2 2 accounting methods for income reco
support vector and kernel methods for pattern recognition
... www.support-vector.net 12 Example: Polynomial Kernels x = ( x1, x ); z = ( z1, z 2) ; x, z = ( x z1 + x z ) = 2 = x 12 z 12 + x z2 + x1z1 x z = 2 = ( x 12 , x2 , x1 x 2) ,( z 12 , z2 , z1z ) = = φ ( x ... K(1,1) K(1 ,2) … K(1,m) K (2, 1) K (2, 2) K (2, 3) … K (2, m) … … … … … K(m,1 ) K= K(1,3) K(m ,2) K(m,3) … K(m,m) www.support-vector.net The Kernel Matrix ! The central structure in kernel machines ! Information ... ) n ( sign x − c− ! Decision function will be: ( 2 sign x + c− − x − c+ 2 ) − x, c− − x − c+ 2 sign c− − c+ + 2( x, c− − x, c+ 1 424 3 2b αi = nclass ( i ) + x, c+ ) ) sign ...
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survey of appearance-based methods for object recognition
... n−1 n ¯ 2 E xj − E [xj x] + E x j=1 n µ + σ − E xj n j=1 n j=1 n j=1 n j=1 n x k + 2 + 2 2 + + n 2 2 + + n = k=1 2 − 2 E xj − n n 2 n = E [xj xk ] j=k 2 n−1 2 − µ µ + 2 − n n n−1 2 2 σ − µ ... growing for the standard approach while the performance of the robust method stays the same even for 25 % occlusion 25 std approach robust approach reconstruction error 20 15 10 0 10 15 20 25 30 ... and Kernel NMF [138] More detailed discussions on kernel methods can be found in [23 , 111, 114] 4.7 .2 2D Subspace Methods For subspace methods usually the input images are vectorized and saved...
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Handbook of Residue Analytical Methods for Agrochemicals VOLUME 1 and VOLUME 2 doc
... 121 3 121 3 121 4 121 4 121 4 121 5 121 5 121 6 121 6 121 6 121 7 121 7 121 7 121 7 121 8 121 9 121 9 121 9 121 9 122 0 122 1 122 1 122 2 122 2 122 2 122 3 122 3 122 3 122 4 122 4 122 5 122 5 122 5 122 6 122 6 122 6 122 7 122 7 xxxiii ... study findings 21 1 21 2 21 3 21 3 21 3 21 5 21 5 21 5 21 6 21 8 21 8 21 9 22 3 22 4 22 4 22 4 22 6 22 7 22 7 22 7 22 8 23 0 23 0 23 1 23 1 23 1 23 2 23 3 23 4 23 5 23 6 23 7 23 8 23 9 23 9 24 0 24 1 24 1 24 2 24 2 24 3 24 5 Contents of ... 122 8 122 8 122 8 122 9 122 9 123 0 123 0 123 0 123 0 123 1 123 1 123 1 123 2 123 2 123 2 123 3 123 3 123 4 123 4 123 5 123 5 123 5 123 5 123 6 123 6 123 6 123 7 123 8 123 8 123 9 124 0 124 1 124 2 124 2 124 3 124 3 124 3 124 3 124 4...
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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
... decomposition type, J− g(i1 )h(i2 )xLL (2n1 − i1 , 2n2 − i2 ), J xHL (n1 , n2 ) = (5) i1 ,i2 J− g(i1 )g(i2 )xLL (2n1 − i1 , 2n2 − i2 ), J xHH (n1 , n2 ) = i1 ,i2 where h(i) represents the low-pass ... −0.0 027 Translation −0.0453 Rotation(1 degree) −0.00 72 Cropping −0.0094 JPEG (QF = 99%) 0 .20 27 considered here is given by [21 ] J− h(i1 )h(i2 )xLL (2n1 − i1 , 2n2 − i2 ), J xLL (n1 , n2 ) = i1 ,i2 ... watermarked image [24 ] For an original image, I(n1 , n2 ) and its watermarked image, W(n1 , n2 ), with 25 5 gray levels, the PSNR is defined as [24 ] (25 5 )2 (9) n2 (I(n1 , n2 ) − W(n1 , n2 )) n1 PSNR...
