Yeah, a rational y wouldn't ever quite satisfy 0 = _2 0 1 p. y
Henry Rich On 2/5/2020 1:04 PM, Devon McCormick wrote:
You especially need guardrails if you try something like this: _2 0 1&p. Newton 1 NB. OK - square root of 2 1.41421 _2 0 1&p. Newton 1x NB. Try extended precision C-c C-c|break NB. After waiting a while... | _2 0 1&p.Newton 1 NB. Failure to terminate... On Wed, Feb 5, 2020 at 7:34 AM Henry Rich <henryhr...@gmail.com> wrote:I misread your function. (^~ - +:) Newton 1.1 0.346323j1.2326e_32 (^~ - +:) Newton 0.5 0.346323 Still need those guardrails! Henry Rich On 2/5/2020 2:21 AM, Skip Cave wrote:In "Fifty Shades of J" chapter 23, the Newton Raphson algorithm is described thusly: Newton =: adverb : ']-u%(u D.1)'(^:_)("0) How would that be defined using the new derivative verbs? Also, what is the replacement for d.? How would I find the roots of (x^x)=2*x using Newton Raphson? Skip Skip Cave Cave Consulting LLC ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
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