February 14

The number of solutions for Brocard's Problem is finite! Proof: Let n! +1 = m^2, and n! = m^2 -1 = (m +1)*(m -1). Let k = m +1, and n! = k*(k -2). If we examine it carefully, we have the following: 4! = 1*2*3*4 = 6*(6 -2) = 6*4 = 1*(2*3)*4 = 24, 5! = 1*2*3*4*5 = 12*(12 -2) = 12*10 = 1*2*3*(2*2)*5 = 1*2*3*4*5 = 120, 6! = 1*2*3*4*5*6 = 30*(30 -2) = 30*28 = 1*2*3*5*(4*7) = 840 <> 720, 7! = 1*2*3*4*5*6*7 = 72*(72 -2) = 72*70 = 1*2*3*4*(3*2)*5*7 = 1*2*3*4*5*6*7 = 5040. Note: 6*1 = 6, 6*2 = 12, 6*5 = 30, 6*12 = 72, 6*27 = 162, and if n = 1, 2, 3, 4, 5, etc., then f(n) = 2^n -n, and if we adjust for n = 4, 5, 6, 7, etc., then we have the following general formula which contains only "n": n! = {6*[2^(n-3) -(n-3)]}*{6*[2^(n-3) -(n-3)] - 2}. Thus, the number of solutions for Brocard's Problem is finite as the LHS doesn't grow as fast as the RHS growing exponentially, vice versa! Do you like the argument??? I thought that you would find the answer interesting. You are probably tell me that it's original research. It's just a very good idea. 108.242.169.13 (talk) 18:32, 14 February 2016 (UTC)[reply]

You found a formula that gives the three known solutions and no others. How do you know there are no other solutions given by some other formula? -- BenRG (talk) 21:12, 14 February 2016 (UTC)[reply]

If I were to finish the proof, then I'd drop the -2 from the RHS such that the RHS would become 36*2^(2*(n-3)). Then,... I'd search for (OR) most likely prove that some other exponential function is easily larger than n!, so we'd have n! < (some proven exponential function that's definitely larger then n!) < 36*2^(2*(n-3)), and we'd have it. Viola! 108.242.169.13 (talk) 19:26, 15 February 2016 (UTC)[reply]

It's simpler than looking for types of curves or new solutions. Just use the information we already have to plot 6*X *(6X -2) VS. X!, verify the three solutions of 4, 5, and 7, and notice that the parabolic curve retreats indefinitely! The current information of the first curve has to at least hold, because it's known to have that form in three places, and we're talking about both curves in the most general of terms without assuming anything new or remotely probable. Inch by inch,... Life's a cinch!108.242.169.13 (talk) 03:04, 16 February 2016 (UTC)[reply]

It is not possible that any solution of this problem could resemble what you've written. --JBL (talk) 04:23, 16 February 2016 (UTC)[reply]

voila not viola which is a musical instrument. --AboutFace 22 (talk) 23:29, 16 February 2016 (UTC)[reply]

Let ${\displaystyle P_{N}}$ denote the Nth prime. I am correct, no, in assuming that the sequence

${\displaystyle (1-1/2)\cdot (1-1/3)\cdot (1-1/5)\cdots (1-1/P_{N})}$

converges to 0 as ${\displaystyle P_{N}}$ goes to infinity?

Question 1. At what rate does this sequence converge?

As ${\displaystyle 1/P_{N}}$? ${\displaystyle 1/(P_{N})^{2}}$? ${\displaystyle 1/\ln(P_{N})}$? ${\displaystyle 1/(\ln(P_{N}))^{2}}$? Or in some other way?

Question 2. Also, is there any constant factor involved, e.g., ${\displaystyle 2/P_{N}}$ or ${\displaystyle 1/(2P_{N})}$?

Bh12 (talk) 23:20, 14 February 2016 (UTC)[reply]

The simplest approach to something like this is to take the logarithm in order to convert an infinite product into an infinite sum. Then you can bound the terms of the sum to get conclusions. For example, the fact that your sum diverges to 0 (this is the usual language for an infinite product; it is analogous behavior is a sum of a series going to negative infinity) follows from the fact that the sum of the reciprocals of the primes diverges. Probably bounds on the latter rate of divergence can give you information about the rate of divergence of the product. --JBL (talk) 02:19, 15 February 2016 (UTC)[reply]

First, thanks to whoever fixed up my math to look like math!

On to business.

You say "For example, the fact that your sum diverges to 0 (this is the usual language for an infinite product; it is analogous behavior is a sum of a series going to negative infinity) follows from the fact that the sum of the reciprocals of the primes diverges."

But the infinte product mentioned above does not diverge: all of its terms are ${\displaystyle >0}$ and ${\displaystyle <1}$. Perhaps I was wrong when I assumed it converges to 0, maybe it converges to some small positive value. But diverge? How?

And taking logarithms to change this infinite product to an infinite sum has a price. It means summing over all ${\displaystyle ln(1-1/P)}$, which for all ${\displaystyle P}$ is always less than ${\displaystyle -1/P}$, so that its divergence is worse than even that of summing the reciprocals of the primes. In other words, it means trading an infinite convergent product for an infinite divergent sum.

