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H: Determine the matrix of a quadratic function
I'm given a quadratic form $\Phi:\mathbb{R}^3\longrightarrow\mathbb{R}$, for which we know that:
$(0,1,0)$ and $(0,1,-1)$ are conjugated by the function
$(1,0,-1)$ belongs to the kernel
$\Phi(0,0,1)=1$
The trace is $0$
From here, I know the matrix must be symmetric, so it will have up to six unique numbers, I have four conditions that I can apply, and I will get four equations that relate the numbers of the matrix, but I still need two to full determine it. Applying the above I get that the matrix must be of the form:
$$A=\pmatrix{2c-1 & b & c\\b&-2c&-2c\\c & -2c & 1}$$
How do I determine $b$ and $c$?
AI: Ok, solved it, the last two equations came from knowing the vector that was in the kernel, so it should be that $f_p[(1,-0,-1),(x,y,z)]=0$, being $f_p$ the polar form of $\Phi$: $f_p=\frac{1}{2}[\Phi(x+y)-\Phi(x)-\Phi(y)]$ |
H: Finding covariance of two negative binomial distributions
Take a set of random variables $X_n$ which indicate number of tosses of a coin ($P(H)=p$) to achieve $n$ heads. What is $\operatorname{cov}(X_1,X_5)$?
I reasoned that the number of tosses till the first heads is not going to affect the number of tosses for the next four. So I can split $X_5$ into $X^{'}_1+X^{'}_4$ which are the number of tosses to get the first and latter 4 respectively of $X_5$. Since those two are independent, $\operatorname{cov}(X_1,X_5)$ becomes $\operatorname{cov}(X_1,X^{'}_1)$. Since $X_1$ and $X^{'}_1$ are the same, $\operatorname{cov}(X_1,X^{'}_1)=\operatorname{var}(X_1)=\frac{1-p}{p^2}=\operatorname{cov}(X_1,X_5)$
Is this reasoning sound?
AI: Community wiki answer so the question can be marked as answered: Yes, the argument is sound. |
H: Does $f(\epsilon)=o(\epsilon\ln(\epsilon))$ imply $\frac{f(\epsilon)}{\epsilon}=o(1)$?
I have the following homework question:
Does $f(\epsilon)=o(\epsilon\ln(\epsilon))$ imply
$\frac{f(\epsilon)}{\epsilon}=o(1)$ ?
It doesn't seem correct to me, using the definition I could only get
$$\frac{f(\epsilon)}{\epsilon}=o(\ln(\epsilon))$$ and it doesn't seem
that if $g(\epsilon)=o(\ln(\epsilon))$ then $g(\epsilon)=o(1)$.
To contradict this I wish to find a function $c(\epsilon)$ s.t $$\lim_{\epsilon\to0}c(\epsilon)=0$$
but $$\lim_{\epsilon\to0}|c(\epsilon)\ln(\epsilon)|>0$$
I am having difficulty finding such a function, I tried a few and
checked them using WA, they all satisfied $$\lim_{\epsilon\to0}c(\epsilon)=0$$
but they all also satisfied that $$\lim_{\epsilon\to0}|c(\epsilon)\ln(\epsilon)|=0$$
Can someone please help me find such $c(\epsilon)$ ? (or surprise
me and show that this statement is in fact true)
AI: The notation
$$f(x)=_a o(g(x))$$
means $$\lim_{x\to a}\frac{f(x)}{g(x)}=0\iff \frac{f(x)}{g(x)}=_a o(1) $$
and to contradict your implication just take $f(\epsilon)=\epsilon$ |
H: Is a linear map of norm $1$ always an isometry?
Let $(E,\|.\|_{E})$ and $(F,\|.\|_{F})$ be two normed spaces and let $f:E\longrightarrow F$ a linear map such that $\|f\|=1$. I don't know if this means that $f$ is an isometry $(\|f(x)\|_{F}=\|x\|_{E}, \,\, \forall x\in E)$. Thanks
AI: It does not. Consider the function $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $f(x,y)=x$, with both $\mathbb{R}^2$ and $\mathbb{R}$ endowed with the $\| \|_\infty$ norm. Then clearly $\|f\|=1$. However $f$ is not an isometry (it is not even invertible : $f(0,y)=0$ for all $y$). |
H: How to solve $e^{ix} = i$
I am taking an on-line course and the following homework problem was posed: $$e^{ix} = i$$ I have no idea how to solve this problem. I have never dealt with solving equations that have imaginary parts. What are the steps to solving such equations? I am familiar with Taylor series and the Euler formula if that is any help.
Thanks.
AI: So, you say that you're okay with Euler's formula:
$$e^{ix}=\cos(x)+i\sin(x),$$ and it should be clear that (for real numbers $a,b,c,d$), we have $$a+bi=c+di\quad\iff\quad a=c\;\text{ and }\;b=d.$$
So, which values of $x$ will satisfy
$$\cos(x)+\sin(x)i=0+1i\quad ?$$ |
H: linear map of linear in/dependent vectors
L a linear map of $\mathbb{R}^m$ in $\mathbb{R}^n$.
1)Does a linear map of two linear dependent vectors $\underline{x}$,$\underline{y} \in \mathbb{R}^m$ to two linear independent vectors $\underline{u}$,$\underline{v} \in \mathbb{R}^n$ exist/is possible?
2)a)Does a linear map of two linear independent vectors $\underline{x}$,$\underline{y} \in \mathbb{R}^m$ to two linear dependent vectors $\underline{u}$,$\underline{v} \in \mathbb{R}^n$ exist/is possible?
b)And if it is what are the consequences for $ker(L)$?
1)No, because if two vectors are dependent:
$\exists k \in \mathbb{R}$,
so that: $k\underline{x}=\underline{y}$
and a linear map to two is linear dependent vectors $\in \mathbb{R}^n$is only possible by two vectors $\in \mathbb{R}^m$.
2)a)Yes, f.ex.: $f:\mathbb{R}^3 -> \mathbb{R}^2$,
with $f(x,y,z)^T = (x, y-z)^T$ for $(0,1,0)^T$ and $(0,0,1)^T$.
b)with the rank-nullity theorem:
$ dim V - dim (im T)= dim (ker T)$
$3 - 2 = 1 = dim (ker T)$ in the example
My questions are:
Is my explication for 1) & 2) sufficient?and if not how can I improve it?
Is 2)b) I don't really know how to interprete that...
AI: Hints:
I'd explain (1) as follows, applying explicitly a linear map:
$$x,y\in\Bbb R^m\;,\;\;a\in\Bbb R\;,\;\;T\;\text{a linear map}\;,\; x=ay\implies Tx= T(ay)=aTy$$
You don't need to worry about (2)-(b) since (1)=(2)-(a) cannot be.
Edit: After the correction to (2)-(a): the zero map is a trivial example of this possibility, and in general, if
$$x,y\in\Bbb R^m\;,\;a\in\Bbb R\;\wedge\;Ty=aTx=T(ax)\implies T(y-ax)=0\implies \ker T\neq 0$$ |
H: What does 'i-th' mean?
I have seen a problem set for the tower of hanoi algorithm that states:
Each integer in the second line is in the range 1 to K where the i-th
integer denotes the peg to which disc of radius i is present in the
initial configuration.
What does i-th actually mean? If i have a line conforming to the format above such as:
4 2 4 3 1 1
How exactly does this translate to the format described above?
AI: The statement that
the $i$-th integer denotes the peg on which the disk of radius $i$ is present in the initial configuration
is shorthand for this statement:
the first integer in the list is the number of the peg on which the disk of radius $1$ is initially placed; the second integer in the list is the number of the peg on which the disk of radius $2$ is initially placed; the third integer in the list is the number of the peg on which the disk of radius $3$ is initially placed; and so on, until the $K$-th and last integer in the list is the number of the peg on which the disk of radius $K$ is initially placed.
The list $4,2,4,3,1,1$ indicates that the disk of radius $1$ starts on peg $4$, the disk of radius $2$ starts on peg $2$, the disk of radius $3$ starts on peg $4$, the disk of radius $4$ starts on peg $3$, and the disks of radius $5$ and $6$ start on peg $1$. The configuration is therefore this one, where the numbers in the array are the radii of the disks.
$$\begin{array}{cc}
\text{Peg }1&\text{Peg }2&\text{Peg }3&\text{Peg }4\\ \hline
5&&&1\\
6&2&4&3
\end{array}$$ |
H: Subtracting roots of unity. Specifically $\omega^3 - \omega^2$
This is question that came up in one of the past papers I have been doing for my exams. Its says that if $\omega=\cos(\pi/5)+i\sin(\pi/5)$. What is $\omega^3-\omega^2$. I can find $\omega^3$ and $\omega^2$ by De Moivre's Theorem. But I cant make much headway into how to subtract these?
The answer they give is $2cos(\dfrac{3\pi}{5})$
Please, some help will be greatly appreciated. I cant seem to find help over the internet either.
AI: Hint: $\cos(3\pi/5)$ and $\cos(2\pi/5)$ are closely related, as are $\sin(3\pi/5)$ and $\sin(2\pi/5)$.
Remark If you are familiar with the complex exponential, note that $\omega^2\omega^3=e^{i\pi}=-1$. So we are looking at $e^{3\pi i/5}+e^{-3\pi i/5}$, which is $2\cos (3\pi/5)$. |
H: How do I prove Binet's Formula?
My initial prompt is as follows:
For $F_{0}=1$, $F_{1}=1$, and for $n\geq 1$, $F_{n+1}=F_{n}+F_{n-1}$.
Prove for all $n\in \mathbb{N}$:
$$F_{n-1}=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt5}{2}\right)^n-\left(\frac{1-\sqrt5}{2}\right)^n\right)$$
Which, to my understanding, is Binet's Formula.
I came up with a proof strategy similar to that described here. But as you can see, it is considering the case of $F_n$, not $F_{n-1}$.
So, my question is: does the same strategy hold for $F_{n-1}$? If not, how is it different?
AI: The problem that you’ve been given uses a less-standard indexing of the Fibonacci numbers. The usual definition has initial values $F_0=0$ and $F_1=1$; your $F_n$ is the usual $F_{n+1}$, and the usual $F_n$ is your $F_{n-1}$. Thus, where your link has $$\varphi^a=F(a)\varphi+F(a-1)\;,$$ you’ll need to write $$\varphi^a=F_{a-1}\varphi+F_{a-2}\;.$$ As a quick check, when $a=2$ that gives you $\varphi^2=F_1\varphi+F_0=\varphi+1$, which you can see from the link is correct.
(I’m assuming here that your proof really does follow pretty much the pattern in the link. You can also just do a straight induction: the induction step is the same whichever indexing you use.) |
H: Quotient group of normal subgroups is cyclic if quotient group of intersection is cyclic
Let $M$, $N$ be normal subgroups of a group $G$.
Prove that if $G/M\cap N$ is cyclic, then $G/M$ and $G/N$ are cyclic. Give a counter example to show that the converse is not always true.
I proved until now that if $G/M\cap N$ is abelian, then $G/M$ and $G/N$ are abelian, I don't know if that helps a lot but it's the only thing I could come up with.
AI: Hint for $\implies$:
Any quotient of a cyclic group is a cyclic group.
The third isomorphism theorem.
Hint for $\;\;\,\not\!\!\!\!\impliedby$:
There is a counterexample to the converse where $G$ has four elements (there are not very many such groups, so it should be easy to work out). |
H: Help me solve this equation: $x^\frac23 - 9x^\frac13+8=0$
Not sure what to call this type of equation so please let me know. I'm having trouble solving it though.
Solve the equation:
$$x^\frac23 - 9x^\frac13+8=0$$
AI: Hint: try the substitution $y = x^{\frac{1}{3}}$. This should give you a quadratic. |
H: What should be proved in the binomial theorem?
I'm following Cambrige mathematics syllabus, from the list of contents of what should be learned:
Induction as a method of proof, including a proof of the binomial theorem with non-negative integral coefficients.
I know what it is, but I'm not sure of what should be proved here. The first thing that comes to mind is the idea of proving it it for $n+1$, but I thought about writing:
$(a+b)^n$
$(a+b)^{n+1}$
But I am missing what premise I should prove. I guess that the proof involves the nature of the coefficients of the expansion of $(a+b)^n$ but from here, I have no idea on how to proceed.
Can you help me?
Edit: I guess I've made some progress. First I evaluated
$$(x+y)^0=1$$
Then I've evaluated it with the summation form
$$\sum_{j=0}^{0}{n \choose j}x^{n-j}y^{j}$$
And confirmed that it's equal to $1$ (I guess this is the base step).
The I did the same for $n+1$:
$$(x+y)^1=x+y$$
Then I've evaluated again for the summation form and checked that the results are the same. Is this all that should be done?
AI: If you know the expansion of $(a+b)^n$, it follows that the expansion of $(a+b)^{n+1}=(a+b)(a+b)^n=a(a+b)^n+b(a+b)^n$ can be found by distributing term by term and collecting coefficients. This is the type of reasoning you should use when doing inductive proofs. |
H: What is the names of $A\vec{x}=\vec{b}$ linear equation system components?
Having $A\vec{x}=\vec{b}$ .
What is the names of $A\vec{x}=\vec{b}$ linear equation system components?
AI: According to Elementary Linear Algebra by Venit and Bishop, we have:
$$A\vec{x}=\vec{b}$$
$A$ is the coefficient matrix of the system
$\vec{x}$ is the (column) vector of unknowns
$\vec{b}$ is the (column) of constants.
Granted, its probably not the most well-known book, but it's the one I have ;) |
H: Simple trigonometry question
I am just wondering how can you get from
$\cos(\pi t)=1/2 $ or $\cos(\pi t)=-1$
for $0<t<4$
to
t = 1/3, 1
t = 5/3, 7/3, 11/3 ,3
I got
$(\pi t) = \pi/3 +2k\pi $ and
$(\pi t) = 5\pi/3 + 2k\pi $
$t = 1/3 + 2k $ and $t = 5/3 +2k$
but i couldn't quite get t = 3, and t =1.... i'm not sure where i got it wrong... so could someone please explain this clearly to me?
AI: There are two values at which $\cos(\pi t)$ is a solution. You've made a dent in finding two of the four solutions to $$\cos(\pi t) = \frac 12, \;\;\text{for}\;\;t \in (0, 4).$$
There are four values at which $t$ gives a solution; they occur when $$0 \lt t \lt 4 \iff 0 \lt \theta \lt 4\pi$$
You found $t = \dfrac 13,\;\dfrac 53$, but also note that $t = \dfrac 73, \dfrac {11}{3} \implies \theta = \frac {7\pi}3, \theta = \frac{11}{\pi} \lt 4\pi = \pi t$
But you also found that $\cos(\pi t) = -1$ is also a solution. This happens when $t$ is an odd integer, so that $\pi t$ is an odd multiple of $\pi$:
$$\cos(\pi t) = -1 \implies \;\;t = 1, \;\;\text{or}\;\; t = 3, \quad0\lt t \lt 4$$ |
H: Finding the value of $c$ that makes this system of equations have a solution
Find the value of $c$ that makes the system of equations below have a solution.
$$\begin{align*}
u + v + 2w &= 2 \\
2u + 3v - w &= 5 \\
3u + 4v + w &= c
\end{align*}$$
I have taken a suggestion from a similar question asked 2 months ago. After row reduction, I have:
$$\begin{bmatrix}
1 & 1 & \hphantom{-}2 & 2 \\
0 & 1 & -3 & 1 \\
0 & 0 & \hphantom{-}1 & (3-c)/2
\end{bmatrix}$$
From here I am confused on how to actually solve for $c$.
Any help would be greatly appreciated!
AI: Adding the left hand sides of the first two equations gives the left hand side of the third equation, so $c=7$ is the only possibility. |
H: Showing existence of an element with order $p$
If a group $G$ has order $p^n$, where $p$ is prime and $n \geq 1$, does there exist some element $a\in G$ s.t. the order of $a$ is $p$?
I happen to know that this is true by Cauchy's theorem, but that theorem has not been presented yet in the book. I only have this so far:
Let $a \in G$, then the order of the cyclic subgroup generated by $a$ divides the order of the group $G$ (by Lagrange's theorem).
But I get stuck because I don't know how to show that the $a$ has order $p$.
AI: To ellaborate on Tobias' comment.
You have that the order of $a$ divides the order of $G$, and the order of $G$ is $p^n$.
This means the order of $a$ can only be $1, p^1, p^2, \cdots, p^n$, as these are the only divisors of $p^n$.
First, you need to pick $a$ such that the order is one of $p^1, \cdots, p^n$ (how?).
Then, you need to pick $k$ such that $a^k$ has order equal to $p$. |
H: Properties of Entropy
When someone writes $H(X_1, X_2, X_3) = H(X_1) + H(X_2\mid X_1) + H(X_3\mid X_2, X_1)$, how should that last term be interpreted/read? As the joint entropy between 2 variables where variable 1 is $X_3\mid X_2$ and variable 2 is $X_1$? Or As the entropy of $X_3$ conditioned on both $X_2$ and $X_1$? In other words is it:
$$H[ (X_3\mid X_2) , (X_1) ]\text{ or }H[ (X_3) \mid (X_2, X_1) ]$$
Are they the same? If so could someone show me how? If not, could someone tell me the correct way to read/interpret/write both possible interpretations?
AI: The intended interpretation in this formula is $H(X_3\mid (X_2,X_1))$. Your other option won't work because $X_3\mid X_2$ is not a random variable. |
H: How can I prove $2\sup(S) = \sup (2S)$?
Let $S$ be a nonempty bounded subset of $\mathbb{R}$ and $T = \{2s : s \in S \}$. Show $\sup T = 2\sup S$
Proof
Consider $2s = s + s \leq \sup S + \sup S = 2\sup S $. $T \subset S$ where T is also bounded, so applying the lub property, we must have $\sup T \leq 2 \sup S$.
On the other hand $2s + s - s \leq \sup T + \sup S - 3\sup S \implies 2\sup S \leq 2s + 2\sup S \leq \sup T $. Which gives the desired result.
Okay I am really worried about my other direction. Especially $2\sup S \leq 2s + 2\sup S$, do I know that $2s$ is positive?
Also in the beginning, how do I know that $\sup S \leq 2 \sup S$? How do I know that the supremum is positive?
AI: You can't assume $s$ is positive, nor can you assume $\sup S$ is positive.
Your proof also assumes a couple of other weird things:
$T \subset S$ is usually not true.
$2s + s - s \le \sup T + \sup S - 3\sup S$ is not necessarily true. Why would $-s \le -3\sup S$?
The first part of your proof is actually correct, ignoring the $T \subset S$ statement.
What you are saying is that any element of $T$, say the element $2s$, is bounded above by $2 \sup S$; thus $2 \sup S$ is an upper bound on $T$; thus $2 \sup S \ge \sup T$ by the least upper bound property.
For the second part of the proof, you need to show that $\sup T \ge 2 \sup S$.
To do this, you need to show that $\frac{\sup T}{2}$ is an upper bound on $S$.
This will imply $\frac{\sup T}{2} \ge \sup S$ by least upper bound property. |
H: Algebraic problem for satisfying a given equation
I'm trying to solve the following exercise:
My backward equation looks like:
$P_{i,j}'(t) = i\lambda P_{i+1,j}(t) - i\lambda P_{i,j}(t) $
So i started with differentiating $P_{i,j}(t)$:
${j-1 \choose i-1}(e^{-i\lambda t}(1-e^{-\lambda t})^{j-i})'$
By using $(uv)' = u'v + uv'$
${j-1 \choose i-1}-i\lambda (e^{-i\lambda t}(1-e^{-\lambda t})^{j-i}) + {j-1 \choose i-1}(j-i)\lambda (e^{-(i+1)\lambda t}(1-e^{-\lambda t})^{j-i-1}) $
where the first term is cleary $-i\lambda P_{i,j}(t) $.
