Thank you Dan!. This is amazing. I will study it further.  
- Bo. 

--- Den tirs 23/2/10 skrev Dan Bron <[email protected]>:

Fra: Dan Bron <[email protected]>
Emne: Re: [Jprogramming] tacitly solving Kepler's equation
Til: "'Programming forum'" <[email protected]>
Dato: tirsdag 23. februar 2010 16.04

Bo Jacoby wrote:
>  I wonder how to translate this explicit verb into a tacit verb?
>     E=. 4 : '(y+x*1&o.)^:_ y'      

How about:

       Et=:  ,~ (p. 1&o.)^:_ ]
    
       0.0167   (E-:E2) 0.46478
    1

The key insight is that  x f^:n y  is  x&f^:n y  .   That is, the 
right-argument to dyadic power is "carried through" all the
iterations (and we need the initial values of both x and y carried through all 
iterations, hence the  ,  ).   I guess a second,
minor insight is that  y+x*u  is just a polynomial evaluated at  u  , which 
allows us to use  p.  to avoid picking apart this fixed
argument (and the coefficient vector for  p.  is always "reversed", hence the  
~  on the  ,  ).

-Dan

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