@robjohn saw your update to the Fourier series problem. like i said in my comment there, it's surprising to me that the 'Fourier series' approach ends up seeming the most obscure
@Semiclassical yes. I believe that the imaginary part of an analytic function on the unit circle determines the analytic function, so I don't think there will be anything other than $\mathrm{Li}_3$ that matches that Fourier Series.
Unless that uniqueness falls apart if there are singularities inside the circle...
@robjohn: Well, the wanton downvoting is continuing. Obviously, I don't care about the rep points, but (as you well know) it would be nice if the person would express a reason ... unless it's just pure revenge. Hmm ... Do I notify mods at some point? [I'm sure @Mike will say there aren't any wontons ...]
@JorgeFernández I have been told the opposite as well. but I don't see how I'm currently being unkind. perhaps the adjective "clearly" was overdoing, and I am not being as helpful as I could be, but I am entertaining myself with other things and only scanning the chatroom very cursorily these days.
well, it's rather straightforward, I don't know if I should call it a trick, but it's "pleasant"
I'd like to share a problem with you I enjoyed, let $G$ be a group of order $2^kl$ with $l$ odd with an element of order $2^kj$. Prove $G$ has a subgroup of index $2$.
consider the action of $G$ on itself by left regular action. then the element of order $2^kj$ induces an odd permutation. Since every subgroup of $S_n$ which has an od permutation has exactly half of it's permutations odd we can consider the elements of $G$ that afford even permutations, these form a subgroup of the desired index.
@TedShifrin I will keep an eye on things, but I can't see anything unless it is bad enough. The community managers and developers are the only ones who can tell anything more.
How can I integrate $1/z$ over a closed curve containing $0$ if $1/z$ is not continuous nor holomorphic at $0$? To use the Cauchy integral formula or the definition of the line integral, I at least need a function that is continuous. In this link <http://en.wikipedia.org/wiki/Methods_of_contour_integration#Example> they seem to ignore these concerns and just integrate anyway.
@TheSubstitute Do you mean $0$ is on the closed curve or is it in the closed curve? If it's the latter (what wiki does) then you should have no problem.
@DanielF Let $(X, d)$ be a metric space, $A \subset X$. Why can't we define $A$ to be convex by the property that for any two $x, y, \in A$, there is a $z$ which is $\neq x $ and $\neq y$ such that $d(x, y) = d(x, z) + d(z, y)$?
You could call $A$ convex if for all $x,y\in A$, for every $t\in [0,1]$, there is a $z_t\in A$ with $d(x,y) = d(x,z_t) + d(z_t,y)$ and $d(x,z_t) = t\cdot d(x,y)$. Maybe that would lead to something useful, no idea.
I have decided to get Gratzer's More Math into LaTeX and Voss's PSTricks for my LaTeX books. Of course, I have decided to use PSTricks to produce graphics.
@MikeMiller He might have upvoted you once and downvoted you four times, for example.
@MikeMiller Accepted answers cannot be deleted except by a moderator, I believe.
I notice that often when my enemy X answers a question and someone else does with a higher number of upvotes, he will be downvoted. It is probably X that downvoted, lol.
In fact, I just saw one such downvote and how the rep of X dropped by 1, so this is very likely, lol.
This is probably the strategy of X to try to get the highest number of upvotes so that her answer will become accepted. Very cunning indeed.
To risk sounding exaggerating, I am sometimes traumatised by the actions of X, lol.
@MikeMiller Could have been that you once edited his/her question, got two points for it, and the question was deleted because of low score when the user was removed.
Ladies and gentlemen, I am very excited as I only need 10 more points to reach 2000 and retire from this site. I will keep my account though as promised.
@ParthKohli I will still come to chat, lol. Hi. You should check if the dictionaries you checked yesterday are authoritative.
@ParthKohli Your take on practice yesterday is something I have never heard before. Chances are that the sources you checked are erroneous.
In American English, practice is noun and verb. In British English, practice is noun and practise is verb. Anyone who disagrees, please correct me citing your sources.
In every area of life the better you become the more envious people surround you. This is also very true for mathematics.
My piece of advice for all these ones would be: don't spend your time with it, it's simply a waste of time, just try to become incredibly good, die trying it. Finally, if you cannot reach that point, appreciate those that can reach the highest peaks, admire them.
The last example that came to mind is Simona Halep that won Seren Williams, she was incredibly good. Some time ago there were some here (in my country) that talked against her when she failed, but I always put trust in her, she has that attitude of one that can even defeat gods (if you know what I mean).
