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00:02
Meaningless concept to me. And who cares if $G$ acts on $M$?
00:13
@BalarkaSen So, consider the set of smooth curves where the endpoints and the tangents there are fixed. Is that set simply connected? Can I do a nontrivial loop?
I feel like the Thurston wiggles from that Inside Out video should do it but I'm not sure
i wonder what application the asker has in mind. or if they're just noodling with formalism.
@leslietownes the example that comes to mind for me are Grassmann numbers in physics
which are very much not numbers in the usual sense. (They're elements of the exterior algebra over $\mathbb{C}$.)
why on earth would you call those numbers
shrug
wikipedia says they've also been called "anticommuting numbers" and "supernumbers"
which is somehow even worse
@Thorgott Do they make you numb
00:19
they call 'em fingers, but i've never seen 'em fing.
"The general element is now [set of equations including $z=z_B+z_S$] where $z_B$ is sometimes referred to as the body and $z_S$ as the soul of the supernumber $z$."
@leslietownes "Oh, wait, there they go"
so Grassmann numbers don't act like numbers and they suffer from the soul-body problem. neat
Remember that $\epsilon$ in $k[\epsilon]/(\epsilon^2)$ is called a dual number.
00:23
that terminology does make my brain numb, fair point
yeah, they mention that one too
@TedShifrin I call that infinitesimal element
what even is supposed to be dual about it
i can understand those in the process of real numbers -> complex numbers -> hypercomplex numbers -> dual numbers -> Grassmann numbers
A homomorphism gives a tangent vector, backwards.
"Dual numbers were introduced in 1873 by William Clifford, and were used at the beginning of the twentieth century by the German mathematician Eduard Study, who used them to represent the dual angle which measures the relative position of two skew lines in space. Study defined a dual angle as $\theta + d\epsilon$, where $\theta$ is the angle between the directions of two lines in three-dimensional space and $d$ is a distance between them."
i hate that notation btw. $d$ as a variable is gross
00:26
Interesting. I wonder why that would be a reasonable choice?
I mean, you're never multiplying angles.
as much as i dislike a lot of modern notation, the stuff in the late 1800s early 1900s shows how bad it could be.
Unless you're doing solid angle stuff…
Having a hard time figuring out how to do solid angles (real or metaphorical) out of skew lines
If you have a knot, by the way, you can (usually) make a Seifert surface using solid angles
yeah, i don't get it either. where is $\epsilon^2=0$ coming in from that
Given a knot and a point, you can look at the solid angle of the knot from the point's perspective
and then look at the set of all points with a given solid angle with the knot
Sometime's it's singular and it's not a surface but it's usually not
This is like the Gauss linking number integral.
00:39
the only proof of that I've ever understood was the "proof by elecromagnetism"
It’s pulling back the area 2-form on the sphere to $S^1\times S^1$ by the unit chord map.
namely, take one loop to be a current-carrying wire. that generates a magnetic field per the Biot-Savart law. if you compute the circulation of said field along the other loop, then Ampere's law tells you it's the linking number (up to irrelevant physical constants)
@TedShifrin unit chord map?
oh, this answer of yours: math.stackexchange.com/a/415176/137524
8 years ago!
so unit chord map = direction vectors from one loop to the other?
00:49
i can see how that connects $S^1\times S^1$ to the sphere at least
The solid-angle shenanigans are mentioned in there somewhere
Yes, Semiclassic.
@Semiclassical Wait I can't
on one side of the map, you've got direction vectors which lie on the sphere
signed area strikes again
00:50
on the other you've got a pair of points, one for each loop. but the loops are each S^1
Using the integral to compute degree of the mapping.
@leslietownes and, occasionally, unstrikes
i think the "magnetism" derivation gets Gauss's version
@Semiclassical Wait is this for doing the solid angle or for finding the linking number
linking number
00:52
"Linguistic ambiguity strikes again!" "What are ambiguity strikes??"
i'm going to assume that if i refuse to acknowlege that, it doesn't exist
01:06
parenting is a pain in the neck.
I have enough pains in the neck and lower back.
it's going to be back pain for me. my daughter somehow broke her leg in day care today. her leg has been immobilized but she will need to be carried around until we can see an orthopedist. probably next week at the earliest.
i'm preparing the lawsuit as we speak.
Ugh. Poor little troublemaker!
