What is the analytical method for finding the composition of a function?

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Homework Help Overview

The discussion revolves around finding an analytical method for determining the composition of a function, specifically the function f(x) = x^2 + x, applied multiple times. Participants explore whether there are alternative methods to the tedious process of manually substituting the function into itself repeatedly.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants suggest that manually plugging in the function is straightforward but tedious, while others question if there are more efficient analytical methods available. There is also mention of using a Computer Algebra System (CAS) for easier computation.

Discussion Status

The conversation is ongoing, with participants sharing their thoughts on the method of substitution and the implications of the function's quadratic nature. Some express skepticism about the feasibility of finding a simpler analytical approach, while others acknowledge the potential for discovering patterns in the compositions.

Contextual Notes

Participants note that the problem is labeled as a CAS problem, implying that it may be designed to be tedious when solved by hand. There is also a recognition of the increasing degree of the function with each composition, which raises questions about the nature of the resulting expressions.

Feldoh
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Homework Statement


Well its not really a problem, I'm just trying to see it there's a different method to solve the problem. My textbooks gives a CAS problem:

Find [tex]f(f(f(f(f(f(x))))))[/tex] if [tex]f(x)= x^2+x[/tex]

Homework Equations


None (or atleast that I know of)

The Attempt at a Solution


Since it's a problem that's supposed to be done on a calculator that's what I did, and got the right answer. I was just wondering if there is an easy analytical way to find the equation, well, other then manually "plugging-in the function" I can do that just fine. I was just wondering if there is any other possible way that I'm not seeing to find the equation. Any help is much appreciated
 
Last edited:
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f(x) = x(x+1)

Does that help?
 
Feldoh said:

Homework Statement


Well its not really a problem, I'm just trying to see it there's a different method to solve the problem. My textbooks gives a CAS problem:

Find [tex]f(f(f(f(f(f(x))))))[/tex] if [tex]f(x)= x^2+x[/tex]


Homework Equations


None (or atleast that I know of)


The Attempt at a Solution


Since it's a problem that's supposed to be done on a calculator that's what I did, and got the right answer. I was just wondering if there is an easy analytical way to find the equation, well, other then manually "plugging-in the function" I can that. I was just wondering if there is any other possible way that I'm not seeing to find the equation. Any help is much appreciated

EnumaElish said:
f(x) = x(x+1)

Does that help?
Not a great deal! :)
Just "manually plugging in". Do it one step at a time:
f(x)= x2+ x so
f(f(x))= (x2+ x)2+ (x2+ x)= [x4+ 4x3+ x2]+ [x2+ x]
= x4+ 4x3+ 2x2+ x

f(f(f(x)))= f(x4+ 4x3+ 2x2+ x)
= (x4+ 4x3+ 2x2+ x)2+ (x4+ 4x3+ 2x2+ x)

and continue. Tedious but straightforward.
 
HallsofIvy said:
Not a great deal! :)
Just "manually plugging in". Do it one step at a time:
f(x)= x2+ x so
f(f(x))= (x2+ x)2+ (x2+ x)= [x4+ 4x3+ x2]+ [x2+ x]
= x4+ 4x3+ 2x2+ x

f(f(f(x)))= f(x4+ 4x3+ 2x2+ x)
= (x4+ 4x3+ 2x2+ x)2+ (x4+ 4x3+ 2x2+ x)

and continue. Tedious but straightforward.

Well, yes I can do that, but as you said it's tedious.. But I guess there's no other way to find the function that you could think of? Just seems like there should be some better way :-/

f(x) = x(x+1)

Does that help?

Well unless you can do something with that other then "manually plugging in" not that much unfortunately. XD
 
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It is a labelled as a CAS (Computer Algebra System) problem. So it's probably meant to be tedious if you do it by hand. With a machine, it's pretty easy. No doubt a clever person could probably come up with expressions for the coefficients, but I don't think it's worth it.
 
HallsofIvy and Feldoh, think again about EnumaElish's post:
f(x) = x(x+1)

What would f(f(x)) be?? [] ( [] + 1 )
(I put boxes where the x's are)
Replace each box with x + 1
Thus, you end up with [x+1] ( [x+1] + 1)
f(f(x))= (x+1)(x+2)

Repeat again to do f( f(f(x)) )
lather, rinse, repeat a few times...
 
drpizza said:
HallsofIvy and Feldoh, think again about EnumaElish's post:
f(x) = x(x+1)

What would f(f(x)) be?? [] ( [] + 1 )
(I put boxes where the x's are)
Replace each box with x + 1
Thus, you end up with [x+1] ( [x+1] + 1)
f(f(x))= (x+1)(x+2)

Repeat again to do f( f(f(x)) )
lather, rinse, repeat a few times...

No, it's not. For one thing, you don't just replace x by x+1, you replace it by the function x(x+1). you would have (x(x+1))(x(x+1)+1). That doesn't seem to me to be an improvement. Obviously, since f(x) is quadratic, f(f(x)) is fourth degree, not another quadratic. Maybe you shouldn't use such a strong shampoo!
 
HallsofIvy said:
No, it's not. For one thing, you don't just replace x by x+1, you replace it by the function x(x+1). you would have (x(x+1))(x(x+1)+1). That doesn't seem to me to be an improvement. Obviously, since f(x) is quadratic, f(f(x)) is fourth degree, not another quadratic. Maybe you shouldn't use such a strong shampoo!

You're correct. Oh my God, that was such a horrible mistake on my part. I don't know why I wanted that to be true. Shame on me.
 
f(x) = x(x+1)
f(f(x)) = x (1 + x) (1 + x + x^2)
f(f(f(x))) = x (1 + x) (1 + x + x^2) (1 + x (1 + x) (1 + x + x^2))

Clearly there is a pattern.
 

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