Difference between revisions of "Lagrange Interpolation Formula"

Revision as of 16:55, 9 June 2020

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.

This article is a stub. Help us out by expanding it.

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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.
-This formula is useful for many olympiad problems, especially since such a polynomial is unique, like the following:
+This formula is useful for many olympiad problems, especially since such a polynomial is unique.
-(United Kingdom) Let <math>{a_1,a_2,\cdots,a_n}</math> be distinct positive integers. Prove that <math>\forall k\in 	\mathbb{Z}</math> so <math>\sum_{i=1}^{n} \frac{a^k}{\prod_{i\neq j}(a_i-a_j) }\in 	\mathbb{Z}</math>
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Difference between revisions of "Lagrange Interpolation Formula"

Revision as of 16:55, 9 June 2020