Dtjytyk

A letter with no particular backstory

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“on average” in the Bombieri-Vinogradov theorem

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TLDR: I don't understand the bit in bold, i.e. where in the formula is the average q? Thanks.

The Bombieri-Vinogradov theorem states the following:

For any $A > 0$ there exists a $B = B(A)$ such that, for $Q= N^{frac{1}{2}}(log N)^{-B}$

$$sum_{q leq Q}max_{substack{a}\{(a,q)=1}}|psi(N;q,a) - frac{N}{varphi(q)}| ll_{A} frac{N}{(log N)^{A}} $$

where $$psi(N;q,a) = sum_{substack{n leq N}\ n equiv a mod q}log n$$

if $n$ is a prime, and $0$ otherwise.

From my understanding, verbally this is addressing the error term of primes in arithmetic progression, for $bf{q leq Q}$ on average

prime-numbers analytic-number-theory

share|cite|improve this question

asked Mar 25 at 16:44

numbertheorylearnernumbertheorylearner

11

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  The true statement is on wiki. It has an equivalent formulation in term of $psi(N,chi) = sum_{p^k le N} chi(p^k) log p$ and the result follows from the orthogonality of characters
  $endgroup$
  – reuns
  Mar 25 at 18:33
  

  
  
  
  


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  @reuns I've looked at the wiki link and am still a bit confused.What is $y$ in the wiki formula? Also how does one arrive at the inequality for $Q$ i.e. $x^{frac{1}{2}}log^{-A}x leq Q leq x^{frac{1}{2}}$ finally do you have a reference to the equivalent formulation you gave? Thanks.
  $endgroup$
  – numbertheorylearner
  Mar 26 at 18:38
  

  
  
  
  


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  @reuns I looked into it more. I see we can write $$psi(x;q,a) = frac{1}{varphi(q)}sum_{chi in X_{q}}bar{chi}(a)psi(x, chi)$$ and we can approximate $psi(x, chi)$ by an integral. So we are averaging over q. The difference in Bomberi Vinogadov is essentially the error term right?
  $endgroup$
  – numbertheorylearner
  Mar 26 at 19:51
  

  
  
  
  

add a comment | 

0

$begingroup$

TLDR: I don't understand the bit in bold, i.e. where in the formula is the average q? Thanks.

The Bombieri-Vinogradov theorem states the following:

For any $A > 0$ there exists a $B = B(A)$ such that, for $Q= N^{frac{1}{2}}(log N)^{-B}$

$$sum_{q leq Q}max_{substack{a}\{(a,q)=1}}|psi(N;q,a) - frac{N}{varphi(q)}| ll_{A} frac{N}{(log N)^{A}} $$

where $$psi(N;q,a) = sum_{substack{n leq N}\ n equiv a mod q}log n$$

if $n$ is a prime, and $0$ otherwise.

From my understanding, verbally this is addressing the error term of primes in arithmetic progression, for $bf{q leq Q}$ on average

prime-numbers analytic-number-theory

share|cite|improve this question

asked Mar 25 at 16:44

numbertheorylearnernumbertheorylearner

11

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The true statement is on wiki. It has an equivalent formulation in term of $psi(N,chi) = sum_{p^k le N} chi(p^k) log p$ and the result follows from the orthogonality of characters
  $endgroup$
  – reuns
  Mar 25 at 18:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @reuns I've looked at the wiki link and am still a bit confused.What is $y$ in the wiki formula? Also how does one arrive at the inequality for $Q$ i.e. $x^{frac{1}{2}}log^{-A}x leq Q leq x^{frac{1}{2}}$ finally do you have a reference to the equivalent formulation you gave? Thanks.
  $endgroup$
  – numbertheorylearner
  Mar 26 at 18:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @reuns I looked into it more. I see we can write $$psi(x;q,a) = frac{1}{varphi(q)}sum_{chi in X_{q}}bar{chi}(a)psi(x, chi)$$ and we can approximate $psi(x, chi)$ by an integral. So we are averaging over q. The difference in Bomberi Vinogadov is essentially the error term right?
  $endgroup$
  – numbertheorylearner
  Mar 26 at 19:51
  

  
  
  
  

add a comment | 

$begingroup$

TLDR: I don't understand the bit in bold, i.e. where in the formula is the average q? Thanks.

The Bombieri-Vinogradov theorem states the following:

For any $A > 0$ there exists a $B = B(A)$ such that, for $Q= N^{frac{1}{2}}(log N)^{-B}$

$$sum_{q leq Q}max_{substack{a}\{(a,q)=1}}|psi(N;q,a) - frac{N}{varphi(q)}| ll_{A} frac{N}{(log N)^{A}} $$

where $$psi(N;q,a) = sum_{substack{n leq N}\ n equiv a mod q}log n$$

if $n$ is a prime, and $0$ otherwise.

From my understanding, verbally this is addressing the error term of primes in arithmetic progression, for $bf{q leq Q}$ on average

prime-numbers analytic-number-theory

share|cite|improve this question

asked Mar 25 at 16:44

numbertheorylearnernumbertheorylearner

11

$endgroup$

share|cite|improve this question

asked Mar 25 at 16:44

numbertheorylearnernumbertheorylearner

11

share|cite|improve this question

asked Mar 25 at 16:44

numbertheorylearnernumbertheorylearner

11

asked Mar 25 at 16:44

numbertheorylearnernumbertheorylearner

11

asked Mar 25 at 16:44

numbertheorylearnernumbertheorylearner

11