Na I meant to quote that post, but it was early morning and I was half asleep before work.
Just intended to continue the conversation.
-
Features
The Math Thread
- Thread starter beero1000
- Start date
Na I meant to quote that post, but it was early morning and I was half asleep before work.
Just intended to continue the conversation.
Kharakov
Quantum Hot Dog
Never mind. All the series other than those generated when x=1 are divergent. Here's a new question for you, that might even be interesting to Loren.... 
Are there any non-simplex related infinite summations using natural numbers that you can divide the harmonic series by and get 0 instead of divergence without rearranging terms?
Simplex related infinite summations are sums of an n:1 to infinity
an=1; 1,2,3...; 1,3,6,10...; 1,4,10,20...; 1,1,2,2,3,3...; 1,1,3,3,6,6...; etc.
When x=1
an=1
\(\frac {1+ \frac {1}{2} +\frac {1}{3} +\frac {1}{4} +\frac {1}{5} +\frac {1}{6} +...}{1 + 1 + 1 + 1 + 1 +1 +...} = 0 \)
You get the sum of the reciprocal of the triagulars/2 (or 1) - the sum of the reciprocal of the tris/2.
1 -( 1/2 +1/6+1/12....)
an=n
\(\frac {1+ \frac {1}{2} +\frac {1}{3} +\frac {1}{4} +\frac {1}{5} +\frac {1}{6} +...}{1 + 2 + 3 + 4 + 5 +6 +...} = 0 \)
You get the sum of the reciprocal of the tetrahedrals/3 -....
1-3/2 = -1/2 + (1/3 +1/12 +1/30 +1/60+1/105....)=0
Next 0 is balanced at the sum of the pentatopes/4. For the other ones, where we double the terms of the simplex related series, we get 0s too:
an= 1,1,3,3,6,6...
1-1/2-13/6+11/12+77/60 = 8/15 =
... -
\(\sum_{n=1}^\infty \frac{n^2+5n+5}{30\frac{n(n+1)(n+2)(n+3)(n+4)(n+5)}{6!}} \)
Are there any non-simplex related infinite summations using natural numbers that you can divide the harmonic series by and get 0 instead of divergence without rearranging terms?
Simplex related infinite summations are sums of an n:1 to infinity
an=1; 1,2,3...; 1,3,6,10...; 1,4,10,20...; 1,1,2,2,3,3...; 1,1,3,3,6,6...; etc.
When x=1
an=1
\(\frac {1+ \frac {1}{2} +\frac {1}{3} +\frac {1}{4} +\frac {1}{5} +\frac {1}{6} +...}{1 + 1 + 1 + 1 + 1 +1 +...} = 0 \)
You get the sum of the reciprocal of the triagulars/2 (or 1) - the sum of the reciprocal of the tris/2.
1 -( 1/2 +1/6+1/12....)
an=n
\(\frac {1+ \frac {1}{2} +\frac {1}{3} +\frac {1}{4} +\frac {1}{5} +\frac {1}{6} +...}{1 + 2 + 3 + 4 + 5 +6 +...} = 0 \)
You get the sum of the reciprocal of the tetrahedrals/3 -....
1-3/2 = -1/2 + (1/3 +1/12 +1/30 +1/60+1/105....)=0
Next 0 is balanced at the sum of the pentatopes/4. For the other ones, where we double the terms of the simplex related series, we get 0s too:
an= 1,1,3,3,6,6...
1-1/2-13/6+11/12+77/60 = 8/15 =
... -
\(\sum_{n=1}^\infty \frac{n^2+5n+5}{30\frac{n(n+1)(n+2)(n+3)(n+4)(n+5)}{6!}} \)
Last edited:
beero1000
Veteran Member
Never mind. All the series other than those generated when x=1 are divergent. Here's a new question for you, that might even be interesting to Loren....
Are there any non-simplex related infinite summations using natural numbers that you can divide the harmonic series by and get 0 instead of divergence without rearranging terms?
Simplex related infinite summations are sums of an n:1 to infinity
an=1; 1,2,3...; 1,3,6,10...; 1,4,10,20...; 1,1,2,2,3,3...; 1,1,3,3,6,6...; etc.
When x=1
an=1
\(\frac {1+ \frac {1}{2} +\frac {1}{3} +\frac {1}{4} +\frac {1}{5} +\frac {1}{6} +...}{1 + 1 + 1 + 1 + 1 +1 +...} = 0 \)
You get the sum of the reciprocal of the triagulars/2 (or 1) - the sum of the reciprocal of the tris/2.
1 -( 1/2 +1/6+1/12....)
an=n
\(\frac {1+ \frac {1}{2} +\frac {1}{3} +\frac {1}{4} +\frac {1}{5} +\frac {1}{6} +...}{1 + 2 + 3 + 4 + 5 +6 +...} = 0 \)
You get the sum of the reciprocal of the tetrahedrals/3 -....
1-3/2 = -1/2 + (1/3 +1/12 +1/30 +1/60+1/105....)=0
Next 0 is balanced at the sum of the pentatopes/4. For the other ones, where we double the terms of the simplex related series, we get 0s too:
an= 1,1,3,3,6,6...
1-1/2-13/6+11/12+77/60 = 8/15 =
... -
\(\sum_{n=1}^\infty \frac{n^2+5n+5}{30\frac{n(n+1)(n+2)(n+3)(n+4)(n+5)}{6!}} \)
You're gonna have to explain what you mean by that division first. If you're taking it as the limit of the ratio of the partial sums, then any series whose partial sums grow asymptotically faster than the harmonic series (about ln n) would work.
Kharakov
Quantum Hot Dog
division method: the reverse of the Cauchy product you mentioned before, as what you mention is the trivial solution, right? Every series I'm dividing by is a greater infinity than the harmonic series, as they consist of non-decreasing sequences of natural numbers (I hope I mentioned that... mehh).
So the revised question would be:
Are there any non-simplex related non decreasing infinite series composed of natural number sequences that you can divide the harmonic series by using the reverse Cauchy product division method to produce a new series that absolutely converges to 0 (no rearranging terms)?
The results I've come across have been divergent, converge to some value, or are exactly equal to 0 {0 in the cases in which one divides the harmonic series by a series based on simplex number progressions (1,1,1,1; 1,2,3,4...; 1,3,6,10...; 1,4,10...; 1,1,3,3,6,6...;}.
Various simplex related ones converge to 0.
Harmonic/ 1+1+1+1... = .........................1 - (1/2 +1/6 + 1/12 + 1/20....) series is reciprocal triangulars/2
Harmonic/ 1+2+3+4... = 1-3/2 +... = ......-1/2 + (1/3 +1/12 +1/30 +1/60...)=0 series is reciprocal tetras/3
Harmonic/ 1+3+6+10... = 1-5/2+11/6 -...= 1/3 - (1/4 + 1/20+1/60+1/140...)= 0 series is reciprocal pentatopes/4
Harmonic/ 1+4+10+20...= ....4 terms.......= -1/4 + (1/5 + 1/30+1/105+1/280...) = 0 series is reciprocal 6 simplex/5
===========================================================================
Continued to simplex infinity....................log(2) - (log(2) - 1 + 1/2 alternating rec. triangulars -1/2 +1/3 alternating recip. tetras-1/3....
so harmonic = alternating triangular reciprocal sum /2 + alternating tetrahedral rec. sum/3 + alternating pentatope reciprocal/4
== 2 log(2)-1 + 4 log (2) -5/2 + 8 log(2) - 16/3 + 16 log(2) - 131/12 + 32 log(2) - 661/30 + 64 log(2) - 2654/60 +128 log(2) -9304/105..
[2 log(2)-1] *2 -1/2 = [4 log (2) -5/2] *2 -1/3 = [8 log(2) - 16/3] *2 - 1/4 = 16 log(2) -131/12...
Am I going in circles?
So the revised question would be:
Are there any non-simplex related non decreasing infinite series composed of natural number sequences that you can divide the harmonic series by using the reverse Cauchy product division method to produce a new series that absolutely converges to 0 (no rearranging terms)?
The results I've come across have been divergent, converge to some value, or are exactly equal to 0 {0 in the cases in which one divides the harmonic series by a series based on simplex number progressions (1,1,1,1; 1,2,3,4...; 1,3,6,10...; 1,4,10...; 1,1,3,3,6,6...;}.
Various simplex related ones converge to 0.
Harmonic/ 1+1+1+1... = .........................1 - (1/2 +1/6 + 1/12 + 1/20....) series is reciprocal triangulars/2
Harmonic/ 1+2+3+4... = 1-3/2 +... = ......-1/2 + (1/3 +1/12 +1/30 +1/60...)=0 series is reciprocal tetras/3
Harmonic/ 1+3+6+10... = 1-5/2+11/6 -...= 1/3 - (1/4 + 1/20+1/60+1/140...)= 0 series is reciprocal pentatopes/4
Harmonic/ 1+4+10+20...= ....4 terms.......= -1/4 + (1/5 + 1/30+1/105+1/280...) = 0 series is reciprocal 6 simplex/5
===========================================================================
Continued to simplex infinity....................log(2) - (log(2) - 1 + 1/2 alternating rec. triangulars -1/2 +1/3 alternating recip. tetras-1/3....
so harmonic = alternating triangular reciprocal sum /2 + alternating tetrahedral rec. sum/3 + alternating pentatope reciprocal/4
== 2 log(2)-1 + 4 log (2) -5/2 + 8 log(2) - 16/3 + 16 log(2) - 131/12 + 32 log(2) - 661/30 + 64 log(2) - 2654/60 +128 log(2) -9304/105..
[2 log(2)-1] *2 -1/2 = [4 log (2) -5/2] *2 -1/3 = [8 log(2) - 16/3] *2 - 1/4 = 16 log(2) -131/12...
Am I going in circles?
beero1000
Veteran Member
I still don't think that the question is well-defined. What do you mean by a series being a 'greater infinity than the harmonic series'? How are you dividing these infinities?
Kharakov
Quantum Hot Dog
So we end up with the harmonic series =
\(\sum_{n=1}^\infty \frac {1}{n} = \sum_{n=1}^\infty 2^n \, log(2) \, - \left [2^{n-1} + \frac{2^{n-2}}{2}+ \frac{2^{n-3}}{3}+ \frac{2^{n-4}}{4} \dots \right] = \sum_{n=1}^\infty 2^n \, log(2) \, - \left [\frac{2^n}{2} + \frac{2^{n}}{2^2 \times 2}+ \frac{2^{n}}{2^3 \times 3}+ \frac{2^{n}}{2^4 \times 4} \dots \right] \)
Which is 2^infinity * (log(2) - series that generates log(2))= infinity, which means that the series itself sums to an infinitesimal less than log 2, if we proceed according to logic.
This means that the previous series of 0s were actually infinitesimals, where infinity*infinitesimal = infinitesimal. So only when we take a real number to the infinitieth power, and multiply it by an infinitesimal do we get infinity.
Is this the hyperreals, or different? This emerges from previous rules (we didn't artificially add rules, we observed rules that exist because of the previous rules).
\(\sum_{n=1}^\infty \frac {1}{n} = \sum_{n=1}^\infty 2^n \, log(2) \, - \left [2^{n-1} + \frac{2^{n-2}}{2}+ \frac{2^{n-3}}{3}+ \frac{2^{n-4}}{4} \dots \right] = \sum_{n=1}^\infty 2^n \, log(2) \, - \left [\frac{2^n}{2} + \frac{2^{n}}{2^2 \times 2}+ \frac{2^{n}}{2^3 \times 3}+ \frac{2^{n}}{2^4 \times 4} \dots \right] \)
Which is 2^infinity * (log(2) - series that generates log(2))= infinity, which means that the series itself sums to an infinitesimal less than log 2, if we proceed according to logic.
This means that the previous series of 0s were actually infinitesimals, where infinity*infinitesimal = infinitesimal. So only when we take a real number to the infinitieth power, and multiply it by an infinitesimal do we get infinity.
Is this the hyperreals, or different? This emerges from previous rules (we didn't artificially add rules, we observed rules that exist because of the previous rules).
Last edited:
Kharakov
Quantum Hot Dog
1+1+1+1+1... > 1 + 1/2 + 1/3 + 1/4...
All of the series are to be composed of non decreasing natural number sequences.
Like I said, reverse Cauchy product- that's the name you gave me, and it looks like what I'm doing.
So the series An/Bn=..
\(\frac {A_n}{B_n} = \frac {A_1}{B_1} + \frac{A_2 B_1- B_2 A_1}{B_1^2} +\frac {B_1^2 A_3 - A_2 B_1 B_2 -B_2^2 A_1-B_3A_1B_1^2}{B_1^3} \,\, ...\)
Last edited:
beero1000
Veteran Member
I think you're playing a little too fast and loose with the rules here.
Kharakov
Quantum Hot Dog
Besides the arithmetic mistake, why do you say that?
