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

[Feldoh]

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

 

[EnumaElish]

f(x) = x(x+1)

Does that help?

 

[HallsofIvy]

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

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.

 

[Feldoh]

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

 

[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.

 

[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...

 

[HallsofIvy]

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!

 

[drpizza]

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.

 

[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.