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00:00
can anyone answer, wolfram alpha freaked out when I asked it this
what would my next step be?
$\text{du} = 2\text{dm}$
okay, yeah I see that
so that means
$\int \frac{du}{u^3}$
i know the integral of $u^3$ is going to be $u^4/4$
what do I do with the dm though?
@Charlie HI.
what nobody knows?
the $\text{dm}$ is gone... you substitued $\text{du}=2\text{dm}$
also, that's not the integral of $u^3$
it's the integral of $u^{-3}$
00:12
@JebediahKerman You have $u^{-3}$, not $u^3$.
I win!
okay
why does the dm go away? that the part that confuses me
all the dm and dx just seem to magically go away
@JebediahKerman What do you mean they go away?
@Mike @PedroTamaroff here
@JebediahKerman Well, it goes away, but is replaced by another differential!
leo
leo
00:17
holle
@PedroTamaroff which one?
I don't get that
leo
leo
have realized that seies is a palindrome?
jebediah do you agree that we have EQUALITY between $du$ and $2dm$? They're the same thing
..yes?
hold on I am gonna read up on this before I bother you more with my ignorance
Sorry if that sounded like you were bothering me, that wasn't intended!
We have $\textrm{du} = 2\textrm{dm}$
So where we used to have $2\textrm{dm}$, we just put $\textrm{du}$ instead, as they're the same thing
00:23
yes
I follow
So $\textrm{dm}$ didn't go away
We just rewrote it
okay
I am still gonna read up on it now though
becuase I only vaguely follow
I need to get a more solid intergration foundation
no rhyme intended
leo
leo
if $F\subseteq E$ are fields, $B_1\subseteq E$ is algebraically independent set and $E$ is algebraic over $F(B_2)$, then $\#(B_1)\leq \#(B_2)$?
it holds if $B_1$ and $B_2$ are both finite
 
2 hours later…
02:27
I really don't wanna study for my discrete math test you guys
i have not even gone to that class for over a week
03:03
how would I find the sum of the residues of \frac{1}{1+z^3}
residue calculator
@Eric Need to have dollar signs ($) around your latex code
@DonLarynx I got you beat. I have an exam coming up in like 3 weeks for a course that I haven't gone to in over a month... Hell, since it started in September, I've gone maybe 10 times at most.
The class is taught by a professor that can barely speak English, so most of the class doesn't bother going at all. We get nothing from him.
I've got an assignment due in that class on Monday, of which I know nothing about... so I have to learn the material over the weekend to do so, taking time away from a group project I'm working on.
leo
leo
03:32
@anon hi
hello
leo
leo
haven't the thing I've wrote above already be asked on main?
it seems pretty standard
I mean, this one:
3 hours ago, by leo
if $F\subseteq E$ are fields, $B_1\subseteq E$ is algebraically independent set and $E$ is algebraic over $F(B_2)$, then $\#(B_1)\leq \#(B_2)$?
I don't know if the question has been asked in that specific form
but it probably has to do with transcendence degree
leo
leo
indeed. It's true when the $B_i$ are both finite. When dealing with finite transcendence basis, it's a lemma which allows conclude that any two finite transcendence basis have the same number of elements. In the infinite case, there's another proof, but I wonder if we can proceed by using this again
I'll ask
but it's a bad time to ask
03:57
I find it really odd and erratic that I'll go on here one day, and one question I frequented will be discussed by my professor the next day.
04:27
Can someone please tell me what my commenter is trying to tell me in my answer to this integration? I am not seeing what he is saying.
I would gladly incorporate additional rigor for the benefit of the OP, but I am not seeing it.
do you agree that the square root of cos(t)^2 is not cos(t), but |cos(t)|?
sure or certainly the absolute value of a is greater than or equal to zero
I am talking about cos(t), not a
Yes I agree absolutely
The integral only makes sense if the value of $t$ is within an interval for which the sign of cos(t) is constant though.
04:37
this is a good point
I am definitely on my feet, and unexpectedly.
so the manipulation is not valid in general, but is valid in context. if you want, you can explain why in your answer, JW
did you read my comments? how would I explain this to my dutiful commenter in any other way than claiming $+c$?
I have some thinking to do
damn
err, rats, strike damn I am not upset, just well...
setting $a\ge0$ by fiat is a rather mundane point
not a serious issue
a constant
of this I am sure
04:41
huh?
Well ok
(note we need a>0 for the formula to work because arcsin is odd)
greater than or equal
well, equal if you want to divide by zero
maybe I hould put that up front. I am just trying to think about where to put it now. Up front or in the end
Ah bleh!
04:42
@KarlKronenfeld I am unable to see how the bound $(4/\pi)^sn^n/n!\le\sqrt{|\Delta|}$ yields Hermite-Minkowski (i.e. there are finitely many number fields of given discriminant). ideas?
yeah don't want to get kicked out of the math union!
does anybody know if there's someone I can email if a Math Reviews link to an article is broken?
the few sources on google search results insinuate ramification is involved, but (a) I don't want to think about ramification and (b) stewart/tall doesn't even seem to mention ramification anywhere, so they can't expect us to use it in an exercise
Let X = {0,1}. List all the elements in ℘(℘(X))
@anon thanks, I think I will put the explicit assumption up front.
04:50
@user109886 where are you having trouble?
I think ℘(X) = {Ø,{0},{1},{0,1}}. But I'm not sure about ℘(℘(X))
well, can you compute ℘(Y) where Y={a,b,c,d}? do this, then set a=Ø,b={0},c={1},d={0,1}
Is my answer for ℘(X) wrong?
It's correct
@anon Yeah, I don't really have any ideas. Of course only finitely many $n$ satisfy that relation, but (given my limited knowledge of any of this stuff) going from there to finitely many extensions seems nontrivial.
04:56
one can't get [finitely many extensions] from [finitely many n], e.g. there are infinitely many quadratic fields.
Yeah, I should have made it clear I know it's necessary to take into account the hypothesis that $\Delta$ is fixed in the latter step.
And I would not know how to use that, esp. without ramification.
05:30
@anon I totally get it.
Now I am just wondering what else I have ignored in my seemingly flawless education.
I think probably to not make assumptions and be more critical of "assumed" reasoning in the future.
05:46
Time to learn boolean algebra in less than 3 hours @agent154
 
3 hours later…
08:56
user image
3
:D
How does one go about proving $\Pi(x)= \sum_{r \ge 1}\frac{\pi(x^{\frac{1}{r}})}{r} \Leftrightarrow \pi(x)= \sum_{r \ge 1} \frac{\mu(r)}{r} \Pi(x^{\frac{1}{r}})$? It's obviously a Mobius inversion but I'm clueless as to how to execute it.
@B.S. :D
 
1 hour later…
10:27
@Alyosha Well, $ \sum_{r\geq 1} \frac{\mu(r)}{r} \Pi\left(x^{1/r}\right) = \sum_{r\geq 1} \frac{\mu(r)}{r} \sum_{j\geq 1} \frac{\pi\left(x^{\frac{1}{r\cdot j}}\right)}{j}$ and the latter is $$\sum_{r,j\geq 1} \frac{\mu(r)}{r\cdot j} \pi\left(x^{\frac{1}{r\cdot j}}\right) = \sum_{r\geq 1} \mu(r) \sum_{r | n} \frac{\pi\left(x^{\frac{1}{n}}\right)}{n} $$
(the second sum is on $n$ of course, not $r$)
(you sum over the multiples of $r$)
(that are not $0$)
try reversing that sum, then, sum first on $n$, and then on $r$, to get a sum over the divisors.
and you should be done
$$ \sum_{r,j\geq 1} \frac{\mu(r)}{r\cdot j} \pi\left(x^{\frac{1}{r\cdot j}}\right) = \sum_{r\geq 1} \mu(r) \sum_{n : r | n} \frac{\pi\left(x^{\frac{1}{n}}\right)}{n} $$
 
2 hours later…
12:16
@B.S. groan...
12:49
Greetings
Sorry, I am new here
I need some urgent help, just to go over certain things with me... is there anyone who would be willing?
I would sincerely appreciate it :)
askaway
I have asked a question regarding continuity, differentiability and existence of partial derivatives.
and I have typed in my steps
I am new to multivariable calculus, so I find this rather difficult
oops, sorry about that
0
Q: Continuity, differentiability and existence of partial derivatives

ArtemisiaHere are a few functions whose continuity, differentiability and existence of partial derivatives are to be checked at the origin. I have given the answers, but I would really appreciate it if someone could check it for me :) $$1. f(x,y)=\sin x\sin(x+y)\sin(x-y)$$ Continuous, differentiable, part...

This is the question
I have shown my steps and my answers... I wanted someone to check them for me so that I know where I am going wrong
Hello...?
Haha sorry :) Anyone who could check the link for me?
