Question

Using the definition of \(e^{z}\) by its power series \([(8.1) \text { of Chapter } 2],\) and the theorem (Chapters 1 and 2) that power series may be differentiated term by term (within the disk of convergence), and the result of Problem \(30,\) show that \((d / d z)\left(e^{z}\right)=e^{z}\).

Short answer

Differentiating the power series term by term confirms that \((d / dz)(e^z) = e^z\).

Step-by-step solution

Recall the Power Series Definition of \(e^z\)

Using the power series definition of the exponential function, \(e^z\) can be expressed as: \[ e^z = \frac{z^0}{0!} + \frac{z^1}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \frac{z^5}{5!} + \frac{z^6}{6!} + \frac{z^7}{7!} + \frac{z^8}{8!} + ... \] In its general form, \[ e^z = \frac{z^n}{n!} \text { for } n = 0, 1, 2, 3, ... \] Note that this series converges for all complex numbers \(z\).

Differentiate the Power Series Term by Term

Using the theorem that power series can be differentiated term by term within the disk of convergence, we differentiate each term of the series for \(e^z\): \[ (d/dz)\bigg(\frac{z^n}{n!}\bigg) = \frac{d}{dz} \bigg( \frac{z^0}{0!} + \frac{z^1}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + ... \bigg)= \frac{z^{0-1}}{0!} + \frac{z^{1-1}}{1!} + \frac{z^{2-1}}{2!} + \frac{z^{3-1}}{3!} + ... \]

Simplify the Differentiated Series

Simplify each differentiated term in the series: \[ (d/dz)\big(e^z\big) = \frac{0}{0!} + \frac{1}{1!} + \frac{2z}{2!} + \frac{3z^2}{3!} + \frac{4z^3}{4!} + ...=0 + 1 + \frac{2z}{2!} + \frac{3z^2}{3!} + \frac{4z^3}{4!} + ... \]

Observe the Resulting Power Series

Notice that the simplified series for the derivative of \(e^z\) now takes the form: \[ (d / dz)\big(e^z\big) = \frac{z^0}{0!} + \frac{z^1}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + ... = e^z\] The power series for \( e^z \) and the original series for \( e^z \) are identical, thus proving the statement.

Key concepts

Power Series

A power series is an infinite sum of the form \[ \text{Power Series} = \frac{z^0}{0!} + \frac{z^1}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \frac{z^5}{5!} + \frac{z^6}{6!} + \frac{z^7}{7!} + \frac{z^8}{8!} + \text{ ... } \] in which a variable 'z' is raised to increasing powers, divided by the factorial of the power number.Power series can represent many mathematical functions. For example, the exponential function, trigonometric functions, and many more can be elegantly expressed as power series.Power series are essential in calculus and complex analysis since they simplify the differentiation and integration of functions.
The convergence of a power series plays a crucial role in defining the interval or disk where the series represents the function reliably. When dealing with power series, always remember that within this interval, operations like differentiation and integration can be performed term by term.

Exponential Function

The exponential function is a fundamental mathematical function denoted as \(e^z\). This function can be described using a power series:\[ e^z = \frac{z^0}{0!} + \frac{z^1}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \text{ ... } \] where 'z' can be any real or complex number.This form is particularly useful in calculus because it simplifies to itself when differentiated or integrated.
The exponential function has interesting properties:

• Its growth rate is proportional to its current value
• The derivative of the function \( (d/dz) e^z = e^z \)
• It is periodic in complex analysis, linking beautifully with trigonometric functions through Euler's formula: \( e^{iz} = \text{cos}(z) + i\text{sin}(z)\)

These characteristics make it an invaluable tool in mathematics and applied sciences.

Complex Analysis

Complex analysis deals with functions of complex variables. It studies complex numbers and their functions, primarily focusing on analytic functions.A major part of complex analysis involves understanding the properties of functions like the exponential function when extended to the complex plane.
In complex analysis, the power series representation is vital because it preserves function properties like differentiability and analyticity within the disk of convergence.This means if a function is represented by a power series, it can be studied through each term of the series, making operations simpler and manageable.Additionally, complex analysis provides tools like contour integration and residue theorem, which are powerful techniques to solve difficult problems involving complex variables.

Convergence

Convergence in the context of a power series refers to the set of values for which the series converges to a particular limit.For the exponential function, the power series \[ e^z = \frac{z^0}{0!} + \frac{z^1}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \text{ ... } \] converges for all complex numbers 'z'.Understanding convergence is crucial because it tells us where the power series is valid.
In the case of differentiation, term-by-term, it ensures that the process of deriving each term will still represent the actual function accurately within the convergence region.Therefore, convergence guarantees that operations like differentiation and integration can be performed safely and accurately within the disk or interval of convergence.

Term-by-Term Differentiation

Term-by-term differentiation is a technique that allows us to differentiate a function represented by a power series by differentiating each term individually.For example, given the power series representation of \( e^z \):\[ e^z = \frac{z^0}{0!} + \frac{z^1}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \text{ ... } \]to find the derivative \( d/dz \big( e^z \big) \), we differentiate each term:\[ (d/dz) \frac{z^n}{n!} = \frac{nz^{n-1}}{n!} = \frac{z^{n-1}}{(n-1)!} \]This gives us another power series:\[ 0 + 1 + \frac{2z}{2!} + \frac{3z^2}{3!} + \text{ ... } \]which simplifies back to \( e^z \).
Term-by-term differentiation is possible only within the disk of convergence, and it tremendously simplifies calculations and proofs in calculus.