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The set of 4-degree polynomials (Linear algebra)

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data

    If P is the set of all 4-degree polynomials, and W is the subset of all 4-degree polynomials such that p(-2) = p(2), find a set S such that W = span(S).

    2. Relevant equations

    3. The attempt at a solution

    My guess is that one set that works is x^4, x^2, and 1. My reasoning is that if x^3 is included, then you could take the coefficients of everything to be zero, and (-2)^3 =/= 2^3. Can't include x for similar reasons.

    Really not sure, I feel like it is possible to construct a polynomial where p(-2) = p(2) but does include an x^3 or an x.
  2. jcsd
  3. Oct 22, 2009 #2


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    Uh, 'I feel like' doesn't really help. Why don't you try and construct a polynomial such that ax^3+bx=p(x) and p(2)=p(-2)? Since you only have one restriction on the five dimensional subspace P, you really would expect W to have dimension 4.
  4. Oct 22, 2009 #3
    ah, it took me a while but i got one.

    x^3 + 3x^2 - 4x - 12

    i guess i don't really understand the question then. how could such a set be constructed?
  5. Oct 22, 2009 #4


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    It took too long because you were working too hard. You already know the span of 1, x^2 and x^4 are in W. The question you asked, and suspected is true, is whether a combination of x and x^3 are in W. If p(x)=ax^3+bx, then p(2)=8a+2b., p(-2)=(-8a-2b). The condition for them to be equal is 4a+b=0, right? So right again, you should include x^3-4x in the spanning vectors. You didn't need to fuss with the even power terms. You already know they work.
  6. Oct 22, 2009 #5
    oh, i see. it makes perfect sense now. oh well, i guess i will just get part marks for that question then... thanks for the explanation.
  7. Oct 23, 2009 #6


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    A general fourth degree polynomial is of the form [itex]p(x)= ax^4+ bx^3+ cx^2+ dx+ e[/itex]. Then the condition that p(2)= p(-2) is that [itex]p(2)= 16a+ 8b+ 4c+ 2d+ e= p(-2)= 16a- 8b+ 4c- 2d+ e[/itex]. a, c, and e cancel immediately leaving [itex]8b+ 2d= -8b- 2d[itex] or [itex]16b+ 4d= 0[/itex]. That reduces to d= -4b while a, c, and e can be anything. Taking a= 1, b= c= d= e= 0 gives [itex]x^4[/itex]. Taking b= 1, a= c= e= 0, d= -4 gives [itex]x^3- 4x[/itex]. Taking c= 1, a= b= d= e= 0 gives [itex]x^2[/itex]. Taking e= 1, a= b= c= d= 0 gives [itex]1[/itex]. A basis for this space is {[itex]x^4[/itex], [itex]x^3- 4x[/itex], [itex]x^2[/itex], 1}.
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