Numerical Methods, Neville's Method Missing Values

Question

I tried to solve question below for over 1 hour, and can't figure it out. Can somebody help me? Question

Answer 1

Given

$$\begin{array}{|c|c|c|c|c|} \hline x_i&y_i& & & \\ \hline x_0 = 1&y_0 = P_0 = a\\ \hline x_1 = 3&y_1 = P_1 = 1.73 & P_{01}= c\\ \hline x_2 = 4&y_2 = P_2 = 2& P_{12} = 1.865 & P_{012} = 1.8729 \\ \hline \hline x_3 = 6&y_3 = P_3 = b & P_{23} = 1.8875 & P_{123} = d & P_{0123} = e\\ \hline \end{array}$$

Using Neville's Algorithm and solving for values in this order, we have

$P_{12}(x) = \dfrac{(x-x_2)P_1 + (x_1-x)P_2}{x_1-x_2} = 1.865$, so $\fbox{x = 3.5}$

$P_{23}(x) = \dfrac{(x-x_3)P_2 + (x_2-x)P_3}{x_2-x_3} = 1.8875$, so $\fbox{b = 2.45}$.

$P_{012} = \dfrac{(x-x_2)P_{01} + (x_0-x)P_{12}}{x_0-x_2} = 1.8729$, so $\fbox{c = 1.9124}$

$P_{123}(x) = \dfrac{(x-x_3)P_{12} + (x_1-x)P_{23}}{x_1-x_3} = d$, so $\fbox{d = 1.86875}$

$P_{0123}(x) = \dfrac{(x-x_3)P_{012} + (x_0 - x)P_{123}}{x_0-x_3} = e$, so $\fbox{e = 1.87083}$

$P_{01}(x) = \dfrac{(x-x_1)P_0 + (x_0-x)P_1}{x_0-x_1} = c$ so $\fbox{a = 1.0004}$