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Yes, differentiating ##f^n(x)## is a crucial step of the induction procedure. ##p_n\left(\frac{1}{x}\right)## indicates ##p_n## is a function of ##\frac{1}{x}##, like how ##f(x)## indicates ##f## is a function of ##x##. You should have ##p_1=\frac{1}{x^2}## because...

Yes, a direct proof at this level is usually shorter and more concise than a proof by induction, and so I think it's better.

You could also manually add up the first few terms of ##R_2## to get an estimate on the error. The terms contribute smaller and smaller amounts as ##n## gets larger, so the more terms you evaluate, the better.

Write out the 'full' Taylor series. The full series with all the terms gives the exact value of ##f##. The first 3 terms make up the 2nd degree Taylor polynomial or ##E_2##. The remaining terms make up the error term or ##R_2##. This is because of the equation $$f(x)=E_2(x)+R_2(x).$$ If this...

The Lagrange form of the remainder is not suitable for this situation because ##c## is not determined. First try writing out the first 3-4 terms of the remainder ##R_2## and see what you get. With any luck, you could be able to find a formula for the infinite sum and maybe even compute the...

$$\lim_{x\to 0^+}e^{-\frac{1}{x}}=\lim_{y\to \infty}e^{-y}=0.$$

Please remind yourself of the correct definition of the chain rule ##\frac{d}{dx}f(g(x))=f'(g(x))g'(x)## and re-check your work for $$\frac{d}{dx}e^{-1/x}.$$ And you haven't yet shown that ##P_k = \frac{1}{x^{k - 1}(-1)^k}## is true for ##k>2##. There are specific requirements for a valid...

The problem statement specifically asks you to use Taylor's theorem, so yes.

I don't know much about the distinction between the two types of duals, but I assume it's the case where ##f## is a linear functional in the continuous dual. If ##f## is allowed to be discontinuous, would it be more trivial to find a discontinuous linear functional to satisfy ##||f||=1## and...

Mr.Lovenstein's cartoons can be so relatably dark

This might be an obvious question, but if ##X_0## corresponds to the closed unit ball, and ##x^*## vanishes on ##X_0##, i.e., ##x^*(x)=0## for all ##x\in X_0##, isn't it a contradiction to ##\sup_{||x||\leq 1}x*(x)=1##?

I considered ##X=\mathbb{R}^n## and quickly realized any linear functional like ##f=a_1x_1+\cdots a_nx_n## would attain a maximum on the boundary. I regret to say that my knowledge of topology is still very limited, and did a lot of experimenting with a pen and paper without fruitful results...

The convergent sequence in the codomain that you give to begin the proof. Sorry if it was a dumb question but I didn't want to take this chance for granted. If the ball is were open, would Bolzano-Weierstraß theorem fail to give a convergent subsequence in the open ball? And if we have a...

Now I see why @FactChecker gave up trying to explain (kidding) 🧐 it's not an simple proof to explain to someone who isn't willing to spend effort and time to learn it. Is the existence of such a convergent sequence guaranteed by ##f## being continuous? It seems a bit unnatural to define the...

I hope you feel better soon and thank you for your efforts to help me understand. Just to make sure I'm understanding , the 'subsequence' mentioned in Bolzano-Weierstraß theorem is necessarily of infinite length, right?