Question

Show from the power series (8.1) that $\frac{\mathbf{d}}{\mathbf{d}\mathbf{z}}{\mathit{e}}^{\mathbf{z}}\mathbf{=}{\mathit{e}}^{\mathbf{z}}$.

Short answer

Answer:

It is proved that $\frac{d}{dz}{e}^z=e^z$.

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 here ${\mathit{a}}_{\mathbf{n}}$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 as:

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

Differentiate the series with respect to z

Differentiate both sides of the series $e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+.........$.

$\frac{d}{dz}{e}^z=\frac{d}{dz}\left(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+.........\right)$

Evaluate the right-hand side of the above series.

$$\begin{gathered} \frac{d}{dz}{e}^z=\frac{d}{dz}1+\frac{d}{dz}\left(z\right)+\frac{d}{dz}\left(\frac{z^3}{2!}\right)+\frac{d}{dz}\left(\frac{z^3}{3!}\right)+\frac{d}{dz}\left(\frac{z^4}{4!}\right)+......... \\ =0+1+\frac{2z}{2}+\frac{3z^3}{3·2·1}+......... \\ =1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+ \\ =e^z \end{gathered}$$

Hence, it is proved that$\frac{d}{dz}{e}^z=e^z$.