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16:00
Hello,moderators?why this post can't comments and can't answers and can't edit and so on? that's why?
Contest questions, for contests that have not yet concluded, are not allowed. Sometimes moderators close questions that look like they might fall into that category, to be safe.
@robjohn,@DanielFischer hello,moderators,
@Australia did you read the link given in the question lock notification? here
sorry, some people say this inequality is 2015 TST,and I believe it,But after survey in fact, before contest problem
I would like to lift this restriction,
morning @Mike.
16:10
morning
This is old problem ,so and is not 2015 TST,
@australia If you are able to find proof of the problem's origin, I expect the restriction will be lifted with no problem. Given the nature of the question however, and your own confusion as to whether the problem does in fact come from a soon to begin contest, the moderators may be inclined to keep it locked until after the contest has ended
Math can wait a few months. If it is something so important that it needs to be done now, then I recommend cracking open a good book to study from.
@DavidWheeler @DavidWheeler : $A_1=\{(x,y)\in \mathbb{R}^2, y>x^2\}, $A_2=\{(x,y)\in\mathbb{R}^2, y<x^2, x<0\}$ and $A_3=\{(x,y)\in \mathbb{R}^2, y<x^2, x>0\}$ if it is true what is the condition on $x$ in $A_1$ ?
@Australia As JMoravitz said, if you can give a publicly available source for the problem, so we can check it's an old and known problem, the question can be unlocked.
Your characterization of $A_2$ and $A_3$ are incomplete you should have: $A_2 = \{(x,y) \in \Bbb R^2: 0 \leq y < x^2, x < 0\}$ and $A_3 = \{(x,y) \in \Bbb R^2: 0 \leq y < x^2, x > 0\}$
16:18
@DavidWheeler that's exactly what i write
@Australia But since you wrote that it is from 2015 TST in the question yourself, it may be decided to leave the question locked nevertheless.
first see on china QQ ,but I can't find origin place,if you can see [tieba.baidu.com/p/3629405125]
@Vrouvrou No, you just wrote $y < x^2$, not $0 \leq y < x^2$
I also heard someone say it is 2015 TST.so I writer in AOPS
@DavidWheeler yes sorry , what about x in A_1 ?
16:20
Ooh, controversy. yawns
@Vrouvrou We don't need to add the condition $y \geq 0$, there, if $y > x^2$ it's already positive, since $x^2$ is always non-negative.
But to prove those sets are connected, you're going to want to assume the contrary, and derive a contradiction.
wel...
1 + 1 =2
@Australia That one is later than your(?) AOPS post. We'd need something older than that.
Sorry, I thought I was smart.
odd only find china QQ
I first see china QQ
@Australia if you don't mind my asking, what is your native language and why are you interested in this problem specifically?
16:26
@ᴇʏᴇs You here?
I fist see in 3.10.benjing time 9.10.am
@Australia That's a different inequality.
No,that's same inequality
\dfrac{a^2+b}{a^2+b^2+1}=1-(b^2+1-b)/(a^2+b^2+1)
Ah, right.
you can see this time, it's old than this two,As far as this blog master like creat some inequality
16:42
@Australia even if it has been posted elsewhere, the moderators may still decide to lock it until the contest is over.
@DavidWheeler Yes thanks for the help. Hard to switch back and forth between windows on a cell phone
No,I interesting this inequality,because It's nice,But I hope someone can solve it
You can think of it this way @ᴇʏᴇs cyclic groups are special because they decompose in just this way. An element of order $d$ (where $d|n$) is a generator of a subgroup of order $d$.
This Euler totient function is pretty useful. Wonder why the professor just didn't teach it to us instead of making everything so confusing
I Hope moderators can lift this restriction,
if this is 2015 TST,I can take all the blame
16:47
It is useful, it's one of the reasons RSA "works"
@DavidWheeler Well the context is finding the roots to a quadratic mod p and cyclic groups came up although I don't know what a group is
@Mike nice results, in case you are interested.
provides an efficient way to compute fundamental group of non-nice orbit spaces.
Cyclic groups do not need the full abstractness of group theory to be explained-they are just the integers, or the integers mod n.
@DanielFischer
16:50
You can think of cyclic groups as things that act like integers on a circle. If the circle is infinitely big, we get the ordinary integers.
The name "cyclic" is supposed to remind you of "cycle".
$$\int_0^{\phi} \frac{d \theta}{\sqrt{1-\cos{\phi}\cos\theta}}=\int_0^{\pi/2} \frac{d \theta}{\sqrt{1+\cos{\phi}\cos\theta}}$$
@BalarkaSen ^^^
17:08
@DavidWheeler we have that $A=A_1\cup A_2\cup A_3$ right
Algebraically, how can I find a equidistant point between two intersecting lines?
@Vrouvrou Ya, sure.
17:40
and yhey are trivially open
so i must prove that they are connected
you need to prove each ONE is connected.
$A_1$ is easy, it's a convex region. $A_2$ is harder, but you don't need to prove $A_3$, because it is homeomorphic to $A_2$.
ooo A_1 is easy
by contradiction if i say that there exists two open disjoint sets U,V in A_1 such that A_1=U\cup V
where is the contradiction ?
17:56
@KalaJ I feel like that question was asked recently on the site, but I cannot find it. The simple explanation is that the point you seek is along a line which bisescts the angle of the two intersecting lines.
If that is not the point you seek, then you will need to clarify, equidistant to what?
@DavidWheeler hey, I gotta quick question: math.stackexchange.com/q/1194949/197705 mind having a look?
@DavidWheeler ?
what i must see please
18:36
0
Q: How can I find the curve?

