Talk:Dixon elliptic functions

Citation problem

I'm relatively new editor, so please forgive me for not including a citation for multiplication formulas. Since I derived them myself (Most probably somebody else derived it earlier, but I have not recearched on who has done it earlier) I can't provide a citation. Is a written proof in talk section, a good citation source, or should I delete that information until somebody publishes a proof on another website?

Proofs:

cm duplication formula (cm(u)not=0):

at first we insert u, -u in difference formula, (cm^2(u)cm(-u)-sm(u)sm^2(-u))/(cm(u)cm^2(-u)-sm^2(u)sm(-u))

then use identities cm(-u)=1/cm(u), sm(-u)=-sm(u)/(cm(u):

(cm^2(u)/cm(u)-sm(u)sm^2(u)/cm^2(u))/(cm(u)/cm^2(u)+sm^2(u)sm(u)/cm(u))

then we multiply both sides by cm^2(u):

(cm^3(u)-sm^3(u))/(cm(u)+sm^3(u)cm(u))

then we use identity cm^3(u)=1-cm^3(u)

(2cm^3(u)-1)/(cm(u)+cm(u)(1-cm^3(u)))

And by opening brackets we get:

(2cm^3(u)-1)/(2cm(u)-cm^4(u))

sm duplication formula (cm(0)not=0):

at first we insert u, -u in difference formula, (sm(u)cm(u)-sm(-u)cm^(-u))/(cm(u)cm^2(-u)-sm^2(u)sm(-u))

then use identities cm(-u)=1/cm(u), sm(-u)=-sm(u)/(cm(u):

(cm(u)sm(u)+sm(u)/cm^2(u))/(cm(u)/cm^2(u)+sm^2(u)sm(u)/cm(u))

then we multiply both sides by cm^2(u)

(cm^3(u)sm(u)+sm(u))/(cm(u)+sm^3(u)cm(u))

we use cm^3(u)+sm^3(u)=1:

((1-sm^3(u))sm(u)+sm(u))/(cm(u)+(1-cm^3(u))cm(u))

By opening brackets we get:

(2sm(u)-sm^4(u))/(2cm(u)-cm^4(u))

For general n, I substituted (u, (n-1)u) into sum formula for cm, and sm.

Proof for triplication I will post in this thread tommorow (standart European time) if this post won't be deleted. Great Cosine (talk) 19:16, 7 January 2023 (UTC)

Please review WP:OR, especially WP:CALC. Short derivations like this are kind of a borderline case, but if there's no source for the formula, it makes me wonder if the formula is important enough to bother including. Apocheir (talk) 01:28, 8 January 2023 (UTC)

Dixon (1890) suggests: $${\begin{aligned}\operatorname {sm} 2u&={\frac {\operatorname {sm} u(1+\operatorname {cm} ^{3}u)}{\operatorname {cm} u(1+\operatorname {sm} ^{3}u)}}\\\operatorname {cm} 2u&={\frac {\operatorname {cm} ^{3}u-\operatorname {sm} ^{3}u}{\operatorname {cm} u(1+\operatorname {sm} ^{3}u)}}{\vphantom {\frac {}{\big |}}}\end{aligned}}$$ And also has triplication formulas on the following page. –jacobolus (t) 02:54, 8 January 2023 (UTC)

Dixon's duplication and triplication formulas can also be found in Robinson (2019). –jacobolus (t) 02:59, 8 January 2023 (UTC)

Thank you for sources, I included citation. Also can you cite a calculator work (If I want to add more specific values, can I cite calculators like WolphramAlpha in doing work, and explaining inputs. Will it count as simple calculation?)? Great Cosine (talk) 08:27, 8 January 2023 (UTC)

Which specific values are you hoping to add? I don’t think these necessarily need a citation (it’s more or less a routine calculation), but they also probably aren’t that valuable for readers. I put a few particular values in so that e.g. people can double-check if they have some code evaluating these functions. A table with more than maybe 10 or 12 entries might start to feel out of scope for the article. You can see that e.g. trigonometric functions includes 7 rows in its table of values, gamma function has 10 specific values listed, and Gudermannian function also has 10. –jacobolus (t) 08:37, 8 January 2023 (UTC)

After reading your reply, I chose not to (maybe creating page specific values, could be a good idea). Also to avoid edit war, I checked that if your cm triplication formula is correct, my cm triplication formula is also correct and it is a bit more elegant than yours.

Proof: let cm^3(u)=c, and sm^3(u)=s,

numerator: cc-s-3cs-ssc (I avoid ^2 notation in some parts to not confuse ^2c with ^(2c))

using identity: s+c=1

cc-(1-c)-3c(1-c)-c(1-c)^2,

cc-1+c-3c+3cc-c(1-2c+cc),

-1-2c+4cc-c+2cc-c^3,

-(c^3)+6cc-3c-1

denominator: c-ss+3cs+ccs,

using s+c=1:

c-((1-c)^2)+3c(1-c)+cc(1-c),

c-(1-2c+cc)+3c-3cc+cc-c^3,

-(c^3)-2cc+4c-1+2c-cc,

-(c^3)-3cc+6c-1.

Then we multiply numerator and denominator by -1, and we get: (ccc-6cc+3c+1)/(ccc+3cc-6c+1)

Finally we replace c with cm^3(u) and s with sm^3(u) to get original formula. (I think it is more elegant because all terms are powers of cm(u), and has same amount of terms), with sn, I suggest to change denominator only (it is the same as cm denominator), and multiply numerator and denominator by -1 (to get rid of a lot of minuses)

For duplication formulas I suggest writing mine nearby, like they did in list of trig identities because some people may prefer to have expressions without brackets. And also because it is useful to have cm duplication formula in cm only. Great Cosine (talk) 10:40, 8 January 2023 (UTC)

Sorry, I wasn't trying to clobber your formulas before, only putting more directly what was in the published sources. I don't really care one way or another about rewriting the expression for cm 3z entirely in terms of cm z; seems fine. I fixed up your sm formula (previously had a typo). I think it should be okay for readers to have it in factored form, and not worth the space to write again expanded out. –jacobolus (t) 04:49, 10 January 2023 (UTC)