Ngày tải lên :
08/08/2014, 01:20
... ν⊢b Substituting in this the expansion s /ν = s /ν , s s = λ⊢a+c s , s s s λ⊢a+c we get hb ∗ s /δ = s , s s s , s s s s µ⊢a ν⊢b λ⊢a+c = s /λ , s s /δ , s s s = µ⊢a ν⊢b λ⊢a+c s /δ ... that equation 1.12 gives h1 s2 1 [X − 1] = det h2 h3 = h2 h1 − h3 − h1 h1 + h1 h1 and equation 3.15 gives U(2,1) = s2 s1 s s + s1 1 s1 s s − (s3 s + s( 2,1) s + s1 3 s ) − s1 s1 s s ... s2 ,1 = s2 ,1 +s1 ,2 +s1 ,1,1 +s0 ,2,1 +s0 ,3 = s2 ,1 s1 ,2 +s1 ,1,1 s1 ,1,1 s2 ,1 = since s1 2 = s1 2 = So we have = here (2) Setting n = in equation 4.16 gives s2 ,2 ∗ s3 ,1 = s3 ,1 + s2 ,2 + s2 ,1,1 + s1 ,2,1...