Let me muse about infinite amounts so that I am not waiting to spill them on a random infinite-adjacent thread.
Someone offers you a choice of 10 fish now or 1 fish a day. Which option lets you eat more fish?
Questions of these kind seem to have properties of:
Even a ludicrous amount of one-off fish will lose to a steady income of fish
Bigger fish streams are bigger. Having 3 fish per day is better than 1 fish per day
Now one could make the case that there is a scheme where each day you grab the new fish and give them a number. For any finite fish per day this can clearly be done. Somebody could come to say that “Hey I can put any fishstream in 1-to-1 correspondence to the natural numbers therefore there is an equal amount of fish in any stream”. Yet it seems if somebody says that their 3 fish per day stream is not an advantage over my 1 fish per day stream, I would not be convinced and think they still have got more.
This makes me think that 1-to-1 correspondence doesn’t really fit with what my brain points to be size. Or some of the intuitive notions come appart when we enter infinity land.
There is also the construction if somebody promises on each day start a new stream of fish. Then for any stream of fixed finite amount of fish per day promised there is a day where the “growing promise” gets you that and then some more afterwards.
I am guessing/feeling that 2 streams of 1 fish vs 1 stream of 2 fish is connected/represented by ω+ω=2ω and the “growing promise” might be connected to ω*ω
Is there/could there be some notion of size that would let me say that ω+ω is smaller than ω*ω and could I deal with people saying that “mappable to integers” is better notion of size?
You can just pretend that ω is finite and plug it into the formula for the partial sum.n∑i=1i=12n2+12n, so ω∑i=1i=12ω2+12ω. If they were to give the ith odd number amount of fish on the ith day (1,3,5,7,9...), then you would have ω2 amount of fish, because n∑i=12i−1=n2. The two links I posted about the handling of infinite divergent series go into greater detail (eg. the question of the starting index).
The links are very on point for my interest thanks for those. Some of it is in rather dense math but alas that is the case when the topic is math.
At one point there is a constuction where in addition to having series of real numbers to define a hyperreal (r1,r2,r3...)=h1 we define a series of hyperreals (h1,h2,h3...)=d1, in order to get a “second tier hyperreal”. So I do wonder whether the “fish gotten per day” is adeqate to distinguish between the scenarios. That is there might be a difference between “each day I get promised an infinite amout of fish” and “each day I get 1 more fish”. That is on day n I have been promised ωn fish and taking it as α∑i=1I am not sure whether α=ω and whether terms like ω2 and ωα refer to the same thing or whether mixing “first-level” and “second level” hyperreals gets you a thing different than mixing just “level 1”s
If you just graph fish vs. time, then the one-time gift is a constant function, the steady income is linear, and the “growing promise” stream-of-streams is a quadratic. The fact that a steady income will eventually surpass any one-time gift is because any positive-slope linear function will eventually exceed any constant-value function. Likewise, any 2nd-order polynomial with a positive x^2 term will eventually exceed any linear function. You could keep going with higher order polynomials if you want.
A similar analogy would be a race where even a large head start will eventually be surpassed by a slightly faster car.
I guess I know that kind of perspective exists but I don’t want to refer to the “up to a time” thing but the total.
For example taking the even and odd integers, that would seem like taking a stream of fish and putting them into alternate buckets every other day. I would like to be able to say something like “even and odd integers are of equal amount and they are half of that of all integers”. But I am unsure whether that is accurate and where the foundations would crumble.
if I do it in a very naive way in graph vs time then I could conclude that the bucket that I start with has a stricly dominating amount of fish (because the graph stay above or matches the other graph). And it feels odd to say that there are more even numbers than odd ones! Althought I guess it wouldn’t be odd that 0+{1,2,3,4,5...| x is even} is more than {1,2,3,4,5...| x is even}. Or is it because the two-bucket arrangement is not the natural choice of odds vs evens? I guess if I received two fish a day and put them on different buckets things could change?
Oh I guess I notice that if I receive two fish a day that seems more than if I receive 2 fish every other day. That seems to map to 2*ω > 2*(ω/2) = ω.
