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00:02
What does it mean for a complex-valued function to be "conjugate-analytic"? Does it mean that the conjugate of the function is analytic?
@Sophie Ah, I see. A group action of $S_3$ on $\mathbb{Z}/p\mathbb{Z}$ is the same data as a homomorphism $S_3\rightarrow\operatorname{Aut}(\mathbb{Z}/p\mathbb{Z})$, but the latter group is cyclic. A subgroup of a cyclic group is cyclic and $S_3$ isn't cyclic, so that homomorphism necessarily has non-trivial kernel. But that means there's a non-identity element of $S_3$ acting like the identity and that gives what we want (for all $a$ simultaneously, even).
because in the example above 1.3.2
the set (-1,1) is open
but if you say we use R as the set X
then (-1,1) would not work
because we want a close set?
yes, but the K from the proposition is [0,1]
(-1,1) is the Y in the proposition
hm what's the use of Tao stating (-1,1) in the example
cause [0,1) is closed in (-1,1), but not closed in R
and that's a situation he wants to give an example of
pedagogically speaking, this is an important thing to illustrate
00:16
erm
but you used [0,1] to be closed in R
but [0,1) is closed in (-1,1)
and then [0,1) is not closed in R
are we supposed to see the relation between [0,1) and [0,1] here?
well, the relevant relation is $[0,1)=[0,1]\cap(-1,1)$
an $[0,1]$ is closed in $\mathbb{R}$, so $[0,1)$ is closed in $(-1,1)$ by prop. 1.3.4
00:37
@Thorgott Oh god I need to study more algebra
we all do
01:02
Thor I see how you choose the Y
but how did you choose the [0,1] as the K
or rather I mean we had [0,1) and (-1,1), so do we just take the intersection and then add the closed brackets to get [0,1]?
the only reason $[0,1)$ isn't closed in $\mathbb{R}$ is that it doesn't contain the limit point $1$, so the choice of $[0,1]$ isn't very farfetched
01:29
0
Q: Equivalent Conditions for Complex Harmonic Functions

user193319 Let $h$ be a smooth complex-valued function. Show that $h$ is harmonic if and only if $\partial h / \partial \overline{z}$ is conjugate-analytic. Recall that $\displaystyle \frac{\partial f}{\partial \overline{z}} = \overline{\frac{\partial \overline{f}}{\partial z}}$ and $\displaystyle \frac{\...

01:58
ok thanks thor
hows your sunday guys
02:39
hey guys can an open ball be empty?
Define open ball
02:56
If a point p inside a triangle ABC such that angle BPC equals 2A then can p be called circumcentre?
03:09
@sheltonBenjamin yes P is the circumcenter because the angle of a segment of a circle seen from the center is twice that seen from another point in the circle
makes sense right? Or do you want a proof?
open ball is defined to be centered at xo with radius r such that B (xo,r) with radius r, and r>0
B(xo,r):{x is an element of X | d(x,xo)<r }
but it doesn't say anything about the points having to be in the ball or not
I think by most definitions the empty set is clopen
Can you name an element of $B(x_0,r)$?
04:09
x_0?
does x_0 have to be in there?
04:38
@BalarkaSen we can start from today.
05:14
@Balarka looooool
it's seriously not bad
machines generate better black metal than I can write
ffs
@Joseph is $d(x_0, x_0) < r$ ?
@Balarka I like how the second half of the last song is just silence, as a good closing song should be
oh no it's literally the whole last song
05:42
yes
but what if there is no such pt
can we still define such a ball
06:03
how do you want to define a ball if you have no point to refer to? An open ball around what?
@JosephRock A ball has a center right? That center is always in the ball.
You could define a notion of constant distance, and assign every point a distance, but then that's not a metric space any longer
I suppose a ball of radius zero would be empty lol
(And at that point you're just mucking about in topology)
Would a ball of radius zero be a point (and necessarily closed, since this is a metric space)?
