Lecture 36:Examples of Taylor SeriesDan SloughterFurman UniversityMathematics 39May 6, 200436.1 Examples of Taylor seriesExample 36.1. Let f(z) = ez. Then f is entire, and so its Maclaurin serieswill converge for all z in the plane. Now f(n)(0) = e0= 1 for n = 0, 1, 2, 3, . . .,and soez=∞Xn=0znn!= 1 + z +z22+z33!+ · · ·for all z ∈ C.Example 36.2. It follows from the previous example thate2z=∞Xn=0(2z)nn!=∞Xn=02nn!znfor all z ∈ C. Later we will prove the uniqueness of power series represen-tations, from which it will follow that the expression above is the Maclaurinseries for e2z.Example 36.3. Similarly,eiz=∞Xn=0inn!znande−iz=∞Xn=0(−1)ninn!zn.1Henceeiz− eiz=∞Xn=0(1 − (−1)n)inn!zn=∞Xn=02i2n+1(2n + 1)!z2n+1=∞Xn=02i(−1)n(2n + 1)!z2n+1.Thus, for all z ∈ C,sin(z) =eiz+ e−iz2i=∞Xn=0(−1)nz2n+1(2n + 1)!= z −z33!+z55!−z77!+ · · · .Example 36.4. We will see later that we may differentiate a power s eries aswe would a polynomial, that is, term by term. From this it will follow thatcos(z) =∞Xn=0(−1)n(2n + 1)z2n(2n + 1)!=∞Xn=0(−1)nz2n(2n)! = z −z22+z44!−z66!+ · · ·for all z ∈ C.Example 36.5. We now havesinh(z) = −i sin(iz) = −i∞Xn=0(−1)ni2n+1z2n+1(2n + 1)!=∞Xn=0(−1)2nz2n+1(2n + 1)!=∞Xn=0z2n+1(2n + 1)!= z +z33!+z55!+z77!+ · · ·andcosh(z) = cos(iz) =∞Xn=0z2n(2n)!= 1 +z22+z44!+z66!+ · · ·for all z ∈ C.Example 36.6. We have seen previously that11 − z=∞Xn=0zn= 1 + z + z2+ z3+ · · ·2when |z| < 1. Hence11 − z2=∞Xn=0z2n= 1 + z2+ z4+ · · ·and11 + z2=∞Xn=0(−1)nz2n= 1 − z2+ z4− z6+ · · ·for all z with |z| < 1.Example 36.7. For another example using the geometric series,1z=11 − (1 − z)=nXn=0(1 − z)n=∞Xn=0(−1)n(z − 1)n= 1 − (z − 1) + (z − 1)2− (z − 1)3+ · · ·for all z with |z − 1| < 1.Example 36.8. We have1z2+ z4=1z2·11 + z2=1z21 − z2+ z4− z6+ · · ·=1z2− 1 + z2− z4+ · · ·for all z with 0 < |z| < 1. Note that this representation is not a Maclaurinseries, but is an example of a Laurent series, which we will consider