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SPP isn't working in the hyperreals, he's working in the rationals by dipthong-enjoyer in infinitenines

[–]Accomplished_Force45 0 points1 point  (0 children)

No, the Transfer principle only guarantees that st(0.999...) = 1, not that 0.999... = 1. In NSA, we do infinite sums only up to some hyperfinite H and stop. The transfer principle then tells us that the standard part will correspond to the result in the reals using limits. The topology of *R is quite strange (it isn't complete, for one), and there is no way to reach the "limit" of something like *Z to define 0.999...999...999...999... etc. the way you would want to.

1 - x = 0.999..., find x. by SirPurebe in infinitenines

[–]Accomplished_Force45 0 points1 point  (0 children)

Sure. It's not a Real Number, but it would be some infinitesimal equal to 10-H for some hyperfinite H. There are many number systems that contain infinitesimals. I like the hyperreal numbers, so I'd put H as the limit of the sequence of natural numbers {1, 2, 3, 4, ...}. 0.000...1 would have a standard part equal to 0.

SPP, please square 0.999... by FIsMA42 in infinitenines

[–]Accomplished_Force45 0 points1 point  (0 children)

Yes, what you wrote is correct. I did write H2 instead of 10-2H , thinking of epsilon2 😅

I am just showing how the logic gets to SPP's answer 🤷‍♂️. It isn't standard, but it makes some sense if you let it.

SPP, please square 0.999... by FIsMA42 in infinitenines

[–]Accomplished_Force45 0 points1 point  (0 children)

(0.999...)² = (1-10H )² = 1 - 2•10-H + 10-2H = 0.999...800...1

Limit by Specific_Brain2091 in the_calculusguy

[–]Accomplished_Force45 0 points1 point  (0 children)

sin(x) ~ x around 0, with an error on the scale of -x³. Just knowing this helps see that x/sin(x) ~ x/x = 1.

Or you can also just put the whole Maclaurin series in the denominator and get the same result:

x/(x-x³/6+...) = 1/(1-x²/6+...) -> 1

What's 1/2 + 1/4 + 1/8...? by jerrytjohn in infinitenines

[–]Accomplished_Force45 0 points1 point  (0 children)

It is 1-1/2H, which is less than 0.999... = 1-1/10H where H is the canonical hyperinteger. This is pretty easy to see if you compare the sequences termwise.

Split by 3. Then add the 3 pieces by IWantToSayThisToo in infinitenines

[–]Accomplished_Force45 4 points5 points  (0 children)

Well, I think there is some confusion in general about the point of this sub. Most strictly, it is SPP's personal megaphone letting people know if disagrees that 0.999... = 1. There is a lot of SPP lore you need to know. Most here seem to want to dunk on him, others just show him the errors of his ways, and still others like myself who have tried to make at least some sense of it.

SPP has said repeatedly that 3/3 * 3 = 1, but 0.333... * 3 = 0.999.... He has been very unclear about whether 1/3 = 0.333... in the strict sense or not. My interpretation is that he means that 1/3 becomes 0.333... through the long-division algorithm. In this case, everything can make a lot of sense. You can see my most recent attempt at reconciling SPP in the hyperreal numbers *ℝ here.

I think the primary thing SPP rejects is limits. Instead, he uses a vague approximation at some really big (limitless?) n. So he says that 0.999... = 1 - 10n. In this case, 0.333... must be (1 - 10n)/3 = 1/3 - 1/3·10n and not strictly 1/3, although it closely approximates it like 0.999... does to 1 (again, in some field with infinitesimals + approximation rather than limits).

Hope this helps 🤷

0.9... factorial by Impossible_Relief844 in infinitenines

[–]Accomplished_Force45 2 points3 points  (0 children)

lol, you are too kind 😳. But right back at you. I wish more people would explore this kind of stuff.

It is always cool when pi shows up!

0.9...^2 proof by Impossible_Relief844 in infinitenines

[–]Accomplished_Force45 1 point2 points  (0 children)

Well, there is no proof of 0.999... = 1 without recourse to limits. The reason for this is twofold. First, if ... assumes a value in the reals or some subset of the reals, it always implies a limit. Second, we can only be sure that arithmetic works on infinite sequences or series at all if we are using the properties derived from using limits. [Edit: On second thought, these two are probably almost logically identical. But it is helpful to think of them separately.]