Metric spaces are Hausdorff and Hausdorff is an awesome, awesome property
Nah I think the radius should be strictly positive, but if you had a ball of radius zero then any point in the ball would violate $d(x,y) \geq 0$
06:08
@Rithaniel $B(a,r) = \{x | d(a,x) < r \}$ and hence for $r=0$ you would have empty
but anyway, radius zero is not allowed
Ah, the definition of open ball, as opposed to just "ball"
I have a question about modules
If $R$ is an integral domain such that all $R-$modules are injective, do you necessarily have that all $R-$modules are projective?
I think the answer is yes, but it's been a while since I've done this stuff
06:12
I think the answer is yes but I can't remember why
Yeah, I always get hung up on the exact split sequences
Do you think it involves constructing the lifting homomorphisms or using split exact sequences?
Err if I were to guess, possibly "if contravariant hom is exact then covariant hom is exact"
but idek if it's true
I think this relies on it being an integral domain, no?
06:34
hang on
if every R-module is injective then every short exact sequence splits
I'm sure I did this as an easy exercise
yeah if every R-module is injective then every short exact sequence splits
lol
so all R-modules injective iff all short exact sequences split iff all R-modules projective
Think that holds for any R
Yeah, I know that one, but how do you get from there to every $R-$module being projective?
if literally every short exact sequence splits then every R-module is projective by definition (if your definition is every exact sequence of the form blah splits)
Oh yeah, that's ridiculously easy
Forest for the trees
yeah hahaha
07:07
@EdwardEvans They're streaming randomly generated AI techdeath rn
so for every nonzero $x$, $R \to R/(x)$ has a section that sends $1 \in R/(x)$ to say $y \in R$, then $y=1 \pmod x$ (since it is a section) and $xy=0$ (since the section is well-defined). from the latter we have $y=0$, so from the former $x$ is a unit
is there any on expert on SVM here?
conclusion: $R$ must be a field @Rithaniel
@SayanChattopadhyay Ya let's start today
@BalarkaSen ffs hahaha
07:14
@EdwardEvans The spirit of metal is pretty on spot
Ah, danke, that was something I was going to do next, Leaky
does anyone know what N means in open ball relative mean
relative topology open ball
@Sayan quick question: need to parse a definition! Suppose I have a representation V of a group G and let K be a subgroup of G. Suppose I have a full decomposition of V into irreducible subreps. Some of those might be isomorphic to each other, so group these together into equivalence classes and sum together the subspaces of V fixed by the action of K wrt these equivalence classes. The sum of those guys is called the K-isotypic decomposition of V, right?
Is that the right construction? Rofl
Sorry @Edward but I do not have a clue, I haven't seen this definition before. Though it would make sense to call this a K-isotypic decomposition
Oh no worries, was just trying to parse the definition
Got lost in the words lol
07:29
@Rithaniel haha I can predict the future
I think in the finite dimensional case you end up with a bunch of weights and then those give you an irreducibility criterion
That might be smth else tho
In the finite dimensional case all you're doing here is taking the sum of all $K$ submodules right? And then you're original representation decomposes into these sums
Sounds reasonable rofl
Well anyway I’m in the infinite dimensional case so meh
08:01
hoy ppl
Hello there!
I am wondering if anyone here is familiar with the setup of the Leray spectral sequence. I am having problems with the differentials because I am getting that many of them are always zero and that ought to be wrong...
I can try to help. Go ahead
Thanks. I already posted it here math.stackexchange.com/questions/3859628/…
the problem basically boils down to the following: I have a map f:X-->Y and a sheaf S on X, which I push forward to Y and resolve with injective sheaves f_S --> I^.
Sorry, f_* S --> I* maybe better notation here in chat.
You can use LaTeX in chat; alternatively wrap your math in between two ` marks to stop italicizing it with all the underscores and stars
f_* S -> I^*
Like that
Great, gimme a moment. f: X\to Y
$f: X\to Y$
08:10
Yup
You need this to read the LaTeX
hmm sorry, does it render for you?
okay, one moment :)
No worries
why doesn't my university teach elementary number theory? Is elementary number theory not undergraduate course?
