线性代数网课代修|ENGO361代写|ENGO361英文辅导|ENGO361

如果你也在线性代数linearalgebra这个学科遇到相关的难题,请随时添加vx号联系我们的代写客服。我们会为你提供专业的服务。 linearalgebra™长期致力于留学生网课服务,涵盖各个网络学科课程:金融学Finance,经济学Economics,数学Mathematics,会计Accounting,文学Literature,艺术Arts等等。除了网课全程托管外,linearalgebra™也可接受单独网课任务。无论遇到了什么网课困难,都能帮你完美解决! Note that the sign of the second normal derivative of \(\tilde{u}\) at a critical point \(x=(\xi,0)\) is exactly that of \(tial_{\nu}u(\xi,0)\), hence \[tial_{\nu}u(\xi,0)>0 \ \Rightarrow\ \operatorname{ind}_{x}\nabla\tilde{u}=\operatorname{ind}_{\xi} \bar{\nabla}h\] \[tial_{\nu}u(\xi,0)0\}}\operatorname{ind}_{\xi} \bar{\nabla}h &-\sum_{\{\xi\intial M:\bar{\nabla}h(\xi)=0,\,tial_{\nu} u(\xi)0\}}\operatorname{ind}_{\xi}\bar{\nabla}h+\sum_{\{\xi\intial M:\bar{ \nabla}h(\xi)=0,\,tial_{\nu}u(\xi)0\}}\operatorname{ind}_ {\xi}\bar{\nabla}h. \tag{11}\]Using\[\chi(\mathscr{D}M)=2\chi(M)-\chi(tial M)\]in (11), we establish (5). The proof is now concluded. We point out that if both \(u\) and \(h\) are _Morse_ functions, formula (5) can be made more precise. In fact, in this case, the \[\sum_{\{x\in M:\nabla u(x)=0\}}\,(-1)^{\mathfrak{m}(x)}=\chi(M)-\sum_{\{\xi\in tial M:\bar{\nabla}h(\xi)=0,\,tial_{\nu}u(\xi)>0\}}.(-1)^{\mathfrak{m} (\xi)} \tag{12}\]In particular, when \(n=2\) (i.e., for Riemannian surfaces), taking into account that all the critical points in the interior are saddle points and on the boundary are maxima or minima, we deduce the following corollary**Corollary**.: _Under the assumptions of .1 we have that_\[\begin{split}\sharp\{\text{critical points of $u$}\}=& \quad\sharp\{\text{maxima of $u$ on $tial M$ with $tial_{\nu}u(\xi)>0$}\} &-\sharp\{\text{minima of $u$ on $tial M$ with $tial_{\nu}u(\xi)>0$}\} &-\chi(M)图片描述

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