+1)!!$ stands for $1\cdot 3\cdot 5\cdot\ldots\cdot (2k+1)$.
|
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|
Trigonometric equation $\sin x+1=\cos x$ $$\sin x+1=\cos x,\quad x\in[-\pi,\pi]$$
How do you solve by squaring both sides? the solution is $x\in\{-\pi/2,0\}$ so the solutions $\pi$ and $-\pi$ are inadmissible, I do not understand how by subbing $-\pi$ back into both sides of the equations makes them unequal, and the same for positive $\pi$. Which equation are you subbing $\pi$ into to check, the original?
|
This is an instance of a “linear equation in sine and cosine”. There are several methods for solving them.
First method.
Write $X=\cos x$, $Y=\sin x$ and consider the system
$$
\begin{cases}
1+Y=X\\
X^2+Y^2=1
\end{cases}
$$
Substitute in the second equation to get
$$
1+2Y+Y^2+Y^2=1
$$
which gives
$$
Y^2+Y=0
$$
so $Y=0$ or $Y=-1$. Thus we get the two solutions
$$
\begin{cases}
X=1\\
Y=0
\end{cases}
\qquad\text{or}\qquad
\begin{cases}
X=0\\
Y=-1
\end{cases}
$$
Solved with respect to $x\in[-\pi,\pi]$, they give $x=0$ or $x=-\pi/2$.
Second method
Rewrite the equation as
$$
\cos x-\sin x=1
$$
and try to rewrite this as $A(\cos x\cos\phi-\sin x\sin\phi)=1$, with $A>0$. Thus we need
$$
A\cos\phi=1,\qquad A\sin\phi=1
$$
so $A^2\cos^2\phi+A^2\sin^2\phi=2$, or $A^2=2$; thus $A=\sqrt{2}$ and
$$
\cos\phi=\frac{1}{\sqrt{2}},\quad\sin\phi=\frac{1}{\sqrt{2}}
$$
that is, $\phi=\pi/4$. Thus the equation becomes
$$
\sqrt{2}\cos\left(x+\frac{\pi}{4}\right)=1
$$
that means
$$
x+\frac{\pi}{4}=\frac{\pi}{4}+2k\pi
\qquad\text{or}
x+\frac{\pi}{4}=-\frac{\pi}{4}+2k\pi
$$
and we get again $x=0$ or $x=-\pi/2$.
Third method
Set $t=\tan(x/2)$ and recall that
$$
\cos x=\frac{1-t^2}{1+t^2},\qquad
\sin x=\frac{2t}{1+t^2}
$$
that transforms the equation into
$$
1+\frac{2t}{1+t^2}=\frac{1-t^2}{1+t^2}
$$
that becomes
$$
1+t^2+2t=1-t^2
$$
or
$$
t^2+t=0
$$
so $t=0$ or $t=-1$. The first solution corresponds to
$$
\frac{x}{2}=k\pi \to x=2k\pi
$$
and the second solution corresponds to
$$
\frac{x}{2}=-\frac{\pi}{4}+k\pi \to x=-\frac{\pi}{2}+2k\pi
$$
and, again, in the given interval we have $x=0$ or $x=-\pi/2$.
You seem to have misunderstood what $x\in[-\pi,\pi]$ means. It means that you have to find all solutions $x$ such that
$$
-\pi\le x\le \pi
$$
|
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|
evaluate $\lim _{x\to \infty} (3x^2-x^3)^{\frac{1}{3}}+x$
$$\lim _{x\to \infty} (3x^2-x^3)^{\frac{1}{3}}+x$$
can I look at $\lim\limits_{x\to \infty} (3^{\frac{1}{3}}x^{\frac{2}{3}}-x+x)$?
|
By factoring, the expression becomes $(x^2(3-x))^{1/3}+x=x^{2/3}(3-x)^{1/3}+x$. Now we take a seemingly random detour and multiply the numerator and denominator by $1/x$, giving $$\lim_{x \to \infty}\frac{x^{-1/3}(3-x)^{1/3}+1}{1/x}= \lim_{x \to \infty}\frac{x^{(-1){1/3}}(3-x)^{1/3}+1}{1/x}=\lim_{x \to \infty}\frac{(\frac{3-x}{x})^{1/3}+1}{1/x}$$
Taking x $\rightarrow$ $\infty$ gives the indeterminate form $\frac{0}{0}$.
Applying L'Hopital's Rule gives
$$\lim_{x \to \infty}\frac{\frac{1}{3}(\frac{3-x}{x})^{-2/3}(\frac{-3}{x^2})}{-1/x^2}.$$
After cancelling like terms, we find
$$\lim_{x \to \infty} (\frac{3}{x}-1)^{2/3}=((-1)^2)^{-1/3}=1.$$
|
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|
Finding a particular angle from a triangle
The way I have seen a solution is very lengthy to find the value of x. Is there any easy way to get the value of x? I have tried different formula and didn't get any shortcut solution.
|
Place point $E$ on $\overline{AD}$ such that $\angle CED = 60^{\circ}$.
Now, by the Law of Sines in $\triangle ABD$,
$$\frac{\sin 15^{\circ}}{1}=\frac{\sin 45^{\circ}}{\overline{AD}}$$
$$\therefore \overline{AD} = 1 + \sqrt{3}$$
Since $\triangle CDE$ is equilateral, $\overline{DE} = 2$.
$$\therefore \overline{AE} = \overline{AD} - \overline{DE} = (1 + \sqrt{3}) - 2 = \sqrt{3} - 1$$
Now notice:
$$\frac{\overline{AE}}{\overline{EC}} = \frac{\overline{BD}}{\overline{DA}}$$
Because:
$$\frac{\sqrt{3} - 1}{2} = \frac{1}{\sqrt{3} + 1}$$
Since that's true, and since $\angle BDA = \angle AEC = 120^{\circ}$, $\therefore \triangle ABD \sim \triangle CAE$.
The rest falls into place.
$\angle CAE = \angle ABD = 45^{\circ}$, so $\angle CAB = 15^{\circ} + 45^{\circ} = 60^{\circ}$, meaning $\angle ACB = \angle x = \color{red}{75^{\circ}}$.
|
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|
Evaluate the Integral: $\int\frac{dx}{\sqrt{x^2+16}}$ I want to evaluate $\int\frac{dx}{\sqrt{x^2+16}}$.
My answer is: $\ln \left| \frac{4+x}{4}+\frac{x}{4} \right|+C$
My work is attached:
|
Notice, your mistake $\sqrt{\text{(Hyp)}^2}=\sqrt{16+x^2}\ne 4+x$
you should have $$\tan\theta=\frac{x}{4}\ \ \text{&} \ \ \sec\theta=\frac{\text{Hyp}}{\text{base}}=\frac{\sqrt{16+x^2}}{4}$$
hence substituting the values of $\sec\theta$ & $\tan\theta$, one should get
$$\ln|\sec\theta+\tan\theta|+C=\ln\left|\frac{\sqrt{16+x^2}}{4}+\frac{x}{4}\right|+C$$
$$=\ln\left|x+\sqrt{16+x^2}\right|-\ln (4)+C=\ln\left|x+\sqrt{16+x^2}\right|+c$$
|
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|
Divisibility Of $(2^{32} +1)$ Here is my problem:
If $2^{32} +1 $ is completely divisible by a whole number. Which of the following numbers is completely divisible by that number :
(A)($2^{16}+1$)
(B)($2^{16}-1$)
(C)$7*2^{23}$
(D)($2^{96}+1$)
I don't know where to start any help is appreciated. Thank you
|
Hint 1: We want to find an option that is divisible by $2^{32} + 1$. Indeed, this is necessary and sufficient: it is necessary since we can choose $2^{32} + 1$ as the factor that divides $2^{32} + 1$ and it is sufficient since if a number is divisible by $r = de$, then it is divisible by $d$ and $e$ as well.
Hint 2: $a^3 + 1 = (a + 1)(a^2 - a + 1)$.
These hints tell us that (d) is the right answer, since $$2^{96} + 1 = (2^{32})^3 + 1 = (2^{32} + 1)(2^{64} - 2^{32} + 1),$$ so $2^{32}+1 \mid 2^{96} + 1$.
We can also rule out the others if we are not given that there are multiple answers: (a) (b) and (c) are all less than $2^{32} + 1$, so $2^{32} + 1$ cannot divide all of them evenly. We may even show that they share no common factors with $2^{32} + 1$ at all.
(a) and (b) are out since their product is $2^{32} - 1$. The only common divisor of $2^{32} - 1$ and $2^{32} + 1$ could be $2$ since their difference is $2$, but they're both odd so $\gcd (2^{32} + 1, 2^{32} - 1) = \gcd(2^{32} + 1, 2^{16} - 1) = \gcd(2^{32} + 1, 2^{16} + 1) = 1$.
(c) can be ruled out since its only divisors are $2$ and $7$. $2$ is not a divisor of $2^{32} + 1$ since it is odd, $7$ is not a divisor since $2^3 \equiv 1 \pmod 7$ and so $2^{32} + 1 \equiv 2^2 (2^3)^{10} +1 1 \equiv 4 + 1 \equiv 5 \pmod 7$.
|
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|
Olympiad Trigonometric Inequality
Let $R$ and $r$ be the circumradius and inradius of $\triangle ABC$.
Prove that $$\frac { \cos { A } }{ { \sin }^{ 2 }A } +\frac { \cos { B } }{ { \sin }^{ 2 }B } +\frac { \cos { C } }{ { \sin }^{ 2 }C } \ge \frac { R }{ r }$$
I am not able to get a solution to this inequality. Any help would be appreciated.
Thank you.
|
It seems the following.
Let $a$, $b$, and $c$ be the respective sides of the triangle $\triangle ABC$, $S$ be its area and $p=(a+b+c)/2$ be its semiperimeter. Then $r=S/p$ and $R=abc/4S$. Hence $R/r=abcp/4S^2$. Moreover,
$4S^2=b^2c^2\sin^2 A=c^2a^2\sin^2 B=a^2b^2\sin^2 C$.
Hence the initial inequality is equivalent to
$b^2c^2\cos A+ c^2a^2\cos B+ a^2b^2\cos C\ge abcp$.
But
$\cos A=\frac{b^2+c^2-a^2}{2bc}$,
$\cos B=\frac{c^2+a^2-b^2}{2ca}$,
$\cos C=\frac{a^2+b^2-c^2}{2ab}$.
