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I want to prove that if $f''$ is continuous at $x_0$, then $$f''(x_0)=\displaystyle\lim_{h\to 0}\dfrac{f(x_0+2h)+f(x_0)-2f(x_0+h)}{h^2}$$

Any hint to prove it?

I can't use l'Hopital or Taylor.

Thanks.

Surtan
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3 Answers3

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Consider that $f'(x)=\lim\limits_{h\to0} \frac{f(x+h)-f(x)}h$ and $f''(x)=\lim\limits_{h\to0} \frac{f'(x+h)-f'(x)}h$. Subsituting the first into the latter should lead you in the right direction.

Pauly B
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  • +1 But be sure that when you substitute the one into the other, you use different limiting variables. From there you have an assumption about $f''$ that allows you to make them both be $h$. – 2'5 9'2 Oct 31 '14 at 07:13
  • Can you explain in more detail how this approach will solve the problem considering the comment by alex.jordan. I tried to continue using your technique but was not able to make the both the limits variables as $h$. – Paramanand Singh Nov 01 '14 at 08:32
  • @alex.jordan: can you please let me know how the continuity of $f''$ helps us to make both the limit variables to be $h$? I was not able to figure that out. – Paramanand Singh Nov 01 '14 at 08:33
  • @ParamanandSingh Show that in this situation, $\lim\limits_{h_1\to0}\lim\limits_{h_2\to0}\text{stuff}=\lim\limits_{(h_1,h_2) \to (0,0)}\text{stuff}$. Then you are free to choose any path to $(0,0)$ that you want to evaluate the limit, including the path $(h,h)$ as $h\to0$. – 2'5 9'2 Nov 01 '14 at 18:05
  • @alex.jordan: now I understand it is same as proof for equality of second order partial derivatives of $F(x,y)=f(x+y)$ at point $(x_{0},0)$. I perhaps had lost touch with double limit theorems. Thanks for reminding. – Paramanand Singh Nov 02 '14 at 01:32
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$$\lim_{h\to 0}\frac{f(x_0+2h)+f(x_0)-2f(x_0+h)}{h^2} \\=\lim_{h\to 0}\frac{\frac{f(x_0+2h)-f(x_0+h)}h-\frac{f(x_0+h)-f(x_0)}h}{h} \\=\lim_{h\to 0}\frac{f'(x_0+h)-f'(x_0)}{h} \\=f''(x_0)$$

RE60K
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We can proceed as follows. Let $$F(h,k)=f(x+h+k)-f(x+h)-f(x+k)+f(x)$$ and we need to show that $$f''(x)=\lim_{h\to 0}\frac{F(h,h)}{h^{2}}$$ Now let $$G(k)=f(x+h+k)-f(x+k)$$ so that $$\begin{aligned}F(h,k)&=G(k)-G(0)\\ &=kG'(ak)\text{ (where } 0<a<1)\\ &= k\{f'(x+h+ak)-f'(x+ak)\}\\ &= hkf''(x+ak+bh)\text{ (where }0<b<1)\end{aligned}$$ It then follows that $$\frac{F(h,h)}{h^{2}}=f''(x+ah+bh)$$ Now by continuity of $f''$ it is clear that RHS of the above equation tends to $f''(x)$ as $h\to 0$.

Paramanand Singh
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