Generalized product rule: Leibniz’s formula

Required math: calculus

Required physics: none

We’ve seen that the product rule for derivatives is, for two functions ${f(x)}$ and ${g(x)}$ :

$\displaystyle (fg)'=f'g+fg'$

This rule can be generalized to give Leibniz’s formula for the ${n^{th}}$ derivative of a product:

$\displaystyle (fg)^{(n)}=\sum_{k=0}^{n}\left(\begin{array}{c} n\\ k\end{array}\right)f^{(k)}g^{(n-k)}$
where the notation ${f^{(n)}\equiv d^{n}f/dx^{n}}$ and ${\left(\begin{array}{c} n\\ k\end{array}\right)=\frac{n!}{k!(n-k)!}}$ is the binomial coefficient.

The proof of this formula is a nice little exercise in proof by mathematical induction. An inductive proof requires two steps. First, we must show that the formula is true for one particular value of ${n}$ , say ${n=0}$ . Second, we assume that the formula is true for a value ${n=m}$ . From this assumption we then must prove that the formula is also true for ${n=m+1}$ . It doesn’t matter what ${m}$ is as long as it is taken to be equal to or greater than the value of ${n}$ used in step 1. The idea is that if we can show that the truth of the formula for a particular value of ${n}$ always implies its truth for the next value of ${n}$ , then as long as we can demonstrate the truth of the formula for some value of ${n}$ (as we do in step 1), the inductive reasoning implies it is true for all ${n}$ greater than or equal to the specific value.

The first step in the proof is called the anchor step and the second step the inductive step.

In our case, taking ${n=0}$ gives us the identity ${(fg)^{(0)}=fg}$ which is clearly true. So now we prove the inductive step. Assuming the formula is true for ${n=m}$ gives us

$\displaystyle (fg)^{(m)}=\sum_{k=0}^{m}\left(\begin{array}{c} m\\ k\end{array}\right)f^{(k)}g^{(m-k)}$

Taking the derivative of this, and applying the regular product rule, gives

$\displaystyle (fg)^{(m+1)}=\sum_{k=0}^{m}\left(\begin{array}{c} m\\ k\end{array}\right)(f^{(k+1)}g^{(m-k)}+f^{(k)}g^{(m-k+1)})$

In the first term, we can shift the summation index by replacing ${k}$ by ${k-1}$ to get

$\displaystyle (fg)^{(m+1)}=\sum_{k=1}^{m+1}\left(\begin{array}{c} m\\ k-1\end{array}\right)f^{(k)}g^{(m-k+1)}+\sum_{k=0}^{m}\left(\begin{array}{c} m\\ k\end{array}\right)f^{(k)}g^{(m-k+1)}$

Note the limits on the two sums. We can now combine the sums in the regions where their indexes overlap to get

$\displaystyle (fg)^{(m+1)}=fg^{(m+1)}+\sum_{k=1}^{m}\left[\left(\begin{array}{c} m\\ k-1\end{array}\right)+\left(\begin{array}{c} m\\ k\end{array}\right)\right]f^{(k)}g^{(m-k+1)}+f^{(m+1)}g\ \ \ \ \ (1)$

We can work out the sum of the two binomial coefficients by expanding them in terms of factorials:

$\displaystyle \left(\begin{array}{c} m\\ k-1\end{array}\right)+\left(\begin{array}{c} m\\ k\end{array}\right)=\frac{m!}{(k-1)!(m-k+1)!}+\frac{m!}{k!(m-k)!}$

Putting over a common denominator:

$\displaystyle$	$\displaystyle =$	$\displaystyle \frac{(k)m!}{k!(m-k+1)!}+\frac{(m-k+1)m!}{k!(m-k+1)!}$
$\displaystyle$	$\displaystyle =$	$\displaystyle \frac{(m+1)!}{k!(m-k+1)!}$

$\displaystyle =\left(\begin{array}{c} m+1\\ k\end{array}\right)$ Since ${\left(\begin{array}{c} m+1\\ 0\end{array}\right)=\left(\begin{array}{c} m+1\\ m+1\end{array}\right)=1}$ we can combine the two outlying terms in 1 to get the final result

$\displaystyle (fg)^{(m+1)}=\sum_{k=0}^{m+1}\left(\begin{array}{c} m+1\\ k\end{array}\right)f^{(k)}g^{(m-k+1)}$

This completes the inductive proof and establishes Leibniz’s formula.

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