Introduction
I was wondering if, you define some interval $[n,m]$ which contains all numbers between $n$ and $m$ (consider either $\mathbb Q$ or $\mathbb R$), what would be the product of all those numbers?
One of my instant guesses was that'll tend to infinity.
Clearly, it would be neat to assume $a\in[n,m], a\ne 0$ since the product will be $0$ otherwise.
In addition, $n,m\ge0$ since multiplying infinitely many negative numbers is undefined.
Q: Is it correct to claim that interval type $[a\gt0,b\gt a]$ in its
product of all of its infinitely many real numbers, tends to $0$ no
matter how big $b$ is, since you can include numbers infinitely close to $0$?A: It seems it can either tend to $0$, $\infty$ or a finite value $x$, which is analyzed below.
Interval of type $[a\gt1,b\gt a]$, would tend to infinity in all cases since we never decrease in value no matter what number we pick to multiply next.
Symmetrical expression to work multiply the numbers
Imagine taking a interval $[n,m]$, and taking the numbers for each $k$ iteration, as averages of previous numbers. Like this:
$$n,m$$
$$n,\frac{n+m}{2},m$$
$$n,\frac{3n+m}{4},\frac{n+m}{2},\frac{n+3m}{4},m$$
$$\dots$$
Which would for $k$ iterations as $k$ tends towards $\infty$, eventually contain all numbers in that interval.
The product of all those numbers for $k$ iterations can be written as:
$$ \prod ^{2^k} _{x=0} \frac{(2^k – x)n+xm}{2^k} $$
If we now take an interval $[n,m]$ as $[2,3]$ for example, we have:
For $k=4$ it is: $5.11501\dots \times 10^6$
For $k=8$ it is: $3.24661\dots \times 10^{101}$
For $k=10$ it is: $7.56107\dots \times 10^{404}$
And we can say it clearly tends towards $\infty$ itself.
And if we pick an interval containing numbers less than 1, we can see it can tend towards $0$;
For example: $[0.1,2],k=14$ then: $1.3926\dots \times 10^{-1062}$
But If we have a greater number, we can still tend towards infinity;
For example: $[0.1,3],k=14$ then: $8.6083\dots \times 10^{1535}$
But then If we choose $[0.1,2.3639556]$ then for $k=14$ we will get:
$1.000756\dots$
Seems it tends to a bit over $1$?Q: Does that mean we can by choosing the right ratio, or the right
values in the interval, make it tend towards any value?A: But then If we increase the $k$, we start tending to either $\infty$ or $0$ very rapidly; for $k=16$ I had to decrease the $m$ a
bit to $2.3638689$ for it to be roughly $1.0076\dots$
This brings up the question;
Can we actually make the product tend to a finite value for the case
where $k$ approaches $\infty$ by choosing the right ratio of $n$ and
$m$ in which $(n \lt 1)$ , $(m\gt 1)$?
Calculating values for finite cases?
I've now tried to find the $m$ for $[0.1,m]$ so that the product of the interval ($x$) :
For $k=2$, $m\approx 3.10745$ for it to be $x\approx 1.00001$
For $k=10$, $m= 2.36569706$ for it to be $x= 1$
For $k=14$, $m\approx 2.363955478620$ for it to be $x\approx 1.00000001474698$
For $k=15$, $m\approx 2.363897553300$ for it to be $x\approx 1.00000169779107$
For $k=16$, $m\approx 2.363868593513$ for it to be $x\approx 1.00000000497575$
$$\dots$$
As you can see, for bigger $k$ we get more precision on the digits of $m$
This brings up another question;
How would one compute $m$ with most precision for choosen $n$,
$(0<n<1)$ where $k$ tends to $\infty$ so that the product of all reals
in $[n,m]$ is a finite value $x\in \mathbb R$?
Do you have any software, any code for a computing algorithm, or a better formula than my symmetrical expression in mind? To preform the calculation where $k$ tends to $\infty$?
