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Thread: Ten is the lovliest number...

  1. #26
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    Registered: Jun 2001
    Location: under God's grace
    Quote Originally Posted by Nicker View Post
    Thanks for the first paragraph, Dema. That makes sense. It doesn't entirely alleviate my confusion but it's one fewer mole to whack.
    Examples might help to see the pattern.

    Binary 10101011.0 is decimal 128 + 32 + 8 + 2 + 1 = 171. Moving the decimal point to the left by one divides all the terms by two: 1010101.10 = 64 + 16 + 4 + 1 + 1/2 = 85.5.

    Code:
    bin        | dec
    ___________|_____________________________________________________________
               |
    10101011.0 | 128 +   0 +  32 +   0 +   8 +   0 +   2 +   1 +   0 = 171
    1010101.10 |  64 +   0 +  16 +   0 +   4 +   0 +   1 + 1/2 +   0 = 85.5
    101010.110 |  32 +   0 +   8 +   0 +   2 +   0 + 1/2 + 1/4 +   0 = 42.75
    10101.0110 |  16 +   0 +   4 +   0 +   1 +   0 + 1/4 + 1/8 +   0 = 21.375
    The real difficulties arise when converting from one base to another. At least for me I'm usually thinking in decimal, even when working with binary or hexadecimal. I convert the value to decimal to "understand" what it is. But some values that can be expressed in some bases cannot be expressed in others. For example decimal 0.3 cannot be finitely expressed in binary. And trinary 0.1 cannot be finitely expressed in decimal.
    Last edited by Qooper; 15th Oct 2023 at 17:05.

  2. #27
    Member
    Registered: Dec 2020
    Something I got interested in recently is 10-adic numbers and the wider p-adic numbers.

    To give an idea, it's math that looks at what happens when you have infinite digits going to the left, not just infinite digits going to the right.

    First think of infinite digits after the decimal point. You can represent this with a geometric series:

    Code:
    0.9999 ... = 9/10 + 9/100 + 9/1000 + 9/10000 ...
    You can work out the sum of an infinite geometric series as

    Code:
    S = a / (1-r)
    ratio r = 0.1 and first term a = 9/10
    S = 9/10 / (0.9) = 0.9 / 0.9 = 1
    There are a number of other proofs that 0.999... = 1, including

    Code:
    S = 0.999 
    S * 10 = 9.99999
    S * 10 - S = 9.99999 - 0.999 
    S * 10 - S = 9
    9 S = 9
    S = 1
    Ok so what do we get when S = ...9999, i.e. infinite 9s going to the LEFT?

    In this case it's still an infinite series:

    Code:
    9 + 90 + 900 + 9000, so a = 9, and r = 10
    We can use the first trick from before:

    Code:
    S = a / (1-r)
    S = 9 / (1-10)
    S = 9 / -9
    S = -1
    So ... fanfare if you add all 9s to the LEFT you get -1

    Let's try the second method:

    Code:
    S = ...999 
    10S = ...9990
    S-10S = 9
    -9S = 9
    S = -1
    Third let's do manual addition. If a number equal -1, then adding 1 to it, should give ZERO

    So we add 1 to the first 9, that becomes 0, carry the one. Then, the 1 carry adds to the next 9, turning it into zero, carry the 1, and so on. In fact adding 1 to the number DOES produce an infinite row of zeros.

    So, three different ways show that infinite 9s to the left actually equal negative one.

    This is a really interesting type of math, because while we might say it's undefined, it's actually well behaved compared to things like 0/0, because you can prove 0/0 equals anything, so it has no clear correct value, whereas these infinite divergent series when you try and compute an actual bound often give you the same real finite result no matter how you approach them.

    So in some sense infinite ...999 can be said to be a way to *represent* the value -1. You'd expect that doing standard addition with it should give you a real result, for example adding it to 14:

    Code:
        14+
    ...999
    ------
    000013
    we add the right-most 9 to 4, and get 13. Write the 3 then carry the 1. The next 9 and the 1 add up to 10, so we add 10 to the 1, which is 11, write the 1, carry the 1. The remaining 9s then cancel each other out, leaving the value 13, the correct result if we assume infinite 9s = -1.
    Last edited by Cipheron; 8th Nov 2023 at 09:45.

