Chapter 1: Problem 1

Show that \(n !>2^{n}\) for all \(n>3\). Hint: Write out a few terms; then consider what you multiply by to go from, say, \(5:\) to \(6 !\) and from \(2^{5}\) to \(2^{6}\).

Short Answer

Expert verified

By induction, \(n! > 2^n\) for all \(n > 3\).

Step by step solution

Base Case

First, check the inequality for small values of n greater than 3. Start with n = 4. Compute both sides: \(4! = 24\) and \(2^4 = 16\). Since 24 > 16, the inequality holds for n = 4.

Check Another Value

Next, consider n = 5. Compute both sides: \(5! = 120\) and \(2^5 = 32\). Again, since 120 > 32, the inequality holds for n = 5.

Generalize for n > 5

Assume the inequality holds for some k > 4, so \(k! > 2^k\). Now consider \((k + 1)!\) and \(2^{k+1}\). Write \((k + 1)! = (k + 1) \times k!\). Because \(k! > 2^k\), this gives \((k + 1)! = (k + 1) \times k! > (k + 1) \times 2^k\).

Simplify the Inequality

Since \(k > 4\), we have \(k + 1 > 2\), making \((k + 1) \times 2^k > 2 \times 2^k = 2^{k + 1}\). Therefore, \((k + 1)! > 2^{k + 1}\). By mathematical induction, the inequality \(n! > 2^n\) holds for all \(n > 3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factorial Comparison

Factorials are products of all positive integers up to a certain integer n, expressed as \(n!\). In this exercise, we compare the factorial function with the exponential function. For instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\), whereas \(2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32\). Notice how quickly the factorial grows compared to exponential functions. When comparing \(n !\) to \(2^n\) for large n, factorials grow significantly faster.
This rapid growth makes proving inequalities involving factorials and exponentials interesting and challenging.

Inequalities

Inequalities show a relationship where one quantity is larger or smaller than another. In this problem, you need to prove that \(n! > 2^n\) for all \(n > 3\). To do this, we:

Start by verifying the inequality for smaller n values like 4 and 5
Use logical steps to generalize it for larger values using mathematical induction

Understanding inequalities requires practice. Always compare terms systematically. Start from known values and slowly build up to general statements. This stepwise approach helps in grasping how inequalities work.

Proof Techniques

Proof techniques provide a systematic way to demonstrate mathematical statements. Here, we use **Mathematical Induction**:

**Base Case:** Verify the statement for an initial value (e.g., \(n = 4\)).
**Induction Hypothesis:** Assume the statement holds for some arbitrary value k greater than the base case.
**Induction Step:** Prove the statement for \(k + 1\) using the assumption from the induction hypothesis.

For \(n! > 2^n\), we verified the base case (e.g., \(4! > 2^4\)). We then assumed \(k! > 2^k\) to hold and proved \((k + 1)! > 2^{(k + 1)}\). The induction step follows similarly by demonstrating that if the statement holds for k, it must hold for \(k + 1\). This systematic method ensures a comprehensive proof valid for all integers greater than 3.

Exponential Functions

Exponential functions are functions of the form \(2^n\), where n is an exponent. These functions grow rapidly, but not as quickly as factorials. In our problem, we compare \(n!\) and \(2^n\).

Exponential functions double their value with each increase in n.
Factorials multiply the existing product by the next integer.

By writing out terms, such as \(2^5 = 32\) and \(5! = 120\), you can directly observe how factorials surpass exponential growth. Recognizing these growth patterns helps in understanding complex inequalities and relationships between different types of mathematical functions.