/*! 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 45 The Taylor series generated by \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The Taylor series generated by \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) is \(\sum_{n=0}^{\infty} a_{n} x^{n}\) A function defined by a power series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) with a radius of convergence \(c>0\) has a Taylor series that con- verges to the function at every point of \((-c, c) .\) Show this by showing that the Taylor series generated by \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) is the series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) itself. An immediate consequence of this is that series like $$x \sin x=x^{2}-\frac{x^{4}}{3 !}+\frac{x^{6}}{5 !}-\frac{x^{8}}{7 !}+\cdots$$ and $$ x^{2} e^{x}=x^{2}+x^{3}+\frac{x^{4}}{2 !}+\frac{x^{5}}{3 !}+\cdots $$ obtained by multiplying Taylor series by powers of \(x,\) as well as series obtained by integration and differentiation of convergent power series, are themselves the Taylor series generated by the functions they represent.

Short Answer

Expert verified
A power series \( \sum_{n=0}^{\infty} a_{n} x^{n} \) with a radius of convergence \((-c, c)\) is its own Taylor series for \( f(x) = \sum_{n=0}^{\infty} a_{n} x^{n} \).

Step by step solution

01

Understand the Power Series Definition

A power series in the form \( \sum_{n=0}^{\infty} a_{n} x^{n} \) is a series where \( a_n \) are coefficients and \( x \) is a variable. The radius of convergence, \( c \), determines the interval \((-c, c)\) in which the series converges.
02

Define the Function by the Power Series

The function \( f(x) \) is defined by the convergent power series \( \sum_{n=0}^{\infty} a_{n} x^{n} \) where the series converges for \( x \) in the radius of convergence \((-c, c)\). This means for each \( x \) in that interval, \( f(x) \) can be exactly represented by this series.
03

Review Taylor Series Definition

A Taylor series generated by a function \( f(x) \) at \( x = 0 \) is given by \( \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n \). If \( f(x) = \sum_{n=0}^{\infty} a_{n} x^{n} \), computing each derivative \( f^{(n)}(0) \) will reproduce the coefficients \( a_n \) of the original series.
04

Show the Taylor Series Equals the Power Series

Consider that \( f(x) = \sum_{n=0}^{\infty} a_{n} x^{n} \). According to the definition of Taylor series, the terms are \( \frac{f^{(n)}(0)}{n!}x^n \). Since \( f^{(n)}(0) = n!a_n \) (demonstrated through differentiation of the power series), the terms simplify to \( a_n x^n \), reproducing the power series. Thus, the Taylor series generated by \( f(x) \) equals the original power series.
05

Discuss Examples

The examples provided, such as \( x \sin x = x^2 - \frac{x^4}{3!} + \cdots \) and \( x^2 e^x = x^2 + x^3 + \frac{x^4}{2!} + \cdots \), demonstrate that multiplying Taylor series or integrating/differentiating power series result in new functions that are still represented by the Taylor series.

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 91Ó°ÊÓ!

