Say I sample $N$ numbers from a mystery discrete distribution and add them all together. My goal is for the sum of these random variables to be uniformly distributed from $[a,b]$ and thus the distribution is paramaterized by $(N, a, b)$. Does such a distribution exist? I can imagine how to calculate the probability of each number being picked here, but I was wondering if this had a name or if anyone had formalized it (My GoogleFu is failing me). I mentioned the discrete case as it's relevant to the problem I'm trying to solve, but I'm curious about the continuous case as well.
Taken from :  Show that it cannot be the case that $U=X+Y$ where $U$ is uniformly distributed on [0,1] and $X$ and $Y$ are independent and identically distributed. You should not assume that X and Y are continuous variables.  A simple proof by contradiction suffices for the case where $X$, $Y$ are assumed discrete by arguing that it always possible to find a $u$ and $u'$ such that $P(U\leq u+u') \geq P(U\leq u)$ while $P(X+Y \leq u) = P(X+Y \leq u+u')$.  However this proof does not extend to $X,Y$ being absolutely continuous or singular continuous. Hints/Comments/Critique?
Time-Stream Crossing and Forgetfulness In The Day of the Doctor, it's explained that the Doctor doesn't remember having interacted with himself, and if he interacts with a future version of himself, the younger version forgets.  But in Time Crash, 10th solves the problem of the TARDISes occupying the same space because he remembers himself doing it from when he was 5th.  If this is true, 5th wouldn't have wondered whether or not he was going to regenerate when he did.  Does the TARDIS occupying the same space with itself re-instate those memories? Or is it just possible that I'm asking a ridiculous question that will never have an answer because Doctor Who.  :D Need to get logfile smaller Solving an equation over the reals: $ x^3 + 1 = 2\sqrt[3]{{2x - 1}}$ Solve the following equation over the reals:$$ x^3  + 1 = 2\sqrt[3]{{2x - 1}} $$ I noticed that 1 is a trivial solution, then I tried raising the equation to the 3rd, then dividing the polynomial by $(x-1)$.. But I can't see the solution, how do I go from here? The inequality is homogeneous, but I am not sure if that helps. Also, I was thinking about considering a concave function and apply Jensen or Karamata on LHS. How to read and write excel file How we can add two date intervals in PHP In .NET, the  is used in a lot of places throughout the .NET base class libraries. Implementing it properly is especially important to find items quickly in a collection or when determining equality.  Is there a standard algorithm or best practice on how to implement GetHashCode for my custom classes so I don't degrade performance? Samsung galaxy S4 Blacked out Screen need to retreive files How can I set the processor affinity of a process on Linux? How can I set the processor affinity of a process on Linux? Why are these estimates to the German tank problem different? How can I use pathauto to generate URLs for deeply nested vocabularies? I am generating breadcrumbs from URL arguments as described .  I have a vocabulary that looks like this:  Arts-> Performing-> Dancing-> Ballet  I want my url to look like this: arts/performing/dancing/ballet  Using the pathauto module, I can do this for taxonomy terms:  [term:parent]/[term:name]   Which, on the Ballet page, gives me the following; not what I want:  dancing/ballet    If I use parents instead of parent,  ([term:parents]/[term:name]); this is close but not quite:  arts-performing-dancing/ballet   How can I get from arts-performing-dancing to arts/performing/dancing? Probability of a 2 pair in poker Switching back to last *used* tab on Chrome Hi I would like to know whether the trace of the inverse of a symmetric positive definite matrix $\mathrm{trace}(S^{-1})$ is convex.  Actually I know that the trace of a symmetric positive definite matrix $S\in M_{m,m}$ is convex since we can find $B\in M_{n,m}$ such that $S=B^T\times B$ then we can write the trace as the sum of scalar quadratic forms, i.e. $\mathrm{trace}(S)=\mathrm{trace}(B^T\times B)=\sum_{j=1}^mb_j^T\times b_j$ where $b_j$ is the $j^{th}$ column of $B$.  