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12:26 AM
So $e_1,...,e_{n-1}$ is an orthonormal frame of $S^{n-1}$ with coframe $\omega_1,...,\omega_{n-1}$. We identify $T_xS^{n-1}\cong\{x\}^{\perp}\subseteq\mathbb{R}^n$ and $dx$ is the vector-valued $1$-form with entries $dx^1,...,dx^n$. So the claim reads as $\sum_{i=1}^ndx^ie_j^i=\omega_j$ ($e_j^i$ being the $i$-th entry of $e_j$), which we can check on a basis: $(\sum_{i=1}^ndx^ixe_j^i)(e_k)=\sum_{i=1}^ndx^i(e_k)e_j^i=\sum_{i=1}^ne_k^ie_j^i=\langle e_k,e_j\rangle=\delta_{ij}=\omega_j(e_k)$.
I don't see where the connection forms come into play
 
12:56 AM
Can anybody tell me how do i find the range of the expression:$x^3-6x^2+11x-6$ ?
 
 
2 hours later…
2:30 AM
@Binod differentiate
 
2:54 AM
When we say “a $k[x]$-module is the same as a $k$-vector space $V$ with a linear map $V\to V$” what do we really mean by “the same as”?
I think it is a categorical equivalence between $k[x]$-modules and the category of “object with endomorphisms” $(V,\phi_V)$.
 
@MaryStar Yes. It is reversed for $0\le a\le1$.
@Binod the range of a cubic polynomial is the whole real line (as long as the lead coefficient is not zero).
 
3:12 AM
-1
Q: Diffeomorphisms that permute angles?

geocalc33I'm new to manifolds. I'm reading "An Intro to Manifolds" by Tu, when I can. A diffeomorphism is a conformal map if the pullback of the metric results in a conformally equivalent metric. Are there any references about diffeomorphisms that permute angles (not conformal)? (If this is impossible, p...

let me know what's unclear
a drew an amazing picture to illustrate my point
 
@robjohn is this true for any odd degree polynomial?
 
@Binod yes
 
@WilliamSun It's not just a categorical equivalence. It's an isomorphism of categories.
 
@robjohn I have one more question. If f(x) =$x^3+x^2+3x +|tan(x)|$ is it a bijective funtion?
 
3:28 AM
@Binod No. $f(-\pi/2)=\infty$, $f(0)=0$, $f(\pi/2)=\infty$. Apply the intermediate value property.
 
When you add two vector field is this the same as thinking of them as fluids colliding and the net effect?
 
@robjohn I see. Thanx.
If there is a polynomial function say f(x) which is one-one . Will f(x) +sin(x) be also one -one ?
 
@Binod $x+\sin(x)$ is one-one
 
Yes, but $f(x) \ne x$
 
3:43 AM
@Binod oh, you're asking will it always be one-one? no. $\frac x2+\sin(x)$ is not one-one
 
@robjohn Can i ask one more ?
 
sure
 
If $1+x^2=\sqrt 3x$ then $\Sigma_{n=1}^{24}(x^n-1/x^n)^2$=?
 
Note that $\left(x^n-1/x^n\right)^2=x^{2n}-2+1/x^{2n}$
 
Yes
 
3:57 AM
also, $x^2+2+1/x^2=3$
 
@robjohn I want to convert roots of this 1+x^2=\sqrt 3x$ into $\omega$ and then substitute this into this.
 
@Binod That is a harder way.
 
Where $\omega$ is the cube root of unity.
@robjohn Yes.
@robjohn Then what do you suggest?
 
You know that $x^2+1/x^2=1$. square and subtract 2 to get $x^4+1/x^4=-1$
square and subtract 2 to get $x^8+1/x^8=-1$
I'm working on $x^3+1/x^3$...
I guess we get $\left(x+1/x\right)^3=x^3+1/x^3+3\left(x+1/x\right)$
$3^{3/2}=x^3+1/x^3+3\sqrt3$, so $x^3+1/x^3=0$
 
@robjohn Isn't this becoming a little complicated?
 
4:10 AM
well, then go ahead and solve for $x$ and plug it in.
 
@robjohn that will take me forever to solve.
 
What, $1+x^2=\sqrt3x$? it's the quadratic equation.
 
@robjohn I am talking about calculating the summation.
 
