characteristics of fixed fractional trading and salutary techniques

... the optical characteristics of ZnO/Al2O3 and ZnO/ZnO core-shell NRAs The nearband-edge (NBE) emission was significantly enhanced, and the deep-level band was suppressed by the Al2O3 and ZnO shells ... Page of Results and discussion Top-viewed and cross-sectional SEM images of asgrown ZnO NRAs are shown in Figure 1a,b, respectively The diameter of ZnO nanorods is in the range of 90 to 100 nm, and ... The ALD condition of ZnO shell layers was the same as that of the ZnO seed layer The thicknesses of ZnO shell layers were 5, 10, and 15 nm, respectively The details of ZnO and Al2 O3 ALD parameters...

Ngày tải lên: 20/06/2014, 22:20

... OUTPUT Enter the no of term –5 Enter the value in form of x– Enter the value of x1 – Enter the value of x2 – Enter the value of x3 – 11 Enter the value of x4 – 13 Enter the value of x5 – 17 Enter ... in the form of y– Enter the value of y1 – 150 597 598 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Enter the value of y2 – 392 Enter the value of y3 – 1452 Enter the value of y4 – 2366 ... Print output End of the program and start of section func 599 600 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Step 13 temp = 1/ (1+(x * x)) Step 14 Return temp Step 15 End of section func...

Ngày tải lên: 04/07/2014, 15:20

... ‘Computer Based Numerical and Statistical Techniques is primarily written according to the unified syllabus of Mathematics for B Tech II year and M.C.A I year students of all Engineering colleges ... feel no difficulty to understand the subject A unique feature of this book is to provide with an algorithm and computer program in Clanguage to understand the steps and methodology used in writing ... presented in a very systematic and logical manner In each chapter, all concepts, definitions and large number of examples in the best possible way have been discussed in detail and lucid manner so that...

Ngày tải lên: 04/07/2014, 15:20

... to COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES be cut-off to a manageable size such as 0.29, 0.286, 0.2857, etc The process of cutting off super-flouts digits and retaining as many digits ... reliability of the numerical result will depend on an error estimate or bound, therefore the analysis of error and the sources of error in numerical methods is also a critically important part of the ... known as rounding off a number or we can say that process of dropping unwanted digits is called rounding-off Number are rounded-off according to the following rules: To round-off the number to...

Ngày tải lên: 04/07/2014, 15:20

... value of 30.5286 = 7.342 = 30.4647 0.241 7.342 lies between 30.4647 – 0.0639 = 30 4008 and 30.4647 + 0.0639 = 0.241 Example 12 Find the product of 346.1 and 865.2 and state how many figures of the ... and (Because Relative Error, Er = 0.0015 = 0.0002 6.61 1 and , correct to decimal places are 0.1429 and 0.0909 11 respectively Find the possible relative error and absolute error in the sum of ... approximation Example Calculate the sum of and relative errors Sol Here 3, and to four significant digits and find its absolute = 1.732, = 2.236, = 2.646 Hence Sum = 6.614 and Absolute Error = Ea = 0.0005...

Ngày tải lên: 04/07/2014, 15:20

... ∆ABC, a = 2.3 cm, b = 5.7 cm and ∠ B = 90o If possible errors in the computed value of b and a are mm and mm respectively, find the possible error in the measurement of angle A Sol Given δb = mm ... relative error of a product of three non-zero numbers does not exceed the sum of the relative errors of the given numbers If y = (0.31x + 2.73)/(x+ 0.35) where the coefficients are round off, find ... ∆ABC, b = 9.5 cm, c = 8.5 cm and A = 45o, find allowable errors in b, c, and A such that the area of ∆ABC may be determined nearest to a square centimeter Sol Let area of the ∆ABC be given by, =...

Ngày tải lên: 04/07/2014, 15:20

... division of mantissa of the numerator by that of the denominator and denominator exponent is subtracted from the numerator exponent The resultant exponent is obtained by adjusting it appropriately and ... value of the number a = 4.568 and b = 6.762 using the four-digit  b − a arithmetic and compare the result by taking c = a +     Sol Since a = 4568el , b = 6762e1 and c be the middle value of ... 0.1823e3] 33 ERRORS AND FLOATING POINT 10 Give example to show that most of the laws of arithmetic fail to hold for floating-point arithmetic 11 Find the root of smaller magnitude of the equation...

