- 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-> Infinity]=1/Log[n] ( from Ulam,

in both books I have by him)

How 3) can be made to contradict 2):

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://home.earthlink.net/~tftn

URL : http://victorian.fortunecity.com/carmelita/435/ - -----BEGIN PGP SIGNED MESSAGE-----

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 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)

> 2) Limit[PrimePi[n],n->Infinity]=n/Log[n] (Hadamard)

> 3) Limit[PrimePi[n]/CompositePi[n],n-> Infinity]=1/Log[n] ( from Ulam,

> in both books I have by him)

>

> How 3) can be made to contradict 2):

> 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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-----END PGP SIGNATURE----- - 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)

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

> 2) Limit[PrimePi[n],n->Infinity]=n/Log[n] (Hadamard)

> 3) Limit[PrimePi[n]/CompositePi[n],n-> Infinity]=1/Log[n] ( from Ulam,

> in both books I have by him)

>

> How 3) can be made to contradict 2):

> 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.

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 density.

I'm sorry.

Roger Bagula wrote:

> 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-> Infinity]=1/Log[n] ( from

> Ulam, in both books I have by him)

>

> How 3) can be made to contradict 2):

> 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://home.earthlink.net/~tftn

URL : http://victorian.fortunecity.com/carmelita/435/