>
>
> I would most appreciate it if anyone could advise me on what to do. Thanks!
>
>
Use SymPy ! :-)
>>> binomial(i,j)
binomial(i, j)
>>> _.subs(i,5)
binomial(5, j)
>>> _.subs(j,2)
10
So your expression is written as
>>> var('lamb mu')
(lamb, mu)
>>> l, i = symbols('l i',integer=True)
>>> f=symbols('f',cls=Function)
>>> f = lamb * ( 1 + (-mu)**(l+i) ) * binomial(l+i,i)
>>> pprint(f)
⎛ i + l ⎞ ⎛i + l⎞
lamb⋅⎝(-μ) + 1⎠⋅⎜ ⎟
⎝ i ⎠
>>> f.subs(((i,5),(l,2)))
21*lamb*(-mu**7 + 1)
Please note that although you are thinking of f as a function, it is (as
you have defined it) actually an expression equal to what is on the right
hand side. If you want f to be a function of l and i (for a fixed mu and
lamb) then you might write
>>> f=lambda l,i:lamb * ( 1 + (-mu)**(l+i) ) * binomial(l+i,i)
>>> f(2, 5)
21*lamb*(-mu**7 + 1)
or write the function as
>>> def f(l, i):
... return lamb * ( 1 + (-mu)**(l+i) ) * binomial(l+i,i)
...
>>> f(5, 2)
21*lamb*(-mu**7 + 1)
>>> f(a,b)
lamb*((-mu)**(a + b) + 1)*binomial(a + b, b)
>>> f(l,i)
lamb*((-mu)**(i + l) + 1)*binomial(i + l, i)
If you want to include an undefined function in an expression, that's the
time to use f as a function:
>>> f = Function('f')
>>> f(1,2) + 3
f(1, 2) + 3
Hope that help,
Chris
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