# How to simplify $f(x)=\sum\limits_{i=0}^{\infty}\frac{x^{i \;\bmod (k-1)}}{i!}$?

$$f(x)=\sum_{i=0}^{\infty}\frac{x^{i \;\bmod (k-1)}}{i!}$$

${i \bmod (k-1)}$ $\quad$ says the $x$ powers can be only $x^0$, $x^1$, ...,$x^{k-2}$

Understand simplify a way to transform this infinity sum in a finite formula.

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This is an expansion of the other answer. There's no point in working modulo $k-1$ instead of $k$.

$$\sum_{m=0}^\infty\frac{x^{(m\mod k)}}{m!}=\sum_{m=0}^{k-1}\left(\frac{1}{m!}+\frac{1}{(m+k)!}+\frac{1}{(m+2k)!}+\cdots\right) x^m.$$

Now if $\omega$ is a primitive $k$-th root of unity we have the property

$$\frac{1}{k}\sum_{l=0}^{k-1}\omega^{jl}=\begin{cases}1&j\equiv0\mod k;\\0&j\not\equiv0\mod k.\end{cases}$$ Thus $$\frac{1}{k}\sum_{l=0}^{k-1}\exp(\omega^l)\omega^{-ml}=\frac{1}{k}\sum_{l=0}^{k-1}\left(\sum_{n=0}^\infty\frac{(\omega^l)^n}{n!}\right)\omega^{-ml}=\sum_{n=0}^\infty\frac{1}{n!}\left(\frac{1}{k}\sum_{l=0}^{k-1}\omega^{(n-m)l}\right).$$

The inner sum above is $1$ when $n\equiv m\mod k$ and $0$ otherwise so this equals

$$\frac{1}{m!}+\frac{1}{(m+k)!}+\frac{1}{(m+2k)!}+\cdots.$$

Therefore our original sum is $$(\bullet)=\frac{1}{k}\sum_{m=0}^{k-1}\sum_{l=0}^{k-1}\exp(\omega^l)(x\omega^{-l})^m.$$

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Very interesting answer. It's what I'm looking for. Interesting question, doesn't it? @anon, I undestood all your process, but it's not clear to me how you arrived in $$\frac{1}{k}\sum_{l=0}^{k-1}\exp(\omega^l)\omega^{-ml}$$. I understood using this you find that sum, but, how you had this guess? I think it will be useful to me in other things. – GarouDan Oct 29 '11 at 19:33
@GarouDan: Sorry I haven't been on in awhile, but that's a very good question. I quoted that expression but didn't explicitly say how it was motivated for space/time's sake, but if you want to know I'd recommend reading that line (and the one after it) backwards, because that provides the a priori direction of reasoning. We know the coefficients of $x^m$ are sums of reciprocals of factorials of numbers in residue class $m$ modulo $k$, and this can be written as a sum of $1/n!$ times ($1$ if $n\equiv m$, $0$ otherwise), and used the formula from the preceding line to create the latter piece. – anon Oct 30 '11 at 1:43
=) My real motivation is I'm searching tools and language to express this another problem and solve it. I think this formula ables me inspect some properties of infinite sums and number theory together. – GarouDan Oct 30 '11 at 13:42

Group them by $0,1,2$ and so on. Your formula is

$$\sum_{j=0}^{k-2} \left( x^j \sum_{n=0}^\infty \frac{1}{(n(k-1)+j)!} \right) \,.$$

You should be able to get a closed formula for inside bracket by calculating the Taylor series of $e^z$ at the $(k-1)$-roots of unity.

P.S. For the last part: adding $e^{w_1}+...+e^{w_{k-1}}$ yields the coefficient of $x^0$. Now if you add $\sum w_ie^{w_i}$ you get the coefficient of $x^1$, and so on.

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It's a good idea, uses the roots of unity. But I didn't understand at all. Can you explain for a small k number? For example 3, and give us the explicit formulas. Thx. – GarouDan Oct 29 '11 at 2:46
Look up multisection of series. – marty cohen Oct 29 '11 at 6:20