4 1 thống kê cho các biến quan sát

... Algorithms 1- 19 widget actual cost amortized cost— P() 1 1 .12 0 .12 1. 12 0. 24 1. 12 0.36 1. 12 0 .48 1. 12 0.60 1. 12 0.72 1. 12 0. 84 1. 12 0.96 1. 12 1. 08 10 1. 12 0.20 11 1. 12 0.32 12 1. 12 0 .44 13 1. 12 0.56 14 ... 14 1. 12 0.68 widget actual cost amortized cost— P() 15 1. 12 0.80 16 1. 12 0.92 17 1. 12 1. 04 18 1. 12 1. 16 19 1. 12 1. 28 20 1. 12 0 .40 21 1 .12 0.52 22 1. 12 0. 64 23 1. 12 0.76 24 1. 12 0.88 25 1. 12 1. 00 ... 64 µs 12 5 µs ms 1s 16 .67 m 11 .57 d 31. 71 y f (n) n4 10 µs 16 0 µs 810 µs 2.56 ms 6.25 ms 10 0 ms 16 .67 m 11 5.7 d 317 1 y 3 .17 ∗ 10 7 y n10 10 s 2. 84 h 6.83 d 12 1 d 3 .1 y 317 1 y 3 .17 ∗ 10 13 y 3 .17 ...

Ngày tải lên: 13/09/2015, 09:45

... 10 07 045 988 544 277770 2 41 7 9 41 8273 944 41 14 8 29990028 01, c = 20 14 0 919 77088555 540 48358836 547 888822965998005599, and the value of M is now M = 25 × 33 × 52 × × 11 × · · · × 10 9 Note that a ≈ 1. 007 × 1 047 ... 1 , α2 , α3 are multiplicatively independent If (i) 4( R1 + 1) (S1 + 1) ≥ T1 + 1, (ii) 4( R1 + 1) (T1 + 1) ≥ S1 + 1, (iii) (R2 + 1) (S2 + 1) (T2 + 1) ≥ 12 (K − 1) 2 (L − 1) , and (iv) (R1 + 1) (S1 + 1) (T1 ... 1/ 2 < |r | < R1 + 1, (R1 + 1) (S1 + 1) T1 + 1/ 2 < |s | < S1 + 1, < |t | < T1 + 1, (S1 + 1) (T1 + 1) R1 + , and 1/ 2 , 10 12 YANN BUGEAUD, MAURICE MIGNOTTE, AND SAMIR SIKSEK |t | < T1 + 1, (R1 + 1) (T1...

Ngày tải lên: 06/03/2014, 08:21

... disclaimer: 4. 1, 4. 5, 4. 13 , 4. 14 , 4. 15 , 4. 20, 4. 22, 4. 27, 4. 28, 4. 48, 5. 24, 5.25, 5.26, 5.27, 5.28, 5.29, 5. 31, 5. 34, 6 .1, 6.2, 6.3, 6 .4, 6.5, 6.6, 6.7, 6.8, 6.9, 6 .13 , 6 .18 , 6. 21, 6.22, 6.23, 6. 24, 6.25, ... Rabbits Month (1) (2) (3) (4) January February March April May June July August September October November December 1 13 21 34 55 89 1 13 21 34 55 1 13 21 34 55 89 13 21 34 55 89 14 4 233 Source: ... ratios 0. 618 , 1. 000, and 1. 618 of the PHI series: FIBONACCI ANALYSIS • 17 • 38.2 percent is the result of the division 0. 618 ÷ 1. 618 • 50.0 percent is the transformed ratio 1. 000 • 61. 8 percent...

