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## Induction for Nth derivative of a product

Posted by peeterjoot on January 2, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Motivation.

I had use for the result $D^n( a b)$, where $D = d/dx$, and felt like doing the induction.

# Guts

Let’s look at the first few cases, starting with $n = 1$. That’s just the chain rule

\begin{aligned}D( a b) = (D a) b + a (D b).\end{aligned} \hspace{\stretch{1}}(2.3)

For $n = 2$ we have

\begin{aligned}D^2( a b) &= D ((D a) b + a (D b)) \\ &= (D^2 a) b + (D a) (D b)(D a) (D b) + a (D^2 b) \\ &= (D^2 a) b + 2 (D a) (D b) + a (D^2 b) \\ \end{aligned}

The induction hypothesis is clearly of the form of a binomial product

\begin{aligned}D^n (a b) = \sum_{k=0}^n \binom{n}{k} (D^k a)(D^{n-k} b).\end{aligned} \hspace{\stretch{1}}(2.3)

Let’s try the induction

\begin{aligned}D &\sum_{k=0}^n \binom{n}{k} (D^k a)(D^{n-k} b) \\ &=\sum_{k=0}^n \binom{n}{k} (D^{k+1} a)(D^{n-k} b)+ (D^{k} a)(D^{n+1-k} b) \\ &=(D^{n+1} a) b+\sum_{k=0}^{n-1} \binom{n}{k} (D^{k+1} a)(D^{n-k} b)+(D^{0} a)(D^{n+1} b) +\sum_{k=1}^n \binom{n}{k} (D^{k} a)(D^{n+1-k} b) \\ &=(D^{n+1} a) b+(D^{0} a)(D^{n+1} b) +\sum_{j=1}^{n} \binom{n}{j-1} (D^{j} a)(D^{n+1-j} b)+\sum_{k=1}^n \binom{n}{k} (D^{k} a)(D^{n+1-k} b) \\ &=(D^{n+1} a) b+(D^{0} a)(D^{n+1} b) +\sum_{j=1}^{n} \left(\binom{n}{j-1} + \binom{n}{j} \right)(D^{j} a)(D^{n+1-j} b) \\ \end{aligned}

Simplification of the binomial coefficients should finish the job

\begin{aligned}\binom{n}{j-1} + \binom{n}{j} &=\frac{n!}{(j-1)!(n+1-j)!}+\frac{n!}{j!(n-j)!} \\ &=\frac{n!}{(j-1)!(n-j)!} \left( \frac{1}{{n+1-j}} + \frac{1}{{j}}\right) \\ &=\frac{(n+1)!}{j!(n+1-j)!} \\ &=\binom{n+1}{j}\end{aligned}

Assembling the terms we have

\begin{aligned}D^{n+1} (a b)=\sum_{j=0}^{n+1} \binom{n+1}{j}(D^{j} a)(D^{n+1-j} b),\end{aligned} \hspace{\stretch{1}}(2.3)

which completes the induction.

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