Can you use Dirichlet Kernel again?

Calculus Level 4

The Cesàro sum of ∑ n = 0 ∞ cos ⁡ n \displaystyle \sum_{n=0}^\infty \cos n is equal to 1 2 \dfrac12 .

Is the Cesàro sum ∑ n = 0 ∞ sin ⁡ n \displaystyle \sum_{n=0}^\infty \sin n also equal to 1 2 \dfrac12 ?

No Yes

Pi Han Goh by Pi Han Goh , 30, Malaysia

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1 solution

Tom Engelsman
Apr 11, 2017

Using the Eulerian form for s i n ( x ) = e i x − e − i x 2 i sin(x) = \frac{e^{ix} - e^{-ix}}{2i} , we now obtain:

∑ n = 0 ∞ sin ⁡ n = ∑ n = 0 ∞ e i n − e − i n 2 i = 1 2 i ⋅ [ 1 1 − e i − 1 1 − e − i ] ; \sum_{n=0}^\infty \sin n = \sum_{n=0}^\infty \frac{e^{in} - e^{-in}}{2i} = \frac{1}{2i} \cdot [ \frac{1}{1-e^{i}} - \frac{1}{1-e^{-i}}];

or 1 2 i ⋅ ( 1 − e − i ) − ( 1 − e i ) ( 1 − e i ) ( 1 − e − i ) ; \frac{1}{2i} \cdot \frac{(1 - e^{-i}) - (1 - e^{i})}{(1 - e^{i})(1 - e^{-i})};

or [ e i − e − i 2 i ] [ 1 2 − e i − e − i ] ; [\frac{e^{i} - e^{-i}}{2i}][\frac{1}{2 - e^{i} - e^{-i}}];

or s i n ( 1 ) 2 − 2 c o s ( 1 ) ; \frac{sin(1)}{2 - 2cos(1)};

or 1 2 ⋅ s i n ( 1 ) 1 − c o s ( 1 ) ; \frac{1}{2} \cdot \frac{sin(1)}{1 - cos(1)};

or 1 2 ⋅ c o t ( 1 2 ) ≠ 1 2 . \frac{1}{2} \cdot cot(\frac{1}{2}) \neq \frac{1}{2}.

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