József Wildt International Mathematical Competition

简介

József Wildt International Mathematical Competition（约瑟夫·维尔德国际数学竞赛）是一个面向高中生的国际数学竞赛，以纪念匈牙利数学家约瑟夫·维尔德（József Wildt）。它通常在罗马尼亚特尔古穆列什举办，旨在鼓励学生对数学的兴趣并挑战他们的解决问题的能力。

各届问题及其解答

2019 年

W1. The Pell numbers $P_n$ satisfy $P_0=0$, $P_1=1$, and $P_n=2P_{n-1}+P_{n-2}$ for $n\ge2$. Find

$$\sum_{n=1}^{\infty}\left(\arctan\frac1{P_{2n}}+\arctan\frac1{P_{2n+2}}\right)\arctan\frac2{P_{2n+1}}.$$

提示

依次证明:
(1). $P_{n}P_{n+2}+\left(-1\right)^{n}=P_{n+1}^{2}$;
(2). $\arctan\frac{1}{P_{2n}}-\arctan\frac{1}{P_{2n+2}}=\arctan\frac{2}{P_{2n+1}}$;
(3). $\sum\limits_{n=1}^{\infty}\left(\arctan\frac{1}{P_{2n}}+\arctan\frac{1}{P_{2n+2}}\right)\arctan\frac{2}{P_{2n+1}}=\arctan^{2}\frac{1}{2}$.

W2. If $0<a\le c\le b$ then:

$$\frac{\left(b^{30}-a^{30}\right)\left(b^{30}-c^{30}\right)}{36b^{10}}\le\frac{\left(b^{25}-a^{25}\right)\left(b^{25}-c^{25}\right)}{25}\le\frac{\left(b^{30}-a^{30}\right)\left(b^{30}-c^{30}\right)}{36\left(ac\right)^{5}}$$

提示

考虑

$$\int_{a}^{b}\int_{c}^{b}t^{24}s^{24}\,ds\,dt\le\frac{1}{(ac)^{5}}\int_{a}^{b}\int_{c}^{b}t^{29}s^{29}\,ds\,dt$$

和

$$\frac{1}{b^{10}}\int_{a}^{b}\int_{c}^{b}t^{29}s^{29}\,ds\,dt\le\int_{a}^{b}\int_{c}^{b}t^{24}s^{24}\,ds\,dt.$$

W3. Compute

$$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\cos x+1-x^{2}}{\left(1+x\sin x\right)\sqrt{1-x^{2}}}dx.$$

提示

$$\begin{align*} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\cos x+1-x^{2}}{\left(1+x\sin x\right)\sqrt{1-x^{2}}}dx & =2\int_{0}^{\frac{\pi}{4}}\frac{\cos x+1-x^{2}}{\left(1+x\sin x\right)\sqrt{1-x^{2}}}dx\\ & =\left.-4\arctan\left(\frac{\cos\left(\frac{\arcsin x-x}{2}\right)}{\sin\left(\frac{\arcsin x+x}{2}\right)}\right)\right|_{0}^{\frac{\pi}{4}}\\ & =2\pi-4\arctan\left(\frac{\cos\left(\frac{\pi-4\arcsin\frac{\pi}{4}}{8}\right)}{\sin\left(\frac{\pi+4\arcsin\frac{\pi}{4}}{8}\right)}\right) \end{align*}$$

W4. If $x,y,z,t>1$ then:

$$\left(\log_{zxt}x\right)^{2}+\left(\log_{xyt}y\right)^{2}+\left(\log_{xyz}z\right)^{2}+\left(\log_{yzt}t\right)^{2}>\frac{1}{4}.$$

提示

$$\begin{align*} \sum_{cyc}\left(\frac{a}{a+c+d}\right)^{2} & >\frac{1}{4}\\ \sum_{cyc}1\sum_{cyc}\left(\frac{a}{a+c+d}\right)^{2} & \ge\left(\sum_{cyc}\frac{a}{a+c+d}\right)^{2}>\left(\sum_{cyc}\frac{a}{a+b+c+d}\right)^{2}=1. \end{align*}$$

W5. Let $n\ge 1$. Find a set of distincts real numbers $(x_j)_{1\le j\le n}$ such that for any bijections $f:\{1,2,\cdots, n\}^2\to\{1,2,\cdots,n\}^2$ the matrix $\left(x_{f(i,j)}\right)_{1\le i,j\le n}$ is invertible.

W6. Compute

$$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{(1+\ln x)\cos x+x\sin x\ln x}{\cos^{2}x+x^{2}\ln^{2}x}\ud x$$

提示

$$\begin{align*} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{(1+\ln x)\cos x+x\sin x\ln x}{\cos^{2}x+x^{2}\ln^{2}x}dx & =\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{(x\ln x)^{\prime}\cos x+\sin x\cdot x\ln x}{\cos^{2}x+x^{2}\ln^{2}x}dx\\ & =\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{1}{1+\left(\frac{x\ln x}{\cos x}\right)^{2}}d\left(\frac{x\ln x}{\cos x}\right)\\ & =\arctan\left(\frac{x\ln x}{\cos x}\right)|_{\pi/6}^{\pi/4}\\ & =\arctan\left(\frac{\sqrt{2}\pi\ln\frac{\pi}{4}}{4}\right)-\arctan\left(\frac{\sqrt{3}\pi\ln\frac{\pi}{6}}{9}\right) \end{align*}$$

