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Magazine Chapter 6: Problem 96
Given that \(\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}\), use term-by-term differentiation or integration to find power series for each function centered at the given point. $$ \ln (1-x) \text { at } x=0 $$
Short Answer
Expert verified
\(\ln(1-x) = -\sum_{n=1}^{\infty} \frac{x^{n}}{n}\) for \(|x| < 1\).
Step by step solution
01
Recognize the power series representation
We are given the power series representation \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^{n} \). This power series holds for \(|x| < 1\). This is a geometric series with a common ratio of \(x\).
02
Rewrite the target function using integration
To find the power series for \( \ln(1-x) \), we notice that the derivative of \( \ln(1-x) \) with respect to \(x\) is \( -\frac{1}{1-x} \). Therefore, we can find \( \ln(1-x) \) by integrating \(-\frac{1}{1-x}\).
03
Integrate the power series term by term
Integrate the power series \( -\frac{1}{1-x} = -\sum_{n=0}^{\infty} x^{n} \) term by term:\[\int -\sum_{n=0}^{\infty} x^{n} \, dx = -\sum_{n=0}^{\infty} \int x^{n} \, dx = -\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} + C\]Here, \(C\) is the constant of integration.
04
Determine the constant of integration
We know that \( \ln(1-x) \) should be zero when \(x = 0\). Substitute \(x = 0\) into the series:\[\ln(1-0) = 0 = -\sum_{n=0}^{\infty} \frac{0^{n+1}}{n+1} + C = 0 + C\]This implies \(C = 0\). So, the constant is zero.
05
State the power series representation
The power series for \( \ln(1-x) \) centered at \(x=0\) is:\[\ln(1-x) = -\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} = -\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots \right)\]Valid for \(|x| < 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Geometric Series
A geometric series is a sum of terms that have a constant ratio between successive terms. The series we are dealing with here is \[ \frac{1}{1-x} = \sum_{n=0}^{\infty} x^{n} \]which can be called a power series representation of the geometric series. In this series, each term is formed by multiplying the previous term by the constant ratio, which in this case is \(x\).
- The standard form for a geometric series looks like: \( a + ar + ar^2 + ar^3 + \cdots \), where \(a\) is the first term and \(r\) is the common ratio.
- For convergence of a geometric series, the absolute value of the ratio, \(|r|\), should be less than 1.
- Geometric series are useful in assessing the convergence and sum of infinite series, allowing us to understand the behavior of functions as series expansions.
Integration
Integration is a fundamental concept in calculus and involves finding a function whose derivative is given. Here, it is used to transform the power series representation of \(\frac{1}{1-x}\) to that of \(\ln(1-x)\).
- Since the derivative of \(\ln(1-x)\) is \(-\frac{1}{1-x}\), we know that by integrating \(-\frac{1}{1-x}\), we can obtain \(\ln(1-x)\).
- When integrating term-by-term, each term of the series \(-\sum_{n=0}^{\infty} x^{n}\) changes by increasing the power of \(x\) by one and dividing by this new power, following the rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \).
Logarithmic Functions
Logarithmic functions are the inverse of exponential functions. The natural logarithm, \(\ln\), is particularly significant because of its properties and its use in calculus.
- The function \(\ln(1-x)\) is important in series expansion because it can be represented as an infinite series that converges for \(|x| < 1\).
- This power series representation is calculated using integration of the geometric series for \(\frac{1}{1-x}\).
- The representation for \(\ln(1-x)\) is \[-(x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots)\],which offers an accessible way to estimate the value of this function within its radius of convergence.
Term-by-Term Differentiation
Differentiating or integrating power series term-by-term allows for flexibility in manipulating series. It's effective because it maintains the convergence properties of the original series.
- When differentiating each term of a power series like \( \sum_{n=0}^{\infty} a_n x^n \), you treat each term independently, making problem-solving more straightforward than dealing with the function analytically.
- Term-by-term differentiation leads directly from the general form of a function to its derivative, aiding the computation of related functions like integrals in our example.
- This approach is valid within the radius of convergence of the series.
