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sine_function [2015/07/03 15:50] nikolaj |
sine_function [2015/07/03 15:50] nikolaj |
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$\sum_{n=a}^{b}\sin(2kn)=\dfrac{\sin (k (a-b-1)) \sin (k (a+b))} {\sin(k)}$ | $\sum_{n=a}^{b}\sin(2kn)=\dfrac{\sin (k (a-b-1)) \sin (k (a+b))} {\sin(k)}$ | ||
- | But that's really just coming from $\sum_{n=a}^{b}{\mathrm e}^{2kn}=\sum_{n=a}^{b}\left{\mathrm e}^{2k}\right)^n=\dots$ | + | But that's really just coming from $\sum_{n=a}^{b}{\mathrm e}^{2kn}=\sum_{n=a}^{b}\left({\mathrm e}^{2k}\right)^n=\dots$ |
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=== Context === | === Context === |