SOLUTIONS TO NON-LINEAR EQUATION
1
a. BISECTION METHOD
Basis of Bisection Method Theorem An equation f(x)=0, where f(x) is a real continuous function, has at least one root between xl and xu if f(xl) f(xu) < 0. f(x)
xď Ź xu
x
Figure 1 At least one root exists between the two points if the function is real, continuous, and changes sign.
Basis of Bisection Method f(x)
xď Ź
xu
x
) Figure 2 If function f ( xdoes not change sign between two points, 0 roots of the equation f ( x ) = may still exist between the two points.
Basis of Bisection Method f(x)
f(x)
x
xu
x
x
xu x
) Figure 3 If the function f ( xdoes not change sign between two ) =0 f ( xbetween points, there may not be any roots for the equation the two points.
Basis of Bisection Method f(x)
x
xu ď Ź
x
Figure 4 If the function f ( x ) changes sign between two points, more than one root for the equation f ( x ) = 0 may exist between the two points.
Step 1 Choose x and xu as two guesses for the root such that f(x) f(xu) < 0, or in other words, f(x) changes sign between x and xu. This was demonstrated in Figure 1. f(x)
x xu
Figure 1
x
Step 2 Estimate the root, xm of the equation f (x) = 0 as the mid point between x and xu as f(x)
x + xu xm = 2 x
xm xu
Figure 5 Estimate of xm
x
Step 3 Now check the following a) If
f ( xm ) = (+ )
b) If
f ( xm ) = (−)
c) If
f ( xm ) = 0
, then x = x ; xu = xm. , then x = xm; xu = xu. , then the root is x m.
Step 4 Find the new estimate of the root x + xu xm = 2
Find the absolute relative approximate error ∈a =
old x new − x m m
x
new m
×100
where xmold = previous estimate of root xmnew = current estimate of root
Example:
Locate the first non-trivial root of sin x = x2. Use Bisection Method with initial interval from 0.5 to 1.0. Perform the εa <= 2% computation until
Solution: f(x) = x2-sin x
x∈(0.5,1)
εa <= 2%
ε a
i
Xl
Xu
Xm
F(xm)
0
0.5
1.0
0.75
-
1
0.75
1.0
0.875
-
14.29
2
0.875
1.0
0.9375
+
6.67
3
0.875
0.9375
0.9063
+
3.44
4
0.875
0.9063
0.8907
+
1.75
-
Example:
Locate the first non-trivial root of sin x = x2. Use Bisection Method with initial interval from 0.5 to 1.0. Perform the εa <= 2% computation until
Solution: f(x) = x2-sin x
x∈(0.5,1)
εa <= 2%
ε a
i
Xl
Xu
Xm
F(xm)
1
0.5000
1.0000
0.7500
-0.1191
2
0.7500
1.0000
0.8750
-0.0019 14.2857
3
0.8750
1.0000
0.9375
0.0728
6.6667
4
0.8750
0.9375
0.9063
0.0341
3.4483
5
0.8750
0.9063
0.8906
0.0157
1.7544
-
Advantages Always convergent ď&#x201A;&#x2014; The root bracket gets halved with each iteration - guaranteed. ď&#x201A;&#x2014;
Drawbacks ď Ž ď Ž
Slow convergence If one of the initial guesses is close to the root, the convergence is slower
Drawbacks ď&#x201A;&#x2014;
