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I often encounter the following statements:

$${D \over e^D - 1} = {\log(\Delta + 1) \over \Delta}$$

$$\int_x^{x+1} f(t)\,dt= {e^D - 1 \over D} [f]$$

$$\Delta = (e^D - 1)\,$$

$$f(a+x)=e^{a D}[f]$$

$$f(a x)=a^{x D}[f]$$

$$f\left(\frac x{1-x}\right)= e^{x^2 D}[f]$$

and so on. Where can I find

  • the complete set of the rules of such manipulations
  • whether the manipulations are applicable to non-linear operators
  • the list of operators in this form (say, convolution operator, integration operator, composition etc)
  • Whether the application of such construct to a function distributive (that is whether ${e^D - 1 \over D} f={e^{Df} - 1 \over Df}$

Any other info is also appreciated.

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1  
The last is a definite "no." Consider the simple case $\frac{D}{D}$ applied to $f$. It should return $f$, but $\frac{Df}{Df}=1$. –  Thomas Andrews Apr 20 '12 at 17:27
    
Why D/D should return f? Where is the rule that says so? –  Anixx Apr 20 '12 at 17:29
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In general, if $p(z)=\sum_{i=0}^\infty a_iz^i$, then $p(D)f = \sum_{i=0}^\infty a_iD^if$. Now, $D/D$ is, as a power series, just $a_0=1$ and $a_i=0$ for $i>0$. And $D^0f = f$, by definition. –  Thomas Andrews Apr 20 '12 at 17:33
    
Is $a e^{x D}=e^{\log a xD}$? I simplified in the question but I am not sure and there is no rules list. –  Anixx Apr 20 '12 at 17:36
    
Is it possible to manipulate such expressions without actually writing down the power series? –  Anixx Apr 20 '12 at 17:42

2 Answers 2

It's better to think of $D$ is being similar to a matrix, in that you can't really define division by $D$, and it is a singular (non-invertible) linear function on some vector space. So you can't in general define $F(D)$ for any $F$ - for example, $\frac{1}{D}$ doesn't make sense, because $D$ is not invertible. (You can see that $D$ is not invertible because $D(f+c)=Df$ for any constant $c$.)

So you are stuck with the kinds of operations you can do with matrices. One of the things you can do with matrices is put them in power series.

For example, if $M$ is a matrix, $e^{M}$ makes sense, when defined using the power series for $e^z$, and it converges for all $M$. $\frac{e^M-1}{M}$ does not strictly make sense, when $M$ is not invertible, but if we define it via the power series for $\frac{e^z-1}{z}$, then it does make sense. So, in this sense, we can only really work with power series, rather than with more general functions.

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As an intro to the operational calculus, you might try looking at Operational Calculus: Based on the two-sided Laplace Transform by Van der Pol and Bremmer and Lectures on Applications-Oriented Mathematics by Friedman. Also, a paper by Lindell, "Heaviside Operational Rules Applicable to Electromagnetic Problems," has a good overview with rules of action of many pseudo-differential operators (ref. from Dead Reckonings website).

Friedman's book gives you a taste of the applications, and many of the formulas you list are covered there, but it is not as rigorous or as systematic as Van der Pol and Bremmer's. These require somewhat heavy study with a good foundation in basic complex analysis.

The first book in the intro explains with a particularly simple operator the necessity of some systematic, consistent method of interpretation:

Which expression is correct,

A: $\displaystyle\frac {1}{1-D} = 1+D+D^2+D^3+\cdots$ or

B: $\displaystyle\frac {1}{1-D} = -\left(\frac {1}{D}+\frac{1}{D^2}+\frac{1}{D^3}+\cdots\right)$ with $\displaystyle\frac{1}{D}H(x) f(x) = H(x)\int^x_0 f(u) \, du$ ?

(H(x) is the Heaviside step function.)

PS: "The series is divergent, therefore we may be able to do something with it." - Heaviside.

Rigor aside, Heaviside often used operator expansions similar to that of A to generate an asymptotic series expansion for functions represented by convergent series generated by expansions similar to B . See Heaviside's Operator Calculus at Dead Reckonings and "The asymptotic solution of an operational equation" by Carson.

Also look at the finite operator calculus, or umbral calculus, associated with Blissard, Bell, Stephen Roman, and Gian-Carlo Rota, among others. Survey articles are available on the Net, e.g., An Introduction to Umbral Calculus by Di Bucchianico, with extensive bibliographies.

Additional references for miscellaneous differential ops and their actions:

H.T. Davis, The Theory of Linear Operators (e.g., p. 89)

K. Jordan, Calculus of Finite Differences

MathOverFlow: MO-107159, Pochhammer symbol of a differential MO-102281, A mysterious Heisenberg algebra identity from Sylvester

MathStackExchange: MSE-116633, MSE-126984, and MSE-169072

OEIS: OEIS-A145271, OEIS-A132440, OEIS-A132681, OEIS-AA094638, OEIS-A021009, OEIS-A218234 (Follow ref. for P. Blasiak and P. Flajolet, G. Dattoli, and W. Lang. Cf. also Merida Lectures--Lie Algebras, Representations, and Semigroups Through Dual Vector Fields by Philip Feinsilver.)

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I thought operational calculus is a completely different thing: it transforms functions into their "images" (other functions) using some integral operator makes some operations on them and then transforms back. How is it connected with the above notation? –  Anixx Apr 21 '12 at 12:52
    
Nothing happens when I click "look inside". –  Anixx Apr 23 '12 at 14:06
    
There is the restriction that $x>0$ in the above expressions. Also, if we regard $DH(x)f(x)=H(x)f{'}(x)+\delta(x)f(x)$ where $H(x)$ is the Heaviside step function and $\delta(x)$ is the Dirac delta function, then $\frac{1}{D}=H(x)\int_{0}^{x}dx$ and $D$ commute when acting on $H(x)f(x)$ and form a pair of inverse operators; i.e., $\frac{1}{D}DH(x)f(x)=D\frac{1}{D}H(x)f(x)=H(x)f(x)$. –  Tom Copeland Apr 24 '12 at 22:22
    
@Tom Copeland, which link? –  Anixx Apr 24 '12 at 23:21
    
As a little technical detail, it's best to consider $\frac{1}{D}=H(x)\int_{0^{-}}^{x}dx$ (i.e., approaching zero from below in the lower limit), so that one doesn't have to worry about a $\delta$ fct. at a boundary of the integral when looking at the commutation argument $[D,\frac{1}{D}]=0.$ –  Tom Copeland May 5 '12 at 23:40

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