Problem 22 - Essential Calculus

Question

Find the Maclaurin series for the function. (Use the table of power series for elementary functions.) $$ g(x)=e^{-3 x} $$

Step-by-Step Solution

Answer

The Maclaurin series for the function \(g(x) = e^{-3x}\) is \(g(x) = \sum_{n=0}^{\infty} \frac{(-1)^n * 3^n * x^n}{n!}\)

Step 1 - Identify the Basic Series

Refer to the table of power series for elementary functions to find the series expression for e^x, which is \(e^{x} = \sum_{n=0}^{\infty} \frac{x^n}{n!}\) where n ranges from 0 to infinity.

Step 2 - Adjust for the Given Function

The given function is \(g(x) = e^{-3x}\). Therefore, we must adjust the series expression so x is replaced by -3x. This is achieved by substituting -3x in place of x in the power series expression giving us \(g(x) = \sum_{n=0}^{\infty} \frac{(-3x)^n}{n!}\)

Step 3 - Simplify the Power of x

The expression \((-3x)^n = (-1)^n * 3^n * x^n\). Substituting this simplification into the series expression, we have \(g(x) = \sum_{n=0}^{\infty} \frac{(-1)^n * 3^n * x^n}{n!}\).

Key Concepts

Power Series

Power series are a wonderful tool in mathematics, allowing us to express functions as infinite sums of terms. This can make complex functions simpler to work with. A power series takes the form:

• \( f(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \ldots \)

Here,

• \( a_0, a_1, a_2, \ldots \) are coefficients.
• \( x \) is the variable.
• The series continues indefinitely.

The key idea is that a power series converges to a specific value within a certain range, known as the radius of convergence. In the context of solving exercises like the one concerning the Maclaurin series, understanding how to adjust basic power series expressions to fit specific functions is crucial.
Power series are especially useful in calculus and algebra because they can preserve relationships and properties of functions that are otherwise challenging to visualize or solve in closed form.

Exponential Functions

Exponential functions are a vital concept in mathematics, showcasing growth or decay. Their key characteristic is that the variable appears in the exponent, such as \( e^x \) where \( e \) is Euler's number, approximately equal to 2.718.

• These functions are ubiquitous, appearing in natural processes such as population growth and radioactive decay.
• Mathematically, exponential functions have the unique property where their derivative is proportional to the function itself, i.e., \( \frac{d}{dx} e^x = e^x \).

In exercises, especially involving series like the Maclaurin series, exponential functions often serve as a starting point for expansion. By understanding how \( e^x \) translates into a series \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \), students can tackle tasks like adjusting for functions such as \( e^{-3x} \) by substituting within the existing series.

Series Expansion

Series expansion refers to expressing a function as a series, revealing its behavior at or around a point. The Maclaurin series is a special case of Taylor series where the expansion is done around zero. For any function that is infinitely differentiable at a point, we can find its series expansion.

• For example, the Maclaurin series for \( e^x \) is \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \).
• In series expansion exercises, replacing variables appropriately is critical, such as substituting \( -3x \) for \( x \) to find the series for \( e^{-3x} \).
• This adaptation requires simplification steps, like recognizing terms such as \( (-3x)^n \) to be \( (-1)^n \, 3^n \, x^n \), to keep computations manageable.

Understanding series expansions enables the breakdown of complex mathematics into comprehensible, manageable parts, aiding in predicting and studying the function's behavior around specific values.