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Chapter 6: Problem 135

Find the smallest value of \(n\) such that the remainder estimate \(\left|R_{n}\right| \leq \frac{M}{(n+1) !}(x-a)^{n+1}\), where \(M\) is the maximum value of \(\left|f^{(n+1)}(z)\right|\) on the interval between \(a\) and the indicated point, yields \(\left|R_{n}\right| \leq \frac{1}{1000}\) on the indicated interval. $$ f(x)=e^{-x} \text { on }[-3,3], a=0 $$

Short Answer

Expert verified
The smallest value of \( n \) is 9.

Step by step solution

01

Identify the Function and Interval

The given function is \( f(x) = e^{-x} \), and the interval is \([-3, 3]\) with \( a = 0 \). We are using the remainder term in the Taylor series approximation to find \( n \) such that the error is less than \( \frac{1}{1000} \).
02

Determine the Derivative

Calculate the \((n+1)\)th derivative of the function. For \( f(x) = e^{-x} \), all derivatives are \((-1)^k e^{-x}\), thus \( f^{(n+1)}(x) = (-1)^{n+1} e^{-x} \).
03

Find Maximum Value of Derivative

The maximum value of \( |f^{(n+1)}(z)| = e^{-z} \) on \([-3, 3]\) is at \( z = -3 \) since \( e^{-x} \) decreases as \( x \) increases. Therefore, \( M = e^3 \).
04

Set Up the Remainder Inequality

The remainder term \( R_n \) is given by \( \left|R_n\right| \leq \frac{M}{(n+1)!}(x-a)^{n+1} = \frac{e^3}{(n+1)!} x^{n+1} \). The condition is \( \left|R_n\right| \leq \frac{1}{1000} \).
05

Solve for Smallest n

Find \( n \) such that for \( -3 \leq x \leq 3 \), the inequality \( \frac{e^3 \cdot 3^{n+1}}{(n+1)!} \leq \frac{1}{1000} \) holds. Test values of \( n \) to find the smallest valid \( n \).
06

Calculate Trial Values

Calculate trial values for several positive integers. For \( n = 6 \):\[ \frac{e^3 \, 3^7}{7!} = \frac{20.0855 \, 2187}{5040} = \frac{43917.9585}{5040} \approx 8.71 \]For \( n = 7 \):\[ \frac{e^3 \, 3^8}{8!} = \frac{20.0855 \, 6561}{40320} = \frac{131763.2755}{40320} \approx 3.27 \]For \( n = 8 \):\[ \frac{e^3 \, 3^9}{9!} = \frac{20.0855 \, 19683}{362880} = \frac{396995.7795}{362880} \approx 1.093 \]
07

Verify the Smallest n

When \( n = 9 \):\[ \frac{e^3 \, 3^{10}}{10!} = \frac{20.0855 \, 59049}{3628800} = \frac{1189948.1245}{3628800} \approx 0.328 \]This satisfies the condition \( \leq \frac{1}{1000} \). Thus, the smallest \( n \) is 9.

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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 method for approximating functions using a sum of terms. These terms are derived from the function鈥檚 derivatives at a single point. This technique is quite powerful for understanding and approximating the behavior of functions around a specific point.
  • The Taylor series for a function \(f(x)\) about the point \(a\) is expressed as: \[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \dots + \frac{f^n(a)}{n!}(x-a)^n + R_n\]
  • The remainder term \(R_n\) signifies the error between the true function and the polynomial approximation.
  • By increasing the number of terms \(n\), the approximation becomes more accurate within a region around \(a\).
Understanding Taylor series allows one to approximate complex functions like the exponential function with simpler polynomial forms around a point \(a\).
remainder estimate
When approximating a function using a Taylor series, an important aspect to understand is the remainder estimate. This estimate helps us to understand the error involved in truncating the infinite series after a finite number of terms.
  • The remainder term for a Taylor series after \(n\) terms, denoted as \(R_n\), is given by:\[\left|R_{n}\right| \leq \frac{M}{(n+1) !}(x-a)^{n+1}\]
  • Here, \(M\) is the maximum value of the \((n+1)\)th derivative on the interval between \(a\) and the point \(x\).
  • This inequality helps to determine how closely the approximation mirrors the actual function.
In practice, the remainder estimate is essential for predicting how many terms are needed to achieve a desired level of accuracy.
maximum value of derivatives
To effectively apply the Taylor Remainder Theorem, it is crucial to identify the maximum value of the derivatives of a function. This value, used as \(M\) in the remainder estimate, influences the accuracy of the polynomial approximation.
  • The value \(M\) is determined by finding the largest value of \(|f^{(n+1)}(z)|\) within the interval of interest.
  • In our case, for \(f(x) = e^{-x}\) evaluated on \([-3,3]\), the maximum of \(e^{-z}\) occurs at \(z = -3\).
  • So, \(M = e^3\). This step is essential in ensuring the remainder estimate is precise.
Understanding how to compute \(M\) allows for effective control over the approximation error in the Taylor series, leading to more reliable results.
exponential function
The exponential function \(e^x\) is a fundamental mathematical function characterized by its unique properties. This function is often used in Taylor series due to its simple and repetitive derivatives that maintain its form.
  • The derivative of \(e^{-x}\) is \((-1)^k e^{-x}\), where \(k\) is the order of the derivative.
  • This characteristic makes \(f(x) = e^{-x}\) advantageous for approximations as all derivatives are easily computed and identical in structure except for coefficients.
  • When expressed as a Taylor series, \(e^{x}\) is infinite, yet it is one of the few functions with a well-known, easy-to-derive Taylor series.
Understanding the properties of exponential functions is powerful in many areas such as calculus, differential equations, and particularly in approximating functions efficiently and with high precision using Taylor series.

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Most popular questions from this chapter

Find the Maclaurin series of \(\sinh x=\frac{e^{x}-e^{-x}}{2}\).

Find the Taylor series of the given function centered at the indicated point. $$ \frac{1}{(x-1)^{3}} \text { at } a=0 $$

Find the Maclaurin series of each function. $$ f(x)=e^{x^{3}} $$

Compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of \(f\). $$ f(x)=\frac{\tan \sqrt{x}}{\sqrt{x}} \text { (see expansion for } \left.\tan x\right) $$

An equivalent formula for the period of a pendulum with amplitude \(\theta_{\max } \quad\) is \(T\left(\theta_{\max }\right)=2 \sqrt{2} \sqrt{\frac{L}{g}} \int_{0}^{\theta_{\max }} \frac{d \theta}{\sqrt{\cos \theta}-\cos \left(\theta_{\max }\right)}\) where \(L\) is the pendulum length and \(g\) is the gravitational acceleration constant. When \(\theta_{\max }=\frac{\pi}{3} \quad\) we \(\frac{1}{\sqrt{\cos t-1 / 2}} \approx \sqrt{2}\left(1+\frac{t^{2}}{2}+\frac{t^{4}}{3}+\frac{181 t^{6}}{720}\right)\). Integrate this approximation to estimate \(T\left(\frac{\pi}{3}\right)\) in terms of \(L\) and \(g\). Assuming \(g=9.806\) meters per second squared, find an approximate length \(L\) such that \(T\left(\frac{\pi}{3}\right)=2\) seconds.
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