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Chapter 10: Problem 59

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} \begin{equation} \begin{array}{l}{\text { to generate the first four nonzero terms of the Taylor series }} \\ {\text { for sin }^{-1} x . \text { What is the radius of convergence? }}\end{array} \end{equation} b. Series for \(\cos ^{-1} x\) Use your result in part (a) to find the first five nonzero terms of the Taylor series for \(\cos ^{-1} x .\)

Short Answer

Expert verified
Taylor series for \( \sin^{-1} x \) is \( x + \frac{x^3}{6} \), radius 1. \( \cos^{-1} x \) series is \( \frac{\pi}{2} - x - \frac{x^3}{6} \). Radius also 1.

Step by step solution

01

Taylor Series for Inverse Sine

The Taylor series expansion of a function \( f(x) \) about \( x = 0 \) is given by \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots \). For \( \sin^{-1} x \), we start by finding \( f(0) \), which is \( \sin^{-1}(0) = 0 \). Now, find the derivatives and evaluate at \( x = 0 \).
02

First Derivative

The first derivative is \( \frac{d}{dx} \sin^{-1}x = (1-x^2)^{-1/2} \). Evaluating at \( x = 0 \) gives \( f'(0) = 1 \).
03

Second Derivative

Using the first derivative, apply the product and chain rule to find \( f''(x) = \frac{x}{(1-x^2)^{3/2}} \). Evaluating at \( x = 0 \) gives \( f''(0) = 0 \).
04

Third Derivative

Take the derivative of \( f''(x) \) to get \( f'''(x) = \frac{2x^2 + 1}{(1-x^2)^{5/2}} \). Evaluating at \( x = 0 \) gives \( f'''(0) = 2 \).
05

Fourth Derivative

Find \( f^{(4)}(x) = \frac{6x(x^2 + 2)}{(1-x^2)^{7/2}} \) and evaluate at \( x = 0 \), resulting in \( f^{(4)}(0) = 0 \).
06

Find Nonzero Taylor Series Terms

The Taylor series is \( \sin^{-1} x = 0 + 1 \cdot x + 0 \cdot \frac{x^2}{2!} + 2 \cdot \frac{x^3}{3!} + 0 \cdot \frac{x^4}{4!} + \cdots = x + \frac{x^3}{6} + \cdots \).
07

Radius of Convergence

The series \( \frac{d}{dx} \sin^{-1}x = (1-x^2)^{-1/2} \) converges for \( |x| < 1 \), implying the radius of convergence for \( \sin^{-1} x \) is 1.
08

Series for Inverse Cosine

Since \( \cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x \), the Taylor series is obtained by subtracting the series for \( \sin^{-1} x \) from \( \frac{\pi}{2} \).
09

Find Nonzero Terms for Inverse Cosine

Using the known series, \( \cos^{-1} x = \frac{\pi}{2} - \left( x + \frac{x^3}{6} + \cdots \right) = \frac{\pi}{2} - x - \frac{x^3}{6} + \cdots \), expanding until the fifth term.

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Key Concepts

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

Binomial Series
The binomial series allows us to expand expressions of the form \((1 + x)^n\) into an infinite series. This is especially useful when working on problems that require manipulation of derivatives of functions like inverse trigonometric functions. The general expression for the binomial series is given by
  • \((1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots\)
Here, \(n\) is not necessarily an integer; it can be any real number. Furthermore, the binomial series helps in simplifying expressions when substituted into functions, often used in Taylor series expansions.
In the context of the inverse sine function, expressing \((1-x^2)^{-1/2}\) using the binomial series becomes pivotal. By substituting for \(x^2\) and expanding via binomial coefficients, the derivatives simplify, thus easing the calculation of nonzero terms in the Taylor series.
Inverse Trigonometric Functions
Inverse trigonometric functions, such as \(\sin^{-1} x\) and \(\cos^{-1} x\), allow us to find angles associated with a given trigonometric value. These functions provide a crucial link between algebraic expressions and geometric interpretations.
For these functions, Taylor series expansions are particularly useful as they form the foundation for representing them near a point, usually around \(x = 0\).
To derive the Taylor series for \(\sin^{-1} x\), we make use of its derivative \(\frac{d}{dx} \sin^{-1}x = (1-x^2)^{-1/2}\). By monitoring the sequence of derivatives and evaluating them at \(x = 0\), the Taylor series unfolds. Each of these derivatives reveals alternating patterns of 0s and nonzero terms, hence leading to the simplified Taylor series:
  • \(\sin^{-1} x = x + \frac{x^3}{6} + \cdots\)
Notably, by leveraging the Taylor series of \(\sin^{-1} x\), we can also deduce the expansion of \(\cos^{-1} x\) using the relationship \(\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x\). This enables similar stepwise derivations for \(\cos^{-1} x\), thereby producing its own sequence of non-zero terms.
Radius of Convergence
The radius of convergence is a critical part of understanding any power series, including those representing Taylor series. It defines the interval within which the series converges to the function.
Determining the radius of convergence involves identifying the values for which the series representation remains valid and does not diverge.
For a function like \(\sin^{-1} x\), the convergence is informed by its derivative \((1-x^2)^{-1/2}\). Since the expansion is contingent upon the factor \((1-x^2)\) being positive, the series converges when \(|x| < 1\).
Therefore, the radius of convergence for the Taylor series of \(\sin^{-1} x\) is 1. This means that within the interval \( -1 < x < 1\), the series provides a close approximation of the function itself. Such an understanding is crucial when applying these functions to numerical computations, as it ensures accuracy and dependability of the series' predictive capability across its interval of validity.

