Mixing
Two Representations of the Prime Counting Function
$begingroup$
The bounty for the best work out of Greg's answer, especially the
"solving for $pi^*(x;q,a)$ in terms of all $Pi^*$ functions (tedious but possible)"
part is over. Since Raymond's contributions might be very helpful to recall the necessary math, upvoting his answers is highly appreciated...
I posted my attempt of a workout here. Comments welcome...
Original question:
I have two representations of $pi(x)$:
The Prime Counting Function $pi(x)$ is given
$$
pi(x) = operatorname{R}(x^1) - sum_{rho}operatorname{R}(x^{rho}) tag{1}
$$
with $ operatorname{R}(z) = sum_{n=1}^{infty} frac{ mu (n)}{n} operatorname{li}(z^{1/n})$ and $rho$ running over all the zeros of $zeta$ function.
This formula, while
widely believed to be correct, has not yet been proved.
$$
pi(x) approx intlimits_2^x{frac{dt}{ln t}} - frac{sqrt x}{ln x}left( 1 + 2sum_{gamma} frac{sin(gammaln x)}{gamma}right) tag{2},
$$
with $gamma=text{Im}({rho})$ being imaginary part of the roots of the $zeta$ function.
Now I have two questions:
- Does the truth of $(2)$ depend on Riemann's Hypothesis or is it "just" what Wikipedia says, that The amplitude of the "noisy" part is heuristically about $sqrt x/ln x$?
- How to show the equivalence between $(1)$ and $(2)$? The integral logarithm is easily found in both representations, but how do the $rho$-parts fit together? How do I get $sin$s from $text{li}(z^{1/n})$s? Does this invoke Gram's series:
$$
operatorname{R}(z) = sum_{n=1}^{infty} frac{ mu (n)}{n} operatorname{li}(z^{1/n}) = 1 + sum_{k=1}^infty frac{(ln z)^k}{k! k zeta(k+1)} ?
$$
We can rewrite $displaystyle frac{sin(gammaln x)}{gamma}=frac{x^{igamma}-x^{-igamma}}{igamma}$ and I remember that I've seen a similar expression at Wikipedia:
$$
psi_0(x) = x - sum_rho frac{x^rho}{rho} - ln 2pi - frac12 ln(1-x^{-2}) ,
$$
But could this help, if at all? ($psi_0(x)$ is the normalization of the Chebyshev function, see here)
prime-numbers riemann-zeta
asked Jan 3 '13 at 23:11
draks ...draks ...
11.5k645131
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|
show 3 more comments
$begingroup$
The bounty for the best work out of Greg's answer, especially the
"solving for $pi^*(x;q,a)$ in terms of all $Pi^*$ functions (tedious but possible)"
part is over. Since Raymond's contributions might be very helpful to recall the necessary math, upvoting his answers is highly appreciated...
I posted my attempt of a workout here. Comments welcome...
Original question:
I have two representations of $pi(x)$:
The Prime Counting Function $pi(x)$ is given
$$
pi(x) = operatorname{R}(x^1) - sum_{rho}operatorname{R}(x^{rho}) tag{1}
$$
with $ operatorname{R}(z) = sum_{n=1}^{infty} frac{ mu (n)}{n} operatorname{li}(z^{1/n})$ and $rho$ running over all the zeros of $zeta$ function.
This formula, while
widely believed to be correct, has not yet been proved.
$$
pi(x) approx intlimits_2^x{frac{dt}{ln t}} - frac{sqrt x}{ln x}left( 1 + 2sum_{gamma} frac{sin(gammaln x)}{gamma}right) tag{2},
$$
with $gamma=text{Im}({rho})$ being imaginary part of the roots of the $zeta$ function.
Now I have two questions:
- Does the truth of $(2)$ depend on Riemann's Hypothesis or is it "just" what Wikipedia says, that The amplitude of the "noisy" part is heuristically about $sqrt x/ln x$?
- How to show the equivalence between $(1)$ and $(2)$? The integral logarithm is easily found in both representations, but how do the $rho$-parts fit together? How do I get $sin$s from $text{li}(z^{1/n})$s? Does this invoke Gram's series:
$$
operatorname{R}(z) = sum_{n=1}^{infty} frac{ mu (n)}{n} operatorname{li}(z^{1/n}) = 1 + sum_{k=1}^infty frac{(ln z)^k}{k! k zeta(k+1)} ?
$$
We can rewrite $displaystyle frac{sin(gammaln x)}{gamma}=frac{x^{igamma}-x^{-igamma}}{igamma}$ and I remember that I've seen a similar expression at Wikipedia:
$$
psi_0(x) = x - sum_rho frac{x^rho}{rho} - ln 2pi - frac12 ln(1-x^{-2}) ,
$$
But could this help, if at all? ($psi_0(x)$ is the normalization of the Chebyshev function, see here)
prime-numbers riemann-zeta
asked Jan 3 '13 at 23:11
draks ...draks ...
11.5k645131
$endgroup$
$begingroup$
+1: interesting question but where did you find the formula $(2)$ ? I would bet that it is only an approximation just from the lower bound $2$ of the first integral ( and... from the $approx$ sign :-) ) ?
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– Raymond Manzoni
Jan 4 '13 at 21:42
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It is from the linked paper here : math.stackexchange.com/q/149755/19341 . Sorry but i post from my tablet. I hate that...
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– draks ...
Jan 4 '13 at 23:28
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Just some preliminary numerical results : I compared $(2)$ to the actual $pi(n)$ function in the interval $(2,100)$ !pi_100 : $(2)$ (darker and on the top) has all the features of a function growing 'by steps' and is very similar to the real $pi(n)$ function but, as seen in $(3,1000)$ !pi_1000, it is starting too low and growing too fast (this effect is real : more than $10^4$ zeros were used).
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– Raymond Manzoni
Jan 4 '13 at 23:42
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From your ref. the zeros seem to be the imaginary parts of the zeros of the Dirichlet beta function. I have Montgomery and Vaughan's book (I'll see tomorrow...).
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– Raymond Manzoni
Jan 4 '13 at 23:43
$begingroup$
But do your plots compare to Figure 4 in Prime Number Races. Beta functions are used in the case $pi_{4n+3}-pi_{4n+1}$ (p.19)...
$endgroup$
– draks ...
Jan 5 '13 at 9:20
|
show 3 more comments
$begingroup$
The bounty for the best work out of Greg's answer, especially the
"solving for $pi^*(x;q,a)$ in terms of all $Pi^*$ functions (tedious but possible)"
part is over. Since Raymond's contributions might be very helpful to recall the necessary math, upvoting his answers is highly appreciated...
I posted my attempt of a workout here. Comments welcome...
Original question:
I have two representations of $pi(x)$:
The Prime Counting Function $pi(x)$ is given
$$
pi(x) = operatorname{R}(x^1) - sum_{rho}operatorname{R}(x^{rho}) tag{1}
$$
with $ operatorname{R}(z) = sum_{n=1}^{infty} frac{ mu (n)}{n} operatorname{li}(z^{1/n})$ and $rho$ running over all the zeros of $zeta$ function.
This formula, while
widely believed to be correct, has not yet been proved.
$$
pi(x) approx intlimits_2^x{frac{dt}{ln t}} - frac{sqrt x}{ln x}left( 1 + 2sum_{gamma} frac{sin(gammaln x)}{gamma}right) tag{2},
$$
with $gamma=text{Im}({rho})$ being imaginary part of the roots of the $zeta$ function.
Now I have two questions:
- Does the truth of $(2)$ depend on Riemann's Hypothesis or is it "just" what Wikipedia says, that The amplitude of the "noisy" part is heuristically about $sqrt x/ln x$?
- How to show the equivalence between $(1)$ and $(2)$? The integral logarithm is easily found in both representations, but how do the $rho$-parts fit together? How do I get $sin$s from $text{li}(z^{1/n})$s? Does this invoke Gram's series:
$$
operatorname{R}(z) = sum_{n=1}^{infty} frac{ mu (n)}{n} operatorname{li}(z^{1/n}) = 1 + sum_{k=1}^infty frac{(ln z)^k}{k! k zeta(k+1)} ?
$$
We can rewrite $displaystyle frac{sin(gammaln x)}{gamma}=frac{x^{igamma}-x^{-igamma}}{igamma}$ and I remember that I've seen a similar expression at Wikipedia:
$$
psi_0(x) = x - sum_rho frac{x^rho}{rho} - ln 2pi - frac12 ln(1-x^{-2}) ,
$$
But could this help, if at all? ($psi_0(x)$ is the normalization of the Chebyshev function, see here)
prime-numbers riemann-zeta
asked Jan 3 '13 at 23:11
draks ...draks ...
11.5k645131
$endgroup$
The bounty for the best work out of Greg's answer, especially the
"solving for $pi^*(x;q,a)$ in terms of all $Pi^*$ functions (tedious but possible)"
part is over. Since Raymond's contributions might be very helpful to recall the necessary math, upvoting his answers is highly appreciated...
I posted my attempt of a workout here. Comments welcome...
Original question:
I have two representations of $pi(x)$:
The Prime Counting Function $pi(x)$ is given
$$
pi(x) = operatorname{R}(x^1) - sum_{rho}operatorname{R}(x^{rho}) tag{1}
$$
with $ operatorname{R}(z) = sum_{n=1}^{infty} frac{ mu (n)}{n} operatorname{li}(z^{1/n})$ and $rho$ running over all the zeros of $zeta$ function.
This formula, while
widely believed to be correct, has not yet been proved.
$$
pi(x) approx intlimits_2^x{frac{dt}{ln t}} - frac{sqrt x}{ln x}left( 1 + 2sum_{gamma} frac{sin(gammaln x)}{gamma}right) tag{2},
$$
with $gamma=text{Im}({rho})$ being imaginary part of the roots of the $zeta$ function.
Now I have two questions:
- Does the truth of $(2)$ depend on Riemann's Hypothesis or is it "just" what Wikipedia says, that The amplitude of the "noisy" part is heuristically about $sqrt x/ln x$?
- How to show the equivalence between $(1)$ and $(2)$? The integral logarithm is easily found in both representations, but how do the $rho$-parts fit together? How do I get $sin$s from $text{li}(z^{1/n})$s? Does this invoke Gram's series:
$$
operatorname{R}(z) = sum_{n=1}^{infty} frac{ mu (n)}{n} operatorname{li}(z^{1/n}) = 1 + sum_{k=1}^infty frac{(ln z)^k}{k! k zeta(k+1)} ?
$$
We can rewrite $displaystyle frac{sin(gammaln x)}{gamma}=frac{x^{igamma}-x^{-igamma}}{igamma}$ and I remember that I've seen a similar expression at Wikipedia:
$$
psi_0(x) = x - sum_rho frac{x^rho}{rho} - ln 2pi - frac12 ln(1-x^{-2}) ,
$$
But could this help, if at all? ($psi_0(x)$ is the normalization of the Chebyshev function, see here)
prime-numbers riemann-zeta
prime-numbers riemann-zeta
asked Jan 3 '13 at 23:11
draks ...draks ...
11.5k645131
asked Jan 3 '13 at 23:11
draks ...draks ...
11.5k645131
edited Jan 24 at 21:59
draks ...
asked Jan 3 '13 at 23:11
draks ...draks ...
11.5k645131
asked Jan 3 '13 at 23:11
draks ...draks ...
11.5k645131
asked Jan 3 '13 at 23:11
draks ...draks ...
11.5k645131
11.5k645131
$begingroup$
+1: interesting question but where did you find the formula $(2)$ ? I would bet that it is only an approximation just from the lower bound $2$ of the first integral ( and... from the $approx$ sign :-) ) ?
$endgroup$
– Raymond Manzoni
Jan 4 '13 at 21:42
$begingroup$
It is from the linked paper here : math.stackexchange.com/q/149755/19341 . Sorry but i post from my tablet. I hate that...
$endgroup$
– draks ...
Jan 4 '13 at 23:28
$begingroup$
Just some preliminary numerical results : I compared $(2)$ to the actual $pi(n)$ function in the interval $(2,100)$ !pi_100 : $(2)$ (darker and on the top) has all the features of a function growing 'by steps' and is very similar to the real $pi(n)$ function but, as seen in $(3,1000)$ !pi_1000, it is starting too low and growing too fast (this effect is real : more than $10^4$ zeros were used).
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– Raymond Manzoni
Jan 4 '13 at 23:42
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From your ref. the zeros seem to be the imaginary parts of the zeros of the Dirichlet beta function. I have Montgomery and Vaughan's book (I'll see tomorrow...).
$endgroup$
– Raymond Manzoni
Jan 4 '13 at 23:43
$begingroup$
But do your plots compare to Figure 4 in Prime Number Races. Beta functions are used in the case $pi_{4n+3}-pi_{4n+1}$ (p.19)...
$endgroup$
– draks ...
Jan 5 '13 at 9:20
|
show 3 more comments
$begingroup$
+1: interesting question but where did you find the formula $(2)$ ? I would bet that it is only an approximation just from the lower bound $2$ of the first integral ( and... from the $approx$ sign :-) ) ?
$endgroup$
– Raymond Manzoni
Jan 4 '13 at 21:42
$begingroup$
It is from the linked paper here : math.stackexchange.com/q/149755/19341 . Sorry but i post from my tablet. I hate that...
$endgroup$
– draks ...
Jan 4 '13 at 23:28
$begingroup$
Just some preliminary numerical results : I compared $(2)$ to the actual $pi(n)$ function in the interval $(2,100)$ !pi_100 : $(2)$ (darker and on the top) has all the features of a function growing 'by steps' and is very similar to the real $pi(n)$ function but, as seen in $(3,1000)$ !pi_1000, it is starting too low and growing too fast (this effect is real : more than $10^4$ zeros were used).
$endgroup$
– Raymond Manzoni
Jan 4 '13 at 23:42
$begingroup$
From your ref. the zeros seem to be the imaginary parts of the zeros of the Dirichlet beta function. I have Montgomery and Vaughan's book (I'll see tomorrow...).
$endgroup$
– Raymond Manzoni
Jan 4 '13 at 23:43
$begingroup$
But do your plots compare to Figure 4 in Prime Number Races. Beta functions are used in the case $pi_{4n+3}-pi_{4n+1}$ (p.19)...
$endgroup$
– draks ...
Jan 5 '13 at 9:20
$begingroup$
+1: interesting question but where did you find the formula $(2)$ ? I would bet that it is only an approximation just from the lower bound $2$ of the first integral ( and... from the $approx$ sign :-) ) ?
$endgroup$
– Raymond Manzoni
Jan 4 '13 at 21:42
$begingroup$
+1: interesting question but where did you find the formula $(2)$ ? I would bet that it is only an approximation just from the lower bound $2$ of the first integral ( and... from the $approx$ sign :-) ) ?
$endgroup$
– Raymond Manzoni
Jan 4 '13 at 21:42
$begingroup$
It is from the linked paper here : math.stackexchange.com/q/149755/19341 . Sorry but i post from my tablet. I hate that...
$endgroup$
– draks ...
Jan 4 '13 at 23:28
$begingroup$
It is from the linked paper here : math.stackexchange.com/q/149755/19341 . Sorry but i post from my tablet. I hate that...
$endgroup$
– draks ...
Jan 4 '13 at 23:28
$begingroup$
Just some preliminary numerical results : I compared $(2)$ to the actual $pi(n)$ function in the interval $(2,100)$ !pi_100 : $(2)$ (darker and on the top) has all the features of a function growing 'by steps' and is very similar to the real $pi(n)$ function but, as seen in $(3,1000)$ !pi_1000, it is starting too low and growing too fast (this effect is real : more than $10^4$ zeros were used).
$endgroup$
– Raymond Manzoni
Jan 4 '13 at 23:42
$begingroup$
Just some preliminary numerical results : I compared $(2)$ to the actual $pi(n)$ function in the interval $(2,100)$ !pi_100 : $(2)$ (darker and on the top) has all the features of a function growing 'by steps' and is very similar to the real $pi(n)$ function but, as seen in $(3,1000)$ !pi_1000, it is starting too low and growing too fast (this effect is real : more than $10^4$ zeros were used).
$endgroup$
– Raymond Manzoni
Jan 4 '13 at 23:42
$begingroup$
From your ref. the zeros seem to be the imaginary parts of the zeros of the Dirichlet beta function. I have Montgomery and Vaughan's book (I'll see tomorrow...).
$endgroup$
– Raymond Manzoni
Jan 4 '13 at 23:43
$begingroup$
From your ref. the zeros seem to be the imaginary parts of the zeros of the Dirichlet beta function. I have Montgomery and Vaughan's book (I'll see tomorrow...).
$endgroup$
– Raymond Manzoni
Jan 4 '13 at 23:43
$begingroup$
But do your plots compare to Figure 4 in Prime Number Races. Beta functions are used in the case $pi_{4n+3}-pi_{4n+1}$ (p.19)...
$endgroup$
– draks ...
Jan 5 '13 at 9:20
$begingroup$
But do your plots compare to Figure 4 in Prime Number Races. Beta functions are used in the case $pi_{4n+3}-pi_{4n+1}$ (p.19)...
$endgroup$
– draks ...
Jan 5 '13 at 9:20
|
show 3 more comments
3 Answers
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Let's start with the nice paper 'Prime Number Races' of Granville and Martin from your comments. It contains the sentence :
"The precise formula he (Riemann) proposed is a bit too technical for
this article, but we can get a good sense of it from the following
approximation when x is large. This formula, while widely believed to
be correct, has not yet been proved."
$$
tag{1}boxed{displaystylefrac{intlimits_2^x{frac{dt}{ln t}}-pi(x)}{sqrt x/ln x} approx 1 + 2sum_{gamma} frac{sin(gammaln x)}{gamma}}
$$
I can only conjecture that the formula referenced in their last sentence is your formula $(1)$ (I don't know a proof of convergence of this formula either) derived from Riemann's explicit formula (proved by von Mangoldt) and not the approximate formula I reproduce above equivalent to your $(2)$ . Note that they add that this approximation is valid 'when $x$ is large'.
Let's add that this approximation doesn't include the trivial zeros from your other thread.
Concerning the $displaystyle frac{sin(gammaln x)}{gamma}$ term it is a good approximation of Riesel's approximation $(2.30)$ from his fine book 'Prime Numbers and Computer Methods for Factorization' :
$$tag{2}frac{cos(gammaln x-arg rho)}{|rho|},quadtext{with} rho=frac 12+igamma$$
since $ gammagg 1 $ we have indeed $ |rho|approx gamma $ and $ arg,rhoapprox frac{pi}2$
(note that $gamma > 14.13 $ so that $ 1<frac {|rho|}{gamma}<1.00063$)
Riesel's method is to approximate the oscillating term (see the next page from Riesel's book) :
$$tag{3}operatorname{R}(x^{rho}) = sum_{n=1}^{infty} frac{ mu (n)}{n}operatorname{li}(x^{rho/n})$$
with the first term :
$$tag{4}operatorname{R}(x^{rho}) approxleft[operatorname{li}(x^{rho})=operatorname{li}(e^{rho;ln x})=operatorname{Ei}(rho;ln x)right]$$
and to use the asymptotic expansion for $operatorname{Ei}$ :
$$tag{5}operatorname{Ei}(z)sim frac{e^z}{z}left(1+frac {1!}z+frac{2!}{z^2}+cdotsright)$$
to get :
$$operatorname{Ei}left(left(frac 12+igammaright);ln xright)approx frac{x^{1/2+igamma}}{(1/2+igamma);ln x}$$
so that combining the two 'mirror' zeros :
$$operatorname{Ei}left(left(frac 12+igammaright)+;ln xright)+operatorname{Ei}left(left(frac 12-igammaright)+;ln xright)approx frac{sqrt{x}}{ln x}left(frac{e^{igammaln x}}{1/2+igamma}+frac{e^{-igammaln x}}{1/2-igamma}right)$$
$$approxfrac{sqrt{x}}{ln x}frac{e^{igammaln x-iarg{rho}}+e^{-igammaln x+iarg{rho}}}{|rho|}quadtext{since} rho=frac 12+igamma$$
we get Riesel's approximation :
$$tag{6}operatorname{R}(x^{rho})+operatorname{R}(x^{overline{rho}})approxfrac{2;sqrt{x};cos(gammaln x-arg{rho})}{|rho|;ln x}$$
while $operatorname{R}(x)$ was approximated by (using $(3), (4), (5)$) :
$$tag{7}operatorname{R}(x)approx operatorname{li}(x)-frac 12operatorname{li}bigl(x^{1/2}bigr)quad text{with} operatorname{li}bigl(x^{1/2}bigr)approx frac{2sqrt{x}}{ln x}$$
this allows too to get better approximations if wished ($operatorname{R}(x)approx operatorname{li}(x)-frac 12operatorname{li}bigl(x^{1/2}bigr)-frac 13operatorname{li}bigl(x^{1/3}bigr) $ for example, adding the trivial zeros contribution and so on...)
