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Magazine Chapter 10: Problem 31
Add up the series. Problem 34 comes from (18). \(1+\frac{1}{2}(.01)+\frac{1}{3}(.001)+\cdots\)
Short Answer
Expert verified
The sum of the series is approximately 1.0536.
Step by step solution
01
Identify the Pattern
Let's observe the given series: \(1 + \frac{1}{2}(0.01) + \frac{1}{3}(0.001) + \cdots\). Notice that each term takes the form \(\frac{1}{n}(0.1)^{n-1}\), where \(n\) starts from 1.
02
Express the Series Formally
We can express this series as \(\sum_{n=1}^{\infty} \frac{1}{n}(0.1)^{n-1}\). This is an infinite series where we are summing terms in the form \(\frac{1}{n}x^{n-1}\) with \(x = 0.1\).
03
Recognize the Series Type
The given series resembles the power series derived from the basic properties of logarithms and can be related to the Taylor series for \(-\ln(1-x)\) when \(|x| < 1\), i.e., \(-\ln(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}\). However, here we have \(x^{n-1}\) instead of \(x^n\).
04
Modify the Expression
Rewrite the series using its properties: \(\sum_{n=1}^{\infty} \frac{1}{n}x^{n-1} = \frac{1}{x}\sum_{n=1}^{\infty} \frac{x^n}{n}\). For our series, \(x = 0.1\).
05
Apply the Logarithmic Series
Substitute into the logarithmic expansion: \(\sum_{n=1}^{\infty} \frac{x^n}{n} = -\ln(1-x)\). Therefore, \(\sum_{n=1}^{\infty} \frac{1}{n}x^{n-1} = \frac{1}{x}(-\ln(1-x))\).
06
Calculate Using \(x=0.1\)
Plug in \(x = 0.1\) and calculate: \(\sum_{n=1}^{\infty} \frac{1}{n}(0.1)^{n-1} = \frac{1}{0.1}(-\ln(0.9)) = 10(-\ln(0.9))\).
07
Simplify the Result
Calculate \(-\ln(0.9)\). Using logarithm tables or a calculator, \(-\ln(0.9) \approx 0.10536\). Thus, the series evaluates to \(10 \times 0.10536 = 1.0536\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Infinite Series
An infinite series is a sum of an endless sequence of numbers. Essentially, it seeks to add up all these numbers to approach a specific value, even though there are infinitely many terms. Infinite series are fundamental in calculus and analysis.
Some important points about infinite series are:
Some important points about infinite series are:
- Convergence: An infinite series converges if the sum approaches a certain finite value as more terms are considered. If the series do not converge to a finite value, we say it diverges.
- Partial Sums: One way to check for convergence is by examining the partial sums. Partial sums are simply the sums of the first few terms. If these sums stabilize at a particular value as more terms are added, the series may converge.
- Examples: Some famous examples include geometric series, harmonic series, and alternating series. Each has different conditions for convergence.
Power Series
A power series is a particular type of infinite series where each term involves a variable raised to increasing powers, multiplied by coefficients. It looks like this: \( \sum_{n=0}^{\infty} a_n x^n \). Here, \(a_n\) represents the coefficients and \(x^n\) represents the variable raised to the nth power.
- Radius of Convergence: In a power series, it's crucial to determine the values of \(x\) for which the series converges, known as the radius of convergence. The series only converges (i.e., the sum reaches a finite number) when \(x\) is within this range.
- Term by Term Differentiation & Integration: Power series can be differentiated or integrated term by term, which is a major utility in calculus.
- Applications: Power series are useful because they can be used to represent functions in terms of series, making them easier to work with analytically.
Taylor Series
The Taylor series is a specific kind of power series used to approximate functions. It represents a function as an infinite sum of terms calculated from its derivatives at a single point. The series takes the form:
\[ f(x) = \sum_{n=0}^{\infty} \frac{f^n(a)}{n!}(x-a)^n \] where \(f^n(a)\) is the nth derivative of the function evaluated at point \(a\).
\[ f(x) = \sum_{n=0}^{\infty} \frac{f^n(a)}{n!}(x-a)^n \] where \(f^n(a)\) is the nth derivative of the function evaluated at point \(a\).
- Center of Expansion: The point \(a\) where the series is centered is crucial as the approximation is most accurate near this point.
- Finite Approximations: Often, only the first few terms are used to approximate complex functions, which is very common in engineering and physics.
- Maclaurin Series: This is a special case of the Taylor series centered at zero, widely used when dealing with standard functions.
Logarithmic Series
Logarithmic series arise when logarithm functions are expanded into an infinite series. A common example is the series for the natural logarithm:
\( \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^n}{n} \) where this series converges when \(|x| < 1\).
\( \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^n}{n} \) where this series converges when \(|x| < 1\).
- Relationship with Logarithm: The series expansion of the logarithmic function showcases how complex continuous functions can be represented as a series.
- Convergence Restrictions: The logarithmic series only converge for
\(|x| < 1\). This is crucial when working with series to ensure correct results. - Applications: Logarithmic series are often used in situations where direct computation of logarithms is challenging. They are also key in calculus when integrating or solving differential equations involving logs.