08:12
Okay, let's try again: I have a map $f:X\to Y$ and a sheaf of abelian groups $S$ on $X$.
talk to your university
Working now, right?
Ok, cool. So I forget how the story is phrased; you take an injective resolution $S \to I$, apply $f_*$, and take hypercohomology of this complex of sheaves?
That's $H(Y; Rf_* S)$
08:15
And the spectral sequence you speak of is a corollary of the general hypercohomology spectral sequence, so you're basically asking how to set that one up. OK, let me read the rest of the post.
I am not that familiar with hypercohomology but that definitely appears in the chapters that I am reading right now.
Gotcha.
@ChristianSander Remind me how to get the horizontal differentials for $I_B \oplus I_H \oplus I_B[1]$ again?
I thought it is just maps like $(0,0,c) \mapsto (c,0,0)$ in this case.
Because as I understand it you use the horseshoe lemma and there you use these kind of maps.
Hm, I see.
I'd need some time to think about the details. What is the paradox, in the end?
From there on I want to use John McCleary's Book on spectral sequences. The Leray spectral sequence is the one obtained from this double complex as I understand it.
08:28
@BalarkaSen have you heard of mathtrainer.org
@ChristianSander Yes, that's right. I typically don't think about the details with the Cartan-Eilenberg resolution, but you can take any injective resolution of the complex $f_* S \to f_* I$, apply $\Gamma$, and compute spectral sequence of the double complex.
Yes, I think that is correct :).
But with this definition I know what the horizontal differentials do.
Or so I thought...
The problem arises when I want to compute the differentials on the r-th page of the spectral sequence.
Because the differential there is kind of assembled from the vertical and horizontal differentials of my double complex, right?
Right, the ladders coming from snake lemma
Sorry, what ladders?
The differentials on the $r$-th page of a spectral sequence of a double complex is given by going along an $r$-step ladder in the double complex by snake lemma.
You have to solve a system of the form $a_0 = da_1$, $d' a_2 = d a_3$, etc
08:34
Ah that is what you mean.
Yeah, sorry, my language for communicating homological algebra is too visual.
Yes, in my double complex with the filtration I used I had to go diagonally from bottom right to top left.
Here is where I am getting my problem.
I have not, @LeakyNun.
When I go up that ladder I have to apply the differentials (of the double complex) to the elements that I get at the end of that ladder, right?
Hm, yeah
08:40
Should I clarify what I mean?
Yes, please go ahead.
McCleary writes that $E_r^{p,q}$ is a quotient of $Z_r^{p,q}$. Do you happen to have a copy of his book @BalarkaSen?
Ah, no, but I remember the description, yes
Okay, let me quickly refresh what those modules were: Given a filtered differential graded module $A$ with filtration $F^\bullet$.
$Z_r^{p,q} = F^p A^{p+q} \cap d^{-1}(F^{p+r} A^{p+q+1})$
$B_r^{p,q} = F^p A^{p+q} \cap d(F^{p-r} A^{p+q-1})$
Okay, I want to take the filtration that cuts threw the collums and keeps the rows of the double complex intact (I refer to that as the vertical filtration)
Yep OK
08:49
Now the $r$-th differential in the spectral sequence is coming from the map $Z_r^{p,q}\xrightarrow{d} Z_r^{p+r,q-r+1}$ where $d$ is the differential of the double complex.
$d = d_h + d_v$
Okay, I am getting to the point :).
I think of $Z_r^{p,q}$ as elements on the diagonal with total degree $p+q$ that map to something nontrivial only above the $p+r$-th row under my differential.
I'm with you.
So in order to find these I (visually) start in the bottom right with an element $a_0$ sitting in the spot $(p,q)$ and apply $d_h$ and $d_v$. $d_h$ must be $0$ in this case leaving me with one element $d_v(a_1)$ which I have to kill with an element sitting in the $(p-1,q+1)$-st spot.
I gotta continue this process up to the top left making sure that I kill all elements along the way that are below the $p+r$-th line on the diagonal with total degree $p+q+1$, right?
