how to calculate the derivative of a finite series in sympy?

Question

I am new to sympy and I want to calculate the derivative of a finite series with variables in two periods. For example, I tried the following code:

from sympy import symbols, Sum, diff, IndexedBase, Idx

T = symbols('T', interger=True)
t = symbols('t', interger=True)
β, σ = symbols('β σ')
a = symbols('a', cls=IndexedBase)

L = Sum(β * a[t] + σ * a[t + 1], (t, 0, T))
print(diff(L, a[t]))

I got the results like Sum(β, (t, 0, T)). But in fact, my expectation is β+σ. Why? Because a[t] appears two times in this series. One is β * a[t] (when time=t) and another is σ * a[t] (when time=t-1).

How could I get the expected result with sympy? Thanks

Answer 1

In the context of a sum the summation variable is a bound dummy variable i.e. the Sum as a whole is not really a function of t just like Integral(x, (x, 0, 1)) = 1/2 is not really a function of x. It does not make sense to use the same symbol t outside of the sum when it is not free in L.

Instead use a different symbol:

In [25]: L
Out[25]: 
  T                        
 ___                       
 ╲                         
  ╲                        
  ╱   (β⋅a[t] + σ⋅a[t + 1])
 ╱                         
 ‾‾‾                       
t = 0                      

In [26]: L.diff(a[n])
Out[26]: 
  T                        
 ___                       
 ╲                         
  ╲   ⎛β⋅δ    + σ⋅δ       ⎞
  ╱   ⎝   n,t      n,t + 1⎠
 ╱                         
 ‾‾‾                       
t = 0                      

In [27]: L.diff(a[n]).doit()
Out[27]: 
⎧β + σ  for T ≥ n ∧ T ≥ n - 1 ∧ n ≥ 0 ∧ n ≥ 1
⎪                                            
⎪  β              for T ≥ n ∧ n ≥ 0          
⎨                                            
⎪  σ            for T ≥ n - 1 ∧ n ≥ 1        
⎪                                            
⎩  0                  otherwise