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How do you find the equation of a Cubic function given 5 points? (no zeros)?

  1. Jan 8, 2012 #1
    1. The problem statement, all variables and given/known data

    what the questions asks is that i need to find the equation of a polynomial with these given points:
    1,1
    2,-3
    3,5
    4,37
    5,105


    i know that one way to solve is by creating 5 equations then solve for ax^3+bx^2+cx+d using the elimination/substitution method.
    however is there another, much easier way of doing this question?

    2. Relevant equations

    ax^3+bx^2+cx+d

    3. The attempt at a solution

    1= a+b+c+d
    -3=8a+4b+2c+d
    5= 27a+9b+3c+d
    37=64a+16b+4c+d
    108=125a+25b+5c+d

    fixed, yes, my mistake
     
    Last edited: Jan 8, 2012
  2. jcsd
  3. Jan 8, 2012 #2

    eumyang

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    Three of the equations are wrong. They should be
    1= a+b+c+d
    -3=8a+4b+2c+d
    5= 27a+9b+3c+d
    37=64a+16b+4c+d
    108=125a+25b+5c+d

    Also, you don't need the last equation, because there are 4 unknowns.

    As for other methods, there's the finite difference method, but I don't think it will help for this particular problem (because you already told us that this is a cubic).
     
    Last edited: Jan 8, 2012
  4. Jan 8, 2012 #3
    fixed, i typed the equations too fast
     
  5. Jan 8, 2012 #4

    SammyS

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    True, but I got the coefficients fairly quickly using a difference method.

    Actually, after playing around a bit with this, I got the result with two different difference methods.
     
  6. Jan 8, 2012 #5
    what exactly is the difference method?
    i ask because i've tried this question, and i kept getting it wrong.
     
  7. Jan 8, 2012 #6

    eumyang

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  8. Jan 8, 2012 #7

    SammyS

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    Make a table of differences

    [tex]\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline
    \quad i \quad & \quad x_i \quad & \quad f(x_i) \quad &\quad (\Delta^1)_i \quad &\quad (\Delta^2)_i \quad & \quad (\Delta^3)_i \quad & \quad (\Delta^4)_i \quad \\
    \hline
    & & & & & & \\
    1 & 1 & 1 & -4 & 12 & 12 & 0 \\
    & & & & & & \\
    2 & 2 & -3 & 8 & 24 & & -- \\
    & & & & & & \\
    3 & 3 & 5 & & & -- & -- \\
    & & & & & & \\
    4 & 4 & 37 & & -- & -- & -- \\
    & & & & & & \\
    5 &5 & 105 & -- & -- & -- & -- \\
    & & & & & & \\
    \hline \end{array}[/tex]

    Where: [itex](\Delta^1)_i=f(x_{i+1})-f(x_{i})\,,[/itex]

    [itex](\Delta^2)_i=(\Delta^1)_{i+1}-(\Delta^1)_i[/itex]

    etc.

    See if you can fill in the rest.

    If f(x) is truly a cubic function then the Δ3 column will all be the same.

    Fill out a similar Table for g(x) = x3 . The Δ3 column will all be 6's.

    What do you suppose that means about the x3 coefficient of f(x) ?
     
  9. Jan 9, 2012 #8
    OMG! i love you forever, i had no idea this method existed, i already solved through almost an hour of writing matrices, with this i solved it in 3 minutes. thank you!
     
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