A function $f:\mathbb{R} \rightarrow \mathbb{R}$ is $C^3$ with $$f(a+h)=f(a)+f'\left(a+\dfrac{1}{2}h\right)h$$ whenever $a \in \mathbb{R}$ and $h \geq 0$. By applying Taylors Theorem to $f$ and to $f'$ or otherwise, show that the third derivative $f'''$, of $f$ is identically zero.
My Attempt
I am confused as to how to apply taylors theorem here, as I can't seem to make out what $f(x)$ is... and what value would I apply Taylor theorem around?
Any help?
On
By Taylor's theorem $$f(x+h)=f(x)+hf'(x)+h^2/2!f''(x)+h^3/3!f'''(x)+o(h^3)\\ f'(x+h/2)=f'(x)+h/2f''(x)+h^2/4 f'''(x)+o(h^2)$$By the given condition then you have $$f'(x)+h/2f''(x)+h^2/4 f'''(x)+o(h^2)=f'(x)+h/2!f''(x)+h^2/3!f'''(x)+o(h^3)/h\\ \Rightarrow \frac{h^2}{12}f'''(x)=o(h^3)/h-o(h^2)\Rightarrow f'''(x)=\frac{o(h^3)}{h^3}-\frac{o(h^2)}{h^2}\ \forall h>0$$
Consider the Taylor expansion of $f$ in any point $a$ yielding $$ f(a+h)=f(a)+f'(a)h+\frac{f''(a)}{2}h^2+\frac{f'''(a)}{3!}h^3+O(h^4) $$ From the identity you stated, it follows that $$ f'(a)h+\frac{f''(a)}{2}h^2+\frac{f'''(a)}{3!}h^3+O(h^4)=f'\left(a+\frac{h}{2}\right)h $$ Now if you expand the right side you find $$ f'\left(a+\frac{h}{2}\right)h=f'(a)h+f''(a)\frac{h^2}{2}+\frac{f'''(a)}{4}h^3+O(h^4) $$
Comparing the coefficients shows that $f'''$ must vanish identically.