詹森不等式据/h1>
有关……据/h4>
- 代数据/span>>据/span>
詹森的不等式据/strong>是涉及函数凸起的不等式。我们首先进行以下定义:据/p>
函数是据strong>凸据/strong>在一个间隔据span class="katex">
一世据/span>如果图上任意两点之间的线段据span class="katex">
(据/span>在据span class="katex">
一世据/span>)据/span>在图表上方。凸函数的一个例子是据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>X据/span>2据/span>.据/p> 函数是据strong>凹据/strong>在一个间隔据span class="katex">
一世据/span>如果图形上任意两点之间的线段据span class="katex">
(据/span>在据span class="katex">
一世据/span>)据/span>在图表下方。凹形函数的一个例子是据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>-据/span>X据/span>2据/span>.据/p>
Jensen不等式的表述据/h2>
Jensen不等式是这样的据/p>
詹森不等式据/strong>
设实值函数据span class="katex"> F据/span>在间隔上被凸据span class="katex"> 一世据/span>.让据span class="katex"> X据/span>1据/span>那据/span>.据/span>.据/span>.据/span>那据/span>X据/span>N据/span>∈据/span>一世据/span>和据span class="katex"> ω据/span>1据/span>那据/span>.据/span>.据/span>.据/span>那据/span>ω据/span>N据/span>≥据/span>0.据/span>.然后我们有据/p>
ω据/span>1据/span>+据/span>ω据/span>2据/span>+据/span>⋯据/span>+据/span>ω据/span>N据/span>ω据/span>1据/span>F据/span>(据/span>X据/span>1据/span>)据/span>+据/span>ω据/span>2据/span>F据/span>(据/span>X据/span>2据/span>)据/span>+据/span>⋯据/span>+据/span>ω据/span>N据/span>F据/span>(据/span>X据/span>N据/span>)据/span>≥据/span>F据/span>(据/span>ω据/span>1据/span>+据/span>ω据/span>2据/span>+据/span>⋯据/span>+据/span>ω据/span>N据/span>ω据/span>1据/span>X据/span>1据/span>+据/span>ω据/span>2据/span>X据/span>2据/span>+据/span>⋯据/span>+据/span>ω据/span>N据/span>X据/span>N据/span>)据/span>.据/span>
如果据span class="katex"> F据/span>为凹,则不等式方向翻转。据/p>
特别是如果我们取权重据span class="katex"> ω据/span>1据/span>=据/span>ω据/span>2据/span>=据/span>⋯据/span>=据/span>ω据/span>N据/span>=据/span>1据/span>,我们得到了不平等据/p>
N据/span>F据/span>(据/span>X据/span>1据/span>)据/span>+据/span>F据/span>(据/span>X据/span>2据/span>)据/span>+据/span>⋯据/span>+据/span>F据/span>(据/span>X据/span>N据/span>)据/span>≥据/span>F据/span>(据/span>N据/span>X据/span>1据/span>+据/span>X据/span>2据/span>+据/span>⋯据/span>+据/span>X据/span>N据/span>)据/span>.据/span>□据/span>
这个证据很长。由于解决问题的方面没有多少帮助,您可能会跳过它。据/p>
注意,我们只需要证明凸函数的陈述。这是由于以下结果:据/p>
如果据span class="katex"> G据/span>那是一个凹形函数据span class="katex"> -据/span>G据/span>是一个凸函数。据/p>
我们首先证明以下声明:据/p>
P.据/span>⟺据/span>问:据/span>那据/span>在哪里据span class="katex"> P.据/span>和据span class="katex"> 问:据/span>如下面所述:据/p>
P.据/span>:据span class="katex"> ∀据/span>X据/span>1据/span>那据/span>.据/span>.据/span>.据/span>那据/span>X据/span>N据/span>∈据/span>一世据/span>和据span class="katex"> ∀据/span>ω据/span>1据/span>那据/span>.据/span>.据/span>.据/span>那据/span>ω据/span>N据/span>≥据/span>0.据/span>以下持有:据/p>
ω据/span>1据/span>+据/span>ω据/span>2据/span>+据/span>⋯据/span>+据/span>ω据/span>N据/span>ω据/span>1据/span>F据/span>(据/span>X据/span>1据/span>)据/span>+据/span>ω据/span>2据/span>F据/span>(据/span>X据/span>2据/span>)据/span>+据/span>⋯据/span>+据/span>ω据/span>N据/span>F据/span>(据/span>X据/span>N据/span>)据/span>≥据/span>F据/span>(据/span>ω据/span>1据/span>+据/span>ω据/span>2据/span>+据/span>⋯据/span>+据/span>ω据/span>N据/span>ω据/span>1据/span>X据/span>1据/span>+据/span>ω据/span>2据/span>X据/span>2据/span>+据/span>⋯据/span>+据/span>ω据/span>N据/span>X据/span>N据/span>)据/span>.据/span>
