4 ms·
I was thinking that any function like f(x)=e^(-x(1+o(1/x^2)) would work, provided that it can be expended into an infinite series (say, f(x)=e^(-x(1+e^(-x))) )
by S4M 10y ago
I was thinking that any function like f(x)=e^(-x(1+o(1/x^2)) would work, provided that it can be expended into an infinite series (say, f(x)=e^(-x(1+e^(-x))) )