Let x 0 , x 1 , x 2 , … be a sequence of real numbers satisfying the recursion,
x n = x n − 1 + x n + 1
for n = 1 , 2 , 3 , … .
If x 0 = 2 3 , what is x 3 0 0 3 ?
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From the linear recurrent relation x n = x n − 1 + x n + 1 , the characteristic equation is as follows:
r 2 − r + 1 ⟹ r = 0 = 2 1 ± 3 i = e ± 3 π i the 6th root of unity
⟹ a n a 0 = a 1 e 3 n π i + a 2 e − 3 n π i = a 1 + a 2
⟹ a 1 + a 2 = 2 3
⟹ a 3 0 0 3 = a 1 e 3 3 0 0 3 π i + a 2 e − 3 3 0 0 3 π i = a 1 e 1 0 0 1 π i + a 2 e − 1 0 0 1 π i = a 1 e π i + a 2 e − π i = − a 1 − a 2 = − 2 3
Nice approach! :)
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If we let x 1 = y , then
And the sequence repeats with a period of 6 .
So, in general, for n divisible by 3 , x n = ( − 1 ) n / 3 × 2 3
So, x 3 0 0 3 = − 2 3