Every other day, the number is the same. On the days in between, there’s a difference. Even taking that notion of ‘dominating’ into account, the difference is finite, and small: 1. If you had a choice between:
a) getting 1⁄3 a fish on day one, then 1⁄9 a fish on day two and 1/27th a fish on day three...(1/3^n on day n)
and
b) getting 1 fish on day one, then 1⁄2 a fish on day two, and 1⁄4 a fish on day three...(1/2^(n-1) on day n)
Then b is better than a...by more than the difference between 0+{1,2,3,4,5...| x is even} and {1,2,3,4,5...| x is even}. But it also seems like a bigger deal in those cases, a and b, because getting what comes out to a finite amount of fish over an infinite period of time isn’t as good as getting an infinite amount.
And it feels odd to say that there are more even numbers than odd ones!
Flip a coin. If it comes up heads put the first fish in the red bucket, and the second fish in the green bucket.
If it comes up tails do the reverse. (With the integers as a whole, where do you start counting from? The answer to that, determines which measure “seems” greater, on odd time/count steps.)
I would like to be able to say something like “even and odd integers are of equal amount and they are half of that of all integers”. But I am unsure whether that is accurate and where the foundations would crumble.
You can color the positive integers, 1, 2, 3, 4,… using this pattern: red, green, red, green,...
You are obviously partitioning them into two collections (sets, because there’s not repetition).
Are there more odd numbers than even? Well if you count how many odd numbers (which are positive integers) are less than or equal to n, and call this function o, o(n) > e(n) for odd n. But for even n, they’re equal. So in the limit as n increases without bound: f(n) = o(n)/(o(n)+e(n) = 1⁄2.* But if you take steps of 2, starting at an even part, like 0, then the two sets are always of equal size. This seems to formalize the intuition well. Using this stream approach, one can articulate that: a randomly chosen integer is as likely to be even as it is to be odd**, and that this doesn’t change over time (as one iterates further through the integers), but the as you iterate out the integers, the frequency with which you see prime numbers goes down.
**This can be generalized further: a randomly chosen integer, for a given radix r, will have its final digit d, drawn from the a distribution indistinguishable from a uniform distribution over [0, r-1].
(Every instance of the word random, I made above is based on a uniform distribution across the integers. If you have a fair coin, and you flip it, until it comes up heads, then the probability that the number of times you flipped it is an odd number is not 1⁄2. The distribution there is: 1⁄2 for 0, 1⁄4 for 1, etc. The probability a red number (even) is chosen is twice that of a green number (odd) being chosen. Using ‘the sum of probabilities is 1’, it can be deduced that the probability the number is odd is 1⁄3, and the probability it is even is 2⁄3.)
*
o(n) + e(n) = n
for even n, o(n) = e(n). f(n) = k/2k = 1⁄2. (k is n/2, and an integer.)
for odd n, o(n) = e(n) + 1. f(n) = (k+1)/(2k) (n is 2k+1, and k is an integer) The limit as n, and thus k goes to infinity for this case is:
1⁄2 + 1/n = 1⁄2. (The limit from above. The other gets infinitely small.)
Every instance of the word random, I made above is based on a uniform distribution across the integers
Strictly speaking, there is no such thing as uniform distribution across the integers, or any other set that can be matched up 1:1 with integers. There are things that aren’t probability distributions that do have some similar properties, such as “asymptotic density”, but they fail to have the properties we require of probabilities. The “stream” idea seems to be another way to describe asymptotic density.
For example we might like to say that there are as many natural numbers of even length in decimal notation as those of odd length, but the stream idea doesn’t support that. The fraction of even-length numbers oscillates between 11% and 89% infinitely often, and there are other properties that behave even more weirdly.
One paradox that illustrates why there is no such thing as a uniform distribution on natural numbers:
Suppose we both pick natural numbers uniformly at random, independently of one another. You reveal your number. What is the probability that my number is greater than yours? No matter what number you chose, there are infinitely many larger numbers and only finitely many numbers that are not larger. Since my number was chosen uniformly at random, my number is larger with probability 1.
