Problem 46 - Calculus Early Transcendentals

Question

(a) Show that \(e^{x} \geq 1+x\) if \(x \geq 0\) (b) Show that \(e^{x} \geq 1+x+\frac{1}{2} x^{2}\) if \(x \geq 0\) (c) Confirm the inequalities in parts (a) and (b) with a graphing utility.

Step-by-Step Solution

Answer

(a) and (b) use calculus/Taylor series; graph confirms visually for \(x \geq 0\).

Step 1: Understand the Problem

We need to prove the inequalities for the exponential function. These relate to the tangent line and parabola approximations of the exponential function at 0, often derived using Taylor series.

Step 2: Prove Inequality (a) Using Calculus

To show that \(e^x \geq 1+x\), consider the function \(f(x) = e^x - x - 1\). Compute the derivative: \(f'(x) = e^x - 1\). Since \(e^x \geq 1\) for \(x \geq 0\), it follows that \(f'(x) \geq 0\). This means \(f(x)\) is non-decreasing for \(x \geq 0\). Since \(f(0) = 0\), it implies \(f(x) \geq 0\) for \(x \geq 0\), thus \(e^x \geq 1 + x\).

Step 3: Prove Inequality (b) Using Taylor Series

Consider the Taylor series expansion of \(e^x\) around 0: \(e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots\). Truncate the series to \(1 + x + \frac{x^2}{2}\) and observe that the remainder is \(\frac{x^3}{6} + \cdots\), which is always non-negative for \(x \geq 0\). Hence, \(e^x \geq 1 + x + \frac{x^2}{2}\) for \(x \geq 0\).

Step 4: Confirm Inequalities with Graphing Utility

Graph \(y = e^x\), \(y = 1 + x\), and \(y = 1 + x + \frac{x^2}{2}\) on the same axes using a graphing utility. Observe that for \(x \geq 0\), \(e^x\) is above both \(1 + x\) and \(1 + x + \frac{x^2}{2}\), confirming the inequalities visually.

Key Concepts

Taylor Series

The Taylor series is a powerful tool in calculus that allows us to approximate more complex functions with polynomials. It represents a function as an infinite sum of terms, calculated from the values of its derivatives at a single point. For the exponential function, the Taylor series expansion around 0 (Maclaurin series) is given by \[ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \]Each term corresponds to the nth derivative of the function evaluated at the specific point, divided by \(n!\), and multiplied by \(x^n\). This formula is especially useful because it helps us understand and approximate the behavior of \(e^x\) in a localized manner by using polynomials, which are simpler to work with.
For the inequalities in the exercise, we truncated the series and considered only a few initial terms against linear and quadratic expressions. This truncation forms the basis for approximating and comparing the function's behavior for small values of \(x\). By using Taylor series, we ensure that our approximations of \(e^x\) are accurate in relation to the expressions \(1 + x\) and \(1 + x + \frac{x^2}{2}\).

Exponential Function

The exponential function \(e^x\) is a fundamental function in mathematics with many unique properties. One key characteristic is that it increases exponentially as its input \(x\) increases. This function is defined as the sum of its Taylor series \[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]which serves to express it precisely as an infinite series of terms. One of the extraordinary features of \(e^x\) is its derivative, \(\frac{d}{dx}e^x = e^x\), which means that the function's rate of change is proportional to its current value, making it a self-replicating function with a uniform rate of growth.
These properties are immensely practical in many fields, including economics, biology, and physics, to model growth processes. The inequalities in our exercise exhibit the gradual growth of \(e^x\) as it surpasses both linear and quadratic terms when \(x \geq 0\), affirming its rapid escalation relative to polynomial growth.

Graphical Analysis

Graphical analysis involves using plots to visually assess and confirm mathematical concepts. In this exercise, graphing the functions \(y = e^x\), \(y = 1 + x\), and \(y = 1 + x + \frac{x^2}{2}\) on the same set of axes provides a clear visualization of their behavior. By plotting these functions, one can easily verify the inequalities from the exercise.
When viewed on a graph, the exponential function \(e^x\) appears as a curve moving steadily upward, whereas the expressions \(1 + x\) and \(1 + x + \frac{x^2}{2}\) appear as a straight line and a parabola, respectively.

• For \(x \geq 0\), \(e^x\) consistently resides above both the line \(1 + x\) and the parabola \(1 + x + \frac{x^2}{2}\).
• This visual interpretation helps reinforce the validity of the algebraic findings, supporting the analytical arguments made using calculus and Taylor series.

Graphical confirmation is an excellent approach to understanding mathematical relationships, as it allows for immediate visual evidence and intuition of how the inequalities hold.

Derivative Analysis

Derivative analysis is a crucial method used to understand the behavior of functions, particularly in verifying inequalities such as those presented in our exercise. By examining the rate of change of a function, we can determine where the function is increasing or decreasing.
For the inequality \(e^x \geq 1 + x\), we introduced the function \(f(x) = e^x - x - 1\), and its derivative \(f'(x) = e^x - 1\). Given that \(e^x \geq 1\) for \(x \geq 0\), it follows that \(f'(x)\) is non-negative, indicating that \(f(x)\) is non-decreasing.

• Since \(f(0) = 0\), it assures us that \(f(x) \geq 0\) for all \(x \geq 0\), hence proving the inequality.
• For the second inequality \(e^x \geq 1 + x + \frac{x^2}{2}\), similar logic is applied using Taylor series with the derivative analysis to confirm that \(e^x\) grows more rapidly than the quadratic expression.

By using derivative analysis, we gain a robust understanding of how function values relate to each other over specific intervals, solidifying the inequalities' proofs.