@Artemisia Your argument for the first one seems wrong. Can you try to rephrase it here?
13:03
@N3buchadnezzar : Sure :) It is a composition of trigonometric functions and hence is continuous... partial derivatives exist and are equal
Yes the function is continous, and differentiable. But you do not justify that in your post.
The partial derivatives are lim(h -> 0)1/h(sinh sinh sin(-h) on both sides
Oh I just assumed they are :) because of continuity of trigonometric functions
Well also you only find that the derivative at $f(0,0)$ is zero, this has unfortunately nothing to do with if the function is continous or not.
@N3buchadnezzar oh... In the first one, you mean?
13:09
Ah
@N3buchadnezzar but since it is a product of trigonometric functions... shouldn't it be differentiable as well?
Indeed! And it is, but your argument does not say that.
@N3buchadnezzar Haha oops :) sorry :)
A sum of continous functions is continous, and a product of continous functions is continious.
@N3buchadnezzar :) that's what I used as my reasoning.
Also to prove that a multivariable function is continous one has to show that its value is independant of the path you aproach the point. Luckilly as long as your function does not contain any singularities, you are fine.
13:13
@N3buchadnezzar To be honest... part 1 was a bit of a hand-wave. I am not sure of the rest of them.
Ah :)
Singularity implies a point where function value is 0?
Something like $$ f(x,y) = \frac{x^2 y}{x^4 + y^2} $$
Oh ok :) I understand
You would have a problem where $x=0,y=0$, or $x^4 = - y^2$ (is the last case really a problem?).
Sorry... I didn't understand what you mean by the last case
Mathematical singularity, a point at which a given mathematical object is not defined or not "well-behaved", for example infinite or not differentiable. Here you would end up dividing by zero, which never is a good thing.
13:17
OH! haha ok I understand :)
@N3buchadnezzar That means 2. and 4. have points of singularity
Great :)
Are the rest of them fine? I know my reasoning is skimpy at places :)
@N3buchadnezzar Oh I have watched this :)
Well you are not using it!
13:23
Oh... I thought that's what I did...
I used that to understand why some functions are discontinuous
Limits are all about the journey towards a point or place. Limits do not care the slightest about what actually happens at the point, only how the function is behaving towards the point.
Yes indeed :)
So to prove that a function is continous you have to prove that the limit, is the same as the value at the point.
Yes. Absolutely.
To do this you have to show that no matter which path you choose to get to the poibnt, you always obtain the same value as at the point.
I can not see any argument like this at (4) ?
13:26
Oh someone has already put that in the comments because I forgot to put it on the original post.
And then I edited it to put in my major doubts and queries
Do just like they did in the video
watch it again if you can not =)
Should I type it in? I calculated it that way for all the functions except 5.
Like the comments say post a new question, where you show your steps to prove conti
@N3buchadnezzar Ok :) I will do that for 2. and 4. What about 5. and 6.?
13:33
Sorry? I don't understand what you mean.
Well again you have no arguments for either 5 or 6.
@N3buchadnezzar for 5. I just looked at the plot and drew conclusions
For 6. I have given my reasoning...
Again you only look at $x=0,y=0$. You focus on a single point
To show that a function is differentiable, it must be differentiable for all $x$ and $y$.
Oh... I thought the question asked for differentiability at the origin
We are very strict on the difference between being differentiable in a point, and being differentiable on a domain or region. Ah, the question states to check origo explicitly.
13:41
Oh. But I think the question asks for just at the origin
Even so your argument only finds the directional derivate at the origin.
Which is not the same as checking if it is differentiable at $(0,0)$.
But if a function is continuous and has continuous partial derivatives at a point, then isn't it differentiable at the point?
In that case yes. But from your post it seems that your argument goes like this: Hey I found the directional derivative at origo, hence my function is differentiable here. That is wrong, eg computing the explicit expression at origion is somewhat pointless.
13:45
No no haha sorry about that.
True
It is enough like you say to say that the function is continous, and so forth.
Ah :) ok :)
That's the reasoning I used :)
But remember that you have to check the two cases.
Cases? Oh directional derivatives
test [ a+b ]
test \[ a+b \]
\( a+b \)
13:49
@N3buchadnezzar What about 6.? It is continuous and has continuous partial derivatives and therefore it is differentiable :)
Part 5 I don't really understand, just that partial derivatives are defined
Being differentiable and having continous partial derivatives are not the same thing.