Ngày tải lên: 03/05/2014, 12:58

... is repeated Define φ((λ, µ)) = (( 1 , 1 , λ2 , ), (µ2 , µ3, )), if 1 < 1 or 1 = 1 = 2s + 1, ((λ2 , λ3 , ), ( 1 , 1 , µ2 , )), if 1 > 1 or 1 = 1 = 2s It is straightforward to verify ... of (1. 2) may be written as 1 (az, q; qa; q, b) (1 − a) (1 .4) By the Heine’s transformation (III .1) in Gasper and Rahman [7, p 2 41 ], (1 .4) is equal to (q, abz; q)∞ 1 (a, b; abz; q, q), (1 − ... odd parts Theorem 2 .1 was established by Chapman [5] in his proof of the q-identity ∞ −q 2n 1 + q 2n n − q 2n n =1 n 1 j =1 ∞ − q 2j 1 − q 2j 1 = − q 2j − q 2j j =1 ∞ ( 1) d d =1 qd , − qd which is...

Ngày tải lên: 07/08/2014, 07:21

... of 10 counted by B (10 ), C (10 ) and D (10 ) Those counted by B (10 ) are {1, 1, 1, 1, 1, 1, 1, 1, 1, 1} , {4, 1, 1, 1, 1, 1, 1} , {4, 4, 1, 1} , {6, 1, 1, 1, 1} , {6, 4} , {9, 1} , those counted by C (10 ) ... sum ′ a1 ≥0 q ma1 m +1 k+a1 ) k =1 (1 + bq a2 ≥a1 q ma2 m +1 m+k+a2 ) k =1 (1 + bq ′ ··· an 1 ≥an−2 q man 1 m +1 (n−2)m+k+an 1 ) k =1 (1 + bq ′ an ≥an 1 q man (3.6) m +1 (1 + bq (n 1) m+k+an ) k =1 Next, ... combinatorics 18 (2 011 ), #P60 Proof We rewrite the left side of (3 .1) as the nested sum q ma1 m +1 k+a1 ) k =1 (1 + bq a2 ≥a1 a1 ≥0 q ma2 m +1 (m +1) +k+a2 ) k =1 (1 + bq q man 1 m +1 (n−2)(m +1) +k+an 1 ) k =1 (1...

Ngày tải lên: 08/08/2014, 14:22

... ReheapDown(child, lastPosition) return 10 Insert new element into min-heap Insert 14 : 14 14 14 The new element is put to the last position, and ReheapUp is called for 11 that position  InsertHeap ... success 13 End InsertHeap Delete minimum element from min-heap 31 31 31 31 The element in the last position is put to the position of the root, and ReheapDown is called for that position 14 Delete ... complete binary tree • If the height is h, the number of nodes n is between 2h -1 and (2h -1) • Complete tree: n = 2h -1 when last level is full • Nearly complete: All nodes in the last level are...

Ngày tải lên: 15/03/2014, 17:20

... kcat, s )1 (%) Km for 1, 2-propanediol (mM) kcat ⁄ Km · 10 )6 (s )1 M )1) kinact (min )1) kcat ⁄ kinact · 10 )4 Wild-type Sa224A Sa224N Kb135R Kb135A Kb135Q Kb135E 336 64 17 2 54 19 6 211 7.7 0 .15 0 .15 1. 90 ... (lM) ± ± ± ± ± ± ± 0 .11 0.05 0.2 0. 04 0 .4 0.6 11 Ki for CN-Cbl (lM) 3.6 1. 8 0.05 10 .2 0.89 0.39 0.000 61 1.5 0.63 2.3 0. 51 1 .4 4 .4 463 ± ± ± ± ± ± ± 0 .1 0.00 0 .1 0.08 0.2 0.5 41 catalysis With the ... 1. 90 0 .12 0.39 0.39 0 .40 2.2 0 .43 0.009 2 .1 0.50 0. 54 0. 019 0.027 0 .46 0.070 0.027 0.059 0.068 0.0085 75 0.8 1. 5 56 20 19 5 .4 (10 0) (19 ) (5) (76) (58) (63) (2) ± ± ± ± ± ± ± 0.02 0. 01 0. 01 0. 01 0.02...

