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6:00 PM
$$\Huge \text{WHAT AN AMAZING RESULT I GOT!!!!!}$$
 
The derivative of a differentiable function is continuous-I need a counter example
 
Huy
$$\Huge \text{HAVEN'T SEEN ONE OF THOSE IN A WHILE!!!!!}$$
 
:-)))))))))))))
 
Huy
@UserX something like $x^2 \sin(1/x)$ is pretty standard. $0$ at $x=0$
 
Thanks
Reached an MSE answer giving that example about 10 seconds before your answer :P
 
Huy
6:02 PM
yeah it's the standard one
:D
 
6:13 PM
@Chris'ssistheartist any interesting result ?
 
@Agawa001 yeah, a very nice result I got!
 
congrats :D
thats just before my interesting result too
 
I wonder if someone is flooding me ...
I trust robjohn anytime, say, but in general no one should know anything about users here, including moderators, e-mails, ips and so on (without personal acceptance).
I'm clearly flooded, my internet works very hard and it's not a problem from my ISP, nor a problem related to my computer.
 
@DanielFischer Could I ask you something? I want to verify that $(2 \cos t, 5 \sin t)$ is a parametrization of $\frac{x^2}{4}+\frac{y^2}{25}=1$.

-For $x=2 \cos t, y=5 \sin t$, $\frac{x^2}{4}+\frac{y^2}{25}=\cos^2 t+ \sin^2 t=1$.

- Now let $\frac{x^2}{4}+\frac{y^2}{25}=1$. We can a pick a $t$ such that $x=2 \cos t$ since $2 \cos t$ is surjective from $\mathbb{R}$ to $[-2,2]$.
Then, $\frac{y^2}{25}=\sin^2 t \Rightarrow y= \pm 5 \sin t$.

How can we reject the case $y=- \sin t$ ?
 
@evinda You can't a priori reject that possibility. If it is, you have picked the wrong $t$, and take a different argument with the same cosine. Depending on the parameter interval, that would be $-t$ or $2\pi - t$.
 
Huy
6:27 PM
if you draw a vertical line on a circle it will intersect it twice...
 
Actually robjohn is the only moderator I trust here. Hope he will remain here for long.
 
Huy
the one time it pays off not to be a moderator
 
@DanielFischer So can't we pick the parametrization $(2 \cos t, 5 \sin t)$ ? Do we have to pick this one: $(2 \cos t, \pm 5 \sin t)$ ?
 
@evinda $x=2\sin(t)$ covers both since $-\sin(t)=\sin(-t)=\sin(2\pi-t)$
 
Huy
you should draw the two using the two parametrizations
 
6:32 PM
@DanielFischer ah, I see you already covered that.
 
@robjohn Could you explain to me why we reject the case $x=- 2 \sin t$ because of the fact that $-\sin(t)=\sin(-t)=\sin(2\pi-t)$ ?
 
I'm so badly flooded I cannot even listen to music on youtube, all loads incredibly hard. Soon I won't even be able to post anything here.
let me run some ping in cmd with yahoo and youtube
 
@evinda we don't reject it, we just don't need it since it is covered by the $x=\sin(t)$ case.
 
@DanielFischer Given a Hilbert space $\mathcal{H}$. If $M_{k}(\mathcal{B}(\mathcal{H}))$ is a $k \times k$ matrix with entries in $\mathcal{B}(\mathcal{H})$, is it clear why $M_{k}(\mathcal{B}(\mathcal{H})) = \mathcal{B}(\mathcal{H}^{k})$?
 
If you have to ask if it is clear, how can it be clear?
 
6:45 PM
This ruins my day ...
 
@Krijn Clarity is subjective...
 
@Moses Kind of. It's clear how each such matrix corresponds to an element of $\mathcal{B}(\mathcal{H}^k)$ I suppose. In the other direction, you have the injections into and projections from $\mathcal{H}^k$, and if you compose an element of $\mathcal{B}(\mathcal{H}^k)$ with an injection and a projection, you get an element of $\mathcal{B}(\mathcal{H})$. That is an entry of the corresponding matrix. Verify that these maps are inverses of each other.
 
