Chapter 2.5

Functions of Large Numbers

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Problem 2.195

Let p1,p2,…,pl,a1,a2,…,alp_{1}, p_{2}, \ldots , p_{l}, a_{1}, a_{2}, \ldots , a_{l} be arbitrary positive numbers. Then

lim⁡n→∞p1a1n+p2a2n+⋯+plalnn \lim _{n \to \infty } \sqrt[n]{p_{1}a_{1}^{n} + p_{2}a_{2}^{n} + \cdots + p_{l}a_{l}^{n}}

exists and it is equal to the largest among the numbers a1,a2,…,ala_{1}, a_{2}, \ldots , a_{l}.

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Problem 2.196

Under the same hypotheses as in Problem 195

lim⁡n→∞p1a1n+1+p2a2n+1+⋯+plaln+1p1a1n+p2a2n+⋯+plaln=max⁡(a1,a2,…,al). \lim _{n \to \infty } \frac{p_{1}a_{1}^{n+1} + p_{2}a_{2}^{n+1} + \cdots + p_{l}a_{l}^{n+1}}{p_{1}a_{1}^{n} + p_{2}a_{2}^{n} + \cdots + p_{l}a_{l}^{n}} = \max (a_{1}, a_{2}, \ldots , a_{l}) .
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Problem 2.197

Let f(x)f(x) be an arbitrary polynomial whose zeros are all real and positive and for which

−f′(x)f(x)=c0+c1x+c2x2+⋯+cnxn⋯ . -\frac{f'(x)}{f(x)} = c_{0} + c_{1}x + c_{2}x^{2} + \cdots + c_{n}x^{n} \cdots .

Show that

lim⁡n→∞1cnn=lim⁡n→∞cn−1cn \lim _{n \to \infty } \frac{1}{\sqrt[n]{c_{n}}} = \lim _{n \to \infty } \frac{c_{n-1}}{c_{n}}

exists and that it is equal to the smallest zero of f(x)f(x).

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Problem 2.198

The two functions φ(x)\varphi (x) and f(x)f(x) are continuous and positive on the interval [a,b]\left[a,b\right]. Then

lim⁡n→∞∫abφ(x)[f(x)]n dxn \lim _{n \to \infty } \sqrt[n]{\int _{a}^{b} \varphi (x) \left[f(x)\right]^{n} \, dx}

exists and is equal to the maximum of f(x)f(x) on [a,b]\left[a,b\right].

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Problem 2.199

Under the same hypothesis as in Problem 198

lim⁡n→∞∫abφ(x)[f(x)]n+1 dx∫abφ(x)[f(x)]n dx=max⁡f(x). \lim _{n \to \infty } \frac{\displaystyle \int _{a}^{b} \varphi (x) \left[f(x)\right]^{n+1} \, dx}{\displaystyle \int _{a}^{b} \varphi (x) \left[f(x)\right]^{n} \, dx} = \max f(x) .
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Problem 2.200

Let kk be a positive constant and a<ξ<ba < \xi < b. Show that for a,b,ξ,ka, b, \xi , k fixed and n→∞n \to \infty

∫abe−kn(x−ξ)2 dx∼πkn. \int _{a}^{b} e^{-kn(x-\xi )^{2}} \, dx \sim \sqrt{\frac{\pi }{kn}} .
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Problem 2.201

The functions φ(x)\varphi (x), h(x)h(x) and f(x)=eh(x)f(x)=e^{h(x)} are defined on the finite or infinite interval [a,b]\left[a,b\right] and satisfy the following conditions:

(1) φ(x)[f(x)]n=φ(x)enh(x)\varphi (x)\left[f(x)\right]^{n} = \varphi (x)e^{nh(x)} is absolutely integrable over [a,b]\left[a,b\right]; n=0,1,2,…n=0,1,2,\ldots

(2) The function h(x)h(x) attains its maximum only at the point ξ\xi in (a,b)(a,b); moreover, the least upper bound of h(x)h(x) is smaller than h(ξ)h(\xi ) on any closed interval that does not contain ξ\xi; there is, furthermore, a neighbourhood of ξ\xi where h′′(x)h''(x) exists and is continuous; finally h′′(ξ)<0h''(\xi ) < 0,

(3) φ(x)\varphi (x) is continuous at x=ξx=\xi, φ(ξ)≠0\varphi (\xi ) \neq 0. Then the following asymptotic formula holds as n→∞n \to \infty

∫abφ(x)[f(x)]n dx∼φ(ξ)[f(ξ)]n+1/2−2πnf′′(ξ)=φ(ξ)enh(ξ)−2πnh′′(ξ). \int _{a}^{b} \varphi (x) \left[f(x)\right]^{n} \, dx \sim \varphi (\xi ) \left[f(\xi )\right]^{n+1/2} \sqrt{-\frac{2\pi }{nf''(\xi )}} = \varphi (\xi ) e^{nh(\xi )} \sqrt{-\frac{2\pi }{nh''(\xi )}} .
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Remark.
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We consider only a neighbourhood of ξ\xi and expand h(x)h(x) in powers of (x−ξ)(x-\xi ) up to terms of the second order.

Problem 2.202

Let nn be an integer, n→+∞n \to +\infty. Using the fact that

∫0π2sin⁡2nx dx=∫0π2cos⁡2nx dx=1⋅3⋯(2n−1)2⋅4⋯2n⋅π2, \int _{0}^{\frac{\pi }{2}} \sin ^{2n} x \, dx = \int _{0}^{\frac{\pi }{2}} \cos ^{2n} x \, dx = \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots 2n} \cdot \frac{\pi }{2},

prove that

1⋅3⋯(2n−1)2⋅4⋯2n∼1nπ. \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots 2n} \sim \frac{1}{\sqrt{n\pi }} .
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Problem 2.203

We assume that λ\lambda is real, λ>1\lambda > 1; Pn(x)P_{n}(x) denotes the nn-th Legendre polynomial. As n→∞n \to \infty

Pn(λ)∼12nπ(λ+λ2−1)n+1/2λ2−14. P_{n}(\lambda ) \sim \frac{1}{\sqrt{2n\pi }} \frac{(\lambda +\sqrt{\lambda ^{2}-1})^{n+1/2}}{\sqrt[4]{\lambda ^{2}-1}} .

The positive value of the roots must be used.

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Remark.
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Problem 86 of Part VI.

Problem 2.204

The Bessel function Jv(t)J_{v}(t) can be defined by Hansen's expansion

eitcos⁡x=J0(t)+2∑v=1∞ivJv(t)cos⁡vx. e^{it\cos x} = J_{0}(t) + 2\sum _{v=1}^{\infty } i^{v} J_{v}(t) \cos vx .

Derive the following asymptotic formula:

Jv(it)∼ivet2πt,t→+∞, v=0,1,2,… J_{v}(it) \sim i^{v} \frac{e^{t}}{\sqrt{2\pi t}}, \qquad t \to +\infty , \ v = 0, 1, 2, \ldots
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Problem 2.205

Show that for positive nn, n→+∞n \to +\infty,

Γ(n+1)=∫0∞e−xxn dx∼(ne)n2πn \Gamma (n+1) = \int _{0}^{\infty } e^{-x} x^{n} \, dx \sim \left(\frac{n}{e}\right)^{n} \sqrt{2\pi n}

and, more accurately,

(en)nΓ(n+1)=2πn+O(1n). \left(\frac{e}{n}\right)^{n} \Gamma (n+1) = \sqrt{2\pi n} + O\left(\frac{1}{\sqrt{n}}\right) .
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Remark.
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Problem 18 of this Part; Problem 167 of Part I.

Problem 2.206

Let kk and ll be real numbers, k>1k > 1. Prove that for n→+∞n \to +\infty

(nk+ln)∼(k−1)n2πn(kk−1)nk+l+12. \binom {nk+l}{n} \sim \frac{(k-1)^{n}}{\sqrt{2\pi n}} \left(\frac{k}{k-1}\right)^{nk+l+\frac{1}{2}} .
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Problem 2.207

Assume that α\alpha is real and that tt is positive and increases to infinity. Then

∫1∞xα(tex)xdxx∼2π tα−12et. \int _{1}^{\infty } x^{\alpha } \left(\frac{te}{x}\right)^{x} \frac{dx}{x} \sim \sqrt{2\pi } \, t^{\alpha -\frac{1}{2}} e^{t} .
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Problem 2.208

Let 0<α<10 < \alpha < 1. The following approximation is justified as τ→+0\tau \to +0:

∫0∞exp⁡(xαα−τx) dx∼2π1−α τ−α2(1−α)−1exp⁡(1−αατ−α1−α). \int _{0}^{\infty } \exp \left(\frac{x^{\alpha }}{\alpha } - \tau x\right) \, dx \sim \sqrt{\frac{2\pi }{1-\alpha }} \, \tau ^{-\frac{\alpha }{2(1-\alpha )}-1} \exp \left(\frac{1-\alpha }{\alpha } \tau ^{-\frac{\alpha }{1-\alpha }}\right) .
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Problem 2.209

Let α>0\alpha > 0. As t→∞t \to \infty we obtain

∫0∞x−αxtx dx∼2πeα t12αexp⁡(e−1αtα). \int _{0}^{\infty } x^{-\alpha x} t^{x} \, dx \sim \sqrt{\frac{2\pi }{e\alpha }} \, t^{\frac{1}{2\alpha }} \exp \left(e^{-1} \alpha t^{\alpha }\right) .
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Problem 2.210

Let α\alpha and β\beta be two real constants. Then the relation

1n!∫0n+αn+βe−xxn dx=A+Bn+o(1n) \frac{1}{n!} \int _{0}^{n+\alpha \sqrt{n}+\beta } e^{-x} x^{n} \, dx = A + \frac{B}{\sqrt{n}} + o\left(\frac{1}{\sqrt{n}}\right)

holds where

A=12π∫−∞αe−t22 dt,B=12π(β−α2+23)e−α22. A = \frac{1}{\sqrt{2\pi }} \int _{-\infty }^{\alpha } e^{-\frac{t^{2}}{2}} \, dt, \qquad B = \frac{1}{\sqrt{2\pi }} \left(\beta - \frac{\alpha ^{2}+2}{3}\right) e^{-\frac{\alpha ^{2}}{2}} .
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Problem 2.211

We denote by λ\lambda a positive proper fraction and by xnx_{n} the only positive root of the transcendental equation

1+x1!+x22!+⋯+xnn!=λex 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \cdots + \frac{x^{n}}{n!} = \lambda e^{x}

As n→∞n \to \infty the root is given by

xn=n+αn+β+o(1), x_{n} = n + \alpha \sqrt{n} + \beta + o(1),

where α\alpha and β\beta satisfy the equations:

12π∫α∞e−t22 dt=λ,β=α2+23. \frac{1}{\sqrt{2\pi }} \int _{\alpha }^{\infty } e^{-\frac{t^{2}}{2}} \, dt = \lambda , \qquad \beta = \frac{\alpha ^{2}+2}{3} .
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Remark.
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Problem 42 of Part V.

Problem 2.212

(Continuation of Problem 201). Let α\alpha denote a real constant. Then, for n→∞n \to \infty,

∫aξ+αnφ(x)[f(x)]n dx∼φ(ξ)[f(ξ)]n+1/21−nf′′(ξ)∫−∞αce−t22 dt \int _{a}^{\xi +\frac{\alpha }{\sqrt{n}}} \varphi (x) \left[f(x)\right]^{n} \, dx \sim \varphi (\xi ) \left[f(\xi )\right]^{n+1/2} \frac{1}{\sqrt{-nf''(\xi )}} \int _{-\infty }^{\alpha c} e^{-\frac{t^{2}}{2}} \, dt =φ(ξ)enh(ξ)1−nh′′(ξ)∫−∞αce−t22 dt,c=−f′′(ξ)f(ξ)=−h′′(ξ). = \varphi (\xi ) e^{nh(\xi )} \frac{1}{\sqrt{-nh''(\xi )}} \int _{-\infty }^{\alpha c} e^{-\frac{t^{2}}{2}} \, dt, \qquad c = \sqrt{-\frac{f''(\xi )}{f(\xi )}} = \sqrt{-h''(\xi )} .
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Problem 2.213

