2008/12/30 Danny <[email protected]>:
>
> Hi Ondrej,
>
> Thanks for the quick response. I took a look at the Sum class tests,
> but there is still one part I'm not clear on.
>
> In defining a summation, I would like to have the sum over an array of
> symbols, indexed by an iterate.
>
> For example, In the sum
>
> \sum_{i=0}^4 tau_i,
>
> I'd like each tau[i] to itself be a symbol, since I later want to
> differentiate with respect to these symbols.
>
> I'd like to do this:
> In [17]: tau = []
>
> In [18]: for i in range(10):
> ....: tau.append(Symbol('tau_%s' % i))
>
> In [19]: Sum(tau[i]**2, (i, 0, 9))
> ---------------------------------------------------------------------------
> ValueError Traceback (most recent call
> last)
>
> or
>
> In [20]: Sum(i**2, (i, tau))
> ---------------------------------------------------------------------------
> ValueError Traceback (most recent call
> last)
>
>
> Ideally, this would give some compact representation of
> tau[0]**2 + tau[1]**2 + ... + tau[9]**2
One way of doing it is this:
In [1]: var("i")
Out[1]: i
In [2]: tau = Function("tau")
In [3]: Sum(tau(i)**2, (i, 0, 9))
Out[3]: Sum(tau(i)**2, (i, 0, 9))
In [4]: Sum(tau(i)**2, (i, 0, 9)).doit()
Out[4]:
2 2 2 2 2 2 2 2 2 2
τ (0) + τ (1) + τ (2) + τ (3) + τ (4) + τ (5) + τ (6) + τ (7) + τ (8) + τ (9)
You can differentiate it like this:
In [7]: Sum(tau(i)**2, (i, 0, 9)).doit().subs(tau(6), x).diff(x).subs(x, tau(6))
Out[7]: 2⋅τ(6)
(One needs to substitute the function tau(i) for a symbol in order to
differentiate.) If you don't call .doit(), it will fail:
In [8]: Sum(tau(i)**2, (i, 0, 9)).subs(tau(6), x).diff(x).subs(x, tau(6))
Out[8]: 0
becuase there is no tau(6) in the Sum expression. This could probably be fixed.
>
> Is there something I'm missing?
Let me know if the above helps, or if we should try to find a
different solution.
Ondrej
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