  3. #28
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    Registered: Jan 2003
    Location: NeoTokyo
    That's great!
    I always got a kick out of the proof that 1+2+3+4+... out to infinity = -1/12, known as Ramaujan's summation, or the most famous exemplar of his summation method.



    It seems surprising or unintuitive, to say the least.
    Well while this thread is going, let's look at it!

    Here's the quick and dirty proof. When you take a sum out to infinity, you're not really adding all of the numbers up to any point. You're taking the amount the sum is averaging towards at the limit of infinity. So the classic quick proof uses three sums:
    S1 = 1-1+1-1...
    S2 = 1-2+3-4+5... and finally
    S3 = 1+2+3+4+...

    So for S1, the sum is going back and forth between 1 and 0. So at the limit of infinity you just take the average of the two. So S1= 1/2.

    For S2, the proof asks us to double S2 to get 2S2, but the trick is to offset the addition one term like this...
    1 - 2 + 3 - 4 + 5 - 6 ...
    +.. 1 - 2 + 3 - 4 + 5 ...

    Then go term by term and you'll get 1-1+1-1+... which is S1, and we already know S1=1/2. Then:
    2S2=1/2
    S2=1/4

    Now for S3, the proof asks us to subtract S2 (1/4) and it looks like this...
    ...1 + 2 + 3 + 4 + 5 +...
    - [1 - 2 + 3 - 4 + 5 +... [Note that the - in front of the bracket reverses all of these signs.]

    So S3-S2 = 0+4+0+8+0+12+... = 4+8+12+16+...
    If you factor out 4, then you get 4(1+2+3+4+...), which is just 4S3.

    Now we have a simple equation S3 - S2 = 4S3
    S3 - 1/4 = 4S3
    -3S3 = 1/4
    S3 = -1/12

    So that's one proof that the sum of all positive integers is -1/12. There are other more complicated proofs.

    You could be forgiven for thinking that this all seems like a pretty arbitrary trick. But it's actually used to get the right answer in physics problems all the time, especially quantum physics and string theory. I mean it has to be -1/12 or you get the wrong answer. I wanted to know how that could be... The problems typically appear to involve processes with infinite contributions at every value that somehow cancel themselves all out to leave some remainder at the limit of infinity that must be -1/12 when it's summing integers from 1 to infinity. So that was my hint.

    I think the easiest way to see why it's at least not completely arbitrary is to make a function that generates the sum and graph that function. When you do that, you'll get a curve that actually has a structure where you can see the -1/12 pop out in a suggestive way vis-a-vis our hint.

    So you can graph it with a generating function: G(x) = x/2 (x+1)

    Let's check that the function works really quickly.
    The actual sum would be n(1)=1, n(2)=1+2=3, n(3)=3+3=6, n(4)=6+4=10...
    And for our generating function, G(1)=.5*2=1, G(2)=1*3=3, G(3)=1.5*4=6, G(4)=2*5=10...

    So it works. When you actually graph the generating function and include its continuation leftwards into negative numbers, it looks like this:



    You'll see there's a bit that drops below the x-axis.

    If you take the integral of that part, that is if you measure the area of the space between the curve and the x-axis, lo and behold...



    The integral is -1/12.

    I don't know if this is entirely logically kosher, but the way I think about it, this whole framework is about parts averaging and canceling out other parts at the limit of infinity. So I'm thinking when these infinite contributions are canceling each other out, it's as if the contributions go to infinity off to the right and actually come back around from infinity off to the left, with the left-side contributions cancelling out the right-side contributions, except for this little extra part that remains, which is the integral of the part dropping below the x-axis, that doesn't cancel out the positive contributions. But you're not left with a positive value, you're left with a negative value contribution. Something along those lines. It somehow seems the right way to think about these physical processes with infinite contributions that still leave us with a finite and negative(!) answer.

    Anyway, that's at least one way I can get some intuition for how 1+2+3+4+... = -1/12, and you know, it kind of makes a weird sense to me when I look at it that way.
    Last edited by demagogue; 8th Nov 2023 at 17:52.

  4. #29
    Member
    Registered: Sep 2001
    Location: The other Derry
    If you stated it like this:
    rsum(1,2,3,...) = -1/12
    where rsum(...) is the Ramanujan sum of the number series
    Then people would be like Ramanujan who?

