Interpolation and Regression → Explore with me!

Given $n+1$ data points $(x_{0},y_{0} ),(x_1,y_1 ),…….,(x_{n-1},y_{n-1}),(x_n,y_n)$ interpolation is the process of finding an equation $y=f(x)$ that passes through above $n+1$ data points and using this equation to find value of $y at x,x_0$ .

Errors in Polynomials

(eq 1) $E_n(x)=f(x)-P_n(x)$

Let us consider $P_n (x)$ => $n^{th}$ degree polynomial through the $n+1$ points $x_1,x_2,……,x_n$

Since both $f(x)$ and $P_n(x)$ have the same value at $x_i, where i=0,1,……n$

using eq 1

(eq 2) $\begin{split} E_n(x) & =f(x)-P_n(x) \\ & = (x-x_0)(x-x_1)(x-x_2)..........(x-x_n)g(x) \end{split}$

$g(x)$ represting error at the non-tabulated points

using eq 1

(eq 3) $f(x) - P_n(x) - E_n(x)=0$

assigning eq 2 in eq 3

$f(x) - P_n(x) - (x-x_0)(x-x_1)(x-x_2).....(x-x_n)g(x)=0$

let us define a function $W(t)$ , which is the function of $t$ and $x$

$W(t) = f(x) - P_n(x) - (x-x_0)(x-x_1)(x-x_2).....(x-x_n)g(x)=0$

at $t=x_i,$ where $i=0,1,.....,n$

$W(t)$ is zero and also at $t=x$ ,
There are about $n+2$ zero’s at $W(t)$
If we impose the law of mean value on $W(t)$ , then $W(t)$ must be continuous and differentiable. Also there exists a root of $W` (t)$ between each of the $(n + 2)$ zeros of $W(t)$ that is there and a total of $(n + 1)$ zeros of $W(t)$ . If there exists $W``(t)$ then there are $n$ zeros and similarly if the $(n + 1)$ derivative exists then there exist at least a zero in the interval of $(x_0, x_n)$
let us call this as $t = y$