So I was given the following table:
| $x$ | -2 | -1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| $y$ | 2 | 14 | 4 | 2 | 38 |
I'm supposed to find a polynomial of degree $3$ and $4$. I was able to find a polynomial of degree 3 using Newton's Method for polynomial interpolation (that is, $p_3(x)=5x^3+4x^2-11x+4$). But I'm completely lost on how to find a polynomial of degree $4$. Please Help!
On
Newton's method uses "divided differences". You are told that f(-2)= 2, f(-1)= 14,f(0)= 4,f(1)= 2, and f(2)= 38. The "differences"are 38-2= 36, 2- 4= -2, 4- 14= -10, an 14- 2= 12. (If the x values were different distances apart we would "divide" by those distances, but here then are all 1). The "second differences" are -2- 36= -38, -10-(-2)= -8, and 12- (-10)= 22. The "third differences are -8- (-38)= 30, and 22- (-8)= 30. Finally, the "fourth difference" is 30- 30= 0. (The fact that the fourth difference is 0 tells us that these points can be interpolated by a third degree polynomial as well as a fourth degree polynomial.)
The interpolating polynomial, by Newton's method, is $2+ 36(x+ 2)- (38/2)(x+ 2)^2+ 30/3!(x+ 2)^3+ (0/4!)(x+ 2)^4= 2+ 36(x+ 2)- 19(x+ 2)^2+ 5(x+ 2)^3= 2+ 36x+ 72- 19(x^2+ 4x+ 4)+ 5(x^3+ 6x^2+ 12x+ 8)= 5x^3+ 11x^2+ 20x+ 38$.
You can easily show that if $x_0< x_1 < \cdots < x_n$, there is one and only one polynomial of degree $\leq n$ such that $p(x_i) = y _i, i = 0, \cdots, n$. In this case, you have $n=4$ and the polynomial has degree $3 \leq 4$, which is fine.