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Calculating limit of function To find limit of $\lim_{x\to 0}\frac {\cos(\sin x) - \cos x}{x^4} $.
I differentiated it using L Hospital's rule. I got
$$\frac{-\sin(\sin x)\cos x + \sin x}{4x^3}\text{.}$$ I divided and multiplied by $\sin x$.
Since $\lim_{x\to 0}\frac{\sin x}{x} = 1$, thus I got
$\frac{1-\cos x}{4x^2}$.On applying standard limits, I get answer $\frac18$. But correct answer is $\frac16$. Please help.
|
Using Prosthaphaeresis Formulas,
$$\cos(\sin x)-\cos x=2\sin\frac{x-\sin x}2\sin\frac{x+\sin x}2$$
So, $$\frac{\cos(\sin x)-\cos x}{x^4}=2\frac{\sin\frac{x-\sin x}2}{\frac{x-\sin x}2}\frac{\sin\frac{x+\sin x}2}{\frac{x+\sin x}2}\cdot\frac{x-\sin x}{x^3}\cdot\frac{x+\sin x}x\cdot\frac14$$
We know, $\lim_{h\to0}\frac{\sin h}h=1$
$$\text{Apply L'Hospital's Rule on }\lim_{x\to0}\frac{x-\sin x}{x^3}$$
$$\text{and we get }\lim_{x\to0}\frac{x+\sin x}x=1+\lim_{x\to0}\frac{\sin x}x$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$ I need a hand in showing that $$ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$$
Thanks in advance for any help.
|
For $m=2$, it amounts to proving $\binom{2}{2} = \binom{3}{3}$, which is true since both equal $1$.
Induction step: let's assume the formula is true for a given $m$,
$$\sum_{n=2}^m \binom{n}{2}=\binom{m+1}{3}$$
Then,
$$\sum_{n=2}^{m+1} \binom{n}{2} = \sum_{n=2}^{m} \binom{n}{2} + \binom{m+1}{2}$$
$$=\binom{m+1}{3}+\binom{m+1}{2}=\binom{m+2}{3}$$
And you are done, by induction.
By the way, the same reasoning would give you for any $k \ge 0$,
$$\sum_{n=k}^m \binom{n}{k} = \binom{m+1}{k+1}$$
For a better understanding of what it means, I suggest you draw Pascal's triangle and see the sum of binomial coefficients in a column, then the sum is at the bottom of this column, one step downward and one step to the right, as in the following:
$$ \begin{array}{cccccccc}
1 & & & & & & &\\
1 & 1 & & & & & &\\
1 & 2 & \color{red}{1} & & & & &\\
1 & 3 & \color{red}{3} & 1 & & & &\\
1 & 4 & \color{red}{6} & 4 & 1 & & &\\
1 & 5 & \color{red}{10} & 10 & 5 & 1& & &\\
1 & 6 & \color{red}{15} & 20 & 15 & 6& 1& &\\
1 & 7 & 21 & \color{blue}{35} & 35 & 21& 7& 1&\end{array}$$
|
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|
Convex Quadrilateral: $ \dfrac {\tan A + \tan B + \tan C + \tan D}{\tan A \tan B \tan C \tan D} = \cot A + \cot B + \cot C + \cot D $ Problem
Let $ABCD$ be a convex quadrilateral with no right angles. Show that $$ \dfrac {\tan A + \tan B + \tan C + \tan D}{\tan A \tan B \tan C \tan D} = \cot A + \cot B + \cot C + \cot D. $$
Source: Geometry Unbound by Kiran Kedlaya.
Attempt: Well, all we really know about a convex quadrilateral are that $ \angle A + \angle B + \angle C + \angle D = 360^\circ $ and that the polygon is convex. Well, that's obvious. But any starts to a proof of this useful fact would be helpful!
|
$$A+B+C+D=2\pi\iff A+B=2\pi-(C+D)$$
$$\implies\tan(A+B)=\tan\{2\pi-(C+D)\}=-\tan(C+D)$$
Using Addition formula, $$\frac{\tan A+\tan B}{1-\tan A\tan B}=-\frac{\tan C+\tan D}{1-\tan C\tan D}$$
On rearrangement, $$\sum \tan A=\sum \tan A\tan B\tan C=\prod\tan A\left(\sum\frac1{\tan A}\right)$$
$$\sum \tan A=\left(\prod\tan A\right)\left(\sum\cot A\right)$$
Divide either sides by $\prod\tan A$ which is non-zero and finite as no angle is multiple of $\displaystyle\frac\pi2$
Interestingly, $\displaystyle A+B+C+D=n\pi$ (where $n$ is any integer) will satisfy this relation
|
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|
Limit of a fraction of double factorials How can we show that
$\begin{align*}
\lim_{n\rightarrow\infty} U_n = 0
\end{align*}$
where
$\begin{align*}
U_n = \frac{(n-1)!!}{n!!}=\frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4}\cdots
\end{align*}$
terminates at $\displaystyle\frac{2}{3}$ (odd) or $\displaystyle\frac{1}{2}$ (even).
At first glance I thought it was 1 because each individual multiplied fraction goes to 1 as $n\rightarrow\infty$. Mahthematica says this is not case, however.
Does it help if I write it in recurrence relation?
$\begin{align*}
U_{n+2} = \frac{n+1}{n+2}U_n\\
U_{n+1} = \frac{1}{n+1}\frac{1}{U_n}
\end{align*}$
|
First observe the following elementary inequality
\begin{gather*}
\frac{m}{m+1}<\sqrt{\frac{m}{m+2}},\qquad \forall m\in\mathbb{N}
\end{gather*}
where $\mathbb{N}$ denotes the set of all the positive integers.
In the case of $n$ is even, we have, by using the above inequality,
\begin{align*}
0<U_n&=\frac{1}{2}\cdot \frac{3}{4}\cdot\frac{5}{6}\cdot\cdots\cdot\frac{2n-3}{2n-2}\cdot\frac{2n-1}{2n}\\
&<\sqrt{\frac{1}{3}}\cdot \sqrt{\frac{3}{5}}\cdot \sqrt{\frac{5}{7}}\cdot\cdots\cdot \sqrt{\frac{2n-3}{2n-1}}\cdot \sqrt{\frac{2n-1}{2n+1}}\\
&=\frac{1}{\sqrt{2n+1}}<\frac{1}{\sqrt{n}},
\end{align*}
and in the case of $n$ is odd, we have similarly
\begin{align*}
0<U_n&=\frac{2}{3}\cdot \frac{4}{5}\cdot\frac{6}{7}\cdot\cdots\cdot\frac{2n-2}{2n-1}\cdot\frac{2n}{2n+1}\\
&<\sqrt{\frac{2}{4}}\cdot \sqrt{\frac{4}{6}}\cdot \sqrt{\frac{6}{8}}\cdot\cdots\cdot \sqrt{\frac{2n-2}{2n }}\cdot \sqrt{\frac{2n }{2n+2}}\\
&=\frac{\sqrt{2}}{\sqrt{2n+2}}<\frac{1}{\sqrt{n}}.
\end{align*}
In summary, we have verified that
\begin{gather*}
0<U_n<\frac{1}{\sqrt{n}},\qquad \forall n\in\mathbb{N}.
\end{gather*}
Then by utilizing the squeezing test and the fact that
\begin{gather*}
\lim_{n\to\infty}\frac{1}{\sqrt{n}}=0,
\end{gather*}
we can get that
\begin{gather*}
\lim_{n\to\infty}U_n=0.
\end{gather*}
|
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|
exponential equation with a sum of exponents I'm trying to solve the following exponential equation:
$e^{2x} - e^{x+3} - e^{x + 1} + e^4 = 0$
According to the the text I am using the answer should be $x = 1,3$ but I can't derive the appropriate quadratic $x^2 -4x + 3$ from the above equation using any of the methods I know. Can someone point me in the right direction? Is there a substitution I'm not seeing?
|
$$e^{2x} - (e^3 +e)e^x + e^4$$
Let $m = e^x$
$$m^2- (e^3 + e)m + e^4$$
$a = 1$, $b=-(e^3 + e)$, $c =e^4$
$$m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{(e^3 + e) \pm \sqrt{{(-(e^3 + e))}^2 -4(1)(e^4)}}{2(1)}$$
$$= \frac{(e^3 + e) \pm \sqrt{e^6 +e^2 + 2e^4 - 4e^4}}{2}$$
$$= \frac{(e^3 + e) \pm \sqrt{e^4 +1 + 2e^2 - 4e^2} \sqrt{e^2}} {2}$$
$$= \frac{(e^3 + e) \pm \sqrt{e^4 - 2e^2 + 1} \sqrt{e^2}} {2}$$
$$= \frac{(e^3 + e) \pm \sqrt{(e^2 - 1)^2} \cdot e} {2}$$
$$=\frac{e^3 + e \pm e^2 - 1 \cdot e}{2}$$
$$=\frac{e^3 + e \pm (e^3 -e)}{2}$$
$$m = \frac{e^3 +e + e^3 - e}{2}, \frac{e^3 + e - e^3 + e}{2}$$
$$\implies m = \frac{2e^3}{2}, \frac{2e}{2}$$
$$m = e^3, e$$
But $m= e^x$
$$e^x = e^3, e^x =e$$
Since the bases are the same, you can equate $x = 1,3$
|
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"timestamp": "2023-03-29T00:00:00",
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|
evaluation of $\int\frac{x^5}{x^5+x+1}dx$ $\displaystyle \int\frac{x^5}{x^5+x+1}dx$
$\bf{My\; Try::}$ $\displaystyle \int\frac{x^5}{x^5+x+1}dx = \int\frac{\left(x^5+x+1\right)-(x+1)}{x^5+x+1}dx = x-\int\frac{x+1}{x^5+x+1}dx$
Now Let $\displaystyle I = \int\frac{x+1}{x^5+x+1}dx = \int \frac{x+1}{(x^2+x+1)\cdot (x^3-x^2+1)}$
Now I Did not understand how can i solve after that
Help Required
Thanks
|
A very nasty solution:
You have observed that $x^5 + x + 1 = (x^2 + x + 1)(x^3 - x^2 + 1)$. This means that you can factor $x^5 + x + 1$ into linear factors :$x^5 + x + 1 = \prod_{i=1}^5 (x-\alpha_i)$, where $\alpha_i$ can be computed. Thus, for any polynomial $p(x)$ or degree less than $5$ you can find unique constants $c_i$ such that:
$$ \frac{p(x)}{x^5 + x + 1 } = \sum_{i=1}^5 \frac{c_i}{x - \alpha_i}$$
Now, each of these terms can be integrated:
$$ \int \frac{p(x)}{x^5 + x + 1 } dx = \sum_{i=1}^5 c_i \ln(x - \alpha_i).$$
So, in principle - problem solved! But please, don't ask me to do the actual computations. Wolfram can do it, though.
|
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|
How do I prove that $\lim_{n\to+\infty}\frac{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}}{\sqrt{n}}=?$
let sequence $\{a_{n}\}$ such $a_{1}=1$,and
$$a_{n+1}a_{n}=n,n\ge 1$$
show that
$$2\sqrt{n}-1\le\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}<\dfrac{5}{2}\sqrt{n}-1$$
(2): I consider we can find this limit
$$\lim_{n\to+\infty}\dfrac{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}}{\sqrt{n}}=?$$
My try:since
$$a_{n+2}a_{n+1}-a_{n+1}a_{n}=n+1-n=1$$
so
$$a_{n+2}=\dfrac{1}{a_{n+1}}+a_{n}$$
so
$$\dfrac{1}{a_{n+1}}=a_{n+2}-a_{n}$$
so
\begin{align*}\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}&=a_{1}+(a_{3}-a_{1})+(a_{4}-a_{2})+(a_{5}-a_{3})+\cdots+(a_{n+1}-a_{n-1})\\
&=a_{1}+a_{n+1}+a_{n}-a_{1}-a_{2}\\
&=a_{n+1}+a_{n}-a_{2}
\end{align*}
since $$a_{1}=1,a_{1}a_{2}=1\Longrightarrow a_{2}=1$$
so
$$a_{n+1}+a_{n}-a_{2}\ge2\sqrt{a_{n+1}a_{n}}-1\ge 2\sqrt{n}-1$$
so left hand inequality is prove it.Now consider Right hand inequality,we only prove this
$$a_{n}+a_{n+1}<\dfrac{5}{2}\sqrt{n}$$
since
$$a_{n}a_{n+1}=n\Longrightarrow a_{n}+\dfrac{n}{a_{n}}<\dfrac{5}{2}\sqrt{n}$$
so maybe we can $a_{n}<2\sqrt{n}$?
becasue $$\dfrac{5}{2}\sqrt{n}=2\sqrt{n}+\dfrac{n}{2\sqrt{n}}$$
so let $$f(x)=x+\dfrac{n}{x}$$
if we can prove $a_{n}<2\sqrt{n}$,Then
$$f(x)\le f(2\sqrt{n})=\dfrac{5}{2}\sqrt{n}$$
Thank you very much
|
For even indices
$$
\frac1{a_{2n}}=\frac{(2n-2)!!}{(2n-1)!!}=\frac1{2n}\frac{4^n}{\binom{2n}{n}}\sim\frac{\sqrt{\pi n}}{2n}
$$
For odd indices
$$
\frac1{a_{2n+1}}=\frac{(2n-1)!!}{(2n)!!}=\frac{\binom{2n}{n}}{4^n}\sim\frac1{\sqrt{\pi n}}
$$
Therefore,
$$
\begin{align}
\frac1{\sqrt{2n}}\sum_{k=1}^{2n}\frac1{a_k}
&\sim\left(\sqrt\pi+\frac2{\sqrt\pi}\right)\frac{\sqrt{n}}{\sqrt{2n}}\\
&=\sqrt{\frac\pi2}+\sqrt{\frac2\pi}\\[9pt]
&\doteq2.05119869811837
\end{align}
$$
|
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|
Evaluate integral $\int_0^\frac{\pi}{2}x\ln(\sin x)~dx$ I've corrected typing error in the integral.
I apologize for my mistake.
Reedited question:
Can anybody solve integral:
$$\int_0^\frac{\pi}{2}x\ln(\sin x)~dx$$
I'm just trying to guess some simple formula for $\zeta(3)$. My "strategy" is simple: Find some conjectures and check them numerically.
Value of similar integral is well known:
$$\int_0^\pi x\ln(\sin x)~dx=-\dfrac{\pi^2}{2}\ln(2) $$
Is there any idea for antiderivative of $x\ln(\sin x)$? It's too difficult to solve it for me.
Any ideas?
|
Note
\begin{align}\int_0^\frac{\pi}{2}x\ln(2\sin x)~dx
=\frac12 \int_0^\frac{\pi}{2}x\ln(2\sin 2x)~dx
+ \frac12 \int_0^\frac{\pi}{2}x\ln\frac{\sin x}{\cos x}~dx\tag1\\
\end{align}
where
\begin{align}
&\int_0^\frac{\pi}{2} \overset{2x\to x} {x\ln(2\sin 2x)}~dx
=\frac14 \int_0^{\pi} \overset{x\to\pi-x} {x\ln(2\sin x)}~{dx}
=\frac\pi8 \int_0^{\pi}\ln(2\sin x)~dx =0
\end{align}
and
\begin{align}
&\int_0^\frac{\pi}{2}x\ln\frac{\sin x}{\cos x} ~dx
\overset{t=\tan x}= \int_0^\infty \frac{\ln t\tan^{-1}t}{1+t^2} dt \\
=& \int_0^\infty \frac{\ln t}{1+t^2} \int_0^1 \frac t{1+ t^2 y^2}dy ~dt
\overset{t^2\to t}=\frac14\int_0^1 \int_0^\infty \frac{\ln t}{(1+t)(1+ t y^2)} \overset{t\to 1/(y^2t)}{dt ~dy}\\
=& \frac14\int_0^1 \int_0^\infty \frac{-\ln t -\ln y^2}{(1+t)(1+ t y^2)}dt ~dy = -\frac14\int_0^1 \int_0^\infty \frac{\ln y}{(1+t)(1+ t y^2)}dt ~dy\\
=& \frac12\int_0^1 \frac{\ln^2 y}{1-y^2}dy
= \frac12\int_0^1 \frac{\ln^2 y}{1-y}dy - \frac12\int_0^1 \frac{y \ln^2 y}{1-y^2}\overset{y^2\to y}{dy}\\
=& \frac7{16}\int_0^1 \frac{\ln^2 y}{1-y}dy
= \frac7{16}\cdot 2\zeta(3)=\frac 78\zeta(3)
\end{align}
Substitute above results into (1) to get
$\int_0^\frac{\pi}{2}x\ln(2\sin x)~dx= \frac7{16}\zeta(3)$, which leads to
$$ \int_0^\frac{\pi}{2}x\ln(\sin x)~dx= \frac7{16}\zeta(3)-\frac{\pi^2}8\ln2$$
|
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|
simplify $\sqrt{3+2\sqrt{2}}-\sqrt{4-2\sqrt{3}}$ Simplify $\sqrt{3+2\sqrt{2}}-\sqrt{4-2\sqrt{3}}$
To do it I have see it that we have basically $\sqrt{2}$ and $\sqrt{3}$ that is we can write it as,
$$\sqrt{\sqrt{3}\sqrt{3}+\sqrt{2}\sqrt{2}\sqrt{2}}-\sqrt{\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{2}-\sqrt{2}\sqrt{2}\sqrt{3}}$$
But even with that I don't get that result.
|
Hints:
$$3+2\sqrt2=(\sqrt2)^2+2\sqrt2+1\qquad 4-2\sqrt3=(\sqrt3)^2-2\sqrt3+1$$
|
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|
Establish $\int_0^{\infty} \frac{x^a}{x^2 + b^2}dx = \frac{\pi b^{a-1}}{2 \cos(\pi a /2)}$ when $-1 < a < 1$ My attempt at a solution: (this is homework, btw)
Let $f(z) = \frac{z^a}{z^2 + b^2}dz$ then the singularities of $f$ occur at $\pm ib$.
$$
Res(f; ib) = \frac{z^a}{z + ib} \biggr |_{ib} = \frac{(ib)^a}{2ib}
$$
$$
Res(f; -ib) = \frac{z^a}{z - ib} \biggr |_{-ib} = \frac{(-ib)^a}{-2ib}
$$
We sum the residues and multiply by $2 \pi i$ to give the value of a contour integral containing the two poles.
$$
2 \pi i \cdot \biggl ( \frac{(ib)^a}{2ib} - \frac{(-ib)^a}{2ib} \biggr ) = \pi b^{a-1} (i^a - (-i)^a) = \pi b^{a-1} (e^{\pi i a/2} - e^{- \pi i a/2}) = \pi b^{a-1} 2 i \sin (\pi i a/2)
$$
We use a circle contour centered at the origin with the origin cut out (since $a$ can be negative)
Also we make the branch cut for $x^a$ along the positive real axis.
$\gamma = [r, R]$
$\mu_R = Re^{i \theta}$, where $0 < \theta < 2\pi$
$\kappa = [-R, -r]$
$\mu_r = re^{i \theta}$, where $0 < \theta < 2\pi$
$$
\pi b^{a-1} 2 i \sin (\pi i a/2) = \int_r^R \frac{x^a}{x^2 + b^2}dx + \int_0^{2\pi} \frac{(Re^{i \theta})^a}{(Re^{i \theta})^2 + b^2}i R e^{i \theta} d \theta + \int_{R}^{r} -\frac{x^a}{x^2 + b^2}dx + \int_{2\pi}^0 \frac{(re^{i \theta})^a}{(re^{i \theta})^2 + b^2}i r e^{i \theta} d \theta
$$
As $r \rightarrow 0$ and $R \rightarrow \infty$ we have that the second and fourth integrals go to zero. Thus,
$$
\pi b^{a-1} 2 i \sin (\pi i a/2) = 2 \int_0^{\infty} \frac{x^a}{x^2 + b^2}dx \implies \int_0^{\infty} \frac{x^a}{x^2 + b^2}dx = \frac{\pi b^{a-1} i \sin (\pi i a/2)}{2}
$$
So where did I go wrong?
|
Setting $m=2$ in this answer, it is shown that
$$
\frac{\pi}{2}\csc\left(\pi\frac{a+1}{2}\right)=\int_0^\infty\frac{x^a}{1+x^2}\,\mathrm{d}x
$$
Thus,
$$
\begin{align}
\int_0^\infty\frac{x^a}{b^2+x^2}\,\mathrm{d}x
&=\frac{\pi}{2}\csc\left(\pi\frac{a+1}{2}\right)b^{1-a}\\
&=\frac{\pi\,b^{1-a}}{2\cos\left(\frac{\pi a}{2}\right)}
\end{align}
$$
|
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|
Is it possible to express $\sin \frac{\pi}{9}$ in terms of radicals? So, yes, this is a math homework question. I've done some research on it and I know that the actual value for $\sin \frac{\pi}{9}$ cannot be expressed without using imaginary numbers.
http://intmstat.com/blog/2011/06/exact-values-sin-degrees.pdf
But, this isn't what the question is asking. It is simply asking if it is possible to do so and for me to prove it. I know that $\frac{3\pi}{9}$ can be simplified to $\frac{\pi}{3}$ and that exact values for the sine and cosine of it can be expressed cleanly and other multiples that can be reduced down to $\frac{\pi}{6}$, $\frac{\pi}{4}$, $\frac{\pi}{3}$ etc.
But how can I prove that $\frac{\pi}{9}$ itself can be expressed as an exact value?
I'm in grade 12 advanced functions and am taking calculus next semester, but I'm totally open to learning new things so if you post very advanced concepts I'll do my best to understand them.
Any ideas where I could start?
Unit circle:
|
$$\sum_{k=1}^{\infty} \frac{1}{6}(-\frac{1}{2})^{k} = \frac{1}{9}\\$$
The basic unit circle goes all the way up to integral multiples of $\frac{\pi}{6}$; thus, it is reasonable to assume that $\sin{\frac{\pi}{9}}$ can reasonably be written as an infinite nested root through half-angle formulas. The resulting numerators of the above stated geometric sequence follow the pattern such that, assuming $a_{0}$ is 1, and ${k}$ follows suit with ${n}$:
\begin{align*}
a_{n+1} = 2a_{n}-1 \ \ \textrm{if} \ \ {n}\mod{2} = 0\\
a_{n+1} = 2a_{n}+1 \ \ \textrm{if} \ \ {n}\mod{2} = 1
\end{align*}
Because of this, every third term ($a_{2},a_{5},...$) can be simplified, and thus have a slightly different final nested root.
Starting the half-angle chains using the sums produced from the equation above, it looks something like this:
\begin{gather*}
\sin{\frac{\pi}{6}}=\frac{1}{2}\\
\sin({\frac{\pi}{6}-\frac{\pi}{12}})=\sin{\frac{\pi}{12}}=\sqrt{\frac{1-\cos{\frac{\pi}{6}}}{2}}=\frac{\sqrt{2-\sqrt{3}}}{2}\\
\sin({\frac{\pi}{6}-\frac{\pi}{12}+\frac{\pi}{24}})=\sin{\frac{3\pi}{24}}=\sqrt{\frac{1-\sqrt{\frac{1+\cos{\frac{3\pi}{6}}}{2}}}{2}}=\frac{\sqrt{2-\sqrt{2}}}{2}\\
\sin({\frac{\pi}{6}-\frac{\pi}{12}+\frac{\pi}{24}-\frac{\pi}{48}})=\sin{\frac{5\pi}{48}}=\sqrt{\frac{1-\sqrt{\frac{1+\sqrt{\frac{1+\cos{\frac{5\pi}{6}}}{2}}}{2}}}{2}}=\frac{\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{3}}}}}{2}\\
\end{gather*}
$\frac{3\pi}{24}$ is left as is to visibly see the divisibility of the fraction, and the correlation with both the terms before and after it regarding the nested roots. Every 3 terms after this can also be simplified; thus, after the following terms of the sequence, things will start to repeat...
$$\sin({\frac{\pi}{6}-\frac{\pi}{12}+\frac{\pi}{24}-\frac{\pi}{48}+\frac{\pi}{96}})=\sin{\frac{11\pi}{96}}=\sqrt{\frac{1-\sqrt{\frac{1+\sqrt{\frac{1+\sqrt{\frac{1+\cos{\frac{11\pi}{6}}}{2}}}{2}}}{2}}}{2}}=\frac{\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{3}}}}}}{2}$$
After this term follows another value, $\frac{21\pi}{192}$, which can be simplified. Just like its predecessor, its final nested root is $\sqrt{2}$, which is then further nested similarly to the last occurrence. Since this is a recurring pattern, the value of $\sin{\frac{\pi}{9}}$ is thus defined to be
$$\frac{\sqrt{2-\Big[{\sqrt{2+{\sqrt{2-{\sqrt{2+...}}}}}}\Big]}}{2}$$
where the bracketed expression is self-recursive (that is, the expression fills in the ellipsis and leaves another ellipsis to be filled again). This value is approximately equal to $0.34202014333$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
2 is a primitive root mod $3^h$ for any positive integer $h$ It's easy to verify that 2 is a primitive root mod $3^2$. But then why does it follow that 2 is a primitive root mod $3^h$ for any positive integer $h$?
