# 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 $Q_x x^3$?

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Whenever in want for $f(D)$ performed on a specific function, where $D$ is an operator, I think it's generally a good idea to expand $f$ as a power series in $D$. For instance, $$Qx^3=x^3-3k^2x$$ (with aid from W|A). –  anon Sep 27 '11 at 11:26
Fractional powers of differential operators have been studied since the beginnings of calculus. One possible interpretation of your operator $Q_x$ could be rooted using fractional calculus: en.wikipedia.org/wiki/Fractional_derivative –  Bill Cook Sep 27 '11 at 19:00

$1+kD^2$ 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.

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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.

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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.

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