Generalized Quasilinearization versus Newton s Method for Convex-Concave Functions

Μέγεθος: px
Εμφάνιση ξεκινά από τη σελίδα:

Download "Generalized Quasilinearization versus Newton s Method for Convex-Concave Functions"

Transcript

1 Journal of Mathematics Research Vol., No. 3; August 010 Generalized Quasilinearization versus Newton s Method for Convex-Concave Functions Cesar Martínez-Garza (Corresponding author) Division of Science The Pennsylvania State University - Berks Campus Tulpehocken Rd., Reading, PA 19610, USA Tel: cxm58@psu.edu Abstract In this paper we use the Method of Generalized Quasilinearization to obtain monotone Newton-like comparative schemes to solve the equation F(x) = 0, where F(x) C[Ω, R], Ω R. Here, F(x) admits the decomposition F(x) = f (x) + g(x), where f (x) and g(x) are convex and concave functions, in C[Ω, R], respectively. The monotone sequences of iterates are shown to converge quadratically. Four cases are explored in this manuscript. Keywords: Newton s Method, Generalized Quasilinearization, Monotone Iterative Technique, Convex Functions 1. Introduction Lakshmikantham and Vatsala (004) showed that the method of Generalized Quasilinearization can be successfully employed to solve the equation 0 = f (x), (1) where f C[R, R], by comparing it to the initial value problem (IVP) x = f (x), x(0) = x 0, () on [0, T]. By exploiting the properties of the method of Generalized Quasilinearization (Lakshmikantham and Vatsala, 1998) they constructed a comparable framework to Newton s method to solve for an isolated zero, x 0 of (1). Newton s method simplicity and rapid convergence renders it as a first choice to solve (1), but its limitations have given rise to a broad body of work that continues to this day (Bellman and Kalaba,1965; Kelley, 1995; Ortega and Rheinboldt (1970); Potra (1987)). The generation of Newton-like schemes for the solution to (1) from the method of Generalized Quasilinearization is a logical choice, as Quasilinearization follows a similar methodology to Newton s method albeit for a seemingly different purpose (Bellman and Kalaba,1965). The main iteration formula for Newton s Method is x n+1 = x n f (x n), n 0, (3) f x (x n ) which requires that f x (x) be continuous and also nonzero in a neighborhood of a simple root of (1). This in turn requires knowledge of the monotone character of f (x). The similarities between Newton s method and standard Quasilinearization are enhanced when expressing (3) in the form 0 = f (x n ) + f x (x n )(x n+1 x n ), (4) and comparing it with the iterative scheme afforded by Quasilinearization to approximate (): x n+1 = f (x n) + f x (x n )(x n+1 x n ), x n+1 (0) = x 0. (5) The similarities between the method of Generalized Quasilinearization and Newton s method come to light when we look at the Newton-Fourier method as is shown next: let f be defined as in (1), and let D = [α 0,β 0 ] be a neighborhood of the simple zero of (1). If 0 < f (α 0 ), 0 < f (β 0 ), f x (x) > 0, and f xx (x) > 0, meaning that f is monotone increasing in D, then there exist monotone sequences {α n } and {β n } that converge quadratically to the simple zero of (1) in D. These sequences are generated by the iterative scheme 0 = f (β n ) + f x (β n )(β n+1 β n ), (6) 0 = f (α n ) + f x (β n )(α n+1 α n ), (7) (Lakshmikantham, V. 1998). Although two sequences are required, the Newton-Fourier method is not more expensive than the traditional Newton s method since the derivative computations are the same for both iterates (Potra, 1987). Comparatively, consider the IVP () with the following conditions: f x (x) 0, f xx (x) 0, α 0 f (β 0), β 0 f (α 0), Published by Canadian Center of Science and Education 63

2 Journal of Mathematics Research Vol., No. 3; August 010 where α 0 (t) β 0 (t), t [0, T], and α 0 (t), β 0 (t) are the upper and lower solutions of (), see Ladde, Lakshmikantham, and Vatsala (1985) and Lakshmikantham & Vatsala (1998, 005). The corresponding iterative scheme using Generalized Quasilinearization is: α n+1 = f (β n) + f x (β n )(α n+1 α n ), α n+1 (0) = x 0, (8) β n+1 = f (α n) + f x (β n )(β n+1 β n ), β n+1 (0) = x 0, (9) where α 0 x β 0 on [0, T]. These schemes give rise to sequences of iterates that converge uniformly and quadratically to x(t), the unique solution of () on [0,T]. For complete details refer to Lakshmikantham and Vatsala (1998). If one considers the case where f is monotone decreasing, then the Newton-Fourier method is no longer applicable, while the solution of a comparable IVP with f monotone decreasing can be approximated with Generalized Quasilinearization. Below we present four cases where iterative schemes are developed to approximate (1) where f can be decomposed into a convex and a concave function. The corresponding Generalized Quasilinearization theorems and proofs can be found in Lakshmikantham and Vatsala (1998, section 1.3).. Statement of the problem Consider the scalar equation 0 = F(x) = f (x) + g(x), (10) where, f, g C[Ω, R], Ω R, and [α 0,β 0 ] = D Ω. Assume that (??) has a simple root x = r in D. Let f and g be convex and concave functions respectively such that, f x, g x, f xx, g xx exist, are continuous and satisfy f xx 0, g xx 0, x Ω. (11) The purpose of this paper is to produce iterative schemes that will generate two monotone sequences {α n } and {β n } that converge from both left and right to r. We will show that this convergence is quadratic in the four following cases. 3. Main Results Theorem 1. Assume that (10), and (11) hold true, and Then, there exist monotone sequences {α n } and {β n } such that 0 < f (α 0 ) + g(α 0 ), (1) 0 > f (β 0 ) + g(β 0 ), (13) f x (β 0 ) + g x (α 0 ) < 0, (14) α 0 x β 0. α 0 <α 1 < <α n < r <β n < <β 1 <β 0 (15) which converge quadratically to r. The sequences {α n } and {β n } are generated by the following iterative scheme: Proof. Let p = α 0 α 1 ; by (1) and (16) with n = 0, 0 = f (α n ) + f x (α n )(α n+1 α n ) + g(α n ) + g x (β n )(α n+1 α n ), (16) 0 = f (β n ) + f x (α n )(β n+1 β n ) + g(β n ) + g x (β n )(β n+1 β n ). (17) 0 < [ f (α 0 ) + g(α 0 )] [ f (α 0 ) + f x (α 0 )(α 1 α 0 ) + g(α 0 ) + g x (β 0 )(α 1 α 0 )] = [ f x (α 0 )(α 1 α 0 ) + g x (β 0 )(α 1 α 0 )] 0 < [ f x (α 0 ) + g x (β 0 )]p By condition (11) f x (α 0 ) < f x (β 0 ) and g x (β 0 ) < g x (α 0 ), so 0 < [ f x (α 0 ) + g x (β 0 )]p < [ f x (β 0 ) + g x (α 0 )]p. Then, condition (14) allows us to conclude that p < 0, thus α 0 <α 1. Now, show that β 0 >β 1 ; let p = β 1 β 0. From (13) and (17) with n = 0, we have 0 < [ f (β 0 ) + f x (α 0 )(β 1 β 0 ) + g(β 0 ) + g x (β 0 )(β 1 β 0 )] [ f (α 0 ) + g(β 0 )] = [ f x (α 0 )(β 1 β 0 ) + g x (β 0 )(β 1 β 0 )] 0 < [ f x (α 0 ) + g x (β 0 )]p 64 ISSN E-ISSN

