/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Find a power series representati... [FREE SOLUTION] | 91Ó°ÊÓ

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Find a power series representation for the function and determine the interval of convergence. $$ f(x)=\frac{x^{2}}{x^{4}+16} $$

Short Answer

Expert verified
The power series is \(\sum_{n=0}^{\infty} (-1)^n \frac{x^{4n+2}}{16^{n+1}}\) with interval of convergence \((-2, 2)\).

Step by step solution

01

Rewrite the Function

Start by rewriting the function in a form that suggests how a power series can be obtained. We have:\[f(x) = \frac{x^2}{x^4 + 16}\]Rewriting the denominator as a difference of squares gives:\[f(x) = \frac{x^2}{16(1 - (-x^4)/16)}\]This transformation will allow us to use the geometric series formula.
02

Identify the Geometric Series

Recall the formula for the geometric series:\[\sum_{n=0}^{\infty} a r^n = \frac{a}{1 - r} \quad \text{if} \quad |r| < 1\]In our case, we can match this to the form:\[\frac{1}{1 - (-x^4)/16}\]which suggests using a geometric series with \(r = -\frac{x^4}{16}\) and \(a = 1\).
03

Apply the Geometric Series Expansion

Substitute \(r = -\frac{x^4}{16}\) into the geometric series expansion:\[\frac{1}{1 + \frac{x^4}{16}} = \sum_{n=0}^{\infty} \left(-\frac{x^4}{16}\right)^n\]This makes the series:\[\sum_{n=0}^{\infty} \left(-1\right)^n \frac{x^{4n}}{16^n}\]
04

Multiply by \(x^2\)

The function \(f(x)\) has an \(x^2\) factor in the numerator. Multiply the series from Step 3 by \(x^2\):\[\frac{x^2}{16} \sum_{n=0}^{\infty} \left(-1\right)^n \frac{x^{4n}}{16^n} = \sum_{n=0}^{\infty} \left(-1\right)^n \frac{x^{4n+2}}{16^{n+1}}\]
05

Determine the Interval of Convergence

The radius of convergence for a geometric series \(\sum_{n=0}^{\infty} r^n\) is obtained from \(|r| < 1\). Here, \(r = \frac{x^4}{16}\) must satisfy:\[\left|\frac{x^4}{16}\right| < 1\]Which simplifies to:\[|x| < 2\]Thus, the interval of convergence is \((-2, 2)\).

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Key Concepts

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

Interval of Convergence
When dealing with power series, determining the interval of convergence is a crucial step. The interval of convergence is the set of input values (for our variable, say \(x\)) where the power series sums up to a finite number. In simpler terms, it's where the series "works" and converges to a real number.
To find this interval, we often rely on conditions from recognized series types. In our particular case with geometric series, the condition to keep in mind is \(|r| < 1\), where \(r\) is the common ratio of the series.
This ratio tells us how each term in the series relates to the previous one. For example, in our exercise, the common ratio was identified as \(r = \frac{-x^4}{16}\).
  • By setting \(|r| < 1\), we derived \(|x| < 2\), indicating that the interval of convergence for our series is \((-2, 2)\).
  • It’s this process that ensures we pinpoint the exact range of \(x\)-values where our power series faithfully represents the function.
Geometric Series
A geometric series is a specific type of series with a distinct feature: each term is a constant multiple, known as the common ratio, of the previous term. It can be expressed as \(a + ar + ar^2 + \, ...\), and so on, where \(a\) is the first term and \(r\) is the common ratio.
This series is especially useful if the absolute value of \(r\) is less than 1 because the series converges or sums to a finite value. To show it more formally, this series is written in the form \(\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}\) for \(|r|<1\).

In the context of our problem, recognizing the structure \(\frac{1}{1-(-x^4)/16}\) as a geometric series allowed us to expand it effectively. This recognition is pivotal because it transforms what might otherwise be a complex expression into a more manageable sum of terms.
  • By setting \(a = 1\) and \(r = -\frac{x^4}{16}\), we could directly apply the geometric series formula.
  • This allowed for an easy expression of the function as a power series, facilitating further mathematical manipulation and understanding, which is particularly useful for integrals and differential equations.
Function Representation
Representing functions as power series is an important technique in calculus and analysis, often used to simplify functions into an infinite sum of terms. This approach gives several benefits:
  • It makes it easier to perform operations like differentiation and integration.
  • Additionally, power series can approximate functions well within their interval of convergence.
In our exercise, the function \(f(x) = \frac{x^2}{x^4 + 16}\) was expressed as a power series by first rewriting it to utilize the geometric series representation.
Starting with the rewritten denominator, \(16(1 - (-x^4)/16)\), enabled us to visualize and transform the function into a series.
This skill is practical because power series can represent a wide range of functions with just simple polynomials.
Furthermore, once a power series representation is obtained, it enables one to estimate function values, solve differential equations, or find anti-derivatives in situations where doing so symbolically or numerically might be complicated.
This flexibility showcases the beauty and utility of power series in mathematical analysis.

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Most popular questions from this chapter

The resistivity \(\rho\) of a conducting wire is the reciprocal of the conductivity and is measured in units of ohm-meters \((\Omega-\mathrm{m}) .\) The resistivity of a given metal depends on the temperature according to the equation \(\rho(t)=\rho_{20} e^{\alpha(t-2 \omega)}\) where \(t\) is the temperature in \(^{\circ} \mathrm{C}\). There are tables that list the values of \(\alpha\) (called the temperature coefficient) and \(\rho_{20}\) (the resistivity at \(20^{\circ} \mathrm{C}\) ) for various metals. Except at very low temperatures, the resistivity varies almost linearly with temperature and so it is common to approximate the expression for \(\rho(t)\) by its first-or second-degree Taylor polynomial at \(t=20 .\) $$ \begin{array}{l}{\text { (a) Find expressions for these linear and quadratic }} \\ {\text { approximations. }} \\ {\text { (b) For copper, the tables give } \alpha=0.0039 /^{\circ} \mathrm{C} \text { and }} \\ {\rho_{20}=1.7 \times 10^{-8} \Omega \text { -m. Graph the resistivity of copper }} \\\ {\text { and the linear and quadratic approximations for }} \\ {-250^{\circ} \mathrm{C} \leqslant t \leqslant 1000^{\circ} \mathrm{C} \text { . }} \\\ {\text { (c) For what values of } t \text { does the linear approximation agree }} \\ {\text { with the exponential expression to within one percent? }}\end{array} $$

Find the Taylor series for \(f(x)\) centered at the given value of \(a\). [ Assume that \(f\) has a power series expansion. Do not show that \(\left.R_{n}(x) \rightarrow 0 .\right]\) Also find the associated radius of convergence. $$ f(x)=e^{2 x}, \quad a=3 $$

Find the Taylor polynomial \(T_{3}(x)\) for the function \(f\) centered at the number \(a\). Graph \(f\) and \(T_{3}\) on the same screen. $$ f(x)=x e^{-2 x}, \quad a=0 $$

Find the Maclaurin series for \(f(x)\) using the definition of a Maclaurin series. [Assume that \(f\) has a power series expansion. Do not show that \(\left.R_{n}(x) \rightarrow 0 .\right]\) Also find the associated radius of convergence. $$ f(x)=\ln (1+x) $$

Use series to evaluate the limit. $$ \lim _{x \rightarrow 0} \frac{x-\ln (1+x)}{x^{2}} $$

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