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

= February 8 =

Second derivative
Given a differentiable twice function $$f$$, prove
 * $$\lim_{h\to0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}=f''(x)$$

יהודה שמחה ולדמן (talk) 14:04, 8 February 2018 (UTC)
 * See L'Hôpital's rule. (You can use the definition directly, too).  –Deacon Vorbis (carbon &bull; videos) 14:22, 8 February 2018 (UTC)
 * You can use the definition and assume $$h=\nu$$
 * $$f''(x)=\lim\limits_{h\to0}\lim\limits_{\nu\to0}\frac{1}{h}\left(\frac{f(h+x)-f(h+x-\nu)}{\nu}-\frac{f(x)-f(x-\nu)}{\nu}\right)=\lim_{h\to0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}$$
 * Ruslik_ Zero 20:38, 8 February 2018 (UTC)
 * This proof doesn't really work. It is not true in general that $$\lim\limits_{h\to0}\lim\limits_{\nu\to0}g(h,\nu)=\lim_{t\to0}g(t,t)$$. -- Meni Rosenfeld (talk) 09:44, 9 February 2018 (UTC)
 * But if $$\lim\limits_{h\to0}\lim\limits_{\nu\to0}g(h,\nu)\ne \lim_{t\to0}g(t,t)$$ 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 $$\lim\limits_{h\to0}\lim\limits_{\nu\to0}g(h,\nu)$$ 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 $$\lim\limits_{h\to0}\lim\limits_{\nu\to0}g(h,\nu)=\lim\limits_{\nu\to0}\lim\limits_{h\to0}g(h,\nu)$$ . Still not a proof that $$\lim\limits_{h\to0}\lim\limits_{\nu\to0}g(h,\nu)=\lim\limits_{t\to0}g(t,t)$$ . Bo Jacoby (talk) 17:30, 9 February 2018 (UTC).