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In each of Exercises 49-54, use Taylor series to calculate the given limit. $$ \lim _{x \rightarrow 0} \frac{\arctan (x)-x}{x^{3}} $$

Short Answer

Expert verified
\(-\frac{1}{3}\)

Step by step solution

01

Identify the Taylor Series

The first step is to identify the Taylor series for each function involved in the limit. The Taylor series for \( \arctan(x) \) around \( x = 0 \) is given by: \[ \arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots \] We'll truncate this to terms up to \( x^3 \) since higher-order terms will vanish as \( x \to 0 \) in the limit calculation.
02

Substitute Taylor Series Into Limit

Replace \( \arctan(x) \) in the limit expression with its Taylor series approximation:\[ \lim _{x \rightarrow 0} \frac{\arctan (x)-x}{x^{3}} = \lim _{x \rightarrow 0} \frac{\left( x - \frac{x^3}{3} + O(x^5) \right) - x}{x^3} \]Simplify the numerator:\[ = \lim _{x \rightarrow 0} \frac{-\frac{x^3}{3} + O(x^5)}{x^3} \]
03

Simplify and Evaluate the Limit

Now simplify the expression further by dividing the terms in the numerator by \( x^3 \):\[ = \lim _{x \rightarrow 0} \left( -\frac{1}{3} + O(x^2) \right) \]As \( x \rightarrow 0 \), the \( O(x^2) \) term approaches zero, so:\[ = -\frac{1}{3} \]

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Key Concepts

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

Limit Calculation
The core idea behind limit calculation in mathematical analysis is to determine how a function behaves as its input approaches a certain point. In this case, we are interested in calculating the limit as \( x \) approaches zero for the expression involving \( \arctan(x) \) and \( x^3 \). Understanding limits helps us find instantaneous rates of change, such as derivatives, and it's a foundational concept in calculus.When calculating limits involving expressions like \( \frac{\arctan(x)-x}{x^3} \), it's useful to apply known expansions, like the Taylor Series, to simplify the expression. By doing so, we can pinpoint the behavior of each term in the expression as the variable approaches the given point, in this case, zero. Here’s a quick recap of the process:
  • Identify the Taylor series representation for involved functions up to the necessary order.
  • Substitute this series into the limit expression.
  • Simplify the expression by factoring out the common terms.
The substitution and simplification allow us to directly evaluate the limit by finding the dominant term and demonstrating how non-dominant terms behave as \( x \rightarrow 0 \). Thus, understanding and applying limits enables us to unlock deeper insights into the behavior of functions near specific points of interest.
Arctangent Function
The arctangent function, often denoted as \( \arctan(x) \), is one of the inverse trigonometric functions. It provides the angle whose tangent gives the number \( x \). The relevance of \( \arctan(x) \) in calculus extends beyond simple trigonometric calculations. Its Taylor series expansion is important for examining and approximating the function near specific values, typically around zero.The series representation for \( \arctan(x) \) around \( x = 0 \) is:\[\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots\]This representation shows that as \( x \) becomes very small, \( \arctan(x) \) approximates \( x \) with further corrections given by higher powers of \( x \). These terms allow us to mitigate approximation errors by choosing how many terms to include, depending on the level of precision we need.In practical calculations, particularly those involving limits, we can truncate this series to a few terms, mainly if the problem centers on understanding behavior near zero, since higher-order terms become insignificant. This approach simplifies calculations and enhances understanding, revealing the nuanced behavior around essential points without excessive complexity.
Higher Order Terms
Higher order terms in series expansions refer to terms involving higher powers of \( x \), such as \( x^5, x^7, \) and so on in the context of the \( \arctan(x) \) Taylor series. These terms become crucial when determining the accuracy of approximations near a given point. As \( x \rightarrow 0 \), these terms diminish much faster than lower-order terms. To simplify the solution of our specific problem, these higher-order terms can often be denoted using the big-\( O \) notation. For example, \( O(x^5) \) suggests all terms involving \( x^5 \) and higher. Using this notation effectively allows us to condense the series expansion without dealing with every term individually, providing a clearer path to evaluate the behavior of the function at small \( x \).For our example problem, understanding that higher-order terms like \( x^5 \) have negligible contributions as \( x \rightarrow 0 \) allows us to focus only on terms up to \( x^3 \). By ignoring these higher-order terms, calculating limits becomes more straightforward, outright simplifying what might otherwise seem a complex expression. This strategic omission not only makes the calculation feasible but also ensures the result retains necessary precision for analysis around \( x = 0 \).

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Most popular questions from this chapter

It is known that \(\sum_{n=1}^{\infty} \frac{n}{2^{n}}=2\) and \(\sum_{n=1}^{\infty} \frac{1}{n 2^{n}}=\ln (2)\). Supposing that \(a, b,\) and \(c\) are constants, evaluate \(\sum_{n=1}^{\infty} \frac{a n^{2}+c n+b}{n 2^{n}}\)

Consider the initial value problem $$ \frac{d y}{d x}=2-x-y, \quad y(0)=1 $$ \(\begin{array}{llll}\text { a. Calculate the power series } & \text { expansion }\end{array}\) \(y(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) of the solution up to the \(x^{7}\) term. b. Using the coefficients you have calculated, plot \(S_{3}(x)=\sum_{n=0}^{3} a_{n} x^{n}\) in the viewing rectangle \([-2,2] \times\) [-10,1.7] c. The exact solution to the initial value problem is \(y(x)=3-x-2 e^{-x},\) as can be determined using the methods of Section 7.7 (in Chapter 7 ). Add the plot of the exact solution to the viewing window. From the two plots, we see that the approximation is fairly accurate for \(-1 \leq x \leq 1\), but the accuracy decreases outside this subinterval. d. To see the improvement in accuracy that results from using more terms in a partial sum, replace the graph of \(S_{3}(x)\) with that of \(S_{7}(x)\)

Find the sum of the given series in closed form. State the radius of convergence \(R\). \(\sum_{n=1}^{\infty} n x^{n+1}\)

In each of Exercises 91-94 a function \(f\), a base point \(c\), and a point \(x_{0}\) are given. Plot \(y=\left|f^{(3)}(t)\right|\) for \(t\) between \(c\) and \(x_{0}\). Use your plot to estimate the quantity \(M\) of Theorem 2 . Then use your value of \(M\) to obtain an upper bound for the absolute error \(\left|R_{2}\left(x_{0}\right)\right|=\left|f\left(x_{0}\right)-T_{2}\left(x_{0}\right)\right|\) that results when \(f\left(x_{0}\right)\) is approximated by the order 2 Taylor polynomial with base point \(c\). $$ f(x)=e^{\sin (x)} \quad c=0 \quad x_{0}=0.4 $$

In each of Exercises \(93-97\), a convergent series is given. Estimate the value \(\ell\) of the series by calculating its partial sums \(S_{N}\) for \(N=1,2,3, \ldots\) Round your evaluations to four decimal places and stop when three consecutive rounded partial sums agree. (This procedure does not ensure that the last partial sum calculated agrees with \(\ell\) to four decimal places. The error that results when a partial sum is used to approximate an infinite series is called a truncation error. Methods of estimating truncation errors will be discussed in later sections.) $$ \sum_{n=1}^{\infty} \frac{e^{-n}}{n} $$

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