    But stating it like this:
    1+2+3+... = -1/12
    is obviously incorrect per normal definitions, so it's a hook

    And I guess I took it. Is there a better proof? This one is invalid because S1 can never equal 1/2. Never. By definition it can only possibly hold two values and will forever oscillate. The mean of S1 will converge to the value of 1/2, but S1 will not. S1 is not a convergent series and therefore you can't assign it a value in the limit.

    I get that Ramanujan was probably trying to find a way to assign finite values to divergent series, but I'm lost as to what purpose it serves. Does his summation method produce finite results for other series besides 1,2,3...? I suppose it does for any positive linear sequence which is equivalent to 1,2,3... multiplied by a scalar. But what about a non-linear series or a series with negative members? Basically, I'm wondering whether it's useful, or just an oddity.

  5. #30
    Member
    Registered: Dec 2020
    But it's no more invalid to say the first result has a sum that to say the final result has a sum.

    S = 1 - 1 + 1 - 1 + 1 - 1 ... is known as Grandi's Series:

    https://en.wikipedia.org/wiki/Grandi%27s_series

    S = 1 − 1 + 1 − 1 + ..., so
    1 − S = 1 − (1 − 1 + 1 − 1 + ...) = 1 − 1 + 1 − 1 + ... = S
    1 − S = S
    1 = 2S
    S = 1/2

    That's the step Demagogue left out but implied, it's to show that 1 - Grandi's Series equals Grandi's Series, so S = 1/2

    That above method is referred to in Wikipedia as "Treating Grandi's series as a divergent geometric series and using the same algebraic methods that evaluate convergent geometric series".

    Ok then we should also be able to try that using the formula for convergent series. Here starting a = 1, and r = -1

    Code:
    S = a / (1-r)
    r = -1 and first term a = 1
    S = 1 / (1 - -1) = 1/2
    There are other methods to estimate an answer but they conflict with the geometric series proof above, so they should be viewed as different contexts for the interpretation.
    Last edited by Cipheron; 8th Nov 2023 at 22:19.

  6. #31
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    Registered: Jan 2003
    Location: NeoTokyo
    But to answer the question -- well back up, I'm no mathematician or even in a STEM field, so I'm not really qualified, although I did major in philosophy with a big focus on pure logic in undergrad, so maybe I'm more qualified than I'm giving myself credit for -- I tend to be functionalist on these things. There are some situations where seemingly-divergent series do actually converge as they go to infinity where you'd want to use these kinds of techniques just to do the job.

    Just speaking of Rama's Summnation, it gets used in quantum physics where you have infinite contributions of discrete sequential elements giving you finite results, cf. the whole problem of renormalization. I can't remember the exact details, but I somehow recall the Casimir effect was an example.

    Of course the famous example is the original brand of String Theory for Bosons. In order to get the zero-point energy of one excitation to vanish, you have to have an infinite series of sequential contributions add up to exactly -1/12 -- which in two states calls for 24 dimensions but that's another thing -- and once you find that, like magic modeling a relativistic string will exactly match the physical properties of actual Bosons. It's one of those cases where your hand is forced to have a sum of integers add up to -1/12 or you don't get the right answer. Fortunately for them, they found the theory that lets them have it. One of the founders wrote an article or book chapter explaining how he cracked that problem while driving home one day that was a fascinating story, but I can't recall it just now.

    While it's the famous example, the Casimir force example is probably better just because it's settled consensus theory.

  7. #32
    Member
    Registered: Aug 2004
    Just so we're clear, you are aware that you can't just apply normal arithmetic to divergent series... Right?

    https://en.wikipedia.org/wiki/Absolu...nce#Background

  8. #33
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    Registered: Jan 2003
    Location: NeoTokyo
    Sure, it's a special regime rule. But if Super String or M Theory is right and we actually live in a 10 dimensional spacetime, with all but 3 of the spatial dimensions curled up, and there turns out to be some experiment that could actually tease that out with 5 sigma confidence, then I'd also recognize that the universe cares more about manifesting abnormal arithmetic to weasel a finite value out of a divergent series than the so-called normal alternative. In that case, forward your complaint to That Guy.

