Let-f-be-defined-in-the-neighborhood-of-x-and-that-f-x-exists-Prove-that-lim-h-0-f-x-h-f-x-h-2f-x-h-2-f-x-

Question Number 62679 by ajfour last updated on 24/Jun/19

L e t f b e d e f i n e d i n t h e n e i g h b o r h o o d

o f x a n d t h a t f^{″} (x) e x i s t s .

P r o v e t h a t

\lim_{h \to 0} \frac{f (x + h) + f (x - h) - 2 f (x)}{h^{2}} = f^{″} (x) .

Commented by kaivan.ahmadi last updated on 24/Jun/19

h i s i r, c h e c k t h e q u o e s t i o n p l e a s e

Commented by ajfour last updated on 24/Jun/19

I t s c o r r e c t, S i r .

Commented by MJS last updated on 24/Jun/19

I think it should be h \to 0 ?

Commented by kaivan.ahmadi last updated on 24/Jun/19

i f h \to 0

\frac{0}{0}

\overset{H o p}{=} {l i m}_{h \to 0} \frac{f^{'} (x + h) - f^{'} (x - h)}{2 h} \overset{H o p}{=}

{l i m}_{h \to 0} \frac{f^{″} (x + h) + f^{″} (x - h)}{2} = \frac{2 f^{″} (x)}{2} = f^{″} (x)

t h i s w a y i s t r u e

i h a d a m i s t a k e i n p r e v i o u s s o l u t i o n

(\frac{d}{d h} (2 f (x)) = 0)

Commented by Mr X pcx last updated on 24/Jun/19

h o s p i t a l t h e o r e m l e t

u (h) = f (x + h) + f (x + h) - 2 f (x)

v (h) = h^{2}

{l i m}_{h \to 0} u (h) = {l i m}_{h \to 0} v (h) = 0

u^{^{'}} (h) = f^{^{'}} (x + h) + f^{^{'}} (x + h) \int \to 2 f^{^{'}} (x)

u^{^{″}} (h) = f^{^{″}} (x + h) + f^{^{″}} (x + h) \to 2 f^{^{″}} (x)

v^{^{'}} (h) = 2 h \to 0

v^{^{″}} (h) = 2 \Rightarrow

{l i m}_{h \to 0} \frac{f (x + h) + f (x + h) - 2 f (x)}{h^{2}}

= \frac{2 f^{^{″}} (x)}{2} = f^{^{″}} (x) .

Answered by mr W last updated on 24/Jun/19

f^{″} (x) = \lim_{h \to 0} \frac{f^{'} (x + \frac{h}{2}) - f^{'} (x - \frac{h}{2})}{h}

f^{″} (x) = \lim_{h \to 0} \frac{\frac{f (x + h) - f (x)}{h} - \frac{f (x) - f (x - h)}{h}}{h}

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

Commented by ajfour last updated on 24/Jun/19

T h a n k s S i r, n i c e w a y!

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