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

Let $r(t)$ be a real-valued function of $t$. Let $v(t)$ be the derivative of $r(t)$. Then $$v(t) = \frac{dr(t)}{dt} = \lim_{\Delta t \to 0} \frac{r(t + \Delta t) - r(t)}{\Delta t}$$ so $$v(t) = \frac{dr(t)}{dt} \approx \frac{r(t + \Delta t) - r(t)}{\Delta t} \text{ for small }\Delta t$$

My question is, is there another way to approximate $v(t) = \dfrac{dr(t)}{dt}$?

For example, I am reading the book Understanding Molecular Simulation by Frenkel and Smit (Second Edition). On page 71 (some pages are available on Google Books here), the authors write $$v(t) = \frac{r(t + \Delta t) - r(t - \Delta t)}{2 \Delta t} + \mathcal{O}(\Delta t^2)$$ or, in other words, $$v(t) \approx \frac{r(t + \Delta t) - r(t - \Delta t)}{2 \Delta t}$$

Basically, then, it seems that there are two ways to express $v(t) = \dfrac{dr(t)}{dt}$:

$$v(t) = \frac{dr(t)}{dt} = \lim_{\Delta t \to 0} \frac{r(t + \Delta t) - r(t)}{\Delta t} \textbf{ (1)}$$

$$v(t) = \frac{dr(t)}{dt} = \lim_{\Delta t \to 0} \frac{r(t + \Delta t) - r(t - \Delta t)}{2 \Delta t} \textbf{ (2)}$$

Are equations (1) and (2) equivalent? Equation (1) Is the definition of derivative that I remember from high school calculus; I do not remember (2). Is (2) an alternative definition of the derivative? Or, what is the relationship between (1) and (2)?

share|cite|improve this question
i can think about one thing,they are same according to following situation,consider difference between $5$ and $4$ difference is $1$ while difference between $5$ and $3$ is $2$ or $2*1$,i think the same principle is here – dato datuashvili Aug 7 '12 at 20:33
up vote 3 down vote accepted

There are many ways to write a modified version of the definition of the derivative that have in fact higher convergence rates and are thus often desirable to use in numerical computation.

Finite difference coefficients can be used for this. For example, we have



$$f'(x_{0}) \approx \displaystyle \frac{-\frac{11}{6}f(x_{0}) + 3f(x_{+1}) -\frac{3}{2}f(x_{+2}) +\frac{1}{3}f(x_{+3}) }{h_{x}} + O\left(h_{x}^3 \right)$$

Addendum: Here is a giant list of high-accuracy derivatives using these coefficients.

share|cite|improve this answer
Thanks. Does this mean that equation (2) in the original post is a correct definition of the derivative? – Andrew Aug 7 '12 at 21:14
@Andrew Yes, very much so! This is a basic three-point estimate. In fact, this derivative formula is mentioned here:… - the error seems to be $$\frac{f^{(3)}(c)h^2}{6}$$ where $$c \in [x-h, x+h]$$ – Argon Aug 7 '12 at 21:17
Thanks! So, based on the link you provided, it seems that equation (2), a three-point estimate, is an actually more accurate approximation of the derivative than equation (1), a two-point estimate. Is this true? – Andrew Aug 7 '12 at 21:24
@Andrew Yes, it is true. Instead of having a bias to the side that the limit is taken, this method has a balance. – Argon Aug 7 '12 at 21:27
"...thus often desirable to use in numerical computation." - not always. To borrow a mantra from a certain popular book on numerics, high order does not necessarily imply high accuracy. One must remember that the basis of the high-order approximations is the use of an interpolating polynomial at equispaced points, and it is well-known how unruly high-order interpolating polynomials can be... – J. M. Aug 7 '12 at 23:37

If the limit (1) exists, the limit (2) exists. To see this, write $$ \frac{r(t+h)-r(t-h)}{2h}=\frac12\left(\frac{r(t+h)-r(t)}{h}+\frac{r(t+(-h))-r(t)}{(-h)}\right). $$ The other implication is wrong, otherwise every even function would be differentiable at $t=0$ with derivative zero (but $t\mapsto|t|$ is a counterexample).

