Suppose f ( x ) = q ( x ) p ( x ) , where p and q are polynomials with real coefficients, f is defined for all real numbers x = 1 , and f ( f ( f ( x ) ) ) = x , for all real x for which f ( f ( f ( x ) ) ) is defined.
The smallest possible positive value of f ( 0 ) for such a function f can be written as b a , where a and b are coprime positive integers. What is the value of a + b ?
Details and assumptions
You are not given if f ( x ) is defined at x = 1 , or at any complex, non-real value.
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This solution provides insight as to how complex analysis can help us understand polynomials. The richer structure allows us to obtain much more information about these functions.
Nice solution. I used the idea of extending f meromorphically onto the Riemann sphere, but wasn't experienced enough in complex analysis to find the step that proves degree(q) < 2.
We begin by proving some preliminary lemmas about the form of our function f ( x ) . Our goal will be to show that both p ( x ) and q ( x ) have degree at most 1 (assuming, of course, that the polynomials p ( x ) and q ( x ) are relatively prime).
Lemma 1 : If f ( f ( f ( x ) ) ) = x for all real x for which f ( f ( f ( x ) ) ) is defined, then f ( f ( f ( x ) ) ) = x for all complex values of x for which f ( f ( f ( x ) ) ) is defined.
Proof : Note that f ( f ( f ( x ) ) ) is a rational function of x , so we can write f ( f ( f ( x ) ) = Q ( x ) P ( x ) for some P ( x ) and Q ( x ) . Then, the statement f ( f ( f ( x ) ) ) = x becomes the polynomial identity P ( x ) = x Q ( x ) . Since this is true for infinitely many real values (i.e. all real values x where f ( f ( f ( x ) ) ) is defined), it must be true for all complex values x .
Lemma 2 : p ( x ) must have degree 0 or 1.
Proof : Assume that p ( x ) has degree at least 2. Then we claim that there is some value λ such that p ( x ) = λ q ( x ) has at least two distinct roots which are not roots of q ( x ) . To see this, first note that since p ( x ) has degree at least 2, p ( x ) − λ q ( x ) will also almost always have degree at least 2 (with the possible exception of when λ times the leading term of q ( x ) equals the leading term of p ( x ) ), so p ( x ) − λ q ( x ) will almost always have at least two roots. Now, if r was a root both of p ( x ) − λ q ( x ) and of q ( x ) , it would also be a root of p ( x ) ; but this implies x − r divides both p ( x ) and q ( x ) , contradicting that p ( x ) and q ( x ) are relatively prime. Therefore, it also follows that these two roots will not be roots of q ( x ) . Finally, to make sure that these roots are distinct, we can simply perturb λ slightly if they aren't.
Call these two roots r 1 and r 2 . Note that q ( r 1 ) p ( r 1 ) = q ( r 2 ) p ( r 2 ) = λ , so in particular, f ( f ( f ( r 1 ) ) ) = f ( f ( f ( r 2 ) ) ) = f ( f ( λ ) ) . But since f ( f ( f ( x ) ) ) = x , this implies r 1 = r 2 , a contradiction. Therefore our initial assumption was wrong, and p ( x ) has degree at most 1.
Lemma 3 : q ( x ) must have degree 0 or 1.
Proof : Similar to the proof of lemma 2 above (we again consider roots of p ( x ) = λ q ( x ) and use the lack of injectivity to our favor).
We have now therefore shown that we can write f ( x ) as
f ( x ) = c x + d a x + b
These are known as Mobius transformations and importantly, compositions of Mobius functions are given by multiplication of the corresponding 2-by-2 matrices (this is straightforward to check). Therefore, in order for f ( f ( f ( x ) ) ) = x , we must have:
[ a c b d ] 3 = [ 1 0 0 1 ]
Equivalently, all eigenvalues of this matrix (which we will henceforth call M ) must be cube roots of unity. In order for M to have real entries, its eigenvalues can either both be 1 (giving rise to the identity matrix and the function f ( x ) = x ) or they can be e ± 2 π i / 3 . In this case, the characteristic polynomial of our matrix will be x 2 + x + 1 , so we must have t r ( M ) = − 1 and det ( M ) = 1 . This gives rise to the system
a + d = − 1 a d − b c = 1
Moreover, since we know f ( x ) is defined everywhere except for x = 1 , we know that c = − d . Substituting this and a = − ( d + 1 ) into the second equation, we have that
b = − d d 2 + d + 1
Finally, f ( 0 ) = d b , which equals
d b = d 2 d 2 + d + 1
This is minimized at when d = − 2 , where d b = 4 3 . Substituting back, we see this minimum is attained by the function
f ( x ) = 4 ( x − 1 ) 2 x − 3
Our desired answer is therefore 3 + 4 = 7 .
This proof is also correct
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The function f ( z ) = p ( z ) / q ( z ) extends to a meromorphic function on the complex numbers with 1 removed, or to a well defined analytic function on the extended (with ∞ ) complex numbers. Since f ( f ( f ( x ) ) ) = x for all real x = 1 , we deduce that f ( f ( f ( z ) ) ) = z for all z . Thus f ( z ) is a bijection of the extended complex plane.
Suppose that either p ( z ) or q ( z ) has degree greater than 1 . If p ( z ) and q ( z ) have distinct degrees, choose a nonzero complex number t . If p ( z ) and q ( z ) have the same degree, choose a complex number t such that a − t b = 0 where a , b are the leading coefficients of p ( z ) , q ( z ) respectively. Then p ( z ) − t q ( z ) is a polynomial of degree at least 2 . If there are two distinct roots of the equation p ( z ) − t q ( z ) = 0 , then there are two distinct solutions of the equation f ( z ) = t in the extended complex plane, and hence f is not a bijection. Thus all the roots of p ( z ) − t q ( z ) = 0 must be the same. If this root is w , then p ( z ) = t q ( z ) + a ( z − w ) n for some nonzero complex number a and integer n ≥ 2 . But then f ′ ( w ) = 0 , which again contradicts the bijectivity of f .
Thus we deduce that both p ( z ) and q ( z ) have degree no greater than 1 . Since f ( x ) is defined for all real numbers except 1 , q ( x ) is either constant or a multiple of x − 1 .
If q ( x ) is constant, then f ( x ) = c x + d for real constants c , d . Then f ( f ( f ( x ) ) ) = c 3 x + ( c 2 + c + 1 ) d and hence c = 1 , d = 0 , so that f ( z ) = x . In this case f ( 0 ) = 0 is not positive.
If q ( x ) is not constant, then we deduce that f ( z ) = z − 1 c z − d c = d for real constants c , d . But f ( 1 ) = ∞ , f ( ∞ ) = c , and hence we must have f ( c ) = 1 , so we deduce that c 2 − d = c − 1 , or d = c 2 − c + 1 . The requirement that c = d tells us that c = 1 . We see that f ( f ( z ) ) = z − c z − d f ( f ( f ( z ) ) ) = z as required. But now f ( 0 ) = d = c 2 − c + 1 = ( c − 1 / 2 ) 2 + 3 / 4 ≥ 3 / 4 , giving us a + b = 7 .