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难以渐近（递归）函数（算法分析）(difficult asymptotic(recurrence) function (algorithm analysis))

 我被困在这一个，我不知道如何解决它，无论我尝试我只是无法找到一种方式来玩这个功能，所以我可以用一种方式来表示它，让我可以找到ag （n），使得g（n）是T（n）∈Θ（g（n））  
 我遇到麻烦的功能是：  
 $ T(n)=4n^4T(\sqrt n) +(n^6lgn+3lg^7n)(2n^2lgn+lg^3n) $  
 另外，如果你可以 - 请你检查我是否在正确的道路上：  
 $ T(n)=T(n-1)+\frac{1}{n}+\frac{1}{n^2} $  
 解决它我尝试使用：$ T(n)-T(n-1)=\frac{1}{n}+\frac{1}{n^2} $ iff $ (T(n)-T(n-1))+(T(n-1)-T(n-2))+\ldots+(T(2)-T(1))=\frac{1}{n}+\frac{1}{n-1}+...+\frac{1}{n^2}+\frac{1}{\left(n-1\right)^2}+.... $ iff $ (T(n)-T(n-1))+(T(n-1)-T(n-2))+\ldots+(T(2)-T(1))=T(n)=T(1)+\sum_{k=2}^n\frac{1}{n}+\sum_{k=2}^n\frac{1}{n^2} $然后使用谐波级数公式。 但是我不知道如何继续并完成它并找到解决它的渐近边界  
 我希望在第二个我正确的道路上。 但是我不知道如何解决第一个问题。 如果我犯了什么错误，请告诉我正确的方法，以便改善我的错误。  
 非常感谢您的帮助  
 对不起，由于某种原因，数学在这里显示不正确 

i am stuck on this one and i don't know how to solve it, no matter what i try i just can't find a way to play with the function so i can represent it in a way that will allow me to find a g(n), so that g(n) is T(n)∈Θ(g(n)) 
the function i am having trouble with is: 
$T(n)=4n^4T(\sqrt n) +(n^6lgn+3lg^7n)(2n^2lgn+lg^3n)$ 
additionally, if you can - could you please check if i am on the right path with: 
$T(n)=T(n-1)+\frac{1}{n}+\frac{1}{n^2}$ 
to solve it i tried to use: $T(n)-T(n-1)=\frac{1}{n}+\frac{1}{n^2}$ iff $(T(n)-T(n-1))+(T(n-1)-T(n-2))+\ldots+(T(2)-T(1))=\frac{1}{n}+\frac{1}{n-1}+...+\frac{1}{n^2}+\frac{1}{\left(n-1\right)^2}+....$ iff $(T(n)-T(n-1))+(T(n-1)-T(n-2))+\ldots+(T(2)-T(1))=T(n)=T(1)+\sum_{k=2}^n\frac{1}{n}+\sum_{k=2}^n\frac{1}{n^2}$ and then using the harmonic series formula. however i don't know how to continue and finish it and find the asymptotic boundaries to solve it 
i hope that on the second i am on the right path. however i don't know how to solve the first one at all. if i've done any mistakes, please show me the right way so i can improve my mistakes. 
thank you very much for your help 
sorry that for some reason math doesn't show correctly here

原文：https://stackoverflow.com/questions/49990600

更新时间：2022-03-05 12:03

最满意答案

 在第二行GetFileByServerRelativeUrl方法中，它需要文件的相对url，请查看下面的工作演示：  
 ClientContext context = new ClientContext(@"http://sp2016/sites/test");

    Microsoft.SharePoint.Client.File file = context.Web.GetFileByServerRelativeUrl(@"/sites/test/Documents1/folder2/test.txt");

    ListItem lstitem = file.ListItemAllFields;
    context.Load(lstitem);
    context.ExecuteQuery();

    lstitem["Title"] = "Mercedes";
    lstitem.Update();
    context.ExecuteQuery();
 
  

In Line two GetFileByServerRelativeUrl method, it needs the file relative url, please check the working demo below: 
 ClientContext context = new ClientContext(@"http://sp2016/sites/test");

    Microsoft.SharePoint.Client.File file = context.Web.GetFileByServerRelativeUrl(@"/sites/test/Documents1/folder2/test.txt");

    ListItem lstitem = file.ListItemAllFields;
    context.Load(lstitem);
    context.ExecuteQuery();

    lstitem["Title"] = "Mercedes";
    lstitem.Update();
    context.ExecuteQuery();

难以渐近（递归）函数（算法分析）(difficult asymptotic(recurrence) function (algorithm analysis))

最满意答案

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