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Math Help - Taylor's Theorem

  1. #1
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    Taylor's Theorem

    Let r be defined on V (x0) = {x : |x − x0| < a}. If r is n-times differentiable at x0 and r(x0) = r'(x0) = r''(x0) = = r(n)(x0) = 0 (nth derivative),
    prove that the limit as x approaches x0 of r(x)/[(x-x0)^n] = 0.
    I use Taylor's Theorem, but all of the terms go away since the r(x0)'s all equal zero. It seems I'm left only with the remainder:
    r(n+1)(c) * (x-x0) / (n+1)! . Somehow I need to show that the limit of the remainder over (x-x0)^n+1 is zero.

    I'm trying to prove it by induction. I can show the base case easily. If I assume the claim is true for all n, then I'm left showing that the limit of the Lagrange Remainder as x goes to x0
    r(n+1) (c) *(x-x0)^(n+1)
    ----------------------------
    (n+1)!
    -------------------------
    (x-x0)^(n+1)
    equals zero right? The r(n+1) (c) is the n+1th derivative of r at a point c between x0 and x. How do I do this? I'm pretty sure this problem needs to incorporate the remainder somehow.
    If I assume the claim for (n-1) and try to show that the claim holds for n, it seems trivial since r(n) (x0) = 0. I'm seriously missing something.
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  2. #2
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    Taylor's theorem works on the entire set V=\{|x-x_0|<a\}, meaning r is differenciable n times at every single point in V and r(x_0)=r'(x_0)=...=r^{(n)}(x_0)=0 then r(x)=0. I can prove that. Is that what you are asking?
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  3. #3
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    No, I need to show that the limit as x goes to x0 of
    r(x) / (x-x0)^n equals 0 with the conditions I gave. I want to use Taylor's Theorem and mathematical induction, but again I'm stuck working it out.
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  4. #4
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    Quote Originally Posted by agentZERO View Post
    I want to use Taylor's Theorem and mathematical induction, but again I'm stuck working it out.
    But you cannot use Taylor's theorem. That is what I am saying, because Taylor's theorem only works given that r is differenciable n times on (x_0-a,x_0+a). It is a condition you need to use the theorem.

    Note: I am not saying what you are saying is false I am saying Taylor's theorem cannot be used over here. You need a different approach.
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  5. #5
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    I don't know, the hint says to use Taylor's Theorem with the Lagrange Remainder so I'm assuming it can be used. This teacher makes mistakes every once in a while, so if it doesn't adhere to the theorem because of one small part, that may be it. Anyway, I'm going to assume I can use Taylor's Theorem. Will you help me apply it with the work I did or show me another way?
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  6. #6
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    Quote Originally Posted by agentZERO View Post
    Will you help me apply it with the work I did or show me another way?
    Okay, say f is differenciable n-times on (x_0-a,x_0+a) then we can write, f(x) = \sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k+R_n(x) where R_n(x)/(x-x_0)^n \to 0 \mbox{ as }x\to x_0. That is Taylor's theorem what I just wrote above.

    Thus, if r(x) is differenciable n-times on (x_0-a,x_0+a) then the Taylor polynomial \sum_{k=0}^n \frac{r^{(k)}(x_0)}{k!}(x-x_0)^k = 0 because all its derivatices vanish at x_0. Thus, r(x) = 0 + R_n(x) since R_n(x)/(x-x_0)^n \to 0 thus, r(x)/(x-x_0)^n \to 0.
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  7. #7
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    Thank you very much for helping. The one thing I don't get is why Rn(x)/(x-x0)^n goes to zero (which you seem to assume)? Why is its limit as x goes to x0 zero?
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  8. #8
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    Well that is Taylor's theorem. There are several versions of Taylor's theorem. The strongest one is referred to as the Integral Form of Remainder.

    Using the integral form of the remainder we can prove what I stated. If f^{(n)} exists on (x_0-a,x_0+a) (and it is integrable) then the remainder term R_n(x) goes to zero faster than (x-x_0)^n as x\to x_0.

    I am not using the Lagrange remainder but a different remainder. What I posted is basically Taylor's theorem.

    --------
    Here is how to maybe prove it without Taylor. Say that r(x_0)=r'(x_0)=0 then since r(x) is differenciable at x_0 we can write r(x)=r(x_0)+r'(x_0)(x-x_0)+E(x) = 0+E(x) and where E(x)/(x-x_0)\to 0 as x\to x_0. That is just basic property of differenciation. Maybe you can induct that somehow.
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  9. #9
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    It seems I could use L'Hopital's Rule and mathematical induction to solve it. Does anyone know how to do this?
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