Problem 59 - Thomas Calculus

Question

a. Use the binomial series and the fact that \begin{equation}\frac{d}{d x} \sin ^{-1} x=\left(1-x^{2}\right)^{-1 / 2}\end{equation} to generate the first four nonzero terms of the Taylor series for \(\sin ^{-1} x .\) What is the radius of convergence? b. Series for \(\cos ^{-1} x\) Use your result in part (a) to find the first five nonzero terms of the Taylor series for \(\cos ^{-1} x .\)

Step-by-Step Solution

Answer

(a) \( \sin^{-1} x = x + \frac{1}{6}x^3 + \frac{3}{40}x^5 + \frac{5}{112}x^7 + \cdots \) with radius of convergence 1. (b) \( \cos^{-1} x = \frac{\pi}{2} - x - \frac{1}{6}x^3 - \frac{3}{40}x^5 - \frac{5}{112}x^7 + \cdots \).

Step 1: Identify the function for the derivative

Given the derivative \( \frac{d}{dx} \sin^{-1} x = (1-x^2)^{-1/2}, \) it is equivalent to \( (1-x^2)^{-1/2} = \sum_{n=0}^{\infty} \binom{1/2}{n} (-x^2)^n. \)

Step 2: Expand using binomial series

The binomial series for \((1-z)^{-\frac{1}{2}}\) is given by \( \sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} (-z)^n \). Setting \(z = x^2\), this becomes \[ \left(1-x^2\right)^{-\frac{1}{2}} = \sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} (-x^2)^n. \]

Step 3: Find the first four nonzero terms

Calculate the first four coefficients:- \( n=0 \): \( \binom{-\frac{1}{2}}{0} (-x^2)^0 = 1 \)- \( n=1 \): \( \binom{-\frac{1}{2}}{1} (-x^2)^1 = \frac{1}{2} x^2 \)- \( n=2 \): \( \binom{-\frac{1}{2}}{2} (-x^2)^2 = \frac{3}{8} x^4 \)- \( n=3 \): \( \binom{-\frac{1}{2}}{3} (-x^2)^3 = \frac{5}{16} x^6 \)Thus, the expansion for the antiderivative gives \[ x + \frac{1}{6}x^3 + \frac{3}{40}x^5 + \frac{5}{112}x^7 + \cdots \]

Step 4: Determine the radius of convergence

The radius of convergence for a binomial expansion of the form \((1-x^2)^{-1/2}\) is when \(|x| < 1\), so the radius of convergence is 1.

Step 5: Relation between \(\sin^{-1} x\) and \(\cos^{-1} x\)

Recall the identity: \(\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x\). To find \(\cos^{-1} x\) Taylor series, substitute the series for \(\sin^{-1} x\) from Step 3 into this identity.

Step 6: Calculate the series for \(\cos^{-1} x\)

From the identity \(\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x\), replace \(\sin^{-1} x\) with its series:\[ \cos^{-1} x = \frac{\pi}{2} - \left(x + \frac{1}{6}x^3 + \frac{3}{40}x^5 + \frac{5}{112}x^7 + \cdots\right). \]Thus, the first five nonzero terms for \(\cos^{-1} x\) series are \[ \frac{\pi}{2} - x - \frac{1}{6}x^3 - \frac{3}{40}x^5 - \frac{5}{112}x^7 + \cdots \]

Key Concepts

Binomial Series

The Binomial Series is a powerful tool for expanding expressions of the form \((1 + x)^n\), where \(n\) can be any number, not just a positive integer. This series is particularly useful for functions related to fractional powers and for generating Taylor series in calculus.

To derive the binomial series, we utilize the binomial theorem. For fractional \(n\), the generalized form of the binomial series becomes:

• \((1 + x)^n = \sum_{k=0}^{\infty} \binom{n}{k}x^k\)

Here, \(\binom{n}{k}\) denotes the binomial coefficient, which can be calculated using:

• \(\binom{n}{k} = \frac{n(n-1)(n-2)...(n-k+1)}{k!}\)

For the particular case of \((1-x^2)^{-1/2}\), which is related to the inverse trigonometric functions, we set \(n = -\frac{1}{2}\) and each \(x\) is replaced by \(-x^2\). This version of the binomial series provides a method to find successive terms of a Taylor expansion, facilitating calculations of derivatives and integrals.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \(\sin^{-1} x\) and \(\cos^{-1} x\), play a crucial role in calculus and are common in mathematical problems dealing with angles and their measures.

The derivative of \(\sin^{-1} x\) is given by \(\frac{d}{dx} \sin^{-1} x = (1-x^2)^{-1/2}\). This derivative is essential for deriving the Taylor series for inverse trigonometric functions.

For instance, to find the Taylor series for \(\sin^{-1} x\), we integrate its derivative, allowing us to expand it into a series using the binomial series for \((1-x^2)^{-1/2}\). This approach yields a series expansion, such as:

• \(\sin^{-1} x = x + \frac{1}{6}x^3 + \frac{3}{40}x^5 + \frac{5}{112}x^7 + \cdots\)

With these expansions, one can explore additional trigonometric inverse functions, like \(\cos^{-1} x\), using the relationship \(\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x\). This relation helps transform one series into another seamlessly.

Radius of Convergence

Understanding the radius of convergence is paramount for any power series, including the Taylor and Binomial series. It defines the set of \(x\)-values for which the series will converge to a finite value.

For any given series of form \((1-x^2)^{-1/2}\) obtained through a Binomial expansion, the series converges when the absolute value of \(x\) is within a certain radius around zero:

• The radius of convergence here is \(|x| < 1\).

This means that the series is guaranteed to converge (i.e., provide a meaningful result) for all \(x\) within the interval \((-1, 1)\). Beyond this interval, the series either becomes divergent or does not provide an accurate representation of the function.

Checking for the radius of convergence is essential when working with power series to ensure the results obtained accurately reflect the original function within the specified range.