# Composite functions

1. Mar 12, 2013

### Saitama

1. The problem statement, all variables and given/known data
Suppose f(x) and g(x) are non-zero polynomials with real coefficients, such that f(g(x))=f(x)⋅g(x). If g(2)=37, what is g(3)?

2. Relevant equations

3. The attempt at a solution
I have no idea where to begin this with.
At x=2,
f(g(2))=f(2).g(2)
f(37)=37f(2)

What should I do with this equation? I cannot even guess any function which would satisfy the above condition.

Any help is appreciated. Thanks!

2. Mar 12, 2013

### Simon Bridge

You know they are polynomials ... what are the properties of polynomials?

Note: if $f=x^2+1$ and $g=x$, both polynomials,
then $f(g)=x^2+1$ and $f.g = x^3+x$ so $f(g)\neq f.g$ so these are not correct.

You have to try to figure out what sort of polynomial will fit in there.
Start by revising the properties of polynimials and then experiment with different short polynomials to get a feel for how they behave.

3. Mar 13, 2013

### Saitama

Does that mean that there's no proper way of doing it and I have to guess the polynomials? D:

4. Mar 13, 2013

### Simon Bridge

The proper way to do it is to use your understanding of the properties of polynomials.

Since you have yet to gain that understanding, you will have to make educated guesses ... refine your guesses as you go: do not make blind guesses - there are infinite possibilities to choose from. You are supposed to use your knowledge of polynomials to guide your choices.

You know things like that $g=\sum_{i=0}^n a_n x^n$ and $g(2)=\sum_{i=0}^n a_n 2^n = 37$ ... and you know how to do powers of 2.

Fortunately (you should have realized) f and g must be quite simple polynomials ... they won't be order n=1045 or anything like that. You know this because your instructor knows your competence level and has set the problem in keeping with it ;) Try guessing just simple powers of x ... what happens if f=x^2 and g=x^3 ? Do f and g have to be different?

eg. f=g=x^2, f.g=x^2, f(g)=x^4 ... so f(g)=fg ... looks good: unfortunately g(2)=4 so it's not right.
g = x^5 + x^2 + 1 works ... but so does: g=x+35
Get the idea?

You can also work with the relationships... eg.
if f(3)=2f(2) then what is f(x) likely to be?

That kind of thing. In your case you know that f(37)=37f(2).

There will also be clues in your coursework so far.

Last edited: Mar 13, 2013
5. Mar 13, 2013

### ehild

What can be the order of the polynomials?

ehild

6. Mar 15, 2013

### Saitama

Sorry for being late.

I don't think that I do have time to revise everything about polynomials as it is almost the end of session and I am done with my coursework. :)

I too guessed those two same polynomials (and some more) for g(x) but not even one for f(x). I could only deduce that both f(x) and g(x) shouldn't be of same degree. For example, had they been both a linear polynomial, the LHS would be a linear one and the RHS would consist of a term containing x^2.

7. Mar 15, 2013

### ehild

Are you sure?

If f is of n degree and g is of m degree what is the degree of f(g(x))? and what is the degree of f(x)g(x)?

Starting with the highest degree term,

f(x) =anxn+...
g(x)=bmxm+...

f(g(x))=an(bmxm+.... )n+....=(anxn+...)(bmxm+...)

What is the relation between n and m?

ehild

8. Mar 15, 2013

### eddybob123

I don't think that there is only one solution. Since we only want g(x), and f(x) is allowed to be anything we want, then just guess a polynomial for g(x) such that g(2) = 37, find the corresponding f(x), then solve for g(3).

9. Mar 15, 2013

### Saitama

f(g(x)) has the degree mn and f(x).g(x) has the degree m+n.

Is the relation $m=\frac{n}{n-1}$ ?

If this is the relation, I get only one solution, which is m=2 and n=2.

Last edited: Mar 15, 2013
10. Mar 15, 2013

### Saitama

Ooohoo, looks like I got it.