Substituting, we obtain that the initial inequality is equivalent to
$a^3b+b^3a+a^3c+c^3a+b^3c+c^3b\ge 2(a^2bc+ab^2c+abc^2),$
which is true by Muirhead inequality, or directly by an equivalent inequality
$(a-b)^2(a+b)c+(b-c)^2(b+c)a+(c-a)^2 (c+a) b\ge 0$.
|
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|
An interesting Sum involving Binomial Coefficients How would you evaluate
$$\sum _{ k=1 }^{ n } k\left( \begin{matrix} 2n \\ n+k \end{matrix} \right) $$
I tried using Vandermonde identity but I can't seem to nail it down.
|
The following proof uses complex variable techniques and improves the
elementary one I posted earlier. It serves to demonstrate the method
even though it requires somewhat more of an effort.
Suppose we seek to evaluate
$$\sum_{k=1}^n k {2n\choose n+k}.$$
Introduce
$${2n\choose n+k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+1}}
\frac{1}{(1-z)^{n+k+1}} \; dz.$$
Observe that this is zero when $k\gt n$ so we may extend
$k$ to infinity to obtain for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\frac{1}{(1-z)^{n+1}}
\sum_{k\ge 1} k \frac{z^k}{(1-z)^k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\frac{1}{(1-z)^{n+1}}
\frac{z/(1-z)}{(1-z/(1-z))^2}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n}}
\frac{1}{(1-z)^{n}}
\frac{1}{(1-2z)^2}
\; dz.$$
Now put $z(1-z)=w$ so that (observe that with $w=z+\cdots$ the image of $|z|=\epsilon$ with $\epsilon$ small is another closed circle-like contour which makes one turn and which we may certainly deform to obtain another circle $|w|=\gamma$)
$$z = \frac{1-\sqrt{1-4w}}{2}
\quad\text{and}\quad
(1-2z)^2 = 1-4w$$
and furthermore
$$dz = -\frac{1}{2}
\times \frac{1}{2} \times (-4) \times (1-4w)^{-1/2} \; dw
= (1-4w)^{-1/2} \; dw$$
to get for the integral
$$\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^n} \frac{1}{1-4w}
(1-4w)^{-1/2} \; dw
= \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^n} \frac{1}{(1-4w)^{3/2}} \; dw.$$
This evaluates by inspection to
$$4^{n-1} {n-1+1/2\choose n-1}
= 4^{n-1} {n-1/2\choose n-1}
= \frac{4^{n-1}}{(n-1)!}
\prod_{q=0}^{n-2} (n-1/2-q)
\\ = \frac{2^{n-1}}{(n-1)!}
\prod_{q=0}^{n-2} (2n-2q-1)
= \frac{2^{n-1}}{(n-1)!}
\frac{(2n-1)!}{2^{n-1} (n-1)!}
\\ = \frac{n^2}{2n} {2n\choose n}
= \frac{1}{2} n {2n\choose n}.$$
Here the mapping from $z=0$ to $w=0$ determines the choice of square
root. For the conditions on $\epsilon$ and $\gamma$ we have that for the series to converge we require $|z/(1-z)|\lt 1$ or $\epsilon/(1-\epsilon) \lt 1$ or $\epsilon \lt 1/2.$ The closest that the image contour of $|z|=\epsilon$ comes to the origin is $\epsilon-\epsilon^2$ so we choose $\gamma \lt \epsilon-\epsilon^2$ for example $\gamma = \epsilon^2-\epsilon^3.$ This also ensures that $\gamma \lt 1/4$ so $|w|=\gamma$ does not intersect the branch cut $[1/4,\infty)$ (and is contained in the image of $|z|=\epsilon$). For example $\epsilon = 1/3$ and $\gamma = 2/27$ will work.
|
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|
Computing an infinite trigonometric sum $\sum \frac{2}{n^2 \pi^2} \cos(n \pi x)\cos(n \pi y)$ Let $G(x,y) = \sum_{n=1}^\infty \frac{2}{n^2 \pi^2} \cos(n \pi x)\cos(n \pi y)$
I'm trying to compute this sum by understanding it as an integral kernel. This question comes from Dym and Mckean Fourier Series and Integrals Ex 1.7.14:
Consider the cosine basis on $L^2[0,1]$ defined by $f_n(x) = \sqrt 2 \cos(n \pi x)$ for $n \ge 1$ and $f_0 = 1$. Define the operator $F$ by $F f_n = \frac{1}{n^2\pi^2} f_n$ acting only on the subspace $n \ge 1$.
It turns out that $F f(x) = \int_0^1 G(x, y) f(y) dy$ for $f \in L^2[0,1]$.
Now let $u = f_n''$ for some unspecified $n \ge 1$ .
$$F (f_n'') = -f_n $$
$$(F(f_n''))'' = -f_n''$$
$$(F u)'' = -u$$
$$(Fu)'(x) =-\int_0^x u(y)dy + K_1$$
$K_1 = 0$ because $(F u)'(0) = C \sin(0) = 0$
$$(Fu)(x) = -\int_0^x \int_0^y u(z)dz\,dy + K_2$$
$K_2 = \int_0^1\int_0^x \int_0^y u(z)dz\,dy\,dx$ because $\int_0^1 (Fu)(x) dx = \int_0^1 C \cos (2 \pi n x) dx = 0$
Now changing order of integration
$$ (Fu)(x) = - \int_0^x (x - z)u(z) dy + K_2$$
$$ = - \int_0^x (x - z)u(z) dy + \int_0^1\int_0^x (x - z)u(z) dz\,dx$$
$$ =... + \int_0^1\int_z^1(x -z)u(z)dx\,dz$$
$$ =... + \int_0^1u(z) [(1 - z^2)/2 - z(1-z)] dz $$
$$ = ... + \int_0^1 u(z) (z - 1)^2/2 dz$$
$$ = - \int_0^x (x - z)u(z) dz + \int_0^1 u(z) (z - 1)^2/2 dz$$
$$ = - \int_0^x (x - z)u(z) dz + \int_0^x u(z) (z - 1)^2/2 dz+ \int_x^1 u(z) (z - 1)^2/2 dz$$
$$ = \int_0^1 T(x,z) u(z) dz$$
where $T(x,y) = (y^2 + 1)/2 - x $ for $y < x$ and $(y-1)^2/2$ for $y > x$
and by necessity $T(x,y) = G(x,y)$
Unfortunately, I don't think this is the right function and it doesn't even look symmetric in $x,y$. I am looking for help for the correct derivation which should lead to $G(x,y) = \frac{x^2 - x + y^2 -y - |x-y|}{2} + 1/3$
Edit:
Even though $T(x,y)$ gives rise to the same operator $F$, it may not equal to $G(x,y)$ since they can differ by any function $\Delta(x)$ so that $T(x,y) + \Delta(x) = G(x,y)$, since $\int_0^1\Delta(x) f_n(y) dy = 0$. Since $\int G(x,y) dy = 0$, we should require $\Delta(x) = -\int_0^1T(x,y)dy = -1/6 + x^2/2$ so that $G(x,y) = (y^2 + 1)/2 - x - x^2/2 + 1/6 $ for $y < x$ and $(y-1)^2/2 + x^2/2 + 1/6$ for $y > x$ and this is the book's answer.
|
Note:
I am finally correcting my mistake
pointed out by Mark.
For
$G(x,y)
= \sum_{n=1}^\infty \frac{2}{n^2 \pi^2} \cos(n \pi x)\cos(n \pi y)
$,
since
(here's where my mistake was:
I had cos-cos
instead of the correct
cos+cos)
$\cos(a)\cos(b)
=\frac12(\cos(a-b)+\cos(a+b))
$,
$\begin{array}\\
G(x,y)
&= \frac12\sum_{n=1}^\infty \frac{2}{n^2 \pi^2} (\cos(n \pi (x-y))+\cos(n \pi (x+y)))\\
&= \frac1{\pi^2}\left(\sum_{n=1}^\infty \frac{1}{n^2} \cos(n \pi (x-y))
+ \sum_{n=1}^\infty \frac{1}{n^2} \cos(n \pi (x+y))\right)\\
&= g(x-y)+g(x+y)\\
\end{array}
$
where
$g(z)
=\frac1{\pi^2}\sum_{n=1}^\infty \frac{1}{n^2} \cos(n \pi z)
$.
$g'(z)
=-\frac1{\pi}\sum_{n=1}^\infty \frac{1}{n} \sin(n \pi z)
$
and,
getting into areas of dubious convergence,
$\begin{array}\\
g''(z)
&=-\sum_{n=1}^\infty \cos(n \pi z)\\
&=-\Re\sum_{n=1}^\infty \exp(in \pi z)\\
&=-\Re\frac{\exp(i \pi z)}{1-\exp(i \pi z)}\\
&=-\Re\frac{\cos( \pi z)+i\sin( \pi z)}{1-\cos( \pi z)-i\sin( \pi z)}\\
&=-\Re\frac{\cos( \pi z)+i\sin( \pi z)}{1-\cos( \pi z)-i\sin( \pi z)}
\frac{1-\cos( \pi z)+i\sin( \pi z)}{1-\cos( \pi z)+i\sin( \pi z)}\\
&=-\Re\frac{\cos( \pi z)+i\sin( \pi z)}{1-\cos( \pi z)-i\sin( \pi z)}
\frac{1-\cos( \pi z)+i\sin( \pi z)}{1-\cos( \pi z)+i\sin( \pi z)}\\
&=-\Re\frac{(\cos( \pi z)+i\sin( \pi z))(1-\cos( \pi z)+i\sin( \pi z))}{(1-\cos( \pi z))^2+\sin^2( \pi z)}\\
&=-\Re\frac{\cos( \pi z)(1-\cos( \pi z)-\sin^2( \pi z))+i(....)}{2-2\cos( \pi z)}\\
&=-\frac{\cos( \pi z)-\cos^2( \pi z)-\sin^2( \pi z)}{2(1-\cos( \pi z))}\\
&=-\frac{\cos( \pi z)-1}{2(1-\cos( \pi z))}\\
&=\frac12\\
\end{array}
$
This seems to imply that
$g(z)
=\frac14 z^2+az+b
$
for some $a$ and $b$,
which I find quite surprising.
I will continue under
the assumption that
this is correct.
If $z=0$,
$b
=g(0)
=\frac1{\pi^2}\sum_{n=1}^\infty \frac1{n^2}
=\frac1{\pi^2}\zeta(2)
=\frac{1}{6}
$.
If
$z=1$,
$\begin{array}\\
g(1)
&=\frac14+a+b\\
&=\frac14+a+\frac1{6}\\
&=\frac{5}{12}+a\\
&=\frac1{\pi^2}\sum_{n=1}^\infty \frac1{n^2}\cos(n\pi)\\
&=\frac1{\pi^2}\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\\
&=-\frac1{\pi^2}\frac12\zeta(2)\\
&=-\frac1{12}\\
\end{array}
$
so
$a
=-\frac12
$.
As a check,
if
$z=2$,
$\begin{array}\\
g(2)
&=1+2a+b\\
&=1+2a+\frac1{6}\\
&=\frac{7}{6}+a\\
&=\frac1{\pi^2}\sum_{n=1}^\infty \frac1{n^2}\cos(2n\pi)\\
&=\frac1{\pi^2}\sum_{n=1}^\infty \frac{1}{n^2}\\
&=\frac1{\pi^2}\zeta(2)\\
&=\frac1{6}\\
\end{array}
$
so
$2a+\frac76
=\frac16
$
or
$a=
-\frac12
$.
Therefore
$\begin{array}\\
G(x, y)
&=g(x-y)+g(x+y)\\
&=\frac14((x-y)^2+(x+y)^2))-\frac12((x-y)+(x+y))+2\frac1{6}\\
&=\frac12(x^2+y^2)-x+\frac13\\
\end{array}
$.
|
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|
Complex number in quadratic equation Find $a,b$ given that a root of $x= 1+2i$ and the equation $ x^2+(a+bi)x+2i-1=0$
I tried finding it by $\Delta$,
which I got $\Delta=a^2+2abi-b^2-8i+4$
I tried substituting the root into the equation but still can't continue.
Can you help me?
|
If you specifically want to do this via the quadratic method (which is not the easiest method as you could use the sum and product of roots to get two expressions that can be solved simultaneously).
But continuing with your method anyway:
Using the quadratic formula for a general quadratic $ax^2+bx+c$ which is
$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$x=\frac{-a-bi\pm \sqrt{(a+bi)^2-4\cdot 1 \cdot (2i-1)}}{2\cdot 1}$$
Since $x=1+2i$ is a solution
$$\frac{-a-bi\pm \sqrt{(a^2+2abi -b^2-8i+4)}}{2}=1+2i$$
$$\implies{\pm \sqrt{a^2+2abi -b^2-8i+4}}=2+a+(4+b)i$$
$$\implies{{a^2+2abi -b^2-8i+4}}=\left(2+a+(4+b)i\right)^2\tag{1}$$
Now simplify ($1$) and equate for the real and imaginary components to find the value of $a$ and $b$.
|
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|
Nice way to solve $\int\int \frac{1}{1-(xy)^2} dydx$? This is something I've been thinking about lately;
$$\int_0^1 \int_0^1 \frac{1}{1-(xy)^2} dydx$$
Solutions I've read involve making the substitutions: $x= \frac{sin(u)}{cos(v)}$ and $y= \frac{sin(v)}{cos(u)}$. This reduces the integral to the area of a right triangle with both legs of length $\frac{\pi}{2}$. My problem is that coming up with this substitution is not at all obvious to me, and realizing how the substitution distorts the unit square into a right triangle seems to require a lot of reflection. My approach without fancy tricks involves letting $u = xy$ and then the integral "simplifies" accordingly:
$\begin{align*} \int_0^1 \int_0^1 \frac{1}{1-(xy)^2} dydx &= \int_0^1\frac{1}{x}\int_0^x \frac{1}{1-u^2}dudx\\
&= \int_0^1\frac{1}{2x}\int_0^x \frac{1}{1-u}+\frac{1}{1+u}dudx\\
&= \int_0^1\frac{1}{2x}ln\left(\frac{1+x}{1-x}\right)dx
\end{align*}$
If I've done everything right this should be $\frac{\pi^2}{8}$ but I haven't figured out how to solve it.
|
You can also do this, and this one does not involve any fancy tricks:
Consider the double integral: $$I=\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{(1+x)(x+y^2)}dydx.$$
Integrate with respect to $y$ first and recognize this is :
$$I=\int_{0}^{\infty} \frac{1}{1+x}\lim_{y \rightarrow \infty} \frac{\arctan{\frac{y}{\sqrt{x}}}}{\sqrt {x}} dx=\int_{0}^{\infty}\frac{\frac{\pi}{2}}{\sqrt{x}{(1+x)}}dx.$$ Now if you apply the transformation $u=\sqrt{x}, du=\frac{1}{2\sqrt{x}}dx,$ you get $$I=\int_{0}^{\infty} \frac{\pi}{(1+u^2)}du=\frac{\pi^2}{2}.$$
On the other hand, reverse the order of integration in $I$ as such: $$I=\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{(1+x)(x+y^2)}dxdy.$$
Now integrate with respect to $x$ using partial fractions as such:
$$\frac{1}{(1+x)(x+y^2)}=\frac{1}{y^2-1} \left(\frac{1}{1+x}-\frac{1}{x+y^2}\right).$$
$$I=\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{y^2-1} \left(\frac{1}{1+x}-\frac{1}{x+y^2}\right)dxdy=\int_{0}^{\infty}\lim_{x\rightarrow \infty}\frac{\ln(1+x)-\ln(x+y^2)}{y^2-1}dy=\int_{0}^{\infty}\frac{\ln(y^2)}{y^2-1}dy=\int_{0}^{\infty}\frac{2\ln(y)}{y^2-1}dy.$$
Now consider $$J=\int_{0}^{1}\frac{\ln(y)}{y^2-1}dy.$$ A u-substitution $y=\frac{1}{z}, dy=\frac{-1}{z^2} dz$ tells us:
$$J=\int_{1}^{\infty}\frac{\ln(z)}{z^2-1}dz.$$ This tells us: $$J=\frac{I}{4}=\frac{\pi^2}{8}.$$ Now let us make the substitution $z=\frac{1-t}{1+t}, dz=\frac{-2}{(t+1)^2}dt.$ We have:
$$J=\frac{\pi^2}{8}=\int_{0}^{1} \frac{\ln(1+t)-\ln(1-t)}{2t} dt,$$ the integral you arrived at.
|
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|
How to prove $2\sqrt{2+\sqrt{3}}=\sqrt{2}+\sqrt{6}$? My calculator and I were arguing one day about the cosine of some number.
The calculator said "$\cos(\frac x2)=\sqrt{2}+\sqrt{6}$".
I said "That's absurd because $\cos(\frac x2)=\sqrt{\frac{1+\cos(x)}2}$, which evaluates to $2\sqrt{2+\sqrt{3}}$ for this particular $x$!"
So I finally decided to do this the non-radical way and found the decimal approximations to be equivalent.
Which is weird and probably not a coincidence. So why is this, and does this have other values that work out like this?
|
$$\color{gray}{2+\sqrt 3=\frac 1 2+\sqrt 3+\frac 3 2=\frac{1+2\sqrt3 +3}{2}=\frac{1+2\sqrt 3+(\sqrt 3)^2}{2}=\frac{(1+\sqrt3)^2}{2}}\\2\sqrt{\frac{(1+\sqrt 3)^2}{2}}\\ \color{gray}{\sqrt{\frac 1 2(1+\sqrt 3)^2}=\frac{\sqrt{(1+\sqrt 3)}}{\sqrt 2}}\\ 2\frac{\sqrt{(1+\sqrt 3)^2}}{\sqrt 2}\\ \color{gray}{\sqrt{(1+\sqrt 3)^2}=1+\sqrt 3:}\\ \frac{2}{\sqrt 2}1+\sqrt 3\\ \color{gray}{\frac{2(1+\sqrt 3)}{\sqrt 2}=\frac{2(1+\sqrt 3)}{\sqrt 2}\times \frac{\sqrt 2}{\sqrt 2}=\frac{1(1+\sqrt 3)\sqrt 2}{2}:}\\ \frac{2(1+\sqrt 3)\sqrt 2}{2}\\ \color{gray}{\frac{2(1+\sqrt 3)\sqrt 2}{2}=\frac 2 2\times(1+\sqrt 3)\sqrt 2=(1+\sqrt3)\sqrt 2}\\ (1+\sqrt 3)\sqrt 2\\ \color{gray}{\sqrt 2(1+\sqrt 3)}=\boxed{\color{blue}{\sqrt 2+\sqrt 6}}$$
|
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|
minimum $x^2 + y^2$ on $\frac{(x-12)^2}{16} + \frac{(y+5)^2}{25} = 1 $ ellipse Given $\frac{(x-12)^2}{16} + \frac{(y+5)^2}{25} = 1$.
Then minimum value of $x^2 + y^2 = ?$
P.S. My solution: Suppose that $x = 4\cos{\theta}+12$and $y = 5\sin{\theta}-5$
and expand $x^2 + y^2$ to find minimum value, but stuck in the end.
Thank you for every comment.
|
HINT: use the Lagrange Multiplier Method
$$F(x,y,\lambda)=x^2+y^2+\lambda((x-12)^2/16+(y+5)^2/25-1)$$
solve the system
$$2\,x+\lambda\, \left( x/8-3/2 \right) =0$$
$$2\,y+\lambda\, \left( {\frac {2}{25}}\,y+2/5 \right)=0 $$
$$1/16\, \left( x-12 \right) ^{2}+1/25\, \left( y+5 \right) ^{2}-1=0$$
|
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|
Find $\frac{AC \times BC}{AD \times BD}$
$AC$ is $2004$. $CD$ bisects angle $C$. If the perimeter of $ABC$ is $6012$, find $\dfrac{AC \times BC}{AD \times BD}$.
Attempt
Let $c = AD+BD$.
We have that $\dfrac{AC}{AD} = \dfrac{BC}{BD}$. Thus, $$\dfrac{BD}{AD}+1 = \dfrac{BC}{AC}+1 \implies \dfrac{BD+AD}{AD} = \dfrac{AC+BC}{AC} = \dfrac{c}{AD} \implies AD = \dfrac{AC \cdot c}{AC+BC}.$$ Likewise it is easy to see by the angle bisector theorem that $BD = \dfrac{BC \cdot c}{AC+BC}$. Thus, we have that $\dfrac{AC \times BC}{AD \times BD} = \dfrac{AC \times BC}{\dfrac{AC \cdot BC \cdot c^2}{(AC+BC)^2}} = \dfrac{(AC+BC)^2}{c^2} = \dfrac{(2004+BC)^2}{(4008-BC)^2}$. But this answer seems to depend on $BC$, which makes no sense as the question indicates it is constant. What have I done wrong?
|
If instead of numerical value of $AC$, numerical value of $AB$ was given then the required expression would have a unique answer.
Let $a, b, c$ represent $BC, AC, AB$ respectively.
By, ange bisector theorem we have, $\frac{AD}{BD}=\frac{b}{a}$.
So, $AD=\frac{bc}{a+b}$ and $\frac{ca}{a+b}$.
$\frac{AC \times BC}{AD \times BD}=\left(\frac{a}{c}+\frac{b}{c}\right)^2=\left(\frac{2004}{c}-1 \right)^2$.
So, knowing $c$ is enough.
|
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|
Prove that for all real numbers $x,y$ and $z$ that $x^2+y^2+z^2 \geq x\sqrt{y^2+z^2}+y\sqrt{x^2+z^2}$
Prove that for all real numbers $x,y$ and $z$ that $x^2+y^2+z^2 \geq x\sqrt{y^2+z^2}+y\sqrt{x^2+z^2}$.
