Proofs of Universal Statements: Mathematical Induction · Base of Induction
Lesson 2
import matplotlib.pyplot as plt
import numpy as np
n = np.linspace(2, 13)
plt.plot(n, n ** 3, label='$n^3$')
plt.plot(n, 2 ** n, label='$2^n$')
plt.legend()
plt.savefig('polyexp.png') 
The graph shows that \(2^{n}\) starts to surpass \(n^{3}\) right around \(n=10\). As \(n\) grows, the first function will move further away from the second, and this will be evident in our proof of the inequality. Below, we prove that \(2^{n} \ge n^{3}\), for all integers \(n \ge 10\), by induction on \(n\).
- Base \(n=10\). It is directly verified: \(1024 > 1000\).
- Step \(n \to n+1\). When transitioning from \(n\) to \(n+1\), the left part is multiplied by two, whereas the right part is multiplied by \[\frac{(n+1)^3}{n^3}=\left(1+\frac{1}{n} \right)^{3} \ .\] For \(n \ge 10\), this number is no more than \(1.1^{3} < 2\).
For the curious 🤓
In general, any exponential function grows faster than any polynomial function. In terms of \(O\)-notation, this is written as follows: for any \(a > 1\) and \(b \ge 0\), it holds that \(n^{b} = O(a^{n})\). We will soon prove this rigorously.