Approximations To Roots (1) Halving The Interval
12 x1 t04 07 approximations to roots (2012)
Jun 23, 2015
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Approximations To Roots(1) Halving The Interval
Approximations To Roots(1) Halving The Interval
y
x
If y = f(x) is a continuous function over the interval , and f(a) and f(b) are opposite in sign,
bxa
xfy
Approximations To Roots(1) Halving The Interval
y
x
If y = f(x) is a continuous function over the interval , and f(a) and f(b) are opposite in sign,
bxa
xfy af
a bf
b
Approximations To Roots(1) Halving The Interval
y
x
If y = f(x) is a continuous function over the interval , and f(a) and f(b) are opposite in sign,
bxa
xfy af
a bf
b
then at least one root of the equation f(x) = 0 lies in the interval bxa
e.g Find an approximation to two decimal places for a root of in the interval01924 xx 31 x
e.g Find an approximation to two decimal places for a root of in the interval01924 xx
01619211 4
f
31 x
1924 xxxf 068
193233 4
f
e.g Find an approximation to two decimal places for a root of in the interval01924 xx
01619211 4
f
22
311
x
31 x
1924 xxxf 068
193233 4
f
01
192222 4
f
1 2 3
e.g Find an approximation to two decimal places for a root of in the interval01924 xx
01619211 4
f
21intervalin liessolution x
22
311
x
31 x
1924 xxxf 068
193233 4
f
01
192222 4
f
1 2 3
e.g Find an approximation to two decimal places for a root of in the interval01924 xx
01619211 4
f
21intervalin liessolution x
22
311
x
31 x
1924 xxxf 068
193233 4
f
01
192222 4
f
1 2 3
5.12
212
x
09.10195.125.15.1 4
f
1 1.5 2
e.g Find an approximation to two decimal places for a root of in the interval01924 xx
01619211 4
f
21intervalin liessolution x
22
311
x
31 x
1924 xxxf 068
193233 4
f
01
192222 4
f
1 2 3
21.5intervalin liessolution x
5.12
212
x
09.10195.125.15.1 4
f
1 1.5 2
75.12
25.13
x
012.61975.1275.175.1 4
f
1.5 1.75 2
21.75intervalin liessolution x
75.12
25.13
x
012.61975.1275.175.1 4
f
1.5 1.75 2
21.75intervalin liessolution x
75.12
25.13
x
012.61975.1275.175.1 4
f
1.5 1.75 2
88.12
275.14
x
075.21988.1288.188.1 4
f
1.75 1.88 2
21.75intervalin liessolution x
75.12
25.13
x
012.61975.1275.175.1 4
f
1.5 1.75 2
21.88intervalin liessolution x
88.12
275.14
x
075.21988.1288.188.1 4
f
1.75 1.88 2
21.75intervalin liessolution x
75.12
25.13
x
012.61975.1275.175.1 4
f
1.5 1.75 2
21.88intervalin liessolution x
88.12
275.14
x
075.21988.1288.188.1 4
f
1.75 1.88 2
94.12
288.15
x
096.01994.1294.194.1 4
f
1.88 1.94 2
21.75intervalin liessolution x
75.12
25.13
x
012.61975.1275.175.1 4
f
1.5 1.75 2
21.88intervalin liessolution x
88.12
275.14
x
075.21988.1288.188.1 4
f
1.75 1.88 2
21.94intervalin liessolution x
94.12
288.15
x
096.01994.1294.194.1 4
f
1.88 1.94 2
97.12
294.16
x
0001.01997.1297.197.1 4
f
1.94 1.97 2
97.11.94intervalin liessolution x
97.12
294.16
x
0001.01997.1297.197.1 4
f
1.94 1.97 2
97.11.94intervalin liessolution x
97.12
294.16
x
0001.01997.1297.197.1 4
f
1.94 1.97 2
96.12
97.194.17
x
032.0
1996.1296.196.1 4
f 1.94 1.96 1.97
97.11.94intervalin liessolution x
97.12
294.16
x
0001.01997.1297.197.1 4
f
1.94 1.97 2
97.11.96intervalin liessolution x
96.12
97.194.17
x
032.0
1996.1296.196.1 4
f 1.94 1.96 1.97
97.11.94intervalin liessolution x
