Q2P - Mathematical Methods in Physical Sciences

Question

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

 $\mathit{z}\mathbf{-}\frac{{\mathbf{z}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{z}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{-}\frac{{\mathbf{z}}^{\mathbf{4}}}{\mathbf{4}}\mathbf{+}\mathbf{\dots }$

Step-by-Step Solution

Answer

The disc of convergence is $\left|z\right|<1$.

Step 1: Given Information.

The given power series, i.e.,   ${\mathrm{S}}_{\mathrm{n}}=\mathrm{z}-\frac{{\mathrm{Z}}^2}{2}+\frac{{\mathrm{Z}}^3}{3}-\frac{{\mathrm{z}}^4}{4}+\dots$

Step 2: Definition of Disc 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 general term of the series.

Use the series to find general terms.

 

$$\begin{gathered} \begin{array}{l}{\mathrm{S}}_{\mathrm{n}}=\mathrm{z}-\frac{{\mathrm{z}}^2}{2}+\frac{{\mathrm{z}}^3}{3}-\frac{{\mathrm{z}}^4}{4}+... ... \left(1\right)\end{array} \\ {\mathrm{S}}_{\mathrm{n}}=\underset{ }{{\displaystyle \sum _{\mathrm{n}=1}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}+1}{\mathrm{z}}^{\mathrm{n}}}{\mathrm{n}}}} ... \left(2\right) \end{gathered}$$ 

Step 4: Find the disc of convergence.

Use the ratio test. 

$$\begin{gathered} {\mathrm{\rho }}_{\mathrm{n}}=\frac{{\mathrm{a}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}}} \\ {\mathrm{\rho }}_{\mathrm{n}}=\left|\frac{{\displaystyle \frac{{\left(-1\right)}^{\mathrm{n}+2}{\mathrm{z}}^{\mathrm{n}+1}}{\mathrm{n}+1}}}{{\displaystyle \frac{{\left(-1\right)}^{\mathrm{n}+1}{\mathrm{z}}^{\mathrm{n}}}{\mathrm{n}}}}\right| \\ =\left|\left(-1\right)\mathrm{z}\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right)\right| \\ =\left|\left(-1\right)\mathrm{z}\left(\frac{1}{1+1/\mathrm{n}}\right)\right| \end{gathered}$$ 

 

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

$$\begin{gathered} \mathrm{\rho }=\underset{\mathrm{n}\to \infty }{\lim }{\mathrm{\rho }}_{\mathrm{n}} \\ =\underset{\mathrm{n}\to \infty }{\lim }\left|\left(-1\right)\mathrm{z}\left(\frac{1}{1+{\displaystyle \frac{1}{\mathrm{n}}}}\right)\right| \\ =\left|\mathrm{z}\right| \end{gathered}$$ 

 

$S_n$is convergent for$\mathrm{\rho }<1$, i.e., $\left|\mathrm{z}\right|<1$.

 

Hence, the disc of convergence is $\left|\mathrm{z}\right|<1$.