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12:14 AM
What ???
0
Q: Look and say type constant

mickI have been fascinated by integer sequences such as the look and say sequence and its many variant. USUALLY such sequences grow like $ a n^b $ or like $ c n ^ d $ where $a,b,c$ are algebraic Numbers or algebraic number raised to an algrbraic power ( like $ \sqrt 2 ^{\sqrt3} $). Very often we me...

 
12:42 AM
Hi
 
Say I fix a list of complex numbers, and want to prove an if and only if theorem statement. If one direction of the statement is ... "if and only if the list of numbers is real", have I done anything incorrect?

Seems a bit strange that I fix a list of complex numbers, and then show that the theorem holds only if the list is reduced to the reals.
What do you think, @robjohn?
Actually, nevermind - I think I got it ... thanks @robjohn
 
12:57 AM
I need to improve my combinatorics skills. I find the following exercise, calculate the probability to obtain a king and a queen of the same suit in a standard 52 deck of cards in a hand of 6 cards. If we choose some suits its completely determined the king and the queen, so we have 4 possible option, and thus 13 choose 4 cards times 4^4 suits for the remaining cards, i.e., 4 (13 choose 4) 4^4 possibles ways. and this divide by 52 choose 4.
I not completely sure, I'm very bad in this kind of problems
No ir's wrong
 
@ForeverMozart I have a book to trade.
New, unused copy of Weinberg's Gravitation and Cosmology.
Amazon sent me two copies for some reason.
Looking for: Spivak 3-5, Steenrod, Milnor-Stasheff, Choquet-Bruhat
Spanier
MacDuff-Salamon
 
1:42 AM
Hi All, why do we say Ring "over" a Field, but Relation 'on' a set? Is that just left over from the German texts?
meaning, why not Ring On a Field? Or does that have some other implication? (Noob here - amature math, but interested in the linguistic aspects.)
 
2:13 AM
Any physicists/ people who know a little quantum mechanics here want to help me figure out what my professor means in this question: s18.postimg.io/xvsbt6yqh/image.png ? Does he just want me to explicitly compute $\Psi(x,t)$? If so, doesn't that require a Hamiltonian?
 
 
7 hours later…
user227867
9:42 AM
@0celo7 You should return it then.
 
10:14 AM
Hi, can anyone give me a hand with implicit differentiation?
 
11:04 AM
@user1618033 ping
 
 
2 hours later…
12:46 PM
@JasperLoy No, I know the reason.
 
1:21 PM
@Agawa001 hey, in the middle of some important research.
 
@user1618033 oooo k' keep goin never look back
 
@Agawa001 Looking back you see the hell image. :-)
 
later autumn/winter /early spring is the most opportunate season for researches
@user1618033 naaaaah aint a hell, this is my paradise
 
@Agawa001 Particularly autumn and winter here
 
@user1618033 i thought u talking about my backround image
 
1:26 PM
@Agawa001 no. But looking back might also mean to consider all the negativity you met on your struggle, your way to reaching your goals.
Looking back is not an option. That's why I called that the hell image.
 
@user1618033 you live in some beautiful place contrarily as you mentioned
as for me, the best spot of earth wher i can find my souls is not a noisy city, or a desert
 
@Agawa001 It depends on the details you considered. In terms of possiblities to find a good job, say, then you're not that lucky, or following some university, or doing many other things, you don't have the resources you had in the city.
@Agawa001 On the other hand, the silence here is magnificent and helpful in the creational process.
 
yes, but as a mathematician, one dosnt need as much and expensive ressources
computer / noteback that would suffices
im confusing french with english again (shh jasper isnt around)
resources*
 
@Agawa001 In general my costs are pretty low here compared with friends and other people I know, because I don't simply need to spend more.
The satisfaction primarily comes from mathematics, I don't buy it. You know, some eat more to be happy, or buy all kind of expensive things.
The happiness comes from inside, there is all you need.
 
all my happiness resides within my cup of coffee
and i'm serious
 
1:37 PM
@Agawa001 hehe, good! :-)
 
capuccino-milk
with chocolate flavor
 
@Agawa001 Then I might say that mine comes from some hot chocolate combined with a bit of capuccino and milk. :D
I don't know a thing more beautiful, more amazing than discovering yourself, by ambition, by passion, by very hard work, and then seeing the universe as you've never seen before.
 
user227867
@user1618033 Well said.
 