Mary StarFind the curve $\overrightarrow{\sigma}(t)$ that describes the following curve or trajectory. Make a graph. $$\{(x, y) \mid 4x^2+y^2=1\}$$ How can I find such a curve??

Any idea about how to prove $$\sum_{\gcd(a,b)=1}{\frac{1}{(ab)^2} }=\frac{5}{2}$$ ?
@StanShunpike I don't know what you mean by "in order"
@Vrouvrou try showing $A_1, A_2$ are path-connected
19:00
19
A: How to evaluate $\sum_{\gcd (p,q)=1} \frac{1}{p^2q^2}$?

Daniel FischerLet's look at the more general case, $$\sum_{\gcd (p,q) = d} \frac{1}{p^2q^2}.$$ $\gcd (p,q) = d$ means we have $p = d\cdot a$ and $q = d\cdot b$ with $\gcd(a,b) = 1$, so we obtain $$\sum_{\gcd(p,q) = d} \frac{1}{p^2q^2} = \sum_{\gcd(a,b) = 1} \frac{1}{(da)^2(db)^2} = \frac{1}{d^4}\sum_{\gcd(a...

@DavidWheeler i can't use path-connectedness
because...?
it is in the end of the chapter
xD
if it is the only way why not
but what is the mane idea of path connected ?
please @DavidWheeler
19:08
you show there is a path entirely in the set connecting any two points
@DanielFischer Hi, I am working on my course on complex analysis and we have a theorem : if $f:D\rightarrow \Bbb{C}$ is continuous, and $D$ is a domain, and if the integral is equal to zero for any closed path $C^1$ piecewise then $f$ admits an antiderivative.
We first proved that the existence of closed path piecewise $C^1$. I was thinking on why it's not immediate ? Can we have the non existence of a such path ?
@Gato Right. Is the proof you have unclear?
@Gato Sounds odd. The existence of piecewise $C^1$ closed paths in a non-empty domain is pretty immediate.
@DanielFischer Ok. It's just because we have a domain ?
Good night @Mike: I'm confused. If $A\in O(n+1)$, isn't $x\rightsquigarrow Ax$ a diffeomorphism of the sphere?
salut @Gato; guten Abend, @DanielF.
@TedShifrin Salut Ted.
19:16
@DavidWheeler the method I cited from Shilov asked me to permute a set of indices in an order like 123456. This could be permuted into 612345 561234 4561234 345612 234561.
Similarly I can do the reverse, 654321 165432 216543 321654 432165 543216. Let's suppose for each term in the determinant I have $a_{a1}a_{b2}a_{c3}a_{d4}a_{e5}a_{f6}$. For a determinant of order 6, there will be 12 terms each with one of the permutations mentioned above. But the term he asked me to find doesn't fit that model. The terms are "out of order" they do not represent a permutation of 123456
hi @Stan
"Salut Ted", I saluted.
Couldn't resist :).
Hey Ted! I got a new book called Exterior Algebra by Suhubi
Its awesome.
@TedShifrin ^
Sorry
I don't know that book :)
Exterior Analysis
That's the title
19:17
heya @Lord
Yes, @Ted. But that's not what I said.
I thought you asked how to canonically associate to an element of $O(n+1)$ a homeomorphism of the sphere?
Hello @TedShifrin the medicine has calmed me down a bit, but that's only the first step.
'llo @Ted. How's life?
Good, @Jasper.
19:18
Other way around.
This is what you get for your disdain for English syntax, @Mike.
No, it's not.
when simplifying expressions does making the numbers smaller count as a simplifying ? such as if both sides can be divided by 2
So you want some map from homeos to the isometry group? No further property?
Sure, @Dave, but you probably should simplify as much as possible.
well for example 10x = 10y becomes 1x=1y
19:20
Sure, @Dave.
okay, no idea why my book doesn't suggest it
@Gato You need the openness so that you can "wiggle" any path a little bit to become piecewise $C^1$. But with room to wiggle, there's no problem.
