inverse of a function with product terms I am trying to find inverse of this function:
$F(x) = \prod_{i=1}^{n} (1-e^{{(- \lambda x)}^{\alpha_i}})$
I am confused what will happen to product symbol if I take inverse of this function. The RHS is $(1-e^{{(- \lambda x)}^{\alpha_1}}) (1-e^{{(- \lambda x)}^{\alpha_2}}) ... (1-e^{{(- \lambda x)}^{\alpha_n}})$. How can it be simplified further so that I can get its inverse?
Should I take inverse of individual terms and add them up as each term is increasing in $\mathbb{R}$?
 A: Let's first change a bit the notation you adopted (better reserve the $i$ for $\sqrt{-1}$)
$$
F(x) = \prod\limits_{1\, \le \,k\, \le \,n} {\left( {1 - e^{\left( { - \lambda \,x} \right)^{\,a_{\,k} } } } \right)}  = \prod\limits_{1\, \le \,k\, \le \,n} {\left( {1 - e^{ - \lambda \,\,a_{\,k} x} } \right)}  = \prod\limits_{0\, \le \,k\, \le \,n - 1} {\left( {1 - e^{ - 2b_{\,k} x} } \right)} 
$$
In the case that the coefficients $a_k$ / $b_k$ have generic real values, you end up with the sum of $n$ exponentials
which do not allow to express the inverse by any algebraic formula.
The only approach I can suggest is to rewrite the single terms as
$$
1 - e^{ - 2b_{\,k} x}  = e^{\, - b_{\,k} x + b_{\,k} x}  - e^{\, - b_{\,k} x - b_{\,k} x}  = e^{\, - b_{\,k} x} \left( {e^{\,b_{\,k} x}  - e^{\, - b_{\,k} x} } \right) = 2e^{\, - b_{\,k} x} \sinh \left( {\,b_{\,k} x} \right)
$$
or equivalently as:
$$
\eqalign{
  & 1 - e^{ - 2b_{\,k} x}  = \left( {1 - e^{ - b_{\,k} x} } \right)\left( {1 + e^{ - b_{\,k} x} } \right) = e^{\, - b_{\,k} x} \left( {e^{\,b_{\,k} x/2}  - e^{\, - b_{\,k} x/2} } \right)\left( {e^{\,b_{\,k} x/2}  + e^{\, - b_{\,k} x/2} } \right) =   \cr 
  &  = 4e^{\, - b_{\,k} x} \cosh \left( {\,{{b_{\,k} } \over 2}x} \right)\sinh \left( {\,{{b_{\,k} } \over 2}x} \right) \cr} 
$$
so that $F(x)$ can be written as:
$$
F(x) = \prod\limits_{0\, \le \,k\, \le \,n - 1} {\left( {1 - e^{ - 2b_{\,k} x} } \right)}  = 2^{\,n} e^{\, - \left( {\sum\limits_{0\, \le \,k\, \le \,n - 1} {b_{\,k} } } \right)\;x} \prod\limits_{0\, \le \,k\, \le \,n - 1} {\sinh \left( {\,b_{\,k} x} \right)} 
$$
which tells you something about its behaviour.
In particular, for $x \to 0$, you can write asymptotically:
$$
\eqalign{
  & \left. {F(x)\;} \right|_{0\; \leftarrow \;x}  \approx 2^{\,n} e^{\, - \left( {\sum\limits_{0\, \le \,k\, \le \,n - 1} {b_{\,k} } } \right)\;x} \prod\limits_{0\, \le \,k\, \le \,n - 1} {b_{\,k} x\left( {1 + O(x^{\,2} )} \right)}  =   \cr 
  &  = 2^{\,n} \left( {\prod\limits_{0\, \le \,k\, \le \,n - 1} {b_{\,k} } } \right)x^{\,n} e^{\, - \left( {\sum\limits_{0\, \le \,k\, \le \,n - 1} {b_{\,k} } } \right)\;x} \left( {1 + O(x^{\,2} )} \right) =   \cr 
  &  \approx 2^{\,n} \left( {\prod\limits_{0\, \le \,k\, \le \,n - 1} {b_{\,k} } } \right)x^{\,n} \left( {1 - \left( {\sum\limits_{0\, \le \,k\, \le \,n - 1} {b_{\,k} } } \right)\;x} \right) + O(x^{\,n + 2} ) \cr} 
$$
A: You could try taking logarithms. Thus, you would get sums instead of a product. When you've finished calculating that inverse, the last step is taking exponentials. 
But that only works if $F>0$, so you would need some restrictions. 
