Wirsing translation - Revision history

Smh82 at 00:50, 3 February 2010

2010-02-03T00:50:56Z

← Older revision		Revision as of 17:50, 2 February 2010
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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

Smh82 at 00:38, 3 February 2010

2010-02-03T00:38:44Z

← Older revision		Revision as of 17:38, 2 February 2010
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	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):		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{ ~~bzw~~.}\sim\prod_{p\leq x}\left(1+\frac{\lambda^{}(p)}{p}+\cdots\right)\left(1+\frac{\lambda(p)}{p}+\cdots\right)^{-1},</math>		:<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.		depending on the product does not tend to 0 or.

	...		...

Smh82 at 00:34, 3 February 2010

2010-02-03T00:34:59Z

← Older revision		Revision as of 17:34, 2 February 2010
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	:<math>\sum_{n\leq x}\lambda^{*}(n)=o(\sum_{n\leq x}\lambda(n))</math>		:<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>:		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>		:<math>(1.5) \lim_{x\to\infty}(\sum_{n\leq x}\lambda^{*}(n))(\sum_{n\leq x}\lambda(n))^{-1}</math>

Smh82 at 00:34, 3 February 2010

2010-02-03T00:34:08Z

← Older revision		Revision as of 17:34, 2 February 2010
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	:<math>\sum_{n\leq x}\lambda^{*}(n)=o(\sum_{n\leq x}\lambda(n))</math>		:<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>:		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>		:<math>(1.5) \lim_{x\to\infty}(\sum_{n\leq x}\lambda^{*}(n))(\sum_{n\leq x}\lambda(n))^{-1}</math>

Smh82 at 00:26, 3 February 2010

2010-02-03T00:26:19Z

Smh82 at 00:22, 3 February 2010

2010-02-03T00:22:11Z

← Older revision		Revision as of 17:22, 2 February 2010
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	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].		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:		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>		<math>\lambda^{*}(n)=\epsilon(n)\lambda(n),\|\epsilon(n)\|\leq1.</math>

Smh82 at 00:20, 3 February 2010

2010-02-03T00:20:09Z

← Older revision		Revision as of 17:20, 2 February 2010
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	<math>\sum_{n\leq x}\lambda^{*}(n)=o(\sum_{n\leq x}\lambda(n))</math>		<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>:		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>		<math>(1.5) \lim_{x\to\infty}(\sum_{n\leq x}\lambda^{*}(n))(\sum_{n\leq x}\lambda(n))^{-1}</math>

Smh82 at 00:14, 3 February 2010

2010-02-03T00:14:53Z

← Older revision		Revision as of 17:14, 2 February 2010
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	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):		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{ bzw.}\sim\prod_{p\leq x}\left(1+\frac{\lambda^{}(p)}{p}+\cdots\right\left(1+\frac{\lambda(p)}{p}+\cdots\right)^{-1},</math>		<math>(1.7) (\sum_{n\leq x}\lambda^{}(n))(\sum_{n\leq x}\lambda(n))^{-1}\to0\mbox{ bzw.}\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.		depending on the product does not tend to 0 or.

	...		...

Smh82 at 00:13, 3 February 2010

2010-02-03T00:13:59Z

← Older revision		Revision as of 17:13, 2 February 2010
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	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].		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)\~~simx~~^{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:		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>		<math>\lambda^{*}(n)=\epsilon(n)\lambda(n),\|\epsilon(n)\|\leq1.</math>

Smh82 at 00:13, 3 February 2010

2010-02-03T00:13:14Z

@@ Line 24: / Line 24: @@
 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>
-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].
 ...

← Older revision		Revision as of 17:26, 2 February 2010
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	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).
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	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>		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>		:<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.		then (1.2) (with <math>\tau=1</math>). Here <math>\lambda</math> will be restricted to the bottom.
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	If <math>\lambda^{}</math> is another multiplicative function <math>\|\lambda^{}\|\leq\lambda</math>, so we could in I with the conditions (1.1) and		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>		:<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		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>		:<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>:		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>		:<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		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>		:<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].		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].
Line 52:		Line 52:
	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:		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>		:<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):		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{ bzw.}\sim\prod_{p\leq x}\left(1+\frac{\lambda^{}(p)}{p}+\cdots\right)\left(1+\frac{\lambda(p)}{p}+\cdots\right)^{-1},</math>		:<math>(1.7) (\sum_{n\leq x}\lambda^{}(n))(\sum_{n\leq x}\lambda(n))^{-1}\to0\mbox{ bzw.}\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.		depending on the product does not tend to 0 or.

	...		...