Problem 67 - Calculus Early Transcendentals

Question

What is the minimum order of the Taylor polynomial required to approximate the following quantities with an absolute error no greater than \(10^{-3}\) ? (The answer depends on your choice of a center.) $$e^{-0.5}$$

Step-by-Step Solution

Answer

Answer: The minimum order needed is 4.

Step 1: Write down the Taylor series for \(e^x\)

The Taylor expansion of \(e^x\) around \(a=0\) (Maclaurin series) is given by: $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...$$

Step 2: Determine the corresponding Taylor polynomial for \(e^{-0.5}\)

Since we want to approximate \(e^{-0.5}\), we will substitute \(x = -0.5\) in the Taylor series: $$e^{-0.5} = \sum_{n=0}^{\infty} \frac{(-0.5)^n}{n!} = 1 - 0.5 + \frac{(-0.5)^2}{2!} + \frac{(-0.5)^3}{3!} + \frac{(-0.5)^4}{4!} + ...$$

Step 3: Apply the remainder term formula to estimate the error

We know that the absolute error is the remainder term, given by the Lagrange formula: $$R_n(x) = \frac{e^{\xi} x^{n+1}}{(n+1)!},$$ where \(0\le\xi \le x\). In this case, x = -0.5, and we want to find minimum n such that the absolute error \(R_n(x) \le 10^{-3}\). $$\frac{e^{\xi}(-0.5)^{n+1}}{(n+1)!} \le 10^{-3}$$ We do not know the value of \(\xi\), but we know that \(0\le \xi \le -0.5\). Since \(e^x\) is a monotonic increasing function, its maximum value in the interval would be at \(\xi = -0.5\). Therefore, to make a conservative estimation of error: $$\frac{e^{-0.5}(-0.5)^{n+1}}{(n+1)!} \le 10^{-3}$$

Step 4: Solve for the minimum order n

Now, we need to find the minimum value of n that satisfies the inequality above. To do this, we will test successive values of n until the inequality holds true. For n = 0: $$\frac{e^{-0.5}(-0.5)^{1}}{(1)!} \approx 0.303 > 10^{-3}$$ For n = 1: $$\frac{e^{-0.5}(-0.5)^{2}}{(2)!} \approx 0.075 > 10^{-3}$$ For n = 2: $$\frac{e^{-0.5}(-0.5)^{3}}{(3)!} \approx 0.012 > 10^{-3}$$ For n = 3: $$\frac{e^{-0.5}(-0.5)^{4}}{(4)!} \approx 0.002 > 10^{-3}$$ Finally, for n = 4: $$\frac{e^{-0.5}(-0.5)^{5}}{(5)!} \approx 0.000238 \le 10^{-3}$$ The minimum order n = 4 satisfies the given conditions. Thus, we need at least a 4th order Taylor polynomial to approximate \(e^{-0.5}\) with the desired accuracy.

Key Concepts

Maclaurin series

The Maclaurin series is a special case of the Taylor series, where the function is expanded around zero. It's named after the Scottish mathematician Colin Maclaurin. This series provides a way to express many functions as infinite sums of powers of their input variable. The general formula for a Maclaurin series is

• \[ f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots \]

For example, the Maclaurin series for the exponential function \( e^x \) is

• \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]

Using this form, complex functions can be approximated using polynomials, which are much easier to compute.
For instance, approximating \( e^{-0.5} \) involves using the Maclaurin series with \( x = -0.5 \). Each term added improves the approximation.

Lagrange remainder

The Lagrange remainder term is an essential concept when dealing with approximations using Taylor or Maclaurin series. It helps us estimate the error involved in truncating the series. When using a Taylor polynomial of degree \( n \), the error, or remainder \( R_n(x) \), is given by:

• \[ R_n(x) = \frac{f^{(n+1)}(\xi) x^{n+1}}{(n+1)!} \]

Where \( \xi \) is some value between 0 and \( x \).
In our exercise, we substitute \( x = -0.5 \) to find how closely a polynomial can approximate \( e^{-0.5} \).
Maximizing \( e^{\xi} \) over the interval ensures we account for the worst-case scenario, leading us to find that a 4th degree polynomial minimizes error within our constraints.

Taylor polynomial approximation

Taylor polynomial approximation is a method used to express functions as polynomials, which are simpler to work with than more complex functions for both computation and understanding. For example, while you can't easily compute \( e^{-0.5} \) without a calculator, approximating it using a polynomial simplifies the task substantially.
In the step-by-step process outlined before, the polynomial grows with each term, beginning with a constant and progressively including terms with higher powers of \( x \).
In our exercise, we discovered that only when we include terms up to the 4th degree, does the Taylor polynomial provide a good approximation with an error of no more than \( 10^{-3} \). This process effectively captures the essence of the function while maintaining usability. By increasing the degree of the polynomial, the approximation becomes more accurate, showcasing the power of Taylor polynomials.