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Dubious step in 'order of mobius' proof which may lead to an interesting fact
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Let $mu(n)$ be the Moebius function, let $M(x)=sum_nleq x mu(x)$ be the Mertens function and let $A(x)=sum_nleq xtfracmu(n)n$ be the truncation of the Dirichlet series expansion of $1/zeta(s)$ at $s=1$.
Question: is it true that $A(x)geq tfracM(x)x$ for all integer $x$?
I provide some facts and context.
By partial summation [3] the difference is $A(x)-tfracM(x)x = tfrac 1xint_0^x A(t) dt$. In an elementary [3] way $|A(x)|leq 1$ for all $x$. It is also elementary [1] [2] to prove the equivalences
$$A(x)=o(1)iff M(x)=o(x)iff textPrime Number Theorem.$$
In fact, I was reading on ProofWiki [2] a proof of the fact that $M(x)=o(x)$. At a certain point of the proof it is claimed that clearly $A(x)geq tfracM(x)x$.
My considerations are: (i) it is not clear to me; (ii) but maybe I'm missing something obvious? (iii) you could think it were obvious if $mu(n)$ didn't change sign; (iv) the statement is true for $x<1000$.
[1] Diamond, H. G. (1982). Elementary methods in the study of the distribution of prime numbers. Bulletin of the American Mathematical Society, 7(3), 553-589.
[2] https://proofwiki.org/wiki/Order_of_M%C3%B6bius_Function
[3] Apostol, T. M. (2013). Introduction to analytic number theory. Springer Science & Business Media.
proof-explanation analytic-number-theory mobius-function
asked Jul 17 at 1:12
Luca Ghidelli
295111
add a comment |Â
up vote
1
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Let $mu(n)$ be the Moebius function, let $M(x)=sum_nleq x mu(x)$ be the Mertens function and let $A(x)=sum_nleq xtfracmu(n)n$ be the truncation of the Dirichlet series expansion of $1/zeta(s)$ at $s=1$.
Question: is it true that $A(x)geq tfracM(x)x$ for all integer $x$?
I provide some facts and context.
By partial summation [3] the difference is $A(x)-tfracM(x)x = tfrac 1xint_0^x A(t) dt$. In an elementary [3] way $|A(x)|leq 1$ for all $x$. It is also elementary [1] [2] to prove the equivalences
$$A(x)=o(1)iff M(x)=o(x)iff textPrime Number Theorem.$$
In fact, I was reading on ProofWiki [2] a proof of the fact that $M(x)=o(x)$. At a certain point of the proof it is claimed that clearly $A(x)geq tfracM(x)x$.
My considerations are: (i) it is not clear to me; (ii) but maybe I'm missing something obvious? (iii) you could think it were obvious if $mu(n)$ didn't change sign; (iv) the statement is true for $x<1000$.
[1] Diamond, H. G. (1982). Elementary methods in the study of the distribution of prime numbers. Bulletin of the American Mathematical Society, 7(3), 553-589.
[2] https://proofwiki.org/wiki/Order_of_M%C3%B6bius_Function
[3] Apostol, T. M. (2013). Introduction to analytic number theory. Springer Science & Business Media.
proof-explanation analytic-number-theory mobius-function
asked Jul 17 at 1:12
Luca Ghidelli
295111
2
That proofwiki article looks really sloppy to me and I would be a little dubious of trusting it for anything.
– Steven Stadnicki
Jul 17 at 1:25
In any case it raises an interesting question in the spirit of "things that (don't?) change sign infinitely often".
– Luca Ghidelli
Jul 17 at 1:46
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $mu(n)$ be the Moebius function, let $M(x)=sum_nleq x mu(x)$ be the Mertens function and let $A(x)=sum_nleq xtfracmu(n)n$ be the truncation of the Dirichlet series expansion of $1/zeta(s)$ at $s=1$.
Question: is it true that $A(x)geq tfracM(x)x$ for all integer $x$?
I provide some facts and context.
