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How to convert $\frac{1}{(1-x)(1-x^3)}$ into a sum of multiple fractions? What I mean is, that one can convert $\frac{1}{(1-x)^2(1-x^2)}$ into the following sum:
$\frac{1}{8}(\frac{1}{1+x}+\frac{1}{1-x} +\frac{2}{(1-x)^2} + \frac{4}{(1-x^3)})$
But I can't seem to do the same here, because when I try to simplify $1-x^3$ I get $(1-x)(1+x+x^2)$, and the second term can't be further simplified.
For context, this is part of another problem in combinatorics, which I'm trying to solve using generating functions.
|
To get the exact answer requested first use a difference of two squares;
$$\frac1{(1-x)^2(1-x^2)}=\frac1{(1+x)(1-x)^3}
$$The standard partial fraction treatment for repeated factors gives
$$\frac1{(1+x)(1-x)^3}=\frac{A}{(1+x)}+\frac{B}{(1-x)}+\frac{C}{(1-x)^2}+\frac{D}{(1-x)^3}
$$
$$1=A(1-x)^3+B(1+x)(1-x)^2+C(1+x)(1-x)+D(1+x)
$$
$x$ = 1 gives $D$ = $\frac{1}{2}$, $x$ = -1 gives $A$ = $\frac{1}{8}$,
matching coefficients of $x^3$ gives $B$ =$\frac{1}{8}$, and $x$ = 0 gives $C$ = $\frac{1}{4}$
\begin{equation}
\frac1{(1+x)(1-x)^3}=\frac{1}{8} \bigg\{
\frac{1}{(1+x)}+\frac{1}{(1-x)}+\frac{2}{(1-x)^2}+\frac{4}{(1-x)^3}
\bigg\}
\end{equation}
It generates the sequence 1, 2, 4, 6, 9, 12, 16, 20, 25, 30, 36, 42, 49, 56, 64, 72, ...
|
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|
series convergence lang page 26 exercise 7 Show that the series
$$\sum_n \frac{z^{n-1}}{(1-z^n)(1-z^{n+1})},\quad z\in\mathbb{C},$$
converges to $\frac{1}{(1-z)^2}$ for $|z|<1$ and to $\frac{1}{z(1-z)^2}$ for $|z|>1$...
I try prove it so;
I know that $\sum z^n = \frac{1}{1-z}$ when $|z|<1$ thus $\sum (n+1)z^n = \frac{1}{(1-z)^2}$ in this way I would to prove that $ \sum (n+1)z^n = \sum \frac{z^{n-1}}{(1-z^n)(1-z^{n+1})}$
Some recommendation for continue it way or some other idea?
This exercise is in $\textit{Complex Analysis - Serge Lang}$ - Third Edition, page $26$
thank your.
|
Note $$\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})} = \frac{1}{z(1-z)}\left[\frac{1}{1-z^n} - \frac{1}{1-z^{n+1}}\right]$$ and $$\lim_{n\to \infty} \frac{1}{1-z^{n+1}} = \begin{cases}1&\text{if $\lvert z\rvert < 1$}\\ 0 & \text{if $\lvert z\rvert > 1$}\end{cases}$$
So the series telescopes to $$\frac{1}{z(1-z)}\left[\frac{1}{1-z} - 1\right] = \frac{1}{(1-z)^2}$$when $\lvert z\rvert < 1$, and to $$\frac{1}{z(1-z)}\left[\frac{1}{(1-z)}\right] = \frac{1}{z(1-z)^2}$$when $\lvert z\rvert > 1$.
|
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|
Calculate $\lim_{n\rightarrow \infty }\left(\frac{n^{2}+1}{n+1}\right)^{\tfrac{n+1}{n^{2}+1}}$
$$\lim_{n\rightarrow \infty }\left(\frac{n^{2}+1}{n+1}\right)^{\tfrac{n+1}{n^{2}+1}}$$
I tried to use $f^{g}=e^{g \ln f}$ and I got $e^{\tfrac{n+1}{n^{2}+1}\ln \left(\tfrac{n^2+1}{n+1} \right)}$. How to continue ?
|
Making the problem more general, consider
$$a_n=\left(\frac{n^{2}+a}{n+b}\right)^{\tfrac{n+c}{n^{2}+d}}\implies \log(a_n)={\tfrac{n+c}{n^{2}+d}}\log\left(\frac{n^{2}+a}{n+b}\right)$$ Now, use Taylor expansions for large $n$
$$\log\left(\frac{n^{2}+a}{n+b}\right)=\log
\left({n}\right)-\frac{b}{n}+\frac{a+\frac{b^2}{2}}{n^2}+O\left(\frac{1}{n^3}\right)$$
$${\tfrac{n+c}{n^{2}+d}}=\frac{1}{n}+\frac{c}{n^2}+O\left(\frac{1}{n^3}\right)$$
$$\log(a_n)=\frac{\log \left({n}\right)}{n}+\frac{-b+c \log
\left({n}\right)}{n^2}+O\left(\frac{1}{n^3}\right)$$ Continuing with Taylor
$$a_n=e^{\log(a_n)}=1+\frac{\log \left({n}\right)}{n}+\frac{2 \left(-b+c \log
\left({n}\right)\right)+\log ^2\left({n}\right)}{2
n^2}+O\left(\frac{1}{n^3}\right)$$
Then $\cdots$ $\text{ ???}$ $\forall \{a,b,c,d\}$
|
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|
Solve $z^2 + 4|z| + 4 = 0$ in $\mathbb{C}$
How can I solve this equation:
$z^2 + 4|z| + 4 = 0$?
If I rewrite $z$ as $a+ib$, I get 2 real solutions. But they are wrong.
Any help?
My attempt. By letting $z=a+ib$, we have that
$$(a+ib)^2 + 4 \sqrt{a^2+b^2} + 4 = 0 $$
$$ a^2 + 2iab - b^2 + 4 \sqrt{a^2+b^2} + 4 = 0 $$
$$a^2 -b^2 + 4 \sqrt{a^2+b^2} + 4 = 0 $$
Then $2ab = 0 $.
So, for $a = 0 \implies z=2$ and $b = 0\implies z=-2$
|
Hint: $z^2 + 4|z| + 4 = 0$ iff $z^2 =-4 |z| - 4 \le 0$ and so $z=bi$ for some $b$.
|
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|
Proving irrationality of $\sqrt[3]{3}+\sqrt[3]{9}$ I need to prove $$\sqrt[3]{3}+\sqrt[3]{9}$$
is irrational, I assumed
$$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m}{n}$$
I cubed both sides and got
$$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m^3-12n^2}{9n^3}$$
I tried setting $$\frac{m^3-12n^2}{9n^3} = \frac{m}{n}$$ but that led me nowhere. so what can I do?
|
Let $\sqrt[3]3+\sqrt[3]9=r$.
Thus, since for all reals $a$, $b$ and $c$ we have:
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$ we obtain:
$$3+9-r^3+9r=0.$$
Now, let $r=\frac{m}{n},$ where $m$ and $n$ are naturals with $gcd=1.$
Thus, $$m^3-9n^2m-12n^3=0,$$ which says that $m$ is divisible by $3$.
Let $m=3m'$, where $m'$ is a natural number.
Thus, $$9m'^3-9n^2m'=4n^3,$$ which says that $n$ is divisible by $3$, which is a contradiction.
|
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|
The greatest value of $|z|$ if $\Big|z+\frac{1}{z}\Big|=3$ where $z\in\mathbb{C}$
$\bigg|z+\dfrac{1}{z}\bigg|=3$ then the greatest value of $|z|$ is ___________
My Attempt
$$
\bigg|z+\frac{1}{z}\bigg|=\bigg|\dfrac{z^2+1}{z}\bigg|=\frac{|z^2+1|}{|z|}=3\\
\bigg|z+\frac{1}{z}\bigg|=3\leq|z|+\frac{1}{|z|}\implies |z|^2-3|z|+1\ge0\\
|z|=\frac{3\pm\sqrt{5}}{2}\implies\color{red}{|z|\in(-\infty,\frac{3-\sqrt{5}}{2}]\cup[\frac{3+\sqrt{5}}{2},+\infty)}
$$
$$
\bigg|z+\frac{1}{z}\bigg|=3\geq\bigg||z|-\frac{1}{|z|}\bigg|\\
|z|^2-3|z|-1\leq0\text{ or }|z|^2+3|z|-1\geq0\\
|z|=\frac{3\pm\sqrt{13}}{2}\text{ or }|z|=\frac{-3\pm\sqrt{13}}{2}\\
\color{red}{|z|\in[\frac{3-\sqrt{13}}{2},\frac{3+\sqrt{13}}{2}]}\text{ or }\color{red}{|z|\in(-\infty,\frac{-3-\sqrt{13}}{2}]\cup[\frac{-3+\sqrt{13}}{2},+\infty)}
$$
The solution given in my reference is $\dfrac{3+\sqrt{13}}{2}$, why am I not able to find it in my attempt ?
Note: A similar question has been asked before, If $∣z+\frac{1}{z}∣=a$ then find the minimum and maximum value of $|z|$, but it does not have any answer which address why we choose a specific version of the triangle inequality.
|
(First, assume that $|z|>1$; our range for $|z|$ will be mirrored with its reciprocal, but the maximum we're looking for is definitely greater than $1$.)
Your inequality $\left|z+\frac1z\right|\le |z|+\left|\frac1z\right|$ is true, but it's only half the picture; we get equality when $z$ and $\frac1z$ are positive real multiples of each other, which makes $|z|$ as small as possible. Solving the resulting quadratic inequality $|z|^2-3|z|+1 \ge 0$, we get $|z|\ge\frac{3+\sqrt{5}}{2}$ or $|z|\le \frac{3-\sqrt{5}}{2}$.
We should also look at an inequality in the other direction: $\left|z+\frac1z\right|\ge |z|-\left|\frac1z\right|$. This leads to the quadratic inequality $|z|^2-3|z|-1\le 0$, with solutions $\frac{3-\sqrt{13}}{2}\le z\le \frac{3+\sqrt{13}}{2}$.
Combine the two inequalities and our restriction $|z|>1$, and we get the range
$$\frac{3+\sqrt{5}}{2} \le |z|\le \frac{3+\sqrt{13}}{2}$$
with equality achieved at pure real values $\pm\frac{3+\sqrt{5}}{2}$ for the lower bound and pure imaginary values $\pm i\frac{3+\sqrt{13}}{2}$ for the upper bound.
So, then, your work - you solved the quadratic wrong, and got the numbers from the other quadratic. (Edit note - fixed now) Then your reference had to go mix things up with an answer that's just plain wrong. No wonder you're confused here.
...why we choose a specific version of the triangle inequality.
The real answer? We choose all of them. They all apply, restricting different aspects of things. Some of those restrictions won't matter for the specific problem we want to solve, but they're all there.
(In response to the red text added)
There's a reason I added the assumption that $|z|>1$; your solution to the inequality $3\ge \left||z|-\frac1{|z|}\right|$ mixes two cases together into a useless mush. Combine the two parts of what you wrote, and it's no restriction on $|z|$ at all - all positive values are in at least one of those ranges.
If $|z|\ge 1$, then $|z|\ge\frac1{|z|}$ and we can write $\left||z|-\frac1{|z|}\right|=|z|-\frac1{|z|}$. This leads to the inequality $|z|^2-3|z|-1\le 0$ and solutions $\frac{3-\sqrt{13}}{2}\le z\le \frac{3+\sqrt{13}}{2}$. But then, we also assumed $|z|\ge 1$, so that's really $1\le z\le \frac{3+\sqrt{13}}{2}$.
If $|z|\le 1$, then $|z|\le\frac1{|z|}$ and we can write $\left||z|-\frac1{|z|}\right|=\frac1{|z|}-|z|$. This leads to the inequality $|z|^2+3|z|-1\ge 0$, with solutions $|z|\le\frac{-3-\sqrt{13}}{2}$ or $|z|\ge \frac{-3+\sqrt{13}}{2}$. The first of those? We can't have $|z|$ negative. Bye. The second? Combine with the case's condition for $\frac{\sqrt{13}-3}{2}\le |z|\le 1$.
So then, the full implication of that inequality is
$$\frac{2}{3+\sqrt{13}}=\frac{\sqrt{13}-3}{2}\le |z|\le\frac{3+\sqrt{13}}{2}$$
Combine with the other half, and we get two ranges for $z$:
$$\frac{2}{3+\sqrt{13}}=\frac{\sqrt{13}-3}{2}\le |z|\le \frac{3-\sqrt{5}}{2}=\frac{2}{3+\sqrt{5}}\quad\text{or}\quad\frac{3+\sqrt{5}}{2}\le |z|\le \frac{3+\sqrt{13}}{2}$$
|
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How can I simplify this fraction problem? I have the problem $\frac{x^2}{x^2-4} - \frac{x+1}{x+2}$ which should simplify to $\frac{1}{x-2}$
I have simplified $x^2-4$, which becomes:
$\frac{x^2}{(x-2)(x+2)} - \frac{x+1}{x+2}$
However, if I combine the fractions I get, $x^2-x-1$ for the numerator, which can't be factored. That's where I get stuck.
How can I get $\frac{1}{x-2}$ out of this problem?
|
You need to put them over a common denominator,
$$\frac {x+1}{x+2}=\frac {(x+1)(x-2)}{(x+2)(x-2)}=\frac {x^2-x-2}{x^2-4}$$
Now you can subtract the numerators
$$x^2-(x^2-x-2)=x+2$$
and finally divide out the $x+2$ from numerator and denominator
Your $-1$ should be $-2$. You didn't show your work, so I can't see why it happened.
|
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For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer? For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer?
For example, if x = 3, $y = 4(3^2) + 8(3) + 5$ = 65 which is divisible by 13
if x = 8, y = 325 which is divisible by 13
if x = 16, y = 1157 which is divisible by 13
if x = 21, y = 1937 which is divisible by 13
I am guessing that values of x = 13i + 3 or x = 13i + 8 where i is an integer will result in a value of y that is evenly divisible by 13.
How do you prove that x = 13i + 3 or x = 13i + 8 will result in a value of y that is evenly divisible by 13?
Is there a general proof to find values of x that will result in a value of y that is evenly divisible by an odd integer p?
|
Let us try with this approach:
$\begin{align} 4x^2+8x+5 &= 4(x+1)^2+1 &\equiv 0 &\quad(\text{mod} 13)\\
&\Rightarrow 4(x+1)^2 &\equiv 12 &\quad(\text{mod} 13)\\
&\Rightarrow (x+1)^2 &\equiv 3 &\quad(\text{mod} 13)\end{align}\\$
Substituting for $x+1 = f$ we are looking to take the square-root of 3 modulo 13. The following holds true:
$$n^2 \equiv (13-n)^2 \quad (\text{mod} 13)$$
It suffices to look at numbers from 0 to 6:
$\begin{align}
0^2 &\equiv 0 (\text{mod} 13),\\
1^2 &\equiv 1 (\text{mod} 13),\\
2^2 &\equiv 4 (\text{mod} 13),\\
3^2 &\equiv 9 (\text{mod} 13),\\
4^2 &\equiv 3 (\text{mod} 13),\\
5^2 &\equiv 12 (\text{mod} 13),\\
6^2 &\equiv 10 (\text{mod} 13).
\end{align}$
Thus numbers of the form $f = 13m + 4$ solve the problem. Then also $f = 13m + 9$ will solve the problem. Since $x = f-1$ our solutions are numbers of the form: $x \in \{13m + 3, 13m +8, \text{where } m\geq 0\}$.
|
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Simplify $\frac{x^3-x}{x^2+xy+x+y}$ $$\frac{x^3-x}{x^2+xy+x+y}$$
What I did:
$$\frac{x}{xy+x+y}$$
through simplifying the $x$'s.
But it's not right.
What did I do wrong?
|
You can only simplify factors of the numerator with factors of the denominator. Thus first you have to decompose the numerator and the denominator, and then simplify the fraction.
Start with
$$
\frac{x^3-x}{x^2+xy+x+y} = \frac{x(x^2-1)}{x(x+y)+x+y} = \frac{x(x+1)(x-1)}{(x+1)(x+y)},
$$
and now simplify the only common factor, that is $x+1$.
|
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|
value of $x$ in Trigonometric equation Find real $x(0<x<180^\circ)$ in
$\tan(x+100^\circ)=\tan(x-50^\circ)+\tan(x)+\tan(x+50^\circ)$
what i try
$\displaystyle \tan(x+100^\circ)-\tan(x)=\tan(x+50^\circ)+\tan(x-50^\circ)$
$\displaystyle \frac{\sin(100^\circ)}{\cos(x+100^\circ)\cos x}=\frac{\sin(2x)}{\cos(x+50^\circ)\cos(x-50^\circ)}$
$\displaystyle \frac{\sin(100^\circ)}{\cos(2x+100^\circ)+\cos(100^\circ)}=\frac{\sin(2x)}{\cos(2x)+\cos(100^\circ)}$
How do i solve further
Help me please
|
Using the addition theorem for $\tan $ you get
$$
\frac{\tan(x) +\tan(100^\circ)}{1 - \tan(x) \tan(100^\circ)}=\frac{\tan(x) - \tan(50^\circ)}{1 + \tan(x) \tan(50^\circ)}+\tan(x)+\frac{\tan(x) + \tan(50^\circ)}{1 -\tan(x) \tan(50^\circ)}
$$
Let $y= \tan(x)$, then this is
$$
\frac{y +\tan(100^\circ)}{1 - y\tan(100^\circ)}=\frac{y- \tan(50^\circ)}{1 + y \tan(50^\circ)}+y+\frac{y + \tan(50^\circ)}{1 -y \tan(50^\circ)}
$$
Clearing denominators gives
$$
0 = - (y +\tan(100^\circ))(1 -y^2 \tan^2(50^\circ))+(y -\tan(50^\circ))(1 -y \tan(50^\circ))(1 - y\tan(100^\circ))\\
+y (1 -y^2 \tan^2(50^\circ))(1 - y\tan(100^\circ)) + (y +\tan(50^\circ))(1 +y \tan(50^\circ))(1 - y\tan(100^\circ))
$$
Multiplying out gives
$$
0 = -\tan^2(50) - 2 y \tan(100) (1 + \tan^2(50)) + y^2 (-1 - 3 \tan^2(50)) + y^4
$$
With some help of Wolframalpha, the only positive solution to this fourth order equation is $\tan x= y \simeq 1.90326$ or $x \simeq 62.28^\circ$ or $x \simeq 0.346 \pi \simeq 1.087$ (in radians). I wonder if this "fits" to something special.
The only negative solution is $\tan x= y \simeq -1.73205$ which gives $x \simeq 120^\circ$. Now this can be verified to be exactly $x =120^\circ$ since indeed, using the original equation,
$$
\tan(220^\circ)=\tan(70^\circ)+\tan(120^\circ)+\tan(170^\circ)
$$
holds.
Though this treatment is sort of ugly, it has the advantage that it shows the only two solutions to the problem.
|
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Subgroup of general linear group over $\mathbb{F}_{3}$ generated by two matrices I have two matrices: $A=\begin{bmatrix}0 & 2 &1\\1 & 2&1\\1 &2&0\end{bmatrix}$
and $B=\begin{bmatrix}2 & 1 &0\\0 & 1&0\\0 &1&2\end{bmatrix}$. I need to describe subgroup generated by these matrices.I know that group generated by A and B, say G, is isomorphic to $Q_{8}$ or $D_{4}$ or $S_{3}$ or $S_{4}$ or $A_{4}$. Any hints?
|
We have $A^2=\begin{bmatrix}0&0&2\\ 0&2&0\\ 2&0&0\end{bmatrix}$, $A^4=\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}$, $B^2=\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}$ and $BAB=A^{-1}$ since $A^3=A^{-1}=\begin{bmatrix}2&1&0\\ 2&1&2\\\ 0&1&2\end{bmatrix}$ and $BAB=\begin{bmatrix}2&1&0\\ 2&1&2\\\ 0&1&2\end{bmatrix}$.
The dihedral group $D_8$ has presentation $\langle a,b\ |\ a^4=b^2=1, bab=a^{-1}\rangle$, clearly $A$ and $B$ satisfy these relations, so we can conclude $\langle A,B\rangle\cong D_8$.
|
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Find all values of a for which the system has trivial solutions $$\begin{pmatrix}1+a & 1 & 1 & 1\\\ 1 & 1-a & 1 & 1\\1 & 1 & 1 & a\\1 & 2 & 2 & 1\\\end{pmatrix}$$
As the title says, find all values of a for which the system has trivial solutions.
I have tried row reduction but I haven't gotten anywhere with it. So what I have done is calculate the determinant, which is:
2 a^3 - a
From what I understand system has trivial solutions if and only if det(A) != 0.
So the solution for this problem would be every ${\rm I\!R}$ number except where det(A) == 0 ?
|
Your determinant is correct, so we get some values to check
$$\det A = a \left(2 a^2-1\right) \implies a = 0, \pm \dfrac{1}{\sqrt{2}}$$
If $a = 0$, the RREF is
$$\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end{bmatrix}$$
If $a = \dfrac{1}{\sqrt{2}}$, the RREF is
$$\begin{bmatrix}
1 & 0 & 0 & \sqrt{2}-1 \\
0 & 1 & 0 & 1-\sqrt{2} \\
0 & 0 & 1 & \frac{1}{\sqrt{2}} \\
0 & 0 & 0 & 0 \\
\end{bmatrix}$$
If $a = -\dfrac{1}{\sqrt{2}}$, the RREF is
$$\begin{bmatrix}
1 & 0 & 0 & -\sqrt{2}-1 \\
0 & 1 & 0 & \sqrt{2}+1 \\
0 & 0 & 1 & -\frac{1}{\sqrt{2}} \\
0 & 0 & 0 & 0 \\
\end{bmatrix}$$
If $a \ne 0, a \ne \pm \dfrac{1}{\sqrt{2}}$, the RREF is
$$\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}$$
Hopefully you can complete the question now.
|
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Find a Jordan's matrix and basis I have a matrix: $A={\begin{bmatrix}3&-1&0&1\\0&3&4&4\\0&0&-5&-8\\0&0&4&7\end{bmatrix}}$.1) I calculate characteristic polynomial. It is: $p_{A}(\lambda)=(\lambda-3)^{3}(\lambda+1)$ 2) So exist Jordan's matrix: $J={\begin{bmatrix}3&?&0&0\\0&3&?&0\\0&0&3&0\\0&0&0&-1\end{bmatrix}}$ 3)I find own subspace: $V_{3}=\ker(A-3I)=\dots= \operatorname{span}({\begin{bmatrix}1&0&0&0\end{bmatrix}}^{T},{\begin{bmatrix}0&1&-1&1\end{bmatrix}}^{T}$ $V_{-1}=\ker(A+I)=\dots= \operatorname{span}({\begin{bmatrix}0&1&-2&1\end{bmatrix}}^{T}$ 4) Jordan's matrix is: $J={\begin{bmatrix}3&0&0&0\\0&3&1&0\\0&0&3&0\\0&0&0&-1\end{bmatrix}}$
However I have a problem with a basis: I read that in basis are vectors from span $V_{3}$ and $V_{-1}$ and the next I must find one more vector. So I can use my vectors from span and I have for example:${\begin{bmatrix}0&-1&0&1&|&1\\0&0&4&4&|&0\\0&0&-8&-8&|&0\\0&0&4&4&|&0\end{bmatrix}}$ and I have for example $(0,d-1,-d,d)^{T}$ and I change for example $d=5$ so my last vectors in basic is $(0,4,-5,5)^{T}$.Hovewer on my lectures we use vectors from image but I don't understand it and I am afraid that my sollution is not good. Can you help me?
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I'll write down an algorithm to find a basis with eigenvectors / generalized eigenvectors, when the wanted vector will be denoted by $\;w\;$ .
First, we must find out what the eigenvalues are. You already did that, they are $\;-1,3\;$, with the second one of geometric multiplicity $2$ . Now
$$(A-(-1)I)w=0\iff \begin{pmatrix}4&-1&0&1\\0&4&4&4\\0&0&-4&-8\\0&0&4&8\end{pmatrix}\begin{pmatrix}w_1\\w_2\\w_3\\w_4\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}\implies w_1=\begin{pmatrix}0\\1\\-2\\1\end{pmatrix}$$$${}$$
$$(A-3I)w=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}\iff\begin{pmatrix}0&-1&0&1\\0&0&4&4\\0&0&-8&-8\\0&0&4&4\end{pmatrix}\begin{pmatrix}w_1\\w_2\\w_3\\w_4\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}\implies w_2=\begin{pmatrix}1\\0\\0\\0\end{pmatrix}\,,\,\,w_3=\begin{pmatrix}0\\1\\\!\!-1\\1\end{pmatrix}$$$${}$$
with $\;w_1,w_2,w_3\;$ eigenvectors of $\;\lambda=-1\;$ (the first one), and of $\;\lambda=3\;$ (the last two) .
We're missing one vector for a basis for $\;\Bbb R^4\;$, and since there are only two linearly ind. eigenvectors of $\;3\;$ , we're going to calculate a generalized eigenvector of $\;3\;$ as follows:
$$(A-3I)w=w_2\;\text{or}\;w_3$$ (we only need one of these to work), so:
$$(A-3I)w=\begin{pmatrix}1\\0\\0\\0\end{pmatrix}=w_2\iff\begin{pmatrix}0&-1&0&1\\0&0&4&4\\0&0&-8&-8\\0&0&4&4\end{pmatrix}\begin{pmatrix}w_1\\w_2\\w_3\\w_4\end{pmatrix}=\begin{pmatrix}1\\0\\0\\0\end{pmatrix}\implies$$$${}$$
$$\begin{cases}-w_2+w_4=1\\
w_3=-w_4\end{cases}\;\;\implies w_4=\begin{pmatrix}0\\-1\\0\\0\end{pmatrix}$$
and we thus have a basis for $\;\Bbb R^4\;:\;\;\{w_1,w_2,w_3,w_4\}$
In this case is pretty easy to find out the JCF of $\;A\;$ , since $\;\dim V_3=2\implies\;$ there are two blocks for $\;\lambda=3\;$ , and thus the JCF (upt to similarity) of $\;A\;$ is
$$J_A=\begin{pmatrix}3&1&0&0\\0&3&0&0\\0&0&3&0\\0&0&0&\!\!-1\end{pmatrix}$$
But ...If you want to get the above by similarity, take the basis $\;\{w_1,w_2,\color{red}{w_4},w_3\}\;$ and form with it the columns of matrix $\;P\;$ . Then you can check that $\;P^{-1}AP=J_A\;$ . The reason is that we "arrange" the basis with "close" vectors to each other. Thus, first the eigenvector $\;w_2\;$ and then the generalized eigenvector $\;w_4\;$ which was obtained with the help of $\;w_2\;$ , and at the end $\;w_3\;$ .
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|
Evaluating $\frac{2013^3-2\cdot 2013^2\cdot 2014+3\cdot 2013\cdot 2014^2-2014^3+1}{2013\cdot 2014}$ What is the value of $$\frac{2013^3-2\cdot 2013^2\cdot 2014+3\cdot 2013\cdot 2014^2-2014^3+1}{2013\cdot 2014}?$$
What I have tried:
$$\implies\frac{2013^2(2013-2\cdot2014)+2014^2(3\cdot 2013-2014)+1}{2013\cdot 2014}$$
$$\implies\frac{2013^2(-2015)+2014^2(4025)+1}{2013\cdot 2014}$$
I'm not sure what to do next...
Help is appreciated!
Furthermore, if you are nice, could you also help me on this problem(Transferring bases of numbers.) too?
Thanks!
Max0815
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Hint
Choose $2013=a$
$$\dfrac{a^3-2a^2(a+1)+3a(a+1)^2-(a+1)^3+1}{a(a+1)}$$
$$=\dfrac{-3a(a+1)+a(a+1)\{3(a+1)-2a\}}{a(a+1)}$$
$$=-3+3+a$$ as $a(a+1)\ne0$
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|
How can i find a inverse of a polynomial in a quotient ring? I am asked to find the inverse of $\widehat {x^3+1} $ in Q/[x]/I where I = $x^2-2$.
I am used to find the inverse by keep doing the division until a find a irreducible elements but the degree of the polynomial i asked to find the inverse is larger than the degree of polynomial in I.
In previous question, we have calculated the inverse of $\widehat {X} $ which is $1/2$(X). And i am trying to breaking down $X^3 - 1 $ into $X(X^2-2)$ + $(2X+1)$. Is this bit some hints to approach this question or is it a new idea to find the inverse with a polynomial of a larger degree?
Thanks a lot!!
