Difference between revisions of "Catalan's constant using Legendre chi"

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==Theorem==
<strong>[[Catalan's constant using Legendre chi]]:</strong> The following formula holds:
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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.
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<strong>Proof:</strong> █
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==Proof==
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Recall, by definition,
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$$K=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{(2k+1)^2},$$
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and
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$$\chi_2(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{z^{2k+1}}{(2k+1)^{\nu}}.$$
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Since $i^{2k+1}=i^{2k}i=(-1)^ki$, plugging in $i$ into $\chi_2$ yields
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$$\chi_2(i) = \displaystyle\sum_{k=0}^{\infty} \dfrac{i^{2k+1}}{(2k+1)^{\nu}}=i \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{(2k+1)^{\nu}}=iK.$$
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Hence,
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$$-i\chi_2(i)=-i(iK)=-i^2K=-(-1)K=K,$$
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completing the proof. $\blacksquare$
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==References==
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[[Category:Theorem]]
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[[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. $\blacksquare$

References