... q of w, one of the magnitudes ur were > 0 (r≦ p) with k rs > 0. Then, on account of λ > 0, we would have Φ(λ)1,2 Φ1,2 ≧ ( ur + us λur + us )krs , where, as q is approached, the quotient on the right side approaches the value ( ur+ ...
... 1 , ‚ \ up ; Up + 1 , ... , Un ) > lim ( u1 , ... , un ) = ŵ , in contradiction to the definition of ŵ . ( 13 ) It follows therefore that for r≤ p all those ur must vanish for which at least one krs > 0 . As q approaches q , 1,2 → → 1 ...
... Q , B -∞ ( Ω , q , r , s -Xpqrs ( 2 , w1 , w2 ) kb , 5 ( 2 ) } S · ƒ $ ( 2 ) ... krs ( w2 ) ) , while that in Eq . ( 7.138 ) is - Xars ( 2 , w1 , w2 ) kos ... 1 ) LI , r ( @ 2 ) Ed , q ( @ 1 ) E ,, ( w2 ) = Σxparz ( 2 , w1 , w2 ) ƒ ( 2 ) ƒ ...
... Q ( fig . 6 ) , the initial amplitude on the surface at the excitation point Qi is related to that of the incoming ... krs ) ( 1 + Kp8 ) ] pr = EP sp [ ( cos 0 + Kr $ ) ( 1 + Kip $ ) ] Qi ( 41 ) Thus , in order for reciprocity to be ...
... q KRS น Ru = 1 + k2 m2 X1 = kmRu Zu = KRA √i + k2m2 / TAN - ' km IF ZIS CAPACITIVE CS 1S SWITCHED TO P ; Xu- KRg Ru ... one of the two , the sign of the Zm term is reversed and the impedance is 1 1 = Zo = 1 Z1 + Z2 + 2 Zm 1. See ...
13 Bx QP ( b ) 14 BK 5 15 Q × Q 16 ExP + 13 KRS 1 14 Q × P + 15 RX Q 16 K Q1 17 K × R 17 S B 4 18 BXS 18 RX B 19 RX S 19 BX R 20 R K 5 20 BQ 5 21 Rx B 21 BXS 22 PXB 22 KB 2 23 B S 3 Resigns . a ] BR 3 is the usual move giving Black a ...
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