Hi everyone,
say I have a function of both an integer n and a complex z
def f(n,z):
return z**n
For any tuple of integer (a_1,a_2, ..., a_k), (actually, k and a_i are
random), I want to form the function
z - sum_{i=1}^k |f(a_i, z)|^2
The result must still be a function. I guess it is
Helloo everybody !!!
Is there a way to compute the longest common subsequence of two
(binary) words in Sage ? I can't find how, but it looks like something
Sage should be able to do :-)
Thaanks !
Nathann
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I don't think you need to make an explicit class here. You can build a
function from within another function, and return that:
sage: def f(n, z):
: return z**n
:
sage: def maker(tup):
: def g(z):
: return sum(abs(f(a_i,z))**2 for a_i in tup)
: return g
Thank's, indeed I went thought the all thing and now it works!
2013/7/17 P Purkayastha ppu...@gmail.com
On 07/16/2013 05:35 PM, Viviane Pons wrote:
Hi everyone,
I have installed sage-5.11-beta3 and for some reason, I cannot use use
easy_install anymore on the sage shell. It seems to be a
On 2013-7-17 02:07, Laurent Decreusefond wrote:
For any tuple of integer (a_1,a_2, ..., a_k), [...]
I want to form the function
z - sum_{i=1}^k |f(a_i, z)|^2
Won't this work?
def ff(atuple,z):
return sum([ abs(f(n,z))**2 for n in atuple ])
--
*\\* Anton Sherwood
Hi Nathann
Here is a Wikipedia article with what looks very much like the python code
for a solution.
http://en.wikipedia.org/wiki/Longest_common_subsequence_problem
Carl
On Wed, Jul 17, 2013 at 12:11 PM, sage-support@googlegroups.com wrote:
Today's Topic Summary
Group:
I am looking to evaluate $\sum_{k=0}^n 1/\binom{n}{k}$ so I type:
sage: n = var('n')
sage: k = var('k')
sage: sum(1/binomial(n,k),k,0,n)
(n + 1)*2^(-n)
and that answer is wrong.
For example, with n=10 we get
sage: sum(1/binomial(10,k) for k in range(11))
1433/630
but the alleged answer of
