Heaviside’s Operator Calculus | Hacker News Reader

Heaviside’s Operator Calculus

myreckonings.com

20 pointsRod6 comments

This link still works: https://web.archive.org/web/20080201075132/http://myreckonin...

Is there a way to make the operator calculus rigorous? Because it's not at all clear which operations are allowed. If we naively used the sort of operations Heaviside used, we could conclude (where p is the operator d/dt) that:

  x = 1
  =>
  x = pt
  =>
  1/x = (1/p) (1/t)
  =>
  1/x = ln(t)
  =>
  x = 1/ln(t)

Which is ridiculous...

If x is a function, what does 1/x mean? The inverse of operator p is indeed 1/p (assuming that the integrator has zero initial condition) but let us not start inverting everything at will ;-)

But clearly some spooky inversion is allowed, e.g., writing:

  v = Lpi
  =>
  v/i = Lp

I don't even know what that means!

Uhmmm, shouldn't the operator p be operating on some function? I don't know what v/i means either, but it looks like an operator to me: you apply it to function i and you obtain v, LOL.

This stuff is trippy!

There are a number of ways to make this rigorous, at least for some narrow classes of operators.

For many self-adjoint (or normal) operators, you can use the functional calculus: http://en.wikipedia.org/wiki/Functional_calculus

You can also do it pretty explicitly using the Fourier transform, or, more generally, the pseudodifferential calculus. This approach typically only gives an approximation, but is often much more useful in many applications. It is also at the heart of a mathematical field called microlocal analysis. http://en.wikipedia.org/wiki/Pseudo-differential_operator

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