Talk:Square pyramidal number: Difference between revisions

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==Untitled==

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

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::::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.) [[User:Frivolous Consultant|Frivolous Consultant]] ([[User talk:Frivolous Consultant|talk]]) 23:21, 25 October 2012 (UTC)

::::::::If people "don't care enough about this topic" it may not be appropriate for wikipedia, meaning not notable--[[User:345Kai|345Kai]] ([[User talk:345Kai|talk]]) 03:12, 23 June 2015 (UTC)

== Quadrature of the Parabola with the "square pyramidal number" (new proof) ==

[[File:Figuraxy.jpg|thumb|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

== Quadrature of the Parabola with the "square pyramidal number" (new proof) ==

Below we show a summary of the proof.

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 canis readreadable at the following web adress:

https://drive.google.com/file/d/0B4iaQ-gBYTaJMDJFd2FFbkU2TU0/view?usp=sharing

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.) = 6^{<small>.</small>}1 + 5^{<small>.</small>}3 + 4^{<small>.</small>}5 + 3^{<small>.</small>}7 + 2^{<small>.</small>}9 + 1^{<small>.</small>}11 = 91 green triangles. (3)

The sum (3) can be written:

:A(cir.) = 6 + 11 + 15 + 18 + 20 + 21 , that is:

:6+

:6+5+

:6+5+4+

:6+5+4+3+

:6+5+4+3+2

:6+5+4+3+2+1

or rather:

: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 n^{<small>3</small>} green triangles

# A_n(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, ....., P_n/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 P_n.

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'' !

The rest came by itself.--[[User:Ancora Luciano|Ancora Luciano]] ([[User talk:Ancora Luciano|talk]]) 18:47, 14 May 2013 (UTC)

: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 [http://mathworld.wolfram.com/DiagonalParadox.html 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 [[WP:RS|reliable source]] here. Personally maintained web sites are not adequate for this purpose. —[[User:David Eppstein|David Eppstein]] ([[User talk:David Eppstein|talk]]) 22:48, 14 May 2013 (UTC)

:::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 (Σ_n n^m)/n^m+1 = 1/(m+1) for each m of N

^n→∞¹

Note that the inverse limit (Σ fom zero to n) diverges, reproducing the sequence of natural numbers. <small><span class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Ancora Luciano|Ancora Luciano]] ([[User talk:Ancora Luciano|talk]] • [[Special:Contributions/Ancora Luciano|contribs]]) 15:25, 18 August 2013 (UTC)</span></small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->

== Sum of the first ''n'' squares (geometrical proof) ==

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{{edit semi-protected|<!-- Page to be edited -->|answered=yes}}

<!-- FuckBegin merequest crazy-->

Something seems amiss in the sentence starting "Now there". I suggest replacing "Now there" with "There are".

<!-- End request -->

[[User:Bill01568|Bill01568]] ([[User talk:Bill01568|talk]]) 16:29, 30 December 2013 (UTC)

:Seems reasonable, [[File:Yes check.svg|20px|link=|alt=]] '''Done'''<!-- Template:ESp -->. ''<b style="font-variant:small-caps;">[[User:Little Mountain 5|<fontspan colorstyle="color:black;">Little</fontspan>]][[User talk:Little Mountain 5|<fontspan colorstyle="color:red;">Mountain</fontspan>]][[Special:Contribs/Little Mountain 5|<fontspan colorstyle="color:#00008B;">5</fontspan>]]</b>'' 22:35, 30 December 2013 (UTC)

== Simplification of the current proof ==

It seems that current proof given in the article could be simplified by using <math>1^2+2^2+\ldots+n^2=1+2+2+3+3+3+\ldots+n+n+\ldots+n</math> instead of <math>1^2+2^2+\ldots+n^2=(2n-1)+(2n-3)+(2n-3)+\ldots+1+1+\ldots+1</math>, with sum of the column still being <math>2n+1</math>. Unfortunately, I really can't be bothered to find citations for this proof. Fortunately, there are no citations for the current proof as well, so changing it won't make it worse. Opinions?

[[User:MYXOMOPbI4|MYXOMOPbI4]] ([[User talk:MYXOMOPbI4|talk]]) 03:51, 7 September 2017 (UTC)

:I'm not sure I see the point of long algebraic derivations at all, given that it's so straightforward to prove any such formula by induction. If you could replace this with a short conceptual visual (and sourced!) proof, that would be better, I think. —[[User:David Eppstein|David Eppstein]] ([[User talk:David Eppstein|talk]]) 04:06, 7 September 2017 (UTC)

::The proof by induction is indeed not that hard, but you need to know the formula in the first place. This way you get the formula automatically, and the proof isn't long at all (it can be done with one picture, http://forumbgz.ru/user/upload/file580638.jpg (with some unimportant text in russian)). But I'm not sure if some random picture on the internet is considered a source. [[User:MYXOMOPbI4|MYXOMOPbI4]] ([[User talk:MYXOMOPbI4|talk]]) 22:53, 7 September 2017 (UTC)

:::It's more or less obvious that the formula is a cubic polynomial and from that and the first four values you can immediately derive it by finite differences. And no, random pictures on the internet, especially from a site whose address syntax suggests that it's an open wiki, are not reliable sources. —[[User:David Eppstein|David Eppstein]] ([[User talk:David Eppstein|talk]]) 00:27, 8 September 2017 (UTC)

== Link addition ==

A link to Lucas Numbers should be provided for completeness. [[Special:Contributions/199.209.255.246|199.209.255.246]] ([[User talk:199.209.255.246|talk]]) 14:40, 10 September 2018 (UTC)

:Only if the connection can be sourced. Otherwise the unsourced paragraph mentioning Lucas should be removed. —[[User:David Eppstein|David Eppstein]] ([[User talk:David Eppstein|talk]]) 16:33, 10 September 2018 (UTC)

== Archimedes ==

I don't see any mention of Archimedes, who probably gave the first formula for the sum of squares of the first n natural numbers in his book 'On Spirals'. It doesn't at first sight look like the modern formula, and the derivation is horribly complicated, but it is there after all. See the discussion in Heath's edition of Archimedes, especially on page 109 in the Dover edition.[[Special:Contributions/2A00:23C8:7906:1301:A453:48F9:36D5:B594|2A00:23C8:7906:1301:A453:48F9:36D5:B594]] ([[User talk:2A00:23C8:7906:1301:A453:48F9:36D5:B594|talk]]) 21:23, 7 March 2021 (UTC)

:Added. I used the 1897 edition rather than the Dover edition, but the pagination is the same. —[[User:David Eppstein|David Eppstein]] ([[User talk:David Eppstein|talk]]) 00:43, 8 March 2021 (UTC)

{{Talk:Square pyramidal number/GA1}}

==Did you know nomination==

{{Template:Did you know nominations/Square pyramidal number}}