"Lagrange's interpolation formula" meaning in English

See Lagrange's interpolation formula in All languages combined, or Wiktionary

Noun

Etymology: Named after Joseph Louis Lagrange (1736–1813), an Italian Enlightenment Era mathematician and astronomer. Head templates: {{en-noun|-}} Lagrange's interpolation formula (uncountable)
  1. (mathematics) A formula which when given a set of n points (x_i,y_i), gives back the unique polynomial of degree (at most) n − 1 in one variable which describes a function passing through those points. The formula is a sum of products, like so: ∑ᵢⁿy_i∏_(j ne i)x-x_j/x_i-x_j. When x=x_i then all terms in the sum other than the iᵗʰ contain a factor x-x_i in the numerator, which becomes equal to zero, thus all terms in the sum other than the iᵗʰ vanish, and the iᵗʰ term has factors x_i-x_j both in the numerator and denominator, which simplify to yield 1, thus the polynomial should return y_i as the function of x_i for any i in the set 1,...,n. Wikipedia link: Joseph Louis Lagrange, Lagrange polynomial Tags: uncountable Categories (topical): Mathematics

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        "A formula which when given a set of n points (x_i,y_i), gives back the unique polynomial of degree (at most) n − 1 in one variable which describes a function passing through those points. The formula is a sum of products, like so: ∑ᵢⁿy_i∏_(j ne i)x-x_j/x_i-x_j. When x=x_i then all terms in the sum other than the iᵗʰ contain a factor x-x_i in the numerator, which becomes equal to zero, thus all terms in the sum other than the iᵗʰ vanish, and the iᵗʰ term has factors x_i-x_j both in the numerator and denominator, which simplify to yield 1, thus the polynomial should return y_i as the function of x_i for any i in the set 1,...,n."
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        "(mathematics) A formula which when given a set of n points (x_i,y_i), gives back the unique polynomial of degree (at most) n − 1 in one variable which describes a function passing through those points. The formula is a sum of products, like so: ∑ᵢⁿy_i∏_(j ne i)x-x_j/x_i-x_j. When x=x_i then all terms in the sum other than the iᵗʰ contain a factor x-x_i in the numerator, which becomes equal to zero, thus all terms in the sum other than the iᵗʰ vanish, and the iᵗʰ term has factors x_i-x_j both in the numerator and denominator, which simplify to yield 1, thus the polynomial should return y_i as the function of x_i for any i in the set 1,...,n."
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