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I find today that I can no longer mount my Netgear ReadyNas drive because apparently it is using SMB1 and Windows 10 have disabled since release 1709. This is very problemaic because I need to access these files and because I have lost the admin password for the ReadyNas itself so I cannot login in order to update/reconfigure it to use SMB2 instead.  Microsoft refers me  to  which in turn refers me to  if I really want re-enable SMB1  And I successfully did re-enable SMB1 by running  Enable-WindowsOptionalFeature -Online -FeatureName smb1protocol   from Powershell as Adminstrator.  However even after reboot I still cannot re-establish connection to my Netgear from Windows Explorer, it gives the same error  What else do I need to do ?  PS C:\WINDOWS\system32&gt; Get-WindowsOptionalFeature –Online –FeatureName SMB1Protocol   FeatureName      : SMB1Protocol DisplayName      : SMB 1.0/CIFS File Sharing Support Description      : Support for the SMB 1.0/CIFS file sharing protocol and the Computer Browser protocol. RestartRequired  : Possible State            : Enabled CustomProperties :                    ServerComponent\Description : Support for the SMB 1.0/CIFS file sharing protocol and the Computer                    Browser protocol.                    ServerComponent\DisplayName : SMB 1.0/CIFS File Sharing Support                    ServerComponent\Id : 487                    ServerComponent\Type : Feature                    ServerComponent\UniqueName : FS-SMB1                    ServerComponent\Deploys\Update\Name : SMB1Protocol   Solution  Thanks Bob, SMB 1.0/CIFS File Sharing Support was enabled but within it SMB 1.0 Client was not enabled, checking that box and rebooting fixed the issue. I am working on SharePoint 2010, I have a custom list with columns Start date , End date, Created by, Approval Status. I have created a page where I am looking to filter the list and show the result depending on the approval status which is a column in custom list as shown in figure . Can some one help me out on how to approach for this. How to avoid noisy renders in Cycles? I know that this question is a bit old or maybe overdone, but I'm running the latest edition of Blender and when I use the Cycles engine I get really fuzzy results and I can't find a simple, non-compromising solution to the quality.  I don't know exactly what I'm doing wrong here, but I'm sure others share my frustration.  Leaving everything in default settings, I started a new project and switched to the Cycles render.  I duplicated the cube and set colors for two Diffuse BSDF shaders, then went ahead and made walls and a floor all sharing the same diffuse shader.  I made two windows to emit blue light.  The screenshot has gaps between the walls because of the setup:  I'm aware that it's considered very incomplete and many have come to terms with some amount of noise, but is this normal?  Furthermore, with default settings or not, I can't change the amount of noise produced.  Supposedly, using "progressively refine" in the rendering tab should make it refine itself indefinitely, however Cycles only counts to 10 refines and then stops, creating a render which is a pixel-perfect match of the ordinary rendering method when it's completed.  I understand that there are tricks that can be used to reduce noise, but at the cost of extra processing time and usually small quality conflicts, like blurring.  My question essentially boils down to is this Cycles' current limit? Assume a p x n data matrix  (e.g. p independent samples of n variables). The n variables are correlated according to a correlation matrix $\rho$. Now if you consider the data matrix as some state t (t=time)  and based on this state I would like to calculate the next timestep (t+1). Therefore the correlation matrix would be allowed to have values &lt; 1 in the diagonal, which means that the diagonal elements would indicate the influence from each variable $x_{p,i}[t]$ on $x_{p,i}[t+1]$. So I would be interested to find a function f like this: $x_{p,i}[t+1] = f(x_{p,i}[t],\rho)$ so that the p x n data matrix  consisting of the $x_{p,i}[t+1]$ replicates the $\rho_{ij}$ (except the diagonal element). f should be something like this: $x_{i,j}[t+1] = \sum_{(k=1\ldots n)}\rho_{i,k} \cdot x_{kj}[t] + e_{i,j}.