I don't see how i can turn $ {j-1 \choose i-1}(j-i)$ into ${j-1 \choose i}i$ so the second term equals $i\lambda P_{i+1,j}(t) $
AI: You seem to have the wrong definition of $P_{i,j}$ it should be ${j-1 \choose i-1}$, not ${j-i \choose i-1}$. If you do the same thing with this corrected version, then
$$
\begin{align*}
{j-1 \choose i-1}(j-i) &= \frac{(j-1)!}{(i-1)!(j-i)!}(j-i) \\
&= \frac{(j-1)!}{(i-1)!(j-i-1)!} = \frac{(j-1)!}{((i+1)-1)!(j-(i+1))!}i \\
&= {j-1 \choose (i+1)-1}i
\end{align*}
$$ |
H: $f: G→G$ defined by $f(x) =x^2$ is a homomorphism if and only if $G$ is abelian.
The function $f: G→G$ defined by $f(x) =x^2$ is a homomorphism if and only if $G$ is abelian.
Can anyone give me any tips how to work on this question?
AI: Hint When is it true that $(ax)^2=a^2x^2$? That is $$axax=aaxx$$
Further hint Multiplying on the left by $a^{-1}$ and right by $x^{-1}$ gives what? |
H: Nilpotent operator / Orthogonal projection
If you have a nilpotent operator $A \colon V \to V$, $V$ being $A$-cyclic (meaning that $V$ is generated by a single vector $v$ in $V$), is it true that the minimal polynomial of the Gram operator $P = A^{T} A$ is $p(t) = t^2 - t$, i.e., $P$ is the orthogonal projection?
AI: If you grant Jordan canonical form, then in a suitable basis $A$ looks like
$$\begin{bmatrix}0&1&0&\dots&0 \\ 0&0&1&\dots&0 \\ & & \ddots&\ddots \\ 0&0&\dots&0&1 \\ 0&0&\dots &0&0
\end{bmatrix}\,.$$
Then $$A^\top A = \begin{bmatrix}0 & \\ & 1 & \\ & & \ddots \\ & & & 1\end{bmatrix}\,.$$
So 'twould appear you're right.
But .... Note that once you bring the transpose into the game, things are no longer basis-independent. The basis that brings $A$ into this nice form is not likely to be orthonormal, and so, if you're bringing the dot product into the game, it really is not meaningful to describe this in terms of an orthogonal projection. The characteristic polynomial of $A$ is basis-independent, but the characteristic polynomial of $A^\top A$ is not. |
H: Estimate the scale of the power series with Poisson pdf-like terms
Sorry to bother you, but I guess that this question is not appropriate for MO, so I repost it here hoping that someone could give me a clue.
I would like to have an estimate for the series
$$P(t) = \sum\limits_{k = 0}^\infty (e^{-t}\frac{t^k}{k!})^m,$$
where $e$ is the base of natural logarithm, $k!$ is the factorial of the integer $k$, $t$ represents the time and $t>0$, $m$ is a positive integer and $m>1$ (Obviously, $P(t)=1$ when $m=1$ since $P(t)$ is exactly the cdf of Poisson distribution with associated parameter $t$ in this case.).
I am interested in showing that the order of the scale could look something like
$$P(t) = O(t^{-\alpha m}),\alpha > 0.$$
But I find it difficult for me to extend their method to the cases when $m>2$.
I also tried to calculate $P(t)$ in Mathematica and it gave me the result like
$$P(t) = \sum\limits_{k = 0}^\infty (e^{-t}\frac{t^k}{k!})^m = e^{-mt}\text{HypergeometricPFQ}[...,t^m]$$
It seems that the decreasing speed of $P(t)$ is much slower than that of $e^{-mt}$ as $t$ increases. But I could not find the closed-form bound for $P(t)$ when $m>2$ just like $O(1/\sqrt t)$ when $m=2$.
Does anyone know of a scale or a bound of $P(t)$ in the literature? Any comments and answers would be highly appreciated. Many thanks!
AI: The function you have, and this is also the closed form that Mathematica gave you, is related to the hypergeometric function,
$$ P(t) = e^{-m t}\,\,{}_0F_{m-1}(;1,1,\ldots,1; t^m). $$
To calculate its behaviour as $t\to\infty$, at least in this case, is relatively easy. In particular the DLMF gives the formulas necessary, 16.11.9 and 16.11.1-4.
The result is
$$ P(t) = \frac{1}{(2\pi)^{\frac{m-1}{2}}\sqrt{m}}\frac{1}{t^{\frac{m-1}{2}}} + O(t^{\frac{-1-m}{2}}). $$
You can also check this for small values of $m$ using Mathematica's function Series, which understands how to calculate asymptotic behaviour. |
H: Spin group without Clifford algebras
I have to build the spin group $Spin(n)$ without use Clifford algebras. Can I find a complete description of spin group with a topological method? How can I build $Spin(n)$ as the double covering of $SO(n)$?
AI: Define $Spin(n)$ as the universal cover of $SO(n)$. The universal cover of a Lie group is a Lie group (e.g. Theorem 2.5 on p. 10 of these notes). In particular, when $n\geq 3$, the homotopy exact sequence gives $\pi_1(SO(n)) = \mathbb{Z}_2$, so $Spin(n)$ is a double covering of $SO(n)$. |
H: What are the cluster points of this filter?
Let X be a topological space, $A\subset X$ and $\mathcal{F}=\{F\subset X|A\subset F\}$. Then $\mathcal{F}$ is a filter on X. I would like to know what are the cluster points of this filter.
I am preparing for an exam and I need help. Thank you in advance.
AI: HINT: The cluster points of any filter $\mathscr{F}$ on a space $X$ are the points of the set $\bigcap_{F\in\mathscr{F}}\operatorname{cl}F$. What is $\bigcap_{F\in\mathscr{F}}\operatorname{cl}F$ for your particular filter? |
H: Simplifying a radical equation?
I'm lost on how to simplify this.
Simplify:
$$\frac{\sqrt[\large 4]{144x^9y^8}}{\sqrt[\large 4]{9x^5y^{-3}}}$$
OK So I think I got it.
$$\frac{\sqrt[\large 4]{144x^9y^8}}{\sqrt[\large 4]{9x^5y^{-3}}} = \left(\frac{{144x^9y^8}}{{9x^5y^{-3}}}\right )^{1/4}$$
You can break that down like this -I think-:
$$\sqrt[\large 4]{16x^4y^{11}}$$
And then finally:
$$2xy^2\sqrt[\large 4]{y^3}$$
EDIT - You can go even further (Thanks amWhy):
$$2xy^{\large \frac{11}{4}}$$
AI: First, recall that
$$ \frac{\sqrt[\large4]{144x^9y^8}}{\sqrt[\large4]{9x^5y^{-3}}} = \sqrt[\large 4]{\frac{144x^9y^8}{9x^5y^{-3}}} = \left(\frac{144x^9y^8}{9x^5y^{-3}}\right)^{1/4}$$
Now, use the "laws" of exponents to simplify the rational expression, (the fraction inside parentheses), and then take distribute the exponent $1/4$ over the factors remaining after simplifying.
ADDED: You are almost there: you can also write the exponent of $y$ as a fraction. Recall that $\;\;(y^a)^b = y^{ab}.\;$ In your case, $\,a = 3$, and $\;b=1/4$:
$$2xy^2\sqrt[4]{y^3} = 2xy^2\left((y^3)^{1/4}\right) = 2x y^2 \left(y^{3/4}\right) = 2x\left(\color{blue}{\bf y^{\left(2 + \frac 34\right)}}\right)\quad ?$$
$\color{blue}{\bf \left(y^cy^d = y^{(c + d)}\right)}$ |
H: Simple Graph Transformation Question $\rightarrow$ $1/f(x)$
for the graph:
such that the function is : $ y = \frac{a+x}{b+cx} $ where a = -2, b = 1 and c = 1/2
how do you sketch the graph of $ y = |\frac{b+cx}{a+x}| $ ??
i got that the VA of the new graph is "$+2$" and the root of the new graph is "$-2$", but then the answer says that the horizontal asymptote is "$\frac{1}{2}$" which i don't really understand..
Could anyone explain to me what to happen, if i am to transform a graph from $f(x)$ to $\frac{1}{f(x)}$.
i only know that the roots of $f(x)$ will become VA value of $1/f(x)$ and VA values of $f(x)$ becomes the zeros of $\frac{1}{f(x)}$
and that if $f(x) \gt 0$ , then $\frac{1}{f(x)}$ $\gt 0$ such that $f(x) \to 0 \implies \frac{1}{f(x)} \to \infty$ and $f(x) \to \infty \implies \frac{1}{f(x)} \to 0$.
AI: You have a function
$$f(x) = \frac{x-2}{\frac{x}{2} + 1}$$
and you want to sketch $y = \frac{1}{f(x)}$.
The $y$-intercept of the graph $y = f(x)$ is given by $f(0)$ (let this be $y_0$). The $y$-intercept of the graph $y = \frac{1}{f(x)}$ is given by $\frac{1}{f(0)}$, which is the reciprocal of $y_0$.
Similarly, the old horizontal asymptote, $y = \displaystyle\lim_{x\to\infty}f(x) = 2$, will become $y = \displaystyle\lim_{x\to\infty} \frac{1}{f(x)} = \frac{1}{\displaystyle\lim_{x\to\infty}f(x)} = \frac{1}{2}$, since $f(x)$ is being replaced by $\frac{1}{f(x)}$.
In general, the $y$ values of $y = f(x)$ will be the reciprocals of $y = \frac{1}{f(x)}$, i.e.,
$$y_{\text{new}} = \frac{1}{y_{\text{old}}}.$$
To then sketch the absolute value of this function, simply reflect any part below the $x$-axis above it, so there are no negative $y$ values for any $x$.
Pictures speak a thousand words...
Note that for each $x$ value, the $y$ values are reciprocals:
Reflecting the part below the $x$-axis:
Also, technically $f$ is not defined at $x = -2$, so neither should $\frac{1}{f}$ or $\left | \frac{1}{f} \right |$, i.e., there are holes at $x=-2$ on the blue graphs. |
H: Eigenvalue of f and df
Given 1 is not an eigenvalue of $df$ at $x_0$, take a chart $(U,\phi)$ around $x_0.$ Then in this coordinate neighborhood, think of $f$ as a map from open ball in $\mathbb{R}^n$ (say $B$), to itself with $f(0)=0.$ Now consider we have a function $f:B\rightarrow B$ such that $f(0)=0.$ Then $det(f - id)(0)\neq 0$.
But I thought it should be det$(df-id)(0) \neq 0$? Did we assume f is linear map so f = df here?
AI: This question is answered by Ted Shifrin.
Yes this should be det$(df - id)_0 \neq 0$. I can't assume that $f$ is linear. |
H: Strange AP Calculus BC question help?
The question is $\int_0^3\frac{1}{(1-x)^2}$. I got an answer (from u-substitution) however the solution manual says that the integral does not converge. Someone told me that the integral is undefined at $x = 1$, but if we look at $\frac {1}{x}$ that function is undefined at $x=0$ yet we can still integrate it.
Can someone explain what is going on?!
AI: As was pointed out to you, the integrand fails to exist at $x=1$, so the integral must be rewritten as the sum of two improper integrals, one from $0$ to $1$ and the other from $1$ to $3$. When you try to evaluate the latter, you get this:
$$\begin{align*}\int_1^3\frac{dx}{(1-x)^2}&=\lim_{a\to 1^+}\int_a^3\frac{dx}{(1-x)^2}\\\\
&=\lim_{a\to 1^+}\left[\frac1{1-x}\right]_a^3\\\\
&=-\frac12-\lim_{a\to 1^+}\frac1{1-a}\;,
\end{align*}$$
and the limit in the last line clearly does not exist. Thus, the improper integral diverges, as does your original integral.
Note that you cannot integrate the function $f(x)=\frac1x$ over any interval that contains $0$, and the same goes for the function $f(x)=\frac1{x^2}$: in both cases you will get a divergent improper integral. |
H: Find the critical points and say whether they are maxima, minima or saddle points
I have this problem:
Find the critical points and say whether they are maxima, minima or saddle points
$$f(x,y)=x^2y(2−x−y)$$
My answer:
$f_x = xy (4-3 x-2 y) $
$f_y = -x^2 (-2+x+2 y)$ then
$xy (4-3 x-2 y)=0 , -x^2 (-2+x+2 y)=0$ the solutions are:
1)x=0 and $y \in{} \mathbb{R}$
2)x=1 and y=1/2
3)x=2 and y=0
I have three critical points,
$f_{xx} = -2 y (-2+3 x+y) $
$f_{yy} = -2 x^2 $
$f_{xy} = x (4-3 x-4 y) $, using
$D=f_{xx}f_{yy}-(f_{xy})^2 = -16 x^2+24 x^3-9 x^4+24 x^2 y-12 x^3 y-12 x^2 y^2$
x=1 , y=1/2 then D=0 but $f_{xx}=-2y (-2+3 x+y)= -2/3$ concluding (1,1/2) is a maxima point
x=2 , y=0 then D=-16 we say that (2,0) is a saddle points
x=0 , $y \in{} \mathbb{R}$ then D=0 the point (x,a) with $a \in{} \mathbb{R}$ is a saddle points because for example if x=0 , a=-8 $fxx=-160$ (maxima) but if x=0 , a=1 then $f_{xx}=2$ (minima)
Is my exercise correct?
AI: Look at the Hessian matrix and consider its eigenvalues. If the determinant is negative, then you get a saddle point. If the determinant is zero, the test is inconclusive; if the determinant is positive and $f_{xx} > 0$, then you get a minima; and if the determinant is positive and $f_{xx} < 0$ , then you have a maxima.
Now, can you see whether you are correct?
Hessian matrix is just the matrix
\begin{pmatrix}
f_{xx} & f_{xy} \\
f_{yx} & f_{yy} \\
\end{pmatrix} |
H: Infinite linear ordered set which doesn't include infinte well-ordered subsets
Need an example of such set.
Thank you for your time
AI: What about the negative integers, usual order? Verification that this has no infinite well-ordered subset should not be difficult. |
H: Cardinality and Measurability
We can show that $\mathbb{R}$ and $\mathbb{R}^2$ or ($\mathbb{R}^n$) have same cardinality using the following one-to-one and onto mapping:
Say x = (0.123456789....)
Then f(x) = {(0.13579...),(0.2468..)}
My question is can we claim that the Borel sigma algebra in $\mathbb{R}$ has a correspondent sigma algebra (if it is, is it THE Borel sigma algebra in $\mathbb{R}^2$?) by mapping each Borel set D to $\mathbb{R}^2$ by f(D)? Why or why not could you give the argument??
Similarly (I am more certain of this)
When we have N^n and N using this mapping could we claim that the sigma algebra in N has a correspondent sigma algebra in N^n? Does the statement hold for N^infinity and N as well, if N^infinity and N have the same cardinality ( i am not sure about my last statement)?
Thank you!
AI: The Borel sigma-algebra of $\mathbb{R}^2$ is defined as that generated by the subsets of $\mathbb{R}^2$ that are open in the product topology. It is almost, but not quite, correct to say that this is equal to the collection of pointwise images of Borel subsets of $\mathbb{R}$ under your a correspondence given by splitting and interleaving decimal expansions.
The problem, as Ittay pointed out in a comment, is that this correspondence is not quite a bijection between $\mathbb{R}$ and $\mathbb{R}^2$. Nevertheless, by changing the argument a bit, one can still show that $\mathbb{R}$ and $\mathbb{R}^2$ are Borel-isomorphic. One way to do this is to get a Borel injection from $\mathbb{R}^2$ to $\mathbb{R}$ and then use the Borel version of the Cantor–Bernstein–Schroeder Theorem.
Indeed, a Borel injection $f: \mathbb{R}^2 \to \mathbb{R}$ can be defined as follows: Given $(x,y) \in \mathbb{R}^2$, let $x_0 x_1 x_2 \ldots$ and $y_0y_1y_2 \ldots$ be the unique decimal expansions of $x$ and $y$ respectively with the property that the digits are not eventually all 9's from some point on.
Then let $f(x,y)$ be the real number whose decimal expansion is $x_0y_0x_1y_1 \ldots.$ (For simplicity of notation I am ignoring the issue of where the decimal point goes, but this is easy to deal with.)
Unfortunately the range of this function is missing uncountably many real numbers (such as those whose even digits are eventually all 9's and whose odd digits are not) and the easiest way that I see to deal with this issue is to use the Cantor–Bernstein–Schroeder Theorem, although it may be overkill.
The sets $\mathbb{R}$ and $\mathbb{R}^2$ with these Borel structures (and indeed $\mathbb{R}^n$ for any positive integer $n$) are examples of
standard Borel spaces, which are the objects one gets from Polish topological spaces by forgetting the topology and remembering only the Borel structure. The Borel isomorphism we established above proves a special case of a theorem of Kuratowski, which says that every uncountable Polish space is Borel-isomorphic to $\mathbb{R}$, so up to isomorphism there is only one uncountable standard Borel space.
The sets $\mathbb{N}$ and $\mathbb{N}^n$ with the discrete Borel structure (every subset is a union of countable many singletons and is therefore Borel in the usual topology) are Borel-isomorphic to one another. They are examples of the countably infinite standard Borel space, which is also unique up to isomorphism.
The set $\mathbb{N}^\infty$ (also known as $\mathbb{N}^\mathbb{N}$) of integer sequences with the Borel algebra generated by the product topology is an uncountable standard Borel space—it is Borel-isomorphic to $\mathbb{R}$ rather than to $\mathbb{N}$. |
H: AP Calculus multiple choice question
The position of a particle along a line is given by $z(t) = 2t^3 -24t^2 + 90t +7$ for $ t \geq 0$. For what values of $t$ is the speed of the particle increasing?
a) $3 < t < 4$ only
b) $t > 4$ only
c) $t > 5$ only
d) $0 < t < 3$ and $t > 5$
e) $3 < t < 4$ and $t > 5$
For this problem I said d) based on my calculations:
$z'(t) = 6t^2 - 48t + 90 = 6(t-5)(t-3)$. I then did a chart and found out that $z'(0) > 0$ ,$z'(4) < 0$ and $z'(6) > 0$ leading me to conclude d) however the textbook says e). I thought that I did everything right, what is wrong with my conclusion?
AI: First of all, realize that the speed is $|z^{'}(t)|$ and not $z^{'}(t)$. This is your mistake.
Now, graph $|z^{'}(t)|$ and you will see that your textbook is spot on. |
H: $R_n = 3(2^n)-4(5^n)$, $n \geq0$, prove $R_n$ satisfies $R_n = 7R_{n-1}-10R_{n-2}$
So the question is:
$R_n=3(2^n)-4(5^n)$ for $n\ge 0$; prove that $R_n$ satisfies $R_n=7R_{n-1}-10R_{n-2}$.
I don't really know what to do from here. If I substitute
$$R_n = 3(2^n)-4(5^n)$$
into
$$Rn = 7R_{n-1}-10R_{n-2}$$
I end up getting
$$R_n = 7\Big(3(2^{n-1})-4(5^{n-1})\Big)-10\Big(3(2^{n-2})-4(5^{n-2})\Big)$$
Dont know what to do...
