show that for twice differentiable function f,
lim_{h rightarrow 0} (f(a+2h)-2f(a+h)+f(a))/h ^{2} = f'' (a)

Question

show that for twice differentiable function f,
lim_{h rightarrow 0} (f(a+2h)-2f(a+h)+f(a))/h ^{2} = f'' (a)

dumbcow · Answer

its the limit definition of 2nd derivative...
$$f''(a) = \lim_{h \rightarrow 0}\frac{f'(a+h) - f'(a)}{h}$$
where
$$f'(a) = \lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h}$$
thus
$$\large f''(a) = \lim_{h \rightarrow 0}\frac{\frac{f(a+2h) - f(a+h)}{h}-\frac{f(a+h)-f(a)}{h}}{h}= \frac{f(a+2h)-2f(a+h)+f(a)}{h}$$

dumbcow · Answer

sorry the last was cut off and it should have h^2 on bottom