Am Donnerstag, 16. April 2020 13:39:39 UTC+2 schrieb Mihai:
>
> fn = diff(FN(x,n), x)
> mean(n) = integral(fn(x, n), (x, 0, oo))
>
The main problem with this is that `fn` is just an expression, not a
function of `(x, n)`. The call `fn(x, n)` switches the position of x and n,
as apparently the arguments of the expression are listed in alphabetical
order.
sage: fn
n*(1/(e^(-x) + 1))^(n - 1)*e^(-x)/(e^(-x) + 1)^2
sage: fn.arguments()
(n, x)
sage: fn(x, n)
x*(e^(-n) + 1)^(-x + 1)*e^(-n)/(e^(-n) + 1)^2
With that in mind, replacing `fn(x, n)` by just `fn` in the integral leads
to the output that Michael obtained, so apparently Sage (or Maxima) has
difficulties to compute this integral symbolically for all n. However, if
you replace n by an actual integer before integrating, you get better
results:
sage: F(x)=1/(1 + exp(-x))
sage: FN(x,n) = F(x)**n
sage: fn = diff(FN(x,n), x)
sage: mean = lambda k: integral(fn.subs({n: k}), (x, 0, oo))
sage: [mean(k) for k in (1..10)]
[1/2, 3/4, 7/8, 15/16, 31/32, 63/64, 127/128, 255/256, 511/512, 1023/1024]
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