1 1 accounting methods for income reco
support vector and kernel methods for pattern recognition
Ngày tải lên :
24/04/2014, 13:39
... kernels: x1, x ← K ( x1, x ) = φ ( x1), φ ( x ) www.support-vector.net 12 Example: Polynomial Kernels x = ( x1, x ); z = ( z1, z 2); x, z = ( x z1 + x z ) = 2 = x12 z12 + x z2 + x1z1 x z = 2 = ( x12 ... approximate it with a convex problem www.support-vector.net The ideal kernel -1 … -1 1 -1 … -1 -1 YY’= -1 … … … … … -1 -1 … 1 www.support-vector.net 50 Spectral Machines ! Can (approximately) maximize ... Alignment (= similarity between Gram matrices) A( K1, K 2) = K1, K K1, K1 K 2, K www.support-vector.net 47 Kernel Alignment K1, K K1, K1 K 2, K A( K1, K 2) = Where we use the Frobenius inner product:...
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survey of appearance-based methods for object recognition
Ngày tải lên :
24/04/2014, 13:47
... j =1 A measurement for the quality of the reconstruction is the squared reconstruction error: n 1 ˜ = ||x − x|| = = a j uj j=k +1 ¯ aj uj + x − n 1 k j =1 ¯ aj uj + x j =1 n 1 a2 j = = (57) j=k +1 ... [4], Kernel CCA [84], and Kernel NMF [13 8] More detailed discussions on kernel methods can be found in [23, 11 1, 11 4] 4.7.2 2D Subspace Methods For subspace methods usually the input images are ... expected value for s2 is given by E s =E n 1 n (xj − x)2 ¯ (99) j =1 Due to the linearity of the expected value we can re-write (99) to n 1 n ¯2 E (xj − x) = j =1 = n 1 = n 1 = n 1 = n 1 = n 1 n ¯ ¯2...
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Recommendations for Use of Antiretroviral Drugs in Pregnant HIV-1-Infected Women for Maternal Health and Interventions to Reduce Perinatal HIV Transmission in the United States pdf
Ngày tải lên :
05/03/2014, 13:20
... tract Antivir Ther 2 011 ;16 (8) :11 49 -11 67 Available at http://www.ncbi.nlm.nih.gov/pubmed/2 215 5899 Recommendations for Use of Antiretroviral Drugs in Pregnant HIV -1- Infected Women for Maternal Health ... Care 2 012 ;24 (1) :1- 11 Available at http://www.ncbi.nlm.nih.gov/pubmed/ 217 77077 Lampe MA Human immunodeficiency virus -1 and preconception care Matern Child Health J Sep 2006 ;10 (5 Suppl):S193 -19 5 Available ... http://www.ncbi.nlm.nih.gov/pubmed /16 832609 Aaron EZ, Criniti SM Preconception health care for HIV-infected women Top HIV Med Aug-Sep 2007 ;15 (4) :13 7 -14 1 Available at http://www.ncbi.nlm.nih.gov/pubmed /17 7 210 00 10 Centers for...
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Handbook of Residue Analytical Methods for Agrochemicals VOLUME 1 and VOLUME 2 doc
Ngày tải lên :
17/03/2014, 02:20
... Water 10 99 10 99 10 99 11 02 11 03 11 04 11 04 11 04 11 07 11 08 11 10 11 11 111 3 11 15 11 17 11 17 11 27 11 27 11 28 11 28 11 28 11 28 11 30 11 38 11 38 11 39 11 41 114 1 11 42 11 43 11 44 11 44 11 47 11 47 11 48 11 53 11 58 11 59 ... 13 06 13 06 13 07 13 08 13 08 13 09 13 09 13 09 13 10 13 10 13 12 13 13 13 13 13 13 13 13 13 14 13 14 13 16 13 16 13 17 13 17 13 17 13 17 13 17 13 17 13 18 13 18 13 18 13 18 13 19 13 19 13 19 13 19 13 19 13 19 13 20 13 20 13 21 13 21 ... 11 91 119 1 11 92 11 92 11 92 11 93 11 93 11 93 11 94 11 95 11 96 11 96 11 96 11 96 11 97 11 98 11 98 11 99 12 00 12 00 12 00 12 00 12 02 12 02 12 02 12 03 12 03 12 05 12 06 12 06 12 06 12 07 12 07 12 08 12 08 12 08 12 09 12 09 12 10...
