25 ms·

The Remarkable Number 1/89 (2004)

413 points by nonsapreiche 6y ago

burlesona 6y ago
Okay, as a non-mathematician, I see something like this and I think... “neat coincidence?” But the world of numbers seems to be full of these neat coincidences. So do any of the math folks here have a theory or explanation of why?
- deleted 6y ago
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- efdb 6y ago
  In addition to this question I would like to know if you can in general say/proof that for every sequence which has some relation between the successive numbers there is a rational number whose decimal expansion is the same as the sequence.
  - stabbles 6y ago
    A standard technique to find explicit expressions for the `n`th number of a recursion is through generation functions; see e.g. https://en.wikipedia.org/wiki/Generating_function https://en.wikipedia.org/wiki/Generating_function. You can plugin x = 10^-1 there. Not sure if the result is always a rational number.
  - magnio 6y ago
    For a linearly recursive sequence x_0, x_1, x_(n+2) = ax_(n+1) + bx_n, the general formula for the terms is x_n = cα^n + dβ^n, where α, β are the roots of the quadratic x²−ax−b; c, d are solutions to the system c + d = x_0 cα + dβ = x_1. If the series Σ x_n⋅10^n converges then its value is 10c/(10−α) + 10d/(10−β) = ((100−10a)x_0 + 10x_1)/(100 − 10a − b). If a, b, x_0, x_1 are all rational then the above series converges to a rational number, too. This is the case for the Fibonacci sequence, with a=b=1, x_0=0 and x_1=1.
    - deleted 6y ago
      [deleted]
- benrbray 6y ago
  There's a "theorem" about this: > The interesting number paradox is a semi-humorous paradox which arises from the attempt to classify every natural number as either "interesting" or "uninteresting". The paradox states that every natural number is interesting. The "proof" is by contradiction: if there exists a non-empty set of uninteresting natural numbers, there would be a smallest uninteresting number – but the smallest uninteresting number is itself interesting because it is the smallest uninteresting number, thus producing a contradiction.
  - wizzwizz4 6y ago
    That just sounds like another formulation of the Surprise Exam paradox.[0] It falls down when you realise that “the four hundred and seventieth otherwise uninteresting number” is not a particularly interesting number, so there must be a problem with the problem statement. [0]: https://en.wikipedia.org/wiki/Unexpected_hanging_paradox https://en.wikipedia.org/wiki/Unexpected_hanging_paradox
    - alisonkisk 6y ago
      It's a paradox related to "meta-logic", but it's different. Surprise exam is about temporal reasoning -- It's only impossible to be surprised on the last day (or else the premise of having an exam is invalidsted), and reasoning backwards in time from a contradiction is not valid. Uninteresting number is a simpler contradiction in definitions.
  - DeusExMachina 6y ago
    This is not a paradox though, as your copy and paste states. It's just a theorem (as you stated) with a proof by contradiction. A paradox must be self-contradictory under all circumstances.
    - phizy 6y ago
      It is a paradox. It assumes you have a definition of uninteresting number such that you can select the least uninteresting number, and then retroactively defines that number to be interesting by brand new criteria, which contradicts that you would have ever selected it in the first place. Thus the axioms invoked are mutually contradictory: the axioms that allow you to identify the least uninteresting number, and the axioms you invoke to declare it interesting are in conflict.
- bobbylarrybobby 6y ago
  This is no coincidence — it’s because 89=100-10-1, and the ordinary generating function for the Fibonacci sequence is 1/(1-x-x^2) (if you go to wolfram alpha and Taylor expand that expression, you’ll see its coefficients are the Fibonacci numbers).
- impendia 6y ago
  Yes. One has the identity: 1/(1 - x - x^2) = 1 + x + 2x^2 + 3x^3 + ... where the Fibonacci numbers are the coefficients on the right. Try writing it out! The basic idea is that because F_n = F_{n - 1} + F_{n - 2}, everything will neatly cancel out. This is an example of a "generating function". Anyway, plug in x = 0.1, and then divide by 100 to see the behavior described in the post. One also gets 1/9899 = 0.00010102030508132134559046368320032326498... 1/998999 = 0.0000010010020030050080130210340550891442334... and so on.
  - contravariant 6y ago
    More to the point this function trivially statisfies f(x)-1-x = X (f(x)-1) + X^2 f(X). You should be able to convince yourself that this is equivalent to the fact that its power series satisfies the Fibonacci recurrence, and that this power series starts with 1 + x. You can also use this to easily find generating functions that satisfy other starting conditions.
- jerf 6y ago
  Imagine the real number line is a database that you can run SELECT queries against, and you don't have to worry about giving a computational procedure that produces the result, it just magically gets produced. Now, imagine you write a query, like, say, "SELECT number WHERE number = .01 * FIB[1] + .001 * FIB[2] + .0001 * FIB[3]" and so on until you get what the article discusses. It isn't necessarily that surprising that you might find something with an uncountably infinite numbers to pick from. Now, consider all possible "interesting" queries you could run, along with all their results. There result is an inconceivably large sea of queries. Most of them are, in fact, utterly pointless; SELECT statements that return no values, SELECT statements that return all values (equally pointless), SELECT statements that return complicated sets of values but have essentially no mathematical interest because there is no practical way to represent them as anything smaller or more interesting, etc. In this massive sea of results, you should expect a lot of interesting things to exist. Finding them is tricky; in percentage terms they make up 0% of the results, but we have mechanisms for finding some of them. Basically, there are so infinitely many mathematical statements that there can't help but be a large supply of "interesting" statements like this. For an interesting view on that, see https://en.wikipedia.org/wiki/Mathematical_coincidence https://en.wikipedia.org/wiki/Mathematical_coincidence . These are true statements or almost true statements (near equalities) about a wide variety of numbers that are essentially meaningless... it's just there's so many ways of putting things together that there are inevitably large numbers of these things (the wiki page is just a sampling). You can even generate these mathematical coincidences yourself. Create a program that will systematically iterate over abstract syntax trees of mathematical expressions involving whatever combination of mathematical operators (+-×/, sqrt, log, sin, whatever) and numbers you like (the first ten integers, e, i, pi, whatever else you like), store up a table of results and emit any two expressions that are, say, within .01% of each other. You will rapidly find a ton of results, because it turns out that even with modest numbers of operators, there are far more mathematical expressions than there are small numbers for them to result in separated by more than .01%. If you think about it, this program can't help but emit a lot of results. Some of them will be humanly "interesting". A few of them will even be mathematically interesting (e.g., this procedure will generate the famous Euler identity relatively quickly if you included the relevant operators and numbers). On a larger scale, this is also known as the Strong Law of Small Numbers: https://en.wikipedia.org/wiki/Strong_Law_of_Small_Numbers https://en.wikipedia.org/wiki/Strong_Law_of_Small_Numbers The previous paragraph is a very bite-sized example of why this holds that you can code up yourself if you are interested.
- alisonkisk 6y ago
  You might also enjoy tan(1 degree/55555555555)
  - huhtenberg 6y ago
    3.1415926536212091649988555782837470430464925667776193389556...e-13 Nice
  - prezjordan 6y ago
    Fun video about this one on Numberphile https://www.youtube.com/watch?v=IMY2_yzDm9I https://www.youtube.com/watch?v=IMY2_yzDm9I
    - dylan604 6y ago
      "they say there's a fine line between a numerator and a denominator" --? another one of those maths jokes that makes me smile and confuses a lot of other people of why i'm laughing
  - paulpauper 6y ago
    not impressive given that more 5s are needed than correct decimals produced
- JackFr 6y ago
  The Penguin Book of Curious and Interesting Numbers https://www.amazon.com/Penguin-Book-Curious-Interesting-Numbers/dp/0140261494/ref=tmm_pap_swatch_0?_encoding=UTF8&qid=&sr= https://www.amazon.com/Penguin-Book-Curious-Interesting-Numb... The caveat here is "for some value of 'interesting'."
doodpants 6y ago
"The successive ratios of the terms, i.e. 1/1, 2/1, 3/2, 5/3 ... tend to a number called the Golden Ratio by the Greeks." Fun fact: take any two numbers (e.g. chosen randomly), and use them as the seeds for a Fibonacci-like sequence by summing the last two terms to generate the next term. The ratio of any two consecutive terms in that series will tend towards the golden ratio.
- cscheid 6y ago
  If you read this and you’re curious, there’s a proof that is fairly easy to follow if you understand eigenvectors. Write the operation that takes the two last elements of the sequence and produces the following two, notice it’s linear, then analyze the eigenvalues of the associated matrix and relate the original operation to the power method.
  - redvenom 6y ago
    You actually don't need eigenvectors for this proof. Sketch: Write x = lim a_n/a_{n-1} where a_n is the nth term of Fibonacci. Replace a_n with a_{n-1} + a_{n-2}, which will give you a quadratic equation in x. Solving this quadratic equation gives you the golden ratio.
    - cscheid 6y ago
      I believe that gives you the result starting from Fibonacci, but not that any starting pair of numbers eventually lands at the golden ratio.
      - redvenom 6y ago
        No, it works for any pair of numbers, as long as they are not both zero. I am only using the recursive relation of the Fibonacci sequence, not the starting terms.
- cleansingfire 6y ago
  This seems like a pretty basic calculus problem. Constants become irrelevant at the limit, so that the relationship (ratio) is all that's left.
jurajvt 6y ago
1/89 = .01123595505..., not .01123595595..., small typo
wumms 6y ago
> As you can easily check, 1/89 = .01123595595... Bizarre, eh? Typo: 1/89 = .01123595505...
ache7 6y ago
Any number is remarkable if you're smart enough to find why it's remarkable.
- eythian 6y ago
  And if it's not remarkable, then that is in itself remarkable.
  - forinti 6y ago
    Aren't most transcendental numbers unremarkable?
    - grahamlee 6y ago
      No number is unremarkable. Let's construct the set of all unremarkable numbers. Now, let's construct the sequence of those numbers in order. The first member of that sequence has the remarkable property that it is the smallest unremarkable number. That is remarkable, so remove it from the set. By induction, the set must be empty.
      - jlokier 6y ago
        > Now, let's construct the sequence of those numbers in order. The first member of that sequence Not happening. Sets of numbers do not always have a first in order. Consider the set of unremarkable real numbers > 0 under the regular arithmetic ordering. Which one is first?
      - gnulinux 6y ago
        That doesn't work because real numbers are not enumerable, so you cannot induce over them. That joke "proof" only works for natural numbers and goes like this: Theorem: all natural numbers are interesting * Base case: 0 is interesting because it is the smallest natural number, as well as the identity element of + operation. * Inductive case: Assume the theorem holds for all m, m<n. Take n. If it is not interesting, then n is the smallest non-interesting number. But that's interesting because it's the smallest such number. Therefore it cannot be non-interesting. Therefore theorem holds for n. By induction, we conclude all natural numbers are interesting. QED.
wumms 6y ago
From an archived talk (2011) on wikipedia [0]: "The linked page misleadingly suggests that a certain Cody Birsner discovered the relationship between the series and the fraction, whereas it had been known for a considerable time before". Günter Köhler, 1983 (published in the The Fibonacci Quarterly, 1985; who cites earlier papers from 1977 and 1981): https://www.fq.math.ca/Scanned/23-1/kohler.pdf https://www.fq.math.ca/Scanned/23-1/kohler.pdf [0] https://en.wikipedia.org/wiki/Talk%3AFibonacci_number%2FArchive_3#1/89 https://en.wikipedia.org/wiki/Talk%3AFibonacci_number%2FArch...
- alisonkisk 6y ago
  It's a bit silly to chase down original authorship of an idea that is a minor detail visible to many people who work in a field. It's like asking who was the first person to discover that all multiples of 11 have the same parity in the respective sums of their odd and even digits.
  - kazinator 6y ago
    Since there had to be a first person who discovered that all multiples of 11 have the same parity in the respective sums of their odd and even digits, that question is obviously interesting to some historians. Due to what circumstances was that person first? What hindered others before that person? How long was the lag before use of decimal and the discovery?
    - munificent 6y ago
      It's important to remember that while technically there is a chronologically first person to discover X for all X, that doesn't imply that that person is the only person to discover X. For sufficiently obvious X, there are likely to be many independent discoverers and highlighting the chronologically first one heaps praise somewhat arbitrarily on one of them.
      - chowned 6y ago
        "History isn't perfect" is not a good reason to avoid doing history. Historians are all well aware of these facts.
ckemere 6y ago
Presumably in a different number base, the number would be different. Would it still be in the aeries?
- Cullinet 6y ago
  Without any basis whatsoever, and I really must one day put this idea out of its misery with some studying, I long suspected that quantum mechanics involved parallel universes where different bases more aptly fit with that other reality. My thought is surely crackpot but I'll explain how my idea arose : A fraction eg 1/3 describes a decimal number to infinite accuracy but creates a challenge for base 10 calculations. I thought about the precision necessary for learning the math of the quantum particles and the experimental way we're trying to figure out how they behave to infer their properties. (Higgs was the other way around but I am optimistic that we'll predict much more in future instead of this convoluted observations rigmarole) and I started wondering if you could approximate extremely high precision decimals to fractions in non decimal bases and from there simplify calculations with far greater resolution. How much more capable would Nyquist - Shannon sampling, if we could clock with the precision of infinite decimal fp digits but handle only a short simple "one third" input or do visor? My silly mind wandered off to imagine particles jumping between different base based universes just as a sequential progression through the precision of their infinitessimal steps through space and time.
