My Math Forum My new formula for pi(x)

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 October 13th, 2012, 02:46 PM #1 Newbie   Joined: Oct 2012 Posts: 22 Thanks: 0 My new formula for pi(x) pi(x) is the count of primers less that x Here is my new formula. $\pi (n)=n\prod_{p=2}^{P_{k}}{(1-\frac{1}{P})}+\pi (P_{k}) while P_{k} <\sqrt{n} It gives very precise results I proved it by evidance in algebrial way. ex. pi(1000) $\pi (1000)=1000(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{5})(1-\frac{1}{7})(1-\frac{1}{11})(1-\frac{1}{13})(1-\frac{1}{17})(1-\frac{1}{19})(1-\frac{1}{23})(1-\frac{1}{29})(1-\frac{1}{31}) +\pi (31) \\$ $\pi (1000)=\frac{1000*30656102400}{200560490130} +11=152.85+11=163.85 \sim 168\\$
 October 13th, 2012, 05:12 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: My new formula for pi(x) You seem to mean $\pi (n)\approx \pi (\sqrt n)+n\prod_{p\le\sqrt n}1-\frac1p$ which has an atrocious amount of cancellation and a consequently huge error term. Can you even prove that the ratio of this to pi(n) is 1? (It seems to be true...)
 October 13th, 2012, 05:30 PM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: My new formula for pi(x) The pi(sqrt(n)) part doesn't seem to help, either. Look at 500,000: your formula is too high by 1136.8..., but it would 'only' be off by 1010.8... if you dropped that term. Similarly at 10^9 it worsens the error from 3,319,288.3... to 3,322,689.3....
 October 13th, 2012, 11:13 PM #4 Newbie   Joined: Oct 2012 Posts: 22 Thanks: 0 Re: My new formula for pi(x) thank you for your helpful feedback. I will applead a word file which contains the proof so wiat for me your feedback is very important to me.
 October 14th, 2012, 09:03 AM #5 Newbie   Joined: Oct 2012 Posts: 22 Thanks: 0 Re: My new formula for pi(x) this is a pdf file contains the proof I change it from word pdf is better than word in reading, I think so. download the file from this link http://depositfiles.com/files/awg3hvlri size 284 kb it is a bit more than attachment maximum allowed size which is 250 Kb zip file does not help still more than 250 kb
 October 14th, 2012, 12:27 PM #6 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: My new formula for pi(x) Your result is incorrect, as shown in your first post: equality does not hold. It may be that the two are approximately equal, but you have not proved this.
October 15th, 2012, 12:21 AM   #7
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Re: My new formula for pi(x)

Quote:
 Originally Posted by CRGreathouse Your result is incorrect, as shown in your first post: equality does not hold. It may be that the two are approximately equal, but you have not proved this.
incorrect means no equality and not approximately equal.

my proof does not mean that the result will be exact

because (for example) when I remove multiples of 5 as 1/5 from the remain it does not exactly 1/5 from it.

the primers will become more and more in the remains while removing the multiples of primers less than sequare root of (n)

and this is what makes the approximation become far from pi(n).

It seems to me I have to consider the method again..

thank you CRGreathouse you gave me very important notes.

October 15th, 2012, 05:36 AM   #8
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Re: My new formula for pi(x)

Quote:
 Originally Posted by nsrmsm incorrect means no equality and not approximately equal.
No, that's not what it means. You wrote an = sign, and it's not an equality.

It may be an approximate equality, but your 'proof' does not show that.

Quote:
 Originally Posted by nsrmsm the primers will become more and more in the remains while removing the multiples of primers less than sequare root of (n) and this is what makes the approximation become far from pi(n).
If you can show that it must stay close to pi(x) that would be interesting.

I see an error of 2,666,913,530,087 at 10^15; that's much further than the error of the logarithmic integral at the same point (1,052,61.

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