Summing Everything up
When calculating the infinite product of all real numbers in the interval $[n,m]$, $(n\lt m)$, We have a few cases we can look at individually:
$1.$ When $n\lt0$ the result is "undefined" since we have $\infty$ many negative numbers of which we can extract $(-1)$, which can then be written as $\lim\limits_{n\to\infty}(-1)^n$ which is undefined.
$2.$ When $a\in[n,m], a=0$ The result is clearly $0$
$3.$ When $0<n<m<1$ then the result tends to $0$
$4.$ When $n>1,m>n$ the result tends to $\infty$
$5.$ When $0<n<1$ and $m>1$ The result can either tend to $0$, $\infty$ or a finite value $x$, which can be calculated by using a symmetric expression I used in my approach.
Better Calculations
I use WolframAlpha while increasing $k$ but it runs up to $20$ since bigger $k$ exceed its computational time allowed.
When applying calculations proposed by String in his answer (using Newton-Raphson method), and feeding them to WolframAlpha, the most precision we get is (for $x=1$):
$n=\frac{1}{2}$, $m\approx 1.6030164899169670747912206652529016572070546450201637$
$n=\frac{1}{3}$, $m\approx 1.8699324270643973162008471123760292568887231725646945$
$n=\frac{1}{4}$, $m\approx 2.0245006935913233776633741232407065465057011574013821$
$n=\frac{1}{5}$, $m\approx 2.1267864647345386651244066521323021196320323179442441$
$$\dots$$
The $m$ seems to tend to $e$ as $n\to\infty$.
I might get back to this if I get a chance, but I doubt it'll be soon enough.
Best Answer
Definition of the interval
One possible definition given a strictly positive interval $[n,m]\subseteq\mathbb R^+$ could be: $$ \prod_{x\in [n,m]}x:=\exp\left(\int_n^m\ln(x)\ dx\right) =\frac{m^m\cdot n^{-n}} {\operatorname e^{m-n}} $$ for an interval $[n,m]$ containing uncountably many elements. The countable and finite versions could then read $$ \prod_{k=1}^{\infty} x_k:=\exp\left(\sum_{k=1}^{\infty}\ln(x_k)\right) \quad\text{and}\quad \prod_{k=1}^n x_k:=\exp\left(\sum_{k=1}^n\ln(x_k)\right)=x_1\cdot x_2\cdot ...\cdot x_n $$
We could even extend this definition to $$ \prod_{x\in [n,m]}f(x):=\exp\left(\int_n^m\ln(f(x))\ dx\right) $$ One nice property is that with this defition we have $$ \prod_{x\in[n,m]} x^a=\left(\prod_{x\in[n,m]} x\right)^a $$ so it appears to follow some nice rules of powers of conventional finite products.
To define it without direct use of integration, my definition should be equivalent to the defining: $$ S_k:=\prod_{x=0}^{2^k}\left(\frac{(2^k-x)n+xm}{2^k}\right)^{(m-n)/2^k} $$ and recognize the product $\prod_{x\in [n,m]}x$ as the limit of those $S_k$'s as $k$ tends to infinity. So it is like multiplying together $2^k+1$ evenly spread out factors over the interval $[n,m]$, but adjusting the exponent of each factor to match the distance between the factors, namely $(m-n)/2^k$. These exponents tend to zero as the number of factors tends to infinity. It can then be shown that $S_k\to m^m\cdot n^{-n}/\operatorname e^{m-n}$.
So how does this relate to your suggested symmetrical expression? Well, you are considering the sequence of the form: $$ T_k:=(S_k)^{2^k/(m-n)}=\prod_{x=0}^{2^k}\frac{(2^k-x)n+xm}{2^k} $$ Now clearly, if $S_k\to a>1$ we will have $T_k\to\infty$ whereas for $S_k\to a\in[0,1)$ we must have $T_k\to0$. So the difficult cases are $S_k\to 1$ or $S_k\to a<0$. The latter I doubt we can make any sense of.
