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Difference operators

For many purposes, it is convenient to think of the symbols $ \triangle,$ $ \nabla$ and $ \delta,$ defined earlier, as operators, which transform a given function $ f(x)$ into related functions, according to the laws:

$\displaystyle \triangle f(x) = f(x+h)-f(x)$

$\displaystyle \nabla f(x)=f(x)-f(x-h)$

$\displaystyle \delta f(x)=f\left(x+\frac{h}{2}\right)- f\left(x-\frac{h}{2}\right)$


In addition, we also define the following:
Averaging operator , denoted by $ \mu$ and defined as

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

Shift operator, denoted by $ E$, and defined as

$\displaystyle E f(x)=f(x+h)$

and the differential operator, denoted by $ D$ and defined as

$\displaystyle D f(x)=f'(x)$

In all these operators except D, the spacing h is implied. Positive integral powers of these operators are defined by iteration. Also we define the zeroeth power of any operator as the identity operator I, which leaves any function unchanged. For the operator $ E$, the power $ E^{\alpha}$ is defined for any real $ \alpha$ so that

$\displaystyle E^{\alpha} f(x)=f(x+\alpha h)$

assuming the existence of $ f(x+\alpha h)$. In view of the above, we also have
$\displaystyle \Delta f(x)$ $\displaystyle =$ $\displaystyle f(x+h)-f(x)$  
  $\displaystyle =$ $\displaystyle Ef(x)-f(x)$  
  $\displaystyle =$ $\displaystyle (E-I)f(x)$  

so that      $ \Delta =E-I$
Again
$\displaystyle \nabla f(x)$ $\displaystyle =$ $\displaystyle f(x)-f(x-h)$  
  $\displaystyle =$ $\displaystyle f(x)-E^{-1}f(x)$  
  $\displaystyle =$ $\displaystyle (I-E^{-1})f(x)$  

giving      $ \nabla = I-E^{-1}$
Also
$\displaystyle \delta f(x)$ $\displaystyle =$ $\displaystyle f\left(x+\frac{h}{2}\right)-f\left(x-\frac{h}{2}\right)$  
  $\displaystyle =$ $\displaystyle E^{\frac{1}{2}}f(x)-E^{\frac{-1}{2}}f(x)$  
  $\displaystyle =$ $\displaystyle (E^{\frac{1}{2}}-E^{\frac{-1}{2}})f(x)$  

and thus      $ \delta = E^{\frac{1}{2}}-E^{\frac{-1}{2}}$
Moreover,
$\displaystyle =$ $\displaystyle \frac{1}{2}\left[f\left(x+\frac{h}{2}\right)+f\left(x-\frac{h}{2}\right)\right]$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}\left[E^{\frac{1}{2}}f(x)+E^{\frac{-1}{2}}f(x)\right]$  

and thus

$\displaystyle \mu=\frac{1}{2}\left(E^{\frac{1}{2}}+ E ^{\frac{-1}{2}}\right)=E^{\frac{1}{2}}-\frac{1}{2}\delta$

Hence the operators $ \triangle$, $ \nabla$, $ \delta$ and $ \mu$, are simply expressed in terms of $ E$. From the above relations, we may deduce the relations

$\displaystyle E-\Delta =I$

$\displaystyle E(I-\nabla)=I$

$\displaystyle \left(E^{\frac{1}{2}}-\frac{1}{2}\delta\right)^{2}-\frac{1}{4}\delta^{2}=I$

$\displaystyle E^{\frac{1}{2}}-\frac{1}{2}\delta-\mu=0$

after which the formal symbolism of elementary algebra suggests the forms

$\displaystyle E=I+\Delta$

$\displaystyle E=\frac{1}{I-\nabla}$

$\displaystyle E^{\frac{1}{2}}={(I+\frac{1}{4}\delta^{2})}^{\frac{1}{2}}+\frac{1}{2}\delta $

and

$\displaystyle \mu=(I+\frac{1}{4}\delta^{2})^{\frac{1}{2}}$

while the first form requires no explanation, the form $ \displaystyle \frac{1}{(I-\nabla)}$ can be interpreted at this stage only as representing the inverse of operator $ (I-\nabla)$, that is, as an alternative notation of the operator $ (I-\nabla)^{-1}$ such that

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

whereas the derivation of the third form shows that $ \displaystyle (I+\frac{1}{4}\delta^{2})^{\frac{1}{2}}$ is to represent an operator such that its iterate is the operator $ (I+\frac{1}{4}\delta^{2})$,   $ \left[\left(I+\frac{1}{4}\delta^{2}\right)^{\frac{1}{2}}\right]^{2}=I+\frac{1}{4}\delta^{2}$

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