"Shew that" Limit (z*(q-1);q)_oo q->1 = 1/Limit (z*(1-q);q)_oo q->1 = e^z . I thought I once did by directly manipulating the infinite product(s), but can't recall how. The usual proof uses identities for equivalent summations whose termwise limits -> (+-z)^n/n! . But recall 26 Jan 11, Re Plouffe's 2 surprising approximations: rwg>Yes, it's immediate. Rewrite eta as its infinite product; log(prod) -> sum(logs); log -> powerseries; sumswap; closed form inner sum. (c1) [%E^(%PI*K*N) = Q^-N,%E^-(%PI*K) = Q,%PI*K = -LOG(Q)]; %pi k n 1 - %pi k (d1) [%e = --, %e = q, %pi k = - log(q)] n q (c2) SUBST(%,SUM(1/(N*(%E^(%PI*K*N)-1)),N,1,INF) = -LOG(ETA(%E^-(%PI*K)))-%PI*K/24); inf ==== \ 1 log(q) (d2) > ---------- = ------ - log(eta(q)) / 1 24 ==== n (-- - 1) n = 1 n q [...]<rwg This easily generalizes to (c1) POQIFY(LOG(%E^-SUM(Z^I/(I*(1-Q^I)),I,1,INF) = PRODUCT(1-Q^I*Z,I,0,INF))); inf ==== i \ z (d1) - > ---------- = log(qpoch(z, q, inf)) / i ==== i (1 - q ) i = 1 which Macsyma can test directly: (c3) BLOCK([SIMPSUM : TRUE],TAYLOR(SUBST((Q-1)*Z,Z,MAKEPROD(D1)),Z,0,3)); 2 2 3 (q - 1) z (q - 2 q + 1) z (d3)/T/ z + ---------- + ----------------- + . . . = 2 q + 2 2 3 q + 3 q + 3 2 2 3 (q - 1) z (q - 2 q + 1) z z + ---------- + ----------------- + . . . 2 q + 2 2 3 q + 3 q + 3 But we only need the first term for the q->1 limit. The two usual q-exponential(z) functions are the cases p=1 and -1 of the continuum ((p*z*(q-1);q)_oo)^p . Based on q-trig, Julian proposes yet another candidate, qexp(pi z) := (theta[4](z ln q)-theta[1](z ln q))/theta[4](0)/q^z^2 = 1 - 2 qpi z ln q/(1-q^2) + z^2 ln q (theta''[4](0) ln q/2/theta[4] - 1) ... (where qpi := (1+q)* "q Wallis's product) but I don't know what nice properties it might have. Presumably one can recast q-trig double angle formulas etc in terms of it. --rwg Mike Hirschhorn privately remarked that the QPochhammer reflection formula can't be new. I agree, and conjecture that BHS omitted it to skirt the gigantic subject of theta functions.