How to calculate the infimum of a quantified sequence ratio with integer constraints?

Questions
,
1

If f:\mathbb{N}\to\mathbb{R} and g:\mathbb{N}\to\mathbb{R} are arbitrary functions, I want to calculate this equation with Mathematica:

\begin{equation} c=\inf\left\{|1-\mathbf{c_1}|:\forall(\epsilon>0)\exists(\mathbf{c_1}>0)\forall(r\in\mathbb{N})\exists(v\in\mathbb{N})\left(\left|\frac{f(r)}{g(v)}-\mathbf{c_1}\right|<\varepsilon\right)\right\}, \end{equation}

Here is what it does:

To obtain c, we want \mathbf{c_1} must satisfy the following:

  1. \mathbf{c_1} is positive
  2. \mathbf{c_1} satisfies 1. and the quantified statement in
    Equation
  3. \mathbf{c_1} satisfies 1. and 2., and has the smallest absolute
    difference from 1.

Here is what I tried:

Clear["Global`*"]
F[r_] := F[r] = r! + 1 
G[v_] := G[v] = 2 v! + 1 (* G can be any arbitrary function *)
c[r_] := FindMinimum[{N[1 - RealAbs[1 - F[r]/G[v]]], 
   Between[v, {1, 10000}] && v \[Element] Integers}, {v}]
Limit[c[r], r -> Infinity]

However, I get the following error message:

During evaluation of In[552]:= FindMinimum::eqineq: Constraints in {v\[Element]\[DoubleStruckCapitalZ],1<=v,v<=10000} are not all equality or inequality constraints. With the exception of integer domain constraints for linear programming, domain constraints or constraints with Unequal (!=) are not supported.

Out[556]= 
\!\(\*UnderscriptBox[\(\[Limit]\), \(r \[Rule] \[Infinity]\)]\) 
 FindMinimum[{N[1 - RealAbs[1 - F[r]/G[v]]], 
   Between[v, {1, 10000}] && v \[Element] Integers}, {v}]
2

With AI I have (I don’t know is this correct:)

Beztytułussss.png
Beztytułussss.png991×173 24.1 KB

Open in Wolfram Cloud ↗
3

This version does not give error messages:

 c[r_] := 
     Minimize[{1 - Abs[1 - F[r]/G[v]], 1 <= v <= 10000}, 
      Element[v, Integers]]

or else this

c[r_] := Min[Table[1 - Abs[1 - F[r]/G[v]], {v, 1, 10000}]]

but it is very much unlikely that you are going to calculate a symbolic Limit of this function.

4

Define h(v) = |g(v)| and suppose h has some maximum value h_\text{max} = h(v_\text{max}) over the natural numbers.

If f(r) \leq h_{\text{max}} then v_\text{max} will be the minimizer of your objective function, since

f(r) \leq h_{\text{max}} \implies 1-\frac{f(r)}{g(v)} \geq 0

and now we can remove the Abs from the objective function and we see we’re just maximizing h(v):

1- \left| 1-\frac{f(r)}{g(v)}\right| = 1 - \left( 1-\frac{f(r)}{g(v)}\right) =\frac{f(r)}{g(v)}

A similar method can applied for the r where f(r) ≥ h_{\text{max}}

In your case, g(v) grows to infinity, so 1- \left| 1-\frac{f(r)}{g(v)}\right| will continue to shrink as v grows