# The umbral calculus proof of the higher order product rule

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.

• You can use induction Commented Sep 21, 2021 at 21:03
• Well, but that wouldn't be very elegant, would it be? I don't want to repeat the proof of the binomial theorem, I want an umbral calculus proof! Commented Sep 21, 2021 at 21:34
• I want to help, but am not familiar enough with umbral calculus. What, for example, would an umbral calculus proof of the falling factorial version of the binomial theorem, $(x+y)_n=\sum_{k=0}^n \binom nk (x)_k (y)_{n-k}$, look like? Here, $(x)_n=x(x-1)\cdots (x-n+1)$. Commented Sep 21, 2021 at 21:37
• @MikeEarnest I made some silly mistakes in my last comment. Please forgive. Commented Sep 21, 2021 at 21:54
• Seems that the umbral calculus is not so easy to learn. I tried reading the Wikipedia article, and sciencedirect.com/science/article/pii/0001870878900877, but I think it will take too much time for me to really grok. (+1) and best of luck. Commented Sep 21, 2021 at 22:24

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.

• Nice answer! One of the applications of umbral calculus is to classical invariant theory, see the review by Kung and Rota in BAMS. A important trick in that context is the use of Cayley's operator and that typically involves introducing a copy $y$ of the variables $x$, applying differentiations in both $x$ and $y$, and at the end, setting $y$ equal to $x$, to erase one's footsteps so to speak. Commented Jan 7 at 15:26
• Thanks a lot! I KNEW there had to be some way of doing this, but hadn't been able to find it. Commented Jan 7 at 17:30
• By the way: For me, this proof is sufficiently umbral, because for me, umbral calculus is operational calculus. (If you want to get people into operational calculus, better call it umbral calculus!) Commented Jan 7 at 18:29
• A year ago, I also happen to have had this feeling that there had to be a way to prove this with operators. After I got more experienced with Umbral/Operational Calc, I stumbled on your question and it made me think about it again, after which I found the answer haha. Commented Jan 8 at 15:10
• I understand your point, but I prefer to see them as two separate, but complementary, fields. The reason is that Operational and Umbral Calc are very distinct in the sens that their proofs are but, still, there are some Operational results that seem completely out ouf reach, but are provable with Umbral Calc and vice-versa. Commented Jan 8 at 15:16