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Finding W perp of inner product space

  1. Nov 4, 2012 #1
    Consider P5(R) together with innner product < p ,q > = ∫p(x)q(x) dx. Find W-perp if W = {p(x) [itex]\in[/itex] P5(R) : p(0) = p'(0) = p''(0) = 0}

    Attempt: Im having trouble with the condition. I always have trouble with these conditions. SO as of now I am going to let q(x) be the standard basis of P5(R). Now I don't know how to apply the condition to p(x).

    After I do apply the condition I would take the inner product and have it set equal to 0. I should have a set of equations that I can solve in matrix form. This should produce some free variables from where I can obtain vectors for W-perp. So the concept is understood......I just can't seem to use the conditions appropriately.....

    thanks.
     
  2. jcsd
  3. Nov 4, 2012 #2

    Dick

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    Write p(x) out as a polynomial. Your conditions are giving you information about the coefficients of the polynomial. What does p(0)=0 tell you?
     
  4. Nov 4, 2012 #3

    LCKurtz

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    If ##p(x) = a + bx + cx^2 + dx^3 + ex^4 +fx^5##, set ##p(0)=0,p'(0)=0,p''(0)=0##. What does that tell you?
     
  5. Nov 4, 2012 #4


    Well I set it up the other way so my "a" was with ax5. Well after doing that it turns out that d=e=f= 0 in order for the polynomial to satisfy the condition.

    Doing that it turns into p(x) = ax5+bx4+cx3

    now I have to multiply this by q(x) and then integrate. Is there any substitution I could possibly do or am I just going to have to literally multiply it out and integrate each component.......if that's the case wouldn't that just be mean?
     
  6. Nov 4, 2012 #5

    Dick

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    It might seem a little more manageable if you notice a basis for W is {x^5,x^4,x^3}, so q(x) is in W-perp if it's perpendicular to each vector in the basis.
     
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