$ This means specifically, I not only want to calculate a data matrix, that replicates the correlation matrix as such, but also to let each observation be depending on its own state and linear dependent on the previous state of all the intercorrelated variables of its own state. For ease one can assume that the p variables are distributed normal and $e_{ij}$ are independent and distributed with mean$(e_{i,j}) = 0 $. Then for a numerical simulation it would be interesting  to find a formula to calculate the variance/standard deviation of each of the j variables (sd($e_{[.,j]} = ?).$ Thank you for any help! Can't redirect cut output How did Luke know where to land on Dagobah? Explain why $E(X) = \int_0^\infty (1-F_X (t)) \, dt$ for every nonnegative random variable $X$ Let $X$ be a non-negative random variable and $F_{X}$ the corresponding CDF. Show,   $$E(X) = \int_0^\infty (1-F_X (t)) \, dt$$   when $X$ has : a) a discrete distribution, b) a continuous distribution.   I assumed that for the case of a continuous distribution, since $F_X (t) = \mathbb{P}(X\leq t)$, then $1-F_X (t) = 1- \mathbb{P}(X\leq t) = \mathbb{P}(X&gt;  t)$. Although how useful integrating that is, I really have no idea. System log full of session and slice messages How can I block incoming SMS from certain senders? How do I cash a USD cheque in Europe, quickly? I'll receive a USD cheque shortly. I prefered payment by money transfer, but all attempts to do so - i.e. SWIFT/IBAN, Payoneer, transferwise - failed. Now I will receive a cheque by mail.  I contacted my bank and they say that they can cash a cheque for me, given a (prob hefty) fee. The bank employee was so kind to warn me, that in case of a cheque the funds are cashed, but reserved until the legal reimbursement period ends, which apparently is between 3 and 6 months.   Is there really no quicker way to receive payments by cheque and have access to the funds quickly? amsmath align environment: row spacing How can I stretch the spacing between consecutive lines in an align* environment? I have a few lines with \fracs in which both the numerator and the denominator consist of textual phrases, not math symbols, and the result looks too cluttered.  \begin{align*} X &amp;= \frac{\textit{Some text here}}{\textit{And some text here}}\\ Y &amp;= \frac{\textit{Will make everything}}{\textit{Look too close together}} \end{align*} I can't run this latex script Latex create a second keywordstyle using listing doesn't work The number $2^{3^n}+1$ is divisible by $3^{n+1}$ and not divisible by $3^{n+2}$. Prove that for every $n\in \mathbb N$, $2^{3^n}+1$ is divisible by $3^{n+1}$ and not divisible by $3^{n+2}$.   I was able to prove that $2^{3^n}+1$ is divisible by $3^{n+1}$ using induction. First, I check whether the statement is true for $n=1$. The statement holds since: $$2^{3^1}+1=9$$ $$3^{1+1}=9$$ Second, I assume that the statement is true for every $n\in \mathbb N$. Finally, I take that $n=n+1$. From that, the original expression becomes $(2^{3^n})^3+1$, which is a sum of cubes, and the same as: $$(2^{3^n}+1)((2^{3^n})^2-2^{3^n}+1)$$ As I assumed that $2^{3^n}+1$ is divisible by $3^{n+1}$, and one of the factors is $2^{3^n}+1$, the entire expression must be divisible by $3^{n+1}$, and so the statement is true for every $n\in \mathbb N$.  Now, I don't know how to prove that $2^{3^n}+1$ isn't divisible by $3^{n+2}$. Please explain how I can do so step-by-step, without using complicated terms. How to get the caret column (not pixels) position in a textarea, in characters, from the start? How do you get the caret position in a &lt;textarea&gt; using JavaScript?  For example: This is| a text  This should return 7.  How would you get it to return the strings surrounding the cursor / selection?  E.g.: 'This is', '', ' a text'.  If the word “is” is highlighted, then it would return 'This ', 'is', ' a text'. Find the value of $P(1)$ Let $P (x) = x^2 + bx + c$, where $b$ and $c$ are integer.       If $P(x)$ is a factor of both $x^4 + 6x^2 + 25$ and $3x^4 + 4x^2 + 28x  + 5$, find the value of $P(1)$.   I am not being able to solve this.Some hints/suggestions?  Tell me if I can edit to improve this question.I could'nt try much.Got stuck. PHPExcel Sumif and Skip How was Dumbledore able to defeat Grindelwald, since Grindelwald had the Elder Wand? When to use self over $this? In PHP 5, what is the difference between using self and $this?  When is each appropriate? When incident light passes through a hydrogen gas, for example, does it have 50% chance (since it's a 1:1 ratio of protons to electrons) of getting absorbed by the proton? Any chance at all? If no, why not? Does a proton have a "bandgap"? If yes, what happens when a photon is absorbed by a proton?