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

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Homework Help Overview

The discussion revolves around a problem from the review section of the Stewart Calculus text, specifically focusing on the recursive definition of functions f_n(x) where f_0(x) = x^2 and f_0(f_n(x)) = f_{n+1}(x). Participants are attempting to derive a general expression for f_n(x).

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between f_n(x) and f_{n+1}(x), with one suggesting to start from the equation (f_n(x))^2 = f_{n+1}(x). Others provide specific calculations for f1(x), f2(x), and f3(x) based on the recursive definition.

Discussion Status

The discussion includes various attempts to derive a formula for f_n(x), with some participants expressing uncertainty about their results. One participant notes a potential discrepancy with the expected outcome, indicating a need for further verification of the problem's requirements.

Contextual Notes

There is mention of needing to check the problem's requirements to ensure that the derived expressions align with the initial conditions provided, particularly regarding f_0(x).

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
[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]
 
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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
 
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 [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.
 
Yes it does. Never mind.

Thanks Viet Dao!
 

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