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Interpolation and Regression


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

    \[ \begin{split} W^{n+1}(y)=0 & ={d^{n+1}\over{ (dt)^{n+1}}}  [f(t)-P_n (t)-(t-x_0 )(t-x_1 )………(t-x_n )g(x)] \\ & = f^{n+1}(y)-0-(n+1!)g(x) \end{split} \]

    \[ g(x) = {1 \over{(n+1)!}}f^{n+1}(y)  \]

    \[ E_n(x)={{(x-x_0)(x-x_1)(x-x_2).....(x-x_n)f^{n+1}(y)} \over {(n+1)!}} \]


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