Kogasa

For an alternative that doesn’t sound insane, try turkey a la king. Turkey in a creamy sauce on puff pastry or toast.

She didn’t step on it, she apparently used her thumb and damaged one of the buns in the pack

Not really, you need to have a basic understanding at least

You might be thinking of a [connection of an affine bundle](https://en.wikipedia.org/wiki/Connection_(affine_bundle). You could learn it through classes (math grad programs usually have a sequence including general topology, differential topology/smooth manifolds, and differential geometry) or just read some books to get the parts you need to know.

Manifolds and differential forms are foundational concepts of differential topology, and connections are a foundational concept of differential geometry. They are mathematical building blocks used in modern physics, essentially enabling the transfer of multivariable calculus to arbitrary curved surfaces (without relying on an explicit embedding into Euclidean space). I think the joke is that physics students don’t typically learn the details of these building blocks, rather just the relevant results, and get confused when they’re emphasized.

For a tl;dr about the concepts mentioned:

• A manifold is a curve, surface, or higher-dimensional object which locally resembles Euclidean space around each point (e.g. the surface of a sphere is a 2D manifold; tiny person standing on a big sphere perceives the area around them to resemble a flat 2D plane).

• Differential forms are “things that can be integrated over a manifold of the corresponding dimension.” In ordinary calculus of 1 variable, that’s a suitably regular function (e.g. a continuous function), and we view such a function f(x) as a differential form by writing it as “f(x) dx.”

• A connection is a way of translating local tangent vectors from one point on a manifold to another in a parallel manner, i.e. literally connecting the local geometries of different points on the manifold. The existence of a connection on a manifold enables one to reason consistently about geometric concepts on the whole manifold.

No, they’re not sure. You’re correct.

Definitely not.

All people. 320kbps mp3 is completely audibly transparent under all normal listening conditions. It’s a low-tier audiophile meme to claim otherwise but they will never pass a double-blind test.

$7 Taco Bell? What’s your order, a tortilla? :'(

Dawg you’re unhinged

It’s not the “where” specifically I’m correcting, it’s the “when.” The model is trained, then the query is run against the trained model. The query doesn’t involve any kind of internet search.

It doesn’t search the internet for cats, it is pre-trained on a large set of labelled images and learns how to predict images from labels. The fact that there are lots of cats (most of which have tails) and not many examples of things “with no tail” is pretty much why it doesn’t work, though.

That’s what the /etc/foo.conf.d/ is for :DDDDD

It means they admit they were wrong and you were correct. As in, “I have been corrected.”

Not the same issue

If you take immortality, you also probably need to take healing. Being mortally wounded and unable to die sounds, uh, bad.

The argument describes an algorithm that can be translated into code.

1/(1-x)^(2) at 0 is 1

(1/(1-x)^(2) - 1)/x = (1 - 1 + 2x - x^(2))/x = 2 - x at 0 is 2

(1/(1-x)^(2) - 1 - 2x)/x^(2) = ((1 - 1 + 2x - x^(2) - 2x + 4x^(2) - 2x^(3))/x(2) = 3 - 2x at 0 is 3

and so on

If your binding is Alt+X, try pressing it down, tapping shift (press and release), then releasing X. The release keybind shouldn’t trigger.

I tried the workaround with all modifier combinations but this still happens, even if you bind Shift+(Your Binding). Pressing “shift” while the a chord is held prevents the release event from being handled.

Let f(x) = 1/((x-1)^(2)). Given an integer n, compute the nth derivative of f as f^((n))(x) = (-1)^{(n)(n+1)!/((x-1)}(n+2)), which lets us write f as the Taylor series about x=0 whose nth coefficient is f^((n))(0)/n! = (-1)^(-2)(n+1)!/n! = n+1. We now compute the nth coefficient with a simple recursion. To show this process works, we make an inductive argument: the 0th coefficient is f(0) = 1, and the nth coefficient is (f(x) - (1 + 2x + 3x^(2) + … + nx^(n-1)))/x(n) evaluated at x=0. Note that each coefficient appearing in the previous expression is an integer between 0 and n, so by inductive hypothesis we can represent it by incrementing 0 repeatedly. Unfortunately, the expression we’ve written isn’t well-defined at x=0 since we can’t divide by 0, but as we’d expect, the limit as x->0 is defined and equal to n+1 (exercise: prove this). To compute the limit, we can evaluate at a sufficiently small value of x and argue by monotonicity or squeezing that n+1 is the nearest integer. (exercise: determine an upper bound for |x| that makes this argument work and fill in the details). Finally, evaluate our expression at the appropriate value of x for each k from 1 to n, using each result to compute the next, until we are able to write each coefficient. Evaluate one more time and conclude by rounding to the value of n+1. This increments n.

Rant incoming, because I just happened to be dealing with this today.

I tried the push to talk binding method in sway, and found it was extremely fragile because if you press any modifier keys before releasing the button, the release binding won’t trigger and your mic will stay open. This is especially problematic for my use case as I use both my push to talk hotkey and my modifier buttons while gaming.

I just use a toggle mute hotkey instead of push-to-talk though. Just gotta remember to re-mute, and have a visual indicator for your mic status so you don’t accidentally leave it in the wrong state.

Another workaround is binding all possible combinations of modifier keys explicitly, and passing in --device-id so that the binding doesn’t consume the keypress and instead forwards it on to the application. This should perfectly emulate the behavior of an X-style global hotkey. With 4 mods (super, ctrl, shift, alt) and 2 bindings per combination there are 2^5 = 32 total bindings for each such hotkey. However there’s still no telling if the key-release event could be missed somehow and leave your mic open. I don’t know how reliable this method is.