HW FIVE

1. 5.1

   (a) By using methods we learned in class, we obtain the polynomial of degree two:

   $$y(x)=\frac{x+x^2}{2}.$$

   (b) Using Lagrange interpolation for the two subintervals, we obtain
   

   $$y(x) = \begin{bmatrix}x, & 0\le x\le 1,\\ 2 x-1, & \le x\le 2\end{bmatrix}$$ (35)

   
   

   (c) by repeating calculations that we have done in class, we obtain

   
   

   $$s(x) = \begin{bmatrix}\frac{3}{4}x+ \frac{x^3}{4} & 0\le x\le 1 \\ 1+\frac{3}{2}(x-1) + \frac{3}{4}(x-1)^2 - \frac{1}{4}(x-1)^3 \end{bmatrix}.$$ (36)

   
   

2. (5.4)

   (a) Lagrange Polynomial is

   $$f(x) = 10(x-2.3)(x-2.6) - 20 *(x-2.1)(x-2.6) + (26/3)(x-2.1)(x-2.3).$$
   This gives, for $x = 2.4$,
   $$f(2.4) = 1.26.$$

   (b) Lagrange Polynomial is

   $$f(x) = -(5/3)(x-0.1)x + 5 ( x+0.2)(x-0.1) + (200/3) x (x+0.2),$$
   which gives, for $x=-0.1$
   $$f(-0.1) = -0.6.$$

   (c) Lagrange Polynomial is

   $$f(x) = - (50/12)(x-1.1)(x-1.2) - (50/3)(x-0.8)(x-1.2) + 25(x-0.8)(x-1.1),$$
   which gives for $x=1$,
   $$f(1) = \frac{1}{12}.$$

3. 5.26

   (a)

   Global polynomial interpolation is

   $$y(x) = \frac{10}{3}x - 3 x^2 +\frac{2}{3}x^3.$$
   Then
   $$y(1/2) = 1b.$$
   (b)

   Writting spline as
   

   $$s_1(x) = a_1+b_1 x + c_1 x^2 + d_1 x^3$$

   $$s_2(x) = a_2 + b_2 (x-1) + c_2 (x-1)^2 + d_2(x-1)^3$$

   $$s_3(x) = a_3 + b_3 (x-2)+ c_3(x-2)^2 +d_3 (x-2)^3$$

   
   and follwoing steps we discussed in class, we obtain system of equations
   

   $$a_1=0, c_1=0$$

   $$b_1+d+1=0, a_2=1, 1+b_2 + c_2 + d+2 = 0$$

   $$a_2 = 1, 1+ b_2 +c_2 + d_2= 0, a_3 = 0,$$

   $$b_3+c_3+d_3 = 1 , b_1 + 3 d_1 =b_2, 6 d_1=2c_2 ,$$

   $$b_2 + 2 c_2 + 3 d_2 = b_3, 2 c_2 + 6 d_2 = 2 c_3,$$

   $$2c_3 + 6 d_3 = 0.$$

   
   We solve this system of equation to finally obtain that
   

   $$s_1(x) = \frac{5}{3} x+ \frac{2}{3} x^3,$$

   $$s_2(x) = 1-\frac{1}{3} (x-1) -2 (x-1)^2+\frac{5}{3}(x-1)^3,$$

   $$s_3(x) = -\frac{1}{3} (x-2)+ 2 (x-2)^2 -\frac{2}{3}(x-2)^3.$$

   
   We therefore obtain
   $$s_1(\frac{1}{2}) = \frac{11}{12}.$$
   c

   Result of part (a) does indeed satisfies interpolation and smoothness condition, but result in part (a) is not a natural spline

   (d)To obtain result in (a) one would have to use clamped spline with

   $$s_1'(0) = \frac{10}{3}= s_3'(3).$$