I've come up with a simple way to "derive" Gauss's prime density function using probability heuristics, alone. The "derivation" involves setting up a nonlinear differential equation whose solution happens to agree exactly with Gauss's estimate. I don't know if I've discovered something interesting, or [more likely] old. At any rate, here's my argument:
.. In 1792, when only 15 years old, Gauss proposed that pi(n), the prime density function
pi(n)∼n/(ln n).
Gauss later refined his estimate to pi(n)∼Li(n),
where Li(n) = Integral] (1 / ln x) dx)
..
I have not seen a simple derivation for this estimate, and if it exists, I am surprised why it is not more widely used in expositions on the subject of the distribution of primes. What follows is a very short argument based on a probability model. According to this model, we'll find that the distribution of primes is governed by a non-linear differential equation of the form
Q'(x) = - Q(x) Q( sqrt(x) ) / x
whose solution is
Q(x) = 1 / 2 ln x
It's probably the only solution, but then, I don't know how to prove uniqueness. (Can there be more than one solution to any given non- linear differential equation, anyway? I suspect not.)
Anyway, to me, what's interesting is how using some specious probability arguments about the distribution primes I was able to set up the equation and try a test function inspired by Gauss's estimate to solve it. This hints at something more meaningful in my Clouseau- esque accident.
The setup
The setup, I have come to find, is similar in spirit to Harald Cramer's probabilistic, heuristic arguments for estimating of the distribution primes (the difference here being that we don't use any other results from number theory). Here it is..
Equ. 1: The joint probability of a randomly chosen positive number n not being divisible by two relative primes p and k is (1 - 1/p)(1 - 1/k).
Equ. 2: Define Q(x) = Pi(1 - 1/p) taken over all primes p <= x (Pi stands for the product symbol).
Using Equ.s 1 & 2, vigorous hand waving, and a pinch of salt, we can say
Equ. 3: The probability that a randomly chosen positive number x being either 1 or a prime is Q( sqrt(x) ).
What we're saying here is that for x to be prime, it suffices to show that it is not divisible by any prime less than the square root of x. Equ. 3 says that in the neighborhood of x, the 'average' distance between primes is 1 / Q( sqrt(x) ).
Now the n.d.e. above comes from trying to approximate Q(x) using this probabilistic model. The idea is to use that approximation in order to estimate a prime counting function:
Integral] Q (sqrt(x)) dx
But we don't have an analytic expression that approximates Q, yet. Instead of setting up an integral equation, we try a differential approach. Consider the change in Q as x passes over 2 very large consecutive primes p1 < p2:
Dividing the top equation by the bottom one, you get the n.d.e. I described above.
The solution
I had read somewhere how the 15 year old Gauss had been able to come up with his logarithmic integral for estimating the number of primes less than n. Was his integral inspired by a similar probabilistic argument? Maybe, but googling it, I couldn't find much. So, I plugged in C / ln x and solved for the constant C (=1/2).
Does it mean anything?
I suspect it might, which is why I posted it. I did a bit of cursory reading on the topic, but alas, I'm an amateur. My claim that
Q(x) = Pi( 1 - 1/p) ~ 1 / 2 ln x
does not agree with Merten's asymptotic formula (1874)
Pi( 1 - 1/p) ~ exp(-C) / ln x,
where C is Euler's constant.
Still, there's something here that piques my nose. My result, when plugged into the prime counting integral, agrees exactly with Gauss's estimate:
In article <ba0292e1-4b25-40b5-9d61-7581108bc...@d1g2000hsg.googlegroups.com>,
Babak <farh...@gmail.com> wrote: >we'll find that the distribution of primes is governed by a non-linear >differential equation of the form
>Q'(x) = - Q(x) Q( sqrt(x) ) / x
See
E. M. Wright, A functional equation in the heuristic theory of primes, Math. Gaz. 44 (1960), 15-16.
G. Hoffman de Visme, "The density of prime numbers," Math. Gaz. 45 (1961), 13-14. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
On Jul 16, 4:44 pm, Babak <farh...@gmail.com> wrote:
> I have not seen a simple derivation for this estimate, and if it > exists, I am surprised why it is not more widely used in expositions > on the subject of the distribution of primes. What follows is a very > short argument based on a probability model. According to this model, > we'll find that the distribution of primes is governed by a non-linear > differential equation of the form
> Q'(x) = - Q(x) Q( sqrt(x) ) / x
Strictly speaking, this is a _functional_ differential equation. A DE should depend only on the current value of G(x) and its derivatives. This is a _delay_ differential equation, since the current value of the derivative depend on earlier values of G.