4:22 AM
Let $a_n=x^n+1/x^n$. Then $\sqrt3a_n=\left(x+1/x\right)\left(x^n+1/x^n\right)=x^{n+1}+1/x^{n+1}+x^{n-1}+1/x^{n-1}=a_{n+1}+a_{n-1}$ thus $a_{n+1}=\sqrt3a_n-a_{n-1}$
Or it might be better to let $b_n=x^{2n}+1/x^{2n}$ and derive a recursion for that
let me see....
Then $b_n=b_{n+1}+b_{n-1}$, that is $b_{n+1}=b_n-b_{n-1}$
$b_0=2$ and $b_1=1$
$2,1,\overline{-1,-2,-1,1,2,1},$
 
$b_n=x^{2n}+1/x^{2n}$ from this how did you get the value of $b_0$ and$ b_1$?
 
$b_1=1$
 
@robjohn Yes .
 
$1\cdot b_n=\left(x^2+1/x^2\right)\left(x^{2n}+1/x^{2n}\right)=x^{2n+2}+1/x^{2n+2}+x^{2n-2}+1/x^{2n-2}=b_{n+1}+b_{n-1}$
we already computed that $b_1=x^2+1/x^2=\left(x+1/x\right)^2-2=3-2=1$
Therefore, $b_{n+1}=b_n-b_{n-1}$
$b_0=x^0+1/x^0=2$
so as computed above, $b_n$ repeats every $6$ terms
and the sum of those $6$ terms is $0$
So the sum is $48$
 
4:44 AM
@robjohn We started off with this $x^2+2+1/x^2=3$, right?
 
@Binod that is squaring the original equation, yes
 
@robjohn after this , is this the next step?
 
@Binod that says that $b_1=x^2+1/x^2=1$
 
@robjohn yes.
@robjohn After finding $b_1$ is this the next step?
 
Then $1\cdot b_n=\left(x^2+1/x^2\right)\left(x^{2n}+1/x^{2n}\right)=\left(x^{2n+2}+1/x^{2n+1}\right)+\left(x^{2n-2}+1/x^{2n-2}\right)=b_{n+1}+b_{n-1}$
which gives $b_{n+1}=b_n-b_{n-1}$
Then we start out with $b_0,b_1,\dots=2,1,-1,-2,-1,1,2,1,\dots$
Note that when we have $2,1,\dots$, this repeats.
 
4:51 AM
@robjohn isn't this $b_{n-2} $ instead of $b_{n-1}$?
 
No... $b_n=x^{2n}+1/x^{2n}$ so $b_{n+1}=x^{2n+2}+1/x^{2n+2}$ and $b_{n-1}=x^{2n-2}+1/x^{2n-2}$
 
@robjohn i see.
 
When $a\leq 0$ or $a\geq 1$ then $\left (1+\frac{1}{n}\right )^{\alpha}\geq 1+\frac{a}{n}$ and then $$ n^{\alpha}\left ( \left (1+\frac{1}{n}\right )^{\alpha}-1\right )\geq \frac{a}{n^{1-\alpha}} $$
Taking the limit $n\rightarrow +\infty$ we get $$\lim_{n\rightarrow +\infty}n^{\alpha}\left ( \left (1+\frac{1}{n}\right )^{\alpha}-1\right )\geq \lim_{n\rightarrow +\infty}\frac{a}{n^{1-\alpha}} \Rightarrow \lim_{n\rightarrow +\infty}n^{\alpha}\left ( \left (1+\frac{1}{n}\right )^{\alpha}-1\right )\geq \begin{cases} 0 & \text{ if } 1-\alpha>0 \\ +\infty & \text{ if } 1-\alpha<0 \\ \alpha & \tex
 
@robjohn what is this in this statement sir?
 
@Binod the sequence of $b_n$. it repeats every $6$ terms
 
5:01 AM
@robjohn Sir if sum of each six terms is 0 then how did we arrive at that result that the sum is 48?
 
@Binod Because $\left(x^n+1/x^n\right)^2=x^{2n}+2+1/x^{2n}$
the $x^{2n}+1/x^{2n}$ sum to $0$, so we are left with $24$ times $2$
 
Ah, because of that 2.
@robjohn Thank you so much sir.
 
@MaryStar since you are computing $(n+1)^a-n^a$, it appears that the limit should be $\infty$ if $a\gt1$, $1$ if $a=1$, and $0$ if $a\lt1$.
 