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extended Euc:
$$ \left( x^{3} + 1 \right) $$
$$ \left( x^{2} - 2 \right) $$
$$ \left( x^{3} + 1 \right) = \left( x^{2} - 2 \right) \cdot \color{magenta}{ \left( x \right) } + \left( 2 x + 1 \right) $$
$$ \left( x^{2} - 2 \right) = \left( 2 x + 1 \right) \cdot \color{magenta}{ \left( \frac{ 2 x - 1 }{ 4 } \right) } + \left( \frac{ -7}{4 } \right) $$
$$ \left( 2 x + 1 \right) = \left( \frac{ -7}{4 } \right) \cdot \color{magenta}{ \left( \frac{ - 8 x - 4 }{ 7 } \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( x \right) } \Longrightarrow \Longrightarrow \frac{ \left( x \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ 2 x - 1 }{ 4 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 2 x^{2} - x + 4 }{ 4 } \right) }{ \left( \frac{ 2 x - 1 }{ 4 } \right) } $$
$$ \color{magenta}{ \left( \frac{ - 8 x - 4 }{ 7 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - 4 x^{3} - 4 }{ 7 } \right) }{ \left( \frac{ - 4 x^{2} + 8 }{ 7 } \right) } $$
$$ \left( x^{3} + 1 \right) \left( \frac{ 2 x - 1 }{ 7 } \right) - \left( x^{2} - 2 \right) \left( \frac{ 2 x^{2} - x + 4 }{ 7 } \right) = \left( 1 \right) $$
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Binomial Theorem expansion and proving an interesting identity? In the identity
$$\frac {n!}{x(x+1)(x+2)...(x+n)} = \sum ^n_{k=0}\frac {A_k}{x+k} $$
Prove that $$A_k =(-1)^{k}\:\cdot\: ^{n}C_k$$
Also from this deduce that,
$$ \;^{n}C_0\frac 1{1.2} - \:^{n}C_1\frac1{2.3} +\; ^{n}C_2\frac1{3.4} \; ... \;{(-1)^n}\; ^{n}C_n\frac1{(n+1)(n+2)}\;=\frac1{(n+2)}$$
So I have to tried to use the binomial theorem on,
$(b-a)^n$, and then multiplied both sides by $a^{x-1}$.
Now I integrated both sides with respect to $a$. This gives me the binomially expanded side as same as the right hand side of the identity that we have to prove when I substitute $a=1$. I dont know how to integrate $a^{x-1}\;(b-a)^n$ with respect to $a$. I am not able to prove the identity and solve the deduction. Can someone please help me out ? Thanks a lot.
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We obtain
\begin{align*}
\frac{n!}{x(x+1)\cdots(x+n)}&=\sum_{k=0}^n\frac{A_k}{x+k}\\
n!&=\sum_{k=0}^nA_k\frac{x(x+1)\cdots(x+n)}{x+k}\tag{1}
\end{align*}
Substituting $x=-j,0\leq j\leq n$ in (1) we get
\begin{align*}
n!&=A_j(-j)(-j+1)\cdots (-1)\cdots1\cdot 2\cdots(n-j)\\
&=A_j(-1)^j j!(n-j)!\\
\end{align*}
We get
\begin{align*}
\color{blue}{A_j}&=(-1)^j\frac{n!}{j!(n-j)!}\\
&\,\,\color{blue}{=(-1)^j\binom{n}{j}\qquad\qquad 0\leq j\leq n}\tag{2}
\end{align*}
and the claim follows.
Using (1) and (2) we want to show $\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{(k+1)(k+2)}=\frac{1}{n+2}$.
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n}&\color{blue}{(-1)^k\binom{n}{k}\frac{1}{(k+1)(k+2)}}\\
&=\sum_{k=0}^n(-1)^k\binom{n}{k}\left(\frac{1}{k+1}-\frac{1}{k+2}\right)\\
&=\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{k+1}-\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{k+2}\\
&=\frac{n!}{1\cdot 2\cdots (n+1)}-\frac{n!}{2\cdot 3\cdots (n+2)}\tag{3}\\
&=\frac{1}{n+1}-\frac{1}{(n+1)(n+2)}\\
&\,\,\color{blue}{=\frac{1}{n+2}}
\end{align*}
and the claim follows.
In (3) we apply (1) and (2) with $x=1$ and $x=2$.
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|
Proof by induction: $a_n > n + \frac{1}{3} \space \forall n \in N,\space n\geq 4$ How do I prove this by induction?
Let $(a_n)_{n\in N}$ be the sequence defined by:
$$a_1=1,\space a_2=\frac{3}{2},\space a_{n+2}=a_{n+1}+\frac{2n+1}{n+2}a_n \space \space (n\in N)$$
Prove that $a_n > n + \frac{1}{3} \space \forall n \in N,\space n\geq 4$
Here's what I did, is it ok?
1) I define
$$P(n): a_n > n + \frac{1}{3} \space \forall n \in N,\space n\geq4\\$$
2) Since the base case is true,
$$a_1=1, \ a_2=\frac{3}{2},\ a_3=\frac{5}{2},\ a_4=5 $$
$$P(4):a_4 = 5 > 4+\frac{1}{3}$$
3) I assume $P(n)$ (already defined) and
$$P(n+1):a_{n+1}>n+1+\frac{1}{3} = n + \frac{4}{3}$$
are true, and want to prove that it implies
$$P(n+2):a_{n+2}>n+2+\frac{1}{3}=n+\frac{7}{3}$$
is true.
4)So since:
by the second induction hypothesis,
$$a_{n+2}=a_{n+1}+\frac{2n+1}{n+2}a_n > n+\frac{4}{3}+\frac{2n+1}{n+2}a_n\ $$
and by the first induction hypothesis,
$$n+\frac{4}{3}+\frac{2n+1}{n+2}a_n > n+\frac{4}{3}+\frac{2n+1}{n+2}(n + \frac{1}{3}) = n+\frac{4}{3} + \dfrac{2n^2+n+\dfrac{2n}{3}+\dfrac{1}{3}}{n+2} $$
5)Now as I know that every term is positive I can remove some terms and assure that:
$$n+\frac{4}{3} + \dfrac{2n^2+n+\dfrac{2n}{3}+\dfrac{1}{3}}{n+2} > n+\frac{4}{3}+\frac{2n+1}{n+2}(n + \frac{1}{3}) = n+\frac{4}{3} + \dfrac{2n^2}{n+2}$$
So the only step remaining would be making sure that:
$$\dfrac{4}{3}+\dfrac{2n^2}{n+2} > \dfrac{7}{3}\ $$
then
$$ \dfrac{2n^2}{n+2} > 1 which \ is \ true \ \forall n \in N, n \geq 4$$
Therefore $P(n+2)$ is true, $P(n+1)$ is true and also $P(n)$ is true.
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What you did is correct. A shortcut would be to notice that
$$a_{n+2}=a_{n+1}+\frac{2n+1}{n+2}a_n > n+\frac{4}{3}+ \underbrace{\frac{2n+1}{n+2}a_n}_{\geq 1} > n+\frac{4}{3}+1 = n + \frac{7}{3}$$
since we have that $a_n > 4 + \frac{1}{3} > 1$ and $\frac{2n+1}{n+2} > 1$ are true for all $n \geq 4$. Hence, $\frac{2n+1}{n+2}a_n >1$ for all $n \geq 4$.
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Proving $\left(1+\frac{1}{m}\right)^m < \left(1+\frac{1}{n}\right)^n$ Let $m,n\in \mathbb{N}$. If $m > n$ show that
$$\left(1+\frac{1}{m}\right)^m > \left(1+\frac{1}{n}\right)^n$$
My works: I tried to show if $g(x)=\left(1+\frac{1}{x}\right)^x$ then $g'(x) > 0$.
\begin{align}
g'(x) &= \frac{d e^ { x \ln(1+\frac{1}{x}) }} {dx} \\
&= e^{x\ln(1+1/x)} \left(\ln(1+\frac{1}{x}) - \frac{1}{1+x} \right) >0
\end{align}
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Normally the proof goes something like, let $a_n=(1+\frac{1}{n})^n$ for all $n\in\mathbb Z^+$. Then, $$\frac{a_{n+1}}{a_n}=\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n})^n}=\frac{(\frac{n+2}{n+1})^{n+1}}{(\frac{n+1}{n})^n}=\frac{(n+2)^{n+1}n^n}{(n+1)^n(n+1)^{n+1}}=\frac{(n^2+2n)^n}{(n+1)^{2n}}\frac{n+2}{n+1}=\frac{((n+1)^2-1)^n}{((n+1)^2)^n}\frac{n+2}{n+1}=\bigg(1-\frac{1}{(n+1)^2}\bigg)^n\frac{n+2}{n+1}\geq\bigg(1-\frac{n}{(n+1)^2}\bigg)\frac{n+2}{n+1}=\frac{n^2+n+1}{(n+1)^2}\frac{n+2}{n+1}=\frac{(n^2+n+1)(n+2)}{(n+1)^3}=\frac{n^3+3n^2+3n+2}{n^3+3n^2+3n+1}> 1$$
(the first inequality that appears above comes from Bernoulli's inequality which can be proven by simple induction as seen in the link)
So, $\frac{a_{n+1}}{a_n}>1\implies a_{n+1}>a_n$. This means if $m$ is any integer greater than $n$, $$a_m>a_{m-1}>a_{m-2}>...>a_{n+1}>a_n$$
giving us $a_m>a_n$, i.e., $(1+\frac{1}{m})^m>(1+\frac{1}{n})^n$
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|
How do we factor out $x^2 - x -2$ in this expression? Suppose we're given that
$$x^4 - 2x^3 +x-2$$
How do we factor out $x^2 - x -2$ in this expression?
$$(x^4 -x^3- 2x^2)-(x^3-x^2-2x)+(x^2-x-2) = x^2(x^2 -x-2)-x(x^2-x-2)+(x^2-x-2) = (x^2-x-2)(x^2-x+1)$$
This satisfies with what we want to get. However, I do not seem to have understood what is done there. Could I get your assistance in order to understand it? Perhaps there's better way of factoring.
Regards
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Another way: complete the square in $\rm\color{#c00}{lead\ 2\ terms}$ yielding
$$ \underbrace{\color{#c00}{(x^2\!-\!x)^2} - (\color{#c00}{x^2}\!-\!x)-2}_{\Large \color{#c00}{x^4\ -\ 2x^3}\ +\ x\ -\ 2\!\!\!\!\!\!\!\!\!\!\!\! } = X^2\!-\!X\!-\!2 = (X\!+\!1)(X\!-\!2) = (x^2\!-\!x\!+\!1)\underbrace{(x^2\!-\!x\!-\!2)}_{\Large (x-2)(x+1)}$$
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If If $x^2+ax+b+1=0$ has roots which are positive integers, then $(a^2+b^2)$ can be which of the given choices?
If $x^2+ax+b+1=0$ has roots which are positive integers, then $(a^2+b^2)$ can be
(A) 50
(B) 37
(C) 19
(D) 61
My approach: I first took roots $\alpha$, $\beta$ and then applied sum and product of roots according to Vieta's formula. But I don't think we can get any value with that. Any help will be much appreciation.
Also can we actually find the range of $a^2+b^2$.
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We need that $a$ and $b$ will be naturals and there is a natural $n$ for which $$a^2-4(b+1)=n^2.$$
Thus, $$a^2+b^2=n^2+b^2+4b+4=n^2+(b+2)^2.$$
Can you end it now?
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|
To find the inverse of a special kind of matrix. In a matrix analysis problem, I encountered the following special kind of matrix
$$
\begin{bmatrix}
0 & 1 & a & a & a & a \\
1 & 0 & a& a& a& a \\
a& a &0 & 1& a& a \\
a& a &1 & 0 & a& a \\
a & a & a & a &0 & 1\\
a & a & a & a &1 & 0
\end{bmatrix}
$$
where $a$ is a positive integer.But this matrix is not a circulant matrix. It is not in my knowledge if this is any known form.
We can also search for inverses for the general form of the matrix for even order.
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You can write it as a sum:
$$ \begin{bmatrix}
0 & 1 & a & a & a & a \\
1 & 0 & a & a & a & a \\
a & a & 0 & 1 & a & a \\
a & a & 1 & 0 & a & a \\
a & a & a & a & 0 & 1 \\
a & a & a & a & 1 & 0
\end{bmatrix} =
\begin{bmatrix}
-a & 1-a & 0 & 0 & 0 & 0 \\
1-a & -a & 0 & 0 & 0 & 0 \\
0 & 0 & -a & 1-a & 0 & 0 \\
0 & 0 & 1-a & -a & 0 & 0 \\
0 & 0 & 0 & 0 & -a & 1-a \\
0 & 0 & 0 & 0 & 1-a & -a \end{bmatrix}
+
\begin{bmatrix} a \\ a\\ a \\ a \\ a \\ a \end{bmatrix}.\begin{bmatrix}1 \\ 1 \\ 1 \\ 1\\ 1 \\ 1 \end{bmatrix}^T $$
and then use the Sherman–Morrison formula to invert this sum of two matrices. The dimension of the matrix can be easily adapted.
|
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Is $8n^3+12n^2-2n-3$ divisible by 5, with $n$ congruent with 1,2,3 mod 5? I make some test using python and I find that this is the case for $n = 5k +p$ with $k$ an integer and $p =$1,2,or 3.
I was able to prove for $p = 1$ and $p=2$ but not for $p = 3$.
What I'm doing wrong? This is my work:
Let's assume 5 is a divisor of $8n^3+12n^2-2n-3$. Let's prove it for $n + 1.$
We have $8(n+1)^3+12(n+1)^2-2(n+1)-3=8(n^3+3 n^2+3n+1)+12(n^2+2n+1)-2(n+1)-3=(8n^3+12n^2-2-3)+24n^2+48n+18$
For induction we have $8n^3+12n^2-2n-3$, we need only that $5$ divide $24n^2+48n+18$
Since $n = 5k + p$, $k$ an integer and $p$ with values $1$,$2$ or $3$.
We need to see that $5$ divide $24n^2+48n+18= 24$ with $(5k+p)^2+48(5k+p)+18$,
Using congruence we have $(5k+p)^2$ is congruent with $p^2$ mod 5 and $5k + p$ with $p$ modulo 5.
$f(p)= 24p^2+48p+18$ is divisible with 5 para $p = 1, p = 2$.
So $f(1) = 90$ is divisible for 5 , $f(2)= 210$ and also divisible for 5.
But $f(3)$ is not divisible for 5.
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OP changed the question:
So, another solution:
$$8n^3+12n^2-2n-3 \equiv_5 3n^2 + 2n^2 -2 n - 3 \stackrel{n \equiv_5 k}{\equiv_5} 3k^3 + 2k^2 - 2k -3$$
Now:
*
*$k = 1$: $3+2-2-3 \equiv_5 0$
*$k = 2$: $3\cdot 8 +2\cdot 4 -2\cdot2 -3 \equiv_5 -1 -2-4-3 \equiv_5 0$
*$k = 3$: $3\cdot 27 +2\cdot 9 -2\cdot3 -3 \equiv_5 6 +3-6-3 \equiv_5 0$
.......
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|
A poker hand contains five cards. Find the probability that a poker hand can be.... a) Four of a kind (Contains four cards of equal face value)
So for this one, we want four cards that have the same face value, different suit. And the last card can be any remaining card.
There are 13 ranks, (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K). We have $\binom{13}{1}$ ways to pick 1 rank out of 13. For each of these ranks, we want to pick 4 cards that are all the same rank (and it is forcefully implied that these 4 cards will differ in suits). So $\binom{13}{1}\binom{4}{4}$, and for each of these ways, we have $\binom{48}{1}$ way to pick 1 card out of the remaining deck of 48 cards, because we picked 4 cards already. Thus, $$\frac{\binom{13}{1}\binom{4}{4}\binom{48}{1}}{\binom{52}{5}}$$ is the total number of ways.
b) Full House (Three cards of equal face value, and two others of equal face value). So i.e: 3, 3, 3, 2, 2 would be a full house where the three 3's and two 2's are distinguishable (different suit).
So there are 13 ranks again, $\binom{13}{1}$ ways to pick 1 rank out of 13 total. For each of these ways, we want to pick three cards of the same rank. So $\binom{13}{1}\binom{4}{3}$. Now we want two more cards that are of equal face value that differ from the other three cards picked earlier, so there are 12 ranks left, and $\binom{12}{1}$ ways to pick 1 rank out of 12 remaining. For each of these ways, we have $\binom{4}{2}$ ways to pick 2 cards out of the 4 suits belonging to the same rank, thus $\binom{12}{1}\binom{4}{2}$. The total number of ways is: $$\frac{\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}}{\binom{52}{5}}$$
c) Three of a kind. (Three cards of equal face value, and two cards with face values that differ from each other and the other three).
13 ranks, we want to pick 3 cards of the same rank. $\binom{13}{1}\binom{4}{3}$. Now we want two cards that have different face values from the three picked, and each other. Now here is where I'm a bit confused, the ranks MUST be different, but the suits can be the same. So proceeding, $\binom{12}{1} \binom{4}{1}$ to pick 1 card of a different face value, but could be same suit, and then $\binom{11}{1}\binom{4}{1}$ to pick another card with a different face value, but could be a different suit. Multiply these altogether and divide by the denominator, and we have our total ways.
Is my work correct?
|
You correctly calculated the probabilities of four of a kind and a full house, not the total number of ways. The number of ways these hands can be obtained is given by the numerators of your probabilities.
What is the probability of three of a kind?
As you know, there are $\binom{52}{5}$ possible five-card hands that can be drawn from a standard deck.
However, your count of the favorable cases is not quite right. There are $\binom{13}{1}$ ways to choose the rank from which three cards are drawn and $\binom{4}{3}$ ways to select cards of that rank, as you found. The remaining two cards must come from different ranks. There are $\binom{12}{2}$ ways to select two ranks from which one card is drawn and $\binom{4}{1}$ ways to choose one card from each of the two ranks. Hence, the number of favorable cases is
$$\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}^2$$
Therefore, the probability of three of a kind is
$$\frac{\dbinom{13}{1}\dbinom{4}{3}\dbinom{12}{2}\dbinom{4}{1}^2}{\dbinom{52}{5}}$$
What was your mistake?
You count each hand twice since the order in which the singletons are selected does not matter. For instance, if the hand is $\color{red}{5\heartsuit}, \color{red}{5\diamondsuit}, 5\spadesuit, 7\clubsuit, \color{red}{10\diamondsuit}$, you count it once when you select $7\clubsuit$ as the card you are drawing from the remaining $12$ ranks in the deck and $\color{red}{10\diamondsuit}$ as the card you are drawing from the remaining $11$ ranks in the deck and once when you select $\color{red}{10\diamondsuit}$ as the card you are drawing from the remaining $12$ ranks in the deck and $7\clubsuit$ as the card you are drawing from the remaining $11$ ranks in the deck. However, both selections result in the same hand. Therefore, you need to divide your answer by $2$. Notice that
$$\frac{1}{2}\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{1}\binom{11}{1}\binom{11}{1}\binom{4}{1} = \binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}^2$$
|
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|
Can anybody help me figure out what the author did in this task? In the expression
$$L = \lim_{x\to+\infty}x\bigg(\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\bigg(\frac{\ln(1+xe^{-x})}{x}+1\bigg)\bigg)$$
The author claims in the next step that
$$L = \lim_{x\to+\infty}x\bigg(\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\bigg)$$
Can anybody help me to understand how they got rid of
$$\frac{\ln (1+xe^{-x})}{x+1}?$$
|
$$L = \lim_{x\to+\infty}x\bigg(\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\bigg(\frac{\ln(1+xe^{-x})}{x}+1\bigg)\bigg)$$
This is equal to
$$L = \lim_{x\to+\infty}x\bigg(\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\frac{\ln(1+xe^{-x})}{x}\bigg)$$
Using the laws of limits, you can write
$$L = \lim_{x\to+\infty}x\bigg(\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\bigg)-\lim_{x\to +\infty}\ln(1+xe^{-x})\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}$$
Looking at the last term, it's equal to
$$\lim_{x\to +\infty}\ln(1+xe^{-x})\lim_{x\to +\infty}\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}$$
But $\frac{x}{e^x}=xe^{-x}\to 0$ as $x\to +\infty$ because exponential functions grow faster than polynomials. Hence, the last term is equal to $\ln(1)=0$. Because $\lim_{x\to +\infty}\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}=1$, the last expression vanishes and we have that
$$L = \lim_{x\to+\infty}x\bigg(\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\bigg)$$
|
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|
How to find limits of integral to find volume? Find the volume generated by the plane region, in the first quadrant, bounded by the graph of the function $ y=\sqrt{9-x^2} $ sbout the y-axis.
I know how to solve it using the formula but how do I get the upper and lower limits of the integration.
|
If I understood the question correctly, the volume of the solid they're asking you to find is a hemisphere. If you have to rotate it about the y-axis, use the shell method:
$$
V=2\pi\int_{0}^{3}x\sqrt{9-x^2}\,dx
$$
Your bounds of integration are from $0$ to $3$ because $\sqrt{9-x^2}$ is a circle of radius $3$ centered at the origin.
The integral itself:
$$
\int x\sqrt{9-x^2}\,dx=
-\frac{1}{2}\int\sqrt{9-x^2}\frac{d}{dx}(9-x^2)\,dx=\\
-\frac{1}{2}\int\sqrt{9-x^2}\,d(9-x^2)=
-\frac{1}{2}\frac{2\sqrt{(9-x^2)^3}}{3}=\\
-\frac{\sqrt{(9-x^2)^3}}{3}+C.
$$
The volume:
$$
V=2\pi\int_{0}^{3}x\sqrt{9-x^2}\,dx=
-2\pi\frac{\sqrt{(9-x^2)^3}}{3}\bigg|_{0}^{3}=\\
-2\pi\left(\frac{\sqrt{(9-3^2)^3}}{3}-\frac{\sqrt{(9-0^2)^3}}{3}\right)=\\
-2\pi\left(0-\frac{9\cdot3}{3}\right)=-2\pi(-9)=18\pi\ cubic\ units.
$$
|
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|
When is a matrix equal to its own inverse? When is a matrix equal to its own inverse?
If you have a $2\times2$ matrix and one if the entries is equal to $x$, for what values of $x$ is this matrix equal to its own matrix? And why?
|
The easiest way is just to write that
$$
A^2=\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)=\left(\begin{matrix}a^2+bc & (a+d)b\\(a+d)c & d^2+bc\end{matrix}\right)=I.
$$
Since $(a+d)b=(a+d)c=0$, either $a+d=0$ or $b=c=0$. In the latter case, we must have $a^2=d^2=1$. In the former case, we must have $a^2+bc=1$... either $a^2=1$ and $bc=0$, or else $a^2\neq 1$ and $bc=1-a^2$. So
$$
A=\left(\begin{matrix}\pm 1 & 0 \\ 0 & \pm 1\end{matrix}\right),
$$
or else
$$
A=\left(\begin{matrix}\pm 1 & b \\ 0 & \mp 1\end{matrix}\right) \text{ or } \left(\begin{matrix}\pm 1 & 0 \\ c & \mp 1\end{matrix}\right),
$$
or
$$
A=\left(\begin{matrix} a & b \\ (1-a^2)/b & -a\end{matrix}\right) \qquad (a^2\neq 1, b\neq 0).
$$
|
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|
Find the convergence radius for this power series The Problem: Find the convergence radius of $\sum_{n=0}^{\infty} \frac{n}{5^{n-1}} z^{\frac{(n)(n+1)}{2}}$
My attempts to find a solution I apply either the ratio test and end up with this expression:
$\lim_{n \to \infty} \frac{1}{5} |z|^{(n+1)} =L$
Since I need $L<1$ for the series to converge, my radius of convergence has to be $|z|<1$.
Is this correct? Or should I use other method?
Any help will appreciated
|
Observe we have
\begin{align}
\sum^\infty_{k=0}a_k z^k
\end{align}
where
\begin{align}
a_k =
\begin{cases}
\frac{n}{5^{n-1}}&\text{ if } k = \frac{n(n+1)}{2},\\
0 & \text{ otherwise }
\end{cases}.
\end{align}
Then by Cauchy-Hadamard theorem (i.e. root test), we see that
\begin{align}
\frac{1}{R}=\limsup_{k\rightarrow \infty} \sqrt[k]{|a_k|}.
\end{align}
By direct calculation, we see that
\begin{align}
\sqrt[k]{|a_k|} =
\begin{cases}
\left(\frac{n}{5^{n-1}}\right)^{\frac{2}{n(n+1)}}&\text{ if } k = \frac{n(n+1)}{2},\\
0 & \text{ otherwise }
\end{cases}.
\end{align}
In particular, we see that
\begin{align}
\lim_{n\rightarrow \infty} \left(\frac{n}{5^{n-1}}\right)^{\frac{2}{n(n+1)}} = 1
\end{align}
which means
\begin{align}
\limsup_{k\rightarrow \infty} \sqrt[k]{|a_k|} = 1.
\end{align}
Then $R=1$.
|
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|
Prove: $2^b-1\mid{}2^a-1 \iff b\mid{}a$ I figured we'd have to show $a=qb$ from $2^a-1=r(2^b-1)$ for $q,r\in\mathbb{Z}$ for the first implication and vice versa, how should I proceed from here?
|
Hint $:$ $2^{qb}-1 = (2^b -1) (2^{b(q-1)} + 2^{b(q-2)} + \cdots + 1).$
The above hint is enough to prove that $b \mid a \implies 2^b-1 \mid 2^a - 1.$
To prove that $2^b - 1 \mid 2^a - 1 \implies b \mid a$ use division algorithm to write $a = bq + r,$ where $0 \leq r < b.$ Then we have
$$\begin{align}
2^a - 1 & = 2^{bq+r} - 1. \\ & = 2^{bq+r} - 2^r + 2^r -1. \\ & = 2^r (2^{bq} -1) + 2^r - 1. \end{align}$$
Now if $2^b-1 \mid 2^a-1$ then since $2^b - 1\mid 2^r(2^{bq} -1)$ so we have $2^b - 1 \mid 2^r - 1,$ which is absurd if $0<r < b.$ Therefore we must have $r=0$ and hence we have $a=bq.$ Or in other words $b \mid a,$ as claimed.
QED
|
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|
Find $K=a^2b+b^2c+c^2a$ for roots $a>b>c$ of a cubic. If $a>b>c$ are the roots of the polynomial $P(x)=x^3-2x^2-x+1$ find the value of $K=a^2b+b^2c+c^2a$.
Using Vièta's formulas:
$$a+b+c=2$$
$$ab+bc+ca=-1$$
$$abc=-1$$
Using those I found that
$$a^2+b^2+c^2=6$$
$$a^3+b^3+c^3=11$$
$$a^2b+b^2c+c^2a+ab^2+bc^2+ca^2=1$$
but I can't separate K from it.
|
HINT: Consider the discriminant of your polynomial; what can you say about its square root given that $a>b>c$? Hover over the yellow box for a (more) complete solution.
The discriminant of your polynomial is $\Delta=49$, and because $a>b>c$ you also have
$$a^2b+b^2c+c^2a-a^2c-b^2a-c^2b=(a-b)(a-c)(b-c)=\sqrt{\Delta}=7.$$
You have already found that
$$a^2b+b^2c+c^2a+ab^2+bc^2+ca^2=1,$$
and from these two it is not hard to see that
$$K=a^2b+b^2c+c^2a=4.$$
|
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|
Why is $\frac{\frac{a}{b} - \frac{b}{a}}{\frac{a+b}{ab}}$ = a - b? I'm given the complex rational expression
$$\frac{\frac{a}{b} - \frac{b}{a}}{\frac{a+b}{ab}}$$
and asked to simplify. The solution provided is $a - b$; however I get $\frac{a^2 - b^2}{a+b}$.
My working:
Numerator first:
$$\frac{a}{b} - \frac{b}{a}$$
Least common denominator is $ab$.
Multiplying each part to get the LCD in the denominator on both sides I get:
$$\frac{a}{b} \cdot \frac{a}{a} - \frac{b}{a} \cdot \frac{b}{b}$$
$$= \frac{a^2 - b^2}{ab}$$
Multiplying this expression by the reciprocal in the original problem:
$$\frac{a^2 - b^2}{ab} \cdot \frac{ab}{a+b}$$
$$= \frac{ab(a^2-b^2)}{ab(a+b)}$$
Cancel out common factor $ab$:
$$\frac{a^2 - b^2}{a + b}$$
Where did I go wrong and how can I arrive at $a - b$?
|
You did nothing wrong. Factor the numerator $a^{2} - b^{2}$ into the difference the squares:
$$a^{2} - b^{2} = (a + b)(a - b)$$
This will leave with:
$$\frac{(a+b)(a-b)}{a+b}$$
Finally, cancel the $a+b$ expression from the numerator and denominator to obtain the desired result.
|
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|
A question about INMO 2017, Question 3
Find all triples $(x,a,b)$ where $x$ is a real number, and $a,b$ are integers belonging to $\{1,2,\dots,9\}$ such that $$x^2-a\{x\}+b=0$$ Here $\{x\}$ denotes the fractional part of $x$.
