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Aug
28
comment how to fairly select a leader
@HagenvonEitzen The natural extension would be to permit any convex combination of any set of votes that you are unable to decide between.
Aug
28
comment how to fairly select a leader
@flawr In your suggested method, note that there is no difference between distributing 2,1,0,0 points to the four candidates and distributing 3,2,1,1 points instead. What happens is that these voters are not able to punish their worst candidate as much as the other voters.
Aug
28
comment how to fairly select a leader
@flawr Assuming the Borda count method is used, this would be correct. A voter voting in this way is essentially saying: "I have no idea what to do about $d$, so I put $d$ in the middle and let the others decide where $d$ should end up."
Aug
28
comment how to fairly select a leader
The condition Independence of Irrelevant Alternatives in Arrow's theorem is very strong, and most election methods used in practice do not satisfy it.
Aug
28
answered how to fairly select a leader
Aug
23
comment Directly prove that $2x^2 -4x + 3 > 0$ for all real $x$
The technique is called completing the square.
Aug
23
comment Polydisc is not biholomorphic to any strictly pseudoconvex domain
@Math Make a holomorphic change of coordinates so that $D$ becomes strictly convex near $\phi(0)$, locally given by $y_2<h(x_1,y_1,x_2)$ where $h$ vanishes to second order. Compose $\phi$ with the projection on the second coordinate $z_2$. The result is a one-variable holomorphic map which maps the interior point $0$ to a boundary point of its image, and such a map has to be constant.
Aug
22
revised Polydisc is not biholomorphic to any strictly pseudoconvex domain
deleted 1 character in body
Aug
22
comment History of notation: “!”
@TonyK Cajori agrees with Hilbert that it should be $\underline{\big|n}$. He also says this notation was introduced in 1827, and became somewhat popular only after Todhunter used it in his texts around 1860.
Aug
22
answered Polydisc is not biholomorphic to any strictly pseudoconvex domain
Aug
22
comment Notation problem in integration: write $dx$ or ${\mathrm{d}}x$?
@ChristianBlatter I think not, but I see the Cambridge typesetting standard disagrees with me. From this description of the ISO standard, "italic symbols should be used only to denote those mathematical and physical entities different values," and "Any other symbol that was not dealt with in the preceding section must be set in roman font". It is my interpretation that if something cannot assume different values then it is because it is the name of something.
Aug
21
revised Notation problem in integration: write $dx$ or ${\mathrm{d}}x$?
added 2 characters in body
Aug
21
answered Notation problem in integration: write $dx$ or ${\mathrm{d}}x$?
Aug
20
awarded  Nice Answer
Aug
20
revised Disproving the claim that the numbers 1+2+4, 1+2+4+8, 1+2+4+8+16… alternate between prime and composite
added 180 characters in body
Aug
20
answered Disproving the claim that the numbers 1+2+4, 1+2+4+8, 1+2+4+8+16… alternate between prime and composite
Aug
20
answered Analysis of a Holomorphic function $f$ given $1 \geq |f '(z)|$.
Aug
20
comment Analysis of a Holomorphic function $f$ given $1 \geq |f '(z)|$.
There is an obvious counterexample: The function $f(z)=z$ satisfies your conditions and is not constant.
Aug
19
comment Mathematical literature to lose yourself in
A related quote by André Weil: "In 1947, in Chicago, I felt bored and depressed, and not knowing what to do, I started reading Gauss's two memoirs on biquadratic residues, which I had never read before (....) This led me in turn to conjectures about varieties over finite fields."
Aug
19
answered Mathematical literature to lose yourself in