    Or again cf. the Casimir effect. It's a real force you can measure. It requires getting a finite value out of a divergent series to properly model it. It's not my fault the universe doesn't respect normal math. »\_(公)_/»

    Edit: Anyway, I have to know about the difference for the QM course I'm taking right now because the only Schroedinger Equation solutions that have physical meaning are the ones that converge to zero before or at infinity and that you can normalize, even if, e.g., you need an infinite number of superposed wavelengths to do it for a free particle. So I get it. But I also respect there are special cases that call for special regimes sometimes, and I think it's okay as long as you know when you're in which regime and why.
    Last edited by demagogue; 9th Nov 2023 at 01:13.

  9. #34
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    Registered: Jun 2001
    Location: under God's grace
    Quote Originally Posted by demagogue View Post
    Sure, it's a special regime rule.
    A non-converging alternating series doesn't have a sum, but there are other ways.

    Let S1 = Grandi's series,
    S2 = 1 - 2 + 3 - 4 + ...
    and finally S3 = 1 + 2 + 3 + 4 + ...

    Using CesÓro summation, S1 gives us a value of 1/2. However, S2 isn't CesÓro summable. We can make a generalization of CesÓro sum where, if the plain sum doesn't converge and the arithmetic means don't converge, we just keep going and calculate means of the means until they converge. This way S2 converges to 1/4, and using this generalization doesn't change the previous result of S1, which was 1/2. However, even using this generalization, S3 doesn't yield a sum. So maybe summation isn't the way to go.

    Is there an analytic function that we could use to get each of these series and values for them? Well, S1 is a geometric series, and there's a formula for those:
    S = ar^0 + ar^1 + ar^2 + ar^3 + ... = a / (1 - r), given that |r| < 1.
    In our case a = 1 and r = -1, and thus the formula gives 1 / (1 - (-1)) = 1/2, although this is just outside the valid range for r.

    We could use the Dirichlet eta-function, η(s), which just so happens to be able to produce both S1 and S2 with the parameters 0 and -1, respectively.

    And what are the results? Well:
    η(0) = 1/2
    η(-1) = 1/4

    Now, what about S3? η(s) doesn't produce that sum for any value s. But Riemann's zeta-function, ζ(s) does. ζ(-1) = S3.

    So, what is the value of ζ(-1)? It is a well-known value, the one and only -1/12.

    To be clear, if we want to be mathematically rigorous, we can't get these values by taking an ordinary sum. If one wants to add and subtract the series to then get a finite value for S3, one first needs to define a special sum that gives 1/2 for Grandi's series. And after that, doing adding and subtracting of the different series no longer means quite the same thing. Once you've gotten S1 = 1/2 in a valid way, you're in a different domain.
    Last edited by Qooper; 9th Nov 2023 at 06:11.

  10. #35
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    Registered: Sep 2001
    Location: The other Derry
    Pyrian basically stated my complaint with the "proof". That bit of algebra isn't valid if S is not convergent. S is not a number. It doesn't have a value unless you truncate it. What does it mean to subtract not a number from 1?

    1 − S = 1 − (1 − 1 + 1 − 1 + ...) = 1 − 1 + 1 − 1 + ... = S is one arbitrary definition
    1 − S = 1 − (1 − 1 + 1 − 1 + ...) = (1 − 1) + (1 − 1) + ... = 0 + 0 + ... = 0 is another arbitrary definition
    1 − S = 1 − (1 − 1 + 1 − 1 + ...) = 1 + (− 1 + 1) + (-1 + 1) + ... = 1 + 0 + 0 + ... = 1 is another arbitrary definition

    We're outside of the rules of algebra. Summation should be commutative, but we're getting different values here depending on order of operations.
    So I think it was really an assertion, not a proof.

    Getting back to my point, S doesn't have a value unless you truncate it. So what happens if you truncate it? Then S becomes a finite sequence of length n:
    S(1) = 1
    S(2) = 1-1 = 0
    S(3) = 1-1+1 = 1
    ...
    S = 1,0,1,0,1,0,1,0,1,0,...
    1 - S = 0,1,0,1,0,1,0,1,0,1...

    Note that 1 - S does not equal S for any value of n out to infinity.

    You can treat it as a process with memory: S(n) = 1 - S(n-1). In that case, 1 - S(n) = S(n-1). Or you can treat it as a geometric series of imaginary numbers: S(n) = (1 - i^2n)/2, and in that case 1 - S(n) = (1 + i^2n)/2. But no matter how you slice it, S and 1- S have the same range {0,1} but never have the same value for any finite length of S up to infinity. So what purpose is served by asserting that 1 - S = S in the infinite?

