Numerical analysis and 组合数学 (Fall 2017)/Problem Set 1: Difference between pages

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[[Image:Ybc7289-bw.jpg|thumb|250px|right|Babylonian clay tablet YBC 7289 (c. 1800–1600 BC) with annotations. The approximation of the [[square root of 2]] is four [[sexagesimal]] figures, which is about six [[decimal]] figures. 1 + 24/60 + 51/60<sup>2</sup> + 10/60<sup>3</sup> = 1.41421296...<ref>[http://it.stlawu.edu/%7Edmelvill/mesomath/tablets/YBC7289.html Photograph, illustration, and description of the ''root(2)'' tablet from the Yale Babylonian Collection]</ref>]]'''Numerical analysis''' studies different [[algorithm]]s to get [[approximation]]s for  problems of [[mathematical analysis]]. Approximations are used for the following reasons:
*每道题目的解答都要有<font color="red" size=4>完整的解题过程</font>。中英文不限。
#There are no known ways to solve a problem using calculus. Examples for such problems are the [[Navier–Stokes equations]] or the [[Three-body problem]]
#There is a way to solve a problem using calculus. Getting an exact solution is impractical though, because it requires a long time, or many resources. An example for this is calculating [[Power series]].


One of the earliest known uses of numerical analysis is a [[Babylonian]] [[clay]] [[tablet]], which approximates the [[square root]] of 2. In a [[unit square]], the diagonal has this length. Being able to compute the sides of a triangle is extremely important, for instance, in [[carpentry]] and [[construction]].<ref>The New Zealand Qualification authority specifically mentions this skill in document 13004 version 2, dated 17 October 2003 titled [http://www.nzqa.govt.nz/nqfdocs/units/pdf/13004.pdf CARPENTRY THEORY: Demonstrate knowledge of setting out a building]</ref>
== Problem 1 ==
#有<math>k</math>种不同的明信片,每种明信片有无限多张,寄给<math>n</math>个人,每人一张,有多少种方法?
#有<math>k</math>种不同的明信片,每种明信片有无限多张,寄给<math>n</math>个人,每人一张,每个人必须收到不同种类的明信片,有多少种方法?
#有<math>k</math>种不同的明信片,每种明信片有无限多张,寄给<math>n</math>个人,每人收到<math>r</math>张不同的明信片(但不同的人可以收到相同的明信片),有多少种方法?
#只有一种明信片,共有<math>m</math>张,寄给<math>n</math>个人,全部寄完,每个人可以收多张明信片或者不收明信片,有多少种方法?
#有<math>k</math>种不同的明信片,其中第<math>i</math>种明信片有<math>m_i</math>张,寄给<math>n</math>个人,全部寄完,每个人可以收多张明信片或者不收明信片,有多少种方法?


Numerical analysis continues this long tradition of practical mathematical calculations. Much like the Babylonian approximation of <math>\sqrt{2}</math>, modern numerical analysis does not seek exact answers, because exact answers are often impossible to obtain in practice. Instead, much of numerical analysis is concerned with obtaining approximate solutions while maintaining reasonable bounds on errors.
== Problem 2 ==
Find the number of ways to select <math>2n</math> balls from <math>n</math> identical blue balls, <math>n</math> identical red balls and <math>n</math> identical green balls.
* Give a combinatorial proof for the problem.
* Give an algebraic proof for the problem.


Numerical analysis naturally finds applications in all fields of engineering and the physical sciences, but in the 21st&nbsp;century, the life sciences and even the arts have adopted elements of scientific computations. Ordinary [[differential equation]]s appear in [[Proper motion|star movement]]; [[Mathematical optimization|optimization]] occurs in portfolio management; [[numerical linear algebra]] is important for data analysis; [[Probability theory|stochastic]] [[differential equation]]s and [[Markov chain]]s are essential in simulating living cells for [[medicine]] and [[biology]].
== Problem 3 ==
*一个长度为<math>n</math>的“山峦”是如下由<math>n</math>个"/"和<math>n</math>个"\"组成的,从坐标<math>(0,0)</math>到<math>(0,2n)</math>的折线,但任何时候都不允许低于<math>x</math>轴。例如下图:


[[Computer]]s greatly helped this task. Before there were computers, numerical methods often depended on hand [[interpolation]] in large printed tables. Since the mid 20th century, computers calculate the required functions instead. These same interpolation formulas nevertheless continue to be used as part of the software [[algorithms]] for solving [[differential equations]].
    /\
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  /        \/\/    \/\/\
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:长度为<math>n</math>的“山峦”有多少?


==References==
*一个长度为<math>n</math>的“地貌”是由<math>n</math>个"/"和<math>n</math>个"\"组成的,从坐标<math>(0,0)</math>到<math>(0,2n)</math>的折线,允许低于<math>x</math>轴。长度为<math>n</math>的“地貌”有多少?
{{reflist}}


{{math-stub}}
== Problem 4==
[[Category:Mathematics]]
李雷和韩梅梅竞选学生会主席,韩梅梅获得选票 <math>p</math> 张,李雷获得选票 <math>q</math> 张,<math>p>q</math>。我们将总共的 <math>p+q</math> 张选票一张一张的点数,有多少种选票的排序方式使得在整个点票过程中,韩梅梅的票数一直高于李雷的票数?等价地,假设选票均匀分布的随机排列,以多大概率在整个点票过程中,韩梅梅的票数一直高于李雷的票数。
 
==Problem 5==
A <math>2\times n</math> rectangle is to be paved with <math>1\times 2</math> identical blocks and <math>2\times 2</math> identical blocks. Let <math>f(n)</math> denote the number of ways that can be done. Find a recurrence relation for <math>f(n)</math>, solve the recurrence relation.
 
== Problem 6 ==
* 令<math>s_n</math>表示长度为<math>n</math>,没有2个连续的1的二进制串的数量,即
*:<math>s_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-1, x_ix_{i+1}\neq 11\}|</math>。
:求 <math>s_n</math>。
 
*令<math>t_n</math>表示长度为<math>n</math>,没有3个连续的1的二进制串的数量,即
*:<math>t_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-2, x_ix_{i+1}x_{i+2}\neq 111\}|</math>。
*#给出计算<math>t_n</math>的递归式,并给出足够的初始值。
*#计算<math>t_n</math>的生成函数<math>T(x)=\sum_{n\ge 0}t_n x^n</math>,给出生成函数<math>T(x)</math>的闭合形式。
 
注意:只需解生成函数的闭合形式,无需展开。
 
== Problem 7 ==
Let <math>a_n</math> be a sequence of numbers satisfying the recurrence relation:
:<math>p a_n+q a_{n-1}+r a_{n-2}=0</math>
with initial condition <math>a_0=s</math> and <math>a_1=t</math>, where <math>p,q,r,s,t</math> are constants such that <math>{p}+q+r=0</math>, <math>p\neq 0</math> and <math>s\neq t</math>. Solve the recurrence relation.

Revision as of 13:14, 17 September 2017

  • 每道题目的解答都要有完整的解题过程。中英文不限。

Problem 1

  1. [math]\displaystyle{ k }[/math]种不同的明信片,每种明信片有无限多张,寄给[math]\displaystyle{ n }[/math]个人,每人一张,有多少种方法?
  2. [math]\displaystyle{ k }[/math]种不同的明信片,每种明信片有无限多张,寄给[math]\displaystyle{ n }[/math]个人,每人一张,每个人必须收到不同种类的明信片,有多少种方法?
  3. [math]\displaystyle{ k }[/math]种不同的明信片,每种明信片有无限多张,寄给[math]\displaystyle{ n }[/math]个人,每人收到[math]\displaystyle{ r }[/math]张不同的明信片(但不同的人可以收到相同的明信片),有多少种方法?
  4. 只有一种明信片,共有[math]\displaystyle{ m }[/math]张,寄给[math]\displaystyle{ n }[/math]个人,全部寄完,每个人可以收多张明信片或者不收明信片,有多少种方法?
  5. [math]\displaystyle{ k }[/math]种不同的明信片,其中第[math]\displaystyle{ i }[/math]种明信片有[math]\displaystyle{ m_i }[/math]张,寄给[math]\displaystyle{ n }[/math]个人,全部寄完,每个人可以收多张明信片或者不收明信片,有多少种方法?

