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Finding a Published Research Paper which Meaningfully Averages the Most Pathalogical Functions (V2)

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Bharath Krishnan^*

This version is not peer-reviewed

Submitted:

14 October 2024

Posted:

15 October 2024

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Abstract

I want to meaningfully average a pathalogical function (i.e., an everywhere surjective function whose graph has zero Hausdorff measure in its dimension). In case this impossible, we wish to average a nowhere continuous function defined on the rationals. We do this by taking the most generalized, satisfying extension of the expected value, w.r.t the Hausdorff measure in its dimension, on bounded functions with domains of finite measure which take finite values only. As of now, I’m unable to solve this due to limited knowledge of advanced math and most people are too busy to help. Therefore, I’m wondering if anyone knows a research paper which solves my doubts. Unlike the previous paper, “Finding a Research Paper Which Meaningfully Averages Pathalogical Functions”) we add motivations, examples, and change parts of the original paper.

Keywords:

Subject: Computer Science and Mathematics - Mathematics

1. Intro

Let

n \in N

and suppose function

f : A \subseteq R^{n} \to R

, where A and f are Borel. Let

\dim_{H} (\cdot)

be the Hausdorff dimension, where

H^{\dim_{H} (\cdot)} (\cdot)

is the Hausdorff measure in its dimension on the Borel

σ

-algebra.

1.1. First special case of f

If the graph of f is G, is there an explicit f where:

(1): The function f is everywhere surjective [5]
(2): $H^{\dim_{H} (G)} (G) = 0$

1.1.1. Potential Answer

A = R

, using this MathOverflow post [8], define f such that:

Consider a Cantor set $C \subseteq [0, 1]$ with Hausdorff dimension 0 [9]. Now consider a countable disjoint union $\cup_{m \in N} C_{m}$ such that each $C_{m}$ is the image of C by some affine map and every open set $O \subseteq [0, 1]$ contains $C_{m}$ for some m. Such a countable collection can be obtained by e.g. at letting $C_{m}$ be contained in the biggest connected component of $[0, 1] ∖ (C_{1} \cup \dots \cup C_{m - 1})$ (with the center of $C_{m}$ being the middle point of the component).

Note that $\cup_{m} C_{m}$ has Hausdorff dimension 0, so $(\cup_{m} C_{m}) \times [0, 1] \subseteq R^{2}$ has Hausdorff dimension one [7].

Now, let $g : [0, 1] \to R$ such that ${g |}_{C_{m}}$ is a bijection $C_{m} \to R$ for all m (all of them can be constructed from a single bijection $C \to R$ , which can be obtained without choice, although it may be ugly to define) and outside $\cup_{m} C_{m}$ let g be defined by $g (x) = h (x)$ , where $h : [0, 1] \to R$ has a graph with Hausdorff dimension 2 [21] (this doesn’t require choice either).

Then the function g has a graph with Hausdorff dimension 2 and is everywhere surjective, but its graph has Lebesgue measure 0 because it is a graph (so it admits uncountably many disjoint vertical translates).

Note, we can make the construction with union of $C_{m}$ rather explicit as follows. Split the binary expansion of x as strings of size with a power of two, say $x = 0.1101000010 \dots$ becomes $(s_{0}, s_{1}, s_{2}, \dots) = (1, 10, 1000, \dots)$ . If this sequence eventually contains only strings of the form $0 \dots 0$ or $1 \dots 1$ , say after $s_{k}$ , then send it to $y = \sum_{i > 0} ϵ_{i} 2^{- i}$ , where $s_{k + i} = ϵ_{i} \dots ϵ_{i}$ . Otherwise, send it to the explicit continuous function h given by the linked article [21]. This will give you something from $[0, 1) \to [0, 1)$

Finally, compose an explicit (reasonable) bijection from $[0, 1)$ to $R$ . In your case, the construction can be easily adapted so that the $[0, 1]$ or $[0, 1)$ target space is actually $(0, 1)$ , then compose with $t \mapsto (1 - 2 x) / (x^{2} - x)$ .

In case this function is impossible to average, consider the following example:

1.2. Second Special Case of f

Suppose, we define

A = Q

, where

f : A \to R

, such that:

f (x) = \{\begin{matrix} 1 & x \in \{(2 s + 1) / (2 t) : s \in Z, t \in N, t \neq 0\} \\ 0 & x \notin \{(2 s + 1) / (2 t) : s \in Z, t \in N, t \neq 0\} \end{matrix}

(1)

In the following sections, we shall see why we chose

§

1.1 and this function.

1.3. Attempting to Analyze/Average f

Suppose, the expected value of f is:

E [f] = \frac{1}{H^{\dim_{H} (A)} (A)} \int_{A} f d H^{\dim_{H} (A)}

(2)

Note, using

§

1.1, explicit f is pathological since it’s everywhere surjective and difficult to meaningfully average (i.e., the most generalized, satisfying (§2) extension of

E [f]

is non-finite).

Thus, we want the most generalized, satisfying extension of

E [f]

on bounded f, where the extension takes finite values for all f defined within §1.1. Moreover, suppose:

(1): The sequence of bounded functions is $\vec{f_{s}} = {(f_{r_{s}}^{(s)})}_{r_{s} \in N}$
(2): The sequence of bounded functions converges to f: i.e., $f_{r_{s}}^{(s)} \to f$
(3): The generalized, satisfying extension of $E [f]$ is $E [f_{r_{s}}^{(s)}]$ : i.e., there exists a $s \in N$ , where $E [f_{r_{s}}^{(s)}]$ is finite
(4): There exists $k, v \in N$ where the expected value of $\vec{f_{k}}$ and $\vec{f_{v}}$ are finite and non-equivelant: i.e.,

$- \infty < E [f_{r_{k}}^{(k)}] \neq E [f_{r_{v}}^{(v)}] < + \infty$

(Whenever (4) is true, (3) is non-unique.)

1.3.1. Example Proving $§$ 1.3 (1)-(4) Correct

Using the second case of

f : A \to R

in §1.2, where

A = Q

, and:

f (x) = \{\begin{matrix} 1 & x \in \{(2 s + 1) / (2 t) : s \in Z, t \in N, t \neq 0\} \\ 0 & x \notin \{(2 s + 1) / (2 t) : s \in Z, t \in N, t \neq 0\} \end{matrix}

(3)

suppose:

{(A_{r}^{★})}_{r \in N} = {(\{c / r! : c \in Z, - r \cdot r! \leq c \leq r \cdot r!\})}_{r \in N}

and

{(A_{j}^{★ ★})}_{j \in N} = (\{c / d : c \in Z, d \in N, d \leq j, - d j \leq c \leq d j\})

where for

f_{r}^{★} : A_{r}^{★} \to R

f_{r}^{★} (x) = f (x) for all x \in A_{r}^{★}

(4)

and for

f_{j}^{★ ★} : A_{j}^{★ ★} \to R

f_{j}^{★ ★} (x) = f (x) for all x \in A_{j}^{★ ★}

(5)

Note,

f_{r}^{★}

and

f_{j}^{★ ★}

are bounded since f is bounded (i.e., criteria (1) of §1.3 is satisfied). Also, the set-theoretic limit of

{(A_{r}^{★})}_{r \in N}

and

{(A_{j}^{★ ★})}_{j \in N}

A = Q

: i.e.,

\begin{matrix} \underset{r \to \infty}{lim sup} A_{r}^{★} = ⋂_{r \geq 1} ⋃_{q \geq r} A_{q}^{★} \\ \underset{r \to \infty}{lim inf} A_{r}^{★} = ⋃_{r \geq 1} ⋂_{q \geq r} A_{q}^{★} \end{matrix}

where:

\underset{r \to \infty}{lim sup} A_{r}^{★} = \underset{r \to \infty}{lim inf} A_{r}^{★} = A = Q

(We are not sure how to prove this; however, a mathematician which specializes in rational numbers and set-theoretic limits should be able to verify the former.)

Hence, one can see that

f_{r}^{★} : A_{r}^{★} \to R

and

f_{j}^{★ ★} : A_{j}^{★ ★} \to R

converges to

f : A \to R

(i.e., criteria (2) of §1.3 is satisfied).

Now, suppose we want to average

f_{r}^{★}

and

f_{j}^{★ ★}

which we denote

E [f_{r}^{★}]

and

E [f_{j}^{★ ★}]

. Note, this is the same as computing the following (i.e., the cardinality is

| \cdot |

and the absolute value is

| | \cdot | |

\forall (ϵ > 0) \exists (N \in N) \forall (r \in N) (r \geq N \Rightarrow ||\frac{1}{| A_{r}^{★} |} \int_{A_{r}^{★}} f (x) - E [f_{r}^{★}]|| < ϵ)

(6)

\forall (ϵ > 0) \exists (N \in N) \forall (j \in N) (j \geq N \Rightarrow ||\frac{1}{| A_{j}^{★ ★} |} \int_{A_{j}^{★ ★}} f (x) - E [f_{j}^{★ ★}]|| < ϵ)

(7)

If we assume,

E [f_{r}^{★}] = 1

in eq. 6, then [15]:

The integral $\int_{A^{★}} f (x) d x$ counts the number of fractions with even denominator and odd numerator in set $A^{★}$ , after cancelling all possible factors of 2 in the fraction. Let’s consider the first case. We can write $| | 1 - | A_{r}^{★} |^{- 1} \int_{A_{r}^{★}} f | | = (| A_{r}^{★} | - \int_{A_{r}^{★}} f) / | A_{r}^{★} | = H (r) / | A_{r}^{★} |$ , where $H (r)$ counts the fractions $x = c / r!$ in $A_{r}^{★}$ that are not counted in $\int_{A^{★}} f$ , i.e., for which $f (x) = 0$ . This is the case when the denominator is odd after the cancellation of the factors of 2, i.e., when the numerator c has a number of factors of 2 greater than or equal to that of $r!$ , which we will denote by $V (r) : = v_{2} (r!)$ a.k.a the 2-valuation of $r!$ , $oeis : A 11371 (r) = r - O (ln (r))$ [17]. That means, c must be a multiple of $2^{V (r)}$ . The number of such c with $- r \cdot r! \leq c \leq r \cdot r$ is simply the length of that interval, equal to $| A_{r}^{★} | = 2 r (r!) + 1$ , divided by $2^{V (r)}$ . Thus, $| | 1 - | A_{r}^{★} |^{- 1} \int_{A_{r}^{★}} f | | = [| A_{r}^{★} | / 2^{V (r)}] / |A_{r}^{★}| \sim 1 / 2^{V (r)} = 1 / 2^{n - O (log n)}$ . This obviously tends to zero, proving $E [f_{r}^{★}] = 1$

Since

E [f_{r}^{★}] = 1

is finite, this proves §1.3 criteria (3).

Last, we need to show

E [f_{j}^{★ ★}] = 1 / 3

, where

E [f_{r}^{★}] \neq E [f_{j}^{★ ★}]

proving §1.3 criteria (4).

Concerning the second case [15], it’s again simpler to consider the complementary set of $x \in A_{j}^{★ ★}$ such that the denominator is odd when all possible factors of 2 are canceled. We can see that for $j = 2 p - 1$ , and these include obviously all those we had for smaller j. The “new" elements in $A_{j}^{★ ★}$ with $j = 2 p - 1$ are those which have denominator $d = 2 p - 1$ when written in lowest terms. Their number is equal to the number of $κ < d$ , $gcd (κ, d) = 1$ , which is given by Euler’s $ϕ$ function. Since we also consider negative fractions, we have to multiply this by 2. Including $x = 0$ , we have $G (j) = | \{x \in A_{j}^{★ ★} | f (x) = 0\} | = 1 + 2 \sum_{0 \leq κ \leq j / 2} ϕ (2 κ + 1)$ . There is no simple explicit expression for this (cf. oeis:A99957 [18]), but we know that $G (j) = 1 + 2 \cdot A 99957 (j / 2) \sim 2 \cdot 8 {(j / 2)}^{2} / π^{2} = 4 j^{2} / π^{2}$ [18]. On the other hand, the total number of all elements of $A_{j}^{★ ★}$ is $| A_{j}^{★ ★} | = 1 + 2 \sum_{1 \leq κ \leq j} ϕ (κ)$ , since each time we increase j by 1, we have the additional fractions with the new denominator $d = j$ and numerators being coprime with d, again with + or − sign. From oeis:A002088 [16] we know that $\sum_{1 \leq κ \leq j} ϕ (κ) = 3 j^{2} / π^{2} + O (j log j)$ , so $| A_{j}^{★ ★} | \sim 6 j^{2} / π^{2}$ , which finally gives $| A_{j}^{★ ★} |^{- 1} \int_{A_{j}^{★ ★}} f = (| A_{j}^{★ ★} | - G (j)) / | A_{j}^{★ ★} | \sim (6 - 4) / 6 = 1 / 3$ as desired.

Hence,

E [f_{j}^{★}] = 1 / 3

and

E [f_{r}^{★}] \neq E [f_{j}^{★ ★}]

proving §1.3 criteria (4).

Therefore, in

§

1.3, since:

Theorem 1.

The set of all Borel f, where

E [f]

is finite, forms a shy [19] subset of all Borel measurable functions in

R^{A}

Note 2

(Proof theorem 1 is true). We follow the argument presented in example 3.6 of this paper [19], take

X : = L^{0} (A)

(measurable functions over A), let P denote the one-dimensional sub-space of A consisting of constant functions (assuming the Lebesgue measure on A) and let

F : = L^{0} (A) ∖ L^{1} (A)

(measurable functions over A with no finite integral). Let

λ_{P}

denote the Lebesgue measure over P, for any fixed

f \in F

λ_{P} (\{α \in R | \int_{A} (f + α) d μ < \infty\}) = 0

Meaning P is a one-dimensional, so f is a 1-prevelant set.

Theorem 3.

The set of all Borel f, where the generalized, satisfying extension of

E [f]

is non-unique

(§

1.3, ((4)), forms a prevalent [19] subset of all Borel measurable functions in

R^{A}

Note 4

(Possible method to proving theorem 3 true). Note, we first must prove that only f whose lines of symmetry intersect at one point have the same expected value for any bounded sequence of functions converging to f. Notice, the set of all symmetric Borel functions forms a shy subset of the set of all Borel measurable functions in

R^{A}

. Hence, the set of all functions, where their lines of symmetry intersect at one point also forms a shy subset of the set of all Borel measurable functions in

R^{A}

(i.e., a subset of a shy subset is also shy).

Since thm. 1 and 3 are true, we need to fix the problems the theorems address with the following:

1.3.2. Blockquote

We want to find an unique, satisfying (§3) extension of $E [f]$ , on bounded functions to f which takes finite values only, such that the set of all f with this extension forms:

(1): a prevalent [19] subset of $R^{A}$
(2): If not prevalent then a non-shy (i.e., neither prevalent [19] nor shy [19]) subset of $R^{A}$ .

For the sake of clarity & precision, we describe examples of “extending

E [f]

on all A with positive & finite Hausdorff measure" (§2) and use the examples to define the terms “unique & satisfying" (§3) in the blockquote of this section.

2. Extending the Expected Value w.r.t the Hausdorff Measure

The following are two methods to determining the most generalized, satisfying extension of

E [f]

on all A with a positive and finite Hausdorff measure:

(1): One way is defining a generalized, satisfying extension of the Hausdorff measure on all A with positive & finite measure which takes positive, finite values for all Borel A. This can theoretically be done in the paper “A Multi-Fractal Formalism for New General Fractal Measures"[1] by taking the expected value of f w.r.t the extended Hausdorff measure.
(2): Another way is finding a generalized, satisfying average of all A in the fractal setting. This can be done with the papers “Analogues of the Lebesgue Density Theorem for Fractal Sets of Reals and Integers" [3] and “Ratio Geometry, Rigidity and the Scenery Process for Hyperbolic Cantor Sets" [4] where we take the expected value of f w.r.t the densities in [3,4].

(Note, the methods in these papers could be used in §3.2 to answer the blockquote of

§

1.3.2.)

3. Attempt to Define “Unique and Satisfying" in The Blockquote of §1.3

3.1. Note

Before reading, when §3.2 is unclear, see §5 for clarity. In §5, we define:

(1): “Sequences of bounded functions converging to f" (§5.1)
(2): “Equivalent sequences of bounded functions" (§5.3, def. 7)
(3): “Nonequivalent sequences of bounded functions" (§5.3, def. 9)
(4): The “measure" of a property on a sequence of bounded functions which increases at rate linear or super-linear to that of “non-equivelant" sequences of bounded functions (§5.4.1, §5.4.2)
(5): The “actual" rate of expansion on a sequence of bounded sets (§5.5)

3.2. Leading Question

To define unique and satisfying in the blockquote of the §1.3, we take the expected value of a sequence of bounded functions chosen by a choice function. To find the choice function we ask the leading question...

