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Math Help - taylor series

  1. #1
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    taylor series

    I need to prove that if f is defined for abs(x) < R and if there exists a constant B such that abs(f*(x)) <= B (where f*(x) denotes the n derivatives of f) for all abs(x) < R and n in the natural numbers, then the Taylor Series expansion,
    summation from n = 0 to infinity of [f*(0)/(n!)]*x^n converges to f(x) for abs(x) < R.

    I did the following work: I have a defined function f(x). Since abs(f*(x)) <= B, it follows that abs(R(x)) <= (abs(x)^n)/n! for n in N and x in R where (R(x)) denotes the sequence of remainders. I know that lim as n goes to infinity of (R(x)) = 0 for each x in R. Because the sequence of remainders converges to 0 for each x in some interval {x: abs(x-c) < R}, I can write that f(x) = summation from n = 0 to infinity of [f*(c)/(n!)]*(x-c)^n. Evaluating at c = 0, the Taylor expansion becomes the summation from n = 0 to infinity of [f*(0)/(n!)]*x^n.
    I use the Cauchy-Hadamard Theorem, which states that if R is the radius of convergence of the power series sum(a_n*x^n), then the series is absolutely convergent if abs(x) < R. Since abs(x) < R, I know that the above Taylor series expansion does converge. But how do I show it converges to f(x)?

    Thanks in advance for your help.
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  2. #2
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    Quote Originally Posted by jamesHADDY View Post
    I need to prove that if f is defined for abs(x) < R and if there exists a constant B such that abs(f*(x)) <= B (where f*(x) denotes the n derivatives of f) for all abs(x) < R and n in the natural numbers, then the Taylor Series expansion,
    summation from n = 0 to infinity of [f*(0)/(n!)]*x^n converges to f(x) for abs(x) < R.
    Let x\in (-R,R) where R>0 and x\not = 0 (at x=0 the series converges at there is nothing to prove). Let T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!}x^k and let R_n(x) = f(x) - T_n (x) then by Lagrange remainder we have that \left| R_n(x) \right| = \left| \frac{f^{(n+1)}(y)}{(n+1)!} x^{n+1} \right| where y is between 0 and x, this means that |R_n(x)| \leq \frac{B R^{n+1}}{(n+1)!} but \frac{R^{n+1}}{(n+1)!} \to 0.
    Last edited by ThePerfectHacker; February 21st 2008 at 07:39 PM.
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  3. #3
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    Where does this show that it converges to f(x)?

    BTW thanks for the help.
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  4. #4
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    Quote Originally Posted by jamesHADDY View Post
    Where does this show that it converges to f(x)?

    BTW thanks for the help.
    The remainder term converges to zero (see proof above) therefore the true value of the function and the infinite series value of the function converge to the same number because the remainder (i.e. their difference) is zero.
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