问:据/span>:据span class="katex"> ∀据/span>X据/span>1据/span>那据/span>.据/span>.据/span>.据/span>那据/span>X据/span>N据/span>∈据/span>一世据/span>和据span class="katex"> ∀据/span>λ据/span>1据/span>那据/span>......据/span>那据/span>λ据/span>N据/span>≥据/span>0.据/span>和据/span>一世据/span>=据/span>1据/span>σ.据/span>N据/span>λ据/span>一世据/span>=据/span>1据/span>以下持有:据/p>
一世据/span>=据/span>1据/span>σ.据/span>N据/span>λ据/span>一世据/span>F据/span>(据/span>X据/span>一世据/span>)据/span>≥据/span>F据/span>(据/span>一世据/span>=据/span>1据/span>σ.据/span>N据/span>λ据/span>一世据/span>X据/span>一世据/span>)据/span>.据/span>
证明:据/strong>
简单地说据span class="katex"> λ据/span>一世据/span>=据/span>R.据/span>=据/span>1据/span>σ.据/span>N据/span>ω据/span>R.据/span>ω据/span>一世据/span>那据/span>在哪里据span class="katex"> 一世据/span>=据/span>1据/span>那据/span>2据/span>那据/span>......据/span>N据/span>.据/span>
这意味着我们降低了刚刚表现出来的问题据span class="katex"> 问:据/span>是真的。要做到这一点,我们使用据一种href="//www.parkandroid.com/wiki/induction-introduction/" class="wiki_link" title="感应GydF4y2Ba" target="_blank">感应据/一种>.据/p>
基本情况据span class="katex"> N据/span>=据/span>1据/span>是非常真实的。据/p>
案子据span class="katex"> N据/span>=据/span>2据/span>是凸函数的定义(证明这是感兴趣的读者)。据/p>
为了据span class="katex"> N据/span>≥据/span>3.据/span>,我们将假设据span class="katex"> λ据/span>N据/span>∈据/span>(据/span>0.据/span>那据/span>1据/span>)据/span> (据/span>如果据span class="katex"> λ据/span>N据/span>=据/span>1据/span>那据/span>断言简单地成立,如果据span class="katex"> λ据/span>N据/span>=据/span>0.据/span>那据/span>我们求助于归纳假设)。据/p>
现在,我们继续假设据span class="katex"> K.据/span>=据/span>2据/span>和据span class="katex"> K.据/span>=据/span>N据/span>-据/span>1据/span>是真的:据/p>
F据/span>(据/span>一世据/span>=据/span>1据/span>σ.据/span>N据/span>λ据/span>一世据/span>X据/span>一世据/span>)据/span>=据/span>F据/span>(据/span>(据/span>1据/span>-据/span>λ据/span>N据/span>)据/span>(据/span>一世据/span>=据/span>1据/span>σ.据/span>N据/span>-据/span>1据/span>μ据/span>一世据/span>X据/span>一世据/span>)据/span>+据/span>λ据/span>N据/span>X据/span>N据/span>)据/span>(据/span>在哪里据/span>μ据/span>一世据/span>=据/span>1据/span>-据/span>λ据/span>N据/span>λ据/span>一世据/span>那据/span>一世据/span>=据/span>1据/span>那据/span>2据/span>那据/span>......据/span>那据/span>N据/span>-据/span>1据/span>那据/span>一世据/span>=据/span>1据/span>σ.据/span>N据/span>-据/span>1据/span>μ据/span>一世据/span>=据/span>1据/span>)据/span>≤.据/span>(据/span>1据/span>-据/span>λ据/span>N据/span>)据/span>F据/span>(据/span>一世据/span>=据/span>1据/span>σ.据/span>N据/span>-据/span>1据/span>μ据/span>一世据/span>X据/span>一世据/span>)据/span>+据/span>λ据/span>N据/span>F据/span>(据/span>X据/span>N据/span>)据/span>≤.据/span>(据/span>一世据/span>=据/span>1据/span>σ.据/span>N据/span>-据/span>1据/span>(据/span>1据/span>-据/span>λ据/span>N据/span>)据/span>μ据/span>一世据/span>F据/span>(据/span>X据/span>一世据/span>)据/span>)据/span>+据/span>λ据/span>N据/span>F据/span>(据/span>X据/span>N据/span>)据/span>=据/span>一世据/span>=据/span>1据/span>σ.据/span>N据/span>λ据/span>一世据/span>F据/span>(据/span>X据/span>一世据/span>)据/span>.据/span>
通过归纳法,证明就完成了。据span class="katex"> □据/span>
最后一份表格是在各种奥林匹克水平问题中遇到和使用的表格。据/p>
我们如何检查一个函数是凸的还是凹的?我们不能总是画出图来检验。最好(而且通常更快)的方法是使用微积分。一个函数据span class="katex">
F据/span>在间隔内凸起据span class="katex">
一世据/span>当且仅当据span class="katex">
F据/span>'据/span>'据/span>(据/span>X据/span>)据/span>≥据/span>0.据/span>对所有人据span class="katex">