97.12
294.16
x
0001.01997.1297.197.1 4
f
1.94 1.97 2
97.11.96intervalin liessolution x
96.12
97.194.17
x
032.0
1996.1296.196.1 4
f 1.94 1.96 1.97
97.12
97.196.18
x
97.11.94intervalin liessolution x
97.12
294.16
x
0001.01997.1297.197.1 4
f
1.94 1.97 2
97.11.96intervalin liessolution x
96.12
97.194.17
x
032.0
1996.1296.196.1 4
f 1.94 1.96 1.97
1.97isroot for theion approximatan x97.1
297.196.1
8
x
(2) Newton’s Method of Approximation
(2) Newton’s Method of Approximationy
x
(2) Newton’s Method of Approximationy
x
If is a good first approximation to a root of the equation f(x) = 0, then a closer approximation is given by;
0x
0
001 xf
xfxx
(2) Newton’s Method of Approximationy
x
If is a good first approximation to a root of the equation f(x) = 0, then a closer approximation is given by;
0x
0
001 xf
xfxx
by;given are,,,,ionsapproximat Successive 132 nn xxxx n
nnn xf
xfxx
1
(2) Newton’s Method of Approximationy
x
If is a good first approximation to a root of the equation f(x) = 0, then a closer approximation is given by;
0x
0
001 xf
xfxx
by;given are,,,,ionsapproximat Successive 132 nn xxxx n
nnn xf
xfxx
1
NOTE:must be a good first
approximationNewton’s method finds where the tangent at cuts the x axis
0x
0x
(2) Newton’s Method of Approximationy
x
xfy
If is a good first approximation to a root of the equation f(x) = 0, then a closer approximation is given by;
0x
0
001 xf
xfxx
by;given are,,,,ionsapproximat Successive 132 nn xxxx n
nnn xf
xfxx
1
NOTE:must be a good first
approximationNewton’s method finds where the tangent at cuts the x axis
0x
0x
0x
(2) Newton’s Method of Approximationy
x
xfy
If is a good first approximation to a root of the equation f(x) = 0, then a closer approximation is given by;
0x
0
001 xf
xfxx
by;given are,,,,ionsapproximat Successive 132 nn xxxx n
nnn xf
xfxx
1
NOTE:must be a good first
approximationNewton’s method finds where the tangent at cuts the x axis
0x
0x
0x1x
(2) Newton’s Method of Approximationy
x
xfy
If is a good first approximation to a root of the equation f(x) = 0, then a closer approximation is given by;
0x
0
001 xf
xfxx
by;given are,,,,ionsapproximat Successive 132 nn xxxx n
nnn xf
xfxx
1
NOTE:must be a good first
approximationNewton’s method finds where the tangent at cuts the x axis
0x
0x
0x1x
failwillmethodtheaxis || tangenti.e.0If 0
xxf
e.g Find an approximation to two decimal places for a root of 01924 xx
e.g Find an approximation to two decimal places for a root of 01924 xx
24 3 xxf
1924 xxxf
e.g Find an approximation to two decimal places for a root of 01924 xx
24 3 xxf
1924 xxxf
5.10 x 9375.10
195.125.15.1 4
f
5.1525.145.1 3
f
e.g Find an approximation to two decimal places for a root of 01924 xx
24 3 xxf
1924 xxxf
5.10 x 9375.10
195.125.15.1 4
f
5.1525.145.1 3
f
21.25.15
9375.105.1
0
001
xfxfxx
e.g Find an approximation to two decimal places for a root of 01924 xx
24 3 xxf
1924 xxxf
5.10 x 9375.10
195.125.15.1 4
f
5.1525.145.1 3
f
21.25.15
9375.105.1
0
001
xfxfxx
2744.91921.2221.221.2 4
f
1754.45
221.2421.2 3
f
00.21754.45
2744.921.22
x
00.21754.45
2744.921.22
x 1
192222 4
f
35
2242 3
f
00.21754.45
2744.921.22
x 1
192222 4
f
35
2242 3
f
97.135123
x
00.21754.45
2744.921.22
x 1
192222 4
f
35
