Newtons Method for Systems of Nonlinear Equations and Root Finding Methods Comparison
Learn how to solve systems of nonlinear equations using Newton's method, with examples and solutions provided. Explore the advantages and disadvantages of different root finding methods such as Bisection, Newton, and Secant methods. Understand the pros and cons to choose the right approach for efficient and accurate results.
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Newtons Method for Systems of Non Linear Equations = : initial an Iteration guess of the root of ( ) 0 Given X F x 0 ' Newton s 1 = ( ' ) ( ) X X F X F X + 1 k k k k f f 1 1 x x ( , ,...) f x x 1 2 1 1 2 f f = = ( ) ( , ,...) , ( ' ) 2 2 F X f x x F X 2 1 2 x x 1 2 1
Example Solve the following system of equations: 2 + = 0 5 . 0 y x x 2 = x 5 0 = x xy y = Initial guess , 1 0 y 2 2 x 1 1 x 1 x + x 5 y 0 5 . y x x = = = , ' , F F X 0 2 2 5 5 1 0 y xy 2
Solution Using Newtons Method Iteration : 1 + x 5 y 2 0 5 . 2 x 1 1 x 1 1 x 0 5 . y x x = = = = = , ' F F 2 1 2 5 5 1 2 6 y xy 1 . 1 1 1 1 0 5 . 25 = = X : . 0 1 0 2 6 1 25 Iteration 2 0.0625 1.5 1 = = = , ' F F . 1 - 0.25 25 . 7 25 1 . 1 . 1 25 1.5 1 0.0625 2332 = = X . 0 . 1 2 25 25 . 7 25 - 0.25 2126 . 0 3
Example Try this Solve the following system of equations: 2 + = 1 0 y x x 2 2 = x 2 0 = x y y = Initial guess , 0 0 y x + y 2 2 1 1 y 0 x 1 y x x = = = , ' , F F X 0 2 2 2 4 1 0 2 x y 4
Example Solution 0 1 2 3 4 5 Iteration __________ __________ __________ __________ __________ __________ _ 0 1 6 . 0 0.5287 0.5257 0.5257 X k 0 0 2 . 0 0.1969 0.1980 0.1980 5
Comparison of Root Finding Methods Advantages/disadvantages Examples 6
Summary Method Bisection Pros - Easy, Reliable, Convergent - One function evaluation per iteration - No knowledge of derivative is needed - Fast (if near the root) - Two function evaluations per iteration Cons - Slow - Needs an interval [a,b] containing the root, i.e., f(a)f(b)<0 Newton - May diverge - Needs derivative and an initial guess x0 such that f (x0) is nonzero Secant - Fast (slower than Newton) - One function evaluation per iteration - No knowledge of derivative is needed - May diverge - Needs two initial points guess x0, x1 such that f(x0)- f(x1) is nonzero 7
Example Secant Use = x f method x to find the root of : 6 ( ) 1 x = 5 . 1 = Two initial points 1 x and x 0 1 ( x ) x x = 1 x ( ) i i x x f x + 1 i i i ( ) ( ) f f 1 i i 8
Solution _______________________________ k xk f(xk) _______________________________ 0 1.0000 -1.0000 1 1.5000 8.8906 2 1.0506 -0.7062 3 1.0836 -0.4645 4 1.1472 0.1321 5 1.1331 -0.0165 6 1.1347 -0.0005 9
Example Use Newton' = x Method s to find a root of : 3 ( ) 1 f x x . 1 = Use the initial point : x 0 Stop after + x iterations three k or , if . 0 001 , or x x 1 k f if ( ) 0001 . 0 . k 10
Five Iterations of the Solution k xk f(xk) ______________________________________ 0 1.0000 -1.0000 2.0000 1 1.5000 0.8750 5.7500 0.1522 2 1.3478 0.1007 4.4499 0.0226 3 1.3252 0.0021 4.2685 0.0005 4 1.3247 0.0000 4.2646 0.0000 5 1.3247 0.0000 4.2646 0.0000 f (xk) ERROR 11
Example Use Newton' = e Method s to find a root of : x ( ) f x x = Use the initial point : . 1 x 0 Stop after x iterations three k x or , if . 0 001 , or + 1 x k if ( ) 0001 . 0 . f k 12
Example Use Newton' = Method s to find = a root of : x x ( ) , ( ' ) 1 f x e x f x e ( ) f x ( ) ( ' ) k x f x f x k k k ( ' ) f x k 1.0000 - 0.6321 - 1.3679 0.4621 0.5379 0.0461 - 1.5840 - 0.0291 0.5670 0.0002 - 1.5672 - 0.0002 0.5671 0.0000 - 1.5671 - 0.0000 13
Example Estimates of the root of: x-cos(x)=0. 0.60000000000000 Initial guess 0.74401731944598 1 correct digit 0.73909047688624 4 correct digits 0.73908513322147 10 correct digits 0.73908513321516 14 correct digits 14
Example In estimating the root of: x-cos(x)=0, to get more than 13 correct digits: 4 iterations of Newton (x0=0.8) 43 iterations of Bisection method (initial interval [0.6, 0.8]) 5 iterations of Secant method ( x0=0.6, x1=0.8) 15
