Some functions are their own inverse.
For example: the function f(x)=1/x has f(f(x))=x
There are also functions that when iterated thrice become the identity.
For example: The function f(x)=1/(1-x) has f(f(x))=(x-1)/x and f(f(f(x)))=f3(x)=x.
Are there functions for other values of n (or perhaps all values of n) where fn(x)=x but fm(x)≠x for any m<n?
If f(x) = x+1 when floor(x) mod 5 ~= 4
= x-4 when floor(x) mod 5 = 4, then
for example f(0.3) = 1.3
f(1.3) = 2.3
f(2.3) = 3.3
f(3.3) = 4.3
f(4.3) = 0.3
and
f(5.3) = 6.3
f(6.3) = 7.3
f(7.3) = 8.3
f(8.3) = 9.3
f(9.3) = 5.3
etc.
The mod 5 could be any mod n, and the = or ~=4 would be = or ~= n-1.
The function could be stretched vertically or horizontally with appropriate tweaks.
The hard case would be to come up with a continuous function. Even 1/(1-x) is not continuous and is not defined at x=0.
Fixed typo: = to - at top.
Edited on April 10, 2022, 12:44 pm
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Posted by Charlie
on 2022-04-10 06:21:47 |
a set of solutions
