Problem 37 - Essential Calculus Early Transcendentals

Question

(a) Show that the function $$ f(x)=\sum_{n=0}^{\infty} \frac{x^{n}}{n !} $$ is a solution of the differential equation $$ f^{\prime}(x)=f(x) $$ (b) Show that \(f(x)=e^{x}\)

Step-by-Step Solution

Answer

The series \(f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}\) satisfies \(f'(x) = f(x)\) and is equivalent to \(f(x) = e^x\).

Step 1: Start with the function definition

Consider the function given by the infinite series: \[ f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]

Step 2: Differentiate the function term by term

Differentiate \(f(x)\) term by term. Differentiating each term yields: \[\frac{d}{dx}\left(\frac{x^n}{n!}\right) = \frac{n x^{n-1}}{n!} = \frac{x^{n-1}}{(n-1)!},\]where \(n = 0\) is a special case as its derivative is zero.

Step 3: Rewrite the differentiated series

Shift the index of summation from \(n\) to \(n-1\) for the differentiated series:\[ f'(x) = \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!} = \sum_{m=0}^{\infty} \frac{x^m}{m!} = f(x),\]where \(m = n - 1\). This shows that \(f'(x) = f(x)\).

Step 4: Recognize the series as the exponential function

The series \(\sum_{n=0}^{\infty} \frac{x^n}{n!}\) is the Maclaurin series expansion for the exponential function \(e^x\).

Step 5: Conclusion from series details

From both the differentiated series and recognizing the standard exponential series, we see that the function described by the series is indeed \(f(x) = e^x\). This completes the proof.

Key Concepts

Differential Equation

A differential equation involves an unknown function and its derivatives. In this exercise, we deal with the differential equation \( f'(x) = f(x) \). This is a first-order equation, which simply means it involves the first derivative of the function. Solving such an equation means finding a function \( f(x) \) that satisfies it for all \( x \). Here, we're asked to show that the infinite series representing \( f(x) \) is a solution. The key property of this particular differential equation is that its solution grows at a rate proportional to its current value. This is a defining characteristic of exponential functions.

Maclaurin Series

A Maclaurin Series is a type of Taylor Series centered around 0. It provides a way to express functions as an infinite sum of terms calculated from the function's derivatives at a single point. The given function \( f(x) = \sum_{n=0}^{\infty} \frac{x^{n}}{n!} \) is a classic example of a Maclaurin Series. This series is particularly important because it describes the exponential function \( e^x \). One of the remarkable aspects of Maclaurin Series is that it can approximate complex functions using simple polynomial terms, making it an essential tool in mathematics for both analysis and computation.

Infinite Series

An infinite series, as the name suggests, is the sum of an infinite sequence of terms. The function provided, \( f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} \), is expressed as an infinite series. Each term in this series becomes smaller as \( n \) increases, thanks to the \( n! \) (factorial of \( n \)) in the denominator. This ensures convergence, meaning the series has a finite limit. The study of infinite series is crucial in calculus because it allows mathematicians to solve differential equations and approximate functions that may not have a straightforward algebraic form.

Differentiation

Differentiation is a central concept in calculus that involves finding the derivative of a function, which represents the rate of change of the function. In this exercise, we differentiate the given series \( f(x) = \sum_{n=0}^{\infty} \frac{x^{n}}{n!} \) term by term to find \( f'(x) \). Differentiating each term \( \frac{x^n}{n!} \) gives \( \frac{x^{n-1}}{(n-1)!} \), illustrating how differentiation affects each component of an infinite series. The differentiated series \( f'(x) \) equates back to \( f(x) \), showing that the original series satisfies the differential equation. This result highlights the power of differentiation in solving and understanding the behavior of functions.