Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Looking at a 2nd-order Taylor series approximation of the function $f$, I have this:

$$f(t_1) = f(t_0) + hf'(t_0) + {h^2\over 2}f''(t_0) + O(h^3)$$

Now say I approximate $f''(t0)$ with a 2nd-order central difference method:

$$f''(t) = {1\over 2}{f'(t+h) - f'(t-h)\over h} + O(h^2)$$

What's the resulting error of this method?

The naive approach would be to substitute the central difference equation into the Taylor series, giving something like this:

$$f(t_1) = f(t_0) + hf'(t_0) + {h\over 4}(f'(t_0+h)-f'(t_0-h)) + {1\over 2}O(h^4) + O(h^3)$$

Is that plausible? Would error actually decrease (go from 2nd-order to 4th-order)?

share|cite|improve this question
It looks like $h$ is being used differently in your Taylor polynomial and your approximation of the second derivative (and perhaps should be $t_1$ instead of $h$ in the Taylor polynomial). – Isaac Sep 28 '10 at 22:09
Since you still have the $O(h^3)$ term, the order of the error is the same as in the first formula. (Remember that $O(h^4)+O(h^3)=O(h^3)$ as $h \to 0$. – Hans Lundmark Sep 29 '10 at 6:44
up vote 2 down vote accepted

If we start with the Taylor expansion (I'll change variables here, too many subscripts confuse me):

$$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h^2}{2}f^{\prime\prime}(x)+\frac{h^3}{3!}f^{\prime\prime\prime}(x)+O(h^4)$$

and the derivative of this w.r.t. $h$

$$f^{\prime}(x+h)=f^{\prime}(x)+h f^{\prime\prime}(x)+\frac{h^2}{2}f^{\prime\prime\prime}(x)+O(h^3)$$

and the version of this with $h$ replaced by negative $h$:

$$f^{\prime}(x-h)=f^{\prime}(x)-h f^{\prime\prime}(x)+\frac{h^2}{2}f^{\prime\prime\prime}(x)+O(h^3)$$

subtracting the third expression from the second expression gives

$$f^{\prime}(x+h)-f^{\prime}(x-h)=2h f^{\prime\prime}(x)+O(h^3)$$

and we see that the even powers drop out of this error expansion.

If we solve for $f^{\prime\prime}(x)$ like so:


and substitute in the first expression,

$$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h^2}{2}\left(\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2h}+O(h^2)\right)+\frac{h^3}{3!}f^{\prime\prime\prime}(x)+O(h^4)$$

we can take the $O(h^2)$ within the parentheses out as an $O(h^4)$ term:

$$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h}{2}\left(\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2}\right)+\frac{h^3}{3!}f^{\prime\prime\prime}(x)+O(h^4)$$

the leading term after the replaced portion is $O(h^3)$, thus simplifying to

$$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h}{2}\left(\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2}\right)+O(h^3)$$

and we see that the formula has $O(h^3)$ error: cutting $h$ in half decreases the error by a factor of $2^3=8$.

share|cite|improve this answer
I don't see what this answer has to do with the question. The question wasn't about what the central difference approximation for $f''(x)$ is in terms of $f$, it was what the order of approximation is in the expression $f(t_0+h) \approx f(t_0) + hf'(t_0) + {h\over 4}(f'(t_0+h)-f'(t_0-h))$. – Hans Lundmark Sep 29 '10 at 11:10
...and the answer is $O(h^3)$. – Hans Lundmark Sep 29 '10 at 11:16
@Hans: there, fixed. I seem to have been thinking of something else while I wrote the first version of this answer. :) Thanks for the heads-up! – J. M. Sep 29 '10 at 11:42
In your formula for $f''(x)$, you've forgotten to divide the remainder term by $h$; it should be $O(h^2)$ instead of $O(h^3)$. Also, in the second to last formula $f(x+h)=\ldots+O(h^3)+O(h^4)$, I don't see where the $O(h^3)$ comes from except that $(h^3/3!) f'''(x)=O(x^3)$; I guess you mean the term that arises from $(h^2/2) O(h^3) = O(h^5)$ (although according to my previous sentence that should in fact be $(h^2/2) O(h^2) = O(h^4)$). – Hans Lundmark Sep 29 '10 at 15:25
All fixed, thanks for the corrections @Hans. – J. M. Sep 29 '10 at 15:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.