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Thread: Showing Differentiation

  1. #1
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    Showing Differentiation

    Let $\displaystyle I\subset \mathbb{R}$ be an open interval, let $\displaystyle f:I \rightarrow \mathbb{R}$ be differentiable on $\displaystyle I$, and suppose $\displaystyle f''(a)$ exists at $\displaystyle a\in I$. Show that
    $\displaystyle f''(a) = \lim_{h\to 0}\frac{f(a+h)-2f(a)+f(a-h)}{h^2}$
    Give an example where this limit exists, but the function does not have a second derivative at $\displaystyle a$ .
    Last edited by CrazyCat87; Apr 18th 2010 at 04:43 PM.
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  2. #2
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    Quote Originally Posted by CrazyCat87 View Post
    Let $\displaystyle I\subset \mathbb{R}$ be an open interval, let [tex]f:I \rightarrow \mathbb{R} be differentiable on $\displaystyle I$, and suppose $\displaystyle f''(a)$ exists at $\displaystyle a\in I$. Show that
    $\displaystyle f''(a) = \lim_{h\to 0}\frac{f(a+h)-2f(a)+f(a-h)}{h^2}$
    Give an example where this limit exists, but the function does not have a second derivative at $\displaystyle a$ .

    Use Taylor polynomials of order 2 around $\displaystyle a\,\,\,for\,\,\,f(a+h)\,\,\,and\,\,\,f(a-h)$ :

    $\displaystyle f(a+h)=f(a)+f'(a)h+\frac{f''(a)h^2}{2!}+O(h^3)$

    $\displaystyle f(a-h)=f(a)+f'(a)(-h)+\frac{f''(a)h^2}{2!}+O(h^3)$

    Now add both eq's above and solve for $\displaystyle f''(a)$ and let $\displaystyle h\rightarrow 0$

    Tonio
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    Does $\displaystyle O(h^3)$ represent the remainder term of the function? I'm confused as to why it disappears..
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    Quote Originally Posted by CrazyCat87 View Post
    Does $\displaystyle O(h^3)$ represent the remainder term of the function? I'm confused as to why it disappears..


    It doesn't disappear: when we add the corresponding eq's we get $\displaystyle 2O(h^3)$ , and when $\displaystyle h\rightarrow 0$ this, of course, vanishes in the limit...

    Tonio
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  5. #5
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    ah yes
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