@Ted: Well, I want it to do what I want it to do. (I'm trying to write down a homotopy equivalence $\text{Homeo}(S^n) \to O(n+1)$.) The desired map should be this homotopy equivalence.
@TedShifrin Good afternoon.
Hmm, @Mike, could we do it with diffeomorphisms?
19:22
The idea being: here's the map for $n=1$. Start with a self-homeomorphism $f: S^1 \to S^1$. The map to $O(2)$ sends $f$ to the element that sends 1 to $f(1)$, and is orientation reversing if $f$ is orientation reversing.
No, @Ted, it's totally false for diffeomorphisms.
Wait a tick now. I'm almost there.
@DanielFischer I don't see why the openness give me the piecewise $C^1$. I need to think about it.
Call this element of $O(2)$ $g$. Then $g^{-1}f$ is a map $S^1 \to S^1$, orientation-preserving, fixes the basepoint 1. So we've got a decomposition $\text{Homeo}(S^1) = O(2)\text{Homeo}_0^+(S^1)$, where the subscript and superscript are what I said they are earlier. :P
@StanShunpike I think you are underestimating the number of terms in a 6x6 determinant
There are 720 terms
@DavidWheeler that would explain everything
@Gato You take any path, and the openness gives you the room to pull it straight locally.
19:25
Now the latter thing is easily seen isomorphic to $\text{Homeo}_0(D^1)$, the self-homeomorphisms of $D^1$ that fix the boundary. This is contractible (this is the content of Alexander's trick). This is where things should fail in the smooth case: this group is usually not contractible, and Alexander's trick is an inherently continuous thing. Can't usually be smoothed.)
@DanielFischer Now I see, great explanation. :D
So we see $\text{Homeo}(S^1) = O(2)$ by hand. I want to modify this process - just by changing the map to be $\text{Homeo}(S^n) \to O(n)$ - to use the same trick.
@TedShifrin if I have a multilinear map $T$ on a set of vectors st $T(\vec{v}_1,\vec{v}_2,...\vec{v}_n)$ and I pull out all the components of the vectors giving me $(v_1v_2...v_n)T(\vec{e}_1,\vec{e}_2,...\vec{e}_n)$ such that each $\vec{e}_i$ is a basis for its respective vector space, is there a name for $T(\vec{e}_1,\vec{e}_2,...\vec{e}_n)$? I didn't realize multilinear maps had this property...that I could pull out all the constants, leaving just canonical basis vectors behind.
@Mike: I dunno.
19:27
@Stan: What you're writing doesn't make sense. You need a ridiculous multisum with all the components of each vector appearing.
@TedShifrin See my starred messages on the right, LOL.
@Chris'ssis hi, do you know why in your currency notes, one contains flowers and another music notes ?
@DavidWheeler thanks a lot. I will go over it again and see if I can figure it out. That must be the problem. Maybe I can answer my own question :D
Hello @Lord_Farin !!! Are you familiar with numerical analysis?
The sign of the term is determined by the sgn of the permutation of the indices you're shuffling (the ones "not in order")
19:30
@DavidWheeler right, I just think I must have misunderstood the formulas because I clear don't have enough terms. Presumably once I figure out how to compute it properly (ie with the terms I'm not accounting for some how) it will be clear how to answer the question I posted.
Could you take a look at my question?
0
Q: system of First-Order ODES

evindaI am looking at the following exercise: Consider the initial value problem $\left\{\begin{matrix} x''(t)=x(t)\\ x(0)=a\\ x'(0)=b \end{matrix}\right.$ Write it as a system of First-Order ODES with suitable initial values and show that Euler method can get unstable for a great step $(h)$. Tha...

@TedShifrin what if all the input vector spaces are 1 dimensional, would I still need a multisum?
@Gato Well, I didn't think of it. I usually keep books in hands, not money. ;)
@TedShifrin ah, he cites the multisum stuff on page 20 :p
For example, with the term $a_{23}a_{31}a_{42}a_{56}a_{14} a_{65}$ the permutation is $(1\ 3\ 2\ 4)(5\ 6)$ which has sgn 1, since it's an even permutation
19:33
@evinda Hello. I'm afraid that I'm not. I have tried to steer clear of numbers for years now. It's just not for me.
A ok.. No problem.. :) How is it going on? :)
@Lord_Farin Have you tried 3?
@MikeMiller I only regularly count my legs.
@evinda I'm fine, thanks. How about you?
@Lord_Farin Me too... :)
@StanShunpike it appears that you were hoping everything would be in terms of a 6-cycle, but that doesn't work
19:35
@evinda I'm surprised you remember me. Why is it that I made such a lasting impression?
I'm fond of 4, particularly. 3 is nice too.
@DavidWheeler exactly.
I obviously misunderstood the method.
@Lord_Farin In general I don't forget my discussions..
I remember for example that we were talking about Set theory.. :)
well the 3x3 case (which is all people usually see worked, since even 4x4 determinants are beastly) is kind of misleading
@evinda :)
@MikeMiller My world often starts at $\omega$; at times, I resort to $\epsilon_0$ ;).
19:40
@Lord_Farin I think you should start counting after the first $\epsilon_0$ limit cardinals. Just sayin'
@Lord_Farin hello i'm happy to see you
Hello @Vrouvrou.
Good to see you too.
Hi @terdon what did they flag about me?
Uh-oh, looks like we have a flag on the play
19:50
@DavidWheeler that's the problem then. I didn't go beyond 3x3. Would have been wise to check that.
@StanShunpike are you trying to prove something, or just calculate a determinant?
@JasperLoy Nothing that was remotely offensive. I can't really tell you more I think.
@terdon I think it's probably the sexual things, lol. I will try to cut down on them.
@JasperLoy No, it was a completely innocuous message. I'd originally come in to ask that people only flag things that are actually problems since the flags are seen network-wide.
So yeah, math folks, please only flag when necessary :)
@terdon gee, you're no fun
(don't mind me, i never know when to keep my mouth shut. Ted can attest to that)
20:01
@DavidWheeler Well, you know, I'm this dour science type. Not quite as fun loving as mathema... hang on!
@DavidWheeler Mind? Why should I mind? I took that as a perfectly civil joke.
Oh dear. I meant it as self-deprecating humor. Missed again.
@DavidWheeler Just calculate the determinant. I'm just doing this for practice.
mathematicians can party like nobody's business
@MikeMiller Ah, there we go. I was wondering why none of you rose to that one :)
mathematicians do it on riemann surfaces
20:04
And let's not mention bootstrapping
not the ones I know...
well, we use numbers, that are based on logic, that is encoded by numbers, which are based on logic, which is encoded by numbers, that-
Mathematics is, essentially, philosophical filo dough
I need to put that on my facebook page
please, no run-on sentences in here
20:09
@MikeMiller How about turn-on sentences?
Ugh, horrible. Sorry.
Well, singles' bars-here I come.
i'm not quite sure that qualifies as a sentence, so I'll rephrase: please, only sentences in here$2\pi$
Hello @Ramanewbie
hi @evinda what's up ?
@MikeMiller If, you're going to be a grammar (something), please end your sentences with periods. Of $2\pi$.
20:10
Fine, thanks.. And you? @Ramanewbie
@evinda I'm ok, too thanks
@Ramanewbie News?
@evinda about ?
@Ramanewbie Anything
20:14
@evinda Not really... I'm having a basic standard life as usual -_-
@Ramanewbie Me too...
@evinda Before I zoomed on your profile image a week ago, I was convinced it was just a small amount of black dirty land...
@Ramanewbie Aha :D
@evinda Are the number special or just random ?
How come I always get "warnings" from the site whenever I vote too many answers in a row and it reminds me to vote on some questions as well, but it doesn't say anything when I vote on many questions in a row
20:19
@Ramanewbie I think that they are random.. What's with your profile image? Why did you choose this one?
probably because you've voted so many answers it's not worried about you voting on too many questions
@Ramanewbie Aha!
@evinda did you know abou it ?
@Ramanewbie No, that's the first time I read about it..
20:27
@evinda Now you know... ^^
@Ramanewbie Yes, thanks for the information :D
@ᴇʏᴇs To encourage voting for questions.
Wow I got 200 points already, not bad.
@ᴇʏᴇs I am going to sleep, good night, I hope my miracle comes soon.
Huy
Huy
@JasperLoy: Have a good sleep.
Hell @Huy
Huy
Huy
20:38
hi evinda
@Huy How are you?
@ᴇʏᴇs It's not a warning. They just want that people vote on questions too, so when you vote on "too many" answers without voting on a question in between, the pop-up comes. You can then decide to not vote on questions, or to vote on questions. If you decide not to vote on questions, I think the nagging keeps coming.