 
2 hours later…
6:54 AM
@robjohn I have to prove this inequality
$$
xyz \gt (y+z-x) ~(x+z -y) ~(x+y -z)
$$
I'm missing something, coz I know it's all about applying $ A.M. ~\gt G.M.$.
We can have something like this
$$
y+ z \gt 2 \sqrt{yz} \\
x+z \gt 2 \sqrt{xz} \\
x+y \gt 2 \sqrt{xy}
$$
But I need x, y and z all together.
 
7:40 AM
How do you get the Poincare Dual of a closed/compact $n$-form, and does it always exist ?
By poincare dual of an $m$-form $\eta$ on a $n$-manifold $M$, I mean a $n-m$-dimensional submanifold $N$ (possibly with boundary) such that for any $n-m$-form $\omega$, we have $\int_{N} i^* \omega = \int_{M} \omega \wedge \eta$
Also, is there a nice way to explicitly write down Poincare duals for $1$ dimensional submanifolds of surfaces ? And possibly unrelated, but it seems if you take a=an embedded one dimensional submanifold in a surface, take the dual of it, and integrate that along the submanifold, the result will be an integer ! I only checked it for toy cases, how do show this (or is this false ?) in general ? Can this be used to give explicit duals ?
 
8:23 AM
@Edward Münster, why?
 
8:43 AM
@Alessandro hahaha I thought so; I've also been looking at Münster for a PhD. There are two profs there who do things that I am potentially interested in
 
9:13 AM
Nice! They have the less painful/bureaucratic application process of all the unis I've seen for PhDs so you should definitely apply haha @Edward
 
Oh nice :D It looks quite streamlined lol, when does one have to apply? Like a year in advance or ? (E.g., if I wanted to start in October 2022 or smth)
 
I applied in spring (applications where open from the end of Februry to the end of March iirc) for October of this year
 
I see, thanks
 
@EdwardEvans yeah you just fill in the online application form, get two professors to send a recommendation letter and you're done
 
@Lelouch It exists on oriented (closed, otherwise use compactly supported cohomology) manifolds.
 
9:18 AM
Nice :) Guess I'll spend the rest of my master studies brown nosing
 
An explicit construction of the Poincare dual is as the Thom class to the normal bundle of the submanifold
 
I suggest contacting the professors you'd like to work with before applying though, I know some people who didn't and they didn't get in
I also didn't but it was kinda particular so dunno
 
yeah fair, the two profs are doing similar stuff to the prof I'd like to write my master thesis with in Heidelberg, so I have some more options lol
(On $(\varphi, \Gamma)$-modules, oof)
 
I don't know if I want to know what those are lol
Also will you be applying for 2022? Not 2021?
 
I'll need to extend my master studies, for mental health reasons I kinda screwed my first semester and that had some effects on my second semester
 
9:29 AM
I see
 
am hoping that won't have too much of an effect on applications, but I have backup from doctors if necessary
lol
 
I don't know if it's possible to start a PhD in the summer term in Münster, I know that some places do that too
In Bonn you can even start a masters in the summer term
 
yeah I would probably still start in October
i mean, I'm just researching profs who do stuff I'm interested in at the moment
 
Fair
It's good to start early, to know all the options and don't miss deadlines you didn't even know about :P
 
aye, and I already have some relevant courses under my belt
and yes, and getting the relevant circular stamps
 
9:51 AM
@robjohn Sir, I got this [hint]
(https://math.stackexchange.com/a/2361192/569595) for that question.
And it has helped me. Thank you sir.
 
is there someone here who has failed midway in a PhD course and records of that failure have badly affected their chances of getting a job in academia or the industries?
 
 
2 hours later…
11:59 AM
I took a look at some sane dynamical systems stuff @Balarka, in particular I saw that there's a nifty dynamical proof of Van der Waerden's theorem. To get that you need Birkhoff's recurrence theorem, turns out that the topological dynamics book had a more abstract nonsense version of Birkhoff's theorem as an unnamed lemma...
 
