Q2P Show from the power series (8.1)... [FREE SOLUTION]

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Chapter 2: Q2P (page 61)

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

Short Answer

Expert verified

Answer:

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

Step by step solution

Given information

The power series (8.1) is ${鈭�}_{n = 0}^{鈭�} a_{n} {(x - c)}^{n} = a_{0} + a_{1} (x - c) + a_{2} {(x - c)}^{2} +$ .

Definition of Power series

A power series (8.1) is an infinite series that is shown in step 1 here $a_{n}$ represents the coefficient of the $n t h$ term and c as a constant.

Expand the series

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

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

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

Evaluate the right-hand side of the above series.

$\frac{d}{d z} e^{z} = \frac{d}{d z} 1 + \frac{d}{d z} (z) + \frac{d}{d z} (\frac{z^{3}}{2!}) + \frac{d}{d z} (\frac{z^{3}}{3!}) + \frac{d}{d z} (\frac{z^{4}}{4!}) + . . . . . . . . . = 0 + 1 + \frac{2 z}{2} + \frac{3 z^{3}}{3 路 2 路 1} + . . . . . . . . . = 1 + z + \frac{z^{2}}{2!} + \frac{z^{3}}{3!} + = e^{z}$

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

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91影视

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

Short Answer

Step by step solution

Given information

Definition of Power series

Expand the series

Differentiate the series with respect to z

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91影视

Show from the power series (8.1) that ddzez=ez.

Short Answer

Step by step solution

Given information

Definition of Power series

Expand the series

Differentiate the series with respect to z

One App. One Place for Learning.

Most popular questions from this chapter

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Study anywhere. Anytime. Across all devices.

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