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For the function $f:\mathbb{N}\to\mathbb{N}$ the descrete derivative for $f$ in $n\in \mathbb{N}$ is defined as follows:

$$f'(n) := f(n+1)-f(n)$$

  1. Find the chain rule for the descrete derivative.

  2. Given that $f:\mathbb{N} \to \mathbb{N}$ is bijective. Is it possible to create a rule for descrete derivative of the inverse function of $f$?

Hello, could someone give me a reasonable solution and explain the approach...thx

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Let $h(n) = f(g(n))$. Then $$h'(n) = h(n+1) - h(n) = f(g(n+1)) - f(g(n))=\sum_{k=g(n)}^{g(n+1)} f'(k)$$ where if $g(n+1) < g(n)$, we switch the upper and lower bounds for summation and take the negative.

From this formula it follows that if $h(n) = f^{-1}(f(n)) = n$ then $h'(n) = 1=\sum_{k=f(n)}^{f(n+1)}f^{-1'} (k)$. (The prime notation won't typset right here: the prime should be lower than it is.)

Both of these results are not very useful because they are just telescoping sums. I find it difficult to do more unless given more information about the functions.

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