This was used in the solution of 2009 Putnam B6
http://math.hawaii.edu/home/pdf/putnam/Putnam_2009.pdf
I saw this Primitive roots of odd primes
but unfortunately I don't have access to the book.
|
Lemma: $2^{2\cdot 3^{n-1}}\equiv 1+3^n\pmod{3^{n+1}}$ for all $n\ge 1$.
Proof: We proceed with induction. Base case for $n=1$ is easy to verify. Suppose the lemma holds for some $n=k$. Then, by the inductive hypothesis $$2^{2\cdot 3^{k-1}}=1+3^k+3^{k+1}\ell,$$ for some $\ell\in\mathbf{Z}$.
Thus, $$2^{2\cdot 3^k}=1+3^{k+1}+3^{k+2}j\equiv 1+3^{k+1}\pmod{3^{k+2}},$$ for some $j\in\mathbf{Z}$, so by induction we are done. $\Box$
Now, suppose the original proposition holds for some $n=k$, so $2^{\varphi(3^k)}=2^{2\cdot 3^{k-1}}\equiv 1\pmod{3^k}$, and let $P=\operatorname{ord}_{3^{k+1}}(2)$. Then we have $2^P\equiv 1\pmod{3^{k+1}}$, so $2^P\equiv 1\pmod{3^k}$, so $2\cdot 3^{k-1}|P$. We also know that $P|\varphi(3^{k+1})=2\cdot 3^k$, so $P=2\cdot 3^{k-1}$ or $P=2\cdot 3^k$.
Then, by our lemma, $$2^{2\cdot 3^{k-1}}\equiv 1+3^k\not\equiv 1\pmod{3^{k+1}},$$ so we must have $P=2\cdot 3^k$, and the rest follows by induction. $\Box$
|
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"timestamp": "2023-03-29T00:00:00",
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|
calculation of $\int\frac{1}{\sin^3 x+\cos^3 x}dx$ and $\int\frac{1}{\sin^5 x+\cos^5x}dx$
Solve the following indefinite integrals:
$$
\begin{align}
&(1)\;\;\int\frac{1}{\sin^3 x+\cos^3 x}dx\\
&(2)\;\;\int\frac{1}{\sin^5 x+\cos^5 x}dx
\end{align}
$$
My Attempt for $(1)$:
$$
\begin{align}
I &= \int\frac{1}{\sin^3 x+\cos ^3 x}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^2 x+\cos ^2 x-\sin x \cos x\right)}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x\right)}\;dx\\
&= \frac{1}{3}\int \left(\frac{2}{\left(\sin x+\cos x\right)}+\frac{\left(\sin x+\cos x \right)}{\left(1-\sin x\cos x\right)}\right)\;dx\\
&= \frac{2}{3}\int\frac{1}{\sin x+\cos x}\;dx + \frac{1}{3}\int\frac{\sin x+\cos x}{1-\sin x\cos x}\;dx
\end{align}
$$
Using the identities
$$
\sin x = \frac{2\tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}},\;\cos x = \frac{1-\tan ^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}
$$
we can transform the integral to
$$I = \frac{1}{3}\int\frac{\left(\tan \frac{x}{2}\right)^{'}}{1-\tan^2 \frac{x}{2}+2\tan \frac{x}{2}}\;dx+\frac{2}{3}\int\frac{\left(\sin x- \cos x\right)^{'}}{1+(\sin x-\cos x)^2}\;dx
$$
The integral is easy to calculate from here.
My Attempt for $(2)$:
$$
\begin{align}
J &= \int\frac{1}{\sin^5 x+\cos ^5 x}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^4 x -\sin^3 x\cos x+\sin^2 x\cos^2 x-\sin x\cos^3 x+\cos^4 x\right)}\;dx\\
&= \int\frac{1}{(\sin x+\cos x)(1-2\sin^2 x\cos^2 x-\sin x\cos x+\sin^2 x\cos^2 x)}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x-\left(\sin x\cos x\right)^2\right)}\;dx
\end{align}
$$
How can I solve $(2)$ from this point?
|
By factorisation of
$$
\begin{aligned}
\sin ^5 x+\cos ^5 x= & \left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^3 x+\cos ^3 x\right) -\sin ^2 x \cos ^2 x(\sin x+\cos x) \\
= & (\sin x+\cos x)\left(1-\sin x \cos x-\sin ^2 x \cos ^2 x\right),
\end{aligned}
$$
we have $$
I=\int \frac{1}{\sin ^5 x+\cos ^5 x} d x=\int \frac{\sin x+\cos x}{(\sin x+\cos x)^2\left(1-\sin x \cos x-\sin ^2x \cos ^2 x\right)} d x
$$
Noting that letting $t=\sin x-\cos x$, then $dt=(\cos x+\sin x)dx$ and $\sin x \cos x=\frac{1-t^2}{2}$ yields
$$
\begin{aligned}I&=4 \int \frac{d t}{\left(t^2-2\right)\left(t^4-4 t^2-1\right)}\\&=
\frac{4}{5} \int\left(\frac{-1}{t^2-2}+\frac{1}{t^2+\sqrt{5}-2}+\frac{1}{t^2-\sqrt{5}-2}\right) d t \\&=\frac{4}{5}\left[-\frac{1}{\sqrt{2}} \tanh ^{-1}\left(\frac{t}{\sqrt{2}}\right)+\frac{1}{\sqrt{\sqrt 5-2}} \tan ^{-1}\left(\frac{t}{\sqrt{\sqrt{5}-2}}\right)-\frac{1}{\sqrt{\sqrt{5}+2}} \tanh ^{-1}\left(\frac{t}{\sqrt{\sqrt{5}+2}}\right)\right]+C\\ &\boxed{I=\frac{4}{5}\left[-\frac{1}{\sqrt{2}} \tanh ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{2}}\right)+\frac{1}{\sqrt{\sqrt 5-2}} \tan ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sqrt{5}-2}}\right)\qquad\qquad -\frac{1}{\sqrt{\sqrt{5}+2}} \tanh ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sqrt{5}+2}}\right)\right]+C}\end{aligned}$$
|
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|
Convergence of $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ I want to check, whether $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ converges or diverges.
I tried to use Leibniz's test :
$|a_n|= \frac{n^2}{\sqrt{n^5+1}} = \frac{n^2}{\sqrt{n^4(n+\frac{1}{n^4})}} = \frac{n^2}{n^2\sqrt{n+\frac{1}{n^4}}} = \frac{1}{\sqrt{n+\frac{1}{n^4}}}$
So $\lim\limits_{n \rightarrow \infty}{{\frac{1}{\sqrt{n+\frac{1}{n^4}}} = 0}} $
$1>|\frac{a_{n+1}}{a_n}|= \frac{(n+1)^2}{\sqrt{(n+1)^5+1}} \frac{\sqrt{n^5+1}}{n^2}= \frac{2n+1 \sqrt{n^5+1}}{\sqrt{(n+1)^5+1}}= \frac{n^2+2n+1 \sqrt{n+\frac {1}{n^4}}}{(n+1)^2\sqrt{n+1+1}} = \frac {\sqrt{n+\frac {1}{n^4}}}{\sqrt{n+2}}$
So $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ converges.
Could somebody please check my solution?
|
We have
$$(n^5+1)^{-1/2}=\frac{1}{n^{5/2}}\left(1+\frac{1}{n^5}\right)^{-1/2}=\frac{1}{n^{5/2}}\left(1+O\left(\frac{1}{n^5}\right)\right)$$
hence
$$(-1)^n\frac{n^2}{\sqrt{n^5+1}}=\underbrace{\frac{(-1)^n}{n^{1/2}}}_{=u_n}+\underbrace{O\left(\frac{1}{n^{11/2}}\right)}_{=v_n}$$
the series $\displaystyle\sum_n u_n$ is convergent by the Leibniz theorem and the series $\displaystyle\sum_n v_n$ is also convergent by comparison with the Riemann series. Conclude.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$ Show that $$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$$
$$\sqrt{2}=\mathbf{2}^{1/2}$$
$$\sqrt{2\sqrt{2}}=\mathbf{2}^{1/2+1/2^2}$$
$$\sqrt{2\sqrt{2\sqrt{2}}}=\mathbf{2}^{1/2+1/2^2+1/2^3}$$
Show the limit of $$\mathbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+\dotsb+\frac{1}{2^n}=1$$ when $n\to\infty$
$$\textbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}$$
$$\Rightarrow \frac{1}{2}\textbf{S}_{n}=\frac{1}{2}(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n})$$
$$\Rightarrow \frac{1}{2}\textbf{S}_{n}=(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n+1}})$$
$$\Rightarrow (1)-(2)=\textbf{S}_{n}-\frac{1}{2}\textbf{S}_{n}=\frac{1}{2}-\frac{1}{2^{n+1}}$$
$$\Rightarrow \textbf{S}_{n}(1-\frac{1}{2})=\frac{1}{2}-\frac{1}{2^{n+1}}$$
$$\Rightarrow \frac{1}{2^{n+1}}\rightarrow\textbf{0}\quad\textit{when n}\rightarrow\infty$$
$$\Rightarrow \textbf{S}_{n}\rightarrow\textbf{1}\quad\textit{when n}\rightarrow\infty$$
$$\Rightarrow \lim_{n \to \infty}\textbf{2}^{\textbf{S}_{n}}=2\quad\textit{when n}\rightarrow\infty$$
|
Let S be your general term for a large value of n. If your square it you have S^2 = 2 S, then S = 2.
|
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|
Finding the taylor series of $f(z) = 1/(1+z^2)$. I am working on the following exercise:
Find the Taylor expansion of the function $f(z) = \frac{1}{1+z^2}$ about $z = 3i$.
We had the Taylor Series Theorem in the lecture:
Let $D \subset \mathbb C$ be a domain and $f: D \to \mathbb C$ a differentiable function. Then $f$ is analytic in $D$ and for any ball $B_R(z_0) \subset D$ the power series expansion
\begin{align*}
f(z) = \sum_{n = 0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
\end{align*}
is valid. Further, if $r \in (0,R)$ then
\begin{align*}
f^{(n)}(z_0) = \frac{n!}{2\pi i} \int_{S_r^+(z_0)} \frac{f(w)}{(w-z_0)^{n+1}}dw,
\end{align*}
where $S_r^+(z_0) = z_0+re^{it}$ with $0\le t \le 2\pi$.
The solution that I don't understand goes like this: Represent $f$ as partial fractions:
\begin{align*}
f(z) = \frac{1}{1+z^2} = \frac{1}{2\pi i} \left(\frac{1}{z-i} - \frac{1}{z+i}\right).
\end{align*}
Then we compute
\begin{align*}
f^{(n)}(z) = \left(\frac{1}{1+z^2}\right)^{(n)} = \frac{1}{2\pi i}\left(\frac{(-1)^nn!}{(z-i)^{n+1}}-\frac{(-1)^nn!}{(z+i)^{n+1}}\right).
\end{align*}
Then one can plug this in the formula and is done.
I don't understand why one can compute so easily the $n$th derivative.
|
A naive, pretty useful many times, approach:
$$\frac1{1+z^2}=\frac1{10+6i+(z-3i)^2}=\frac1{10+6i}\frac1{1+\left(\frac{z-3i}{\sqrt{10+6i}}\right)^2}$$
Now, notice that
$$\left|\frac{z-3i}{\sqrt{10+6i}}\right|<1\iff|z-3i|^2<|10+6i|=4\sqrt{10}$$
for which values of $\;z\;$ we can write
$$\frac1{1+\left(\frac{z-3i}{\sqrt{10+6i}}\right)^2}=\sum_{n=0}^\infty(-1)^n\left(\frac{z-3i}{\sqrt{10+6i}}\right)^n$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Limit of series $4\left( \frac {1}{8}+\frac {1}{12}\right) +6\left( \frac {1}{24}+\frac {1}{36}\right) +\ldots$ How to find this serie
$4\left( \dfrac {1}{8}+\dfrac {1}{12}\right) +6\left( \dfrac {1}{24}+\dfrac {1}{36}\right) +8\left( \dfrac {1}{64}+\dfrac {1}{96}\right) +\ldots $
I think it's telescopic, isn't it?
|
The terms in this sum seem to be
$$
2n\left( \dfrac {1}{2^{n-1}2n}+\dfrac {1}{3\cdot2^{n-2}2n}\right)
=
\dfrac {5}{6}\frac1{2^{n-2}}
$$
starting from $n=2$.
This is a geometric series and easy to sum.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How find this $I=\iint_{\Sigma}(x^2+y^2+z^2)^{-\frac{3}{2}}(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4})^{-\frac{1}{2}}dS$ Find this Surface integral
$$I=\iint_{\Sigma}(x^2+y^2+z^2)^{-\frac{3}{2}}\left(\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}+\dfrac{z^2}{c^4}\right)^{-\frac{1}{2}}dS$$
where
$$\Sigma:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1,a>0,b>0,c>0$$
My try: let
$$x=a\sin{\alpha}\cos{\beta},y=b\sin{\alpha}\sin{\beta},z=c\cos{\alpha},\alpha\in[0,\pi],\beta\in[0,2\pi]$$
$$E=x''_{\alpha}+y''_{\alpha}+z''_{\alpha}=a^2\cos^2{\alpha}\cos^2{\beta}+b^2\cos^2{\alpha}\sin^2{\beta}+c^2\sin^2{\alpha}$$
$$G=x''_{\beta}+y''_{\beta}+z''_{\beta}=a^2\sin^2{\alpha}\sin^2{\beta}+b^2\sin^2{\alpha}\cos^2{\beta}$$
$$F=x'_{\alpha}x'_{\beta}+y'_{\alpha}y'_{\beta}+z'_{\alpha}z'_{\beta}=-a^2\sin{\alpha}\cos{\alpha}\sin{\beta}\cos{\beta}+b^2\sin{\alpha}\cos{\alpha}\sin{\beta}\cos{\beta}$$
so
$$EG-F^2=(a^2b^2\cos^2{\alpha}+a^2c^2\sin^2{\alpha}\sin^2{\theta}+b^2c^2\sin{\alpha}\cos^2{\beta})\sin^2{\alpha}$$
then
$$\int_{\Sigma}f(x,y,z)dS=\int_{\Delta}f(a\sin{\alpha}\cos{\beta},b\sin{\alpha}\sin{\beta},c\cos{\alpha})\sqrt{EG-F^2}d\alpha d\beta$$
where $\Delta:0\le\alpha\le \pi,0\le \beta\le 2\pi$
$$f(x,y,z)=(x^2+y^2+z^2)^{-\frac{3}{2}}\left(\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}+\dfrac{z^2}{c^4}\right)^{-\frac{1}{2}}$$
and follow I fell very ugly,Thank you very much
|
Lets write the integral
$$I=\iint_{\Sigma}(x^2+y^2+z^2)^{-\frac{3}{2}}\left(\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}+\dfrac{z^2}{c^4}\right)^{-\frac{1}{2}}dS$$
in the form
$$I=\iint_{\Sigma} \frac{(x, y, z)}{(x^2+y^2+z^2)^{\frac{3}{2}}} \cdot \left(\left(\frac{x}{a^2}, \frac{y}{b^2}, \frac{z}{c^2}\right)\left(\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}+\dfrac{z^2}{c^4}\right)^{-\frac{1}{2}}\right)dS$$
and notice the integral now reads
$$I=\iint_{\Sigma}(\mathbf{F\cdot n})dS$$
with
$$\mathbf{F}= \frac{(x, y, z)}{(x^2+y^2+z^2)^{\frac{3}{2}}}, \mathbf{n}=\left(\left(\frac{x}{a^2}, \frac{y}{b^2}, \frac{z}{c^2}\right)\left(\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}+\dfrac{z^2}{c^4}\right)\right)^{-\frac{1}{2}}$$
Notice how $\nabla\cdot \mathbf{F}=0$ on $\mathbb{R}\backslash\{(0, 0, 0)\}$, which means you can integrate $\mathbb{F}$ over any surface $S$ with the origin in its interior. For simplicity, take $S: \{(x,y,z)\in \mathbb{R}^3|\ x^2+y^2+z^2 = 1\}$. The integral is now simply
$$I=\iint_{S} \frac{(x, y, z)}{(x^2+y^2+z^2)^{\frac{3}{2}}} \cdot \frac{(x, y, z)}{(x^2+y^2+z^2)^{\frac{1}{2}}}dS = \iint_{S} \frac{dS}{r^{2}} $$
Therefore, $$I= \int_{0}^{1} \frac{4\pi r^2 dr}{r^2} \Rightarrow I=4\pi$$
|
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|
If $a$ and $b$ are positve integers, show that the inequality $e <\frac{a}{b}<\frac{87}{32} $ imples that $b \geq 39$. If $a$ and $b$ are positve integers, show that the inequality $e <\frac{a}{b}<\frac{87}{32} $ imples that $b \geq 39$.
Use the continued fraction $e=[2:1,2,1,1,4,1,1,6,...]$
|
Note that $e>\frac{19}7$ and that $\frac{19}7<\frac ab<\frac{87}{32}$ implies that the numerators of the differences $\frac{a}{b}-\frac{19}{7}=\frac{7a-19b}{7b}$ and $\frac{87}{32}-\frac ab=\frac{87b-32a}{32b}$ are integers $\ge 1$.
Therefore
$$b=(7\cdot 87-32\cdot19)b=32(7a-19b)+7(87b-32a)\ge 32+7=39 .$$
Why is $e>\frac{19}7$?
$$\begin{align}[2:1,2,1,1,4,1,1,6,\ldots]>[2:1,2,2] &\iff [1:2,1,1,4,1,1,6,\ldots]<[1:2,2]\\
&\iff[2:1,1,4,1,1,6,\ldots]>[2:2]\\
&\iff [1:1,4,1,1,6,\ldots]<2 \end{align}$$
|
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|
big determinant calculation I have found this exercise in a book, and having troubles solving it:
How to calculate this determinant?
$$\det\begin{pmatrix}
5 & 6 & 0 & 0 & 0 & \cdots & 0 \\
4 & 5 & 2 & 0 & 0 & \cdots & 0 \\
0 & 1 & 3 & 2 & 0 & \cdots & 0 \\
0 & 0 & 1 & 3 & 2 & \cdots & \vdots \\
0 & 0 & 0 & 1 & \ddots & \ddots & 0 \\
& & & & \ddots & 3 & 2 \\
0 & 0 & 0 & \cdots & 0 & 1 & 3 \\
\end{pmatrix}
_{n\times n}$$
Thanks!!
|
In general, when calculating a determinant of tridiagonal matrix of this form
$D_n=
\begin{vmatrix}
c& b & 0 & 0 & \ldots & 0 \\
a& c & b & 0 & \ldots & 0 \\
0& a & c & b & \ldots & 0 \\
\vdots & \ddots & \ddots & \ddots & \vdots \\
0 & \ldots & 0 & a & c & b \\
0 & \ldots & 0 & 0 & a & c
\end{vmatrix}$
you can use Laplace expansion twice to obtain recurrent relation.
$D_n=
\begin{vmatrix}
c& b & 0 & 0 & \ldots & 0 \\
a& c & b & 0 & \ldots & 0 \\
0& a & c & b & \ldots & 0 \\
\vdots & \ddots & \ddots & \ddots & \vdots \\
0 & \ldots & 0 & a & c & ab \\
0 & \ldots & 0 & 0 & a & c
\end{vmatrix}
\overset{(1)}= cD_{n-1} - b
\begin{vmatrix}
a & 0 & 0 & \ldots & 0 \\
a & c & b & \ldots & 0 \\
\vdots & \ddots & \ddots & \ddots & \vdots \\
0 & \ldots & a & c & ab \\
0 & \ldots & 0 & a & c
\end{vmatrix}
\overset{(2)}= cD_{n-1} - ab D_{n-2}$
The equality (1) is Laplace for the first row and the equality (2) is Laplace for the first column.
This is not exactly the situation from the original post, but we can modify it to a very similar situation using reflection w.r.t vertical and horizontal axis. This can be done by reverting the order of rows and then the order of columns. Since this is the same as doing even number of exchanges of rows/columns, this does not change the determinant. So to solve the original question you simply have to solve the following recurrence:
$D_n=
\begin{vmatrix}
3& 1 & 0 & 0 & \ldots & 0 \\
2& 3 & 1 & 0 & \ldots & 0 \\
0& 2 & 3 & 1 & \ldots & 0 \\
\vdots & \ddots & \ddots & \ddots & \vdots \\
0 & \ldots & 0 & 2 & 5 & 4 \\
0 & \ldots & 0 & 0 & 6 & 5
\end{vmatrix}$
$D_3=
\begin{bmatrix}
3 & 1 & 0 \\
2 & 5 & 4 \\
0 & 6 & 5
\end{bmatrix}
$
$D_4=
\begin{bmatrix}
3 & 1 & 0 & 0\\
2 & 3 & 1 & 0 \\
0 & 2 & 5 & 4 \\
0 & 0 & 6 & 5
\end{bmatrix}
$
$D_n=3D_{n-1}-2D_{n-2}$
|
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|
How to find $\lim_{(x,y)\to(1,1)} \frac{x^3-y^3}{x^2 - y^2}$? Suppose $$ f(x,y) = \begin{cases}
\frac{x^3-y^3}{x^2 - y^2}, & \text{if $x^2 \ne y^2$} \\
\alpha, & \text{if $x^2 = y^2$} \\
\end{cases}$$
I need to find $\alpha$ such that $f(x,y)$ will be continous at $(1,1)$.
I started:
If $f(x,y)$ continous at $(1,1)$ $\implies$ $\lim_{(x,y)\to(1,1)} \frac{x^3-y^3}{x^2 - y^2} = \alpha$
Then I need to find $\lim_{(x,y)\to(1,1)} \frac{x^3-y^3}{x^2 - y^2}$ but I stuck here.
Can you please help me find the limit? I'd glad to a detailed solution or hint .. thanks in advance..
|
Hint: Notice that the numerator factors as follows:
$$
x^3 - y^3 = (x - y)(x^2 + xy + y^2)
$$
Now factor the denominator and cancel the $(x-y)$ factor.
|
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|
Number of roots of a quadratic equation modulo 325 Any Help solving this question ?
a) Find ONE solution $\overline x\in\Bbb Z/325\Bbb Z$ such that $x^2\equiv-1\pmod{325}$. (Hint: CRT and lifting.)
b) How many solutions $\overline x$ to the above equation are there, and why?
|
$325=5^2\cdot 13$ so let's solve $x^2\equiv -1\pmod{25}$, $x^2\equiv -1\pmod{13}$.
