Catalan Numbers

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基数词语法课件

13 (thirteen)
45 (forty-five)
33 (thirty-three) 82 (eighty-two) 91 (ninety-one)
14 (fourteen) 102 (one hundred and two) 139 (one hundred and thirty-nine)
Amy 5558 6390 __fi_ve_,_fi_ve_,_fi_v_e,_e_ig_ht_;_si_x,_t_hr_e_e_,n_in_e_,o_
Daniel 5554 3986 __f_iv_e,_f_iv_e,_f_iv_e,_f_ou_r_;t_h_re_e_,n_in_e_,e_i_gh_t_,six
pay RMB Yuan five hundred and ninety-eight
￥598.00
pay RMB Yuan twelve thousand,
, three hundred and twelve ￥12312.00
第十三页，共16页。
Homework
• Write the numbers in English.
_a_ 6. one thousand and twenty-six _c_ 7. one hundred and ten thousand, two hundred and
six
a. 1,026
b. 10,206
c. 110,206
d. 126
e. 100,260 f. 12,026
g. 1,200,026
• 23,813= t_w_e_n_t_y-_t_h_re_e__ thousand,eight h_u_n_d_re_d__an_d__th_i_rt_e_e_n___

2009罗马尼亚大师杯解答

The 2nd Romanian Master of Mathematics Competition –SolutionsBucharest,Saturday,February 28,2009Problem 1.For any positive integers a 1,...,a k ,let n =k i =1a i ,and letna 1,...,a k be the multinomial coefﬁcient n !ki =1(a i!).Let d =gcd(a 1,...,a k )denote the greatest com-mon divisor of a 1,...,a k .Prove thatdn n a 1,...,a kis an integer.Romania,Dan Schwarz [1]Solution.The key idea is the fact that the greatest com-mon divisor is a linear combination with integer coefﬁcients of the numbers involved[2],i.e.there exist u i ∈Z such thatd =ki =1u i a i .Butn a 1,...,a k =n a i n −1a 1,...,a i −1,a i −1,a i +1,...,a k,sod nn a 1,...,a k=ki =1u in −1a 1,...,a i −1,a i −1,a i +1,...,a k,which clearly is an integer,since multinomial coefﬁcients are known (and easy to prove)to be integer. Problem 2.A set S of points in space satisﬁes the property that all pairwise distances between points in S are distinct.Given that all points in S have integer coordinates (x ,y ,z ),where 1≤x ,y ,z ≤n ,show that the number of points in S is less than min (n +2) n /3,n 6 .Romania,Dan Schwarz [3]Solution.The critical idea is to estimate the total number possible T of distinct distances realized by pairs of points (x ,y ,z ),of integer coordinates 1≤x ,y ,z ≤n .However,any such distance is also realized by a pair anchored at (1,1,1),from symmetry considerations.But the number of distinct distances to points with no co-ordinates x ,y ,z equal is at most n 3 =16n (n −1)(n −2);the number of distinct distances to points with two of the three coordinates x ,y ,z equal is at most 2 n 2 =n (n −1);while the number of distinct distances to points with all three coordi-nates x ,y ,z equal is n −1,henceT ≤16n (n −1)(n −2)+n (n −1)+(n −1)<16(n 3+3n 2+2n ).On the other hand,the total number of distinct distancesbetween the N points in S needs be N 2 =12N (N −1)≤T ,yielding(2N −1)2<13(4n 3+12n 2+8n )+1≤13(2n n +3 n )2,hence N <12(2n +3) n /3+1 ≤(n +2) n /3for n ≥3.Onecan easily check that the inequality is true for n =2also,since then[4]T =3.On the other hand,since the squares of the distances canonly take the integer values between 1and the trivial upper bound 3(n −1)2(for the diagonal of the cube),it follows that T ≤3(n −1)2,yielding N <n 6. Problem 3.Given four points A 1,A 2,A 3,A 4in the plane,no three collinear,such thatA 1A 2·A 3A 4=A 1A 3·A 2A 4=A 1A 4·A 2A 3,let us denote by O i the circumcenter of ∆A j A k A ,with {i ,j ,k , }={1,2,3,4}.Assuming A i =O i for all indices i ,prove that the four lines A i O i are concurrent or parallel.Bulgaria,Nikolai Ivanov Beluhov Solution.(D.Schwarz)The given triple equality being in-variated by any permutation in S 4,it is enough to prove that the lines A i O i for 2≤i ≤4are concurrent or parallel.The relations can then be writtenA 1A 2A 1A 3=A 4A 2A 4A 3,A 1A 3A 1A 4=A 2A 3A 2A 4,A 1A 4A 1A 2=A 3A 4A 3A 2.Consider the Apollonius circles Γk of centers ωk ∈A i A j ,for{i ,j ,k }={2,3,4},determined by the point A 1,which there-fore lies on all three,while the points A k lie on Γk .Moreover,the points ωk are collinear,since the point A kwhich is the other meeting point (than A 1,if any)of Γi and Γj fulﬁlls A k A j A k A k=A i A j A i A kandA k A i A k A k=A j A i A j A k,thusA k A i A k A j=A k A iA k A j,therefore A kalso lies on Γk ,hence all three circles Γk share the same meeting point(s),thus their centers are collinear.Now,the circumcenters O i and O j ,as well as the point ωk ,lie on the perpendicular bisector of the segment A 1A k ,for {i ,j ,k }={2,3,4}.It follows that the pairs of lines A i A j ,O i O j meet at the collinear points ωk .Desargues’theorem for the perspective triangles ∆A i A j A k and ∆O i O j O k yields the claim. Alternate Solution.The author’s original solution makes use of inversions of poles A i to reach the same conclusion via Desargues,in a dual-by-inversion to the solution above manner,with a lot more details than concepts.We feel that making use of the well-known properties of the Apollonius circles renders the idea in a more striking way. Remark.There exists a particular (degenerate)case,when the points are the vertices of a kite of π6equal angles,hence one of the associated ratios is 1,so a corresponding Apollonius circle degenerates to the perpendicular bisector.This (together with the use of Desargues)shows the deep projective nature of the problem,better handled through projective methods.Also,there is no converse implication,since the case of concyclic points trivially warrants the conclusion,without fulﬁlling the stated condition (as in conﬂict with Ptolemy’s relation).2Problem 4.For a ﬁnite set X of positive integers,letΣ(X )=x ∈X arctan 1x .Given a ﬁnite set S of positive integers for which Σ(S )<π2,show that there exists at least one ﬁnite set T of positive in-tegers for which S ⊂T and Σ(T )=π2.United Kingdom,Kevin Buzzard Solution.(D.Schwarz)We will step-by-step augment the set S with positive integers t n ,by taking each time t n as the least positive integer larger than max(S ),and not already used,such that Σ(S ∪{t 1,t 2,...,t n })remains at most π2(thisis possible since arctan 1t →0when t →∞).If at some point we get exactly π2we are through,since we have augmented S to a set T as required,so assume the process continues indeﬁnitely.Clearly the sequence (t n )n ≥1is built (strictly)increasing,so for all n ≥1we have t n +1>t n >max(S ).We will make some useful notations.Take S 0=S ,S n +1=S n ∪{t n +1},for n ∈N .Also take x n =tan π2−Σ(S n ).Usingthe well-known formula tan(α+β)=tan α+tan β1−tan αtan βone caneasily prove by simple induction that a lesser than π2sum of arcs of rational tangents is as well an arc of rational tan-gent,therefore x n =p nq n ,with p n ,q n ∈N ∗,(p n ,q n )=1.Sincearctan is increasing,we need take t n +1≥ 1x nin order that we may augment S n with t n +1to obtain S n +1.Assume that for all n ≥1we have 1x n ≤t n .Since we need both t n +1≥ 1x n and t n +1>t n ≥1x n,it follows that t n +1=t n +1(the least available value),so t k +1=t 1+k forall k ≥0.But then π2>Σ({t 1,t 2,...,t n })=n −1 k =0arctan 1t 1+k >12n −1 k =01t 1+k→∞when n →∞,absurd (see Lemma ).Therefore there exists some N ≥1for which 1x N >t N ,so 1x Nis available for t N +1.Moreover,for any n ≥N with t n +1=1x n,we have x n +1=x n −1n +11+x n 1t n +1=x n t n +1−1t n +1+x n <x n t n +1+x n <1t n +1,since t n +1= 1x n implies x n t n +1−1<x n ;and so we can take t n +1= 1x nindeﬁnitely for n ≥N .Now we use the fact that x n =p nq n .Then p n +1q n +1=pnq n −1t n +11+p n q n 1t n +1=p n t n +1−q n q n t n +1+p n ,hence p n +1≤p n t n +1−q n <p n ,since t n +1= q n p n ,and so t n +1<q np n+1.Therefore the sequence (p n )n ≥1of the numerators of x n eventually becomes (strictly)decreasing,absurd for any se-quence of positive integers.Lemma.For x ∈(0,π2)one has arctan x >x2.Proof.We start by proving that under given condition onehas sin x >tan x2,in turn equivalent to 2sin x 2cos x 2>sin x 2cos x 2,2cos 2x 2−1>0,and ﬁnally cos x >0,patently true.Now,arctan is increasing,hence applied to the above,together with the well-known inequality x >sin x ,true forall x >0,yields arctan x >arctansin x >arctantan x 2=x2. As a corollary,arctan 1n >12n ,for all positive integersn ,inequality used to yield the divergence of the series n ≥1arctan 1n in the above solution.Remark.The above solution shows that it is irrelevantthat we start with the arc π2− s ∈S arctan 1s;in fact we maystate the problem like thisProve that for any arc α∈(0,π2)of some rational tangent τ=tan α,and any ﬁnite set S of distinct positive integers,there exists some ﬁnite set T of distinct positive integers such that T ∩S = andt ∈Tarctan1t=α.The problem is strongly reminiscent of a strengthened form of the famous Egyptian fraction [5]theoremProve that for any rational number r ∈(0,1),and any ﬁnite set S of distinct positive integers,there exists a ﬁnite set T of distinct positive integers such that T ∩S = andt ∈T 1t=r .All the ingredients are there:the greedy algorithm,going beyond the largest element of S ,using the divergence of theseries n ≥11n ,and the (Fermat)inﬁnite descent method of a(strictly)decreasing sequence of positive integers.In fact,it is enough to consider a (strictly)increasing func-tion f :Q +→R +with the properties that there exists a func-tion ϕ:Q +×Q +→Q +such that f (r )−f (s )=f (ϕ(r ,s ))for any 0≤s <r in Q ,lim x →0f (x )=0,and lim n →∞n k =1f 1k =∞.Moreover,we need that ϕ(r ,s )has not larger numerator than r −s ,and not lesser denominator.Then the Egyptian fraction method extends perfectly.Or f (x )=arctan x andϕ(x ,y )=x −y1+x y conform to this model.END[1]Based on a property of quasi-Catalan numbers of J.Conway,see [G UY ,R.K.,Unsolved Problems in Number Theory ].[2]Easily proven by induction from the classical Bézout’s relationgcd(M ,N )=uM +vN for some integers u ,v .[3]A 3-dimensional extrapolation of a plane lattice points casestudy of P .Erdös and R.K.Guy.[4]An example of N =3points for n =2is (1,1,1),(2,2,1),(2,2,2);and of N =4points for n =3is (1,1,1),(1,1,2),(2,2,1),(2,3,3).[5]An Egyptian fraction is written as a ﬁnite sum of fractions withall unit numerators and all distinct denominators.Such frac-tions were used by ancient Egyptians,as apparent in the Rhind Papyrus,but their use is discontinued today.。

s, Volume 48, Num

2001 Steele Prizes
The 2001 Leroy P. Steele Prizes were awarded at the 107th Annual Meeting of the AMS in January 2001 in New Orleans. The Steele Prizes were established in 1970 in honor of George David Birkhoff, William Fogg Osgood, and William Caspar Graustein and are endowed under the terms of a bequest from Leroy P. Steele. The prizes are awarded in three categories: for expository writing, for a research paper of fundamental and lasting importance, and for cumulative influence extending over a career. The current award is $4,000 in each category (in case of multiple recipients, the amount is divided equally). The recipients of the 2001 Steele Prizes are RICHARD P. STANLEY for Mathematical Exposition, L ESLIE G REENGARD and V LADIMIR R OKHLIN for a Seminal Contribution to Research (limited this year to applied mathematics), and HARRY KESTEN for Lifetime Achievement. The Steele Prizes are awarded by the AMS Council acting through a selection committee whose members at the time of these selections were: Constantine M. Dafermos, Bertram Kostant, Hugh L. Montgomery, Marc A. Rieffel, Jonathan M. Rosenberg, Barry Simon, François Treves (chair), S. R. S. Varadhan, and Herbert S. Wilf. The text that follows contains, for each prize recipient, the committee’s citation, a brief biographical sketch, and a response from the recipient upon receiving the prize.

组合数学-第十一节：Fibonacci数和Catalan数

4.5 Fibonacci 数和Catalan 数Fibonacci 数列和Catalan 数列是递推关系的典型问题，它们经常出现在组合计数问题中，而这两个数列本身的应用也十分广泛。

4.5.1 Fibonacci 数关于Fibonacci 数列的问题是一个古老的数学问题，它是由意大利著名数学家Fibonacci 于1202年提出来的。

这个问题是：把一对兔子（雌、雄各一只）在某年的开始放到围栏中，每个月这对兔子都生出一对新兔子，其中雌、雄各一只。

从第二个月开始，每对新兔子每个月也生出一对新兔子，也是雌、雄各一只。

问一年后围栏中有多少对兔子？对于1,2,n = ，令()f n 表示第n 个月开始时围栏中的兔子对数。

显然有()()11,22f f ==。

在第n 个月的开始，那些第1n -个月初已经在围栏中的兔子仍然存在，而且每对在第2n -个月初就存在的兔子将在第1n -个月生出一对新兔子，所以有()()()()()()12311,22f n f n f n n f f =-+-≥⎧⎪⎨==⎪⎩ （4.5.1）这是一个带有初值的递推关系，如果我们规定()01f =，则递推关系（4.5.1）就变成()()()()()()12201,11f n f n f n n f f =-+-≥⎧⎪⎨==⎪⎩ （4.5.2）满足递推关系（4.5.2）的数列就叫做Fibonacci 数列，它的项就叫做Fibonacci 数。

下面我们来求解这个递推关系，它的特征方程为210x x --=其特征根为12x x ==所以，通解为()121122n nf n c c ⎛⎛+=+ ⎝⎭⎝⎭代入初值来确定1c 和2c ，得到方程组12121,11122c c c +=⎧⎪⎨++=⎪⎩ 解这个方程组，得121,212cc+==所以，原递推关系的解为()()110,1,2n nf nn++=⎭⎭=Fibonacci数常出现在组合计数问题中。

Catalan数,括号序列和栈

Catalan数,括号序列和栈全是⼊门的⼀些东西.基本全是从别处抄的.栈: ⽀持单端插⼊删除的线性容器. 也就是说,仅允许在其⼀端加⼊⼀个新元素或删除⼀个元素. 允许操作的⼀端也叫栈顶,不允许操作的⼀端也叫栈底.数个箱⼦相叠就可以认为是⼀个栈,只能在最顶端加⼊⼀个新箱⼦或拿⾛⼀个箱⼦.栈中的元素遵循后进先出(last in first out,LILO)的规律.即:更早出栈的元素,应为更早⼊栈者.这是⼀个演⽰:奇数⾏为栈中元素(右端可以进⾏插⼊删除),元素以逗号隔开, EMPTY表⽰栈为空偶数⾏为进⾏的操作EMPTY插⼊1010插⼊2010,20插⼊5110,20,51插⼊1010,20,51,10删除⼀个元素10,20,51删除⼀个元素10,20插⼊3010,20,30删除⼀个元素10,20删除⼀个元素10删除⼀个元素EMPTY栈混洗问题给出三个栈S1,S2,S3,⼀开始S1中含有1到n的n个数字且从栈顶到栈底数字依次为1,2,3,....n-1,n.只有两种允许的操作:A S1⾮空时从S1取出⼀个元素放⼊S2,B S2⾮空时可以从S2取出⼀个元素放⼊S3.最后S3中⾃底向上形成的序列称作⼀个栈混洗.例如,如果S1中⼀开始有1,2,3,4四个元素,那么先进⾏4次A操作再进⾏4次B操作,将得到序列4,3,2,1.如果A操作和B操作交替进⾏,将得到序列1,2,3,4.显然,栈混洗的结果并不唯⼀.⼀个长为n的序列的栈混洗可以表⽰成n次A操作和n次B操作组成的⼀个操作序列. ⽽n次A操作和n次B操作组成的⼀个操作序列也可以表⽰唯⼀的⼀个栈混洗序列.不同的操作序列必然得到不同的栈混洗,不同的栈混洗也必然对应不同的操作序列.并不是所有含有n个A,n个B的序列都是合理的操作序列.例如BBBBAAAA,将导致我们尝试拿出空栈S2中的元素.⼀个序列合理的条件是:对于任意m(1<=m<=2n),该序列的前m个操作中,B操作的数⽬不超过A操作的数⽬.只要满⾜这个条件就能保证任意时刻不会拿出空栈中的元素.如何判断⼀个栈混洗序列是否是可能出现的栈混洗序列例如,对于序列1,2,3,通过栈混洗可以得到[1,2,3],[3,2,1],[1,3,2],[2,1,3],[2,3,1],但是⽆法得到[3,1,2].如果使3最先出栈,那么就必须先令1,2⼊栈,从⽽2会在1之前出栈,只能得到[3,2,1]任意给出⼀个n和⼀个排列,如何判断这个排列能否通过栈混洗得到?例如,n=5,序列为5,4,1,3,2,是否可能?直接的思路是,直接根据我们要得到的序列,尝试进⾏A操作和B操作.例如,n=5时,要使序列的第⼀个元素为5,就必须⼀直进⾏A操作直到5出现在S2的栈顶.之后需要4,4恰好在S2的栈顶,于是弹出4.接下来需要的1不在栈顶,因此这个序列⽆法通过栈混洗得到.时间复杂度显然为O(n).其正确性也是显然的.栈混洗与括号序列⾸先只考虑由⼀种括号组成的括号序列.()()()(),(()())(),((()))都是能够匹配的括号序列.)(,))(,())(都是不能够匹配的括号序列.只要将A操作与左括号"("对应,B操作与右括号")"对应,栈混洗的合法操作序列就可以和能够匹配的括号序列⼀⼀对应.例如,AABB对应(()),ABAB对应()()按照我们之前的理解,如果保证序列中左右括号数⽬相同,那么我们只需要扫描⼀遍序列并维护⼀个计数器,初始为0,遇到左括号+1,遇到右括号-1,只要这个计数器的数值始终⾮负,就说明任意⼀个前缀中左括号的数⽬多于右括号的数⽬,等价于这个序列是能够匹配的括号序列.另⼀种判断括号序列是否合法的⽅法:初始化⼀个空栈S,从左向右扫描序列,遇到⼀个左括号将其⼊栈,遇到⼀个右括号时判断栈中是否有⼀个左括号,如果有,那么这个左括号与当前遇到的右括号相匹配.如果栈为空,那么这个括号序列并不合法.通过这种⽅式,我们除了得知括号序列是否合法,还可以得知每个右括号具体是和哪个左括号匹配,还可以处理序列中出现了多种括号且只有对应种类的括号能匹配的情况.可以⾃⾏尝试,"维护⼀个计数器"的⽅式并不能⽅便地扩展到多种括号的情况.拓展:允许循环移位的合法括号序列允许将括号序列的最左侧元素拿到最右边,问能否通过若⼲次这样的操作使得括号序列合法.⾸先这样的序列仍然需要满⾜左括号和右括号数⽬相同. 但是除此之外还需要满⾜什么条件呢?令⼈惊讶的是,除此之外并不需要满⾜什么条件.只要左括号和右括号数⽬相同,就可以保证能够通过若⼲次循环移位使得括号序列合法.给出⼀个构造⽅式:将左括号视为1,右括号视为-1,求取该序列的前缀和.如果前缀和中的最⼩值为0那么序列本⾝就是合法的.否则我们找到前缀和的最⼩值的出现位置(多个数值相同时取最左侧的),将这个位置之前的⼀段序列移位到右端,得到的即为合法的括号序列.构造⽅式的正确性不难证明.括号序列计数有多少个不同的含有n对括号的合法括号序列?这个问题等价于通过栈混洗可以得到多少个不同序列.也等价于,有多少个含有n个0,n个1的长为2n的字符串使得任意⼀个前缀中1的数⽬不少于0的数⽬.假设答案为f[n].边界条件为f[0]=1,表⽰没有括号/没有元素时也算有⼀种⽅案(空串也是⼀种合法序列).插播两条⾼中数学:分类加法计数原理:做⼀件事有m类⽅法,每类⽅法分别有A1,A2...Am种做法,那么做这件事共有A1+A2+...+Am种⽅法分步乘法计数原理:做⼀件事有m个步骤,每个步骤分别有A1,A2...Am种做法,那么做这件事共有A1∗A2∗...∗Am种⽅法好了,接下来我们来将所学的⽤于实践.考虑最左侧的左括号匹配的右括号在什么位置.假设这个左括号和匹配的右括号之间有2i个括号(这个数⽬必须是偶数),这2i个括号排列成合法括号序列的⽅案数为f[i].匹配的右括号右边还有2n-2i-2个括号,将它们排列为合法括号序列的⽅案数为f[n-i-1],这两部分可以认为是两个步骤,是独⽴的,那么总的⽅案数是f[i]*f[n-i-1].对于不同的i,我们可以认为是做⼀件事的不同种类的⽅法.于是f[1]=f[0]∗f[0],f[2]=f[1]∗f[0]+f[0]∗f[1],f[3]=f[0]∗f[2]+f[1]∗f[1]+f[2]∗f[0],这是⼀个O(n^2)的计算⽅式.这⾥我们得到的f[n]也叫Catalan数(卡特兰数),它还具有很多实际意义.紫书330~331页给出了Catalan数的另⼀个实际意义:多边形三⾓剖分数⽬,并通过另⼀个O(n^2)的递推公式,推倒了O(n)的递推公式.另⼀个常⽤的计算公式是f[n]=C(2n,n)/(n+1)=C(2n,n) - C(2n,n-1),C(2n,n)为组合数.catalan数的⼀个推⼴:n个左括号,m个右括号,任意前缀中左括号不少于右括号的序列数？答案是C(n+m,m)-C(n+m,m-1).Processing math: 100%。

