Recent content by snesnerd

Hmmmm it did not translate my latex writing. I apologize for that.

If you take the Fourier series of a function $f(x)$ where $0 < x < \pi$, then would $a_{0}$, $a_{n}$, and $b_{n}$ be defined as, $a_{0} = \displaystyle\frac{1}{\pi}\int_{0}^{\pi}f(x)dx$ $a_{n} = \displaystyle\frac{2}{\pi}\int_{0}^{\pi}f(x)\cos(nx)dx$ $b_{n} =...

Homework Statement If (X,d) is a metric space. I want to show that the set of all open balls and \emptyset form a base.Homework Equations The Attempt at a Solution I know that we need to show that the union of all these sets (or balls) is the whole set. I feel like this is simple yet, I am...

Proof (Attempt) : Suppose S^1 \approx D. Then \exists a map f : S^1 \rightarrow D that is \approx . Let x \in (0,1) and y = f(x) . Then f : S^1 - \{x\} \rightarrow D - \{y\}. Here S^1 - \{x\} and D - \{y\} are connected. Suppose we remove another point x' \in (0,1), and y' = f(x')...

Well to be honest, the idea sounds good, but I would like to mathematically write it out nicely. As you know in mathematics it is easy to say something is true, but it is only true if you can prove it. Granted the wording still sort of proves it well, but I would like to translate what I said...

Homework Statement I recently read and proved that the real line is not homeomorphic to the circle. It was very interesting and made sense to me. I started pondering to myself and thought of another interesting problem. What if I considered the circle S^1 and D where D is the open disk. The...

But since we have that n and \frac{n}{d} are essentially the same set of numbers then then \frac{n}{c} and c are the same set of numbers so we can commute them and nothing will change/

Because if you consider the set of divisors of n which are cd , and you substitute \frac{n}{d} in for c then you get that n = (\frac{n}{d})d = n .

(f \ast g)(n) = \sum\limits_{d|n}f(d)g(\frac{n}{d}). Note that if d|n then their exists a c such that n = cd . Then d = \frac{n}{c} and c = \frac{n}{d} . Substituting above gives us: \sum\limits_{d|n} f(\frac{n}{c})g(c) Now from here how would I word it to explain why...

Okay if that is the case, would the following proof show that it is commutative: (f \ast g)(n) = \sum\limits_{d|n}f(d)g(\frac{n}{d}) = \sum\limits_{d|n}f(\frac{n}{c})g(c) = \sum\limits_{d|n}g(c)f(\frac{n}{c}) = (g \ast f)(n)

I understand what you mean pasmith, but I don't see how that would help show it is commutative.

If \ast : (f \ast g)(n) = \sum\limits_{d|n}f(d)g(\frac{n}{d}), show that \ast is commutative. Note that d|n says d divides n. Now I was not sure how to do this from an abstract algebra point of view although when I stare at it my though process was to maybe rewrite it somehow, which will then be...

Homework Statement Find the Fourier transform of (1/p)e^{[(-pi*x^2)/p^2]} for any p > 0 Homework Equations The Fourier transform of e^{-pi*x^2} is e^{-pi*u^2}. The scaling property is given to be f(px) ----> (1/p)f(u/p) The Attempt at a Solution Using the information above, I got...

If I have the unit sphere and I mod out its equator, I get two spheres touching at one point. I have been thinking what the bijection between these could be but can not come up with one.

Let x = (2,1+i,i) and y = (2-i,2,1+2i). Find <x,y> So my work is the following: 2(2-i) + (1+i)2 + i(1+2i) = 4+i, but my book says the correct answer is 8+5i. Hmmm what am I doing wrong?