\chapter{Basic concepts} \section{Algorithms} \section{Mathematical Preliminaries} \subsection{Mathematical induction} \begin{myexercise}{4}{18} Let $F_n$ be the $n$-th Fibonacci number, defined by the recursion relation \[ \begin{cases} F_0 = 1, \\ F_1 = 1, \\ F_n = F_{n-1} + F_{n-2}, \quad n \geq 2. \end{cases} \] Let $\phi = \frac{1+\sqrt{5}}{2}$ be the golden ratio. We have \begin{equation} F_n \leq \phi^{n-1} \label{eq:fibo_leq_golden_ratio} \end{equation} $\forall n \in \mathbb{N}$. Prove that, in addition to Eq. \ref{eq:fibo_leq_golden_ratio}, $F_n \geq \phi^{n-2}$. \begin{myanswer} Let $P(n)$ be the assertion that $F_n \geq \phi^{n-2}$. We proceed by recursion on $n$. \begin{equation} 1 + \phi = \phi^2 \label{eq:golden_ratio_squared} \end{equation} \emph{Initiation} For $n = 0$: $\phi^{0-2} = \phi^{-2} = \frac{1}{\phi^2}$. As $\phi^2 > 1$ . $\frac{1}{\phi^2} < 1$, so $P(0)$ is true. For $n = 1$: $\phi^{1-2} = \phi^{-1} = \frac{1}{\phi}$. As $\phi > 1$, $\frac{1}{\phi} < 1$, so $P(1)$ is true. \emph{Heredity} Let $P(n)$ and $P(n-1)$ be true. We have to prove $P(n+1)$. \begin{align*} P(n) \wedge P(n-1) & \Leftrightarrow F_n \geq \phi^{n-2} \wedge F_{n-1} \geq \phi^{n-1-2}\\ & \Leftrightarrow F_{n} + F_{n-1} \geq \phi^{n-2} + \phi^{n-3} \\ & \Leftrightarrow F_n + F_{n-1} \geq \phi^{n-3} (1 + \phi) \\ & \Leftrightarrow F_n + F_{n-1} \geq \phi^{n-3} \phi^2 \qquad \text{using Eq. \ref{eq:golden_ratio_squared}} \\ & \Leftrightarrow F_n + F_{n-1} \geq \phi^{n-1} \\ & \Leftrightarrow F_{n+1} \geq \phi^{n+1-2} \\ & \Leftrightarrow P(n+1) \end{align*} \emph{Conclusion} $P(0)$ and $P(1)$ are true. $P(n) \wedge P(n-1)$ true implies $P(n+1)$ true. So $P(n)$ is true for all $n \in \mathbb{N}$. \end{myanswer} \end{myexercise} \begin{myexercise}{5}{19} A prime number is an integer $> 1$ that has no exact divisors other than $1$ and itself. Using this definition and mathematical induction, prove that every integer $> 1$ can be written as a product of prime numbers or is a prime itself. \begin{myanswer} Let $P(n)$ be the assertion that $n$ can be written as a product of prime numbers or is a prime. \begin{proof} Let $N > 1$ be the smallest integer such that $P(N)$ is false. As $\forall N' < N$, $N' > 1$, $P(N')$ is true. We have either of the following assertions: \begin{itemize} \item $N$ is a prime number; \item there exists $m$ and $n$ below $N$ such that, $m \times n = N$. As $m$ and $n$ are below $N$, they satisfies $P$, so they are either primes or a product of primes. So $mn = N$ is also a product of primes. \end{itemize} We have a contradiction, and $P(N)$ must be true. \end{proof} \end{myanswer} \end{myexercise} \begin{myexercise}{7}{19} Formulate and prove by induction a rule for the sums $1^2$, $2^2 - 1^2$, $3^2 - 2^2 + 1^2$, $4^2 - 3^2 + 2^2 - 1^2$, $5^2 - 4^2 + 3^2 - 2^2 + 1^2$, etc. \begin{myanswer} $1^2 = 1$ $2^2 - 1^2 = 3$ $3^2 - 2^2 + 1^2 = 6$ $4^2 - 3^2 + 2^2 - 1^2 = 10$ $5^2 - 4^2 + 3^2 - 2^2 + 1^2 = 15$ We can rewrite the computed sequence with \[ T_n = \sum_{i=0}^n (-1)^i (n-i)^2 \] With little help from formal computation, we get \begin{equation} T_n = \frac{1}{2} n (n + 1) \label{eq:ex7_half_factor_n_n+1} \end{equation} \begin{proof} Let us prove the formula. Let $P(n)$ be the proposition that Eq. \ref{eq:ex7_half_factor_n_n+1} is correct for $T_n$. By recursion on $n$ \emph{Initiation} For $n = 1$, $1^2 = 1$ and $\frac{1}{2} 1 \times (1+1) = 1$, so $P(n)$ is true. \emph{Heredity} Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true. \begin{align*} P(n) & \Leftrightarrow T_n = \frac{1}{2} n(n+1) \\ & \Leftrightarrow \sum_{i=0}^n (-1)^i (n-i)^2 = \frac{1}{2} n(n+1) \end{align*} \unfinished \end{proof} \end{myanswer} \end{myexercise} \begin{myexercise}{8}{19} (a) Prove the following theorem of Nicomachus by induction: $1^3 = 1$, $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$, $4^3 = 13 + 15 + 17 + 19$, etc. (b) Use this result to prove the remarkable formula $1^3 + 2^3 + \dots + n^3 = (1 + 2 + \dots + n)^2$ \begin{myanswer} (a) The theorem states the following: For all $n \in \mathbb{N}$ we have \[ n^3 = \sum_{i = 1}^n | n (n - 1) - 1 + 2i| \] \begin{proof} We proceed by recursion on $n$. Let $P(n)$ be the proposition ``$n^3 = \sum_{i = 1}^n | n (n - 1) - 1 + 2i|$''. \emph{Initiation} For $n = 1$, $1^3 = 1^2$, so $P(1)$ is true. \emph{Heredity} Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true. \begin{align*} P(n) \Leftrightarrow n^3 &= \sum_{i = 1}^n n (n - 1) - 1 + 2i \\ % &\Leftrightarrow (n+1)^3 &= \sum_{i = 1}^n n^2 - n - 1 + 2i \\ \end{align*} \end{proof} \unfinished (b) \begin{theorem}[Formula of Nicomachus] The sum of the cubes of the first $n$ natural numbers is equal to the square of the sum of the first $n$ natural numbers. \end{theorem} \begin{equation} \sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2 \end{equation} \begin{proof} Let $P(n)$ be the proposition ``$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$''. $P(n) \Longleftrightarrow \sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$ because $\sum_{i=1}^n i = \frac{n(n+1)}{2}$. We want to prove that: $\sum_{k=1}^{n+1} k^3 = \left(\frac{(n+1)(n+2)}{2}\right)^2$ \begin{align*} \sum_{k=1}^{n+1} k^3 &= \frac{n^2 (n+1)^2}{2^2} + (n+1)^3 \\ &= (n+1)^2 \left(\frac{n^2}{4} + n + 1\right) \\ &= (n+1)^2 \cdot \frac{n^2 + 4n + 4}{4} \\ &= \frac{(n+1)^2(n+2)^2}{2^2} \\ & \Updownarrow \\ & P(n+1) \end{align*} \emph{Conclusion} $P(1)$ is true because $1^3 = 1$ and $\left(\frac{1(1+1)}{2}\right)^2 = 1$. $P(n)$ true implies $P(n+1)$ true, so $P(n)$ is true for all $n \in \mathbb{N}_+$. \end{proof} \end{myanswer} \end{myexercise} \begin{myexercise}{9}{19} Prove by induction that if $0 < a < 1$ then $(1 - a)^n \geq 1 - na$. \begin{myanswer} We proceed by recursion on $n$. Let $P(n)$ be the proposition ``$(1 - a)^n \geq 1 - na$''. \emph{Initiation} For $n = 0$, $(1 - a)^n = 1 = 1 - na$, so $P(0)$ is true. \emph{Heredity} Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true. We want to prove that $(1 - a)^{n+1} \geq 1 - (n+1)a$. \begin{align*} P(n) \Leftrightarrow (1-a)^n &\geq 1-na \\ (1-a)^n (1-a) & \geq (1 - na) (1-a) \qquad \text{$1-a > 0 $} \\ (1-a)^{n+1} & \geq 1 - na - a + na \\ & \geq 1 - (n+1)a - a \\ & \geq 1 - (n+1)a \qquad \text{because $a > 0$} & \Rightarrow P(n+1) \end{align*} \emph{Conclusion} $P(1)$ is true. $P(n)$ true implies $P(n+1)$ true, so $P(n)$ is true for all $n \in \mathbb{N}_+$. \end{myanswer} \end{myexercise} % \begin{myexercise}{10}{19} % Prove by induction that if $n \geq 10$ then $2^n > n^3$. % \begin{myanswer} % We proceed by recursion on $n$. % Let $P(n)$ be the proposition ``$2^n > n^3$''. % \emph{Initiation} For $n = 10$, % $2^{10} = 1024 > 1000 = 10^3$, so $P(10)$ is true. % \emph{Heredity} Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true. % We want to prove that $2^{n+1} > (n+1)^3$. % \begin{align*} % P(n) \Leftrightarrow 2^n &> n^3 \\ % 2^n \times 2 &> n^3 \times 2 \qquad \text{because $2 > 0$} \\ % 2^{n+1} &> n^3 + n^3 \\ % \end{align*} % $n^3 = n^2 \times n$ % \emph{Conclusion} % $P(10)$ is true. % $P(n)$ true implies $P(n+1)$ true, so $P(n)$ is true for all integer $n \geq 10$. % \end{myanswer} % \end{myexercise} % Dirty false proof... To be continued