Q1P - Mathematical Methods in Physical Sciences

Question

Find the disk of convergence for each of the following complex power series.a

${\mathbf{e}}^{\mathbf{z}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{z}\mathbf{+}\frac{{\mathbf{z}}^{\mathbf{2}}}{\mathbf{2}\mathbf{!}}\mathbf{+}\frac{{\mathbf{z}}^{\mathbf{3}}}{\mathbf{3}\mathbf{!}}\mathbf{·}\mathbf{·}\mathbf{·}\left[\mathrm{equation} (8.1)\right]$

Step-by-Step Solution

Answer

The entire complex plane is the radius of convergence.

Step 1: Given Information.

The given power series, i.e.,${\mathrm{S}}_{\mathrm{n}}={\mathrm{e}}^{\mathrm{z}}$ 

Step 2: Definition of Radius of convergence.

The interior of the set of points of convergence of a power series is called the disc of convergence. Its radius is known as the series' convergence radius.

Step 3: Find the value of the general term.

Expand the series to find the general term.

 $$\begin{gathered} S_n=e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+... ...\left(1\right) \\ {S}_n=\sum _{ }\frac{{\mathrm{z}}^{\mathrm{n}}}{\mathrm{n}!} \end{gathered}$$

 

Find the value of ${\mathrm{a}}_{\mathrm{n}}$ and ${\mathrm{a}}_{\mathrm{n}+1}$   

 $\begin{array}{l}{\mathrm{a}}_{\mathrm{n}}=\frac{{\mathrm{z}}^{\mathrm{n}}}{\mathrm{n}!}\\ {\mathrm{a}}_{\mathrm{n}+1}=\frac{{\mathrm{z}}^{\mathrm{n}+1}}{(\mathrm{n}+1)!}\end{array}$

Step 4: Find the radius of convergence.

Use the ratio test. 

$$\begin{gathered} {\mathrm{\rho }}_{\mathrm{n}}=\frac{{\mathrm{a}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}}} \\ {\mathrm{\rho }}_{\mathrm{n}}=\left|\frac{{\displaystyle \frac{{\mathrm{z}}^{\mathrm{n}+1}}{\left(\mathrm{n}+1\right)!}}}{{\displaystyle \frac{{\mathrm{z}}^{\mathrm{n}}}{\mathrm{n}!}}}\right| \\ =\left|\frac{\mathrm{z}}{\mathrm{n}+1}\right| \end{gathered}$$ 

 

Calculate the value of $\mathrm{\rho }$,i.e., 

$\begin{array}{l}\mathrm{p}=\underset{\mathrm{n}\to \infty }{\lim }\left|{\mathrm{\rho }}_{\mathrm{n}}\right|\\ =\underset{\mathrm{n}\to \infty }{\lim }\left|\frac{\mathrm{z}}{\mathrm{n}+1}\right|\\ =0\end{array}$

 

${\mathrm{S}}_{\mathrm{n}}={\mathrm{e}}^{\mathrm{z}}$  is convergent for all values of z.

 

Hence, the entire complex plane is the radius of convergence.