The umbral calculus proof of the higher order product rule

Question

unfortunately, I seem to be quite unable to come up with the correct umbral calculus proof of the identity $$ \frac{\mathrm{d}^{n}\left(fg\right)}{\mathrm{d}x^{n}}\left(x\right) = \sum_{k=0}^{n}{\binom{n}{k} f^{\left(k\right)}\left(x\right) g^{\left(n-k\right)}\left(x\right)}. $$ I tried to write $\frac{\mathrm{d}}{\mathrm{d}x}$ as an element of a ring in which $f$ and $g$ might be idempotents, but the problem is that we are multiplying and not adding $f$ and $g$. I then tried and failed to find the proof on the internet. I'd be enourmously grateful for any courteous hints.

Answer 1

The trick is to make a little detour by bivariate calculus.

Let $F(x, y)$ be a bivariate function, and recall the chain rule

$$ D_t F(x(t),y(t)) = \dfrac{\text{d} F}{\text{d} a_1} (x,y) D_t x(t) + \dfrac{\text{d}F}{\text{d} a_2}(x, y) D_t y(t) $$

where $\dfrac{\text{d} F}{\text{d} a_i}(x, y)$ is the derivative of $F$ with respect to its $i$-th variable, evaluated at $x$ and $y$. (I want to avoid any confusion regarding the variables).

This implies that

$$ D_x F(x, x) = \dfrac{\text{d} F}{\text{d} a_1} (x,x)+ \dfrac{\text{d}F}{\text{d} a_2}(x, x). \tag{1} $$

If I define a linear operator by $\mathcal E_{y\to x} F(x, y) = F(x, x)$, then in operator fashion, (1) becomes

$$ D_x \mathcal E_{y\to x} = \mathcal E_{y\to x} (D_x + D_y), \tag{2} $$

which is true for all (totally) differentiable functions. By applying $D_x$ on the left and using (2), we get

$$ D_x^2 \mathcal E_{y\to x} = D_x \mathcal E_{y\to x} (D_x + D_y) = \mathcal E_{y\to x} (D_x + D_y)^2, $$

which can be generalized by induction for integers $n$ to

$$ D_x^n \mathcal E_{y\to x} = \mathcal E_{y\to x} (D_x + D_y)^n. \tag{3} $$

Finally we can apply this equality (3) to $f(x)g(y)$ to get on the left hand side

$$ D_x^n \mathcal E_{y\to x} f(x) g(y) = D_x^n f(x)g(x), $$

and on the right hand side

$$ \begin{align} \mathcal E_{y\to x} (D_x + D_y)^n f(x)g(y) &= \mathcal E_{y\to x} \sum_{k=0}^n \binom{n}{k} D_x^k D_y^{n-k} f(x) g(y) \\ &= \mathcal E_{y\to x} \sum_{k=0}^n \binom{n}{k} f^{(k)}(x) g^{(n-k)}(y) \\ &= \sum_{k=0}^n \binom{n}{k} f^{(k)}(x) g^{(n-k)}(x), \end{align} $$

where we have used the binomial theorem, which works because $D_x$ and $D_y$ commute when applied to the space generated by products of the form $l(x)m(y)$ by the linearity of the derivative. Hence the equality

$$ D_x^n f(x)g(x) = \sum_{k=0}^n \binom{n}{k} f^{(k)}(x) g^{(n-k)}(x). $$

Note: I'd like to mention that this proof is not an "Umbral Calculus" proof, but more of an "Operational Calculus" one, as Umbral Calculus is the study of Sheffer sequences. But both are strongly connected.