Problem 33 - A Graphical Approach to Precalculus with Limits

Question

Prove each statement by mathematical induction. If \(n \geq 4,\) then \(n !>2^{n}\)

Step-by-Step Solution

Answer

The statement is proven by induction for all \( n \geq 4 \).

Step 1: Base Case

Start by checking the statement for the initial value of n, which is 4 in this case. Calculate \( 4! \) and \( 2^4 \). We have \( 4! = 24 \) and \( 2^4 = 16 \). Since \( 24 > 16 \), the statement holds true for \( n = 4 \).

Step 2: Inductive Hypothesis

Assume that the statement \( n! > 2^n \) is true for some arbitrary positive integer \( n = k \), where \( k \geq 4 \). That means we suppose \( k! > 2^k \).

Step 3: Inductive Step

We need to prove that the statement is true for \( n = k + 1 \). Consider \( (k+1)! = (k+1) \times k! \). By the inductive hypothesis, \( k! > 2^k \), so \( (k+1)! = (k+1) \times k! > (k+1) \times 2^k \). To prove the inductive step, we need \( (k+1) \times 2^k > 2^{k+1} \).

Step 4: Simplify the Inductive Step

Observe that \( 2^{k+1} = 2 \times 2^k \). Therefore, we need \( (k+1) \times 2^k > 2 \times 2^k \), which simplifies to \( k + 1 > 2 \). Since \( k \geq 4 \), clearly \( k + 1 \geq 5 \), so \( k+1 > 2 \) holds true. This confirms the statement for \( n = k + 1 \).

Step 5: Conclusion

Since the base case holds and the inductive step has been proved, by mathematical induction, the statement \( n! > 2^n \) is true for all integers \( n \geq 4 \).

Key Concepts

Factorials

Factorials, represented by the symbol "!", are a fundamental concept in mathematics frequently used in combinatorics, algebra, and calculus. A factorial of a non-negative integer \( n \), denoted as \( n! \), is the product of all positive integers less than or equal to \( n \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

Factorials grow extremely fast as \( n \) increases. This rapid increase makes factorials larger than many other functions when \( n \) is sufficiently big. In the context of the problem, the challenge is to show that factorials grow faster than exponential functions for \( n \geq 4 \).

• \( 0! \) is defined as 1.
• \( n! \) grows super-exponentially, which means it outpaces exponential growth for large \( n \).

Exponential Functions

Exponential functions are mathematical functions of the form \( a^x \), where \( a \) is a constant and \( x \) is a variable. They are characterized by their constant ratio of growth. When the base \( a \) is greater than 1, these functions grow faster than linear and polynomial functions, but not as fast as factorial functions.

In this problem, we consider the exponential function \( 2^n \). This represents the growth of doubling as \( n \) increases. While \( 2^n \) grows rapidly, it does not compare to the super-exponential growth of \( n! \) beyond a certain point.

• \( 2^n \) starts relatively small but can grow large for increasing values of \( n \).
• Exponential growth is constant for each increment in \( n \), shown as multiplying by the base \( a \).

Inequalities

Inequalities are mathematical expressions used to compare two values or functions. They show relationships like greater than (\( > \)), less than (\( < \)), greater than or equal to (\( \geq \)), and less than or equal to (\( \leq \)). Understanding inequalities is crucial when comparing growth rates of different types of mathematical functions.

In the problem we are dealing with the inequality \( n! > 2^n \).
Mathematical induction is used to establish this inequality for \( n \geq 4 \).

The process involves:

• Verifying a base case to ensure the smallest valid integer reaches the requirement.
• Assuming the statement is true for an arbitrary integer \( k \).
• Proving that if it’s true for \( k \), it must be true for \( k + 1 \).

This systematic approach reveals how inequalities hold when comparing the explosive growth rates of factorials against exponential functions in specific intervals.