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Báo cáo khoa học: Molecular basis for substrate recognition and drug ˚ resistance from 1.1 to 1.6 A resolution crystal structures of HIV-1 protease mutants with substrate analogs pptx
Ngày tải lên :
30/03/2014, 20:20
... CA-p2 P 212 12 p2-NC P 212 12 p2-NC P 212 12 p2-NC P 212 12 p6pol-PR P 212 12 p6pol-PR P 212 12 p1-p6 P 212 12 p1-p6 P 212 12 57.89 85.96 46 .19 50 1. 54 34 544 6.5 (37 .1) 13 .4 (5.8) 10 1. 54 0 .12 0 .19 19 4 58.00 ... 50 1. 40 44 2 91 10 .1 (45.6) 8.9 (2 .1) 10 1. 40 0 .15 0 .19 18 8 58.02 85.89 46. 61 50 1. 10 95 318 10 .2 (37.7) 10 .4 (2 .1) 10 1. 10 0 .13 0 .17 206 57.80 85.59 46.46 50 1. 30 55 009 7.9 (33.0) 14 .4 (3.2) 10 1. 30 ... 99.2 (93 .1) 90.3 (10 0) 89.2 (96.2) 99.9 (10 0) 0. 010 0.029 0. 011 0.029 0. 015 0.033 0. 013 0.030 0.009 0.029 0. 011 0.0 31 0. 010 0.030 0. 012 0.034 15 .9 21. 9 27.5 31. 5 8.0 12 .2 14 .0 23.7 9.0 13 .7 10 .6...
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formal methods for components and objects 9th international symposium, fmco 2010, graz, austria, november 29-december 1, 2010 revised papers
Ngày tải lên :
31/05/2014, 00:38
... Science P.O Box 9 512 2300 RA Leiden, The Netherlands E-mail: marcello@liacs.nl ISSN 0302-9743 e-ISSN 16 11- 3349 ISBN 978-3-642-25270-9 e-ISBN 978-3-642-252 71- 6 DOI 10 .10 07/978-3-642-252 71- 6 Springer ... Marcello M Bonsangue (Eds.) Formal Methods for Components and Objects 9th International Symposium, FMCO 2 010 Graz, Austria, November 29 - December 1, 2 010 Revised Papers 13 Volume Editors Bernhard ... The 9th Symposium on Formal Methods for Components and Objects (FMCO 2 010 ) was held in Graz, Austria, from November 29 to December 1, 2 010 The venue was Hotel Weitzer FMCO 2 010 was realized as...
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Báo cáo hóa học: "Research Article Time-Frequency and Time-Scale-Based Fragile Watermarking Methods for Image Authentication Braham Barkat1 and Farook Sattar (EURASIP Member)2 1 Department 2 Faculty" pdf
Ngày tải lên :
21/06/2014, 08:20
... extracted chirp signal 1. 3 1. 25 1. 2 12 9 (13 5, 13 8) 1. 15 13 6 1. 1 14 4 16 × authentication block with index 2 81 1.05 0.95 50 10 0 15 0 200 250 300 350 400 450 500 Block index Figure 14 : Magnitudes of watermark ... extracted chirp signal 1. 3 1. 2 Pixel location (16 1, 12 9) 1. 1 Block index 267 283 299 315 3 31 284 300 316 332 (16 1, 16 8) 0.9 0.8 16 × authentication block 0.7 0.6 0.5 0.4 0.3 50 10 0 15 0 200 250 300 350 ... 0.0267 15 .74 80% 35.94 0.06 41 11. 93 70% 34. 41 0.0956 10 .19 60% 33.34 0 .10 61 9.74 50% 32.56 0 .12 27 9 .11 w mag (i) − , Nw i =1 given by 267, 268, 283, 284, 299, 300, 315 , 316 , 3 31, and 332, exactly match...
