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

AI Thread Summary
The discussion revolves around finding the composition of the function f(x) = x^2 + x, specifically f(f(f(f(f(f(x)))))). While one participant initially used a calculator to solve the problem, they sought a more analytical method beyond simply plugging the function in repeatedly. The consensus is that the process is tedious and primarily involves calculating each composition step-by-step, as there isn't a significantly simpler analytical method available. Some participants note that recognizing the pattern in the compositions can help, but ultimately, the task remains complex due to the nature of the quadratic function. The conversation highlights the challenges of manual computation versus using a Computer Algebra System (CAS) for such problems.
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 f(f(f(f(f(f(x)))))) if f(x)= x^2+x

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 f(f(f(f(f(f(x)))))) if f(x)= x^2+x


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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