Ngày tải lên: 04/07/2014, 15:20

... iterations, the root of f ( x) = x3 − x − = up to three places of decimals is 1.325, which is of desired accuracy Example Find the root of the equation x3 – x – = between and to three places of decimal ... ALGEBRAIC AND TRANSCENDENTAL EQUATION Here ei and ei + are the errors in ith and (i + 1)th iterations respectively Comparing the above equation with lim i →∞ ei +1 e k i ≤A We get k = and A = 0.5 ... −1 , which is negative Thus f(0.5) is negative and f(1) is positive Then the root lies between 0.5 and 40 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Second approximation: The second approximation...

Ngày tải lên: 04/07/2014, 15:20

... 0.10203  Since, x3 and x4 are approximately the same upto four places of decimal, hence the required root of the given equation is 1.4036 54 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example ... substituting the value of ei+1 and ei–1 in (5), we get  ei  Ae ik = ei    A or e ik = MA 1/ k M –(1 + 1/k).e (1+1/k) i .(7) Comparing the powers of ei on both sides of (7), we get or k = ... the root is 2.9428 correct to four places of decimal 53 ALGEBRAIC AND TRANSCENDENTAL EQUATION Example Using the method of False Position, find the root of equation x6 – x4 – x3 –1 = up to four...

Ngày tải lên: 04/07/2014, 15:20

... loge – 12 and 2.6137 Therefore, f(3) and f(4) are of opposite signs Therefore, a real root lies between and For the approximation to the root, taking x0 = 3, x1 = 4, f(x0) = – 4.0986 and f(x1) ... digits using of Falsi Position Sol Let Since and f(x) = tan x + tan h x = f(2.35) = – 0.03 f(2.37) = 0.009 Hence the root lies between 2.35 and 2.37 Let x0 = 2.35 and x1 = 2.37 Using method of Falsi ... root of the equations: (a) x = tan x (b) x2 – loge x – 12 = (c) 3x = cos x + Find the rate of convergence of Regula-Falsi method Find real cube root of 18 by Regula-Falsi method Discuss method of...

Ngày tải lên: 04/07/2014, 15:20

... (or Rate) of Convergence of Newton-Raphson Method Let α be the actual root of equation f(x) = i.e., f(a) = Let xn and xn+1 be two successive approximations to the actual root α If en and en+1 ... PROBLEM SET 2.3 Use the method of Iteration to find a positive root between and of the equation xex – [Ans 0.5671477] Find the Iterative method, the real root of the equation 3x – log10 x = correct ... step is proportional to the square of the previous error and as such the convergence is quadratic Example Find the real root of the equation x2 – 5x + = between and by NewtonRaphson’s method .(1)...

Ngày tải lên: 04/07/2014, 15:20

... NUMERICAL AND STATISTICAL TECHNIQUES = 4( 238.977) + 240 = 2.9925 399.627 Hence, the required value of (240)1/5 correct to three places of decimal is 2.993 Example 12 Determine the value of p and q ... lemmas (1) 81 ALGEBRAIC AND TRANSCENDENTAL EQUATION Lemma 2.1: Under the assumptions and notations of the theorem: f ′′( x∞ ) Ek Ek f ′( x∞ ) −1 Proof Using the definition of xk+1, we find Ek + ... find a root of the equation x3 – 3x – = (U.P.T.U 2005) [Ans 2.279] Find the four places of decimal, the smallest root of the equation e–x = sin x [Ans 0.5885] Find the cube root of 10 [Ans 2.15466]...

Ngày tải lên: 04/07/2014, 15:20

... approximated values of p and q Since R and S are both functions of the two parameters p and q then the improved values are given by R(p + ∆p, q + ∆q) = .(4) S(p + ∆p, q + ∆q) = Expand equation (4) ... extract the quadratic factors that are the products of the pairs of 92 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES complex roots, and then complex arithmetic can be avoided because such ... values of p and q First approximation: Let p0 and q0 be the initial approximations, then the first approximation can be obtained by p1 = p0 + ∆p and q1 = q0 + ∆q Because given equation is of the...