  - whatshisface 6y ago
    Base 10 doesn't fit particularly well with our reality, it was picked for historical reasons. Every other base works just as well.
    - philsnow 6y ago
      including fractional bases (rational or irrational) and even negative bases, I think.
  - core-questions 6y ago
    This is my favourite kind of pure speculation. You don't know what you're talking about, really, but you're clearly exploring vast areas of thought at the same time. Remember, integers are integers are integers, because they represent the intrinsic "whole quantity" of something; this is as concrete as logic will get, the idea that there are "ones", it's pretty hard to imagine a universe that doesn't have that. Once you have integers, then you're going to do math in an integer base; and making too high of a base has a diminishing value at at certain point, so it's unlikely we'd see higher than maybe 60. Non-integer bases exist - https://en.wikipedia.org/wiki/Non-integer_base_of_numeration https://en.wikipedia.org/wiki/Non-integer_base_of_numeration - but it's clear to me that I'm too stupid to use them, and so probably most other people are too. This tells me that it's going to be a comparatively rare Many World that chooses to do this. Choosing a number like 12 or 60 with a lot of divisors would have been nice. 1/3 in base 12 is "0.4", which is a lot nicer than 0.333... and would probably help make a lot of younger math education way easier. Combining that with skipping "degrees" entirely and using radians from the start would probably have been wise choices. We'd have been much better equipped to divide things! I imagine a six-fingered being would have had an immediate advantage in that regard, but, alas. Now, would some of those number constants look particularly different? Not really. Pi in base 12 is "3.184809493B91866", for instance, so it doesn't look like that would be much easier. E and other numbers similarly just end up with different expansions. Remember, you can use whatever number base you want to, in this universe. The key is that it's just a way your brain interprets the symbols to represent a quantity; don't confuse the map for the territory. Five, the quality of having five whole entities, exists the same when it's 101 in binary or 10 in base 5 or 11 in base 4; either way it's all still just five, and so the right thing to do is to use the base system that most intuitively works for you so that it becomes transparent.
ckemere 6y ago
Presumably in a different number base (base 12, eg) it would be a different number that had this reciprocal property. Would it still be in the fibonacci series in that base?
- stabbles 6y ago
  Compute the generating function f(x) = sum_n f_n x^n where f_n satisfies your favorite recursion. Compute f(b^-1) where b is your base. For fibonacci f(x) = x / (1 - x - x^2).
- core-questions 6y ago
  This was my thought too. If 1/89 in base 10 works, and 89 in the 10th unique term of the series > 1,1,2,3,5,8,13,21,34,55,89,144,... then does 1/144 work in base 11, and 1/233 in base 12? Assuming we of course translate the base-10 number 144 to the appropriate base-11 number (121) first. I'm too bad at math to do this anymore, maybe someone else can tell us ;)
carlob 6y ago
Fibonacci died in 1242, so 13th century not 15th.
- alisonkisk 6y ago
  8th Fibbonacci century. (0 1 1 2 3 5 8 13)
tetris11 6y ago
oh weird, I accidentally "discovered" this when I was a kid, and assumed it was known for a long time... but 1994?
stabbles 6y ago
For people who are generally interested in these kind of identities relating a fraction to a recursion, please check out the generatingfunctionology book [1]. The basic idea is to define f(x) = sum_n f_n x^n where f_n satisfies some recursion equation. Very often you can find f(x) as p(x) / q(x) where p and q are polynomials in x. Now you can simply evaluate f(0.1) or more generally f(b^-l) where b is a base and l a positive integer, which gives you on the left-hand side your rational number as a fraction, and on the right-hand side the decimal expansion. [1] https://www.math.upenn.edu/~wilf/gfologyLinked2.pdf https://www.math.upenn.edu/~wilf/gfologyLinked2.pdf
chrismorgan 6y ago
On the decimal expansion part, 1⁄7 has always fascinated me, having something very similar going on. Doubling from 7, you get 14, 28, 56; and 1⁄7 is 0.1̅4̅2̅8̅5̅7̅, 2⁄7 is 0.2̅8̅5̅7̅1̅4̅, 3⁄7 is 0.4̅2̅8̅5̅7̅1̅, &c. (just changing which digit you start the recurring sequence with). https://en.wikipedia.org/wiki/142,857 https://en.wikipedia.org/wiki/142,857 talks about it a bit more; the doubling sequence thing is covered in the section 1⁄7 as an infinite sequence (including the reason the recurring decimal has 57 instead of 56—that 56 doubled is 112, so the hundred there overlaps with the six, much as the ten of the 13 adds to the 8 in the 1⁄89 expansion of this article).
- strictnein 6y ago
  That really is rather fascinating.
- NickHoff 6y ago
  I've always liked 1/7 as well, and I never realized that thing about doubling from 7 giving 14, 28, 56. People are always impressed when I can rattle off the digits of x/7. By the way, I appreciate your use of U+0305 combining overline. Did you enter those manually or do you have some neat way of doing it?
  - red_phone 6y ago
    To me, the fact that you’re in an environment where knowing the digits of x/7 is appreciated is the most-impressive thing of all!
  - chrismorgan 6y ago
    Heh, last time I spouted about 1⁄7 here someone asked about the overline too: https://news.ycombinator.com/item?id=24302104 https://news.ycombinator.com/item?id=24302104. I love my Compose key.
    - foobarian 6y ago
      How do you get one of those. Track down an old Sun keyboard? :)
quickthrower2 6y ago
Ah makes sense if you think of it as 1 / (100 - 10 - 1) 89 being a fib itself is quite a coincidence though?
jorgenveisdal 6y ago
So strange. Published another post about the same topic an hour before you
aqme28 6y ago
This is even more clear with larger expansions like `1/99989999 =1.00010002000300050008001300210034005500890144...×10−8` See discussion on math overflow here: https://math.stackexchange.com/questions/656183/why-does-frac1-99989999-generate-the-fibonacci-sequence https://math.stackexchange.com/questions/656183/why-does-fra...
karmakaze 6y ago
An interesting property. Seeing the proof is like seeing how a magic trick is done--ah of course it works like that.
grahamlee 6y ago
Did nobody tell Fibonacci that rabbits are mortal?
- asveikau 6y ago
  I think it would be built into the curve or sequence that the new births replace the dying ones. Obviously though they would have limited resources and couldn't keep growing indefinitely.
anonu 6y ago
This is cool. The proof is a bit over my head though. How does someone go about learning or dissecting the syntax of this proof...?
- qorrect 6y ago
  As a self-studyer, I'm a big fan of this book https://www.people.vcu.edu/~rhammack/BookOfProof/ https://www.people.vcu.edu/~rhammack/BookOfProof/ .
- LolWolf 6y ago