1 number of length 0, 9 numbers of length 1 (and maybe 0), 90 numbers of length 2, 900 numbers of length 3, 9000 numbers of length 4
9*10^(n-1) numbers of length n. For each n the amount of numbers of length just before that is 10 times less and the amount of numbers the next length is 10 times more. If you take a rolling fraction of n odd to all numbers seen it starts to go down when even numbered length is reached and starts to go up when an odd length number is reached.
(Ignoring that most people don’t think of 0 as of being length 0.)
Jump by two orders of magnitude every time and it stays stable:
Starting with nothing:
1 of even length, 9 of odd length.
90 of even length, 900 of odd length.
10% versus 90%.
Start after an even jump:
91 of even length, 9 of odd length.
9,091 of even length, 909 of odd length.
(Starts at 91% even, but drops after a double jump. I don’t know what the limit on this is.)
By comparison, resolving proportion of even numbers versus odd numbers, is much easier, because it’s a simple pattern which oscillates at the same rate, instead of changing.
9*10^(n-1) numbers of length n.
(in base 10)
Well, if a different color is used every time, then the coloring aspect is solved. If you ask about addition though, then things get weird.
Right, that was a math typo. It really oscillates between 9% and 91%.
For example, 909090 of the numbers below a million have even length, i.e. 91%. As you increase the bound toward ten million, this fraction decreases until it hits a minimum of 9%, and then starts increasing again until you reach a hundred million, and so on.
Well, the old solution to what is the limit of: +1, −1, +1, −1, etc. was: (index starts at one, pattern is (-1)^(n+1))
Consider the cases:
a) +1, odd index
b) −1, even index
Average them.
0.
If that was applied directly, it’d be: (9+91)/2% = 50%.
You could argue that it should be broken down differently, because there’s different proportions here though.
You could also declare the answer undefined, and say infinity is about growth, it doesn’t have a value, for x % 2 (or odd or even number as the case may be), and averages are ridiculous. (And once you have a breakdown of cases, and probability what more is there?)
That is one of the puzzle in that 0+0+0+0+0… converges and has a value but +1-1+1-1+1-1… which seems to be like (1-1)+(1-1)+(1-1)+(1-1)… diverges (and the series with and without the paranthesis are not equivalent)
The strram idea gives it a bit more wiggleroom. Getting 1,0,1,0,1.. fish seems equivalent to getting 1⁄2 fish a day but 1,1,1,1,1.. seems twice the fish of 1,0,1,0,1,0,1,0… So which with the other methods are “can’t say anthing” there is maybe hope to capture more cases with this kind of approach.
Too bad its not super formal and I can’t even pinpoint where the painpoints for formalization would be.
That paradox is good in that it cuts to the matter very cleanly.
To my mind it “numbers larger than my number” and “uniform integer” don’t need to be same. There are n smaller numbers and ω-n bigger numbers. (ω-n)/ω is going to be near 1 (infinidesimally so) but not quite up to 1. Maybe crucially (ω-2n)/ω is smaller than (ω-n)/ω ie if I hit a high number my chances are better. I get that standard approach somehow gets into the way of this and I would like to know which axiom I have a bone to pick with.
There is a (from my perspective a problem) that events of 0 probablity can happen and events with probability 1 can fail to happen. The associated verbal language is “almost surely” and “almost never”. Showing me that a thing can almost surely happen doesn’t guarantee it. To my mind this is because some zeroes have rounding to the nearest real and some don’t.
For positive values, there will be some time t where the quadratic overtakes the linear which overtakes the constant. The best advice for dealing with infinity, though, is “don’t”. In some cases, you can deal with the approach to infinity as something else increases, but most of the time you don’t even need to do that.
Also, there is some point in time at which fish don’t help you. You only need to calculate to that point. And there is likely some sense in which earlier fish are more useful than later fish, and you should discount far-future fish to the point they don’t affect your decisions.
Let me muse about infinite amounts so that I am not waiting to spill them on a random infinite-adjacent thread.
Someone offers you a choice of 10 fish now or 1 fish a day. Which option lets you eat more fish?
