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For the following fi denotes the i-fold composition of a function f with itself.
1. No, there is not. We prove this by contradiction and assume f is such a function. First note that f must be a bijection, because y=fn is. It follows that f must be strictly monotonically increasing or decreasing. We can rule out that f is increasing as follows: Consider a x_1 with x_2:=f(x_1)<>x_1 ("<>" stands for "unequal"). Say x_2>x_1, then by applying f we get x_3:=f(x_2)>f(x_1)=x_2>x_1. By induction it follows that x_i:=fi(x_1)>x_1 for all i and we see that f is no identity root after all. If x_2<x_1 it follows similarly that fi(x_1)<x_1 for all i. So now we know that f, if it exists, must be strictly monotonically decreasing. Then the function g(x):=f(f(x)) is monotonically increasing. But its n-fold self-composition is also the identity function. From what we have just proven, this is only possible if g itself is the identity function, i.e. g(x)=x for all x. But this means f is a 2nd identity root, which contradicts n>2.
2. Assume c=0 first. If we consider the unit circle, rather than the (straight line of) real numbers as domain for our function, there is an obvious n-th identity root, namely the rotation by 2*Pi/n. Let's denote this rotation by D. Then if p is a bijection from the real numbers (to which we add a number that we call "infinity", otherwise no continuous bijection is possible) to the unit circle, then f=p-1Dp is an n-th root defined on the real numbers. It is undefined at f-1(infinity).
We choose the unit circle with mid point (0,1). It is easy to find the bijection
p(x):=[u(x),v(x)]:=[sin(2arctan(x+tan(Pi/2-Pi/n))),1-cos(2arctan(x+tan(Pi/2-Pi/n)))].
The term tan(Pi/2-Pi/n) is only to ensure we end up with a pole in 0. After applying the rotation we get
D(p(x))=[sin(2arctan(x+2Pi/n+tan(Pi/2-Pi/n))),1-cos(2arctan(x+2Pi/n+tan(Pi/2-Pi/n)))]
Finally we get
f(x)=p-1(D(p(x)))=(1-cos(2arctan(x+2Pi/n+tan(Pi/2-Pi/n))))/sin(2arctan(x+2Pi/n+tan(Pi/2-Pi/n)))
which is along the lines of Joel's 2nd solution.
Simplification yields
f(x)=-1/(x sin(Pi/n)^2)-2*cot(Pi/n)
which looks more similar to Richard's solution. To shift the pole from 0 to c, a simple transformation t(x):=x-c and g(x):=t-1(D(t(x))) gets us the final solution
g(x)=-1/((x-c)sin(Pi/n)^2)-2*cot(Pi/n)+c
Example: For n=4 and c=1 we get g(x)=(1+x)/(1-x)
There are many, many such roots: If g is a solution to the problem and if t is any bijection of the real numbers (i.e. strictlty monotonic decreasing or increasing) with t(c)=c, then t-1Dt is also a solution. E.g. for t(x)=x^3 and the 3rd root g(x)=1-1/x we get another 3rd root sqrt_3(x3-1)/x. Because there are so many such t's, we get many 3rd identity roots. In terms of set cardinality, there are as many such roots as there are continuous functions. This is no bigger than the set of real numbers, but still pretty big. |
