数值分析-5[1].1 Lagrange Interpolation

See the textbook p.091.
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½n5.1.1 (35Ú5) ®.: Ù¥x , x , · · · , x *dpɧK3õªp (x) ∈ P ÷v¢ ^
0 1 n n n
(x0, f (x0)), (x1, f (x1)), · · · , (xn, f (xn)),
where x0, x1, · · · , xn are distinct. Then there exists unique polynomial pn(x) ∈ Pn, which passes through these points, le Page
Proof
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(x0, f (x0)), (x1, f (x1)), · · · , (xn, f (xn)), pn(xi) = f (xi), for i = 0, 1, 2, · · · , n.
Theorem 5.1.1 (Existence and Uniqueness) Given the points
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Example 5.1.1 Using the data f (−1) = 2, f (1) = 1, f (2) = 1 to find a Lagrange interpolating polynomial for f (x). Solution From the assumption we denote x0 = −1, x1 = 1, x2 = 2; y0 = 2, y1 = 1, y2 = 1, and (5.1.6) gives l0(x) = l1(x) = and l2(x) = (x − 1)(x − 2) 1 (x − x1)(x − x2) = = (x2 − 3x + 2), (x0 − x1)(x0 − x2) (−1 − 1)(−1 − 2) 6
f (xi) = yi,
Kõª ÷v¢^ Ù¥
n
i = 0, 1, · · · , n.
n
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pn(x) = y0l0(x) + y1l1(x) + · · · + ynln(x) =
k=0
yk lk (x)
(5.1.4)
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pn(xi) = yi,
n
i = 0, 1, · · · , n, k = 0, 1, · · · , n.
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pn(xi) = f (xi),
i = 0, 1, 2, · · · , n.
5.1.3 Lagrange Interpolating Polynomial
Theorem 5.1.2 (Lagrange Interpolating Polynomial) Suppose x0, x1, · · · , xn are n + 1 distinct numbers and f is a function whose values are given at these numbers: f (xi) = yi, for i = 0, 1, · · · , n. Then the polynomial
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(x + 1)(x − 2) 1 (x − x0)(x − x2) = = − (x2 − x − 2), (x1 − x0)(x1 − x2) (1 + 1)(1 − 2) 2 (x + 1)(x − 1) 1 2 (x − x0)(x − x1) = = (x − 1). (x2 − x0)(x2 − x1) (2 + 1)(2 − 1) 3
n
pn(x) = y0l0(x) + y1l1(x) + · · · + ynln(x) =
k=0
yk lk (x)
(5.1.4)
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satisfies the interpolating conditions pn(xi) = yi, for each i = 0, 1, · · · , n, where lk (x) =
n
(5.1.1)
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P L«¤kgêØL n êõª|¤8Ü"®
(x0, f (x0)), (x1, f (x1)), · · · , (xn, f (xn)),
n
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Ù¥: x , x , · · · , x *dpɧ¿¡¢!:" ¢ ¯Kµ ¦õªp (x) ∈ P §¦÷v
.KF¢
5.1.1 Problem
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5.1.2 Existence and Uniqueness of the Interpolating Polynomial
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5.1.3 Lagrange Interpolating Polynomial
i=0,i=k n
(5.1.5)
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x − xi , xk − xi
for k = 0, 1, · · · , n.
(5.1.6)
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The polynomial pn(x) defined in (5.1.4) is called the nth Lagrange interpolating polynomial, and l0(x), l1(x), · · · , ln(x) are called Lagrange bases. Proof See the textbook p.092.
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From (5.1.4), we get p2(x) = y0l0(x) + y1l1(x) + y2l2(x) = x + 1, which is a polynomial with degree one.
.KF¢õª .KFØúª
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5.1.4 Lagrange Error Formula
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5.1.1 Problem
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Let Pn be the set of algebraic polynomials of degree at most n, and the points (x0, f (x0)), (x1, f (x1)), · · · , (xn, f (xn)) be given, where x0, x1, · · · , xn are distinct and called the interpolating nodes. The problem is to find a polynomial pn(x) ∈ Pn, which called the interpolating polynomial, to passes through these points, that is, pn(xi) = f (xi), for i = 0, 1, 2, · · · , n, which are called the interpolating conditions.
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l1(x) = and l2(x) =
(x − x0)(x − x2) (x − 0)(x − 2) = = −(x2 − 2x), (x1 − x0)(x1 − x2) (1 − 0)(1 − 2) (x − x0)(x − x1) (x − 0)(x − 1) 1 2 = = (x − x). (x2 − x0)(x2 − x1) (2 − 0)(2 − 1) 2
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From (5.1.4), we have 1 p2(x) = y0l0(x) + y1l1(x) + y2l2(x) = (x2 − 3x + 8), 6 which is the Lagrange interpolating polynomial for f (x) we want.
0 1 n n
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¿¡ p (x) ¢õª§¡^(5.1.1)¢^" ¢ ¢
n
pn(xi) = f (xi),
i = 0, 1, 2, · · · , n,
(5.1.1)
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5.1.2 Existence and Uniqueness of the Interpolating Polynomial
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5.5 Piecewise Polynomial Interpolation
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5.6 Cubic Spline Interpolation
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5.1 Lagrange Interpolation
0 1
(5.1.5) (5.1.6)
n
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lk (x) =
i=0,i=k
¡õª p (x) ng . K F ¢ õ ª§l (x), l (x), · · · , l (x)¡ .KFļê" .
x − xi , xk − xi
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Chapter 5 Interpolation
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5.1 Lagrange Interpolation