I am currently trying to learn advanced mathematics. So I'm currently reading 'The Art of Proof' by Matthias Beck and Ross Geoghegan.
I tried to solve one of the problems but I don't know if I'm actually close to an answer or not. The thing is I'm still trying to wrap my head around the second part of mathematical induction i.e the P(n) implies P(n+1) part.
So here's the problem and my attempt at a solution
Project 4.23 (Leibniz’s formula) Consider an operation denoted by $^\prime$ that is applied to symbols such as u, v, w. Assume that the operation $^\prime$ satisfies the following axioms:
$(u + v)^\prime = u^\prime + v^\prime \\ (uv)^\prime = uv^\prime + u^\prime v\\ (cu)^\prime = cu^\prime$
Define $w^{k}$ recursively by
(i) $w^{0} := w$
(ii) Assuming $w^{n}$ defined (where $n \in Z_{\ge0}$), define $w^{(n+1)} := {(w^{n})}^\prime$
Proof $(uv)^{k} = \sum_{m=0}^{k}\binom{k}{m}u^m v^{k-m}$
Solution
$P(k):(uv)^{k} = \sum_{m=0}^{k}\binom{k}{m}u^m v^{k-m}$
$P(1):$ base case - we use 1 (instead of 0) for the first derivative
$(uv)^{1}$ $= \sum_{m=0}^{1}\binom{1}{m}u^m v^{1-m}\\ = \binom{1}{0}u^0 v^{1-0} + \binom{1}{1}u^1 v^{1-1}\\ = u^0v^1 + v^1u^0\\ = uv^1 + v^1u - - (i)$
$P(n):$ Assumed defined
$(uv)^n = \sum_{m=0}^{n}\binom{n}{m}u^m v^{n-m}$
$P(n+1):$ to be proved
$(uv)^{n+1} = \sum_{m=0}^{n+1}\binom{n+1}{m}u^m v^{n+1-m}$
$(uv)^{n+1}$ $= \sum_{m=0}^{n+1}\binom{n+1}{m}u^m v^{n+1-m}\\ = \sum_{m=0}^{n}\binom{n}{m}u^m v^{n-m} + \binom{n+1}{n+1}u^{n+1} v^{n+1-({n+1})}\\ = \sum_{m=0}^{n}\binom{n}{m}u^m v^{n-m} + u^{n+1} v^0\\ = \sum_{m=0}^{n}\binom{n}{m}u^m v^{n-m} + (u^{n})^1 v\\ = (uv)^n + (u^{n})^1 v$
I know $u^{n}$ is already defined. I also think I could make same case for $(uv)^{n}$. But I don't have the rationale to make that jump.
All and any help will be appreciated. If you don't have time for typing mathjax syntax, just do it on piece of paper and post a snapshot. I'll do the work and post it here
The basic idea with induction is to use: $$(uv)^{n+1} = \dfrac{d}{dx}(uv)^n = \dfrac{d}{dx}\Big( \sum_{k=0}^n\binom{n}{k}u^k v^{n-k}\Big) = \sum_{k=0}^n\binom{n}{k}\dfrac{d}{dx}(u^kv^{n-k}) = ... $$
The rest is simply some algebraic manipulation using the fact that $\binom{n+1}{k} = \binom{n}{k}+\binom{n}{k-1}.$