[HN Gopher] Is the largest root of a random real polynomial more...
___________________________________________________________________
Is the largest root of a random real polynomial more likely real or
complex?
Author : jjgreen
Score : 157 points
Date : 2024-05-10 09:01 UTC (13 hours ago)
(HTM) web link (mathoverflow.net)
(TXT) w3m dump (mathoverflow.net)
| Sebb767 wrote:
| For those not well versed in the Stack Exchange network, what's
| the difference between Math Stack Exchange and Math Overflow?
| SassyBird wrote:
| MSE is for noobs, think HS and uni students. MO is for pros,
| think PhDs, researchers etc.
| lupire wrote:
| Adults, who might or not have graduated or attended high
| school or University, also use MSE.
| OJFord wrote:
| There are probably some very clever people who are not PhDs
| nor professional researchers using MO too, GP was just
| giving examples ('think [...]').
| colechristensen wrote:
| This is not a helpful comment.
| jjgreen wrote:
| MO is for research-level discussions, questions are often
| bumped from one to the other (more often from MO to MSE).
| admissionsguy wrote:
| I believe one was for professionals and the other one for
| students. Don't remember which one, I guess Overflow is for
| professionals since we use it for coding.
| Sharlin wrote:
| Honestly it's Stack Overflow that's really the "noob" level
| programming discussion, with 95% of questions incredibly
| elementary, whether "professional" or not.
| cstheory.stackexchange.com is more analogous to
| mathoverflow.com in that it's about research-level questions.
|
| I think Math Overflow was the first overflow instance after
| SO, so it has a sort of a special status and its own domain.
| *.stackexchange.com came later to meet the growing demand for
| more instances.
| HappyPanacea wrote:
| Math Overflow domain and name are owned by a nonprofit
| organization, see
| https://meta.mathoverflow.net/questions/969/who-owns-mathove...
| melenaboija wrote:
| > The MathOverflow corporation is completely independent from
| Stack Exchange and its mission is to ensure the continued
| operation of the site in a manner that meets the needs and
| expectations of the community.
|
| Does it make it more or less hostile than Stack Exchange?
| rvnx wrote:
| They claim "The MathOverflow corporation is completely
| independent from Stack Exchange"
|
| and then when you click Legal notice, you see that all
| conditions and legal of Stack Exchange applies.
|
| In addition:
|
| > This letter is to confirm that, per our communications,
| that MathOverflow ("MathOverflow") agrees that the website,
| MathOverflow (located at www.mathoverflow.net,
| www.mathoverflow.com and www.mathoverflow. org (the
| "MathOverflow Domains"), the assets of which are owned and
| controlled by MathOverflow, will be upgraded to Stack
| Exchange, Inc.'s ("Stack Exchange") version 2.0 software
| model and, in conjunction with this upgrade, MathOverflow
| accepts Stack Exchange 2.0's Terms of Service
| (http://stackexchange.com/legal/tems-of-service, the
| "Terms"
|
| ===
|
| " _completely_ independent " is perhaps exaggerated claim.
| lupire wrote:
| The corporation is independent, and has the right to
| unilaterally leave the website at any time. The current
| website is not independent.
| The_suffocated wrote:
| MSE is sometimes perceived as hostile to beginners because
| it often closes off newbie questions that lack context or
| are thought to be of low quality. MO is not suffering from
| this because almost all such questions are migrated to MSE.
| In this sense, yes, MO is less hostile than MSE because its
| hostility is outsourced to MSE.
| barrenko wrote:
| > Thievery corporation
| mamediz wrote:
| My experience with MathOverflow was funny, the first time I
| posted a question it was almost immediatly downvoted and
| criticized by the moderators. Later the same day, I was
| surprised when my question was answered by Terence Tao,
| then the moderators quickly changed their opinion and
| people started to upvote the question.
| renewiltord wrote:
| Mathematics even worse than Software Engineering because
| lay practitioner can miss significance of question. If
| you have link to question, would appreciate. Just out of
| curiosity to see if I would realize why question is
| interesting. That in itself is a skill.
| mamediz wrote:
| To be fair the question was kind of naive, about
| language, but T. Tao bother to answer and made my day.
| https://mathoverflow.net/questions/206544/why-do-people-
| use-...
| renewiltord wrote:
| Thank you for responding. Assuaged my curiosity :)
| selimthegrim wrote:
| Looking through the links from here one of the answers by Boris
| Hanin may have resolved a question I had for a while.
