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I'm wondering if the following is correct. The original question(1.1.21, Fa96) asks to prove that \begin{align*} f''(x) = \lim_{h\rightarrow0}\dfrac{f(x+h) - 2 f(x) +f(x-h) }{h^2} \end{align*}

I thought it was straightforward and proposed the following solution:

\begin{align} f'(x) &= \lim_{h\rightarrow0}\dfrac{f(x+h) - f(x)}{h} \\ f'(x-h)&= \lim_{h\rightarrow0}\dfrac{f(x) - f(x-h)}{h}\\ f''(x) &=\lim_{h\rightarrow0} \dfrac{f'(x) - f'(x-h)}{h} \\ \end{align}

Substitute $f'(x)$ and $f'(x-h)$ and we'll end up with our desired identity. This solution is different from the one proposed in the book so I want to make sure it is correct.

  • Consider the Taylor expansion :

    $$ f(x+h) = f(x) + hf'(x) + \frac{h^{2}}{2} f''(x) + \mathop{o} \limits_{h \to 0}\big( h^{2} \big) $$

     – pitchounet Jul 27 '14 at 09:49
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    The second equality in your derivation cannot be right: It has $h$ as a free variable on the left but not on the right. – WimC Jul 27 '14 at 09:59
  • Ah, I see it now. The second equality still equates to $f'(x)$, not $f'(x-h)$. –  Jul 27 '14 at 10:07

3 Answers3

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Strictly speaking no your solution is not correct.

We have a function $$ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} $$ but note if we try to evaluate $f'(x-h)$ then the $h$ inputted and the $h$ in the limit are two different variables (one is being taken to $0$ while the other is being taken to be constant). So really what we have is $$ f'(x-h') = \lim_{h \to 0} \frac{f(x-h'+h) - f(x-h')}{h} $$ and now looking at $f''$ we get $$ f''(x) = \lim_{h' \to 0} \frac{f'(x) - f(x-h')}{h'} = \lim_{h' \to 0} \frac{\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} - \lim_{h \to 0} \frac{f(x-h'+h) - f(x-h')}{h}}{h'} $$ And now assuming sufficient differentiability of $f$ we get $$ f''(x) = \lim_{h' \to 0} \lim_{h \to 0} \frac{f(x+h)-f(x)-f(x-h'+h)+f(x-h')}{h h'} $$ but this requires justification in that you can interchange these limits (or take them both to $0$ at the same rate).

DanZimm
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  • Ah. I knew it was probably wrong to use only one free variable $h$. Thanks for the clarification. Is there a formal way to justify the last claim? –  Jul 27 '14 at 10:10
  • @JunichiKoganemaru not atm unfortunately – DanZimm Jul 27 '14 at 10:18
  • I forgot to add that $f \in C^2$. That takes care of the differentiability assumption. Perhaps an epsilon-delta is required to take care of the latter? –  Jul 27 '14 at 10:19
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    @JunichiKoganemaru perhaps so, in my current state (my mind is juggling a bit) I'm not sure I can come up with the proof. My first reaction would have been to look at the Taylor approximations – DanZimm Jul 27 '14 at 10:23
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Recently came across this in wikipedia : http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule We need to prove that : \begin{align*} f''(x) = \lim_{h\rightarrow0}\dfrac{f(x+h) - 2 f(x) +f(x-h) }{h^2} \end{align*} Now, observe that this limit is of the ${0\over 0}$ form. So, we can apply the l'hopitals rule. So, we differentiate with respect to $h$ to get

\begin{align*} \lim_{h\rightarrow0}\dfrac{f(x+h) - 2 f(x) +f(x-h) }{h^2}= \lim_{h\rightarrow0}\dfrac{f'(x+h) - f'(x-h) }{2h}=f''(x)\end{align*}

Srinivas K
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Hint: Another approach is to use Taylor series as

$$ f(x+h) \approx f(x)+f'(x)h+\frac{f''(x)}{2!}h^2 $$

$$ f(x-h) \approx f(x)-f'(x)h+\frac{f''(x)}{2!}h^2. $$

See a related problem.

Community
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Mhenni Benghorbal
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