@JasperLoy JASPER!! How is it going? :-)
 
user227867
@user1618033 CHRIS
 
1:43 PM
@JasperLoy CHRIS'SIS
@JasperLoy :D
 
user227867
No difference, until you tell me your real name one day. =)
 
@JasperLoy Indeed. ;)
@JasperLoy Did you upload some new songs?
 
user227867
Nope, doing other things these few days.
 
@user1618033 everyone has a way to conceive his own utopia and live in it
 
user227867
@user1618033 I imagine your book will be called Limits, Series, and Integrals.
 
1:44 PM
@Agawa001 hehe, I think so. :-)
@JasperLoy No, it has a pretty different name.
 
and i m sure not all of us can dive hardly dive among a 4x4 paper
 
user227867
@user1618033 'The lost notebooks of Chris'
 
@JasperLoy I'm still around, not the case to consider lost notebooks. ;) Careful here all is kept in good conditions. :D
 
user227867
@user1618033 Aha! I was listening to the 4 songs already on my channel. I think they are very good, better than the ones I deleted. =)
 
@JasperLoy which one listening now?
 
user227867
1:47 PM
@user1618033 I just listened to all 4 a few times, lol.
 
@JasperLoy sole mio?
 
user227867
I think I will do lots of walking in September.
 
user227867
@user1618033 Yes, that too. O sole mio
 
@JasperLoy That is a success.
@JasperLoy Walking is pretty healthy I think (if possible daily). I miss walking lately, but hope to fix that somehow.
I need to do all kind of changes to my book and this takes time.
 
user227867
@user1618033 Singing is like math. The novice cannot tell the difference between a song badly sung and a song well sung. =)
 
1:50 PM
@JasperLoy in this respect, yeah :D
 
user227867
Just like not many people in the world can understand a difficult theorem, not many people in the world can hear the subtleties in the voice. =)
 
@JasperLoy :-)))
 
user227867
@user1618033 Only Chris appreciates good mathematics and good music, LOL.
 
bbl need to replunge into my 1366 x 768 screen and chase some 1's and 0's
 
@JasperLoy lolll, not really.
BBL (visitors at the door)
 
 
1 hour later…
Huy
2:54 PM
@0celo7: help with elementary geometry
 
what is le problem
 
Huy
given two rays with the same starting point and another point $P$ between the rays, how do I construct a line through $P$ such that $P$ halves the segment within the rays?
 
hi chat
@Huy not sure I understand which segment you mean
 
:32004950 does not work.
$P$ need not lie in "the middle"
 
2:57 PM
ah, wait, i think i get it
 
Huy
call the line I want to construct $g$. it intersects the ray in $A, B$. then, the distance $AP = BP$.
 
right.
 
@Huy draw the arc CPD with O as center where O is the "same starting point"
then find the midpoint of that arc and call it as M
then construct a line through M perpen OM
 
That need not even pass through $P$
 
2:59 PM
then draw a line parallel to the previous line passing through P
 
then $P$ need not be the mid-point anymore
 
Oops, I completely misunderstood the problem
 
@Huy pls draw
I do differential geometry, not Euclidean though
 
it's easiest to sketch 'in reverse'
 
Huy
it's just a special case
 
3:02 PM
Touche
 
He's essentially asking you to find an edge of a triangle, already given two of them, such that a given ray is a median.
 
Huy
I've drawn it "in reverse" and am trying to find another characterization of a point on that line
 
Huy
yes, that's the sketch
 
right.
i'm doing it through geogebra myself
 
3:05 PM
is this about constructions
 
we skipped those in high school, too hard
 
Huy
well yeah, I asked how to construct it
 
I can't do this off the top of my head.
 
Huy
3:06 PM
yeah this is some introductory material that I'm supposed to use for my geometry HS class
 
In my title's question, do I put 4 or $4$?
 
Huy
can't solve the problems
kinda problematic
 
@Huy if you solve my Riemannian metric problem I will help.
 
