Question

Show from the power series (8.1) that ${\mathit{e}}^{{\mathbf{z}}_{\mathbf{1}}}\mathbf{·}{\mathit{e}}^{{\mathbf{z}}_{\mathbf{2}}}\mathbf{=}{\mathit{e}}^{{\mathbf{z}}_{\mathbf{1}}\mathbf{+}{\mathbf{z}}_{\mathbf{2}}}$.

Short answer

Answer:

It is proved that $e^{{\mathrm{z}}_1}·e^{{\mathrm{z}}_2}=e^{{\mathrm{z}}_1+{\mathrm{z}}_2}$.

Step-by-step solution

Given information

The power series (8.1) is$\sum _{n=0}^{\infty }{a}_n{\left(x-c\right)}^n=a_0+a_1\left(x-c\right)+a_2{\left(x-c\right)}^2+\boxed{\xcancel{}}$.

Definition of power series

A power series (8.1) is an infinite series that is shown in step 1, where ${\mathit{a}}_{\mathbf{n}}$represents the coefficient of the $\mathit{n}\mathit{t}\mathit{h}$term and c as a constant.

Expand the series

It is known that $e^z=\sum _{n=0}^{\infty }\frac{z^n}{n!}$.

Expand the series.

$e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+.........$

Change the value of the power according to the question.

Put z in place of z.

$e^{z_1}=1+z_1+\frac{z_1^3}{3!}+.........$

Put $z_2$in place of z.

$e^{z_2}=1+{}_{}+\frac{z_2^2}{2!}+\frac{z_2^3}{3!}+.........$

Multiply both the expanded series

Multiply the expanded form of the series.

$$\begin{gathered} e^{z_1}·e^{z_2}=\left(1+z_1+\frac{z_1^2}{2!}+\frac{z_1^3}{3!}+.........\right)\left(1+z_2+\frac{z_2^2}{2!}+\frac{z_2^3}{3!}+.........\right) \\ =\left\{\begin{array}{c}\left(1+z_1+\frac{z_1^3}{2!}+\frac{z_1^3}{3!}+........\right)1+\left(1+z_1+\frac{z_1^2}{2!}+\frac{z_1^3}{3!}+.........\right)z_2\\ +\left(1+z_1+\frac{z_1^2}{2!}+\frac{z_1^3}{3!}+.........\right)\frac{z_2^2}{2!}+\left(1+z_1+\frac{z_1^2}{2!}+\frac{z_1^3}{3!}+........\right)\frac{z_2^3}{3!}+.........\end{array}\right\} \\ =\left\{\begin{array}{c}\left(1+z_1+\frac{z_1^3}{2!}+\frac{z_1^3}{3!}+........\right)+z_2+z_1{z}_2+z_2\frac{z_1^2}{2!}+z_2\frac{z_1^3}{3!}+........\\ +\frac{z_2^2}{2!}+z_1\frac{z_2^2}{2!}\frac{z_2^2}{2!}+\frac{z_2^2}{2!}\frac{z_2^3}{3!}.........\frac{z_2^3}{2!}+z_1\frac{z_2^3}{3!}+\frac{z_1^2}{2!}\frac{z_1^3}{3!}+\frac{z_1^3}{3!}\frac{z_2^3}{3!}+............\end{array}\right\} \end{gathered}$$

Simplify the expression further.

$$\begin{gathered} e^{z_1}·e^{z_2}=1+\left(z_1+z_2\right)+\frac{z_1^2}{2!}+\frac{z_2^2}{2!}+z_1{z}_2+\frac{z_1^3}{3!}+\frac{z_2^3}{3!}+\frac{z_1^{2}2}{2!}{z}_2+\frac{z_2^2}{2!}{z}_1+... \\ =1+\left(z_1+z_2\right)+\frac{1}{2!}\left(z_1^2+z_2^2+2z_1{z}_2\right)+\frac{1}{3!}\left(z_1^3+z_2^3+3z_1^2{z}_2+3z_1{z}_2^2\right)+......... \\ =1+\left(z_1+z_2\right)+\frac{{\left(z_1+z_2\right)}^2}{2!}+\frac{{\left(z_1+z_2\right)}^3}{3!}+......\left(1\right) \end{gathered}$$

The power series states that $e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+..........$.

Replace z by $\left(z_1+z_2\right)$in the above power series as:

$$\begin{gathered} \Rightarrow \left(z_1+z_2\right)+\frac{{\left(z_1+z_2\right)}^2}{2!}+\frac{{\left(z_1+z_2\right)}^2}{3!}+... \\ \Rightarrow e^{z_1+z_2} \end{gathered}$$

Substitute the derived value in the equation (1).

$e^{z_1}·e^{z_2}=e^{z_1+z_2}$

Hence, it is proved that $e^{z_1}·e^{z_2}=e^{z_1+z_2}$by using the power series.