By partial summation [3] the difference is $A(x)-tfracM(x)x = tfrac 1xint_0^x A(t) dt$. In an elementary [3] way $|A(x)|leq 1$ for all $x$. It is also elementary [1] [2] to prove the equivalences
$$A(x)=o(1)iff M(x)=o(x)iff textPrime Number Theorem.$$
In fact, I was reading on ProofWiki [2] a proof of the fact that $M(x)=o(x)$. At a certain point of the proof it is claimed that clearly $A(x)geq tfracM(x)x$.
My considerations are: (i) it is not clear to me; (ii) but maybe I'm missing something obvious? (iii) you could think it were obvious if $mu(n)$ didn't change sign; (iv) the statement is true for $x<1000$.
[1] Diamond, H. G. (1982). Elementary methods in the study of the distribution of prime numbers. Bulletin of the American Mathematical Society, 7(3), 553-589.
[2] https://proofwiki.org/wiki/Order_of_M%C3%B6bius_Function
[3] Apostol, T. M. (2013). Introduction to analytic number theory. Springer Science & Business Media.
proof-explanation analytic-number-theory mobius-function
asked Jul 17 at 1:12
Luca Ghidelli
295111
Let $mu(n)$ be the Moebius function, let $M(x)=sum_nleq x mu(x)$ be the Mertens function and let $A(x)=sum_nleq xtfracmu(n)n$ be the truncation of the Dirichlet series expansion of $1/zeta(s)$ at $s=1$.
Question: is it true that $A(x)geq tfracM(x)x$ for all integer $x$?
I provide some facts and context.
By partial summation [3] the difference is $A(x)-tfracM(x)x = tfrac 1xint_0^x A(t) dt$. In an elementary [3] way $|A(x)|leq 1$ for all $x$. It is also elementary [1] [2] to prove the equivalences
$$A(x)=o(1)iff M(x)=o(x)iff textPrime Number Theorem.$$
In fact, I was reading on ProofWiki [2] a proof of the fact that $M(x)=o(x)$. At a certain point of the proof it is claimed that clearly $A(x)geq tfracM(x)x$.
My considerations are: (i) it is not clear to me; (ii) but maybe I'm missing something obvious? (iii) you could think it were obvious if $mu(n)$ didn't change sign; (iv) the statement is true for $x<1000$.
[1] Diamond, H. G. (1982). Elementary methods in the study of the distribution of prime numbers. Bulletin of the American Mathematical Society, 7(3), 553-589.
[2] https://proofwiki.org/wiki/Order_of_M%C3%B6bius_Function
[3] Apostol, T. M. (2013). Introduction to analytic number theory. Springer Science & Business Media.
proof-explanation analytic-number-theory mobius-function
asked Jul 17 at 1:12
Luca Ghidelli
295111
asked Jul 17 at 1:12
Luca Ghidelli
295111
asked Jul 17 at 1:12
Luca Ghidelli
295111
asked Jul 17 at 1:12
Luca Ghidelli
295111
295111
2
That proofwiki article looks really sloppy to me and I would be a little dubious of trusting it for anything.
– Steven Stadnicki
Jul 17 at 1:25
In any case it raises an interesting question in the spirit of "things that (don't?) change sign infinitely often".
– Luca Ghidelli
Jul 17 at 1:46
add a comment |Â
2
That proofwiki article looks really sloppy to me and I would be a little dubious of trusting it for anything.
– Steven Stadnicki
Jul 17 at 1:25
In any case it raises an interesting question in the spirit of "things that (don't?) change sign infinitely often".
– Luca Ghidelli
Jul 17 at 1:46
2
2
That proofwiki article looks really sloppy to me and I would be a little dubious of trusting it for anything.
– Steven Stadnicki
Jul 17 at 1:25
That proofwiki article looks really sloppy to me and I would be a little dubious of trusting it for anything.
– Steven Stadnicki
Jul 17 at 1:25
In any case it raises an interesting question in the spirit of "things that (don't?) change sign infinitely often".
– Luca Ghidelli
Jul 17 at 1:46
In any case it raises an interesting question in the spirit of "things that (don't?) change sign infinitely often".
