Mathematics - 2020-09-12

3:20 AM

Hmm ... It doesn't have rotational symmetry about the $z$-axis, so it's only a rotating shape.

4:07 AM

Can anybody explain how the normalization coefficients calculated?
I have recently given a paper which had 12 shifts in september and 8 shifts in january....I found
[this](https://jeemain.nta.nic.in/WebInfo/Handler/FileHandler.ashx?i=File&ii=97&iii=Y) . But this looks like BS....it doesnt give substantial information about comparing candidates across different shifts

Anindya Prithvi

4:38 AM

bounty 200 rep^

SarGe

6:31 AM

I need help in solving this question !image

The Testosterone Fanatic

7:26 AM

@SarGe use the formula to express sin(3x) in terms of sin(x) and (sin(x))^3 and you get a telescopic series

Martin Sleziak

7:44 AM

There is this post on the main site: Limit of the given sum: $f(x) = \lim_{n\to \infty} \sum_{r=1}^n 3^{r-1}\sin^3(x/(3^r))$ - as mentioned in the calculus chatroom.

robjohn

@TedShifrin $x^{2/3}+y^{2/3}+z^{2/3}=1$

The Testosterone Fanatic

@MartinSleziak what I said exactly matches with the answer on that page

SarGe

Thank you @TheTestosteroneFanatic and @MartinSleziak for your concern.

robjohn

@TheTestosteroneFanatic yes, it details your suggestion, but SarGe was given that link in another forum is what Martin was trying to point out.

Martin Sleziak

@SarGe As a side note, you can use Approach0 and many other methods to search among the posts on Mathematics.

Relevant post on meta: How to search on this site?

Sayan Chattopadhyay

8:25 AM

Is there a coordinate free way to show that if $(V, \omega)$ is a vector space with a skew symmetric non degenerate bilinear form, then $\wedge^{n} \omega \neq 0$ ?

SarGe

@MartinSleziak Actually, I'm using mobile site and Approach0 isn't supported on mobile. I tried searching on Google but didn't find anything constructive.

Martin Sleziak

That's a bit unfortunate. Still, I guess you'll be able to get to desktop eventually. Can SearchOnMath be used on mobile devices?

Balarka Sen

@SayanChattopadhyay Yes but the coordinate free definition of wedge is just as much work.

Start by writing down a formula for $\omega^n(X_1, Y_1, \cdots, X_n, Y_n)$

It's best to choose $X_1, Y_1, \cdots, X_n, Y_n$ to be a symplectic basis for $\omega$, in which case the end result will be Pfaffian of the matrix of $\omega$ wrt this basis.

That is, the matrix $(\omega(X_i, Y_j))_{i, j}$

And Pfaffian is square-root of the determinant, which is nonzero by nondegeneracy, and thus...

Martin Sleziak

Anyway, further discussion of Approach0 (and searching in general) might be more suitable in the searching chatroom. (I have asked there some details on how things work on mobile devices - I have never used any mobile device, so I do not have any experience with that.)

Sayan Chattopadhyay

@BalarkaSen Okay, let me do that, makes sense. And also on the half lives half die thing, I think I have figured out this commutative diagram using PD, once I am free from this category theory intensive symplectic geometry HW, I will write it down

Balarka Sen

8:38 AM

OK cool

Leaky Nun

@BalarkaSen youtube.com/watch?v=W-nWeRrlGMM

Balarka Sen

Lmfao this guy is 5head

brilliant

Anindya Prithvi

9:00 AM

Can anybody explain how the normalization coefficients calculated?
I have recently given a paper which had 12 shifts in september and 8 shifts in january....I found
[this](https://jeemain.nta.nic.in/WebInfo/Handler/FileHandler.ashx?i=File&ii=97&iii=Y) . But this looks like BS....it doesnt give substantial information about comparing candidates across different shifts

$$\Huge{\mathrm{HELP}}$$

Astyx

9:38 AM

@LeakyNun That's not a wooden shield

Balarka Sen

Lol

Astyx

Is complex geometry as boring as my prof makes is sound ?

Balarka Sen

What are they teaching in complex geometry

Astyx

For now not much, we went over multidimensional holomorphic functions and the Cauchy formula

Balarka Sen

Cool. I actually don't know much about this

Multivariable holomorphicity is quite different from one-variable holomorphicity

Astyx

9:44 AM

Is it ?

Balarka Sen

Yeah, let's see if I can tell you why

$f : \Bbb C^2 \setminus \{(0, 0)\} \to \Bbb C$ be a holomorphic function of two variables. Then I claim $(0, 0)$ is a removable singularity, $f$ always extends to a holomorphic function on all of $\Bbb C^2$.

Astyx

What about $z, z' \mapsto 1/zz'$ ?

Leaky Nun

so the singularities can only be codimension 1?

@Astyx it's singular on x-axis and y-axis

Balarka Sen

@Astyx That has singularities along $z = 0$ or $z' = 0$

@LeakyNun Yes, that's correct

Astyx

Indeed

Balarka Sen

9:47 AM

But anyway let's try to prove this

Leaky Nun

why did I say x-axis and y-axis

@BalarkaSen do or do not, there is no try

Balarka Sen

Lol

Ok, so $f(z, w)$ is a function of two variables. Fix some $w \neq 0$ a complex number, and consider $g(z) = f(z, w)$, the restriction of $f$ to a "horizontal" slice.

This is of course also holomorphic as a function $g : \Bbb C \to \Bbb C$

Hm I am not sure I want this, give me a moment.

Yeah OK. I can write $g(z) = \sum_{n = 0}^\infty a_n z^n$ by Taylor expanding near $z = 0$ (because $w \neq 0$ was fixed). Here the Taylor coefficients $a_n = a_n(w)$ only depends on $w$.

Explicitly, $a_n(w) = \frac{1}{2\pi i} \int_{|z| = 1} f(z, w)/z^{n+1} \, dz$, yes?

Alessandro Codenotti

Is there a simple example of a coalgebra which is not a bialgebra? I only have an example of an algebra which is not a coalgebra

Actually what I'm looking for is a coalgebra without group-like elements (or a proof that no such things exist), and a coalgebra which is not a bialgebra seems like a good place to start from

Astyx

10:04 AM

@BalarkaSen Yep

Balarka Sen

@Astyx I realized this doesn't work, but some modification does. Suppose that $f : \Bbb C^2 \setminus \Bbb D_r(0) \times \Bbb D_r(0) \to \Bbb C$ was a holomorphic function, where $r > 0$ is anything.

Leaky Nun

@BalarkaSen why doesn't it work

Balarka Sen

So instead of removing a point I am removing a polydisk nbhd of the point.

Leaky Nun

@AlessandroCodenotti $K[X]$?

Balarka Sen

@LeakyNun I mean, my argument wasn't working. The fact I am trying to prove still is; my solution is to prove a more general fact.

Alessandro Codenotti

10:11 AM

@LeakyNun Is this an example for the first or the second of my questions? In any case I don't see why it works

Leaky Nun

why doesn't your argument work @BalarkaSen

Balarka Sen

What is my argument

I didn't write it yet

Leaky Nun

your taylor series lol

Balarka Sen

I don't understand

Leaky Nun

it has a taylor series therefore it is holomorphic

Balarka Sen

10:13 AM

Sorry, I do not understand. What is $a_n(0)$?

Leaky Nun

$\frac1{2\pi i} \oint f(z,0) / z \ dz$

Balarka Sen

Why does that exist (you meant $z^{n+1}$)

Leaky Nun

because it only invokes $f$ for values outside the origin

Balarka Sen

Ok, because you're integrating over $|z| = 1$.

You still need to prove $a_n(w)$ is holomorphic, but that's probably easy by Morera's theorem.