Case $1$: $\mod{25}$
If $x^2\equiv -1\pmod{25}$, then $x^2\equiv -1\pmod 5$ so $x\equiv\pm2\pmod 5$
$$(5k+2)^2=25k^2+20k+4\equiv 20k+4\equiv-1\pmod{25}\implies 4k\equiv -1\pmod 5\\\implies k\equiv 1\pmod 5$$
$$(5k-2)^2=25k^2-20k+4\equiv -20k+4\equiv-1\pmod{25}\implies 4k\equiv 1\pmod 5\\\implies k\equiv 4\pmod 5$$
So we have $x^2\equiv-1\pmod{25}\implies x\equiv\pm7\pmod{25}$
Case $2$: $\mod 13$
We see upon inspection that $x^2\equiv -1\pmod{13}\implies x=\pm 5$
See if you can use CRT to find all solutions to $x^2\equiv -1\pmod{325}$
|
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|
Finding a formula for a recursively defined sequence I have a sequence given by:
\begin{align}
r_1 &= 1\\
r_2 &= 0\\
r_3 &= -1\\
r_n &= r_{n-1}r_{n-2} + r_{n-3}\\
R &= \{1, 0, -1, 1, -1, -2, 3, -7, -23, etc...\}
\end{align}
The first four lines were all we were given in order to help study for our final tomorrow. I have a feeling one of the questions will be to find a formula but I don't know where to start. Any suggestions?
My first intuition was to start with $x^3 = x^2\ x + 1$ but that leads to $x^3 - x^3 = 1 \rightarrow 0 = 1$
|
I've not much time at the moment; but just a first approach if you have no other better idea: write the first few iterates, most explicite, beginning with [a,b,c] and try to discern a pattern. Later I'd look, whether there is some simplification, given the special values from the initial problem (like a pattern in the exponents, sums partial geometric series and the like, products with increasing number of factors...).
a
b
c
b*c + a
b*c^2 + a*c + b
b^2*c^3 + 2*b*a*c^2 + (a^2 + (b^2 + 1))*c + b*a
b^3*c^5 + 3*b^2*a*c^4 + (3*b*a^2 + (2*b^3 + b))*c^3 + (a^3 + (4*b^2 + 1)*a)*c^2 + (2*b*a^2 + (b^3 + 2*b))*c + (b^2 + 1)*a
b^5*c^8 + 5*b^4*a*c^7 + (10*b^3*a^2 + (3*b^5 + 2*b^3))*c^6 + (10*b^2*a^3 + (12*b^4 + 6*b^2)*a)*c^5 + (5*b*a^4 + (18*b^3 + 6*b)*a^2 + (3*b^5 + 5*b^3 + b))*c^4 + (a^5 + (12*b^2 + 2)*a^3 + (9*b^4 + 11*b^2 + 1)*a)*c^3 + (3*b*a^4 + (9*b^3 + 7*b)*a^2 + (b^5 + 3*b^3 + 3*b))*c^2 + ((3*b^2 + 1)*a^3 + (2*b^4 + 4*b^2 + 2)*a)*c + ((b^3 + b)*a^2 + b)
|
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|
Show that the sum of the series Show that the sum of the series is greater than 24
$$\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{9}+\sqrt{11}} +\cdots+\frac{1}{\sqrt{9997}+\sqrt{9999}} > 24$$
I see that
$\frac{1}{\sqrt{3}+\sqrt{1}}\cdot \frac{\sqrt{3}-\sqrt{1}}{\sqrt{3}-\sqrt{1}}=\frac{\sqrt{3}-\sqrt{1}}{2}$
In each term, the denominator is 2 so the series becomes $\frac{1}{2}\sum^{2499}_{n=0}[\sqrt{4n+3}-\sqrt{4n+1}]$
|
The simplest way I can think of is to write it as $\frac12(\sum(\sqrt{4n+3} - \sum(\sqrt{4n-1}).$ You can approximate each sum by the integral, and since the function $\sqrt{x}$ is monotonic (and concave up) you can estimate the difference very closely.
A second (maybe even simpler) argument is to use your form, and note that by the mean value theorem each term equals $\frac{1}2 \frac{1}{\sqrt{x}}$ for some $x$ between $4n+1$ and $4n+3.$
|
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|
conditional expectation of squared standard normal Let $A,B$ independent standard normals. What is $E(A^2|A+B)$?
Is the following ok?
$A,B$ iid and hence $(A^2,A+B),(B^2,A+B)$ iid.
Therefore we have $\int_M A^2 dP = \int_M B^2 dP$ for every $A+B$-measurable set $M$ and hence $E(A^2|A+B) = E(B^2|A+B)$.
We obtain $2 \cdot E(A^2|A+B) = E(A^2|A+B) + E(B^2|A+B) = E(A^2+B^2|A+B) = A^2+B^2$ where the last equation holds since $A^2+B^2$ is $A+B$-measurable.
Finally we have $E(A^2|A+B) = \frac{A^2+B^2}{2}$.
|
if $A$ and $B$ are independent, then $A^2$ and $A+B$ are not independent because $A^2$ and $A$ are correlated.
|
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|
Find the solution of the equation Find all real solutions of this equation :
$$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$
|
In order for the equation to hold, we can see that $2-\sqrt{2+x}\geq0$, so $x\leq2$. Clearly, $x\geq0$. Since we are only interested in $\cos t$ and $0\leq\cos t\leq 1$, we may assume $0\leq t\leq\frac\pi2$.
Substitute $x=2\cos t$.
Then:
$$\begin{align*}&2+x=2(1+\cos t)=4\cos^2\frac t2,\hspace{5pt}\cos\frac t2\geq0 \hspace{5pt}\Rightarrow\hspace{5pt} \sqrt{2+x}=2\cos\frac t2\\
&2-\sqrt{2+x}=2-2\cos\frac t2=2\cdot2\sin^2\frac t4,\hspace{5pt} \sin\frac t4\geq 0 \hspace{5pt}\Rightarrow\hspace{5pt} \sqrt{2-\sqrt{2+x}}=2\sin\frac t4\\
&2+\sqrt{2-\sqrt{2+x}}=2+2\sin\frac t4=2\left(1+\cos\left(\frac\pi2-\frac t4\right)\right)=4\cos^2\left(\frac\pi4-\frac t8\right)
\end{align*}$$
So we get $2\cos t=2\cos\left(\frac\pi4-\frac t8\right)$, i.e. $t=\frac\pi4-\frac t8$ (since if $a=\frac\pi4-\frac t8$ then $0\leq a,t\leq\frac\pi2$, so $\cos a=\cos t$ implies $a=t$). We have $t=\frac 89\left(\frac\pi4\right)=\frac{2\pi}9$. So $x=2\cos\frac{2\pi}9$.
|
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|
How can I calculate $\sum_{k=1}^n (3k-1)^{2}$? I was trying to calculate the sum
$$\sum_{k=1}^n (3k-1)^2,$$
but actually I cannot frame the type. In fact is neither a geometric series, because the terms are not raised to the $k$th power, nor harmonic.
Any suggestions?
|
Let $\displaystyle f(k)=Ak^3+Bk^2+Ck+D$ where $A,B,C,D$ are arbitrary constants
$\displaystyle\implies f(k+1)-f(k)=A\{(k+1)^3-k^3\}+B\{(k+1)^2-k^2\}+C\{(k+1)-k\}$$\displaystyle=A(3k^2+3k+1)+B(2k+1)+C$
$\displaystyle\implies f(k+1)-f(k)=3Ak^2+k(3A+2B)+A+B+C$
Comparing with $\displaystyle(3k-1)^2=9k^2-6k+1,$
$\displaystyle 3A=9\implies A=3,3A+2B=-6\implies B=-\frac{15}2;A+B+C=1\implies C=\frac{11}2 $
So, we can write $\displaystyle9k^2-6k+1=f(k+1)-f(k)$ which is a Telescoping series
$\displaystyle\implies\sum_{1\le k\le n}(3k-1)^2=f(n+1)-f(1)$
$\displaystyle=A(n+1)^3+B(n+1)^2+C(n+1)+D-(A+B+C+D) $
$\displaystyle=A\{(n+1)^3-1\}+B\{(n+1)^2-1\}+Cn$
Put the values of $A,B,C$ and optionally simplify
|
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|
Solving the equation $\frac{x^7}{7}=1+10^{1/7}x(x^2-10^{1/7})^2$ $$\frac{x^7}{7}=1+10^{1/7}x(x^2-10^{1/7})^2$$ Find $x$ where $x$ is real.
|
This is really a trick question. Let $\alpha = 10^{1/14}$, the equation we have can be rewritten as
$$P(x) = 0 \quad\text{ where }\quad P(x) = \frac{x^7}{7} - 1 - \alpha^2 x (x^2-\alpha^2)^2$$
Let $x = \alpha y$, we can further simplify the equation
$$P(x) = 0
\quad\iff\quad\alpha^{-7} P(\alpha y) = 0
\quad\iff\quad\frac{y^7}{7} - y^5 + 2y^3 - y - \alpha^{-7} = 0$$
The polynomial in $y$ on the RHS looks familiar. In fact, aside from the constant term,
it is proportional to a Chebyshev polynomial of the first kind to degree 7:
$$\frac{y^7}{7} - y^5 + 2y^3 - y = \frac{2}{7}T_7\left(\frac{y}{2}\right) =
\begin{cases}
\frac{2}{7}\cos\left(7\cos^{-1}\left(\frac{y}{2}\right)\right), & |y| \le 2\\
\frac{2}{7}\cosh\left(7\cosh^{-1}\left(\frac{y}{2}\right)\right),& |y| \ge 2
\end{cases}$$
Since $\alpha^{7} = \sqrt{10}$, we get
$$\begin{align}
& y = 2\cosh\left(\frac17\cosh^{-1}\left(\frac{7}{2\sqrt{10}}\right)\right)\\
\iff & x = 2\times 10^{1/14}\cosh\left(\frac17\cosh^{-1}\left(\frac{7}{2\sqrt{10}}\right)\right) \sim 2.362588464315639
\end{align}$$
Update
I'm missing an obvious simplification.
Let $\theta = \cosh^{-1}\left(\frac{7}{2\sqrt{10}}\right)$, we have
$$
e^\theta = \frac{7 + \sqrt{7^2 - (2\sqrt{10})^2}}{2\sqrt{10}} = \frac{5}{\sqrt{10}}
\;\;\implies\;\;x = 10^{1/14} \left(e^{\frac{\theta}{7}} + e^{-\frac{\theta}{7}}\right) = 5^{1/7} + 2^{1/7}
$$
|
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|
How does mod multiplication work? For example, $10^{10} \equiv 4\pmod{6}$
If I used $\pmod{2}$ and $\pmod{3}$, how does the multiplication process work?
Since $10^{10} \equiv 0 \pmod{2}$ and $10^{10}\equiv 1\pmod{3}$,
$$
10^{10}\equiv (0,1) \pmod{(2,3)}
$$
how do we get the value $4$ at the end? do we list out the possible values of $0\pmod{2}$ and $1\pmod{3}$?
$$
1\pmod{3} = 1, 4, 7, 10
$$
so on.
Since only $4, 10$ and so on satisfy $\pmod{2}$, only values that satisfy both criteria can be used.
In general, can we do this for $\pmod{n}$, $n$ being any integer?
|
Let me try to rephrase what you said.
Let $10^{10} \equiv a \pmod 6$, $a \in \{0, 1, 2, 3, 4, 5\}$.
As $10^{10} \equiv 0 \pmod 2$, we must have $a \equiv 0 \pmod 2$, so $a \in \{0, 2, 4\}$.
As $10^{10} \equiv 1 \pmod 3$, we must have $a \equiv 1 \pmod 3$, so $a \in \{1, 4\}$.
Therefore, $a = 4$.
This does give the correct answer as you've noticed. This works because $2, 3 \mid 6$ and $(2, 3) = 1$.
What you are doing is strongly related to the Chinese Remainder Theorem which states that if $n_1, \dots, n_k$ are pairwise coprime positive integers, then for any integers $a_1, \dots, a_k$, the system of congruences $x \equiv a_1 \pmod{n_1},\dots, x \equiv a_k\pmod{n_k}$ has a solution and it is unique modulo $n_1\dots n_k$. In particular, one way to find this solution is to use the method you used (as can be seen here).
|
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|
Solving $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=c$ Question:
let $a,b,c$ be positive constants. Find $u=u(x,y)$ if is satisfies the partial differential equation
$$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c$$
and the boundary condition
$$u=0,\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.$$
my try: I know this is screened Poisson equation
$$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c$$
I only find this poisson equation one of the solution
$$u(x,y)=-\dfrac{c}{2}\cdot\dfrac{a^2b^2}{a^2+b^2}\left(1-\dfrac{x^2}{a^2}-\dfrac{y^2}{b}\right)$$
because
$$\dfrac{\partial u}{\partial x}=-\dfrac{c}{2}\cdot\dfrac{a^2b^2}{a^2+b^2}\left(-\dfrac{2x}{a^2}\right)$$
$$\dfrac{\partial^2 u}{\partial x^2}=c\cdot\dfrac{b^2}{a^2+b^2}$$
and
$$\dfrac{\partial^2 u}{\partial y^2}=c\cdot\dfrac{a^2}{a^2+b^2}$$
so
$$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c\cdot\left(\dfrac{a^2}{a^2+b^2}+\dfrac{b^2}{a^2+b^2}\right)=c$$
and when
$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\Longrightarrow u=0$$
and this solution is uniqueness? and How prove it? Thank you
Thank you very much!
|
Let
$$v(x,y)=u(x,y)+\dfrac{c}{2}\cdot\dfrac{a^2b^2}{a^2+b^2}\left(1-\dfrac{x^2}{a^2}-\dfrac{y^2}{b}\right)$$
Then $\Delta v=0$ in $\mathbb R^2$ and $v(x,y)=0$ on the boundary of the ellipse $U=\big\{\frac{x^2}{a^2}+\frac{y^2}{b^2}<1\big\}$. Thus, as the Dirichlet problem
$$
\Delta v=0 \,\,\text{in $U$} \quad\text{and} \quad v=0 \,\,\text{on $\partial U$},
$$
has unique solution $v\equiv 0$, then $v$ vanishes in $U$ as well. But $v$ is a harmonic function in $\mathbb R^2$, and harmonic functions are real analytic, and thus it is expressible as a power series around $(0,0)$
$$
v(x,y)=\sum_{m,n=0}^\infty \frac{\partial^{m+n}v(0,0)}{\partial x^m\partial y^n}\frac{x^my^n}{m!n!}=0,
$$
since $\frac{\partial^{m+n}v(0,0)}{\partial x^m\partial y^n}=0$, for all $m,n$.
|
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|
Determine indefinite-integral $\text{I}=\int \sin^{n-1}x\sin\left[(n+1)x\right]\text{d}x$ Determine indefinite-integral:
$$\text{I}=\int \sin^{n-1}x\sin\left[(n+1)x\right]\text{d}x$$
My tried:
$$\text{I}=\int \sin^{n-1}x\sin\left[(n+1)x\right]\text{d}x=-\int\sin^{n-2}x\sin\left[(n+1)x\right]d(\cos x)$$
$$=- \sin^{n-2}x\sin\left[(n+1)x\right]\cos x+\int \left[(n-2)\sin^{n-3}x\sin\left[(n+1)x\right]\right]\cos^2xdx$$
$$+\int \sin^{n-2}x (n+1)\cos\left[(n+1)x\right]\cos xdx$$
$$=- \sin^{n-2}x\sin\left[(n+1)x\right]\cos x+(n-2)\int \sin^{n-3}x\sin\left[(n+1)x\right]dx-(n-2)I$$
$$+(n+1)\int \sin^{n-2}x\cos\left[(n+1)x\right]\cos xdx$$
Come here I don't know how, please help me.
|
Use angle sum expansion on the second term.
$$\int \sin^{n-1}{x}\left(\sin{x}\cos{nx} + \cos{x} \sin{nx}\right)\text{d}x $$
$$\frac{1}{n}\int (n\sin^{n-1}x\cos{x}\sin{nx} + n\sin^{n}{x}\cos{nx})\text{d}x$$
$$\frac{1}{n}\int \text{d}(\sin^{n}{x})\sin{nx}+\sin^{n}{x}\text{d}(\sin{nx})$$
Reverse the product rule
$$\frac{1}{n}\int\text{d}(\sin^n{x}\sin{nx}) = \frac{\sin^n{x}\sin{nx}}{n} + C$$
|
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|
Definite integration problem (trig). I have this definite integral:
$$
\int_0^\Pi \cos{x} \sqrt{\cos{x}+1} \, dx
$$
For finding the indefinite integral, I have tried substitution, integration by parts, but I'm having trouble solving it.
By parts
$$
\int \cos{x} \sqrt{\cos{x}+1} \, dx\ = \sqrt{\cos{x}+1}\sin{x} + \frac{1}{2} \int \frac{\sin^{2}{x}}{\sqrt{\cos{x}+1}} \, dx
$$
$
f(x) = \sqrt{\cos{x}+1} \\ f'(x) = \frac{1}{2} \frac{-\sin{x}}{\sqrt{\cos{x}+1}} \\ g(x) = \sin{x} \\ g'(x) = \cos{x}
$
I don't know how to approach this further because of the $\sin^{2}{x}$.
Substitution
$
\cos{x} + 1 = u \\ -\sin{x} \, dx = du
$
But I have no use for $sin\,x$.
I believe it has something to do with trig manipulations.
WolframAlpha tells me to substitute, but I don't understand how to get the first u-substituted integral like shown:
I would really appreciate any help on this. Thank you.
|
Here is how I would do it: first, let's recall the cosine double-angle identity. $$\cos 2x = \cos^2 x - \sin^2 x = \cos^2 x - (1 - \cos^2 x) = 2\cos^2 x - 1.$$ Thus the corresponding half-angle identity can be written $$\cos x = \sqrt{\frac{1 + \cos 2x}{2}}$$ or equivalently, $$\sqrt{1 + \cos x} = \sqrt{2} \cos \frac{x}{2}, \quad 0 \le x \le \pi.$$ So the integral becomes $$I = \int_{x=0}^\pi \sqrt{2} \cos x \cos \frac{x}{2} \, dx.$$ Now recall the angle addition identity $$\cos(a \pm b) = \cos a \cos b \mp \sin a \sin b,$$ from which we obtain $$\cos (a+b) + \cos (a-b) = 2 \cos a \cos b.$$ Then with $a = x$, $b = x/2$, we easily see the integral is now $$I = \frac{1}{\sqrt{2}} \int_{x=0}^\pi \cos \frac{3x}{2} + \cos \frac{x}{2} \, dx.$$ Now it is a simple matter to integrate each term: $$\begin{align*} I &= \frac{1}{\sqrt{2}} \left[ \frac{2}{3} \sin \frac{3x}{2} + 2 \sin \frac{x}{2} \right]_{x=0}^\pi \\ &= \frac{1}{\sqrt{2}} \left( -\frac{2}{3} + 2 \right) = \frac{2\sqrt{2}}{3}. \end{align*} $$
|
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|
Prove that $ \lim_{n \rightarrow \infty } \frac{n+6}{n^2-6} = 0 $. My attempt:
We prove that $ \lim\limits_{n \rightarrow \infty } \dfrac{n+6}{n^2-6} =0$.
It is sufficient to show that for any $ \epsilon \in\textbf{R}^+ $, there exists an $ K \in \textbf{R}$ such that for any $ n > K $
$ \displaystyle \left| \frac{n+6}{n^2-6} - 0 \right| < \epsilon $.
Suppose $ n>4 $. Then $ n+6 < 7n $ and $ n^2 -6 > \frac{1}{2} n^2 $. So
$ \displaystyle \left| \frac{n+6}{n^2-6} - 0 \right| = \frac{n+6}{n^2-6 } < \frac{7n}{\frac{1}{2}n^2} = \frac{14}{n} $.
Consider $ K = \max\{4,\displaystyle \frac{14}{\epsilon}\} $ and suppose $ n> K $. Then $ n > \displaystyle \frac{14}{\epsilon} $. This implies that $ \epsilon > \displaystyle \frac{14}{n} $. Therefore
$ \displaystyle \left| \frac{n+6}{n^2-6} - 0 \right| = \frac{n+6}{n^2-6 } < \frac{7n}{\frac{1}{2}n^2} = \frac{14}{n} < \epsilon $.
Thus $ \lim\limits_{n \rightarrow \infty } \displaystyle \frac{n+6}{n^2-6}=0 $.
Is this proof correct? What are some other ways of proving this? Thanks!
|
$$\frac{n+6}{n^{2}-6}=r_{n}\times\frac{1+6r_{n}}{1-6r_{n}}$$ for: $$r_{n}=\frac{1}{n}$$
If $n\rightarrow\infty$ then $r_{n}\rightarrow0$ (easy) and consequently:
$$\frac{n+6}{n^{2}-6}\rightarrow0\times\frac{1+0}{1-0}=0$$
|
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|
find the limit: $\lim_{x\to 0} \dfrac{e^x \cos x - (x+1)}{\tan x -\sin x}$
find
$\lim_{x\to 0} \dfrac{e^x \cos x - (x+1)}{\tan x -\sin x}$
i tried using l'hopital's rule, but it just gets very complicated very fast
edit: i made a mistake in the numerator (sorry!) its $(x+1)$
|
$$
\begin{aligned}
\lim _{x\to 0}\left(\frac{e^x\cos(x)-\left(x+1\right)}{\tan(x)\:-\sin(x)}\right)
& = \lim _{x\to 0}\left(\frac{\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+o\left(x^3\right)\right)\left(1-\frac{x^2}{2}+o\left(x^2\right)\right)-\left(x+1\right)}{\left(x+\frac{x^3}{3}+o\left(x^3\right)\right)\:-\left(x-\frac{x^3}{6}+o\left(x^3\right)\right)}\right)
\\& = \lim _{x\to 0}\left(\frac{\frac{-x^5-3x^4-4x^3}{12}+o\left(x^2\right)}{\frac{x^3}{2}+o\left(x^3\right)}\right)
\\& = \color{red}{-\frac{2}{3}}
\end{aligned}
$$
Solved with Taylor expansion
|
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|
If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that $(a+1/a)^2+(b+1/b)^2\ge 25/2$
If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that
$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\frac{1}{b}\bigg)^2\ge \dfrac{25}{2}.$$
My work:
$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge \dfrac{25}{2}\implies a^2+\dfrac{1}{a^2}+b^2+\dfrac{1}{b^2}+4\ge \dfrac{25}{2}$$
Now, we have $a^2+\dfrac{1}{a^2}\ge 2$ and $b^2+\dfrac{1}{b^2}\ge 2$.
Here, I am stuck, I cannot use the information provided, $a+b=1$ to any use. Please help!
|
Hint: Substitute $a=\frac{1}{2}+x$ and $b=\frac{1}{2}-x$, $|x|<\frac{1}{2}$. Then you should only find the minimum of a one variable function $f(x)$.
I'll write the whole solution, maybe someone finds it helpful. After the substitution we get:
$(a+\frac{1}{a})^2+(b+\frac{1}{b})^2=\frac{9}{2}+2x^2+\frac{\frac{1}{2}+2x^2}{(\frac{1}{4}-x^2)^2}=f(x)$.
Since $x^2\geqslant0$, we get that $f(x)\geqslant\frac{9}{2}+\frac{\frac{1}{2}}{\frac{1}{16}}=\frac{25}{2}$. Equality: for $x=0$ or $a=b=\frac{1}{2}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding sum of sequence $\sum\limits_{k=2}^n \frac{1}{k^2-1}$ Here is a problem I need to solve:
$$ \sum_{k=2}^n \frac{1}{k^2-1} $$
It came with another one alreay done in the same task:
$$ \sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^n \frac{(k+1)-k}{k(k+1)} = \sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n \frac{1}{k+1} = \sum_{k=1}^n \frac{1}{k} - \sum_{k=2}^{n+1} \frac{1}{k} = 1 - \frac{1}{n+1} $$
So, the problem have to have similar solution. I can't find a proper change for the numerator. I think the numeratore could be find using factorised dominator:
$$ \sum_{k=2}^n \frac{1}{(k-1)(k+1)} $$
|
$\frac{1}{(k-1)(k+1)}=\frac{1}{2}\frac{1}{k-1}-\frac{1}{2}\frac{1}{k+1}$
|
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|
Prove the following equality using mathematical induction: Prove the following equality using mathematical induction: For any integer $n \ge 1$
$$\sum_{i=1}^{n} \frac{1}{i(i+1)} = \frac{n}{n+1}$$
I understand for the base base I need to have $n=1$. If I substituted $n$ for 1 and $i$ for 1 I would get
\begin{align}
\sum_{i=1}^{1} \frac{1}{1(1+1)} &= \frac{1}{1+1} \\
\sum_{i=1}^{1} \frac{1}{2} &= \frac{1}{2}
\end{align}
As for the inductive step, I inputted $n +1$ but it did not work for me because honestly I did not know what to do with the $i$ and what I should do next.
