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that c ∈DF (b1⟨1,−u1⟩ ⊥ . . . ⊥ br−1⟨1,−ur−1⟩) ∩ −DF (br⟨1,−ur⟩). By Lemma 7.1.15 we can thus assume br−1 = c and br = −c. This implies ψ ≅ b1⟨1,−u1⟩ ⊥ . . . ⊥ br−2⟨1,−ur−2⟩ ⊥ c⟨1,−ur−1,−1, ur⟩, the last [...] Q(x)((t1))⋯((t4)) we consider the forms ψ1 ∶= ⟨x,−(x + 4)⟩ ⊥ −t1⟨1,−(x + 4)⟩, ψ2 ∶= ⟨x,−(x + 1)⟩ ⊥ −2t1⟨1,−(x + 1)⟩ ρ1 ∶= ⟨1,−x,−t1t2(x + 4), t1t2x(x + 4)⟩ = ⟨⟨x, t1t2(x + 4)⟩⟩, ρ2 ∶= ⟨1,−x(x + 1)(x + 4) [...] Witt equivalent to ⟨−1, (x + 1)(x + 4),2t1(x + 1)(x + 2)(x + 4),−2t1(x + 2)⟩ ⊥ t3(⟨1,−(x + 4)⟩ ⊥ −t1⟨1,−(x + 4)⟩ ⊥ −t2(⟨1,−(x + 1)⟩ ⊥ −2t1⟨1,−(x + 1)⟩)) = − ⟨⟨(x + 1)(x + 4),−2t1(x + 2)⟩⟩ ´¹¹¹¹¹¹¹¹¹¹¹ …

2 "genuine" "58" 215 129.6 129.4 8.8 9 141.1 "genuine" "59" 215.1 130.1 129.9 7.9 11 141.3 "genuine" "60" 215.1 130 129.8 8.2 10.3 141.4 "genuine" "61" 215.1 129.6 129.3 8.3 9.9 141.6 "genuine" "62" 215 [...] 130.4 130.3 9.1 10.2 141.1 "genuine" "98" 214.4 129.7 129.5 8 10.3 141.2 "genuine" "99" 215.1 130 129.8 9.1 10.2 141.5 "genuine" "100" 214.7 130 129.4 7.8 10 141.2 "genuine" "101" 214.4 130.1 130.3 9.7 11 [...] "121" 214.8 130.3 130.4 10.1 12.1 139.6 "counterfeit" "122" 215.1 130.6 130.3 12.3 10.2 139.6 "counterfeit" "123" 215.3 130.8 131.1 11.6 10.6 140.2 "counterfeit" "124" 215.1 130.7 130.4 10.5 11.2 139.7 …

Microsoft Word - skript_baleanoptera2

           1 2 3 1 (0)e if h h h h h     Example: 1 2 1 2 35; 20; 0.2; 0.1; 0.3 (0) 6.29x x h h h e       2 pieces:     1 1 1 2 1 1 1 2 10 1 1 1 (0) exp exp xp(- h ·x )· x [...]  Keyfitz entropy 1 1 1 1 1 0 1 1 1 10 ln 1 ( ) ln ( ) ln 1 ( ) ln S S S S l x l x dx S k S k H S S l x dx S k                Maximum age  11 max 1 ln ln 1 1 N SN S x k k       [...] functions Hazard rate or force of mortality: 1 1 2 1 2 3 2 h h x h 0 x x (x) x x x x          Cumulated hazard rates:       1 1 1 2 1 1 2 1 1 2 3 1 1 2 2 1 3 2 2 H h H h x H h x 0 x x H(x) x h …

500 Bayern 5 5 10 10.1 0.83 1 500 Berlin 10 5 25 19.1 1.20 1 196 Brandenburg 3 2 8 6.5 0.59 1 60 Bremen 5 3 10 9.6 1.01 1 132 Hamburg 10 5 20 14.5 0.78 1 120 Hessen 5 3 10 7.3 0.33 1 50 Mecklenburg-Vorpommern [...] g 65* 50 78 64.8 1.66 1 300 69* 50 80 64.9 1.07 1 365 66 50 80 64.8 0.90 1 365 Bayern 60 50 77 62.0 1.48 1 250 60 50 70 59.4 0.81 2 280 60 50 71 60.0 0.71 1 280 Berlin 23 23 30 29.6 1.11 5 130 23 23 25 [...] 35.2 -1.0 88.5 -1.0 94.3 -0.5 95.9 -0.3 96.9 +0.9 Westdeutschland 1,278,631 653,442 656,551 658,611 635,898 206,382 384,057 562,905 615,104 605,115 16.1 -0.4 58.8 -0.8 85.7 -1.1 93.4 -1.1 95.2 +0.1 Ostd …

