Interpolate the function f(x) = 2* over the interval [1,1] by a polynomial (a) Construct the interpolating polynomial p2 (x) of degree n = 2 in the Newton form using uniformly spaced nodes. (b) Find an upper bound for the error |ƒ(x) — p₂(x)| at x = ½. (c) Find an upper bound for the error |f(x) - p₂(x)| on [-1,1]. Note that max |f(x) p₂(x)| ≤ x= [a,b] h³ M3 9√3 2 max f(x) pn(x)| ≤ x= [a,b] where h = (ba)/2 and |f"(x)| ≤ M3 for x = [a, b]. (d) How many equally-spaced nodes are required to interpolate the function f(x) to within € = 0.01? Note that for n > 2, 1 max | f(n+¹) (§) (n + 1)! € [a,b] hn+¹n! 4 where h = (b -a)/n.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
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Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 93E
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3. Interpolate the function f(x) = 2ª over the interval [−1, 1] by a polynomial
(a) Construct the interpolating polynomial p2(x) of degree n = 2 in the Newton form using uniformly
spaced nodes.
1
(b) Find an upper bound for the error |ƒ(x) — p2(x)| at x = 2
(c) Find an upper bound for the error |ƒ(x) — p2(x)| on [−1,1]. Note that
max f(x) p2(x)| ≤
x= [a,b]
h³ M3
9√3
9
max | f(x) — pn(x)| ≤
x= [a,b]
where h = (b − a)/2 and |ƒ""'(x)| ≤ M3 for x € [a, b].
(d) How many equally-spaced nodes are required to interpolate the function f(x) to within € = 0.01?
Note that for n > 2,
1
max | f(n+¹) (§) |
(n + 1)! E[a,b]
hn + 1n!
4
where h = (b − a)/n.
Transcribed Image Text:3. Interpolate the function f(x) = 2ª over the interval [−1, 1] by a polynomial (a) Construct the interpolating polynomial p2(x) of degree n = 2 in the Newton form using uniformly spaced nodes. 1 (b) Find an upper bound for the error |ƒ(x) — p2(x)| at x = 2 (c) Find an upper bound for the error |ƒ(x) — p2(x)| on [−1,1]. Note that max f(x) p2(x)| ≤ x= [a,b] h³ M3 9√3 9 max | f(x) — pn(x)| ≤ x= [a,b] where h = (b − a)/2 and |ƒ""'(x)| ≤ M3 for x € [a, b]. (d) How many equally-spaced nodes are required to interpolate the function f(x) to within € = 0.01? Note that for n > 2, 1 max | f(n+¹) (§) | (n + 1)! E[a,b] hn + 1n! 4 where h = (b − a)/n.
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