The alternating harmonic series log(2) > other series log(2) by an infinitesimal amount. It has to be, if (2^n + 2) * [alt logs log (2) - other log(2)] --> infinity as n-->infinity.
Of course, we also have an infinite amount of series generated log(2) from the alternating simplex series.
What a mess. Ok- so it's not as cut and dried as it looked when I first noticed it.
The alternating harmonic series log(2) > other series log(2) by an infinitesimal amount. It has to be, if (2^n + 2) * [alt logs log (2) - other log(2)] --> infinity as n-->infinity.
Of course, we also have an infinite amount of series generated log(2) from the alternating simplex series.
What a mess. Ok- so it's not as cut and dried as it looked when I first noticed it.
Last edited:
beero1000
Veteran Member
You keep using terms like 'infinity' and 'infinitesimal' without specifying exactly what you mean, so setting things equal and doing all that symbolic manipulation needs to be explained. I'm saying you haven't specified what those strings of symbols mean, so I can't say if you're right or wrong.
Kharakov
Quantum Hot Dog
The harmonic inverse Cauchy products are correct. If you keep on dividing by the next simplex series, you get the following:
Harmonic/ 1+1+1+1... = ......................... 1 - (1/2 + 1/6 + 1/12 + 1/20....) series is reciprocal triangulars/2
Harmonic/ 1+2+3+4... = 1-3/2 +... = ...... -1/2 + (1/3 +1/12 + 1/30 +1/60...)=0 series is reciprocal tetras/3
Harmonic/ 1+3+6+10... = 1-5/2+11/6 -...= 1/3 - (1/4 + 1/20+ 1/60 +1/140...)= 0 series is reciprocal pentatopes/4
Harmonic/ 1+4+10+20...= ....4 terms.......= -1/4 + (1/5 + 1/30+ 1/105 +1/280...) = 0 series is reciprocal 6 simplex/5
===========================================================================
............................................................. log(2) + A + B + C + D +......
A: log(2)................................................-1;
B: 1/2 alternating triangular # reciprocal -1/2
C: 1/3 alternating recip. tetrahedral #s -1/3
D: 1/4 alternating pentatopic recipricols -1/4
You should see the harmonic right above here
And I was wrong: it's the harmonic is equal to 2 log(2)+ 1/2 alternating triangular # reciprocal (2 log(2)-1) + 1/3 alternating recip. tetrahedral #s (4log(2)-5/2) + 1/4 alternating pentatopic recipricols.....
So it's.. 2 log(2) + [ 2 log(2)-1 + 4 log (2) -5/2 + 8 log(2) - 16/3 + 16 log(2) - 131/12 + 32 log(2) - 661/30 ...]
Which is (by parts, without the first 2 log(2)), and sum 2^n = 2^(n+1)-2:
\(\sum_{n=1}^\infty 2^n log(2) = (2^{n+1}-2) log(2)\)
-
\(\sum_{n=1}^\infty \frac{2^n}{2} + \frac{2^n}{2^2*2} + \frac {2^n}{2^3*3} ...= (2^{n+1}-2) log(2) \)
zeroed, but...have to take out first terms above
+
\(\sum_1^1 \frac{2^n}{2^2*2} + \sum_1^2 \frac {2^n}{2^3*3} +\sum_1^3 \frac {2^n}{2^4*4} ...= \frac{2^1-2}{2^1*1} +\frac{2^2-2}{2^2*2} +\frac{2^3-2}{2^3*3} +\frac{2^4-2}{2^4*4} \)
Which is the harmonic series - 2 log(2). Now it's time to fetch our 2 log(2) from above. Now we have the harmonic series as equal to 2 log(2) + 1/2 recip tris + 1/3 recip tetras + 1/4 recip pentas..
Everything is zeroed out perfectly. Which probably looks right to you, ehh?
I still wonder if there are any non-simplex related non-decreasing natural number series you can divide the harmonic series by and get 0s. I tried a bunch. I'll work on the wording of that question tomorrow.
Question 2: What ratio do you change the harmonic series by to keep it equal when you change the first term by multiplying by something like 2log(2) (or a bit more arbitrary)?
Harmonic/ 1+1+1+1... = ......................... 1 - (1/2 + 1/6 + 1/12 + 1/20....) series is reciprocal triangulars/2
Harmonic/ 1+2+3+4... = 1-3/2 +... = ...... -1/2 + (1/3 +1/12 + 1/30 +1/60...)=0 series is reciprocal tetras/3
Harmonic/ 1+3+6+10... = 1-5/2+11/6 -...= 1/3 - (1/4 + 1/20+ 1/60 +1/140...)= 0 series is reciprocal pentatopes/4
Harmonic/ 1+4+10+20...= ....4 terms.......= -1/4 + (1/5 + 1/30+ 1/105 +1/280...) = 0 series is reciprocal 6 simplex/5
===========================================================================
............................................................. log(2) + A + B + C + D +......
A: log(2)................................................-1;
B: 1/2 alternating triangular # reciprocal -1/2
C: 1/3 alternating recip. tetrahedral #s -1/3
D: 1/4 alternating pentatopic recipricols -1/4
You should see the harmonic right above here
And I was wrong: it's the harmonic is equal to 2 log(2)+ 1/2 alternating triangular # reciprocal (2 log(2)-1) + 1/3 alternating recip. tetrahedral #s (4log(2)-5/2) + 1/4 alternating pentatopic recipricols.....
So it's.. 2 log(2) + [ 2 log(2)-1 + 4 log (2) -5/2 + 8 log(2) - 16/3 + 16 log(2) - 131/12 + 32 log(2) - 661/30 ...]
Which is (by parts, without the first 2 log(2)), and sum 2^n = 2^(n+1)-2:
\(\sum_{n=1}^\infty 2^n log(2) = (2^{n+1}-2) log(2)\)
-
\(\sum_{n=1}^\infty \frac{2^n}{2} + \frac{2^n}{2^2*2} + \frac {2^n}{2^3*3} ...= (2^{n+1}-2) log(2) \)
zeroed, but...have to take out first terms above
+
\(\sum_1^1 \frac{2^n}{2^2*2} + \sum_1^2 \frac {2^n}{2^3*3} +\sum_1^3 \frac {2^n}{2^4*4} ...= \frac{2^1-2}{2^1*1} +\frac{2^2-2}{2^2*2} +\frac{2^3-2}{2^3*3} +\frac{2^4-2}{2^4*4} \)
Which is the harmonic series - 2 log(2). Now it's time to fetch our 2 log(2) from above. Now we have the harmonic series as equal to 2 log(2) + 1/2 recip tris + 1/3 recip tetras + 1/4 recip pentas..
Everything is zeroed out perfectly. Which probably looks right to you, ehh?
I still wonder if there are any non-simplex related non-decreasing natural number series you can divide the harmonic series by and get 0s. I tried a bunch. I'll work on the wording of that question tomorrow.
Question 2: What ratio do you change the harmonic series by to keep it equal when you change the first term by multiplying by something like 2log(2) (or a bit more arbitrary)?
Last edited:
Kharakov
Quantum Hot Dog
Ok, so what's the name of this reverse Cauchy product division of the harmonic series that results in 0's for various well defined infinite sums?
H/ (1+1+1+1.. = 0 (first well defined Harmonic Cauchy 0)
log 2 is the series -1 + the series.... although that seems a bit fast and loose:
.../0+0+0+1+1+... //maybe we don't need to specify 0s since the series
.../0+0+1+1+1... // are not being rearranged?
.../0+1+1+1...
H/(1+1+1+1... = H/1+2+3+4... = 0 (2nd infinity division 0)
log(2) things...
To get to the last series, subtract the series- the series seed from itself... you'll go back to zero (since our series are all Harmonic Cauchy zeroes).
H/1+2+3+4
./... -1-2- 3.. = H/1+1+1+1= 0
keep shifting and stacking for the next series that gives you a Harmonic zero.
.../0+0+1+2+3.. (these are the same as adding 1+2+3+4 - (1+1+1+1...) for every shift over, this layer is -2* (1+1+1...)
.../0+1+2+3...
H/1+2+3+4... = H/1+3+6+10... = 0
If you mix series, do you get convergent Harmonic Cauchys? The following look divergent, like they are log(0) or something.
.../.+1+1+1
H/1+2+3+4= H/ 1+3+4+5... looks divergent to negative infinity- don't think you can add the specific Cauchy zeroes to one another, although you can subtract them...
H/1+2+3+4...= H/1+1+2+3+4.... = looks like it diverges to -infinity.
-0-1 -1 -1...
So for something to be a Harmonic reverse Cauchy product division zero, it looks like it has to be directly related to specific, well defined infinite series.
0+0+1+ 3+....
0+1+3+ 6+10...
1+3+6+10...
1+4+10+20... = H/this = Harmonic Cauchy 0 (first is 1,1,1... second is 1,2,3... third is 1,3,6,... 4th is 1,4,10....)
You can also construct other 0 ladders, if you want:
The Cauchy zeros are infinitesimals that can be divided by one another.
If you take the first Cauchy zero I mentioned, and divide it by the second:
zero 1: Harm/ 1+1+1+ = 1- (1/2+1/6+1/12...)
zero 2: Harm/ 1+2+3....= 1- 3/2 +1/3 +1/12 +1/30..
zero1/zero2 = 1+1+1+1+1... because (1+1+1+1...)^2 = 1+2+3+4....
zero2/zero1 = 1-1 = 0
I wonder what zero3/zero1 equals?? I gotta check.
H/ (1+1+1+1.. = 0 (first well defined Harmonic Cauchy 0)
log 2 is the series -1 + the series.... although that seems a bit fast and loose:
.../0+0+0+1+1+... //maybe we don't need to specify 0s since the series
.../0+0+1+1+1... // are not being rearranged?
.../0+1+1+1...
H/(1+1+1+1... = H/1+2+3+4... = 0 (2nd infinity division 0)
log(2) things...
To get to the last series, subtract the series- the series seed from itself... you'll go back to zero (since our series are all Harmonic Cauchy zeroes).
H/1+2+3+4
./... -1-2- 3.. = H/1+1+1+1= 0
keep shifting and stacking for the next series that gives you a Harmonic zero.
.../0+0+1+2+3.. (these are the same as adding 1+2+3+4 - (1+1+1+1...) for every shift over, this layer is -2* (1+1+1...)
.../0+1+2+3...
H/1+2+3+4... = H/1+3+6+10... = 0
If you mix series, do you get convergent Harmonic Cauchys? The following look divergent, like they are log(0) or something.
.../.+1+1+1
H/1+2+3+4= H/ 1+3+4+5... looks divergent to negative infinity- don't think you can add the specific Cauchy zeroes to one another, although you can subtract them...
H/1+2+3+4...= H/1+1+2+3+4.... = looks like it diverges to -infinity.
-0-1 -1 -1...
So for something to be a Harmonic reverse Cauchy product division zero, it looks like it has to be directly related to specific, well defined infinite series.
0+0+1+ 3+....
0+1+3+ 6+10...
1+3+6+10...
1+4+10+20... = H/this = Harmonic Cauchy 0 (first is 1,1,1... second is 1,2,3... third is 1,3,6,... 4th is 1,4,10....)
You can also construct other 0 ladders, if you want:
The Cauchy zeros are infinitesimals that can be divided by one another.
If you take the first Cauchy zero I mentioned, and divide it by the second:
zero 1: Harm/ 1+1+1+ = 1- (1/2+1/6+1/12...)
zero 2: Harm/ 1+2+3....= 1- 3/2 +1/3 +1/12 +1/30..
zero1/zero2 = 1+1+1+1+1... because (1+1+1+1...)^2 = 1+2+3+4....
zero2/zero1 = 1-1 = 0
I wonder what zero3/zero1 equals?? I gotta check.
Last edited:
lpetrich
Contributor
Some more group-theory stuff.
First, homomorphisms or mappings: f(a) = b for a in group G and b in group H. There are two trivial ones: f(a) = a (G to G) and f(a) = e (G to the identity group).
The kernel K of a homomorphism is f(a) = e for a in K, with no other elements giving e. It is easy to show that K is a normal subgroup of G, and that the destination group is the quotient group G/K.
An automorphism is a homomorphism of a group onto itself: f(a) = b for both a and b in G. An "inner" one can be constructed with some g in G such that f(a) = g.a.g-1. An "outer" one is all the others.
Inner automorphisms form a group: Inn(G) = G/Z(G) where Z(G) is the center of G, all of G's elements that commute with every element of G.
Automorphisms in general also form a group: Aut(G), and the outer-automorphism group Out(G) = Aut(G)/Inn(G).
Every abelian group's inner-automorphism group is the identity group, the trivial one. All others are outer ones.
Now for some examples.
A cyclic group can be generated by a single element, and a homomorphism of a cyclic group is thus another cyclic group: f(Z
) = Z(m) where m evenly divides n.
