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Hyperoperations by Hamilton Nudgebanks - Wed, 29 Mar 2017 16:37:10 EST ID:uqJv93qR No.15439 Ignore Report Quick Reply
File: 1490819830015.png -(109711B / 107.14KB, 790x726) Thumbnail displayed, click image for full size. 109711
Does anyone have any ideas about pic related?
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Phineas Beblingstork - Fri, 31 Mar 2017 11:02:27 EST ID:A2j/BW/W No.15440 Ignore Report Quick Reply
Notice how the degree of the polynomials for the sums is always one higher than the exponent on the terms. If the exponent increases for subsequent terms, it makes no sense that such a sum could be expressed as a polynomial. There may be a formula, but I think we can safely rule out polynomial expressions.
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Edward Semmlespear - Fri, 31 Mar 2017 14:34:49 EST ID:MnYRJPQ7 No.15441 Ignore Report Quick Reply
>>15440
Another way to see this is to note that the hypersquare sum grows superexponentially in n, which is faster than any polynomial. The formula must grow at least this fast. The sum of powers is already a polynomial in n, but we can find a very simple polynomial expression which is equal to it. Is it possible that such an expression for this new sum exists in terms of the tetration operation? Can we rule out this possibility somehow?
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George Cudgeshaw - Fri, 31 Mar 2017 21:33:54 EST ID:A2j/BW/W No.15442 Ignore Report Quick Reply
Hmm, according to https://oeis.org/A001923 if the sum is S(n), then S(n+1)/S(n) is asymptotic to e*n. This means the sum grows like (e^n)(n!).
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Phyllis Bunford - Sat, 01 Apr 2017 13:39:24 EST ID:vrOFV9fT No.15443 Ignore Report Quick Reply
1491068364081.png -(6078B / 5.94KB, 329x78) Thumbnail displayed, click image for full size.
>>15439

I think I figured it out. Recently interest has been centered on describing locally elliptic, reducible, pairwise sub-local equations but there are people interested in this too.

These operations are usually super-invertible and standard techniques can demonstrate that this type of hyper operation has relationships to the critical points of Noetherian irreducible rings and anti-unconditionally Hausdorff sub-symmetric stochastic systems.
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Polly Sendlehadging - Sat, 01 Apr 2017 13:43:29 EST ID:KW3RHxlP No.15444 Ignore Report Quick Reply
>>15443
Is that mathgen?
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Isabella Furringmock - Sun, 02 Apr 2017 13:30:15 EST ID:vrOFV9fT No.15445 Ignore Report Quick Reply
>>15444

Indeed
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Wesley Snodforth - Mon, 03 Apr 2017 21:32:12 EST ID:A2j/BW/W No.15446 Ignore Report Quick Reply
1491269532789.gif -(1891B / 1.85KB, 277x87) Thumbnail displayed, click image for full size.
>>15442
>This means the sum grows like (e^n)(n!).
Whoops, I messed this up. I tried to work backwards from the previous statement, but you can't do that. The sum (it's actually called the hypertriangular function) is actually asymptotic to n^n. This means the sum's highest term is so much greater than the other terms that they don't effect the growth rate. And by Stirling's approximation, n^n ~ e^n*n!/sqrt(2*pi*n). From here, you can work out S(n+1)/S(n) ~ e*n. As far as I can tell, there is no known closed-form formula.
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Wesley Snodforth - Mon, 03 Apr 2017 22:23:19 EST ID:A2j/BW/W No.15447 Ignore Report Quick Reply
http://www.ijpam.eu/contents/2007-36-2/9/9.pdf
http://sci-hub.ac/10.2307/2306261
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Simon Hushstone - Tue, 04 Apr 2017 21:03:52 EST ID:ojvh+mCX No.15448 Ignore Report Quick Reply
>>15447
It's nice to see some more serious interest in these things. While the connection with polynomial arithmetic is somewhat geometric, I would like to see a general geometric object that corresponds to these quantities in the same way that one can deduce the sum of squares by arranging rectangles.

