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Thread: 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 $\displaystyle x >0$ and let $\displaystyle f$ be defined on $\displaystyle (a,b)$ where $\displaystyle a < 0 < b$ and suppose the n-th derivative $\displaystyle f^{(n)}$ exists on $\displaystyle (a,b)$. Then for each nonzero $\displaystyle x\in (a,b)$ there is some $\displaystyle y$ between $\displaystyle 0$ and $\displaystyle x$ such that $\displaystyle R_{n}(x)=\frac{f^{(n)}(y)}{n!}x^{n}$.

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

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

    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 $\displaystyle F(0) = F(x)$, 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 $\displaystyle F(0) = f(x) + (0)$ and $\displaystyle F(x) = 2f(x)$. 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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    Senior Member Pinkk's Avatar
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    Er, I plugged in a $\displaystyle k=0$, not $\displaystyle k=1$ first for the series. Okay, so I'm hoping this is correct:

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

    $\displaystyle F(x) = f(x) + 0$ and so both are equal to $\displaystyle f(x)$, yes?
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    Since M is chosen so that $\displaystyle f(x)= \sum_{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}x^{k} + \frac{Mx^{n}}{n!}
    $, 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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