Now to answer your questions :
- We have to suppose the R.H. to be able to write $rho=frac 12+igamma $ for any root $rho$ so that the simplifications proposed apply (in practice we don't need R.H. since $(2)$ requires precomputed zeros and since all the zeros known satisfy the R.H. ... to this date...)
- The implication from $(1)to (2)$ was provided without using the Gram series
Hoping this clarified things,
edited Apr 13 '17 at 12:20
Community♦
1
answered Jan 5 '13 at 21:33
Raymond ManzoniRaymond Manzoni
37.2k563117
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thank you very much. Again :-)
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– draks ...
Jan 6 '13 at 22:14
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You are welcome @draks : this gave me the occasion to compute some zeros (say 331) of Dirichlet beta, see a fine approximation for $pi(n)$ and ways to make it better. Thanks for the journey !
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– Raymond Manzoni
Jan 6 '13 at 23:43
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Wow 331 sounds great. I was just out to find a table where these zeros are listed. Would you mind sharing the zeros of beta with me? And to be honest I still have one more question: Is it possible to write $#{text{primes} 4n+3 le x} - #{text{primes} 4n+1 le x}$(form [here](math.stackexchange.com/q/149755/19341)) in terms like $operatorname{R}(x^1) - sum_{rho}operatorname{R}(x^{rho})$ ? I hope I don't bother you too much...
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– draks ...
Jan 6 '13 at 23:54
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Thanks a lot. Concerning the last question, I thought about using a simple analogy $(pi_{4n+3}-pi_{4n+1})-text{Li}(x)=text{R}(x)-sum_rho text{R}(x^rho)$ with $rho$ being the values you've kindly provided...?
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– draks ...
Jan 7 '13 at 0:21
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@draks: I gave the zeros in your other thread and will erase the values here... Concerning your other question I don't know but will look at it in the coming days (it's late...). I must add that from your formula $(4')$ there is probably a similar (or identical) formula as you suppose !
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– Raymond Manzoni
Jan 7 '13 at 0:30
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show 7 more comments
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Before answering to your question in the comment :
" Is it possible to write $ #{text{primes} 4n+3 le x},-
,#{text{primes} 4n+1 le x}$ (form here) in terms like $,operatorname{R}(x^1) - sum_{rho}operatorname{R}(x^{rho})$ ? "
let's start with a sketch using von Mangoldt's derivation to obtain your equation $(1)$ that will be used for inspiration (for proofs see for example Edwards' chapter 3) :
Riemann's Explicit Formulas
The Euler product formula gives us :
$$tag{1}boxed{displaystylezeta(s)=prod_{p text{prime}}frac 1{1-p^{-s}}}quadtext{for} Re(s)>1$$
so that
$$log zeta(s)=-sum_{p text{prime}}log(1-p^{-s})=sum_psum_{k=1}^infty frac{p^{-ks}}k$$
minus the derivative will be :
$$tag{2}f(s):=-frac{zeta'(s)}{zeta(s)}=sum_psum_{k=1}^infty frac{log,p}{p^{ks}}=sum_{n=1}^infty frac{Lambda(n)}{n^s}quadtext{for} Re(s)>1$$
with $Lambda$ the von Mangoldt function defined by $ Lambda(n):=begin{cases}
log, p & text{if} n=p^k text{and} k>0\
0 & text{else}
end{cases}$
Let's use the definition of the second Chebyshev function : $displaystyle psi(x)=sum_{nleq x}Lambda(n)$ and Abel's sum formula applied to $a(n):=Lambda(n)$ and $phi(n):=n^{-s}$ :
$$sum_{n=1}^infty frac{Lambda(n)}{n^s}=sint_1^infty frac {sum_{nleq x}Lambda(n)}{x^{s+1}};dx$$
to rewrite $f(s)$ as (the lower bound became $0$ since $phi(x)=0$ for $x<1$) :
$$tag{2.1}f(s)=sint_0^{infty}frac{psi(x)}{x^{s+1}}dx$$
But this is a Mellin transform which may be reverted to get Perron's formula (let's observe that the Dirichlet series $displaystyle f(s)=sum_{n=1}^infty frac{Lambda(n)}{n^s}$ is absolutely convergent for $Re(s)>1$ and suppose $c>1$) :
$$tag{3}psi^*(x):=frac1{2pi i}int_{c-iinfty}^{c+iinfty}f(s)frac{x^s}s,ds=sum_{n=1}^infty Lambda(n) frac1{2pi i}int_{c-iinfty}^{c+iinfty}left(frac xnright)^sfrac{ds}s$$
This last integral may be evaluated using Fourier's theorem :
$$ displaystylefrac1{2pi i}int_{c-iinfty}^{c+iinfty}frac{y^s}s ds=
begin{cases}
0 & 0<y<1\
1/2 & quad y=1\
1 & quad y>1\
end{cases}$$
and we obtain (as wished) that : $displaystyle tag{4}psi^*(x)=sum_{nle x}^{*}Lambda(n)$
with $psi^*$ the second Chebyshev function except when $x$ is an integer because in this case the last term of the sum has to be divided by $2$. This will be the meaning of the $^*$ symbols in this article : at a first order discontinuity point (i.e. a jump) the result is the mean value of the limit at the left and the right.
Now $psi^*(x)$ may also be written as :
$$tag{5}psi^*(x)=-frac1{2pi i}int_{c-iinfty}^{c+iinfty}frac{zeta'(s)}{zeta(s)}frac{x^s}s,ds$$
Here the idea is simply to sum the residues of (all) the poles at the left of the vertical line of integration ($c>1$) at least if $x>1$ (for $0<x<1$ we would have to take the poles at the right and obtain $0$). These poles come from the denominator $s$, the pole of $zeta$ at $1$ and the zeros with the correspondence : $0mapsto -frac{zeta'(0)}{zeta(0)}, 1mapsto x^1, rhomapsto -frac{x^rho}{rho}$ for $rho$ any zero of $zeta,$ (from the Weierstrass factorization of the Hadamard product) so that :
$$tag{6}boxed{displaystylepsi^*(x)=x-sum_{rho} frac {x^{rho}}{rho}-frac{zeta'(0)}{zeta(0)}},quad(x>1)$$
(in this post we won't distinguish the trivial from the non-trivial zeros $rho$ ; for convergence the non-trivial roots should be grouped by pairs and sorted by increasing $|Im(rho)|$)
The Riemann prime-counting function is defined by :
begin{align}
tag{7}Pi^*(x):&=sum_{p^kle x}^{*}frac 1k=sum_{nle x}^{*}frac {Lambda(n)}{log,n}\
&=sum_{nle x}^{*} Lambda(n)left(int_n^x frac{dt}{t,log^2 t}+frac 1{log,x}right)\
&=int_2^xfrac{psi^*(t) dt}{t,log^2 t}+frac{psi^*(x)}{log ,x}\
&tag{8}=int_2^xfrac{psi^{*'}(t) dt}{log,t}\
end{align}
But $ displaystyleoperatorname{li}(x)=int_2^x frac{dt}{log,t},$ (Riemann's variant of the logarithmic integral) verifies $ displaystyleoperatorname{li}(x^r)'=frac{r,x^{r-1}}{log x^r}=frac{x^{r-1}}{log,x}$ so that from $(6)$ :
$$tag{9}boxed{displaystylePi^*(x)=operatorname{li}(x)-sum_{rho} operatorname{li}(x^{rho})},quad(x>1)$$
The prime-counting function is defined by $ displaystylepi^*(x):=sum_{ple x}^{*}1 $ while $ displaystylePi^*(x):=sum_{p^kle x}^{*}frac 1k$ will be :
$$tag{10}Pi^*(x)=sum_{k>0} frac{pi^{*}bigl(x^{1/k}bigr)}k$$
Applying the Möbius inversion formula $ displaystylepi^{*}(x):=sum_{n=1}^{infty} frac{mu(k)}k Pi^*bigl(x^{1/k}bigr) $ to $(9)$ we get (with questionable convergence...) :
$$tag{11}boxed{displaystylepi^*(x)=R(x)-sum_{rho} R(x^{rho})},quad(x>1)$$
Where Riemann's $,displaystyle R(x):=sum_{n=1}^{infty} frac{mu(k)}k operatorname{li}bigl(x^{1/k}bigr),$ may be written as a Gram series.
Dirichlet's L-functions
We will try to repeat the previous derivations but with $zeta$ replaced by the Dirichlet L-function
$$L(s,chi)=sum_{k=1}^inftyfrac{chi(k)}{k^s}$$
With $chi$ a Dirichlet character we get following Euler product :
$$tag{1'}boxed{displaystyle L(s,chi)=prod_{p text{prime}}frac 1{1-chi(p)p^{-s}}}quadtext{for} Re(s)>1$$
so that ($chi$ is multiplicative implying $chi(p)^k=chi(p^k)$) :
$$log L(s,chi)=-sum_{p text{prime}}log(1-chi(p)p^{-s})=sum_psum_{k=1}^infty frac{chi(p^k)p^{-ks}}k$$
minus the derivative relatively to $s$ will be :
$$tag{2'}f(s,chi):=-frac{L'(s,chi)}{L(s,chi)}=sum_psum_{k=1}^infty frac{chi(p^k)log,p}{p^{ks}}=sum_{n=1}^infty frac{chi(n)Lambda(n)}{n^s}quadtext{for} Re(s)>1$$
The Perron formula applied to $f(s,chi)$ gives :
$$tag{3'}psi^*(x,chi):=frac1{2pi i}int_{c-iinfty}^{c+iinfty}f(s,chi)frac{x^s}s,ds=sum_{n=1}^infty chi(n)Lambda(n) frac1{2pi i}int_{c-iinfty}^{c+iinfty}left(frac xnright)^sfrac{ds}s$$
$$tag{4'}psi^*(x,chi)=sum_{nle x}^{*}chi(n)Lambda(n)$$
But $psi^*(x,chi)$ may also be written as :
$$tag{5'}psi^*(x,chi)=-frac1{2pi i}int_{c-iinfty}^{c+iinfty}frac{L'(s,chi)}{L(s,chi)}frac{x^s}s,ds$$
Again we compute the sum of the residues at (all) the poles to the left of the vertical line of integration (for $c>1$). The contributions from the different poles will be : $0mapsto -frac{L'(0,chi)}{L(0,chi)}, rhomapsto -frac{x^rho}{rho}$ for $rho$ any zero of $L(s,chi),$ (from Weierstrass factorization) so that :
$$tag{6'}boxed{displaystylepsi^*(x,chi)=-sum_{rho} frac {x^{rho}}{rho}-frac{L'(0,chi)}{L(0,chi)}},quad(x>1)$$
(the sum over $rho$ combines the trivial and the non-trivial zeros and we will suppose that $chi$ is a non principal character so that $,L(s,chi),$ is regular everywhere, for a principal character we should add a $x$ contribution from the pole at $1$)
An analogue to the Riemann prime-counting function will be :
begin{align}
tag{7'}Pi^*(x,chi):&=sum_{p^kle x}^{*}frac {chi(p^k)}k=sum_{nle x}^{*}frac {chi(n)Lambda(n)}{log,n}\
&=sum_{nle x}^{*} chi(n)Lambda(n)left(int_n^x frac{dt}{t,log^2 t}+frac 1{log,x}right)\
&=int_2^xfrac{psi^*(t,chi) dt}{t,log^2 t}+frac{psi^*(x,chi)}{log ,x}\
&tag{8'}=int_2^xfrac{psi^{*'}(t,chi) dt}{log,t}\
end{align}
From $ displaystyleoperatorname{li}(x^r)'=frac{x^{r-1}}{log,x}$ and $(6')$ we get :
$$tag{9'}boxed{displaystylePi^*(x,chi)=-sum_{rho} operatorname{li}(x^{rho})},quad(x>1)$$
(for $chi$ a principal character there is an additional $,operatorname{li}(x)$ term)
The $chi$-prime-counting function is $ displaystylepi^*(x,chi):=sum_{ple x}^{*}chi(p) $ while $ displaystylePi^*(x,chi):=sum_{p^kle x}^{*}frac {chi(p^k)}k$ so that :
$$tag{10'}Pi^*(x,chi)=sum_{k>0} frac{pi^{*}bigl(x^{1/k},chi^kbigr)}k,quad(x>1)$$
Broken analogy...
The problem here is that we have $,pi^{*}bigl(x^{1/k},chi^kbigr)$ in the sum and not $,pi^{*}bigl(x^{1/k},chibigr)$.
For principal characters (when $,chi(p)=0$ or $1$ so that $,chi^k(p)=chi(p),$ for $k$ positive) we may apply the Möbius inversion formula to $(9')$ and get :
$$tag{11'}displaystylepi^*(x,chi)=[R(x)]-sum_{rho} R(x^{rho})$$
(since $chi$ is a principal character we must add the $R(x)$ term from the pole at $1$)
but this doesn't work in general and the equality becomes questionable after the first prime $p$ such that $,chi^k(p)not =chi(p),$ as we will see by considering your specific question.
We want $chi$ to be the Dirichlet character modulo $4$ ($chi_2$ in the link) defined by $ displaystylechi(0)=0, chi(1)=1, chi(2)=0, chi(3)=-1 $ and get :
$$tag{12'}L(s,chi)=sum_{n=0}^inftyfrac {(-1)^n}{(2n+1)^s}=beta(s)$$
with $beta$ the Dirichlet beta function we could hope that :
$$pi^*(x,chi)=displaystylepi_{4,1}^*(x)-pi_{4,3}^*(x)approx -sum_{rho} R(x^{rho})$$
with $pi_{q,a}(x)=#{p:p text{is prime and} ,ple x, text{and} ,pequiv apmod q}$
But, from $(10')$ and $(9')$, we have in fact :
begin{align}
Pi^*(x,chi)&=sum_{k>0} frac{pi_{4,1}^{*}bigl(x^{1/k}bigr)+(-1)^k,pi_{4,3}^{*}bigl(x^{1/k}bigr)}k=-sum_{rho} operatorname{li}(x^{rho})\
&=sum_{k>0} frac{pi_{4,1}^{*}bigl(x^{1/k}bigr)-,pi_{4,3}^{*}bigl(x^{1/k}bigr)}k+2sum_{j>0} frac{pi_{4,3}^{*}left(sqrt{x}^{1/j}right)}{2j}\
end{align}
so that the Möbius transformation will return the exact (with $rho$ any zero of $beta$) :
$$tag{13'}boxed{displaystylepi_{4,1}^{*}(x)-pi_{4,3}^{*}(x)+pi_{4,3}^{*}left(sqrt{x}right)=-sum_{rho} R(x^{rho})}$$
To show the effect of the $,pi_{4,3}^{*}left(sqrt{x}right),$ term let's exhibit an approximation of the expression at the right (taking the $40$ first terms of the sum $R$ and the $300$ first non-trivial zeros of the Dirichlet-beta function) that we compare to the exact (dark) result for $,pi^*(x,chi)=displaystylepi_{4,1}^*(x)-pi_{4,3}^*(x),$ in the interval $,[5,80]$ :
The initial behavior seems right : $-1$ at $3$ followed by $+1$ at $5$ and $-1$ at $7$ but just after that there is a first 'irregularity' at $9=3^2$ because $chi(3^2)=1$ while $chi(3)=-1$. After that the behavior is correct (shifted $+1$ of course) until another irregularity at $49=7^2$ again because $chi(7^2)=1$. The next one will be at $121=11^2$ and at all the following squares of primes of gender $4n+3$...
ADDITION: Draks telescoping
The situation was blocked at this point until draks, using a suggestion of Greg Martin, proposed in february his powerful telescoping method to obtain (the $rho_zeta$ and $rho_beta$ are respectively all the zeros of the $zeta$ and $beta$ function) :
$$tag{14'}pi_{4,3}(x):=sum_{k=0}^infty;2^{-k-1}left(
Rleft(x^{1/2^{k}}right)-sum_{rho_zeta}
Rleft(x^{rho_zeta/2^k}right) +sum_{rho_beta}
Rleft(x^{rho_beta/2^k}right) right)$$
We may obtain this result too (with a minor correction) by rewriting $(11)$ as (for $x>2$) :
$$pi^*(x)=1+pi_{4,1}^{*}(x)+pi_{4,3}^{*}(x)=R(x)-sum_{rho_zeta} R(x^{rho_zeta})$$
($1$ was added since all primes except $2$ are equal to $1$ or $3bmod{4}$)
and using $(13')$ :
$$pi_{4,1}^{*}(x)-pi_{4,3}^{*}(x)+pi_{4,3}^{*}left(sqrt{x}right)=-sum_{rho_beta} R(x^{rho_beta})$$
Subtracting these equations gives :
$$2pi_{4,3}^{*}(x)-pi_{4,3}^{*}left(sqrt{x}right)=R(x)-1-sum_{rho_zeta} R(x^{rho_zeta})+sum_{rho_beta} R(x^{rho_beta})$$
while draks' telescoping sum $ displaystylefrac 12sum_{k=0}^infty frac{fleft(x^{1/2^{k}}right)}{2^k}$ is :
$$sum_{k=0}^infty frac {2,pi_{4,3}^{*}(x^{1/2^{k}})-pi_{4,3}^{*}bigl(x^{1/2^{k+1}}bigr)}{2^{k+1}}=sum_{k=0}^infty frac {Rleft(x^{1/2^{k}}right)-1-sum_{rho_zeta} Rleft(x^{rho_zeta/2^k}right)+sum_{rho_beta} Rleft(x^{rho_beta/2^k}right)}{2^{k+1}}$$
producing draks' formula :
$$tag{15'}boxed{displaystylepi_{4,3}^{*}(x)=sum_{k=0}^infty frac {Rleft(x^{1/2^{k}}right)-sum_{rho_zeta} Rleft(x^{rho_zeta/2^k}right)+sum_{rho_beta} Rleft(x^{rho_beta/2^k}right)}{2^{k+1}}-1}$$
A similar formula may be written for $pi_{4,1}$ and this answers positively the initial question !
Let's finish with a plot of $pi_{4,3}^{*}(x)$ from $4$ to $100$ (more details here with the pari/gp scripts used)
edited Apr 13 '17 at 12:20
Community♦
1
answered Jan 20 '13 at 17:22
Raymond ManzoniRaymond Manzoni
37.2k563117
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Raymond, first of all: Wow! What an answer. "+1". Thank you so much for everything including the link to Edward's book. My copy is buried somewhere in a box because I moved. It'll take me a while to verify all that, but like you, I think that the reasoning by analogy is correct. Specially the last equation thrills me. You might have noticed my attempts with "Summing over General Functions of Primes" ...
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– draks ...
Jan 20 '13 at 22:44
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... or the follow-on question: "An Expression for $logzeta(ns)$ derived from the Limit of the truncated Prime $zeta$ Function". So I think it's possible to combine these result to get an analytic expression for the truncated Prime $zeta$ function with a restriction to primes $4npm 1$. This is ever so cool and I can't leave this unawarded...
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– draks ...
Jan 20 '13 at 22:51
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Thanks @draks... I think that most things are right here except tiny things like the - from the citation, the starting value $n=1$ of the Dirichlet beta function, and perhaps very important things like the transitions from $(9')$ to $(11')$ : these should be (more than!) double checked (because of things like $chi(3^2)=1not =-1$ and of warnings from different authors like Ingham that Riemann's method to obtain $pi$ may not be as easily generalized as $psicdots$ but works only for some specific Dirichlet characters ($chi(p)=0,1$ only?)). But this clear path may at least be checked!
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– Raymond Manzoni
Jan 21 '13 at 0:05
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@draks... I corrected the end of my answer.
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– Raymond Manzoni
Jan 26 '13 at 23:04
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aha, looks like we've gottta add $(mu^2(n)-1)$ to get rid of the squares...interesting, what do you think? And thanks for your thoughts. At least my ${rm li}(x)$ works...
