Question

1. Given the data table with f(x) = yn for a unkown function f, determine the cubic spline interpolation that intersects with the 3 data points. No need to solve for the coefficients. Just set up the eight equations. 1.1 3.5 1.2 3.7 1.3 2.9 2. The fixed point iteration can be used to find the solution of a function f(r) = 0. To use this method, we need to first identify g(x) such that the solution of g(x) = r is the solution of f(x) = 0. (a) Compare fiexed point and the Jacobi iterative method, what is the function (1) in Jacobi iterative method? (b) For this method, one important condition is that the inverse of the dingonal matrix, D-, exists. In this *2 + 2xy = 6 linear sytsem -11-23-2what is the matrix D? What can we do to make the algorithm work? 3r2 - 3 = 5

Answer 1

as we know that 2 + 2 رم1 -0.17 + املا - دارو و درملوو 4 قليلا - اول وملا Sile) a bot bil-x) + balx+x2 گل-لاليط + د کرد

NOW Page No. sjlx ; J = Vj j = 0, 1, 2, Snailan) = YA Date: To for jo Solxo) (11 so, Soltol) = a + a, (1.1-111) + az 111-11172+ ag (1490 1.38 3.5 = ao - so ao 3.5 for j = 1 Sila) - 9 3.1 bot bilat: 21:21+ be ( 112 +112) 2 + b3 ( 112=1218 bor 3.7 bo = 3.7 (2) Now, Tor Iz six2) 3,113) = 2.9

Page No. (Date: TT bot b, (1,3-1,2) + b2 (1.3-1-2/2 + ballos -1,219 22.9 3.7 + bilo.1) + bg (0.01) + b3 (0.001) - 29 so, Oolbi & 0.01 b2 +0.001 bz = -0.8 - (3) Now, Silxg+1) - Sj + 1 l xj +1) for i=o Solxu = s, la) so, sol 12) = 9, (12) ao + 9, 11,2-1, 1) + Q₂ ( 102-11 172 + Og (12-10138 bot bil 12-12) + b2 11,2-112/²+ b3 (1-210213 ja Oil 3.5t oola, torol O2 + 0.001az - 3.7 to Oolat, t000102 +0.0010 - 02 -- Now (4) S; 1 x + 1) Sitil asal

Page No. Date: Solo) = sila) So'l112) = Si (12) a + 204 11:2-1) + 3 agl 1.2-1.2 b, + 2 b2 (112-112) + 368 42-71212 a t 0.2 a2 + 0.03 Az = bi (5) Now, n-2 For i = 0 j= Oil, sola So, 20₂ + 6 a 1 x,-1.1) - 2b2 + 3 x 2 bg {1:2-1-2) 20276 02 (102-101) a 2b2 to 66 202 +6Qg (Dol) = 262 202 + 0.603 - 2 bz Now clamped condition,

Page No. Date: Solxn) afn forno Solo) a fo Strola f (x) sol 1.1) = f'(11) a, +20₂ (111+lel) at Bag (11-101) = a to - flog a NOW sna lan) a to for nar silze) = fal s'ha 3) = te - f(x) = f (trg) bi + 2 bg (1.3-12) + 3 ba (108-11292 $413) bit 0.2b₂ t 0.05 bz = f'(1,2) (8 Hence tight equations ao - 3.5 2. bo = 3.7 3. Oilbi + 0.0162 +0.001 bz = -0.8

4. Oula, +0.0102 +0.001 az = 0.2 5. a, 40120 at 0.05 az bi 6. 2 a 2 + 0.6 az 2b2 7. a = f'll.1) 8. bit 0.2b2 +0.03 63 of 131