Using the above relation, m=2 and n=2, I assumed that f(x) is a quadratic of the form $ax^2+bx+c$.
Since, $f(37)=37 \cdot f(2)$
$$(37)^2a+37b+c=37(4a+2b+c)$$
Rearranging, I end up with
$$37 \times 33a-37b-36c=0$$
The possible values of $a,b,c$ can be $a=1, b=33$ and c=0.
Therefore, $f(x)=x^2+33x$
As $f(g(x))=f(x).g(x)$
$$(g(x))^2+33 \cdot g(x)=(x^2+33x) \cdot g(x)$$
The above relation is a quadratic in g(x) and from here I get two solutions.
$g(x)=0$ and $g(x)=x^2+33x-33$ but g(x) is a non-zero polynomial, hence $g(x)=0$ is not admissible. Both f(x) and g(x) satisfy the given conditions.
Therefore, g(3)=75.

Thanks a lot both of you.

11. Mar 16, 2013

### ehild

Well, that is the solution, but the problem said that the coefficients are real. So the equation 37×33a−37b−36c=0 has infinite number of solutions. The coefficient of the quadratic term is arbitrary, take it a=1.

If f(x)=x2+bx+c and g(x) = Ax2+Bx+C

f(g(x))=(Ax2+Bx+C)2+b(Ax2+Bx+C)+c=(x2+bx+c)(Ax2+Bx+C)

You can rearrange the equation so as one side is quadratic.

b(Ax2+Bx+C)+c=(Ax2+Bx+C)(Ax2+Bx+C-x2-bx-c)

Expand, and compare the coefficients. Start with the highest degree terms. The coefficient of X4 is A(A-1)=0 so A=1. The coefficient of x3 is A(B-b), so B=b....

ehild

12. Mar 16, 2013

### Saitama

Won't this equation give two values for A? And corresponding to each value, there will be a different value for B?

13. Mar 16, 2013

### ehild

A can not be zero as it would make g(x) first order.

ehild

14. Mar 16, 2013

### vio

?

frm which book did ya get dis qstn??

15. Mar 18, 2013

### Saitama

Comparing coefficient of $x^2$,
$$Ab=A(C-c)+B(B-b)+C(A-1)$$
But A=1 and B=b,
$$b=C-c$$

Comparing coefficient of $x$,
$$Bb=B(C-c)+C(B-b)$$
As B=b=C-c, hence
$$b^2=b(C-c)$$
$$b=0,C-c$$

Comparing the constant terms,
$$bC+c=C(C-c)$$
$$bC+c=bC$$
$$c=0$$

From these, $b,c$ and $C$ are zero.

16. Mar 18, 2013

### ehild

I made a sign error in my equation, I am afraid. It should be
b(Ax2+Bx+C)+c=(Ax2+Bx+C)(x2+bx+c-(Ax2+Bx+C)) or maybe not? You have to check what is said to you

ehild

17. Mar 18, 2013

### ehild

Edited:

As it was my fault, I do one more step with the correct formula and delete anything further.

b(Ax2+Bx+C)+c=(Ax2+Bx+C)(x2(1-A)+(b-B)x+(c-C))

We know already that A=1 and B=b.
.............

ehild

Last edited: Mar 18, 2013
18. Mar 18, 2013

### Saitama

Oh ehild, you should have waited for me to do it. I cannot reply instantly now a days because of my finals. I still haven't checked the derivation so you may wish to remove it. :)

19. Mar 19, 2013

### ehild

I removed everything but a hint: before comparing further coefficients replace A=1 and B=b into the equation. Good luck for the finals! :)

ehild

20. Mar 22, 2013

### Saitama

Thanks, my finals just got over today. I did well especially in Physics and Maths.

Getting back onto the question, I replaced A with 1 and B with b.
The equation reduces to
b(x2[/sup]+bx+C)+c=(x2[/sup]+bx+C)(c-C)

Comparing coefficients of x2[/sup],
b=c-C

Comparing coefficients of x,
b2[/sup]=(c-C)b or b2[/sup]=b2[/sup]. This means b=0?

Comparing the constant terms,
bC+c=C(c-C) or bC+c=Cb
Again c=0.

I am ending up with the same results again. :(