2242 3
f
97.135123
x 001.0
1997.1297.197.1 4
f
58.32
297.1497.1 3
f
00.21754.45
2744.921.22
x 1
192222 4
f
35
2242 3
f
97.135123
x 001.0
1997.1297.197.1 4
f
58.32
297.1497.1 3
f
97.158.32
001.097.14
x
00.21754.45
2744.921.22
x 1
192222 4
f
35
2242 3
f
97.135123
x 001.0
1997.1297.197.1 4
f
58.32
297.1497.1 3
f
97.158.32
001.097.14
x
rootfor theion approximatbetter ais1.97 x
( ) Use Newton's Method to obtain an approximation to 23 correct to two decimal placesii
( ) Use Newton's Method to obtain an approximation to 23 correct to two decimal placesii
2 23f x x
( ) Use Newton's Method to obtain an approximation to 23 correct to two decimal placesii
2 23f x x
2f x x
( ) Use Newton's Method to obtain an approximation to 23 correct to two decimal placesii
2 23f x x
2f x x
21
11
21
1
232
232
nn n
n
n
n
xx xx
xx
( ) Use Newton's Method to obtain an approximation to 23 correct to two decimal placesii
2 23f x x
2f x x
21
11
21
1
232
232
nn n
n
n
n
xx xx
xx
0 5x
( ) Use Newton's Method to obtain an approximation to 23 correct to two decimal placesii
2 23f x x
2f x x
21
11
21
1
232
232
nn n
n
n
n
xx xx
xx
0 5x
2
15 232 5
x
1 4.8x
( ) Use Newton's Method to obtain an approximation to 23 correct to two decimal placesii
2 23f x x
2f x x
21
11
21
1
232
232
nn n
n
n
n
xx xx
xx
0 5x
2
15 232 5
x
1 4.8x
2
24.8 232 4.8
x
2 4.795833333x
2 4.80 (to 2 dp)x
( ) Use Newton's Method to obtain an approximation to 23 correct to two decimal placesii
2 23f x x
2f x x
21
11
21
1
232
232
nn n
n
n
n
xx xx
xx
0 5x
2
15 232 5
x
1 4.8x
2
24.8 232 4.8
x
2 4.795833333x
2 4.80 (to 2 dp)x
23 4.80 (to 2 dp)
Other Possible Problems with Newton’s Method
Other Possible Problems with Newton’s MethodApproximations oscillate
Other Possible Problems with Newton’s Method
y
x
Approximations oscillate
Other Possible Problems with Newton’s Method
y
x
Approximations oscillate
want to find this root
Other Possible Problems with Newton’s Method
y
x
Approximations oscillate
1x
want to find this root
2x
Other Possible Problems with Newton’s Method
y
x
Approximations oscillate
1x 2x
want to find this root
Other Possible Problems with Newton’s Method
y
x
Approximations oscillate
1x 2x
wrong side of stationary pointconverges to wrong root
want to find this root
Other Possible Problems with Newton’s Method
y
x
Approximations oscillate
1x 2x
wrong side of stationary pointconverges to wrong rooty
x
want to find this root
Other Possible Problems with Newton’s Method
y
x
Approximations oscillate
1x 2x
wrong side of stationary pointconverges to wrong rooty
x
want to find this root
want to find this root
Other Possible Problems with Newton’s Method
y
x
Approximations oscillate
1x 2x
wrong side of stationary pointconverges to wrong rooty
x1x 2x
want to find this root
want to find this root
Other Possible Problems with Newton’s Method
y
x
Approximations oscillate
1x 2x
wrong side of stationary pointconverges to wrong rooty
x1x 2x
want to find this root
want to find this root
Other Possible Problems with Newton’s Method
y
x
Approximations oscillate
1x 2x
wrong side of stationary pointconverges to wrong rooty
x1x 2x
want to find this root
want to find this root
Exercise 6E; 1, 3ac, 6adf, 8a, 10, 12
Related Documents