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– draks ...
Jan 27 '13 at 16:03
|
show 3 more comments
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You probably know that
$$
Pi^*(x;q,a) = sum^*_{substack{nle x \ nequiv apmod q}} frac{Lambda(n)}{log n} = frac1{phi(q)} sum_{chipmod q} overline{chi(a)}Pi^*(x,chi).
$$
I find that the easiest way to convert from counting prime powers (with weights) to counting only primes is to use the fact that
begin{align*}
Pi^*(x;q,a) &= sum^*_{substack{ple x \ pequiv apmod q}} 1 + sum^*_{substack{p^2le x \ p^2equiv apmod q}} tfrac12 + sum^*_{substack{p^3le x \ p^3equiv apmod q}} tfrac13 + cdots \
&= pi^*(x;q,a) + tfrac12 sum_{substack{bpmod q \ b^2equiv apmod q}} pi^*(x^{1/2};q,b) + tfrac13 sum_{substack{cpmod q \ c^3equiv apmod q}} pi^*(x^{1/3};q,c) + cdots
end{align*}
and solving for $pi^*(x;q,a)$ in terms of all $Pi^*$ functions (tedious but possible).
answered Jan 31 '13 at 7:39
Greg MartinGreg Martin
36.3k23565
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+1 thanks, doesn't it boil down a little/lot when $a=1$? Could you give me a reference hint or something else to read on that topic?
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– draks ...
Jan 31 '13 at 20:43
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I'm not sure it gets much better when $a=1$. Except that all the values $chi(a)$ (which I forgot to put in in the first place - they're there now) equal $1$ and thus disappear. But $a=1$ will yield the largest possible number of terms in the sums over $b$, $c$, etc. in the last displayed equation line.
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– Greg Martin
Feb 1 '13 at 1:25
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hmm, when I look at the second term in the last line and insert $a=1$, the sum's running over $b=1,3,5,7,dots$. But $b>3$ doesn't make sense since it's included already before, right? If this is correct and I continue, I get $x^mequiv 1 bmod 4to cases{x=1,3 text{ if $m$ even }\x=1;;,, text{ if $m$ odd}} $. But doesn't summing $pi(x^{1/m};4,pm 1)$ equal $pi(x^{1/m})$? How to continue from here? Could you kindly sketch the way...?
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– draks ...
Feb 8 '13 at 23:01
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Right, when I write a sum over $bpmod q$, I mean a sum over residue classes - so don't separately include both $b=1$ and $b=5$, since they're the same residue class. And in this particular case, when $m$ is even you do in fact count all odd primes in the sum, yes.
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– Greg Martin
Feb 9 '13 at 6:13
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you are awarded the bounty since it was your prime race paper which gave me new hope. I'm currently studying your arXiv stuff. Could you point me to something that would help me once more?
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– draks ...
Feb 16 '13 at 13:35
|
show 7 more comments
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$begingroup$
Let's start with the nice paper 'Prime Number Races' of Granville and Martin from your comments. It contains the sentence :
"The precise formula he (Riemann) proposed is a bit too technical for
this article, but we can get a good sense of it from the following
approximation when x is large. This formula, while widely believed to
be correct, has not yet been proved."
$$
tag{1}boxed{displaystylefrac{intlimits_2^x{frac{dt}{ln t}}-pi(x)}{sqrt x/ln x} approx 1 + 2sum_{gamma} frac{sin(gammaln x)}{gamma}}
$$
I can only conjecture that the formula referenced in their last sentence is your formula $(1)$ (I don't know a proof of convergence of this formula either) derived from Riemann's explicit formula (proved by von Mangoldt) and not the approximate formula I reproduce above equivalent to your $(2)$ . Note that they add that this approximation is valid 'when $x$ is large'.
Let's add that this approximation doesn't include the trivial zeros from your other thread.
Concerning the $displaystyle frac{sin(gammaln x)}{gamma}$ term it is a good approximation of Riesel's approximation $(2.30)$ from his fine book 'Prime Numbers and Computer Methods for Factorization' :
$$tag{2}frac{cos(gammaln x-arg rho)}{|rho|},quadtext{with} rho=frac 12+igamma$$
since $ gammagg 1 $ we have indeed $ |rho|approx gamma $ and $ arg,rhoapprox frac{pi}2$
(note that $gamma > 14.13 $ so that $ 1<frac {|rho|}{gamma}<1.00063$)
Riesel's method is to approximate the oscillating term (see the next page from Riesel's book) :
$$tag{3}operatorname{R}(x^{rho}) = sum_{n=1}^{infty} frac{ mu (n)}{n}operatorname{li}(x^{rho/n})$$
with the first term :
$$tag{4}operatorname{R}(x^{rho}) approxleft[operatorname{li}(x^{rho})=operatorname{li}(e^{rho;ln x})=operatorname{Ei}(rho;ln x)right]$$
and to use the asymptotic expansion for $operatorname{Ei}$ :
$$tag{5}operatorname{Ei}(z)sim frac{e^z}{z}left(1+frac {1!}z+frac{2!}{z^2}+cdotsright)$$
to get :
$$operatorname{Ei}left(left(frac 12+igammaright);ln xright)approx frac{x^{1/2+igamma}}{(1/2+igamma);ln x}$$
so that combining the two 'mirror' zeros :
$$operatorname{Ei}left(left(frac 12+igammaright)+;ln xright)+operatorname{Ei}left(left(frac 12-igammaright)+;ln xright)approx frac{sqrt{x}}{ln x}left(frac{e^{igammaln x}}{1/2+igamma}+frac{e^{-igammaln x}}{1/2-igamma}right)$$
$$approxfrac{sqrt{x}}{ln x}frac{e^{igammaln x-iarg{rho}}+e^{-igammaln x+iarg{rho}}}{|rho|}quadtext{since} rho=frac 12+igamma$$
we get Riesel's approximation :
$$tag{6}operatorname{R}(x^{rho})+operatorname{R}(x^{overline{rho}})approxfrac{2;sqrt{x};cos(gammaln x-arg{rho})}{|rho|;ln x}$$
while $operatorname{R}(x)$ was approximated by (using $(3), (4), (5)$) :
$$tag{7}operatorname{R}(x)approx operatorname{li}(x)-frac 12operatorname{li}bigl(x^{1/2}bigr)quad text{with} operatorname{li}bigl(x^{1/2}bigr)approx frac{2sqrt{x}}{ln x}$$
this allows too to get better approximations if wished ($operatorname{R}(x)approx operatorname{li}(x)-frac 12operatorname{li}bigl(x^{1/2}bigr)-frac 13operatorname{li}bigl(x^{1/3}bigr) $ for example, adding the trivial zeros contribution and so on...)
Now to answer your questions :
- We have to suppose the R.H. to be able to write $rho=frac 12+igamma $ for any root $rho$ so that the simplifications proposed apply (in practice we don't need R.H. since $(2)$ requires precomputed zeros and since all the zeros known satisfy the R.H. ... to this date...)
- The implication from $(1)to (2)$ was provided without using the Gram series
Hoping this clarified things,
edited Apr 13 '17 at 12:20
Community♦
1
answered Jan 5 '13 at 21:33
Raymond ManzoniRaymond Manzoni
37.2k563117
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$begingroup$
thank you very much. Again :-)
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– draks ...
Jan 6 '13 at 22:14
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You are welcome @draks : this gave me the occasion to compute some zeros (say 331) of Dirichlet beta, see a fine approximation for $pi(n)$ and ways to make it better. Thanks for the journey !
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– Raymond Manzoni
Jan 6 '13 at 23:43
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Wow 331 sounds great. I was just out to find a table where these zeros are listed. Would you mind sharing the zeros of beta with me? And to be honest I still have one more question: Is it possible to write $#{text{primes} 4n+3 le x} - #{text{primes} 4n+1 le x}$(form [here](math.stackexchange.com/q/149755/19341)) in terms like $operatorname{R}(x^1) - sum_{rho}operatorname{R}(x^{rho})$ ? I hope I don't bother you too much...
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– draks ...
Jan 6 '13 at 23:54
$begingroup$
Thanks a lot. Concerning the last question, I thought about using a simple analogy $(pi_{4n+3}-pi_{4n+1})-text{Li}(x)=text{R}(x)-sum_rho text{R}(x^rho)$ with $rho$ being the values you've kindly provided...?
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– draks ...
Jan 7 '13 at 0:21
$begingroup$
@draks: I gave the zeros in your other thread and will erase the values here... Concerning your other question I don't know but will look at it in the coming days (it's late...). I must add that from your formula $(4')$ there is probably a similar (or identical) formula as you suppose !
$endgroup$
– Raymond Manzoni
Jan 7 '13 at 0:30
|
show 7 more comments
$begingroup$
Let's start with the nice paper 'Prime Number Races' of Granville and Martin from your comments. It contains the sentence :
"The precise formula he (Riemann) proposed is a bit too technical for
this article, but we can get a good sense of it from the following
approximation when x is large. This formula, while widely believed to
be correct, has not yet been proved."
$$
tag{1}boxed{displaystylefrac{intlimits_2^x{frac{dt}{ln t}}-pi(x)}{sqrt x/ln x} approx 1 + 2sum_{gamma} frac{sin(gammaln x)}{gamma}}
$$
I can only conjecture that the formula referenced in their last sentence is your formula $(1)$ (I don't know a proof of convergence of this formula either) derived from Riemann's explicit formula (proved by von Mangoldt) and not the approximate formula I reproduce above equivalent to your $(2)$ . Note that they add that this approximation is valid 'when $x$ is large'.
Let's add that this approximation doesn't include the trivial zeros from your other thread.
Concerning the $displaystyle frac{sin(gammaln x)}{gamma}$ term it is a good approximation of Riesel's approximation $(2.30)$ from his fine book 'Prime Numbers and Computer Methods for Factorization' :
$$tag{2}frac{cos(gammaln x-arg rho)}{|rho|},quadtext{with} rho=frac 12+igamma$$
since $ gammagg 1 $ we have indeed $ |rho|approx gamma $ and $ arg,rhoapprox frac{pi}2$
(note that $gamma > 14.13 $ so that $ 1<frac {|rho|}{gamma}<1.00063$)
Riesel's method is to approximate the oscillating term (see the next page from Riesel's book) :
$$tag{3}operatorname{R}(x^{rho}) = sum_{n=1}^{infty} frac{ mu (n)}{n}operatorname{li}(x^{rho/n})$$
with the first term :
$$tag{4}operatorname{R}(x^{rho}) approxleft[operatorname{li}(x^{rho})=operatorname{li}(e^{rho;ln x})=operatorname{Ei}(rho;ln x)right]$$
and to use the asymptotic expansion for $operatorname{Ei}$ :
$$tag{5}operatorname{Ei}(z)sim frac{e^z}{z}left(1+frac {1!}z+frac{2!}{z^2}+cdotsright)$$
to get :
$$operatorname{Ei}left(left(frac 12+igammaright);ln xright)approx frac{x^{1/2+igamma}}{(1/2+igamma);ln x}$$
so that combining the two 'mirror' zeros :
$$operatorname{Ei}left(left(frac 12+igammaright)+;ln xright)+operatorname{Ei}left(left(frac 12-igammaright)+;ln xright)approx frac{sqrt{x}}{ln x}left(frac{e^{igammaln x}}{1/2+igamma}+frac{e^{-igammaln x}}{1/2-igamma}right)$$
$$approxfrac{sqrt{x}}{ln x}frac{e^{igammaln x-iarg{rho}}+e^{-igammaln x+iarg{rho}}}{|rho|}quadtext{since} rho=frac 12+igamma$$
we get Riesel's approximation :
$$tag{6}operatorname{R}(x^{rho})+operatorname{R}(x^{overline{rho}})approxfrac{2;sqrt{x};cos(gammaln x-arg{rho})}{|rho|;ln x}$$
while $operatorname{R}(x)$ was approximated by (using $(3), (4), (5)$) :
$$tag{7}operatorname{R}(x)approx operatorname{li}(x)-frac 12operatorname{li}bigl(x^{1/2}bigr)quad text{with} operatorname{li}bigl(x^{1/2}bigr)approx frac{2sqrt{x}}{ln x}$$
this allows too to get better approximations if wished ($operatorname{R}(x)approx operatorname{li}(x)-frac 12operatorname{li}bigl(x^{1/2}bigr)-frac 13operatorname{li}bigl(x^{1/3}bigr) $ for example, adding the trivial zeros contribution and so on...)
Now to answer your questions :
- We have to suppose the R.H. to be able to write $rho=frac 12+igamma $ for any root $rho$ so that the simplifications proposed apply (in practice we don't need R.H. since $(2)$ requires precomputed zeros and since all the zeros known satisfy the R.H. ... to this date...)
- The implication from $(1)to (2)$ was provided without using the Gram series
Hoping this clarified things,
edited Apr 13 '17 at 12:20
Community♦
1
answered Jan 5 '13 at 21:33
Raymond ManzoniRaymond Manzoni
37.2k563117
$endgroup$
$begingroup$
thank you very much. Again :-)
$endgroup$
– draks ...
Jan 6 '13 at 22:14
$begingroup$
You are welcome @draks : this gave me the occasion to compute some zeros (say 331) of Dirichlet beta, see a fine approximation for $pi(n)$ and ways to make it better. Thanks for the journey !
$endgroup$
– Raymond Manzoni
Jan 6 '13 at 23:43
$begingroup$
Wow 331 sounds great. I was just out to find a table where these zeros are listed. Would you mind sharing the zeros of beta with me? And to be honest I still have one more question: Is it possible to write $#{text{primes} 4n+3 le x} - #{text{primes} 4n+1 le x}$(form [here](math.stackexchange.com/q/149755/19341)) in terms like $operatorname{R}(x^1) - sum_{rho}operatorname{R}(x^{rho})$ ? I hope I don't bother you too much...
$endgroup$
– draks ...
Jan 6 '13 at 23:54
$begingroup$
Thanks a lot. Concerning the last question, I thought about using a simple analogy $(pi_{4n+3}-pi_{4n+1})-text{Li}(x)=text{R}(x)-sum_rho text{R}(x^rho)$ with $rho$ being the values you've kindly provided...?
$endgroup$
– draks ...
Jan 7 '13 at 0:21
$begingroup$
@draks: I gave the zeros in your other thread and will erase the values here... Concerning your other question I don't know but will look at it in the coming days (it's late...). I must add that from your formula $(4')$ there is probably a similar (or identical) formula as you suppose !
$endgroup$
– Raymond Manzoni
Jan 7 '13 at 0:30
|
show 7 more comments
$begingroup$
Let's start with the nice paper 'Prime Number Races' of Granville and Martin from your comments. It contains the sentence :
"The precise formula he (Riemann) proposed is a bit too technical for
this article, but we can get a good sense of it from the following
approximation when x is large. This formula, while widely believed to
be correct, has not yet been proved."
$$
tag{1}boxed{displaystylefrac{intlimits_2^x{frac{dt}{ln t}}-pi(x)}{sqrt x/ln x} approx 1 + 2sum_{gamma} frac{sin(gammaln x)}{gamma}}
$$
I can only conjecture that the formula referenced in their last sentence is your formula $(1)$ (I don't know a proof of convergence of this formula either) derived from Riemann's explicit formula (proved by von Mangoldt) and not the approximate formula I reproduce above equivalent to your $(2)$ . Note that they add that this approximation is valid 'when $x$ is large'.
Let's add that this approximation doesn't include the trivial zeros from your other thread.
Concerning the $displaystyle frac{sin(gammaln x)}{gamma}$ term it is a good approximation of Riesel's approximation $(2.30)$ from his fine book 'Prime Numbers and Computer Methods for Factorization' :
$$tag{2}frac{cos(gammaln x-arg rho)}{|rho|},quadtext{with} rho=frac 12+igamma$$
since $ gammagg 1 $ we have indeed $ |rho|approx gamma $ and $ arg,rhoapprox frac{pi}2$
(note that $gamma > 14.13 $ so that $ 1<frac {|rho|}{gamma}<1.00063$)
Riesel's method is to approximate the oscillating term (see the next page from Riesel's book) :
$$tag{3}operatorname{R}(x^{rho}) = sum_{n=1}^{infty} frac{ mu (n)}{n}operatorname{li}(x^{rho/n})$$
with the first term :
$$tag{4}operatorname{R}(x^{rho}) approxleft[operatorname{li}(x^{rho})=operatorname{li}(e^{rho;ln x})=operatorname{Ei}(rho;ln x)right]$$
and to use the asymptotic expansion for $operatorname{Ei}$ :
$$tag{5}operatorname{Ei}(z)sim frac{e^z}{z}left(1+frac {1!}z+frac{2!}{z^2}+cdotsright)$$
to get :
$$operatorname{Ei}left(left(frac 12+igammaright);ln xright)approx frac{x^{1/2+igamma}}{(1/2+igamma);ln x}$$
so that combining the two 'mirror' zeros :
$$operatorname{Ei}left(left(frac 12+igammaright)+;ln xright)+operatorname{Ei}left(left(frac 12-igammaright)+;ln xright)approx frac{sqrt{x}}{ln x}left(frac{e^{igammaln x}}{1/2+igamma}+frac{e^{-igammaln x}}{1/2-igamma}right)$$
$$approxfrac{sqrt{x}}{ln x}frac{e^{igammaln x-iarg{rho}}+e^{-igammaln x+iarg{rho}}}{|rho|}quadtext{since} rho=frac 12+igamma$$
we get Riesel's approximation :
$$tag{6}operatorname{R}(x^{rho})+operatorname{R}(x^{overline{rho}})approxfrac{2;sqrt{x};cos(gammaln x-arg{rho})}{|rho|;ln x}$$
while $operatorname{R}(x)$ was approximated by (using $(3), (4), (5)$) :
$$tag{7}operatorname{R}(x)approx operatorname{li}(x)-frac 12operatorname{li}bigl(x^{1/2}bigr)quad text{with} operatorname{li}bigl(x^{1/2}bigr)approx frac{2sqrt{x}}{ln x}$$
this allows too to get better approximations if wished ($operatorname{R}(x)approx operatorname{li}(x)-frac 12operatorname{li}bigl(x^{1/2}bigr)-frac 13operatorname{li}bigl(x^{1/3}bigr) $ for example, adding the trivial zeros contribution and so on...)
Now to answer your questions :
- We have to suppose the R.H. to be able to write $rho=frac 12+igamma $ for any root $rho$ so that the simplifications proposed apply (in practice we don't need R.H. since $(2)$ requires precomputed zeros and since all the zeros known satisfy the R.H. ... to this date...)
- The implication from $(1)to (2)$ was provided without using the Gram series
Hoping this clarified things,
edited Apr 13 '17 at 12:20
Community♦
1
answered Jan 5 '13 at 21:33
Raymond ManzoniRaymond Manzoni
37.2k563117
$endgroup$
Let's start with the nice paper 'Prime Number Races' of Granville and Martin from your comments. It contains the sentence :
"The precise formula he (Riemann) proposed is a bit too technical for
this article, but we can get a good sense of it from the following
approximation when x is large. This formula, while widely believed to
be correct, has not yet been proved."
$$
tag{1}boxed{displaystylefrac{intlimits_2^x{frac{dt}{ln t}}-pi(x)}{sqrt x/ln x} approx 1 + 2sum_{gamma} frac{sin(gammaln x)}{gamma}}
$$
I can only conjecture that the formula referenced in their last sentence is your formula $(1)$ (I don't know a proof of convergence of this formula either) derived from Riemann's explicit formula (proved by von Mangoldt) and not the approximate formula I reproduce above equivalent to your $(2)$ . Note that they add that this approximation is valid 'when $x$ is large'.
Let's add that this approximation doesn't include the trivial zeros from your other thread.