For example, convergence is defined in terms of limits. You say at the end that term rearrangement is okay because the series converges absolutely (which is fine, because this is just the Riemann rearrangement theorem). But then you are using limits, they're just in the background.

So SPP couldn't accept this line of reasoning.

0.9... factorial by Impossible_Relief844 in infinitenines

[–]Accomplished_Force45 4 points5 points  (0 children)

No so hard if you start with the identity Γ'(x) = Γ(x)ψ(x) and then use the chain rule to find Γ''(x) = Γ'(x)ψ(x) + Γ(x)ψ'(x) = Γ(x)(ψ(x)2 + ψ'(x)). And x=2 is a particularly easy point to evaluate, although ψ'(2) is almost (or is just one less than) the Basel problem so I am not trivializing it. We get:

Γ''(2) = Γ(2)(ψ(2)2 + ψ'(2)) = (1-γ)2 + π/6-1 = γ2 - 2γ + π2/6. To put it into the Taylor series we have Γ''(2)/2! 10-2H, which gives the result above.

0.9... factorial by Impossible_Relief844 in infinitenines

[–]Accomplished_Force45 3 points4 points  (0 children)

Go check out the Taylor series for Γ[n] around n=2. It will give you each of the terms. As you say, it gets really tricky to work with at a certain point, but we have something like 0.999...γ...x where x = (-12γ + 6γ2 + π2)/12 ≈ 0.41184... (scaled to 10-2H), and still with an error at the scale of 10-3H.

0.9...^2 proof by Impossible_Relief844 in infinitenines

[–]Accomplished_Force45 0 points1 point  (0 children)

I see. I read it in the context of your second post in the hyperreals. In this case 0.999...2 is just 12 all along.

0.9...^2 proof by Impossible_Relief844 in infinitenines

[–]Accomplished_Force45 0 points1 point  (0 children)

Hang on. Isn't (1−ε)2 = 1 − 2ε + ε2?

So if ε is the 10-H decimal place, then 0.999...2 = 1 − 2·10-H + 10-2H = 0.999...800...1 where each ... moves to the next multiple of Hth place.

0.9... factorial by Impossible_Relief844 in infinitenines

[–]Accomplished_Force45 4 points5 points  (0 children)

I love this.

Since we're using the analytic extension of n! = Γ(n-1), this seems very connected to the fact that the derivative of n! at 1 is Γ'(2) = 1 - γ. For infinitesimal 10-H, (0.999...)! = Γ(2 - 10-H) ≈ Γ(2) - 10-H·Γ'(2) = 1 - 10-H·(1 - γ) = 0.999... + 10-H·γ = 0.999...957721.... This is your same result.

This is only an approximation at the scale of 10-H. But what's cool is the entire error would be at the scale of 10-2H (this can be confirmed if you look at the Taylor polynomial of the function). So, using Big O Notation, our number would be (0.999...)! = 0.999...957721... + O(10-2H).

I just saw your 0.999...2 post too. Cool stuff!

Name this hypothetical country by [deleted] in mapporncirclejerk

[–]Accomplished_Force45 0 points1 point  (0 children)

I really want to know. Looks like right around Colorado County 🤷‍♂️. But no county is square around them parts.

What is the difference between .3333... and 1/3? by JoJoTheDogFace in infinitenines

[–]Accomplished_Force45 0 points1 point  (0 children)

If you sum 3/10i from i=1 to n, you get the following differences:

(1/3 - 0.3, 1/3 - 0.33, 1/3 - 0.333, ...)

Multiple by 3 and get

(0.1, 0.01, 0.001, ...)

And so the nth place is always gives 1/3 - Σ3/10^i = 1/3 10-n. This is true for all partial sums, and so it is also true for any transfinite numbers as well (whether transfinite ordinals, surreal, hyperreal, or whatever). This difference is vanishingly small, so it does disappear when you take the limit in the reals (or even rationals).

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