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I've read Wikipedia about it and it says:

In mathematics before the 1970s, the term umbral calculus referred to the surprising similarity between seemingly unrelated polynomial equations and certain shadowy techniques used to 'prove' them.

  • What are these techniques?

These similarities allow one to construct umbral proofs, which, on the surface cannot be correct, but seem to work anyway.

  • What does "seem to work" mean here?
  • It seems that umbral calculus is a mathematical idea with almost no uses, why? (At least it's not so famous as calculus and algebra, for example.)
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Calculus and algebra are massive branches of mathematics encompassing thousands of techniques each. It is far to much to demand of umbral calculus, which is essentially a single technique, to be equally useful. – Alex Becker Sep 6 '12 at 3:50
"What are these techniques?" Aren't several examples given in the very Wikipedia article you're quoting? – Rahul Sep 6 '12 at 4:01
Roman's Advanced Linear Algebra has a nice chapter on this, that might be a good place to read. – James S. Cook Sep 6 '12 at 4:02
The Wikipedia article itself already gives good examples and also has references. – Qiaochu Yuan Sep 6 '12 at 4:14
You can download some of Roman's articles on umbral calculus here: – wj32 Sep 6 '12 at 6:06
up vote 12 down vote accepted

Umbral relations shadow those of the basic binomial transform, revealing underlying connections between diverse areas of math (as Leibnitz himself predicted--see H. Davis "Theory of Linear Operators"):

I) Umbral notation is brief and suggestive (courtesy of Blissard and contemporaries):

$ \displaystyle (a.)^n= a_n \;  \;  \; $ (umbral variable and lowering of superscript).

Expressing binomial convolution simply:

$$  \displaystyle (a. + b.)^{n} = \sum_{k=0}^{n}  \binom{n}{k} a_{k} b_{n-k} \; \;$$ (be careful to evaluate $(a.+b.)^0=a_0b_0$ and $(a.+b.)^1=a_0b_1+a_1b_0$), $$ \displaystyle e^{a. \; x}= \sum_{n \ge 0} a_n \frac{x^n}{n!} \; \; ,$$

$$ \displaystyle e^{a.\;x}\; e^{b.\; x} = e^{(a. + b.)x}\; \; .$$

A more precise notation is to use $\langle a.^n \rangle = a_n$ to clearly specify when the lowering op, or evaluation of an umbral quantity, is to be done. E.g.,

$$\langle a.^n a.^m\rangle=\langle a.^{n+m}\rangle= a_{n+m} \ne a_n a_m= \langle a.^n\rangle\langle a.^m\rangle$$

and $$\langle\exp[\ln(1+a.x)]\rangle=\langle(1+a.x)\rangle=1+a_1x$$

$$\ne \exp[\langle\ln(1+a.x)\rangle]=\exp \left[ \sum_{n \ge 1} \langle\frac{a.^nx^n}{n}\rangle\right]=\exp\left[\sum_{n \ge 1} \frac{a_nx^n}{n}\right]\; .$$

II) Same for umbralized ops, allowing succint specification and derivation of many relations, especially among special functions. A good deal of umbral calculus is about defining these ops for special sequences, such as the falling $(x)_{n}=x!/(x-n)!$ and rising factorials $(x)_{\bar{n}}=(x+n-1)!/(x-1)!$ and Bell polynomials $\phi_n(x)$.


$ (:AB:)^n = A^n B ^n$  (defn. for order preserving exponentiation for any operators )

$$ (xD)^n = (\phi.(:xD:))^n = \phi_n(:xD:) \; \; ,$$

$$ e^{txD} = e^{t \phi.(:xD:)} \; \; .$$

From $xD \; x^{n} = n \; x^{n}$, it's easy to derive

$$e^{t\phi.(x)} = e^{x (e^t-1)} \; \; .$$

(See this MO-Q for the o.g.f.)

III) Umbral compositional inverse pairs allow for easy derivations of combinatorial identities and reveal associations among different reps of operator calculi.

Look at how this connects the distributive operator exponentiation $:xD:^n=x^nD^n$ to umbral lowering of superscripts. The falling factorials and Bell polynomials are an umbral inverse pair, i.e., $\phi_n((x).)=x^n=(\phi.(x))_n$. This is reflected in the functions $\log(1+t)$ and $e^t-1$, defining their e.g.f.s $e^{x\log(1+t)}$ and $e^{x(e^t-1)}$, being regular compositional inverses and to the lower triangular matrices containing the coefficients for the polynomials (the Stirling numbers of the first and second kinds) being multiplicative inverses, so we can move among many reps to find and relate many formulas. For the derivative op rep,


so we have a connection to the umbral lowering of indices


IV) The generalized Taylor series or shift operator is at the heart of umbral calculus:

$$ e^{p.(x)D_y}f(y) = f(p.(x)+y)  \; , $$

(e.g., this entry on A class of differential operators and another on the Bernoulli polynomials) with special cases

$$ e^{:p.(x) D_x:} f(x) = f(p.(x) + x) \; , $$ and

$$ e^{-(1-q.(x))D_y}y^{s-1} \; |_{y=1} = (1-(1-q.(x)))^{s-1} \; ,$$ giving a Gauss-Newton interpolation of $q_n(x)$ (shadows of the binomial relations).