Questions of these kind seem to have properties of:
Even a ludicrous amount of one-off fish will lose to a steady income of fish
Bigger fish streams are bigger. Having 3 fish per day is better than 1 fish per day
Now one could make the case that there is a scheme where each day you grab the new fish and give them a number. For any finite fish per day this can clearly be done. Somebody could come to say that “Hey I can put any fishstream in 1-to-1 correspondence to the natural numbers therefore there is an equal amount of fish in any stream”. Yet it seems if somebody says that their 3 fish per day stream is not an advantage over my 1 fish per day stream, I would not be convinced and think they still have got more.
This makes me think that 1-to-1 correspondence doesn’t really fit with what my brain points to be size. Or some of the intuitive notions come appart when we enter infinity land.
There is also the construction if somebody promises on each day start a new stream of fish. Then for any stream of fixed finite amount of fish per day promised there is a day where the “growing promise” gets you that and then some more afterwards.
I am guessing/feeling that 2 streams of 1 fish vs 1 stream of 2 fish is connected/represented by ω+ω=2ω and the “growing promise” might be connected to ω*ω
Is there/could there be some notion of size that would let me say that ω+ω is smaller than ω*ω and could I deal with people saying that “mappable to integers” is better notion of size?
You can absolutely count your fish that way with the help of hyperreals! (“growing promise” stream would be 12ω2+12ω though)
I think https://en.wikipedia.org/wiki/Hyperreal_number#The_ultrapower_construction is a good introduction. https://math.stackexchange.com/questions/2649573/how-are-infinite-sums-in-nonstandard-analysis-defined and https://papers.ssrn.com/sol3/papers.cfm?abstract_id=3459243 address the handling of infinite divergent series with hyperreals and https://arxiv.org/pdf/1106.1524v1.pdf talks about uniform probability over N (among other things).
Why 12ω2+12ω and not any other? What kind of stream would correspond to ω2 ?
You can just pretend that ω is finite and plug it into the formula for the partial sum.n∑i=1i=12n2+12n, so ω∑i=1i=12ω2+12ω. If they were to give the ith odd number amount of fish on the ith day (1,3,5,7,9...), then you would have ω2 amount of fish, because n∑i=12i−1=n2. The two links I posted about the handling of infinite divergent series go into greater detail (eg. the question of the starting index).
The links are very on point for my interest thanks for those. Some of it is in rather dense math but alas that is the case when the topic is math.
At one point there is a constuction where in addition to having series of real numbers to define a hyperreal (r1,r2,r3...)=h1 we define a series of hyperreals (h1,h2,h3...)=d1, in order to get a “second tier hyperreal”. So I do wonder whether the “fish gotten per day” is adeqate to distinguish between the scenarios. That is there might be a difference between “each day I get promised an infinite amout of fish” and “each day I get 1 more fish”. That is on day n I have been promised ωn fish and taking it as α∑i=1I am not sure whether α=ω and whether terms like ω2 and ωα refer to the same thing or whether mixing “first-level” and “second level” hyperreals gets you a thing different than mixing just “level 1”s
If you just graph fish vs. time, then the one-time gift is a constant function, the steady income is linear, and the “growing promise” stream-of-streams is a quadratic. The fact that a steady income will eventually surpass any one-time gift is because any positive-slope linear function will eventually exceed any constant-value function. Likewise, any 2nd-order polynomial with a positive x^2 term will eventually exceed any linear function. You could keep going with higher order polynomials if you want.
A similar analogy would be a race where even a large head start will eventually be surpassed by a slightly faster car.
I guess I know that kind of perspective exists but I don’t want to refer to the “up to a time” thing but the total.
For example taking the even and odd integers, that would seem like taking a stream of fish and putting them into alternate buckets every other day. I would like to be able to say something like “even and odd integers are of equal amount and they are half of that of all integers”. But I am unsure whether that is accurate and where the foundations would crumble.
if I do it in a very naive way in graph vs time then I could conclude that the bucket that I start with has a stricly dominating amount of fish (because the graph stay above or matches the other graph). And it feels odd to say that there are more even numbers than odd ones! Althought I guess it wouldn’t be odd that 0+{1,2,3,4,5...| x is even} is more than {1,2,3,4,5...| x is even}. Or is it because the two-bucket arrangement is not the natural choice of odds vs evens? I guess if I received two fish a day and put them on different buckets things could change?