$OA\sin\theta=OB\sin\varphi$
 
Huy
no trig
 
3:06 PM
I fear the solution involves Weierstrass approximations
I had a dream where Riemann told me to use them
 
@Huy trig is just right angled triangles
 
er, not Weierstrass
 
eh, i like to use trig for these just to get a feel for how the problem works
 
Whitney?
 
one observation: $\triangle AOP$ and $\triangle POB$ have the same area
 
3:09 PM
Riemann told me to use that
 
@Huy Eh. If you're given $P$ you can draw the circle centered at it and intersect to get the endpoints of the segment, no? Are circles allowed?
 
since 1) they have the same base and 2) if you drop an altitude from $O$ to the line $AB$, they have the same 'height.'
 
Huy
@Semiclassical doesn't that follow from the fact that OP is a median?
 
probably.
my knowledge of triangle stuff is shallow
 
Huy
@BalarkaSen: yes, that's allowed. but how do I get both endpoints on the rays?
 
3:10 PM
Pondering. It's not clear if you can find the radius of the circle so easily.
 
Done
Construct C and D as shown
Observe that PC:PD = OB:OA
then mark a point on the right ray (you can even reuse D)
construct a point E on the left ray such that OE:OD=PD:PC
then ED is parallel to AB
which can be constructed by drawing a line parallel to ED through P
QED.
 
not sure I see why you're comparing with OB and OA, which aren't part of right triangles
 
welcome to the boring world of Niels bohr
 
OB, OA are not given to you.
 
@BalarkaSen I didn't use OB and OA
 
3:14 PM
You have to find those.
 
@Semiclassical see the trigonometric ratios above
 
2 mins ago, by Leaky Nun
Observe that PC:PD = OB:OA
 
@Semiclassical that is just showing how my answer works
you can delete that line
 
i'd rather not, since it's part of the logic
 
@Semiclassical I mean, I did not use OB and OA in the construction
that's just proving why my construction works
 
3:15 PM
No, ok, what he says seems fine.
 
Huy
"construct a point E on the left ray such that OE:OD=PD:PC"
how does that work?
 
Well, why is PC:PD :: OB:OA?
 
@Huy Just cut lengths.
 
@Semiclassical $PC:PD = \sin\theta:\sin\varphi = OB:OA$ as proven above
 
3:16 PM
You need to find E on the left ray such that OE = PD/PC*OD.
You know PD, PC, OD.
So it's a matter of cutting lengths.
 
neither $OA$ nor $OB$ are legs of a right triangle.
did you mean OD instead of OB?
 
@Semiclassical I just proved the latter equality above
10 mins ago, by Leaky Nun
$OA\sin\theta=OB\sin\varphi$
by using sine rule
 
okay, that i can buy
 
The tools used are inherently non-Euclidean though. I wonder if there's a simpler way.
 
Huy
how do you justify the ratio without trig or intercept theorem
this is supposed to be solvable without that stuff
 
3:21 PM
@Huy trig is just right-angled triangle
 
trig is just similarity of right triangles
 
Huy
(because trig and similarity comes much later)
ok let me look at the construction again
 
so it's hardly non-Euclidean. that doesn't mean it's as elementary as possible, though.
 
@Semiclassical And similarity of right triangles is just trig. I do not distinguish the two.
 
okay? my point is merely that calling this non-Euclidean seems a stretch
 
3:23 PM
I mean, by that logic, coordinate geometry is also Euclidean :) Sure, in principle, this can be made into a Euclidean geometry proof but the thought process isn't.
 
I think we're using different meanings of Euclidean, then
 
Perhaps.
 
I think it is fair to say that this isn't an Element-ary construction :)
 
hah, like that.
(is it a pun on what I think it is?)
 
yeah
saw another (much harder, i suspect) geometry problem on main recently: math.stackexchange.com/questions/1906795/…
the 'secondary question' is a neat observation
 
3:27 PM
I am rusty on tricky Euclidean geometry. Haven't done those in a while.
 
ellipses !!!
i got too old for geometry
 
i suspect that it is indeed something the realm of projective geometry
but since i don't know how to actually prove stuff like that...
 