| meindnoch wrote:
| As there's no formula for degree-5 polynomials and above, how do
| you distinguish between a real root vs. a real number +
| epsilon*i?
| jcla1 wrote:
| Just because of the impossibility of an exact formula for the
| roots of a high-degree polynomial that does not rule out the
| possibility of figuring out, say, the distribution of roots of
| such polynomials. The question is not about any polynomial in
| particular (hence Abel's theorem is no barrier).
|
| edit: Think of the following example: take a polynomial a_n x^n
| + ... + a_0, where the coefficients a_i are i.i.d. Bernoulli
| random variables. Even though the degree n might be large (> 4)
| I can say with confidence that such a polynomial has a real
| root (x = 0) with probability 1/2. Similar though more
| sophisticated arguments are at work in the linked question.
| vouaobrasil wrote:
| The estimations are not based on numerics, but rather
| reasoning. For example, since real-valued polynomials are
| stable under complex conjugation, each complex root must have a
| complex conjugate. For polynomials with three distinct roots
| then, one must be real. This is just an EXAMPLE of the kind of
| reasoning that can be used to infer whether a root is purely
| real or complex.
|
| As for formulas: no, there is no GENERAL formula for the
| quintic polynomial or higher. General formulas only exist for
| polynomials of degree four or lower.
|
| Of course, there are SPECIFIC formulas for specific KINDS of
| higher-degree polyomials. But for the general quintic and
| higher, nothing. That's already been proved classically.
| orlp wrote:
| > As for formulas: no, there is no GENERAL formula for the
| quintic polynomial or higher.
|
| Note that there is no general formula for quintics (and
| higher) _in radicals_.
|
| But it is a bit arbitrary to draw the line there. There is
| also no 'general formula' using basic arithmetic operations
| (addition, subtraction, multiplication, division) that gives
| roots for x^2 = 2.
|
| However it is quite useful to be able to talk about those
| roots, so we added a dedicated symbol for them (sqrt), and
| more generally extended the definition of exponentiation to
| fractional powers.
|
| If it were generally useful to talk about the roots of
| quintics we also would've added common notation for their
| roots.
| lanstin wrote:
| People sort of have:
| https://en.wikipedia.org/wiki/Bring_radical
| yorwba wrote:
| You use Budan's theorem
| https://en.wikipedia.org/wiki/Budan%27s_theorem to certify that
| there's exactly one (or exactly zero) real root in the interval
| ( _r_ - e, _r_ + e]. As long as your guess _r_ is within e of a
| root, this allows distinguishing between real and complex roots
| even if you don 't have the exact value. (There are instances
| where Budan's theorem cannot give an answer, e.g. most
| obviously if there's more than one root in the interval.)
| red_trumpet wrote:
| I could imagine that one uses the derivative here. If x+iy and
| x-iy are complex roots, and y is really small, then the
| derivative at x should be small as well (if y = 0, you have a
| double zero, so p'(x)=0). So if p'(x) is far away from zero,
| you have a single real zero. And I guess double zeroes don't
| really happen with random coefficients.
| HappyPanacea wrote:
| If we have polynomial with integer coefficients, we can bound
| the largest and smallest root by an expression that is maximum
| of abs. value of coefficients to some power. There should be
| similar bounds for the smallest imaginary value a root can
| have. If your polynomial have real numbers for coefficients
| having a formula doesn't help - we have the same problem of
| being not able to tell if number is equal zero. For example in
| the quadratic formula, we can have discriminant equal minus
| epsilon. If epsilon was zero there is no imaginary root but if
| it isn't we have an imaginary root.
| bmacho wrote:
| I think the others answered your question.
|
| In addition, I suggest you to forget this "degree-5
| polynomials" thing. They are not different from degree 2
| polynomials in any meaningful way, if you talk about computing
| their roots or reason about their roots.
| LeonardoTolstoy wrote:
| I assume this is one of these things where it is just like ... Oh
| I don't know this kind of math.
|
| Because in my mind I take a "random" polynomial and only consider
| the largest two roots. Then I consider doing two reflections:
| across the bottom/top of the curve, and across the x axis. If
| those top two roots aren't degenerate the combination of the four
| curves using the reflections have two curves with a largest real
| root, and two curves with a largest imaginary root (I think). If
| it is degenerate then the largest root is real.
|
| So I would then conclude that (1) there are more real largest
| roots than imaginary and (2) the "advantage" is vanishingly small
| since there are infinitely more curves with a largest non-
| degenerate root than degenerate root.