– Luca Ghidelli
Jul 17 at 1:46
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
It is false. Using PARI/GP I found:
$M(18798)=0$
$A(18798)*18798 =-0.3589042...$
answered Jul 17 at 14:59
Luca Ghidelli
295111
Bonus question: does the difference change sign infinitely often?
– Luca Ghidelli
Jul 17 at 15:00
I would expect so, but I would not be surprised if that is still unknown.
– Daniel Fischer♦
Jul 17 at 15:17
1
@LucaGhidelli, it can be shown to change sign infinitely often, using the tricks in Chapter 15 of Montgomery and Vaughan.
– Peter Humphries
Jul 20 at 9:35
I will be happy to learn it, thank you for the reference! @Peter Humphries
– Luca Ghidelli
Jul 20 at 9:46
I learnt something, thank you! Lemma 15.1 applied to the function 1/s(s-1)zeta(s) = int_0^infty [A(x)-M(x)/x]x^-s (no need to do better) does the job perfectly, because it is analytic for real s>0 but not on all the semiplane Re s>0 ! Thank you, would you like to write it as an answer?
– Luca Ghidelli
Jul 20 at 19:43
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
It is false. Using PARI/GP I found:
$M(18798)=0$
$A(18798)*18798 =-0.3589042...$
answered Jul 17 at 14:59
Luca Ghidelli
295111
Bonus question: does the difference change sign infinitely often?
– Luca Ghidelli
Jul 17 at 15:00
I would expect so, but I would not be surprised if that is still unknown.
– Daniel Fischer♦
Jul 17 at 15:17
1
@LucaGhidelli, it can be shown to change sign infinitely often, using the tricks in Chapter 15 of Montgomery and Vaughan.
– Peter Humphries
Jul 20 at 9:35
I will be happy to learn it, thank you for the reference! @Peter Humphries
– Luca Ghidelli
Jul 20 at 9:46
I learnt something, thank you! Lemma 15.1 applied to the function 1/s(s-1)zeta(s) = int_0^infty [A(x)-M(x)/x]x^-s (no need to do better) does the job perfectly, because it is analytic for real s>0 but not on all the semiplane Re s>0 ! Thank you, would you like to write it as an answer?
– Luca Ghidelli
Jul 20 at 19:43
add a comment |Â
up vote
1
down vote
It is false. Using PARI/GP I found:
$M(18798)=0$
$A(18798)*18798 =-0.3589042...$
answered Jul 17 at 14:59
Luca Ghidelli
295111
Bonus question: does the difference change sign infinitely often?
– Luca Ghidelli
Jul 17 at 15:00
I would expect so, but I would not be surprised if that is still unknown.
– Daniel Fischer♦
Jul 17 at 15:17
1
@LucaGhidelli, it can be shown to change sign infinitely often, using the tricks in Chapter 15 of Montgomery and Vaughan.
– Peter Humphries
Jul 20 at 9:35
I will be happy to learn it, thank you for the reference! @Peter Humphries
– Luca Ghidelli
Jul 20 at 9:46
I learnt something, thank you! Lemma 15.1 applied to the function 1/s(s-1)zeta(s) = int_0^infty [A(x)-M(x)/x]x^-s (no need to do better) does the job perfectly, because it is analytic for real s>0 but not on all the semiplane Re s>0 ! Thank you, would you like to write it as an answer?
– Luca Ghidelli
Jul 20 at 19:43
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is false. Using PARI/GP I found:
$M(18798)=0$
$A(18798)*18798 =-0.3589042...$
answered Jul 17 at 14:59
Luca Ghidelli
295111
It is false. Using PARI/GP I found:
$M(18798)=0$
$A(18798)*18798 =-0.3589042...$
answered Jul 17 at 14:59
Luca Ghidelli
295111
answered Jul 17 at 14:59
Luca Ghidelli
295111
answered Jul 17 at 14:59
Luca Ghidelli
295111
answered Jul 17 at 14:59
Luca Ghidelli
295111
295111
Bonus question: does the difference change sign infinitely often?