for instance if we have  $trace([\begin{array}{cc} 1 &amp; 2  \\ 3 &amp; 4  \\ \end{array}] \times [\begin{array}{cc} 1 &amp; 3  \\ 2 &amp; 4  \\ \end{array}])= [\begin{array}{cc} 1 &amp; 2  \\ \end{array}]\times [\begin{array}{c} 1   \\ 2   \\ \end{array}]+ [\begin{array}{cc} 3 &amp; 4  \\ \end{array}]\times [\begin{array}{c} 3   \\ 4   \\ \end{array}]=30$  And so I wonder if $\mathrm{trace}(S^{-1})$ is convex too.. Does proving (second countable) $\Rightarrow$ (Lindelöf) require the axiom of choice? Why does the aorta spray blood when atmospheric pressure is higher than blood pressure? I'm trying to solve a question where I need to write $\frac {1}{1+z^2} $ as a power series centered at $z_0=1$  I'm not allowed to use taylor expansion. So my first thought was to rewrite the function in a form where I can apply the basic identity:  $$ \frac{1}{1-x} = \sum_{n=0}^\infty x^n $$  So let's rewrite the original function $f(z) = \frac {1}{1+z^2} $ and use above identity:  $$\frac {1}{1+z^2} = \frac {1}{1-(-z^2)} $$ But observe that I need to center it at $z_0=1$ so I rewrote it to:  $$\frac {1}{1-(-z^2)}=\frac {1}{2z-(-(z-1)^2)} $$  I can factor out $\frac {1}{2z}$ and use the basic identity to get : $$\frac {1}{2z}\frac {1}{1-\frac{(z-1)^2}{2z}} = \frac {1}{2z}\sum_{n=0}^\infty \left(\frac{-(z-1)^2}{2z}\right)^n =\frac {1}{2z}\sum_{n=0}^\infty \frac{(-1)^n(z-1)^{2n}}{(2z)^n} = \sum_{n=0}^\infty \frac{(-1)^n(z-1)^{2n}}{(2z)^{n+1}}$$  The problem is if I ask wolfram alpha: power series$ \frac{1}{1+z^2} $centered at z = 1 it will give me a different answer that doesn't match the sum I found.  Can anyone help me figure this one out? What am I doing wrong? I've already see a  by Marko Riedel which I list it follows:     The standard way to treat these sums is to integrate   $$ f(z) = \frac{1}{(z+\alpha)^2} \pi \cot(\pi z)$$   along a contour consisting of a circle of radius $R$ and with $R$ going to infinity and hence being larger than $\alpha$, where the circle does not pass through the poles on the real axis.   Now along the semicircle in the upper half plane we have   $$|f(z)| \le \frac{1}{(R-|\alpha|)^2}\pi \left|\frac{e^{i\pi R\exp(i\theta)} + e^{-i\pi R\exp(i\theta)}} {e^{i\pi R\exp(i\theta)} - e^{-i\pi R\exp(i\theta)}}\right|= \frac{1}{(R-|\alpha|)^2} \pi  \left|\frac{e^{2i\pi R\exp(i\theta)}+1}{e^{2i\pi R\exp(i\theta)}-1}\right| &lt; \frac{1}{(R-|\alpha|)^2} \pi   \frac{1+e^{-2\pi R\sin(\theta)}e^{2i\pi R\cos(\theta)}} {1-e^{-2\pi R\sin(\theta)}e^{2i\pi R\cos(\theta)}}$$   This last term is clearly $O(1/R^2)$ as $R$ goes to infinity as the quotient of the two exponentials goes to one since $\exp(-R)$ vanishes and there is no singularity when $\theta = 0$ or $\theta = \pi$ as $R\cos\theta = \pm R$, which is not an integer by the assumption that the circle avoids the poles and hence cannot be one.     My question: It is clear that for each $z=Re^{i \theta}$ , $f(z)=O(1/R^2)$ ，but can this guarantee that the last term on RHS is uniform bounded ? Since $\theta$ varies on a compact interval $[0, \pi]$ , I want to show that for each $ \theta \in [0, \pi]$ , and a fixed positive number $M \gt 1 $ there exist an open ball contains $\theta$ and a fixed $R_{\theta}\gt 0$ , for every $R \ge R_{\theta}$ and $x \in $ the open ball $$\frac{1+e^{-2\pi R\sin(x)}e^{2i\pi R\cos(x)}} {1-e^{-2\pi R\sin(x)}e^{2i\pi R\cos(x)}} \le M$$  For $\theta \neq 0,\pi$ , it is easy to find the desired $R_{\theta}$ and the open ball , but if $\theta = 0$ or $ \theta = \pi$ , for every open ball containing $\theta$ , it behave erratically near $\theta$ , I have no idea how to deal with this . What are the differences between rings, groups, and fields? Rings, groups, and fields all feel similar. What are the differences between them, both in definition and in how they are used? custom button visibility list view I have addded a custom button to list views in the Contact object. I have multiple list views but I just need to show the custom button in some of the list views and it should be hidden in the remaining lists views.  Is this possible? Thanks!