DEs tend to be associated with initial conditions that make the solution unique. On the other hand, a DDE is associated with an initial _set_ of values that makes the solution unique.
> whose solution is
> Q(x) = 1 / 2 ln x
> It's probably the only solution, but then, I don't know how to prove > uniqueness. (Can there be more than one solution to any given non- > linear differential equation, anyway? I suspect not.)
Nonlinear differential equations typically have infinitely many solutions. An initial condition typically pins it down to a unique one. A delay differential equation will also typically have infinitely many solutions and an initial _range_ of values will typically pin it down to a unique one.
Given any differentiable function q(x) on any interval of the form [a, a^2] (a > 1) that satisfies the boundary consistency condition
q'(a^2) = -q(a^2)q(a)/(a^2),
there is a uniquely determined solution Q of
Q'(x)= -Q(x)Q(sqrt(x))/x
obtained on [a^2, a^4] by integrating
Q(x)'/Q(x) = -q(sqrt(x))/x .
Whence, by induction, one obtains a solution for all x > a. The solution can also be extended to (1,a): obtain it on (sqrt a, a) by
In article <9f6826b8-f66a-4f3a-a1b2-bf61baeb2...@r66g2000hsg.googlegroups.com>,
Dan <lueck...@uark.edu> wrote: > > Q'(x) = - Q(x) Q( sqrt(x) ) / x
> Strictly speaking, this is a _functional_ differential > equation. A DE should depend only on the current value of > G(x) and its derivatives. This is a _delay_ differential > equation, since the current value of the derivative depend > on earlier values of G.
When x<1, sqrt(x) is a *future* time. When x=1, x=sqrt(x) ... this suggests expanding Q in powers of (x-1) ??
> On Jul 16, 4:44 pm, Babak <farh...@gmail.com> wrote:
> > I have not seen a simple derivation for this estimate, and if it > > exists, I am surprised why it is not more widely used in expositions > > on the subject of the distribution of primes. What follows is a very > > short argument based on a probability model. According to this model, > > we'll find that the distribution of primes is governed by a non-linear > > differential equation of the form
> > Q'(x) = - Q(x) Q( sqrt(x) ) / x
> Strictly speaking, this is a _functional_ differential > equation. A DE should depend only on the current value of > G(x) and its derivatives. This is a _delay_ differential > equation, since the current value of the derivative depend > on earlier values of G.
> DEs tend to be associated with initial conditions that make > the solution unique. On the other hand, a DDE is associated > with an initial _set_ of values that makes the solution unique.
> > whose solution is
> > Q(x) = 1 / 2 ln x
> > It's probably the only solution, but then, I don't know how to prove > > uniqueness. (Can there be more than one solution to any given non- > > linear differential equation, anyway? I suspect not.)
> Nonlinear differential equations typically have infinitely > many solutions. An initial condition typically pins it down > to a unique one. A delay differential equation will also > typically have infinitely many solutions and an initial > _range_ of values will typically pin it down to a unique one.
> Given any differentiable function q(x) on any interval of > the form [a, a^2] (a > 1) that satisfies the boundary > consistency condition
> q'(a^2) = -q(a^2)q(a)/(a^2),
> there is a uniquely determined solution Q of
> Q'(x)= -Q(x)Q(sqrt(x))/x
> obtained on [a^2, a^4] by integrating
> Q(x)'/Q(x) = -q(sqrt(x))/x .