$\int_{|z| = 1, |w| = 1} f(z, w)/z^{n+1} dz dw$

Yeah, that's zero because $f(z, w)/z^{n+1}$ is holomorphic in $w$ for all $z$ along $|z| = 1$, so the integral over $w$ vanishes.

Leaky Nun

@AlessandroCodenotti what's an algebra that's not a coalgebra?

Balarka Sen

10:19 AM

@LeakyNun Ok, thanks. So $\sum_{n = 0}^\infty a_n(w) z^n$ does seem to define an analytic function on $\Bbb C^2$ which agrees with $f$ on $\Bbb C^2 \setminus \{(0, 0)\}$

Leaky Nun

great

Balarka Sen

@Astyx So there you go.

Alessandro Codenotti

Which coalgebra/algebra structure do you have in mind for $K[X]$? Because with the usual algebra structure and the coalgebra given by $\Delta(X^n)=\sum X^i\otimes X^{n-i}$ and $\varepsilon(X^n)=1$ iff $n=0$ and $0$ otherwise it looks like a bialgebra to me

Balarka Sen

Yeah I never remember this proof, I don't know why.

Alessandro Codenotti

@LeakyNun $M_n(D)$ where $D$ is a nonzero division ring which is not a field

Leaky Nun

10:20 AM

@AlessandroCodenotti yeah forget my example

Astyx

Cheers, that makes sense

Balarka Sen

But actually note that this same proof works to show that any holomorphic function $f : \Bbb C^2 \setminus \Bbb D_r(0) \times \Bbb D_r(0) \to \Bbb C$ extends to $f : \Bbb C^2 \to \Bbb C$

Leaky Nun

oh no

all my rings are commutative

Balarka Sen

You do the same argument but with Laurent series, and argue that the negative terms vanish.

Super surprising stuff. This is called Hartog's extension if you want a name to look up

Alessandro Codenotti

@LeakyNun rip

It's easy to see that this is not a coalgebra because if you try to build a counit $\varepsilon\colon M_n(D)\to K$ you run into issues by looking at its kernel

Leaky Nun

10:26 AM

@AlessandroCodenotti so basically if $C$ is a coalgebra then $\operatorname{Hom}(C,V)$ is a monoid for every $V$?

Balarka Sen

@Alessandro Can't you always dualize a finite-dimensional algebra? If $A$ is an algebra over a field $K$, look at $\text{Hom}(A, K)$.

Alessandro Codenotti

@BalarkaSen Yes and that's also a coalgebra but I don't see your point

Balarka Sen

That's automatically a bialgebra is it? Because you can take product of two functionals... Hm.

Alessandro Codenotti

@BalarkaSen yes

@LeakyNun I guess so for nonzero $V$ because you should be able to use the counit of $C$ to build the identity of the Hom set

Leaky Nun

why do you need nonzero

Alessandro Codenotti

10:32 AM

Ok I guess a monoid with one element is still a monoid

Balarka Sen

@Alessandro I have an example, but it might be overkill

Alessandro Codenotti

Go for it

Balarka Sen

Actually I am not sure anymore. Consider the coalgebra defined as follows. $A = \Bbb Q[1] \oplus \Bbb Q[x]$, with co-unit map projecting to the first factor, and co-product defined by $\Delta(x) = 1 \otimes x + x \otimes 1$.

This is a valid coalgebra right? Does it have a compatible algebra structure?

Alessandro Codenotti

Wait $\Bbb Q[1]$ is $\Bbb Q$ or am I drunk?

Balarka Sen

Yeah, I am specifying the generators in every factor.

Alessandro Codenotti

10:39 AM

Oh ok

Let me think about it then

Balarka Sen

It's just $\Bbb Q \oplus \Bbb Q$

$1$ and $x$ are the generators in the respective grades

I didn't mean polynomial rings

Alessandro Codenotti

And the 2020 award for the worst possible notation goes to... :P

Balarka Sen

Sorry :(

I am saying because $\Bbb Q[x]/(x^2)$ feels like the natural algebra structure but it's not compatible with $\Delta$

Alessandro Codenotti

Hm ok let me see

@BalarkaSen Wait what's $\Delta(1)$?

