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Determining polynomials

  1. Jul 28, 2004 #1
    Hi

    I got a Linear Algebra question.

    I'm suppose to find two polynomials p1 and p2 both of highest degree 3, and which satisfies the following:

    p1(-1) = 1

    p1'(-1) = 0

    p2(1) = 3

    p2'(1) = 0

    p1(0) = p2(0)

    p1'(0) = p2'(0)


    I hope that there is somebody out there who can explain to me how I do that ?

    Thanks in advance.

    Fred
     
  2. jcsd
  3. Jul 28, 2004 #2

    arildno

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    Hint:
    If your polynomials are of highest degree 3, they may be written as:
    [tex]p_{1}=a_{1}x^{3}+b_{1}x^{2}+c_{1}x+d_{1}[/tex]
    [tex]p_{2}=a_{2}x^{3}+b_{2}x^{2}+c_{2}x+d_{2}[/tex]

    The letters are the constants you must determine with the aid of the given relations (you will gain a matrix system you''ll need to solve)
     
  4. Jul 28, 2004 #3
    Thanks for Your answer.

    If I understand You correctly.

    I take these two polynomials p1 and p1'.

    I then insert the terms into the polynomial equations? So I get the following

    p1 = a1*(-1)^3 + b2*(-1)^2 + c1*(-1) +d = 1

    p1' = 3*a1*(-1)^2 + 2*b1*(-1) +c1 = 0

    I use a matrix system to solve these two equations so they satisfy the given conditions?

    And then find the variables for p2 ?

    Sincerely
    Fred
     
  5. Jul 28, 2004 #4

    matt grime

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    you have far more than just those two equations. for instance the last two tell you that d_1=d_2 and c_1=c_2
     
  6. Jul 28, 2004 #5
    You mean I have these equations:

    p1 = a1*(-1)^3 + b1*(-1)^2 + c1*(-1) +d1 = 1

    p1' = 3*a2*(-1)^2 + 2*b2*(-1) +c2 = 0

    p2 = a1*(1) ^3 +b1 * (1) ^2 + c1*(1) + d1 = 3

    p2' = 3*a2*(1)^2 + 2*b2*(-1) +c2 = 0

    a1*(0)^3 + b1*(0)^2 + c1*(0) +d1 = a1*(0) ^3 +b1 * (0) ^2 + c1*(0) + d1

    3*a2*(0)^2 + 2*b2*(0) +c2 = 3*a2*(0)^2 + 2*b2*(0) +c2

    and solve these as one system of equations ?

    /Fred
     
  7. Jul 28, 2004 #6

    matt grime

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    not quite since you appear to have several mistakes in there, let us use arildno's notation properly

    1=-a_1+b_1-c_1+d_1
    from (p_1(-1)=1)

    0=3a_1-2b_1+c_1
    from (p_1'(-1)=0)

    3=a_2+b_2+c_2+d_2
    from (p_2(1)=3)

    0=3a_2+2b_2+c_2
    from (p_2'(1)=0)

    d_1=d_2
    from (p_1(0)=p_2(0))

    c_1=d_1
    from (p_1'(0)=p_2'(0))

    note you have 8 unkowns and 6 equations
     
  8. Jul 28, 2004 #7
    okay Thank You.

    Fred
     
  9. Jul 28, 2004 #8
    I get the following solution values:

    a1, a2 = -1/2

    b1, b2 = 3/2

    c1,c2 = 0

    d1, d2 = 2

    Can anybody tell me if they are correct ?

    Thanks again.

    Sincerely

    Fred
     
  10. Jul 29, 2004 #9

    matt grime

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    But you've all these identities that allow you to check your answer by yourself, you don't need us to verify it.
     
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