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Chapter 9: Problem 40

Use the power series \(\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n},|x|<1\) Find the series representation of the function and determine its interval of convergence. $$ f(x)=\frac{x}{(1-x)^{2}} $$

Short Answer

Expert verified
The series representation of the function \(f(x)=\frac{x}{(1-x)^{2}}\) is \(f(x) = \sum_{n=1}^{\infty} nx^{n}\) for \(|x|<1\).

Step by step solution

01

Identify the given power series and the function

We are given a function \(f(x)=\frac{x}{(1-x)^{2}}\) and a power series \(\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}\) when \(|x|<1\). We can use the power series and differentiation to find the series representation of the function.
02

Differentiate the power series

We differentiate the given power series by x, obtaining \(\frac{1}{(1-x)^{2}}= \sum_{n=0}^{\infty} nx^{n-1}\). This is almost similar to our function except for the x in the numerator in our function. The presence of that x changes the exponent on x in the power series from \(n-1\) to n.
03

Multiply by x

Now, we multiply the above series by x. So we get \(f(x) = \frac{x}{(1-x)^{2}}= \sum_{n=0}^{\infty} nx^{n}\). Now we need to handle the lower bound of the series, which cannot remain at n=0 when we multiply by n.
04

Adjust the lower bound

If we keep the lower bound at n = 0, that term contributes nothing to the sum, because 0*x^0 = 0. Therefore, we can safely raise the lower bound to n = 1. Hence, the series representation for the function is, \(f(x) = \sum_{n=1}^{\infty} nx^{n}\) with an adjusted lower bound.
05

Identify the interval of convergence

The interval of convergence remains unchanged. The original series was valid for \(|x|<1\), and the operations performed (differentiation and multiplication) did not change that. Thus, the interval of convergence for the series representation of the function is \(|x|<1\).

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Key Concepts

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

Interval of Convergence
When we talk about the interval of convergence, we're referring to the range of values for which a power series converges to a finite number. This is incredibly important because it determines where the power series representation of a function accurately depicts the behavior of the function.

For our given function, after differentiating the series to find the representation of the derivative, the interval of convergence does not change. It remains the same as that of the original power series, meaning the series representation of the function \(f(x)=\frac{x}{(1-x)^{2}}\) will also converge for \( |x|<1 \).

It's critical to note that operations like differentiation or integration can potentially change the interval of convergence, but in this case, it didn't. Let's examine more closely how different operations can affect this interval.
Series Differentiation
The concept of series differentiation is used when we want to find the power series representation of the derivative of a given function. Differentiating a power series term by term is a valid operation within the interval of convergence of the original series.

Here's a step-by-step differentiation of our given power series:
  • Start with \(\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}\) when \( |x|<1 \).
  • Differentiate the series term by term to find \(\frac{1}{(1-x)^{2}}= \sum_{n=0}^{\infty} nx^{n-1}\).
  • By multiplying each term by \(x\), we alter the power of x to \(n\) and get the series representation for our initial function \(f(x)\).
Without this procedure, finding the coefficients for the power series representation would be much more complex or in some cases, not obvious at all.
Infinite Series
An infinite series is essentially the sum of the terms of an infinite sequence. These series can either converge to a finite value or diverge to infinity.

Understanding how these series work is pivotal for evaluating power series representations of functions. There are various tests to determine if such a series converges, but in our exercise, we are already provided that the original series converges for \( |x|<1 \).

When we multiply the differentiated series by \(x\), we get the power series for our function \(f(x)=\frac{x}{(1-x)^{2}}\), which is an infinite series. While the series appeared to start from \(n=0\), we established that the term at \(n=0\) contributes nothing to the sum. Therefore, the effective series starts from \(n=1\), which is a critical detail for accurately representing the function's behavior through its power series.

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Most popular questions from this chapter

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. The series \(\sum_{n=1}^{\infty} \frac{n}{1000(n+1)}\) diverges.

Use a graphing utility to graph the function. Identify the horizontal asymptote of the graph and determine its relationship to the sum of the series. $$ f(x)=2\left[\frac{1-(0.8)^{x}}{1-0.8}\right] \quad \sum_{n=0}^{\infty} 2\left(\frac{4}{5}\right)^{n} $$

Investigation you found that the interval of convergence of the geometric series \(\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{n}\) is \((-2,2)\). (a) Find the sum of the series when \(x=\frac{3}{4}\). Use a graphing utility to graph the first six terms of the sequence of partial sums and the horizontal line representing the sum of the series. (b) Repeat part (a) for \(x=-\frac{3}{4}\). (c) Write a short paragraph comparing the rate of convergence of the partial sums with the sum of the series in parts (a) and (b). How do the plots of the partial sums differ as they converge toward the sum of the series? (d) Given any positive real number \(M\), there exists a positive integer \(N\) such that the partial sum \(\sum_{n=0}^{N}\left(\frac{3}{2}\right)^{n}>M\) $$ \begin{aligned} &\text { Use a graphing utility to complete the table. }\\\ &\begin{array}{|c|c|c|c|c|} \hline \boldsymbol{M} & 10 & 100 & 1000 & 10,000 \\ \hline \boldsymbol{N} & & & & \\ \hline \end{array} \end{aligned} $$

(a) Let \(f(x)=\sin x\) and \(a_{n}=n \sin 1 / n\). Show that \(\lim _{n \rightarrow \infty} a_{n}-f^{\prime}(0)=1\) (b) Let \(f(x)\) be differentiable on the interval \([0,1]\) and \(f(0)=0 .\) Consider the sequence \(\left\\{a_{n}\right\\}\), where \(a_{n}=n f(1 / n) .\) Show that \(\lim _{n \rightarrow \infty} a_{n}=f^{\prime}(0)\).

Fibonacci Sequence In a study of the progeny of rabbits, Fibonacci (ca. \(1170-\) ca. 1240 ) encountered the sequence now bearing his name. It is defined recursively by \(a_{n+2}=a_{n}+a_{n+1}, \quad\) where \(\quad a_{1}=1\) and \(a_{2}=1\). (a) Write the first 12 terms of the sequence. (b) Write the first 10 terms of the sequence defined by $$ b_{n}=\frac{a_{n+1}}{a_{n}}, \quad n \geq 1 . $$ (c) Using the definition in part (b), show that $$ b_{n}=1+\frac{1}{b_{n-1}} $$ (d) The golden ratio \(\rho\) can be defined by \(\lim _{n \rightarrow \infty} b_{n}=\rho\). Show that \(\rho=1+1 / \rho\) and solve this equation for \(\rho\).
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