3 Journal of Mathematics Research Vol., No. 3; August 010 Again, by (11) and (14) we conclude that p < 0, making β 1 <β 0. To show that α 1 <β 1, we combine (16) and (17), both with n = 0: 0 = [ f (α 0 ) + f x (α 0 )(α 1 α 0 ) + g(α 0 ) + g x (β 0 )(α 1 α 0 )] [ f (β 0 ) + f x (α 0 )(β 1 β 0 ) + g(β 0 ) + g x (β 0 )(β 1 β 0 )] = [ f (α 0 ) f (β 0 )] + [g(α 0 ) g(β 0 )] + [ f x (α 0 ) + g x (β 0 )][α 1 α 0 + β 0 β 1 ]. Using the Mean Value Theorem (MVT) with σ 1,σ (α 0,β 0 ) we next obtain, 0 = [ f x (σ 1 ) + g x (σ )][α 0 β 0 ] + [ f x (α 0 ) + g x (β 0 )][α 1 α 0 + β 0 β 1 ]. Now, (11) states that f x (x) and g x (x) are strictly increasing and strictly decreasing respectively. So, f x (α 0 ) < f x (σ 1 ) < f x (β 0 ) and g x (β 0 ) < g x (σ ) < g x (α 0 ). 0 < [ f x (α 0 ) + g x (β 0 )][α 0 β 0 ] + [ f x (α 0 ) + g x (β 0 )][α 1 α 0 + β 0 β 1 ] = [ f x (α 0 ) + g x (β 0 )]p < [ f x (β 0 ) + g x (α 0 )]p. Then, p < 0 by (14) and α 1 <β 1. The next step is to show that α 1 < r <β 1. Let p = α 1 r. The fact that f (r) + g(r) = 0 together with (16) (with n = 0), yields: 0 = [ f (α 0 ) + f x (α 0 )(α 1 α 0 ) + g(α 0 ) + g x (β 0 )(α 1 α 0 )] [ f (r) + g(r)] = [ f (α 0 ) f (r)] + [g(α 0 ) g(r)] + [ f x (α 0 ) + g x (β 0 )][α 1 α 0 ]. We again use the MVT with γ 1,γ (α 0, r), and (11) to obtain 0 = [ f x (γ 1 ) + g x (γ )][α 0 r] + [ f x (α 0 ) + g x (β 0 )][α 1 α 0 ] < [ f x (β 0 ) g x (α 0 )][α 0 r] + [ f x (α 0 ) + g x (β 0 )][α 1 α 0 ] < [ f x (β 0 ) g x (α 0 )][α 0 r] + [ f x (β 0 ) g x (α 0 )][α 1 α 0 ] = [ f x (β 0 ) + g x (α 0 )][α 0 r + α 1 α 0 ] < [ f x (β 0 ) + g x (α 0 )]p. Where we have again appealed to the strictly increasing and strictly decreasing natures of f x (x) and g x (x), respectively. Condition (14) leads to p < 0, so α 1 < r. By letting p = r β 1 one can easily show that r <β 1. Now that both the base and n = 1 cases have been proven, by induction we see that (15) holds true, where lim n α n = r = lim n β n. To show the quadratic convergence of the current scheme, we let p n+1 = r α n+1 > 0, and q n+1 = β n+1 r > 0. Consider the case p n+1 = r α n > 0. Combining (10) and (16) we have: 0 = [ f (r) + g(r)] [ f (α n ) + f x (α n )(α n+1 α n ) + g(α n ) + g x (β n )(α n+1 α n )] = [ f (r) f (α n )] + [g(r) g(α n )] + [ f x (α n ) + g x (β n )][α n α n+1 ] The Mean Value Theorem with η 1,η (α n, r) produces the next result where we let p n = r α n. We also observe that by (11) f x (η 1 ) < f x (r) and g x (η ) < g x (α n ). 0 = [ f x (η 1 ) + g x (η )][r α n ] + [ f x (α n ) + g x (β n )][α n r + r α n+1 ] < [ f x (r) + g x (α n )]p n + [ f x (α n ) + g x (β n )][p n+1 p n ] < [ f x (r) f x (α n )]p n [g x (β n ) g x (α n )]p n + [ f x (α n ) + g x (β n )]p n+1. Application of the MVT once more yields the results below with ζ 1 (α n, r), ζ (α n,β n ); we also let q n = β n r. 0 < f xx (ζ 1 )p n g xx (ζ )[β n r + r α n ]p n + [ f x (α n ) + g x (β n )]p n+1 = f xx (ζ 1 )p n g xx (ζ )[q n + p n ]p n + [ f x (α n ) + g x (β n )]p n+1. The concave-convex behavior of f and g in C[Ω, R] given by (11) along with condition (14) yield: 0 < f xx (ζ 1 ) p n + g xx (ζ ) [q n + p n ]p n f x (β 0 ) + g x (α 0 ) p n+1 < M 1 p n + M (q n + p n ) p n M 3 p n+1, (18) Published by Canadian Center of Science and Education 65

4 Journal of Mathematics Research Vol., No. 3; August 010 where Now, (18) yields, M 1 > f xx (x), M > g xx (x), M 3 < f x (β 0 ) + g x (α 0 ), for x D. (19) p n+1 < M 1 M 3 p n + M M 3 ( qn p n + p n ) = M 1 p n + M ( q n p n + p ) n ( < M 1 p 3 n + M p n + 1 ) q n, since (p n q n ) = p n p n q n + q n 0 which implies 3p n + q n p n + p n q n. We arrive at the desired result: p n+1 < (M 1 + 3M ) p n + M q n. (0) Similarly, for β n+1 = β n r > 0 we obtain the estimate ( 3M1 q n+1 < ) + M q n + M 1 p n, (1) where, M 1, M are defined in (19). The combination of the results (0) with (1) yields: p n+1 + q n+1 < [(M 1 + 3M ) ] [( ) p n + M q 3M1 n + + M q n + M 1 ( 3M1 = + 3M ) ( p 3M1 n + + 3M ) q n. p n ] Finally, p n+1 + q n+1 = r α n+1 + β n+1 r < K ( p n + q n), () where, K = 3 (M 1 + M ). This completes the proof of the theorem. If F(x) is a decreasing function with an isolated zero x = r D, one can modify Theorem 1 accordingly as follows. Theorem. Assume that (10), and (11) hold true, and 0 > f (α 0 ) + g(α 0 ), (3) 0 < f (β 0 ) + g(β 0 ), (4) f x (α 0 ) + g x (β 0 ) > 0, (5) α 0 x β 0. Then, there exist monotone sequences {α n } and {β n } which satisfy (15), and which converge quadratically to r. The sequences {α n } and {β n } are generated by the following iterative scheme: 0 = f (α n ) + f x (β n )(α n+1 α n ) + g(α n ) + g x (α n )(α n+1 α n ), (6) 0 = f (β n ) + f x (β n )(β n+1 β n ) + g(β n ) + g x (α n )(β n+1 β n ). (7) The mechanics of the proof of Theorem are the same as those of Theorem 1. Here we present the highlights of the proof of Theorem. Letting p = α 0 α 1 and combining (3) with (6) with n=0 yields the same result given for p = β 1 β 0 while combining (4) with (7) with n=0: 0 > [ f x (β 0 ) + g x (α 0 )]p > [ f x (α 0 ) + g x (β 0 )]p, which by (11) and (5) indicates that p < 0 on both cases, thus α 0 <α 1 and β 1 <β 0. Likewise, when showing that α 1 < r <β 1 we obtain 0 > [ f x (β 0 ) + g x (α 0 )]p > [ f x (α 0 ) + g x (β 0 )]p, where we first let p = α 1 r followed by p = r β 1. So, the n-case becomes 0 > [ f x (β n ) + g x (α n )]p > [ f x (α 0 ) + g x (β 0 )]p, which establishes that α n < r <β n, from which (15) follows. Quadratic convergence follows in the same fashion as Theorem ISSN E-ISSN

5 Journal of Mathematics Research Vol., No. 3; August 010 Theorem 3. Assume that (10), and (11) hold true, and 0 < f (β 0 ) + g(α 0 ), (8) 0 > f (α 0 ) + g(β 0 ), (9) f x (β 0 ) < 0, (30) f x (β 0 ) + g x (α 0 ) < 0, (31) α 0 x β 0. Then, there exist monotone sequences {α n } and {β n } which satisfy (15), and which converge quadratically to r, the isolated zero of (10). The sequences {α n } and {β n } are generated by the following iterative scheme: 0 = f (β n ) + f x (β n )(α n+1 α n ) + g(α n ) + g x (β n )(α n+1 α n ), (3) 0 = f (α n ) + f x (β n )(β n+1 β n ) + g(β n ) + g x (β n )(β n+1 β n ). (33) The proof follows the same strategy as the previous proofs, but with some variation in the results which we address below. Proof. Begin by subtracting (3) with n = 0 from (8) 0 < f (β 0 ) + g(α 0 ) [ f (β 0 ) + f x (β 0 )(α 1 α 0 ) + g(α 0 ) + g x (β 0 )(α 1 α 0 )] = f x (β 0 )(α 0 α 1 ) + g x (β 0 )(α 0 α 1 ) < [ f x (β 0 ) + g x (α 0 )](α 0 α 1 ) By (31) we conclude that α 0 <α 1. Next, subtract (33) from (9). 0 < f (α 0 ) + f x (β 0 )(β 1 β 0 ) + g(β 0 ) + g x (β 0 )(β 1 β 0 ) [ f (α 0 ) + g(β 0 )] = f x (β 0 )(β 1 β 0 ) + g x (β 0 )(β 1 β 0 ) < [ f x (β 0 ) + g x (α 0 )](β 1 β 0 ) Again, by (31) we conclude that β 1 <β 0 Now, subtract (33) with n=0 from (3) with n=0: 0 = f (β 0 ) + f x (β 0 )(α 1 α 0 ) + g(α 0 ) + g x (β 0 )(α 1 α 0 ) [ f (α 0 ) + f x (β 0 )(β 1 β 0 ) + g(β 0 ) + g x (β 0 )(β 1 β 0 )] = [ f (β 0 ) f (α 0 )] [g(β 0 ) g(α 0 )] + f x (β 0 )(α 1 α 0 β 1 + β 0 ) + g x (β 0 )(α 1 α 0 β 1 + β 0 ) = f x (σ)(β 0 α 0 ) g x (η)(β 0 α 0 ) + f x (β 0 )(α 1 α 0 β 1 + β 0 ) + g x (β 0 )(α 1 α 0 β 1 + β 0 ) < f x (β 0 )(β 0 α 0 ) g x (β 0 )(β 0 α 0 ) + f x (β 0 )(α 1 α 0 β 1 + β 0 ) + g x (β 0 )(α 1 α 0 β 1 + β 0 ) = f x (β 0 )(β 0 α 0 + α 1 α 0 β 1 + β 0 ) + g x (β 0 )(β 0 α 0 + α 1 α 0 β 1 + β 0 ) = f x (β 0 )(β 0 α 0 ) + f x (β 0 )(α 1 β 1 ) + g x (β 0 )(α 1 β 1 ) < f x (β 0 )(β 0 α 0 ) + f x (β 0 )(α 1 β 1 ) + g x (α 0 )(α 1 β 1 ) By (30), f x (β 0 ) < 0 and since β 0 α 0 > 0, we obtain 0 < f x (β 0 )(β 0 α 0 ) + f x (β 0 )(α 1 β 1 ) + g x (α 0 )(α 1 β 1 ) < ( f x (β 0 ) + g x (α 0 ))(α 1 β 1 ) Which establishes that α 1 <β 1 by virtue of (31). We now establish that α 1 < r <β 1. Subtract (10) with x=r from (3) with n=0: Published by Canadian Center of Science and Education 67

6 Journal of Mathematics Research Vol., No. 3; August = f (β 0 ) + f x (β 0 )(α 1 α 0 ) + g(α 0 ) + g x (β 0 )(α 1 α 0 ) [ f (r) + g(r)] = [ f (β 0 ) f (r)] [g(r) g(α 0 )] + f x (β 0 )(α 1 α 0 ) + g x (β 0 )(α 1 α 0 ) = f x (σ)(β 0 r) g x (η)(r α 0 ) + f x (β 0 )(α 1 α 0 ) + g x (β 0 )(α 1 α 0 ) < f x (β 0 )(β 0 r) + g x (α 0 )(α 0 r) + f x (β 0 )(α 1 α 0 ) + g x (β 0 )(α 1 α 0 ) < f x (β 0 )(β 0 r + α 1 α 0 ) + g x (α 0 )(α 0 r + α 1 α 0 ) = f x (β 0 )(β 0 α 0 ) + ( f x (β 0 ) + g x (α 0 ))(α 1 r) 0 < ( f x (β 0 ) + g x (α 0 ))(α 1 r) Where, r <σ<β 0, and α 0 <η<r by the MVT. Condition (11) implies that f x (β 0 ) > f x (σ) and g x (β 0 ) < g x (η) < g x (α 0 ). We have again appealed to (30) and finally to (31), so we conclude that α 1 r < 0. Similarly, subtract (3) with n=0 from (10) with x=r. 0 = f (r) + g(r) [ f (α 0 ) + f x (β 0 )(β 1 β 0 ) + g(β 0 ) + g x (β 0 )(β 1 β 0 )] = [ f (r) f (α 0 )] [g(β 0 ) g(r)] f x (β 0 )(β 1 β 0 ) g x (β 0 )(β 1 β 0 ) = f x (σ)(r α 0 ) g x (η)(β 0 r) + f x (β 0 )(β 0 β 1 ) + g x (β 0 )(β 0 β 1 ) < f x (β 0 )(r α 0 ) + g x (α 0 )(r β 0 ) + f x (β 0 )(β 0 β 1 ) + g x (α 0 )(β 0 β 1 ) = f x (β 0 )(β 0 α 0 ) + ( f x (β 0 ) + g x (α 0 ))(r β 1 ) < ( f x (β 0 ) + g x (α 0 ))(r β 1 ) So, we now conclude that α 0 <α 1 < r <β 1 <β 0 ; hence, by induction we obtain (15). Quadratic convergence now follows. Let p n+1 = r α n+1 > 0, and q n+1 = β n+1 r > 0. Subtracting (3) from (10) with x=r we obtain the following inequalities. 0 = f (r) + g(r) [ f (β n ) + f x (β n )(α n+1 α n ) + g(α n ) + g x (β n )(α n+1 α n )] = [ f (β n ) f (r)] + [g(r) g(α n )] f x (β n )(α n+1 r + r α n ) g x (β n )(α n+1 r + r α n ) = f x (σ)(β n r) + g x (η)(r α n ) + f x (β n )(p n+1 p n ) + g x (β n )(p n+1 p n ) = f x (σ)(β n α n + α n r) + g x (η)p n + f x (β n )(p n+1 p n ) + g x (β n )(p n+1 p n ) = f x (σ)(β n α n ) + f x (σ)p n + g x (η)p n + f x (β n )(p n+1 p n ) + g x (β n )(p n+1 p n ) = f x (σ)(β n α n ) f x (σ)p n g x (η)p n + f x (β n )(p n p n+1 ) + g x (β n )(p n p n+1 ) < f x (β n )(β n α n ) f x (r)p n g x (r)p n + f x (β n )(p n p n+1 ) + g x (β n )(p n p n+1 ) Again, f x (β n ) < 0 and by the MVT, r <σ<β n and α n <η<r. By (11), f x (β 0 ) > f x (σ), f x (r) > f x (σ),and - g x (r) > g x (η). So, 0 < [ f x (β n ) f x (r)]p n + [g x (β n ) g x (r)]p n [ f x (β n ) + g x (β n )]p n+1 < [ f x (β n ) f x (r)]p n + [g x (β n ) g x (r)]p n [ f x (β 0 ) + g x (α 0 )]p n+1 = f xx (ζ)(β n r)p n + g xx (ξ)(β n r)p n [ f x (β 0 ) + g x (α 0 )]p n+1 = [ f xx (ζ) + g xx (ξ)]p n q n [ f x (β 0 ) + g x (α 0 )]p n+1 Since, p n p n q n + q n 0 it follows that 1 (p n + q n) p n q n. So, And, placing bounds on the derivatives as before, we obtain, ( p [ f x (β 0 ) + g x (α 0 )]p n+1 < [ f xx (ζ) + g xx (ξ)] n + q ) n [ M1 + M Np n+1 < [ M1 + M p n+1 < N ] ( ) p n + q n ) ] ( p n + q n 68 ISSN E-ISSN

7 Journal of Mathematics Research Vol., No. 3; August 010 Now, subtract (10) with x=r from (33). 0 = f (r) + g(r) [ f (α n ) + f x (β n )(β n+1 β n ) + g(β n ) + g x (β n )(β n+1 β n )] =[ f (r) f (α n )] [g(β n ) g(r)] [ f x (β n ) + g x (β n )](β n+1 r + r β n ) = f x (σ)(r β n + β n α n ) g x (η)q n [ f x (β n ) + g x (β n )](q n+1 q n ) = f x (σ)q n + f x (σ)(β n α n ) g x (η)q n [ f x (β n ) + g x (β n )](q n+1 q n ). Where we again used the MVT with α n < σ < r and r < η < β n. By (11) we have that f x (σ) < f x (α n ) and g x (η) < g x (β n ), using condition (30) we produce the following inequalities: 0 < f x (α n )q n + f x (β n )(β n α n ) g x (β n )q n [ f x (β n ) + g x (β n )](q n+1 q n ) < f x (α n )q n g x (β n )q n + g x (β n )q n + f x (β n )q n [ f x (β n ) + g x (β n )]q n+1 < [ f x (β n ) f x (α n )]q n [ f x (β n ) + g x (β n )]q n+1 = f xx (ζ)(β n α n )q n [ f x (β n ) + g x (β n )]q n+1 = f xx (ζ)(β n r + r α n )q n [ f x (β n ) + g x (β n )]q n+1 = f xx (ζ)(q n + p n )q n [ f x (β n ) + g x (β n )]q n+1 = f xx (ζ)(q n + p n q n ) [ f x (β n ) + g x (β n )]q n+1 ( 3q ) f xx (ζ) n + p n [ f x (β n ) + g x (β n )]q n+1. Here, we have used the fact that p n p n q n + q n 0 3q n + p n q n + p n q n. Now, place bounds on the derivatives. Combine results. Finally, [ M1 + M p n+1 + q n+1 < N ( 3q ) 0 < M n 1 + p n Nq n+1 q n+1 < M 1 N p n + 3M 1 N q n ] ( ) p n + q M 1 n + N p n + 3M 1 [ M1 N q n [ M1 = N + M 1 N + M ] p n + N N + 3M 1 N + M N [ ] [ ] M1 + M = p 4M1 + M n + q n N N ] q n where, K = max [ ] M 1 +M N, 4M 1+M N. p n+1 + q n+1 = r α n+1 + β n+1 r < K ( p n + q n), Theorem 4. Assume that (10), and (11) hold true, and 0 > f (β 0 ) + g(α 0 ), 0 < f (α 0 ) + g(β 0 ), f x (α 0 ) < 0, f x (α 0 ) + g x (β 0 ) > 0, α 0 x β 0. Published by Canadian Center of Science and Education 69

8 Journal of Mathematics Research Vol., No. 3; August 010 Then, there exist monotone sequences {α n } and {β n } which satisfy (15), and which converge quadratically to r, the isolated zero of (10). The sequences {α n } and {β n } are generated by the following iterative scheme: 0 = f (α n ) + f x (α n )(α n+1 α n ) + g(β n ) + g x (α n )(α n+1 α n ), (34) 0 = f (β n ) + f x (α n )(β n+1 β n ) + g(α n ) + g x (α n )(β n+1 β n ). (35) The proof of this theorem is not presented as it follows the patterns established in the proof of the previous theorem. The results above are found to be in agreement, when f is either convex or concave accordingly, with those presented in (Lakshmikantham, V. 1998). Acknowledgments The author wishes to express his utmost gratitude to Dr. V. Lakshmikantham at the Florida Institute of Technology for the original idea and his initial guidance, and to Dr. Ali Alikhani at Penn State Berks for his revisions and suggestions to this manuscript. Further recognition is rendered to the reviewer of the document whose comments were essential to bring this article to its final form. References Bellman, R.E. & Kalaba, R.E. (1965). Quasilinearization and nonlinear boundary value problems. New York: Elsevier. Kelley, C.T. (1995). Iterative methods for linear and nonlinear equations, Philadelphia: SIAM. Ladde, G.S., Lakshmikantham, V. & Vatsala, A.S. (1985). Monotone Iterative Technique for Nonlinear Differential Equations. Boston: Pitman. Lakshmikantham, V. & Vatsala, A.S. (1998). Generalized Quasilinearization for Nonlinear Problems. Dordrecht, The Netherlands: Kluwer Academic Publishers. Lakshmikantham, V. & Vatsala, A.S. (005). Generalized quasilinearization versus Newton s method. Applied Mathematics and Computation, 164, Ortega, J.M. & Rheinboldt, W.C. (1970). Iterative Solution of Nonlinear Equations in Several Variables. New York: Academic Press. Potra, F.A. (1987). On a monotone Newton-like method, Computing, 39, ISSN E-ISSN

2 Composition. Invertible Mappings

2 Composition. Invertible Mappings Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan Composition. Invertible Mappings In this section we discuss two procedures for creating new mappings from old ones, namely,

Διαβάστε περισσότερα

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS CHAPTER 5 SOLVING EQUATIONS BY ITERATIVE METHODS EXERCISE 104 Page 8 1. Find the positive root of the equation x + 3x 5 = 0, correct to 3 significant figures, using the method of bisection. Let f(x) =

Διαβάστε περισσότερα

C.S. 430 Assignment 6, Sample Solutions

C.S. 430 Assignment 6, Sample Solutions C.S. 430 Assignment 6, Sample Solutions Paul Liu November 15, 2007 Note that these are sample solutions only; in many cases there were many acceptable answers. 1 Reynolds Problem 10.1 1.1 Normal-order

Διαβάστε περισσότερα

Example Sheet 3 Solutions

Example Sheet 3 Solutions Example Sheet 3 Solutions. i Regular Sturm-Liouville. ii Singular Sturm-Liouville mixed boundary conditions. iii Not Sturm-Liouville ODE is not in Sturm-Liouville form. iv Regular Sturm-Liouville note

Διαβάστε περισσότερα

Concrete Mathematics Exercises from 30 September 2016

Concrete Mathematics Exercises from 30 September 2016 Concrete Mathematics Exercises from 30 September 2016 Silvio Capobianco Exercise 1.7 Let H(n) = J(n + 1) J(n). Equation (1.8) tells us that H(2n) = 2, and H(2n+1) = J(2n+2) J(2n+1) = (2J(n+1) 1) (2J(n)+1)

Διαβάστε περισσότερα

SCITECH Volume 13, Issue 2 RESEARCH ORGANISATION Published online: March 29, 2018

SCITECH Volume 13, Issue 2 RESEARCH ORGANISATION Published online: March 29, 2018 Journal of rogressive Research in Mathematics(JRM) ISSN: 2395-028 SCITECH Volume 3, Issue 2 RESEARCH ORGANISATION ublished online: March 29, 208 Journal of rogressive Research in Mathematics www.scitecresearch.com/journals

Διαβάστε περισσότερα

EE512: Error Control Coding

EE512: Error Control Coding EE512: Error Control Coding Solution for Assignment on Finite Fields February 16, 2007 1. (a) Addition and Multiplication tables for GF (5) and GF (7) are shown in Tables 1 and 2. + 0 1 2 3 4 0 0 1 2 3

Διαβάστε περισσότερα

Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)

Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required) Phys460.nb 81 ψ n (t) is still the (same) eigenstate of H But for tdependent H. The answer is NO. 5.5.5. Solution for the tdependent Schrodinger s equation If we assume that at time t 0, the electron starts

Διαβάστε περισσότερα

Homework 3 Solutions

Homework 3 Solutions Homework 3 Solutions Igor Yanovsky (Math 151A TA) Problem 1: Compute the absolute error and relative error in approximations of p by p. (Use calculator!) a) p π, p 22/7; b) p π, p 3.141. Solution: For

Διαβάστε περισσότερα

Second Order Partial Differential Equations

Second Order Partial Differential Equations Chapter 7 Second Order Partial Differential Equations 7.1 Introduction A second order linear PDE in two independent variables (x, y Ω can be written as A(x, y u x + B(x, y u xy + C(x, y u u u + D(x, y

Διαβάστε περισσότερα

4.6 Autoregressive Moving Average Model ARMA(1,1)

4.6 Autoregressive Moving Average Model ARMA(1,1) 84 CHAPTER 4. STATIONARY TS MODELS 4.6 Autoregressive Moving Average Model ARMA(,) This section is an introduction to a wide class of models ARMA(p,q) which we will consider in more detail later in this

Διαβάστε περισσότερα

HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:

HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch: HOMEWORK 4 Problem a For the fast loading case, we want to derive the relationship between P zz and λ z. We know that the nominal stress is expressed as: P zz = ψ λ z where λ z = λ λ z. Therefore, applying

Διαβάστε περισσότερα

Section 8.3 Trigonometric Equations

Section 8.3 Trigonometric Equations 99 Section 8. Trigonometric Equations Objective 1: Solve Equations Involving One Trigonometric Function. In this section and the next, we will exple how to solving equations involving trigonometric functions.