Concerning the $displaystyle frac{sin(gammaln x)}{gamma}$ term it is a good approximation of Riesel's approximation $(2.30)$ from his fine book 'Prime Numbers and Computer Methods for Factorization' :
$$tag{2}frac{cos(gammaln x-arg rho)}{|rho|},quadtext{with} rho=frac 12+igamma$$
since $ gammagg 1 $ we have indeed $ |rho|approx gamma $ and $ arg,rhoapprox frac{pi}2$
(note that $gamma > 14.13 $ so that $ 1<frac {|rho|}{gamma}<1.00063$)
Riesel's method is to approximate the oscillating term (see the next page from Riesel's book) :
$$tag{3}operatorname{R}(x^{rho}) = sum_{n=1}^{infty} frac{ mu (n)}{n}operatorname{li}(x^{rho/n})$$
with the first term :
$$tag{4}operatorname{R}(x^{rho}) approxleft[operatorname{li}(x^{rho})=operatorname{li}(e^{rho;ln x})=operatorname{Ei}(rho;ln x)right]$$
and to use the asymptotic expansion for $operatorname{Ei}$ :
$$tag{5}operatorname{Ei}(z)sim frac{e^z}{z}left(1+frac {1!}z+frac{2!}{z^2}+cdotsright)$$
to get :
$$operatorname{Ei}left(left(frac 12+igammaright);ln xright)approx frac{x^{1/2+igamma}}{(1/2+igamma);ln x}$$
so that combining the two 'mirror' zeros :
$$operatorname{Ei}left(left(frac 12+igammaright)+;ln xright)+operatorname{Ei}left(left(frac 12-igammaright)+;ln xright)approx frac{sqrt{x}}{ln x}left(frac{e^{igammaln x}}{1/2+igamma}+frac{e^{-igammaln x}}{1/2-igamma}right)$$
$$approxfrac{sqrt{x}}{ln x}frac{e^{igammaln x-iarg{rho}}+e^{-igammaln x+iarg{rho}}}{|rho|}quadtext{since} rho=frac 12+igamma$$
we get Riesel's approximation :
$$tag{6}operatorname{R}(x^{rho})+operatorname{R}(x^{overline{rho}})approxfrac{2;sqrt{x};cos(gammaln x-arg{rho})}{|rho|;ln x}$$
while $operatorname{R}(x)$ was approximated by (using $(3), (4), (5)$) :
$$tag{7}operatorname{R}(x)approx operatorname{li}(x)-frac 12operatorname{li}bigl(x^{1/2}bigr)quad text{with} operatorname{li}bigl(x^{1/2}bigr)approx frac{2sqrt{x}}{ln x}$$
this allows too to get better approximations if wished ($operatorname{R}(x)approx operatorname{li}(x)-frac 12operatorname{li}bigl(x^{1/2}bigr)-frac 13operatorname{li}bigl(x^{1/3}bigr) $ for example, adding the trivial zeros contribution and so on...)
Now to answer your questions :
- We have to suppose the R.H. to be able to write $rho=frac 12+igamma $ for any root $rho$ so that the simplifications proposed apply (in practice we don't need R.H. since $(2)$ requires precomputed zeros and since all the zeros known satisfy the R.H. ... to this date...)
- The implication from $(1)to (2)$ was provided without using the Gram series
Hoping this clarified things,
edited Apr 13 '17 at 12:20
Community♦
1
answered Jan 5 '13 at 21:33
Raymond ManzoniRaymond Manzoni
37.2k563117
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Jan 5 '13 at 21:33
Raymond ManzoniRaymond Manzoni
37.2k563117
answered Jan 5 '13 at 21:33
Raymond ManzoniRaymond Manzoni
37.2k563117
answered Jan 5 '13 at 21:33
Raymond ManzoniRaymond Manzoni
37.2k563117
37.2k563117
$begingroup$
thank you very much. Again :-)
$endgroup$
– draks ...
Jan 6 '13 at 22:14
$begingroup$
You are welcome @draks : this gave me the occasion to compute some zeros (say 331) of Dirichlet beta, see a fine approximation for $pi(n)$ and ways to make it better. Thanks for the journey !
$endgroup$
– Raymond Manzoni
Jan 6 '13 at 23:43
$begingroup$
Wow 331 sounds great. I was just out to find a table where these zeros are listed. Would you mind sharing the zeros of beta with me? And to be honest I still have one more question: Is it possible to write $#{text{primes} 4n+3 le x} - #{text{primes} 4n+1 le x}$(form [here](math.stackexchange.com/q/149755/19341)) in terms like $operatorname{R}(x^1) - sum_{rho}operatorname{R}(x^{rho})$ ? I hope I don't bother you too much...
$endgroup$
– draks ...
Jan 6 '13 at 23:54
$begingroup$
Thanks a lot. Concerning the last question, I thought about using a simple analogy $(pi_{4n+3}-pi_{4n+1})-text{Li}(x)=text{R}(x)-sum_rho text{R}(x^rho)$ with $rho$ being the values you've kindly provided...?
$endgroup$
– draks ...
Jan 7 '13 at 0:21
$begingroup$
@draks: I gave the zeros in your other thread and will erase the values here... Concerning your other question I don't know but will look at it in the coming days (it's late...). I must add that from your formula $(4')$ there is probably a similar (or identical) formula as you suppose !
$endgroup$
– Raymond Manzoni
Jan 7 '13 at 0:30
|
show 7 more comments
$begingroup$
thank you very much. Again :-)
$endgroup$
– draks ...
Jan 6 '13 at 22:14
$begingroup$
You are welcome @draks : this gave me the occasion to compute some zeros (say 331) of Dirichlet beta, see a fine approximation for $pi(n)$ and ways to make it better. Thanks for the journey !
$endgroup$
– Raymond Manzoni
Jan 6 '13 at 23:43
$begingroup$
Wow 331 sounds great. I was just out to find a table where these zeros are listed. Would you mind sharing the zeros of beta with me? And to be honest I still have one more question: Is it possible to write $#{text{primes} 4n+3 le x} - #{text{primes} 4n+1 le x}$(form [here](math.stackexchange.com/q/149755/19341)) in terms like $operatorname{R}(x^1) - sum_{rho}operatorname{R}(x^{rho})$ ? I hope I don't bother you too much...
$endgroup$
– draks ...
Jan 6 '13 at 23:54
$begingroup$
Thanks a lot. Concerning the last question, I thought about using a simple analogy $(pi_{4n+3}-pi_{4n+1})-text{Li}(x)=text{R}(x)-sum_rho text{R}(x^rho)$ with $rho$ being the values you've kindly provided...?
$endgroup$
– draks ...
Jan 7 '13 at 0:21
$begingroup$
@draks: I gave the zeros in your other thread and will erase the values here... Concerning your other question I don't know but will look at it in the coming days (it's late...). I must add that from your formula $(4')$ there is probably a similar (or identical) formula as you suppose !
$endgroup$
– Raymond Manzoni
Jan 7 '13 at 0:30
$begingroup$
thank you very much. Again :-)
$endgroup$
– draks ...
Jan 6 '13 at 22:14
$begingroup$
thank you very much. Again :-)
$endgroup$
– draks ...
Jan 6 '13 at 22:14
$begingroup$
You are welcome @draks : this gave me the occasion to compute some zeros (say 331) of Dirichlet beta, see a fine approximation for $pi(n)$ and ways to make it better. Thanks for the journey !
$endgroup$
– Raymond Manzoni
Jan 6 '13 at 23:43
$begingroup$
You are welcome @draks : this gave me the occasion to compute some zeros (say 331) of Dirichlet beta, see a fine approximation for $pi(n)$ and ways to make it better. Thanks for the journey !
$endgroup$
– Raymond Manzoni
Jan 6 '13 at 23:43
$begingroup$
Wow 331 sounds great. I was just out to find a table where these zeros are listed. Would you mind sharing the zeros of beta with me? And to be honest I still have one more question: Is it possible to write $#{text{primes} 4n+3 le x} - #{text{primes} 4n+1 le x}$(form [here](math.stackexchange.com/q/149755/19341)) in terms like $operatorname{R}(x^1) - sum_{rho}operatorname{R}(x^{rho})$ ? I hope I don't bother you too much...
$endgroup$
– draks ...
Jan 6 '13 at 23:54
$begingroup$
Wow 331 sounds great. I was just out to find a table where these zeros are listed. Would you mind sharing the zeros of beta with me? And to be honest I still have one more question: Is it possible to write $#{text{primes} 4n+3 le x} - #{text{primes} 4n+1 le x}$(form [here](math.stackexchange.com/q/149755/19341)) in terms like $operatorname{R}(x^1) - sum_{rho}operatorname{R}(x^{rho})$ ? I hope I don't bother you too much...
$endgroup$
– draks ...
Jan 6 '13 at 23:54
$begingroup$
Thanks a lot. Concerning the last question, I thought about using a simple analogy $(pi_{4n+3}-pi_{4n+1})-text{Li}(x)=text{R}(x)-sum_rho text{R}(x^rho)$ with $rho$ being the values you've kindly provided...?
$endgroup$
– draks ...
Jan 7 '13 at 0:21
$begingroup$
Thanks a lot. Concerning the last question, I thought about using a simple analogy $(pi_{4n+3}-pi_{4n+1})-text{Li}(x)=text{R}(x)-sum_rho text{R}(x^rho)$ with $rho$ being the values you've kindly provided...?
$endgroup$
– draks ...
Jan 7 '13 at 0:21
$begingroup$
@draks: I gave the zeros in your other thread and will erase the values here... Concerning your other question I don't know but will look at it in the coming days (it's late...). I must add that from your formula $(4')$ there is probably a similar (or identical) formula as you suppose !
$endgroup$
– Raymond Manzoni
Jan 7 '13 at 0:30
$begingroup$
@draks: I gave the zeros in your other thread and will erase the values here... Concerning your other question I don't know but will look at it in the coming days (it's late...). I must add that from your formula $(4')$ there is probably a similar (or identical) formula as you suppose !
$endgroup$
– Raymond Manzoni
Jan 7 '13 at 0:30
|
show 7 more comments
$begingroup$
Before answering to your question in the comment :
" Is it possible to write $ #{text{primes} 4n+3 le x},-
,#{text{primes} 4n+1 le x}$ (form here) in terms like $,operatorname{R}(x^1) - sum_{rho}operatorname{R}(x^{rho})$ ? "
let's start with a sketch using von Mangoldt's derivation to obtain your equation $(1)$ that will be used for inspiration (for proofs see for example Edwards' chapter 3) :
Riemann's Explicit Formulas
The Euler product formula gives us :
$$tag{1}boxed{displaystylezeta(s)=prod_{p text{prime}}frac 1{1-p^{-s}}}quadtext{for} Re(s)>1$$
so that
$$log zeta(s)=-sum_{p text{prime}}log(1-p^{-s})=sum_psum_{k=1}^infty frac{p^{-ks}}k$$
minus the derivative will be :
$$tag{2}f(s):=-frac{zeta'(s)}{zeta(s)}=sum_psum_{k=1}^infty frac{log,p}{p^{ks}}=sum_{n=1}^infty frac{Lambda(n)}{n^s}quadtext{for} Re(s)>1$$
with $Lambda$ the von Mangoldt function defined by $ Lambda(n):=begin{cases}
log, p & text{if} n=p^k text{and} k>0\
0 & text{else}
end{cases}$
Let's use the definition of the second Chebyshev function : $displaystyle psi(x)=sum_{nleq x}Lambda(n)$ and Abel's sum formula applied to $a(n):=Lambda(n)$ and $phi(n):=n^{-s}$ :
$$sum_{n=1}^infty frac{Lambda(n)}{n^s}=sint_1^infty frac {sum_{nleq x}Lambda(n)}{x^{s+1}};dx$$
to rewrite $f(s)$ as (the lower bound became $0$ since $phi(x)=0$ for $x<1$) :
$$tag{2.1}f(s)=sint_0^{infty}frac{psi(x)}{x^{s+1}}dx$$
But this is a Mellin transform which may be reverted to get Perron's formula (let's observe that the Dirichlet series $displaystyle f(s)=sum_{n=1}^infty frac{Lambda(n)}{n^s}$ is absolutely convergent for $Re(s)>1$ and suppose $c>1$) :
$$tag{3}psi^*(x):=frac1{2pi i}int_{c-iinfty}^{c+iinfty}f(s)frac{x^s}s,ds=sum_{n=1}^infty Lambda(n) frac1{2pi i}int_{c-iinfty}^{c+iinfty}left(frac xnright)^sfrac{ds}s$$
This last integral may be evaluated using Fourier's theorem :
$$ displaystylefrac1{2pi i}int_{c-iinfty}^{c+iinfty}frac{y^s}s ds=
begin{cases}
0 & 0<y<1\
1/2 & quad y=1\
1 & quad y>1\
end{cases}$$
and we obtain (as wished) that : $displaystyle tag{4}psi^*(x)=sum_{nle x}^{*}Lambda(n)$
with $psi^*$ the second Chebyshev function except when $x$ is an integer because in this case the last term of the sum has to be divided by $2$. This will be the meaning of the $^*$ symbols in this article : at a first order discontinuity point (i.e. a jump) the result is the mean value of the limit at the left and the right.
Now $psi^*(x)$ may also be written as :
$$tag{5}psi^*(x)=-frac1{2pi i}int_{c-iinfty}^{c+iinfty}frac{zeta'(s)}{zeta(s)}frac{x^s}s,ds$$
Here the idea is simply to sum the residues of (all) the poles at the left of the vertical line of integration ($c>1$) at least if $x>1$ (for $0<x<1$ we would have to take the poles at the right and obtain $0$). These poles come from the denominator $s$, the pole of $zeta$ at $1$ and the zeros with the correspondence : $0mapsto -frac{zeta'(0)}{zeta(0)}, 1mapsto x^1, rhomapsto -frac{x^rho}{rho}$ for $rho$ any zero of $zeta,$ (from the Weierstrass factorization of the Hadamard product) so that :
$$tag{6}boxed{displaystylepsi^*(x)=x-sum_{rho} frac {x^{rho}}{rho}-frac{zeta'(0)}{zeta(0)}},quad(x>1)$$
(in this post we won't distinguish the trivial from the non-trivial zeros $rho$ ; for convergence the non-trivial roots should be grouped by pairs and sorted by increasing $|Im(rho)|$)
The Riemann prime-counting function is defined by :
begin{align}
tag{7}Pi^*(x):&=sum_{p^kle x}^{*}frac 1k=sum_{nle x}^{*}frac {Lambda(n)}{log,n}\
&=sum_{nle x}^{*} Lambda(n)left(int_n^x frac{dt}{t,log^2 t}+frac 1{log,x}right)\
&=int_2^xfrac{psi^*(t) dt}{t,log^2 t}+frac{psi^*(x)}{log ,x}\
&tag{8}=int_2^xfrac{psi^{*'}(t) dt}{log,t}\
end{align}
But $ displaystyleoperatorname{li}(x)=int_2^x frac{dt}{log,t},$ (Riemann's variant of the logarithmic integral) verifies $ displaystyleoperatorname{li}(x^r)'=frac{r,x^{r-1}}{log x^r}=frac{x^{r-1}}{log,x}$ so that from $(6)$ :
$$tag{9}boxed{displaystylePi^*(x)=operatorname{li}(x)-sum_{rho} operatorname{li}(x^{rho})},quad(x>1)$$
The prime-counting function is defined by $ displaystylepi^*(x):=sum_{ple x}^{*}1 $ while $ displaystylePi^*(x):=sum_{p^kle x}^{*}frac 1k$ will be :
$$tag{10}Pi^*(x)=sum_{k>0} frac{pi^{*}bigl(x^{1/k}bigr)}k$$
Applying the Möbius inversion formula $ displaystylepi^{*}(x):=sum_{n=1}^{infty} frac{mu(k)}k Pi^*bigl(x^{1/k}bigr) $ to $(9)$ we get (with questionable convergence...) :
$$tag{11}boxed{displaystylepi^*(x)=R(x)-sum_{rho} R(x^{rho})},quad(x>1)$$
Where Riemann's $,displaystyle R(x):=sum_{n=1}^{infty} frac{mu(k)}k operatorname{li}bigl(x^{1/k}bigr),$ may be written as a Gram series.
Dirichlet's L-functions
We will try to repeat the previous derivations but with $zeta$ replaced by the Dirichlet L-function
$$L(s,chi)=sum_{k=1}^inftyfrac{chi(k)}{k^s}$$
With $chi$ a Dirichlet character we get following Euler product :
$$tag{1'}boxed{displaystyle L(s,chi)=prod_{p text{prime}}frac 1{1-chi(p)p^{-s}}}quadtext{for} Re(s)>1$$
so that ($chi$ is multiplicative implying $chi(p)^k=chi(p^k)$) :
$$log L(s,chi)=-sum_{p text{prime}}log(1-chi(p)p^{-s})=sum_psum_{k=1}^infty frac{chi(p^k)p^{-ks}}k$$
minus the derivative relatively to $s$ will be :
$$tag{2'}f(s,chi):=-frac{L'(s,chi)}{L(s,chi)}=sum_psum_{k=1}^infty frac{chi(p^k)log,p}{p^{ks}}=sum_{n=1}^infty frac{chi(n)Lambda(n)}{n^s}quadtext{for} Re(s)>1$$
The Perron formula applied to $f(s,chi)$ gives :
$$tag{3'}psi^*(x,chi):=frac1{2pi i}int_{c-iinfty}^{c+iinfty}f(s,chi)frac{x^s}s,ds=sum_{n=1}^infty chi(n)Lambda(n) frac1{2pi i}int_{c-iinfty}^{c+iinfty}left(frac xnright)^sfrac{ds}s$$
$$tag{4'}psi^*(x,chi)=sum_{nle x}^{*}chi(n)Lambda(n)$$
But $psi^*(x,chi)$ may also be written as :
$$tag{5'}psi^*(x,chi)=-frac1{2pi i}int_{c-iinfty}^{c+iinfty}frac{L'(s,chi)}{L(s,chi)}frac{x^s}s,ds$$
Again we compute the sum of the residues at (all) the poles to the left of the vertical line of integration (for $c>1$). The contributions from the different poles will be : $0mapsto -frac{L'(0,chi)}{L(0,chi)}, rhomapsto -frac{x^rho}{rho}$ for $rho$ any zero of $L(s,chi),$ (from Weierstrass factorization) so that :
$$tag{6'}boxed{displaystylepsi^*(x,chi)=-sum_{rho} frac {x^{rho}}{rho}-frac{L'(0,chi)}{L(0,chi)}},quad(x>1)$$
(the sum over $rho$ combines the trivial and the non-trivial zeros and we will suppose that $chi$ is a non principal character so that $,L(s,chi),$ is regular everywhere, for a principal character we should add a $x$ contribution from the pole at $1$)
An analogue to the Riemann prime-counting function will be :
begin{align}
tag{7'}Pi^*(x,chi):&=sum_{p^kle x}^{*}frac {chi(p^k)}k=sum_{nle x}^{*}frac {chi(n)Lambda(n)}{log,n}\
&=sum_{nle x}^{*} chi(n)Lambda(n)left(int_n^x frac{dt}{t,log^2 t}+frac 1{log,x}right)\
&=int_2^xfrac{psi^*(t,chi) dt}{t,log^2 t}+frac{psi^*(x,chi)}{log ,x}\
&tag{8'}=int_2^xfrac{psi^{*'}(t,chi) dt}{log,t}\
end{align}
From $ displaystyleoperatorname{li}(x^r)'=frac{x^{r-1}}{log,x}$ and $(6')$ we get :
$$tag{9'}boxed{displaystylePi^*(x,chi)=-sum_{rho} operatorname{li}(x^{rho})},quad(x>1)$$
(for $chi$ a principal character there is an additional $,operatorname{li}(x)$ term)
The $chi$-prime-counting function is $ displaystylepi^*(x,chi):=sum_{ple x}^{*}chi(p) $ while $ displaystylePi^*(x,chi):=sum_{p^kle x}^{*}frac {chi(p^k)}k$ so that :
$$tag{10'}Pi^*(x,chi)=sum_{k>0} frac{pi^{*}bigl(x^{1/k},chi^kbigr)}k,quad(x>1)$$
Broken analogy...
The problem here is that we have $,pi^{*}bigl(x^{1/k},chi^kbigr)$ in the sum and not $,pi^{*}bigl(x^{1/k},chibigr)$.
For principal characters (when $,chi(p)=0$ or $1$ so that $,chi^k(p)=chi(p),$ for $k$ positive) we may apply the Möbius inversion formula to $(9')$ and get :
$$tag{11'}displaystylepi^*(x,chi)=[R(x)]-sum_{rho} R(x^{rho})$$
(since $chi$ is a principal character we must add the $R(x)$ term from the pole at $1$)
but this doesn't work in general and the equality becomes questionable after the first prime $p$ such that $,chi^k(p)not =chi(p),$ as we will see by considering your specific question.
