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Newton Interpolating Polynomial

Tool to find the equation of a curve via Newton's algorithm. Newtonian Interpolating algorithm is a polynomial interpolation/approximation allowing to obtain the Lagrange polynomial as equation of the curve by knowing its points.

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Newton Interpolating Polynomial -

Tag(s) : Functions

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# Newton Interpolating Polynomial

## Newtonian Interpolation Calculator

### Extrapolation

Tool to find the equation of a curve via Newton's algorithm. Newtonian Interpolating algorithm is a polynomial interpolation/approximation allowing to obtain the Lagrange polynomial as equation of the curve by knowing its points.

### How to find the equation of a curve using Newton algorithm?

dCode allows to use Newton's method for Polynomial Interpolation in order to find the equation of the polynomial (identical to Lagrange) in the Newton form from the already known values of the function..

From $n + 1$ known points $(x_i, y_i)$, the Newton form of the polynomial is equal to $$P(x)= [y_0] + [y_0,y_1] (x-x_0) + \ldots + [y_0,\ldots ,y_n] (x-x_0) \ldots (x-x_{n-1})$$

with the notation $[y_i]$ for divided difference.

Example: Curve whose points (1,3) and (2,5) are known. $$P(x) = [y_0] + [y_0,y_1] (x-x_0) \\ = 3 + \left(\frac{3}{1-2}+\frac{5}{2-1}\right) (x-1) = 3+2(x-1) = 2x+1$$

### What are the limits for Interpolating with Newton?

Newton Divided Differences are noted $[y_i]$ and computed by the formula $$[y_0,\dots ,y_k]=\sum_{j=0}^k {\frac{y_j}{\prod_{0\leq i\leq k,\,i\neq j}(x_j-x_i)}}$$

NB: If $k = 0$, then the product $\prod(x_j-x_i) = 1$ (empty product)

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