Difference between revisions of "Liouville lambda"

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m (Tom moved page Liouville function to Liouville lambda)
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=Properties=
 
=Properties=
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<strong>Theorem:</strong> For every $n \geq 1$
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$$\displaystyle\sum_{d | n} \lambda(d) = \left\{ \begin{array}{ll}
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1 &; n \mathrm{\hspace{2pt}is \hspace{2pt} a \hspace{2pt} square} \\
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0 &; \mathrm{otherwise},
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\end{array} \right.$$
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where $d | n$ denotes that the sum is over all [[divisor|divisors]] $d$ of $n$.
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<strong>Proof:</strong> █
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<strong>Theorem:</strong> The following formula holds:
 
<strong>Theorem:</strong> The following formula holds:
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where $\zeta$ denotes the [[Riemann zeta function]].
 
where $\zeta$ denotes the [[Riemann zeta function]].
 
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<strong>Proof:</strong> proof goes here █  
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<strong>Proof:</strong> █  
 
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Revision as of 04:10, 30 April 2015

The Liouville function is defined by the formula $$\lambda(n) = (-1)^{\Omega(n)},$$ where $\Omega(n)$ indicates the number of prime factors of $n$, counted with multiplicity.

Properties

Theorem: For every $n \geq 1$ $$\displaystyle\sum_{d | n} \lambda(d) = \left\{ \begin{array}{ll} 1 &; n \mathrm{\hspace{2pt}is \hspace{2pt} a \hspace{2pt} square} \\ 0 &; \mathrm{otherwise}, \end{array} \right.$$ where $d | n$ denotes that the sum is over all divisors $d$ of $n$.

Proof:

Theorem: The following formula holds: $$\dfrac{\zeta(2s)}{\zeta(s)}=\displaystyle\sum_{n=1}^{\infty} \dfrac{\lambda(n)}{n^s},$$ where $\zeta$ denotes the Riemann zeta function.

Proof: