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Lagrange Interpolating Polynomial

Tool to find the equation of a function. Lagrange Interpolating Polynomial is a method for finding the equation corresponding to a curve having some dots coordinates of it.

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Lagrange Interpolating Polynomial -

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# Lagrange Interpolating Polynomial

## Lagrange Interpolation Calculator

### Extrapolation

Tool to find the equation of a function. Lagrange Interpolating Polynomial is a method for finding the equation corresponding to a curve having some dots coordinates of it.

### How to find the equation of a curve using Lagrange?

Lagrange polynomials are computed using the formula :

$$P(X)=\sum_{j=0}^n y_j \left(\prod_{i=0,i\neq j}^n \frac{X-x_i}{x_j-x_i} \right)$$

with $P(X)$ the Lagrange polynomial and the dots $(x_0, y_0),\dots,(x_n, y_n)$ and $x_i$ distinct.

From the points whose coordinates are known, the lagrange polynomial calculator can thus predict other points based on the assumption that the curve formed by these points is derived from a polynomial equation.

dCode allow to use the Lagrangian method for interpolating a Polynomial and finds back the original equation using known points (x,y) values.

Example: By the knowledge of the points $(x,y)$ : $(0,0),(2,4),(4,16)$ the Polynomial Lagrangian Interpolation method allow to find back the équation $y = x^2$. Once deducted, the interpolating function $f(x) = x^2$ allow to estimate the value for $x = 3$, here $f(x) = 9$.

The Lagrange interpolation method allows a good approximation of polynomial functions.

There are others interpolation formulas (rather than Lagrange/Rechner) such as Neville interpolation also available online on dCode.

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

Since the complexity of the calculations increases with the number of points, the program is limited to 25 coordinates (with distinct x-values in the rational number set Q).

### How to calculate/anticipate another value?

From a list of numbers, the Lagrange interpolation allows to find an equation for $f(x)$. Using this equation with a new value of $x$, it is possible to calculate the image of $x$ by $f$.

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