【Good系列】高中大学数学公式表

DeﬁnitionsSeriesf (n )=O (g (n ))iﬀ∃positive c,n 0such that0≤f (n )≤cg (n )∀n ≥n 0.ni =1i =n (n +1)2,ni =1i 2=n (n +1)(2n +1)6,ni =1i 3=n 2(n +1)24.In general:n i =1i m =1m +1 (n +1)m +1−1−n i =1(i +1)m +1−i m +1−(m +1)i mn −1 i =1i m=1m +1m k =0m +1k B k n m +1−k .Geometric series:n i =0c i =c n +1−1c −1,c =1,∞i =0c i=11−c ,∞ i =1c i =c1−c ,|c |<1,ni =0ic i=nc n +2−(n +1)c n +1+c (c −1)2,c =1,∞ i =0ic i =c(1−c )2,|c |<1.Harmonic series:H n =n i =11i ,ni =1iH i =n (n +1)2H n −n (n −1)4.n i =1H i =(n +1)H n −n,n i =1i m H i = n +1m +1 H n +1−1m +1 .f (n )=Ω(g (n ))iﬀ∃positive c,n 0such that f (n )≥cg (n )≥0∀n ≥n 0.f (n )=Θ(g (n ))iﬀf (n )=O (g (n ))and f (n )=Ω(g (n )).f (n )=o (g (n ))iﬀlim n →∞f (n )/g (n )=0.lim n →∞a n =a iﬀ∀ >0,∃n 0such that |a n −a |< ,∀n ≥n 0.sup S leastb ∈R such that b ≥s ,∀s ∈S .inf S greatest b ∈R such that b ≤s ,∀s ∈S .lim inf n →∞a nlim n →∞inf {a i |i ≥n,i ∈N }.lim sup n →∞a n lim n →∞sup {a i |i ≥n,i ∈N }. n kCombinations:Size k sub-sets of a size n set. n kStirling numbers (1st kind):Arrangements of an n ele-ment set into k cycles.1. n k =n !(n −k )!k !,2.n k =0 n k=2n , 3. n k = n n −k ,4. n k =n k n −1k −1 , 5. n k = n −1k + n −1k −1 ,6. n m m k = n k n −km −k ,7.n k =0r +k k = r +n +1n ,8.n k =0 k m = n +1m +1 ,9.n k =0 r k s n −k = r +sn ,10. n k=(−1)k k −n −1k ,11. n 1 = n n =1,12. n 2=2n −1−1,13. n k =k n −1k + n −1k −1 , n kStirling numbers (2nd kind):Partitions of an n element set into k non-empty sets.n k 1st order Eulerian numbers:Permutations π1π2...πn on{1,2,...,n }with k ascents. n k2nd order Eulerian numbers.C n Catalan Numbers:Binary trees with n +1vertices.14. n 1 =(n −1)!,15. n 2 =(n −1)!H n −1,16. n n =1,17. n k ≥ n k,18. n k =(n −1) n −1k + n −1k −1 ,19. n n −1 = n n −1 = n2,20.n k =0 n k =n !,21.C n =1n +1 2n n ,22. n 0 = n n −1 =1,23. n k = n n −1−k ,24. n k =(k +1) n −1k +(n −k ) n −1k −1,25. 0k = 1if k =0,0otherwise26. n 1=2n −n −1,27. n 2 =3n −(n +1)2n+ n +12 ,28.x n =n k =0 n k x +k n ,29. n m=m k =0 n +1k (m +1−k )n (−1)k ,30.m ! n m =n k =0n k kn −m ,31. n m=n k =0 n kn −k m (−1)n −k −mk !,32. n 0 =1,33. n n =0for n =0,34. n k =(k +1) n −1k +(2n −1−k ) n −1k −1,35.n k =0n k =(2n )n2n ,36. xx −n=n k =0n k x +n −1−k 2n ,37. n +1m +1 = kn k k m =n k =0k m (m +1)n −k ,Identities Cont.Trees 38.n +1m +1= k n k k m =n k =0 k m n n −k =n !n k =01k ! km ,39. x x −n =n k =0 n k x +k 2n,40. nm=kn k k +1m +1 (−1)n −k ,41. n m = kn +1k +1 k m (−1)m −k ,42. m +n +1m=m k =0k n +k k ,43. m +n +1m =m k =0k (n +k ) n +k k ,44. n m = k n +1k +1 k m (−1)m −k ,45.(n −m )! n m = k n +1k +1 km(−1)m −k ,for n ≥m ,46. n n −m = k m −n m +k m +n n +k m +k k ,47. n n −m= k m −n m +k m +n n +k m +k k ,48.n +m +m=kkn −k m n k ,49. n +m +m = kk n −k m n k .Every tree with nvertices has n −1edges.Kraft inequal-ity:If the depths of the leaves of a binary tree are d 1,...,d n :ni =12−d i ≤1,and equality holds only if every in-ternal node has 2sons.RecurrencesMaster method:T (n )=aT (n/b )+f (n ),a ≥1,b >1If ∃ >0such that f (n )=O (n log b a − )thenT (n )=Θ(n log b a ).If f (n )=Θ(n log b a )thenT (n )=Θ(n log b a log 2n ).If ∃ >0such that f (n )=Ω(n log b a + ),and ∃c <1such that af (n/b )≤cf (n )for large n ,thenT (n )=Θ(f (n )).Substitution (example):Consider the following recurrenceT i +1=22i·T 2i ,T 1=2.Note that T i is always a power of two.Let t i =log 2T i .Then we havet i +1=2i +2t i ,t 1=1.Let u i =t i /2i .Dividing both sides of the previous equation by 2i +1we gett i +12i +1=2i 2i +1+t i 2i.Substituting we ﬁndu i +1=12+u i ,u 1=12,which is simply u i =i/2.So we ﬁndthat T i has the closed form T i =2i 2i −1.Summing factors (example):Consider the following recurrenceT (n )=3T (n/2)+n,T (1)=1.Rewrite so that all terms involving T are on the left sideT (n )−3T (n/2)=n.Now expand the recurrence,and choose a factor which makes the left side “tele-scope”1 T (n )−3T (n/2)=n 3 T (n/2)−3T (n/4)=n/2.........3log 2n −1 T (2)−3T (1)=2Let m =log 2n .Summing the left side we get T (n )−3m T (1)=T (n )−3m =T (n )−n k where k =log 23≈1.58496.Summing the right side we getm −1 i =0n 2i 3i=n m −1 i =0 32 i .Let c =32.Then we have nm −1 i =0c i =n c m −1c −1 =2n (clog 2n−1)=2n (c (k −1)log c n −1)=2n k −2n,and so T (n )=3n k −2n .Full history re-currences can often be changed to limited history ones (example):ConsiderT i =1+i −1j =0T j ,T 0=1.Note thatT i +1=1+i j =0T j .Subtracting we ﬁndT i +1−T i =1+i j =0T j −1−i −1 j =0T j =T i .And so T i +1=2T i =2i +1.Generating functions:1.Multiply both sides of the equa-tion by x i .2.Sum both sides over all i for which the equation is valid.3.Choose a generating functionG (x ).Usually G (x )= ∞i =0x i g i .3.Rewrite the equation in terms of the generating function G (x ).4.Solve for G (x ).5.The coeﬃcient of x i in G (x )is g i .Example:g i +1=2g i +1,g 0=0.Multiply and sum:i ≥0g i +1x i = i ≥02g i x i+ i ≥0x i .We choose G (x )=i ≥0x i g i .Rewrite in terms of G (x ):G (x )−g 0x=2G (x )+ i ≥0x i .Simplify:G (x )x =2G (x )+11−x .Solve for G (x ):G (x )=x(1−x )(1−2x ).Expand this using partial fractions:G (x )=x21−2x −11−x=x2i ≥02i x i − i ≥0x i =i ≥0(2i +1−1)x i +1.So g i =2i −1.