If we make sure to:

(A): See $§$ 3.1 and (C)-(E) when something is unclear
(B): Take all sequences of bounded functions which converge to f
(C): Define C to be chosen center point of $R^{n + 1}$
(D): Define E to be the chosen, fixed rate of expansion of a sequence on the graph of bounded functions
(E): Define $E$ to be actual rate of expansion of a sequence on the graph of bounded functions ( $§$ )5.5

Does there exist a unique choice function which chooses a unique set of equivalent sequences of bounded functions where:

(1)

The chosen, equivelant sequences of bounded functions should satisfy (B).

(2)

The “measure" of the graph of all chosen, equivalent sequences of bounded functions which satisfy (B) should increase at a rate linear or superlinear to that of non-equivelant sequences of bounded functions satisfying (B)

(3)

The expected values, defined in the papers of §2, for all equivalent sequences of bounded functions are equivalent and finite

(4)

For the chosen, equivalent sequences of bounded functions satisfying (1), (2) and (3), when f is unbounded (i.e, skip when f is bounded):

The absolute difference between criteria (3) and the $(n + 1)$ -th coordinate of C is the less than or equal to that of non-equivalent sequences of bounded functions satisfying (1), (2), and (3)
The “rate of divergence" [20] of $||E - E||$ , using the absolute value $| | \cdot | |$ , is less than or equal to that of non-equivalent sequences of bounded functions which satisfy (a), (b), and (c)

(5)

When set

Q \subseteq R^{A}

is the set of all

f \in R^{A}

, where the choice function chooses all equivalent sequences of bounded functions satisfying (1), (2), (3) and (4), then Q is

(a): a prevelant [19] subset of $R^{A}$
(b): If not (a) then a non-shy (i.e., neither prevelant [19] nor shy [19]) subset of $R^{A}$ .

(6)

Out of all choice functions which satisfy (1), (2), (3), (4) and (5), we choose the one with the simplest form, meaning for each choice function fully expanded, we take the one with the fewest variables/numbers?

(In case this is unclear, see §5.)

3.2.1. Explaining Motivation Behind $§$ 3.2

(1): When defining “the measure" (§5.4.1,§5.4.2) of a function, we want a bounded sequence of functions with a “high" entropic density (i.e., we aren’t sure if this is infact what the “measure" measures.) For example, when $A = R$ and f is everywhere surjective [5], the “measure" chooses bounded sequence of functions whose lines of symmetry intersect at one point rather than non-symmetrical functions §5.4.1-§6, §8.
(2): Note, the later criteria doesn’t apply to bounded functions: e.g., using §1.3.1, when $A = Q$ and $f : A \to R$ , where:

$f (x) = \{\begin{matrix} 1 & x \in \{(2 s + 1) / (2 t) : s \in Z, t \in N, t \neq 0\} \\ 0 & x \notin \{(2 s + 1) / (2 t) : s \in Z, t \in N, t \neq 0\} \end{matrix}$

(8)

depending on the sequence of bounded function $(f_{r}^{★})$ chosen which converge to f: $E [f_{r}^{★}]$ can be any number in $[0, 1]$ . Since $[0, 1]$ is bounded, we need only §3.2.1 crit. (1) to solve the blockquote of §1.3.2.
(3): Using an unbounded function in §1.1 (i.e., an everywhere surjective f whose graph has zero Hausdorff measure in its dimension) depending on the sequence of bounded functions ${(f_{r}^{★})}_{r \in N}$ chosen which converge tof: $E [f_{r}^{★}]$ can be any real number (when it exists). To fix this, take all ${(f_{r}^{★})}_{r \in N}$ , where the $E [f_{r}^{★}]$ has smallest absolute difference from the $(n + 1)$ -th coordinate of a reference point (i.e., the center point $C \in R^{n}$ ). The problem is there exists f, where the expected value of non-equivalent sequences (§5.3, def. 9) of bounded functions have the same minimum absolute difference from $(n + 1)$ -th coordinate of C.
(4): Thus, we take the sequence of functions whose actual rate of expansion $E$ from C ( $§$ 5.5) “diverges" [20] pp.275-322 at the smallest rate from the chosen, fixed rate of expansion E from C (i.e., the “rate of divergence of $||E - E||$ , using the absolute value $| | \cdot | |$ , is less than or equal to that of all the non-equivalent sequences of bounded functions which satisfy §3.2 criteria (1), (2), and (3)).
(5): Finally, since there might still be non-equivalent sequences (§5.3, def. 9) of bounded functions which satisfy $§$ 3.2.1 criteria (1), (3) and (4), but their graphs are congruent with different $E [f_{r}^{★}]$ , we use equation T in §6.3.1 eq. 144 to choose a unique set of all equivalent sequences of bounded sets with the same expected value.

I’m convinced the expected values of the sequences of bounded functions chosen by a choice function which answers the leading question aren’t unique nor satisfying enough to answer the blockquote of §1.3.2. Still, adjustments are possible by changing the criteria or by adding new criteria to the question.

4. Question Regarding My Work

Most don’t have time to address everything in my research, hence I ask the following:

Is there a research paper which already solves the ideas I’m working on? (Non-published papers, such as mine [12], don’t count.)

Using AI, papers that might answer this question are “Prediction of dynamical systems from time-delayed measurements with self-intersections" [2] and “A Hausdorff measure boundary element method for acoustic scattering by fractal screens" [6].

Does either of these papers solve the blockquote of

§

1.3.2?

5. Clarifying §3

Suppose

{(f_{r})}_{r \in N}

is a sequence of bounded functions converging to f and

{(G_{r})}_{r \in N}

is a sequence of the graph of each

f_{r}

. Let

\dim_{H} (\cdot)

be the Hausdorff dimension and

H^{\dim_{H} (\cdot)} (\cdot)

be the Hausdorff measure in its dimension on the Borel

σ

-algebra.

See §3.2 once reading §5, and consider the following:

Is there a simpler version of the definitions below?

5.1. Defining Sequences of Bounded Functions Converging to f

The sequence of bounded functions

{(f_{r})}_{r \in N}

, where

f_{r} : A_{r} \to R

and

{(A_{r})}_{r \in N}

is a sequence of bounded sets, converges to function

f : A \to R

when:

For any $x \in A$ there exists a sequence $x \in A_{r}$ s.t. $x \to (x_{1}, \cdot \cdot \cdot, x_{n})$ and $f_{r} (x) \to f (x_{1}, \cdot \cdot \cdot, x_{n})$ (see [10] for info).

Example 5

(Example of §5.1). If

A = R

and

f : A \to R

, where

f (x) = 1 / x

, then an example of

{(f_{r})}_{r \in N}

, such that

f_{r} : A_{r} \to R

is:

(1): ${(A_{r})}_{r \in N} = {([- r, - 1 / r] \cup [1 / r, r])}_{r \in N}$
(2): $f_{r} (x) = 1 / x$ for $x \in A_{r}$

Example 6

(More Complex Example). If

A = R

and

f : A \to R

, where

f (x) = x

, then an example of

{(f_{r})}_{r \in N}

, such that

f_{r} : A_{r} \to R

is:

(1): ${(A_{r})}_{r \in N} = {([- r, r])}_{r \in N}$
(2): $f_{r} (x) = x + (1 / r) sin (x)$ for $x \in A_{r}$

5.2. Expected Value of Bounded Sequence of Functions

(f_{r})

converges to f (§5.1), the expected value of f w.r.t

{(f_{r})}_{r \in N}

E [f_{r}]

(when it exists), where the absolute value is

| | \cdot | |

, such that

E [f_{r}]

satisfies:

\begin{matrix} \forall (ϵ > 0) \exists (N \in N) \forall (r \in N) (r \geq N \Rightarrow ||\frac{1}{H^{\dim_{H} (A_{r})} (A_{r})} \int_{A_{r}} f_{r} d H^{\dim_{H} (A_{r})} - E [f_{r}]|| < ϵ) \end{matrix}

(9)

Note,

E [f_{r}]

can be extended by using

§

5.2.1. Example

Using example 5, when

{(f_{r})}_{r \in N} = {(\{(x, 1 / x) : x \in [- r, - 1 / r] \cup [1 / r, r]\})}_{r \in N}

where:

(1): ${(A_{r})}_{r \in N} = {([- r, - 1 / r] \cup [1 / r, r])}_{r \in N}$
(2): $f_{r} (x) = 1 / x$ for $x \in A_{r}$

If we assume

E [f_{r}] = 0

\begin{matrix} \forall (ϵ > 0) \exists (N \in N) \forall (r \in N) (r \geq N \Rightarrow ||\frac{1}{H^{\dim_{H} (A_{r})} (A_{r})} \int_{A_{r}} f_{r} d H^{\dim_{H} (A_{r})} - E [f_{r}]|| < ϵ) = \end{matrix}

(10)

\begin{matrix} \forall (ϵ > 0) \exists (N \in N) \forall (r \in N) \end{matrix}

(11)

\begin{matrix} (r \geq N \Rightarrow ||\frac{1}{H^{\dim_{H} ([- r, - 1 / r] \cup [1 / r, r])} ([- r, - 1 / r] \cup [1 / r, r])} \int_{[- r, - 1 / r] \cup [1 / r, r]} 1 / x d H^{\dim_{H} ([- r, - 1 / r] \cup [1 / r, r])} - 0|| < ϵ) = \end{matrix}

(12)

\begin{matrix} \forall (ϵ > 0) \exists (N \in N) \forall (r \in N) (r \geq N \Rightarrow ||\frac{1}{H^{1} ([- r, - 1 / r] \cup [1 / r, r])} \int_{[- r, - 1 / r] \cup [1 / r, r]} 1 / x d H^{1}|| < ϵ) = \end{matrix}

(13)

\begin{matrix} \forall (ϵ > 0) \exists (N \in N) \forall (r \in N) (r \geq N \Rightarrow ||\frac{1}{(- 1 / r - (- r)) + (r - 1 / r)} (\int_{- r}^{- 1 / r} 1 / x d x + \int_{1 / r}^{r} 1 / x d x)|| < ϵ) = \end{matrix}

(14)

\begin{matrix} \forall (ϵ > 0) \exists (N \in N) \forall (r \in N) (r \geq N \Rightarrow ||\frac{1}{(r - 1 / r) + (- 1 / r + r)} (ln (| | x | |) + C |_{- r}^{- 1 / r} + ln (| | x | |) + C |_{1 / r}^{r})|| < ϵ) = \end{matrix}

(15)

\begin{matrix} \forall (ϵ > 0) \exists (N \in N) \forall (r \in N) (r \geq N \Rightarrow ||\frac{1}{(r - 1 / r) + (- 1 / r + r)} (ln (| | - r | |) - ln (| | - 1 / r | |) + ln (| | r | |) - ln (| | 1 / r | |))|| < ϵ) = \end{matrix}

(16)

\begin{matrix} \forall (ϵ > 0) \exists (N \in N) \forall (r \in N) (r \geq N \Rightarrow ||\frac{1}{2 r - 2 / r} \cdot 4 ln (r)|| < ϵ) = \end{matrix}

(17)

To prove eq. 17 is true, recall:

\begin{matrix} r ≪ e^{r / 2}, e^{1 / r} ≪ e^{r} \end{matrix}

(18)

\begin{matrix} r ≪ e^{r / 2}, e^{1 / (2 r)} ≪ e^{r / 2} \end{matrix}

(19)

\begin{matrix} r e^{1 / (2 r)} ≪ e^{r / 2} \end{matrix}

(20)

\begin{matrix} r ≪ e^{r / 2} / e^{1 / (2 r)} \end{matrix}

(21)

\begin{matrix} r ≪ e^{r / 2 - 1 / (2 r)} \end{matrix}

(22)

\begin{matrix} ln (r) ≪ r / 2 - 1 / (2 r) \end{matrix}

(23)

\begin{matrix} 4 ln (r) ≪ 2 r - 2 / r \end{matrix}

(24)

Hence, for all

ε > 0

\begin{matrix} 4 ln (r) < ε (2 r - 2 / r) \end{matrix}

(25)

\begin{matrix} \frac{4 ln (r)}{2 r - 2 / r} < ε \end{matrix}

(26)

\begin{matrix} ||\frac{4 ln (r)}{2 r - 2 / r}|| < ε \end{matrix}

(27)

Since eq. 17 is true,

E [f_{r}] = 0

. Note, if we simply took the average of f from

(- \infty, \infty)

, using the improper integral, the expected value:

\begin{matrix} lim_{(x_{1}, x_{2}, x_{3}, x_{4}) \to (- \infty, 0^{-}, 0^{+}, + \infty)} \frac{1}{(x_{4} - x_{3}) + (x_{2} - x_{1})} (\int_{x_{1}}^{x_{2}} \frac{1}{x} d x + \int_{x_{3}}^{x_{4}} \frac{1}{x} d x) = \end{matrix}

(28)

\begin{matrix} lim_{(x_{1}, x_{2}, x_{3}, x_{4}) \to (- \infty, 0^{-}, 0^{+}, + \infty)} \frac{1}{(x_{4} - x_{3}) + (x_{2} - x_{1})} (ln (| | x | |) + C |_{x_{1}}^{x_{2}} + ln (| | x | |) + C |_{x_{3}}^{x_{4}}) = \end{matrix}

(29)

\begin{matrix} lim_{(x_{1}, x_{2}, x_{3}, x_{4}) \to (- \infty, 0^{-}, 0^{+}, + \infty)} \frac{1}{(x_{4} - x_{3}) + (x_{2} - x_{1})} (ln (| | x_{2} | |) - ln (| | x_{1} | |) + ln (| | x_{4} | |) - ln (| | x_{3} | |)) \end{matrix}

(30)

+ \infty

(when

x_{2} = 1 / x_{1}

x_{3} = 1 / x_{4}

, and

x_{1} = exp (x_{4}^{2})

) or

- \infty

(when

x_{2} = 1 / x_{1}

x_{3} = 1 / x_{4}

, and

x_{4} = - exp (x_{1}^{2})

), making

E [f]

undefined. (However, using eq. 10-17, we get the

E [f_{r}] = 0

instead of an undefined value.)

5.3. Defining Equivelant and Non-Equivelant Sequences of Bounded Functions

Let

S \subseteq N

be an arbitrary set and define the following sequence of functions:

\vec{f_{1}} = {f_{r_{1}}^{(1)}}_{r_{1} \in N}, \vec{f_{2}} = {f_{r_{2}}^{(2)}}_{r_{2} \in N}, \cdot \cdot \cdot, \vec{f_{s}} = {f_{r_{s}}^{(s)}}_{r_{s} \in N}

Note, the sequences of bounded functions in

\{\vec{f_{s}} : s \in N\}

converges to f and the sequences of the graphs of all functions in each former sequence are:

\begin{matrix} \vec{G_{1}} & = {(graph (f_{r_{1}}^{(1)}))}_{r_{1} \in N} & = {(G_{r_{1}}^{(1)})}_{r_{1} \in N}, \\ \vec{G_{2}} & = {(graph (f_{r_{2}}^{(2)}))}_{r_{2} \in N} & = {(G_{r_{2}}^{(2)})}_{r_{2} \in N}, \\ ⋮ \\ \vec{G_{s}} & = {(graph (f_{r_{s}}^{(s)}))}_{r_{s} \in N} & = {(G_{r_{s}}^{(s)})}_{r_{s} \in N} \end{matrix}

(31)

Definition 7

(Equivelant Sequences of functions). Suppose

S \subseteq N

is an arbitrary set. The sequences of bounded functions in:

\{\vec{f_{s}} : s \in S\}

are equivalent, if for all

k, v \in S

, where

k \neq v

\vec{f_{k}}

and

\vec{f_{v}}

are equivelant: i.e., there exists a

N^{'} \in N

, such for all

r_{k} \geq N^{'}

, there is a

r_{v} \in N

, where:

H^{\dim_{H} (G_{r_{k}}^{(k)})} (G_{r_{k}}^{(k)} Δ G_{r_{v}}^{(v)}) = 0

and for all

r_{v} \geq N^{'}

, there is a

r_{k} \in N

, where:

H^{\dim_{H} (G_{r_{v}}^{(v)})} (G_{r_{k}}^{(k)} Δ G_{r_{v}}^{(v)}) = 0

More, for each

s \in N

, we denote all equivalent sequences of bounded functions to

{\{f_{r_{s}}^{(s)}\}}_{r_{s} \in N}

using the notation

\sim {f_{r_{s}}^{(s)}}_{r_{s} \in N}

5.3.1. Explanation

We define

\vec{f_{k}}

and

\vec{f_{v}}

as equivalent, where

\vec{A_{k}}

and

\vec{A_{v}}

are constant in

f_{r_{k}}^{(k)} : A_{r_{k}}^{(k)} \to R

and

f_{r_{v}}^{(v)} : A_{r_{v}}^{(v)} \to R

, such for all Borel measurable

f \in R^{A}

, when both

E [f_{r_{k}}^{(k)}]

and

E [f_{r_{v}}^{(v)}]

exist:

E [f_{r_{k}}^{(k)}] = E [f_{r_{v}}^{(v)}] (§ 5.2)

Hence, consider the following:

Theorem 8.

