Functional Square Root

[tommy1729]

On the topic of function composition. Is the successor function the only function f(x) such that f(f(x)) = f(x)+1?

Let f(x) = v(x) + x 

then f(x) + 1 = f(f(x)) = f( v(x) + x ) = v( v(x) + x ) + v(x) + x = v(x) + x + 1.

so v( v(x) + x ) = 1. since this implies that v is independant from x , we get that v(x) is a constant function. 

Hence v(x) = 1. 
And thus f(x) = x + 1.

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notice that equation

g(x+1) = g(x) + 1 has solution g(x) = x + 1-periodic(x).

this is cruxial for superfunctions !

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also fun :

h(x+1) = h(h(x))

let t(x) be the inverse of h :

t h(x+1) = t h(h(x))

x + 1 = h(x).


I could give more similar looking equations and solve them ... but why ?


regards

tommy1729