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Proof Taylor Series

  1. Oct 26, 2007 #1

    rock.freak667

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    How does one prove taylor series? Is it proven the same way as Maclaurin's Series(Which i know is a special case of taylor series)

    [tex]f(x)=A_0+A_1x+A_2x^2+A_3x^3+...[/tex]
    [tex]f(\alpha)=A_0+A_1\alpha+A_2(\alpha)^2+A_3(\alpha)^3+...[/tex]

    this kinda doesnt seem like a good way to prove it...as that is how I know how to prove maclaurin series
     
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  3. Oct 26, 2007 #2

    mathman

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    Can you state precisely what is "it" that you are trying to prove?
     
  4. Oct 26, 2007 #3
    If I understand your problem, f(g(x))=f(x-a) is just as differentiable as f is. Although, the logical approach taught in analysis courses is to prove Taylor's theorem and then the result about Maclaurin series holds at a=0.

    I'm not even sure if you prove the result for Maclaurin series without implicitly proving Taylor series.
     
    Last edited: Oct 26, 2007
  5. Oct 26, 2007 #4

    rock.freak667

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    Well I am trying to show that if a function is differentiable at x=a
    then f(x)=f(a)+f'(a)(x-a)+f''(a)(x-a)^2/2!+...
    I never learned Taylor series...but I was yet shown Maclaurin's series and how to prove it without know that maclaurin's series is just a special case of Taylor series
     
  6. Oct 26, 2007 #5
    The Taylor series is not so easy as it looks. The general result is the following.

    First, a class [tex]C^n[/tex] function on [tex](a,b)[/tex] (with [tex]a<0<b[/tex]) is continously differentiable [tex]n[/tex] times. Then the polynomial [tex]T_n(x) = \sum_{j=0}^{n} \frac{f^{(j)}(0)}{j!}x^j[/tex] has the property that the error, i.e. [tex]R_n(x) = f(x) - T_n(x)[/tex] satisfies [tex]\lim_{x\to 0}\frac{R_n(x)}{x^n} = 0[/tex]. This is a result which becomes of theoretic important.
     
  7. Oct 26, 2007 #6

    Gib Z

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    Wikipedia's article on Taylor Series has several different proofs, I'm sure you can read them there. The simplest one by memory was the integration by parts one.
     
  8. Oct 26, 2007 #7

    rock.freak667

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    I didnt see any proofs when i checked it

    EDIT:I see it ...thanks
     
  9. Nov 5, 2007 #8
    Well Taylor series is a simple consequence of Cauchy's Integral formula. That is if you allow a complex analysis based proof(real ones aren't all that elegant).
    For any analytic function f(z), represent with the integral formula for any suitable disc centred at say a and expand the denominator in a geometric series such that the intgration variable isn't caught up in it. Seperate the integral inside the sum and there's your Taylor coefficient. The fact that this is a higher derivative of the original function is again a simple consequence of the Cauchy Integral Formula(which isn't terribly difficult to prove using Cauchy's theorem once you understand the basic notions of holomorphism)
     
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