Difference between revisions of "Lagrange Interpolation Formula"
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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 | + | This formula is useful for many olympiad problems, especially since such a polynomial is unique, like the following: |
+ | (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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Revision as of 16:42, 9 June 2020
For any distinct complex numbers and any complex numbers , there exists a unique polynomial of degree less than or equal to such that for all integers , , 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 that goes through the points and for . 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:
(United Kingdom) Let be distinct positive integers. Prove that so
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