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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.

  • 1
    The last is a definite "no." Consider the simple case $\frac{D}{D}$ applied to $f$. It should return $f$, but $\frac{Df}{Df}=1$.2012-04-20
  • 0
    Why D/D should return f? Where is the rule that says so?2012-04-20
  • 1
    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.2012-04-20
  • 0
    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.2012-04-20
  • 0
    Is it possible to manipulate such expressions without actually writing down the power series?2012-04-20
  • 0
    @Anixx : $D/D=1$, so $(D/D)f = 1f = f$.2012-04-20
  • 0
    @Anixx : The rule is $e^{cxd/dx}f(x)=f(e^{c}x)$ or $a^{xd/dx}f(x)=f(ax)$ with $c=ln(a)$.2012-04-21
  • 0
    @Michael Hardy in $(D/D)f$ do you mean multiplication or application of operator? I used square brackets to indicate application of operator to a function.2014-04-15
  • 0
    @Anixx : I meant application of an operator.2014-04-15

3 Answers 3