    Let me compare it with Schr÷dinger's cat. We all know that per the design of the experiment, when the box is opened, the cat can only be in one of two states, dead or alive. During the experiment, we model the state of the cat as a two-valued random variable whose pdf varies over time. Prior to ending the experiment, the cat exists in the superposition of both states with different probabilities. Opening the box is equivalent to sampling the random process and that's when state variable acquires a value. But what if we never open the box? If we never observe it, is the state of the cat in the box still constrained by the two possible states it can be observed in? Or can we split the difference and say it's maimed?

    I'm somewhat familiar with renormalization. It's been 30 years since quantum mechanics in school, but I recall renormalization techniques being useful for dealing with infinities and singularities. I'm not yet seeing the parallel with these sequences. It's not obvious these special summations can be generalized to other oscillating and divergent series, or what properties a series has to have in order for the method to work.

    Strangely interesting topic.
    Last edited by heywood; 9th Nov 2023 at 14:26.

  11. #36
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    Location: NeoTokyo
    Just a quick clarification that as far as I understand it renormalization doesn't use these techniques, unless you interpret the String Theory zero-point energy problem as a kind of renormalization problem, but that's not the textbook type even if you did (I think).

    I just meant more to handwave towards that kind of world. They both have that flavor of asserting a value by fiat when you have some divergent function, if we're putting it that way. That part I'd agree with. I think the punchline is that it isn't as arbitrary as one might at first think, but I don't know how to defend that intuition well. I'm more banking on the authority of math or physics professors talking about it in that way and thinking they should have a good grounds to do that.

  12. #37
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    Registered: May 2004
    Location: Canuckistan GWN
    Quote Originally Posted by Qooper View Post
    The real difficulties arise when converting from one base to another. At least for me I'm usually thinking in decimal, even when working with binary or hexadecimal. I convert the value to decimal to "understand" what it is. But some values that can be expressed in some bases cannot be expressed in others. For example decimal 0.3 cannot be finitely expressed in binary. And trinary 0.1 cannot be finitely expressed in decimal.
    Is there a point where a base becomes 'universal'? By which I mean, it can express values in all the preceding and proceeding bases?

    Or does the problem of limitations in expression continue as the base increases?



    And thanks to those still contributing here. I am following along and trying to keep up.

  13. #38
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    Registered: Aug 2004
    Quote Originally Posted by Nicker View Post
    Is there a point where a base becomes 'universal'? By which I mean, it can express values in all the preceding and proceeding bases?
    This question simplifies to "Is there a 3+ integer whose prime divisors include the prime divisors of all preceding and proceeding positive integers?", to which the answer is "lol no". Like, you can't even do preceding, nevermind proceeding. 3 isn't divisible by 2. 6 is divisible by 2 & 3, but not 5. And it gets worse as you go higher.

    Consider the most familiar base to us, 10. It's divisible by 5 and 2. You can take any rational number of the form 1/n, and figure out how many digits it will take to express in decimal by the greater of how many 2's and/or 5's the n divides into; and any other prime multiples makes it infinite. 1/2? 0.5, one digit. 1/8=1/(2*2*2)? 0.125, three digits. And so on. Throw in any other prime, e.g., 1/3 or 1/7, (nevermind 1/11) and it goes infinite.

    This holds for any other base. Base 6 can handle 1/2 and 1/3, but not 1/5. Base 30 can handle 1/2, 1/3, and 1/5, but not 1/7. Base 15 can handle 1/3 and 1/5, but not 1/2. No base can handle any prime number greater than that base.
    Last edited by Pyrian; 12th Nov 2023 at 05:46.

  14. #39
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    I especially get a kick out of base e, which is useful for exponential functions & natural log but I doubt it's clean in any other base you'd want to use. But as a base it's far from arbitrary.

  15. #40
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    Registered: Jun 2001
    Location: under God's grace
    Quote Originally Posted by heywood View Post
    That bit of algebra isn't valid if S is not convergent. S is not a number. It doesn't have a value unless you truncate it. What does it mean to subtract not a number from 1?

