riemann Sum.据/h1>
已经有一个帐户?据一种HR.E.F="//www.parkandroid.com/account/login/?next=/wiki/riemann-sums/" class="ax-click" data-ax-id="clicked_signup_modal_login" data-ax-type="link">这里登录。据/a>
测验据/h4>
有关……据/h4>
- 结石据/span>>据/span>
一种据S.trong>黎曼和据/strong>是通过添加该区域的多个简化切片的区域而获得的区域区域的近似值。它以微积分应用以形式化疲惫的方法,用于确定区域的面积。此过程产生积分,其精确地计算了该区域的值。据/p>
定义据/h2>
让我们分解给定闭区间据S.pan class="katex">
[据/span>一种据/span>那据/span>B.据/span>]据/span>进入据S.pan class="katex">
N据/span>插入子内部据S.pan class="katex">
N据/span>-据/span>1据/span>点据S.pan class="katex">
X据/span>1据/span>那据/span>X据/span>2据/span>那据/span>......据/span>那据/span>X据/span>N据/span>-据/span>1据/span>这样据/p>
一种据/span>=据/span>X据/span>0.据/span>据据/span>X据/span>1据/span>据据/span>X据/span>2据/span>据据/span>......据/span>据据/span>X据/span>K.据/span>据据/span>......据/span>据据/span>X据/span>N据/span>=据/span>B.据/span>。据/span> 点的这样一个集合据S.pan class="katex">
{据/span>X据/span>0.据/span>那据/span>X据/span>1据/span>那据/span>......据/span>那据/span>X据/span>N据/span>}据/span>被称为据E.m>划分据/em>的据S.pan class="katex">
[据/span>一种据/span>那据/span>B.据/span>]据/span>,这个分区决定了以下内容据S.pan class="katex">
N据/span>封闭的子区间:据/p>
[据/span>X据/span>0.据/span>那据/span>X据/span>1据/span>]据/span>那据/span>[据/span>X据/span>1据/span>那据/span>X据/span>2据/span>]据/span>那据/span>......据/span>那据/span>[据/span>X据/span>N据/span>-据/span>1据/span>那据/span>X据/span>N据/span>]据/span>。据/span> 让据S.pan class="katex">
[据/span>X据/span>K.据/span>-据/span>1据/span>那据/span>X据/span>K.据/span>]据/span>表示据S.pan class="katex">
K.据/span>TH.据/span>闭子区间的划分,并令据S.pan class="katex">
δ.据/span>X据/span>K.据/span>表示的长度据S.pan class="katex">
K.据/span>TH.据/span>subinterval。注意据S.pan class="katex">
δ.据/span>X据/span>K.据/span>不必为每个子一样。据/p>
如果据S.pan class="katex">
F据/span>在封闭间隔内定义据S.pan class="katex">
[据/span>一种据/span>那据/span>B.据/span>]据/span>和据S.pan class="katex">
C据/span>K.据/span>在任何点据S.pan class="katex">
[据/span>X据/span>K.据/span>-据/span>1据/span>那据/span>X据/span>K.据/span>]据/span>,那么A.据S.trong>黎曼和据/strong>被定义为据/p>
K.据/span>=据/span>1据/span>σ.据/span>N据/span>F据/span>(据/span>C据/span>K.据/span>)据/span>δ.据/span>X据/span>K.据/span>。据/span> 甲黎曼和可以被可视化为(大约)的区域的曲线下的分割据S.pan class="katex">
F据/span>(据/span>X据/span>)据/span>在据S.pan class="katex">
[据/span>一种据/span>那据/span>B.据/span>]据/span>进入据S.pan class="katex">
N据/span>相邻的矩形跨越区间,其中据S.pan class="katex">
K.据/span>TH.据/span>矩形的宽度据S.pan class="katex">
δ.据/span>X据/span>K.据/span>和身高据S.pan class="katex">
F据/span>(据/span>C据/span>K.据/span>)据/span>。每个矩形的区域都是据S.pan class="katex">
F据/span>(据/span>C据/span>K.据/span>)据/span>δ.据/span>X据/span>K.据/span>(高度倍宽度)。所有矩形的面积之和接近下的实际面积据S.pan class="katex">
F据/span>在据S.pan class="katex">
[据/span>一种据/span>那据/span>B.据/span>]据/span>并且等于黎曼总和。注意据S.pan class="katex">
