The  is an 80 year-old open Math problem. Recently, I made a remarkable discovery of the long sought visual pattern in the conjecture. This never-before-seen visual pattern led me to two significant results: (a) a natural proof that the Collatz conjecture has no non-trivial cycles, and (2) a direct link between the Collatz conjecture and prime numbers. My question: If you have proven a major unresolved question in mathematics but you are out of touch with university professors, or someone who can endorse you on arXiv, how can you get your paper published and noticed by professional mathematicians? My paper can be found . I'm currently not interested in publishing in a journal, unless someone wants to pick up that challenge with me.
I am interested in the situation where you have a very interesting result. For instance, you have solved a very important open problem. However, you are not known in the field and do not have any remarkable publications. Your supervisor thinks the work is good and you submit the work to a high profile journal, but you get rejected.  The thing is that the contribution is very strong. It breaks what most people believe or what they have already proven: e.g., you solve the P vs. NP problem or any other well known open problem.   The reviewers strongly reject your work with no justification and they do not state why the result is wrong. Examples of reviewer comments include:   "The proof must be wrong."  "You cannot achieve such a result."  "You do not understand well the notion of ..."   My question is what to do in this situation? Where to go? If your advisor accepts the work, but the reviewers from the top journal reject the work without even explaining the mistakes,  what should you do?
Why Session object destruction failed ASCII art generator Can't reach wi-fi signal on Ubuntu, but can do it on other OS/devices (ath9k) I have two laptops, both running Ubuntu 13.10, as well as a different OS (Windows 7), dual boot on both.  My wi-fi router is on my bedroom and I want to get the wi-fi signal from the living room. I can reach the wi-fi signal from both pcs when running windows, with a relative good signal, but can't reach it on both of them when running Ubuntu, can't even connect to it.  Is there something that I can do to amplify the wi-fi signal? What is the Python equivalent of Perl's chomp function, which removes the last character of a string if it is a newline? Dealing with questions of nefarious intent The help pages in R assume I know what those numbers mean, but I don't. I'm trying to really intuitively understand every number here. I will just post the output and comment on what I found out. There might (will) be mistakes, as I'll just write what I assume. Mainly I'd like to know what the t-value in the coefficients mean, and why they print the residual standard error.   Call: lm(formula = iris$Sepal.Width ~ iris$Petal.Width)  Residuals:      Min       1Q   Median       3Q      Max  -1.09907 -0.23626 -0.01064  0.23345  1.17532    This is a 5-point-summary of the residuals (their mean is always 0, right?). The numbers can be used (I'm guessing here) to quickly see if there are any big outliers. Also you can already see it here if the residuals are far from normally distributed (they should be normally distributed).  Coefficients:                  Estimate Std. Error t value Pr(&gt;|t|)     (Intercept)       3.30843    0.06210  53.278  &lt; 2e-16 *** iris$Petal.Width -0.20936    0.04374  -4.786 4.07e-06 *** --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1    Estimates $\hat{\beta_i}$, computed by least squares regression. Also, the standard error is $\sigma_{\beta_i}$. I'd like to know how this is calculated.  I have no idea where the t-value and the corresponding p-value come from. I know $\hat{\beta}$ should be normal distributed, but how is the t-value calculated?  Residual standard error: 0.407 on 148 degrees of freedom   $\sqrt{ \frac{1}{n-p} \epsilon^T\epsilon }$, I guess. But why do we calculate that, and what does it tell us?  Multiple R-squared: 0.134,  Adjusted R-squared: 0.1282    $ R^2 = \frac{s_\hat{y}^2}{s_y^2} $, which is $ \frac{\sum_{i=1}^n (\hat{y_i}-\bar{y})^2}{\sum_{i=1}^n (y_i-\bar{y})^2} $. The ratio is close to 1 if the points lie on a straight line, and 0 if they are random.  What is the adjusted R-squared?  F-statistic: 22.91 on 1 and 148 DF,  p-value: 4.073e-06    F and p for the whole model, not only for single $\beta_i$s as previous. The F value is $ \frac{s^2_{\hat{y}}}{\sum\epsilon_i} $. The bigger it grows, the more unlikely it is that the $\beta$'s do not have any effect at all. Determining possible minimal polynomials for a rank one linear operator I have come across a question about determining possible minimal polynomials for a rank one linear operator and I am wondering if I am using the correct proof method.  I think that the facts needed to solve this problem come from the section on Nilpotent operators from Hoffman and Kunze's "Linear Algebra".     Question: Let $V$ be a vector space of dimension $n$ over the field $F$ and consider a linear operator $T$ on $V$ such that $\mathrm{rank}(T) = 1$.  List all possible minimal polynomials for $T$.   Sketch of Proof:   If $T$ is nilpotent then the minimal polynomial of $T$ is $x^k$ for some $k\leq n$.  So suppose $T$ is not nilpotent, then we can argue that $T$ is diagonalizable based on the fact that $T$ must have one nonzero eigenvalue otherwise it would be nilpotent (I am leaving details of the proof of diagonalization but it is the observation that the characteristic space of the nonzero eigenvalue is the range of T and has dimension $1$).  Thus the minimal polynomial of $T$ is just a linear term $x-c$.    Did I make a mistake assuming that $T$ can have only one nonzero eigenvalue?  