410 lines
11 KiB
TeX
Executable File
410 lines
11 KiB
TeX
Executable File
\chapter{Basic concepts}
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\section{Algorithms}
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\section{Mathematical Preliminaries}
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\subsection{Mathematical induction}
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\begin{myexercise}{4}{18}
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Let $F_n$ be the $n$-th Fibonacci number, defined by the recursion relation
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\[
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\begin{cases}
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F_0 = 1, \\
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F_1 = 1, \\
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F_n = F_{n-1} + F_{n-2}, \quad n \geq 2.
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\end{cases}
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\]
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Let $\phi = \frac{1+\sqrt{5}}{2}$ be the golden ratio.
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We have
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\begin{equation}
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F_n \leq \phi^{n-1}
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\label{eq:fibo_leq_golden_ratio}
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\end{equation}
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$\forall n \in \mathbb{N}$.
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Prove that, in addition to Eq. \ref{eq:fibo_leq_golden_ratio}, $F_n \geq \phi^{n-2}$.
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\begin{myanswer}
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Let $P(n)$ be the assertion that $F_n \geq \phi^{n-2}$.
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We proceed by recursion on $n$.
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\begin{equation}
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1 + \phi = \phi^2
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\label{eq:golden_ratio_squared}
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\end{equation}
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\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.
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For $n = 1$:
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$\phi^{1-2} = \phi^{-1} = \frac{1}{\phi}$. As $\phi > 1$, $\frac{1}{\phi} < 1$, so $P(1)$ is true.
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\emph{Heredity}
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Let $P(n)$ and $P(n-1)$ be true. We have to prove $P(n+1)$.
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\begin{align*}
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P(n) \wedge P(n-1) & \Leftrightarrow F_n \geq \phi^{n-2} \wedge F_{n-1} \geq \phi^{n-1-2}\\
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& \Leftrightarrow F_{n} + F_{n-1} \geq \phi^{n-2} + \phi^{n-3} \\
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& \Leftrightarrow F_n + F_{n-1} \geq \phi^{n-3} (1 + \phi) \\
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& \Leftrightarrow F_n + F_{n-1} \geq \phi^{n-3} \phi^2 \qquad \text{using Eq. \ref{eq:golden_ratio_squared}} \\
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& \Leftrightarrow F_n + F_{n-1} \geq \phi^{n-1} \\
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& \Leftrightarrow F_{n+1} \geq \phi^{n+1-2} \\
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& \Leftrightarrow P(n+1)
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\end{align*}
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\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 $n \in \mathbb{N}$.
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\end{myanswer}
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\end{myexercise}
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\begin{myexercise}{5}{19}
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A prime number is an integer $> 1$ that has no exact divisors other than $1$ and itself.
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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.
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\begin{myanswer}
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Let $P(n)$ be the assertion that $n$ can be written as a product of prime numbers or is a prime.
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\begin{proof}
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Let $N > 1$ be the smallest integer such that $P(N)$ is false.
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As $\forall N' < N$, $N' > 1$, $P(N')$ is true.
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We have either of the following assertions:
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\begin{itemize}
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\item $N$ is a prime number;
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\item there exists $m$ and $n$ below $N$ such that, $m \times n = N$.
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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.
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\end{itemize}
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We have a contradiction, and $P(N)$ must be true.
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\end{proof}
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\end{myanswer}
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\end{myexercise}
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\begin{myexercise}{7}{19}
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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.
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\begin{myanswer}
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$1^2 = 1$
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$2^2 - 1^2 = 3$
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$3^2 - 2^2 + 1^2 = 6$
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$4^2 - 3^2 + 2^2 - 1^2 = 10$
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$5^2 - 4^2 + 3^2 - 2^2 + 1^2 = 15$
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We can rewrite the computed sequence with
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\[
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T_n = \sum_{i=0}^n (-1)^i (n-i)^2
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\]
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With little help from formal computation, we get
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\begin{equation}
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T_n = \frac{1}{2} n (n + 1)
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\label{eq:ex7_half_factor_n_n+1}
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\end{equation}
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\begin{proof}
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Let us prove the formula.
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Let $P(n)$ be the proposition that Eq. \ref{eq:ex7_half_factor_n_n+1} is correct for $T_n$.
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By recursion on $n$
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\emph{Initiation}
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For $n = 1$, $1^2 = 1$ and $\frac{1}{2} 1 \times (1+1) = 1$, so $P(n)$ is true.
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\emph{Heredity}
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Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true.
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\begin{align*}
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P(n) & \Leftrightarrow T_n = \frac{1}{2} n(n+1) \\
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& \Leftrightarrow \sum_{i=0}^n (-1)^i (n-i)^2 = \frac{1}{2} n(n+1)
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\end{align*}
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\unfinished
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\end{proof}
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\end{myanswer}
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\end{myexercise}
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\begin{myexercise}{8}{19}
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(a) Prove the following theorem of Nicomachus by induction:
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$1^3 = 1$, $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$, $4^3 = 13 + 15 + 17 + 19$, etc.
