# Find the positive root of correct to four decimal places ... · PDF file = 0.75, 0.75 = 0.0183 = + The root lies between and 1 i.e. 0.5 and 0.75, put = 1 2 = 0.5 + 0.75 2 = 0.625,

• View
3

0

Embed Size (px)

### Text of Find the positive root of correct to four decimal places ... · PDF file = 0.75, 0.75 = 0.0183...

• The Bisection method or BOLZANOβs method or Interval halving method:

Find the positive root of ππ β π = π correct to four decimal places by bisection method.

Solution:

Let π π₯ = π₯3 β π₯ β 1

Here π 0 = β1 = βπ£π, π 1 = βπ£π, π 2 = 5 = +π£π

Hence a root lies between 1 and 2.

Here π = 1 πππ π = 2

π»πππ π₯0 = π + π

2 =

1 + 2

2 = 1.5, π π₯0 = π(1.5) = 0.875 = +π£π

Since π π is negative and π(π₯0) is positive

The root lies between π and π₯0 i.e. 1 and 1.5, put π = π₯0

π₯1 = 1 + 1.5

2 = 1.25, π 1.25 = β0.2969 = βπ£π

The root lies between π₯1 and π i.e. 1.25 and 1.5, put π = π₯1

π₯2 = 1.25 + 1.5

2 = 1.375, π 1.375 = 0.2246 = +π£π

Therefore the root lies between π and π₯2 i.e. 1.25 and 1.375

Proceeding like this and these values are put in a tabular form as follows

π a b π₯π π(π₯π)

0 1 2 1.5 0.875

1 1 1.5 1.25 -0.2969

2 1.25 1.5 1.375 0.2246

3 1.25 1.375 1.3125 -0.0515

4 1.3125 1.375 1.3438 0.0828

5 1.3125 1.3438 1.3282 0.0149

6 1.3125 1.3282 1.3204 -0.0183

7 1.3204 1.3282 1.3243 -0.0018

8 1.3243 1.3282 1.3263 0.0068

9 1.3243 1.3263 1.3253 0.0025

10 1.3243 1.3253 1.3248 0.0003

11 1.3243 1.3248 1.3246 -0.0005

12 1.3246 1.3248 1.3247 -0.0001

13 1.3247 1.3248 1.3248 0.0003

14 1.3247 1.3248 1.3248 0.0003

Since the values in 13th and 14th iteration are same, the root is 1.3248.

• Find the positive root of π β ππ¨π¬π = π by bisection method.

Solution:

Let π π₯ = π₯ β cos π₯

Here π 0 = β1 = βπ£π, π 1 = 1 β cos 1 = 0.4597 = +π£π

Hence a root lies between 0 and 1.

Here π = 0 πππ π = 1

π»πππ π₯0 = π + π

2 =

0 + 1

2 = 0.5, π π₯0 = π 0.5 = β0.3776 = βπ£π

Since π π₯0 is negative and π(π) is positive

The root lies between π₯0 and π i.e. 0.5 and 1, put π = π₯0

π₯1 = 0.5 + 1

2 = 0.75, π 0.75 = 0.0183 = +π£π

The root lies between π and π₯1 i.e. 0.5 and 0.75, put π = π₯1

π₯2 = 0.5 + 0.75

2 = 0.625, π 0.625 = β0.186 = βπ£π

Therefore the root lies between π₯2 and π i.e. 0.625 and 0.75

Proceeding like this and these values are put in a tabular form as follows

Since the value of the last column is zero in last column of 11th iteration, the root is 0.7391

Regula Falsi method or False position method:

Solve for a positive root of ππ β ππ + π = π by Regula Falsi method.

Solution: Let π π₯ = π₯3 β 4π₯ + 1

Here π 0 = 1 = +π£π, π 1 = β2 = βπ£π

β΄ A root lies between 0 and 1. Here π = 0 and π = 1

π a b π₯π π(π₯π)

0 0 1 0.5 -0.3776

1 0.5 1 0.75 0.0183

2 0.5 0.75 0.625 -0.186

3 0.625 0.75 0.6875 -0.0853

4 0.6875 0.75 0.7188 -0.0338

5 0.7188 0.75 0.7344 -0.0078

6 0.7344 0.75 0.7422 0.0052

7 0.7344 0.7422 0.7383 -0.0013

8 0.7383 0.7422 0.7403 0.002

9 0.7383 0.7403 0.7393 0.0004

10 0.7383 0.7393 0.7388 -0.0005

11 0.7388 0.7393 0.7391 0

• π₯1 = ππ π β ππ π

π π β π π =

0 Γ π 1 β 1 Γ π 0

π 1 β π 0 = β

1

β2 β 1 = 0.3333

π π₯1 = π 0.3333 = β0.2963 = βπ£π

Hence the root lies between 0 and 0.3333 i.e. π = 0 πππ π = 0.3333

(πππππ π 0 = +π£π πππ π 0.3333 = βπ£π)

π₯2 = ππ π β ππ π

π π β π π =

0 Γ π 0.3333 β 1 Γ π 0

π 0.3333 β π 0 = 0.2571

π 0.2571 = β0.0114 = βπ£π

Hence the root lies between 0 and 0.3333 i.e. π = 0 πππ π = 0.2571

(πππππ π 0 = +π£π πππ π 0.2571 = βπ£π)

Continuing in the same way and the values are put in a table below

π π π π(π) π(π) π₯π π(π₯π)

1 0 1 1 -2 0.3333 -0.2962

2 0 0.3333 1 -0.2962 0.2571 -0.0114

3 0 0.2571 1 -0.0114 0.2542 -0.0004

4 0 0.2542 1 -0.0004 0.2541 0

Since the value of the last column of 4th iteration is zero, the root is 0.2541

Find a positive root of πππ = π by the method of false position.

