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What if I re-index so that n\in \mathbb{N} so that d_0=e_0, d_1=e_1, d_2=e_{-1},\dots. Then by Bessel's inequality we have \sum_{n=0}^{\infty}|(f,d_n)|^2\le \parallel f\parallel ^{2}<\infty. Hence \sum_{n=0}^{\infty}|(f,d_n)|^2 converges absolutely and whence (f,d_n)\to 0 and thus (f,e_n)\to\ 0...

What should I be careful with? Does L^2-convergence not imply absolute convergence? Sorry, just a little confused by your statement.

Yep we have. Could I say that: Since \{e_n\} is complete it follows that f=\lim_{n\to\infty} \sum_{m=-n}^{n}(f,e_m)e_m. Thus, the latter sum converges and hence \lim_{n\to\infty}\sum_{m=-n}^{n}|(f,e_m)|^{2}<\infty. Thus |(f,e_m)|^2 \to 0 so |(f,e_m)|\to 0 and whence (f,e_m)\to 0, as required...

Wait, if {e_n} is complete, then \lim_{n\to\infty}\sum_{m=-n}^{n}(f,e_m)e_m converges to f absolutely, so the coefficients necessarily converge to zero. Would this work?

Homework Statement Let e_{n}(t)= \frac{1}{ \sqrt{2\pi}}\cdot e^{int} for n\in\mathbb{Z} and -\pi\le t\le\pi. Show that for any f\in L^{2}[-\pi,\pi] we have that (f,e_{n})=\int_{-\pi}^{\pi}f(t)\cdot e^{-int}dt\to0 as |n|\to \infty. The Attempt at a Solution I want to use dominant convergence...

I know that thinking a R-module is simply a ring R acting on a set (following the usual axioms), would it be safe to think of an R-algebra as the ring R acting on another ring? This may seem convoluted, but I'm just having a little trouble getting through all the different definitions of an...

Thanks for the help! I got it.

We have 0= 1\otimes 0 = 1\otimes 12= 12(1\otimes 1)=12\otimes 1=(2\otimes 1)+ (10\otimes 1)=(2\otimes 1)+0=2\otimes 1. Hence, m\otimes 1 = k(2\otimes 1)=0.

Actually, I guess that doesn't really prove anything since it isn't assumed that \phi is injective. Hrmm...

Well it is trivial if m=0, so suppose m\neq 0 even. Then it follows that m=2k hence m \otimes 1 = 2k\otimes 1 = 2(k\otimes 1) Hence, for any morphism of \mathbb{Z}-modules \phi : (\mathbb{Z}_{10}\otimes_{\mathbb{Z}}\mathbb{Z}_{12})\to \mathbb{Z}_{2}, it follows that...

yepp 1 \otimes 10 = 10(1\otimes 1)=10\otimes 1 = 0\otimes 1=0.

Homework Statement Show that \mathbb{Z}_{10}\otimes_{\mathbb{Z}}\mathbb{Z}_{12} \cong \mathbb{Z}_{2} The Attempt at a Solution Clearly, for any 0\neq m\in\mathbb{Z}_{10} and 0\neq n \in \mathbb{Z}_{12} we have that m\otimes n = mn(1\otimes 1), and if either m=0 or n=0 we have that m\otimes n...

omg how do i call myself a math major. thank you.

Homework Statement Let H be a Hilbert space. Prove \Vert x \Vert = \sup_{0\neq y\in H}\frac{\vert (x,y) \vert}{\Vert y \Vert} The Attempt at a Solution First suppose x = 0. Then we have \sup_{0\neq y\in H}\frac{\vert (x,y) \vert}{\Vert y \Vert} = \sup_{0\neq y\in H}\frac{\vert (0,y)...

Algebraic Topology Algebraic Geometry I Linear Algebraic Groups Functional Analysis II Rings and Modules Field Theory Commutative Algebra Lie Algebras Differential Equations Algebraic Geometry II I bet you wouldn't guess that I'm applying to an algebra group for graduate studies...

Homework Statement Let \sigma (E)=\{(x,y):x-y\in E\} for any E\subseteq\mathbb{R}. If E has measure zero, then \sigma (E) has measure zero. The Attempt at a Solution I'm trying to show that if \sigma (E) is not of measure zero, then there exists a point in E such that \sigma (\{e\}) that...

Nevermind, I'll just use Monotone Convergence on a sequence of simple functions.

So my question is if I'm allowed to say this: \int f\, d\nu = \sup \{\int \phi\, d\nu :0\le\phi\le f,\, \phi \,simple\} =\sup \{\int \phi g\,d\mu :0\le\phi\le f,\,\phi\,simple\} = \int fg\,d\mu?

I'm using Royden, where the integral of a non-negative function is defined as \int f\, d\,\mu =\sup \{\int \phi \, d\mu :0\le\phi\le f,\,\phi \, simple\}

Hey, thanks for the reply! I've already done this for simple functions, I'm just stuck on how to show that any non-negative measurable function satisfies the equality. Should i take approximations from above and below? or should what I gave in my first reply be sufficient?

Should I do some inequalities for simple functions above and below? I feel like that last equality should be an inequality. Hrmf.

