Newton Gregory Forward Interpolation Questions and Answers - Sanfoundry PDF

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3/9/2021

Newton Gregory Forward Interpolation Questions and Answers - Sanfoundry

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This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) focuses on “NewtonGregory Forward Interpolation Formula”. 1. Newton- Gregory Forward interpolation formula can be used _____________ a) only for equally spaced intervals b) only for unequally spaced intervals c) for both equally and unequally spaced intervals d) for unequally intervals View Answer Answer: a Explanation: Newton – Gregory Forward Interpolation formula is given by f(x) = y + nΔy + n(n-1)Δ2y /2! + n(n-1)(n-2) Δ3 y /3! + …..

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2. Find n for the following data if f(0.2) is asked.

x

0

1

2

3

4

5

6

f(x)

176

185

194

203

212

220

229

a) 0.4 b) 0.2 c) 1 d) 0.1 View Answer Answer: b Explanation: The formula is x = x 0 + nh. Here x 0 is 0 as 0 is the rst element and h is 1. Since in the question it is given that we have to nd f(0.2), x= 0.2. So, substituting the values in the formula we get, 0.2 = 0 + n(1) . Hence, n= 0.2. 3. Find n for the following data if f(1.8) is asked.

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a) 2.4 b) 3.4 c) 2.6 d) 3.6 View Answer Answer: d Explanation: Here, x 0 is 0, h is 0.5, x is 1.8. Substituting the values in the formula x = x 0 + nh, 1.8 = 0 + n(0.5) n = 3.6. advertisement

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4. Find the polynomial for the following data.

x

4

6

8

10

f(x)

1

3

8

16

a)

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d) View Answer Answer: a Explanation: Here,

x

y

Δy

Δ2 y

Δ3 y

4

1

2

3

0

6

3

5

3

8

8

8

10

16

y 0 is 1 since it is forward interpolation formula. Δy 0 = 2 Δ 2y 0 = 3 Δ 3y 0 = 0 x = x 0 + nh, x = 4 + n(2) Hence n = (x-4)/2 Substituting these values in the formula, f(x) = y 0 + nΔy0 + n(n-1)Δ2y0 /2! + n(n-1)(n-2) Δ3 y0 /3!,

5. Using Newton’s Forward formula, nd sin(0.1604) from the following table.

x

0.160

0.161

0.162

f(x)

0.1593182066

0.1603053541

0.1612923412

a) 0.169713084 b) 0.159713084 c) 0.158713084

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Answer: b Explanation: Here, x 0 = 0.160 x = 0.1604 h = 0.001 x = x 0 + nh, 0.1604 = 0.160 + n(0.001) n = 0.4 Δ2 y

x

y

Δy

0.160

0.1593182066

9.871475*10

0.161

0.1603053541

9.869871*10

0.162

0.1612923412

-4

-7

-1.604*10

-4

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y 0 is 0.1593182066 since it is forward interpolation formula. -4

Δy 0 = 9.871475*10 2 -7 Δ y 0 = -1.604*10

Substituting in the formula, f(x) = y 0 + nΔy0 + n(n-1)Δ2y0 /2! + n(n-1)(n-2) Δ3 y0 /3! , f(0.1604) = 0.1593182066+(0.4)(9.871475*10-4 )+(0.4)(-0.6)

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Newton Gregory Forward Interpolation Questions and Answers - Sanfoundry

x

0

2

4

6

8

f(x)

4

26

58

112

466

a) 71.109375 b) 61.103975 c) 70.103957 d) 71.103957 View Answer Answer: a Explanation: Here, x0 = 0 x=5 h=2 x = x 0 + nh, 5 = 0 + n(2) n = 2.5

x

y

Δy

Δ2 y

Δ3 y

Δ4 y

0

4

22

10

12

266

2

26

32

22

278

4

58

54

300

6

112

354

8

466

y 0 is 4 since it is forward interpolation formula. Δy 0 = 22 Δ 2y 0 = 10 Δ 3y 0 = 12 Δ 4y 0 = 266 Substituting in the formula, f(x) = y 0 + nΔy0 + n(n-1)Δ2y0 /2! + n(n-1)(n-2) Δ3 y0 /3! +…. f(5)=4+(2.5)(22)+