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báo cáo hóa học:" Research Article Towards Automation 2.0: A Neurocognitive Model for Environment Recognition, Decision-Making, and Action Execution" docx
... perceptual system of the brain for the purpose of sensor EURASIP Journal on Embedded Systems [6] [7] [8] [9] [10] [11] [ 12] [13] [14] [15] [16] [17] [18] [19] [20 ] [21 ] [22 ] [23 ] [24 ] fusion,” in Human-Computer ... York, NY, USA, 20 02 [26 ] C Roesener, Adaptive behavior arbitration for mobile service robots in building automation, Ph.D thesis, Vienna University of Technology, Vienna, Austria, 20 07 [27 ] B Palensky, ... D Care (a) Internal States of Agent A that lead to the Formulation of a Request 100 100 100 100 75 75 75 75 50 50 50 50 25 25 25 25 A B C D Basic emotions a b c A B C D a b c Complex emotions...
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Quantitative Methods for Business chapter 2 doc
... chartered) 120 100 80 Cargo 60 Passenger 40 20 0 20 40 60 x (Lotkas chartered) 80 100 80 100 Figure 2. 22 The feasible region in Example 2. 23 y (Soodnas chartered) 120 100 80 60 40 A 20 0 20 40 60 ... by elimination: (a) 3x ϩ 2y ϭ xϩ yϭ3 (c) 2x ϩ 7y ϭ 4x ϩ 3y ϭ 17 (d) 6x ϩ 6y ϭ 27 4x ϩ 5y ϭ 22 2. 4 (b) 5x ϩ 3y ϭ 19 2x Ϫ y ϭ (e) 3x Ϫ 2y ϭ x ϩ 4y ϭ 26 (f) 12x ϩ 4y ϭ 2x Ϫ y ϭ Following a crash, ... y ϭ 20 0 ϩ 0.1x Subtract 0.1x from both sides: y Ϫ 0.1x ϭ 20 0 ϩ 0.1x Ϫ 0.1x to get: y Ϫ 0.1x ϭ 20 0 For the second equation: y ϭ ϩ 0.2x Subtract 0.2x from both sides: y Ϫ 0.2x ϭ ϩ 0.2x Ϫ 0.2x to...
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Quantitative Methods for Ecology and Evolutionary Biology (Cambridge, 2006) - Chapter 2 pptx
... Expected reproductive success 120 30 Size (cm) 25 20 15 10 0 10 15 Age (yr) 20 25 30 100 80 60 40 20 0 10 15 Age (yr) 20 25 Figure 2. 2 (a) von Bertalanffy growth for an organism with asymptotic ... dL ¼ 3L2 dt dt (2: 10) and if we use this equation in Eq (2. 9), we see that 3L2 dL ¼ L2 À cL3 dt (2: 11) so that now if we divide through by 3L2, we obtain dL c ¼ À L dt 3 3 (2: 12) and we ... 120 N(500) 100 80 60 40 20 1 .2 1.4 1.6 1.8 r 2. 2 2. 4 2. 6 2. 8 How we understand what is happening? To begin we rewrite Eq (2. 29) as N ðt þ 1Þ ¼ ð1 þ rÞNðtÞ À rN ðt 2 K and investigate this as a...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 2 ppt
... + a2 j + a3 k) × (b1 i + b2 j + b3 k) = a1 i × (b1 i + b2 j + b3 k) + a2 j × (b1 i + b2 j + b3 k) + a3 k × (b1 i + b2 j + b3 k) = a1 b2 k + a1 b3 (−j) + a2 b1 (−k) + a2 b3 i + a3 b1 j + a3 b2 ... b2 (−i) = (a2 b3 − a3 b2 )i − (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 ... ax2 x→∞ x2 + βx + χ 2ax = lim x→∞ 2x + β 2a = lim x→∞ =a lim f (x) = lim x→∞ f (x) = 2x2 x2 + βx + χ Now we use the fact that f (x) is even to conclude that q(x) is even and thus β = f (x) = 2x2...