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Computational Methods for Protein Structure Prediction and Modeling Volume 1: Basic Characterization pot
Ngày tải lên :
27/06/2014, 10:20
... 2006 10 :1 Ying Xu, Dong Xu, and Jie Liang (Eds.) Computational Methods for Protein Structure Prediction and Modeling Volume 1: Basic Characterization iii SVNY330-Xu-Vol-I November 4, 2006 10 :1 Dong ... methods CASP deserves special recognition in any consideration of the role of modeling/computational methods for biology, since the meeting/process has transformed the level of recognition (for ... (positive-inside rule) (Heijne, 19 86), the hydrophobic/hydrophilic patterns of TM regions (Kyte and Doolittle, 19 82), and the minimum length of TM regions 1. 3 .1. 1 Methods for Topology Model Prediction...
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Quantitative Methods for Business chapter 1 ppsx
Ngày tải lên :
06/07/2014, 00:20
... really use correlation and regression? Review questions 99 10 2 10 8 10 9 11 1 11 1 11 6 12 3 12 8 13 5 13 6 14 3 15 7 16 0 16 2 16 9 17 0 17 7 17 8 19 4 211 213 215 223 224 237 244 247 248 Contents Chapter vii Counting ... 4.5 * 10 1 4.5 * 10 0 4.5 * 10 1/2 4.5 * 10 Ϫ3 4.5 * 10 5 Match the operations on the left below to the answers on the right (i) (ii) (iii) (iv) (v) (vi) (vii) 1. 4 14 or Ϫ8 1. 5 or 1. 5 48 20 18 12 or ... service 30 Quantitative methods for business 1. 13 1. 14 1. 15 Chapter An energy company charges households a quarterly fee of £9.20 for providing an electricity supply and 7 .11 pence for every unit of...
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Quantitative Methods for Ecology and Evolutionary Biology (Cambridge, 2006) - Chapter 1 potx
Ngày tải lên :
06/07/2014, 13:20
... assumption, the only things that the forager does is search and handle prey items, so that T ¼ S þ H or T ¼ S þ h1 l1 S ¼ S 1 þ l1 h1 Þ (1: 1) We now solve this equation for the time spent searching, ... tackle for the Pittsburgh Steelers (played 19 68 19 81) , although he might provide an excellent metaphor too However, I mean the great composer of opera Giuseppe Verdi (lived 18 13 19 01; Figure 1. 7) ... 0 10 0.08 0.8 0.06 0.6 Gain 0 .1 R(t) 25 t (d) (c) Rate of gain, 20 15 Residence time, 0.04 0.4 0.02 0.2 Optimal residence time 0 10 20 15 Residence time, t 25 –5 10 15 Residence time Figure 1. 3...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 1 pdf
Ngày tải lên :
06/08/2014, 01:21
