- #1
eljose
- 492
- 0
let be e an small parameter e<<<1 then if we want to find a solution to the equation:
[tex] e\ddot x + f(t)x=0 [/tex]
then we could write a solution to it in the form:
[tex] x(t)=exp(i \int dt f(t)^{1/2}/e)[a_{0}(t)+ea_{1}(t)+e^{2}a_{2}(t)+...] [/tex]
My question is if we could apply Borel resummation (or other technique) to give a "sum" for a divergent series in the form:
[tex]a_{0}(t)+ea_{1}(t)+e^{2}a_{2}(t)+...\rightarrow \int_{0}^{\infty}dxe^{-x}B(t,x,e)dx [/tex]
With [tex] B(x,t,e)= \sum_{n=0}^{\infty} \frac{a_{n}(t)e^{n} x^{n}}{n!} [/tex]
the generating function of the coefficient..so we can extend the domain of convergence for the solution not only to the case e--->0 but to every value of e or at least valid when e-->1.:tongue2:
[tex] e\ddot x + f(t)x=0 [/tex]
then we could write a solution to it in the form:
[tex] x(t)=exp(i \int dt f(t)^{1/2}/e)[a_{0}(t)+ea_{1}(t)+e^{2}a_{2}(t)+...] [/tex]
My question is if we could apply Borel resummation (or other technique) to give a "sum" for a divergent series in the form:
[tex]a_{0}(t)+ea_{1}(t)+e^{2}a_{2}(t)+...\rightarrow \int_{0}^{\infty}dxe^{-x}B(t,x,e)dx [/tex]
With [tex] B(x,t,e)= \sum_{n=0}^{\infty} \frac{a_{n}(t)e^{n} x^{n}}{n!} [/tex]
the generating function of the coefficient..so we can extend the domain of convergence for the solution not only to the case e--->0 but to every value of e or at least valid when e-->1.:tongue2: