Guess Extended Rate

[Ralf Hemmecke]

Suppose I have

aList(n)==[n*factorial(n+2*k-1)/(factorial(n + k)*factorial(k))*(-1)^k_
for k in 0..14]
res := guessAlg(aList 1)

The result shows as:

n 2 2 3 4
[[[x ]f(x): x f(x) + f(x) - 1 = 0, f(x) = 1 - x + 2 x - 5 x + O(x )]]

with type List(Expression(Integer)).

I can do

(7) -> k := first rest tower res.1

n 2 2 3 4
[[x ]f(x): x f(x) + f(x) - 1 = 0, f(x) = 1 - x + 2 x - 5 x + O(x )]
Type: Kernel(Expression(Integer))
(8) -> operator k

(8) rootOfADE
Type: BasicOperator
(9) -> argument k

(9) [n, %infoser0()]
Type: List(Expression(Integer))

But if I want the equation? How do I get it from res?

In general it would also for the other guess functions to be able to do
something with the result other than displaying it.

Ralf

[Ralf Hemmecke]

Meanwhile, I found one answer myself.
https://fricas.github.io/api/RecurrenceOperator

ROZE==>RecurrenceOperator(Integer, Expression Integer)
getEq(res)$ROZE

I do not like this abbreviated name. That goes against the usual rule of
having fully expanded names. We have ==> to abbreviate.



Other question...

aList(n)==[n*factorial(n+2*k-1)/(factorial(n + k)*factorial(k))*(-1)^k_
for k in 0..14];
g := guessPRec(aList 1)

[[f(n): (n+2)*f(n+1)+(4*n+2)*f(n)=0, f(0)=1]]

rec := getEq(first g)$ROZE

(n+2)*f(n+1)+(4*n+2)*f(n)=0

Such a recurrence cries for turning it into a program that can compute
the elements of the sequence.

Of course, it doesn't work to simply define

fexpr := subst(f(n+1)-rec/(n+2),n=n-1)
f(0) == 1
f(n) == fexpr

because then f(1) would return exactly the expression involving n.

also

f(k) == subst(fexpr, n=k)

doesn't work, since it does not trigger recursion.

(1) Is it actually possible to turn an expression into a program?
(2) How do I transform rec to an expression of the form
f(n) = ... f(n-1) + ... f(n-2)
but NOT doing by hand.
(3) Can I extract the initial values from the result of guessPRec?

Thank you
Ralf

[Martin R]

I wrote a utility seriesSolve to (try to) compute the sequence given a differential equation, but it is very ugly to use. Some (untested) examples (possibly bitrotten):

f := operator 'f
SSOLVE ==> EXPRSOL(INT, EXPR INT, UFPS EXPR INT, UFPS SMPEXPR EXPR INT)
binTree := seriesSolve(-2*f z + 2*z + f z^2+f(z^2), f, z, [0])$SSOLVE
alcohol := seriesSolve(2*z*f(z^3) + z*f(z)^3  - 3*f(z) + 3, f, z, [])$SSOLVE

)lib UTSSOL EXPRSOL SUPEXPR
G := operator 'G
s := seriesSolve(D(G x, x)-exp(-G(-x)), G, x, [log 2])$EXPRSOL(INT, EXPR INT, UFPS EXPR INT, UFPS SUPEXPR EXPR INT)

fm: UFPS SMPEXPR EXPR INT -> UFPS SMPEXPR EXPR INT
z := monomial(1,1)$UFPS SMPEXPR EXPR INT
fm f == f * (1-q*z) - f.(q*z)*(y*q*z+y)- f.(q*z)*q*z*f - y*q*z

res := seriesSolve(fm, [0])$UTSSOL(EXPR INT, UFPS EXPR INT, UFPS SMPEXPR EXPR INT);

Good Luck,

Martin