Taylor Series

SOLUTION OFNON-LINEAR EQUATIONS

• All equations used in horizontal adjustment are non-linear.

• Solution involves approximating solution using 1'st order Taylor series expansion, and

• Then solving system for corrections to approximate solution.

• Repeat solving system of linearized equations for corrections until corrections become small.

• This process of solving equations is known as: ITERATING

Taylor’s SeriesGiven a function, L = f(x,y)

!2!1

!2!1),(),(

2

02

2

0

2

02

2

000

dyy

LdyyL

dxx

LdxxL

yxfyxfL

Taylor’s Series

• The series is also non-linear (unknowns are the dx’s, dy’s, and higher order terms)

• Therefore, truncate the series after only the first order terms, which makes the equation an approximation

• Initial approximations generally need to be reasonably close in order for the solution to converge

dyyL

dxxL

yxfyxfL00

00 ),(),(

Solution

• Determine initial approximations (closer is better)

• Form the (first order) equations

• Solve for corrections, dx and dy

• Add corrections to approximations to get improved values

• Iterate until the solution converges

Example C.1Solve the following pair of non-linear equations. Use initial approximation of 1 (one) for both x and y.

8),(

42),(22

2

yxyxG

yyxyxF

First, determine the partial derivatives

1

xF

yyF

41

xxG

2 y

yG

2

Partials

Write the Linearized Equations

8)1(2)1(2)1()1(),(

4)]1(41[)1(211),(22

2

dydxyxG

dydxyxF

Simplify

622

43

dydx

dydx

Solve

75.1

25.1

8)6(1)4(2

8)6(3)4(2

6

4

12

32

81

6

4

22

31

dy

dx

dy

dx

dy

dx

New approximations:75.275.11

25.225.11

dy

dx

Linearized Equations – Iteration 2

8)75.2(2)25.2(2)75.2()25.2(),(

4)]75.2(41[)75.2(275.225.2),(22

2

dydxyxG

dydxyxF

Simplify

625.45.55.4

125.610

dydx

dydx

Solve – Iteration 2

64.0

25.0

5.50)625.4(1)125.6(5.4

5.50)625.4(10)125.6(5.5

625.4

125.6

15.4

105.5

5.501

625.4

125.6

5.55.4

101

dy

dx

dy

dx

dy

dx

New approximations:11.264.075.2

00.225.025.2

dy

dx

Iteration 3

Same procedure yields: dx = 0.00 and dy = -0.11

This results in new approximations of x = 2.00 and y = 2.00

Further iterations are negligible

General Matrix Form

• The coefficient matrix formed by the partial derivatives of the functions with respect to the variables is the Jacobian matrix

• It can be used directly in a general matrix form

y

Gx

Gy

Fx

FJ

General Form for Example

00

0

22

411

yx

yJ

dy

dxX

JX = K

),(8

),(4

00

00

yxG

yxFK

Circle Example

222 )()( rkyhx

Determine the equation of a circle that passes through the points (9.4, 5.6), (7.6, 7.2), and (3.8, 4.8).

Initial approximations for unknown and circle equation:Center point: (7, 4.5), Radius: 3

Linearizing

0)()(),,( 222 rkyhxrkhCRearranged

Set Up General Matrix Form

Substitute the Values and Solve

Continue Until Converged