Balarka Sen

That has to be $1$, right?

I could be dead wrong about this example, I am getting confused.

Alessandro Codenotti

10:47 AM

I'm getting confused by the notation I think

No ok it has to be $1$

Balarka Sen

Yeah because projection to $1$ is the counit. It's very confusing to write.

Alessandro Codenotti

So $\varepsilon(x)=x$?

I'm a bit suspicious because in finite dimension the dual of a coalgebra is an algebra (whose dual is the starting coalgebra), and I thought that by identifying the coalgebra with its dual you get a bialgebra in this way, but now that I think about it I don't know if the operations are compatible

Balarka Sen

I think $\varepsilon(a + bx) = a$. Does that work? Let me see, one moment.

So $\varepsilon(x) = 0$

$\Delta(x) = 1 \otimes x + x \otimes 1$, $\text{id} \otimes \varepsilon$ of that is $0 + x = x$, like we want

Yeah seems to work

$\varepsilon : \Bbb Q_1 \oplus \Bbb Q_x \to \Bbb Q_1$ is projection to first factor, like I said.

Alessandro Codenotti

Can we write $1$ as $(1,0)$ and $x$ as $(0,1)$ please

Balarka Sen

Ok haha sorry

That gets too cumbersome. I think the reason nobody cares about coalgebras is how its equally natural as algebras but so much harder to write compared to them

Alessandro Codenotti

10:57 AM

"nobody cares about coalgebras" doesn't sound very accurate (unfortunately)

Anyway why do you say your example doesn't have an algebra structure?

Balarka Sen

So I don't know, under an algebra structure, what would $x^2$ be? $\Delta(x^2) = (1 \otimes x + x \otimes 1)^2 = (1 \otimes x)(1 \otimes x) + (1 \otimes x)(x \otimes 1) + (x \otimes 1)(1 \otimes x) + (x \otimes 1)^2$ and $(a \otimes b)(c \otimes d) = (ac) \otimes (bd)$ is the forced algebra structure on $A \otimes A$, right?

So I get $\Delta(x^2) = 1 \otimes x^2 + 2x \otimes x + x^2 \otimes 1$

Konformist Liberal

Could you suggest anything to search which is (maybe even slightly) related to "symmetries of a space determines the space"

Balarka Sen

Note that $x^2 = 0$, the natural algebra structure on $\Bbb Q_1 \oplus \Bbb Q_x = \Bbb Q[x]/(x^2)$, does not work because $x \otimes x$ is not a torsion element.

What are the possibilities?

We have to try $x^2 = a + bx$ in general

I can always cook up a $y$ such that $y^2 = c$ some constant, by completing the square

Eg $y = x - b/2$

Then $\Delta(y^2) = \Delta(c) = c$ whereas $\Delta(y^2) = 1 \otimes c + 2y \otimes y + c \otimes 1$?

Um

No what is $\Delta(1)$ again? $1 \otimes 1$, right?

That's what it has to be

Alessandro Codenotti

@BalarkaSen Hm wait I'm not convinced here

Why do you have $\Delta(x^2)=(\Delta(x))^2$

Balarka Sen

Because we assumed the algebra structure is compatible with the coalgebra structure, no?

Isn't that what it means

Alessandro Codenotti

11:08 AM

We want $\Delta(xy)=\sum x_1y_1\otimes x_2y_2$

Where $\Delta(x)=x_1\otimes x_2$ is the awful notation everybody uses for $\Delta(x)=\sum_{i=0}^n x_{i1}\otimes x_{i2}$

Balarka Sen

Comultiplication $A \to A \otimes A$ is a homomorphism of algebras, that's what you're saying, yeah?

That implies $\Delta(x^2) = (\Delta(x))^2$, no?

