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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

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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.

Answers to Questions

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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