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Math Help - Taylor Series Problem

  1. #1
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    Taylor Series Problem

    Find the first five terms of the Taylor series for f(x) = lnx about x = 1.


    Okay, so I know the general formula for a Taylor series, and roughly how to do this, but I've seen conflicting stuff in my notes and whenever I've looked at examples (it's not really "conflicting", I just don't understand, I'm sure).

    Could someone walk me through this one so I can make sure I understand the proper steps and everything? I really appreciate it.
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  2. #2
    Super Member Quacky's Avatar
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    Re: Taylor Series Problem

    f(x)=ln(x)

    f'(x)=\frac{1}{x}

    f''(x)=\frac{-1}{x^2}

    f^3(x)=\frac{2}{x^3}

    f^4(x)=\frac{-6}{x^4}

    f^5(x)=\frac{24}{x^5}

    Substitute x=1 into all of these to get (respectively) 0, 1, -1, 2, -6, -24

    Then, just substitute into the formula:

    f(x)\approx f(a)+f'(a)(x-a)+\frac{f''(a)(x-a)^2}{2!}+\frac{f^3(a)(x-a)^3}{3!}+\frac{f^4(a)(x-a)^4}{4!}+\frac{f^5(a)(x-a)^5}{5!}

    So, for example, for this term:

    \frac{f^4(a)(x-a)^4}{4!}

    We'd substitute in as:

    \frac{-6(x-1)^4}{4!}

    =\frac{-1}{4}(x-1)^4

    Hope this helps.
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  3. #3
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    Re: Taylor Series Problem

    Thank you! That clears up a lot of my confusion.
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    MHF Contributor chisigma's Avatar
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    Re: Taylor Series Problem

    Quote Originally Posted by bobsanchez View Post
    Find the first five terms of the Taylor series for f(x) = lnx about x = 1.


    Okay, so I know the general formula for a Taylor series, and roughly how to do this, but I've seen conflicting stuff in my notes and whenever I've looked at examples (it's not really "conflicting", I just don't understand, I'm sure).

    Could someone walk me through this one so I can make sure I understand the proper steps and everything? I really appreciate it.
    A fast solution starts from the well known series expansion...

    \ln (1+\xi)= \xi - \frac{\xi^{2}}{2} + \frac{\xi^{3}}{3} - \frac{\xi^{4}}{4} + ... (1)

    ... and now You set in (1) x=1 + \xi \implies \xi= x-1 ...

    Kind regards

    \chi \sigma
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