    1 − S = 1 − (1 − 1 + 1 − 1 + ...) = 1 − 1 + 1 − 1 + ... = S is one arbitrary definition
    1 − S = 1 − (1 − 1 + 1 − 1 + ...) = (1 − 1) + (1 − 1) + ... = 0 + 0 + ... = 0 is another arbitrary definition
    1 − S = 1 − (1 − 1 + 1 − 1 + ...) = 1 + (− 1 + 1) + (-1 + 1) + ... = 1 + 0 + 0 + ... = 1 is another arbitrary definition
    Very true, S is not a number. But I think we can assign mathematical meaning to such algebra in certain cases, as long as the path is well defined.

    Since in this case everything depends on Grandi's series, we must get a proper value to it in some way. The Dirichlet eta-function gives us that value at 0: η(0) = 1/2. The eta-function is defined for all complex values, so it's also defined at 0. To be clear, this does not mean that the sum of Grandi's series is 1/2, it just means we have a value for it. If we use this value while doing algebra with other series, it means we're no longer in the classical domain of series summation, we're somewhere else. So using this to get a value for 1 - 2 + 3 - 4 + ... means that this value is also not a sum, but something else. It's a valid value, just not a sum.

    The real problem is that getting -1/12 for the sum of all positive integers using Riemann's zeta-function goes outside the defined range of the function. ζ(s) is defined only if Re(s) > 1, and the sum of all positive integers is ζ(-1). However, there is a way! Analytic continuation. ζ(s) can be defined in terms of η(s), and since η(s) was defined everywhere, so is this analytic continuation for ζ(s).

    Let me compare it with Schr÷dinger's cat. We all know that per the design of the experiment, when the box is opened, the cat can only be in one of two states, dead or alive. During the experiment, we model the state of the cat as a two-valued random variable whose pdf varies over time. Prior to ending the experiment, the cat exists in the superposition of both states with different probabilities. Opening the box is equivalent to sampling the random process and that's when state variable acquires a value. But what if we never open the box? If we never observe it, is the state of the cat in the box still constrained by the two possible states it can be observed in? Or can we split the difference and say it's maimed?
    I'd say before observation we can't say it's one or the other, and it definitely cannot be a mixture of the two where it's a little bit an alive cat and a little bit dead. The paths of reality are discrete and do not mix, and before observation both possibilities exist distinctly without mixing into each other. The analogy works really well for Grandi's series, since it is a type of on/off/which-one-is-it-neener-neener -type of series. But what about for example 1 - 2 + 3 - 4 + ... = 1/4, what does this mean? Does the analogy end here, or does 1/4 also encode some aspect of a quantum system? Maybe not superposition, but something else?

    Strangely interesting topic.
    I find it fascinating too. I'm feeling a bit poetic:

    1 is where infinity and the finite kiss
    1 is where we try so hard yet miss
    1 is where infinity comes to tell us whether the cat is alive or dead
    1 is where we get a glimpse of infinity but dare not further tread


    Okay it's a lousy poem. Shoot me.

    Quote Originally Posted by Pyrian View Post
    Base 15 can handle1/3 and 1/7, but not 1/2.
    I think you meant 1/5 instead of 1/7?

  16. #41
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    Registered: Aug 2004
    Quote Originally Posted by demagogue View Post
    I especially get a kick out of base e, which is useful for exponential functions & natural log but I doubt it's clean in any other base you'd want to use. But as a base it's far from arbitrary.
    How do you even use a non-integer as a base in the first place? Like... Base refers to the number of symbols. You can't have a 0.71th symbol or whatever.

    Quote Originally Posted by Qooper View Post
    Very true, S is not a number. But I think we can assign mathematical meaning to such algebra in certain cases, as long as the path is well defined.

    Since in this case everything depends on Grandi's series, we must get a proper value to it in some way.
    The problem with an indeterminacy isn't that you can't assign it a value, it's that you can assign it two or more different values. Any given value is therefore wrong. They can't be rigorously simplified. If you're doing Physics and you encounter something like that, it's another matter; it can be approximated easily enough as 1/2 +/- 1/2.

    Quote Originally Posted by Qooper View Post
    I think you meant 1/5 instead of 1/7?
    Yes, thank you for the correction.

  17. #42
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    Location: NeoTokyo
    Quote Originally Posted by Pyrian View Post
    How do you even use a non-integer as a base in the first place? Like... Base refers to the number of symbols. You can't have a 0.71th symbol or whatever.
    But when you use it as the base for log, things magically clear up: ln e = 1, ln e^x = x, etc.
    That's what I was referring to.

  18. #43
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    Registered: Jun 2001
    Location: under God's grace
    Quote Originally Posted by Pyrian View Post
    How do you even use a non-integer as a base in the first place? Like... Base refers to the number of symbols. You can't have a 0.71th symbol or whatever.
    A more general meaning of 'base' is a value b to the power of the digit's index (the first digit left of the decimal point is index 0 and the first one to the right is index -1) that each digit gets multiplied by. So for example 45100 would be 4*b^4 + 5*b^3 + 1*b^2 + 0*b^1 + 0*b^0. For base e we could have the symbols 0, 1 and 2, and as such, 10 = e. So 122 would be 1*e^2 + 2*e^1 + 2*e^0 = e^2 + 2e + 2. Though, having an irrational base means you can't express almost any value easily

    By the way, base doesn't necessarily have to be the number of symbols. We could have a base-10 writing system where we have 20 symbols, spaced 0.5 apart: 0, a, 1, b, 2, c, ..., i, 9 and finally j. This would mean that there's redundancy in expression, for example a0 is the same as 5, but the system works. It's just not a very useful or interesting system.

    The problem with an indeterminacy isn't that you can't assign it a value, it's that you can assign it two or more different values. Any given value is therefore wrong. They can't be rigorously simplified.
    Unless it is gotten analytically, like from the eta-function. And that can then be used in other divergent series for getting a value. You can use η(0)=1/2 to get a valid value for 1 - 2 + 3 - 4 + ..., which would be the same as directly evaluating η(-1), is all I'm sayin'.

    If you're doing Physics and you encounter something like that, it's another matter; it can be approximated easily enough as 1/2 +/- 1/2.
    But the calculations in physics still follow from rigorous math, even if physicists aren't concerned with the proofs. The math has to work, and it can only work if it's true. Approximations are of course on purpose not precise, but they usually come only at the very end of a general calculation. A precise equation is used for a system, and then a specific case is finally approximated to sufficient precision.
    Last edited by Qooper; 12th Nov 2023 at 12:05.

  19. #44
    Member
    Registered: Dec 2020
    Quote Originally Posted by Pyrian View Post
    How do you even use a non-integer as a base in the first place? Like... Base refers to the number of symbols. You can't have a 0.71th symbol or whatever.
    Actually not quite. What about base -10 for example? Presumably that would need -10 symbols?


    With base -10, each odd power of the base is negative so we need to use a system of subtraction similar to Roman Numerals to get all numbers. However, 10 symbols is both sufficient and required to make everything:

    50 in base -10 = 1 * (-10)^2 + 5 * (-10)^1 + 0 * (-10)^0 = "150"

    950 is trickier, since you want 1000-50, but the 1000s column is negative so you need to go one along:

    950 in base -10 = 1 * (-10)^4 + 9 * (-10)^3 + 0 * (-10)^2 + 5 * (-10)^1 + 0 * (-10)^0 = "19050"

    So the system works, and needs 10 symbols so at the very least it's now |b| symbols not just b symbols


    As for fractional bases, consider what "base 0.5" actually means. Each position going to the left is halved, each position going to the right is doubled. So, it would basically be binary, reflected around the units column. The number of symbols needed is still 2. You could pull the same trick and prove that base 0.1 would be mirror-image base 10, using 10 symbols.


    This just shows that the "rule" that base-X needs X symbols is a simplification that's not true once we look at alternative bases. The number of symbols needed depends on the relative size of each power of the base, in both directions.

    As for bases that aren't some neat multiple or divisor, they work, but you'll probably get an infinite expansion when trying to encode integers; just like base 10 cannot neatly encode 1/3, base 0.71 would not be able to neatly encode ANY integers. However we can use logic to work out how many symbols is needed.


    Take base 1/7 for example, that can use 7 symbols and is just mirror-image base-7. But base 7/2 or 2/7 is different. none of the places to either left or right of the decimal are whole numbers, so it can only express integers as decimal expansions.

