Chapter 9: Problem 8
Find the first four nonzero terms of the Maclaurin series for the function by making an appropriate substitution in a known Maclaurin series and performing any algebraic operations that are required. State the radius of convergence of the series. $$ \text { (a) } \frac{x}{x-1} \quad \text { (b) } 3 \cosh \left(x^{2}\right) \quad \text { (c) } \frac{x}{(1+2 x)^{3}} $$
Short Answer
Step by step solution
Identify the Known Series
Derive Series for \( \frac{x}{x-1} \)
State Radius of Convergence for \( \frac{x}{x-1} \)
Use Known Series for \( \cosh(x^2) \)
Multiply by 3 for \( 3 \cosh(x^2) \)
State Radius of Convergence for \( 3 \cosh(x^2) \)
Identify the Known Series for \( \frac{1}{(1-x)^n} \)
Expand \( \frac{x}{(1+2x)^3} \)
State Radius of Convergence for \( \frac{x}{(1+2x)^3} \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Radius of Convergence
For example, the series \(\frac{1}{1-x}\) has a radius of convergence of 1, meaning it converges for all \(|x| < 1\). When finding the Maclaurin series for \(\frac{x}{x-1}\), which can be rewritten as \(-\frac{x}{1-x}\), the radius of convergence remains 1. This implies that the series representation of this function is valid as long as the value of \(x\) is within 1 unit on either side of the center 0.
Similarly, for the expression \(3 \cosh(x^2)\), since the hyperbolic cosine function \(\cosh(x)\) converges for all real numbers, the radius of convergence is infinite. This highlights how substitution of variables, such as substituting \(x^2\) in \(\cosh(x)\), doesn't affect the convergence range.
Lastly, in the case of \(\frac{x}{(1+2x)^3}\), using the binomial series leads to a radius of convergence of \(\frac{1}{2}\). This is calculated based on the condition \(|2x| < 1\), implying \(|x| < \frac{1}{2}\). Understanding this concept is crucial when utilizing power series to approximate functions.
Taylor Series
The Maclaurin series is essentially a Taylor series centered at zero. Both series take derivatives of the function, evaluated at the center point, to build polynomial terms. Each term in a Taylor series involves a factorial denominator and a ratio of the function's derivatives.
The series is represented by the formula:
\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots\]
A key insight from using Taylor series is that complicated functions can be approximated to great accuracy with only a few terms, especially when \(x\) is near the center \(a\).
To find the series for expressions like \(\cosh(x^2)\) or \(\frac{x}{(1+2x)^3}\), we can leverage known series and perform substitutions or algebraic manipulations. It's also possible to determine further terms of a series to increase approximation precision, as we did by calculating additional nonzero terms in the provided exercise.
Binomial Series
\[ (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots\]
To handle functions such as \(\frac{x}{(1+2x)^3}\), we use the binomial series for \((1+2x)^{-3}\).
Here's a brief overview of the steps:
- Use binomial coefficient notation to expand the series \((1+2x)^{-3}\).
- Calculate individual terms by substituting \(-3\) for \(n\) and \(2x\) for \(x\).
- Apply each term derived to the expression \(\frac{x}{(1+2x)^3}\).