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COMP 361 or COMP 5611 complete Assignment 4 fall Concordia University
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COMP 361 or COMP 5611 complete Assignment 4 fall Concordia University
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COMP 361 or COMP 5611 complete Assignment 4fall 2024-2025 Concordia University
Solutions Problem 1. (20%)
(a) If we approximate f (x) = 11+x
on [0, 1] with local polynomial interpolation using 3 equally
spaced interpolation points per interval, then how many intervals of equal size are
needed to ensure that the maximum error is less than 10−4?
(b) Let N be the number of intervals found in part (a). Using local polynomial
interpolation with N intervals and 3 equally spaced interpolation points per interval,
what approximation do we get for f (0.3)?
(c) Repeat part (a) for arbitrary local interpolation points.
Solution
(a) Let N be the number of intervals, h = 1/N and n = 2. The maximum error |f (x) −
(n+1)
(ξ) n+1
pj (x)| is f(n+1)! 2h Cn for some ξ ∈ [0, 1]. By differentiating f (x), we have f (n+1)
(x) =
(— 1) (n + 1)!(1 + x)
n+1 −(n+1)
. For x in [0, 1], | f (n+1)(x) is at most (n + 1)!. So a
|
bound on the error is
.. f 3!(ξ)..
(3)
3
2h C2
≤. 3 C2
2
.. 3!. h
3!
1
=
(2N )3
C2 2 = 0.38490 from
4
1/3
2
To make this less than 10−4, we need N8 > 10 C . Using the
value of C
page 183 of the lecture notes, this is 7.835 so N = 8 guarantees the error is less than 10−4.
(b) Using N = 8 intervals, 0.3 lies in the third interval,
8 8
[ 2 , 3 ]. The three interpolation points
are x4,0 = 2 , x4,1 = 5 and x4,2 = 3 . The interpolation polynomial p4(x) is
8 16 8
p4(x) = f (x4,0)l0(x) + f (x4,1)l1(x) + f (x4,2)l2(x).
1
, Then p4(0.3) = 0.76925.
2
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