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Chapter 4: Problem 3

For which natural numbers \(n\) is \(n !>3^{n} ?\) Justify your conclusion.

Short Answer

Expert verified
The inequality \(n! > 3^n\) holds true for all natural numbers \(n \geq 5\).

Step by step solution

01

Understand the inequality

We have to compare the factorial function, \(n!\), with the exponential function, \(3^n\), and determine for which natural number values of \(n\), the factorial function is greater than the exponential function.
02

Test for small values of n

Let's start by testing the inequality for small values of n: For n = 1: \(1! = 1\) and \(3^1 = 3\), so the inequality doesn't hold. For n = 2: \(2! = 2\) and \(3^2 = 9\), so the inequality doesn't hold. For n = 3: \(3! = 6\) and \(3^3 = 27\), so the inequality doesn't hold. For n = 4: \(4! = 24\) and \(3^4 = 81\), so the inequality doesn't hold. For n = 5: \(5! = 120\) and \(3^5 = 243\), so the inequality holds. For n = 6: \(6! = 720\) and \(3^6 = 729\), so the inequality holds.
03

Analyze the pattern

Notice that the inequality doesn't hold for n = 1, 2, 3, and 4; however, it starts to hold for n = 5 and 6. Let's analyze the pattern of the ratio between the functions: For n = 1: \(R_1 = \frac{1}{3}\) For n = 2: \(R_2 = \frac{2}{9}\) For n = 3: \(R_3 = \frac{6}{27}\) For n = 4: \(R_4 = \frac{24}{81}\) For n = 5: \(R_5 = \frac{120}{243}\) We can see that from n = 5, the value of \(n!\) starts to exceed the value of \(3^n\). Let's verify if the inequality holds for greater values of n: For n = 7: \(7! = 5,040\) and \(3^7 = 2,187\), so the inequality holds. For n = 8: \(8! = 40,320\) and \(3^8 = 6,561\), so the inequality holds. For n = 9: \(9! = 362,880\) and \(3^9 = 19,683\), so the inequality holds. As n increases, the difference between the values of \(n!\) and \(3^n\) grows. This is due to the fact that the factorial function grows much faster than the exponential function for large values of n, especially after crossing a certain threshold.
04

Conclusion

Based on our analysis, we can conclude that for all natural numbers \(n \geq 5\), the inequality \(n! > 3^n\) holds true.

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Key Concepts

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

Natural Numbers
Natural numbers are the most basic counting numbers that we use in everyday life. These are the numbers that start from 1 and go on indefinitely: 1, 2, 3, 4, 5, and so on. We use them to count discrete, indivisible quantities, such as the number of apples in a basket or the number of books on a shelf.

In mathematics, the set of natural numbers is often denoted by the symbol \( \mathbb{N} \). This set includes all positive integers but does not include zero, fractions, or negative numbers. When it comes to inequalities like \( n! > 3^{n} \), it is important to select values from this set to check the conditions where the inequality holds true or not, just as demonstrated in the exercise. Recognizing the properties of natural numbers can help students understand foundational concepts and can be critical in higher-level mathematics, including number theory and combinatorics.
Factorial Function
The factorial function is a mathematical operation that multiplies a given natural number by all the natural numbers less than it, down to 1. This function is represented by an exclamation mark \( n! \). For example, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \). Factorials grow very rapidly with increasing values of \( n \).

Understanding how the factorial function operates is crucial for grasping more complex concepts in probability, combinatorics, and algebra. While \( n! \), for small values of \( n \), may not seem much larger than respective values of an exponential function, very soon after a certain point, the growth rate of factorial overtakes that of the exponential function due to its multiplicative nature.
Exponential Function
An exponential function is another important mathematical concept characterized by a constant base raised to a variable exponent. The form of an exponential function is \( b^{n} \), where \( b \) is the base, and \( n \) is the exponent. For our exercise, the base is 3, and \( n \) is a natural number. Examples include \( 3^{1} = 3 \), \( 3^{2} = 9 \), and \( 3^{3} = 27 \).

Unlike the factorial function, the exponential function increases in size by repeatedly multiplying by a constant number, the base. Exponentials are used in a wide variety of ways in real-world applications, including computing compound interest in finance, modeling population growth in biology, and describing nuclear decay in physics. While it may initially grow faster than polynomial functions for small values of \( n \), it eventually lags behind the astronomical growth rate of factorials once \( n \) reaches higher values, as shown in the inequality \( n! > 3^{n} \) for \( n \) equal or greater than 5.

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Most popular questions from this chapter

Use mathematical induction to prove each of the following: (a) For each natural number \(n, 3\) divides \(\left(4^{n}-1\right)\). (b) For each natural number \(n, 6\) divides \(\left(n^{3}-n\right)\).

The Lucas numbers are a sequence of natural numbers \(L_{1}, L_{2}, L_{3}, \ldots, L_{n}, \ldots\) which are defined recursively as follows: \- \(L_{1}=1\) and \(L_{2}=3,\) and \- For each natural number \(n, L_{n+2}=L_{n+1}+L_{n}\). List the first 10 Lucas numbers and the first ten Fibonacci numbers and then prove each of the following propositions. The Second Principle of Mathematical Induction may be needed to prove some of these propositions. (a) For each natural number \(n, L_{n}=2 f_{n+1}-f_{n}\) (b) For each \(n \in \mathbb{N}\) with \(n \geq 2,5 f_{n}=L_{n-1}+L_{n+1}\). (c) For each \(n \in \mathbb{N}\) with \(n \geq 3, L_{n}=f_{n+2}-f_{n-2}\).

The quadratic formula can be used to show that \(\alpha=\frac{1+\sqrt{5}}{2}\) and \(\beta=\frac{1-\sqrt{5}}{2}\) are the two real number solutions of the quadratic equation \(x^{2}-x-1=0\). Notice that this implies that $$ \begin{array}{l} \alpha^{2}=\alpha+1, \text { and } \\ \beta^{2}=\beta+1 \end{array} $$ It may be surprising to find out that these two irrational numbers are closely related to the Fibonacci numbers. (a) Verify that \(f_{1}=\frac{\alpha^{1}-\beta^{1}}{\alpha-\beta}\) and that \(f_{2}=\frac{\alpha^{2}-\beta^{2}}{\alpha-\beta}\). (b) (This part is optional, but it may help with the induction proof in part (c).) Work with the relation \(f_{3}=f_{2}+f_{1}\) and substitute the expressions for \(f_{1}\) and \(f_{2}\) from part (a). Rewrite the expression as a single fraction and then in the numerator use \(\alpha^{2}+\alpha=\alpha(\alpha+1)\) and a similar equation involving \(\beta .\) Now prove that \(f_{3}=\frac{\alpha^{3}-\beta^{3}}{\alpha-\beta}\).(c) Use induction to prove that for each natural number \(n,\) if \(\alpha=\frac{1+\sqrt{5}}{2}\) and \(\beta=\frac{1-\sqrt{5}}{2},\) then \(f_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta} .\) Note: This formula for the \(n^{t h}\) Fibonacci number is known as Binet's formula, named after the French mathematician Jacques Binet ( \(1786-1856\) ).

Prove that for each natural number \(n\), any set with \(n\) elements has \(\frac{n(n-1)}{2}\) two-element subsets.

For the sequence \(a_{1}, a_{2}, \ldots, a_{n}, \ldots,\) assume that \(a_{1}=1, a_{2}=5,\) and that for each \(n \in \mathbb{N}, a_{n+1}=a_{n}+2 a_{n-1}\). Prove that for each natural number \(n\), \(a_{n}=2^{n}+(-1)^{n}\)
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