Ten is the lovliest number... - by Nicker
Tocky on 14/10/2023 at 03:29
It was motherfuckers like you who designed the alien algebra crap in Omicron that took so long to figure out wasn't it?
Nicker on 14/10/2023 at 19:16
I think I am having a similar difficulty to SD trying to understand this.
Moving the point in base 10 is the same as dividing or multiplying by 10.
But moving the point doesn't provide the same elegant effect in other bases, does it?
Nicker on 14/10/2023 at 19:26
On a side note...
Quote:
"The Nobel Prize in mathematics was awarded to a California professor who has discovered a new number! The number is bleen, which he claims belongs between 6 and 7." --George Carlin
Given the almost universal authority granted to Carlin on all matters, I would like to know why bleen is still being suppressed.
demagogue on 14/10/2023 at 19:27
Moving the point would be the same as dividing or multiplying by the base number, which is what you'd want if you're working in a different base system. It'd be the elegant thing in that base.
The thing that I want to wrap my head around is not only are the complex numbers you map to a 2D plane, but there's quaternions numbers you map to 4D, and octonions numbers you map to 8D, and there are all kinds of crazy rules to apply in those contexts.
Nicker on 14/10/2023 at 21:48
Thanks for the first paragraph, Dema. That makes sense. It doesn't entirely alleviate my confusion but it's one fewer mole to whack.
After that you are just flexing over by the free weights while I plod on the treadmill though. LOL
I am finding the discussion fascinating and inspiring, never the less.
Qooper on 15/10/2023 at 20:59
Quote Posted by Nicker
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.
Cipheron on 8/11/2023 at 14:39
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.
demagogue on 8/11/2023 at 15:51
That's great!
I always got a kick out of the proof that 1+2+3+4+... out to infinity = -1/12, known as (
https://en.wikipedia.org/wiki/Ramanujan_summation) Ramaujan's summation, or the most famous exemplar of his summation method.
Inline Image:
https://i.ibb.co/qs78d44/RamaSum.jpgIt 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:
Inline Image:
https://i.ibb.co/LgKn2Qd/main-qimg-788ab3d7dff8d763dac1c1d7ae9d3b62.webpYou'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...
Inline Image:
https://i.ibb.co/2n4T9s0/main-qimg-923cd83fea56ab233671178355bdd91c.webpThe 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.
heywood on 9/11/2023 at 01:37
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.
Cipheron on 9/11/2023 at 03:13
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.
