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University of Florence


17h
comment Can a convex polygon inside a square with edge length 1 have a perimeter > 4?
The projection map is the map which sends every point of the plane into the closest point on the convex set.
2d
comment (Dis)continuity of function in $R^2$
You have to carefully identify what is the boundary of the set $\{(x,y) \colon x^2>2+x \wedge y<6 \}$. In this case you find that this boundary is composed by four half-lines.
2d
comment (Dis)continuity of function in $R^2$
Ok, so $\wedge$ is an and not a min as I thought!... :-)
Jul
20
comment (Dis)continuity of function in $R^2$
I find it strange to write something like $A > B < C$... are you sure?
Jul
14
comment What will be probability of this problem?
@Did: you are right. Hope it is corrected now.
Jul
13
comment Pythagorean triple means
you mean H^2 + G^2 = A^2
Jul
12
comment Is it differentiable?
@user157130 that's not the definition of derivative.
Jul
12
comment Is it differentiable?
To have a chord passing through $(0,f(0))$ you need to know what is $f(0)$.
Jul
11
comment Spherical Sampling of Projected Disk
Are you speaking of a 2-dimensional disk in 3-dimensional space?
Jul
10
comment $f(x,y,z)$ must be equal at $(0,0,a)$ and $(0,0,-a)$ when $\nabla f(x,y,z)$ is always parallel to $(x,y,z)$
the gradient is in the title but not in the exercise
Jun
25
comment What is the domain of the function $F(x)=\int_{0}^{x}\frac{\operatorname{arctan}(t)}{t}dt$?
@Paul: thanks i've corrected my answer. In fact it is indeed not clear how the improper integral could be possibly defined on the empty interval $(a,a)$... I think that it should be $0$ by definition.
Jun
25
comment What is the domain of the function $F(x)=\int_{0}^{x}\frac{\operatorname{arctan}(t)}{t}dt$?
It much depends on the definitions you are given. Notice however that if you decide it is not defined for $x=0$ it is not defined for any value of $x$. Hence you have two possibilities: either the function is defined for all $x$ or for no $x$.
Jun
12
comment probability in roulette!
The best thing you can is not to play.
Jun
11
comment Solve set problems without Venn diagrams
@MauroALLEGRANZA my position is that your solution is using the inclusion-exclusion principle when you say: in order to avoid counting twice. Once you assume that the equation given by this principle is true, then the problem becomes algebraic. Notice that this principle is about finite sets and can be understood without knowning anything about general set theory (let alone Cantor).
Jun
11
comment $l_{p}$ metric on $\mathbb{R}^{n}$ and its open balls
1. cannot read indices in the answer. 2. what you mean with ??
Jun
11
comment Solve set problems without Venn diagrams
@AsafKaragila I don't know category theory. However you could explain your position by solving the stated problem using combinatorics without set theory.
Jun
11
comment Solve set problems without Venn diagrams
@AsafKaragila I would say that combinatorics is based on set theory... I cannot imagine an axiomatization of combinatorics which does not use sets. Instead I can imagine an axiomatization of numbers (real or natural) without sets.
Jun
11
comment Solve set problems without Venn diagrams
I would say no: because the statement of the problem itself can only be modeled by using sets. A purely algebraic solution cannot explain why the formula used is true.
Jun
11
comment How to find the number of possible permutations in a composition function
You found $khk = h^3$. Multiply both sides of equation by $k$ on the right to obtain: $khkk = h^3 k$. But you know that $kk=1$ hence you obtain $kh = h^3 k$. About $kh^3 = kh$ you can multiply on the left by $k$ and obtain $h^3 = h$ which becomes $h^2=1$ which is false.
Jun
11
comment How to find the number of possible permutations in a composition function
I don't use any theorem, just some algebraic computation. What is not clear?