Kip, Raul,Thanks for replying.
The matrix way would probably be better.
As a follow up question, I am having trouble creating an adverb/conjunction to
build solutions from.
if I define the conjunction
F =: 2 : '^@:(n&*)'
and I give, as my noun argument, the solutions to some polynomial (e.g. >1{ p.
_2 1 1)
Then I need to do
'' F _2 1
^@:(_2 1&*) NB. the two solutions (without constants)
Which gives me my verb "solution". Please note that I had to give '' as the
left argument to F to return my verb.
Is there some other way of doing this?
i.e. I want some "part of speech" (conjunction or adverb I suppose) that takes
a noun and returns a verb.
It seems conjunctions take verbs and nouns and return verbs
Conj: V x N --> V
but since, in my case for F, the verb is superfluous, is there any "part of
speech" that takes a noun and returns a verb?
I hope my question made sense. My motivation is, I don't see why I should need
to pass an empty verb '' to F to
return a verb. I only want to give F my noun and from that, I want to get a
verb.
Regards,
Jon
> From: [email protected]
> Date: Sun, 15 Mar 2015 14:46:35 -0400
> To: [email protected]
> Subject: Re: [Jprogramming] Solving Differential Eqns with J
>
> Oops, I should have seen that.
>
> Thanks,
>
> --
> Raul
>
> On Sun, Mar 15, 2015 at 2:45 PM, Kip Murray <[email protected]> wrote:
> > Raul, where you have _1 * y you need _2 * y
> >
> > --Kip
> >
> > On Sunday, March 15, 2015, Raul Miller <[email protected]> wrote:
> >
> >> On Sat, Mar 14, 2015 at 7:11 AM, Kip Murray <[email protected]
> >> <javascript:;>> wrote:
> >> > I recommend a matrix approach involving eigenvalues and eigenvectors,
> >> > and assume you have a way to find eigenvalues and eigenvectors for an n
> >> by
> >> > n matrix which has n linearly independent eigenvectors. I confine my
> >> > attention to linear differential equations with constant coefficients and
> >> > right hand side 0, for example
> >> >
> >> > u''' - u'' - 4 u' + 4 u = 0 with initial conditions u(0) = 2 , u'(0) =
> >> _1
> >> > , and u''(0) = 5
> >> >
> >> > This has the solution u(t) = ( ^ _2 * t ) + ( ^ t ) where I use a mix
> >> of
> >> > conventional and J notation.
> >>
> >> I decided to try converting your notes here to J, and I ran into a snag.
> >>
> >> First, I tried expressing your initial constraints in J:
> >>
> >> constraint=: 1 :0
> >> if. 0 = y do.
> >> assert. 2 = u y
> >> assert. _1 = u D. 1 y
> >> assert. 5 = u D. 2 y
> >> end.
> >> assert. 0= (u D.3 + (_1*u D.2) + (_4 * u D. 1) + 4 * u) y
> >> )
> >>
> >> And then I expressed your solution in J:
> >>
> >> solution=:3 :0
> >> (^ _1 * y) + ^ y
> >> )
> >>
> >> Or, alternatively:
> >>
> >> tacitsolution=: +&^ -
> >>
> >> And then I tested my code to see if I had gotten it right:
> >>
> >> solution constraint 0
> >> |assertion failure
> >> | _1=u D.1 y
> >>
> >> Looking at this:
> >> solution D.1 (0)
> >> 9.76996e_8
> >> tacitsolution D.1 (0)
> >> 9.76996e_8
> >>
> >> working this through manually,
> >>
> >> dsolution=: 3 :0
> >> (_1 * ^ _1*y) + ^ y
> >> )
> >>
> >> dsolution 0
> >> 0
> >>
> >> So J's implementation of D. isn't perfect, but I'm wondering if you
> >> might not also have a typo somewhere in your presentation here?
> >>
> >> Thanks,
> >>
> >> --
> >> Raul
> >> ----------------------------------------------------------------------
> >> For information about J forums see http://www.jsoftware.com/forums.htm
> >>
> >
> >
> > --
> > Sent from Gmail Mobile
> > ----------------------------------------------------------------------
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> ----------------------------------------------------------------------
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