# Square Root of a Function

So I came across the
Quantum::Usrn Perl module, and it got me to thinking about square roots of other functions. I’m sure there’s some way, as there is with integrating, to find these, but so far I’ve had to use different methods for each one, along with quite a bit of mathematical intuition.

For instance, sqrt(λx.-x) is λx.ix, and in general, sqrt(λx.ax) is λx.sqrt(a)x. Also, quite obviously, sqrt(λx.x+a) is λx.x+a/2. I found sqrt(λx.ax+b), and it is significantly more complicated, but an interesting property is that it is in the form αx+β, and α is sqrt(a), which leads me to believe that there is some way of methodically finding these, at least for polynomials.

I have a hunch that the solution of sqrt(f) lies somehow in the scalar eigenvalues of f (that is, x where f(x) = x).

This is all fun stuff. I don’t know what practical use it has, but I’m sure the dynamical systems folks have done something with it at some point.

## 3 thoughts on “Square Root of a Function”

1. Fergal Daly says:

I saw a problem a long time ago: what’s the square root of f(z) = z*z + 1 (in the complex plain) and is it analytic? I never got the solution. Finding the square root of z*z and trying to add terms with coefficients didn’t do me any good (give it a try). Playing further I found that some 4th degree polynomials have quadratic sqaure roots but the problem that the space of 4th degree polys is 5-d and the space of quadratics is 3-d so there are a whole bunch of 4th degree polys that don’t have a polynomial square root.

I’d really love to the answer to the z*z + 1 question.

2. Just as a quick clarification: do you mean z multiplied by its conjugate, or just z squared?

3. Anonymous says:

Just z squared. Actually the problem was posed as a “find or prove there is no function” but I couldn’t find a contradiction either.