She jumped off a 4 ft high jungle gym?
indoors. it sounded like she was moving around very rapidly. per my daughter, she was 'spinning around' in a room with a lot of low-lying kid-sized furniture. her foot probably got caught on something, and she crashed into furniture on the way down.
she's at the age where she can move extremely fast, but not old enough to control her movement very well or appreciate risk.
I’m at that age, too.
01:10
haha
she's in high spirits. she demanded a blanket, a snack, and the cat when she got home.
What does the cat say?
she's on an arm of the sofa. she is not going to risk the lap.
i hope my daughter doesn't take the wrong lesson from this. break a leg, have gobs of attention and entertainment and no day care.
She’s in charge, but you knew that.
 
2 hours later…
03:09
@leslietownes I guess the non-catchy version would be: the completion of an algebraic tensor product is not a categorical tensor product.
yeah, something like that. completions of tensored spaces/normed algebras is an interesting subject. grothendieck was an early contributor to the field although i understand he is better known for other things.
and without dissing it, his work in functional analysis turned out to be less impactful than his work in other areas. he was wise to switch.
judging from the bit about nuclear spaces, i should probably be more careful about what i mean by "completion"
completion with respect to Hilbert-Schmidt norm, vs completion with respect to trace norm
@leslietownes sorry to hear about your daughter.
she's happily in bed now. she does seem to appreciate that she shouldn't put weight on her leg.
it is tough at that age to have casts.
03:19
she was picking at her splint. we put her in a tight pair of pajamas to hopefully limit this behavior. i will scream if we cannot get a cast tomorrow.
semiclassical an interesting and fairly introductory treatment of aspects of some of the theory a monograph by raymond ryan, titled something like 'tensor products of banach spaces,' from springer.
he's apparently irish but you can trust his math.
neat
i need to pick up some expertise on Banach algebras/spaces and C* algebras
is there any function f:R->R with domain (0,1) and range [0,1], where f is also continuous, 1-to-1 and onto?
seems like the conditions would force it to be monotonic, in which case i can't see how you'd get it to work
but my imagination for analysis counterexamples is...not great
no. see if you can prove that f^{-1} (which exists by your hypotheses) is continuous and use the fact that the continuous image of a compact set is compact.
or get there the same way using, i dunno, sequences or something.
or click there.
brilliant
04:01
on the other hand i am irish but you cannot trust my maths.
Only your spuds.
he can also cut and stack turf
greetings ticks
I just realized that bjarne added a feature in c++ so that it always throws errors in the first compilation
120
Q: Generate the longest error message in C++

Elazar LeibovichWrite a short program, that would generate the longest possible error message, in a standard C++ compiler (gcc, cl.exe, icc, or clang). The score of each entry is the number of characters in the longest error message the compiler emitted. Types included in your source code and quoted by the comp...

04:24
Theodore John Kaczynski
let $f:\mathbb{R}^2 \to \mathbb{R}$. author gives a definition of differentiability using $f(x_0 + h, y_0 + k) - f(x_0, y_0) = f_x(x_0,y_0)h + f_y(x_0,y_0)k + \epsilon(h,k)\sqrt{h^2+k^2}$
why would it not be sufficient to write $\epsilon(h,k)$ without multiplying by $\sqrt{h^2 + k^2}$?
@Semiclassical i fear more than one option have values b/w 500-750.
you could, as long as you then quantified the desired property of epsilon accordingly
do you have the complete original problem? @kunalCh.
04:28
It’s a question of what limit condition $\epsilon$ must satisfy.
oh! the author does add $\epsilon(h,k) \to 0$ as $\sqrt{h^2 + k^2} \to 0$, which i'm guessing then has to be interpreted as: the error goes to zero as the change in x,y goes to 0
@leslietownes i do find it amusing that the business of trace class vs Hilbert-Schmidt class (and thus whether the Hilbert-Schmidt class is a categorical tensor product) hinges on "$\sum_k 1/k^2$ converges but $\sum_k 1/k$ diverges"
No, the error goes to zero faster than $\|(h,k)\|$. That’s the whole point.
(i mean, it's really coming down to $\sum_k a_k^2 converges doesn't imply $\sum_k a_k$ converges. but that's the obvious counterexample)
@TedShifrin oh! now it makes sense. thanks a lot!