W7. If

$$\Omega_n=\sum_{k=1}^n\left(\int_{-\frac{1}{k}}^{\frac{1}{k}}\left(2 x^{10}+3 x^8+1\right) \cdot \cos ^{-1}(k x) d x\right)$$

then find:

$$\Omega=\lim _{n \rightarrow \infty}\left(\Omega_n-\pi \cdot H_n\right)$$

提示

分部积分得到:

$$\int_{-\frac{1}{k}}^{\frac{1}{k}}x^{2n}\arccos\left(kx\right)\ud x=\frac{\pi}{(2n+1)k^{2n+1}},$$

代入,

$$\Omega_{n}=\sum_{k=1}^{n}\left(\int_{-\frac{1}{k}}^{\frac{1}{k}}\left(2x^{10}+3x^{8}+1\right)\cdot\arccos(kx)dx\right)=\sum_{k=1}^{n}\left(\frac{2\pi}{11k^{11}}+\frac{\pi}{3k^{9}}+\frac{\pi}{k}\right),$$

所以

$$\Omega=\frac{2\pi}{11}\zeta(11)+\frac{\pi}{3}\zeta(9).$$

W8. Let $(a_n)_{n \geq 1}$ be a positive real sequence given by $a_n=\sum_{k=1}^n \frac{1}{k}$. Compute

$$\lim _{n \to \infty} e^{-2 a_n} \sum_{k=1}^n\left[(\sqrt[2 k]{k!}+\sqrt[2(k+1)]{(k+1)!})^2\right]$$

where we denote by $[x]$ the integer part of $x$.

提示

由 Stirling 公式,

$$n!=\sqrt{2\pi n}\left(\frac{n}{\ue}\right)^{n}\ue^{O\left(\frac{1}{n}\right)},$$

所以

$$\sqrt[n]{n!}=\frac{n}{\ue}\cdot\ue^{\frac{\ln\left(2\pi n\right)}{2n}+O\left(\frac{1}{n^{2}}\right)}=\frac{n}{\ue}\left(1+\frac{\ln\left(2\pi n\right)}{2n}+o\left(\frac{1}{n}\right)\right),$$

又由于不等式

$$\sqrt[k]{k!}<\sqrt[k+1]{(k+1)!}\Longrightarrow\sqrt[2k]{k!}<\sqrt[2(k+1)]{(k+1)!},$$

所以有自然的不等式

$$4\sqrt[k]{k!}<\left[\left(\sqrt[2k]{k!}+\sqrt[2(k+1)]{(k+1)!}\right)^{2}\right]<4\sqrt[k+1]{(k+1)!}.$$

所以所求极限化为

$$\lim_{n\to\infty}4\ue^{-2H_{n}}\sum_{k=1}^{n}\sqrt[k]{k!}<\lim_{n\to\infty}\ue^{-2H_{n}}\sum_{k=1}^{n}\left[\left(\sqrt[2k]{k!}+\sqrt[2(k+1)]{(k+1)!}\right)^{2}\right]<\lim_{n\to\infty}4\ue^{-2H_{n}}\sum_{k=1}^{n}\sqrt[k+1]{(k+1)!}.$$

由于 $\ue^{-2H_{n}}=\ue^{-2(H_{n}-\ln n)-2\ln n}\to\ue^{-2\gamma}\cdot n^{-2}$,
所以

$$\begin{align*} \lim_{n\to\infty}4\ue^{-2H_{n}}\sum_{k=1}^{n}\sqrt[k]{k!} & =\lim_{n\to\infty}\frac{4}{\ue^{2\gamma}}\cdot\frac{1}{n}\sum_{k=1}^{n}\frac{\sqrt[k]{k!}}{n}\\ & =\lim_{n\to\infty}\frac{4}{\ue^{2\gamma}}\cdot\frac{1}{n}\sum_{k=1}^{n}\frac{\frac{k}{\ue}\left(1+\frac{\ln\left(2\pi k\right)}{2k}+o\left(\frac{1}{k}\right)\right)}{n}\\ & =\frac{4}{\ue^{2\gamma}}\cdot\int_{0}^{1}\frac{x}{\ue}\ud x+\lim_{n\to\infty}\frac{4}{\ue^{2\gamma}}\cdot\frac{1}{n}\sum_{k=1}^{n}\left(\frac{\ln\left(2\pi k\right)}{2kn}+o\left(\frac{1}{kn}\right)\right)\\ & =\frac{2}{\ue^{2\gamma+1}}. \end{align*}$$

同理可证:

$$\lim_{n\to\infty}4\ue^{-2H_{n}}\sum_{k=1}^{n}\sqrt[k+1]{(k+1)!}=\frac{2}{\ue^{2\gamma+1}}.$$

所以

$$\lim_{n\to\infty}\ue^{-2H_{n}}\sum_{k=1}^{n}\left[\left(\sqrt[2k]{k!}+\sqrt[2(k+1)]{(k+1)!}\right)^{2}\right]=\frac{2}{\ue^{2\gamma+1}}.$$