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... j! d j 1 dz j 1 ∞ k 1 z k 1 d j 1 dz j 1 z 1 − z 1 j 1 ! 1 − z −j−2 for 0 <z< ;1; j 0, 1, ,nm 1 . 4.22 Thus, it follows that μ 0 ≤ ∞ k0 nm 1 1 j0 kk!ρ k−j 0 v j 0 k − j!j! 3/2 ≤ nm 1 1 j0 j ... Moreover, sup k 1, 2, xk C nm 1 ≤ M w0 C nm 1 1 − qθ 0 1 , 4 .15 where w0ϕ0,ϕ 1 , ,ϕ−m. Proof. By 4 .11 ,weobtain β 0 ≤ sup k 1, 2, nm 1 1 j0 k!ρ k−j 0 v j 0 k ... Medina 13 On the other hand, μ 0 ≤ ∞ k0 nm 1 1 j0 kk!ρ k−j 0 v j 0 k − j!j! 3/2 . 4. 21 But ∞ k 1 kk!z k−j k − j! ≤ ∞ k 1 k 1 !z k−j k − j! d j 1 dz j 1 ∞ k 1 z k 1 d j 1 dz j 1 z1...
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... always work the first time. 02 0430 CH 01 5/22/ 01 10 :19 AM Page 11 01 0430 PT 01 5/22/ 01 10:09 AM Page 2 4 Chapter 1 Getting Started 1. 1 Editing with Emacs An editor is the program that you use ... CH 01 5/22/ 01 10 :19 AM Page 10 Advanced UNIX Programming with Linux I 1 Getting Started 2 Writing Good GNU/Linux Software 3 Processes 4 Threads 5 Interprocess Communication 01 0430 PT 01 5/22/ 01 ... rule for this target uses the rm command to remove the files. 02 0430 CH 01 5/22/ 01 10 :19 AM Page 9 5 1. 1 Editing with Emacs 1. 1.2 Automatic Formatting If you’re accustomed to programming in an Integrated...
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... 39.2 51. 9 49.9 37.9 34.9 62.5 15 .4 21. 5 32.0 30.0 47.4 45.4 31. 9 28.9 10 0 19 .8 20 .1 24.5 22.5 44.3 42.3 27.8 24.8 200 29.0 18 .0 10 .8 8.8 39.8 37.8 21. 8 18 .8 250 32.8 17 .3 5.5 3.5 38.3 36.3 19 .8 ... Power Sum ELFEXT (MHz) (dB /10 0m) (dB /10 0m) (dB /10 0m) (dB /10 0m) (dB100m) (dB /10 0m) (dB /10 0m) (dB /10 0m) 1 2 20.0 72.3 70.3 74.3 72.3 67.8 64.8 4 3.8 23.0 61. 5 59.5 65.3 63.3 55.8 52.8 8 ... 250 32.8 17 .3 5.5 3.5 38.3 36.3 19 .8 16 .8 300 36.6 16 .8 - - 37.2 35.2 18 .2 15 .2 350 40.0 16 .3 - - 36.2 34.2 16 .9 13 .9 400 43.2 15 .9 - - 35.3 33.3 15 .7 12 .7 Note: The above listed discrete...
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Tài liệu Báo cáo "REMARKS ON LOCAL DIMENSION OF FRACTAL MEASURE ASSOCIATED WITH THE (0, 1, 9) - PROBLEM " pdf
... Let x 0 = (1, 1, 1, 1, 0, 0 , 1 , 1, 1, 1, 1, 0, 0 , 1 ) x 1 = (1, 1, 1, 1, 1, 0, 0 , 1 , 1, 1, 1, 1, 0, 0 , 1 , ) x 2 = (1, 1, 1, 1, 1, 1, 0, 0 , 1 , 1, 1, 1, 1, 0, 0 , 1 , ) x 3 = (1, ... ) x 3 = (1, 1, 0, 1, 1, 1, 1, 0, 0 , 1 , 1, 1, 1, 1, 0, 0 , 1 , ) x 4 = (1, 1, 0, 0, 1, 1, 1, 1, 0, 0 , 1 , 1, 1, 1, 1, 0, 0 , 1 , ) x 5 = (1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0 , 1 , 1, 1, 1, 1, ... by s = S(x), where x = (1, 1, 1, 1, 0, 0 , 1 k 1 =6 0, 0, ,0 , 1 k 2 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0 , 1 k 3 =18 0, 0, ,0, , 1 k 4 ). (17 ) Note that, for i ∈...
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Báo cáo Y học: Differential scanning calorimetric study of myosin subfragment 1 with tryptic cleavage at the N-terminal region of the heavy chain pdf
... DH cal by 18 %, from 12 30 to 14 50 kJặmol )1 . In the case of the complex t-S1ặADPặBeF x ,thethermaltransitionoft-S1 shiftedto55.1C and its enthalpy (14 00 kJặmol )1 ) increased by only 14 % in comparison ... subfrag- ment 1- nucleotide complexes. Biophys. J. 81, 11 01 11 14. 13 . Levitsky, D.I., Shnyrov, V.L., Khvorov, N.V., Bukatina, A.E., Vedenkina, N.S., Permyakov, E.A., Nikolaeva, O.P. & Poglazov, B.F. (19 92) ... complexes with t-S1 (B) or Nt-S1 (C). Conditions: 26 l M F-actin, 50 l M phalloidin, 13 l M t-S1 or Nt-S1 in 15 m M Hepes, pH 7.3, 2 m M MgCl 2 ,0.5m M ADP, and twice-diluted G-buffer. Heating rate 1...
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... (M3) F109-Q123 vs. homologue Inactive ± ± PAI-2 sequence b and E128G Mutant 4 (M4) F109-Q123 vs. AAAA Inactive ± ± Mutant 5 (D109 -11 2) D F109-H 112 10 000 + NT Mutant 6 (A 114 -11 5) F 114 A and R 115 A ... NT Mutant 7 (A 114 11 8) F 114 -R 118 vs. AAAAA 20 800 + + Mutant 8 (M8) F 114 -R 118 vs. AAAAA and D68G 7 700 + + Mutant 9 (M9) V284-G294 vs. homologue PAI-2 sequence c Inactive ± ± Mutant 10 (P73A) P73A ... Mutant A 114 11 8 displayed a slower association to (on-rate: 5.35 ´ 10 5 M )1 ás )1 )andafaster dissociation from Vn (off-rate: 1. 0 ´ 10 )3 s )1 ) c ompared t o wild-type PAI -1 (on-rate: 1. 0 ´ 10 6 M )1 ás )1 ;...
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