π≈3.14159,e ≈2.71828,γ≈0.57721,φ=1+√52≈1.61803,ˆφ=1−√52≈−.61803i 2i p i GeneralProbability122Bernoulli Numbers (B i =0,odd i =1):B 0=1,B 1=−12,B 2=16,B 4=−130,B 6=142,B 8=−130,B 10=566.Change of base,quadratic formula:log b x =log a x log a b ,−b ±√b 2−4ac 2a.Euler’s number e :e =1+12+16+124+1120+···lim n →∞ 1+x nn=e x . 1+1nn <e < 1+1n n +1. 1+1n n=e −e 2n +11e 24n 2−O 1n 3.Harmonic numbers:1,32,116,2512,13760,4920,363140,761280,71292520,...ln n <H n <ln n +1,H n =ln n +γ+O 1n.Factorial,Stirling’s approximation:1,2,6,24,120,720,5040,40320,362880,...n !=√2πnn e n 1+Θ 1n.Ackermann’s function and inverse:a (i,j )=2ji =1a (i −1,2)j =1a (i −1,a (i,j −1))i,j ≥2α(i )=min {j |a (j,j )≥i }.Continuous distributions:IfPr[a <X <b ]= bap (x )dx,then p is the probability density function ofX .IfPr[X <a ]=P (a ),then P is the distribution function of X .If P and p both exist thenP (a )=a−∞p (x )dx.Expectation:If X is discreteE [g (X )]=xg (x )Pr[X =x ].If X continuous thenE [g (X )]=∞−∞g (x )p (x )dx = ∞−∞g (x )dP (x ).Variance,standard deviation:VAR[X ]=E [X 2]−E [X ]2,σ=VAR[X ].For events A and B :Pr[A ∨B ]=Pr[A ]+Pr[B ]−Pr[A ∧B ]Pr[A ∧B ]=Pr[A ]·Pr[B ],iﬀA and B are independent.Pr[A |B ]=Pr[A ∧B ]Pr[B ]For random variables X and Y :E [X ·Y ]=E [X ]·E [Y ],if X and Y are independent.E [X +Y ]=E [X ]+E [Y ],E [cX ]=c E [X ].Bayes’theorem:Pr[A i |B ]=Pr[B |A i ]Pr[A i ]nj =1Pr[A j ]Pr[B |A j ].Inclusion-exclusion:Prni =1X i =n i =1Pr[X i ]+nk =2(−1)k +1i i <···<i kPrk j =1X i j.Moment inequalities:Pr |X |≥λE [X ] ≤1λ,Pr X −E [X ] ≥λ·σ ≤1λ2.Geometric distribution:Pr[X =k ]=pq k −1,q =1−p,E [X ]=∞ k =1kpq k −1=1p.24338541675321166413712817825619951223101,02429112,04831124,09637138,192411416,384431532,768471665,5365317131,0725918262,1446119524,28867201,048,57671212,097,15273224,194,30479238,388,608832416,777,216892533,554,432972667,108,86410127134,217,72810328268,435,456107Binomial distribution:Pr[X =k ]= n kp k qn −k,q =1−p,E [X ]=n k =1k n kp k q n −k =np.Poisson distribution:Pr[X =k ]=e −λλkk !,E [X ]=λ.Normal (Gaussian)distribution:p (x )=1√2πσe −(x −µ)2/2σ2,E [X ]=µ.The “coupon collector”:We are given a random coupon each day,and there are n diﬀerent types of coupons.The distribu-tion of coupons is uniform.The expected number of days to pass before we to col-lect all n types isnH n .29536,870,912109301,073,741,824113312,147,483,648127324,294,967,296131Pascal’s Triangle111121133114641151010511615201561172135352171182856705628811936841261268436911104512021025221012045101πCalculusWallis’identity:π=2·2·2·4·4·6·6···1·3·3·5·5·7···Brouncker’s continued fraction expansion:π4=1+122+322+522+722+···Gregrory’s series:π4=1−13+15−17+19−···Newton’s series:π6=12+12·3·23+1·32·4·5·25+···Sharp’s series:π6=1√31−131·3+132·5−133·7+··· Euler’s series:π26=112+122+132+142+152+···π28=112+132+152+172+192+···π212=112−122+132−142+152−···Derivatives:1.d (cu )dx =c du dx, 2.d (u +v )dx =du dx +dv dx , 3.d (uv )dx =u dv dx +v dudx,4.d (u n )dx =nu n −1du dx ,5.d (u/v )dx =v du dx −u dv dx v 2,6.d (e cu )dx =ce cu du dx,7.d (c u )dx =(ln c )c u du dx ,8.d (ln u )dx =1u dudx,9.d (sin u )dx =cos u du dx ,10.d (cos u )dx =−sin u dudx ,11.d (tan u )dx =sec 2u du dx,12.d (cot u )dx =csc 2u dudx,13.d (sec u )dx =tan u sec u du dx ,14.d (csc u )dx =−cot u csc u dudx,15.d (arcsin u )dx =1√1−u 2du dx,16.d (arccos u )dx =−1√1−u 2dudx ,17.d (arctan u )dx =11+u 2du dx ,18.d (arccot u )dx =−11+u 2du dx ,19.d (arcsec u )dx =1u √1−u 2du dx,20.d (arccsc u )dx =−1u √1−u 2dudx ,21.d (sinh u )dx =cosh u du dx ,22.d (cosh u )dx =sinh u du dx ,23.d (tanh u )dx =sech 2u du dx ,24.d (coth u )dx =−csch 2u du dx,25.d (sech u )dx =−sech u tanh u du dx,26.d (csch u )dx =−csch u coth u dudx ,27.d (arcsinh u )dx =1√1+u 2du dx,28.d (arccosh u )dx =1√u 2−1dudx ,29.d (arctanh u )dx =11−u 2du dx ,30.d (arccoth u )dx =1u 2−1du dx ,31.d (arcsech u )dx =−1u √1−u 2du dx,32.d (arccsch u )dx =−1|u |√1+u 2dudx.Integrals:1. cu dx =c u dx, 2. (u +v )dx =u dx +v dx,3. x n dx =1n +1x n +1,n =−1,4.1xdx =ln x, 5.e x dx =e x ,6. dx1+x 2=arctan x,7.u dvdx dx =uv − v du dxdx,8.sin x dx =−cos x,9. cos x dx =sin x,10.tan x dx =−ln |cos x |,11. cot x dx =ln |cos x |,12.sec x dx =ln |sec x +tan x |,13.csc x dx =ln |csc x +cot x |,14.arcsin x a dx =arcsin xa +a 2−x 2,a >0,Partial FractionsLet N (x )and D (x )be polynomial func-tions of x .We can break down N (x )/D (x )using partial fraction expan-sion.First,if the degree of N is greater than or equal to the degree of D ,divide N by D ,obtainingN (x )D (x )=Q (x )+N (x )D (x ),where the degree of N is less than that ofD .Second,factor D (x ).Use the follow-ing rules:For a non-repeated factor:N (x )(x −a )D (x )=A x −a +N (x )D (x ),whereA =N (x )D (x )x =a.For a repeated factor:N (x )(x −a )m D (x )=m −1 k =0A k (x −a )m −k +N (x )D (x ),whereA k =1k ! dkdx kN (x )D (x )x =a.The reasonable man adapts himself to the world;the unreasonable persists in trying to adapt the world to himself.Therefore all progress depends on the unreasonable.–George Bernard ShawCalculus Cont.15.arccos x a dx =arccos xa −a 2−x 2,a >0,16.arctan x a dx =x arctan xa −a2ln(a 2+x 2),a >0,17.sin 2(ax )dx =12aax −sin(ax )cos(ax ),18.cos 2(ax )dx =12aax +sin(ax )cos(ax ),19.sec 2x dx =tan x,20.csc 2x dx =−cot x,21.sin nx dx =−sin n −1x cos x n +n −1n sin n −2x dx,22.cos nx dx =cos n −1x sin x n +n −1ncos n −2x dx,23.tan n x dx =tan n −1x n −1− tan n −2x dx,n =1,24.cot nx dx =−cot n −1x n −1− cot n −2x dx,n =1,25.sec nx dx =tan x sec n −1x n −1+n −2n −1sec n −2x dx,n =1,26.csc n x dx =−cot x cscn −1x n −1+n −2n −1csc n −2x dx,n =1,27. sinh x dx =cosh x,28. cosh x dx =sinh x,29.tanh x dx =ln |cosh x |,30. coth x dx =ln |sinh x |,31. sech x dx =arctan sinh x,32.csch x dx =lntanh x 2 ,33.sinh 2x dx =14sinh(2x )−12x,34.cosh 2x dx =14sinh(2x )+12x,35.sech 2x dx =tanh x,36.arcsinhx a dx=x arcsinhxa−x 2+a 2,a >0,37.arctanh x a dx =x arctanh xa +a2ln |a 2−x 2|,38. arccosh xa dx = x arccosh x a −x 2+a 2,if arccosh xa >0and a >0,x arccosh x a+x 2+a 2,if arccosh x a <0and a >0,39.dx √a 2+x 2=ln x + a 2+x 2 ,a >0,40. dx a 2+x 2=1a arctan x a ,a >0,41. a 2−x 2dx =x 2 a 2−x 2+a 22arcsin x a ,a >0,42. (a 2−x 2)3/2dx =x 8(5a 2−2x 2) a 2−x 2+3a 48arcsin x a,a >0,43. dx √a 2−x 2=arcsin x a ,a >0,44. dx a 2−x 2=12a ln a +x a −x ,45. dx (a 2−x 2)3/2=x a 2√a 2−x 2,46.a 2±x 2dx =x 2 a 2±x 2±a 22ln x + a 2±x 2 ,47. dx √x 2−a 2=ln x + x 2−a 2 ,a >0,48.dx ax 2+bx =1a ln x a +bx ,49. x √a +bx dx =2(3bx −2a )(a +bx )3/215b 2,50. √a +bx x dx =2√a +bx +a 1x √a +bx dx,51. x √a +bx dx =1√2ln √a +bx −√a √a +bx +√a ,a >0,52. √a 2−x 2x dx = a 2−x 2−a lna +√a 2−x 2x,53. x a 2−x 2dx =−13(a 2−x 2)3/2,54. x 2 a 2−x 2dx =x 8(2x 2−a 2) a 2−x 2+a 48arcsin x a ,a >0,55. dx √a 2−x 2=−1a ln a +√a 2−x 2x ,56. x dx √a 2−x 2=− a 2−x 2,57. x 2dx √a 2−x 2=−x 2 a 2−x 2+a 22arcsin x a,a >0,58. √a 2+x 2x dx = a 2+x 2−a ln a +√a 2+x 2x ,59. √x 2−a 2xdx = x 2−a 2−a arccos a |x |,a >0,60. xx 2±a 2dx =13(x 2±a 2)3/2,61. dx x √x 2+a 2=1a ln x a +√a 2+x 2,Calculus Cont.Finite Calculus62.dx x √x 2−a 2=1a arccos a|x |,a >0,63.dx x 2√x 2±a 2=∓√x 2±a 2a 2x ,64.x dx√x 2±a 2= x 2±a 2,65. √x 2±a 2x 4dx =∓(x 2+a 2)3/23a 2x 3,66.dx ax 2+bx +c =1√b 2−4ac ln 2ax +b −√b 2−4ac 2ax +b +√b 2−4ac,if b 2>4ac ,2√4ac −b 2arctan 2ax +b √4ac −b2,if b 2<4ac ,67.dx√ax 2+bx +c =1√a ln 2ax +b +2√a ax 2+bx +c ,if a >0,1√−a arcsin −2ax −b √b 2−4ac,if a <0,68.ax 2+bx +c dx =2ax +b 4a ax 2+bx +c +4ax −b28a dx √ax 2+bx +c,69.x dx√ax 2+bx +c=√ax 2+bx +c a −b 2adx√ax 2+bx +c,70.dxx √ax 2+bx +c = −1√c ln 2√c √ax 2+bx +c +bx +2c x ,if c >0,1√−c arcsin bx +2c |x |√b 2−4ac,if c <0,71. x 3x 2+a 2dx =(13x 2−215a 2)(x 2+a 2)3/2,72.x nsin(ax )dx =−1a xncos(ax )+n axn −1cos(ax )dx,73.x n cos(ax )dx =1ax nsin(ax )−nax n −1sin(ax )dx,74.x n e axdx =x n e ax a −n ax n −1e ax dx,75.x n ln(ax )dx =x n +1ln(ax )n +1−1(n +1)2,76.x n (ln ax )m dx =x n +1n +1(ln ax )m −m n +1x n (ln ax )m −1dx.Diﬀerence,shift operators:∆f (x )=f (x +1)−f (x ),E f (x )=f (x +1).Fundamental Theorem:f (x )=∆F (x )⇔f (x )δx =F (x )+C.b af (x )δx =b −1 i =af (i ).Diﬀerences:∆(cu )=c ∆u,∆(u +v )=∆u +∆v,∆(uv )=u ∆v +E v ∆u,∆(x n )=nx n −1,∆(H x )=x −1,∆(2x )=2x ,∆(c x )=(c −1)c x ,∆ x m = x m −1 .Sums: cu δx =cu δx, (u +v )δx = u δx + v δx, u ∆v δx =uv −E v ∆u δx,x nδx =x n +1m +1, x −1δx =H x , c x δx =c xc −1, x m δx = x m +1 .Falling Factorial Powers:x n =x (x −1)···(x −n +1),n >0,x 0=1,x n =1(x +1)···(x +|n |),n <0,x n +m =x m (x −m )n .Rising Factorial Powers:x n =x (x +1)···(x +n −1),n >0,x 0=1,x n =1(x −1)···(x −|n |),n <0,x n +m =x m (x +m )n .Conversion:x n =(−1)n (−x )n =(x −n +1)n=1/(x +1)−n ,x n =(−1)n (−x )n =(x +n −1)n=1/(x −1)−n ,x n =n k =1 n k x k =n k =1 nk (−1)n −k x k ,x n =n k =1nk (−1)n −k x k ,x n =n k =1n kx k.x 1=x 1=x 1x 2=x 2+x 1=x 2−x 1x 3=x 3+3x 2+x 1=x 3−3x 2+x 1x 4=x 4+6x 3+7x 2+x 1=x 4−6x 3+7x 2−x 1x 5=x 5+15x 4+25x 3+10x 2+x 1=x 5−15x 4+25x 3−10x 2+x 1x 1=x 1x 1=x 1x 2=x 2+x 1x 2=x 2−x 1x 3=x 3+3x 2+2x 1x 3=x 3−3x 2+2x 1x 4=x 4+6x 3+11x 2+6x 1x 4=x 4−6x 3+11x 2−6x 1x 5=x 5+10x 4+35x 3+50x 2+24x 1x 5=x 5−10x 4+35x 3−50x 2+24x 1Series Taylor’s series:f(x)=f(a)+(x−a)f (a)+(x−a)22f (a)+···=∞i=0(x−a)ii!f(i)(a).Expansions:1 1−x =1+x+x2+x3+x4+···=∞i=0x i,11−cx =1+cx+c2x2+c3x3+···=∞i=0c i x i,11−x n =1+x n+x2n+x3n+···=∞i=0x ni,x (1−x)2=x+2x2+3x3+4x4+···=∞i=0ix i,x kd ndx n11−x=x+2n x2+3n x3+4n x4+···=∞i=0i n x i,e x=1+x+12x2+16x3+···=∞i=0x ii!,ln(1+x)=x−12x2+13x3−14x4−···=∞i=1(−1)i+1x ii,ln11−x=x+12x2+13x3+14x4+···=∞i=1x ii,sin x=x−13!x3+15!x5−17!x7+···=∞i=0(−1)ix2i+1(2i+1)!,cos x=1−12!x2+14!x4−16!x6+···=∞i=0(−1)ix2i(2i)!,tan−1x=x−13x3+15x5−17x7+···=∞i=0(−1)ix2i+1(2i+1),(1+x)n=1+nx+n(n−1)2x2+···=∞i=0nix i, 1(1−x)n+1=1+(n+1)x+n+22x2+···=∞i=0i+nix i, xe x−1=1−12x+112x2−1720x4+···=∞i=0B i x ii!,1 2x (1−√1−4x)=1+x+2x2+5x3+···=∞i=01i+12iix i,1√1−4x=1+x+2x2+6x3+···=∞i=02iix i,1√1−4x1−√1−4x2xn=1+(2+n)x+4+n2x2+···=∞i=02i+nix i,1 1−x ln11−x=x+32x2+116x3+2512x4+···=∞i=1H i x i,1 2ln11−x2=12x2+34x3+1124x4+···=∞i=2H i−1x ii, x1−x−x2=x+x2+2x3+3x4+···=∞i=0F i x i,F n x1−(F n−1+F n+1)x−(−1)n x2=F n x+F2n x2+F3n x3+···=∞i=0F ni x i.Ordinary power series:A(x)=∞i=0a i x i.Exponential power series:A(x)=∞i=0a ix ii!.Dirichlet power series:A(x)=∞i=1a ii x.Binomial theorem:(x+y)n=nk=0nkx n−k y k.Diﬀerence of like powers:x n−y n=(x−y)n−1k=0x n−1−k y k.For ordinary power series:αA(x)+βB(x)=∞i=0(αa i+βb i)x i,x k A(x)=∞i=ka i−k x i,A(x)−k−1i=0a i x ix k=∞i=0a i+k x i,A(cx)=∞i=0c i a i x i,A (x)=∞i=0(i+1)a i+1x i,xA (x)=∞i=1ia i x i,A(x)dx=∞i=1a i−1ix i,A(x)+A(−x)2=∞i=0a2i x2i,A(x)−A(−x)2=∞i=0a2i+1x2i+1.Summation:If b i=ij=0a i thenB(x)=11−xA(x).Convolution:A(x)B(x)=∞i=0ij=0a jb i−jx i.God made the natural numbers;all the rest is the work of man.–Leopold KroneckerSeriesEscher’s KnotExpansions:1(1−x )n +1ln11−x=∞i =0(H n +i −H n ) n +i i x i , 1x −n =∞ i =0i n x i ,x n=∞ i =0 n i x i ,(e x −1)n=∞ i =0i n n !x i i !, ln11−x n=∞ i =0 i n n !x i i !,x cot x =∞ i =0(−4)i B 2i x 2i(2i )!,tan x =∞ i =1(−1)i −122i (22i −1)B 2i x 2i −1(2i )!,ζ(x )=∞ i =11i x ,1ζ(x )=∞ i =1µ(i )i x ,ζ(x −1)ζ(x )=∞ i =1φ(i )i x ,Stieltjes Integration ζ(x )= p11−p −x ,ζ2(x )=∞ i =1d (i )x iwhere d (n )= d |n 1,ζ(x )ζ(x −1)=∞ i =1S (i )x iwhere S (n )=d |n d,ζ(2n )=22n −1|B 2n |(2n )!π2n,n ∈N ,x sin x =∞ i =0(−1)i −1(4i −2)B 2i x 2i (2i )!,1−√1−4x2xn =∞ i =0n (2i +n −1)!i !(n +i )!x i,e x sin x =∞ i =12i/2sin iπ4i !x i,1−√1−xx =∞ i =0(4i )!16i √2(2i )!(2i +1)!x i ,arcsin x x2=∞ i =04i i !2(i +1)(2i +1)!x 2i .If G is continuous in the interval [a,b ]and F is nondecreasing then b aG (x )dF (x )exists.If a ≤b ≤c thencaG (x )dF (x )=baG (x )dF (x )+cbG (x )dF (x ).If the integrals involved existba G (x )+H (x )dF (x )=baG (x )dF (x )+ba H (x )dF (x ),b a G (x )d F (x )+H (x ) =baG (x )dF (x )+ba G (x )dH (x ),bac ·G (x )dF (x )=baG (x )d c ·F (x ) =cba G (x )dF (x ),baG (x )dF (x )=G (b )F (b )−G (a )F (a )−baF (x )dG (x ).If the integrals involved exist,and F possesses a derivative F at everypoint in [a,b ]thenbaG (x )dF (x )= b aG (x )F (x )dx.Cramer’s Rule00471876299385346152861157287039944502639580226738714956130459968133074872602415736990824417580135266874099183552712463037087519928466235041142536405162037788992132435465061089977842536405162031987987Fibonacci Numbers If we have equations:a 1,1x 1+a 1,2x 2+···+a 1,n x n =b 1a 2,1x 1+a 2,2x 2+···+a 2,n x n =b 2.........a n,1x 1+a n,2x 2+···+a n,n x n =b nLet A =(a i,j )and B be the column matrix (b i ).Then there is a unique solution iﬀdet A =0.Let A i be A with column i replaced by B .Thenx i =det A idet A .1,1,2,3,5,8,13,21,34,55,89,...Deﬁnitions:F i =F i −1+F i −2,F 0=F 1=1,F −i =(−1)i −1F i ,F i =1√5 φi −ˆφi ,Cassini’s identity:for i >0:F i +1F i −1−F 2i =(−1)i .Additive rule:F n +k =F k F n +1+F k −1F n ,F 2n =F n F n +1+F n −1F n .Calculation by matrices:F n −2F n −1F n −1F n =0111n .The Fibonacci number system:Every integer n has a unique representationn =F k 1+F k 2+···+F k m ,where k i ≥k i +1+2for all i ,1≤i <m and k m ≥2.Improvement makes strait roads,but the crooked roads without Improvement,are roads of Genius.–William Blake (The Marriage of Heaven and Hell)。