All possible tuples $a_i$ that I obtain in this way make up my $Z_r^{p,q}$.
Correct.
09:00
Okay, great.
After applying the differential $d$ of the double complex to those elements I land in $Z_r^{p+r,q-r+1}$
And in order to get an element in $E_r^{p+r,q-r+1}$ I need to mod out the group $Z_{r-1}^{p+r+1,q-r} + B_{r-1}^{p+r,q-r+1}$
Sorry for the many indices...
I now claim that all elements in $Z_r^{p,q}$ land in the latter module $B_{r-1}^{p+r,q-r+1}$ when I apply the differential $d$ of the double complex.
I think at least from $r=2$ on and this has imho to do with the horizontal differentials of the double complex.
Ok, tell me how (I checked your computation of the differentials of the Cartan-Eilenberg resolution, and they are correct)
Okay. Thanks so far.
Sorry, I need to quickly get the door. I'll be right back (10 mins tops).
09:37
Okay, I'm back.
$B_{r-1}^{p+r,q-r+1}$ is the image of $d$ of all elements on the diagonal with total degree $p+q$ above the $p+1$-st row that also land above the $p+r$-th row, right?
10:02
@BalarkaSen Or let me rephrase the last question: In order to understand $B_{r-1}^{p+r,q-r+1}$ I need to do that "going up the latter" just as it was done for $Z_r^{p,q}$ but instead of starting in the spot $(q,p)$ I start at the next (up and left) spot $(q-1, p+1)$.
Sorry, I was away from keyboard. It's hard for me to follow symbols, but maybe you could skip it because I know the description you're trying to recall. Could you point out why your differentials are becoming zero?
So far, everything in the question you linked seems correct. So if there is some inconsistency it's in this part -- certainly in the Leray spectral sequence not all the differentials beyond page 2 are zero in general so we know something you're doing is wrong.
Yes, I agree with last part :).
I'll try to keep it as simple as possible: Take an element in $E_r^{p,q}$. This is represented by an element in $Z_r^{p,q}$.
To compute the image of that under the differential $d_r$ we have to do that "going up the ladder".
Let's say the module at the bottom right on our diagonal with total degree $p+q$ is $A_0 \oplus B_0\oplus C_0$.
this is where we start that ladder process. Since we want to land above the $p+r$-th row, we cannot take something in $C_0$, right?
Because the horizontal differential is sending that not to zero.
So th eprocess starts with an element like $(a,b,0)$. We apply the vertical differential and get $(d_va,d_vb,0)$ and this we can only kill with an element next to it on the left if $d_vb=0$. In this case an element that kills it via the horizontal differential looks like $(x,y,d_va)$.
And here I could freely choose $x$ and $y$. Am I talking sense up to here?
10:21
Let's just do $r = 2$. $E_2^{p, q}$ is represented by an element $\alpha \in K^{p, q}$ such that $d_v \alpha = 0$, $d_h \alpha = - d_v \alpha'$ where $\alpha' \in K^{p-1, q-1}$. Am I correct?
You are taking the other filtration right?
Where you take whole columns?
Yes, indeed. You've got the reverse one, so let me fix that
$d_h \alpha = 0$, $d_v \alpha = -d_h \alpha'$
If $\alpha = (a, b, c)$ in coordinates, then as $d_h \alpha = (c, 0, 0) = 0$, we are forced to have $c = 0$
$d_v \alpha = -d_h \alpha' = -(c', 0, 0)$ if $\alpha' = (a', b', c')$.
But we have no clue what $d_v$ looks like right? It need not act componentwise
I don't understand how you write $d_v (a, b, 0) = (d_v a, d_v b, 0)$
I thought it acts componentwise...
Hmm, let me quickly check that.
10:27
The horseshoe lemma doesn't seem to say anything about the vertical differentials.
What do you mean by $d_v a$ anyway?
The components of $I_B \oplus I_H \oplus I_B[1]$ are not double complexes, they are just chain complexes. What is $d_v$ of $I_B$?
