Extend Euler Product Convergence Over Primes: Basic Qs

In summary, the Euler product convergence over primes is a mathematical concept that states the infinite product of (1+1/p) for all primes p, where p is a prime number, converges to a non-zero value. Extending this convergence is important because it allows us to use it in more complex equations and understand the behavior of prime numbers. The basic steps for extending the convergence are identifying the original Euler product, defining the new function or equation, manipulating it using the properties of the original product, and proving its convergence. The Euler product convergence over primes has real-life applications in various fields of mathematics and physics, but it has limitations as it can only be extended to certain types of functions or equations.
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benorin
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I would like to extend the convergence of the Euler product over primes, and I tried to do so in the exact manor it was done for the Dirichlet series, namely, given a completely multiplicative sequence ##a( {kj} ) =a(k) \cdot a(j)\text{ and }a(1)=1##, the Dirichlet series ##\xi (s) := \sum_{k=1}^{\infty} \tfrac{a(k)}{k^s}## can be shown (by adding the alternating version of itself to itself and simplifying) to be equal to ## \hat \xi (s) := \left( 1-2^{1-s} a(2) \right) ^{-1} \sum_{k=1}^{\infty} (-1)^k \tfrac{a(k)}{k^s}##, and this series is an analytic continuation of the former.
Tried to do this to the Euler product

$$\sum_{k=1}^{\infty} \tfrac{a(k)}{k^s} = \prod_{k=1}^{\infty} \left( 1+a( {p_k} ) p_k^{-s} + a ( {p_k^2} ) p_k^{-2s}+\cdots \right)=\prod_{k=1}^{\infty} \tfrac{1}{1-a( {p_k} ) p_k^{-s}}$$

(where ##p_k## is the ##k^{th}## prime and the later equality holds only for completely multiplicative sequences ##a(k)## taking a cue from the analytic continuation of the Dirichlet series I set ##a(k) :=(-1)^{k-1}a^{\prime}(k)## where ##a^{\prime}(k)## is a completely multiplicative sequence and the product becomes

$$\sum_{k=1}^{\infty} (-1)^{k-1}\tfrac{a^{\prime}(k)}{k^s} =\prod_{k=1}^{\infty} \tfrac{1}{1-(-1)^{p_k -1}a^{\prime}( {p_k} ) p_k^{-s}} = \tfrac{1}{1+a^{\prime}( {2} ) 2^{-s}}\prod_{k=2}^{\infty} \tfrac{1}{1-a^{\prime}( {p_k} ) p_k^{-s}}$$

I was hoping for an analytic continuation of the product over primes but this product differs from the original Euler product by only one term hence converges in the same region. I had hoped to follow this up with the rest of the steps to globally analytically continue the zeta function and wind up with a product over primes converging for all complex ##s\neq 1## but effecting the convergence of the Dirichlet series the Euler product is equal to didn't effect the convergence of the product itself. I think my problem may be that I need to be working with absolute convergence? Do you understand what I'm trying to do? How can I do that?

Edit: sorry if this is an easy one but I just started on these and I have no text on it.
 
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Hello,

Thank you for sharing your ideas on extending the convergence of the Euler product over primes. It seems like you have a good understanding of the topic and have made some interesting observations.

To answer your question, I think you are correct in thinking that absolute convergence may be necessary for your method to work. In order to globally analytically continue the zeta function, it is important to have absolute convergence for all values of s. This means that the series must converge for all complex numbers s, not just for certain values.

One way to ensure absolute convergence is to use the absolute values of the terms in the product, rather than alternating them as you have done. This would give you a product of the form

$$\prod_{k=1}^{\infty} \left( \frac{1}{1+a( {p_k} ) |p_k|^{-s}} \right)$$

where |p_k| denotes the absolute value of the kth prime. This product should converge absolutely for all complex s, and you can then continue with the rest of your steps to globally analytically continue the zeta function.

I hope this helps and good luck with your research!
 

FAQ: Extend Euler Product Convergence Over Primes: Basic Qs

What is the Euler product convergence over primes?

The Euler product convergence over primes is a mathematical concept that states that if an infinite product of numbers converges, then the corresponding infinite sum also converges. This concept was introduced by the mathematician Leonhard Euler in the 18th century.

How does the Euler product convergence over primes work?

The Euler product convergence over primes works by breaking down an infinite product into a sum of terms, each of which corresponds to a prime number. This allows for easier analysis and calculation of the convergence of the product.

What is the significance of the Euler product convergence over primes?

The Euler product convergence over primes is significant because it provides a powerful tool for analyzing the convergence of infinite products. It has been used in various areas of mathematics, including number theory, complex analysis, and probability theory.

What are some applications of the Euler product convergence over primes?

Some applications of the Euler product convergence over primes include the proof of the infinitude of primes, the Riemann zeta function, and the Dirichlet series. It has also been used in the study of the distribution of prime numbers and in the theory of modular forms.

Are there any limitations to the Euler product convergence over primes?

Yes, there are some limitations to the Euler product convergence over primes. It only applies to infinite products that converge absolutely, meaning that the terms in the product decrease in magnitude as the index increases. Additionally, it does not work for all types of infinite products, such as those with alternating signs.

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