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bio website math.brown.edu/~lubinj
location Minnesota
age 78
visits member for 3 years, 2 months
seen 13 hours ago

Aged mathematician


Nov
24
comment given a point cloud of n points, create a convex shape that defines their outer limits
As a total outsider to the field, I’d say that this is one of the core problems of linear programming. But in what format is your set going to be described? By a bunch of linear inequalities, perhaps?
Nov
24
comment The property a(bc)=(ab)c for linear operators
Indeed. For matrices of finite size at least, it’s easier to use this obviously true principle to show associativity of matrix multiplication, than to get yourself tangled up in the mess of indices.
Nov
24
comment Find a complex number for the splitting field
e-mail me if you’d like to discuss this further.
Nov
23
comment Factorize x^3+3
GF(7) is $\mathbb F_7=\mathbb Z/(7)$
Nov
23
comment Solve $z^{1+i}=4$
This is certainly the right method, just take account of the many-valuedness.
Nov
23
comment Find a complex number for the splitting field
Well, almost any combination of the two separate things will give you a primitive element. My guess is that your process gets only some of these.
Nov
22
comment Find a complex number for the splitting field
Yes, in the sense that for $c$ equal to either of the quoted elements, $E = \mathbb Q(c)$. But not via the procedure you were following!
Nov
22
comment Find a complex number for the splitting field
Both are primitive elements. That is, either of them will generate your sextic extension of $\mathbb Q$
Nov
22
comment Find a complex number for the splitting field
I made a comment that the sum $\root3\of2+\omega$ should work, but that was based on experience, not on using the method you were told to use. With the knowledge of the special properties of this field, though, I found that $(2\omega+1)/(\root3\of2+1)$ has $\mathbb Q$-polynomial $X^6-3X^4+3X^2+3$, a very pretty Eisenstein sextic, more in line with the kind of thing you were looking for.
Nov
21
answered Laurent series for $f(z)= \frac{1}{ (z-i)(z+2i)}$
Nov
21
comment Cyclotomic polynomial identity
This comes from the Möbius Inversion Theorem, which you should look up. Also, in your exponent, you really just wanted ordinary division: $X^{n/d}$.
Nov
21
comment How do I get the expansion of $(2x-1)^{-1} $ to the $x^3$ term
@ClaudeLeibovici, I wish long division of power series was taught in all US schools…
Nov
21
comment Ratio of an isosceles triangle to its altitude is 3:4. find the measures of the angles
When you say, “ratio of an isosceles triangle to its altitude”, are you talking about the base of the triangle versus the altitude? If so, could you edit?
Nov
21
comment How to evaluate $(0.9)^4$ without calculator
Just a matter of terminology: one solves an equation, here you want to evaluate your expression.
Nov
20
comment Is the map, $ f:(0,1)⊂ \mathbb{R}$ → $(1,∞)⊂ \mathbb{R}$ : $x ↦ 1/x $continuous?
If there were a point of discontinuity, what would it be? And what is the inverse function of $f(x)=1/x$?
Nov
20
comment Can $\pi$ be a ratio of angles?
Any real number can be the ratio between two angles.
Nov
20
comment Is there a formula for calculating the area of 2d shapes on a sphere?
multiply by $r^2$
Nov
20
comment Is there a formula for calculating the area of 2d shapes on a sphere?
For a spherical triangle (bounded by arcs of great circles: these are the geodesics on the sphere), it’s easy, and has been known for centuries: the area is proportional to the “spherical excess”, the amount by which the sum of the angles exceeds a straight angle (of $180^\circ$). So, for a sphere of unit radius, if the sum of the angles is $S$ radians, the area is precisely $S-\pi$.
Nov
20
comment Fixed fields,cyclic groups and Galois theory
Much of this is covered in the Fundamental /theorem of Galois Theory, which you should look into. For your question on $\mathbb Z$, this group is not a Galois group as this is ordinarily defined. Galois groups are profinite: compact topological groups with a neighborhood basis of open subgroups.
Nov
19
answered $\sum_{\zeta^p=1}(\zeta-1)^n$