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[函数] 一道竞赛题目

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theonly Posted 2020-3-30 07:42 |Read mode
Last edited by theonly 2020-3-30 11:55\begin{align*}
u+u^2+......+u^8+10u^9=v+.....+v^{10}+10v^{11}=8
\end{align*}
试比较u,v的大小,证明之


另外坛主的减压群能不能带我一个please~

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kuing Posted 2020-3-30 08:10
减压群已经被封了好久了

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 Author| theonly Posted 2020-3-30 11:14
现在还有群吗

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青青子衿 Posted 2020-3-30 11:58
Last edited by 青青子衿 2020-3-30 12:09回复 3# theonly
减压群2019年夏天就没有了。

QQ群的话就只有一个“人教群”
群号等于按如下步骤相乘所得数:
二乘以十九,再乘以二十九万一千三百五十九

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kuing Posted 2020-3-30 14:52
回复 4# 青青子衿

给个群号还搞乘法……真有闲情……撸题吧……

首先 `u`, `v` 显然都是正的,令
\begin{align*}
f(x)&=x+x^2+\cdots+x^8+10x^9,\\
g(x)&=x+x^2+\cdots+x^{10}+10x^{11},
\end{align*}其中 `x>0`,显然它们都是单增,所以 `u`, `v` 唯一确定,作差得
\[f(x)-g(x)=9x^9-x^{10}-10x^{11}=x^9(1+x)(9-10x),\]所以当 `x\in(0,0.9)` 时 `f(x)>g(x)`,且 `f(0.9)=g(0.9)`,来算算该函数值是多少,由
\[f(x)=\frac{x-x^9}{1-x}+10x^9,\]得
\[f(0.9)=10(0.9-0.9^9)+10\times0.9^9=9,\]也就是 `f(0.9)=g(0.9)=9`,那么 `f(u)=g(v)=8` 的话就是 `u<v` 了。

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 Author| theonly Posted 2020-3-30 15:59
答得漂亮,减压群为啥被封了?

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kuing Posted 2020-3-30 18:33
回复 6# theonly

太H太XX呗,难道你以为那是数学群

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 Author| theonly Posted 2020-3-30 20:08
我真的以为是数学群

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