EDIT: Thanks to Zev, what I did was:
$$3(2^n)-4(5^n)=7\bigg[3(2^{n-1})-4(5^{n-1})\bigg]-10\bigg[3(2^{n-2})-4(5^{n-2})\bigg].$$
$$\begin{align*}
3(2^n)-4(5^n)&=21(2^{n-1})-28(5^{n-1})-30(2^{n-2})+40(5^{n-2})\\\\
3(2^n)-4(5^n)&=21(2^{n})(2^{-1})-28(5^{n})(5^{-1})-30(2^{n})(2^{-2})+40(5^{n})(5^{-2})\\\\
3(2^n)-4(5^n)&=21/2(2^{n})-28/5(5^{n})-30/2(2^{n})+40/5(5^{n})\\\\
3(2^n)-4(5^n)&=(2^{n})[21/2-30/4]+(5^{n})[40/25-28/25]\\\\
3(2^n)-4(5^n)&=(2^{n})[3]+(5^{n})[-4]\\\\
3(2^n)-4(5^n)&=3(2^{n})-4(5^{n})
\end{align*}$$
AI: You're on the right track so far; you've used the definition of $R_n$ to express the right side of the equation. Now just do this for the left side as well. You want to show that for any $n\geq 0$,
$$3(2^n)-4(5^n)=7\bigg[3(2^{n-1})-4(5^{n-1})\bigg]-10\bigg[3(2^{n-2})-4(5^{n-2})\bigg].$$
This can be done directly:
$$\begin{align*}
3(2^n)-4(5^n)&=21(2^{n-1})-28(5^{n-1})-30(2^{n-2})+40(5^{n-2})\\\\
12(2^{n-2})-100(5^{n-2})&=42(2^{n-2})-140(5^{n-2})-30(2^{n-2})+40(5^{n-2})
\end{align*}$$ |
H: Solving For Variables In Simultaneous Equations
I'm doing some work in linear algebra and these came up and I realized I don't know how to solve them as they have quadratics in them. I'm sure I've done this before but if someone could give me a crash course on how to find the values of the variables, it would be most appreciated. For the following questions, let $a,b,c,d \in \mathbb{R}$
1)
$\begin{cases}
a^2+b^2=a\\
(a+d)b=b\\
b^2+d^2=d
\end{cases}$
2)
$\begin{cases}
a^2+bc=a\\
(a+d)b=b\\
(a+d)c=c\\
bc+d^2=d
\end{cases}$
AI: 1)
$\begin{cases}
a^2+b^2=a\\
(a+d)b=b\\
b^2+d^2=d
\end{cases}$
in equation $2$ divide both side by b
$\begin{cases}
a^2+b^2=a\\
a+d=1\\
b^2+d^2=d
\end{cases}$
from eqn $2$ and $3$
$$a=1-d$$
$$b^2=d-d^2$$
put these value in eqn $1$
$$(1-d)^2+d-d^2=1-d\implies d=0$$
it will give $a=1,b=0$
2)
$\begin{cases}
a^2+bc=a\\
(a+d)b=b\\
(a+d)c=c\\
bc+d^2=d
\end{cases}$
in eqn $2,3$ divide both side by b and c
$
a^2+bc=a\\
a+d=1\\
a+d=1\\
bc+d^2=d
$
$
a^2+bc=a\\
a+d=1\\
bc+d^2=d
$
from eqn $1$ and $2$
$$bc=a-a^2$$
$$d=1-a$$
put these value in $3$
$$a-a^2+(1-a)^2=1-a\implies a=0$$
that will give $d=1$ and either of b and c will 0.
$$$$ |
H: Flux across surface
Find A uniform fluid that flows vertically downward is described by the vector field F (x, y, z) = (0, 0, −1).
Find the flux through the cone z =
$z= \sqrt{x^2 + y^2}$,
$x^2 + y^2 \le 1$.
I attempted this question with spherical coordinates and i don't know why it didn't work out. I used that $\sin(\theta) = \frac{\pi}{4}$ and evaluated the cross product T$\phi$ x Tr
This came to $(r\sin(\theta)\cos(\theta)\cos(\phi),-r\sin(\theta)\cos(\theta)\sin(\phi),r\sin^2(\theta))$
I then integrated
$\int_0^{2\pi}\int_0^1(0,0,-1).(r\sin(\theta)\cos(\theta)\cos(\phi),-r\sin(\theta)\cos(\theta)\sin(\phi),r\sin^2(\theta))~dr~d\phi$
=$-\int_0^{2\pi}\int_0^1r\sin^2(\theta)~dr~d\phi$
=$-\frac{\pi}{2}$ as $\theta = \frac{\pi}{4}$
Because I didn't get the right answer I checked my lecture notes, and I thought 'hey there is a $\hat n$ at the end of this integral and i didn't divide by magnitude. But to no avail.
I ended up getting -2$\pi$ by dividing by the magnitude and the answer is $\pi$
I now know that I can do it better with cylindrical coordinates and parametrization. But this way i don't divide by the magnitude
Do I divide by the normals magnitude to get $\hat n$ or not?
AI: If you know Divergence Theorem, then this problem looks like a standard "applying the Divergence theorem" exercise.
Denote the cone surface as $S=\{(x,y,z):z = \sqrt{x^2+y^2},\;z\leq 1\}$, the region surrounded by this surface is $D = \{(x,y,z):\sqrt{x^2+y^2}< z< 1\}$.
This region has a "cap": $C = \{(x,y,z): x^2+y^2\leq 1, z= 1\}$ such that $\partial D = S\cup C$. Apply the divergence theorem for $F = (0,0,-1)$:
$$
\iiint_D \nabla\cdot F\,dxdydz = \iint_{\partial D} F\cdot \nu\,dS = \iint_{S} F\cdot \nu\,dS + \iint_{C} F\cdot \nu\,dS,
$$
where $\nu$ are the unit surface normal pointing outward. The divergence of $F$ is zero, what we wanna compute is the flux across the cone surface $S$:
$$
\iint_{S} F\cdot \nu\,dS = - \iint_{C} F\cdot \nu\,dS.\tag{1}
$$
Basically this identity from divergence theorem tells us that the amount of fluid flowing across the cone surface outward away from the cone, is the same with the amount of the fluid flowing into the cone from the cap. Flowing in/out yields the sign difference.
Now back to evaluating (1): $\nu$ on the cap is $(0,0,1)$, thus $F\cdot \nu = -1$, the right side is just the area of the cap $C$, which is $\pi$. Therefore:
$$
\iint_{S} F\cdot \nu\,dS = \pi,
$$
and we do not have to deal with the integral on a parametrized surface which is significantly messier than this approach. |
H: Group acting on a set
Let $A$ be a set, and let $G$ be any subgroup of $S_A$. $G$ is a group of permutations of $A$; we say it is a group acting on the set $A$. Assume here that $G$ is a finite group. If $u \in A$, the orbit of $u$ (with respect to $G$) is the set
$$O(u)=\{g(u): g \in G\}.$$
Prove that $G_u = \{g \in G: g(u)=u\}$ is a subgroup of $G$.
Any advice? Thanks a lot.
AI: Let $H$ be a group, whose operation is denoted by $\star$, and whose identity element is $e$.
By definition, a subset $K\subseteq H$ is a subgroup of $H$ when
$e\in K$.
For any $a,b\in K$, we also have $(a\star b)\in K$.
For any $a\in K$, we also have $a^{-1}\in K$.
Prove that $G(u)$ satisfies these requirements. Remember, the group operation of your group $G$ is composition of permutations, so the identity element is the trivial permutation (i.e. the one that does nothing). Also, remember that (by the definition of a group action), for any $g_1,g_2\in G$ and any $u\in A$, we have that $g_1(g_2(a))=(g_1\circ g_2)(a)$. |
H: Can you recommend some books on elliptic function?
I plan to study elliptic function. Can you recommend some books? What is the relationship between elliptic function and elliptic curve？Many thanks in advance!
AI: McKean and Moll have written the nice book Elliptic Curves: Function Theory, Geometry, Arithmetic that cleanly illustrates the connection between elliptic curves and elliptic/modular functions. If you haven't seen the book already, you should.
As for elliptic functions proper, my suggested books tend to be a bit on the old side, so pardon me if I don't know the newer treatments. Anyway, I quite liked Lawden's Elliptic Functions and Applications and Akhiezer's Elements of the Theory of Elliptic Functions. An oldie but goodie is Greenhill's classic, The Applications of Elliptic Functions; the notation is a bit antiquated, but I have yet to see another book that has a wide discussion of the applications of elliptic functions to physical problems.
At one time... every young mathematician was familiar with $\mathrm{sn}\,u$, $\mathrm{cn}\,u$, and $\mathrm{dn}\,u$, and algebraic identities between these functions figured in every examination.
– E.H. Neville
Finally, I would be remiss if I did not mention the venerable Abramowitz and Stegun, and the successor work, the DLMF. The chapters on the Jacobi and Weierstrass elliptic functions give a nice overview of the most useful identities, and also point to other fine work on the subject. |
H: How to calculate $\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$?
I need to calculate the sum $\displaystyle S=\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$,
where $\displaystyle H_n=\sum\limits_{m=1}^n\frac1m$.
Using a CAS I found that $S=\lim\limits_{k\to\infty}s_k$ where $s_k$ satisfies the recurrence relation
\begin{align}
& s_{1}=-1,\hspace{5mm} s_{2}=\frac18,\hspace{5mm} s_{3}=-\frac{215}{216},\hspace{5mm} s_{4}=\frac{155}{1728},\hspace{5mm} \text{for all} \quad k>4, \\ s_{k} &=\frac1{k^3(2k-3)}\left(\left(-4k^4+18k^3-25k^2+12k-2\right)s_{k-1}+\left(12k^3-39k^2+38k-10\right)s_{k-2} \right.\\
& \hspace{5mm} \left. +\left(4k^4-18k^3+25k^2-10k\right)s_{k-3}\\+\left(2k^4-15k^3+39k^2-40k+12\right)s_{k-4}\right),
\end{align}
but it could not express $S$ or $s_k$ in a closed form.
Can you suggest any ideas how to calculate $S$?
AI: Write down the function
$$ g(z) = \sum_{n\geq1} \frac{z^n}{n}H_n^2, $$
so that $S=g(-1)$ and $g$ can be reduced to
$$ zg'(z) = \sum_{n\geq1} z^n H_n^2 = h(z). $$
Now, using $H_n = H_{n-1} + \frac1n$ ($n\geq2$), we can get a closed form for $h(z)$:
$$h(z) = z + \sum_{n\geq2}\frac{z^n}{n^2} + \sum_{n\geq 2}z^n H_{n-1}^2 + \sum_{n\geq 2} 2\frac{z^n}{n}H_{n-1}. $$
Now, the first and third sums Mathematica can evaluate itself in closed form (the third one evaluates to the function $p(z)$ below, the first one is $\text{Li}_2(z)-z$), and the middle sum is $z h(z)$.
Substituting this into the expression for $g(z)$, we get
$$g(z) = \int \frac{\text{Li}_2(z) + p(z)}{z(1-z)}\,dz, $$
$$p(z) = -\frac{\pi^2}{3} + 2\log^2(1-z)-2\log(1-z)\log(z)+2\text{Li}_2((1-z)^{-1}) - 2\text{Li}_2(z). $$
Mathematica can also evaluate this integral, giving (up to a constant of integration)
\begin{align}
g(z) &= \frac{1}{3} \left(-2 \log(1-z^3+3 \log(1-z)^2 \log(-z)+\log(-1+z)^2 (\log(-1+z)+3 \log(-z) \right. \\
& \hspace{5mm} \left. -3 \log(z))+\pi ^2 (\log(-z)-2 \log(z))+\log(1-z) \left(\pi^2 - 3 \log(-1+z)^2 \right. \right.\\
& \hspace{5mm} \left.\left. +6 (\log(-1+z)-\log(-z)) \log(z)\right)-6 (\log(-1+z)-\log(z)) \left(\text{Li}_{2}\left(\frac{1}{1-z}\right)-\text{Li}_{2}(z)\right) \right.\\
& \hspace{10mm} \left. -3 \log(1-z) \text{Li}_{2}(z)+3 \text{Li}_{3}(z)\right).
\end{align}
The constant of integration is fixed by requiring $g(0)=0$.
Some care needs to be taken, because the function
is multi-valued, when evaluating $g(-1)$. The answer is
$$ \frac{1}{12}(\pi^2\log2-4(\log 2)^3-9\zeta(3)). $$ |
H: Inner product polynomials
Let $V$ be the vector space of real polynomial $\mathbb{R}[x]$ endowed
with the inner product
$\langle f,g \rangle = \displaystyle\int_{-\infty}^{\infty}
e^{-|x|}f(x)g(x) \ dx$
By considering the sequence of subspaces $\{V_n\}$ where
$V_n = \{f(x) \in \mathbb{R}[x] : \deg f \leq n \}$
or otherwise, show that there exist unique monic polynomials
$\phi_n(x)$ for $n \geq 0$ such that
$\displaystyle\int_{-\infty}^{\infty} e^{-|x|}\phi_n(x)g(x) \ dx = 0$
whenever $\deg g < n$, and find $\phi_n(x)$ for $n = 0,1,2.$
What is the coefficient of $x^{2000}$ in $\phi_{2007}(x)$?
I'm having trouble even knowing where to begin with this question, any help appreciated!
AI: $n = 0$: $\phi_n$ being monic implies $\phi_0 = 1$, if $\mathrm{deg}(g)<0$, then $g = 0$, and
$$
\int_{-\infty}^{\infty} e^{-|x|}\phi_0(x)g(x) \, dx = 0
$$
is trivial.
$n= 1$: $\phi_1$ has to be $x$ for being monic, and $g = c$ is a constant, by symmetry of the integral:
$$
\int_{-\infty}^{\infty} e^{-|x|}\phi_1(x) c\, dx = 0
$$
for $e^{-|x|}\phi_1(x)$ is odd.
$n= 2$: let $\phi_2(x) = x^2 + a_1 x +a_2$, $g\in \mathrm{span}\{1,x\}$, by symmetry
$$
\int_{-\infty}^{\infty} e^{-|x|}(x^2 + a_1 x +a_2) x\, dx = 0
$$
implies $a_1= 0$, and
$$
\int_{-\infty}^{\infty} e^{-|x|}(x^2 + a_2)\, dx = 0
$$
yields $a_2 = -2$. Hence $\phi_2(x) = x^2 - 2$ which does not have a degree 1 term.
$n= 2007$: $g\in \mathrm{span}\{1,x,\ldots,x^{2006}\}$, choosing certain $g$ will let you find out certain degree terms in $\phi_{2007}$ is zero by a symmetry argument like above (the integration of an odd function is zero from $-\infty$ to $\infty$, given that it is integrable), the rest is left for you to try.
General result for $\phi_n$, discuss the cases for $n$ being odd and even like above. |
H: Problem related with solving ODE
I was solving old exam papers and am stuck on the following problem:
Consider the system of ODE
$\frac {d}{dx}Y=AY,Y(0)=\begin{pmatrix}
2\\
-1
\end{pmatrix}$ where $A=\begin{pmatrix}
1 &2 \\
0&-1
\end{pmatrix},Y=\begin{pmatrix}
y_1(x)\\
y_2(x)
\end{pmatrix}$. Then I have to determine which of the following options hold good.
$y_1(x) \to \infty,y_2(x) \to 0\,\,$ as $\,\,x \to \infty$
$y_1(x) \to 0,y_2(x) \to 0\,\,$ as $\,\,x \to \infty$
$y_1(x) \to \infty,y_2(x) \to -\infty \,\,$ as $\,\,x \to -\infty$
$y_1(x),y_2(x) \to -\infty \,\,$ as $\,\,x \to -\infty$ .
My Attempt: Using the eigenvalues of $A$,I get the ODE of the form:
$y(x)=c_1e^x+c_2e^{-x},$ where $c_1,c_2$ being arbitrary constants to be determined .
Here,I am stuck.I could not use the fact that $Y(0)=\begin{pmatrix}
2\\
-1
\end{pmatrix}$ for finding $c_1,c_2$.
Can someone point me in the right direction with some explanation? Thanks in advance for your time.
AI: If you want to use the eigenvalue approach, you would generate eigenvectors with them.
These are shown below as column vectors (is it clear how to generate the eigenvectors). You would write:
$$y(x) = c_1 e^x \begin{pmatrix}1 \\ 0 \end{pmatrix} + c_2 e^{-x}\begin{pmatrix}-1 \\ 1 \end{pmatrix}$$
From this, we have two solutions, $y_1(x)$ and $y_2(x)$.
Now, you have two equations, two unknowns and two initial conditions to use for finding $c_1$ and $c_2$.
So, we have:
$$y_1(x) = c_1e^x - c_2e^{-x}$$
$$y_2(x) = c_2e^{-x} $$
From the given ICs, we have:
$$y_1(0) = c_1 - c_2 = 2$$
$$y_2(0) = c_2 = -1$$
So, $c_2 = -1$ and $c_1 = 1$, thus:
$$y_1(x) = e^x + e^{-x}$$
$$y_2(x) = - e^{-x} $$
Can you now answer the four questions? |
H: intuition for the closed form of the fibonacci sequence
I'm trying to picture this closed form from Wikipedia visually:
The idea is, if you take $\phi^n / \sqrt{5}$ and round it to the nearest integer, you'll get the $n$th Fibonacci number. I see how it works out on paper, but is there an intuitive way to understand this? Specifically I'm trying to wrap my brain around how $\phi^n$ would lead to the correct answer... if I want the 8th Fibonacci number, for example, how does it help to multiply $\phi$ by itself 8 times?
AI: The fact that $F_n$ is the integer nearest to $\dfrac{\varphi^n}{\sqrt5}$ follows from the closed form for the Fibonacci numbers known as the Binet formula:
$$F_n=\frac{\varphi^n-\widehat\varphi^n}{\sqrt5}=\frac{\varphi^n}{\sqrt5}-\frac{\widehat\varphi^n}{\sqrt5}\;,$$
where $\widehat\varphi=\dfrac{1-\sqrt5}2$. Note that $\widehat\varphi\approx-0.618$, so $|\widehat\varphi|<1$, and $|\widehat\varphi^n|$ decreases rapidly as $n$ increases. It turns out that even for small $n$ the correction $\dfrac{\widehat\varphi^n}{\sqrt5}$ is small enough so that $F_n$ is the integer nearest to $\dfrac{\varphi^n}{\sqrt5}$.
The $\sqrt5$ in the Binet formula ultimately comes from the initial conditions $F_0=0$ and $F_1=1$; a sequence with the same recurrence $x_n=x_{n-1}+x_{n-2}$ but different initial values would still grow approximately proportionally to $\varphi^n$ (or in exceptional cases $\widehat\varphi^n$), but the coefficient of approximate proportionality would be different.
Added: Specifically, each such sequence has a closed form $x_n=\alpha\varphi^n+\beta\widehat\varphi^n$, where $\alpha$ and $\beta$ are chosen to give the correct values when $n=0$ and $n=1$. Suppose that $x_0=a$ and $x_1=b$. Then from $n=0$ we must have $\alpha+\beta=a$, and from $n=1$ we must have $\alpha\varphi+\beta\widehat\varphi=b$. This pair of linear equations can then be solved for $\alpha$ and $\beta$, and provided that $\alpha\ne0$, we’ll have $x_n\approx\alpha\varphi^n$ for sufficiently large $n$. (Just how large sufficiently large actually is will depend on $\alpha$ and $\beta$.) |
H: Find all polynomials $P(x)$ satisfying this functional equation
Find all polynomials $P(x)$ which have the property
$$P[F(x)]=F[P(x)], \quad P(0) = 0$$
where $F(x)$ is a given function with the property $F(x)>x$ for all $x\geq 0$.
