Recent content by schniefen

Thank you for replying. Ok, I guess in computing ##r## the author assumed ##x\neq c##. Because if ##R=0##, then, as you write, $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim_{n\to\infty}\frac1{\left|\frac{a_{n}}{a_{n+1}}\right|}=\infty.$$ And so ##r=[|x-c|\cdot\infty]=\infty##, i.e...

In these lecture notes, there is the following theorem and proof: I'm confused about "...the power series converges if ##0\leq r<1##, or ##|x-c|<R##...". In other words, why is ##|x-c|<R## equivalent to ##0\leq r<1##? I guess the author reasons as follows. If $$R=\lim _{n\to \infty...

Ok, I think this clarified it. Thanks a lot! To summarize; by the definition of the supremum, we have ##|x| \lt |x_0| \leq R##. Now, ##\sum a_nx^n## converges for ##x=x_0## and so for all ##x\in\mathbb R## with ##|x|<|x_0|##. As @FactChecker pointed out, we can find an ##|x_1|## such that...

I thought maybe the statement ##|x|<|x_0|< R## is not incorrect after all if the set $$\left\{\left|x\right|\ge 0:\sum a_nx^n \text{ converges}\right\}$$ is an interval, but this is not something we know a priori. Besides I am really not sure how to show the set is an interval -- probably...

I am reading the following passage in these lecture notes (chapter 10, in the proof of theorem 10.3) on power series (and have seen similar statements in other texts): I'm confused about ##|x_0|<R##. If ##M=\sup (A)##, then for every ##M'<M##, there exists an ##x\in A## such that ##x>M'##...

Hmm, I'm confused. Are you claiming there is a sequence ##(h_n)_1^\infty## (independent of ##k##) such that $$\sum_{k = 1}^{\infty}\frac{\arctan(kh_n)}{h_nk^2}$$ diverges? I don't see how that would be possible.

We'll always have uniform convergence of $$\sum_{k = 1}^{\infty}\frac{\arctan(kh)}{hk^2}$$ no matter how we fix ##h##, be it as ##h_n=1/n## or some other sequence tending to ##0## as ##n\to\infty##.

Ok, I think I also may have a solution. Grateful for any feedback. Take ##h_k=1/k^2##, then we have ##h_k\to 0## as ##n\to\infty##. Also, $$\frac{f(h_k)}{h_k}=\sum_{k=1}^\infty \arctan (1/k).$$ This series diverges according to the limit comparison test with ##1/k##, i.e. we have...

Ok, thanks for clarifying things @PeroK. I had to remind myself of the sequential characterization of the limit and the aim is clear now. Have you been able to find a sequence such that $$\forall n: \frac{f(h_n)}{h_n} > n?$$

You are right! I forgot. So ##x\neq 0##. That said, I'm still unsure what exactly it is I need to check in order for the exchange of limits to be valid.

The statement that a limit function ##f## of a sequence ##(f_n)_1^\infty## (of continuous functions) is continuous at a point ##a## means that ##\lim_{x\to a} f(x)=f(a)##, i.e. ##\lim_{x\to a}(\lim_{n \to \infty}f_n(x))=\lim_{n \to \infty}(\lim_{x\to a} f_n(x)).## This exchange is permitted...

Can we exchange those limits? That is, does ##\lim_{x \to 0} \lim_{K \to \infty}\ldots=\lim_{K \to \infty} \lim_{x \to 0}\ldots##?

Definitely.

I have previously shown that the function series is differentiable at ##x\neq 0##. The series converges uniformly (thus pointwise) on ##\mathbb R## and the term wise differentiated series is uniformly convergent on any interval ##d\leq |x|##, where ##d>0##. Moreover, the terms are continuously...

You were right, fixed it. As @FinBurger pointed out, I just evaluate ##V## at the equilibrium distance for b). However, why at the equilibrium distance? Is this where the kinetic energy is ##0## and thus we obtain the total energy of the system?