If the sequence of functions in:

\{\vec{f_{s}} : s \in S\}

are equivalent, then for all

k, v \in S

, where

k \neq v

E [f_{r_{k}}^{(k)}] = E [f_{r_{v}}^{(v)}]

Note, this explains criteria ((3)) in

§

5.3.2. Example of Equivalent Sequences of Bounded Functions

Suppose we define

\vec{f_{1}} = {(f_{r_{1}}^{(1)})}_{r_{1} \in N}

, where

f_{r_{1}}^{(1)} : A_{r_{1}}^{(1)} \to R

and

\vec{f_{2}} = {(f_{r_{2}}^{(2)})}_{r_{2} \in N}

, where

f_{r_{2}}^{(2)} : A_{r_{2}}^{(2)} \to R

such that:

(1): $f_{r_{1}}^{(1)} (x) = x$
(2): $A_{r_{1}}^{(1)} = [- r_{1} - 2, r_{1} + 2]$
(3): $G_{r_{1}}^{(1)} = \{(x, x) : x \in [- r_{1} - 2, r_{1} + 2]\}$

and

(1): $f_{r_{2}}^{(2)} (x) = x$
(2): $A_{r_{2}}^{(1)} = [- r_{2}, r_{2}] \cup (Q \cap [- r_{2} - 1, r_{2} + 1])$
(3): $G_{r_{2}}^{(2)} = \{(x, x) : x \in [- r_{2}, r_{2}] \cup (Q \cap [- r_{2} - 1, r_{2} + 1])\}$

Note, using def. 7,

S = {1, 2}

k = 1

, and

v = 2

. In other words,

r_{k} = r_{1}

and

r_{v} = r_{2}

. Now, suppose

N = 3

, where

r_{1} \in N

and

r_{2} \in N

. Then,

For all

r_{1} \geq N = 3

, there exists a

5 \leq r_{2} = : r_{1} + 2 \in N

, where:

H^{\dim_{H} (G_{r_{1}}^{(1)})} (G_{r_{1}}^{(1)} Δ G_{r_{2}}^{(2)}) = 0

(32)

We show this with the following:

In eq. 32, since

\{(x, x) : x \in [- r_{1} - 2, r_{1} + 2]\}

is a 1-d interval,

\dim_{H} (G_{r_{1}}^{(1)}) = 1

. Hence,

\begin{matrix} H^{1} (\{(x, x) : x \in [- r_{1} - 2, r_{1} + 2]\} Δ \{(x, x) : x \in [- r_{2}, r_{2}] \cup (Q \cap [- r_{2} - 1, r_{2} + 1])\}) = \end{matrix}

(33)

\begin{matrix} H^{1} (\{(x, x) : x \in [- r_{1} - 2, r_{1} + 2] Δ ([- (r_{1} + 2), (r_{1} + 2)] \cup (Q \cap [- (r_{1} + 2) - 1, (r_{1} + 2) + 1])\}) = \end{matrix}

(34)

\begin{matrix} H^{1} (\{(x, x) : x \in [- r_{1} - 2, r_{1} + 2] Δ ([- (r_{1} + 2), (r_{1} + 2)] \cup (Q \cap [- r_{1} - 3, r_{1} + 3]))\}) = \end{matrix}

(35)

\begin{matrix} H^{1} (\{(x, x) : x \in Q \cap ([r_{1} + 2, r_{1} + 3]) \cup (Q \cap [- r_{1} - 3, r_{1} - 2])\}) = \end{matrix}

(36)

\begin{matrix} 0 \end{matrix}

(37)

We also show:

(2)

For all

r_{2} \geq N = 3

, there exists a

1 \leq r_{2} - 2 = : r_{1} \in N

, where:

H^{\dim_{H} (G_{r_{2}}^{(2)})} (G_{r_{1}}^{(1)} Δ G_{r_{2}}^{(2)}) = 0

(38)

We show this with the following:

In eq. 38, since

\dim_{H} (G_{r_{2}}^{(2)}) = \dim_{H} ([- r_{2}, r_{2}] \cup (Q \cap [- r_{2} - . 5, r_{2} + 1])) = 1

\begin{matrix} H^{1} (\{(x, x) : [- r_{1} - 2, r_{1} + 2] Δ ([- r_{2}, r_{2}] \cup (Q \cap [- r_{2} - 1, r_{2} + 1]))\}) = \end{matrix}

(39)

\begin{matrix} H^{1} (\{(x, x) : [- (r_{2} - 2) - 2, (r_{2} - 2) + 2] Δ ([- r_{2}, r_{2}] \cup (Q \cap [- r_{2} - 1, r_{2} + 1])\}) = \end{matrix}

(40)

\begin{matrix} H^{1} (\{(x, x) : [- r_{2}, r_{2}] Δ ([- r_{2}, r_{2}] \cup (Q \cap [- r_{2} - 1, r_{2} + 1])\}) = \end{matrix}

(41)

\begin{matrix} H^{1} ((Q \cap [- r_{2} - 1, - r_{2}]) \cup (Q \cap [r_{2}, r_{2} + 1])) = \end{matrix}

(42)

\begin{matrix} 0 \end{matrix}

(43)

Since crit. ((1)) and ((2)) is true, using def. 7, we have shown

f_{r_{1}}^{(1)}

and

f_{r_{2}}^{(2)}

are equivalent.

Definition 9

(Non-Equivalent Sequences of functions). Again,

S \subseteq N

is an arbitrary set. Therefore, all sequences of bounded functions in:

\{\vec{f_{s}} : s \in S\}

are non-equivalent, if def. 7 is false, meaning for some

k, v \in S

, where

k \neq v

\vec{f_{k}}

and

\vec{f_{v}}

are non-equivelant: there is a

N^{'} \in N

, where for all

r_{k} \geq N^{'}

, there is either a

r_{v} \in N

, where:

H^{\dim_{H} (G_{r_{k}}^{(k)})} (G_{r_{k}}^{(k)} Δ G_{r_{v}}^{(v)}) \neq 0

or for all

r_{v} \geq N^{'}

, there is a

r_{k} \in N

, where

H^{\dim_{H} (G_{r_{v}}^{(v)})} (G_{r_{k}}^{(k)} Δ G_{r_{v}}^{(v)}) \neq 0

5.3.3. Explanation

We define

\vec{f_{k}}

and

\vec{f_{v}}

as equivalent, where

\vec{A_{k}}

and

\vec{A_{v}}

are constant in

f_{r_{k}}^{(k)} : A_{r_{k}}^{(k)} \to R

and

f_{r_{v}}^{(v)} : A_{r_{v}}^{(v)} \to R

, such for all Borel measurable

f \in R^{A}

, when either

E [f_{r_{k}}^{(k)}]

E [f_{r_{v}}^{(v)}]

exist:

E [f_{r_{k}}^{(k)}] \neq E [f_{r_{v}}^{(v)}] (§ 5.2)

Hence, consider the following:

5.3.4. Example of Non-Equivalent Bounded Sequences of Functions

Suppose we define

\vec{f_{1}} = {(f_{r_{1}}^{(1)})}_{r_{1} \in N}

, where

f_{r_{1}}^{(1)} : A_{r_{1}}^{(1)} \to R

and

\vec{f_{2}} = {(f_{r_{2}}^{(2)})}_{r_{2} \in N}

, where

f_{r_{2}}^{(2)} : A_{r_{2}}^{(2)} \to R

such that:

(1): $f_{r_{1}}^{(1)} (x) = x$
(2): $A_{r_{1}}^{(1)} = [- r_{1}, r_{1}]$
(3): $G_{r_{1}}^{(1)} = \{(x, x) : x \in [- r_{1}, r_{1}]\}$

and

(1): $f_{r_{2}}^{(2)} (x) = x + (1 / r_{2}) sin (x)$
(2): $A_{r_{2}}^{(1)} = [- 2 r_{2}, 2 r_{2}]$
(3): $G_{r_{2}}^{(2)} = \{(x, x + (1 / r_{2}) sin (x)) : x \in [- r_{2}, r_{2}] \cup (Q \cap [- r_{2} - 1, r_{2} + 1])\}$

Note, using def. 7,

S = {1, 2}

k = 1

, and

v = 2

. In other words,

r_{k} = r_{1}

and

r_{v} = r_{2}

. Now, suppose

N : = 1

, where

r_{1} \in N

and

r_{2} \in N

. Then,

(1): For all $r_{2} \geq N = 1$ , there exists a $3 \leq 2 r_{2} + 1 = : r_{2} \in N$ , where:

$H^{\dim_{H} (G_{r_{2}}^{(2)})} (G_{r_{1}}^{(1)} Δ G_{r_{2}}^{(2)}) \neq 0$

We show this with the following:

Since $G_{r_{2}}^{(2)} = [- 2 r_{2}, 2 r_{2}]$ is a 1-d interval, $\dim_{H} (G_{r_{1}}^{(1)}) = 1$ . Hence:

$\begin{matrix} H^{1} (\{(x, x) : x \in [- r_{1}, r_{1}]\} Δ \{(x, x + (1 / r_{2}) sin (x)) : x \in [- 2 r_{2}, 2 r_{2}]\}) = \end{matrix}$

(44)

$\begin{matrix} H^{1} (\{(x, x) : x \in [- (2 r_{2} + 1), (2 r_{2} + 1)]\} Δ \{(x, x + (1 / r_{2}) sin (x)) : x \in [- 2 r_{2}, 2 r_{2}]\}) = \end{matrix}$

(45)

$\begin{matrix} H^{1} (\{(x, x) : x \in [- 2 r_{2} - 1, 2 r_{2} + 1]\} Δ \{(x, x + (1 / r_{2}) sin (x)) : x \in [- 2 r_{2}, 2 r_{2}]\}) = \end{matrix}$

(46)

$\begin{matrix} \neq & 0 \end{matrix}$

(47)

Which is true, since $x \neq (1 / r_{2}) sin (x)$ for $x \in ([- 2 r_{2} - 1, - 2 r_{2}] \cup [2 r_{2}, 2 r_{2} + 1]) ∖ {π m : m \in N, - r_{2} \leq π m \leq r_{2}}$ .

Now, considering the examples in

§

5.3.2 and

§

5.3.2, here are some shortcuts to definitions of equivelant and non-equivelant sequences of bounded functions:

5.3.5. Shortcuts to Determining Equivelant and Non-equivelant Sequences of Bounded Functions

Suppose we define

\vec{f_{1}} = {(f_{r_{1}}^{(1)})}_{r_{1} \in N}

, where

f_{r_{1}}^{(1)} : A_{r_{1}}^{(1)} \to R

and

\vec{f_{2}} = {(f_{r_{2}}^{(2)})}_{r_{2} \in N}

, where

f_{r_{2}}^{(2)} : A_{r_{2}}^{(2)} \to R

(1)

\vec{f_{1}}

and

\vec{f_{2}}

cannot be equivalent, unless:

\begin{matrix} H^{\dim_{H} (A_{r_{1}}^{(1)})} (\{(x_{1}, \cdot \cdot \cdot, x_{n}) : f_{r_{1}}^{(1)} (x_{1}, \cdot \cdot \cdot, x_{n}) \neq f_{r_{2}}^{(2)} (x_{1}, \cdot \cdot \cdot, x_{n})\}) = \end{matrix}

(48)

\begin{matrix} H^{\dim_{H} (A_{r_{2}}^{(2)})} (\{(x_{1}, \cdot \cdot \cdot, x_{n}) : f_{r_{1}}^{(1)} (x_{1}, \cdot \cdot \cdot, x_{n}) \neq f_{r_{2}}^{(2)} (x_{1}, \cdot \cdot \cdot, x_{n})\}) = 0 \end{matrix}

(49)

(2)

\vec{f_{1}}

and

\vec{f_{2}}

is non-equivelant, when either:

(1): $H^{\dim_{H} (A_{r_{1}}^{(1)})} (\{(x_{1}, \cdot \cdot \cdot, x_{n}) : f_{r_{1}}^{(1)} (x_{1}, \cdot \cdot \cdot, x_{n}) \neq f_{r_{2}}^{(2)} (x_{1}, \cdot \cdot \cdot, x_{n})\}) > 0$
(2): $H^{\dim_{H} (A_{r_{2}}^{(2)})} (\{(x_{1}, \cdot \cdot \cdot, x_{n}) : f_{r_{1}}^{(1)} (x_{1}, \cdot \cdot \cdot, x_{n}) \neq f_{r_{2}}^{(2)} (x_{1}, \cdot \cdot \cdot, x_{n})\}) > 0$

5.4. Defining the “Measure"

5.4.1. Preliminaries

We define the “measure" of

{(f_{r})}_{r \in N}

§

5.4.2, where

{(G_{r})}_{r \in N}

is a sequence of the graph of each

f_{r}

. To understand this “measure", continue reading.

For every

r \in N

, “over-cover"

G_{r}

with minimal, pairwise disjoint sets of equal

H^{\dim_{H} (G_{r})}

measure. (We denote the equal measures

ε

, where the former sentence is defined

C (ε, G_{r}, ω)

: i.e.,

ω \in Ω_{ε, r}

enumerates all collections of these sets covering

G_{r}

. In case this step is unclear, see §8.1.)

(2)

For every

ε

, r and

ω

, take a sample point from each set in

C (ε, G_{r}, ω)

. The set of these points is “the sample" which we define

S (C (ε, G_{r}, ω), ψ)

: i.e.,

ψ \in Ψ_{ε, r, ω}

enumerates all possible samples of

C (ε, G_{r}, ω)

. (If this is unclear, see §8.2.)

(3)

For every

ε

, r,

ω

and

ψ

(a): Take a “pathway” of line segments: we start with a line segment from arbitrary point $x_{0}$ of $S (C (ε, G_{r}, ω), ψ)$ to the sample point with the smallest $(n + 1)$ -dimensional Euclidean distance to $x_{0}$ (i.e., when more than one sample point has the smallest $(n + 1)$ -dimensional Euclidean distance to $x_{0}$ , take either of those points). Next, repeat this process until the “pathway” intersects with every sample point once. (In case this is unclear, see §8.3.1.)
(b): Take the set of the length of all segments in (a), except for lengths that are outliers (i.e., for any constant $C > 0$ , the outliers are more than C times the interquartile range of the length of all line segments as $r \to \infty$ ). Define this $L (x_{0}, S (C (ε, G_{r}, ω), ψ))$ . (If this is unclear, see §8.3.2.)
(c): Multiply remaining lengths in the pathway by a constant so they add up to one (i.e., a probability distribution). This will be denoted $P (L (x_{0}, S (C (ε, G_{r}, ω), ψ)))$ . (In case this is unclear, see §8.3.3)
(d): Take the shannon entropy [14] of step (c). We define this:

$E (P (L (x_{0}, S (C (ε, G_{r}, ω), ψ)))) = \sum_{x \in P (L (x_{0}, S (C (ε, G_{r}, ω), ψ)))} - x {log}_{2} x$

which will be shortened to $E (L (x_{0}, S (C (ε, G_{r}, ω), ψ)))$ . (If this is unclear, see §8.3.4.)
(e): Maximize the entropy w.r.t all "pathways". This we will denote:

$E (L (S (C (ε, G_{r}, ω), ψ))) = sup_{x_{0} \in S (C (ε, G_{r}, ω), ψ)} E (L (x_{0}, S (C (ε, G_{r}, ω), ψ)))$

(In case this is unclear, see §8.3.5.)

