Make unallocated space into a partition with no OS

Kinda messed up and deleted C drive instead of formatting it because a site said the install would recognize it and make a gpt partition out of it (did not happen). So now I have just one partition and I don’t want to install win on that one since I use it a lot, I’d like it and my windows install separate.

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What is an asteroid called if it is in deep space?

Pretty much every definition I’ve read about asteroids is something like “Asteroids are rocky worlds revolving around the sun that are too small to be called planets.”

What would an asteroid be called if it was in deep space, not orbiting any sun?

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Why vector space only requires a finite sum?


I know this question might be a bit weird. More out of curosity.

I know from the definition that one of the criteria for $V$ to be a vector space is that $forall v_1,v_2 in V, v_1 + v_2 in V$.

But how about countable infinite sum? From my viewpoint, it sounds more like a vector space when infinite countable sum is still contained in $V$.


I have this question because when I look at the $l^{infty}$, i.e., the bounded sequence space. I feel like if I can recursively sum up two proper sequence in $l^{infty}$, I can make resulting sequence larger than any given bound. So it sounds like, it might be out of the vector space that I am interested in.

I know the definition of vector space only requires “two element sum still contained in the space”. Just curious if there is a reason for not allowing infinite sum.

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Absolute convergence of a random variable in a countable space

I am reading Jacod and Protter’s Probability Essentials and I am struggling to understand the following string of inequalities:

enter image description here

where $L^1$ is the space of real valued random variables on ($Omega$, $A$, $P$) for $sigma$-algebra $A$ and probability measure $P$. Note that in this chapter, $A$ is defined s.t. $A$=$2^Omega$ for a countable $Omega$.

Specifically, I do not understand why the expecation would be “bifurcated” in to the sum of the two subsets of the image of $omega$ such that $lvert X( omega) rvert <1$ and $lvert X( omega) rvert ge 1$.

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Sample space vs Support.

Can anyone give me a good and clear example (or, examples) where a value could be found in the Support of an RV, but, not in the Sample-space of the experiment, and vice-versa?

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Continuity constraints between observed and latent space

Given a generative model as shown below, I am interested in the analysis of the to observable variables $x,y$ given the latent space $z,w$.

enter image description here

Do you know if there exists a general continuity constraint for the manifolds so that I can say something like $Delta x Leftrightarrow Delta z Leftrightarrow Delta y$ (where $Delta$ denotes a change or small pertubation)?
Further, is there something like a discontinuity constraint like: If $Delta xnot{propto} Delta y$, then there exist some latent $w$ which is not shared by the observations.

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Consider the equation $ax + by = c$ in 2-space and the slope for that equation where $a$ and $b$ are real numbers.

Consider the equation $ax + by = c$ in 2-space and the slope for that equation where a and b
are real numbers.

Explain using multiple representations why it is equivalent to say that a linear equation moves vertically $−a/b$ units for every $1$ unit of movement horizontally as it is to say that linear equations move vertically $-a$ units for every b units of horizontal movement.

I am having a hard time showing this. Should I plug in values and show rise/run graphically and maybe use a table? How do I explain this?

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Can a set of vectors be linearly independent in one vector space, but be linearly dependent in another vector space?

For example,

let S = {$(x_1, x_1)| x_1 in mathbb R$} be a subspace of $mathbb R^2$.

By definition, dim(S) = 1, and dim($mathbb R^2$) = 2.

Then the set {(1, 1)} only has one vector, so is it linearly independent in S, but is linearly dependent in $mathbb R^2$?

I know this doesn’t make any sense, but we learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is in.

Since (1,1) is in both S and $mathbb R^2$, but the set {(1, 1)} only spans S, how come it is not only linearly independent in S and linearly dependent in $mathbb R^2$ (since {(1,1)} does not span $mathbb R^2$).

Sorry for this stupid question

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In Powershell, how to check drive space on remote server with non-admin account?

I work in an active directory environment with many servers. I have a user account that I would like to use to check the status of other servers, without giving this account full administrative access to these other servers. Specifically, I want to check the drive space on these servers, and I’d like to do it with Powershell.

I have executed Enable-PSRemoting on the target server, and I can successfully invoke Get-PSDrive on them remotely using administrator credentials. The results are returned almost immediately, and include information about the used / free space on all drives.

However, when I run the same command (Invoke-Command -computer remoteserver1 {Get-PSDrive -PSProvider FileSystem}) as the non-administrative user, the results come back very slowly (takes about 30 seconds), and none of the drives have any information about their used / free space.

What I have done so far:

  • I have added the non-administrative user account to the Remote Management Users group on the target server.
  • Edited SDDL for scmanager (on the target server) to add the same DiscretionaryACL for Remote Management Users as Built-in Administrators have.
  • Per this post, I have granted this user WMI permissions in wmimgmt.exe > WIM Control (Local) > (right click) > Properties > Security tab > Expand ‘Root’ > click on SECURITY node > click ‘Security’ button > added non-admin user with full access.
  • Added user to the Distributed COM Users group on the target server.

Some also suggested trying Invoke-Command -computer remoteserver1 {Get-WmiObject -Class Win32_MappedLogicalDisk} to troubleshoot, but it comes back ‘Access is denied.’ I believe if I could get Get-WmiObject working successfully for this limited user, it would solve my issue.

What should I do to get this limited user account the access they need to check drive space on other servers? without giving the account admin rights, and preferably without having to map and unmap any drives?

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Lens space and its generalization, homotopy/homology groups

The three-dimensional lens spaces $L(p;q)$ are quotients of $S^3$ by $mathbb{Z}/p$-actions. More precisely, let$p$ and $q$ be coprime integers and consider $S^3$ as the unit sphere in $mathbb C^2$. Then the $mathbb{Z}/p$-action on $S^3$ generated by the homeomorphism
:$$(z_1,z_2) mapsto (e^{2pi i/p} cdot z_1, e^{2pi i q/p}cdot z_2)$$
is free. The resulting quotient space is called the ”’lens space”’ $L(p;q)$.

This can be generalized to higher dimensions as follows: Let $p,q_1,ldots,q_n$ be integers such that the q_i are coprime to $p$ and consider $S^{2n-1}$ as the unit sphere in $mathbb C^n$. The lens space $$L(p;q_1,ldots q_n)$$ is the quotient of $S^{2n-1}$ by the free $mathbb Z/p$-action generated by
: $$(z_1,ldots,z_n) mapsto (e^{2pi iq_1/p} cdot z_1,ldots, e^{2pi i q_n/p}cdot z_n).$$

my inquiries:

  1. In Wikipedia it says: In three dimensions we have $L(p;q)=L(p;1,q).$ Why is that? Isnt that one is in 3 real and the other in 5 real dimensions?

  2. How to show: The fundamental group of all the lens spaces $L(p;q_1,ldots, q_n)$ is $mathbb Z/pmathbb Z$ independent of the $q_i$?

  3. $L ( 5 ; 1 )$ and $L ( 5 ; 2 )$ were not homeomorphic even though they have isomorphic fundamental groups and the same homology, though they do not have the same homotopy type. What are the homotopy groups of $L ( 5 ; 1 )$ and $L ( 5 ; 2 )$, respectively? And their homology groups?

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