# Where to find reference about dealing with operators in form of formal power series?

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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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
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
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 $c$.)
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.