# Discrete Derivative

Standard

I came across the following interesting question in the book “Math Girls” by Hiroshi Yuki :

Develop a definition for the differential operator $\Delta$ in discrete space,corresponding to the definition of the differential operator D in the continuous space.

We know that derivative of a function f at point x is the rate at which f changes at the point x . Geometrically, derivative at a point x is the slope of the tangent to the function f at x where a tangent is the limit of the secant lines as shown below :

But this happens only in the continuous world where x “glides smoothly” from one point to another. But this is not the case in a discrete world. In the discrete world there is nothing like “being close to each other”. Hence we cannot use the earlier definition of bringing h arbitrarily close to x. In a discrete world we cannot talk about getting “close” to something but instead we can talk about being “next” to each other.

We can talk about the change in x as it moves from x to x+1 while f changes from f(x) to f(x+1). We do not need limits here, so the definition of “difference operator” (analogous to differential operator ) will be :

$\Delta f(x) = \frac{ f(x+1) - f(x) }{(x+1) - x} = f(x+1) - f(x)$

Hence to find derivative of a function, say $g(x) = x^2$ , it is easy to verify that $Dg(x) = 2x$ but $\Delta g(x) = 2x + 1$ (using definitions mentioned above)

Now, when will we be able to get the same derivative in both discrete and continuous worlds? I read a little about this question in math girls and a little more in “An introduction to the calculus of finite differences” by C.H.Richardson.

Calculus of differences is the study of the relations that exist between the values assumed by the function whenever the independent variable takes on a series of values in arithmetic progression.

Let us write f(x) as $f_x$ instead from now onwards. So $f(x+1) - f(x) = \Delta f(x) = f_{x+1} - f_x$. Using above definition we can prove the following for functions $U_x$ and $V_x$ :

1) $\Delta^{k+1} U_x = \Delta^{k} U_{x+1} - \Delta^{k} U_x$

2) $\Delta (U_x + V_x) = \Delta U_x + \Delta V_x$ (or) $\Delta^k (U_x + V_x) =\Delta^k U_x + \Delta^k V_x$

3) $\Delta^k (cU_x) = c \Delta^k U_x$

Theorem. $\Delta^n x^n = n!$

Proof. $\Delta x^n = (x+1)^n - x^n = n\cdot x^{n-1} + \text{terms of degree lower than} (n - 1)$. Each repetition of the process of differencing reduces the degree by one and also adds one factor to the succession $n(n - 1) (n - 2) \cdots$. Repeating the process $n$ times we have $\Delta^k x^n = n!$.

Corollary 1. $\Delta^n ax^n = a\cdot n!$

Corollary 2. $\Delta^{n+1} x^n = 0$

Corollary 3. If $U_x$ is a polynomial of degree n i.e. $U_x= a_0+ a_1 x + a_3 x + \ldots + a_n x^n$ , then $\Delta^n U_x = a_n\cdot n!$.

We call the continued products $U_x^{|n|} = U_x\cdot U_{x+1}\cdot U_{x+2} \cdots U_{x+(n-1)}$ and $U_x^{(n)} = U_x \cdot U_{x-1}\cdot U_{x-2}\cdots U_{x-(n-1)}$ as factorial expressions.

If $U_x$ is the function ax+b for some real numbers a and b, then the factorial forms we get by replacing $U_x$ by ax+b is $(ax+b)^{|n|} = (ax+b)\cdot(a(x+1)+b)\cdot (a(x+2)+b)\cdots (a(x+n-1)+b)$ and $(ax+b)^{(n)} =(ax+b)\cdot (a(x-1)+b)\cdot (a(x-2)+b)\cdots (a(x-(n-1))+b)$.

We define $(ax+b)^{|0|}$ and $(ax+b)^{(0)}$ as 1.

Using the above definition of factorial we can show the following :

(i) $\Delta (ax+b)^{(n)} = a\cdot n \cdot (ax+b)^{(n-1)}$

(ii) $\displaystyle{\Delta \frac{1}{(ax+b)^{|n|}} = \frac{-an}{(ax+b)^{|n+1|}}}$

When we consider the special case of a=1 and b=0, the factorial representations are called raising and falling factorials :

$x^{|n|} = x \cdot (x+1)\cdot (x+2)\cdots (x+n-1)$ – rising factorial

$x^{(n)} =x\cdot (x-1) \cdot (x-2) \cdots (x-n+1)$ – falling factorial.

Substituting a=1 and b=0 in (i) and (ii) above , we get that

$\Delta x^{(n)} = n\cdot x^{(n-1)}$ , $\Delta^n x^{(n)} = n!$ and $\displaystyle{\Delta \frac{1}{x^{|n|}} = - \frac{n}{x^{|n+1|}}}$.

Summary:

source: Richardson, C. H. An introduction to the calculus of finite differences. pp. 10.

Due to the fact that $x^{(n)}$ plays in the calculus of finite differences a role similar to that played by $x^n$ in the infinitesimal calculus, for many purposes in finite differences it is advisable to express a given polynomial in a series of factorials. A method of accomplishing this is contained in Newton’s Theorem.

source: Richardson, C. H. An introduction to the calculus of finite differences. pp. 10.

Since these differences and $U_x$ are identities, they are true for all values of x, and consequently must hold for x = 0. Setting x = 0 in the given function and the differences, we have the required values for all $a_i$ and theorem is proved.