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Lagrange Interpolation Polynomial Pdf Lagrange interpolation polynomial pdf Continue Lagrange Interpolation Polynomials If we want to describe all the ups and downs in the data set, and hit every point, we use what is called polynomial interpolation. This method is associated with Lagrange. Suppose the dataset consists of N data points: (x1, y1), (x2, y2), (x3, y3), ..., (xN, yN) Polynomial Interpolation will have a degree N - 1 . It is given: P (x) j1 (x) y1 y2 (x) y2 y3 (x) y3 ... It's easier than it looks. The main thing to note is that the ji numerator (x) contains the entire sequence of factors (x - x1), (x - x2), (x - x3), ... (x - xN), except for one factor (x - xi). Similarly, the denominator contains the entire sequence of factors (xi - x1), (xi - x2), (xi - x3), ... (xi - xN), except for one factor (xi - xi). Now note: ji (xi) No 1 (number and denominator), but ji (xj) 0 (number 0, as it contains the factor (xj - xj) ) for any j is not i. This means P (xi) and yi, which is exactly what we want! If this seems like smoke and mirrors, consider a simple example. Here's the data for g again: x g(x) 0 -250 10 0 20 50 30 -100 In this case there is a N 4 data point, so we will create a polynomial degree N - 1 and 3 . We have x1 y 0, x2 x 10, x3 x 20, x4 y 30, and y1 - 250, y2 y 0, y3 y 50, y4 and 100. So: Multiplying each of them appropriate yi (i q 1, 2, 3, 4) and adding together conditions of similar power, then gives: cubic conditions are canceled, and we come to a simple square description of the data. A quick data story along with a polynomic shows that it actually passes through each of the data points: For an interactive demonstration of Lagrange's polynomialization, showing how changes in data points affect the resulting curve, go here. This image shows four dots ((No9, 5), (No4, 2), (No1, No2), (7, 9)), (cubic) interpolation of polynomial L (x) (dashed, black), which is the sum of the scaled polynomial bases y0l0 (x), y1l1 (x), y2l2 (x) and y3l3 (x). Polynomial interpolation passes through all four checkpoints, and each scaled polynomial base passes through the appropriate checkpoint and is 0, where x corresponds to the other three checkpoints. In numerical analysis, Lagrange polynomials are used for polynomial interpolation. For this set of dots (x j, y j) (display (x_ j',y_'j)) without two x j (display x_ j'j) values are equal, Lagrange-polynomial is the least polynomial, which assumes at each value x j 'displaystyle x_ j' appropriate value y j 'displaystyle y_'j' Although named after Joseph-Louis Lagrange, who in 1795, the method was first discovered in 1779 by Edward Waring, which is also a simple consequence of a formula published in 1783 by Leonhard Euler. The use of Lagrange polynomials includes the Newton-Cote numerical integration method and Shamir's secret cryptography exchange scheme. Lagrange interpolation is subject to the phenomenon of large swings of Runge. Because changing the x j'displaystyle points x_ requires a recalculation of the entire interpolant, it is often easier to use Newton's polynomials instead. Definition Here we are building Lagrange basic functions of the 1st, 2nd and 3rd order on a two-unit domain. The linear combinations of Lagrange's base functions are used to construct the polynomial interpolation lagrange. Lagrange's basic functions are commonly used in the analysis of the end elements as the basis for the function of the form of the element. In addition, two domain units are typically used as a natural space to determine the final item. Given the set of k 1 data points (x 0, y 0), ... , (x j, y j), ... ( x k, y k y_{0} x_{0}) (x_, y_ jj), ldots, (x_'y_'k) where there are no two x 'displaystyle x_ j'j), Interpolation polynomial in the form of Lagrange is a linear combination of L (x) : ∑ j y 0 k y j j ( x) : 'sum zj'y_ 'j' (x) of Lagrange basis polynomials l j (x) := ∏ 0 ≤ m ≤ k m ≠ j x − x m x