04:32
Yup.
actually, come to think of it. is there another 'obvious' example of a sequence $\{a_k\}$ such that $\sum_k a_k^2$ converges but not $\sum_k a_k$? I'm sure one can do some variations on the harmonic series but i can't think of another kind of example
(again, failure of imagination for analysis pathologies)
@Semiclassical you could do $\sum_{n=1}^\infty\frac1{n^{3/4}}$ which diverges and $\sum_{n=1}^\infty\frac1{n^{3/2}}$ which converges, or did you mean something other than $p$-series?
Variations? Just the harmonic series?
hmm, that's a good point
any 1/2<p<1 will work, yeah
$\le 1$
04:37
woops
given that that was the whole point of the above, i should've noticed that
so yeah, p-series give generic examples. i wonder if there's others
I missed the early discussion; sorry.
Lots of others.
yeah, i buy that
How about a series where $\sum\limits_{n=1}^\infty a_n$ converges, but $\sum\limits_{n=1}^\infty a_n^2$ diverges?
i wondered about that, and I don't think that can happen
$a_n$ would need to converge to zero as a sequence
but then eventually $a_n^2<|a_n|$
$\sum\limits_{n=1}^\infty\frac{(-1)^{n-1}}{\sqrt{n}}$
04:41
oh.
yeah
doesn't converge absolutely but does converge
it had to be something sneaky like that
given the trace class context, i should've insisted on $|a_n|$ i.e. absolute convergence for $\sum_n a_n$ anyways
Hi I want to use { bracket. As of now I am using it like this: \{. But its size is small. I am not able to use \left{, the way we can use for \left( to increase the size of (. What is the code for bigger {? Thanks.
\big, \Big, etc.
@TedShifrin in that case, how to end }? because \big} is not working.
04:52
\}
otherwise it's just interpreted as yet another latex bracket
similarly \{
@Semiclassical but this way, the size is small. the content inside is bigger
sure? you can still do \Big\{
e.g. $\Big\{\big\}$
the entire "\Big" is the modifier. it's not Big modifying the \
@Semiclassical even with \big, I am getting small size of the bracket. Maybe after I post it, the formatting may look better. If not I'll post the link here. I may take more than an hour to finish. Thanks for your help.
well, if you want to do LaTeX in chat, use the link in the upper right
i am afraid that Kaczynski's path and my own had some linking coefficient.
I want to return to a happy place where meat and 3 veg. means 3 kinds of spuds :-)
05:17
@Semiclassical the answer can be found by using this formula median$=l + \frac {\frac{s}{2}-c}{f}×i$ where[here we are taking specific values of 3rd entry cz 6/2=3, (even) no of entries = 6] $l=500$ lower limit of 3rd entry $f=$frequency of 3rd term i.e. 23 $c=$cumultative frequency i.e. 12+10=32 $s=$total frequency 100 $i=$ limit width 750-500= 250. So median in this way = 500+ [{100/2-32}/23]250 or 500 + 0.782608696×250 or 695.652174
@Semiclassical that was the original question btw which i copy pasted.
06:51
How can I determine the convergence of $\sum_{n=1}^\infty \frac{(2n)!}{4^n(n!)^2}$ ?
it's a bit delicate. suppose you insert a factor of $x^n$ into that series. then for $-1<x<1$ the series definitely converges per the root test, and indeed sums (via the binomial theorem) to $(1-x)^{-1/2}$
so $x=1$ lies right on the radius of convergence.
at the very least, the ratio test is not going to do a bit of good here
should have "ratio test" above, not "root test" (though that's also not helpful)
07:18
@aarbee $\left\{\frac{\frac54}{\frac45}\right\}$ gives $\left\{\frac{\frac54}{\frac45}\right\}$
@barista it diverges.
@robjohn how can you tell?
Look at $(9)$ from this answer
and $\binom{2n}{n}=\frac{(2n)!}{n!^2}$
so your terms are like $\frac1{\sqrt{\pi n}}$
Oh it's harder than I thought
When two planes intersect, the intersection is a line. The cross product of the normal vectors of both planes will lie on that line of intersection. I understand that there is several ways in which 3 planes can intersect. When 3 planes intersect at a point, will the direction vector of the line (when two planes intersect) be parallel to all 3 planes?