This is an exercise from my homework. I would appreciate any kind explanations. Thanks!
AI: Denote $x_0=0$ and $x_{n+1}=F(x_n)$ for every $n\ge 0$. Since $F(x)>x$ for every $x\ge 0$, $x_{n+1}>x_n$ for every $n\ge 0$. Note that $P(x_0)=x_0$ and
$$P(x_{n+1})=P(F(x_n))=F(P(x_n)),\quad\forall n\ge 0.$$
Then by induction, it is easy to see that
$$P(x_n)=x_n,\quad\forall n\ge 0.\tag{1}$$
$(1)$ implies that the polynomial $P(x)-x$ has infinitely many roots, i.e. $P(x)\equiv x$. |
H: Help with Combination
Guys need help to solve this one..
How will we arrange Red balls in '$N$' places , so that if you choose any '$M$' consecutive places, there should be at least '$K$' Red balls among this '$M$' chosen places.And we should use minimum number of Red balls.
Now, If $N = 6$ and $M = 3$ and $K = 2$ then, one combination is like '$1\,\; R\,\; R\,\; 1\; R\; R\; 1$' Here if we choose any $3$ consecutive places, there will be $2$ Red Balls. So like that how many ways we can do it..
There could be a larger numbers also.
Please help..
AI: Completely revised to answer the general question:
Assume first that $n=3m$, so that we may segment the string into $m$ blocks of three balls. Each block must be of one of the forms RRW, RWR, and WRR, and a string of $m$ RRW blocks is certainly acceptable, so there must be $2m$ red and $m$ white balls.
Let $s_m$ be the number of acceptable strings of $3m$ balls. Suppose that I have such a string. If it ends with a WRR block, I can append any of the three kinds of block to get an acceptable string of length $3(m+1)$. If it ends with an RWR block, I can append either an RWR or an RRW block, but not a WRR block. And if it ends with an RRW block, I can only append another RRW block. To turn it around, a WRR block can only follow another WRR block; an RWR block can follow a WRR block or another RWR block; and an RRW block can follow anything.
Let $r_m$ be the number of acceptable strings of length $3m$ that end with an RRW block, $c_m$ the number that end with an RWR block, and $\ell_m$ the number that end with a WRR block. Then from the last paragraph we see that
$$\begin{align*}
s_m&=r_m+c_m+\ell_m\;,\\
r_m&=s_{m-1}\;,\\
c_m&=c_{m-1}+\ell_{m-1}\;,\text{ and}\\
\ell_m&=\ell_{m-1}\;.
\end{align*}\tag{1}$$
Clearly $r_1=c_1=\ell_1=1$, so $\ell_m=1$ for all $m\ge 1$, and $(1)$ reduces to
$$\begin{align*}
s_m&=r_m+c_m+1\;,\\
r_m&=s_{m-1}\;,\text{ and}\\
c_m&=c_{m-1}+1\;.\\
\end{align*}\tag{2}$$
From the last line of $(2)$ it’s clear that $c_m=m$ for all $m\ge 1$, so $$s_m=s_{m-1}+m+1\;.\tag{3}$$ If we set $s_0=1$ and $r_0=c_0=0$, the recurrences are all valid for $m\ge 1$, and it’s not hard to verify that
$$s_m=\sum_{k=1}^{m+1}=\frac12(m+1)(m+2)=\binom{m+2}2\;.$$
This completely answers the question when $n$ is a multiple of $3$.
Now suppose that $n=3m+1$. There is an acceptable string of length $3m$ ending in WRR, namely, a string of $m$ WRR blocks; we can append a white ball to this string, so there is an acceptable string of length $n$ with just $2m$ red balls. However, this is the only such string, since, as we saw before, a WRR block must follow another WRR block, and so on back to the beginning of the string. Thus, there is only one acceptable string of length $n=3m+1$ that uses the minimum possible number of red balls: $(\text{WRR})^m\text{W}$.
Finally, suppose that $n=3m+2$. Clearly at least one of the last two balls must be red, and one suffices, since we can append a red ball to the string $(\text{WRR})^m\text{W}$, so the minimum possible number of red balls in this case is $2m+1$. We can append WR to an acceptable string of length $3m$ if and only if the string is $(\text{WRR})^m$; that accounts for one acceptable string of length $n$. We can append RW to any acceptable string of length $3m$ that ends in RWR or WRR. There is just one of the latter, and there are $c_m=m$ of the former, so this case accounts for another $m+1$ acceptable strings of length $n$. Thus, there are $m+2$ acceptable strings of length $n=3m+2$. |
H: the partial derivatives computing
Given that:
$u_{xx}^{''}=u_{yy}^{''} \;, \;
u(x,2x)=x \;, \;
u_{x}^{'}(x,2x)=x^{2}\;, \;
$
How to find the following values?
$u_{xx}^{''}(x,2x)=?\;
u_{xy}^{''}(x,2x)=? \;
u_{yy}^{''}(x,2x)=?\; $
Thanks a lot.
AI: As we have $u(x, 2x) = x$, the chain rule gives
$$ u_x(x,2x) + 2u_y(x,2x) = 1 $$
As $u_x(x,2x) = x^2$, we have $u_y(x, 2x) = \frac{1-x^2}{2}$. Starting with $u_x(x,2x) = x^2$, the chain rule gives
$$ u_{xx}(x,2x) + 2u_{xy}(x,2x) = 2x \tag 1 $$
and starting with $u_y(x, 2x) = \frac{1-x^2}{2}$, we get
$$ u_{xy}(x,2x) + 2u_{yy}(x,2x) = -x \tag 2$$
Subtracting (2) twice from the equation (1), we get
$$ u_{xx}(x,2x) - 4u_{yy}(x,2x) = 4x $$
As $u_{xx} = u_{yy}$, we have
$$ u_{xx}(x,2x) = u_{yy}(x,2x) = -\frac 43x $$
And hence, by (2)
$$ u_{xy}(x,2x) = -x + \frac 83 x = \frac 53 x. $$ |
H: Spherical coordinate system
I can easily write $z$ axis value is $r\cos\theta$ but what will be for $x$ and $y$ axis, explain a bit please.
From the above how can I write the area element as $d\vec{a} = r^2\sin\theta d\theta d\phi\hat{r}$?
AI: The piece of the sphere with radius between $r$ and $\mathrm{d}r$ etc. has a shape tends towards a cuboid as the size of the piece cut decreases, with height $\mathrm{d}r$, width $\mathrm{d}\theta$ and length $r\sin\theta\mathrm{d}\phi$. Its volume is $r^2\sin\theta \mathrm{d}\theta \ \mathrm{d}\phi\mathrm{d}r$. |
H: A few questions on nonstandard analysis
I know that nonstandard analysis is analysis plus the existence of infinitesimal numbers. Does it mean that nonstandard analysis is the same theory as $ZF+\exists$infinitesimal numbers?
From what I read about it on Wikipedia there seem to be a few different approaches to it. There is "Robinson's method" which is "based on studying models". There is "Nelson's method" which is "Internal set theory" which is an extension of ZF. Are they the same? Or are there theorems that are true in one but not in the other?
If IST is an extension of ZF how much stronger than ZF is the "theory of nonstandard analysis"?
AI: To answer your first question, one can indeed think of nonstandard analysis, in a first approximation, as entailing the existence of infinitesimal numbers as you write, but to be more precise one would have to elaborate further. I like to think of it in terms of three approaches (other editors may disagree), as follows:
(1) the most straightforward approach (in my opinion) is through the construction of a proper field extension $\mathbb R^\star$ (notation varies, but let's stick with this one which is used in Keisler's book, see http://www.math.wisc.edu/~keisler/calc.html) of the real field $\mathbb R$. This field extension can be constructed in a way somewhat similar to constructing $\mathbb R$ out of the rationals $\mathbb Q$. Namely, one starts with sequences of real numbers, introduces a suitable equivalence, and obtains the hyperreal field $\mathbb R^\star$ as the quotient. This construction is called the ultrapower construction. A serious undergraduate algebra course provides enough background to understand this construction; namely what is required is the existence of a certain maximal ideal in a suitable ring.
(2) A more sophisticated route (and the one taken by Robinson in his 1966 book, see http://www.google.co.il/books?id=OkONWa4ToH4C&source=gbs_navlinks_s) is to invoke the compactness theorem from mathematical logic (more specifically, model theory) so as to prove the existence of such an $\mathbb R^\star$ (in Robinson's book this is denoted $^\star\mathbb R$).
(3) Edward Nelson's approach, called IST (internal set theory) is a reformulation of Robinson's approach where, instead of extending the field $\mathbb R$, Nelson takes the "syntactic" route. This means that the language of ordinary set theory is enriched by the addition of a unary predicate "Standard". Then infinitesimals are found within the "ordinary" $\mathbb R$ itself; so to speak they have been there all along, it's just that we haven't noticed (because the ordinary syntax of set theory is insufficiently rich). An infinitesimal number is NOT "Standard".
All three approaches are equivalent (at least at a basic level), so one proves the same theorems in all of them. No new axioms are needed beyond those of ZFC.
A more detailed summary by Joel David Hamkins appears here.
Since ZF was mentioned in your question, I add the following. Recently it turned out that one can do axiomatic nonstandard analysis conservatively over ZF (for a discussion of the meaning of conservativity see https://mathoverflow.net/a/401076/28128). The relevant publications can be consulted at https://u.math.biu.ac.il/~katzmik/spot.html . |
H: I would like to know an intuitive way to understand a Cauchy sequence and the Cauchy criterion.
My understanding from the definition in my book (Rudin) is this.
A seq. $\{p_n\}$ in a metric space $X$ (I only really know $\mathbb R^k$) is said to be a Cauchy sequence if for any given $\epsilon > 0$, $\exists N\in \mathbb N$ such that $\forall n,m\ge N$, $d(p_n,p_m)<\epsilon$.
(1) I see it as, given any tiny value $\epsilon$, we can find a natural number $N$ large enough so that the distance between $p_n$ and $p_m$ is less than $\epsilon$. Am I right ?
The reason I'm asking this is because I was trying to understand the proof of how
$$\sum a_nb_n$$
can converge, and the book said this
$$\left\lvert \sum_{n=p}^{q}a_nb_n\right\rvert \leq \epsilon$$
satisfies the Cauchy criterion and therefore it converges.
I read other questions and answers about the Cauchy sequence, but it didn't really help me…
Can someone explain me what's going on?
Edit:
Suppose
a) the partial sums of $A_n = \Sigma a_n$ form a bounded sequence
b) $b_0 \geq b_1 \geq \dotsb$
c) $\lim_{b \to \infty} b_n = 0$
Using the partial summation formula, algebraically the equation in the bottom is proved
$$\left\lvert \sum_{n=p}^{q}a_nb_n \right\rvert \leq \epsilon$$
Algebraically I had no problem, but I don't know why this proves convergence. I thought to show that a sequence is Cauchy, we need to find the distance between two terms in a sequence. That's where I'm confused.
AI: Since you asked specifically how to understand Cauchy sequences "intuitively" (rather than how to do $\epsilon,\delta$ proofs with them), I would say that the best way to understand them is as Cauchy himself might have understood them. Namely, for all infinite indices $n$ and $m$, the difference $p_n-p_m$ is infinitesmal. Such formalisations exist, for instance, in the context of the hyperreal extension of the field of real numbers.
As far as the particular series you asked about, what is going on is that the book is considering the sequence $p_n$ of partial sums of the series, and applying the Cauchy criterion to this sequence. Then the difference $p_n-p_m$ is the expression $\sum_m^n$ that you wrote down (up to a slight shift in index).
Some thoughts on Cauchy can be found here. |
H: Fourier series for $f(x)=(\pi -x)/2$
I need to find the Fourier series for $$f(x)=\frac{\pi -x}{2}, 0<x<2\pi$$
Since the interval isn't symmetric over $0$, I guess I need to consider $f$'s periodic extension to $\mathbb R$. let's call it $g$. Then $$g(x)=\begin{cases}\frac{\pi -x}{2}, \text{ if } 0<x<\pi\\ \frac{-\pi -x}{2}, \text{ if }{-\pi <x<0}\end{cases}$$
Due to $g$ being odd, the fourier coefficients $a_n$ are all $0$.
And $$b_n=\frac{2}{\pi}\int _0^\pi g(x)\sin(nx)dx=\frac{1}{\pi}\int _0^\pi (x-\pi)\sin (nx)dx$$
According to wolfram alpha $b_n=1/n$ so the Fourier series is
$$\sum _{n=1}^{+\infty}\frac{\sin(kx)}{n}$$
But when I check the result of $x=1/2$ I get this which seems to diverge. It also fails for $x=\pi /2$ and a few other $x$'s.
What am I doing wrong?
I just saw this question. it looks like my answer is correct. am I doing anything wrong in my verification?
EDIT: I realised I forgot to divide by $\pi$. I fixed that but it still doesn't work. I also updated the links.
AI: You forgot to divide by $\pi$. WolramAlpha's result is correct, but the coefficients $b_n$ are given by
$$b_n=\frac{1}{n}$$
Your second formula for the series in WolframAlpha is wrong, you should divide by $k$, not multiply. Then everything should be fine: WolframAlpha |
H: homomorphisms of infinite groups
Prove that each of the following is a homomorphism, and describe its kernel:
the function $f: \mathbb{R}^*\to\mathbb{R}_{>0}$ defined by $f(x)=|x|$
My proof step: The kernel of $f$ is the set $$k=\{x\in\mathbb{R}^*:f(x)=e\}.$$
Let $a,b \in k$, $f(a)f(b)=f(ab)=ee=e$. I think I'm missing a lot of steps.
AI: HINT: To find $\ker f$ you need to answer the following questions:
What is the identity in $\Bbb R_{>0}$? In other words, what is the element $e$ of $\Bbb R_{>0}$?
What elements of $\Bbb R^*$ are mapped to that identity by $f$?
To show that $f$ is a homomorphism, you need to show that it has the homomorphism property: for all $x,y\in\Bbb R^*$, $f(xy)=f(x)f(y)$.
All of this requires using the definition of $f$. |
H: Why is this polynomial a function of $X^3$?
In studying that recent question, I noticed that curious (or perhaps
not so curious) property : if $x,y$ are rational numbers and $a$ is the real
part of a cubic root of $x+iy$, then $Q(a^3)=0$ where $Q$ is a polynomial
of degree three with rational coefficients.
This is a purely algebraic property, and I obtained it by calculating some
resultants with my computer.
Now, can we show a priori that $a$ is annihilated by a
polynomial of this form, without computer-made computations ? Note that
this property does not hold for the imaginary part of a cubic root.
AI: That's because the solutions to the system $(a+b)^3 = x+y, (a-b)^3 = x-y$, are the $9$ couples that you obtain by choosing a complex cube root $u$ of $x+y$, a complex cube root $v$ of $x-y$, and set $a=(u+v)/2, b = (u-v)/2$ (this is clear after a change of variables)
Since you can multiply both $u$ and $v$ by the same cube root of $1$ to obtain new solutions,
if $(a,b)$ is a solution, so are $(\zeta_3a,\zeta_3b)$ and $(\zeta_3^2 a, \zeta_3^2 b)$, hence the degree $9$ polynomial for $a$ is in fact a polynomial in $a^3$ (and of course the same goes with the one for $b$) |
H: $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$?
Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?
I thought it would be 12 this as per pemdas rule:
$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$
Wanted to confirm the right answer from you guys. Thanks for your help.
AI: Your answer of $12$ is numerically correct, but you’ve applied the precedence rules incorrectly. In the absence of parentheses multiplication is done before addition, so
$$1+1+1+1+1+1+1+1+1+1+1+1\cdot 0+1$$
is to be evaluated as
$$1+1+1+1+1+1+1+1+1+1+1+(1\cdot 0)+1\;,$$
not as
$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot(0+1)\;.$$
Since $1\cdot0=0$, this simplifies to
$$1+1+1+1+1+1+1+1+1+1+1+0+1=12\;.$$ |
H: Finding a palindromic number which is the difference of two palindromic numbers
Let $X$ and $Y$ be two $4$-digit palindromes and $Z$ be a $3$-digit palindrome. They are related in the way $X-Y=Z$. How can we figure out $Z$?
AI: Hint: Write $X = 1001a + 110b$, $Y = 1001c + 110d$ with $a,b,c,d \in \{0,\ldots, 9\}$, without loss of generality we may assume $a \ge c$. Then
$$ Z= X-Y = 1001(a-c) + 110(b-d) $$
Now note: Can $a-c > 1$ hold? Can $a - c = 0$ hold? Now check for which value of $k := d-b$ we have that $1001 - 110k$ is a palindrome. |
H: Proof of the Riesz Representation Theorem
Theorem: Let $F$ be a continuous linear functional on the Hilbert space $H$, then $\exists !$ (exists one and only one) $y \in H$ such that $F(x) = (x,y)$ for $x\in H$.
Proof:
Uniqueness: $$F(x)=(x,y)=(x,y') \Rightarrow (x,y-y')=0 \space \forall x \in H \Rightarrow ||y-y'||^2=0$$
Existence: if for any $x\in H, \space F(x) =(x,y)= 0 $ we'll set $y=0$. Let that $F \neq 0. $ and define a closed subspace $H_0 = \{x \in H: F(x)=0\}$. We know that $H \neq H_0$ because $H_0 = F^{-1}(\{0\})$.
Q1: "$H_0$ is a subspace by the linearity of $F$, closed by continuity of $F$." What does this mean? I understand what's a subspace, but don't understand what does it have to do with linearity and continuity.
$\exists z\in H$ such that $||z||=1$. For $u \in H_0, \space F(u)=(u,z) =0$.
Q2: Does this imply that $u \perp z \space \forall u \in H_0, \forall z \in H$?
Let's define $u=F(x)z - F(z)x, \space x \in H$
$$F(u)=(F(x)z - F(z)x,z)=F(x)F(z)-F(z)F(x)=0=F(x)(z,z)-F(z)(x,z)$$
Since we know that $||z|| = \sqrt{(z,z)}=1$, then
$$0=F(x)(z,z)-F(z)(x,z) \Rightarrow F(x) =F(z)(x,z) = (x,\bar{F(z)z)} $$
By defining $y= \bar{F(z)}z$ we've completed the proof.
By the way, the bar on the top denotes the complex conjugate, as is common in Hilbert spaces. i.e. $\alpha (x,y)= (\alpha x,y)=(x,\bar{\alpha} y) $
Corollary: If $H=L^2(\mu)$ and $F$ is a continuous linear functional on $L^2(\mu)$, then $\exists ! \space g \in L^2(\mu)$ such that $$F(f)=\int_X f \bar{g} d\mu = (f,g)$$
In the context of the Radon-Nikodym theorem, $g$ is the R-N derivative.
AI: Question 1: You defined $H_0 = F^{-1}(\{0\})$, where $F \in H^*$ (the continuous linear functionals on $H$). In order for $H_0$ to be a subspace, you have to check that $\lambda x + \mu y \in H_0$ for all $x,y \in H_0$, $\lambda, \mu \in \mathbb K$, right? But
$$ \lambda x + \mu y \in H_0 \iff F(\lambda x + \mu y) = 0 $$
As $F$ is linear (and $x,y \in H_0$), we have
$$ F(\lambda x + \mu y) = \lambda F(x) + \mu F(y) = 0 $$
So $H_0$ being a subspace follows from the linearity of $F$. In order for $H_0$ to be closed, we must show that for all sequences $(x_n)$ in $H_0$ converging to some $x \in H$ we must have $x\in H_0$. But, as $F$ is continuous, we have that $x_n \to x$, implies $F(x_n) \to F(x)$. As $F(x_n) = 0$ for every $n$, we get $F(x) = 0$, so $x \in H_0$. So the closedness of $H_0$ follows from the continuity of $F$.
