4 ms·

I can give examples of each. If you try to build Calculus on top of the complex numbers, you get complex analysis where every differentiable function winds up

by btilly 5y ago

I can give examples of each. If you try to build Calculus on top of the complex numbers, you get complex analysis where every differentiable function winds up locally a power series, and things just work out over and over again unreasonably nicely. Build Calculus on top of the real numbers, and as you dive in you wind up with menagerie of counterexamples. For example Weierstrass produced a function which is everywhere differentiable and nowhere twice differentiable. These are not mere curiosities, for example wavelets (heavily used in data analysis) always are only differentiable a finite number of times. We can ask whether infinitely differentiable functions can be represented by a power series. e^(-1/x^2) is the classic example of one which can't be at 0. https://en.wikipedia.org/wiki/Non-analytic_smooth_function#A_smooth_function_which_is_nowhere_real_analytic https://en.wikipedia.org/wiki/Non-analytic_smooth_function#A... offers one which is nowhere. And so on. But now let's go from the real numbers to the natural numbers. The real numbers have an axiom system which is known to be consistent and complete. See https://math.stackexchange.com/a/362840/6708 https://math.stackexchange.com/a/362840/6708 for some of the details. However the natural numbers famously do not. (That's Gödel's Incompleteness Theorem.) What goes wrong? Philosophically it turns out that the natural numbers can encode computation, computation can encode reasoning, and reasoning about reasoning allows us to encode the liar's paradox. Everything blows up, hopelessly, after that. At a conceptual level what is happening is this. The complex numbers have a tremendous amount of structure. As you go towards having less and less structure, we can use that freedom to write down more logically complex things. Logically complex things give ways for stuff to go wrong. And the more depth you study these things in, the more that turns out to matter.

hyperpallium2 5y ago
Could you elaborate on the "tremendous amount of structure" of complex numbers please?
- ssivark 5y ago
  Think of the following: there are many many paths to get from one location to another on a plane. If you imagine a function on the plane, it’s behavior as you walk along each part would have a sanity constraint. That (along with the constraints of behavior under complex conjugation, should you choose to impose “analyticity”) adds up to a lot of constraints, which is the “structure” in complex numbers — which makes complex analysis simple.
  - btilly 5y ago
    This is the right idea. For example if you draw a loop, specify a continuous function on the loop, and say it has to be differentiable inside of that loop, you can prove from the 2-d structure of differentiable functions that the value of the function at every point must be given by a specific integral around that loop. And from that you can start to recover a lot of the structure. Here is a trivial example. The value of a differentiable function inside of the loop has reach its maximum absolute value at the boundary. Now consider a polynomial p(z) which doesn't have 0 as a root. 1/p(z) is differentiable everywhere that p(z) is not zero. If we draw a large enough circle around the origin, the value of 1/p(z) on that boundary is arbitrarily close to 0. But it isn't arbitrarily close to 0 at the origin, and therefore somewhere in that circle it must not be differentiable. And that point is a root of p(z). This is the first proof that I saw that every polynomial over the complex numbers has a root. And it heavily uses the geometric structure of differentiable functions over the complex numbers. Note, that the reals lack this 2-D structure, and it is likewise easy to come up with polynomials over the reals that do not have real roots.
    - hyperpallium2 5y ago
      Sorry, this trivial eg is still way over my head. Is it related to (high school) polnomials always having complex roots, but not necessarily real roots? But that constraint on reals seems like "structure" to me, and complex numbers lack this structure...? (It may well just require a few more years math on ny part.)
      - btilly 5y ago
        Yes. Polynomials over the complex numbers can always be factored completely based on the roots. Polynomials over the real numbers cannot. The result is easy to prove using complex analysis, which uses the 2-D structure of the complex numbers to prove it.
itronitron 5y ago
Interesting, so you're telling me there is a mathematical basis for why things always work out over and over again in the imaginary world but not in the real world?
jesuslop 5y ago
hmm, I've deleted a long reply elaborating the distinction between syntactic and semantic completeness (the terminology is a mess). Just wanted to add that your interesting point shines here: https://math.stackexchange.com/a/117914/150078 https://math.stackexchange.com/a/117914/150078

by btilly 5y ago

hyperpallium2 5y ago

Could you elaborate on the "tremendous amount of structure" of complex numbers please?

ssivark 5y ago
Think of the following: there are many many paths to get from one location to another on a plane. If you imagine a function on the plane, it’s behavior as you walk along each part would have a sanity constraint. That (along with the constraints of behavior under complex conjugation, should you choose to impose “analyticity”) adds up to a lot of constraints, which is the “structure” in complex numbers — which makes complex analysis simple.
- btilly 5y ago
  This is the right idea. For example if you draw a loop, specify a continuous function on the loop, and say it has to be differentiable inside of that loop, you can prove from the 2-d structure of differentiable functions that the value of the function at every point must be given by a specific integral around that loop. And from that you can start to recover a lot of the structure. Here is a trivial example. The value of a differentiable function inside of the loop has reach its maximum absolute value at the boundary. Now consider a polynomial p(z) which doesn't have 0 as a root. 1/p(z) is differentiable everywhere that p(z) is not zero. If we draw a large enough circle around the origin, the value of 1/p(z) on that boundary is arbitrarily close to 0. But it isn't arbitrarily close to 0 at the origin, and therefore somewhere in that circle it must not be differentiable. And that point is a root of p(z). This is the first proof that I saw that every polynomial over the complex numbers has a root. And it heavily uses the geometric structure of differentiable functions over the complex numbers. Note, that the reals lack this 2-D structure, and it is likewise easy to come up with polynomials over the reals that do not have real roots.
  - hyperpallium2 5y ago
    Sorry, this trivial eg is still way over my head. Is it related to (high school) polnomials always having complex roots, but not necessarily real roots? But that constraint on reals seems like "structure" to me, and complex numbers lack this structure...? (It may well just require a few more years math on ny part.)
    - btilly 5y ago
      Yes. Polynomials over the complex numbers can always be factored completely based on the roots. Polynomials over the real numbers cannot. The result is easy to prove using complex analysis, which uses the 2-D structure of the complex numbers to prove it.

itronitron 5y ago

Interesting, so you're telling me there is a mathematical basis for why things always work out over and over again in the imaginary world but not in the real world?

jesuslop 5y ago

hmm, I've deleted a long reply elaborating the distinction between syntactic and semantic completeness (the terminology is a mess). Just wanted to add that your interesting point shines here: https://math.stackexchange.com/a/117914/150078 https://math.stackexchange.com/a/117914/150078