许以超习题

egin{example}
设域$F$上多项式$f(x)$被$x-1,x-2,x-3$除后,余式分别为$4,8,16$.试求$f(x)$被$(x-1)(x-2)(x-3)$除后的余式.
end{example}
egin{solution}
$12$.
end{solution}

egin{example}
给定正整数$n$,试证:存在正整数$m$,使得域$mathbb{F}$上多项式
[
left( 1+x ight) left( 1+x^2 ight) cdots left( 1+x^{2^n} ight) =1+x+x^2+cdots +x^m.
]
end{example}
egin{solution}
注意到
egin{align*}
left( 1+x ight) left( 1+x^2 ight) cdots left( 1+x^{2^n} ight) &=frac{left( 1-x ight) left( 1+x ight) left( 1+x^2 ight) cdots left( 1+x^{2^n} ight)}{1-x}
\
&=frac{left( 1-x^2 ight) left( 1+x^2 ight) cdots left( 1+x^{2^n} ight)}{1-x}=cdots =frac{1-x^{2^{n+1}}}{1-x}
\
&=1+x+x^2+cdots +x^{2^{n+1}-1},
end{align*}
即$m=2^{n+1}-1$.
end{solution}

egin{example}
试计算复多项式
[
x^n+left( a+b ight) x^{n-1}+left( a^2+ab+b^2 ight) x^{n-2}+cdots +left( a^n+a^{n-1}b+cdots +ab^{n-1}+b^n ight)
]
的根的方幂和$S_1,S_2,cdots,S_n$.
end{example}
egin{solution}
等价于$x^n+frac{a^2-b^2}{a-b}x^{n-1}+frac{a^3-b^3}{a-b}x^{n-2}+cdots +frac{a^n-b^n}{a-b}$.
end{solution}

egin{example}
对任意$x_jin (0,1/2],j=1,2,cdots,n$和正整数$n$,证明不等式
[
frac{prod_{j=1}^n{x_j}}{left( sum_{j=1}^n{x_j} ight) ^n}le frac{prod_{j=1}^n{left( 1-x_j ight)}}{left( sum_{j=1}^n{left( 1-x_j ight)} ight) ^n}.]
end{example}
egin{solution}
等价于证明
[
frac{prod_{j=1}^n{x_j}}{left( sum_{j=1}^n{x_j} ight) ^n}le frac{prod_{j=1}^n{left( 1-x_j ight)}}{left( n-sum_{j=1}^n{x_j} ight) ^n}Leftrightarrow frac{left( n-sum_{j=1}^n{x_j} ight) ^n}{left( sum_{j=1}^n{x_j} ight) ^n}le frac{prod_{j=1}^n{left( 1-x_j ight)}}{prod_{j=1}^n{x_j}},
]
即
[
left( frac{n}{sum_{j=1}^n{x_j}}-1 ight) ^nle prod_{j=1}^n{left( frac{1}{x_j}-1 ight)}Leftrightarrow ln left( frac{n}{sum_{j=1}^n{x_j}}-1 ight) le frac{sum_{j=1}^n{ln left( frac{1}{x_j}-1 ight)}}{n}.
]

考虑辅助函数$fleft( x ight) =ln left( frac{1}{x}-1 ight)$,由Jensen不等式即可.
end{solution}

egin{example}
计算
[
sum_{j=1}^{n-1}{frac{1}{1-exp left{ frac{2pi ij}{n} ight}}}.
]
end{example}
egin{solution}
(2011年清华金秋营)设$varepsilon_n=e^{frac{2pi i}{n}}$,试求: $sum_{k=0}^{n-1}frac{1}{1-varepsilon_n^kt},
sum_{k=1}^{n-1}frac{1}{1-varepsilon_n^k},
sum_{k=1}^{n-1}frac{1}{(1-varepsilon_n^k)(1-varepsilon_n^{-k})}$.

记$varepsilon_n=e^{frac{2pi i}{n}}$,则$1,varepsilon_n,varepsilon_n^2,cdots,varepsilon_n^{n-1}$为$x^n=1$的$n$个根.