C据/span>K.据/span>可能是任何一点据S.pan class="katex">
K.据/span>TH.据/span>子区间,这意味着定义在选择每个的高度时留有一定的自由度据S.pan class="katex">
N据/span>矩形。据/p>
关系定积分据/h2>
有关明确积分的更多信息,请参阅据一种HR.E.F="//www.parkandroid.com/wiki/definite-integrals/" class="wiki_link" title="明确的积分" target="_blank">明确的积分据/a>。据!-- end-meta -->
函数的黎曼和与定积分的关系如下:据/p>
N据/span>→据/span>∞据/span>林据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>F据/span>(据/span>C据/span>K.据/span>)据/span>δ.据/span>X据/span>K.据/span>=据/span>∫据/span>一种据/span>B.据/span>F据/span>(据/span>X据/span>)据/span>D.据/span>X据/span>。据/span>
应用和实例据/h2>
评价黎曼和如下图所示,这是该区域的下左手近似据/p>
F据/span>(据/span>X据/span>)据/span>=据/span>1据/span>6.据/span>-据/span>X据/span>2据/span> 那据/span>0.据/span>≤.据/span>X据/span>≤.据/span>4.据/span>。据/span>
由于我们使用的4个矩形左边端点上的间隔据S.pan class="katex"> [据/span>0.据/span>那据/span>4.据/span>]据/span>那据/span>我们可以首先找到价值据S.pan class="katex"> F据/span>(据/span>X据/span>)据/span>在范围内的每个左端点:据/p>
F据/span>(据/span>0.据/span>)据/span>F据/span>(据/span>1据/span>)据/span>F据/span>(据/span>2据/span>)据/span>F据/span>(据/span>3.据/span>)据/span>=据/span>4.据/span>=据/span>1据/span>5.据/span> =据/span>1据/span>2据/span> =据/span>7.据/span> 。据/span>
我们还知道间隔是长度4的,我们将其分成4个相同宽度的矩形。所以每个矩形的宽度为1,或据S.pan class="katex"> δ.据/span>X据/span>=据/span>1据/span>。据/p>
我们现在剩下的长方形的面积的总和找到上面的图片在总蓝色区域的面积:据/p>
一世据/span>=据/span>0.据/span>σ.据/span>3.据/span>F据/span>(据/span>X据/span>一世据/span>)据/span>δ.据/span>X据/span>=据/span>[据/span>F据/span>(据/span>X据/span>0.据/span>)据/span>+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>F据/span>(据/span>X据/span>3.据/span>)据/span>]据/span>δ.据/span>X据/span>=据/span>[据/span>(据/span>F据/span>(据/span>0.据/span>)据/span>+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>F据/span>(据/span>3.据/span>)据/span>]据/span>δ.据/span>X据/span>=据/span>(据/span>4.据/span>+据/span>1据/span>5.据/span> +据/span>1据/span>2据/span> +据/span>7.据/span> )据/span>⋅据/span>1据/span>≈据/span>1据/span>3.据/span>。据/span>9.据/span>8.据/span>2据/span>那据/span>
如我们从图中看到的那样,随着函数的时间间隔降低,曲线下的区域略微高估。据S.pan class="katex"> □据/span>
让据S.pan class="katex"> F据/span>(据/span>X据/span>)据/span>是从下面界限的连续功能据S.pan class="katex"> X据/span>-轴。找到的高估和有界区域的面积在闭区间的低估据S.pan class="katex"> [据/span>一种据/span>那据/span>B.据/span>]据/span>使用riemann um。据/p>
为简单地,我们划分了间隔据S.pan class="katex"> [据/span>一种据/span>那据/span>B.据/span>]据/span>进入据S.pan class="katex"> N据/span>长度相等的子区间,据S.pan class="katex"> δ.据/span>X据/span>=据/span>N据/span>B.据/span>-据/span>一种据/span>。通过极值定理据S.pan class="katex"> F据/span>(据/span>X据/span>)据/span>在每个子区间上都有一个最小值和一个最大值,对于据S.pan class="katex"> K.据/span>TH.据/span>Subinterval我们将表示他们据S.pan class="katex"> F据/span>(据/span>m据/span>K.据/span>)据/span>和据S.pan class="katex"> F据/span>(据/span>m据/span>K.据/span>)据/span>那据/span>分别。在据S.pan class="katex"> m据/span>K.据/span>我们构建具有高度的矩形据S.pan class="katex"> F据/span>(据/span>m据/span>K.据/span>)据/span>哪个在曲线下面,在一点上据S.pan class="katex"> m据/span>K.据/span>我们构建具有高度的矩形据S.pan class="katex"> F据/span>(据/span>m据/span>K.据/span>)据/span>部分位于曲线上方。据/p>