We want $chi$ to be the Dirichlet character modulo $4$ ($chi_2$ in the link) defined by $ displaystylechi(0)=0, chi(1)=1, chi(2)=0, chi(3)=-1 $ and get :
$$tag{12'}L(s,chi)=sum_{n=0}^inftyfrac {(-1)^n}{(2n+1)^s}=beta(s)$$
with $beta$ the Dirichlet beta function we could hope that :
$$pi^*(x,chi)=displaystylepi_{4,1}^*(x)-pi_{4,3}^*(x)approx -sum_{rho} R(x^{rho})$$
with $pi_{q,a}(x)=#{p:p text{is prime and} ,ple x, text{and} ,pequiv apmod q}$
But, from $(10')$ and $(9')$, we have in fact :
begin{align}
Pi^*(x,chi)&=sum_{k>0} frac{pi_{4,1}^{*}bigl(x^{1/k}bigr)+(-1)^k,pi_{4,3}^{*}bigl(x^{1/k}bigr)}k=-sum_{rho} operatorname{li}(x^{rho})\
&=sum_{k>0} frac{pi_{4,1}^{*}bigl(x^{1/k}bigr)-,pi_{4,3}^{*}bigl(x^{1/k}bigr)}k+2sum_{j>0} frac{pi_{4,3}^{*}left(sqrt{x}^{1/j}right)}{2j}\
end{align}
so that the Möbius transformation will return the exact (with $rho$ any zero of $beta$) :
$$tag{13'}boxed{displaystylepi_{4,1}^{*}(x)-pi_{4,3}^{*}(x)+pi_{4,3}^{*}left(sqrt{x}right)=-sum_{rho} R(x^{rho})}$$
To show the effect of the $,pi_{4,3}^{*}left(sqrt{x}right),$ term let's exhibit an approximation of the expression at the right (taking the $40$ first terms of the sum $R$ and the $300$ first non-trivial zeros of the Dirichlet-beta function) that we compare to the exact (dark) result for $,pi^*(x,chi)=displaystylepi_{4,1}^*(x)-pi_{4,3}^*(x),$ in the interval $,[5,80]$ :
The initial behavior seems right : $-1$ at $3$ followed by $+1$ at $5$ and $-1$ at $7$ but just after that there is a first 'irregularity' at $9=3^2$ because $chi(3^2)=1$ while $chi(3)=-1$. After that the behavior is correct (shifted $+1$ of course) until another irregularity at $49=7^2$ again because $chi(7^2)=1$. The next one will be at $121=11^2$ and at all the following squares of primes of gender $4n+3$...
ADDITION: Draks telescoping
The situation was blocked at this point until draks, using a suggestion of Greg Martin, proposed in february his powerful telescoping method to obtain (the $rho_zeta$ and $rho_beta$ are respectively all the zeros of the $zeta$ and $beta$ function) :
$$tag{14'}pi_{4,3}(x):=sum_{k=0}^infty;2^{-k-1}left(
Rleft(x^{1/2^{k}}right)-sum_{rho_zeta}
Rleft(x^{rho_zeta/2^k}right) +sum_{rho_beta}
Rleft(x^{rho_beta/2^k}right) right)$$
We may obtain this result too (with a minor correction) by rewriting $(11)$ as (for $x>2$) :
$$pi^*(x)=1+pi_{4,1}^{*}(x)+pi_{4,3}^{*}(x)=R(x)-sum_{rho_zeta} R(x^{rho_zeta})$$
($1$ was added since all primes except $2$ are equal to $1$ or $3bmod{4}$)
and using $(13')$ :
$$pi_{4,1}^{*}(x)-pi_{4,3}^{*}(x)+pi_{4,3}^{*}left(sqrt{x}right)=-sum_{rho_beta} R(x^{rho_beta})$$
Subtracting these equations gives :
$$2pi_{4,3}^{*}(x)-pi_{4,3}^{*}left(sqrt{x}right)=R(x)-1-sum_{rho_zeta} R(x^{rho_zeta})+sum_{rho_beta} R(x^{rho_beta})$$
while draks' telescoping sum $ displaystylefrac 12sum_{k=0}^infty frac{fleft(x^{1/2^{k}}right)}{2^k}$ is :
$$sum_{k=0}^infty frac {2,pi_{4,3}^{*}(x^{1/2^{k}})-pi_{4,3}^{*}bigl(x^{1/2^{k+1}}bigr)}{2^{k+1}}=sum_{k=0}^infty frac {Rleft(x^{1/2^{k}}right)-1-sum_{rho_zeta} Rleft(x^{rho_zeta/2^k}right)+sum_{rho_beta} Rleft(x^{rho_beta/2^k}right)}{2^{k+1}}$$
producing draks' formula :
$$tag{15'}boxed{displaystylepi_{4,3}^{*}(x)=sum_{k=0}^infty frac {Rleft(x^{1/2^{k}}right)-sum_{rho_zeta} Rleft(x^{rho_zeta/2^k}right)+sum_{rho_beta} Rleft(x^{rho_beta/2^k}right)}{2^{k+1}}-1}$$
A similar formula may be written for $pi_{4,1}$ and this answers positively the initial question !
Let's finish with a plot of $pi_{4,3}^{*}(x)$ from $4$ to $100$ (more details here with the pari/gp scripts used)
edited Apr 13 '17 at 12:20
Community♦
1
answered Jan 20 '13 at 17:22
Raymond ManzoniRaymond Manzoni
37.2k563117
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Raymond, first of all: Wow! What an answer. "+1". Thank you so much for everything including the link to Edward's book. My copy is buried somewhere in a box because I moved. It'll take me a while to verify all that, but like you, I think that the reasoning by analogy is correct. Specially the last equation thrills me. You might have noticed my attempts with "Summing over General Functions of Primes" ...
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– draks ...
Jan 20 '13 at 22:44
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... or the follow-on question: "An Expression for $logzeta(ns)$ derived from the Limit of the truncated Prime $zeta$ Function". So I think it's possible to combine these result to get an analytic expression for the truncated Prime $zeta$ function with a restriction to primes $4npm 1$. This is ever so cool and I can't leave this unawarded...
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– draks ...
Jan 20 '13 at 22:51
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Thanks @draks... I think that most things are right here except tiny things like the - from the citation, the starting value $n=1$ of the Dirichlet beta function, and perhaps very important things like the transitions from $(9')$ to $(11')$ : these should be (more than!) double checked (because of things like $chi(3^2)=1not =-1$ and of warnings from different authors like Ingham that Riemann's method to obtain $pi$ may not be as easily generalized as $psicdots$ but works only for some specific Dirichlet characters ($chi(p)=0,1$ only?)). But this clear path may at least be checked!
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– Raymond Manzoni
Jan 21 '13 at 0:05
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@draks... I corrected the end of my answer.
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– Raymond Manzoni
Jan 26 '13 at 23:04
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aha, looks like we've gottta add $(mu^2(n)-1)$ to get rid of the squares...interesting, what do you think? And thanks for your thoughts. At least my ${rm li}(x)$ works...
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– draks ...
Jan 27 '13 at 16:03
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show 3 more comments
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Before answering to your question in the comment :
" Is it possible to write $ #{text{primes} 4n+3 le x},-
,#{text{primes} 4n+1 le x}$ (form here) in terms like $,operatorname{R}(x^1) - sum_{rho}operatorname{R}(x^{rho})$ ? "
let's start with a sketch using von Mangoldt's derivation to obtain your equation $(1)$ that will be used for inspiration (for proofs see for example Edwards' chapter 3) :
Riemann's Explicit Formulas
The Euler product formula gives us :
$$tag{1}boxed{displaystylezeta(s)=prod_{p text{prime}}frac 1{1-p^{-s}}}quadtext{for} Re(s)>1$$
so that
$$log zeta(s)=-sum_{p text{prime}}log(1-p^{-s})=sum_psum_{k=1}^infty frac{p^{-ks}}k$$
minus the derivative will be :
$$tag{2}f(s):=-frac{zeta'(s)}{zeta(s)}=sum_psum_{k=1}^infty frac{log,p}{p^{ks}}=sum_{n=1}^infty frac{Lambda(n)}{n^s}quadtext{for} Re(s)>1$$
with $Lambda$ the von Mangoldt function defined by $ Lambda(n):=begin{cases}
log, p & text{if} n=p^k text{and} k>0\
0 & text{else}
end{cases}$
Let's use the definition of the second Chebyshev function : $displaystyle psi(x)=sum_{nleq x}Lambda(n)$ and Abel's sum formula applied to $a(n):=Lambda(n)$ and $phi(n):=n^{-s}$ :
$$sum_{n=1}^infty frac{Lambda(n)}{n^s}=sint_1^infty frac {sum_{nleq x}Lambda(n)}{x^{s+1}};dx$$
to rewrite $f(s)$ as (the lower bound became $0$ since $phi(x)=0$ for $x<1$) :
$$tag{2.1}f(s)=sint_0^{infty}frac{psi(x)}{x^{s+1}}dx$$
But this is a Mellin transform which may be reverted to get Perron's formula (let's observe that the Dirichlet series $displaystyle f(s)=sum_{n=1}^infty frac{Lambda(n)}{n^s}$ is absolutely convergent for $Re(s)>1$ and suppose $c>1$) :
$$tag{3}psi^*(x):=frac1{2pi i}int_{c-iinfty}^{c+iinfty}f(s)frac{x^s}s,ds=sum_{n=1}^infty Lambda(n) frac1{2pi i}int_{c-iinfty}^{c+iinfty}left(frac xnright)^sfrac{ds}s$$
This last integral may be evaluated using Fourier's theorem :
$$ displaystylefrac1{2pi i}int_{c-iinfty}^{c+iinfty}frac{y^s}s ds=
begin{cases}
0 & 0<y<1\
1/2 & quad y=1\
1 & quad y>1\
end{cases}$$
and we obtain (as wished) that : $displaystyle tag{4}psi^*(x)=sum_{nle x}^{*}Lambda(n)$
with $psi^*$ the second Chebyshev function except when $x$ is an integer because in this case the last term of the sum has to be divided by $2$. This will be the meaning of the $^*$ symbols in this article : at a first order discontinuity point (i.e. a jump) the result is the mean value of the limit at the left and the right.
Now $psi^*(x)$ may also be written as :
$$tag{5}psi^*(x)=-frac1{2pi i}int_{c-iinfty}^{c+iinfty}frac{zeta'(s)}{zeta(s)}frac{x^s}s,ds$$
Here the idea is simply to sum the residues of (all) the poles at the left of the vertical line of integration ($c>1$) at least if $x>1$ (for $0<x<1$ we would have to take the poles at the right and obtain $0$). These poles come from the denominator $s$, the pole of $zeta$ at $1$ and the zeros with the correspondence : $0mapsto -frac{zeta'(0)}{zeta(0)}, 1mapsto x^1, rhomapsto -frac{x^rho}{rho}$ for $rho$ any zero of $zeta,$ (from the Weierstrass factorization of the Hadamard product) so that :
$$tag{6}boxed{displaystylepsi^*(x)=x-sum_{rho} frac {x^{rho}}{rho}-frac{zeta'(0)}{zeta(0)}},quad(x>1)$$
(in this post we won't distinguish the trivial from the non-trivial zeros $rho$ ; for convergence the non-trivial roots should be grouped by pairs and sorted by increasing $|Im(rho)|$)
The Riemann prime-counting function is defined by :
begin{align}
tag{7}Pi^*(x):&=sum_{p^kle x}^{*}frac 1k=sum_{nle x}^{*}frac {Lambda(n)}{log,n}\
&=sum_{nle x}^{*} Lambda(n)left(int_n^x frac{dt}{t,log^2 t}+frac 1{log,x}right)\
&=int_2^xfrac{psi^*(t) dt}{t,log^2 t}+frac{psi^*(x)}{log ,x}\
&tag{8}=int_2^xfrac{psi^{*'}(t) dt}{log,t}\
end{align}
But $ displaystyleoperatorname{li}(x)=int_2^x frac{dt}{log,t},$ (Riemann's variant of the logarithmic integral) verifies $ displaystyleoperatorname{li}(x^r)'=frac{r,x^{r-1}}{log x^r}=frac{x^{r-1}}{log,x}$ so that from $(6)$ :
$$tag{9}boxed{displaystylePi^*(x)=operatorname{li}(x)-sum_{rho} operatorname{li}(x^{rho})},quad(x>1)$$
The prime-counting function is defined by $ displaystylepi^*(x):=sum_{ple x}^{*}1 $ while $ displaystylePi^*(x):=sum_{p^kle x}^{*}frac 1k$ will be :
$$tag{10}Pi^*(x)=sum_{k>0} frac{pi^{*}bigl(x^{1/k}bigr)}k$$
Applying the Möbius inversion formula $ displaystylepi^{*}(x):=sum_{n=1}^{infty} frac{mu(k)}k Pi^*bigl(x^{1/k}bigr) $ to $(9)$ we get (with questionable convergence...) :
$$tag{11}boxed{displaystylepi^*(x)=R(x)-sum_{rho} R(x^{rho})},quad(x>1)$$
Where Riemann's $,displaystyle R(x):=sum_{n=1}^{infty} frac{mu(k)}k operatorname{li}bigl(x^{1/k}bigr),$ may be written as a Gram series.
Dirichlet's L-functions
We will try to repeat the previous derivations but with $zeta$ replaced by the Dirichlet L-function
$$L(s,chi)=sum_{k=1}^inftyfrac{chi(k)}{k^s}$$
With $chi$ a Dirichlet character we get following Euler product :
$$tag{1'}boxed{displaystyle L(s,chi)=prod_{p text{prime}}frac 1{1-chi(p)p^{-s}}}quadtext{for} Re(s)>1$$
so that ($chi$ is multiplicative implying $chi(p)^k=chi(p^k)$) :
$$log L(s,chi)=-sum_{p text{prime}}log(1-chi(p)p^{-s})=sum_psum_{k=1}^infty frac{chi(p^k)p^{-ks}}k$$
minus the derivative relatively to $s$ will be :
$$tag{2'}f(s,chi):=-frac{L'(s,chi)}{L(s,chi)}=sum_psum_{k=1}^infty frac{chi(p^k)log,p}{p^{ks}}=sum_{n=1}^infty frac{chi(n)Lambda(n)}{n^s}quadtext{for} Re(s)>1$$
The Perron formula applied to $f(s,chi)$ gives :
$$tag{3'}psi^*(x,chi):=frac1{2pi i}int_{c-iinfty}^{c+iinfty}f(s,chi)frac{x^s}s,ds=sum_{n=1}^infty chi(n)Lambda(n) frac1{2pi i}int_{c-iinfty}^{c+iinfty}left(frac xnright)^sfrac{ds}s$$
$$tag{4'}psi^*(x,chi)=sum_{nle x}^{*}chi(n)Lambda(n)$$
But $psi^*(x,chi)$ may also be written as :
$$tag{5'}psi^*(x,chi)=-frac1{2pi i}int_{c-iinfty}^{c+iinfty}frac{L'(s,chi)}{L(s,chi)}frac{x^s}s,ds$$
Again we compute the sum of the residues at (all) the poles to the left of the vertical line of integration (for $c>1$). The contributions from the different poles will be : $0mapsto -frac{L'(0,chi)}{L(0,chi)}, rhomapsto -frac{x^rho}{rho}$ for $rho$ any zero of $L(s,chi),$ (from Weierstrass factorization) so that :
$$tag{6'}boxed{displaystylepsi^*(x,chi)=-sum_{rho} frac {x^{rho}}{rho}-frac{L'(0,chi)}{L(0,chi)}},quad(x>1)$$
(the sum over $rho$ combines the trivial and the non-trivial zeros and we will suppose that $chi$ is a non principal character so that $,L(s,chi),$ is regular everywhere, for a principal character we should add a $x$ contribution from the pole at $1$)
An analogue to the Riemann prime-counting function will be :
begin{align}
tag{7'}Pi^*(x,chi):&=sum_{p^kle x}^{*}frac {chi(p^k)}k=sum_{nle x}^{*}frac {chi(n)Lambda(n)}{log,n}\
&=sum_{nle x}^{*} chi(n)Lambda(n)left(int_n^x frac{dt}{t,log^2 t}+frac 1{log,x}right)\
&=int_2^xfrac{psi^*(t,chi) dt}{t,log^2 t}+frac{psi^*(x,chi)}{log ,x}\
&tag{8'}=int_2^xfrac{psi^{*'}(t,chi) dt}{log,t}\
end{align}
From $ displaystyleoperatorname{li}(x^r)'=frac{x^{r-1}}{log,x}$ and $(6')$ we get :
$$tag{9'}boxed{displaystylePi^*(x,chi)=-sum_{rho} operatorname{li}(x^{rho})},quad(x>1)$$
(for $chi$ a principal character there is an additional $,operatorname{li}(x)$ term)
The $chi$-prime-counting function is $ displaystylepi^*(x,chi):=sum_{ple x}^{*}chi(p) $ while $ displaystylePi^*(x,chi):=sum_{p^kle x}^{*}frac {chi(p^k)}k$ so that :
$$tag{10'}Pi^*(x,chi)=sum_{k>0} frac{pi^{*}bigl(x^{1/k},chi^kbigr)}k,quad(x>1)$$
Broken analogy...
The problem here is that we have $,pi^{*}bigl(x^{1/k},chi^kbigr)$ in the sum and not $,pi^{*}bigl(x^{1/k},chibigr)$.
For principal characters (when $,chi(p)=0$ or $1$ so that $,chi^k(p)=chi(p),$ for $k$ positive) we may apply the Möbius inversion formula to $(9')$ and get :
$$tag{11'}displaystylepi^*(x,chi)=[R(x)]-sum_{rho} R(x^{rho})$$
(since $chi$ is a principal character we must add the $R(x)$ term from the pole at $1$)
but this doesn't work in general and the equality becomes questionable after the first prime $p$ such that $,chi^k(p)not =chi(p),$ as we will see by considering your specific question.
We want $chi$ to be the Dirichlet character modulo $4$ ($chi_2$ in the link) defined by $ displaystylechi(0)=0, chi(1)=1, chi(2)=0, chi(3)=-1 $ and get :
$$tag{12'}L(s,chi)=sum_{n=0}^inftyfrac {(-1)^n}{(2n+1)^s}=beta(s)$$
with $beta$ the Dirichlet beta function we could hope that :
$$pi^*(x,chi)=displaystylepi_{4,1}^*(x)-pi_{4,3}^*(x)approx -sum_{rho} R(x^{rho})$$
with $pi_{q,a}(x)=#{p:p text{is prime and} ,ple x, text{and} ,pequiv apmod q}$
But, from $(10')$ and $(9')$, we have in fact :
begin{align}
Pi^*(x,chi)&=sum_{k>0} frac{pi_{4,1}^{*}bigl(x^{1/k}bigr)+(-1)^k,pi_{4,3}^{*}bigl(x^{1/k}bigr)}k=-sum_{rho} operatorname{li}(x^{rho})\
&=sum_{k>0} frac{pi_{4,1}^{*}bigl(x^{1/k}bigr)-,pi_{4,3}^{*}bigl(x^{1/k}bigr)}k+2sum_{j>0} frac{pi_{4,3}^{*}left(sqrt{x}^{1/j}right)}{2j}\
end{align}
so that the Möbius transformation will return the exact (with $rho$ any zero of $beta$) :
$$tag{13'}boxed{displaystylepi_{4,1}^{*}(x)-pi_{4,3}^{*}(x)+pi_{4,3}^{*}left(sqrt{x}right)=-sum_{rho} R(x^{rho})}$$
To show the effect of the $,pi_{4,3}^{*}left(sqrt{x}right),$ term let's exhibit an approximation of the expression at the right (taking the $40$ first terms of the sum $R$ and the $300$ first non-trivial zeros of the Dirichlet-beta function) that we compare to the exact (dark) result for $,pi^*(x,chi)=displaystylepi_{4,1}^*(x)-pi_{4,3}^*(x),$ in the interval $,[5,80]$ :
The initial behavior seems right : $-1$ at $3$ followed by $+1$ at $5$ and $-1$ at $7$ but just after that there is a first 'irregularity' at $9=3^2$ because $chi(3^2)=1$ while $chi(3)=-1$. After that the behavior is correct (shifted $+1$ of course) until another irregularity at $49=7^2$ again because $chi(7^2)=1$. The next one will be at $121=11^2$ and at all the following squares of primes of gender $4n+3$...