How would I go about doing the next step and an explanation for $i$ would be helpful.
|
Hint:
Write $$\frac{n}{n+1} = \frac{n+1-1}{n+1} = 1-\frac{1}{n+1}$$
And note that the sum on the left side is a telescop sum. Thus you do not even need induction to proof this.
However, for the induction step, you could write
$$\begin{array}{rcl}
\sum_{i=1}^{n+1} \frac{1}{i(i+1)} &=& \frac{1}{(n+1)(n+2)}+\sum_{i=1}^{n} \frac{1}{i(i+1)} \\
&=& 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} \\
&=& 1 - \frac{n+2}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)} \\
&=& 1 - \frac{n+2-1}{(n+1)(n+2)}\\
&=& 1 - \frac{1}{n+2}\\
\end{array}$$
|
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|
If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
My work:
$(\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b})=\frac{1}{a+b}$
By squaring both sides, we get,
$\frac{\sin^8 x}{a^2}+\frac{\cos^8 x}{b^2}+2\frac{\sin^4 x \cos^4 x}{ab}=\frac{1}{(a+b)^2}$
$\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
So, now, we have to prove that,
$-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=0$
I cannot do this. Please help!
|
HINT:
Use $$\cos^2x=1-\sin^2x$$ to form a Quadratic Equation in $\displaystyle\sin^2x$
writing $\displaystyle\sin^2x=p$ we get $$\frac{p^2}a+\frac{(1-p)^2}b=\frac1{a+b}$$
$$\implies b p^2+ a(1+p^2-2p)=\frac{ab}{a+b}$$
$$\implies (a+b)\{(a+b)p^2-2ap+a\}=ab$$
$$\implies (a+b)^2p^2-2a\cdot (a+b)p+a^2=0\implies \left[p(a+b)-a\right]^2\implies p=\frac a{a+b}$$
Can you take it from here?
|
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|
Calculate a hard limit Calculate: $$
\lim_{x\to+\infty} x\left( \frac{1}{x^2+1^2}+\frac{1}{x^2+2^2}+\dots+\frac{1}{x^2+x^2}\right)$$
|
$$
\begin{align}
& \phantom{=} \lim_{x\to+\infty} x\left( \frac{1}{x^2+1^2} + \frac{1}{x^2+2^2} + \dots + \frac{1}{x^2+x^2}\right) \\[8pt]
& = \lim_{x\to\infty} \frac 1 x \left( \frac{1}{1+\frac{1^2}{x^2}} + \frac{1}{1+\frac{2^2}{x^2}} +\cdots+\frac1{1+\frac{x^2}{x^2}} \right) \\[8pt]
& = \int_0^1 \frac{1}{1+w^2} \, dw = \frac \pi 4.
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Sum of consecutive square roots inside a square root $$\large\sqrt{1+\sqrt{1+2+\sqrt{1+2+3+\sqrt{1+2+3+4+\cdots}}}}$$
I saw this somewhere in the internet but, the website didn't provide me any further information. What is the sum of the equation above? What is it called?
|
As Gerry Myerson said this type of problems are called nested radical.
Let us try the following method: Writing the sequence of partials $$D_n:=
\sqrt {1 + \sqrt {1 + 2 + \sqrt {1 + 2 + 3 + \sqrt {1 + 2 + 3 + 4 + \ldots + \sqrt { \ldots + \sqrt { \ldots + \sqrt {1 + 2 + 3 + \ldots + n} } } } } } }
$$
Labeling the terms of the series beginning from the last one up to the first one; denoted by $y_1^2,y_1^2,\cdots,y_{n-1}^2,y_n^2$, i.e.,
\begin{align}
&y_n^{2}=1 + 2 + 3 + \ldots + n = \sum\limits_{k = 1}^n {k}=\frac{n(n+1)}{2}
\\
&y_{n-1}^{2}=1 + 2 + 3 + \ldots + (n-1) = \sum\limits_{k = 1}^{n-1} {k}=\frac{n(n-1)}{2}
\\
&y_{n-2}^{2}=1 + 2 + 3 + \ldots + (n-2) = \sum\limits_{k = 1}^{n-2} {k}=\frac{(n-1)(n-2)}{2}
\\
&\vdots
\\
&y_{3}^{2}=1 + 2 + 3 = \sum\limits_{k = 1}^{3} {k}=\frac{(3)(4)}{2}=6
\\
&y_{2}^{2}=1 + 2 = \sum\limits_{k = 1}^{2} {k}=\frac{(2)(3)}{2}=3
\\
&y_{1}^{2}=1 = \sum\limits_{k = 1}^{1} {k}=\frac{(1)(2)}{2}=1
\end{align}
Thus,
\begin{align}
D_n =
\sqrt {y_1 + \sqrt {y_2 + \sqrt {y_3 + \ldots + \sqrt { \ldots + \sqrt { \ldots + \sqrt {y_{n - 2} + \sqrt {y_{n - 1} + \sqrt {y_n } } } } } } } }
\end{align}
Now, let us divide $D_n$ into the following subsequences
\begin{align}
& S_{n,1}^2 = y_n \\
& S_{n,2}^2 = y_{n - 1} + S_{n,1} \\
&S_{n,3}^2 = y_{n - 2} + S_{n,2} \\
&S_{n,4}^2 = y_{n - 3} + S_{n,3} \\
&\vdots \\
&S_{n,n - 2}^2 = y_2 + S_{n,n - 1} \\
&S_{n,n - 1}^2 = y_1 + S_{n,n}
\end{align}
Finally, we have
\begin{align}
D_n &= S_{n,1}^2 + S_{n,2}^2 + S_{n,3}^2 + S_{n,4}^2 + \ldots + S_{n,n - 2}^2 + S_{n,n - 1}^2
\\
&= \sum\limits_{j = 1}^{n - 1} {S_{n,j}^2 }
\\
&= \sum\limits_{j = 1}^n {y_j } + \sum\limits_{j = 1}^n {S_{n,j} }
\end{align}
I think this enough to give an indication.
|
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|
If $p$, $q$ are naturals, solve $p^3-q^5=(p+q)^2$. In If $p,q$ are prime, solve $p^3-q^5=(p+q)^2$., the author asks to solve the equation $p^3-q^5=(p+q)^2$ for primes $p$ and $q$. A proof is given that $p=7, q=3$ is the only solution.
In this "followup", I would like to ask for a proof that does not depend on $p$ and $q$ being primes but allows arbitrary positive integers.
I do have an elementary proof for the case in which $p$ and $q$ are relatively prime, but the case in which they aren't is giving me a hard time. Anyone here who can help?
9/12 update:
As I commented, I may have been wrong when I claimed I had a proof for the case where $p$ and $q$ are relatively prime, but it should be possible to prove with the additional condition $(p,q+1)=1$ (not $q-1$ as I had written in the comment). Here's the proof:
Assume that $p$ and $q$ are positive integers such that
$$p^3-q^5=(p+q)^2\text{,}\tag{1} $$
$$(p,q)=1,\tag{2}$$
$$(p,q+1)=1.\tag{3}$$
Note that (1) implies that
$$q<p.\tag{4}$$
Evaluating (1) modulo $q$ gives $p^3\equiv p^2\pmod{q}$, so by (2), $p\equiv1\pmod{q}$, i.e. there exists $a\in\mathbb{N}$ such that
$$p=aq+1.\tag{5}$$
Likewise, if we evaluate (1) modulo $p$, we get $-q^5\equiv q^2\pmod{p}$, so $p$ divides
$q^5+q^2=q^2(q+1)(q^2-q+1)$ and thus by (2) and (3),
$$p\;|\;q^2-q+1.\tag{6}$$
Combining (5) and (6), we get that $p$ divides $q^2-q+1-aq-1=q(q-a-1)$ and therefore by (2),
$$p\;|\;q-a-1\tag{7}.$$
Note that, since the right-hand side in (6) is positive, $q-a-1$ must not be negative. On the other hand, (4) implies that it cannot be positive either, so it is $0$ and we have $a=q-1$ and therefore
$$p=q^2-q+1.\tag{8}$$
Now, substituting (8) in (1) and evaluating modulo $q^2$ gives $-3q+1\equiv1\pmod{q^2}$, i.e. $q^2$ divides $3q$ which forces $q=3$ and $p=7$.
|
I came up with a solution for $\gcd(p, q)=1$. Note that $p>q$. Looking mod $p$ gives $q^2(q^3+1) \equiv 0 \pmod p$, thus $p|q^3+1$. Take modulo $q$ of both sides of the equation to get $p^2(p-1) \equiv 0 \pmod q$, hence $q|p-1$. So we have $p=qr+1$ for some positive integer $r$. It follows that$$p|q^3+1-p=q^3-qr=q(q^2-r)$$Hence there is some non-negative integer $s$ for which $q^2-r=sp=s(qr+1)$ and consequently$$q^2-rsq-(r+s)=0$$In order to get integer values for $q$ discriminant of this quadratic must be a perfect square, but notice that if $r>1$ and $s>1$ $$(rs)^2<r^2s^2+4(r+s)<(rs+2)^2$$and we must have $r^2s^2+4(r+s)=(rs+1)^2$, which leads to$$r=\frac{4s-1}{2s-4},$$Which is never an integer. Contradiction. Therefore, we must have $r =1$ or $s \le1$.
$r=1$ gives $p=q+1$ and equation becomes $(q+1)^3-q^5=(2q+1)^2$. One can simplify this and get $-q^4+q^2-q-1=0$, so $q|1$ and $q=1$, which fails to satisfy equation.
$s=0$ gives $p=q^3+1$ and equation becomes $(q^3+1)^3-q^5=(q^3+q+1)^2$. So $q$ divides the constant term of polynomial equation, that is $q|2$, and $q=1, 2$, which again fail to satisfy equation.
$s=1$ gives $p=q^2-q+1$ and equation becomes $(q^2-q+1)^3-q^5=(q^2+1)^2$. So $q$ divides the constant term of polynomial equation, that is $q|3$, and $q=1, 3$. $q=1$ fails to satisfy equation. $q=3$ satisfies the equation and gives $p=7$. Therefore $(p, q)=(7, 3)$ is the only solution when $\gcd(p, q)=1$.
|
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|
if $A+B+C+D=\pi$, what is $\min(\cos{A}+\cos{B}+\cos{C}+\cos{D})$ Let $A,B,C,D \in R ~|~ A+B+C+D=\pi$, what is the minimum of the following function
$$f(A,B,C,D)=\cos{A}+\cos{B}+\cos{C}+\cos{D}$$
I found a post about a similar problem, butI think an even number of variables is harder than an odd number of variables, because for three variables,
$A+B+C=\pi~|~A,B,C\in R$ then it follows that $\cos{A}+\cos{B}+\cos{C}=1+4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}\to 1$, but for four variables, we can't.
My try: $$\cos{A}+\cos{B}+\cos{C}+\cos{D}=2\cos{\dfrac{A+B}{2}}\cos{\dfrac{A-B}{2}}+2\cos{\dfrac{C+D}{2}}\cos{\dfrac{C-D}{2}}=2\sin{\dfrac{C+D}{2}}\cos{\dfrac{A-B}{2}}+2\cos{\dfrac{C+D}{2}}\cos{\dfrac{C-D}{2}}$$
|
An alternative approach:
It is known that
$$
\cos A + \cos B + \cos C + \cos D = 2 \cos\frac{A+B}{2} \cos\frac{A-B}{2} +2 \cos\frac{C+D}{2} \cos\frac{C-D}{2}.
$$
By introducing the new variables
$$
A' = \frac{A+B}{2} \quad B' = \frac{A-B}{2} \quad C' = \frac{C+D}{2} \quad D' = \frac{C-D}{2},
$$
hence
$$
A = A'+B' \quad B=A'-B' \quad C = C' + D' \quad D = C'-D',
$$
we see that $A + B + C + D = \pi$ becomes $2 A' + 2 C' = \pi$. Hence we have to minimize the four functions
$$
2 ( \pm \cos A' \pm \cos(\frac{\pi}{2} - A')) = 2( \pm \cos A' \mp \sin A')
$$
It can be shown analytically (I used the computer) that these all have a minimum of $- 2 \sqrt{2}$. For example for both $`+'$ we have $A' = 2 \tan^{-1}(1+\sqrt{2})$. Hence $C' = \frac{\pi}{2} - 2 \tan^{-1}(1+\sqrt{2})$. Furthermore have we chosen $B' = 0$ and $C' = 0$. Using this we find that:
$$
A = \tan^{-1}(1+\sqrt{2}) \\
B = \tan^{-1}(1+\sqrt{2}) \\
C = \frac{\pi}{4} - \tan^{-1}(1+\sqrt{2}) \\
D = \frac{\pi}{4} - \tan^{-1}(1+\sqrt{2}).
$$
|
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|
Proof of an equality involving cosine $\sqrt{2 + \sqrt{2 + \cdots + \sqrt{2 + \sqrt{2}}}}\ =\ 2\cos (\pi/2^{n+1})$ so I stumbled upon this equation/formula, and I have no idea how to prove it. I don't know how should I approach it:
$$
\sqrt{2 + \sqrt{2 + \cdots + \sqrt{2 + \sqrt{\vphantom{\large A}2\,}\,}\,}\,}\
=\
2\cos\left(\vphantom{\Large A}\pi \over 2^{n + 1}\right)
$$
where $n\in\mathbb N$ and the square root sign appears $n$-times.
I thought about using sequences and limits, to express the LHS as a recurrence relation but I didn't get anywhere.
edit: Solved, thanks for your answers and comments.
|
I know it is a bit old, but I'd like to elaborate a bit the answer here going step by step.
As commented above, the best hint is to use the cosine half-angle formula ($*=\cos\left(\frac{\alpha}{2}\right)=\pm\sqrt{\frac{1+\cos(\alpha)}{2}}$) and work by induction. Therefore, let's get to it:
Base case:
The first partial sum, $S(1)$, will be:
$$S(1)=\sqrt{2}$$
Using the provided formula, one sees that
$$\sqrt{2}=\sqrt{2}\frac{\sqrt{2}}{\sqrt{2}}=\frac{2}{\sqrt{2}}=2\frac{1}{\sqrt{2}}=2\cos\left(\frac{\pi}{4}\right)=2\cos\left(\frac{\pi}{2^2}\right)=2\cos\left(\frac{1}{2}\cdot\frac{\pi}{2}\right)\stackrel{*}{=}2\sqrt{\frac{1+\cos\left(\frac{\pi}{2}\right)}{2}}=2\sqrt{\frac{1}{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$$
If we start directly from the cosine and try to get to the partial sum, it is actually nicer imho:
$$2\cos\left(\frac{\pi}{4}\right)=2\cos\left(\frac{\pi}{2^2}\right)=2\cos\left(\frac{1}{2}\cdot\frac{\pi}{2}\right)\stackrel{*}{=}2\sqrt{\frac{1+\cos\left(\frac{\pi}{2}\right)}{2}}=2\sqrt{\frac{1}{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$$
If we go for the second partial sum, we have:
$$S(2)=\sqrt{2+\sqrt{2}}$$
Which is:
$$2\cos\left(\frac{\pi}{8}\right)=2\cos\left(\frac{\pi}{2^3}\right)=2\cos\left(\frac{1}{2}\cdot\frac{\pi}{4}\right)\stackrel{*}{=}2\sqrt{\frac{1+\cos\left(\frac{\pi}{4}\right)}{2}}=2\sqrt{\frac{1}{2}+\frac{1}{2}\cos\left(\frac{\pi}{4}\right)}=2\sqrt{\frac{1}{2}+\frac{1}{2}\frac{\sqrt{2}}{2}}=2\sqrt{\frac{1}{2}+\frac{1}{4}\sqrt{2}}=2\sqrt{\frac{2+\sqrt{2}}{4}}=\sqrt{2+\sqrt{2}}$$
So now you can already develop some intuition that eventually you will have a factor $\frac{1}{2}$ inside the sum that will be cancelled by the outer 2.
Induction step:
Assume $S(n)$ works, i.e.:
$$\sqrt{2+\sqrt{2+\cdots+\sqrt{2+\sqrt{2}}}}=2\cos\left(\pi\over2^{n + 1}\right)$$
which clearly implies:
$$**=\frac{1}{2}S(n)=\frac{1}{2}\sqrt{2+\sqrt{2+\cdots+\sqrt{2+\sqrt{2}}}}=\cos\left(\frac{\pi}{2^{n+1}}\right)$$
Let's find out if $$S(n+1)=2\cos\left(\frac{\pi}{2^{n+2}}\right)$$
$$2\cos\left(\frac{\pi}{2^{n+2}}\right)=2\cos\left(\frac{1}{2}\cdot\frac{\pi}{2^{n+1}}\right)\stackrel{*}{=}2\sqrt{\frac{1+\cos\left(\frac{\pi}{2^{n+1}}\right)}{2}}=2\sqrt{\frac{1}{2}+\frac{1}{2}\cos\left(\frac{\pi}{2^{n+1}}\right)}\stackrel{**}{=}2\sqrt{\frac{1}{2}+\frac{1}{2}\frac{1}{2}S(n)}=2\sqrt{\frac{1}{2}+\frac{1}{4}S(n)}=2\sqrt{\frac{2+S(n)}{4}}=\sqrt{2+S(n)}$$
Bottom line for other cases:
To me, if I would encounter the $S(n)$, trying to get a nice closed-form solution would be the most complicated thing to do. Once you realise it goes in the direction of cosines is quite straight forward, as you can see. In this case, the trick was to observe that the cosine formula they relate it to is the half angle: see how $\cos\left(\frac{\alpha}{2^{n+1}}\right)$ has this $2^{n+1}$. It should pop up as an alarm as saying "hey, you will eventually have to split this denominator".
|
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|
Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$
For $a\ge 0$ let's define $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)dx.$$ Find explicit formula for $I(a)$.
My attempt: Let
$$\begin{align*}
f_n(x)
&= \frac{\ln\left(1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2\right)-\ln\left(1-2a\cos x+a^2\right)}{\frac{1}{n}}\\
&=\frac{\ln\left(\displaystyle\frac{1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2}{1-2a\cos x+a^2}\right)}{\frac{1}{n}}\\
&=\frac{\ln\left(1+\dfrac{1}{n}\left(\displaystyle\frac{2a-2\cos x+\frac{1}{n}}{1-2a\cos x+a^2}\right)\right)}{\frac{1}{n}}.
\end{align*}$$
Now it is easy to see that $f_n(x) \to \frac{2a-2\cos x}{1-2a\cos x+a^2}$ as $n \to \infty$. $|f_n(x)|\le \frac{2a+2}{(1-a)^2}$ RHS is integrable so $\lim_{n\to\infty}\int_0^\pi f_n(x)dx = \int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2} dx=I'(a)$. But
$$\int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2}=\int_0^\pi\left(1-\frac{(1-a)^2}{1-2a\cos x+a^2}\right)dx.$$ Consider
$$\int_0^\pi\frac{dx}{1-2a\cos x+a^2}=\int_0^\infty\frac{\frac{dy}{1+t^2}}{1-2a\frac{1-t^2}{1+t^2}+a^2}=\int_0^\infty\frac{dt}{1+t^2-2a(1-t^2)+a^2(1+t^2)}=\int_0^\infty\frac{dt}{(1-a)^2+\left((1+a)t\right)^2}\stackrel{(*)}{=}\frac{1}{(1-a)^2}\int_0^\infty\frac{dt}{1+\left(\frac{1+a}{1-a}t\right)^2}=\frac{1}{(1-a)(1+a)}\int_0^\infty\frac{du}{1+u^2}=\frac{1}{(1-a)(1+a)}\frac{\pi}{2}.$$
So $$I'(a)=\frac{\pi}{2}\left(2-\frac{1-a}{1+a}\right)\Rightarrow I(a)=\frac{\pi}{2}\left(3a-2\ln\left(a+1\right)\right).$$
It looks too easy, is there any crucial lack?
$(*)$ — we have to check $a=1$ here by hand and actually consider $[0,1), (1,\infty)$ but result on these two intervals may differ only by constant - it may be important but in my opinion not crucial for this proof.
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For some reason, this problem appeared on the main page for me even though the problem was posted quite a while ago. Let me present an alternative approach which does not really require any integrating.
Let us consider your integral:
$$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos(x)+a^2\right)\ \text{d}x,\;\ a>1$$
Consider a triangle $ABC$ with angles $\alpha,\beta,\gamma$ opposite to the sides $a,b,c$, respectively:
The law of cosines states
$$c^2=a^2+b^2-2ab\cos(\gamma)$$
To keep with this notation, let $b=1$ and $\gamma=x$.
Our integral becomes
$$\begin{align}
I(a) & = \int_{0}^{\pi}\ln\left(c^2\right)\ \text{d}\gamma \\
& = 2\int_{0}^{\pi}\ln\left(c\right)\ \text{d}\gamma \\
\end{align}$$
The law of sines states
$$\frac{\sin(\alpha)}{a}=\frac{\sin(\beta)}{b}=\frac{\sin(\gamma)}{c}$$
Our integral becomes
$$\begin{align}
I(a) & = 2\int_{0}^{\pi}\ln\left(a\ \frac{\sin(\gamma)}{\sin(\alpha)}\right)\ \text{d}\gamma \\
& = 2\pi \ln(a)+2\int_{0}^{\pi}\ln\left(\sin(\gamma)\right)\ \text{d}\gamma-2\int_{0}^{\pi}\ln\left(\sin(\alpha)\right)\ \text{d}\gamma \\
\end{align}$$
Consider the right-hand integral. In a triangle, the three angles $\alpha,\beta,\gamma$ add up to $\pi$ radians. Let $\gamma=\pi-\alpha-\beta$ such that $\text{d}\gamma=-\text{d}\alpha-\text{d}\beta$.
When $\gamma\rightarrow 0:\alpha\rightarrow \pi,\beta\rightarrow 0$. When $\gamma\rightarrow \pi:\alpha\rightarrow 0,\ \beta\rightarrow 0$.
Our integral becomes
$$\begin{align}
I(a) & = 2\pi \ln(a)+2\int_{0}^{\pi}\ln\left(\sin(\gamma)\right)\ \text{d}\gamma\\
& -2\left[\int_{\pi}^{0}\ln\left(\sin(\alpha)\right)(-\text{d}\alpha)+\int_{0}^{0}\ln\left(\sin(\alpha)\right)(-\text{d}\beta)\right] \\
& = 2\pi \ln(a)+2\int_{0}^{\pi}\ln\left(\sin(\gamma)\right)\ \text{d}\gamma-2\int_{0}^{\pi}\ln\left(\sin(\alpha)\right)\ \text{d}\alpha \\
\end{align}$$
Recognize that the integrals evaluate to the same value and therefore will cancel out each other.
$$I(a)=2\pi\ln(a),\;\ a>1$$
|
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Induction: show that $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + ... + \frac{1}{\sqrt{n}} < 2\sqrt{n}$ The question:
Induction: show that:
$$1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + ... + \frac{1}{\sqrt{n}} < 2\sqrt{n}$$
for $n \geq 1$
My attempt at a solution:
First we test the base case: $n=1$ this gives us:
$$1 < 2$$
Which works.