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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 1 pps
... = (nπ)1 /2 , n ∈ Z± z2 2z = lim 2) z→0 sin (z z→0 2z cos (z ) lim = lim z→0 cos (z ) =1 379 − 4z sin (z ) lim z→(nπ)1 /2 z − (nπ)1 /2 = lim 2) 1 /2 2z cos (z ) sin (z z→(nπ) = 1 /2 (−1)n 2( nπ) Example ... are ux = vy and uy = −vx −(x − 1) + y −(x − 1 )2 + y 2( x − 1)y 2( x − 1)y = and = ((x − 1 )2 + y )2 ((x − 1 )2 + y )2 ((x − 1 )2 + y )2 ((x − 1 )2 + y )2 2 The Cauchy-Riemann equations are each identities ... x cosh y − ı cos x sinh y x2 − y + x + ı(2xy − y) Hint, Solution Exercise 8.5 f (z) is analytic for all z, (|z| < ∞) f (z1 + z2 ) = f (z1 ) f (z2 ) for all z1 and z2 (This is known as a functional...
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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 2 pptx
... function Recall that for analytic functions f , f = fx = −ıfy So that fx + ıfy = 2 2 + = ∂u2 ∂v dw dz −1 dw dz 2 2 dw + = 2 ∂u ∂v dz −1 2 2 2 + ∂x2 ∂y 2 2 + ∂x2 ∂y Solution 8.16 ... analytic 423 We calculate the first partial derivatives of u and v u x = ex −y uy = 2 e (x cos(2xy) − y sin(2xy)) x2 −y (y cos(2xy) + x sin(2xy)) vx = e x2 −y (y cos(2xy) + x sin(2xy)) vy = e x2 −y ... f (z) = ux Note that d 2u(z /2, −ız /2) dz z z z z , −ı − ıuy , −ı 2 2 = ux (z /2, −ız /2) − ıuy (z /2, −ız /2) We integrate the equation to obtain: f (z) = 2u z z + c , −ı 2 We know that the real...
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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 3 ppt
... θ0 + π] z2 − dz ≤ z2 + z2 − |dz| C z +1 θ0 +π R2 e 2 −1 ≤ |R dθ| R2 e 2 +1 θ0 θ0 +π R2 + dθ R2 − θ0 R2 + = πr R −1 ≤R 490 Solution 10.6 f (z) dz = C √ π eıθ /2 ı eıθ dθ + r dr + 2 = + − −ı ... 2 ) 2 eınθ ı eıθ dθ (z − z0 )n dz = C = 2 eı(n+1)θ n+1 [ıθ ]2 for n = −1 = for n = −1 2 for n = −1 for n = −1 We parameterize the contour and the integration z − z0 = 2 + eıθ , 2 ... of the integral z = eıθ , θ ∈ [−π /2 π /2] 489 Log z dz ≤ |Log z| |dz| C C π /2 | ln + ıθ |2 dθ = −π /2 π /2 (ln + |θ|) dθ 2 −π /2 π /2 =4 (ln + θ) dθ π (π + ln 2) = We parameterize the contour and...
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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 4 ppsx
... , C2 and C2 C z dz = z3 − C1 z− + √ C2 z− C3 z− + = 2 + 2 + 2 z− √ √ √ z− z− z √ dz e 2 /3 z − e− 2 /3 z √ √ dz 2 /3 z − 9e z − e− 2 /3 z √ √ dz z − e 2 /3 z − e− 2 /3 z− 9 √ z e 2 /3 ... Integral Formula ezt dz 2 C z (z + a ) ezt ı ezt ı ezt ω= + − 2 C a2 z 2a (z − ıa) 2a (z + ıa) d ezt ı eıat ı e−ıat ω= − + dz a2 z=0 2a3 2a3 sin(at) t ω= 2 a a3 at − sin(at) ω= a3 ω= 2 dz Solution ... fractions z2 a= z z+ı z a b = + +1 z−ı z+ı z=ı = , b= z z−ı = z=−ı Now we can the integral with Cauchy’s formula C z2 z dz = +1 1 /2 dz + C z −ı 1 = 2 + 2 2 = 2 C 1 /2 dz z+ı C z2 + dz = z...