The group of automorphisms of the cyclic group Z is the group Z(*,n), the group under multiplication modulo n of all numbers 1...(n-1) that are relatively prime to n.
Thus, Aut(Z2) = {e}, Aut(Z3) = Z2, Aut(Z5) = Z4, Aut(Z7) = Z6, ...
Products of groups may include interchanges of those groups among their automorphisms. Thus, Aut(Z2*Z2) = Sym(3), the group of all interchanges of the group's non-identity elements.
But Aut(Sym(3)) = Sym(3) and is all inner automorphisms.
First, homomorphisms or mappings: f(a) = b for a in group G and b in group H. There are two trivial ones: f(a) = a (G to G) and f(a) = e (G to the identity group).
The kernel K of a homomorphism is f(a) = e for a in K, with no other elements giving e. It is easy to show that K is a normal subgroup of G, and that the destination group is the quotient group G/K.
An automorphism is a homomorphism of a group onto itself: f(a) = b for both a and b in G. An "inner" one can be constructed with some g in G such that f(a) = g.a.g-1. An "outer" one is all the others.
Inner automorphisms form a group: Inn(G) = G/Z(G) where Z(G) is the center of G, all of G's elements that commute with every element of G.
Automorphisms in general also form a group: Aut(G), and the outer-automorphism group Out(G) = Aut(G)/Inn(G).
Every abelian group's inner-automorphism group is the identity group, the trivial one. All others are outer ones.
Now for some examples.
A cyclic group can be generated by a single element, and a homomorphism of a cyclic group is thus another cyclic group: f(Z
) = Z(m) where m evenly divides n.The group of automorphisms of the cyclic group Z
is the group Z(*,n), the group under multiplication modulo n of all numbers 1...(n-1) that are relatively prime to n.Thus, Aut(Z2) = {e}, Aut(Z3) = Z2, Aut(Z5) = Z4, Aut(Z7) = Z6, ...
Products of groups may include interchanges of those groups among their automorphisms. Thus, Aut(Z2*Z2) = Sym(3), the group of all interchanges of the group's non-identity elements.
But Aut(Sym(3)) = Sym(3) and is all inner automorphisms.
lpetrich
Contributor
Now Lie algebras. A group homomorphism f(I + e*L + ...) = I' + e*L' + ... reduces to f(L) = L' where f must be linear: T.L = L'
Finding the commutator gives Tia*Tjb*fabk = f'ijc*Tck where f and f' are the structure constants of the source and destination algebras: [Li,Lj] = fijk*Lk
It's hard for me to proceed further in this general case, so I will turn to automorphisms: Tia*Tjb*fabk = fijc*Tck (same f on both sides)
This is still rather difficult, so I will expand T around the identity matrix: T -> I + e*W + ... . I get Wia*fajk + Wjb*fibk = fijcWck
There is a simple solution: Wi[/sup]k[/sup] = Va*faik for parameters V. It can be verified with the Jacobi identity, and it is the inner automorphisms. The outer ones are the remaining ones.
Now for what makes these ones inner ones. Consider a group inner automorphism: (I + e*M + ...).(I + e*L + ...).(I - e*M ...) It is (I + e*L + e2*[M,L] + ...) giving L' = L + e*[M,L] + ... Expanding out the commutator gives the inner-automorphism part of W.
I have attempted to find solutions for some Lie algebras, and here is my results:
U(1): only one automorphism, an outer one: L -> T*L where T is nonzero.
I have found that SU and SO have only inner automorphisms, with a few exceptions.
These are: SU(1) and SO(1) are the empty algebras, SO(2) ~ U(1) and has only one automorphism, an outer one, and SO(4) ~ SU(2)*SU(2), meaning that it has an outer automorphism in addition to its inner ones: interchange of the two SU(2)'s.
I haven't yet done the symplectic algebras, Sp(2n), or the exceptional simple ones: G2, F4, E6, E7, and E8.
Finding the commutator gives Tia*Tjb*fabk = f'ijc*Tck where f and f' are the structure constants of the source and destination algebras: [Li,Lj] = fijk*Lk
It's hard for me to proceed further in this general case, so I will turn to automorphisms: Tia*Tjb*fabk = fijc*Tck (same f on both sides)
This is still rather difficult, so I will expand T around the identity matrix: T -> I + e*W + ... . I get Wia*fajk + Wjb*fibk = fijcWck
There is a simple solution: Wi[/sup]k[/sup] = Va*faik for parameters V. It can be verified with the Jacobi identity, and it is the inner automorphisms. The outer ones are the remaining ones.
Now for what makes these ones inner ones. Consider a group inner automorphism: (I + e*M + ...).(I + e*L + ...).(I - e*M ...) It is (I + e*L + e2*[M,L] + ...) giving L' = L + e*[M,L] + ... Expanding out the commutator gives the inner-automorphism part of W.
I have attempted to find solutions for some Lie algebras, and here is my results:
U(1): only one automorphism, an outer one: L -> T*L where T is nonzero.
I have found that SU
and SO have only inner automorphisms, with a few exceptions.These are: SU(1) and SO(1) are the empty algebras, SO(2) ~ U(1) and has only one automorphism, an outer one, and SO(4) ~ SU(2)*SU(2), meaning that it has an outer automorphism in addition to its inner ones: interchange of the two SU(2)'s.
I haven't yet done the symplectic algebras, Sp(2n), or the exceptional simple ones: G2, F4, E6, E7, and E8.
lpetrich
Contributor
Oof. My results are incomplete. Outer automorphisms of simple Lie Algebras - MathOverflow has the solutions that I was trying to get.
The inner-automorphism parts of the T's are matrix groups constructed from the algebra's adjoint representation, the one that's most directly related to its generators.
The outer automorphisms are related to the symmetries of the algebras' Dynkin dyagrams.
For SU or A(n-1), the Dynkin diagram is o - o - o - ... - o with (n-1) o's. It has end-to-end reflection symmetry, and that means for SU and its relative SL the negative transpose.
For SO(2n+1) or B, the Dynkin diagram is o - o - o - ... - o = x and for Sp(2n) or C, the Dynkin diagram is x - x - x - ... - x = o, both with n o's and x's, and with o being the long root and x the short one (usually a darkened circle). Both with no symmetry.
For SO(2n) or D, the Dynkin diagram is o - o - o - ... - o (- o) - o with n o's. It has a symmetry of reflecting the two short branches, corresponding to using an O(2n) matrix with determinant -1 to be applied to the algebra generators.
