2 2 accounting methods for income reco

... 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 ...

Ngày tải lên: 24/04/2014, 13:39

... 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...

Ngày tải lên: 24/04/2014, 13:47

Ngày tải lên: 26/03/2015, 11:04

... 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...

Ngày tải lên: 17/03/2014, 02:20

... 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...

Ngày tải lên: 21/06/2014, 08:20

... 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...

Ngày tải lên: 21/06/2014, 11:20

... 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...

Ngày tải lên: 06/07/2014, 00:20

... 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...

Ngày tải lên: 06/07/2014, 13:20

... + 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...

Ngày tải lên: 06/08/2014, 01:21

... = (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...

Ngày tải lên: 06/08/2014, 01:21

... 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...

Ngày tải lên: 06/08/2014, 01:21

... 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...

Ngày tải lên: 06/08/2014, 01:21

... , 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...

Ngày tải lên: 06/08/2014, 01:21

... 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...

Ngày tải lên: 06/08/2014, 01:21

... = = 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...

Ngày tải lên: 06/08/2014, 01:21

... 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...

Ngày tải lên: 06/08/2014, 01:21

... (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,...

Ngày tải lên: 06/08/2014, 01:21

... 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...

Ngày tải lên: 06/08/2014, 01:21

... 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...

Ngày tải lên: 06/08/2014, 01:21