... 12 16 12 16 12 19 12 24 12 25 12 48 12 49 12 51 12 51 12 55 12 63 12 70 12 72 12 72 12 78 12 78 12 80 12 81 12 83 12 83 12 84 12 87 12 88 12 94 12 97 13 02 13 03 13 04 25 .14 Solutions ... 10 25 10 27 10 27 10 29 10 32 10 34 10 35 10 41 10 41 10 43 10 45 10 46 10 48 10 50 10 52 10 59 10 59 10 61 10 65 10 65 10 68 10 71 10 74 10 74 10 76 10 77 10 79 10 82 21. 7 .1 Green ... 11 84 11 84 11 88 11 98 12 01 12 03 12 06 10 92 10 95 10 98 11 00 11 04 11 09 11 17 11 23 11 26 11 64 11 65 23.3 23.4 23.5 23.6 23.7 23.8 23.9 Irregular...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 2 ppt
Ngày tải lên :
06/08/2014, 01:21
... = 10 9 = x0 11 10 = x0 x0 = 10 9 At 7:00 pm the number of bacteria is 10 11 10 60 = 11 60 ≈ 3.04 × 10 11 51 10 At 3:00 pm the number of bacteria was 10 9 11 10 18 0 = 18 10 189 ≈ 35.4 11 180 Figure 1. 13: ... (a1 b3 − a3 b1 )j + (a1 b2 − a2 b1 )k Next we evaluate the determinant i j k a a a a a a a1 a2 a3 = i − j + k b2 b3 b1 b3 b1 b2 b1 b2 b3 = (a2 b3 − a3 b2 )i − (a1 b3 − a3 b1 )j + (a1 b2 − a2 b1 ... plane 1, 1, −2 × 2, 1, 1 = −3, −3, −3 We see that the plane is orthogonal to the vector 1, 1, and passes through the point (1, 2, 3) The equation of the plane is 1, 1, · x, y, z = 1, 1, · 1, 2,...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 3 pptx
Ngày tải lên :
06/08/2014, 01:21
... = ( 1) n 1 (n − 1) !, for n ≥ By Taylor’s theorem of the mean we have, ln x = (x − 1) − (x − 1) 2 (x − 1) 3 (x − 1) 4 (x − 1) n (x − 1) n +1 + − + · · · + ( 1) n 1 + ( 1) n n n + ξ n +1 72 -1 0.5 1. 5 2.5 ... in Figure 3 .11 2.5 1. 5 0.5 -1 -0.5 0.5 2.5 1. 5 0.5 -1 -0.5 2.5 1. 5 0.5 -1 -0.5 0.5 0.5 2.5 1. 5 0.5 -1 -0.5 0.5 Figure 3 .11 : Four Finite Taylor Series Approximations of ex Note that for the range ... 6! (2(n − 1) )! (2n)! Here are graphs of the one, two, three and four term approximations 0.5 0.5 -3 -2 -1 -0.5 -1 -3 -2 -1 -0.5 -1 0.5 -3 -2 -1 -0.5 -1 0.5 -3 -2 -1 -0.5 -1 Figure 3 .12 : Taylor...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 4 pptx
Ngày tải lên :
06/08/2014, 01:21
... − 1) 2 The expansion has the form a0 a1 b0 b1 + + + 2 x x (x − 1) x 1 128 The coefficients are 1 + x + x2 = 1, 0! (x − 1) 2 x=0 d + x + x2 = a1 = 1! dx (x − 1) 2 x=0 1 + x + x2 = 3, b0 = 0! x2 x =1 ... − x→0 10 9 x =0 c ln lim x→+∞ 1+ x x = lim x→+∞ = lim x→+∞ = lim x→+∞ = lim ln + 1/ x 1+ x→+∞ 1+ =1 Thus we have lim x→+∞ 1+ 11 0 x x 1 x − x2 1/ x2 x→+∞ = lim x ln 1+ x x ln + x x = e x 1 d It ... 4.4 .1 Consider the partial fraction expansion of + x + x2 (x − 1) 3 The expansion has the form a0 a1 a2 + + (x − 1) (x − 1) x 1 127 The coefficients are (1 + x + x2 )|x =1 = 3, 0! d a1 = (1 + x...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 5 pdf
Ngày tải lên :
06/08/2014, 01:21
