Changing Variables in Discrete Calculus

Question

In discrete calculus one soon meets 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 the second formula from the first? Is there something like the change of variable formula in calculus? Maybe some version of 'dimensional analysis' that tells you where to insert $h$'s? As another example, given that $$\Delta[\log x] = \log(1 + \frac{1}{x})$$ I'd like to be able to see immediately that $$\Delta_h[\log x] = \log(1 + \frac{h}{x})$$

Any help or references would be welcome.

Answer 1

In the context of discrete calculus define the $h$-difference operator

$$\Delta_h[f(x)] := f(x+h) - f(x). $$

Notice that $\,\Delta = \Delta_1\,$ where $\,\Delta\,$ is defined as the difference operator

$$ \Delta[f(x)] := f(x+1) - f(x). $$

You asked the question

What I'd like is a uniform procedure for recovering the "h-version" from such results.

Sadly, the answer, in general, is no. Consider any function that has period $1$. That is, $\,f(x+1) = f(x)\,$ for all $\,x.\,$ This is equivalent to $\,\Delta[f(x)] = 0.\,$ Thus, all information about the function is lost. There is no way to compute $\,\Delta_h[f(x)]\,$ from the $\,\Delta_1\,$ result.

Again, consider any function $\,f(x)\,$ and suppose that the first difference $\,g(x) := \Delta[f(x)]\,$ is known. Then the $2$-difference is $\,\Delta_2[f(x)] = f(x+2)-f(x) = g(x)+g(x+1).\,$ now, in general, these is no easy way to express this in terms of $\,g(x).\,$

This is only a special case of the general problem. Suppose given any function of two variables $\,f(h,x)\,$ where we only know the explicit expression for $\,f(1,x).\,$ It is impossible, in general, to recover the expression for $\,f(h,x).\,$

However, given more information, it may be possible to get results.

For the $h$-difference example, suppose the expression for the function $$ F_r(x) := \Delta[f(rx)] = f(rx+r) - f(rx) $$ is known. Then $$ \Delta_h[f(rx)] = f(r(x+h)) - f(rx) = f(rx+rh) - f(rx) = F_{rh}(x/h). $$

For the indefinite sum example, suppose the expression for the function $$ G_r(x) := \Delta^{-1}[f(rx)] $$ is known where $\,\Delta[G_r(x)] = G_r(x+1) - G_r(x) = f(rx).\,$ Then similarly $$ \Delta_h^{-1}[f(rx)] = G_{rh}(x/h) $$ because $$ \Delta_h[G_{rh}(x/h)] = G_{rh}\left(\frac{x+h}h\right) - G_{rh}\left(\frac{x}h\right) \\ = G_{rh}\left(\frac{x}h+1\right) - G_{rh}\left(\frac{x}h\right) = f(rx). $$