数学代写|离散数学作业代写discrete mathematics代考|Math 1030Q

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• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|离散数学作业代写discrete mathematics代考|The Axiom of Mathematical Induction

The Axiom of Mathematical Induction in equivalent form. Let $S$ be some set of natural numbers. Let $0 \in S$ and if some natural number $k \in S$, then also the following natural number $k+1 \in S$. Then $S$ is the set of all natural numbers ${0,1,2, \ldots}$.

Problem 8 Prove that in every problem above one can apply any of the equivalent forms of the axiom of induction.

Problem 9 Prove that $n-3$ diagonals divide an $n$ – gon, that is, the polygon with $n$ sides (not necessarily convex) into $n-2$ parts.

Proof. It is convenient in this problem to start at $n=3$. Thus, a polygon is a triangle, which has no diagonal, and consists of $3-2=1$ part, that establishes the basis of induction. Now assume that for all the $k-$ gons the statement is true, and consider any polygon $P$ with $k+l$ sides. Any its diagonal $d$ splits $P$ into two smaller polygons with a $k_{1}$ and a $k_{2}$ sides, respectively, where $k_{1}+k_{2}$ counts $d$ twice. Thus, $k_{1}+k_{2}=k+1-1=k$. On the other hand, the total number of parts is $k_{1}-1+k_{2}-1+1=k_{1}+k_{2}-1=k-1=(k+1)-2$.
The following example shows that the induction can be employed to prove inequalities as well.
Problem 10 Prove that $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}>\sqrt{n}$.
Problem 11 The next is called The Strong Mathematical Induction Principle. Let $S$ be some set of natural numbers and $0 \in S$. Moreover, if for any natural number $k$, the natural numbers $0,1,2, \ldots, k-1, k \in S$, then also the following natural number $k+1 \in S$. Then $S$ is the set of all natural numbers ${0,1,2, \ldots}$.
The assumptions here seem to be stronger than in the standard principle above, since we assume something not only about $k$, but also about all smaller natural numbers $j \leq k$. Thus the statement looks weaker. But both principles are equivalent.

Indeed, prove that the Strong Principle of Mathematical Induction is equivalen to the Principle of Mulheтulical Induction in stundurl form.
The natural numbers satisfy the Axiom of Mathematical Induction. Moreover, they have a more general property, the Well-Ordering Principle; it considered in more detail, for example, in [31, p. 20].

数学代写|离散数学作业代写discrete mathematics代考|FACTORIALS AND THE STIRLING FORMULA

In the previous computation, we had to multiply consecutive natural numbers. This procedure occurs so often that it deserved its own name and symbol.

Definition 2 The product of $n$ consecutive natural numbers from 1 through $n$ inclusive is called the $n$ factorial and is denoted as $n !$.

For example, $1 !=1,2 !=2 \times 1=2,3 !=3 \times 2 \times 1=6$; if $k<n$, then $\frac{n !}{k !}=(k+1)(k+2) \cdots n$. If we want to preserve this property for $k=0$, it is natural to define $0 !=1$.

The symbol $n$ ! does not look very impressive for small $n$ but let us study it more carefully. Already $10 !=3,628,800$, and we observe that the factorials grow very fast. James Stirling (1692-1770), the younger contemporary of Newton (1643-1727), showed the asymptotic formula to be proved in Lemma 2,
$$n ! \approx\left(\frac{n}{e}\right)^{n} \sqrt{2 \pi n}, n \rightarrow \infty$$
The letter $e$ is a standard symbol for the famous real number $e \approx 2.7$, which will be discussed below. The symbol $\approx$ and name here mean that the ratio of the left- and right-hand sides of the formula tend to 1 as $n \rightarrow \infty$. This formula, without the precise value of the constant, was known to Abraham de Moivre (1667-1754) even before Stirling.

Problem 12 Approximate 10 ! by formula (1.4) and compare with the exact value given above.

Solution. By Stirling’s formula, $10 ! \approx 3.6 \times 10^{6}$, with the relative error less than $1 \%$. If $n$ is increasing, the relative error is even smaller.

Problem 13 Compute $\frac{(n+1) !}{(n-1) !}$. For what natural numbers is this expression defined?
Solution. For $n-1 \geq 0$, hence $n \geq 1$.
Problem 14 How many digits are in the number $100 !$ ?

数学代写|离散数学作业代写discrete mathematics代考| FACTORIALS AND THE STIRLING FORMULA

$$n ! \approx\left(\frac{n}{e}\right)^{n} \sqrt{2 \pi n}, n \rightarrow \infty$$

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