Primal-Trigo Summation tan

Geometry Level 3

∑ n = 1 2015 tan ⁡ 5 ( n π 403 ) = ? \large{\sum _{ n=1 }^{ 2015 }{ \tan ^{ 5 }{ \left( \frac { n\pi }{ 403 } \right) } } =?}


The answer is 0.

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Department 8 by Department 8 , 27, Israel

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1 solution

Otto Bretscher
Oct 25, 2015

By symmetry tan ⁡ ( 5 π − t ) = tan ⁡ ( π − t ) = − tan ⁡ ( t ) \tan(5\pi-t)=\tan(\pi-t)=-\tan(t) , the summands with indices n n and 2015 − n 2015-n will cancel out, so that ∑ n = 1 2014 tan ⁡ 5 ( n π 403 ) = 0 \sum_{n=1}^{2014}\tan^5\left(\frac{n\pi}{403}\right)=0 . Adding all the way to n = 2015 n=2015 makes the sum tan ⁡ 5 ( 5 π ) = 0 \tan^5(5\pi)=\boxed{0} still.

(I must confess that I just copied my solution from here , mutatis mutandis.)

4 Helpful 0 Interesting 0 Brilliant 0 Confused

thats really easy. i took t a n ( x ) = − i e 2 i x − 1 e 2 i x + 1 tan(x)=-i\dfrac{e^{2ix}-1}{e^{2ix}+1} and used roots of unity to solve.

Aareyan Manzoor - 5 years, 7 months ago

Wow! I learned a new term: mutatis mutandis. Thanks Comrade Otto.

Pi Han Goh - 5 years, 4 months ago

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