# Thread: Taylor series and remainder

1. ## Taylor series and remainder

Let f be a continuous function and its third derivative exists. I need to show that [ -f(x+2h) + 4f(x+h) - 3f(x)] / 2h = f’(x) + O[h^2].

I know how to show that -f(x+2h) + 4f(x+h) - 3f(x) = 2h f’(x) by using the Taylor expansions:

f(x+2h) = f(x) + 2h f’(x) + [(2h)^2] f’’(x) / 2! + O[(2h)^3],

f(x+h) = f(x) + h f’(x) + [h^2] f’’(x) / 2! + O[h^3].

Note that O[h^(n+1)] is the remainder of the Taylor expansion, and O[h^(n+1)] = f^(n+1)(z_n) h^(n+1) / (n+1)!, where z_n is some number between x and c (since h = x-c).

How can I prove that we get O[h^2] in the first equation above? I cannot prove it by directly using the definition above. Please help me answer this question. If you know any links that may help me, please give me.

2. It seems you are looked at divided differences.

3. This link is similar to the problem. However, it does not deal with remainders (a.k.a. orders) O[h^(n+1)]. I'm struggling with proving that there is O[h^2]... Any help?

4. In general if $\displaystyle f$ is defined in the neighborhood of $\displaystyle a$ then $\displaystyle n$-th divided difference is $\displaystyle \sum_{k=0}^n \frac{{n\choose k}f(a+kh)(-1)^{n-k}}{h^n}$. So for $\displaystyle n=2$ this expression is (in numerator) $\displaystyle f(a+2h) - 2f(a+h)+f(a)$. This means if $\displaystyle f$ is (infinitly) differenciable in the neighborhood of $\displaystyle a$ then $\displaystyle f(a+2h) = f(a) + 2hf'(a)+2h^2f''(a)+E_1(h)$ where $\displaystyle E_1(h)/h^3\to 0$ as $\displaystyle h\to 0$ by Taylor's theorem. Furthermore, $\displaystyle 2f(a+h) = 2f(a)+2hf'(a)+h^2f''(a)+E_2(h)$ where $\displaystyle E_2(h)/h^3\to 0$ as $\displaystyle h\to 0$ by Taylor's theorem. This means this difference is simply $\displaystyle d(h) = f''(a)h^2 + E_1(h) - E_2(h)$. Thus $\displaystyle d(h)/h^2 \to f''(a)$.

5. I figured the problem out. Thank you.