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Prove that if $f''(a)$ exists, then $$f''(a)=\lim_{h\to0}\frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}.$$

I really have no idea on this one. Am I supposed to apply the mean value theorem?

Ernie060
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Lincoln77
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3 Answers3

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One way would be to observe that

$f''(a)=\lim_{h\to0}\frac{f'(a+h)-f'(a-h)}{2h}$

and so the result follows by a simple application of L'Hospital'sRule on

$\lim_{h\to0}\frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}$

Remark:

The first item is proved by noting that

$f''(a)=\lim_{h\to0}\frac{f'(a+h)-f'(a)}{h}$

but also

$f''(a)=\lim_{h\to0}\frac{f'(a-h)-f'(a)}{-h}$

so averaging these we get the first formula.

J.G.
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Matematleta
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    I think this should be the answer. Using Taylor to prove the result seems intellectually dishonest to me (as in, using a hammer to crack a nut). – Jonathan H Feb 21 '17 at 15:05
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    @JonathanH It may be a bigger hammer than necessary, but hardly intellectually dishonest. Using Taylor (whose proof does not depend on L'Hôpital, not that this is in any way relevant to the question) makes the first order cancellation obvious, which is, to some extent, the whole point of the formula in the question. – copper.hat Oct 09 '22 at 19:39
  • One may also add that using Taylor, you can come up with approximations for $f''(a)$ that converge better than linearly in $h^2$. – Diger Oct 26 '22 at 16:08
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Taylor's theorem shows that there exists some $\eta$ such that $f(a+h) = f(a) + f'(a)h + {1 \over 2} f''(a) h^2 + \eta(h)$, where $\lim_{h \to 0} {\eta(h) \over h^2} = 0$.

Then ${1 \over h^2} (f(a+h)+f(a-h)-2f(a)) = {1 \over 2} (f''(a)+f''(a)+{\eta(h) \over h^2} + {\eta(-h) \over h^2} )$ from which the result follows.

Aside: Note that with $f(x) = x |x|$, we see that the limit $\lim_{h \to 0} {f(h)+f(-h)-2f(0) \over h^2} = 0$ but $f$ is not twice differentiable at $h=0$.

copper.hat
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    The hypothesis does not allow the usage of Taylor's theorem. – Aloizio Macedo Jun 02 '16 at 03:48
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    @AloizioMacedo: Please elaborate, the above corresponds to the most basic form of Taylor's theorem for the second derivative. – copper.hat Jun 02 '16 at 03:52
  • What form of Taylor's theorem are you using? If you want to have a remainder of order $o(h^n)$, you must have $f^{(n+1)}$ defined in an interval. At least, that is the case I am aware of... which is not in the hypothesis of this exercise. (not even $f''$ is defined in an interval) – Aloizio Macedo Jun 02 '16 at 03:57
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    For reference, Rudin has this exact same exercise, to which it hints: "Use Theorem 5.13" - namely, L'Hopital rule. – Aloizio Macedo Jun 02 '16 at 03:58
  • For more reference, there is a proof on wikipedia which proves the statement under the hypothesis of this exercise (and not on the restrictive hypothesis I mentioned, which usually entails some sort of explicit remainder). However, the proof of such fact (Taylor's theorem) on the less restrictive assumption is also based on L'hopital Rule, and goes exactly as this exercise would. – Aloizio Macedo Jun 02 '16 at 04:15
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    Well, to some extent L'Hôpital and the expression in terms of 'little o' are equivalent. In any case, the intent was to provide intuition for the OP. Taylor's theorem comes in many forms, stronger assumptions yield stronger results. – copper.hat Jun 02 '16 at 04:26
  • If one is willing to admit Lebesgue integration to the toolbox, then the basic result can be proved without recourse to L'Hôpital. However, this is a sledgehammer approach... – copper.hat Jun 02 '16 at 04:33
  • This is a sledgehammer approach, and your form of Taylor's theorem is a circular approach in my toolbox (since the proof computes (essentially) the very limit you are trying to compute here). For instance, I don't know how to prove your version of Taylor's theorem without L'hopital, and I sincerely do not consider it "basic". I would be interested in a proof which does not use L'hopital (also sincerely: I don't like relying on L'hopital). – Aloizio Macedo Jun 02 '16 at 04:38
  • I would need to think about how to do this without the fundamental theorem of calculus (for which I need Lebesgue integration so I can get $\int_0^h f'(a+t) dt = f(a+h)-f(a)$). – copper.hat Jun 02 '16 at 05:06
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    @AloizioMacedo: The form of Taylor series used by copper.hat here is not so famous (known as Taylor theorem with Peano's remainder) and one of its proofs is based on L'Hospital's Rule and that's why some consider it to be equivalent to LHR. But there is a simple proof using mean value theorem for this form of Taylor series also. let me know if you are interested to know this. There is no circularity involved here. – Paramanand Singh Jun 02 '16 at 05:15
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    @AloizioMacedo: See the elementary proof of Taylor's Theorem used here at http://math.stackexchange.com/a/1809307/72031 – Paramanand Singh Jun 02 '16 at 05:56
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Let us make the problem more general trying to approximate the second derivative based on three points $x+ah$, $x+bh$, $x+ch$.

By Taylor, we have $$f(x+kh)=f(x)+h k f'(x)+\frac{1}{2} h^2 k^2 f''(x)+\frac{1}{6} h^3 k^3 f^{(3)}(x)+\frac{1}{24} h^4 k^4 f^{(4)}(x)+O\left(h^5\right)$$ Now, consider $$F=Af(x+ah)+Bf(x+bh)+Cf(x+ch)$$ and apply the above formula for each of the terms and group terms; you should get $$F=f(x) (A+B+C)+h f'(x) (a A+b B+c C)+\frac{1}{2} h^2 f''(x) \left(a^2 A+b^2 B+c^2 C\right)+\frac{1}{6} h^3 f^{(3)}(x) \left(a^3 A+b^3 B+c^3 C\right)+\frac{1}{24} h^4 f^{(4)}(x) \left(a^4 A+b^4 B+c^4 C\right)+O\left(h^5\right)$$ What we then want is to only get the term with $f''(x)$; so we need to cancel the previous terms. Then, this implies $$A+B+C=0\tag 1$$ $$a A+b B+c C=0\tag 2$$ $$\frac{1}{2} \left(a^2 A+b^2 B+c^2 C\right)=1\tag 3$$ So, three linear equations in $A,B,C$ to be solved. This gives $$A=\frac{2}{(a-b) (a-c)}\qquad B=\frac{2}{(b-a) (b-c)}\qquad C=\frac{2}{(c-a) (c-b)}$$ For your case $a=1,b-1,c=0$ which gives $A=1,B=1,C=-2$.

So, we have $$f(x+h)+f(x-h)-2f(x)=h^2 f''(x)+\frac{1}{12} h^4 f^{(4)}(x)+O\left(h^5\right)$$

The same procedure can apply to the derivative of any order.

Claude Leibovici
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