- #1
eljose
- 492
- 0
"strange" sums...
Let be the next 2 sums in the form:
[tex] f(x)+f(x+1)+f(x+2)+... [/tex] (1) and
[tex] f(x)+f(2x)+f(3x)+... [/tex] (2)
how would you calculate them?..well i used a "non-rigorous" but i think correct method to derive their sums..let be the infinitesimal generator
[tex] D=d/dx [/tex] (traslation) [tex] xD=x(d/dx) [/tex] (dilatation)
the 2 series above can be "summed" as:
[tex] (1+e^{D}+e^{2D}+e^{3D}+...)f(x) [/tex]
[tex](1+e^{xD}+e^{2xD}+e^{3xD}+...)f(x) [/tex]
now we put [tex] (1/D)f(x)=F(x) [/tex] and [tex](1/xD)f(x)=F(x)/x [/tex]
where F(x) is the "primitive" of f(x) then we would have..
[tex] \sum_{n=0}^{\infty}\frac{B_{n}}{n!}F^{n}(x) [/tex]
[tex] \sum_{n=0}^{\infty}\frac{B_{n}}{n!}x^{n-1}F^{n}(x) [/tex]
these 2 expressions above would be the sum for (1) and (2) respectively where the B are the Bernoulli,s number and (n) means n-th derivative of F.
Let be the next 2 sums in the form:
[tex] f(x)+f(x+1)+f(x+2)+... [/tex] (1) and
[tex] f(x)+f(2x)+f(3x)+... [/tex] (2)
how would you calculate them?..well i used a "non-rigorous" but i think correct method to derive their sums..let be the infinitesimal generator
[tex] D=d/dx [/tex] (traslation) [tex] xD=x(d/dx) [/tex] (dilatation)
the 2 series above can be "summed" as:
[tex] (1+e^{D}+e^{2D}+e^{3D}+...)f(x) [/tex]
[tex](1+e^{xD}+e^{2xD}+e^{3xD}+...)f(x) [/tex]
now we put [tex] (1/D)f(x)=F(x) [/tex] and [tex](1/xD)f(x)=F(x)/x [/tex]
where F(x) is the "primitive" of f(x) then we would have..
[tex] \sum_{n=0}^{\infty}\frac{B_{n}}{n!}F^{n}(x) [/tex]
[tex] \sum_{n=0}^{\infty}\frac{B_{n}}{n!}x^{n-1}F^{n}(x) [/tex]
these 2 expressions above would be the sum for (1) and (2) respectively where the B are the Bernoulli,s number and (n) means n-th derivative of F.