  Here's a simple way of looking at it (and we'll build up to the matrix notation). The nth and (n+1)st fibonacci numbers can be written as a system of two equations in the nth and (n-1)st numbers: F(n) = F(n) F(n+1) = F(n) + F(n-1) where the base cases are F(0) = 0 and F(1) = 1. The question is then to show that (note that the indices under F are actually wrong in the OP...): 1/89 = .01F(2) + .001F(3) + .0001F(4) + ... + 10^(-n)F(n) + ... One beautiful (and mathematically simple) way of analyzing this system is by the use of linear algebra. The idea is that, because F(n+1) depends linearly on F(n) and F(n-1) (see the above definition), then we can write the previous system with the following (linear-algebraic) notation [ F(n) ] = [ 0 1 ] [ F(n-1) ] [ F(n+1) ] = [ 1 1 ] [ F(n) ]. If we write x(n+1) as the vector (F(n), F(n+1)), i.e., first entry is F(n) and the second entry is F(n+1), and the matrix as A, then x(n+1) = Ax(n). (Note that A is the matrix whose entries are exactly the coefficients of the linear equation we gave above!) In other words, we've reduced the problem down to the question of investigating the properties of the matrix A ! Now, the sum we were looking at, originally, can be written (in terms of x(n)) as the first entry of (note that x is a vector as we've defined it!) x(2) + .1x(3) + .01x(4) + ... = x(2) + .1Ax(2) + .01 A Ax(3) + ... = x(2) + (.1A)x(2) + (.1A)^2 x(3) + ... There's a slick proof (see [0]) that, if this sequence converges, then its result is given by the first entry of (I - .1A)^(-1)x(2), which is exactly what the result gives. (I have changed the normalization a little bit for convenience, but it is the same proof :) ----- [0] If y + By + B^2y + ... = z converges, then we can multiply both sides by B to get By + B^2y + ... = Bz But, here's the magic! Let's subtract the first equation from the second to get y = z - Bz = (I - B)z, so, multiplying by the inverse of B on both sides, we get: y + By + B^2y + ... = z = (I - B)^(-1)y, as required!
sabujp 6y ago
fibonacci retracements are used heavily in trading, i wonder how prices/averages line up so nicely around those points
sabujp 6y ago
btw, there's a typo on the page for the value of 1/89, it should be : .01123595505617977528 but the math does work of course, it converges : >>> .01 + .001 + .0002 + .00003 + .000005 + .0000008 + .00000013 + .000000021 + .0000000034 + .00000000055 + .000000000089 + .0000000000144 + .00000000000233 + .000000000000377 0.011235955056107002
adenadel 6y ago
I think I found a typo in the final line (not impactful for the result, however). 11/89 should be 11/8900 This can be verified by using the following Sage code (simply go to https://sagecell.sagemath.org/ https://sagecell.sagemath.org/ to avoid installing sage) b = 10 A = Matrix([[0, 1],[1/(b**2), 1/b]]) I = matrix.identity(2) show((I - A).inverse() * Matrix([[1/b**2],[1/b**3]]))
adenadel 6y ago
In case anyone is interested, I ran this procedure using integer bases besides 10 and obtained the following sequence 1, 5, 11, 19, 29, 41, 55, 71, 89, ... Searching it on the OEIS (a great resource for mathematics) gave http://oeis.org/search?q=1%2C5%2C+11%2C+19%2C+29%2C+41%2C+55%2C+71%2C+89&language=english&go=Search http://oeis.org/search?q=1%2C5%2C+11%2C+19%2C+29%2C+41%2C+55... It turns out that these are precisely the first values of the Fibonacci polynomial n^2 - n - 1 I haven't verified this fact, but it seems like it comes from an application of the generating function of the Fibonacci numbers. I posted my code in another comment https://news.ycombinator.com/item?id=24933085 https://news.ycombinator.com/item?id=24933085
justinlloyd 6y ago
I went to college in '88 or '89 and one of my teachers showed me the 1/89 trick, along with a few others, e.g. 1/7, and so forth. My teacher claimed to have been shown the various tricks by one of his teachers back in the '60s. In the same period I "discovered" an error detection technique which is commonly known as Hamming codes. I would never dare to claim I invented them or discovered them. In '92 and '93, I was heavily in to the BBS scene, some people had 56K modems, others had 14.4K. With verifiable evidence of written notes and digital artifacts (a BBS door and protocol for AmiBBS and Citadel and a couple of others) , I created a technique whereby multiple peers with low-bandwidth connections could transfer small fragments of a larger dataset to a peer with a lot of bandwidth. If you were heavily in to the warez scene and/or part of Fate, it is probably you made use of this protocol to transfer pirated software between FTP sites. A warehousing server would tell each peer who had what part of a piece of data, and any peer could make requests for any piece of the data from any other peer who happened to have a copy of that data. Today, a very similar protocl is commonly known as BitTorrent. I would not say I "discovered" or "invented" the protocol as my work was based on the various X-, Y- & Z- modem protocols. There was a TCP/IP packet to -Modem packet translator so that a BBS talking over that new fangled internet thing could take advantage of a T1 (1.5Mbps) connection for instance, which really helped with spreading the warez around the various FTP sites by the couriers. I doubt the veracity of the claim by the author to have "discovered it as original" in 1994 before anybody else.
hellokaushik 6y ago
Nice
Jeff_Brown 6y ago
It had to add up to something. The surprise to me is that it's a rational number.
markc 6y ago
Question (perhaps naive): Given the decimal expansion of 1/89, is there any way to directly retrieve the Fibonacci sequence from it? (that is, not using any knowledge of the sequence itself) I'm assuming it can't be done because different sets of fractions can sum to 1/89, but maybe I'm missing something.
mjcohen 6y ago
Another surprising fraction: 987654312/123456789 = 8 And this works in any base.
Konohamaru 6y ago
What's more interesting is that the sum is periodic.
syrrim 6y ago
I noticed a while ago that the powers of 1001 encode the successive rows of pascal's triangle/the binomial coefficients. This is not so surprising, since we are effectively taking powers of the polynomial (1+x), but replacing x with 1000 (clearly it works the same if we use any other power of ten instead). I wonder if we can find a relationship to that here. We might start by looking at this relation: 1/(1-x) = 1 + x + x^2 + x^3 + ... Hopefully this will yield an operator x whose succesive powers are the fibonacci numbers. Take .01/(1-x) = 1/89, then x = 0.11. Actually, the powers of x, just like 1001 above, will yield rows of pascal's triangle. So the taylor expansion above tells us that F(k) = Σ(n=0..k-1) B(n, k), in other words that each fibonacci number is the sum of a diagonal of pascal's triangle (like here: https://cdn1.byjus.com/wp-content/uploads/2018/11/maths/2016/08/01092801/PascalTriangleFibonacci.png https://cdn1.byjus.com/wp-content/uploads/2018/11/maths/2016...) More generally, we can compute numbers with decimal expansions of the fibonacci numbers with 10^-2n / (1 - 10^-n - 10^-2n). Notice that this is just the z-transform of the recurrence relation of the fibonacci series, with 10^n replacing z: Z(f(n)) = Z(f(n-1) + f(n-2) + δ(n-2)) (δ is the kronecker delta) F(z) = z^-2 / (1 - z^-1 - z^-2) The taylor expansion of this expression has coefficients equal to the terms of the fibonacci sequence - which makes sense, because that's the definition of the z-transform. We can, with a little rearranging, get an explicit formula for the fibonacci sequence from it too: Take φ± = (1 ± √5)/2 Then F(z) = z^2/(z - φ+)/(z - φ-) = ( φ+ z/(z - φ+) - φ- z/(z - φ-) )/√5 (by partial fraction decomposition) Z^-1(F(z)) = f(n) = (φ+^(n+1) - φ-^(n+1))/√5

