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A question about double zeta values

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Post Mon May 08, 2017 4:31 am

Posts: 105
By some simple calculations, we can deduce the following results
\begin{align*}
&\zeta \left( {\bar 2,2} \right) + 2\zeta \left( {\bar 3,1} \right) = \frac{5}{{16}}\zeta \left( 4 \right), \\
&\zeta \left( {\bar 4,2} \right) + 4\zeta \left( {\bar 5,1} \right) = \frac{1}{4}\zeta \left( {4,2} \right) - \frac{{37}}{{64}}\zeta \left( 6 \right) + \frac{1}{2}{\zeta ^2}\left( 3 \right), \\
&\zeta \left( {\bar 6,2} \right) + 6\zeta \left( {\bar 7,1} \right) = - \frac{9}{8}\zeta \left( {6,2} \right) - \frac{{1887}}{{256}}\zeta \left( 8 \right) + 6\zeta \left( 3 \right)\zeta \left( 5 \right).
\end{align*}
Based on the above three results, I conjecture that
\begin{align*}
\zeta \left( {\bar 8,2} \right) + 8\zeta \left( {\bar 9,1} \right) =& {a_1}\zeta \left( {8,2} \right) + {a_2}\zeta \left( 2 \right)\zeta \left( {6,2} \right) + {a_3}\zeta \left( {10} \right) + {a_4}{\zeta ^2}\left( 5 \right) \\
&+ {a_5}\zeta \left( 3 \right)\zeta \left( 7 \right) + {a_6}\zeta \left( 2 \right)\zeta \left( 3 \right)\zeta \left( 5 \right) + {a_7}{\zeta ^2}\left( 3 \right)\zeta \left( 4 \right)\quad ?.
\end{align*}
Here the coefficients $a_i\ (i=1,2,3,4,5,6,7)$ are rational numbers.

In general, the combined alternating mzvs
\[\zeta \left( {{\overline {2m}},2} \right) + 2m\zeta \left( {{\overline {2m + 1}},1} \right)=?\quad (m\in \mathbb{N}).\]
It can be expressed in terms of non-alternating mzvs and zeta values?

Here the double zeta values are defined by
\begin{align*}
\zeta \left( {p,q} \right): = \sum\limits_{{n_1} > {n_2} > 0} {\frac{1}{{n_1^pn_2^q}}} ,\zeta \left( {\bar p,q} \right): = \sum\limits_{{n_1} > {n_2} > 0} {\frac{{{{\left( { - 1} \right)}^{{n_1}}}}}{{n_1^pn_2^q}}} \quad (p,q\in \mathbb{N}).
\end{align*}

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