To find polynomial roots, p. is the primitive of choice. Given the coefficients, p. finds the roots; given the roots and a multiplier, p. finds the coefficients.
p. <1+i.20 NB. coefficients from roots and multiplier (here 1) 2432902008176640000 _8752948036761600000 13803759753640704000 _12870931245150988800 8037811822645051776 _3599979517947607200 1206647803780373360 _311333643161390640 63030812099294896 _10142299865511450 1307535010540395 _135585182899530 11310276995381 _7561... p. p. <1+i.20 NB. roots and multiplier from coefficients ┌─┬──────────────────────────────────────────────────┐ │1│20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1│ └─┴──────────────────────────────────────────────────┘ The polynomial above is Wilkinson's polynomial <https://en.wikipedia.org/wiki/Wilkinson%27s_polynomial>. On Wed, Feb 5, 2020 at 1:57 PM Lippu Esa <esa.li...@varma.fi> wrote: > If I remember correctly you should deflate the original equation by > dividing it by x-x1 where x1 is the first root found in order to find x2 > the second root. > > Esa > > > > > > Sent from my Samsung Galaxy smartphone. > > > -------- Original message -------- > From: Skip Cave <s...@caveconsulting.com> > Date: 2/5/20 21:49 (GMT+02:00) > To: "programm...@jsoftware.com" <programm...@jsoftware.com> > Subject: Re: [Jprogramming] Derivatives > > Won't (x^x)-(2*x) = 0 have two roots? A real one, and a complex one? > > Will Newton Raphson find both? > > Skip > > > > > On Wed, Feb 5, 2020 at 12:08 PM Henry Rich <henryhr...@gmail.com> wrote: > > > 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 > https://eur01.safelinks.protection.outlook.com/?url=http%3A%2F%2Fwww.jsoftware.com%2Fforums.htm&data=01%7C01%7C%7Cf73805d19a944a8ea1e608d7aa747a61%7C5090e269dbea4e98a9aa3e70be5890f7%7C0&sdata=LCHRYwGvI5EXRoKUQbKA%2BWG8v1vyKu9MDopz6xEBcSE%3D&reserved=0 > > >> ---------------------------------------------------------------------- > > >> For information about J forums see > https://eur01.safelinks.protection.outlook.com/?url=http%3A%2F%2Fwww.jsoftware.com%2Fforums.htm&data=01%7C01%7C%7Cf73805d19a944a8ea1e608d7aa747a61%7C5090e269dbea4e98a9aa3e70be5890f7%7C0&sdata=LCHRYwGvI5EXRoKUQbKA%2BWG8v1vyKu9MDopz6xEBcSE%3D&reserved=0 > > >> > > > > > > > ---------------------------------------------------------------------- > > For information about J forums see > https://eur01.safelinks.protection.outlook.com/?url=http%3A%2F%2Fwww.jsoftware.com%2Fforums.htm&data=01%7C01%7C%7Cf73805d19a944a8ea1e608d7aa747a61%7C5090e269dbea4e98a9aa3e70be5890f7%7C0&sdata=LCHRYwGvI5EXRoKUQbKA%2BWG8v1vyKu9MDopz6xEBcSE%3D&reserved=0 > > > ---------------------------------------------------------------------- > For information about J forums see > https://eur01.safelinks.protection.outlook.com/?url=http%3A%2F%2Fwww.jsoftware.com%2Fforums.htm&data=01%7C01%7C%7Cf73805d19a944a8ea1e608d7aa747a61%7C5090e269dbea4e98a9aa3e70be5890f7%7C0&sdata=LCHRYwGvI5EXRoKUQbKA%2BWG8v1vyKu9MDopz6xEBcSE%3D&reserved=0 > > > T?m?n viestin sis?lt? liitteineen on luottamuksellinen ja tarkoitettu vain > sen vastaanottajalle. 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