Huy
Huy
tired, as always, you ?
Hi @Huy.
Huy
Huy
hey, @DanielFischer
watching tomorrow's solar eclipse, @DanielFischer?
20:42
@Huy Don't think so.
Huy
Huy
:(
did you see the one in 1999?
@Huy Sort of. I didn't do anything big, but I went outside.
Huy
Huy
I did too, but we barely saw anything. It was raining.
Ah, we had a sunny day, so it became noticeably darker.
@Huy I am not that tired..
Hey @DanielFischer!!! Do you have about 20 Celsius? I heard so.. :)
20:46
@evinda Hallo!
@evinda If we add the temperatures of morning and afternoon, it comes out to about 20.
@Alessandro Hallo!!! Wie geht es dir?
@DanielFischer Aha :D
@evinda Gut, danke, und dir?
@Alessandro Ganz ok.. Was gibt es so neues?
@evinda nicht besonderes :( Heute habe ich den Genitiv im Kurs gelernt
20:52
Und? Hast du ihn verstanden? @Alessandro
Hello @AndrewSzymczak
@evinda Ja, ich wusste schon ungefähr wie man ihn benutzt
@Alessandro Gut!!! Bist du auch spazieren gegangen? Oder war es in deiner Gegend nicht sonnig?
@evinda Das wetter war richtig schön heute, deshalb bin ich mit meinem Hund im Wald spazieren gegangen :) was hast du heute gemacht?
Achso :) @Alessandro Ich habe 2 Vorlesungen besucht und jetzt lerne ich ein bisschen und bin im Internet.. Nichts außergewöhnliches..
21:27
Wir haben Deutsch und Francais hier
So @Evinda & @Alessandro speak German, and @Hippalectryon & @iwriteonbananas speak French. That's pretty cool :)
21:47
@Committing: Some of us are multilingual, but also Ramanewb is French, DanielF is German, Gato, vrouvrou speak French, and plenty more ...
@TedShifrin: looks like I finally need to use that French I claimed to know.
We know it's a vacuous claim, @Mike.
But good night.
Apparently something I want to know is proved in "sur les diffeomorphisms de la sphere de dimension trois"...
par qui, @Mike?
(oh, and you misspelled diffeomorphismes :) )
I just answered my own question on here :)
Feels good
21:50
LOL, so you gave yourself lots of votes, @Committing?
@TedShifrin If I could :)
Typical politician.
I get no rep for answering my own question
Ah, of course I did. :P
That's so unfair lol
21:51
@Ted: It contains a proof (depending on Smale's conjecture, now Hatcher's theorem) of the question I posted the other day.
You're into territory I know nothing about, @Mike.
Sure. I was just saying that I had to read some French. :P
Don't come whining to me.
Do I ever?
22:04
@Ted: Can you figure out what this quesiton is trying to ask, at all?
Thank you kind upvoter(if you are here)
No, @Mike, I decided not to meddle. Originally, it was full of "manifold" because someone told him that word in his first question.
I have 101 helpful flags and 999 account views :), two milestones in one day
(assuming I get 1 more view today)
Oh dear, @Ted.
I'll make sure not to look at your profile, @Committing.
22:12
Oh dear?
I'm never looking at you, @Committing.
Aren't you supposed to be on your commute to school, @Committing?
@Ted: Oh dear that someone is giving him words that are probably unhelpful.
@TedShifrin I am at uni
Yeah, "shape without dimensions" = manifold???
@TedShifrin I got here at 6:20am
ah, @Committing
poor dear, @Committing
22:13
And I got good sleep :)
No it is a good thing
I love being at uni, so much easier to do work
except you waste so much time in here, @Committing
@TedShifrin I have been working for ~2 hours and I have posted 13 comments
It's 8:15am atm
right, MSE contributions ≠ homework done
@TedShifrin I answered my own question lol(which was relevant to class)
And now I am decoding some vigenere cipher for class
I dropped my 5th class, so I am doing normal load now
(Dropped operations research, my only non-pure math course)
(Though I am doing non-pures next semester)
Bah! I just spent far too long trying to prove the hypothesis of a question true, when in fact it was false. Well, at least it is easy to see how to extend my counterexample to generate a lot of other counterexamples. >8(
22:26
@robjohn Usually when that happens for me it is actually one of the best ways to learn the content, but that is possibly only true at a low level
@Committingtoachallenge It can help anyone. I think all the same learning steps can happen at all levels.
I wonder if my professor is trolling the class..he gave all the exact problems he claimed will be the midterm problems
So you already knew the questions prior?
hmmm, I developed another integral by Ramanujan ...
(maybe I prepare a new question to post)
@Committingtoachallenge He gave us a set of problems, and he said these will be the exact problems on the midterm
22:30
@Chris'ssis I enveloped a piece of chicken by KFC
@ᴇʏᴇs Maybe he meant they are the same with changes in variables or something(I don't know the problems)
@Committingtoachallenge Even so, I think that's still giving too much away
Hello @Committingtoachallenge
@Committingtoachallenge aha ...
Does someone have an idea for:
0
Q: Product rule for curves