Problem 2

Find the number of ways to select [math]\displaystyle{ 2n }[/math] balls from [math]\displaystyle{ n }[/math] identical blue balls, [math]\displaystyle{ n }[/math] identical red balls and [math]\displaystyle{ n }[/math] identical green balls.

  • Give a combinatorial proof for the problem.
  • Give an algebraic proof for the problem.

Problem 3

  • 一个长度为[math]\displaystyle{ n }[/math]的“山峦”是如下由[math]\displaystyle{ n }[/math]个"/"和[math]\displaystyle{ n }[/math]个"\"组成的,从坐标[math]\displaystyle{ (0,0) }[/math][math]\displaystyle{ (0,2n) }[/math]的折线,但任何时候都不允许低于[math]\displaystyle{ x }[/math]轴。例如下图:
   /\
  /  \/\/\    /\/\
 /        \/\/    \/\/\
 ----------------------
长度为[math]\displaystyle{ n }[/math]的“山峦”有多少?
  • 一个长度为[math]\displaystyle{ n }[/math]的“地貌”是由[math]\displaystyle{ n }[/math]个"/"和[math]\displaystyle{ n }[/math]个"\"组成的,从坐标[math]\displaystyle{ (0,0) }[/math][math]\displaystyle{ (0,2n) }[/math]的折线,允许低于[math]\displaystyle{ x }[/math]轴。长度为[math]\displaystyle{ n }[/math]的“地貌”有多少?

Problem 4

李雷和韩梅梅竞选学生会主席,韩梅梅获得选票 [math]\displaystyle{ p }[/math] 张,李雷获得选票 [math]\displaystyle{ q }[/math] 张,[math]\displaystyle{ p\gt q }[/math]。我们将总共的 [math]\displaystyle{ p+q }[/math] 张选票一张一张的点数,有多少种选票的排序方式使得在整个点票过程中,韩梅梅的票数一直高于李雷的票数?等价地,假设选票均匀分布的随机排列,以多大概率在整个点票过程中,韩梅梅的票数一直高于李雷的票数。

Problem 5

A [math]\displaystyle{ 2\times n }[/math] rectangle is to be paved with [math]\displaystyle{ 1\times 2 }[/math] identical blocks and [math]\displaystyle{ 2\times 2 }[/math] identical blocks. Let [math]\displaystyle{ f(n) }[/math] denote the number of ways that can be done. Find a recurrence relation for [math]\displaystyle{ f(n) }[/math], solve the recurrence relation.

Problem 6

  • [math]\displaystyle{ s_n }[/math]表示长度为[math]\displaystyle{ n }[/math],没有2个连续的1的二进制串的数量,即
    [math]\displaystyle{ s_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-1, x_ix_{i+1}\neq 11\}| }[/math]
[math]\displaystyle{ s_n }[/math]
  • [math]\displaystyle{ t_n }[/math]表示长度为[math]\displaystyle{ n }[/math],没有3个连续的1的二进制串的数量,即
    [math]\displaystyle{ t_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-2, x_ix_{i+1}x_{i+2}\neq 111\}| }[/math]
    1. 给出计算[math]\displaystyle{ t_n }[/math]的递归式,并给出足够的初始值。
    2. 计算[math]\displaystyle{ t_n }[/math]的生成函数[math]\displaystyle{ T(x)=\sum_{n\ge 0}t_n x^n }[/math],给出生成函数[math]\displaystyle{ T(x) }[/math]的闭合形式。

注意:只需解生成函数的闭合形式,无需展开。

Problem 7

Let [math]\displaystyle{ a_n }[/math] be a sequence of numbers satisfying the recurrence relation:

[math]\displaystyle{ p a_n+q a_{n-1}+r a_{n-2}=0 }[/math]

with initial condition [math]\displaystyle{ a_0=s }[/math] and [math]\displaystyle{ a_1=t }[/math], where [math]\displaystyle{ p,q,r,s,t }[/math] are constants such that [math]\displaystyle{ {p}+q+r=0 }[/math], [math]\displaystyle{ p\neq 0 }[/math] and [math]\displaystyle{ s\neq t }[/math]. Solve the recurrence relation.