Hi Fred,
thanks for reporting this. I made some improvements in the power functions, see SVN 393. /// Jürgen On 07/23/2014 06:16 PM, Frederick H. Pitts wrote:
Jurgen, The last comment in the listing below summarizes the issue. The input file for the listing is attached. Regards Fred Retired Chemical Engineer⎕PP ← 16 ⍝ Compute first 5, 14th, 15th, 70th and 71th Fibonacci numbers using ⍝ Binet's formula (http://www.math.rutgers.edu/~erowland/fibonacci.html) ⍝ Desired results: ⍝ 1 1 2 3 5 377 610 190392490709135 308061521170129R ← 1 2 3 4 5 14 15 70 71R5 ← 5 ⋆ .5 P5 ← 1 + R5 M5 ← 1 - R5 ⍝ Using repeated multiplication to compute the powers gives valid answers ⍝ up to the 70th number after rounding. Maybe this result is the best one ⍝ can expect from double precision arithmetic. ( ( ( ×/ )¨ R ⍴¨ P5 ) - ( ( ×/ )¨ R ⍴¨ M5 ) ) ÷ R5 × ( ×/ )¨ R ⍴¨ 2 1 1 2 3 5 377.0000000000001 610.0000000000003 190392490709135.4 308061521170129.7 ⍝ Using the APL ⋆ operator to compute the powers gives valid answers only ⍝ up to the 14th number. The 15th number is no where near correct. ⍝ Shouldn't the ⋆ operator results match the repeated multiplication ⍝ results more closely? ( ( P5 ⋆ R ) - M5 ⋆ R ) ÷ R5 × 2 ⋆ R 1 1 2 3 5 377.0000000000002 0.001639344262295081 5.252307989014705E¯15 3.246104856593697E¯15
