Problem 12 - Essential Calculus

Question

Prove that the Maclaurin series for the function converges to the function for all \(x\). $$ f(x)=e^{-2 x} $$

Step-by-Step Solution

Answer

Through calculations, it is shown that the nth derivative of the function \(e^{-2x}\) is \(f^{(n)}(0) = (-2)^n\). The nth term of the Maclaurin series is thus \(\frac{(-2)^n}{n!} x^n\). As the limit of this expression as \(n\to\infty\), for all \(x\), is 0, the Maclaurin series for \(e^{-2x}\) converges to the function for all \(x\).

Step 1: Understand the Problem

The task is to prove the convergence of the Maclaurin series for the given function \(e^{-2x}\). The Maclaurin series of a function \(f(x)\) is represented by \(f(x) = f(0) + f'(0)\cdot x + f''(0)\cdot \frac{x^2}{2!} + f'''(0)\cdot \frac{x^3}{3!} + \cdots\)

Step 2: Find the nth derivative of the function

Each term in the Maclaurin series involves taking successive derivatives of the function. For the function \(e^{-2x}\), the nth derivative \(f^{(n)}(x) = (-2)^n e^{-2x}\). At \(x=0\), the nth derivative can be given as \(f^{(n)}(0) = (-2)^n\)

Step 3: Find the general term of the Maclaurin series

Using the formula for the Maclaurin series, the nth term of the series can be written as \(\frac{f^{(n)}(0)}{n!} x^n = \frac{(-2)^n}{n!} x^n\)

Step 4: Determine and Prove the Convergence

The Maclaurin series converges to the function if \(\lim_{n\to\infty} \frac{f^{(n)}(0)}{n!} x^n = 0\). As we calculated, \(\lim_{n\to\infty} \frac{(-2)^n}{n!} x^n = 0\) for all \(x\). Hence we have proved that the Maclaurin series for \(e^{-2x}\) converges to the function for all \(x\).

Key Concepts

Calculus

Calculus is a branch of mathematics focusing on limits, functions, derivatives, integrals, and infinite series. It's a tool that allows us to study change and motion, providing a framework for modeling and understanding the physical universe. In the context of our exercise, calculus is used to determine the series expansion of functions through a process referred to as Maclaurin series. This particular series is a special case of the Taylor series, where the expansion is about the point zero. The Maclaurin series is instrumental in approximating functions to a polynomial form that is easier to analyze and apply. As calculus deals with infinite processes, understanding series and their convergence is an essential part of mastering this field.

Power Series

A power series is an infinite series of the form \( \sum_{n=0}^{\infty} a_n (x-c)^n \), where \( a_n \) are the coefficients of the series, \( x \) is the variable, and \( c \) is the center of the series expansion. When \( c = 0 \) the series is called a Maclaurin series. These series are extremely useful for functions that can be hard to solve otherwise, such as exponential, trigonometric, and logarithmic functions.

Understanding the concept of a power series is crucial for mathematical computations and solving complex problems. It allows us to represent functions in a form that can be easily manipulated for purposes like integration, differentiation, and solving differential equations. For the function \( e^{-2x} \), the power series representation allows us to work with a polynomial approximation of the exponential function.

Convergence of Series

The convergence of a series is a fundamental concept in calculus that deals with whether the infinite sum of the terms of the series approaches a finite value or not. For a Maclaurin series to be a valid representation of a function, it must converge to the function for all values within a certain interval. There are various tests for convergence, but one of the most common ones used in the context of power series is the ratio test.

To ensure that we can safely use the series to represent the function \( e^{-2x} \) for all \( x \) within its domain, we must prove that it converges. The given solution uses the fact that factorial growth \( n! \) in the denominator outpaces the growth of the power \( (-2)^n \) in the numerator, leading to the conclusion that the terms in the series get smaller and smaller and approach zero as \( n \) approaches infinity.

Exponential Functions

Exponential functions are a type of mathematical function of the form \( f(x) = a^x \), where \( a \) is a constant base and \( x \) is the exponent. In the case of the exercise, the exponential function in question is \( e^{-2x} \), where \( e \) is Euler's number, approximately equal to 2.71828, and \( -2x \) is the exponent.

Exponential functions model growth or decay processes and have unique properties that make them valuable across various disciplines, from finance to physics. One significant property is that the rate of change of an exponential function is directly proportional to its value. This makes the analysis of exponential functions and their representations, such as the Maclaurin series, particularly important for both theoretical and practical applications.