Question 2: No, it doesn't. As $H_0$ is a proper closed subspace of $H$ ($H_0 \subsetneq H$ as $F \ne 0$), there is some $z \in H$, with $z \mathrel\bot H_0$ and $\|z\| = 1$. This means that for this $z$, not for all we have
$$ (z, u) = 0, \quad \text{all } u \in H_0. $$ |
H: On Absolutely Continuous Functions
I would like to know if we can extend the concept of absolute continuity to functions $f:[a,b]\to X$, where $X$ is a topological vector space. I browsed some books on Topological Vector Spaces but can't find the definition of absolutely continuous functions defined on $[a,b]$ and take values on $X$. I would be greatful if someone can provide me a definition (if possible) of absolute continuous function $f:[a,b]\to X$. Thanks in advance...
AI: For general Topological Vector Spaces: no.
There are two definitions of absolute continuity: one is for functions, the other for measures.
Definition (AC for functions): Let $I$ be an interval in $\mathbb{R}$. A function $f:\mathbb{R}\to\mathbb{R}$ is said to be absolutely continuous if for every $\epsilon > 0$ there exists $\delta > 0$ such that if $(a_k,b_k) = I_k \subset I$, $k \in \{1\ldots n\}$ are pairwise disjoint subintervals of $I$ satisfying
$$ \sum_1^n |I_k| = \sum_1^n |b_k - a_k| < \delta $$
then
$$ \sum_1^n |f(b_k) - f(a_k)| < \epsilon.$$
This definition only requires that we can evaluate
$$ \sum_1^n |f(b_k) - f(a_k)| $$
and naturally we can extend this to the case of the codomain being any metric space $(X,d)$ by replacing
$$ \sum_1^n |f(b_k)- f(a_k)| \implies \sum_1^n d(f(b_k),f(a_k)) $$
A general TVS may not even be metrizable, and I don't see how you can have a reasonable generalisation of the notion of absolute continuity if you don't have a reasonable replacement for $\sum_1^n|f(b_k) - f(a_k)|$.
See also the EOM entry.
Definition (AC for measures): Let $(\Omega,\mathcal{S},\mu)$ be a measure space (in the interval case we usually take $\Omega = [a,b]$, $\mathcal{S},\mu$ corresponding to the Lebesgue measure). A measure $\lambda$ is said to be absolutely continuous with respect to $\mu$ if every $\mu$-null set is $\lambda$-null.
The usual generalisation of this concept is to the case where $\lambda$ is a vector measure (a measure "taking values" in a Banach space). See, e.g., Chapter 29 of Schecter's HAF. |
H: Can $\frac{n!}{(n-r)!r!}$ be simplified?
I'm trying to calculate in a program the number of possible unique subsets of a set of unique numbers, given the subset size, using the following formula:
$\dfrac{n!}{(n-r)!r!}$
The trouble is, on the face of it, you may need an enormous structure to hold the dividend (at least). Is there a way of simplifying this calculation, so that it can be calculated using something smaller like $64$-bit integers, or are you really going to have to store numbers like $60!$?
Alternatively, is there a formula that is more suited to computing the aforementioned value?
AI: As mentioned by others, the binomial coefficient
$$\binom{n}{r}=\frac{n!}{r!(n-r)!}$$
is already so fundamental a quantity, that it is itself considered a canonical form. Nevertheless, you seem to be asking how one might compute this, while avoiding unsavory events like integer overflow.
Since this is for programming purposes, if you can stand to use a two-dimensional array, you might be interested in the doubly-indexed recurrence relation
$$\binom{n}{r}=\binom{n-1}{r}+\binom{n-1}{r-1}$$
along with the initial conditions $\tbinom{n}{0}=\tbinom{n}{n}=1$, or the singly-indexed recurrences
$$\begin{align*}
\binom{n}{r}&=\frac{n}{n-r}\binom{n-1}{r}\\
\binom{n}{r}&=\frac{n-r+1}{r}\binom{n}{r-1}
\end{align*}$$
One other possibility is to instead deal with the logarithms of the factorials, since $\log(ab)=\log\,a+\log\,b$ and $\log(a/b)=\log\,a-\log\,b$. You might want to look into Stirling's formula if you want to take this route, but this is only intended for the cases where $n$ and $r$ are large. |
H: Second Derivative of a matrix
Pardon me for not knowing LateX representation,
I have following function, where $\mu$ and $\Sigma$ are both Matrices.
$$ h = \mu^T \Sigma \mu $$
which is a function of $\alpha$, such that its derivative can be written as
$$ \def\p#1#2{\frac{\partial #1}{\partial #2}}
\p h\alpha = \left(\p \mu\alpha\right)^T\Sigma^{-1}\mu - \mu^T\Sigma^{-1}\p\Sigma\alpha \Sigma^{-1}\mu + \mu^T \Sigma \p \mu\alpha
$$
I need to find out the Second derivative wrt $\alpha$. Please help, I searched through net, but couldn't fond any such property.
AI: Assuming that $h = \mu^T \Sigma^{-1}\mu$, as suggested by the derivative, we have: Taking another derivative we apply the product rule as we did for the first and use the chain rule for $\Sigma^{-1}$, in using $\def\pd#1#2{\frac{\partial #1}{\partial #2}}\def\pdd#1#2{\frac{\partial^2 #1}{\partial #2^2}}$
$$ \pd{\Sigma^{-1}}{\alpha} = -\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1} $$
So for the first term of $\pd h\alpha$ we have
$$ \pd{}\alpha\left(\pd{\mu^T}\alpha \Sigma^{-1}\mu \right)
= \pdd{\mu^T}\alpha\Sigma^{-1}\mu - \pd{\mu^T}\alpha \Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\mu + \pd{\mu^T}\alpha \Sigma^{-1}\pd\mu\alpha
$$
For the second term
\begin{align*}
&\!\!\!\!\!\!\!\pd{}\alpha\left( \mu^T\Sigma^{-1}\pd\Sigma\alpha \Sigma^{-1}\mu\right)\\
&= \pd{\mu^T}\alpha \Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\mu
- \mu^T\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\mu
+ \mu^T\Sigma^{-1}\pdd\Sigma\alpha\Sigma^{-1}\mu\\& \ \ {} - \mu^T\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\mu + \mu^T\Sigma^{-1}\pd\Sigma\alpha \Sigma^{-1}\pd \mu\alpha
\end{align*}
And for the third term
$$ \pd{}\alpha\left(\mu^T \Sigma^{-1}\pd\mu\alpha \right)
= \pd{\mu^T}\alpha\Sigma^{-1}\pd\mu\alpha - {\mu^T}\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\pd\mu\alpha + {\mu^T} \Sigma^{-1}\pdd\mu\alpha
$$
Adding everything, we should have
\begin{align*} \pdd h \alpha &=
\pdd{\mu^T}\alpha \Sigma^{-1}\mu - \mu^T\Sigma^{-1}\pdd\Sigma\alpha\Sigma^{-1}\mu + \mu^T\Sigma^{-1}\pdd\mu\alpha\\
&{}+ 2\left(\pd{\mu^T}\alpha \Sigma^{-1}\pd\mu\alpha - \pd{\mu^T}\alpha\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\mu-\mu^T\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\pd\mu\alpha + \mu^T\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\mu\right)
\end{align*} |
H: Is it possible to find function that contains every given point?
Let say we have a arbitrary number of given points and there is at least one function, for which every point lies on its graph. Is it possible to find that function using only X and Y coordinates of every given point?
Example:
We are given points $A(0,1)$; $B(0.27;0)$ and $C (3.73;0)$. All of these points lie on the graph of the function $f(x) = x^2 - 4x + 1$. Is it possible to find this function using only X and Y coordinates of the point A,B and C.
Note: The number of point is can be really big and also the exponent of the function.
AI: There are many such functions so in general it's not possible to recover the one you started with.
For example if you had a set of points $\{(n,n): n\in\mathbb N\}$ both the functions $f(x) = x$ and $g(x) = x+sin(\pi x)$ would pass through all the points.
If you have a finite number of points you can find a polynomial that passes through them all.
One way to find it would be the following algorithm.
Let the coordinates of the points be $(x_1,y_1), (x_2,y_2),\dots, (x_N,y_N)$
First we find a function $f_1$ that passes through $(x_1,y_1)$ that's easy. $F_1(x) = y_1$.
Now suppose we've constructed $F_n$ that passes through $(x_1,y_1), \dots,(x_n, y_n)$ (so far we've only done this for $n=1$).
Let $$g_n(x) = \prod_{i=1}^n (x-x_i)$$
then $g_n(x_i) = 0$ for every $i\leq n$. Therefore for every value of $\lambda$ the function
$F_n(x_i) + \lambda g_n(x_i) = y_i$ for every $i\leq n$.
So set $$\lambda_n = \frac{y_{n+1} - F_n(x_{n+1})}{g_n(x_{n+1})}$$ and $$F_{n+1}(x) = F_n(x) + \lambda_n g_n(x).$$
Then $F_{n+1}$ passes through $(x_i,y_i)$ for every $i\leq n+1$.
So apply this step repeatedly until you get $F_N$, which passes through every point.
Let's try it with your equation I'm going to choose slightly different points so that we can use whole numbers. Set
$$\begin{array}{rlrl}
x_1 &=0 & y_1 &=1 \\
x_2 &=2 & y_2 &=-3 \\
x_3 &=5 & y_3 &=6
\end{array}$$
It's easy to check that $x^2-4x+1$ passes through all three points. So let's follow the recipe.
First we set $F_1(x) = y_1 = 1$ and $g_1(x) = (x-x_0) = x$.
Then we can calculate $$\begin{array}{rl}\lambda_1 &= \frac{y_2 - F_1(x_2)}{g_1(x_2)} \\&= -2\end{array}$$
Hence $$\begin{array}{rl}F_2(x) &= F_1(x) + \lambda_1 g_1(x) \\ &= 1 -2x \end{array}$$
We can check that $F_2(0)=1$ and $F_2(2)=-3$.
Next we have $g_2(x) = x(x-2) = x^2-2x$
and we can work out $$\begin{array}{rl}\lambda_2 &= \frac{y_3 - F_2(x_3)}{g_2(x_3)} \\ &= \frac{6 - (-3)}{40} \\&= 1\end{array}$$
So $\lambda_2 = -\frac 3{15}$ and we have
Then we have $$\begin{array}{rl}F_3(x) &= F_2(x) + \lambda_2 g_2(x) \\
&= (1-2x) + (x^2 - 2x) \\
&= x^2-4x+1
\end{array}$$
Then as we've run out of points we set $f(x) = F_3(x)$ and we are done. |
H: Tangential Space of a differentiable manifold is always $\mathbb R^n$?
Let $\mathcal M$ be a differential manifold with a point $p$. Let U be an open set, $p\in U$, on $\mathcal M$ and let $\phi,\psi:U\to \mathbb R^n$ be a charts on $\mathcal M$. I'm having diffculties arranging all the concepts of a differential manifold. The "manifold" of required spaces gives me a headache. At the moment im utterly confused with the construction of the tangent space at $p$.
I'll start describing my thoughts from the start: $\require{AMScd}$
\begin{CD}
U @>\phi>> \mathbb R_\phi^n \\
@V \psi VV @.\\
\mathbb R_\psi^n
\end{CD}
That's the open set U with charts $\psi,\phi$. I denoted $\mathbb R^n$ with the chart pointing to it, so we can make sure to differentiate between the different $\mathbb R^n$ -spaces we will encounter.
The tangent space of $\mathcal M$ or $U$ in $p$ is a $n$-dimensional $\mathbb R$-vector space, which has isomorpisms pointing to $T_{\phi(p)} \mathbb R^n$ and $T_{\psi(p)} \mathbb R^n$.
Where $T_{\phi(p)} \mathbb R^n$ and $T_{\psi(p)} \mathbb R^n$ are vector spaces we're inducing by the choice of our bases $B_\psi , B_\phi$ depending on our charts as follows after the diagram:
\begin{CD}
T_p \mathcal M @> \cong >> T_{\phi(p)} \mathbb R^n \\
@V \cong VV @.\\
T_ {\psi(p)} \mathbb R^n
\end{CD}
$T_{\phi(p)} \mathbb R^n$ and $T_{\phi(p)} \mathbb R^n$ are induced by the vector space bases $B_\psi , B_\phi$ using isomorphisms from the images of our charts ($\phi(U)\subset \mathbb R_{\phi}^n, \psi(U)\subset\mathbb R_{\psi}^n)$ to the tangent space $T_{\phi(p)} \mathbb R^n$.
The first base vector of the base $B_\phi$ is given as $\frac{\partial}{\partial x_i}|_p := \phi^-1 (\phi(p)+ t e_i)$, the other base vectors of $B_\phi$ follow with $i \in {2,....,n}$.
Let's call the function that takes $p$ and gives us $\frac{\partial}{\partial x_i}|_p$ and $\frac{\partial}{\partial y_i}|_p$ by names $\bar{\phi}_p$ and $\bar{\psi}_p$. In a diagram we now got two choices for the base of our ismorphic tangent spaces depending on $\phi,\psi$:
\begin{CD}
T_{\phi(p)} \mathbb R^n @. T_{\psi(p)} \mathbb R^n \\
@A \bar{\phi}_p AA @A \bar{\psi}_p AA\\
\mathbb R_\phi^n @. \mathbb R_\psi^n
\end{CD}
First question: Why is $T_p \mathcal M$ an $\mathbb R$-vector space, i don't know why the underlying field has to be $\mathbb R$. Or is it, that we just identify $T_p M$ with $T_{\phi(p)}\mathbb R^n$, if so, why can we do that?
I have the notion that $\mathcal M$ is of a abstract nature, so how comes it's tangential space is so $\mathbb R^n$ related?
Second question:
Assuming i understood why $T_p \mathcal M$ is a $\mathbb R$-vector space, i can now ask myself how i can change bases.
So we get to linear algebra and the pushforward as linear function between vector spaces.
The image of the coordinate vector $(1,0,...,0)^t$,of the first base vector $\frac{\partial}{\partial x_i}|_p$, is the first row of the matrix $D((\psi)\circ\psi^-1)$ giving us the coordinate vector expressed in the base $B_\psi$.
\begin{CD}
T_{\phi(p)} \mathbb R^n @>D((\psi)\circ\phi^-1)>> T_{\psi(p)} \mathbb R^n
\end{CD}
Third question: The "last" notion that gives me a headache right now is that of the function:
$$\phi_*: T_p \mathcal M \to T_{\phi(p)} \mathbb R^n\text{ defined as }
\phi_*(\gamma) := \frac{d}{dt}(\phi \circ \gamma )|_{t=0}$$
where $\gamma$ is a tangent vector, for example a curve(geometric defintion of a tangent vector).
Is that definition base independend? If so taking the first base vector of $B_\phi$
namely $\frac{\partial}{\partial x_i}|_p := \phi^-1 (\phi(p)+ t e_i)$ and using $\phi_*$ on it, gives us another tangent vector in the tangent space $T_{\phi_*} \mathbb R^n$, or is it that same as $\mathbb R_\phi^n$ or $T_{\phi(p)}R^n$?
Thanks for any answers!
AI: First question:
Congratulations, you have discovered algebraic varieties and their Zariski tangent space. For classical differential geometry, the reason that the tangent space is an $\mathbb{R}$ vector space is by definition: the local model of a differentiable manifold is $\mathbb{R}^n$. The whole motivation of classical differential geometry is to study stuff that, at sufficiently small neighborhoods, look like $\mathbb{R}^n$. It would be strange if its tangent space does not look like that of $\mathbb{R}^n$, but be over some other field, no? If you model the local geometry by an affine variety associated to a different base field, you will get appropriately different vector spaces as the tangent space.
Second question:
Looks like you more or less got it, except you probably meant $\psi\circ \phi^{-1}$ rather than $\psi \circ \psi^{-1}$. Again, locally you are just working with functions defined over $\mathbb{R}^n$ (which may take $\mathbb{R}^k$ values).
Third question:
$\phi_*$ pushes forward to $T_{\phi(p)}\mathbb{R}^n$, not $T_p\mathbb{R}^n$. Since you didn't explain what $\gamma$ is, I have no idea what your second equation means. But the derivative with respect to $t$ makes me wonder if you intended $\gamma:\mathbb{R}\to M$ and you are actually thinking about the expression
$$ \phi_*(\frac{\mathrm{d}}{\mathrm{d}t}{\gamma}(t)) = \frac{\mathrm{d}}{\mathrm{d}t}( \phi\circ \gamma)(t)~? $$
In that case, this definition is base independent. You can explicitly verify it by composing with the smooth maps associated to changes of charts.
Your expression $\partial/\partial x_i |_p = \phi^{-1}(\phi(p) + t e_i)$ is wrong. The left hand side is an object living in $T_p M$ and the right hand side is an object living in $M$. You appear to have a type error. |
H: Principal value of $\int_0^\infty \frac{x^{-p}}{x-1}dx$ for $|p|<1$
If $|p|<1$, how to find the Cauchy Principal Value of $$\int_0^\infty \frac{x^{-p}}{x-1}dx$$
I tried spliting the integration from $0\to 1$ and $1 \to \infty$ and switching $x = 1/u$, but no luck getting desired result which the book says is $ \pi \cot p \pi$. Please help.
ADDED:: using above method I got $\displaystyle \sum_{k=0}^\infty \left( \frac{1}{k + p} + \frac{1}{p - (k+1)} \right) $, it turned out to be $p \cot p\pi$ from Mathematica, anyone any ideas?
ADDED:: It seems that $$\displaystyle \sum_{k=0}^\infty \left( \frac{1}{k + p} + \frac{1}{p - (k+1)} \right) = \sum_{k=-\infty}^\infty \frac{1}{k +p} = - \text{Res}[ \pi \cot(z \pi)/(p+z), z=-p] = \pi \cot p\pi$$
Also I was having trouble doing this via contour. Is this the shape of contour?
AI: Note: we must restrict $p$ to $p \in (0,1)$ for the Cauchy PV of the integral to converge at infinity.
The idea is to consider an integral in the complex plane and use Cauchy's theorem. That is, if we define a contour $C$ within which there are no poles, we have
$$\oint_C dz \frac{z^{-p}}{z-1} = 0$$
$C$ will take the form of a keyhole contour, except that we will indent with a semicircle about the singularity at $z=1$ each time we make a pass along the real axis. Thus, the above contour integral is broken up into $8$ pieces:
$$\oint_C dz \frac{z^{-p}}{z-1} = \int_{\epsilon}^{1-\epsilon} dx \frac{x^{-p}}{x-1} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{(1+\epsilon e^{i \phi})^{-p}}{\epsilon e^{i \phi}} + \int_{1+\epsilon}^{R} dx \frac{x^{-p}}{x-1} + \\i R \int_{0}^{2 \pi} d\phi \, e^{i \phi} \frac{R^{-p} e^{-i p \phi}}{R e^{i \phi}-1} + e^{-i 2 \pi p}\int_{R}^{1+\epsilon} dx \frac{x^{-p}}{x-1} + \\ i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{(e^{i 2 \pi}+\epsilon e^{i \phi})^{-p}}{\epsilon e^{i \phi}} + e^{-i 2 \pi p} \int_{1-\epsilon}^{\epsilon} dx \frac{x^{-p}}{x-1} + \\ i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\epsilon^{-p} e^{-i p \phi}}{\epsilon e^{i \phi}-1} $$
Note that when we make the second pass along the real axis, we exploit the fact that we have traversed $2 \pi$ in argument, so we parametrize the real line using $z=x\, e^{i 2 \pi}$. The multivaluedness of the integrand produces the nontrivial result we seek.