(4)

Therefore, the maximum entropy, using (1) and (2) is:

E_{\max} (ε, r) = sup_{ω \in Ω_{ε, r}} sup_{ψ \in Ψ_{ε, r, ω}} E (L (S (C (ε, G_{r}, ω), ψ)))

5.4.2. What Am I Measuring?

Suppose we define two sequences of the graph of the bounded functions converging to the graph of f: e.g.,

{(G_{r}^{★})}_{r \in N}

and

{(G_{j}^{★ ★})}_{j \in N}

, where for constant

ε

and cardinality

| \cdot |

Using (2) and (3(e)) of Section 5.4.1, suppose:

\begin{matrix} \underset{̲}{|S (C (ε, G_{r}^{★}, ω), ψ)|} = \\ sup \{|S (C (ε, G_{j}^{★ ★}, ω^{'}), ψ^{'})| : j \in N, ω^{'} \in Ω_{ε, j}, ψ^{'} \in Ψ_{ε, j, ω}, E (L (S (C (ε, G_{j}^{★ ★}, ω^{'}), ψ^{'}))) \leq E (L (S (C (ε, G_{r}^{★}, ω), ψ)))\} \end{matrix}

then (using

\underset{̲}{|S (C (ϵ, G_{r}^{★}, ω), ψ)|}

) we get

\underset{̲}{α} (ε, r, ω, ψ) = \underset{̲}{|S (C (ε, G_{r}^{★}, ω), ψ)|} / |S (C (ε, G_{r}^{★}, ω), ψ))|

(b)

Also, using ((2)) and (3(e)) of Section 5.4.1, suppose:

\begin{matrix} \bar{|S (C (ε, G_{r}^{★}, ω), ψ)|} = \\ inf \{|S (C (ε, G_{j}^{★ ★}, ω^{'}), ψ^{'})| : j \in N, ω^{'} \in Ω_{ε, j}, ψ^{'} \in Ψ_{ε, j, ω}, E (L (S (C (ε, G_{j}^{★ ★}, ω^{'}), ψ^{'}))) \geq E (L (S (C (ε, G_{r}^{★}, ω), ψ)))\} \end{matrix}

then (using

\bar{|S (C (ε, G_{r}^{★}, ω), ψ)|}

) we also get:

\bar{α} (ε, r, ω, ψ) = \bar{|S (C (ε, G_{r}^{★}, ω), ψ)|} / |S (C (ε, G_{r}^{★}, ω), ψ))|

(1)

If using

\bar{α} (ϵ, r, ω, ψ)

and

\underset{̲}{α} (ϵ, r, ω, ψ)

we have:

1 < \underset{ε \to 0}{lim sup} \underset{r \to \infty}{lim sup} sup_{ω \in Ω_{ε, r}} sup_{ψ \in Ψ_{ε, r, ω}} \bar{α} (ε, r, ω, ψ), \underset{ε \to 0}{lim inf} = \underset{r \to \infty}{lim inf} inf_{ω \in Ω_{ε, r}} inf_{ψ \in Ψ_{ε, r, ω}} \underset{̲}{α} (ε, r, ω, ψ) < + \infty

then what I’m measuring from

{(G_{r}^{★})}_{r \in N}

increases at a rate superlinear to that of

{(G_{j}^{★ ★})}_{j \in N}

(2)

If using equations

\bar{α} (ε, j, ω, ψ)

and

\underset{̲}{α} (ε, j, ω, ψ)

(swapping

r \in N

and

{(G_{r}^{★})}_{r \in N}

, in

\bar{α} (ϵ, r, ω, ψ)

and

\underset{̲}{α} (ϵ, r, ω, ψ)

, with

j \in N

and

{(G_{j}^{★ ★})}_{j \in N}

) we get:

1 < \underset{ε \to 0}{lim sup} \underset{j \to \infty}{lim sup} sup_{ω \in Ω_{ε, j}} sup_{ψ \in Ψ_{ε, j, ω}} \bar{α} (ε, j, ω, ψ), \underset{ε \to 0}{lim inf} \underset{j \to \infty}{lim inf} inf_{ω \in Ω_{ε, j}} inf_{ψ \in Ψ_{ε, j, ω}} \underset{̲}{α} (ε, j, ω, ψ) < + \infty

then what I’m measuring from

{(G_{r}^{★})}_{r \in N}

increases at a rate sublinear to that of

{(G_{j}^{★ ★})}_{j \in N}

(3)

If using equations

\bar{α} (ε, r, ω, ψ)

\underset{̲}{α} (ε, r, ω, ψ)

\bar{α} (ε, j, ω, ψ)

, and

\underset{̲}{α} (ε, j, ω, ψ)

, we both have:

(a): $\underset{ε \to 0}{lim sup} \underset{r \to \infty}{lim sup} sup_{ω \in Ω_{ε, r}} sup_{ψ \in Ψ_{ε, r, ω}} \bar{α} (ε, r, ω, ψ)$ or $\underset{ε \to 0}{lim inf} \underset{r \to \infty}{lim inf} inf_{ω \in Ω_{ε, r}} inf_{ψ \in Ψ_{ε, r, ω}} \underset{̲}{α} (ε, r, ω, ψ)$ are equal to zero, one or $+ \infty$
(b): $\underset{ε \to 0}{lim sup} \underset{j \to \infty}{lim sup} sup_{ω \in Ω_{ε, j}} sup_{ψ \in Ψ_{ε, j, ω}} \bar{α} (ε, j, ω, ψ)$ or $\underset{ε \to 0}{lim inf} \underset{j \to \infty}{lim inf} inf_{ω \in Ω_{ε, j}} inf_{ψ \in Ψ_{ε, j, ω}} \underset{̲}{α} (ε, j, ω, ψ)$ are equal to zero, one or $+ \infty$

then what I’m measuring from

{(G_{r}^{★})}_{r \in N}

increases at a rate linear to that of

{(G_{j}^{★ ★})}_{j \in N}

5.4.3. Example of The “Measure" of $(G_{r}^{★})$ Increasing at Rate Super-linear to that of $(G_{j}^{★ ★})$

Suppose, we have function

f : A \to R

, where

A = Q \cap [0, 1]

, and:

f (x) = \{\begin{matrix} 1 & x \in \{(2 s + 1) / (2 t) : s \in Z, t \in N, t \neq 0\} \cap [0, 1] \\ 0 & x \notin \{(2 s + 1) / (2 t) : s \in Z, t \in N, t \neq 0\} \cap [0, 1] \end{matrix}

(50)

such that:

{(A_{r}^{★})}_{r \in N} = {(\{c / r! : c \in Z, 0 \leq c \leq r!\})}_{r \in N}

and

{(A_{j}^{★ ★})}_{j \in N} = {(\{c / d : c \in Z, d \in N, d \leq j, 0 \leq c \leq j\})}_{j \in N}

where for

f_{r}^{★} : A_{r}^{★} \to R

f_{r}^{★} (x) = f (x) for all x \in A_{r}^{★}

(51)

and

f_{j}^{★ ★} : A_{j}^{★ ★} \to R

f_{j}^{★ ★} (x) = f (x) for all x \in A_{j}^{★ ★}

(52)

Hence, when

{(G_{r}^{★})}_{r \in N}

is:

{(G_{r}^{★})}_{r \in N} = {(\{(x, f (x)) : x \in \{c / r! : c \in Z, 0 \leq c \leq r!\}\})}_{r \in N}

(53)

and

{(G_{j}^{★ ★})}_{j \in N}

is:

{(G_{j}^{★ ★})}_{j \in N} = {(\{(x, f (x)) : x \in \{c / d : c \in Z, d \in N, d \leq j, 0 \leq c \leq j\}\})}_{j \in N}

(54)

Note, the following:

Since

ε > 0

and

A = Q \cap [0, 1]

is countably infinite, there exists minimum

ε

which is 1. Therefore, we don’t need

ε \to 0

. Also, we maximize

E (L (S (C (ε, G_{r}^{★}, ω), ψ)))

by the following procedure:

(1)

For every

r \in N

, group

x \in G_{r}^{★}

into elements with an even numerator when simplified: i.e.,

x \in \{(2 s + 1) / (2 t) : s \in Z, t \in N, t \neq 0\}

which we call

S_{1, r}

, and group

x \in G_{r}^{★}

into elements with an odd denominator when simplified: i.e.,

x \in Q ∖ \{(2 s + 1) / (2 t) : s \in Z, t \in N, t \neq 0\}

which we call

S_{2, r}

(2)

Arrange the points in

S_{1, r}

from least to greatest and take the 2-d Euclidean distance between each pair of consecutive points in

S_{1, r}

. In this case, since all points lie on

y = 1

, take the absolute difference between the x-coordinates of

S_{1, r}

then call this

D_{1, r}

. (Note, this is similar to

§

5.4.1 step (3)(a)).

(3)

Repeat step (2) for

S_{2, r}

, then call this

D_{2, r}

. (Note, all point of

S_{2, r}

lie on

y = 0

(4)

Remove any outliers from

D_{r} = D_{1, r} \cup D_{2, r} \cup {d ((\frac{n! - 1}{n!}, 1), (1, 0))}

(i.e., d is the 2-d Euclidean distance between points

(\frac{n! - 1}{n!}, 1)

and

(1, 0)

). Note, in this case,

D_{2, r}

and

{d ((\frac{n! - 1}{n!}, 1), (1, 0))}

should be outliers (i.e., for any

C > 0

, the lengths of

D_{2, r}

are more than C times the interquartile range of the lengths of

D_{r}

) leaving us with

D_{1, r}

(5)

Multiply the remaining lengths in the pathway by a constant so they add up to one. (See P[r] of code 1 for an example)

(6)

Take the entropy of the probability distribution. (See entropy[r] of code 1 for an example.)

We can illustrate this process with the following code:

Code 1: Illustration of step ((1))-((6))

Taking Table[{r,entropy[r]},{r,3,8}], we get:

Code 2: Output of Table[{r,entropy[r]},{r,3,8}]

and notice when:

(1): $c (r) = (r!) / 2 - 1$
(2): $\{b (4) \mapsto 9, b (5) \mapsto 45, b (6) \mapsto 315, b (7) \mapsto 2205, b (8) \mapsto 19845\}$
(3): $a (r) + b (r) = c (r)$

the output of code can be defined:

\frac{a (r) {log}_{2} (c (r))}{c (r)} + \frac{b (r) log (2 c (r))}{c (r)} = \frac{a (r) {log}_{2} (c (r)) + b (r) log (2 c (r))}{c (r)}

(55)

Hence, since

a (r) = c (r) - b (r) = (r!) / 2 - 1 - b (r)

\begin{matrix} \frac{a (r) {log}_{2} (c (r)) + b (r) log (2 c (r))}{c (r)} = \end{matrix}

(56)

\begin{matrix} \frac{(r! / 2 - 1 - b (r)) {log}_{2} (c (r)) + b (r) {log}_{2} (2 c (r))}{c (r)} = \end{matrix}

(57)

\begin{matrix} \frac{(r! / 2) {log}_{2} (c (r)) - {log}_{2} (c (r)) - b (r) {log}_{2} (r) + b (r) {log}_{2} (c (r)) + b (r) {log}_{2} (2)}{c (r)} = \end{matrix}

(58)

\begin{matrix} \frac{(r! / 2) {log}_{2} (c (r)) - {log}_{2} (c (r)) + b (r)}{c (r)} = \end{matrix}

(59)

\begin{matrix} \frac{(r! / 2 - 1) {log}_{2} (c (r)) + b (r)}{c (r)} = \end{matrix}

(60)

\begin{matrix} \frac{(r! / 2 - 1) {log}_{2} (r! / 2 - 1) + b (r)}{r! / 2 - 1} = \end{matrix}

(61)

\begin{matrix} {log}_{2} (r! / 2 - 1) + \frac{b (r)}{r! / 2 - 1} = \end{matrix}

(62)

and

{lim}_{r \to \infty} b (r) / c (r) = 1

(I need help proving this):

\begin{matrix} {log}_{2} (r! / 2 - 1) + \frac{b (r)}{r! / 2 - 1} \sim {log}_{2} (r! / 2 - 1) + 1 \end{matrix}

(63)

\begin{matrix} {log}_{2} (r! / 2 - 1) + {log}_{2} (2) = \end{matrix}

(64)

\begin{matrix} {log}_{2} (2 (r! / 2 - 1)) \end{matrix}

(65)

\begin{matrix} {log}_{2} (r! - 2) \sim {log}_{2} (r!) \end{matrix}

(66)

Hence, entropy[r] is the same as:

\begin{matrix} E (L (S (C (1, G_{r}^{★}, ω), ψ))) \sim {log}_{2} (r!) \end{matrix}

(67)

Now, repeat code with:

{(A_{j}^{★ ★})}_{j \in N} = (\{c / d : c \in Z, d \in N, d \leq j, 0 \leq c \leq j\})

Code 3: Illustration of step ((1))-((6)) on (A_j^**)

Using this post [22], we assume an approximation of Table[entropy[j],{j,3,Infinity}] or

E (L (S (C (1, G_{j}^{★ ★}, ω^{'}), ψ^{'})))

is:

\begin{matrix} E (L (S (C (1, G_{j}^{★ ★}, ω^{'}), ψ^{'}))) \sim 2 {log}_{2} (j) + 1 - {log}_{2} (3 π) \end{matrix}

(68)

Hence, using §5.4.2 ((a)) and §5.4.2 ((1)), take

|S (C (ε, G_{j}^{★ ★}, ω^{'}), ψ^{'})| = \sum_{M = 1}^{j} ϕ (M) \approx \frac{3}{π^{2}} j^{2}

(where

ϕ

is Euler’s Totient function) computing the following:

\begin{matrix} \underset{̲}{|S (C (ε, G_{r}^{★}, ω), ψ)|} = \\ sup \{|S (C (ε, G_{j}^{★ ★}, ω^{'}), ψ^{'})| : j \in N, ω^{'} \in Ω_{ε, j}, ψ^{'} \in Ψ_{ε, j, ω}, E (L (S (C (ε, G_{j}^{★ ★}, ω^{'}), ψ^{'}))) \leq E (L (S (C (ε, G_{r}^{★}, ω), ψ)))\} = \\ sup \{\frac{3}{π^{2}} j^{2} : j \in N, ω^{'} \in Ω_{ε, j}, ψ^{'} \in Ψ_{ε, j, ω}, 2 {log}_{2} (j) + 1 - {log}_{2} (3 π) \leq {log}_{2} (r!)\} = \end{matrix}

(69)

where:

(1): For every $r \in N$ , we find a $j \in N$ , where $2 {log}_{2} (j) + 1 - {log}_{2} (3 π) \leq {log}_{2} (r!)$ , but the absolute value of ${log}_{2} (r!) - (2 {log}_{2} (j) + 1 - {log}_{2} (3 π))$ is minimized. In other words, for every $r \in N$ , we want $j \in N$ where:

$\begin{matrix} 2 {log}_{2} (j) + 1 - {log}_{2} (3 π) \leq {log}_{2} (r!) \end{matrix}$

(70)

$\begin{matrix} 2 {log}_{2} (j) \leq {log}_{2} (r!) - 1 + {log}_{2} (3 π) \end{matrix}$

(71)

$\begin{matrix} {(2^{{log}_{2} (j)})}^{2} \leq 2^{{log}_{2} (r!) - 1 + {log}_{2} (3 π)} \end{matrix}$

(72)

$\begin{matrix} j^{2} \leq (2^{{log}_{2} (r!)} 2^{{log}_{2} (3 π)}) / 2 \end{matrix}$

(73)

$\begin{matrix} j \leq \sqrt{\frac{r! (3 π)}{2}} \end{matrix}$

(74)

$\begin{matrix} j = ⌊\sqrt{\frac{3 π r!}{2}}⌋ \end{matrix}$

(75)

$\begin{matrix} \frac{3}{π^{2}} j^{2} = \frac{3}{π^{2}} {(⌊\sqrt{\frac{3 π r!}{2}}⌋)}^{2} \sim \underset{̲}{|S (C (1, G_{r}^{★}, ω), ψ)|} \end{matrix}$

(76)

Finally, since

|S (C (ε, G_{r}^{★}, ω), ψ)| = r!