It looks like AM-GM should be used here but the square roots make it difficult. So maybe Cauchy-Schwarz works?
|
This is simply 2 applications of $ab\leq\frac{1}{2}(a^2+b^2)$:
$$
x\sqrt{y^2+z^2}\leq\frac{1}{2}(x^2+y^2+z^2),\quad y\sqrt{x^2+z^2}\leq\frac{1}{2}(y^2+x^2+z^2)
$$
Now, sum the above 2 inequalities.
|
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|
Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \left (\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right) \geq \frac{9}{a+b+c}$
Let $a,b,$ and $c$ be positive real numbers, prove that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq 2 \left (\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a} \right) \geq \dfrac{9}{a+b+c}$.
Should I use AM-GM for the expression in the middle of the inequality? We have $a+b \geq 2\sqrt{ab}$ etc.?
|
Hints: For the first inequality, note that $\frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}$ because this is equivalent to $(a+b)^2\geq 4ab$. For the second, apply AM-GM to
$$
(a+b) + (b+c) + (c+a)\quad\text{and}\quad\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}
$$
and then multiply the resulting two inequalities.
|
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|
The cubic equation $x^3 - 4 x^2 + x + 1 =0$
*
*The cubic polynomial
$P(x) = x^{3} - 4x^{2} + x + 1$ has discriminant $\Delta = 169 = 13^{2}$ which tells us that the extension $\mathbb{Q}(a)/\mathbb{Q}$ is normal, where $a$ is any root of the equation $P(x) = 0$.
*Therefore, given one root $a$, one can find the other as polynomial expressions in $a$.
*For instance, in this case it is not hard to check that the other roots are $a^{2} - 4a + 2$ and $-a^{2} + 3 a + 2$. But what if we didn't know these expressions? Is there a way to get them?
|
Since $P(x) = x^3 - 4 x^2 + x + 1$ is normal, $P(x)$ splits in $\mathbb Q(\alpha)$.
This means that the polynomial will factor into linear factors over $\mathbb Q(\alpha)$. One of the factors will be $(x - \alpha)$. The others will be $(x - (\alpha^2 - 4 \alpha + 2))$ and $(x - (-\alpha^2 + 3 \alpha + 2))$
You can calculate these polynomials quickly using factorization modulo an ideal, for example in pari/gp:
? f(x) = x^3 - 4*x^2 + x + 1
%1 = (x)->x^3-4*x^2+x+1
? factor(Mod(f(x),f(a)))
%2 =
[ x + Mod(-a, a^3 - 4*a^2 + a + 1) 1]
[x + Mod(-a^2 + 4*a - 2, a^3 - 4*a^2 + a + 1) 1]
[ x + Mod(a^2 - 3*a - 2, a^3 - 4*a^2 + a + 1) 1]
If you were just working with pencil and paper you could perform the long division of $P(x)/(x-\alpha)$ modulo $P(\alpha)$.
This gives $x^2 + (\alpha - 4) x + (\alpha^2 - 4\alpha + 1)$ which you can now try to factorize into two linear factors.
|
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|
Solving equation involving complex numbers
(a) Find real numbers $a$ and $b$ such that $(a+bi)^2 = -3-4i.$
(b) Hence solve the equation: $z^2+i\sqrt{3}z+i = 0$.
Original Image
In the above question, I have solved part (a), with $a=\pm1$ and with $b=\mp2$, but I am not sure how to use this information to solve (b). I know they are relevant, because of the key word "hence," which dictates that I should use the information derived from part (a).
Please advise. Sorry in advance for any mistakes in labelling of the title and tags (I'm not very good at those, but am trying to improve on it)
|
Notice, $$(a+bi)^2=-3-4i$$
$$(a^2-b^2)+2iab=-3-4i$$
comparing real & imaginary parts, one should get $$a^2-b^2=-3\tag 1$$ $$2ab=-4\tag 2$$
solve (1) & (2) for the values of $a$ & $b$.
b) $$z^2+i\sqrt 3z+i=0$$
$$\left(z^2+2i\frac{\sqrt3}{2}z+\left(\frac{i\sqrt{3}}{2}\right)^2\right)-\left(\frac{i\sqrt{3}}{2}\right)^2+i=0$$
$$\left(z+\frac{i\sqrt{3}}{2}\right)^2=-\frac{3}{4}-i=\frac{-3-4i}{4}$$
$$\left(2z+i\sqrt 3\right)^2=-3-4i$$
Alternative Method: use quadratic formula to find the roots of $z^2+i\sqrt 3z+i=0$ as follows $$z^2+i\sqrt 3z+i=0$$
$$z=\frac{-i\sqrt 3\pm\sqrt{(-i\sqrt 3)^2-4(1)(i)}}{2(1)}=\frac{-i\sqrt 3\pm i\sqrt{3+4i}}{2}$$
|
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|
Prove that $5^n + 2\cdot3^{n-1} + 1$ is multiple of $8$ Prove that $5^n + 2\cdot3^{n-1}+ 1$ is multiple of $8$.
I've tried using induction (it isn't):
For $n=1$:
$$5^1 + 2\cdot3^{n-1} + 1 = 8$$
If it is true for $n$, then $n+1$?
\begin{align}
5^{n+1} + 2\cdot3^n + 1
=
&(4+1)^n\cdot(4+1)+ 2\cdot(2+1)^n + 1
\\
=& (4^n + n4^{n-1} + 1)\cdot(4+1) + 2\cdot(2^n + n2^{n-1} + 1) + 1
\\
=
&
(4k+1)\cdot(4+1) + 2(2r+1) + 1
\\
= &16k+4k+4 +1+4r+2+1
\\
=
&20k + 4r + 8 = 4(5+r+2)
\end{align}
But i've only proved it is multiple of $4$.
|
If you want to use induction:
$5^{n+1} + 2\cdot3^n +1 = 5^n\cdot5 + 2\cdot3^{n-1}\cdot3 + 1 = 5^n + 3^{n-1} + 1 + 4\cdot5^n + 4\cdot3^{n-1} = 8K + (4\cdot5^n + 4\cdot3^{n-1})$.
Suffices to show $4\cdot5^n + 4\cdot3^{n-1}$ is divisible by 8. It's clearly divisible by 4. So it suffices to show $5^n + 3^{n-1}$ is even. Which we can do by, heh heh, induction (yes, you can do induction within induction).
$n = 1; 5^1 + 3^0 = 6$ Even. Induction: $5^{n+1} + 3^n = 5^n + 3^{n-1} + (4\cdot5^n + 2\cdot3^{n-1})$ It's even.
|
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|
Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$ Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$
First inequality using MVT:
$\frac{1}{a+1}<\ln \frac{a+1}{a}:$
$f(a)=\frac{1}{a+1}-\ln \frac{a+1}{a}$
$f(1)=\frac{1-2\ln 2}{2},f^{'}(a)=\frac{1}{a(a+1)^2}>f(1)\Rightarrow f(a)>f(1)$
$\frac{1}{a+1}-\ln \frac{a+1}{a}-\frac{1-2\ln 2}{2}>0$
This is not the starting inequality.
Is there something wrong in this method?
|
Hint: use $f(x)=\ln x$ in $[a,a+1]$.
|
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|
Multiplying the denominator $$\frac{2}{3} \gt -4y - \frac{25}{3}$$
My question is to solve this problem we need to multiply both sides by $3$. The result will be
$$2 \gt -12y - 25.$$
Why doesn't the numerator of both $\frac{2}{3}$ and $\frac{25}{3}$ get multiplied?
|
As an expansion of what I said in the comments:
When we multiply the equation $\frac{2}{3} > -4y - \frac{25}{3}$ by $3$ we get the following equation
$\frac{2\times3}{3} > (-4y)\times3 - \frac{25\times3}{3}$. Simplifying the following we get
$\frac{6}{3} > -12y - \frac{75}{3}$.
Now $\frac{6}{3} =2$ and $\frac{75}{3} = 25$. Thus utilizing these known facts we can simplify the above equation giving us
$2 > -12y - 25$. So the numerators got multiplied by $3$, but then we divided by $3$, which amounts to returning to the original numerator. This is because multiplication and division are inverse operators.
From your comment, if you changed the $3$ in the denominator to a $6$ you would not get the same result. Instead of ending up with $25$, you would get $\frac{25}{2}$.
|
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|
Compute $\int\frac{x}{2x^2+x+3}dx$ $$\int\frac{x}{2x^2+x+3}\,dx$$
Well, to approach this kind of exercises I know that I need to check the derivative of the denominator. which is $4x + 1$.
Then, I can re-write the integral: $\int\frac{0.25(4x+1) - 0.25}{2x^2+x+3}\,dx$.
Then, I get:
$$0.25\int\frac{(4x+1)}{2x^2+x+3}\,dx-0.25\int\frac{1}{2x^2+x+3}\,dx.$$
The left one is pretty straight forward with $\ln|\cdot|$,
Problem: does anyone have some "technique" to solve the right integral? hints would be appreciated too.
Edit: maybe somehow: $$0.25\int\frac{1}{2(2x^2/2+x/2+3/2)}\,dx = 0.25\int\frac{1}{(x+0.25)^2 + \frac{23}{16}}\,dx$$
|
$$\int\frac{1}{2x^2+x+3}dx = \int\frac{1}{2\left(x+\frac{1}{4}\right)^2+\frac{23}{8}}dx= \frac{2}{\sqrt{23}}\tan^{-1}\left(\frac{4x+1}{\sqrt{23}}\right)+c$$
|
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|
Solve irrational equation $x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$ Solve irrational equation
$$x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$$
Here is what I tried
$t^3 = 35-x^3 \implies x = \sqrt[3]{35-t^3} $
which takes me to nowhere.
|
See you need $6.5$ in reduced form . So we can just manipulate an get solution . So we need a cubic term to get real number so $x\sqrt{35-x^3}=6$ and $x=3$ gives you $6$ also $x+\sqrt{35-x^3}=5$ where $3$ is the solution thus we have one real answer which is 3 other way is solving by Vieta method.
|
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|
Given matrix A. To find $A^{2010}$
Let $\theta = 2\pi/67$. Now consider the matrix
$$
A =
\begin{pmatrix}
\phantom{-}\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{pmatrix}.
$$
Then the matrix $A^{2010}$ is
\begin{align*}
&\text{(A)}\;
\begin{pmatrix}
\phantom{-}\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{pmatrix},
&
&\text{(B)}\;
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}, \\
&\text{(C)}\;
\begin{pmatrix}
\phantom{-}\cos^{30} \theta & \sin^{30} \theta \\
-\sin^{30} \theta & \cos^{30} \theta
\end{pmatrix},
&
&\text{(D)}\;
\begin{pmatrix}
\phantom{-}0 & 1 \\
-1 & 0
\end{pmatrix}.