@barista not really... just a sec
07:37
if $a_n=\frac{(2n)!}{4^nn!^2}$, then $a_0=1$ and
$$
\begin{align}
a_n
&=\frac{2n(2n-1)}{4n^2}a_{n-1}\\
&=\frac{2n-1}{2n}a_{n-1}
\end{align}
$$
Therefore,
$$
a_n=\underbrace{\frac{2n-1}{2n}\frac{2n-3}{2n-2}\frac{2n-5}{2n-4}\cdots\frac34}_{\substack{\text{we bound the squares}\\\text{of these terms from below below}}}\frac12
$$
cross-multiplying shows that $\frac{2n-1}{2n}\ge\frac{2n-2}{2n-1}$; therefore, $\left(\frac{2n-1}{2n}\right)^2\ge\frac{2n-2}{2n-1}\frac{2n-1}{2n}=\frac{n-1}{n}$. Therefore,
@barista This shows that $\frac{(2n)!}{4^nn!^2}\ge\frac1{\sqrt{4n}}$
@robjohn Thanks :)
The other answer just refines the $\frac12$ to $\frac1{\sqrt\pi}$ and bounds it from above as well
 
5 hours later…
12:56
Hey guys! I've almost solved this interesting question on second-order nonlinear ODE, but I can't seem to get the last part. My question of Math SE details the question and my current solution; I would really appreciate if anyone here could take look and give me any advice on how to continue
0
Q: Newton's Second Law and Second-order Nonlinear Ordinary Differential Equation

David ChoiThe following question is from Courant's Differential and Integral Calculus, Vol. 1: A particle of unit mass moves along the x-axis and is acted upon by a force $f(x) = -\sin x$. (a) Determine the motion of the point if at time $t=0$ it is at the point $x=0$ and has velocity $v_0=2$. Show that a...

13:08
@DavidChoi this is equivalent to the ODE for a mathematical pendulum. the integrals actually can be carried out...well, at least if you count "expressed as elliptic integrals/functions"
the details get gory pretty fast tho
well i mean elliptic integrals are just integrals we can't express in elementary functions
so we just go
u be a function lmao
anywhooooo
did u get to the end of my post?
yeah. main thing is that there's a loooooot that's known about said functions
i feel like there's no logical error in my reasoning
that said, the "simple pendulum" version gives the intuition immediately: If you hit a simple pendulum hard enough (thereby giving it an initial speed) it'll pass over the top and swing back down
and since no energy is lost, it'll just keep repeating that motion forever
if it's not hit hard enough, by contrast, it'll just swing back and forth
right i see that, but i wanted to be able to show it mathematically
since well i got the question from a math textbook lmao
13:11
fair
omitting the details, i get this expression $\frac{dt}{dx} = \frac{1}{\sqrt{2\cos x + v_0^2 -2}}$, where t is time and x is displacement
what i'd be inclined to do is look at $\frac12 \dot{x}^2+1-\cos x(t)=\frac12 v_0^2$
hmmmmmm
why that expression specifically?
one way to understand it is that it's saying that the trajectory of $(x,\dot{x})$ is restricted to that curve
to the curve of $\frac{1}{2}v_0^2$ correct?
13:16
so suppose if we look at the level sets of such. i guess there's no reason to have the $1$ on the left-hand side tho
not quite: the curve is $H(x,\dot x) = \frac12 \dot{x}^2-\cos x=\frac 12 v_0^2-1$ (using your version)
i.e. levels sets of $H(x,\dot x)$.
i have no idea what level sets are
can you give me a quick run down of what it is?
sets on which a function $f(x,y,\cdots)$ assume a particular value
think of $z=f(x,y)$ as some 2D surface above the xy plane. then the level sets are just horizontal cross-sections
i.e., sets of identical altitude/level
ahhhh i seeeeee
with that in mind, suppose we plot $H(x,\dot{x})$:
woahhh okk
13:22
what you can notice is that, if you pick $H(x,\dot{x})=E$ with small enough $E>0$, then the level sets are all closed curves
in particular, the level set in the nbhd of the origin is such
right i see that
so evidently if you pick a value of $E=v_0^2/2-1\in (0,1)$, then your trajectory is trapped on such a closed curve for all time
but if you pick $E>1$, then the curve you get isn't trapped
it just wiggles back and forth for all $x$
i got you i got you
the case of $E=1$ (equivalently $v_0=2$) is known as the separatrix
can i trouble you with one more question
13:25
one way to understand it is to notice that there's saddle points at $(x,\dot{x})=(\pm \pi, 0)$
@barista Raabe's test will work
what do you mean by saddle points?
there's a saddle point right smack dab in the middle of that graph
ok ok ok
so essentially if i can somehow show $H(x, \dot x) = E$
can be represented by a circle
for $E < 1$
then i'm done
not a circle, specifically, but homeomorphic to such
(as $E\to 0^+$ they do become circles)
13:29
right sorry the important point is just that the curve is closed
damn there's so much i don't know lmaoo
more analytically, near $x=\pi$ we have $\cos x\approx 1-\frac12 (x-\pi)^2$ so $H\approx \frac12 \dot{x}^2-\frac12 (x-\pi)^2+1$
i haven't even learned ODEs properly yet
which is a saddle point, because $H$ increases in the $\dot{x}$ direction but decreases in the $x$ direction
all of this said: one should be able to get this directly from the integral
u mean from $\int \frac{1}{\sqrt{2\cos x + v_0^2 -2}}$
13:32
i dont see that at all
to clean that up a bit, note that $2\cos x-2=4\cos^{2}\frac{x}{2}$
so the integral becomes $$t=\int_0^{x(t)}\frac{1}{\sqrt{v_0^2+4\cos^2(x/2)}}\,dx$$
oops
$2\cos x-2=-4\sin^2(x/2)$
$$t=\int_0^{x(t)}\frac{1}{\sqrt{v_0^2-4\cos^2(x/2)}}\,dx$$
which starts to make the role of $v_0$ more apparent. if $v_0\geq 2$, then the denominator is well-defined for all $x$
but if $v_0<2$, then you seemingly have a problem when $\cos(x/2)=v_0/2<1$
that was essentially my argument in my post
that's why i didn't have trouble showing that the particle goes to infinity as $t \to \infty$ for $v_0 > 2$
but for $v_0 < 2$, my line of reasoning lead me to the conclusion that $x \to \cos^{-1} (\frac{2-v_0^2}{2})$ as $t \to \infty$
where $\cos^{-1} (\frac{2-v_0^2}{2})$ is a vertical tangent of the graph $\frac{1}{\sqrt{2\cos x + v_0^2 -2}}$
that said, recall that the denominator is really just $\dot x$
so if you think about the loops, what has to happen is that $\dot{x}$ goes from being positive to negative
(i.e., the pendulum reaches a certain height and then starts to swing back down)
to make this a bit more precise, let $x_0=2\cos^{-1}(v_0/2)$ for $0<v_0<2$
then the integral makes sense up to some value $T=\int_0^{x_0}$ (i'm too lazy to write out the integrand again)
Excuse me, but do I need some sort of common algebra for working with boolean algebra and other algebras like elementary algebra? I'm trying to figure out why wolfram here can't recognize something like wolframalpha.com/input/…
So it takes a time $T$ for the particle to reach the $x=x_0$ "turning point"
13:44
im sorry, where did you get $x_0=2\cos^{-1}(v_0/2)$ ?
@AMDG is a + b intended as mod 2, i.e., A XOR B?
@DavidChoi solving $\dot{x}=\sqrt{v_0^2-4\cos^2(x_0/2)}=0$ for $x\in(0,\pi)$
@Semiclassical a + b here denotes arithmetic sum in elementary algebra. I'm using two different algebras here in the same equation.
in terms of the loops, it's the horizontal 'radius'
These are the kinds of things I would like to analyze, yet it would seem even Wolfram cannot help me do that.
i'm not surprised. || is logical OR and applies to propositions
a+b applies to numbers
wolfram can't read your mind and know that you're using a non-standard definition for ||
13:47
Propositions? Logical OR is an operation that applies a truth table to an input. In both cases, a and b are numbers.
Oh, but that would explain why you mentioned mod 2 I suppose. I intend to work with two base 2 numbers.
right
i don't know what Mathematica itself does to solve boolean systems, come to think of it
So how would I properly phrase this? In the case of binary addition where there is no carry required, $A + B = A \lor B$.
that said, while WolframAlpha can't do it, Mathematica can. the `Solve` function doesn't owrk here, but `FindInstance` does:
`FindInstance[a||b==a+b,{a,b},Integers]` yields `{{a->0,b->0}}`
hmm. i must be forgetting how to format code in chat
test
ok i dunno
The relevance of solving something like $a + b = a \lor b$ as specified in my query for both $a$ and $b$ solution sets gives a nice pattern matching function on the basis that a number $n = m(a + b)$ to help me find a substring within a string.
