3.4 Hermite Interpolation 3.5 Cubic Spline Interpolation
[Pages:14]3.4 Hermite Interpolation 3.5 Cubic Spline Interpolation
1
Hermite Polynomial
Definition. Suppose 1[, ]. Let 0, ... , be distinct numbers in [, ], the Hermite polynomial
() approximating is that:
1. = , for = 0, ... ,
2.
= ,
for = 0, ... ,
Remark: () and () agree not only function values but also 1st derivative values at , = 0, ... , .
2
Theorem. If 1 , and 0, ... , , distinct
numbers, the Hermite polynomial is:
2+1 = ,() + ,()
=0
=0
Where
, = [1 - 2( - ),()]2,()
, = - 2,
Moreover, if 2+2 , , then
= 2+1
+
- 0 2 ... - 2 + 2 !
2
2+2
(())
for some , .
Remark:
1. 2+1 is a polynomial of degree at most 2 + 1.
2. ,() is jth Lagrange basis polynomial of degree .
3.
-0 2... - 2+2 !
2
2+2
(())
is
the
error
term.
3
3rd Degree Hermite Polynomial
? Given distinct 0, 1 and values of and at these
numbers.
3 =
1
+
2
1
- 0 - 0
1 - 1 - 0
2
0
+ - 0
1 - 1 - 0
2
0
+
1
+
2
1 - 1 - 0
0 - 0 - 1
2
1
+ - 1
0 - 0 - 1
2
1
4
Hermite Polynomial by Divided Differences
Suppose 0, ... , and , are given at these numbers.
Define 0, ... , 2+1 by
2 = 2+1 = ,
for = 0, ... ,
Construct divided difference table, but use 0 , 1 , . . ,
to set the following undefined divided difference:
0, 1 , 2, 3 , ... , 2, 2+1 .
The Hermite polynomial is
2+1
2+1 = 0 +
0, ... , - 0 ... ( - -1)
=1
5
Divided Difference Notation for Hermite Interpolation
? Divided difference notation: 3 = 0 + 0 - 0 + 0, 0, 1 - 0 2 + [0, 0, 1, 1] - 0 2( - 1)
6
Problems with High Order Polynomial Interpolation
? 21 equal-spaced numbers to interpolate
=
1 1+162
.
The
interpolating
polynomial
oscillates between interpolation points.
7
Cubic Splines
? Idea: Use piecewise polynomial interpolation, i.e, divide the interval into smaller sub-intervals, and construct different low degree polynomial approximations (with small oscillations) on the sub-intervals.
? Challenge: If () are not known, can we still generate interpolating polynomial with continuous derivatives?
8
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