Taylor's theorem works on the entire set , meaning is differenciable n times at every single point in and then . I can prove that. Is that what you are asking?
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)
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(n+1)!
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(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.
But you cannot use Taylor's theorem. That is what I am saying, because Taylor's theorem only works given that is differenciable n times on . 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.
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?
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 exists on (and it is integrable) then the remainder term goes to zero faster than as .
I am not using the Lagrange remainder but a different remainder. What I posted is basically Taylor's theorem.
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Here is how to maybe prove it without Taylor. Say that then since is differenciable at we can write and where as . That is just basic property of differenciation. Maybe you can induct that somehow.