Then we do the inductive assumption that it holds true for $k=n$, this gives us:
$$1 + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n}} < 2\sqrt{n}$$
If we can prove that the following is true, we have solved the problem:
$$1 + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n+1}} < 2\sqrt{n+1}$$
we then have to show that $$2\sqrt{n} + \frac{1}{\sqrt{n+1}} \leq 2\sqrt{n+1}$$
Which can be written as: $$2\sqrt{n}\sqrt{n+1} + 1 \leq 2(n+1)$$
$$1 \leq 2(n+1) - 2\sqrt{n}\sqrt{n+1}$$
$$2\sqrt{n}\sqrt{n+1} \leq 2(n+1) - 1$$
$$2\sqrt{n}\sqrt{n+1} \leq 2n+1$$
$$(2\sqrt{n}\sqrt{n+1})^2 \leq (2n+1)^2$$
$$4n^2 + 4n \leq 4n^2 + 4n + 1$$
Can anyone please help me to continue from here or does this mean that I have proved the inductive assumption?
Thank you!
|
You have proved it but if you dislike switching to (essentially) showing a new inequality is true you could approach this slightly differently.
$\begin{align}
1 + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n+1}}
&< 2\sqrt{n} + \frac{1}{\sqrt{n+1}} \mathrm{(by\ inductive\ hypothesis)} \\
&= \frac{2\sqrt{n}(n+1)}{n+1} + \frac{\sqrt{n+1}}{n+1} \\
&= \frac{(2\sqrt{n}\sqrt{n+1}+1)\sqrt{n+1}}{n+1} \\
&= \frac{(2\sqrt{n^2+n}+1)\sqrt{n+1}}{n+1} \\
&< \frac{(2\sqrt{n^2+n+\frac{1}{4}}+1)\sqrt{n+1}}{n+1} \\
&= \frac{(2\sqrt{(n+\frac{1}{2})^2}+1)\sqrt{n+1}}{n+1} \\
&= \frac{(2(n+\frac{1}{2})+1)\sqrt{n+1}}{n+1} \\
&= \frac{(2n+1+1)\sqrt{n+1}}{n+1} \\
&= \frac{(2n+2)\sqrt{n+1}}{n+1} \\
&= \frac{2(n+1)\sqrt{n+1}}{n+1} \\
&= 2\sqrt{n+1} \\
\end{align}$
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Solve Recurrence Equation with Induction Question: Given the recurrence equation for the recursive Fibonacci sequence program:
$T(n) = T(n-1) + T(n-2) + b$
$T(0) = T(1) = a$
Using induction, show that $T(n) \leq f(n)$, where $f(n) = c2^n, \forall n \geq 0$. Find a value for $c$ in the process.
Attempt at a Solution:
Base case for n=0: $f(0) = c$, so you have the constraint $a \leq c$
Base case for n=1: $f(1) = 2c$, so you have the constraint $a \leq 2c$
Inductive step:
*
*$T(n) = T(n-1) + T(n-2) + b$
*$T(n+1) = T(n) + T(n-1) + b$
*$T(n+1) = c2^n + c2^{n-1} + b$
I'm not really sure where to go after this, nor am I really sure if what I've done so far is of any value whatsoever.
|
Calculate. We get $T(2)=2a+b$, $T(3)=3a+2b$, $T(4)=5a+4b$, $T(5)=8a+7b$, $T(6)=13a+12b$, and so on. The coefficients grow fast, but not too fast.
For simplicity assume that $a$ and $b$ are non-negative.
We show by induction that $T(n)\le 2^n a+2^n b=(a+b)2^n$. For the induction step, suppose that $T(k-1)\le 2^{k-1}(a+b)$ and $T(k-2)\le 2^{k-2}(a+b)$. We need to show that
$T(k)\le 2^k(a+b)$.
To do this it is enough to show that $2^{k-1}+2^{k-2}+1\le 2^k$. This is not difficult, since $1+2+\cdots +2^{k-1}=2^k-1$.
Detail: Let $c=a+b$. Let $A(n)$ be the assertion that $T(n)\le (a+b)2^n$ and $T(n+1)\le
(a+b)2^{n+1}$. It is clear that $A(0)$ holds. We show that if $A(k)$ holds, then $A(k+1)$ holds. So we need to show that $T(k+1)\le (a+b)2^{k+1}$ and $T(k+2)\le (a+b)2^{k+2}$.
It is built into $A(k)$ that $T(k+1)\le (a+b)2^{k+1}$. So it is enough to show that $T(k+2)\le (a+b)2^{k+2}$.
We have $T(k+2)=T(k+1)+T(k)+b$. By the induction hypothesis, $T(k+1)\le (a+b)2^{k+1}$ and $T(k)\le (a+b)2^k$. Thus
$$T(k+2)\le (a+b)2^{k+1}+(a+b)2^k +b\le (a+b)2^{k+1}+(a+b)2^k +a+b.\tag{1}$$
Thus from (1) we get
$$T(k+2)\le (a+b)(2^{k+1}+2^k+1).$$
But $2^{k+1}+2^k+1 \le 2^{k+2}$. It follows that $T(k+2)\le (a+b)2^{k+2}$. This completes the induction step.
|
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problem requires condition probability Q. A robot fires 3 shots at a moving target. For the first shot, the probability of hitting the moving target is 1/3. For subsequent shots beyond the first shot, the probability of hitting the moving target is 1/2 if the previous shot is a hit (for example, the probability of hitting the moving target on the 3rd shot is 1/2 if the 2nd shot is a hit) and the probability of hitting the moving target is 1/4 if the previous shot is a miss. What is the mean and variance of the number of hits?
I understand that we can draw a probability tree diagram and find out the possibility of each case, but what i don't understand is how to calculate mean for the number of hits.
some kind of hint would be really helpful
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Let $X$ be the number of hits. Then $X$ can take on values $0$, $1$, $2$, or $3$.
We want to find the probability of $k$ hits, that is, $\Pr(X=k)$, for $k=0,1,2,3$.
To save space, write for example $hmm$ for hit then miss then miss.
The probability that $X=0$ is the probability of $mmm$. This is $\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{3}{4}$, that is, $\frac{9}{24}$.
There are three ways we can have $1$ hit. They are $hmm$, $mhm$, and $mmh$.
The probabilities are $\frac{1}{3}\cdot\frac{1}{2}\cdot\frac{3}{4}$, $\frac{2}{3}\cdot \frac{1}{4}\cdot\frac{1}{2}$ and $\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{1}{4}$. These add up to $\frac{8}{24}$.
The probability that $X=3$ is $\frac{1}{3}\cdot\frac{1}{2}\cdot\frac{1}{2}$, which is $\frac{2}{24}$.
We skipped $X=2$, because the probabilities must add up to $1$. So the probability that $X=2$ is $\frac{5}{24}$. It might be a good idea to find $\Pr(X=2)$ the long way, by listing cases, as a check.
In summary, $\Pr(X=0)=\frac{9}{24}$ and $\Pr(X=1)=\frac{8}{24}$ and $\Pr(X=2)=\frac{5}{24}$ and $\Pr(X=3)=\frac{2}{24}$.
Now finding the mean is simple:
$$E(X)=\frac{9}{24}\cdot 0+\frac{8}{24}\cdot 1+\frac{5}{24}\cdot 2+\frac{2}{24}\cdot 3.$$
For the variance, if is easiest to use the fact that $\text{Var}(X)=E(X^2)-(E(X))^2$. And we have
$$E(X^2)=\frac{9}{24}\cdot 0^2+\frac{8}{24}\cdot 1^2+\frac{5}{24}\cdot 2^2+\frac{2}{24}\cdot 3^2.$$
|
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$2017$ as the sum of two squares Write the prime $2017$ as the sum of two squares
$2017$ can be written as the sum of two squares because it is a prime of the form $p\equiv 1\ ($mod $4)$
Using an appropriate algorithm find the two numbers that, when squared, add to the total of $2017$
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About 200 years ago, Jacobi would have known, and shown, that the number of representing 2017 as a sum of two squares is exactly 8.
More generally, for a positive integer $n$, let $r_2(n)$ denote the number of representations of $n$ as a sum of two squares of integers; for example, $r_2(5) = 8$, since $5 = 2^2 + 1^2 = (-2)^2 + 1^2 = 2^2 + (-1)^2 = (-2)^2 + (-1)^2 = 1^2 + 2^2 = 1^2 + (-2)^2 = (-1)^2 + 2^2 = (-1)^2 + (-2)^2.$
Let also $d_1(n)$ denote the number of positive divisors of $n$ of the form $4k+1$ and $d_3(n)$ the number of positive divisors of $n$ of the form $4k+3$. Jacobi's Formula is
$$
r_2(n) = 4(d_1(n) - d_3(n)) .
$$
Since 2017 is prime, and equal to $4 \times 504 + 1$, we have immediately $r_2(2017) = 4(2-0) = 8$. Easy to check:
$2017 = (44)^2 + 9^2 = (-44)^2 + 9^2 = \cdots$.
One out of many possible references for a proof of the formula of Jacobi: Theorem 3.2.1 in
Bruce C. Berndt, Number Theory in the Spirit of Ramanujan, Student Mathematical Library 34, Amer. Math Soc. 2006.
|
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Simplify the surd expression. Simplify the surd.
$(2\sqrt 3 + 3\sqrt 2)^2$
I know I should us this formula: $(a^2+2ab+b^2)$
But this gets complicated later. Please explain. :(
|
Alternatively, with a little simplification:
$$(2\sqrt{3}+3\sqrt{2})=(\sqrt{2}\sqrt{2}\sqrt{3}+\sqrt{3}\sqrt{3}\sqrt{2})=(\sqrt{2}(\sqrt{2}\sqrt{3}+\sqrt{3}\sqrt{3}))=\sqrt{2}(\sqrt{6}+3)$$
we can find the result:
$$\begin{align}[\sqrt{2}(\sqrt{6}+3)]^2&=2(\sqrt{6}+3)^2\\ & =2(6+2\sqrt{6}\cdot3+3^2) \\ &=2(15+6\sqrt{6}) \\ &=30+12\sqrt{6}
\end{align}$$
|
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Prove that $\sum_{k=0}^{\infty} (k-1)/2^k = 0$ How to prove that this series converges, and that the limit is 0 ?
|
You don't need calculus at all.
The sum is equal to $-1+ 0 + \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \dots$
In essence, we want to show that $\frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \dots = 1$
We can decompose this into an infinite number of infinite geometric series as follows:
$$S_1 = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{1}{2}$$
$$S_2 = \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \dots = \frac{\frac{1}{8}}{1 - \frac{1}{2}} = \frac{1}{4}$$
$$S_3 = \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \dots = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{1}{8}$$
And so forth.
Furthermore, notice that $$S = S_1 + S_2 + S_3 + \dots = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$$ is itself geometric! In addition, it's sum is $1$! Thus, we have just shown that $\displaystyle \sum_{k = 0}^{\infty} \frac{(k-1)}{2^k} = 0$
|
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Computing this limit $$\lim_{n\to\infty}{\sqrt[3]{n^3+5n^2+6}-\sqrt{n^2+3n+4}}$$
At first glance, we see that it's an indeterminate form ($\infty-\infty$). Here are my tries:
I.) I tried to form $a^2-b^2$ in the numerator, where $\cdots$ represents something that $\to 0$:
$$\frac{n^2(\sqrt[3]{1+\cdots}-1)-3n-4}{n(\sqrt[3]{1+\cdots}+\sqrt{1+\cdots})}$$
II.) I tried to form $a^3-b^3$ in the numerator, same as I.):
$$\frac{n^3(1-\sqrt[3]{1+\cdots})+5n^2+6}{n^2(\sqrt{1+\cdots}+\sqrt{1+\cdots}\sqrt{1+\cdots}+1)+3n+4}$$
What method can I apply here?
Thank you.
|
Try this:
$$\sqrt[3]{n^3+5n^2+6}-\sqrt{n^2+3n+4} = n\cdot \left(\sqrt[3]{1+\frac{5}{n}+\frac{6}{n^3}} - \sqrt{1+\frac{3}{n}+\frac{4}{n^2}}\right).$$
L'Hospital rules should help from here on.
|
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For a positive integer $n$ both $5n+1$ and $7n+1$ are perfect squares. Show that $n$ is divisible by 24. My try:
$5n + 1 = k^2$
$7n +1 = \frac{7k^2-2}5$
Just don't know how to proceed after this. Please help.
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Given, both $5n+1$ and $7n+1$ are perfect squares, We have to prove, $24\mid n$
Consider the sum: $(5n+1)+(7n+1)=12n+2$. This number leaves the remainder $2$ when divided by $3$. This is possible only under three situations:
a) $5n+1\equiv 2\mod3$ and $7n+1\equiv 0\mod3$
b) $5n+1\equiv 1\mod3$ and $7n+1\equiv 1\mod3$
c) $5n+1\equiv 0\mod3$ and $7n+1\equiv 2\mod3$
However, a perfect square never leaves remainder $2$ when divided by $3$. Hence, cases (a) and (c) cannot exist. The only choice, is thus case (b). Hence, both $5n+1$ and $7n+1$ leave remainder $1$ when divided by $3$. Hence, both $5n$ and $7n$ are divisible by $3$. As _gcd_$(5,3)=1$ and _gcd_$(7,3)=1$, we conclude that $3$ divides $n$.
Exactly similar argument holds for divisibility by $8$: The sum $12n+2$ leaves remainder $2$ or $6$ when divided by $8$.
But, $k^2\equiv 0,1,4 \mod 8$.
Thus, the only way two perfect squares can add up to a number that leaves remainder $2\mod 8$ is when each of them leaves remainder $1 \mod 8$. Hence, both $5n$ and $7n$ are divisible by $8$, and as $5$ and $7$ both are coprime with $8$, hence $8\mid n$.
Finally, because $8$ and $3$ are coprime, hence their product will also divide $n$. Hence $24\mid n$. Thus proved.
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Solving $\sqrt{x+2}+\sqrt{x-3}=\sqrt{3x+4}$ I try to solve this equation:
$$\sqrt{x+2}+\sqrt{x-3}=\sqrt{3x+4}$$
So what i did was:
$$x+2+2*\sqrt{x+2}*\sqrt{x-3}+x-3=3x+4$$
$$2*\sqrt{x+2}*\sqrt{x-3}=x+5$$
$$4*{(x+2)}*(x-3)=x^2+25+10x$$
$$4x^2-4x-24=x^2+25+10x$$
$$3x^2-14x-49$$
But this seems to be wrong! What did i wrong?
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Note: the original question read $= \sqrt{3x + 5}$ instead of $= \sqrt{3x + 4}$. There is no problem with the fixed question. Maybe just that it's unfinished. The final line should read $3x^2 - 14x + 49 = 0$ rather than just $x^2 - 14x + 49$. After that, solve the quadratic; only one of the solutions ($7$) to the quadratic gives a solution to the original equation (the other ($-7/3$) was introduced by the squaring operations).
Original response: the $3x + 4$ instead of $3x + 5$ on the first line seems a genuine mistake, rather than a typo; you continue with it.
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Divisibility problem: $ \frac{3^{m}}{2^{n} - 3^r} $
Is divisible a power of 3 for a difference of powers of 2 and 3? That
is, can result, this division, in an integer? $$ \frac{3^{m}}{2^{n} -
3^r} $$ where $n,m,r$ natural number.
Edit: $n>r$, $r=m+1$.
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HINT:
We need $\displaystyle 2^n-3^r=\pm3^s$ where $s\le m$
$\displaystyle 2^n=3^r\pm3^s$
Case $1:$ Taking the $+$ sign, $2^n=3^r+3^s=3^{r-s}(3^s+1)$ (WLOG $r\ge s$)
As $(2,3)=1,$ we need $3^{r-s}=1\iff r=s$ and $2^n=3^r+1$
Case $2:$ Taking the $-$ sign, $2^n=3^r-3^s=3^{r-s}(3^s-1)$ (we need $r\ge s$)
Like the previous case, we must have $r=s\implies 2^n=3^r-1$
See Catalan's Conjecture
|
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If $x+\frac1x=5$ find $x^5+\frac1{x^5}$. If $x>0$ and $\,x+\dfrac{1}{x}=5,\,$ find $\,x^5+\dfrac{1}{x^5}$.
Is there any other way find it?
$$
\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=23\cdot 110.
$$
Thanks
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Observe the recurrence relation $$x^{n+1} + x^{-(n+1)} = (x+x^{-1})(x^n+x^{-n}) - (x^{n-1} + x^{-(n-1)}).$$
This immediately gives us the specific recurrence $$f_{n+1} = 5f_n - f_{n-1}, \quad f_0 = 2, \quad f_1 = 5,$$ where $f_n = x^n + x^{-n}$.
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Proof of Divisibility of $n(n^2+20)$ by 48. This is a question from Bangladesh National Math Olympiad 2013 - Junior Category that still haunts me a lot. I want to find an answer to this question. Please prove this.
If $n$ is an even integer, prove that $48$ divides $n(n^2+20)$.
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${} \bmod 16$, we have $n(n^2+20) \equiv n(n^2+4)=2k(4k^2+4) = 8k(k^2+1)$. If $k$ is even then clearly $8k(k^2+1)\equiv 0$. If $k$ is odd, then $k^2+1$ is even and again $8k(k^2+1)\equiv 0$.
${} \bmod 3$, we have $n(n^2+20) \equiv n(n^2-1)=(n-1)n(n+1)\equiv 0$, since given three consecutive numbers, exactly one of them is a multiple of $3$.
Therefore, $16$ and $3$ divide $n(n^2+20)$ and so does $48=16\cdot 3=lcm (16,3)$.
|
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|
$\displaystyle \frac{6}{2n-1} - \frac{1}{n} = \frac{p}{2^i5^j}$ $\displaystyle \frac{6}{2n-1} - \frac{1}{n} = \frac{p}{2^i5^j}$
For which $n$ is this expression true. $n$ and $p$ are integers. $i$ and $j$ are positive integers or zero.
|
$$\frac{p}{2^i5^j} = \frac{6}{2n-1} - \frac{1}{n} = \frac{4n+1}{n(2n-1)}.$$
$n=1, 2$ are clearly solutions, and $3$ is not a solutions. Henceforth let $n\geq 4$.
Since $\gcd(n, 4n+1) = 1$, hence we must have $n \mid 2^i 5^j$. Since $\gcd(n, 2n-1) = 1$, they do not share any common factors. Note that $ \gcd( 2n-1, 4n+1) = \gcd(2n-1, 3 )$, hence is either 3 or 1. We consider cases as follows
Case 1: $\gcd(2n-1, 4n+1) = 1$.
Then one of $n$ and $2n-1$ is a power of 2 and the other is a power of 5. Since $2n-1$ is odd and greater than 7, it is the power of 5, and we have $ n = 2^i, 2n-1 = 5^j$. This gives us $$2^{i+1} - 1 = 5^j.$$
Consider mod 4. Since $n \geq 4$, hence $i\geq 2$, and so the LHS is $3 \pmod 4$, and the RHS is 1 $\pmod 4$,which is a contradiction. Thus, there are no solutions.
Case 2: $\gcd(2n-1, 4n+1) = 3$.
Since $2n-1$ is odd and greater than 7, we must have $n = 2^i$ and $ 2n-1 = 3 \times 5^j$. It is easy to check that $j=0, 1$ lead to the solutions $n=2, 8$. Henceforth, let $j \geq 2$.
Since $3 \mid 2^{i+1} - 1$, this tells us that $i$ is odd. Let $i = 2k + 1 $. This gives us
$$ 4^{k+1} -1 = 3 \times 5^j.$$
Dividing by $4 - 1 = 3$, we get $ 4^{k} + 4^{k-1} + \ldots + 1 = 5^j$. Since $j \geq 2$, we work mod 5 to get that $k+1$ must be even. Let $k = 2l + 1$. then dividing throughout by $4+1 = 5$, we get that
$$ 16^l + 16^{l-1} + \ldots + 1 = 5^{j-1}$$
Since $j-1 \geq 1 $, we work mod 5 to get that $l+1$ must be a multiple of 5. But this implies that $ 1 + 16 + 16^2 + 16^3 + 16^4 \mid 5^{j-1}$, which is clearly not possible since 69905 is not a power of 5.
Hence, the only solutions in this case are $n=2, 8$.
In conclusion, the only solutions are $n=1, 2, 8$.
I believe Case 2 has a simpler approach. I just bashed my way through it.
|
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|
Inverse of the Joukowski map $\phi(z) = z + \frac{1}{z}$ We know the Joukowski map $$\phi(z) = z + \frac{1}{z}$$ which maps the upper semidisc of radius $1$ in the lower half plane, and the lower semidisc of radius $1$ in the upper half plane.
What is the inverse of this function ? We obtain $z ^{2}-zy + 1 = 0$ and this equation has $2$ solutions, which is the right one ?
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We obtain $z^2−zy+1=0$ and this equation has $2$ solutions, which is the right one?
The solutions are given by
$$z = \frac{y - \sqrt{y^2-4}}{2},$$
where the different solutions correspond to the different choices of the square root.
We want a holomorphic inverse, so we need a holomorphic branch of $\sqrt{y^2-4}$ on the image of $\phi$, and the choice of the square root at one point determines whether we land inside the unit circle or outside. Since $\phi(\frac12) = \frac52$, we want
$$1 = \frac{5}{2} - \sqrt{\left(\frac{5}{2}\right)^2-4} = \frac{5}{2} - \sqrt{\frac{25-16}{4}},$$
so we need the branch of $\sqrt{y^2-4}$ on $\mathbb{C}\setminus [-2,2]$ that takes the value $\frac{3}{2}$ in $y = \frac{5}{2}$.
|
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Random variable transformation function I am stuck with a random variable transformation problem ($Y=\phi(X)$). The random variable $X$ has a uniform distribution $U(-1,1)$, and I want to transform it into $Y$ which is also an uniform distribution $U(\frac{s}{2},2s)$, where $s$ is some scalar quantity. Does anyone know a transformation function $\phi$ which transforms $X$ into $Y$ with these constraints?
Presently, I am using the function $Y=s2^X$, which unfortunately gives me an exponential distribution of $Y$ and not uniform distribution. Thank you in advance
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$2s-\frac{s}{2}= \frac{3s}{2} = \frac{3s}{4} \times (1-(-1))$
$\frac{3s}{4} \times 1 = \frac{3s}{4} = 2s - \frac{5s}{4}$ and $\frac{3s}{4} \times (-1) = -\frac{3s}{4} = \frac{s}{2} - \frac{5s}{4}$
So $Y=\frac{3s}{4} X +\frac{5s}{4} = \frac{s}{4} (3x+5) $ is a possible answer.
$Y=-\frac{3s}{4} X +\frac{5s}{4} = \frac{s}{4} (5-3x) $ is another and there are many more non-linear possibilities, though they will be more complicated.
|
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If $(a,b) = 1$, show that $(a-b,a^2+ab+b^2)=1 \text{ or } 3$ I have a question...We know that $(a,b)=1$ and we want to show that $(a-b,a^2+ab+b^2)=1 \text{ or } 3$.How can I show this??
I thought that we could suppose that $(a-b,a^2+ab+b^2)=d$.Then we know that $d|a-b$ and $d|,a^2+ab+b^2$.But how can I continue??
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$d|a-b$,$d|a^2+ab+b^2$ and $a^2+ab+b^2=(a-b)^2+3ab$ imply that
$$d|3ab,d|3a(a^2+ab+b^2)=3a^3+3a^2b+3ab^2,d|3ab(a-b)=3a^2b-3ab^2,d|2a(3ab).$$
Hence, $$d|3a^3+3a^2b+3ab^2+3a^2b-3ab^2-6a^2b=3a^3$$ and similarly $d|3b^3$ and now
$$d\le(3a^3,3b^3)=3(a^3,b^3)=3.$$ so $d=1\,\text{or}\,2\,\text{or}\,3$, but $d$ can't be $2$, since in this case $a^3$ and $b^3$ must be two even numbers and $(a^3,b^3)\neq1$. (Note that $(a,b)=1$ implies $(a^3,b^3)=1$)
|
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If $x,y,z\in(0;1)$, prove that $(x+1)(y+1)(z+1)\ge \sqrt{8(x+y)(y+z)(z+x)}$.
