好题合辑

Posted 2022-05-05 eufisky

1、AOPS论坛

2、匈牙利语版本

3、英文版本

4、其它试题

Suppose that $f: \mathbbR^+ \to \mathbbR^+$ is a continuous function such that for all positive real numbers $x,y$ the following is true :

$$(f(x)-f(y)) \left ( f \left ( \fracx+y2 \right ) - f ( \sqrtxy ) \right )=0.$$

Is it true that the only solution to this is the constant function ?

---

陶哲轩解答：Yes.  If $f$ were not constant, then (since $\bf R^+$ is connected) it could not be locally constant, thus there exists $x_0 \in \bf R^+$ such that $f$ is not constant in any neighbourhood of $x_0$.  By rescaling (replacing $f(x)$ with $f(x_0 x)$) we may assume without loss of generality that $x_0=1$.

 

For any $y \in \bf R^+$, there thus exists $x$ arbitrarily close to $1$ for which $f(x) \neq f(y)$, hence $f((x+y)/2) = f(\sqrtxy)$.  By continuity, this implies that $f((1+y)/2) = f(\sqrty)$ for all $y \in \bf R^+$.  Making the substitution $z := (1+y)/2$, we conclude that $f(z) = f(g(z))$ for all $z \in \bf R^+$, where $g(z) := \sqrt2z-1$.  The function $g$ has the fixed point $z=1$ as an attractor, so on iteration and by using the continuity of $f$ we conclude that $f(z)=f(1)$ for all $z \in \bf R^+$, so $f$ is indeed constant.

来源：这里

---

(1950年)令$a>0,d>0$,设$$ f(x)=\frac1a+\fracxa(a+d)+\cdots+\fracx^na(a+d)\cdots(a+nd)+\cdots$$给出$f(x)$的封闭解.

---

也就是求\[\sum\limits_n = 0^\infty  \fracx^n\prod\limits_k = 0^n \left( a + kd \right) .\]

解.首先有$$\prod_k=0^n\frac1a+kd=\frac\Gamma \left( \fracad \right)d^n+1\Gamma \left( \fracad+n+1 \right),$$

又因为$$\gamma \left( s,x \right) =\sum_k=0^\infty\fracx^se^-xx^ks\left( s+1 \right) ...\left( s+k \right)=x^s\,\Gamma \left( s \right) \,e^-x\sum_k=0^\infty\fracx^k\Gamma \left( s+k+1 \right),$$我们有

\beginalign*\sum_n=0^\infty\fracx^n\prod\limits_k=0^n\left( a+kd \right)&=\frac\Gamma \left( \fracad \right)d\sum_n=0^\infty\frac\left( x/d \right) ^n\Gamma \left( \fracad+n+1 \right)\\&=\frac\Gamma \left( \fracad \right)d\gamma \left( \fracad,\fracxd \right) \left( \fracdx \right) ^a/d\frace^x/d\Gamma \left( \fracad \right)=\left( \fracdx \right) ^a/d\frace^x/dd\gamma \left( \fracad,\fracxd \right) ,\endalign*

其中$\displaystyle\Gamma(s,x) = \int_x^\infty t^s-1\,e^-t\,\rm dt$为the upper incomplete gamma function,而$\displaystyle\gamma(s,x) = \int_0^x t^s-1\,e^-t\,\rm dt$为the lower incomplete gamma function.参考这里.

---

$g(x) = x^a f(x^d)$ satifies $g‘(x) = x^a-1 + x^d-1 g(x)$. Solve the associated differential equation and conclude.

---

令$a\in (0,\pi)$,设$n$为正整数.证明$$\int_0^\pi\frac\cos \left( nx \right) -\cos \left( na \right)\cos x-\cos adx=\pi \frac\sin \left( na \right)\sin a.$$

求$$\int_1^\frac\sqrt5+12\left( \frac\arctan x\arctan x-x \right) ^2dx,$$

$$\int_0^1\frac\arctan xx\sqrt1-x^2dx.$$

---

Let $n$ be a positive integer. Prove that, for $0<x<\frac\pin+1$,

$$\sinx-\frac\sin2x2+\cdots+(-1)^n+1\frac\sinnxn-\fracx2$$

is positive if $n$ is odd and negative if $n$ is even.

---

Since

\beginalign*f_n(x) &= \sinx - \frac \sin2x2 + \cdots + ( - 1)^n + 1\frac \sinnxn - \frac x2,\\f_n‘(x) &= - \mboxRe\left(\sum_n = 1^nz^n\right) - \frac12.\endalign*

 

After some simplifications we get

$$ f_n‘(x) = \frac ( - 1)^n + 12((1 - \cos(x))\frac \sin((n + 1)x)\sin(x) + \cos((n + 1)x))$$

and $$ f_n‘‘(x) = \frac ( - 1)^n2\frac (n + 1)\sin(nx) + n\sin((n + 1)x)1 + \cos(x).$$

The formula for $ f_n‘‘$ shows that $ ( - 1)^n f$ is convex for $ 0 < x < \frac \pin + 1$. Since $ f_n(0) = 0$ and $ f_n‘(0) = \frac ( - 1)^n + 12$.We are ready when we can show that $ ( - 1)^n + 1f_n(\frac \pin + 1) > 0$.

 

We have to distinct between two different, but very similar cases, namely $ n$ is odd, and $ n$ is even.

Let‘s restrict to the case $ n$ is even.

We prove $ f_2n(\frac \pi2n + 1) < 0$.

 

\beginalign*f_2n\left( \frac\pi2n+1 \right) &=\sum_k=1^2n\left( -1 \right)^k+1\frac\sin \left( \frack\pi2n+1 \right)k-\frac\pi2\left( 2n+1 \right)\\&=\frac\pi2n+1\left( \sum_k=1^n\frac\sin \left( \frac\left( 2k-1 \right) \pi2n+1 \right)\frac\left( 2k-1 \right) \pi2n+1-\sum_k=1^n\frac\sin \left( \frac2k\pi2n+1 \right)\frac2k\pi2n+1 \right) -\frac\pi2\left( 2n+1 \right).\endalign*

 

The function $ x \mapsto \frac \sin(x)x$ is descending on $ [0,\pi]$, thus

both sums lay between $ a$ and $ a + \frac 2\pi2n + 1$, where $ a = \int_0^\pi\frac \sin(x)x\,dx$.

Thus $$ f_2n\left(\frac \pi2n + 1\right) < \frac \pi2n + 1\cdot\frac 2\pi2n + 1 - \frac \pi2(2n + 1) < 0.$$

 \[\int e^x\frac1+\sin x1+\cos xdx,\quad \int_\frac12^2\left(1+x-\frac1x\right) e^x+\frac1x d x\]

日本高考题