Editing Polynomial interpolation (section)

==== Newton formula ====
Taking negative slope transversal from <math>y_0 </math> to <math>\Delta^n y_0 </math> gives the interpolation formula of all the <math>n+1 </math> consecutively arranged points, equivalent to Newton's forward interpolation formula:

<math>\begin{aligned}
y(s)  &=y_0+C(s, 1) \Delta y_0+C(s, 2) \Delta^2 y_0+C(s, 3) \Delta^3 y_0+\cdots \\
& =y_0+s \Delta y_0+\frac{s(s-1)}{2} \Delta^2 y_0+\frac{s(s-1)(s-2)}{3 !} \Delta^3 y_0+\frac{s(s-1)(s-2)(s-3)}{4 !} \Delta^4 y_0+\cdots      
\end{aligned}    </math>

whereas, taking positive slope transversal from <math>y_n </math> to <math>\nabla^n y_n = \Delta^n y_0 </math>, gives the interpolation formula of all the <math>n+1 </math> consecutively arranged points, equivalent to Newton's backward interpolation formula:

<math>\begin{aligned}
y(u) & =  y_k+C(u-k, 1) \Delta y_{k-1}+C(u-k+1,2) \Delta^2 y_{k-2} +C(u - k+2,3) \Delta^3 y_{k-3}+\cdots \\           
& = y_k+(u-k) \Delta y_{k-1} +\frac{(u-k+1) (u-k)}{2} \Delta^2 y_{k-2}+\frac{(u-k+2)(u-k+1)(u-k)}{3 !} \Delta^3 y_{k-3}+\cdots   \\   
y(k+s) & = y_k+(s) \nabla y_{k} +\frac{(s+1) s}{2} \nabla^2 y_{k}+\frac{(s+2)(s+1) s}{3 !} \nabla^3 y_{k}+\frac{(s+3)(s+2)(s+1) s}{4 !} \nabla^4 y_{k}+\cdots \\   
\end{aligned}    </math>

where <math>s=u-k </math> is the number corresponding to that introduced in Newton interpolation.