|
| As one can tell I barely know the proper nomenclature. I assume
| I'm mainly missing some consideration about the uniformity of
| random coefficients not resulting in a uniform distribution in
| space? Or possibly my reflections can violate one of the
| conditions (real coefficients?)? Or I'm just plain wrong.
| red_trumpet wrote:
| I don't follow your reflections. Reflecting the graph of a
| polynomial p(x) along the x-axis means replacing p(x) by p(-x).
| What do you mean by reflecting "across the bottom/top of the
| curve"? Reflecting along the y-axis? This would mean p(x) is
| replaced by -p(x).
| r0uv3n wrote:
| What do you mean with reflection across the bottom/top of the
| curve?
|
| With combination you presumably mean taking
| (polynomial+reflected polynomial)/2?
| LeonardoTolstoy wrote:
| Ah there we go. At the very least I was only considering the
| subspace where the polynomial's derivative has only real
| roots?
|
| Think of a simple quadratic. The reflection is across the
| min(f(x)) axis. But for higher order polynomials that doesn't
| always work so that is certainly an issue. Thanks
| keepamovin wrote:
| Wow that is crazy that it is between chance and 1/phi.
|
| _In the linked MSE post, it has now been proved that the
| probability that the largest root is real is at least 6.2%_
|
| And! at least 1/10th of 1/phi. Hahaha! :)
|
| I think the link of phi with number theory is natural in the
| sense that the primes are not _random_ (as is commonly mistaken)
| but arise recursively from prior primes: they are exactly the
| 'gaps' not hit by multiples of prior primes, so there's a kind of
| 'natural growth' pattern that we would expect to reflect some
| manner of _e_ and _phi_
|
| It's a very fundamental pattern of magnitude. Truth beauty et all
| dr_dshiv wrote:
| Beautiful! Phi always seems to show up when there is crowding.
| keepamovin wrote:
| Thanks! Let's collab. We should start a "primes
| investigation" github repo!!!! :)
| dr_dshiv wrote:
| I love number theory in the context of waves. Here's an
| example:
|
| https://arxiv.org/pdf/1910.10751
| keepamovin wrote:
| Thanks. When you mention waves I'm wondering how we can
| leverage FFT to make a function of the next prime? But it
| has to include all primes, I guess? Or is this Reimann,
| already??
| Analemma_ wrote:
| Yes, high-level descriptions of the RH are often vague
| about the connection between the zeta function and the
| prime numbers, but the short version is that, for every
| point on on the complex plane, there is an associated
| real-valued "Riemann harmonic function"; if you sum up
| the Riemann harmonic functions of the non-trivial zeros
| of the zeta function, you get the exact prime-counting
| step function, and this function is much more predictable
| if the RH is true.
| RockofStrength wrote:
| You get e from the primes by averaging the set size of only
| growing gaps (2,3,5)(7,11)(13,17)... or only non-shrinking gaps
| (2,3,5,7,11)(13,17)(19,23,29)...
|
| Not a property of primes per se, just a property of growth. It
| works better with a set of randoms.
| lupire wrote:
| I thinking that works for primes because primes are
| approximately randomly distributed.
|
| But primes are more directly "e" if you look at p(N), the
| count of primes less than N. That is ~n/ ln(n)
|
| https://en.m.wikipedia.org/wiki/Prime-counting_function
| kzrdude wrote:
| This reminds me of another of those random polynomial questions
| where a good answer showed why the question asker's chosen
| distribution of coefficients lead to the result they were seeing.
| Anyone recognize that and have a link? It seems like it's
| relevant here too.
| evertedsphere wrote:
| https://math.stackexchange.com/questions/206890/the-egg-biza...
| ?
| kzrdude wrote:
| There's many of these then!
|
| I think I found the one I was thinking of:
|
| https://mathoverflow.net/questions/182412/why-do-roots-of-
| po...
|
| See answer by Will Jagy
| Aardwolf wrote:
| Does random polynomial mean with real or complex coefficients,
| though?
| cool_dude85 wrote:
| The question posed specifies real coefficients, looks like
| experimentally tested on random coefficients selected from [-1,
| 1] from uniform distribution and a sort of scaled normal
| distribution.
| VyseofArcadia wrote:
| This stuck out to me, and might be a place where our
| intuition breaks down. As a human if you tell me, "give me a
| polynomial with degree n with coefficients drawn uniformly
| from [-1, 1]," I will likely produce one with a lot of zero
| coefficients, just because a lot of the polynomials you
| encounter in the wild have a lot of zero coefficients.