Ngày tải lên: 04/07/2014, 15:20

... the complex roots and multiple roots of polynomials and also for determining the Eigen values of a matrix An important feature of the method is that it gives approximate values of all the roots ... 10.480733 After substituting the values of bi and Ci in equations (1) and (2), we get ∆p = 0.018582 ∆q = – 0.0002186 100 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Therefore the third approximation ... value of x in the given range However if the function f is not known, the value of y can be obtained, when a set of values of x is given The method to find out such values is based on principle of...

Ngày tải lên: 04/07/2014, 15:20

... function is given by h Example Find the first term of the series whose second and subsequent terms are 8, 3, 0, –1, and Sol If the interval of differencing is unity, then f(1) = E–1f(2) = (1+ ... subtract from each value of y except y0 , the previous value of y, we get y1 – y0, y2 –y1, y3 – y2, yn – yn–1 These differences are called first backward differences of y and are denoted by ∇y ... operator D of differential calculus In finite differences, we deal with ratio of simultaneous increments of mutually dependent quantities where as in differential calculus, we find the limit of such...

Ngày tải lên: 04/07/2014, 15:20

... n) = (–1)m ∆m   n 118 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES  ∆2  Ee x Example 17 Show that ex =   ex x ; the interval of differencing being h  E ∆ e Sol Let f(x) = ex, ... (x)  (a) µ   = 1  g(x)  g(x − )g(x + ) 2 Here, interval of differencing being unity 120 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES (b) ∇ = h2 D2 – h3 D3 + 4 – h D 12 (c) ∇ – ∆ = ... 1 = f (x + ) g(x – ) – f(x – ) g(x + ) 2 2 = 124 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Therefore right hand side as 1 1 f (x + ) g (x − ) − f (x − ) g (x + ) 2 2 = 1 g( x − ) g (...

Ngày tải lên: 04/07/2014, 15:20

... DIFFERENCES OF POLYNOMIAL Statement: If f(x) be the nth degree polynomial in x, then the nth difference of f(x) is constant and ∆n+1f(x) and all higher differences are zero when the values of the ... value of y lies In odd difference columns of ∆1y, ∆3y, , the maximum error lies in the two middle terms and the incorrect value of y lies between these two middle terms Example Find the error and ... Sum of all values in column of fourth difference is –0.004 which is very small as compared to sum of values in other columns ∴ ∆4y = Errors in this column are e, –4e, 6e, –4e and e Term of Maximum...

Ngày tải lên: 04/07/2014, 15:20

... 137 CALCULUS OF FINITE DIFFERENCES y= f(x) = f(x0), f(x1), f(x2) , f(a + nh) i.e., f(xn) Let one of the value of f(x) is missing Say it f(i) To determine this missing value of f(x), assume ... COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES i.e., 4f(3) = 124 i.e., f(3) = 31 (function values are 3n type and this is not a polynomial) Example Find the missing value of the data: x f (x ... f(5) + f(3) = 152 and 10f(5) + 3f(3) = 1331 On solving these two, we get f(3) = 27 and f(5) = 125 Example 10 Assuming that the following values of y belong to a polynomial of degree 4, compute...

Ngày tải lên: 04/07/2014, 15:20

... of different powers of x in order beginning with the coefficients of higher power of x Put in the left hand side column and write zero below the coefficients of highest power of x (in this case ... written below The remainder –10 is the value of D Add the terms of corresponding columns of (a) and get 2, –1 and of (b) Now again apply the steps (1) and (3): in this way we get (2 × 2) + (0 × ... below –1 The remainder of (b) is equal to C Apply step (4) on (b) and get and of (c) Again apply the steps (2), (3) and (4) to get which will be equal to A and remainder of (c) which will be equal...

Ngày tải lên: 04/07/2014, 15:20

... the arithmetic mean of q and r and B is arithmetic between the arguments at q and r is A + 24 mean of 3q – 2p – s and 3r – 2s – p Sol Given A is the arithmetic mean of q and r q+r q + r = 2A ... the marks obtained by 492 candidates in a certain examination: Marks − 40 40 − 45 45 − 50 50 − 55 55 − 60 60 − 65 No of Candidates 210 43 54 74 32 79 Find out (a) No of candidates, if they secure ... COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES The following are the numbers of deaths in four successive ten year age groups Find the number of deaths at 45-50 and 50-55 Age 25 − 35 35...

Ngày tải lên: 04/07/2014, 15:20