why $\begin{pmatrix}2 & 0 \\ 1 & 2\end{pmatrix}^n=\begin{pmatrix}\text{not important}& \text{not important} \\ n2^{n-1} & \text{not important}\end{pmatrix}$? I've multiplied it three times, and I do not see the idea.
Hmm, I still cannot quite see in general why $2^{\Bbb{R}}$ breaks it while $\Bbb{R}$ don't:
In $\Bbb{R}$ a binary expansion is something of the form: ...bbbbbbbb.bbbbbbb... So moving to the left, it can become unbounded, while moving to the right the importance of the digits tends to zero There are $\aleph_1$ many possible diverging sequences and $\aleph_1$ may possible sequences that converges to zero, yet for all reals we have no problem telling which is bigger than which
For $2^{\Bbb{R}}$, a "binary expansion" should look something like this:
Do defining lexicographic order need well ordering, or it is the binary representation of some element in the set is the one that need to be well ordered?
@MikeMiller Thanks for the book tip. It seems like much of it is actually explained fairly well in the second lecture of Wednesday of qgm.au.dk/video/mc/soergelkl but it might be interesting to take a closer look at the "real" results at some point
@Secret Lexicographic order has nothing to do with well-order. It works for the product of any two ordered sets
If I tried to define the lexicographic ordering by finding the first pair of elemnts that differ and then x < y, then I ran into the following counterexample:
yes, @LeakyNun, and $\begin{pmatrix}a_{n+1}\\a_n\end{pmatrix}=\begin{pmatrix} 4 & -4 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} a_n \\ a_{n-1}\end{pmatrix}$ was the sequence I was originally solving.
Can't you start with $A_0=\omega_2$, for any pair of elements with no third element in between add a third, call the result $A_1$, keep iterating this process, $A_\omega$ should work
Hi. In the context of probability theory, a random element is a generalization of a random variable. Is there some further generalization of a random element?
Then you will get $\begin{pmatrix}a_n \\ b_n\end{pmatrix}=\begin{pmatrix}2 & 0 \\ 1 & 2\end{pmatrix}^n \cdot \begin{pmatrix}1 \\ 0\end{pmatrix}$. But why do they have $n2^{n-1}$? @LeakyNun
they expand the function I get $$\frac{1}{(1-2x)^2}=(\sum_{n=0}^{\infty}2^nx^n)^2=\sum_{n=0}^{\infty}(\sum_{k=0}^{n}2^k2^{n-k})x^n=\sum_{n=0}^{\infty}(n+1)2^nx^n$$, @LeakyNun
The theorem is that A smooth complex projective variety comes with a "polarized hodge structure" on the abelian groups $H^k(X;\Bbb Z)$. A Hodge structure of weight k on an abelian group is...
So far I've only needed Kahler. If the Kahler 2-form omega is also integral then you get a further decomposition of the V^{p,q} into "primitive pieces" on which a certain intersection form is definite
@TedShifrin the question about the change of basis was an essential question that I do not need directly for tomorrow. But the whole concept of terms like $S^{-1}AS$ is there.
@LeakyNun as said, I am lost, sorry. The function I gave you should be correct, but the one from the solution refers to the other function, so the calculation above doesn't refer to our $b_n$.
they expand the function I get $$\frac{1}{(1-2x)^2}=(\sum_{n=0}^{\infty}2^nx^n)^2=\sum_{n=0}^{\infty}(\sum_{k=0}^{n}2^k2^{n-k})x^n=\sum_{n=0}^{\infty}(n+1)2^nx^n$$, @LeakyNun
@Danu Unfortunately I was not in China for the conference where Williamson again created a stir, that time by announcing a disproof of a 30-year old conjecture.