Ngày tải lên: 23/03/2014, 06:20

Ngày tải lên: 23/04/2014, 22:35

... variables α0 ,α 1 , ,α−r +1 : V1 = a0 α0 + a1 α 1 + · · · + ar 1 α−r +1 , V2 = a0 V1 + a1 α0 + a2 α 1 + · · · + ar 1 α−r+2 + a0 α−r +1 , V3 = a0 V2 + a1 V1 + a2 α0 + a3 α 1 + · · · + ar 1 α−r+3 + a0 ... coefficient sequence (1. 3), we have n= Vk +1 = a0 Vk + a1 Vk 1 + · · · + ar −2 Vk−r+2 + (ar 1 + 1) Vk−r +1 = a0 Vk + a1 Vk 1 + · · · + ar −2 Vk−r+2 + ar 1 Vk−r +1 + a0 Vk−r + a1 Vk−r 1 + · · · ∞ = i=0 ... sequence (1. 3), for every n ∈ Z and N > 0, we have Wn +1 = a0 Wn + a1 Wn 1 + · · · + ar −2 Wn−r+2 + ar 1 + Wn−r +1 = a0 Wn + a1 Wn 1 + · · · + ar −2 Wn−r+2 + ar 1 Wn−r +1 + Wn−r +1 = a0 Wn + a1 Wn 1 +...

Ngày tải lên: 22/06/2014, 22:20

Ngày tải lên: 02/07/2014, 04:20

Ngày tải lên: 02/07/2014, 04:20

Ngày tải lên: 02/07/2014, 04:20

Ngày tải lên: 02/07/2014, 04:20

Ngày tải lên: 02/07/2014, 04:20

Ngày tải lên: 02/07/2014, 04:20

Ngày tải lên: 02/07/2014, 04:20

Ngày tải lên: 02/07/2014, 04:20

Ngày tải lên: 02/07/2014, 04:20

... = K[x, y, z, t, T1 , T2 , T3 , T4 ]/J , where the ideal J is generated by z T1 + wT2 + xT3 , y T1 + xT2 + wT4 , xw2 T1 + 2 z T2 + y T3 , x2 wT1 + y T2 + z T4 , xwT1 + T2 + T3 T4 From this it ... Theorem 2 .1 The Rees ring R(I) is equal to K[x, T ]/(L1 , L2 , L3 , L4 , Q), i.e J = (L1 , L2 , L3 , L4 , Q) Denote by A the ideal (L1 , L2 , L3 , L4 , Q) Let us first remark that xs L1 − xt L2 ... Lsat = 1 1 1 −3 1 More precisely, the ideal ILsat is the definition ideal of the curve (s4 , s3 t, st3 , t4 ), and IL = (xz − y w, y z − x2 w2 , z − xw3 , y − x3 w) Applying Theorem 2 .1, we...

Ngày tải lên: 06/08/2014, 05:20

... 5,6 12 2,3 22 13 14 16 15 17 29 23 18 24 26 25 27 30 28 31 33 32 34 35 Figure 3: The tree F2 n a0 a1 a2 d0 d1 d2 p0 p1 p2 1 1 1 1 1 2 1 0 2 1 2 1 2 0 11 14 12 15 4 11 14 17 1 12 15 18 4 1 16 ... 20 24 10 4 0 17 21 25 11 1 19 23 27 12 1 20 24 28 13 0 23 27 31 14 1 24 28 32 15 8 1 26 30 34 16 0 27 31 35 17 10 1 32 37 42 18 10 8 0 33 38 43 19 11 8 0 35 40 45 20 12 1 36 41 46 Table 1: The ... since D0 = E∞ By induction En +1 = En En = 11 0r1 11 0r2 · · · 11 0r2n 1 110 r1 11 0r2 · · · 11 0r2n 1 = 11 0r1 11 0r2 · · · 11 0r2n 1 110 r1+2n 1 110 r2+2n 1 · · · 11 0r2n 1 110 n +1 , as required The second...

Ngày tải lên: 07/08/2014, 13:21