/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 Use the series definition (2.72)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 50

Use the series definition (2.72) of \(e^{z}\) to prove that \(^{12} d e^{z} / d z=e^{z}\).

Short Answer

Expert verified
The derivative of \( e^{z} \) is \( e^{z} \).

Step by step solution

01

Recall the series definition of exponential function

The exponential function for a complex number \( z \) is defined by the series:\[ e^{z} = \sum_{n=0}^{\infty} \frac{z^n}{n!}. \] This series represents the exponential function \( e^{z} \) in terms of the series expansion.
02

Differentiate the series term by term

To differentiate \( e^{z} \) with respect to \( z \), we differentiate each term of the series:\[ \frac{d}{dz} \left( \sum_{n=0}^{\infty} \frac{z^n}{n!} \right) = \sum_{n=0}^{\infty} \frac{d}{dz} \left( \frac{z^n}{n!} \right). \]The derivative of each term \( \frac{z^n}{n!} \) is \( \frac{n z^{n-1}}{n!} \). For \( n = 0 \), the derivative is zero since \( z^0 \) is a constant.
03

Simplify the differentiated series

After differentiating, the series becomes:\[ \sum_{n=1}^{\infty} \frac{n z^{n-1}}{n!} = \sum_{n=1}^{\infty} \frac{z^{n-1}}{(n-1)!}. \]This is obtained because \( n/n! = 1/(n-1)! \). By shifting the index of summation \( m = n-1 \), the series becomes:\[ \sum_{m=0}^{\infty} \frac{z^m}{m!}. \]
04

Recognize the resulting series as the original series

The series \( \sum_{m=0}^{\infty} \frac{z^m}{m!} \) is the original definition of the exponential function \( e^{z} \). Therefore, the derivative of \( e^{z} \) is \( e^{z} \) itself.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Series Definition and Exponential Functions
The series definition is a powerful tool for understanding complex functions, such as the exponential function. In mathematics, an exponential function like \( e^{z} \) can be expressed as an infinite series:
  • \( e^{z} = \sum_{n=0}^{\infty} \frac{z^n}{n!} \).
This representation allows us to break down the function into simpler pieces, making the function more approachable. By breaking down the exponential into a series, each term contributes a piece of the puzzle, where:
  • \( n! \) is the factorial of \( n \).
  • \( z^n \) represents \( z \) raised to the power of \( n \).
Using such a series, we can perform operations like differentiation more straightforwardly, even with complex numbers. This definition provides a systematic way to expand and manipulate functions, crucial for solving mathematical problems.
Term-by-Term Differentiation of Series
When dealing with series like the one for the exponential function, term-by-term differentiation is a fundamental technique. This method involves differentiating each term of the series individually:
  • Differentiate \( \frac{z^n}{n!} \) to obtain \( \frac{n z^{n-1}}{n!} \).
  • For the first term when \( n = 0 \), the derivative is zero because it's a constant.
The beauty of term-by-term differentiation lies in its simplicity and directness. It relies on basic calculus rules applied systematically:
  • We don't differentiate the entire series at once, rather each term on its own.
  • It's a method that can often simplify solving differential equations.
This process leads to a new series that, for exponential functions, interestingly resembles the original function itself. This property is what makes functions like \( e^{z} \) so extraordinary and elegant in calculus and analysis.
Complex Analysis and Exponential Functions
Complex analysis introduces us to the fascinating world of functions of complex numbers. Here, the exponential function \( e^{z} \) becomes even more interesting. In complex analysis:
  • \( z \) represents a complex number, often written as \( z = x + yi \) where \( x \) and \( y \) are real numbers.
  • The function \( e^{z} \) extends its properties from real numbers to this complex plane.
An exponential function in complex analysis retains its identity under differentiation:
  • The derivative of \( e^{z} \) is \( e^{z} \), maintaining its form even for complex numbers.
This property highlights the harmony and predictability of exponential functions, whether in real or complex domains. Complex analysis uses these properties to explore deeper relationships and solve more intricate problems, revealing patterns that are both elegant and powerful.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

