3 methods are functions and message expressions are function calls

... whether the clips are bits of a completed work or a work-in-progress So 35 Virilio, War and Cinema : The Logistics of Perception, 66 36 Cubitt, The Cinema Effect, 250 37 Ibid 33 footageheads edit ... Kittler and Lev Manovich to examine how the automatisms of both the hardware and software of digital cinema technologies encourage new forms, contents and participants From an extrinsic standpoint, ... Screens 37 Perpendicular Cinema 39 Ubiquity and Art 43 The Dissipating Aura of the Cinematic Art Object 45 III How Cinema is Digital: How Cinema Technologies are Changing How Movies Are Produced,...

Ngày tải lên: 07/03/2014, 15:20

... (407 ,37 2,266) 14,220,410 (39 3,151,856) 31 7,485 ,33 5 30 , 535 ,685 1,562,115 2,091, 733 3, 066,777 43, 5 23, 519 94, 837 (21, 839 ,779) — — — — — — — 3, 645,000 31 7,485 ,33 5 30 , 535 ,685 1,562,115 2,091, 733 3, 066,777 ... 4 73, 436 ,888 11,650,441 6,449,596 6,661 ,30 3 634 ,37 3 33 0,814 46,805 — 49,082 — — — — 11,650,441 6,498,678 6,661 ,30 3 634 ,37 3 33 0,814 46,805 25,7 73, 332 49,082 25,822,414 11,842 ,31 6 — 11,842 ,31 6 37 ,615,648 ... maturities (Note 3) 14,741,227 122,805, 238 Total current assets 2 ,30 9,277 3, 4 13, 469 — 11, 233 ,0 23 2,214,542 Total receivables 2,261,687 851,517 39 5,000 11, 233 ,0 23 277 ,34 1 ,32 3 — — — — 277 ,34 1 ,32 3 27,277,052...

Ngày tải lên: 18/06/2014, 20:20

... this study were ACH [30 ] and K30 [31 ] The K30 -3' /5681 and K30 -3' / 4089 constructs were derived from K30 DNA by subcloning 3' segments (positions 5681 to 90 43 and 4089 to 90 43, respectively) in ... GGCUGUUU (unspliced) HBZ 3 LTR 1P Ja s0 YB 03 YB 09 YB 13 YB 16 YB 17 YB 18 YB 27 YB 34 M T4 D YB 35 J1 + K 30 J M M C 81 66 -4 C K 30 -3 ’/5 68 A C H 293T * Figure HBZ transcripts are alternatively ... purposes) Retrovirology 2006, 3: 15 29 30 31 32 33 34 35 http://www.retrovirology.com/content /3/ 1/15 heterologous strain replication J Gen Virol 1997, 78 ( Pt 10):25 03- 2511 Kim JH, McLinden RJ,...

Ngày tải lên: 13/08/2014, 09:21

... n n holds, and ( 23) contradicts (a2) in Theorem 3. 6 This completes the proof that E cu is volume-expanding 3. 6 Proof of Theorem F As already said, let us assume Theorems 3. 6 and 3. 7 and show how ... Th 3. 8]) and (b) the nonwandering set outside a neighborhood of the singularities is hyperbolic (Lemma 4 .3) We next extend this splitting to all of Λ, obtaining Theorem C Theorems 3. 6 and 3. 7 are ... , ≤ i ≤ 3, are the eigenvalues of L at σ, then λi , ≤ i ≤ 3, are real and satisfy λ2 < 3 < < − 3 < λ1 [12] Inspired by this property we introduce the following definition Definition 1 .3 A singularity...

Ngày tải lên: 28/03/2014, 22:20

... Language of Functions and Graphs: An Examination Module for Secondary Schools, 1985, Shell Centre for Mathematical Education 1.2 What Are Functions? Basic Vocabulary and Notation 1.2 WHAT ARE FUNCTIONS? ... −→ f 1.2 What Are Functions? Basic Vocabulary and Notation A is a function with domain {1, 2, 3, 4, 5} and range {c, d, g, o, s}.4 B is a function with domain {1, 2, 3, 4, 5, 6} and range {c, ... of the following rules is a function? Are any of the functions 1-to-1? Q R −→ 13 −→ π −→ −→ π −→ 3 −→ 14 −→ π −→ 11 −→ π −→ 16 S −→ ±1 −→ ±8 Answer Q and R are both functions, since every number...

