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Find a basis for polynomials with complex coefficients

  1. Oct 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Let V=Pn(C) (polynomials of nth degree with complex coefficients), where n >=1. Find a basis for W=V s.t. every f(x) belonging to the basis satisfies f(0)f(1)= -1. (Demonstrate that the set you find is linearly independent and spans W.)


    2. Relevant equations
    For example, for V=Pn(F) and n>=1:
    S=(x^n, x^n + x^n-1, x^n + x^n-2, x^n + x^n-3,..., x^n +x^2, x^n + x, x^n +1 ) is a basis for Pn(F).

    3. The attempt at a solution
    I a bit confused at how f(0)f(1) = -1 is satisfied, because f(0) = 0 for functions without a constant. Does that mean that we have a constant function here?
    e.g. f(x) = -1 so f(0) = -1
    f(x) = x so f(1) = 1
    So f(0)f(1) = -1?
    Or is there something about complex number coefficients that I'm missing?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 12, 2008 #2
    Each f in the basis is a different polynomial, but f(0)f(1)=-1 has to hold for the same polynomial; that is, the example you gave is f1(x)=-1, f2(x)=x, f1(0)f2(1)=-1, not both f1 or both f2.

    Start with a concrete case. Let n=1. We need two basis vectors. Is there a constant basis vector f(x) such that f(0)f(1)=-1? if there is, then that means f(x)^2=-1(remember, you're are in Complex numbers). Then what is the degree-1 basis vector? Call it ax+b. Then f(0)f(1)=b(a+b)=-1. Pick a friendly non-zero a and see what b has to be.

    Is this a good method for n>1? Maybe. However, you have given another basis, can you work with it to ensure that for each basis element f, you have f(0)f(1)=-1?
     
  4. Oct 12, 2008 #3

    HallsofIvy

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    Science Advisor

    No, it simply means that you CANNOT have f(0)= 0 which means the constant term cannot be 0. And, by the way, f(x)= -1 is not in the set because f(0)f(1)= (-1)(-1)= 1 not -1. The constant function f(x)= i is in the set because f(0)f(1)= (i)(i)= -1.
    f(x) = x so f(1) = 1
    So f(0)f(1) = -1?

    I have given you one function you can put in the basis, f(x)= i. The set of all polynomials of degree n, with complex coefficients, is a space of dimension n+ 1. One possible basis is {x^n, x^{n-1}, ..., x^2, x, 1} but that does not satisfy the condition that f(0)f(1)= -1.
    To construct a first degree polynomial that satisfies that let p(x)= ax+ b. Then f(0)= b and f(1)= a+ b. Find two numbers, a and b, that satisfy b(a+b)= -1 with a not 0. Of course, there are an infinite number of such functions. You might, for example, choose b to be some convenient number, say 1, and solve for a.
    Now do the same with p(x)= ax^2+ bx+ c, etc.
     
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