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

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To solve for f_n(x) in the Stewart Calculus review section, the relationship f_0(f_n(x)) = f_{n+1}(x) is established, starting with f_0(x) = x^2. The discussion reveals that f_1(x) = x^4, f_2(x) = x^8, and continues with a pattern suggesting f_n(x) = x^{2^{n+1}}. However, there is confusion about whether this holds true for f_0(x) = x^2, leading to a realization that the initial conditions may have been misinterpreted. Ultimately, the conclusion is that the derived formula needs to be consistent with the base case.
Jeff Ford
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I'm going through the review section of the Stewart Calculus text and I'm stuck on this problem.

Given
<br /> f_0(x) = x^2<br />
and
<br /> f_0(f_n(x)) = f_{n+1}(x) , n = 0,1,2...<br />
how do you solve for
<br /> f_n(x)<br />
 
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The only place I can think to start is making (f_n(x))^2 = f_{n+1}(x), but that's as far as I got
 
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,
 
From that I get f_n(x) = x^{2^{n+1}} but that doesn't work for f_0(x) = x^2. 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.
 
Yes it does. Never mind.

Thanks Viet Dao!
 
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