Talk:Triangular number

Overcomplicated Result notation

Please, could you edit the result formula, removing the extremely unpopular "binomial coefficient" by the more popular n * (n+1) / 2 "as the Last Result" ? if you want, you can add the reference to the binomial coefficient in a side note, because when the vast majority of the people read "open parenthesis" (n+1) / 2 "close parenthesis", they will just discard the parenthesis because it means nothing when used as the first and last characters in algebraic notation (which will be the CONTEXT employed to read this result, since up to that point, just simple algebra was used. Please, take this info into account (if the objective is give people answers instead of letting them confused, trying to find why the "n" was dropped in the final result, it is just "(n+1)/2" ? -- Please.

Popular Culture

Personally, I wouldn't consider the bible being "pop culture" — Preceding unsigned comment added by 211.29.185.4 (talk) 06:08, 15 June 2007 (UTC)

Yeah, it’s religious, and been around for hundreds of years. So saying that it’s popular culture doesn’t make sense. — Preceding unsigned comment added by 2601:140:8a00:2059:fde6:9b72:9a53:1d4 (talk) 13:01, 1 March 2019 (UTC)

Alternate notation

Does anyone else support the inclusion of other, perhaps less common, forms of notation? I was able to dig up discussion here as well as some additional support for the n? notation here. 67.171.222.203 (talk) 22:53, 15 April 2014 (UTC)

Exponential functions of Triangular numbers

These formulas apply to any x > 0

T(x²) = T(x-1)² + T(x)²

T(1) = 0² + 1² = 1
T(4) = 1² + 3² = 10
T(9) = 3² + 6² = 45

PROOF
T(x²) = [x² * (x² + 1)] / 2 = [x⁴ + x²] / 2

T(x-1)^2 = [x⁴ - 2x³ + x²] / 4
T(x)^2 = [x⁴ + 2x³ + x²] / 4

T(x)² + T(x-1)²
= [2x⁴ + 2x²] / 4
= [x⁴ + x²] / 2

edit: Whoops, just noticed that this one is redundant. My bad.

Next, any T(x)² is equal to the sum of cubes from 1 to x
T(1)² = 1³ = 1 (1²)
T(2)² = 1³ + 2³ = 9 (3²)
T(3)² = 1³ + 2³ + 3³ = 1 + 8 + 27 = 36 (6²)
1 + 8 + 27 + 64 = 100 (10²)

I have not found a formula to express this (as is expected with iterative functions) but it can be verified in about 20 lines of java code.

Finally, 6*T(x) + 1 = (x+1)³ - x³
2³ - 1³ = 8 - 1 = 7 = 6 * T(1) + 1
3³ - 2³ = 27 - 8 = 19 = 6 * T(2) + 1

PROOF
(x+1)³ = x³ + 3x² + 3x + 1
-x³ = 3x² + 3x + 1

6 * T(x) + 1 = 6 * (x² + x) / 2 + 1 = 3x² + 3x + 1

Hopefully these can be put to use. They seem to be similar in structure to existing algebraic formulas, but a bit more straightforward. Jozbornn (talk) 06:50, 2 March 2015 (UTC)

The sum of cubes fact that you state is well known — see squared triangular number. But if you want any of this to be incorporated into the article (and why else would you be posting it here?) you'll need to find reliable sources that already state these facts — otherwise it would be original research. —David Eppstein (talk) 07:24, 2 March 2015 (UTC)

Assessment comment

The comment(s) below were originally left at Talk:Triangular number/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Comment(s)