Alessandro Codenotti

Hm ok I guess it does

Makes sense

Balarka Sen

Or put $x = y$ in your notation

$x_{01} = 1, x_{11} = x, x_{02} = x, x_{12} = 1$

So basically you have to prove that $x^2 = c$ and $\Delta(x^2) = 1 \otimes x^2 + 2x\otimes x + x^2 \otimes 1$ are not consistent conditions

Because the only algebra structures on our beast are $\Bbb Q[x]/(x^2 - c)$, upto isomorphism.

Alessandro Codenotti

Hm ok, your example does seem to work to me now, but I'm still suspicious for some reason

Balarka Sen

I claim that my example is, give or take, the reason that $S^2$ is not a $H$-space :)

Alessandro Codenotti

11:17 AM

So if we take this coalgebra and put the standard algebra structure on its dual (and then identify it with the dual), what do we get?

Balarka Sen

I was confused at the beginning because graded bialgebra is different from just bialgebra (and the graded info is what distinguishes $S^2$ from $S^3$ say)

But this is ok

Mike Miller

@BalarkaSen Curiously I believe there are places where the coalgebra is much easier to work with

Balarka Sen

@AlessandroCodenotti Thats a good question. So we need to understand where $x^2$ goes to in the dual

@MikeMiller I'd listen if you have examples off the top of your head

I find it very difficult to write coalgebras

Alessandro Codenotti

Ok so $x$ goes to $x^\ast$ as usual

Balarka Sen

So let's call $A = (\Bbb Q_1 \oplus \Bbb Q_x, \varepsilon, \Delta)$ as we defined it. $m : A^* \times A^* \to (A \times A)^* \to A^*$ gives $m(x^*, x^*)(a) = (x, x)^*(\Delta(a))$

for any $a \in A$

Alessandro Codenotti

11:26 AM

Right

Well this is a thing in $\Bbb Q\otimes\Bbb Q$ as written to be very pedantic but then we just multiply the two pieces

Balarka Sen

So if $a = \alpha + \beta x$, $\Delta(a) = 1 \otimes (\alpha + \beta x) + (\alpha + \beta x) \otimes 1 = 1\otimes\alpha + \alpha \otimes 1 + 1 \otimes \beta x + \beta x \otimes 1$

$(x, x)^*$ of this thing is uh

Alessandro Codenotti

Uhm

Mike Miller

@BalarkaSen Steenrod algebra is famously better written in terms of the coproduct. Milnor has a basis in which the dual algebra to Steenrod is polynomial

Since you usually don't want to pass to the dual algebra you instead talk about the coproduct of things in the Milnor basis

Whereas the Steenrod algebra itself is complicated

Balarka Sen

Pretty sure it's $2\beta$ or something like this, and $x^*(a) = \beta$, so $(x^*)^2 = 2x^*$

Something of this sort

Alessandro Codenotti

Yeah that seems sensible

Balarka Sen

11:31 AM

@MikeMiller Hrm ok, I am not familiar with this though.

Alessandro Codenotti

Unfortunately this example still has a group-like element

@Mike do you happen to know off the top of your head whether any coalgebra has a group-like element? This is obvious for bialgebras, so I was trying to look at examples of coalgebras that are not bialgebras, but with little success so far

Mike Miller

@BalarkaSen Just think of it like this. The Steenrod algebra has all these Adem relations with binomials etc. Seems pretty complicated, even if you write it in terms of the generators Sq^{2^i}. But surprisingly the dual of this complicated algebra is free, so the coproduct is easy to write down (in the right basis).

@AlessandroCodenotti Why is this obvious for bialgebras?

Alessandro Codenotti

Isn't the unit of the algebra always group like?

Mike Miller

Oh sure

Maybe try a cofree coalgebra

Alessandro Codenotti

11:46 AM

Uhm I need to look up what that is, let me see

Mike Miller

As an example of something which isn't a bialgebra I mean.

It probably has grouplikes

You could also look in Sweedler's book

Alessandro Codenotti

So if I'm understanding correctly the comultiplication of the cofree coalgebra extends that of the tensor coalgebra? In that case there is at least a group like

@MikeMiller The one just called "Hopf Algebras"?