|
| In reality, if you grab a coefficient uniformly from [-1,1],
| the probability that it is 0 is 0.
| madcaptenor wrote:
| And how random sparse polynomials behave would _also_ be an
| interesting question!
| Someone wrote:
| Likely (can't find the test code) tested on coefficients
| selected from the IEEE doubles in _[-1, +1]_ , so from a
| subset of numbers of the form _m /2^n_ with _m_ and _n_
| integers and m between 0 and 2^53 or thereabouts.
|
| For this problem, I don't see how that would give different
| results as picking any number in _[-1, +1]_ , but that's not
| a priori guaranteed. Mathematicians are well-trained as
| nitpickers.
| eternauta3k wrote:
| Aren't roots continuous functions of the coefficients?
| Smaug123 wrote:
| The first line of the question says "real".
| Aardwolf wrote:
| But that first line is a statement about something related
| but different (the question is about largest root, this
| statement is about any root), and the term "random
| polynomial" without specifying anything is used throughout
| the rest of the content
| outop wrote:
| If you allow complex coefficients then the distribution is
| rotationally symmetric in the complex plane. So a root is no
| more likely to be in the real line than in any other line
| through the origin. There would be vanishingly few polynomials
| with largest root real.
| dailykoder wrote:
| Sligthly OT, but I always have fun reading these maths posts
| here: I enjoyed maths in University very much (I did CE) and the
| joy from my professors always inspired me. I'd love to learn more
| and dip into solving things with it (maybe with a hang towards
| numerics?).
|
| But been out of Uni for 2 years now and haven't done much, so
| probably gotta relearn quite a bit. Does anyone have some good
| ideas where to start and find fun ideas? Not numerics, but I did
| quite a few projecteuler problems during my bachelors - maybe
| there is more like this? Any ideas welcome.
|
| Or should I redo all tasks in my textbooks first? ;-)
| belter wrote:
| A bit unrelated to your question, but if you love Maths you
| can't miss The Math Sorcerer:
| https://www.youtube.com/@TheMathSorcerer
| jjgreen wrote:
| I went to University to do maths 6-7 years after leaving school
| (with decent maths results there) so had forgotten everything.
| Nothing beats doing all of the exercises that you did before,
| and actually you find that you've not quite forgotten
| everything. Good luck!
| jsbg wrote:
| Mathematics and Its History
| (https://link.springer.com/book/10.1007/978-1-4419-6053-5) is a
| really fun book for someone who already knows some university-
| level math and is just curious.
|
| You may also enjoy Proofs from THE BOOK
| (https://link.springer.com/book/10.1007/978-3-662-57265-8).
| ptero wrote:
| Two questions that immediately jump at me are:
|
| 1. What is random? They seem to have run numerical experiments,
| which suggests they used (something equivalent to) uniform
| integer coefficients from a bounded set. And that could affect
| findings a lot!
|
| 2. Did they look at odd degrees (which always guarantees a real
| root), even or both?
|
| Not trying to knock this fun post.
| lupire wrote:
| Right. The problem as originally stated is paradoxical. You
| can't uniformly sample an unbounded set. I assume OP used a
| random floating point, which are bounded.
|
| But the important part of the question can be asked for any
| possible distribution over the reals.
| Sesse__ wrote:
| > You can't uniformly sample an unbounded set.
|
| You definitely can, as long as you are accepting the use of
| limits (and without that, sampling uniform over any
| subsegment over the reals would also be pretty much
| impossible). "Sample a uniformly chosen real" just is
| shorthand for "Sample over a real from [-N, +N], then we are
| interested in the answer then N goes to infinity". In this
| case, then just divide all the coefficients by N, which
| obviously doesn't affect the answer. So uniformly over
| [-1,+1] is as good as uniformly over R.
|
| This is very common. See e.g. number theory, where you can
| ask questions like "what is the probability that two randomly
| chosen natural numbers are coprime"; the set of natural
| numbers is also unbounded, but it doesn't prevent this
| question from being well-defined, and the answer is 6/p2
| (about 0.608).
| mturmon wrote:
| This comment opens perhaps a bigger can of worms than
| intended.
|
| The original statement:
|
| >> You can't uniformly sample an unbounded set. [of real
| numbers]
|
| is true, provided that by "sample" we mean "sample in a way
| that obeys Kolmogorov's axioms". (I'm adding the qualifier
| "of real numbers" to keep this somewhat grounded, so that
| we don't get distracted with abstract metric spaces, or
| worse.)
