**The problem:**

If the infinite sum of a function is known, how to find:

$$\begin{align*} \sum_{i\equiv 0 \mod m}f(x_0+i)=\\\\ f(x_0)+f(x_0+m)+f(x_0+2m)+f(x_0+3m)+\ldots \end{align*}$$

And if the finite sum of a function is known, how to find:

$$\begin{align*} \sum_{i\equiv 0 \mod m}^{i = {(x_0+\lfloor \frac{x-x_0+1}{m}\rfloor m)}}f(x_0+i)=\\\\ f(x_0)+f(x_0+m)+f(x_0+2m)+f(x_0+3m) &\quad +\ldots+f\left(x_0+\left\lfloor \frac{x-x_0+1}{m}\right\rfloor m\right) \end{align*}$$

**Details:**

I had posted this question in Math.StackExchange too (about one day before). It's in this link.

If we know a function $f$ and we can find the sum of its terms (defined as $S_f$), how to find the sum, but jumping some factors (defined as $MS_f$, where M representes modular)?

What's the relation with the sum function ($S_f$)? (I think this uses the root of the unity, but don't know how.)

For example, if:

$$S_f=\displaystyle\sum_{i=1}^{\infty}f(i)=f(1)+f(2)+\ldots$$

with infinite terms, how to find

$$\begin{align*} MS_f(x_0,m)&=\sum_{i\equiv 0 \mod m}f(x_0+i)\\\\ & = f(x_0)+f(x_0+m)+f(x_0+2m)+f(x_0+3m)+\ldots \end{align*}$$

And if:

$$S_f(x)=\displaystyle\sum_{i=1}^{x}f(i)=f(1)+\ldots+f(x-1)+f(x),$$

how to find

$$\begin{align*} MS_f(x,x_0,m)&=\sum_{i\equiv 0 \mod m}^{i = {(x_0+\lfloor \frac{x-x_0+1}{m}\rfloor m)}}f(x_0+i)\\\\ & = f(x_0)+f(x_0+m)+f(x_0+2m)+f(x_0+3m) \\\\ &\quad +\ldots+f\left(x_0+\left\lfloor \frac{x-x_0+1}{m}\right\rfloor m\right) \end{align*}$$

where $(x_0+\lfloor \frac{x-x_0+1}{m}\rfloor m)$ is the ultimate term of the arithmetic progression $x_0+k\times m$ which not exceeds $x$.

Edited:

As Jacques Carette said, I think the answer is using something like:

$MS_f(x,x_0,m)=\displaystyle\sum_{i=0}^{m-1}a_iS_f(w^ix)$ or $\displaystyle\sum_{i=0}^{m-1}a_iS_f(w^i(x+x0))$

but I don't know exactly.

**Example:**

$$S_f=\sum_{i=1}^{\infty}\frac{x^{i-1}}{(i-1)!}=e^x, \quad f(i)=\frac{x^{i-1}}{(i-1)!}$$ $$\begin{align*} MS_f(x_0,m)=\sum_{i\equiv 0 \mod m}f(x_0+i)=\sum_{i\equiv 0 \mod m}\frac{x^{(x_0+i)-1}}{((x_0+i)-1)!}\implies\\\\ MS_f(3,2)=\sum_{i\equiv 0 \mod 2}\frac{x^{(3+i)-1}}{((3+i)-1)!}=\sum_{j=0}^{\infty}\frac{x^{3+2j-1}}{(3+2j-1)!}=\cosh (x)-1 \end{align*}$$

firston tea.mathoverflow.net if the question fits MO? – Mariano Suárez-Alvarez♦ Nov 28 '11 at 18:48