Wikipedia:Reference desk/Archives/Mathematics/2013 May 1

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May 1

The closeness of the hyperreals

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There was a contribution in the infinitesimal article asserting that 0.999... can be reinterpreted from its standard meaning as a real number so that it is not equal to 1. Upon further reading I came to understand that this might be the case if 0.999... represents a set of numbers less than 1 but greater than that of a given rank and this was being suggested as an alternative interpretation of 0.999... proposed by Karin Katz and Mikhail Katz which is described in our 0.999... article in the infinitesimal section here:

"The standard definition of the number 0.999... is the limit of the sequence 0.9, 0.99, 0.999, ... A different definition considers the equivalence class [(0.9, 0.99, 0.999, ...)] of this sequence in the ultrapower construction, which corresponds to a number that is infinitesimally smaller than 1. More generally, the hyperreal number uH=0.999...;...999000..., with last digit 9 at infinite hypernatural rank H, satisfies a strict inequality uH < 1. Accordingly, Karin Katz and Mikhail Katz have proposed an alternative interpretation of "0.999...These authors propose we define 0.999... as:"

${\underset {H}{0.\underbrace {999\ldots } }}\;=1\;-\;{\frac {1}{10^{H}}}.$

My problem with this definition which I posted on the infinitesimal talkpage is as follows: Suppose that H has infinite rank, then consider that a standard number is 3/10 + 3/100 + ... = 0.333... = 1/3 (this should be true whether we are working with reals or hyperreals, since hyperreals include all reals). Therefore: 0.999... = 9/10 + 9/100 + ... = 3*(3/10 + 3/100 + ...) = 3*(1/3) = 1 = 1 - 0 <> 1 - |1/x| for all x because the reciprocal function f(x)=1/x<>0 can not be zero for any and all x including all possible infinitely small numbers given by the infinitesimals such as the specific one given by Katz&Katz: 1/10^H. Thus, 0.999... <> 1 - 1/10^H. QED.

Therefore, the real number .999... = 1, but if the rank of 0.999... is infinite, I don't understand how there could be a set of infinitimal numbers .999... < 1 that would still be close, let alone infinitely close, to 1? -Modocc (talk) 02:11, 1 May 2013 (UTC)

[Correction: my initial post was missing part of Katzes' definition. But its fixed now I hope (and I hope our article isn't incomplete). -Modocc (talk) 06:33, 1 May 2013 (UTC)

Note that the article being discussed there is the one by Ely as well as the article by Norton and Baldwin. Tkuvho (talk) 14:41, 1 May 2013 (UTC)

The problem with the Katzes' take on this (assuming that you've reported it correctly and I've understood you correctly) is that there is just no reason to think 0.999... should be interpreted as meaning 9s up to a position given by some nonstandard natural H (or is it H-1 in their example?) followed by all 0s. Where are the 0s? You'd need to write it as 0.999...;...999000... or something like that. Also we want a given numeral to denote a specific number, and there's no obvious canonical choice for H.

So basically I agree with you. This is a red herring when talking about infinitesimals. You can use nonstandard models if you want, but the natural interpretation of 0.999... in that context is that you have 9s through all decimal positions (standard and nonstandard), and then you get 1 again.

(You might ask, what if I take 9s just through the standard positions, and 0s for all nonstandard ones? That might be fairly natural, but such a numeral can't exist in the model. If it did, the model would be able to tell which of its natural numbers were standard, and that's not allowed.) --Trovatore (talk) 03:28, 1 May 2013 (UTC)

Er.... its late here and I've corrected my omission. --Modocc (talk) 06:33, 1 May 2013 (UTC)

I'm not familiar with the abstract algebras, nevertheless I can't see how any conical position for the rank H can work, because any and all nonzero numbers must be excluded, and its because .999... is the natural number 1. --Modocc (talk) 03:44, 1 May 2013 (UTC)

Since this concerns material dealt with in detail in the article 0.999... the natural place to deal with this is at Talk:0.999... where I created a thread. Tkuvho (talk) 10:20, 1 May 2013 (UTC)

