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r9m
12:00 AM
@mixedmath ya .. but I have to wait for a long time before it finally becomes available for free
 
@r9m Is there something in particular that you're looking for?
 
if you're affiliated with an institution (e.g., are either an undergraduate or graduate student at a university) they may very well have access to this journal
 
r9m
@mixedmath Crux with Mayhem vol. 36 (5) pages 321–324 :)
@MikeMiller I'm 1200 km away from my insti at this moment (summer vacations going on)
 
ah
well, if you like walking...
 
r9m
@MikeMiller :P .. it will be the end of my vacation by the time I reach there by foot :P .. I can just wait in that case till the last day and get a flight :P
 
12:17 AM
Oh, well, I'd been thinking about becoming a member for a long time, and I'm now a subscriber to Crux... but they're going to snail mail me my online access...
so I'm sure that in a mere 2 weeks, I'll have access
 
@mixedmath Hah!
 
Test
How do I change my name on this chat? My stackexchange does not have this name :c
 
@mito_562condrio How long have you been a user?
 
@mixedmath for months!
 
@mito_562condrio Wow, your profile and background make absolutely no sense to me and my moderator powers
I've got nothing - I'm sorry
can you link me to any of your parent users on one of the SE sites?
oh wait, that's weird
are you Lucas?
 
12:23 AM
@mixedmath yes!
 
@LucasZanella Ok, I've done what I can do. I don't actually quite know what will happen. I foresee three possibilities: maybe if you logout and login, your chat name will be correct now; or maybe it will update within 24 hours or so, as many things sort of work like that; or maybe I didn't do anything successfully and you should click the "contact us" link at the bottom of one of the main sites to talk to the real SE mods
oohh, but I'm optimistic, as responding to your post used your name
so there's that
 
@mixedmath Thank you so much :)
I'm gonna log in again
 
ok
 
:D
@mixedmath Thank you so much, again, now it works
$\frac{1}{2}$ testing LaTeX
LaTeX is not accepted here?
 
Not natively. But if you use the super special workaround from other mod robjohn, it works fine
then it'll display for you
 
r9m
12:29 AM
@mixedmath can you give me a sneak peek at those 3 pages after you get subscription ? :D
 
Ok, thanks by all, see you guys another time!
 
MJD
12:40 AM
Suppose $P$ is a permutation that consists of a single cycle, and is not a product of two or more smaller disjoint cycles. Is it normal to say that $P$ is “cyclic”?
 
@MJD That doesn't feel natural to me at all. I'm much more inclined to call $P$ a "cycle"
@r9m Sure. If you pop me an email, I'll actually remember, too
 
MJD
Thanks. I see Wikipedia is iffy about it also; it has an entry for cyclic permutation which says it sometimes means what I said, and sometimes means something more general.
 
r9m
@MJD well a cyclic permutation is a single cycle
@mixedmath if I ping you in this room .. will you get notified ?
 
@r9m umm... that's a good question
probably - I'm here often enough for that to probably work
 
r9m
@mixedmath okay .. ur blog had your mail address .. :) .. thanks :D
 
12:51 AM
@r9m yes indeed!
 
 
1 hour later…
2:07 AM
@r9m there is probably a VPN that you could use to access it remotely
 
Mew
how can we show the fourier transform of e^(2pikt) is the dirac delta of (k1-k)
 
@AlexanderGruber I was surprised that my university doesn't have electronic access to the new Crux stuff
I've sort of been taking it for granted that I have access to almost everything I want math journal wise
 
r9m
2:23 AM
@AlexanderGruber my institute doesn't have Crux subscription :(
 
2:38 AM
@mixedmath They don't? Mine does, curiously enough.
 
