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Taylor's Theorem

  1. Oct 29, 2007 #1
    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.
     
  2. jcsd
  3. Oct 29, 2007 #2

    mathwonk

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    as i recall its just integration by parts.
     
  4. Oct 29, 2007 #3
    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.
     
  5. Oct 29, 2007 #4

    mathwonk

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    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.
     
  6. Oct 29, 2007 #5

    mathwonk

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    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.
     
  7. Oct 29, 2007 #6
    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.
     
  8. Oct 29, 2007 #7

    mathwonk

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    is it true for n=1? n=2?....
     
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