In en.wikipedia.org/wiki/Horseshoe_lemma indeed they do not say anything about that. But in Weibel's book he says that it acts componentwise if I understand it correctly.
You have to tell me what acting componentwise means in this context
$d_va$ was admittedly abuse of notation.
Right, you mean $d_v(a, 0, 0) + d_v(0, b, 0)$. It is additive but why does that prevent this element to have nonzero third component?
It need not be of the form $(blah, blah, 0)$ as far as I'm concerned
We don't know anything about that. At least I don't.
10:32
Why?
this is what I would expect.
Why would I expect that?
I don't see it.
From the horseshoe lemma again. Let me find an online reference.
I have Weibel, where does he say this?
page 37
The last part under the last SES.
Right before the proof.
10:38
@BalarkaSen try it out
Have I misunderstood something here?
@Christian I'm not following. How does this lemma say anything about $d_v$? I may be being stupid.
In Weibel's case we are talking about the (horizontal) differential of the complex $P = P' + P''$, right?
Hm? Aren't those what give you your $d_h$, not $d_v$?
Oh, no, I see.
No, they give me my $d_h$.
Sorry for the confusion, horizontal and vertical is changing a lot here :(.
10:47
Right, I think I got confused because $v$ and $h$ were switched in my earlier computation.
So are you on board with the middle resolution having differentials $d'\oplus d''$?
I am still using Weibel's notation here.
Hm, seems fine.
But then I see your paradox.
Oh yes :)?
Meaning: You see how the differentials in the spectral sequence end up to be zero?
Yeah, that's fine. I dunno, something must be wrong, let's see.
Must be a stupid algebra mistake
I first was suspecting $\Gamma$.
10:52
Yeah but that acts componentwise on sums of sheaves so I don't see the problem there, as you said
Yeah, at least for finite direct sums I don't see a problem either.
Hmm I admit that I am thinking about this for two weeks now. I couldn't find anything but I agree: It is probably just a stupid mistake that I made :-/.
I can't spot an error either. This is bizarre.
It is bizarre :-/.
I guess that applies to many other spectral sequences too...
11:08
If this calculation was correct it'd be correct for every hypercohomology spectral sequence, heh.
Let me try to say the contradiction explicitly. From our calculation we have $\alpha = (a, b, 0)$ where $d_v b = 0$
But then you claim this is also a boundary?
What is a boundary, $\alpha$?
Sorry, you want to see $d_2 = 0$, right? How do you see that, again?
This is a one step ladder. We start with $\alpha=(a,b,0)$ and apply $d_v+d_h$ to get $d_v(a,b,0) = (d_va,0,0)$.
I abused notation in the last step again.
I gotta kill $d_va$ and can take an element $(x,y,-d_va)$ where I get to choose $x,y$ freely.
11:16
Yes
Ahm sorry not quite, I forgot a minus.
And $d_v(x,y,-d_va) = (d_vx,d_vy,0)$ represents the image $d_2(\alpha)$, correct?
I am claiming this is an element of $B_{r-1}^{p+r,q-r+1}$.
Just $B^{p+1, q}_1$ in our case
Ah yes, sorry.
11:23
But $D(x, y, 0) = (d_v x, d_v y, 0)$?
Wait, I am a bit confused.
Yes, I think that is it...
$D = d_h+d_v$ I assume.
Yes
This is ridiculous. What is happening?
Sigh
Tell me
I am almost sure I have made a mistake in the whole setup...:(
why don't you just try an example
11:31
I don't know what kind of example that would be :(...
@MikeMiller: Was that directed at me?
This isn't an example question, he's trying to understand some specific resolution (that I am not understanding either).
11:44
@Christian Sorry, I was right.
The differential in horseshoe lemma is not $d' \oplus d''$
The line in Weibel just says $P_n = P_n' \oplus P_n''$. That's a group decomposition, not a chain complex decomposition.
Work out the differential by hand, you'll see it.
This solves the issue I think
Hmm...it probably would. Let me reread his statement and then the proof on page 37.