My contention, which I know is wrong, is that the equation has no solutions.
Reason: If $x$ has $0$ fractional part, this clearly has no solutions. Let us suppose now that $x$ has a fractional part that runs upto $n$ decimals. Then $\{x\}$ also has a fractional part which runs up to $n$ decimals. Consequently, $x^2$ has a fractional part that runs up to $2n$ decimals, and $a\{x\}$ has a fractional part that runs only up to $n$ decimal places (as $a$ is an integer). Hence, $x^2-a\{x\}$ has a non-zero fractional part (at least in the $n+1$ to $2n$ decimal places), which implies $x^2-a\{x\}+b$ can never be $0$ as it has a fractional part.
Where in this argument am I going wrong?
|
As for where in your argument you went wrong, as stated in the comments, you can't necessarily assume that $x$ is a terminating decimal.
As to determine what values will work, let
$$x = c + r \text{ where } c \in \mathbb{Z} \text{ and } 0 \le r \lt 1 \tag{1}\label{eq1}$$
This gives
$$\left(c + r\right)^2 - ar + b = 0 \tag{2}\label{eq2}$$
Expanding this and collecting into powers of $r$ terms gives
$$r^2 + \left(2c - a\right)r + \left(c^2 + b\right) = 0 \tag{3}\label{eq3}$$
First, note that as $r \ge 0$, $c^2 \ge 0$ and $b \gt 0$, this means that
$$2c - a \lt 0 \tag{4}\label{eq4}$$
Solving for $r$ using the quadratic formula gives
$$r = \frac{-\left(2c - a\right) \pm \sqrt{\left(2c - a\right)^2 - 4\left(c^2 + b\right)}}{2} \tag{5}\label{eq5}$$
The $r$ inequality in \eqref{eq1} gives that
$$0 \le -\left(2c - a\right) \pm \sqrt{\left(2c - a\right)^2 - 4\left(c^2 + b\right)} \lt 2 \tag{6}\label{eq6}$$
Due to \eqref{eq4}, plus that $c^2 + b \gt 0$, the left side of \eqref{eq6} will always be true. As such, consider the right side. From \eqref{eq4}, to be able to add the square root means that $2c - a = -1$. However, this makes the discriminant $1 - 4\left(c^2 + b\right) \lt 0$, so $r$ won't be real. Thus, only subtracting the square root can possibly work. Thus, moving the square root term to the right side and the $2$ to the left gives
$$-\left(2c - a\right) - 2 \lt \sqrt{\left(2c - a\right)^2 - 4\left(c^2 + b\right)} \tag{7}\label{eq7}$$
Next, since $2c - a < -1$, as discussed above, then both sides are non-negative so we can square each side to get
$$\left(2c - a\right)^2 + 8c - 4a + 4 \lt \left(2c - a\right)^2 - 4\left(c^2 + b\right) \tag{8}\label{eq8}$$
Simplifying by removing the common first term on both sides and dividing by $4$ gives
$$2c - a + 1 \lt -c^2 - b \tag{9}\label{eq9}$$
This becomes
$$c^2 + 2c + 1 \lt a - b \Rightarrow \left(c + 1\right)^2 \lt a - b \tag{10}\label{eq10}$$
As $\left(c + 1\right)^2 \ge 0$, this means $a \ge b$ and it limits the possible values of $c$ as $-2 \le c + 1 \le 2$ since $a - b \lt 9$, so $c$ can only be one of $-3, -2, -1, 0, 1$. You should be able to finish this now by checking these few values of $c$ to determine the appropriate values for $r$, thus $x$, plus $a$ and $b$, where the value $r$ obtained in \eqref{eq5} is real, with this requiring that $a^2 - 4ac - 4b \ge 0$.
|
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What is the number of pairs of $(a,b)$ of positive real numbers satisfying $a^4+b^4<1$ and $a^2+b^2>1$?
The number of pairs of $(a,b)$ of positive real numbers satisfying
$a^4+b^4<1$ and $a^2+b^2>1$ is -
$(i)$0
$(ii)$1
$(iii)$2
$(iv)$ more than 2
Solution:We have $a,b>0$,
According to the given situation,$0<a^4+b^4<1<a^2+b^2\implies a^4+b^4<a^2+b^2\implies 0<a^2(a^2-1)<b^2(1-b^2)\implies b^2(1-b)(1+b)<0 ;a^2(a-1)(a+1)>0\implies a\in(-\infty,-1)\cup (0,1);b\in(-1,0)\cup(1,\infty)$
But, in particular, if we choose $a=-1/2,b=1/2$,then $a^4+b^4=1/16+1/16=1/8<1$ and $a^2+b^2=1/4+1/4=1/2<1$.Which contradicts the given condition $a^2+b^2>1$
Where is the mistake in my approach? How should I approach this problem(means how to think ?). Can Triangle inequality be used here?
Please provide some hint...
|
Let $a^3+b^3=1$, there are uncountably many such pairs with $a\in (0,1)$ as we can set $a^3=\cos^2t, b^3=\sin^2t$.
Now for all these pairs, $a^4+b^4<a^3+b^3=1<a^2+b^2$ is obvious.
|
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|
Solving the linear first order differential equation? I was given the equation
$y' = -xy$, where $y(0) = 1$.
My solution was as follows:
$$\frac{dy}{dx} = -xy $$
$$dy = -xy \ dx $$
$$\int {dy} = \int {-xy dx} $$
$$y = -\frac{x^2}{2}y + c $$
$$y + \frac{x^2}{2}y = c $$
$$y(1 + \frac{x^2}{2}) = c $$
$$y = \frac{c}{1 + \frac{x^2}{2}} $$
I tried plugging in $0$ and $1$ respectively to find $c$, and got $c = 0$.
I just know that what I did is wrong. Where did I go wrong, please? Thanks!
|
(Too long for a comment.)
No one else has mentioned this, so I will...
Remember: $y$ is a function of $x$, not something that is constant with respect to variation of $x$. (Forgetting this is a fairly common error when first performing implicit differentiation.) It can help to write "$y(x)$" rather than just "$y$".
Your error is believing $\int -x y(x) \, \mathrm{d}x = \frac{-1}{2}x^2 y$. But $y$ is not some constant in this integral; it is a function that depends on $x$. Otherwise, what is written could be the absurdity (when, say $y(x) = \frac{1}{x^2}$), "$\int -x \cdot \frac{1}{x^2} \,\mathrm{d}x = \frac{-1}{2} x^2 \cdot \frac{1}{x^2} + C_1 = C$" rather than the correct
$$ \int -x \cdot \frac{1}{x^2} \,\mathrm{d}x = \ln|x| + C \text{.} $$
Quick way to verify this: Implicitly differentiate your antiderivative to see if you get the integrand and differential element back:\begin{align*}
\left( \frac{-1}{2}x^2 y(x) \right)'
&= \frac{-1}{2}x^2 (y(x))' + \left( \frac{-1}{2}x^2 \right)' y(x) \\
&= \frac{-1}{2}x^2 \,\mathrm{d}y(x) + \left( \frac{-1}{2} \cdot 2 x \,\mathrm{d}x \right) y(x) \\
&= \frac{-1}{2}x^2 \,\mathrm{d}y(x) - x y(x) \,\mathrm{d}x \text{,}
\end{align*}
which isn't quite "$- x y\,\mathrm{d}x$".
(However, we have shown \begin{align*}
\int \left( \frac{-1}{2}x^2 \frac{\mathrm{d}y(x)}{\mathrm{d}x} - x y(x) \right) \,\mathrm{d}x
&= \int \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{-1}{2}x^2 y(x) \right) \, \mathrm{d} x \\
&= \frac{-1}{2}x^2 y(x) + C \text{.}
\end{align*}
Familiarity with this kind of manipulation could be useful in the future.)
|
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|
How many four-digit integers are there that contain exactly one $8$? My book gives:
$$1 \times 9 \times 9 \times 9 + 3 \times 8 \times 9 \times 9 = 2673$$ integers
I don't understand where the $3$ comes from in the second part.
I understand the first part, $1 \times 9 \times 9 \times 9$. I believe it's saying we have $1$ choice to put $8$ in the beginning, $9$ choices of $\{0\} + \{1,2,3,4,5,6,7,9\}$ for the remaining three positions.
|
You made a good start. Indeed, if you fix the first spot as $8$, the second, third and fourth spot can be filled with any digit in $\{0,1,2,3,4,5,6,7,9\}$, hence you have $9^3$ possible choices.
It remains to count how many number you could have when the first spot is not $8$. Notice that, as Peter suggests above, usually the first digit is supposed to be not zero.
Fix the second spot to be $8$. Then you have only $8$ possible choices for the first digit, infact you can choose any number in $\{1,2,3,4,5,6,7,9\}$. The third and fourth spot can be filled with any number different to $8$, that is any number in $\{0,1,2,3,4,5,6,7,9\}$. Thus you have $8\times 9^2$ choices fixing the second spot to be $8$. The same reasoning can be done also when you fix the third spot to be $8$ and when you fix the fourth spot to be $8$. In other words, when the first spot in not $8$ you have $8\times 9^2 + 8\times 9^2+8\times 9^2=3\times 8\times 9^2$ possible choices.
Hence the total amount is $9^3+3\times 8\times 9^2$ numbers.
|
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|
integrate sin(x)cos(x) using trig identity. Book tells me the answer is:
$$ \int \sin(x)\cos(x) dx = \frac{1}{2} \sin^{2}(x) + C $$
however, I get the result:
$$
\sin(A)\cos(B) = \frac{1}{2} \sin(A-B)+\frac{1}{2}\sin(A+B)
$$
$$
\begin{split}
\int \sin(x)\cos(x) dx
&= \int \left(\frac{1}{2}\sin(x-x) + \frac{1}{2}\sin(x+x)\right) dx \\
&= \int \left(\frac{1}{2}\sin(0) + \frac{1}{2}\sin(2x)\right) dx \\
&= \int \left(\frac{1}{2}\sin(2x)\right) dx \\
&= -\frac{1}{2} \frac{1}{2}\cos(2x) +C\\
&= -\frac{1}{4} \cos(2x) +C
\end{split}
$$
How did the book arrive at the answer $\frac{1}{2}\sin^2(x)$?
|
Note that
$$\cos(2x)=1-2\sin^2x$$
From here you get
$$-\frac{1}{4}\cos(2x) + C = -\frac{1}{4} + \frac{1}{2}\sin^2 x+C = \frac{1}{2}\sin^2x + C_1$$
where $C_1 = -\frac{1}{4}+C$ is a new constant.
|
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|
What is the $n^{th}$ term derivative of $f(x) = (x^2-x-1)(\ln(1-x))$? I have the first three terms but am struggling with finding the $n^{th}$ term derivative of the function. Here is my work:
$$\\$$
$$f(x) = (x^2-x-1)(\ln(1-x)) $$
$$f'(x) = (2x-1)(\ln(1-x))-\left(\dfrac{x^2-x-1}{1-x}\right)$$
$$f''(x) = \dfrac{3x^2-5x+2(x-1)^2 \ln(1-x)+3}{(1-x)^2}$$
$$f'''(x) = \dfrac{2x^2-5x+1}{(x-1)^3}$$
$$f^{(n)}(x) = \ ?$$
|
It might be useful here to make a small substitution and to expand the parentheses:
*
*$y = 1-x \Rightarrow y' = -1$
*$\Rightarrow f(x) = g(y(x)) = -(1+x(1-x))\ln (1-x) = -(1+(1-y)y)\ln y$
Now differentiate $\boxed{g(y) = -\ln y - y\ln y + y^2 \ln y}$ with respect to $x$ and have in mind that $\color{blue}{y'(x) = -1}$ (which leads to the changing signs below):
\begin{eqnarray*}
f'(x) & = & \frac{1}{y} + (\ln y + 1) - (2y\ln y + y) \\
f''(x) & = & \frac{1}{y^2} - \frac{1}{y} + (2\ln y + 2) +1 \\
f^{(3)}(x) & = & \frac{2}{y^3} - \frac{1}{y^2} - \frac{2}{y}\\
& \ldots & \\
f^{(n)} & = & \ldots \\
\end{eqnarray*}
For example replacing $y = 1-x$ gives $f^{(3)}(x) = \frac{2}{(1-x)^3} - \frac{1}{(1-x)^2} - \frac{2}{1-x}$.
|
{
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|
Vector Fields on Lie Groups in a Coordinate Parameterisation My question is about how to write out vector fields on Lie groups in terms of a coordinate basis when a coordinate parametrisation of the group is given.
Consider the group $E(2)$ of Euclidian transformations. A general element can be written in coordinates as
$$g = \begin{bmatrix}\cos z & -\sin z & x \\ \sin z & \cos z & y \\ 0 & 0 & 1 \end{bmatrix} $$
where $(x,y,z) \in \mathbb{R}^2 \times S^1$. This gives an identification of the group with $\mathbb{R}^2 \times S^1$. We can get a general left-invariant vector field on the group by multiplying any element of the Lie algebra on the left by the above matrix. The Lie algebra is spanned by
$$e_1 = \begin{bmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \ \ e_2 = \begin{bmatrix}0 & 0 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}, \ \ e_3 = \begin{bmatrix}0 & 0 & 1 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{bmatrix} $$
So for instance there is a vector field
$$X = ge_1 = \begin{bmatrix}-\sin z & -\cos z & 0 \\ \cos z & -\sin z & 0 \\ 0 & 0 & 0 \end{bmatrix} $$
I would like to picture this vector field by plotting it as a vector field in $\mathbb{R}^2 \times S^1$. How can I write it out in a coordinate basis, i.e. $X = X_x \partial_x + X_y \partial_y + X_z \partial_z$?
|
Careful: you mean $SE(2)$, the group of proper rigid transformations.
If $g = \begin{bmatrix}\cos c & -\sin c & a \\ \sin c & \cos c & b \\ 0 & 0 & 1 \end{bmatrix}$ is a fixed group element and $h = \begin{bmatrix}\cos z & -\sin z & x \\ \sin z & \cos z & y \\ 0 & 0 & 1 \end{bmatrix}$ then
$$L_g(h) = \begin{bmatrix}\cos (c+z) & -\sin (c+z) & a + x\cos(c) - y\sin(c) \\ \sin (c+z) & \cos (c+z) & b + y\cos(c) + x \sin(c) \\ 0 & 0 & 1 \end{bmatrix},$$
or by identifying the group with $\mathbb R^2 \times S^1$, $$L_g(x,y,z) = (a+x \cos c - y\sin c, b + y\cos c + x \sin c, c+z).$$
Hence by differentiating, we see that the pushforward is $$dL_g = \begin{bmatrix}\cos c & -\sin c & 0
\\ \sin c & \cos c & 0\\ 0 & 0 & 1 \end{bmatrix}.$$
Now your $e_1 = \partial_z$, so
$$(dL_g)_I(e_1) = \begin{bmatrix}\cos c & -\sin c & 0
\\ \sin c & \cos c & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix} = \partial_z.$$
Compare e.g. with $e_2 = \partial_x + \partial_y$, where
$$\begin{align}(dL_g)_I(e_2) &= \begin{bmatrix}\cos c & -\sin c & 0
\\ \sin c & \cos c & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix} \\&= \begin{bmatrix} \cos c - \sin c \\ \sin c + \cos c \\ 0\end{bmatrix} \\&= (\cos c - \sin c)\partial_x + (\sin c + \cos c)\partial_y.\end{align}$$
|
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|
What is the value of $2x+3y$ if $x+y=7$ and $x^2-y^2=21$? If $x+y=7$ and $x^2-y^2=21$, then what is $2x+3y$?
I solved it like this:
\begin{align}
y& =7-x \\
x^2-(7-x)^2-21&=0 \\
x^2-49+14x-x^2-21&=0 \\
14x&=70 \\
x&=5
\end{align}
Then I solved for $y$ and I got $2$. I plugged in the values of $x$ and $y$ to $2x+3y$, and I got $16$.
Is this correct?
|
Hint: A slightly easier way.
$$x^2-y^2=(x+y)(x-y)=7(x-y)=21 \implies x-y=3$$
But your solution is correct.
|
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|
Proof of exponential law using limit definition of exponential function? For fun, I tried to prove the well-known exponential property $e^{a+b} = e^a e^b$ using the limit definition of the exponential function, below.
Definition. The exponential function is defined as follows.
$$e^x := \lim_{\epsilon \rightarrow 0} \left( 1 + \epsilon x \right)^{1/\epsilon}$$
I was able to outline the majority of the proof, however I do not have sufficient justification to go from line $\eqref 1$ to step $\eqref 2$. What limit properties might be used to fill in the blanks? I'd prefer not to use the binomial theorem or calculus-based argument, if possible (though if an expansion like that seems necessary, that is OK!).
Proof.
$$e^{a+b} = \lim_{\epsilon \rightarrow 0} \left( 1 + \epsilon (a+b) \right)^{1/\epsilon}$$
$$e^{a+b} = \lim_{\epsilon \rightarrow 0} \left( 1 + \epsilon a + \epsilon b \right)^{1/\epsilon} \tag 1 \label 1$$
$$\vdots$$
$$e^{a+b} = \lim_{\epsilon \rightarrow 0} \left( 1 + \epsilon a + \epsilon b
+ \epsilon^2 ab\right)^{1/\epsilon} \tag 2 \label 2$$
$$e^{a+b} = \lim_{\epsilon \rightarrow 0} \left( (1 + \epsilon a)(1 + \epsilon b)\right)^{1/\epsilon}$$
$$e^{a+b} = \lim_{\epsilon \rightarrow 0} \left(1 + \epsilon a \right)^{1/\epsilon} \left(1 + \epsilon b\right)^{1/\epsilon}$$
$$e^{a+b} = e^a e^b$$
|
Taking for granted that $e^x$ is well-defined in this way for all real $x$, it follows that for all real $x$:
$$
e^x = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n.
$$
For all real $a$ and $b$, and every positive integer $n$,
\begin{align*}
& \phantom{={}} \left\lvert\left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^n \!\! - \left(1+\frac{a}{n}+\frac{b}{n}\right)^n\right\rvert \\
& = \frac{|ab|}{n^2}\left\lvert\left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^{n-1} \!\! + \left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^{n-2}\left(1+\frac{a}{n}+\frac{b}{n}\right) + \cdots \right. \\
& \phantom{={}} \left. \cdots + \left(1+\frac{a}{n}+\frac{b}{n}\right)^{n-1}\right\rvert \\
& \leqslant \frac{|ab|}{n}\left(1+\frac{|a|}{n}+\frac{|b|}{n}+\frac{|ab|}{n^2}\right)^{n-1} = \frac{|ab|}{n}\left(1+\frac{|a|}{n}\right)^{n-1} \left(1+\frac{|b|}{n}\right)^{n-1} \\
& \leqslant \frac{|ab|}{n}\left(1+\frac{|a|}{n}\right)^n \left(1+\frac{|b|}{n}\right)^n,
\end{align*}
and this tends to zero as $n$ tends to infinity, because:
$$
\lim_{n\to\infty}\left(1+\frac{|a|}{n}\right)^n = e^{|a|}
\quad\text{and}\quad
\lim_{n\to\infty}\left(1+\frac{|b|}{n}\right)^n = e^{|b|}.
$$
Therefore:
\begin{align*}
e^ae^b & = \lim_{n\to\infty}\left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^n \\
& = \lim_{n\to\infty}\left[\left(\left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^n \!\! - \left(1+\frac{a}{n}+\frac{b}{n}\right)^n\right) + \left(1+\frac{a}{n}+\frac{b}{n}\right)^n\right] \\
& = \lim_{n\to\infty}\left(\left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^n \!\! - \left(1+\frac{a}{n}+\frac{b}{n}\right)^n\right) + \lim_{n\to\infty}\left(1+\frac{a}{n}+\frac{b}{n}\right)^n \\
& = 0 + e^{a+b} = e^{a+b}.
\end{align*}
|
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|
solve $\frac{\sqrt{5}(\cos \theta - \sin \theta)}{3\sqrt{2}}=\tan \theta$ I came across a question from another forum -
find the $x$ in the following diagram:
I managed to deduce an equation from the following diagram:
which is:
$\dfrac{\sqrt{5}(\cos \theta - \sin \theta)}{3\sqrt{2}}=\tan \theta$
and I know the answer (from WolframAlpha) is:
$\cos \theta= \dfrac{3}{\sqrt{10}}$
but I'm not able to deduce the answer myself, any ideas?
By the way, $x=2\sqrt{5}$, which can be easily deduced by:
$\dfrac{3\sqrt{2}}{x}=\cos\theta$
I also tried to solve the original question geometrically:
Somehow, I managed to figure out that $y=3$ in the above diagram, but I can't prove it either.
|
Defining $c := \cos\theta$ and $s := \sin\theta$, we can write
$$c^2\sqrt{5} = s\left(3 \sqrt{2} + c\sqrt{5}\right) \tag{1}$$
Squaring, re-writing $s^2 = 1-c^2$, and re-arranging,
$$10 c^4 + 6 c^3 \sqrt{10} + 13 c^2 - 6c \sqrt{10} - 18 = 0 \tag{2}$$
At this point, if we had the presence of mind to identify $10$ as $\sqrt{10}^2$ and $13$ as $-27 + 4\sqrt{10}^2$, then (defining $r:=\sqrt{10}$) we could gather terms and factor
$$\begin{align}
0 &= r^2 c^4 +\left(-3r + 9 r\right)c^3 + (-27+4r^2)c^2 +(-12r+ 6r)c - 18 \\[4pt]
&= \left(r^2 c^4 - 3rc^3\right) + \left(9 rc^3 -27c^2\right)+\left(4r^2c^2 -12rc\right)+ \left(6rc - 18\right)\\[4pt]
&= \left(r c - 3 \right) \left( rc^3 + 9 c^2 +4 r c + 6 \right) \tag{3}
\end{align}$$
(That is, we have factored over $\mathbb{Q}\left[\sqrt{10}\right]$.) The first factor yields the target root, $\cos\theta = 3/\sqrt{10}$. (Note that the second factor obviously has no positive solutions.)
Without such intuition, but with a suspicion that the $\sqrt{10}$s were impeding progress to a reasonably-nice solution, we could write $(2)$ as
$$10 c^4 + 13 c^2-18 = 6c\sqrt{10}\left(1-c^2\right) \tag{4}$$
Now, squaring will eliminate the pesky $\sqrt{10}$, and we have
$$100 c^8 - 100 c^6 + 529 c^4 - 828 c^2 + 324 = 0 \tag{5}$$
From here, old-fashioned factoring gives
$$\left(10 c^2 - 9\right) (10 c^6 - c^4 + 52c^2 -36 ) = 0\tag{6}$$
Again, the first factor gives the target root, $\cos\theta=3/\sqrt{10}$ (as well as a newly-introduced extraneous root, $\cos\theta=-3/\sqrt{10}$). It's not clear that the second factor has no valid roots; indeed, Mathematica gives the positive solution $\cos\theta = 0.80501\ldots$ (amid otherwise negative or non-real candidates), but we can check that it doesn't satisfy $(1)$.
|
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|
Proving $\lim\limits_{n{\rightarrow}\infty} \int_{0}^{n} \frac{{\sqrt{x}}\ln x}{1+x^2}dx=\pi$
Show that $$\lim\limits_{n{\rightarrow}\infty} \int_{0}^{n} \frac{x^{\frac{1}{2}}\ln x}{1+x^2}dx=\pi$$
After a number of transformations I ended having $$4\int_{0}^{\frac{1}{n^2}}\frac{\ln(x)}{x^4+1}dx$$ From here on I have
no idea on how to continue.
|
Here is an approach that will make use of the derivative of the Beta function $\operatorname{B} (m,n)$ of the form
$$\operatorname{B} (m,n) = \int_0^\infty \frac{x^{m - 1}}{(1 + x)^{m + n}} \, dx, \quad m,n > 0.$$
Let
$$I = \int_0^\infty \frac{\sqrt{x} \, \ln x}{1 + x^2} \, dx.$$
Enforcing a substitution of $x \mapsto \sqrt{x}$ leads to
$$I = \frac{1}{4} \int_0^\infty \frac{x^{-1/4} \ln x}{1 + x} \, dx.$$
Now consider
$$J(s) = \frac{1}{4} \int_0^\infty \frac{x^s}{1 + x} \, dx, \qquad - 1 < s < 0.$$
On differentiating with respect to $s$ we have
$$J'(s) = \frac{1}{4} \int_0^\infty \frac{x^s \ln x}{1 + x} \, dx.$$
For our integral we see that $I = J'(-1/4)$.
Now
\begin{align}
J(s) &= \frac{1}{4} \int_0^\infty \frac{x^{(s + 1) - 1}}{(1 + x)^{(s + 1) + (-s)}} \, dx\\
&= \frac{1}{4} \operatorname{B} (s + 1, -s)\\
&= \frac{1}{4} \Gamma (s + 1) \Gamma (-s).
\end{align}
From the reflexion formula for the Gamma function, namely
$$\Gamma (1 - x) \Gamma (x) = \frac{\pi}{\sin (\pi x)},$$
replacing $x$ with $-s$ gives
$$\Gamma (s + 1) \Gamma (-s) = - \pi \operatorname{cosec} (\pi s).$$
Thus
$$J(s) = -\frac{\pi}{4} \operatorname{cosec} (\pi s).$$
Differentiating with respect to $s$ gives
$$J'(s) = \frac{\pi^2}{4} \cot (\pi s) \operatorname{cosec} (\pi s).$$
Setting $s = -1/4$ gives
$$J \left (-\frac{1}{4} \right ) = \frac{\pi^2}{4} \cot \left (-\frac{\pi}{4} \right ) \operatorname{cosec} \left (-\frac{\pi}{4} \right ),$$
or
$$\int_0^\infty \frac{\sqrt{x} \, \ln x}{1 + x^2} \, dx = \frac{\pi^2}{2 \sqrt{2}}.$$
|
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|
How to evaluate a binomial sum with $2n$ in the exponent. The question is to evaluate the sum $$\sum_{k=0}^n\binom nk 3^{2n-k}$$
have tried fitting into the binomial form of $\binom nk \times x^k\times y^{n-k}$
but I can't seem to bring it to the correct form.
|
Using @lulu's hint:
\begin{align}
\sum_{k = 0}^{n} \binom{n}{k} 3^{2n - k}
& = \sum_{k = 0}^{n} \binom{n}{k} 3^{n + n - k}
= 3^{n} \sum_{k = 0}^{n} \binom{n}{k} 3^{n - k} \\
& = 3^n \sum_{k = 0}^{n} \binom{n}{k} 1^k 3^{n - k}
= 3^n (1 + 3)^n
= 3^n 4^n
= 12^n.
\end{align}
Alternatively, one could pull out the $3^{2n}$:
\begin{align}
\sum_{k = 0}^{n} \binom{n}{k} 3^{2n - k}
& = 3^{2n} \sum_{k = 0}^{n} \binom{n}{k} 3^{- k}
= 9^{n} \sum_{k = 0}^{n} \binom{n}{k} 1^{n - k} \left(\frac{1}{3}\right)^{k} \\
& = 9^n \left(1 + \frac{1}{3}\right)^n
= 9^n \frac{4^n}{3^n}
= 3^n 4^n
= 12^n.
\end{align}
|
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|
Evaluate $\iint_{\{(x,y)\mid (x-1)^2+y^2\leq1\}}\sqrt{x^2+y^2} dxdy$ [Background]: I'm trying to find the volume of the region bounded by the $xy$-plane, the cone $z^2=x^2+y^2$ and the cylinder $(x-1)^2+y^2=1$.
[Attempt]: I tried to use the polar coordinate:
\begin{align*}
\iint_{\{(x,y)\mid (x-1)^2+y^2\leq1\}}\sqrt{x^2+y^2} dxdy &= \int_0^{2\pi}\int_0^1 \sqrt{(1+r\cos{\theta})^2+(r\sin{\theta})^2}rdrd\theta\\
&= \int_0^{2\pi}\int_0^1 \sqrt{1+2r\cos\theta+r^2}r drd\theta
\end{align*}
Then I cannot continue. Could you give me a hint?
|
Hint Writing the domain in polar coordinates gives
$$(r \cos \theta - 1)^2 + (r \sin \theta)^2 \leq 1 .$$
Expanding and using the Pythagorean identity gives
$$r^2 - 2 r \cos \theta + 1 \leq 1,$$
rearranging gives
$$r^2 \leq 2 r \cos \theta ,$$
and for $r \neq 0$, dividing by $r$ gives
$$r \leq 2 \cos \theta .$$
So, in polar coordinates the integral becomes
$$\int_{-\pi / 2}^{\pi / 2} \int_0^{2 \cos \theta} \underbrace{r}_{\sqrt{x^2 + y^2}} \cdot \underbrace{r \,dr \,d\theta}_{dx\,dy} = \int_{-\pi / 2}^{\pi / 2} \int_0^{2 \cos \theta} r^2 \,dr \,d\theta .$$
|
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|
Locus of orthocenter of triangle inscribed in ellipse While messing around with ellipses in Geogebra, I found the following interesting result:
Let $\alpha$ be an ellipse. Let $AB$ be a fixed chord, and let $P$ be a point that moves freely on $\alpha$. Then as $P$ traces the ellipse, the orthocenter of triangle $PAB$ traces another ellipse(which passes through $AB$).