Oh I guess I notice that if I receive two fish a day that seems more than if I receive 2 fish every other day. That seems to map to 2*ω > 2*(ω/2) = ω.
Every other day, the number is the same. On the days in between, there’s a difference. Even taking that notion of ‘dominating’ into account, the difference is finite, and small: 1. If you had a choice between:
a) getting 1⁄3 a fish on day one, then 1⁄9 a fish on day two and 1/27th a fish on day three...(1/3^n on day n)
and
b) getting 1 fish on day one, then 1⁄2 a fish on day two, and 1⁄4 a fish on day three...(1/2^(n-1) on day n)
Then b is better than a...by more than the difference between 0+{1,2,3,4,5...| x is even} and {1,2,3,4,5...| x is even}. But it also seems like a bigger deal in those cases, a and b, because getting what comes out to a finite amount of fish over an infinite period of time isn’t as good as getting an infinite amount.
Flip a coin. If it comes up heads put the first fish in the red bucket, and the second fish in the green bucket.
If it comes up tails do the reverse. (With the integers as a whole, where do you start counting from? The answer to that, determines which measure “seems” greater, on odd time/count steps.)
You can color the positive integers, 1, 2, 3, 4,… using this pattern: red, green, red, green,...
You are obviously partitioning them into two collections (sets, because there’s not repetition).
Are there more odd numbers than even? Well if you count how many odd numbers (which are positive integers) are less than or equal to n, and call this function o, o(n) > e(n) for odd n. But for even n, they’re equal. So in the limit as n increases without bound: f(n) = o(n)/(o(n)+e(n) = 1⁄2.* But if you take steps of 2, starting at an even part, like 0, then the two sets are always of equal size. This seems to formalize the intuition well. Using this stream approach, one can articulate that: a randomly chosen integer is as likely to be even as it is to be odd**, and that this doesn’t change over time (as one iterates further through the integers), but the as you iterate out the integers, the frequency with which you see prime numbers goes down.
**This can be generalized further: a randomly chosen integer, for a given radix r, will have its final digit d, drawn from the a distribution indistinguishable from a uniform distribution over [0, r-1].
(Every instance of the word random, I made above is based on a uniform distribution across the integers. If you have a fair coin, and you flip it, until it comes up heads, then the probability that the number of times you flipped it is an odd number is not 1⁄2. The distribution there is: 1⁄2 for 0, 1⁄4 for 1, etc. The probability a red number (even) is chosen is twice that of a green number (odd) being chosen. Using ‘the sum of probabilities is 1’, it can be deduced that the probability the number is odd is 1⁄3, and the probability it is even is 2⁄3.)
*
o(n) + e(n) = n
for even n, o(n) = e(n). f(n) = k/2k = 1⁄2. (k is n/2, and an integer.)
for odd n, o(n) = e(n) + 1. f(n) = (k+1)/(2k) (n is 2k+1, and k is an integer) The limit as n, and thus k goes to infinity for this case is:
1⁄2 + 1/n = 1⁄2. (The limit from above. The other gets infinitely small.)
Strictly speaking, there is no such thing as uniform distribution across the integers, or any other set that can be matched up 1:1 with integers. There are things that aren’t probability distributions that do have some similar properties, such as “asymptotic density”, but they fail to have the properties we require of probabilities. The “stream” idea seems to be another way to describe asymptotic density.
For example we might like to say that there are as many natural numbers of even length in decimal notation as those of odd length, but the stream idea doesn’t support that. The fraction of even-length numbers oscillates between 11% and 89% infinitely often, and there are other properties that behave even more weirdly.
One paradox that illustrates why there is no such thing as a uniform distribution on natural numbers:
Suppose we both pick natural numbers uniformly at random, independently of one another. You reveal your number. What is the probability that my number is greater than yours? No matter what number you chose, there are infinitely many larger numbers and only finitely many numbers that are not larger. Since my number was chosen uniformly at random, my number is larger with probability 1.
