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Let α be the positive root of the equation 8 x 3 − 6 x − 1 = 0 and f ( x ) be a function defined on real numbers satisfying f ( x + 1 ) + f ( x − 1 ) = 2 α f ( x ) for all real x .
Find the period of f ( x ) .
The answer is 18.
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1 solution
Isn't this solution incomplete sir?
Since you have although shown that there exists a function with such a property and that has period 18, but doesn't it needs to be shown that all functions which satisfy the functional equation must have period 18?
Or either you show that this is the only function possible or it could be shown that all functions satisfying the functional equation have the same period.
And even for this solution, it needs to be shown why the substitution x = cos θ is allowed (although that is obvious).
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For any 0 ≤ x < 1 define the sequence u x , n = f ( x + n ) n ∈ Z Then we have the recurrence relation u x , n + 1 + u x , n − 1 = 2 α u x , n n ∈ Z for any 0 ≤ x < 1 . This recurrence relation is easy to solve, and we obtain that u x , n = A x cos ( 9 1 π n + ε x ) n ∈ Z for some constants A x , ε x . Thus we deduce that f ( n + x ) = A x cos ( 9 1 π ( n + x ) + η x ) n ∈ Z for any 0 ≤ x < 1 , where η x = ε x − 9 1 π x .
Thus the general solution of this functional equation involves choosing two functions A : [ 0 , 1 ) → R and η : [ 0 , 1 ) → R . All these functions have period 1 8 .
The substitution x = cos θ is "allowed" because it gives the answer. It is justified by the outcome.
Writing x = cos θ , the equation 8 x 3 − 6 x − 1 = 0 becomes 2 cos 3 θ − 1 = 0 , and hence α = cos 9 1 π . A function with the required property is therefore f ( x ) = cos ( 9 1 π x ) x ∈ R and this function f has period 1 8 .