Recent content by autre

Homework Statement Show by example that an infinite union of closed sets in \mathbb{C} need not be closed. The Attempt at a Solution In \mathbb{R} I know that an infinite union of the closed sets A_{n}=[1/n,1-1/n] is open. Not sure if it works in \mathbb{C} as well.

I meant that "there are finitely many such x\in(x_{n})". How do I get started proving that there are infinitely many x\in(x_{n}) s.t. x<r+\epsilon?

Homework Statement \{x_{n}\}\in\mathbb{R^{+}} is a bounded sequence and r=\lim\sup_{n\rightarrow\infty}x_{n}. Show that \forall\epsilon>0,\exists finitely many x_{n}>r+\epsilon and infinitely many x_{n}<r+\epsilon. The Attempt at a Solution By definition of limit superior, r\in\mathbb{R}...

Oh, sorry -- I want to prove that \sum\frac{1}{n^{p}} diverges if \sum\frac{1}{n^{1-p}} diverges using the Comparison Test.

Okay, so going back to k=sk' and n=sn', can I use the fact that for s>1, n/s \leq n but z^{n/s}=e^{2\pi i(k/s)}=1? Or am I still using fractional powers improperly?

Sorry, typo -- p<1.

e^\frac{2 i \pi}{3}\neq{(e^{2 i \pi})}^\frac{1}{3}. That was some sloppy algebra on my part -- x^{2/3}=(x^{2})^{1/3} iff x>0.

Ah, the problem seems to be that I should have s\neq k .

Do I need to exclude 2 somehow? It seems like 2 is the only prime number where the proof would fail.

Homework Statement \sum\frac{1}{n^{p}} converges for p>1 and diverges for p<1, p\geq0. The Attempt at a Solution (1) Diverges: I want to prove it diverges for 1-p and using the comparison test show it also diverges for p. \sum\frac{1}{n^{1-p}}=\sum\frac{1}{n^{1}n^{-p}}=\sum...

How about this? Let z\in\mathbb{C} be a primitive n-th root of unity. Then, for n\in\mathbb{N} and k\in\mathbb{Z} s.t. 0<k<n ,z^{n}=1 and z^{k}\neq1. Suppose k and n are not coprime. Then, \exists s\in\mathbb{N} s.t. n=sn' and k=sk' for some k'\in\mathbb{Z},n'\in\mathbb{N}. Then...

How's this for starters? Let z\in\mathbb{C} be a primitive n-th root of unity. Then, for n\in\mathbb{N} and k\in\mathbb{Z} s.t. 0<k<n ,z^{n}=1 and z^{k}\neq1. Suppose n=jk for some j\in\mathbb{Z}. Then, since z is a root of unity, z=e^{2k\pi i/n} and z^{k}=z^{n/j}=e^{(2\pi...

We never covered it in class so I've had to make do with various online sources and have had some difficulty coming up with a formal definition to help me with the proof (though I basically understand what's going on). Do you have a moreformal definition that I could use as a starting point? Thanks!

A primitive n-th root has the smallest such n that z^n = 1. So if k and n aren't coprime then they would have a common factor except 1, because if they did, you could have a smaller n s.t. z^n = 1.

Homework Statement Show that primitive n-th roots of unity have the form e^{i2\pi k/n} for k\in\mathbb{Z},n\in\mathbb{N}, k and n coprime. The attempt at a solution So the n-th roots of unity z have the property z^{n}=1. I have previously shown that (e^{2\pi ik/n})^{n}=e^{2\pi ik}=(e^{2\pi...