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Chapter 9: Problem 26

Using a Power Series In Exercises \(17-26,\) use the power series $$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n} x^{n}$$ to determine a power series, centered at \(0,\) for the function. Identify the interval of convergence. $$ f(x)=\arctan 2 x $$

Short Answer

Expert verified
The power series representation for \(f(x) = \arctan 2x\) is \(\sum_{n=0}^{\infty}(-1)^n 4^n x^{2n}\) and its interval of convergence is \(-0.5 < x < 0.5\).

Step by step solution

01

Recognize the Power Series

Recognize that the power series given \(\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n} x^{n}\) is the Maclaurin series representation for \(\frac{1}{1+x}\).
02

Convert Function Into Form of Power Series

Notice that \(\arctan 2x\) can be written as \(\frac{\pi}{4} - \frac{1}{1 + (2x)^2}\). We can easily find the power series of \(\frac{1}{1 + u}\) where \(u = (2x)^2\), by replacing \(x\) in the given power series with \(u\), resulting with \(\frac{1}{1 + u}=\sum_{n=0}^{\infty}(-1)^{n} u^{n}\).
03

Apply Power Series

Now, the power series representation for \(\arctan 2x\) can be found by replacing \(u\) in the derived power series in the previous step with \((2x)^2\). So, the power series representation will be \(\sum_{n=0}^{\infty}(-1)^n (2x)^{2n}\). So, the power series for \(\arctan 2x\) would be \(\sum_{n=0}^{\infty}(-1)^n 4^n x^{2n}\).
04

Determine Interval of Convergence

Recall that the interval of convergence for the original function \(\frac{1}{1 + x}\) is \(-1 < x < 1\) as given by the standard theory for power series. When \(x\) is replaced with \((2x)^{2}\), \(\frac{1}{1 + (2x)^2}\) will converge when \(-1 < (2x)^2 < 1\), which simplifies to \(-1 < 2x < 1\). Then the interval would be \(-0.5 < x < 0.5\). That's the interval of convergence.
05

Identify Coefficients

Identify the coefficients for the power series as \( (-1)^n 4^n \), for the variable \( x^{2n} \).

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Key Concepts

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

Maclaurin series
The Maclaurin series is a special kind of power series. It represents a function as an infinite sum of terms calculated from the derivatives of the function at a single point, usually zero. In simple terms, it's a way to express a function as an infinite polynomial centered at zero. For the series given in the exercise,
  • The function \(\frac{1}{1+x}\) is expanded as \(\sum_{n=0}^{\infty}(-1)^{n} x^{n}\).
  • This series showcases how \(1 + x\) can be decomposed into an infinite series that approaches the function for small values of \(x\).
This Maclaurin series provides a powerful tool for approximating functions and performing calculations in contexts where finding an exact solution is challenging.
interval of convergence
The interval of convergence is a crucial concept when dealing with power series. It tells us the range of values for which the series actually represents the function. For a power series to work correctly, it must converge within a certain interval, otherwise, the infinite sum would not approach a finite result. In this exercise, the initial interval of convergence for
  • The Maclaurin series \(\frac{1}{1+x}\) is \(-1 < x < 1\).
  • After substituting \(x\) with \((2x)^2\), this fundamental interval shifts.
  • We deduced that \(-0.5 < x < 0.5\) is valid for \(arctan(2x)\).
Think of it like this: the interval of convergence is the boundary where the series behaves well, representing the desired function accurately.
arctangent function
The arctangent function, often written as \(\arctan(x)\), is the inverse of the tangent function. It is used to determine the angle whose tangent is a given value. In trigonometry, this function is essential for converting between angle measures and tangent values. Here, the goal was to find a power series representation for \(\arctan(2x)\).The Maclaurin series helps by transforming \(\arctan(2x)\) into a form that can be expanded as an infinite series. This can be done by recognizing patterns from known series:
  • For instance, we already used the power series for \(\frac{1}{1+u}\), where \(u\) is \(((2x)^2\), allowing us to alter the existing series rather than creating one from scratch.
  • By manipulating it, we're seeing more of a structured approach to expressing trigonometric and inverse trigonometric functions.
Thus, using known patterns, we can simplify complex functions like \(\arctan(2x)\) into a manageable series form.
coefficient identification
Coefficient identification is the heart of breaking down a power series into useful parts. Each term in the series has a coefficient that modifies the impact of the term in the polynomial representation of the function. In this exercise:
  • We identified the coefficients in the series for \(arctan(2x)\) as \((-1)^n 4^n\).
  • These coefficients are derived directly from the process of substitution and manipulation of the basic power series terms.
It's like assigning the right weight to each term in the series to ensure the function approximates correctly. Always remember, identifying these coefficients correctly is essential since they determine how each term in the infinite polynomial contributes to the final representation of the function.

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Most popular questions from this chapter

Verifying a Sum In Exercises \(55-58\) , verify the sum. Then use a graphing utility to approximate the sum with an error of less than \(0.0001 .\) $$ \sum_{n=0}^{\infty}(-1)^{n}\left[\frac{1}{(2 n+1) !}\right]=\sin 1 $$

Using a Binomial Series In Exercises \(17-26,\) use the binomial series to find the Maclaurin series for the function. $$ f(x)=\frac{1}{(2+x)^{3}} $$

the terms of a series \(\sum_{n=1}^{\infty} a_{n}\) are defined recursively. Determine the convergence or divergence of the series. Explain your reasoning. \(a_{1}=1, a_{n+1}=\frac{\sin n+1}{\sqrt{n}} a_{n}\)

Finding a Maclaurin Series In Exercises \(27-40\) , find the Maclaurin series for the function. Use the table of power series for elementary functions on page 670 . $$ f(x)=\cos ^{2} x $$

Probability In Exercises 73 and \(74,\) approximate the normal probability with an error of less than 0.0001 , where the probability is given by $$P(a < x < b)=\frac{1}{\sqrt{2 \pi}} \int_{a}^{b} e^{-x^{2} / 2} d x$$ $$ P(1 < x < 2) $$
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