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Suppose that $f: mathbb{R}^+ o mathbb{R}^+$ is a continuous function such that for all positive real numbers $x,y$ the following is true :
$$(f(x)-f(y)) left ( f left ( frac{x+y}{2}
ight ) - f ( sqrt{xy} )
ight )=0.$$
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 ${f R}^+$ is connected) it could not be locally constant, thus there exists $x_0 in {f R}^+$ 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 $y in {f R}^+$, there thus exists $x$ arbitrarily close to $1$ for which $f(x)
eq f(y)$, hence $f((x+y)/2) = f(sqrt{xy})$. By continuity, this implies that $f((1+y)/2) = f(sqrt{y})$ for all $y in {f R}^+$. Making the substitution $z := (1+y)/2$, we conclude that $f(z) = f(g(z))$ for all $z in {f R}^+$, where $g(z) := sqrt{2z-1}$. 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 $z in {f R}^+$, so $f$ is indeed constant.
来源：这里
(1950年)令$a>0,d>0$,设$$ f(x)=frac{1}{a}+frac{x}{a(a+d)}+cdots+frac{x^n}{a(a+d)cdots(a+nd)}+cdots$$给出$f(x)$的封闭解.
也就是求[sumlimits_{n = 0}^infty {frac{{{x^n}}}{{prodlimits_{k = 0}^n {left( {a + kd} ight)} }}} .]
解.首先有$$prod_{k=0}^n{frac{1}{a+kd}}=frac{Gamma left( frac{a}{d} ight)}{d^{n+1}Gamma left( frac{a}{d}+n+1 ight)},$$
又因为$$gamma left( s,x ight) =sum_{k=0}^{infty}{frac{x^se^{-x}x^k}{sleft( s+1 ight) ...left( s+k ight)}}=x^s\,Gamma left( s ight) \,e^{-x}sum_{k=0}^{infty}{frac{x^k}{Gamma left( s+k+1 ight)}},$$我们有
egin{align*}sum_{n=0}^{infty}{frac{x^n}{prodlimits_{k=0}^n{left( a+kd
ight)}}}&=frac{Gamma left( frac{a}{d}
ight)}{d}sum_{n=0}^{infty}{frac{left( x/d
ight) ^n}{Gamma left( frac{a}{d}+n+1
ight)}}\&=frac{Gamma left( frac{a}{d}
ight)}{d}gamma left( frac{a}{d},frac{x}{d}
ight) left( frac{d}{x}
ight) ^{a/d}frac{e^{x/d}}{Gamma left( frac{a}{d}
ight)}=left( frac{d}{x}
ight) ^{a/d}frac{e^{x/d}}{d}gamma left( frac{a}{d},frac{x}{d}
ight) ,end{align*}
其中$displaystyleGamma(s,x) = int_x^{infty} t^{s-1}\,e^{-t}\,{
m d}t$为the upper incomplete gamma function,而$displaystylegamma(s,x) = int_0^x t^{s-1}\,e^{-t}\,{
m d}t$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$ain (0,pi)$,设$n$为正整数.证明$$int_0^{pi}{frac{cos left( nx ight) -cos left( na ight)}{cos x-cos a}dx}=pi frac{sin left( na ight)}{sin a}.$$
求$$int_1^{frac{sqrt{5}+1}{2}}{left( frac{arctan x}{arctan x-x}
ight) ^2dx},$$
$$int_0^1{frac{arctan x}{xsqrt{1-x^2}}dx}.$$
Let $n$ be a positive integer. Prove that, for $0<x<fracpi{n+1}$,
$$sin{x}-frac{sin{2x}}{2}+cdots+(-1)^{n+1}frac{sin{nx}}{n}-frac{x}{2}$$
is positive if $n$ is odd and negative if $n$ is even.
Since
egin{align*}f_n(x) &= sin{x} - frac {sin{2x}}{2} + cdots + ( - 1)^{n + 1}frac {sin{nx}}{n} - frac {x}{2},\f_n'(x) &= - mbox{Re}left(sum_{n = 1}^{n}z^n
ight) - frac12.end{align*}
After some simplifications we get
$$ f_n'(x) = frac {( - 1)^{n + 1}}{2}((1 - cos(x))frac {sin((n + 1)x)}{sin(x)} + cos((n + 1)x))$$
and $$ f_n''(x) = frac {( - 1)^{n}}{2}frac {(n + 1)sin(nx) + nsin((n + 1)x)}{1 + cos(x)}.$$
The formula for $ f_n''$ shows that $ ( - 1)^n f$ is convex for $ 0 < x < frac {pi}{n + 1}$. Since $ f_n(0) = 0$ and $ f_n'(0) = frac {( - 1)^{n + 1}}{2}$.We are ready when we can show that $ ( - 1)^{n + 1}f_n(frac {pi}{n + 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 {pi}{2n + 1}) < 0$.
egin{align*}f_{2n}left( frac{pi}{2n+1}
ight) &=sum_{k=1}^{2n}{left( -1
ight)}^{k+1}frac{sin left( frac{kpi}{2n+1}
ight)}{k}-frac{pi}{2left( 2n+1
ight)}\&=frac{pi}{2n+1}left( sum_{k=1}^n{frac{sin left( frac{left( 2k-1
ight) pi}{2n+1}
ight)}{frac{left( 2k-1
ight) pi}{2n+1}}}-sum_{k=1}^n{frac{sin left( frac{2kpi}{2n+1}
ight)}{frac{2kpi}{2n+1}}}
ight) -frac{pi}{2left( 2n+1
ight)}.end{align*}
The function $ x mapsto frac {sin(x)}{x}$ is descending on $ [0,pi]$, thus
both sums lay between $ a$ and $ a + frac {2pi}{2n + 1}$, where $ a = int_0^{pi}frac {sin(x)}{x}\,dx$.
Thus $$ f_{2n}left(frac {pi}{2n + 1} ight) < frac {pi}{2n + 1}cdotfrac {2pi}{2n + 1} - frac {pi}{2(2n + 1)} < 0.$$
[int e^xfrac{1+sin x}{1+cos x}dx,quad int_{frac{1}{2}}^{2}left(1+x-frac{1}{x}
ight) e^{x+frac{1}{x}} d x]