In the last identity,EllipticK[ -m^2/(4(1-m)) ] == (1-m)^(1/2) EllipticK[(2-m)m], let m=2.  We get EllipticK[ 1 ] == (-1)^(1/2) EllipticK[0]. But the LHS is infinite, while the RHS is finite.   --  Gene From: Warren D Smith <warren.wds@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Sunday, May 24, 2015 4:38 PM Subject: [math-fun] (no subject) Bill Gosper: Then I bashed on it so hard that it got unnervingly ungrotesque: (*)  EllipticK[1/4 (2 + (-2 + t)/Sqrt[1 - t])] == (1 - t)^(1/4) EllipticK[t] How could anything this simple be new? WDS: Let s=Sqrt[1-t].  Then  (*) is rewritten as EllipticK[ 1/2 - (1+s^2)/(4s) ] == EllipticK[ -(1-s)^2/(4s) ] == s^(1/2) EllipticK[1-s^2] Now let s=1-m.  Then this is EllipticK[ -m^2/(4(1-m)) ] == (1-m)^(1/2) EllipticK[(2-m)m]. These look kind of like the known identities here http://functions.wolfram.com/EllipticIntegrals/EllipticK/17/01/ but not the same.  Perhaps yours arises by composing one of them, with a quadratic transformation. The final form of it I wrote down makes it clear how the argument (if m is small) gets to be quadratically way small.  Thus, this offers a very fast way to compute EllipticK.  Also, to compute pi/2=EllipticK[0]  if we happened to know some other EllipticK special value. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)