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... earth’s surface is sufficient, even though such a frame is noninertial to the extent of its taking part in the daily rotation of the earth about its axis and also its yearly rotation about the sun ... PROBLEM 2.25 2.14.2 Example of Spherical Shell Subjected to Internal Pressure 2.58 2.14.3 Example of Cylindrical Shell Subjected to Internal Pressure 2.58 2.14.4 Discontinuity Analysis 2.58 2.15 ... yearly rotation about the sun When more precise results are required, a set of axes fixed at the center of the earth may be used Such a set of axes is subject only to the orbital motion of the earth...

Ngày tải lên: 06/05/2015, 07:48

... basedon such a factorization A matrix can be partitioned by groupingthe elements of the matrix into submatrices Thesesubmatricescan then be treated as elements of a smaller matrix To ensure that ... more than 900 FORTRAN subroutines for use in general mathematics and statistical data analysis Also available in C and Java Visual Numerics, Inc., 1300W.Sam Houston Parkway S., Suite 150, Houston ... generally identified by a double subscripted lowercase letter, for example, ai,j, where the first subscript i identifies the row of the matrix and the secondsubscriptj identifies the column of...

Ngày tải lên: 02/04/2014, 15:44

... heavy sun By this we mean that, although the sun exerts a gravitational attraction on the planet, we regard the corresponding attraction of the planet on the sun as negligible, and that the sun ... φ(z)) ≤ sup exp(−K(x − a)) x∈[a,b] a x ≤ sup exp(−K(x − a)) x∈[a,b] f (s, y(s)) − f (s, z(s)) ds f (s, y(s)) − f (s, z(s)) ds a x ≤ L sup exp(−K(x − a)) x∈[a,b] y(s) − z(s) ds a x ≤ Lρ(y, z) sup ... k < 1, such that, for any η, ζ ∈ M , ρ(φ(η), φ(ζ)) ≤ kρ(η, ζ) Then there exists a unique ξ ∈ M such that φ(ξ) = ξ DIFFERENTIAL AND DIFFERENCE EQUATIONS 23 We can now state our main result Theorem...

Ngày tải lên: 13/02/2014, 20:20

... the same directional move of every common risk factor However, the problem with nonsubadditive risk measures such as VaR is that there happen to be cases in which the hedging benefit turns out ... of subadditivity violations of VaR are available in the literature [5, 8], although these may appear to be fictitious and unrealistic It may be surprising to learn, however, that examples of subadditivity ... COHERENT MEASURES OF RISK INTO EVERYDAY MARKET PRACTICE both standard normals, leading to the conclusion that it is never sufficient to study the marginals to ward off a VaR violation of subadditivity,...

Ngày tải lên: 07/03/2014, 15:20

... the same directional move of every common risk factor However, the problem with nonsubadditive risk measures such as VaR is that there happen to be cases in which the hedging benefit turns out ... of subadditivity violations of VaR are available in the literature [5, 8], although these may appear to be fictitious and unrealistic It may be surprising to learn, however, that examples of subadditivity ... COHERENT MEASURES OF RISK INTO EVERYDAY MARKET PRACTICE both standard normals, leading to the conclusion that it is never sufficient to study the marginals to ward off a VaR violation of subadditivity,...

Ngày tải lên: 18/03/2014, 00:20

... of calculations Out of such differences, I have usually managed to learn something of value in improving my subsequent work, either by accepting new ideas or by being reassured that what I was ... = 1, successive doubling will give some number q = 2k A starting point such that (q + 1) and q are represented identically By then setting r to successive integers 2, 3, , , a value such that ... non-orthogonality measure (3.17), it can be shown that no subsequent rotation can exchange them Suppose that the algorithm has proceeded so far that the non-orthogonality measure Z satisfies the...

Ngày tải lên: 31/03/2014, 15:02

... Contents Chapter Theoretical Results 1.0 Foreword 1.1 Introduction and preliminaries 1.2 Constitutive and momentum equations 13 1.3 A brief survey of theoretical results 15 1.4 Splitting the two-dimensional ... (see Remark 1.3.3), and therefore, we shall not study this case here 1.3 A brief survey of theoretical results The results presented here are for homogeneous boundary conditions The theory of the ... using (1.3.8) and (1.3.7), Section 1.3 Theoretical Results 17 Then by substituting this bound into the left-hand side of (1.3.11), and by substituting the estimate deduced from (1.3.6) to bound...