But if the function is continuous, then it should be differentiable provided partial derivatives are continuous... ?
I mean the opposite, sorry. Continous partial derivatives implies differentiability.
However differentiability does not imply continous partial derivatives.
An example is this mathinsight.org/…
Yes that's how I understood it :)
@N3buchadnezzar Any other mistakes you spot in the answers that I have given?
If you could point out my mistakes, I can check the concepts once again :)
14:06
I do not see many "mistakes", since you mostly point out facts without stating why they hold.
Whys is this continous, why is it differentiable eg.
Thats the only "mistake" I see
Oh... but are the facts right? Sometimes I have some very weird intuitions haha. Like 5. was just guesswork in some sense :)
Yeah, it is correct. You just have to justify it =)
Haha ok :) Thank you for all your help :)
And all the useful pointers :)
Read the articles on mathinsights, they are really helpfull =)
I surely will :)
I actually have an exam on groups and symmetries round the corner... any advice ? :)
@N3buchadnezzar Apparently the answers are wrong. I was trying this online assignment and that says that my answers are incorrect... I don't know which one is wrong, though.
14:22
Dear @DanielFischer, please, save my soul and my mind!!! math.stackexchange.com/questions/556977/…
Norwegian Magnus Carlsen is the new world chess champion!
@nullgeppetto $S^n_{++}$ means symmetric positive definite?
@N3buchadnezzar I tried it again... I don't see why I am getting it wrong :(
@N3buchadnezzar \0/
14:39
I saw this inequality a while ago, stated as an obvious fact
$$ \frac{x - y}{\log x - \log y} \leq \sqrt{xy} $$
Is there any obvious argument as to why this holds?
It looks similar to the Lipshitz condition, or perhaps Jensens Inequality. But it is not quite the same.
@DanielFischer, exactly!
@N3buchadnezzar fix $y$ and let $x\to \infty$, does it make sense?
@PabloRotondo No?
@N3buchadnezzar Exactly. That couldn't have been the inequality, unless the source was wrong.
Hi :) How can I start a private chat with someone?
14:52
@PabloRotondo Silly me
$$ \frac{x - y}{\log x - \log y} > \sqrt{xy} \ , \qquad x>y$$
I posted it as a question instead =)
15:41
Anyone knows what would be the purpose of calculation of the cube of the adjacency matrix of the complete graph?
@theUg finding trips of length 3?
So, there is the practical purpose, besides purely computational masturbation?
I need to find A^3 via matrix tree theorem without actually computing it. And I don’t know where to start. :/
16:06
Greetings people!
@Chris'ssis Hi ChrisSis!
@RonGordon Hello! How are you? Long time I haven't seen you around! :-) (referring to chatroom)
Oh, yeah...I don;t think to come here often
I guess I should be more social
@RonGordon Yeah, that could be a good idea :D
@RonGordon We miss you long time!
16:10
Really? You get plenty of me on the main site...
I certainly know what most of you are up to
I always look at your answers and think "Hah! The physicist strikes again". In regard to your silly placement of symbols..
@N3buchadnezzar :)
Because of the people who frequent the chat, the topic often sway towards integration, sums and problem solving. So being absent from chat has a few downsides, including missing out on cake and soda.
Mmmmmm....cake...
I feel like I have been acting a little arrogant the past week or so. I apologize to anyone who has been offended by anything I have said or done.
^^
$$ \int_{0}^{\infty} \frac{x^{29}}{\left( 5x^2 + 49 \right)^{17}} \,\mathrm{d}x = \frac{14!}{2 \cdot 49^2 \cdot 5^{15} \cdot 16!} $$
Old present! (No beta, or gamma allowed)
16:21
Residue theorem...involving a pair of 16th derivatives
Oh fun!
Because it is one-sided, you need to introduce a log term or something like it
Easy to solve with a few elementary subs
I was wondering if there is some hope for computing this by real methods $$\int_0^{\infty} e^{-nx} (\cot(x)+\coth(x))\sin(nx) \ dx =\frac{\pi}{2}\left(\frac{1+e^{-n\pi}}{1-e^{-n\pi}}\right)^{(-1)^{n}}$$
@robjohn Why would it be so? I mean, I am trying to understand the sense of it, so I know where to look for the solution. Let’s say I have a complete graph K_5, then there would be no trips of length 3, would there? They all would be longer, because all vertices are connected.
