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Math Help - Taylor's Theorem problem 2

  1. #1
    Senior Member Pinkk's Avatar
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    Taylor's Theorem problem 2

    This involves a proof to Taylor's Theorem. Assume x >0 and let f be defined on (a,b) where a < 0 < b and suppose the n-th derivative f^{(n)} exists on (a,b). Then for each nonzero x\in (a,b) there is some y between 0 and x such that R_{n}(x)=\frac{f^{(n)}(y)}{n!}x^{n}.

    Let M be the unique solution of f(x)= \sum_{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}x^{k} + \frac{Mx^{n}}{n!} and let F(t) = f(t) + \sum_{k=0}^{n-1}\frac{(x-t)^{k}}{k!}f^{(k)}(t) + M\frac{(x-t)^{n}}{n!}

    Show that F is differentiable on [0,x] (I think this part is straightforward since f is differentiable on (a,b) and the other terms are simply polynomials). Show that F'(t)= \frac{(x-t)^{n-1}}{(n-1)!}[f^{(n)}(t) - M] (not really sure how this works out, not great with combining series), show that F(0)=F(x) (again, don't know), and finally show that there is y\in (0,x) such that f^{(n)}(y) = M.

    My professor showed an alternate proof in class which seems at least somewhat easier, but I am absolutely lost here. Thanks.
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  2. #2
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    When you differentiate, the sum in F telescopes (use the product rule). To show F(0) = F(x), just, uh, plug those in.
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  3. #3
    Senior Member Pinkk's Avatar
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    Okay, I forgot the first time of the series doesn't cancel out when you plug in x, so that was throwing me off. Like I said, infinite series was my weakest area in basic calculus and now in analysis.

    Ugh, nevermind, I'm not getting equal answers after all. I'm getting F(0) = f(x) + (0) and F(x) = 2f(x). I don't know what I'm doing wrong....
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    How did you get F(x)=2f(x)?
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  5. #5
    Senior Member Pinkk's Avatar
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    Er, I plugged in a k=0, not k=1 first for the series. Okay, so I'm hoping this is correct:

    F(0) = f(0) + f(x) - f(0); the last two terms are pretty much f(x) except the series is from 1 to n-1, not 0 to n-1, so we have to subtract a f(0) to compensate.

    F(x) = f(x) + 0 and so both are equal to f(x), yes?
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  6. #6
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    Since M is chosen so that f(x)= \sum_{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}x^{k} + \frac{Mx^{n}}{n!}<br />
, just see that F(0) is the RHS of this.

    So, ja, they're both f(x).
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  7. #7
    Senior Member Pinkk's Avatar
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    Okay, thanks after that I can take it from there with the Rolle's and all that. For some reason when it comes to series my mind hits a brick wall.
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