由韦达定理可知
[sum_{i=1}^{n}prod varepsilon_n^{k_1}cdots varepsilon_n^{k_i}=0,quad 1leq i<n.]
由$
x^n-1=left( x-1 ight) left( x-varepsilon _n ight) left( x-varepsilon _{n}^{2} ight) cdots left( x-varepsilon _{n}^{n-1} ight)
$可得
[
frac{1}{x^n}-1=left( frac{1}{x}-1 ight) left( frac{1}{x}-varepsilon _n ight) left( frac{1}{x}-varepsilon _{n}^{2} ight) cdots left( frac{1}{x}-varepsilon _{n}^{n-1} ight),
]
于是
[
1-x^n=left( 1-x ight) left( 1-varepsilon _n x ight) left( 1-varepsilon _{n}^{2}x^2 ight) cdots left( 1-varepsilon _{n}^{n-1}x^{n-1} ight).
]
因此
egin{align*}
sum_{k=1}^{n-1}{frac{1}{1-varepsilon _{n}^{k}t}} &=frac{n}{1-t^n}-frac{1}{1-t}=frac{n-left( 1+t+t^2+cdots +t^{n-1} ight)}{1-t^n}
\
&=frac{left( 1-t ight) +left( 1-t^2 ight) +cdots +left( 1-t^{n-1} ight)}{1-t^n}
\
&=frac{left( n-1 ight) +left( n-2 ight) t+left( n-3 ight) t^2+cdots +t^{n-2}}{1+t+t^2+cdots +t^{n-1}}.
end{align*}

令$t=1$,有
[
sum_{k=1}^{n-1}{frac{1}{1-varepsilon _{n}^{k}}}=frac{left( n-1 ight) +left( n-2 ight) +cdots +1}{n}=frac{n-1}{2}.
]

而
egin{align*}
sum_{k=1}^{n-1}{frac{1}{left( 1-varepsilon _{n}^{k} ight) left( 1-varepsilon _{n}^{-k} ight)}} &=sum_{k=1}^{n-1}{frac{1}{left( 1-cos frac{2kpi}{n}-isin frac{2kpi}{n} ight) left( 1-cos frac{2kpi}{n}+isin frac{2kpi}{n} ight)}}
\
&=sum_{k=1}^{n-1}{frac{1}{left( 1-cos frac{2kpi}{n} ight) ^2+sin ^2frac{2kpi}{n}}}=sum_{k=1}^{n-1}{frac{1}{2-2cos frac{2kpi}{n}}}
\
&=sum_{k=1}^{n-1}{frac{1}{4sin ^2frac{kpi}{n}}}=frac{n-1}{4}+frac{1}{4}sum_{k=1}^{n-1}{cot ^2frac{kpi}{n}},
end{align*}

又因为
[
left( cos frac{kpi}{n}+isin frac{kpi}{n} ight) ^n=left( -1 ight) ^k=sum_{j=0}^n{C_{n}^{j}left( cos frac{kpi}{n} ight) ^{n-j}left( isin frac{kpi}{n} ight) ^j},
]
当$n=2m$时,
egin{align*}
&sum_{j=1}^m{C_{2m}^{2j-1}left( cos frac{kpi}{n} ight) ^{2m-2j+1}left( isin frac{kpi}{n} ight) ^{2j-1}}=0,
\
&sum_{j=1}^m{C_{2m}^{2j-1}left( cot frac{kpi}{n} ight) ^{2m-2j}left( -1 ight) ^j}=0,
end{align*}
所以$cot ^2frac{pi}{n},cot ^2frac{2pi}{n},cdots,cot ^2frac{left( m-1 ight) pi}{n}$为多项式$sum_{j=1}^m{C_{2m}^{2j-1}x^{m-j}left( -1 ight) ^j}=0$的根,因此
[
sum_{k=1}^{m-1}{cot ^2frac{kpi}{n}}=frac{C_{2m}^{3}}{C_{2m}^{1}}=frac{left( m-1 ight) left( 2m-1 ight)}{3},
]
故
[
sum_{k=1}^{n-1}{cot ^2frac{kpi}{n}}=frac{left( 2m-2 ight) left( 2m-1 ight)}{3}=frac{left( n-2 ight) left( n-1 ight)}{3},
]
当$n=2m+1$时,类似可得
[
sum_{k=1}^{n-1}{cot ^2frac{kpi}{n}}=2frac{C_{2m+1}^{3}}{C_{2m+1}^{1}}=frac{left( 2m-1 ight) 2m}{3}=frac{left( n-2 ight) left( n-1 ight)}{3},
]
因此
egin{align*}
sum_{k=1}^{n-1}{frac{1}{left( 1-varepsilon _{n}^{k} ight) left( 1-varepsilon _{n}^{-k} ight)}} &=frac{n-1}{4}+frac{1}{4}sum_{k=1}^{n-1}{cot ^2frac{kpi}{n}}
\
&=frac{n-1}{4}+frac{1}{4}cdot frac{left( n-2 ight) left( n-1 ight)}{3}=frac{n^2-1}{12}.
end{align*}
end{solution}