You get maps $P_n''[1] \to P_n'$ in the process of lifting the chain map, by projectivity. These contribute to the total differential
Judging from his statement I would say that it is a direct sum of complexes because of him writing "...the right hand column lifts to an exact sequence of complexes $0\to P'\xrightarrow{i} P\xrightarrow{\pi} P''\to 0$.
It's simply not true, just work it out!
Even better look at Exercise 2.2.4. He writes down the differential
I'll try, what is $P_n''[1]$ for you here?
11:52
$P''_n$ but grade-shifted up by $1$
Okay thanks. And in that exercise there is a typo right? After i.e. it should be a $d$ and not $d'$, right?
I will work that out. @BalarkaSen: Thank you very much for your patience and time up to here!
It's all good, I learnt something as well. Good to see that I wasn't 100% off as to where the issue was.
I never remember this horseshoe lemma because it's an immediate corollary of the definition of injectivity/projectivity. So I never remember the differentials either
@Christian Here's the takeaway, which is what was surprising me about your confusion. You can never expect to know everything about a spectral sequence; if you know $d_h$, you must not know what $d_v$ is, at least very explicitly. And that's right, the differentials in the horseshoe lemma aren't super-trivial - the off-diagonal term you need to compute.
If you know everything about a spectral sequence you either did a easy computation or you're wrong.
Which I was, hehe :)
So in this case the middle differential doesn't fulfill $d(a,b) = d(a,0)\oplus d(0,b)$. So it is kind of not linear...
12:02
Well, no, it is linear. It's just that $d(a, 0)$ is not $(da, 0)$
@ChristianSander That's okay, the only person among us who can compute using spectral sequences is @MikeMiller
Ah yes, of course, your right. It is linear :).
I think I need a break and then try to work out why the differentials are as they claim in the paper :).
12:19
hi, i read this def in my notes: $ f \in \mathbb{C}[x,y]$ has no repeated factors over $\mathbb{C}$ iff it is not a product of hte form $g^2 h$ where $g$ is non-constant. Equivalently, $f = P_1 ... P_k$ where $P_i$ are distinct irreducible polynomials. I don't think this is quite right, I think what they mean to say is 'Equivalently, $f = P_1...P_k$ where $P_i$ are mutually non-associate irreducible polynomials'. Which one is correct?
@porridgemathematics yeah you're right
@Leak
@LeakyNun thanks!
Hi
I need help here
Good morning! Are there sine and cosine functions for a parabola available?
12:39
Hello @AMDG, can you clarify a bit?
maybe they see "hyperbolic sine" and they want "parabolic sine"
@ChristianSander for any given quadratic of the form $ax^{2} + bx + c$, what is the point along the parabola at some arbitrary distance along the parabola and away from the vertex of the parabola towards either negative or positive infinity, and consequently, what are the functions of x and y that describe this parabola parametrically as $(\operatorname{cosp}(\theta), \operatorname{sinp}(\theta))$?
@LeakyNun lol
@AMDG You are given a parabola as above and a distance that you want to walk along this parabola starting at the vertex?
@BalarkaSen it wasnt a joke
12:52
And you want to know at what point you end up?
@ChristianSander yes
After that, I'd like a generalized formula for all linear equations :D
Do you know how to measure the length of a curve?
In this case that curve is given by your parabola.
@LeakyNun evidently not
I know it requires calculus, I think by some form of integration.
@BalarkaSen let's play chess
12:53
And to answer your first question: There aren't any parabolic sine/cosine versions I know of.
fog of war @BalarkaSen
65
Q: Do "Parabolic Trigonometric Functions" exist?

ArgonThe parametric equation $$\begin{align*} x(t) &= \cos t\\ y(t) &= \sin t \end{align*}$$ traces the unit circle centered at the origin ($x^2+y^2=1$). Similarly, $$\begin{align*} x(t) &= \cosh t\\ y(t) &= \sinh t \end{align*}$$ draws the right part of a regular hyperbola ($x^2-y^2=1$). The ...

