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Chapter 9: Problem 27

(a) Use the Maclaurin series for \(1 /(1-x)\) to find the Maclaurin series for $$f(x)=\frac{x}{1-x^{2}}$$ (b) Use the Maclaurin series obtained in part (a) to find \(f^{(5)}(0)\) and \(f^{(6)}(0)\). (c) What can you say about the value of \(f^{(n)}(0) ?\)

Short Answer

Expert verified
(a) \(f(x) = x + x^3 + x^5 + \cdots\). (b) \(f^{(5)}(0) = 120\), \(f^{(6)}(0) = 0\). (c) \(f^{(n)}(0) = 0\) for even \(n\), \(n!\) for odd \(n\).

Step by step solution

01

Recall the Maclaurin series for \(\frac{1}{1-x}\)

The Maclaurin series expansion for \(\frac{1}{1-x}\) is given by:\[\frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + \cdots = \sum_{n=0}^{\infty} x^n\]
02

Substitute \(x^2\) for \(x\) in the series

Since we have \(\frac{1}{1-x^2}\) in \(f(x)\), we substitute \(x^2\) for \(x\) in the Maclaurin series of \(\frac{1}{1-x}\):\[\frac{1}{1-x^2} = 1 + x^2 + x^4 + x^6 + \cdots = \sum_{n=0}^{\infty} x^{2n}.\]
03

Multiply the series by \(x\) to find \(f(x)\)

Given \(f(x) = \frac{x}{1-x^2}\), multiply the series from Step 2 by \(x\):\[f(x) = x(1 + x^2 + x^4 + x^6 + \cdots) = x + x^3 + x^5 + x^7 + \cdots = \sum_{n=0}^{\infty} x^{2n+1}.\]
04

Identify coefficients for derivatives

The coefficients in the series \(x + x^3 + x^5 + x^7 + \cdots\) represent the values of the derivatives at \(x=0\). Specifically, the coefficient of \(x^n\) corresponds to \(\frac{f^{(n)}(0)}{n!}\).
05

Calculate \(f^{(5)}(0)\) and \(f^{(6)}(0)\)

In the series, the coefficient of \(x^5\) is 1, so the fifth derivative \(f^{(5)}(0)\) is determined by \(\frac{f^{(5)}(0)}{5!} = 1\), thus \(f^{(5)}(0) = 5! = 120\).The coefficient of \(x^6\) is 0, as \(x^6\) does not appear in the series, thus \(f^{(6)}(0) = 0\).
06

Analyze the derivative pattern \(f^{(n)}(0)\)

From the series, observe that non-zero coefficients occur only at odd powers of \(x\). Thus, \(f^{(n)}(0)\) is zero for even \(n\) and \(n!\) for odd \(n\) when \(n = 2m+1\), and the coefficient of \(x^{2m+1}\) is 1.

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Key Concepts

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

Derivatives at Zero
Derivatives at zero play a crucial role in finding the Maclaurin series of a function. The Maclaurin series lets us express functions as a power series centered at zero. This involves taking derivatives at zero. For a function \( f(x) \), its Maclaurin series is:
  • \( f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \cdots \)
The coefficients \( \frac{f^{(n)}(0)}{n!} \) are derived from the function's value and its derivatives at zero. Understanding the pattern of these derivatives helps uncover more about the function's behavior and structure. This approach helps in making predictions about the function at non-zero values.
Power Series
A power series is a representation of a function as an infinite sum of terms involving powers of the variable. The Maclaurin series is a particular type of power series centered at zero. For example, the power series for \( \frac{1}{1-x} \) is given by:
  • \( \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \)
  • General form: \( \sum_{n=0}^{\infty} x^n \)
Power series have both a theoretical and practical applications. They simplify complex functions into infinite polynomials, making analytical and numerical computations easier.
The Maclaurin series trick of substituting variables (such as replacing \( x \) with \( x^2 \) in this case) allows us to find power series for more complex functions.
Function Expansion
Function expansion refers to expressing a function as a sum of simpler terms, most commonly using the Maclaurin series. In our example, the function \( f(x) = \frac{x}{1-x^2} \) was expressed as an infinite series by expanding the denominator's series:
  • Start with \( \frac{1}{1-x^2} = 1 + x^2 + x^4 + \cdots \)
  • Multiply by \( x \) to find \( f(x) = x + x^3 + x^5 + \cdots \)
This process involves creating an infinite polynomial where each term represents increasingly higher powers of \( x \). The ability to expand functions into series is powerful in calculus and analysis. It helps solve otherwise complicated problems by transforming them into simpler algebraic problems.
Higher-Order Derivatives
Higher-order derivatives refer to the successive derivatives of a function, beyond the first derivative. They are essential in understanding the curvature and behavior within the vicinity of a point, specifically at zero for Maclaurin series.
For the function \( f(x) = \frac{x}{1-x^2} \), the pattern of higher-order derivatives becomes clear:
  • Odd derivatives \( (f^{(5)}(0) = 120) \) take values based on factorials due to presence in the series.
  • Even derivatives \( f^{(6)}(0) = 0 \) are zero because their corresponding coefficients are absent in the odd series.
Observing these patterns simplifies differentiation and prediction. Understanding the role of higher-order derivatives in function expansion allows for precise approximations within proximity to \( x = 0 \).

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Most popular questions from this chapter

Determine whether the statement is true or false. Explain your answer. The Maclaurin series for a polynomial function has radius of convergence \(+\infty .\)

For each given \(p\) -series, identify \(p\) and determine whether the series converges. $$ \text { (a) } \sum_{k=1}^{\infty} k^{-4 / 3} \text { (b) } \sum_{k=1}^{\infty} \frac{1}{\sqrt[4]{k}} \quad \text { (c) } \sum_{k=1}^{\infty} \frac{1}{\sqrt[3]{k^{5}}} \quad \text { (d) } \sum_{k=1}^{\infty} \frac{1}{k^{\pi}} $$

Given an infinite series, discuss a strategy for deciding what convergence test to use.

Show that if \(p\) is a positive integer, then the power series $$ \sum_{k=0}^{\infty} \frac{(p k) !}{(k !)^{p}} x^{k} $$ has a radius of convergence of \(1 / p^{p}\)

Use the Maclaurin series for \(\sinh x\) and \(\cosh x\) to obtain the first four nonzero terms in the Maclaurin series for tanh \(x .\)
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