Yeah, well semester here starts in October, and they want to already get my degree certificate before even drafting up a contract, so that'll take another month
@Kirill you asked for an explanation of the bottom-left corner of $\begin{bmatrix}2&0\\1&2\end{bmatrix}^n$, which is essentially the bottom entry of $\begin{bmatrix}2&0\\1&2\end{bmatrix}^n \begin{bmatrix}1\\0\end{bmatrix}$. I basically transformed your problem to a sequence, with definition being $\begin{bmatrix}a_{n+1}\\b_{n+1}\end{bmatrix} = \begin{bmatrix}2&0\\1&2\end{bmatrix} \begin{bmatrix}a_n\\b_n\end{bmatrix}$ such that $\begin{bmatrix}a_n\\b_n\end{bmatrix} = \begin{bmatrix}2&0\\1&2\end{bmatrix}^n \begin{bmatrix}a_0\\b_0\end{bmatrix}$.
@Kirill @Leaky: There's an easy way to compute the matrix powers directly. Perhaps you've already discussed this. Write the matrix as $A=2I + J$ where $J=\begin{bmatrix} 0 &0\\1&0\end{bmatrix}$. Then $J^2=0$ and you can easily write down the formula for $A^n$ (because $I$ and $J$ commute, of course).
@LeakyNun that was good! In the solution they simply say $\begin{pmatrix}2 & 0 \\ 1 & 2\end{pmatrix}^n = \begin{pmatrix}2^n & 0 \\ n2^{n-1} & 2^n\end{pmatrix}$, and I haven't thought it would need an hour to understand why, to be honest.
I have no idea. I muddled up something I'd never taught before in the very last lecture of the first semester, where I was presenting the isoperimetric inequality (diff geo) done with Fourier analysis.
But that's hardly an essential part of the course.
When I've looked back at a few, I've seen mistakes that no students pointed out immediately, and I was sitting there saying, "No, no, no," but eventually in the lecture I caught it. I just wished they'd caught it fast. :P
> I wonder how many professors stand up there not ever thinking that they would have ever gone through to graduate school, and much less become a tenured faculty member only to give lectures to young kids who are in the same position that they themselves were in years ago. It's a little nostalgic if you really think about it.
if you substitute a generating function $f$ with $\sum_{n=0}^{\infty}a_nx^n$, is it also allowed to replace $f$ back in terms like $\sum_{n=0}^{\infty}a_{n+1}x^{n+1}$ or $\sum_{n=1}^{\infty}a_{n-1}x^{n-1}$? Or, should $f$ be replaced back strictly when the term $\sum_{n=0}^{\infty}a_{n}x^{n}$ is there?
@robjohn it was really about, if there such a term :) e.g. you call a matrix $A$ as degenerated in Russian, if $\det(A)=0$.
That's what someone else said to me in some other situation yesterday in some other place.
Anonymous
What's the correct way to approach these kind of questions: Show that $\forall x>0$ $x-x^2/2+x^3/(3x+3) < \log (1+x) < x-x^2/2+x^3/3$. I apparently can't use Maclaurin expansion for this as it won't converge for x>1.
Anonymous
Maybe I could write out the difference of two functions and then use derivative test....
Let $a,b \in \mathbb{Z}$ be integers such that $\sqrt{a} \notin \mathbb{Z}$ and $
\sqrt[3]{b} \notin \mathbb{Z}$ (the number $a$ is allowed to be negative). I need to prove that
$$ \mathbb{Q}(\sqrt{a}+\sqrt[3]{b}) = \mathbb{Q}(\sqrt{a}, \sqrt[3]{b})$$
but I cannot use any Galois theory to do s...
Let $a,b \in \mathbb{Z}$ be integers such that $\sqrt{a} \notin \mathbb{Z}$ and $
\sqrt[3]{b} \notin \mathbb{Z}$ (the number $a$ is allowed to be negative). I need to prove that
$$ \mathbb{Q}(\sqrt{a}+\sqrt[3]{b}) = \mathbb{Q}(\sqrt{a}, \sqrt[3]{b})$$
but I cannot use any Galois theory to do s...
