NONLINEAR EQUATIONS AND TAYLOR S THEOREM

Transcription

1 APPENDIX C NONLINEAR EQUATIONS AND TAYLOR S THEOREM C.1 INTRODUCTION In adjustment computations it is frequently necessary to deal with nonlinear equations. For example, some observation equations relate observed quantities to unknown parameters through the transcendental functions of sine, cosine, or tangent, while others relate them through terms raised to second- and higherorder powers. The task of solving a system of nonlinear equations is formidable. To facilitate the solution, a first-order Taylor series approximation can be used to create a set of linear equations. The equations can then be solved by the matrix methods discussed in Appendix B. C.2 TAYLOR SERIES LINEARIZATION OF NONLINEAR EQUATIONS Suppose that the following equation relates a observed value L to its unknown parameters x and y through nonlinear coefficients as L = f (x, y) (C.1) By Taylor s theorem, the equation is represented as L = f (x, y) = f (x, y ) + ( L/ x) ( 2 L/ x 2) dx + dx ( n L/ x n ) dx n 1! 2! n! + ( L/ y) ( 2 L/ y 2) dy + dy ( n L/ y n ) dy n + R 1! 2! n! 576 Adjustment Computations: Spatial Data Analysis, Fifth Edition, Charles D. Ghilani 21 by John Wiley & Sons, Inc. Published by John Wiley & Sons, Inc. (C.2)

2 C.3 NUMERICAL EXAMPLE 577 In Equation (C.2), x and y are approximations for x and y; f (x, y ) is the nonlinear function evaluated at these approximations; R is the remainder, and dx and dy are corrections to the approximations, such that x = x + dx (C.3) y = y + dy A more exact Taylor series approximation is obtained by increasing the value of n in Equation (C.2). However, as the order of each successive term increases, its significance in the overall expression decreases. If all terms containing derivatives higher than the first are dropped, the following linear expression is obtained: ( ) ( ) L L L = f (x, y) = f (x, y ) + dx + dy (C.4) x y Once the initial approximations are selected, the only unknowns in Equation (C.4) are the corrections dx and dy. Of course, by dropping the higher-order terms from the Taylor series, Equation (C.4) becomes only a good approximation of the original equation. However, an iterative procedure can be followed in the solution that yields accurate results. This iterative procedure uses the following steps: Step 1: Determine initial approximations for the unknowns. They may be obtained by guessing or from observations. It should be understood that the closer the initial approximations are to the final solution, the faster the final solution will be obtained. For some problems, initial approximations are obtained from graphical solutions or computed from available data or observations. For others, the determination of initial approximations can involve considerable computational effort. Step 2: Substitute the initial approximations into Equation (C.4) and solve for the corrections dx and dy. Step 3: Calculate revised values of x and y using Equations (C.3). Step 4: Using these newly revised values for x and y, repeat steps 2 and 3. Step 5: Continue the procedure until the corrections dx and dy are small enough to bring x and y within tolerable accuracy. When this occurs, the solution is said to have converged. C.3 NUMERICAL EXAMPLE To clarify this procedure further, a numerical example will be solved. Example C.1 Linearize the following pair of nonlinear equations containing the two unknowns x and y and solve for the unknown parameters x and y. F: x + y 2y 2 = 4 G: x 2 + y 2 = 8

3 578 NONLINEAR EQUATIONS AND TAYLOR S THEOREM SOLUTION Determine the partial derivative for each equation with respect to each unknown. x = 1 x = 2x y = 1 4y y = 2y Compute an approximate solution. An estimate of x = 1andy = 1 is used for the approximations initially. First iteration: Write the linearized equations in the form of Equation (C.4). F: dx + [1 4(1)] dy = 4 [ (1) 2] G: 2(1) dx + 2(1) dy = 8 ( ) From the two equations above, solve for the unknowns dx and dy according to Equation (C.3): dx = 1.25 and dy = 1.75 Using this solution, determine updated values for x and y: x = x + dx = = 2.25 y = y + dy = = 2.75 Second iteration: Continue the procedure demonstrated for the first iteration. F: dx + [1 4(2.75)] dy = 4 [ (2.75) 2] G: 2(2.25)dx + 2(2.75) dy = 8 ( ) From the two equations above, dx =.25 and dy =.64, from which x = x + dx = = 2. y = y + dy = = 2.11 Third iteration: Substitute the new approximations into the equations and solve the equations for the unknown parameters. F: dx + [1 4(2.11)] dy = 4 [ (2.11) 2] G: 2(2.)dx + 2(2.11) dy = 8 ( )