I've shown for non-negative simple functions that \int \phi\,d\nu=\int \phi g \,d\mu. Now I wish to show it in general for non-negative measurable functions. So I say let f be a non-negative measurable function on X. Fix \phi as a simple function such that 0\le\phi \le f. Hence we have \int...

Again, thanks all! This helped so much!

I'm still confused as to where I'm supposed to go with all of this. Let p be an odd prime. For each d|p-1, let f(d) be the amount of elements of \mathbb{Z}_{p}^{\times} such that those elements have order d. Clearly \sum_{d|p-1}f(d)=p-1. Fix a \in\mathbb{Z}_{p}^{\times} such that ord(a)=d, for...

Then a^{p-1}\equiv 1 \mod (p^{2}) Why is this a problem?

Homework Statement Let (X,\mathcal{B},\mu) be a measure space and g be a nonnegative measurable function on X. Set \nu (E)=\int_{E}g\,d\mu. Prove that \nu is a measure and \int f\, d \nu =\int fg\,d\mu for all nonnegative measurable functions f on X. The Attempt at a Solution I'm basically...

Yep, we had to use the fact that a+p\in\mathbb{Z}_{p^{2}}^{\times} is a generator, hence showing that it is cyclic. My original question was the starting point to this proof. I'm still confused on how exactly we showed that \gcd(k,p)=1. Also, thank you so much for your help otherwise!

Yep that's the question exactly. What is it exactly that you're trying to show that makes the claim untrue? I'll take it up with my prof if it indeed isn't a true claim.

OOOOOH Since p odd prime it follows that it cannot be that both p|a^{m}-1 and p|a^{m}+1. I understand why we want to do this, but what if p^2|a^{m}+1?

I'm still stuck on your last hint :(

No, for any odd prime p it is not the case for both p|n and p|n+2, for any n\in\mathbb{Z}.

Either p|(a^{m}+1) or p|(a^{m}-1). If it is the latter, then p|(a^{m}-1) implies that a^{m}\equiv 1\mod p, but that would mean that the order of a is m, which contradicts the fact that a is a generator of \mathbb{Z}_{p}^{\times}. Hence, it must be that a^{m}\equiv -1\mod p.

Well we can say kp=(a^{m})^2-1=(a^{m}-1)(a^{m}+1), or is there something that I'm missing?

Well if 1\le k<p, then clearly \gcd(k,n)=1. But if p<k, then its not so clear.

We can take d=p-1 and hence there exists \phi(p-1) many elements of \mathbb{Z}_{p}^{\times} whose order is p-1. But since the order of \mathbb{Z}_{p}^{\times} is p-1, it follows that these elements are exactly the generators of \mathbb{Z}_{p}^{\times}. Hence, we've show that there exists at...

Yep since f(d) is either \phi(d) or 0; both of which are less than or equal to \phi(d). And since \sum_{d|p-1}f(d)=\sum_{d|p-1}\phi(d), it follows from the previous sentence that f(d)=\phi(d). So now we know that \phi(d) is exactly the number of elements in \mathbb{Z}_{p}^{\times} that have...

Yep :) \sum_{d|p-1}\phi(d)=p-1

For every d|p-1, it follows that every f(d) is either \phi(d) or 0, because we haven't verified the fact that there exists an element of order d for each d|p-1.

If we take d=p-1, then there are only \phi (p-1) many generators of \mathbb{Z}_{p}^{\times}. Is this right? Since d|p-1, it follows that \phi(d)|\phi(p-1). Hence there exists k\in\mathbb{Z} such that \phi(p-1)=k\phi(d).

\phi (d) many.

The size of \mathbb{Z}_{p}^{\times} is \phi (p)=p-1

Any element a^{r} such that \gcd(r,d)=1 is a generator.

Well I know its obvious that a^{p-1}\equiv 1 (\mod p) since [a]\in\mathbb{Z}_{p}^{\times}. Hence there exists k\in\mathbb{Z} such that a^{p-1}=1+kp. How do I show that \gcd (k,p)=1?

Homework Statement Let p be an odd prime. Show that there exists a\in\mathbb{Z} such that [a]\in\mathbb{Z}^{\times}_{p} is a generator and a^{p-1}=1+cp for some c coprime to p. Homework Equations The Attempt at a Solution I honestly have no idea where to even start with this. Any help will...

Well if you're trying to show that it is not Cauchy, state what it means for a sequence to not be Cauchy. That is where I would start :)

If order didn't matter, I'd look up Bell numbers, they arise as a way to count the amount of partitions there are to any given set. Since order does matter, you know that your number you will want to count should be less than this. I hope that helps.

Is undergraduate QM a little hand-wavy? Is there a good text i could supplement with that treats it a little more axiomatically maybe?

Just a side note, I'm 3rd year pursuing a math degree. I plan on taking QM courses either in supplement to my Master's (I'm hoping to get into quantum computing through non-commutative geometry). Would undergraduate QM courses be relatively more accessible with a background in linear algebra...

Group Theory General Topology Measure Theory Cryptography Combinatorics YAAAAAY.

I'll be in my 4th year of a Pure Math degree next year, and I'm think of what courses to fill up my timetable with. I'll be taking an overload, so I'll be taking 6.0 credits worth of courses (3.0 per semester). I'm only required to take 3.0 more courses in Pure Math to get my degree, so I have...