Question

The sequence \(\\{n !\\}\) ultimately grows faster than the sequence \(\left\\{b^{n}\right\\},\) for any \(b > 1,\) as \(n \rightarrow \infty .\) However, \(b^{n}\) is generally greater than \(n !\) for small values of \(n\). Use a calculator to determine the smallest value of \(n\) such that \(n ! > b^{n}\) for each of the cases \(b=2, b=e,\) and \(b=10\).

Short answer

Answer: For b=2, the smallest value of n is 4. For b=e, the smallest value of n is 5. For b=10, the smallest value of n is 13.

Step-by-step solution

Case 1: b=2

For this case, we want to find the smallest value of \(n\) such that \(n! > 2^n\). Calculate factorials and powers of 2 for increasing values of n until the inequality is satisfied: For \(n=1\), \(1! = 1 \not\gt 2^1 = 2\) For \(n=2\), \(2! = 2 \not\gt 2^2 = 4\) For \(n=3\), \(3! = 6 \not\gt 2^3 = 8\) For \(n=4\), \(4! = 24 > 2^4 = 16\) The smallest value of \(n\) such that \(n! > 2^n\) is \(\boxed{n=4}\).

Case 2: b=e

For this case, we want to find the smallest value of \(n\) such that \(n! > e^n\). Since \(e\approx2.718\), we can start checking from \(n=4\). Calculate factorials and powers of e for increasing values of n until the inequality is satisfied: For \(n=4\), \(4! = 24 \not\gt e^4 \approx 54.6\) For \(n=5\), \(5! = 120 > e^5 \approx 148.4\) The smallest value of \(n\) such that \(n! > e^n\) is \(\boxed{n=5}\).

Case 3: b=10

For this case, we want to find the smallest value of \(n\) such that \(n! > 10^n\). We can start checking from \(n=5\) since the previous case showed that \(5! > b^n\) for smaller values of \(b\). Calculate factorials and powers of 10 for increasing values of n until the inequality is satisfied: For \(n=5\), \(5! = 120 \not\gt 10^5 = 100000\) For \(n=6\), \(6! = 720 \not\gt 10^6 = 1000000\) For \(n=7\), \(7! = 5040 \not\gt 10^7 = 10000000\) For \(n=8\), \(8! = 40320 \not\gt 10^8 = 100000000\) For \(n=9\), \(9! = 362880 \not\gt 10^9 = 1000000000\) For \(n=10\), \(10! = 3628800 \not\gt 10^{10} = 10000000000\) For \(n=11\), \(11! = 39916800 \not\gt 10^{11} = 100000000000\) For \(n=12\), \(12! = 479001600 \not\gt 10^{12} = 1000000000000\) For \(n=13\), \(13! = 6227020800 > 10^{13} = 10000000000000\) The smallest value of \(n\) such that \(n! > 10^n\) is \(\boxed{n=13}\).

Key concepts

Asymptotic Behavior

Asymptotic behavior is the analysis of the growth patterns of mathematical functions or sequences as the input size grows towards infinity. In many mathematical scenarios, understanding how a function behaves asymptotically helps to simplify complex calculations.

Consider the sequence of factorials, denoted as \(n!\), which represents the product of all positive integers up to \(n\). Another common sequence encountered in mathematics is \(b^n\) where \(b\) is a constant. When analyzing these sequences, particularly for large values of \(n\), it's crucial to understand which sequence eventually dominates the other.

• The factorial sequence, \(n!\), tends to grow very rapidly because for each step \(n\), it multiplies by a larger factor.
• The exponential sequence, \(b^n\), grows at a rate determined by the base \(b\).

For instance, for any \(b > 1\), \(n!\) grows faster than \(b^n\) due to factorial's multiplicative nature, as \(n\) approaches infinity.

Exponential Growth

Exponential growth is a process that increases quantity over time at a rate proportional to the current value. It is expressed as \(b^n\), where \(b\) is a base greater than one and \(n\) is the exponent. This concept is frequently observed in real-world scenarios such as population growth, interest calculations, and more.

In our exercise, exponential growth is represented by sequences like \(2^n\), \(e^n\), and \(10^n\). These sequences initially grow fast, quickly surpassing factorials for smaller values of \(n\).

• The base value \(b\) dictates the initial growth rate.
• For \(b = 2\), the exponential doubling effect is evident, making \(2^n\) quite prominent for small \(n\).
• Similarly, with \(b = 10\), each increase in \(n\) results in a tenfold increase, making it grow very swiftly.

As \(n\) increases further, however, the factorial growth's multiplicative nature begins to catch up and eventually surpasses the exponential sequence.

Sequence Comparison

Sequence comparison involves evaluating two or more sequences to determine which grows faster or dominates in large timeframes. In mathematical analysis, this helps us comprehend their long-term behaviors and relative growth patterns.

In the exercise, comparing \(n!\) with \(b^n\) exposes insights into when the factorial surpasses the exponential sequence:

• Case \(b=2\): For smaller values, \(2^n\) is greater. However, for \(n=4\), factorial begins to grow larger than the power of two.
• Case \(b=e\): As \(e\) is slightly greater than 2.7, \(e^n\) stays ahead until \(n=5\), beyond which the factorial takes over.
• Case \(b=10\): This large base results in a \(10^n\) that surpasses factorials for larger ranges, but eventually, from \(n=13\) onwards, factorials dominate.

Understanding such comparisons allows mathematicians and students to predict long-term outcomes and make informed decisions based on sequence behavior.