, we wish to prove

1 < \underset{ε \to 0}{lim inf} \underset{r \to \infty}{lim inf} inf_{ω \in Ω_{ε, r}} inf_{ψ \in Ψ_{ε, r, ω}} \underset{̲}{α} (ε, r, ω, ψ) < + \infty

within §5.4.2 crit. (1):

\begin{matrix} \underset{ε \to 0}{lim inf} \underset{r \to \infty}{lim inf} inf_{ω \in Ω_{ε, r}} inf_{ψ \in Ψ_{ε, r, ω}} \underset{̲}{α} (ε, r, ω, ψ) = \underset{r \to \infty}{lim inf} inf_{ω \in Ω_{ε, r}} inf_{ψ \in Ψ_{ε, r, ω}} \frac{\underset{̲}{|S (C (ε, F_{r}^{★}, ω), ψ)|}}{|S (C (1, G_{r}^{★}, ω), ψ))|} \end{matrix}

(77)

\begin{matrix} = lim_{r \to \infty} \frac{\frac{3}{π^{2}} {(⌊\sqrt{\frac{3 π r!}{2}}⌋)}^{2}}{r!} \end{matrix}

(78)

where using mathematica, we get the limit is greater than one:

Code 4: Limit of eq. 78

Also, using §5.4.2 ((b)) and §5.4.2 ((1)), take

|S (C (ε, G_{j}^{★ ★}, ω^{'}), ψ^{'})| = \sum_{M = 1}^{j} ϕ (M) \approx \frac{3}{π^{2}} j^{2}

(where

ϕ

is Euler’s Totient function) to compute the following:

\begin{matrix} \underset{̲}{|S (C (ε, G_{r}^{★}, ω), ψ)|} = \\ inf \{|S (C (ε, G_{j}^{★ ★}, ω^{'}), ψ^{'})| : j \in N, ω^{'} \in Ω_{ε, j}, ψ^{'} \in Ψ_{ε, j, ω}, E (L (S (C (ε, G_{j}^{★ ★}, ω^{'}), ψ^{'}))) \geq E (L (S (C (ε, G_{r}^{★}, ω), ψ)))\} = \\ inf \{\frac{3}{π^{2}} j^{2} : j \in N, ω^{'} \in Ω_{ε, j}, ψ^{'} \in Ψ_{ε, j, ω}, 2 {log}_{2} (j) + 1 - {log}_{2} (3 π) \geq {log}_{2} (r!)\} = \end{matrix}

(79)

where:

(1): For every $r \in N$ , we find a $j \in N$ , where $2 {log}_{2} (j) + 1 - {log}_{2} (3 π) \geq {log}_{2} (r!)$ , but the absolute value of $(2 {log}_{2} (j) + 1 - {log}_{2} (3 π)) - {log}_{2} (r!)$ is minimized. In other words,

for every $r \in N$ , we want $j \in N$ where:

$\begin{matrix} 2 {log}_{2} (j) + 1 - {log}_{2} (3 π) \geq {log}_{2} (r!) \end{matrix}$

(80)

$\begin{matrix} 2^{2 {log}_{2} (j)} \geq {log}_{2} (r!) - 1 + {log}_{2} (3 π) \end{matrix}$

(81)

$\begin{matrix} {(2^{{log}_{2} (j)})}^{2} \geq 2^{{log}_{2} (r!) - 1 + {log}_{2} (3 π)} \end{matrix}$

(82)

$\begin{matrix} j^{2} \geq (2^{{log}_{2} (r!)} 2^{{log}_{2} (3 π)}) / 2 \end{matrix}$

(83)

$\begin{matrix} j \geq \sqrt{\frac{r! (3 π)}{2}} \end{matrix}$

(84)

$\begin{matrix} j = ⌈\sqrt{\frac{3 π r!}{2}}⌉ \end{matrix}$

(85)

$\begin{matrix} \frac{3}{π^{2}} j^{2} = \frac{3}{π^{2}} {(⌈\sqrt{\frac{3 π r!}{2}}⌉)}^{2} \sim \bar{|S (C (1, G_{r}^{★}, ω), ψ)|} \end{matrix}$

(86)

Finally, since

|S (C (1, G_{r}^{★}, ω), ψ)| = r!

, we wish to prove

1 < \underset{ε \to 0}{lim sup} \underset{r \to \infty}{lim sup} sup_{ω \in Ω_{ε, r}} sup_{ψ \in Ψ_{ε, r, ω}} \bar{α} (ε, r, ω, ψ) < + \infty

within §5.4.2 crit. (1):

\begin{matrix} \underset{ε \to 0}{lim sup} \underset{r \to \infty}{lim sup} sup_{ω \in Ω_{ε, r}} sup_{ψ \in Ψ_{ε, r, ω}} \bar{α} (ε, r, ω, ψ) = \underset{r \to \infty}{lim sup} sup_{ω \in Ω_{ε, r}} sup_{ψ \in Ψ_{ε, r, ω}} \frac{\bar{|S (C (1, G_{r}^{★}, ω), ψ)|}}{|S (C (1, G_{r}^{★}, ω), ψ))|} \end{matrix}

(87)

\begin{matrix} = lim_{r \to \infty} \frac{\frac{3}{π^{2}} {(⌈\sqrt{\frac{3 π r!}{2}}⌉)}^{2}}{r!} \end{matrix}

(88)

where using mathematica, we get the limit is greater than one:

Code 5: Limit of eq. 88

Hence, since the limits in eq. 77 and eq. 87 are greater than one and less than

+ \infty

: i.e.,

1 < \underset{ε \to 0}{lim inf} \underset{r \to \infty}{lim inf} inf_{ω \in Ω_{ε, r}} inf_{ψ \in Ψ_{ε, r, ω}} \underset{̲}{α} (ε, r, ω, ψ) = \underset{ε \to 0}{lim sup} \underset{r \to \infty}{lim sup} sup_{ω \in Ω_{ε, r}} sup_{ψ \in Ψ_{ε, r, ω}} \bar{α} (ε, r, ω, ψ) < + \infty

(89)

what we’re measuring from

{(G_{r}^{★})}_{r \in N}

increases at a rate superlinear to that of

{(G_{j}^{★ ★})}_{j \in N}

(i.e., Section 5.4.2 crit. (1)).

5.4.4. Example of The “Measure" from ${(G_{r}^{★})}_{r \in N}$ Increasing at a Rate Sub-Linear to that of ${(G_{j}^{★ ★})}_{j \in N}$

Using our previous example, we can use the following theorem:

Theorem 10.

If what we’re measuring from

{(G_{r}^{★})}_{r \in N}

increases at a rate superlinear to that of

{(G_{j}^{★ ★})}_{j \in N}

, then what we’re measuring from

{(G_{j}^{★ ★})}_{r \in N}

increases at a ratesublinearto that of

{(G_{r}^{★})}_{r \in N}

Hence, in our definition of super-linear (§5.4.2 crit. (1)), swap

G_{r}^{★}

and

r \in N

for

(G_{j}^{★ ★})

and

j \in N

regarding

\bar{α} (ϵ, r, ω, ψ)

and

\underset{̲}{α} (ϵ, r, ω, ψ)

(i.e.,

\bar{α} (ϵ, j, ω, ψ)

and

\underset{̲}{α} (ϵ, j, ω, ψ)

) and notice thm. 10 is true when:

1 < \underset{ε \to 0}{lim sup} \underset{j \to \infty}{lim sup} sup_{ω \in Ω_{ε, j}} sup_{ψ \in Ψ_{ε, j, ω}} \bar{α} (ε, j, ω, ψ), \underset{ε \to 0}{lim inf} \underset{j \to \infty}{lim inf} inf_{ω \in Ω_{ε, j}} inf_{ψ \in Ψ_{ε, j, ω}} \underset{̲}{α} (ε, j, ω, ψ) < + \infty

5.4.5. Example of The “Measure" from ${(G_{r}^{★})}_{r \in N}$ Increasing at a Rate Linear to that of ${(G_{j}^{★ ★})}_{j \in N}$

Suppose, we have function

f : A \to R

, where

A = Q \cap [0, 1]

, and:

f (x) = \{\begin{matrix} 1 & x \in \{(2 s + 1) / (2 t) : s \in Z, t \in N, t \neq 0\} \cap [0, 1] \\ 0 & x \notin \{(2 s + 1) / (2 t) : s \in Z, t \in N, t \neq 0\} \cap [0, 1] \end{matrix}

(90)

such that:

{(A_{r}^{★})}_{r \in N} = {(\{c / r! : c \in Z, 0 \leq c \leq r!\})}_{r \in N}

and

{(A_{j}^{★ ★})}_{j \in N} = (\{c / ({(j!)}^{2}) : c \in N, 1 \leq c \leq {(j!)}^{2}\})

where for

f_{r}^{★} : A_{r}^{★} \to R

f_{r}^{★} (x) = f (x) for all x \in A_{r}^{★}

(91)

and

f_{j}^{★ ★} : A_{j}^{★ ★} \to R

f_{j}^{★ ★} (x) = f (x) for all x \in A_{j}^{★ ★}

(92)

Hence, when

{(G_{r}^{★})}_{r \in N}

is:

{(G_{r}^{★})}_{r \in N} = {(\{(x, f (x)) : x \in \{c / r! : c \in Z, 0 \leq c \leq r!\}\})}_{r \in N}

(93)

and

{(G_{j}^{★ ★})}_{j \in N}

is:

{(G_{j}^{★ ★})}_{j \in N} = {(\{(x, f (x)) : x \in \{c / ({(j!)}^{2}) : c \in Z, 0 \leq c \leq {(j!)}^{2}\}\})}_{j \in N}

(94)

We already know, using §5.4.3:

E (L (S (C (1, G_{r}^{★}, ω), ψ))) \sim {log}_{2} (r! - 2) \sim {log}_{2} (r!)

(95)

Also, using §5.4.3 steps (1)-(6) on

{(A_{j}^{★ ★})}_{j \in N}

Code 6: Illustration of step ((1))-((6)) on (A_j^**)

where the output is

Code 7: Output of Code 6

Notice when:

(1): $c (j) = {(j!)}^{2} / 2 - 1$
(2): $\{b (4) \mapsto 9, b (5) \mapsto 279, b (6) \mapsto 6975, b (7) \mapsto 257175, b (8) \mapsto 19845\}$
(3): $a (j) + b (j) = c (j)$

the output of code can be defined:

\frac{a (j) {log}_{2} (c (j))}{c (j)} + \frac{b (j) log (2 c (j))}{c (j)} = \frac{a (j) {log}_{2} (c (j)) + b (j) log (2 c (j))}{c (j)}

(96)

Hence, since

a (j) = c (j) - b (j) = {(j!)}^{2} / 2 - 1 - b (j)

\begin{matrix} \frac{a (j) {log}_{2} (c (j)) + b (j) log (2 c (j))}{c (j)} = \end{matrix}

(97)

\begin{matrix} \frac{({(j!)}^{2} / 2 - 1 - b (j)) {log}_{2} (c (j)) + b (j) {log}_{2} (2 c (j))}{c (j)} = \end{matrix}

(98)

\begin{matrix} \frac{({(j!)}^{2} / 2) {log}_{2} (c (j)) - {log}_{2} (c (j)) - b (j) {log}_{2} (j) + b (j) {log}_{2} (c (j)) + b (j) {log}_{2} (2)}{c (j)} = \end{matrix}

(99)

\begin{matrix} \frac{({(j!)}^{2} / 2) {log}_{2} (c (j)) - {log}_{2} (c (j)) + b (j)}{c (j)} = \end{matrix}

(100)

\begin{matrix} \frac{({(j!)}^{2} / 2 - 1) {log}_{2} (c (j)) + b (j)}{c (j)} = \end{matrix}

(101)

\begin{matrix} \frac{({(j!)}^{2} / 2 - 1) {log}_{2} ({(j!)}^{2} / 2 - 1) + b (j)}{{(j!)}^{2} / 2 - 1} = \end{matrix}

(102)

\begin{matrix} {log}_{2} ({(j!)}^{2} / 2 - 1) + \frac{b (j)}{{(j!)}^{2} / 2 - 1} = \end{matrix}

(103)

since

{lim}_{r \to \infty} b (r) / c (r) = 1

(this is proven in [23]):

\begin{matrix} {log}_{2} ({(j!)}^{2} / 2 - 1) + \frac{b (j)}{{(j!)}^{2} / 2 - 1} \sim {log}_{2} ({(j!)}^{2} / 2 - 1) + 1 \end{matrix}

(104)

\begin{matrix} {log}_{2} ({(j!)}^{2} / 2 - 1) + {log}_{2} (2) = \end{matrix}

(105)

\begin{matrix} {log}_{2} ({(j!)}^{2} - 2)) \sim \end{matrix}

(106)

\begin{matrix} {log}_{2} ({(j!)}^{2}) = \end{matrix}

(107)

\begin{matrix} 2 {log}_{2} (j!) \end{matrix}

(108)

Hence, entropy[r] is the same as:

\begin{matrix} E (L (S (C (1, G_{r}^{★}, ω), ψ))) \sim \end{matrix}

(109)

\begin{matrix} 2 {log}_{2} (j!) \end{matrix}

(110)

Therefore, using §5.4.2 ((a)) and §5.4.2 ((3)(a)), take

|S (C (ε, G_{j}^{★ ★}, ω^{'}), ψ^{'})| = {(j!)}^{2}

to compute the following:

\begin{matrix} \underset{̲}{|S (C (ε, G_{r}^{★}, ω), ψ)|} = \\ sup \{|S (C (ε, G_{j}^{★ ★}, ω^{'}), ψ^{'})| : j \in N, ω^{'} \in Ω_{ε, j}, ψ^{'} \in Ψ_{ε, j, ω}, E (L (S (C (ε, G_{j}^{★ ★}, ω^{'}), ψ^{'}))) \leq E (L (S (C (ε, G_{r}^{★}, ω), ψ)))\} = \\ sup \{{(j!)}^{2} : j \in N, ω^{'} \in Ω_{ε, j}, ψ^{'} \in Ψ_{ε, j, ω}, 2 {log}_{2} (j!) \leq {log}_{2} (r!)\} = \end{matrix}

(111)

where:

(1): For every $r \in N$ , we find a $j \in N$ , where $2 {log}_{2} (j!) \leq {log}_{2} (r!)$ , but the absolute value of ${log}_{2} (r!) - 2 {log}_{2} (j!)$ is minimized. In other words, for every $r \in N$ , we want $j \in N$ where:

$\begin{matrix} 2 {log}_{2} (j!) \leq {log}_{2} (r!) \end{matrix}$

(112)

$\begin{matrix} 2^{2 {log}_{2} (j!)} \leq 2^{{log}_{2} (r!)} \end{matrix}$

(113)

$\begin{matrix} {(2^{{log}_{2} (j!)})}^{2} \leq r! \end{matrix}$

(114)

$\begin{matrix} {(j!)}^{2} \leq r! \end{matrix}$

(115)

$\begin{matrix} {(j!)}^{2} = ⌊ r! ⌋ \end{matrix}$

(116)

To solve for j, we try the following code:

Code 8: Code for j in eq. 116

Note, the output is:

Code 9: Output for code 8

Figure 1. Plot of loweralphr.