If a function f(x) is such that it just touches the x-axis it will be unable to find the lower and upper guesses. f(x)
f ( x) = x x
2
Drawbacks ď Ž
Function changes sign but root does not exist f(x)
1 f ( x) = x x
NEXT TOPIC:
b. REGULA-FALSI METHOD (FALSE POSITION METHOD)
f ( xu ) − f ( xl ) = xu −x r xr − xl
xr f ( xu ) − xl f ( xu ) = − xu f ( x l ) + xr f ( xl ) xr [ f ( xu ) − f ( xl )] = xl f ( xu ) − xu f ( xl ) xl f ( xu ) − xu f ( xl ) xr = f ( xu ) − f ( xl )
Example:
Locate the first non-trivial root of sin x = x2. Use Bisection Method with initial interval from 0.5 to 1.0. Perform the computation until εa <= 2%
Solution: f(x) = x2-sin x
x∈(0.5,1)
εa <= 2%
xl f ( xu ) − xu f ( xl ) xr = f ( xu ) − f ( xl ) A
B
C
E= D F(xu)
AD-BC D-B
E
F
Xr
F(xr)
ε a
i
Xl
F(xl)
Xu
1
0.5
-0.2294
1.0
0.1585 0.7957 -0.0812
2
0.7957 -0.0812
1.0
0.1585 0.8649 -0.0130
3
0.8649 -0.0130
1.0
0.1585 0.8751 -0.0018 1.17
-
8
Example:
Locate the first non-trivial root of sin x = x2. Use Bisection Method with initial interval from 0.5 to 1.0. Perform the εa <= 2% computation until
Solution: f(x) = x2-sin x i
A Xl
x∈(0.5,1) B F(xl)
C Xu
εa <= 2% D F(xu)
E Xm
F F(xm)
ε a
1
0.5000 -0.2294 1.0000 0.1585 0.7957 -0.0812
2
0.7957 -0.0812 1.0000 0.1585 0.8649 -0.0130
3
0.8649 -0.0130 1.0000 0.1585 0.8751 -0.0018 1.17%
xl f ( xu ) − xu f ( xl ) xr = f ( xu ) − f ( xl )
Xm=
AD-BC D-BC
-
8%
NEXT TOPIC:
c. SIMPLE FIXED-POINT ITERATION
SFPI - also known as Open Method
f ( x) = 0 â&#x2021;&#x201D; x = g ( x)
Example:
Locate the first non-trivial root of sin x = x2. Use Bisection Method with initial interval from 0.5 to 1.0. Perform the εa <= 2% computation until
Solution: f(x) = x2-sin x
x∈(0.5,1)
εa <= 2%
x = sin x sin x x= x −1 2 x = sin x
(1) (2) (3)
If x = sin x
At x = 0.5 i 0 1 2 3 4
xi 0.5 0.6924 0.7990 0.8466 0.8655
x (i+1) 0.6924 0.7990 0.8466 0.8655 0.8726
Ea 13.34 5.62 2.18 0.81
xi 1 0.9173
x (i+1) 0.9173 0.8911
Ea 2.95
At x = 1.0 i 0 1 2
If x = sin x
At x = 0.5 i 0 1 2 3 4
xi 0.5 0.6924 0.7990 0.8466 0.8654
x (i+1) 0.6924 0.7990 0.8466 0.8654 0.8726
Ea 13.34 5.62 2.18 0.82
xi 1 0.9173 0.8911
x (i+1) 0.9173 0.8911 0.8819
Ea 2.95 1.04
At x = 1.0 i 0 1 2
sin x if x = x
At x = 0.5 i 0 1 2 3
xi 0.5 0.9589
x (i+1) 0.9589 0.8537
Ea 12.32
xi 1 0.8415
x (i+1) 0.8415 0.8861
Ea 5.04
At x = 1.0 i 0 1 2
sin x if x = x
At x = 0.5 i 0 1 2 3
xi 0.5 0.9589 0.8537 0.8829
x (i+1) 0.9589 0.8537 0.8829 0.8751
Ea 12.32 3.31 0.90
xi 1 0.8415 0.8861
x (i+1) 0.8415 0.8861 0.8742
Ea 5.04 1.36
At x = 1.0 i 0 1 2
Assignment: To be submitted on or before July 12, 2011
1.
Solve V1,V2,V3 of the circuit below using any of the iterative method.
2.
Solve for the roots of f(x) = with an interval of (0.5,1). Use BM, RFM, SFPI.
sin x â&#x2C6;&#x2019; x