Dilogarithm of Reciprocal of Golden Mean

From ProofWiki
Jump to navigation Jump to search

Theorem

$\map {\Li_2} {\dfrac 1 \phi} = \dfrac 3 5 \map \zeta 2 - \paren {\map \ln \phi}^2$


where:

$\map {\Li_2} x$ is the dilogarithm function of $x$
$\map \zeta 2$ is the Riemann $\zeta$ function of $2$
$\phi$ denotes the golden mean.


Proof

We first note the following:

\(\ds -\frac 1 \phi\) \(=\) \(\ds 1 - \phi\) Reciprocal Form of One Minus Golden Mean
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \frac 1 {\phi^2}\) \(=\) \(\ds 1 - \dfrac 1 \phi\) dividing through by $-\phi$ and rearranging


We now note:

\(\ds \map {\Li_2} z + \map {\Li_2} {-z}\) \(=\) \(\ds \dfrac 1 2 \map {\Li_2} {z^2}\) Dilogarithm of Square
\(\ds \leadsto \ \ \) \(\ds \map {\Li_2} {\frac 1 \phi} + \map {\Li_2} {-\frac 1 \phi}\) \(=\) \(\ds \dfrac 1 2 \map {\Li_2} {\frac 1 {\phi^2} }\) setting $z := \dfrac 1 \phi$
\(\ds \leadsto \ \ \) \(\ds \map {\Li_2} {\dfrac 1 \phi}\) \(=\) \(\ds \dfrac 1 2 \map {\Li_2} {1 - \dfrac 1 \phi} - \map {\Li_2} {-\frac 1 \phi}\) substituting $\paren {1}$ above
\(\ds \) \(=\) \(\ds \dfrac 1 2 \paren {\dfrac 2 5 \map \zeta 2 - \paren {\map \ln \phi}^2} - \paren {-\dfrac 2 5 \map \zeta 2 + \dfrac 1 2 \paren {\map \ln \phi}^2}\) Dilogarithm of One Minus Reciprocal of Golden Mean and Dilogarithm of Minus Reciprocal of Golden Mean
\(\ds \leadsto \ \ \) \(\ds \map {\Li_2} {\dfrac 1 \phi}\) \(=\) \(\ds \dfrac 3 5 \map \zeta 2 - \paren {\map \ln \phi}^2\)

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Dilogarithm_of_Reciprocal_of_Golden_Mean&oldid=678563"