Yeah I have seen this proof before
Not really my cup of tea though
 
Birkhoff says that if $X$ is compact and $f$ an automorphism then there is an $x\in X$ and a sequence $n_k\in\Bbb Z$ with $f^{n_k}(x)\to x$ (there is a recurrent point in dynamics terminology)
 
Yeah I think this follows from Poincare recurrence theorem
 
But in fact you can get a sequence $n_k$ with bounded gaps
(there is an almost periodic point)
 
You can get an $f$-invariant probability measure on $X$, and then you just do Poincare recurrence on this
 
12:04 PM
And it's actually true whenever a topological group $G$ acts on a compact space $X$, there is an almost periodic point $x$, meaning that for all nbhds $U$ of $x$ there is a syndetic $A\subset G$ with $xA\subseteq U$ (syndetic is a generalization of quasi-dense in discrete groups, it means that there is a compact $K\subseteq G$ with $G=AK$)
 
Hm OK
Believable
@Lelouch Did you follow the weird local geodesic stuff? That's too computational for me
 
The idea is that compact spaces always have a minimal subspace (this is an easy Zorn's lemma argument) and every point in a minimal subspace is almost periodic (this part is more painful but not too bad either)
 
Just do Poincare recurrence man
Don't do set theory
 
@BalarkaSen But that's ergodic theory
 
Yeah I'm saying you can just put this in ergodic theory context
You can always cook up an $f$-invariant probability measure on $X$
 
12:11 PM
Can you?
That's interesting, I didn't know, what should I google?
 
I mean take any probability measure on $X$ and average it along $f$
You can always pass to a subsequence which weak* converges to an $f$-invariant probability measure
 
What do you mean with average? It looks like I need a locally compact group or something like that
 
No, what? You can do $1/n \sum_{k = 0}^n (f^k)^* \mu$ or whatever
 
Also can you get an invariant measure with good regularity properties? Like a Borel measure at least I'd hope
I was thinking about a generic topological group as acting group
 
I was just doing it for one transformation
I dunno the general stuff
 
12:17 PM
I can integrate against the Haar measure along an orbit if my group is locally compact
Fair, the Z thing makes sense, and I believe it can be generalized. I'll look this up, thanks
 
But how do you use Poincare recurrence?
for a general group
 
I don't know if there's a general version
There's probably some version for many commuting maps? Like there is one for Birkhoff
 
yeah I dunno
 
I guess I'll read some ergodic theory too eventually
There are characterizations of amenability (for topological not necessarily discrete groups) in terms on invariant measures
 
@BalarkaSen part 3 of the Lemma ?
 
12:27 PM
yah
 
well I just took notes of all the theorems, I need to prove most of them (didn't follow this one, the incircle bound one 100%)
 
I'll have to read it from Brideson-Haefliger
too hard man
 
tbh it seems like just some olympiad combo problem, use the $\delta$-hyperbolicity condition to prove these theorems and you don't need to use anything except triangle inequality
 
yeah haha its like Euclidean geometry all over again
 
I'm finding it to be a little boring tbh, though tha'ts probably because I lack the prereqs to appreciate it
 
12:30 PM
I have to read to "get it" as well
I am not convinced by his answer to my last question, though I see his point
 
I thought he was going to do some diff geo/curvature stuff
 
haha
 
@BalarkaSen What was your question regarding the projection parametrization btw ?
Didn't get you fully
 
Oh that was just saying the following: Take any two geodesics which are Hausdorff close, then they are close in the pointwise sense, under the uniform norm, as well
So since in lemma we're dealing with local geodesics, his kind of example where you go forward and come back a lot shouldnt pop up
 
What do you mean by "close in the pointwise sense" ? Didn't understand that part
 
12:33 PM
As in, $f, g$ be the curves, say they are pointwise $\epsilon$-close if $d(f(t), g(t)) < \epsilon$
for all $t$
 
oh ok. I see
 
So if your local geodesic is Hausdorff close to an actual geodesic, you zoom in and you have two honest geodesics which are Hausdorff close
so they should be pointwise close and thus the local guy should be a quasigeod
Maybe that doesnt work for some stupid reason I'd have to think
 
@BalarkaSen By "zooming in", presumably you mean using the second part to find a geodesic which is $3 \delta$ close to it (or something else) ?
 