If $x,y,z\in(0;1)$, prove that $$(x+1)(y+1)(z+1)\ge \sqrt{8(x+y)(y+z)(z+x)}$$
Both sides of the inequality are positive, so I could square them: $$(x+1)^2(y+1)^2(z+1)^2\ge8(x+y)(y+z)(z+x)$$
Even if there's a solution using brute force, we'd have to prove this:
$$x^2 y^2 z^2+2 x^2 y^2 z+x^2 y^2+2 x^2 y z^2+4 x^2 y z-6 x^2 y+x^2 z^2-6 x^2 z+x^2+2 x y^2 z^2+4 x y^2 z-6 x y^2+4 x y z^2-8 x y z+4 x y-6 x z^2+4 x z+2 x+y^2 z^2-6 y^2 z+y^2-6 y z^2+4 y z+2 y+z^2+2 z+1\ge0$$
I'm sure there's a more elegant approach. But I can't find it. Some ideas would be great. Thanks.
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You can prove that
$$
(x+1)(y+1)\geq 2(x+y)
$$
From that, you can multiply the three inequalities you get by permuting the variable to get what you want.
To prove that, first expand both sides
$$
xy+x+y+1\geq 2x+2y
$$
simplifying yields
$$
xy+1\geq x+y\\
xy-x-y+1\geq 0\\
(x-1)(y-1)\geq 0\\
(1-x)(1-y)\geq 0
$$
But since $0< x,y< 1$, we know the last statement is true, and because these are all two-way steps, we now have proven the theorem. Since the initial inequalities are strict, we also know that equality cannot happen.
|
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|
How can we convert the number $(0.\overline{101})_2$ written in dual-system into the decimal system?
How can we convert the number $(0.\overline{101})_2$ written in dual-system into the decimal system ?
$0.\overline{101}=\underbrace{\dfrac{1}{2}+\dfrac{0}{4}+\dfrac{1}{8}}_{\dfrac{5}{8}}+\underbrace{\dfrac{1}{16}+\dfrac{0}{32}+\dfrac{1}{64}}_{\dfrac{5}{64}}+...$,
then i get:
$\sum_{k\ge0}\dfrac{5}{8}\cdot (\dfrac{1}{8})^{k}=\dfrac{5}{7}$
but must it not be $(0.\overline{101})_2=\dfrac{101}{111}=(0.\overline{909})_{10}$
Thanks in advance
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There is no such thing as a characeterwie substitution to obtain base ten periods from base two periods. Just thinkn of $\frac13=0.\overline 3_{10}=0.\overline{01}_2$ and especially, $\frac15=0.2_{10}=0.\overline{0011}_2$.
The actual analogy is that $$0.\overline{909}_{10}=\frac{909}{10^3-1}=\frac{909}{999}=\frac{101}{111}$$ and $$0.\overline{101}_2 = \frac{101_2}{2^3-1}=\frac 57.$$.
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Formal Limit Proofs for Limits Involving Factorials How does one use the definition of the limit to formally prove that
$$
\lim_{n \to \infty} \frac{n^4}{n^2 + n!} = 0?
$$
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Well, you know that $n! \ge n(n-1)(n-2)(n-3)(n-4) \ge (n-4)^5$ (for $n >4$), so
you have $0 \le{n^4 \over n^2+n!} \le {n^4 \over n!} \le {n^4 \over (n-4)^5} = {{1 \over n} \over (1-{4 \over n})^5 }$. Hence for $n>8$ we have $(1-{4 \over n}) \ge {1 \over 2}$, and so ${n^4 \over n^2+n!} \le 2^5 {1 \over n}$.
Now choose $n > \max(8, {2^5 \over \epsilon})$, then ${n^4 \over n^2+n!} < \epsilon$.
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Find the maximum or minimum value of the quadratic function. Find the maximum or minimum value of the quadratic function by completing the squares. Also, state the value of $x$ at which the function is maximum or minimum.
$y=2x^2-4x+7$
$x^2$ has a coefficient of $2$, how should I complete the squares?
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$y=2x^2-4x+7$ therefore $y=2(x^2-2x+7/2)$
completing the square now gives $y=2[(x-1)^2-1+7/2]$
which rearranges to $y=2(x-1)^2+5$
For completing the square, I always divide by a number that ensures I have $1$ as the coefficient of $x^2$.
To get the minimum value there are two options:
1) By looking at the equation after we have completed the square we can see that we will always have the $+5$ term, so we need to minimize the $2(x-1)^2$ term. Because it's square it will never be negative, however, we can make it equal to zero. So, $2(x-1)^2=0$, therefore $(x-1)^2=0$ and $x-1=0$ so $x=1$ is the minimum point.
2) Using differentiation. $y'=4x-4$ then set this equal to zero so $4x-4=0$, $4x=4$ and $x=1$.
|
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Prove that $n^3 + 5n$ is divisible by $6$ by using induction Prove that for integers $n > 0$, $n^3 + 5n$ is divisible by $6$.
Here is what I have done:
Base Step: $n=1$, $1^3+5(1)=6$
Inductive Step:
$p(k)=k^3 + 5k =6m$, $m$ is some integer
$p(k+1)=(k+1)^3+5(k+1)=6m$ $m$ is some integer
Since both are equal to $6m$ I set them equal to each other.
$k^3 + 5k=(k+1)^3+5(k+1)$
$k^3+5k=(k+1)[(k+1)^2+5]$
Proven that $p(k)=p(k+1)$
I do not think this is the correct way of proving this problem but I couldn't think of anything else.
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Hint: $(k+1)^3+5(k+1) = (k^3 + 5k) + 3k^2 + 3k + 6$
Since $(k^3 + 5k)$ is divisible by 6, all you have to prove is $3k^2 + 3k + 6$ is divisible by 6 too. Since adding $6$ does not change divisibility, you just have to proove $3k(k+1)$ is divisible by 6. Think about even and odd numbers.
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|
Expected number of steps between states in a Markov Chain Suppose I am given a state space $S=\{0,1,2,3\}$ with transition probability matrix
$\mathbf{P}= \begin{bmatrix}
\frac{2}{3} & \frac{1}{3} & 0 & 0 \\[0.3em]
\frac{2}{3} & 0 & \frac{1}{3} & 0\\[0.3em]
\frac{2}{3} & 0 & 0 & \frac{1}{3}\\[0.3em]
0 & 0 & 0 & 1
\end{bmatrix}$
and I want the expected number of steps from states $0 \rightarrow 3$ which I will denote $E_0(N(3))$.
Attempt at solving: First I write the transient states $\{0,1,2\}$ and recurrent state $\{3\}$ which I got from drawing the chain. I now want to write $\mathbf{P}$ in canonical form, i.e. with state space $S=\{3,0,1,2\}$ as so:
$\mathbf{P}=\begin{bmatrix}
1 & 0 & 0 & 0\\[0.3em]
0 & \frac{2}{3} & \frac{1}{3} & 0 \\[0.3em]
0 & \frac{2}{3} & 0 & \frac{1}{3}\\[0.3em]
\frac{2}{3} & 0 & 0 & \frac{1}{3}
\end{bmatrix}$
It's clear that the transient matrix is
$\mathbf{Q}=
\begin{bmatrix}
\frac{2}{3} & \frac{1}{3} & 0 \\[0.3em]
\frac{2}{3} & 0 & \frac{1}{3}\\[0.3em]
0 & 0 & \frac{1}{3}
\end{bmatrix}$
Now I can get the matrix I want for computing expected steps (calculated with Mathematica):
$\mathbf{M}=(\mathbf{I}-\mathbf{Q})^{-1}=\begin{bmatrix}
9 & 3 & 3\\[0.3em]
6 & 3 & 3\\[0.3em]
0 & 0 & \frac{3}{2}
\end{bmatrix}$
From this, we get $E_0(N(3))=9+3+3=15$. Is this correct? I am sort of weak in finding the "canonical form" of a matrix.
Note: although this looks like a homework question, it's simply a preparation problem for an upcoming exam, so a complete solution/correction of my work is appreciated.
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The distribution for the number of time steps to move between marked states in a discrete time Markov chain is the discrete phase-type distribution. You made a mistake in reorganising the row and column vectors and your transient matrix should be
$$\mathbf{Q}=
\begin{bmatrix}
\frac{2}{3} & \frac{1}{3} & 0 \\
\frac{2}{3} & 0 & \frac{1}{3}\\
\frac{2}{3} & 0 & 0
\end{bmatrix}$$
which you can then continue to find
$$\mathbf M = (\mathbf I - \mathbf Q)^{-1} =
\begin{bmatrix}
27 & 9 & 3\\
24 & 9 & 3 \\
18 & 6 & 3\end{bmatrix}$$
and summing the first row gives you (as you require) 39.
Generally when I have seen these distributions written the 'exit' state is put last, which would mean your matrix with elements in the order $\{0,1,2,3\}$ was already in canonical form and you needed to extract the top left portion as your transient matrix,
$$\mathbf P =
\left(
\begin{array}{ccc|c}
\frac{2}{3} & \frac{1}{3} & 0 & 0\\
\frac{2}{3} & 0 & \frac{1}{3} & 0 \\
\frac{2}{3} & 0 & 0 & \frac{1}{3}\\
\hline\\
0 & 0 & 0 & 1
\end{array}
\right)
$$
For further moments you might be interested in this paper:
*
*Dayar, Tuǧrul. "On moments of discrete phase-type distributions." Formal Techniques for Computer Systems and Business Processes. Springer Berlin Heidelberg, 2005. 51-63. doi:10.1007/11549970_5.
|
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|
What is the coefficient of the term $x^4 y^5$ in $(x+y+2)^{12}$? What is the coefficient of the term $x^4 y^5$ in $(x+y+2)^{12}$?
How can we calculate this expression ?
I've applied the binomial theorem formula and got $91$ terms but I am not sure if it is right or wrong.
|
The trick is to apply trinomial formula here, the coefficient is
$${12\choose4,5,3}2^3=\frac{12!2^3}{4!5!3!}$$
To see this, first consider $(x+y)$ as a whole body and then use binomial formula. Then the coefficient of term $(x+y)^9$ is
$\displaystyle{12\choose 9}2^3=\frac{12!2^3}{9!3!}$ (Note we don't need to consider terms like $(x+y)^k$ other than $k=9$ here). And then consider the term $x^4y^5$ in $(x+y)^9$, whose coefficient shall be $\displaystyle{9\choose4}=\frac{9!}{4!5!}$. Combine two things, we obtain
$${12\choose 9}2^3{9\choose4}=\frac{12!2^3}{9!3!}\frac{9!}{4!5!}=\frac{12!2^3}{4!5!3!}$$
|
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|
FINDING PMF probability theory A die is tossed until first 6 occurred. Let X be Random variable that number of one in the experiment. Find the PMF of X. And E(X).
I noticed PMF of X is geometric distribution but I don't know why.
|
The distribution of $X$ is indeed geometric, under one of the definitions of geometric distribution.
Note that $X$ is the number of $1$s until the first $6$. In particular, for this game, anything other than $1$ or $6$ does not matter, and need not even be recorded. All that matters is $1$ or $6$. Given that anything else is irrelevant, and $1$ and $6$ are equally likely, each has probability $\frac{1}{2}$.
Thus $\Pr(X=0)=\frac{1}{2}$ (a $6$ comes before any $1$). Similarly, $\Pr(X=1)=\frac{1}{4}$ (among recorded tosses, there is a $1$ and then a $6$). In general, $\Pr(X=n)=\frac{1}{2^n}$.
Thus we have
$$E(X)=1\cdot \frac{1}{2^2}+2\cdot \frac{1}{2^3}+3\cdot\frac{1}{2^4}+\cdots.\tag{1}$$
Sums of this type have been repeatedly calculated on MSE. There are many ways to compute. One way is to note that for $|x|\lt 1$, we have
$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots.$$
Differentiate. We get
$$\frac{1}{(1-x)^2}=1+2x+3x^2+\cdots.$$
Set $x=\frac{1}{2}$, and multiply through by $\frac{1}{2^2}$. We get Expression (1), which is therefore equal to $1$. We conclude that $E(X)=1$.
There are many other ways to evaluate (1). Recall that the more common geometric distribution, in which we count the number of trials until the first success, has expectation $\frac{1}{p}$, where $p$ is the probability of success on any trial. In our case, the number of $1$s and/or $6$s until the first $6$ has expectation $\frac{1}{1/2}=2$. Thus the number of failures, that is, $1$s, until the first $6$ has expectation $2-1$, that is, $1$.
|
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Solving a curve integral around part of an elipse I'm having trouble calculating a curve integral in a vector field:
$\int_C y (18x + 1)\ dx + 2y^2\ dy$
where $C$ is the curve along the ellipse
$9x^2 + y^2 = 64$
going counterclockwise from the point ( $-\frac{4}{3}\sqrt{3} $ , $4$) to the point (-$\frac{4}{3}$ , $4\sqrt{3} $)
Thats almost one "lap" around the ellipse..
When making a parametrisation I come up with:
x = $\sqrt{ \frac{64}{9}\ } \cos t $
y = $\sqrt{ 64 } \sin t $
$- \frac{\Pi}{3} \le t < \arctan(3 \sqrt{3}) $
But the integral created from this parametrization give an answer involving arctan. Any ideas to get a rational answer?
I had an idea to split the curve into multiple curves and integrating them piecewise, but that gets really messy aswell.
This is my calculations when making a variable change:
First the variable change:
$ u= 3x $
$ du = 3dx $
$ \ u^2 + y^2 =64 $
$ \int y(18x + 1)dx + 2y^2dy = \int y(6u + 1)\frac{du}{3} + 2y^2dy $
$ u = \sqrt{64}\cos t $
$ y = \sqrt{64}\sin t $
$ du = -\sqrt{64}\sin tdt $
$ dy = \sqrt{64}\cos tdt $
The startpoint after making variable change:
$ ( -4\sqrt{3},4)$
$ arctan( \frac{4}{-4\sqrt{3}}) = -\Pi/6$
Startangle: $ \frac{5\Pi}{6} $
The endpoint after making variable change: $ ( -4,4\sqrt{3})$
$ arctan( \frac{-4\sqrt{3}}{4}) = -\Pi/3$
End angle: $\frac{2\Pi}{3} $
$\int y(6u + 1)\frac{du}{3} + 2y^2dy = \int -\frac{1}{3}\sqrt{64}\sin t(6\sqrt{64}\cos t + 1)\sqrt{64}\sin tdt + 128\sin^2t\sqrt{64}\cos tdt = $
$\int -\frac{64}{3}\sin^2tdt $
$ \frac{5\Pi}{6} \le t \le \frac{2\Pi}{3} $
The answer becomes: $\frac{16}{9}\Pi $
|
Draw a figure! The parametrization of the ellipse is
$$t\mapsto \bigl(x(t),y(t)\bigr):=\left({8\over3}\cos t,8\sin t\right)\qquad(t\in{\mathbb R})\ .\tag{1}$$
Now we have to find the boundary values which are relevant for the integral $J$ in question. Both endpoints $p$ and $q$ of the arc lie in the second quadrant. At the starting point we have $8\sin t=4$, or $\sin t={1\over2}$, which amounts to $t=-\pi-{\pi\over6}$, and at the endpoint we have $8\sin t=4\sqrt{3}$, which amounts to $t={2\pi\over3}$.
Therefore we obtain
$$J=\int_{-7\pi/6}^{2\pi/3} \left(y(t)(18 x(t)+1)\dot x(t) + 2y^2(t)\dot y(t)\right)\ dt\ ,$$
where you now have to plug in the parametrization $(1)$. The computation can be simplified somewhat by observing that
$$\int\nolimits_\gamma 2y^2\ dy={2\over3}y^3\biggr|_p^q\quad .$$
|
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|
Why doesn't this approach work for $\int \sec^4 x\,dx$? I been trying to integrate $\sec^4$ , without much luck. But I don't entirely understand why my result is invalid and would like some feedback if possible.
I'm attacking the issue in the following way
$$
\int (\sec^2{x})^2dx = \int (\tan^2+1)^2dx
$$
then, I put $u=\tan^2+1$ which means $x = \arctan(\sqrt{u-1})$, which allows me to do the following backwards substitution
$$
\int (\sec^2{x})^2dx = \int (\tan^2+1)^2dx = \int u^2 \frac{1}{2u\sqrt{u-1}} dx = \frac{1}{2}\int u (u-1)^{\frac{-1}{2}} dx
$$
Now using integration by parts I get
$$\begin{align*}
\int u (u-1)^{\frac{-1}{2}} dx &= \frac{1}{2}(2u(u-1)^{\frac{1}{2}} - \int 2(u-1)^{\frac{1}{2}})\\\\
&= \frac{1}{2}(2u(u-1)^{\frac{1}{2}}-\frac{4}{3}(u-1)^{3/2}) \\\\
&= \frac{1}{2}(2(\tan^2{x}+1)(\tan^2{x})^{\frac{1}{2}}-\frac{4}{3}(\tan^2{x})^{3/2})
\end{align*}$$
This however, seems to be incorrect. How come?
|
Your result is almost valid.
Taking a different approach, using Integration by Parts from the start:
Note that $\sec^2x = \frac{d}{dx} (\tan x)$.
We can use integration by parts:
$u = \sec^2 x \implies\,du = 2\sec^2 x \tan x\,dx$
$dv = \sec^2 x \,dx\implies \, v = \tan x$.
$$\int \sec^4 x \,dx = \sec^2x\tan x - 2\int \tan^2 x \sec^2 x$$
Now, the remaining integral can be easily solved by substitution: $w = \tan x,\;\implies dw = \sec^2 x$: $$2\int \tan^2 x \sec^2 x \,dx = 2\int w^2 \,dw = \dfrac 23 w + c$$
Putting the above together gives us:
$$\int \sec^4 x \,dx = \sec^2x\tan x - \frac 23 \tan^3 x + C$$
Note: Your answer is very, very close, and can be manipulated algebraically to closely match the above:
$$\frac{1}{2}\Big(2(\underbrace{\tan^2{x}+1}_{\sec^2 x})\underbrace{(\tan^2{x})^{\frac{1}{2}}}_{|\tan x|}-\frac{4}{3}\underbrace{(\tan^2{x})^{3/2}}_{|\tan^3 x|}\Big)
= \sec^2 |\tan x| - \frac 23 |\tan^3 x| + C$$
|
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|
If $k$ cannot equal $0$ and $A$ is as given below, what is $A$ inverse? $$
A =
\pmatrix{
1 & 0 & 2k\\
0 & 1 & k\\
0 & 0 & k
}
$$
I don't know what method to use to solve this problem as I haven't encountered and variable before when solving for the inverse of a matrix. I've tried to use the identity method to solve it. It didn't seem right. I also, tried solving for each row and setting them equal to 0, this also didn't seem right.
|
Hint: We can write the matrix as the product of elementary matrices:
$$
\begin{bmatrix}
1 & 0 & 2k \\
0 & 1 & k \\
0 & 0 & k \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & k \\
\end{bmatrix}.
$$
Perhaps you'll find it easier to find $$
\left(\begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & k \\
\end{bmatrix}\right)^{-1}=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & k \\
\end{bmatrix}^{-1}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{bmatrix}^{-1}
\begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}^{-1}.
$$
Their inverses are described on the linked wikipedia page.
|
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|
Determine the value of the integral $I=\int_{0}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$ Determine the value of the integral $$I(a)=\int_{0}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$$
My try:
$\to I'(a)=\int_{0}^{\infty}\frac{2a}{(a^2+x^2)(b^2+x^2)}dx=\frac{\pi}{b(a+b)}$
Hence $I(a)=\frac{\pi}{b}\ln(a+b)+C$
Question: Find C???
Thank you!
|
Since integrand is even function, so
$\int_{0}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$=$\frac{1}{2}\int_{-\infty}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$
so solving the R.H.S using residues--
$\int_{C} f(z)dz=\int_{Cr} f(z)dz +\int_{-R}^{R}\frac{1}{2} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$
where C is closed loop in upper half of plane and f(z)= $ \frac{1}{2}\frac{\ln\left(a^2+z^2\right)}{b^2+z^2}$
and f(z) becomes $\frac{1}{2}\frac{\ln\left(a^2+z^2\right)}{(z+bi)(z-bi)}$ by factorizing denominator
there for f(z) has singularities at z=+bi,-bi, but only z=bi lie in upper half of plane
so,$\int_{C} f(z)dz=2(pi)(i)$[residue of f(z) at $bi$]
and as R goes to infinity $\int_{Cr} f(z)dz-->0$
we get
$\int_{0}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx=\int_{C} f(z)dz=2(\pi)(i)$[residue of f(z) at $bi$]
=$\frac{\pi}{2b}\ln(a^2-b^2)$
|
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|
Proof that there are at the most two numbers of exactly six digits that squared end with the same six digits Written in a more formal way, proof that there are at the most $2$ numbers $n$ of six digits, that
$$n^2 \equiv n \mod 10^6$$
Research effort:
if $n^2 \equiv n \mod 10^6$ this means $10^6\mid n^2-n \rightarrow10^6\mid n(n-1)$
Given that $5^6\mid 10^6$ then $5^6\mid n(n-1)$ whitch means $5^6\mid n$ or $5^6\mid n-1$
Same for $2^6$, Given that $2^6\mid 10^6$ then $2^6\mid n(n-1)$ whitch means $2^6\mid n$ or $2^6\mid n-1$
By the chinese remainder theorem I got assured $4$ solutions, but I tried to solve the systems and it's not a very happy thing to do, so I think I did something wrong somewhere.
What do you think?
|
If $\displaystyle2^6|(n-a)$ and $5^6|(n-a),$
$\displaystyle$ lcm$(2^6,5^6)|(n-a)$
As $(2,5)=1,(2^6,5^6)=(2,5)^6=1$
$\implies$ lcm$(2^6,5^6)=(2^6\cdot5^6)=10^6\iff n\equiv a\pmod{10^6}$
Here $a=0,1$
|
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|
Show that there is a unique sequence of positive integers $(a_n)$ satisfying $a_1=1,a_2=2,a_4=12,a_{n+1}a_{n-1}=a_n^2\pm 1 $ Show that there is a unique sequence of positive integers $(a_n)$ satisfying the following conditions.
$$a_1=1,a_2=2,a_4=12,a_{n+1}a_{n-1}=a_n^2\pm 1$$
I approached the problem to find out,
$a_3=4 \pm 1$
$a_2a_4=a_3^2\pm1 \implies 24\mp1=a_3^2 \implies a_3=5$ as $a_n \in \mathbb{N}$
I could not do anything more. Even, I could not understand what should I show to prove that the sequence is unique. Please help.
|
Suppose that after $a_n$, that both possible values of $a_{n+1}$ , $\frac{a_n\,^2 - 1} {a_{n-1}}$ and $\frac{a_n\,^2 + 1} {a_{n-1}}$, are both integers.
Then $$\begin{cases} a_{n-1} | a_n\,^2 - 1 \\ a_{n-1} | a_n\,^2 + 1 \end{cases}$$
So $a_{n-1}$ can only be $1$ or $2$ if the sequence forks into to valid sequences after $a_n$. So if the sequence is increasing, then it is unique for $a_{n-1} > 2$.
For existence, just look at the sequence mechanically first:
$$\begin{array} {c|c} n & a_n \\ \hline 1 & 1 \\ 2 & 2 \\ 3 & 5 \\ 4 & 12 \\ 5 & 29 \\ 6 & 70 \\ 7 & 169 \end{array}$$
It seems to be geometric with a ratio of $\frac {169} {70} \text{ is about } 2.414 \text{ is about } 1 + \sqrt{2}$. Checking $a_n = x(1 + \sqrt{2})^n + y (1 - \sqrt{2})^n$ gets $x = 2^{-3/2}, y = -2^{-3/2}$.
