Difference between revisions of "2021 WSMO Speed Round/Problem 1"

(Solution (bash))
(Solution (bash))
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<cmath>=f(f(f(f(f((1-1)^2)))))=f(f(f(f(f(0)))))=f(f(f(f((0-1)^2))))=f(f(f(f(1))))=f(f(f((1-1)^2)))</cmath>
 
<cmath>=f(f(f(f(f((1-1)^2)))))=f(f(f(f(f(0)))))=f(f(f(f((0-1)^2))))=f(f(f(f(1))))=f(f(f((1-1)^2)))</cmath>
 
<cmath>=f(f(f(0)))=f(f((0-1)^2))=f(f(1))=f((1-1)^2)=f(0)=(0-1)^2=\boxed{1}.</cmath>
 
<cmath>=f(f(f(0)))=f(f((0-1)^2))=f(f(1))=f((1-1)^2)=f(0)=(0-1)^2=\boxed{1}.</cmath>
 +
 
  ~pinkpig
 
  ~pinkpig

Revision as of 16:35, 22 December 2021

Problem

Let $f^1(x)=(x-1)^2$, and let $f^n(x)=f^1(f^{n-1}(x))$. Find the value of $|f^7(2)|$.

Solution (bash)

Note that \[|f^7(2)|=f(f(f(f(f(f(f(2)))))))=f(f(f(f(f(f((2-1)^2))))))=f(f(f(f(f(f(1))))))\] \[=f(f(f(f(f((1-1)^2)))))=f(f(f(f(f(0)))))=f(f(f(f((0-1)^2))))=f(f(f(f(1))))=f(f(f((1-1)^2)))\] \[=f(f(f(0)))=f(f((0-1)^2))=f(f(1))=f((1-1)^2)=f(0)=(0-1)^2=\boxed{1}.\]

~pinkpig