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The first differences, (differences between consecutive terms of a single function $f(x)$) can easily be computed for any function $f(x)$ in polynomial form by simplifying $f(x)-f(x-1)$.

For any given function $g(x)$, is there a known way to find a function $f(x)$ such that $f(x)-f(x-1) = g(x)$ evaluated at the same term $x$?

For instance, $f(x)=x^2+x+1$, and $(x^2+x+1)-((x-1)^2+(x-1)^2+1) = 2x$

Suppose we started the function, $g(x)=2x$ and wanted to find $f(x)$ such that $f(x)-f(x-1) = g(x)$? Someone please explain the theory, if any on this. Thanks.

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    The first difference is a kind of discrete version of differentiation, so what you want is a discrete version of the integral. I don't know which one works, but that's the direction you should search in. – 2017-02-28

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  • Define your function $f(x)$ on $[0,1)$ with any values.

  • Then note that it is straightforwardly defined on $[1,2)$ as $f(x) = g(x)+f(x-1)$. And so on for $[n,n+1)$ iteratively.