|
| In the above comment, by placing a limit on the set ([-N,
| +N]), you have made the set bounded -- thereby
| contradicting the premise of the original statement!
|
| *
|
| Your last paragraph is interesting. Suffice to say, the
| situation is more complex than your summary would indicate,
| precisely because of the above issue -- that is, "what does
| _sample_ mean in this context ".
|
| It led me to this paper by Lei and Kadane (the latter, the
| well-known Bayesian theorist) --
|
| https://arxiv.org/pdf/1806.00053
|
| It is well worth a read, if you're into such things.
|
| See especially section 1, and section 7. In essence, the
| above statement ("You can't sample uniformly") is true, if
| you interpret "sample" as "following a scheme that obeys
| Kolmogorov's well-known axioms."
|
| However, if you are willing to abandon countable additivity
| of your measure, and drop down to finitely-additive
| measures, there are several classes to choose from that
| support a notion of uniformity.
|
| (Again, just thinking about measures on integers, not real
| numbers. But these are not measures in Kolmogorov's sense.)
|
| These measures support a notion of uniformity, but they may
| not have some properties that you might hope for, such as
| that the measure of a set A of natural numbers is the same
| as A + i, where adding "i" shifts the set by "i" units left
| or right.
|
| For some of these measures, the result about co-prime
| numbers is as you say, and for others, the result is
| indeterminate. That is, the limit exists, but it can be any
| number between 0 and $6/\pi/\pi$.
| selimthegrim wrote:
| >It led me to this paper by Lei and Kadane (the latter,
| the well-known Bayesian theorist)
|
| Not to be confused with J-P Kahane, who also studied this
| area (of the posted article) via Fourier series
| ptero wrote:
| They are also discrete, coming from a finite set. I remember
| at the undergrad differential equation class the teacher
| asked for some arbitrary coefficients so we can classify the
| result.
|
| She was very surprised that 3 differential equations she got
| all turned out to be parabolical, which form a set of
| probability 0 in the full space. But of course the
| probabilities for equations with small integer coefficients
| (which is what people wrote) are totally different.
| owyn wrote:
| > You can't uniformly sample an unbounded set.
|
| I was asked to do this as an interview question at google! It
| was about log file processing not math (sample an infinite
| log file), and I thought my answer was ok? But that's why I
| don't work there so I am really enjoying this whole thread.
| CamperBob2 wrote:
| What's the correct answer (and what was yours?)
| racional wrote:
| It was definitely a dorky question, and a glaring sign of
| the interviewer's lack of maturity to blithely burn through
| your precious time in this manner.
|
| But if you're going to play Google's game, this what you're
| going to get apparently. So what was your answer? One
| guesses that they weren't so much interested in a correct
| answer (since it's a bullshit question anyway) but that you
| were willing to give it a college try (and reference some
| concepts about limits, and perhaps the geometric layout of
| this "file system" in the Doctor Who universe they were
| apparently hoping to roll out this system in, the speed of
| light, etc).
|
| In order to, you know, pass their dipshit filter.
| pedrosorio wrote:
| Given the description about "sampling from an infinite"
| something, it's likely this was intended as a variation
| on reservoir sampling.
|
| https://en.wikipedia.org/wiki/Reservoir_sampling
| hcks wrote:
| "I refuse categorically to entertain fun maths problems."
| ykonstant wrote:
| As stated in the definition, the distribution is that of
| independent uniform coefficients in (-1,1).
| thaumasiotes wrote:
| True, but they give a justification for that distribution,
| and the justification is incorrect.
|
| > The number of real roots of a random polynomial with real
| coefficients is much smaller than the number of complex
| roots. Assume that the coefficients are independently and
| uniformly random in (-1,1) for if not then we can divide each
| coefficient by the coefficient with the largest absolute[]
| value to scale each coefficient to (-1,1).
|
| It's true that, if you have a known polynomial, you can
| normalize the coefficients in this way... assuming that (-1,
| 1) is a typo for [-1, 1].
|
| But it's definitely not true that you can produce a random
| polynomial as the input to this idea. That can't be done. The
| concept of "a random polynomial with real coefficients" is
| known not to exist; you would have to specify some kind of
| nonuniform distribution over the reals for the coefficients
| to be drawn from. Once you have done this, the distribution
| of coefficients in the normalized equivalents that have
| coefficients in [-1, 1] will not be uniform.
