12 x1 t04 07 approximations to roots (2013)

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