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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 5 pps
... z +2 z + 5z + 1 √ =√ + z/3 + z /2 −1 /2 z −1 /2 z = √ 1+ + + ··· 3 z 3z z z2 = √ 1− + + ··· 1− + + ··· 24 32 17 = √ − z + z2 + · · · 12 96 12. 6 1+ −1 /2 z −1 /2 + 2 z 2 + ··· Laurent Series Result 12. 6.1 ... n)n ∞ (−1)n ln 11 n =2 ∞ 12 n =2 ∞ 13 n =2 n (n! )2 (2n)! 3n + 4n + 5n − 4n − 5 62 ∞ 14 n =2 ∞ 15 n =2 ∞ 16 n=1 ∞ 17 n=1 ∞ 18 n=1 ∞ 19 n=1 ∞ 20 n =2 n! (ln n)n en ln(n!) (n! )2 (n2 )! n8 + 4n4 + 3n9 − ... converge ∞ n =2 ∞ n =2 n ln(n) ln (nn ) ∞ ln √ n ln n n =2 561 ∞ n(ln n)(ln(ln n)) n=10 ∞ n=1 ∞ n=0 ∞ n=0 ln (2n ) ln (3n ) + 1 ln(n + 20 ) 4n + 3n − ∞ (Logπ 2) n n=0 ∞ n =2 ∞ 10 n =2 n2 − n4 − n2 (ln n)n...
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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 6 doc
... = = for x = 2 k for x = 2 k 1−eınx 1−eıx for x = 2 k e−ıx /2 − eı(N −1 /2) x e−ıx /2 − eıx /2 for x = 2 k e−ıx /2 − eı(N −1 /2) x − 2 sin(x /2) = N −1 sin(nx) = n=1 for x = 2 k for x = 2 k for x = 2 k ... function Hint 12. 22 cos z = − cos(z − π) sin z = − sin(z − π) Hint 12. 23 CONTINUE Hint 12. 24 CONTINUE Hint 12. 25 Hint 12. 26 Hint 12. 27 Hint 12. 28 Hint 12. 29 Hint 12. 30 CONTINUE 581 12. 9 Solutions ... Solution 12. 4 ∞ n=1 (−1)n+1 = n ∞ n=1 ∞ = n=1 ∞ < 1 − 2n − 2n (2n − 1)(2n) (2n − 1 )2 n=1 ∞ < = n=1 π 12 Thus the series is convergent 587 n2 Solution 12. 5 Since 2n−1 |S2n − Sn | = j=n 2n−1 ≥ j=n j 2n...
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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 7 pdf
... 1)(−1)n 2 n z n , n=0 ∞ n=0 − ı /2 − ı /2 = (z − 2) z2 − ı /2 = z2 2 n n+1 n z , 2n 1− z ∞ 2 n n=0 ∞ for |z| < for |z| < 2 − z n , for |2/ z| < (−1)n (n + 1)(−1)n 2n z −n 2 , = (2 − ı /2) for |z| ... z 2 − 2/ z =− z ∞ n=0 ∞ z n , 2n z −n−1 , =− for |2/ z| < for |z| > n=0 −1 2 n−1 z n , =− n=−∞ 620 for |z| > 2 − ı /2 = (2 − ı /2) (1 − z /2) 2 (z − 2) 4−ı = = = 4−ı 4−ı ∞ n=0 ∞ − z n , for |z /2| < ... − ı /2 z − (z − 2) 2 |z|=3 (z − ı /2) + ı /2 (z − 2) + c(z − 2) + 2c − + dz = z − ı /2 z 2 (z − 2) 2 ı 2 2+ c =0 ı c =2 z |z|=3 Thus we see that the function is f (z) = 1 − ı /2 − + + d, z − ı /2 z...
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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 8 pot
... (z + ı5) z =2 ı (z + (ı7 − 2) z − − ı 12) eız e 2 + −ı = 2 (z + ı5 )2 (z − 2) 58 116 z=0 ı e 2 = 2 − + + −ı 25 20 58 116 π 6π = − + π cos − π sin + ı − + π cos + π sin 10 58 29 25 29 58 687 We ... cosh y − ı sin x sinh y sin x cosh y + ı cos x sinh y = cos2 x cosh2 y + sin2 x sinh2 y sin2 x cosh2 y + cos2 x sinh2 y ≤ cosh2 y sinh2 y = | coth(y)| The hyperbolic cotangent, coth(y), has a ... 1/(ız) dz = + a(z − z −1 )/ (2 ) C z2 2/ a dz + ( 2/ a)z − We factor the denominator of the integrand z1 = ı 2/ a dz f (a) = C (z − z1 )(z − z2 ) √ √ −1 + − a2 −1 − − a2 , z2 = ı a a Because |a| < 1,...
