@AkivaWeinberger There are many... The first is a Taylor series about $0$. The second is a Taylor series about $\infty$. Probably not what the teacher would mean on an exam.
@LittleRookie I'm not sure how you deduce that from the given formulas on slide 1 and 2, f(beta) and f(x) are not the same thing, those terms do not cancel
well the way i reason through why that could be true is because h(x) is kind of an error term, and the error at the endpoints can always be set to zero
equivalent to shifting the origin around
h isn't the error term, to be clear
on slide one, it is the final term with the (n+1)'th power
the original expression is the taylor series, here in taylor theorem, u are proving that u can change the infinite series to a summation of finite number of terms, with the expression of taylor polynomial + remainder.
@AkivaWeinberger There is the Taylor series, then there are two Laurent series that converge either in an anulus between $|z|=1$ and $|z|=2$ or outside of $|z|=2$
How to make sense of complex differentiable functions at a point z_0 , i mean in real it was tangent line what is the case in complex intuitivly speaking
I already had to postpone oral surgery (dental implant), and I can't postpone again, because I'm leaving for a month of travel in Europe for June. Ugh.
@BAYMAX: So, based on what we did a few days ago, can't you give me a continuous function on the square missing the origin that won't extend to a continuous function on the entire square?
@TedShifrin so the whole idea of cauchy integral formula is that we can find the value of any point z inside the contour by knowning the integral arount the countour yeah ? :D
@A---B: You will have to design that yourself. It doesn't exist in LaTeX, so far as I know. You need to design a vertical line of the right length and then shift it on top of the rightarrow.
I'm trying to produce an arrow similar to $\nrightarrow $ only with the smaller line being vertical rather than horizontal. I went over the big list of latex symbols several times and could not find it. Any help would be appreciated (I'm using xypic, so extended codes are fine if xypic supports t...
One of which was on your midterm, 2 of which were on your pset, 2 of which I couldn't find
But I only have 5 psets of yours
So it's possible (though I think unlikely) that she assigned the remaining 2 problems on one of the psets I don't have, unless she really did assign you only 5
the only problem I have left is showing that every geodesic in a lens space is closed, and that there are lens spaces with simple closed geodesics of different length
Like, today she talked about topologies on trees that correspond to games, so how the binary game was homeomorphic to the Cantor set, the countable one to $\mathbb{R}\setminus\mathbb{Q}$ (the verification of which would be on the homework)
Also today we proved that the Borel hierarchy does not terminate in countably many steps, using a method I only 30% understand
What is the necessity of having a local entropy definition and a classical entropy definition?
I guess this is a little open-ended, let me elaborate
Using the partition function $\int_{x'} exp(f(x'|\theta)) dx'$, we get that the local entropy is log(partition function)
We take the log to find the probability density around the $\theta$ parameters of our space, as opposed to $E[log(f(x'|\theta))]$ which is classical entropy.
Then you define $A\subset \mathbb{R}^2$ to be universal with respect to a property $P$ if it has that property, and if for any $B\subset \mathbb{R}$, there's a horizontal level of the set which projects onto $B$
Now, in the Borel hierarchy, if you're at some level and you can find a set in its complement which isn't there (e.g. an $F_{\sigma\delta}$ set which isn't $G_{\delta\sigma}$), it means that moving to this step must have added something new, because any set in the hierarchy contains all previous lines
Now, the idea is to show that you can construct a universal set $A$ such that $\{x:(x,x)\in A\}$ has this property $P$ if you've only gone countably many steps in the hierarchy, by indexing things through $\mathbb{N}^{\mathbb{N}}$
You do stuff I only partially followed and it works
No, Marianna said that it does, by saying that if you chose countably many levels of the hierarchy when you've reached $\omega_1$, there's still stuff afterwards