413 points by nonsapreiche 6y ago

Okay, as a non-mathematician, I see something like this and I think... “neat coincidence?” But the world of numbers seems to be full of these neat coincidences. So do any of the math folks here have a theory or explanation of why?

deleted 6y ago
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efdb 6y ago
In addition to this question I would like to know if you can in general say/proof that for every sequence which has some relation between the successive numbers there is a rational number whose decimal expansion is the same as the sequence.
- stabbles 6y ago
  A standard technique to find explicit expressions for the `n`th number of a recursion is through generation functions; see e.g. https://en.wikipedia.org/wiki/Generating_function https://en.wikipedia.org/wiki/Generating_function. You can plugin x = 10^-1 there. Not sure if the result is always a rational number.
- magnio 6y ago
  For a linearly recursive sequence x_0, x_1, x_(n+2) = ax_(n+1) + bx_n, the general formula for the terms is x_n = cα^n + dβ^n, where α, β are the roots of the quadratic x²−ax−b; c, d are solutions to the system c + d = x_0 cα + dβ = x_1. If the series Σ x_n⋅10^n converges then its value is 10c/(10−α) + 10d/(10−β) = ((100−10a)x_0 + 10x_1)/(100 − 10a − b). If a, b, x_0, x_1 are all rational then the above series converges to a rational number, too. This is the case for the Fibonacci sequence, with a=b=1, x_0=0 and x_1=1.
  - deleted 6y ago
    [deleted]
benrbray 6y ago
There's a "theorem" about this: > The interesting number paradox is a semi-humorous paradox which arises from the attempt to classify every natural number as either "interesting" or "uninteresting". The paradox states that every natural number is interesting. The "proof" is by contradiction: if there exists a non-empty set of uninteresting natural numbers, there would be a smallest uninteresting number – but the smallest uninteresting number is itself interesting because it is the smallest uninteresting number, thus producing a contradiction.
- wizzwizz4 6y ago
  That just sounds like another formulation of the Surprise Exam paradox.[0] It falls down when you realise that “the four hundred and seventieth otherwise uninteresting number” is not a particularly interesting number, so there must be a problem with the problem statement. [0]: https://en.wikipedia.org/wiki/Unexpected_hanging_paradox https://en.wikipedia.org/wiki/Unexpected_hanging_paradox
  - alisonkisk 6y ago
    It's a paradox related to "meta-logic", but it's different. Surprise exam is about temporal reasoning -- It's only impossible to be surprised on the last day (or else the premise of having an exam is invalidsted), and reasoning backwards in time from a contradiction is not valid. Uninteresting number is a simpler contradiction in definitions.
- DeusExMachina 6y ago
  This is not a paradox though, as your copy and paste states. It's just a theorem (as you stated) with a proof by contradiction. A paradox must be self-contradictory under all circumstances.
  - phizy 6y ago
    It is a paradox. It assumes you have a definition of uninteresting number such that you can select the least uninteresting number, and then retroactively defines that number to be interesting by brand new criteria, which contradicts that you would have ever selected it in the first place. Thus the axioms invoked are mutually contradictory: the axioms that allow you to identify the least uninteresting number, and the axioms you invoke to declare it interesting are in conflict.
bobbylarrybobby 6y ago
This is no coincidence — it’s because 89=100-10-1, and the ordinary generating function for the Fibonacci sequence is 1/(1-x-x^2) (if you go to wolfram alpha and Taylor expand that expression, you’ll see its coefficients are the Fibonacci numbers).
impendia 6y ago
Yes. One has the identity: 1/(1 - x - x^2) = 1 + x + 2x^2 + 3x^3 + ... where the Fibonacci numbers are the coefficients on the right. Try writing it out! The basic idea is that because F_n = F_{n - 1} + F_{n - 2}, everything will neatly cancel out. This is an example of a "generating function". Anyway, plug in x = 0.1, and then divide by 100 to see the behavior described in the post. One also gets 1/9899 = 0.00010102030508132134559046368320032326498... 1/998999 = 0.0000010010020030050080130210340550891442334... and so on.
- contravariant 6y ago
  More to the point this function trivially statisfies f(x)-1-x = X (f(x)-1) + X^2 f(X). You should be able to convince yourself that this is equivalent to the fact that its power series satisfies the Fibonacci recurrence, and that this power series starts with 1 + x. You can also use this to easily find generating functions that satisfy other starting conditions.
jerf 6y ago
Imagine the real number line is a database that you can run SELECT queries against, and you don't have to worry about giving a computational procedure that produces the result, it just magically gets produced. Now, imagine you write a query, like, say, "SELECT number WHERE number = .01 * FIB[1] + .001 * FIB[2] + .0001 * FIB[3]" and so on until you get what the article discusses. It isn't necessarily that surprising that you might find something with an uncountably infinite numbers to pick from. Now, consider all possible "interesting" queries you could run, along with all their results. There result is an inconceivably large sea of queries. Most of them are, in fact, utterly pointless; SELECT statements that return no values, SELECT statements that return all values (equally pointless), SELECT statements that return complicated sets of values but have essentially no mathematical interest because there is no practical way to represent them as anything smaller or more interesting, etc. In this massive sea of results, you should expect a lot of interesting things to exist. Finding them is tricky; in percentage terms they make up 0% of the results, but we have mechanisms for finding some of them. Basically, there are so infinitely many mathematical statements that there can't help but be a large supply of "interesting" statements like this. For an interesting view on that, see https://en.wikipedia.org/wiki/Mathematical_coincidence https://en.wikipedia.org/wiki/Mathematical_coincidence . These are true statements or almost true statements (near equalities) about a wide variety of numbers that are essentially meaningless... it's just there's so many ways of putting things together that there are inevitably large numbers of these things (the wiki page is just a sampling). You can even generate these mathematical coincidences yourself. Create a program that will systematically iterate over abstract syntax trees of mathematical expressions involving whatever combination of mathematical operators (+-×/, sqrt, log, sin, whatever) and numbers you like (the first ten integers, e, i, pi, whatever else you like), store up a table of results and emit any two expressions that are, say, within .01% of each other. You will rapidly find a ton of results, because it turns out that even with modest numbers of operators, there are far more mathematical expressions than there are small numbers for them to result in separated by more than .01%. If you think about it, this program can't help but emit a lot of results. Some of them will be humanly "interesting". A few of them will even be mathematically interesting (e.g., this procedure will generate the famous Euler identity relatively quickly if you included the relevant operators and numbers). On a larger scale, this is also known as the Strong Law of Small Numbers: https://en.wikipedia.org/wiki/Strong_Law_of_Small_Numbers https://en.wikipedia.org/wiki/Strong_Law_of_Small_Numbers The previous paragraph is a very bite-sized example of why this holds that you can code up yourself if you are interested.
alisonkisk 6y ago
You might also enjoy tan(1 degree/55555555555)
- huhtenberg 6y ago
  3.1415926536212091649988555782837470430464925667776193389556...e-13 Nice
- prezjordan 6y ago
  Fun video about this one on Numberphile https://www.youtube.com/watch?v=IMY2_yzDm9I https://www.youtube.com/watch?v=IMY2_yzDm9I
  - dylan604 6y ago
    "they say there's a fine line between a numerator and a denominator" --? another one of those maths jokes that makes me smile and confuses a lot of other people of why i'm laughing
- paulpauper 6y ago
  not impressive given that more 5s are needed than correct decimals produced
JackFr 6y ago
The Penguin Book of Curious and Interesting Numbers https://www.amazon.com/Penguin-Book-Curious-Interesting-Numbers/dp/0140261494/ref=tmm_pap_swatch_0?_encoding=UTF8&qid=&sr= https://www.amazon.com/Penguin-Book-Curious-Interesting-Numb... The caveat here is "for some value of 'interesting'."

