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Math Help - Find one & all polynomial of degree 4

  1. #1
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    Find one & all polynomial of degree 4

    If someone can advise me how to approach the following problem it would be greatly appreciated.

    Q: Find one & all polynomials of degree 4 such that p(-1)=1, p(0)=0, p(1)=2, p(2)=1

    I tried...

    p(x)=ax^4+bx^3+cx^2+dx+e
    p(-1)=a-b+c-d+e=1
    p(0)=e=0
    p(1)=a+b+c+d+e=2
    p(2)=16a+8b+4c+2d+e=1

    Is it correct to place it into a matrix and reduce to echelon form? But then I'm not sure what to do with it.

    [1 1 1 1 2] [1 1 1 1 2]
    [1 -1 1 -1 1] ~ [0 -2 0 -2 -1]
    [16 8 4 2 1] [0 0 1 1/2 -9/4]
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  2. #2
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    Sure, you can go to row echelon form, and then do back substitution. I would, incidentally, leave out the last column there, as you already know that e = 0. You can save yourself some computations. If the system is consistent (and it better be, because having a polynomial of degree 4 that goes through those four points is definitely possible - just think about Lagrange interpolating polynomials to see that), then I would say you should end up with a 1-parameter family of solutions. What do you get?
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  3. #3
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    I did as you suggested. I went to echelon form then back substituted to row canonical form.

    [1 1 1 1 2]
    [1 -1 1 -1 1]
    [16 8 4 2 1]

    [1 1 1 1 2]
    [0 -2 0 -2 -1]
    [0 0 1 1/2 -9/4]

    [1 0 0 -1/2 15/4]
    [0 1 0 1 1/2]
    [0 0 1 1/2 -9/4]

    I'm not sure what I can do with this matrix though. Am I on the right track? I represented a, b, c, d and the solution in the above matrices given that e=0.
    Last edited by chipwager; March 23rd 2011 at 01:15 PM.
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  4. #4
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    You might want to check your accuracy here. It looks like your last column might have some errors in it.

    What you do now is back substitution by letting one variable, say, d, be an arbitrary parameter. I'll do the first one, and you do the rest. Let the parameter t = d. Then, by the third row, we know that c + d/2 = 9/4, or c = 9/4-t/2. And continue in this manner, solving for b and a. What do you get?
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  5. #5
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    I see where you're going with this. I get a = (t/2-3/4), b = (1/2 - t), c= (9/4 - t/2) then solve for d given that e=0. Can I assume that the value for "a" is to the degree of 4 and "b"=3, etc?
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  6. #6
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    Not quite. You're not going to be able to eliminate t, because you have more unknowns than equations. However, you can write the parameters a through d as linear functions of t. Then you plug those into the original polynomial, along with the fact that e = 0, and you've found all the polynomials that satisfy your conditions.
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  7. #7
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    thank you so much for walking me through this! much appreciated.
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  8. #8
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    What did you get for your final answer?
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