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Most popular questions from this chapter

The sequence \(\\{n /(n+1)\\}\) has a least upper bound of 1 Show that if \(M\) is a number less than \(1,\) then the terms of 1 \(\\{n /(n+1)\\}\) eventually exceed \(M .\) That is, if \(M<1\) there is an integer \(N\) such that \(n /(n+1)>M\) whenever \(n>N .\) since \(n /(n+1)<1\) for every \(n,\) this proves that 1 is a least upper bound for \(\\{n /(n+1)\\} .\)

According to the Alternating Series Estimation Theorem, how many terms of the Taylor series for tan \(^{-1} 1\) would you have to add to be sure of finding \(\pi / 4\) with an error of magnitude less than \(10^{-3}\) . Give reasons for your answer.

Proof of Theorem 21 Assume that \(a=0\) in Theorem 21 and that \(f(x)=\sum_{n=0}^{\infty} c_{n} x^{n}\) converges for \(- R < x < R .\) Let \(g(x)=\sum_{n=1}^{\infty} n c_{n} x^{n-1} .\) This exercise will prove that \(f^{\prime}(x)=g(x)\) that is, \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=g(x)\) $$ \begin{array}{c}{\text { a. Use the Ratio Test to show that } g(x) \text { converges for }} \\ {-R < x < R .} \\ {\text { b. Use the Mean Value Theorem to show that }} \\ {\frac{(x+h)^{n}-x^{n}}{h}=n c_{n}^{n-1}} \\ {\text { for some } c_{n} \text { between } x \text { and } x+h \text { for } n=1,2,3, \ldots}\\\\{\text { c. Show that }} \\ {\quad g(x)-\frac{f(x+h)-f(x)}{h}|=| \sum_{n=2}^{\infty} n a_{n}\left(x^{n-1}-c_{n}^{n-1}\right) |} \\ {\text { d. Use the Mean Value Theorem to show that }} \\\ {\frac{x^{n-1}-c_{n}^{n-1}}{x-c_{n}}=(n-1) d_{n-1}^{n-2}} \\ {\text { for some } d_{n-1} \text { between } x \text { and } c_{n} \text { for } n=2,3,4, \ldots}\\\\{\text { e. Explain why }\left|x-c_{n}\right| < h \text { and why }} \\ {\left|d_{n-1}\right| \leq \alpha=\max \\{|x|,|x+h|\\}} \\ {\text { f. Show that }} \\ {\left|g(x)-\frac{f(x+h)-f(x)}{h}\right| \leq|h| \sum_{n=2}^{\infty}\left|n(n-1) a_{n} \alpha^{n-2}\right|}\\\\{\text { g. Show that } \sum_{n=2}^{\infty} n(n-1) \alpha^{n-2} \text { converges for }- R < x < R \text { . }} \\ {\text { h. Let } h \rightarrow 0 \text { in part (f) to conclude that }} \\ {\quad \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=g(x)}\end{array} $$

Which of the sequences converge, and which diverge? Give reasons for your answers. $$ a_{n}=\frac{2^{n}-1}{2^{n}} $$

Use the following steps to prove that the binomial series in Equation \((1)\) converges to \((1+x)^{m}\). \begin{equation}\begin{array}{l} \\ {\text { a. Differentiate the series }}\end{array}\end{equation} \begin{equation}\quad \quad \quad f(x)=1+\sum_{k=1}^{\infty} \left( \begin{array}{l}{m} \\ {k}\end{array}\right) x^{k}\end{equation} to show that \begin{equation}f^{\prime}(x)=\frac{m f(x)}{1+x}, \quad-1< x < 1.\end{equation} \begin{equation} \begin{array}{l}{\text { b. Define } g(x)=(1+x)^{-m} f(x) \text { and show that } g^{\prime}(x)=0} \\ {\text { c. From part (b), show that }}\end{array}\end{equation} \begin{equation}f(x)=(1+x)^{m}\end{equation}
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