Harmonic transients in time signals - Signal Processing in MRS with Biomedical Applications - page 168

where m > 0 (m = 1, 2, 3,...). If (4.56) is compared with the identity
∞

∞

c n+s u −n−1 = c s

1
u

c n+s+1 u −n−1

u +

(4.58)

n=0 n=0

the following relation is obtained
∞

c n+s+1 u −n−1

n=0

q (s)

r+1 e (s)

q (s)
1
u−q (s)
1

q (s 2 e (s)

r

1

= c s

−e (s 1 − −e (s 2 −· · ·−

r+1 −· · ·

u−q (s)
2

u−q (s)
r+1

−e (s)

q (s)
1
u−γ (s)
1

[δ (s 1 ] 2
u−γ (s)
2

[δ (s r ] 2
u−γ (s)

= c s (4.59)

− −· · ·−

r+1 −· · ·

where

= q (s) γ (s)

n

= q (s)
n

+ e (s)
n

[δ (s)
n

] 2

n+1 e (s n .

(4.60)

On the other hand, by employing (4.21) and (4.27) we arrive at

+ e (s+1)
n−1

= α (s+1)
n−1

q (s)
n

+ e (s)
n

= q (s+1)
n

(4.61)

q (s)

n+1 e (s)

= q (s+1)
n

e (s+1)
n

= [β (s+1)
n

] 2 (4.62)

n

= α (s+1)
n−1

γ (s)
n

[δ (s n ] 2 = [β (s+1) ] 2 (4.63)

n

so that
∞

c n+s+1 u −n−1

n=0

q (s)

r+1 e (s)

q (s 2 e (s) c s+1

u−q (s)
1

r

1

=

−e (s 1 − −e (s 2 −· · ·−

r+1 −· · ·

u−q (s)
2

u−q (s)
r+1

−e (s)

[β (s+1 1 ] 2
u−α (s+1)
1

[β (s+1 r ] 2
u−α (s+1)

c s+1
u−α (s+1)
0

= . (4.64)

− −· · ·− −· · ·

r

Here, the second line of (4.64) is identical to the second line of (4.53), provided
that s + 1 is used instead of s, as it ought to be. As such, (4.64) is, in fact,
the proof that (4.56) is correct. Returning now to (4.57) for the odd part of
G CCF(s)

(u)

n

(u) = G CF(s) G LCF(s)

o,n

(u) = G CCF(s)
o,n

2n+1 (u).

(4.65)

Hence, we see from (4.55) and (4.65) that the even and odd parts of the delayed
Lanczos approximants G LCF(s) (u) and G LCF(s)

o,n (u) of order n (n = 1, 2, 3,...)
coincide with the delayed continued fractions G CF(s)
2n

e,n

(u) and G CF(s)

2n+1 (u) of or

ders 2n and 2n + 1, respectively.