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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.

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  • $\begingroup$ 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 $x=0$. $\endgroup$ – Brian M. Scott May 22 at 23:48
  • $\begingroup$ Should I use the same reasoning as hamam_Abdallah gave me in his hint for a)? $\endgroup$ – SocietyViper May 23 at 0:30
  • $\begingroup$ I don’t see any need to use anything but the definitions of limit and derivative. $\endgroup$ – Brian M. Scott May 23 at 0:37
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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 $$f''(0) \text{ doesn't existe}$$

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  • $\begingroup$ 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? $\endgroup$ – SocietyViper May 23 at 0:57
  • $\begingroup$ @SocietyViper Look now, i just added some lines for $b)$. $\endgroup$ – hamam_Abdallah May 23 at 1:05
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Fot this kind of problems, start with the general formula $$F_n=f(a+n h)=f(a)+ n f'(a)h+\frac{1}{2} n^2 f''(a)h^2+\frac{1}{6} n^3 f'''(a)h^3+O\left(h^4\right)$$ $$F_2-2F_1+F_0=h^2 f''(a)+h^3 f'''(a)+O\left(h^4\right)$$ $$\frac{f(a+2h)-2f(a+h)+f(a)}{h^2}=f''(a)+h f'''(a)+O\left(h^2\right) \to f''(a)$$

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