Thanks, but to be clear, I'm not asking anyone here how that article should be edited. But I would like help understanding how the maths involved with this proposed definition can be contrary to my understanding of the reals and basic algebra. -Modocc (talk) 12:19, 1 May 2013 (UTC)

It is not contrary to your understanding of the reals and algebra. Judging from your earlier comments, I have the impression that it may be contrary to your somewhat too literal understanding of the concept of an "infinite sum". I can try to elaborate. Tkuvho (talk) 13:22, 1 May 2013 (UTC)

At the moment where I live I've some more gardening to do and more seeds to sow and I've thus far only read the student journal article and the Wikipedia articles, thus I've only a precursory understanding of K&K's research and what is being taught. I'll see what I can find out within the next few days, but I'm going to be disconnected from the internet during most of this period. In the meantime, any elaborations, clarifications, insights or reference materials posted here by you guys regarding this very recent proposal would be appreciated. My concern is primarily with any alternation of the meaning of the ... convention. I can see its meaning being modified in limited contexts when that context is accepted and known, but changing meanings or conventions can confuse rather than enlighten. If I tell you that my research results showed that 0.333... of my patients scored well on the exam, does that mean 1/3 did so or does it mean 1/3-1/3*10^H did? If the former then 0.999... = 1, if the latter meaning is meant, then we only have instead the latter result of 0.999... = "(1 - 1/10^H)/3" (K&K). Conventionally, its my understanding that the ... notation means to include (or continue) an entire sequence of some set (or operation), not just a portion of it. That is why I should be able to write that the summation of the individual members of the set {3/10, 3/100,...} equals 1/3. That is my understanding of the ... notation and what is meant by an infinite sum. Yes, my interpretation is definitely limited to a literal meaning, but that is how we communicate with maths and this isn't the humanities desk. :-) But few math conventions are set-in-stone!  ;-) In fact, I do have a real sword-yanking demonstration to give when I get back. Modocc (talk) 14:05, 2 May 2013 (UTC).

Again, note that the article being discussed at Infinitesimal is the one by Ely as well as the article by Norton and Baldwin. Ely's text does not even cite the article you mentioned. Your misconception about infinite sums has nothing to do with the particular proper extension of R chosen; it can be illustrated in the context of the dual numbers for example. Tkuvho (talk) 15:16, 2 May 2013 (UTC)

I'm not very familiar with the extensions of R (and I don't generally enter caves such as these without a well-lit torch, so I'm somewhat prepared I hope, for the beast of a...), so I took a look at dual numbers. It adds ε which is nilpotent where ε² = 0 and has the property xⁿ = 0 where n is positive. Within the realm of basic algebra, only ε=x=0 can satisfy those conditions, so I'm circling back upon the standard algebra of reals because we added an element that is already in R. -Modocc (talk) 16:14, 2 May 2013 (UTC)

If you don't like nilpotent elements there are certainly simple-minded infinitesimal systems without them (I was merely referring you to the simplest example of an extension). Now I don't quite understand your "circling back". If you want to stick with the real numbers, that's certainly fine, but there are NO infinitesimals in the real numbers. Why are you interested in the infinitesimal page then? Tkuvho (talk) 16:34, 2 May 2013 (UTC)

I've numerous interests, including the historical infinitesimal which is 1/infinity <> 0, and these infinitesimals exist because 1/x <> 0 is a property of the reciprocal function. Without this property, we would not have an article on infinitesimals. By "circling back", I mean that if in order to extend R to elements that do not satisfy a standard algebra, then those elements are useful only within the context of the algebra that these were extended to. -Modocc (talk) 17:35, 2 May 2013 (UTC)

Good thing you like standard algebra. The hyperreal numbers satisfy it by the transfer principle. Tkuvho (talk) 08:30, 3 May 2013 (UTC)

I've not studied cardinal numbers, but you keep saying that I take infinite sums too literally. Perhaps you can explain whats wrong with either my understanding here or my objections: For the proposed definition I get 0.999... <≠ 1. Tradition vs proposed! The new definition is strictly less than 1 thus these are consistent statements once these are given the required context. My objection to this which I thought I made explicit above, was that although its fine within the context used, its not fine as a standard! What does the ... notation suppose to mean for repeating decimals in general? -Modocc (talk) 09:40, 3 May 2013 (UTC)

To respond to your question "Perhaps you can explain whats wrong with either my understanding here or my objections": first, what is the meaning of your notation "<>" ? Furthermore, your remarks concerning "infinitesimals existing because of the properties of the reciprocal function" are vague, and don't indicate any awareness of modern foundational issues. Tkuvho (talk) 08:52, 5 May 2013 (UTC)

<> means "not equal" (see http://en.wikipedia.org/wiki/List_of_mathematical_symbols), and since this thread is now archived, I'll be more explicit regarding my views and understanding at another time. I've clarified the infinitesimal article with this edit [1] accordingly. Note also that I started a discussion about the appropriateness of the references on the infinitesimal talkpage. Thanks for your input on this, and a special thank you to Trovatore too. :-) -Modocc (talk) 18:25, 5 May 2013 (UTC)

(Un)folded paper pattern

I've seen a video about this before, but can't remember the name of the concept/sequence - basically, by continually folding a strip of paper in half left-over-right, then unfolding it, the direction of the folds forms a pattern. For example, with only 1 fold you get a "down" fold, after 2 folds you get "up-down-down", 3 folds is "up-up-down-down-up-down-down", etc. This sequence follows a specific pattern (just playing with paper I've determined some of the rules myself), and I know that there is a specific name for it but I can't recall it. Any ideas? Organics^LRO 12:58, 1 May 2013 (UTC)

See Regular paperfolding sequence, I think that may both give answers and some cool new directions to think about.Naraht (talk) 13:31, 1 May 2013 (UTC)

The eventual fractal is called a Dragon curve. Dmcq (talk) 13:39, 1 May 2013 (UTC)

Ah, Dragon curve was the name I was looking for, however both articles give some excellent info. Thanks to you both! Organics^LRO 14:12, 1 May 2013 (UTC)

Resolved

What fraction of moonlight is at night?

I'd like to know what fraction of moonlight is at night. I'm making a *large* number of assumptions here to move this to math...

1. Circular moon Orbit
2. Constant Planetary Day/night length
3. Ignore eclipses/ earth atmosphere.
4. No craters etc.
5. Using a *32* day month for illustration

If I deal with only the first half of the orbit (day 0 =New to day 16=Full), then on day 0 the moon gives zero moonlight to either day or night. As the orbit goes on , the percentage of the time that the moon spends in the night increases linearly. However even though on day 8, the moon's face is exactly half lit, is it true that on day 4 that the crescent that shows makes up exactly one quarter lit and on day 12 that the percentage of the crescent is exactly three quarters lit? If that increase is also linear then the amount of moonlight at night during the first half is the integral of x^2 from 0 to 1 while the amount of moonlight at night is the integral of x(1-x) from 0 to 1. The first integral gives 1/3, the second 1/6. So the fraction of moonlight that occurs at night is 2/3 and the fraction of moonlight that occurs during the day is 1/3. Can people please check my logic after the assumptions?Naraht (talk) 13:27, 1 May 2013 (UTC)

Even with your simplifications you are missing at least one important factor. A full moon is far more than twice as bright as a half moon, because the average angle of sunlight striking the lunar surface is much higher. Essentially you need to know the phase curve of the moon, or at least have an approximation of it. Looie496 (talk) 15:04, 1 May 2013 (UTC)

Well, it also depends on what they really want. Your point is very apt if the goal is to quantify the amount of moonlight, but the OP seems to be on the right track for quantifying the amount of time, during which moonlight occurs at night. SemanticMantis (talk) 18:40, 1 May 2013 (UTC)

