/*! 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 43 By recognizing each series in Pr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

By recognizing each series in Problems \(43-51\) as a Taylor series evaluated at a particular value of \(x,\) find the sum of each of the following convergent series. $$1+\frac{2}{1 !}+\frac{4}{2 !}+\frac{8}{3 !}+\dots+\frac{2^{n}}{n !}+\dots$$

Short Answer

Expert verified
The sum of the series is \(e^2\).

Step by step solution

01

Recognize the Taylor Series Pattern

The given series is \[1 + \frac{2}{1!} + \frac{4}{2!} + \frac{8}{3!} + \dots + \frac{2^n}{n!} + \dots\] which can be rewritten as \[\sum_{n=0}^{\infty} \frac{2^n}{n!}\]. This series has a form that resembles a Taylor series expansion of the exponential function \(e^x\).
02

Recall the Taylor Series for Exponential Function

The Taylor series for \(e^x\) about \(x = 0\) is given by \(e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\).
03

Compare and Identify x Value

By comparing the series \(\sum_{n=0}^{\infty} \frac{2^n}{n!}\) with the Taylor series \(\sum_{n=0}^{\infty} \frac{x^n}{n!}\), we can identify that \(x = 2\). This means our series is a special case of the exponential function, specifically \(e^2\).
04

Conclude the Sum of the Series

Since the series \(\sum_{n=0}^{\infty} \frac{2^n}{n!}\) is the Taylor series for \(e^x\) with \(x = 2\), the sum of the series is \(e^2\).

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.

Exponential Function
The exponential function is one of the most fascinating and fundamental functions in mathematics. It's often denoted as \( e^x \), where \( e \) is a special mathematical constant approximately equal to 2.71828. This function is everywhere continuous and differentiable, making it incredibly important in calculus and real-world applications. An exponential function models growth or decay processes, such as population growth or radioactive decay.
When we talk about the Taylor series for exponential functions, it provides a way to approximate \( e^x \) using an infinite sum of terms based on derivatives at a single point. The ability to express \( e^x \) as a Taylor series allows mathematicians and scientists to understand and use exponential behaviors in various fields without evaluating complex expressions directly. Importantly, this series is always convergent for any real number \( x \).
Convergent Series
A convergent series is an infinite series that approaches a specific value as more terms are added. For a series \( \sum_{n=0}^{\infty} a_n \) to be convergent, the sum must get closer and closer to a finite limit. This is crucial because it means the series can be relied upon to represent real-world phenomena accurately.
In the context of the Taylor series, convergence is particularly significant. Since the series converges to \( e^x \) for any \( x \), it helps us calculate exponential values accurately. Convergent series provide powerful tools for approximations and calculations across physics, engineering, and finance, where precise values are essential.
  • An important test for convergence is the ratio test, which can be used to determine whether a series will converge based on the behavior of its terms as \( n \) increases.
  • Convergent series ensure that mathematical models remain valid over infinite processes.
Series Evaluation
Evaluating a series refers to determining its sum. With Taylor series, especially when dealing with exponentially convergent series, evaluation often means recognizing patterns that match standard expansions like the exponential function's series.
In the exercise provided, the series \( \sum_{n=0}^{\infty} \frac{2^n}{n!} \) needed to be identified and interpreted correctly. Recognizing it as a form of the exponential function's Taylor series allowed us to evaluate it as \( e^2 \). This kind of pattern recognition and substitution is crucial in mathematics, making complex series manageable and solvable.
Series evaluation techniques can extend to other functions by understanding and applying their respective Taylor series. Whether dealing with trigonometric functions, logarithmic functions, or exponentials, evaluating series accurately requires practice and familiarity with function expansions.

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

Find the value of the integer \(a_{i}\) given that the Taylor series of \(f(x)=\cos \sqrt{x}\) is: $$a_{0}+\frac{1}{a_{1}} x+\frac{1}{a_{2}} x^{2}+\frac{1}{a_{3}} x^{3}+\cdots$$ $$a_{2}$$

Decide if the statements are true or false. Give an explanation for your answer. The quadratic approximation to \(f(x)\) near \(x=-1\) shows that if \(f(-1)=g(-1), f^{\prime}(-1)=g^{\prime}(-1),\) and \(f^{\prime \prime}(-1)

By recognizing each series in Problems \(43-51\) as a Taylor series evaluated at a particular value of \(x,\) find the sum of each of the following convergent series. $$1-\frac{1}{3 !}+\frac{1}{5 !}-\frac{1}{7 !}+\dots+\frac{(-1)^{n}}{(2 n+1) !}+\cdots$$

Find a simplified formula for \(P_{5}(x),\) the fifth-degree Taylor polynomial approximating \(f\) near \(x=0\) Use the values in the table. $$\begin{array}{c|c|c|c|c|c}\hline f(0) & f^{\prime}(0) & f^{\prime \prime}(0) & f^{\prime \prime \prime}(0) & f^{(4)}(0) & f^{(5)}(0) \\\\\hline-3 & 5 & -2 & 0 & -1 & 4 \\\\\hline\end{array}$$

The theory of relativity predicts that when an object moves at speeds close to the speed of light, the object appears heavier. The apparent, or relativistic, mass, \(m\) of the object when it is moving at speed \(v\) is given by the formula $$m=\frac{m_{0}}{\sqrt{1-v^{2} / c^{2}}}$$, where \(c\) is the speed of light and \(m_{0}\) is the mass of the object when it is at rest. (a) Use the formula for \(m\) to decide what values of \(v\) are possible. (b) Sketch a rough graph of \(m\) against \(v,\) labeling intercepts and asymptotes. (c) Write the first three nonzero terms of the Taylor series for \(m\) in terms of \(v\) (d) For what values of \(v\) do you expect the series to converge?

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.