Ngày tải lên: 02/04/2014, 14:54

... heavy sun By this we mean that, although the sun exerts a gravitational attraction on the planet, we regard the corresponding attraction of the planet on the sun as negligible, and that the sun ... φ(z)) ≤ sup exp(−K(x − a)) x∈[a,b] a x ≤ sup exp(−K(x − a)) x∈[a,b] f (s, y(s)) − f (s, z(s)) ds f (s, y(s)) − f (s, z(s)) ds a x ≤ L sup exp(−K(x − a)) x∈[a,b] y(s) − z(s) ds a x ≤ Lρ(y, z) sup ... k < 1, such that, for any η, ζ ∈ M , ρ(φ(η), φ(ζ)) ≤ kρ(η, ζ) Then there exists a unique ξ ∈ M such that φ(ξ) = ξ DIFFERENTIAL AND DIFFERENCE EQUATIONS 23 We can now state our main result Theorem...

Ngày tải lên: 27/06/2014, 05:20

... heavy sun By this we mean that, although the sun exerts a gravitational attraction on the planet, we regard the corresponding attraction of the planet on the sun as negligible, and that the sun ... φ(z)) ≤ sup exp(−K(x − a)) x∈[a,b] a x ≤ sup exp(−K(x − a)) x∈[a,b] f (s, y(s)) − f (s, z(s)) ds f (s, y(s)) − f (s, z(s)) ds a x ≤ L sup exp(−K(x − a)) x∈[a,b] y(s) − z(s) ds a x ≤ Lρ(y, z) sup ... k < 1, such that, for any η, ζ ∈ M , ρ(φ(η), φ(ζ)) ≤ kρ(η, ζ) Then there exists a unique ξ ∈ M such that φ(ξ) = ξ DIFFERENTIAL AND DIFFERENCE EQUATIONS 23 We can now state our main result Theorem...

Ngày tải lên: 27/06/2014, 08:20

... heavy sun By this we mean that, although the sun exerts a gravitational attraction on the planet, we regard the corresponding attraction of the planet on the sun as negligible, and that the sun ... φ(z)) ≤ sup exp(−K(x − a)) x∈[a,b] a x ≤ sup exp(−K(x − a)) x∈[a,b] f (s, y(s)) − f (s, z(s)) ds f (s, y(s)) − f (s, z(s)) ds a x ≤ L sup exp(−K(x − a)) x∈[a,b] y(s) − z(s) ds a x ≤ Lρ(y, z) sup ... k < 1, such that, for any η, ζ ∈ M , ρ(φ(η), φ(ζ)) ≤ kρ(η, ζ) Then there exists a unique ξ ∈ M such that φ(ξ) = ξ DIFFERENTIAL AND DIFFERENCE EQUATIONS 23 We can now state our main result Theorem...

Ngày tải lên: 27/06/2014, 18:20

... of closed convex nonempty subsets of V with Kh a Vh, V h (in general, we not assume Kh a K) such that {Kh}h satisfies the following two conditions: (i) If {vh}h is such that vhe Kh,V h and {vH}h ... P^x - y)C(y) dy, (2.45) then 5B e @(R2), Supp(Cn) < Supp(y) + Supp(pJ, = lim = v strongly in (2.46) H^R ) n~* oo Hence from (2.41) and (2.46), we have Supp( | »„ |) < 12 for n large enough = (2.47) ... {Vh}h of closed subspaces of V (in practice, we will take Vh to be finite dimensional and h to vary over a sequence), we suppose that {Vh}h satisfies: (i) There exists U c V such that U = V,...

Ngày tải lên: 27/06/2014, 18:20

... heavy sun By this we mean that, although the sun exerts a gravitational attraction on the planet, we regard the corresponding attraction of the planet on the sun as negligible, and that the sun ... to differentiate (103j) and substitute from (103f) and (103g) to give the result y1 y3 + y2 y4 = (103o) The second step is to differentiate (103o) and to make various substitutions from (103f)–(103i) ... predictor–corrector methods Order and Stability Barriers 440 Survey of barrier results 441 Maximum order for a convergent k-step method 442 Order...

Ngày tải lên: 13/08/2014, 05:21

... φ(z)) ≤ sup exp(−K(x − a)) x∈[a,b] a x ≤ sup exp(−K(x − a)) x∈[a,b] f (s, y(s)) − f (s, z(s)) ds f (s, y(s)) − f (s, z(s)) ds a x ≤ L sup exp(−K(x − a)) x∈[a,b] y(s) − z(s) ds a x ≤ Lρ(y, z) sup ... k < 1, such that, for any η, ζ ∈ M , ρ(φ(η), φ(ζ)) ≤ kρ(η, ζ) Then there exists a unique ξ ∈ M such that φ(ξ) = ξ DIFFERENTIAL AND DIFFERENCE EQUATIONS 23 We can now state our main result Theorem ... examples of stiff problems, in Subsections 102 and 104, and of course there are many more such examples The concept of the ‘one-sided Lipschitz condition’ was mentioned in Subsection 110 without any...

