Relations between Difference operators

  1. We note that

    $\displaystyle Ef(x)=f(x+h) \;\; = [f(x+h)-f(x)]+f(x) \;\; = \Delta f(x)+f(x)=
(\Delta +1)f(x).$

    Thus,

    $\displaystyle \framebox[1in][c]{ $E \equiv 1 + \Delta$ } \;\; {\mbox{ or }} \;\;
\Delta \equiv E-1. $

  2. Further, $ \nabla (E(f(x)) = \nabla(f(x+h)) = f(x+h) - f(x).$ Thus,

    $\displaystyle (1-\nabla) Ef(x)= E(f(x)) - \nabla(E(f(x)) = f(x+h) - [f(x+h)-f(x)] = f(x).$

    Thus $ E \equiv 1 + \Delta,$ gives us

    $\displaystyle (1-\nabla)(1+\Delta) f(x)=f(x)\; {\mbox{ for all }} \; x.$

    So we write,

    $\displaystyle (1+\Delta)^{-1}=1-\nabla \,\,\,{\mbox{ or }} \;\;
\framebox[1.5in][c]{$ \nabla=1-(1+\Delta)^{-1},$} \;\; {\mbox{ and }}$

    $\displaystyle (1-\nabla)^{-1}=1+\Delta =E.$

    Similarly,

    $\displaystyle \Delta=(1-\nabla )^{-1}-1.$

  3. Let us denote by $ E^{\frac{1}{2}}f(x)=f(x+ \frac{h}{2}).$ Then, we see that

    $\displaystyle \delta f(x)=f(x+ \frac{h}{2})-f(x-\frac{h}{2}) = E^{\frac{1}{2}}f(x)-
E^{-\frac{1}{2}}f(x). $

    Thus,

    $\displaystyle \;\; \framebox[1.4in][c]{$\delta =
E^{\frac{1}{2}}-E^{-\frac{1}{2}}.$}$

    Recall,

    $\displaystyle \delta ^2 f(x)=f(x+h)-2f(x)+f(x-h) = [f(x+h)+2f(x)+f(x-h)]-4f(x)
= 4(\mu^2-1)f(x).$

    So, we have,

    $\displaystyle \framebox[1in][c]{$\mu^2\equiv \frac{\delta^2}{4}+1$} \;\; {\mbox{ or }}
\;\; \framebox[1in][c]{$\mu \equiv \sqrt{1+\frac{\delta^2}{4}}$}.$

    That is, the action of $ \displaystyle{\sqrt{1+\frac{\delta^2}{4}}}$ is same as that of $ \mu.$
  4. We further note that,
    $\displaystyle \Delta f(x)$ $\displaystyle =$ $\displaystyle f(x+h)-f(x) = \frac{1}{2}\bigl[f(x+h)-2f(x)+f(x-h)
\bigr]+ \frac{1}{2}\bigl[f(x+h)-f(x-h)\bigr]$  
      $\displaystyle =$ $\displaystyle \frac{1}{2} \delta^2 (f(x)) +
\frac{1}{2}\bigl[f(x+h)-f(x-h)\bigr]$  

    and
    $\displaystyle \delta \mu f(x)$ $\displaystyle =$ $\displaystyle \delta\left[\frac{1}{2}\left\{f(x+ \frac{h}{2})+
f(x-\frac{h}{2})\right\} \right] =
\frac{1}{2}\bigl[\{f(x+h)-f(x)\}+\{f(x)-f(x-h)\}\bigr]$  
      $\displaystyle =$ $\displaystyle \frac{1}{2}\left[f(x+h)-f(x-h)\right].$  

    Thus,

    $\displaystyle \Delta f(x) = \left[\frac{1}{2}\delta^2+\delta
\mu\right]f(x),$

    i.e.,

    $\displaystyle \Delta \equiv \frac{1}{2} \delta
^2 +\delta \mu \equiv \frac{1}{2} \delta ^2
+\delta\sqrt{1+\frac{\delta^2}{4}}.$

    In view of the above discussion, we have the following table showing the relations between various difference operators:

      E $ \Delta$ $ \nabla$ $ \delta$
    E E $ \Delta+1$ $ (1-\nabla)^{-1}$ $ \frac{1}{2} \delta ^2 +\delta\sqrt{1+\frac{\delta^2}{4}}+1 $
    $ \Delta$ $ E-1$ $ \Delta$ $ (1-\nabla)^{-1}-1$ $ \frac{1}{2} \delta ^2 +\delta\sqrt{1+\frac{1}{4}\delta^2} $
    $ \nabla$ $ 1-E^{-1}$ $ 1-(1+\nabla)^{-1}$ $ \nabla$ $ -\frac{1}{2} \delta ^2 +\delta\sqrt{1+\frac{1}{4}\delta^2} $
    $ \delta$ $ E^{1/2}-E^{-1/2} $ $ \Delta(1+\Delta)^{-1/2}$ $ \nabla(1-\nabla)^{-1/2}$ $ \delta$

EXERCISE 11.3.1  
  1. Verify the validity of the above table.
  2. Obtain the relations between the averaging operator and other difference operators.
  3. Find $ \Delta^2 y_2, \; \nabla^2 y_2, \; \delta^2 y_2$ and $ \mu^2 y_2$ for the following tabular values:
    $ i$ 0 1 2 3 4
    $ x_i$ 93.0 96.5 100.0 103.5 107.0
    $ y_i$ 11.3 12.5 14.0 15.2 16.0

A K Lal 2007-09-12