Resolving when $S_k$ tends to $1$
If $S_k$ tends to $1$ for some $n\in(0,1)$ and an appropriate matching $m>1$ we then know that $$ \int_n^m\ln(x)\ dx=\lim_{k\to\infty}\ln(S_k)=\ln(1)=0 $$ Considering the graph of $\ln(x)$ it can be shown that $$ \left[\ln m -q_k\cdot(\ln m -\ln n)\right]\cdot\Delta x\leq\ln(S_k)\leq\left[\ln n +(1-q_k)\cdot(\ln m -\ln n)\right]\cdot\Delta x $$ where $0.5\leq q_k\leq 1$ is a sequence tending to $0.5$, and $\Delta x$ is short hand for the distance $(m-n)/2^k$. Since $T_k$ is equal to $S_k$ except for raising to the reciprocal of $\Delta x$ we get $\ln(T_k)=\ln(S_k)/\Delta x$ and therefore $$ \left[\ln m-q_k\cdot(\ln m -\ln n)\right]\leq\ln(T_k)\leq\left[\ln n +(1-q_k)\cdot(\ln m -\ln n)\right] $$ As $k$ tends to infinity both bounds tend to $(\ln m+\ln n )/2$ showing us that $$ \lim_{k\to\infty} T_k=\exp\left(\frac{\ln m+\ln n}2\right)=\sqrt{n\cdot m} $$
Now we should able to do the following:
Resolving the computation method
Given $n\in(0,1)$ one can solve $$ \int_n^m\ln(x)\ dx=m(\ln m-1)-n(\ln n-1)=0 $$ for $m>1$ in order to find the corresponding $m$ so that $T_k$ converges. One way to do this is to use the Newton-Raphson method on the function $$ f(x)=x(\ln x-1)-n(\ln n-1) $$ with initial guess $x_0=2$. Then $m=\lim_{k\to\infty}x_k$ where the $x_k$'s are defined recursively as $$ x_{k+1}:=x_k-\frac{x_k(\ln x_k-1)-n(\ln n-1)}{\ln x_k} $$ It turns out that in fact $1<m<\operatorname e\approx 2.7182818$. For $n=0.5$ it takes only a few iterations before one has $$ m\approx x_5=1.603016489916967074791... $$ and it can be verified that the digits listed above do not change for future iterations so we already have $m$ to a very high precision. So we have $$ \lim_{k\to\infty}\prod ^{2^k} _{x=0} \frac{(2^k - x)0.5+x\cdot 1.603016489916967074791...}{2^k}\\=\sqrt{0.5\cdot 1.603016489916967074791...}\approx 0.8952699285458456285 $$ But this is a very unstable result anyway! My earlier computations showed that if $m$ is either the slightest bit larger than the actual solution to $f(x)=0$ the product tends to infinity, or if it is the slightest bit less then it tends to zero.
Some general remarks to conclude
The infinite symmetrical product you defined is very unstable in more than one respect. If we change the definition even slightly we may get an entirely different result: $$ \prod ^{2^k} _{x=1} \frac{(2^k - x)n+xm}{2^k} $$ removes the first factor $n$ whereas $$ \prod ^{2^k-1} _{x=0} \frac{(2^k - x)n+xm}{2^k} $$ removes the last factor $m$. Whereas these would both lead to the same values of $S_k$ removing only a negligible contribution at that level, they affect the value of $T_k$ by a non-negligible factor. Also if we distributed the factors in the interval $[n,m]$ slightly different the product $T_k$ could change a lot whereas $S_k$ would not. So overall $S_k$ is a much more stable value. And $S_k$ tends to represent an actually uncountable product, whereas $T_k$ tends to something I would characterize as product of a countable subset of the factors in question. This might be another reason it is so unstable - there are infinitely many other ways we could have defined $T_k$ that would yield totally different results whereas all definitions of $S_k$ by partitioning $[n,m]$ into subintervals that decrease toward zero width as $k$ tends to infinity would all point to the same limit value of $S_k$. This in a sense addresses your question 1 as an infinite product of the kind you are suggesting would or would not tend to zero depending on the distribution you use to select factors from $[n,m]$.