Oilers' product? help please.

  • Thread starter Eshez
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Main Question or Discussion Point

Hello.
Let me first apologize, I've been reading this Math book in hebrew and since my Math knowledge is pretty basic I'm not firmiliar with the terminology in the English lanauge, I promise to do my best tho.
The book I'm reading is called "The music of the prime numbers"(thats the hebrew name atleast, should be pretty popular one).
Maybe it is just me but this book has a tendency to sometimes show some stuff without really explaning them.
Anyway, I got to this part which I don't really understand. Oiler took the Zetha(?) function and used the fact that every number can be written as the product of prime numbers to basically write the Zetha function in a different way.
firstly the book shows an example on how you can write 1/60 using primes--> 1/2x1/2x1/3x1/5.
Than is more or less jumps to this:
Zehtha(x)= 1/1^x + 1/2^x + 1/3^x + .... + 1/n^x + ....
hope I wrote this correctly, basically its a sequence of 1/1 +1/2 + 1/3 each at the power of x.
So this is the Zetha function, and with Oilers' improvment they jump to this:
=(1+1/2^x + 1/4^x + ....)x(1 + 1/3^x + 1/9^x+...)x....x(1+1/p^x+1/(p^2)^x)+....)x...
I hope this is clearly written, mind the the Xs between the brackets are not the variable x but simply the times icon.

I have no clue if I was suppost to understand why this is true from the book or not, but I definitly don't understand this and would be happy to.

Thanks in advanced.
 

Answers and Replies

188
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the easiest proof is that since EVERY natural number can be written as a product of primes then

[tex] \sum_{n=1}^{\infty} n^{-s} = \prod _{p} \sum_{k=0}^{\infty}p^{-ks} [/tex]

and [tex] \sum_{k=0}^{\infty} p^{-ks}= (1-p^{-s})^{-1}) [/tex]

and this last completes the proof, is the easiest method to obtain Euler's Product (they call it Oiler in pronounciation but is written Euler )
 
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I'm sorry, im not sure what some of this signs mean. Where can I get some info on that manner?
 
122
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http://en.wikipedia.org/wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function

The proof under the title "another proof" is written in simple terms with none of those signs =] It's a very beautiful result.
It does look beautiful, but, I can't quite follow it in my head. One mistake in the proof is where the author writes

"When ζ(i) > 1, ..."

This should have been

"When Re(i) > 1, ..."

There might be more mistakes later on in the proof, but, if there are, I'm sure sure that they are as easy to fix as this one.

DJ
 
122
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the easiest proof is that since EVERY natural number can be written as a product of primes then

[tex] \sum_{n=1}^{\infty} n^{-s} = \prod _{p} \sum_{k=0}^{\infty}p^{-ks} [/tex]

and [tex] \sum_{k=0}^{\infty} p^{-ks}= (1-p^{-s})^{-1}) [/tex]

and this last completes the proof, is the easiest method to obtain Euler's Product. ...

Mhill,

I can't quite follow your first step.

I tried to see if the product operation commutes with the sum operation, but it does not look to me like it does.

Even then, I don't quite see how the prime factorization of "n" is used, but, I think it has to be.

I would like to be able to follow it. The only proof that I know right now of the Euler product formula for the zeta function is first prove the Euler product formula for the reciprocal of the zeta function and then to use the Mobius inversion formula and the fact that mu^(-1) = 1 to derive the Euler product for the zeta function.

Eshez: I got the idea for this proof from looking at Borek's thread "primes to 1" that is just about 10 threads below this one in this number theory forum. There's a link:

https://www.physicsforums.com/showthread.php?t=240100

DJ
 
Last edited:
122
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Zeta(x)= 1/1^x + 1/2^x + 1/3^x + .... + 1/n^x + ....
So this is the Zeta function, and with Eulers' improvment they jump to this:
=(1+1/2^x + 1/4^x + ....)x(1 + 1/3^x + 1/9^x+...)x....x(1+1/p^x+1/(p^2)^x)+....)x...
Thanks in advance.
That looks right to me. Except when you wrote "with Eulers' improvment" I think you meant to write
"applying Euler's equation" or "applying Euler's product formula" or something like that.

The missing step is that one then applies the formula for the infinite geometric series,

[tex] \sum_{k=0}^{\infty} p^{-ks}= (1-p^{-s})^{-1} [/tex]

to the Euler product. When you do that, you get the expression you wrote above.

DJ
 
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For me, this is one of the most beautiful identities in number theory. I think a book by T. Apostol is very easy to follow, I think it's called introduction to analytic number theory. If you want to know more, I recommend it! :)
 
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For me, this is one of the most beautiful identities in number theory. I think a book by T. Apostol is very easy to follow, I think it's called introduction to analytic number theory. If you want to know more, I recommend it! :)
Thanks, I have a copy. I will check it out. I have never looked at his words on Euler products.

Apostol is one of my favorite math authors. He is one of the few who succeds in showing how simple it really is.

John
 

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