5 ms·

I am not sure why the point that reals appear to be more likely would be counterintuitive. If instead of uniform random real coefficients in an interval one pi

by pama 2y ago

I am not sure why the point that reals appear to be more likely would be counterintuitive. If instead of uniform random real coefficients in an interval one picked uniform random roots in a circular disk centered at the origin of the complex plane as the way to construct polynomials, there is a fat chance that one almost never gets a polynomial with real coefficients back, whereas if the random roots were real then the polynomial is guaranteed real. So a priori neither of the possible answers would be surprising and getting reals to be more likely seems a bit more intuitive to me, though not necessary.

eigenket 2y ago
A polynomial of order n has n roots (real and complex). A random polynomial has roughly log(n) of those be real and n - log(n) be nonreal. For large n, log(n) is tiny compared to n, so it's very surprising that over half the time the largest root is one of the tiny number of real roots
- pama 2y ago
  Thanks. Finding the right combo to balance the largest root if it were non-real and end up with real coefficients seems harder than balancing a smaller root, ie the polynomial with the larger real root also has an n-1 degree polynomial with real coefficients and smaller maximal root as a factor, whereas for the one with a non-real largest root this n-1 polynomial factor has to have non-real coefficients in a way that exactly balance the largest term to make all coefficients real again and now has less space in the complex plane in order to do so. I am not explaining it well as I am away from a keyboard and I am sure it is very complicated to prove but I am not a priori very surprised either way.
  - eigenket 2y ago
    I think you're thinking about it in a slightly weird way, if I am understanding you correctly. Picking random roots would be a very different thing. Here were only dealing with polynomials whose complex roots come in conjugate pairs (I.e. if a+ib is a root then so is a-ib) this gives you real coefficients "for free" since (x - (a+ib))(x-(a-ib)) = x^2 -2ax + a^2 + b^2 Theres no coincidental balancing act in choosing the roots to obtain real coefficients. It happens because of the type of polynomials we've chosen.
    - pama 2y ago
      Say that it is true that for degree N-1 polynomials the largest root is real with more than 0.5 probability. Then for degree N, either you multiply a degree N-1 with real coefficients by a linear (first order) polynomial with a real root that may also be larger than the one in the N-1,, or you multiply it with a first order polynomial of a complex root that has to be conjugate to an existing one in the (non-real coefficient) N-1 degree which can only be as large as the one already in the degree N-1; that particular degree N-1 polynomial has to have a distribution that is similar to those of the ones with real coefficients except for one additional unbalanced root. For a degree 1 polynomial with real coefficients the probability that the root is real is trivially 1, so my natural guess is that with a careful framing of a (double?) induction one could work out a proof that the probability remains larger than 0.5 for all N. (The second induction would be over polynomials that are products of those with real coefficients times a complex first order polynomial). I am guessing it is not that simple because somebody would have blurted it out, but my own (perhaps slightly weird) way to think about it doesn’t lead to any prior surprise.
      - eigenket 2y ago
        This reasoning doesn't quite work. If you're multiplying a polynomial P by a linear factor L = (x-a) with a being a nonreal complex number then at least one of these statements is true 1. P does not have real coefficients 2. PL does not have real coefficients I think you try to avoid this in what you say by multiplying P by a linear term with a complex root whose conjugate is already in P, but if the conjugate is in P then P wasn't real to start with. The term with the complex root has to "balance" with its conjugate for things to work out. E.g. (x-a)^n (x-a*)^k has real coefficients only when n=k or a is real.
        pama 2y ago
        That’s why you have to do a double induction (I’m not a mathematician so my terminology may be wrong) and also keep proving that those special polynomials that are a product of a complex linear term with a polynomial of real coefficients also have a higher (or equal) to 50% chance of having the largest root be real. For degree 2, I think one can prove this directly, and the rest is also an induction.

by pama 2y ago

eigenket 2y ago

A polynomial of order n has n roots (real and complex). A random polynomial has roughly log(n) of those be real and n - log(n) be nonreal. For large n, log(n) is tiny compared to n, so it's very surprising that over half the time the largest root is one of the tiny number of real roots

pama 2y ago
Thanks. Finding the right combo to balance the largest root if it were non-real and end up with real coefficients seems harder than balancing a smaller root, ie the polynomial with the larger real root also has an n-1 degree polynomial with real coefficients and smaller maximal root as a factor, whereas for the one with a non-real largest root this n-1 polynomial factor has to have non-real coefficients in a way that exactly balance the largest term to make all coefficients real again and now has less space in the complex plane in order to do so. I am not explaining it well as I am away from a keyboard and I am sure it is very complicated to prove but I am not a priori very surprised either way.
- eigenket 2y ago
  I think you're thinking about it in a slightly weird way, if I am understanding you correctly. Picking random roots would be a very different thing. Here were only dealing with polynomials whose complex roots come in conjugate pairs (I.e. if a+ib is a root then so is a-ib) this gives you real coefficients "for free" since (x - (a+ib))(x-(a-ib)) = x^2 -2ax + a^2 + b^2 Theres no coincidental balancing act in choosing the roots to obtain real coefficients. It happens because of the type of polynomials we've chosen.
  - pama 2y ago
    Say that it is true that for degree N-1 polynomials the largest root is real with more than 0.5 probability. Then for degree N, either you multiply a degree N-1 with real coefficients by a linear (first order) polynomial with a real root that may also be larger than the one in the N-1,, or you multiply it with a first order polynomial of a complex root that has to be conjugate to an existing one in the (non-real coefficient) N-1 degree which can only be as large as the one already in the degree N-1; that particular degree N-1 polynomial has to have a distribution that is similar to those of the ones with real coefficients except for one additional unbalanced root. For a degree 1 polynomial with real coefficients the probability that the root is real is trivially 1, so my natural guess is that with a careful framing of a (double?) induction one could work out a proof that the probability remains larger than 0.5 for all N. (The second induction would be over polynomials that are products of those with real coefficients times a complex first order polynomial). I am guessing it is not that simple because somebody would have blurted it out, but my own (perhaps slightly weird) way to think about it doesn’t lead to any prior surprise.
    - eigenket 2y ago
      This reasoning doesn't quite work. If you're multiplying a polynomial P by a linear factor L = (x-a) with a being a nonreal complex number then at least one of these statements is true 1. P does not have real coefficients 2. PL does not have real coefficients I think you try to avoid this in what you say by multiplying P by a linear term with a complex root whose conjugate is already in P, but if the conjugate is in P then P wasn't real to start with. The term with the complex root has to "balance" with its conjugate for things to work out. E.g. (x-a)^n (x-a*)^k has real coefficients only when n=k or a is real.
      - pama 2y ago
        That’s why you have to do a double induction (I’m not a mathematician so my terminology may be wrong) and also keep proving that those special polynomials that are a product of a complex linear term with a polynomial of real coefficients also have a higher (or equal) to 50% chance of having the largest root be real. For degree 2, I think one can prove this directly, and the rest is also an induction.