Difference between revisions of "2001 SMT/Algebra Problems/Problem 5"
(Created page with "==Problem== Let <math>f \colon \mathbb{N} \to \mathbb{N}</math> be defined by <math>f(n) =\begin{cases}2 & \text{if }x = 0, \\(f(x-1))^2 & \text{if }x \neq 0 \end{cases}</ma...") |
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==Solution== | ==Solution== | ||
| − | Notice that each time we increase <math>x</math> by <math>1</math> starting fro <math>0</math>, we square some power of <math>2</math>. By playing around with the recursion, we can come up with an explicit formula <math>f(x)=2^{2^x}</math>. Then <math>f(11)=2^{2^11}=2^2048</math>, so <math>\log_2f(11)=\boxed{2048}</math>. | + | Notice that each time we increase <math>x</math> by <math>1</math> starting fro <math>0</math>, we square some power of <math>2</math>. By playing around with the recursion, we can come up with an explicit formula <math>f(x)=2^{2^x}</math>. Then <math>f(11)=2^{2^{11}}=2^{2048}</math>, so <math>\log_2f(11)=\boxed{2048}</math>. |
~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406] | ~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406] | ||
Latest revision as of 21:02, 19 March 2025
Problem
Let 
be defined by 
. What is 
?
Solution
Notice that each time we increase 
by 
starting fro 
, we square some power of 
. By playing around with the recursion, we can come up with an explicit formula 
. Then 
, so 
.