要计算每个矩形的区域,我们发现相应长度和高度的乘积:据/p>
(据/span>内部矩形的面积据/span>)据/span>=据/span>δ.据/span>X据/span>F据/span>(据/span>m据/span>K.据/span>)据/span>≤.据/span>(据/span>区下据/span>F据/span>在据/span>K.据/span>TH.据/span>子区间据/span>)据/span>≤.据/span>δ.据/span>X据/span>F据/span>(据/span>m据/span>K.据/span>)据/span>=据/span>(据/span>外矩形区域据/span>)据/span>。据/span>
(我们省略了这一说法,并就目前依靠检测证明。)据/p>
让我们用据S.pan class="katex"> S.据/span>N据/span>所有内部矩形的区域的总和,以及据S.pan class="katex"> S.据/span>N据/span>所有外矩形的区域的总和。然后我们获得公式据/p>
(据/span>近似于下方据/span>)据/span>(据/span>逼近从上面据/span>)据/span>=据/span>S.据/span>N据/span>=据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>F据/span>(据/span>m据/span>K.据/span>)据/span>δ.据/span>X据/span>=据/span>(据/span>内矩形的总面积据/span>)据/span>=据/span>S.据/span>N据/span>=据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>F据/span>(据/span>m据/span>K.据/span>)据/span>δ.据/span>X据/span>=据/span>(据/span>外矩形的总面积据/span>)据/span>。据/span>
有界函数的曲线下面积据S.pan class="katex"> F据/span>(据/span>X据/span>)据/span>大于内矩形的总面积,但小于外矩形的总面积。也就是说,应用不等式的性质如果据S.pan class="katex"> 一种据/span>据据/span>B.据/span>和据S.pan class="katex"> C据/span>据据/span>D.据/span>, 然后据S.pan class="katex"> 一种据/span>+据/span>C据/span>据据/span>B.据/span>+据/span>D.据/span>上面描述的不等式据S.pan class="katex"> K.据/span>TH.据/span>subinterval,我们获得据/p>
S.据/span>N据/span>=据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>F据/span>(据/span>m据/span>K.据/span>)据/span>δ.据/span>X据/span>≤.据/span>(据/span>地区据/span>)据/span>≤.据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>F据/span>(据/span>m据/span>K.据/span>)据/span>δ.据/span>X据/span>=据/span>S.据/span>N据/span>那据/span>
制作据S.pan class="katex"> S.据/span>N据/span>低估和据S.pan class="katex"> S.据/span>N据/span>高估。据S.pan class="katex"> □据/span>
笔记:据/strong>
- 如果据S.pan class="katex">
m据/span>K.据/span>=据/span>X据/span>K.据/span>-据/span>1据/span>, 然后据S.pan class="katex">
S.据/span>N据/span>=据/span>σ.据/span>K.据/span>=据/span>1据/span>N据/span>F据/span>(据/span>X据/span>K.据/span>-据/span>1据/span>)据/span>δ.据/span>X据/span>我们称此为左侧黎曼和近似。据/li>
- 如果据S.pan class="katex"> m据/span>K.据/span>=据/span>X据/span>K.据/span>, 然后据S.pan class="katex"> S.据/span>N据/span>=据/span>σ.据/span>K.据/span>=据/span>1据/span>N据/span>F据/span>(据/span>X据/span>K.据/span>)据/span>δ.据/span>X据/span>我们称此为右手黎曼和近似。据/li>
评估的黎曼和据S.pan class="katex"> F据/span>(据/span>X据/span>)据/span>=据/span>X据/span>2据/span>在间隔上据S.pan class="katex"> [据/span>0.据/span>那据/span>4.据/span>]据/span>,它使用左端点的每一个据/p>
- 一)10级相等的子区间据/li>
- B) 100个相等的子区间。据/li>
我们划分了间隔据S.pan class="katex"> [据/span>一种据/span>那据/span>B.据/span>]据/span>=据/span>[据/span>0.据/span>那据/span>4.据/span>]据/span>进入据S.pan class="katex"> N据/span>长度相等的子区间:据/p>
δ.据/span>X据/span>K.据/span>=据/span>δ.据/span>X据/span>=据/span>N据/span>B.据/span>-据/span>一种据/span>=据/span>N据/span>4.据/span>-据/span>0.据/span>=据/span>N据/span>4.据/span>。据/span>
这个分区据S.pan class="katex"> [据/span>0.据/span>那据/span>4.据/span>]据/span>包括以下几点:据/p>
0.据/span>=据/span>一种据/span>=据/span>X据/span>0.据/span>据据/span>X据/span>0.据/span>+据/span>δ.据/span>X据/span>据据/span>X据/span>0.据/span>+据/span>2据/span>δ.据/span>X据/span>据据/span>⋯据/span>据据/span>X据/span>0.据/span>+据/span>K.据/span>δ.据/span>X据/span>据据/span>⋯据/span>据据/span>X据/span>0.据/span>+据/span>N据/span>δ.据/span>X据/span>=据/span>B.据/span>=据/span>4.据/span>。据/span>
替代据S.pan class="katex"> X据/span>0.据/span>和据S.pan class="katex"> δ.据/span>X据/span>, 我们获得据/p>
0.据/span>据据/span>N据/span>4.据/span>据据/span>2据/span>×据/span>N据/span>4.据/span>据据/span>⋯据/span>据据/span>K.据/span>×据/span>N据/span>4.据/span>据据/span>⋯据/span>据据/span>N据/span>×据/span>N据/span>4.据/span>=据/span>4.据/span>。据/span>
然后据/p>
X据/span>K.据/span>=据/span>K.据/span>×据/span>N据/span>4.据/span>。据/span>
接下来,我们发现左手riemann和近似:据/p>
S.据/span>N据/span>=据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>F据/span>(据/span>X据/span>K.据/span>-据/span>1据/span>)据/span>δ.据/span>X据/span>=据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>(据/span>X据/span>K.据/span>-据/span>1据/span>2据/span>)据/span>(据/span>N据/span>4.据/span>)据/span>=据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>[据/span>(据/span>K.据/span>-据/span>1据/span>)据/span>(据/span>N据/span>4.据/span>)据/span>]据/span>2据/span>(据/span>N据/span>4.据/span>)据/span>=据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>(据/span>N据/span>3.据/span>6.据/span>4.据/span>)据/span>(据/span>K.据/span>2据/span>-据/span>2据/span>K.据/span>+据/span>1据/span>)据/span>=据/span>N据/span>3.据/span>6.据/span>4.据/span>[据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>K.据/span>2据/span>-据/span>2据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>K.据/span>+据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>1据/span>]据/span>=据/span>(据/span>N据/span>3.据/span>6.据/span>4.据/span>)据/span>[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>-据/span>2据/span>×据/span>2据/span>N据/span>(据/span>N据/span>+据/span>1据/span>)据/span>+据/span>N据/span>]据/span>=据/span>(据/span>3.据/span>N据/span>3.据/span>3.据/span>2据/span>)据/span>(据/span>2据/span>N据/span>3.据/span>+据/span>3.据/span>N据/span>2据/span>+据/span>N据/span>-据/span>6.据/span>N据/span>2据/span>-据/span>6.据/span>N据/span>+据/span>6.据/span>N据/span>)据/span>=据/span>(据/span>3.据/span>N据/span>2据/span>3.据/span>2据/span>)据/span>(据/span>2据/span>N据/span>2据/span>-据/span>3.据/span>N据/span>+据/span>1据/span>)据/span>那据/span>
