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Unit II : Interpolation and Approxiamtion
Part A
1. Construct a linear interpolating polynomial given the points (x0,y0) and (x1,y1).
2. Obtain the interpolation quadratic polynomial for the given data by using
Newton’s forward difference formula.
X : 0 2 4 6
Y : -3 5 21 45
3. Obtain the divided difference table for the following data.
X : -1 0 2 3
Y : -8 3 1 12
4. Find the polynomial which takes the following values.
X : 0 1 2
Y : 1 2 1
5. Define forward, backward, central differences and divided differences.
6. Evaluate 10 (1-x) (1-2x) (1-3x)--------(1-10x), by taking h=1.
7. Show that the divided difference operator is linear.
8. State the order of convergence of cubic spline.
9. What are the natural or free conditions in cubic spline.
10. Find the cubic spline for the following data
X : 0 2 4 6
Y : 1 9 21 41
11. State the properties of divided differences.
12. Show that
bcd a abcd
1
)
1
( 3 .
13. Find the divided differences of f(x) = x3 + x + 2 for the arguments 1,3,6,11.
14. State Newton’s forward and backward interpolating formula.
15. Using Lagranges find y at x = 2 for the following
X : 0 1 3 4 5
Y : 0 1 81 256 625
Part B
1. Using Lagranges interpolation formula find y(10) given that y(5) = 12, y(6) = 13,
y(9) = 14 and y(11) = 16.
2. Find the missing term in the following table
x : 0 1 2 3 4
y : 1 3 9 - 81
3. From the data given below find the number of students whose weight is between
60 to 70.
Wt (x) : 0-40 40-60 60-80 80-100 100-120
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No of
students : 250 120 100 70 50
4. From the following table find y(1.5) and y’(1) using cubic spline.
X : 1 2 3
Y : -8 -1 18
5. Given sin 450 = 0.7071, sin 500 = 0.7660, sin 550 = 0.8192, sin 600 = 0.8660, find
sin 520 using Newton’s forward interpolating formula.
6. Given log 10 654 = 2.8156, log 10 658 = 2.8182, log 10 659 = 2.8189, log 10 661 =
2.8202, find using Lagrange’s formula the value of log 10 656.
7. Fit a Lagrangian interpolating polynomial y = f(x) and find f(5)
x : 1 3 4 6
y : -3 0 30 132
8. Find y(12) using Newton’ forward interpolation formula given
x : 10 20 30 40 50
y : 46 66 81 93 101
9. Obtain the root of f(x) = 0 by Lagrange’s inverse interpolation given that f(30) = -30,
f(34) = -13, f(38) = 3, f(42) = 18.
10. Fit a natural cubic spline for the following data
x : 0 1 2 3
y : 1 4 0 -2
11. Derive Newton’s divided difference formula.
12. The following data are taken from the steam table:
Temp0 c : 140 150 160 170 180
Pressure : 3.685 4.854 6.502 8.076 10.225
Find the pressure at temperature t = 1420 and at t = 1750
13. Find the sixth term of the sequence 8,12,19,29,42.
14. From the following table of half yearly premium for policies maturing at different ages, estimate the premium for policies maturing at the age of 46.
Age x : 45 50 55 60 65
Premium y : 114.84 96.16 83.32 74.48 68.48
15. Form the divided difference table for the following data
x : -2 0 3 5 7 8
y : -792 108 -72 48 -144 -252
Unit II : Interpolation and Approxiamtion
Part A
1. Construct a linear interpolating polynomial given the points (x0,y0) and (x1,y1).
2. Obtain the interpolation quadratic polynomial for the given data by using
Newton’s forward difference formula.
X : 0 2 4 6
Y : -3 5 21 45
3. Obtain the divided difference table for the following data.
X : -1 0 2 3
Y : -8 3 1 12
4. Find the polynomial which takes the following values.
X : 0 1 2
Y : 1 2 1
5. Define forward, backward, central differences and divided differences.
6. Evaluate 10 (1-x) (1-2x) (1-3x)--------(1-10x), by taking h=1.
7. Show that the divided difference operator is linear.
8. State the order of convergence of cubic spline.
9. What are the natural or free conditions in cubic spline.
10. Find the cubic spline for the following data
X : 0 2 4 6
Y : 1 9 21 41
11. State the properties of divided differences.
12. Show that
bcd a abcd
1
)
1
( 3 .
13. Find the divided differences of f(x) = x3 + x + 2 for the arguments 1,3,6,11.
14. State Newton’s forward and backward interpolating formula.
15. Using Lagranges find y at x = 2 for the following
X : 0 1 3 4 5
Y : 0 1 81 256 625
Part B
1. Using Lagranges interpolation formula find y(10) given that y(5) = 12, y(6) = 13,
y(9) = 14 and y(11) = 16.
2. Find the missing term in the following table
x : 0 1 2 3 4
y : 1 3 9 - 81
3. From the data given below find the number of students whose weight is between
60 to 70.
Wt (x) : 0-40 40-60 60-80 80-100 100-120
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No of
students : 250 120 100 70 50
4. From the following table find y(1.5) and y’(1) using cubic spline.
X : 1 2 3
Y : -8 -1 18
5. Given sin 450 = 0.7071, sin 500 = 0.7660, sin 550 = 0.8192, sin 600 = 0.8660, find
sin 520 using Newton’s forward interpolating formula.
6. Given log 10 654 = 2.8156, log 10 658 = 2.8182, log 10 659 = 2.8189, log 10 661 =
2.8202, find using Lagrange’s formula the value of log 10 656.
7. Fit a Lagrangian interpolating polynomial y = f(x) and find f(5)
x : 1 3 4 6
y : -3 0 30 132
8. Find y(12) using Newton’ forward interpolation formula given
x : 10 20 30 40 50
y : 46 66 81 93 101
9. Obtain the root of f(x) = 0 by Lagrange’s inverse interpolation given that f(30) = -30,
f(34) = -13, f(38) = 3, f(42) = 18.
10. Fit a natural cubic spline for the following data
x : 0 1 2 3
y : 1 4 0 -2
11. Derive Newton’s divided difference formula.
12. The following data are taken from the steam table:
Temp0 c : 140 150 160 170 180
Pressure : 3.685 4.854 6.502 8.076 10.225
Find the pressure at temperature t = 1420 and at t = 1750
13. Find the sixth term of the sequence 8,12,19,29,42.
14. From the following table of half yearly premium for policies maturing at different ages, estimate the premium for policies maturing at the age of 46.
Age x : 45 50 55 60 65
Premium y : 114.84 96.16 83.32 74.48 68.48
15. Form the divided difference table for the following data
x : -2 0 3 5 7 8
y : -792 108 -72 48 -144 -252
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