Question

QUESTION 4 (a) A natural cubic spline that fits the data given by (10) 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) z+ y = 9, 1² + y2 = 25, 2, y > 0. Perform one iteration of Newton's method to approximate the solution, starting with (2, 4) as the initial solution.

its a numerical analysis question

Answer 1

0 second region: (301,-3-6674) to (3.22.2176) Let Cubic spline in this region in S. C): 0, tb, (2-3.1) 4 (-3.13+d(x-3.153 Giveni f(3-0) -- 5.6790 ,A(3,1)= -3.6674, f(3.2)= -2:2178 First divide the region into given part. • First region : (3,-516790) to (3.1, -3.66 74) Let cubic spline for this region be 50 (X) = Qot bo (x-3) + (x-3)? + do (x-33. From Cubic so We know that it is passed tterough (3.-5.679) So (3) -5.6790 -5.6790= lotbo (3-3) & Co (3-3,² tdo (3-373 90-5.7690 Similarity se passes througle (3.1, -3,6 674) So (3-D =-3,6674. -3.66745-5.6790+ 0.1 bo +0.01 C6+0.001 do 100 bo+ 10 Cot do = 20116 - 0 simileerdly s, pagses terorghe (3,2, -22178) S (32) = -2.2178 -2.2178= -3.6674+0.1 b, to-01 0,1+ 0.001d, 1449. 6 = 100 b. tlocitd. At point (3.1,-3.6674) Ithe slope of bothi the spline are equal Now, we com fety tent si co= S(x) at *23.) Só (X) = bot 2 co(x-3) +3d.(3-3)? si 012 6,+2C,CX-3.)) +3d. Col-3.13 Equate botte equations and put +23.) we get bo +(0.2160+60-03)do: b, 3 Also double derivative at x =3.1 will be same Só (x) = sila at $223.1 50" (x)= 26 +6 do (x-3) S," QUE 2 (1+6d, CX-3.1) In the above equation, put x23-1 2 Co to. 60 do =26,

2 Now, we have 5 equations to solve 5 conetants. bo 22.457. do--234. 166, b= 15-432, GE-7.025, d, -23.416, we know that boundary condition Si (3) si (3:2)-0 S." (3) - 220-0 We have the value & Co 20 Put Coro in all the above equations 5" (3-2)- 26+0.6 d.=0 - Equations are: 0.6 do=2c, 20,-0.62,50 By solving these equations, we get 202-5.6790, 9,3-3.6674, Coco x 3-3.) la botdo= 2011.6 100b-410cistd.: 1449.6 boot 0.03do=bi S(): -5.6796 +22.45+(x-3)?--234. 1667(x-3) and between x 3.1 to 3,2 SC1): -3.6674 +15,432(2-3-1)-7.02(x-314-23-416 Q2-3.1? nig zo. X0=2, 40-4 x² + y² = 25 964,972 x y*=25 b) Given functions: x²+4=9 fox, y)= x+y=9 fcxo, yo) = f(2,4)=-1 xkt = xk Ji f (2k, Yk) For first iteration put k =0 (f(x6,9) -Yo (g(x, yo find If Its Jacobian matrie is 22 ſoffox (oglar zgloy) zy JK! 4y-2x ray-1 offay 7 - (23 24] =>JË', the la J Dk Determinent of Jk+ 4xy-2x Come zis] JE at (xo, Yo) put x=2, 924 in the above equation Jk' = 4xy-2x [-2x 2 x )

من (24) لمه JE [] 29 ال After first iteration x=2.106, og = 4.868 D: 32 -4 0 = 2S 4 4 6- 28 -0.63S1 60142 -6042 6042 4:56 0142 . الا ] - :: 60285 -6003S ][] 2،164): اد (2)