doodpants 6y ago

"The successive ratios of the terms, i.e. 1/1, 2/1, 3/2, 5/3 ... tend to a number called the Golden Ratio by the Greeks." Fun fact: take any two numbers (e.g. chosen randomly), and use them as the seeds for a Fibonacci-like sequence by summing the last two terms to generate the next term. The ratio of any two consecutive terms in that series will tend towards the golden ratio.

cscheid 6y ago
If you read this and you’re curious, there’s a proof that is fairly easy to follow if you understand eigenvectors. Write the operation that takes the two last elements of the sequence and produces the following two, notice it’s linear, then analyze the eigenvalues of the associated matrix and relate the original operation to the power method.
- redvenom 6y ago
  You actually don't need eigenvectors for this proof. Sketch: Write x = lim a_n/a_{n-1} where a_n is the nth term of Fibonacci. Replace a_n with a_{n-1} + a_{n-2}, which will give you a quadratic equation in x. Solving this quadratic equation gives you the golden ratio.
  - cscheid 6y ago
    I believe that gives you the result starting from Fibonacci, but not that any starting pair of numbers eventually lands at the golden ratio.
    - redvenom 6y ago
      No, it works for any pair of numbers, as long as they are not both zero. I am only using the recursive relation of the Fibonacci sequence, not the starting terms.
cleansingfire 6y ago
This seems like a pretty basic calculus problem. Constants become irrelevant at the limit, so that the relationship (ratio) is all that's left.

jurajvt 6y ago

1/89 = .01123595505..., not .01123595595..., small typo

wumms 6y ago

> As you can easily check, 1/89 = .01123595595... Bizarre, eh? Typo: 1/89 = .01123595505...

ache7 6y ago

Any number is remarkable if you're smart enough to find why it's remarkable.

eythian 6y ago
And if it's not remarkable, then that is in itself remarkable.
- forinti 6y ago
  Aren't most transcendental numbers unremarkable?
  - grahamlee 6y ago
    No number is unremarkable. Let's construct the set of all unremarkable numbers. Now, let's construct the sequence of those numbers in order. The first member of that sequence has the remarkable property that it is the smallest unremarkable number. That is remarkable, so remove it from the set. By induction, the set must be empty.
    - jlokier 6y ago
      > Now, let's construct the sequence of those numbers in order. The first member of that sequence Not happening. Sets of numbers do not always have a first in order. Consider the set of unremarkable real numbers > 0 under the regular arithmetic ordering. Which one is first?
    - gnulinux 6y ago
      That doesn't work because real numbers are not enumerable, so you cannot induce over them. That joke "proof" only works for natural numbers and goes like this: Theorem: all natural numbers are interesting * Base case: 0 is interesting because it is the smallest natural number, as well as the identity element of + operation. * Inductive case: Assume the theorem holds for all m, m<n. Take n. If it is not interesting, then n is the smallest non-interesting number. But that's interesting because it's the smallest such number. Therefore it cannot be non-interesting. Therefore theorem holds for n. By induction, we conclude all natural numbers are interesting. QED.

wumms 6y ago

From an archived talk (2011) on wikipedia [0]: "The linked page misleadingly suggests that a certain Cody Birsner discovered the relationship between the series and the fraction, whereas it had been known for a considerable time before". Günter Köhler, 1983 (published in the The Fibonacci Quarterly, 1985; who cites earlier papers from 1977 and 1981): https://www.fq.math.ca/Scanned/23-1/kohler.pdf https://www.fq.math.ca/Scanned/23-1/kohler.pdf [0] https://en.wikipedia.org/wiki/Talk%3AFibonacci_number%2FArchive_3#1/89 https://en.wikipedia.org/wiki/Talk%3AFibonacci_number%2FArch...

alisonkisk 6y ago
It's a bit silly to chase down original authorship of an idea that is a minor detail visible to many people who work in a field. It's like asking who was the first person to discover that all multiples of 11 have the same parity in the respective sums of their odd and even digits.
- kazinator 6y ago
  Since there had to be a first person who discovered that all multiples of 11 have the same parity in the respective sums of their odd and even digits, that question is obviously interesting to some historians. Due to what circumstances was that person first? What hindered others before that person? How long was the lag before use of decimal and the discovery?
  - munificent 6y ago
    It's important to remember that while technically there is a chronologically first person to discover X for all X, that doesn't imply that that person is the only person to discover X. For sufficiently obvious X, there are likely to be many independent discoverers and highlighting the chronologically first one heaps praise somewhat arbitrarily on one of them.
    - chowned 6y ago
      "History isn't perfect" is not a good reason to avoid doing history. Historians are all well aware of these facts.

ckemere 6y ago

Presumably in a different number base, the number would be different. Would it still be in the aeries?

