Problem 11 - Thomas Calculus

Question

Find the Maclaurin series for the functions. \(e^{-x}\)

Step-by-Step Solution

Answer

The Maclaurin series for \(e^{-x}\) is \(\sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n!}\).

Step 1: Recall the Maclaurin series formula

The Maclaurin series is a special case of the Taylor series expansion of a function about zero. The formula is given by:\[f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots\]This series represents the sum of the derivatives of the function evaluated at zero, multiplied by powers of \(x\) divided by the factorial of the degree.

Step 2: Compute the derivatives of \(e^{-x}\)

We need to find the first few derivatives of \(e^{-x}\):1. \(f(x) = e^{-x}\)2. \(f'(x) = -e^{-x}\)3. \(f''(x) = e^{-x}\)4. \(f'''(x) = -e^{-x}\)5. Continue this pattern... Notice the derivatives alternate in sign and repeat in a cycle.

Step 3: Evaluate the derivatives at \(x = 0\)

Calculate each derivative at \(x = 0\):1. \(f(0) = e^{0} = 1\)2. \(f'(0) = -e^{0} = -1\)3. \(f''(0) = e^{0} = 1\)4. \(f'''(0) = -e^{0} = -1\)This pattern persists for higher degree derivatives.

Step 4: Write the Maclaurin series for \(e^{-x}\)

Substitute these values into the Maclaurin series formula:\[e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n!}\]Each term is calculated by multiplying the derivative evaluated at zero by \(x^n\) and dividing by \(n!\). The alternation in sign is due to the pattern observed in the derivatives.

Key Concepts

Taylor series

The Taylor series is an important tool in calculus for approximating functions. It allows us to express a function as an infinite sum of terms calculated from its derivatives. This series is written in terms of its derivatives evaluated at a specific point, typically represented by the formula \[f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)(x - a)^2}{2!} + \frac{f'''(a)(x - a)^3}{3!} + \cdots\]

• "\(f(a)\)" is the function value at the point \(a\).
• "\(f'(a)\)" is the first derivative evaluated at \(a\).
• "\((x - a)\)" accounts for the distance from \(a\).

A Taylor series centered at zero is called a Maclaurin series, giving it a special place in mathematics because it simplifies calculations when expanding functions at zero. Understanding Taylor series helps grasp function behaviors near certain points and improve predictions in analysis.

series expansion

Series expansion is a method of expressing a function as a sum of elements calculated through systematic procedures, often involving derivatives. A common type of series expansion is the Taylor series, where a function is expanded around a particular point.In the case of the Maclaurin series, the expansion occurs around the point zero. This means we don't need shifts in our expansion formula and simplifies the process:

• Only derivatives at zero are needed.
• The resulting series combines these derivatives with powers and factorials of \(x\).

The beauty of series expansions is their ability to capture the essence of a function with remarkable accuracy, especially near the point of expansion. These expansions enable a deeper insight into the function’s behavior and can approximate complex calculations in fields like physics and engineering.

derivatives of functions

Derivatives are fundamental in calculus and play a crucial role in series expansions, such as Taylor and Maclaurin series. They measure how a function changes as its input changes, providing insights into the function’s slope and rate of change.When calculating a series like the Maclaurin series, derivatives become essential because each term involves a derivative evaluated at the expansion point:

• First, the function itself is evaluated at zero.
• Next, its successive derivatives are found: first, second, third, etc.
• Each derivative contributes to one term in the series.

The sign and magnitude of these derivatives affect the series' terms, showing both patterns and periodic behavior. In the exercise of expanding \(e^{-x}\), each derivative oscillates between positive and negative, which leads to alternating signs in the resulting series. Understanding these patterns in derivatives makes creating series expansions more intuitive and accessible.