Mathematics - 2017-05-09 (page 2 of 5)

Sha Vuklia

16:00

((((lol))))

Daminark

@Eric I wonder if today's difftop class drove a wedge between us all

SteamyRoot

Badum tsss

If we're doing wedge jokes: "Old MacDonald had a form, $e_i \wedge e_i = 0$"

15

Daminark

Oh that's beautiful

I shall treasure it

Hey @EricS!

And @Akiva

Sha Vuklia

Guys, still about this $k$-regular proof; they say "By Hall's marriage theorem, $G'$ has a decomposition into $k$ edge-disjoint perfect matchings." Do they mean there are $k$ different perfect matchings we can make for this graph?

Daminark

Sounds like it, yeah

Hey @Riker!

Riker

16:17

o/

I popped in because I was going to ask a quick math question, but I solved it already

Daminark

Lel

Eric

pls no @Daminark

Riker

@Daminark my brother... I have found you

other users always comment on my 'lel's and 'kek's

Semiclassical

The definition of the cross product cited in this problem makes my head hurt: math.stackexchange.com/q/2273362/137524

Riker

yes

> What does the book means? Please explain.

sounds about right

Daminark

16:20

@Riker kek

Sha Vuklia

omgoodness @Semi

what is wrong with those people

oh wait

they say "can be considered as"

you confused me by calling it a definition :P

Semiclassical

Yeah.

Daminark

Still... wai tho

Semiclassical

Yeah, I guess that's a bit misleading.

But that construction is just...ow.

Daminark

And @Eric kek, you know I live for puns

Riker

16:22

@Daminark lol

I have massively more keks and lels than you though :P

I have 273 lels and you have 10

and 49 to 35 keks

Daminark

Yeah I'm gonna need to up my game a good bit

Semiclassical

Demonark's just a lel-ophyte, it would seem.

Riker

@Daminark ..... kek

Daminark

(Also I only just now learned how to create hyperlinks in response to yours so thank you)

Hey @Astyx!

Riker

@Daminark fwiw, if you're interested: the problem I was working on was "express |a+b|-|b| without absolute value signs"

Astyx

16:27

Hello

Riker

treating each case separately

Daminark

Ah, I see

Astyx

What's up ?

Riker

not much here, what about you?

Daminark

Not too much, finally seeing forms in the proper generality (and also I know why determinant is unique)

You?

Astyx

16:29

Exams exams exams

And being tired of them too

Why is the determinant unique ?

Daminark

Oh also, to anyone who knows this, one of the problems on my manifolds pset defines $OX = \{(x,v)\in NX : \|v\| = 1\}$ where $X^n\subset \mathbb{R}^{n+1}$, then talks about the projection map, and says to show that $X$ is orientable if and only if $OX$ is connected. Shouldn't this be disconnected? My idea is that you take a path $\gamma$ between $(x,v)$ and $(x,-v)$, then $\pi(\gamma)$ is a loop that reverses orientation

Astyx

How do I get mathjax on my phone ?

Danu

In the representation theory of Lie algebras, what does "isotropic subspace" mean?

Astyx

Back to graphs are we @Sha ?

Daminark

@Astyx Good luck, I don't know if you can get mathjax on your phone, and the idea is to define $\Lambda^r(V)^*$ to be the space of r-linear alternating forms on the vector space $V$

(Form meaning maps to $\mathbb{R}$)

Now, the dimension of this is the dimension of $V$ choose $r$

So that if $r = \dim(V)$, it's a one-dimensional space

Astyx

16:35

So $\lambda \det$ is unique :)

Daminark

Thus, the determinant (which is an alternating, n-linear form) is unique up to a constant, so you declare the standard basis to map to 1, so that you... determine the mapping

Astyx

I've seen some nice applications of this result in differential equation theory

Concerning the Wronskien if I'm not mistaken (not sure about terminology, at all)

Shaun

Hello all :)

Astyx

Hi

Shaun

Hello, @Astyx

Daminark

16:42

I don't think I've heard of the Wronskien @Astyx

Sha Vuklia

@Astyx lol yea someone asked me this question, and I couldn't let go of it:P

but I haven't solved it:(

I've contacted my teacher about it

like, I'm not supposed to be able to prove Peterson and Hall's Marriage.

that's impossible

I don't really understand wikipedia's definition of a parametric equation?

I came across the term here

but I don't really see what a parametric equation really is

like, what's the different between a parametric equation and a regular... i dunno, function?

Astyx

@Daminark when you has a homogeneous differential equation on a vector space of dimension n, you have a set of n solutions $s_1, \dots, s_n$. If you take the determinant of these, you'll find by differentiating it that it's derivative is a n - linear 1 n to symmetric for of the solutions, so you deduce it is equal to the Wronskien, thus it's an exponential @Dami

Daminark

Interesting

Astyx

"That's impossible" ? @ShaVuklia

(Sorry if my maths typing is messed up, I am struggling with my phone)

Sha Vuklia

@Astyx it's not realistic:l like, we don't have the tools to prove this.

and if we really were supposed to be using those theorems, then it would have been subquestions

Fargle

16:48

@ShaVuklia Normally, you'd have y in terms of x, or z in terms of x and y. In a parametric equation, instead of y = f(x) or z = f(x,y), you have x, y, and z all in terms of some other variable or variables, in this case s and t.

Sha Vuklia

with the actual proof being the last subquestion

Astyx

What is it you're trying to prove again ?

Fargle

So, for example, $x = t, y = t^2$ would be a parametric representation of $y = x^2$.

Shaun

Do you think this question is too long?

Sha Vuklia

@Astyx I posted it on the forum: math.stackexchange.com/questions/2273260/…

Fargle

16:50

It's called a parametric equation because you give all the coordinates in terms of some independent parameter(s)

Sha Vuklia

@Fargle so technically, in a parametric equation, everything depends on a certain set of variables?

ahh

it's the independence that's key here

i see

right, it makes a lot more sense now

Astyx

No @Shaun

Mary Star

Hello!! For the deinition of the derivative we use the slope. When we have a linear function to find the slope we draw a triangle, right? What do we when we have a quadratic function?

Shaun

Okay, good.

Astyx

You differentiate @MaryStar

Mary Star

16:51

@Astyx What do you mean?

Astyx

Or you draw the tangent to the point of which you want the slope and apply what you know on linear functions

Mary Star

Is it possible not to use the tangent? Or is this the only way? @Astyx

Semiclassical

I would not be surprised if I've said something goofy amid my answer here: math.stackexchange.com/a/2273402/137524

Astyx

Do you have an mathematical expression for you function ? @MaryStar

Sha Vuklia

@Fargle So in the case of the equation for a circle, $r^2=x^2+y^2$, we have two variables that are depended of each other. But say $r$ is fixed, then we might as well say that $x$ is our only variable.. so is this really a good example?

Semiclassical

16:55

Eh. Keep in mind that unless $x=1,-1$ there's not just one value of $y$.

Sha Vuklia

oh right

Fargle

@ShaVuklia What I think you mean to say is that there's in some sense only one degree of freedom, and I agree.

Semiclassical

Knowing the angle uniquely specifies the point, but the x-coordinate almost never does.

On the other hand, knowing the x-coordinate and the sign of the y-coordinate is enough.

Fargle

This fact falls out when you use the parametric representation $x = \cos t, y = \sin t$, and notice that this representation is one-dimensional.

Sha Vuklia

@Semiclassical So $y$ depends on $x$ in the sense that it is "restricted" now

Mary Star

16:57

@Astyx At the point x_0 the res function has a specific slope. Is there a possibility to draw an other quadratic function (i.e. the green one) and find in that way the slope?

Semiclassical

Right.

Mary Star

Sha Vuklia

ohh, riight, thanks!

Semiclassical

another way to put the point is that there's two functions for y(x), one with y>0 and the other with y<0

Astyx

@semi are you not integrating 1/t around 0 there ?

Semiclassical

16:58

And if you want to go smoothly around the circle, you'll have to change which one you use when you pass through (1,0) or (-1,0)

In my possibly goofy answer?

Keep in mind that t is a parameter, not the integration variable. (or should have been)

Mary Star

Do you have an idea @Astyx ?

Semiclassical

Fixed. @Astyx

(also, I find it funny for a question by "TheQuantumMan" to be answered by "Semiclassical.")

Sha Vuklia

Does a function always have independent variables tho?

so I'm not just talking about expression/equations

Astyx

Not really @MaryStar

Semiclassical

it's not much of a function if it doesn't have a domain.

Sha Vuklia

17:01

yea, I'm just talking about functions

like, are they always parametrised expressions technically?

Semiclassical

not sure what you mean by that.

Astyx

@Semiclassical Haha I hadn't noticed the username :p

Sha Vuklia

Is a function a parametric equation?

Semiclassical

I guess I'd say that a parametric equation is just a function whose domain and codomain aren't R.

If it's got n parameters, then the domain is R^n.

Sha Vuklia

why can't the domain be $R$?

Semiclassical

17:03

If there are m equations, then the codomain is R^m.

Hmm. I think it's just a matter of terminology.

Astyx

It depends what you mean by parametric equation

Semiclassical

No point referring to it as 'parametric' if it just maps one variable to another.

Sha Vuklia

@Astyx lol I'm trying to figure out what a parametric equation $is$

but it's technically a parametric equation, right? @Semi

Astyx

You can write $y_0 =y (x, t)$ to designate the set of $(x, t)$ verifying the equation

Fargle

@ShaVuklia I tried to give the definition earlier--a representation of a set where the coordinates are given in terms of some non-coordinate quantities, instead of in terms of the other coordinates.

Sha Vuklia

17:05

just so that I know I understood it

Semiclassical

Yeah, I guess.

That may be the better one, though.

One way to get a surface, for instance, is as $z=f(x,y)$.

Is that a parametric surface or not?

Sha Vuklia

lol

i know it's not

because I just read that :P

Semiclassical

lol

Sha Vuklia

but alright let me think

Semiclassical

Well, the tricky thing is that you can cast it as something that's 'obviously' parametric.

Namely, $(x,y,z)(u,v)=(u,v,f(u,v))$.

Sha Vuklia

17:08

huh, I don't understand that expression

oh wait

I see something

:P

Semiclassical

It's intended as $x(u,v)=u$, $y(u,v)=v$, $z(u,v)=f(u,v)$.

Astyx

For instance if you have two orthogonal nonzero vectors u, v in the plane, the orthogonal space of u is $\{w, \langle u, w\rangle =0\}$ and $\{\lambda v, \lambda \in \Bbb R\}$, the first being a parametrised definition, the second an explicit one

Sha Vuklia

yea

I see it

Astyx

(Took way too long to write)

Semiclassical

It's just a rewriting, really, but one is explicitly parametric and the other isn't.

Astyx

17:10

In fact any vector subspace has both parametric equations and a explicit definition as the span of something

Semiclassical

I'd tend to avoid calling $z=f(x,y)$ a parametric surface, but I think that's actually just personal taste.

Sha Vuklia

Okay, I won't bother you too long with this, but I'll try one last time:

Astyx

(Which, I find, is a nice fact)

Sha Vuklia

(sorry @Astyx I don't see why the one is parametrised and the other is not)

so say we have $z(x,y)=x^2+4y^2$

is this one parametrised or not?

Semiclassical

I think I'd go with yes, at this point, though part of me wants to say no.

Astyx

17:13

A parametric equation is an equation of the form $f (x) =c $ where the unknown is x

Sha Vuklia

lol well, I think if even you are unsure about it, I'll just let it go

Semiclassical

In saying that, though, I'm being influenced by Wiki's definition here: en.wikipedia.org/wiki/Parametric_surface

Sha Vuklia

I'll try to act like a physicist, and just do math intuitively :P

Semiclassical

including the first example

Astyx

You mean pretend to do maths

Sha Vuklia

17:14

lolll

Semiclassical

Namely: "The simplest type of parametric surfaces is given by the graphs of functions of two variable."

pfffffft

Sha Vuklia

A physicist with a strong background in math is love :P

Astyx

I'll leave before semi lynches me

Semiclassical

you do that.

Sha Vuklia

honestly, even the best of our teachers (like Erik Verlinde) don't have a good enough background in math (if you ask me)

Semiclassical

17:15

Though, I would be curious if there's anything obviously silly now about that one answer I gave.

It depends on how one defines 'a good background in math', is part of the problem.

Sha Vuklia

lol yea, that's why I added "if you ask me" :P

like, it's good enough for him to do his work

but it's not good enough for me to have the math explained :P and that's my no 1.,2,3,4,5,6,... frustration in physics

Semiclassical

I think most mathematicians would tend to discount a lot of the math you see in a physics course. (e.g. bessel functions, separation of variables, etc.)

Sha Vuklia

lol of course!

Semiclassical

As just being computational and classical.

I find that to be a bit annoying.