@robjohn the discovery I did this evening is mind-blowing, just let me prepare some stuff with it.
 
@DanielFischer How is it clear how each such matrix corresponds to an element of $\mathcal{B}(\mathcal{H}^k)$?
 
@Moses Matrix multiplication. Where "multiplying" a linear map with a vector is applying it.
 
6:52 PM
@DanielFischer Are you taking $\mathcal{H}^{k}$ to be a tensor product?
 
@Chris'ssistheartist they arp-spoof your computer
 
@Moses No, a Cartesian product.
 
If $\alpha^3 = \alpha + 2$, then the ideal $(2,\alpha) = (\alpha)$ right? Because $2 = \alpha^3 - \alpha$
(Should add that this is an ideal of $\mathbb{Z}[\alpha]$)
 
@DanielFischer Oh yeah, I understand.
 
@Krijn Whatever the ring. $(\alpha) \subset (2,\alpha)$ is always true, and $\alpha^3 - \alpha = 2$ shows $2\in (\alpha)$, whence $(2,\alpha) \subset (\alpha)$.
 
7:01 PM
Yeah, but only in this ring can we multiply by $\alpha^2$ I thought
 
@Krijn Whenever a ring contains $\alpha$, it also contains all powers of $\alpha$.
 
Ah, yes!
 
@Agawa001 Think so.
 
@Chris'ssistheartist raymond.cc/blog/…
 
@Agawa001 thanks! Good stuff.
 
7:07 PM
@Chris'ssistheartist i think i got a grasp on these congruent number-sequences
gonna post it tomorrow, now im sleepyyy
gd night
 
@Agawa001 hope on that blog the stuff there do not contain trojans.
 
@Chris'ssistheartist Mma 10 is working on it now.
 
@Chris'ssistheartist cant guarantee, copy the name of software and get it from different server :)
 
What is the right way to talk about $x^t A x$. Say that $A$ is symmetric. Do we say that this is the quadratic form of $A$ with respect to the vector $x$?
 
7:18 PM
@DanielFischer Have you heard of Stinespring Dilation Theorem?
 
@Moses Not until now.
 
@DanielFischer Does it seem interesting?
 
@Agawa001 Xarp installed, and no attack so far. :-)
 
@Moses It looks like it's useful when investigating $C^\ast$-algebras.
 
@Chris'ssistheartist Ick...
`-(1/96) I (2 \[Pi]^3 -
4 I \[Pi]^2 Log[
15] + \[Pi] (6 Log[2]^2 + 24 Log[3]^2 + 6 Log[5]^2 -
12 Log[5] Log[9] - 6 Log[9]^2 - 12 Log[10]^2 +
6 Log[9] Log[16] + 6 Log[10] Log[16] + 6 Log[5] Log[25] -
6 Log[12] Log[25] - 6 Log[10] Log[36] + 12 Log[5] Log[54] +
Log[64]^2 + 6 Log[3] Log[640000/6561] + 6 Log[9] Log[108] -
6 Log[16] Log[108] - 6 Log[2] Log[2916] +
24 PolyLog[2, -(1/2)] + 24 PolyLog[2, -(1/3)] -
12 PolyLog[2, -(1/9)] - 12 PolyLog[2, 1/10] +
12 PolyLog[2, 2/5] + 12 PolyLog[2, 3/5] - 12 PolyLog[2, 3/4]) +
 
7:27 PM
@robjohn ooo, this look horrible ... :-)
 
@Chris'ssistheartist I am running a FullSimplify on it to see if it can be made simpler.
 
@DanielFischer Yeah. Do you think that $V^{*}$ has to be some mapping from $K$ to $H$, maybe the inverse of $V$?
 
@robjohn it should make it far nicer.
 
hi, can somebody help me?
i am a bit confused
 
Askaway
:-)
 
7:31 PM
so metric tensors define the coeffecients of x_i x_j, correct?
 
@user3502615 It helps if you tell us what you are confused about :-)
 
sorry i was typing it
didnt know how to word it
 
@user3502615 yes. You mean the $g_{i,j}$
 
why do you have to define for all i / j ?
cant you just define x_1 to x_n
then, for say $g_{a,b}$ just do x_a * x_b?
 
@Moses It's the adjoint of $V$.
 
7:33 PM
@user3502615 no. because the axes are not always perpendicular
 
what do you mean by that?
 
@DanielFischer Oh okay. Is it from $K \to H$?
 
@user3502615 Say that your coordinates are related to the Cartesian coordinates by some matrix $M$, then $g$ is going to be $M^TM$
You can't just use the diagonal
 
Hello all what can I don I do not understand mathematical statistics?
 
@user3502615 But then there are far more complicated embeddings, where the coordinates are curvy
 
7:36 PM
specially tranformations of random variables
 
@Moses Yes, $V \colon H \to K$ means $V^\ast \colon K \to H$.
 
why do the axes have to be perpendicular for basic math rules (in this case, the distributive property) to exist?
 
@Chris'ssistheartist Still simplifying
 
@robjohn OK
@robjohn A series coming from my research today
$$\sum _{n=1}^{\infty} \frac{(-1)^{n+1} }{2 n+1}\cdot\frac{3\cdot 5 \cdot 7 \cdots (2n+1)}{2\cdot4\cdot 6\cdots 2n}\cdot\Phi (-1,1,n+1)$$
It can be rewritten in many other forms.
 
@user3502615 the distributive property is visualized in terms of the area of rectangles whose angles are perpendicular, right?
 
7:48 PM
i guess
 
Draw a sketch.
 
well yes
i understand what you mean
oh wait i realize
$x_i$ is not expressed linearly
 
Now, that^ makes a huge difference :-)
 
thank you :)
 
np pal
:-)
 
7:55 PM
I mean a slightly different series though, that is $$\log (2)+\frac{1}{\sqrt{\pi }}\sum _{n=1}^{\infty } \frac{\Phi (-1,1,n+1) \Gamma \left(n+\frac{1}{2}\right)}{(2 n+1) \ \Gamma (n+1)}$$
 
@Chris'ssistheartist What is $\Phi$?
 
@Chris'ssistheartist Oh, Lerch Transcendant
(from the Addams Family)
 
@robjohn A generalization of Hurwitz zeta function.
@robjohn :-)
I think I also wanna put at work the Legendre duplication formula.
 
@Chris'ssistheartist: Mathematica is still working on simplifying that huge formula.
 
8:06 PM
yo
 
Working on this one $$\int_0^1 \int_0^1 \int_0^1\frac{dx \ dy \ dz}{(1+x) (1+y) (1+z) (1+x+y+z+x y+x z+yz+9 x y z)}$$
right?
 
@Chris'ssistheartist working on its closed form of that one to make it simpler, yes
often, it can't do much
 
@robjohn OK. The closed form is incredibly simple, you'll see. :-)
 
@Chris'ssistheartist if it can actually simplify it
 
Hey @anon what do you think of the following argument.
The question is: If A and B are normal subgroups such that G/A and G/B are both abelian prove that $G/(A \cap B)$ is abelian.
Consider $\tau : G/A \rightarrow G/B$ $gA \rightarrow gB$
 
8:08 PM
Is it okay to ask here for people to check my answer to my own question on stackexchange?
 
@robjohn $\displaystyle \frac{3}{32} \zeta (3)$ :D
 
$Ker(\tau) = A \cap B$ By 1st isomorphism theorem $G/(A \cap B) \equiv \phi(G/A) \leq G/B$
since G/B is abelian so is $G/(A \cap B)$
 
@Chris'ssistheartist With so many polylogs floating around, I wonder if it will be able to simplify it.
 