\end{align*}
I think answer is B. But I am not sure.
|
Note that
$$
A=PDP^{-1}
$$
where
\begin{align*}
P &=
\begin{bmatrix}
i & -i \\
1 & 1
\end{bmatrix}
&
D &=
\begin{bmatrix}
\cos\theta+i\sin\theta & 0 \\
0 & \cos\theta-i\sin\theta
\end{bmatrix}
\end{align*}
De Moivre's formula then implies
\begin{align*}
A^{2010}
&= PD^{2010} P^{-1} \\
&= P
\begin{bmatrix}
\cos(2010\,\theta)+i\sin(2010\,\theta) & 0 \\
0 & \cos(2010\,\theta)-i\sin(2010\,\theta)
\end{bmatrix}P^{-1}\\
&=
P
\begin{bmatrix}
\cos(60\,\pi) & \sin(60\,\pi) \\
-\sin(60\,\pi) & \cos(60\,\pi)
\end{bmatrix}
P^{-1} \\
&=
P
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
P^{-1} \\
&= PIP^{-1} \\
&= PP^{-1} \\
&= I
\end{align*}
|
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|
How do I evaluate the sum $\sum_{k=1}^\infty\left(\ln\big(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\big)\right)$ How do I evaluate the sum
$$\sum_{k=1}^\infty\left(\ln\Big(1+\frac{1}{k+a}
\Big)-\ln\Big(1+\frac{1}{k+b}\Big)\right)$$
where $0 <a<b<1$?
Hints will be appreciated
Thanks
|
Let
$$S_n = \ln\prod\limits_{k=1}^n\dfrac{k+a+1}{k+a} - \ln\prod\limits_{k=1}^n\dfrac{k+b+1}{k+b}= \ln\dfrac{k+a+1}{a+1} - \ln\dfrac{k+b+1}{b+1}$$$$ = \ln\dfrac{b+1}{a+1} + \ln\dfrac{k+b+1}{k+a+1}.$$
$$\sum_{k=1}^{\infty} \ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right) = \lim\limits_{n\to \infty} S_n = \ln\dfrac{b+1}{a+1} +\ln1 =\ln\dfrac{b+1}{a+1}$$
|
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|
How to evaluate the following integral, $\int\frac{x \, dx}{x^2+2x+17}$? I am new to integration. This function is kinda tricky for me : $$\int\frac{x \, dx}{x^2+2x+17}$$
I came up with following three approaches:
*
*Partial fraction decomposition, but I can't factor the denominator into different parts.
*Substitution: I tried $u = x^{2}$ and $u = x^2+2x+17$, but both of them seem not to be helpful.
*I also tried dividing both denominator and numerator by $x$, then the fraction became $\frac{1}{x+\frac{17}{x}+2}$ . It is more complex.
Any suggestions or hints? Thank you so much!
|
Variations on this one come up here from time to time.
There is a standard technique in algebra for reducing a problem involving a quadratic polynomial with a first-degree term to a problem involving a quadratic polynomial without a first-degree term.
$$
u = x^2+2x+17,\qquad du = (2x+2)\,dx, \qquad \frac{du} 2 = (x+1)\,dx
$$
$$
\frac{x}{x^2+2x+17} = \underbrace{\frac{x+1}{x^2+2x+17}} + \frac{-1}{x^2+2x+17}
$$
The integral of the function over the $\underbrace{\text{underbrace}}$ is done via the substitution above. Then next term has to be treated differently.
$$
\underbrace{x^2+2x+17 = (x^2 + 2x + 1) + 16 = (x+1)^2 + 16}_\text{completing the square}
$$
$$
\int \frac 1 {(x+1)^2 + 16} \,dx = \int \frac{1/16}{\left( \frac{x+1} 4 \right)^2 + 1} \, dx = \int \frac{1/4}{\left( \frac{x+1} 4 \right)^2 + 1} \, \frac{dx} 4 = \frac 1 4 \int \frac{du}{u^2+1}
$$
etc.
There is a standard technique in algebra for reducing a problem involving a quadratic polynomial with a first-degree term to a problem involving a quadratic polynomial without a first-degree term.
|
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|
Show that $(\frac{S_1}{S_n+1},\frac{S_2}{S_n+1},...\frac{S_n}{S_n+1})=_d (U_{(1)},U_{(2)},...,U_{(n)})$. Let $(X_1, X_2,...,X_n) \in \mathbb R^n$ have density function $p(x)$.
(1) Find the density of $(U_{(1)},U_{(2)},...,U_{(n)})$, the order statistics from a sample of iid $\mathbb U[0,1]$ (uniform distributions) variables.
(2) Let $E_1, E_2, ..., E_n$ be i.i.d. exponential with density $p(x)=e^{-x}$, where $x>0$ and for $k=1,2,3,...,n+1$, set $S_k=\Sigma_{i=1}^{k} E_i$. Show that $\left(\frac{S_1}{S_{n+1}},\frac{S_2}{S_{n+1}},...\frac{S_n}{S_{n+1}}\right)=_d (U_{(1)},U_{(2)},...,U_{(n)})$.
(3) For $m$ an integer such that $\frac{m}{n} \rightarrow \alpha$ as $n \rightarrow \infty$, take $U_{(m)}$ as the estimate of the $\alpha$ quantile $x_{\alpha}$; for $\mathbb U[0,1]$, we have $x_{\alpha}=\alpha$. The error made in estimating $x_{\alpha}$ is $U_{(m)}-x_{\alpha}$. Use part (2) to find the asymptotic distribution of $\sqrt n (U_{(m)}-x_{\alpha})$.
Progress: Part (1) is easy and I just plugged in the formula of density function of order statistics. However, I have no idea about part (2) and part (3).
|
If we let
$$ \begin{align*} Y_k & = \frac {S_k} {S_{n+1}}, k = 1, 2, \ldots, n \\
Y_{n+1} & = S_{n+1}
\end{align*}$$
Then it is not hard to find out the inverse transform:
$$ \begin{align*} E_1 & = Y_1Y_{n+1} \\
E_k &= (Y_k - Y_{k-1})Y_{n+1}, k = 2, 3, \ldots, n \\
E_{n+1} & = Y_{n+1} - Y_nY_{n+1} \end{align*}$$
And thus we have the Jacobian Matrix:
$$ \begin{bmatrix} y_{n+1} & 0 & 0 & \ldots & 0 & y_1 \\
-y_{n+1} & y_{n+1} & 0 & \ldots & 0 & y_2-y_1 \\
0 & -y_{n+1} & y_{n+1} & \ldots & 0 & y_3-y_2 \\
\vdots & \vdots & \vdots & \ddots & \vdots &\vdots \\
0 & 0 & 0 & \ldots & y_{n+1} & y_n - y_{n-1} \\
0 & 0 & 0 & \ldots & -y_{n+1} & 1 - y_n \\
\end{bmatrix}$$
Since the determinant is unchanged under row addition, the determinant of the above matrix is the same as an upper triangular matrix by successive row additions, and thus the determinant is just the product of the diagonal entries of the resulting matrix, which is
$$y_{n+1}^n$$
Eventually, since the joint pdf of $(E_1, E_2, \ldots, E_n, E_{n+1})$ is
$$ f_{E_1, E_2, \ldots, E_n, E_{n+1}}(v_1, v_2, \ldots, v_n, v_{n+1})
= \exp\left\{-\sum_{i=1}^{n+1} v_i\right\}, v_i > 0, i = 1, 2, 3, \ldots, n $$
The joint pdf of $(Y_1, Y_2, \ldots, Y_n, Y_{n+1})$ is
$$ \begin{align*} & f_{Y_1, Y_2, \ldots, Y_n, Y_{n+1}}(y_1, y_2, \ldots, y_n, y_{n+1}) \\
= ~& \exp\left\{-\left(y_1y_{n+1} + \sum_{k=2}^n(y_k - y_{k-1})y_{n+1} + y_{n+1}
- y_ny_{n+1} \right)\right\}y_{n+1}^n \\
= ~& \exp\left\{-y_{n+1}\right\}y_{n+1}^n, 0 < y_1 < y_2 < \ldots < y_n < 1, y_{n+1} > 0 \end{align*}$$
By integrating this joint pdf with respect to $y_{n+1}$ (actually this auxiliary random variable is independent of the remaining), we have
$$ \begin{align*} f_{Y_1, Y_2, \ldots, Y_n}(y_1, y_2, \ldots, y_n) &= \int_0^{+\infty} \exp\left\{-y_{n+1}\right\}y_{n+1}^n dy_{n+1} \\
& = \Gamma(n+1) \\
& = n!, 0 < y_1 < y_2 < \ldots < y_n < 1
\end{align*} $$
which has the same joint pdf as the ordered uniforms. For the last part, we can set
$$ m = \lfloor n\alpha\rfloor$$
(ceiling is also ok) It is well known that
$$ 0 \leq n\alpha - \lfloor\alpha n\rfloor < 1$$
Dividing the inequality by $n$, by squeezing principle, we have
$$ \lim_{n\to+\infty} \frac {\lfloor\alpha n\rfloor} {n} = \alpha $$
Let
$$ W = S_m = \sum_{i=1}^{\lfloor n\alpha\rfloor}E_i,
V = \sum_{i=\lfloor n\alpha\rfloor+1}^{n+1}E_i$$
Note that
$$ E[W] = \lfloor n\alpha\rfloor, E[V] = n + 1 - \lfloor n\alpha\rfloor$$
and $U, V$ are independent. Since we have shown that $U_{(m)}$ has the same distribution of $Y_m$, we can consider the asymptotic of $Y_m$ instead. Note
$$ Y_m = \frac {U} {U + V}$$
(it is well known that this function of gamma random variables has a beta distribution) and note
$$ y = \frac {w} {w+v}, \frac {\partial y} {\partial w} = \frac {v} {(w+v)^2}, \frac {\partial y} {\partial v} = -\frac {w} {(w+v)^2} $$
Therefore we can use the old trick: Taylor expansion about the mean.
