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Magazine Chapter 8: Problem 13
(a) [BB] Show that \(2^{n} \prec n !\). (b) Show that \(b^{n} \prec n !\) for any \(b>1\).
Short Answer
Expert verified
Both \(2^n\) and \(b^n\) are outpaced by \(n!\) for large \(n\).
Step by step solution
01
Understand the Problem
We need to show that the growth rate of the function \(2^n\) is eventually outpaced by the growth rate of \(n!\), which means there exists an \(N\) such that for all \(n > N\), \(n! > 2^n\). Similarly, for part (b), we need to prove the same for \(b^n\) where \(b > 1\).
02
Base Case for \(2^n \prec n!\)
Check small values of \(n\) to establish a base case. For \(n = 0, 1, 2, 3\), calculate and compare. We find that starting from \(n = 4\), \(4! = 24 > 2^4 = 16\). This suggests \(4! > 2^4\).
03
Inductive Step for \(2^n \prec n!\)
Assume \(2^k < k!\) for some \(k \geq 4\). We need to show \(2^{k+1} < (k+1)!\). Start with \((k+1)! = (k+1) \cdot k!\). Since \(k! > 2^k\), we have \((k+1)\cdot k! > (k+1) \cdot 2^k > 2^{k+1}\) for \(k \geq 4\). Thus \(2^{k+1} < (k+1)!\).
04
Conclusion for \(2^n \prec n!\)
By mathematical induction, for \(n \geq 4\), the inequality \(n! > 2^n\) holds. We have verified the base case and the inductive step.
05
Initial Steps for \(b^n \prec n!\)
Fix \(b > 1\) and repeat the process similarly to part (a). Calculate initial values and use induction starting with \(n = b + 1\).
06
Base Case for \(b^n \prec n!\)
Start with the smallest integer greater than \(b\). If \(b=2\), start with \(n = 3\), and show that \(3! > 2^3\). Generally, establish \((b+1)! > b^{b+1}\).
07
Inductive Step for \(b^n \prec n!\)
Assume \(b^k < k!\) for \(k\) greater than some \(b\), and prove \(b^{k+1} < (k+1)!\). Follow the logic: \((k+1)! = (k+1)k! > (k+1)b^k > b \cdot b^k = b^{k+1}\), if \(k + 1 > b\).
08
Final Conclusion for \(b^n \prec n!\)
By induction, \(n! > b^n\) holds for all \(n \geq b+1\). With initial computation and induction, the inequality is confirmed.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mathematical Induction
Mathematical induction is a powerful tool in mathematics used to prove that a statement holds true for all natural numbers. It is particularly useful for establishing the truth of an equation or inequality that involves a variable, like proving growth comparisons. The process of mathematical induction involves two key steps:
- Base Case: Verify that the statement holds for the initial value, usually starting with the smallest natural number, often 0 or 1. In our case, for example, proving that for a certain value of \( n \), such as 4, \( 4! > 2^4 \).
- Inductive Step: Assume the statement is true for an arbitrary positive integer \( k \), and then prove it true for \( k+1 \). This involves showing that if \( n! > 2^n \) is true for any \( n = k \), then \( (k+1)! > 2^{k+1} \) must also be true.
Factorials
Factorials, denoted with an exclamation point (!), represent the product of all positive integers up to a certain number. For example, \( n! = n \times (n-1) \times (n-2) \times ...\times 2 \times 1 \). They are essential in both combinatorics and analysis.
- Factorial growth is incredibly fast. Even for moderately large \( n \), the values of \( n! \) escalate rapidly compared to exponential functions like \( 2^n \) or \( b^n \).
- Understanding factorial growth helps in comparing it with other functions. For instance, it is evident that if you calculate five factorial, \( 5! = 120 \), it's already much larger than \( 2^5 = 32 \).
Exponential Growth
Exponential growth occurs when numbers increase at a consistent rate per unit interval, commonly represented as \( b^n \). This form of growth is fast, but factorial growth is even faster in the long run.
- For smaller values of \( n \), exponential growth might show larger outputs compared to \( n! \). However, as \( n \) increases, especially beyond certain thresholds, factorials become significantly larger.
- This comparison is crucial in fields like algorithms and computational complexity, where understanding which functions dominate others can inform the efficiency and structure of processes.
