Here is the problem. A nonconstant rational function over the real numbers(a rational function is a

hovudverkocym6

hovudverkocym6

Answered question

2022-04-10

Here is the problem.
A nonconstant rational function over the real numbers(a rational function is a function that can be expressed as p ( x ) q ( x ) , with p ( x ), q ( x ) as polynomial function) f ( x ), is defined such that:
( f ( x ) ) 2 a = f ( x 2 ) for all x ( a is a constant value)
Prove that f ( x ) must be of the form x k for some constant k.
I have an idea of how to prove it, but I can't do so rigorously. I can prove that any polynomial with more than 2 terms, or monomials, would not work. But I can't prove it won't work for any rational function.

Answer & Explanation

Madalynn Acosta

Madalynn Acosta

Beginner2022-04-11Added 15 answers

If q ( z ) is nonconstant, then q ( z 2 ) has at least one complex zero that q ( z ) doesn't have. (Choose, for example, the zero z 1 of q ( z 2 ) with minimal nonzero argument.) This means that q ( z ) must be constant, for otherwise the right-hand side is undefined at a point where the left-hand side is defined. This reduces to the case where f is a polynomial, which you have (almost?) solved.
Edit: wnoise points out that this argument fails if q ( z ) is a power of z. Perhaps the argument for polynomials extends to Laurent polynomials (where negative powers of z are allowed)?

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