Final verification of existence, put value of $a_n$ into:
$$a_n = \frac{a_{n-1}\,^2 - (-1)^n} {a_{n-2}}$$
|
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|
Proof this curious trigonometric identity Proof that $$\cos^2{10^\circ} + \cos^2{50^\circ} - \sin{40^\circ}\sin{80^\circ} = \frac{3}{4}$$
I notice that $10^\circ + 80^\circ = 90^\circ$, and $50^\circ +40^\circ = 90^\circ$. I tried doing some manipulation but my efforts were futile. Any hints?
|
Generalization :
$$I=\cos^2A+\cos^2B-\cos A\cos B$$
$$=\cos^2A-\sin^2B+1-\frac{\cos(A-B)+\cos(A-B)}2$$
$$=\cos(A-B)\cos(A+B)+1-\frac{\cos(A-B)+\cos(A+B)}2$$
$$I=\cos(A-B)\left(\cos(A+B)-\frac12\right)+1-\frac{\cos(A+B)}2$$
If $\displaystyle\cos(A+B)=\frac12=\cos60^\circ\iff A+B=2n\pi\pm60^\circ$ where $n$ is any integer
$I$ becomes $\displaystyle1-\frac{\dfrac12}2=\frac34$
$$\text{Again, }I=\cos(A+B)\left(\cos(A-B)-\frac12\right)+1-\frac{\cos(A-B)}2$$
What if $\displaystyle\cos(A-B)=\frac12 ?$
|
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|
Using residues to evaluate an improper integral
Use residues to evaluate the improper integral
\begin{align}
\int_{0}^{\infty} \frac{1}{(x^2+1)^2} \, dx
\end{align}
First, I said $f(z) = \frac{1}{(z^2+1)^2}$. My only question so far is how do I establish the region $C$ (from the given real limits of $0$ to $\infty$) so I can do countour integration and find residues in $C$?
|
In response to @Cameron Williams' hint and comments, I am going to attempt the solution.
We have $f(z) = \frac{1}{(z^2+1)^2}$. Let $C$ be the half circle as described by @Cameron Williams. Now, we have $z = i$ to be the singularity point inside $C$.
In finding the residue,
\begin{align}
\text{Res}_{z = i} f(z) &= \text{Res}_{z = i} \frac{1}{(z^2+1)^2} = \text{Res}_{z = i} \frac{1}{(z+i)^2(z-i)^2} \\
&= \frac{d}{dz} \frac{1}{(z+i)^2} \Bigg\vert_{z=i} = -\frac{2}{(z+i)^3} \Bigg\vert_{z=i} = \frac{1}{4i}
\end{align}
For the horizontal line and half-circle arc, we have $z = x$ and $z=Re^{i \theta}$ respectively. Employing the residue theorem for integrals, we have
\begin{align}
\int_C f(z) \, dz = \int_{-R}^{R} \frac{1}{(x^2+1)^2} \, dx + \int_{0}^{\pi} \frac{1}{(Re^{i \theta}+1)^2} (iRe^{i \theta} \, d\theta) = 2\pi i \, \text{Res}_{z = i} f(z) = \frac{\pi}{2}
\end{align}
Taking the limit as $R \rightarrow \infty$, we get
\begin{align}
\int_{-\infty}^{\infty} \frac{1}{(x^2+1)^2} \, dx + 0 = \frac{\pi}{2}
\end{align}
Therefore,
\begin{align}
\int_{0}^{\infty} \frac{1}{(x^2+1)^2} \, dx = \frac{\pi}{4}
\end{align}
|
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|
Find the area of the indicated surface Find the surface area of the part of the sphere $x^2 + y^2 + z^2 = a^2$ inside the circular cylinder $x^2 + y^2 = ay$ ($r = a\sin(\theta)$ in polar coordinates), with $a > 0$.
First time posting on this website, sorry for the lack of details on my attempts but I am really not sure where to start on this problem.
A formula that is useful is $ \displaystyle A(G) = \iint \sqrt{f_x^2 + f_y^2 + 1} \, dA.$
$f_x$ is the partial derivative with respect to $x,$
$f_y$ is the partial derivative with respect to $y$
I know that I need to find an equation which should be $x^2 + y^2 + z^2 = a^2$, and I need to find the limits which is where I am really struggling.
Also according to my professor, I shouldn't have to use any polar coordinate conversions in order to complete this problem, which was my initial approach.
|
I'm going to assume that, since the problem asks for the surface area of the sphere found within the cylinder, the total for both hemispheres is what is asked for. So we'll work out the expression for the "upper" hemisphere and double that.
The function we will work with for that hemisphere is $ \ f(x,y) \ = \ z \ = \ \sqrt{a^2 - x^2 - y^2} \ . $ Using direct or implicit differentiation, we find that $ \ f_x \ = \ -\frac{x}{z} \ $ and $ \ f_y \ = \ -\frac{y}{z} \ , $ which gives us the "surface area projection factor",
$$ \ \sqrt{f_x^2 + f_y^2 + 1} \ = \ \sqrt{\frac{x^2}{z^2} + \frac{y^2}{z^2} + 1} \ = \ \frac{\sqrt{x^2 + y^2 + z^2}}{z} \ = \ \frac{a}{\sqrt{a^2 - x^2 - y^2}} \ \ . $$
It is this last result which perhaps led your professor to suggest that polar coordinates were not needed (opinions concerning this may differ...).
To obtain the surface area for the selected portion of the hemisphere, we will want to integrate this function over the circular cross-section of the cylinder in the $ \ xy-$ plane. In Cartesian coordinates, the circle is
$$ x^2 \ + \ y^2 \ = \ ay \ \ \Rightarrow \ \ x^2 \ + \ (y^2 \ - \ ay \ + \ \frac{a^2}{4}) \ = \ \frac{a^2}{4} \ \ \Rightarrow \ \ x^2 \ + \ (y \ - \ \frac{a}{2})^2 \ = \ \frac{a^2}{4} \ , $$
hence it is centered at $ \ ( 0 , \frac{a}{2} ) $ and has radius $ \frac{a}{2} \ . $ (You likely already knew that, since you also gave the polar function for this circle.) The limits of integration are less messy if we integrate in the $ \ x-$ direction first; the equation of the circle yields
$$ x^2 \ = \ \frac{a^2}{4} \ - \ (y^2 \ - \ ay \ + \ \frac{a^2}{4}) \ \ \Rightarrow \ \ x \ = \ \pm \sqrt{ay - y^2} \ . $$
Since the circle is symmetric about the $ \ y-$ axis, we can use just the positive square root and double the result. So the limits of integration in the $ \ x-$ direction are $ \ 0 \ \text{to} \ \sqrt{ay - y^2} \ $ ; in the $ \ y-$ direction, they are $ \ 0 \ \text{to} \ a \ . $
The surface area integral covered both halves of the cross-sectional circle and both hemispheres is then
$$ 2 \ \cdot \ 2 \ \int_0^a \ \int_0^{\sqrt{ay - y^2}} \ \frac{a}{\sqrt{a^2 - x^2 - y^2}} \ \ dx \ dy \ \ . $$
This is not a horrible integral, but also a not too pleasant one, in rectangular coordinates. In polar coordinates, we have
$$ 2 \ \cdot \ 2 \ \int_0^{\pi/2} \ \int_0^{a \sin \theta} \ \frac{a}{\sqrt{a^2 - r^2}} \ \ r \ dr \ d\theta \ \ . $$
Note that this is not a "spherical cap", since the curve of intersection on the spherical surface is not a circle; so there is no convenient shortcut for checking the result.
$$ \ \ $$
This approach does give the correct result for spherical caps. If we pick the one for a circle of radius $ \ \frac{a}{2} \ $ centered on the origin, we can use limits of integration $ \ 0 \ \le \ x \ \le \ \sqrt{\frac{a^2}{4} - y^2} \ $ and $ \ 0 \ \le \ y \ \le \ \ \frac{a}{2} \ . $ We are then integrating over a quarter of the circle, so the area of the full cap is given by
$$ 4 \ \int_0^{a/2} \ \int_0^{\sqrt{\frac{a^2}{4} - y^2}} \ \frac{a}{\sqrt{a^2 - x^2 - y^2}} \ \ dx \ dy \ \ . $$
Using polar coordinates, we have
$$ 4 \ \int_0^{\pi/2} \ \int_0^{a/2} \ \ \frac{a}{\sqrt{a^2 - r^2}} \ \ r \ dr \ d\theta \ \ = \ \ 4a \ \int_0^{\pi/2} d\theta \ \ \int_0^{a/2} \frac{r \ dr}{\sqrt{a^2 - r^2}} \ \ \ $$
$$ = \ \pi a \ \int_{\frac{3}{4} a^2}^{a^2} \ \frac{du}{\sqrt{u}} \ = \ \pi a \ \left( \ 2 \ u^{1/2} \ \vert_{\frac{3}{4} a^2}^{a^2} \ \right) \ = \ 2 \pi a \ ( \ a \ - \ \frac{\sqrt{3}}{2} a ) $$
$$ = \ ( \ 2 \ - \ \sqrt{3} ) \ \pi a^2 \ \ , $$
which agrees with the result from the standard formula for the surface area of a spherical cap.
|
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|
Elementary Symmetric Polynomials, Roots of cubic polynomials I'm given $a_1, a_2, a_3$ as roots of the equation $x^3 + 7x^2 - 8x + 3$ and need to find the cubic polynomials with roots $a_1^2, a_2^2, a_3^3$ and $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}$. It's in a chapter about Galois Theory and is preceded by a problem about elementary symmetric polynomials. I'm guessing I need to use elementary symmetric polynomials to solve for the two desired functions, but I really don't know where to start. Hints would be much appreciated.
|
$a$, $b$ and $c$ are the roots of $x^3 + px^2 + qx + r$ if and only if $abc = -r$, $ab+bc+ca=q$ and $-(a+b+c)=p$.
We have $\frac{1}{a_1} \cdot \frac{1}{a_2} \cdot\frac{1}{a_3} = \frac{1}{a_1a_2a_3} = -1/3$, $\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} = \frac{a_1a_2 + a_2a_3 + a_3a_1}{a_1a_2a_3} = \frac{-8}{-3}$ and $\frac{1}{a_1a_2} + \frac{1}{a_2a_3} + \frac{1}{a_3a_1} = \frac{a_1+a_2+a_3}{a_1a_2a_3} = \frac{-7}{-3}$. It follows that $\frac{1}{a_1} $, $\frac{1}{a_2} $ and $\frac{1}{a_3} $ are the roots of $X^3 - \frac{8X^2}{3} + \frac{7X}{3} + \frac{1}{3}$. The problem with $a_1^2$, $a_2^2$, $a_3^2$ can be solved similarly.
|
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|
Inequality problem $(a+b+c)^5\geq27(ab+bc+ca)(ab^2+bc^2+ca^2)$ While solving one inequality, I arrived at a much simpler, but still nontrivial inequality
$$(a+b+c)^5\geq27(ab+bc+ca)(ab^2+bc^2+ca^2)$$
where $a,b,c$ are positive real numbers.
It apparently holds, but I can't seem to find a proof.
The problem is it's not symmetric and applying inequalities $(a+b+c)^2\geq3(ab+bc+ca)$ or $a^3+b^3+c^3\geq ab^2+bc^2+ca^2$ won't work.
Any ideas?
|
Without loss of generality, let $a+b+c=3$. Then we only have to prove
$$(a^2b+b^2c+c^2a)(ab+bc+ac)\le 9$$
Assume $b=\text{mid}(a,b,c)$, then we have
$$c(b-a)(b-c)\le 0\Longleftrightarrow a^2b+b^2c+c^2a\le b(a^2+ac+c^2)$$
So
$$(a^2b+b^2c+c^2a)(ab+bc+ac)\le b(a^2+ac+c^2)(ab+bc+ac)$$
Using AM-GM inequality, we get
$\begin{align*}
b(a^2+ac+b^2)(ab+bc+ca)&\le\,b\cdot\dfrac{(a^2+ac+c^2+ab+bc+ac)^2}{4}=
b\cdot\dfrac{(a+c)^2(a+b+c)^2}{4}\\
&=\dfrac{9}{8}\cdot 2b\cdot(a+c)\cdot(a+c)\\
&\le\dfrac{9}{8}\cdot\dfrac{(2b+a+c+a+c)^3}{27}\\
&=\,9
\end{align*}$
Done!
|
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|
Can $\frac {100-100}{100-100}=2$? \begin{align*}
\frac{0}{0} &= \frac{100-100}{100-100} \\
&= \frac{10^2-10^2}{10(10-10)} \\
&= \frac{(10+10)(10-10)}{10(10-10)} \\
&= \frac{10+10}{10} \\
&= \frac{20}{10} \\
&= 2
\end{align*}
I know that $0/0$ isn't equal to $2$, the what is wrong in this proof? I wasn't satisfied after searching it on Google, any help would be appreciated.
|
$1.$ You cannot start with $\frac00$. This is not a number.
$2.$ You are trying to simplify the fraction by dividing numerator and denominator with $(10-10)$. This cannot be done.
Assume beginning from the end going towards the start. The wrong part is when you multiply with $(10-10)$. Because then denominator would be $0.$
With your logic $\frac00=\frac{0*a}{0*b}=\frac{a}b$.
In general: In almost all these kind of problems where someone proves that $1=2$ (or something similar), most of the times, somewhere the denominator will be $0.$
|
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|
Induction proof of $n^{(n+1) }> n(n+1)^{(n-1)}$ The question statement from my homework booklet goes:
Prove by mathematical induction that $n^{n+1} > n(n+1)^{n-1}$
is true for all integers $n \geq 2$.
I've managed to come up with this for the induction step (the base case is trivial), but I am not sure what to do from here:
Assume true for n=k. For n=k+1,
\begin{align*}
&k^{k+1} > k(k+1)^{k+1} \\
&(k+1)k^{k+1} > k(k+1)^k \\
&(k+1)^{k+2} < (k+1)^3 k^k < (k+1)^3 (k+2)^k \\
&(k+1)^{k-1} < (k+2)^k
\end{align*}
I would greatly appreciate any help with how to solve this. Thanks in advance.
|
We are required to prove that: $(k+1)^{k+2}>(k+1)(k+2)^k$.
The assumption is that: $k^{k+1} > k(k+1)^{k-1}$. Dividng both sides by $k^k$ gives: $k> \left(\frac{k+1}{k}\right)^{k-1}$, thus $k>\left(1+\frac{1}{k}\right)^{k-1}$
Starting from this,
\begin{eqnarray}
k&>&\left(1+\frac{1}{k}\right)^{k-1}\\
\frac{1}{\left(1+\frac{1}{k}\right)^{k-1}}&>&\frac{1}{k}\\
1+\frac{1}{\left(1+\frac{1}{k}\right)^{k-1}}&>&1+\frac{1}{k}\\
\left(1+\frac{1}{k}\right)^{k-1}+1&>&\left(1+\frac{1}{k}\right)^k\\
\end{eqnarray}
Now, since $k>\left(1+\frac{1}{k}\right)^{k-1}$, then:
\begin{eqnarray}
k+1&>&\left(1+\frac{1}{k}\right)^k\\
(k+1)^{k+2}&>&(k+1)^{k+1}\left(1+\frac{1}{k}\right)^k
\end{eqnarray}
Also, $\frac{1}{k}>\frac{1}{k+1}$, therefore:
\begin{eqnarray}
(k+1)^{k+2}&>&(k+1)^{k+1}\left(1+\frac{1}{k+1}\right)^k\\
&>&(k+1)(k+2)^k
\end{eqnarray}
Therefore, $(k+1)^{k+2}>(k+1)(k+2)^k$. $\Box$
|
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|
Probability of a certain ball drawn from one box given that other balls were drawn Box 1 contains 2 green and 3 red balls, 2 has 4 green and 2 red, and 3 has 3 green and 3 red. Only one ball is drawn from each of the 3 boxes. What is the probability that a green ball was drawn from box 1 given that two green balls were drawn?
So in total there were exactly 2 green balls and 1 red ball drawn, from a different combinations of the 3 boxes. We could have selected the 2 greens from the first 2 boxes and a red from the last box, 2 greens from the last 2 boxes, or 2 greens from box 1 and 3. I get $\frac{2}{5} \frac{4}{6} \frac{3}{6} + \frac{2}{5} \frac{2}{6} \frac{3}{6} + \frac{3}{5} \frac{4}{6} \frac{3}{6} = \frac{2}{5}$. Now this is the probability of drawing 2 green balls. What do I do from here?
|
E1 = Green from Box 1
E2 = 2 Greens Drawn
E1 and E2 = Green from Box 1 and 2 Greens Drawn
E1:
2 Green in Box 1
5 Total in Box 1
P(E1) = 2/5
E2:
Possible combinations are GGR and GRG. Add these two combinations together.
P(GGR) = (2/5) * (4/6) * (3/6) = 2/15
P(GRG) = (2/5) * (2/6) * (3/6) = 1/15
P(E1 and E2) = P(GGR) + P(GRG) = 3/15 = 1/5
P(E1 | E2)
P(E1 | E2) = P(E1 and E2) / P(E2)
P(E1 | E2) = (1/5) / (2/5) = 1/2
|
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|
Proving divisibility by using induction: $133 \mid (11^{n+2} + 12^{2n+1})$ If $n > 0$, then prove the following by using induction: $$133|(11^{n+2} + 12^{2n+1}).$$
|
Inductive step:
$$\begin{align}11^{(n+1) + 2} + 12^{2(n+1)+1} &= 11^{n+3} + 12^{2n+3}\\
&=11\cdot11^{n+2} + 144\cdot12^{2n+1}\\
&= 11\cdot11^{n+2} + 11\cdot12^{2n+1} + 133\cdot12^{2n+1}\\
&=11\cdot(11^{n+2} + 12^{2n+1}) + 133\cdot12^{2n+1}\end{align}$$
But $133 | (11^{n+2} + 12^{2n+1})$. This is the inductive hypothesis. Hence $133$ must divide the above.
|
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|
Solving a combintorical problem using generating function
In how many ways $n$ balls can be divided into groups where each group may contain only one ball or two.
Each ball can be a singleton, or coupled with one of the $n-1$ options. Hence,
$a_0 = 1$
$a_1 = 1$
$a_n = a_{n-1} + (n-1)a_{n-2}$
The correspond generating function:
$$F(x) = 1 + x + \sum\nolimits_{n \ge 2} {\left( {{a_{n - 1}} + (n - 1) \cdot {a_{n - 2}}} \right)} \cdot {x^n}$$
$$F(x) = 1 + x + \sum\nolimits_{n \ge 2} {{a_{n - 1}}} {x^n} + \sum\nolimits_{n \ge 2} {(n - 1)} \cdot {a_{n - 2}} \cdot {x^n}$$
$$F(x) = 1 + x + x \cdot \sum\nolimits_{n \ge 1} {{a_n}} {x^n} + {x^2} \cdot \sum\nolimits_{n \ge 0} {(n - 1)} \cdot {a_n} \cdot {x^n}$$
$$F(x) = 1 + x + x\left( {F(x) - 1} \right) + {x^2}\left( ? \right)$$
How to treat the last expression? $$\sum\nolimits_{n \ge 0} {(n - 1)} \cdot {a_n} \cdot {x^n}$$
|
To solve the recurrence that you have given,
$$
\begin{align}
f(x)
&=\sum_{k=0}^\infty a_kx^k\\
&=1+x+\sum_{k=2}^\infty (\color{#C00000}{a_{k-1}}+\color{#00A000}{(k-1)a_{k-2}})x^k\\
&=1+x+\color{#C00000}{x(f(x)-1)}+\color{#00A000}{x^2(xf'(x)+f(x))}\\[10pt]
-1
&=x^3f'(x)+f(x)(x^2+x-1)
\end{align}
$$
This equation can be solved using the integrating factor
$$
\begin{align}
g(x)&=xe^{-\frac1{x^{\vphantom{2}}}+\frac1{2x^2}}\\
g'(x)&=\left(1+\frac1x-\frac1{x^2}\right)e^{-\frac1{x^{\vphantom{2}}}+\frac1{2x^2}}
\end{align}
$$
which yields
$$
-\frac1{x^2}e^{-\frac1{x^{\vphantom{2}}}+\frac1{2x^2}}
=(gf)'(x)
$$
giving
$$
\begin{align}
f(x)
&=\left(\mathrm{erfi}\left(\frac{1-x}{x\sqrt2}\right)+i\right)\frac{\sqrt{2\pi}}{2x}e^{-\frac{(1-x)^2}{2x^2}}\\[6pt]
&=1+x+2x^2+4x^3+10x^4+26x^5+76x^6+232x^7+764x^8+\dots
\end{align}
$$
The series above was verified in Mathematica 8 using
Series[(Erfi[(1-x)/x/Sqrt[2]]+I)Sqrt[Pi/2]/x Exp[-(1-x)^2/2/x^2],{x,0,8},Assumptions->Element[x,Reals]&&x>=0]
To get the exponential generating function,
$$
\begin{align}
g(x)
&=\sum_{k=0}^\infty a_k\frac{x^k}{k!}\\
g'(x)
&=1+\sum_{k=2}^\infty(\color{#C00000}{a_{k-1}}+\color{#00A000}{(k-1)a_{k-2}})\frac{x^{k-1}}{(k-1)!}\\
&=1+\color{#C00000}{g(x)-1}+\color{#00A000}{xg(x)}\\[9pt]
&=(1+x)g(x)
\end{align}
$$
which is easily solved by
$$
\begin{align}
g(x)
&=e^{x+x^2/2}\\
&=1+x+2\frac{x^2}{2!}+4\frac{x^3}{3!}+10\frac{x^4}{4!}+26\frac{x^5}{5!}+76\frac{x^6}{6!}+232\frac{x^7}{7!}+764\frac{x^8}{8!}+\dots
\end{align}
$$
The series above was verified with Mathematica 8 using
CoefficientList[Series[Exp[x+x^2/2],{x, 0, 8}],x]Table[n!,{n, 0, 8}]
|
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|
Matrix linear transformation from $\mathbb{R}^3$ to $\mathbb{R}^4$. Find the matrix representation Say that $T : \mathbb{R}^3 \to \mathbb{R}^4$ is linear and satisfies $T(1,1,1) = (3,2,0,1)$, $T(1,1,0) = (2,1,3, −1)$ and $T(1,0,0) =(5, −2,1,0)$. Find $T(x, y, z)$ and then find the matrix representation for $T$.
I am not sure how to proceed. I thought of two approaches, one the matrix $A$ from $Ax=b$ could be expressed as
$$A=\begin{pmatrix}
3 & 2 & 5\\
2 & 1 & -2\\
0 & 3 & 1\\
1 & -1 & 0
\end{pmatrix}$$
Then possibly solve for $Ax=0$?
Or since $T(1,0,0)=v_3$ etc, we can also write $T(x,y,z)=x\cdot v_1+y \cdot v_2+z \cdot v_3$, but I am not sure how to find $v_1$, $v_2$ or $v_3$.
Appreciate your help.
|
$T(0,0,1) = T(1,1,1) - T(1,1,0) = (1, 1, -3, 2)$, and $T(0,1,0) = T(1,1,0) - T(1,0,0) = (-3, 3, 2, -1)$, and $T(1,0,0) = (5, -2, 1, 0)$. So
\begin{align*}T(x,y,z) &= xT(1,0,0) + yT(0,1,0) + zT(0,0,1) \\
&= x(5,-2,1,0) + y(-3,3,2,-1) + z(1,1,-3,2) \\
&= (5x-3y+z, -2x+3y+z, x+2y-3z, -y+2z).\end{align*} And from this the matrix $A$ is:
\begin{pmatrix}
5 & -3 & 1 \\
-2 & 3 & 1 \\
1 & 2 & -3 \\
0 & -1 & 2 \\
\end{pmatrix}
You can continue.
|
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|
Using the polar form of $1 + i$ and $\sqrt3 + i$ to deduce $\cos (\frac{\pi}{12}), \sin(\frac{\pi}{12})$ I have been beating my head against the following problem and would like a gentle nudge in the right direction.
