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--- means_._note [2015/06/20 17:03]
nikolaj
+++ means_._note [2015/06/20 17:05]
nikolaj
@@ Line 14: / Line 14: @@
 == Note ==
-I use this in the context of [[Minus twelve . Note]]. If $f(k):= k\,z^k$, then for $z\in(0,1)$, we find
+I use this in the context of [[Minus twelve . Note]]. For $z\in(0,1)$, we find
-$\sum_{k=0}^\infty \left(k\,z^k-\langle f\rangle_{[k,k+1]}\right)=\dfrac{z}{(z-1)^2}-\dfrac{1}{\ln(z)^2}=-\dfrac{1}{12}+{\mathcal O}\left((z-1)^1\right)$
+$\sum_{k=0}^\infty \langle q\mapsto q\,z^q\rangle_{[k,k+1]}=\dfrac{1}{\ln(z)^2}$,
-See also [[Natural logarithm of complex numbers]].
+i.e. (see [[Natural logarithm of complex numbers]])
+$\sum_{k=0}^\infty \left(k\,z^k-\langle q\mapsto q\,z^q\rangle_{[k,k+1]}\right)=\dfrac{z}{(z-1)^2}-\dfrac{1}{\ln(z)^2}=-\dfrac{1}{12}+{\mathcal O}\left((z-1)^1\right)$
 -----
 === Requirements ===
 [[Function integral]]