/*! 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 11 In Problems \(9-12\), the indica... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 11

In Problems \(9-12\), the indicated number is a zero of the given function. Use a Maclaurin or Taylor series to determine the order of the zero. $$ f(z)=1-e^{z-1} ; z=1 $$

Short Answer

Expert verified
The order of the zero at \( z=1 \) is 1.

Step by step solution

01

Identify the Function and Point

We have the function \( f(z) = 1 - e^{z-1} \) and we are given that \( z = 1 \) is a zero of the function. This means when \( z = 1 \), \( f(1) = 0 \). Our goal is to determine the order of this zero by expanding the function around this point.
02

Expand Using Taylor Series

The Taylor series for \( e^{x} \) around \( x = 0 \) is given by \( e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \). We will use this to expand \( e^{z-1} \) around \( z=1 \). Substituting \( x = z-1 \), the expansion becomes:\[e^{z-1} = e^0 + (z-1) + \frac{(z-1)^2}{2!} + \frac{(z-1)^3}{3!} + \cdots = 1 + (z-1) + \frac{(z-1)^2}{2} + \frac{(z-1)^3}{6} + \cdots \]
03

Substitute into the Function

Substitute the Taylor series expansion of \( e^{z-1} \) into \( f(z) \):\[f(z) = 1 - \left( 1 + (z-1) + \frac{(z-1)^2}{2} + \frac{(z-1)^3}{6} + \cdots \right) = 1 - 1 - (z-1) - \frac{(z-1)^2}{2} - \frac{(z-1)^3}{6} - \cdots \]Simplifying, we get:\[f(z) = -(z-1) - \frac{(z-1)^2}{2} - \frac{(z-1)^3}{6} - \cdots \]
04

Determine the Order of the Zero

From the expression \( f(z) = -(z-1) - \frac{(z-1)^2}{2} - \frac{(z-1)^3}{6} - \cdots \), we observe that the lowest power term in the expansion is \( (z-1) \). Thus, the order of the zero at \( z=1 \) is 1, since the term \( (z-1)^1 \) is the first non-zero term in the expansion.

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.

Order of Zero
Understanding the order of a zero is crucial when dealing with functions and their expansions using series. The 'order of zero' refers to the smallest positive integer, say \( n \), such that \( (z-z_0)^n \) is the first term in the series that does not vanish. In simpler terms, it indicates how many times the graph of the function "touches" or "crosses" the x-axis at a zero point.
  • The order of a zero is 1 if the function linearly passes through the zero.
  • If the zero is of order 2, the graph "bounces" at that point and does not cross the axis.
In our exercise, the function \( f(z) = 1 - e^{z-1} \) at \( z = 1 \) is such that the linear term \(- (z-1)\) is the first non-zero term when expanded, hence the order of the zero at \( z = 1 \) is 1.
Maclaurin Series
The Maclaurin series is a special case of the Taylor series and is centered at zero. It expresses a function as an infinite sum of its derivatives at this center point. Mathematically, it is given by: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots\]
  • It is used for approximating functions.
  • Maclaurin series help simplify complex functions at points close to zero.
  • The series applies to functions with derivatives of all orders.
In the given exercise, though centered at \( z = 1 \), it serves as a foundation for understanding series expansions around different points, resembling a Maclaurin series when \( z_0 = 0 \). By substituting \( (z-1) \), this transforms our series into a more conventional Taylor expansion with the same principle of using derivatives and their sums.
Exponential Function Expansion
One of the most common expansions involves the exponential function \( e^x \). The Taylor series expansion for \( e^x \), centered around zero, shows:\[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\]
  • This series is simple to remember and crucial for various calculations.
  • The exponential function is unique due to its infinite radius of convergence.
  • This expansion helps transform complex exponential terms into simple polynomial-like expressions.
In the exercise, expanding \( e^{z-1} \) using this concept allows us to simplify \( f(z) = 1 - e^{z-1} \) by breaking it down into terms that vividly illustrate the function's behavior around \( z = 1 \). Understanding how to expand exponential functions is key to analyzing and solving many equations in calculus and differential equations.

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

Evaluate the given trigonometric integral. \(\int_{0}^{2 \pi} \frac{\sin ^{2} \theta}{5+4 \cos \theta} d \theta\)

Evaluate the Cauchy principal value of the given improper integral. \(\int_{-\infty}^{\infty} \frac{\cos x}{\left(x^{2}+1\right)\left(x^{2}+9\right)} d x\)

Use Cauchy's residue theorem to evaluate the given integral along the indicated contour. \(\oint_{c} \frac{2 z-1}{z^{2}\left(z^{3}+1\right)} d z, C\) is the rectangle defined by \(x=-2, x=1\), \(y=-\frac{1}{2}, y=1\)

Use Cauchy's residue theorem to evaluate the given integral along the indicated contour. \(\oint_{C} \frac{\cos z}{(z-1)^{2}\left(z^{2}+9\right)} d z, C:|z-1|=1\)

Evaluate the Cauchy principal value of the given improper integral. \(\int_{-\infty}^{\infty} \frac{x^{2}}{\left(x^{2}+1\right)^{2}} d x\)
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Nuclear Physics

Read Explanation

Astrophysics

Read Explanation

Radiation

Read Explanation

Rotational Dynamics

Read Explanation

Further Mechanics and Thermal Physics

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.