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E. Wirsing, "Das asymptotische verhalten von summen über multiplikative funktionen. II."
E. Wirsing, "Das asymptotische verhalten von summen über multiplikative funktionen. II."
Acta Mathematica Academiae Scientiarum Hungaricae
Acta Mathematica Academiae Scientiarum Hungaricae
Tomus 18 (3-4), 1978, pp. 411-467.
Tomus 18 (3-4), 1967, pp. 411-467.


English Translation by: Google Translator
English Translation by: Google Translator
Line 11: Line 11:
multiplicative functions <math> \lambda</math> essentially under the condition
multiplicative functions <math> \lambda</math> essentially under the condition


<math> (1.1)  \sum_{p\leq x}\lambda(p)\log(p)\sim\tau x \mbox{ (p prime)}</math>
:<math> (1.1)  \sum_{p\leq x}\lambda(p)\log(p)\sim\tau x \mbox{ (p prime)}</math>


: Determine
Determine:


<math> (1.2) \sum_{n\leq x}\lambda(n)\sim\frac{e^{-ct}}{\Gamma(\tau)}\frac{x}{\log x}\prod_{p\leq x}\left(1+\frac{\lambda(p)}{p}+\frac{\lambda(p^{2})}{p^{2}}+\cdots\right)</math>
:<math> (1.2) \sum_{n\leq x}\lambda(n)\sim\frac{e^{-ct}}{\Gamma(\tau)}\frac{x}{\log x}\prod_{p\leq x}\left(1+\frac{\lambda(p)}{p}+\frac{\lambda(p^{2})}{p^{2}}+\cdots\right)</math>


(<math> c</math> is the Euler-) constant. Special rates are the same type Delange [3]. The same result (1.2) is here under the much weaker assumption
(<math> c</math> is the Euler-) constant. Special rates are the same type Delange [3]. The same result (1.2) is here under the much weaker assumption


<math> (1.3) \sum_{p\leq x}\lambda(p)\frac{\log p}{p}\sim\tau\log x</math>
:<math> (1.3) \sum_{p\leq x}\lambda(p)\frac{\log p}{p}\sim\tau\log x</math>


However, with the additional. Call <math> \lambda(p)= O(1)</math> and only for <math>\tau>0</math> are shown (Theorem 1.1).
However, with the additional. Call <math> \lambda(p)= O(1)</math> and only for <math>\tau>0</math> are shown (Theorem 1.1).
The terms of <math> \lambda(p^{v}) (v\geq2)</math> are thieves than I, but we want them in the introduction
The terms of <math> \lambda(p^{v}) (v\geq2)</math> are thieves than I, but we want them in the introduction
. neglect The same result for complex-function <math> \lambda</math>, we get only if <math> \lambda</math>
. neglect The same result for complex-function <math> \lambda</math>, we get only if <math> \lambda</math>
by <math> |\lambda|</math> not significantly different, namely, if
by <math> |\lambda|</math> not significantly different, namely, if <math>\sum\frac{1}{p}(|\lambda(p)-Re\lambda(p)|)</math> converges (Theorem 1.1.1). The special case <math>\tau=1,|\lambda|\leq1,\sum\frac{1}{p}(1-\lambda(p))</math> is convergent was proved by Delange [4] easier to by Rényi [8].
 
An interesting counterpart to give Erdos and Rényi [7]: convergence <math>\sum\frac{1}{p}(1-\lambda(p)),\sum\frac{1}{p^{2}}\lambda(p)^{2}</math> and <math>\sum_{p}\sum_{v\geq2}\frac{1}{p^{v}}\lambda(p^{v})</math> and for each <math>\epsilon>0</math>
 
:<math>\liminf_{x\to\infty}\sum_{x<p\leq(1+\epsilon)x}\lambda(p)\frac{\log p}{p}>0,</math>
 
then (1.2) (with <math>\tau=1</math>). Here <math>\lambda</math> will be restricted to the bottom.
 
If <math>\lambda^{*}</math> is another multiplicative function <math>|\lambda^{*}|\leq\lambda</math>, so we could in I with the conditions (1.1) and
 
:<math>(1.4)  \sum_{p\leq x}\lambda^{*}(p)\log p\sim\tau^{*}x</math>
 
n is the sum <math>\sum_{n\leq x}\lambda^{*}(n)</math> up to <math>o(\sum_{n\leq x}\lambda(n))</math> identify and, in particular
 
:<math>\sum_{n\leq x}\lambda^{*}(n)=o(\sum_{n\leq x}\lambda(n))</math>
 
show if <math>\sum_{p}\frac{1}{p}(\lambda(p)-Re\lambda^{*}(p))</math> diverges. In the event <math>\lambda=1</math> see Delange [3]. In the present paper we obtain such results without (1.4), some of which (1.1), some of them already with (1.3), if the range of values of <math>\lambda^{*}</math> is suitably restricted. In particular, we prove (Theorem 1.2.2): Do (1.3), <math>\lambda(p)=O (1)</math>, <math>|\lambda^{*}|\leq\lambda</math> is <math>\lambda^{*}</math> real-valued, the average of <math>\lambda^{*}</math> exists regarding <math>\lambda</math>:
 