Hmmm - would it help to sum ${\displaystyle (ln(1-1/P))+1/P}$ for all ${\displaystyle P}$? I suspect that sum is convergent.

Bh12 (talk) 09:16, 15 February 2016 (UTC)[reply]

It is called "diverging to zero" in the language of infinite products. Infinite products (of positive numbers) are called convergent if the limit of finite partial products is a positive real number. If the limit exists and is zero, it is called "divergent". The reason is that infinite products are essentially always dealt with by taking logarithms, and the sum of the logarithms of your terms diverges to ${\displaystyle -\infty }$. Replacing ${\displaystyle \log(1+x)}$ by x is not making a huge mistake, as the error is of order ${\displaystyle x^{2}}$ (and using Taylor series, you can get it smaller if you like). —Kusma (t·c) 10:33, 15 February 2016 (UTC)[reply]

OK, I accept all that you say, and thank all of you for your help - and patience.

So, I've replaced the infinite divergent product by an infinite divergent logarithmic sum, and replaced the latter by an infinite divergent sum over ${\displaystyle -1/P}$ with a second order error of ${\displaystyle 1/P^{2}}$.

And if need be I can add to each ${\displaystyle ln(1-1/P)}$ the term ${\displaystyle 1/P}$, to get a convergent sum ${\displaystyle [(1/P)+ln(1-1/P)]}$ with the same error term as before.

But - I have no idea what comes next...Bh12 (talk) 11:40, 15 February 2016 (UTC)[reply]

The sequence tends to zero. More generally, consider the infinite product

${\displaystyle {\frac {1}{\zeta (s)}}=\prod _{n=1}^{\infty }\left(1-{\frac {1}{p_{n}^{s}}}\right)}$

which is the reciprocal of the Riemann zeta function. The zeta function has a pole at ${\displaystyle s=1}$, so this is a zero of the infinite product. The sequence converges very slowly. One should be able to estimate the rate of convergence (aka "divergence to zero") by using the prime number theorem. Sławomir
Biały 12:00, 15 February 2016 (UTC)[reply]

I assumed at the start that the sequence tends to zero, which you have confirmed with a reason (the sequence is the reciprocal of a function that has a pole at ${\displaystyle s=1}$).

And you have suggested a way to estimate the rate of convergence - use the prime number theorem.

So, how do I use this theorem?

(You'd be surprised how little I know about number theory; then again, probably not...)

Bh12 (talk) 18:23, 15 February 2016 (UTC)[reply]

Belatedly, you are welcome for the LaTeXing. See Divergence of the sum of the reciprocals of the primes. The discussion above establishes more or less that your product is of the order

${\displaystyle \exp \left(-\sum _{i=1}^{N}{\frac {1}{P_{i}}}\right)}$; the linked article gives the rate of divergence of this series. So that should give you the rate at which your product goes to 0. To get the exact constant is probably trickier. --JBL (talk) 18:35, 15 February 2016 (UTC)[reply]

You will also note the appearance of your product in the linked article! --JBL (talk) 18:40, 15 February 2016 (UTC)[reply]

Please note that I am now using log in place of ln (even though my computer's calculator disagrees).

I went to your linked article (Divergence of the sum of the reciprocals of the primes) and saw there that

${\displaystyle \sum _{\scriptstyle p{\text{ prime }} \atop \scriptstyle p\leq n}{\frac {1}{p}}\geq \log \log(n+1)-\log {\frac {\pi ^{2}}{6}}}$

Now ${\displaystyle -log(1-1/P)}$ is always greater than ${\displaystyle 1/P}$. I assume that this difference (summed from ${\displaystyle P_{1}}$ to large enough ${\displaystyle P_{N}}$) at least equals the constant term (${\displaystyle \log {\frac {\pi ^{2}}{6}}}$, a bit less than 0.5), which can then be considered cancelled out.

Also, multiply both sides by ${\displaystyle -1}$, reversing the inequality.

Altogether, this gives

${\displaystyle \sum _{\scriptstyle p{\text{ prime }} \atop \scriptstyle p\leq n}\log(1-1/p)\leq -\log \log(n+1)}$

To get back the original infinite product, take the exponential of both sides.

${\displaystyle \prod _{\scriptstyle p{\text{ prime }} \atop \scriptstyle p\leq n}(1-1/p)\leq 1/\log(n+1)}$

Considering just primes, and noting that ${\displaystyle 1/log(P+1)<1/log(P)}$, we have as ${\displaystyle P}$ approaches infinity

${\displaystyle \prod (1-1/P)<1/\log(P)}$

It would seem the answer to my question is that, as ${\displaystyle P_{N}}$ approaches infinity, the sequence ${\displaystyle (1-1/2)\cdot (1-1/3)\cdot (1-1/5)\cdots (1-1/P_{N})}$ diverges to 0 at a rate less than ${\displaystyle 1/log(P_{N})}$.

Or, maybe the correct way to put it is that, as ${\displaystyle P_{N}}$ approaches infinity, the aforesaid sequence approaches 0 faster than ${\displaystyle 1/log(P_{N})}$ approaches 0?

Criticism invited.

Bh12 (talk) 09:15, 16 February 2016 (UTC)[reply]

Nice job. I didn't notice any errors. Sławomir
Biały 11:10, 16 February 2016 (UTC)[reply]

Resolved