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Question: Let $f(x)$ be a twice differentiable function on $\mathbb{R}$. Show that for any $h>0$ and any $x\in\mathbb{R}$ there exists a $\zeta$ between $x-h$ and $x+h$ such that $$ \frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(\zeta). $$

My Attempt: For $a \in (x-h,x) $ and $b \in (x,x+h)$ by Mean Value Theorem: $$ f'(a)=\frac{f(x)-f(x-h)}{h} $$ $$ f'(b)=\frac{f(x+h)-f(x)}{h} $$ Then, because $f$ is twice differentiable $\exists \zeta \in(a,b) \subset (x-h,x+h)$ s.t. $$ f''(\zeta)=\frac{f'(b)-f'(a)}{b-a} =\frac{\frac{f(x+h)-f(x)}{h}-\frac{f(x)-f(x-h)}{h}}{b-a}=\frac{f(x+h)-2f(x)+f(x-h)}{h(b-a)}$$ can we now, choose $a$ and $b$ to be $h$ apart? I don't think Mean Value Theorem allows us to choose it this way...

If I use the derivative definition I'm not sure how to get rid of the limit. I also think the "linearization" formula may be useful something like $f(a)+f'(a)(x-a)+1/2 f''(\zeta)(x-a)^2$ ?

elcharlosmaster
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  • You're definitely correct that the MVT does not allow you to choose this way. Not sure how to fix it yet though – MathTrain Aug 27 '19 at 22:40

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By Lagrange Remainder Theorem for Taylor polinomial you have:

$$\exists a\in [x,x+h] \mid f(x+h)=f(x)+f’(x)h+\frac{f’’(a)}2h^2\iff\frac{f(x+h)-f(x)-f’(x)h}{h^2}=\frac{f’’(a)}2$$ And in the same way $$\exists b\in [x-h,x] \mid f(x-h)=f(x)-f’(x)h+\frac{f’’(b)}2h^2\iff\frac{f(x-h)-f(x)+f’(x)h}{h^2}=\frac{f’’(b)}2$$ Summing these last two expression we have $$\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=\frac{f’’(a)+f’’(b)}2$$ But $\frac{f’’(a)+f’’(b)}2$ stays between $f’’(a)$ and $f’’(b)$ so for Intermediate Value Theorem $\exists \zeta\in [a,b]\subset [x-h,x+h]$ so that $$f’’(\zeta)=\frac{f’’(a)+f’’(b)}2=\frac{f(x+h)-2f(x)+f(x-h)}{h^2}$$

Alessandro Cigna
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