- #1
Karlisbad
- 131
- 0
"Borel resummation" is useful?
My question is if this is nothing but a "math tool" but not valid for realistic example, for example if we wish to calculate the divergent (but Borel summable) series:
[tex] a(0)+a(1)+a(2)+..... =S [/tex]
then you take the expression : [tex] B(x)=\sum_{n=0}^{\infty}\frac{a(n). x^{n} }{n!} [/tex] ,
so the sum of the series is just "defined":
tex] a(0)+a(1)+a(2)+.....=S=\int_{0}^{\infty}dxB(x)e^{-x} [/tex]
Of course if [tex] a(n)=(-1)^{n} [/tex] or [tex] a(n)=n! [/tex] then it's very easy to get B(x), but in a "realistic" situation that you don't even know the general term a(n) or it's very complicated there's no way to obtain its Borel sum
My question is if this is nothing but a "math tool" but not valid for realistic example, for example if we wish to calculate the divergent (but Borel summable) series:
[tex] a(0)+a(1)+a(2)+..... =S [/tex]
then you take the expression : [tex] B(x)=\sum_{n=0}^{\infty}\frac{a(n). x^{n} }{n!} [/tex] ,
so the sum of the series is just "defined":
tex] a(0)+a(1)+a(2)+.....=S=\int_{0}^{\infty}dxB(x)e^{-x} [/tex]
Of course if [tex] a(n)=(-1)^{n} [/tex] or [tex] a(n)=n! [/tex] then it's very easy to get B(x), but in a "realistic" situation that you don't even know the general term a(n) or it's very complicated there's no way to obtain its Borel sum