4 C.4 USING MATRICES TO SOLVE NONLINEAR EQUATIONS 579 From the two equations above, dx =. and dy =.11, from which x = x + dx = = 2. y = y + dy = = 2. Fourth iteration: Continue the iteration procedure. F: dx + [1 4(2.)] dy = 4 [ (2.) 2] G: 2(2.)dx + 2(2.) dy = 8 ( ) Using the two preceding equations, the corrections to x and y are zero to the nearest hundredth. Thus, the solution has converged and the values of x = 2. and y = 2. are the desired unknowns. Note that the initial values for the approximations were relatively poor and four iterations were required to find the final solution. However, had better estimates been made (say, x = 2.1 and y = 1.9), the solution would have converged in one or two iterations and saved computational effort. Fortunately, there are accepted computational procedures to determine close approximations in most surveying problems. C.4 USING MATRICES TO SOLVE NONLINEAR EQUATIONS The example of Section C.3 could be solved using matrix methods. However, as in the algebraic approach, the equations must be linearized using Taylor s series. To facilitate linearization using Taylor s theorem, a Jacobian matrix (a matrix consisting of the partial derivatives taken with respect to the unknown variables) is formed. This is the coefficient matrix of the linearized equations. The Jacobian matrix for the example of Section C.3 is J = x x In the preceding Jacobian matrix, the first column contains the partial derivative for each equation with respect to x, and the second column contains each partial derivative of each equation with respect to y. The linearized form of the equations can then be expressed in matrix notation as y y JX = K (C.5) In Equation (C.5), J is the Jacobian matrix, X the matrix of unknown corrections dx and dy, and K the matrix of constants. Specifically, for the example of

5 58 NONLINEAR EQUATIONS AND TAYLOR S THEOREM Section C.3, these matrices are [ ] 1 1 4y J = 2x 2y [ ] dx X = dy [ ] 4 F(x, y ) K = 8 G(x, y ) where F(x, y ) and G(x, y ) are the equations F and G solved at the approximations of x and y. Beginning with a set of approximations x and y,thej and K matrices of Equation (C.5) are formed. X is computed using the matrix methods presented in Appendix B. Having updated the unknowns according to Equations (C.3), the J and K matrices are formed again and the solution for X computed again. This procedure is iterated until convergence is achieved. C.5 SIMPLE MATRIX EXAMPLE Example C.2 Find the solution of the nonlinear system of equations shown below using matrix methods. F: x 2 + 3xy 4y 2 = 6 G: x + xy y 2 = 3 SOLUTION The partial derivatives of functions F and G with respect to the unknown s, x and y, are = 2x + 3y x x = 1 + y y = 3x 8y y = x 2y Thus, the Jacobian matrix is J = x x y y [ ] 2x = + 3y 3x 8y 1 + y x 2y The system of equations to solve is [ ][ ] [ ] 2x + 3y 3x 8y dx 6 F(x, y ) = 1 + y x 2y dy 3 G(x, y )