|
| In other words, the question says "assume that the
| coefficients [of a random polynomial] are independently and
| uniformly random in (-1, 1), for if not, we can exhibit
| equivalent polynomials whose coefficients are dependently and
| nonuniformly random in [-1, 1]". But this observation makes
| no sense.
| ykonstant wrote:
| >The concept of "a random polynomial with real
| coefficients" is known not to exist;
|
| The concept of "a random polynomial with real coefficients"
| means what we want it to mean in each particular context;
| mathematicians don't care too much about semantics, but
| interesting questions. And this is a very interesting
| question.
| eternauta3k wrote:
| He's right that this question isn't fundamentally about
| the space of polynomials but rather about this particular
| arbitrary-looking distribution. But yeah, the question is
| interesting nonetheless.
| scotty79 wrote:
| I don't get your objection. (-1,1) uniform distribution
| is equivalent to uniform distribution over (-inf, +inf)
| when it comes to roots. Because you can map one to one
| between those intervals without affecting roots.
| SassyBird wrote:
| There is no such thing as a uniform distribution over
| (-inf, +inf) for the simple reason that the Lebesgue
| measure of (-inf, +inf) is +inf and you can't make the
| integral of the whole set be equal to 1. For it to be a
| valid probabilistic distribution, the integral of 1 over
| the whole space relative to the distribution should be 1.
| But it won't be. In U(a,b) a and b need to be bounded, so
| that you can divide the Lebesgue measure by (b - a).
| jprete wrote:
| It's impossible to have a uniform distribution over
| (-inf, +inf). Since the distribution is uniform, each
| digit has an equal chance of being zero through nine, but
| that's true of all digits or else it's no longer uniform.
| At that point 100% of the values drawn in any sampling of
| the distribution are infinite.
|
| If scaling the coefficients down to [-1, +1] gives a
| uniform distribution, then the original distribution must
| have been uniform in [-x, +x] where x is the largest
| coefficient in the original polynomial. That, too, is a
| contradiction! It means that either -x or +x are
| guaranteed to be one of the random coefficients, which is
| very much not a uniform distribution.
|
| (That said, I think the (-1, +1) version of the problem
| still has interest!)
| zeroonetwothree wrote:
| Typically you would do a uniform (-x, x) and then take
| the limit.
|
| This isn't necessarily the same as the (-1, 1) version
| though
| eigenket wrote:
| Its the same because the uniform distribution over (-x,x)
| can be obtained by taking a random variable uniformly
| distributed over (-1,1) and multiplying it by x.
|
| This doesn't change anything for the question about roots
| because the roots of a polynomial don't change when you
| multiply it by x.
| shiandow wrote:
| There are ways to 'approximate' a uniform distribution by
| just choosing greater and greater ranges and then
| normalising the result.
|
| One problem with this is that there's no one way to do it.
| You could argue equally well that the result should be a
| uniformly random point on the S100 sphere.
|
| What it definitely cannot be is a random point on the
| [-1,1]^100 cube because if you divide by the greatest
| absolute value then one of the coefficients will be -1 or 1
| andrewla wrote:
| If you're looking to run some numerical experiments with this,
| there is built-in support in R for this sort of thing:
| plot(polyroot(runif(101,-1,1)))
|
| will give you a visualization of the roots of a polynomial of
| degree 100.
| dahart wrote:
| > Assume that the coefficients are independently and uniformly
| random in (-1,1) for if not then we can divide each coefficient
| by the coefficient with the largest absolutely value to scale
| each coefficient to (-1,1).
|
| Not sure about my intuition here, but won't scaling this way
| create a non-uniform distribution for all but the largest coeffs,
| since the scaling is a divide?
| klyrs wrote:
| Yeah, but you can't pick a random real number with uniform
| distribution. Something's gotta give.
| axblount wrote:
| The important thing is that multplying a polynomial by a
| constant doesn't change its roots.
| dahart wrote:
| Ah right, of course. The original coefficient space would
| still be uniform. I guess this means that a bi-unit [-1,1]
| interval for the random numbers is equivalent to any space of
| coefficients [-n,n] when it comes to answering the
| probability question?
| charlieyu1 wrote:
| Now it surprised me as a guy who studied a lot of math. I would
| think the complex plane is much larger than the real line so the
| probability would go towards 0.
| guyomes wrote:
| That would probably be true if all the coefficients were
| complex numbers. The biais here comes from the fact that all
| the coefficients are real numbers.
___________________________________________________________________
(page generated 2024-05-10 23:00 UTC)