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c. Derive, from first principles, the answers to parts (a) and (b). (Hint:
Look up in a standard integral table the sine integral function.)
NOTE An important goal of filter theory is to find methods to smooth these
kinds of oscillations.

8.7.6 Interpolating the Coefficients of an (n – 1)-degree Polynomial from
n Points
The problem at hand can be posed as follows:

Given the coordinates of n points: (x , y ), (x , y ), …, (x , y ), we want to find
2
2
1
1
n
n
the polynomial of degree (n – 1), denoted by p n–1 (x), whose curve passes
through these points.
Let us assume that the polynomial has the following form:
x =
2
p () a + a x + a x +…+ a x n−1 (8.22)
n−1 1 2 3 n
From a knowledge of the column vectors X and Y, we can formulate this prob-
lem in the standard linear system form. In particular, in matrix form, we can
write:
a
1 x x 2 L x n −1    y 1 
1
 1 1 2 1 n −1     
a
 1 x 2 x 2 L x 2     y 2 
2
M
VA* =  M M M O M    =  M  = Y (8.23)
     
 M M M L M     M 
M
 2 n −1     y 
a
 1 x n x n L x n     n 
n
Knowing the matrix V and the column Y, it is then a trivial matter to deduce
the column A:
A = V * Y (8.24)
−1
What remains to be done is to generate in an efficient manner the matrix V
using the column vector X as input. We note the following recursion relation
for the elements of V:

V(k, j) = x(k) * V(k, j – 1) (8.25)

Furthermore, the first column of V has all its elements equal to 1.
The following routine computes A:

© 2001 by CRC Press LLC