Balarka Sen

@MikeMiller Oh ok bizarre

Alessandro Codenotti

I have to leave for lunch, be back soon

Mike Miller

@AlessandroCodenotti Yeah but I thought you wanted examples of coalgebras which aren't bialgebras

Balarka Sen

11:53 AM

I gave him one for that already I think

He wants no grouplike elements

To which, shrug

Alessandro Codenotti

Yeah the thing is that I want an example of a coalgebra with no group-like elements (or a proof that this is impossible), so coalgebras which are not bialgebras seemed like a natural place to start from, but the two examples I've seen so far both have group-like elements

Balarka Sen

I remember the time coalgebras really clicked for me is when I was thinking about topology of divisor complements in $E \times E$ where $E$ is an elliptic curve

Something silly, but yeah

Mike Miller

Meh/10

orientablesurface

2:38 PM

is the following true

Let $M$ be a Riemannian manifold

Let $\omega \in \Omega^1(M)$

Then there exists a unique smooth vector field $Y$ on $M$

sucht that for any smooth vector field $X$

we get $\omega(X)=<X,Y>$

How can I prove it without using coordinates

Edward Evans

"The natural-medicine guru Andrew Weil claims that plant extracts are effective because they operate on multiple systems..." apparently Weil's american counterpart dabbles in natural-medicine and not in algebraic geometry

Mike Miller

Smoothness you should check in coordinates

Existence of a unique (possibly discontinuous) vector field Y with that property follows from linear algebra

If you know that a vector field is smooth iff <X, Y> is smooth for all smooth X, then this follows from the fact that omega(X) is smooth for all smooth X

Balarka Sen

@EdwardEvans Curing germs by sheaves of stalks

Edward Evans

ha

actually this is about the effectiveness of Saint-John's-Wort

Balarka Sen

wort is that

Edward Evans

2:51 PM

some natural remedy to depressive symptoms

also nice puns

Balarka Sen

Hm I see

Edward Evans

ANT 2 and Modular Forms will be running concurrently, but ANT 2 will be asynchronously run as an online lecture course, while Modular Forms will be run at university with standard lectures

which is the dream

change of topic

maybe this time I won't fail modular forms rofl

Tobias Kildetoft

@EdwardEvans You better not fail them this time. Those forms count on you.

Edward Evans

ha

The lecturer this year has some weird space of modular forms named after him and proved some stuff with Gross and Zagier

orientablesurface

3:21 PM

@MikeMiller how would you prove the converse of a vector field $Y$ is smooth on $M$ if and only if $<X,Y>$ is smooth for all $X$

Mike Miller

In charts

In the end a map is smooth iff it's smooth in every chart. That's the definition. So you shouldn't be surprised that you have to talk about charts somewhere to check something is smooth...

orientablesurface

5:46 PM

if $f,g:M\rightarrow \mathbb{R}$ and $f$ is smooth, then $g$ is smooth

right?

Thorgott

you forgot to tell us how $f$ and $g$ are related

orientablesurface

Sorry, I meant to add that $fg=c$, where c is a constant

if f is non-zero then surely it holds

Mike Miller

If $c=0$ it's false

Thorgott

If $c\neq0$, then $f$ is everywhere non-zero and $g=c/f$ is smooth

Ted Shifrin

G'day, @Mike, @Thor, @orientablesurface.

Mike Miller

5:51 PM

hi

Thorgott

hi Ted

orientablesurface

okay, the second question is: given a riemannian manifold $(M,g)$, then expressing the metric tensor locally i.e. $g_{ij}$, we may invert the matrix $[g_{ij}]^{-1}=[g^{ij}]$, each $g^{ij}:M\rightarrow \mathbb{R}$ is smooth, right?

hello @TedShifrin

Ted Shifrin

@orientablesurface: How do you show the entries of $A^{-1}$ are smooth functions in the entries of $A$?

