# Composition of a function

1. Sep 5, 2007

### Feldoh

1. The problem statement, all variables and given/known data
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 $$f(x)= x^2+x$$

2. Relevant equations
None (or atleast that I know of)

3. 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: Sep 6, 2007
2. Sep 5, 2007

### EnumaElish

f(x) = x(x+1)

Does that help?

3. Sep 6, 2007

### HallsofIvy

Staff Emeritus
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.

4. Sep 6, 2007

### Feldoh

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 :-/

Well unless you can do something with that other then "manually plugging in" not that much unfortunately. XD

Last edited: Sep 6, 2007
5. Sep 6, 2007

### Dick

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.

6. Sep 6, 2007

### drpizza

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...

7. Sep 6, 2007

### HallsofIvy

Staff Emeritus
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!

8. Sep 7, 2007

### drpizza

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.

9. Sep 7, 2007

### EnumaElish

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.