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@nbruin I thought it assumed $X(i) = i$ from the expansion to $(-1)^n ... X(0)X(1)...$ but now that I see it again, it might have come to mind from some other tests I run (but can't replicate now). The next thing I'll do is take the logarithm of $L$, so differentiating shouldn't be a problem so long as it understands the product will turn into a sum (not sure it will since log of the product simply puts a log around the whole expression even when simplified). Perhaps you could expand on an answer? Thanks

That is very strange... However you found an issue with my code. $a$ and $b$ should be greater than or equal to 0, not strictly greater. In that case the result remains the same.

No, x has variable length, so that's not really a possibility. It shouldn't be that hard to generalize this case. And even if it doesn't go straight to my problem I'd have guessed there should be a way to work with variable-length variables... Perhaps I'm thinking too much like a programmer.

I need to work symbolically with expressions such as this

$L(x; a,b) = \prod_{i=1}^n{abx_i^{a-1}(1-x_i^a)^{b-1}}$

where $x$ would be a random sample of size $n$.

I don't know of any way to express the indexing of the sample $x$ by each element...

The closest I got was defining the variables $n$ and $i$ and representing $x$ as a function

var('a','b','x','n','i')
assume(x>0,a>0,b>0,i>0,n>0)
X = function('X',nargs=1)
L = product(a*b*X(i)^(a-1)*(1-X(i)^a)^(b-1), i, 1, n)

But this seems to inmediately assume that $X(i) = i$ and $L$ is represented as:

$-\frac{\left(-1\right)^{n} a^{n} b^{n} X\left(0\right) X\left(-1\right) X\left(-2\right) X\left(-3\right) X\left(-4\right) X\left(-5\right) X\left(-6\right) X\left(-7\right) X\left(-8\right) X\left(-9\right) {\prod_{i=1}^{n} {\left(-X\left(i\right)^{a} + 1\right)}^{b}} {\prod_{i=1}^{n} X\left(i\right)^{a}}}{X\left(n - 1\right) X\left(n - 2\right) X\left(n - 3\right) X\left(n - 4\right) X\left(n - 5\right) X\left(n - 6\right) X\left(n - 7\right) X\left(n - 8\right) X\left(n - 9\right) X\left(n\right) {\prod_{i=1}^{n} X\left(i\right)^{a} - 1} {\prod_{i=1}^{n} X\left(i - 10\right)}}$

I don't know how to deal with this expression, and it seems to me like it should be straight forward.

In case it's meaningful, after defining the expression, I will be differentiating it with respect to both $a$ and $b$.

I need to work symbolically with expressions such as this

$L(x; a,b) = \prod_{i=1}^n{abx_i^{a-1}(1-x_i^a)^{b-1}}$

where $x$ would be a random sample of size $n$.

I don't know of any way to express the indexing of the sample $x$ by each element...

The closest I got was defining the variables $n$ and $i$ and representing $x$ as a function

var('a','b','x','n','i')
assume(x>0,a>0,b>0,i>0,n>0)
X = function('X',nargs=1)
L = product(a*b*X(i)^(a-1)*(1-X(i)^a)^(b-1), i, 1, n)

But this seems to inmediately assume that $X(i) = i$ and $L$ is represented as:

$-\frac{\left(-1\right)^{n} a^{n} b^{n} X\left(0\right) X\left(-1\right) X\left(-2\right) X\left(-3\right) X\left(-4\right) X\left(-5\right) X\left(-6\right) X\left(-7\right) X\left(-8\right) X\left(-9\right) {\prod_{i=1}^{n} {\left(-X\left(i\right)^{a} + 1\right)}^{b}} {\prod_{i=1}^{n} X\left(i\right)^{a}}}{X\left(n - 1\right) X\left(n - 2\right) X\left(n - 3\right) X\left(n - 4\right) X\left(n - 5\right) X\left(n - 6\right) X\left(n - 7\right) X\left(n - 8\right) X\left(n - 9\right) X\left(n\right) {\prod_{i=1}^{n} X\left(i\right)^{a} - 1} {\prod_{i=1}^{n} X\left(i - 10\right)}}$

I don't know how to deal with this expression, and it seems to me like it should be straight forward.