We take the limit as $\epsilon \to 0$ and $R \to \infty$. The integrals over $x$ combine to form a factor times the Cauchy PV we seek. The fourth and eighth integrals vanish in this limit because of the restriction over the value of $p$. We are then left with the integrals over the bumps which do not vanish. We thus have, after some simplification:
$$\left ( 1-e^{-i 2 \pi p}\right ) PV \int_0^{\infty} dx \frac{x^{-p}}{x-1} - i \pi \left ( 1+e^{-i 2 \pi p}\right ) = 0$$
Note that the second term comes from evaluating the second and sixth integrals and combining, while the first term comes from combining the first, third, fifth, and seventh integrals.
Solving for the Cauchy principal value, we get
$$PV \int_0^{\infty} dx \frac{x^{-p}}{x-1} = i \pi \frac{1+e^{-i 2 \pi p}}{1-e^{-i 2 \pi p}} = \pi \cot{\pi p}$$ |
H: generalized derivative of Wiener process
Defined a standard Wiener process $W = (W_t , \mathcal F_t)_{t≥0}$ and a deterministic, continuously differentiable function $f : [0, ∞) → \mathbb R$. Prove that
$$f(t)W_t=\int_0^tW_sf'(s)ds+\int_0^tf(s)dW_s$$
AI: If you know Ito Lemma, then you just consider $F(t,x) = f_t\cdot x$ and the differential of $F$ is
$$
\mathrm dF(t,W_t) = f'_tW_t\mathrm dt + f_t\mathrm dW_t + \mathrm d[f,W]_t
$$
and since $f$ is deterministic and continuous, the latter term is zero. Hence
$$
\mathrm dF(t,W_t) = f'_tW_t\mathrm dt + f_t\mathrm dW_t \quad\Leftrightarrow \quad F(T,W_T) = F(0,W_0) + \int_0^T f'_tW_t\mathrm dt + \int_0^Tf_t\mathrm dW_t.
$$ |
H: Pigeon Hole Principle; 3 know each other, or 3 don't know each other
I found another question in my text book, it seems simple, but the hardest part is to prove it. Here the question
There are six persons in a party. Prove that either 3 of them recognize each other or 3 of them don't recognize each other.
I heard the answer use Pigeon Hole Principle, but i have no idea in using it. Could somebody please tell me the way to solve it? Thanks for the attention and sorry for the bad English and my messy post
AI: NOTE: in my answer I assume that the relation "know someone" is symmetric (i.e., A knows B if and only if B knows A). If this relation is not symmetric for you then, I did not really check it but I believe the statement is not true.\\\
Choose a person A at the party. The following two situations are possible:
(CASE 1) A knows at least three people, say B, C and D, at the party;
(CASE 2) A doesn't know at least three people at the party.
In (CASE 1), if at least a pair among {B,C}, {C,D} or {D,B} is formed by people that know each other, then you have three people that know each other. If there is no such pair among these three, then B, C and D are three people that do not know each other.
In (CASE 2) proceed similarly. |
H: Projective general linear groups of order 2 = projective special linear group of order 2
If $q\geq 5$ is a prime and we consider the group PGL(2,q) the projective general linear group of order 2 and the group PSL(2,q) the projective special linear group of order 2.
Can I then conclude PSL(2,q) = PGL(2,q)?
AI: The cokernel of the embedding $\mathrm{PSL}(2,q) \to \mathrm{PGL}(2,q)$ is isomorphic to $\mathbb F_q^*/(\mathbb F_q^*)^2$, that is you will have equality exactly iff every element of $\mathbb F_q$ is a square. For $q = 5$ for example, we have $\mathbb F_5^* = \{1,2,3,4\}$ and $(\mathbb F_5^*)^2 = \{1,4\}$, so $\mathrm{PSL}(2,5) < \mathrm{PGL}(2,5)$. |
H: Find derivatives of functions with respect to $ x$
Can someone help me with these.
Find the derivatives of the following functions with respect to $x$: here $a$ is an arbitrary
(fixed) real number.
$(a)$ $\displaystyle\int_{a}^{x^4} t^3\ \mathrm dt$
$(b)$ $\displaystyle \int_{-x^3}^{a^3} \dfrac{\mathrm dt}{1+t^2}$
AI: For example, let $\,F(x)\,$ be a function s.t. $\,F'(x)=x^3\,$ (why such a a function exists?), so that (assuming you've a typo in the first integral's upper limit):
$$g(x):=\int\limits_a^{x^4}t^3\,dt=F(x^4)-F(a)\implies g'(x)\stackrel{\text{chain rule}}=4x^3F'(x^4)=4x^3(x^3)^4=4x^{15}$$
Now you try in a similar fashion the other one. |
H: Matrix operations - equivalent operation for a given operation
This is the given problem, I need to write a code for this:
$(M*Q) \circ (N*Q) $
where $M,Q,N$ are known matrices, "$\ast$" denotes matrix multiplication and "$\circ$" denotes elementwise division.
Dimensions of the matrices:
\begin{align*}
M: &a\times l,\\
N: &a\times l,\\
Q: &l\times1.
\end{align*}
By element-wise division, I mean this:
$A \circ B = C$ where $C_{ij} = A_{ij}/B_{ij} \ \ \forall\ i,j$.
I want to convert this above problem to this:
Find $K$ such that $K*Q = (M*Q) \circ (N*Q)$.
AI: In general, this is impossible because $(M*Q) \circ (N*Q)$ is a rational function in the entries of $Q$ but $K*Q$ is linear. For instance, consider $M=(1,0),N=(0,1)$ and $Q=\pmatrix{x\\ y}$. Then $(M*Q) \circ (N*Q)=\frac xy$. Surely it is not a linear function in $x$ and $y$. |
H: Proof of the Lebesgue-Radon-Nikodym Theorem
Theorem: Let $\lambda, \mu$ be $\sigma$-finite measures defined on the $\sigma$-algebra $\mathcal{A}$ of the space $X$. Then,
a) Lebesgue decomposition: $\lambda=\lambda_a+\lambda_s$ where $\lambda_a \ll \mu$ and $\lambda_s \perp \mu$
b) $\exists h \in L^1(\mu)$ such that $\lambda_a = \int h d\mu$
Proof:
Define $\phi=\lambda + \mu \space \phi(X)<\infty, \space f \in L^2(\phi) $. Then $f \to \int f d\lambda $ is a continuous linear functional on $L^2(\phi)$ Because:
$$\int_X |f| d\lambda \le \int |f| d\phi \le \Bigg( \int_X |f|^2 d\phi \Bigg)^{\frac12}\Bigg( \int_X d\phi \Bigg)^{\frac12} $$
Q1: How does this prove the aforementioned claim? Don't we already have $f \to \int f d\lambda $ is a continuous linear functional on $L^2(\phi)$ from the fact that $f \in L^2(\phi) $?
Moving on, from the Riesz Representation theorem we know, that $\exists ! \space g \in L^2(\phi) $ such that $$\int_X f d\lambda = \int_X fg d\phi $$.
Choosing $f=\chi_E, \space E \in \mathcal{A}, \space \phi(E)<\infty$ we obtain $\lambda(E)=\int_E g d\phi$ and I'm told, that
$$\lambda(E)=\int_E g d\phi \Rightarrow 0 \le \frac{1}{\phi(E)}\int_E g d\phi = \frac{\lambda(E)}{\phi(E)} \le 1$$
Q2 Why does the above implication hold?
From the above equality we know, that $g(x) \in [0,1]$, an absolutely crucial fact for the rest of the proof. We define $A=\{x:g(x)<1\} \text{ and } B=\{x: g(x)=1\}$. Now,
$$\lambda(E) = \lambda(E\cap A) + \lambda(E \cap B) = \lambda_a (E)+\lambda_s(E)$$
$$\int_X f d\lambda = \int_X fg d (\lambda+\mu) \Rightarrow \int_X f(1-g) d \lambda = \int_X fg d \mu $$.
Choosing $f=\chi_B$ we obtain $\mu(B)=0 \Rightarrow \mu \perp \lambda_s$
Choosing $f=\chi_E \sum_{k=0}^n g^k$ we obtain
$$\int_E (1-g^{n+1}) d \lambda = \int_E \sum_{k=1}^n g^k d \mu \\ \lim_{n \to \infty} \int_E (1-g^{n+1}) d \lambda = \lim_{n \to \infty} \int_E \sum_{k=1}^n g^k d \mu$$.
and since $(1-g^{n+1}) \nearrow 1$ and $\sum_{k=1}^n g^k \nearrow h$ we can use the monotone convergence theorem to get
$$\lambda(E\cap A)=\lambda_a(E) = \int_E h d \mu $$
Remark: for $x \in B$ we get 0 measure because $g(x)=1$.
AI: Question 1: We define a functional $\ell\colon L^2(\phi) \to \mathbb K$ by
$$ \ell(f) = \int_X f \, d\lambda, \quad f \in L^2(\phi). $$
So to each $f \in L^2(\phi)$ we associate a number, namely the integral over $X$ with respect to $\lambda$. To know that $f$ is linear and continuous, we have to prove it. Linearity is easy, so we concentrate on continuity, we have for all $f \in L^2(\phi)\def\abs#1{\left|#1\right|}\def\norm#1{\left\|#1\right\|}$
\begin{align*}
\abs{\ell(f)} &= \abs{\int_X f\, d\lambda}\\
&\le \int_X \abs f\, d\lambda\\
&\le \int_X \abs f \, d(\lambda + \mu) \text{ note that $\int_X \abs f \, d\mu \ge 0$}\\
&= \int_X \abs f\cdot 1 \, d\phi\\
&\le \left(\int_X \abs{f}^2\, d\phi\right)^{1/2} \left(\int_X 1\, d\phi\right)^{1/2} \text{ this is Cauchy-Schwarz}\\
&= \phi(X)^{1/2} \cdot \norm f_{L^2(\phi)}
\end{align*}
So $\ell$ is a bounded, hence continuous linear functional on $L^2(\phi)$.
Question 2: As $\lambda$ and $\phi$ are positive measures, we obviously have $\lambda(E)$, $\phi(E)\ge 0$, moreover note, that by defintion
$$ \phi(E) = \lambda(E) + \mu(E) \ge \lambda(E) $$
So for $\phi(E) > 0$, we have
$$ 0 \le \frac{\lambda(E)}{\phi(E)} = \frac{\lambda(E)}{\lambda(E) + \mu(E)} \le 1. $$ |
H: Extrapolating an abstract algebra proof, arriving upon an incorrect conclusion.
Could you kindly point out what is wrong with my reasoning?
EDIT: What I have unintendedly proven through my reasoning is that every field can only have one automorphism- the identity mapping. Hope this helps in navigating the mess below
Let $k_{1},k_{2},\dots k_{n}$ be an arbitrary number of distinct, non-zero elements $\in\mathbb{K}$. $\mathbb{K}$ is a field of $0$ characteristic.
Also, $\sigma_{1},\sigma_{2},\dots \sigma_{m}$ are distinct automorphisms of $\mathbb{K}$.
I will show that $m\leq n$.
Let us assume that $m>n$. We can form the following system of linear equations:
$$\sigma_{1}(k_{1})x_{1}+\sigma_{2}(k_{1})x_{2}+\dots \sigma_{m}(k_{1})x_{m}=0$$ $$\sigma_{1}(k_{2})x_{1}+\sigma_{2}(k_{2})x_{2}+\dots \sigma_{m}(k_{2})x_{m}=0$$ $$\dots$$$$\dots$$$$\sigma_{1}(k_{n})x_{1}+\sigma_{2}(k_{n})x_{2}+\dots \sigma_{m}(k_{n})x_{m}=0$$
As $m>n$, we are assured a non-zero solution for $(x_{1},x_{2},\dots x_{m})$. Let it be $(a_{1}, a_{2},\dots a_{m})$, not all zero.
Then, taking the first linear equation into consideration, $$\sigma_{1}(k_{1})a_{1}+\sigma_{2}(k_{1})a_{2}+\dots \sigma_{m}(k_{1})a_{m}=0, a_{i}\in\mathbb{K}$$
This is not possible (I can give a proof, but it would unnecessarily lengthen the discussion).
Hence, on selecting an arbitrary ($n$) number of elements from $\mathbb{K}$, this result shows that the number of automorphisms $\mathbb{K}$ can have $\leq n$. This directly implies that the number of automorphisms $\mathbb{K}$ has is $1$- the identity automorphism.
It is quite obvious that this conclusion is wrong. What is erroneous in my thought process though? Thanks in advance!
AI: The mistake must lie in the assertion that
$$\sigma_1(k_1)a_1 + \cdots +\sigma_m(k_1)a_m = 0$$
is impossible.
Namely, if we let $k_1 = 1$, it is trivial to see that there will be $a_1\ldots a_m$ making this true: just let $a_m = -(a_1+\cdots + a_{m-1})$. This is in direct contradiction with your observation.
Without your purported proof of the above statement, not much more can be said. |
H: Spliting Field over $\mathbb{F}_3$
How to find the splitting field of $f(x)=x^3-x+1$ and $g(x)=x^3-x-1$ over $\mathbb{F}_3$ and how to construct a isomorphism between them?
AI: I'll try to help you with one, you do the other one and try to build an isomorphism between the corresponding fields (knowing it exists must help...).
Since $\,f(x)=x^3-x+1\in\Bbb F_3[x]\,$ is irreducible, we get that $\,\Bbb F_{3^3}:=\Bbb F_3[x]/\langle f(x)\rangle\,$ is a field with $\,3^{\deg f}=27\,$ elements. Let $\,w\in \overline{\Bbb F_3}\,$ be a root of $\,f\,$ , so :
$$(**)\;\;\;\;w^3=w-1=w+2\pmod 3\implies$$
$$\implies \Bbb F_{27}=\{0,1,2,w, 2w,...\}=\{q(w)\;;\;q(x)\in\Bbb F_3[x]\;,\;\deg q\le 2\} $$
with the sum and product operations done modulo three and with the relation (**) taken into account.
Thus, for example, we have that
$$(2w+1)(w^2+w+1)=2w^3+\color{red}{2w^2}+\color{blue}{2w}+\color{red}{w^2}+\color{blue}w+1=2(w-1)+1=2w-1=2w+2$$
or also
$$w(aw^2+bw+c)=1\;,\;\;a,b,c\in\Bbb F_3\;\implies aw-a+bw^2+cw=1\implies$$
$$\implies a = 2\,,\,c=1\,,\,b=0\implies\;\text{in the field}\;\;\Bbb F_{27}\;,\;\;w^{-1}=2w^2+1$$
Well, try to do the same with the other polynomial and then try to come up with some isomorphism between the respective fields.
Further hint: each one of the above fields has an element (a primitive element) s.t. every non-zero element in the field is the power of that elements, which is just a lengthy way to say that the multiplicative groups of the non-zero elements of each field is a cyclic group of order $\;3^3-1=26\;$ .
If you understand how the product is done from the examples above then that'll help you to find those primitive elements in each field... |
H: 1D Green's function: from interval to infinite line
Let's consider two problems for diffusion equation.
The first one:
$$
u_t = a^2u_{xx},\qquad 0<x<l,\quad 0<t\leq T
$$
$$
u(x,0) = \phi(x), \qquad 0 \leq x \leq l
$$
\begin{equation}
u(0,t)=0,\quad u(l,t)=0, \quad 0 \leq t \leq T
\end{equation}
and the second one:
$$
u_t = a^2u_{xx},\qquad -\infty <x<+\infty,\quad t>0
$$
$$
u(x,0) = \phi(x), \qquad -\infty < x < +\infty
$$
For both these cases there are well known expressions for Green's function:
$$
G_1(x,\xi,t) = \frac{2}{l}\sum_{n=1}^{\infty} \exp\left({-\left(\frac{\pi n}{l}\right)^2}a^2t\right)\sin\frac{\pi nx}{l}\sin\frac{\pi n\xi}{l}
$$
$$
G_2(x,\xi,t) = \frac{1}{\sqrt{4\pi a^2 t}}\exp{\left(-\cfrac{(x-\xi)^2}{4a^2t}\right)}
$$
Thus we obtain solutions for first (finite) problem:
\begin{equation}
u(x,t)=\int\limits_0^lG_1(x,\xi,t)\phi(\xi)\,d\xi
\end{equation}
and for the second (infinite):
\begin{equation}
u(x,t)=\int\limits_{-\infty}^{+\infty}G_2(x,\xi,t)\phi(\xi)\,d\xi
\end{equation}
Is there any way to obtain Green's function for infinite case from Green's function for range $[0,l]$ (for example by stating $\quad l\rightarrow+\infty$, but I have not achieved any success with this idea)?
Or may be second solution from first?
AI: Yes, $G_2$ can be obtained as a limit of $G_1$ but I will do a slightly easier limit first.
Write
$${G}_1(x,\xi,t)=\frac{2}{l}\sum_{q} \exp\left(-q^2a^2t\right)\sin q x\sin q\xi,$$
where $q=\frac{\pi n}{l}$, $n=1,2,\ldots$ Now since $q_{j+1}-q_j=\frac{\pi}{l}$, for $l\rightarrow\infty$ we have
$$\frac{\pi}{l}\sum_{q}\rightarrow \int_0^{\infty}dq,$$
and therefore
\begin{align}\lim_{l\rightarrow\infty}{G}_1(x,\xi,t)&=\frac{2}{\pi}\int_0^{\infty}\exp\left(-q^2a^2t\right)\sin q x\sin q\xi\;dq=\\
&=\frac{1}{\sqrt{4\pi a^2t}}\left[\exp{\left(-\cfrac{(x-\xi)^2}{4a^2t}\right)}-
\exp{\left(-\cfrac{(x+\xi)^2}{4a^2t}\right)}\right].
\end{align}
As you may guess, this is the Green function for the heat equation on the half-line. Now the modifications for the whole line will be as follows:
Instead of $G_1(x,\xi,t)$ one should start with translated Green function $$\tilde{G}_1(x,\xi,t)=G_1(x-l/2,\xi-l/2,t).$$ It corresponds to the solution of the heat equation on $[-l/2,l/2]$.
Similarly transform the sum over $n$ into an integral over $q$ (you will need to consider odd and even values of $n$ separately). Evaluating two resulting gaussian integrals, you will recover $G_2(x,\xi,t)$. |
H: derivative of a numerical expression
I have a function $$g(y) = \int{f(y, t)dt}$$ and I am currently able to evaluate the integral numerically (doesn't seem like it is possible to find an analytical solution...), which I have also plotted for various values of $g(y)$. All good so far. However, I am also interested in the quantity $$\frac{dg(y)}{dy}$$ Is there a way to obtain this derivative?
AI: In common cases it is possible to swap the integral and the derivative so that:
$$\dfrac{\partial g(y)}{\partial y} = \int \dfrac{\partial f(y, t)}{\partial y}dt$$
For this equality to be true you need an inversion result, a sufficient condition is that the partial derivative of $f$ is continuous w.r.t. both $y$ and $t$ and bounded uniformly w.r.t. $y$ on the domain of integration. The last condition means that there exist a function $h$ on the domain of $t$ such that for all $x, t$: $\|\dfrac{\partial f(y,t)}{\partial y}\| \leq h(t)$.
