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Finding a Basis

  1. Jan 21, 2014 #1
    1. The problem statement, all variables and given/known data
    The problem asks, find a basis for the P2 subspace that consists of polynomials, p(x) such that p'(5)=0.



    3. The attempt at a solution

    I know that a set of vectors is a basis if it's linear independent and spans the vector space.

    So I let p(x) = ax2 + bx +c .

    Then p'(x) = 2ax + b.
    p'(2) = 4a + b

    I set that equal to 0. so 4a+b = 0. then 4a=-b, and a= =1/4*b.

    Now I am stuck as to where to go from here when finding a basis (prof hasn't really taught this yet)

    I was wondering what the next step would be?

    Thank you.
     
  2. jcsd
  3. Jan 21, 2014 #2

    tiny-tim

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    Hi Nexttime35! :smile:

    (do you mean p'(2) = 0 ?)

    Hints:

    i] is 1 in the subspace??

    ii] how many elements do you think the basis has? :wink:
     
  4. Jan 21, 2014 #3
    I think the basis has two elements, a and b.

    However, I am unsure how to apply your first hint to this question... Do you mean, can 4a+b=1?
     
  5. Jan 21, 2014 #4

    tiny-tim

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    Hi Nexttime35! :smile:
    No, I mean is p = 1 an element of P2 such that p'(2) = 0 ? :wink:
     
  6. Jan 21, 2014 #5

    Mark44

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    What tiny-tim is asking about is the inconsistency between what you're given, p'(5) = 0, and what your work shows, p'(2) = 0.
     
  7. Jan 21, 2014 #6
    Ahh, ok.

    Then p=1 is not an element of p2 such that p'(2) = 0.
     
  8. Jan 21, 2014 #7
    So a basis for this problem must consist of a function with only a degree 1, such as p(x) = x, or p(x) = 2x?
     
  9. Jan 21, 2014 #8
    Oh sorry, I meant p'(2) = 0 in the problem header. Sorry about that.
     
  10. Jan 21, 2014 #9

    tiny-tim

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    i don't mind whether it's 2 or 5, it makes little difference, i was just checking :wink:
    yes it is!!

    p = 1 is a polynomial, and its derivative is 0 everywhere!
     
  11. Jan 21, 2014 #10
    Duh! My mistake! For some reason I understood that differently. So p(x) = 1 for some x, and then obviously p'(x)=0 . I am confused about how that helps me find a basis, though. Thoughts?
     
  12. Jan 21, 2014 #11

    vela

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    An element of P2 is a polynomial. We can think of these polynomials as vectors because with the appropriate definitions of vector addition and scalar multiplication, P2 has all the properties required of a vector space.

    A basis is a set of linearly independent vectors that span the vector space. So in this problem, the vectors are polynomials, so you're looking for a set of polynomials that are linearly independent and that span the given subspace. Clearly ##a## and ##b## (the way you used them) aren't elements of the subspace, so they can't be elements of the basis.
     
  13. Jan 21, 2014 #12
    So vela,

    there isn't just one basis. There are essentially many, many bases that can be formed? So I could make a set of vectors {v1,v2,v3} and that would be a basis if each of those vectors (or polynomials in this case) satisfy p'(2)=0, and that are linearly independent?
     
  14. Jan 21, 2014 #13

    tiny-tim

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    well, p = 1 is in the subspace, so is p = 2, p = √2, etc

    so does that help with one of the elements of the basis? :smile:
     
  15. Jan 21, 2014 #14

    vela

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    Yes, there are many possible bases for a vector space.

    You have a subset S of P2 such that any element in S satisfies p'(2) = 0. A basis for S should be a set of linearly independent vectors that you can express any element of S as a linear combination of the basis vectors. This is close to what you said but not exactly. You need to make sure you can express every element of S in terms of the basis vectors.
     
  16. Jan 21, 2014 #15
    Yes!

    So one element of the basis could be a constant, since that derivative of a constant is equal to 0. One element of the basis could be the function p(x) = x , because that satisfies the p'(2) = 0 condition. So, could {1, x} be a basis for the subspace of P2 in this case?
     
  17. Jan 21, 2014 #16

    tiny-tim

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    Hi Nexttime35! :wink:
    Yes!! :smile:

    (obviously, any constant will do, but equally obviously most people would choose 1 !)
    erm :redface:nooo! :rolleyes:

    you had that part correct at the beginning …

    do you need some sleep? :zzz:​
     
  18. Jan 21, 2014 #17

    vela

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    If p(x)=x, then p'(x)=1. That doesn't satisfy p'(2)=0.

    Suppose for the moment that p(x)=x did satisfy the condition. If {1, x} was a basis, that means any polynomial satisfying p'(2)=0 can be written as a linear combination of 1 and x. You need to prove that is the case for whatever prospective basis you come up with.
     
  19. Jan 21, 2014 #18
    Oh gosh. I think I do need some sleep...

    So one element of the basis will be a constant. Say p(x) = 4x^2. Then plugging in x=2, we get p(2) = 16. p'(2) = 0. OK I understand. I had a brain fart... Say p(x) = 4x. Then plugging in x=2, we get p(2) = 8. p'(2) = 0. So, could a basis be {1,x,x^2}? Or am I just overthinking this?
     
  20. Jan 21, 2014 #19

    tiny-tim

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    Sorry, I'm calling a time-out

    GET SOME SLEEP! :zzz:

    NOW!!
     
  21. Jan 21, 2014 #20
    I sincerely apologize for my lapse in mathematical judgement.

    The only derivative that can result in 0 is the derivative of a constant. Thus, is the basis for this subspace of P2 a one-element vector, a constant?
     
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