Solve this using python$/$Jupyter Notebook

Input: A set of values $y_1,y_2,$cdots,$y_n$ and the corresponding

coordinate vectors $x_1,x_2,$cdots,$x_n.$

Input: A coordinate vector of a point of interest $c_j,$ for which an

estimated value will be computed.

Output: The interpolated value hat$(y{)}_j^R,$ computed for the location $c_j.$

nlarr the number of input points;

for all input points iin$\{1,2,$cdots,$n\}$ do

$d_{i,j}$larr Euclidean distance between $x_i$ and $c_j$;

for all input points iin$\{1,2,$cdots,$n\}$ do

$w_{i,j}$larr$\frac{d_{i,j}^{-2}}{{\displaystyle \underset{k=1}{\overset{n}{}}}{d}_{k,j}^{-2}}$

$6$hat$(y{)}_j^{\text{I}\text{D}\text{W}}$larr${\displaystyle \underset{i=1}{\overset{n}{}}}{w}_{i,j}{y}_i$

$7$hat$(y{)}_j^R$larrhat$(y{)}_j^{\text{I}\text{D}\text{W}}+n\frac{{\displaystyle \underset{i=1}{\overset{n}{}}}{y}_i-\text{n}\text{h}\text{a}\text{t}(y{)}_j^{\text{I}\text{D}\text{W}}}{n^2-{\displaystyle \underset{i=1}{\overset{n}{}}}{d}_{i,j}^{-2}{\displaystyle \underset{i=1}{\overset{n}{}}}{d}_{i,j}^2}$

return hat$(y{)}_j^R$

Note that $x_n$ represents a coordinate vector, i$.$e$.,x=x$ in $2$D or $y=f(x),$ and $x=(x_1,x_2)$ in $3$D or $y=f(x_1,x_2).$

Part A

Demonstrate how the algorithm described above works for the following set of coordinates $(x,Y)$

$x=\left[\begin{array}{c}0\mathrm{.}0\\ 0\mathrm{.}05\\ 0\mathrm{.}01\\ 0\mathrm{.}2\\ 0\mathrm{.}8\\ 0\mathrm{.}85\\ 0\mathrm{.}9\\ 1\mathrm{.}0\end{array}\right],Y=\left[\begin{array}{c}0\mathrm{.}0\\ 0\mathrm{.}7\\ 1\mathrm{.}0\\ 1\mathrm{.}0\\ 0\mathrm{.}3\\ 0\mathrm{.}05\\ 0\mathrm{.}1\\ 1\mathrm{.}0\end{array}\right]$

and show that the interpolated values for $x=\{0\mathrm{.}02,0\mathrm{.}18,0\mathrm{.}4,0\mathrm{.}6,0\mathrm{.}89\}$ are $y=\{0\mathrm{.}7925,0\mathrm{.}9848,0\mathrm{.}9959,0\mathrm{.}1680,0\mathrm{.}1057\}$ respectively.

For Part A show a detailed solution. A step by step process, do not use a 'def' function

yet.