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February 8

Second derivative

Given a differentiable twice function $f$ , prove

$\underset{h\to 0}{\lim }\frac{f(x+h)+f(x-h)-2f(x)}{h^2}=f^{″}(x)$

יהודה שמחה ולדמן (talk) 14:04, 8 February 2018 (UTC)

See L'Hôpital's rule#Examples. (You can use the definition directly, too). –Deacon Vorbis (carbon • videos) 14:22, 8 February 2018 (UTC)

You can use the definition and assume $h=\nu$

$f^{″}(x)=\lim \underset{}{h\to 0}\underset{\nu \to 0}{\lim }\frac{1}{h}(\frac{f(h+x)-f(h+x-\nu )}{\nu }-\frac{f(x)-f(x-\nu )}{\nu })=\underset{h\to 0}{\lim }\frac{f(x+h)+f(x-h)-2f(x)}{h^2}$

Ruslik_Zero 20:38, 8 February 2018 (UTC)

This proof doesn't really work. It is not true in general that $\underset{h\to 0}{\lim }\underset{\nu \to 0}{\lim }g(h,\nu )=\underset{t\to 0}{\lim }g(t,t)$. -- Meni Rosenfeld (talk) 09:44, 9 February 2018 (UTC)

But if $\underset{h\to 0}{\lim }\underset{\nu \to 0}{\lim }g(h,\nu )\ne \underset{t\to 0}{\lim }g(t,t)$ where $g(h,\nu )=\frac{\frac{f(x+h+\nu )-f(x+h)}{\nu }-\frac{f(x+\nu )-f(x)}{\nu }}{h}$ then the function $f$ is not twice differentiable. Bo Jacoby (talk) 10:01, 9 February 2018 (UTC).

Ok, but this needs to be proven. The only thing you can directly deduce from "f is twice differentiable" is that $\underset{h\to 0}{\lim }\underset{\nu \to 0}{\lim }g(h,\nu )$ exists, you need to actually show that it equals the RHS in this case. -- Meni Rosenfeld (talk) 14:42, 9 February 2018 (UTC)

Right! Obviously $g(h,\nu )=g(\nu ,h)$ , so $\underset{h\to 0}{\lim }\underset{\nu \to 0}{\lim }g(h,\nu )=\underset{\nu \to 0}{\lim }\underset{h\to 0}{\lim }g(h,\nu )$ . Still not a proof that $\underset{h\to 0}{\lim }\underset{\nu \to 0}{\lim }g(h,\nu )=\underset{t\to 0}{\lim }g(t,t)$ . Bo Jacoby (talk) 17:30, 9 February 2018 (UTC).