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Lagrange Interpolating Polynomial

Tool for calculating the Lagrange interpolating polynomial from a list of points. Lagrange polynomial interpolation allows us to find the equation of a polynomial function and perform extrapolation.

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Lagrange Interpolating Polynomial -

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Lagrange Interpolating Polynomial

Lagrange Interpolation Calculator


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Find the equation of a curve

Answers to Questions (FAQ)

What is the Lagrange Interpolation? (Definition)

Lagrange interpolation is a method for determining the polynomial $ P $ of degree $ n $ that passes through $ n+1 $ given distinct points $ \{ (x_0,y_0), (x_1,y_1), \dots, (x_n,y_n) \} $

The Lagrange interpolating polynomial can be written in the form $$ P(X) = \sum_{j=0}^n y_j \left(\prod_{i=0,i\neq j}^n \frac{X-x_i}{x_j-x_i} \right) $$

It checks that $ P(x_j) = y_j $ for all $ j $, which ensures that the curve passes exactly through the given points.

For distinct abscissas, there exists a unique Lagrange interpolating polynomial.

How to find the equation of a curve using Lagrange?

From a set of points with known coordinates, Lagrange interpolation allows us to explicitly construct the polynomial $ P(x) $. This polynomial can then predict other points (by extrapolation) based on the assumption that the curve formed by these points comes from a polynomial-type equation.

Example: By the knowledge of the points $ (x,y) $ : $ (0,0),(2,4),(4,16) $ the Polynomial Lagrangian Interpolation method allows to find back the équation $ y = x^2 $. Calculations details step by step: $$ P(x) = 0 \times \frac{(x-2)}{(0-2)} \frac{(x-4)}{(0-4)} + 4 \times \frac{(x-0)}{(2-0)} \frac{(x-4)}{(2-4)} + 16 \times \frac{(x-0)}{(4-0)} \frac{(x-2)}{(4-2)} \\ = 4 \times \frac{x}{2}\frac{(x-4)}{(-2)} + 16 \times \frac{x}{4}\frac{(x-2)}{2} \\ = -x(x-4)+2x(x-2) \\ = -x^2+4x+2x^2-4x \\ = x^2
$$ Once deducted, the interpolating function $ f(x) = x^2 $ allows to estimate the value for $ x = 3 $, here $ f(x) = 9 $.

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

What are the limits for interpolating with Lagrange?

Lagrange interpolation requires that the x-coordinates $ x_i $ be distinct in pairs, a necessary condition to guarantee the existence and uniqueness of the interpolating polynomial.

As the number of points increases, the degree of the polynomial also increases, making the expression more complex and increasing the computational cost.

From a numerical perspective, the method can become unstable with many points or when they are poorly distributed, as rounding errors can be amplified. Furthermore, for equidistant points and a high degree, the polynomial can exhibit strong oscillations at the ends of the interpolation interval, a phenomenon known as the Runge phenomenon.

How to calculate/anticipate another value?

For estimating a new value $ x $, evaluate $ P(x) $:

— if $ x $ belongs to the interval $ [\min x_i,\max x_i] $, it is an interpolation;

— otherwise, it is an extrapolation, generally less reliable.

The quality of the estimation depends heavily on the number of points used and their distribution.

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