Ngày tải lên: 13/08/2014, 05:21

... stepsize, because of its evident success in Subsection 202 218 Rounding error The mathematical analysis of the behaviour of a numerical method, such as the Euler method, is usually idealized to exact ... results is difficult, because we are no longer able to take limits, as stepsizes tend to zero Hence, our attention will move towards qualitative questions, such as whether or not a computed result ... result computed by the Euler method and a result computed, at least for the current step, using a method which has a higher order As we will see in Section 22, there are many ways in which such...

Ngày tải lên: 13/08/2014, 05:21

... xn−1 and yn−1 substituted for x and y, respectively The results for these experiments are shown in Figure 223(i) In each case the error is plotted, where we note that the exact result is exp( ... b3 where a21 = c2 and a31 + a32 = c3 The conditions for order 3, taken from results that will be summarized in Subsection 234, are b1 + b2 + b3 = 1, b2 c2 + b3 c3 = , b2 c2 + b3 c2 = , 3 b3 ... for the fourth order method (237c), the results were approximately six decimal places better This suggests that, even though the cost of evaluating the result in each step of an implicit method...

Ngày tải lên: 13/08/2014, 05:21

... two-step result and the double-step result as an error estimator 27.3 Denote the starting point in Exercise 27.2 as x−1 so that the results are computed at x0 = x−1 + h and x1 = x0 + h Find a suitable ... (|t|)(y(x)) dx ∗ is the sum of F (|u|)(y(x)) over all u ∈ TS such that the subtree formed by removing s from the set of vertices is t Proof If S = {s0 , s}, then the result is equivalent to d ... rooted trees tu and ut occur, where r(t) = r(u), we need to subtract half the number of such trees unless t = u, in which case no subtraction is performed 146 NUMERICAL METHODS FOR ORDINARY...

Ngày tải lên: 13/08/2014, 05:21

... We discuss in Subsection 325 the construction of fifth order methods These usually satisfy the so-called D(1) condition, which we introduce in Subsection 321 This simplifying assumption has, as ... integer such that 2q + ≥ p and suppose that for any i ∈ S ⊂ {1, 2, , s − 1}, the method has internal order q If there exists b ∈ Rs , with bs = such that s bi = 1, (323g) i=1 and such that ... O(hn+1 ), σ(t) i = 1, 2, , s [n] Calculate Yi using (313c) and the preliminary result follows Assume that h is su ciently small to guarantee convergence of the sequence [0] [1] [2] (Yi , Yi ,...

Ngày tải lên: 13/08/2014, 05:21

... with s = p = such that c = [0, , 1] 32.3 Find a method with s = p = such that b1 = and c2 = 32.4 Find a method with s = p = such that b2 = and c2 = 32.5 Find a method with s = p = such that b1 ... have at least nine stages It is possible to construct such a method using the principles of Subsection 323, extending the approach used in Subsection 326 The abscissa vector is chosen as c = [0 ... Richardson extrapolation as in Subsection 331, is it feasible to modify the method so that a proportion of the estimated error (331c) is subtracted from the result yn ? 33.2 Find a problem for...

Ngày tải lên: 13/08/2014, 05:21

... cases with m − l > suggest that these members of the Pad´ table are not A-stable e For the third sub-diagonal, this was proved by Ehle (1969), and for the fourth and fifth sub-diagonals by Nørsett ... an up arrow for which |z| is su ciently large to ensure that it is not possible that z is a pole or that z is real with (d/dz)(R(z) exp(−z)) = In this case we can assume without loss of generality ... consequences of assuming a method is irreducible This result was published in Dahlquist and Jeltsch (1979): Corollary 356D Under the same conditions of Theorem 356C, with the additional assumption that...

Ngày tải lên: 13/08/2014, 05:21

... assumption that the method is irreducible Suppose it were possible that two stages, Yi and Yj , say, give identical results for any Lipschitz continuous differential equation, provided h > is su ciently ... is possible to construct β such that (αβ)(t) = or (βα)(t) = 0, for all t ∈ T Because (αβ)(τ ) = (βα)(τ ) = α(τ ) + β(τ ), the result is clear for order Suppose the result has been proved for all ... corresponding term in (384a) is zero Hence, we sum only over I defined as the subset of I from which such i are omitted For any i ∈ I , define R S = (V, E) such that all the vertices associated with...

Ngày tải lên: 13/08/2014, 05:21