$$ \begin{align*} Y_m = ~& \frac {\lfloor n\alpha\rfloor}
{\lfloor n\alpha\rfloor + n+1 - \lfloor n\alpha\rfloor}
+ (W - \lfloor n\alpha\rfloor) \frac {n + 1 - \lfloor n\alpha\rfloor} {(n+1)^2} - (V - (n+1 - \lfloor n\alpha\rfloor)) \frac {\lfloor n\alpha\rfloor} {(n+1)^2} \\ & + R \end{align*}$$
where $R$ is the remaining term. Rearranging a little bit, we have
$$ \begin{align*} & \sqrt{n}(Y_m - \alpha) \\
= ~& \sqrt{n}\left(\frac {\lfloor n\alpha\rfloor} {n+1} - \alpha \right) + \sqrt{\lfloor n\alpha\rfloor}
\left(\frac {W} {\lfloor n\alpha\rfloor} - 1\right)
\frac {(n+1-\lfloor n\alpha\rfloor)\lfloor n\alpha\rfloor} {(n+1)^2}
\sqrt{\frac {n} {\lfloor n\alpha\rfloor}} \\
~& - \sqrt{n + 1 - \lfloor n\alpha\rfloor}
\left(\frac {V} {n + 1 - \lfloor n\alpha\rfloor} - 1\right)
\frac {(n+1-\lfloor n\alpha\rfloor)\lfloor n\alpha\rfloor} {(n+1)^2}
\sqrt{\frac {n} {n + 1 - \lfloor n\alpha\rfloor}} \\
~& + \sqrt{n}R
\end{align*}\\ $$
Eventually we can bound it term by term. For the first term, again since
$$ \frac {\lfloor n\alpha\rfloor} {n+1} - \alpha =
\frac {\lfloor n\alpha\rfloor - n\alpha - \alpha} {n+1}$$
and $ -1 < \lfloor n\alpha\rfloor - n\alpha \leq 0$, therefore
$$ -\sqrt{n}\frac {1 + \alpha} {n+1} < \sqrt{n}\left(\frac {\lfloor n\alpha\rfloor} {n+1} - \alpha \right) \leq -\sqrt{n}\frac {\alpha} {n+1}$$
And by squeezing principle again, this goes to zero as $n \to +\infty$. For the second and third term, note by Central Limit Theorem (CLT),
$$ \sqrt{\lfloor n\alpha\rfloor}
\left(\frac {W} {\lfloor n\alpha\rfloor} - 1\right) \text{ and }
\sqrt{n + 1 - \lfloor n\alpha\rfloor}
\left(\frac {V} {n + 1 - \lfloor n\alpha\rfloor} - 1\right) $$
both converges to the standard normal distribution. From the previous limit result,
$$ \lim_{n\to+\infty} \frac {(n+1-\lfloor n\alpha\rfloor)\lfloor n\alpha\rfloor} {(n+1)^2}
= (1 - \alpha)\alpha$$
$$\lim_{n\to+\infty} \sqrt{\frac {n} {\lfloor n\alpha\rfloor}} = \sqrt{\frac {1} {\alpha}}, \lim_{n\to+\infty} \sqrt{\frac {n} {n+1 - \lfloor n\alpha\rfloor}} = \sqrt{\frac {1} {1 - \alpha}} $$
For the remainder term R, it contains the higher power term which contain at least two of the standardized $W, V$, so by CLT they are at least $O_p(n)$, $o_p(\sqrt{n})$ or in other words goes to zero in probability as $n \to +\infty$. Finally, since $W, V$ are independent, we conclude that $\sqrt{n}(Y_m - \alpha)$ also has a asymtotic normal distribution, with mean zero, and variance
$$ (1 - \alpha)^2 \alpha + (1 - \alpha)\alpha^2 = (1 - \alpha)\alpha $$
which agrees with the well known result of the sample quantile.
|
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|
Testing the convergence of cube root of some function of n I have to test the convergence of the following series:-
$$\sum_{n=1}^\infty\sqrt[3]{n^3+1}-n$$
My approach is as follows :-
$$n^3+1>1=\sqrt[3]{n^3+1}>1=\sqrt[3]{n^3+1}-n>1-n$$
Now since$\sum 1-n$ diverges, the series under consideration diverges.
Is this right or wrong?
|
Hint
Let $a=\sqrt[3]{n^3+1}$ and $b=n$. Then,
$$a-b=\frac{a^3-b^3}{a^2+ab+b^2}=\frac{1}{a^2+ab+b^2}.$$
Observe that $a \geq b$, therefore
$$\sqrt[3]{n^3+1}-n=a-b \leq \frac{1}{3b^2}=\frac{1}{3n^2}.$$
Now use comparison to claim convergence.
|
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|
Difficult Integral I have a problem with solving this integral:
$$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx$$
I tried to use substitution but I got stuck. Can anyone help me?
|
You can also use one of the Euler's substitutions:
$$\begin{align}
\sqrt{x^2+1} &= t-x \\
x &= \frac{1}{2}\left(t - \frac{1}{t}\right) \\
dx &= \frac{1}{2}\left(1 + \frac{1}{t^2}\right)dt \\
\sqrt{x^2+1} &= \frac{1}{2}\left(t + \frac{1}{t}\right) \\
\frac{dx}{\sqrt{x^2+1}}&=\frac{dt}{t}\\
2x^2+3x+1&= \frac{t^4+3t^3-3t+1}{2t^2}
\end{align}
$$
Integration is now straightforward
$$\begin{align}\int\frac{t^4+3t^3-3t+1}{2t^3}dt&=\frac{-t + 6t + 6 t^3 + t^4}{4 t^2}+C\\&=\frac{(t^2+1)(t^2+6 t-1)}{4 t^2}+C \\
&=\frac{1}{2}\left(1+\frac{1}{t}\right)\left[\frac{1}{2}\left(t-\frac{1}{t}\right)+3\right]\\
&=\sqrt{x^2+1}(x+3)
\end{align}$$
As it often happens with the Euler's method, the algebra is tedious, but it gets you there.
|
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|
Prove that that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \geq \frac{(x+y+z)^2}{a+b+c}.$
Prove that that $\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c} \geq \dfrac{(x+y+z)^2}{a+b+c}.$ with $a,b,c$ positive real numbers.
Attempt
I tried using Cauchy-Schwarz, but I can't find the correct $a_i$ and $b_i$. How would you solve this using Cauchy-Schwarz?
|
HINT: the left-hand side minus the right-hand side is equal to
$${z}^{2}{a}^{2}b+{y}^{2}{a}^{2}c+{z}^{2}a{b}^{2}-2\,abcxy-2\,abcxz-2\,a
bcyz+{y}^{2}a{c}^{2}+{x}^{2}{b}^{2}c+{x}^{2}b{c}^{2}
\geq 0$$
can you proceed? (sum of squares!)
|
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|
Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$. Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$.
That's what I've tried:
Let a Cauchy-Schwarz Inequality be :
\begin{array}
(((\sqrt{a} )^2+(\sqrt{b})^2+(\sqrt{c})^2 )\left(\left(\cfrac{ab}{\sqrt{a}}\right)^2 +\left(\cfrac{ac} {\sqrt{b}}\right)^2+\left(\cfrac{bc}{\sqrt{c}}\right)^2\right)& \ge (ab+ac+bc)^2 \\
(a+b+c)(ab^2+\cfrac{a^2 c^2}{b}+b^2 c) &\ge (ab+ac+bc)^2 \\
\end{array}
However how should I now factorize the left hand side of the inequality as $(a^2+b^2+c^2) \cdot Something $ ?
I think I've made the problem harder than it needs to be .
|
I have given a detailed proof using an easier method. Please click on this link to see the proof
|
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|
Prove that $(a^2+2)(b^2+2)(c^2+2)\geq 3(a+b+c)^2$
For the non-negative real numbers $a, b, c$ prove that $$(a^2+2)(b^2+2)(c^2+2)\geq 3(a+b+c)^2$$
What I did is applying Holder's inequality in LHS:$$(a^2+(\sqrt{2})^2)(b^2+(\sqrt{2})^2)(c^2+(\sqrt{2})^2) \geq (abc + 2\sqrt{2})^2$$
Then it suffices to prove that $$(abc+2\sqrt2)^2 \geq 3(a+b+c)^2 \\ \Rightarrow abc+2\sqrt2 \geq \sqrt3(a+b+c)$$
But I don't know how to proceed. I also think I applied Holder's Inequality incorrectly.
|
Among numbers $a-1, b-1, c-1$ there are two having the same sign, say $a-1, b-1$. In other words, $(a-1)(b-1)\ge 0.$ Multiplying both sides by a positive number $(a+1)(b+1)$ we get
\begin{align*}
\left(a^2-1\right)\left(b^2-1\right)&\ge 0 \\
a^2b^2-a^2-b^2+1&\ge 0 \\
a^2b^2+2a^2+2b^2+4&\ge 3\left(a^2+b^2+1\right) \\
\left(a^2+2\right)\left(b^2+2\right) &\ge 3\left(a^2+b^2+1\right) \\
\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right) &\ge 3\left(a^2+b^2+1\right)\left(c^2+2\right).
\end{align*}
Using Schwarz inequality we find
$$\left(a^2+b^2+1\right)\left(c^2+2\right)=\left(a^2+b^2+1\right)\left(1+1+c^2\right) \ge (a+b+c)^2.$$
Therefore
$$\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right) \ge 3\left(a^2+b^2+1\right)\left(c^2+2\right) \ge 3(a+b+c)^2.$$
|
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|
Why this system have one solution Let $b\in (1,2),x\in (0,\frac{\pi}{2})$,if such
$$\begin{cases}
2b^2+b-4=2\sqrt{4-b^2}\cos{x}\\
2b^2-4=2b\cos{(x+\frac{\pi}{18})}-2\sqrt{4-b^2}\cos{\frac{5\pi}{18}}
\end{cases}$$
show that:$$x=\dfrac{\pi}{6}$$
Here is what I already got.
First of all, one should notice equation $x=\dfrac{\pi}{6}$,
$$2b^2+b-4=\sqrt{12-3b^2}$$
then $b$ such
$$b^3-3b+1=0$$
But this kind of proof does not fit my appetite as it not only involves some additional theorem but also not very nice as the simple nice form of the question of itself.
|
In the spirit as Ian Miller's answer, the first equation leads to $$\cos(x)=\frac{2 b^2+b-4}{2 \sqrt{4-b^2}}$$ In the second equation, replace $$\cos{(x+\frac{\pi}{18})}=\cos \left(\frac{\pi }{18}\right) \cos (x)-\sin \left(\frac{\pi }{18}\right) \sin (x)$$ which allows to extract $\sin(x)$ given by $$-\frac{\csc \left(\frac{\pi }{18}\right) \left(-4 \sqrt{4-b^2}-b \left(-2
\sqrt{4-b^2} b+\left(2 b^2+b-4\right) \cos \left(\frac{\pi }{18}\right)+2 b \sin
\left(\frac{2 \pi }{9}\right)\right)+8 \sin \left(\frac{2 \pi
}{9}\right)\right)}{2 b \sqrt{4-b^2}}$$ (nice monster !). Now (do not try to write it !) consider $$f(b)=\sin^2(x)+\cos^2(x)-1=0$$ and plot the function. You will notice that, in the range $1\leq b \leq 2$, there are two roots to the equation. Solving numerically $$b_1\approx 1.41421356237310$$ $$b_2\approx 1.53208888623796$$ The first one is clearly $b_1=\sqrt 2$; for the second root, an inverse symbolic calculator reports that this is a solution of the cubic $x^3-3x+1=0$. Using Cardano method, this equation presents three real roots. So, using the trigonometric method for solving cubic equations, the only acceptable root is given by $$b_2=\sqrt{3} \sin \left(\frac{\pi }{9}\right)+\cos \left(\frac{\pi }{9}\right)=2 \cos \left(\frac{2 \pi }{9}\right)$$ Back to the definition of $\cos(x)$ as a function of $b$, we then find $$b_1=\sqrt 2\implies x_1=\frac{\pi }{3}$$ $$b_2=2 \cos \left(\frac{2 \pi }{9}\right)\implies x_2=\frac{\pi }{6}$$ However, $x_1$ satifies the first equation but not the second (this probably corresponds to a false root introduced by the multiple quaring processes).
