1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Finding a basis for a set of polynomials and functions

  1. Sep 21, 2012 #1
    Find a basis for and the dimension of the subspaces defined for each of the following sets of conditions:

    {p [itex]\in[/itex] P3(R) | p(2) = p(-1) = 0 }

    { f[itex]\in[/itex]Span{ex, e2x, e3x} | f(0) = f'(0) = 0}

    Attempt: Having trouble getting started...

    So I think my issue is interpreting what those sets are and setting it up. So I think the sets are: i) the set of all polynomials s.t P(2) = p(-1) = 0 and ii) the set of exp functions where at 0 equal 0.

    So how do I put these each into a matrix form to find the basis and dimension?
  2. jcsd
  3. Sep 21, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I doubt this. It is probably the set of all polynomials with degree <= 3, such that p(2) = p(-1) = 0.
    Not just any exp functions. They have to be in [itex]\text{span}(e^x, e^{2x}, e^{3x})[/itex].

    I suggest that you start by finding the dimension of these two spaces: [itex]P_3(\mathbb{R})[/itex] and [itex]\text{span}(e^x, e^{2x}, e^{3x})[/itex]. Also, what is the form of a general element for each of these two spaces?
  4. Sep 21, 2012 #3


    User Avatar
    Science Advisor

    Personally, I wouldn't use a matrix, I would use the basic definition. First, I am going to assume that P3 is the vector space of polynomials of degree 3 or less, which has dimension 4 (some texts use that to mean the space of polynomials 2 or less which has dimension 3- the same ideas will apply but it is simpler). Any such polynomial canbe written [itex]p(x)= ax^3+ bx^2+ cx+ d[/itex]. The condition that p(2)= 0 means that we must have [itex]p(2)= 8a+ 4b+ 2c+ d= 0[/itex] or [itex]d= -(8a+ 4b+ 2c)[/itex]. The condition that p(-1)= 0 means that [itex]-a+ b- 2c+ d= 0[/tex] or [tex]d= -(a- b+ 2c)[/tex]. Then d= -(8a+ 4b+ 2c)= -(a- b+ 2c) so that 8a+ 4b+ 2c= a- b+ 2c which reduces to 7a= -5b. Sp we can replace a by -5b/7 which means d= -(-(5/7)b- b+ 2c)= -(12/7)b- 2c. Using those, [itex]ax^3+ bx^2+ cx+ d= -(5/7)bx^3+ bx^2+ cx- (12/7)b- 2c= (-(5/7)x^3+ x^2- 12/7)b+ c(x- 2)[/itex]

    Now, do you see what a basis is and what the dimension is?
    (You could have made a quick "guess" at what the dimension is by the fact that the basic space has dimension 4 and there are 2 conditions put on it.)

    For the second one, any f in the span of ex, e2x, and e3x, can be written as f(x)= aex+ be2x+ ce3x, and f'(x)= aex+ 2bex+ 3cex.

    The condition that f(0)= 0 gives a+ b+ c= 0 and f'(0)= 0 gives a+ 2b+ 3c= 0. We can subtract the first equations from the second to get b+ 2c= 0 or b= -2c. Putting that into the first equation a- 2c+ c= a- c= 0 so a= c. That is, we can write aex+ bex+ cex= cex[/sup- 2ce2x+ ce3x= c(ex- 2e2x+ e3x. Now, what is the dimension and what is a basis?
    (Here, the basic space has dimension three and there are two conditions.)
  5. Sep 21, 2012 #4

    That's what I intend on doing, but my issue is setting it up in order ot find those dimensions. So here's how I'm trying to piece it together:

    I know the general form for P3(R) is: ax3+bx2+cx+d, now the condition is that p(2) = P(-1) = 0. So I have to some how write out a set of vectors that satisfy that form.

    As for ii) a function would be f(x) = ex-2e2x+e3x, but I'm utterly clueless as to how this is line independent and how I could even find this vector if I set up a matrix
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook