considering the flowing code: ———————————————————————— from sage.symbolic.assumptions import GenericDeclaration; var('lamda,n,k,p'); decl1 = GenericDeclaration(k, 'integer'); decl1.assume(); assume(k>0,k>1,n>k,lamda>0,lamda<n); ep="binomial(n,k)<em">p^k(1-p)^(n-k); ep=ep.subs_expr(p==lamda/n); ep=ep.subs_expr(binomial(n,k)==factorial(n)/factorial(k)/factorial(n-k)); ep=limit(ep,n=oo); ep.simplify_full(); ———————————————————————— the result is: lamda^klimit((-(lamda - n)/n)^nfactorial(n)/((-lamda + n)^kfactorial(-k + n)), n, +Infinity)/factorial(k) ———————————————————————— it is not the simplest form. In fact , sage will give me the simplest form as below when I paste the result above to sage again. ———————————————————————— lamda^ke^(-lamda)/factorial(k) ———————————————————————— the question is how can i get the simplest form directly?
Revision history [back]
 | 1 | initial version |
sage cannot get the simplest form on the limit expression with binomial and exponential operations
| | 2 | No.2 Revision |
sage cannot get the simplest form on the limit expression with binomial and exponential operations
considering the flowing code:
————————————————————————
code:
————————————————————————
from sage.symbolic.assumptions import GenericDeclaration;
var('lamda,n,k,p');
GenericDeclaration;
var('lamda,n,k,p');
decl1 = GenericDeclaration(k, 'integer');
decl1.assume();
assume(k>0,k>1,n>k,lamda>0,lamda<n); ep="binomial(n,k)<em">p^k(1-p)^(n-k);
ep=ep.subs_expr(p==lamda/n);
ep=ep.subs_expr(binomial(n,k)==factorial(n)/factorial(k)/factorial(n-k));
ep=limit(ep,n=oo);
ep.simplify_full();
————————————————————————
'integer');
decl1.assume();
assume(k>0,k>1,n>k,lamda>0,lamda<n);< p="">
ep=binomial(n,k)p^k(1-p)^(n-k);
ep=ep.subs_expr(p==lamda/n);
ep=ep.subs_expr(binomial(n,k)==factorial(n)/factorial(k)/factorial(n-k));
ep=limit(ep,n=oo);
ep.simplify_full();
————————————————————————
the result is:
is:
lamda^klimit((-(lamda - n)/n)^nfactorial(n)/((-lamda + n)^kfactorial(-k n)^k*factorial(-k + n)), n, +Infinity)/factorial(k)
————————————————————————
+Infinity)/factorial(k)
————————————————————————
it is not the simplest form. In fact , sage will give me the simplest form as below when I paste the result above to sage again.
————————————————————————
lamda^ke^(-lamda)/factorial(k) again.
————————————————————————
lamda^k*e^(-lamda)/factorial(k)
————————————————————————
————————————————————————
the question is how can i get the simplest form directly?
| | 3 | No.3 Revision |
sage cannot get the simplest form on the limit expression with binomial and exponential operations
considering the flowing code:
————————————————————————
from sage.symbolic.assumptions import GenericDeclaration;
var('lamda,n,k,p');
decl1 = GenericDeclaration(k, 'integer');
decl1.assume();
assume(k>0,k>1,n>k,lamda>0,lamda<n);< p="">
ep=binomial(n,k)p^k(1-p)^(n-k);
ep=ep.subs_expr(p==lamda/n);
ep=ep.subs_expr(binomial(n,k)==factorial(n)/factorial(k)/factorial(n-k));
ep=limit(ep,n=oo);
ep.simplify_full();
————————————————————————
the result is:
lamda^klimit((-(lamda - n)/n)^nfactorial(n)/((-lamda + n)^k*factorial(-k + n)), n, +Infinity)/factorial(k)
————————————————————————
it is not the simplest form. In fact , sage will give me the simplest form as below when I paste the result above to sage again.
————————————————————————
lamda^k*e^(-lamda)/factorial(k)
————————————————————————the question is how can i get the simplest form directly?
| | 4 | No.4 Revision |
sage cannot get the simplest form on the limit expression with binomial and exponential operations
considering the flowing code:
————————————————————————
var('lamda,n,k,p');
decl1.assume();
assume(k>0,k>1,n>k,lamda>0,lamda<n);< p="">
ep=binomial(n,k)p^k(1-p)^(n-k);
ep=ep.subs_expr(p==lamda/n);
ep=ep.subs_expr(binomial(n,k)==factorial(n)/factorial(k)/factorial(n-k));
ep=limit(ep,n=oo);
ep.simplify_full();
————————————————————————
the result is:
lamda^klimit((-(lamda - n)/n)^nfactorial(n)/((-lamda + n)^k*factorial(-k + n)), n, +Infinity)/factorial(k)
————————————————————————
it is not the simplest form. In fact , sage will give me the simplest form as below when I paste the result above to sage again.
————————————————————————
lamda^k*e^(-lamda)/factorial(k)
lamda^k*e^(-lamda)/factorial(k)
————————————————————————
the question is how can i get the simplest form directly?
| | 5 | No.5 Revision |
sage cannot get the simplest form on the limit expression with binomial and exponential operations
considering the flowing code:
————————————————————————
from sage.symbolic.assumptions import GenericDeclaration;
var('lamda,n,k,p');
decl1 = GenericDeclaration(k, 'integer');
decl1.assume();
assume(k>0,k>1,n>k,lamda>0,lamda<n);
ep=binomial(n,k)*p^k*(1-p)^(n-k);
ep=ep.subs_expr(p==lamda/n);
ep=ep.subs_expr(binomial(n,k)==factorial(n)/factorial(k)/factorial(n-k));
ep=limit(ep,n=oo);
ep.simplify_full();
————————————————————————
the result is:
lamda^klimit((-(lamda lamda^k*limit((-(lamda - n)/n)^nfactorial(n)/((-lamda n)/n)^n*factorial(n)/((-lamda + n)^k*factorial(-k + n)), n, +Infinity)/factorial(k)
————————————————————————
it is not the simplest form. In fact , sage will give me the simplest form as below when I paste the result above to sage again.
————————————————————————
lamda^k*e^(-lamda)/factorial(k)
————————————————————————
the question is how can i get the simplest form directly?
| | 6 | No.6 Revision |
sage cannot get the simplest form on the limit expression with binomial and exponential operations
considering the flowing code:
————————————————————————
from sage.symbolic.assumptions import GenericDeclaration;
var('lamda,n,k,p');
decl1 = GenericDeclaration(k, 'integer');
decl1.assume();
assume(k>0,k>1,n>k,lamda>0,lamda<n);
ep=binomial(n,k)*p^k*(1-p)^(n-k);
ep=ep.subs_expr(p==lamda/n);
ep=ep.subs_expr(binomial(n,k)==factorial(n)/factorial(k)/factorial(n-k));
ep=limit(ep,n=oo);
ep.simplify_full();
————————————————————————
the result is:
lamda^k*limit((-(lamda - n)/n)^n*factorial(n)/((-lamda + n)^k*factorial(-k + n)), n, +Infinity)/factorial(k)+Infinity)/factorial(k)
————————————————————————
it is not the simplest form. In fact , sage will give me the simplest form as below when I paste the result above to sage again.
————————————————————————
lamda^k*e^(-lamda)/factorial(k)
————————————————————————
the question is how can i get the simplest form directly?
| | 7 | retagged |
sage cannot get the simplest form on the limit expression with binomial and exponential operations
considering the flowing code:
————————————————————————
from sage.symbolic.assumptions import GenericDeclaration;
var('lamda,n,k,p');
decl1 = GenericDeclaration(k, 'integer');
decl1.assume();
assume(k>0,k>1,n>k,lamda>0,lamda<n);
ep=binomial(n,k)*p^k*(1-p)^(n-k);
ep=ep.subs_expr(p==lamda/n);
ep=ep.subs_expr(binomial(n,k)==factorial(n)/factorial(k)/factorial(n-k));
ep=limit(ep,n=oo);
ep.simplify_full();
————————————————————————
the result is:
lamda^k*limit((-(lamda - n)/n)^n*factorial(n)/((-lamda + n)^k*factorial(-k + n)), n, +Infinity)/factorial(k)
————————————————————————
it is not the simplest form. In fact , sage will give me the simplest form as below when I paste the result above to sage again.
————————————————————————
lamda^k*e^(-lamda)/factorial(k)
————————————————————————
the question is how can i get the simplest form directly?
| | 8 | retagged |
sage cannot get the simplest form on the limit expression with binomial and exponential operations
considering the flowing code:
————————————————————————
from sage.symbolic.assumptions import GenericDeclaration;
var('lamda,n,k,p');
decl1 = GenericDeclaration(k, 'integer');
decl1.assume();
assume(k>0,k>1,n>k,lamda>0,lamda<n);
ep=binomial(n,k)*p^k*(1-p)^(n-k);
ep=ep.subs_expr(p==lamda/n);
ep=ep.subs_expr(binomial(n,k)==factorial(n)/factorial(k)/factorial(n-k));
ep=limit(ep,n=oo);
ep.simplify_full();
————————————————————————
the result is:
lamda^k*limit((-(lamda - n)/n)^n*factorial(n)/((-lamda + n)^k*factorial(-k + n)), n, +Infinity)/factorial(k)
————————————————————————
it is not the simplest form. In fact , sage will give me the simplest form as below when I paste the result above to sage again.
————————————————————————
lamda^k*e^(-lamda)/factorial(k)
————————————————————————
the question is how can i get the simplest form directly?
