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Linear Algebra - Find a Polynomial

  1. Jan 18, 2013 #1
    1. The problem statement, all variables and given/known data


    Find all polynomials of the form a + bx + cx^2 that:

    Goes through the points (1,1) and (3,3)

    and such that f'(2) = 1


    2. Relevant equations

    a + bx + cx^2
    f'(x) = x+2cx
    f'(2) = 2 + 4c

    polynomial through (1,1) = a + b1 + c1 = 1
    polynomial through (3,3) = a + b3+ c3^2 = 3

    3. The attempt at a solution

    I have the general idea that this should result in a series of equations that I need to do gauss Jordan on. Similar problems like this resulted in 3 similar equations and were quite simple.

    My problem here is that, since I have taken the derivative of f(x) I have lost my constant a. So I'm not sure what my matrix should look like. I've tried:

    1 1 1 1
    1 3 9 3

    But for the one with the derivative, my f"(2) isn't in the same form - it's 2+4c = 1

    so I"m not sure whether to use

    1 2 4 1
    0 2 4 1

    or something more general. If I use something more general though like:

    a 2 4 1

    I can't get a pivot in my first column...

    'elp!

    -Dave K
     
  2. jcsd
  3. Jan 18, 2013 #2

    CompuChip

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    You can rewrite 2+4c = 1 to 4c = -1.
    That's of the form ... a + ... b + ... c = ....
     
  4. Jan 18, 2013 #3
    That helps, but I'm still not sure whether my a's and b's for that equation should be 0s or 1's. I can try both, I suppose, but clearly my understanding of the problem is muddled or I wouldn't have to guess...
     
  5. Jan 18, 2013 #4

    CompuChip

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    Is [itex]1 \cdot a + 1 \cdot b + 4 \cdot c = -1[/itex] the same as [itex]4c = -1[/itex]?
    Or is [itex]0 \cdot a + 0 \cdot b + 4 \cdot c = -1[/itex] the same as [itex]4c = -1[/itex]?
     
  6. Jan 18, 2013 #5
    Thanks for that clarification. I'm still getting some weird stuff (fractions I shouldn't be getting) but there must be another problem, since I'm pretty sure I'm not supposed to get those and that I should wind up with a free variable instead.

    I'll check back later if I'm still working on it. Thanks again!
    -Dave K
     
  7. Jan 18, 2013 #6

    lurflurf

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    1 1 1 1
    1 3 9 3
    0 2 4 1
    do you know why?

    Since x is such a polynomial it is easier to let
    f(x)=a + bx + cx^2
    g(x)=f(x)-x
    f(1)=1
    f(3)=3
    f'(2)=1
    imply
    g(1)=0
    g(3)=0
    g'(2)=0
    the matrix is
    1 1 1 0
    1 3 9 0
    0 2 4 0
     
  8. Jan 18, 2013 #7

    Ray Vickson

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    Your expression for f '(2) is incorrect.
     
    Last edited: Jan 18, 2013
  9. Jan 21, 2013 #8

    CompuChip

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    Good catch Ray! It took three people to see that :)
     
  10. Jan 22, 2013 #9
    Yes, indeed. That was a problem too. But your question "Is 1⋅a+1⋅b+4⋅c=−1 the same as 4c=−1?" was also instructive.

    Thanks to the whole team. :)
     
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