The functions φ(x)\varphi (x), h(x)h(x) and f(x)=eh(x)f(x)=e^{h(x)} are defined on the finite or infinite interval [a,b]\left[a,b\right] and satisfy the following conditions:

(1) φ(x)[f(x)]n=φ(x)enh(x)\varphi (x)\left[f(x)\right]^{n} = \varphi (x)e^{nh(x)} is absolutely integrable over [a,b]\left[a,b\right], n=0,1,2,…n=0,1,2,\ldots

(2) The value of the function h(x)h(x) at a point ξ\xi of (a,b)(a,b) is larger than its least upper bound in any closed interval to the left of ξ\xi which does not contain ξ\xi. Moreover there is a neighbourhood of ξ\xi in which h′′(x)h''(x) exists and is bounded. Finally h′(ξ)>0h'(\xi ) > 0.

(3) φ(x)\varphi (x) is continuous at x=ξx=\xi, φ(ξ)≠0\varphi (\xi ) \neq 0.

Prove, for n→∞n \to \infty, the asymptotic formula

∫aξ+αlog⁡nn+βnφ(x)[f(x)]n dx∼φ(ξ)h′(ξ)eβh′(ξ)⋅nαh′(ξ)−1⋅enh(ξ), \int _{a}^{\xi +\frac{\alpha \log n}{n}+\frac{\beta }{n}} \varphi (x) \left[f(x)\right]^{n} \, dx \sim \frac{\varphi (\xi )}{h'(\xi )} e^{\beta h'(\xi )} \cdot n^{\alpha h'(\xi )-1} \cdot e^{nh(\xi )},

where α\alpha and β\beta stand for real constants.

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Problem 2.214

Let ξ\xi denote the only real root of the transcendental equation e1+ξξ=1e^{1+\xi }\xi = 1. Then we have for n→∞n \to \infty

1n!∫0ξn+αlog⁡n+βexxn dx∼nAB, \frac{1}{n!} \int _{0}^{\xi n + \alpha \log n + \beta } e^{x} x^{n} \, dx \sim n^{A} B,

where α\alpha and β\beta are real constants and

A=α1+ξξ−12,B=12πξ1+ξeβ1+ξξ. A = \alpha \frac{1+\xi }{\xi } - \frac{1}{2}, \qquad B = \frac{1}{\sqrt{2\pi }} \frac{\xi }{1+\xi } e^{\beta \frac{1+\xi }{\xi }} .
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Problem 2.215

Suppose that nn is odd and let −xn-x_{n} denote the only real root of the equation

1+x1!+x22!+⋯+xnn!=0 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \cdots + \frac{x^{n}}{n!} = 0

As n→∞n \to \infty, xnx_{n} is asymptotically given by

xn=ξn+αlog⁡n+β+o(1), x_{n} = \xi n + \alpha \log n + \beta + o(1),

where ξ\xi is the only real root of the equation e1+ξξ=1e^{1+\xi }\xi = 1 and α\alpha and β\beta are given by

α=12ξ1+ξ,β=ξ1+ξlog⁡(2π1+ξξ). \alpha = \frac{1}{2}\frac{\xi }{1+\xi }, \qquad \beta = \frac{\xi }{1+\xi } \log \left(\sqrt{2\pi }\frac{1+\xi }{\xi }\right) .
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Remark.
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Problem 74 of Part V.

Problem 2.216

Assume that the function g(x)g(x) is monotone increasing for positive xx and that

lim⁡x→+∞g(x)=+∞,lim⁡x→+∞g(x)x=0. \lim _{x \to +\infty } g(x) = +\infty , \qquad \lim _{x \to +\infty } \frac{g(x)}{x} = 0 .

We define

an=1n!∫0∞e−x+g(x)xn dx. a_{n} = \frac{1}{n!} \int _{0}^{\infty } e^{-x+g(x)} x^{n} \, dx .

