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Math Help - Differentiation proof

  1. #1
    Super Member Showcase_22's Avatar
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    Differentiation proof

    Suppose f: \mathbb{R} \rightarrow \mathbb{R} is twwice differentiable and a \in \mathbb{R}. Prove that \forall \ h>0 \ \exists \ \xi \in(a-h,a+h) with

    f(a+h)-2f(a)+f(a-h)=h^2f''(\xi)

    Hint: consider the function \phi given by

    \phi(t)=f(a+t)-2f(a)+f(a-t)-\left( \frac{t}{h} \right)^2(f(a+h)-2f(a)+f(a-h)).
    I decided to use the mean value theorem.

    Let \xi_1 \in(0,h] and \xi_2 \in [-h,0).
    Then \xi=\xi_1 \cup \xi_2.
    (this is done so that \xi \neq 0).

    MVT gives: \phi'(t_0)=\frac{f(a+h)+f(a-h)-f(a+h)-f(a-h)-f(a)+2f(a)-f(a)}{h}=0 using the interval (0,h].

    I have no idea why this helps me (if it does!).
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  2. #2
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    Hi

    what does "Then \xi=\xi_1 \cup \xi_2" means?
    What is t_0?
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    I decided to use the mean value theorem.

    Let \xi_1 \in(0,h] and \xi_2 \in [-h,0).
    Then \xi=\xi_1 \cup \xi_2.
    (this is done so that \xi \neq 0).

    MVT gives: \phi'(t_0)=\frac{f(a+h)+f(a-h)-f(a+h)-f(a-h)-f(a)+2f(a)-f(a)}{h}=0 using the interval (0,h].

    I have no idea why this helps me (if it does!).
    I would probably do something like this:

    We want to show that \frac{f(a+h)+f(a-h)-2f(a)}{h^2}=f''(\xi), for some \xi\in[a-h,a+h]

    Factor out h and break up the remaining fraction:

    \frac{1}{h}\left(\frac{f(a+h)-f(a)}{h}-\frac{f(a)-f(a-h)}{h}\right)

    By the MVT, we now have \frac{1}{h}[f'(t_2)-f'(t_1)], for some t_1,t_2\in[a-h,a+h].

    Distribute the \frac{1}{h} through to get \frac{f'(t_2)-f'(t_1)}{h}.

    Using the MVT again achieves the desired result.
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  4. #4
    Super Member Showcase_22's Avatar
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    but then can't \xi=0?

    Either way, that's a pretty good proof!
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  5. #5
    Super Member redsoxfan325's Avatar
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    Why can't \xi=0? You're not dividing by it.

    Also, another interesting fact that follows from above is that if you take

    \lim_{h\to 0}\frac{f(a+h)+f(a-h)-2f(a)}{h^2}

    you get out f''(a). The reason is that we know \xi\in[a-h,a+h], so as h\to 0, then [a-h,a+h]\to[a,a], and if \xi\in[a,a], then \xi=a.
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  6. #6
    Super Member Showcase_22's Avatar
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    I have a "comments on the test" sheet where one of my lecturers has written:

    "..and in the problem apply MVT to \phi on the interval [-h,h] which gives a \xi which could turn out to be 0. The latter can lead to complications and it is easier to use the interval [0,h]".
    I'm going to ask my supervisor on Tuesday why \xi \neq 0 and exactly what complications arise if you do. I expected that when I applied MVT to \phi some reason would reveal itself. Unfortunately this hasn't happened!
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