N∑ i=1 αi = 1 2 N∑ i=1 N∑ j=1 αiαjyiyj 〈 ~xi , ~xj 〉 − N∑ i=1 N∑ j=1 αiαjyiyj 〈 ~xi , ~xj 〉 + N∑ i=1 αi =+ N∑ i=1 αi − 1 2 N∑ i=1 N∑ j=1 αiαjyiyj 〈 ~xi , ~xj 〉 unter den Nebenbedingungen 0 = N∑ i=1 αiyi [...] Problem 1 2 ||~β||2 − N∑ i=1 αi [ yi (〈 ~xi , ~β 〉 + β0 ) − 1 ] = 1 2 ||~β||2 − N∑ i=1 αi yi (〈 ~xi , ~β 〉 + β0 ) + N∑ i=1 αi = 1 2 ||~β||2 − N∑ i=1 αiyi 〈 ~xi , ~β 〉 − N∑ i=1 αiyiβ0 + N∑ i=1 αi (10) = [...] } H1 H2 + + + + + − − − − − − − Der Abstand der mittleren Ebene H∗ zum Ursprung beträgt d(~0,H∗) = β0 ‖~β‖ Der Abstand zwischen den Ebenen H1 und H2 ist d(H1,H2) = β0+1 ‖~β‖ − β0−1 ‖~β‖ = β0−β0+1+1 ‖~β‖ …

N∑ i=1 αiyi~xi führt zu 1 2 ||~β||2 − N∑ i=1 αiyi 〈 ~xi , ~β 〉 + N∑ i=1 αi = 1 2 N∑ i=1 N∑ j=1 αiαjyiyj 〈~xi , ~xj〉 − N∑ i=1 N∑ j=1 αiαjyiyj 〈~xi , ~xj〉 + N∑ i=1 αi =+ N∑ i=1 αi − 1 2 N∑ i=1 N∑ j=1 αiαjyiyj [...] Problem 1 2 ||~β||2 − N∑ i=1 αi [ yi (〈 ~xi , ~β 〉 + β0 ) − 1 ] = 1 2 ||~β||2 − N∑ i=1 αi yi (〈 ~xi , ~β 〉 + β0 ) + N∑ i=1 αi = 1 2 ||~β||2 − N∑ i=1 αiyi 〈 ~xi , ~β 〉 − N∑ i=1 αiyiβ0 + N∑ i=1 αi (12) = [...] =   1 2 3   dann ist die Ebenengleichung nicht in Hesse Normalform, weil ‖~β′‖ = √ 14 6= 1. Wir normalisieren ~β = ~β′ ‖~β′‖ 1√ 14   1 2 3   〈 ~β , ~x 〉 − β0 = 0 1√ 14 x1 + 1√ 14 x2 + 1√ 14 x3 …

Vorlesung Wissensentdeckung - Stützvektormethode

N∑ i=1 αiyi~xi führt zu 1 2 ||~β||2 − N∑ i=1 αiyi 〈 ~xi , ~β 〉 + N∑ i=1 αi = 1 2 N∑ i=1 N∑ j=1 αiαjyiyj 〈~xi , ~xj〉 − N∑ i=1 N∑ j=1 αiαjyiyj 〈~xi , ~xj〉 + N∑ i=1 αi =+ N∑ i=1 αi − 1 2 N∑ i=1 N∑ j=1 αiαjyiyj [...] Problem 1 2 ||~β||2 − N∑ i=1 αi [ yi (〈 ~xi , ~β 〉 + β0 ) − 1 ] = 1 2 ||~β||2 − N∑ i=1 αi yi (〈 ~xi , ~β 〉 + β0 ) + N∑ i=1 αi = 1 2 ||~β||2 − N∑ i=1 αiyi 〈 ~xi , ~β 〉 − N∑ i=1 αiyiβ0 + N∑ i=1 αi (12) = [...] ~β′ =  1 2 3  dann ist die Ebenengleichung nicht in Hesse Normalform, weil ‖~β′‖ = √ 14 6= 1. Wir normalisieren ~β = ~β′ ‖~β′‖ 1√ 14  1 2 3 〈 ~β , ~x 〉 − β0 = 0 1√ 14 x1 + 1√ 14 x2 + 1√ 14 x3 − …

Microsoft Word - Prachayasittikul_thiotetrahydropyridine_proof_060509.doc

(CO) cm-1; UV(95 % ethanol)λmax nm (logε): 202 (4.33), 243 (4.07); 1H-NMR (300 MHz, CDCl3): δ6.12 (d, J = 4.1 Hz, 1H, H-5), 5.95 (d, J = 4.1 Hz, 1H, H-6), 5.64 (d, J = 1.2 Hz, 1H, H-2), 3.95 (d, J = 1.2 Hz [...] phenyl)-1,2,3,4-tetrahydro- pyridines (15a-b), 1-acetyl-2-(1-adaman- tylthio)-3-hydroxy-3-methyl-4-acetoxy- 1,2,3,4-tetrahydropyridine (16), 1-acetyl-2- (1-adamantylthio)-3-acetoxy-4-phenyl-6- hydroxy-1,2,3 [...] isomeric forms; 1,2,3,6-, 1,2,3,4- and 2,3,4,5-tetrahydro- pyridines. 1-Methyl-4-phenyl–1,2,3,6–tetra- hydropyridines (1) is a potent neurotoxin in dopaminergic system. Tetrahydropyridine 1 was a byproduct …