Cool Functional Equation
f ( f ( x ) ) = f ( x ) + x
Let f : R → R be a continuous function such that the above equation is satisfied for all x ∈ R , and for all x > 0 , we have f ( x ) > 0 .
If f ( 1 ) = 2 0 1 4 , find f ( 2 0 1 5 ) .
This is a original problem.
The answer is 4029.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
Moderator note:
The reason why we have a unique solution under the given conditions, is because the equation x 2 − x − 1 = 0 has a unique positive real root. Can you fill in the rest of the details for uniqueness?
You have not demonstrated that such a function exists.
If such a function exists, then yes the value must be equal to 4029. But if no such function exists, then there is no value.
Log in to reply
Sir, it is given in the question that such a function exists, and so, we get 4029 as the answer, but yes, if there does not exist such a function, then obviously, we ha no solution!!
Log in to reply
The problem was edited since I made my comment.
In the previous version, there were certain conditions given, which made this problem much more interesting, because we could actually prove uniqueness of the functional equation.
Log in to reply
@Calvin Lin – Yes sir I have edited this after your comment
The reason why we have a unique solution under the given conditions, is because the equation x 2 − x − 1 = 0 has a unique positive real root. Can you fill in the rest of the details for uniqueness?
Log in to reply
Yes sir, I think this we can get by defining a recursion as taking a n as the n th iterate of f, and then writing its characteristic equation.
Log in to reply
Great! Actually, after working out the details, the solution isn't unique as I previously thought.
To elaborate, set a n = f n ( 1 ) . We have a n + 2 = a n + 1 + a n . By the theory of linear recurrence relations , we solve the characteristic equation of x 2 − x − 1 = 0 , which has roots α = 2 1 + 5 , β = 2 1 − 5 . Then, we may set a n = A α n + B β n .
In order to get the uniqueness of the solution, I would have needed β < 0 < α and ∣ β ∣ > ∣ α ∣ , in order to conclude that for a large enough n , we must have B = 0 to satisfy the given positive condition.
Since, f ( f ( x ) = f ( x ) + x
Replacing x by f ( x ) we get,
f ( f ( x ) + x ) =2 f ( x ) + x .
Substituting x =1 , f ( f ( 1 ) + 1 ) =2 f ( 1 ) +1 .
Substituting value of f ( 1 ) =2015 ,
f ( 2 0 1 5 ) =2 f ( 1 ) +1
i . e . f ( 2 0 1 5 ) = 4 0 2 9
It is hard to follow what you are saying. It is best to show your working step by step.
IE By substituting in x = f ( x ) , we get f ( f ( f ( x ) ) = f ( f ( x ) ) + f ( x ) . Next, since f ( f ( x ) ) = f ( x ) + x , we get that f ( f ( x ) + x ) = 2 f ( x ) + x .
You have not demonstrated that such a function exists.
If such a function exists, then yes the value must be equal to 4029. But if no such function exists, then there is no value.
Since f(f(x)) = f(x) + x and f(1) = 2014, then f(f(1)) = f(1) + 1 = 2014 + 1 = 2015. From this f(f(1)) = f(2014) = 2015. Then f(f(2014)) = f(2015) = f(2014) + 2014 = 2015 + 2014 = 4029, so f(2015) = 4029