It can often be used to easily reveal interesting combinatorial relations among operators. A simple example:

$$ e^{txD} f(x) = e^{t\phi.(:xD:)} f(x) = e^{(e^t-1):xD:} f(x) = f(e^{t}x) \; .$$ You could even umbralize $t$ to obtain the Faa di Bruno formula. Try discovering some op relations with the Laguerre polynomials (hint--look at $:Dx:^n= D^nx^n$).

As another example (added May 2015) of the interplay between differential operators, umbral calculus, and finite differences, note the relations for the Bell polynomials

$$\phi_{n}(:xD_x:)= \sum_{k=0}^n S(n,k)x^kD_x^k = (xD_x)^n=\sum_{j=0}^\infty j^n \frac{x^jD^j_{x=0}}{j!}=\sum_{j=0}^\infty (-1)^j \left[\sum_{k=0}^j(-1)^k \binom{j}{k}k^n\right] \frac{x^jD_x^j}{j!} \;$$

and apply these operators on $x^m$, $e^{x}$, and $x^s$. (The S(n,k) are the Stirling numbers of the second kind.)

I've used the power monomials $x^n$ and their associated raising and lowering ops, $x$ and $D_x$, but these relations are shadowed by the raising and lowering ops of all umbral sequences $p_n(x)$ such that $R \; p_n(x) = p_{n+1}(x)$ and $L \; p_n(x)  = n \;  p_{n-1}(x)$. (Shadows of Lie and quantum mechanics here also.)

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I'm having introductory lectures in combinatorics. I'm amazed that Umbral Calculus is a lot similar to some of the things I've seen there, I thought it was nearer to real analysis. Although, as you pointed in the start of the text: It might have a lot of connections. – Voyska Jan 2 '15 at 12:19
Could you recommend some really elementary introduction to umbral calculus? It seems really interesting, but I still don't have much background. – Voyska Jan 2 '15 at 12:23
Master basic combinatorics first (binomials, Stirling numbers, Chu-Vandermonde identities--keeping an eye on their apps in analysis--check OEIS entries also) and linear algebra (linear functionals, dual spaces). Then you should be ready for Rota, Roman, et al., as Michael suggests. – Tom Copeland Jan 2 '15 at 19:29
Can't wait? Start studying the Appell polynomials, in particular, the Hermite and Bernoulli polynomials, and the three binomial Sheffer sequences the rising and falling factorials and Bell / Touchard polynomials. In parallel, look at finite differences and their differential operator reps. (Wiki refs again). I think after that you can easily make your own choices about what relevant operator calculi and integral transforms to explore. Use the umbral notation to simplify expressions and suggest formulas. – Tom Copeland Jan 2 '15 at 19:50
Yep, the intersections inform. Also, you can often contribute something new yourself. Look at noncrossing partitions A134264. – Tom Copeland Jan 3 '15 at 10:32

From the Wikipedia article: "The combinatorialist John Riordan in his book Combinatorial Identities published in the 1960s, used techniques of this sort extensively." There you see the classical umbral calculus. Basically he pretends subscripts are exponents, and somehow it works. Take a look at that book. It doesn't require a lot of apparatus.

The 1978 paper by Roman and Rota, cited in the article, is the beginning of a technique for making the classical umbral calculus rigorous.

The 1975 paper by Rota, Kahaner, and Odlyzko appears to be a paper about Sheffer sequences, which are certain sequences of polynomials (see the Wikipedia article titled "Sheffer sequence"). If you lay that paper and the 1978 paper side-by-side, you can see that they're really two different ways of looking at the same thing.

In the mean time, look at the concrete examples in the Wikipedia article that you cited.

Is it useful? I think one could argue about that. But I don't want to try to make the case for its utility in research.

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What you mean with useful? I've read a few stuff on the differences of pure and applied mathematics, considering the view of pure mathematics, this definition seems to don't exist. – Voyska Sep 6 '12 at 6:18
@Voyska : In English one normally uses the preposition "by", saying "What do you mean by 'useful'?" (I suspect those who use "with" are relying on the fact that the German preposition "mit" would usually be translated as "with".) $\qquad$ – Michael Hardy Jun 16 at 17:48

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