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Newton Gregory Forward Interpolation Questions and Answers - Sanfoundry

x

0

0.1

0.2

0.3

0.4

f(x)

1

1.052

1.2214

1.3499

1.4918

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a) 1.18878784 b) 1.8878784 c) 1.9878785 d) 0.8878784 View Answer Answer: a Explanation: Here, x0 = 0 x = 0.18 h = 0.1 x = x 0 + nh, 0.18 = 0 + n(0.1) n = 1.8

x

y

Δy

Δ2 y

Δ3 y

Δ4 y

0

1

0.052

0.1174

-0.1583

0.2126

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Newton Gregory Forward Interpolation Questions and Answers - Sanfoundry

0.3

1.3499

0.4

1.4918

0.1419

y 0 is 1 since it is forward interpolation formula. Δy 0 = 0.052 2

Δ y 0 = 0.1174 3 Δ y 0 = -0.1583 4

Δ y 0 = 0.2126 Substituting in the formula, 2

3

f(x) = y 0 + nΔy0 + n(n-1)Δ y0 /2! + n(n-1)(n-2) Δ y0 /3! +….

f(0.18) = 1.18878784. 8. Find f(2.75) using Newton’s Forward interpolation formula from the following table.

x

1.5

2

2.5

3

3.5

4

y

3.375

7

13.625

24

38.875

59

a) 1.8296875 b) 18.296875 c) 22.296875 d) 24.296875 View Answer Answer: a Explanation: Here, x 0 = 1.5 x = 2.75 h = 0.5 x = x 0 + nh, 2.75 = 1.5 + n(0.5) n = 2.5

x

y

Δy

Δ2 y

Δ3 y

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Δ4 y

Δ5 y

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2.5

13.625

10.375

4.5

3

24

14.875

5.25

3.5

38.875

20.125

4

59

0.75

y 0 is 3.375 since it is forward interpolation formula. Δy 0 = 3.625 Δ 2y 0 = 3 Δ 3y 0 = 0.75 Δ 4y 0 = 0 Δ 5y 0 = 0 Substituting in the formula, f(x) = y 0 + nΔy0 + n(n-1)Δ2y0 /2! + n(n-1)(n-2) Δ3 y0 /3! +…. f(2.75)=3.375+(2.5)(3.625)+

+0+0

f(2.75) = 18.296875. 9. Find n if x0 = 0.75825, x = 0.759 and h = 0.00005. a) 1.5 b) 15 c) 2.5 d) 25 View Answer Answer: b Explanation: Given x 0 = 0.75825 x = 0.759 h = 0.00005 Substituting in the formula, x = x 0 + nh, 0.759 = 0.75825 + n(0.00005) Therefore, n = 15. 10. Find x if x0 = 0.6, n = 2.6 and h = 0.2. a) 12 b) 1.2

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Answer: c Explanation: Given x 0 = 0.6 n = 2.6 h = 0.2 Substituting in the formula, x = x 0 + nh x = 0.6+(0.2)(2.6) x = 1.12. Sanfoundry Global Education & Learning Series – Numerical Methods. To practice all areas of Numerical Methods, here is complete set of 1000+ Multiple Choice Questions and Answers. Participate in the Sanfoundry Certication contest to get free Certicate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs! Telegram | Youtube | LinkedIn | Instagram | Facebook | Twitter | Pinterest « Prev - Matrix Inversion Questions and Answers – Jacobi’s Iteration Method » Next - Numerical Analysis Questions and Answers – Approximation of Functions using Least Square Method advertisement

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