MATH529: L23 Taylor series

Overview

• Taylor series

• Laurent series

• Singularities, zeros, and poles

Taylor series

• A power series ${\sum }_{k=0}^{\infty }{a}_k{(z-z_k)}^k$ represents a continuous function $f:ℂ\to ℂ$ within its radius of convergence $|z-z_0|<R,R\ne 0$

• The power series can be differentiated and integrated term-by-term within $|z-z_0|<R,R\ne 0$

• $f:D\to ℂ$ analytic, $z_0\in D$ can be represented by the Taylor series

  ${\displaystyle f\left(z\right)=\sum _{k=0}^{\infty }{a}_k{(z-z_0)}^k=\sum _{k=0}^{\infty }\frac{f^{\left(k\right)}\left(z_0\right)}{k!}{(z-z_0)}^k}$ ${\displaystyle f\left(z_0\right)=a_0,f^{\text{'}}\left(z_0\right)=a_1,f^{\text{'}\text{'}}\left(z_0\right)=2a_2,\dots }$

Taylor series in $ℂ$ proof (basic facts)

• Algebraic identity

  ${\displaystyle \frac{1-t^n}{1-t}=1+t+t^2+\cdots +t^{n-1}\Rightarrow \frac{1}{1-t}=1+\cdots +t^{n-1}+\frac{t^n}{1-t}}$

• Cauchy's integral formulas

  ${\displaystyle f^{\left(n\right)}\left(z_0\right)=\frac{n!}{2\pi i}\oint \frac{f\left(z\right)dz}{{(z-z_0)}^{n+1}}}$

Taylor series in $ℂ$ (proof)

• $f\left(z\right)={\sum }_{k=0}^{\infty }{a}_k{(z-z_0)}^k={\sum }_{k=0}^{\infty }\frac{f^{\left(k\right)}}{k!}{(z-z_0)}^k$

• Start from Cauchy-Goursat

  ${\displaystyle f\left(z\right)=\frac{1}{2\pi i}\oint \frac{f\left(s\right)ds}{s-z}=\frac{1}{2\pi i}\oint \frac{f\left(s\right)ds}{s-z_0-(z-z_0)}=}$ ${\displaystyle =\frac{1}{2\pi i}\oint \frac{f\left(s\right)ds}{(s-z_0)[1-\frac{z-z_0}{s-z_0}]}}$ ${\displaystyle =\frac{1}{2\pi i}\oint \frac{f\left(s\right)}{(s-z_0)}[1+\frac{z-z_0}{s-z_0}+\cdots +{\left(\frac{z-z_0}{s-z_0}\right)}^{n-1}+\frac{{\left(\frac{z-z_0}{s-z_0}\right)}^n}{1-\frac{z-z_0}{s-z_0}}]ds}$

• Recall geometric series, $\frac{1}{1-t}=1+\cdots +t^{n-1}+\frac{t^n}{1-t}$, $t=$$\frac{z-z_0}{s-z_0}$

Taylor series proof (cont)

• After expansion using polynomial factorization identity

  ${\displaystyle f\left(z\right)=\frac{1}{2\pi i}\oint \frac{f\left(s\right)}{(s-z_0)}[1+\frac{z-z_0}{s-z_0}+\cdots +{\left(\frac{z-z_0}{s-z_0}\right)}^{n-1}+\frac{{\left(\frac{z-z_0}{s-z_0}\right)}^n}{1-\frac{z-z_0}{s-z_0}}]ds\Rightarrow }$ ${\displaystyle 2\pi if\left(z\right)=\oint \frac{f\left(s\right)ds}{(s-z_0)}+(z-z_0)\oint \frac{f\left(s\right)ds}{{(s-z_0)}^2}+\cdots +{(z-z_0)}^{n-1}\oint \frac{f\left(s\right)ds}{{(s-z_0)}^n}+2\pi i{R}_n}$ ${\displaystyle f\left(z\right)=f\left(z_0\right)+(z-z_0)+\cdots +{(z-z_0)}^{n-1}+\dots }$ ${\displaystyle R_n=\frac{1}{2\pi i}\oint \cdot ds=\frac{1}{2\pi i}\oint \frac{f\left(s\right)}{(s-z)}{\left(\frac{z-z_0}{s-z_0}\right)}^{n}ds\to 0,\mathrm{as}n\to \infty }$

