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Chapter 9: Problem 21

Confirm the derivative formula by differentiating the appropriate Maclaurin series term by term. $$ \text { (a) } \frac{d}{d x}[\cos x]=-\sin x \quad \text { (b) } \frac{d}{d x}[\ln (1+x)]=\frac{1}{1+x} $$

Short Answer

Expert verified
Derivative formulas are confirmed by term-by-term differentiation of their Maclaurin series.

Step by step solution

01

Recall Maclaurin Series

The Maclaurin series for a function is the Taylor series centered at zero. For \(\cos x\), the Maclaurin series is \(\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\). For \(\ln(1+x)\), the series is \(\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\).
02

Differentiate the Series of \(\cos x\) Term by Term

Differentiate each term of the Maclaurin series for \(\cos x\):- The derivative of \(1\) is \(0\).- The derivative of \(-\frac{x^2}{2!}\) is \(-x\).- The derivative of \(\frac{x^4}{4!}\) is \(\frac{x^3}{3!}\).- Continue differentiating each term in a similar manner leading to the series for \(-\sin x\):\(-\sin x = -x + \frac{x^3}{3!} - \frac{x^5}{5!} + \cdots\).
03

Differentiate the Series of \(\ln(1+x)\) Term by Term

Differentiate each term of the Maclaurin series for \(\ln(1+x)\):- The derivative of \(x\) is \(1\).- The derivative of \(-\frac{x^2}{2}\) is \(-x\).- The derivative of \(\frac{x^3}{3}\) is \(x^2\).- Continue on which leads to the series for \(\frac{1}{1+x}\):\(\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots\).
04

Conclusion

Both differentiated series yield the expected results:- Differentiating the Maclaurin series for \(\cos x\) term by term yields \(-\sin x\).- Differentiating the Maclaurin series for \(\ln(1+x)\) term by term yields \(\frac{1}{1+x}\). Hence, the derivative formulas are confirmed.

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Key Concepts

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

Maclaurin series
The Maclaurin series is a special case of the Taylor series. It focuses on expanding functions about zero. This means for any function, you can express it as a power series centered at zero using its derivatives.
  • The Maclaurin series is written as \[ f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots \]
  • This is helpful in approximating functions like \ \cos x\ and \ \ln(1+x)\ near zero.
  • For example, \ \cos x\ can be expanded into the series: \[\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \]
  • The function \ \ln(1+x)\ uses: \[\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \]
The beauty of this series lies in its ability to approximate complicated functions using polynomials, making calculations much simpler. Understanding how to expand these functions helps in various calculus problems, such as differentiation and integration using series.
derivatives
Derivatives help us understand the rate of change of a function. When working with series like the Maclaurin series, differentiating term by term gives us insight into new series.
  • For a given function, its derivative at a point shows how the function value changes with respect to changes in input.
  • In the exercise, the series for \ \cos x\ is differentiated term by term to confirm the formula \ \frac{d}{dx}[\cos x] = -\sin x\.
    • The initial term of 1 becomes 0.
    • The term \( -\frac{x^2}{2!} \) turns into \( -x \), and so forth.
  • Similarly, the series for \ \ln(1+x)\ is differentiated to reach \ \frac{1}{1+x}\.
    • Here, \( x \) becomes 1 and \( -\frac{x^2}{2} \) results in \( -x \).
Term-by-term differentiation maintains the accuracy of the series expansion, and is a crucial technique that showcases how derivatives reveal underlying changes in functions, especially near zero for the Maclaurin series.
Taylor series
A Taylor series gives a more general way to expand a function about any point, not just zero. Unlike the Maclaurin series, it can center on any value \( a \), developing a deeper understanding of function behavior around specific points.
  • The Taylor series is expressed as \[ f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots \]
  • Choosing \( a = 0 \) simplifies this to the Maclaurin series. Hence, every Maclaurin series is a Taylor series, just specifically centered at zero.
  • Taylor series are beneficial for functions where behavior and changes need inspection around particular \( a \) values.
  • They offer flexibility when solving more complex problems in mathematics, physics, and engineering by approximating non-polynomial functions.
Taylor series and its specific case, the Maclaurin series, provide powerful methods for making infinite series approximations of functions, helping us solve calculus-related problems effectively.

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Most popular questions from this chapter

Determine whether the series converges. $$ \sum_{k=1}^{\infty} \frac{k}{\ln (k+1)} $$

(a) Find examples to show that if \(\sum a_{k}\) converges, then \(\sum a_{k}^{2}\) may diverge or converge. (b) Find examples to show that if \(\sum a_{k}^{2}\) converges, then \(\quad \sum a_{k}\) may diverge or converge.

Consider the series $$ \sum_{k=1}^{\infty} \frac{(-3)^{k}}{k} x^{k} $$ Determine the intervals of convergence for this series and for the series obtained by integrating this series term by term.

Find the first five nonzero terms of the Maclaurin series for the function by using partial fractions and a known Maclaurin series. $$\frac{x^{3}+x^{2}+2 x-2}{x^{2}-1}$$

In each part, find upper and lower bounds on the error that results if the sum of the series is approximated by the 10 th partial sum. $$ \text { (a) } \sum_{k=1}^{\infty} \frac{1}{(2 k+1)^{2}} \text { (b) } \sum_{k=1}^{\infty} \frac{1}{k^{2}+1} \quad \text { (c) } \sum_{k=1}^{\infty} \frac{k}{e^{k}} $$
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