@robjohn It has a hard time because probably it didn't perform the integration in a clever way.
 
would an abstract coordinate system be considered in euclidean space?
or not
 
r9m
8:30 PM
0
Q: An Inequality for sides and diagonal of convex quadrilateral from AMM

r9mLet $ABCD$ be vertices of a convex quadrilateral. If $AC$ and $BD$ have mid-points $E$ and $F$ respectively, show that: $$\overline{AB} + \overline{BC} +\overline{CD} + \overline{DA} \ge \overline{AC}+\overline{BD}+2\overline{EF}$$ where, $\overline{XY}$ denotes the length of the line segment $...

anyone interested? :-)
@Chris'ssistheartist @robjohn @DanielFischer ^^
 
good evening everybody
 
@r9m Great. Just out of curiousity, do you ever receive the problems I ask you about here? ;)
 
r9m
@Chris'ssistheartist ya ... :) but I am not clever enough to attend those problems :)
especially the most recent ones seem to be Out of this World difficult ..
 
@r9m It's not much about cleverness, but about the particular corner of mathematics we attend. With enough practice one can excel in any corner, you know well my philosophy.
@r9m They are actually very easy. You didn't see yet my hard problems.
 
r9m
@Chris'ssistheartist for example .. the last problem involved inverse tan integrals .. and I was mostly clueless because I am not used to them yet ,,
 
8:39 PM
@r9m What I was saying, you didn't play in that particular corner but that doesn't mean you couldn't excel there.
 
Everyone can excel in mathematics, you just have to be clever from time to time.
 
r9m
@Chris'ssistheartist I could scratch my head all day but I still would't know how to respond to this :P
@Krijn That sounds like 'Everyone can get a medal in Olympic, you just have to be an athlete from time to time'
 
Well, yeah, thats true, is it not
 
@DanielFischer @robjohn I want to show that $r(t)=\left (\cos^2 t-\frac{1}{2}, \sin t\cos t, \sin t\right )$ is a parametrization of the curve
of intersection of the circular cylinder of radius $\frac{1}{2}$ and axis the $z$-axis with the sphere of radius $1$ and centre $\left (-\frac{1}{2}, 0, 0\right )$.

The circular cylinder of radius $\frac{1}{2}$ and axis the $z$-axis has the form: $x^2+y^2=\frac{1}{4}$.

The sphere of radius $1$ and centre $\left (-\frac{1}{2}, 0, 0\right )$ has the form $\left( x+ \frac{1}{2}\right)^2+y^2+z^2=1 \Rightarrow x^2+x+\frac{1}{4}+y^2+z^2=1$.
 
r9m
@Krijn cleverness just like being an athlete is a state of being .. I don't think part timing (time to time) is gonna cut it .. :) just my humble opinion :-)
 
8:48 PM
@Chris'ssistheartist It simplified it, but not completely:
`1/96 (-2 I \[Pi]^3 - 2 \[Pi]^2 Log[1440] +
6 I \[Pi] (7 Log[2]^2 - Log[5] Log[20] +
2 (Log[3]^2 - 2 PolyLog[2, -(1/3)] + PolyLog[2, -(1/9)] +
2 PolyLog[2, 3/4] - PolyLog[2, 9/10])) +
6 (-3 Log[27/25] Log[2]^2 + 6 Log[2]^3 +
Log[3] (ArcCoth[4] Log[25] - 4 PolyLog[2, -(2/3)]) +
2 Log[6/5] PolyLog[2, -(1/3)] - 2 Log[6] PolyLog[2, -(1/5)] +
Log[81] PolyLog[2, 3/5] +
Log[2] (Log[9/5] Log[81] + 8 PolyLog[2, 2/3]) +
Log[100/9] PolyLog[2, 3/4] + Log[36] PolyLog[2, 5/6] +
Log[4] (-PolyLog[2, -(1/9)] + PolyLog[2, 9/10])) + 9 Zeta[3])`
 
@robjohn It still looks horrible. :-)
 
You know how we can represent a string of yes/no decisions in binary, and thus as a number quite nicely. I just realised that my problem is different and is in fact permutations without replacement (so factorial) is there a nice way to assign numbers to permutations?
 
@AlecTeal There are ways to assign a unique integer from $1$ to $n!$ to each permutation of $n$ items.
 
Yes I know there are, but is there a nice one.
 
@AlecTeal what do you mean by "nice"?
 
