If we have two functions, $a(x)$ and $b(x)$ and those function are differentiable inifitely many times. What is a closed form to $$\frac{d^n}{dx^n} (ab)$$
How can I use induction here?
I don't really know how to start this task.
Many thanks in advance.
On
Hint: suppose you have $r_1$ and $r_2$, two real numbers. Than, how you prove the binormal formula $$ (r_1+r_2)^n=\sum_{i=0}^n {n\choose {i}} r_1^ir_2^{n-i} \tag 1$$ Can you use induction to prove $(1)$?
Then your problem will be proven by using exactly same method and finally you have
$$\frac{d^n}{dx^n}(ab)=\sum_{k=0}^n {n\choose {i}} \frac{d^i}{dx^i}a \,\frac{d^{n-i}}{dx^{n-i}}b$$
On
Start with $$\frac{d}{dx}(ab) = ab' + a'b = \sum_{i=0}^{1}\binom{1}{i}a^{(i)}b^{(1-i)}$$
$$\frac{d^2}{dx^2}(ab) = ab'' + 2a'b' + a''b = \sum_{i=0}^{2}\binom{2}{i}a^{(i)}b^{(2-i)} $$
Notice a patern? Use induction over $n$ to find
$$\frac{d^n}{dx^n}(ab) = \sum_{i=1}^{n}\binom{n}{i}a^{(i)}b^{(n-i)} $$
Here is the inductive step
$$\begin{align}\frac{d^{n+1}}{dx^{n+1}}(ab) &= \frac{d}{dx}\Bigg(\frac{d^{n}}{dx^n}(ab)\Bigg) = \frac{d}{dx}\Bigg(\sum_{i=0}^{n}\binom{n}{i}a^{(i)}b^{(n-i)}\Bigg)\end{align}$$
use linearity of $\frac{d}{dx}$, that is, $\frac{d}{dx}(A + B) =\frac{d}{dx}A + \frac{d}{dx}B$ and $\frac{d}{dx} cA = c \frac{d}{dx}A$ to get
$$\begin{align}\frac{d}{dx}\Bigg(\sum_{i=0}^{n}\binom{n}{i}a^{(i)}b^{(n-i)}\Bigg) &=\sum_{i=0}^{n}\binom{n}{i}\frac{d}{dx}\Big(a^{(i)}b^{(n-i)}\Big) \\&= \sum_{i=0}^{n}\binom{n}{i}\Bigg(a^{(i)}b^{(n-i+1)} + a^{i+1}b^{n-i}\Bigg)\\&=\sum_{i=0}^{n}\binom{n}{i}a^{(i)}b^{((n+1)-i)}+\sum_{i=0}^{n}\binom{n}{i+1}a^{i}b^{n-i}\\&=ab^{n+1} + \sum_{i=0}^{n}\Bigg(\binom{n}{i-1} + \binom{n}{i}\Bigg)a^{(i+1)}b^{((n+1)-i)}+a^{n+1}b\\&=ab^{n+1} + \sum_{i=0}^{n}\binom{n+1}{i}a^{(i+1)}b^{((n+1)-i)}+a^{n+1}b \tag{1} \\&=\sum_{i=0}^{n+1}\binom{n+1}{i}a^{(i)}b^{(n+1)-i}\end{align}$$
where in $(1)$ we have used Pascal's Rule, and there you have it.
$$(ab)^{(n)}=\sum_{k=0}^n {n\choose {k}} a^{(k)} b^{(n-k)}$$