# Strange sums

1. Apr 14, 2006

### eljose

"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.

2. Apr 15, 2006

### matt grime

What on earth allows you to assume that the f are differentiable? And even then what does raising e to the power d/dx do? Let what be an infinitesimal generator of what? What's the/a primitive of f? Why invent a name for it that you don't ever use again? Not that the primitive makes much sense (inverting d/dx?).

You really need to focus on explaining what you mean properly (not rigorously; you have a completely unnecessary issue with rigour).

If by F^n you mean F raised to the power n (F(x)^n), then it cannot be a correct formula, for obvious reasons which I'll let you figure out (hint what if F(x)=0 for some x).

If it means F applied n times, then my gut reaction is that it still will be false, not that it is clear what F actually is.

Your second sum is also not defined at x=0.

Last edited: Apr 15, 2006
3. Apr 15, 2006

### eljose

-No.... $$F(x)=\int{dx}f(x)$$ and $$\frac{d^{n}F}{dx^{n}}=F^(n)(x)$$ by the way you say is not correct Euler himself used (1) in the same manner i do to prove that:

$$1+2+3+4+5+6+...........=(-1/12)$$ just put f(x)=x and apply the differential operator..by the way is a known result that:

$$e^{aD}f(x)=f(x+a)$$ or $$e^{xD}=f(2x)$$ just take a look at "Wikipedia" (english version)..chao.

4. Apr 15, 2006

### matt grime

There you go again. I didn't say it was not a correct method I said that what you wrote doesn't make sense. You have a very odd sense of what is 'universally known and accepted' terminology, and a poor way of presenting things; partly this is a language problem, partly it is not.

If you ask a question, why not explain the terms properly? However, I'll take a guess that e^D is short hand for the operator

$$\sum \frac{1}{n!}\frac{d^n}{dx^n}$$