Given some difference operator of e.g. the form $$\Delta [f(x)] = f(x+1)-f(x),$$how would I go about finding its inverse $\Delta^{-1}$, so some type of "discrete antiderivative" so that $$\Delta^{-1}[\Delta [f(x)]] = f(x).$$
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For discrete $x$, we can use symbolic summation. Ask here if you need this case. – IV_ Apr 13 '23 at 16:24
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@IV_ This is indeed the case, $x$ is supposed to be discrete. – jfeis Apr 15 '23 at 14:44
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Technically, the inverse of the difference operator is the summation operator. If we assume that $x$ runs over the positive integers, then we have the following telescoping sum:
$$\begin{eqnarray}f(x) & = & f(0) + [f(1) - f(0)] + [f(2) - f(1)] + \ldots + [f(x) - f(x-1)] \\ \\ & = & f(0) + \Delta[f(0)] + \Delta[f(1)] + \ldots + \Delta[f(x-1)] \\ & = & f(0) + \sum_{i = 1}^x \Delta[f(i - 1)] \end{eqnarray}$$
Notice that the $f(0)$ term out the front performs a similar role to the $+C$ in the indefinite integral - in general a discrete difference equation will define a family of functions.
And also just like for the continuous integral, there's no single method to find a closed form for $f(x)$ given a difference equation $\Delta[f(x)] = g(x, f(x))$, but for specific forms of $g(x)$ there are - for example, if $g$ is polynomial in $x$ then $f$ will also be polynomial, if $g$ is a linear function of $f(x)$ then $f$ will involve some kind of exponential term, etc.
ConMan
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1.)
$\Delta\left[f(x)\right]$ is a forward difference.
The inverse operator of the forward difference operator is the indefinite sum or antidifference operator.
The inverse difference operator is treated in
Károly Jordán, Charles Jordan: Calculus of Finite Differences. American Mathematical Soc., 1965, 100 ff.: Inverse Operation of Differences and Means. Sums
2.)
If $x$ is discrete, we can use the concepts of Symbolic Summation:
$$\Delta\left[f(k)\right]=f(k+1)-f(k)$$ $$F(k)=\Delta\left[f(k)\right]$$
The goal of Symbolic Summation is to find $f(k)$ for a given $F(k)$:
$$\sum_{k=m}^nF(k)=f(n+1)-f(m)$$
see e.g.:
Are there some techniques which can be used to show that a sum "does not have a closed form"?
IV_
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