n
gn(x) = e "5ix∑ teixt
1 e -0.5iχdy,(eix)t = 1
i dx t= 1 i
( ʌ
ixd e,χ 1 eβ∙xx
dx 1 -eix
\ /
( . ■ „ \ A
1 -e +e -°-i ix( 1-( x+1) einx)
e -0.5ix_e 0.5ix}2 e ~o∙5ix-e 0'5ix
_ cos(0.5x)-isin(0.5x)ɑ_etnx^ + 2insin(0.5x)eιxx
-4sin2(0.5 x ) -4sin2(0.5 x )
(A.54)
Thus,
cos(0.5x)(1 - cos(xx) - (2x— 1)sin(0.5x)sin(xx)
-4sin2(0.5 x )
i (2x— 1)sin(0.5x)cos(xx) - cos(0.5x)sin(xx) + sin(0.5x)
-4sin2(0.5 x )
gx(x ) + gx( “x ) =
cos(0.5x)(1 - cos(xx) - (2x-1)sin(0.5x)sin(xx)
-2sin2(0.5 x )
(A.55)
Since cos(2kπ) = 1 and sin(2kπ) = 0, the second equality in (38) follows. The proof of the first
equality goes similarly, and (39) is trivial. Q.E.D.
Proof of Lemma 9: Observe that
K
∑ expi (xt + У) =
t≡ 1
exp( i (y + 0.5 x ))-------1----exp( iKx )------
exp(-0.5 ix) - exp(0.5ix)
_ (cos(y + 0.5x) + i sin(y + 0.5 x))(1 - cos(Kx) - i sin(Kx))
-2 i sin(0.5 x )
_ i cos(y + 0.5x)(1 - cos(Kx)) + sin(y + 0.5 x)sin(Kx)
2sin(0.5 x )
cos(y + 0.5x)sin(Kx) - sin(y + 0.5x)(1 - cos(Kx))
2sin(0.5 x ) ’
(A.56)
hence
48
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