Question

QUESTION 4 (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 x2 + y = 9, 22 + y2 = 25, x, y > 0. Perform one iteration of Newton's method to approximate the solution, starting with (2, 4) as the initial solution.

Answer 1

(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

آه - - - 2 จร » - «ها - 21 da ع So I h al (20%) put na 2, J =4 in above equation Dk Dk ها - 34 - 08 Ik at (2,4) اٹل | ع با ما 1 6.385 | -0.6351 ووه ۵.۱ - - [3] = ابو 0، 49 6. 8 -6 ،5351 و ها) .0- و با ۵.۱ ]] او مها، و) 568 دا ل دکمه Ib6 .و 568 : ها | 1 - 3.106 568 : ها : 8 1 X

-3.66447-5.6790 +0.1bot 0.0160701001 do slechton Let us first divide the region into given part ,S? region = (3, -5,6790) to (3.1, -3.6694) Let the cubic spline for this region be So (X) = 20+ b (0-3) + Cos(x-3597 d,(6-3 gegion = (3.1, -3.6674) to (3.4,-2.2178 ) Let the cobic spline in this region is 5,() = , + b, (– 3. 1) +6,60–3.994 d. (x-3-1) 1 from cebic So we know that, Il passed through 13, -5.6790) S. (3) ---5-6790 -5.6790 = 9+5063 - 3) + 6(3-3)+ d3(3-3)* 3 19 = 5.7690 we know si passes through (301,-3.6574) Similarly $(3.1) = -3.6674. -700 100 bot 106 tdo = gollie Similarly wo know & passes through 13.8,-8-2+3) 5.03.2) = -2,2178

1449-6 = 1005, + 10C, we know that at polut (3.1, -3.6674) the slope of both the spline are equal we can say that s(x) = s, (x) at x = 3.1 26(2-3 +3 do (N-31 s, (x) = 6, +26, (x-3.1)+3d, (x-3.1) put 2=3.1 and equate both the equation we bo +0.296+ (0.03) dob-73 s! (x)=bo . "Con 26t6do (x-3) -2.2178 =-3-6674 +0.1b, +0.01 +0.001d, td, Q get also dos double derivate at a=3.) will be Same so Cay - 11 ES." (x) at 8=3.1 S. Si put Casade, +60, (X- 3.1) x=3,1 in above equation 20+ 0.6 do =20, ~7 we know that boundary condition S''(g) = 5'(3.2)=0 "(3) = 266-0 No coso put Coto in above equations

S: (3.c) 3.2) -2, +0.6d1-O So now we have to hand constants lobot do 2011.6 100 b, +106, +4, +1449.6 bo +0.0%3d, 26, 0.6 do =26 20+0.644-07 By Soluto solving above equations bo = 29.457 -234.166 b,= 15.432 Ciz-7.025 dia-23.416 a=-5,6790 a = -3.6674 Co to X (3, 3.1) 3(X) = -5.6790 +22.45718-32-234.1667 (2-3)3 and between aa3.1 to 312

5(X) = -3.6674 +15.432 GX-3.1) -7.02 (x-3.11'- 23,446 (a -3,133 2 5(x) = -3.6674 +15.432 (x - 3,19 –7002 (-3.1) - 23.41668–3.1) Given that floup) =2 ty=9 gloney) = x + ug" =25 for %) = f(2,4)=-1 kt & k =x-The franada) First Pregration put kao Roto) ما - *= (3-si k --- JE tid 24 l can 2x Dk "(%18 L gooi Let us find the I co If ran da dy dg da Jx dy (28 al Jina 4ky-an 1-2n an - Determinent of Jk =484 - 24 ſ : Jacobian mothin? J