Though some new regulations on public websites requiring them to be accessible to people of all handicaps will change that unless we can get some specialized UI people to do it for us
I do dread if we need to look closed into the front end though
Seeing as the back-end stuff is in some places a bit of a mess, so there is the possibility that the front-end will be as well
And there are only so many times we can tell the customer that something will take many times the usual amount of time due to the poor quality of the system we have taken over
@s.harp: I was going to say it's from Latin, although the context isn't quite right. But I looked at the link you gave. The OP is abbreviating Professor, I'm pretty sure.
@TedShifrin ohhh, my secondary interpretation was that people were basically using pro as the slang for professional or "really-good-at-this-thing guy", which also made sense and made these conversations seem rather endearing
im guessing that if it is not empty ore comprised of isolated points then it (at least) admits a smooth regular path $(-\epsilon, \epsilon)\to f^{-1}(\{0\})$, ie we dont have osmething like not locally path-connected
@s.harp Any closed subset of $\Bbb R^n$ is level set of a differentiable function.
You just have to smoothen out the distance function well enough. I can draw a picture of a smooth function $\Bbb R \to \Bbb R$ whose zero locus is the Cantor set.
This came up studying fully connected networks like Hopfield networks. Suppose something like below for a function space S:
$$F \subseteq S, \ \forall{f} {\ \epsilon \ } F: f(F) \rightarrow f', f' {\ \epsilon \ } F', F' \subseteq S.$$
What does the formal nomenclature look like for such a set...
Let f be a map from $\mathbb{R}^2$ to $\mathbb{R}^2$ be defined as $f(x,y)=(x^2,y^2)$ let T from $\mathbb{R}^2$ to $\mathbb{R}^2$ be the map $T(x,y)=(2x,4y)$ then to show that f is differentiable at $(1,2)$ with derivative T, the following change of variables is made:
$$\lim_{(x,y) \rightarrow...
@LegionMammal978: Way too hard. Consider $0a=(0+0)a$.
@topologicalmagician The point is that you can always set $x=a+h$ (whether with real numbers or with vectors), so letting $x\to a$ is equivalent to letting $h\to 0$. If you look at my YouTube lectures, you'll see that I always use $h\to 0$ for this because it saves algebra/steps.
(Since you seem to be lacking confidence, note that the definition of $x\to a$ is that $\|x-a\|\to 0$ (here I'm assuming vectors). But by definition $x-a = h$, so $\|h\|\to 0$, which means $h\to 0$. And all vice versa. OK?)
@TedShifrin I'm not exactly sure how it can be simplified further, using strictly the axioms. To cancel out the original $0a$, we need to introduce a $-0a$, and then we use $0a+0a=(0+0)a=0a$ to remove the additional $0a$ we introduced, which ultimately takes 6 steps.
I suppose $-0a$ in particular isn't necessary, but I fail to see how it would make the chain any shorter.
@LegionMammal978: Did you try doing what I said? I take three more steps from where I am. $0a=(0+0)a = 0a + 0a$. Now add $-0a$ to both sides, and I'm done.
This came up studying fully connected networks like Hopfield networks. Suppose something like below for a function space S:
$$F \subseteq S, \ \forall{f} {\ \epsilon \ } F: f(F) \rightarrow f', f' {\ \epsilon \ } F', F' \subseteq S.$$
What does the formal nomenclature look like for such a set...
Let f be a map from $\mathbb{R}^2$ to $\mathbb{R}^2$ be defined as $f(x,y)=(x^2,y^2)$ let T from $\mathbb{R}^2$ to $\mathbb{R}^2$ be the map $T(x,y)=(2x,4y)$ then to show that f is differentiable at $(1,2)$ with derivative T, the following change of variables is made:
$$\lim_{(x,y) \rightarrow...