ADDITION: Draks telescoping
The situation was blocked at this point until draks, using a suggestion of Greg Martin, proposed in february his powerful telescoping method to obtain (the $rho_zeta$ and $rho_beta$ are respectively all the zeros of the $zeta$ and $beta$ function) :
$$tag{14'}pi_{4,3}(x):=sum_{k=0}^infty;2^{-k-1}left(
Rleft(x^{1/2^{k}}right)-sum_{rho_zeta}
Rleft(x^{rho_zeta/2^k}right) +sum_{rho_beta}
Rleft(x^{rho_beta/2^k}right) right)$$
We may obtain this result too (with a minor correction) by rewriting $(11)$ as (for $x>2$) :
$$pi^*(x)=1+pi_{4,1}^{*}(x)+pi_{4,3}^{*}(x)=R(x)-sum_{rho_zeta} R(x^{rho_zeta})$$
($1$ was added since all primes except $2$ are equal to $1$ or $3bmod{4}$)
and using $(13')$ :
$$pi_{4,1}^{*}(x)-pi_{4,3}^{*}(x)+pi_{4,3}^{*}left(sqrt{x}right)=-sum_{rho_beta} R(x^{rho_beta})$$
Subtracting these equations gives :
$$2pi_{4,3}^{*}(x)-pi_{4,3}^{*}left(sqrt{x}right)=R(x)-1-sum_{rho_zeta} R(x^{rho_zeta})+sum_{rho_beta} R(x^{rho_beta})$$
while draks' telescoping sum $ displaystylefrac 12sum_{k=0}^infty frac{fleft(x^{1/2^{k}}right)}{2^k}$ is :
$$sum_{k=0}^infty frac {2,pi_{4,3}^{*}(x^{1/2^{k}})-pi_{4,3}^{*}bigl(x^{1/2^{k+1}}bigr)}{2^{k+1}}=sum_{k=0}^infty frac {Rleft(x^{1/2^{k}}right)-1-sum_{rho_zeta} Rleft(x^{rho_zeta/2^k}right)+sum_{rho_beta} Rleft(x^{rho_beta/2^k}right)}{2^{k+1}}$$
producing draks' formula :
$$tag{15'}boxed{displaystylepi_{4,3}^{*}(x)=sum_{k=0}^infty frac {Rleft(x^{1/2^{k}}right)-sum_{rho_zeta} Rleft(x^{rho_zeta/2^k}right)+sum_{rho_beta} Rleft(x^{rho_beta/2^k}right)}{2^{k+1}}-1}$$
A similar formula may be written for $pi_{4,1}$ and this answers positively the initial question !
Let's finish with a plot of $pi_{4,3}^{*}(x)$ from $4$ to $100$ (more details here with the pari/gp scripts used)
edited Apr 13 '17 at 12:20
Community♦
1
answered Jan 20 '13 at 17:22
Raymond ManzoniRaymond Manzoni
37.2k563117
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Raymond, first of all: Wow! What an answer. "+1". Thank you so much for everything including the link to Edward's book. My copy is buried somewhere in a box because I moved. It'll take me a while to verify all that, but like you, I think that the reasoning by analogy is correct. Specially the last equation thrills me. You might have noticed my attempts with "Summing over General Functions of Primes" ...
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– draks ...
Jan 20 '13 at 22:44
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... or the follow-on question: "An Expression for $logzeta(ns)$ derived from the Limit of the truncated Prime $zeta$ Function". So I think it's possible to combine these result to get an analytic expression for the truncated Prime $zeta$ function with a restriction to primes $4npm 1$. This is ever so cool and I can't leave this unawarded...
$endgroup$
– draks ...
Jan 20 '13 at 22:51
$begingroup$
Thanks @draks... I think that most things are right here except tiny things like the - from the citation, the starting value $n=1$ of the Dirichlet beta function, and perhaps very important things like the transitions from $(9')$ to $(11')$ : these should be (more than!) double checked (because of things like $chi(3^2)=1not =-1$ and of warnings from different authors like Ingham that Riemann's method to obtain $pi$ may not be as easily generalized as $psicdots$ but works only for some specific Dirichlet characters ($chi(p)=0,1$ only?)). But this clear path may at least be checked!
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– Raymond Manzoni
Jan 21 '13 at 0:05
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@draks... I corrected the end of my answer.
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– Raymond Manzoni
Jan 26 '13 at 23:04
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aha, looks like we've gottta add $(mu^2(n)-1)$ to get rid of the squares...interesting, what do you think? And thanks for your thoughts. At least my ${rm li}(x)$ works...
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– draks ...
Jan 27 '13 at 16:03
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show 3 more comments
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Before answering to your question in the comment :
" Is it possible to write $ #{text{primes} 4n+3 le x},-
,#{text{primes} 4n+1 le x}$ (form here) in terms like $,operatorname{R}(x^1) - sum_{rho}operatorname{R}(x^{rho})$ ? "
let's start with a sketch using von Mangoldt's derivation to obtain your equation $(1)$ that will be used for inspiration (for proofs see for example Edwards' chapter 3) :
Riemann's Explicit Formulas
The Euler product formula gives us :
$$tag{1}boxed{displaystylezeta(s)=prod_{p text{prime}}frac 1{1-p^{-s}}}quadtext{for} Re(s)>1$$
so that
$$log zeta(s)=-sum_{p text{prime}}log(1-p^{-s})=sum_psum_{k=1}^infty frac{p^{-ks}}k$$
minus the derivative will be :
$$tag{2}f(s):=-frac{zeta'(s)}{zeta(s)}=sum_psum_{k=1}^infty frac{log,p}{p^{ks}}=sum_{n=1}^infty frac{Lambda(n)}{n^s}quadtext{for} Re(s)>1$$
with $Lambda$ the von Mangoldt function defined by $ Lambda(n):=begin{cases}
log, p & text{if} n=p^k text{and} k>0\
0 & text{else}
end{cases}$
Let's use the definition of the second Chebyshev function : $displaystyle psi(x)=sum_{nleq x}Lambda(n)$ and Abel's sum formula applied to $a(n):=Lambda(n)$ and $phi(n):=n^{-s}$ :
$$sum_{n=1}^infty frac{Lambda(n)}{n^s}=sint_1^infty frac {sum_{nleq x}Lambda(n)}{x^{s+1}};dx$$
to rewrite $f(s)$ as (the lower bound became $0$ since $phi(x)=0$ for $x<1$) :
$$tag{2.1}f(s)=sint_0^{infty}frac{psi(x)}{x^{s+1}}dx$$
But this is a Mellin transform which may be reverted to get Perron's formula (let's observe that the Dirichlet series $displaystyle f(s)=sum_{n=1}^infty frac{Lambda(n)}{n^s}$ is absolutely convergent for $Re(s)>1$ and suppose $c>1$) :
$$tag{3}psi^*(x):=frac1{2pi i}int_{c-iinfty}^{c+iinfty}f(s)frac{x^s}s,ds=sum_{n=1}^infty Lambda(n) frac1{2pi i}int_{c-iinfty}^{c+iinfty}left(frac xnright)^sfrac{ds}s$$
This last integral may be evaluated using Fourier's theorem :
$$ displaystylefrac1{2pi i}int_{c-iinfty}^{c+iinfty}frac{y^s}s ds=
begin{cases}
0 & 0<y<1\
1/2 & quad y=1\
1 & quad y>1\
end{cases}$$
and we obtain (as wished) that : $displaystyle tag{4}psi^*(x)=sum_{nle x}^{*}Lambda(n)$
with $psi^*$ the second Chebyshev function except when $x$ is an integer because in this case the last term of the sum has to be divided by $2$. This will be the meaning of the $^*$ symbols in this article : at a first order discontinuity point (i.e. a jump) the result is the mean value of the limit at the left and the right.
Now $psi^*(x)$ may also be written as :
$$tag{5}psi^*(x)=-frac1{2pi i}int_{c-iinfty}^{c+iinfty}frac{zeta'(s)}{zeta(s)}frac{x^s}s,ds$$
Here the idea is simply to sum the residues of (all) the poles at the left of the vertical line of integration ($c>1$) at least if $x>1$ (for $0<x<1$ we would have to take the poles at the right and obtain $0$). These poles come from the denominator $s$, the pole of $zeta$ at $1$ and the zeros with the correspondence : $0mapsto -frac{zeta'(0)}{zeta(0)}, 1mapsto x^1, rhomapsto -frac{x^rho}{rho}$ for $rho$ any zero of $zeta,$ (from the Weierstrass factorization of the Hadamard product) so that :
$$tag{6}boxed{displaystylepsi^*(x)=x-sum_{rho} frac {x^{rho}}{rho}-frac{zeta'(0)}{zeta(0)}},quad(x>1)$$
(in this post we won't distinguish the trivial from the non-trivial zeros $rho$ ; for convergence the non-trivial roots should be grouped by pairs and sorted by increasing $|Im(rho)|$)
The Riemann prime-counting function is defined by :
begin{align}
tag{7}Pi^*(x):&=sum_{p^kle x}^{*}frac 1k=sum_{nle x}^{*}frac {Lambda(n)}{log,n}\
&=sum_{nle x}^{*} Lambda(n)left(int_n^x frac{dt}{t,log^2 t}+frac 1{log,x}right)\
&=int_2^xfrac{psi^*(t) dt}{t,log^2 t}+frac{psi^*(x)}{log ,x}\
&tag{8}=int_2^xfrac{psi^{*'}(t) dt}{log,t}\
end{align}
But $ displaystyleoperatorname{li}(x)=int_2^x frac{dt}{log,t},$ (Riemann's variant of the logarithmic integral) verifies $ displaystyleoperatorname{li}(x^r)'=frac{r,x^{r-1}}{log x^r}=frac{x^{r-1}}{log,x}$ so that from $(6)$ :
$$tag{9}boxed{displaystylePi^*(x)=operatorname{li}(x)-sum_{rho} operatorname{li}(x^{rho})},quad(x>1)$$
The prime-counting function is defined by $ displaystylepi^*(x):=sum_{ple x}^{*}1 $ while $ displaystylePi^*(x):=sum_{p^kle x}^{*}frac 1k$ will be :
$$tag{10}Pi^*(x)=sum_{k>0} frac{pi^{*}bigl(x^{1/k}bigr)}k$$
Applying the Möbius inversion formula $ displaystylepi^{*}(x):=sum_{n=1}^{infty} frac{mu(k)}k Pi^*bigl(x^{1/k}bigr) $ to $(9)$ we get (with questionable convergence...) :
$$tag{11}boxed{displaystylepi^*(x)=R(x)-sum_{rho} R(x^{rho})},quad(x>1)$$
Where Riemann's $,displaystyle R(x):=sum_{n=1}^{infty} frac{mu(k)}k operatorname{li}bigl(x^{1/k}bigr),$ may be written as a Gram series.
Dirichlet's L-functions
We will try to repeat the previous derivations but with $zeta$ replaced by the Dirichlet L-function
$$L(s,chi)=sum_{k=1}^inftyfrac{chi(k)}{k^s}$$
With $chi$ a Dirichlet character we get following Euler product :
$$tag{1'}boxed{displaystyle L(s,chi)=prod_{p text{prime}}frac 1{1-chi(p)p^{-s}}}quadtext{for} Re(s)>1$$
so that ($chi$ is multiplicative implying $chi(p)^k=chi(p^k)$) :
$$log L(s,chi)=-sum_{p text{prime}}log(1-chi(p)p^{-s})=sum_psum_{k=1}^infty frac{chi(p^k)p^{-ks}}k$$
minus the derivative relatively to $s$ will be :
$$tag{2'}f(s,chi):=-frac{L'(s,chi)}{L(s,chi)}=sum_psum_{k=1}^infty frac{chi(p^k)log,p}{p^{ks}}=sum_{n=1}^infty frac{chi(n)Lambda(n)}{n^s}quadtext{for} Re(s)>1$$
The Perron formula applied to $f(s,chi)$ gives :
$$tag{3'}psi^*(x,chi):=frac1{2pi i}int_{c-iinfty}^{c+iinfty}f(s,chi)frac{x^s}s,ds=sum_{n=1}^infty chi(n)Lambda(n) frac1{2pi i}int_{c-iinfty}^{c+iinfty}left(frac xnright)^sfrac{ds}s$$
$$tag{4'}psi^*(x,chi)=sum_{nle x}^{*}chi(n)Lambda(n)$$
But $psi^*(x,chi)$ may also be written as :
$$tag{5'}psi^*(x,chi)=-frac1{2pi i}int_{c-iinfty}^{c+iinfty}frac{L'(s,chi)}{L(s,chi)}frac{x^s}s,ds$$
Again we compute the sum of the residues at (all) the poles to the left of the vertical line of integration (for $c>1$). The contributions from the different poles will be : $0mapsto -frac{L'(0,chi)}{L(0,chi)}, rhomapsto -frac{x^rho}{rho}$ for $rho$ any zero of $L(s,chi),$ (from Weierstrass factorization) so that :
$$tag{6'}boxed{displaystylepsi^*(x,chi)=-sum_{rho} frac {x^{rho}}{rho}-frac{L'(0,chi)}{L(0,chi)}},quad(x>1)$$
(the sum over $rho$ combines the trivial and the non-trivial zeros and we will suppose that $chi$ is a non principal character so that $,L(s,chi),$ is regular everywhere, for a principal character we should add a $x$ contribution from the pole at $1$)
An analogue to the Riemann prime-counting function will be :
begin{align}
tag{7'}Pi^*(x,chi):&=sum_{p^kle x}^{*}frac {chi(p^k)}k=sum_{nle x}^{*}frac {chi(n)Lambda(n)}{log,n}\
&=sum_{nle x}^{*} chi(n)Lambda(n)left(int_n^x frac{dt}{t,log^2 t}+frac 1{log,x}right)\
&=int_2^xfrac{psi^*(t,chi) dt}{t,log^2 t}+frac{psi^*(x,chi)}{log ,x}\
&tag{8'}=int_2^xfrac{psi^{*'}(t,chi) dt}{log,t}\
end{align}
From $ displaystyleoperatorname{li}(x^r)'=frac{x^{r-1}}{log,x}$ and $(6')$ we get :
$$tag{9'}boxed{displaystylePi^*(x,chi)=-sum_{rho} operatorname{li}(x^{rho})},quad(x>1)$$
(for $chi$ a principal character there is an additional $,operatorname{li}(x)$ term)
The $chi$-prime-counting function is $ displaystylepi^*(x,chi):=sum_{ple x}^{*}chi(p) $ while $ displaystylePi^*(x,chi):=sum_{p^kle x}^{*}frac {chi(p^k)}k$ so that :
$$tag{10'}Pi^*(x,chi)=sum_{k>0} frac{pi^{*}bigl(x^{1/k},chi^kbigr)}k,quad(x>1)$$
Broken analogy...
The problem here is that we have $,pi^{*}bigl(x^{1/k},chi^kbigr)$ in the sum and not $,pi^{*}bigl(x^{1/k},chibigr)$.
For principal characters (when $,chi(p)=0$ or $1$ so that $,chi^k(p)=chi(p),$ for $k$ positive) we may apply the Möbius inversion formula to $(9')$ and get :
$$tag{11'}displaystylepi^*(x,chi)=[R(x)]-sum_{rho} R(x^{rho})$$
(since $chi$ is a principal character we must add the $R(x)$ term from the pole at $1$)
but this doesn't work in general and the equality becomes questionable after the first prime $p$ such that $,chi^k(p)not =chi(p),$ as we will see by considering your specific question.
We want $chi$ to be the Dirichlet character modulo $4$ ($chi_2$ in the link) defined by $ displaystylechi(0)=0, chi(1)=1, chi(2)=0, chi(3)=-1 $ and get :
$$tag{12'}L(s,chi)=sum_{n=0}^inftyfrac {(-1)^n}{(2n+1)^s}=beta(s)$$
with $beta$ the Dirichlet beta function we could hope that :
$$pi^*(x,chi)=displaystylepi_{4,1}^*(x)-pi_{4,3}^*(x)approx -sum_{rho} R(x^{rho})$$
with $pi_{q,a}(x)=#{p:p text{is prime and} ,ple x, text{and} ,pequiv apmod q}$
But, from $(10')$ and $(9')$, we have in fact :
begin{align}
Pi^*(x,chi)&=sum_{k>0} frac{pi_{4,1}^{*}bigl(x^{1/k}bigr)+(-1)^k,pi_{4,3}^{*}bigl(x^{1/k}bigr)}k=-sum_{rho} operatorname{li}(x^{rho})\
&=sum_{k>0} frac{pi_{4,1}^{*}bigl(x^{1/k}bigr)-,pi_{4,3}^{*}bigl(x^{1/k}bigr)}k+2sum_{j>0} frac{pi_{4,3}^{*}left(sqrt{x}^{1/j}right)}{2j}\
end{align}
so that the Möbius transformation will return the exact (with $rho$ any zero of $beta$) :
$$tag{13'}boxed{displaystylepi_{4,1}^{*}(x)-pi_{4,3}^{*}(x)+pi_{4,3}^{*}left(sqrt{x}right)=-sum_{rho} R(x^{rho})}$$
To show the effect of the $,pi_{4,3}^{*}left(sqrt{x}right),$ term let's exhibit an approximation of the expression at the right (taking the $40$ first terms of the sum $R$ and the $300$ first non-trivial zeros of the Dirichlet-beta function) that we compare to the exact (dark) result for $,pi^*(x,chi)=displaystylepi_{4,1}^*(x)-pi_{4,3}^*(x),$ in the interval $,[5,80]$ :
The initial behavior seems right : $-1$ at $3$ followed by $+1$ at $5$ and $-1$ at $7$ but just after that there is a first 'irregularity' at $9=3^2$ because $chi(3^2)=1$ while $chi(3)=-1$. After that the behavior is correct (shifted $+1$ of course) until another irregularity at $49=7^2$ again because $chi(7^2)=1$. The next one will be at $121=11^2$ and at all the following squares of primes of gender $4n+3$...
ADDITION: Draks telescoping
The situation was blocked at this point until draks, using a suggestion of Greg Martin, proposed in february his powerful telescoping method to obtain (the $rho_zeta$ and $rho_beta$ are respectively all the zeros of the $zeta$ and $beta$ function) :
$$tag{14'}pi_{4,3}(x):=sum_{k=0}^infty;2^{-k-1}left(
Rleft(x^{1/2^{k}}right)-sum_{rho_zeta}
Rleft(x^{rho_zeta/2^k}right) +sum_{rho_beta}
Rleft(x^{rho_beta/2^k}right) right)$$
We may obtain this result too (with a minor correction) by rewriting $(11)$ as (for $x>2$) :
$$pi^*(x)=1+pi_{4,1}^{*}(x)+pi_{4,3}^{*}(x)=R(x)-sum_{rho_zeta} R(x^{rho_zeta})$$
($1$ was added since all primes except $2$ are equal to $1$ or $3bmod{4}$)
and using $(13')$ :
$$pi_{4,1}^{*}(x)-pi_{4,3}^{*}(x)+pi_{4,3}^{*}left(sqrt{x}right)=-sum_{rho_beta} R(x^{rho_beta})$$
Subtracting these equations gives :
$$2pi_{4,3}^{*}(x)-pi_{4,3}^{*}left(sqrt{x}right)=R(x)-1-sum_{rho_zeta} R(x^{rho_zeta})+sum_{rho_beta} R(x^{rho_beta})$$
while draks' telescoping sum $ displaystylefrac 12sum_{k=0}^infty frac{fleft(x^{1/2^{k}}right)}{2^k}$ is :
$$sum_{k=0}^infty frac {2,pi_{4,3}^{*}(x^{1/2^{k}})-pi_{4,3}^{*}bigl(x^{1/2^{k+1}}bigr)}{2^{k+1}}=sum_{k=0}^infty frac {Rleft(x^{1/2^{k}}right)-1-sum_{rho_zeta} Rleft(x^{rho_zeta/2^k}right)+sum_{rho_beta} Rleft(x^{rho_beta/2^k}right)}{2^{k+1}}$$
producing draks' formula :
$$tag{15'}boxed{displaystylepi_{4,3}^{*}(x)=sum_{k=0}^infty frac {Rleft(x^{1/2^{k}}right)-sum_{rho_zeta} Rleft(x^{rho_zeta/2^k}right)+sum_{rho_beta} Rleft(x^{rho_beta/2^k}right)}{2^{k+1}}-1}$$
A similar formula may be written for $pi_{4,1}$ and this answers positively the initial question !
