Asymptotics in combinatorics

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Using Stirling's formula, approximately compute: C10040C_{100}^{40}.

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Using Stirling's formula, approximately compute: 100!20!⋅30!⋅50!\dfrac {100!}{20! \cdot 30! \cdot 50!}.

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If nn is very large, approximately compute C2nnC_{2n}^{n}.

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Let the sequence of natural numbers {kn}\left\{ k_{n}\right\} be such that knn→p,kn⩽n\frac{k_{n}}{n} \rightarrow p, k_{n} \leqslant n. Compute the limit lim⁡1nln⁡Cnkn\lim \frac{1}{n} \ln C_{n}^{k_{n}}.

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Let α>0\alpha >0. Prove that as n→+∞(n∈N)n \rightarrow +\infty (n \in \mathbb {N})

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(a)

∑1⩽k⩽nkαk/n∼n1+ααln⁡n\sum_{1 \leqslant k \leqslant n} k^{\alpha k / n} \sim \frac{n^{1+\alpha }}{\alpha \ln n};

(b)

∑1⩽k⩽nk−αk/n∼nαln⁡n\sum_{1 \leqslant k \leqslant n} k^{-\alpha k / n} \sim \frac{n}{\alpha \ln n};

(c)

∑1⩽k⩽n(k!)−α/n∼nαln⁡n\sum_{1 \leqslant k \leqslant n}(k!)^{-\alpha / n} \sim \frac{n}{\alpha \ln n};

(d)

∑1⩽k⩽nkαn∼nαn1−e−α\sum_{1 \leqslant k \leqslant n} k^{\alpha n} \sim \frac{n^{\alpha n}}{1-e^{-\alpha }}.

Prove that ∑1⩽k⩽nnkk!∼12en\sum_{1 \leqslant k \leqslant n} \frac{n^{k}}{k!} \sim \frac{1}{2} e^{n}.

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Prove that

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(a)

∑1⩽k⩽nCnk∼2πn42n/2\sum_{1 \leqslant k \leqslant n} \sqrt{C_{n}^{k}} \sim \sqrt[4]{2 \pi n} 2^{n / 2};

(b)

∑1⩽k⩽nCnk∣2k−n∣∼Γ(34)π2n42n\sum_{1 \leqslant k \leqslant n} C_{n}^{k} \sqrt{\left|2 k-n\right|} \sim \frac{\Gamma \left(\frac{3}{4}\right)}{\sqrt{\pi }} \sqrt[4]{2 n} 2^{n}.

Let a1,a2,…,ala_1, a_2, \ldots , a_l be positive integers without a common factor different from 1 and AnA_n be the number of non-negative integral solutions of

a1x1+a2x2+a3x3+⋯+alxl=n. a_1 x_1 + a_2 x_2 + a_3 x_3 + \cdots + a_l x_l = n .

Then we have

lim⁡n→∞Annl−1=1a1a2⋯al(l−1)!. \lim _{n \rightarrow \infty } \frac{A_n}{n^{l-1}} = \frac{1}{a_1 a_2 \cdots a_l (l-1)!} .
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Remark.
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Cf. Laguerre: Oeuvres, Vol. 1. Paris: Gauthier-Villars 1898, pp. 218-220.

Let kk be a fixed positive number and nn an integer increasing to infinity. Then

∑v=0n(nv)k∼2knk(2πn)k−12 \sum _{v=0}^{n} \binom {n}{v}^{k} \sim \frac{2^{kn}}{\sqrt{k}} \left(\frac{2}{\pi n}\right)^{\frac{k-1}{2}}

(Problem 58). Observe the particular cases k=1k = 1, k=2k = 2.

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Let AnA_{n} denote the arithmetic, and GnG_{n} the geometric, mean of the binomial coefficients

(n0),(n1),(n2),…,(nn). \binom {n}{0}, \binom {n}{1}, \binom {n}{2}, \ldots , \binom {n}{n} .

Show that

lim⁡n→∞Ann=2,lim⁡n→∞Gnn=e. \lim _{n \to \infty } \sqrt[n]{A_{n}} = 2, \qquad \lim _{n \to \infty } \sqrt[n]{G_{n}} = \sqrt{e} .
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Let nn and vv be integers, 0<v<n0 < v < n. If nn and vv increase to infinity in such a way that

lim⁡n→∞v−n2n=λ, \lim _{n \to \infty } \frac{v - \dfrac {n}{2}}{\sqrt{n}} = \lambda ,

then

lim⁡n→∞n2n(nv)=2π e−2λ2. \lim _{n \to \infty } \frac{\sqrt{n}}{2^{n}} \binom {n}{v} = \sqrt{\frac{2}{\pi }} \, e^{-2\lambda ^{2}} .
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In the expansion of a symmetric nnth order determinant (with arbitrary elements) let sn′s_{n}' be the number of distinct positive terms and sn′′s_{n}'' the number of distinct negative terms. For sn=sn′+sn′′s_{n} = s_{n}' + s_{n}'' we have already in the works of Cayley the recurrence relation

sn+1=(n+1)sn−(n2)sn−2. s_{n+1} = (n+1) s_{n} - \binom {n}{2} s_{n-2}.

Show that for dn=sn′−sn′′d_{n} = s_{n}' - s_{n}'' we have the recurrence relation

dn+1=−(n−1)dn−(n2)dn−2, d_{n+1} = -(n-1) d_{n} - \binom {n}{2} d_{n-2},

and that

lim⁡n→∞n1/2snn!=e3/4π,lim⁡n→∞(−1)n−1n3/2dnn!=e−3/42π. \lim _{n \rightarrow \infty } \frac{n^{1/2} s_{n}}{n!} = \frac{e^{3/4}}{\sqrt{\pi }}, \qquad \lim _{n \rightarrow \infty } \frac{(-1)^{n-1} n^{3/2} d_{n}}{n!} = \frac{e^{-3/4}}{2\sqrt{\pi }}.
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Show that

Tn1/n→4(1+a)1+a(1−a)1−a T_{n}^{1 / n} \rightarrow \frac{4}{\sqrt{(1+a)^{1+a}(1-a)^{1-a}}}

as n→∞n \rightarrow \infty, where 0<a<10 < a < 1 and

Tn=∑k:∣k−12n∣>12an(nk) T_{n} = \sum _{\substack {k: \\ \left|k-\frac{1}{2} n\right| > \frac{1}{2} a n}}\binom {n}{k}

Find the asymptotic behaviour of Tn1/nT_{n}^{1 / n} where

Tn=∑k⋅k>n(1+a)nkk!, where a>0 T_{n} = \sum _{\substack {k \cdot \\ k > n(1+a)}} \frac{n^{k}}{k!}, \quad \text{ where } a > 0
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Find the asymptotics of

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(1)

the sums from Problem 1.1.6 (1);

(2)

the sums from Problem 1.1.6 (2);

(3)

the sums from Problem 1.1.6 (3);

(4)

the number AnA_n of subsets of the set {1,2,…,n}\left\{ 1, 2, \ldots , n\right\} containing no two consecutive numbers;

(5)

the same as in part (4), for three consecutive numbers.

In the answer you may use the function xP(a,b)x_P(a,b) which, given numbers a,ba, b and a polynomial PP having a unique root on the interval [a,b][a,b], returns this root.

Find the asymptotics of the maximum number of edges in a graph with nn vertices containing no kk-clique. Here k=kn=o(n)k = k_n = o(n).

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Prove the following relations, assuming in the asymptotics that n→∞n \to \infty and kk is fixed. (The number \exH(n)\ex_H(n) denotes the maximum number of edges in a graph on nn vertices that does not contain HH as a subgraph; it is defined in Problem 2.7.6.)