Cullinet 6y ago
Without any basis whatsoever, and I really must one day put this idea out of its misery with some studying, I long suspected that quantum mechanics involved parallel universes where different bases more aptly fit with that other reality. My thought is surely crackpot but I'll explain how my idea arose : A fraction eg 1/3 describes a decimal number to infinite accuracy but creates a challenge for base 10 calculations. I thought about the precision necessary for learning the math of the quantum particles and the experimental way we're trying to figure out how they behave to infer their properties. (Higgs was the other way around but I am optimistic that we'll predict much more in future instead of this convoluted observations rigmarole) and I started wondering if you could approximate extremely high precision decimals to fractions in non decimal bases and from there simplify calculations with far greater resolution. How much more capable would Nyquist - Shannon sampling, if we could clock with the precision of infinite decimal fp digits but handle only a short simple "one third" input or do visor? My silly mind wandered off to imagine particles jumping between different base based universes just as a sequential progression through the precision of their infinitessimal steps through space and time.
- whatshisface 6y ago
  Base 10 doesn't fit particularly well with our reality, it was picked for historical reasons. Every other base works just as well.
  - philsnow 6y ago
    including fractional bases (rational or irrational) and even negative bases, I think.
- core-questions 6y ago
  This is my favourite kind of pure speculation. You don't know what you're talking about, really, but you're clearly exploring vast areas of thought at the same time. Remember, integers are integers are integers, because they represent the intrinsic "whole quantity" of something; this is as concrete as logic will get, the idea that there are "ones", it's pretty hard to imagine a universe that doesn't have that. Once you have integers, then you're going to do math in an integer base; and making too high of a base has a diminishing value at at certain point, so it's unlikely we'd see higher than maybe 60. Non-integer bases exist - https://en.wikipedia.org/wiki/Non-integer_base_of_numeration https://en.wikipedia.org/wiki/Non-integer_base_of_numeration - but it's clear to me that I'm too stupid to use them, and so probably most other people are too. This tells me that it's going to be a comparatively rare Many World that chooses to do this. Choosing a number like 12 or 60 with a lot of divisors would have been nice. 1/3 in base 12 is "0.4", which is a lot nicer than 0.333... and would probably help make a lot of younger math education way easier. Combining that with skipping "degrees" entirely and using radians from the start would probably have been wise choices. We'd have been much better equipped to divide things! I imagine a six-fingered being would have had an immediate advantage in that regard, but, alas. Now, would some of those number constants look particularly different? Not really. Pi in base 12 is "3.184809493B91866", for instance, so it doesn't look like that would be much easier. E and other numbers similarly just end up with different expansions. Remember, you can use whatever number base you want to, in this universe. The key is that it's just a way your brain interprets the symbols to represent a quantity; don't confuse the map for the territory. Five, the quality of having five whole entities, exists the same when it's 101 in binary or 10 in base 5 or 11 in base 4; either way it's all still just five, and so the right thing to do is to use the base system that most intuitively works for you so that it becomes transparent.

ckemere 6y ago

Presumably in a different number base (base 12, eg) it would be a different number that had this reciprocal property. Would it still be in the fibonacci series in that base?

stabbles 6y ago
Compute the generating function f(x) = sum_n f_n x^n where f_n satisfies your favorite recursion. Compute f(b^-1) where b is your base. For fibonacci f(x) = x / (1 - x - x^2).
core-questions 6y ago
This was my thought too. If 1/89 in base 10 works, and 89 in the 10th unique term of the series > 1,1,2,3,5,8,13,21,34,55,89,144,... then does 1/144 work in base 11, and 1/233 in base 12? Assuming we of course translate the base-10 number 144 to the appropriate base-11 number (121) first. I'm too bad at math to do this anymore, maybe someone else can tell us ;)

carlob 6y ago

Fibonacci died in 1242, so 13th century not 15th.

alisonkisk 6y ago
8th Fibbonacci century. (0 1 1 2 3 5 8 13)

tetris11 6y ago

oh weird, I accidentally "discovered" this when I was a kid, and assumed it was known for a long time... but 1994?

stabbles 6y ago

For people who are generally interested in these kind of identities relating a fraction to a recursion, please check out the generatingfunctionology book [1]. The basic idea is to define f(x) = sum_n f_n x^n where f_n satisfies some recursion equation. Very often you can find f(x) as p(x) / q(x) where p and q are polynomials in x. Now you can simply evaluate f(0.1) or more generally f(b^-l) where b is a base and l a positive integer, which gives you on the left-hand side your rational number as a fraction, and on the right-hand side the decimal expansion. [1] https://www.math.upenn.edu/~wilf/gfologyLinked2.pdf https://www.math.upenn.edu/~wilf/gfologyLinked2.pdf

chrismorgan 6y ago

On the decimal expansion part, 1⁄7 has always fascinated me, having something very similar going on. Doubling from 7, you get 14, 28, 56; and 1⁄7 is 0.1̅4̅2̅8̅5̅7̅, 2⁄7 is 0.2̅8̅5̅7̅1̅4̅, 3⁄7 is 0.4̅2̅8̅5̅7̅1̅, &c. (just changing which digit you start the recurring sequence with). https://en.wikipedia.org/wiki/142,857 https://en.wikipedia.org/wiki/142,857 talks about it a bit more; the doubling sequence thing is covered in the section 1⁄7 as an infinite sequence (including the reason the recurring decimal has 57 instead of 56—that 56 doubled is 112, so the hundred there overlaps with the six, much as the ten of the 13 adds to the 8 in the 1⁄89 expansion of this article).

strictnein 6y ago
That really is rather fascinating.
NickHoff 6y ago
I've always liked 1/7 as well, and I never realized that thing about doubling from 7 giving 14, 28, 56. People are always impressed when I can rattle off the digits of x/7. By the way, I appreciate your use of U+0305 combining overline. Did you enter those manually or do you have some neat way of doing it?
- red_phone 6y ago
  To me, the fact that you’re in an environment where knowing the digits of x/7 is appreciated is the most-impressive thing of all!
- chrismorgan 6y ago
  Heh, last time I spouted about 1⁄7 here someone asked about the overline too: https://news.ycombinator.com/item?id=24302104 https://news.ycombinator.com/item?id=24302104. I love my Compose key.
  - foobarian 6y ago
    How do you get one of those. Track down an old Sun keyboard? :)

quickthrower2 6y ago

Ah makes sense if you think of it as 1 / (100 - 10 - 1) 89 being a fib itself is quite a coincidence though?

jorgenveisdal 6y ago

So strange. Published another post about the same topic an hour before you

aqme28 6y ago

This is even more clear with larger expansions like `1/99989999 =1.00010002000300050008001300210034005500890144...×10−8` See discussion on math overflow here: https://math.stackexchange.com/questions/656183/why-does-frac1-99989999-generate-the-fibonacci-sequence https://math.stackexchange.com/questions/656183/why-does-fra...

karmakaze 6y ago

An interesting property. Seeing the proof is like seeing how a magic trick is done--ah of course it works like that.

grahamlee 6y ago

Did nobody tell Fibonacci that rabbits are mortal?

asveikau 6y ago
I think it would be built into the curve or sequence that the new births replace the dying ones. Obviously though they would have limited resources and couldn't keep growing indefinitely.

anonu 6y ago

This is cool. The proof is a bit over my head though. How does someone go about learning or dissecting the syntax of this proof...?

