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Thread: 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 $\displaystyle y = f\left( {x} \right) = 1.293{x^3} + 31.27{x^2} + 4.149x + 1688$

    then in 1925 year the population is $\displaystyle {y_{1925}} = f\left( {2.5} \right) = 1.293 \cdot {2.5^3} + 31.27 \cdot {2.5^2} + 4.149 \cdot 2.5 + 1688 \approx 1914$
    Last edited by DeMath; Aug 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 $\displaystyle P(x) = Ax^{3}+Bx^{2}+Cx + D $ where $\displaystyle P(x)$ is the population in year $\displaystyle x$

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

    $\displaystyle 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 $\displaystyle P(x) = Ax^{3}+Bx^{2}+Cx + D $ where $\displaystyle P(x)$ is the population in year $\displaystyle x$

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

    $\displaystyle 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??

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

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

    Then we have to solve the standard system of linear equations

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

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

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

    $\displaystyle \frac{{\partial S}}
    {{\partial d}} = a\sum\limits_{i = 1}^n {x_i^3} + b\sum\limits_{i = 1}^n {x_i^2} + c\sum\limits_{i = 1}^n {{x_i}} + nd - \sum\limits_{i = 1}^n {{y_i}} = 0.$
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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.

    $\displaystyle A^{T}Ax = A^{T}b $


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