How to solve for f_n(x) in Stewart Calculus text's review section?

In summary, the conversation discusses the problem of solving for f_n(x) using the given equations involving f_0(x) and f_{n+1}(x). The conversation also includes attempts at solving the problem and finding the pattern for f_n(x), with the final conclusion that the solution is f_n(x) = x^{2^{n+1}}, which satisfies the given requirements.
  • #1
Jeff Ford
155
2
I'm going through the review section of the Stewart Calculus text and I'm stuck on this problem.

Given
[itex]
f_0(x) = x^2
[/itex]
and
[itex]
f_0(f_n(x)) = f_{n+1}(x) , n = 0,1,2...
[/itex]
how do you solve for
[itex]
f_n(x)
[/itex]
 
Last edited:
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  • #2
The only place I can think to start is making [itex] (f_n(x))^2 = f_{n+1}(x) [/itex], but that's as far as I got
 
  • #3
Now, f1(x) = f0(f0(x)) = f0(x2) = x4
f2(x) = f0(f1(x)) = f0(x4) = x8
f3(x) = f0(f2(x)) = ...
f4(x) = f0(f3(x)) = ...
So what's fn(x)?
Can you go from here?
Viet Dao,
 
  • #4
From that I get [itex] f_n(x) = x^{2^{n+1}} [/itex] but that doesn't work for [itex] f_0(x) = x^2 [/itex]. That's the same answer the book has, so maybe I wrote the requirements down wrong. I'll have to check when I get home if this had to work for 0.
 
  • #5
Yes it does. Never mind.

Thanks Viet Dao!
 

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