/*! 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 62 Use the Taylor series for 1\(/\l... [FREE SOLUTION] | 91Ó°ÊÓ

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Use the Taylor series for 1\(/\left(1-x^{2}\right)\) to obtain a series for 2\(x /\left(1-x^{2}\right)^{2}.\)

Short Answer

Expert verified
The series is \( \sum_{n=1}^{\infty} 4n x^{2n} \).

Step by step solution

01

Identify the Taylor series for \( \frac{1}{1-x^2} \)

We start by finding the Taylor series for the function \( \frac{1}{1-x^2} \). The geometric series formula states that for \(-1 < x < 1\), \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \). Substitute \(x^2\) for \(x\) in this series:\[ \frac{1}{1-x^2} = \sum_{n=0}^{\infty} x^{2n}. \]
02

Derive the series for \( \frac{1}{(1-x^2)^2} \)

We take the derivative of the series \( \frac{1}{1-x^2} \) with respect to \(x\) to find the series for \( \frac{1}{(1-x^2)^2} \). The derivative of \( \frac{1}{1-x^2} \) is \( \frac{2x}{(1-x^2)^2} \).Differentiating the series term by term:\[ \frac{d}{dx} \left( \sum_{n=0}^{\infty} x^{2n} \right) = \sum_{n=1}^{\infty} 2n x^{2n-1}. \]Thus, \( \frac{1}{(1-x^2)^2} = \sum_{n=1}^{\infty} 2n x^{2n-1}. \)
03

Multiply the series by \( 2x \)

To find the series for \( \frac{2x}{(1-x^2)^2} \), we multiply the series obtained from Step 2 by \(2x\):\[ 2x \cdot \sum_{n=1}^{\infty} 2n x^{2n-1} = \sum_{n=1}^{\infty} 4n x^{2n}. \]Now we have the series representation for \( \frac{2x}{(1-x^2)^2} \), which is \( \sum_{n=1}^{\infty} 4n x^{2n}. \)

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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 one of the fundamental concepts in mathematics.Understanding this helps unravel the Taylor series, like in the exercise given.A geometric series is a sum of terms where each term is a constant multiple (called the common ratio) of the previous.For instance, the series: \( a, ar, ar^2, ar^3, \dots \) forms a geometric series.This type of series can be summed using the formula:
  • \( S = \frac{a}{1 - r} \)
where \(a\) is the first term and \(r\) is the common ratio.
In the context of the Taylor series, we use the geometric series formula as follows:
  • For \(-1 < x < 1\), the geometric series for \( \frac{1}{1-x} \) is \( \sum_{n=0}^{\infty} x^n \)
By substituting \(x^2\) for \(x\), we achieve the series for \( \frac{1}{1-x^2} \):
  • \( \frac{1}{1-x^2} = \sum_{n=0}^{\infty} x^{2n} \)
This manipulation is crucial because it lays the foundation for more complex series manipulations.
Derivative of a Series
Once we have a series, finding its derivative can offer deeper insights and new formulas.In this exercise, the series \( \frac{1}{1-x^2} = \sum_{n=0}^{\infty} x^{2n} \) is differentiated with respect to \(x\).By doing so, we aren't just developing a new series, but also applying calculus to see how the function behaves.
The differentiation is straightforward:
  • Differentiate term by term to get \( \sum_{n=1}^{\infty} 2n x^{2n-1} \)
This gives the series for the derivative \( \frac{d}{dx}( \frac{1}{1-x^2}) = \frac{2x}{(1-x^2)^2} \).
In essence, taking the derivative transforms our original problem into a new form, helping us create a series for a more complex function like \( \frac{1}{(1-x^2)^2} \).This is key in developing the new series representation through calculus.
Power Series Representation
Power series representation is a way of expressing a function as an infinite sum of terms with powers of a variable.This concept is critical in the exercise as it breaks down functions into manageable series.Having derived \( \frac{1}{(1-x^2)^2} \) as \( \sum_{n=1}^{\infty} 2n x^{2n-1} \), the next step is crucial.
We find the power series representation for \( \frac{2x}{(1-x^2)^2} \) by multiplying our series by \(2x\).This process involves only simple multiplication:
  • Multiply each term by \(2x\) to get \( \sum_{n=1}^{\infty} 4n x^{2n} \)
Power series representation allows functions to be expressed in an infinite sum, facilitating easy computation and estimation.
Functions distilled into a power series are more straightforward to explore, giving insights into their behavior at different points. This makes power series representation a powerful tool for mathematicians and students alike.

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Most popular questions from this chapter

Which of the series in Exercises 13 46 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{n}{n^{2}+1} $$

Estimate the value of \(\sum_{n=1}^{\infty}\left(1 / n^{3}\right)\) to within 0.01 of its exact value.

Pythagorean triples A triple of positive integers \(a, b,\) and \(c\) is called a Pythagorean triple if \(a^{2}+b^{2}=c^{2} .\) Let \(a\) be an odd positive integer and let $$ b=\left\lfloor\frac{a^{2}}{2}\right\rfloor \text { and } c=\left\lceil\frac{a^{2}}{2}\right\rceil $$ be, respectively, the integer floor and ceiling for \(a^{2} / 2\) a. Show that \(a^{2}+b^{2}=c^{2} .\) (Hint: Let \(a=2 n+1\) and express \(b\) and \(c\) in terms of \(n .\) b. By direct calculation, or by appealing to the accompanying figure, find $$ \lim _{a \rightarrow \infty} \frac{\left\lfloor\frac{a^{2}}{2}\right\rfloor}{ | \frac{a^{2}}{2} \rceil} $$

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L\) ? b. If the sequence converges, find an integer \(N\) such that \(\quad\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{1}=1, \quad a_{n+1}=a_{n}+\frac{1}{5^{n}} $$

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L\) ? b. If the sequence converges, find an integer \(N\) such that \(\quad\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=\frac{n^{41}}{19^{n}} $$

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