割圜密率捷法卡特兰数

割圜密率捷法卡特兰数割圆密率（Catalan numbers）是一种组合数学中的数列，以比利时数学家欧仁·查理·卡特兰（Eugène Charles Catalan）命名，该数列的前几项依次为1、1、2、5、14、42、132、429等。

在组合数学中，割圆密率数的计算和应用有很多，下面将介绍一些与割圆密率数相关的内容。

1. 割圆密率的递归定义：割圆密率数可以通过以下的递归公式定义：C(0) = 1C(n+1) = ∑[i=0 to n] C(i) * C(n-i) (for n≥0)2. 割圆密率的计算公式：割圆密率数还有一种常见的计算公式，即用阶乘的形式表示：C(n) = (2n)! / (n+1)! * n!3. 割圆密率数的性质：割圆密率数具有一些有趣的性质，例如：- 割圆密率数是一个递增的数列。

- 割圆密率数满足递推关系：C(n) = C(0)*C(n-1) + C(1)*C(n-2) + ... + C(n-1)*C(0)。

- 割圆密率数的生成函数为：C(z) = 1 / (1 - z - z^2)。

4. 割圆密率数的应用：割圆密率数在组合数学、图论、计算几何等领域都有广泛的应用。

以下是一些经典的应用例子：- 引理1：在一个二叉树中，如果叶子节点的数量为n，则该二叉树的种类数为C(n)。

- 引理2：在一个凸多边形中，通过n条不相交对角线将该多边形分割成若干个三角形，则共有C(n)种分割方法。

- 引理3：在一个n*n的方格图中，从左下角移动到右上角，每次只能向上或向右移动，且不能超过对角线，共有C(n)种路径方法。

- 引理4：在一个由n+1个点形成的多边形中，每个内角都是锐角，共有C(n)种三角剖分方法。

以上只是割圆密率数的一些应用案例，实际上割圆密率数还有很多其他有趣的应用领域。

总结：割圆密率数作为一种常见的数列，在组合数学中扮演着重要的角色。

通过递归公式、计算公式和生成函数等方法，可以计算出任意项的割圆密率数。

克里斯托数列

克里斯托数列摘要：一、克里斯托数列的定义与起源二、克里斯托数列的性质与应用三、克里斯托数列的扩展与变体四、克里斯托数列在实际问题中的解决方案五、总结与展望正文：克里斯托数列（Catalan numbers）是一种二项式系数，具有重要的数学地位和广泛的应用。

它们被命名为西班牙数学家Catalan 的名字，起源于他在19 世纪对组合数学的研究。

克里斯托数列的定义如下：C(n) = (2n)! / (n!(n+1)!)其中n 为正整数。

根据这个定义，我们可以计算出前几项克里斯托数列的值：C(0) = 1，C(1) = 1，C(2) = 1/2，C(3) = 1/2 + 1/6 = 1/2，C(4) = 1/2 + 1/6 + 1/24 = 1/3。

克里斯托数列具有以下性质：1.递归性：C(0) = 1，C(n) = C(n-1) + C(n-2)，n > 0。

2.组合数性质：C(n) 表示从n+1 个元素中取2 个元素的组合数。

3.生成函数：T(x) = 1 + x + x^2 + ...+ x^n，求导得T"(x) = C(n) *x^(n-1)。

克里斯托数列在组合数学、图论、概率论等领域具有广泛的应用。

例如，在图论中，克里斯托数列表示有限无向图的顶点着色数的上界。

此外，克里斯托数列还与Stirling 数、Binomial 系数、Pochhammer 符号等数学符号密切相关。

克里斯托数列的扩展与变体包括：1.超克里斯托数列：定义为C(n+1, k) = C(n, k) + C(n, k-1)，其中n 和k 为非负整数。

2.反克里斯托数列：定义为C(n, k) = C(n-1, k-1) + C(n-1, k)，其中n 和k 为非负整数。

在实际问题中，克里斯托数列及其扩展可以帮助我们解决各种组合优化问题、概率问题等。

例如，在组合优化中，我们可以利用克里斯托数列计算二进制排列数，从而找到最优解。

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Catalan Numbers
Q1
问题描述：
有n个字符X和n个字符Y，字符X之间没有差别，字符Y之间也没有差别，将所有的X和Y进行排列，要求在任意的前i(i=1,2,…,n)个字符中X的个数不得小于Y的个数，满足要求的排列的个数。

(X=(，Y=)是一种特殊情况)
分析：
令f(n)表示n个X和n个Y的排列个数，定义配对规则：Y与前面离它最近的X为一对。

由定义可知，在一对XY之间的XY个数相等，即可以一一配对。

由要求可知，第一个字符一定为X，在包含这个X 的XY对之间的其他XY对个数可能为0、1、…、n-1，对应外面的XY 对个数为n-1、n-2、…、0。

每种情况是一个子问题，子问题的排列
个数为f(i)f(n-1-i)，f(n)与子问题之间的关系为
f(n)=f(0)f(n-1)+f(1)f(n-2)+…+f(n-1)f(0)
对照Catalan Numbers的表示形式可知f(n)即Catalan Numbers。

Q2
问题描述：
将边数为n的凸多边形划分成三角形，有多少种分法。

分析：
边数为n的凸多边形分法为f(n)，f(2)=f(3)=1。

多边形的顶点分别为a1、a2、…、an，边分别为a1a2、a2a3、…、ana1。

边a1a2所在的三角形由3个顶点组成，a1、a2和ax，x可以取3、…、n，当x确定以后，该种情况下产生两个子凸多边形，大小分别为x-1、n-x+2，所以
f(n)=f(2)f(n-1)+f(3)f(n-2)+…+f(n-1)f(2)
将f(j)与Ci做如下映射
f(2)、f(3)、f(4)、f(5)、…、f(n)
C0 、C1 、C2 、C3 、…、Cn-2
可以发现f(n)=Cn-2(n>=2)。