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Chapter 8: Problem 28

In each of Exercises 23-34, derive the Maclaurin series of the given function \(f(x)\) by using a known Maclaurin series. $$ f(x)=e^{1-x} $$

Short Answer

Expert verified
The Maclaurin series for \( f(x) = e^{1-x} \) is \( e - ex + \frac{ex^2}{2} - \frac{ex^3}{6} + \cdots \).

Step by step solution

01

Recall the Maclaurin Series for a Known Function

The Maclaurin series for the exponential function \( e^x \) is given by: \[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \] This series will be used as a basis to derive the series for \( e^{1-x} \).
02

Substitute & Express \(f(x)\) Appropriately

We have \( f(x) = e^{1-x} = e^1 \cdot e^{-x} \). This can be seen as the product of \( e^1 \) and \( e^{-x} \). Recall that \( e^1 = e \) is a constant multiplier.
03

Derive the Maclaurin Series for \(e^{-x}\)

Replace \( x \) with \( -x \) in the Maclaurin series of \( e^x \), which gives:\[ e^{-x} = \sum_{n=0}^{\infty} \frac{(-x)^n}{n!} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots \]
04

Combine Results to Write the Series for \(f(x)\)

Since \( f(x) = e \cdot e^{-x} \), and \( e \) is simply a constant multiplier, the series for \( f(x) \) will be:\[ f(x) = e \left(1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots \right) \]This expansion means that each term in the series is multiplied by \( e \).
05

Write the Final Maclaurin Series

Therefore, the Maclaurin series for \( f(x) = e^{1-x} \) is:\[ f(x) = e - ex + \frac{ex^2}{2} - \frac{ex^3}{6} + \cdots \]Each term is the corresponding term from \( e^{-x} \) multiplied by \( e \).

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Key Concepts

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

Understanding the Exponential Function
The exponential function, typically expressed as \( e^x \), is a mathematical function showing exponential growth. Its base, \( e \), is a special constant approximately equal to 2.71828. The function has unique properties that make it an essential part of calculus and mathematical analysis.
  • It's continuous and differentiable everywhere.
  • The derivative and integral of \( e^x \) are itself \( e^x \).
  • It is characterized by a constant rate of growth proportional to its current value.
In this exercise, we are using the exponential function to find the Maclaurin series of a transformation, \( e^{1-x} \). This slight change modifies how we express its series, yet it retains its core characteristics of continuous and steady growth.
Exploring Series Expansion
Series expansion involves representing a function as an infinite sum of terms calculated from the values of its derivatives at a single point. The Maclaurin series is one common type of series expansion.

For the exponential function \( e^x \), the series expansion starts from \( n = 0 \) and goes to infinity:
\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\]
This series expansion allows a complex function to be expressed in simpler terms, making it easier to analyze or approximate values.
  • Each term is derived from the function's derivatives at zero.
  • It provides a polynomial approximation for functions.
The Maclaurin series helps in breaking down and reconstructing \( e^{1-x} \) by tweaking the series for \( e^x \) to account for the shift in variable.
The Role of Derivation in Series
Derivation is the technique in calculus used to determine the rate at which a function is changing. With series expansion, derivation plays a critical role in forming the terms of the series.

For the Maclaurin series, each coefficient in the polynomial series is determined by the derivatives of the function evaluated at zero:
\[ f^{(n)}(0) = \text{Coefficient for } \frac{x^n}{n!}\]
  • The first derivative gives the linear component.
  • The second derivative contributes to the quadratic component.
  • And so on, for higher-order derivatives.
Derivation thus enables the transformation of \( e^x \) to \( e^{1-x} \) by adjusting the base series through the substitution of \(-x\), tweaking each derivative鈥檚 contribution in the series.
Calculus and Its Applications
Calculus is a branch of mathematics focusing on limits, functions, derivatives, integrals, and infinite series. Its tools, such as derivation and integration, help us understand changes and find areas under curves.

  • It builds upon foundational concepts like limits and functions.
  • It provides ways to model dynamics and change.
  • Tools from calculus are widely used in physics, engineering, economics, and beyond.
In the specific context of the Maclaurin series, calculus allows us to express complex functions like \( e^{1-x} \) in simpler, more understandable polynomial terms. This helps in quick approximations and deeper understanding of behavior near the chosen point, usually zero in Maclaurin series.

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Most popular questions from this chapter

If \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) and \(g(x)=\sum_{n=0}^{\infty} b_{n} x^{n}\) converge on \((-R, R),\) then we may formally multiply the series as though they were polynomials. That is, if \(h(x)=f(x) g(x),\) then $$ h(x)=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n} a_{k} b_{n-k}\right) x^{n} $$ The product series, which is called the Cauchy product, also converges on \((-R, R) .\) Exercises \(59-62\) concern the Cauchy product. Suppose \(|x|<1 .\) Calculate the power series of \(h(x)=\) \(1 /(1-x)^{2}\) with base point 0 by using the method of Exercise 43\. Using \(f(x)=g(x)=1 /(1-x)=\sum_{n=0}^{\infty} x^{n},\) verify the Cauchy product formula for \(h=f \cdot g\) up to the \(x^{6}\) term.

It is known that \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{1}{6} \pi^{2}\) and \(\sum_{n=1}^{\infty} \frac{1}{n^{4}}=\frac{1}{90} \pi^{4}\). Let \(f(x, y)=\sum_{n=1}^{\infty} \frac{x+y \cdot(\pi n)^{2}}{n^{4}} .\) Describe the solution set \(\left\\{(x, y) \in \mathbb{R}^{2}: f(x, y)=0\right\\}\).

Prove that \(\sum_{n=1}^{\infty} \ln (1 / n)\) diverges.

Let \(\left\\{a_{n}\right\\}\) be a sequence of positive numbers. In a course on mathematical analysis, one learns that if the two limits \(\lim _{n \rightarrow \infty} a_{n+1} / a_{n}\) and \(\lim _{n \rightarrow \infty} a_{n}^{1 / n}\) exist, then they are equal. In each of Exercises \(65-68\), produce a plot that illustrates the equality of these two limits. Your plot should include a horizontal line that is the asymptote of the points \(\left\\{\left(n, a_{n+1} / a_{n}\right)\right\\}\) and \(\left\\{\left(n, a_{n}^{1 / n}\right)\right\\}\). \(a_{n}=e^{n} / \ln (1+n)\)

If \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) and \(g(x)=\sum_{n=0}^{\infty} b_{n} x^{n}\) converge on \((-R, R),\) then we may formally multiply the series as though they were polynomials. That is, if \(h(x)=f(x) g(x),\) then $$ h(x)=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n} a_{k} b_{n-k}\right) x^{n} $$ The product series, which is called the Cauchy product, also converges on \((-R, R) .\) Exercises \(59-62\) concern the Cauchy product. Suppose that the series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) converges on \((-R, R)\) to a function \(f(x)\) and that \(|f(x)| \geq k>0\) on that interval for some positive constant \(k\). Then, \(1 / f(x)\) also has a convergent power series expansion on \((-R, R) .\) Compute its coefficients in terms of the \(a_{n}\) 's. Hint: Set $$ \frac{1}{f(x)}=g(x)=\sum_{n=0}^{\infty} b_{n} x^{n} $$ Use the equation \(f(x) \cdot g(x)=1\) to solve for the \(b_{n}\) 's.
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