8:57 PM
By nice I mean "easy" or "obvious choice" or "natural"
 
@AlecTeal The one I like enumerates all the permutations of the first $k$ items before moving any of the rest.
 
@robjohn rather than clues, is there more than one scheme? Because honestly, I'm thinking of just "being bothered" to work out the algorithm that generates them in lexographic order as n goes from 0 to n! and working out the inverse.
 
@AlecTeal I am not trying to simply give clues, I am trying to find out what you would like to use. There are many ways of counting the permutations.
 
@robjohn I want to know the "most natural" scheme to say here are n distinct things, here's a permutation "oh that's permutation number k" or something, and going vice-versa
I'm pretty sure that lexiographic ordering is the most obvious, but would be surprised if this lacks a name when Dijikstra's algorithm has a name.
Sorry for being short @robjohn I just wanted to not reinvent the wheel, thinking about it, lexiographic order is the only "sensible" orders
 
"factorial base"
 
r9m
9:08 PM
@Chris'ssistheartist Mother of God!! That was neatly crafted :D
 
count how many things less than $k$ follow $k$ in the permutation; multiply that by $(k-1)!$ and sum.
 
@robjohn got a link? I don't see how that'd work.
 
r9m
@Chris'ssistheartist that Phi being Lerch transcendent?
 
@AlecTeal It is very easy to compute, and not to difficult to reproduce.
 
9:11 PM
@r9m yeap
 
r9m
@Chris'ssistheartist looks Badass!
 
@robjohn I'm not entirely sure what you mean, and yes, it probably is. as generating permutations isn't!
 
@r9m Yeap. :-)
 
$f(2,4,6,3,1,5)=1\cdot1!+1\cdot2!+2\cdot3!+0\cdot4!+3\cdot5!=375$
 
@robjohn please do explain (sum of the digits?, choice of k?)
Oooh! Okay, @robjohn sorry to be a pest (got a name for this) but what about going the other way, from 375 to whatever it came from?
 
r9m
9:16 PM
@Chris'ssistheartist I think I'm gonna lose my sanity for good after I get a hold of your book!! :P
 
$f(2,4,6,3,1,5)= \overbrace{1\cdot1!}^{\text{$1$ after $2$}}+\overbrace{1\cdot2!}^{\text{$1$ after $3$}}+\overbrace{2\cdot3!}^{\text{$1,3$ after $4$}}+\overbrace{0\cdot4!}^{\text{nothing after $5$}}+\overbrace{3\cdot5!}^{\text{$1,3,5$ after $6$}}=375$
 
@r9m lol, my mom was saying this about me if I wouldn't take longer breaks from doing math. :D
 
@AlecTeal you can write integers in what is called the factorial base, where digit $k$ represents a $k!$.
$375$ would be written $30211$
 
@r9m When people I usually know have fun, dance, take breaks, sleep, I'm working hard and very hard. :-)
 
r9m
@Chris'ssistheartist seems she knows what she is saying :P ... but anyway .. Sanity is Boring :P
 
9:20 PM
@r9m TRUE! :D
 
In combinatorics, the factorial number system, also called factoradic, is a mixed radix numeral system adapted to numbering permutations. It is also called factorial base, although factorials do not function as base, but as place value of digits. By converting a number less than n! to factorial representation, one obtains a sequence of n digits that can be converted to a permutation of n in a straightforward way, either using them as Lehmer code or as inversion table representation; in the former case the resulting map from integers to permutations of n lists them in lexicographical order. General...
 
r9m
@Chris'ssistheartist I think while people you know are having fun, dancing, taking breaks, sleeping, your'e having more fun than what their 'puny' minds can comprehend :P
 
@robjohn I'm trying (on paper) but I keep mixing what your "k follow k" means, given a vector $(a_1,\ldots,a_n)$ and a $k$ I think you mean the coefficient of $k!$ is the count of the $a_i$ such that $i>k$ and $a_k\ge k$
Okay that's wrong, but you know what I mean. Also found that page when you mentioned "factorial base".
 
