Question

Let f(x) = xlnx. Approximate f(2) by the Hermite interpolating polynomial using x0 = 1 and x1 = e and compare the error.(e ≈ 2.7...)

Answer 1

f(x) = xlna f'(x) = x1 + enx = 14 dna 20=1 , 91=e. To compute: f(2) fee, 2 ln 2. Hermite Interpolation for mula is; pens = È [1- 2 L CER) Cx- ap)] [ Lp (2) ² frak) + ² (x-Xp ] [ Lk (833² f cale) kao so, first we evaluate the values of Lp (ak) & 2 car and then putting values in p(x) we get the required Polynomial. Loca) = x-x1 = 1-€ | f(x) = xlna do-a l-e f(ao) = 1. ln (1) = 0 Lo'cxo) = the L1 (%) = 1-20 = 21 f'(ao) = 0+1=1 f(21)= e ence) = e. f'(a)= Itence) =lt1 = 2. Now, pcx) = R20 { CX-ART kzo pw. ŚLI- 2. Laiky(7-782] [ LA TO DATA) + Eco-XR] [ 4p73]® flek) = 1-2 10 (90) (*-* 33 C 1060]°71%) + 1-24, 40, -3)? CScanned with canh (01) 412) + (-30) [ Lg603] ?F'(%) + (x-2) [L, C23 ]² flex)

_~) = 1 1+oX EX (7-1){re xo + {1- 2 x L x (1-1)} x (x-1) re. + (x-1) [42.] 1 + (9-e) (24)?. 2 + = 11 - (2-Uca-e12 Cheiatex-e)(a-2 (en) 2 þ (2) = f(2)= 2002 2 + 2(2-6) (2712 : (ey) 2 - 2 = (1-262-0) (2-1) + (2-443-03? 4 2016 (2 (2-1 (2-0)² 612 Cle2 12-e) e-1 (2-e) 2 + 2.(2-e) - 3e-5xe e-l (e-)2 72 (les2 ce-112 . = 1.690397 + 0.174743 + (-0486559) If (2) = 1.37 8.58] i r obtained by Hermite enter palation formula. the value of f(2) in general is Now f(2)= 2112 1412) = 1.38629) so the approximade error is :- LEE 0:00772] CScanned with CamScanner
firstly we obtained the value of f(2) by Hermite Interpolation and then the exact value and by comparing the two we found the approximate error.