How is Leibniz's rule for the derivative of a product related to the binomial formula?

Question

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“Binomial theorem”-like identities

The binomial formula describes the expansion of the $n$th power of the sum $(a+b)$:

$$(a+b)^n = \sum_{k = 0}^n {n\choose k}a^kb^{n-k}$$

In calculus, there is a generalization of the product rule called Leibniz's rule, which describes the expansion of the $n$th derivative of the product of two functions $f$ and $g$:

$$(fg)^{(n)} = \sum_{k = 0}^n {n\choose k}f^{(k)}g^{(n-k)}$$

It's not difficult to prove using the product rule and induction, what I'm really trying understand here is why the formula is true. Is there a way to prove it using the binomial formula? How are these two formulas related? Is there a combinatorial proof for Leibniz's rule like there is for the binomial formula?

I understand this question is a bit vague, but I hope you'll agree that these two formulas look so similar it just feels like there should be some relation between the two. Of course this is not necessarily true, but I'm hoping it's not just a crazy coincidence.

Answer 1

There's a formal correspondence between the two rules. Both can be understood as special cases of:

Abstract binomial theorem. Assume that we have:

1. A vector space (or even a module) $V$.
2. A family of "general terms" $(T_{i,j}\in V)_{i,j\in\mathbb N}$.
3. A linear operatior $F: V\to V$, such that $F(T_{i,j}) = T_{(i+1),j} + T_{i,(j+1)}$.

Then $F^n(T_{0,0}) = \sum_{i=0}^n \binom{n}{i} T_{i,(n-i)}$.

The ordinary binomial theorem results if we take $F$ to be multiplication by $(a+b)$ and $T_{i,j}=a^ib^j$ -- in particular $T_{0,0}=1$.

The Leibniz rule results if we take $F$ to be $\frac{d}{dx}$ and $T_{i,j}=f^{(i)}g^{(j)}$ -- in particular $T_{0,0}=fg$.

Answer 2

Hint $\ $ Let $\rm\: f(x) = {\it e}^{ax},\ g(x) = {\it e}^{bx}$

To understand the innate duality from a more advanced standpoint look up the Umbral calculus, e.g. see Steven Roman's book by that name, or the Rota-Odlyzko exposition.