... ( Xo ( 1 ) Xo ( 2 ) · · · Xo ( j ) ) d ( Xo ( j + 1 ) ) Xo ( j + 2 ) · · · Xo ( n ) ) · ... : + p ) ƒ ( x1 , . . . , Xn ) — √ ( ƒ ( X1 , . . . , Xn ) ) p + nƒd ( x1 , . . . , Xn ) ( ( c + ( n ) ) + n Σoes , α ( 10 ) Σ ...

... xo 4 T ( 1 + Mo ) ▽ Moto + o + π MOL arc sin 1 ΔΡ 2 ( 1 - x ) -to ( 1 + Mo ) to ( 1 - Mo ) ! ( AP ) , α = 8 -- M = arc sin 2xo - to ( 1 - Mo ) ( 1 + Mo ) to + ( A6c ) ( A6d ) to - xo π ; —EF ( 4 , k ' ) — 2 90/4 Marc sin 16 xo - 1 + Moto ...

... xo + 2k - xo + 4f ( xo ) 4ƒ ( xo ) + 2 ( xo − 1 ) ā - ( 64 ) Since 4f ( xo ) + 2 ( xo - −xo ( 4ƒ ( xo ) + 2 ( xo − 1 ) ã ) + 4ƒ ( xo ) - 4f ( xo ) + 2 ( xo − 1 ) ā - 1 ) ā < 0 , to prove ( 61 ) , we only need to prove -xo ( 4f ( xo ) + ...

... 1 , Σ ( Z , xo ) * = G1 , n − 1 , Σ ( z , Xo ) , Fn , ( 2 , 0 ) G1 , n - 1 , Σ ( z , Xo ) = G2 , n − 1 , Σ ( z , Xo ) Fn , Σ ( z , Xo ) , Hn + 1 , ( 2 , 0 ) G2 , n − 1 , ( z , xo ) = G1 , n − 1 , Σ ( z , xo ) Hn + 1 ...

... 1 cos xo = cos xo = cos 60o 2 = = cosec 2A = cosec 30o A 2A = 30o = 15o xo = 60o (ii) sin2 xo + cos2 xo = sin2 60o + cos2 60o and cot 3A – 1 = 0 3 2 1 2 4 = 1 = = 1 3 4 4 A = 15o cot 3A = 1 cot 3A = cot 45o 3A = 45o Hence, A = 15 3. If ...

... ( XO - 1 ) SHOP EQUIPMENT , AN / TSM - 92 ( XO - 1 ) STANDARD , CALIBRATOR TEST KIT MK - 1055 ( XO - 1 ) TRANSPORTABILITY GUIDANCE HAWK GUIDED MISSILE SYSTEMS ( TOWED ) EQUIPMENT SERVICEABILITY HAWK GUIDED MISSILE SYTEMS COUNTER ...

... ( 1 ) i = 1 mn + 2gabe ble ; 1 ) ele ; 2 ) ] 771 169 The third term is ( y2 - y , ) - dependent , so that , as argued ... xo ) ] 2 ( 4.3 ) = - ( 4c45 ) 2 ; and then performing Q and S transformations with independent parameters ( -o ( 1 ) ...

... xo of in a neighborhood of to . Then ( 3.104 ) does not hold , that is , r + 1. On the other hand , there exist weight functions w having an isolated zero at xo of ( exact ) order r such that ( 3.104 ) holds , i.e. , w is in W ( p , J ) ...

... 1 . ( i ) Let Xo = -1 . Then = f ( x ) – f ( xo ) = lim XXO 0 – 0 lim x → xo – X – ( -1 ) = - : 0 , X – Xo f ( x ) – f ( xo ) lim XX0 + X – Xo = ( 1 + x ) - 0 x – ( -1 ) lim x → X0 + = = 1 . = = Hence , f ' ( - 1 ) = 0 and f ( -1 ) = 1 .

... 1 0 : " + 1g → CA ( Ug ) 0 ( хо ^ X1 ^ ^ xn ) = xo & e ( x1 ^ • • • ^ Xn ) . Because we are working modulo the image of 1 —t , 0 is well defined . The following result is implicit in the proof of [ 7 , 10.2.4 ] for r = 1 , A = Ug ...