Why does $n$-time differentiation of product have the same structure as raising sum to $n$th power?

A formula for differentiating a product is well known:

$$(ab)'=a'b+ab'.$$

At first sight it doesn't resemble anything interesting. But what if we differentiate twice? We'll get

$$(ab)''=a''b+2a'b'+ab''.$$

Now this does resemble something:

$$(a+b)^2=a^2b^0+2a^1b^1+a^0b^2.$$

Of course, going to the next level now gives us the expected result:

$$(ab)'''=a'''b+3a''b'+3a'b''+ab''',$$

in full accordance to known binomial expansion:

$$(a+b)^3=a^3b^0+3a^2b^1+3a^1b^2+a^0b^3.$$

I know that binomial expansion stems from distributive law of multiplication over addition. It's easy because integer powers can be interpreted as iterated multiplication. But I can't seem to see any simiar distributive law of differentiation over product.

Observing this, I wonder: how to explain this striking similarity between seemingly unrelated things? Can this be explained by some generalized form of distributive law?

Solution 1:

This can be easily shown using the bialgebras.

First, introduce a something called tensor product, denoted $\otimes$, which has property \begin{equation} (a\otimes b)(c\otimes d)=ac\otimes bd. \end{equation}

Then, introduce something called multiplication map, denoted by $m$, such that, for example: \begin{equation} m(a\otimes b + c\otimes d+...) = ab+cd+... \end{equation} i.e. $m$ turns each tensor product into normal product.

And let's denote derivative operator with $\partial$.

Then, Leibnitz rule "simply" states: \begin{equation} \partial(ab)=m[(\partial\otimes1+1\otimes\partial)(a\otimes b)] \end{equation} It's easy to see that this must hold. Left hand side is simply $(ab)'$. Right hand side is: \begin{align} m[(\partial\otimes1+1\otimes\partial)(a\otimes b)] &=m(\partial a \otimes b + a \otimes \partial b)\\ &=m(a' \otimes b + a \otimes b')\\ &=a'b+ab' \end{align}

So far so good! Oh and btw. $\Delta\partial\equiv\partial\otimes1+1\otimes\partial$ is something called coproduct. It has a property that $(\Delta h)^n=\Delta h^n$, which leads to the property $f(\Delta h)=\Delta f(h)$ and it's used to see how does some operator act on a product, e.g.:

\begin{equation} \partial(ab)=m[\Delta\partial(a\otimes b)] \end{equation}

Now, we want to act on the product with derivative operator $n$ times: \begin{equation} (ab)^{(n)}=\partial^n(ab) =m[\Delta\partial^n(a\otimes b)] =m[(\Delta\partial)^n(a\otimes b)] \end{equation}

So we only need to calculate $(\Delta\partial)^n$. But this is simply a binomial raised to the $n$-th power: \begin{equation} (\Delta\partial)^n=(\partial\otimes1+1\otimes\partial)^n \end{equation}

And this is where all those binomial coefficients come from.

Solution 2:

Well, there are many places in Calculus or Algebra that this similarity appears. I don't understood exactly what you mean't by "explanation of this similarity", but this similarity can be easily proven by induction over n (n being the number of differentiations).

We want to prove that $(ab)^{(n)} = \sum^{n}_{i=0}\binom{n}{i}a^{(n-i)}b^{(i)}$.

Well, it works for $n=2$. Now let's suppose it works for $n$ and prove that it works for $n+1$. Just take the derivative of $(ab)^{(n)}$.

$(ab)^{(n+1)} = \sum^{n}_{i=0}\binom{n}{i}(a^{(n-i)}b^{(i)})' = \sum^{n}_{i=0}\binom{n}{i}(a^{(n+1-i)}b^{(i)} + a^{(n-i)}b^{(i+1)} )$

$(ab)^{(n+1)} = \sum^{n}_{i=0}\binom{n}{i}a^{(n+1-i)}b^{(i)} + \sum^{n}_{i=0}\binom{n}{i}a^{(n-i)}b^{(i+1)}$

$(ab)^{(n+1)} = a^{(n+1)}b + \sum^{n}_{i=1}\binom{n}{i}a^{(n+1-i)}b^{(i)} + \sum^{n+1}_{i=1}\binom{n}{i-1}a^{(n+1-i)}b^{(i)} $

$(ab)^{(n+1)} = a^{(n+1)}b + \sum^{n}_{i=1}\binom{n}{i}a^{(n+1-i)}b^{(i)} + \sum^{n}_{i=1}\binom{n}{i-1}a^{(n+1-i)}b^{(i)} + ab^{(n+1)}$

Here, you apply that rule that says that $\binom{n}{i}+\binom{n}{i-1}=\binom{n+1}{i} $. Thus, giving you that:

$(ab)^{(n+1)} = a^{(n+1)}b + \sum^{n}_{i=1}\binom{n+1}{i}a^{(n+1-i)}b^{(i)} + ab^{(n+1)}$

$(ab)^{(n+1)} = \sum^{n+1}_{i=0}\binom{n+1}{i}a^{(n+1-i)}b^{(i)} $

Which is exactly what we wanted to prove.