Problem 51 - Calculus Early Transcendentals

Question

Evaluating an infinite series Let \(f(x)=\left(e^{x}-1\right) / x,\) for \(x \neq 0\) and \(f(0)=1 .\) Use the Taylor series for \(f\) about 0 and evaluate \(f(1)\) to find the value of \(\sum_{k=0}^{\infty} \frac{1}{(k+1) !}\)

Step-by-Step Solution

Answer

Answer: The value of the infinite series is equal to the value of \(f(1)\), where \(f(x) = \frac{e^x - 1}{x}\).

Step 1: Find the Taylor series for f(x) about 0

Recall that the Taylor series of a function \(f(x)\) around 0, also called the Maclaurin series, is given by: $$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n $$ Where \(f^{(n)}(0)\) denotes the \(n\)-th derivative of the function evaluated at 0. So, we first need to find the derivatives of \(f(x)\) at 0: \(f(x) = \frac{e^x - 1}{x}\) \(f'(x) = \frac{e^x}{x} - e^x \frac{1}{x^2}\) \(f''(x) = \frac{e^x - 1}{x^3} - 2e^x \frac{1}{x^3}\) Now, we evaluate these derivatives at x=0 (excluding f(x)): \(f'(0) = \frac{e^0}{0} - e^0 \frac{1}{0^2} = 1\) \(f''(0) = \frac{e^0 - 1}{0^3} - 2e^0 \frac{1}{0^3} = \frac{1}{2}\) We can now use these derivatives to find the Taylor series for \(f(x)\): \(f(x) = 1 + \frac{1}{1!}x + \frac{1}{2}\frac{1}{2!}x^2 + \cdots = \sum_{k=0}^{\infty} \frac{1}{(k+1)!} x^k\)

Step 2: Evaluate f(1) using the Taylor series

We have found the Taylor series for \(f(x)\). Now, we substitute \(x=1\) to evaluate \(f(1)\): $$ f(1) = \sum_{k=0}^{\infty} \frac{1}{(k+1)!} $$

Step 3: Determine the value of the infinite series

We can see that the value of the infinite series is equal to the value of \(f(1)\), which we found in step 2. Thus, $$ \sum_{k=0}^{\infty} \frac{1}{(k+1)!} = f(1) $$ The value of the infinite series is equal to \(f(1)\).

Key Concepts

Taylor Series

The Taylor series is a powerful tool in calculus used to represent a function as an infinite sum of terms. These terms are derived from the function's derivatives evaluated at a particular point. The general formula for a Taylor series around the point zero, also known as a Maclaurin series, is:\[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\]Here, the function is expressed as an infinite sum involving the derivatives of the function at zero. The coefficient of each term is determined by the value of the function's derivatives at this point, divided by the factorial of the term's order.If a function is infinitely differentiable at a point, its Taylor series can provide an accurate approximation of the function near that point. This property is particularly useful in various fields of science and engineering.

Maclaurin Series

The Maclaurin series is a special case of the Taylor series that is centered at zero. This series is particularly useful for several common functions, where the function and all its derivatives evaluated at zero are known and easily calculated.Common functions with well-known Maclaurin series include the exponential function, logarithmic functions, and trigonometric functions, among others. For example, the function \(f(x) = e^x\) has a Maclaurin series given by:\[e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\]Expressing functions as Maclaurin series can greatly simplify complex calculations, especially when approximating functions with numerical methods or in physics problems where such simplifications lead to convenient results.

Calculus

Calculus is the branch of mathematics that studies continuous change. It consists of differential calculus, which deals with rates of change and slopes of curves, and integral calculus, which deals with the accumulation of quantities and the areas under and between curves. In the context of infinite series, calculus provides tools to study the behavior of functions as they are expressed as sums of their derivatives. The study of how functions behave near certain points, and how they can be expanded using series, is a key part of calculus. Taylor and Maclaurin series are essential techniques taught in calculus to understand function approximations, error estimations, and theoretical aspects of mathematical analysis. They are part of the foundation upon which more complex calculus ideas are built.

Derivative Evaluation

Derivative evaluation is a critical step in forming Taylor or Maclaurin series. The derivatives of a function at a specific point provide the necessary coefficients for the series expansion.To evaluate derivatives for a series:

• Find the first few derivatives of the function.
• Evaluate each derivative at the point of expansion (often zero for Maclaurin series).
• Use these evaluated values to construct the terms of the series.

For example, if we are expanding \(f(x) = \frac{e^x - 1}{x}\) around zero, we first find its derivatives, such as \(f'(x)\) and \(f''(x)\), and then evaluate them at zero. This process gives us the coefficients we need for the series terms.This systematic approach enables us to approximate functions effectively. Evaluating derivatives is thus a crucial skill in solving problems involving Taylor or Maclaurin series in calculus.