eljose
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"strange" sums...
Let be the next 2 sums in the form:
f(x)+f(x+1)+f(x+2)+... (1) and
f(x)+f(2x)+f(3x)+... (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
D=d/dx (traslation) xD=x(d/dx) (dilatation)
the 2 series above can be "summed" as:
(1+e^{D}+e^{2D}+e^{3D}+...)f(x)
(1+e^{xD}+e^{2xD}+e^{3xD}+...)f(x)
now we put (1/D)f(x)=F(x) and (1/xD)f(x)=F(x)/x
where F(x) is the "primitive" of f(x) then we would have..
\sum_{n=0}^{\infty}\frac{B_{n}}{n!}F^{n}(x)
\sum_{n=0}^{\infty}\frac{B_{n}}{n!}x^{n-1}F^{n}(x)
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:
f(x)+f(x+1)+f(x+2)+... (1) and
f(x)+f(2x)+f(3x)+... (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
D=d/dx (traslation) xD=x(d/dx) (dilatation)
the 2 series above can be "summed" as:
(1+e^{D}+e^{2D}+e^{3D}+...)f(x)
(1+e^{xD}+e^{2xD}+e^{3xD}+...)f(x)
now we put (1/D)f(x)=F(x) and (1/xD)f(x)=F(x)/x
where F(x) is the "primitive" of f(x) then we would have..
\sum_{n=0}^{\infty}\frac{B_{n}}{n!}F^{n}(x)
\sum_{n=0}^{\infty}\frac{B_{n}}{n!}x^{n-1}F^{n}(x)
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.