hello
The question is :
How to substitute n floating values for the n variables of a
polynomial in Sage, where n is a variable ?
For example,
Let P
P=x3*x6+x1*x6+x4*x5+x2*x5+x1*x4+x2*
and
f=x^6+2
with approximate roots
rr=gp.polroots(f)
=[0.9720806486198328151427283823 + 0.5612310241546864907167665248*I,
0.9720806486198328151427283823 - 0.5612310241546864907167665248*I,
0.E-36 + 1.122462048309372981433533050*I, 0.E-36 -
1.122462048309372981433533050*I, -0.9720806486198328151427283823 +
0.5612310241546864907167665248*I, -0.9720806486198328151427283823 -
0.5612310241546864907167665248*I]
We want to substitute in P the n=6 values contained in rr for the
variables xi.
P(rr[0],rr[1],rr[2],rr[3],rr[4],rr[5])
is not convenient because we want
P(rr[0],rr[1],rr[2],...,rr[n])
and to know how to construct the series rr[0],rr[1],rr[2],...,rr[n] ?
The substitution may be performed as follows :
************************
n=6
for i in range(1,n+1):
p2=P.subs({x[i]:rr[i]})
result:
(x5 + (0.9720806486198328151427283823
+0.5612310241546864907167665248*I))*x4 +
((x5 + x3)*x2 + (x3 +(0.9720806486198328151427283823 +
0.5612310241546864907167665248*I))*x6)
**********************************
But just the first variable is substituted. A type problem w.r.t. the
coefficients ?
When the degree n of f is know, We can also substitute using the
following command
RESULT1= P.subs({x[1]:rr[0],x[2]:rr[1],x[3]:rr[2],x[4]:rr[3],x[5]:rr
[4],x[6]:rr[5]})
1.259921049894873164767210607 - 2.524354897 E-29*I
But in a program the degree of f is n and not necessarily 6. The
problem is to construct
x[1]:rr[0],x[2]:rr[1],x[3]:rr[2],x[4]:...,x[n]:rr[n-1]
when n is a variable.
Moreover, We have used the map fonction of maxima in order to
construct a substitution list for the maxima.subst fonction:
******************************
n=6
v=maxima.map('lambda([rac,i],concat(x,i)=rac)',rr,'makelist(i,i,
1,n)');v
RESULT2 = gp(maxima.subst(v,P));
1.259921049894873327847796133 + 2.524354897 E-29*I
*************************
However, result RESULT2 is not the desired result RESULT1 computed
with the subs fonction of Sage
(derived from the subst fonction of maxima) because maxima troncates
the float:
***********************
rr[0]
0.9720806486198328151427283823 + 0.5612310241546864907167665248*I
with maxima rr[0] is troncated:
maxima(rr[0])
.5612310241546865*I+.9720806486198328
*********************************************
The only solution is to extend the precision of the floats of maxima ?
And how in Sage (not in maxima) ?
It is very complicated !
We think that it is not a nice solution.
Can you help us ?
Best regards, Faical Bouazizi and Annick Valibouze
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