Ngày tải lên: 05/07/2014, 18:20

... C: 45, 32 , 12, 23, 26, 27, 39 1401, 138 8, 136 8, 137 9, 138 2, 138 3, 139 5 225, 160, 50, 115, 130 , 135 , 195 Ask at least two of your friends to, by inspection, identify the most variable and least ... Eq (3. 1) to this situation (Figure 3. 1a) we must have PrfAjBg ¼ Area common to A and B Area of B (3: 4) and if we divide numerator and denominator by the area of S, the right hand side of Eq (3. 4) ... eÀlt (3: 39) and before going on, I ask that you compare this equation with the first line of Eq (3. 33) Are these two descriptions inconsistent with each other? The answer is no From Eq (3. 39) the...

Ngày tải lên: 06/07/2014, 13:20

... Figure 3. 3 Properties of Continuous Functions Arithmetic If u(x) and v(x) are continuous at x = ξ then u(x) ± v(x) and u(x)v(x) are continuous at x = ξ is continuous at x = ξ if v(ξ) = u(x) v(x) Function ... Figure 3. 7 Example 3. 5 .3 Consider y = x3 and the point x = The function is differentiable The derivative, y = 3x2 is positive for x < and positive for < x Since y is not identically zero and the ... xx = e0 = x→0 80 3. 8 3. 8.1 Exercises Limits of Functions Exercise 3. 1 Does x lim sin x→0 exist? Hint, Solution Exercise 3. 2 Does lim x sin x→0 x exist? Hint, Solution Exercise 3. 3 Evaluate the...

Ngày tải lên: 06/08/2014, 01:21

... Consider two functions, f (x, y) and g(x, y) They are said to be functionally dependent if there is a an h(g) such that f (x, y) = h(g(x, y)) f and g will be functionally dependent if and only if ... a primitive in functions of several variables Consider a function f (x) F (x) is an integral of f (x) if and only if dF = f dx Now we move to functions of x and y Let P (x, y) and Q(x, y) be ... y), and find the analytic function f (z) = u(x, y) + ıv(x, y) as a function of z x3 − 3xy − 2xy + y ex sinh y 454 ex (sin x cos y cosh y − cos x sin y sinh y) Exercise 9 .3 For an analytic function, ...

Ngày tải lên: 06/08/2014, 01:21

... cotangent of the angle x/y (See Figure 14 .3. ) Example 14 .3. 3 Consider the one-parameter family of functions: y(x) = f (x) + cg(x), where f (x) and g(x) are known functions The derivative is y = f + ... homogeneous if it has no terms that are functions of the independent variable alone Thus an inhomogeneous equation is one in which there are terms that are functions of the independent variables ... eα(t−t0 ) 776 14 .3 One Parameter Families of Functions Consider the equation: F (x, y(x), c) = 0, (14.1) which implicitly defines a one-parameter family of functions y(x; c) Here y is a function of...

Ngày tải lên: 06/08/2014, 01:21

... = (u2 + 2u − 3) dx + x(u − 2) du = dx u−2 + du = x (u + 3) (u − 1) dx 5/4 1/4 + − du = x u +3 u−1 ln(x) + ln(u + 3) − ln(u − 1) = c 4 x4 (u + 3) 5 =c u−1 x4 (y/x + 3) 5 =c y/x − (y + 3x)5 =c y−x Solution ... +9 We factor the denominator of the fraction to see that z = 3 and z = − 3 are regular singular points dw + w=0 dz (z − 3) (z + 3) We make the transformation z = 1/ζ to examine the point at ... solution of the differential equation is defined on the interval (−1.67 835 0.768 039 ) 836 -5 -4 -3 -2 -1 -1 -2 -3 Figure 14.11: The function 2(1 − x) ex −1 Solution 14.11 We consider the differential...

Ngày tải lên: 06/08/2014, 01:21

... 3 −2 −5 − λ −4 = − 3 + 3 2 − 3 + = −(λ − 1 )3 −4 3 λ λ = is an eigenvalue of multiplicity The rank of the null space of A − I is (The second and third rows are multiples of the first.)   3 ... xi1 and xi2 Thus we can take the first two components of η to be zero        3 −2  −6 −4   = c1 0 + c2   3 3 −4 −2 3 = c1 , −4 3 = 2c2 , c1 = c2 , 888 2 3 = 2c1 − 3c2 c1 3 = ... canonical form of the matrix is J= 3 3 882 The solution of the initial value problem is x = eAt x0 x = eAt x0 = S eJt S−1 x0 = e−3t t e−3t e−3t 1/4 x= −4 3 + 4t −3t e + 4t Solution 15.9 We consider...

