Understanding Taylor's Theorem: Rudin Problem & Analyticity vs C^infinity

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Discussion Overview

The discussion revolves around understanding Taylor's Theorem, particularly in the context of a problem from Rudin's analysis text. Participants explore the theorem's implications, the nature of the remainder in Taylor expansions, and the distinction between analyticity and \( C^\infty \) functions.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in grasping Taylor's Theorem and presents a specific problem from Rudin involving the nth derivative and a Taylor polynomial.
  • Another participant suggests using integration by parts as a method to approach the problem, although the original poster questions the applicability of this method given the context.
  • A further reply attempts to provide an intuitive understanding of the theorem by relating it to the integral of the derivative of a function.
  • Another participant proposes a weaker version of the remainder, linking it to the Mean Value Theorem and discussing the computation of a constant \( K \) related to the remainder.
  • There is a request for clarification on the specific form of the remainder that is required, emphasizing that \( Q(t) \) is defined in the context of the problem.
  • A question is raised about the validity of the theorem for specific values of \( n \), such as \( n=1 \) and \( n=2 \).

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to prove the theorem or the nature of the remainder. Multiple approaches and interpretations are presented, indicating ongoing debate and exploration of the topic.

Contextual Notes

There are limitations in the discussion regarding the assumptions underlying the problem and the definitions of terms like analyticity and \( C^\infty \). The participants do not resolve these aspects, leaving them open for further exploration.

SiddharthM
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I find the idea behind this theorem is somewhat difficult to grasp.

Anyhow, there is a related problem in Rudin that I can't figure out.

So let f be continuously differentiable n-1 times on [a,b] where the nth derivative exists on (a,b). z and a are two distinct points of the interval. Put

Q(t)=(f(t)-f(z))/(t-z)

Let P(t) be the (n-1)th degree taylor polynomial of f about a. Show that

f(z)=P(z) + [Q^(n-1)(a)(z-a)^n]/(n-1)!

I've tried expanding this every way I can think of but am not getting the result. I can't even see why it's true.

My 2nd question regards analyticity vs. C^infinity. So e^(-1/x) and 0 at 0 shows these aren't the same concepts. What does it mean to say that r(h) (the remainder function) goes to zero faster than h does ie what does r(h)/h goes to zero as h goes to zero show/mean?

Taylor's theorem was not emphasised in my singe variable analysis class and we took it as given in my complex analysis class - so it's relatively new to me and I don't have an intuitive understanding of the concept.
 
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as i recall its just integration by parts.
 
can you be more specific?

if you are talking about the proof, then although that may help there has to be a more elementary method because integration has not been defined as yet.
 
you asked for an intuitive understanidng. i am givinmg one.

f(b) -f(a) = integral of f' from a to b.

so f(b) = f(a) + integral of f' from a to b. now integrate by parts, with u = f'(x), and dv = -(b-x)dx...

this gives the [preferred] integral form of the remainder.
 
a weaker version, apparently the one you want, gives the remainder as a corollary of the mean valkue theorem./

choose a constant K such that f(b) = Pn(b) + K(b-a)^(n+1), where Pn is the nth taylor, polynomial for f.

The idea is to compute K. since the function f(x) - Pn(x) - K(x-a)^(n+1) has all derivatives at a equal to zero up to order n, and equals zero also at b, applying the rolle theorem n+1 times, gives a point c between a and b where this function has n+1st derivative equal to zero.

i.e. since the n+1st derivative of Pn is dead zero, this says that f^(n+1)(c) = K(n+1)!. dividing we have "computed" K = f^(n+1)(c)/(n+1)! as desired.

this is not as precise as the integral form given above since the point c is unknown.
 
yes the theorem with that remainder is the proof given in Rudin, but I'm supposed to find another version of the remainder.

Q^(n-1)(a)(z-a)^n]/(n-1)!

where Q(t)=(f(t)-f(z))/(t-z) and a IS known.
 
is it true for n=1? n=2?...
 

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