I feel like I'd prefer to attend a series this beautiful evening. Maybe this one $$\sum_{n=1}^{\infty}\frac{1}{3^n}\left(\frac{\pi }{\pi^2-3\pi+3}\right)^{1/3^n}$$
16:29
@theUg Not round trips. The cube of the adjacency matrix would show where you could go in three hops.
@robjohn So, how exactly would I interpret this: wolframalpha.com/input/…
This crap is totally over my head. Graph theory was easy up until now.
@theUg You have three nodes, each connected to the other two. There are two paths to get back to the same node (e.g. 1->2->1 and 1->3->1 are the ways to get from 1 back to 1 in two hops) and only one way to get to another node (e.g. 1->3->2 is the only way to get from 1 to 2 in two hops)
And by 1, 2 and 3 in those → schemes you mean v_1, v_2 etc?
@theUg for that one, you can get from 1 to 1 in three hops in two ways: 1->2->3->1 and 1->3->2->1. To get from 1 to 2, there are 3 ways in three hops to get from 1 to 2: 1->2->3->2, 1->2->1->2, and 1->3->1->2.
@robjohn Okay, that makes sense now, but does not help me relate it to matrix tree theorem. I just do not understand relation fundamentally. And just looking at two pages of dense text of proof of this theorem just scares me.
And it kinda sucks too, because I hardly ever did matrices before (I have trouble with basic properties), and now this is on exam all of a sudden.
Oh, by the way, it is not for K_m, it is for K_4, so, at least, it is finite.
16:41
@theUg using matrices for graphs does not really require that you've done a lot of matrices in the past. As long as you know how to multiply matrices, that is usually good enough to get started.
Things I suppose I could use are adjacency, degree, incidence, what else?
@theUg I am not sure what you are asking. You mean what kind of matrices are used with graphs?
@robjohn Well, I get how to multiply them, but it is hard to keep track of operations (all the dot-products), since I had no practise with it.
@robjohn I am asking what can I use to solve my problem. What notions out of matrix tree theorem I should look at.
@theUg I am not sure what you mean by the matrix tree theorem, so I don't know what to say.
@robjohn Like I said, there is literally two pages of proof, and it boggles me. And all this just for a 1/4 point problem. So it should be something obvious, but I do not see it.
Kirghoff's theorem
In the mathematical field of graph theory Kirchhoff's theorem or Kirchhoff's matrix tree theorem named after Gustav Kirchhoff is a theorem about the number of spanning trees in a graph, showing that this number can be computed in polynomial time as the determinant of a matrix derived from the graph. It is a generalization of Cayley's formula which provides the number of spanning trees in a complete graph. Kirchhoff's theorem Kirchhoff's theorem relies on the notion of the Laplacian matrix of a graph that is equal to the difference between the graph's degree matrix (a diagonal matrix with ...
16:46
@Chris'ssis where do you get these sums from?
@RonGordon These ones I received from some kids that have fun challenging me. I usually teach them some things.
@Chris'ssis Nice.
@robjohn Degree matrix for K_4 is just:
@theUg Perhaps someone who is more acquainted with this may be of more help. I would need to do a bit of research to help more. Sorry.
3 0 0 0
0 3 0 0
0 0 3 0
0 0 0 3
16:48
@Chris'ssis I assume you know that this has a nice answer?
@robjohn Yes.
@theUg That makes sense.
@robjohn So I wonder if there is something from making Laplacian matrix out of those [D - A]
Though, I suppose it would be just:
3 1 1 1
1 3 1 1
1 1 3 1
1 1 1 3
Rather, -1s
3 -1 -1 -1
-1 3 -1 -1
-1 -1 3 -1
-1 -1 -1 3
@robjohn Then there's this. Is this where I should look?
0
A: How to calculate square matrix to power n?

IuliOne method is induction. Another way to calculate $A^{n}$ for a $2 \times2$ matrix generally is Hamilton-Cayley Theorem: $A^{2}-Tr(A)\cdot A +\det{A} \cdot I_{2}=0$. This is a very useful theorem which can be applyed for any $n \times n$ matrix. for example if you have a $2 \times 2$ matrix wit...

