All you have to do is show that if $f_n$ is a sequence of continuous functions vanishing at endpoints, its limit (which you know is continuous) also vanishes at endpoints.
Theorem: Suppose that the function $f:[a,b]\rightarrow\mathbb{R}$ is bounded. Then $f$ is integrable iff for each $\epsilon>0$ there exists a $\delta>0$ such that $U(f,P)-L(f,P)<\epsilon$ whenever $P$ is a partition of $[a,b]$ such that $||P||<\delta$.
Proof: ($\rightarrow$). Let $\epsilon>0...
@PedroTamaroff I really want to challenge myself, you see...
(I need to take 12 units worth of classes to stay a full-time student, but it's possible they won't let me just take the one course I need to graduate, and then 8 units of independent studies. If not, I waste time with something like that.)
wikipedia says schemes are formed by gluing together parts of spectra. I know how the spectrum of a ring is endowed with a sheaf, but I want to know what this "gluing" is exactly. presumably the space associated to a scheme is a disjoint union of open sets from various spectra, but now we have to put the sheaves on these spaces together in some way. @mike @ted
@PedroTamaroff You should see a doctor. It's not safe to have fire to your fingers so long.
@TedShifrin I need to stay above 12 units... and I'm not sure yet whether I can get away with doing that by having 8 independent study units. If I can't, I'm going to be wasting my time somehow.
So I might as well waste as little as possible.
My alternative is to sign up for a class for four weeks and drop it mid-quarter, after the aid has been disbursed
Hi guys, i was wondering if anyone knows a bit of basic stats here: I am looking to get the PDF of the following RV: $$Y = \frac{1}{k} \sum_{i=1}^k X_i $$ where each $X_i$~ Exp($\lambda$)
I know, if it was just the sum, Y would have a PDF given by the k fold convolution of the exponential distribution
@Pedro: Spell Stokes correctly :) Also, please don't confound vector field $F$ and associated $1$-form. We can take d of the latter, not of the former. That's all. :D
But i've told you 99% vector calc students don't know forms. So what you're writing is doubly bad. It's sloppy (i.e., wrong) and they don't know what you're talking about.
@anon I want to show an finite noncyclic abelian group contains a subgroup that is iso to Z_p x Z_p for some prime p. I have done the following: cyclic iff unique subgroup of order d for each d | |G|. Thus, for some d | |G| there are two subgroups of order d. I claim that for some prime factor of d, I can find two distinct elts of order p. Right or left?
It is very similar to what Lang does which is quite clean, but it gives something stronger so the proof is very abstract, one doesn't get hands dirty like Hungerford.
If S_{n-1} embeds, consider the coset space A_n/S_{n-1} of size n/2. Then A_n acts on it by permutations and we can totally assume n>4, and since A_n is simple for n>4, we get an injection, so n!/2 $\geqslant $ (n/2)!, impossibru.
"If $G$ is abelian and $x$ is an element of maximal order, $|x|=\exp G$".
Proof. Write $|x|=p_1^{e_1}\cdots p_s^{e_s}$, $|h|=p_1^{f_1}\cdots p_s^{f_s}$. If $h^{|x|}\neq 1$, some $f_i>e_i$, wlog $i=1$. Let $x'=x^{p_1^{e_1}}$ and $h'=h^{|h|/p_1^{f_1}}$ then $|x'h'|=|x'||h'|>|x|$ contra maximality of $x$.
(Note $(|x'|,|h'|)=1$)
@Mike In fact $G$ finite abelian is cyclic iff $\exp G=|G|$. This gives a slick proof that $F^\times$ is cyclic for a finite field.
Theorem: Suppose that the function $f:[a,b]\rightarrow\mathbb{R}$ is bounded. Then $f$ is integrable iff for each $\epsilon>0$ there exists a $\delta>0$ such that $U(f,P)-L(f,P)<\epsilon$ whenever $P$ is a partition of $[a,b]$ such that $||P||<\delta$.
Proof: ($\rightarrow$). Let $\epsilon>0...
