Lecture 35:Taylor SeriesDan SloughterFurman UniversityMathematics 39May 5, 200435.1 Taylor seriesDefinition 35.1. If f is analytic at a point z0∈ C, we call the power series∞Xn=0f(n)(z0)n!(z − z0)nthe Taylor series of f about z0. When z0= 0, we call∞Xn=0f(n)(0)n!znthe Maclaurin series of f.The following fundamental theorem is known as Taylor’s theorem.Theorem 35.1. If R0> 0, z0∈ C, and f is analytic in the diskD = {z ∈ C : |z − z0| < R0},thenf(z) =∞Xn=0f(n)(z0)n!(z − z0)nfor all z ∈ D.1Proof. We first assume z0= 0. Let z ∈ D, let r = |z|, let r < r0< R0, andlet C0be the positively oriented circle of radius r0centered at the origin. Bythe Cauchy integral formula, we havef(z) =12πiZC0f(s)s − zds.Now1s − z=1s·11 −zs=1s N −1Xn=0zsn+zsN1 −zs!=N −1Xn=0znsn+1+zN(s − z)sN.Hencef(z) =12πiN −1Xn=0znZC0f(s)sn+1ds +zN2πiZC0f(s)(s − z)sNds=N −1Xn=0f(n)(0)n!zn+ ρN(z),whereρN(z) =zN2πiZC0f(s)(s − z)sNds.It remains to show that limN →∞ρN(z) = 0. Let M be the maximum valueof |f(s)| on C0and note that|s − z| ≥ ||s| − |z|| = r0− r.Thenf(s)(s − z)sN≤M(r0− r)rN0,and so|ρN(z)| ≤rN2π·M(r0− r)rN0· 2πr0=Mr0r0− rrr0N.2Sincerr0< 1, we have limN →∞ρN(z) = 0.Finally, if z06= 0, let g(z) = f(z + z0). Then g is analytic when|(z + z0) − z0| < R0,that is, when |z| < R0. Hencef(z + z0) = g(z) =∞Xn=0g(n)(0)n!zn=∞Xn=0f(n)(z0)n!znwhen |z| < R0. Thus if |z − z0| < R0,f(z) = f((z − z0) + z0) =∞Xn=0f(n)(z0)n!(z −