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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.

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# Strange sums

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