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Math Help - Cubic function

  1. #1
    s3a
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    Cubic function

    I have never studied the cubic function before so it would be great if someone could explain it to me so that I can answer #25.

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by s3a View Post
    I have never studied the cubic function before so it would be great if someone could explain it to me so that I can answer #25.

    Thanks in advance!
    you need to enter the data into your calculator and run a cubic regression. I set the year 1900 as 0. graph and equation screenshots follow ...
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by s3a View Post
    I have never studied the cubic function before so it would be great if someone could explain it to me so that I can answer #25.

    Thanks in advance!
    MS Excel shown



    So we have a equation of population y = f\left( {x} \right) = 1.293{x^3} + 31.27{x^2} + 4.149x + 1688

    then in 1925 year the population is {y_{1925}} = f\left( {2.5} \right) = 1.293 \cdot {2.5^3} + 31.27 \cdot {2.5^2} + 4.149 \cdot 2.5 + 1688 \approx 1914
    Last edited by DeMath; August 30th 2009 at 09:37 PM. Reason: typo
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  4. #4
    Super Member Random Variable's Avatar
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    let  P(x) = Ax^{3}+Bx^{2}+Cx + D where P(x) is the population in year x

    then 1650 = 1900^{3}A+ 1900^{2}B + 1900C + D

     1750 = 1910^{3}A+1910^{2}B+1910C + D

    and so on

    What you have is an overdetermined linear system that can be "solved" for the coefficients A, B, C, and D by the method of least squares (which can be done by a program like Maple or Mathematica).
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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Random Variable View Post
    let  P(x) = Ax^{3}+Bx^{2}+Cx + D where P(x) is the population in year x

    then 1650 = 1900^{3}A+ 1900^{2}B + 1900C + D

     1750 = 1910^{3}A+1910^{2}B+1910C + D

    and so on

    What you have is an overdetermined linear system that can be "solved" for the coefficients A, B, C, and D by the method of least squares (which can be done by a program like Maple or Mathematica).
    Maybe this??

    {{\hat y}_t} = a{x^3} + b{x^2} + cx + d

    S = f\left( {a,b,c,d} \right) = \sum\limits_{i = 1}^n {{{\left( {{y_i} - {{\hat y}_t}} \right)}^2}}  = \sum\limits_{i = 1}^n {{{\left[ {{y_i} - \left( {ax_i^3 + bx_i^2 + c{x_i} + d} \right)} \right]}^2}}  \to \min

    Then we have to solve the standard system of linear equations

    \frac{{\partial S}}<br />
{{\partial a}} = a\sum\limits_{i = 1}^n {x_i^6}  + b\sum\limits_{i = 1}^n {x_i^5}  + c\sum\limits_{i = 1}^n {x_i^4}  + d\sum\limits_{i = 1}^n {x_i^3}  - \sum\limits_{i = 1}^n {x_i^3{y_i}}  = 0,

    \frac{{\partial S}}<br />
{{\partial b}} = a\sum\limits_{i = 1}^n {x_i^5}  + b\sum\limits_{i = 1}^n {x_i^4}  + c\sum\limits_{i = 1}^n {x_i^3}  + d\sum\limits_{i = 1}^n {x_i^2}  - \sum\limits_{i = 1}^n {x_i^2{y_i}}  = 0,

    \frac{{\partial S}}<br />
{{\partial c}} = a\sum\limits_{i = 1}^n {x_i^4}  + b\sum\limits_{i = 1}^n {x_i^3}  + c\sum\limits_{i = 1}^n {x_i^2}  + d\sum\limits_{i = 1}^n {{x_i}}  - \sum\limits_{i = 1}^n {{x_i}{y_i}}  = 0,

    \frac{{\partial S}}<br />
{{\partial d}} = a\sum\limits_{i = 1}^n {x_i^3}  + b\sum\limits_{i = 1}^n {x_i^2}  + c\sum\limits_{i = 1}^n {{x_i}}  + nd - \sum\limits_{i = 1}^n {{y_i}}  = 0.
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  6. #6
    Super Member Random Variable's Avatar
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    I would set up the system in matrix form and use the normal equation.

     A^{T}Ax = A^{T}b


    But it would probably be best to first factorize matrix A using QR decomposition.
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