@r9m rofl rofl rofl, I'M ON THE FLOOR LAUGHING!!!! :-)))))))))))
 
@AlecTeal just count how many things follow $4$ that are less than $4$ in the permutation. In standard order, all these counts are $0$.
 
9:24 PM
@r9m You said it damn well!!!!:-))))))))
 
r9m
@Chris'ssistheartist :P I know what I am saying alright :P
 
@robjohn did you typo in your 375 example? For k=1 there is only one thing less than 2 after the first term, that is the 1. For k=2 there are 2 things less than 4 - so the second term should be 2.2! right?
 
@AlecTeal nothing will be to the right of $1$ that should be to the left of $1$. Only $1$ can be to the right of $2$ when it should be to the left of $2$. So for $k$ there are $k-1$ possibilities of things that can be to the right that should be to the left.
 
If you're using the 1 thing less than 2 for k=2, then the first digit should be 0.
 
9:27 PM
the more mathematics i learn
the more subjects of it i see
 
@AlecTeal hang on... I have to deal with something outside...
 
like as a kid i thought everything was just spooky scary 5x+6
 
@r9m It'd bad that here on MSE problems like the one you posted are not appealing to the users. At any rate, MSE do not seem to embrace pleasantly such problems.
 
@robjohn sure, I'm sorry for being so slow. I cannot work out where the 3 (coefficient of 5!) came from for example.
 
r9m
@Chris'ssistheartist seems that way ... doesn't matter :-) I just hope Blue notices it before more of the (close) count reaches 5 :))
 
9:34 PM
@AlecTeal That is because $1,3,5$ follow $6$.
 
But 6 is the third term? Oh okay, so it's the number of things that follow k where k actually is in the sequence?
 
@r9m hehe! In my research I found some interesting ways of calculating tough integrals by using geometry. I should also do some more geometry, much more.
 
But you can only go as high as k=5 (as k goes as high as n-1, and you don't have a 6! despite having 6 terms)
 
r9m
@Chris'ssistheartist :D :D !!
 
@r9m btw, are you better now with your condition?
 
9:37 PM
@AlecTeal the reverse permutation should give $6!-1$
 
r9m
@Chris'ssistheartist ya .. I feel well enough! :) Thanks ,, but the symptoms are quickly reappearing if I miss my meds even a single time in a day!
 
I don't want to sound ungrateful but I cannot work out what the algorithm is from what it maps to the end points and something about $k$. All I know is as $k$ goes from 1 to 5, it is neither the count of the things after k (so from k+1 to n) that are less than $a_k$ or less than $k$ @robjohn (provided your example is right)
 
$f(6,5,4,3,2,1)=1\cdot1!+2\cdot2!+3\cdot3!+4\cdot4!+5\cdot5!$
 
@r9m Judging you after the sense of humour you have today, I'd say you're almost healed. That is you're funny as before. :D
 
OR "less than or equal to" as the relation (that's 4 cases tested)
23 mins ago, by robjohn
$f(2,4,6,3,1,5)= \overbrace{1\cdot1!}^{\text{$1$ after $2$}}+\overbrace{1\cdot2!}^{\text{$1$ after $3$}}+\overbrace{2\cdot3!}^{\text{$1,3$ after $4$}}+\overbrace{0\cdot4!}^{\text{nothing after $5$}}+\overbrace{3\cdot5!}^{\text{$1,3,5$ after $6$}}=375$
(for reference)
For k=1, what did you do there? and for k=2 and such?
 
9:41 PM
$k=2$ contributes $1\cdot1!$
$k=3$ contributes $1\cdot2!$
$k=4$ contributes $2\cdot3!$
 
r9m
@Chris'ssistheartist I feel reassured ... I am almost heeled! :D
 
$k=5$ contributes $0\cdot4!$
 
Okay there, for $k=4$ what did you do, the 4th term is a $3$ but there's only one thing less than (or equal to) $a_k$ or $k$ after the $k$th term,
 
$k=6$ contributes $3\cdot5!$
 
9:45 PM
@AlecTeal don't look at the $k^{\text{th}}$ position, look at what follows $k$
$1,3$ follow $4$
 
r9m
@Chris'ssistheartist something like this
 
@robjohn are you around in a bit, I want to try using what I know, eg: "given 6 things, there are 6 places for the first to go. So I can do "position of first thing x (k-1)!"" sort of logic. I'd like to pester you if I get stuck.
Okay, so if $k=2$ there is only one thing after 2 in the sequence smaller than 2. If we pick $k=4$, there are 3 things after it smaller than it?
 
