Recent content by DPMachine

EDIT: Nevermind, that's not right. Sorry about not using latex, by the way. I was writing from a mobile device.

I'm trying to show: If f: S -> S' is an isomorphism of <S, *> with <S', *'>, then f^(-1) is homomorphic. My take: So I have to show that f^(-1)(x' *' y') = f^(-1)(x') * f^(-1)(y'). Since f is bijective (onto, more precisely) I know that f^(-1)(x') = x and f^(-1)(y') = y. So...

p preserves addition, but it's definitely not a bijection... I don't think it's onto. i.e. p(f)(x)=\int^{x}_{0} f(t) dt \neq x^2 for any f in F even though x^2 is in R... Does that make sense?

I fixed #4. Sorry about that. For #3, p(f) = f(t), so p is an identity, right? So p(f+g) = f(t) + g(t) = p(f) + p(g).

Homework Statement Let F be the set of all functions f mapping R into R that have derivatives of all orders. Determine whether p is an isomorphism of the first binary structure with the second. 1. <F, +> with <R, +> where p(f) = f'(0) 2. <F, +> with <F, +> where p(f)(x) = \int^{x}_{0}...

In general, would it be true that if a set is bounded, there must also be a supremum for the set? Too obvious, perhaps?

Can \epsilon = \frac{l+\epsilon}{n}?...

If \epsilon = \frac{l}{2}, we have \left| na_{n} - l \right| < \epsilon = \frac{l}{2} which = -\frac{l}{2} < na_{n} - l < \frac{l}{2} = -\frac{l}{2n} < a_{n} - l < \frac{l}{2n} = -\frac{l}{2n} + l < a_{n} < \frac{l}{2n} + l = \frac{l}{2n} < a_{n} < \frac{l}{2n} + l

|an-l/n| approaches 0 doesn't it? I'm not sure if this was what you were getting at... I understand that an is some form of 1/n, but I want to prove that by using the e def of limit and without providing a specific an that works. How about if e = l/2? Then I would have l/2 < an < l/2 + l...

ok, thank you. I think I proved it by letting epsilon = 2l (which is >0 because a_n >0) Then -2l < na_{n}-l < 2l implies -l < na_{n} < 3l, which implies -l/n < a_{n} Since ∑ -l/n = (-1)*∑ 1/n which doesn't converge, by comparison test a_n doesn't converge.

I made a mistake on the original post. The goal is to prove that \sum a_{n} diverges. Sorry about that... Does my explanation make more sense now?

Homework Statement Given that a_{n} > 0 and lim(na_{n}) = l with l\neq0, prove that \sum a_{n} diverges.Homework Equations The Attempt at a Solution lim(na_n)=l (with =/= 0), so I can safely say that: \left|na_{n}-l\right| < \epsilon by the definition of limit. Then isn't it also true that...

Right, so \frac{1}{ln(e^{n}+e^{-n})} > \frac{1}{ln(e^{n}+e^{n})} = \frac{1}{ln(2)+ln(e^{n})} = \frac{1}{ln(2)+n} \sum \frac{1}{ln(2)+n} certainly looks like it diverges... but how would I prove that? comparison test wouldn't work since it's not quite true that \sum \frac{1}{ln(2)+n} > 1/n

Oh wow... okay. I don't know why that was so hard to figure out. Thank you!

Homework Statement I was looking over my notes and there was a part that didn't make sense. It's basically using the Euler's formula (e^{ix}=cos(x)+isin(x)) and the fact that sin(x)=Im(e^{ix}) to find what Σ sin n sums to. It starts out like this: But this part isn't relevant to my...