where the primes denote successive derivatives of \(f(x)\). (Depending on the function this series may converge for any increment \(\delta\) or only for values of \(\delta\) less than some nonzero "radius of convergence.") This theorem is enormously useful, especially for small values of \(\delta\), when the first one or two terms of the series are often an excellent approximation. \(^{11}\) (a) Find the Taylor series for \(\ln (1+\delta)\). (b) Do the same for cos \(\delta\). (c) Likewise sin \(\delta\). (d) And \(e^{\delta}\).

2.16 \(\star\) A golfer hits his ball with speed \(v_{\mathrm{o}}\) at an angle \(\theta\) above the horizontal ground. Assuming that the angle \(\theta\) is fixed and that air resistance can be neglected, what is the minimum speed \(v_{\mathrm{o}}(\min )\) for which the ball will clear a wall of height \(h\), a distance \(d\) away? Your solution should get into trouble if the angle \(\tan \theta < h / d .\) Explain. What is \(v_{\mathrm{o}}(\min )\) if \(\theta=25^{\circ}, d=50 \mathrm{m},\) and \(h=2 \mathrm{m} ?\)

A gun can fire shells in any direction with the same speed \(v_{\mathrm{o}}\). Ignoring air resistance and using cylindrical polar coordinates with the gun at the origin and \(z\) measured vertically up, show that the gun can hit any object inside the surface $$z=\frac{v_{0}^{2}}{2 g}-\frac{g}{2 v_{0}^{2}} \rho^{2}$$ Describe this surface and comment on its dimensions.

Problem 2.7 is about a class of one-dimensional problems that can always be reduced to doing an integral. Here is another. Show that if the net force on a one-dimensional particle depends only on position, \(F=F(x),\) then Newton's second law can be solved to find \(v\) as a function of \(x\) given by $$v^{2}=v_{\mathrm{o}}^{2}+\frac{2}{m} \int_{x_{0}}^{x} F\left(x^{\prime}\right) d x^{\prime}$$ [Hint: Use the chain rule to prove the following handy relation, which we could call the " \(v\) dv/d \(x\) rule": If you regard \(v\) as a function of \(x,\) then $$\dot{v}=v \frac{d v}{d x}=\frac{1}{2} \frac{d v^{2}}{d x}$$ Use this to rewrite Newton's second law in the separated form \(m d\left(v^{2}\right)=2 F(x) d x\) and then integrate from \(x_{\mathrm{o}}\) to \(x .\) ] Comment on your result for the case that \(F(x)\) is actually a constant. (You may recognise your solution as a statement about kinetic energy and work, both of which we shall be discussing in Chapter 4.)

For any complex number \(z=x+i y,\) the real and imaginary parts are defined as the real numbers \(\operatorname{Re}(z)=x\) and \(\operatorname{Im}(z)=y .\) The modulus or absolute value is \(|z|=\sqrt{x^{2}+y^{2}}\) and the phase or angle is the value of \(\theta\) when \(z\) is expressed as \(z=r e^{i \theta} .\) The complex conjugate is \(z^{*}=x-i y\) (This last is the notation used by most physicists; most mathematicians use \(\bar{z}\).) For each of the following complex numbers, find the real and imaginary parts, the modulus and phase, and the complex conjugate, and sketch \(z\) and \(z^{*}\) in the complex plane: (a) \(z=1+i \quad\) (b) \(z=1-i \sqrt{3}\) (d) \(z=5 e^{i \omega t}\) (c) \(z=\sqrt{2} e^{-i \pi / 4}\) In part (d), \(\omega\) is a constant and \(t\) is the time.
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Dynamics

Read Explanation

Torque and Rotational Motion

Read Explanation

Famous Physicists

Read Explanation

Rotational Dynamics

Read Explanation

Thermodynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.