Yeah, shame on my university. I also have enjoyed every copy of the Mathematical Gazette that I've read (from the UK), but as far as I can tell, it is not distributed electronically at all
which is too bad
 
r9m
3:19 AM
@Chris'ssis Have you seen these two ?! :D

$$\displaystyle\sum\limits_{m=1}^{\infty} \sum\limits_{n=1}^{\infty} (-1)^{m+n}\left(\ln 2 - \sum\limits_{k=1}^{m+n}\frac{1}{m+n+k}\right)$$
and $$\displaystyle \lim\limits_{n \to \infty} (-1)^n \dfrac{n^2}{n!} \sum\limits_{k=2}^{n} {n \choose k} (-1)^k k^{n-1} \ln k $$
The second one is interesting .. :)
 
 
2 hours later…
4:56 AM
anyone here?
 
r9m
@user52932 :)
 
I have a question regarding closed, open stuff.
one sec. I'll type it
 
@user52932 still typing? :D
 
Define $C=\{f:[0,1]\to[0,1]\}$ such that elements of C are continuous. Then define a metric on $C$ of the form $d(f,g)=\max_{x\in[0,1]}|f(x)-g(x)|$. Let $C_i$ and $C_s$ be the set of injective and surjective elements of $C$. Prove or disprove that $C_i$ is closed and $C_s$ is closed.
sorry, 2nd one is correct
 
All good, you could always edit instead of rewrite :P
What have you tried to do?
Also, you're using the usual topology for a metric space I'm guessing?
 
5:02 AM
my question is that how do I approach such questions in general. I know the definitions of open and closed. The way I see it there are two ways. Either I can go about saying $C$ is a subset of larger space of continuous functions and try and show that $C_i=D\cap C$ where $D$ is closed in larger space
Or I directly go into trying and proving that complement of $C_i$ and $C_s$ are open by selecting a point and trying to construct an open ball around it that it lies completely in the complement.
This is a sample question from one of the past exams. My exam is on monday. The issue is that I don't understand how to approach such problems in general spaces (ball approach or set approach) and I do not even know whether I will just be wasting time or not because the thing might be false
 
r9m
@user52932 how about the sequence of functions $f_n(x)=x^n$ ?
 
this is like one of the 10 qs to be done in 3 hours
 
I personally always shoot for the ball approach first
@r9m raises a nice point, that sequence lives in that set, but the limit is definitely not in the set so it cannot be sequentially compact $\implies$ its not compact in general so it cannot be closed
 
@r9m are you implying that this sequence of functions will have a limit that will not be injective?
 
r9m
@user52932 its a sequence of functions in $C_i \cap C_s$ .. ie bijective (injective and surjective) functions that does not have limit point in $C$ .. so neither $C_i$ or $C_s$ are closed .. I guess
 
5:09 AM
@r9m nice catch!
so theres a third approach for you, look at sequences!
 
r9m
:D
 
@r9m wait why is the limit not continuous?
sorry for being dumb...i wish you guys were giving this exam
 
@user52932 the limit is the function $f(x) = 0, x \in [0,1), f(x) = 1, x = 1$
 
oh yeah...this is that classic case
true...nice
@r9m how did you come up with that so quickly
 
presumably you're not in a topology course, otherwise you would need to take into consideration a couple of other factors (you would need to make sure that it's hausdroff)
 
5:12 AM
No I am in an analysis course
 
very nice! I'm out of school currently, studying on my own, but I love those sorts of questions!
 
yeah I'll be graduating soon and studying math on my own too. I want to do some measure theory and probability theory. I am not that good at proofs though :(
Like I can understand them when they are given in the book and they seem ridiculous
but i lack the creativity to come up with own proofs
:(
Do you guys think this set $C$ is connected?
 
Intuitively I would say yes, but I don't know a proof for that right now
so it might not be true
Assume there exists two disjoint subsets and draw a contradiction from the continuity?
 
I have another general question too. Sometimes on exams you get questions where they give you weird sequences of functions and then they ask you to evaluate limit of the sequence etc...for example consider this sequence
$f_n=\frac{sin(\frac{x}{n})}{x\frac{1}{n}}$ and then question asks what happens to $f_n'$ and $f_n$ as $n$ gets large.
and what would be the limits...
is there a way to do this stuff analytically and directly
like how to approach this
 
r9m
@user52932 well a convex combination of two elements $f,g \in C$, $tf+(1-t)g \in C$, so $C$ is path-connected I guess ..
 
