# Results for: “Manish Goyal”

1 eBook

## ALLC8-1 |
Manish Goyal | Laxmi Publications | |||||

8 The Eigen Value Problem 8.1. INTRODUCTION Computation of eigen values and the corresponding eigen vectors of a matrix is of practical importance. For example, in solid mechanics, when we consider an element in a continuum, subjected to normal and shear stresses, we usually find principal stresses which are the maximum and minimum stresses in an element. Consider the wedge of unit thickness, subjected to normal and shear stresses. If it has to be in equilibrium, the following system of equations, written in matrix notation, must be satisfied. LMσ Nσ x − σθ xy σ xy σ y − σθ OP Lcos θO = L0O Q MNsin θPQ MN0PQ For non-trivial solution to exist, the determinant of the coefficient matrix must be zero. σ x − σθ σ xy σ xy =0 σ y − σθ ⇒ σθ2 – σθ (σx + σy) + (σx σy – σxy2) = 0 This is a quadratic equation whose roots give two eigen values corresponding to principal stresses. In this chapter, we will describe power method for finding largest eigen value of a matrix and inverse power method for finding smallest eigen value of a matrix. Several methods will be discussed for finding eigen values of symmetric tridiagonal matrices. See All Chapters |
|||||||

## ALLC9-1 |
Manish Goyal | Laxmi Publications | |||||

9 Numerical Solution of Ordinary Differential Equations 9.1. INTRODUCTION A physical situation that concerns with the rate of change of one quantity with respect to another gives rise to a differential equation. Consider the first order ordinary differential equation dy = f (x, y) ...(1) dx with the initial condition ...(2) y(x0) = y0 Many analytical techniques exist for solving such equations. But these methods can be applied to solve only a selected class of differential equations. However, a majority of differential equations appearing in physical problems cannot be solved analytically. Thus it becomes imperative to discuss their solution by numerical methods. In numerical methods, we donot proceed in the hope of finding a relation between variables but we find the numerical values of the dependent variable for certain values of independent variable. It must be noted that even the differential equations which are solvable by analytical methods, can be solved numerically also. 9.2. INITIAL-VALUE AND BOUNDARY-VALUE PROBLEMS See All Chapters |
|||||||

## ALLC-IND |
Manish Goyal | Laxmi Publications | |||||

Index A A general error formula 36 Acceptance region 716 Acceptance sampling 701 Accuracy of numbers 28 Adams-Moulton (or Adams-Bashforth) formula 546 Advantages of statistical quality control 701 Advantages/features of ‘C’ language 6 Aitken’s method 308 Aitken’s ∆2 method 150 Algorithmic errors 27 Alternative hypothesis 715 Analysis of data 701 Analysis of time-series 755 Analysis of variance (ANOVA) 745 Angle between two lines of regression 684 Annihilation of off-diagonal elements 465 ANOVA table 745 Applications of t-distribution 726 Approximate numbers 28 Approximation of functions 616 Area diagrams 678 Arithmetic operations with normalized floating point numbers 50 Arrays and string 15 Assignable causes 702 Assumptions for interpolation 165 Automatic error monitoring 550 Averaging operator 169 B Backward differences 168 Bar diagrams 678 Basic structure of ‘C’ program 10 Bender-Schmidt explicit finite difference scheme 658 Bender-Schmidt 657 Bessel’s interpolation formula 253 Bisection method 63 Bolzano method 64 Boole’s rule 336 See All Chapters |
|||||||

## ALLC4-7 |
Manish Goyal | Laxmi Publications | |||||

260 NUMERICAL METHODS AND STATISTICAL TECHNIQUES USING ‘C’ Using Bessel’s formula, RS f (0) + f (1) UV + FG u − 1IJ ∆f (0) + u (u − 1) RS ∆ 2 T 2 W H 2K T 2 f(u) = FG H (u − 1) u − + ⇒ f(.25) = UV W f (− 1) + ∆2 f (0) 2 3! IJ K 1 u 2 ∆3 f (− 1) FG 32 + 35IJ + (.25 – .5) (3) + (.25) (.25 − 1) RS − 5 + 2 UV H 2 K 2 T 2 W + (.25 − 1) (.25 − .5) (.25) (7) 3! = 32.9453125 Hence y25 = 32.9453125. Example 2. Apply Bessel’s formula to find the value of f(27.4) from the table : x : 25 26 27 28 29 30 f(x) : 4.000 3.846 3.704 3.571 3.448 3.333. Sol. Taking origin at 27 and h = 1 x = a + uh ⇒ 27.4 = 27 + u × 1 ∴ u = 0.4 The difference table is as follows : u 103f(u) –2 4000 –1 3847 103 ∆f(u) 103 ∆2f(u) 103 ∆3f(u) 103 ∆4f(u) 103 ∆5f(u) – 154 12 – 142 0 –3 3704 9 – 133 1 3571 2 3448 3 3333 4 1 –7 10 – 123 –3 –2 8 – 115 Bessel’s formula is RS f (0) + f (1) UV + FG u − 1IJ ∆f (0) + u (u − 1) RS ∆ f (0) + ∆ f (− 1) UV 2! 2 T 2 W H 2K W T F 1I (u − 1) G u − J u H 2 K ∆ f (− 1) + (u + 1) u (u − 1) (u − 2) RS ∆ f (− 1) + ∆ f (− 2) UV + 3! 4! 2 W T F 1I (u − 2) (u − 1) G u − J u (u + 1) H 2K ∆ f (− 2) + 2 2 f(u) = See All Chapters |
|||||||

## ALLC2-2 |
Manish Goyal | Laxmi Publications | |||||

43 ERROR ANALYSIS AND ESTIMATION Rounding off all other numbers to two decimal digits, we have 0.15, 15.45, 0.00, 8.12, 0.02, 0.64 and 0.17. The sum S is given by S = 305.1 + 143.3 + 0.15 + 15.45 + 0.00 + 8.12 + 0.02 + 0.64 + 0.17 = 472.59 = 472.6. To determine absolute error, we note that first two numbers have each an absolute error of 0.05 and remaining seven numbers have an absolute error of 0.005 each. ∴ Absolute error in all 9 numbers = 2(0.05) + 7(0.005) = 0.1 + 0.035 = 0.135 ≈ 0.14. In addition to the above absolute error, we have to take into account rounding error which is 0.01. Hence the total absolute error in S = 0.14 + 0.01 = 0.15 Thus, S = 472.6 ± 0.15. Example 17. 5.5 = 2.345 and 6.1 = 2.470 correct to four significant figures. Find the relative error in taking the difference of these numbers. 1 × 0.001 = 0.0005 2 ∆x1 ∆x2 ∆x1 0.0005 + ∴ Relative error < =2 =2 = 0.008 . X X 0.125 X Example 18. 10 = 3.162 and e ~ – 2.718 correct to three decimal places. Find the percentage error in their difference. 0.0005 See All Chapters |