The exceptional ones are G2: o = x (three dashes in the =), no symmetry, F4: o - o = x - x, no symmetry, E6: o - o - o (- o) - o - o, end-to-end symmetry of reflecting the two long branches, E7: o - o - o - o (- o) - o - o, no symmetry, E8: o - o - o - o - o (- o) - o - o, no symmetry.
The inner-automorphism parts of the T's are matrix groups constructed from the algebra's adjoint representation, the one that's most directly related to its generators.
The outer automorphisms are related to the symmetries of the algebras' Dynkin dyagrams.
For SU
or A(n-1), the Dynkin diagram is o - o - o - ... - o with (n-1) o's. It has end-to-end reflection symmetry, and that means for SU and its relative SL the negative transpose.For SO(2n+1) or B
, the Dynkin diagram is o - o - o - ... - o = x and for Sp(2n) or C, the Dynkin diagram is x - x - x - ... - x = o, both with n o's and x's, and with o being the long root and x the short one (usually a darkened circle). Both with no symmetry.For SO(2n) or D
, the Dynkin diagram is o - o - o - ... - o (- o) - o with n o's. It has a symmetry of reflecting the two short branches, corresponding to using an O(2n) matrix with determinant -1 to be applied to the algebra generators.The exceptional ones are G2: o = x (three dashes in the =), no symmetry, F4: o - o = x - x, no symmetry, E6: o - o - o (- o) - o - o, end-to-end symmetry of reflecting the two long branches, E7: o - o - o - o (- o) - o - o, no symmetry, E8: o - o - o - o - o (- o) - o - o, no symmetry.
Kharakov
Quantum Hot Dog
Simplex convolution series division, results are r_whatever:
\(\begin{tabular}{ l | c c c c c } & a & b & c & d & e ... \\ \div & A & B & C & D & E ... \\ \hline r_0 & r_0 A & r_0 B & r_0 C & r_0 D & r_0 E \\ r_1 & r_1 A & r_1 B & r_1 C & r_1 D & \\ r_2 & r_2 A & r_2 B & r_2 C & & \\ r_3 & r_3 A & r_3 B & & & \\ r_4 & r_4 A & & & & \\ \end{tabular} r_0=\frac{a}{A} r_1=\frac{b-B r_0}{A} r_2=\frac{c- C r_0 - B r_1}{A} .... \)
Square convolution series division, results are r_whatever:
\(\begin{tabular}{ l | c c c c c } & a & b & c & d & e ... \\ \div & A & B & C & D & E ... \\ \hline r_0 & r_0 A & r_0 B & r_0 C & r_0 D & r_0 E \\ r_1 & r_1 A & r_1 B & r_1 C & r_1 D & r_1 E \\ r_2 & r_2 A & r_2 B & r_2 C & r_2 D & r_2 E \\ r_3 & r_3 A & r_3 B & r_3 C & r_3 D & r_3 E \\ r_4 & r_4 A & r_4 B & r_4 C & r_4 D & r_4 E\\ \end{tabular} r_0=\frac{a}{A} r_1=\frac{b-B r_0}{A + B} r_2=\frac{c- C (r_0 + r_1)}{A + B + C} r_3=\frac{d- D (r_0 + r_1 + r_2)}{A + B + C + D} .... \)
Square looks easier to code, but has "messier" results. I will code something for it tomorrow. For now, what I've been doing is dividing the harmonic series by other infinities generated by natural numbers.
Simplex convolution multiplication is Cauchy, I think. Square is a bit more complex. Infinite series have square roots. If you want to try a couple out:
Simplex infinity root (multiplications are simplex convolution, you might recognize these as the coefficients of a Taylor series for square roots):
\(\frac{(2n)!}{4^n n!^2} = 1 + \frac {1}{2} +\frac {3}{8} + \frac {5}{16} ....\)
SimplexRoot * SimplexRoot = 1+1+1+1...
(sqrt(A) * SR) * (sqrt(A) * SR) = A + A + A + A....
(1+1+1...) (1+1+1...) = 1+2+3+4....
sqrt(A) + sqrt(A) +sqrt(A)... = A + 2 A + 3A....
(1+2+3...)^2 = 1 +4+ 10...
(sqrt(2) + 2 sqrt(2) + 3 sqrt(2)...)^2 = 2 + 8 + 20...
Square infinity root for 1+1+1+1...:
\(1 +(\sqrt{2} - \sqr{1}) + (\sqrt{3} - \sqr{2}) + (\sqrt{4} - \sqr{3}) ... \)
Same operations on series as above, except use square convolution. I haven't checked square convolution Harmonic Zeroes yet.
I justify the "larger infinities" of the same Cantor set because. well, magic. There are Harmonic zeros that pop at at the infinities. Which is magic... not logic! Just joking. Same size infinities, but they aren't. They are distinct. They have behaviors near the Harmonic zeroes.
Each infinity is higher dimensional than the last. An infinite line (1+1+1+1...) is a smaller infinity than an infinite plane ([(1,1)=1]+[(2,1) (2,2)=2] +3+4...) or infinite volume. The "y" axis of the (1+1+1...)^2 is the position in the simplex diagonal... gnight.
Question about the zeroes still stands.
\(\begin{tabular}{ l | c c c c c } & a & b & c & d & e ... \\ \div & A & B & C & D & E ... \\ \hline r_0 & r_0 A & r_0 B & r_0 C & r_0 D & r_0 E \\ r_1 & r_1 A & r_1 B & r_1 C & r_1 D & \\ r_2 & r_2 A & r_2 B & r_2 C & & \\ r_3 & r_3 A & r_3 B & & & \\ r_4 & r_4 A & & & & \\ \end{tabular} r_0=\frac{a}{A} r_1=\frac{b-B r_0}{A} r_2=\frac{c- C r_0 - B r_1}{A} .... \)
Square convolution series division, results are r_whatever:
\(\begin{tabular}{ l | c c c c c } & a & b & c & d & e ... \\ \div & A & B & C & D & E ... \\ \hline r_0 & r_0 A & r_0 B & r_0 C & r_0 D & r_0 E \\ r_1 & r_1 A & r_1 B & r_1 C & r_1 D & r_1 E \\ r_2 & r_2 A & r_2 B & r_2 C & r_2 D & r_2 E \\ r_3 & r_3 A & r_3 B & r_3 C & r_3 D & r_3 E \\ r_4 & r_4 A & r_4 B & r_4 C & r_4 D & r_4 E\\ \end{tabular} r_0=\frac{a}{A} r_1=\frac{b-B r_0}{A + B} r_2=\frac{c- C (r_0 + r_1)}{A + B + C} r_3=\frac{d- D (r_0 + r_1 + r_2)}{A + B + C + D} .... \)
Square looks easier to code, but has "messier" results. I will code something for it tomorrow. For now, what I've been doing is dividing the harmonic series by other infinities generated by natural numbers.
