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Chapter 9: Problem 32

Let \(f(x)=(1+x)^{1 / 2}+(1-x)^{1 / 2} .\) Find the Maclaurin series for \(f\) and use it to find \(f^{(4)}(0)\) and \(f^{(51)}(0)\).

Short Answer

Expert verified
\(f^{(4)}(0) = -3\), \(f^{(51)}(0) = 0\).

Step by step solution

01

Recognize the General Maclaurin Series Formula

The Maclaurin series of a function is the expansion of the function around 0, given by \( f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \cdots \).
02

Expand Each Component Using the Binomial Series

Recognize that both functions \((1+x)^{1/2}\) and \((1-x)^{1/2}\) can be expanded using the binomial series: \( (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots \).Apply this to: \((1+x)^{1/2} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \cdots \) and \((1-x)^{1/2} = 1 - \frac{1}{2}x - \frac{1}{8}x^2 - \frac{1}{16}x^3 + \cdots \).
03

Sum the Series to Form the Maclaurin Series for \(f(x)\)

Add the series expansions from Step 2: \(f(x) = ((1+x)^{1/2} + (1-x)^{1/2}) = (1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \cdots) + (1 - \frac{1}{2}x - \frac{1}{8}x^2 - \frac{1}{16}x^3 + \cdots) = 2 - \frac{1}{4}x^2 - \frac{1}{8}x^4 - \cdots \).
04

Identify Coefficients for Derivatives at 0

In the expansion from Step 3, note the coefficients of \(x^n\) give the factorial of the derivatives. The term with \(x^0\) is for \(f(0)\), the term \(x^2\) gives \(f''(0) = -\frac{1}{4} \times 2!\), and the term \(x^4\) gives \(f^{(4)}(0) = -\frac{1}{8} \times 4!\).
05

Compute \(f^{(4)}(0)\)

Using the series, \(f''(0) = -\frac{1}{4} \times 2! = -\frac{1}{4} \times 2 = -\frac{1}{2}\).For \(f^{(4)}(0) = -\frac{1}{8} \times 4! = -\frac{1}{8} \times 24 = -3\).
06

Determine \(f^{(51)}(0)\)

The Maclaurin series terms alternate signs and powers of even x due to the symmetry of roots.Since terms with odd powers of x are zero, \(f^{(51)}(0)\) does not exist as a term in our series expansion is zero by pattern observation.Therefore, it must be 0.

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Key Concepts

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

Binomial Series
The binomial series is a fundamental concept in calculus, providing a way to expand functions into infinite series based on binomials raised to any real number power. For a function of the form \[(1+x)^n,\] it can be expanded into a series as follows:
  • Start with the term 1, which doesn't depend on x.
  • Add the linear term: \(nx\).
  • Continue with quadratic term: \(\frac{n(n-1)}{2!}x^2\).
  • Subsequent terms follow the pattern \(\frac{n(n-1)(n-2)}{3!}x^3 + \cdots\)
Notice that the coefficients involve factorials and the powers of x increase successively. In our exercise, we expanded \((1+x)^{1/2}\) and \((1-x)^{1/2}\) using this series to facilitate the creation of a combined Maclaurin series for the function.
Derivatives
Derivatives are a key component in series expansions like the Maclaurin series. They represent the rates of change and are used to construct the coefficients in these series. For a function鈥檚 Taylor or Maclaurin series, the derivative coefficients are critically linked:
  • For the constant term, it's simple: use the function evaluated at 0, denoted as \(f(0)\).
  • The first derivative term, \(f'(0)\), gives the coefficient of \(x\).
  • Following this, the second derivative, \(f''(0)\), helps define the \(x^2\) term.
  • This pattern continues, with each derivative evaluated at 0 providing the coefficient for a corresponding term in the series.
In our exercise, the calculated derivatives like \(f''(0)\) and \(f^{(4)}(0)\) were extracted from the series terms to give exact results, showcasing the practical utility of derivatives in these expansions.
Series Expansion
Series expansion involves rewriting functions as infinite sums of terms. In calculus, they allow complex functions to be expressed in simpler polynomial-like forms, often making them easier to analyze or compute. The Maclaurin series, a special case of Taylor series centered at 0, is such an expansion:
  • It starts with the value of the function at 0.
  • Then it adds terms involving derivatives multiplied by powers of x, all evaluated at 0.
  • The format follows: \[f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \cdots\]
In our task, the function \(f(x)\) was simplified by converting its components via series expansion, showing how larger functions can be handled in a straightforward, step-by-step approach.
Taylor Series
The Taylor series is an extension of the Maclaurin series, providing a method to represent functions around any point using derivatives. While Maclaurin series are specifically centered around 0, Taylor series generalize this:
  • To construct a Taylor series, calculate derivatives of the function at a specific point \(a\).
  • Use these derivatives to form the series: \[f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots\]
  • The terms progressively fit the function鈥檚 behavior as \(x\) moves away from \(a\).
In practical applications, including our original exercise, the Taylor and Maclaurin series provide powerful tools to approximate functions. They are especially useful when direct computation is complex or impossible. Understanding the principles of these series is crucial for problem-solving in calculus.

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Most popular questions from this chapter

Find the terms through \(x^{5}\) in the Maclaurin series for \(f(x) .\) Hint: It may be easiest to use known Maclaurin series and then perform multiplications, divisions, and so on. For example, \(\tan x=(\sin x) /(\cos x)\). $$ f(x)=\frac{\cos x-1+x^{2} / 2}{x^{4}} $$

Find the Taylor polynomial of order 4 based at 2 for \(f(x)=x^{4}\) and show that it represents \(f(x)\) exactly.

Find the radius of convergence of $$ \sum_{n=1}^{\infty} \frac{1 \cdot 2 \cdot 3 \cdots n}{1 \cdot 3 \cdot 5 \cdots(2 n-1)} x^{2 n+1} $$

The author of a biology text claimed that the smallest positive solution to \(x=1-e^{-(1+k) x}\) is approximately \(x=2 k\), provided \(k\) is very small. Show how she reached this conclusion and check it for \(k=0.01\).

Determine the order \(n\) of the Maclaurin polynomial for \(e^{x}\) that is required to approximate \(e\) to five decimal places, that is, so that \(\left|R_{n}(1)\right| \leq 0.000005\).
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