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 suggests: $${\begin{aligned}\operatorname {sm} 2u&={\frac {\operatorname {sm} u(1+\operatorname {cm} ^{3}u)}{\operatorname {sm} u(1+\operatorname {cm} ^{3}u)}}\\\operatorname {cm} 2u&={\frac {\operatorname {cm} ^{3}u-\operatorname {sm} ^{3}u}{\operatorname {sm} u(1+\operatorname {cm} ^{3}u)}}{\vphantom {\frac {}{\big |}}}\end{aligned}}$$ –jacobolus (t) 02:54, 8 January 2023 (UTC)