@Sushma Hi there
I am running the 35 version on pari/gp right now - but it might be slow
and it has occurred to me that there might not be any solution at all - depends on quadratic residues, I recall
16:59
Sir, I tried the same code in pari/gp 2.5.5 but for n=35, it didn't give
@Sushma That might mean that a solution does not exist :(
ok, quite possible, but that prime is cong. to 1 mod 35, so I thought, a soln might exist
I think it might depend on whether 35 is a square mod p
ok I will check
35 seems to be a square mod p
17:03
yes
and 21 is also -
yes, for that I got
as 442263555, 97162880914
I meant for 21
ah yes - just seen that
I think 23 is going to fail - it is not a square mod p :(
ok, thanks
and it looks like 35 is taking some time on my machine here - but I can leave it running for a long time if necessary
17:06
i was using nzmath, for python developed in tokyo metropolin university, for these type of problem,
yes sir, don't take trouble please
@Sushma It sounds as if GAP might be a better method from one of the other comments on the question
@Sushma Oh - it is no trouble :)
ok, so should I code it
in Gap
or a command exists
ok, I can install Gap and check
@Sushma Just re-read the GAP comment - unfortunately, gap only seems to give $x^2+y^2$ representation
yes I think so
pity
17:09
why pity?
it is a pity that it does not seem to have a quick way to do representations in the form $x^2+dy^2$
no, when I was without any answer I got your answer, so will wait
last year, I had run the command in nzmath to get a representation of a prime no. with125 digits, but this is not a huge number I do not know why this time it is taking such a long time
There might be another reason why a solution is impossible ... did you know a guy called Cox wrote a whole book on "Primes of the form $x^2+dy^2$"?
I have a copy - and it is a hard subject!
yes I read that book, this is the side effect of that);
@Sushma I haven't finished it - don't know enough algebra, as I am basically an analyst
17:18
I didn't finish the full book; the last chapter is still remaining, of complex multiplication, rest I finished, but initially I didn't understand, now slowly knowing that subject
@OldJohn Hey Old man =)
@N3buchadnezzar Hi there young fella
I also read a paper by Lenstra and stevenhagen on"Mersenne primes and artin reciprocity", which is nice,
@Sushma sounds hard
no, it is a joy, gothrough once
17:20
@Sushma excellent
@OldJohn yes really
@OldJohn Do you remember how to add spoilers for answers? I tried <! but alas.
@Sushma I will have to go shortly - will be back later, and I can leave my program running
@OldJohn, have a good day(night), thanks
@N3buchadnezzar not sure off-hand, but I am sure Makoto Kato did something similar recently, so it might be worth looking at his answers to his own questions
@Sushma Thanks - and if you want to contact me, my email is on my profile page
17:23
@Chris'ssis I am trying something that looks promising, but just to check, I ran this through Mathematica and it gets an answer of $0.4886437056940325001$ and the ISC finds nothing for that. So the nice answer is not nice enough for the ISC :-)
@OldJohn, thank you nice of you, Bye will in next chat
@Sushma Bye for now - speak again sometime
@theUg I am sorry. I just am not familiar enough with this to make a suggestion.
@robjohn Oh, the answer looks nice $$\frac{1}{\log(\pi-2)}+\frac{1}{3-\pi}$$ :-)
@Chris'ssis Hmmm... That doesn't match what I got on Mma
17:33
$S \otimes_R R \approx S$, where $f : R \to S$ is a given hom of rings
@Chris'ssis Do you have a proof that this is the sum, or is this what you were told is the sum?
@robjohn I was told this is the sum. I'm sure this must be the sum. (well, not 100%)
@Chris'ssis The series converges fairly quickly, so I believe Mma's approximation, but it only matches the answer you give above to 3 places.
@robjohn Yeah, I noticed that. I don't know what to say at the moment. I'll ask for the answer again.
Am I on a wild goose chase trying to separate $\frac{1}{x+x^2}$ into something "regular"
17:46
@MickLH Do you have ChatJax installed?
not on this computer, let me grab the bookmarklet
@MickLH $\frac1{x+x^2}=\frac1{x}-\frac1{x+1}$ Is that what you are looking for?
I believe so, I am trying to encode polynomials for a computer program, let me get back to you in a moment
derp I don't know how I didn't notice that, thanks!
@Chris'ssis If we let $\pi$ be a variable and let $\pi\to3$, then that function limits to $\frac12$ which agrees with the sum if we substitute $3$ for $\pi$...
@MickLH anytime :-)
@robjohn I see.
17:58
@Chris'ssis: I got an upvote today for my answer to one of your questions. It is 9 months old, so it was a bit surprising.
@robjohn Let me see. I missed that.
@Chris'ssis The high-school level answer, not the EMS one :-)
@robjohn Oh, this is an amazing answer! math.stackexchange.com/questions/307117/… I like that very much! Indeed! It's an elementary approach.
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