@Committingtoachallenge I was just saying some things that affected some of my math projects, I mean they were delayed to some extent (this is painful to me), but I didn't ask for your opinion, I don't care it.
I work hard to get everything in place such that I'm able to publish my book as soon as possible.
The plane $Π_1$ has equation
$$r=i+2j+k+ \theta (2j-k) + \phi (3i+2j-2k) $$
Which is $$2x+3y+6z=14$$
The line $l$ has equation
$$r = 3i + 8j + 2k + t(4i + 6j + 5k)$$
The point on $l$ where $t = λ$ is denoted by P. Find the set of values of λ for which the perpendicular...
@Chris'ssis You don't care about your narcissistic qualities? I could easily generate a list of $30$ examples from the transcript with their classifications under narcissism within a week if you wish me to. I personally feel everyone should work on eliminating such qualities as they are hardly conducive to a healthy life(or a realistic depiction of reality). Anyway I shall sleep now, goodnight.
@Committingtoachallenge I have the right to talk about my work in a very appreciative way. Do you know why? Because, look, while you slept like 10-12 hours per night (maybe), in the last years I slept like 3 or 4 hours and assigned my time to work, to become very good, and I got thousand of questions and solutions, maybe over 10000 questions.
Did I say "very good"? Yeah, exactly, and you know what? I have great plans, ooo yeah, I wanna become like Ramanujan one day in terms of integrals, series and limit, and this has nothing to do with the narcissism.
@Committingtoachallenge and I'd really like you to be right about me, and thus the world we live in to be much better than I perceive it. Hope one day you publish a paper with some work of yours (I mean your original work) and then someone suggests some of your work comes from another paper, that you took if from there. Then you'll see exactly my point.
I have a question that I feel is super trivial - if two polynomials of x are equal then they are also equal as polynomials of arbitrary powers of x right? i.e. f(x)=g(x) => f(x^n)=g(x^n)
@Committingtoachallenge One of the things I said last days that seem intriguing and called as narcissistic? Well, yeah, I said I have the best proof to Au-yeung series in the world. I mean I'm very proud about it, yeah, it's an achievement I'm not willing to hide it, I worked d*amn hard for it, I suffered to get that you know?
@Max Write out the polynomials as $\sum a_ix^i$ and $\sum b_i x^i$. That the polynomials are equal means the coefficients are equal, so subbing in anything for $x$ gives two still equal polynomials.
how would that make proving equality harder, though. if anything i'd expect easier since taking the quotient maps zero to zero, but also maps other things to zero
@Max Just so you know, not everyone agrees on the term "polynomial ring"... Some (me) take it to mean "ring of polynomials over another ring", some "quotient of rings in the previous sense", some require the coefficient ring to be a field, etc
@Max We can first factor out $p$ - that is a prime, I hope, then we have a nice polynomial ring $(\mathbb{Z}/p\mathbb{Z})[X] \cong \mathbb{Z}[X]/(p)$. Then you want to see whether $f(X) \equiv g(X) \pmod{X^r-1} \implies f(X^n) \equiv g(X^n) \pmod{X^r-1}$, where $f,g \in (\mathbb{Z}/p\mathbb{Z})[X]$.
@Max Let $R$ be a commutative ring (with $1$), and $m \in R[X]$. Then you want to see whether $f(X) \equiv g(X)\pmod{(m)} \implies f(X^n) \equiv g(X^n)\pmod{(m)}$, right?
@Max Okay, so you have $f(X) = g(X) + m(X)\cdot q(X)$, whence $f(X^n) = g(X^n) + m(X^n)\cdot q(X^n)$. For that to always hold, whatever $f,g$, you need $m(X^n) \equiv 0 \pmod{(m)}$, as Mike said.
@Chris'ssis Despite what some people might say about me, I have very little confidence in my mathematical abilities and I'm always questioning myself. I wish I had just a small fraction of the confidence that you have.
If you don't have the $f(x^n)\in I$ condition you automatically have a counterexample like that; if you do, suppose $f \equiv g$, that is $f(x)=g(x)+i(x)$. Then $f(x^n)=g(x^n)+i(x^n)$ and by our condition $f(x^n)\equiv g(x^n)$.
The plane $Π_1$ has equation
$$r=i+2j+k+ \theta (2j-k) + \phi (3i+2j-2k) $$
Which is $$2x+3y+6z=14$$
The line $l$ has equation
$$r = 3i + 8j + 2k + t(4i + 6j + 5k)$$
The point on $l$ where $t = λ$ is denoted by P. Find the set of values of λ for which the perpendicular...