maybe some other time
Yes @AMDG: Let's do a simple example.
To make it easy, let's consider $f(x) = x^2$.
12:56
@LeakyNun yeah I was looking at that. I asked here first though :P
IIRC the length of the graph of that function between $a<b$ is given by $\int_a^b\sqrt((f'(x)^2+1)dx$.
is that another way of asking how to transform the parabola into polar coordinates
Starting from the vertex we have to take $a=0$, right?
Yes
@LeakyNun No, but since what I'm doing requires that all things be waves, and norms, it makes sense to find a wave function that describes a parabola and work up to all polynomials eventually as a generalization of sinp and cosp.
norms
13:00
Or just absolute value if you like
$f'(x) = 2x$ so if $d$ is your given distance you have to find $b$ such that $d = \int_0^b \sqrt{4x^2+1}dx$.
Is that what you were looking for?
Sounds about right.
Nice, so this will give $f(x)$, but what is the parametric form?
I thought you were looking for the point $(b,b^2)$ depending on your given distance $d$ were you not?
13:06
Also the infinite series shows this is surprisingly close to $\arctan(x)$. Hm. Interesting. (Now how to remove the coefficients in the numerator I wonder?)
@SayanChattopadhyay Let's go?
@ChristianSander
> and consequently, what are the functions of x and y that describe this parabola parametrically as (cosp(θ),sinp(θ))?
@Balarka Yeah let's do it.
come to garbo
i mean quanto
I don't know neither $\operatorname{cosp}$ nor $\operatorname{sinp}$
13:10
As parabolic sine and cosine.
What is their definition?
That's what I'm asking about. :)
Yeah as I said: I don't know them. So from here on I'd have to do the same as you would have to: Using google :(.
:(
Couldn't u find something?
13:19
In the question that LeakyNun posted here which I was also looking at, it would appear that maybe $\operatorname{sinp}(t)=t$ and $\operatorname{cosp}(t)=1+\frac{1}{2}t^{2}$ in the accepted answer, however, he doesn't state that as the definition, but more as a guess.
58
A: Do "Parabolic Trigonometric Functions" exist?

hmakholm left over MonicaIndeed there are, but they're not usually called that. What ordinary trigonometric and hyperbolic functions have in common is that they are solutions to the differential equation $$f''(t) = af(t)$$ When $a$ is negative, the solutions are ordinary sines and cosines, scaled horizontally by a factor...

One of the limitations I have put on myself as well in this area that I am researching (I don't know what it would be called) is no calculus of infinitesimals, so I can't get something normally via integration or differentiation. It has to be through composition or applying some other function, and in a closed-form expression.
The benefits will be enormous. I find that while it is easy to compose functions using elementary particles of addition and subtraction fundamentally, it is difficult to decompose functions into their elementary parts and describe them all the same using a closed-form expression.
But I also don't want the fruits of my labor to be a long "list of identities for common functions" that you have to memorize, but rather something very terse that can be universally applied and easily remembered to "split" a function into parts which can be manipulated and rearranged.
The simple questions I'm still contemplating the answers to are "What is every $x$?", "What is every function?", and "What are all axes?" They'll be the substance of my study.
Anyways, thanks for the help, @ChristianSander
But you get that in a closed form expression, right?
But that expression uses the inverse of sinh...
What specifically?
Hyperbolic functions I consider to have closed-form definitions that can be trivially computed since they are real functions of x and do not strictly require imaginary numbers.
If I could trivially compute complex-valued functions, well then that would be a different story.
I've been looking for a cheaply computable form of the circular functions, you see, but I would say that is only a catalyst rather than the motivation for what I'm now studying, as this is actually relevant to a project of mine which is non-trivial in computer science and applicable to all sciences once complete.
13:55
The formula we derived above. Is that something you consider a closed form?
Yes, I would consider that a closed form.
And I don't understand the part about complex numbers. It is not so different from the reals.
sure, but... well, you tell me how long it takes to compute $e^{ix}$ for arbitrary x, accurately, and I'll go brew some coffee. :)
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