qorrect 6y ago
As a self-studyer, I'm a big fan of this book https://www.people.vcu.edu/~rhammack/BookOfProof/ https://www.people.vcu.edu/~rhammack/BookOfProof/ .
LolWolf 6y ago
Here's a simple way of looking at it (and we'll build up to the matrix notation). The nth and (n+1)st fibonacci numbers can be written as a system of two equations in the nth and (n-1)st numbers: F(n) = F(n) F(n+1) = F(n) + F(n-1) where the base cases are F(0) = 0 and F(1) = 1. The question is then to show that (note that the indices under F are actually wrong in the OP...): 1/89 = .01F(2) + .001F(3) + .0001F(4) + ... + 10^(-n)F(n) + ... One beautiful (and mathematically simple) way of analyzing this system is by the use of linear algebra. The idea is that, because F(n+1) depends linearly on F(n) and F(n-1) (see the above definition), then we can write the previous system with the following (linear-algebraic) notation [ F(n) ] = [ 0 1 ] [ F(n-1) ] [ F(n+1) ] = [ 1 1 ] [ F(n) ]. If we write x(n+1) as the vector (F(n), F(n+1)), i.e., first entry is F(n) and the second entry is F(n+1), and the matrix as A, then x(n+1) = Ax(n). (Note that A is the matrix whose entries are exactly the coefficients of the linear equation we gave above!) In other words, we've reduced the problem down to the question of investigating the properties of the matrix A ! Now, the sum we were looking at, originally, can be written (in terms of x(n)) as the first entry of (note that x is a vector as we've defined it!) x(2) + .1x(3) + .01x(4) + ... = x(2) + .1Ax(2) + .01 A Ax(3) + ... = x(2) + (.1A)x(2) + (.1A)^2 x(3) + ... There's a slick proof (see [0]) that, if this sequence converges, then its result is given by the first entry of (I - .1A)^(-1)x(2), which is exactly what the result gives. (I have changed the normalization a little bit for convenience, but it is the same proof :) ----- [0] If y + By + B^2y + ... = z converges, then we can multiply both sides by B to get By + B^2y + ... = Bz But, here's the magic! Let's subtract the first equation from the second to get y = z - Bz = (I - B)z, so, multiplying by the inverse of B on both sides, we get: y + By + B^2y + ... = z = (I - B)^(-1)y, as required!

sabujp 6y ago

fibonacci retracements are used heavily in trading, i wonder how prices/averages line up so nicely around those points

sabujp 6y ago

btw, there's a typo on the page for the value of 1/89, it should be : .01123595505617977528 but the math does work of course, it converges : >>> .01 + .001 + .0002 + .00003 + .000005 + .0000008 + .00000013 + .000000021 + .0000000034 + .00000000055 + .000000000089 + .0000000000144 + .00000000000233 + .000000000000377 0.011235955056107002

adenadel 6y ago

I think I found a typo in the final line (not impactful for the result, however). 11/89 should be 11/8900 This can be verified by using the following Sage code (simply go to https://sagecell.sagemath.org/ https://sagecell.sagemath.org/ to avoid installing sage) b = 10 A = Matrix([[0, 1],[1/(b**2), 1/b]]) I = matrix.identity(2) show((I - A).inverse() * Matrix([[1/b**2],[1/b**3]]))

adenadel 6y ago

In case anyone is interested, I ran this procedure using integer bases besides 10 and obtained the following sequence 1, 5, 11, 19, 29, 41, 55, 71, 89, ... Searching it on the OEIS (a great resource for mathematics) gave http://oeis.org/search?q=1%2C5%2C+11%2C+19%2C+29%2C+41%2C+55%2C+71%2C+89&language=english&go=Search http://oeis.org/search?q=1%2C5%2C+11%2C+19%2C+29%2C+41%2C+55... It turns out that these are precisely the first values of the Fibonacci polynomial n^2 - n - 1 I haven't verified this fact, but it seems like it comes from an application of the generating function of the Fibonacci numbers. I posted my code in another comment https://news.ycombinator.com/item?id=24933085 https://news.ycombinator.com/item?id=24933085

justinlloyd 6y ago

I went to college in '88 or '89 and one of my teachers showed me the 1/89 trick, along with a few others, e.g. 1/7, and so forth. My teacher claimed to have been shown the various tricks by one of his teachers back in the '60s. In the same period I "discovered" an error detection technique which is commonly known as Hamming codes. I would never dare to claim I invented them or discovered them. In '92 and '93, I was heavily in to the BBS scene, some people had 56K modems, others had 14.4K. With verifiable evidence of written notes and digital artifacts (a BBS door and protocol for AmiBBS and Citadel and a couple of others) , I created a technique whereby multiple peers with low-bandwidth connections could transfer small fragments of a larger dataset to a peer with a lot of bandwidth. If you were heavily in to the warez scene and/or part of Fate, it is probably you made use of this protocol to transfer pirated software between FTP sites. A warehousing server would tell each peer who had what part of a piece of data, and any peer could make requests for any piece of the data from any other peer who happened to have a copy of that data. Today, a very similar protocl is commonly known as BitTorrent. I would not say I "discovered" or "invented" the protocol as my work was based on the various X-, Y- & Z- modem protocols. There was a TCP/IP packet to -Modem packet translator so that a BBS talking over that new fangled internet thing could take advantage of a T1 (1.5Mbps) connection for instance, which really helped with spreading the warez around the various FTP sites by the couriers. I doubt the veracity of the claim by the author to have "discovered it as original" in 1994 before anybody else.

hellokaushik 6y ago

Nice

Jeff_Brown 6y ago

It had to add up to something. The surprise to me is that it's a rational number.

markc 6y ago

Question (perhaps naive): Given the decimal expansion of 1/89, is there any way to directly retrieve the Fibonacci sequence from it? (that is, not using any knowledge of the sequence itself) I'm assuming it can't be done because different sets of fractions can sum to 1/89, but maybe I'm missing something.

mjcohen 6y ago

Another surprising fraction: 987654312/123456789 = 8 And this works in any base.

Konohamaru 6y ago

What's more interesting is that the sum is periodic.

syrrim 6y ago

I noticed a while ago that the powers of 1001 encode the successive rows of pascal's triangle/the binomial coefficients. This is not so surprising, since we are effectively taking powers of the polynomial (1+x), but replacing x with 1000 (clearly it works the same if we use any other power of ten instead). I wonder if we can find a relationship to that here. We might start by looking at this relation: 1/(1-x) = 1 + x + x^2 + x^3 + ... Hopefully this will yield an operator x whose succesive powers are the fibonacci numbers. Take .01/(1-x) = 1/89, then x = 0.11. Actually, the powers of x, just like 1001 above, will yield rows of pascal's triangle. So the taylor expansion above tells us that F(k) = Σ(n=0..k-1) B(n, k), in other words that each fibonacci number is the sum of a diagonal of pascal's triangle (like here: https://cdn1.byjus.com/wp-content/uploads/2018/11/maths/2016/08/01092801/PascalTriangleFibonacci.png https://cdn1.byjus.com/wp-content/uploads/2018/11/maths/2016...) More generally, we can compute numbers with decimal expansions of the fibonacci numbers with 10^-2n / (1 - 10^-n - 10^-2n). Notice that this is just the z-transform of the recurrence relation of the fibonacci series, with 10^n replacing z: Z(f(n)) = Z(f(n-1) + f(n-2) + δ(n-2)) (δ is the kronecker delta) F(z) = z^-2 / (1 - z^-1 - z^-2) The taylor expansion of this expression has coefficients equal to the terms of the fibonacci sequence - which makes sense, because that's the definition of the z-transform. We can, with a little rearranging, get an explicit formula for the fibonacci sequence from it too: Take φ± = (1 ± √5)/2 Then F(z) = z^2/(z - φ+)/(z - φ-) = ( φ+ z/(z - φ+) - φ- z/(z - φ-) )/√5 (by partial fraction decomposition) Z^-1(F(z)) = f(n) = (φ+^(n+1) - φ-^(n+1))/√5