@Semiclassical More problematic is that I have only been doing coordinate geometry for the last few weeks in school.
 
yeah
it's hard to move from one geometry mindset to another
 
3:38 PM
Agreed.
@LeakyNun OK, no, I don't see any other proof. Your proof is good, nice thinking.
 
@BalarkaSen thank you
 
Didn't think to fiddle with the angles.
 
3:56 PM
@Semiclassical I studied some statics.
Moment of forces is a nice concept.
 
yeah, leverage and such
though leverage is the very simplest version of that
 
What d'you mean by leverage? I haven't heard that terminology.
 
4:09 PM
well, technically what i'm getting at is the first moment of force, i.e. torque
which in its simplest form (a force $F$ applied perpendicular to some arm at a certain distance $d$) is just $\tau = Fd$
you can increase that either by increasing the applied force or by increasing the distance from the center of rotation. in the latter case, one is increasing one's leverage
 
That's what I meant by "moment". What other moment are there? Curious.
 
well, moment of inertia for one is a 2nd moment
In physics, a moment is an expression involving the product of a distance and a physical quantity, and in this way it accounts for how the physical quantity is located or arranged. Moments are usually defined with respect to a fixed reference point; they deal with physical quantities as measured at some distance from that reference point. For example, the moment of force acting on an object, often called torque, is the product of the force and the distance from a reference point. In principle, any physical quantity can be multiplied by distance to produce a moment; commonly used quantities include...
 
ah, haven't yet studied that
 
in the context of electromagnetism one also has monopole, dipole, quadrupole etc moments
which are fun visually, if a bit of a pain in the arse algebraically
 
yeah, I have encountered dipole moments in chemistry.
 
4:12 PM
yeah
to get a monopole, put a single charge down. then you just have field lines radiating from it. moreover, if you have an overall charge at all, then regardless of how that charge is distributed it looks like a single net charge from far enough away
by contrast, if you have a positive charge and a negative charge, you'll get a dipole configuration
with the dipole moment being $qd$, the product of the charge and their separation
 
what's the monopole moment?
that is to say, how is it defined?
 
$q$
it's just the charge itself
 
oh.
 
yeah. boring but useful
 
not very interesting :(
 
4:16 PM
note that it's also a scalar quantity. by contrast, the dipole moment is technically a vector quantity: $\vec{p}=q\vec{d}$ where the vector points from the negative charge to the positive charge (i think that's the convention)
 
agree
 
now, suppose you had two dipoles of equal moment facing in opposite directions
then the net moment is $\vec{p}+(-\vec{p})=0$, so there's no net dipole or monopole moments
 
sure
 
but you still will have a field. in that case, it'll be a quadrupole field and have a quadrupole moment
 
ah
how does on define that?
 
4:18 PM
it's a tensor quantity. let me pull up the precise definition
 
makes sense; it should be a tensor
 
$$Q_{ij}=\int (3r_i r_j-|r|^2 \delta_{ij})\rho(\vec{r})d^3 r$$
so the diagonal elements would be stuff like $Q_{33}=\int (2z^2-x^2-y^2)\rho(\vec{r})d^3 r$ whereas the off-diagonals would be e.g. $Q_{12}\int 3xy\,\rho(\vec{r})d^3 r$
one can also define higher multipole moments, octopole being the next one
 
huh
 
but you can see that it's getting to be a pain
 
right
 
4:23 PM
what this leads to is being able to write the electric field in terms of a multipole expansion
 
e.g. a monopole term that falls off like $1/r^2$, a dipole term that falls off like $1/r^3$, and so forth
 
that's pretty interesting, at least theoretically, i suppose
going off-topic slightly, have you heard of molecular resonance?
 
in typical applications one doesn't keep all the moments, of course. which is good, because the higher you go in the expansion the more involved it all becomes. (the 2^nth moment would be a rank-n tensor i think)
i've probably heard the term, but i don't really know about it
i know what resonance is in a general sense, of course, but i haven't studied its molecular forms
 
the essential idea is that if you have a molecule, you can place the free electrons, or electrons in pi-bonds in different places to get different structures of the molecule, all of which are probable
 
Huy
ok
I found out a much more elementary proof imo
 
the last image is of a quadrupole
 
the reason I asked is because it seems a lot like the idea of Feynman diagrams to me.
 
could not tell you, tbh
 
but I don't really know much of anything of the latter.
 