Sha Vuklia

dude, I refused to do physics for 5 months because of the lousy maths

but I finally understand the real math now, and now I'm doing physics again :P

and I'm LOVING it :P

Semiclassical

17:17

lol

It is true that the more math you have, the more you appreciate physics.

Sha Vuklia

definitely

Semiclassical

To a certain point, at least.

Sha Vuklia

yea of course

Secret

For me, the maths tells me what are those magic steps is being used to go from one step in the equation to another

and it also makes quantum mechanics easier because it is basically linear algebra

Sha Vuklia

true. like, the concepts in physics are already difficult enough. if anything, the math should not be the problem

Astyx

17:18

Seems fine to me @semi

Semiclassical

@Secret Mostly.

Sha Vuklia

like, if you're struggling with the math, you're not really doing the physics you're supposed to:/

Semiclassical

If you're doing finite-dimensional problems, that's all it is.

Astyx

(Your answer)

Semiclassical

@ShaVuklia Eh, depends on the problem. Struggling with Bessel functions is the nature of the beast

Secret

17:19

yes, for the continuum, you need functional analysis (which I still have to catch up because (obviously) my uni undergrad cut that off completely, thus I have to self study for it)

and also more complicated things like spinors, group representations etc.

Semiclassical

There are some computations that, simply put, are just a pain in the arse.

Sha Vuklia

okay true, but as far as the first-year courses go at uni, it holds :P

Semiclassical

Sure.

And even in those problems, there's usually two steps.

1) Convert the problem into an integral. 2) Simplify said integral as much as possible.

Usually the first step is not too hard.

Sha Vuklia

lol true

Semiclassical

Figuring out how far one can go with the second step is what's hard.

Sha Vuklia

17:21

except for classical mechanics though

you don't always need the integral

because the problems are so easy

but with electromagnetism, it's literally just that

Semiclassical

Tell that to action-angle variables, lol.

Mary Star

Is it possible to say something about the slope of a quadratic function (we don't know the formula) when we know an other quadratic function?

Fargle

I've found that the more advanced a subject in math or physics is, the more likely it is to use $\chi$. It's like the gatekeeper of all forbidden knowledge, and I flee in terror when I see it.

Sha Vuklia

and with quantum too, as far as I've seen

Semiclassical

@Fargle Come to think of it, you're right.

Sha Vuklia

17:22

and vibrations and waves seems to be all about solving differential equations?

Astyx

That's why we have MSE isn't it ? To make others solve our integrals :p

Secret

I struggle with electromagnetism because of the byzatine number of assumptions I need to put into the formulae to get the answers they want or to even progress

Semiclassical

$\chi$ in stats means "nope, I don't understand what's going on"

@Secret Yeah. That's especially true in stuff like radiation problems.

Far-field versus near-field, Fraunhofer diffraction vs. Fresnel

A right pain.

Hippalectryon

@Waiting Great, how are you ?

Secret

and I actually think more like a mathematician (which is why I actually struggle a bit in physics) I don't like assumptions because they make things inaccurate

Astyx

17:23

Hi @Hippalectryon

Semiclassical

$\chi$ in physics means fluctuation-dissipation theorem, for me, and I hate that.

Secret

so what I lack is a physicist's mind to do physics

Sha Vuklia

lol @Hippalectryon you and Waiting are really having a "fractured conversation" :P like, one day one person says hi how are you, the other day the other responds by "hi I'm good you"

Hippalectryon

@Astyx o/

Semiclassical

($\chi$ can also just mean susceptibility, but I don't find that terribly intuitive either)

Astyx

17:24

To me $\chi$ means what you wrote your answer Semi

Semiclassical

Yeah, that's the math one

Hippalectryon

@ShaVuklia I know, it's because my computer is often on even though I'm not in front of it :-)

Semiclassical

You don't see that in physics much.

Sha Vuklia

haha

Semiclassical

You're more likely to see the Heaviside step function $H(x)=1$ if $x>0$ and $0$ otherwise.

Secret

17:24

I recall when I first learnt about waves reflected by potential barriers, I treat the whole thing as if it is an ordinary ODE and solve it step by step. The problem:

Astyx

I learned about Airy's function today in my exams

Secret

It turns out I need to introduce a physical assumption that stuff don't blow up to infinity to kill off two of the arbitrary constants

Semiclassical

oh lawd, airy functions

Astyx

Are they famous ?

Semiclassical

@secret yep. an incident wave produces a transmitted wave and a reflected wave

eh, I've run into them before.