So $x_2=\frac{\pi }{6}$ is the only solution.
|
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|
What is the correct way to approach $\int\frac{\cos x}{6+2\sin x-\cos^2x}dx$? I have $$\int\frac{\cos x}{6+2\sin x-\cos^2x}dx .$$
Online solvers have problems with it so what I need are some general guidelines on how to proceed with such as to whether do I need to apply some goniometric identity, or do I need to split it to partial fractions or such. Thanks very much for any help.
|
$$\int\frac{\cos(x)}{6+2\sin(x)-\cos^2(x)}\space\text{d}x=$$
Use $\cos^2(x)=1-\sin^2(x)$:
$$\int\frac{\cos(x)}{5+2\sin(x)+\sin^2(x)}\space\text{d}x=$$
Substitute $u=\sin(x)$ and $\text{d}u=\cos(x)\space\text{d}x$:
$$\int\frac{1}{u^2+2u+5}\space\text{d}u=$$
$$\int\frac{1}{(u+1)^2+4}\space\text{d}u=$$
Substitute $s=u+1$ and $\text{d}s=\text{d}u$:
$$\int\frac{1}{s^2+4}\space\text{d}s=$$
$$\int\frac{1}{4\left(\frac{s^2}{4}+1\right)}\space\text{d}s=$$
$$\frac{1}{4}\int\frac{1}{\frac{s^2}{4}+1}\space\text{d}s=$$
Substitute $p=\frac{s}{2}$ and $\text{d}p=\frac{1}{2}\space\text{d}s$:
$$\frac{1}{2}\int\frac{1}{p^2+1}\space\text{d}p=$$
$$\frac{\arctan\left(p\right)}{2}+\text{C}=$$
$$\frac{\arctan\left(\frac{s}{2}\right)}{2}+\text{C}=$$
$$\frac{\arctan\left(\frac{u+1}{2}\right)}{2}+\text{C}=$$
$$\frac{\arctan\left(\frac{\sin(x)+1}{2}\right)}{2}+\text{C}$$
|
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|
Prove that the equations of common tangents to the two hyperbolas $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ are Prove that the equations of common tangents to the two hyperbolas $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ are $y=x+\sqrt{a^2-b^2},y=x-\sqrt{a^2-b^2},y=-x+\sqrt{a^2-b^2},y=-x-\sqrt{a^2-b^2}$.
I tried to solve the two equations and find the points of intersections of the hyperbolas but these hyperbolas do not intersect each other.
Let the point of tangency be $(x_1,y_1)$ on the $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $(x_2,y_2)$ on the $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$For $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{dy}{dx}=\frac{b^2x}{a^2y}$
For $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\frac{dy}{dx}=\frac{a^2x}{b^2y}$
$\frac{b^2x_1}{a^2y_1}=\frac{a^2x_2}{b^2y_2}$
$b^4x_1y_2=a^4x_2y_1$
I do not know how to take it further.
|
Hint: Implicitly differentiate each, then set the derivatives equal to one another.
|
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|
General technique for finding minimal polynomial? I always have a lot of trouble with these problems, "find the minimal polynomial of {number} over {field}"
What are the general procedures for solving problems of this format?
Thank you for your help
|
I'll assume you're working over $\mathbb Q$.
As an example, take $\alpha = 1+\sqrt{2}$. Notice that $\alpha^2 = 3+2\sqrt{2}$, and so we can write
$$\left[\begin{array}{c} \alpha \\ \alpha^2\end{array}\right] =
\left[\begin{array}{cc} 1 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{c} 1 \\ \sqrt{2}\end{array}\right]$$
Inverting the 2-by-2 matrix on the right gives
$$\left[\begin{array}{cc} -2 & 1 \\ 3 & -1 \end{array}\right]\left[\begin{array}{c} \alpha \\ \alpha^2\end{array}\right]=\left[\begin{array}{c} 1 \\ \sqrt{2}\end{array}\right]$$
Expanding the the first row of the left hand side gives $-2\alpha + \alpha^2 = 1$, i.e.
$$\alpha^2 - 2\alpha-1=0$$
You can do a similar thing with numbers like $\beta = \sqrt{2}+\sqrt{3}$. We note that
$$\begin{eqnarray*}\beta &=& {\bf 0} + {\bf 1}\sqrt{2}+{\bf 1}\sqrt{3}+{\bf 0}\sqrt{6} \\ \\
\beta^2 &=& {\bf 5} + {\bf 0}\sqrt{2}+{\bf 0}\sqrt{3}+{\bf 2}\sqrt{6} \\ \\
\beta^3 &=& {\bf 0} + {\bf 11}\sqrt{2}+{\bf 9}\sqrt{3}+{\bf 0}\sqrt{6} \\ \\
\beta^4 &=& {\bf 49} + {\bf 0}\sqrt{2}+{\bf 0}\sqrt{3}+{\bf 20}\sqrt{6}
\end{eqnarray*}$$
Putting this into a matrix equation, we get
$$\left[\begin{array}{c} \beta \\ \beta^2 \\ \beta^3 \\ \beta^4 \end{array}\right] = \left[ \begin{array}{cccc} 0 & 1 & 1 & 0 \\
5 & 0 & 0 & 2 \\
0 & 11 & 9 & 0 \\
49 & 0 & 0 & 20 \end{array}\right]\left[\begin{array}{c} 1 \\ \sqrt{2} \\ \sqrt{3} \\ \sqrt{6}\end{array}\right]$$
If you invert the 4-by-4 matrix then you get
$$\left[\begin{array}{cccc} 0 & 10 & 0 & -1 \\ -9/2 & 0 & 1/2 & 0 \\
11/2 & 0 & -1/2 & 0 \\ 0 & -49/2 & 0 & 5/2 \end{array}\right]\left[\begin{array}{c} \beta \\ \beta^2 \\ \beta^3 \\ \beta^4 \end{array}\right]=\left[\begin{array}{c} 1 \\ \sqrt{2} \\ \sqrt{3} \\ \sqrt{6}\end{array}\right]$$
Reading off the first row gives
$$\beta^4-10\beta^2+1=0$$
|
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Evaluate $ \lim_{x \rightarrow - \infty} \frac{\sqrt{9x^6-x}}{x^3+1} $ I have started learning limits in calculus and I came across this question:
Evaluate $ \displaystyle \lim_{x \rightarrow \infty} \dfrac{\sqrt{9x^6-x}}{x^3+1} $ .
I rewrite the above as $ \displaystyle \lim_{x \rightarrow \infty} \dfrac{\sqrt{9-\dfrac{1}{x^5}}}{1+\dfrac{1}{x^3}} = \lim_{x \rightarrow \infty} \dfrac{\sqrt{9}}{1} = \boxed{3} $
But now I am asked to compute $ \displaystyle \lim_{x \rightarrow -\infty} \dfrac{\sqrt{9x^6-x}}{x^3+1} $
How to solve for minus infinity? Where to put minus sign , I am getting confused , please help. Thanks.
|
Notice, one can easily change limit as $x\to +\infty$ as follows $$\lim_{x\to -\infty}\frac{\sqrt{9x^6-x}}{x^3+1}$$
$$=\lim_{x\to +\infty}\frac{\sqrt{9(-x)^6-(-x)}}{(-x)^3+1}$$
$$=\lim_{x\to +\infty}\frac{\sqrt{9x^6+x}}{1-x^3}$$
$$=\lim_{x\to +\infty}\frac{|3x^3|\sqrt{1+\frac{1}{9x^5}}}{x^3\left(\frac{1}{x^3}-1\right)}$$
$$=\lim_{x\to +\infty}\frac{3x^3\sqrt{1+\frac{1}{9x^5}}}{x^3\left(\frac{1}{x^3}-1\right)}$$
$$=3\lim_{x\to +\infty}\frac{\sqrt{1+\frac{1}{9x^5}}}{\frac{1}{x^3}-1}$$
$$=3\cdot \frac{\sqrt{1+0}}{0-1}=\color{red}{-3}$$
|
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|
What is the inverse of $f(x) = \dfrac{x}{x^2 - 1}$ I need to find a continuous inverse of $$f(x) = \dfrac{x}{x^2 - 1}$$
Let $y = f(x) = \dfrac{x}{x^2 - 1} \Rightarrow y(x^2-1) = x \Rightarrow yx^2-x = y$
How should I proceed from here?
|
Let $f(x) = y = \frac{x}{x^{2} - 1}.$ To find the inverse of $f(x),$ we switch $x$ and $y$ in the equation and solve for $y.$ This yields $x = \frac{y}{y^{2} - 1},$ which we can rearrange to get $xy^{2} - y - x = 0.$ This is a quadratic with respect to $y.$ Using the quadratic equation, we get that $y = \boxed{f(x) = \frac{1 \pm \sqrt{1 + 4x^{2}}}{2x}}.$
|
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|
Prove that $ \sum\limits_{cyc} \sqrt{\cot{A}+\cot{B}} \ge 2\sqrt2$
Let $ \triangle ABC$ be an acute-angled triangle. Prove that
$ \sum\limits_\text{cyc} \sqrt{\cot{A}+\cot{B}} \ge 2\sqrt2 $
Attempt
Since $\triangle{ABC}$ is acute, we may say that $A,B,C \in (0, \frac{\pi}{2})$. Now, we have that by a result for triangles $\displaystyle \sum_{cyc} \cot{A} \cdot \cot{B} = 1$. Then see that $$ \sum_{cyc} \sqrt{\cot{A}+\cot{B}} = \sqrt{\cot{A}+\cot{B}}+\sqrt{\cot{B}+\cot{C}}+\sqrt{\cot{A}+\cot{C}}.$$ Now see that by Cauchy-Schwarz $\sqrt{(\cot{A}+\cot{B})(2)} \geq (\sqrt{\cot{A}}+\sqrt{\cot{B}})$ and thus $\sqrt{\cot{A}+\cot{B}}+\sqrt{\cot{B}+\cot{C}}+\sqrt{\cot{A}+\cot{C}} \geq \sqrt{2}(\sqrt{\cot{A}}+\sqrt{\cot{B}}+\sqrt{\cot{C}})$. Now since $$\cot{A}\cot{B}+\cot{B}\cot{C}+\cot{A}\cot{C} = 1.$$
I get stuck here.
|
Let $\cot\alpha=a$, $\cot\beta=b$ and $\cot\gamma=c$.