Simplex convolution multiplication is Cauchy, I think. Square is a bit more complex. Infinite series have square roots. If you want to try a couple out:
Simplex infinity root (multiplications are simplex convolution, you might recognize these as the coefficients of a Taylor series for square roots):
\(\frac{(2n)!}{4^n n!^2} = 1 + \frac {1}{2} +\frac {3}{8} + \frac {5}{16} ....\)
SimplexRoot * SimplexRoot = 1+1+1+1...
(sqrt(A) * SR) * (sqrt(A) * SR) = A + A + A + A....
(1+1+1...) (1+1+1...) = 1+2+3+4....
sqrt(A) + sqrt(A) +sqrt(A)... = A + 2 A + 3A....
(1+2+3...)^2 = 1 +4+ 10...
(sqrt(2) + 2 sqrt(2) + 3 sqrt(2)...)^2 = 2 + 8 + 20...
Square infinity root for 1+1+1+1...:
\(1 +(\sqrt{2} - \sqr{1}) + (\sqrt{3} - \sqr{2}) + (\sqrt{4} - \sqr{3}) ... \)
Same operations on series as above, except use square convolution. I haven't checked square convolution Harmonic Zeroes yet.
I justify the "larger infinities" of the same Cantor set because. well, magic. There are Harmonic zeros that pop at at the infinities. Which is magic... not logic!
Just joking. Same size infinities, but they aren't. They are distinct. They have behaviors near the Harmonic zeroes.Each infinity is higher dimensional than the last. An infinite line (1+1+1+1...) is a smaller infinity than an infinite plane ([(1,1)=1]+[(2,1) (2,2)=2] +3+4...) or infinite volume. The "y" axis of the (1+1+1...)^2 is the position in the simplex diagonal... gnight.
Question about the zeroes still stands.
lpetrich
Contributor
I can't figure out what that's getting at.
But back to Lie-algebra automorphisms.
For SO, Tij,kl = (1/2)*(Aik*Ajl - Ail*Ajk)
where A is orthonormal: A.AT = AT.A = I. Thus, A is in the group O.
for j>i and l>k, det(T) = det(A)n-1.
For SO(2n), det(T) is -1 if det(A) = -1 (A is a reflection or a rotoreflection)
For SO(2n+1), det(T) is always 1 and det(A) is always 1
det(A) = 1: inner automorphism because A here is generated by SO
det(A) = -1: outer automorphism
The SO(2n) outer-automorphism group is thus Z2, and it corresponds to exchanging the algebra's two spinor reps. SO(2n+1) has only one spinor rep, and its outer-automorphism group is the identity group.
SO(8) is weird. Its vector rep and its two spinor reps have dimension 8, so they can all be interchanged.
-
Turning to SU and its relatives U, SL, and GL, I find Tij,kl = Aik*(A-1,T)jl with a crossed one: - Ail*(A-1,T)jk where A is a member of that group.
SO has that form also, but with the crossed form (i-l,j-k) equivalent to the straight form (i-k,j-l) because of its antisymmetry.
Sp(2n) is a bit tricky. Its group's matrices A satisfy A-1 = J.AT.J where J is the antisymmetric form {{0,-i*I},{i*I,0}}. J satisfies J.J = I. It is easy to show that J is in Sp(2n) Its algebra's generators L satisfy LT = - J.L.J
The straight form of T is thus Aik*(J.A.J)jl.
The crossed form of T is thus - Ail*(J.A.J)jk. Interchange k and l: (A.J)ik*(J.A)jl. Using J.(A.J).J = J.A shows that it is the straight form with A -> A.J Since J is in the group, A.J is also in the group.
I'm not sure about the exceptional algebras. G2, F4, E7, and E8 likely have crossed = straight, while E6 likely has crossed != straight.
But back to Lie-algebra automorphisms.
For SO
, Tij,kl = (1/2)*(Aik*Ajl - Ail*Ajk)where A is orthonormal: A.AT = AT.A = I. Thus, A is in the group O
.for j>i and l>k, det(T) = det(A)n-1.
For SO(2n), det(T) is -1 if det(A) = -1 (A is a reflection or a rotoreflection)
For SO(2n+1), det(T) is always 1 and det(A) is always 1
det(A) = 1: inner automorphism because A here is generated by SO
det(A) = -1: outer automorphism
The SO(2n) outer-automorphism group is thus Z2, and it corresponds to exchanging the algebra's two spinor reps. SO(2n+1) has only one spinor rep, and its outer-automorphism group is the identity group.
SO(8) is weird. Its vector rep and its two spinor reps have dimension 8, so they can all be interchanged.
-
Turning to SU
and its relatives U, SL, and GL, I find Tij,kl = Aik*(A-1,T)jl with a crossed one: - Ail*(A-1,T)jk where A is a member of that group.SO
has that form also, but with the crossed form (i-l,j-k) equivalent to the straight form (i-k,j-l) because of its antisymmetry.Sp(2n) is a bit tricky. Its group's matrices A satisfy A-1 = J.AT.J where J is the antisymmetric form {{0,-i*I},{i*I,0}}. J satisfies J.J = I. It is easy to show that J is in Sp(2n) Its algebra's generators L satisfy LT = - J.L.J
The straight form of T is thus Aik*(J.A.J)jl.
The crossed form of T is thus - Ail*(J.A.J)jk. Interchange k and l: (A.J)ik*(J.A)jl. Using J.(A.J).J = J.A shows that it is the straight form with A -> A.J Since J is in the group, A.J is also in the group.
I'm not sure about the exceptional algebras. G2, F4, E7, and E8 likely have crossed = straight, while E6 likely has crossed != straight.
Kharakov
Quantum Hot Dog
Before I explained the "infinite series division" method I was using, I mentioned that there is a set of specific sequences of non decreasing natural numbers that you could use to get zero sums (that I call "harmonic 0s") when dividing the harmonic series by the series formed by the non-decreasing natural number sequences.
Here's an example of a harmonic 0: (1 + 1/2 + 1/3...) / (1+1+1...) = 1 -1/2 -1/6 -1/12.... = 0
I asked if there were any other non-decreasing natural number sequences, NOT related to the simplex number sequences (1+1+1...)^n that generate harmonic 0s.