Let's finish with a plot of $pi_{4,3}^{*}(x)$ from $4$ to $100$ (more details here with the pari/gp scripts used)
edited Apr 13 '17 at 12:20
Community♦
1
answered Jan 20 '13 at 17:22
Raymond ManzoniRaymond Manzoni
37.2k563117
$endgroup$
Before answering to your question in the comment :
" Is it possible to write $ #{text{primes} 4n+3 le x},-
,#{text{primes} 4n+1 le x}$ (form here) in terms like $,operatorname{R}(x^1) - sum_{rho}operatorname{R}(x^{rho})$ ? "
let's start with a sketch using von Mangoldt's derivation to obtain your equation $(1)$ that will be used for inspiration (for proofs see for example Edwards' chapter 3) :
Riemann's Explicit Formulas
The Euler product formula gives us :
$$tag{1}boxed{displaystylezeta(s)=prod_{p text{prime}}frac 1{1-p^{-s}}}quadtext{for} Re(s)>1$$
so that
$$log zeta(s)=-sum_{p text{prime}}log(1-p^{-s})=sum_psum_{k=1}^infty frac{p^{-ks}}k$$
minus the derivative will be :
$$tag{2}f(s):=-frac{zeta'(s)}{zeta(s)}=sum_psum_{k=1}^infty frac{log,p}{p^{ks}}=sum_{n=1}^infty frac{Lambda(n)}{n^s}quadtext{for} Re(s)>1$$
with $Lambda$ the von Mangoldt function defined by $ Lambda(n):=begin{cases}
log, p & text{if} n=p^k text{and} k>0\
0 & text{else}
end{cases}$
Let's use the definition of the second Chebyshev function : $displaystyle psi(x)=sum_{nleq x}Lambda(n)$ and Abel's sum formula applied to $a(n):=Lambda(n)$ and $phi(n):=n^{-s}$ :
$$sum_{n=1}^infty frac{Lambda(n)}{n^s}=sint_1^infty frac {sum_{nleq x}Lambda(n)}{x^{s+1}};dx$$
to rewrite $f(s)$ as (the lower bound became $0$ since $phi(x)=0$ for $x<1$) :
$$tag{2.1}f(s)=sint_0^{infty}frac{psi(x)}{x^{s+1}}dx$$
But this is a Mellin transform which may be reverted to get Perron's formula (let's observe that the Dirichlet series $displaystyle f(s)=sum_{n=1}^infty frac{Lambda(n)}{n^s}$ is absolutely convergent for $Re(s)>1$ and suppose $c>1$) :
$$tag{3}psi^*(x):=frac1{2pi i}int_{c-iinfty}^{c+iinfty}f(s)frac{x^s}s,ds=sum_{n=1}^infty Lambda(n) frac1{2pi i}int_{c-iinfty}^{c+iinfty}left(frac xnright)^sfrac{ds}s$$
This last integral may be evaluated using Fourier's theorem :
$$ displaystylefrac1{2pi i}int_{c-iinfty}^{c+iinfty}frac{y^s}s ds=
begin{cases}
0 & 0<y<1\
1/2 & quad y=1\
1 & quad y>1\
end{cases}$$
and we obtain (as wished) that : $displaystyle tag{4}psi^*(x)=sum_{nle x}^{*}Lambda(n)$
with $psi^*$ the second Chebyshev function except when $x$ is an integer because in this case the last term of the sum has to be divided by $2$. This will be the meaning of the $^*$ symbols in this article : at a first order discontinuity point (i.e. a jump) the result is the mean value of the limit at the left and the right.
Now $psi^*(x)$ may also be written as :
$$tag{5}psi^*(x)=-frac1{2pi i}int_{c-iinfty}^{c+iinfty}frac{zeta'(s)}{zeta(s)}frac{x^s}s,ds$$
Here the idea is simply to sum the residues of (all) the poles at the left of the vertical line of integration ($c>1$) at least if $x>1$ (for $0<x<1$ we would have to take the poles at the right and obtain $0$). These poles come from the denominator $s$, the pole of $zeta$ at $1$ and the zeros with the correspondence : $0mapsto -frac{zeta'(0)}{zeta(0)}, 1mapsto x^1, rhomapsto -frac{x^rho}{rho}$ for $rho$ any zero of $zeta,$ (from the Weierstrass factorization of the Hadamard product) so that :
$$tag{6}boxed{displaystylepsi^*(x)=x-sum_{rho} frac {x^{rho}}{rho}-frac{zeta'(0)}{zeta(0)}},quad(x>1)$$
(in this post we won't distinguish the trivial from the non-trivial zeros $rho$ ; for convergence the non-trivial roots should be grouped by pairs and sorted by increasing $|Im(rho)|$)
The Riemann prime-counting function is defined by :
begin{align}
tag{7}Pi^*(x):&=sum_{p^kle x}^{*}frac 1k=sum_{nle x}^{*}frac {Lambda(n)}{log,n}\
&=sum_{nle x}^{*} Lambda(n)left(int_n^x frac{dt}{t,log^2 t}+frac 1{log,x}right)\
&=int_2^xfrac{psi^*(t) dt}{t,log^2 t}+frac{psi^*(x)}{log ,x}\
&tag{8}=int_2^xfrac{psi^{*'}(t) dt}{log,t}\
end{align}
But $ displaystyleoperatorname{li}(x)=int_2^x frac{dt}{log,t},$ (Riemann's variant of the logarithmic integral) verifies $ displaystyleoperatorname{li}(x^r)'=frac{r,x^{r-1}}{log x^r}=frac{x^{r-1}}{log,x}$ so that from $(6)$ :
$$tag{9}boxed{displaystylePi^*(x)=operatorname{li}(x)-sum_{rho} operatorname{li}(x^{rho})},quad(x>1)$$
The prime-counting function is defined by $ displaystylepi^*(x):=sum_{ple x}^{*}1 $ while $ displaystylePi^*(x):=sum_{p^kle x}^{*}frac 1k$ will be :
$$tag{10}Pi^*(x)=sum_{k>0} frac{pi^{*}bigl(x^{1/k}bigr)}k$$
Applying the Möbius inversion formula $ displaystylepi^{*}(x):=sum_{n=1}^{infty} frac{mu(k)}k Pi^*bigl(x^{1/k}bigr) $ to $(9)$ we get (with questionable convergence...) :
$$tag{11}boxed{displaystylepi^*(x)=R(x)-sum_{rho} R(x^{rho})},quad(x>1)$$
Where Riemann's $,displaystyle R(x):=sum_{n=1}^{infty} frac{mu(k)}k operatorname{li}bigl(x^{1/k}bigr),$ may be written as a Gram series.
Dirichlet's L-functions
We will try to repeat the previous derivations but with $zeta$ replaced by the Dirichlet L-function
$$L(s,chi)=sum_{k=1}^inftyfrac{chi(k)}{k^s}$$
With $chi$ a Dirichlet character we get following Euler product :
$$tag{1'}boxed{displaystyle L(s,chi)=prod_{p text{prime}}frac 1{1-chi(p)p^{-s}}}quadtext{for} Re(s)>1$$
so that ($chi$ is multiplicative implying $chi(p)^k=chi(p^k)$) :
$$log L(s,chi)=-sum_{p text{prime}}log(1-chi(p)p^{-s})=sum_psum_{k=1}^infty frac{chi(p^k)p^{-ks}}k$$
minus the derivative relatively to $s$ will be :
$$tag{2'}f(s,chi):=-frac{L'(s,chi)}{L(s,chi)}=sum_psum_{k=1}^infty frac{chi(p^k)log,p}{p^{ks}}=sum_{n=1}^infty frac{chi(n)Lambda(n)}{n^s}quadtext{for} Re(s)>1$$
The Perron formula applied to $f(s,chi)$ gives :
$$tag{3'}psi^*(x,chi):=frac1{2pi i}int_{c-iinfty}^{c+iinfty}f(s,chi)frac{x^s}s,ds=sum_{n=1}^infty chi(n)Lambda(n) frac1{2pi i}int_{c-iinfty}^{c+iinfty}left(frac xnright)^sfrac{ds}s$$
$$tag{4'}psi^*(x,chi)=sum_{nle x}^{*}chi(n)Lambda(n)$$
But $psi^*(x,chi)$ may also be written as :
$$tag{5'}psi^*(x,chi)=-frac1{2pi i}int_{c-iinfty}^{c+iinfty}frac{L'(s,chi)}{L(s,chi)}frac{x^s}s,ds$$
Again we compute the sum of the residues at (all) the poles to the left of the vertical line of integration (for $c>1$). The contributions from the different poles will be : $0mapsto -frac{L'(0,chi)}{L(0,chi)}, rhomapsto -frac{x^rho}{rho}$ for $rho$ any zero of $L(s,chi),$ (from Weierstrass factorization) so that :
$$tag{6'}boxed{displaystylepsi^*(x,chi)=-sum_{rho} frac {x^{rho}}{rho}-frac{L'(0,chi)}{L(0,chi)}},quad(x>1)$$
(the sum over $rho$ combines the trivial and the non-trivial zeros and we will suppose that $chi$ is a non principal character so that $,L(s,chi),$ is regular everywhere, for a principal character we should add a $x$ contribution from the pole at $1$)
An analogue to the Riemann prime-counting function will be :
begin{align}
tag{7'}Pi^*(x,chi):&=sum_{p^kle x}^{*}frac {chi(p^k)}k=sum_{nle x}^{*}frac {chi(n)Lambda(n)}{log,n}\
&=sum_{nle x}^{*} chi(n)Lambda(n)left(int_n^x frac{dt}{t,log^2 t}+frac 1{log,x}right)\
&=int_2^xfrac{psi^*(t,chi) dt}{t,log^2 t}+frac{psi^*(x,chi)}{log ,x}\
&tag{8'}=int_2^xfrac{psi^{*'}(t,chi) dt}{log,t}\
end{align}
From $ displaystyleoperatorname{li}(x^r)'=frac{x^{r-1}}{log,x}$ and $(6')$ we get :
$$tag{9'}boxed{displaystylePi^*(x,chi)=-sum_{rho} operatorname{li}(x^{rho})},quad(x>1)$$
(for $chi$ a principal character there is an additional $,operatorname{li}(x)$ term)
The $chi$-prime-counting function is $ displaystylepi^*(x,chi):=sum_{ple x}^{*}chi(p) $ while $ displaystylePi^*(x,chi):=sum_{p^kle x}^{*}frac {chi(p^k)}k$ so that :
$$tag{10'}Pi^*(x,chi)=sum_{k>0} frac{pi^{*}bigl(x^{1/k},chi^kbigr)}k,quad(x>1)$$
Broken analogy...
The problem here is that we have $,pi^{*}bigl(x^{1/k},chi^kbigr)$ in the sum and not $,pi^{*}bigl(x^{1/k},chibigr)$.
For principal characters (when $,chi(p)=0$ or $1$ so that $,chi^k(p)=chi(p),$ for $k$ positive) we may apply the Möbius inversion formula to $(9')$ and get :
$$tag{11'}displaystylepi^*(x,chi)=[R(x)]-sum_{rho} R(x^{rho})$$
(since $chi$ is a principal character we must add the $R(x)$ term from the pole at $1$)
but this doesn't work in general and the equality becomes questionable after the first prime $p$ such that $,chi^k(p)not =chi(p),$ as we will see by considering your specific question.
We want $chi$ to be the Dirichlet character modulo $4$ ($chi_2$ in the link) defined by $ displaystylechi(0)=0, chi(1)=1, chi(2)=0, chi(3)=-1 $ and get :
$$tag{12'}L(s,chi)=sum_{n=0}^inftyfrac {(-1)^n}{(2n+1)^s}=beta(s)$$
with $beta$ the Dirichlet beta function we could hope that :
$$pi^*(x,chi)=displaystylepi_{4,1}^*(x)-pi_{4,3}^*(x)approx -sum_{rho} R(x^{rho})$$
with $pi_{q,a}(x)=#{p:p text{is prime and} ,ple x, text{and} ,pequiv apmod q}$
But, from $(10')$ and $(9')$, we have in fact :
begin{align}
Pi^*(x,chi)&=sum_{k>0} frac{pi_{4,1}^{*}bigl(x^{1/k}bigr)+(-1)^k,pi_{4,3}^{*}bigl(x^{1/k}bigr)}k=-sum_{rho} operatorname{li}(x^{rho})\
&=sum_{k>0} frac{pi_{4,1}^{*}bigl(x^{1/k}bigr)-,pi_{4,3}^{*}bigl(x^{1/k}bigr)}k+2sum_{j>0} frac{pi_{4,3}^{*}left(sqrt{x}^{1/j}right)}{2j}\
end{align}
so that the Möbius transformation will return the exact (with $rho$ any zero of $beta$) :
$$tag{13'}boxed{displaystylepi_{4,1}^{*}(x)-pi_{4,3}^{*}(x)+pi_{4,3}^{*}left(sqrt{x}right)=-sum_{rho} R(x^{rho})}$$
To show the effect of the $,pi_{4,3}^{*}left(sqrt{x}right),$ term let's exhibit an approximation of the expression at the right (taking the $40$ first terms of the sum $R$ and the $300$ first non-trivial zeros of the Dirichlet-beta function) that we compare to the exact (dark) result for $,pi^*(x,chi)=displaystylepi_{4,1}^*(x)-pi_{4,3}^*(x),$ in the interval $,[5,80]$ :
The initial behavior seems right : $-1$ at $3$ followed by $+1$ at $5$ and $-1$ at $7$ but just after that there is a first 'irregularity' at $9=3^2$ because $chi(3^2)=1$ while $chi(3)=-1$. After that the behavior is correct (shifted $+1$ of course) until another irregularity at $49=7^2$ again because $chi(7^2)=1$. The next one will be at $121=11^2$ and at all the following squares of primes of gender $4n+3$...
ADDITION: Draks telescoping
The situation was blocked at this point until draks, using a suggestion of Greg Martin, proposed in february his powerful telescoping method to obtain (the $rho_zeta$ and $rho_beta$ are respectively all the zeros of the $zeta$ and $beta$ function) :
$$tag{14'}pi_{4,3}(x):=sum_{k=0}^infty;2^{-k-1}left(
Rleft(x^{1/2^{k}}right)-sum_{rho_zeta}
Rleft(x^{rho_zeta/2^k}right) +sum_{rho_beta}
Rleft(x^{rho_beta/2^k}right) right)$$
We may obtain this result too (with a minor correction) by rewriting $(11)$ as (for $x>2$) :
$$pi^*(x)=1+pi_{4,1}^{*}(x)+pi_{4,3}^{*}(x)=R(x)-sum_{rho_zeta} R(x^{rho_zeta})$$
($1$ was added since all primes except $2$ are equal to $1$ or $3bmod{4}$)
and using $(13')$ :
$$pi_{4,1}^{*}(x)-pi_{4,3}^{*}(x)+pi_{4,3}^{*}left(sqrt{x}right)=-sum_{rho_beta} R(x^{rho_beta})$$
Subtracting these equations gives :
$$2pi_{4,3}^{*}(x)-pi_{4,3}^{*}left(sqrt{x}right)=R(x)-1-sum_{rho_zeta} R(x^{rho_zeta})+sum_{rho_beta} R(x^{rho_beta})$$
while draks' telescoping sum $ displaystylefrac 12sum_{k=0}^infty frac{fleft(x^{1/2^{k}}right)}{2^k}$ is :
$$sum_{k=0}^infty frac {2,pi_{4,3}^{*}(x^{1/2^{k}})-pi_{4,3}^{*}bigl(x^{1/2^{k+1}}bigr)}{2^{k+1}}=sum_{k=0}^infty frac {Rleft(x^{1/2^{k}}right)-1-sum_{rho_zeta} Rleft(x^{rho_zeta/2^k}right)+sum_{rho_beta} Rleft(x^{rho_beta/2^k}right)}{2^{k+1}}$$
producing draks' formula :
$$tag{15'}boxed{displaystylepi_{4,3}^{*}(x)=sum_{k=0}^infty frac {Rleft(x^{1/2^{k}}right)-sum_{rho_zeta} Rleft(x^{rho_zeta/2^k}right)+sum_{rho_beta} Rleft(x^{rho_beta/2^k}right)}{2^{k+1}}-1}$$
A similar formula may be written for $pi_{4,1}$ and this answers positively the initial question !
Let's finish with a plot of $pi_{4,3}^{*}(x)$ from $4$ to $100$ (more details here with the pari/gp scripts used)
edited Apr 13 '17 at 12:20
Community♦
1
answered Jan 20 '13 at 17:22
Raymond ManzoniRaymond Manzoni
37.2k563117
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Jan 20 '13 at 17:22
Raymond ManzoniRaymond Manzoni
37.2k563117
answered Jan 20 '13 at 17:22
Raymond ManzoniRaymond Manzoni
37.2k563117
answered Jan 20 '13 at 17:22
Raymond ManzoniRaymond Manzoni
37.2k563117
37.2k563117
$begingroup$
Raymond, first of all: Wow! What an answer. "+1". Thank you so much for everything including the link to Edward's book. My copy is buried somewhere in a box because I moved. It'll take me a while to verify all that, but like you, I think that the reasoning by analogy is correct. Specially the last equation thrills me. You might have noticed my attempts with "Summing over General Functions of Primes" ...
$endgroup$
– draks ...
Jan 20 '13 at 22:44
$begingroup$
... or the follow-on question: "An Expression for $logzeta(ns)$ derived from the Limit of the truncated Prime $zeta$ Function". So I think it's possible to combine these result to get an analytic expression for the truncated Prime $zeta$ function with a restriction to primes $4npm 1$. This is ever so cool and I can't leave this unawarded...
$endgroup$
– draks ...
Jan 20 '13 at 22:51
$begingroup$
Thanks @draks... I think that most things are right here except tiny things like the - from the citation, the starting value $n=1$ of the Dirichlet beta function, and perhaps very important things like the transitions from $(9')$ to $(11')$ : these should be (more than!) double checked (because of things like $chi(3^2)=1not =-1$ and of warnings from different authors like Ingham that Riemann's method to obtain $pi$ may not be as easily generalized as $psicdots$ but works only for some specific Dirichlet characters ($chi(p)=0,1$ only?)). But this clear path may at least be checked!
$endgroup$
– Raymond Manzoni
Jan 21 '13 at 0:05
$begingroup$
@draks... I corrected the end of my answer.
$endgroup$
– Raymond Manzoni
Jan 26 '13 at 23:04
$begingroup$
aha, looks like we've gottta add $(mu^2(n)-1)$ to get rid of the squares...interesting, what do you think? And thanks for your thoughts. At least my ${rm li}(x)$ works...
$endgroup$
– draks ...
Jan 27 '13 at 16:03
|
show 3 more comments
$begingroup$
Raymond, first of all: Wow! What an answer. "+1". Thank you so much for everything including the link to Edward's book. My copy is buried somewhere in a box because I moved. It'll take me a while to verify all that, but like you, I think that the reasoning by analogy is correct. Specially the last equation thrills me. You might have noticed my attempts with "Summing over General Functions of Primes" ...
$endgroup$
– draks ...
Jan 20 '13 at 22:44
$begingroup$
... or the follow-on question: "An Expression for $logzeta(ns)$ derived from the Limit of the truncated Prime $zeta$ Function". So I think it's possible to combine these result to get an analytic expression for the truncated Prime $zeta$ function with a restriction to primes $4npm 1$. This is ever so cool and I can't leave this unawarded...
$endgroup$
– draks ...
Jan 20 '13 at 22:51
$begingroup$
Thanks @draks... I think that most things are right here except tiny things like the - from the citation, the starting value $n=1$ of the Dirichlet beta function, and perhaps very important things like the transitions from $(9')$ to $(11')$ : these should be (more than!) double checked (because of things like $chi(3^2)=1not =-1$ and of warnings from different authors like Ingham that Riemann's method to obtain $pi$ may not be as easily generalized as $psicdots$ but works only for some specific Dirichlet characters ($chi(p)=0,1$ only?)). But this clear path may at least be checked!
$endgroup$
– Raymond Manzoni
Jan 21 '13 at 0:05
$begingroup$
@draks... I corrected the end of my answer.