在这里我们使用了求和公式扩大的款项。据/p>
因此,答案如下:据/p>
- 一个)据S.pan class="katex"> N据/span>=据/span>1据/span>0.据/span>那据S.pan class="katex"> S.据/span>N据/span>=据/span>1据/span>8.据/span>。据/span>2据/span>4.据/span>。据/span>
- b)for.据S.pan class="katex"> N据/span>=据/span>1据/span>0.据/span>0.据/span>那据S.pan class="katex"> S.据/span>N据/span>=据/span>2据/span>1据/span>。据/span>0.据/span>1据/span>4.据/span>4.据/span>。据/span> □据/span>
笔记据/strong>那据S.pan class="katex"> ∫据/span>0.据/span>4.据/span>X据/span>2据/span>D.据/span>X据/span>=据/span>2据/span>1据/span>。据/span>3.据/span>是实际的曲线下面积,并通过增加更多的矩形,我们会取得更好的近似。据!-- end-example -->
评估的黎曼和据S.pan class="katex"> F据/span>(据/span>X据/span>)据/span>=据/span>1据/span>-据/span>X据/span>2据/span>在间隔上据S.pan class="katex"> [据/span>0.据/span>那据/span>1据/span>]据/span>,它对每个无穷个子区间使用正确的端点。据/p>
让我们先划分据S.pan class="katex"> X据/span>- XIS宽度的许多间隔据S.pan class="katex"> δ.据/span>X据/span>=据/span>N据/span>1据/span>,所以间隔会据/p>
[据/span>0.据/span>那据/span>N据/span>1据/span>]据/span>那据/span>[据/span>N据/span>1据/span>那据/span>N据/span>2据/span>]据/span>那据/span>......据/span>那据/span>[据/span>N据/span>N据/span>-据/span>1据/span>那据/span>1据/span>]据/span>。据/span>
然后,我们让每个宽度的子区间据S.pan class="katex"> N据/span>1据/span>和身高:据/p>
[据/span>F据/span>(据/span>N据/span>1据/span>)据/span>]据/span>那据/span>[据/span>F据/span>(据/span>N据/span>2据/span>)据/span>]据/span>那据/span>......据/span>那据/span>[据/span>F据/span>(据/span>N据/span>K.据/span>)据/span>]据/span>那据/span>......据/span>那据/span>[据/span>F据/span>(据/span>N据/span>N据/span>)据/span>]据/span>。据/span>
因此,所有子区间的面积将由以下求和(我们称之为雷曼和)给出:据/p>
K.据/span>=据/span>1据/span>σ.据/span>N据/span>F据/span>(据/span>N据/span>K.据/span>)据/span>(据/span>N据/span>1据/span>)据/span>=据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>(据/span>1据/span>-据/span>(据/span>N据/span>K.据/span>)据/span>2据/span>)据/span>(据/span>N据/span>1据/span>)据/span>=据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>(据/span>N据/span>1据/span>-据/span>N据/span>3.据/span>K.据/span>2据/span>)据/span>=据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>N据/span>1据/span>-据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>N据/span>3.据/span>K.据/span>2据/span>=据/span>N据/span>×据/span>N据/span>1据/span>-据/span>N据/span>3.据/span>1据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>K.据/span>2据/span>=据/span>1据/span>-据/span>(据/span>N据/span>3.据/span>1据/span>)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>=据/span>1据/span>-据/span>6.据/span>N据/span>3.据/span>2据/span>N据/span>3.据/span>+据/span>3.据/span>N据/span>2据/span>+据/span>N据/span>。据/span>
因此,面积将是极限据S.pan class="katex"> N据/span>→据/span>∞据/span>上述表达式,其等于据/p>
