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Thread: Taylors theorem question.

  1. #1
    C.E
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    Taylors theorem question.

    Hi, I have been stuck on the following for ages, any guidance would be welcome.

    A function f: R --> R is C^3 and,

    f(a+h)= f(a) + f'(a + 0.5h)h

    for real a and h 0

    by applying Taylors theorem to f and f' show that the third

    derivative of f is identically 0.

    I can see that f'(a + 0.5h)h is the Lagrange error term

    and I know that similarly f'(a+05h)=f'(a)+0.5f''$\displaystyle \delta$h

    Therefore f(x)= f(a) + (f'(a) + 0.5 f''(a))(x-a).

    Is this even the right way to attempt the question? If so, how does the above imply that the third derivative of f is always zero?
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  2. #2
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    Taylor's theorem for f says $\displaystyle f(a+h) = f(a) + hf'(a) + \tfrac12h^2f''(a) + \tfrac16f'''(a+\theta h)$ ( for some $\displaystyle \theta$ between 0 and 1).

    Taylor's theorem for f' says $\displaystyle f'(a+
    \tfrac h2) = f'(a) + \tfrac h2f''(a) + \tfrac12\bigl(\tfrac h2\bigr)^2f'''(a+\phi h)$ ( for some $\displaystyle \phi$ between 0 and 1/2). Multiply that equation by h, subtract it from the previous one, and use the information that $\displaystyle f(a+h) = f(a) + hf'(a+\tfrac h2)$. You'll end up with the equation $\displaystyle 4f'''(a+\theta h) = 3f'''(a+\phi h)$. Now let $\displaystyle h\to0$ to see that $\displaystyle 4f'''(a) = 3f'''(a)$, from which obviously $\displaystyle f'''(a)=0$.
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