Problem 58 - Essential Calculus Early Transcendentals

Question

(a) Show that \(e^{x} \geqslant 1+x\) for \(x \geqslant 0\) . (b) Deduce that \(e^{x} \geqslant 1+x+\frac{1}{2} x^{2}\) for \(x \geqslant 0\) . (c) Use mathematical induction to prove that for \(x \geqslant 0\) and any positive integer \(n\) $$e^{x} \geqslant 1+x+\frac{x^{2}}{2 !}+\cdots+\frac{x^{n}}{n !}$$

Step-by-Step Solution

Answer

Using mathematical induction and the Taylor series, it is shown that \(e^x \geq 1+x+\frac{x^{2}}{2 !}+\cdots+\frac{x^{n}}{n !}\) for \(x \geq 0\), \(n \in \mathbb{N}\).

Step 1: Understanding the Function

The function we need to analyze is the exponential function, particularly focusing on how it compares to polynomial approximations. Considerations on the behavior when evaluated at zero and in its derivative form will help in establishing inequalities like these.

Step 2: Proof for e^x ">=" 1 + x (Taylor Expansion)

Using the Taylor series expansion of the exponential function:\[e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\]For any \(x \geq 0\), the exponential function \(e^x\) is equal to or exceeds \(1 + x\) since all higher-order positive terms contribute non-negatively. Thus, \(e^{x} \geq 1 + x\).

Step 3: Starting From Basic Inequality

Using the result from Step 2, consider how additional terms from the Taylor expansion affect the inequality.For any additional term \(\frac{x^2}{2!}\) added to \(1 + x\), because \(\frac{x^2}{2!} = \frac{x^2}{2} > 0\), the inequality \(e^{x} \geq 1 + x + \frac{x^2}{2}\) holds for \(x \geq 0\).

Step 4: Verify the Base Case for Mathematical Induction

Set the base case n = 2. Show:\[e^{x} \geq 1 + x + \frac{x^2}{2!}\]This was addressed in Step 3, so the base case holds.

Step 5: Inductive Step

Assume that the inequality holds for some integer \(n = k\):\[e^x \geq 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^k}{k!}\]Prove it holds for \(n = k + 1\):Using Taylor series:\[e^x = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^k}{k!} + \frac{x^{k+1}}{(k+1)!} + \cdots\]All terms in the exponential expansion beyond \(\frac{x^k}{k!}\) are positive for \(x \geq 0\). Hence,\[e^x \geq 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^k}{k!} + \frac{x^{k+1}}{(k+1)!}\]

Step 6: Conclusion for Mathematical Induction

By the principle of mathematical induction, since the inequality holds for the base case and the inductive step has been proven, the inequality:\[e^{x} \geqslant 1+x+\frac{x^{2}}{2 !}+\cdots+\frac{x^{n}}{n !}\]holds true for all positive integers \(n\) and for any \(x \geq 0\).

Key Concepts

Taylor Series

The Taylor series is a powerful mathematical tool used to approximate functions. It breaks down a function into an infinite sum of terms calculated from the values of its derivatives at a single point. For the exponential function, the expression for its Taylor series, centered at zero (known as the Maclaurin series), is:

• \[ e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]

Here, each term \( \frac{x^n}{n!} \) is derived from the successive derivatives of \( e^x \) evaluated at zero.
This series allows us to approximate \( e^x \) with any desired level of accuracy by including more terms.
When \( x \geq 0 \), each of these additional terms \( \frac{x^n}{n!} \) is positive, reinforcing the approximation to be always greater than or equal to the truncated sum. This property is crucial when proving inequalities involving the exponential function as it provides a straightforward way to compare \( e^x \) with polynomial expressions.

Mathematical Induction

Mathematical induction is a technique used to prove statements or formulas related to integers. It involves two key steps. First, the base case, where the statement is proven for the initial integer (usually \( n = 0 \) or \( n = 1 \)). Second, the inductive step, where we assume the statement holds for some integer \( k \) and we prove it holds for \( k + 1 \).
Induction is often likened to dominoes falling; once the base case is proven, and you show the inductive step works, all subsequent numbers "fall" because they follow logically from the assumptions.

• Base Case: Establish the truth of the statement for an initial value.
• Inductive Step: Assuming the proposition is true for some \( n = k \), prove it holds for \( n = k + 1 \).

In the context of the exponential inequality, by confirming the base case with fewer terms from the Taylor series, and then proving the inductive step, it assures us that the inequality holds for any integer \( n \).

Polynomial Approximation

Polynomial approximation involves expressing complex functions with simpler polynomial expressions. In many cases, functions can be approximated by polynomials which capture their essential characteristics over a specific range.
For the exponential function, a natural polynomial approximation is provided by utilizing the Taylor series. By truncating the series after a finite number of terms, you derive polynomial approximations, such as:

• \( P_1(x) = 1 + x \)
• \( P_2(x) = 1 + x + \frac{x^2}{2!} \)

These serve as successive approximations of \( e^x \) for \( x \geq 0 \).
Because each term added through the Taylor series represents a higher degree of accuracy, these polynomial forms approach the behavior of \( e^x \) more closely as more terms are included.
Polynomial approximations are practical for calculations when the exact value of \( e^x \) is cumbersome or unnecessary at high precision.

Exponential Function

The exponential function, represented as \( e^x \), is a fundamental mathematical function with the base of the natural logarithm \( e \), approximately equal to 2.71828. It exhibits unique properties crucial in various branches of mathematics and applied sciences.
Characterized by its smooth and continuous nature, \( e^x \) is continuously increasing for all real \( x \), and has a derivative equal to itself, meaning that \( \frac{d}{dx} [e^x] = e^x \).

• Grows rapidly as \( x \) increases.
• \( e^0 = 1 \), illustrating its fixed point at zero.
• Increases infinitely without asymptotes in the positive direction.

Analyzing \( e^x \) against polynomial approximations like \( 1 + x \) demonstrates its growth by considering additional positive terms from its Taylor series expansion.
This behavior is crucial in proving exponential inequalities, where minimizing expressions with lower power terms underestimates the true value of \( e^x \) but never surpasses it, hence comparisons like \( e^x \geq 1+x \) hold true.