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(1)

\exPk(n)≳k−22⋅n\ex_{P_k}(n) \gtrsim \dfrac {k-2}{2} \cdot n.

(2)

\exC2k+1(n)≳n24\ex_{C_{2k+1}}(n) \gtrsim \dfrac {n^2}{4}.

(3)

\exK1,k(n)∼k−12⋅n\ex_{K_{1,k}}(n) \sim \dfrac {k-1}{2} \cdot n.

Remark.
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The famous Erdős--Stone--Simonovits theorem states that for any fixed HH such that χ(H)>2\chi (H) > 2, as n→∞n \to \infty we have \exH(n)∼n22χ(H)−2χ(H)−1\ex_H(n) \sim \dfrac {n^2}{2}\dfrac {\chi (H)-2}{\chi (H)-1}. (For bipartite HH it is only known that \exH(n)=o(n2)\ex_H(n) = o(n^2).) That is, if we forbid a graph to contain some fixed subgraph HH, then the fraction of edges that can be drawn, among all possible edges, is determined by the chromatic number of the graph HH. It is surprising that the chromatic number arises in this problem! (This theorem may not be used in solving Problem 6.1.3 (2).)

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(1)

n2(2+1n)n=(2+o(1))nn^2\left(2+\dfrac {1}{n}\right)^n = (2+o(1))^n. By definition, this means that there exists a function ψ(n)=o(1)\psi (n) = o(1) such that

n2(2+1n)n=(2+ψ(n))n; n^2\left(2+\frac{1}{n}\right)^n = (2+\psi (n))^n;

or, equivalently,

n2(2+1n)nn−2=o(1). \sqrt[n]{n^2\left(2+\frac{1}{n}\right)^n} - 2 = o(1).
(2)

3n(2+1n)n=(2+o(1))n3^{\sqrt{n}}\left(2+\dfrac {1}{n}\right)^n = (2+o(1))^n.

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(1)

Find the asymptotics of (n[n/2])n\sqrt[n]{\dbinom {n}{[n/2]}} (cf. Problems 1.4.3 (2) and 6.1.10 (2)).

(2)

2nn+1<(n[n/2])<2n\dfrac {2^n}{n+1} < \dbinom {n}{[n/2]} < 2^n.

(3)

Find the asymptotics of (3mm)m\sqrt[m]{\dbinom {3m}{m}}.

(4)

33m3m+1<22m(3mm)<33m\dfrac {3^{3m}}{3m+1} < 2^{2m}\dbinom {3m}{m} < 3^{3m}.

(5)

(n[an])=(a−a(1−a)a−1+o(1))n\dbinom {n}{[an]} = (a^{-a}(1-a)^{a-1}+o(1))^n, where 0<a<10 < a < 1.

(6)

n![a1n]!…[asn]!=(e−a1ln⁡a1−…−asln⁡as+o(1))n\dfrac {n!}{[a_1n]!\ldots [a_sn]!} = (e^{-a_1\ln a_1 - \ldots - a_s\ln a_s}+o(1))^n, where a1+…+as=1a_1+\ldots +a_s=1, 0<ak<10 < a_k < 1.

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(1)

Find the asymptotics of ln⁡(n!)\ln (n!).

(2)

Find the asymptotics of n!n\sqrt[n]{n!}.

(3)

nne−n+1⩽n!⩽nn+1e−n+1n^ne^{-n+1} \leqslant n! \leqslant n^{n+1}e^{-n+1};

(4)

n!⩽nne−n+1nn! \leqslant n^ne^{-n+1}\sqrt{n}.

(5)

Stirling’s formula. n!∼nne−n2πnn! \sim n^ne^{-n}\sqrt{2\pi n}, i.e., lim⁡n→∞n!nne−n2πn=1\lim_{n \to \infty } \dfrac {n!}{n^ne^{-n}\sqrt{2\pi n}} = 1.

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(1)

(nk)<nkk!\dbinom {n}{k} < \dfrac {n^k}{k!}.

(2)

ln⁡(n−1)(n−2)…(n−k)nk=−k(k+1)2n(1+O(kn))\ln \dfrac {(n-1)(n-2)\ldots (n-k)}{n^k} = -\dfrac {k(k+1)}{2n}\left(1+O\left(\dfrac {k}{n}\right)\right) for k=kn<n2k=k_n < \dfrac {n}{2}.

This means that there exists a function ψ(n)=O(kn)\psi (n) = O\left(\dfrac {k}{n}\right) such that

ln⁡(n−1)(n−2)…(n−k)nk=−k(k+1)2n(1+ψ(n)); \ln \frac{(n-1)(n-2)\ldots (n-k)}{n^k} = -\frac{k(k+1)}{2n}(1+\psi (n));

or, equivalently,

−1−2nk(k+1)ln⁡(n−1)(n−2)…(n−k)nk=O(kn). -1 - \frac{2n}{k(k+1)}\ln \frac{(n-1)(n-2)\ldots (n-k)}{n^k} = O\left(\frac{k}{n}\right).
(3)

(nk)=nkk!e−k(k−1)2n+O(k3/n2)\dbinom {n}{k} = \dfrac {n^k}{k!}e^{-\frac{k(k-1)}{2n}+O(k^3/n^2)} for k=kn<n2k=k_n < \dfrac {n}{2}.

(Formulate yourself what e−k(k−1)2n+O(k3/n2)e^{-\frac{k(k-1)}{2n}+O(k^3/n^2)} means here.)

(4)

(nk)∼nkk!\dbinom {n}{k} \sim \dfrac {n^k}{k!} for k=kn=o(n)k=k_n=o(\sqrt{n}).

(5)

(2nn−k)/(2nn)=e−k2n(1+o(1))\dbinom {2n}{n-k}\Big/\dbinom {2n}{n} = e^{-\frac{k^2}{n}(1+o(1))} for k=kn=o(n)k=k_n=o(n).

(Formulate yourself what e−k2n(1+o(1))e^{-\frac{k^2}{n}(1+o(1))} means here.)

Informally, part (4) means that for k≪nk \ll \sqrt{n} the probability of getting exactly kk heads in nn coin tosses is approximately equal to nkk!2−n\dfrac {n^k}{k!}2^{-n}. For the informal remarks on this and the next part, an intuitive understanding of what probability is suffices.

Informally, part (5) means that for k≪nk \ll n the probability PkP_k of getting exactly n−kn-k heads in 2n2n coin tosses is approximately equal to P0e−k2/nP_0e^{-k^2/n} (the normal distribution).

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(1)

Is it true that the notations eo(n)e^{o(n)} and o(en)o(e^n) are ``equivalent''?

That is, is it true that for any function f:Z→(0,+∞)f : \mathbb {Z} \to (0,+\infty ) the conditions lim⁡n→∞ln⁡f(n)n=0\lim_{n\to \infty }\dfrac {\ln f(n)}{n}=0 and lim⁡n→∞f(n)e−n=0\lim_{n\to \infty }f(n)e^{-n}=0 are equivalent?

(2)

Find functions f,g:Z→(0,+∞)f,g : \mathbb {Z} \to (0,+\infty ) such that f(n)∼g(n)f(n)\sim g(n), but ef(n)≠O(eg(n))e^{f(n)} \neq O(e^{g(n)}).

(3)

Can functions f,g:Z→(0,+∞)f,g : \mathbb {Z} \to (0,+\infty ) simultaneously satisfy the relations f(n)=o(g(n))f(n)=o(g(n)) and g(n)=o(f(n))g(n)=o(f(n))?

(4)

Can functions f,g:Z→(0,+∞)f,g : \mathbb {Z} \to (0,+\infty ) simultaneously satisfy the relations f(n)=O(g(n))f(n)=O(g(n)) and g(n)=O(f(n))g(n)=O(f(n))?

(5)

Do the two relations of part (4) imply that f(n)∼g(n)f(n)\sim g(n)?

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(1)

Which function grows faster: x(xx)x^{(x^x)} or (x!)(2x)(x!)^{(2^x)}?

That is, find lim⁡x→∞x(xx)(x!)(−2x)\lim_{x\to \infty } x^{(x^x)}(x!)^{(-2^x)}.

(2)

Does there exist a function ψ(n)=o(1)\psi (n) = o(1) such that (2+ψ(n))n×2−ne−n→∞(2+\psi (n))^n \times 2^{-n}e^{-\sqrt{n}} \to \infty?

(As in any mathematical problem, the answer must be justified: give an example of such a function, or prove its existence, or prove that no such function exists.)

In Problems 6.1.10 (2, 3, 4, 5, 6, 7), unlike the other problems, one may use Stirling’s formula 6.1.6 (5) without proof.

Find the asymptotics of

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(1)

ln⁡(n2n)\ln \dbinom {n^2}{n};

(2)

(n[n/2])\dbinom {n}{[n/2]};

(3)

(n2n)\dbinom {n^2}{n};

(4)

(n[nα])\dbinom {n}{[n^\alpha ]}, α∈(0,1)\alpha \in (0,1);

(5)

(2n−1)!!:=(2n−1)⋅(2n−3)⋅…⋅3⋅1(2n-1)!! := (2n-1)\cdot (2n-3)\cdot \ldots \cdot 3\cdot 1;

(6)

∑k=0n(nk)2\displaystyle \sum_{k=0}^{n}\dbinom {n}{k}^2;

(7)

∑k=0n(nk)4\displaystyle \sum_{k=0}^{n}\dbinom {n}{k}^4.

Find the asymptotics of the function s=s(n)s = s(n) defined by

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(1)

ss=ns^s = n;

(2)

ss3=ns^{s^3} = n;

(3)

s(n):=max⁡{k:k!⩽n}s(n) := \max \left\{ k : k! \leqslant n\right\};

(4)

s(n):=min⁡{m∈N:(nm)<2(m2)}s(n) := \min \left\{ m \in \mathbb {N} : \dbinom {n}{m} < 2^{\binom {m}{2}}\right\} (cf. Problem 4.1.5);

(5)

s(n):=min⁡{m∈N:2mm>n}s(n) := \min \left\{ m \in \mathbb {N} : \dfrac {2^m}{m} > n\right\} (the function 2m/m2^m/m arises as the complexity of realizing Boolean functions);

(6)

s(n):=min⁡{m∈N:(m[m/2])>n}s(n) := \min \left\{ m \in \mathbb {N} : \dbinom {m}{[m/2]} > n\right\} (cf. Problem 1.4.3 (2)).

In the answers you may use constants given as sums of series. Find the asymptotics of

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(1)

the number of linear subspaces of Z2n\mathbb {Z}_2^n (see Problem 1.4.7 and the definition preceding it);

(2)

the number of unicyclic graphs with nn vertices (see Problem 2.2.5 (2) and the definition preceding it).

References

[Dem97] Demidovich B.P. Сборник задач и упражнений по математическому анализу. 13-е изд., Издательство Московского университета / Издательство "ЧеРо", 1997.
[МГЛ+04] Макаров Б.М., Голузина М.Г., Лодкин А.А., Подкорытов А.Н. Избранные задачи по вещественному анализу. 2-е изд., Невский Диалект; БХВ-Петербург, 2004.
[PóSz98] Pólya G., Szegő G. Problems and Theorems in Analysis. Springer, 1998.
[GrSt20] Grimmett G.R., Stirzaker D.R. Probability and Random Processes. 4th ed., Oxford University Press, 2020.
[ГДИ+16] Глибичук А.А., Дайняк А.Б., Ильинский Д.Г., Купавский А.Б., Райгородский А.М., Скопенков А.Б., Чернов А.А. Элементы дискретной математики в задачах. МЦНМО, 2016.