# 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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From Numerical Analysis 8th Edition by Richard Burden, Section 3.4

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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:

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

Step 3

Using the above conditions and ...

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