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?
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}...
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.
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
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...