– Luca Ghidelli
Jul 17 at 15:00
I would expect so, but I would not be surprised if that is still unknown.
– Daniel Fischer♦
Jul 17 at 15:17
1
@LucaGhidelli, it can be shown to change sign infinitely often, using the tricks in Chapter 15 of Montgomery and Vaughan.
– Peter Humphries
Jul 20 at 9:35
I will be happy to learn it, thank you for the reference! @Peter Humphries
– Luca Ghidelli
Jul 20 at 9:46
I learnt something, thank you! Lemma 15.1 applied to the function 1/s(s-1)zeta(s) = int_0^infty [A(x)-M(x)/x]x^-s (no need to do better) does the job perfectly, because it is analytic for real s>0 but not on all the semiplane Re s>0 ! Thank you, would you like to write it as an answer?
– Luca Ghidelli
Jul 20 at 19:43
add a comment |Â
Bonus question: does the difference change sign infinitely often?
– Luca Ghidelli
Jul 17 at 15:00
I would expect so, but I would not be surprised if that is still unknown.
– Daniel Fischer♦
Jul 17 at 15:17
1
@LucaGhidelli, it can be shown to change sign infinitely often, using the tricks in Chapter 15 of Montgomery and Vaughan.
– Peter Humphries
Jul 20 at 9:35
I will be happy to learn it, thank you for the reference! @Peter Humphries
– Luca Ghidelli
Jul 20 at 9:46
I learnt something, thank you! Lemma 15.1 applied to the function 1/s(s-1)zeta(s) = int_0^infty [A(x)-M(x)/x]x^-s (no need to do better) does the job perfectly, because it is analytic for real s>0 but not on all the semiplane Re s>0 ! Thank you, would you like to write it as an answer?
– Luca Ghidelli
Jul 20 at 19:43
Bonus question: does the difference change sign infinitely often?
– Luca Ghidelli
Jul 17 at 15:00
Bonus question: does the difference change sign infinitely often?
– Luca Ghidelli
Jul 17 at 15:00
I would expect so, but I would not be surprised if that is still unknown.
– Daniel Fischer♦
Jul 17 at 15:17
I would expect so, but I would not be surprised if that is still unknown.
– Daniel Fischer♦
Jul 17 at 15:17
1
1
@LucaGhidelli, it can be shown to change sign infinitely often, using the tricks in Chapter 15 of Montgomery and Vaughan.
– Peter Humphries
Jul 20 at 9:35
@LucaGhidelli, it can be shown to change sign infinitely often, using the tricks in Chapter 15 of Montgomery and Vaughan.
– Peter Humphries
Jul 20 at 9:35
I will be happy to learn it, thank you for the reference! @Peter Humphries
– Luca Ghidelli
Jul 20 at 9:46
I will be happy to learn it, thank you for the reference! @Peter Humphries
– Luca Ghidelli
Jul 20 at 9:46
I learnt something, thank you! Lemma 15.1 applied to the function 1/s(s-1)zeta(s) = int_0^infty [A(x)-M(x)/x]x^-s (no need to do better) does the job perfectly, because it is analytic for real s>0 but not on all the semiplane Re s>0 ! Thank you, would you like to write it as an answer?
– Luca Ghidelli
Jul 20 at 19:43
I learnt something, thank you! Lemma 15.1 applied to the function 1/s(s-1)zeta(s) = int_0^infty [A(x)-M(x)/x]x^-s (no need to do better) does the job perfectly, because it is analytic for real s>0 but not on all the semiplane Re s>0 ! Thank you, would you like to write it as an answer?
– Luca Ghidelli
Jul 20 at 19:43
add a comment |Â
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2
That proofwiki article looks really sloppy to me and I would be a little dubious of trusting it for anything.
– Steven Stadnicki
Jul 17 at 1:25
In any case it raises an interesting question in the spirit of "things that (don't?) change sign infinitely often".
– Luca Ghidelli
Jul 17 at 1:46