using maxima output from solve_rec

[Ken Levasseur]

Hello,

I've been trying to solve recurrence relations and managed to find the solve_rec in maxima.  For example, when I evaluate 

sol=maxima('solve_rec(a[n+2]+a[n+1]-2*a[n]=0,a[n],a[0]=1,a[1]=3)')

the value of sol is

a[n]=(-2)^(n+1)/3+5/3

Now what I'd like to do is define a Sage function that matches this output.  I haven't been about to do so.  Anyone know how?

Ken Levasseur

UMass Lowell

[Ken Levasseur]

Kent:

Thanks, I wasn't aware of the rhs function in maxima.

Ken

On Wednesday, April 3, 2013 3:42:12 PM UTC-4, Kent Morrison wrote:

You can define a Sage function by grabbing the right hand side of sol

a(n) = maxima('rhs(sol)')

 

[kcrisman]

On Thursday, April 4, 2013 8:35:28 AM UTC-4, Ken Levasseur wrote:

Kent:

Thanks, I wasn't aware of the rhs function in maxima.

Ken

There is also a .rhs() method for most symbolic things in Sage, just in case that helps some other place.

In this case, this leads to a problem.

First, implicit in Ken's post is that we need to load that package.

sage: maxima('load(solve_rec)')

"/Users/.../sage-5.7.rc0/local/share/maxima/5.29.1/share/solve_rec/solve_rec.mac"

That's fine.  But next,

sage: sol=maxima('solve_rec(a[n+2]+a[n+1]-2*a[n]=0,a[n],a[0]=1,a[1]=3)')

sage: sol

a[n]=(-2)^(n+1)/3+5/3

sage: type(sol)

sage.interfaces.maxima.MaximaElement

sage: a(n) = maxima('rhs(sol)')

sage: a

n |--> 0

This is because Maxima doesn't know what sol is, only Sage does.  In general Kent's solution would work, but here that little quirk prevents it from working.  Try instead

sage: sol.rhs()  # a Maxima method, so still a Maxima element

(-2)^(n+1)/3+5/3

sage: sol.rhs().sage()  # convert it to Sage

1/3*(-2)^(n + 1) + 5/3

sage: a(n) = sol.rhs().sage()

sage: a  # Hooray!

n |--> 1/3*(-2)^(n + 1) + 5/3

sage: a(55)  # is this right?

24019198012642647

This has the advantage of staying object-oriented, if one cares about such things (which I probably don't...)

- kcrisman