... dx2 = (1 + 2x) x= 1 = 1 x= 1 = x= 1 (2) x= 1 =1 Then we can the integration + x + x2 dx = (x + 1) 3 1 − + (x + 1) (x + 1) x +1 1 + + ln |x + 1| =− 2(x + 1) 2 x + x + 1/ 2 + ln |x + 1| = (x + 1) 2 dx ... (ξ), a for some ξ ∈ [a b] Solution 4.6 a b c d = + + + x(x − 1) (x − 2)(x − 3) x x 1 x−2 x−3 15 2 1 =− (0 − 1) (0 − 2)(0 − 3) 1 = b= (1) (1 − 2) (1 − 3) 1 c= =− (2)(2 − 1) (2 − 3) 1 = d= (3)(3 − 1) (3 ... = −3 to solve for a, b and c Hint 4 .17 dx = lim δ→0+ (x − 1) 2 1 δ dx + lim →0+ (x − 1) 2 Hint 4 .18 1 √ dx = lim →0+ x 1 √ dx x Hint 4 .19 x2 1 x dx = arctan +a a a 14 0 1+ dx (x − 1) 2 4.8 Solutions...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 6 pps
Ngày tải lên :
06/08/2014, 01:21
... more that s values For instance, (12 ) 11 /2 = 1 and 11 /2 = ( 1) 2 = Example 6.6.2 Consider 21/ 5 , (1 + ı )1/ 3 and (2 + ı)5/6 √ 21/ 5 = eı2πk/5 , for k = 0, 1, 2, 3, 19 9 = √ (1 + ı )1/ 3 = = (2 + ı)5/6 ... arctan(3, 1) ) − ı sin(9 arctan(3, 1) )) 10 000 10 (x − y) √ (cos(9 arctan(3, 1) ) + sin(9 arctan(3, 1) )) = 10 000 10 (x − y) √ (cos(9 arctan(3, 1) ) − sin(9 arctan(3, 1) )) +ı 10 000 10 = (1 + ı)(x − y) 212 ... = √ = 12 √ 1/ 3 eıπ/4 eıπ /12 eı2πk/3 , eı Arctan(2 ,1) for k = 0, 1, 5/6 55 eı5 Arctan(2 ,1) 1/ 6 55 eı Arctan(2 ,1) eıπk/3 , for k = 0, 1, 2, 3, 4, Example 6.6.3 We find the roots of z + (−4 )1/ 5 =...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 7 ppt
Ngày tải lên :
06/08/2014, 01:21
... Cartesian form √ 1+ ı 1 = √ 1+ ı √ −2 + ı2 1 = √ −2 + ı2 √ −2 + ı2 √ −8 − ı8 1 = = √ −2 + ı2 √ 12 8 + 12 8 10 √ 1+ ı √ = − 512 − ı 512 1 1 √ 512 + ı √ 1 − ı √ √ = 512 + ı − ı √ =− +ı 2048 2048 = 214 ... 6 .16 225 -1 -1 Figure 6 .15 : ( 1) −3/4 Solution 6 .13 ( 1) 1/ 4 = (( 1) 1 )1/ 4 = ( 1) 1/4 = (eıπ )1/ 4 = eıπ/4 11 /4 = eıπ/4 eıkπ/2 , k = 0, 1, 2, = eıπ/4 , eı3π/4 , eı5π/4 , eı7π/4 + ı 1 + ı 1 − ı − ... 1/ 3 eıπ /12 11 /3 eıπ /12 eı2πk/3 , k = 0, 1, √ ıπ /12 √ ı3π/4 √ 17 π /12 6 = , 2e , 2e 2e = The principal root is √ 1+ ı= √ eıπ /12 The roots are depicted in Figure 6.9 1/ 4 1/ 4 = eıπ/2 = eıπ/8 11 /4...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 8 ppt
Ngày tải lên :
06/08/2014, 01:21
... draw a 2 71 -1 -2 y -1 y -1 -1 x -1 -2 -1 x -2 -2 -1 -2 01 x -2 -1 y -1 -2 01 x -1 -2 12 10-2 -1 y -1 -1 y -1 1 210 -2 -1 y Figure 7.24: Plots of x -1 -2 -1 z 1/ 2 (left) and 272 x -1 -2 z 1/ 2 (right) ... arguments: log( 1) = log 1 = log (1) − log( 1) = − log( 1) , therefore, log( 1) = = 11 /2 = (( 1) ( 1) )1/ 2 = ( 1) 1/2 ( 1) 1/2 = ıı = 1, therefore, = 1 Hint, Solution Exercise 7 .11 Write the following ... (z) √ 1/ 3 then what is f (−2)? f (z) = z 1/ 3 (z − 1) 1/3 (z + 1) 1/3 There are branch points at z = 1, 0, We consider the point at infinity f ζ = = ζ 1/ 3 1 ζ 1/ 3 +1 ζ 1/ 3 (1 − ζ )1/ 3 (1 + ζ )1/ 3...
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