If you take the $\alpha$ to be of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, then interestingly the foci of the ellipse drawn by the orthocenter is parallel to the $y$-axis. The two ellipses also seem to have the same eccentricity.
I believe a proof could be along the lines of applying the linear transformation given by the matrix $\begin{bmatrix} \frac{1}{a} && 1 \\ 1 && \frac{1}{b} \end{bmatrix}$ which takes the ellipse to the unit circle, solving the locus in the circle case (which turns out to be just the unit circle reflected after $AB$) and then applying the reverse transformation, but it does not seem to work very well for some reason.
I also thought about parametrizing the points and using complex numbers; I obtained a formula for the orthocenter in terms of an angle $\phi$ (where $P = (a\cos(\phi), b\sin(\phi))$ but it seems very unwieldly to show that the points form an ellipse, especially since I haven't been able to guess the real part of the foci.
Any ideas would be welcome.
image
|
With the help of Mathematica, I was able to confirm your suspicion.
Let the ellipse be parameterized by $(a \cos\theta, b \sin\theta)$, and let $A$ and $B$ be points corresponding to $\theta = 2\alpha$ and $\theta = 2\beta$. After some symbol-crunching, we find that the orthocenter $(x,y)$ satisfies
$$\begin{align}
ax &= \left(\; a^2\sin^2(\alpha+\beta) + b^2 \cos^2(\alpha + \beta) \;\right)\cos\theta \\[4pt]
&+\left(a^2-b^2\right)\cos(\alpha+\beta)\sin(\alpha+\beta) \sin\theta \\[4pt]
&+\left(a^2 + b^2\right) \cos(\alpha-\beta)\cos(\alpha+\beta) \\[18pt]
-by&=\left(a^2 - b^2\right) \cos(\alpha+\beta) \sin(\alpha+\beta) \cos\theta \\[4pt]
&- \left(\; a^2 \sin^2(\alpha+\beta)+ b^2 \cos^2(\alpha + \beta) \;\right)\sin\theta \\[4pt]
&- \left(a^2 + b^2\right) \cos(\alpha-\beta) \sin(\alpha+\beta)
\end{align} \tag{1}$$
Solving system $(1)$ for $\cos\theta$ and $\sin\theta$, substituting into $\cos^2\theta+\sin^2\theta=1$, and simplifying, we obtain the equation of a new ellipse:
$$\begin{align}
&\phantom{+\;\;}a^2 \left(x - \frac{a^2 + b^2}{a} \cos(\alpha-\beta) \cos(\alpha+\beta)\right)^2 \\[4pt]
&+b^2 \left(y - \frac{a^2 + b^2}{b} \cos(\alpha-\beta) \sin(\alpha+\beta)\right)^2 \\[4pt]
&=a^4 \sin^2(\alpha+\beta) + b^4 \cos^2(\alpha+\beta)
\end{align} \tag{$\star$}$$
Since the horizontal and vertical radii are proportional to $1/a$ and $1/b$, we see that this ellipse's major and minor axes are perpendicular to the original's axes, respectively. Moreover, for $a\geq b$, the eccentricity is $\sqrt{a^2-b^2}/a$, which matches that of the original ellipse.
More generally, we can take a conic of eccentricity $e$, parameterized by
$$\frac{p}{1+e \cos\theta}\left(\cos\theta,\sin\theta\right) $$
whose focus is at the origin and whose major/transverse axis coincides with the $x$-axis. Taking $A$ and $B$ to correspond to $\theta=\alpha$ and $\theta = \beta$, we can perform the same kind of analysis as above to obtain a comparable conic:
$$\begin{align}
&\phantom{+\;\;}
x^2 \phantom{\left( 1 - e^2 \right) }( 1 + e \cos\alpha ) ( 1 + e \cos\beta ) \\
&+ y^2 \left( 1 - e^2 \right) ( 1 + e \cos\alpha ) ( 1 + e \cos\beta ) \\
&- x p \left( \left( 2 + e^2 \right) \left( \cos\alpha + \cos\beta \right) + 2 e \left(1 + 2 \cos\alpha \cos\beta \right) \right) \\
&- y p \left( 2 - e^2 \right) ( \sin\alpha + \sin\beta + e \sin(\alpha+\beta) ) \\
= &- p^2 \left( 1 + 2\cos(\alpha-\beta) + e (\cos\alpha + \cos\beta ) - e^2 \sin\alpha \sin\beta \right)
\end{align}$$
For $e=1$, the original conic is a parabola; we see that the new conic is, too, since its $y^2$ term vanishes. Otherwise, the horizontal and vertical radii of the new conic are proportional to $1$ and $1/|1-e^2|$, respectively, and we deduce that the new eccentricity is also $e$. In all cases, we see that the new conic is rotated $90^\circ$ with respect to the original.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Showing that $\arctan x = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)$ Since I don't have the answer to this one, I want to make sure I've done this correctly.
Show that
$$\arctan x = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)$$
Since
$$\arctan x = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)$$
we have
$$\tan\left(\arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)\right) = x$$
Applying the Pythagorean theorem, we learn that $b = 1$, so we have $\tan(x/1) = x$.
It may be a bit short, but I really want to be sure I have this right before I continue on. :)
|
Let's consider the definitions of arcsine and arctangent.
Let $\arctan x = \theta$. Then $\theta$ is the unique angle in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\tan\theta = x$.
Let
$$\arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right) = \varphi$$
Then $\varphi$ is the unique angle in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ such that
$$\sin\varphi = \frac{x}{\sqrt{1 + x^2}}$$
We need to show that $\theta = \varphi$.
Observe that
$$\left(\frac{x}{\sqrt{1 + x^2}}\right)^2 = \frac{x^2}{1 + x^2} < 1$$
for every real number $x$ since
$$\frac{x^2}{1 + x^2} < 1 \iff x^2 < 1 + x^2 \iff 0 < 1$$
Hence,
$$-1 < \frac{x}{\sqrt{1 + x^2}} < 1 \implies -\frac{\pi}{2} < \varphi = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right) < \frac{\pi}{2}$$
By the Pythagorean identity $\sin^2\varphi + \cos^2\varphi = 1$,
\begin{align*}
\cos^2\varphi & = 1 - \sin^2\varphi\\
& = 1 - \left(\frac{x}{\sqrt{1 + x^2}}\right)^2\\
& = 1 - \frac{x^2}{1 + x^2}\\
& = \frac{1 + x^2 - x^2}{1 + x^2}\\
& = \frac{1}{1 + x^2}
\end{align*}
Since $-\frac{\pi}{2} < \varphi < \frac{\pi}{2}$, $\cos\varphi > 0$, so we take the positive square root. Thus,
$$\cos\varphi = \frac{1}{\sqrt{1 + x^2}}$$
Hence,
\begin{align*}
\tan\varphi & = \frac{\sin\varphi}{\cos\varphi}\\
& = \frac{\frac{x}{\sqrt{1 + x^2}}}{\frac{1}{\sqrt{1 + x^2}}}\\
& = x\\
& = \tan\theta
\end{align*}
Since $\theta$ is the unique angle in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\tan\theta = x$, $\tan\varphi = x$, and $-\frac{\pi}{2} < \varphi < \frac{\pi}{2}$, we may conclude that $\theta = \varphi$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3137453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Pattern for Induction?
Find and prove a formula for the sum
$$\frac{1^3}{1^4+4} - \frac{3^3}{3^4+4}+...+\frac{(-1)^n(2n+1)^3}{(2n+1)^4 + 4}$$ where $n$ is an integer.
I tried listing out the partial sums of the sequence to see if there was a pattern, however the only thing I can make out is that the denominators of the sum always either end in 5 or 9.
I would just like a hint as to how to solve this.
Thanks.
|
Extended hint:
Writing down 4-5 first partial sums reveals after reducing the fractions the general pattern:
$$
S_n=(-1)^n\frac{n+1}{4(n+1)^2+1}.
$$
It remains to apply induction or telescoping. The following facts will be useful:
$$\begin{array}{ll}
(2n+1)^4+4&=(4n^2+1)(4n^2+8n+5),\\
4n^2+8n+5&=4(n+1)^2+1,\\
(2n+1)^3-n(4n^2+8n+1)&=(n+1)(4n^2+1).
\end{array}
$$
$\frac{(2n+1)^3}{(2n+1)^4+4}=\frac{n}{4n^2+1\vphantom{)^2}}+\frac{n+1}{4(n+1)^2+1}.$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3137706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Eigenvalue by Gaussian elimination? Let $A=\begin{bmatrix}-1&4&0\\-2&4&-1\\2&-5&0 \end{bmatrix}$
I want to find out the eigenvalue of matrix my Gaussian elimination. My idea was to make the matrix triangular and by that being able to write an expression for the determinant that equals 0. Like this: $(aλ+a_1)(bλ+b_1)(cλ+c_1)$
My try:
$ \det \begin{vmatrix} λ+1&-4&0\\2&λ-4&1\\-2&5&λ \end{vmatrix} = \begin{vmatrix} λ+1&0&0\\2&λ-4+\frac{8}{(λ+1)}&1\\-2&5-\frac{8}{(λ+1)}&λ \end{vmatrix}=
\begin{vmatrix} λ+1&0&0\\2&λ-4+\frac{8}{(λ+1)}&0\\-2&5-\frac{8}{(λ+1)}&\frac{5-\frac{8}{(λ+1)}}{-4+\frac{8}{(λ+1)}} \end{vmatrix}$,
which would give the equation: $ (λ+1)(λ-4+\frac{8}{(λ+1)})(\frac{5-\frac{8}{(λ+1)}}{-4+\frac{8}{(λ+1)}})=0 $
Now solving each of the parentheses for λ, should give me the right Eigenvalues if I understand the concept of eigenvalues. Am I doing anything wrong or do i simply not understand eigenvalues yet?
|
Here is a simple way to obtain the characteristic polynomial:
\begin{align}
&\begin{vmatrix}
\lambda+1&-4&0\\ 2&\lambda- 4&\phantom{-}\\-2&5&\lambda
\end{vmatrix} =
\begin{vmatrix}
\lambda+1&-4&0\\ 0&\lambda+1&\lambda+1\\-2&5&\lambda
\end{vmatrix} =
(\lambda+1)\begin{vmatrix}
\lambda+1&-4&0\\ 0 & 1 & 1\\-2&5&\lambda
\end{vmatrix}\\[1ex]
={}
(\lambda+1)&\begin{vmatrix}
\lambda+1&-4&4\\ 0 & 1 & 0\\-2&5&\lambda -5
\end{vmatrix} = (\lambda+1)\begin{vmatrix}
\lambda+1&4 \\-2&\lambda -5
\end{vmatrix}=(\lambda+1)\bigl((\lambda+1)(\lambda-5)+8\bigr)\\[1ex]
=(\lambda+1)&(\lambda^2-4\lambda+3)=(\lambda+1)(\lambda-1)(\lambda-3).
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3137831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Help finding the derivative of $f(x) = \cos{\left(\sqrt{e^{x^5} \sin{x}}\right)}$ I am trying to find the derivative of $f(x) = \cos(\sqrt{(e^{x^5} \sin(x)})$.
I keep getting the wrong answer, and I'm not sure what I'm doing wrong.
$$\frac{d}{dx} e^{x^5} = e^{x^5} \cdot 5x^4$$
$$\frac{d}{dx} \sin(x) = \cos(x)$$
$$\frac{d}{dx} (e^{x ^5} \cdot \sin(x)) = [(e^{x^5} \cdot 5x^4) \cdot \sin(x)] + [\cos(x) \cdot e^{x^5}]$$
$$\frac{d}{dx} \sqrt{x} = \frac{1}{2\sqrt{x}}$$
$$\frac{d}{dx} \sqrt{(e^{x^5} \sin(x))} = \frac{[(e^{x^5} \cdot 5x^4) \cdot \sin(x)] + [\cos(x) \cdot e^{x^5}]}{2 \sqrt{e^{x^5} \sin(x)}}$$
Therefore, since $\frac{d}{dx} \cos(x) = -\sin(x)$, I have
$$ f'(x) = -\sin(\sqrt{e^{x^5} \sin(x)}) \cdot \frac{[(e^{x^5} \cdot 5x^4) \cdot \sin(x)] + [\cos(x) \cdot e^{x^5}]}{2 \sqrt{e^{x^5} \sin(x)}}$$
However, the website I'm using, "WeBWorK", says this is incorrect.
|
I differentiated your function and did not look at your result. As you can see, they're identical:
$$\begin{align}
\frac{d}{dx}\left[\cos\left(\sqrt{e^{x^5}\cdot\sin{x}}\right)\right]
&=-\sin{\left(\sqrt{e^{x^5}\cdot\sin{x}}\right)}\left(\sqrt{e^{x^5}\cdot\sin{x}}\right)'\\
&=-\sin{\left(\sqrt{e^{x^5}\cdot\sin{x}}\right)}\frac{1}{2\sqrt{e^{x^5}\cdot\sin{x}}}\left(e^{x^5}\cdot\sin{x}\right)'\\
&=-\sin{\left(\sqrt{e^{x^5}\cdot\sin{x}}\right)}\frac{1}{2\sqrt{e^{x^5}\cdot\sin{x}}}\left[\left(e^{x^5}\right)'\cdot\sin{x}+e^{x^5}\cdot\left(\sin{x}\right)'\right]\\
&=-\sin{\left(\sqrt{e^{x^5}\cdot\sin{x}}\right)}\frac{1}{2\sqrt{e^{x^5}\cdot\sin{x}}}\left[e^{x^5}\left(x^5\right)'\cdot\sin{x}+e^{x^5}\cdot\cos{x}\right]\\
&=-\sin{\left(\sqrt{e^{x^5}\cdot\sin{x}}\right)}\frac{1}{2\sqrt{e^{x^5}\cdot\sin{x}}}\left[5e^{x^5}x^4\cdot\sin{x}+e^{x^5}\cdot\cos{x}\right]
\end{align}$$
Your differentiation skills are fine. It's the problem with the website that you're suing.
|
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"url": "https://math.stackexchange.com/questions/3138417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
A simpler method to prove $\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c} = 1?$ This is the way I think about it:
$
1 = \log_aa = \log_bb = \log_cc \\~\\
\textbf{Using the ‘change of base rule':} \\
\log_{a}{b} = \frac{\log_{b}{b}}{\log_{b}{a}}, \hspace{0.3cm} \log_{b}{c} = \frac{\log_{c}{c}}{\log_{c}{b}}, \hspace{0.3cm} \log_{c}{a} = \frac{\log_{a}{a}}{\log_{a}{c}}\\~\\
\rightarrow \log_{b}{a} = \frac{\log_{b}{b}}{\log_{a}{b}}, \hspace{0.3cm} \log_{c}{b} = \frac{\log_{c}{c}}{\log_{b}{c}}, \hspace{0.3cm} \log_{a}{c} = \frac{\log_{a}{a}}{\log_{c}{a}}\\~\\
\rightarrow \large\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}\rightarrow\\~\\
\rightarrow\frac{\log_{b}{b}}{\log_{a}{b}} \times \frac{\log_{c}{c}}{\log_{b}{c}} \times \frac{\log_{a}{a}}{\log_{c}{a}} \overset? = 1\\
\rightarrow \frac{1}{\log_{a}{b} \times \log_{b}{c} \times \log_{c}{a}} \overset? = 1\\
\rightarrow \log_{a}{b} \times \log_{b}{c} \times \log_{c}{a} = 1\\
\rightarrow \log_{a}{b} \times \log_{b}{c} \times \log_{c}{a} \overset? = \log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}\\
\rightarrow \frac{\log_{a}{b} \times \log_{b}{c} \times \log_{c}{a}}{\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}} = 1 \\
\rightarrow \frac{\log_{a}{b}}{\log_{b}{a}} \times \frac{\log_{b}{c}}{\log_{c}{b}} \times \frac{\log_{c}{a}}{\log_{a}{c}} = 1\\~\\
\small \text{------ Using the ‘change of base rule' again ------} \\~\\
\Large \therefore \hspace{0.2cm}\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c} = 1
$
Any other methods?
Thanks in advance.
|
Well it is much simpler. By definition
$$\log_b(a) = \frac{\ln(a)}{\ln(b)}, \quad \log_c(b) = \frac{\ln(b)}{\ln(c)}, \quad log_a(c) = \frac{\ln(c)}{\ln(a)}$$
where $\ln$ is the logarithm in base $e$.
Your identity directly follows.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate integral $\int \sin^4(t)\cos^3(t)dt$ $$\int \sin^4(t)\cos^3(t)dt = \int \sin^4(t)(1-\sin^2(t))\cos(t) dt $$
$$u = \sin(t) \\ du = \cos(t)dt$$
$$ \int \sin^4(t)\cos^3(t)dt = \int u^4(1-u^2) du \\
= u^4 - u^6 = \frac{1}{5}u^5 - \frac{1}{7}u^7 + C \\
= \frac{1}{5}\sin^5(t) - \frac{1}{7}\sin^7(t) + C $$
This seems like a simple enough trig substitution integral problem to me. However, when I check my answer with wolphram alpha, it gives:
This looks like a simplified version of my answer, but it is not entirely clear to me how it gets reduced down. The furthest I can get is this:
$$ \frac{1}{5}\sin^5(t) - \frac{1}{7}\sin^7(t) = \sin^5(t) \bigg( \frac{1}{5} - \frac{1}{7} sin^2(t) \bigg) \\
= \frac{1}{5}\sin^5(t) \bigg(\frac{1}{5} - \frac{1}{7} \cdot \frac{1}{2} (1 - \cos 2x) \bigg) \\
= \frac{1}{5}\sin^5(t) \bigg(\frac{1}{5} - \frac{1}{14} - \frac{1}{14}\cos 2x\bigg) \\
= \frac{1}{5}\sin^5(t) \bigg(\frac{14}{90} - \frac{5}{90} - \frac{5}{90} \cos 2x \bigg)$$
and I feel like my simplification is not really going anywhere meaningful...
|
Use the formula $$\int \cos^m(t)\sin^n(t)dt=-\frac{\cos^{m+1}(t)\sin^{n-1}(t)}{m+n}+\frac{n-1}{m+n}\int\cos^m(t)\sin^{n-2}(t)dt$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3146621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Closed form representation of alternating series I consider the following series
\begin{align}
\sum_{n=0}^\infty{\frac{(-1)^{n+2}}{n+2}} \stackrel{?}{=} 1-\ln(2)
\end{align}
Wolfram tells me that it is equal to $1-\ln(2)$.
I know the following
\begin{align}
&\sum_{k=1}^\infty{\frac{(-1)^{k+1}}{k}} = \ln(2)\\
&\sum_{n=0}^\infty{\frac{(-1)^{n+2}}{n+2}} = \sum_{k=1}^\infty{\frac{(-1)^{k+1}}{k+1}}
\end{align}
I thus need to show that the following is true, but I would not know how to do it.
\begin{align}
\sum_{k=1}^\infty{\frac{(-1)^{k+1}}{k+1}} + \sum_{k=1}^\infty{\frac{(-1)^{k+1}}{k}} = 1
\end{align}
|
\begin{align}
\sum_{k=1}^\infty{\frac{(-1)^{k+1}}{k+1}} + \sum_{k=1}^\infty{\frac{(-1)^{k+1}}{k}}
&=
\sum_{k=1}^\infty{\frac{(-1)^{k+1}}{k+1}} + \sum_{k=0}^\infty{\frac{(-1)^{k}}{k+1}}
\\ &=
\sum_{k=1}^\infty{\frac{(-1)^{k+1}}{k+1}} + \sum_{k=1}^\infty{\frac{(-1)^{k}}{k+1}}+1
\\ &=1
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Closed form of the sum of two binomial expansions I would like to know if there is a simple closed form for the following expression:
$(x+y)^n + (x+z)^n$
Expanding the above I get
$(y^n + z^n) + nx(y^{n-1}+z^{n-1}) + \frac{(n-1)n}{2}x^2(y^{n-2} + z^{n-2}) + \mathcal{O}(x^3) $,
but it isn't clear to me if this can factor into something of the form $(x+w)^n + u$ where $w$ and $u$ depend on $y$ and $z$ (in general I mean, it is fairly easy to get something of this form for $n=1$ or $n=2$).
|
Maybe you are looking for a symmetric formulation like this
$$
\eqalign{
& \left( {x + y} \right)^{\,n} + \left( {x + z} \right)^{\,n} = \cr
& = \left( {x + \left( {{{z + y} \over 2}} \right) - \left( {{{z - y} \over 2}} \right)} \right)^{\,n}
+ \left( {x + \left( {{{z + y} \over 2}} \right) + \left( {{{z - y} \over 2}} \right)} \right)^{\,n} = \cr
& = \left( {x + \left( {{{z + y} \over 2}} \right)} \right)^{\,n}
\left( {\left( {1 - \left( {{{z - y} \over {2x + y + z}}} \right)} \right)^{\,n} + \left( {1 + \left( {{{z - y} \over {2x + y + z}}} \right)} \right)^{\,n} } \right) \cr}
$$
|
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|
Find the number of roots of the equation, $x^3 + x^2 +2x +\sin x = 0$ in $[-2\pi , 2\pi]$.
Find the number of roots of the equation,
$$x^3 + x^2 +2x +\sin x = 0$$
in $[-2\pi , 2\pi]$.
What I have tried:
$$x^3 + x^2 +2x = -\sin x$$
$$x^2 +x +2 = \frac{-\sin x }{x}$$
$$(x + \frac{1}{2})^2 + \frac{7}{4} = \frac{-\sin x }{x}$$
I am getting somewhere from here, but I don't know how to continue. Please help!
The answer is 1
|
For $x<0, \sin x=x-\frac{x^3}{3!}+O(x^5)>x$:
$$x^3+x^2+2x-\sin x <x^3+x^2+2x-x=x(x^2+x+1)<0,$$
for $x>0, \sin x<x$:
$$x^3+x^2+2x-\sin x>x^3+x^2+2x-x=x^3+x^2+x>0.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How is the partial sum of a geometric sequence calculated? I have been asked to calculate a general formula for $1 + b + 2b^2 + 3b^3 + ... + Nb^N$
I think that a part of this is a geometric sequence, and I have rewritten this as
$f(x) = 1 + \sum_{i=1}^n i\cdot x^i$
(I couldn't figure out a way to make the first term be 1 with the progression so I just removed it from the series instead)
When I look on Wolfram Alpha it says that the partial sum formula for $ \sum_{i=1}^n i\cdot x^i$ is:
$$\sum_{i=1}^n i\cdot x^i = \frac{(nx-n-1)x^{n+1}+x}{(1-x)^2}$$
On this question, an answer said that the general formula for the sum of a finite geometric series is:
$$\sum_{k=0}^{n-1}x^k = \frac{1-x^n}{1-x}$$
But if I substitute my ($i\cdot x^i$) into the formula mentioned above I don't get the same value as the one Wolfram gives me.
Two questions:
*
*Is it correct to pull the first term out of the series so it becomes a geometric series, or is there another way?
*How did Wolfram Alpha calculate that expression?
|
Start with $$S=1 + b + 2b^2 + 3b^3 + \cdots + Nb^N$$
Multiply by $b$ and then subtract to give
$$bS=b + b^2 + 2b^3 + 3b^4 + \cdots +(N-1)b^N + Nb^{N+1}$$
$$(b-1)S = -1 - b^2 - b^3 - b^4 - \cdots -b^N + Nb^{N+1}$$
Do that again
$$b(b-1)S = -b - b^3 - b^4 - b^5 - \cdots -b^N-b^{N+1} - Nb^{N+2}$$
$$(b-1)^2S = 1-b +b^2 -(N+1)b^{N+1} + Nb^{N+2}$$
Finally simplify
$$S = \dfrac{1-b +b^2 -(N+1)b^{N+1} + Nb^{N+2}}{(b-1)^2}.$$
As an example, with $b=2$ and $N=3$ the first line gives $S=1+2+8+24=35$ while the final line gives $S=\frac{1-2+4-64+96}{1^2}= 35$
|
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|
How to prove that function $f(p)= \frac{1}{(p^2 - 1)} \sum_{q=3}^p \frac{q^2-3}{q}[q\text{ is prime}]$ is less than 0.3 for all $p$? How to prove that function $f(p)= \frac{1}{(p^2 - 1)} \sum_{q=3}^p \frac{q^2-3}{q}[q\text{ is prime}]$ is less than 0.3 for all $p$?
Let's suppose that we have a function defined as follows:
$f(p) = \frac{1}{(p^2 - 1)} \sum_{q=3}^p \frac{q^2-3}{q}[q\text{ is prime}]$
where $p$ is a prime number and $[...]$ are Iverson brackets.
Here are the first few values of $f(p)$:
$f(3) = \frac{1}{(3^2 - 1)} \times \frac {(3^2-3)}{3}$
$f(3) = \frac{1}{8} \times \frac{6}{3}$
$f(3) = 0.25$
$f(5) = \frac{1}{(5^2 - 1)} \times [ \frac{(3^2-3)}{3} +\frac{(5^2-3)}{5} ]$
$f(5) = \frac{1}{24} \times [ \frac{6}{3} +\frac{22}{5} ]$
$f(5) = 0.26667$
$f(7) = \frac{1}{(7^2 - 1)} \times [ \frac{(3^2-3)}{3} +\frac{(5^2-3)}{5} +\frac{(7^2-3)}{7}]$
$f(7) = \frac{1}{48} \times [ \frac{6}{3} +\frac{22}{5} +\frac{46}{7}]$
$f(7) = 0.273214286$
$f(11) = \frac{1}{(11^2 - 1)} \times [ \frac{(3^2-3)}{3} +\frac{(5^2-3)}{5} +\frac{(7^2-3)}{7}+\frac{(11^2-3)}{11}]$
$f(11) = \frac{1}{120} \times [ \frac{6}{3} +\frac{22}{5} +\frac{46}{7}+\frac{118}{11}]$
$f(11) = 0.198679654$
How to prove that $f(p)$ is always less than 0.3 for all prime numbers $p$?
I wrote a program to calculate $f(p)$ for the first $1000$ primes up to $p = 7,927$ and graphically it appears to approach $0$ with a maximum value of $0.2732$ at $p=7$. From the graph, it looks like it should be easy, but for some reason, I cannot figure it out.
Notice that $f(3) < f(5) < f(7)$, but $f(7) > f(11)$ so a proof by induction may not work.
|
$$f(p)=\frac{1}{p^2-1}\sum_{q=3}^p\frac{q^2-3}{q}$$
If we use the fact that:
$$\sum_{q=3}^p\frac{q^2-3}{q}=\sum_{q=3}^pq-3\sum_{q=3}^p\frac1q$$
Now we know that:
$$\sum_{q=3}^pq=\sum_{q=1}^pq-\sum_{q=1}^2q=\frac{p(p+1)}{2}-3$$
$$-3\sum_{q=3}^p\frac1q=-3\left[\sum_{q=1}^p\frac1q-\sum_{q=1}^2\frac1q\right]=-3\left[\ln(p)+\gamma+\frac{1}{2p}-\frac32\right]$$
If we add these together we get:
$$f(p)=\frac{1}{p^2-1}\left[\frac{p(p+1)}{2}-3\left(\ln(p)+\gamma+\frac{1}{2p}-\frac12\right)\right]$$
Now if you find $f'(p)$ you can find a maximum value and find what this is
|
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|
Find the number of $k$ satisfying $\frac{n}{3} - 1 < k \leq \frac{n}{2}$ for $n > 9$. Show that the number of $k \in \mathbb{N}$ satisfying $\frac{n}{3} - 1 < k \leq \frac{n}{2}$ for $n \in \mathbb{N}, n> 9$ is greater than $\frac{n}{6}$.
