Action of an anti-difference operator

[flower321]

hi, i want to ask you that how to perform an anti-difference operator on the product of two functions? i.e.
D^{-1}(f_{n} g_{n})

 

[member 587159]

hi, i want to ask you that how to perform an anti-difference operator on the product of two functions? i.e.
D^{-1}(f_{n} g_{n})

What do you mean, find the antiderivative of a function $(fg)(x)$? There is no standard formula for this, but you might want to google: integration by parts.

 

[flower321]

What do you mean, find the antiderivative of a function $(fg)(x)$? There is no standard formula for this, but you might want to google: integration by parts.

no this is not a usual calculas, this is discrete calculus, i.e. Df_{n} = f_{n+1}-f_{n}, where D is the forward difference operator, then how an anti-difference act on the product of two discrete functions?

 

[Svein]

An anti-difference operator is a sum (operator).

 

[flower321]

An anti-difference operator is a sum (operator).

ok, if it is a sum operator then how it is act?

 

[Stephen Tashi]

hi, i want to ask you that how to perform an anti-difference operator on the product of two functions? i.e.
D^{-1}(f_{n} g_{n})

As math_QED said, there is no general formula.

In the special case where $g(n) = \triangle h(n)$ we can derive a formula. I suspect it is just a special case of the formula for "summation by parts".

Integration by parts is derived from the formula for differentiating a product.

By analogy, try:

$\triangle (f(k) h(k)) = f(k+1) h(k+1) - f(k) h(k)$
$= f(k+1) h(k+1) - f(k)h(k+1) + f(k)h(k+1) - f(k) h(k)$
$= h(k+1)\triangle f(k) + f(k)\triangle h(k)$

$f(k) \triangle h(k) = \triangle (f(k) h(k)) - h(k+1) \triangle f(k)$

$\triangle^{-1} ( f(k) \triangle h(k) ) = \triangle^{-1} \triangle ( f(k) h(k)) - \triangle^{-1} ( h(k+1)\triangle f(k) )$
$\triangle^{-1} ( f(k) \triangle h(k)) = f(k) h(k) - \triangle^{-1} ( h(k+1) \triangle f(k) )$So if $g(k) = \triangle h(k)$

$\triangle^{-1} ( f(k) g(k) ) = f(k) h(k) - \triangle^{-1} ( h(k+1) \triangle f(k) )$