2

Suppose $f$ is differentiable and $\exists f''(0)$. I want to show $$\lim _{h \rightarrow 0} \frac{f(h)-2f(0)+f(-h)}{h^2}=f''(0)$$ not using the L'hospital rule.

I know how to calculate with L'hospital rule, but I cannot see how otherwise.

Martin Sleziak
  • 53,687
Focus
  • 1,204
  • 4
    As a sidenote, $\exists f''(0)$ is a horrible jargon of notation – Hagen von Eitzen Jun 01 '18 at 12:48
  • Use Lagrange theorem – Jakobian Jun 01 '18 at 12:51
  • @HagenvonEitzen How can I write it neatly with the result making sense? – Focus Jun 01 '18 at 13:23
  • 1
    @izimath you just say “and $f’’(x)$ exists.” Typically when you use $\exists$ you are talking about the existence of some variable satisfying an expression. The use of the symbol here doesn’t make sense. Although it may seem like what I suggested and what you wrote are the same thing, it goes against the standard existence notation – Teh Rod Jun 01 '18 at 15:54

3 Answers3

1

Since $f''(0)$ does exist, $$f(h) = f(0) + h \cdot f'(0) + \frac{h^2 \cdot f''(0)}{2!} + o(h^2)$$

$$f(-h) = f(0) - h \cdot f'(0) + \frac{h^2 \cdot f''(0)}{2!} + o(h^2)$$

Upon adding we get ,

$$f(h)+f(-h)-2f(0) = h^2 \cdot f''(0)+o(h^2)$$

$$\frac {f(h)+f(-h)-2f(0)}{h^2} = f''(0) + o(h^2)/h^2$$

Let $h\to 0$ and the desired result is achieved.

Mohammad Riazi-Kermani
  • 68,728
1

Here is one proof which is an adaptation of the proof of Taylor's theorem as given here.

Let $$g(x) =f(x) - \frac{x^2}{2}f''(0)-xf'(0)-f(0)\tag{1}$$ so that $$g(0)=g'(0)=g''(0)=0\tag{2}$$ and then $$g(h) - 2g(0)+g(-h)=f(h)-2f(0)+f(-h)-h^2f''(0)$$ so that our job is done if we show that $$\lim_{h\to 0}\frac{g(h)-2g(0)+g(-h)}{h^2}=0\tag{3}$$ Let's note that by definition of derivative we have $$g'(x) =g'(0)+xg''(0)+o(x)=o(x)$$ and by mean value theorem we have $$g(h) =g(0)+hg'(\xi)=hg'(\xi) $$ for some $\xi$ between $0$ and $h$. And using previous equation we can see that $$g(h) =ho(\xi) =o(h^2)$$ as $0<|\xi|<|h|$. And changing sign of $h$ we get $$g(-h) =o(h^2)$$ Using these we can easily see that $(3)$ holds and we are done.

Paramanand Singh
  • 87,309
0

Use two Taylor series for $f'$ at $0$ using steps $h$ and $-h$.

$$f(h) = f(0) + h \cdot f'(0) + \frac{h^2 \cdot f''(0)}{2!} + \frac{h^3 \cdot f'''(0)}{3!} + \frac{h^4 \cdot f''''(0)}{4!} ...$$ $$f(-h) = f(0) - h \cdot f'(0) + \frac{h^2 \cdot f''(0)}{2!} - \frac{h^3 \cdot f'''(0)}{3!} + \frac{h^4 \cdot f''''(0)}{4!} ...$$

Next, lets add them together.

$$f(h) + f(-h) = 2f(0) + 2\frac{h^2 \cdot f''(0)}{2!} + 2\frac{h^4 \cdot f''''(0)}{4!} ...$$

Note the odd powers of $h$ cancel. Solve for $f''(0)$:

$$f''(0) = \frac{f(h) - 2 f(0) + f(-h)}{h^2} + 2\frac{h^2 \cdot f''''(0)}{4!} ...$$

What we have so far is the derivation to the numerical approximation to the second derivative use in computational science. But, just take your limit and the second and all following terms disappear because h is in the numerator.

Dan Sp.
  • 334