@Araske did I not tell you? He was with Calegari and was like "Oh yeah the guy is a good lecturer, you should take a class of his, even though he doesn't know much math"
Basically, he gave an integrated treatment of more vector space/linear transformation-y stuff and matrix stuff, focused on making sure we understood how to bounce between one and the other
For what it's worth I wasn't quite going for seeing that theorems were true in hindsight, almost smelling their presence before you know for sure that you should use them
So, define $Mf(x) = \frac{1}{\mu(B(x,r))}\int_{B(x,r)} |f|d\mu$. Then you want to show that $f\mapsto Mf$ is a bounded operator on $Lp$ for $p > 1$
One of the lemmas in this proof is that $\mu(\{x: Mf(x) > t\}) \le \frac{c}{t}\int_{\{x:|f(x)| > \frac{t}{2}\}} |f| d\mu$
Proving it is fine, you just define $f_1$ to be $f$ where it's less than $\frac{t}{2}$ and $f_2$ elsewhere, so that $f = f_1 + f_2$
Then $t < Mf(x) \le Mf_1(x) + Mf_2(x)$, but since $\|f_1\|_{\infty} \le \frac{t}{2}$, we have that $Mf_2(x) > \frac{t}{2}$ (both statements are almost everywhere), you get that except on for a null set, $\{x:Mf(x) > t\} \subset \{x:Mf_2(x) > \frac{t}{2}\}$
Then you use the weak 1-1 estimate and get $\mu(\{x:Mf(x) > t\}) \le \mu(\{x:Mf_2(x) > \frac{t}{2}\}) \le \frac{2c}{t}\int_{x:|f(x)| > \frac{t}{2}} |f|d\mu$
I'm trying to approximate E[X] where X ~ Geo(1/2) with two methods: once I just "throw the coin", yields ~2, good. But the second method using CDF gives me 1.44
using -Math.log2(1-Math.random())
I don't see what I'm doing wrong
CDF is: 1-(1/2)^k
for Geo(1/2)
the value of k should be >= 1, but the code does not guarantee that
Though I think I've more or less got it down. It's only on the stuff from the last few weeks and I think I've materialized most of the main proofs except for the dual of $L_p$
My original plan was to just stay up all night studying for analysis, then go to sleep after add-dropping into bio, logic, and complex analysis until it gets closer to time for math
That's cutting it a bit close, though, so instead I'll just sleep the next 2-3 hours, wake up in time for add-drop, then go back to sleep, so this gives me panic review time before the exam
Part of it is that I'm fasting, no food or water all day. Taking a 6:30PM exam like that is merp. At least shifting sleep like this will help lessen that blow somewhat
Being given a 9 digit number $N$ in which digits MAY be repeated. Consider another single digit no. $X$. Now if we remove any digit from $N$ and multiply it by $X$, the sum of digits of the product is the same in all possible cases. Also whatsoever digit we remove, on the same position in the pro...