Sequence

Algebra Level 5

The sequences { a n } n = 1 \{a_n\}_{n=1}^\infty and { b n } n = 1 \{b_n\}_{n=1}^\infty are defined through a 1 = α a_1=\alpha , b 1 = β b_1=\beta , a n + 1 = α a n β b n a_{n+1}=\alpha a_n-\beta b_n , and b n + 1 = β a n + α b n b_{n+1}=\beta a_n+\alpha b_n for all n 2 n\ge2 . How many pairs of real numbers ( α , β ) (\alpha,\beta) are there such that a 2014 = b 1 a_{2014}=b_1 and b 2014 = a 1 b_{2014}=a_1 ?


The answer is 2016.

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2 solutions

Jack D'Aurizio
Apr 14, 2014

By setting z n = a n + i b n z_n = a_n + i\cdot b_n we can see that a n = ( z n ) , b n = ( z n ) a_n = \Re(z_n), b_n=\Im(z_n) and z n = ( α + i β ) n z_n = (\alpha+i\cdot \beta)^n . In order to have z 2014 = β + i α z_{2014}=\beta+i\cdot\alpha we must have z 1 = 0 z_1 = 0 or z 1 = e i θ z_1 = e^{i\theta} . In the second case, 2014 θ + 2 k π = π 2 θ 2014\theta +2k\pi = \frac{\pi}{2}-\theta must hold, so θ = 2 π m 2015 + π 4030 \theta = \frac{2\pi m}{2015}+\frac{\pi}{4030} with m [ 1 , 2015 ] m\in[1,2015] . In conclusion, we have 2016 2016 solutions.

Cody Johnson
Apr 13, 2014

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