Διαβάστε περισσότερα

3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β

3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β 3.4 SUM AND DIFFERENCE FORMULAS Page Theorem cos(αβ cos α cos β -sin α cos(α-β cos α cos β sin α NOTE: cos(αβ cos α cos β cos(α-β cos α -cos β Proof of cos(α-β cos α cos β sin α Let s use a unit circle

Διαβάστε περισσότερα

Every set of first-order formulas is equivalent to an independent set

Every set of first-order formulas is equivalent to an independent set Every set of first-order formulas is equivalent to an independent set May 6, 2008 Abstract A set of first-order formulas, whatever the cardinality of the set of symbols, is equivalent to an independent

Διαβάστε περισσότερα

Tridiagonal matrices. Gérard MEURANT. October, 2008

Tridiagonal matrices. Gérard MEURANT. October, 2008 Tridiagonal matrices Gérard MEURANT October, 2008 1 Similarity 2 Cholesy factorizations 3 Eigenvalues 4 Inverse Similarity Let α 1 ω 1 β 1 α 2 ω 2 T =......... β 2 α 1 ω 1 β 1 α and β i ω i, i = 1,...,

Διαβάστε περισσότερα

Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit

Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit Ting Zhang Stanford May 11, 2001 Stanford, 5/11/2001 1 Outline Ordinal Classification Ordinal Addition Ordinal Multiplication Ordinal

Διαβάστε περισσότερα

6.3 Forecasting ARMA processes

6.3 Forecasting ARMA processes 122 CHAPTER 6. ARMA MODELS 6.3 Forecasting ARMA processes The purpose of forecasting is to predict future values of a TS based on the data collected to the present. In this section we will discuss a linear

Διαβάστε περισσότερα

Math 446 Homework 3 Solutions. (1). (i): Reverse triangle inequality for metrics: Let (X, d) be a metric space and let x, y, z X.

Math 446 Homework 3 Solutions. (1). (i): Reverse triangle inequality for metrics: Let (X, d) be a metric space and let x, y, z X. Math 446 Homework 3 Solutions. (1). (i): Reverse triangle inequalit for metrics: Let (X, d) be a metric space and let x,, z X. Prove that d(x, z) d(, z) d(x, ). (ii): Reverse triangle inequalit for norms:

Διαβάστε περισσότερα

Uniform Convergence of Fourier Series Michael Taylor

Uniform Convergence of Fourier Series Michael Taylor Uniform Convergence of Fourier Series Michael Taylor Given f L 1 T 1 ), we consider the partial sums of the Fourier series of f: N 1) S N fθ) = ˆfk)e ikθ. k= N A calculation gives the Dirichlet formula

Διαβάστε περισσότερα

D Alembert s Solution to the Wave Equation

D Alembert s Solution to the Wave Equation D Alembert s Solution to the Wave Equation MATH 467 Partial Differential Equations J. Robert Buchanan Department of Mathematics Fall 2018 Objectives In this lesson we will learn: a change of variable technique

Διαβάστε περισσότερα

Coefficient Inequalities for a New Subclass of K-uniformly Convex Functions

Coefficient Inequalities for a New Subclass of K-uniformly Convex Functions International Journal of Computational Science and Mathematics. ISSN 0974-89 Volume, Number (00), pp. 67--75 International Research Publication House http://www.irphouse.com Coefficient Inequalities for

Διαβάστε περισσότερα

Second Order RLC Filters

Second Order RLC Filters ECEN 60 Circuits/Electronics Spring 007-0-07 P. Mathys Second Order RLC Filters RLC Lowpass Filter A passive RLC lowpass filter (LPF) circuit is shown in the following schematic. R L C v O (t) Using phasor

Διαβάστε περισσότερα

Math221: HW# 1 solutions

Math221: HW# 1 solutions Math: HW# solutions Andy Royston October, 5 7.5.7, 3 rd Ed. We have a n = b n = a = fxdx = xdx =, x cos nxdx = x sin nx n sin nxdx n = cos nx n = n n, x sin nxdx = x cos nx n + cos nxdx n cos n = + sin

Διαβάστε περισσότερα

Problem Set 3: Solutions

Problem Set 3: Solutions CMPSCI 69GG Applied Information Theory Fall 006 Problem Set 3: Solutions. [Cover and Thomas 7.] a Define the following notation, C I p xx; Y max X; Y C I p xx; Ỹ max I X; Ỹ We would like to show that C

Διαβάστε περισσότερα

Section 7.6 Double and Half Angle Formulas

Section 7.6 Double and Half Angle Formulas 09 Section 7. Double and Half Angle Fmulas To derive the double-angles fmulas, we will use the sum of two angles fmulas that we developed in the last section. We will let α θ and β θ: cos(θ) cos(θ + θ)

Διαβάστε περισσότερα

Inverse trigonometric functions & General Solution of Trigonometric Equations. ------------------ ----------------------------- -----------------

Inverse trigonometric functions & General Solution of Trigonometric Equations. ------------------ ----------------------------- ----------------- Inverse trigonometric functions & General Solution of Trigonometric Equations. 1. Sin ( ) = a) b) c) d) Ans b. Solution : Method 1. Ans a: 17 > 1 a) is rejected. w.k.t Sin ( sin ) = d is rejected. If sin

Διαβάστε περισσότερα

Solutions to Exercise Sheet 5

Solutions to Exercise Sheet 5 Solutions to Eercise Sheet 5 jacques@ucsd.edu. Let X and Y be random variables with joint pdf f(, y) = 3y( + y) where and y. Determine each of the following probabilities. Solutions. a. P (X ). b. P (X

Διαβάστε περισσότερα

Areas and Lengths in Polar Coordinates

Areas and Lengths in Polar Coordinates Kiryl Tsishchanka Areas and Lengths in Polar Coordinates In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the

Διαβάστε περισσότερα

Areas and Lengths in Polar Coordinates

Areas and Lengths in Polar Coordinates Kiryl Tsishchanka Areas and Lengths in Polar Coordinates In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the

Διαβάστε περισσότερα

Chapter 6: Systems of Linear Differential. be continuous functions on the interval

Chapter 6: Systems of Linear Differential. be continuous functions on the interval Chapter 6: Systems of Linear Differential Equations Let a (t), a 2 (t),..., a nn (t), b (t), b 2 (t),..., b n (t) be continuous functions on the interval I. The system of n first-order differential equations

Διαβάστε περισσότερα

Lecture 2. Soundness and completeness of propositional logic

Lecture 2. Soundness and completeness of propositional logic Lecture 2 Soundness and completeness of propositional logic February 9, 2004 1 Overview Review of natural deduction. Soundness and completeness. Semantics of propositional formulas. Soundness proof. Completeness

Διαβάστε περισσότερα

9.09. # 1. Area inside the oval limaçon r = cos θ. To graph, start with θ = 0 so r = 6. Compute dr

9.09. # 1. Area inside the oval limaçon r = cos θ. To graph, start with θ = 0 so r = 6. Compute dr 9.9 #. Area inside the oval limaçon r = + cos. To graph, start with = so r =. Compute d = sin. Interesting points are where d vanishes, or at =,,, etc. For these values of we compute r:,,, and the values

Διαβάστε περισσότερα

Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013

Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013 Notes on Average Scattering imes and Hall Factors Jesse Maassen and Mar Lundstrom Purdue University November 5, 13 I. Introduction 1 II. Solution of the BE 1 III. Exercises: Woring out average scattering

Διαβάστε περισσότερα

Approximation of distance between locations on earth given by latitude and longitude

Approximation of distance between locations on earth given by latitude and longitude Approximation of distance between locations on earth given by latitude and longitude Jan Behrens 2012-12-31 In this paper we shall provide a method to approximate distances between two points on earth

Διαβάστε περισσότερα

Exercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.

Exercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1. Exercises 0 More exercises are available in Elementary Differential Equations. If you have a problem to solve any of them, feel free to come to office hour. Problem Find a fundamental matrix of the given

Διαβάστε περισσότερα

ST5224: Advanced Statistical Theory II

ST5224: Advanced Statistical Theory II ST5224: Advanced Statistical Theory II 2014/2015: Semester II Tutorial 7 1. Let X be a sample from a population P and consider testing hypotheses H 0 : P = P 0 versus H 1 : P = P 1, where P j is a known

Διαβάστε περισσότερα

CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS

CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS EXERCISE 01 Page 545 1. Use matrices to solve: 3x + 4y x + 5y + 7 3x + 4y x + 5y 7 Hence, 3 4 x 0 5 y 7 The inverse of 3 4 5 is: 1 5 4 1 5 4 15 8 3

Διαβάστε περισσότερα

Partial Differential Equations in Biology The boundary element method. March 26, 2013

Partial Differential Equations in Biology The boundary element method. March 26, 2013 The boundary element method March 26, 203 Introduction and notation The problem: u = f in D R d u = ϕ in Γ D u n = g on Γ N, where D = Γ D Γ N, Γ D Γ N = (possibly, Γ D = [Neumann problem] or Γ N = [Dirichlet

Διαβάστε περισσότερα

Appendix to On the stability of a compressible axisymmetric rotating flow in a pipe. By Z. Rusak & J. H. Lee

Appendix to On the stability of a compressible axisymmetric rotating flow in a pipe. By Z. Rusak & J. H. Lee Appendi to On the stability of a compressible aisymmetric rotating flow in a pipe By Z. Rusak & J. H. Lee Journal of Fluid Mechanics, vol. 5 4, pp. 5 4 This material has not been copy-edited or typeset

Διαβάστε περισσότερα

Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.

Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ. Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο The time integral of a force is referred to as impulse, is determined by and is obtained from: Newton s 2 nd Law of motion states that the action

Διαβάστε περισσότερα

derivation of the Laplacian from rectangular to spherical coordinates

derivation of the Laplacian from rectangular to spherical coordinates derivation of the Laplacian from rectangular to spherical coordinates swapnizzle 03-03- :5:43 We begin by recognizing the familiar conversion from rectangular to spherical coordinates (note that φ is used

Διαβάστε περισσότερα

Congruence Classes of Invertible Matrices of Order 3 over F 2

Congruence Classes of Invertible Matrices of Order 3 over F 2 International Journal of Algebra, Vol. 8, 24, no. 5, 239-246 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/.2988/ija.24.422 Congruence Classes of Invertible Matrices of Order 3 over F 2 Ligong An and

Διαβάστε περισσότερα

Homomorphism in Intuitionistic Fuzzy Automata

Homomorphism in Intuitionistic Fuzzy Automata International Journal of Fuzzy Mathematics Systems. ISSN 2248-9940 Volume 3, Number 1 (2013), pp. 39-45 Research India Publications http://www.ripublication.com/ijfms.htm Homomorphism in Intuitionistic

Διαβάστε περισσότερα

k A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R +

k A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R + Chapter 3. Fuzzy Arithmetic 3- Fuzzy arithmetic: ~Addition(+) and subtraction (-): Let A = [a and B = [b, b in R If x [a and y [b, b than x+y [a +b +b Symbolically,we write A(+)B = [a (+)[b, b = [a +b

Διαβάστε περισσότερα

Matrices and Determinants

Matrices and Determinants Matrices and Determinants SUBJECTIVE PROBLEMS: Q 1. For what value of k do the following system of equations possess a non-trivial (i.e., not all zero) solution over the set of rationals Q? x + ky + 3z

Διαβάστε περισσότερα

Other Test Constructions: Likelihood Ratio & Bayes Tests

Other Test Constructions: Likelihood Ratio & Bayes Tests Other Test Constructions: Likelihood Ratio & Bayes Tests Side-Note: So far we have seen a few approaches for creating tests such as Neyman-Pearson Lemma ( most powerful tests of H 0 : θ = θ 0 vs H 1 :

Διαβάστε περισσότερα

( y) Partial Differential Equations

( y) Partial Differential Equations Partial Dierential Equations Linear P.D.Es. contains no owers roducts o the deendent variables / an o its derivatives can occasionall be solved. Consider eamle ( ) a (sometimes written as a ) we can integrate

Διαβάστε περισσότερα

DESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.

DESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0. DESIGN OF MACHINERY SOLUTION MANUAL -7-1! PROBLEM -7 Statement: Design a double-dwell cam to move a follower from to 25 6, dwell for 12, fall 25 and dwell for the remader The total cycle must take 4 sec

Διαβάστε περισσότερα

Homework 8 Model Solution Section

Homework 8 Model Solution Section MATH 004 Homework Solution Homework 8 Model Solution Section 14.5 14.6. 14.5. Use the Chain Rule to find dz where z cosx + 4y), x 5t 4, y 1 t. dz dx + dy y sinx + 4y)0t + 4) sinx + 4y) 1t ) 0t + 4t ) sinx

Διαβάστε περισσότερα

ANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?

ANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =? Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) 3 00 000 www.tekoclasses.com ANSWERSHEET (TOPIC DIFFERENTIAL CALCULUS) COLLECTION # Question Type A.Single Correct Type Q. (A) Sol least

Διαβάστε περισσότερα

Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in

Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in : tail in X, head in A nowhere-zero Γ-flow is a Γ-circulation such that

Διαβάστε περισσότερα

forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with

forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with Week 03: C lassification of S econd- Order L inear Equations In last week s lectures we have illustrated how to obtain the general solutions of first order PDEs using the method of characteristics. We

Διαβάστε περισσότερα

A Note on Intuitionistic Fuzzy. Equivalence Relation

A Note on Intuitionistic Fuzzy. Equivalence Relation International Mathematical Forum, 5, 2010, no. 67, 3301-3307 A Note on Intuitionistic Fuzzy Equivalence Relation D. K. Basnet Dept. of Mathematics, Assam University Silchar-788011, Assam, India dkbasnet@rediffmail.com

Διαβάστε περισσότερα

Strain gauge and rosettes

Strain gauge and rosettes Strain gauge and rosettes Introduction A strain gauge is a device which is used to measure strain (deformation) on an object subjected to forces. Strain can be measured using various types of devices classified

Διαβάστε περισσότερα

SCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions

SCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions SCHOOL OF MATHEMATICAL SCIENCES GLMA Linear Mathematics 00- Examination Solutions. (a) i. ( + 5i)( i) = (6 + 5) + (5 )i = + i. Real part is, imaginary part is. (b) ii. + 5i i ( + 5i)( + i) = ( i)( + i)

Διαβάστε περισσότερα

Quadratic Expressions

Quadratic Expressions Quadratic Expressions. The standard form of a quadratic equation is ax + bx + c = 0 where a, b, c R and a 0. The roots of ax + bx + c = 0 are b ± b a 4ac. 3. For the equation ax +bx+c = 0, sum of the roots

Διαβάστε περισσότερα

1 String with massive end-points

1 String with massive end-points 1 String with massive end-points Πρόβλημα 5.11:Θεωρείστε μια χορδή μήκους, τάσης T, με δύο σημειακά σωματίδια στα άκρα της, το ένα μάζας m, και το άλλο μάζας m. α) Μελετώντας την κίνηση των άκρων βρείτε

Διαβάστε περισσότερα

CHAPTER 101 FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD

CHAPTER 101 FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD CHAPTER FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD EXERCISE 36 Page 66. Determine the Fourier series for the periodic function: f(x), when x +, when x which is periodic outside this rge of period.

Διαβάστε περισσότερα

Srednicki Chapter 55

Srednicki Chapter 55 Srednicki Chapter 55 QFT Problems & Solutions A. George August 3, 03 Srednicki 55.. Use equations 55.3-55.0 and A i, A j ] = Π i, Π j ] = 0 (at equal times) to verify equations 55.-55.3. This is our third

Διαβάστε περισσότερα

Chapter 6: Systems of Linear Differential. be continuous functions on the interval

Chapter 6: Systems of Linear Differential. be continuous functions on the interval Chapter 6: Systems of Linear Differential Equations Let a (t), a 2 (t),..., a nn (t), b (t), b 2 (t),..., b n (t) be continuous functions on the interval I. The system of n first-order differential equations

Διαβάστε περισσότερα

DiracDelta. Notations. Primary definition. Specific values. General characteristics. Traditional name. Traditional notation

DiracDelta. Notations. Primary definition. Specific values. General characteristics. Traditional name. Traditional notation DiracDelta Notations Traditional name Dirac delta function Traditional notation x Mathematica StandardForm notation DiracDeltax Primary definition 4.03.02.000.0 x Π lim ε ; x ε0 x 2 2 ε Specific values

Διαβάστε περισσότερα

A Two-Sided Laplace Inversion Algorithm with Computable Error Bounds and Its Applications in Financial Engineering

A Two-Sided Laplace Inversion Algorithm with Computable Error Bounds and Its Applications in Financial Engineering Electronic Companion A Two-Sie Laplace Inversion Algorithm with Computable Error Bouns an Its Applications in Financial Engineering Ning Cai, S. G. Kou, Zongjian Liu HKUST an Columbia University Appenix

Διαβάστε περισσότερα

( ) 2 and compare to M.