Mike Miller

$g_{ij}$ is not a function on $M$

but this is a detail you can fix

orientablesurface

$g_{ij}$ is a function on $M$

Ted Shifrin

5:54 PM

Not literally, @orientablesurface. What's the correct statement?

Mike Miller

"expressing the metric tensor locally" doesn't take place on M

Ted Shifrin

Huh? @MikeM

orientablesurface

$g_{ij}(p)= <\frac{\partial}{\partial{x}^i|_p},\frac{partial}{\partial{x}^j}|_p>=$ $<\frac{\partial}{\partial{x}^i},\frac{partial}{\partial{x}^j}>(p)$

Thorgott

what are $x^i,x^j$?

orientablesurface

coordinates on a chart

Mike Miller

5:58 PM

and those coordinate functions naturally live on R^n, not M. :)

Thorgott

so their domain is?

orientablesurface

open subset of a manifold

@Thorgott

Mike Miller

whatever

Thorgott

yeah, so it only makes sense to talk about $g_{ij}$ on that open subset as well

Ted Shifrin

I can't compete with Mike's pedantry.

orientablesurface

5:59 PM

oh yeah, ofcourse

would $g^{ij}$ be a function $U\rightarrow \mathbb{R}$ as well?

Ted Shifrin

Well, Mike's point is that once you choose a chart, everything is a function on an open subset of $\Bbb R^n$, not the open subset of the manifold. :D

orientablesurface

oh, I see

Write a book on Riemannian Geometry, @TedShifrin please :P

Ted Shifrin

Sorry, @orientablesurface ... just the curves and surfaces book. I'm not about to make my lecture notes for the graduate stuff accessible to the public.

orientablesurface

ahh I see, was worth a try anyways :P

Leaky Nun

I always thought the $x_i$'s are functions from an open subset of $M$

Ted Shifrin

6:04 PM

Right.

They are @LeakyNun

$x^i$ are

Hi, a @Balarka!

Hi @Ted

Gotta get dinner, be back soon

SK19

8:48 PM

quick question: I know in English $\chi'$ is pronounced "chi prime". But what about $\chi''$? Is it just "chi two prime" or something else? I'm not a native speaker.

Leaky Nun

1. it's "primed" not "prime"; 2. I would say "chi double primed" or "chi primed primed"

Balarka Sen

chi tick tick

Tobias Kildetoft

@LeakyNun I have always said "prime"

Balarka Sen

Me too

I think I alternate between primed and prime

Lol

Leaky Nun

Prime (symbol)

The prime symbol ′, double prime symbol ″, triple prime symbol ‴, and quadruple prime symbol ⁗ are used to designate units and for other purposes in mathematics, science, linguistics and music. Although the characters differ little in appearance from those of the apostrophe (or the single or double quotation mark), the appropriate uses of the prime symbol are different. However, an apostrophe is often used in place of the prime (and a double quote in place of the double prime), due to the lack of the prime and double prime symbols either on commodity keyboards. These substitutions would not normally...

I stand corrected

but now you make me curious: where did I get this "primed" from if you also sometimes say "primed"?

Tobias Kildetoft

8:55 PM

I have never said primed

But I assume it comes from thinking of it as the symbol being primed

(not that this actually makes much sense in this context)

Leaky Nun

and what sense does "prime" make?

> The name "prime" is something of a metonymy. Through the early part of the 20th century, the notation x′ was read as "x prime" not because it was an x followed by a "prime symbol", but because it was the first in the series that continued with x″ ("x second") and x‴ ("x third"). It was only later, in the 1950s and 1960s, that the term "prime" began to be applied to the apostrophe-like symbol itself. Although it is now more common to pronounce x″ and x‴ as "x double prime" and "x triple prime", these are still sometimes pronounced in the old manner as "x second" and "x third".[citation nee…

SK19

9:29 PM

thanks for the feedback!

Ted Shifrin

9:54 PM

I've never heard of ”primed” in a mathematical application. Painting, yes.

Transcript for

$Mathematics$ Mathematics