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Magazine Chapter 9: Problem 10
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, \(\tan x=(\sin x) /(\cos x) .\) \(f(x)=\frac{1}{1+x} \ln \left(\frac{1}{1+x}\right)=\frac{-\ln (1+x)}{1+x}\)
Short Answer
Expert verified
The terms are: \(-x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5}\).
Step by step solution
01
Understand Maclaurin Series
The Maclaurin series for a function is a representation of the function as an infinite sum of terms, calculated from the derivatives of the function at zero. For any function \(f(x)\), its Maclaurin expansion is given by \(f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \).
02
Recall Known Series
Recall the Maclaurin series for \(\ln(1+x)\) and \(\frac{1}{1+x}\). They are:\[\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\]\[\frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - \cdots\]
03
Compute Maclaurin Series for \(-\ln(1+x)\)
Multiply the series for \(\ln(1+x)\) by \(-1\) to find the series expansion for \(-\ln(1+x)\):\[-\ln(1+x) = -x + \frac{x^2}{2} - \frac{x^3}{3} + \frac{x^4}{4} - \cdots\]
04
Divide Series by \(1+x\)
To find \(\frac{-\ln(1+x)}{1+x}\), divide the series for \(-\ln(1+x)\) by the series for \(\frac{1}{1+x}\). This involves polynomial division: \([-x + \frac{x^2}{2} - \frac{x^3}{3} + \frac{x^4}{4}] \div [1-x + \dots]\).
05
Perform Division Step-by-Step
Start dividing term by term: - Divide \(-x\) by 1 to get \(-x\). - Multiply \(-x\) by \(1-x\) to get \(-x + x^2\). Subtract this from the first terms: \(-x + \frac{x^2}{2}\) becomes \(-\frac{x^2}{2}\). - Divide \(-\frac{x^2}{2}\) by 1 to get \(-\frac{x^2}{2}\). - Repeat the process for further terms to find terms up to \(x^5\).
06
Collect Terms
Collect and combine the coefficients for each power of \(x\) after dividing:\[f(x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} + \cdots\]
07
Conclusion
The terms of the Maclaurin series for \(f(x)\) through \(x^5\) are:\[-x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5}\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Division
Polynomial division is a method used to divide one polynomial by another, similar to long division with numbers. In this context, it's utilized to simplify the expression of a function as a series. Consider dividing the polynomial \(-x + \frac{x^2}{2} - \frac{x^3}{3} + \frac{x^4}{4}\) by \(1-x\).
- Identify the leading term in both polynomials: For the numerator, it is \(-x\), and for the denominator, it is \(1\).
- Divide the leading term of the numerator by the leading term of the denominator to get the first term of the quotient.
- Multiply the entire divisor by this first term and subtract from the original polynomial to find the remainder.
- Repeat the process with the new polynomial (the remainder) until the desired degree is reached.
Series Expansion
Series expansion involves expressing a function as an infinite sum of terms, based on its Taylor or Maclaurin series. The idea is to approximate complex functions with simpler polynomial forms by using derivatives evaluated at a specific point.
- Maclaurin series are special cases of Taylor series, expanded at zero.
- Series expansion simplifies functions such as \(-\ln(1+x)\) and complex quotients like \(\frac{-\ln(1+x)}{1+x}\).
- By expanding these functions, you can compute terms up to desired powers of \(x\), such as terms through \(x^5\).
Derivatives
Derivatives are at the core of calculating a function's series expansion. A derivative provides a measure of how a function changes as its input changes, crucial for forming Taylor or Maclaurin series. Let's delve into why they matter:
- The Maclaurin series is developed from the derivatives of \(f(x)\) at zero: \(f(0), f'(0), f''(0), \ldots\).
- These derivatives give the coefficients for the powers of \(x\) in the series.
- For instance, the first derivative of \(\ln(1+x)\) reflects the gradient, while the second, third, etc., tell how the rate of change itself changes.
Power Series
A power series is an infinite series of the form \(a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots\), used to express functions in terms of simple polynomials that have potentially infinite terms. This concept is foundational in calculus:
- Power series enable representation of functions such as \(\ln(1+x)\ and \frac{1}{1+x}\) using basic operations.
- They can converge to functional forms over certain intervals of \(x\), allowing complex functions to be understood as simpler series.
- Power series are not only useful in approximating functions but are also pivotal in solving differential equations and evaluating integrals, where exact solutions are challenging.
