Problem 37 - 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 \( f'(x) = f(x) \) (b) Show that \( f(x) = e^x. \)

Step-by-Step Solution

Answer

(a) The function satisfies the equation \( f'(x) = f(x) \). (b) \( f(x) = e^x \).

Step 1: Differentiate the function

To prove that \( f(x) = \sum_{n = 0}^{\infty} \frac{x^n}{n!} \) satisfies the differential equation \( f'(x) = f(x) \), let's first find the derivative of \( f(x) \).The derivative of a power series \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \) is given by differentiating term by term:\[ f'(x) = \sum_{n=0}^{\infty} \frac{d}{dx}\left(\frac{x^n}{n!}\right) = \sum_{n=1}^{\infty} \frac{nx^{n-1}}{n!} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!}. \]

Step 2: Re-index the summation

Next, we will re-index the series from Step 1 to make it comparable to \( f(x) \).Let \( m = n - 1 \), then \( n = m + 1 \).Hence,\[ f'(x) = \sum_{m=0}^{\infty} \frac{x^m}{m!}. \]Notice that this is the same as \( f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} \), hence we have shown that \( f'(x) = f(x) \).

Step 3: Recognize the power series as the exponential function

To show that \( f(x) = e^x \), use the definition of the exponential function. The function \( e^x \) is defined by the Taylor series \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \).Since \( f(x) \) is expressed precisely by this series, we conclude that \( f(x) = e^x \).

Key Concepts

Power Series

Power series are infinite sums of the form \[ extstyle S(x) = rac{a_0}{0!} + \frac{a_1 x}{1!} + \frac{a_2 x^2}{2!} + \ldots = extstyle \ \sum_{n=0}^{\infty} \frac{a_n x^n}{n!}, \] where each term is a product of a coefficient and a power of the variable.
In this exercise, we're given a power series \[ f(x) = extstyle \ sum_{n=0}^{\infty} \frac{x^n}{n!}, \] which represents an infinite series. Here, 'n!' represents the factorial of 'n', which is the product of all positive integers up to 'n'.
Power series are useful tools in calculus for expressing more complex functions in an infinite polynomial form. This allows for operations like differentiation and integration to be handled term-by-term.
For instance, differentiating a power series means simply differentiating each term separately, as done in this exercise.

Exponential Function

The exponential function, denoted \( e^x \), is a fundamental mathematical function crucial to many areas of mathematics, especially in calculus.
It is represented by the power series\[ extstyle e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}, \] which is the same series as our given function \( f(x) \) in the exercise.
This series converges for all real numbers \( x \), meaning it reliably evaluates to a finite number as you add more terms.
The exponential function has unique properties, such as

• Being its own derivative. This means that \( \frac{d}{dx} e^x = e^x \). This property made it straightforward to show that \( f'(x) = f(x) \).
• The base \( e \) (~2.718) is an irrational number and is the constant such that the function \( e^x \) satisfies the above derivative property.

Understanding these properties helps illuminate why \( f(x) \) matches \( e^x \) exactly.

Taylor Series

A Taylor series is a way of approximating functions using polynomials.
It is represented as
\[ extstyle T(x) = \ sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n, \] where \( f^{(n)}(a) \) is the nth derivative of the function evaluated at a certain point \( a \).
In the context of our exercise, the power series \[ f(x) = extstyle \ sum_{n=0}^{\infty} \frac{x^n}{n!}, \] is actually the Taylor series for \( e^x \) centered at \( x = 0 \), because all the derivatives of \( e^x \) are \( e^x \) and thus equal to 1 at \( x = 0 \).

• By writing a function this way, it becomes easier to handle functions that are otherwise complex to integrate or differentiate directly.
• The Taylor series provides insight into the behavior and properties of functions by expressing them as an infinite sum of simpler polynomial terms.

Recognizing \( f(x) \) as the Taylor series of \( e^x \) not only confirms its identity but also illustrates the power of such series in simplifying complex mathematical tasks.