Finding Polynomial Equations from Table of Values

[F.B]

I need help with these sort of questions where you have given a set of
points and you must determine the equation, linear ones are easy but ones
that are quadratic or cubic. So could go about and tell me how to determine
equations of polynomials. Heres the set of points i am given.

x...y
-3...-110
-2...-21
-1...4
0...7
1...6
2...-5
3...-56
The dots separate the columns because they are supposed to be in a table of values.

So can anyone help me determine an equation for this. I know i have to find the differences in the y-column, so the fourth difference of the y-column are constant at -24, but what do i do now. I know the equation is in the form y=ax^4 + bx^3 + cx^2 + dx + e, but what do i do with the 24 and how do i solve the rest.

 

[arildno]

you are to find an interpolating polynomial, right?

Remember that you can always interpolate with a polynomial of degree one less than the number of points you've got.
that is:
You may use a sixth-degree polynomial here, if it is actually possible to interpolate with a polynomial of less degree, that will become apparent when solving the equations for the sixth-degree polynomial

 

[Fermat]

... but what do i do now. I know the equation is in the form y=ax^4 + bx^3 + cx^2 + dx + e, but what do i do with the 24 and how do i solve the rest.

You don't bother with the 24 now. You just used that to determine the order of the polynomial.

You have a 4th order polynomial with 5 unknowns, a,b,c,d,e.

Use 5 points from your data set, 5 (x,y)-values, to create 5 simultaneous eqns in a,b,c,d,e.
Then use matrix methods to solve for the unknowns, i.e. a,b,c,d,e.

 

[HallsofIvy]

Since you talk about the "fourth difference" apparently you know about "Newton's divided difference formula". It is essentially the same as Taylors series but works with finite differences:
P(x)= y(x_0)+ \Delta_1(x- x_0)+ \frac{\Delta_2}{2}(x-x_0)(x-x_1)+ ...+ \frac{\Delta_n}{n!}(x- x_0)(x-x_1)...(x- x_n)
x₀, x₁,etc are the x values starting at some point (in your case, x₀= -3, x₁= -2,etc.) The terms \Delta_1, etc are the divided differences (since your x-values are all unit steps apart, that's just the differences). Notice that you do not have powers of (x-x₀) but products of (x-x<sup>0[/sub](x- x₁) etc.

The computation involved is a bit easier than solving 4 equations for 4 unknowns. Also this has the advantage that it is easier to program.</sup>