I looked at the distance $|\frac{n}{3} - 1 - (\frac{n}{2})| = 1 + \frac{n}{6}$ which is $> \frac{n}{6}.$ Is this correct or am I missing a proper justification? I'm not sure how to use the $n > 9$ fact either. Thank you.
|
Let $n = 6m + r$ for some non-negative integer $m$ and integer $0 \le r \le 5$. Thus, the question asks to prove there are more than $\frac{n}{6} = m + \frac{r}{6} \lt m + 1$ values of $k$, i.e., that there are at least $m + 1$ such values.
The inequality becomes
$$2m + \frac{r}{3} - 1 \lt k \le 3m + \frac{r}{2} \tag{1}\label{eq1}$$
In general for large divisors, you would normally look at various spans of values of $r$. However, since there aren't too many here, you can instead fairly easily manually check the results of the $6$ possible values of $r$ to get:
$$r = 0 \; \Rightarrow \; 2m - 1 \lt k \le 3m \tag{2}\label{eq2}$$
$$r = 1 \; \Rightarrow \; 2m - \frac{2}{3} \lt k \le 3m + \frac{1}{2}\tag{3}\label{eq3}$$
$$r = 2 \; \Rightarrow \; 2m - \frac{1}{3} \lt k \le 3m + 1 \tag{4}\label{eq4}$$
$$r = 3 \; \Rightarrow \; 2m \lt k \le 3m + \frac{3}{2} \tag{5}\label{eq5}$$
$$r = 4 \; \Rightarrow \; 2m + \frac{1}{3} \lt k \le 3m + 2 \tag{6}\label{eq6}$$
$$r = 5 \; \Rightarrow \; 2m + \frac{2}{3} \lt k \le 3m + \frac{5}{2} \tag{7}\label{eq7}$$
With \eqref{eq2} and \eqref{eq3}, $k$ goes from $2m$ to $3m$ inclusive, so there are $m + 1$ values which work. Similarly with \eqref{eq4}, $k$ goes from $2m$ to $3m + 1$, so there are actually $m + 2$ values which work. With \eqref{eq5}, $k$ goes from $2m + 1$ to $3m + 1$, for $m + 1$ values. Finally, \eqref{eq6} & \eqref{eq7} give $k$ going from $2m + 1$ to $3m + 2$, for $m + 2$ values.
As such, this works for all $n \in \mathbb{N}$, including for $n \gt 9$. I'm not sure why this restriction is used in the question. Are there perhaps any other unstated conditions?
|
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|
$\iint_{\Sigma} \frac{d\sigma}{\sqrt{x^2+y^2+(z+R)^2}}$ with $\Sigma$ the upper half of the sphere $x^2+y^2+z^2=R^2$ My attempt:
$K = \{(r,\theta): 0 \le r \le R, 0 \le \theta \le 2\pi \}$. I chose following parameterization: $$ \vec{\varphi}(r,\theta)=(r\cos\theta, r\sin\theta,\sqrt{R^2-r^2}).$$ And after further calculations, I got $$ \left\|\frac{\partial{\vec{\varphi}}}{\partial r} \times\frac{\partial{\vec{\varphi}}}{\partial \theta}\right\| = \frac{rR}{\sqrt{R^2-r^2}}.$$
And now, $$\iint_{\Sigma} \frac{d\sigma}{\sqrt{x^2+y^2+(z+R)^2}} = \int_0^{2\pi}d\theta\int_0^R \frac{r^2R}{\sqrt{(2R^2+2R\sqrt{R^2-r^2})(R^2-r^2)}}dr.$$ Applying the substitution $t = \sqrt{R^2-r^2}$, we get $$ 2\pi R\int_R^0\frac{R^2-t^2}{\sqrt{(2R^2+2Rt)t^2}}\frac{-t}{\sqrt{R^2-t^2}}dt=\sqrt{2R}\pi\int_0^R\sqrt{R-t}dt=\frac23\sqrt2\pi R^2.$$
The correct answer, however, should be $2\pi R(2-\sqrt2)$. Can somebody spot my mistake?
|
When you wrote $\int_0^R \frac{r^2}{\sqrt{(2 R^2+2 R\sqrt{R^2-r^2} ) (R^2-r^2)}}$ the numerator should have been $rR$ rather than $r^2$. You can evaluate the integral using your substitution $t=\sqrt{R^2-r^2}$ as follows:
$$\int_0^{2 \pi}d\theta\int_0^R \frac{rR}{\sqrt{(2 R^2+2 R\sqrt{R^2-r^2} ) (R^2-r^2)}}dr$$
$$
= 2 \pi \int_R^0 \frac{R \sqrt{R^2-t^2}}{\sqrt{(2 R^2+2 Rt ) t^2}} \frac{-t}{\sqrt{R^2-t^2}} dt
$$
$$
= 2 \pi \int_R^0 \frac{R }{\sqrt{(2 R^2+2 Rt ) t^2}} (-t) dt
$$
$$
= 2 \pi \int_0^R \frac{R }{\sqrt{2 R^2+2 Rt }} dt
$$
$$
= 2\sqrt2 \pi \int_0^R \frac{R }{2\sqrt{R^2+ Rt }} dt
$$
$$
= 2\sqrt2 \pi (\sqrt{R^2+ R^2 } - R)
$$
$$
= 2\sqrt2 \pi R (\sqrt{2} - 1)
$$
$$
= 2 \pi R (2 - \sqrt2).
$$
|
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|
On the transport equation $x \cdot \nabla u = |x|^2$ I have no clue about how to solve the following system
\begin{equation*}
\begin{cases}
x\cdot\nabla u=|x|^2, \quad x\in\mathbb{R}^n, \\
u|_{x_1=1}=3x_n.
\end{cases}
\end{equation*}
Study the domain where the function is defined.
Can you help me with some hints?
|
Use the method of characteristics to decompose the problem. Observe, we have the string of equations:
$$\frac{dx_1}{x_1} = \ldots = \frac{dx_n}{x_n} = \frac{dz}{x_1^2 + \ldots x_n^2}$$
Solving for the constant characteristics for the first n-1 equations, we get:
$$\phi_i = \frac{x_{i-1}}{x_i}$$
So the last equation becomes:
$$\frac{dx_n}{x_n} = \frac{dz}{x_n^2 \phi_{n-1}^2(1 + \phi_{n-2}^2(\ldots (1 + \phi_1^2) \ldots )))}$$
Which, if we allow $\psi := \phi_{n-1}^2(1 + \phi_{n-2}^2(\ldots (1 + \phi_1^2) \ldots )))$, then this boils down to:
$$\frac{dz}{\psi} = x_n dx_n$$
$$\implies \frac{z}{\psi} = \frac{1}{2}x_n^2 + \eta(\phi_1, \ldots, \phi_{n-1})$$
$$\implies z = \psi(\frac{1}{2}x_n^2 + \eta)$$
where we defined $\psi$ already, which we know equals:
$$\psi = \frac{|x|^2}{x_n^2}$$
and $\eta$ is a function on the constants $\phi_1, \ldots, \phi_{n-1}$.
This leads to:
$$u = \frac{1}{2} x \cdot x + \eta(\phi_1, \ldots, \phi_{n-1})\frac{|x|^2}{x_n^2}$$
Now considering the initial conditions:
$$u(x_1 = 1, \cdot ) = 3x_n = \frac{1}{2}\sum_{i=1}^n x_i^2 + \eta(1, \phi_2, \ldots, \phi_{n-1}) \frac{1}{x_n^2} \sum_{i=1}^n x_i^2$$
$$\implies 3x_n^3 - \frac{x_n^2}{2}\sum_{i=1}^n x_i^2 = \eta(1, \phi_2, \ldots, \phi_{n-1}) \sum_{i=1}^n x_i^2$$
$$\implies \frac{3x_n^3}{|x|^2} - \frac{x_n^2}{2} = \eta(1, \phi_2, \ldots, \phi_{n-1}) $$
Of course we can easily translate the variables $\phi_2 = x_2, \phi_3 = \frac{x_3}{x_2}, \ldots$ which instantly makes $\eta$ a function of $x_2, \ldots, x_n$. Therefore,
$$ \implies \eta = \frac{3x_n^3}{1 + \sum_{i=2}^n x_i^2} - \frac{x_n^2}{2}$$
Thus, we get:
$$u = \frac{1}{2} x \cdot x + |x|^2 \left ( \frac{3x_n}{1 + \sum_{i=2}^n x_i^2} - \frac{1}{2} \right )$$
Which simplifies to:
$$u = x \cdot x \left ( \frac{3x_n}{1 + \sum_{i=2}^n x_i^2} \right )$$
|
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|
Using Cardano's method to find an algebraic equation whose root is $\sqrt{2} +\sqrt[3]{3}$ $\sqrt{2} +\sqrt[3]{3}$ solution of an algebraic equation $ a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{0}=0 $
How to find this equation?
I tried Cardano's method, noting that
$$\sqrt{2} +\sqrt[3]{3} = \sqrt[3]{\sqrt{8}} +\sqrt[3]{3}$$
It means
$$\begin{align}
-\frac{q}{2}+\sqrt{ \frac{q^{2}}{4}+\frac{p^{3}}{27}}&=3 \\[4pt]
-\frac{q}{2}-\sqrt{ \frac{q^{2}}{4}+\frac{p^{3}}{27}}&=\sqrt{8}
\end{align}$$
but this system doesn't have natural solutions.
Maybe it has rational $p$ and $q$?
|
Let $x=\sqrt 2 + \sqrt[3]{3}$. Then,
$$
x-\sqrt 2 = \sqrt[3]{3}
$$
Cubing both sides and rearranging we have,
$$
x^3+6x-3=\sqrt 2 (3x^2+2)
$$
Squaring both sides and rearranging we have,
$$
x^6-6x^4-6x^3+12x^2-36x+1=0
$$
|
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|
Sum of all values that satisfy $\frac{x^{x-3}}{x}=\frac{x}{x^\frac{4}{x}}$.
What is the sum of all values of $x$ that satisfy the equation
$\frac{x^{x-3}}{x}=\frac{x}{x^\frac{4}{x}}$?
I start off by cross multiplying.
$$x^2=x^{x-3+\frac{4}{x}}$$
Taking the square root of both sides gives me:
$$x=\pm x^{\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}}$$
I start with the positive side first:
$$1=\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}$$$$x^2-5x+4=0$$$$x=4, 1$$
Now, I start with the negative side:
$$x=-x^{\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}}$$$$x^\frac{x^2-5x+4}{2x}=-1$$
I can't log both sides because of the $-1$, so I raise both sides by a power of $2x$ to get rid of the fraction. Because $2x$ is even, the RHS becomes a $1$.
$$x^{x^2-5x+4}=1$$
Now, taking the log of both sides, I have
$$(x^2-5x+4)\cdot\operatorname{log}(x)=0$$
Dividing both sides by $\operatorname{log}(x)$, I get $x^2-5x+4=0$ again, which should give me $x=4, 1$ again. So thus, the answer should be $4+1=5$.
However, this is wrong, as the answer key says that the answer is $-1+1+4=\boxed{4}$. I have checked $-1$ as a solution on all of my equations, and all of them work. How can I derive the $-1$ out from my equations? Or, if my approach or any of my equations are wrong, how do I get the answer $x=-1$?
|
In your second case, after you render
$x^{\frac{x^2-5x+4}{2x}}=-1$
forget about logarithms. Taking absolute values of the equation $a^b=-1$ and remembering that anything to a zero power can only be $+1$, conclude that the only way to get a value of $-1$ using real variables is to have $a=x=-1$. So put $x=-1$ into the equation you derived and see whether, in fact, it's consistent with the power having a value of $-1$.
|
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|
nth element in the given sequence Given two constants, a and b, and a sequence constructed as follows:
1, 1, 1, 1...(a - 1 times), 2, 2, 2, 2,...(a-2) times,......a-1
For example, for a = 6,
1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5
Ques: Determine the bth term in the sequence
*
*This is opposite to the sequence given here: Finding the nth element of a sequence
I am not working for a proof here, any pointers to the explanation would help loads. Thank you
|
Let's flip the sequence to give,
5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1
Now if we perform, $a-t_n$ for all the terms it gives us,
1, 2, 2, 3, 3, 3.....5
We know the term $t_n = \lfloor \sqrt{2n} + \frac{1}{2} \rfloor$. Our term for the flipped sequence is just $t_n = a-\lfloor \sqrt{2n} + \frac{1}{2} \rfloor$.
The index of the original sequence w.r.t the flipped sequence is $\frac{a(a-1)}{2} - n$. So putting it together,
$$
t_{\left(\frac{a(a-1)}{2} - n\right) } = a-\lfloor \sqrt{2n} + \frac{1}{2} \rfloor $$
Replacing $n$ with $\frac{a(a-1)}{2} - n$,
$$
t_{n} = a-\lfloor \sqrt{2\left(\frac{a(a-1)}{2} - n\right) } + \frac{1}{2} \rfloor $$
|
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|
Prove that the conic $x^2 - 4xy + y^2 -2x -20y -11 = 0$ is a hyperbola and find the centre $(h,k)$
I have to prove that the conic $$x^2 - 4xy + y^2 -2x -20y -11 = 0$$ is a hyperbola and find the centre $(h,k)$.
I proved it is a hyperbola using discriminant $b^2-4ac $ and the answer was greater than zero hence a hyperbola.
But I cannot seem to change the equation into the form
$(x-h)^2/a^2 - (y-k)^2/b^2=1$ so as to find the centre...
I could finally solve it with everyone's Help
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Write first the terms containing $x$ as the beginning of the square of an affine function in $x$ and $y$:
$$x^2-4xy-2x= (x-2y-1)^2-(4y+4y^2+1),$$
so that the equation becomes
\begin{align}
x^2 - 4xy + y^2 -2x -20y -11 &= (x-2y-1)^2-(4y+4y^2+1)+y^2-20y-11 \\
&= (x-2y-1)^2-3(y^2+8y+4)\\
&= (x-2y-1)^2-3\bigl((y+4)^2-16\bigr).
\end{align}
Can you end the calculations?
|
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|
Isn't there any divisor $k$ of $n^4$ such that $n^2-nI did some experiment with my Python script to find a number which could divide $n^4$ in this interval ($n^2-n$, $n^2$).
I watched the form of prim factors of the numbers in this ($n^2-n$, $n^2$) interval, but I hadn't was any idea for proofing it.
Can anyone to help with some start idea to proof that if $n^2-n < k < n^2$ then $k$ can't divide $n^4$.
|
Answer to the Question
Let $k=n^2-a$. Since $n^2-a\mid n^4$ and
$$
\frac{n^4}{n^2-a}=n^2+a+\frac{a^2}{n^2-a}
$$
we must have $d=\frac{a^2}{n^2-a}\in\mathbb{Z}$. However, since $\color{#C00}{1\le a\le n-1}$ and $\frac{a^2}{n^2-a}$ is increasing in $a$,
$$
0\lt\overbrace{\ \frac1{n^2-1}\ }^{\large\color{#C00}{a=1}}\le d\le\overbrace{\frac{n^2-2n+1}{n^2-n+1}}^{\large\color{#C00}{a=n-1}}\lt1
$$
which is impossible because there is no integer between $0$ and $1$.
Bounding the Greatest Factor of $\boldsymbol{n^4}$ less than $\boldsymbol{n^2}$
Case: $\boldsymbol{d=1}$
Since $n^2=a^2+a$, we have $n-1\lt a\lt n$, which is impossible.
Case: $\boldsymbol{d=2}$
Since $n^2=\frac{a(a+2)}2$, $\sqrt2\,n-1\lt a\lt\sqrt2\,n$. In fact, $2n^2+1=(a+1)^2$. That means that
$$
\left(\frac{a+1}n\right)^2=2+\frac1{n^2}\lt\left(\sqrt2+\frac1{2n^2}\right)^2
$$
Which forces $\frac{a+1}n$ to be a continued fraction overestimate of $\sqrt2$ . This gives the first pairs $(n,a)$ to be
$$
\{(0,0),(2,2),(12,16),(70,98),(408,576),(2378,3362),(13860,19600)\}
$$
The recursion for $n_k$ is $n_k=6n_{k-1}-n_{k-2}$ and $a_k=\left\lfloor\sqrt2\,n\right\rfloor$.
In any case, we have
The largest factor of $n^4$ less than $n^2$ must be less than $n^2-\sqrt2\,n+1$ and there are an infinite number of $n$ so that the largest factor of $n^4$ less than $n^2$ is greater than $n^2-\sqrt2\,n$.
|
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|
Evaluate definite integral using limit of a sum definition
From the definition of a definite integral as the limit of a sum, evaluate
$$\int_a^b\frac{1}{\sqrt x}dx$$
My progress so far:
$\begin{align}
\int_a^b\frac{1}{\sqrt x}dx & =\lim_{n\to\infty}h\times\bigg(f(a)+f(a+h)+\cdots+f\big(a+(n-1)h\big)\bigg) \\
& = \lim_{n\to\infty}h\times\bigg(\frac1{\sqrt a}+\frac1{\sqrt {a+h}} +\cdots+\frac1{\sqrt {a+(n-1)h}}\bigg)
\end{align}$
After this point, I cant't think of a way to combine the reciprocals of square roots into a neat form. Usually a convenient summation formula is applicable but how would I proceed in this case?
|
We assume $0\leq a\leq b$. In order to cope with the square root function $\frac{1}{\sqrt{x}}$ it is convenient to use variable length intervals with length $j^2\frac{b-a}{n^2}$. When taking square roots we can factor out $j$ and summation is expected to become simpler. The corresponding Riemann sum is
\begin{align*}
\int_{a}^{b}\frac{1}{\sqrt{x}}\,dx=\lim_{n\to \infty}\sum_{j=1}^n\frac{1}{\sqrt{\color{blue}{a}+ j^2\frac{b-a}{n^2}}}\cdot\left(j^2\frac{b-a}{n^2}-(j-1)^2\frac{b-a}{n^2}\right)\tag{1}
\end{align*}
But it's still not easy due to the constant $\color{blue}{a}$. To overcome this difficulty we write the integral as difference of two improper integrals
\begin{align*}
\int_{a}^{b}\frac{1}{\sqrt{x}}\,dx=\int_{0}^{b}\frac{1}{\sqrt{x}}\,dx-\int_{0}^{a}\frac{1}{\sqrt{x}}\,dx\tag{2}
\end{align*}
If the limit of the Riemann sums of the improper integrals exists we have found the wanted Riemann integral and we are done.
We subdivide the interval $[0,b]$ in
\begin{align*}
[0,b]=\bigcup_{j=1}^n\left[(j-1)^2\frac{b}{n^2},j^2\frac{b}{n^2}\right]
\end{align*}
with length
\begin{align*}
j^2\frac{b}{n^2}-(j-1)^2\frac{b}{n^2}=\frac{(2j-1)b}{n^2}\qquad\qquad1\leq j\leq n
\end{align*}
and obtain the Riemann sum
\begin{align*}
\color{blue}{\int_{0}^b\frac{1}{\sqrt{x}}\,dx}&=\lim_{n\to\infty}\sum_{j=1}^n\frac{1}{\sqrt{\frac{j^2b}{n^2}}}\cdot\frac{(2j-1)b}{n^2}\tag{3}\\
&=\sqrt{b}\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^n\left(2-\frac{1}{j}\right)\tag{4}\\
&=\sqrt{b}\lim_{n\to\infty}\frac{1}{n}\left(2n-H_n\right)\tag{5}\\
&\,\,\color{blue}{=2\sqrt{b}}\tag{6}
\end{align*}
We finally conclude from (2) and (6)
\begin{align*}
\color{blue}{\int_{a}^b\frac{1}{\sqrt{x}}\,dx=2\left(\sqrt{b}-\sqrt{a}\right)}
\end{align*}
Comment:
*
*In (3) we have the now convenient representation (1) evaluated at $a=0$.
*In (4) we do some simplifications and factor out terms not dependent on $n$ resp. $j$.
*In (5) we note the Harmonic numbers $H_n=\sum_{j=1}^n \frac{1}{j}$ grow asymptotically with $\log n$, so that
\begin{align*}
\lim_{n\to\infty}\frac{H_n}{n}=\lim_{n\to\infty}\frac{\log n}{n}=0.
\end{align*}
|
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|
Evaluating $\sum\limits_{i=1}^8 \cos\frac{\pi t_i}{4}$ and $\sum\limits_{i=1}^8 \cos^2\frac{\pi t_i}{4}$ for $t\in\{-7,-5,-3,-1,1,3,5,7\}$ I am dealing with a series$$\sum_{i=1}^8 \cos\frac{\pi t_i}{4}$$
with $t\in\{-7,-5,-3,-1,1,3,5,7\}$. How can I determine a global solution for this summation? Would it be the same logic for$$\sum_{i=1}^8 \cos^2\frac{\pi t_i}{4}?$$Many thanks for your help!
|
You have a formula for the sum of sines or cosines of arcs in an arithmetic progression:
$$ \cos a + \cos(a + \theta )+\cos(a + 2 \theta )+ \dots + \cos(a + n \theta ) = \frac{\sin \frac{(n +1) \theta }{2}}{\sin \frac{ \theta }{2}}\,\cos\Bigl (a + \frac{n \theta }{2}\Bigr)$$
$$\sin a+ \sin(a + \theta )+ \sin(a + 2 \theta )+ \dots + \sin(a + n \theta ) = \frac{\sin \frac{(n + 1)\theta }{2}}{\sin \frac{\theta }{2}}\,\sin\Bigl (a + \frac{n \theta }{2}\Bigr),$$
that you can prove calculating the sum of the geometric progression
$$\sum_{k=0}^n\mathrm e^{i(a+k\theta)}=\mathrm e^{ia}\sum_{k=0}^n\mathrm e^{ik\theta}$$
and identifying the real and imaginary parts.
|
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|
About the probability in rolling dice If A and B each roll two dice, how do we calculate the probability that the maximum number of A is greater than the maximum of B and that the minimum number of A is also greater than the minimum of B?
|
Hint: Start by computing the probability mass function of the maximum for player $A$, i.e., compute $Pr(\max=x)$ for each $x$.
By the way, if you are able to solve the problem for the maximum, it will be easy to get the solution for the minimum.
Edit: Since it's solved, I will have some fun adding the formulas for the cases:
Let $m$ be the maximum of the roll of two dices $d_1$ and $d_2$. Assuming that the dice rolls are independent, we can compute the probability of the maximum being $x \in \{1,2,3,4,5,6\}$ by
$$
Pr(m=x) = \\
\ \ \
Pr(d_1=x \wedge d_2<x) \\
+ Pr(d_1<x \wedge d_2=x) \\
+ Pr(d_1=x \wedge d_2=x)
$$
Which means that either the first dice $d_1$ is the maximum, or the second dice $d_2$ is the maximum or both are the maximum.
This gives:
$$
Pr(M=x) = \frac{1}{6} \cdot \frac{x-1}{6}
+ \frac{x-1}{6} \cdot \frac{1}{6}
+ \frac{1}{6} \cdot \frac{1}{6}
= \frac{2x-1}{36} \ .
$$
Alternatively, you could compute
$$
Pr(m=x) = Pr(d_1 \leq x \wedge d_2 \leq x)
- Pr(d_1 \leq x-1 \wedge d_2 \leq x-1)
$$
Which means the probability of at least one of the dice being less than or equal to $x$ but not both.
This gives:
$$
Pr(M=x) = \frac{x}{6} \cdot \frac{x}{6}
- \frac{x-1}{6} \cdot \frac{x-1}{6}
= \frac{x^2 - (x-1)^2}{36} \ .
$$
And simplifying we get the same result as before.
The computation for the minimum $m$ is similar, with the inequality signs switched. The end result will be symmetrical, i.e.:
$$
Pr(m=x) = Pr(M=7-x) = \frac{13-2x}{36}\ .
$$
If you make the computations for each $x$ you'll get the results presented by HJ_beginner.
While we are at it, we may take the opportunity to compute the cumulative probability distribution of the maximum $M$ for each $x \in \{1,2,3,4,5,6\}$:
$$
F(M=x) = Pr(M \leq x) = \sum_{j=1}^x Pr(M = x) = \frac{x^2}{36} \ .
$$
Similarly, for the minimum $m$:
$$
F(m=x) = Pr(m \leq x) = \sum_{j=1}^x Pr(m = x) = \frac{12x-x^2}{36} \ .
$$
Now, for the question, if we want to know the probability of player $A$ having his/her maximum above player $B$, i.e., $M_A > M_B$, we will have:
$$
Pr(M_A > M_B) = \sum_{j=2}^6 Pr(M_A = j \wedge M_B < j)
$$
Assuming that the rolls of $A$ and $B$ are independent, we get
$$
Pr(M_A > M_B) = \sum_{j=2}^6 Pr(M_A = j) \times Pr(M_B < j)
$$
Noting that $Pr(M_B < j) = Pr(M_B \leq j-1)$, we get
$$
Pr(M_A > M_B) = \sum_{j=2}^6 Pr(M_A = j) \times Pr(M_B \leq j-1)
$$
Using the formulas for each probability we have:
$$
Pr(M_A > M_B) = \sum_{j=2}^6 \frac{2j-1}{36} \times \frac{(j-1)^2}{36}
$$
$$
Pr(M_A > M_B) = \sum_{j=1}^5 \frac{(2j+1)j^2}{36}
$$
Using Faulhaber's formulas we obtain:
$$
Pr(M_A > M_B) = \frac{2(5^4 + 2 \times 5^3 + 5^2)/4 + (2 \times 5^3 + 3 \times 5^2 + 5)/6}{6^4}
$$
$$
Pr(M_A > M_B) = \frac{505}{6^4} = 0.38966
$$
Of course, we could also use the symmetry argument by HJ_beginner to get the same result:
$$
Pr(M_A > M_B) = \frac{1}{2} \times Pr(M_A \neq M_B) = \frac{1}{2} \times (1 - Pr(M_A = M_B))
$$
And $Pr(M_A = M_B)$ is:
$$
Pr(M_A = M_B) = \sum_{j=1}^6 Pr(M_A = j) \times Pr(M_B = j)
$$
$$
Pr(M_A = M_B) = \sum_{j=1}^6 \frac{2j-1}{36} \times \frac{2j-1}{36} = \frac{1}{6^4} \times \sum_{j=1}^6 4j^2-4j+1
$$
$$
Pr(M_A = M_B) = \frac{1}{6^4} \times
\left(
4 \times \frac{\color{red}{6} \times (6+1) \times (2 \times 6 + 1)}{\color{red}{6}}
- 4 \times \frac{6 \times (6+1)}{2} + 6 \right)
$$
$$
Pr(M_A = M_B) = \frac{286}{6^4} = 0.22068
$$
Hence
$$
Pr(M_A > M_B) = \frac{1}{2} \times \left(1 - \frac{286}{6^4} \right) = \frac{505}{6^4}
$$
As for the minimum, $Pr(m_A > m_B)$ the result is the same by symmetry.
|
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|
Suppose we roll a die repeatedly and let Tk be the number of rolls until some number appears k times in a row. Suppose we roll a die repeatedly and let Tk be the number of rolls until some number
appears k times in a row. (a) Find P(T2 = k) for each integer k ≥ 2. (b) Find E[T2].
I am thinking that P(T2=2) = (1/6)(1/6) ; P(T2=3) = (5/6)(1/6)^3 . So I was trying to develop the pattern : P(T2=k) = (5/6)^(k-2) * (1/6)^k . However, I run into the issue that I have with P(T2=4) : The first role has 5 options, the second has 4 (because it cannot be the previous OR the next number). So I cannot figure out the general formula.
** I also originally thought that for all k>3, P(T2+k) would be zero. Is that correct or is the above reasoning better?
Thanks!
|
We can answer the question for a die with $n$ faces, rolling until
some number appears $k$ times in row. We have for the probability that
we need $m$ rolls where $m\ge k$ from first principles
$$\mathrm{P}[T=m]
= \frac{1}{n^m} \frac{n}{n-1}
[z^m] (n-1) z^k \sum_{q\ge 0} (z+\cdots+z^{k-1})^q (n-1)^q
\\ = \frac{1}{n^m} \frac{n}{n-1}
[z^m] (n-1) z^k \sum_{q\ge 0} z^q (1+\cdots+z^{k-2})^q (n-1)^q
\\ = n \frac{1}{n^m}
[z^m] z^k \sum_{q\ge 0} z^q (1-z^{k-1})^q (n-1)^q / (1-z)^q
\\ = n \frac{1}{n^m}
[z^m] z^k \frac{1}{1-(n-1)z(1-z^{k-1})/(1-z)}
\\ = n \frac{1}{n^m}
[z^m] z^k \frac{1-z}{1-z-(n-1)z+(n-1)z^{k}}
\\ = n \frac{1}{n^m}
[z^m] z^k \frac{1-z}{1-nz+(n-1)z^{k}}.$$
Let us verify that this is a probability distribution.
