Recent content by Shoelace Thm.

SammyS: My apologies, it was not twenty-four hours. Micromass: I am aware of that theorem, and in fact used it to prove the theorem in the original post. The problem is that I do not know how to show that the set of discontinuities of f \circ \psi are of measure 0. Clearly it is...

Any suggestions?

Homework Statement Let ψ(x) = x sin 1/x for 0 < x ≤ 1 and ψ(0) = 0. (a) If f : [-1,1] → ℝ is Riemann integrable, prove that f \circ ψ is Riemann integrable. (b) What happens for ψ*(x) = √x sin 1/x? Homework Equations I've proven that if ψ : [c,d] → [a,b] is continuous and for every set...

I doubt it has a simple formula, just by calculating the first few coefficients. Is there not some general way of proving the analyticity of a quotient? Neither of SammyS's links are helpful in this regard.

Does anyone have a good way of proving this? Any help is appreciated.

That won't work. Say the Taylor series for \frac{ x }{ e^x -1 } about zero is \sum_{n=0}^{\infty} \frac{B_n}{n!} x^n . We know nothing about \lim_{ n \to \infty } \sqrt[n]{ \frac{B_n}{n!}} because we know nothing about \lim_{n \to \infty} \frac{B_{n+1}}{B_n} . For all we know, the...

Well no, and I can't even find any mention of it online.

How do I prove that if f(x) and g(x) are real analytic, then f(x)/g(x) is real analytic where g(0)≠0?

Any thoughts?

Alright its simple enough. On a side note (although it is not necessary because it is given in the problem statement), if we were not given that g was given by its taylor series in a neighborhood of zero, how to prove that it is? The series manipulations require us knowing g is given by its...

Homework Statement How can I show that \frac{x}{e^x-1} is real analytic in a neighborhood of zero, excluding zero?Homework Equations The Attempt at a Solution

I meant what I wrote in #16. Also, I think what I wrote in #4 is incorrect because of what Dick said, i.e. the power series i had written is not a power series in x.

I don't quite understand what you've written; could you explain? Is what I've written correct?

Ok here is an argument: Because f is analytic, f(c+x) = \sum_{n=0}^{\infty} a_n x^n . Then g(0+x) = f(c+x) = f(x+c) = \sum_{n=0}^{\infty} a_n x^n , so g is analytic in a neighborhood of 0. Then g is analytic in (-R,R), R being the radius of convergence of the series. Can someone confirm...

No I mean for a general function f (not necessarily the one I've specified).