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Wyatt's Notes — University

Taylor and Laurent Series

7.1 Taylor Series

Theorem 7.1. If ff is analytic on ∣z−z0∣<R|z - z_0| \lt RThen

f(z)=∑n=0∞f(n)(z0)n!(z−z0)nf(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z - z_0)^n

And the series converges uniformly on compact subsets of ∣z−z0∣<R|z - z_0| \lt R.

Proof. For ∣z−z0∣<r<R|z - z_0| \lt r \lt RApply Cauchy”s integral formula on ∣ζ−z0∣=r|\zeta - z_0| = r:

f(z)=12πi∫∣ζ−z0∣=rf(ζ)ζ−z dζf(z) = \frac{1}{2\pi i}\int_{|\zeta - z_0| = r} \frac{f(\zeta)}{\zeta - z}\, d\zeta

Write 1ζ−z=1(ζ−z0)−(z−z0)=1ζ−z0⋅11−(z−z0)/(ζ−z0)\frac{1}{\zeta - z} = \frac{1}{(\zeta - z_0) - (z - z_0)} = \frac{1}{\zeta - z_0} \cdot \frac{1}{1 - (z - z_0)/(\zeta - z_0)} =∑n=0∞(z−z0)n(ζ−z0)n+1= \sum_{n=0}^{\infty} \frac{(z - z_0)^n}{(\zeta - z_0)^{n+1}} (geometric series, convergent since ∣z−z0∣/∣ζ−z0∣<1|z - z_0|/|\zeta - z_0| \lt 1).

Substituting and integrating term by term gives the Taylor series. ■\blacksquare

Remark. The radius of convergence RR is the distance from z0z_0 to the nearest singularity of ff.

7.2 Common Taylor Series

ez=∑n=0∞znn!=1+z+z22!+⋯e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!} = 1 + z + \frac{z^2}{2!} + \cdots

sin⁡z=∑n=0∞(−1)nz2n+1(2n+1)!\sin z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!}

cos⁡z=∑n=0∞(−1)nz2n(2n)!\cos z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!}

11−z=∑n=0∞zn,∣z∣<1\frac{1}{1 - z} = \sum_{n=0}^{\infty} z^n, \quad |z| \lt 1

ln⁡(1+z)=∑n=1∞(−1)n+1znn,∣z∣<1\ln(1 + z) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} z^n}{n}, \quad |z| \lt 1

7.3 Worked Examples: Taylor Series

Solution

Problem. Find the Taylor series of f(z)=1zf(z) = \frac{1}{z} centered at z0=1z_0 = 1.

1z=11+(z−1)=∑n=0∞(−1)n(z−1)n\frac{1}{z} = \frac{1}{1 + (z - 1)} = \sum_{n=0}^{\infty} (-1)^n (z - 1)^n for ∣z−1∣<1|z - 1| \lt 1.

Radius of convergence: distance from z0=1z_0 = 1 to the singularity at z=0z = 0Which is 11.

Problem. Find the Taylor series of f(z)=1(1−z)2f(z) = \frac{1}{(1 - z)^2} centered at z0=0z_0 = 0.

1(1−z)2=ddz[11−z]=ddz∑n=0∞zn=∑n=1∞nzn−1=∑n=0∞(n+1)zn\frac{1}{(1-z)^2} = \frac{d}{dz}\left[\frac{1}{1 - z}\right] = \frac{d}{dz}\sum_{n=0}^{\infty} z^n = \sum_{n=1}^{\infty} nz^{n-1} = \sum_{n=0}^{\infty} (n+1)z^n for ∣z∣<1|z| \lt 1.

Problem. Find the Taylor series of f(z)=ezsin⁡zf(z) = e^z \sin z up to the z4z^4 term.

ez=1+z+z2/2+z3/6+z4/24+⋯e^z = 1 + z + z^2/2 + z^3/6 + z^4/24 + \cdots sin⁡z=z−z3/6+z5/120−⋯\sin z = z - z^3/6 + z^5/120 - \cdots

ezsin⁡z=(1+z+z2/2+z3/6+z4/24+⋯ )(z−z3/6+⋯ )e^z \sin z = (1 + z + z^2/2 + z^3/6 + z^4/24 + \cdots)(z - z^3/6 + \cdots)

=z+z2+z3/2+z4/6+⋯−z3/6−z4/6+⋯= z + z^2 + z^3/2 + z^4/6 + \cdots - z^3/6 - z^4/6 + \cdots =z+z2+z3/3−z4/30+⋯= z + z^2 + z^3/3 - z^4/30 + \cdots

7.4 Laurent Series

Theorem 7.2 (Laurent Series). If ff is analytic on the annulus r<∣z−z0∣<Rr \lt |z - z_0| \lt R Then

f(z)=∑n=−∞∞an(z−z0)n=⋯+a−2(z−z0)2+a−1z−z0+a0+a1(z−z0)+⋯f(z) = \sum_{n=-\infty}^{\infty} a_n(z - z_0)^n = \cdots + \frac{a_{-2}}{(z - z_0)^2} + \frac{a_{-1}}{z - z_0} + a_0 + a_1(z - z_0) + \cdots

Where

an=12πi∫γf(z)(z−z0)n+1 dza_n = \frac{1}{2\pi i}\int_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\, dz

For any simple closed contour γ\gamma in the annulus encircling z0z_0.

The principal part is ∑n=−∞−1an(z−z0)n\sum_{n=-\infty}^{-1} a_n(z - z_0)^n (negative powers). The analytic Part is ∑n=0∞an(z−z0)n\sum_{n=0}^{\infty} a_n(z - z_0)^n (non-negative powers).

7.5 Classification of Laurent Series

The Laurent series expansion depends on the annulus of convergence. A function may have different Laurent expansions in different annuli.

Proposition 7.3. The Laurent series expansion of ff in an annulus is unique.

7.6 Worked Examples: Laurent Series

Solution

Problem. Find the Laurent series of f(z)=1z(z−1)f(z) = \frac{1}{z(z-1)} in 0<∣z∣<10 \lt |z| \lt 1.

Solution. Using partial fractions: 1z(z−1)=1z−1−1z\frac{1}{z(z-1)} = \frac{1}{z-1} - \frac{1}{z}.

In ∣z∣<1|z| \lt 1: 1z−1=−11−z=−∑n=0∞zn\frac{1}{z - 1} = -\frac{1}{1 - z} = -\sum_{n=0}^{\infty} z^n.

So f(z)=−∑n=0∞zn−1z=⋯−z2−z−1−1zf(z) = -\sum_{n=0}^{\infty} z^n - \frac{1}{z} = \cdots - z^2 - z - 1 - \frac{1}{z}.

The principal part is −1/z-1/zSo z=0z = 0 is a simple pole. ■\blacksquare

Problem. Find the Laurent series of f(z)=1z(z−1)f(z) = \frac{1}{z(z-1)} in 1<∣z∣<∞1 \lt |z| \lt \infty.

In ∣z∣>1|z| \gt 1: 1z−1=1z⋅11−1/z=∑n=2∞z−n\frac{1}{z - 1} = \frac{1}{z} \cdot \frac{1}{1 - 1/z} = \sum_{n=2}^{\infty} z^{-n}.

f(z)=∑n=2∞z−n−1z=1z2+1z3+⋯f(z) = \sum_{n=2}^{\infty} z^{-n} - \frac{1}{z} = \frac{1}{z^2} + \frac{1}{z^3} + \cdots

Problem. Find the Laurent series of f(z)=ezz2f(z) = \frac{e^z}{z^2} in 0<∣z∣<∞0 \lt |z| \lt \infty.

ez=∑n=0∞znn!e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}So f(z)=∑n=0∞zn−2n!=1z2+1z+12+z6+⋯f(z) = \sum_{n=0}^{\infty} \frac{z^{n-2}}{n!} = \frac{1}{z^2} + \frac{1}{z} + \frac{1}{2} + \frac{z}{6} + \cdots

Residue at z=0z = 0: a−1=1a_{-1} = 1.

Problem. Find the Laurent series of f(z)=1z2(z−3)f(z) = \frac{1}{z^2(z - 3)} in 0<∣z∣<30 \lt |z| \lt 3.

1z−3=−13∑n=0∞zn3n\frac{1}{z - 3} = -\frac{1}{3}\sum_{n=0}^{\infty} \frac{z^n}{3^n}.

f(z)=−∑n=0∞zn−23n+1=−13z2−19z−127−z81−⋯f(z) = -\sum_{n=0}^{\infty} \frac{z^{n-2}}{3^{n+1}} = -\frac{1}{3z^2} - \frac{1}{9z} - \frac{1}{27} - \frac{z}{81} - \cdots

Residue at z=0z = 0: a−1=−19a_{-1} = -\frac{1}{9}.

7.7 Residue at Infinity

Definition. The residue at infinity of ff is defined as

Res(f,∞)=−12πi∫∣z∣=Rf(z) dz\mathrm{Res}(f, \infty) = -\frac{1}{2\pi i}\int_{|z|=R} f(z)\, dz

For sufficiently large RR (enclosing all finite singularities).

Proposition 7.4. For a function ff with finitely many singularities in C\mathbb{C}:

∑all finite zkRes(f,zk)+Res(f,∞)=0\sum_{\mathrm{all\ finite\ } z_k} \mathrm{Res}(f, z_k) + \mathrm{Res}(f, \infty) = 0

Proof. By the residue theorem applied to ∣z∣=R|z| = R enclosing all finite singularities:

∫∣z∣=Rf dz=2πi∑finiteRes(f,zk)\int_{|z|=R} f\, dz = 2\pi i \sum_{\mathrm{finite} \mathrm{Res}(f, z_k)}.

But Res(f,∞)=−12πi∫∣z∣=Rf dz\mathrm{Res}(f, \infty) = -\frac{1}{2\pi i}\int_{|z|=R} f\, dzSo the sum is zero. ■\blacksquare