Wikipedia:Reference desk/Archives/Mathematics/2018 February 8

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February 8

Second derivative

Given a differentiable twice function $f$ , prove

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

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 $h=\nu$

$f''(x)=\lim \limits _{h\to 0}\lim \limits _{\nu \to 0}{\frac {1}{h}}\left({\frac {f(h+x)-f(h+x-\nu )}{\nu }}-{\frac {f(x)-f(x-\nu )}{\nu }}\right)=\lim _{h\to 0}{\frac {f(x+h)+f(x-h)-2f(x)}{h^{2}}}$

Ruslik_Zero 20:38, 8 February 2018 (UTC)

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

But if $\lim \limits _{h\to 0}\lim \limits _{\nu \to 0}g(h,\nu )\neq \lim _{t\to 0}g(t,t)$ 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 $\lim \limits _{h\to 0}\lim \limits _{\nu \to 0}g(h,\nu )$ 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 $\lim \limits _{h\to 0}\lim \limits _{\nu \to 0}g(h,\nu )=\lim \limits _{\nu \to 0}\lim \limits _{h\to 0}g(h,\nu )$ . Still not a proof that $\lim \limits _{h\to 0}\lim \limits _{\nu \to 0}g(h,\nu )=\lim \limits _{t\to 0}g(t,t)$ . Bo Jacoby (talk) 17:30, 9 February 2018 (UTC).