Difference between revisions of "Catalan's constant using Legendre chi"
From specialfunctionswiki
| (One intermediate revision by the same user not shown) | |||
| Line 1: | Line 1: | ||
| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
$$K=-i\chi_2(i),$$ | $$K=-i\chi_2(i),$$ | ||
where $K$ is [[Catalan's constant]] and $\chi$ denotes the [[Legendre chi]] function. | where $K$ is [[Catalan's constant]] and $\chi$ denotes the [[Legendre chi]] function. | ||
| − | + | ||
| − | + | ==Proof== | |
| − | + | Recall, by definition, | |
| − | + | $$K=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{(2k+1)^2},$$ | |
| + | and | ||
| + | $$\chi_2(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{z^{2k+1}}{(2k+1)^{\nu}}.$$ | ||
| + | Since $i^{2k+1}=i^{2k}i=(-1)^ki$, plugging in $i$ into $\chi_2$ yields | ||
| + | $$\chi_2(i) = \displaystyle\sum_{k=0}^{\infty} \dfrac{i^{2k+1}}{(2k+1)^{\nu}}=i \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{(2k+1)^{\nu}}=iK.$$ | ||
| + | Hence, | ||
| + | $$-i\chi_2(i)=-i(iK)=-i^2K=-(-1)K=K,$$ | ||
| + | completing the proof. $\blacksquare$ | ||
| + | |||
| + | ==References== | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
Latest revision as of 15:46, 25 February 2018
Theorem
The following formula holds: $$K=-i\chi_2(i),$$ where $K$ is Catalan's constant and $\chi$ denotes the Legendre chi function.
Proof
Recall, by definition, $$K=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{(2k+1)^2},$$ and $$\chi_2(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{z^{2k+1}}{(2k+1)^{\nu}}.$$ Since $i^{2k+1}=i^{2k}i=(-1)^ki$, plugging in $i$ into $\chi_2$ yields $$\chi_2(i) = \displaystyle\sum_{k=0}^{\infty} \dfrac{i^{2k+1}}{(2k+1)^{\nu}}=i \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{(2k+1)^{\nu}}=iK.$$ Hence, $$-i\chi_2(i)=-i(iK)=-i^2K=-(-1)K=K,$$ completing the proof. $\blacksquare$
