And then you come up with that bound now that you know this is what you need to do
I've seen in some books that they throw a bunch of seemingly random bounds at the beginning of the proof and then invoke them like black magic once they're needed
I taught my Spivak students to write these sorts of proofs in two passes. "Scratchwork" where you discover what's the right bounds ... and then "official proof" where you basically reorganize and write a polished sentence.
Well, @PVAL, you may end up teaching such things in the not-too-distant future ... Then you'll read ... :P
I find myself often picking up books and enjoying them, but feeling as though I have to read them and ultimately enjoying the experience less. Further exemplified by my latest reading of Anna Karenina, where I would enjoy it every time I picked it up, but felt that I was not reading it fast enough. So, I read it less often out of discontent.
We also used rectangles for Morera. I'm still confused though, but it's 1AM and I'm too tired to think about it, maybe it'll make more sense after some sleep
He goes and he says okay, if integrals on loops are zero, then you have path independence, so the notion of $F(z) = \int_{z_0}^z f(\xi)d\xi$ makes sense. Now, $f$ is continuous, so if only you had FTC and could say that $f$ being continuous implies $F$ is analytic and $F' = f$, then you're done because then $f$ is as well. But oh look here you DO have it! :D
Like, assume you're writing it in order of coefficients of $1,x,x^2,\ldots$, you get $(a_0,a_1,a_2,\ldots)\in\mathbb{R}^{\omega}$ fully contains the information
Now, the idea is that you think about a power series as such
Also, fun fact, there are some languages in Uganda that have eleven grammatical genders. (Here we should think of these as "classes of nouns" rather than anything relating to human biology.) The verbs and adjectives all change depending on the gender of the noun, and they all have different pluralization rules
Similarly, polynomials $\mathbb{R}[t]$ are ones where all but finitely many of the $a_i$ are 0
And formally you don't think about them as functions, just as sequences. It's relevant in the power series case when worrying about convergence (and other things I'm sure)
Find the number of solutions : x_1+x_2 +....+x_5 = r , where x_1 >= 3 , 2=<x_2 <= 4 , x_3 <= 2 , x_4 is even , x_5 is odd @TedShifrin @AkivaWeinberger @Daminark
@KasmirKhaan for example, (1+z+z^2)(z^5+z^7+z^9) will be the generating function whose z^k coefficient is the number of solutions to u_1+u_2=k where u_1 is in {0,1,2} and u_2 is in {5,7,9}. does that make sense?
for example, when you multiply out (1+z+z^2)(z^5+z^7+z^9), the terms will be of the form (z^a)(z^b), i.e. z^(a+b), where a is from {0,1,2} and b is from {5,7,9}. does that make sense?
(1+z+z^2) can be thought of as (z^0+z^1+z^2) by the way
for example, the z^7 coefficient is 2 since it is (z^0)(z^7) and (z^2)(z^5) combined together, corresponding to the solutions 0+7=7 and 2+5=7. (specifically, solutions to the system u_1+u_2=7 where u_1 is in {0,1,2} and u_2 is in {5,7,9})
@Ted when you see this, the very next thing to happen in the notes was using Morera to prove that uniform limits of analytic functions are analytic, which is rather convenient
