"that has at least 1 root in L"

so, for example, let's take Q(i) as an extension of Q

$x^3 - 2$ is an irreducible polynomial over Q which has no root in Q(i), so it doesn't need to split in Q(i)

whereas x^2+1, for example, has a root in Q(i) (namely i), so it splits as (x+i)(x-i)

Q(i) is actually a normal extension of Q, but Q(i) is not algebraically closed because x^3-2 doesn't have a root in Q(i)

should we include the polynomial associated to the normal extension then?

*the min polynomial

what do you mean by "polynomial associated to the normal extension"?

i mean just for your example Q(i) over Q

yes? what's the "polynomial associated to the normal extension"?

x^2 + 1

why is it associated to it?

because its roots split in Q(i)

but x-1 is also such a polynomial

I think what you're missing is the word "every"

"little detail"

i just read ur example a couple of times to understand now.

Hi @Ted

Rehi, a @Balarka :)

Finally we're in the same time zone!

Oh? How is that?

Haha, as in online. I think we missed each other consistently during the last few weeks

Did you see my radioactivity example? I thought you'd like that one.

Oh, no. I said hello to you just a few hours ago when we were both here :P

I think that was a rare event in the past few days. I worked on assignments throughout the evening and slept early.

Yeah, I saw what you wrote, although I did take time to ponder too thoroughly.

OK :)

Hey, can I ask a question to someone who has a bit more experience with MSE?

It's a very cute computation, let me know if I should send it to you sometime (or not, it's not that big a deal).

I remember that when I taught correlation in probability there were some surprising examples of correlation, zero and nonzero. Of course, that was 6 years ago (yikes!), so I'd have to look at my notes to remember.

I've got a series of downvotes on older answers, is there anything i can do about it?

Depends what the question is, @supinf.

Aha yes I do remember you gave me one example out of your notes, regarding independence!

Were they malicious or were your answers not good, in retrospect?

@Balarka: I stole it from some textbook, in truth.

@TedShifrin I once saw some notes where positively correlated is drawn as a straight line with positive slope, negatively correlated with negative slope, and 0 correlation with 0 slope

@Leaky: It's all about dot products.

Agreed.

(the joke is that 0 correlation isn't 0 slope)

@TedShifrin it seems suspicious to me. I will look for an example. I got -38 reputation today.

I directed a masters thesis in which the author fleshed out tons of connections between statistics and the geometry of linear algebra.

@LeakyNun Missed that completely. You're right of course.

Statistics: A Geometrical Approach

@TedShifrin math.stackexchange.com/questions/3820186/… one example, 1 downvote an hour ago. All downvotes today were 1-2 hours ago i think.

@supinf: I'm no moderator, but I know that they run a computer check that's supposed to catch malicious mass downvoting or its opposite.

it's all fake internet points

A lot of manipulation with Gaussian variables, independence, etc is about linear algebra.

@supinf: I would not downvote, but I certainly don't like your talking about upper and lower bounds of $F^\perp$. What in the world does that mean?

@LeakyNun you are not wrong about it, but it does not seem to be appropriate. If the answers were bad some comments would have been nice, but I got no comments.

I don't like your answer.

A random theorem I learnt 2 days ago in my statistics class: If $P$ is a symmetric projection matrix and $Q$ is a symmetric positive semidefinite matrix then $I - P - Q$ is positive semidefinite iff $PQ = QP = 0$

@TedShifrin that is a valid point. Upper bound on $F^\perp$ means a set $Z$ such that $F^\perp\subset Z$. I was just trying to give a hint there, not a complete solution.

@TedShifrin I also got two downvotes today on this otherwise very popular question of mine: math.stackexchange.com/questions/1818520/…

@BalarkaSen $P=0$, $Q=2I$

Since when is $-I$ positive semidefinite?

exactly

Oh I wrote iff, I meant implies.

Other direction is not true.

So we're talking about three downvotes total? That's not huge numbers. That question is interesting, but you haven't convinced me to switch from the usual definition.

@BalarkaSen alright

@TedShifrin Its 19 downvotes total today, I was just giving examples.

If $P$ is an orthoprojection, $Q$ some self-adjoint positive guy, $P + Q$ will be dominated by the identity operator only if $P$ is projection onto the kernel of $Q$, or something of this sort.

You can put a question in the moderator room asking them please to check up on this.

Very hard to interpret, very easy to prove.

@TedShifrin Thank you for your advice. I did not know about that room.

I know that sometimes a person will, having seen a question or answer of yours, look to see what other questions or answers there are. This often results in a bunch of votes.

@TedShifrin true, but then usually they would leave some comments and tell me what I did wrong?

I have a problem that I think is interesting. I want to design a function whose gradients will lead directly to the local minima from anywhere on the function in $k$ steps.

However, I feel like such a function may trivially already exist since I'm at the beginning of my research. Does anyone know of one?

Heya @Axoren. Long time!

@TedShifrin How goes it! It's good to see you're still around

@Axoren what do you mean by "steps", do you mean gradient descent?

That may or may not be the case :P

@supinf Gradient Descent, yes.