Ngày tải lên: 06/08/2014, 01:21

... determinant is 5x4 − 4x3 To check the theorem, d x x2 d ∆[A(x)] = dx dx x2 x4 2x x x2 = + 2x 4x3 x x = x4 − 2x3 + 4x4 − 2x3 = 5x4 − 4x3 16.4.2 The Wronskian of a Set of Functions A set of functions {y1 ... solutions, L[yk ] = If the coefficient functions and inhomogeneity are real-valued, then the general solution can be written in terms of real-valued functions 904 16 .3 Transformation to a First Order ... implies that the the set of functions is linearly dependent Result 16.4.2 The Wronskian of a set of functions vanishes identically over an interval if and only if the set of functions is linearly dependent...

Ngày tải lên: 06/08/2014, 01:21

... + 13 eλx = λ2 − 4λ + 13 = λ = ± 3i Thus two linearly independent solutions are e(2+3i)x , and 966 e(2−3i)x Noting that e(2+3i)x = e2x [cos(3x) + ı sin(3x)] e(2−3i)x = e2x [cos(3x) − ı sin(3x)], ... exp( x2 dx) to make this an exact equation d 3 ex /3 y = c1 ex /3 dx ex /3 ex y = c1 y = c1 e−x /3 ex /3 /3 945 dx + c2 dx + c2 e−x /3 Result 17 .3. 1 If you can write a differential equation in ... solutions are x2+ 3 , x2− 3 We can put this in a more understandable form x2+ 3 = x2 e 3 ln x = x2 cos (3 ln x) + x2 sin (3 ln x) We can write the general solution in terms of real-valued functions y...

Ngày tải lên: 06/08/2014, 01:21

... have a first order equation for x d 3 ey /3 x = y ey /3 dy x = e−y /3 y ey /3 dy + c e−y Example 18 .3. 2 Consider the equation y + 2x Interchanging the dependent and independent variables yields ... 18 .3. 1 Consider the equation y = y3 − xy 990 Instead of considering y to be a function of x, consider x to be a function of y That is, x = x(y), x = dy = dx y − xy dx = y − xy dy x + y2x = y3 ... 3 + = λ = 1, The general solution is then y = c1 x + c2 x2 974 Solution 17. 13 We note that xy + y + y = x is an Euler equation The substitution y = xλ yields 3 − 3 2 + 2λ + λ2 − λ + λ = λ3...

Ngày tải lên: 06/08/2014, 01:21

... for x(y) x − y 1/2 x = y 1/2 d 2y 3/ 2 2y 3/ 2 x exp − = y 1/2 exp − dy 3 2y 3/ 2 2y 3/ 2 x exp − = − exp − + c1 3 2y 3/ 2 x = −1 + c1 exp 3/ 2 x+1 2y = exp c1 x+1 = y 3/ 2 log c1 y= y= log c+ x+1 c1 log(x ... p(x) dx 3 1 p(x) dx y = u − pu + (p2 − 3p )u exp − 1 y = u − pu + (p2 − 3p )u + (9p − 9p − p3 )u exp − 27 p(x) dx yields the differential equation 1 u + (3q − 3p − p2 )u + (27r − 9pq − 9p + 2p3 )u ... Equations 1017 2 /3 2 /3 Chapter 19 Transformations and Canonical Forms Prize intensity more than extent Excellence resides in quality not in quantity The best is always few and rare abundance lowers...

Ngày tải lên: 06/08/2014, 01:21

... 1 = − cos(x) sin(3x) − sin(x) dx + sin(x) cos(3x) + cos(x) dx 2 1 1 = − cos(x) − cos(3x) + cos(x) + sin(x) sin(3x) + sin(x) 3 1 = sin2 (x) − cos2 (x) + cos(3x) cos(x) + sin(3x) sin(x) 1 = − cos(2x) ... x=ξ 0.1 0.1 -0.1 -0.2 -0 .3 0.5 -0.1 -0.2 -0 .3 0.1 0.5 0.1 0.5 -0.1 -0.2 -0 .3 -0.1 -0.2 -0 .3 Figure 21 .3: Plot of G(x|0.05),G(x|0.25),G(x|0.5) and G(x|0.75) Thus the Green function is (ξ − 1)x ξ(x ... the behavior of the ramp function, the Heaviside function, the delta function, and the derivative of the delta function We write the differential equation for the Green function G (x|ξ) + p(x)G...

