With regards to part c) how would you guys have done it? I did it already, I got an orthonormal basis for the set of polynomials and obtained the adjoint matrix
but the other way of doing it is to just get the coefficients directly. WHich would be a more efficient way to go about it?
i don't think the tradeoff matters, generally. just understand how it can be done without computing an orthonormal basis (T^* x is the thing making <T*x,y> = <x, Ty> true for all y).
outside of computing an ONB you could just use that and y = [any basis] to deduce a system of equations satisfied by the coefficients of T*x in [that basis] and solve it.
it is a huuuuuuuuuuuuuuuuuuuuge waste of time to try to tune your strategy so you can complete homework exercise x in minimal lines although that is what a lot of math.SE answers are ultimately about.
computing the ONB allows you to compute T*[anything] for any input [anything]. so it's useful for that purpose. if you only have a fixed vector (as you might in homework but maybe not elsewhere) you can shorten the work just a little bit, but not by much.
anything that reduces to solving one system of n equations in n unknowns with n less than 3 billion is basically equivalent to anything else of that sort. in the grand scheme of things.
As an aside, I really dislike the notation $\mathbf1_X$. The important bit, $X$, is shoved into a subscript. I would use something like $[X]$, except that if $X$ were an interval (say $X=(0,1]$) then I'd be stuck writing $[(0,1]]$, and nobody wants that. \\ EDIT: Another alternative, I suppose, is to simply identify sets with $\{0,1\}$-valued functions and just write $X$. I'm not sure there is much opportunity for ambiguity. It also lets us write fun things like $X\cap Y=XY$ and $X\cup Y=X-XY+Y$. — Akiva Weinberger7 mins ago
I once explained the concept of topology to a kid as, basically, the thing that doesn't change with the first and third arrows in that picture but does with that second arrow
(except with words 'cause I didn't have the image with me)
"When you bite far enough into a donut, the way it's connected to itself changes. The way something is connected up is called its topology"
The fundamental group doesn't just count holes - it's a group, not just a number. The fundamental group of a knot complement in general can be quite complicated
it really was a number before it was a group, akiva. your hisotrical questions lately were about this. betti numbers and torsion coefficients came first.
but I'm thinking maybe the best equivalent definition, for intuition, is that a set is connected if you can't split it into to subsets, neither of which contains a limit point of the other.
:-). the daughter of a irish high school friend is at a conference, i wanted to at least get her a nice breakfast, i don't think albany would be very exciting...
I read a comment above saying that you're not fine recently, so I hope you will get better first, then share a little time to help me with a little combinatorics question.
@CalvinKhor The algorithms are created by Google Engineers and those (creepy!) Psychologists. Those algorithms are very offended for the users. They want you to watch as many videos as possible!
The foundations of these algorithms are statistics. If you can resist them, you're abnormal. Because 80% of the users will watch the next video in the recommendations. The only way you will stop watching them is when you are tired (or more rude word: you're feeling sick!).
@linear_combinatori_probabi After watching 3b1b videos, Calvins brain won't have the mental energy anymore to do his/her work.On the other hand, Numberphile videos (specifically the -1/12 videos) will make calvins brain say "I have better things to do than watch these kinds of videos). And Calvin will leave youtube and do some work.
Not to be offensive. But some of the videos of the PBS infinite series are explained by a girl. I remembered I watched all of them in 1 day. Each of topic itself is very interesting for sure.
The bad thing is the channel stopped uploading any more videos 3 years ago.
@linear_combinatori_probabi I found a big bug in my method: (1) After watching bad math videos, The user could start watching good videos again. But we want the user to leave youtube. (2) Good math videos can be brain intensive sometimes, which makes it a bit stressful. But we want to relax.
PBE infinite series. I'm watching the SOMA game in your link too because I feel like my mind is decaying after too much video on theories I cannot understand.
I will be pretty happy if my mind is bent by Bézier Curves
@Koro I think the variable $\epsilon$ here is not important right? You can change it to whatever you like since the behaviour of any variable goes to $0^+$ is the same.(maybe)
Hello Everyone, I wanted to ask if $f(z)=\sum_{k=0}^n \frac{1}{k!z^k}$ is a polynomial. Because I would like to say something about the zeros of $f$ thus I thought maybe we can say that it has $n$ zeros in $\Bbb{C}$ if it is a polynomial. Is this true?
@CalvinKhor Hmm okey, so I need to show that there are no zeros in $|z|\geq 2$ so i thought if I can show that there are $n$ zeros in total and i can show that in $|z|<2$ there are $n$ zeros I would be done right?
@CalvinKhor sorry to disturb again, I somehow got a bit stuck. If I can show that $f(1/z)$ has n zeros in $|z|<2$ does this then also hold for $f(z)$ so I mean does $f$ has the same number of zeros in $|z|<2$?
Because I get in trouble with $z=0$ while applying Rouche.
Is there something stupid about this formula for the Dirichlet divisor problem: $$\sigma _0(n)=\frac{i}{2 \pi } \sum _{k=1}^n \log \left(-\exp \left(\frac{2 \pi i (n+1)}{k}\right)\right)+(n+1) H_n-\frac{n}{2}$$
so you mean if I show f(1/z) has n zeros in |z|<2 then f(z) has n zeros in |z|>1/2? Okey so this will not help me to get further because I should use |z|\geq 2
@robjohn: Sorry for the ping, but I think you are the right person that can help me with my little combinatorics question: math.stackexchange.com/q/4445719/390226
hey. i am having problems understanding how such a sgn function, basically i had the following $\sigma_I \wedge \sigma_J = sgn (1,...,n, I,J) \sigma_1 \wedge ... \wedge \sigma_n$ how can i understand such signum function?
@Shub what exactly do you mean by approximated? They are both continuous functions, they vary from $1$ at $k\pi+\pi/2$ to $0$ at $k\pi$, so their derivatives are zero at all of those extrema. They won't be too far apart, but I don't know if one would call them approximations of each other.
@linear_combinatori_probabi I need to re-read your question, then.
@TedShifrin Good morning. Getting ready for Mother's Day celebrations. Unfortunately, my wife has a fever, so she might not be going. We'll see how she feels later today.
I worked out a couple of bounds for $\operatorname{W}_0(x)$ and $\operatorname{W}_{-1}(-1/x)$. I am looking to see if I can find other references for these bounds. (Lambert W).
