# Derivative and integral operators

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 9 295 - 366 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 9.31a

The operator $f_{t}=[-1,\;+1]$ is called the “derivative operator”; explain why.

### Solution

In terms of finite differences (see problem 9.29), the derivative of a function is approximately equal to the change in values between two adjacent points divided by the distance between the two points. Convolving $f_{t}$ with $g_{t}$ gives values of the finite-difference derivative of $g_{t}$ for different values of $t$ . To show this, we find the $(n+1)^{th}$ term of $h_{t}=f_{t}*g_{t}$ . From equation (9.2b), we have

{\begin{aligned}h_{n}=\mathop {\sum } \limits _{k=0}^{}f_{k}g_{n-k}=[-g_{0},\;(-g_{1}+g_{0}),\;.\;.\;.\;(-g_{n}+g_{n-1})].\end{aligned}} Each term (except the first) equals $-\Delta g$ at $=n\Delta$ . Taking $\Delta =1$ , we see that $h_{n}$ is minus the finite difference derivative at $t=n\Delta$ .

## Problem 9.31b

What is the integral operator?

### Solution

The integral, $\int _{g_{0}}^{g_{n}}\ g(t)\,\mathrm {d} t$ , for continuous functions becomes a summation for digital functions, the equivalent of the above integral being $\Delta \mathop {\sum } \limits {_{g_{0}}^{g_{n}}}g_{t}$ . If we take the operator $f_{t}=[1,1,1,\ldots ..,1]$ where both $f_{t}$ and $g_{t}$ have $(n+1)$ terms, we obtain the following for the terms of $f_{t}*g_{t}$ (omitting $\Delta$ ):

{\begin{aligned}&[(g_{0}),(g_{0}+g_{1}),...,(g_{0}+g_{1}+\cdots +g_{r}),(g_{0}+g_{1}+\cdots +g_{n}),\\&\quad (g_{1}+g_{2}+\cdots +g_{n}),(g_{2}+g_{3}+\cdots +g_{n}),...,(g_{n-1}+g_{n}),(g_{n}).\end{aligned}} The terms are equivalent to the integrals between 0 and $r$ for $0\leq r\leq n$ . After the value $r=n$ is reached, the upper limit remains $n$ while the lower limit is successively 1, 2, . . . , $n$ . Thus we conclude that the operator [1, 1, 1, . . . , 1] is an integral operator. By taking $f_{t}$ with fewer terms than $g_{t}$ , we can get the equivalent of an integral over selected parts of $g_{t}$ .