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Chapter 10: Problem 55

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.

Short Answer

Expert verified
499 terms are needed to ensure an error less than \(10^{-3}\).

Step by step solution

01

Understand the Taylor Series for \(\tan^{-1}x\)

The Taylor series for \(\tan^{-1}x\) is given by:\[\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \ldots\]This series is an alternating series where each term is given by \((-1)^{n}\frac{x^{2n+1}}{2n+1}\) for \(n = 0, 1, 2, 3, ...\). Here, \(x = 1\).
02

Apply the Alternating Series Estimation Theorem

The Alternating Series Estimation Theorem states that the error in approximating the sum of an alternating series by its \(n\)-th partial sum is less than the absolute value of the first omitted term. The error bound for the Taylor series approximation is given as:\[|R_n| < |a_{n+1}| = \left| \frac{x^{2n+3}}{2n+3} \right|\]For this problem, we need \(|R_n| < 10^{-3}\).
03

Set Up the Inequality for Error Bound

Since we need \(|R_n| < 10^{-3}\), we'll use the inequality:\[\left| \frac{1^{2n+3}}{2n+3} \right| < 10^{-3}\]This simplifies to:\[\frac{1}{2n+3} < 10^{-3}\].
04

Solve the Inequality for \(n\)

Solving \(\frac{1}{2n+3} < 10^{-3}\) leads to:\[2n+3 > 1000\]Solving for \(n\), we get:\[2n > 997 > \frac{997}{2} > 498.5\]Thus, rounding up, we find \(n > 498\).
05

Determine the Number of Terms

Since \(n\) represents the number of term indices (starting from 0), you need to include terms from \(n=0\) to \(n=498\). Therefore, the number of terms required is \(499\).

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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 allows us to approximate complex functions using a polynomial expression. It is expressed through an infinite sum of terms based on the derivatives of a function at a single point. For instance, the Taylor series for the function \( an^{-1} x\) at \(-1 \leq x \leq 1\) is given by:\[\tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \ldots\]In this series, each term alternates in sign, contributing to the concept of an "alternating series." The series becomes especially important when finding approximations because it allows us to express the relationship between functions and their polynomial representations.
  • The Taylor series is used to approximate continuous functions around a certain point.
  • It uses derivatives to determine the precise nature and magnitude of each term in the polynomial expansion.
  • Convergence of a Taylor series, which is the process of the series approaching a limit, depends on the function and the point chosen for development.
approximation error
When using Taylor series to approximate functions, achieving precision is important. The approximation error tells us how far off an approximation is from the exact function value. In the context of Taylor series, this error is often depicted with the term \(|R_n|\), representing the remainder or error after \(n\) terms.The Alternating Series Estimation Theorem helps assess this error for alternating series. The theorem assures us that the absolute error \(|R_n|\) is less than the absolute value of the next term \(a_{n+1}\). Hence, when we approximate \(\tan^{-1} x\) and seek a small error such as less than \(10^{-3}\), we ensure our result is close to the true value.
  • Approximation error helps in determining how many terms are needed in a series to achieve a desired accuracy.
  • Smaller approximation errors signify closer matches to the real function, which is the goal of solving calculus problems.
  • Understanding and handling errors are crucial as they guide decisions on the number of terms to compute in practical applications.
inverse tangent function
The inverse tangent function, denoted as \(\tan^{-1} x\) or \( ext{arctan}(x)\), is a fundamental component of trigonometry and calculus. It finds the angle whose tangent is \(x\), and is often used when analyzing periodic behavior or solving inverse trigonometric problems.In the exercise, we're interested in the Taylor series approximation of \(\tan^{-1} x\) at \(x = 1\), which leads to an approximation of \(\pi/4\). The inverse tangent function is significant because it connects angles through trigonometric ratios and can be represented through its Taylor series expansion.
  • \(\tan^{-1} x\) helps transition from trigonometric results back to angle measures.
  • The function is periodic and arcs through the Cartesian plane, useful in various mathematical and engineering fields.
  • It is especially effective in problems involving circular arcs or oscillatory systems.
calculus problem solving
Solving calculus problems often calls for creativity and understanding of multiple concepts. In the given problem, we apply the Alternating Series Estimation Theorem, which is a strategy used to measure convergence and error in series approximations.The task involves ensuring the Taylor series approximation for \(\tan^{-1} 1\) achieves an error less than \(10^{-3}\). Through the development of the series, setting up an inequality to determine how many terms are needed becomes the goal.
  • Calibrating problem-solving skills involves understanding the series, their errors, and theorem applications.
  • Each step, such as transforming inequalities to measure error, requires precision and concept clarity.
  • Calculus connects seemingly unrelated mathematical ideas into cohesive solutions, often demanding interpretation and symbolic manipulation.
Thus, effective problem solving in calculus is about diligence, methodical thinking, and layering of mathematical techniques to arrive at the most accurate results possible.

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Most popular questions from this chapter

Use the identity \(\sin ^{2} x=(1-\cos 2 x) / 2\) to obtain the Maclaurin series for \(\sin ^{2} x\). Then differentiate this series to obtain the Maclaurin series for \(2 \sin x \cos x .\) Check that this is the series for \(\sin 2 x\).

Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals. Step 1: Plot the function over the specified interval. Step 2: Find the Taylor polynomials \(P_{1}(x), P_{2}(x),\) and \(P_{3}(x)\) at \(\bar{x}=0\) 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. $$f(x)=\frac{1}{\sqrt{1+x}}, \quad|x| \leq \frac{3}{4}$$

Which of the sequences converge, and which diverge? Give reasons for your answers. The first term of a sequence is \(x_{1}=\cos (1) .\) The next terms are \(x_{2}=x_{1}\) or \(\cos (2),\) whichever is larger; and \(x_{3}=x_{2}\) or \(\cos (3)\) whichever is larger (farther to the right). In general, $$x_{n+1}=\max \left\\{x_{n}, \cos (n+1)\right\\}$$.

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 \(L ?\) \(a_{n}=\frac{8^{n}}{n !}\)

To construct this set, we begin with the closed interval \([0,1] .\) From that interval, remove the middle open interval (1/3, 2/3). leaving the two closed intervals [ 0, 1/3 ] and [2/3, 1 ]. At the second step we remove the open middle third interval from each of those remaining. From \([0.1 / 3]\) we remove the open interval \((1 / 9,2 / 9),\) and from \([2 / 3,1]\) we remove \((7 / 9,8 / 9),\) leaving behind the four closed intervals \([0,1 / 9]\) \([2 / 9,1 / 3],[2 / 3,7 / 9],\) and \([8 / 9,1] .\) At the next step, we remove the middle open third interval from each closed interval left behind, so \((1 / 27,2 / 27)\) is removed from \([0,1 / 9],\) leaving the closed intervals \([0,1 / 27]\) and \([2 / 27,1 / 9] ;(7 / 27,8 / 27)\) is removed from \([2 / 9,1 / 3]\), leaving behind \([2 / 9,7 / 27]\) and \([8 / 27,1 / 3],\) and so forth. We continue this process repeatedly without stopping, at each step removing the open third interval from every closed interval remaining behind from the preceding step. The numbers remaining in the interval \([0,1],\) after all open middle third intervals have been removed, are the points in the Cantor set (named after Georg Cantor, \(1845-1918\) ). The set has some interesting properties. a. The Cantor set contains infinitely many numbers in [0,1] List 12 numbers that belong to the Cantor set. b. Show, by summing an appropriate geometric series, that the total length of all the open middle third intervals that have been removed from [0,1] is equal to 1
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