We sum the probabilities to obtain
$$n\sum_{m\ge k} \frac{1}{n^m}
[z^m] z^k \frac{1-z}{1-nz+(n-1)z^{k}}
= n \frac{1}{n^k} \frac{1-1/n}{(n-1)/n^{k}}
= \frac{n-1}{n-1} = 1$$
and the sanity check goes through. We continue with the expectation.
$$\mathrm{E}[T] =
n\sum_{m\ge k} \frac{1}{n^m} m
[z^m] z^k \frac{1-z}{1-nz+(n-1)z^{k}}
\\ = n\sum_{m\ge k} \frac{1}{n^m}
[z^{m-1}] \left(z^k \frac{1-z}{1-nz+(n-1)z^{k}}\right)'
\\ = \sum_{m\ge k} \frac{1}{n^{m-1}}
[z^{m-1}]
\left(\frac{kz^{k-1}-(k+1)z^k}{1-nz+(n-1)z^{k}}
- \frac{(z^k-z^{k+1})((n-1)kz^{k-1}-n)}{(1-nz+(n-1)z^k)^2}
\right)
\\ = \frac{k/n^{k-1}-(k+1)/n^k}{(n-1)/n^{k}}
- \frac{(1/n^k-1/n^{k+1})((n-1)k/n^{k-1}-n)}{(n-1)^2/(n^k)^2}
\\ = \frac{nk-(k+1)}{n-1}
- \frac{(1-1/n)((n-1)k/n^{k-1}-n)}{(n-1)^2/n^k}
\\ = \frac{(n-1)k-1}{n-1}
- \frac{1}{n} \frac{((n-1)k/n^{k-1}-n)}{(n-1)/n^k}
\\ = k-\frac{1}{n-1}
- \frac{1}{n} \frac{n(n-1)k-n^{k+1}}{n-1}
= k-\frac{1}{n-1}
- \frac{(n-1)k-n^{k}}{n-1}.$$
This yields the desired answer
$$\bbox[5px,border:2px solid #00A000]{
\mathrm{E}[T] = \frac{n^k-1}{n-1}.}$$
|
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|
Long division of polynomials: $(2x^4 - 5x^3 - 15x^2 + 10x +8) \div (x^2-x-2)$ I've been self-studying from Stroud & Booth's amazing "Engineering Mathematics". I'm currently stuck on an aspect of long division of polynomials, when the denominator itself is a polynomial. So, I know how to do long division when the deniminator is something like $(x+1)$, but when it's a polynomial, I'm not sure what the mechanics are, like in the below example:
$$(2x^4 - 5x^3 - 15x^2 + 10x +8) \div (x^2-x-2)$$
Can anybody shed some light here, please? Thank you!
|
Usually in these problems, I will first analyze the denominator polynomial $g(x)=x^2-x-2$.
The roots of this polynomial are: $$g(x)=(x-2)(x+1)=0$$
giving, $x=-1,2$. Now the second thing I do is I check whether the numerator polynomial is having a common root. So we check $f(x)=2x^4-5x^3-15x^2+10x+8$, for $x=-1,2$.
$f(-1)=-10$ and $f(2)=-40$, so none of the roots are common telling that this division will lead to a remainder.
The next thing I would do is just take $(x-2)$ common from $f(x)$ like this (take common whatever is remaining you just add to make it equal to $f(x)$):
$$f(x)=2x^3(x-2)-x^2(x-2)-17x(x-2)-24(x-2)-40$$
$$\frac{f(x)}{g(x)}=\frac{2x^3(x-2)-x^2(x-2)-17x(x-2)-24(x-2)-40}{(x-2)(x+1)}$$
$$\frac{f(x)}{g(x)}=\frac{2x^3-x^2-17x-24}{x+1}-\frac{40}{(x-2)(x+1)}$$
For the first part do the same thing again. Take $(x+1)$ common to get:
$$2x^2(x+1)-3x(x+1)-14(x+1)-10$$
$$\frac{f(x)}{g(x)}=\frac{(2x^2-3x-14)(x+1)}{x+1}-\frac{10}{x+1}+\frac{40}{(x-2)(x+1)}$$
$$\frac{f(x)}{g(x)}=2x^2-3x-14-\frac{10}{x+1}-\frac{40}{(x-2)(x+1)}$$
This is the final answer for your division. You can also club the last two terms so that you get a common denominator function.
$$\frac{f(x)}{g(x)}=2x^2-3x-14-\frac{10(x+2)}{(x-2)(x+1)}$$
You can also try synthetic division which is far more easier but for that you need to have a good hold on the concept of long division.
Hope this helps.....
|
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|
Find number of ways to split $1$ dollar into $5$, $10$, $20$, $50$ cents
Find number of ways to split $1$ dollar into $5$, $10$, $20$, $50$ cents
I am going to use generating functions:
$$n = [x^{100}] (1+x^5+x^{10}+\cdots)(1+x^{10}+x^{20}+\cdots)(1+x^{20}+x^{40}+\cdots)(1+x^{50}+x^{100}+\cdots) = \\
[x^{100}]\frac{1}{1-x^5}\frac{1}{1-x^{10}}\frac{1}{1-x^{20}}\frac{1}{1-x^{50}}
$$
Ok, I know that computers are able to solve that but how can I easily get coefficient at $x^{100}$?
|
Let $$\begin{align}f(y)&=\frac{1}{1-y}\frac{1}{1-y^2}\frac{1}{1-y^4}\\&=\frac{(1+y+y^2+y^3)(1+y^2)}{(1-y^4)^3}\\&=\frac{1+y+2y^2+2y^3+y^4+y^5}{(1-y^4)^3}\end{align}$$
Letting $p(y)=1+y+2y^2+2y^3+y^4+y^5$, then $$[y^{20}]f(y)\frac{1}{1-y^{10}} = [y^0]f(y)+[y^{10}]f(y)+[y^{20}]f(y)$$
Now use that $$\frac{1}{(1-y^4)^3}=\sum_{j=0}^{\infty} \binom{j+2}{2}y^{4j}$$
So $$\begin{align}[y^0]f(y)&=1\\ [y^{10}]f(y)&=\binom{0+2}{2}[y^{10}]p(y) + \binom{1+2}{2}[y^{6}]p(y)+\binom{2+2}{2}[y^2]p(y)\\&=0+0+6\cdot2\\&=12\\
[y^{20}]f(y)&=\binom{4+2}{2}[y^4]p(y)+\binom{5+2}{2}[y^0]p(y)\\
&=15\cdot 1 + 21\cdot 1\\
&=36\end{align}$$
So your result is $1+12+36=49.$
You have in general that
$$\begin{align}[y^{4j}]f(y)&=\binom{j+1}{2}+\binom{j+2}{2}=(j+1)^2\\
[y^{4j+1}]f(y)&=\binom{j+1}{2}+\binom{j+2}{2}=(j+1)^2\\
[y^{4j+2}]f(y)&=2\binom{j+2}{2}=(j+2)(j+1)\\
[y^{4j+3}]f(y)&=2\binom{j+2}{2}=(j+2)(j+1)
\end{align}$$
|
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|
Finding the value of $x-y$. Given two equations \begin{align} & x^4+y^4=\dfrac {-7}{9} \\ & x^3-y^3=3. \end{align} From these equations find the value of $(x-y).$
I have just factorize $x^3-y^3=(x-y)(x^2+xy+y^2)=3$. But don't know how to proceed from here. One way to find the values of $x$ and $y$ and then compute possible values of $x-y$. But that is a too long way. I think there is any simple way to find the value of $x-y$. Please help me to solve this.
|
Multiply equation (2) by $x-y$, giving $(x^4-x^3 y +xy^3 + y^4) = 3(x-y)$.
Next, (1) = -7/9, so
$-x^3 y +xy^3 - 7/9 = 3(x-y)$.
Factorise.
$xy(-x^2 + y^2) - 7/9 = 3(x-y)$.
$(y-x)(y+x)xy - 7/9 = 3(x-y)$.
$(x-y)(3+(y+x)xy) = - 7/9$
Using your idea, $x^2+xy+y^2 = 3/(x-y) = - 7/3*(3+(y+x)xy)$
Congrats, you now have a quadratic in $x$, or $y$, depending on how you feel.
Next we solve this to get $y$ in terms of $x$.
$3x^2 +3xy + 3y^2 = -21-7xy^2-7x^2y$
|
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|
Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation eliminates $4y$ to solve strictly for $x$; i.e. $6x - 5x = 72 - 64 \Rightarrow x = 8$. Substituting $x=8$ into $(2)$ reveals that $y=6$.
I could also subtract $(1)$ from $(2)$ and divide by $2$, yielding $x+y=14$. Let $$\begin{align}3x+3y - y &= 36 \tag{1a}\\ 5x + 5y - y &= 64\tag{1b}\end{align}$$ then expand brackets, and it follows that $42 - y = 36$ and $70 - y = 64$, thus revealing $y=6$ and so $x = 14 - 6 = 8$.
I can even use matrices!
$(1)$ and $(2)$ could be written in matrix form:
$$\begin{align}\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}&=\begin{bmatrix}36 \\ 64\end{bmatrix}\tag3 \\ \begin{bmatrix} x \\ y\end{bmatrix} &= {\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}}^{-1}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix}4 &-2 \\ -5 &3\end{bmatrix}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &=\frac12\begin{bmatrix} 16 \\ 12\end{bmatrix} \\ &= \begin{bmatrix} 8 \\ 6\end{bmatrix} \\ \\ \therefore x&=8 \\ \therefore y&= 6\end{align}$$
Question
Are there any other methods to solve for both $x$ and $y$?
|
Another method to solve simultaneous equations in two dimensions, is by plotting graphs of the equations on a cartesian plane, and finding the point of intersection.
|
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|
Finding Angle using Geometry In an equilateral triangle $ ABC $ the point $ D $ and $ E $ are on sides $ AC $ and $AB$ respectively, such that $ BD $ and $ CE $ intersect at $P$ , and the area of the quadrilateral $ ADPE $ is equal to area of $ \Delta BPC $ find $ \angle BPE $.
This question when I first tried looked easy and I was also able to guess the answer but when I tried to proof, was not able to work it out. I want some help. Thank you. No image was provided in question. I am attaching my drawing.
|
Let $a$ be the side length of the equilateral $\triangle ABC$,
$\angle BPE=\theta$, $\angle PBC=\phi$,
and let $[\cdot]$ denote the area.
Note that
\begin{align}
[AEPD]&=[ABC]-[BCD]-[BCE]+[BCP]
,
\end{align}
and since $[AEPD]=[BCP]$,
we must have
\begin{align}
[BCD]+[BCE]&=[ABC]
.
\end{align}
Using the three-angles-and-a-side expression for the area,
we get
\begin{align}
\frac{\sqrt3}4\cdot a^2\cdot \frac{\sin(\phi)}{\sin(120^\circ-\phi)}
+
\frac{\sqrt3}4\cdot a^2\cdot \frac{\sin(\theta-\phi)}{\sin(120^\circ+\phi-\theta)}
&=
\frac{\sqrt3}4\cdot a^2
,\\
\frac{\sin(\phi)}{\sin(120^\circ-\phi)}
+
\cdot \frac{\sin(\theta-\phi)}{\sin(120^\circ+\phi-\theta)}
-1
&=0
.
\end{align}
After expanding and refactoring we get
\begin{align}
\frac{(\sin^2\phi+\cos^2\phi)(\sqrt3\sin\theta-3\cos\theta)}{
(\sqrt3\cos\phi+\sin\phi)
(\sqrt3\cos\theta\cos\phi
+\sqrt3\sin\theta\sin\phi+\sin\theta\cos\phi
-\cos\theta\sin\phi)
}
&=0
,
\end{align}
and it follows that $\theta=60^\circ$.
|
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|
Is there some strategy to find the general term of a recursive sequence? Given $(a_{n})_{n \in \mathbb{N}}$,
$a_{0} = 2, a_{1} = 4, a_{n+2} = 4a_{n+1} - 3a_{n}$,
is there any way to find the general term? I reckoned that every $a_{i}$ is a factor of 2, and then the series becomes 1, 2, 5, 14, 41, ..., and obviously that's $a_{i} = 3(a_{i-1}) - 1$ (if $i≠0$) but even then I just can't get rid of these damn recursive clauses.
Any help? Is there some way to tackle this kind of problems?
|
Writing each pair of consecutive terms as a linear combination of the previous pair of consecutive terms,
$$\pmatrix{a_{n}\\a_{n-1}} = \pmatrix{4&-3\\1&0}\pmatrix{a_{n-1}\\a_{n-2}}= \cdots = \pmatrix{4&-3\\1&0}^{n-1}\pmatrix{a_{1}\\a_{0}}$$
The question becomes how to represent the matrix power in close form, which is by diagonalisation.
$$\begin{align*}
\pmatrix{4-\lambda&-3\\1&0-\lambda} &= 0\\
(4-\lambda)(-\lambda) +3 &= 0\\
(\lambda-3)(\lambda-1) &= 0
\end{align*}$$
(This characteristic equation has the same roots as the ones given in the other answers.)
For $\lambda = 3$, $$\pmatrix{4&-3\\1&0}\pmatrix{3\\1} = 3\pmatrix{3\\1}$$
For $\lambda = 1$, $$\pmatrix{4&-3\\1&0}\pmatrix{1\\1} = \pmatrix{1\\1}$$
So
$$\begin{align*}
\pmatrix{4&-3\\1&0}^{n-1} &= \pmatrix{3&1\\1&1}\pmatrix{3^{n-1}&0\\0&1}\pmatrix{3&1\\1&1}^{-1}\\
&= \frac12\pmatrix{3&1\\1&1}\pmatrix{3^{n-1}&0\\0&1}\pmatrix{1&-1\\-1&3}\\
a_n &= \pmatrix{1&0}\pmatrix{a_{n}\\a_{n-1}}\\
&= \pmatrix{1&0}\pmatrix{4&-3\\1&0}^{n-1}\pmatrix{a_{1}\\a_{0}}\\
&= \pmatrix{1&0}\pmatrix{3&1\\1&1}\pmatrix{3^{n-1}&0\\0&1}\pmatrix{3&1\\1&1}^{-1}\pmatrix{4\\2}\\
&= \pmatrix{1&0}\pmatrix{3&1\\1&1}\pmatrix{3^{n-1}&0\\0&1}\pmatrix{1&-1\\-1&3}\pmatrix{2\\1}\\
&= \pmatrix{3&1}\pmatrix{3^{n-1}&0\\0&1}\pmatrix{1\\1}\\
&= 3^n+1
\end{align*}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3181668",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
If $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}$. Find $\lim_{n \to \infty} nt_n$ If $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}$
Find $\lim_{n \to \infty} nt_n$
First attempt: $t_n$ is positive(grouping two terms and performing subtraction we will get it) so is $nt_n$. Now can we prove it is monotonically decreasing? If so then $\lim_{n \to \infty} nt_n=\lim_{n \to \infty}[(\frac{1}{2+1/n}-\frac{1}{2+2/n})+(\frac{1}{2+3/n}-\frac{1}{2+4/n})+\cdots +(\frac{1}{4-1/n}-\frac{1}{4})]$ and each of these terms will go to zero so is the limit.
Second attempt: I was trying to use Riemann summation $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}=
\frac{1}{2n+1}+\frac{1}{2n+2}+\frac{1}{2n+3}+\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}+\frac{1}{4n}-2[\frac{1}{2n+2}+\frac{1}{2n+4}+\cdots \frac{1}{4n}]\Rightarrow \lim \frac 1n [nt_n]=\int_0^2\frac{dx}{2+x}-\int_0^1\frac{dx}{1+x}=\ln4-\ln 2-\ln 2=0$
So $\lim t_n=0$
So what will happen with $\lim nt_n$
Edit: As I got the answer is not $0$ because of the flaw. So can we have different approaches even with Riemann Sum to have the answer?
|
Let $$H_n = 1 + \frac{1}{2} + \frac{1}{3} + \dotsb + \frac{1}{n}$$ be the $n$th harmonic number. Then, as you noted, $$t_n = H_{4n} - H_{2n} - [H_{2n} - H_n] = H_{4n} - 2H_{2n} + H_n.$$
By the Euler-Maclaurin summation formula,
$$H_n = \log n + \gamma + \frac{1}{2n} + O\left(\frac{1}{n^2}\right)$$
so, after the smoke clears, $$t_n = \frac{1}{8n} + O\left(\frac{1}{n^2}\right)$$ whence $n t_n \to 1/8$.
|
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"url": "https://math.stackexchange.com/questions/3181789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Quickest way to find $a^5+b^5+c^5$ given that $a+b+c=1$, $a^2+b^2+c^2=2$ and $a^3+b^3+c^3=3$
$$\text{If}\ \cases{a+b+c=1 \\ a^2+b^2+c^2=2 \\a^3+b^3+c^3=3}
\text{then}\ a^5+b^5+c^5= \ ?$$
A YouTuber solved this problem recently and, though he spent some time explaining it, took over 40 minutes to solve it.
Like the video, the best I can do with this is relying on expansion formulas and substitution. As trivial a problem this is, the numerous trinomials and binomials with mixed terms makes it very, very tedious.
What is the quickest/shortest approach to this problem (meaning it doesn't need to be solved algebraically)? You don't have to type the entire solution out, I think if I'm given a good hint then I can take it from there.
|
Fun video!
Much time was spent on finding $abc=1/6$.
Alternative method for this:
$$\begin{align}a^2+b^2&=2-c^2 \Rightarrow \\
(a+b)^2-2ab&=2-c^2 \Rightarrow \\
(1-c)^2-2ab&=2-c^2 \Rightarrow \\
ab&=c^2-c-\frac12 \Rightarrow \\
abc&=c^3-c^2-\frac c2 \end{align}$$
Similarly:
$$abc=a^3-a^2-\frac a2\\
abc=b^3-b^2-\frac b2$$
Now adding them up:
$$3abc=(a^3+b^3+c^3)-(a^2+b^2+c^2)-\frac12(a+b+c)=3-2-\frac12 \Rightarrow abc=\frac16.$$
In fact, you can find other terms as well:
$$ab+bc+ca=(a^2+b^2+c^2)-(a+b+c)-\frac32=2-1-\frac32=-\frac12;\\
a^2b^2+b^2c^2+c^2a^2=ab(c^2-c-\frac12)+bc(a^2-a-\frac12)+ca(b^2-b-\frac12)=\\
abc(a+b+c)-3abc-\frac12(ab+bc+ca)=\\
\frac16-\frac12+\frac14=-\frac1{12}$$
Hence:
$$a^5+b^5+c^5=(a^2+b^2+c^2)(a^3+b^3+c^3)-(a^2b^2+b^2c^2+c^2a^2)+abc(ab+bc+ca)=\\
2\cdot 3-(-\frac1{12})+\frac16\cdot (-\frac12)=6.$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
An apparently harmless exercise concerning induction Let $b \in \mathbb{R}, b \ge 2$. Prove by induction that $$(b^n - 1)(b^n - b)(b^n -b^2)\cdots(b^n - b^{n-2}) \ge b^{n(n-1)}-b^{n(n-1)-1}$$ for all $n \in \mathbb{N}, n \ge 1$.
For the case $n = 2$, I have to show that $$b^2 - 1 \ge b^2 - b$$ which is true for $b \ge 1$.
Suppose the result holds for $k \ge 2$, I want to show that it holds for $k + 1$ too.
I know that $$(b^k - 1)(b^k - b)(b^k -b^2)\cdots(b^k - b^{k-2}) \ge b^{k(k-1)}-b^{k(k-1)-1}$$ thanks to the inductive hypothesis, and I want to prove that $$(b^{k+1} - 1)(b^{k+1} - b)(b^{k+1} -b^2)\cdots(b^{k+1} - b^{k-1}) \ge b^{(k+1)k}-b^{(k+1)k-1}.$$
The LHS is $$(b^{k+1} - 1)(b^{k+1} - b)(b^{k+1} -b^2)\cdots(b^{k+1} - b^{k-1}) =$$$$ (b^{k+1} - 1)b(b^k-1)b(b^k-b)b(b^k-b^2)\cdots b(b^k-b^{k-2})=$$$$ b^{k-1}(b^{k+1} - 1)(b^k-1)(b^k-b)(b^k-b^2)\cdots(b^k-b^{k-2}) \ge$$$$ b^{k-1}(b^{k+1} - 1)(b^{k(k-1)}-b^{k(k-1)-1}) =$$$$ b^{k^2+k}-b^{k^2+k-1}-b^{k^2-1}+b^{k^2-2}.$$
The RHS is $$b^{(k+1)k}-b^{(k+1)k-1}= b^{k^2 + k} - b^{k^2 + k - 1}.$$
So I am left to prove that $$b^{k^2+k}-b^{k^2+k-1}-b^{k^2-1}+b^{k^2-2} \ge b^{k^2 + k} - b^{k^2 + k - 1} \Leftrightarrow b^{k^2-2} \ge b^{k^2-1}$$ which is unfortunately not true.
|
Change variable to $c = \frac1b$. The inequality at hand (the version for $b \ge 2)$ is equivalent to
$$\prod_{k=2}^n (1-c^k) \ge 1 - c\quad\text{ for }\;n \ge 2, c \in \left(0,\frac12\right]$$
To prove this by induction, we will use a stronger form of induction statement.
For $n \ge 1$ and $c \in (0,\frac12]$, let $\mathcal{S}_n$ be the statement
$$\mathcal{S}_n :\quad \prod_{k=2}^n (1-c^k) \ge \frac{1-c}{1-c^n}$$
When $n = 1$, LHS is an empty product and evaluates to $1 = $ RHS, so $\mathcal{S}_1$ is true.
Assume $\mathcal{S}_{n}$ is true, we have
$$\prod_{k=2}^{n+1} (1-c^k) = (1-c^{n+1})\prod_{k=2}^n (1-c^k)
\ge (1-c^{n+1})\frac{1-c}{1-c^n}$$
Since $c \in (0,\frac12]$, we have
$$(1-c^{n+1})^2 > 1 - 2c^{n+1} \ge 1 - c^n
\implies \frac{1-c^{n+1}}{1-c^n} > \frac{1}{1-c^{n+1}}$$
This leads to
$$\prod_{k=2}^{n+1} (1-c^k) > \frac{1-c}{1-c^{n+1}}$$
and hence $\mathcal{S}_n \implies \mathcal{S}_{n+1}$. By induction, $\mathcal{S}_n$ is true for all $n \ge 1$.
As a result,
$$\prod_{k=2}^n (1-c^k) \ge \frac{1-c}{1-c^n} > 1 - c\quad\text{ for }\; n \ge 1, c \in \left(0,\frac12\right]$$
As a side note, the conjecture about $b \ge 1$ doesn't work in general.
For example, the inequality for $n = 3$ fails when $b <
\frac{\sqrt[3]{9+\sqrt{69}} + \sqrt[3]{9-\sqrt{69}}}{\sqrt[3]{18}} \approx 1.324717957244746$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
What am I doing wrong when testing for concavity with $f(x)=5x^{2/3}-2x^{5/3}$? I can't figure out what I'm doing wrong when testing for concavity with $f(x)=5x^{2/3}-2x^{5/3}$. I can't find the right intervals for where the graph is concave-down and concave-up.
My Steps:
find $f'(x)$:
$$f'(x)=\frac{10}{3}x^{-1/3}-\frac{10}{3}x^{2/3}=\frac{10}{3\sqrt[3]{x}}-\frac{10\sqrt[3]{x^2}}{3}$$
find $f''(x)$:
$$\frac{d}{dx} 3\sqrt[3]{x} = 0 + \frac{1}{3\sqrt[3]{x^2}} \cdot 3 = \frac{1}{\sqrt[3]{x^2}}$$
$$\frac{d}{dx}\frac{10}{3\sqrt[3]{x}}=\frac{0-[\frac{1}{\sqrt[3]{x^2}} \cdot 10]}{(3\sqrt[3]{x})^2}=\frac{-10}{(3\sqrt[3]{x})^2(\sqrt[3]{x^2})}=\frac{-10}{9\sqrt[3]{x^4}}$$
$$\frac{d}{dx}10\sqrt[3]{x^2}= 0 +10 \cdot \frac{d}{dx}x^{2/3}=\frac{10}{3\sqrt[3]{x}}$$
$$\frac{d}{dx}\frac{10\sqrt[3]{x^2}}{3}=\frac{0 - [\frac{10}{3\sqrt[3]{x}} \cdot 3]}{9} = \frac{-30}{9(3\sqrt[3]{x})}=\frac{-10}{9\sqrt[3]{x}}$$
$$f''(x)=-\frac{10}{9\sqrt[3]{x^4}}-\frac{-10}{9\sqrt[3]{x}}= \frac{10}{9\sqrt[3]{x}}-\frac{10}{9\sqrt[3]{x^4}}$$
set $f''(x) = 0$ and solve for $x$.
$$\frac{10}{9\sqrt[3]{x}}-\frac{10}{9\sqrt[3]{x^4}}=0 \to x=1$$
find $f''(-1)$.
$$\frac{10}{9\sqrt[3]{(-1)}}-\frac{10}{9\sqrt[3]{(-1)^4}}=\frac{-20}{9}$$
find $f''(2)$
$$\frac{10}{9\sqrt[3]{(2)}}-\frac{10}{9\sqrt[3]{(2)^4}}=\frac{5\sqrt[3]{4}}{18}$$
So the graph should be concave-down on $(-\infty, 1)$ and concave-up on $(1, \infty)$.
However, this is wrong.
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It is a bit easier if you factor out the smallest fractional power of $x$ at each step.
\begin{eqnarray}
f(x)&=&5x^{2/3}-2x^{5/3}=x^{2/3}(5-2x)\\
f^\prime(x)&=&\frac{10}{3}x^{-1/3}-\frac{10}{3}x^{2/3}=\frac{10}{3}x^{-1/3}(1-x)\\
f^{\prime\prime}(x)&=&-\frac{10}{9}x^{-4/3}-\frac{20}{9}x^{-1/3}=-\frac{10}{9}x^{-4/3}(1+2x)
\end{eqnarray}
ADDENDUM:
Note that since $f^\prime(x)$ is undefined at $x=0$ and $f^{\prime\prime}(x)=0$ at $x=-\frac{1}{2}$ you should check the concavity on the intervals $\left(-\infty,-\frac{1}{2}\right)$, $\left(-\frac{1}{2},0\right)$ and $(0,\infty)$
Here is the graph with the visually indiscernible inflection at $\left(-\frac{1}{2}, 3\sqrt[3]{2}\right)$.
|
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"url": "https://math.stackexchange.com/questions/3186799",
"timestamp": "2023-03-29T00:00:00",
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|
System of Distributional Differential Equations
Here are the problem and my attempt to the solution. Is it correct?
|
Let
$$
A = \begin{pmatrix}2 & -1 \\ 3 & -2\end{pmatrix},
\quad
y = \begin{pmatrix}y_1 \\ y_2\end{pmatrix},
\quad
b = \begin{pmatrix}0 \\ 2\end{pmatrix} \delta(x).
$$
The equation can then be written
$y' = Ay + b.$
This can be rewritten as
$y' - Ay = b.$
After multiplication from the left with $e^{-Ax}$ we get
$e^{-Ax}y' - e^{-Ax}Ay = e^{-Ax}b.$
The left hand side can now be written as $(e^{-A}y)'$ giving
$(e^{-Ax}y)' = e^{-Ax}b,$
i.e. $e^{-Ax}y = \int e^{-Ax}b + c,$
where $c$ is a constant vector and $\int$ denotes taking antiderivative.
Thus the solution is given by
$$y = e^{Ax} \int e^{-Ax}b + e^{Ax} c.$$
Now $A^2 = I$ (check it!) so
$$
e^{Ax}
= \sum_{n=0}^{\infty} \frac{1}{n!} (Ax)^n
= \sum_{n\text{ even}} \frac{1}{n!} x^n I + \sum_{n\text{ odd}} \frac{1}{n!} x^n A
= (\cosh x) I + (\sinh x) A
$$
and
$$
e^{-Ax} = (\cosh x) I - (\sinh x) A
.
$$
This makes
$$\begin{align}
e^{-Ax} b
&= \left( (\cosh x) I - (\sinh x) A \right) b \\
&=
\begin{pmatrix}
\cosh x - 2 \sinh x & \sinh x \\
-3 \sinh x & \cosh x + 2 \sinh x
\end{pmatrix}
\begin{pmatrix}
0 \\
2
\end{pmatrix} \delta(x) \\
&=
\begin{pmatrix}
2 \sinh x \\
2 \cosh x + 4 \sinh x
\end{pmatrix}
\delta(x) \\
&=
\begin{pmatrix}
0 \\
2
\end{pmatrix}
\delta(x),
\end{align}$$
where the last identity holds since $f(x) \, \delta(x) = f(0) \, \delta(x).$
Now,
$$
\int e^{-Ax} b = \begin{pmatrix}0 \\ 2\end{pmatrix} H(x),
$$
where $H(x)$ is the Heaviside function.