You need the asymptotics of the Airy function for the textbook presentation of the WKB method.

Secret

17:26

Assumptions are never straightforward to me unless I do enough problems, they don't have that nice structure as mathematical objects that you can derive one assumption from a given one in some systematic fashion

Semiclassical

And it's really annoying. (Hence why I don't like that approach.)

Secret

In fact, to me, thinking what assumptions I need to use in a physics question is as nontrivial as remembering all the steps you need to e.g. renew a car license

Astyx

Wkb ?

Semiclassical

The issue being that the Airy function satisfies an ODE with an irregular singular point, and that makes it weirder than the usual examples.

Werner-Keller-Brillouin, I think? Hang on

Astyx

Yeah so I figured in my exam :p

Semiclassical

17:28

nuts. Wentzel–Kramers–Brillouin

Secret

The issue I have with these thinking methods is that there is no pattern that logically connect step i to step i+1, thus you cannot derive the whole series from just information in one step

Astyx

Almost what you said

Semiclassical

From the Wikipedia description: In mathematical physics, the WKB approximation or WKB method is a method for finding approximate solutions to linear differential equations with spatially varying coefficients. It is typically used for a semiclassical calculation in quantum mechanics in which the wavefunction is recast as an exponential function, semiclassically expanded, and then either the amplitude or the phase is taken to be slowly changing.

The linkage to the Airy asymptotics also appears on that page: en.wikipedia.org/wiki/WKB_approximation

Secret

e.g. Say you need to travel, if you forget to bring your passport, nothing among the things that you have remember to bring will let you derive that you forget to bring your passport

Semiclassical

In my own work I've tended to rely not so much on the WKB approximation per se but rather a certain result that you can derive from it.

Secret

17:30

But in maths, if there is any missing things, the existence of that missing thing is easily inferred from the underlying rules

Semiclassical

Mostly because I hate having to deal with the actual approximation.

Fargle

@Secret Or the impossibility thereof.

Astyx

Fair enough

Semiclassical

Hence I should probably emphasize the 'classical' part of my name :P

Astyx

Put it in bold maybe ?

Secret

17:32

Yeah, so maths fits me more because everything including the steps between are well explained, but reality itself, all the daily life skills that we all need to survive, while have a strict ordering, has no (obvious) logic between the steps

Semiclassical

heh.

Astyx

Semiclassical

Not sure that would work as a username though

Secret

i.e. there's no master mathematical object that allow you to derive all daily life skills, otherwise, life will be MUCH easier

Semiclassical

yeah.

and it's not worth it regardless.

Mark McClure

I'm not sure if I've ever seen so many bad yet highly upvoted answers as there are to this question.

Astyx

17:33

Evidently :p

These are fine are they not @MarkMcClure ?

Semiclassical

@MarkMcClure Which of those answers would you accept, out of curiousity? (if any)

Mark McClure

@Astyx See my comment on the accepted answer.

@Semiclassical I like this answer the best.

Astyx

Ah I see, yes these are not expliciting it's a convention when they should. However they are giving good arguments

Secret

While this might provide some reasonable level of intuition for beginners, this answer is ultimately completely incorrect. How do you define $g$, for example, if $f(x)=\sin(\sin(x))$? Perhaps by $g(x)=\int_0^x f(t)\,dt$. So, what is $g(0)$? — Mark McClure 24 mins ago

Actually do all continuous functions must have an antiderivative (even if it is nonelementary)?

Semiclassical

Continuous -> Integrable

Secret

17:39

oops, yes

Mark McClure

@Secret Correct, by the Fundamental Theorem of Calculus - that's the point.

Semiclassical

The point, as I take it, is that one shouldn't need to know anything about $f(x)$ away from $x=a$ to declare that $\int_a^a f(x)\,dx=0$.

Astyx

@MarkMcClure The very last answer is good though

Or at least it does specify it's a convention

Mark McClure

@Semiclassical I agree with you. Also, though, you can't decide that $\int_a^a f(x) dx=0$, because it equals $F(a)-F(a)$, when you've defined $F(x)=\int_a^x f(t) dt$.

Astyx

This one @MarkMcClure

Semiclassical

17:43

not following you.

Mark McClure

@Astyx I'm not sure what the last answer is. The answers don't always appear in the same order.

Semiclassical

@Astyx That's basically the same as one of the comments, though. (Not clear which one came first, mind.)