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Currently one long unstructured section consisting of a sequence of unrelated facts. Needs organization and editing. —David Eppstein 01:44, 24 April 2007 (UTC) Add Fermat's Theorem on triangular numbers (first proved by Gauss). I think that a mention of the theorem that every positive integer can be represented as the sum of at most 3 triangular numbers, stated by Fermat and first proved by Gauss in Disquisitiones Arithmeticae, should be added to this page. Perhaps also a reference to its generalization, also stated by Fermat and proved by Cauchy (?) after work by others that every positive integer was a sum of at most 3 triangular numbers, 4 square numbers, 5 pentagonal numbers, etc. Yes I know the history of the 4-square theorem, but I don't think that would be appropriate in the page on triangular numbers. Achava 00:44, 3 May 2007 (UTC) I would consider myself an expert on this topic, and just wanted to note that think this article is extremely well written. I would love to see its author(s) contribute to more content like it. — Preceding unsigned comment added by 24.19.214.183 (talk) 20:26, 10 February 2013 (UTC)

Last edited at 20:27, 10 February 2013 (UTC). Substituted at 02:39, 5 May 2016 (UTC)

Easy formula

Why isn't there a easy formula? Like f(x)=(x/2)(x+1) works very well. 178.121.66.129 (talk) 17:39, 11 March 2019 (UTC)

That's the second-to-last of the terms in the big formula at the start of the "Formula" section. It's also repeated a couple more times in the same section. —David Eppstein (talk) 17:56, 11 March 2019 (UTC)

It is a little confusing, because it looks like just one big equation, rather than three different ways to write the same thing. Sometimes we put them all on separate lines, but perhaps its just done this way to save space since two of the three identities are just one term. —Soap— 03:45, 7 April 2020 (UTC)

Easier proof

Is there a reason why there isn't an easier proof on the page? I mean, it can be solved by anyone who saw $\sum _{x=1}^{n}x=n(n+1)/2$ in school; by induction.

Extended content

Obviously 1³ = 1². if applies for $n$, then it applies for $(n+1)$ as well; so $(\sum _{x=1}^{n+1}x)^{2}=\sum _{x=1}^{n+1}x^{3}$ seperate $n+1$ $((\sum _{x=1}^{n}x)+(n+1))^{2}=(\sum _{x=1}^{n}x^{3})+(n+1)^{3}$ considering we assume it applies for $n$, we can substitute $\sum _{x=1}^{n}x^{3}=(\sum _{x=1}^{n}x)^{2}$ $((\sum _{x=1}^{n}x)+(n+1))^{2}=(\sum _{x=1}^{n}x)^{2}+(n+1)^{3}$ solving the quadratic equation $(\sum _{x=1}^{n}x)^{2}+2(\sum _{x=1}^{n}x)(n+1)+(n+1)^{2}=(\sum _{x=1}^{n}x)^{2}+(n+1)^{3}$ remove $(\sum _{x=1}^{n}x)^{2}$ from both sides $2(\sum _{x=1}^{n}x)(n+1)+(n+1)^{2}=(n+1)^{3}$ considering $(\sum _{x=1}^{n}x)=n(n+1)/2$ $2n(n+1)(n+1)/2+(n+1)^{2}=(n+1)^{3}$ as $2*a*b*b/2=a*b^{2}$ $n(n+1)^{2}+(n+1)^{2}=(n+1)^{3}$ next, we single out $(n+1)^{2}$ as $a*b^{2}+b^{2}=b^{2}(a+1)$ $(n+1)^{2}(n+1)=(n+1)^{3}$ which proves the theorem

Qube0 (talk) 17:18, 5 January 2020 (UTC)

@Qube0: Easier than what? Other than a graphical illustration, there's no proof in this article at all, which is perfectly reasonable. Inductive proofs of identities like this aren't generally very enlightening and just tend to clutter up the article. On a side note, your proof is full of errors – most seriously, starting by writing what you're trying to ultimately prove, rather than winding up with that as the final result. –Deacon Vorbis (carbon • videos) 17:44, 5 January 2020 (UTC)

non-integer triangular numbers

Since the ${\frac {n(n+1)}{2}}$ formula can be evaluated for all real numbers, both positive and negative, integer or non-integer, is there any special term for the numbers that lie along the resulting graph? x^2 evaluates to the same value regardless of sign, so the resulting curve is symmetrical across the axis of x=-1/2. But Im more interested in the fractional terms and if there is anything special about them. e.g. the π^th "triangular number" is approximately 6.50559852734. If there's anything to this, it certainly would be worth mentioning, but I couldnt find anything about "generalized triangular numbers" anywhere. Thank you, —Soap— 03:42, 7 April 2020 (UTC)

