Infinity Boost **

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Complex infinity at a point and division by zero

Edit: I’ve found out that the book was written for Mathematica 7, which was a pretty long time ago. It boils down to changes in syntax most probably, but simple renaming to lower lettercase does not work.

Following Stan Wagon’s Mathematica in Action, chapter 19, subsection 19.2, I’ve run into problems.

  1. I cannot redefine ReIm[z] as is done on p.496, Mathematica just states that Tag ReIm in ReIm[z_] is protected.
  2. Trying the same procedure on that page and over the course of the following two pages with a function that I’ve called reim[z], it’s not possible to get the hyperbolic triangle.
  3. I’m then left to use ReIm[z] for the triangle, which will work regardless of whether I “redefine” or not.
  4. Those four definitions of LFT functions and turning off the division-by-zero message off, nothing happens once again.
  5. And then, regardless of what I do in the previous 4 steps, I can not get the tessellation shown on p.498. Instead, I get one of two shown below.

What I think happens is that at the time of writing the book, ReIm was not a legitimate function in Mathematica. It was probably implemented sometime afterwards and now it inadvertently affects this code as well. Is it possible to “add on” to a predefined definition in Mathematica? Or to somehow bypass these errors with a new function?

enter image description here

The problematic ReIm[z] part:

ReIm[z_]:=N[{Re[z], Im[z]}];  ReIm[ComplexInfinity]={0,1000};  Attributes[ReIm]=Listable; 

The LFT (Linear Fractional Transformation) involving ReIm[z] which seems to do nothing, together with the turning off of the errors:

LFT[mat_List][z_?NumericQ] := reim[Divide @@ (mat - {z, 1})]; Off[Power::infy, General::dbyz, Divide::infy]; 

The most problematic part of the code:

polys = Table[{FaceForm[Hue[Random[], 0.6]],      Polygon[LFT[w][triangle[]]]}, {w, G}]; 

^This lists out errors of the type: “\emph{Indeterminate expression $ \frac{0}{0}$ encountered}.” While thisˇ gives one of the two attached pictures:

Graphics[{EdgeForm[Black], polys}, PlotRange -> {{-3, 4}, {0, 2.4}},   Frame -> True, FrameTicks -> False] 

Find a sequence of computable functions that its compute time in WC is growing to infinity in a sense

Define $ T:\mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ by:

$ T(k,n) = 2^{T(k-1,n)}$

$ T(0,n) = 1$

$ T(1,n) = 2^n$

$ T(2,n) = 2^{2^n}$

And denote the WC time of computing $ f(n)$ as $ WC(f(n))$

Can we find computable functions $ f_i: \mathbb{N} \rightarrow \mathbb{N}$ s.t computing $ f_i(n)$ takes $ \Omega(T(i,n))$ W.C and such that for every other computable function $ f \equiv f_i$ $ \lim_{n \rightarrow \infty} \frac{WC(f_i(n))}{WC(f(n))} \neq 0$ ?

I am in a sense trying to find a concrete example of functions that takes at least X time to compute where X can be as large (asymptotically) as we want.

I was thinking maybe $ f_i(n) =$ the $ T(i,n)$ ‘th digit of $ \pi$

but not quite sure this fulfills the second condition

Would Ultimate L really “[reduce] all questions of set theory to axioms of strong infinity”?

According to these slides, the axiom $ V = \mathrm{Ultimate} \,L$ has the following consequences (p. 55):

  1. It implies the Continuum Hypothesis.

  2. It reduces all questions of set theory to axioms of strong infinity (which are with intrinsic justification).

  3. Provides an axiomatic foundation for set theory which is immune to independence by Cohen’s method.

If true, this sounds totally amazing. However, I’m skeptical that (2) is really an accurate description of one of the consequences of $ V = \mathrm{Ultimate} \,L.$ In particular, I was under the impression that there’s no currently-accepted definition of what makes an axiom a large cardinal axiom.

Question. Does anyone here understand enough of the technical details of Woodin’s proposal to be able to comment on the accuracy of the above statements, especially (2)?

Homotopy fibers of infinity functors

Let $ F: C \to D$ be an infinity functor. Is it true that the homotopy fiber at $ y$ can be described as $ C \times_D D^{\simeq}_{/y}$ ? If not, is there a simple formula resembling this one?

Beside the infinity structure, its points are pairs $ (x \in C, s: F(x) \to y \text{ equivalence})$ . I heard this in a class but never seen a proof. I came now with a weird fact that seems a consequence of this fact and I would like to have a confirm.

Thanks, Andrea

How to solve a Poisson’s differential equation with a boundary condition at infinity?

Context: This question is relevant to the physical problem of calculating potential for a set of p-n-p junctions. We have to solve a Poisson’s differential equation for a p-n-p junction with potential equal zero outside it on the left and right sides. For simplification and due to symmetry law we analyze only right side from 0 to some delta (from which potential is the same as for Infinity) and do not analyze left side. Boundary conditions are that in point on Infinity function and its 1st derivative is equal 0. Derivative in x=0 is equal 0. Alpha is a random very small number for Fermi step. In code bcd are boundary conditions

α = 0.00001; bcd1 = ϕ'[0] == 0; bcd2 = ϕ'[Infinity] == 0; bcd3 = ϕ[Infinity] == 0; eqn = Div[ Grad  [ϕ[x], x], x] == -((1/(exp ((x - 1)/α) + 1)) - (1/(exp (((-x - 1)/α) + 1))) + exp (-ϕ[x]) - exp (ϕ[x])); DSolve[{eqn, bcd1, bcd2, bcd}, ϕ, {x, 0, Infinity}] 

i have tried to use numbers(some delta from which Phi is 0) instead of Infinity or set boundary conditions like

ϕ'[x == 0] == 0 ϕ[x == -Infinity] == 0 ϕ'[x == -Infinity] == 0 

and put it directly into eqn but it does not seem to work. And I obtain as a result

DSolve[{Div[Grad[ϕ[x],x],x] == 1/(exp (1 + 100000. (-1 - x))) - 1/(1 + 100000. exp (-1 + x)) + 2 exp ϕ[x], Derivative[1][ϕ][0] == 0,    Derivative[1][ϕ][∞] == 0, ϕ[∞] ==  0}, ϕ, {x, 0, ∞}] 

If I try to vary boundary conditions or use more complex version of equation I obtain this

DSolve::dsvar: ∞ (-∞..) cannot be used as a variable. 

Thank you for your time.