n^n is the number of functions from a set to itself, but how do I turn this into a shape in the "correct" way?
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Phoebe Hondlepun - Wed, 05 Apr 2017 11:25:05 EST ID:A2j/BW/W No.15449 Ignore Report Quick Reply
>>15448
>one can deduce the sum of squares by arranging rectangles
Show what you mean by this.
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Simon Hushstone - Wed, 05 Apr 2017 19:31:58 EST ID:ojvh+mCX No.15450 Ignore Report Quick Reply
>>15449
If you try looking up what I mentioned you will find this.

http://www.maa.org/sites/default/files/Kanim.MathMag.2001.pdf

It is not what I meant. I don't have time to think about it right now, but the method I was shown was cleaner. I can produce an image later, say in a week when I'm less busy, but basically you just place a square of area 1 next to a square of area 4 next to a square of area 9 and so on so that their bottom edges line up. Consider the smallest rectangle you can form which encloses these squares. The "rows" in the rectangle are the successive triangular numbers. Measure the area of the rectangle in two different ways to obtain the formula.
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Molly Crenkinshit - Thu, 06 Apr 2017 12:54:30 EST ID:Ku9rvp0b No.15451 Ignore Report Quick Reply
1491497670945.png -(80820B / 78.93KB, 739x680) Thumbnail displayed, click image for full size.
Found it in David Burton's History of Mathematics. He just makes the argument without giving a source.
>>
Frederick Sondleham - Fri, 07 Apr 2017 02:02:02 EST ID:A2j/BW/W No.15454 Ignore Report Quick Reply
>>15450
>>15451
Interesting. I learned to sum series like that algebraically, never seen these geometric approaches. I like the derivation on Wikipedia: https://en.wikipedia.org/wiki/Square_pyramidal_number#Derivation_of_the_summation_formula

With summing squares, all the terms are squares; with summing cubes, all the terms are cubes; and so on. With these sums, you have a common currency: area for the sum of squares, volume for the sum of cubes, etc. But with the hypertriangular sum, you have a line of length 1, added to a square of length 2, added to a cube of length 3, added to a hypercube of length 4, etc. This approach doesn't seem fruitful.
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Rebecca Sibbersine - Fri, 07 Apr 2017 09:30:04 EST ID:oAD+Ltu+ No.15455 Ignore Report Quick Reply
>>15454
You could make the same objection when summing squares.

>With summing multiples of 2, all the terms are multiples of 2; with summing multiples of 3, all the terms are multiples of 3; and so on. With these sums, you have a common currency: 2 for the sum of multiples of 2, 3 for the sum of multiples of 3, etc. But with the sum of squares, you have a line of length 1, added to a line of length 4, added to a line of length 9, added to a line of length 16, etc.

Obviously this is only a problem for someone who doesn't understand the geometric import of squaring. It sounds quaint to us. We may think that we understand hypersquares, but this problem suggests that thinking of them as cubes may be as parochial as thinking of 3^2 as a line of length 9.

