Square root of differential operator If $D_x$ is the differential operator. eg. $D_x x^3=3 x^2$.
How can I find out what the operator $Q_x=(1+(k D_x)^2)^{(-1/2)}$ does to a (differentiable) function $f(x)$? ($k$ is a real number)
For instance what is $Q_x x^3$?  
 A: It probably means this: Expand the expression $(1+(kt)^2)^{-1/2}$ as a power series in $t$, getting
$$
a_0 + a_1 t + a_2 t^2 + a_3 t^3 + \cdots,
$$
and then put $D_x$ where $t$ was:
$$
a_0 + a_1 D_x + a_2 D_x^2 + a_3 D_x^3 + \cdots
$$
and then apply that operator to $x^3$:
$$
a_0 x^3 + a_1 D_x x^3 + a_2 D_x^2 x^3 + a_3 D_x^3 x^3 + \cdots.
$$
All of the terms beyond the ones shown here will vanish, so you won't have an infinite series.
A: $1+kD^2$ has positive spectrum but this is not enough for the existence of a square root operator on the same space of functions.  This is because derivative is an unbounded operator (so that the expansion of $Q_x$ as a power series in $D$ may not converge), and the square root function is multi-valued.  
The restriction to functions that are polynomials does give a well-defined $Q$ using the power series.
A: Dirac won a Nobel Prize for finding the square root of a pesky differential operator while working on the relativistic Schrodinger equation. http://www.youtube.com/watch?v=zM-Lc16nyho. <--- That video  gives a clear overview of the hacks Dirac used. 