The question states, by writing $1 + i$ and $\sqrt3 + i$ in polar form, deduce that
$$\cos (\frac{\pi}{12}) = \frac{\sqrt3 + 1}{2\sqrt2}, \sin(\frac{\pi}{12}) = \frac{\sqrt3 - 1}{2\sqrt2}$$
so I have written them in polar form
EDIT (as polar forms were incorrect):
$$1 + i = \sqrt2e^{i\pi/4}, \sqrt3 + i = 2e^{i\pi/6}$$
another part of the question also asks you to put $\frac{1 + i}{\sqrt3 + i}$ into form $x + yi$ which I figured is
$$\frac{\sqrt3 + 1}{4} + \frac{1 - \sqrt3}{4}i$$
I just can't seem to connect it all together unfortunately so any help would be greatfully received.
|
Notice
$$1 + i = \sqrt2e^{i\frac{\pi}{4}}$$
$$\sqrt{3} - i = 2e^{-i\frac{\pi}{6}}$$
Multiply both and we get
$$(1 + i)(\sqrt{3} - i) = \sqrt2e^{i(\frac{\pi}{4} - \frac{\pi}{6})}$$
$$(\sqrt3 + 1) + (\sqrt3-1)i= 2\sqrt{2}e^{i\frac{\pi}{12}}$$
$$(\sqrt3 + 1) + (\sqrt3-1)i= 2\sqrt2\left(\sin{\frac{\pi}{12}} + i\cos\frac{\pi}{12}\right)$$
$$\frac{\sqrt3 + 1}{2\sqrt2} + \frac{\sqrt3 - 1}{2\sqrt2}i = \cos{\frac{\pi}{12}} + i\sin\frac{\pi}{12}$$
Now, compare both real and imaginary parts to deduce:
$$\sin{\frac{\pi}{12}} = \frac{\sqrt3 + 1}{2\sqrt2}$$
$$\cos{\frac{\pi}{12}} = \frac{\sqrt3 - 1}{2\sqrt2}$$
|
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|
Getting incorrect result on limit of square root I have a limit problem:
$\lim_{x \to \infty} \sqrt{x^2 + x} - x$
According to Wolfram|Alpha the answer is $\frac{1}{2}$
However, my calculation gives $1$. Please help me understand what I'm doing wrong. The process is:
$\lim_{x \to \infty} \sqrt{x^2 + x} - x = \lim_{x \to \infty} \sqrt{x^2(1 + \frac{1}{x})} - x = \lim_{x \to \infty} x\sqrt{1 + \frac{1}{x}} - x = \lim_{x \to \infty} x(\sqrt{1 + \frac{1}{x}} - 1)$
Let $t = \frac{1}{x}$
$\lim_{x \to 0} \frac{\sqrt{1 + t} - 1}{t} \Rightarrow (l'Hopital) \lim_{x \to 0} \frac{\frac{1}{\sqrt{1 + t}}}{1} = \lim_{x \to 0} \frac{1}{\sqrt{1+t}} = 1$
|
$\lim_{x \to \infty} \sqrt{x^2 + x} - x$
approach:
taking $t = \frac{1}{x}$ hence t tends to 0
So,
$\lim_{t \to 0} \sqrt{\frac{1}{t^2} + \frac{1}{t}} - \frac{1}{t}$
$= \lim_{t \to 0} \frac{\sqrt{1+t}}{t} - \frac{1}{t}$
$= \lim_{t \to 0} \frac{1}{t} (\sqrt{1+t} - 1)$
multiplying by $\frac{\sqrt{1+t} + 1}{\sqrt{1+t} + 1}$
$= \lim_{t \to 0} \frac{1}{\sqrt{1+t} + 1}$
Now apply limits
|
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|
How to solve system of two equation with two unknown and using substitution? $$\begin{align}
a^2 - b^2 = 3\\
a \cdot b = 2
\end{align}$$
In aforementioned equations, we can mentally find out the value of $a = 2, b = 1$. But what is the general way to solve this system algebraically?
I tried to use substitution but I got stuck. Rearranging two equations to $a^2=b^2+3$ and $a^2 \cdot b^2 = 2^2$ we will get:
$$ (b^2 + 3) \cdot b^2 = 4\\
\rightarrow b^4 + 3b^2 = 4\\
\rightarrow b^2(b^2 + 3) = 4\\
$$
Then what?
Alternatively can we solve by elimination?
|
HINT:
$$\implies b^4+3b^2-4=0\iff (b^2+4)(b^2-1)=0$$
Then from the second equation, find corresponding value of $a$ for each value of $b$
Then, check whether each pair of $(a,b)$ satisfy the first
Alternatively, $$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2$$
|
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|
Prove by induction on $n\geq k$ that $n^3 \lt 3^n$. What is the value $k$? Prove by induction on $n\geq k$ that $n^3 \lt 3^n$. What is the value $k$?
It looks like $k$ should be $0$ after trying random values but obviously that is a terrible way of doing things.
Proof:
P(n): n^3<3^n
Assume k holds
Prove k+1
P(k+1): (k+1)^3<3^(k+1)
At this point, I am not entirely sure where to go. Any advice would be great.
Thanks,
|
First we compute a little, systematically. We have
$0^3\lt 3^0$;
$1^3 \lt 3^1$;
$2^3\lt 3^2$;
$3^3=3^3$, so there inequality $n^3\lt 3^n$ is not true at $n=3$;
$4^3=64$, $3^4=81$, so $4^3\lt 3^4$;
$5^3=125$, and $3^5=243$;
$6^3=216$, and $3^6=729$;
$7^3-343$, and $3^7=2187$.
If this were a horse race, it looks as if the horse $3^n$ is now well ahead of the horse $n^3$, and is getting further and further ahead as time goes on.
We will prove that $n^3\lt 3^n$ if $n\ge 4$.
The result is certainly true at $n=4$. We show that for any $k\ge 4$, if $k^3\lt 3^k$, then $(k+1)^3\lt 3^{k+1}$. There are several ways to do this. We mention a couple.
(i) We have $3^{k+1}=3\cdot 3^k$. By the induction assumption, $3^k\gt k^3$. So it is enough to show that $3k^3\ge (k+1)^3$, or equivalently that $3\ge \left(1+\frac{1}{k}\right)^3$. Note that $\left(1+\frac{1}{4}\right)^3\lt 3$, and if $k\gt 4$, then $\left(1+\frac{1}{k}\right)^3\lt \left(1+\frac{1}{4}\right)^3$. This completes the proof. The idea behind this proof is that if we increment $n$ by $1$, then $3^n$ jumps by a factor of $3$, but $n^3$ only jumps by a factor of $\left(\frac{n+1}{n}\right)^3$, which very soon is $\lt 3$.
(ii) We want to show that if $k\ge 4$, then $3k^3\ge (k+1)^3$, or, expanding, that $2k^3\gt 3k^2+3k+1$. Note that if $k\ge 4$, then $2k^3\ge 8k^2$. But $3k^2+3k+1\lt 3k^2+3k^2+k^2=7k^2$.
Alternately, we could look at the function $2x^3-3x^2-3x-1$, and use techniques from the calculus or elsewhere to find out when it is positive.
|
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|
Prove variance in Uniform distribution (continuous) I read in wikipedia article, variance is $\frac{1}{12}(b-a)^2$ , can anyone prove or show how can I derive this?
|
By the definition of the variance, $\operatorname{Var} X = \mathbb{E}[X^2] - (\mathbb{E} X)^2$.
Since here $\mathbb{E} X = \frac{1}{b-a}\int_{[a,b]}x dx = \frac{a+b}{2}$, and $\mathbb{E} X^2 = \frac{1}{b-a}\int_{[a,b]}x^2 dx = \frac{b^3-a^3}{3(b-a)}=\frac{a^2+ab+b^2}{3}$, it follows that
$$
\operatorname{Var} X = \frac{a^2+ab+b^2}{3} - \frac{a^2+2ab+b^2}{4} = \frac{a^2-2ab+b^2}{12} =\frac{(b-a)^2}{12}
$$
|
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|
$ \frac{x}{4 \ \sqrt{x^2+1}} \ = \ \frac{y}{5 \ \sqrt{y^2+1}} \ = \ \frac{z}{6 \ \sqrt{z^2+1}} $ and $ \ x+y+z \ = \ xyz \ $ Consider the system of equations in real numbers $ \ x,y,z \ $ satisfying
$$ \frac{x}{4 \ \sqrt{x^2+1}} \ = \ \frac{y}{5 \ \sqrt{y^2+1}} \ = \ \frac{z}{6 \ \sqrt{z^2+1}} $$
and $ \ x+y+z \ = \ xyz \ $. Find $ \ x,y,z \ . $
I substituted $ \ x,y,z \ $ as $ \ \tan(\theta_1) \ , \ \tan(\theta_2) \ , \ \tan(\theta_3) \ $ , where
$$ \theta_1 \ + \ \theta_2 \ + \ \theta_3 \ = \ 180º \ \ . $$
|
This is a comment that’s too long for the usual format.
LAcarguy’s method generalizes to any values $a,b,c$ instead
of $\frac14,\frac15,\frac16$.
If we put
$$
\phi(a,b,c)=ab+ac-bc, \ \psi(a,b,c)=ab+ac+bc \tag{1}
$$
and
$$
G(a,b,c)=\phi(a,b,c)\phi(b,c,a)\phi(c,a,b)\psi(a,b,c) \tag{2}
$$
then LAcarguy’s method leads to the following explicit formulas for the
solutions :
$$
x=\frac{\varepsilon}{\phi(a^2,b^2,c^2)}\sqrt{G(a,b,c)},\
y=\frac{\varepsilon}{\phi(b^2,c^2,a^2)}\sqrt{G(a,b,c)},\
z=\frac{\varepsilon}{\phi(c^2,a^2,b^2)}\sqrt{G(a,b,c)} \tag{3}
$$
where $\varepsilon$ is $+1$ or $-1$.
In your initial question where $a=\frac14,b=\frac15,c=\frac16$, one finds
$G=\frac{7}{921600}$,
$$
x=\varepsilon\frac{\sqrt{7}}{3}, \
y=\varepsilon\frac{5\sqrt{7}}{9}, \
z=3\varepsilon\sqrt{7} \tag{4}
$$
|
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|
Find the angle of intersection of circles $x^2+y^2-6x+4=0 \ \&\ x^2+y^2-2x-2y-8=0$ Find the angle of intersection of circles $$x^2+y^2-6x+4=0 \\ x^2+y^2-2x-2y-8=0$$
my answer is : 41.14 degrees. but i'm not sure if it's right. please help me.
|
Write the equations as
$$
(x-3)^2+y^2=5
$$
and
$$
(x-1)^2+(y-1)^2=10
$$
Thus, we have a triangle with sides $\sqrt{5}$, $\sqrt{10}$, $\sqrt{5}$:
$\hspace{3cm}$
The Law of Cosines says
$$
5=5+10-2\sqrt{5}\sqrt{10}\cos(\alpha)
$$
which implies
$$
\cos(\alpha)=\frac1{\sqrt2}
$$
We could also recognize the $45{-}45{-}90$ triangle.
|
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|
Complex Modulus Is Less Than One Show that $|\frac{-a}{b}+\frac{\sqrt{a^2-b^2}}{b}|<1$ where $a>|b|>0$
This is just a minor part of a more full complex trigonometric integral proof that I'm working on. I'm pretty much set once I can show this inequality is true. Please Help!
What I was thinking was:
$|\frac{-a}{b}+\frac{\sqrt{a^2-b^2}}{b}|=|\frac{-a}{b}+\frac{a}{b}\sqrt{1-\frac{b^2}{a^2}}|$
Then $\sqrt{1-\frac{b^2}{a^2}}<1$, Which implies $\frac{a}{b}\sqrt{1-\frac{b^2}{a^2}}<\frac{a}{b}$
The problem is, this doesn't guarantee that first inequality is true, in less I'm missing something.
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Hint:
$$
\sqrt{a^2-b^2}-a = \frac{a^2-b^2-a^2}{\sqrt{a^2-b^2}+a}$$
details:
$$
0\le a-\sqrt{a^2-b^2}
= \frac{b^2}{\sqrt{a^2-b^2}+a}\le |b| \frac {|b|}a\le |b|$$
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Integral over a sphere How to calculate $$I=\int\int_S\frac{d^2S}{r^n}$$ where S is a surface of the sphere $S=\{x^2+y^2+z^2=R^2\}$ and $r=dist((x,y,z),(0,0,c))$, $c>R,c\in\mathbb{R}$ ?
I believe $r=\sqrt{x^2+y^2+(\sqrt{R^2-x^2-y^2}-c)^2}$ so $$I=\int\int_{x^2+y^2<R^2}\frac{1}{\Big(x^2+y^2+(\sqrt{R^2-x^2-y^2}-c)^2\Big)^\frac{n}{2}}*\sqrt{G}$$ where $G = \left| \begin{array}{cc}
1 & 0 \\
0 & 1 \\
\frac{-x}{\sqrt{R^2-x^2-y^2}} & \frac{-y}{\sqrt{R^2-x^2-y^2}} \end{array} \right|$
I made few calculations using $x=tcos\alpha,y=tsin\alpha$ and then $t=Rsin\beta$ substitutions but I'm stuck with some ugly looking integral, so I'm not sure if I got it all right.
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Work in spherical coordinates $(R, \theta, \phi)$. Note that
$$r = \sqrt{x^2 + y^2 + (z-c)^2} = \sqrt{x^2 + y^2 + z^2 - 2cz + c^2} = \sqrt{R^2 - 2Rc \cos \phi + c^2}$$ and $$d^2 S = R^2 \sin \phi\, d\theta\, d\phi$$
so
$$\iint_S \frac{d^2S}{r^n} = \int_0^\pi \int_0^{2\pi} \frac{R^2 \sin \phi}{(R^2 - 2Rc\cos \phi + c^2)^{n/2}}\, d\theta\, d\phi$$
You should be able to calculate this.
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Need help to solve taylor series of $e^{\sin x}$ How to derive the taylor series of $e^{\sin x}$, up to $x^5$?
i just don't know how to get the answer
$$f(x) = 1 + x + \frac{x^2}{2} - \frac{x^4}{8} -\frac{x^5}{15}$$
really need some help. Thanks
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Let $f(x) = e^{\sin x}$.
\begin{equation*}
\begin{split}
f'(x) &= e^{\sin x} \cos x = f(x) \cos x\\
f''(x) &= f'(x) \cos x - f(x) \sin x \\
f'''(x) &= f''(x) \cos x - f'(x)\sin x - (f(x)\cos x + f'(x) \sin x)\\
&= f''(x) \cos x - 2f'(x) \sin x - f'(x)\\
f^4 (x) &= f'''(x) \cos x - f''(x) \sin x - 2(f'(x)\cos x + f''(x)\sin x) - f''(x)\\
&= f'''(x) \cos x - 3f''(x) \sin x - 2f'(x) \cos x - f''(x)
\end{split}
\end{equation*}
Calculating higher order derivatives at $x=0$,
\begin{equation*}
\begin{split}
f(0) &= 1\\
f'(0) &= 1 \times 1 = 1\\
f''(0) &= 1 - 0 = 1\\
f'''(0) &= 1-0-1 = 0 \\
f^4 (0) &= 0 - 0 -2 -1 = -3\\
f^5 (0) &= f^4 (0) \cos 0 - 3 f''(0) \cos 0 - 2f''(0) \cos 0- f'''(0)\\
&= - 3 -3-2-0 =-8
\end{split}
\end{equation*}
Using Maclaurin's expansion for infinite series,
\begin{equation*}
\begin{split}
f(x) &= f(0) + xf'(0) + \frac{x^2}{2!}f''(0) + \frac{x^3}{3!}f'''(0) + \\
& \frac{x^4}{4!}f^4 (0) + \frac{x^5}{5!}f^5 (0) + \frac{x^6}{6!}f^6 (0) + \ldots\\
e^{\sin x} &= 1 + x + \frac{x^2}{2!}\times 1 + 0 + \frac{x^4}{4!}\times (-3) + \frac{x^5}{5!} \times (-8) \ldots\\
&= 1 + x + \frac{x^2}{2} - \frac{x^4}{8} - \frac{x^5}{15} \ldots\\
\end{split}
\end{equation*}
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Taylor expansion of $\frac{1}{2-z-z^2}$ The problem is:
Find the Taylor expansion of $f(z):= \dfrac{1}{2-z-z^2}$ on the disc $|z| < 1$
So far I have used partial fractions to obtain $f(z) = \dfrac{1}{3}\left(\dfrac{1}{1-z} + \dfrac{1}{2+z}\right)$ which I then rewrite as $f(z) = \dfrac{1}{3}\left(\dfrac{1}{1-z} + \dfrac{1}{1-((-1)(1+z))}\right)$.
My problem is that we are told specifically to find the expansion on the disc $|z| < 1$, and whilst $\dfrac{1}{1-z}$ is valid for $|z| < 1$, $\dfrac{1}{1-((-1)(1+z))}$ is valid for $|1+z| < 1$. I am a guessing that we cannot just say $f(z) = \dfrac{1}{3}\left(\displaystyle\sum_{n=0}^\infty z^n + (-1)^n(1+z)^n\right)$ but am not sure as my notes don't have any examples of this sort of difficulty.
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Hint: $\frac{1}{1-z}=1+z+z^2+\ldots$ as infinitely descending geometric progression. And $\frac{1}{2+z}=\frac{1}{2}\cdot\frac{1}{1-(-z/2)}=\ldots$.
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consecutive prime power I'm interesting on consecutive prime power numbers. I see that there is the Mersenne primes and the Fermat Primes that give solutions and $(8,9)$. In Sloane collection it is referred on A006549 and it is written:
David W. Wilson and Eric Rains found a simple proof that in this case of Catalan's conjecture either n or n+1 must be a power of 2 and the other number must be a prime, except for n=8.
But I can't find this proof on the web (and I don't find it myself...).
I tried the first case. I get a proof for $2^k+1=3^q$, but found it complicated. If someone has something simple?
Simple: I see that it's enough to consider the case $(2^k, q^n)$ because... ok it's clear
I see that it's enough to show that $2^k=-1$ mod $q$ or $q^2$ is impossible. But for exemple that give nothing for $2^k+1=5^q$ because $2$ is primite modulo $5$ and $25$ and so modulo all the $5^q$.
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There is indeed a relatively easy proof of the fact that if $n$ and $n+1$ are consecutive prime powers, then
1) Either $n$ or $n+1$ must be a power of 2.
2) Apart from $n=8$, the other number must be a prime.
Note: Here, the proof is relatively easy because we exploit the fact that the numbers are prime powers, and not just perfect powers. The proof of the full statement of Catalan's conjecture/Mihailescu's theorem is much harder.
Proof:
1) is trivial: One of $n, n+1$ must be even, hence must a power of $2$ since it is a prime power.
2) We have two cases ($n$ even or $n+1$ even)
Case 1: $n$ even. We have $n=2^a, n+1=p^b$, $p$ an odd prime, $a, b \in \mathbb{Z}^+$. Thus $2^a=p^b-1$.
If $b$ is even, then write $b=2c$.
$$2^a=p^b-1=(p^c-1)(p^c+1)$$
$p^c \pm 1$ are two powers of $2$ which differ by $2$, so must be $2$ and $4$. Thus $n=2^a=(2)(4)=8$. (the only exception where the other number is not prime)
If $b$ is odd, then
$$2^a=p^b-1=(p-1)(p^{b-1}+ \ldots +p+1)$$
$p^{b-1}+ \ldots +p+1 \equiv b \equiv 1 \pmod{2}$, so $p^{b-1}+ \ldots +p+1 \mid 2^a$ implies $p^{b-1}+ \ldots +p+1=1$. Thus $b=1$, so $n+1$ is a prime and we are done.
Case 2: $n$ odd. Then $n+1=2^a, n=p^b$, $p$ an odd prime, $a, b \in \mathbb{Z}^+$. Thus $2^a=p^b+1$.
If $b$ is even, then $p^b \equiv 1 \pmod{4}$. Thus $2^a=p^b+1 \equiv 2 \pmod{4}$. Therefore $a=1$, so $n+1=2^a=2$, so $n=1$, a contradiction.
If $b$ is odd, then
$$2^a=p^b+1=(1+p)(p^{b-1}-p^{b-2}+ \ldots +p^2-p+1)$$
We have $(p^{b-1}-p^{b-2}+ \ldots +p^2-p+1) \equiv b \equiv 1 \pmod{2}$, and $(p^{b-1}-p^{b-2}+ \ldots +p^2-p+1)$ is a power of $2$, so $(p^{b-1}-p^{b-2}+ \ldots +p^2-p+1)=1$. Thus $b=1$, so $n$ is a prime and we are done.
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how to factor $x^4+2x^3+4x^2+3x+2$ I'm trying my hand on these types of expressions.
How to factorize $x^4+2x^3+4x^2+3x+2$ into two (or more) polynomials with rational coefficients. please write step by step solution.
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Let $P(x)=x^4+2x^3+4x^2+3x+2$. If $P$ factors over the rationals, it must factor over the integers. Since $P$ is monic (i.e., its lead coefficient is $1$), the only possible roots (which would correspond to linear factors, $x-r$) are $\pm1$ and $\pm2$. We can dismiss $1$ and $2$ out of hand, since $P(x)$ is clearly positive for $x\ge0$. But $P(-1)=1-2+4-3+2=2$ and $P(-2)=16-16+16-6+2=12$ rules those out as well. So it remains to see if $P$ can be factored into a pair of quadratics, $P(x)=(x^2+ax+b)(x^2+cx+d)$.
If it can, then $b$ and $d$ are necessarily positive, since otherwise $P$ would have a positive real root, which we know is not the case since, as we already noted, $P(x)$ is positive for $x\ge0$. (E.g., if $b$ were negative, then the factor $x^2+ax+b$ would be negative at $x=0$ but eventually positive, once $x$ is sufficiently large.) Thus, from $bd=P(0)=2$, we can restrict ourselves to looking at factorizations of the form
$$P(x)=(x^2+ax+2)(x^2+cx+1)$$
Expanding this out as
$$P(x)=x^4+(a+c)x^3+(ac+3)x^2+(a+2c)x+2$$
we see that we need to have $a+c=2$, $ac+3=4$, and $a+2c=3$. In general, three equations in two unknowns can not be solved (which is to say, most quartics do not factor into two quadratics), but in this case the solution is pretty clearly $a=c=1$, and that gives the factorization
$$x^4+2x^3+4x^2+3x+2=(x^2+x+2)(x^2+x+1)$$
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How to evaluate $\int\frac{x}{\sqrt{x^2+x+1}} \, dx$ using trigonometric substitution? I am pretty sure that my answer is correct but given answer for the exercise from textbook Calculus James Steward was slightly different. Any idea to solve this:
$$\int\frac{x}{\sqrt{x^2+x+1}} \, dx$$
The given answers: $\sqrt{x^2+x+1}-\frac{1}{2}\ln(\sqrt{x^2+x+1}+x+\frac{1}{2})+c$
Thanks in advance.
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$$x^2+x+1 = \left ( x+\frac12 \right )^2+\frac{3}{4} $$
$$\begin{align} \int dx \frac{x}{\sqrt{x^2+x+1}} &= \int dx \frac{x+\frac12}{\sqrt{\left ( x+\frac12 \right )^2+\frac{3}{4}}}-\frac12 \int dx \frac{1}{\sqrt{\left ( x+\frac12 \right )^2+\frac{3}{4}}}\end{align}$$
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How do I find the inverse of this exponential function? $x=-3(3^{-x})+9$
I know the steps up until a certain point.
$x=-3(3^{-y})+9$
$x-9=-3(3^{-y})$
$\frac{(x-9)}{-3} = 3^y$
$ln (\frac{x-9}{-3}) = -y * ln 3$
Not sure what to do from here. I know I have to get y by itself but thats it. Can anyone help please?
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We want to solve for $y$, after switching $x$ and $y$.
$y=-3(3^{-x})+9$
becomes
$x=-3(3^{-y})+9\\
\implies x-9=-3(3^{-y})\\
\implies \dfrac{x-9}{-3}=3^{-y}\\
\implies -\dfrac{1}{3}x+3=3^{-y}\\
\implies \ln\left(-\dfrac{1}{3}x+3\right)=\ln (3^{-y})\\
\implies \ln\left(-\dfrac{1}{3}x+3\right)=-y\cdot \ln 3\\
\implies -\dfrac{ \ln\left(-\dfrac{1}{3}x+3\right)}{\ln 3}=y$
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