Key Concepts

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

Power Series
A power series is essentially an infinite sum in the form \[ \sum_{n=0}^{\infty} a_{n} x^{n} \]where each term consists of a coefficient \( a_n \) and a power of a variable \( x \). You can think of a power series as a polynomial of infinite degree. Unlike in a regular polynomial, where you only have a finite number of terms, a power series keeps going
  • Power series are versatile and form the basis for many mathematical functions.
  • They are key to understanding complex functions in an easy-to-grasp way.
If you plug different values of \( x \) into the series, the series may converge (sum up to a finite number) or diverge (go off to infinity). The power series can represent a real-valued function whenever it converges for the input \( x \). This gives us a handy tool to work with functions because we are able to treat them as infinite-degree polynomials, using calculus techniques like differentiation and integration more effectively.
Convergence
Convergence in the context of power series tells us where the series sums up to a finite number, making it behave nicely. Imagine a balance beam that only stays balanced on a specific section of a board. Similarly, a series is said to converge when its terms approach a certain value as you go to infinity.
  • This means you can rely on the series to accurately represent a function within this interval.
  • For a series to converge, the infinite sum of its terms must result in a finite number within its radius of convergence.
A convergent series paints a clear and coherent picture, akin to connecting dots smoothly between known points. When dealing with a power series like\[ \sum_{n=0}^{\infty} a_{n} x^{n} \]understanding where it converges allows us to know where the series realistically expresses the actual function values.
Radius of Convergence
The radius of convergence provides a barometer of how far from the center (usually zero), a power series will converge. Think of it as a safety zone beyond which the series becomes untrustworthy. It's defined as the distance from the center point \(x = 0\) up to where the series can be counted on to converge.
  • The radius of convergence \( c \) determines the interval \((-c, c)\) where the power series converges.
  • Beyond this interval, the series may diverge or not make sense.
Finding the radius of convergence is vital because it helps us know the valid range of \( x \) for which the series and its corresponding function agree. In practical terms, this is done through several tests, like the ratio test or the root test, to show where exactly a series holds its ground.
Differentiation of Series
Differentiation of power series is a powerful tool. It allows us to find new functions by taking derivatives term by term, just like you would with polynomials. When you differentiate a power series, you get another power series, which means you can use calculus techniques with ease.
  • For a series \( \sum_{n=0}^{\infty} a_{n} x^{n} \), differentiating term by term gives \( \sum_{n=1}^{\infty} na_{n} x^{n-1} \).
  • This retains the function within its radius of convergence.
Differentiation doesn't just stop at creating new series. When a power series converges to a function, differentiation can yield derivatives that are themselves represented as power series. This ensures continuity between algebraic and calculus operations, providing a deeper insight into the behavior of complex mathematical functions and their representations.

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

Uniqueness of least upper bounds Show that if \(M_{1}\) and \(M_{2}\) are least upper bounds for the sequence \(\left\\{a_{n}\right\\},\) then \(M_{1}=M_{2} .\) That is, a sequence cannot have two different least upper bounds.

Which of the series converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series’ convergence or divergence.) $$ \sum_{n=3}^{\infty} \frac{(1 / n)}{(\ln n) \sqrt{\ln ^{2} n-1}} $$

Sequences generated by Newton's method Newton's method, applied to a differentiable function \(f(x),\) begins with a starting value \(x_{0}\) and constructs from it a sequence of numbers \(\left\\{x_{n}\right\\}\) that under favorable circumstances converges to a zero of \(f .\) The recursion formula for the sequence is $$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} $$ a. Show that the recursion formula for \(f(x)=x^{2}-a, a>0\) can be written as \(x_{n+1}=\left(x_{n}+a / x_{n}\right) / 2\) b. Starting with \(x_{0}=1\) and \(a=3\) , calculate successive terms of the sequence until the display begins to repeat. What number is being approximated? Explain.

Logistic difference equation The recursive relation $$ a_{n+1}=r a_{n}\left(1-a_{n}\right) $$ is called the logistic difference equation, and when the initial value \(a_{0}\) is given the equation defines the logistic sequence \(\left\\{a_{n}\right\\} .\) Throughout this exercise we choose \(a_{0}\) in the interval \(03.57\) . Choose \(r=3.65\) and calculate and plot the first 300 terms of \(\left\\{a_{n}\right\\} .\) Observe how the terms wander around in an unpredictable, chaotic fashion. You cannot predict the value of \(a_{n+1}\) from previous values of the sequence. g. For \(r=3.65\) choose two starting values of \(a_{0}\) that are close together, say, \(a_{0}=0.3\) and \(a_{0}=0.301 .\) Calculate and plot the first 300 values of the sequences determined by each starting value. Compare the behaviors observed in your plots. How far out do you go before the corresponding terms of your two sequences appear to depart from each other? Repeat the exploration for \(r=3.75 .\) Can you see how the plots look different depending on your choice of \(a_{0} ?\) We say that the logistic sequence is sensitive to the initial condition a_{0} .

Approximate the sums in Exercises 49 and 50 with an error of magnitude less than \(5 \times 10^{-6}\) . $$ \sum_{n=0}^{\infty}(-1)^{n} \frac{1}{n !} $$

See all solutions

Recommended explanations on Math Textbooks

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.