If there is a positive number γ\gamma such that

lim⁡x→+∞g(αx)g(x) \lim _{x \to +\infty } \frac{g(\alpha x)}{g(x)}

exists and is a continuous function of α\alpha for 1−γ≤α≤1+γ1-\gamma \leq \alpha \leq 1+\gamma then

lim⁡n→∞log⁡ang(n)=1. \lim _{n \to \infty } \frac{\log a_{n}}{g(n)} = 1 .
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Problem 2.217
lim⁡n→∞∫−ππn! 22ncos⁡ϑ∣(2neiϑ−1)(2neiϑ−2)(2neiϑ−3)⋯(2neiϑ−n)∣ dϑ=2π. \lim _{n \to \infty } \int _{-\pi }^{\pi } \frac{n! \, 2^{2n\cos \vartheta }}{\left|(2ne^{i\vartheta }-1)(2ne^{i\vartheta }-2)(2ne^{i\vartheta }-3)\cdots (2ne^{i\vartheta }-n)\right|} \, d\vartheta = 2\pi .
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Remark.
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Put ϑ=xn\vartheta = \dfrac {x}{\sqrt{n}} and recall Problems 59 and 115.

Problem 2.217.1

Analogy to Problem 201 suggests sufficient conditions under which for n→+∞n \to +\infty

∬Rφ(x,y)enh(x,y) dx dy∼φ(ξ,η)enh(ξ,η)2πnhxxhyy−hxy2 \iint _{\mathfrak {R}} \varphi (x,y) e^{nh(x,y)} \, dx \, dy \sim \varphi (\xi ,\eta ) e^{nh(\xi ,\eta )} \frac{2\pi }{n\sqrt{h_{xx}h_{yy}-h_{xy}^{2}}}

where the partial derivatives of second order hxx,hyyh_{xx}, h_{yy}, and hxyh_{xy} are taken at the point (ξ,η)(\xi ,\eta ). Give a full statement and a proof.

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Problem 2.217.1

Analogy to Problem 201 suggests sufficient conditions under which for n→+∞n \to +\infty

∬Rφ(x,y)enh(x,y) dx dy∼φ(ξ,η)enh(ξ,η)2πnhxxhyy−hxy2 \iint _{\mathfrak {R}} \varphi (x,y) e^{nh(x,y)} \, dx \, dy \sim \varphi (\xi ,\eta ) e^{nh(\xi ,\eta )} \frac{2\pi }{n\sqrt{h_{xx}h_{yy}-h_{xy}^{2}}}

where the partial derivatives of second order hxx,hyyh_{xx}, h_{yy}, and hxyh_{xy} are taken at the point (ξ,η)(\xi ,\eta ). Give a full statement and a proof.

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Problem 2.218

The function

x(x−1)(x−2)⋯(x−n) a−x, \sqrt{x(x-1)(x-2)\cdots (x-n)} \, a^{-x},

where a>1a > 1, has the maximum MnM_{n} on the interval (n,+∞)(n, +\infty ). It can be approximated by

Mnn!∼12π1(a−1)n+1/2. \frac{M_{n}}{n!} \sim \frac{1}{\sqrt{2\pi }} \frac{1}{(a-1)^{n+1/2}} .
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Remark.
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Problem 16.

Problem 2.219

The function

x(x2−12)(x2−22)⋯(x2−n2) a−x, x(x^{2}-1^{2})(x^{2}-2^{2})\cdots (x^{2}-n^{2}) \, a^{-x},

where a>1a > 1, has the maximum MnM_{n} on the interval (n,+∞)(n, +\infty ). It can be approximated by

Mnn!2∼12π(2aa−1)2n+1. \frac{M_{n}}{n!^{2}} \sim \frac{1}{2\pi } \left(\frac{2\sqrt{a}}{a-1}\right)^{2n+1} .
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Remark.
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Problem 17.

