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Math Help - Taylor's Theorem

  1. #1
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    Taylor's Theorem





    I tried to write out the taylor's theorem formula , but from there onwards i have no idea how to continue to show 1\leq f(\frac{1}{2}) \leq \frac{5}{4}

    Can anyone guide me along? Thanks in advance.
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  2. #2
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    Quote Originally Posted by xcluded View Post




    I tried to write out the taylor's theorem formula , but from there onwards i have no idea how to continue to show 1\leq f(\frac{1}{2}) \leq \frac{5}{4}

    Can anyone guide me along? Thanks in advance.
    [font=arial][size=4] From the given info it seems like they expect you to write out a MacClaurin series = a Taylor series around x = 0, but then I don't get it:

    f(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\  frac{f'''(0)}{3!}x^3 +....= 2+0+\frac{1}{2}x^2+\frac{f'''(0)}{6}x^3+...

    As all the visible terms are non-negative f(\frac{1}{2}) \geq 2...

    Tonio
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  3. #3
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    yeah. it is kind of weird.

    anyone with different views on it ?
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  4. #4
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    Quote Originally Posted by xcluded View Post
    yeah. it is kind of weird.

    anyone with different views on it ?
    Looks to me like you are expected to use the LaGrange estimate of the error:
    If you truncate a Taylor's series at the n^{th} power term, then the error is less than \frac{M}{(n+1)!}(x- a)^{n+1} where M is an upper bound on the n+1 derivative.
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