Guest wrote:The answer for Landau's Fourth Problem is affirmative!
Here are two important hints for our proof:
1. The gap between two consecutive primes can be arbitrarily long.
2. The probability there is a negative answer for Landau's Fourth Problem when [tex]n \ge i[/tex] is
[tex]\prod_{n=i}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})}{(\pi(\sqrt{(2n)^2+1})+1)}= 0.[/tex]
Guest wrote:Guest wrote:The answer for Landau's Fourth Problem is affirmative!
Here are two important hints for our proof:
1. The gap between two consecutive primes can be arbitrarily long.
2. The probability there is a negative answer for Landau's Fourth Problem when [tex]n \ge i[/tex] is
[tex]\prod_{n=i}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})}{\pi(\sqrt{(2n)^2+1})+1}= 0.[/tex]
"Please observe [tex]\pi(x)[/tex] is the prime-counting function." -- Dave the antworker.
Guest wrote:Guest wrote:Guest wrote:The answer for Landau's Fourth Problem is affirmative!
Here are two important hints for our proof:
1. The gap between two consecutive primes can be arbitrarily long.
2. The probability there is a negative answer for Landau's Fourth Problem when [tex]n \ge i[/tex] is
[tex]\prod_{n=i}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})}{\pi(\sqrt{(2n)^2+1})+1}= 0.[/tex]
"Please observe [tex]\pi(x)[/tex] is the prime-counting function." -- Dave the antworker.
We also observe that
[tex]k_{n }p_{n}[/tex] = [tex](2n)^{2}[/tex] + 1
where prime, [tex]p_{n}[/tex] is:
[tex]3 \le p_{n}\le\sqrt{k_{n}p_{n}} =\sqrt{(2n)^2+1}\le k_{n }.[/tex]
Note: [tex]k_{n }[/tex] = 1 is a violation when [tex]n \ge i.[/tex]
Dave the antworker
Guest wrote:The answer for Landau's Fourth Problem is affirmative!
Here are two important hints for our proof:
1. The gap between two consecutive primes can be arbitrarily long.
2. The probability there is a negative answer for Landau's Fourth Problem when [tex]n \ge i[/tex] is
[tex]\prod_{n=i}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})}{\pi(\sqrt{(2n)^2+1})+1}= 0.[/tex]
"Please observe [tex]\pi(x)[/tex] is the prime-counting function." -- Dave the antworker.
Guest wrote:Please compute:
Prob(1000: [tex]\infty[/tex])
= [tex]\prod_{n=1000}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})-1}{\pi(\sqrt{(2n)^2+1})}[/tex]
to account for prime two.
Prob(1000: 10,000) = .000138989;
Prob(1000: 11,000) = .0000909983;
Prob(1000: 12,000) = 0000616167;
Prob(1000: 13,000) = 0000429001;
Prob(1000: 14,000) = 0000305871;
Prob(1000: 15,000) = 0000222699;
Prob(1000: 16,000) = .0000165075;
...
Computational Tool Link:
[url]http://m.wolframalpha.com/input/?i=Prod ... qrt%28%282×i%29%5E2%2B1%29%29+-1.+%29%2FPi%28sqrt%28%282×i%29%5E2%2B1%29%29+from+i%3D1000+to+i%3D15000[/url]
Dave the antworker.
Guest wrote:Please compute:
Prob([tex]10^{10^{10^{10000}}}[/tex]: [tex]\infty[/tex])
=
[tex]\prod_{n=10^{10^{10^{10000}}}}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})-1}{\pi(\sqrt{(2n)^2+1})}[/tex]
to account for prime two.
Dave the antworker.
Guest wrote:Please compute:
[tex]\prod_{m=1}^{\infty}(\sum_{n=1}^{10^{100000} }9×10^{-n})[/tex]
Guest wrote:Guest wrote:Please compute:
[tex]\prod_{m=1}^{\infty}(\sum_{n=1}^{10^{100000} }9×10^{-n})[/tex]
Guest wrote:Guest wrote:Guest wrote:Please compute:
[tex]\prod_{m=1}^{\infty}(\sum_{n=1}^{10^{10^{10^{10^{100000}}}} }9×10^{-n})[/tex]
Guest wrote:Relevant Reference Link:
[tex].9999999999^{10^{13}}[/tex] = 5.07550973604988 × [tex]10^{-435}[/tex]
http://m.wolframalpha.com/input/?i=%28Sum+9×10.%5E%28-n%29+from+n%3D1+to+n%3D10%29%5E%28
10%5E13%29
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