X据/span>∈据/span>一世据/span>和凹据span class="katex">
F据/span>'据/span>'据/span>(据/span>X据/span>)据/span>≤.据/span>0.据/span>对所有人据span class="katex">
X据/span>∈据/span>一世据/span>.以下给出了确切的陈述:据/p>
让据span class="katex">
F据/span>:据/span>一世据/span>→据/span>R.据/span>是一个两次可分辨动的功能。据/p>
然后据/p>
可以找到此证明据一种target="_blank" rel="nofollow" href="//www.parkandroid.com/wiki/convex-functions/">在这里据/一种>.据/p>
例如,对于据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>X据/span>2据/span>我们有据span class="katex">
F据/span>'据/span>(据/span>X据/span>)据/span>=据/span>2据/span>X据/span>那据/span>所以据span class="katex">
F据/span>'据/span>'据/span>(据/span>X据/span>)据/span>=据/span>2据/span>>据/span>0.据/span>对所有人据span class="katex">
X据/span>∈据/span>R.据/span>.所以这个函数在图上处处是凸的。同样的据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>-据/span>X据/span>2据/span>我们有据span class="katex">
F据/span>'据/span>'据/span>(据/span>X据/span>)据/span>=据/span>-据/span>2据/span>据据/span>0.据/span>对所有人据span class="katex">
X据/span>∈据/span>R.据/span>.所以该函数在图中无处不在。据/p>
由于我们一直在谈论据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>X据/span>2据/span>,让我们申请jensen。我们已经证明它是凸显的。我们选择真实据span class="katex">
X据/span>1据/span>=据/span>1据/span>那据/span>X据/span>2据/span>=据/span>2据/span>那据/span>.据/span>.据/span>.据/span>那据/span>X据/span>N据/span>=据/span>N据/span>.应用jensen我们得到了据/p>
N据/span>F据/span>(据/span>X据/span>1据/span>)据/span>+据/span>F据/span>(据/span>X据/span>2据/span>)据/span>+据/span>⋯据/span>+据/span>F据/span>(据/span>X据/span>N据/span>)据/span>N据/span>F据/span>(据/span>1据/span>)据/span>+据/span>F据/span>(据/span>2据/span>)据/span>+据/span>⋯据/span>+据/span>F据/span>(据/span>N据/span>)据/span>N据/span>1据/span>2据/span>+据/span>2据/span>2据/span>+据/span>⋯据/span>+据/span>N据/span>2据/span>N据/span>6.据/span>N据/span>(据/span>N据/span>+据/span>1据/span>)据/span>(据/span>2据/span>N据/span>+据/span>1据/span>)据/span>6.据/span>(据/span>N据/span>+据/span>1据/span>)据/span>(据/span>2据/span>N据/span>+据/span>1据/span>)据/span>N据/span>≥据/span>F据/span>(据/span>N据/span>X据/span>1据/span>+据/span>X据/span>2据/span>+据/span>⋯据/span>+据/span>X据/span>N据/span>)据/span>≥据/span>F据/span>(据/span>N据/span>1据/span>+据/span>2据/span>+据/span>⋯据/span>+据/span>N据/span>)据/span>≥据/span>F据/span>(据/span>N据/span>2据/span>N据/span>(据/span>N据/span>+据/span>1据/span>)据/span>)据/span>≥据/span>F据/span>(据/span>2据/span>N据/span>+据/span>1据/span>)据/span>≥据/span>(据/span>2据/span>N据/span>+据/span>1据/span>)据/span>2据/span>≥据/span>1据/span>.据/span>
Jensen不等式的特殊情况据/h2>
Jensen不等式的应用据/h2>
一种据/span>1据/span>+据/span>B.据/span>1据/span>+据/span>C据/span>1据/span>=据/span>一种据/span>+据/span>B.据/span>+据/span>C据/span>
让据span class="katex"> 一种据/span>那据/span>B.据/span>那据/span>C据/span>为正实数,以满足上述方程。如果下面表达式的最大值为据span class="katex"> N据/span>m据/span>那据/span>在哪里据span class="katex"> m据/span>那据/span>N据/span>是coprime正整数,是什么据span class="katex"> m据/span>+据/span>N据/span>还是据/span>
(据/span>2据/span>一种据/span>+据/span>B.据/span>+据/span>C据/span>)据/span>2据/span>1据/span>+据/span>(据/span>2据/span>B.据/span>+据/span>C据/span>+据/span>一种据/span>)据/span>2据/span>1据/span>+据/span>(据/span>2据/span>C据/span>+据/span>B.据/span>+据/span>一种据/span>)据/span>2据/span>1据/span>