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means_._note [2015/06/20 17:03] nikolaj |
means_._note [2015/06/20 17:04] nikolaj |
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| I use this in the context of [[Minus twelve . Note]]. For $z\in(0,1)$, we find | I use this in the context of [[Minus twelve . Note]]. For $z\in(0,1)$, we find | ||
| - | $\sum_{k=0}^\infty \left(k\,z^k-\langle k\mapsto k\,z^k\rangle_{[k,k+1]}\right)=\dfrac{z}{(z-1)^2}-\dfrac{1}{\ln(z)^2}=-\dfrac{1}{12}+{\mathcal O}\left((z-1)^1\right)$ | + | $\sum_{k=0}^\infty \langle q\mapsto q\,z^q\rangle_{[k,k+1]}=\dfrac{1}{\ln(z)^2}$, |
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| + | i.e. | ||
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| + | $\sum_{k=0}^\infty \left(k\,z^k-\langle q\mapsto q\,z^q\rangle_{[k,k+1]}\right)=\dfrac{z}{(z-1)^2}-\dfrac{1}{\ln(z)^2}=-\dfrac{1}{12}+{\mathcal O}\left((z-1)^1\right)$ | ||
| See also [[Natural logarithm of complex numbers]]. | See also [[Natural logarithm of complex numbers]]. | ||