@AlecTeal no. only $1$ and $3$ follow $4$ in that permutation (of the things that should be before $4$)
 
Can we use A, B, C, D, E, F for the permutation please? I sense the numbers are not helping
 
9:49 PM
@r9m lol, 2 users voted to close your question. I really don't understand what some users do here (with their powers.)
 
Also why did you start with $k=2$ - thus ending at $k=6$ - nothing can follow the last. Well 3 things do. Oh I may have seen it.
 
Is it such a bad thing to have a geometry problem on MSE?
 
r9m
@Chris'ssistheartist what superman or any other super hero do with it ,, 'MISUSE' :P
 
hehe :D
@r9m I'm not aware of anything giving such a great time, such much fun like math, it's bad for those that miss this point, I don't say they have some kind of fault, but I don't think in life one can meet something more fantastic than mathematics.
It makes you live your life in the most fantastic possible dimension, surrounded by rivers of pleasure and mountains of amazing results that feeds your soul and spirit every day.
 
@AlecTeal $f(B,D,F,C,A,E)= \overbrace{1\cdot1!}^{\text{$A$ after $B$}}+\overbrace{1\cdot2!}^{\text{$A$ after $C$}}+\overbrace{2\cdot3!}^{\text{$A,C$ after $D$}}+\overbrace{0\cdot4!}^{\text{nothing after $E$}}+\overbrace{3\cdot5!}^{\text{$A,C,E$ after $F$}}=375$
 
r9m
9:56 PM
@Chris'ssistheartist you haven't met my future self yet .. :P
 
@r9m :-))))))))
:24530348 oh, don't say it, you can get banned even for less. The last time I got banned was for defending here the Paul Nahin's book. Perhaps if this event happened some centuries ago I would have risked to be burned alive. :-)
 
@robjohn I think I've got it, and I see why we start at k=2. I just tried something and I found the last term in mine would always be zero, so you've got to shift one - this probably makes no sense to you though, thanks!
To go backwards, it's like dividing in your head, you just see how many 5!s you can remove, then 4!s and so forth. It tells you what position they're in in a list, where you .... yes it's hard to say.
 
r9m
@Chris'ssistheartist I don't think he needed defending ... for his book is a wolf in sheeps fur :D
 
@AlecTeal that is why the location of the things to the right of $2$ only contribute $1!$ and those to the right of $3$ contribute $2!$, etc
 
@r9m I like very much his book, all my respect for what he did!
 
r9m
10:03 PM
@Chris'ssistheartist ME 2 :)
 
@AlecTeal The nice thing about this counting is that it gives the same index no matter how many items you have if only the first 6 are permuted.
 
r9m
@PVAL 4321 :) AWESOME rep!
 
eh
 
@r9m that is the combined rep from all sites
 
r9m
@robjohn ya :-) .. will I get mod super powers if my total rep goes 10k?
 
10:06 PM
@r9m it is hard to even find that usually. I hover over the rep in the chat profile. Can you find it elsewhere? (also if they've typed enough lines in chat)
@r9m no, just when your rep on one site goes that high, and only on that site.
 
r9m
@robjohn I see :)
 
@r9m You'll be there soon in math
 
Joel Hamkins has so much rep
 
r9m
@robjohn I get that from hovering too .. idk if the there is a place which shown total rep other than that
 
If
you
type
enough
 
10:09 PM
@robjohn what about going the other way? Given (1,1,2,0,3) how do you get the order back out. Something like "start with (A)" and build back, but I'm not sure (picking your brains now, incase you're not around later)
 
Wow @robjohn I didn't realize just how much rep you had until just now, dunno why
 
@AlecTeal BDFCAE
 
r9m
@robjohn I left click on the avatars and hover the pointer over the name .. that shows the total rep in chat
 
If that's the case then we have $f^{-1}(f(F,E,D,C,B,A))=(B,D,F,C,A,E)$
 
@r9m Ah. so it does
@AlecTeal wait... what do you mean by 1,1,2,0,3, then?
 