See http://en.wikipedia.org/wiki/Leibniz_integral_rule. |
H: About the connection of $L^2$-convergence and convergence in distribution.
Let $(T_{1,n})_{n\in\mathbb{N}}$ and $(T_{2,n})_{n\in\mathbb{N}}$ two sequences of real valued random variables in $L^2(\Omega,\mathbb{A},P)$. Suppose that
$\|T_{1,n}\|_{L^2(\Omega)}\xrightarrow{n\rightarrow\infty} 1$,
$\|T_{2,n}\|_{L^2(\Omega)}\xrightarrow{n\rightarrow\infty} 1$ and
$\|T_{1,n}-T_{2,n}\|_{L^2(\Omega)}\xrightarrow{n\rightarrow\infty} 0$.
Suppose further that $T\in L^2(\Omega,\mathbb{A},P)$ with continuous distribution function and $T_{1,n}\xrightarrow{d} T$ (convergence in distribution).
Does it hold that $T_{2,n}\xrightarrow{d} T$?
AI: We actually just need that $T_{1,n}-T_{2,n}$ converges to $0$ in probability. We use portmanteau theorem.
Let $f\colon\Bbb R\to\Bbb R$ be a uniformly continuous bounded function. Then:
$$|E(f(T_{2,n}))-E(f(T))|\leqslant |E(f(T_{2,n}))-E(f(T_{1,n}))|+|E(f(T_{1,n}))-E(f(T))|,$$
and if $\varepsilon$ is fixed, and $\delta$ the corresponding number in the definition of uniform continuity,
\begin{align}|E(f(T_{2,n}))-E(f(T_{1,n}))|&\leqslant\int_{\{|T_{1,n}-T_{2,n}|>\delta\}}|E(f(T_{2,n}))-E(f(T_{1,n}))|d\mu\\
&+\int_{\{|T_{1,n}-T_{2,n}|\leqslant\delta\}}|E(f(T_{2,n}))-E(f(T_{1,n}))|d\mu\\
&\leqslant 2\sup |f|\cdot \mu\{|T_{1,n}-T_{2,n}|>\delta\}+\varepsilon.
\end{align} |
H: Roots of a polynomial with integer coefficients
Let $f(x)$ be a polynomial in the ring $\mathbb{Z}[x]$. Suppose that $f(i)=0$, where $i^2=-1$.
Can I conclude that $x^2+1$ is a factor of $f(x)$? If so, how can be proven?
AI: Hint: Apply the remainder factor theorem.
Hint: Apply the conjugate root theorem. |
H: Half-line - open or closed set
Half-line is given by
$$A=\{x\in R^n \mid x=x_0 + t(a-x_0), \forall t\geq 0\}$$
Is it open or closed set?
AI: Clearly the complement of $A$ is open (since the point $x_0$ is not in the complement of $A$). This $A$ is closed.
On the other hand, the set $A$ is not open, so it must be closed (not clopen). |
H: Deferred Annuity not working
A simple financial math problem:
Mack obtains $500\ 000$ repayable over $20$ years. If interest is compounded monthly at $9.25\%$ per annum, determine the monthly repayments if the repayment begins in $6$ months time.
I used the formula:
$$P_v = x[(1-(1+i)^{-n})/i]$$
but I'm not getting the right value.
What am I doing wrong?
AI: Because repayment begins in $6$ months, the effective principal is
$$P = 500,000\, \left ( 1 + \frac{i}{12} \right )^6$$
where $i=0.0925$ is the annual interest rate. The monthly payment is then
$$m = \frac{P \, (i/12)}{1-\left [ 1 + (i/12) \right ]^{-240}}$$
Plugging in the numbers, I get a monthly payment of about $\$4795.25$. |
H: Wirtinger derivative of composition of functions
So I have a very basic question : let $h : \mathbb{R} \rightarrow \mathbb{R}$ be a $C^1$ function, and let $g : \mathbb{C} \rightarrow \mathbb{R}$ be defined by $g(z)=h(z \overline{z})$. I want to compute $\frac{\partial g}{\partial \overline{z}}$ (which is defined because $g$ is defined on $\mathbb{C}$). I know the chain rule for $\overline{\partial}$. However, it involves terms like $\frac{\partial h}{\partial z}$ and $\frac{\partial h}{\partial \overline{z}}$, which confuses me because $h$ is only defined on $\mathbb{R}$.
Any help appreciated !
AI: You're absolutely right to be skeptical of the expressions $\partial h/\partial z$ and $\partial h/\partial\overline{z}$. Just like you say, since $h$ is only defined on $\mathbb{R}$, it isn't possible to take these derivatives. There are a couple of things you might do here in order to solve your problem.
(1) You can extend $h$ to a function $\tilde{h}$ on $\mathbb{C}$ in some way, and then define $g(z) = \tilde{h}(z\overline{z})$. This of course doesn't change the function $g$ in any way, but now you'll be able to take derivatives $\partial\tilde{h}/\partial z$ and $\partial\tilde{h}/\partial \overline{z}$. The most natural extension $\tilde{h}$ is the one given by $\tilde{h}(z) := h(\Re z)$.
(2) You can use the ordinary real chain rule as follows. Let $f(z) = z\overline{z}$, or, in terms of $x$ and $y$, this is $f(x,y) = x^2 + y^2$. Then $g = h\circ f$, so $$\frac{\partial g}{\partial x} = (h'\circ f)\frac{\partial f}{\partial x}\,\,\,\,\,\mbox{ and }\,\,\,\,\,\frac{\partial g}{\partial y} = (h'\circ f)\frac{\partial f}{\partial y}.$$ You can then use these expressions to compute $\partial g/\partial z$ and $\partial g/\partial\overline{z}$. |
H: Ordinal arithmetic $3+\omega^2 = \omega^2$
Which one of the following equalities is false:
a) $2\cdot \omega = 3\cdot\omega$
b) $3+\omega+\omega^2 = \omega+3+\omega^2$
c) $\omega^2 + 3 = 3+\omega^2$
d) $12\cdot(5+\omega)=60 \cdot\omega$
I think the wrong one is c):
In a) and d) both sides are $\omega$. And for b) I found that:
$\omega^2$ can be viewed as $\{0,1,2,\dots,\omega,\omega+1,\dots,\omega\cdot 2,\omega\cdot 2+1,\omega\cdot 2+2,\dots,\omega\cdot 3, \omega\cdot 3+1,\dots\}$.
Thus in b) LHS we have $3+\omega = \omega$ and RHS $3+\omega^2=\omega^2$.
But in c) we get $\omega^2+3\not=\omega^2$.
Is that right?
AI: In c) you have limit ordinal on one side of the equation and non-limit on the other side. The ordinal sum $\alpha+\beta$ is limit if and only if $\beta$ is limit. The same is true for non-limit (successor) ordinals.
So if you have been told that exactly one of them is incorrect, this must be the one.
The other computations you have given seem to be correct.
a) You have $n\cdot \omega=\omega$ for every $n<\omega$. (This can be visualized as follows: If we replace in the usual ordering of $\omega=\{0,1,2,3,\dots\}$ each number by finitely many copies, we get an order-isomorphic set.)
b) You are correct in saying that $3+\omega=\omega$ and $3+\omega^2=\omega^2$. Then you have $\omega+\omega^2$ on both sides. This can be further simplified as
$\omega+\omega^2=1\cdot\omega+\omega\cdot \omega=(1+\omega)\cdot\omega=\omega\cdot\omega=\omega^2$.
d) $5+\omega=\omega$; so you get $12\cdot\omega=60\cdot\omega=\omega$ for the same reasons as in a). |
H: Cardinality of a power set (cartesian product)
$A = \{0,1,2\}$ and $C = \{1,2\}$
$|P(A \times C)| = ?$
The answer states $|P(A \times C)| = 2^{3×2} = 2^6 = 64$
What formula/logic is used to obtain this answer please?
AI: There are two theorems at play here:
Theorem 1 If $|A|=n$ and $|B|=m$ then $|A \times B|= n\cdot m$.
Theorem 2 If $|C|=n$ then $|\mathcal{P}(C)| = 2^n$.
In your particular example, as $|A|=3$ and $|C|=2$, then by Theorem 1 we have $|A \times C| = 6$. Then, by Theorem 2, we have that $|\mathcal{P}(A \times C)| = 2^6=64.$ |
H: Minimize $(x+y)(x+z)$ with constraint without calculus
Let $x,y,z \in \mathbb R^+$ such that $xyz\cdot(x+y+z) = 1$
Find $\min\{(x+y)(x+z)\}.$
Using calculus, and Lagrange multipliers, I get:
$(x+y)(x+z) \ge2$ (with the equality occurring if and only if $y=z=1,\ x=\sqrt{2} - 1$).
But I want to solve it in an easy way, which doesn't need calculus.
How can I do it?
Thanks.
AI: Hint: Your expression is $x(x+y+z)+yz= 1/yz+yz$ |
H: Confirmation of correctness in proof regarding norm preserving operator
I just want to know if my solutions are correct for the following problem (euclidean norm assumed):
A linear transformation $T:\Bbb R^n \to \Bbb R^n$ is norm preserving if $|T(x)| = |x|$ for all $x \in \Bbb R^n$, and inner product preserving if $\left\langle T(x),T(y)\right\rangle = \left\langle x, y\right\rangle$.
$(a)$ Prove that $T$ is norm preserving if and only if $T$ is inner product preserving.
$(b)$ Prove that such a linear transformation is $1$-$1$ and $T^{-1}$ is of the same sort.
My solution for $(a)$ was: suppose $T$ is inner product preserving, then since $|x|=\sqrt{\left\langle x,x \right\rangle}$ we have that $|T(x)| = \sqrt{\left\langle T(x),T(x)\right\rangle}=\sqrt{\left\langle x, x\right\rangle} = |x|$ so that $T$ is norm preserving. Suppose now that $T$ is norm preserving, so we have the polarization identity:
$$\left\langle T(x), T(y)\right\rangle=\frac{|T(x)+T(y)|^2 - |T(x)-T(y)|^2}{4}$$
But by virtue of linearity $T(x)+T(y) = T(x+y)$ and $T(x)-T(y)=T(x-y)$ and so since $T$ preserves norms we have $\left\langle T(x),T(y)\right\rangle = \left\langle x,y\right\rangle$ so that $T$ is inner product preserving also.
My solution for $(b)$ was: Suppose that $T(x) = 0$, then we have $|T(x)| = 0$, but since $T$ is norm preserving $|x| = 0$ and this implies $x = 0$ so that $T$ is injective. Also by the rank-nullity theorem $T$ is surjective because $\dim \ker T = 0$ and so $\dim \operatorname{Im} T = \dim \Bbb R^n$. Now, because of that, given $y \in \Bbb R^n$ there's some $x \in \Bbb R^n$ such that $ y = T(x)$. In that case, consider the norm of $T^{-1}(y)$, because of the analysis that there's some $x$ such that $y= T(x)$ we will have:
$$T^{-1}(y) = x$$
Now recalling that $T$ preserves norms, we have that $|x| = |T(x)|$, but $T(x) = y$ so that $|x| = |y|$, so taking the norm on the above equation we finally have:
$$\left|T^{-1}(y)\right| = |x| = |T(x)| = |y|$$
Are my solutions correct? Thanks very much in advance!
AI: Just so that this question has an answer, I would make this comment into an answer.
You are absolutely correct in all your arguments. Where is your doubt? |
H: In a integral domain every prime element is irreducible
I'm trying to understand a proof of Hungerford's book which says that in a integral domain every prime element is irreducible:
I didn't understand why this implication $p=ab\implies p|a$ or $p|b$, is not the contrary $p=ab\implies a|p$ and $b|p$ ?
I'm a little confused
Thanks in advance
AI: If $p = ab$, then in particular $p$ divides $ab$, because $ab = p \cdot 1$. Since $p$ is prime, it has to divide either $a$ or $b$. |
H: Prime ideals in quotients of polynomial ring over finite field
Reading a book, i found this argument ($\mathbb{F}_2$ is the field with 2 elements):
consider the quotient ring:
$$\mathbb{F}_2[x]/(x+1)^2$$
Then it has only one prime ideal, namely the following:
$$(x+1)\mathbb{F}_2[x]/(x+1)^2$$
I've some questions:
what does the notation $(x+1)\mathbb{F}_2[x]/(x+1)^2$ means, i.e. what are the elements of this ideal?
how can i see that this ideal is prime?
how can i exclude that other primes are there, different from that?
Thanks for any help
AI: This is one of those cases where knowing less makes everything a lot simpler. Let $R$ be a commutative ring, and $I$ an ideal in $R$.
If $J$ is another ideal in $R$, the ideal $JR/I$ means the image of the ideal $J$ in $R/I$, given by the natural projection $p:R \to R/I$.
As rschwieb has pointed out below, $JR/I$ is actually just $(J+I)/I$.
In this case, $J=(x+1), R = \mathbb{F}_2[x], I = (x+1)^2$.
There is a one-to-one correspondence between ideals in $R$ containing $I$ and ideals of $R/I$, sending $A \to p(A)$ for $A$ an ideal in $R$, and $B \to p^{-1}(B)$, for $B$ an ideal in $R/I$. This is also a correspondence between prime ideals. Have you seen this result?
Thus what remains to be checked is that the ideal $(x+1)$ is the only prime ideal in $\mathbb{F}_2[x]$ containing $(x+1)^2$. Luckily, $\mathbb{F}_2[x]$ doesn't contain that many elements of small degree, so this can be checked manually. |
H: Is there an image to the point at infinity through this map?
I encountered a conformal mapping on the complex plane:$$z\rightarrow e^{i\pi z}$$
and I am not sure about where it does send the point at infinity. If I could say something along the lines: $$\text{Im}(\infty) = \infty$$
Then it would map it to the origin but there is still a voice in my head saying that this equality is non-sense. And from the usual definition of the imaginary part:$$\text{Im}(z)=\frac{z+\bar{z}}{2i}$$ it makes even less sense.
I'd appreciate some enlightenment.
AI: If we allow $z$ to approach the point at infinity along the positive imaginary axis, we find that $e^{i\pi z}$ tends to $0$. Approaching along the negative imaginary axis, we find that $e^{i\pi z}$ gets big without bound. It turns out that we can make $e^{i\pi z}$ tend toward anything we like, just by allowing $z$ to approach the point at infinity along an appropriate path, and along some paths (such as the positive real axis) it won't tend toward anything at all. Thus, there's no nice way to extend the map to the point at infinity. |
H: $\mathbb N$ a Banach space?
Is $\mathbb N$ a Banach space with the norm $|x-y|$ from $\mathbb R$? I think is Banach space because there is no convergent sequence that is not constant after some $N$. Then all limit points are in the space. But I am not sure.
AI: It surely is a complete metric space, your proof is correct. Another justification of this would be that it is a closed subspace of the Banach space $(\Bbb R,|\cdot|)$. But it has no linear structure, so it's not a Banach space. |
H: $p$ is irreducible if and only if the only divisors of $p$ are the associates of $p$ and the unit elements of $R$
Let $R$ be an integral domain and $p ∈ R$ be such that $p$ is nonzero and a nonunit. Then $p$ is irreducible if and only if the only divisors of $p$ are the associates of $p$ and the unit elements of $R$.
Proof. Suppose the only divisors of p are the associates of $p$ and the unit elements of $R$. Let $p = ab$ for some $a, b ∈ R$. Suppose $a$ is not a unit. Then $a$ is an associate of $p$. Therefore, $a = pu$ for some unit $u ∈ R$. Now
$p = pub$. Since $R$ is an integral domain, it follows that $ub = 1$. Hence, $b$ is a unit and so $p$ is irreducible. We leave the converse as an exercise.
this is a theorem from a book and I have tried to proof the converse in the following way:-
suppose $p$ is irreducible and $a|p$. Then there exist $b\in R$ such that $p=ab$ . Since $p$ is irreducible so one of $a$ & $b$ is unit.Let $a$ is unit.Then we need to show that $b$ is associate of $p$. Since $a$ is unit there exist $u \in R$ such that $au=1$. hence $aup=p \implies aup=ab \implies up=b $.So the result follows.
Am I correct?
AI: Yes. The key idea is this: $ $ a factor $\,b\,$ of $\,\color{#c00}{p = ab}\,$ is a unit $\iff$ its cofactor $\,a\,$ is associate to $\,p,\,$ by
$$\,b\ \ {\rm unit} \iff {\overbrace{b\mid 1 \iff\!\!\ p\mid a}^{\Large\color{#c00}{ 1/b\ \ \ \ =\ \ \ \ a/p}}}\!\overset{\color{#c00}{\large a\,\mid\, p}}\iff a\ \ {\rm associate\ to}\ \ p\qquad\ $$ |
H: A branch cut problem
In Ahlfors' Complex Analysis text, chapter 3, section 4 the transformation $z=\zeta+\frac{1}{\zeta}$ is discussed. The author notes that for every $z$, there exists 2 solutions for $\zeta$ and they are inverses of each other. In order to get a unique $\zeta$ he suggests the restriction $|\zeta|<1$, and he says that in that case the interval $(-2,2)$ must be removed from the $z$ plane.
My question is: why must that interval be removed? In previous chapters it was shown that in order to get an analytic branch for $\sqrt{w}$ one might use a plane with a ray removed (say the negative real axis) - not an interval.
The inverse function is
$$\zeta=\frac{z \pm \sqrt{z^2-4}}{2}$$
and requiring that the argument of the square root is not negative doesn't get me to the interval $(-2,2)$.
AI: This function has two branch points at $\pm2$ as your inversion formula shows. By contrast, $\sqrt z$ alone has only one finite branch point at $0$; however, formally, it also has one at infinity. (Set $w=1/z$, or look at the Riemann sphere to make sense of this.)
Branch cuts always join two branch points, hence the line segment he chose. (An alternative would be two line segments from $2\to\infty$ and $-2\to-\infty$, which is secretly one line going via infinity.)
I can explain more if you like, but this is the reason.
Take $f(z)$ to be mostly two-valued, like $f=\sqrt{g(z)}$ for polynomial $z$. Now suppose you try to define $f$ by starting at some $z_0$ with one of the branches and smoothly (analytically) extending it. When does this go wrong? When you come back to a point $z_1$ you've seen before but with a different value.
Now imagine drawing the two curves you followed from $z_0$ to $z_1$; they form a loop. Start tightening it by bringing them closer together. Since it must be impossible for the two paths to be smoothly deformed onto each other (since they have different values at the $z_1$ end), there must be some point $z_\star$ (along the line you almost formed) such that you can't give a single consistent value to $f$ in any neighbourhood of $z_\star$. $z_\star$ is a branch point.
For example, with $\sqrt{z}$ this happens at the origin. Intuitively, if you go around it in an arbitrarily tiny circle, you get an inconsistent result.
It also happens at infinity; taking a small circle in the variable $1/z$ or a large circle in $z$ makes it clear that you have the same problem out here.
By contrast, $\sqrt{z^2-4}$ has two branch points at $\pm 2$. At infinity, for example, because $$z=Re^{i\theta} \implies (z^2-4)^{1/2}\sim (R^2e^{2i\theta})^{1/2}=Re^{i\theta}$$ there is no inconsistency on going round by a circle, as $\theta\sim\theta+2\pi$. (The $z^2$ goes round twice as fast to compensate for the way that $\sqrt z$ lags at half speed.)
Anyway, the point is that if you want to make it impossible to follow a pair of contradictory paths like those described above, then it seems you can't include any region containing a branch point, or indeed any region enclosing a branch point since the discontinuity will be felt further way. The exception is that when you enclose two branch points, it might be that going around both of them doesn't lead to a contradiction. In fact, we saw this for $\sqrt{z^2-4}$ above already. (It's an equivalent calculation in this case to seeing that infinity isn't a branch point.)
