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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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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
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1 Answer

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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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