Change of variable formulas in discrete calculus?

Question

Crossposted from MSE.

In discrete calculus one defines the $h$-difference operator $$\Delta_h[f(x)] = f(x+h) - f(x)$$ and we often define $\Delta = \Delta_1.$ We can similarly define the indefinite sums $\Delta_h^{-1}$ and set $\Delta^{-1} = \Delta_1^{-1}.$

Most books on discrete calculus only include results for $h=1.$ For example, the above link gives that $$\Delta^{-1}\sin rx = \frac{-\cos(r[x-\frac{1}{2}])}{2\sin\frac{r}{2}}$$

Through an ugly computation I managed to work out that $$\Delta_h^{-1}[\sin rx] = \frac{-\cos(r[x-\frac{h}{2}])}{2\sin\frac{rh}{2}}$$

What I'd like is a uniform procedure for recovering the "$h$-version" from such results. How could I have derived (or even just guessed) the second formula from the first? Is there something like the change of variable formula in calculus? Maybe some version of the 'dimensional analysis' heuristics in ordinary calculus that tells you where to insert the $h$'s? As a test, given a formula at random from the previous link like $\Delta[\log x] = \log(1 + \frac{1}{x})$ I'd like to be able to see immediately that $\Delta_h[\log x] = \log(1 + \frac{h}{x})$ without carrying out the actual calculation.

A search trough the standard references didn't turn up much, so any help or references would be welcome.

Answer 1

In terms of the differential operator $\partial_x\equiv d/dx$ one has $f(x+h)=e^{h\partial_x}f(x)$, hence $$\Delta_h=e^{h\partial_x}-1.$$ Upon Fourier transformation, $\hat{f}(k)=\int_{-\infty}^\infty e^{ikx}f(x)\,dx$, one has $\partial_x\mapsto -ik$, hence $$\hat{\Delta}_h=e^{-ihk}-1\Leftrightarrow \hat{\Delta}_h^{-1}=\frac{e^{ihk}}{1-e^{ihk}}.$$

In this way you can calculate, for any function $f(x)$ with Fourier transform $\hat{f}(k)$: $$\Delta_h^{-1}f(x)=\int_{-\infty}^\infty\frac{e^{i(h-x)k}}{1-e^{ihk}}\hat{f}(k)\,\frac{dk}{2\pi}=\int_{-\infty}^\infty F_h(x'-x)f(x')\,dx'$$ with Dirac-delta-function kernel $$F_h(x)=\int_{-\infty}^\infty\frac{e^{i(h+x)k}}{1-e^{ihk}}\,\frac{dk}{2\pi}=\sum_{n=1}^\infty\delta(x+nh).$$ The relation $F_h(x)=(1/h)F_1(x/h)$ gives the scaling with $h$ requested by the OP:

$\Delta_h^{-1}f(x)$ follows from $\Delta_1^{-1}f(rx)$ by rescaling $r\mapsto rh$ and $x\mapsto x/h$.

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The "test cases" mentioned by the OP include: $\Delta_1^{-1}\log rx=x\log r+\log \Gamma(x)$, hence $$\Delta_h^{-1}\log rx=(x/h)\log(rh)+\log\Gamma(x/h).$$ Similarly, $\Delta_1^{-1}a^{rx}=a^{rx}/(a^r-1)$, hence $$\Delta_h^{-1} a^{rx}=a^{rx}/(a^{rh}-1),$$ and $\Delta_1^{-1}(rx)^a=r^aB_{a+1}(x)/(a+1)$, hence $$\Delta_h^{-1}(rx)^a=(rh)^aB_{a+1}(x/h)/(a+1).$$

Answer 2

$\newcommand\De\Delta$Elementarily: The relation $f=\De_h^{-1}g$ means that $f(x+h)-f(x)=g(x)$ for all $x$.

For a function $g$ and each real $r$, we know $f_r:=\De_1^{-1}g_r$, where $g_r(x):=g(rx)$, so that $$f_r(x+1)-f_r(x)=g(rx) \tag{10}\label{10}$$

For real $h\ne0$, real $r$, and all real $x$, let now $$f_{r,h}(x):=f_{rh}(x/h). \tag{20}\label{20}$$ Then $$f_{r,h}(x+h)-f_{r,h}(x)=f_{rh}(x/h+1)-f_{rh}(x/h)=g((rh)(x/h))=g(rx)$$ for all real $x$. That is, $f_{r,h}=\De_h^{-1}g_r$. $\quad\Box$

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Note that we do not need or impose any conditions on the functions $g$ and $f_r$ (such as measurability, continuity, growth, etc.) except for \eqref{10}.

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To illustrate these simple considerations, consider the example from your post, with $g_r(x)=\sin rx$ and $f_r(x)=\dfrac{-\cos(r(x-1/2))}{2\sin(r/2)}$. Then, by \eqref{20}, for $\De_h^{-1}g_r=f_{r,h}$ we have $$(\De_h^{-1}g_r)(x)=f_{r,h}(x)=\dfrac{-\cos((rh)(x/h-1/2))}{2\sin(rh/2)} =\dfrac{-\cos(r(x-h/2))}{2\sin(rh/2)}. \quad\Box$$