Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.