Hence (2) is not a definition of the derivative. The definition of the derivative is (1). But as soon as (1) holds, the derivative coincides with the limit $$ \lim\limits_{h\to0}\frac{r(t+ah)-r(t-bh)}{(a+b)h}, $$ for every $(a,b)$ such that $a+b\ne0$, and in particular with the limit in (2).

share|cite|improve this answer

As a complement to other fine answers (did's for example) let's suppose that $f$ admits a Taylor expansion at $t$ then :

$$f(t+h)=f(t)+hf'(t)+\frac {h^2}2 f''(t)+\frac {h^3}6 f'''(t) +O\left(h^4 f''''(t)\right)$$

so that : $$\tag{1}\frac{f(t+h)-f(t)}h=f'(t)+\frac {h}2 f''(t)+\frac {h^2}6 f'''(t) +O\left(h^3 f''''(t)\right)$$ while : $$\tag{2}\frac{f(t+h)-f(t-h)}2=f'(t)+\frac {h^2}6 f'''(t) +O\left(h^3 f''''(t)\right)$$

That is why the second method is more precise : the $f''$ term disappeared (well all the even terms in fact!) so that $f'$ will be evaluated with more precision (with an $h^2$ error instead of $h$). Because of the precision obtained Feynman proposed this second method to evaluate derivatives in his famous Physics Lectures (Vol I 9-6).

But when there is no convergence ($f(t)=\frac 1t$ for example) you'll get a limit in the second case ($0$) without any problem and that's clearly different of the first case.

The second method is much used too when you need a 'discretized' version of a differential equation (respecting time symmetry, energy conservation and so on). Ed Fredkin for example proposed following equation in his article "Feynman, Barton and the reversible Schrödinger difference equation" : $$\frac{C_{x,t+1}-C_{x,t-1}}2=ik\left(C_{x-1,t}-2C_{x,t}+C_{x+1,t}\right)$$

share|cite|improve this answer
@Raymound:These techniques are used in numerical methods. – Mhenni Benghorbal Aug 7 '12 at 23:11
@MhenniBenghorbal: yes you are right (I insisted on the symmetric part of the second method), – Raymond Manzoni Aug 7 '12 at 23:13
Oh, the central difference is very useful in numerics indeed. See this. – J. M. Aug 7 '12 at 23:35
Thanks for the link(s) @J.M. ! – Raymond Manzoni Aug 7 '12 at 23:37

Consider the linear terms in the Taylor series of $r(t+h)$ and $r(t-h)$, where $h = \Delta t$ and $r'(t)=v(t)$, $$ r(t+h) \approx r(t) + r'(t)\,h $$ and $$ r(t-h) \approx r(t) - r'(t)\,h $$ Multiplying the second equation by $-1$ and adding $$ 2h r'(t) = r(t+h) + r(t-h) \Rightarrow v(t) \approx \frac{r(t+h) + r(t-h)}{2h}\,, $$

which implies

$$ v(t) = \lim_{h \to 0} \frac{r(t+h) + r(t-h)}{2h} $$

share|cite|improve this answer

Here is yet another equivalent definition: $f'(x) = \lim_{r \to 1} \frac{f(rx) - f(x)}{(r-1)x}$, if $x \ne 0$. The differentiation rule $\frac{d}{dx}x^n = n x^{n-1}$ is especially easy to derive with this definition.

share|cite|improve this answer
Thanks. Are you implying that equation (2) in the original post is indeed a correct definition of a derivative? The answer by did seems to imply it is not. – Andrew Aug 7 '12 at 20:54
@Andrew, it isn't a good definition. See this. – J. M. Aug 7 '12 at 23:42
@Andrew: When you have questions about some parts of an answer, you might as well ask directly the author. – Did Aug 24 '12 at 15:45

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.