6 C.6 PRACTICAL EXAMPLE 581 First iteration: Using the approximate solution of x = 3andy = yields [ ][ ] [ ] [ ] 6 9 dx = = 1 3 dy 3 3 The determinant for the Jacobian matrix above is 3(6) 9(1) = 9, and thus the matrix solution is [ ] dx = 1 [ ][ ] [ ] = dy Applying Equation (C.3), the new approximations for the unknowns used in the second iteration are [ ] [ ] [ ] [ ] [ ] [ ] x x dx = + = + = y dy.3.3 y Second iteration: [ ][ ] [ ] [ ] dx = = dy From this dx and dy are found to be.45 and.77, respectively. This makes the approximations for the unknowns of x and y for the third iteration 1.55 and 1.7, respectively. These procedures are followed until the final solution for x and y is found to be 2. and 1., respectively. Again, fewer iterations would have been required if the initial approximations had been closer to the final values. C.6 PRACTICAL EXAMPLE Example C.3 Assume that the x and y coordinates of three points on a circle have been observed. Their coordinates are (9.4, 5.6), (7.6, 7.2), and (3.8, 4.8), respectively. The equation for a circle with center (h, k) and radius r is (x h) 2 + (y k) 2 = r 2. Determine the coordinates of the center of the circle and its radius. SOLUTION The equation of a circle is rewritten as C (h, k, r) = (x h) 2 + (y k) 2 r 2 =. The partial derivatives with respect to the unknowns h, k, and r are C h = 2 (x h) C k = 2 (y k) C r = 2r

7 582 NONLINEAR EQUATIONS AND TAYLOR S THEOREM For each point observed, one equation is written, resulting in a system of three equations and three unknowns. The general linearized form of these equations expressed using matrices is C 1 h C 2 h C 3 h C 1 k C 2 k C 2 k C 1 r C 2 r C 3 r dh [ (x 1 h ) 2 + (y 1 k ) 2 r 2 ] dk = [ (x 2 h ) 2 + (y 2 k ) 2 r 2 ] dr [ (x 3 h ) 2 + (y 3 k ) 2 r 2 ] After taking partial derivatives, Equation (C.6) becomes (C.6) 2 (x 1 h ) 2 (y 1 k ) 2r dh [ (x 1 h ) 2 + (y 1 k ) 2 r 2 ] 2 (x 2 h ) 2 (y 2 k ) 2r dk = [ (x 2 h ) 2 + (y 2 k ) 2 r 2 ] 2 (x 3 h ) 2 (y 3 k ) 2r dr [ (x 3 h ) 2 + (y 3 k ) 2 r 2 ] (C.7) Equations (C.7) can be simplified by multiplying each side by 1/2. The resulting equations are (x 1 h ) (y 1 k ) r (x 2 h ) (y 2 k ) r dh.5 [ (x 1 h ) 2 + (y 1 k ) 2 r 2 ] dk =.5 [ (x 2 h ) 2 + (y 2 k ) 2 r 2 ] (x 3 h ) (y 3 k ) r dr.5 [ (x 3 h ) 2 + (y 3 k ) 2 r 2 ] (C.8) Assuming approximate values for h, k, andr as 7, 4.5, and 3, respectively, Equations (C.8) are (9.4 7) ( ) 3 (7.6 7) ( ) 3 dh.5 [ (9.4 7) 2 + ( ) 2 3 2] dk =.5 [ (7.6 7) 2 + ( ) 2 3 2] (3.8 7) ( ) 3 dr.5 [ (3.8 7) 2 + ( ) 2 3 2] (C.9) Simplifying Equations (C.9) yields dh dk = dr.665

8 C.7 CONCLUDING REMARKS 583 Solving this system gives the results dh dk = dr.7115 After applying these changes to the approximation values for the unknowns, updated values for h, k, and r of , , and , respectively, are obtained. The second iteration results in corrections of,, and Since the correction for r is still comparatively large, the iteration process must continue. After the third iteration, suitable convergence was achieved. The final values for h, k, andr are 6.72, 4.39, and 2.94, respectively, which are within.1 of a perfect solution. Sometimes, more than one method is available for solving a problem. For example, in Example C.3 an alternative linear form of the equation of a circle could be used. That equation is x 2 + y 2 + 2dx + 2ey + f =, where the center ofthecircleisat( d, e) and the circle s radius is found as d 2 + e 2 f.note that the equation is linear in terms of its unknowns (d, e, f ), and thus iterations are not necessary in solving for the unknowns. Writing a rearranged form of this equation for each of three observed sets of (x, y) coordinates yields 2dx 1 + 2ey 1 + f = ( x1 2 + ) y2 1 2dx 2 + 2ey 2 + f = ( x2 2 + ) y2 2 (C.1) 2dx 3 + 2ey 3 + f = ( x3 2 + ) y2 3 Equations (C.1) can, in turn, be represented in matrix notation as 2x 1 2y 1 1 2x 2 2y 2 1 d ( x1 2 e + ) y2 1 = ( x2 2 + ) y2 2 2x 3 2y 3 1 f ( x3 2 + ) y2 3 Solving this matrix system, the center of the circle is again found to be (6.72, 4.39) and its radius is determined to be C.7 CONCLUDING REMARKS In this appendix, the Taylor series has been applied to solve for the unknowns in nonlinear equations. Many equations in surveying, geodesy, and photogrammetry are nonlinear. In surveying, examples include the distance and angle formulas that are nonlinear in terms of station coordinates. The Taylor series is used to linearize these equations and find least squares solutions. Thus, when performing least squares adjustments of plane observations, the techniques presented in this appendix must be used in the solutions.