( ) 2 and compare to M. Problems and Solutions for Section 4.2 4.9 through 4.33) 4.9 Calculate the square root of the matrix 3!0 M!0 8 Hint: Let M / 2 a!b ; calculate M / 2!b c ) 2 and compare to M. Solution: Given: 3!0 M!0 8

Διαβάστε περισσότερα

w o = R 1 p. (1) R = p =. = 1

w o = R 1 p. (1) R = p =. = 1 Πανεπιστήµιο Κρήτης - Τµήµα Επιστήµης Υπολογιστών ΗΥ-570: Στατιστική Επεξεργασία Σήµατος 205 ιδάσκων : Α. Μουχτάρης Τριτη Σειρά Ασκήσεων Λύσεις Ασκηση 3. 5.2 (a) From the Wiener-Hopf equation we have:

Διαβάστε περισσότερα

The Simply Typed Lambda Calculus

The Simply Typed Lambda Calculus Type Inference Instead of writing type annotations, can we use an algorithm to infer what the type annotations should be? That depends on the type system. For simple type systems the answer is yes, and

Διαβάστε περισσότερα

Statistical Inference I Locally most powerful tests

Statistical Inference I Locally most powerful tests Statistical Inference I Locally most powerful tests Shirsendu Mukherjee Department of Statistics, Asutosh College, Kolkata, India. shirsendu st@yahoo.co.in So far we have treated the testing of one-sided

Διαβάστε περισσότερα

Practice Exam 2. Conceptual Questions. 1. State a Basic identity and then verify it. (a) Identity: Solution: One identity is csc(θ) = 1

Practice Exam 2. Conceptual Questions. 1. State a Basic identity and then verify it. (a) Identity: Solution: One identity is csc(θ) = 1 Conceptual Questions. State a Basic identity and then verify it. a) Identity: Solution: One identity is cscθ) = sinθ) Practice Exam b) Verification: Solution: Given the point of intersection x, y) of the

Διαβάστε περισσότερα

Finite Field Problems: Solutions

Finite Field Problems: Solutions Finite Field Problems: Solutions 1. Let f = x 2 +1 Z 11 [x] and let F = Z 11 [x]/(f), a field. Let Solution: F =11 2 = 121, so F = 121 1 = 120. The possible orders are the divisors of 120. Solution: The

Διαβάστε περισσότερα

Testing for Indeterminacy: An Application to U.S. Monetary Policy. Technical Appendix

Testing for Indeterminacy: An Application to U.S. Monetary Policy. Technical Appendix Testing for Indeterminacy: An Application to U.S. Monetary Policy Technical Appendix Thomas A. Lubik Department of Economics Johns Hopkins University Frank Schorfheide Department of Economics University

Διαβάστε περισσότερα

ORDINAL ARITHMETIC JULIAN J. SCHLÖDER

ORDINAL ARITHMETIC JULIAN J. SCHLÖDER ORDINAL ARITHMETIC JULIAN J. SCHLÖDER Abstract. We define ordinal arithmetic and show laws of Left- Monotonicity, Associativity, Distributivity, some minor related properties and the Cantor Normal Form.

Διαβάστε περισσότερα

Fourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics

Fourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics Fourier Series MATH 211, Calculus II J. Robert Buchanan Department of Mathematics Spring 2018 Introduction Not all functions can be represented by Taylor series. f (k) (c) A Taylor series f (x) = (x c)

Διαβάστε περισσότερα

SOME PROPERTIES OF FUZZY REAL NUMBERS

SOME PROPERTIES OF FUZZY REAL NUMBERS Sahand Communications in Mathematical Analysis (SCMA) Vol. 3 No. 1 (2016), 21-27 http://scma.maragheh.ac.ir SOME PROPERTIES OF FUZZY REAL NUMBERS BAYAZ DARABY 1 AND JAVAD JAFARI 2 Abstract. In the mathematical

Διαβάστε περισσότερα

On a four-dimensional hyperbolic manifold with finite volume

On a four-dimensional hyperbolic manifold with finite volume BULETINUL ACADEMIEI DE ŞTIINŢE A REPUBLICII MOLDOVA. MATEMATICA Numbers 2(72) 3(73), 2013, Pages 80 89 ISSN 1024 7696 On a four-dimensional hyperbolic manifold with finite volume I.S.Gutsul Abstract. In

Διαβάστε περισσότερα

Optimal Parameter in Hermitian and Skew-Hermitian Splitting Method for Certain Two-by-Two Block Matrices

Optimal Parameter in Hermitian and Skew-Hermitian Splitting Method for Certain Two-by-Two Block Matrices Optimal Parameter in Hermitian and Skew-Hermitian Splitting Method for Certain Two-by-Two Block Matrices Chi-Kwong Li Department of Mathematics The College of William and Mary Williamsburg, Virginia 23187-8795

Διαβάστε περισσότερα

Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1

Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8  questions or comments to Dan Fetter 1 Eon : Fall 8 Suggested Solutions to Problem Set 8 Email questions or omments to Dan Fetter Problem. Let X be a salar with density f(x, θ) (θx + θ) [ x ] with θ. (a) Find the most powerful level α test

Διαβάστε περισσότερα

ΗΜΥ 220: ΣΗΜΑΤΑ ΚΑΙ ΣΥΣΤΗΜΑΤΑ Ι Ακαδημαϊκό έτος Εαρινό Εξάμηνο Κατ οίκον εργασία αρ. 2

ΗΜΥ 220: ΣΗΜΑΤΑ ΚΑΙ ΣΥΣΤΗΜΑΤΑ Ι Ακαδημαϊκό έτος Εαρινό Εξάμηνο Κατ οίκον εργασία αρ. 2 ΤΜΗΜΑ ΗΛΕΚΤΡΟΛΟΓΩΝ ΜΗΧΑΝΙΚΩΝ ΚΑΙ ΜΗΧΑΝΙΚΩΝ ΥΠΟΛΟΓΙΣΤΩΝ ΠΑΝΕΠΙΣΤΗΜΙΟ ΚΥΠΡΟΥ ΗΜΥ 220: ΣΗΜΑΤΑ ΚΑΙ ΣΥΣΤΗΜΑΤΑ Ι Ακαδημαϊκό έτος 2007-08 -- Εαρινό Εξάμηνο Κατ οίκον εργασία αρ. 2 Ημερομηνία Παραδόσεως: Παρασκευή

Διαβάστε περισσότερα

Overview. Transition Semantics. Configurations and the transition relation. Executions and computation

Overview. Transition Semantics. Configurations and the transition relation. Executions and computation Overview Transition Semantics Configurations and the transition relation Executions and computation Inference rules for small-step structural operational semantics for the simple imperative language Transition

Διαβάστε περισσότερα

MATH423 String Theory Solutions 4. = 0 τ = f(s). (1) dτ ds = dxµ dτ f (s) (2) dτ 2 [f (s)] 2 + dxµ. dτ f (s) (3)

MATH423 String Theory Solutions 4. = 0 τ = f(s). (1) dτ ds = dxµ dτ f (s) (2) dτ 2 [f (s)] 2 + dxµ. dτ f (s) (3) 1. MATH43 String Theory Solutions 4 x = 0 τ = fs). 1) = = f s) ) x = x [f s)] + f s) 3) equation of motion is x = 0 if an only if f s) = 0 i.e. fs) = As + B with A, B constants. i.e. allowe reparametrisations

