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Yes compactness includes a boundedness condition so it now looks more plausible. I don't have a proof yet, neither for the existence nor for the uniqueness. I'll need to think about it.

Alan and Barbara play a game in which they take turns filling entries of an initially empty $2008 \times 2008$ array. Alan plays first. At each turn, a player chooses a real number and places it in a vacant entry. The game ends when all the entries are filled. Alan wins if the determinant of the res...

Hi Tolis,

I believe we need some boundedness condition. For example the result is not true for $\mathbb{R}$.

This is immediate by the little Picard theorem which says that the range of every non-constant entire function is either the whole of the complex plane, or the complex plane minus one point. Here however the range of $f$ does not contain any point with imaginary value less than or equal to $-2$. Sin...

Let us write $D_n$ for the determinant of the $n \times n $ matrix. It is immediate that $D_1=1$ and $D_2=2$. Taking cofactor expansion along the first column we have \[D_{n+2} = \left|{\begin{array}{rrrrrrr} 1 & 1 & 0 & \cdots & 0 & 0 & 0\\ -1 & 1 & 1 & \cdots &a...

No there isn't even if $f$ is allowed to be unbounded. So let $f$ be a monotonic function. We will show that it has at most countably many discontinuities. Let us also suppose without loss of generality that $f$ is monotone increasing. Since $f$ is monotonic, then it has left and right limits at eac...

We have \[\sum_{n=0}^{\infty} \frac{(-1)^{n+1}(2n+1)!}{(n+2)!n!4^{2n+3}} = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{2(n+2)4^{2n+3}} \binom{2n+2}{n+1} = \frac{1}{8}\sum_{n=0}^{\infty} \frac{1}{n+2} \binom{2n+2}{n+1}\left(-\frac{1}{16} \right)^{n+1}.\] It is well-known that the generating function of the...