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Lagrange Interpolation Formula: Difference between revisions

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For any distinct [[complex number]]s <math> x_0, \ldots , x_n </math> and any complex numbers <math> y_0, \ldots, y_n </math>, there exists a unique [[polynomial]] <math> \displaystyle P(x) </math> of [[degree of a polynomial | degree]] less than or equal to <math> \displaystyle n </math> such that for all [[integer]]s <math> 0 \le i \le n </math>, <math> P(x_i) = y_i </math>, and this polynomial is
For any distinct [[complex number]]s <math> x_0, \ldots , x_n </math> and any complex numbers <math> y_0, \ldots, y_n </math>, there exists a unique [[polynomial]] <math>P(x) </math> of [[degree of a polynomial | degree]] less than or equal to <math>n </math> such that for all [[integer]]s <math> 0 \le i \le n </math>, <math> P(x_i) = y_i </math>, and this polynomial is
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While this formula may appear intimidating, it's actually not so difficult to see what is going on: for each term in the sum, we are finding a polynomial of degree <math>n</math> that goes through the points <math>(x_i,y_i)</math> and <math>(x_k,0)</math> for <math>k\neq i</math>. When we add them all together, we end up with a polynomial that interpolates the desired points.
While this formula may appear intimidating, it's actually not so difficult to see what is going on: for each term in the sum, we are finding a polynomial of degree <math>n</math> that goes through the points <math>(x_i,y_i)</math> and <math>(x_k,0)</math> for <math>k\neq i</math>. When we add them all together, we end up with a polynomial that interpolates the desired points.
This formula is useful for many olympiad problems, especially since such a polynomial is unique.




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Revision as of 07:27, 18 April 2012

For any distinct complex numbers $x_0, \ldots , x_n$ and any complex numbers $y_0, \ldots, y_n$, there exists a unique polynomial $P(x)$ of degree less than or equal to $n$ such that for all integers $0 \le i \le n$, $P(x_i) = y_i$, and this polynomial is

$P(x) = \sum_{i=0}^{n}y_i \frac{(x-x_0) \cdots (x-x_{i-1}) (x-x_{i+1}) \cdots (x-x_n)}{(x_i-x_0) \cdots (x_i-x_{i-1}) (x_i - x_{i+1}) \cdots (x_i - x_n)}$.

While this formula may appear intimidating, it's actually not so difficult to see what is going on: for each term in the sum, we are finding a polynomial of degree $n$ that goes through the points $(x_i,y_i)$ and $(x_k,0)$ for $k\neq i$. When we add them all together, we end up with a polynomial that interpolates the desired points.

This formula is useful for many olympiad problems, especially since such a polynomial is unique.


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