10:13 PM
they're the coefficients in the $F,E,D,C,B,A$ example
 
@AlecTeal no, that is just the permutation we were working with
where did you get $f^{-1}(f(F,E,D,C,B,A))=(B,D,F,C,A,E)$?
 
Well (1,1,2,0,3) are the coefficients of the factorials in $375=f(F,E,D,C,B,A)$
 
$f(F,E,D,C,B,A)=719 =1,2,3,4,5$
 
That explains why it wasn't working.
 
Once you've written your index in the factorial base, it is easy to generate the permutation
 
10:19 PM
For some reason I just can't seem to get it right, I'm wondering if it's me mixing things up or the method I'm applying
 
@AlecTeal which permutation are you looking at?
 
(1,1,2,0,3) which should be BDFCAE
 
yes
 
I keep getting AFDECB
 
there is one thing to the right of B so you start BA
there is one thing to the right of C, so you get BCA
there are 2 things to the right of D so we get BDCA
there is nothing to the right of E, so BDCAE
there are 3 things to the right of F, so BDFCAE
@TedShifrin: we are counting permutations... wanna count?
 
10:25 PM
My fingers broke years ago.
 
I should say indexing permutations
 
Okay I tried again but slowly and this time I got BDAFCE - I'm getting closer, thanks @robjohn I shouldn't need any more guides.
 
I am procrastinating on writing a review of a book for SIAM. I don't like the book.
 
@AlecTeal Did you see how I did it above?
 
@robjohn I promise you that's what I was trying to do. I did see though, I also see how it works (honestly, but I cannot convey it in a sentence) thanks for taking the time.
 
10:27 PM
@AlecTeal any time
 
Well, @robjohn, I did get rained on in my 2-mile hike. But now it's cloudy and sorta sunny.
Is mr eyeglasses really lurking?
 
@TedShifrin It is looking gloomy here, but the chance of rain is negligible.
@TedShifrin who is that?
 
my name for @morphic, who changes names more often than I change socks.
 
lol
 
1 hour ago, by r9m
@Chris'ssistheartist I think while people you know are having fun, dancing, taking breaks, sleeping, your'e having more fun than what their 'puny' minds can comprehend :P
Read that as "white" not "while" - 3 times. I'm having a bad day!
AND added a "the"
 
10:50 PM
@KarimMansour (a) how is that supposed to prove G/(A cap B) is abelian? (b) that map isn't well-defined
 
yeah
the map I should have used is $\phi : G \rightarrow G/A x G/B$
 
right
 
ye
@anon I need to prove that a finite abelian group is the direct product of its sylow subgroup for this I show that $G = P_1 ... P_i$, where $P_i$ is the sylow p subgroup of the ith prime of the order of G.
We can show by induction that $P_1 ... P_i$ is a group.
Since the group is abelian so any subgroup must be normal so we only have 1 sylow p subgroup for each prime
 
howdy @Karim @anon
 
But why is the $|P_1...P_n| = |P_1| |P_2| ... |P_n|$ ?
Hi @TedShifrin
 
10:56 PM
Hi @TedShifrin
 
oh, mr eyeglasses finally speaks other than "lol" :D
 
lol
 
Well, that didn't last long.
 
I'm kind of failing algebra
 
hi
 
10:57 PM
kind of, mr eyeglasses?
 
don't remind of algebra
 
Yay, I just passed 1k
 
oh stop your whining, @Karim
 
fucken prof gives us 30 question each assignment
I got 99.6 but I spent like 50 hr on the assignment
 
@TedShifrin I really wish I could take an undergrad course in algebra but there's none available to me
 
10:59 PM
@TedShifrin
Hi
did you assign your students too 30 questions as prof ?
 

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