App D on sum REVD
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Appendix D of Phil's Spectral Theory book, in the Aug 2013 update folder; a note says it was installed into FT on Aug 8, 2013. It proves that a sum over z^-s with z=e^{ik}, weighted by a linear factor, equals 2π times a delta-like function from Appendix A. The method splits the sum, differentiates a sine sum, and has Maple confirm the remainder is zero. A comment suggests a Gradshteyn and Ryzhik alternative. Sum and delta symbols are garbled in the extraction.
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This was installed into FT on Aug 8,2013
Appendix D: Calculation of a Sum which appears in (35.17)
We want to show that (k = ωT1, z = eik)
S ≡ !Syntax Error, I z-s = (D.1)
where the right side is 2πδ6(k,N) of (A.20). There are no doubt elegant ways to verify this identity, but we shall use the tried and true brute force method. We start off:
S = !Syntax Error, I z-s = !Syntax Error, I z-s - !Syntax Error, I|s| z-s = S1 - S2 . (D.2)
The sum S1 we know from (A.30) and (A.15) is
S1 = !Syntax Error, I z-s = 2π δ5(k,2N) = . (D.3)
So we work then on S2 :
S2 = !Syntax Error, I|s| z-s = !Syntax Error, I|s| z-s + !Syntax Error, I|s| z-s + 0
= !Syntax Error, I(-s) z-s + !Syntax Error, I(s) z-s = !Syntax Error, I (s) zs + !Syntax Error, I(s) z-s = !Syntax Error, Is (zs + z-s)
= 2 !Syntax Error, Is cos(ks) . (D.4)
Suppose we define
S3 ≡ !Syntax Error, I sin(ks) . (D.5)
Then (here ∂k means d/dk)
S2 = 2 ∂k S3. (D.6)
So we now work on S3 :
S3 = !Syntax Error, I sin(ks) = (1/2i) !Syntax Error, I[ zs - z-s] = (1/2i) [ !Syntax Error, I zs - c.c. ]
= (1/2i) [ S4 - c.c. ] (D.7)
where c.c means complex conjugate. We next work on
S4 = !Syntax Error, I zs = !Syntax Error, Izs - 1 . (D.8)
Change summation variable to r = s-N so this becomes, again using (A.30),
S4 = !Syntax Error, I z(r+N) - 1 = zN !Syntax Error, Izr - 1 = zN 2π δ5(k,N) - 1 (D.9)
Then
[ S4 - c.c. ] = { zN 2π δ5(k,N) - 1} - { z-N 2π δ5(k,N) - 1}
= (zN - z-N) 2π δ5(k,N) . (D.10)
Backtracking now we find for (D.7) that
S3 = (1/2i) [ S4 - c.c. ] = (1/2i) (zN - z-N) 2π δ5(k,N) = sin(Nk) 2π δ5(k,N) . (D.11)
which gives a result we could have looked up (see Comment below),
!Syntax Error, I sin(ks) = sin(Nk) . (D.11)'
It remains to compute S2 according to (D.6),
S2 = 2 ∂k S3 = 2 ∂k[sin(Nk) 2π δ5(k,N)] . (D.12)
Backing up more we then have from (D.2),
S = S1 – S2
= 2π δ5(k,2N) – 2 ∂k[sin(Nk) 2π δ5(k,N)] (D.13)
so that
(2N+1)S = (2N+1) 2π δ5(k,2N) – 2∂k[sin(Nk) 2π δ5(k,N)] . (D.14)
Anticipating denominators of powers up to sin2(k/2) we rewrite this as
sin2(k/2) (2N+1)S = (2N+1) sin2(k/2)[ 2π δ5(k,2N)] – 2 sin2(k/2) ∂k[sin(Nk) 2π δ5(k,N)]
= T1 – T2
Then let
D5(k,N) ≡ [2π δ5(k,N)] = (A.15)
to get
T1 = (2N+1) sin2(k/2) D5(k,2N)
T2 = 2 sin2(k/2) ∂k[sin(Nk) D5(k,N)] . (D.15)
The statement (D.1) which we are trying to prove is now this:
S = 2πδ6(n,K) = (D.1)
and adding our factors to both sides this becomes
sin2(k/2) (2N+1)S = sin2(k/2) (2N+1) 2πδ6(n,K)
= sin2(k/2) (2N+1) { }
= sin2[(N+1/2)k] . (D.16).
If we then define this last factor as
T3 ≡ sin2[(N+1/2)k] (D.17)
our task is then to show that
T1 - T2 = T3
or
Q ≡ T1 - T2 - T3 = 0. (D.18)
where
T1 = (2N+1) sin2(k/2) D5(k,2N)
T2 = 2 sin2(k/2) ∂k[sin(Nk) D5(k,N)]
D5(k,N) ≡
T3 ≡ sin2[(N+1/2)k] . (D.19)
This is a task for Maple. We enter the quantities T1,T2,T3 and the function D5(k,N) :
The Maple value command causes the differentiation to be carried out, so we continue :
:
Thus we have shown that Q = 0 so (D.1) is then verified. QED
Comment: Gradshteyn and Ryzhik provide the following summation formulas :
An alternate method of verifying (D.1) would be to use two of these sums in an approach that begins this way
S ≡ !Syntax Error, I z-s = !Syntax Error, I (z-s + zs )/2 = !Syntax Error, I cos(ks)
= 1 + 2 !Syntax Error, I cos(ks) = 1 + 2 !Syntax Error, I cos(ks) – 2 !Syntax Error, Is cos(ks)