Technische Universität München
Fakultät für Mathematik
Prof. Dr. M. Mehl
B. Gatzhammer
WT 2012/13
November 22, 2012
Numerical Programming I (for CSE)
Tutorial 6: Interpolation I
I 1) The Scheme of Aitken and Neville
Let p : R → R be a polynomial interpolant with support points Pi = (xi , yi ), i = 0, . . . , n.
Compute the value p(2) of the cubic polynomial with support points
P0 = (−2, −7), P1 = (−1, 0), P2 = (0, 1), P3 = (1, 2).
Use the scheme of Aitken and Neville!
I 2) Newton’s Interpolation Formula
Let p : R → R be a polynomial interpolant with support points Pi = (xi , yi ), i = 0, . . . , n.
Given the support points
P0 = (−1, −3), P1 = (1, 1), P2 = (3, −3),
find a closed representation of p that is based on
(i) Lagrange polynomials,
(ii) divided differences (Newton’s interpolation formula).
Now, another support point P3 = (0, 0) is added. Find again a closed representation of the polynomial interpolant. Is it possible to use the previous results?
I 3) Estimation of the Interpolation Error
Let p be a polynomial that interpolates the function
f : [0, 1] → R,
x 7→ cos(πx)
and xk = k/5, k = 0, . . . , 5 the equidistant support abscissas. Estimate the interpolation error at
the points x̃1 = 0.1, x̃2 = 0.5, and x̃3 = 0.6.
You can use your program of exercise 4 to verify your results.
P 4) Implementation of Newton’s Interpolation Formula
Develop a program for polynomial interpolation based on Newton’s interpolation formula.
(i) Write a function
c = interp newton(x,y)
that returns the coefficients c = (c0 , c1 , . . . , cn ) of the polynomial interpolant with support
abscissas x = (x0 , . . . , xn ) and support ordinates y = (y0 , . . . , yn ).
(ii) Write a function
yp = eval newton(c,x,xp)
that evaluates Newton’s interpolation formula for given coefficients c = (c0 , c1 , . . . , cn ) and
support abscissas x = (x0 , . . . , xn ) at a point xp and returns the result yp = p(xp).
Think about how to implement the evaluation of p efficiently with only O(n) additions and
multiplications.
(iii) Test your program with the Runge function
f1 : [a, b] → R,
x 7→
1
1 + x2
and n + 1 support points x0 , . . . , xn that are
(1) uniformly distributed on [a, b],
(2) equal to the Chebyshev points transformed on [a, b].
Run your program for a = −5, b = 5, and n = 20 and interpolate at the nodes xp=[a:1e-3:b].
Plot the results and the interpolation error for both distributions (1) and (2) of the support
points.
Remark: The Chebyshev points on [−1, 1] are defined by
2k + 1
xk = cos
π , k = 0, . . . , n
2(n + 1)
and lead to an almost optimal approximation. Use a linear transformation to get them for an
arbitrary interval [a, b].
5) Hermite Interpolation
Hermite interpolation is a method of piecewise interpolation that allows us to consider given derivatives y00 , . . . , yn0 at the support points (x0 , y0 ), . . . , (xn , yn ) as well as the support points themselves.
In every subinterval [xi , xi+1 ], the function f is approximated by a cubic polynomial p such that
p(xi ) = yi ,
0
p (xi ) =
p(xi+1 ) = yi+1 ,
yi0 ,
0
p (xi+1 ) =
(1)
0
yi+1
.
(2)
Thus, p(x) is continuously differentiable at all points (p ∈ C 1 ).
a) Find the Hermite interpolant p(x) in the interval [xi , xi+1 ].
Hint: Use the ansatz p(x) = pi (x), x ∈ [xi , xi+1 ] with
pi (x) = ai0 + ai1
x − xi
+ ai2
hi
x − xi
hi
2
+ ai3
x − xi
hi
3
,
hi = xi+1 − xi
and determine the coefficients ai0 , . . . , ai3 using the interpolation conditions (1), (2).
0
i
b) By sorting the result of a) by yi , yi+1 , yi0 , yi+1
and with substitution t = x−x
hi , we get another
representation of the interpolant p:
0
pi (t) = yi · H0 (t) + yi+1 · H1 (t) + yi0 · hi H2 (t) + yi+1
· hi H3 (t)
(3)
Find the Hermite functions H0 (t), . . . , H3 (t).
P c) Write a function
yp = interp hermite(xi,yi,yprime,xp)
that computes the value yp = p(xp) of the Hermite interpolant
p(x) = pi (x),
x ∈ [xi , xi+1 ],
given the support abscissas xi = (x0 , . . . , xn ), the support ordinates yi = (y0 , . . . , yn ), and the
derivatives yprime = (y00 , . . . , yn0 ).
Use equation (3) with Hermite functions
H0 (t) = 1 − 3t2 + 2t3 ,
H2 (t) = t − 2t2 + t3 ,
H1 (t) = 3t2 − 2t3 ,
H3 (t) = −t2 + t3 .
Test your program with the interpolation problem of exercise 4.