Question

why

EXAMPLE 2. One can use expansion (1) and the definition (Sec. 34) eiz sin z = 2i = sin z. To give the details, to find the M aclaurin series for the entire function f (z) we refer to expansion (1) and write (-iz)" (iz)" Σ sin z = 2i ΣΣ1--1y" 1 "- п! п! n! *(0 > |2|) n=0 n=0 B ut 1– (-1)" series: O when n is even, and so we can replace n by 2n +1 in this last ¡2n+1,2n+1 - (-1)2n+1] sin z (2n + 1)! 2i n=0 (∞ > |2|) Inasmuch as = 2 and i2n+1 = (i?)"i = (-1)"i, 1- (-1)2n+1 this reduces to z2n+1 sinzΣ-1 . (2) (|z| < ∞). (2n + 1)! n=0

why did they not include the 2

Answer 1

soll e we know that sin z = Cizred? - using formula ez we get - 02 h azh and 2 - 2 2 - 12) substitute these naives in eq? ), we get C1Z1600) So, we det - E-c-1"] hope o [1- (-1)] chain - Sinza But 1-1-1,1 =0 for every even value of n. so we consides only odd value of n .so we can replace n by anti in eam (?) , we ged Sinz = [1-C-1) 7 & 8949 229+1 (QNH)! Cizko But note from odo o that for any values of n cont = 1 which implies 1-C-13717!=1-(-1) =>

and é pnt = è Me= cine = (-ilé (as e8=4) - substitute this two values èe Incydent = 2 and panth= (-m é in last equation , we get e anti anti (2011! Sinza e [l-caganti ja - = (4157 62294 anti! zamani con zana Can+1)! ant)! so finaly we get Sinz= . (2nt)! cyjn zant (IZKO n=0 CANS)