Thus is turns out that excluding the region $(-2,2)$, which prevents us from making any path which encircles just one of the branch points, is sufficient to allow a smooth definition everywhere. Hooray! |
H: How prove this $\sum_{k=1}^{m-1}\dfrac{(-1)^{k+1}}{C_{m}^{k}}=\dfrac{(-1)^{m}+1^m}{m+2}$
prove that
$$\sum_{k=1}^{m-1}\dfrac{(-1)^{k+1}}{C_{m}^{k}}=\dfrac{(-1)^{m}+1^m}{m+2}$$
where $C_{m}^{k}=\dfrac{m!}{(m-k)!k!}$
This problem is my frend ask me,I think we can prove by case$:m=2n,m=2n+1$
Thank you everyone can nice methods.
AI: Since
$$\frac{1}{C^m_k}=(m+1)\frac{\Gamma(m-k+1)\Gamma(k+1)}{\Gamma(m+2)}=(m+1)B(m-k+1,k+1),$$
the sum can be rewritten as
\begin{align}
&(m+1)\int_0^1\left(\sum_{k=1}^{m-1}(-1)^{k+1}\left(\frac{x}{1-x}\right)^k\right)(1-x)^mdx=\\
=&(m+1)\int_0^1\left[x(1-x)^m+(-1)^m(1-x)x^m\right]dx=\\=&(m+1)\left(1+(-1)^m\right)B(2,m+1)=\\
=&\frac{1+(-1)^m}{m+2}.
\end{align}
Here $B(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$ denotes Euler beta function, and the summation over $k$ was performed using geometric series. |
H: expanding convoluted integrand
I have a function on the form $$g(y) = \int_{-\infty}^{\infty}{e^{-v^2}f(y-v)dv}$$ I know that $g(y)$ is linear around $0$, $g(y\approx 0)\approx yG$, and I am interested in finding this gradient $G$.
For this reason I thought that the integrand could be expanded around $y=0$. In general, am I allowed to do that for a convolution? Or should I expand around $y-v=0$?
AI: Yes, you should expand around $y=0$ if you're interested near there! With suitable reasonable restrictions the expansion, which is really just a derivative with respect to $y$ at 0, commutes with the integral (see http://en.m.wikipedia.org/wiki/Differentiation_under_the_integral_sign for example) so everything works out.
$$G= \int e^\cdots \partial_y f(y-v) \text{ at }y=0$$
Incidentally a trick which is sometimes useful is using $\partial_x f(x-y)=-\partial_y f(x-y)$ so that you can integrate by parts, or similar. |
H: How to prove there are no more positive integers that are products of 2 and 3 consecutive numbers?
$6$ and $210$ share the property that both are the products of both two and three consecutive numbers. $6$ is $2\times3$ and $1\times2\times3$ and $210$ is $14\times15$ and $5\times6\times7$. It was easy enough to write a program to search for more numbers with this property, I found that there were no more up to at least $1{,}000{,}000{,}000{,}000$ ($1$ Trillion). But it is beyond me to prove that there are either no more numbers like this or to find the next one. Any ideas?
AI: In his book Diophantine equation,(page $257-258$) L.J.Mordell proved that the equation $$y(y+1)=x(x+1)(x+2)$$ has only the integer solutions $x=-1,-2,0,1,5.$ |
H: A definite integration problem on law of large number
The problem is given as follows:
If $g(x),h(x)$ are continuous function on $[0,1]$, satisfying $0\le g(x) <M h(x)$, where $M$ is a nonzero constant. Prove that
$$\lim_{n\to \infty} \int_{0}^1 \int_0^1\cdots\int_0^1\frac{g(x_1)+g(x_2)+\cdots +g(x_n)}{h(x_1)+h(x_2)+\cdots +h(x_n)}dx_1dx_2\cdots dx_n=\frac{\int_{0}^1g(x)dx}{\int_0^1h(x)dx}$$
I have some idea of the solution, but I think there must be some problem. Here is my solution:
Let $T_i$ be the random variables with density function $\frac{g(x)}{\int_{0}^1 g(t)dt}$ , where $x\in[0,1]$.Similarly, we define $S_i$, with density function $\frac{h(x)}{\int_{0}^1 h(t)dt}$.
By the law of large numbers, $$\lim_{n\to\infty} A_n=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n T_i=1 \mbox{ a.s,}\lim_{n\to\infty} B_n=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n S_i=1 \mbox{ a.s}$$
But I got stuck here, because I cannot combine the two result to get the solution, it does not mean anything since the density function cannot be divided directly. Is there any method to tackle this problem?
Any hint of a solution is welcomed, thanks!
AI: (This was already posted on the site.) Consider an i.i.d. sequence $(U_n)$ uniform on $(0,1)$, then the $n$th integral on the LHS is $E[W_n]$, where
$$
W_n=\frac{g(U_1)+\cdots+g(U_n)}{h(U_1)+\cdots+h(U_n)}.
$$
You already know that $W_n\to w$ almost surely, where $w$ is the RHS. Furthermore, $0\leqslant W_n\leqslant M$ almost surely hence $W_n\to w$ in $L^1$, which implies that $E[W_n]\to w$. |
H: Show that $n^2\log\left(1+\frac{1}{n}\right)$ does not converge to $1$
How to show that $n^2\log\left(1+\dfrac{1}{n}\right)\to 1$ is false?
I have to show that $\left(1+\dfrac{1}{n}\right)^{n^2}$ doesn't tend to $e.$
AI: $$\log\left(1+\frac1n\right)=-\log\left(1-\frac1{n+1}\right)\geqslant\frac1{n+1}$$ |
H: Matrix equation involving a Pauli matrix
I should solve the following problem:
find the matrix $A$ that satisfies the following equation:
$$\sin(\pi A)+\cos(\pi A)^2= \left( \begin{array}{ccc}
0 & 1 \\
1 & 0 \end{array} \right)$$
How can I solve the problem?
Thanks in advance.
AI: Note: diagonalization is essential here. One possibility is to change the basis back and forth. That's what I do below. A more elegant way is to simply work with the spectral projections without ever writing the four coefficients of the matrices. That's Muphrid's approach.
First diagonalize the rhs matrix by $P$ orthogonal:
$$
PSP^*=P\pmatrix{0&1\\1&0}P^*=\pmatrix{1&0\\0&-1}=D\qquad \mbox{where}\quad P=\pmatrix{\frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2}\\-\frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}}.
$$
Then observe, via the series representations of $\sin$ and $\cos $ and $PB^nP^*=(PBP^*)^n$, that
$$
\sin (\pi A)+\cos(\pi A)^2=S\iff P(\sin (\pi A)+\cos(\pi A)^2)P^*=D
$$
$$
\iff \sin (\pi PAP^*)+\cos(\pi PAP^*)^2=D.
$$
So we need to solve $(E): \;\sin (B)+\cos(B)^2=D$ and make $A=\frac{1}{\pi}P^*BP$ in the end.
If $B$ is a solution of $(E)$, $D$ and $B$ commute. In particular, $B$ must leave the eigenspaces of $D$ invariant. Since they are one-dimensional, this forces $B$ to be diagonal. Whence $(E)$ is equivalent to
$$
B=\pmatrix{\lambda&0\\0&\mu}\qquad \pmatrix{\sin \lambda+\cos^2\lambda&0\\0&\sin \mu+\cos^2\mu}=\pmatrix{1&0\\0&-1}
$$
This yields $\sin \lambda=0$ or $\sin\lambda =1$, and $\sin\mu=-1$. So the solutions of $(E)$ are
$$
B=\pmatrix{k\pi&0\\0&-\frac{\pi}{2}+2l\pi}\qquad B=\pmatrix{\frac{\pi}{2}+2k\pi&0\\0&-\frac{\pi}{2}+2l\pi}.
$$
To obtain the solutions of your initial equation, just use the formula $A=\frac{1}{\pi}P^*BP$:
$$
A=\pmatrix{\frac{-1+2k+4l}{4} & \frac{1+2k-4l}{4}\\ \frac{1+2k-4l}{4}&\frac{-1+2k+4l}{4} }\qquad A=\pmatrix{k+l & \frac{1}{2}+k-l\\\frac{1}{2}+k-l&k+l }
$$
where $k$ and $l$ range over $\mathbb{Z}$.
Conclusion: this can be rewritten
$$
A=\left(\frac{-1+2k+4l}{4}\right) I_2+\left(\frac{1+2k-4l}{4}\right) S\qquad A= (k+l)I_2+\left(\frac{1}{2}+k-l \right)S
$$
where $S$ is your original Pauli matrix on the rhs of your equation. |
H: Calculate interior, closure and boundary
As part of an exercise, I want to calculate the interior, closure and boundary of the following sets in $\mathbb{R}^2$ (with the standard topology).
1. $\mathbb{Z} \times \mathbb{Z}$
2. $\mathbb{Q} \times (\mathbb{Q}\cap]0,+\infty[)$
I found the following solution and would like to verify whether those are correct.
1. Interior: $\emptyset$; Closure: $\mathbb{Z} \times \mathbb{Z}$; Boundary: $\mathbb{Z} \times \mathbb{Z}$
2. Interior: $\emptyset$; Closure: $\mathbb{R} \times (\mathbb{R}\cap[0,+\infty[)$; Boundary: $\mathbb{R} \times (\mathbb{R}\cap[0,+\infty[)$
AI: You are correct. You're basically done after interior and closure: the boundary follows easily as the difference between the closure and the interior. So empty interior implies boundary = closure, as we have here. |
H: Choosing a bound when it can be plus or minus? I.e. $\sqrt{4}$
My textbook glossed over how to choose integral bounds when using substitution and the value is sign-agnostic. Or I missed it!
Consider the definite integral: $$ \int_1^4\! \frac{6^{-\sqrt{x}}}{\sqrt x} dx $$
Let $ u = -\sqrt{x} $ such that $$ du = - \frac{1}{2\sqrt{x}} dx $$
Now, if one wishes to alter the bounds of the integral so as to avoid substituting $ - \sqrt{x} $ back in for $ u $, how is the sign of the integral's bounds determined?
Because: $ u(1) = -\sqrt 1 = -(\pm 1) = \pm 1 $ and $ u(4) = -\sqrt{4} = -(\pm2) = \pm2 $
How does one determine the correct bound? My textbook selected $ -1 $ and $-2 $ without explaining the choices.
AI: It is convention that $\sqrt{x} = + \sqrt{x}$. Thus, you set $u(1) = -\sqrt{1}=-1$ and $u(4) = -\sqrt{4}=-2$.
The only situation where you introduce the $\pm$ signs is when you are finding the root of a quadratic such as $y^2=x$ in which case both $y=+\sqrt{x}$ and $y=-\sqrt{x}$ satisfy the original equation. |
H: Modulo question about equality
True or false?
$$24 \equiv 77 \mod 16 $$
$1.$ $77/16 = 4.8125 $
$2.$ $4.8125 - 4 = .8125$
$3.$ $0.8125 \times 16 = 13$
$4.$ $24 != 13$
So the answer is false? Am I right?
AI: A simpler approach: $a\equiv b\pmod{n}$ if and only if $n$ divides $(b-a)$.
In this problem, $77-24=53$, which is not a multiple of $16$ ($53/16=3.3125$), so $77\not \equiv 24\pmod{16}$. |
H: Isomorphism between two finite fields
We have $k_1:= \mathbb F_7(\alpha)$ and $k_2 := \mathbb F_7(\beta)$ where $\alpha^2 = 3$ and $\beta^2 = -1$ in $\mathbb F_7$. I have to show that these two are isomorphic.
Let $\phi:k_1 \rightarrow k_2$ be a homomorphism which preserves $1 \in k_1$. Then
$$\phi(\alpha^2)= \phi(3) = 3 = \phi(\alpha)^2$$
where
$$\phi(\alpha) = x + y\beta\;,\;\;x,y \in \mathbb F_7$$
Thus
$$(x+y\beta)^2 = \phi(\alpha)^2 = x^2 + 2xy\beta +y ^2 \beta^2 = x^2 +2xy\beta -y^2=3$$
So $x$ or $y$ must be $0$. But $y$ can't be zero because $3$ has no root in $\mathbb F_7$. So $x = 0$ such that
$$-y^2 = 3 \rightarrow y \in \{-2, 2\}$$
Is the function $\phi$ with $\phi(x) = x\;$ for $\;x \in \mathbb F_7\;$ and $\;\phi(\alpha) = 2\beta\;$ then an isomorphism ?
AI: Well, you have a candidate; why don't you check whether $\phi((a+b\alpha)(c+d\alpha)) = \phi(a+b\alpha)\phi(c+d\alpha)$, for all $a,b,c,d \in \mathbb{F}_7$? For sums it's trivial, as you already define it as a linear map over the base field. |
H: Is it possible to isolate p in $"a = 2bpq + 2apq + a(p^2) + 2aqq + bqq"?$
Is it possible to simplify this equation so that $p$ is isolated on one side instead of $a$? I tried factoring out $p$ on the right side, but I get stuck with the $ap^2$.
$$a = 2bpq + 2apq + ap^2 + 2aq^2 + bq^2$$
AI: By solving p in this equation you get:
$p = \frac{\sqrt{a^2(-q)^2+a^2+abq^2+b^2q^2} -aq - bq}{a}$
or
$p = \frac{-\sqrt{a^2(-q)^2+a^2+abq^2+b^2q^2} -aq - bq}{a}$
EDIT:
Steps as requested:
$ap^2+2apq+2bpq+2aq^2+bq^2=a$
$p^2 + \frac{p(2aq+2bq)}{a }+ \frac{2aq^2+bq^2}{a} = 1$
$p^2 + \frac{p(2aq+2bq)}{a} = 1 - \frac{2aq^2+bq^2}{a}$
$p^2 + \frac{p(2aq+2bq)}{a} + \frac{(2aq+2bq)^2}{4a^2} = 1 + \frac{(2aq+2bq)^2}{4a^2} - \frac{2aq^2+bq^2}{a}$
$(p + \frac{(2aq+2bq)}{2a})^2 = 1 + \frac{(2aq+2bq)^2}{4a^2} - \frac{2aq^2+bq^2}{a}$
Can you take it from here? |
H: addition with a variable (mod)
Given $2+x \equiv 7 \pmod 3$.
$2 + 0 = 2$
$2 + 1 = 3$
$2 + 2 = 4$
.
.
.
$2 + 5 = 7$
so, the answer will be $x = 5, 8, 11, 14, 17,\dots$
Is this correct? Because somebody told me the answer should be $x = 2, 5, 8, 11, 14, 17,\dots$
AI: Hint: $\ x\equiv 7-2 \equiv \color{#c00}5 \pmod 3 = \color{#c00}5 + 3\,\Bbb Z = \{\,\ldots,-4, -1, 2, \color{#c00}5, 8, 11,14,\ldots\}$
Or: $\ x\equiv 5\pmod{3}\!\iff\! 3\mid x\!-\!5\!\iff\! 3n = x\!-\!5 \!\iff\! x = 5\! +\! 3n,\ $ for some $\, n\in \Bbb Z$ |
H: What is the Idea of a Relative Open Set?
My real-analysis text gave the following defintion:
Let U be a subset of E. U is open relative to E if
for $\forall t \in U$, $\exists \epsilon$ such that $N_\epsilon(t) \cap E \subset U$.
Although the idea that U is open in $\mathbb R$ follows the definition,
I normally do not think about the intersection of $N_\epsilon(t) \cap \mathbb R$.
U is open if for $\forall t \in U$, $\exists \epsilon$ such that $N_\epsilon(t) \subset U$; every t is an interior point of U.
Intuitively, each point, t, is contained in a "bubble".
So, what is the significance of specifying the intersection?
Apparently, I can't think of a relative open set in terms like interior points and "bubbles".
AI: As you said, a subset $U$ of $\mathbb{R}$ is open, if for every point $x$ of $U$, there is some $\epsilon>0$ such that $N_\epsilon(x) \subset U$, which means "every point of $\mathbb{R}$ that is closer to $x$ than $\epsilon$ lies in $U$".
Now we want to talk more generally about open subsets of $E$, where $E \subset \mathbb{R}$, but using the same distance function; i.e. keeping as much the same as possible.
So we have $U \subset E$ and we want to define what it means to be open in $E$. The easiest way is to say: for every $x \in U$ there is some $\epsilon > 0$ such that "every point of $E$ that is closer to $x$ than $\epsilon$ lies in $U$". The statement between quotes is just $E \cap N_\epsilon(x) \subset U$, as your text's definition. It makes no sense to consider points not in $E$; we consider $E$ to be the new "universe of discourse", as we only talk about its subsets, no longer about all subsets of $\mathbb{R}$. |
H: How does one derive $O(n \log{n}) =O(n^2)$?
I was studying time complexity where I found that time complexity for sorting is $O(n\log n)=O(n^2)$. Now, I am confused how they found out the right-hand value. According to this $\log n=n$. So, can anyone tell me how they got that value?
Here is the link where I found out the result.
AI: $f(x)=\mathcal{O}(g(x))$ means that $f$ is asymptotically smaller or equal to $g$. This means that $|f(x)|\le c|g(x)|\quad \forall x$ for some constant c.
Now in your example, $$n\log n \le n^2$$
This does not say that $n=\log n$, but instead, that $\log n\le n$. In other words, that $\log n$ is asymptotically smaller or equal to $n$, or, in Big-O notation, $\log n = \mathcal{O}(n)$. |
H: If I have three points, is there an easy way to tell if they are collinear?
Points $(a,b)$, $(m,n)$, and $(x,y)$ are selected at random. What is the quickest/easiest way to tell if they are collinear? At first I thought it was a matter of comparing slopes but that doesn't appear to be enough.
AI: "At first I thought it was a matter of comparing slopes"... and you were right!
If the line segments AB and BC have the same slope, then A, B, C are necessarily collinear. Note that there are some corner cases having to do with whether B is the "middle" point or not (in which case the slopes will still be equal), and one having to do with vertical lines (where some formula you use to compute slope might divide by 0).
Putting all this together, the points $(a, b)$, $(m, n)$ and $(x, y)$ are collinear if and only if $$(n-b)(x-m) = (y-n)(m-a)$$
(comes from $\frac{n-b}{m-a} = \frac{y-n}{x-m}$, but not writing it in fraction form to avoid division by $0$). |
H: Why $v_1 \in L$ and $v_2 \in L^\perp$?
Let $L$ be a one dimensional subspace of $R^2$ (we may view $L$ as a line in the plane through the origin).
Suppose $\alpha$ is the angle from the positive x axis to $L$.
Let $v_1=(\cos \alpha, \sin \alpha)$ and $v_2=(-\sin \alpha, \cos \alpha)$.
Then $v_1 \in L$ and $v_2 \in L^\perp$.
I can't understand the last line. Why $v_1 \in L$ and $v_2 \in L^\perp$?
AI: The vector $ \ \overrightarrow{v_1} \ $ is a unit vector which lies in the line L , so it would be considered an element of the vector subspace $ \ L \ $ , which is the set of all vectors pointing along the line L (in either direction). Since $ \ \overrightarrow{v_1} \cdot \overrightarrow{v_2} \ = \ 0 $ , the second vector is perpendicular to the first, and so must lie in the vector subspace in $ \ \mathbb{R}^2 $ which is designated as $ \ L^{\bot} \ $ ; it is also a unit vector, so $ \ \overrightarrow{v_2} \ $ would be one element of $ \ L^{\bot} \ . $ |
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