How to prove the following statements using the Taylor polynomial.

a) I have to prove that if $$f''(a)$$ exists, then $$f''(a)=\lim_{h \to 0}\frac{f(a+2h)-2f(a+h)+f(a)}{h^2}$$ and the hint that they gave me is to use Taylor polynomial of grade 2 at $$a$$, with $$x = a + h$$ and $$x = a - h$$. First, I know that if $$f''(a)$$ exists, then, by definition, $$f''(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$ Then, I tried to do the Taylor polynomial of $$f(a+h) = f(x)$$, so $$P_{2,a}(x) = \frac{f(a)}{0!} + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 = f(a) + f'(a)(x-a) + \frac{1}{2}f''(a)(x-a)^2$$ And I got stuck in this point, because of the fact that $$x = a + h$$ and $$x = a - h$$. How do I continue?

b) And the second statement is: Let $$f(x) = x^2$$ for $$x \geq 0$$, and $$-x^2$$ for $$x \leq 0$$. Prove that $$\lim_{h \to 0} \frac{f(0 + h) + f(0 - h) - 2f(0)}{h^2}$$ exists, while $$f''(0)$$ doesn't exists. In this one, I don't really know where to start, so, I'm just looking for a hint if it's possible.

• For the second one, the first step is to simplify the displayed limit, figure out what it is, and then prove it. Then find $f'$ and show that it’s not differentiable at $x=0$. – Brian M. Scott May 22 at 23:48
• Should I use the same reasoning as hamam_Abdallah gave me in his hint for a)? – SocietyViper May 23 at 0:30
• I don’t see any need to use anything but the definitions of limit and derivative. – Brian M. Scott May 23 at 0:37

hint

If $$f''(a)$$ exists, the by Taylor-Young formula,

$$f(a+h)=f(a)\color{red}{+}hf'(a)+\frac{h^2}{2}f''(a)+h^2\epsilon(h)$$

$$f(a-h)=f(a)\color{red}{-}hf(a)+\frac{h^2}{2}f''(a)+h^2\epsilon(h)$$

$$f(a+h)+f(a-h)=...$$

with $$\lim_{h\to 0}\epsilon(h)=0$$

For $$b)$$,

$$\lim_{h\to0}\frac{f(h)+f(-h)-2f(0)}{h^2}=$$ $$\lim_{h\to0}\frac{h^2+(-h^2)}{h^2}=0$$

but for $$x\ge 0$$, $$f'(x)=2x$$ and

for $$x\le 0$$, $$f'(x)=-2x$$, thus

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

$$\ne \lim_{x\to 0^-}\frac{f'(x)-f'(0)}{x-0}=-2$$ therefore $$f''(0) \text{ doesn't existe}$$

• This worked! Now, I used the Taylor Polynomial for b), and I obtained that the limit is equal to 2, but there is no terms f′′(0). So then I could say that f′′(0) doesn't exist. Is that correct? – SocietyViper May 23 at 0:57
• @SocietyViper Look now, i just added some lines for $b)$. – hamam_Abdallah May 23 at 1:05

Fot this kind of problems, start with the general formula $$F_n=f(a+n h)=f(a)+ n f'(a)h+\frac{1}{2} n^2 f''(a)h^2+\frac{1}{6} n^3 f'''(a)h^3+O\left(h^4\right)$$ $$F_2-2F_1+F_0=h^2 f''(a)+h^3 f'''(a)+O\left(h^4\right)$$ $$\frac{f(a+2h)-2f(a+h)+f(a)}{h^2}=f''(a)+h f'''(a)+O\left(h^2\right) \to f''(a)$$