Διαβάστε περισσότερα

Commutative Monoids in Intuitionistic Fuzzy Sets

Commutative Monoids in Intuitionistic Fuzzy Sets Commutative Monoids in Intuitionistic Fuzzy Sets S K Mala #1, Dr. MM Shanmugapriya *2 1 PhD Scholar in Mathematics, Karpagam University, Coimbatore, Tamilnadu- 641021 Assistant Professor of Mathematics,

Διαβάστε περισσότερα

Lecture 26: Circular domains

Lecture 26: Circular domains Introductory lecture notes on Partial Differential Equations - c Anthony Peirce. Not to be copied, used, or revised without eplicit written permission from the copyright owner. 1 Lecture 6: Circular domains

Διαβάστε περισσότερα

Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3

Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3 Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3 1 State vector space and the dual space Space of wavefunctions The space of wavefunctions is the set of all

Διαβάστε περισσότερα

Section 9.2 Polar Equations and Graphs

Section 9.2 Polar Equations and Graphs 180 Section 9. Polar Equations and Graphs In this section, we will be graphing polar equations on a polar grid. In the first few examples, we will write the polar equation in rectangular form to help identify

Διαβάστε περισσότερα

Appendix S1 1. ( z) α βc. dβ β δ β

Appendix S1 1. ( z) α βc. dβ β δ β Appendix S1 1 Proof of Lemma 1. Taking first and second partial derivatives of the expected profit function, as expressed in Eq. (7), with respect to l: Π Π ( z, λ, l) l θ + s ( s + h ) g ( t) dt λ Ω(

Διαβάστε περισσότερα

Problem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.

Problem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ. Chemistry 362 Dr Jean M Standard Problem Set 9 Solutions The ˆ L 2 operator is defined as Verify that the angular wavefunction Y θ,φ) Also verify that the eigenvalue is given by 2! 2 & L ˆ 2! 2 2 θ 2 +

Διαβάστε περισσότερα

CRASH COURSE IN PRECALCULUS

CRASH COURSE IN PRECALCULUS CRASH COURSE IN PRECALCULUS Shiah-Sen Wang The graphs are prepared by Chien-Lun Lai Based on : Precalculus: Mathematics for Calculus by J. Stuwart, L. Redin & S. Watson, 6th edition, 01, Brooks/Cole Chapter

Διαβάστε περισσότερα

6.1. Dirac Equation. Hamiltonian. Dirac Eq.

6.1. Dirac Equation. Hamiltonian. Dirac Eq. 6.1. Dirac Equation Ref: M.Kaku, Quantum Field Theory, Oxford Univ Press (1993) η μν = η μν = diag(1, -1, -1, -1) p 0 = p 0 p = p i = -p i p μ p μ = p 0 p 0 + p i p i = E c 2 - p 2 = (m c) 2 H = c p 2

Διαβάστε περισσότερα

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007 Οδηγίες: Να απαντηθούν όλες οι ερωτήσεις. Αν κάπου κάνετε κάποιες υποθέσεις να αναφερθούν στη σχετική ερώτηση. Όλα τα αρχεία που αναφέρονται στα προβλήματα βρίσκονται στον ίδιο φάκελο με το εκτελέσιμο

Διαβάστε περισσότερα

Solution Series 9. i=1 x i and i=1 x i.

Solution Series 9. i=1 x i and i=1 x i. Lecturer: Prof. Dr. Mete SONER Coordinator: Yilin WANG Solution Series 9 Q1. Let α, β >, the p.d.f. of a beta distribution with parameters α and β is { Γ(α+β) Γ(α)Γ(β) f(x α, β) xα 1 (1 x) β 1 for < x

Διαβάστε περισσότερα

Fractional Colorings and Zykov Products of graphs

Fractional Colorings and Zykov Products of graphs Fractional Colorings and Zykov Products of graphs Who? Nichole Schimanski When? July 27, 2011 Graphs A graph, G, consists of a vertex set, V (G), and an edge set, E(G). V (G) is any finite set E(G) is

Διαβάστε περισσότερα

Arithmetical applications of lagrangian interpolation. Tanguy Rivoal. Institut Fourier CNRS and Université de Grenoble 1

Arithmetical applications of lagrangian interpolation. Tanguy Rivoal. Institut Fourier CNRS and Université de Grenoble 1 Arithmetical applications of lagrangian interpolation Tanguy Rivoal Institut Fourier CNRS and Université de Grenoble Conference Diophantine and Analytic Problems in Number Theory, The 00th anniversary

Διαβάστε περισσότερα

EE101: Resonance in RLC circuits

EE101: Resonance in RLC circuits EE11: Resonance in RLC circuits M. B. Patil mbatil@ee.iitb.ac.in www.ee.iitb.ac.in/~sequel Deartment of Electrical Engineering Indian Institute of Technology Bombay I V R V L V C I = I m = R + jωl + 1/jωC

Διαβάστε περισσότερα

A summation formula ramified with hypergeometric function and involving recurrence relation

A summation formula ramified with hypergeometric function and involving recurrence relation South Asian Journal of Mathematics 017, Vol. 7 ( 1): 1 4 www.sajm-online.com ISSN 51-151 RESEARCH ARTICLE A summation formula ramified with hypergeometric function and involving recurrence relation Salahuddin

Διαβάστε περισσότερα

: Monte Carlo EM 313, Louis (1982) EM, EM Newton-Raphson, /. EM, 2 Monte Carlo EM Newton-Raphson, Monte Carlo EM, Monte Carlo EM, /. 3, Monte Carlo EM

: Monte Carlo EM 313, Louis (1982) EM, EM Newton-Raphson, /. EM, 2 Monte Carlo EM Newton-Raphson, Monte Carlo EM, Monte Carlo EM, /. 3, Monte Carlo EM 2008 6 Chinese Journal of Applied Probability and Statistics Vol.24 No.3 Jun. 2008 Monte Carlo EM 1,2 ( 1,, 200241; 2,, 310018) EM, E,,. Monte Carlo EM, EM E Monte Carlo,. EM, Monte Carlo EM,,,,. Newton-Raphson.

Διαβάστε περισσότερα

2. THEORY OF EQUATIONS. PREVIOUS EAMCET Bits.

2. THEORY OF EQUATIONS. PREVIOUS EAMCET Bits. EAMCET-. THEORY OF EQUATIONS PREVIOUS EAMCET Bits. Each of the roots of the equation x 6x + 6x 5= are increased by k so that the new transformed equation does not contain term. Then k =... - 4. - Sol.

Διαβάστε περισσότερα

Potential Dividers. 46 minutes. 46 marks. Page 1 of 11

Potential Dividers. 46 minutes. 46 marks. Page 1 of 11 Potential Dividers 46 minutes 46 marks Page 1 of 11 Q1. In the circuit shown in the figure below, the battery, of negligible internal resistance, has an emf of 30 V. The pd across the lamp is 6.0 V and

Διαβάστε περισσότερα

A General Note on δ-quasi Monotone and Increasing Sequence

A General Note on δ-quasi Monotone and Increasing Sequence International Mathematical Forum, 4, 2009, no. 3, 143-149 A General Note on δ-quasi Monotone and Increasing Sequence Santosh Kr. Saxena H. N. 419, Jawaharpuri, Badaun, U.P., India Presently working in

Διαβάστε περισσότερα

Math 6 SL Probability Distributions Practice Test Mark Scheme

Math 6 SL Probability Distributions Practice Test Mark Scheme Math 6 SL Probability Distributions Practice Test Mark Scheme. (a) Note: Award A for vertical line to right of mean, A for shading to right of their vertical line. AA N (b) evidence of recognizing symmetry

Διαβάστε περισσότερα

SOLVING CUBICS AND QUARTICS BY RADICALS

SOLVING CUBICS AND QUARTICS BY RADICALS SOLVING CUBICS AND QUARTICS BY RADICALS The purpose of this handout is to record the classical formulas expressing the roots of degree three and degree four polynomials in terms of radicals. We begin with

Διαβάστε περισσότερα

On New Subclasses of Analytic Functions with Respect to Conjugate and Symmetric Conjugate Points

On New Subclasses of Analytic Functions with Respect to Conjugate and Symmetric Conjugate Points Global Journal of Pure Applied Mathematics. ISSN 0973-768 Volume, Number 3 06, pp. 849 865 Research India Publications http://www.ripublication.com/gjpam.htm On New Subclasses of Analytic Functions with

Διαβάστε περισσότερα