Difference between revisions of "Ramanujan tau"

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$$\displaystyle\sum_{n=1}^{\infty} \tau(n)q^n = q \prod_{n=1}^{\infty} (1-q^n)^{24} = \eta(z)^{24}=\Delta(z),$$
 
$$\displaystyle\sum_{n=1}^{\infty} \tau(n)q^n = q \prod_{n=1}^{\infty} (1-q^n)^{24} = \eta(z)^{24}=\Delta(z),$$
 
where $q=e^{2\pi i z}$ with $\mathrm{Re}(z)>0$, $\eta$ denotes the [[Dedekind eta function]], and $\Delta$ denotes the [[discriminant modular form]].
 
where $q=e^{2\pi i z}$ with $\mathrm{Re}(z)>0$, $\eta$ denotes the [[Dedekind eta function]], and $\Delta$ denotes the [[discriminant modular form]].
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File:Ramanujantau.png|Plot of $\tau(n)$ for $n=0,1,\ldots,250$.
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=Properties=
 
=Properties=
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[[Ramanujan tau is multiplicative]]<br />
<strong>Theorem:</strong> $\tau(mn)=\tau(m)\tau(n)$ if [[Greatest common divisor|$\gcd$]]$(m,n)=1$
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[[Ramanujan tau of a power of a prime]]<br />
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[[Ramanujan tau inequality]]<br />
<strong>Proof:</strong>
 
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=References=
<strong>Theorem:</strong> $\tau(p^{r+1})=\tau(p)\tau(p^r)-p^{11}\tau(p^{r-1})$ whenever $p$ is prime and $r>0$
 
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<strong>Proof:</strong> █
 
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[[Category:SpecialFunction]]
<strong>Theorem:</strong> $|tau(p)| \leq 2p^{\frac{11}{2}}$ for all primes $p$
 
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<strong>Proof:</strong> █
 
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Latest revision as of 00:53, 23 December 2016

The Ramanujan tau function $\tau \colon \mathbb{N} \rightarrow \mathbb{Z}$ is defined by the formulas $$\displaystyle\sum_{n=1}^{\infty} \tau(n)q^n = q \prod_{n=1}^{\infty} (1-q^n)^{24} = \eta(z)^{24}=\Delta(z),$$ where $q=e^{2\pi i z}$ with $\mathrm{Re}(z)>0$, $\eta$ denotes the Dedekind eta function, and $\Delta$ denotes the discriminant modular form.

Properties

Ramanujan tau is multiplicative
Ramanujan tau of a power of a prime
Ramanujan tau inequality

References