And "lead directly" means you actually get there in precisely $k$ steps?

There's a single global minimum?

Or in the neighborhood thereabouts

That's way too vague for a mathematician.

"the neighborhood thereabouts"?

then on a smaller neighbourhood it will only take $1$ step

but then the function must be $|x|$

which cannot be differentiable at $0$

I truly do not understand.

This is somewhat of an XY Problem. Let me explain further somewhat.

I have a vague notion of a function whose properties involve a global minima at -1, 1 and a global maxima at 1, 1 and it's continuous.

I'll draw an example

one moment

Are we in $\Bbb R^1$ or in $\Bbb R^2$?

A global MINIMUM at $(-1,1)$ and a global MAXIMUM at $(1,1)$?

The function is $f: \mathbb R \to \mathbb R$

I was giving (x,y) coordinates

Ohhhh. Damn.

Haha.

But the $y$ values are the same?

I made a mistake and it's too late to edit

(-1, -1) and (1, 1)

OK.

And the drawing was upside down

LOL

Coulda just flipped it along the y axis

So what happens if you start at $0$?

The idea is that if you take the gradient from a point outside of $[-1, 1]$, you can be arbitrarily far away and still get to at least within $[-1, 1]$ in finite steps; ideally to the nearest extrema.

I don't see why that's true. Since the slope is constant, your step sizes will always be the same.

This is a rough drawing. Assume continuous and curvy

That's the vague shape

Oh, right, it has to change concavity

To your question about 0, it depends on if we're doing a gradient ascent or descent.

if you are far away then the gradient is small. And then gradient descent takes a long time. If you are closer it is quicker.

So it seems to me that we overshoot in both cases.

my argument still works

Which was?

So to refine what I'm talking about. I want to, while in $(-\infty, -1)$ searching for the minimum, land in $[-1, 1]$ in $k$ steps. While in $(1, \infty)$ searching for the maximum, land in $[-1, 1]$ in $k$ steps.

And your steps are precisely what?

Essentially, the land between -1 and 1 is "home"

$x\mapsto x-f'(x)$?

Yes.

Negative or Positive thereof depending on if looking for min/max

(this is WLOG assuming the minimum is $f(0) = 0$)

I am not convinced at all. But I will think about it.

If you're talking about my situation, then no. The minimum can't be at $f(0) = 0$

I'm sorry if I interrupted a different convo

I said WLOG

No, it was your convo, but before your function. He needs to modify it.

I can just shift your graph to the right and upwards

the situation is the same

@Leaky Nun if you use gradient descent (without line search and scaling) and you are near $0$, then you are near $-1$ or $1$ in the next step.

I don't care if I get "home" faster.

I just mustn't get "home" in more than $k$ steps. My bad

the point is that near the minimum the graph will be straight lines

because that's the only way you can get "home" in 1 step

but then it won't be differentiable at that point

I'm already home if I'm at an extremum.

But you're talking about in a neighborhood around it, right?

straight line is the point

That does actually make sense.

@Leaky Nun I do not get why it should be a straight line.

that's the only way you can get home in 1 step

However, it would only be straight on one side of the extremum. We haven't put any constraints on the inner side.

So it could still be differentiable.

someone knows a thing or two about complex analysis?

it must be a straight line pointing downwards to the minimum

if you want it to be differentiable, it must go downwards still

which makes it not the minimum

Leaky Nun that does not convince me. It could also be the function $x\mapsto 2x^2$.

what is one step?

is it $x \mapsto x - f'(x)$ or $x \mapsto x - xf'(x)$?

I thought gradient descent was the first.

I can understand straight lines if you are thinking about the second "step rule" that you mention.

For me, I'm using the first.

so you want $x - f'(x) = -1$ for $x$ sufficiently close to $-1$

xf'(x) is another way of solving the problem I'm trying to solve if I wasn't in control of the function. It makes gradients farther away travel faster towards the center.

which means $f'(x) = x+1$, so $f(x) = (x+1)^2-1$

@Axoren what about the function $f(x)=x^2/2$, you always get to $0$ in $1$ step (i think).

you mean $0.5x^2$

@Keep_On_Cruising depends on what it is

@supinf Asymptote at $y = 0$ wasn't mentioned, but that's something I want this function to have.

@LeakyNun Is this for a single step towards the extermum?

towards the stipulated minimum $f(-1) = -1$

Hmm... that would make having the asymptote impossible without truncating the function to some (-u, u).

This might very well be bust.

I was kind of hoping for an activation function which punished inputs outside of the neighborhood, as opposed to tapering off normally towards -1 and 1 on the respective sides.

But with the properties I've mentioned, all that leaves is something like this

Red dotted lines would be parabola whereas the function I was hoping to have was in solid black

I can easily hack an application together that just does descent by the parabolic sections instead of the underlying function, but I worry that will have more issues.

@LeakyNun @TedShifrin @supinf thanks for humoring me on this

you're welcome

@LeakyNun determining the order of poles of an analytic function.

@Keep_On_Cruising an analytic function has no poles; you mean meromorphic.