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amWhy
amWhy
12:56 AM
What do users here follow, wrt questions: New tab, or Active tab. I have always followed "new" questions. But it seems like many follow "active" questions, and hence the concern about bumping posts via edits or a community bot. Poll? It seems that a lot of users follow "active questions". I'd consider changing my preference for new, if I understood better, the value of "active". Or at least I'd consider using both, in any given day. @TedShifrin What are your thoughts?
 
Ted Shifrin
Ted Shifrin
1:19 AM
I just have the default setting. I dunno.
 
amWhy
amWhy
@TedShifrin Yeah. I think default, aka "front page" is of currently active posts.
 
Ted Shifrin
Ted Shifrin
1:35 AM
I actually want sincerely active ones, not just new. Ugh.
 
 
4 hours later…
copper.hat
copper.hat
5:35 AM
to me, this highlights why real analysis can be tough: math.stackexchange.com/a/4315730/27978
@Koro did you sort out the $\|Px\| \le \|x\|$ issue?
 
leslie townes
leslie townes
that's a classic.
koro's problem also a classic. great stuff. happy thanksgiving, everybody.
stuff like that feels harder if you have to give 'formulas' to see it. fairly intuitive that a curve can bounce between two curves that share a common non-horizontal tangent at a point without 'settling down' enough to make the curve 1-1 on a neighborhood of the point.
a lot of that stuff is in the 'after a bit of thought, no reason to expect it to be true' ballpark and then you just have to find formulas if you want them. which tend to suck and involve lots of arbitrary choices.
 
copper.hat
copper.hat
the fact that it is true when the hypotheses hold globally is almost magical to me
Happy Thanksgiving to you & all!
 
leslie townes
leslie townes
i want to open tomorrow's wine today. but i won't.
 
copper.hat
copper.hat
i will hit the wine shortly. my consumption has been unusually light lately. something is even more wrong than usual. my usual accomplice was traveling for a while which induced the break.
i like a drink every now and then, but its much more enjoyable with company.
i mean company that pretends to listen.
Reading Terrence Tao is depressing for me. I know he's a wunderkind, but still, it reminds me of a comment Norbert Wiener made
99% of mathematics are done by 1%
 
Rover
Rover
5:54 AM
Any suggestions for books covering Single Variable calculus and Multivariable calculus?
 
copper.hat
copper.hat
sorry, i'm not really sure what those entail.
 
leslie townes
leslie townes
yeah, if you're trying to fit it into a program of study, more detail might help those with opinions. i don't have strong opinions, because to me most of those books (interpreting those topics broadly) are kind of the same
 
Rover
Rover
well it is in our course, mainly mathematics in engineering
I have some books and was looking for books which cover it in depth
Kind of trying to see which book suits me best, so I want to know some books covering those topics.
 
copper.hat
copper.hat
i hate to suggest rudin's "principles of mathematical analysis", that has more an analysis focus than calculus.
 
Rover
Rover
Okay
 
copper.hat
copper.hat
6:04 AM
i suspect there are better texts, but i am not aware of many in the calculus realm
 
Rover
Rover
, how is Introduction to Real Analysis by R.G. Bartle and donald R. Sherbert
@copper.hat Ok
 
copper.hat
copper.hat
when i say i am not aware, i mean aware of other books period, not just better ones
 
Rover
Rover
Okk
 
copper.hat
copper.hat
my big weakness is abstract algebra
so i have accumulated 5+ books on it
Hi Koro
 
Koro
Koro
Happy thanksgiving everybody :)
 
copper.hat
copper.hat
6:07 AM
You too!
 
Koro
Koro
@copper.hat yes, copper. I could figure that out.
 
Rover
Rover
@copper.hat Ohhhk
 
Koro
Koro
@Rover you may look up Apostle's also.
 
Rover
Rover
Ok
 
Koro
Koro
copper, the thing that I was missing was that: if $x$ is in range P then $P(x)=x$
using this idea, it is possible to show that every vector in null P is orthogonal to every vector in range P.
:)
 
copper.hat
copper.hat
6:14 AM
If $x = Py$ then $Px = PPy = Py = x$.
sorry, i misread what you wrote, i thought you were asking!
 