:<math>(1.5)  \lim_{x\to\infty}(\sum_{n\leq x}\lambda^{*}(n))(\sum_{n\leq x}\lambda(n))^{-1}</math>
 
and has the value
 
:<math>(1.6)  \lim_{x\to\infty}\prod_{p\leq x}\left(1+\frac{\lambda^{*}(p)}{p}+\frac{\lambda^{*}(p^2)}{p^2}+\cdots\right)\left(1+\frac{\lambda(p)}{p}+\frac{\lambda(p^2)}{p^2}+\cdots\right)^{-1}.</math>
 
If one specifically for <math>\lambda</math> is the constant 1 and allowed <math>\lambda^{*}</math> only values <math>\pm1</math>, then this is the solution of a known problem of Wintner [9] (there) with a supposed proof, see Erdos [6].
 
For complex-valued functions <math>\lambda^{*}</math>, the situation is more complicated. The example <math>\lambda(n)=1,\lambda^{*}(n)=n^{i}</math> shows that the existence of (1.6), the foltg of (1.5) does not if only (1.1), <math>\lambda(p)=O(1)</math> and <math>|\lambda^{*}|\leq\lambda</math> requires. It is then that is <math>\sum_{n\leq x}\lambda^{*}(n)\sim x^{1+i}(1+i)^{-1}</math> while (1.6) has the value 0. But if <math>|\lambda^{*}|\leq\lambda</math> vershärft to the following claim:
 
:<math>\lambda^{*}(n)=\epsilon(n)\lambda(n),|\epsilon(n)|\leq1.</math>
 
There are a number <math>e^{i\phi}</math> of magnitude 1, which is "not an accumulation point of the sequence <math>(\epsilon(p))</math>, then (Theorem 1.2.1 follows):
 
:<math>(1.7) (\sum_{n\leq x}\lambda^{*}(n))(\sum_{n\leq x}\lambda(n))^{-1}\to0\mbox{, resp.}\sim\prod_{p\leq x}\left(1+\frac{\lambda^{*}(p)}{p}+\cdots\right)\left(1+\frac{\lambda(p)}{p}+\cdots\right)^{-1},</math>
 
depending on the product does not tend to 0 or.


...
...

Latest revision as of 16:50, 2 February 2010

E. Wirsing, "Das asymptotische verhalten von summen über multiplikative funktionen. II." Acta Mathematica Academiae Scientiarum Hungaricae Tomus 18 (3-4), 1967, pp. 411-467.

English Translation by: Google Translator



In I we have the asymptotic behavior of the sum [math]\displaystyle{ \sum_{n \leq x} \lambda (n) }[/math] for nonnegative multiplicative functions [math]\displaystyle{ \lambda }[/math] essentially under the condition

[math]\displaystyle{ (1.1) \sum_{p\leq x}\lambda(p)\log(p)\sim\tau x \mbox{ (p prime)} }[/math]

Determine:

[math]\displaystyle{ (1.2) \sum_{n\leq x}\lambda(n)\sim\frac{e^{-ct}}{\Gamma(\tau)}\frac{x}{\log x}\prod_{p\leq x}\left(1+\frac{\lambda(p)}{p}+\frac{\lambda(p^{2})}{p^{2}}+\cdots\right) }[/math]

([math]\displaystyle{ c }[/math] is the Euler-) constant. Special rates are the same type Delange [3]. The same result (1.2) is here under the much weaker assumption

[math]\displaystyle{ (1.3) \sum_{p\leq x}\lambda(p)\frac{\log p}{p}\sim\tau\log x }[/math]

However, with the additional. Call [math]\displaystyle{ \lambda(p)= O(1) }[/math] and only for [math]\displaystyle{ \tau\gt 0 }[/math] are shown (Theorem 1.1). The terms of [math]\displaystyle{ \lambda(p^{v}) (v\geq2) }[/math] are thieves than I, but we want them in the introduction . neglect The same result for complex-function [math]\displaystyle{ \lambda }[/math], we get only if [math]\displaystyle{ \lambda }[/math] by [math]\displaystyle{ |\lambda| }[/math] not significantly different, namely, if [math]\displaystyle{ \sum\frac{1}{p}(|\lambda(p)-Re\lambda(p)|) }[/math] converges (Theorem 1.1.1). The special case [math]\displaystyle{ \tau=1,|\lambda|\leq1,\sum\frac{1}{p}(1-\lambda(p)) }[/math] is convergent was proved by Delange [4] easier to by Rényi [8].