I would also advise people to not give up so easily. If you pursue the n-dimensional version of Burton's method you will come up with a formula. It won't be closed form and it will require you to do n-dimensional geometry, but you can come up with something. What I hope is that a better understanding of what these numbers should represent will give an easier answer.
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Frederick Sondleham - Fri, 07 Apr 2017 12:06:45 EST ID:A2j/BW/W No.15456 Ignore Report Quick Reply
>>15455
>If you pursue the n-dimensional version of Burton's method you will come up with a formula.
Show me. How do you do this for the 3D case for instance?
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Rebecca Sibbersine - Fri, 07 Apr 2017 12:15:12 EST ID:oAD+Ltu+ No.15457 Ignore Report Quick Reply
>>15456
The 3D case isn't that hard. I'll leave it for a little while for someone else to try and post, since I do have actual work that I'm supposed to be doing right now.
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Reuben Hecklewill - Fri, 21 Apr 2017 01:44:04 EST ID:A2j/BW/W No.15478 Ignore Report Quick Reply
1492753444473.png -(6748B / 6.59KB, 809x145) Thumbnail displayed, click image for full size.
I thought I found a pattern, but when you get to the fourth line in pic related, it messes up. Where is the additional 4 term coming from?
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Doris Claystock - Fri, 21 Apr 2017 10:06:14 EST ID:BwZBkGcf No.15479 Ignore Report Quick Reply
>>15478
Did you find this pattern by doing something geometric or just by messing around? I would guess the latter since your third line is stated a little weirdly for someone who is counting cubes in the way I'm thinking. However you did it, the problem is likely that at each stage there are more different kinds of lower-dimensional "holes" to fill in when counting the volume of the Burton figure in two different ways. In the 2D case you just have to add in the "row" areas, but in the 3D case there are 2D "faces" and 1D "rows". To make matters worse, you are actually stacking parallelpipeds which are not all cubes in order to compute the sum we are interested in.
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Walter Nushmet - Sat, 22 Apr 2017 04:17:08 EST ID:A2j/BW/W No.15480 Ignore Report Quick Reply
>>15479
I was trying to apply the same reasoning behind Burton's method to the hypertriangular sum. It seems to only work for dimensions up to 3.
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Walter Nushmet - Sat, 22 Apr 2017 06:08:13 EST ID:A2j/BW/W No.15481 Ignore Report Quick Reply
1492855693219.png -(7598B / 7.42KB, 808x148) Thumbnail displayed, click image for full size.
Never mind, I figured it out.
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Jenny Ferrynirk - Sat, 22 Apr 2017 11:22:20 EST ID:u3qow64e No.15482 Ignore Report Quick Reply
>>15481
I've tried doing an analogue of Burton's method in two different ways but I'm not sure how you arrived at that sum. Can you prove that it works in general? I guess next steps are to check it for a bunch of values and then attempt a rigorous proof if it lools good.
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Jenny Ferrynirk - Sat, 22 Apr 2017 11:22:43 EST ID:u3qow64e No.15483 Ignore Report Quick Reply
>>15481>>15481
I've tried doing an analogue of Burton's method in two different ways but I'm not sure how you arrived at that sum. Can you prove that it works in general? I guess next steps are to check it for a bunch of values and then attempt a rigorous proof if it looks good.
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Jenny Ferrynirk - Sat, 22 Apr 2017 11:23:11 EST ID:u3qow64e No.15484 Ignore Report Quick Reply
>>15481
I've tried doing an analogue of Burton's method in two different ways but I'm not sure how you arrived at that sum. Can you prove that it works in general? I guess next steps are to check it for a bunch of values and then attempt a rigorous proof if it looks good.
>>
Basil Bardwell - Sat, 22 Apr 2017 12:12:54 EST ID:jKEmS4eI No.15485 Ignore Report Quick Reply
1492877574393.png -(42764B / 41.76KB, 1010x210) Thumbnail displayed, click image for full size.
I checked the first thousand values of the proposed formula. It looks good.
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Basil Bardwell - Sat, 22 Apr 2017 12:18:30 EST ID:jKEmS4eI No.15486 Ignore Report Quick Reply
>>15485
Now that I've taken a second to just expand the formula I see that it obviously is true, and I have a complaint. I would still like to see how it's geometric, but this formula is not ideal in the sense that you are pretty much just summing the 4-hypersquares in disguise. The nice thing about the sum of squares formula is that it reduces a long computation to a short one, but this formula doesn't really do that. Arguably it takes more steps than the original expression.
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Samuel Lightfuck - Sun, 23 Apr 2017 16:05:01 EST ID:A2j/BW/W No.15487 Ignore Report Quick Reply
>>15482
>I've tried doing an analogue of Burton's method in two different ways but I'm not sure how you arrived at that sum.
Show me what you did first.
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David Blythestock - Sun, 23 Apr 2017 19:46:31 EST ID:KPDi1EOJ No.15488 Ignore Report Quick Reply
>>15487
Why so secretive? I'll draw you a picture of each of the ways I tried some time later this week. Can't be assed to describe it in words right now, and in any case I wasn't able to finish the double-counting either way I did it.


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