In all articles found the formula for the square pyramidal number is proofed by using the inductive methode. Isn't there a direct proof?
Jon van den Helder
I thought inductive was direct. An empirical proof (e.g., having a computer calculate the first thousand values using each formula and then having it compare that the lists are all the same) would convince more people, but symbological snobs would look down on it. Anton Mravcek17:49, 14 July 2006 (UTC)Reply
Latest comment: 12 years ago5 comments3 people in discussion
I have an immense amount of experimentally discovered things I could add to "Squares in a Square" which involve different grid shapes and altogether different shapes such as triangles. So much so that I could very easily create a whole other Wikipedia article on it. I would enjoy doing so, but I thought I would get a second opinion. In addition, although my stuff is definitely true, as I said, I figured out this stuff experimentally, so I have nothing to reference but myself. I know that isn't a very good answer, but I could really improve the depth of understanding and knowledge of that topic. I hope you will comment. Frivolous Consultant (talk) 22:56, 17 October 2012 (UTC)Reply
You need to find a published source for this material if you want to include it here; see Wikipedia:Verifiability and Wikipedia:No original research. In addition, if the solution doesn't involve the same sequence of numbers, it may be a bit off-topic for this article, because this article is not about square-in-square type problems, it's about the sequence of numbers. That doesn't mean it can't be mentioned at all (after all, we do mention the problem of rectangles in a square grid) but it means that it would probably be inappropriate to include much detail. —David Eppstein (talk) 23:01, 17 October 2012 (UTC)Reply
I sort-of expected this answer, but I don't know what to do. People don't care enough about this topic, so no one makes anything about it. I've searched all over the Internet, but I haven't found anything. Although I haven't searched any books, I wouldn't know where to start, and besides, I would be exceptionally surprised if they had anything of value in them. If anyone can find anything, I'll defintely cite it, but as of now, I have nothing. I've thought about creating a formal website so that I have something to source, but I don't know how to do that. This isn't the first time something like this has happened to me before either. I could expand on that, but I'm already taking up too much space. I could add a formula to the "Squares in a square" part to improve it that wouldn't take up too much space and go off topic, but it still wouldn't have a reference. If you ask, I could figure out how to put it on here so you can look it over. That would be only a measely fraction of the unreferenceable knowledge I have on the topic. Also, I was suggesting creating a whole other page on Wikipedia for putting the rest and/or moving "Squares in a square" there, not adding a bunch to this article. I tried putting the suggestion on Wikipedia: Articles for creation today, but I couldn't get it to show up. I would appreciate help. Frivolous Consultant (talk) 00:20, 19 October 2012 (UTC)Reply
Have you tried the Online Encyclopedia of Integer Sequences? You should be able to generate some sequences of numbers from your results and plug them into OEIS. I think it's pretty likely that you'll find some of your results there. You may also find references, maybe enough to write an article. Ozob (talk) 01:50, 21 October 2012 (UTC)Reply
That website helped a lot. There is so much information on there about what I was looking for that it gave me a slight inferiority complex because of how many hours and hours I spent working on this stuff when it took me under a second to get a full sequence along with formulas. To my relief though, there was a few things that it didn't know, so I don't feel as bad, but it knows enough to give a good resorce. Be prepared for a new article on the subject, but there are two more things. First, I don't know what I should title it; I've recently been refering to the topic as tessellation conglomerates for lack of a better term, but that name is completely made up by me. Also, when I finish, it might be good to move "Squares in a square" to the page. (I realize what I've been typing takes up a lot of room. I won't be offended if you delete my previous entries.) Frivolous Consultant (talk) 23:21, 25 October 2012 (UTC)Reply
Quadrature of the Parabola with the "square pyramidal number" (new proof)
Latest comment: 12 years ago6 comments2 people in discussion
Quadrature of the Parabola
I found that the "square pyramidal number" can be used to prove the Archimedes' theorem on the area of parabolic segment. The entire proof, carried out without the use of "mathematical analysis", one can read at the following web adress:
https://sites.google.com/site/leggendoarchimede/12---quadratura
Below we show a summary of the proof.
Proposition: The area of parabolic segment is a third of the triangle ABC.
Divide AB and BC into 6 equal parts and use the green triangle as measurement unit of the areas.
The triangle ABC contains:
(1+3+5+7+9+11).6 = 62.6 = 63 green triangles.
The parabola circumscribed figure (in red) contains:
A(cir.) = sum of the squares of first 6 natural numbers !
Generally, for any number n of divisions of AB and BC, it is:
The triangle ABC contains n3 green triangles
An(cir.) = sum of the squares of first n natural numbers
So, the saw-tooth figure that circumscribes the parabolic segment can be expressed with the "square pyramidal number" of number theory!
For the principle of mathematical induction, this circumstance (which was well hidden in (3)) we can reduce the proof to the simple check of the following statement:
the sequence of the areas ratio: 1, 5/8, 14/27, 30/64, ....., Pn/n3, ..... tends at number 1/3, as n tends to infinity (4a)
where the numerator of the sequence terms is the nth square pyramidal number Pn.
But (4a) state that: the area (measured in green triangles) of the circumscribed figure is one-third the area of the triangle ABC, at the limit of n = infinity. End of proof
This proof is very beautiful! Notice its three essential steps:
Choice of equivalent triangles for measuring areas.
With this choice, the area of triangle ABC measure n3 triangles.
Counting the number of triangles in the saw-tooth figure that encloses the parabolic segment and discovery that, for each number n of divisions, this number is the square pyramidal number !
That does indeed seem to be a valid and nice proof (or at least something that could be made into a proof with a little more care about why the limit of the saw-tooth areas converges to the parabola area (in contrast to situations like this one for which the convergence argument doesn't work). But either it's not new or it doesn't (yet) belong here; see WP:NOR. So if you think it's new, the appropriate thing to do would be to get it properly published elsewhere first, so that the publication can be used as a reliable source here. Personally maintained web sites are not adequate for this purpose. —David Eppstein (talk) 22:48, 14 May 2013 (UTC)Reply
In the article in pdf format contained in the first figure, to step 2 of the discussion, is written the following general formula, derived with Excel:
n
lim (Σn nm)/nm+1 = 1/(m+1) for each m of N
n→∞ 1
We represent the square pyramidal number P6 = 91 with cubes of unit volume, as shown, and inscribe in building a pyramid (in red). Let V6 the volume of the inscribed pyramid. To obtain P6 you may add to V6 the excess external volume to the red pyramid. Such excess is: 2/3 for each cube placed on the central edge, and 1/2 for the cubes forming the steps of the building (enlarge for a better look of highlighted part). Then, calculating one has:
P6 = V6+(2/3)*6+(1+2+3+4+5)
For the induction principle, will be:
Pn = Vn+(2n)/3+Σn(from 1 to n-1)n
Pn = n3/3+2n/3+(n2+n)/2-n
Pn = (2n3+3n2+n)/6
Geometric representation of the square pyramidal number using cubes, instead of sferes, is more useful.
George Polya, in his book "Mathematical discovery", Volume 2, 1968, presents this solution saying it "rained from the sky", as obtained algebraically with a trick, like a rabbit drawn out from the hat.
Returning to the introduction to this talk page, seems that this (realistic) geometrical derivation is a little more direct than the algebraic presented in the article.