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(b) Use this result to prove the remarkable formula $1^3 + 2^3 + \dots + n^3 = (1 + 2 + \dots + n)^2$
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\begin{myanswer}
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(a) The theorem states the following:
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For all $n \in \mathbb{N}$ we have
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\[
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n^3 = \sum_{i = 1}^n | n (n - 1) - 1 + 2i|
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\]
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\begin{proof}
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We proceed by recursion on $n$.
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Let $P(n)$ be the proposition ``$n^3 = \sum_{i = 1}^n | n (n - 1) - 1 + 2i|$''.
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\emph{Initiation} For $n = 1$, $1^3 = 1^2$, so $P(1)$ is true.
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\emph{Heredity} Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true.
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\begin{align*}
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P(n) \Leftrightarrow n^3 &= \sum_{i = 1}^n n (n - 1) - 1 + 2i \\
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% &\Leftrightarrow (n+1)^3 &= \sum_{i = 1}^n n^2 - n - 1 + 2i \\
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\end{align*}
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\end{proof}
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\unfinished
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(b)
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\begin{theorem}[Formula of Nicomachus]
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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.
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\end{theorem}
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\begin{equation}
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\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2
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\end{equation}
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\begin{proof}
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Let $P(n)$ be the proposition ``$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$''.
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$P(n) \Longleftrightarrow \sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$
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because $\sum_{i=1}^n i = \frac{n(n+1)}{2}$.
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We want to prove that: $\sum_{k=1}^{n+1} k^3 = \left(\frac{(n+1)(n+2)}{2}\right)^2$
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\begin{align*}
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\sum_{k=1}^{n+1} k^3 &= \frac{n^2 (n+1)^2}{2^2} + (n+1)^3 \\
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&= (n+1)^2 \left(\frac{n^2}{4} + n + 1\right) \\
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&= (n+1)^2 \cdot \frac{n^2 + 4n + 4}{4} \\
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&= \frac{(n+1)^2(n+2)^2}{2^2} \\
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& \Updownarrow \\
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& P(n+1)
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\end{align*}
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\emph{Conclusion}
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$P(1)$ is true because $1^3 = 1$ and $\left(\frac{1(1+1)}{2}\right)^2 = 1$.
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$P(n)$ true implies $P(n+1)$ true, so $P(n)$ is true for all $n \in \mathbb{N}_+$.
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\end{proof}
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\end{myanswer}
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\end{myexercise}
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\begin{myexercise}{9}{19}
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Prove by induction that if $0 < a < 1$ then $(1 - a)^n \geq 1 - na$.
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\begin{myanswer}
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We proceed by recursion on $n$.
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Let $P(n)$ be the proposition ``$(1 - a)^n \geq 1 - na$''.
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\emph{Initiation} For $n = 0$,
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$(1 - a)^n = 1 = 1 - na$, so $P(0)$ is true.
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\emph{Heredity} Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true.
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We want to prove that $(1 - a)^{n+1} \geq 1 - (n+1)a$.
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\begin{align*}
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P(n) \Leftrightarrow (1-a)^n &\geq 1-na \\
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(1-a)^n (1-a) & \geq (1 - na) (1-a) \qquad \text{$1-a > 0 $} \\
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(1-a)^{n+1} & \geq 1 - na - a + na \\
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& \geq 1 - (n+1)a - a \\
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& \geq 1 - (n+1)a \qquad \text{because $a > 0$}
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& \Rightarrow P(n+1)
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\end{align*}
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\emph{Conclusion}
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$P(1)$ is true.
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$P(n)$ true implies $P(n+1)$ true, so $P(n)$ is true for all $n \in \mathbb{N}_+$.
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\end{myanswer}
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\end{myexercise}
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% \begin{myexercise}{10}{19}
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% Prove by induction that if $n \geq 10$ then $2^n > n^3$.
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% \begin{myanswer}
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% We proceed by recursion on $n$.
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% Let $P(n)$ be the proposition ``$2^n > n^3$''.
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% \emph{Initiation} For $n = 10$,
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% $2^{10} = 1024 > 1000 = 10^3$, so $P(10)$ is true.
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% \emph{Heredity} Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true.
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% We want to prove that $2^{n+1} > (n+1)^3$.
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% \begin{align*}
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% P(n) \Leftrightarrow 2^n &> n^3 \\
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% 2^n \times 2 &> n^3 \times 2 \qquad \text{because $2 > 0$} \\
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% 2^{n+1} &> n^3 + n^3 \\
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% \end{align*}
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% $n^3 = n^2 \times n$
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% \emph{Conclusion}
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% $P(10)$ is true.
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% $P(n)$ true implies $P(n+1)$ true, so $P(n)$ is true for all integer $n \geq 10$.
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% \end{myanswer}
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% \end{myexercise}
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% Dirty false proof... To be continued
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\subsection{Numbers, Powers and Logarithms}
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\begin{myexercise}{1}{25}
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What is the smallest positive rational number?
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\begin{myanswer}
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The smallest positive rational number is 0.
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\end{myanswer}
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\end{myexercise}
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\begin{myexercise}{2}{25}
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Is $1 + 0.23999999999...$ a decimal expansion?
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\begin{myanswer}
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0.23999999999... is not a decimal number as it has an infinite decimal expansion.
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The decimal expansion of $1 + 0.23999999999...$ is in the form
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\[
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1 \cdot 10^0 + 2 \cdot 10^{-1} + 3 \cdot 10^{-2} + 9 \cdot 10^{-3} + \sum_{i=1}^{\infty} 9 \cdot 10^{-3-i}
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\]
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\textit{Nota bene}: This surely does not answer the question...
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\end{myanswer}
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\end{myexercise}
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\begin{myexercise}{3}{25}
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What is $(-3)^{-3}$?
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\begin{myanswer}
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\[
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(-3)^{-3} = \frac{1}{(-3)^3} = \frac{1}{(-3)(-3)(-3)} = -\frac{1}{27}
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\]
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\end{myanswer}
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\end{myexercise}
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\begin{myexercise}{4}{25}
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What is $(0.125)^{-2/3}$?
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\begin{equation}
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b^{p/q} = \sqrt[q]{b^p}
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\end{equation}
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\begin{myanswer}
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\begin{align*}
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0.125^{-2/3} &= \left(\frac{1}{8}\right)^{-2/3} \\
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&= 8^{2/3} \\
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&= \sqrt[3]{8^2} \\
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&= \sqrt[3]{64} \\
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&= 4
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\end{align*}
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\end{myanswer}
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\end{myexercise}
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\begin{myexercise}{5}{25}
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Defining real numbers in terms of binary expansion.
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\begin{myanswer}
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\[
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x = \sum_{i=0}^{\infty} a_i \cdot 2^{-i}
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\]
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\end{myanswer}
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\end{myexercise}
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\begin{myexercise}{6}{25}
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Let $x = m + 0.d_1d_2\dots$ and $y = n + 0.e_1e_2\dots$.
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Give a rule for determining whether $x = y$, $x < y$, $x > y$ based on the decimal representation.
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\begin{myanswer}
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\begin{algorithm}[H]
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\begin{algorithmic}[1]
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\Procedure {LowerGreaterEqual} {$X$, $Y$}
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\Comment{$X$ and $Y$ are arrays of decimal of numbers $x$ and $y$ respectively.}
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\State $i \gets 0$
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\While {$i < X.length$ and $i < Y.length$}
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\If{$X[i] == Y[i]$}
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\State $i \gets i+1$
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\ElsIf {$X[i] > Y[i]$}
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\State \Return \texttt{GREATER}
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\Else
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\State \Return \texttt{LOWER}
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\EndIf
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\EndWhile
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\If{$X.length > Y.length$}
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\State \Return \texttt{GREATER}
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\ElsIf{$X.length < Y.length$}
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\State \Return \texttt{LOWER}
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\Else
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\State \Return \texttt{EQUALS}
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\EndIf
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\EndProcedure
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\end{algorithmic}
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\caption{Greater, Lower or Equal Decimal representation}
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\end{algorithm}
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\end{myanswer}
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\end{myexercise}
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\begin{myexercise}{7}{25}
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Given that $x$ and $y$ are integers. \textit{Prove} the laws of exponents, from definition given by Eq. \ref{eq:rule-of-exponents-2} from Eq. \ref{eq:rule-of-exponents}.
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\begin{eqnarray}
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b^0 = 1, & b^n = b^{n-1} b \quad \text{ if $n > 0$}, & b^n = b^{n+1} / b \quad \text{ if $n < 0$}.
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\label{eq:rule-of-exponents}
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\end{eqnarray}
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\begin{eqnarray}
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b^{x+y} = b^x b^y, & (b^x)^y = b^{xy},
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\label{eq:rule-of-exponents-2}
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\end{eqnarray}
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\begin{myanswer}
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We will prove this in $\Reals$
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Let us recall the following formula:
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\begin{equation}
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b^x = e^{x \ln b}
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\end{equation}
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and
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\begin{equation}
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\exp(x + y) = \exp(x) \exp(y)
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\end{equation}
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Thus it comes:
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\begin{align*}
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b^{x + y} &= e^{(x + y) \ln b} \\
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&= e^{x \ln b + y \ln b} \\
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&= e^{x \ln b} e^{y \ln b} \\
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&= b^x b^y
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\end{align*}
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Similarly for $(a^x)^y$:
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\begin{align*}
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(a^x)^y &= e^{y \ln (a^x)} \\
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&= e^{xy \ln a} \\
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&= a^{xy}
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\end{align*}
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\end{myanswer}
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\end{myexercise}
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