Finally, since the lower bound of loweralphr is zero, we have shown:

\underset{ε \to 0}{lim inf} \underset{r \to \infty}{lim inf} inf_{ω \in Ω_{ε, r}} inf_{ψ \in Ψ_{ε, r, ω}} \underset{̲}{α} (ε, r, ω, ψ) = 0

(117)

Next, using §5.4.2 ((b)) and §5.4.2 ((3)(b)), take

|S (C (ε, G_{r}^{★}, ω^{'}), ψ^{'})| = r!

and swap

r \in N

and

{(G_{r}^{★})}_{r \in N}

with

j \in N

and

{(G_{j}^{★ ★})}_{j \in N}

, to compute the following:

\begin{matrix} \underset{̲}{|S (C (ε, G_{j}^{★ ★}, ω), ψ)|} = \\ inf \{|S (C (ε, G_{r}^{★}, ω^{'}), ψ^{'})| : r \in N, ω^{'} \in Ω_{ε, r}, ψ^{'} \in Ψ_{ε, r, ω}, E (L (S (C (ε, G_{r}^{★}, ω^{'}), ψ^{'}))) \geq E (L (S (C (ε, G_{j}^{★ ★}, ω), ψ)))\} = \\ inf \{r! : r \in N, ω^{'} \in Ω_{ε, r}, ψ^{'} \in Ψ_{ε, j, ω}, {log}_{2} (r!) \geq 2 {log}_{2} (j!)\} = \end{matrix}

(118)

where:

(1): For every $j \in N$ , we find a $r \in N$ , where ${log}_{2} (r!) \leq 2 {log}_{2} (j!)$ , but the absolute value of $2 {log}_{2} (j!) - {log}_{2} (r!)$ is minimized. In other words, for every $j \in N$ , we want $r \in N$ where:

$\begin{matrix} {log}_{2} (r!) \leq 2 {log}_{2} (j!) \end{matrix}$

(119)

$\begin{matrix} 2^{{log}_{2} (r!)} \leq 2^{2 {log}_{2} (j!)} \end{matrix}$

(120)

$\begin{matrix} r! \leq {(2^{{log}_{2} (j!)})}^{2} \end{matrix}$

(121)

$\begin{matrix} r! \leq {(j!)}^{2} \end{matrix}$

(122)

$\begin{matrix} r! = {(j!)}^{2} \end{matrix}$

(123)

To solve r, we try the following code:

Code 10: Code for r in eq. 123

Note, the output is:

Code 11: Output for code 10

Figure 2. Plot of loweralphj.

since the lower bound of loweralphj is zero, we have shown:

\underset{ε \to 0}{lim inf} \underset{j \to \infty}{lim inf} inf_{ω \in Ω_{ε, j}} inf_{ψ \in Ψ_{ε, j, ω}} \underset{̲}{α} (ε, j, ω, ψ) = 0

(124)

Hence, using eq. 117 and 124, since both:

(1): $\underset{ε \to 0}{lim sup} \underset{r \to \infty}{lim sup} sup_{ω \in Ω_{ε, r}} sup_{ψ \in Ψ_{ε, r, ω}} \bar{α} (ε, r, ω, ψ)$ or $\underset{ε \to 0}{lim inf} \underset{r \to \infty}{lim inf} inf_{ω \in Ω_{ε, r}} inf_{ψ \in Ψ_{ε, r, ω}} \underset{̲}{α} (ε, r, ω, ψ)$ are equal to zero, one or $+ \infty$
(2): $\underset{ε \to 0}{lim sup} \underset{j \to \infty}{lim sup} sup_{ω \in Ω_{ε, j}} sup_{ψ \in Ψ_{ε, j, ω}} \bar{α} (ε, j, ω, ψ)$ or $\underset{ε \to 0}{lim inf} \underset{j \to \infty}{lim inf} inf_{ω \in Ω_{ε, j}} inf_{ψ \in Ψ_{ε, j, ω}} \underset{̲}{α} (ε, j, ω, ψ)$ are equal to zero, one or $+ \infty$

then what I’m measuring from

{(G_{r}^{★})}_{r \in N}

increases at a rate linear to that of

{(G_{j}^{★ ★})}_{j \in N}

5.5. Defining The Actual Rate of Expansion of Sequence of Bounded Sets

5.5.1. Definition of Actual Rate of Expansion of Sequence of Bounded Sets

Suppose

{(f_{r})}_{r \in N}

is a sequence of bounded functions converging to f, where

{(G_{r})}_{r \in N}

is a sequence of the graph on each

f_{r}

, and

d (Q, R)

is the Euclidean distance between points

Q, R \in R^{n}

. Therefore, using the “chosen" center point

C \in R^{n + 1}

, when:

G (C, G_{r}) = sup \{d (C, y) : y \in G_{r}\}

the actual rate of expansion is:

E (C, G_{r}) = G (C, G_{r + 1}) - G (C, G_{r})

Note, there are cases of

{(G_{r})}_{r \in N}

when

E

isn’t fixed and

E \neq E

(i.e., the chosen, fixed rate of expansion).

5.5.2. Example

Suppose, we have

f : A \to R

, where

A = R

and

f (x) = x

, such that

{(A_{r}^{★})}_{r \in N} = {([- r, r])}_{r \in N}

and for

f_{r}^{★} : A_{r}^{★} \to R

f_{r}^{★} (x) = f (x) for all x \in A_{r}^{★}

Hence, when

{(G_{r}^{★})}_{r \in N}

is:

{(G_{r}^{★})}_{r \in N} = {(\{(x, x) : x \in [- r, r]\})}_{r \in N}

such that

C = (0, 0)

, note the farthest point of

G_{r}^{★}

from C is either

(- r, - r)

(r, r)

. Hence, to compute

G (C, G_{r}^{★})

, we can take

d ((0, 0), (r, r))

d ((0, 0), (- r, r))

\begin{matrix} G (C, G_{r}^{★}) = & sup \{d (C, y) : y \in G_{r}^{★}\} = \end{matrix}

(125)

\begin{matrix} d ((0, 0), (r, r)) = \end{matrix}

(126)

\begin{matrix} \sqrt{{(0 - r)}^{2} + {(0 - r)}^{2}} = \end{matrix}

(127)

\begin{matrix} \sqrt{r^{2} + r^{2}} = \end{matrix}

(128)

\begin{matrix} \sqrt{2 r^{2}} = \end{matrix}

(129)

\begin{matrix} \sqrt{2 r^{2}} = \end{matrix}

(130)

\begin{matrix} \sqrt{2} | r | \end{matrix}

(131)

\begin{matrix} \sqrt{2} r \sin ce r > 0 \end{matrix}

(132)

and the actual rate of expansion is:

\begin{matrix} E (C, G_{r}^{★}) = & G (C, G_{r + 1}^{★}) - G (C, G_{r}^{★}) = \end{matrix}

(133)

\begin{matrix} \sqrt{2} (r + 1) - \sqrt{2} r = \end{matrix}

(134)

\begin{matrix} \sqrt{2} (r + 1) - \sqrt{2} r = \end{matrix}

(135)

\begin{matrix} \sqrt{2} r + \sqrt{2} - \sqrt{2} r = \end{matrix}

(136)

\begin{matrix} \sqrt{2} \end{matrix}

(137)

5.6. Reminder

See if §3.2 is easier to understand.

6. My Attempt At Answering The Blockquote of §1.3.2

6.1. Choice Function

Suppose we define the following:

(1): ${(f_{k}^{★ ★ ★})}_{k \in N}$ is the sequence of bounded functions which satisfies ((1)), ((2)), ((3)), ((4)) and ((5)) of the leading question in $§$ 3.2
(2): $S^{'} (f)$ is all sequences of bounded functions satisfying ((1)) of the leading question where the expected values, defined in the papers of $§$ 2, is finite.
(3): ${(f_{j}^{★ ★})}_{j \in N}$ is an element $S^{'} (f)$ but not an element in the set of equivelant sequences of bounded functions to that of ${(f_{k}^{★ ★ ★})}_{k \in N}$ (def. 7), where using the end of def. 7, we represent this criteria as:

${(f_{j}^{★ ★})}_{j \in N} \in S^{'} (f) ∖ \sim {(f_{k}^{★ ★ ★})}_{k \in N}$

Further note, from §5.4.2 ((b)), if we take:

\begin{matrix} \bar{|S (C (ε, G_{k}^{★ ★ ★}, ω), ψ)|} = \\ inf \{|S (C (ε, G_{j}^{★ ★}, ω^{'}), ψ^{'})| : j \in N, ω^{'} \in Ω_{ε, j}, ψ^{'} \in Ψ_{ε, j, ω}, E (L (S (C (ε, G_{j}^{★ ★}, ω^{'}), ψ^{'}))) \geq E (L (S (C (ε, G_{k}^{★ ★ ★}, ω), ψ)))\} \end{matrix}

(138)

and from §5.4.2 ((a)), we take:

\begin{matrix} \underset{̲}{|S (C (ε, G_{k}^{★ ★ ★}, ω), ψ)|} = \\ sup \{|S (C (ε, G_{j}^{★ ★}, ω^{'}), ψ^{'})| : j \in N, ω^{'} \in Ω_{ε, j}, ψ^{'} \in Ψ_{ε, j, ω}, E (L (S (C (ε, G_{j}^{★ ★}, ω^{'}), ψ^{'}))) \leq E (L (S (C (ε, G_{k}^{★ ★ ★}, ω), ψ)))\} \end{matrix}

(139)

Then, §5.4.1 ((2)), eq. 138, and eq. 139 is:

\begin{matrix} sup_{ω \in Ω_{ε, k}} sup_{ψ \in Ψ_{ε, k, ω}} |S (C (ε, G_{k}^{★ ★ ★}, ω), ψ)| = |S^{'} (ε, G_{k}^{★ ★ ★})| = |S^{'}| \end{matrix}

(140)

\begin{matrix} sup_{ω \in Ω_{ε, k}} sup_{ψ \in Ψ_{ε, k, ω}} \bar{|S (C (ε, G_{k}^{★ ★ ★}, ω), ψ)|} = \bar{|S^{'} (ε, G_{k}^{★ ★ ★})|} = \bar{|S^{'}|} \end{matrix}

(141)

\begin{matrix} sup_{ω \in Ω_{ε, k}} sup_{ψ \in Ψ_{ε, k, ω}} \underset{̲}{|S (C (ε, G_{k}^{★ ★ ★}, ω), ψ)|} = \underset{̲}{|S^{'} (ε, G_{k}^{★ ★ ★})|} = \underset{̲}{|S^{'}|} \end{matrix}

(142)

6.2. Approach

We manipulate the definitions of §5.4.2 ((a)) and §5.4.2 ((b)) to solve (((1)), ((2)), ((3)), ((4)) and ((5)) of the leading question in

§

3.2

6.3. Potential Answer

6.3.1. Preliminaries (Definition of T in Case of §3.2.1 ((5)))

When the difference of point

X = (x_{1}, \cdot \cdot \cdot, x_{n})

and

Y = (y_{1}, \cdot \cdot \cdot, y_{n})

is:

X - Y = (x_{1} - y_{1}, x_{2} - y_{2}, \cdot \cdot \cdot, x_{n} - y_{n})

the average of

G_{r}^{★}

for every

r \in N

is:

Avg (G_{r}^{★}) = \frac{1}{H^{\dim_{H} (G_{r}^{★})} (G_{r}^{★})} \int_{G_{r}^{★}} (x_{1}, \cdot \cdot \cdot, x_{n}) d H^{\dim_{H} (G_{r}^{★})}

(143)

and

d (P, Q)

is the n-d Euclidean distance between points

P, Q \in R^{n}

, we define an explicit injective

F : R^{n} \to R

such that:

(1): If $d (Avg (G_{r}^{★}), C) < d (Avg (G_{j}^{★ ★}), C)$ , then $F (Avg (G_{r}^{★}) - C) < F (Avg (G_{j}^{★ ★}) - C)$
(2): If $d (Avg (G_{r}^{★}), C) > d (Avg (G_{j}^{★ ★}), C)$ , then $F (Avg (G_{r}^{★}) - C) > F (Avg (G_{j}^{★ ★}) - C)$
(3): If $d (Avg (G_{r}^{★}), C) = d (Avg (F_{j}^{★ ★}), C)$ , then $F (Avg (G_{r}^{★}) - C) \neq F (Avg (G_{j}^{★ ★}) - C)$

where using “chosen" center point

C \in R^{n}

T (C, G_{r}^{★}) = F (Avg (G_{r}^{★}) - C)

(144)

6.3.2. Question

Does T exist? If so, how do we define it?

Hence, using

|S^{'}|

\bar{|S^{'}|}

\underset{̲}{|S^{'}|}

, E,

E (C, G_{k}^{★ ★ ★})

(§5.5), and

T (C, F_{k}^{★ ★ ★})

, such that with the absolute value function

| | \cdot | |

, ceiling function

⌈ \cdot ⌉

, and nearest integer function

[\cdot]

, we define:

\begin{matrix} K (ε, G_{k}^{★ ★ ★}) = \\ (1 + ||E - E (C, G_{k}^{★ ★ ★})||) 5 (5 | 5 | \frac{|S^{'}| (1 + ⌈\frac{|S^{'}| (\underset{̲}{|S^{'}|} + 2 |S^{'}|)}{(\underset{̲}{|S^{'}|} + |S^{'}|) (\underset{̲}{|S^{'}|} + |S^{'}| + \bar{|S^{'}|})}⌉) (1 + [\underset{̲}{|S^{'}|} / |S^{'}|])}{(1 + [|S^{'}| / \bar{|S^{'}|}]) (1 + [\underset{̲}{|S^{'}|} / \bar{|S^{'}|}])} - |S^{'}| 5 | 5 | + |S^{'}| 5) - T (C, G_{k}^{★ ★ ★}) E (C, G_{k}^{★ ★ ★}) \end{matrix}

(145)

where

E

, E, and T are “removed" when

E, E = 0

, the choice function which answers the leading question in

§

3.2 could be the following, s.t.we explain the reason behind choosing the choice function in §Section 6.4:

Theorem 11.

If we define:

M (ε, G_{k}^{★ ★ ★}) = | S^{'} (ε, G_{k}^{★ ★ ★}) | (K (ε, G_{k}^{★ ★ ★}) - | S^{'} (ε, G_{k}^{★ ★ ★}) |)

M (ε, G_{j}^{★ ★}) = | S^{'} (ε, G_{j}^{★ ★}) | (K (ε, G_{j}^{★ ★}) - | S^{'} (ε, G_{j}^{★ ★}) |)

where for

M (ε, G_{k}^{★ ★ ★})

, we define

M (ε, G_{k}^{★ ★ ★})

to be the same as

M (ε, G_{j}^{★ ★})

when swapping “

j \in N

" with “

k \in N

" (for eq. 138 & 139) and sets

G_{k}^{★ ★ ★}

with

G_{j}^{★ ★}

(for eq. 138–145), then for constant

v > 0

and variable

v^{*} > 0

, if:

\bar{S} (ε, k, v^{*}, G_{j}^{★ ★}) = inf (\{| S^{'} (ε, G_{j}^{★ ★}) | : j \in N, M (ε, G_{j}^{★ ★}) \geq M (ε, G_{k}^{★ ★ ★}) \geq v^{*}\} \cup {v^{*}}) + v

(146)

and:

\underset{̲}{S} (ε, k, v^{*}, G_{j}^{★ ★}) = sup (\{| S^{'} (ε, G_{j}^{★ ★}) | : j \in N, v^{*} \leq M (ε, G_{j}^{★ ★}) \leq M (ε, G_{k}^{★ ★ ★})\} \cup {- v^{*}}) + v

(147)

then for all

{(f_{j}^{★ ★})}_{j \in N} \in S^{'} (f) ∖ \sim {(f_{k}^{★ ★ ★})}_{k \in N}

(def. 7), if:

inf \{| | 1 - c | | : \forall (ϵ > 0) \exists (c > 0) \forall (k \in N) \exists (j \in N) (||\frac{|S^{'} (ε, G_{k}^{★ ★ ★})|}{|S^{'} (ε, G_{j}^{★ ★})|} - c|| < ε)\}

(148)

where

⌈ \cdot ⌉

is the ceiling function, E is the fixed rate of expansion, Γ is the gamma function, n is the dimension of

R^{n}

\dim_{H} (G_{k}^{★ ★ ★})

is the Hausdorff dimension of set

G_{k}^{★ ★ ★} \subseteq R^{n + 1}

, and

A_{k}

is area of the smallest

(n + 1)

-dimensional box that contains

A_{k}^{★ ★ ★}

, then:

\begin{matrix} V (ε, G_{k}^{★ ★ ★}, n) = \\ ⌈ & (A_{k}^{1 - sign (E)} (E - sign (E) + 1) (\frac{exp (n ln (π) / 2)}{Γ (n / 2 + 1)}) ({k!}^{(n - \dim_{H} (G_{k}^{★ ★ ★}))}) (k^{sign (E) (\dim_{H} (G_{k}^{★ ★ ★}) - sign (\dim_{H} (G_{k}^{★ ★ ★})) + 1)}) + \\ (1 - sign (\dim_{H} (G_{k}^{★ ★ ★})))) / ε ⌉ / | S^{'} (ε, G_{k}^{★ ★ ★}) | \end{matrix}

(149)

the choice function is:

\begin{matrix} \underset{ε \to 0}{lim sup} lim_{v^{*} \to \infty} \underset{k \to \infty}{lim sup} & (\frac{sign (M (ε, G_{k}^{★ ★ ★})) \bar{S} (ε, k, v^{*}, G_{j}^{★ ★})}{| S^{'} (ε, G_{k}^{★ ★ ★}) | + v} - c^{- V (ε, G_{k}^{★ ★ ★}, n)}) \\ (\frac{sign (M (ε, G_{k}^{★ ★ ★})) \underset{̲}{S} (ε, k, v^{*}, G_{j}^{★ ★})}{| S^{'} (ε, G_{k}^{★ ★ ★}) | + v} - c^{- V (ε, G_{k}^{★ ★ ★}, n)}) = \end{matrix}

(150)

\begin{matrix} \underset{ε \to 0}{lim inf} lim_{v^{*} \to \infty} \underset{k \to \infty}{lim inf} & (\frac{sign (M (ε, G_{k}^{★ ★ ★})) \bar{S} (ε, k, v^{*}, G_{j}^{★ ★})}{| S^{'} (ε, G_{k}^{★ ★ ★}) | + v} - c^{- V (ε, G_{k}^{★ ★ ★}, n)}) \\ (\frac{sign (M (ε, G_{k}^{★ ★ ★})) \underset{̲}{S} (ε, k, v^{*}, G_{j}^{★ ★})}{| S^{'} (ε, G_{k}^{★ ★ ★}) | + v} - c^{- V (ε, G_{k}^{★ ★ ★}, n)}) = 0 \end{matrix}

(151)

such that

{(G_{k}^{★ ★ ★})}_{k \in N}

satisfies eq. 150 & eq. . (Note, we want

sup \emptyset = - \infty

inf \emptyset = + \infty

, and

{(f_{k}^{★ ★ ★})}_{k \in N}

to answer the leading question of

§

3.2) where the answer to the blockquote of

§

1.3.2 is

E [f_{k}^{★ ★ ★}]

(when it exists).

