Question

Suppose that 𝑓 is differentiable and 𝑓 ′′(𝑎) exists. Prove that 𝑓 ′′(𝑎) = lim ℎ→𝑎 (𝑓(𝑎+ℎ)−2𝑓(𝑎)+𝑓(𝑎−ℎ))/ ℎ^2. Give an example where the above limit exists, but 𝑓 ′′(𝑎) does not exist.

Answer 1

$f\left( {x + h} \right) = f\left( x \right) + hf'\left( x \right) + \frac{{{h^2}}}{{2!}}f''\left( x \right) + O\left( {{h^3}} \right) \qquad (i) \\ f\left( {x - h} \right) = f\left( x \right) - hf'\left( x \right) + \frac{{{h^2}}}{{2!}}f''\left( x \right) -O\left( {{h^3}} \right) \qquad (ii) \\ \\ \text{adding } (i) ~\text{and } (ii) f\left( {x + h} \right) + f\left( {x - h} \right) = 2f\left( x \right) + \frac{{{h^2}}}{{2!}}f''\left( x \right) \\ \text{or} f''\left( x \right) = \frac{{f\left( {x + h} \right) - 2f\left( x \right) + f\left( {x - h} \right)}}{{{h^2}}} \\ \text{Taking the Limit as } h \to 0 \\ f''\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - 2f\left( x \right) + f\left( {x - h} \right)}}{{{h^2}}} \\ \Rightarrow f''\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - 2f\left( a \right) + f\left( {a - h} \right)}}{{{h^2}}} \\ \text{We need to find }f \text{ such that } f'\left( x \right) = \left| x \right|. \\ \Rightarrow f\left( x \right) = \int_0^x {\left| t \right|dt} = \frac{{{x^2}{\mathop{\rm sgn}} \left( x \right)}}{2} \\ \text{Since } \mathop {\lim }\limits_{h \to 0} f\left( {0 + h} \right) - 2f\left( 0 \right) + f\left( {0 - h} \right) = \frac{{{h^2}}}{2} - \frac{{{h^2}}}{2} = 0 \\ \text{and } \mathop {\lim }\limits_{h \to 0} {h^2} = 0, \\ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - 2f\left( a \right) + f\left( {a - h} \right)}}{{{h^2}}}\underline{\underline {L.H}} \mathop {\lim }\limits_{h \to 0} \frac{{\frac{d}{{dh}}\left( {f\left( {a + h} \right) - 2f\left( a \right) + f\left( {a - h} \right)} \right)}}{{\frac{d}{{dx}}\left( {{h^2}} \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{{f'\left( h \right) + f\left( { - h} \right)}}{{2h}} \\ = \mathop {\lim }\limits_{h \to 0} \frac{{f'\left( h \right) - f\left( { - h} \right)}}{{2h}} = \mathop {\lim }\limits_{h \to 0} \frac{{\left| h \right| + \left| { - h} \right|}}{{2h}} = 0 \\ \text{Clearly, } f''\left( 0 \right) \text{ does not exist since } \left| x \right| \text{ is not differentiable at point 0}$