Question

6. Let f(z) = z² sin z. (a) (5%) Find the Taylor series expansion of f(z) about zo = 0. Where does the series converge? (b) (5%) Find f(?)0) and f(*)(0).

Answer 1

solution : Here f(2)= z² sin z Let g (2) = sinz So: 23 g (2) f(2)= about Zoro find the Taylor series Expansion of fres where does the series Converge ? het sina 2 g (2) ³ g (o) = 0 (2) Cos z CO) go (2) 11 - sinz → g 0 = 0 (3) 9 (2) - Cosz ܩܘ -1 sin 2 => 0 11 go (z (2) (2) go) Cosz ( (2) = - sinz CO) O şion, By Mac laurin's expan. g(z) = g(0) + z glos 23 + g 3) 0 sin z = 0 + zit (-1) + .(0) +.....

sin z = z- 23 2 + + 31 s! onla 7! or in sigma notation: 2k+1 sin (z) = z k=o (@k+1)! 3 multiply And Lo, if we desired series: by Z we get the f(2)= z sinz 3 7 2 z = 23 (z- + + 3! s! 깃 9! 6 lo - Z Z 12 z 9! 31 71 On 2k +1+3 f(2)= (2k+1)! k=0 akty f (2) = Ź (-1) Z Answer kro (2K+1)!

Now, we check the series series for for conver convergence, Hence, series es (-1) kro het an (2k+1) ! 1. an is positive and Monotone decreasing le lim kt an lim fotos (@k+1) sol By the alterneting series test criteria we have given, series Et is (2k+i)! k=0 Convergent. gt Convergent and at 0.

e = 62 (6) find (o) and f of (a) = 23 sina het u= sinz and Differentiating uz? Differentiating w. z. t. 2 W.. + Un sin(2+ in vi = 32 2 Un-l= sin (z + (n-1) II ) Un-a = sin (z +(2-2) Un-3 = sin sin (z + (n-3) II) y = 0 Leibnitz's Theorem, we using have f (2) = Un vt nc, Unni ", + nea Un-2 ₂ t.... + fWcz) = sin (2+ **). 2? + "c, (sin 2 + (-1) ).3z? sin (2+(2-2) II). 62 + ng sin (2 + (0 - 3) / ). 6 to ng Sin (ot(n-3) 1 - 6 fco C₂ (sin 21). 6 = 0 $ xlb, (0) = 6

similarly [rsin at = o flos, (0) sin ( 1 ) . 6 11 8! xlx 6 51X3! 8 X 7 X 6 X6 3! 8x7 x 6 &x42 336 Hence, Foto (0) (8) f (o) = 336 Answer yo