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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 9 ppt
... max z∈CR z2 (z + 1 )2 = lim R→∞ R R2 (R2 − 1 )2 =0 we can apply Result 13.4.1 ∞ −∞ Res x2 dx = 2 Res (x2 + 1 )2 z2 ,z = ı (z + 1 )2 ∞ −∞ z2 ,z = ı (z + 1 )2 d z2 (z − ı )2 z→ı dz (z + 1 )2 d z2 = lim ... x2 ∞ ıaπ /2 a e eıa3π /2 x dx = 2 + − e 2 a + x2 2 − 2 ∞ eıaπ /2 − eıa3π /2 xa dx = π + x2 − eı2aπ ∞ eıaπ /2 (1 − eıaπ ) xa dx = π + x2 (1 + eıaπ )(1 − eıaπ ) ∞ π xa dx = −ıaπ /2 e 1+x + eıaπ /2 ... angle range −π /2 < θ < 3π /2 We consider the integral of log2 z/(1 + z ) on this contour C log2 z dz = 2 Res + z2 log2 z ,z = ı + z2 log2 z = 2 lim z→ı z + ı (ıπ /2) 2 = 2 2 π3 =− 720 Let CR be...
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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 10 ppt
... x + 2 ) dx = 2 Res (x + 1 )2 ∞ 0 x1 /2 d 1 /2 dx = 2 lim (z log z) z→−1 dz (x + 1 )2 x1 /2 dx = 2 lim z→−1 (x + 1 )2 x1 /2 log x dx + 2 (x + 1 )2 ∞ z 1 /2 log z , −1 (z + 1 )2 x1 /2 dx = 2 (x ... −ı 2 −ı Res (z − 1 )2 √ √ , + Res z(z − + 2) (z − − 2) 768 √ (z − 1 )2 √ √ ,z = − 2 z(z − + 2) (z − − 2) π lim z→0 (z − 1 )2 √ √ (z − + 2) (z − − 2) + lim√ z→3 2 (z − 1 )2 √ z(z − − 2) √ x2 (2 − 2) π ... Res , z = −c − c2 − (z + c + c2 − 1 )2 (z + c − c2 − 1 )2 d z √ = 4π lim √ z→−c− c2 −1 dz (z + c − c2 − 1 )2 2z √ √ − = 4π lim √ z→−c− c2 −1 (z + c − c2 − 1 )2 (z + c − c2 − 1)3 √ c − c2 − − z √ = 4π...
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Tìm thêm: hệ việt nam nhật bản và sức hấp dẫn của tiếng nhật tại việt nam xác định các nguyên tắc biên soạn khảo sát các chuẩn giảng dạy tiếng nhật từ góc độ lí thuyết và thực tiễn khảo sát chương trình đào tạo của các đơn vị đào tạo tại nhật bản khảo sát chương trình đào tạo gắn với các giáo trình cụ thể xác định thời lượng học về mặt lí thuyết và thực tế tiến hành xây dựng chương trình đào tạo dành cho đối tượng không chuyên ngữ tại việt nam điều tra với đối tượng sinh viên học tiếng nhật không chuyên ngữ1 khảo sát các chương trình đào tạo theo những bộ giáo trình tiêu biểu phát huy những thành tựu công nghệ mới nhất được áp dụng vào công tác dạy và học ngoại ngữ mở máy động cơ lồng sóc mở máy động cơ rôto dây quấn các đặc tính của động cơ điện không đồng bộ hệ số công suất cosp fi p2 đặc tuyến hiệu suất h fi p2 đặc tuyến tốc độ rôto n fi p2 đặc tuyến dòng điện stato i1 fi p2 động cơ điện không đồng bộ một pha sự cần thiết phải đầu tư xây dựng nhà máy chỉ tiêu chất lượng theo chất lượng phẩm chất sản phẩm khô từ gạo của bộ y tế năm 2008