Thanks for your help How to add a confirm dialog to a command button? I'd think this would work just fine, but it seems to prevent the action function from firing regardless of the confirm response  &lt;apex:commandButton value="Remove All" action="{!removeAll}" rerender="form"      status="LoadingStatus" onclick="return confirm('Are you sure?');"/&gt; Frattini subgroup $Φ ( G )$ is not trivial Frattini subgroup $Φ ( G )$ of a group G is the intersection of all maximal subgroups of $G$. The question is  Let $G$ be an abelian group.What is the sufficient and necessary condition of $Φ ( G ) ≠｛e｝$？  I think if $G$ has no torsion part, we cannot say $Φ ( G ) ≠｛e｝$,because $\Bbb Z ×\Bbb Z×・・・ ×\Bbb Z$ has a maximal subgroup $p\Bbb Z ×\Bbb Z×・・・×\Bbb Z$, with intersectioning distinct primes, we gain just a trivial group. So, we need torsion part. But I cannot proceed from here.I only know the answer, $G$ has an element of order $p^2$, that  is, $G$ has theree subgroups. I want to deduce this consequence from steady discussion. P.S  According to this answer, in a ring that is a finite direct product of quotient rings of $\mathbf{Z}$, we can say rattini subgroup is the same as the Jacobson radical of the corresponding ring. Has anyone ever booked a flight with ?  I am just wondering where they get their prices and data from, and if I might find a cheaper flight by using something like Expedia or Travelocity. Since I updated the system to 10.11.5, the Facebook message notification is not showing message text.   Is this a bug in this software upgrade of OSX? Find all natural numbers $x,y$ such that   $$3^x=2y^2+1$$  solutions are $(1,1)$, $(2,2)$, $(5,11)$. I found that parity of both is same and If $x$ Is odd it is of the form $4k+1$. After hitting 80% LTV, my mortgage lender refuses to remove PMI. Any recourse? Please add documentation about the redaction tool to the moderator help How can I install Sun/Oracle's proprietary Java JDK 6/7/8 or JRE? I am trying to help my daughter on her math homework and I am having some trouble on some equation solving steps. My current major concern relies on understanding why $\sqrt{8}/2$ equal to $\sqrt{2}$.  Thanks in advance Keep only essential packages Opt in "mentor" privilege at 7.5k I am suggesting a Mentor privilege at 7.5k. It would be on a per-user per-site opt in basis.  At some reputation level (perhaps 7.5k) there could be an opt in "mentor" privilege.  If someone opts in then when a new user signs up, they become their Mentor. The new user is the Mentee.  They will get notified of any posts or suggested edits they make, and can help and support them.  While this would by no means help everyone (I'm sure there are more new people signing up than there are 7.5k people willing to help them), it could help mitigate the "new user's bad posts" and such like.  By no means would being someone's mentor mean you are the only person who can edit and help them, and you aren't obliged to do so - but if you have opted in then you should help.  How would this work?   Any Mentor would have no more than 5 active Mentees at once. Once the Mentee gets to 125 rep, the downvote privilege (or any other level deemed sensible) they are no longer a Mentee. If they aren't active for a month, they are no longer a Mentee. If the Mentor has 5 Mentees (i.e. cannot mentor anyone else) then any Mentee not active for 2 weeks is no longer a Mentee. The Mentee can opt out at any time, the Mentor can opt out, but if they opt out a lot (say 5 people in 2 months) they are blocked from the privilege for 2 weeks. The Mentor gets no extra rep. They can still get rep from answering their Mentee's questions and similar. Once the new user gets 20 rep on the site, a chat room is created for them to talk. No user with the association bonus will become a Mentee. This would (maybe) be opt in for sites. For example, this might not be at all useful for this site, or for , but on  or  I imagine this could be quite good.    appears to be asking the same thing, based on the title, but that is asking about people who know each other offline mentoring each other.    Every single one of these suggestions (especially the numbers) is open to discussion. Maybe 125 is to low to stop being mentored, or maybe it's too high. What do you think? Why is gksu no longer installed by default? I have always recommended pressing ALT+F2 and typing for example gksu nautilus or gksudo nautilus both are equivalent; to run graphical programs with root privileges.  However, I have just installed raring on a test system and this no longer works.  gksu and gksudo are not installed as part of a default raring install; though they are in the repositories. In 12.04 at least gksudo was just a symbolic link to gksu  The reasons I have always done this are because while it is often safe to use sudo under some circumstances it can cause problems with graphical applications because gksu has you running the program as root while with sudo you are running as yourself but with elevated privileges.  This can mean you inadvertently change the owner of critical system files.  This can cause problems.  This reasoning is explained better here:         Why was command removed from a fresh install and what should we be using instead? What is the term for accusing a person of mischief through invented, twisting and changeable terminology? What is the term for accusing a person of invented wrongdoing in such a way that he cannot respond or even clearly understand the accusations? For example, frequently reversing the definitions of the terminology used, or basing argumentation on hints and roundabout phrasings, metaphors, invented-on-the-fly floating definitions and rules to fit with the accusations.    Update: I think inquisitorial is the most proper term for context in which the result does not depend on the argumentation of one of the sides.    Though, jargonaut was completely unheard by me and incrementally more useful.