    With base 7/2 = 3.5, you'd need 4 symbols, because you'd need to express either 0,1,2,3 times any power of the base. you don't need more than that because 3.5 * 4 would move it over a column, then you could move the remainder down. So yeah, because it's not an integer multiple between powers of the base, you'd have remainders to redistribute any time you add 1 to a column and it carries over.

    As a rule, basically anything base greater than 1 up to 2 would need 2 symbols, then anything greater than to up to 3 would need 3 symbols etc. The inverse would hold true, that any fractional base less than 1 down to 1/2 needs 2 symbols, then 3 symbols below 1/2 down to 1/3 and so on.
    Last edited by Cipheron; 13th Nov 2023 at 11:51.

  20. #45
    Member
    Registered: Jun 2001
    Location: under God's grace
    Quote Originally Posted by Cipheron View Post
    for example - base sqrt(2). If you try and express normal integers in base-sqrt(2) then you can just note every even column is just the powers of 2, so you realize you can do it with exactly 2 symbols: by leaving all the in-between values as 0 and filling in the normal binary digits every 2nd column. Job complete.
    Interesting find! What about base e, can we find a way to make it work? My example of using 0, 1 and 2 for base e obviously won't work because you wouldn't even be able to add 2 + 2. Of course using only 0 and e works, but that's trivial and has the downside of not being able to express any integers expect 0.

  21. #46
    Member
    Registered: Dec 2020
    Quote Originally Posted by Qooper View Post
    Interesting find! What about base e, can we find a way to make it work? My example of using 0, 1 and 2 for base e obviously won't work because you wouldn't even be able to add 2 + 2. Of course using only 0 and e works, but that's trivial and has the downside of not being able to express any integers expect 0.
    Heh I edited that out because the area is just too big and I didn't want it to ramble too much from the main point about fractional bases. Sqrt-bases are kind of cheating, because they have that property of hitting every whole number power of the squared base.

    but keep in mind what all these tricks mean. Negative bases work. Inverse bases work, and square roots of bases work. So you can construct a base of "negative one over square root of 10" and get it to work.

    As for base e, you're right, that you can express 2 by having a 2 in the 1's column, but then if you add 2, you now have 4, which is some value "k" minus e. You then have a 1 in the e^1 column, then a 1 in the e^0 column, then you have some fractional part which you have to compute and subtract powers of 1/e from. You can do that using standard long division. However you're going to have to estimate it as it's an infinite expansion.

    The issue is just as e has no clean representation in base 10, integers greater than e have no clean representation in base-e
    Last edited by Cipheron; 13th Nov 2023 at 12:11.

  22. #47
    Moderator
    Registered: Jan 2003
    Location: NeoTokyo
    Posting a meme basically comparing Ramanujan's summation business vs. the however many other ways you can arbitrarily factor a geometric sum to probably come up with a lot of different answers.

    I guess the punchline is that there's a lot more behind the former than just the cheap proof I gave above, but on the other hand, it's no less arbitrary if you're looking at it as a one special regime out of a lot of others you could choose from. (All this keeping in mind the points already made before that strictly speaking you're not supposed to do algebra with divergent series to begin with.)

    Well it's a cheap meme. I don't really have to explain it much.


  23. #48
    Member
    Registered: Aug 2002
    Location: Maupertuis
    Ramanajan summation would cause far less havoc among enthusiasts if it used a colon instead of an equals sign. Next week I'll ask some of my fellow grad students about Ramanajan summation. I vaguely remember them expressing frustration with its misinterpretation.

    (Yes, I'm in graduate school again, this time in math. I belong in academia. This is my way back in.)

    Cipheron: the p-adic numbers sure are mind-bending, aren't they! There's an analogy to be made with Laurent series in complex analysis.
    Last edited by Anarchic Fox; 3rd Dec 2023 at 21:09.

  24. #49
    Member
    Registered: Aug 2002
    Location: Maupertuis
    I asked three classmates. The consensus was that it's bad notation. Two said the equals sign should not be used, one said the summation symbol should not be used.

  25. #50
    Member
    Registered: Dec 2020
    Thought about where to post this, maybe it should go in the math thread

    https://twitter.com/realstewpeters/s...36101555327317

    Stew Peters (the dollar store Alex Jones) posts a video by a guy who's proved 1x1=3, says we should listen to that guy because he's blown everyone out of the water proving everything we're taught is a lie.

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