Taylor series

The Taylor series of an analytic function $f$ at $z_0$ is the Power series

${\sum }_{n=0}^{\infty }\frac{f^{(n)}(z_0)}{n!}(z-z_0{)}^n.$

Theorem (Taylor’s theorem in $\mathbb{C}$). If $f$ is analytic on the open disk $|z-z_0|<R$, then on that disk,

$f(z)={\sum }_{n=0}^{\infty }\frac{f^{(n)}(z_0)}{n!}(z-z_0{)}^n.$

The series converges to $f$ throughout the disk.

Every analytic function equals its Taylor series on every disk where it’s analytic. There’s no distinction between “analytic” and “represented by its Taylor series,” which doesn’t hold in the real case.

Sketch of proof

For any $z$ inside a disk of analyticity around $z_0$, choose a circle $C$ around $z_0$ with $z$ inside $C$ and $C$ still inside the disk. By the Cauchy integral formula,

$f(z)=\frac{1}{2\pi i}{\oint }_C\frac{f(w)}{w-z}dw.$

Expand $1/(w-z)$ as a geometric series in $(z-z_0)/(w-z_0)$ (valid because $|z-z_0|<|w-z_0|$):

$\frac{1}{w-z}=\frac{1}{(w-z_0)-(z-z_0)}=\frac{1}{w-z_0}{\sum }_{n=0}^{\infty }{\left(\frac{z-z_0}{w-z_0}\right)}^n.$

Substitute back and integrate term by term. The integral $\oint f(w)/(w-z_0{)}^{n+1}dw$ is $2\pi i{f}^{(n)}(z_0)/n!$ by the generalized CIF. Collecting:

$f(z)={\sum }_{n=0}^{\infty }\frac{f^{(n)}(z_0)}{n!}(z-z_0{)}^n.■$

Radius of convergence = distance to nearest singularity

A consequence of the proof: the largest disk around $z_0$ on which the Taylor series converges to $f$ is the largest disk around $z_0$ on which $f$ is analytic. So the radius of convergence equals the distance from $z_0$ to the nearest singularity of $f$.

Example. $f(z)=1/(1+z^2)$ has singularities at $z=\pm i$. Taylor series around $z_0=0$: the distance to the nearest singularity is $1$, so $R=1$. The series $\sum (-1{)}^n{z}^{2n}$ converges on $|z|<1$ to $1/(1+z^2)$.

$1/(1+x^2)$ is perfectly smooth for all real $x$, yet its real Taylor series at $0$ only converges for $|x|<1$. The complex picture explains why: the singularities at $\pm i$ constrain the radius of convergence even on the real line.

The four core Taylor expansions

Memorize these. They’re the building blocks for almost every series computation.

$\frac{1}{1-z}={\sum }_{n=0}^{\infty }{z}^n=1+z+z^2+z^3+\cdots ,|z|<1.$

$e^z={\sum }_{n=0}^{\infty }\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots ,\text{all }z.$

$\sin z={\sum }_{n=0}^{\infty }\frac{(-1{)}^n{z}^{2n+1}}{(2n+1)!}=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots ,\text{all }z.$

$\cos z={\sum }_{n=0}^{\infty }\frac{(-1{)}^n{z}^{2n}}{(2n)!}=1-\frac{z^2}{2!}+\frac{z^4}{4!}-\cdots ,\text{all }z.$

Anything else, build it from these with the manipulation techniques below.

Euler’s formula, finally derived

Back in Chapter 1 we took $e^{i\theta }=\cos \theta +i\sin \theta$ as a definition. The Taylor series for $e^z$ now justifies it. Substitute $z=i\theta$:

$e^{i\theta }={\sum }_{n=0}^{\infty }\frac{(i\theta {)}^n}{n!}.$

Split by parity of $n$. For even $n=2k$: $i^{2k}=(i^2{)}^k=(-1{)}^k$. For odd $n=2k+1$: $i^{2k+1}=i\cdot (-1{)}^k$. So

$e^{i\theta }=\underset{\cos \theta }{\underbrace{\sum _{k=0}^{\infty }\frac{(-1{)}^k{\theta }^{2k}}{(2k)!}}}+i\underset{\sin \theta }{\underbrace{\sum _{k=0}^{\infty }\frac{(-1{)}^k{\theta }^{2k+1}}{(2k+1)!}}}=\cos \theta +i\sin \theta .$

The rearrangement is legal because the $e^z$ series converges absolutely on all of $\mathbb{C}$, and absolutely convergent series can be rearranged freely.

So Euler’s formula is just the $e^z$ series evaluated at $i\theta$, splitting by parity into the cosine and sine series. See Euler’s formula.

Three manipulation techniques

1. Substitution. Replace $z$ in a known expansion with $g(z)$.

Example. $1/(1+z^2)$ around $0$: use $1/(1-w)$ with $w=-z^2$: $1/(1+z^2)=\sum (-z^2{)}^n=\sum (-1{)}^n{z}^{2n}$ for $|z|<1$.

Example. $e^{z^2}$: replace $z$ in $e^z$: $\sum z^{2n}/n!$.

2. Differentiation. Differentiate a known series term by term (legal inside the disk of convergence).

Example. $1/(1-z{)}^2$: differentiate $\sum z^n$: $\sum n{z}^{n-1}=1+2z+3z^2+\cdots$.

Example. From $\sin z=z-z^3/3!+\cdots$, differentiating gives $\cos z=1-z^2/2!+z^4/4!-\cdots$.

3. Integration. Integrate a known series term by term.

Example. $\log (1+z)$ around $0$: $d/dz[\log (1+z)]=1/(1+z)=\sum (-z{)}^n$. Integrate (with constant zero since $\log 1=0$):

$\log (1+z)=z-z^2/2+z^3/3-z^4/4+\cdots ={\sum }_{n=1}^{\infty }(-1{)}^{n+1}{z}^n/n,|z|<1.$

Example. $\arctan z$: ${\arctan }^{\prime }(z)=1/(1+z^2)=\sum (-1{)}^n{z}^{2n}$. Integrate:

$\arctan z=z-z^3/3+z^5/5-z^7/7+\cdots ,|z|<1.$

(Setting $z=1$ gives Leibniz’s formula for $\pi /4$.)

Often you mix techniques. Expand $z^2{e}^{z^3}$: start with $e^z=\sum z^n/n!$, substitute $z\mapsto z^3$: $e^{z^3}=\sum z^{3n}/n!$. Multiply by $z^2$: $z^2{e}^{z^3}=\sum z^{3n+2}/n!$.

Where they show up

• Computing residues at simple poles when $f(z)=g(z)/(z-z_0)$ with $g$ analytic.
• Reading off the behavior of $f$ near a point.
• Numerical work: polynomial approximations of transcendental functions.
• Base case for Laurent series (Taylor with possibly-negative powers).