I am not at all sure how this would work if you work with bitwise strings tho
13:52
What does {{a->0,b->0}} mean?
Also, I don't have mathematica.
yeah, thtat's mathematica speak. basically it means it's found a solution (a,b)=(0,0)
That makes no sense to me. Why would (0,0) be the solution? Maybe can you specify the constraint that a or b but not both are constant at evaluation?
"a solution", not "the solution"
Me want solution set
0+0 = 0 = 0 || 0
13:54
Yes, of course.
i suppose the only case which fails is (1,1)
all of this is a bit academic if you don't have mathematica, ofc
But... ${10}_{two} + {1}_{two} = {10}_{two} \lor {01}_{two}$.
ahh
that's fair
Yeah, so there are infinitely many solutions.
in that case Integers is the wrong domain
13:55
wat
for mathematica I mean
hmm
Oh. Man, don't you love when software is straightforward like mathematica? Target domain: integers. Your software: solution is (0, 0).
Mathematica's solutions don't make a lot of sense to me in this context tho
One day, software will be intuitive.
if I force it to generate more than one solution, the others i get are all nonsense
13:59
Hm, I can probably help by specifying which particular constraints the equation I mentioned is useful for.
Mathematica does have operations on bits so i'm a bit perplexed what's going on
for reference, what is 10+11 in your context? (as bit strings not integers)
101
ahh.
maybe i know how to deal with this, then
at least on the mathematica side
Yep. Like I said, a + b = a or b if the sum does not require any carries.
okay, the fix on the mathematica side is to use BitOr instead of Or
14:03
I'm sure you can see how this helps for finding substrings. Evaluating the solution set as a boolean function tells us whether or not a substring exists in the string at all. Evaluating the solution set as a set of numbers and computing a function that maps all a to all b can give us an occurrence of that substring within the string assuming it is a true substring.
the issue is that Or expects arguments of either True or False
BitOr correctly interprets everything in terms of binary representations of integers
let's see if that works in WolframAlpha. (i am not confident)
so forcing bitwise OR allows WolframAlpha to handle it correctly
(how you'd do bitwise or without knowing about BitOr? heck if i know)
So correction: not a function that maps from a to b. Rather, a function $f : a\to n$ for $n = a + b$ and $n$ satisfying $a + b = a \lor b$. That is the full set of constraints.
@Semiclassical Nice, that's better than nothing
right. so for a given $a$, find $n$ such that there's a $b$ which works
Yes, or vice versa
getting WolframAlpha to do what you say can be kind of an arse at times
14:08
Everything is pain and never works when you want it to.
and sometimes it just sorta shrugs
Looks like I'll have to speed up development to make my "Libre Mathematica :tm:".
Advertisement: "It can use multiple algebras and solve them in any target algebra, but intuitively."
Or: "It's mathematica, but intuitive"
an example I was seeing this morning: compare this with this
Yes, wolfram can be retarded sometimes. Severely retarded. Sometimes it parses latex correctly; other times, "Wolfram|Alpha doesn't understand your query".
if i pick numbers a,b with -1<a<b<1, then WA is perfectly happy to do that integral numerically
but if i pick a=-1, b=1? "nope, i don't get what you're asking"
not "i can't compute this" but "nah, i dunno what you mean"
14:13
Amazing. Look at Wolfram's massive brain every time I try to add constraints. wolframalpha.com/input/…
do naturals mean positive integers or nonnegative? i hate that that's an ambiguity that exists
Positive integers. How is that ambiguous?
Z = integers. N = naturals.
there's a lot of sources which include 0 in the natural numbers
129
Q: Is $0$ a natural number?

brynIs there a consensus in the mathematical community, or some accepted authority, to determine whether zero should be classified as a natural number? It seems as though formerly $0$ was considered in the set of natural numbers, but now it seems more common to see definitions saying that the natura...

for an ancient debate on the subject
Yes, well here I mean N and 0
it's the math equivalent of "do arrays start at 0"
it does a bit better if you implement them as inequalities: wolframalpha.com/input/…
14:17
Also, I determined long ago that zero is a "neutral" number so-called (because there is no such thing as neutral in reality) with its own sign in my eyes because it lacks the properties of positive and negative numbers, and it cannot be both negative and positive at the same time.
for my tastes i just insist on only ever using the phrases "positive integers" and "nonnegative integers"
just so i never have to debate whether 0 is natural
That's fair. You can't lose with that sort of terminology.
(that said, i do think there are contexts where 0 is natural and contexts where it isn't. for power series, you start with the x^0 term so including 0 makes sense. but 0 is quite unnatural when it comes to multiplication.)
(if all you're doing is addition, 0 is natural as the identity element. but if you're also doing multiplication, it's not.)
0 cannot sometimes be natural and sometimes not be natural. It is either always natural, or never natural.
meh. "natural" is just a convention
i've long given up on expecting people to come up with consistent conventions
14:21
I always took natural to be a definitive property accepted by all mathematicians as the way to describe positive integers because that is the accepted name of the set of naturals, and the numbers in that set.
(c.f., French definition of continuity vs. American definition)
@Semiclassical It is impossible that anyone with a disordered foundation and understanding of reality can come up with consistent conventions that are coherent, and worse, is still capable of making internally consistent systems that are erroneous as a result.
meh. humans create mathematical formalisms
what's important is not that everyone use the same conventions, because different conventions are legitimately more useful in some contexts vs others
it's that everyone be able to convert between them
Yes, well that in summary means that it is best to have a single, common, objective standard because without that standard, no conversion can take place between the subjective models.
setting aside irritating questions of whether {0,1,2,..} and {1,2,3,...} exist in a Platonic sense or not, the name 'natural' is not an objective property
14:25
Conventions belong to the subjective order.
@Semiclassical Agreed.
I partly come at this from a physics mindset. In special relativity, there's no preferred reference frame: each one is just as good as another. You could insist that one of them is the "true" reference frame, with all others being defined relative to it, but there would be no empirical content to that
i.e., I'd get exactly the same predictions based on whether I said that my reference frame is the "correct" one, or if I said that yours was
there's an Einstein quote i'm struggling to remember
nope, i don't have a good enough grasp on it
@Semiclassical There always exists object and subject. Without the objective, the subjective is meaningless. Einstein's (special) relativity is nothing more than the realization that object and subject are always present in reality, but the philosophical concepts of object and subject are universal, generally speaking.
the common example of this is in electrodynamics. suppose you hold a coil of wire in one hand and a magnet in the other.
if you hold the coil in place and move the magnet through it, an electric current is generated
if you hold the magnet in place and move the coil, you get the same electric current
but if you look at Maxwell's equations, the mechanisms by which those currents are generated seem different
The objective perspective or "correct" frame of reference is one that would transcend both the coil and the magnet themselves such that the coil and magnet become subjects.
right, well, that's not what a reference frame means
14:35
Otherwise, the object and subject are determined by the circumstances, in this case, the object is what is stationary.
not really. what's stationary depends on whether you're taking the coil's point of view or the magnet's
nevertheless, the emperical predictions are exactly the same
which suggests that the question of "which reference frame is correct" is empirically meaningless
there's no experiment one could run to settle the question
@Semiclassical That is incorrect. Whether or not an object is stationary depends on whether or not the object is in fact stationary, not whether the object appears to move according to some observer. Hence why I said the objective perspective is one that transcends both the coil and the magnet as such an observer understands what is stationary and what is moving with absolute certainty regardless of other observers.
(which is a pity, because having a definite answer would make QM interpretation a lot easier)
@AMDG You may say it's incorrect, but modern physics does not agree with you at all
Modern physics does not even consent to the idea that the causal order of two events is necessarily fixed
(There are physicists who resist that, but they are by far the minority.)
@Semiclassical Precisely because we do not transcend the confines of reality being that we are both rational spirit and material body, and reality being purely material with the instruments we use being material. A spiritual entity can easily observe these things, however. Thus, in the Catholic model of reality, even things such as Godel's incompleteness are not contradicted because there is always an outside observer or outside means of observation.
...right
14:42
@Semiclassical Well that's because science has abandoned reason. It has become a cesspool of pride and ignorance. Harsh, for sure, but that is reality.
And then they claim that we who believe are pseudoscientists. How ironic.

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