Hence, $ab+ac+bc=1$ and we need to prove that $\sum\limits_{cyc}\sqrt{a+b}\geq2\sqrt2$ or
$a+b+c+\sum\limits_{cyc}\sqrt{a^2+1}\geq4$, which is true because by C-S
$a+b+c+\sum\limits_{cyc}\sqrt{a^2+1}=a+b+c+\frac{1}{2}\sum\limits_{cyc}\sqrt{(1+3)(a^2+1)}\geq$
$\geq a+b+c+\frac{1}{2}\sum\limits_{cyc}(a+\sqrt3)=\frac{3}{2}(a+b+c)+\frac{3\sqrt{3}}{2}\geq\frac{3\sqrt{3}}{2}+\frac{3\sqrt{3}}{2}=3\sqrt3>4$
|
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|
Integers represented by $x^2 + 3 y^2$ vs. integers represented by $x^2 + x y + y^2$. How does one show that the quadratic forms $x^2 + 3 y^2$ and $x^2 + x y + y^2$ represent the same set of integers?
I think it relates to a classical result of Euler about primes of form $6k+1$. In fact a positive integer $n$ is of the form $x^2 + 3 y^2$ if and only if the the primes $\equiv -1 \mod 3$ have an even exponent in $n$. Similarly for $x^2 + x y + y^2$.
However somebody told me that this can be shown without so much number theoretical background. Any ideas would be appreciated!
|
Others have already answered well; the principal form $x^2 + xy + k y^2$ always represents a superset of the numbers represented by $x^2 + (4k-1)y^2.$ For $k=1,$ the two sets agree. Same for $k = -1,$ so $x^2 + xy - y^2$ represents the same numbers as $x^2 - 5 y^2.$
It is always the case that $x^2 + xy + 2k y^2$ always represents the same ODD numbers represented by $x^2 + (8k-1)y^2.$ So, $x^2 + xy + 2y^2$ and $x^2 + 7 y^2$ represent the same odd numbers. Same with $x^2 + xy -4y^2$ and $x^2 - 17 y^2$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $3 \le a+b+c \le 2\sqrt{3}$ in a triangle
Let $a,b,$ and $c$ be the lengths of the sides of a triangle satisfying $ab+bc+ca = 3.$ Prove that $3 \le a+b+c \le 2\sqrt{3}$.
The idea I had was $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) = a^2+b^2+c^2+6 \geq 9$ by rearrangement. That takes care of the first inequality. How do I show the other inequality?
|
Why is $a^2+b^2+c^2\geq 3$?
I think you need to apply somewhere that you are dealing with a triangle.
Maybe something like this?
If you sum up the following inequalities,
$$ab+bc=b(a+c)> b^2$$
$$ab+ac=a(b+c)> a^2$$
$$ac+bc=c(a+b)> c^2$$
you obtain $a^2+b^2+c^2\leq 6$ which by your argument gives you
$(a+b+c)^2<12.$ This implies $a+b+c\leq 2< 3.$
However, the left-hand side of the inequality follows from Cauchy-Schwarz inequality for triples $(a,b,c)$ and $(b,c,a)$ and
$$ab+bc+ca\leq \sqrt{a^2+b^2+c^2} \cdot \sqrt{b^2+c^2+a^2}=a^2+b^2+c^2.$$
This finally implies $a^2+b^2+c^2\geq 3$ as you assumed before.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $a^4 + 4^b$ is prime, then $a$ is odd and $b$ is even. We say an integer $p>1$ is prime when its only positive divisors are $1$ and $p$. Let $a$ and $b$ be natural number not both $1$. Prove that if $a^4+4^b$ is prime, then $a$ is odd and $b$ is even.
I'm also given a hint:
Consider the expression $(x^2-2xy+2y^2)(x^2+2xy+2y^2)$.
What I have so far is:
Let $x=a^2, y=2^b$ and substitute it into the expression.
But I am not sure how to proceed. Can anyone help please?
|
It's clear that $a$ must be odd, as otherwise the number would be divisible by $2$.
Suppose $b$ is odd. Then we can write our number as $a^4 + 4\cdot 4^{2n} = a^4 + 4 \cdot 2^{4n}$ for some $n$. You were given the key factorization,
$$ (a^4 + 4 b^4) = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2).$$
here, that means
$$ a^4 + 4 \cdot 2^{4n} = (a^2 + 2 a 2^n + 2\cdot 2^{2n})(a^2 - 2 a 2^n + 2\cdot 2^{2n}),$$
which is a nontrivial factorization. Thus $b$ cannot be odd.
Aside:
This factorization identity is called Sophie Germain's identity, after the French mathematian and general polymath who noticed it while exploring number theory. She corresponded with Gauss and Legendre, and attended lectures at the newly founded École Polytechnique. To avoid ridicule, as it was extremely uncommon for women to pursue math (or indeed, to be allowed to pursue a great many professions), she used a male pen name for her initial correspondence.
|
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|
Dealing with a difficult sum of binomial coefficients, $\sum_{l=0}^{n}\binom{n}{l}^{2}\sum_{j=0}^{2l-n}\binom{l}{j} $
I am interested in finding an upper bound for the sum
$$F(n)= \sum_{l=0}^{n}\binom{n}{l}^{2}\;\sum_{j=0}^{2l-n}\binom{l}{j}$$
Ideally it should be possible to evaluate it exactly using some combinatorial identities or generating functions. Note that the $\binom{n}{l}$ term is squared, and that when $2l-n<0$, we define the inner to be zero.
Simple upper bound: By noticing that $2l-n\leq n$, we have that $$\sum_{j=0}^{2l-n}\binom{l}{j}\leq 2^n,$$ and so $$F(n)\leq 2^n \sum_{l=0}^{n}\binom{n}{l}^{2}=2^n\binom{2n}{n}.$$ (See this question for a proof that $\sum_{l=0}^{n}\binom{n}{l}^{2}=\binom{2n}{n}$) Modifying the above based on when the inner sum is zero, that is when $2l\leq n$, we can replace this upper bound by $$F(n)\leq 2^{n-1}\binom{2n}{n}.$$ However it should be possible to do significantly better than this.
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By way of an intermittent progress report I submit several integral
representations of the sum
$$\sum_{k=0}^n {n\choose k}^2 \times
\sum_{j=0}^{2k-n} {k\choose j}.$$
They were obtained using the Egorychev method.
The reader is cordially invited to comment, verify and prove these.
Saddle point asymptotics could be attempted if these were univariate
integrals. I present them in the hope that perhaps a univariate
generating function can be found where the catalog of asymptotics of
coefficient extractor integrals could be applied. Sometimes there
exist substitutions which produce radical cancellation /
simplification in these binomial sum integrals.
First representation:
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{1-w} \frac{1}{w^{n+1}}
(1 + w + w^2 z)^n
\; dw\; dz.$$
Second representation:
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+2z)^n (1+z)^n
\; dz
\\ +
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{2w-1} \frac{1}{(1-w)^{n+1}}
\frac{1}{w^n}
((1-w)^2 z + w)^n
\; dw\; dz.$$
Third representation:
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n+1}}\frac{(1+w)^n}{1-w}
\frac{1}{z-w/(1-w)}
((1+z)^2 w^2 + z)^n
\; dw\; dz.$$
Fourth representation:
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n+1}}
\frac{1}{1-w}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
(1 + z + wz)^n
(1 + w^2 z)^n
\; dz\; dw.$$
Fifth representation:
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n+1}} \frac{1}{1-w}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\frac{1}{(1-z)^{n+1}}
(1 + w - wz)^n
(z + w^2 - w^2 z)^n
\; dz\; dw $$
Sixth representation:
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n+1}}
\frac{1}{1-w}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} \frac{1}{(1-z)^{n+1}}
(w+1-z)^n
\; dz\; dw
\\ - \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{2n}}
\frac{1}{1-w}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} \frac{1}{(1-z)^{n+1}}
(w+1-z)^n (1-z + w^2 z)^n
\; dz\; dw .$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
double integral getting different results I am trying to calculate the double integral $$\lim_{b \to 0^+} \int_{b}^1 \int_b^1 \frac{y-x}{(y+x)^3}dydx$$ If you plug this into wolfram, you get $-\frac{1}{2}$ and if you plug it into symbolab you get $\frac{1}{2}$ I will show you my steps, I just want to make sure I got the right answer.
$$\lim_{b \to 0^+} \int_{b}^1 \int_b^1 \frac{y-x}{(y+x)^3}dydx=\lim_{b \to 0^+} \int_{b}^1 \int_b^1 \frac{y+x}{(y+x)^3}-\frac{2x}{(y+x)^3}dydx$$
$$=\lim_{b \to 0^+} \int_{b}^1 \frac{-1}{(1+x)^2}dx=\lim_{b \to 0^+} \frac{1}{1+x}\Big|_b^1=\frac{-1}{2}$$ I just wanted to verify because these two different websites are giving me different answers.
|
I may be wrong but I think the correct result is $0$.
Where I think you got it wrong is in the second equality, where you exchange limit and the integral for one term.
Here I sketched some more detailed computations:
\begin{align}
\lim_{b \to 0}
\int_b^1 \int_b^1 \frac{y-x}{(y+x)^3}dydx
{}={} &
\lim_{b \to 0}
\int_b^1 \int_b^1
\left(\frac{y+x}{(y+x)^3}-\frac{2x}{(y+x)^3}\right)dydx
\\
{}={} &
\lim_{b \to 0}
\int_b^1
\left(
-\frac{1}{y+x}\Bigg|_{y=b}^{y=1}
+
\frac{2x}{2(y+x)^2}\Bigg|_{y=b}^{y=1}
\right)dx
\\
{}={} &
\lim_{b \to 0}
\int_b^1
\left(
\frac{1}{b+x}
-\frac{1}{1+x}
+
\frac{x}{(1+x)^2}
-\frac{x}{(b+x)^2}
\right)dx
\\
{}={} &
\lim_{b \to 0}
\int_b^1
\left(
\frac{-1}{(1+x)^2}
+\frac{b}{(b+x)^2}
\right)dx
\\
{}={} &
\lim_{b\to0}
\left(
\frac{1}{1+x}
-\frac{b}{b+x}
\Bigg|_{x=b}^{x=1}
\right)
\\
{}={} &
\lim_{b\to0}
\left(
\frac{1}{2}
-\frac{1}{1+b}
+\frac{b}{2b}
-\frac{b}{b+1}
\right)
\\
{}={} &
0
\end{align}
|
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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.