with differential global optimization one of the roots is found:
In [10]: differential_evolution(f,((-2,2), (-2,2), (-2,2), (-2,2), (-2,2),
(-2,2
: ), (-2,2), (-2,2), (-2,2), (-2,2), (-2,2), (-2,2), (-2,2), (-2,2),
(-2,
: 2), (-2,2),(-2,2), (-2,2),(-2,2), (-2,2), (-2,2)))
Out[10]:
something like this:
~ $ ptipython
Python 3.5.1 |Anaconda 4.0.0 (64-bit)| (default, Dec 7 2015, 11:16:01)
Type "copyright", "credits" or "license" for more information.
IPython 4.1.2 -- An enhanced Interactive Python.
? -> Introduction and overview of IPython's features.
%quickref -> Q
Have you tried to use *Idx* instead of *Symbol* for the index?
k = Idx("k", *4*)
4 is the range.
In this case, *k *ranges from 0 to 4-1 = 3.
On Thursday, 14 April 2016 09:31:38 UTC+2, Nico Schlömer wrote:
>
> Once again, atoms to the rescue.
> ```
> y.atoms(IndexedBase)
> ```
> will give you al
Sorry for being passive these days. I was busy with some course projects
and am done. I have gone through the two classes again to recall
everything. I have a small doubt, a conceptual one. Can you please explain
me the the difference between argument qs in LagrangesMethod and q_ind &
q_dep in
Once again, atoms to the rescue.
```
y.atoms(IndexedBase)
```
will give you all IndexedBase object from an expression.
On Thursday, April 14, 2016 at 9:25:09 AM UTC+2, Nico Schlömer wrote:
>
> It seems that, unfortunately, the information of whether or not a variable
> is an IndexedBase object i
It seems that, unfortunately, the information of whether or not a variable
is an IndexedBase object is discarded when using it in an expression. Check
```
from sympy import *
u = IndexedBase('u')
k = Symbol('k')
y = u[k]
print(isinstance(u, IndexedBase))
for s in y.free_symbols:
print(s, is
>From an object like `sin(u[k]) + u0[k]` I would like to get the
corresponding C code as a string. Since `k` is a variable, I cannot use
`MatrixSymbol`, but there's always `IndexedBase` of course. With
```
u = IndexedBase('u')
u0 = IndexedBase('u0')
k = Symbol('k')
y = sin(u[k]) + u0[k]
```
thing
