Difference between revisions of "2021 WSMO Speed Round/Problem 1"
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Let <math>f^1(x)=(x-1)^2</math>, and let <math>f^n(x)=f^1(f^{n-1}(x))</math>. Find the value of <math>|f^7(2)|</math>. | Let <math>f^1(x)=(x-1)^2</math>, and let <math>f^n(x)=f^1(f^{n-1}(x))</math>. Find the value of <math>|f^7(2)|</math>. | ||
| − | ==Solution | + | <math>\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24</math> |
| + | |||
| + | ==Solution 1== | ||
Note that | Note that | ||
<cmath>|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))))))</cmath> | <cmath>|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))))))</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(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 | ||
Revision as of 17:39, 22 December 2021
Problem
Let 
, and let 
. Find the value of 
.
Solution 1
Note that
![\[|f^7(2)|=f(f(f(f(f(f(f(2)))))))=f(f(f(f(f(f((2-1)^2))))))=f(f(f(f(f(f(1))))))\]](http://latex.artofproblemsolving.com/3/3/d/33d9637d2533ffb876684aa542996fda25518c30.png)
![\[=f(f(f(f(f((1-1)^2)))))=f(f(f(f(f(0)))))=f(f(f(f((0-1)^2))))=f(f(f(f(1))))=f(f(f((1-1)^2)))\]](http://latex.artofproblemsolving.com/1/b/5/1b5fa488afbec394528cda2caed09fec8730b4ed.png)
![\[=f(f(f(0)))=f(f((0-1)^2))=f(f(1))=f((1-1)^2)=f(0)=(0-1)^2=\boxed{1}.\]](http://latex.artofproblemsolving.com/d/7/a/d7a4ba936b0fc5c7024af9f646b9b8adb31c971f.png)
~pinkpig
