Subset as arithmetic mean of geometric means. Not really?

Question

Let $\,M \le N\,$ and $X$ be a finite set of $N$ reals $x_i\,$: $X = \{x_1,x_2, .. ,x_i, .. ,x_N\}\,$, $Y$ be a finite set of $M$ reals $y_j\,$: $Y = \{y_1,y_2, .. ,y_j, .. ,y_M\}\,$. In order to determine whether $X$ is a subset of $Y$, we could compare the elements: $X$ is a subset of $Y$, $\,X \subseteq Y\,$, if and only if for each of the $\,x_i\,$ one of them is equal to $\,y_j\,$ : $$ \quad ((x_1 = y_1) \vee (x_1 = y_2) \vee \cdots \vee (x_1 = y_j) \vee \cdots \vee (x_1 = y_M)) \\ \wedge ((x_2 = y_1) \vee (x_2 = y_2) \vee \cdots \vee (x_2 = y_j) \vee \cdots \vee (x_2 = y_M)) \\ \wedge \cdots \cdots \wedge \quad \\ \quad ((x_i = y_1) \vee (x_i = y_2) \vee \cdots \vee (x_i = y_j) \vee \cdots \vee (x_i = y_M)) \\ \wedge \cdots \cdots \wedge \quad \\ \quad ((x_N = y_1) \vee (x_N = y_2) \vee \cdots \vee (x_N = y_j) \vee \cdots \vee (x_N = y_M)) $$ $$ (X \subseteq Y) \Longleftrightarrow \bigwedge_{i=1}^N \left[ \bigvee_{j=1}^M \left(x_i=y_j\right) \right] $$ Because all $\,x_i\,$ and $\,y_j\,$ are real numbers, we can alsow write this as follows; a product of terms is zero, if and only if one (or more) of the factors is zero. $$ (X \subseteq Y) \Longleftrightarrow \bigwedge_{i=1}^N \left[ \prod_{j=1}^M \left(x_i-y_j\right) = 0\right] $$ And each of the terms is zero if and only if the sum of the squares of all of these terms is zero: $$ (X \subseteq Y) \Longleftrightarrow \sum_{i=1}^N \left[ \prod_{j=1}^M \left(x_i-y_j\right) \right]^2 = 0 $$ One can make this somewhat more computation friendly by implementing products as geometric means: take the $M$-th root of each of the terms. For the same reason, we shall implement the sum as an arithmetic mean. Furthermore the above can easily be generalized to vectors $(\vec{x}_i,\vec{y}_j)$. The we have at last, for the discrete case: $$ (X \subseteq Y) \Longleftrightarrow \sum_{i=1}^N \frac{1}{N} \left[ \prod_{j=1}^M \left|\vec{x}_i-\vec{y}_j\right|^2 \right]^{1/M} = 0 $$ In Wikipedia a definition is found of the Geometric Mean. And there is a relationship with the arithmetic mean of logarithms at that page: $$ {\displaystyle \left(\prod _{i=1}^{n}a_{i}\right)^{\frac {1}{n}}=\exp \left[{\frac {1}{n}}\sum _{i=1}^{n}\ln a_{i}\right];} \quad a_i > 0 $$ Together with the above then we have: $$ (X \subseteq Y) \Longleftrightarrow \sum_{i=1}^N \frac{1}{N} \exp\left[\frac{1}{M}\sum_{j=1}^M \ln(\left|\vec{x}_i-\vec{y}_j\right|^2)\right] = 0 $$ The following essential reading is needed, in order to be able to proceed.

Product integral

Namely for converting the discrete into the continuous: $$ (X \subseteq Y) \Longleftrightarrow \int_0^1 \exp\left(\int_0^1 \ln(\left|\vec{x}(t)-\vec{y}(u)\right|^2)\,dt\right)\,du = 0 $$ where a minor detail is restriction to the integration interval $[0,1]$ by parameter transformation.
So far so good. Let's try now the simplest example possible, namely $\;\vec{x}(t)=t \; ; \; 0 \le t\le 1\;$ and $\;\vec{y}(u)=u \; ; \; 0 \le u\le 1\;$.
Then it is clear that $(X=Y$ and so $(X \subseteq Y)$ and $(Y \subseteq X)$. And our calculation effort is defined by proving that: $$ \int_0^1 \exp\left(\int_0^1 \ln\left(|t-u\right|^2)\,dt\right)\,du = \int_0^1 \exp\left(\int_0^1 \ln(\left|t-u\right|^2)\,du\right)\,dt = 0 \\ = \exp(-2)\int_0^1 u^{2u}(1-u)^{2(1-u)}\,du \approx 0.05378539284 $$ This is not by far the accuracy we did expect! I think it has something to do with evaluating the logarithm for $t=u$ but I'm not sure. What precisely is going on?