2+2+2... or 1/2+ 1/2 +1/2... give you harmonic 0s as well, as do other a*(1+1+1...)^n. The question excludes zeroes by series comparison between the harmonic series and the non decreasing natural number sequence series (because the one series "approaches" infinity much faster). These 0s are based on invalid reasoning about division by infinite series.
Division method:
results of division are in left hand columns:
\( \begin{tabular}{ r | r r r r r r} & 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & ... \\ \div & 1 & 1 & 1 & 1 & 1& ... \\ \hline 1 & 1 & 1 & 1 & 1 & 1 \\ -\frac {1}{2} & -\frac {1}{2} & -\frac {1}{2} & -\frac {1}{2} & -\frac {1}{2} & \\ -\frac {1}{6} & -\frac {1}{6} & -\frac {1}{6} &-\frac {1}{6} & & \\ -\frac {1}{12} & -\frac {1}{12} & -\frac {1}{12} & & & \\ -\frac {1}{20} & -\frac {1}{20} & & & & \\ \end{tabular} \) ................ \(\begin{tabular}{ r | r r r r r r } & 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5}& ... \\ \div & 1 & 2 & 3 & 4 & 5 & ... \\ \hline 1 & 1 & 2 & 3 & 4 & 5 \\ -\frac {3}{2} & -\frac {3}{2} & -3 & -\frac {9}{2} & -6 & \\ +\frac {1}{3} & \frac {1}{3} & \frac {2}{3} & 1 & & \\ +\frac {1}{12} & \frac {1}{12} & \frac {1}{6} & & & \\ +\frac {1}{30} & \frac {1}{30} & & & & \\ \end{tabular} \)
So it's 1-1 for the first division (by 1+1+1+...) and 1/2-1/2 for the second (by 1+2+3...).
Here are a couple of the series:
The square convolution I mention for (1+1+1...)^2 isn't technically correct- I don't know if this is something that Cauchy proved, which is why the simplex product is called the Cauchy product, despite the fact that this leads people away from valuable information: that the product is directly related to simplexes and simplex number sequences.
(1+1+1...)^2 = does = 1+2+3+4.... because you're multiplying the first term in the series by (1+1+1+1...), then the second term, then the third, so you have:
\(\begin{tabular}{ l | c c c c c } & 1 & 1 & 1 & 1 & 1 ... \\ \times & 1 & 1 & 1 & 1 & 1 ... \\ \hline & 1 & 1 & 1 & 1 & 1... \\ + & & 1 & 1 & 1 & 1 ...\\ +& & & 1 & 1 & 1 ..\\ +& & & & 1 & 1... \\ +& & & & & 1 ...\\ \hline & 1 & 2 & 3 & 4 & 5... \\ \end{tabular} \) |||||||||| \( \begin{tabular}{ l | c c c c c } &1 & 1 & 1 & 1 & 1 ... \\ \times & 2 & 1 & 1 & 1 & 1 ... \\ \hline & 2 & 2 & 2 & 2 & 2... \\ + & & 1 & 1 & 1 & 1 ...\\ +& & & 1 & 1 & 1 ..\\ +& & & & 1 & 1... \\ +& & & & & 1 ...\\ \hline & 2 & 3 & 4 & 5 & 6... \\ \end{tabular} \)|||||||||| \( \begin{tabular}{ l | c c c c c } &2 & 1 & 1 & 1 & 1 ... \\ \times & 1 & 1 & 1 & 1 & 1 ... \\ \hline & 2 & 1 & 1 & 1 & 1... \\ + & &2 & 1 & 1 & 1 ...\\ +& & & 2 & 1 & 1 ..\\ +& & & & 2 & 1... \\ +& & & & & 2 ...\\ \hline & 2 & 3 & 4 & 5 & 6... \\ \end{tabular} \) ||||||| \( \begin{tabular}{ l | r r r r r r} &1 & \frac{1}{2} & \frac{3}{8} & \frac{5}{16} &\frac{35}{128} & ... \\ \times &1 & \frac{1}{2} & \frac{3}{8} & \frac{5}{16} &\frac{35}{128} & ... \\ \hline & 1 & \frac{1}{2} & \frac{3}{8} & \frac{5}{16} &\frac{35}{128} & ... \\ + & &\frac{1}{2} & \frac{1}{4} & \frac{3}{16} &\frac{5}{32} & ... \\ +& & & \frac{3}{8} & \frac{3}{16} &\frac{9}{64} & ... \\ +& & & & \frac{5}{16} &\frac{5}{32} & ... \\ +& & & & &\frac{35}{128} & ... \\ \hline & 1 & 1 & 1 & 1 & 1 & ... \\ \end{tabular} \)
Some properties of the specific harmonic 0s:
Dividing a smaller harmonic 0 by a larger harmonic 0 results in 0.
Dividing a larger harmonic 0 by a smaller harmonic 0 results in a specific infinity related to relative position of harmonic 0s.
First four harmonic 0s:
H0_1/ H0_2 = 0
H0_2/ H0_1 = 1+1+1....
H0_2/ H0_4 = 0
H0_4/H0_2 = (1+1+1...)^2 =1+2+3+4... (the division is below)
In the following division, the diagonals sum to the original series (the numerator), like mentioned at the top of this post. The results are in the first column. The second column is the first column times 1, which is the first value in the denominator (divisor).
\( \begin{tabular}{ r | r r r r r r} & 1 &-\frac{3}{2} & \frac{1}{3} & \frac{1}{12}& \frac{1}{30}& ... \\ \div & 1 & - \frac{7}{2} & \frac{13}{3} & -\frac{25}{12} & \frac{1}{5} & ... \\ \hline 1 & 1 & - \frac{7}{2} & \frac{13}{3} & -\frac{25}{12} & \frac{1}{5} & ... \\ 2 & 2 & -7 & \frac {26}{3} & - \frac{25}{6} & \\ 3 & 3 & -\frac {21}{2} & 13 & & \\ 4 & 4 & -14 & & & \\ 5 & 5 & & & & \\ \end{tabular} \)
Some questions:
It doesn't seem that Ramanujan summation works, if you are strict. Are they justified, or are they fast and loose solutions that appear to get us to the right answer the wrong way?
What about the physical case with 1+2+3+4... = -1/12? Are there other numbers associated with the physic's properties associated with that series? Are the claims about the physical properties bullshit?
Last, and first question: Are there any non simplex (not a*(1+1+1...)^n) series that result in harmonic 0s when you divide the harmonic series by the non simplex related series?
A voice in my head said "Ho Ho", and I thought "Ho" and laughed. Am I sane?