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  1. #1
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    Is there a cubic polynomial that takes...

    Question : Is there a cubic polynomial that takes these values

    \begin{array}{|c|c|c|c|c|c|c|}x & 1 & -2 & 0 & 3 & -1 & 7 \\ \hline<br />
y & -2 & -56 & -2 & 4 & 16 & 376 \end{array}
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  2. #2
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    Hello,

    Just try it out !

    A cubic polynomial is in the form ax^3+bx^2+cx+d
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    but what should i try to do .Do i stubstitute the values in the table, but where should i substitute
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    You substitute the x with the values your given.
    Then you get an equation with each of the first 4 values, which gives you a system of 4 equations with 4 unknowns (a,b,c,d).
    Then see if the remaining values are such that the equations are satisfied.

    I suggest you first consider 0,1,-1 because they're nice values to deal with
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    but what about the y values what are those for???
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    Quote Originally Posted by zorro View Post
    but what about the y values what are those for???
    y=ax^3+bx^2+cx+d<br />

    They're the values of the polynomial taken at a given x.
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  7. #7
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    Quote Originally Posted by Moo View Post
    You substitute the x with the values your given.
    Then you get an equation with each of the first 4 values, which gives you a system of 4 equations with 4 unknowns (a,b,c,d).
    Then see if the remaining values are such that the equations are satisfied.

    I suggest you first consider 0,1,-1 because they're nice values to deal with
    Putting x=1 in the eq
    a+b+c+d=0

    Putting x=0 in the eq
    d=0

    Putting x=-1 in the eq
    -a+b-c+d=0

    Is this what u meant ..........but i dont knw what are we doing
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  8. #8
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    Hello, zorro!

    You don't understand a thing we're saying, do you?


    Is there a cubic polynomial that takes these values?

    . . \begin{array}{|c||c|c|c|c|c|c|}x & \text{-}2 & \text{-}1 & 0 & 1 & 3 & 7\\ \hline<br />
y & \text{-}56 & 16 & \text{-}2 & \text{-}2 & 4 & 376 \end{array}

    The general cubic polynomial is: . f(x) \:=\:ax^3 + bx^2 + cx + d
    . . and we must determine the four coefficients: a,b,c,d.

    We can use the first four values of the function and create a system of equations.

    . . \begin{array}{ccccccccc}<br />
f(\text{-}2) = 56: & \text{-}8a + 4b - 2c + d &=& \text{-}56 \\<br />
f(\text{-}1) = 16: & \text{-}a + b - c + d &=& 16 \\<br />
f(0) = 0: & 0 + 0 + 0 + d &=& \text{-}2 \\<br />
f(1) = \text{-}2: & a + b + c + d &=& \text{-}2 \end{array}

    Solve the system of equations: . a = 18,\;b = 9,\;c = -27,\;d = -2

    Hence, the cubic is: . f(x) \;=\;18x^3 + 9x^2 - 27x - 2
    . . which goes through the first four points.

    But: . f(3) \:=\:18(3^3) + 9(3^2) - 27(3) - 2 \;=\;322
    . . It does not go through the fifth point.

    Therefore, there is no cubic polynomial through the given six points.

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  9. #9
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    Quote Originally Posted by zorro View Post
    Question : Is there a cubic polynomial that takes these values

    \begin{array}{|c|c|c|c|c|c|c|}x & 1 & -2 & 0 & 3 & -1 & 7 \\ \hline<br />
y & -2 & -56 & -2 & 4 & 16 & 376 \end{array}

    we only need four points to constuct a cubic polynomial so

    I choose those points which are located at  x = -2 , -1 , 0 , 1

    To find the cubic polynomial which passes through those points

    we need to find the first , second and the third difference ,


     -56 , 16 , -2 , -2

     \delta = 16 - (-56) , -2 - 16 , -2 - (-2) = 72 , -18 , 0

     \delta^2 = - 90 , 18

     \delta^3 = 108

    Therefore , by using Newton's Interpolation Formula ,

    the cubic polynomial is :

     -56 + 72 \binom{x+2}{1} - 90\binom{x+2}{2} + 108\binom{x+2}{3}


    To check whether there is a cubic polynomial passing through the other points either , we just sub.  x = 3 , x = 7
    and see if the values the polynomial gives us equal to  4 and  376 respectively .
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  10. #10
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    thanks moo and simplependulum for ur post.........
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