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Math Help - taylor theorem proof - extra term?

  1. #1
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    Nov 2008
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    taylor theorem proof - extra term?

    I am working through the proof of taylor's theorem going from this step:

    f(a)+f'(a)(x-a)+\int_{a}^{x}(x-t)f''(t)dt

    to this step:

    f(a)+f'(a)(x-a)+\frac{1}{2}(x-a)^2f''(a)+\frac{1}{2}\int_{a}^{x}(x-t)^2f'''(t)dt

    for the integral in the first equation I have
    u=f''(t),  u'=f'''(t)
    v=-\frac{1}{2}(x-t)^2+\frac{x^2}{2},  v'=(x-t)dt

    so for uv I get the 0.5*(x-a)^2*f''(a) just fine.

    but for integral of u'v I get:
    \frac{1}{2}\int_{a}^{x}(x-t)^2f'''(t) + \frac{x^2}{2} dt
    and if i integrate and evaluate the extra \frac{x^2}{2} term, it does not seem to help.
    Which step/s are incorrect?
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  2. #2
    Member
    Joined
    Feb 2010
    Posts
    130
    To show:
    \int_{a}^{x}(x-t)f''(t)dt = \frac{1}{2}(x-a)^2f''(a)+\frac{1}{2}\int_{a}^{x}(x-t)^2f'''(t)dt

    Set u(t) = f''(t)
    Set dv = (x-t)dt

    Then v = \frac{-(x-t)^2}{2} and
    \int_{a}^{x}(x-t)f''(t)dt = \int_{a}^{x}udv = \frac{1}{2}(x-a)^2f''(a)+\frac{1}{2}\int_{a}^{x}(x-t)^2f'''(t)dt
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