Question : Is there a cubic polynomial that takes these values

$\displaystyle \begin{array}{|c|c|c|c|c|c|c|}x & 1 & -2 & 0 & 3 & -1 & 7 \\ \hline

y & -2 & -56 & -2 & 4 & 16 & 376 \end{array}$

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- Dec 29th 2009, 11:06 PMzorroIs there a cubic polynomial that takes...
Question : Is there a cubic polynomial that takes these values

$\displaystyle \begin{array}{|c|c|c|c|c|c|c|}x & 1 & -2 & 0 & 3 & -1 & 7 \\ \hline

y & -2 & -56 & -2 & 4 & 16 & 376 \end{array}$ - Dec 29th 2009, 11:34 PMMoo
Hello,

Just try it out !

A cubic polynomial is in the form $\displaystyle ax^3+bx^2+cx+d$ - Dec 29th 2009, 11:50 PMzorro
but what should i try to do .Do i stubstitute the values in the table, but where should i substitute

- Dec 29th 2009, 11:52 PMMoo
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 ;) - Dec 29th 2009, 11:56 PMzorro
but what about the y values what are those for???

- Dec 29th 2009, 11:57 PMMoo
- Dec 31st 2009, 04:30 PMzorro
- Dec 31st 2009, 05:50 PMSoroban
Hello, zorro!

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

Quote:

Is there a cubic polynomial that takes these values?

. . $\displaystyle \begin{array}{|c||c|c|c|c|c|c|}x & \text{-}2 & \text{-}1 & 0 & 1 & 3 & 7\\ \hline

y & \text{-}56 & 16 & \text{-}2 & \text{-}2 & 4 & 376 \end{array}$

The general cubic polynomial is: .$\displaystyle f(x) \:=\:ax^3 + bx^2 + cx + d $

. . and we must determine the four coefficients: $\displaystyle a,b,c,d.$

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

. . $\displaystyle \begin{array}{ccccccccc}

f(\text{-}2) = 56: & \text{-}8a + 4b - 2c + d &=& \text{-}56 \\

f(\text{-}1) = 16: & \text{-}a + b - c + d &=& 16 \\

f(0) = 0: & 0 + 0 + 0 + d &=& \text{-}2 \\

f(1) = \text{-}2: & a + b + c + d &=& \text{-}2 \end{array}$

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

Hence, the cubic is: .$\displaystyle f(x) \;=\;18x^3 + 9x^2 - 27x - 2$

. . which goes through the first four points.

But: .$\displaystyle f(3) \:=\:18(3^3) + 9(3^2) - 27(3) - 2 \;=\;322$

. . It doesgo through the fifth point.*not*

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

- Dec 31st 2009, 08:39 PMsimplependulum

we only need four points to constuct a cubic polynomial so

I choose those points which are located at $\displaystyle 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 ,

$\displaystyle -56 , 16 , -2 , -2 $

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

$\displaystyle \delta^2 = - 90 , 18 $

$\displaystyle \delta^3 = 108 $

Therefore , by using Newton's Interpolation Formula ,

the cubic polynomial is :

$\displaystyle -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. $\displaystyle x = 3 , x = 7 $

and see if the values the polynomial gives us equal to $\displaystyle 4 $ and $\displaystyle 376 $ respectively . - Jan 1st 2010, 01:26 AMzorro
thanks moo and simplependulum for ur post.........