Solution: Let π π₯ = π₯ππ₯ β 2

Here π 0 = β2 = βπ£π, π 1 = 0.7183 = +π£π

β΄ A root lies between 0 and 1. Here π = 0 and π = 1

π₯1 = ππ π β ππ π

π π β π π =

0 Γ π 1 β 1 Γ π 0

π 1 β π 0 =

2

0.7183 + 2 = 0.3333

π π₯1 = π 0.3333 = β0.2963 = βπ£π

Hence the root lies between 0 and 0.3333 i.e. π = 0 πππ π = 0.3333

(πππππ π 0 = +π£π πππ π 0.3333 = βπ£π)

π₯2 = ππ π β ππ π

π π β π π =

0 Γ π 0.3333 β 1 Γ π 0

π 0.3333 β π 0 = 0.2571

• π 0.2571 = β0.0114 = βπ£π

Hence the root lies between 0 and 0.3333 i.e. π = 0 πππ π = 0.2571

(πππππ π 0 = +π£π πππ π 0.2571 = βπ£π)

Continuing in the same way and the values are put in a table below

π π π π(π) π(π) π₯π π(π₯π)

1 0 1 -2 0.7183 0.7358 -0.4643

2 0.7358 1 -0.4643 0.7183 0.8395 -1.7664

3 0.8395 1 -0.0564 0.7183 0.8512 -1.7881

4 0.8512 1 -0.0061 0.7183 0.8525 -1.7904

5 0.8525 1 -0.0005 0.7183 0.8526 -1.7906

6 0.8526 1 0 0.7183 0.8526 -1.7906

Since the value of the π₯π column of 5 th and 6th iteration are same, the root is 0.8526 up to four decimal

places.

Gauss Jacobi method:

Solve by Gauss Jacobi method, the system

12π₯ + π¦ + π§ = 31,2 π₯ + 8π¦ β π§ = 24, 3π₯ + 4π¦ + 10π§ = 58

Solution:

1 + 1 < 12 , 2 + β1 < |8|, |3| + |4| < |10|

Therefore the given equations are diagonally dominant, so we can use Gauss Jacobi method.

Proceeding in the same manner and writing the solutions in the tabular form we get

π π₯ π¦ π§

1 2.5833 3 5.8

2 1.85 3.0792 3.825

3 2.008 3.0156 4.0133

4 1.9976 2.9997 3.9914

• 5 2.0007 2.9995 4.0008

6 2 2.9999 4

7 2 3 4

8 2 3 4

Since the values in 7th and 8th iteration are same, the solution is π₯ = 2, π¦ = 3, π§ = 4 up to four

decimal places.

Solve by Gauss Jacobi method, the system

9π₯ β π¦ + 2π§ = 9, π₯ + 10π¦ β 2π§ = 15, 2π₯ β 2π¦ β 13π§ = β17

Solution:

β1 + 2 = 3 < 9 , 1 + β2 < 10 , 2 + β2 < β13

Therefore the given equations are diagonally dominant, so we can use Gauss Jacobi method.

π₯ = 1

9 (9 + π¦ β 2π§)

π¦ = 1

10 (15 β π₯ + 2π§)

π§ = 1

13 (17 + 2π₯ β 2π¦)

1st iteration

π₯(1) = 1

9 9 + 0 β 2 0 = 1

π¦(1) = 1

10 15 β 0 + 2 0 = 1.5

π§(1) = 1

13 17 + 2(0) β 2(0) = 1.3077

2nd iteration

π₯(2) = 1

9 9 + 1.5 β 2 1.3077 = 0.8761

• π¦(2) = 1

10 15 β 1.5 + 2 1.3077 = 1.6615

π§(2) = 1

13 17 + 2(1) β 2(1.5) = 1.2308

Proceeding in the same manner and writing the solutions in the tabular form we get

π π₯ π¦ π§

1 1 1.5 1.3077

2 0.8761 1.6615 1.2308

3 0.9111 1.6586 1.1869

4 0.9205 1.6463 1.1927

5 0.9179 1.6465 1.196

6 0.9172 1.6474 1.1956

7 0.9174 1.6474 1.1954

8 0.9174 1.6473 1.1954

9 0.9174 1.6473 1.1954

Since the values in 8th and 9th iteration are same, the solution is π₯ = 0.9174, π¦ = 1.6473, π§ = 1.1954

up to four decimal places.

Gauss Seidel method

Solve by Gauss Seidel method, the system

12π₯ + π¦ + π§ = 31,2 π₯ + 8π¦ β π§ = 24, 3π₯ + 4π¦ + 10π§ = 58

Solution:

1 + 1 < 12 , 2 + β1 < |8|, |3| + |4| < |10|

Therefore the given equations are diagonally dominant, so we can use Gauss Seidel method.

π₯ = 1

12 31 β π¦ β π§

π¦ = 1

8 24 β 2π₯ + π§

• π§ = 1

10 58 β 3π₯ β 4π¦

1st iteration

π₯ = 1

12 31 β 0 β 0 = 2.5833

π¦ = 1

8 24 β 2 2.5833 + (0) = 2.3542

π§ = 1

10 58 β 3(2.5833) β 4(2.3542) = 4.0833

2nd iteration

π₯ = 1

12 31 β 2.3542 β 4.0833 = 2.0469

π¦ = 1

8 24 β 2(2.0469) + 4.0833 = 2.9987

π§ = 1

Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents