None of these cute ways is very accurate in tough cases:
0j15 ": (2*a) %~ (-b) (+,-) %: (b^2) - 4*a*c [ 'a b c' =. 1e_6 1e6 1e_6
0.000000000000000 _1000000000000.000000000000000
But p. does better:
0j15 ": 1 {:: p. c,b,a
_1000000000000.000000000000000 _0.000000000001000
Henry Rich
On 12/11/2012 6:29 PM, Roger Hui wrote:
There are some cheeky (or is it cheesy?) versions:
(2*a) %~ (-b) (+,-) %: (b^2) - 4*a*c NB. Kip Murray
(2*a) %~ - b (+,-) %: (b^2) - 4*a*c
(+:a) %~ - b (+,-) %: (*:b) - 4*a*c
-: a %~ - b (+,-) %: (*:b) - 4*a*c
On Tue, Dec 11, 2012 at 11:37 AM, km <k...@math.uh.edu> wrote:
It appears this could be translated into J as the rather cool
(2*a) %~ (-b) (+,-) %: (b^2) - 4*a*c
Sent from my iPad
On Dec 11, 2012, at 12:59 PM, Roger Hui <rogerhui.can...@gmail.com> wrote:
Example from the Iverson and McDonnell *Phrasal
Forms*<http://www.jsoftware.com/papers/fork.htm>paper (which
introduced fork):
(-b)(+,-)√((b*2)-4×a×c)÷2×a
√ is a postulated APL primitive, spelled %: in J.
On Tue, Dec 11, 2012 at 10:49 AM, km <k...@math.uh.edu> wrote:
What is the coolest way of programming the quadratic formula in J? We
are
finding the roots of polynomial c + x*(b + x*a) without using p. . I
offer
roots
3 : 0
'a b c' =. y
q =. %: (b^2) - 4*a*c
(2*a) %~ (-b) + q,-q
)
roots 1 3 2
_1 _2
roots 1 0 1
0j1 0j_1
roots 1 _2 1
1 1
partly as problem definition. I am looking for cool roots verbs!
Kip Murray
Sent from my iPad
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