- #1
Sangoku
- 20
- 0
can we find for a well-behaved f(x) a Rational (Padé approximation) so
[tex] \int_{0}^{\infty}dx f(x) - \int_{0}^{\infty}dx Q(x) \approx 0 [/tex] ??
Where Q(x) is a rational function, the main idea is that the integral for Q(x) can be performed exactly , whereas the initial integral of f(x) not.
[tex] \int_{0}^{\infty}dx f(x) - \int_{0}^{\infty}dx Q(x) \approx 0 [/tex] ??
Where Q(x) is a rational function, the main idea is that the integral for Q(x) can be performed exactly , whereas the initial integral of f(x) not.