I have the following sequence: 1 , 2 cos &#x2061;<!-- ⁡ --> <mrow class="MJX-Te

Jackson Duncan

Jackson Duncan

Answered question

2022-06-10

I have the following sequence:
1 , 2 cos ( x ) ,   1 + 2 cos ( 2 x ) ,   2 cos ( x ) + 2 cos ( 3 x ) .
I need to find the n th term, considering the 1 s t term as 1, 2 n d as 2 cos ( x ) and so on.
We can represent it by a recurrence function as:
f ( n ) = f ( n 2 ) + 2 cos ( ( n 1 ) x ) , f ( 1 ) = 1 , f ( 2 ) = 2 cos ( x )
So what is the non-recursive version of the function, or how to approach to find one?

Answer & Explanation

Samantha Reid

Samantha Reid

Beginner2022-06-11Added 22 answers

For odd n = 2 k + 1 , we have
f n = f 1 + j = 1 k ( f 2 j + 1 f 2 j 1 ) = 1 + 2 j = 1 k cos ( 2 j x ) .
For even n = 2 k , we have
f n = f 2 + j = 2 k ( f 2 j f 2 j 2 ) = 2 j = 1 k cos ( ( 2 j 1 ) x ) .

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