3.4 (Graded) Cubic splines2. Determine the clamped cubic spline s that interpolates the dataf (0) 0, f(1) 1, f(2) 2and satisfies s' (0) = s'(2) = 1Note: this can be done effectively by hand.4c. Construct the natural cubic spline for the following data.f (x)х-0.290049960.1-0.560797340.2-0.814019720.3Note: this can be done effectively with the aid of software - avoid ugly numbers by hand.8c. Construct the clamped cubic spline using the data of Exercise 4 and the fact thatf'(0.1) =-2.801998 and f'(0.3) = -2.453395.Note: Book has f(0.1) listed incorrectly. Also this can be done effectively with the aidof software avoid ugly numbers by hand.12. A clamped cubic splines for a function f is defined on [1,3] byJso()3(-1)+2( -1)2 - (x - 1)3, if 1 < 2S1()a+b(x - 2)+ c( - 2)2 + d(x - 2)3,s(x) =if 2

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Asked Oct 31, 2019
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From Numerical Analysis 8th Edition by Richard Burden, Section 3.4

3.4 (Graded) Cubic splines
2. Determine the clamped cubic spline s that interpolates the data
f (0) 0, f(1) 1, f(2) 2
and satisfies s' (0) = s'(2) = 1
Note: this can be done effectively by hand.
4c. Construct the natural cubic spline for the following data.
f (x)
х
-0.29004996
0.1
-0.56079734
0.2
-0.81401972
0.3
Note: this can be done effectively with the aid of software - avoid ugly numbers by hand.
8c. Construct the clamped cubic spline using the data of Exercise 4 and the fact that
f'(0.1) =-2.801998 and f'(0.3) = -2.453395.
Note: Book has f(0.1) listed incorrectly. Also this can be done effectively with the aid
of software avoid ugly numbers by hand.
12. A clamped cubic splines for a function f is defined on [1,3] by
Jso()3(-1)+2( -1)2 - (x - 1)3, if 1 < 2
S1()a+b(x - 2)+ c( - 2)2 + d(x - 2)3,
s(x) =
if 2 <I< 3.
Given f'(1) f'(3), find a, b, c, and d.
Note: this can be done effectively by hand.
16. Construct a natural cubic spline to approximate f(x) = e by using the values given by
f(x) at x 0, 0.25, 0.75, and 1.0.
1
1
e dx 1-
the result to
(a) Integrate the spline over [0, 1], and
compare
е
(b) Use the derivatives of the spline to approximate f (0.5) and f"(0.5). Compare the
approximations to the actual values.
Note: this can be done effectively with the aid of software.
19. Suppose that f(x) is a polynomial of degree 3. Show that f(x) is its own clamped cubic
spline, but that it cannot be its own natural cubic spline.
29. (Bonus/graduate) A car traveling along a straight road is clocked at a number of points.
The data from the observations are given in the following table, where the time is in
seconds, the distance is in feet, and the speed is in feet per second.
13
8.
0
3
5
Time
Distance 0
Speed
225 383 623 993
72
80
74
75
77
(a) Use a clamped cubic spline to predict the position of the car and its speed when t = 10 s.
(b) Use the derivative of the spline to determine whether the car ever exceeds a 55
speed limit on the road; if so, what is the first time the car exceeds this speed?
(c) What is the predicted maximum speed for the car?
Note: this can be done effectively with the aid of software.
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3.4 (Graded) Cubic splines 2. Determine the clamped cubic spline s that interpolates the data f (0) 0, f(1) 1, f(2) 2 and satisfies s' (0) = s'(2) = 1 Note: this can be done effectively by hand. 4c. Construct the natural cubic spline for the following data. f (x) х -0.29004996 0.1 -0.56079734 0.2 -0.81401972 0.3 Note: this can be done effectively with the aid of software - avoid ugly numbers by hand. 8c. Construct the clamped cubic spline using the data of Exercise 4 and the fact that f'(0.1) =-2.801998 and f'(0.3) = -2.453395. Note: Book has f(0.1) listed incorrectly. Also this can be done effectively with the aid of software avoid ugly numbers by hand. 12. A clamped cubic splines for a function f is defined on [1,3] by Jso()3(-1)+2( -1)2 - (x - 1)3, if 1 < 2 S1()a+b(x - 2)+ c( - 2)2 + d(x - 2)3, s(x) = if 2 <I< 3. Given f'(1) f'(3), find a, b, c, and d. Note: this can be done effectively by hand. 16. Construct a natural cubic spline to approximate f(x) = e by using the values given by f(x) at x 0, 0.25, 0.75, and 1.0. 1 1 e dx 1- the result to (a) Integrate the spline over [0, 1], and compare е (b) Use the derivatives of the spline to approximate f (0.5) and f"(0.5). Compare the approximations to the actual values. Note: this can be done effectively with the aid of software. 19. Suppose that f(x) is a polynomial of degree 3. Show that f(x) is its own clamped cubic spline, but that it cannot be its own natural cubic spline. 29. (Bonus/graduate) A car traveling along a straight road is clocked at a number of points. The data from the observations are given in the following table, where the time is in seconds, the distance is in feet, and the speed is in feet per second. 13 8. 0 3 5 Time Distance 0 Speed 225 383 623 993 72 80 74 75 77 (a) Use a clamped cubic spline to predict the position of the car and its speed when t = 10 s. (b) Use the derivative of the spline to determine whether the car ever exceeds a 55 speed limit on the road; if so, what is the first time the car exceeds this speed? (c) What is the predicted maximum speed for the car? Note: this can be done effectively with the aid of software.