j − x m = ( x − x 0 ) ( x j − x 0 ) ⋯ ( x − x j − 1 ) ( x j − x j − 1 ) ( x − x j + 1 ) ( x j − x j + 1 ) ⋯ ( x − x k ) ( x j − x k ) , displayell j'j (x): Prod start smallmatrix0'leq m'leq k'meq j'end'smallmatrix frac (x-x_'m'm'm'x_ yoya-x_mhamemerrak (x-x_{0}) (x_ jj x_{0} x_) (x_-y-x_ j x_-x_ x_1) frak (x-x_'k) (x_ j'-x_'k), where 0 ≤ jstyle ≤ k display 0leq j'leq k. I don't know Note that, given the initial assumption that there are no two x j'displaystyle x_ j'j' are the same, then (when m ≠ j 'displaystyle meq j) x j x m ≠ 0 displaystyle x_ j -x_'m'eq 0 so this expression is always clearly defined. The reason par x i x display style x_i'x_'j'yo y i'≠ y j display style y_i'eq y_ J e is not allowed in that no interpolation function L displaystyle L is such that i and L (x i) displaystyle y_ I L (x_'i) will exist; the feature can only get one value for each argument x i displaystyle x_ i. On the other hand, if also I and y j 'displaystyle y_ 'i'y_'j' then these two points would actually be one point. For all, ≠ j 'displaystyle ieq j', l j (x) (display style (x)) includes the term (x - x i) in the numerator (x-x_), so the whole product will be zero on x x i display style x x_ i: ∀ (j ≠ i) : l (x i ≠ ∏) м.з. (х я х х 0) ( х х х х 0) ⋯ ( х я х х х х х х х х х 0 ) ( х х х х х х я) ( х й х х я ) ⋯ ( х я х к) ( х х х х к) 0. Displaystyle forall jeq i): ell j (x_i)prod meq j'frac (x_x_'m'x_ j'-x_'m'frac (x_'-x_{0}) (x_ (x_{0}) (x_-I-x_) (x_ j'-x_'i) (x_-x_'k) (x_ j x_'k). С другой стороны, l j ( x j ) ∏ м ≠ j x j х x j х м х х х 1 «дисплей»ell j'j'(x_'j)): »prod »meq j'frac »x_ j»-x_'m'x_'j'j x_'m все базисные полиномиалы являются нулевыми на х х х я «displaystyle x»x_'i» , за исключением l j ( x ) »displaystyle »элл йо j»(x)» for which he believes that l j (x j) - 1 x_ (display style) because it lacks (x x j) (x j) (x-x_)) term. It follows that y j l j (x j) - y j 'displaystyle y_ j'ell j'j' (x_ j'j) - y_ j'j' , so at each point x j 'displaystyle x_ j ' , L (x j) j - 0 - 0 - ⋯ - 0 - y j 'displaystyle L (x_'j)) y_j'0'0'dots (0'y_'j' showing that L-displaystyle L' interpolates function accurately. Proof of function L(x) is sought is polynomial in x least that interpolates this data set. that is, it takes the value of yj on the corresponding xj for all data points J: L (x j) and y j y j 0 , ... , k. Displaystyle L (x_ j) y_ j'j'qquad j'0, 'ls,k.) Note this: In l j (x) displaystyle (el yo (x) there are k factors in the product, and each factor contains one x, so that L (x) (which is the sum of these k-degree polynomials) should be a polynomial degree no more than k. l j (x i ≠ ∏) Displaystyle ' ell j (x_) start smallmatrixm'0'meq j'end'smallmatrix-k'frac (x_'i'-x_'m'x_'j'j'x_. , где м й j, если я й J, то все термины, которые появляются являются х j х х х х х х х 1 дисплей стилем frac x_j'-x_'m'x_ jj'-x_'m'1 . Also, if I ≠ j, then one term in the product will be (for m and i), x i x i x i x i x i 0 display style frak x_ x_ I-x_ x_ So L j (x i) - δ j i 1 , x_ if j and i 0 if j ≠ i, cases No1, text if j i'0 , text if jeq iend where δ i j displaystyle delta ij is the Kroneker Delta. So: L (x i) - ∑ j 0 k y j j (x i) - ∑ j 0 k y j δ j i and i i. (L (x_) amount (j'0'k) y_ j'j'ell (x_'i) amount y_ (j'0'k'y_ j'delta So L/x function is polynomial with no more than k and where L (xi) and yi. , polynomial interpolation is unique, as evidenced by the monotheration theorem in the article of polynomial interpolation. It is also true that: ∑ j y 0 k j (x) 1 ∀ x 'display style 'j'0'k'ell j'j'(x) 1qquad 'forall x' it should be a polynomial degree no more than to and passes through all these k 1 data points: ( x 0, 1) , (x j, 1) , ... ( x k, 1) displaystyle (x_{0}.1), ldots,' (x_ j',1), ldots, (x_'k,1) resulting in a lineline. The prospect of linear algebra Solution to the problem of interpolation leads to a problem of linear algebra, equivalent to matrix inversion.
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