Problem 2.220

We define x=Q0(x)\sqrt{x} = Q_{0}(x),

x(1−x1)(1−x2)⋯(1−xn)=Qn(x),n=1,2,3,… \sqrt{x}\left(1-\frac{x}{1}\right)\left(1-\frac{x}{2}\right)\cdots \left(1-\frac{x}{n}\right) = Q_{n}(x), \qquad n=1,2,3,\ldots

The sequence of functions

Q1(x)a−x,Q2(x)a−x,…,Qn(x)a−x,… Q_{1}(x)a^{-x}, \quad Q_{2}(x)a^{-x}, \quad \ldots , \quad Q_{n}(x)a^{-x}, \quad \ldots

is uniformly bounded for x>0x>0 if a≥2a \geq 2; it is not uniformly bounded if 0<a<20 < a < 2.

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Problem 2.221

We define x=P0(x)x = P_{0}(x),

x(1−x21)(1−x24)(1−x29)⋯(1−x2n2)=Pn(x),n=1,2,3,… x\left(1-\frac{x^{2}}{1}\right)\left(1-\frac{x^{2}}{4}\right)\left(1-\frac{x^{2}}{9}\right)\cdots \left(1-\frac{x^{2}}{n^{2}}\right) = P_{n}(x), \qquad n=1,2,3,\ldots

The sequence of functions

P1(x)a−x,P2(x)a−x,…,Pn(x)a−x,… P_{1}(x)a^{-x}, \quad P_{2}(x)a^{-x}, \quad \ldots , \quad P_{n}(x)a^{-x}, \quad \ldots

is uniformly bounded for x>0x>0 if a≥3+8a \geq 3+\sqrt{8}; it is not uniformly bounded if 0<a<3+80 < a < 3+\sqrt{8}.

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Problem 2.222

Assume that a>0a>0, 0<μ<10<\mu <1 and that MnM_{n} is the maximum of e−(x+axμ)xne^{-(x+ax^{\mu })} x^{n} in the interval (0,+∞)(0,+\infty ). We find

lim⁡n→∞(Mnn!)n−μ=e−a. \lim _{n \to \infty } \left(\frac{M_{n}}{n!}\right)^{n^{-\mu }} = e^{-a} .
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Problem 2.223

The function f(x,y)f(x,y) is continuous in the rectangle

a≤x≤a′,b≤y≤b′. a \leq x \leq a', \qquad b \leq y \leq b' .

Then its maximum for a given xx and b≤y≤b′b \leq y \leq b' (along a segment parallel to the yy-axis)

max⁡yf(x,y)=φ(x) \max _{y} f(x,y) = \varphi (x)

is a continuous function of xx.

We interpret the surface

z=f(x,y) z = f(x,y)

in a rectangular coordinate system x,y,zx,y,z with vertical zz-axis as a topographical surface in a mountainous region. Then the curve

z=φ(x) z = \varphi (x)

(in the x,zx,z-plane) is the skyline of the range (as it appears when seen from a faraway point of the yy-axis) and the minimax

min⁡xφ(x)=min⁡xmax⁡yf(x,y) \min _{x} \varphi (x) = \min _{x} \max _{y} f(x,y)

(the minimum of the maxima) refers to the lowest point in the skyline.

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Problem 2.224

(continued). Show that

max⁡ymin⁡xf(x,y)≤min⁡xmax⁡yf(x,y). \max _{y} \min _{x} f(x,y) \leq \min _{x} \max _{y} f(x,y) .
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Problem 2.225

(continued). Which one of the two signs, << and ==, is valid in the example

f(x,y)=1−(x−y+1)2, f(x,y) = 1 - (x-y+1)^{2},

a=0a=0, a′=2a'=2, b=0b=0, b′=4b'=4?

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Problem 2.226

Add to the assumptions of Problem 223 that f(x,y)>0f(x,y) > 0. Then

lim⁡n→∞[∫aa′(∫bb′[f(x,y)]n dy)−1dx]−1n=min⁡xmax⁡yf(x,y). \lim _{n \to \infty } \left[\int _{a}^{a'} \left(\int _{b}^{b'} \left[f(x,y)\right]^{n} \, dy\right)^{-1} dx\right]^{-\frac{1}{n}} = \min _{x} \max _{y} f(x,y) .
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