This gives
$$
e^{Ax} \int e^{-Ax} b
= \left( (\cosh x) I + (\sinh x) A \right)
\begin{pmatrix}0 \\ 2\end{pmatrix} H(x)
= \begin{pmatrix}-2 \sinh x \\ 2 \cosh x - 4 \sinh x\end{pmatrix} H(x).
$$
Thus the general solution is
$$
y(x) =
\begin{pmatrix}
-2 \sinh x\\
2 \cosh x - 4 \sinh x
\end{pmatrix} H(x)
+ \begin{pmatrix}
c_1 \cosh x + 2 c_1 \sinh x - c_2 \sinh x \\
c_2 \cosh x + 3 c_1 \sinh x - 2 c_2 \sinh x
\end{pmatrix},
$$
where $c_1$ and $c_2$ are constants.
Solving the equation using Fourier transform, just like you I get
$$
\begin{pmatrix}\hat{y}_1 \\ \hat{y}_2\end{pmatrix}
= \frac{-1}{1+\xi^2} \begin{pmatrix}i\xi+2 & -1 \\ 3 & i\xi-2\end{pmatrix}
\begin{pmatrix}0 \\ 2\end{pmatrix}
$$
but then you make some mistakes. First you make a sign error. I get
$$
\begin{pmatrix}\hat{y}_1 \\ \hat{y}_2\end{pmatrix}
= \frac{2}{1+\xi^2} \begin{pmatrix}1 \\ 2-i\xi\end{pmatrix}.
$$
The right hand side is the Fourier transform of the convolution
$$
e^{-|x|} * \begin{pmatrix}\delta \\ 2\delta-\delta' \end{pmatrix}
= \begin{pmatrix}e^{-|x|} * \delta \\ e^{-|x|}*(2\delta-\delta') \end{pmatrix}
= \begin{pmatrix}e^{-|x|} \\ 2 e^{-|x|} - (e^{-|x|})' \end{pmatrix} \\
= \begin{pmatrix}e^{-|x|} \\ 2 e^{-|x|} + \operatorname{sign}(x) e^{-|x|} \end{pmatrix}
= e^{-|x|} \begin{pmatrix}1 \\ 2 + \operatorname{sign}(x) \end{pmatrix}.
$$
Thus,
$$
\begin{pmatrix}{y}_1 \\ {y}_2\end{pmatrix}
= e^{-|x|} \begin{pmatrix}1 \\ 2 + \operatorname{sign}(x) \end{pmatrix}
+ \text{solutions to the homogeneous equation}.
$$
I should add something about the solutions to the homogeneous equation and also show that the two solutions are equal.
|
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|
Find the third expression of Taylor's for $f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}$ around $x=0$
Find the third expression of Taylor's for $f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}$ around $x=0$
My try:Let $y=x+1$. Then: $$f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}=\frac{1}{(2y-1)(4-\frac{1}{y^2})^2}=-\frac{1}{2}\cdot\frac{1}{1-2y}\cdot\frac{1}{(1-\frac{1}{4y^2})^2}=$$ $$=-\frac{1}{2}\sum(2y)^n\sum(\frac{1}{4y^2})^n$$That is why I have $$f^{(3)}=3!\frac{-1}{2}\cdot2\cdot\frac{1}{4}=\frac{1}{4}$$Why is it not correct solution?
|
Just for the fun of it !
Starting from @marty cohen's answer and continuing the process
$$f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}=\sum_{n=0}^\infty 2^{n-8} \left[27 (6 n+7)+(-1)^n (2 n^3+13n+67)\right]\,x^n$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving ${\lim\limits_ {n\to\infty}}\frac{6n^3+5n-1}{2n^3+2n+8} = 3$ I'm trying to show that $\exists \,\varepsilon >0\mid\forall n>N\in\mathbb{N}$ such that:
$$\left|\frac{6n^3+5n-1}{2n^3+2n+8}-3\right| < \varepsilon$$
Let's take $\varepsilon = 1/2$:
$$\left|\frac{6n^3+5n-1}{2n^3+2n+8}-3\right| = \left|\frac{6n^3+5n-1 - 6n^3 -6n -24}{2n^3+2n+8}\right| < \left|\frac{-n -25}{2n^3+2n+8}\right| = \left|-\left(\frac{n +25}{2n^3+2n+8}\right)\right|$$
Since $|-x| = |x|$:
$$\left|-\left(\frac{n +25}{2n^3+2n+8}\right)\right| = \left|\frac{n +25}{2n^3+2n+8}\right|<\varepsilon $$
This is the part which I have trouble with. Here, I always end up with my minimum-required index to be smaller than zero, which seems peculiar to me:
$$\left|\frac{n +25}{2n^3+2n+8}\right| < |n+25| = n+ 25 < \varepsilon = 1/2$$
From here, I will get that my $N$ will be less than zero, which means that all the elements of the sequence are inside of my given epsilon environment, but I know that's not true since $a_1 = 5/6 < 3 - \varepsilon $, so what I did do wrong? Is it because I completely removed the denominator? If so, why does that break the inequality?
|
I do it like this:
$\forall n \ge 1, \; \dfrac{6n^3 + 5n - 1}{2n^3 + 2n + 8} = \dfrac{6 + 5n^{-2} - n^{-3}}{2 + 2n^{-2} + 8n^{-3}}; \tag 1$
since it is easy to see that
$\displaystyle \lim_{n \to \infty} (6 + 5n^{-2} - n^{-3}) = 6, \tag 2$
and
$\displaystyle \lim_{n \to \infty} (2 + 2n^{-2} + 8n^{-3} ) = 2, \tag 3$
we have
$\displaystyle \lim_{n \to \infty} \dfrac{6n^3 + 5n - 1}{2n^3 + 2n + 8} = \lim_{n \to \infty} \dfrac{6 + 5n^{-2} - n^{-3}}{2 + 2n^{-2} + 8n^{-3}} = \dfrac{\lim_{n \to \infty}( 6 + 5n^{-2} - n^{-3})}{\lim_{n \to \infty} ( 2 + 2n^{-2} + 8n^{-3}) } = \dfrac{6}{2} = 3. \tag 4$
Of course, such an answer, though perfectly rigorous and on point, is bound to be less than satisfactory to readers who want, as the text of the question indicates, to see the $\epsilon$-$N$ mechanism operate in detail; what I have done here is simply invoke standard and elementary results on the behavior of limits; we may also cast things into $\epsilon$-$N$ form if we write
$\dfrac{6 + 5n^{-2} - n^{-3}}{2 + 2n^{-2} + 8n^{-3}} - 3$
$= \dfrac{ 6 + 5n^{-2} - n^{-3} - 3( 2 + 2n^{-2} + 8n^{-3})}{2 + 2n^{-2} + 8n^{-3}}$
$=\dfrac{5n^{-2} - n^{-3} - 6n^{-2} - 24n^{-3}}{2 + 2n^{-2} + 8n^{-3}} = \dfrac{-n^{-2} - 25n^{-3}}{2 + 2n^{-2} + 8n^{-3}} ; \tag 5$
$\left \vert \dfrac{6 + 5n^{-2} - n^{-3}}{2 + 2n^{-2} + 8n^{-3}} - 3 \right \vert = \left \vert \dfrac{-n^{-2} - 25n^{-3}}{2 + 2n^{-2} + 8n^{-3}} \right \vert; \tag 6$
it is evident via inspection of the right-hand side of this equation that, given any $\epsilon > 0$ there exists a sufficiently large $N \in \Bbb N$ that
$n > N \Longrightarrow \left \vert \dfrac{-n^{-2} - 25n^{-3}}{2 + 2n^{-2} + 8n^{-3}} \right \vert < \epsilon, \tag 7$
which in light of (6) by definition implies
$\displaystyle \lim_{n \to \infty} \dfrac{6 + 5n^{-2} - n^{-3}}{2 + 2n^{-2} + 8n^{-3}} = 3, \tag 8$
completing an $\epsilon$-$N$ demonstration of the requisite limit.
It appears to me that our OP Fatima Ens' principal error lies in
confusing the roles of $\exists$ and $\forall$ in the definition of limit; the quantifying symbol string should read
$\forall \epsilon \exists N \mid \forall n > N \; \text{and so forth}; \tag 9$
if this correction is accepted and the remaining statements bought into accord with this (corrected) version, the chances of completing a successful proof are greatly enhanced.
|
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|
Сonvert trigonometric equation to quadratic equation I am struggling to know how to transform this equation as it involves more than one type of trigonometric function, I know how to do it with one repeated function.
Question:
$\sin^2 \theta/2$ + $\sin \theta$ + $1$ = $0$
In must be transform in pure quadratic form:
$t^2$ + $t$ + $1$ = $0$
|
This is what is sometime called a "trick question" since it does not require you to solve a quadratic equation.
Notice that if $\sin^2\left(\dfrac{\theta}{2}\right)+\sin\theta+1=0$, then it follows that $\sin^2\left(\dfrac{\theta}{2}\right)=-\sin\theta-1\ge0$. So it must be the case that $\sin\theta\le-1$. So $\sin\theta=-1$.
Therefore one would think that $\theta=-\dfrac{\pi}{2}+2\pi n$. But this will contain extraneous solutions. For example, $-\dfrac{\pi}{2}$ is not a solution since
$$ \sin^2\left(-\frac{\pi}{4}\right)+\sin\left(-\frac{\pi}{2}\right)+1\ne0 $$
This raises the question "Are there any values of $n$ which yield a solution?"
\begin{eqnarray}
&&\sin^2\left(-\frac{\pi}{4}+\pi n\right)+\sin\left(-\frac{\pi}{2}+2\pi n\right)+1\\
&=&-\frac{\sqrt{2}}{2}\cos(\pi n)+0-\cos(2\pi n)+0\\
&=&-\frac{\sqrt{2}}{2}\cos(\pi n)-1
\end{eqnarray}
But, since $\cos(\pi n)=\pm 1$ there are no values of $n$ which will give $0$.
So the equation has no solutions.
ADDENDUM
If it is the case that OP intended to write $\dfrac{\sin^2\theta}{2}+\sin\theta+1=0$, that also has no solutions since
\begin{eqnarray}
&&\sin^2\theta+2\sin\theta+2\\
&=&(\sin\theta+1)^2+1\ge1
\end{eqnarray}
SECOND ADDENDUM
Here is an even faster way to show that there are no solutions.
Rewrite the equation using the identity $\sin^2\left(\dfrac{\theta}{2}\right)=\dfrac{1-\cos\theta}{2}$ to obtain
$$ 2\sin\theta-\cos\theta+3=0 $$
$$ \sqrt{5}\left(\frac{2}{\sqrt{5}}\sin\theta-\frac{1}{\sqrt{5}}\cos\theta\right)+3=0$$
Letting $\sin\psi=\dfrac{1}{\sqrt{5}}$ this can be re-written
$$ \sqrt{5}\sin(\theta-\psi)+3=0 $$
But $\sqrt{5}\sin(\theta-\psi)+3\ge3-\sqrt{5}>0$. So there are no solutions.
|
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"url": "https://math.stackexchange.com/questions/3193934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Matrixes with common parameters to result in no inverse I've been given three matrices $A, B \ \& \ C$ which are defined as follows:
$$ A = {
\left[
\begin{array}{ccc}
b & 5 & 8 \\
c & 1 & 3 \\
a & 4 & 3 \\
\end{array}
\right]
},\ B = {
\left[
\begin{array}{ccc}
7 & 4 & 2 \\
5 & 5 & -1 \\
-2 & -a & -b \\
\end{array}
\right]
}, \ C = {
\left[
\begin{array}{ccc}
3 & c & 5 \\
1 & b & 4 \\
2 & a & 1 \\
\end{array}
\right]
} $$
First, I've been asked to find the determinant of each matrix in terms of the parameters, and I've done it like this:
$ |A| = b \begin{vmatrix}
1 & 3 \\
4 & 3 \\
\end{vmatrix} - 5 \begin{vmatrix}
c & 3 \\
a & 3 \\
\end{vmatrix} + 8 \begin{vmatrix}
c & 1 \\
a & 4 \\
\end{vmatrix} = \ b(3-12)-5(3c-3a)+8(4c-a) = \\ -9b - 15c + 15a + 32c - 8a = \mathbf{7a - 9b + 17c} $
$ |B| = 7 \begin{vmatrix}
5 & -1 \\
-a & -b \\
\end{vmatrix} - 4 \begin{vmatrix}
5 & -1 \\
-2 & -b \\
\end{vmatrix} + 2 \begin{vmatrix}
5 & 5 \\
-2 & -a \\
\end{vmatrix} = 7(-5b-a)-4(-5b-2)+2(-5a+(-10)) = -35b - 7a + 20b + 8 - 10a + 20 = \mathbf{-17a -15b + 28} $
$ |C| = 3 \begin{vmatrix}
b & 4 \\
a & 1 \\
\end{vmatrix} - c \begin{vmatrix}
1 & 4 \\
2 & 1 \\
\end{vmatrix} + 5 \begin{vmatrix}
1 & b \\
2 & a \\
\end{vmatrix} = \ 3(b - 4a) - c(1 - 8) + 5(a - 2b) = \\ 3b - 12a – c + 8c + 5a - 10b = \mathbf{-7a -7b + 7c} $
So after this, I have to calculate the value of the parameters to make all three of the matrices not invertible, so I assume $|M| = 0$ has to be true for each matrix. But I don't know how to proceed.
I thought of defining a system of linear equations using the three of them equaled to $ 0 $ but as $|B|$ has no $c$ term I don't know how to acomplish this.
|
Your determinants are right. Next you set them equal to $0$. You do not ignore the number 28. You put it on the RHS as mentioned in the comments.
$$\left(
\begin{array}{ccc|c}
7 & -9 & 17 & 0\\
-17 & -15 & 0 & -28\\
-7 & -7 & 7 &0 \\
\end{array} \right)$$
Now you can use the Gaussian elimination algorithm to obtain the solution. Here are the two steps.
Step 1
Multiply row 1 by $17/7$ and add to row 2.
Add row 1 to row 3.
$$\left(
\begin{array}{ccc|c}
7 & -9 & 17 & 0\\
0 & -258/7 & 289/7 & -28\\
0 & -16 & 24 &0 \\
\end{array} \right)$$
Step 2
Multiply row 2 by $-56/129$ and add to row 3.
$$\left(
\begin{array}{ccc|c}
7 & -9 & 17 & 0\\
0 & -258/7 & 289/7 & -28\\
0 & 0 & 784/129 &1568/129 \\
\end{array} \right)$$
|
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|
Advanced Complex numbers Let $\omega$ be a nonreal root of $z^3 = 1.$ Let $a_1,$ $a_2,$ $\dots,$ $a_n$ be real numbers such that
$$\frac{1}{a_1 + \omega} + \frac{1}{a_2 + \omega} + \dots + \frac{1}{a_n + \omega} = 2 + 5i.$$Compute
$$\frac{2a_1 - 1}{a_1^2 - a_1 + 1} + \frac{2a_2 - 1}{a_2^2 - a_2 + 1} + \dots + \frac{2a_n - 1}{a_n^2 - a_n + 1}.$$
I have no clue how to do this. Can someone help?
|
Hint: $\overline \omega$ is another non-real root of $z^3=1$,
$$\sum_{j=1}^n\frac1{a_j+\overline\omega} = \overline{2+5i}$$
$1,\omega, \overline\omega$ are roots of $z^3-1 = 0$, so
$$\omega\overline\omega = 1,\quad \omega+\overline\omega + 1 = 0$$
Consider the following sum,
$$\begin{align*}
\sum_{j=1}^n\frac1{a_j+\omega} + \sum_{j=1}^n\frac1{a_j+\overline\omega}
&= \sum_{j=1}^n\frac{(a_j+\omega)+(a_j+\overline\omega)}{(a_j+\omega)(a_j+\overline\omega)}\\
&= \sum_{j=1}^n\frac{2a_j+\omega+\overline\omega}{a_j^2 + \omega a_j +\overline\omega a_j + \omega\overline\omega}\\
&= \sum_{j=1}^n\frac{2a_j-1}{a_j^2 -a_j + 1}\\
\end{align*}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Central forces and conservation problem An object A moves under the influence of a planet B's gravity, a central force of magnitude $\frac{1}{r^2}$. At some point, its velocity is $u$e$_{\theta}$ and $r=a$. It is given that $au^2 < 2$. I am looking to find an upper limit to the distance A gets from B.
Integrating the force I found the potential energy to be V = $-\frac{1}{r}$.
From conservation of angular momentum I found that:
L = $au$k = $r^2\dot{\theta}$k $\implies$ $\dot{\theta} = \frac{au}{r^2}$.
From conservation of energy I found that:
\begin{equation*}
\frac{1}{2}\Big(\dot{r}^2 + r^2\dot{\theta}^2\Big) - \frac{1}{r} = \frac{\dot{r}^2}{2} + \frac{a^2u^2}{2r^2} - \frac{1}{r} = \frac{u^2}{2} - \frac{1}{a}.
\end{equation*}
If I use that $\frac{\dot{r}^2}{2} \geq 0$, simplify everything and use $au^2 < 2$ then I end up with:
\begin{equation*}
au^2(r^2 -a^2) -2r^2 +2ar \geq 0 \implies 2a(r-a) \geq 0.
\end{equation*}
So at this point I only have $r \geq a$, but that only gives me a lower bound for this distance, right? If someone could point out where I'm going wrong that would be great.
|
Note that the condition that at $r=a$ the velocity is $u$e$_{\theta}$ means that when $r=a$ we are either at a minimum distance or at a maximum distance.
This is because the condition that marks that we are at an extreme distance is $\dot{r} = 0$, i.e., no radial velocity.
So, from the equation
\begin{equation*}
\frac{\dot{r}^2}{2} + \frac{a^2u^2}{2r^2} - \frac{1}{r} = \frac{u^2}{2} - \frac{1}{a}
\end{equation*}
Using $\dot{r} = 0$ and after multiplying by $2 a r^2$ gives
\begin{equation*}
a^3u^2 - 2 a r = (a u^2 - 2 ) r^2
\end{equation*}
Which has as solutions
\begin{equation*}
r = a \frac{1 \pm \sqrt{1 - (2 - a u^2)a u^2}}{2 - a u^2}
\end{equation*}
Hence
\begin{equation*}
r = a \frac{1 \pm \sqrt{(1 - a u^2)^2}}{2 - a u^2}
\end{equation*}
And, finally
\begin{equation*}
r = a \vee r = \frac{a^2 u^2}{2 - a u^2} > a
\end{equation*}
So, the lower bound for $r$ is $a$ and the upper bound for $r$ is
\begin{equation*}
\frac{a^2 u^2}{2 - a u^2}
\end{equation*}
|
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|
Answer to - proving $\cos 207^o$ is irrational. At first, $\cos(207^o) = \cos(180^o+27^o) = - \cos(27^o)$
We have, $(\sin 27^o + \cos 27^o)^2 = \sin^2 27^o + \cos^2 27^o + 2 \sin 27^o \cos 27^o$
$$ = 1 + \sin 2.27^o = 1 + \sin 54^o = 1 + \sin (90^o - 36^o) = 1 + \cos 36^o$$
$$(\sin 27^o + \cos 27^o)^2 = 1 + \frac{\sqrt{5}+1}{4} = \frac{1}{4}(5+\sqrt 5)$$
$$\sin 27^o + \cos 27^o = \frac{1}{2}{\sqrt{5+\sqrt 5}} ...(a)$$
Similarly, we have,
$$(\sin 27^o- \cos 27^o)^2 = 1 - \cos 36° = 1 -\frac{\sqrt 5 +1}{4} = \frac{1}{4}(3-\sqrt 5)$$
$$ \sin 27^o - \cos 27^o = \pm\frac{1}{2}\sqrt{3-\sqrt 5}$$
Until $45^o$ in the first quadrant, $\sin x < \cos x$.
Thus,
$$ \sin 27^o - \cos 27^o = -\frac{1}{2}\sqrt{3-\sqrt 5}...(b)$$
On subtracting (a) from (b)
$$\cos 207^o = -\cos 27^o = - \frac{1}{4}(\sqrt{5+\sqrt5} + \sqrt{3 - \sqrt 5})$$
As $\sqrt{5+\sqrt5}$ and $\sqrt{3 - \sqrt 5}$ are irrational, $\cos 207^o$ is irrational.
|
As you noticed we have to prove that:
$$\cos(27°)\notin \Bbb{Q}$$
First of all:
$$ \cos(54°)=2\cos^2(27°)-1$$
So: $$\cos(54°)\notin \Bbb{Q}\Rightarrow \cos(27°)\notin \Bbb{Q}$$
Notice that:
$$\cos(54°)=\sin(36°)=\sqrt{\frac{5-\sqrt{5}}{8}} \notin \Bbb{Q}$$
:)
|
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|
If $x = 2$ is a root of $\det\left[\begin{smallmatrix}x&-6&-1\\2&-3x&x-3\\-3&2x&x+2\end{smallmatrix}\right]=0$, find other two roots
If $x = 2$ is a root of equation
$$ \begin{vmatrix}
x & -6 & -1 \\
2 & -3x & x-3\\
-3 & 2x & x+2
\end{vmatrix} = 0 $$
Then find the other two roots.
I solved it and got a cubic equation, and then I divided it by $(x-2)$ to get the other two roots. But this is a long method to do.
Please help me with some shorter approach to this question.
|
$\begin{vmatrix}
x & -6 & -1 \\
2 & -3x & x-3\\
-3 & 2x & x+2
\end{vmatrix} = 0$
$\begin{vmatrix}
x-2 & 3x-6 & 2-x\\
2 & -3x & x-3\\
-3 & 2x & x+2
\end{vmatrix} = 0$
$R_1 \rightarrow R_1-R_2$
$(x-2)\begin{vmatrix}
1 & 3 & -1\\
2 & -3x & x-3\\
-3 & 2x & x+2
\end{vmatrix} = 0$
$\begin{vmatrix}
1 & 3 & -1\\
0 & -3x-6 & x-1\\
0 & 2x+9 & x-1
\end{vmatrix} = 0$
$R_2 \rightarrow R_2-2R_1$,
$R_3 \rightarrow R_3+3R_1$
Now open the determinant using $C_1$
clearly,one factor is (x-1). You get -3x-6= 2x+9, x=-3.
|
{
"language": "en",
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|
Eliminate $\theta$ from $\lambda\cos2\theta=\cos(\theta + \alpha) \space$ and $\space \space\lambda \sin2\theta=2\sin(\theta + \alpha)$ Eliminate $\theta$ from $\lambda \cos2\theta=\cos(\theta + \alpha)$ and $\lambda\sin2\theta=2\sin(\theta + \alpha)$
My approach:
Dividing the RHS and LHS of both equations by $\lambda$, then squaring and adding them, we get,
$$\frac{\cos^2(\theta+\alpha)}{\lambda^2}+\frac{4\sin^2(\theta+\alpha)}{\lambda^2}=\cos^22\theta + \sin^22\theta=1$$
$$\Rightarrow \sin^2(\theta+\alpha)=\frac{\lambda^2-1}{3}$$
I am unable to proceed.
|
We have
\begin{align*}
\tan2\theta&=2\tan(\theta+\alpha)\\
\frac{2\tan\theta}{1-\tan^2\theta}&=\frac{2(\tan\theta+\tan\alpha)}{1-\tan\theta\tan\alpha}\\
\tan\theta-\tan^2\theta\tan\alpha&=\tan\theta(1-\tan^2\theta)+\tan\alpha(1-\tan^2\theta)\\
\tan^3\theta&=\tan\alpha
\end{align*}
From $\lambda\sin2\theta=2\sin(\theta+\alpha)$,
\begin{align*}
2\lambda\sin\theta\cos\theta&=2(\sin\theta\cos\alpha+\cos\theta\sin\alpha)\\
\lambda&=\frac{\cos\alpha}{\cos\theta}+\frac{\sin\alpha}{\sin\theta}\\
&=\frac{\cos\alpha}{\cos\theta}\left(1+\frac{\tan\alpha}{\tan\theta}\right)\\
\lambda^2&=\frac{\sec^2\theta}{\sec^2\alpha}\left(1+\frac{\tan\alpha}{\tan\theta}\right)^2\\
&=\left(\frac{1+\tan^2\theta}{1+\tan^2\alpha}\right)(1+\tan^2\theta)^2\\
&=\frac{(1+\tan^2\theta)^3}{1+\tan^6\theta}\\
&=\frac{1+2\tan^2\theta+\tan^4\theta}{1-\tan^2\theta+\tan^4\theta}\\
(\lambda^2-1)(1+\tan^4\theta)&=(\lambda^2+2)\tan^2\theta\\
\tan^2\theta+\frac{1}{\tan^2\theta}&=\frac{\lambda^2+2}{\lambda^2-1}
\end{align*}
Note that
$$\tan^2\alpha+\frac{1}{\tan^2\alpha}=\tan^6\theta+\frac{1}{\tan^6\theta}=\left(\tan^2\theta+\frac{1}{\tan^2\theta}\right)^3-3\left(\tan^2\theta+\frac{1}{\tan^2\theta}\right)$$
Therefore, $\displaystyle \tan^2\alpha+\frac{1}{\tan^2\alpha}=\left(\frac{\lambda^2+2}{\lambda^2-1}\right)^3-3\left(\frac{\lambda^2+2}{\lambda^2-1}\right)$.
|
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|
If $(2^b-1+c^b)^a = (2^a-1+c^a)^b,$ then is it true that $a=b?$
Question: Assume that $a,b,c$ are real numbers such that
$$1\leq a\leq b, \quad\text{and}\quad 0<c<1.$$
If we have
$$(2^b-1+c^b)^a = (2^a-1+c^a)^b,$$
then is it true that
$$a=b?$$
I can only prove that the statement is true if $a$ is an integer.
In particular, since
$$(2^b-1+c^b)^a = (2^a-1+c^a)^b,$$
by manipulation, we have
$$2^a = (2^b-1+c^b)^{\frac{a}{b}} -c^a+1. $$
Since LHS is an even natural number, so is RHS.
This forces $a=b.$
However, I have no idea how to prove it for real number $a.$
|
Let
$$
f(a,b,c)=a\cdot\ln\Big(2^b-1+c^b\Big)-b\cdot\ln\Big(2^a-1+c^a\Big)
$$
Obviously your statement is equal to finding roots of $f$.
Now, its easy to see that $f(a,b,1)=0$, so we take the derivative:
$$
\frac{\partial f}{\partial c}= \frac{ab}{c(2^b-1+c^b)(2^a-1+c^a)}(2^ac^b-2^bc^a+c^a-c^b)
$$
Notice that the last term can be estimated as
$$
2^ac^b-2^bc^a+c^a-c^b\leq2^bc^b-2^bc^a+c^a-c^b=(2^b-1)(c^b-c^a)<0
$$
due to $1\leq a<b$ and $0\leq c \leq 1$. Thus, $f$ is strictly decreasing in $c$ and hence has no other roots other than $c=1$, for $a\neq b$.
|
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|
Solving limit without L'Hôpital's rule: $\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}$ How can I solve this limit without L'Hôpital's rule?
$$\begin{align}\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}&=\lim\limits_{x \to 0} \frac{\tan x-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}\\&=\lim\limits_{x \to 0} \frac{\frac{\tan x-\sin x}{x^3}}{\sqrt{1+\tan x}+\sqrt{1+\sin x}}\end{align}$$
I can't proceed anymore from here.
|
$$\frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}
=\frac{\tan x-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}.$$
From Maclaurin series $\sin x=x-x^3/6+O(x^5)$ and $\tan x=x+x^3/3+O(x^5)$.
Therefore
$$\frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}
\sim\frac{x^3/2}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}$$
as $x\to0$, and so the limit is $1/4$.
|
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|
Integration by Partial Fractions $\int \sec^3xdx$ I have looked at multiple ways to do partial fractions to integrate $\sec^3x$, but there is a part where I keep getting stuck when I see the partial fractions get split up.
$$\int \frac{\cos x}{\cos^4 x} = \int \frac{1}{(1-y^2)^2}$$
The next step I have seen is shown like this on Wikipedia:
$$\frac{1}{(1-u^2)^2}=\frac{1/4}{1-u}+\frac{1/4}{(1-u)^2}+\frac{1/4}{1+u}+\frac{1/4}{(1+u)^2}.$$
Where does the $\frac{1}{4}$ come from in the numerator?
