Zelotes T80 Mouse Driver 44

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Category:Computer miceQ:

Why did $E$’s eigenvalues approach 1/2 as the mass increased?

Consider a frame of reference where the $E$’s eigenvalues approach 0 at infinite distance.
Let’s say that $\rho$ is a differentiable function so that $E’$ is invertible, then we can say that $E'(x)$ is an $n\times n$ positive definite matrix with unit determinant so that
$$E'(x) = \frac{1}{2} (I + 2\rho'(x)v\otimes v)$$
where $I$ is the $n\times n$ identity matrix and $\otimes$ denotes the Kronecker product.
Now let $M$ be any $n\times n$ positive definite matrix so that $\rho'(x)M(x)$ is also positive definite. Then
$$E'(x) = \frac{1}{2} (I + 2\rho'(x)M(x)v\otimes v)$$
Thus
$$E(x) = \frac{1}{2} (I + 2\rho(x)M(x)v\otimes v)$$
So the eigenvalues of $E$ become
$$\frac{1}{2} (1 + \rho(x)M(x))\pm\frac{1}{2} (\rho(x)M(x))$$
So if we have a $n\times n$ symmetric matrix $M(x)$ with $m_{ii}(x)$ and $m_{ij}(x)$ as the diagonal and off-diagonal elements of $M(x)$ respectively, then we have
$$\frac{1}{2} (1 + \rho(x)M(x)) + \frac{1}{2} (\rho(x)M(x)) = 1 + \frac{1}{2} (\rho(x)M(x))$$
or
$$\frac{1}{2} (1 + \rho(x)M(x)) + \frac{1}{2} (\rho(x)M(x)) = 1$$
and
$$\

 

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Category:Computer miceQ:

What is the difference between shell and shell script?

What is the difference between shell and shell script?

A:

A shell script is a shell program.
Shell scripts are often written in a scripting language like sh, bash or ksh.
The most common use of shell scripts is to automate a command sequence, which might look like this:
sh -c ‘rm /usr/bin/du -sh’

The above is a single line in a script, that deletes /usr/bin/du -sh.
Typically such a command sequence would be executed via the shell in an interactive shell session.
An example of an shell script that is not a shell script, is:
ls

This will list the current directory.
One difference between shell scripts and programs in general, is that shell scripts are usually written as if there is only one user, and do not require a terminal.

U.S. Army to cut by more than 100,000 personnel

The United States Army is planning on cutting more than 100,000 active-duty troops, becoming the first service to start shrinking after the wars in Iraq and Afghanistan ended.

The army is targeting to reduce the size of its active-duty force to 440,000 by early 2023, from about 570,000 today.

“We have no plans to go back to a pre-Iraq-and-Afghanistan, pre-sequestered size,” acting Army chief of staff Gen. Mark Milley said Tuesday. “We are where we are with the force structure that we have at this time.”

In addition to the cuts, the Army has also announced that it plans to shrink the size of its infantry, which will lose more than a third of its troops to the Marines.

To meet its target, the Army is planning to cut its ranks by 100,000 through early next year.

“Over the next few months, if you’re serving today, your best guess is you’ll be out the door at the end of this fiscal year,” he said.

To help close the gap, the service has announced it will expand the traditional three-year
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