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Magazine Chapter 6: Problem 15
Use an appropriate series in (2) to find the Maclaurin series of the given function. Write your answer in summation notation. $$\ln (1-x)$$
Short Answer
Expert verified
\(\ln(1-x) = -\sum_{n=1}^{\infty} \frac{x^n}{n}\).
Step by step solution
01
Identify the Known Series Expansion
Recall the Maclaurin series for the natural logarithm function. The series for the function \(-\ln(1-x)\) is given by the geometric series: \(-x - \frac{x^2}{2} - \frac{x^3}{3} - \cdots - \frac{x^n}{n}\). This is equivalent to \(\sum_{n=1}^{\infty} \frac{x^n}{n}\) for \(x\) in the interval \(-1 < x \leq 1\).
02
Adjust Series for the Target Function
Since we want \(\ln(1-x)\), we note that \(-\ln(1-x)\) was actually \(-1\) times the series we need. Thus, multiplying the series by \(-1\), we obtain:\(\ln(1-x) = - \left( x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots + \frac{x^n}{n} \right)\).
03
Express as Summation Notation
The Maclaurin series can be written in summation notation as:\[\ln(1-x) = - \sum_{n=1}^{\infty} \frac{x^n}{n},\]where each term \( \frac{x^n}{n} \) corresponds to the components of the original expansion.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Series Expansion
Series expansion is a powerful mathematical tool that helps us to represent complex functions in a simpler form using an infinite sum of terms. This is particularly useful for functions that are not easily manageable, like the logarithm or exponential functions.
To perform a series expansion, we break a function down into a series of simpler components. In mathematics, one of the most popular series expansions is the Maclaurin series. The Maclaurin series is a type of Taylor series centered at zero. It approximates functions by using polynomial terms derived from a function's derivatives at zero.
To perform a series expansion, we break a function down into a series of simpler components. In mathematics, one of the most popular series expansions is the Maclaurin series. The Maclaurin series is a type of Taylor series centered at zero. It approximates functions by using polynomial terms derived from a function's derivatives at zero.
- The terms of the series gradually increase in degree, allowing for a more accurate approximation of the function as more terms are added.
- This can provide a way to understand the behavior of functions in different intervals.
- Series expansion is particularly valuable in calculus, physics, and engineering calculations.
Natural Logarithm Function
The natural logarithm function, denoted as \( \ln(x) \), is a fundamental function in mathematics with a wide range of applications, especially in exponential growth and decay modeling. It is the inverse of the exponential function, \( e^x \), where \( e \) is Euler's number, an irrational constant approximately equal to 2.718.
The function \( \ln(1-x) \) specifically represents the natural logarithm of \( 1-x \), and it showcases interesting properties, particularly when combined with series expansions.
When analyzing \( \ln(1-x) \), remember that:
The function \( \ln(1-x) \) specifically represents the natural logarithm of \( 1-x \), and it showcases interesting properties, particularly when combined with series expansions.
When analyzing \( \ln(1-x) \), remember that:
- It is only defined for \( x \leq 1 \) (but \( x < 1 \) if expressed as \( \ln(1-x) \)).
- The term \( \ln(1-x) \) indicates shrinking as \( x \) approaches 1.
- It is often expanded into a series to be solvable in polynomial terms, especially for small values of \( x \).
Geometric Series
A geometric series is a series where each term after the first is found by multiplying the previous term by a constant, known as the common ratio. This kind of series can be expressed as:
\[ S = a + ar + ar^2 + ar^3 + \, \cdots \]
where \( a \) is the first term and \( r \) is the common ratio.
Understanding the geometric series is crucial in tackling series expansions like the Maclaurin series for \( \ln(1-x) \). When \( -\ln(1-x) \) is encoded as a geometric series, we derive that:
\[ S = a + ar + ar^2 + ar^3 + \, \cdots \]
where \( a \) is the first term and \( r \) is the common ratio.
Understanding the geometric series is crucial in tackling series expansions like the Maclaurin series for \( \ln(1-x) \). When \( -\ln(1-x) \) is encoded as a geometric series, we derive that:
- Each term \( \frac{x^n}{n} \) comes from integrating a geometric series \( x + x^2 + x^3 + \cdots \).
- The terms are structured to correspond directly to the coefficients in front of the \( x \) terms in the series.
- The reversal from \( -\ln(1-x) \) to \( \ln(1-x) \) straightens out the original but adjusted series, reflecting terms without alternating signs.
Summation Notation
Summation notation, also known as sigma notation, provides a concise way of expressing a sum of terms. It is represented by the Greek letter sigma (\( \Sigma \)) and is especially useful when dealing with series.
For \( \ln(1-x) \) in the Maclaurin series, summation notation helps display the series efficiently and clearly, as:\[ \ln(1-x) = - \sum_{n=1}^{\infty} \frac{x^n}{n} \]
For \( \ln(1-x) \) in the Maclaurin series, summation notation helps display the series efficiently and clearly, as:\[ \ln(1-x) = - \sum_{n=1}^{\infty} \frac{x^n}{n} \]
- The symbol \( \sum \) signifies that a sum of terms will follow.
- The subscripts and superscripts define the range of the sum, with \( n=1 \) as the lower limit and \( \infty \) representing an indefinite or "infinite" upper limit.
- The expression within the summation denotes what every term is in the series, in this case, \( \frac{x^n}{n} \), indicating how each term in the expansion is calculated.
Using summation notation is handy to avoid writing infinitely repeating terms and is a preferred format in mathematical expressions.