I mean $c$ be the local geodesic, we know $c$ is some $3\delta$-Hausdorff close to $[c(a), c(b)]$ (I think part $1$ says $\text{im}(c)$ is in the $2\delta$-nbhd of $[c(a), c(b)]$ and part $2$ says $[c(a), c(b)]$ is in the $3\delta$-nbhd of $\text{im}(c)$, but whatever). Now pick a point on $c$, look at the arc of length $k$ around that point -- that is an honest geodesic segment which is $3\delta$-close to some segment of $[c(a), c(b)]$
and Hausdorff-close geodesics track each other in the parametrized sense as well aka they are pointwise close... so
Anyway I don't know seems like an annoying fact
@Lelouch I'm really intrigued by the 4-point condition and I'm disappointed he doesn't like my idea. Here's what I really want to note: Fix a basepoint $o \in X$ in your $\delta$-hyperbolic space, and say the distance between two points $x, y \in X$ - imagine them as being very very far from $o$ - is $e^{-(x, y)_o}$.
Why? because $(x, y)_o$ is measuring how long the geodesics $[o, x]$ and $[o, y]$ stay close to each other (think of the tripod picture), if you do $e^{-that}$ that tells you if they are nearby as points at infinity
 
12:49 PM
@BalarkaSen I think I have a counterexample for why this don't work in $\mathbb{H}^2$, though I'll need to make sure it works
 
Then $(x, y)_o \geq \min\{(x, z)_o, (z, y)_o\} - \delta$ is really saying that this metric, let's call it $v(x, y) = e^{-(x, y)_o}$, satisfies $v(x, y) \leq e^\delta \max\{v(x, z), v(z, y)\}$
That's nearly an ultrametric condition
For trees it's exactly an ultrametric, and says that the "boundary" of the tree is a Cantor set with the usual ultrametric
@Lelouch OK tell me
 
Sorry, nevermind, I forgot the $k \geq 8 \delta$ condition must also hold
 
Ah yeah, that condition should say the zooming should be comparable with the $\delta$-thinness of your ambient
If you zoom so small that you can't even apply thinness then something is bound to go wrong
I dunno whatever man this sucks
 
@BalarkaSen Yes, exactly. Otherwise you get a counterexample in the circle graph too: Just consider a loop
 
Yeah
Anyway I'm thoroughly annoyed. Do hyperbolic geometers think in terms of inequalities?
 
12:56 PM
@BalarkaSen By the way, I don't understand why is $\langle v, w \rangle_x$ is referred as the inner product here ?
 
That's garbage
 
He mentioned something like you get an inner product structure even when there's no tangent space, but I don't understand what he meant by you get an inner product structure
 
Meh that's some analogy. $(v, w)_x$ is like the angle at $x$ appended by the geodesics $[x, v]$ and $[x, w]$
because it tells you how far those geodesics track each other
That's what it is for a tree, right? it's just distance of $x$ to the tripod point in the triangle spanned by $x, v, w$
that's how much the geodesics $[x, v]$ and $[x, w]$ track each other aka are SAME in $\delta = 0$
 
@BalarkaSen Yes. So the rough picture to keep in mind is that $(v,w)_x$ is large will imply that the geodesics $[x,v]$ and $[x,w]$ are very close to each other for most of the time (and vice versa), right ?
 
Yup!
 
1:01 PM
ok, now I understand your original comment regarding the 4-point condition and sending the points at infinity
that's nice
 
But he thinks it's unhelpful so maybe this is a useless point
 
It also tells us how much it decays
 
I mean I get what he's saying; (1) there's no preferred basepoint (2) by sending the points to infinity I am not really understanding the $\delta$-hyperbolicity of the space, because the condition is for all $o, x, y, z$
Doesn't it irritate you that the inequality has to be algebraically manipulated to make sense of it using the tetrahedron condition?
That's GARBAGE man
 
@BalarkaSen lol
it's probably instructive (and maybe trivial) to find a space for which the inner product condition holds for one particular base point, but it's not $\delta$-hyperblic
 
yeah your approach to this is good, we should write down lots of examples
i know 0 examples
 
1:05 PM
Maybe some $p$-adic space will give an counterexample ?
 
Too hard, maybe you can just take like a tree, $o$ be a basepoint, and then very very very very far from $o$ append a massive complete graph of a bunch of vertices
then its mostly delta-hyperbolic but very very very very far away its not
so at that basepoint the inner product thing should work out
But maybe that doesn't work
You need some substantial set of vertices I mean, otherwise its just quasi isometric to the tree
 
I think you can modify yours with a slight change: Set $d(o, x) = 100$, $d(x , x') = 1$.
For all $x, x'$
so basically a cone
 
Haha yeah
That works
so I wonder if $\delta$-hyperbolicity is the same as "4-point condition, but at a basepoint and at infinity, for all basepoints"
if that question can be made precise
It should be true. Anyway I don't know these are curiosities, probably not useful
 
1:35 PM
@Knight Sorry I couldn't respond earlier. So you can now show that $\frac12\sqrt{(a+b+c)(b+c-a)(a-b+c)(a+b-c)}\le\frac{\sqrt3}2(abc)^{2/3}$
The quantity on the left being the area of a triangle with sides $a,b,c$
 
2:06 PM
@robjohn sir what’s the RHS? How you got it?
 
