[VA] SRC#002- Almost integers and other beasties

[Gerson W. Barbosa]

(02-13-2019 01:49 PM)Albert Chan Wrote:  Your code might be shortened and faster by removing (-1)^k factor.
Since it pre-calculated iterations required, summing backwards may be more accurate.

Sure I am aware of these, but initially I was not interested in optimizing the code for speed, only for size. Anyway, since the summation can be evaluated without resorting to exponentiation, as you've pointed out, there's no reason to use it, at least inside the loop:

enter k;
sign := -2*mod(k, 2) + 1;
sum := 0;
a := 2^k - 1;
for i = k to 1 step -1;
  sum := sum + sign/(a*i);
  sign := -sign;
  a := a div 2
next i;
display sum


The RPL code, however, is now slightly slower, because either it's not properly optimized yet or the LongFloat FY^X function is very efficient for integer arguments.

%%HP: T(3)A(R)F(.);
\<< RCLF SWAP -105 CF -3 CF DUP 1 + 'DIGITS' STO 10 LN * 2 LN SWAP DUP2 / LN + SWAP / CEIL 2 OVER ^ 1 - OVER 2 MOD -2 * 1 + SWAP 0 4 ROLL 1
  FOR i OVER i FMULT 4 PICK FMULT FINV FADD ROT NEG ROT 2 FDIV FIP ROT -1
  STEP UNROT DROP2 DUP FADD 2 FLN FDIV 12 FINV FADD FPI FSQ 6 FDIV 2 FLN FSQ FDIV FADD ZZ\<-\->F NEG SWAP \->STR DUP SIZE ROT \=/ -51 FC? { "." } { "," } IFTE UNROT { DUP TAIL SWAP HEAD } { "0" } IFTE UNROT + + DUP SIZE " " REPL " " "" SREPL DROP SWAP STOF
\>>

# 1A8h, 428.5 bytes

141.34 seconds on my HP-50g, for 100 (previously, 132.59 seconds)

  1 -> 0.9

  2 -> 1.0

  3 -> 0.999

  4 -> 0.9999

  5 -> 0.99998

  6 -> 1.00000

  7 -> 0.9999997

 12 -> 1.00000000000

 20 -> 1.0000000000012374125

 40 -> 1.000000000001237412575736110228719610646

 50 -> 1.0000000000012374125757361102287196106466728742977

100 -> 1.000000000001237412575736110228719610646672874297732048196548443844171825640530​428850913885586193525

Regards,

Gerson.