6.4. Explaining The Choice Function and Evidence The Choice Function Is Credible

Notice, before reading the programming in code 12, without the “c"-terms in eq. 150 and eq. :

(1)

The choice function in eq. 150 and eq. 151 is zero, when what I’m measuring from

{(G_{k}^{★ ★ ★})}_{k \in N}

(§5.4.2 criteria (1)) increases at a rate superlinear to that of

{(G_{j}^{★ ★})}_{j \in N}

, where

sign (M (ε, G_{k}^{★ ★ ★})) = 0

(2)

The choice function in eq. 150 and eq. 151 is zero, when for a given

{(G_{k}^{★ ★ ★})}_{k \in N}

and

{(G_{j}^{★ ★})}_{j \in N}

there doesn’t exist c where eq. 148 is satisfied or

c = 0

(3)

When c does exist, suppose:

\{J (k) : k \in N, \frac{| S^{'} (ε, G_{k}^{★ ★ ★}) |}{| S^{'} (ε, G_{J (k)}^{★ ★}) |} \approx c\}

(152)

(a): When $| S^{'} (ε, G_{k}^{★ ★ ★}) | < | S^{'} (ε, G_{J (k)}^{★ ★}) |$ , then:

$\begin{matrix} \underset{ε \to 0}{lim sup} lim_{v^{*} \to \infty} \underset{k \to \infty}{lim sup} \frac{sign (M (ε, G_{k}^{★ ★ ★})) \bar{S} (ε, k, v^{*}, G_{j}^{★ ★})}{| S^{'} (ε, G_{k}^{★ ★ ★}) | + v} = c \end{matrix}$

(153)

$\begin{matrix} \underset{ε \to 0}{lim inf} lim_{v^{*} \to \infty} \underset{k \to \infty}{lim inf} \frac{sign (M (ε, G_{k}^{★ ★ ★})) \underset{̲}{S} (ε, k, v^{*}, G_{j}^{★ ★})}{| S^{'} (ε, G_{k}^{★ ★ ★}) | + v} = 0 \end{matrix}$

(154)
(b): When $| S^{'} (ε, G_{k}^{★ ★ ★}) | > | S^{'} (ε, G_{J (k)}^{★ ★}) |$ , then:

$\begin{matrix} \underset{ε \to 0}{lim sup} lim_{v^{*} \to \infty} \underset{k \to \infty}{lim sup} \frac{sign (M (ε, G_{k}^{★ ★ ★})) \bar{S} (ε, k, v^{*}, G_{j}^{★ ★})}{| S^{'} (ε, G_{k}^{★ ★ ★}) | + v} = + \infty \end{matrix}$

(155)

$\begin{matrix} \underset{ε \to 0}{lim inf} lim_{v^{*} \to \infty} \underset{k \to \infty}{lim inf} \frac{sign (M (ε, G_{k}^{★ ★ ★})) \underset{̲}{S} (ε, k, v^{*}, G_{j}^{★ ★})}{| S^{'} (ε, G_{k}^{★ ★ ★}) | + v} = 1 / c \end{matrix}$

(156)

Hence, for each sub-criteria under crit. ((3)), if we subtract one of their limits by their limit value, then eq. 150 and eq. 151 is zero. (We do this using the “c"-term in eq. 150 and 151). However, when the exponents of the “c"-terms aren’t equal to

- 1

, the limits of eq. 150 and 151 aren’t equal to zero. We want this, infact, whenever we swap

S^{'} (ε, G_{k}^{★ ★ ★})

with

S^{'} (ε, G_{j}^{★ ★})

. Moreover, we define function

V (ε, G_{k}^{★ ★ ★}, n)

(i.e., eq. 149), where:

(i)

When

S^{'} (ε, G_{k}^{★ ★ ★}) ≫ Numerator (V (ε, G_{k}^{★ ★ ★}, n))

, then eq. 150 and without the “c"-terms are zero. (The “c"-terms approach zero and still allow eq. 150 and to equal zero.)

(ii)

When

S^{'} (ε, G_{k}^{★ ★ ★}) ≪ Numerator (V (ε, G_{k}^{★ ★ ★}, n))

, then

sign (M (ε, G_{k}^{★ ★ ★}))

is zero which makes eq. 150 and equal zero.

(iii)

Here are some examples of the numerator of

V (ε, G_{k}^{★ ★ ★}, n)

(eq. 149):

(A): When $E = 0$ , $n = 1$ , and $\dim_{H} (A) = 0$ , the numerator of $V (ε, G_{k}^{★ ★ ★}, n)$ is $⌈ (A k! + 1) / ε ⌉$
(B): When $E = z$ , $n = 1$ , and $\dim_{H} (A) = 0$ , the numerator of $V (ε, G_{k}^{★ ★ ★}, n)$ is $⌈ (2 z k \cdot k! + 1) / ε ⌉$
(C): When $E = 0$ , $n = z_{2}$ , and $\dim_{H} (A) = z_{2}$ , the numerator of $V (ε, G_{k}^{★ ★ ★}, n)$ is ceiling of constant $A$ times the volume of an n-dimensional ball with finite radius: i.e.,

$⌈\frac{A z_{1} exp (z_{2} ln (π) / 2)}{Γ (z_{2} / 2 + 1)} / ε⌉$
(D): When $E = z_{1}$ , $n = z_{2}$ , and $\dim_{H} (A) = z_{2}$ , the numerator of $V (ε, G_{k}^{★ ★ ★}, n)$ is ceiling of the volume of the n-dimensional ball: i.e.,

$⌈\frac{z_{1} exp (z_{2} ln (π) / 2)}{Γ (z_{2} / 2 + 1)} k^{z_{2}} / ε⌉$

Now, consider the code for eq. 150 and eq. 151. (Note, the set theoretic limit of

G_{k}^{★ ★ ★}

is the graph of function

f : A \to R

.) In this example,

A = Q \cap [0, 1]

, and:

f (x) = \{\begin{matrix} 1 & x \in \{(2 s + 1) / (2 t) : s \in Z, t \in N, t \neq 0\} \cap [0, 1] \\ 0 & x \notin \{(2 s + 1) / (2 t) : s \in Z, t \in N, t \neq 0\} \cap [0, 1] \end{matrix}

(157)

such that:

{(A_{k}^{★ ★ ★})}_{r \in N} = {(\{c / k! : c \in Z, 0 \leq c \leq k!\})}_{k \in N}

the ceiling function is

⌈ \cdot ⌉

, and:

{(A_{j}^{★ ★})}_{j \in N} = {(\{c / ⌈ j! / 3 ⌉ : c \in Z, 0 \leq c \leq ⌈ j! / 3 ⌉\})}_{j \in N}

such for

f_{k}^{★} : A_{k}^{★ ★ ★} \to R

f_{k}^{★ ★ ★} (x) = f (x) for all x \in A_{k}^{★ ★ ★}

(158)

and

f_{j}^{★ ★} : A_{j}^{★ ★} \to R

f_{j}^{★ ★} (x) = f (x) for all x \in A_{j}^{★ ★}

(159)

Hence, when

{(G_{k}^{★ ★ ★})}_{r \in N}

is:

{(G_{k}^{★ ★ ★})}_{k \in N} = {(\{(x, f (x)) : x \in \{c / k! : c \in Z, 0 \leq c \leq k!\}\})}_{k \in N}

(160)

and

{(G_{j}^{★ ★})}_{j \in N}

is:

{(G_{j}^{★ ★})}_{j \in N} = {(\{(x, f (x)) : x \in \{c / ⌈ j! / 3 ⌉ : c \in Z, 0 \leq c \leq ⌈ j! / 3 ⌉\}\})}_{j \in N}

(161)

Note, the following (we leave this to mathematicians to figure LengthS1, LengthS2, Entropy1 and

Entropy 2

for other A and f in code 12).

6.4.1. Evidence with Programming

Code 12: Code for eq. 150 and 151 to eq. 160 and eq. 161

7. Questions

(1): Does $§$ 6 answer the $leading question$ in $§$ 3.2
(2): Using thm. 11, when f is defined in $§$ 1.1, does $E [f_{k}^{★ ★ ★}]$ have a finite value?
(3): Using thm. 11, when f is defined in $§$ 1.2, does $E [f_{k}^{★ ★ ★}]$ have a finite value?
(4): If there’s no time to check questions (1), (2) and (3), see $§$ .

8. Appendix of §5.4.1

8.1. Example of §5.4.1, Step (1)

Suppose

(1): $A = R$
(2): When defining $f : A \to R$ :

$f (x) = \{\begin{matrix} 1 & x < 0 \\ - 1 & 0 \leq x < 0.5 \\ 0.5 & 0.5 \leq x \end{matrix}$

(162)
(3): ${(G_{r}^{★})}_{r \in N} = {(\{(x, f (x)) : - r \leq x \leq r\})}_{r \in N}$

Then one example of

C (\sqrt{2} / 6, G_{1}^{★}, 1)

, using §5.4.1 step (1), (where

G_{1}^{★} = {(\{(x, f (x)) : - 1 \leq x \leq 1\})}_{r \in N}

) is:

\begin{matrix} { & \{(x, f (x)) : - 1 \leq x \leq \frac{\sqrt{2} - 6}{6}\}, \{(x, f (x)) : \frac{\sqrt{2} - 6}{6} \leq x \leq \frac{2 \sqrt{2} - 6}{6}\}, \{(x, f (x)) : \frac{2 \sqrt{2} - 6}{6} \leq x \leq \frac{3 \sqrt{2} - 6}{6}\} \\ \{(x, f (x)) : \frac{3 \sqrt{2} - 6}{6} \leq x \leq \frac{4 \sqrt{2} - 6}{6}\}, \{(x, f (x)) : \frac{4 \sqrt{2} - 6}{6} \leq x \leq \frac{5 \sqrt{2} - 6}{6}\}, \{(x, f (x)) : \frac{5 \sqrt{2} - 6}{6} \leq x \leq \frac{6 \sqrt{2} - 6}{6}\} \\ \{(x, f (x)) : \frac{6 \sqrt{2} - 6}{6} \leq x \leq \frac{7 \sqrt{2} - 6}{6}\}, \{(x, f (x)) : \frac{7 \sqrt{2} - 6}{6} \leq x \leq \frac{8 \sqrt{2} - 6}{6}\}, \{(x, f (x)) : \frac{8 \sqrt{2} - 6}{6} \leq x \leq \frac{9 \sqrt{2} - 6}{6}\}} \end{matrix}

(163)

Note, the length of each partition is

\sqrt{2} / 6

, where the borders could be approximated as:

\begin{matrix} { & \{(x, f (x)) : - 1 \leq x \leq - . 764\}, \{(x, f (x)) : - . 764 \leq x \leq - . 528\}, \{(x, f (x)) : - . 528 \leq x \leq - . 293\} \\ \{(x, f (x)) : - . 293 \leq x \leq - . 057\}, \{(x, f (x)) : - . 057 \leq x \leq . 178\}, \{(x, f (x)) : . 178 \leq x \leq . 414\} \\ \{(x, f (x)) : . 414 \leq x \leq . 65\}, \{(x, f (x)) : . 65 \leq x \leq . 886\}, \{(x, f (x)) : . 886 \leq x \leq 1.121\}} \end{matrix}

(164)

which is illustrated using alternating orange/black lines of equal length covering

G_{1}^{★}

(i.e., the black vertical lines are the smallest and largest x-cooridinates of

G_{1}^{★}

Figure 3. The alternating orange & black lines are the “covers" and the vertical lines are the boundaries of

G_{1}^{★}

Figure 3. The alternating orange & black lines are the “covers" and the vertical lines are the boundaries of

G_{1}^{★}

(Note, the alternating covers in Figure 3 satisfy step ((1)) of §5.4.1, because the Hausdorff measure in its dimension of the covers is

\sqrt{2} / 6

and there are 9 covers over-covering

G_{1}^{★}

: i.e.,

Definition 12

(Minimum Covers of Measure

ε = \sqrt{2} / 6

covering

G_{1}^{★}

). We can compute the minimum covers of

C (\sqrt{2} / 6, G_{1}^{★}, 1)

, using the formula:

⌈ H^{\dim_{H} (G_{1}^{★})} (G_{1}^{★}) / (\sqrt{2} / 6) ⌉

where

⌈ H^{\dim_{H} (G_{1}^{★})} (G_{1}^{★}) / (\sqrt{2} / 6) ⌉ = ⌈ Length ([- 1, 1]) / (\sqrt{2} / 6) ⌉ = ⌈ 2 / (\sqrt{2} / 6) ⌉ = ⌈ 6 \sqrt{2} ⌉ = ⌈ 6 (1.4) ⌉ = ⌈ 8 + . 4 ⌉ = 9

Note there are other examples of

C (\sqrt{2} / 6, G_{1}^{★}, ω)

for different

ω

. Here is another case:

Figure 4. This is similar to Figure 3, except the start-points of the covers are shifted all the way to the left.

which can be defined (see eq. 163 for comparison):

\begin{matrix} { & \{(x, f (x)) : \frac{6 - 9 \sqrt{2}}{6} \leq x \leq \frac{6 - 8 \sqrt{2}}{6}\}, \{(x, f (x)) : \frac{6 - 8 \sqrt{2}}{6} \leq x \leq \frac{6 - 7 \sqrt{2}}{6}\}, \{(x, f (x)) : \frac{6 - 7 \sqrt{2}}{6} \leq x \leq \frac{6 - 6 \sqrt{2}}{6}\} \\ \{(x, f (x)) : \frac{6 - 6 \sqrt{2}}{6} \leq x \leq \frac{6 - 5 \sqrt{2}}{6}\}, \{(x, f (x)) : \frac{6 - 5 \sqrt{2}}{6} \leq x \leq \frac{6 - 4 \sqrt{2}}{6}\}, \{(x, f (x)) : \frac{6 - 4 \sqrt{2}}{6} \leq x \leq \frac{6 - 3 \sqrt{2}}{6}\} \\ \{(x, f (x)) : \frac{6 - 3 \sqrt{2}}{6} \leq x \leq \frac{6 - 2 \sqrt{2}}{6}\}, \{(x, f (x)) : \frac{6 - 2 \sqrt{2}}{6} \leq x \leq \frac{6 - \sqrt{2}}{6}\}, \{(x, f (x)) : \frac{6 - \sqrt{2}}{6} \leq x \leq 1\}} \end{matrix}

(165)

