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Maybe this question is dumb but I dont know how to approach it. In a book it is stated that, for $f\in C^1(\Bbb R,\Bbb R)$ two times differentiable at $c$ then it holds that

$$f''(c)=\lim_{h\to 0^+}\frac{f(c+2h)-2f(c+h)+f(c)}{h^2}\tag1$$

But I'm unable to justify the above formula from the natural one

$$f''(c)=\lim_{h\to 0}\lim_{s\to 0}\frac{f(c+h+s)-2f(c+s)+f(c)}{hs}\tag2$$

Can someone show me (or give a reference) about how to prove $(1)$? I'm also very intrigued by the reasons about why it is stated the limit in $(1)$ as just a lateral limit. Thank you.

Masacroso
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  • (1) is just a special case of (2) under the assumption that the derivate exists. In that case any way you approach $c$ will result in the same value (uniqueness of the limit). If it is not known that the limit exists (i.e. if the statement reads that $f$ is twice differentiable if the limit (1) exists than additional work is required). – Thomas Sep 12 '18 at 12:13
  • @Thomas I dont see why it is the same a double limit that a simple one. Ok about the uniqueness of the limit but I cant see why it must hold. Maybe it is a dumb question but I dont grasp it. – Masacroso Sep 12 '18 at 12:15

1 Answers1

1

Hint Do you know derivative of a function at a point can be approximated by

$$f(a+h)-f(a)-f'(a)h=h\epsilon\tag 1$$ where $\epsilon$ is a error term

Write the similar expression for second derivative,

$$f'(a+h)-f'(a)-f''(a)h=h\epsilon'$$

Some more work,

Now $$hf'(a+h)-hf'(a)-f''(a+h)h^2=h^2\epsilon$$ Replace $$hf'(a+h)=f(a+2h)-f(a+h)-h\epsilon$$ we get $$f(a+2h)-f(a+h)-h\epsilon -f(a+h)+f(a)+h\epsilon-f''(a)h^2=h^2\epsilon'$$ $$f(a+2h)-2f(a+h)+f(a)-f''(a)h^2=h^2\epsilon'$$

Dividing by $h^2$, you get the desired expression.

Shweta Aggrawal
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