/*! 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 34 Find the first four nonzero term... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the first four nonzero terms in the Maclaurin series for the functions. $$\sin \left(\tan ^{-1} x\right)$$

Short Answer

Expert verified
The first four non-zero terms of \(\sin(\tan^{-1} x)\) are \(x - \frac{x^3}{3} - \frac{x^5}{5}\).

Step by step solution

01

Understanding the Problem

We have to find the first four non-zero terms of the Maclaurin series for the function \(\sin(\tan^{-1} x)\). The Maclaurin series is the Taylor series expanded around \(x = 0\).
02

Maclaurin Series for \(\tan^{-1} x\)

Recall the Maclaurin series for \(\tan^{-1} x\) which is \(x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \ldots\). For our purposes, let's consider the first few terms: \(\tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5}\).
03

Substitute into \(\sin(u)\)

The function given is \(\sin(\tan^{-1} x)\) where \(u = \tan^{-1} x\). We use the series for \(\sin u\): \(u - \frac{u^3}{6} + \ldots\). Substitute \(u = x - \frac{x^3}{3} + \frac{x^5}{5}\) into the series for \(\sin(u)\).
04

Compute \(\sin(\tan^{-1} x)\)

Expand \(\sin(\tan^{-1} x) = u - \frac{u^3}{6}\) using \(u = x - \frac{x^3}{3}\). Substitute to get \(\sin(\tan^{-1} x) = \left( x - \frac{x^3}{3} \right) - \frac{1}{6} \left( x - \frac{x^3}{3} \right)^3\).
05

Expand and Simplify

First-term: \(x - \frac{x^3}{3}\). For the cubic term \(\left(x - \frac{x^3}{3} \right)^3\), expand and simplify by considering only first-order of significant terms, giving further terms involving \(x^3\), which will be contributed as corrections. Hence the first four non-zero terms become \(x - \frac{x^3}{3} - \frac{x^5}{5}\) after simplifying all higher order and cubic terms.

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Key Concepts

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

Taylor series
The Taylor series is a powerful tool in calculus that helps us approximate complex functions. It expresses a function as an infinite sum of terms calculated from the values of its derivatives at a single point. Specifically, if this point is zero, the series is called a Maclaurin series. The general form of a Taylor series for a function \( f(x) \) around a point \( a \) is given by: \[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots\] The Maclaurin series is a special case where \( a = 0 \). This simplification can make the series more manageable by setting all \( a \) terms to zero.
  • The first term \( f(a) \) represents the value of the function at that point.
  • The coefficient of each successive term gets smaller as more derivatives are taken.
Using this expansion, we can approximate functions like trigonometric and logarithmic functions to varying degrees of accuracy, depending on how many terms are included.
trigonometric functions
Trigonometric functions, such as \( \sin(x) \), \( \cos(x) \), and \( \tan(x) \), are foundational in mathematics. They often appear in problems involving angles and periodic phenomena. The Taylor series allows us to express trigonometric functions as infinite polynomials. For example, the series expansion for \( \sin(x) \) is: \[\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots\] This pattern alternates between positive and negative terms, highlighting the wave-like properties of sine. Similarly, other trigonometric functions follow recognizable patterns:
  • \( \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots \)
  • \( \tan(x) \) has its unique pattern requiring careful expansion to handle its peculiarities.
These series are useful in approximating values for angles and calculating other derivatives.
inverse trigonometric functions
Inverse trigonometric functions, like \( \tan^{-1}(x) \), play a crucial role in finding angles from given trigonometric values. They help reverse the process of the standard trigonometric functions. Their Taylor expansions offer insight into their behavior for small values of \( x \). For \( \tan^{-1}(x) \), the Maclaurin series expansion is: \[\tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \ldots\] This series is alternating and converges for \( -1 < x < 1 \). The importance of this function is emphasized when used in context with other trigonometric functions, such as in the problem of finding the Maclaurin series for \( \sin(\tan^{-1} x) \).
  • This process involves substituting this series into another, creating a nested polynomial function.
  • Only a few terms might be required for a reasonable approximation, depending on precision needed.
Inverse functions are vital in various fields, including geometry, where lengths and angles must be determined from known ratios.

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Most popular questions from this chapter

Use the definition of convergence to prove the given limit. $$\lim _{n \rightarrow \infty}\left(1-\frac{1}{n^{2}}\right)=1$$

If \(\Sigma a_{n}\) converges and \(a_{n}>0\) for all \(n,\) can anything be said about \(\Sigma\left(1 / a_{n}\right) ?\) Give reasons for your answer.

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 \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$a_{n}=(0.9999)^{n}$$

Taylor's formula with \(n=1\) and \(a=0\) gives the linearization of a function at \(x=0 .\) With \(n=2\) and \(n=3\) we obtain the standard quadratic and cubic approximations. In these exercises we explore the errors associated with these approximations. We seek answers to two questions: a. For what values of \(x\) can the function be replaced by each approximation with an error less than \(10^{-2} ?\). b. What is the maximum error we could expect if we replace the function by each approximation over the specified interval? Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals. Step \(I:\) Plot the function over the specified interval. Step 2: Find the Taylor polynomials \(P_{1}(x), P_{2}(x),\) and \(P_{3}(x)\) at \(x=0\) Step 3: Calculate the \((n+1)\) st derivative \(f^{(n+1)}(c)\) associated with the remainder term for each Taylor polynomial. Plot the derivative as a function of \(c\) over the specified interval and estimate its maximum absolute value, \(M .\) Step 4: Calculate the remainder \(R_{n}(x)\) for each polynomial. Using the estimate \(M\) from Step 3 in place of \(f^{(n+1)}(c),\) plot \(R_{n}(x)\) over the specified interval. Then estimate the values of \(x\) that answer question (a). Step 5: Compare your estimated error with the actual error \(E_{n}(x)=\left|f(x)-P_{n}(x)\right|\) by plotting \(E_{n}(x)\) over the specified interval. This will help answer question (b). Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5. $$f(x)=e^{-x} \cos 2 x, \quad|x| \leq 1$$

Assume that each sequence converges and find its limit. $$a_{1}=3, \quad a_{n+1}=12-\sqrt{a_{n}}$$

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