Laurent series

• $f:D\to ℂ$ analytic in $D:r<|z-z_0|<R$ has the series representation

  ${\displaystyle f\left(z\right)=\sum _{k=-\infty }^{\infty }{a}_k{(z-z_0)}^k}$

  where coefficients are defined as

  ${\displaystyle a_k=\frac{1}{2\pi i}\underset{C}{\oint }\frac{f\left(z\right)dz}{{(z-z_0)}^{k+1}},k=0,\pm 1,\dots ,}$

  with $C$ a contour within $D$.

Laurent series proof

1. Integral over $C_2$ is treated similarly to Taylor series proof

   ${\displaystyle \frac{1}{2\pi i}\underset{C_2}{\oint }\frac{f\left(s\right)ds}{s-z}=\frac{1}{2\pi i}\underset{C_2}{\oint }\frac{f\left(s\right)ds}{s-z_0-(z-z_0)}=}$ ${\displaystyle \frac{1}{2\pi i}\underset{C_2}{\oint }=\sum _{k=0}^{\infty }{a}_k{(z-z_0)}^k(\mathrm{as}\mathrm{in}{\mathrm{Taylor}}^{\text{'}}s\mathrm{theorem})}$

2. Integral over $C_1$ is analogous, but you factor out a different term

   ${\displaystyle -\frac{1}{2\pi i}\underset{C_2}{\oint }\frac{f\left(s\right)ds}{s-z}=\frac{1}{2\pi i}\underset{C_2}{\oint }\frac{f\left(s\right)ds}{z-z_0-(s-z_0)}=}$ ${\displaystyle \frac{1}{2\pi i}\underset{C_2}{\oint }=\sum _{k=1}^{\infty }{a}_{-k}{(z-z_0)}^{-k}}$

Singularities

• Points at which $f:ℂ\to ℂ$ is not analytic are singularities of the function, e.g.

  ${\displaystyle f_1\left(z\right)=\frac{1}{z^2+1}\mathrm{is}\mathrm{singular}\mathrm{at}\pm i,f_2\left(z\right)=\log z\mathrm{is}\mathrm{singular}\mathrm{for}\mathrm{Im}z=0,\mathrm{Re}z⩽0}$

• A singularity $z_0$ is isolated if there exists $R$ such that $f$ analytic for

  ${\displaystyle 0<|z-z_0|.<R}$

• A singularity $z_0$ is not isolated if every neighborhood contains another singularity

  ${\displaystyle z_0=0\mathrm{is}\mathrm{not}\mathrm{an}\mathrm{isolated}\mathrm{singularity}\mathrm{of}f\left(z\right)=\log z}$

• A series representation is possible for $f$ with isolated singularities

Zeros and poles

• $f\left(z\right)=0$, then $z$ is a zero of $f$

• The function $f\left(z\right)=p\left(z\right)/q\left(z\right)$ has a pole at $z$ if $q\left(z\right)$ has a zero at $z$

• The order of the pole is the number of time $z$ is a repeated root

• Laurent series of functions with $k^{\mathrm{th}}$ order poles have terms up to the power $z^{-k}$

• Define two parts of the Laurent series: $f\left(z\right)={\sum }_{k=-\infty }^{\infty }{a}_k{(z-z_0)}^k$

  • Principal part: ${\sum }_{k=-\infty }^{-1}{a}_k{(z-z_0)}^k={\sum }_{k=1}^{\infty }{a}_{-k}{(z-z_0)}^{-k}$

  • Taylor part: ${\sum }_{k=0}^{\infty }{a}_k{(z-z_0)}^k$

Isolated singularity classification

• If principal part is zero, $z_0$ is a removable singularity

  ${\displaystyle =1-+-\dots }$

• If principal part has a finite number of terms $n$, $z_0$ is a pole of order n

  ${\displaystyle }$

• A pole of order 1 is a simple pole

  ${\displaystyle =-+-\dots }$

• If the principal part contains infinitely many terms, $z_0$ is an essential singularity