@TedShifrin This only works for fixed polynomials, or? But the statement is true for every irreducible polynomial over a perfect field.

The gcd of two polynials is 1 iff the determinant of the sylvester matrix is not equal to zero, but one has to compute the determiant....

I was asking because I've proved that in a real-closed field $f$ is square-free iff $\gcd(f, f') = 1$ using that the irreducible polynomials in a real-closed field have at most degree $2$, so they can't have repeated roots. Then I looked at wikipedia and found about perfect fields, and they said it holds for any field of characteristic $0$.

Because $(x-a)^2 = x^2-2ax+a^2$, so it can't be irreducible, $a$ in algebraic closure

is the p-adic heat equation well understood?

@XanderHenderson But if we do that and make a lot of money we can throw an MSE-exclusive pizza party or something

@geocalc33 Mostly, depending on what you mean by it. But there are still open problems.

e.g. link.springer.com/article/10.1007/s11253-018-1496-x

(again, this relies heavily on what you mean by "the $p$-adic heat equation", as $p$-adic analysis is quite different from real analysis; no least upper bound property, the space is totally disconnected, etc)

I think that the usual approach is to use the (inverse) Fourier transform to turn multiplicative operators into pseudo-differential operators, via the spectral theorem. And yes, I am throwing around a lot of big words. I used to understand what they all mean and how they all fit together, but I am using them today just to feel smart.

But I am pretty sure that what I have written is no entirely nonsense.

@XanderHenderson I see - and what about the region of convergence of composite p-adic functions? Do you think it is trickier than in real analysis?

@Jakobian If the function $f:[a,b]\to\mathbb{R}$, then how would one compute $f(a^-)$?

@geocalc33 I don't think I understand the question. What do you mean by "composite $p$-adic functions"? What is the region of convergence of such an object?

@robjohn sorry, I meant $f(a^+)$

got it mixed up

@XanderHenderson Yeah I guess it doesn't have to be composite (two functions composed together) I guess what I'm asking is do you think it is more laborious to determine region of convergence of p-adic functions, than for real functions?

What is the "region of convergence"?

Maybe you are thinking of a function given by a power series, and asking where that power series converges? If so: no, I don't think that the problem is very different in the $p$-adic when compared to, for example, the reals or the complex numbers. This is because you will ultimately be looking at series like $$ \sum \|a_n x^n\|_p, $$ which are series of real (well, rational) numbers.

But I'm not sure if that is the question you are asking.

You’ve never watched Food Wars, have you? @XanderHenderson

@Jakobian Over a fixed field every irreducible polynomial is the minimal polynomial of all its roots. In characteristic zero this means an irreducible polynomial and its derivative cannot share a common factor for degree greater than one because this would contradict the property to be the minimal polynomial. Only in characteristic non zero the derivative can be the zero polynomial with every element as a root and it doesn't yield a contradiction.

Probably this is the detailed proof that Ted Shifrin had in mind.

I realize, that's why I commented that it doesn't work for characteristic $p$ because $f'$ could be zero

thanks for your comment though

@XanderHenderson Did Beethoven ever compose a pizza? I have often composed myself.

@Jakobian Yes.

@robjohn You know, a group of archaeologists recently exhumed Beethoven. When they opened his coffin, they found him madly erasing his works. When asked what he was doing, he responded, "Decomposing."

<groan>

And I thought I’m unfunny

That wasn't unfunny, just very punny.

I thought it was funny

@robjohn do you have any ideas about the problem I sent a while ago?

@冥王Hades which one is that?

This @robjohn

@冥王Hades Hadn't seen that until now.

i'm going to need to see more vast expanses of white space around the pointer and at the bottom of that image before i can give that the proper amount of attention

@robjohn I didn’t write on it since I think it’s obvious but MN is parallel to AB

@leslietownes yeah but then it’ll destroy my eyes

Oh hey I see a small triangle in there

@leslietownes Better?

wonderful. just enough of a margin for a proof of some number theory triviality i've been working on.

See? You can be funny if you try Xander

@AkivaWeinberger I made a HTML / JavaScript thing you can use to make diagrams for your domino puzzle. It can dump the diagram to SVG (and copy it to the clipboard). Click on a cell edge to place a domino that straddles that edge. Click a domino to delete it.

sagecell.sagemath.org/…

It doesn't bother to check for overlaps, but hopefully that's not a problem. ;)

I also updated my polyomino exact cover code to do SVG output. I haven't done a version (yet) for the domino puzzles, but here's an example of a tromino cover:

@冥王Hades since M and N are midpoints, the triangles are similar.

So, yes, the lines are parallel

@robjohn have you looked into it yet?

I mean always true if K is not an immaginary quadratic extension.

Any ideas on that would be helpful.

Furthermore, it is meant that statement one is always false under these conditions.

And K contains no imaginary quadratic subfield instead of K is not an imaginary quadratic subfield, sorry.

There is only one natural way to measure the mean in the normal distribution

@geocalc33 There is only one way to measure the mean of any distribution.

Integrate $x f_X(x)$ over the reals.

Or whatever notation you prefer---here, $f_X$ is the pdf.

@XanderHenderson the mean can be viewed as the "balancing point" of the distribution

@geocalc33 I know what the mean is.

I don't understand your comment.

I'm just saying, integrating $xf_{X}(x)$ over the reals assumes the the balancing point is a vertical line

No.

It assumes that $f_X$ is the probability density function.

so you're against my claim?

More generally, $E[X] = \int_{\Omega} x f_X(x)\,\mathrm{d}x$, where $\Omega$ is your sample space.

What does it mean for there to be "only one natural way to measure the mean in the normal distribution"? Are there unnatural ways? Is the normal distribution somehow unique? What do you mean?

@XanderHenderson Well my claim is about divvying up the density for density functions by halving it in 2 distinct ways. For example, the normal distribution has that reflection symmetry about a vertical line. That's clearly the natural way to find the mean (or balancing point)

and doing $\int_{\Omega} xf_{X}(x)~dx$ clearly supports that

You have used the word "clearly". I believe that I have make my opinion about that word quite... clear.

What you are observing is that the mean / expected value coincides with the center of mass. This is a more generally true result, if you consider $\Omega$ to be a physical object and the pdf to represent an actual density.

@XanderHenderson If $f_{X}(x)$ is a self inverse function then you could halve the area by slicing with the identity function. And because of this I think you could define the mean differently than the classical definition