N据/span>→据/span>∞据/span>林据/span>1据/span>-据/span>6.据/span>N据/span>3.据/span>2据/span>N据/span>3.据/span>+据/span>3.据/span>N据/span>2据/span>+据/span>N据/span>=据/span>1据/span>-据/span>6.据/span>2据/span>=据/span>3.据/span>2据/span>。据/span>□据/span>
明确的积分作为总和的限制据/h2>
我们可以将明确的积分表示为一定数量的术语总和的限制。让据S.pan class="katex">
F据/span>(据/span>X据/span>)据/span>在间隔中是一个连续的功能据S.pan class="katex">
[据/span>一种据/span>那据/span>B.据/span>]据/span>。划分据S.pan class="katex">
一种据/span>-据/span>B.据/span>进入据S.pan class="katex">
N据/span>相等的部分,使得每个部分的宽度为据S.pan class="katex">
H据/span>。然后据/p>
N据/span>H据/span>=据/span>B.据/span>-据/span>一种据/span>。据/span> 函数的定积分据S.pan class="katex">
F据/span>(据/span>X据/span>)据/span>在间隔据S.pan class="katex">
[据/span>一种据/span>那据/span>B.据/span>]据/span>可以定义为据/p>
∫据/span>一种据/span>B.据/span>F据/span>(据/span>X据/span>)据/span>D.据/span>X据/span>=据/span>N据/span>→据/span>∞据/span>那据/span>H据/span>→据/span>0.据/span>林据/span>H据/span>[据/span>F据/span>(据/span>一种据/span>+据/span>H据/span>)据/span>+据/span>F据/span>(据/span>一种据/span>+据/span>2据/span>H据/span>)据/span>+据/span>⋯据/span>+据/span>F据/span>(据/span>一种据/span>+据/span>N据/span>H据/span>)据/span>]据/span>=据/span>N据/span>→据/span>∞据/span>那据/span>H据/span>→据/span>0.据/span>林据/span>H据/span>R.据/span>=据/span>1据/span>σ.据/span>N据/span>F据/span>(据/span>一种据/span>+据/span>R.据/span>H据/span>)据/span>。据/span> 在此公式的帮助下,我们可以评估一些简单的确定积分。使用上述公式找到明确积分的过程称为据S.trong>定积分的总和的极限据/strong>。据/p>
用定积分求级数的和据/h2>
考虑前一节中所定义的“总和的极限”式的,即据/p>
∫据/span>一种据/span>B.据/span>F据/span>(据/span>X据/span>)据/span>D.据/span>X据/span>=据/span>N据/span>→据/span>∞据/span>那据/span>H据/span>→据/span>0.据/span>林据/span>H据/span>[据/span>F据/span>(据/span>一种据/span>+据/span>H据/span>)据/span>+据/span>F据/span>(据/span>一种据/span>+据/span>2据/span>H据/span>)据/span>+据/span>⋯据/span>+据/span>F据/span>(据/span>一种据/span>+据/span>N据/span>H据/span>)据/span>]据/span>。据/span>
现在,我们得到了一节中的概念,我们将寻找工作规则。据/p>
工作规则:据/p>
- 首先,快递的形式给定系列据S.pan class="katex">
N据/span>1据/span>σ.据/span>F据/span>(据/span>N据/span>R.据/span>)据/span>。据/li>
- 代替据S.pan class="katex"> σ.据/span>对于积分符号据S.pan class="katex"> ∫据/span>那据/span> N据/span>R.据/span>为了据S.pan class="katex"> X据/span>那据/span>和据S.pan class="katex"> N据/span>1据/span>为了据S.pan class="katex"> D.据/span>X据/span>。据/span>
- 整合的下限和上限是值的据S.pan class="katex"> N据/span>→据/span>∞据/span>林据/span>N据/span>R.据/span>和据S.pan class="katex"> R.据/span>那据/span>分别。据/li>
注意这个表达式据S.pan class="katex">
N据/span>→据/span>∞据/span>那据/span>H据/span>→据/span>0.据/span>林据/span>H据/span>R.据/span>=据/span>1据/span>σ.据/span>N据/span>F据/span>(据/span>一种据/span>+据/span>R.据/span>H据/span>)据/span>也是表单据S.pan class="katex">
N据/span>1据/span>σ.据/span>F据/span>(据/span>N据/span>R.据/span>)据/span>因为据S.pan class="katex">