And for the denominator, why are there two additional positive fraction decompositions $1+u$ and $(1+u)^2$?
|
And for the denominator, why are there two additional positive
fraction decompositions
Because $1-u^2 = (1-u)(1+u).$ It's the "difference of squares" factorization. And so, squaring both sides, we also have $(1-u^2)^2 = (1-u)^2(1+u)^2$.
In partial fractions decomposition, all irreducible factors must appear as denominators, to the highest power that appears, as well as to all lower powers. So that means we need a denominator for $(1-u)$ and $(1-u)^2,$ but also for $(1+u)$ and $(1+u)^2.$
Where does the $\frac{1}{4}$ come from in the numerator?
For partial fraction decomposition, the numerators must be chosen so that if you were to give every term common denominator and combine the numerators, you would get the starting fraction back. It takes a bit of algebra to find what numerators do the job, but that is where the $\dfrac{1}{4}$ comes from.
But let's do the algebra. The normal way to find these numerators is to introduce variables for them and solve. Like so:
$$
\dfrac{1}{(1-u^2)^2} = \dfrac{1}{(1+u)^2(1-u)^2} = \dfrac{A}{1+u} + \dfrac{B}{(1+u)^2} + \dfrac{C}{1-u} + \dfrac{D}{(1-u)^2}
$$
here the "variable" numerators are called $A, B, C,$ and $D.$
We solve for them by clearing denominators:
$$
1 = A(1-u)^2(1+u) + B(1-u)^2 + C(1-u)(1+u)^2 + D(1+u)^2
$$
Then in general you may have to expand all those binomials and match like terms of the resulting polynomials in $u$. But a shortcut can be to sub in the values of the roots in $u$. For example, at $u=1,$ the above equation becomes
$$
1 = A(1-1)^2(1+1) + B(1-1)^2 + C(1-1)(1+1)^2+ D(1+1)^2 = D(1+1)^2.
$$
So we have $D=1/4.$
Similarly, by next subbing $u=-1,$
$$
1 = B(1-(-1))^2
$$
So we have $B=1/4.$
To figure out that $A$ and $C$ are also $1/4$, we can either sub in some other, non-root values of $u$, or bite the bullet and do the algebra. (But if this were a partial fractions problem with no powers of irreducibles, we'd be done).
I'll do the algebra, I guess. Expand, and gather like terms in $u$, treating the coefficients $A, B, C,$ and $D$ as variables.
$$
1 = A(1-u-u^2+u^3) + B(1-2u+u^2) + C(1+u-u^2-u^3) + D(1+2u+u^2) \\
= (A+B+C+D) + (-A-2B+C+2D)u + (-A+B-C+D)u^2 + (A-C)u^3.
$$
Now for two polynomials in $u$ to be equal for all values of $u$, every coefficient must be equal. On the left-hand side we have the polynomial $1$, which has a constant coefficient of $1$ and all higher coefficients are $0$.
So this gives us the following equations:
$$
1 = A+B+C+D\\
0 = -A-2B+C+2D\\
0 = -A+B-C+D\\
0 = A-C.
$$
We could solve this system of four linear equations in four unknowns, using Gaussian elimination or substitution or matrix methods. But first let's remember that we already know that $B=D=1/4.$ So we don't have four unknowns any more, our only unknowns left are $A$ and $C$. By the fourth equation, $A=C$, so our only unknown is really $A$. The first equation becomes
$$
1 = A+B+C+D = A + \dfrac{1}{4} + A + \dfrac{1}{4}.
$$
So $2A=1-\dfrac{2}{4} = \dfrac{1}{2}.$ And so $A=B=C=D=\dfrac{1}{4}.$
|
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|
Compute $\sum_{k≥1,n≥1}\frac{(-1)^{n+k}}{k(k+1+n)^2}$ How can Compute in closed form this double summation :
$$\sum_{k≥1,n≥1}\frac{(-1)^{n+k}}{k(k+1+n)^2}$$
I think here can use harmonic series
Actually I don't have any ideas to approach it
|
The idea is to replace the double sum by a double integral which then hopefully can be solved. The hope is justified and we find the following
Result
The closed form is
$$s = \sum_{k≥1,n≥1}\frac{(-1)^{n+k}}{k(k+1+n)^2}\\=\frac{1}{2}\zeta(2) - 2 (1-\log(2))-\frac{1}{8} \zeta(3)\simeq 0.0585043\tag{1}$$
Derivation
The replacement starts with
$$\frac{1}{(n+k+1)^2} = \int_{0}^1 \frac{1}{y} \left(\int_0^y x^{n+k}\,dx \right)\,dy\tag{2}$$
Now we do the double sum under the integral
$$\sum_{n=1}^\infty (\sum_{k=1}^\infty \frac{1}{k} (-x)^{n+k}) =\frac{x \log (x+1)}{x+1}\tag{3}$$
then the $x$-integral
$$\int_0^y \frac{x \log (x+1)}{x+1} \, dx = -y-\frac{1}{2} \log ^2(y+1)+(y+1) \log (y+1)\tag{4}$$
and finally the remaining $y$-integral
$$\int_0^1 \frac{1}{y}(-y-\frac{1}{2} \log ^2(y+1)+(y+1) \log (y+1)) \, dy\tag{5}$$
Observing that
$$\int_0^1 \frac{1}{y} (\log(1+y))^2 = \frac{1}{4} \zeta(3)\tag{6a}$$
$$\int_0^1 \frac{1}{y} \log(1+y) = \frac{1}{2} \zeta(2)\tag{6b}$$
$$\int_0^1 \log(1+y) = -1+\log(4)\tag{6c}$$
$(5)$ gives $(1)$.
Q.E.D.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the equation of the plane through the point $(2,1,4)$ Find the equation of the plane through the point $(2,1,4)$ and perpendicular to each of the planes $9x-7y+6z+48=0$ and $x+y+z=0$.
My attempt:
The equation of the plane passing through the point $(2,1,4)$ is given by
$$A(x-2)+B(y-1)+C(z-4)=0$$
Here, $A,B,C$ represents the direction ratios of normal to the plane.
Since, it is perpendicular to the plane $9x-7y+6z+48=0$,
$$9A-7B+6C=0$$
Also, the plane is perpendicular to another plane $x+y+z=0$,
$$A+B+C=0$$
|
By solving the second and third equations, we have
\begin{align}
B&=\frac{3}{13}A\\
C&=-\frac{16}{13}A
\end{align}
So the plane is
\begin{align}
A(x-2)+3(y-1)A/13-16(z-4)A/13 &=0\\
13(x-2)+3(y-1)-16(z-4) &=0
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Decomposing a quartic polynomial into a difference of squares Given quartic polynomial $1 + x + x^2 +x^3 + x^4$, I want to find polynomials $p, q \in \mathbb R[x]$ such that
$$1 + x + x^2 +x^3 + x^4 = p^2 (x) - q^2 (x)$$
|
$$1 + x + x^2 +x^3 + x^4 = \frac 12 \begin{bmatrix} 1\\ x\\ x^2\end{bmatrix}^\top \begin{bmatrix} 2 & 1 & t\\ 1 & 2-2t & 1\\ t & 1 & 2\end{bmatrix} \begin{bmatrix} 1\\ x\\ x^2\end{bmatrix}$$
Since we want the matrix to be rank-$2$,
$$\det \begin{bmatrix} 2 & 1 & t\\ 1 & 2-2t & 1\\ t & 1 & 2\end{bmatrix} = 2 (t - 2) (t^{2} + t - 1) = 0$$
For $\color{blue}{t = 2}$, the matrix has one positive eigenvalue, one negative eigenvalue, and one zero eigenvalue. Using the spectral decomposition, eventually we obtain
$$\begin{aligned} p (x) &:= \left(\frac{\sqrt{\sqrt{11} + 1}}{4 \sqrt{11 + 3 \sqrt{11}}}\right) \left( \left(\sqrt{22} + 3 \sqrt{2}\right) x^{2} + 2 \sqrt{2} x + \left( \sqrt{22} + 3 \sqrt{2} \right)\right)\\ q (x) &:= \left(\frac{\sqrt{\sqrt{11} - 1}}{4 \sqrt{11 - 3 \sqrt{11}}} \right) \left( \left(\sqrt{22} - 3 \sqrt{2}\right) x^{2} - 2 \sqrt{2} x + \left( \sqrt{22} - 3 \sqrt{2} \right)\right)\end{aligned}$$
such that
$$1 + x + x^2 +x^3 + x^4 = p^2 (x) - q^2 (x)$$
SymPy code
from sympy import *
t = Symbol('t', real=True)
x = Symbol('x', real=True)
M0 = Matrix([[ 2, 1, 0],
[ 1, 2, 1],
[ 0, 1, 2]])
M1 = Matrix([[ 0, 0, 1],
[ 0,-2, 0],
[ 1, 0, 0]])
M = M0 + t * M1
# find roots of the determinant
roots = solve(M.det(), t)
for r in roots:
print "Root: ", r.evalf()
print "Determinant:", M.subs(t,r).det()
print "Eigenvalues:", M.subs(t,r).eigenvals()
print "\n"
# compute eigendecomposition
V, D = M.subs(t,roots[0]).diagonalize()
# normalize eigenvectors
Q = V * ( diag(V[:,0].norm(), V[:,1].norm(), V[:,2].norm()) )**-1
p = collect(simplify(sqrt( D[1,1] / 2) * (Q[:,1].T * Matrix([1,x,x**2]))[0,0]), x)
q = collect(simplify(sqrt(-D[2,2] / 2) * (Q[:,2].T * Matrix([1,x,x**2]))[0,0]), x)
print "p: ", p
print "q: ", q
print "Difference of squares:", simplify(p**2 - q**2)
|
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|
Find $P$ where $P^{-1}AP$ for a given matrix $A$ I am doing a past paper and I have been given a matrix A:
\begin{bmatrix}
4 & -1 & -3 & 2 \\
4 & -2 & -4 & 4 \\
-4 & 4 & 6 & -4 \\
-6 & 5 & 7 & -4 \\
\end{bmatrix}
and I need to find a matrix $P$ such that $P^{-1}AP$ =
\begin{bmatrix}
2 & 0 & 0 & 0 \\
0 & 2 & 1 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & -2 \\
\end{bmatrix}
I'm really not sure how to begin with this question, would anyone be able to help out?
|
From the given information, we know that matrix A had eigenvalues 2, -2, and 2 is an eigenvalue of multiplicity.
We want to find eigenvectors that are associated with each of these eigenvalues.
find $v$ such that
$(A-\lambda I) v = 0$
$\lambda = 2$
$\begin{bmatrix}
2 & -1 & -3 & 2 \\
4 & -4 & -4 & 4 \\
-4 & 4 & 4 & -4 \\
-6 & 5 & 7 & -6 \\
\end{bmatrix}v = 0$
With a little bit of trial and error we find
$\begin{bmatrix} 1\\1\\1\\1 \end{bmatrix}, \begin{bmatrix} -1\\0\\0\\1 \end{bmatrix} $
You might have found different eigenvectors, any two independent vectors in the eigenspace will do.
for the remaining vector we are looking for some vector $w$ such that
$Aw = 2w + v$ where $v$ is one of the the eigenvectors that is associated with the eigenvalue 2.
if such a vector exists then
$(A-2I) w = v\\
(A-2I)^2 w = (A-2I)v = 0$
$(A-2I)^2 = \begin{bmatrix} 0&0&0&0\\-16&16&16&-16\\16&-16&-16&16\\16&-16&-16&16\end{bmatrix}$
Which means that $w$ must be some linear combination of
$\begin{bmatrix} 1\\1\\0\\0\end{bmatrix},\begin{bmatrix} 1\\0\\1\\0\end{bmatrix},\begin{bmatrix} 1\\0\\0\\1\end{bmatrix}$
find some combination of these such that $(A-2I)w$ equals one of the eigenvectors you already have.
$(A-2I)\begin{bmatrix} 1\\0\\1\\0\end{bmatrix} =\begin{bmatrix} -1\\0\\0\\1 \end{bmatrix}$
$A \begin{bmatrix} 1&-1&1\\1&0&0\\1&0&1\\1&1&0\end{bmatrix}= \begin{bmatrix} 1&-1&1\\1&0&0\\1&0&1\\1&1&0\end{bmatrix}\begin{bmatrix} 2&0&0\\0&2&1\\0&0&2\end{bmatrix}$
And I will leave it to you to find the remaining eigenvector and complete the matrix.
|
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|
Differential equation $y'' - y + 2\sin(x)=0$ I need help and explanation with this differential equation. Actually I really don't know how to solve just this type of equations. So the problem: $$y'' - y + 2\sin(x)=0$$
In my opinion first of all we solve homogeneous equation $y''-y=0$ and the solution of this is $y=c_1e^x+c_2e^{-x}$. And after that to solve it with $2\sin(x)$. From this point I need help.
|
The given differential equation is
$y'' - y + 2\sin\ x=0$
$\implies y'' - y =- 2\sin x \implies (D^2 -1)y=- 2\sin x$
where $D \equiv \frac{d}{dx} $
I think you have an idea about how to find the Complementary Function (i.e., C.F.), (for your case, which is nothing but the solution of the homogeneous differential equation $y'' - y =0$).
Here C.F. is $c_1e^{x}+c_2e^{-x}$.
Now for the Particular Integral (i.e., P.I.) there are some general rules
If $f(D)$ can be expressed as $\phi(D^2)$ and $\phi(-a^2)\neq 0$, then
$1.$ $\frac{1}{f(D)} \sin ax=\frac{1}{\phi(D^2)} \sin ax = \frac{1}{\phi(-a^2)} \sin ax$
$2.$ $\frac{1}{f(D)} \cos ax=\frac{1}{\phi(D^2)} \cos ax = \frac{1}{\phi(-a^2)} \cos ax$
Note: If $f(D)$ can be expressed as $\phi(D^2)=D^2+a^2$, then $\phi(-a^2)= 0$.
$1.$ $\frac{1}{f(D)} \sin ax =\frac{1}{\phi(D^2)} \sin ax=x\frac{1}{\phi'(D^2)} \sin ax= x \frac{1}{2D} \sin ax= -\frac{x}{2a} \cos ax$.
$2.$ $\frac{1}{f(D)} \cos ax =\frac{1}{\phi(D^2)} \cos ax=x\frac{1}{\phi'(D^2)} \cos ax= x \frac{1}{2D} \cos ax= \frac{x}{2a} \sin ax$.
where $\phi'(D^2)\equiv\frac{d}{dD}\phi(D^2)$
So for your problem, P.I. is $\frac{1}{D^2 -1} (-2\sin x)=-2[\frac{1}{-1^2 -1} \sin x]= \sin x$
Hence the general solution is $y=$ C.F. $+$ P.I. $ = c_1e^{x}+c_2e^{-x}+\sin\, x$
|
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|
Proof that $ p^2 \equiv 1 \mod{30} \vee p^2 \equiv 19 \mod{30} $ for $p>5$ Proof that $$ p^2 \equiv 1 \mod{30} \vee p^2 \equiv 19 \mod{30} $$
for $p>5$ and $p$ is prime.
$\newcommand{\Mod}[1]{\ (\mathrm{mod}\ #1)}$
My try
Let show that
$$p^2 - 1 \equiv 0 \Mod{30} \vee p^2 - 1\equiv 18 \Mod{30}$$
Let check $$p^2 -1 = (p-1)(p+1) $$
We know that (for example from here) that this is dividable by $2$ and by $3$ so by $6$
Let consider $5$ cases:
$$\exists_k p=5k \rightarrow \mbox{false because p is prime}$$
$$\exists_k p=5k+1 \rightarrow p^2 - 1 = 5k(5k+2) \rightarrow \mbox{ dividable by 6 and by 5 so we have rest 0}$$
$$\exists_k p=5k+2 \rightarrow p^2 - 1 = (5k+1)(5k+3) \rightarrow \mbox{ ??? }$$
$$\exists_k p=5k+3 \rightarrow p^2 - 1 = (5k+2)(5k+4) \rightarrow \mbox{ ??? }$$
$$\exists_k p=5k+4 \rightarrow p^2 - 1 = (5k+3)(5k+5) \rightarrow \mbox{ dividable by 6 and by 5 so we have rest 0}$$
I have stucked with $???$ cases...
|
$p^2-1$ is divisible by $2$ and $3$ and leaves remainder $0$ or $3$ when divided by $5$
[since if $p\not\equiv0\pmod{5}$ then $p^2\equiv1$ or $4\pmod5$].
In the former case, $p^2-1\equiv0\pmod{30}$; in the latter, $p^2-1\equiv18\pmod {30}.$
|
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|
Evaluate $\lim\limits_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}$ Solution
Expanding $(1+x)^{\frac{1}{x}}$ at $x=0$ by Taylor's Formula，we obtain
\begin{align*}
(1+x)^{\frac{1}{x}}&=\exp\left[\frac{\ln(1+x)}{x}\right]=\exp \left(\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots }{x}\right)\\
&=\exp \left(1-\frac{x}{2}+\frac{x^2}{3}+\cdots \right)=e\cdot\exp \left(-\frac{x}{2}+\frac{x^2}{3}+\cdots \right) \\
&=e\left[1+\left(-\dfrac{x}{2}+\dfrac{x^2}{3}+\cdots \right)+\frac{1}{2!}\left(-\dfrac{x}{2}+\dfrac{x^2}{3}+\cdots \right)^2+\cdots\right]\\
&=e\left(1-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)\\
&=e-\frac{ex}{2}+\frac{11}{24}ex^2+\cdots
\end{align*}
Likewise, expanding $e^{(1+x)^{\frac{1}{x}}}$ at $x=0$, we obtain
\begin{align*}
e^{(1+x)^{\frac{1}{x}}}&=(e^e)^{1-\frac{x}{2}+\frac{11}{24}x^2-\cdots}=e^e\cdot (e^e)^{-\frac{x}{2}+\frac{11}{24}x^2+\cdots}\\
&=e^e\cdot\left[1+\left(-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)\ln e^e+\frac{1}{2!}\left(-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)^2\ln^2 e^e+\cdots\right]\\
&=e^e \cdot\left[1-\frac{ex}{2}+\frac{1}{24}(11e+3e^2)x^2+\cdots\right]
\end{align*}
Expanding $(1+x)^{\frac{e}{x}}$ at $x=0$, it follows that
\begin{align*}
(1+x)^{\frac{e}{x}}&=\exp\left[\frac{e\ln(1+x)}{x}\right]=\exp \left(e\cdot\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots }{x}\right)\\
&=\exp \left(e-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)=e^e\cdot\exp \left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right) \\
&=e^e\left[1+\left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)+\frac{1}{2!}\left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)^2+\cdots\right]\\
&=e^e\left[1-\frac{ex}{2}+\frac{1}{24}e(8+3e)x^2+\cdots\right]
\end{align*}
Therefore
\begin{align*}
&\lim_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}\\
=&\lim_{x \to 0}\frac{e^e\cdot\left[1-\dfrac{ex}{2}+\dfrac{1}{24}e(11+3e)x^2+\cdots\right]-e^e\cdot\left[1-\dfrac{ex}{2}+\dfrac{1}{24}e(8+3e)x^2+\cdots\right]}{x^2}\\
=&e^e\cdot\frac{1}{8}e\\
=&\frac{1}{8}e^{e+1}
\end{align*}
Please check. Is there any more simpler solution?
|
Your result is correct. A simpler method is to decompose this limit into a product of simpler limits:
Let $f(x)=(1+x)^{1/x}$. We have $\lim_{x\rightarrow 0} f(x) = e$. Note that your expression can be written as
$$ \lim_{x\rightarrow 0} \frac{e^{f(x)}- f(x)^e}{x^2} = \lim_{x\rightarrow 0} \frac{e^{f(x)}- f(x)^e}{(f(x)-e)^2}\frac{(f(x)-e)^2}{x^2} = \lim_{y\rightarrow e} \frac{e^y - y^e}{(y-e)^2} \cdot \left(\lim_{x\rightarrow 0} \frac{f(x)-e}{x}\right)^2 $$
From l'Hôpital's rule we have
$$ \lim_{y\rightarrow e} \frac{e^y - y^e}{(y-e)^2} =^H \lim_{y\rightarrow e} \frac{e^y - ey^{e-1}}{2(y-e)} =^H \lim_{y\rightarrow e} \frac{e^y - e(e-1)y^{e-2}}{2} = \frac12 e^{e-1}$$
$$ \lim_{x\rightarrow 0} \frac{f(x)-e}{x} = \lim_{x\rightarrow 0} \frac{e^{\ln(1+x)/x}- e}{\ln(1+x)/x - 1}\frac{\ln(1+x)/x - 1}{x} = \lim_{y\rightarrow 1}\frac{e^y -e}{y-1} \lim_{x\rightarrow 0}\frac{\ln(1+x)-x}{x^2} $$
$$ \lim_{y\rightarrow 1}\frac{e^y -e}{y-1} =^H \lim_{y\rightarrow 1} \frac{e^y}{1} = e $$
$$ \lim_{x\rightarrow 0}\frac{\ln(1+x)-x}{x^2} =^H \lim_{x\rightarrow 0}\frac{\frac{1}{1+x}-1}{2x} = \lim_{x\rightarrow 0}\frac{-1}{2(1+x)} = -\frac12$$
In total
$$ \lim_{x\rightarrow 0} \frac{e^{f(x)}- f(x)^e}{x^2} = \frac12 e^{e-1} \cdot\left(-\frac{e}{2}\right)^2 = \frac{e^{e+1}}{8}$$
|
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|
To find minimum distance between two curves Let $P(x, y, 1)$ and $Q(x, y, z)$ lie on the curves $$\frac{x^2}{9}+\frac{y^2}{4}=4$$ and $$\frac{x+2}{1}=\frac{y-\sqrt{3}}{\sqrt{3}}=\frac{z-1}{2}$$ respectively. Then find the square of the minimum distance between $P$ and $Q$.
My Attempt is:
I tried to find minimum distance between the points $(-2,\sqrt{3})$ and $(6\cos \theta,4\sin \theta)$.
|
Starting from @Christian Blatter's answer, using $s=2 \tan ^{-1}(x)$ and expanding, we end with
$$2 \sqrt{3}\, x^4+70 \,x^3+72 \sqrt{3} \,x^2-274\, x-26 \sqrt{3}=0$$
Let $x=t-\frac{35}{4 \sqrt{3}}$ to get the depressed quartic
$$t^4-\frac{937 }{8}t^2+\frac{24467}{24 \sqrt{3}} t-\frac{166043}{256}=0$$ which can be exactly solved using radicals.
Following the steps given here, we have
$$\Delta=\frac{386701126204}{27}\quad P=-937\quad Q=\frac{24467}{3 \sqrt{3}}\quad \Delta_0=5935\quad D=-261003$$ So, four real roots with
$$p=-\frac{937}{8}\quad q=\frac{24467}{24 \sqrt{3}}$$
Just finish to get the exact values of $(t_1,t_2,t_3,t_4)$ from which $(x_1,x_2,x_3,x_4)$ and finally $(s_1,s_2,s_3,s_4)$ in terms of messy radicals.
|
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|
Arrange $x,y,z$ in ascending order. $c>1, c \in \mathbb{R}$
$x = \frac{\sqrt{c+2} - \sqrt{c+1}}{\sqrt{c} - \sqrt{c-1}}
= (\sqrt{c+2} - \sqrt{c+1})(\sqrt{c} + \sqrt{c-1})$
$y = \frac{\sqrt{c+2} - \sqrt{c+1}}{\sqrt{c + 1} - \sqrt{c}}
= (\sqrt{c+2} - \sqrt{c+1})(\sqrt{c+1} + \sqrt{c})$
$z = \frac{\sqrt{c} - \sqrt{c-1}}{\sqrt{c+2} - \sqrt{c+1}}
= (\sqrt{c} - \sqrt{c-1})(\sqrt{c+2} + \sqrt{c+1})$
It is pretty easy to prove that $x<y$. I have already done that. By experimental substitution, I have found out that $y<z$. But every time I try out prove this generically, I end up with LHS and RHS both looking terrible and yielding unwanted false results and conclusions like $ \sqrt x < 0$. Can someone please help me out in proving that $ y<z $??? Please. Even a definitive good hint is helpful.
|
Hint One way to look at it is that $t\mapsto \sqrt t$ is concave, so successive secants have lower slope. Now can you relate $z,x$ by considering them as ratios of slopes?
|
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|
Three ways to find the normal of a hyperboloid? In the hyperboloid $x^2 + y^2 − z^2 = 4$, where $z \ge 0$, I have found three ways to get the normal vector, but my problem is they do not seem equivalent.
The first is as I have been taught that you can find the gradient of the function, and that will be the normal:
$$\langle 2x,2y,-2z \rangle$$
The unit normal should be
$$\langle {2x\over \sqrt{4x^2+4y^2+4z^2} },{2y\over \sqrt{4x^2+4y^2+4z^2}},{-2z\over \sqrt{4x^2+4y^2+4z^2}} \rangle$$
$$=\langle {2x\over \sqrt{4x^2+4y^2+4(x^2+y^2-4)} },{2y\over \sqrt{4x^2+4y^2+4(x^2+y^2-4)}},{-2\sqrt{x^2+y^2-4}\over \sqrt{4x^2+4y^2+4(x^2+y^2-4)}} \rangle$$
$$=\langle {2x\over 2\sqrt{2x^2+2y^2-4} },{2y\over 2\sqrt{2x^2+2y^2-4}},{-2\sqrt{x^2+y^2-4}\over 2\sqrt{2x^2+2y^2-4}} \rangle$$
A second way would be to rewrite the hyperboloid as $z=\sqrt{x^2+y^2-4}$
and then use the identity that the normal is $\langle f_x,f_y,-1 \rangle$
Then the normal is $$\langle {x\over \sqrt{x^2+y^2-4} },{y\over \sqrt{x^2+y^2-4}},-1 \rangle$$
The unit normal is then
$$\langle {x \sqrt{x^2+y^2-4} \over 2x^2+2y^2-4 },{y \sqrt{x^2+y^2-4} \over 2x^2+2y^2-4},-\big( {x^2+y^2-4\over 2x^2+2y^2-4} \big) \rangle$$
The third is to use the parametrization $x=r \cos t, y=r\sin t ,z= r^2$. Then the normal is given by $$r_r \times r_t= \langle -2r^2\cos t, -2r^2\sin t, r \rangle$$
The unit normal should be
$$\langle {-2r\cos t\over \sqrt{4r^2+1}}, {-2r\sin t \over \sqrt{4r^2+1}}, {1\over \sqrt{4r^2+1}} \rangle$$
However, it does not seem that these are equivalent, so which ones are incorrect and why?
|
Ignoring normalization ...
*
*Your first vector, divided-through by $2$, is $(x,y,−z)$.
*Your second vector, multiplied-through by $\sqrt{x^2+y^2−4}$, is $(x,y,−\sqrt{x^2+y^2−4})=(x,y,−z)$
So, those vectors are equivalent.
Your third vector is problematic, which isn't surprising, since your parameterization doesn't satisfy the surface's equation:
$$(r\cos t)^2 + ( r\sin t)^2 - ( r^2 )^2 = 4 \quad\to\quad r^2\cos^2t+r^2\sin^2t-r^4=4\quad\to\quad r^2-r^4=4$$
which, for arbitrary $r$, generally isn't true.
One correct parameterization would be
$$x = r \cos t \qquad y = r \sin t \qquad z = \sqrt{r^2-4}$$
(Verifying that this satisfies the surface's equation is left as an exercise to the reader.)
If you don't care for the square root, you can make the substitution $r=2\sec s$, so that the parameterization becomes
$$x = 2\sec s \cos t \qquad y = 2 \sec s \sin t \qquad z = \sqrt{4\sec^2s-4}=\sqrt{4\tan^2s}=2\tan s$$
In the latter case,
$$\begin{align}
\frac{\partial}{\partial s}(x,y,z) \times \frac{\partial}{\partial t}(x,y,z) &= ( 2 \sec s \tan s \cos t, 2 \sec s \tan s \sin t, 2 \sec^2 s ) \tag{1}\\
&\quad\times (-2 \sec s \sin t, 2 \sec s \cos t, 0) \\[4pt]
&\propto ( \sin s \cos t, \sin s \sin t, 1 ) \times (-\sin t, \cos t, 0) \tag{2}\\[4pt]
&= (-\cos t, -\sin t, \sin s) \tag{3}\\
&\propto (2\sec s\cos t, 2\sec s\sin t, -2\tan s) \tag{4}\\
&=(x,y,-z) \tag{5}
\end{align}$$
In $(3)$, I scaled the vectors by dividing-through by $2\sec^2s$ and $2\sec s$, respectively, to simplify the calculation of their cross-product. In $(4)$, I multiplied-through by $-2\sec s$ to make the equivalence with $(x,y,-z)$ apparent.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3228455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
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Subsets and Splits
Fractions in Questions and Answers
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