Question

QUESTION 4 (10) (a) A natural cubic spline that fits the data given by f(3.0) = -5.6790, f(3.1) = -3.6674, f(3.2) = -2.2178 is to be constructed. Write down explicitly the system of equations that need to be used to construct the required natural cubic spline. (b) Consider the nonlinear system (10) 2+y=9, 22 + y2 = 25, 2,y> 0. Perform one iteration of Newton's method to approximate the solution, starting with (2, 4) as the initial solution. [20]

Answer 1

a) Let us 3 us first divide the region into given part. First region= (3-5.6790) to ( 3.1, -3,6674) Let the cubic spline for this region be S, (a) = ax + b.(X+3)+ cos ( (21–3] + d. (x-3) Second region (3.1, -3, 6674),to (3,2,-2.2178) Let the cubic spline in this region is S, (x)= a, + b con-3,1)+ G (X-3.5+d, con-3,13 from cebic So we know that it ropassed through (3, 5, 6790) S. (3) = -5.6790 -5.6790 = atb, (3-3)+6. (3-3)+ d.(3-3) 3 ao = -5.7690 Similarly we know s, passes through (3,1, -3,6674) S.C3.0 = -3,6674 -3,6674=-5.679040, lb.to. 01 Coat 0.00) do 100b. + 10+ do - 2011. 6 Similarly we know s, passes through (3,2, -2,278) $, (3,2)= -2,2178 -2.2178 = -3,6674+0.16 +0.019 +0,00 d, 1449,6 oob, toc, td, loc 2

2= 3.1 we know that at point (3,1, -3,6674) the slope of both the spline are equal. we can say that S (x) = s, (a) at s,c)= 6+2 6,04-1)+3d,(04-3)? S' (x)= b,+2 c, (5-3,1+3d0-3.132 Put X=3,3 and equate both the equálio we bo+ (0.2)67 10.03)d, zb, 3 we also double derivation at x=3, I will be same s" (x) = 5" (2) at x = 3.1 so C) = 26+6 d. 20-3) 5", (X) = 25+ 6d, C91-3. Put a=3,) in above equation 268+0.6 d. = 24 We know that boundary conclition s', (3) - 5"(3, 2) =6 S" (3)=26=0. we have the value of 670 ped 670 in all the above equations S",(3, 2) > 24,+0,6d,=0 G So J

5 So libe now we have five fine equations to Solue Constants Equations are lobot do=2011.6 loob, +105,+d, =1449,6 bo+0,030,- b, 0.6d-ac, 26, +0.6d, co. By solving the equations, we gert bo=22,457, do = -234.166, 6 = 15.432 4,=-7,025, d. = -23,416, 2,5-5,6790 a,=-3,6674 Co=o. & 2= 3-3,) Soul--5.6790+22.457 (21-33- 234.1667631-3) and between s= 3,1 to 3,2 s 20 = -3,6674 +15.43263-3.) – 7,02 (X-3.15-23.416 (4-3.190 S(M= –3, 6674+15.432(X-3.1)-7.02 (9–3.1723.41661-28

b) Given functions fica, y)= x+y=q, gx,y)= a tý=25 %=2, y=4 f (x, y) = f (2,4)= -1 sekt' - J F CZY For first iteration put kao f(x, y] 1. (%) 5 534,99 (269) let us us find the I K Its Jacobian matrix es 2 JE oflox affay 2%%* aglay *Jl. 1/24 2x ay k Dk/-2 DK -21 22 - ay - Jul Tx - 404-22 -27 28 다.

Dk - Determinent of Jx = 48ly-28 2y ( -2x 2011 I' 414-221 So I i at ca. y.) put 01=2, you, in above equation Dr= 32-4728 Dk=28 JK at (2, 4) = 133 연 -4 is 0,285 -0.0357 -0,142 0.142 0.285 -0,035 1612 -0,142 Joe x= [4]-( 20 After first iteration Olu2 2.106 ll 4.568 2 = 2.106 y=4.568 2