An interesting counterpart to give Erdos and Rényi [7]: convergence [math]\displaystyle{ \sum\frac{1}{p}(1-\lambda(p)),\sum\frac{1}{p^{2}}\lambda(p)^{2} }[/math] and [math]\displaystyle{ \sum_{p}\sum_{v\geq2}\frac{1}{p^{v}}\lambda(p^{v}) }[/math] and for each [math]\displaystyle{ \epsilon\gt 0 }[/math]

[math]\displaystyle{ \liminf_{x\to\infty}\sum_{x\lt p\leq(1+\epsilon)x}\lambda(p)\frac{\log p}{p}\gt 0, }[/math]

then (1.2) (with [math]\displaystyle{ \tau=1 }[/math]). Here [math]\displaystyle{ \lambda }[/math] will be restricted to the bottom.

If [math]\displaystyle{ \lambda^{*} }[/math] is another multiplicative function [math]\displaystyle{ |\lambda^{*}|\leq\lambda }[/math], so we could in I with the conditions (1.1) and

[math]\displaystyle{ (1.4) \sum_{p\leq x}\lambda^{*}(p)\log p\sim\tau^{*}x }[/math]

n is the sum [math]\displaystyle{ \sum_{n\leq x}\lambda^{*}(n) }[/math] up to [math]\displaystyle{ o(\sum_{n\leq x}\lambda(n)) }[/math] identify and, in particular

[math]\displaystyle{ \sum_{n\leq x}\lambda^{*}(n)=o(\sum_{n\leq x}\lambda(n)) }[/math]

show if [math]\displaystyle{ \sum_{p}\frac{1}{p}(\lambda(p)-Re\lambda^{*}(p)) }[/math] diverges. In the event [math]\displaystyle{ \lambda=1 }[/math] see Delange [3]. In the present paper we obtain such results without (1.4), some of which (1.1), some of them already with (1.3), if the range of values of [math]\displaystyle{ \lambda^{*} }[/math] is suitably restricted. In particular, we prove (Theorem 1.2.2): Do (1.3), [math]\displaystyle{ \lambda(p)=O (1) }[/math], [math]\displaystyle{ |\lambda^{*}|\leq\lambda }[/math] is [math]\displaystyle{ \lambda^{*} }[/math] real-valued, the average of [math]\displaystyle{ \lambda^{*} }[/math] exists regarding [math]\displaystyle{ \lambda }[/math]:

[math]\displaystyle{ (1.5) \lim_{x\to\infty}(\sum_{n\leq x}\lambda^{*}(n))(\sum_{n\leq x}\lambda(n))^{-1} }[/math]

and has the value

[math]\displaystyle{ (1.6) \lim_{x\to\infty}\prod_{p\leq x}\left(1+\frac{\lambda^{*}(p)}{p}+\frac{\lambda^{*}(p^2)}{p^2}+\cdots\right)\left(1+\frac{\lambda(p)}{p}+\frac{\lambda(p^2)}{p^2}+\cdots\right)^{-1}. }[/math]

If one specifically for [math]\displaystyle{ \lambda }[/math] is the constant 1 and allowed [math]\displaystyle{ \lambda^{*} }[/math] only values [math]\displaystyle{ \pm1 }[/math], then this is the solution of a known problem of Wintner [9] (there) with a supposed proof, see Erdos [6].

For complex-valued functions [math]\displaystyle{ \lambda^{*} }[/math], the situation is more complicated. The example [math]\displaystyle{ \lambda(n)=1,\lambda^{*}(n)=n^{i} }[/math] shows that the existence of (1.6), the foltg of (1.5) does not if only (1.1), [math]\displaystyle{ \lambda(p)=O(1) }[/math] and [math]\displaystyle{ |\lambda^{*}|\leq\lambda }[/math] requires. It is then that is [math]\displaystyle{ \sum_{n\leq x}\lambda^{*}(n)\sim x^{1+i}(1+i)^{-1} }[/math] while (1.6) has the value 0. But if [math]\displaystyle{ |\lambda^{*}|\leq\lambda }[/math] vershärft to the following claim:

[math]\displaystyle{ \lambda^{*}(n)=\epsilon(n)\lambda(n),|\epsilon(n)|\leq1. }[/math]

There are a number [math]\displaystyle{ e^{i\phi} }[/math] of magnitude 1, which is "not an accumulation point of the sequence [math]\displaystyle{ (\epsilon(p)) }[/math], then (Theorem 1.2.1 follows):

[math]\displaystyle{ (1.7) (\sum_{n\leq x}\lambda^{*}(n))(\sum_{n\leq x}\lambda(n))^{-1}\to0\mbox{, resp.}\sim\prod_{p\leq x}\left(1+\frac{\lambda^{*}(p)}{p}+\cdots\right)\left(1+\frac{\lambda(p)}{p}+\cdots\right)^{-1}, }[/math]

depending on the product does not tend to 0 or.

...