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Some other kind of root extraction (Posted on 2006-11-07) Difficulty: 4 of 5
Let's call a real-valued continuous function an n-th identity root when it generates the identity function after exactly n compositions with itself. For example f(x)=-x is a 2nd identity root because f(f(x))=x.

1. The function f(x)=1-1/x is a 3rd identity root. Unfortunately it is undefined at x=0. Are there identity roots for n>2 which are defined for all real numbers?

2. For a given real number c and n>1, give an example of an n-th identity root which is defined for all real numbers except c. How many such roots exist?

See The Solution Submitted by JLo    
Rating: 4.2500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Curious | Comment 9 of 10 |
(In reply to re: Curious by Richard)

So you thought of projective mappings of the line of real numbers, represented them as 2x2-matrices, took a subset of these that can be considered as complex numbers under matrix multiplication and finally you chose those matrices that represent complex unity roots. Following all backwards would give you the formula. Very neat!!! Thanks for this nice solution! I agree there is a strong connection between your construction and mine, although I must admit that yours is certainly much more elegant and "math-like".
  Posted by JLo on 2006-11-10 08:36:32

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