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. This is one of the deepest payoffs of complex analysis: there is no distinction between “analytic” and “represented by its Taylor series.”

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 $1/(1+z^2)$.

Striking: $1/(1+x^2)$ is perfectly smooth for all real $x$, yet its real Taylor series at $0$ only converges for $|x|<1$. The complex perspective explains why — singularities at $\pm i$ are what constrain the radius of convergence on the real line. Real-analysis mystery solved by complex analysis.

The four core Taylor expansions

Memorize these — they’re the building blocks for almost every series computation:

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

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

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

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

These four cover an enormous fraction of practical needs. Anything more complex, use one of three manipulation techniques.

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 $\mathbb{C}$ — absolutely convergent series can be rearranged freely.

Euler’s formula is the statement that the $e^z$ series, evaluated at $i\theta$, splits exactly 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 $\pi /4$.)

Combinations. Often 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 $z^2$: $z^2{e}^{z^3}=\sum z^{3n+2}/n!$.

In context

Taylor series provide the most computational tool for working with analytic functions. They’re used:

• To compute residues at simple poles when $f(z)=g(z)/(z-z_0)$ with $g$ analytic.
• To extract behavior of $f$ near specific points.
• For numerical computation: polynomial approximations of transcendental functions.
• As the foundation for Laurent series (Taylor with possibly-negative powers).