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Expert Answer

Step 1

Consider the function f(x) = ex and x0 = 0, x1 = 0.25, x2 = 0.75, x3 = 1.0  

 

Let the polynomial passing through the above points:

 

.(x)a,+b(x-0) + c, (x - 0)' +d,(x-0) for x e [0,0.25]
S(x)={S (x) ab, (x- 0.25) + c, (x - 0.25) +d(x-0.25) for x e[0.25,0.75
s. (x) a,+b (x-0.75) + c, (x - 0.75) + d (x-075) for x e [0.75, 1]
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.(x)a,+b(x-0) + c, (x - 0)' +d,(x-0) for x e [0,0.25] S(x)={S (x) ab, (x- 0.25) + c, (x - 0.25) +d(x-0.25) for x e[0.25,0.75 s. (x) a,+b (x-0.75) + c, (x - 0.75) + d (x-075) for x e [0.75, 1]

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Step 2

Then

 

 a0 = f(x0) = f (0) = e0 = 1

 

a1 = f(x1) = f (0.25) = e.25 = 1.2840

 

a2 = f(x2) = f (0.75) = e0.75 = 2.1170

 

a3 = f(x3) = f (1) = e 1 = 2.7182

 

 

we have n = 3

 

h0 = x1 – x0 = 0.25

 

h1 = x2 – x1 = 0.50

 

h2 = x3 – x2 = 0.25

use conditions
f.(x)y
f.(x)y
f(x)
f.(x.)y
f.(x
+1
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use conditions f.(x)y f.(x)y f(x) f.(x.)y f.(x +1

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Step 3

Using the above conditions and ...

1+1.0828 x-.9416 10 x2 +8.5 12.x2 for xe[0,0.25]
S(x)1.00681.0015.r 3.2520 10 x2 4.1760-10 r for xE (0.25,0.75]
1.8944-2.5490x +5.0592.x2-1.6864x2 for x e (0.75, 1]
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1+1.0828 x-.9416 10 x2 +8.5 12.x2 for xe[0,0.25] S(x)1.00681.0015.r 3.2520 10 x2 4.1760-10 r for xE (0.25,0.75] 1.8944-2.5490x +5.0592.x2-1.6864x2 for x e (0.75, 1]

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