H据/span>=据/span>N据/span>B.据/span>-据/span>一种据/span>。据/p>
发现该系列的总和据S.pan class="katex">
N据/span>→据/span>∞据/span>林据/span>(据/span>N据/span>1据/span>+据/span>N据/span>+据/span>1据/span>1据/span>+据/span>N据/span>+据/span>2据/span>1据/span>+据/span>⋯据/span>+据/span>6.据/span>N据/span>1据/span>)据/span>。据/span> 首先,我们将以表格表达据S.pan class="katex">
N据/span>1据/span>σ.据/span>F据/span>(据/span>N据/span>R.据/span>)据/span>:据/span>
S.据/span>=据/span>N据/span>→据/span>∞据/span>林据/span>(据/span>N据/span>1据/span>+据/span>N据/span>+据/span>1据/span>1据/span>+据/span>N据/span>+据/span>2据/span>1据/span>+据/span>⋯据/span>+据/span>6.据/span>N据/span>1据/span>)据/span>=据/span>N据/span>→据/span>∞据/span>林据/span>N据/span>1据/span>(据/span>1据/span>+据/span>1据/span>+据/span>N据/span>1据/span>1据/span>+据/span>1据/span>+据/span>N据/span>2据/span>1据/span>+据/span>⋯据/span>+据/span>1据/span>+据/span>N据/span>5.据/span>N据/span>1据/span>)据/span>=据/span>N据/span>→据/span>∞据/span>林据/span>N据/span>1据/span>R.据/span>=据/span>0.据/span>σ.据/span>5.据/span>N据/span>1据/span>+据/span>N据/span>R.据/span>1据/span>。据/span> 自据/p>
(据/span>下限据/span>)据/span>=据/span>一种据/span>=据/span>N据/span>→据/span>∞据/span>林据/span>N据/span>R.据/span>=据/span>N据/span>→据/span>∞据/span>林据/span>N据/span>0.据/span>=据/span>0.据/span>那据/span>(据/span>上限据/span>)据/span>=据/span>B.据/span>=据/span>N据/span>→据/span>∞据/span>林据/span>N据/span>R.据/span>=据/span>N据/span>→据/span>∞据/span>林据/span>N据/span>5.据/span>N据/span>=据/span>5.据/span>那据/span> 我们有据/p>
S.据/span>=据/span>N据/span>→据/span>∞据/span>林据/span>N据/span>1据/span>R.据/span>=据/span>1据/span>σ.据/span>5.据/span>N据/span>1据/span>+据/span>N据/span>R.据/span>1据/span>=据/span>∫据/span>0.据/span>5.据/span>1据/span>+据/span>X据/span>D.据/span>X据/span>=据/span>[据/span>LN.据/span>|据/span>1据/span>+据/span>X据/span>|据/span>]据/span>0.据/span>5.据/span>=据/span>LN.据/span>6.据/span>。据/span>□据/span>
找到据/p>
N据/span>→据/span>∞据/span>林据/span>N据/span>N据/span>
1据/span>
+据/span>2据/span>
+据/span>3.据/span>
+据/span>4.据/span>
+据/span>⋯据/span>+据/span>N据/span>
。据/span> 这可以重写为据/p>
N据/span>→据/span>∞据/span>林据/span>N据/span>1据/span>K.据/span>=据/span>1据/span>σ.据/span>N据/span>N据/span>K.据/span>
=据/span>∫据/span>0.据/span>1据/span>X据/span>
D.据/span>X据/span>=据/span>3.据/span>2据/span>。据/span>□据/span>
评估据/p>
N据/span>→据/span>∞据/span>林据/span>(据/span>2据/span>N据/span>1据/span>+据/span>2据/span>N据/span>+据/span>1据/span>1据/span>+据/span>2据/span>N据/span>+据/span>2据/span>1据/span>+据/span>2据/span>N据/span>+据/span>3.据/span>1据/span>+据/span>⋯据/span>+据/span>3.据/span>N据/span>1据/span>)据/span>。据/span> 答案在于形式据S.pan class="katex">
LN.据/span>B.据/span>一种据/span>那据/span>在哪里据S.pan class="katex">
一种据/span>和据S.pan class="katex">
B.据/span>为互质的正整数。据/p>
找到据S.pan class="katex">
一种据/span>+据/span>B.据/span>。据/p>