9 584 NONLINEAR EQUATIONS AND TAYLOR S THEOREM FIGURE C.1 Spreadsheet for Example C.3. The MATRIX software can be used in conjunction with a spreadsheet to solve these nonlinear problems. As an example, matrices for Example C.3 are formulated using a spreadsheet as shown in Figure C.1. The final values for h, k, andr are formed by summing the columns to their left. These values are used to formulate the J and K matrices. Thus, as the updates to the approximate values for the unknowns are entered into the appropriate cells, the matrices are updated automatically. These updated matrices can then be cut and pasted directly into the MATRIX editor, saved, and the problem solved. Of course, for small problems such as this, the matrix routines available in most spreadsheets can also be used. This was done in Figure C.1. However, as the problem grows, the MATRIX software can easily handle their solution. These procedures are demonstrated in the spreadsheet Example C-3.xls at the book s companion web site ( PROBLEMS C.1 Solve for the unknowns x and y in the following nonlinear equations using Taylor s theorem. (Use x = 5andy = 5 for initial approximations.) x 2 y 3x 2 = 75 x 2 y = 19 C.2 Solve for the unknown values of x, y, andz in the following three nonlinear equations using the Taylor series. (Use x = y = z = 2 for initial approximations.) x 2 y 2 + 2xy + z = 4 x + y + z = 4 2x 2 y + z 3 = 23

10 PROGRAMMING PROBLEMS 585 C.3 Use the MATRIX software to solve Problem C.1. C.4 Use the MATRIX software to solve Problem C.2. C.5 Find the center and radius of a circle using the equation (x h) 2 + (y k) 2 = r 2, given the coordinates of points A, B, andc on the circle. Follow the procedures discussed in Section C.5. Use initial approximations of h = 5, k = 4, and r = 2 for the first iteration. A: (7.2, 5.2) B: (4., 6.4) C : (4., 2.4) C.6 Repeat Problem C.5 using the linear equation x 2 + y 2 + 2dx + 2ey + f =. C.7 Repeat Problem C.5 using the points A: (.5,.7), B: (1.,.), C : (.7,.7). Use initial approximations of h =, k =, and r = 1. C.8 Repeat Problem C.7 using the linear equation x 2 + y 2 + 2dx + 2ey + f =. C.9 Use the ADJUST software to solve Problem C.5. C.1 Use the ADJUST software to solve Problem C.7. C.11 The distance formula between two stations i and j is (xj ) 2 ( ) 2 D ij = x i + yj y i Write the linearized form of this equation in terms of the variables x i, y i, x j,andy j. C.12 The azimuth formula between two stations i and j is α ij = tan 1 x j x i y j y i Write the linearized form of this equation in terms of the variables x i, y i, x j,andy j. C.13 The formula for an angle jik is α ik α ij, where α is defined in Problem C.12. Write the linearized form of this equation in terms of the variables x i, y i, x j, y j, x k,andy k. PROGRAMMING PROBLEMS C.14 Create a programmed package that solves Problem C.6. C.15 Create a programmed package that solves Problem C.5.