Problem 2 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the Maclaurin polynomial of order 4 for \(f(x)\) and use it to approximate \(f(0.12) .\) $$ f(x)=e^{-3 x} $$

Step-by-Step Solution

Answer

The approximate value of \( f(0.12) \) is 0.7095.

Step 1: Identify the function and its derivatives

We need the Maclaurin series, which is a Taylor series expansion about 0. For the function \( f(x) = e^{-3x} \), we will compute derivatives up to the fourth order.

Step 2: Compute the function derivatives

First, compute the derivatives of \( f(x) = e^{-3x} \):1. \( f(x) = e^{-3x} \)2. \( f'(x) = -3e^{-3x} \)3. \( f''(x) = 9e^{-3x} \)4. \( f'''(x) = -27e^{-3x} \)5. \( f^{(4)}(x) = 81e^{-3x} \)

Step 3: Evaluate derivatives at x = 0

Substitute \( x = 0 \) into the derivatives:- \( f(0) = 1 \) - \( f'(0) = -3 \) - \( f''(0) = 9 \)- \( f'''(0) = -27 \)- \( f^{(4)}(0) = 81 \)

Step 4: Write Maclaurin polynomial

The Maclaurin polynomial of order 4 is given by:\[ P_4(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 \]Substitute the evaluated derivatives:\[ P_4(x) = 1 - 3x + \frac{9}{2} x^2 - \frac{27}{6} x^3 + \frac{81}{24} x^4 \]This simplifies to:\[ P_4(x) = 1 - 3x + \frac{9}{2} x^2 - \frac{9}{2} x^3 + \frac{81}{24} x^4 \]

Step 5: Substitute x = 0.12 to approximate f(0.12)

Substitute \( x = 0.12 \) into the polynomial:\[ P_4(0.12) = 1 - 3(0.12) + \frac{9}{2}(0.12)^2 - \frac{9}{2}(0.12)^3 + \frac{81}{24}(0.12)^4 \]Calculate each term and sum them:- \(-3(0.12) = -0.36\)- \(\frac{9}{2}(0.12)^2 = 0.0648\)- \(\frac{9}{2}(0.12)^3 = 0.007776\)- \(\frac{81}{24}(0.12)^4 = 0.0004672\)Thus, \( P_4(0.12) \approx 0.7094912 \).

Step 6: Conclusion – Approximate value

From the calculation steps, the approximate value of \( f(0.12) \) using the Maclaurin polynomial of order 4 is approximately 0.7095.

Key Concepts

Taylor series

The Taylor series is a powerful tool in mathematics used for approximating functions. It represents a function as an infinite sum of terms, each derived from the function's derivatives at a certain point.
When we center the series at zero, it becomes known as the Maclaurin series. This is essentially a special case of the Taylor series. In this exercise, we're dealing with a function centered at zero, making it a Maclaurin series.

• This series is expressed as:
  \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots \]

This series expansion can approximate complex functions, making calculations more manageable. However, in practice, we often use only a finite number of terms. This offers a polynomial approximation to the function, which brings us to our next key concept.

Polynomial approximation

Polynomial approximation simplifies complicated functions by expressing them as polynomials.
These polynomials are often derived from Taylor or Maclaurin series. They make calculations tractable and are especially useful when dealing with non-linear or transcendental functions.

• In the exercise, a Maclaurin polynomial of order 4 was used to approximate the exponential function \(e^{-3x}\).
  

The polynomial approximation lets us evaluate the function at given points without complex calculations. For instance, to approximate \(f(0.12)\) for \(f(x) = e^{-3x}\), only polynomial terms up to \(x^4\) were considered.
While the approximation becomes more accurate as we include higher-order terms, it also becomes more complex. Thus, there's often a trade-off between simplicity and accuracy.

Exponential function

The exponential function is a fundamental mathematical expression characterized by its constant growth rate. In the exercise, the function in focus is \(f(x) = e^{-3x}\),
a modified exponential function where \(-3\) dictates the rate of growth or decay.

• Exponential functions are widely utilized in various fields, including science, finance, and engineering, due to their natural occurrence in growth and decay processes.
  

In the context of Taylor and Maclaurin series, exponential functions stand out because their derivatives are simple to compute, making them ideal for such approximations.
Specifically, the derivatives of an exponential function like \(e^{kx}\) are consistent in form, only scaled by coefficients, as seen in this exercise. This feature simplifies constructing a Taylor series and helps illustrate the concept of polynomial approximation quite effectively.

Higher-order derivatives

Higher-order derivatives are crucial when constructing Taylor or Maclaurin series. These derivatives help determine the polynomial's accuracy.

• In the calculation, derivatives up to the fourth order were used to form the Maclaurin polynomial of function \(f(x) = e^{-3x}\).
  
• The higher the order of the derivative used, the closer the polynomial will approximate the function.

Each derivative corresponds to a term in the polynomial, with higher derivatives capturing more nuanced aspects of the function's behavior.
Computing these higher-order derivatives, as shown in the solution, involves differentiating the function repeatedly.
This iterative process results in terms like \(f''(x), f'''(x),\) and \(f^{(4)}(x)\), which are then evaluated at the expansion point (usually \(x = 0\) for a Maclaurin series). By understanding both the computational aspect and the theoretical foundation, students can better grasp the usefulness and potential limitations of polynomial approximations.