How to solve recurrence relation with one real root and two complex roots ?

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SUMMARY

The discussion focuses on solving the recurrence relation defined by the equation a_n - a_{n-1} - a_{n-3} = 0, with initial conditions a_1 = 1, a_2 = 1, and a_3 = 2. The characteristic equation derived is r^3 - r^2 - 1 = 0, which leads to one real root and two complex roots. A key insight provided is that the recurrence can be simplified to a_n = -a_{n-2}, allowing for the calculation of subsequent terms in the sequence. This approach reveals a clear pattern in the values of a_n, facilitating further analysis.

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How to solve recurrence relation with one real root and two complex roots ?

The Example is ;

Solve the recurrence relation a n-1 + a n-3 = 0 where n ≥ 3 and a 1 = 1 a 2 = 1 a 3 = 2
a n = nth order
a n-1 = (n-1)th order.
a n-3 = (n-3)th order.

I've started the solving ;
a n = r^n
so
the equation will be ;

r^n - r^(n-1) - r^(n-3)
r^3 - r^2 - 1 = 0
I could'nt do anything after find the roots ?
What should i do ?
Thanks for helping.
 
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I think you are making this harder than it is. Hint:

you can rewrite you recurrence relation as

<br /> a_{n} = - a_{n-2}.<br />

So what is a_4? How about a_5? See the pattern?

jason
 

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