Ngày tải lên: 06/08/2014, 01:21

... sin − 3) cos x e − cos − sin + (2e − cos − 3) sin x + (3 − cos − sin 1) ex Fredholm Alternative Theorem Orthogonality Two real vectors, u and v are orthogonal if u · v = Consider two functions, ... u y + u3 y = u1 y1 + u2 y2 + u3 y3 = To avoid some messy algebra when solving for uj , use Kramer’s rule Green Functions Hint 21.9 Hint 21.10 Hint 21.11 Hint 21.12 Hint 21. 13 cosh(x) and sinh(x ... b where c1 and c2 are arbitrary constants and a and b are any conveniently chosen points Using the result of part (a) show that the solution satisfying the initial conditions y(0) = and y (0)...

Ngày tải lên: 06/08/2014, 01:21

... y2 + u2 y2 + u3 y3 + u3 y3 + p2 (u1 y1 + u2 y2 + u3 y3 ) + p1 (u1 y1 + u2 y2 + u3 y3 ) + p0 (u1 y1 + u2 y2 + u3 y3 ) = f (x) u1 y1 + u2 y2 + u3 y3 + u1 L[y1 ] + u2 L[y2 ] + u3 L[y3 ] = f (x) u1 ... solution is yp = y1 (y2 y3 − y2 y3 )f (x) dx + y2 W (x) (y3 y1 − y3 y1 )f (x) dx + y3 W (x) (y1 y2 − y1 y2 )f (x) dx W (x) Green Functions Solution 21.9 We consider the Green function problem G = ... Kramer’s rule u1 = (y2 y3 − y2 y3 )f (x) , W (x) u2 = − (y1 y3 − y1 y3 )f (x) , W (x) u3 = (y1 y2 − y1 y2 )f (x) W (x) Here W (x) is the Wronskian of {y1 , y2 , y3 } Integrating the expressions for uj...

Ngày tải lên: 06/08/2014, 01:21

... functions is defined b φ1 |φ2 = φ1 (x)φ2 (x) dx a φ|φ The two functions are orthogonal if φ1 |φ2 = The L2 norm of a function is defined φ = Let {φ1 , φ2 , 3 , } be a set of complex valued functions ... factors The first few are φ1 = x − P0 (x) = P1 (x) = x 3x2 − P2 (x) = 5x − 3x P3 (x) = 35 x4 − 30 x2 + P4 (x) = Expanding cos(πx) in Legendre polynomials cos(πx) ≈ cn Pn (x), n=0 and calculating the ... m then the functions are orthogonal with respect to the weighting function σ(x) If the functions satisfy φn |σ|φm = δnm then the set is orthonormal with respect to σ(x) Example 25.8 .3 We know...

Ngày tải lên: 06/08/2014, 01:21

... ∂ξi + ∂x2 ∂ξi + ∂x3 ∂ξi The gradient, divergence, etc., follow a1 ∂u a2 ∂u a3 ∂u + + h1 ∂ξ1 h2 ∂ξ2 h3 ∂ 3 ∂ ∂ ∂ (h2 h3 v1 ) + (h3 h1 v2 ) + (h1 h2 v3 ) ∂ξ1 ∂ξ2 ∂ 3 h2 h3 ∂u ∂ h3 h1 ∂u ∂ h1 h2 ... [Iν , Kν ] from the value of W [Iν , I−ν ] Hint 34 .9 Hint 34 .10 1655 Hint 34 .11 Hint 34 .12 Hint 34 . 13 Hint 34 .14 1656 34 .10 Solutions Solution 34 .1 Bessel’s equation is L[y] ≡ z y + zy + z − ... (z)ın = eız (34 .3) (−1)n Jn (z)ın = e−ız (34 .4) J0 (z) + n=1 Next we substitute t = −ı into the generating function ∞ J0 (z) + n=1 1667 Dividing the sum of Equation 34 .3 and Equation 34 .4 by gives...

Ngày tải lên: 06/08/2014, 01:21