Problem 15 - Calculus Early Transcendentals

Question

a. Find the first four nonzero terms of the Maclaurin series for the given function. b. Write the power series using summation notation. c. Determine the interval of convergence of the series. $$f(x)=\tan ^{-1} x$$

Step-by-Step Solution

Answer

Question: Find the Maclaurin series for the function \(f(x)=\tan^{-1}(x)\) and determine its interval of convergence. Solution: a) The first 4 non-zero terms of the Maclaurin series for the given function are: $$x - \frac{1}{3}x^3$$. b) The power series representation of the function using summation notation is: $$f(x) = \sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{(2n-1)}x^{2n-1}$$ c) The interval of convergence for this power series is: $$-1 < x < 1$$

Step 1: Part a: Find the first four nonzero terms of the Maclaurin series

We know that the Maclaurin series representation of a function is given by: $$f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$$ So, we need to find the derivatives of \(f(x) = \tan^{-1}(x)\), followed by evaluating them at \(x=0\). 1. First, find the first four derivatives of the given function: $$f(x) = \tan^{-1}(x)$$ $$f'(x) = \frac{1}{1+x^2}$$ $$f''(x) = \frac{-2x}{(1+x^2)^2}$$ $$f'''(x) = \frac{2(3x^2-1)}{(1+x^2)^3}$$ $$f^{(4)}(x) = \frac{24x(1-x^2)}{(1+x^2)^4}$$ 2. Then, evaluate these derivatives at \(x=0\): $$f(0) = \tan^{-1}(0) = 0$$ $$f'(0) = \frac{1}{1+0^2} = 1$$ $$f''(0) = \frac{-2(0)}{(1+0^2)^2} = 0$$ $$f'''(0) = \frac{2(3(0)^2-1)}{(1+0^2)^3} = -2$$ $$f^{(4)}(0) = \frac{24(0)(1-0^2)}{(1+0^2)^4} = 0$$ Now, we can plug these values back into the Maclaurin series equation: $$f(x) = 0 + 1x + \frac{0}{2!}x^2 - \frac{2}{3!}x^3 + \cdots$$ The first four nonzero terms are \(x - \frac{1}{3}x^3\).

Step 2: Part b: Write the power series using summation notation

Let's find the pattern for the generalized term of the Maclaurin series. Notice that only the odd-order terms are nonzero. For the kth term \(a_kx^k\), we have: $$a_1x^1 = x$$ $$a_3x^3 = -\frac{1}{3}x^3$$ So, the general term can be expressed as: $$a_{2n-1}x^{2n-1} = (-1)^{n-1}\frac{1}{(2n-1)}x^{2n-1}$$ Using summation notation, we can write the power series as: $$f(x) = \sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{(2n-1)}x^{2n-1}$$

Step 3: Part c: Determine the interval of convergence

To determine the interval of convergence, we use the Ratio Test: $$\lim_{n\to\infty} \left |\frac{a_{n+1}}{a_n}\right| = L$$ If \(L < 1\), the series converges, if \(L > 1\), the series diverges, and if \(L = 1\), the test is inconclusive. By plugging the expression of the terms into the limit ratio: $$\lim_{n\to\infty} \left|\frac{(-1)^{n}\frac{1}{(2(n+1)-1)}x^{2(n+1)-1}}{(-1)^{n-1}\frac{1}{(2n-1)}x^{2n-1}}\right|= \lim_{n\to\infty} \frac{2n-1}{2n+1}|x|$$ We can simplify and get: $$L = \lim_{n\to\infty} (1-\frac{2}{2n+1})|x| = |x|$$ So, we have convergence when \(|x| < 1\), which leads to the interval of convergence: $$-1 < x < 1$$

Key Concepts

Power Series

A power series is a way to represent a function as an infinite sum of terms that are powers of a variable. In mathematical notation, it can be written as:

• \[\sum_{n=0}^{\infty} a_n x^n\]

Here, each term is expressed as \(a_n x^n\), where \(a_n\) represents the coefficient of the \(n\)-th term, and \(x\) is the variable. This type of series is very flexible and can be used to approximate many functions.

Power series are fundamental in mathematics because they allow us to express complex functions in simpler forms that can be easily manipulated and computed. The Maclaurin series is a special type of power series that is particularly useful for evaluating functions at or around zero.

Interval of Convergence

The interval of convergence is the range of values for which a power series converges to a function. To determine this, we need to test input values to see if substituting them leads to a series that converges or not.

In the case of our series for \(\tan^{-1}(x)\), we've found that it converges when the absolute value of \(x\) is less than 1. Mathematically, this is expressed as:

• \[-1 < x < 1\]

Finding the interval of convergence is crucial as it tells us precisely where the power series accurately represents the function. Beyond these bounds, the series diverges, meaning it doesn't settle to a particular value, and thus, cannot be used for practical approximations of the original function.

Ratio Test

The Ratio Test is a common method used to find the interval of convergence for a power series. This test helps determine whether the series converges or diverges by examining the limit of the ratio of consecutive terms.

To apply the Ratio Test, consider two consecutive terms in the series: \(a_n\) and \(a_{n+1}\). You compute the limit:

• \[L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|\]

The test gives the following rules:

• If \(L < 1\), the series converges absolutely.
• If \(L > 1\), the series diverges.
• If \(L = 1\), the test is inconclusive, and you may need a different approach.

For the given series representation of \(\tan^{-1}(x)\), it was shown that:

• \[L = |x|\]

This indicates convergence when \(|x| < 1\), helping to establish the interval \(-1 < x < 1\). The Ratio Test is particularly useful in handling series with factorial or exponential-like growth and provides a straightforward way to assess convergence.