Mary StarShow the following rule for differentiable curves in $\mathbb{R}^3$: $$\frac{d}{dt}\left \{\overrightarrow{\sigma}(t) \cdot \left [\overrightarrow{\rho}(t)\times \overrightarrow{\tau}(t)\right ]\right \}=\frac{d\overrightarrow{\sigma}}{dt}\cdot \left [\overrightarrow{\rho(t)}\times \overrightar...

@Committingtoachallenge Most of them are proofs, so I don't know
22:31
??
@Chris'ssis I didn't actually, KFC is too unhealthy for me :P (When studying)
@ᴇʏᴇs Oh well learning proofs can be hard, so that is fair enough
@Committingtoachallenge Do you also take smart pills?
If you don't understand them well, you won't write them down in time
@Chris'ssis I take no medication or drug of any sort, other than coffee
I assume you refer to adderall?
@Committingtoachallenge That's good.
@Chris'ssis Have you ever experimented with nootropics
22:34
I think taking them seems absurd. If you don't have the passion to work all day on it, you shouldn't be studying it anyway
@Committingtoachallenge nootropics in general.
@ᴇʏᴇs I took once, many years ago when it was available on market. Now it's not available anymore. I only took it for a short period of time.
My friend who is a PhD candidate at Harvard purchased like 4 pounds of aniracetam or something
The best drug should be the passion, and the good results should come as a result of a hard work.
@Chris'ssis Exactly
@Committingtoachallenge People sometimes try the alternatives thinking some miracles might produce, but no, there is no such a miracle.
22:37
@Chris'ssis I think people who take those drugs in Math should change degree to something that doesn't require them
2
Nothing can replace the hard work.
@Committingtoachallenge Agree 100%.
I always feel better when people tell me I can succeed in math with hard work since I'm not very smart but I don't know if it's true that someone that's a bit below average can succeed in math with hard work alone
2
Q: Solution of $y''+xy=0$

evindaThe differential equation $y''+xy=0$ is given. Find the solution of the differential equation, using the power series method. That's what I have tried: We are looking for a solution of the form $y(x)= \sum_{n=0}^{\infty} a_n x^n$ with radius of convergence of the power series $R>0$. Then: $$...

Could you take a look at my question?
@robjohn I'm also able to compute $$\sum_{n=1}^{\infty} (-1)^{n+1}\frac{1^{(2011)}+2^{(2011)}+ \cdots + n^{(2011)}}{n^{2014}}$$
I'm going to add these series in my book, not in the particular case, but considering the general case.
@ᴇʏᴇs I am not sure if this matters to you, but there was a very very successful researcher in the field of genetics(if I recall correctly, but can't find via google) who had a tested IQ of under 80, who succeeded solely on passion and hard work
Unfortunately it is hard to search for genetics and IQ in the same google search and expect anything other than studies on IQ and genetics haha
22:53
I think the discussion on IQ is very complex in general, much more complex than it seems at first sight. I also think that when one puts much passion into something and much hard work, the brain simply works differently in that area.
I'll just do my best since that's all I can do
I'd be curious to know how many people can excel in some areas without any passion. My philosophy is that one cannot excel being separated by passion. I never ever met in the real life someone like robjohn, with his skills.
Why do I say that?
Well, because I never ever met someone such passionate about doing math either (in the real life). The fact that he is here doing math all day long says a lot. I bet no one would do that without a crazy amount of passion.

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