OP, here. I didn't know about the phase curve. :( "The Moon's apparent magnitude at full phase is −12.7[29] while at quarter phase it is less than 10 percent as bright." Given that, I'm quite sure that the fraction of moonlight at night is an even higher fraction, but the calculations even with the graph are *ugly*.Naraht (talk) 20:40, 1 May 2013 (UTC)

The Polylogarithm and Palindromic polynomials

$\sum _{n=0}^{\infty }{n^{k} \over a^{n}}\ =\ \operatorname {Li} _{\ -k}\left({1 \over a}\right)\ =\ {a\cdot P_{k-1}(a) \over (a-1)^{k+1}}\ ,\ \qquad \qquad k\in \mathbb {N}$

where

P_k-1 (a) is a palindromic polynomial of degree k-1 in a, with the coefficients of a^k-1 and a⁰ being equal to 1.

My question would be:

How do we to deduce the rest of its coefficients ? Are there any formulas that describe them ? — 79.113.215.147 (talk) 18:42, 1 May 2013 (UTC)

How do you get the sum of the series $nx^{n}$? List of mathematical series#Low-order polylogarithms might be what you want. Dmcq (talk) 14:38, 2 May 2013 (UTC)

The recursive relation ${\frac {d}{dz}}\operatorname {Li} _{n}(z)={\frac {\operatorname {Li} _{n-1}(z)}{z}}\,\!$ is indeed pertinent, but I was sort of hoping to turn it into a more `obvious` one, involving the coefficients directly, kinda like the one in Pascal's triangle... and I did indeed find one (recursive) that gives me the coefficients of the second and next-to-last terms in a... But that's pretty much how far I'm mentally able to go... I'm really challenged when it comes to induction... kinda like a golf handicap, only that my weakspots are algebra and "classic" (non-Cartesian) geometry... Heck, I can't even understand how those formulas (which I'm now trying to `simplify`) are actually computed... I saw a couple of them recently in Ramanujan's journals (for k = 5 and a = 2 and 3)... but how he (or anyone else, for that matter) was able to calculate them in the first place, is simply beyond me... I'm just wandering alone in the dark, all lost, with no one to guide me, despite having a degree in engineering and a mathematician for a father. I'm afraid I will die in this darkness. — 79.113.234.47 (talk) 19:25, 2 May 2013 (UTC)

The original equation you've got here is a bit wrong. The lower limit should be 1 not 0. May be what went wrong. Dmcq (talk) 07:46, 4 May 2013 (UTC)

It makes no difference. 0^k / a⁰ = 0 / 1 = 0. — In case you were refering to a previous version of my last comment, what went wrong there was the fact that I've replaced z with a instead of ¹/_a : silly mistake! :-) — 79.113.208.131 (talk) 16:35, 4 May 2013 (UTC)

It is slightly different, for k=0 the usual convention for lots of series is that 0⁰ is 1 and of course one can't stick in negative k. Dmcq (talk) 18:40, 4 May 2013 (UTC)

I still don't know how to transform the recursive formula involving calculus into one involving only coefficients [be it either directly or recursively]... :-( The only thing I've come up so far is b_{k + 1} = 2 b_k + k - 2, where the b`s are the coefficents for powers 1 and k - 2 of a. Where are Euler and Ramanujan when you need them? :-( — 79.113.208.131 (talk) 19:30, 4 May 2013 (UTC)

I just had a look and starting with

$\operatorname {Li} _{-n}(x)={{\sum p(n,i)x^{i}} \over {(1-x)^{n+1}}}$

Applying

$\operatorname {Li} _{-(n+1)}(x)=x{\frac {d}{dx}}\operatorname {Li} _{-n}(x)$

I get

$p(n+1,i)=ip(n,i)+(n-i+2)p(n,i-1)$

Does that work for you? Dmcq (talk) 21:59, 4 May 2013 (UTC)

I think it just might... Thank you! I'd wish you a happy Easter as well, but I doubt you're from Eastern Europe... You've definitely made my Holidays happier... :-) — 79.113.208.71 (talk) 23:39, 4 May 2013 (UTC)

Resolved