## the three limits of classical prime theory-> a contradiction

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• 1) Limit[Prime[n], n-- Infinity]=n*Log[n] ( Euler, I think) 2) Limit[PrimePi[n],n- Infinity]=n/Log[n] (Hadamard) 3) Limit[PrimePi[n]/CompositePi[n],n-
Message 1 of 4 , Nov 3, 2003
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1) Limit[Prime[n], n--> Infinity]=n*Log[n] ( Euler, I think)
3) Limit[PrimePi[n]/CompositePi[n],n-> Infinity]=1/Log[n] ( from Ulam,
in both books I have by him)

PrimePi[n]+CompositePi[n]=n
PrimePi[n]/CompositePi[n]=PrimePi[n]/(n-PrimePi[n])=1/Log[n]
Solving for
PrimePi[n]=n/(1+Log[n])

I'm puzzled by this which I found this morning.

--
Respectfully, Roger L. Bagula
tftn@..., 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
URL : http://victorian.fortunecity.com/carmelita/435/
• ... Hash: SHA1 Please, man, spare us the laughters (at your cost) and go learn some basic asymptotics. I suggest you look up in particular L Hôspital s rule
Message 2 of 4 , Nov 3, 2003
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Please, man, spare us the laughters (at your cost) and go learn some basic
asymptotics. I suggest you look up in particular L'Hôspital's rule for
limits.

Remember: when we get tired of laughing at you, you'll just get moderated and
then you'll lose your primary source of fun, which is trying to troll us.

So at least be more intelligent in your attempts.

Décio

On Monday 03 November 2003 15:56, Roger Bagula wrote:
> 1) Limit[Prime[n], n--> Infinity]=n*Log[n] ( Euler, I think)
> 3) Limit[PrimePi[n]/CompositePi[n],n-> Infinity]=1/Log[n] ( from Ulam,
> in both books I have by him)
>
> PrimePi[n]+CompositePi[n]=n
> PrimePi[n]/CompositePi[n]=PrimePi[n]/(n-PrimePi[n])=1/Log[n]
> Solving for
> PrimePi[n]=n/(1+Log[n])
>
> I'm puzzled by this which I found this morning.
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• ... n/(log n) is asymptotic to n/(1+log n). Think about it. And 1) is certainly not due to Euler. 1) is an equivalent form of the prime number theorem, and so
Message 3 of 4 , Nov 3, 2003
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On Mon, Nov 03, 2003 at 09:56:54AM -0800, Roger Bagula wrote:
> 1) Limit[Prime[n], n--> Infinity]=n*Log[n] ( Euler, I think)
> 3) Limit[PrimePi[n]/CompositePi[n],n-> Infinity]=1/Log[n] ( from Ulam,
> in both books I have by him)
>
> PrimePi[n]+CompositePi[n]=n
> PrimePi[n]/CompositePi[n]=PrimePi[n]/(n-PrimePi[n])=1/Log[n]
> Solving for
> PrimePi[n]=n/(1+Log[n])
>
> I'm puzzled by this which I found this morning.

n/(log n) is asymptotic to n/(1+log n). Think about it.

And 1) is certainly not due to Euler. 1) is an equivalent form of the
prime number theorem, and so wasn't proved until Hadamard (and de la
Vallee Poussin independently) proved it in 1896. 3) follows painfully
obviously from 2), and so doesn't need to be quoted from anywhere.

Before the PNT was proved, it was only known that pi(x) was of order
x/log x, which is weaker than it being asymptotically x/log x.

Andy
• My mistake here was that: PrimePi[n]/CompositePi[n]=(1/Log[n])/(1-1/Log[n])=1/(Log[n]-1) That checks to give: PrimePi[n]=n/Log[n] I had the ratio not the
Message 4 of 4 , Nov 4, 2003
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My mistake here was that:
PrimePi[n]/CompositePi[n]=(1/Log[n])/(1-1/Log[n])=1/(Log[n]-1)
That checks to
give:
PrimePi[n]=n/Log[n]
I had the ratio not the density.
I'm sorry.

Roger Bagula wrote:

> 1) Limit[Prime[n], n--> Infinity]=n*Log[n] ( Euler, I think)
> 3) Limit[PrimePi[n]/CompositePi[n],n-> Infinity]=1/Log[n] ( from
> Ulam, in both books I have by him)
>
> PrimePi[n]+CompositePi[n]=n
> PrimePi[n]/CompositePi[n]=PrimePi[n]/(n-PrimePi[n])=1/Log[n]
> Solving for
> PrimePi[n]=n/(1+Log[n])
>
> I'm puzzled by this which I found this morning.
>

--
Respectfully, Roger L. Bagula
tftn@..., 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :