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Use power series operations to find the Taylor series at \(x=0\) for the functions. $$e^{x}+\frac{1}{1+x}$$

Short Answer

Expert verified
The Taylor series is: \[ 1 + \sum_{n=1}^{\infty} \left( \frac{1}{n!} + (-1)^n \right) x^n. \]

Step by step solution

01

Recall Taylor Series for Basic Functions

The Taylor series for basic functions around the point \(x=0\) are important for solving the problem. The Taylor series for \(e^x\) is \[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}. \] Similarly, the Taylor series for \(\frac{1}{1+x}\) is \[ \frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n. \] We will use these series to find the Taylor series of \(e^x + \frac{1}{1+x}\).
02

Sum the Series

To find the Taylor series for \(e^x + \frac{1}{1+x}\), we simply add the Taylor series for each separate function. This gives: \[\sum_{n=0}^{\infty} \frac{x^n}{n!} + \sum_{n=0}^{\infty} (-1)^n x^n.\]
03

Combine into One Series

Combine the series into one expression which results in: \[ 1 + \sum_{n=1}^{\infty} \left( \frac{1}{n!} + (-1)^n \right) x^n. \] Here, the constant terms were combined, and the rest of the terms were combined into a single series.
04

Conclusion

The Taylor series of the function \(e^x + \frac{1}{1+x}\) at \(x=0\) is: \[ 1 + \sum_{n=1}^{\infty} \left( \frac{1}{n!} + (-1)^n \right) x^n. \] Each term of \(x^n\) is a sum of the coefficients from each individual series.

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Key Concepts

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

Understanding Power Series
A power series is a type of infinite series where each term involves a power of a variable. In our exercise, the variable is \(x\). The general form of a power series can be expressed as
  • \( \sum_{n=0}^{http://http://http://http://http://fulltexthttp://infty} a_n x^n \),
where \(a_n\) is the coefficient for each term and \(x^n\) is the variable raised to the power \(n\). Power series are particularly useful for representing functions in a form that is easy to work with, especially for finding derivatives and integrals. They can be considered as a polynomial of infinite degree.
It allows us to approximate functions using polynomials, making complex calculations simpler.
For example, the Taylor series is a specific type of power series that provides an approximation of functions using derivatives. In the problem, the established Taylor series for basic functions like \(e^x\) and \(\frac{1}{1+x}\) are power series that expand these functions into an infinite sum.
Exploring Exponential Function
The exponential function, \(e^x\), is one of the most important functions in mathematics. It is characterized by the constant \(e\), which is an irrational number approximately equal to 2.71828. The function itself is defined for all real numbers and has a unique property: its rate of growth is proportional to its current value.
The Taylor series expansion of \(e^x\) is a classic example of a power series, given by:
  • \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \).
Here, each term in the series represents the derivative of \(e^x\) at zero divided by \(n!\), which is the factorial of \(n\). This property allows us to express \(e^x\) as an infinite polynomial, where each additional term increases the approximation's accuracy.
By understanding this power series representation, students can derive useful approximations and analyze the behavior of the exponential growth in various mathematical models.
Grasping Series Convergence
Series convergence is a critical concept when dealing with infinite series like power series. It refers to the ability of a series to approach a specific value as more terms are added.
In other words, as \(n\) becomes very large, the sum of the series approaches a particular value, which we call the limit.
To check the convergence of a series, mathematicians use various tests. For our task, since we are dealing with a known power series representation of \(e^x\) and \(\frac{1}{1+x}\), we can rely on their well-known convergence properties:
  • The series for \(e^x\) converges for any real number \(x\) due to its rapid decrease in the factorial \(n!\) in the denominator.
  • The series for \(\frac{1}{1+x}\) converges for \(|x| < 1\) because each term \((-1)^n x^n\) diminishes quickly as \(|x|\) is less than 1.
When we added these two series in our exercise, the convergence properties allow us to guarantee that the resulting series effectively represents the original function for acceptable ranges of \(x\).
Understanding series convergence helps students grasp why certain series work and how they approximate functions reliably.

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Most popular questions from this chapter

Prove that \(\lim _{n \rightarrow \infty} x^{1 / n}=1,(x>0)\).

A patient takes a 300 mg tablet for the control of high blood pressure every morning at the same time. The concentration of the drug in the patient's system decays exponentially at a constant hourly rate of \(k=0.12\) a. How many milligrams of the drug are in the patient's system just before the second tablet is taken? Just before the third tablet is taken? b. In the long run, after taking the medication for at least six months, what quantity of drug is in the patient's body just before taking the next regularly scheduled morning tablet?

a. Show that the improper integral $$ \int_{2}^{\infty} \frac{d x}{x(\ln x)^{p}} \quad(p \text { a positive constant }) $$ converges if and only if \(p>1\) b. What implications does the fact in part (a) have for the convergence of the series $$ \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}} ? $$ Give reasons for your answer.

Use the identity \(\sin ^{2} x=(1-\cos 2 x) / 2\) to obtain the Maclaurin series for \(\sin ^{2} x\). Then differentiate this series to obtain the Maclaurin series for \(2 \sin x \cos x .\) Check that this is the series for \(\sin 2 x\).

a. Find the interval of convergence of the power series $$\sum_{n=0}^{\infty} \frac{8}{4^{n+2}} x^{n}$$ b. Represent the power series in part (a) as a power series about \(x=3\) and identify the interval of convergence of the new series. (Later in the chapter you will understand why the new interval of convergence does not necessarily include all of the numbers in the original interval of convergence.)

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