Post on 18-Jul-2015
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
so is the solution closer to 1.96 or 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
so is the solution closer to 1.96 or 1.97?
1.96 1.965 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
so is the solution closer to 1.96 or 1.97?
1.96 1.965 1.97
41.965 1.965 2 1.965 190.16 0
f
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
so is the solution closer to 1.96 or 1.97?
1.96 1.965 1.97
41.965 1.965 2 1.965 190.16 0
f
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 x
so is the solution closer to 1.96 or 1.97?
1.96 1.965 1.97
41.965 1.965 2 1.965 190.16 0
f
(2) Newton’s Method of Approximationy
x
xfy
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
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
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
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
(2) Newton’s Method of Approximationy
x
xfy
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
Using the tangent at to find0x 1x
(2) Newton’s Method of Approximationy
x
xfy
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
Using the tangent at to find0x 1x
0
0 1
0slope of tangent
f xx x
(2) Newton’s Method of Approximationy
x
xfy
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
Using the tangent at to find0x 1x
0
0 1
0slope of tangent
f xx x
00
0 1
0f xf x
x x
0 1 0 0
00 1
0
x x f x f x
f xx x
f 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
0 1 0 0
00 1
0
x x f x f x
f xx x
f 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
0 1 0 0
00 1
0
x x f x f x
f xx x
f x
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