Proposed merge from Termial

Termials and triangular numbers are the same thing. Since Wikipedia articles are about concepts not names (WP:NOTDICT), they should have together only one article. Therefore, I am proposing a merge. I think that termial is also a rather obscure term for what is much more widely known as a triangular number, so the merge should go from termial to triangular number. Any other opinions? —David Eppstein (talk) 19:26, 20 August 2020 (UTC)

Support I wouldn't say they are the same thing. One is a sequence while the other a series, and termial can apparently apply to non-integers. However, they are certainly intertwined, and Termial might be better as a section of the Triangular Number article. The term "termial" doesn't seem to appear in much mathematical literature though, so I question its notability in the first place. Kstern (talk) 20:54, 3 September 2020 (UTC)

Support They're the same thing! If someone wants to know more about termials, and they read the article about triangular numbers, that will serve their purpose just fine. — Preceding unsigned comment added by 73.241.189.0 (talk) 17:00, 24 September 2020 (UTC)

• Support it doesn't make much difference whether the concept is presented as a sequence or a series, the two are clearly functionally equivalent and this article does already note that triangular numbers can be represented as a series. Triangular number is definitely the primary usage and I suggest we just note somewhere that they are sometimes known as termials (The Art of Computer Programming is very influential). Termial doesn't actually specify how it can be calculated for non-integers. Hut 8.5 19:11, 24 September 2020 (UTC)
• This MUST be merged with triangular numbers. It has no independent conceptual difference, and much less content, and it's only used by Knuth, while triangular number has been used since the pythagoreans at least

Termial means exactly the same as triangular number. The idea is to simplify knowledge, not to create for each synonim a different new article. And this is obviously a term only used by no one.

Santropedro (talk) 03:13, 4 January 2021 (UTC)

Seeing support and no opposition, I have gone ahead and done the merge. —David Eppstein (talk) 05:08, 4 January 2021 (UTC)

Rotations in n-dimensional geometry

The amount of rotations in any n-dimensional geometry is the triangular number of n-1. Unfortunately, I haven't found anything on this on the internet so it would count as OR. Is there any way to get around this?

You can play around with it yourself if you want: https://www.mathgoodies.com/calculators/triangular-numbers

75.129.231.4 (talk) 23:19, 28 October 2020 (UTC)

What exactly do you mean by "amount of rotations?" Do you mean possible axes of rotation? In three dimensions an infinite number of rotational axes exist.—Anita5192 (talk) 00:00, 29 October 2020 (UTC)

Edward Waring

While looking at Edward Waring's Meditationes Algebraicae (3rd ed, 1782) for another purpose, I noticed that he states (in Latin, of course), without proof, that every whole number is the sum of 1, 2 or 3 triangular numbers. (This doesn't exclude the possibility that it is also the sum of 4, 5, 6... etc such numbers). (op. cit. p 349 prop. 3.) Waring goes on to state the corresponding propositions for pentagonal and hexagonal numbers (prop. 4) and squares (prop. 5). (This is on the same page as the proposition often known as Waring's Theorem.) I was wondering if this was already known before Waring. Looking at the article here, I see that it was later rediscovered by Gauss, who was excited enough to call it his Eureka theorem. However, I see that it is a special case of a conjecture of Fermat, known as his Polygonal Number Theorem, but characteristically stated without proof. Quite possibly Waring found it in Fermat. As a conjecture, without proof, it is therefore certainly earlier than Gauss, though Gauss may have been the first to prove it for the triangular case.2A00:23C8:7907:4B01:6415:B77:5411:4DE (talk) 14:06, 17 June 2022 (UTC)