Answer 1

In the discrete case we have $\sum_{i=1}^n\log((x_i-y)^2)=-\infty$ whenever $y\in\{x_1,\dots,x_n\}$.

But in the continuous case, we do not have $\int_a^b\log((t-y)^2)\,\mathrm dt=-\infty$ when $y\in[a,b]$. The log function goes to $\pm\infty$ too "slowly" for the integral to diverge.

Answer 2

In Wikipedia a definition is found of the Geometric Mean. And there is a relationship with the arithmetic mean of logarithms at that page: $$ {\displaystyle \left(\prod _{i=1}^{n}a_{i}\right)^{\frac {1}{n}}=\exp \left[{\frac {1}{n}}\sum _{i=1}^{n}\ln a_{i}\right];} \quad a_i > 0 $$ Together with the above then we have: $$ (X \subseteq Y) \Longleftrightarrow \sum_{i=1}^N \frac{1}{N} \exp\left[\frac{1}{M}\sum_{j=1}^M \ln\left|\vec{x}_i-\vec{y}_j\right|^2\right] = 0 $$

No, the conclusion is not true. The condition $a_i>0$ is not met if you have $a_i$ equal to the sum of differences of squares. So the equality does not hold.

In particular, if $X\subseteq Y$, then the expression $$\sum_{i=1}^N\frac{1}{N}\left[\Pi_{j=1}^M\left|\vec x_i - \vec x_j\right|^2\right]^{\frac1M}$$ is equal to $0$, while the expression

$$ \sum_{i=1}^N \frac{1}{N} \exp\left[\frac{1}{M}\sum_{j=1}^M \ln\left|\vec{x}_i-\vec{y}_j\right|^2\right]$$

is undefined.

Answer 3

Too long for a comment, in view of the (+1) answers by user937523 and 5xum.
While preparing the question, these were some of my notes: $$ \int_0^1\ln(|y-x|^2)\,dx = 2\lim_{\delta\downarrow 0}\left[\int_0^{y-\delta} \ln(y-x)\,dx + \int_{y+\delta}^1 \ln(x-y)\,dx\right] = \\ 2\lim_{\delta\downarrow 0} \left[-(y-x)\ln(y-x)+(y-x)\right]_{y-x=y}^{y-x=\delta} + \\ 2\lim_{\delta\downarrow 0} \left[(x-y)\ln(x-y)-(x-y)\right]_{x-y=\delta}^{x-y=1-y} = \\ 4\lim_{\delta\downarrow 0}\left[\,-\delta\ln(\delta)+\delta\,\right] + 2y\ln(y)-2y + 2(1-y)\ln(1-y)-2(1-y) $$ And the limit between square brackets is zero, just because, indeed, the The log function goes to $\pm \infty$ too "slowly" for the integral to diverge. Furthermore, because of the condition $a_i > 0$, in the continuous case, we would have to consider instead the Cauchy principal value for the singularity at zero in the product integral / logarithm: $$ \int_0^1 \exp\left(-\!\!\!\!\!\!\int_0^1 \ln(\left|\vec{x}(t)-\vec{y}(u)\right|^2)\,dt\right)\,du $$ But the meaning of the latter is simply that we cannot calculate for being a subset in this way, because zero is essential as the outcome and can never be obtained.

CGM of a straight line segment

It has been established that we cannot calculate for being a subset, so the integral representing the arithmetic mean can safely be discarded. Then we are left with the Cauchy principal value of a Continuous Geometric Mean: $$ \operatorname{CGM}(\vec{r}) = \exp\left(-\!\!\!\!\!\!\int_0^1 \ln(\left|\vec{p}(t)-\vec{r}\right|^2)\,dt\right) $$ where for example $\,\vec{p}(t)\,$ is a curve in the plane and $\,\vec{r}\,$ is a point in the plane. For our straight line segment, we substitute: $$ \vec{p}(t) = (t,0) \quad ; \quad \vec{r} = (x,y) \\ \operatorname{CGM}(x,y) = \exp\left(-\!\!\!\!\!\!\int_0^1 \ln\left|(t-x)^2+y^2\right| dt \right) $$ A computer algebra system (MAPLE) has been employed for the purpose:

> int(ln((t-x)^2+y^2),t=0..1,continuous);

Giving that $\,\operatorname{CGM}(x,y) = e^{\Pi(x,y)}$ , with: $$ \Pi(x,y) = (1-x)\ln(1-2x+x^2+y^2)+x\ln(x^2+y^2)-2 \\ + 2y\left[\arctan\left(\frac{1-x}{y}\right)+\arctan\left(\frac{x}{y}\right)\right] $$ The result has been programmed and when visualized it looks as follows:

The lowest function values ($\approx 0.034$) are nearby the red line segment.
Window: xmin := -0.25; xmax := +1.25; ymin := -0.75; ymax := +0.75;