In the case of

G_{1}^{★}

, there are uncountable different covers

C (\sqrt{2} / 6, G_{1}^{★}, ω)

which can be used. For instance, when

0 \leq α \leq (12 - 9 \sqrt{2}) / 6

(i.e.,

ω = α - (12 - 9 \sqrt{2}) / 6 + 1

) consider:

\begin{matrix} { & \{(x, f (x)) : α - 1 + α \leq x \leq α + \frac{\sqrt{2} - 6}{6}\}, \{(x, f (x)) : α + \frac{\sqrt{2} - 6}{6} \leq x \leq α + \frac{2 \sqrt{2} - 6}{6}\}, \{(x, f (x)) : α + \frac{2 \sqrt{2} - 6}{6} \leq x \leq α + \frac{3 \sqrt{2} - 6}{6}\} \\ \{(x, f (x)) : α + \frac{3 \sqrt{2} - 6}{6} \leq x \leq α + \frac{4 \sqrt{2} - 6}{6}\}, \{(x, f (x)) : α + \frac{4 \sqrt{2} - 6}{6} \leq x \leq α + \frac{5 \sqrt{2} - 6}{6}\}, \\ \{(x, f (x)) : α + \frac{5 \sqrt{2} - 6}{6} \leq x \leq α + \frac{6 \sqrt{2} - 6}{6}\}, \{(x, f (x)) : α + \frac{6 \sqrt{2} - 6}{6} \leq x \leq α + \frac{7 \sqrt{2} - 6}{6}\} \\ \{(x, f (x)) : α + \frac{7 \sqrt{2} - 6}{6} \leq x \leq α + \frac{8 \sqrt{2} - 6}{6}\}, \{(x, f (x)) : α + \frac{8 \sqrt{2} - 6}{6} \leq x \leq α + \frac{9 \sqrt{2} - 6}{6}\}} \end{matrix}

(166)

When

α = 0

and

ω = (9 \sqrt{2} - 6) / 6

, we get Figure 4 and when

α = (12 - 9 \sqrt{2}) / 6

and

ω = 1

, we get Figure 3

8.2. Example of §5.4.1, Step (2)

Suppose:

(1): $A = R$
(2): When defining $f : A \to R$ : i.e.,

$f (x) = \{\begin{matrix} 1 & x < 0 \\ - 1 & 0 \leq x < 0.5 \\ 0.5 & 0.5 \leq x \end{matrix}$

(167)
(3): ${(G_{r}^{★})}_{r \in N} = {(\{(x, f (x)) : - r \leq x \leq r\})}_{r \in N}$
(4): $G_{1}^{★} = \{(x, f (x)) : - 1 \leq x \leq 1\}$
(5): $C (\sqrt{2} / 6, G_{1}^{★}, 1)$ , using eq. 164 and fig. Figure 3, which is approximately

$\begin{matrix} { & \{(x, f (x)) : - 1 \leq x \leq - . 764\}, \{(x, f (x)) : - . 764 \leq x \leq - . 528\}, \{(x, f (x)) : - . 528 \leq x \leq - . 293\} \\ \{(x, f (x)) : - . 293 \leq x \leq - . 057\}, \{(x, f (x)) : - . 057 \leq x \leq . 178\}, \{(x, f (x)) : . 178 \leq x \leq . 414\} \\ \{(x, f (x)) : . 414 \leq x \leq . 65\}, \{(x, f (x)) : . 65 \leq x \leq . 886\}, \{(x, f (x)) : . 886 \leq x \leq 1.121\}} \end{matrix}$

(168)

Then, an example of

S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1)

is:

{(- . 9, 1), (- . 65, 1), (- . 4, 1), (- . 2, 1), (. 1, - 1), (. 3, - 1), (. 55, . 5), (. 75, . 5), (1, . 5)}

(169)

Below, we illustrate the sample: i.e., the set of all blue points in each orange and black line of

C (\sqrt{2} / 6, G_{1}^{★}, 1)

covering

G_{1}^{★}

Figure 5. The blue points are the “sample points", the alternative black and orange lines are the “covers", and the red lines are the smallest & largest x-coordinates

G_{1}^{★}

Figure 5. The blue points are the “sample points", the alternative black and orange lines are the “covers", and the red lines are the smallest & largest x-coordinates

G_{1}^{★}

Note, there are multiple samples that can be taken, as long as one sample point is taken from each cover in

C (\sqrt{2} / 6, G_{1}^{★}, 1)

8.3. Example of §5.4.1, Step (3)

Suppose

(1): $A = R$
(2): When defining $f : A \to R$ :

$f (x) = \{\begin{matrix} 1 & x < 0 \\ - 1 & 0 \leq x < 0.5 \\ 0.5 & 0.5 \leq x \end{matrix}$

(170)
(3): ${(G_{r}^{★})}_{r \in N} = {(\{(x, f (x)) : - r \leq x \leq r\})}_{r \in N}$
(4): $G_{1}^{★} = \{(x, f (x)) : - 1 \leq x \leq 1\}$
(5): $C (\sqrt{2} / 6, G_{1}^{★}, 1)$ , using eq. 164 and fig. Figure 3, is approx.

$\begin{matrix} { & \{(x, f (x)) : - 1 \leq x \leq - . 764\}, \{(x, f (x)) : - . 764 \leq x \leq - . 528\}, \{(x, f (x)) : - . 528 \leq x \leq - . 293\} \\ \{(x, f (x)) : - . 293 \leq x \leq - . 057\}, \{(x, f (x)) : - . 057 \leq x \leq . 178\}, \{(x, f (x)) : . 178 \leq x \leq . 414\} \\ \{(x, f (x)) : . 414 \leq x \leq . 65\}, \{(x, f (x)) : . 65 \leq x \leq . 886\}, \{(x, f (x)) : . 886 \leq x \leq 1.121\}} \end{matrix}$

(171)
(6): $S (C (13 / 6, G_{1}^{★}, 1), 1)$ , using eq. 169, is:

${(- . 9, 1), (- . 65, 1), (- . 4, 1), (- . 2, 1), (. 1, - 1), (. 3, - 1), (. 55, . 5), (. 75, . 5), (1, . 5)}$

(172)

Therefore, consider the following process:

8.3.1. Step (3)(a)

S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1)

is:

{(- . 9, 1), (- . 65, 1), (- . 4, 1), (- . 2, 1), (. 1, - 1), (. 3, - 1), (. 55, . 5), (. 75, . 5), (1, . 5)}

(173)

suppose

x_{0} = (- . 9, 1)

. Note, the following:

(1): $x_{1} = (- . 65, 1)$ is the next point in the “pathway" since it’s a point in $S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1)$ with the smallest 2-d Euclidean distance to $x_{0}$ instead of $x_{0}$ .
(2): $x_{2} = (- . 4, 1)$ is the third point since it’s a point in $S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1)$ with the smallest 2-d Euclidean distance to $x_{1}$ instead of $x_{0}$ and $x_{1}$ .
(3): $x_{3} = (- . 2, 1)$ is the fourth point since it’s a point in $S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1)$ with the smallest 2-d Euclidean distance to $x_{2}$ instead of $x_{0}$ , $x_{1}$ , and $x_{2}$ .
(4): we continue this process, where the “pathway" of $S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1)$ is:

(- . 9, 1) \to (- . 65, 1) \to (- . 4, 1) \to (- . 2, 1) \to (. 55, . 5) \to (. 75, . 5) \to (1, . 5) \to (. 3, - 1) \to (. 1, - 1)

(174)

Note 13.

If more than one point has the minimum 2-d Euclidean distance from

x_{0}

x_{1}

x_{2}

, etc. take all potential pathways: e.g., using the sample in eq. 173, if

x_{0} = (- . 65, 1)

, then since

(- . 9, 1)

and

(- . 4, 1)

have the smallest Euclidean distance to

(- . 65, 1)

, taketwopathways:

(- . 65, 1) \to (- . 9, 1) \to (- . 4, 1) \to (- . 2, 1) \to (. 55, . 5) \to (. 75, . 5) \to (1, . 5) \to (. 3, - 1) \to (. 1, - 1)

and also:

(- . 65, 1) \to (- . 4, 1) \to (- . 2, 1) \to (- . 9, 1) \to (. 55, . 5) \to (. 75, . 5) \to (1, . 5) \to (. 3, - 1) \to (. 1, - 1)

8.3.2. Step (3)(b)

Next, take the length of all line segments in each pathway. In other words, suppose

d (P, Q)

is the n-th dim.Euclidean distance between points

P, Q \in R^{n}

. Using the pathway in eq. 174, we want:

\begin{matrix} { & d ((- . 9, 1), (- . 65, 1)), d ((- . 65, 1), (- . 4, 1)), d ((- . 4, 1), (- . 2, 1)), d ((- . 2, 1), (. 55, . 5)), \\ d ((. 55, . 5), (. 75, . 5)), d ((. 75, . 5), (1, . 5)), d ((1, . 5), (. 3, - 1)), d ((. 3, - 1), (. 1, - 1))} \end{matrix}

(175)

Whose distances can be approximated as:

\begin{matrix} { & . 25, . 25, . 2, . 901389, . 2, . 25, 1.655295, . 2} \end{matrix}

Also, we see the outliers [11] are

. 901389

and

1.655295

(i.e., notice that the outliers are more prominent for

ε ≪ \sqrt{2} / 6

). Therefore, remove

. 901389

and

1.655295

from our set of lengths:

\begin{matrix} { & . 25, . 25, . 2, . 2, . 25, . 2} \end{matrix}

This is illustrated using:

Figure 6. The black arrows are the “pathways" whose lengths aren’t outliers. The length of the red arrows in the pathway are outliers.

Hence, when

x_{0} = (- . 9, 1)

, using §5.4.1 step (3)(b) & eq. 173, we note:

L ((- . 9, 1), S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1)) = {. 25, . 25, . 2, . 2, . 25, . 2}

(176)

8.3.3. Step (3)(c)

To convert the set of distances in eq. 176 into a probability distribution, we take:

\sum_{x \in {. 25, . 25, . 2, . 2, . 25, . 2}} x = . 25 + . 25 + . 2 + . 2 + . 25 + . 2 = 1.35

(177)

Then divide each element in

{. 25, . 25, . 2, . 2, . 25, . 2}

by 1.35

{. 25 / (1.35), . 25 / (1.35), . 2 / (1.35), . 2 / (1.35), . 25 / (1.35), . 2 / (1.35)}

which gives us the probability distribution:

{5 / 27, 5 / 27, 4 / 27, 4 / 27, 5 / 27, 4 / 27}

Hence,

P (L ((- . 9, 1), S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1))) = {5 / 27, 5 / 27, 4 / 27, 4 / 27, 5 / 27, 4 / 27}

(178)

8.3.4. Step (3)(d)

Take the shannon entropy of eq. 178:

\begin{matrix} E (P (L ((- . 9, 1), S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1)))) = \\ \sum_{x \in P (L ((- . 9, 1), S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1)))} - x {log}_{2} x = \sum_{x \in {5 / 27, 5 / 27, 4 / 27, 4 / 27, 5 / 27, 4 / 27}} - x {log}_{2} x = \\ - (5 / 27) {log}_{2} (5 / 27) - (5 / 27) {log}_{2} (5 / 27) - (4 / 27) {log}_{2} (4 / 27) - (4 / 27) {log}_{2} (4 / 27) - (5 / 27) {log}_{2} (5 / 27) - (4 / 27) {log}_{2} (5 / 27) = \\ - (15 / 27) {log}_{2} (5 / 27) - (12 / 27) {log}_{2} (4 / 27) \approx 2.57604 \end{matrix}

We shorten

E (P (L ((- . 9, 1), S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1))))

E (L ((- . 9, 1), S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1)))

, giving us:

E (L ((- . 9, 1), S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1))) \approx 2.57604

(179)

8.3.5. Step (3)(e)

Take the entropy, w.r.t all pathways, of the sample:

{(- . 9, 1), (- . 65, 1), (- . 4, 1), (- . 2, 1), (. 1, - 1), (. 3, - 1), (. 55, . 5), (. 75, . 5), (1, . 5)}

(180)

In other words, we’ll compute:

E (L (S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1))) = sup_{x_{0} \in S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1)} E (L (x_{0}, S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1)))

We do this by repeating §8.3.1-§Appendix 8.3.4 for different

x_{0} \in S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1)))

(i.e., in the equation with multiple values, see note 13)

\begin{matrix} E (L ((- . 9, 1), S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1))) \approx 2.57604 \end{matrix}

(181)

\begin{matrix} E (L ((- . 65, 1), S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1))) \approx 2.3131, 2.377604 \end{matrix}

(182)

\begin{matrix} E (L ((- . 4, 1), S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1))) \approx 2.3131 \end{matrix}

(183)

\begin{matrix} E (L ((- . 2, - 1), S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1))) \approx 2.57604 \end{matrix}

(184)

\begin{matrix} E (L ((- . 1, - 1), S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1))) \approx 1.86094 \end{matrix}

(185)

\begin{matrix} E (L ((- . 3, - 1), S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1))) \approx 1.85289 \end{matrix}

(186)

\begin{matrix} E (L ((. 55, . 5), S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1))) \approx 2.08327 \end{matrix}

(187)

\begin{matrix} E (L ((. 75, . 5), S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1))) \approx 2.31185 \end{matrix}

(188)

\begin{matrix} E (L ((1, . 5), S (C (\sqrt{2} / 6, G_{1}^{★}, 1), 1))) \approx 2.2622 \end{matrix}

(189)

Hence, since the largest value out of eq. 181-189 is

2.57604

E (L (S (C (13 / 6, G_{1}^{★}, 1), 1))) = sup_{x_{0} \in S (C (ε, G_{1}^{★}, 3), 1)} E (L (x_{0}, S (C (ε, G_{1}^{★}, 1), 1))) \approx 2.57604

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Finding a Published Research Paper which Meaningfully Averages the Most Pathalogical Functions (V2)

Abstract

1. Intro

1.1. First special case of f

1.1.1. Potential Answer

1.2. Second Special Case of f

1.3. Attempting to Analyze/Average f

1.3.1. Example Proving § 1.3 (1)-(4) Correct

1.3.2. Blockquote

2. Extending the Expected Value w.r.t the Hausdorff Measure

3. Attempt to Define “Unique and Satisfying" in The Blockquote of §1.3

3.1. Note

3.2. Leading Question

3.2.1. Explaining Motivation Behind § 3.2

4. Question Regarding My Work

5. Clarifying §3

5.1. Defining Sequences of Bounded Functions Converging to f

5.2. Expected Value of Bounded Sequence of Functions

5.2.1. Example

5.3. Defining Equivelant and Non-Equivelant Sequences of Bounded Functions

5.3.1. Explanation

5.3.2. Example of Equivalent Sequences of Bounded Functions

5.3.3. Explanation

5.3.4. Example of Non-Equivalent Bounded Sequences of Functions

5.3.5. Shortcuts to Determining Equivelant and Non-equivelant Sequences of Bounded Functions

5.4. Defining the “Measure"

5.4.1. Preliminaries

5.4.2. What Am I Measuring?

5.4.3. Example of The “Measure" of ( G r ★ ) Increasing at Rate Super-linear to that of ( G j ★ ★ )

5.4.4. Example of The “Measure" from ( G r ★ ) r ∈ N Increasing at a Rate Sub-Linear to that of ( G j ★ ★ ) j ∈ N

5.4.5. Example of The “Measure" from ( G r ★ ) r ∈ N Increasing at a Rate Linear to that of ( G j ★ ★ ) j ∈ N

5.5. Defining The Actual Rate of Expansion of Sequence of Bounded Sets

5.5.1. Definition of Actual Rate of Expansion of Sequence of Bounded Sets

5.5.2. Example

5.6. Reminder

6. My Attempt At Answering The Blockquote of §1.3.2

6.1. Choice Function

6.2. Approach

6.3. Potential Answer

6.3.1. Preliminaries (Definition of T in Case of §3.2.1 ((5)))

6.3.2. Question

6.4. Explaining The Choice Function and Evidence The Choice Function Is Credible

6.4.1. Evidence with Programming

7. Questions

8. Appendix of §5.4.1

8.1. Example of §5.4.1, Step (1)

8.2. Example of §5.4.1, Step (2)

8.3. Example of §5.4.1, Step (3)

8.3.1. Step (3)(a)

8.3.2. Step (3)(b)

8.3.3. Step (3)(c)

8.3.4. Step (3)(d)

8.3.5. Step (3)(e)

References

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1.3.1. Example Proving $§$ 1.3 (1)-(4) Correct

3.2.1. Explaining Motivation Behind $§$ 3.2

5.4.3. Example of The “Measure" of $(G_{r}^{★})$ Increasing at Rate Super-linear to that of $(G_{j}^{★ ★})$

5.4.4. Example of The “Measure" from ${(G_{r}^{★})}_{r \in N}$ Increasing at a Rate Sub-Linear to that of ${(G_{j}^{★ ★})}_{j \in N}$

5.4.5. Example of The “Measure" from ${(G_{r}^{★})}_{r \in N}$ Increasing at a Rate Linear to that of ${(G_{j}^{★ ★})}_{j \in N}$