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If \(\cos x\) is replaced by \(1-\left(x^{2} / 2\right)\) and \(|x|<0.5,\) what estimate can be made of the error? Does \(1-\left(x^{2} / 2\right)\) tend to be too large, or too small? Give reasons for your answer.

Short Answer

Expert verified
The error is approximately 0.0026, and 1 - (x^2/2) tends to underestimate \(\cos x\).

Step by step solution

01

Understand the Taylor Series for Cosine

The Taylor series expansion for \(\cos x\) around \(x = 0\) is given by \(\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \ldots\) If we truncate the series to \(1 - \frac{x^2}{2}\), the remaining terms form the error.
02

Define the Remainder (Error) Term

Since we're truncating after the \(x^2\) term, the error starts from the \(x^4\) term. For the Taylor series of \(\cos x\), the error can be approximated by the next term: \(R_4(x) = \frac{x^4}{24}\).
03

Calculate the Maximum Error Estimate

Given \(|x| < 0.5\), substitute \(x = 0.5\) into the error term to find the maximum potential error: \(R_4(0.5) = \frac{(0.5)^4}{24} = \frac{0.0625}{24} \approx 0.0026\). This represents the maximum error in the approximation.
04

Analyze the Direction of the Approximation Error

Since the Taylor expansion term \(\frac{x^4}{24}\) is positive for all \(|x| < 0.5\), \(1 - \frac{x^2}{2}\) underestimates \(\cos x\). Hence, it tends to be smaller than the true value of \(\cos x\).

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Key Concepts

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

Truncation Error
When we talk about truncation error, we're essentially discussing the inaccuracy that comes from stopping a series expansion too early. In this scenario, we're using a Taylor Series, which is a powerful tool for approximating complicated functions with simpler polynomial expressions. However, when we choose to truncate the series and keep only the terms up to a certain degree, we exclude the rest. This exclusion is what contributes to truncation error.

The Taylor Series for the cosine function is infinite, but practically, we cannot compute infinite terms. Hence, approximating \(\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \ldots\)by truncating it to \(1 - \frac{x^2}{2}\) involves truncation error, which starts with the term \(\frac{x^4}{24}\).

Understanding where you stop in the series helps in estimating how much error you are introducing. The size of the next term gives us a good approximation of the error magnitude.
Cosine Function
The cosine function is a fundamental trigonometric function, often represented as \(\cos x\). It is an even function, meaning \(\cos(-x) = \cos(x)\), and is periodic with a period of \(2\pi\). Cosine is commonly used in physics, engineering, and various fields that involve wave analysis.

For small values of \(x\), such as \(|x| < 0.5\), the cosine function can be efficiently approximated using its Taylor Series expansion around \(x = 0\).
  • The series begins with the term 1, providing the value of \(\cos(0)\), which is 1.
  • As more terms are added, the approximation becomes more accurate.
Understanding this behavior is key to knowing how the function acts for various values of \(x\). Using only the first two terms like \(1 - \frac{x^2}{2}\) gives a simple but approximate representation over small intervals.
Error Estimation
Estimating the error is a crucial part of using series expansions in practical applications. When we truncate the Taylor series, we must understand how much error is potentially introduced. In this case, by considering only the term \(1 - \frac{x^2}{2}\), the error is estimated by looking at the next term in the series.

The error term \(R_4(x) = \frac{x^4}{24}\)provides a way to quantify this error. Since this term is positive, it contributes a positive value to the cosine calculation, which means that our approximation is consistently below the actual value of \(\cos x\). In essence, \(1 - \frac{x^2}{2}\) tends to underestimate \(\cos x\).

For \(|x| < 0.5\), the maximum error can be computed directly using the substitute \(x = 0.5\), yielding an error value of approximately 0.0026. This value indicates how close the approximation is to the real value and offers insight into the reliability of the simplified model over the specified interval.

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Most popular questions from this chapter

Assume that each sequence converges and find its limit. \(a_{1}=-1, \quad a_{n+1}=\frac{a_{n}+6}{a_{n}+2}\)

Prove that a sequence \(\left\\{a_{n}\right\\}\) converges to 0 if and only if the sequence of absolute values \(\left\\{\left|a_{n}\right|\right\\}\) converges to 0 .

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 \(L ?\) \(a_{n}=\sin n\)

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

Use a geometric series to represent each of the given functions as a power series about \(x=0,\) and find their intervals of convergence. a. \(f(x)=\frac{5}{3-x}\) b. \(\quad g(x)=\frac{3}{x-2}\)

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