I’m sorry about that Raul, that should read “y”. ----
Thanks Roger. You were right, after a quick search, I’ve found that indeed I would need to perform a matrix division. Several documents give this as a multi-variable Newton method: v_n+1 = v_n - J_f (v_n) ^-1 f(v_n) In J, this would be: multinewt=: 2 : ‘(- u %. u D.1)^:n’ However, if I take a paraboloid z =: +/@:*: x y , and look for its zeros with +/@:*: multinewt (<5) 3 4 I get this: 3 4 _0.5 0.5 _0.5 0.5 _0.5 0.5 _0.5 0.5 which is obviously wrong. Would I need some sort of transposition? This could just be due to precision errors. Or because I wrote an erroneous conjunction? Cheers, Louis > On 13 Feb 2016, at 20:52, Raul Miller <[email protected]> wrote: > > Something is wrong here. > > You say "where A is the original argument, or the result of the > previous iteration." > > ... but I can find no use of A in the expression you are describing. > > Can you look over what you wrote and decide if it is what you meant to say? > > Thanks, > > -- > Raul > > On Sat, Feb 13, 2016 at 10:50 AM, Louis de Forcrand <[email protected]> wrote: >> I’ve been trying to write a conjunction that will find the zeros to a >> function using the >> Newton-Raphson method. The simplest way to do this is probably: >> >> English: >> x_n+1 = x_n - f ( x_n ) / f ‘ ( x_n ) >> >> J: >> eznewt=: 2 : ‘ ( - u % u D.1 ) ^: n ‘ >> >> This works fine for scalar -> scalar functions, but won’t work if the rank >> of the >> result is AND the rank of the arguments are above 1. >> Probably the most evident situation where this would be a problem is if one >> were searching for a minimum instead of a zero, in which case the algorithm >> would be applied to the derivative of the function: +/@:*: D.1 newt 30 ] 3 4 >> As I understand it, this would give a result shape at each iteration (with a >> function u) of: >> >> ( $y ) , $ u eznewt 1 y >> >> where A is the original argument, or the result of the previous iteration. >> What we would >> want is a result with shape $y . >> >> First, let’s get some rules clear: >> - the syntax should be: (verb) newt (# of iterations) (argument) >> - the argument can be of any rank >> - the shape of the argument matches the shape of the result at any iteration >> - this implies that $ u y matches i. 0 or $ y >> >> The hard part is getting u % u D.1 to have the shape of the argument. >> If y is a vector, and u y is a scalar, then u D.1 y will be a vector, and >> u D.1 D.1 y will be a $y by $y matrix. All that is needed then is to take >> its diagonal with (<0 1)&|: . >> >> But what if y is a matrix? Since $ u D.1 y -: ( $ y ) , $ u y , I though >> maybe running >> y's axis together with |: might work >> >> newt=: 2 : '(- u diag_and_divide u D.1)^:n’ >> diag_and_divide=: [ % ] |:~ -@#@$@[ <@}. i.@#@$@] >> >> but something about it doesn’t. >> My head is $pinning now, and I figured I’d send this and then take a break. >> >> Thanks in advance! >> Louis >> ---------------------------------------------------------------------- >> For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