Astyx

Sure. It's still a correct answer

Mark McClure

@Astyx Ahh.

Astyx

I'm not so keen on leaving answers in comments

Semiclassical

17:45

But eh.

I find questions like this just kinda tedious.

Astyx

For posterity

I agree, the OP is asking to mathematically justify a convention, there are no better answer than : it works

Semiclassical

I think there are reasonable arguments to be made, I just don't find it interesting.

Secret

@Astyx That's a cool answer that making use of only the symmetries of the function itself and no other considerations

I always like these abstract algebraic approaches to calculus, because they are simpler to understand due to no need to assume anything except the bare minimum about the nature of the functions

Astyx

I agree with that @Secret

SoumyoB

18:05

sooo I gave the online copy of a book to be printed at a xerox shop, and that book was Brownian Motion and Stochastic Calculus

and they only printed the first 30-ish pages of that book and mistakenly replaced the remainder of that book by some Algebra book's pages someone else must have given to be printed

I've paid almost four dollars for printing this entire book and this is so annoying

Felix.C

18:16

Hey small beginner question here...I want to show that a convergent series is a null set, now my idea was to find a k such that I can put my first k-1 points into their own cubes, and the rest in the k-th cube, of course I had to watch out that they don't intersect, but I could easily make an estimate for it. What bothers me is that I get something like this:
$k*s^n < \varepsilon$ with $s > 2* \Vert x - x_k \Vert$ e.g. s is the length of my cube, but now I have a like two variables dependant from k....Any hints how to express s and k?

Balarka Sen

18:35

morning, @MikeMiller

Astyx

@Felix.C What's a null set ?

Fargle

@Astyx A set with measure $0$, I believe.

Astyx

Oh so he means the set of the terms of a convergent series

It's countable

And there are only finitely many $\gt \epsilon $ for any $\epsilon \gt 0$

If I understand what you are going for @Felix.C

Fargle

@Felix.C Yeah, you can just use its countability. Because a single point has measure zero, and because measure is countably additive, you have that any countable collection of points has measure zero.

Felix.C

@Astyx In german it's called a Jordan Nullmenge e.g. a set where you can find cubes so that your set is a subset of those cubes and you can make the volume of those cubes as small as you want

I'll quickly write down the definition we used

Fargle

18:47

That corresponds exactly with the notion of a null set w.r.t. Lebesgue measure, but I see you don't have the machinery of Lebesgue measure here.

Astyx

See what I wrote above, if it helps you @Felix.C

Ans what Fargle said too of course

Alessandro Codenotti

@Fargle He's using the Jordan (outer) measure

(which isn't actually a measure)

Felix.C

I had a quick look on the Lebesgue measure, what I never encountered so far, and it seems to me as it uses a infinite number of cubes, I need a finite amount

Fargle

Ahh. Remind me not to take myself too seriously when I've just woken up.

You're right, @Felix, that's my bad.

Astyx

You can fix a small cube (as small as you want) around 0, and then there are only finitely many points remaining outside that cube so you can just add one small cube for every point @Felix.C

DarthVader1056

18:53

Does someone knows product of 25 positive consecutive integers which is a perfect square?

Fargle

@DarthVader1056 $0 \times 1 \times \cdots \times 24$, but I guess you're not looking for that...

Felix.C

@Astyx Not really...I need to show that if I have a convergent series $(x_n) \subset \mathbb{R}^n$ then we can find a finite number of cubes $W_i$ such that $(x_n) \subset \bigcup_{i=1}^n W_i$ and $\sum_{n=1}^n W_i < \varepsilon$ for any $\varepsilon$

Fargle

@DarthVader1056 Oh, you said positive. I don't think this can happen if the top integer is $\geq 29$.

Astyx

Yes, so use what I said and that should work, I haven't done all the work for you @Felix.C

Fargle

And it can't happen below that, because any interval containing 23 would have to contain 46 as well, which means it has to contain 29, and it fails.

Felix.C

18:56

ok, thank you!

Astyx

Glad to help, don't hesitate to ask if you misunderstood something

Fargle

Well, now that I think about it, it could happen if you find 25 consecutive numbers which are all composite and have between them an even amount of each prime factor.

DarthVader1056

@Fargle Yes any product with a prime such that p<=n+24<2p will not be so.

Astyx

Can you not build such a sequence with factorials ? (Just an idea, I don't have any paper to work with so I might be saying nonsense)

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$Mathematics$ Mathematics