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.
Lagrange Interpolating Polynomial - dCode
Tag(s) : Functions
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Lagrange Interpolating Polynomial
Lagrange Interpolation Calculator
Answers to Questions (FAQ)
What is the Lagrange Interpolation? (Definition)
The Lagrangian interpolation (known as Lagrange/Rechner) is a method which makes it possible to find the equation of a polynomial function which passes through a series of $ n $ given points $ \{ (x_0,y_0), (x_1,y_1), \dots, (x_n,y_n) \} $.
The Lagrange polynomial is calculated by the formula $$ P(X) = \sum_{j=0}^n y_j \left(\prod_{i=0,i\neq j}^n \frac{X-x_i}{x_j-x_i} \right) $$
How to find the equation of a curve using Lagrange?
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 allows 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 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 $.
The Lagrange interpolation method allows a good approximation of polynomial functions.
There are other 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 automatically limited (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 $ by extrapolation.
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