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sine_function [2015/07/03 15:50] nikolaj |
sine_function [2015/07/03 15:51] nikolaj |
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| == Theorem == | == Theorem == | ||
| + | From | ||
| + | $\sum_{n=a}^{b}{\mathrm e}^{2kn}=\sum_{n=a}^{b}\left({\mathrm e}^{2k}\right)^n=\dots$ | ||
| + | we get | ||
| $\sum_{n=a}^{b}\sin(2kn)=\dfrac{\sin (k (a-b-1)) \sin (k (a+b))} {\sin(k)}$ | $\sum_{n=a}^{b}\sin(2kn)=\dfrac{\sin (k (a-b-1)) \sin (k (a+b))} {\sin(k)}$ | ||
| - | But that's really just coming from $\sum_{n=a}^{b}{\mathrm e}^{2kn}=\sum_{n=a}^{b}\left{\mathrm e}^{2k}\right)^n=\dots$ | ||
| ----- | ----- | ||
| === Context === | === Context === | ||
