Re: [Bug-apl] Unexpected results for the power function

[Juergen Sauermann]

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 308061521170129
>        
>        R ← 1 2 3 4 5 14 15 70 71
>        R5 ← 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