5:36 AM
@r9m I see. So in general to check connectedness, it is good to check if the set is convex. Becuase that will guarantee path connectedness and hence connectedness (i.e. $C$ cannot be union of two disjoint sets)
 
r9m
@user52932 :)
 
For that sequence you can use l'hopitals rule and come to see that the limit is 1 I believe
(For both f,f')
Sorry for lack of detail, on iPad as my laptop decided to stop charging
 
thats fine. ill try l'hopital. its so hard to know all these theorems
 
r9m
5:54 AM
@robjohn is it true that $C_0([0,\infty)) = \{f: f \text{is continuous} and \lim\limits_{x \to infty} f(x) = 0\} = \overline{V}$, where $V=\{e^{-x}P(x): P\text{is a polynomial}\}$ ?
 
r9m
6:07 AM
@robjohn is it true that $C_0([0,\infty)) = \{f: f:[0,\infty) \to \mathbb{R} \textrm {is continuous and} \lim\limits_{x \to \infty} f(x) = 0\} = \overline{V}$ where $V=\{e^{-x}P(x) :P \textrm{is a polynomial}\}$ ?
sorry for multiple pings ..
 
6:59 AM
yeah, l'Hopital on that :D
 
@r9m looks right. Look on compact sets and use Stone-Weierstrass.
 
Hi @robjohn :-)
Have you ever heard what "reverse-math" is about?
 
Techmology wots it ol abowt?
 
user116900
7:48 AM
I just answered 2 lhfs.
 
8:18 AM
@skullpatrol nope. never heard of it.
 
@robjohn Thanks for responding :-)
 
oh how leet, a math chat, with latex even. greetings all ;)
 
so what's the difference between this and the math overflow chat? my first guess is that this room gets more homework questions and drive-by shootings and such
 
@DavidKirby Hello, there are usually no homework question asked here, rather questions from people self-studying a given topic
 
8:32 AM
lovely.
well then, i'm sure i have questions; but none at this moment... it's 4:37 am in georgia, i'm listening to minimal techno, and reading this recent paper on gamma error bounds : arxiv.org/pdf/1310.0166v2.pdf ;)
 
Hi all! If you like, please take a look!
 
8:52 AM
 
@karp2345, nice!
 
@nullgeppetto Can you guess what it is? ;)
 
@karp2345, maybe a \LaTeX exercise? For very bad students?
 
expand expand expand expand ....
 
@nullgeppetto Nah, its the angle X in terms of w,x,y,z in the figure here:
 
8:55 AM
@karp2345, it seems legit! Really!
 
:D
 
9:11 AM
:D
 
@karp2345 the two angles marked "x" are radically different
 
I use to be able to handle the drive-by shootings @David Kirby until I took an arrow to the knee :D
 
9:31 AM
@MikeMiller You here?
Inspired by the coverings of compactified modular curves, I was wondering if similar can be done in general, i.e., can galois groups be realized as galois coverings of compact Riemann surfaces?
 
A wild @Sab appears
 
9:47 AM
@robjohn One is capital and the other one is small x :-)
 
9:59 AM
@robjohn Heya
 
@N3buchadnezzar howdy
 
You showed me that $$ \int_0^{T} f(nx)\,\mathrm{d}x = \int_0^{T} f(x) \,\mathrm{d}x$$
Given that $f(x+T) = f(x)$.
What can one say about the sine case?
$$
\int_0^{T/4} f(\sin n x) \,\mathrm{d}x
$$
The $T$ case is clear, but what abut $T/2$ or $T/4$?
It is already know that
$$
\int_0^{\pi/2} \log (\sin 2x) \,\mathrm{d}x = \int_0^{\pi/2} \log( \sin x)\,\mathrm{d}x
$$
So it must hold in some cases, even when not integrating over a whole period :p
 
10:16 AM
Hello everyone
 
hola
 
What approach should I use to compute this equation? arcsinx+(x-1)/(sqrt(1-x*x))=0. I need only hint, not solution, please.
 
Greetings
 
Greetings
 
10:18 AM
hi
 
@r9m I did the first one with the tools of the high school. The second one was studied by Gosper.
 
kirill: hint, a constant
 
@DavidKirby So I should do something with zero...
@DavidKirby Thanks, let me think.
Still can't get it
 
I brought cake to uni today =)
 
kirill ... think, harmoniously.
 
10:31 AM
@N3buchadnezzar here, the period of $f(\sin(nx))$ is $2\pi/n$, not $T$.
 
@DavidKirby it's something not traditional, right?
 
@N3buchadnezzar so I doubt there is much you can say about the integral over $[0,T/4]$
 
@robjohn I was more thinking of trying to figure out when
$$
\int_0^{\pi/2} f(\sin nx ) \mathrm{d}x = \int_0^{\pi/2} f(\sin x)\,\mathrm{d}x
$$
And the case $\pi$ instead of $\pi/2$
 
by traditional I assume you mean rational?
 