> Whence, by induction, one obtains a solution for all x > a. > The solution can also be extended to (1,a): obtain it on > (sqrt a, a) by
Thanks for pointing out de Visme's heuristic argument along a similar line of argument. One thing I noted in my original post was that though my claim that
Q(x) = Pi( 1 - 1/p) ~ 1 / 2 ln x
does not agree with Merten's asymptotic formula
Pi( 1 - 1/p) ~ exp(-C) / ln x,
where C is Euler's constant, my result *surprisingly* yielded the correct prime counting integral:
Integral] Q(sqrt(x)) dx = Integral] (1 / ln x) dx
That observation puzzled me, and while attempting to fix my heuristic model, I discovered that in fact *any* model claiming that the probability of x being prime is roughly equal to
Pi(1 - 1/p)
taken over all primes p < x^a, a > 0, yields a d.d.e of the form
Q'(x) + Q(x)Q(x^a)/x = 0
which has a solution of the form
Q(x) = a / ln x .
and which when plugged into the prime counting integral gives
Integral] Q(x^a) dx ~ Integral] (1 / ln x) dx
So one surprise is replaced with another: it doesn't matter whether the model looks into the past (a < 1), or into the future (a > 1)--the resulting prime counting integral is the same.
> In article <ba0292e1-4b25-40b5-9d61-7581108bc...@d1g2000hsg.googlegroups.com>,
> Babak <farh...@gmail.com> wrote: > >we'll find that the distribution of primes is governed by a non-linear > >differential equation of the form
> >Q'(x) = - Q(x) Q( sqrt(x) ) / x
> See
> E. M. Wright, A functional equation in the heuristic theory of primes, > Math. Gaz. 44 (1960), 15-16.
> G. Hoffman de Visme, "The density of prime numbers," Math. Gaz. 45 > (1961), 13-14. > -- > Tim Chow tchow-at-alum-dot-mit-dot-edu > The range of our projectiles---even ... the artillery---however great, will > never exceed four of those miles of which as many thousand separate us from > the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
> Thanks for pointing out that this is in fact a *delay* diff equ. > Re your method of solution..
> > Given any differentiable function q(x) on any interval of > > the form [a, a^2] (a > 1) that satisfies the boundary > > consistency condition
> > q'(a^2) = -q(a^2)q(a)/(a^2),
> > there is a uniquely determined solution Q of
> > Q'(x)= -Q(x)Q(sqrt(x))/x
> > obtained on [a^2, a^4] by integrating
> > Q(x)'/Q(x) = -q(sqrt(x))/x .
> > Whence, by induction, one obtains a solution for all x > a.
> I'm not following the argument. I can't see how one can obtain > a solution by integrating
> Q(x)'/Q(x) = -q(sqrt(x))/x .
If a^2 <= x <= a^4, the a <= sqrt(x) <= a^2. We let Q(x) = q(x) for a <= x <= a^2 and for a^2 <= x <- a^4 we obtain Q(x) by integrating the above equation and using the obvious initial condition. Q(a^2) = q(a^2). Clearly then Q'(x)/Q(x) = - Q(sqrt(x))/x for x in [a^2,a^4] and so we have Q defined on [a,a^4]. Perform the same steps for x in [a^4,a^8] using the values obtained on [[a^2,a^4], etc.
> Let me try a counter example. Consider
> q(x) = - x^2 / 2
> Now > q'(a^2) = -q(a^2)q(a)/(a^2)
> is satisfied for a = 2. However, integrating
> Q'/Q = - x / 2
> does not yield a solution to the original delay diff equ.
But -q(sqrt(x))/x = 1/2 so one needs to integrate 1/2. With this correction it yields ln Q(x) = x/2 + C or
Q(x) = C exp(x/2).
Using the initial condition Q(4) = q(4) = -8 we get C = -8 exp(-2) whence
Q(x) = -x^2/2, on [2,4] Q(x) = -8 exp (x/2 - 2), on [4,16].
which clearly satisfies the equation for all x in [4,16]. for x in the interval [16,256] we would integrate Q'(x)/Q(x) = 8 exp (sqrt(x)/2 - 2) / x and use the initial condition Q(16) = -8 exp(6).
In order to get the equation satisfied on [2,4] we need Q defined on [sqrt(2), 2] one again uses the equation to define it: Q(sqrt(x)) = -x q'(x)/q(x), or
Q(x) = - x^2 q'(x^2)/q(x^2) = - 2, x in [sqrt(2),2].
Unfortunately, Q fails to be differentiable at 2. It also seems we cannot continue to [2^{1/4},2^{1/2}] without sacrificing continuity. However, getting a solution on [4,infinity) is no problem and the proceedure is standard for this type of delay differential equation.