Foundations of Data Science

Let C e = (c 0 , c 2 , . . . , c d−2 ), C o = (c 1 , c 3 , . . . , c d−1 ), B e = (b 0 , b 2 , . . . , b d−2 ), and B o =
(b 1 , b 3 , . . . , b d−1 ). Equations 12.1, 12.2, and 11.3 can be expressed as convolutions 35 of
these sequences. Equation 12.1 is C e ∗ Be R + C o ∗ Bo R = 0, 12.2 is C e ∗ Ce R + C o ∗ Co R = δ(k),
and 11.3 is B e ∗ Be R + B o ∗ Bo R = δ(k), where the superscript R stands for reversal of the
sequence. These equations can be written in matrix format as
( )
Ce C o
B e B o
( C
R
∗ e Be
R
Co
R Bo
R
)
=
Taking the Fourier or z-transform yields
( ) ( F (Ce ) F (C o ) F (C
R
e ) F (Be R )
F (B e ) F (B o ) F (Co R ) F (Bo R )
( 2δ 0
0 2δ
)
=
)
( 2 0
0 2
where F denotes the transform. Taking the determinant yields
(
)(
)
F (C e )F (B o ) − F (B e )F (C o ) F (C e )F (B o ) − F (C o )F (B e ) = 4
Thus F (C e )F (B o ) − F (C o )F (B e ) = 2 and the inverse transform yields
Convolution by C R e
Now d−1 ∑
j=0
yields
C e ∗ B o − C o ∗ B e = 2δ(k).
C R e ∗ C e ∗ B o − C R e ∗ B e ∗ C o = C R e ∗ 2δ(k)
c j b j−2k = 0 so −C R e ∗ B e = C R o ∗ B o . Thus
C R e ∗ C e ∗ B o + C R o ∗ B o ∗ C o = 2C R e ∗ δ(k)
(C R e ∗ C e + C R o ∗ C o ) ∗ B o = 2C R e ∗ δ(k)
2δ(k) ∗ B o = 2C R e
C e = B R o
∗ δ(k)
)
.
Thus, c i = 2b d−1−i for even i. By a similar argument, convolution by C R 0
C R 0 ∗ C e ∗ B 0 − C R 0 ∗ C 0 ∗ B e = 2C R 0 δ(k)
yields
Since C R )
∗ B 0 = −C R 0 ∗ B e
−C R e ∗ C R e ∗ B e − C R 0 ∗ C 0 ∗ B e = 2C R 0 δ(k)
−(C e ∗ C R e + C R 0 ∗ C 0 ) ∗ B e = 2C R 0 δ(k)
−2δ(k)B e = 2C R 0 δ(k)
−B e = C R 0
Thus, c i = −2b d−1−i for all odd i and hence c i = (−1) i 2b d−1−i for all i.
35 The convolution of (a 0 , a 1 , . . . , a d−1 ) and (b 0 , b 1 , . . . , b d−1 ) denoted
(a 0 , a 1 , . . . , a d−1 ) ∗ (b 0 , b 1 , . . . , b d−1 ) is the sequence
(a 0 b d−1 , a 0 b d−2 + a 1 b d−1 , a 0 b d−3 + a 1 b d−2 + a 3 b d−1 . . . , a d−1 b 0 ).
366