What?
1 number of length 0, 9 numbers of length 1
(and maybe 0), 90 numbers of length 2, 900 numbers of length 3, 9000 numbers of length 49*10^(n-1) numbers of length n. For each n the amount of numbers of length just before that is 10 times less and the amount of numbers the next length is 10 times more. If you take a rolling fraction of n odd to all numbers seen it starts to go down when even numbered length is reached and starts to go up when an odd length number is reached.
(Ignoring that most people don’t think of 0 as of being length 0.)
Jump by two orders of magnitude every time and it stays stable:
Starting with nothing:
1 of even length, 9 of odd length.
90 of even length, 900 of odd length.
10% versus 90%.
Start after an even jump:
91 of even length, 9 of odd length.
9,091 of even length, 909 of odd length.
(Starts at 91% even, but drops after a double jump. I don’t know what the limit on this is.)
By comparison, resolving proportion of even numbers versus odd numbers, is much easier, because it’s a simple pattern which oscillates at the same rate, instead of changing.
(in base 10)
Well, if a different color is used every time, then the coloring aspect is solved. If you ask about addition though, then things get weird.
Right, that was a math typo. It really oscillates between 9% and 91%.
For example, 909090 of the numbers below a million have even length, i.e. 91%. As you increase the bound toward ten million, this fraction decreases until it hits a minimum of 9%, and then starts increasing again until you reach a hundred million, and so on.
Well, the old solution to what is the limit of: +1, −1, +1, −1, etc. was: (index starts at one, pattern is (-1)^(n+1))
Consider the cases:
a) +1, odd index
b) −1, even index
Average them.
0.
If that was applied directly, it’d be: (9+91)/2% = 50%.
You could argue that it should be broken down differently, because there’s different proportions here though.
You could also declare the answer undefined, and say infinity is about growth, it doesn’t have a value, for x % 2 (or odd or even number as the case may be), and averages are ridiculous. (And once you have a breakdown of cases, and probability what more is there?)
That is one of the puzzle in that 0+0+0+0+0… converges and has a value but +1-1+1-1+1-1… which seems to be like (1-1)+(1-1)+(1-1)+(1-1)… diverges (and the series with and without the paranthesis are not equivalent)
The strram idea gives it a bit more wiggleroom. Getting 1,0,1,0,1.. fish seems equivalent to getting 1⁄2 fish a day but 1,1,1,1,1.. seems twice the fish of 1,0,1,0,1,0,1,0… So which with the other methods are “can’t say anthing” there is maybe hope to capture more cases with this kind of approach.
Too bad its not super formal and I can’t even pinpoint where the painpoints for formalization would be.
That paradox is good in that it cuts to the matter very cleanly.
To my mind it “numbers larger than my number” and “uniform integer” don’t need to be same. There are n smaller numbers and ω-n bigger numbers. (ω-n)/ω is going to be near 1 (infinidesimally so) but not quite up to 1. Maybe crucially (ω-2n)/ω is smaller than (ω-n)/ω ie if I hit a high number my chances are better. I get that standard approach somehow gets into the way of this and I would like to know which axiom I have a bone to pick with.
There is a (from my perspective a problem) that events of 0 probablity can happen and events with probability 1 can fail to happen. The associated verbal language is “almost surely” and “almost never”. Showing me that a thing can almost surely happen doesn’t guarantee it. To my mind this is because some zeroes have rounding to the nearest real and some don’t.
There are two bones you might pick. One is that probabilities are real numbers. The other is that probabilities are countably additive.
For positive values, there will be some time t where the quadratic overtakes the linear which overtakes the constant. The best advice for dealing with infinity, though, is “don’t”. In some cases, you can deal with the approach to infinity as something else increases, but most of the time you don’t even need to do that.
Also, there is some point in time at which fish don’t help you. You only need to calculate to that point. And there is likely some sense in which earlier fish are more useful than later fish, and you should discount far-future fish to the point they don’t affect your decisions.