10:37 AM
@DavidKirby right C:
@DavidKirby Are you about 0=arcsin(0)?
 
@N3buchadnezzar For a general $f$, it's true for $n=1$ and $n=2$, but not for any other $n$.
 
Yeah, thats what I saw
$$
\int_0^{\pi/2} f(\sin 2x) dx
= \frac{1}{2} \int_0^{\pi} f(\sin x) dx
= \int_0^{\pi/2} f(\sin x) dx
$$
 
@N3buchadnezzar if $f$ is even, then it is true for all $n\in\mathbb{Z}$
 
@robjohn Even about the origin $f(x) = f(-x)$ ?
 
@N3buchadnezzar If $f$ is odd, then for $n=-1$ and $n=-2$ as well
 
10:45 AM
Can I have one more little hint?
 
kirill, it's not known whether it's irrational
 
@robjohn Sure about this? I checked a few integrals in maple, and they do not agree numerically. I checked $f(x) = \log(1+x^2)$ for a simple example
 
@DavidKirby what do you mean?
 
well, if it's rational the denominator is huge...
a quarter of a million zeros at least
 
@N3buchadnezzar I just checked it with that function and $n\in\{1,2,3,4\}$ all give $0.591331$
 
10:50 AM
@DavidKirby When you said constant you meant 0 or 1?
 
oh, no, a named constant.. you're looking for the root right?
 
Mmm strange I get
 
@DavidKirby Exactly
 
That's $f=\log(2+x^2)$, but it should be the same
I get $1.4233571556713879977$ with that $f$ for all $n$
 
10:58 AM
@robjohn I agree, I checked with another program. Strange that maple makes such errors.
 
I found a marvelous algorithm from comparing size of power towers.
It's really quite cleverly done, especially, the trick $\log(a+b)=\log(a)+\log(1+b/a)$
 
kirill, ok, here's another hint. look for the root between 0.5 and 0.55
 
@DavidKirby I know the root, but I need way to find it
@DavidKirby thank you anyway. I keep trying.
 
@robjohn How did you figure out the even general case?
 
@robjohn could you delete my answer above?
 
11:06 AM
@Chris'ssis ah, good morning. I didn't see you here... which answers?
 
@robjohn Hello
The one above to @r9m
 
@Chris'ssis Why ? =)
 
@Chris'ssis done
 
@robjohn Yes
@N3buchadnezzar Thanks!
@N3buchadnezzar I think I missed something ...
 
kirill: well, it seems unlikely that euler-mascheroni is algebraic, but by all means -- keep trying. when you run out of meth continued fractions will be waiting for you.
;)
 
11:16 AM
Is $\text{Res}[f(z), \infty]= \text{Res}[f\left(\frac{1}{z}\right), 0]$, assuming that $f(z)$ converges for $|z|>R$ for some $R$? I have an example that seems to indicate not, but I don't know why.
 
@robjohn hehe, it was correct though :-)
@r9m for the first series it's $$1/8(\log(32)-\pi)$$ The second one comes from the Gosper's work.
 
kirill: just curious -- what's the application? is it school, or for software, or simply the joy of discovery? the method, especially in this case, really depends on the precision needed for the application.
 
it's college problem
Part of it
 
I think I got it =)
wop
 
And probably I am using wring approach to solve it, may be I don't need to find the root of this equation
@DavidKirby So, I have function f(x)=(x-1)arcsinx on [-1,1], and I need to verify somehow that the zero(s) of f represent points where function has tangent lines
 
11:33 AM
ah, now that's a great deal easier.
 