@Threnody yes, and you soon find out that the arctangent that comes from integrating $\frac1{1+x^2}$ can be written as $\frac1{2i}\log\left(\frac{x-i}{x+i}\right)+\frac\pi2$
@Knight Think about the inequality you just proved.
 
@robjohn Do you mean $$ (a+b-c) (b+c-a)(a+c-b) \lt abc $$??
 
3:09 PM
@Lelouch I asked Mahan this point, and his response was "what you are saying is right. and is the right perspective at infinity. but for finite points, better think tetrahedra" -- so maybe not entirely useless.
So we can maybe spend some time thinking about this interpretation and won't be a waste of time
 
@Astyx Hello! Seeing you after a time
 
hi
 
Will you people have normal classroom education this year in France?
 
Heyo
@Knight Hi
@Knight What all is this?
 
@abhas_RewCie I was really missing you pal!
(I was trying to understand Robbie sir’s proof)
Thanks for pinging me! It feels like we are still like before
 
3:24 PM
@Knight I messaged you on facebook, but you didn't reply. I thought you were busy with life or so...
haha :-)
 
@abhas_RewCie I think I missed it.
 
@Knight ah okay... What are you doing these days?
@Knight Talking you back is feeling like finding a missing brother after a long time!
 
@abhas_RewCie Nothing, everything is as it is as before
 
@Knight I messed with my life a bit... Recovery time...
 
@Knight I don't know
 
3:31 PM
Okay!
 
4:21 PM
prime and maximal ideals coincide in Zn?
 
@robjohn Sir I want to ask something very directly from you, can I?
(it’s not a mathematics problem so “don’t ask to ask” doesn’t apply here)
 
@ManolisLyviakis yes
Well apart from (0) of course
 
same goes for any PID
 
4:39 PM
$s=14i$
$n=7$
$0.5 + 14.1347i = s-n \left(\lim_{c\to 1} \, \frac{\sum _{k=1}^{n} \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta ((c-1) (k-1)+s)}}{\zeta (c) \sum _{k=1}^{n+1} \frac{(-1)^{k-1} \binom{n}{k-1}}{\zeta ((c-1) (k-1)+s)}}\right)$
Is this a bad formula for some reason? Ideally $n \rightarrow \infty$
 
@JackOhara Hi
How you doin’ ?
 
@Thorgott right, I wrote for any ring of dim=1 in my removed messaged, but then I realized there might be many chains with different smallwr element switnessing dim=1. Of course if (0) is prime that doesn't happen
 
@MatsGranvik After inspecting the thing for 5 min, I quit. It's not my piece of cake....
@AlessandroCodenotti Hey! Are you free for 5 min?
 
It seems to be no ones piece of cake. I guess I will have to eat it myself.
 
@MatsGranvik Either you are enjoying Latex a bit or a bit of mathematics....
or both, possibly!
Okay... there's zeta function too....
 
4:52 PM
@abhas_RewCie Do you like chocolate cake?
 
@MatsGranvik Yes... when cooking... No, when eating :P
@MatsGranvik Have you tried using Mathematica to ask that?
I've seen people believing more in mathematica than Mathematician when it comes to mathematics
 
@abhas_RewCie Yes I found this by doing FullSimplify in Mathematica on nested limits that give derivatives of the reciprocal of the Riemann zeta function.
And then testing it numerically.
 
@Knight yes
@Knight what?
 
@MatsGranvik Bad formula, good formula doesn't matter. If mathematica says, it is!
 
@robjohn Sir, high school algebra is much about : adding and subtracting the same thing to get a good expression, dividing both num and deno with same number to get something familiar. In high school, all the indefinite integrals are more about algebra than the Calculus.
In high school we manipulate more and solve less.
 
4:59 PM
@Knight Is that a problem?
 
My question is: Is higher mathematics similar to this? I don’t think so, because whenever I see @Thorgott discussing something of his level, he almost never writes any expression, they all discuss it like philosophy
 
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