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Estimate the error in the approximation \(\sinh x=x+\left(x^{3} / 3 !\right)\) when \(|x|<0.5 .\left(\text {Hint} : \text { Use } R_{4}, \text { not } R_{3} .\right)\)

Short Answer

Expert verified
The error is approximately \( 2.604 \times 10^{-4} \).

Step by step solution

01

Understanding the Function Series

The hyperbolic sine function can be expanded into an infinite series: \( \sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \). This is a Taylor series centered at 0 (Maclaurin series). Our task is to approximate it up to the \( x^3 \) term and estimate the error using the next term.
02

Identifying the Relevant Error Term

The remainder term \( R_n \) represents the error in approximating the series to the nth term. We want \( R_4 \), which involves the \( x^5 \) term in the series: \( \frac{x^5}{5!} \). This considers terms after the approximation of \( x + \frac{x^3}{3!} \).
03

Applying the Error Term

Since the absolute value of \( x \) is less than 0.5, the error term \( R_4 \) for this function can be expressed as approximately \( \left| \frac{x^5}{5!} \right| \leq \frac{(0.5)^5}{120} \). Calculating this gives us \( \frac{0.03125}{120} \approx 2.604 \times 10^{-4} \).
04

Conclusion

The error in approximating \( \sinh x = x + \frac{x^3}{3!} \) for \( |x| < 0.5 \) is approximately \( 2.604 \times 10^{-4} \).

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Key Concepts

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

Taylor series
Taylor series allow us to approximate complex functions using polynomials. They are incredibly useful because polynomials are much easier to work with, especially for calculus-related tasks like integration and differentiation. A Taylor series represents a function as an infinite sum of terms calculated from the values of its derivatives at a single point. When the point is 0, it's called a Maclaurin series.
For example, the hyperbolic sine function, \( \sinh x \), can be expanded in a Maclaurin series as: \(\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \). The more terms included, the more accurate the approximation. The beauty of Taylor series lies in their ability to approximate functions to any desired degree of accuracy by choosing a sufficient number of terms.
hyperbolic functions
Hyperbolic functions are analogs of trigonometric functions but for the hyperbola, not the circle. Just like sine and cosine are for trigonometry, \( \sinh \) (hyperbolic sine) and \( \cosh \) (hyperbolic cosine) are for hyperbolic geometry.
  • Definition: The hyperbolic sine function is defined as: \( \sinh x = \frac{e^x - e^{-x}}{2} \).
  • Connections: These functions naturally occur in several areas of mathematics, including calculus and complex analysis. They model real-world phenomena like the shape of a hanging cable or chain (catenary curve).
  • Properties: Hyperbolic functions have properties similar to their trigonometric counterparts, like identities and periodicity, but they differ in that \( \sinh x \) is not bounded in the same way as sine.
Understanding these functions is important, as they permit new perspectives on solving mathematical problems and modeling real-life scenarios.
approximation error
Approximation error is the difference between the exact value and the approximation. When we approximate mathematical functions using Taylor or Maclaurin series, calculating the error gives us insight into how good our approximation is. This concept is critical in mathematical modeling, engineering, and sciences where exact solutions may be difficult to obtain.
For functions like \( \sinh x \), the error related to its Taylor series can be represented by the remainder term \( R_n \). This remainder term estimates how much the finite polynomial differs from the actual function. When approximating \( \sinh x = x + \frac{x^3}{3!} \), the relevant error term \( R_4 \) is \( \frac{x^5}{5!} \).
  • Significance: Estimating the error involves knowing not just the quality of the approximation but also the conditions under which it holds. For example, when \( |x| < 0.5 \), the error is relatively small.
  • Practicality: Knowing that the error is approximately \( 2.604 \times 10^{-4} \) assures us that the approximation is precise enough for most practical purposes within this range.
Understanding approximation error helps in assessing the precision of solutions and ensuring reliability in various applications.

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Most popular questions from this chapter

Uniqueness of limits Prove that limits of sequences are unique. That is, show that if \(L_{1}\) and \(L_{2}\) are numbers such that \(a_{n} \rightarrow L_{1}\) and \(a_{n} \rightarrow L_{2},\) then \(L_{1}=L_{2} .\)

Logistic difference equation The recursive relation $$ a_{n+1}=r a_{n}\left(1-a_{n}\right) $$ is called the logistic difference equation, and when the initial value \(a_{0}\) is given the equation defines the logistic sequence \(\left\\{a_{n}\right\\} .\) Throughout this exercise we choose \(a_{0}\) in the interval \(03.57\) . Choose \(r=3.65\) and calculate and plot the first 300 terms of \(\left\\{a_{n}\right\\} .\) Observe how the terms wander around in an unpredictable, chaotic fashion. You cannot predict the value of \(a_{n+1}\) from previous values of the sequence. g. For \(r=3.65\) choose two starting values of \(a_{0}\) that are close together, say, \(a_{0}=0.3\) and \(a_{0}=0.301 .\) Calculate and plot the first 300 values of the sequences determined by each starting value. Compare the behaviors observed in your plots. How far out do you go before the corresponding terms of your two sequences appear to depart from each other? Repeat the exploration for \(r=3.75 .\) Can you see how the plots look different depending on your choice of \(a_{0} ?\) We say that the logistic sequence is sensitive to the initial condition a_{0} .

Estimating Pi The English mathematician Wallis discovered the formula $$ \frac{\pi}{4}=\frac{2 \cdot 4 \cdot 4 \cdot 6 \cdot 6 \cdot 8 \cdot \cdots}{3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdot 7 \cdot \cdots} $$ Find \(\pi\) to two decimal places with this formula.

Use a CAS to perform the following steps for the sequences in Exercises \(129-140 .\) 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 \(L ?\) $$ a_{n}=(0.9999)^{n} $$

Prove that \(\lim _{n \rightarrow \infty} \sqrt[n]{n}=1\).

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