Uniplanar Algebra: Being Part I of a Propædeutic to the Higher Mathematical AnalysisBerkeley Press, 1893 - 141페이지 |
도서 본문에서
100개의 결과 중 1 - 5개
8 페이지
... cosk preserves Kan morphisms . Also cosk ( 4 ) = Am if k > m . Thus cosk ( e ) = e ko , where e is the I , k > 0 , where I one - point simplicial set , and cosk ( I ) is the one - simplex . If F - > E > B p is a sequence of pointed ...
... cosk preserves Kan morphisms . Also cosk ( 4 ) = Am if k > m . Thus cosk ( e ) = e ko , where e is the I , k > 0 , where I one - point simplicial set , and cosk ( I ) is the one - simplex . If F - > E > B p is a sequence of pointed ...
3 페이지
... cosk , cosk , -cosk , cosk , ) + ( 1 - cosk cosky cosk ) - ( m2 / 2a ) sink , sink , ( o + 7 cosk , ) sink , sink , ( o + Tсosk , ) sink , sink , ( a + cosk2 ) sink , sink , ( o + cosk , ) -0 , 1 - cosky + 0 ( 2 -cosk , cosk , -cosk ...
... cosk , cosk , -cosk , cosk , ) + ( 1 - cosk cosky cosk ) - ( m2 / 2a ) sink , sink , ( o + 7 cosk , ) sink , sink , ( o + Tсosk , ) sink , sink , ( a + cosk2 ) sink , sink , ( o + cosk , ) -0 , 1 - cosky + 0 ( 2 -cosk , cosk , -cosk ...
페이지
... cosk ( t ) = cos ( kt ) ifk > 0,1ifk = 0 , cosh ( -kt ) ifk < 0 , tank ( t ) = sink ( t ) cosk ( t ) = 1 kt ifk > 0 , tifk = 0,1 - ktanh ( -kt ) ifk < 0 , cotk ( t ) = cosk ( t ) sink ( t ) = kcot ( kt ) ifk > 0,1tifk = 0 , -Kcoth ( -kt ) ...
... cosk ( t ) = cos ( kt ) ifk > 0,1ifk = 0 , cosh ( -kt ) ifk < 0 , tank ( t ) = sink ( t ) cosk ( t ) = 1 kt ifk > 0 , tifk = 0,1 - ktanh ( -kt ) ifk < 0 , cotk ( t ) = cosk ( t ) sink ( t ) = kcot ( kt ) ifk > 0,1tifk = 0 , -Kcoth ( -kt ) ...
13 페이지
... cosk x2 ] G + B2 [ cosk_x1 - cosk cosk * , ] dễ } n Since , we chose to write ỹ ( x ) = { __ ~ _cosk_x ; o < x < L = M = 0 m m and y1 y ( x1 ) , y2 = ỹ ( x2 ) equation ( 13 ) is equivalent to the matrix equation ( 13a ) ( 13b ) 1 1 1 g ...
... cosk x2 ] G + B2 [ cosk_x1 - cosk cosk * , ] dễ } n Since , we chose to write ỹ ( x ) = { __ ~ _cosk_x ; o < x < L = M = 0 m m and y1 y ( x1 ) , y2 = ỹ ( x2 ) equation ( 13 ) is equivalent to the matrix equation ( 13a ) ( 13b ) 1 1 1 g ...
20 페이지
... cosk [ _ dɅ σ , ( ^ ) . 2πp1 ( k ) = cosk So dk p1 ( k ) + 8u Ви u2 + 16 ( sink -A ) , u2 + 16 ( sink - A ) 2 - 2153 € ( A 。) = − 2t so 。 p1 ( k , A。) cosk dk + E。( Q ) – E。( Qo ) · ( 4.12 ) We have explicitly indicated the ...
... cosk [ _ dɅ σ , ( ^ ) . 2πp1 ( k ) = cosk So dk p1 ( k ) + 8u Ви u2 + 16 ( sink -A ) , u2 + 16 ( sink - A ) 2 - 2153 € ( A 。) = − 2t so 。 p1 ( k , A。) cosk dk + E。( Q ) – E。( Qo ) · ( 4.12 ) We have explicitly indicated the ...
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a+ib addition and subtraction Addition Theorem affix Agenda amplitude angle AOQ arc AVQ arc-ratio assumed base CALIFORNIA circular sector co-ordinates commutative law complex quantities corresponding cosh COSK csch defined definition direction distance equal equation equilateral hyperbola expm exponential expressed factors formula functions geometric addition Goniometric Ratios Hence hyperbolic functions Hyperbolic Ratios hyperbolic sector imaginary indeterminate form integer inverse law of indices law of involution law of metathesis length logarithmic spiral logm metathesis modular normal modulus natural logarithms negative nth roots orthomorphosis parallel path plane polynomial positive Prop proportion PROPOSITION Prove the following quotient radius real axis real magnitudes real quantities reciprocal represent respectively roots sech sector sinh speed straight line tanh tensor tion triangle unit circle x+iy z-plane zero