$endgroup$
– Raymond Manzoni
Jan 26 '13 at 23:04
$begingroup$
aha, looks like we've gottta add $(mu^2(n)-1)$ to get rid of the squares...interesting, what do you think? And thanks for your thoughts. At least my ${rm li}(x)$ works...
$endgroup$
– draks ...
Jan 27 '13 at 16:03
$begingroup$
Raymond, first of all: Wow! What an answer. "+1". Thank you so much for everything including the link to Edward's book. My copy is buried somewhere in a box because I moved. It'll take me a while to verify all that, but like you, I think that the reasoning by analogy is correct. Specially the last equation thrills me. You might have noticed my attempts with "Summing over General Functions of Primes" ...
$endgroup$
– draks ...
Jan 20 '13 at 22:44
$begingroup$
Raymond, first of all: Wow! What an answer. "+1". Thank you so much for everything including the link to Edward's book. My copy is buried somewhere in a box because I moved. It'll take me a while to verify all that, but like you, I think that the reasoning by analogy is correct. Specially the last equation thrills me. You might have noticed my attempts with "Summing over General Functions of Primes" ...
$endgroup$
– draks ...
Jan 20 '13 at 22:44
$begingroup$
... or the follow-on question: "An Expression for $logzeta(ns)$ derived from the Limit of the truncated Prime $zeta$ Function". So I think it's possible to combine these result to get an analytic expression for the truncated Prime $zeta$ function with a restriction to primes $4npm 1$. This is ever so cool and I can't leave this unawarded...
$endgroup$
– draks ...
Jan 20 '13 at 22:51
$begingroup$
... or the follow-on question: "An Expression for $logzeta(ns)$ derived from the Limit of the truncated Prime $zeta$ Function". So I think it's possible to combine these result to get an analytic expression for the truncated Prime $zeta$ function with a restriction to primes $4npm 1$. This is ever so cool and I can't leave this unawarded...
$endgroup$
– draks ...
Jan 20 '13 at 22:51
$begingroup$
Thanks @draks... I think that most things are right here except tiny things like the - from the citation, the starting value $n=1$ of the Dirichlet beta function, and perhaps very important things like the transitions from $(9')$ to $(11')$ : these should be (more than!) double checked (because of things like $chi(3^2)=1not =-1$ and of warnings from different authors like Ingham that Riemann's method to obtain $pi$ may not be as easily generalized as $psicdots$ but works only for some specific Dirichlet characters ($chi(p)=0,1$ only?)). But this clear path may at least be checked!
$endgroup$
– Raymond Manzoni
Jan 21 '13 at 0:05
$begingroup$
Thanks @draks... I think that most things are right here except tiny things like the - from the citation, the starting value $n=1$ of the Dirichlet beta function, and perhaps very important things like the transitions from $(9')$ to $(11')$ : these should be (more than!) double checked (because of things like $chi(3^2)=1not =-1$ and of warnings from different authors like Ingham that Riemann's method to obtain $pi$ may not be as easily generalized as $psicdots$ but works only for some specific Dirichlet characters ($chi(p)=0,1$ only?)). But this clear path may at least be checked!
$endgroup$
– Raymond Manzoni
Jan 21 '13 at 0:05
$begingroup$
@draks... I corrected the end of my answer.
$endgroup$
– Raymond Manzoni
Jan 26 '13 at 23:04
$begingroup$
@draks... I corrected the end of my answer.
$endgroup$
– Raymond Manzoni
Jan 26 '13 at 23:04
$begingroup$
aha, looks like we've gottta add $(mu^2(n)-1)$ to get rid of the squares...interesting, what do you think? And thanks for your thoughts. At least my ${rm li}(x)$ works...
$endgroup$
– draks ...
Jan 27 '13 at 16:03
$begingroup$
aha, looks like we've gottta add $(mu^2(n)-1)$ to get rid of the squares...interesting, what do you think? And thanks for your thoughts. At least my ${rm li}(x)$ works...
$endgroup$
– draks ...
Jan 27 '13 at 16:03
|
show 3 more comments
$begingroup$
You probably know that
$$
Pi^*(x;q,a) = sum^*_{substack{nle x \ nequiv apmod q}} frac{Lambda(n)}{log n} = frac1{phi(q)} sum_{chipmod q} overline{chi(a)}Pi^*(x,chi).
$$
I find that the easiest way to convert from counting prime powers (with weights) to counting only primes is to use the fact that
begin{align*}
Pi^*(x;q,a) &= sum^*_{substack{ple x \ pequiv apmod q}} 1 + sum^*_{substack{p^2le x \ p^2equiv apmod q}} tfrac12 + sum^*_{substack{p^3le x \ p^3equiv apmod q}} tfrac13 + cdots \
&= pi^*(x;q,a) + tfrac12 sum_{substack{bpmod q \ b^2equiv apmod q}} pi^*(x^{1/2};q,b) + tfrac13 sum_{substack{cpmod q \ c^3equiv apmod q}} pi^*(x^{1/3};q,c) + cdots
end{align*}
and solving for $pi^*(x;q,a)$ in terms of all $Pi^*$ functions (tedious but possible).
answered Jan 31 '13 at 7:39
Greg MartinGreg Martin
36.3k23565
$endgroup$
$begingroup$
+1 thanks, doesn't it boil down a little/lot when $a=1$? Could you give me a reference hint or something else to read on that topic?
$endgroup$
– draks ...
Jan 31 '13 at 20:43
$begingroup$
I'm not sure it gets much better when $a=1$. Except that all the values $chi(a)$ (which I forgot to put in in the first place - they're there now) equal $1$ and thus disappear. But $a=1$ will yield the largest possible number of terms in the sums over $b$, $c$, etc. in the last displayed equation line.
$endgroup$
– Greg Martin
Feb 1 '13 at 1:25
$begingroup$
hmm, when I look at the second term in the last line and insert $a=1$, the sum's running over $b=1,3,5,7,dots$. But $b>3$ doesn't make sense since it's included already before, right? If this is correct and I continue, I get $x^mequiv 1 bmod 4to cases{x=1,3 text{ if $m$ even }\x=1;;,, text{ if $m$ odd}} $. But doesn't summing $pi(x^{1/m};4,pm 1)$ equal $pi(x^{1/m})$? How to continue from here? Could you kindly sketch the way...?
$endgroup$
– draks ...
Feb 8 '13 at 23:01
$begingroup$
Right, when I write a sum over $bpmod q$, I mean a sum over residue classes - so don't separately include both $b=1$ and $b=5$, since they're the same residue class. And in this particular case, when $m$ is even you do in fact count all odd primes in the sum, yes.
$endgroup$
– Greg Martin
Feb 9 '13 at 6:13
$begingroup$
you are awarded the bounty since it was your prime race paper which gave me new hope. I'm currently studying your arXiv stuff. Could you point me to something that would help me once more?
$endgroup$
– draks ...
Feb 16 '13 at 13:35
|
show 7 more comments
$begingroup$
You probably know that
$$
Pi^*(x;q,a) = sum^*_{substack{nle x \ nequiv apmod q}} frac{Lambda(n)}{log n} = frac1{phi(q)} sum_{chipmod q} overline{chi(a)}Pi^*(x,chi).
$$
I find that the easiest way to convert from counting prime powers (with weights) to counting only primes is to use the fact that
begin{align*}
Pi^*(x;q,a) &= sum^*_{substack{ple x \ pequiv apmod q}} 1 + sum^*_{substack{p^2le x \ p^2equiv apmod q}} tfrac12 + sum^*_{substack{p^3le x \ p^3equiv apmod q}} tfrac13 + cdots \
&= pi^*(x;q,a) + tfrac12 sum_{substack{bpmod q \ b^2equiv apmod q}} pi^*(x^{1/2};q,b) + tfrac13 sum_{substack{cpmod q \ c^3equiv apmod q}} pi^*(x^{1/3};q,c) + cdots
end{align*}
and solving for $pi^*(x;q,a)$ in terms of all $Pi^*$ functions (tedious but possible).
answered Jan 31 '13 at 7:39
Greg MartinGreg Martin
36.3k23565
$endgroup$
$begingroup$
+1 thanks, doesn't it boil down a little/lot when $a=1$? Could you give me a reference hint or something else to read on that topic?
$endgroup$
– draks ...
Jan 31 '13 at 20:43
$begingroup$
I'm not sure it gets much better when $a=1$. Except that all the values $chi(a)$ (which I forgot to put in in the first place - they're there now) equal $1$ and thus disappear. But $a=1$ will yield the largest possible number of terms in the sums over $b$, $c$, etc. in the last displayed equation line.
$endgroup$
– Greg Martin
Feb 1 '13 at 1:25
$begingroup$
hmm, when I look at the second term in the last line and insert $a=1$, the sum's running over $b=1,3,5,7,dots$. But $b>3$ doesn't make sense since it's included already before, right? If this is correct and I continue, I get $x^mequiv 1 bmod 4to cases{x=1,3 text{ if $m$ even }\x=1;;,, text{ if $m$ odd}} $. But doesn't summing $pi(x^{1/m};4,pm 1)$ equal $pi(x^{1/m})$? How to continue from here? Could you kindly sketch the way...?
$endgroup$
– draks ...
Feb 8 '13 at 23:01
$begingroup$
Right, when I write a sum over $bpmod q$, I mean a sum over residue classes - so don't separately include both $b=1$ and $b=5$, since they're the same residue class. And in this particular case, when $m$ is even you do in fact count all odd primes in the sum, yes.
$endgroup$
– Greg Martin
Feb 9 '13 at 6:13
$begingroup$
you are awarded the bounty since it was your prime race paper which gave me new hope. I'm currently studying your arXiv stuff. Could you point me to something that would help me once more?
$endgroup$
– draks ...
Feb 16 '13 at 13:35
|
show 7 more comments
$begingroup$
You probably know that
$$
Pi^*(x;q,a) = sum^*_{substack{nle x \ nequiv apmod q}} frac{Lambda(n)}{log n} = frac1{phi(q)} sum_{chipmod q} overline{chi(a)}Pi^*(x,chi).
$$
I find that the easiest way to convert from counting prime powers (with weights) to counting only primes is to use the fact that
begin{align*}
Pi^*(x;q,a) &= sum^*_{substack{ple x \ pequiv apmod q}} 1 + sum^*_{substack{p^2le x \ p^2equiv apmod q}} tfrac12 + sum^*_{substack{p^3le x \ p^3equiv apmod q}} tfrac13 + cdots \
&= pi^*(x;q,a) + tfrac12 sum_{substack{bpmod q \ b^2equiv apmod q}} pi^*(x^{1/2};q,b) + tfrac13 sum_{substack{cpmod q \ c^3equiv apmod q}} pi^*(x^{1/3};q,c) + cdots
end{align*}
and solving for $pi^*(x;q,a)$ in terms of all $Pi^*$ functions (tedious but possible).
answered Jan 31 '13 at 7:39
Greg MartinGreg Martin
36.3k23565
$endgroup$
You probably know that
$$
Pi^*(x;q,a) = sum^*_{substack{nle x \ nequiv apmod q}} frac{Lambda(n)}{log n} = frac1{phi(q)} sum_{chipmod q} overline{chi(a)}Pi^*(x,chi).
$$
I find that the easiest way to convert from counting prime powers (with weights) to counting only primes is to use the fact that
begin{align*}
Pi^*(x;q,a) &= sum^*_{substack{ple x \ pequiv apmod q}} 1 + sum^*_{substack{p^2le x \ p^2equiv apmod q}} tfrac12 + sum^*_{substack{p^3le x \ p^3equiv apmod q}} tfrac13 + cdots \
&= pi^*(x;q,a) + tfrac12 sum_{substack{bpmod q \ b^2equiv apmod q}} pi^*(x^{1/2};q,b) + tfrac13 sum_{substack{cpmod q \ c^3equiv apmod q}} pi^*(x^{1/3};q,c) + cdots
end{align*}
and solving for $pi^*(x;q,a)$ in terms of all $Pi^*$ functions (tedious but possible).
answered Jan 31 '13 at 7:39
Greg MartinGreg Martin
36.3k23565
edited Feb 1 '13 at 1:24
answered Jan 31 '13 at 7:39
Greg MartinGreg Martin
36.3k23565
answered Jan 31 '13 at 7:39
Greg MartinGreg Martin
36.3k23565
answered Jan 31 '13 at 7:39
Greg MartinGreg Martin
36.3k23565
36.3k23565
$begingroup$
+1 thanks, doesn't it boil down a little/lot when $a=1$? Could you give me a reference hint or something else to read on that topic?
$endgroup$
– draks ...
Jan 31 '13 at 20:43
$begingroup$
I'm not sure it gets much better when $a=1$. Except that all the values $chi(a)$ (which I forgot to put in in the first place - they're there now) equal $1$ and thus disappear. But $a=1$ will yield the largest possible number of terms in the sums over $b$, $c$, etc. in the last displayed equation line.
$endgroup$
– Greg Martin
Feb 1 '13 at 1:25
$begingroup$
hmm, when I look at the second term in the last line and insert $a=1$, the sum's running over $b=1,3,5,7,dots$. But $b>3$ doesn't make sense since it's included already before, right? If this is correct and I continue, I get $x^mequiv 1 bmod 4to cases{x=1,3 text{ if $m$ even }\x=1;;,, text{ if $m$ odd}} $. But doesn't summing $pi(x^{1/m};4,pm 1)$ equal $pi(x^{1/m})$? How to continue from here? Could you kindly sketch the way...?
$endgroup$
– draks ...
Feb 8 '13 at 23:01
$begingroup$
Right, when I write a sum over $bpmod q$, I mean a sum over residue classes - so don't separately include both $b=1$ and $b=5$, since they're the same residue class. And in this particular case, when $m$ is even you do in fact count all odd primes in the sum, yes.
$endgroup$
– Greg Martin
Feb 9 '13 at 6:13
$begingroup$
you are awarded the bounty since it was your prime race paper which gave me new hope. I'm currently studying your arXiv stuff. Could you point me to something that would help me once more?
$endgroup$
– draks ...
Feb 16 '13 at 13:35
|
show 7 more comments
$begingroup$
+1 thanks, doesn't it boil down a little/lot when $a=1$? Could you give me a reference hint or something else to read on that topic?
$endgroup$
– draks ...
Jan 31 '13 at 20:43
$begingroup$
I'm not sure it gets much better when $a=1$. Except that all the values $chi(a)$ (which I forgot to put in in the first place - they're there now) equal $1$ and thus disappear. But $a=1$ will yield the largest possible number of terms in the sums over $b$, $c$, etc. in the last displayed equation line.
$endgroup$
– Greg Martin
Feb 1 '13 at 1:25
$begingroup$
hmm, when I look at the second term in the last line and insert $a=1$, the sum's running over $b=1,3,5,7,dots$. But $b>3$ doesn't make sense since it's included already before, right? If this is correct and I continue, I get $x^mequiv 1 bmod 4to cases{x=1,3 text{ if $m$ even }\x=1;;,, text{ if $m$ odd}} $. But doesn't summing $pi(x^{1/m};4,pm 1)$ equal $pi(x^{1/m})$? How to continue from here? Could you kindly sketch the way...?
$endgroup$
– draks ...
Feb 8 '13 at 23:01
$begingroup$
Right, when I write a sum over $bpmod q$, I mean a sum over residue classes - so don't separately include both $b=1$ and $b=5$, since they're the same residue class. And in this particular case, when $m$ is even you do in fact count all odd primes in the sum, yes.
$endgroup$
– Greg Martin
Feb 9 '13 at 6:13
$begingroup$
you are awarded the bounty since it was your prime race paper which gave me new hope. I'm currently studying your arXiv stuff. Could you point me to something that would help me once more?
$endgroup$
– draks ...
Feb 16 '13 at 13:35
$begingroup$
+1 thanks, doesn't it boil down a little/lot when $a=1$? Could you give me a reference hint or something else to read on that topic?
$endgroup$
– draks ...
Jan 31 '13 at 20:43
$begingroup$
+1 thanks, doesn't it boil down a little/lot when $a=1$? Could you give me a reference hint or something else to read on that topic?
$endgroup$
– draks ...
Jan 31 '13 at 20:43
$begingroup$
I'm not sure it gets much better when $a=1$. Except that all the values $chi(a)$ (which I forgot to put in in the first place - they're there now) equal $1$ and thus disappear. But $a=1$ will yield the largest possible number of terms in the sums over $b$, $c$, etc. in the last displayed equation line.
$endgroup$
– Greg Martin
Feb 1 '13 at 1:25
$begingroup$
I'm not sure it gets much better when $a=1$. Except that all the values $chi(a)$ (which I forgot to put in in the first place - they're there now) equal $1$ and thus disappear. But $a=1$ will yield the largest possible number of terms in the sums over $b$, $c$, etc. in the last displayed equation line.
$endgroup$
– Greg Martin
Feb 1 '13 at 1:25
$begingroup$
hmm, when I look at the second term in the last line and insert $a=1$, the sum's running over $b=1,3,5,7,dots$. But $b>3$ doesn't make sense since it's included already before, right? If this is correct and I continue, I get $x^mequiv 1 bmod 4to cases{x=1,3 text{ if $m$ even }\x=1;;,, text{ if $m$ odd}} $. But doesn't summing $pi(x^{1/m};4,pm 1)$ equal $pi(x^{1/m})$? How to continue from here? Could you kindly sketch the way...?
$endgroup$
– draks ...
Feb 8 '13 at 23:01
$begingroup$
hmm, when I look at the second term in the last line and insert $a=1$, the sum's running over $b=1,3,5,7,dots$. But $b>3$ doesn't make sense since it's included already before, right? If this is correct and I continue, I get $x^mequiv 1 bmod 4to cases{x=1,3 text{ if $m$ even }\x=1;;,, text{ if $m$ odd}} $. But doesn't summing $pi(x^{1/m};4,pm 1)$ equal $pi(x^{1/m})$? How to continue from here? Could you kindly sketch the way...?
$endgroup$
– draks ...
Feb 8 '13 at 23:01
$begingroup$
Right, when I write a sum over $bpmod q$, I mean a sum over residue classes - so don't separately include both $b=1$ and $b=5$, since they're the same residue class. And in this particular case, when $m$ is even you do in fact count all odd primes in the sum, yes.
$endgroup$
– Greg Martin
Feb 9 '13 at 6:13
$begingroup$
Right, when I write a sum over $bpmod q$, I mean a sum over residue classes - so don't separately include both $b=1$ and $b=5$, since they're the same residue class. And in this particular case, when $m$ is even you do in fact count all odd primes in the sum, yes.
$endgroup$
– Greg Martin
Feb 9 '13 at 6:13
$begingroup$
you are awarded the bounty since it was your prime race paper which gave me new hope. I'm currently studying your arXiv stuff. Could you point me to something that would help me once more?
$endgroup$
– draks ...
Feb 16 '13 at 13:35
$begingroup$
you are awarded the bounty since it was your prime race paper which gave me new hope. I'm currently studying your arXiv stuff. Could you point me to something that would help me once more?
$endgroup$
– draks ...
Feb 16 '13 at 13:35
|
show 7 more comments
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$begingroup$
+1: interesting question but where did you find the formula $(2)$ ? I would bet that it is only an approximation just from the lower bound $2$ of the first integral ( and... from the $approx$ sign :-) ) ?
$endgroup$
– Raymond Manzoni
Jan 4 '13 at 21:42
$begingroup$
It is from the linked paper here : math.stackexchange.com/q/149755/19341 . Sorry but i post from my tablet. I hate that...
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– draks ...
Jan 4 '13 at 23:28
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Just some preliminary numerical results : I compared $(2)$ to the actual $pi(n)$ function in the interval $(2,100)$ !pi_100 : $(2)$ (darker and on the top) has all the features of a function growing 'by steps' and is very similar to the real $pi(n)$ function but, as seen in $(3,1000)$ !pi_1000, it is starting too low and growing too fast (this effect is real : more than $10^4$ zeros were used).
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– Raymond Manzoni
Jan 4 '13 at 23:42
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From your ref. the zeros seem to be the imaginary parts of the zeros of the Dirichlet beta function. I have Montgomery and Vaughan's book (I'll see tomorrow...).
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– Raymond Manzoni
Jan 4 '13 at 23:43
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But do your plots compare to Figure 4 in Prime Number Races. Beta functions are used in the case $pi_{4n+3}-pi_{4n+1}$ (p.19)...
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– draks ...
Jan 5 '13 at 9:20