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Composition and polynomial

  1. Oct 31, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the function g(x), if there are [itex]h=g \circ f[/itex]

    f(x)=x+1, [itex]x \in \mathbb{R}[/itex] and [itex]h(x)=x^3+3x+1[/itex]

    2. Relevant equations

    [tex](g \circ f)(x)=g(f(x))=h(x)[/tex]

    [tex]f \circ g \neq g \circ f [/tex]

    3. The attempt at a solution

    [tex]h=g \circ f[/tex]

    [tex]h(x)=g(f(x))[/tex]

    [tex]x^3+3x+1=g(x+1)[/tex]

    Is there any way that I will directly find the result of g(x), or I should guess and try some things? I tried something at home, but useless. Please help me!
     
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  3. Oct 31, 2008 #2

    Office_Shredder

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    Re: Composition

    My guess is g is going to be a polynomial. You can use the degree of f and h to find what the degree of g is going to be, then plug f into a generic polynomial of that degree and see what conditions the coefficients have to satisfy
     
  4. Oct 31, 2008 #3

    Dick

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    Re: Composition

    Well, you can figure g ought to be a polynomial function of degree 3, right? So put g(x+1)=A(x+1)^3+B(x+1)^2+C(x+1)+D. If you set that equal to x^3+3x+1, can you find A,B,C and D? You should get four equations in four unknowns if you equate the powers of x. Of course, you can also be clever and pick some special values of x that might make the job easier (like x=-1).
     
  5. Oct 31, 2008 #4
    Re: Composition

    Thanks for the posts.

    I found A=1, B=-3, C=6, D=-3.
    [tex]g(x+1)=(x+1)^3-3(x+1)^2+6(x+1)-3[/tex]
    So
    [tex]g(x)=x^3-3x^2+6x-3[/tex]

    Yes it is correct.

    How will x=-1 make the job easier?
     
  6. Oct 31, 2008 #5

    Dick

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    Re: Composition

    If you put x=-1, you get D=-3 directly. You can also see A must be 1 without doing any any algebra. So you really only need to solve for B and C.
     
  7. Oct 31, 2008 #6

    gabbagabbahey

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    Re: Composition

    An alternative method would be to just make the substitution u=x+1 (and hence x=u-1) into this equation and find g(u), then rename u to x and your done....You wouldn't even need to expand the polynomial, you could just leave it as a polynomial in powers of (x-1)
     
  8. Oct 31, 2008 #7
    Re: Composition

    Thanks for the replys.
    gabbagabbahey, do you think like [itex](u-1)^3+3(u-1)+1=g(u)[/itex] ?

    I got the same result [itex]u^3-3u^2+6u-3=g(u)[/itex], so if I write [itex]g(x)=u^3-3u^2+6u-3[/itex]. Thanks again.
     
  9. Oct 31, 2008 #8

    gabbagabbahey

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    Re: Composition

    Yes, [itex](u-1)^3+3(u-1)+1=g(u)[/itex] so automatically, [itex]g(x)=(x-1)^3+3(x-1)+1[/itex] and you could leave the answer just like that. (or you could expand it in powers of x and get [itex]g(x)=x^3-3x^2+6x-3[/itex])
     
  10. Oct 31, 2008 #9

    Dick

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    Re: Composition

    That's even more direct. Good observation.
     
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