Re: Smarter way to calculate middle-right terms of continued fraction partial sums

*To*: mathgroup at smc.vnet.net*Subject*: [mg70080] Re: Smarter way to calculate middle-right terms of continued fraction partial sums*From*: ab_def at prontomail.com*Date*: Mon, 2 Oct 2006 00:34:11 -0400 (EDT)*References*: <eflefm$dg8$1@smc.vnet.net><efnui4$7m8$1@smc.vnet.net>

Diana wrote: > > Note that the signs of a1 alternate between positive and positive. > Likewise, all the variables alternate in sign. > > That would be "the signs of a1 alternate between positive and > negative." > > > Diana M. I think you forgot to mention that d[n] was defined as d[n_] := Product[a[k]^2^(n - k), {k, n}] You can generate the polynomial convergents automatically, as was discussed on this list quite recently: pcf[e_] := Drop[#, -2]&@ NestWhile[ Flatten@ {Drop[#, -2], PolynomialReduce @@ #[[{-1, -2}]], #[[-2]]}&, {Denominator@ #, Numerator@ #}&@ Together@ e, !TrueQ[#[[-2]] == 0]&] In[3]:= pcf[Sum[1/d[k], {k, 4}]] Out[3]= {0, a[1], -a[2], -a[1], -a[3], a[1], a[2], -a[1], -a[4], a[1], -a[2], -a[1], a[3], a[1], a[2], -a[1]} Then you can extract the indices and look up the sequence, e.g., in Sloane's Encyclopedia of Integer Sequences: In[4]:= % /. a -> (#&) Out[4]= {0, 1, -2, -1, -3, 1, 2, -1, -4, 1, -2, -1, 3, 1, 2, -1} Without the signs this is the ruler function. To account for signs, consider that +/-n appears at positions (2*k + 1)*2^(n - 1) + 1 and the signs are alternating (between positive and...well, not positive), so the sign depends on Mod[k, 2]: lcf[n_] := Module[ {L = Range[2^n - 1], Lpow}, Lpow = IntegerExponent[L, 2]; Prepend[(-1)^(Quotient[Quotient[L, 2^Lpow] - 1, 2] + Mod[L, 2] + 1)* a /@ (Lpow + 1), 0] ] In[6]:= lcf[6] === pcf[Sum[1/d[k], {k, 6}]] Out[6]= True Note that the 33rd term is -a[6], not +a[6]. Maxim Rytin m.r at inbox.ru ***********************************