On Fri, Jul 27, 2012 at 6:05 PM, fulio <[email protected]> wrote:
> Hi all,
>
> is it possible to increase the index of a symbolic variable.
>
> I tried this.
>
> var('b x')
> var('i n', integer=True)
> print(var('x:5'))
> print(summation(b_i*x(n-i),(i, 0, 4)))
>
>
> The output should look like this:
> b4*x(n - 4) + b3*x(n - 3) + b2*x(n - 2) + b1x(n - 1) + b0x(n)
>
>
I'm showing the steps I used to give an answer:
>>> [w for w in dir() if 'Index' in w]
['Indexed', 'IndexedBase']
>>> help(Indexed)
Help on class Indexed in module sympy.tensor.indexed:
class Indexed(sympy.core.expr.Expr)
| Represents a mathematical object with indices.
|
| >>> from sympy.tensor import Indexed, IndexedBase, Idx
| >>> from sympy import symbols
| >>> i, j = map(Idx, ['i', 'j'])
| >>> Indexed('A', i, j)
| A[i, j]
|
| It is recommended that Indexed objects are created via IndexedBase:
|
| >>> A = IndexedBase('A')
| >>> Indexed('A', i, j) == A[i, j]
| True
|
>>> b=IndexedBase('b')
>>> summation(b[i]*f(n-i),(i,0,4))
f(n)*b[0] + f(n - 4)*b[4] + f(n - 3)*b[3] + f(n - 2)*b[2] + f(n - 1)*b[1]
You can also create the symbols as needed and use Add rather than summation:
>>> Add(*(Symbol('b'+str(i))*f(n-i) for i in range(5)))
b0*f(n) + b1*f(n - 1) + b2*f(n - 2) + b3*f(n - 3) + b4*f(n - 4)
What do you want to do after you have the expression? The Indexed doesn't
respond to subs -- seems to me that this is an error -- while the symbol
does:
>>> summation(b[i]*f(n-i),(i,0,4))
f(n)*b[0] + f(n - 4)*b[4] + f(n - 3)*b[3] + f(n - 2)*b[2] + f(n - 1)*b[1]
>>> _.subs(b[0],1)
f(n)*b[0] + f(n - 4)*b[4] + f(n - 3)*b[3] + f(n - 2)*b[2] + f(n - 1)*b[1]
-------> notice that the b[0] is still there
>>> Add(*(Symbol('b'+str(i))*f(n-i) for i in range(5)))
b0*f(n) + b1*f(n - 1) + b2*f(n - 2) + b3*f(n - 3) + b4*f(n - 4)
>>> _.subs('b0',1)
b1*f(n - 1) + b2*f(n - 2) + b3*f(n - 3) + b4*f(n - 4) + f(n)
-------> the b0 is now 1
HTH,
Chris
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