@DavidKirby Sorry for confusion.
 
split into two equations
arcsin(x)=0 and x-1=0
exactly
 
Why they should be equal to zero?
 
solve for x, so you have x=1 and arcsin(x)=0
 
Yep, where x = 0
 
11:36 AM
now eliminate the inverse sine
then take the sine of both sides
 
Sorry, how should I eliminate that? Can't get it
x(x-1)=0 ?
 
correct
there are two solutions, x=0 and x=1
 
So what we have now? I am lost
I will just find the roots for y = 0
 
@robjohn Seems like the same result holds for cosine as well
 
kirill, right, at y=0,x=0 and y=0, x=1 --- the arc of the sine passes through these points
 
11:44 AM
So...
?
I need point of tangent line
@DavidKirby Ok, thanks. I will do try to do it myself
 
ok, so then the next step would be to find the domain and range
you already know the domain -1 <= x <= 1
 
12:07 PM
In my second solution to the above question I met another series you might like ...
$$\sum_{n=1}^{\infty} (-1)^{n+1} \left(\log(2)-\frac{1}{n+2}-\frac{1}{n+3}-\cdots -\frac{1}{2(n+1)}\right)$$
 
kirill, i'm assuming you've got it now.. ;) d/dx((x-1) sin^(-1)(x)) = (x-1)/sqrt(1-x^2)+sin^(-1)(x)
 
Since everything goes perfect here I'm singing a song ... "I'm feeling like a star"
:-)
$$\sum_{n=1}^{\infty} (-1)^{n+1} \left(\log(2)-\frac{1}{n+2}-\frac{1}{n+3}-\cdots -\frac{1}{2(n+1)}\right)=\frac{1}{8} (2 \log(2)-4+\pi)$$
 
user116900
12:22 PM
;) is often used by the user I hate most on this site.
 
:-) ?
smiley semantics, oodles of fun
 
@JasperLoy don't use the word "hate" pal, it is too extreme
 
@N3buchadnezzar for even $f$, yes. Do you see why? The integrals should also be the same.
 
user116900
@skullpatrol I am full of hatred, it's true. I am not a saint.
 
12:24 PM
@JasperLoy no one is a saint
@JasperLoy hate destroys the hater
 
user116900
Among my two lhfs today, the more sophisticated answer got 0 votes compared to the other one which got 2, lol.
 
@JasperLoy Hate not only destroys people around us; it also erodes our own character. Never let hate engulf your mind.
 
user116900
@skullpatrol OK.
 
:-)
 
on the topic of hate, ... log(2)
 
12:31 PM
@DavidKirby What's to hate about that?
 
oh well, it's that one constant that i always forget to try, and inevitably it's the constant that i needed.
 
user116900
I demand an upvote for my answer at math.stackexchange.com/questions/799993/…
 
@JasperLoy you are very demanding
 
user116900
@skullpatrol Yes. I rather demand than delete the answer. I have a habit of deleting answers at 0 votes after a while.
 
why?
 
user116900
12:41 PM
Just a habit.
 
um, how very log2 of you
so you guys all answer questions? i can't imagine ever having the time... it takes a while to put those beautiful answers together i'd assume
 
user116900
I only answer low hanging fruits @david
 
watch out what you say in here pal,
you could be the next victim of a drive by
 
oh no i've already been shot and i almost got shot earlier when that absurd expansion of a step function was posted as an answer to a problem with no explicit answers
from now on i'm just throwing fruit and making topology jokes
 
12:48 PM
@Alyosha As I understand residues in $\Bbb C \cup \{\infty\}$, residue of $f(z)$ at $z = \infty$ is the residue of $-1/z^2f(1/z)$ at $z = 0$. This is explained if you think of the contour-integral definition of residues. [$f(1/z)d(1/z) = -1/z^2f(1/z)dz$]
 
@BalarkaSen i thought you said you were going to take some time to do some serious math?
 
@skullpatrol I am answering math.
 
@BalarkaSen you said away from here pal
 
I believe that falls into the category of "doing math"?
@skullpatrol Away from general room, I think.
 
12:51 PM
And, I also said I won't chat useless. So being silent a bit.
 
yes
 
'doing math' --- my girlfriend calls it 'playing math', which has always sort of bothered me... if it's the weekend and i'm working on math, she rolls her eyes and says, 'oh, you're playing math, call me next month', and i'm like ok awesome, meanie.
'playing serious math'
alright, here's a bad math joke and then i actually do need to go do some math :
Q: Why did the cat fall of the roof?
A: Because he lost his mu.
 
that's why mathematicians have both a wife and a girlfriend
 
teeeheee
 
1:08 PM
because people are mean
 
Is it wrong?
 
@blue Downvotes are the most random thing. So don't worry about it.
 
@Sawarnik You think there is no reason for the downvote?
 
@blue If they don't leave any comment, and there is apparently nothing obviously wrong, then yes.
 
That's weird.
Why did you delete it?
 
