Transcript Approximate Solution of Equations Solve the equation Graphical Method
Approximate Solution of Equations Graphical Method
Example :
Solve the equation x
2
+ 2x - 1 = 0
Method 1 : Draw the graph y = x 2 + 2x - 1
-3 y = x 2 + 2x - 1 2 3 y -2 -1 1 -1 -2 1 2 3 x
x -3 2 y -2 -1 -1 -2 0 -1 1 2 2 7 3 14
-3
0.8
0.6
0.4
0.2
0 -0.2
-0.4
-0.6
-0.8
-2.8
-2.6
-2.4
x = -2.4
-2.2
x x -3 2.0
y -2.8 1.2
-2.6 0.6
-2.4 0.0
-2.2 -0.6
-2 -1.0
Method 1 : Draw the graph y = x 2 + 2x - 1
3 y 2 y = x 2 + 2x - 1 1 3 x -3 -2 -1 1 2 -1 -2 -3
0.8
0.6
0.4
0.2
0 -0.2
-0.4
-0.6
-0.8
0.2
0.4
0.6
x = 0.4
0.8
x x 0 y -1.0
0.2 -0.6
0.4 0.0
0.6 0.6
0.8 1.2
1 2.0
Method 2 : Use of the given graph y = x 2 •Rewrite the equation as x 2 = -2x + 1 •Add the line y = -2x + 1 on the same graph paper •Note the x -coordinates of the points of intersection
-4 -3 -2 -1 -1 -2 2 1 5 4 3 7 y 6 y = x 2 1 2 3 4 x 5 y = -2x+1
Drawing Software :
Winplot
http://math.exeter.edu/rparris
Graphmatica
http://www.pair.com/ksoft
Answer to Worksheet Exercise 1: Use the graph y = 2x 2 x 2 – x – 1 = 0 to solve Rewrite the equation as x 2 2x 2 = x + 1 = 2x + 2 Draw the line y = 2x + 2
y = 2x 2
(-0.6, 0.7)
-2 2 1 6
y y
(1.6, 5.3)
5 4 3 -1 -0.6
0 y = 2x+2 1 1.6
2
x x
Answer to Worksheet Exercise 1: Use the graph y = 2x 2 4x 2 + x – 6 = 0 to solve Rewrite the equation as 4x 2 2x 2 = -x + 6 = -x/2 + 3 Draw the line y = -x/2 + 3
y = 2x 2 -2 -1.4
-1 2 1 6
y y
5 4 3 0 y = -x/2+3 1 1.1
2
x x
Approximate Solution of Equations Method of Bisection
Example :
Solve the equation x
3
- 3x
2
+ 5 = 0
Step 1 : Locate the root
y = x 3 - 3x 2 + 5 5 y -3 -2 -1 4 3 2 1 -1 1 2 3 x 4
x -2 -1 0 1 2
-2
The root lies between -2 and -1 y -15 1 5 3 1
Step 2 : Find the mid-point of the interval that contains the root
-2 -1
Mid-point = 1 (-2 + -1) = -1.5
2
Step 3 : Choose the half-interval that contains the root
-2 -1.5
-1
f(x) = x 3 - 3x 2 + 5
x f(x)
-2 -1.5
-15 -5.13 -1 1
-1.5
-1.25
-1 x f(x)
-1.5
-5.13 -1.25
-1.64 -1 1
-1.25
-1.125
-1 x f(x)
-1.25
-1.641 -1.125
-0.221 -1 1
-1.125
-1.0625
-1 x f(x)
-1.125
-0.221 -1.0625
0.4138
-1 1 Root of x 3 - 3x 2 + 5 = 0 is
-1.1
(1 d.p.)
-1.125
-1.0938
-1.0625
x f(x)
-1.125 -1.09375 -1.063
-0.221 0.1027 0.4138
Method of Bisection
Find the interval that contains the root No Find the mid-point Answer Yes Precise enough ?
Choose the half-interval that contains the root
a -
-2
-2 -1.5
-1.25
-1.125
-1.125
+ -1
b -1 -1 -1 -1 -1.0625
f(-2) = -15
a
b
2 -1.5
-1.25
-1.125
-1.0625
f(-1) = 1
b f
2 -5.13
-1.64
-0.22
0.42
Root of x 3 - 3x 2 + 5 = 0 is –1.1
(1d.p.)
Method of Bisection Use of Excel Spreadsheet
a
-2 -1.5
-1.25
-1.125
-1.125
-1.125
-1.1094
b
-1 -1 -1 -1 -1.0625
-1.0938
-1.0938
(a+b)/2
-1.5
-1.25
-1.125
-1.0625
-1.0938
-1.1094
-1.1016
f(a)
-15 -5.125
-1.6406
f(b)
1 1
f(mid-value)
-5.125
-1.640625
1 -0.2207031
-0.2207
1 -0.2207 0.41382
0.4138184
0.1026917
-0.2207 0.10269 -0.0574608
-0.0575 0.10269
0.0230002
Answer to Worksheet Exercise 3: Find a root of 2x 4 – 3x – 5 = 0
x f(x)
0 -5 1 -6 2 21 Bracketing interval : 1 < x 0 < 2 3 148
a
1 1 1.25
1.375
1.4375
1.46875
b
2 1.5
1.5
1.5
1.5
1.5
a
b
2
1.5
1.25
1.375
1.4375
1.46875
f b
2
0.625
-3.87
-1.98
-1.98
-0.77
The root is 1.5
(2 sig. fig.)
Answer to Worksheet Exercise 4: Find a root of x 3 – 7x +2 = 0
a
-3 -3 -3 -2.875
-2.8125
-2.78125
b
-2 -2.5
-2.75
-2.75
-2.75
-2.75
-2.78125 -2.765625
-2.78125 -2.773438
-2.78125 -2.777344
(a+b)/2
-2.5
-2.75
-2.875
f(a)
-4 -4
f(b)
3.875
8
f(mid-value)
3.875
0.453125
-4 0.45313 -1.638672
-2.8125 -1.6387 0.45313 -0.559814
-2.78125 -0.5598 0.45313 -0.045197
-2.76563 -0.0452 0.45313 0.2059898
-2.77344 -0.0452 0.20599 0.0809045
-2.77734 -0.0452
0.0809 0.0179811
-2.7793 -0.0452 0.01798 -0.013576
Useful Websites for Method of Bisection http://www.rohan.sdsu.edu/faculty/ symbol/bisect.html
http://www.krellinst.org/UCES/archive/ resources/roots/node2.html