4:28 PM
typically you get Feynman diagrams when there's some notion of a scattering process
 
seems optical triicky
 
or more generally a sequence of possible interactions
 
it lists down all possible interactions between two particles, not?
 
something like that
 
vaguely, that's what molecular resonance does too.
 
4:30 PM
i think it's not entirely implausible to apply that to states of molecules, but i'm not 100% sure
 
lists down all possible positions of the weaker-energy electrons
@Semiclassical the analogy might be superficial tho
it's not about interacts of molecules... just about positions of the free electrons
 
@Semiclassical wheel the first image up and down hastly and focus on the center
 
The kind of thing which resonance explains is e.g., the bonds the C's form in benzene say.
 
(checking my eyes deficiency or efficiency)
 
according to the usual theory, two consecutive C's either form a single bond or a double bond.
but experimentally all the bonds are found to be of the same nature. same bond energy/length i mean.
 
Huy
4:34 PM
@BalarkaSen @Semiclassical @LeakyNun:
Construct a parallel line of one of the rays through P. It intersects the other ray uniquely. The intersection is the midpoint of one side of the desired triangle.

(this works because for any triangle $ABC$, the quadrilateral $A M_c M_a M_b$ is a parallelogram)
 
@Huy :o beautiful
 
This is the reason
 
Huy
 
@Huy what do you mean by "one of the rays through P"?
 
Huy
parallel line through P
 
4:37 PM
Oh
 
@Semiclassical it works for second image also, focus on the arrows, you see them moving
 
@huy nice
 
Huy
now I need to elementarily justify why the quadrilateral is always a parallelogram :P
 
isn't that trivial? each pair of sides is parallel by construction
 
Huy
hm
 
4:39 PM
$OM_2$ is parallel to $M_3 M_4$, for instance
 
Huy
I think either you're misunderstanding me or I'm misunderstanding you
 
I did mean the quadrilateral $OM_2 P M_4$, though
that's the one I think is obviously a parallelogram
 
Huy
here, yes, by construction
 
should've been $M_3$ in my last expression
you've also got $M_2 M_3$ parallel to the line through $P$ by construction
and i think that's enough to ensure that any other quadrilaterals you get will be parallelograms
 
 
2 hours later…
6:19 PM
hey @ted
 
 
2 hours later…
7:54 PM
zbogom
 
8:53 PM
Thanks for your feedback in my questions @user1952009 and Daniel. Ths site isn't possible without user as you. Well night.
 
 
2 hours later…
10:26 PM
Which devil voted to close ?? :)
2
Q: Find $ ? = \sqrt[3] {1 + \sqrt[3] {1 + 2 \sqrt[3] {1 + 3 \sqrt[3] \cdots}}} $

mickI wonder about a closed form for $ ? = \sqrt[3] {1 + \sqrt[3] {1 + 2 \sqrt[3] {1 + 3 \sqrt[3] \cdots}}} $ I tried solving $ f(x) ^3 = 1 + (x+y) f(x+1) $ for various fixed values integer $y$ , but I failed.

 
user227867
10:40 PM
@user1618033 Seems that you spent the whole night with the visitors. Must be Monica!
 
11:14 PM
how do those infinite root things work o.o
 
Depends on the definition of work.
 
user227867
@deepak Ignore the stupid sarcastic comments. Keep posting good answers. We need more users like you and less of these stupid comments. — Jasper Loy 4 mins ago
 
user227867
Thank you for your attention. ^
 
@BalarkaSen how is pi equal to an infinite product of repeated square roots :O
 
@0celo7 Are you asking for a derivation or expressing surprise? I don't feel surprised by it: they are cute but not at all surprising.
 
user227867
11:27 PM
@0celo7 How do you pronounce your username?
 
11:49 PM
@JasperLoy oh-cello-"Seven" (cello like the instrument), or like the cat
both are acceptable
@BalarkaSen very surprising to me
@BalarkaSen Did you know this? Let $g:M\to\Bbb R$ be a positive cont. function on a manifold $M$. Then there is a positive smooth function $f:M\to\Bbb R$ such that $f(x)<g(x)$ for all $x$.
 

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