1:18 PM
:D
 
:D
 
:D
 
:-D
 
@blue Is it true that if $f(x)$ is continuously differentiable, then $e^{f(x)}$ is differentiable?
 
Yes, there is a theorem that if $f,g$ are differentiable at $x$ then $g \circ f$ (the composition) is differentiable at $x$.
of course $e^x$ is differentiable.
 
1:21 PM
Hmm, is it easy to prove?
 
Not as easy as the theorems for $f + g$ and $f / g$ but it's not impossibly difficult.
 
Ooh ok.
 
The proof is a calculation of the derivative: it is $g' (f) \cdot f'$.
It is called the chain rule.
There is probably more than one way of proving it.
 
does anyone know how $\sum\limits_{n=0}^{\infty}\frac{a^n}{n!} = e^{-a} $ ,$a$is a integer
how do i enable latex
 
@Danny I don't know but maybe you can show that the series converges and then that the derivative also converges and that the derivative equals the original sum. Then maybe apply the theorem that gives uniqueness of solution of a differential equation.
@Danny I don't know.
 
1:29 PM
ok
 
@BalarkaSen Okay.
Is there a way of formalising that a circular contour as its radius goes to infinity is effectively a contour around infinity?
In the sense that $|z|=1$ is a contour around $0$, that is.
@Danny i.gyazo.com/7e871aba5da827f44e8803f74d20522c.png click 'LaTeX support for chat'.
 
yup, blue is right, you can use the cauchy ratio test to show that the series converges
 
ok
 
@blue By any chance, are you interested in Euclidean geometry?
 
1:48 PM
@DanielFischer @robjohn please help out with this problem. I am stuck on this for quite some time, math.stackexchange.com/questions/800327/…
@Sawarnik Hi!
 
@Hawk Have you shown it for all triangles?
 
@DanielFischer No, I couldn't show anything, I just verified it for a few cases, and found to be true...
 
@Hawk hey!
@Hawk After so long..
 
@Sawarnik Yes...was reading a few things...
@DanielFischer I think, I need to use step function over here...
 
@Sawarnik "May be related" is an understatement.
 
1:53 PM
@DanielFischer Yeah, I had just seen the title, so I thought of linking this here, before reading the question.
 
@Hawk Show that if the formula holds for two polygons that have a common side, it also holds for the union of these polygons. That allows you to reduce the general case to the case of triangles.
 
@DanielFischer you mean, triangulation of a polygon?
 
@Hawk If you want to give a big name to it, yes.
 
c c
2:16 PM
What is a space with a negative dimension? what does it mean?
 
@Sawarnik I'm sorry I don't know what it is. Is it geometry with triangles and circles and so on?
 
@blue Yea.
 
@Sawarnik I don't know much about it. Do you have a question about it?
 
@blue No, actually there is another user name Blue here, and his answers, many in geometry, are really awesome.
:D
 
2:37 PM
@cc does it really matter what it means IRL? can you generalize to $n$ dimensions and plug in a negative $n$ ?
 
@Hawk as ronno says, this looks like Pick's Theorem, but in your question, you never say that $Z$ is the number of interior lattice points and $B$ is the number of boundary lattice points.
 
2:59 PM
@Alyosha No, that'd result in the whole complex plane. But we use an annulus instead, with external radius tending to infinity.
 
3:21 PM
@BalarkaSen Did you see this one? $$\sum_{n=1}^{\infty} (-1)^{n+1} \left(\log(2)-\frac{1}{n+2}-\frac{1}{n+3}-\cdots -\frac{1}{2(n+1)}\right)=\frac{1}{8} (2 \log(2)-4+\pi)$$
You might like it.
 
@Chris'ssis Ah, I think I have an idea for that...
 
@robjohn Related to integrals?
 
@Chris'ssis No, just series.
 
@robjohn Great then.
 
@Chris'ssis I find it less interesting than the series you usually create =)
 
3:25 PM
@BalarkaSen This is just a small thing to compute in a much more complex question. I think is pretty enjoyable.
 
@Chris'ssis Like?
 
@BalarkaSen see @r9m's questions to me above.
@robjohn If you thought of the summation by parts, then it should work too.
 
3:47 PM
Is there a general form for the solution of a Quintic with elementary functions?
 
@Alizter Depends on what you mean by "elementary".
 

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