Koro
Koro
:)
 
copper.hat
copper.hat
one key when dealing with such things is the derivative of $\|Px-y\|^2$.
 
Koro
Koro
I never thought about that.
 
copper.hat
copper.hat
i like things convex, and linear is a special case :-)
 
Koro
Koro
derivative of norm is very new for me :)
 
leslie townes
leslie townes
6:16 AM
norm squared. easier that way :)
 
copper.hat
copper.hat
since you have an inner product, the derivative is easy to estimate and check.
a nice result from rudin of all places is that $x \bot y$ iff $\|y\| \le \| t x +y\| $ for all scalars $t$.
 
Koro
Koro
in particular, I liked how $\sin x $ was approximated by a polynomial of degree $5$ using inner products with better accuracy than Taylor's polynomial of the same degree on $[-\pi, \pi]$
 
leslie townes
leslie townes
copper - that's also in axler, oddly enough. :)
i wonder where he got it from.
koro wow did you actually do that exercise? oof.
i mean, software can. i remember working out the basis, or at least some of it, and giving up. too many factors of pi for me to continue by hand.
 
Koro
Koro
yes, Leslie.
Oh you mean approximating sine ?
 
copper.hat
copper.hat
i used to like calculations, now i make too many mistakes to be fun
 
Leaky Nun
Leaky Nun
6:20 AM
@Koro when I see your name I see Кого
 
copper.hat
copper.hat
hmmm?
 
Leaky Nun
Leaky Nun
which means "whose" in russian
 
Koro
Koro
Leslie: Actually I was referring to a solved example in Axler's wherein he approximates $\sin x$ by projecting it on space of polynomials of degree 5 using an inner product.
 
leslie townes
leslie townes
doesn't he do it numerically? and leave a symbolic calculation as an exercise?
cool application but horrible exercise, in my view
 
copper.hat
copper.hat
i like chebychev etc, or however you like to spell it
 
Koro
Koro
6:23 AM
Yes, you are right. But in the exercise, he also tells to use a symbolic calculator to retain $\pi$'s etc. which in the solved example he replaced by numerical approximations.
 
copper.hat
copper.hat
strangely i know very few Russians
one recruited me to the local high school math council
 
leslie townes
leslie townes
nice of him to suggest using software.
 
Koro
Koro
:)
 
leslie townes
leslie townes
is that what they're calling it these days
 
copper.hat
copper.hat
she just wanted an aggravator there
but i blew my chance by snorting when the high school principle announced to the council that they were removing the model of a ww 1 fokker from the local studio
i asked if they realised that ww 1 & ww 2 were separated by a few years in addition to some other minor details?
oh, but there are guns in the plane.
toy guns.
 
leslie townes
leslie townes
6:27 AM
you were snorting what when the high school principal announced that?
:)
 
copper.hat
copper.hat
well, i started laughing initially, i said that's funny, followed by going quite quiet and asking "are you serious???"
 
Koro
Koro
@LeakyNun I didn't know that. :D
 
copper.hat
copper.hat
i have many eastern bloc friends, all of whom speak Russian, but no Russians unfortunately
one of my aunts used to travel to Russian regularly
not a spy as far as we know
i want to visit Владивосток someday
apparently Cyrillic derived in part from Greek
used to be there were lots of Аэрофло́т flights to.from ireland
 
copper.hat
copper.hat
7:03 AM
$\kappa \alpha \lambda \eta \nu \upsilon \kappa \tau \alpha$
 
leslie townes
leslie townes
that stuff goes great with tzatziki
a little ouzo wouldn't hurt either
 
wklm
wklm
8:08 AM
Hi! I got two densities from the exponential family. Does their joint density come from the exponential family as well?
I got to show that the joint density of normal and inverse gamma comes from the exponential family and was hoping that is going to do the trick :S
 
 
2 hours later…
Martin Sleziak
Martin Sleziak
9:45 AM
1
Q: Chat room for ProofAssistants proposal

Guy CoderProposal: Proof Assistants A StackExchange chat room has been setup at https://chat.stackexchange.com/rooms/131732/proofassistants-temp

science
 
 
2 hours later…
love_sodam
love_sodam
12:01 PM
Is every fibration a Serre fibration? Is that really true?
 
 
1 hour later…
1010011010
1010011010
1:30 PM
Goodday
In hopes somebody will know something (source appreciated!):
Could anyone point towards the reasoning as to why the Heaviside cover-up method works?
Is there a derivation?
I tried something with induction but it's pretty horrible to be fair.
So hopefully my chat message here will lead to an alternative venue:-)
Oh and BTW, I need it for generalisations of the "order 1" case (order of the pole I guess), for which I already have the formula
I'll leave this up for a bit -- if it is deemed too complex I'll head over to the Q&A
 
EnthusiastiC
EnthusiastiC
1:52 PM
just one simple question,
Does ln(x) continuous at infinity?
According to the condition that states a function is continuous if both the limit and its value at that point equal and exist
but ln(infty) is infty
Does that mean it exists?
 
robjohn
robjohn
@EnthusiastiC no, infty is not real, nor is log defined there (since it is not real). the limits are equal, but that does not extend the log function to non-reals.
 
EnthusiastiC
EnthusiastiC
good to know, convincing is a little fuzzy
we can say that ln(infty) doesn't have a value
since infty is not a value either
 
robjohn
robjohn
@EnthusiastiC You can define an extension to the log function anyway you want, as long as you don't expect it to obey the relations that log obeys.
if you want to talk about an extension to the log function that is defined on an extension of the reals, you need to define those extensions. You can work with the two-point compactification of the positive reals, and define $\log(0)=-\infty$ and $\log(\infty)=\infty$. Then $\log$ is continuous on the compactification.
So with the proper extensions, the answer to your question is yes.
 
EnthusiastiC
EnthusiastiC
but infty is not real?
 
robjohn
robjohn
it is not a real number
 
Adil Mohammed
Adil Mohammed
2:07 PM
@robjohn do you remember the name of the shortcut rule for finding determinants
 
robjohn
robjohn
@AdilMohammed I am not sure what you mean. Are you thinking of a particular subset of matrices? One can expand on a row or column.
@EnthusiastiC a "neighborhood of infty" is the set of all reals greater than a given lower real bound.
 
EnthusiastiC
EnthusiastiC
it makes sense
 
robjohn
robjohn
for any neighborhood of infinity $V$, one can find another neighborhood of infinity $U$ so that if $x\in U$, then $\log(x)\in V$.
that is what is usually considered continuity at infinity without having to extend the reals
 
love_sodam
love_sodam
2:55 PM
Suppose I write one of 1~7 on each tile on infinite tile. Is it possible to write 1~7 so that every time I see 3X3 subtile, all numbers 1 to 7 appear on the boundary 8 tiles.
 
Adil Mohammed
Adil Mohammed
@robjohn oh I was talking about sarrus rule but I found it now
 
robjohn
robjohn
@AdilMohammed oh, that is not really a short cut, as you are computing all the same products and adding them. It is just a way to remember the signs used. And it is only for 3x3 matrices.
 
EnthusiastiC
EnthusiastiC
3:23 PM
@robjohn Sorry for late reply but appreciated
 
marshblocker
marshblocker
which is a more numerically stable: a(b + c + d + ...) or ab + ac + ad + ... ?
I'm thinking the former since we only multiply once
 
AMDG
AMDG
3:35 PM
Define numerically stable. From the perspective of mathematics, the semantics are the same, and the result is the same. I assume you're referring to some limited subset of the model of numbers such as floating point arithmetic.
Anyways, so has it been considered whether or not division by zero manifests a number line perpendicular to the reals as well as the imaginary numbers?
 
ypercubeᵀᴹ
ypercubeᵀᴹ
@AMDG are you referring to possible compactifications of the complex numbers?
 
AMDG
AMDG
What is a compactification?
 
ypercubeᵀᴹ
ypercubeᵀᴹ
Compactification (mathematics)
In mathematics, in general topology, compactification is the process or result of making a topological space into a compact space. A compact space is a space in which every open cover of the space contains a finite subcover. The methods of compactification are various, but each is a way of controlling points from "going off to infinity" by in some way adding "points at infinity" or preventing such an "escape". == An example == Consider the real line with its ordinary topology. This space is not compact; in a sense, points can go off to infinity to the left or to the right. It is possible to turn...
 
AMDG
AMDG
3:51 PM
So compactification is just another way of saying that a topological space is converted from one of infinite space for every dimension to one of bounded indefinite space?
Regardless, I don't know enough about complex analysis to give a proper answer to your question. I'm just considering division by zero as an operation and its immediate conclusion as a result of further contemplation.
If we go back to first principles and consider division as asking "how many times can I subtract one number from another before it equals zero?", then for $\frac{0}{0}$, we could say that there are infinitely many answers, and generalize or build upon this as a foundation for numerators other than zero such that you do not end up with the contradiction that every number is equal to every number.
So any real number greater than or equal to zero must intuitively be a valid solution to $\frac{0}{0}$ since I can substract zero from itself any arbitrary quantity and end up once more at zero, including a total of zero subtractions. I suppose zero would in some way here be considered a "minimal" or "optimal" answer, but we would have to extend the idea of functions to operations on sets which yields sets and state that all reals greater than or equal to zero is the answer to $\frac{0}{0}$.
 
Adil Mohammed
Adil Mohammed
4:13 PM
@robjohn Haha maybe at your level it may not be a shortcut but at my level it's like using l'hopital for differentiation to save time😂
Also Rob, I have a HS doubt, if f(x)=4x³-18x²+27x-7 and f'(x)= 3(2x-3)² would you say the function is increasing throughout the domain? Or throughout the domain except 3/2?
 
robjohn
robjohn
4:50 PM
@AdilMohammed increasing means that if $x\gt y$, then $f(x)\gt f(y)$. Can you find any $x,y$ for which that doesn’t hold?
@AMDG once you utter “division by zero”, most people take exception.
 
AMDG
AMDG
What do you mean by that?
 
robjohn
robjohn
There is no good way to handle division by zero, so as soon as most people see someone suggesting it, they think “crank”.
 
AMDG
AMDG
Why is there no good way to handle division by zero?
I've seen all the videos on it, heard all the usual things about it, but it just feels like all the avenues haven't been considered.
Or at least, if they have, something's been missed because someone thought it was interesting for a second and then dismissed a possibility out of bias.
 
robjohn
robjohn
No matter how you try to define it, you get contradictions. That is why in almost all, if not all, computer languages, division by zero raises an exception.
 
Lobic
Lobic
$\forall x\forall y\forall z([(x,y)\in f\wedge(x,z)\in f]\Rightarrow y=z)$ is this a correct definition of a function i think not?r
Am I right
 
robjohn
robjohn
5:01 PM
@Lobic that is more a definition of “well-defined“
 
Lobic
Lobic
What do you mean by well defined ?
I said not because {1} satisfies it vacuously but is clearly not a function
@robjohn
 
robjohn
robjohn
Well-defined expression
In mathematics, a well-defined expression or unambiguous expression is an expression whose definition assigns it a unique interpretation or value. Otherwise, the expression is said to be not well defined, ill defined or ambiguous. A function is well defined if it gives the same result when the representation of the input is changed without changing the value of the input. For instance, if f takes real numbers as input, and if f(0.5) does not equal f(1/2) then f is not well defined (and thus not a function). The term well defined can also be used to indicate that a logical expression is unambiguous...
 
Ted Shifrin
Ted Shifrin
Happy Thanksgiving, @robjohn, @leslie, @copper, @amWhy, and all others who celebrate!
 
robjohn
robjohn
@TedShifrin Happy T-Day!
 
Lobic
Lobic
So it’s not a correct def of a function for the reason I gave correct?
 
Ted Shifrin
Ted Shifrin
5:04 PM
Out of bias? How about out of the laws of basic algebra?
 
robjohn
robjohn
I just got a Costco pumpkin pie yesterday. They are huge!
 
Ted Shifrin
Ted Shifrin
In any ring (in particular, the real numbers), $0$ times any number is always $0$
i hate pumpkin pie. I’m now making an apple tart with Calvados custard :)
 
robjohn
robjohn
@Lobic $\{(x,1)\}$ is a function
 
Lobic
Lobic
But {1} is not
And satisfies the def
 
AMDG
AMDG
Bruh, zero is an anomaly, all things considered. All this time and it still has no definition for division by zero. You would think something so fundamental to the set of reals would have a definition under the operation of division.
 
robjohn
robjohn
5:07 PM
@TedShifrin them’s fighting words. My two favorite pies are pumpkin and apple.
 
Ted Shifrin
Ted Shifrin
You’re hopeless.
Not you, robjohn.
 
robjohn
robjohn
@TedShifrin the tart does sound scrumptious, however.
 
AMDG
AMDG
For starters, what is the formal definition of zero as a real number?
 
Lobic
Lobic
@AMDG No it’s not ; it’s really as simple as 0+0….+0=0
 
Ted Shifrin
Ted Shifrin
It is. I’ve made it numerous times. I’m not a big custard fan, but it’s discreet and so yummy,
 
robjohn
robjohn
5:10 PM
It’s the additive identity. That leads to $0\cdot x=0$ for all $x$ and therein lies the problem with division by zero.
 
Lobic
Lobic
@robjohn am I missing something trivial
 
Ted Shifrin
Ted Shifrin
That PLUS the distributive law. Probably AMDG doesn’t believe in distributivity.
 
AMDG
AMDG
Let's try to keep things objective eh
 
Ted Shifrin
Ted Shifrin
You are one to talk. Seriously.
 
AMDG
AMDG
I'm not sure what that's supposed to imply, but whatever.
Is there something wrong with wanting something to be defined?
 
Lobic
Lobic
5:13 PM
If you want you can define division by 0 but it’s completely useless in most cases and makes us loose some important properties
 
robjohn
robjohn
@Lobic a function is a set of pairs, so how does $\{1\}$ fit in? When I see that, I assume you mean $\{(x,1)\}$
 
Lobic
Lobic
That’s exactly my point @robjohn if you plug f={1} in the def it holds vacuously
 
robjohn
robjohn
Any set that has no pairs would satisfy that “definition”. It is not a definition of a function from one set to another.
$\{(1,1,1)\}$ also satisfies it
 
Lobic
Lobic
Exactly so that’s a wrong def of a function
correct@robjohn
 
robjohn
robjohn
Yes
 
Lobic
Lobic
5:19 PM
because i say it in a post today
By a reputed user
 
robjohn
robjohn
If it’s specified as a set of pairs from $X\times Y$, then it might work
 
Ted Shifrin
Ted Shifrin
One would have assumed that $f$ was that from the formulation Lobic wrote above. But I came in late.
I took $f$ to be an arbitrary subset of $X\times Y$ with that additional property.
 
robjohn
robjohn
@Ted: I am pinned by a cat on my lap, so I am chatting on my phone. I hate the mobile interface.
Just as I wrote that, my cat released me :-)
 
Ted Shifrin
Ted Shifrin
Yes, sometimes it gets very glitchy.
I hate typing on the phone, but I do use my iPad.
 
robjohn
robjohn
i have a laptop and an iPhone, so I don't have an iPad
It seems a bit overkill
 
Lobic
Lobic
5:28 PM
Then it would be a function with domain a subset of X @TedShifrin
 
Ted Shifrin
Ted Shifrin
Correct, @Lobic. As I say, I came in in the middle. If you want the domain to be all of $X$, then you have to add that condition (for all $x$, there exists $(x,y)\in f$).
@robjohn I got a mini iPad long before I ever got an iPhone.
The phone is too small for reading and typing. I use it only when I'm out and about.
 
Lobic
Lobic
Still the user never specified that it was a subset of the Cartesian product of 2 sets so the answer was wrong !
 
BannedUser
BannedUser
a lot of physicist couldn't answer this question so I came to mathematician and fake my question to be mathematical physics lol
 
Ted Shifrin
Ted Shifrin
@Lobic I have not seen the post, so I can't comment.
 
BannedUser
BannedUser
I saw this weird question in physics textbook for electric distribution
I wonder why is at last stage q=Q/4?
 
Ted Shifrin
Ted Shifrin
5:38 PM
What does any of that mean, @BannedUser?
 
BannedUser
BannedUser
@TedShifrin it ask how much electric charge at last stage is at location A
The line is wire.
 
robjohn
robjohn
@TedShifrin I have a stylus and using Notes I can make hand drawings on my iPhone. It saves me lots of paper.
 
BannedUser
BannedUser
Physically question doesn't makes sense to me but I don't know bout abstractly.
 
Ted Shifrin
Ted Shifrin
Again, I would prefer the iPad, but I totally get it, @robjohn.
I need words explaining how the diagrams are related. It just seems like a bunch of symbols with no connection, @Banned.
 
BannedUser
BannedUser
@TedShifrin At first you can see there is a line connecting from point a to point b where point both a and b contains 0 charge and c got charge $Q^+$ but c is not connected to any of them. You can distribute charge from the line.
At second all of the position is connected by line which means charge can be distributed.
At third stage a and c is disconnected.
 
Ted Shifrin
Ted Shifrin
5:46 PM
So the ABC in diagram 4 are the same as in diagram 1, just after a succession of "moves."
So what happens in 2)?
 
BannedUser
BannedUser
@TedShifrin all of the position is connected by line which means charge can be distributed
@TedShifrin ?
 
Ted Shifrin
Ted Shifrin
Yes, so what happens?
Presumably A becomes Q/2 and B and C are each Q/4.
Then in 3) A and B would become 3Q/8. So I'm not getting their answer.
 
BannedUser
BannedUser
Well I have thought about this in terms of protons which can move lol
 
Ted Shifrin
Ted Shifrin
The key is to figure out 2).
 
leslie townes
leslie townes
i'm thankful for not having that on my plate. happy thanksgiving, ted.
 
Ted Shifrin
Ted Shifrin
5:51 PM
Give my regards to the munchkin and the ducks, @leslie.
 
BannedUser
BannedUser
I think it depends on amount of protons. If I got 1 proton then nothing will happen. If I got 2 proton then there will be Q/2 on C and Q/2 on B....
Charge doesn't makes sense to me without using real life stuff like proton and electron
 
Ted Shifrin
Ted Shifrin
It's just a matter of figuring out the balancing act. If they had three lines in diagram 2), then all of them would end up with Q/3.
 
leslie townes
leslie townes
if you're beaming protons from one place to another, make sure there's a wall of lead between you and the other side.
 
BannedUser
BannedUser
@TedShifrin by "them" are you saying point?
 
Ted Shifrin
Ted Shifrin
Yes, the charges at all three would be equal, in that case. But we have only two lines.
 
BannedUser
BannedUser
5:56 PM
then Q/2
 
Koro
Koro
I think you should use the fact that charge flows from higher potential to lower potential.
 
Ted Shifrin
Ted Shifrin
I think we're just looking at equilibrium state, Koro. So what do you say happens after equilibrium in position 2)?
 
Koro
Koro
C will do something more at A than at B so you shouldn't have equal charges but I don't remember the details. I think I did these kind of problems in my class 12.
 
Ted Shifrin
Ted Shifrin
With one wire joining two point charges, at equilibrium they have the same charge.
@Koro: My claim was that A had to balance B+C, and by symmetry B,C should be equal. So A=Q/2, B=C=Q/4. But that may be wrong. Does it matter if all the charge starts off at C or at A?
 
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