Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

${i \bmod (k-1)}$ $\quad$ 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.

share|improve this question

2 Answers 2

up vote 1 down vote accepted

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

share|improve this answer
    
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.

share|improve this answer
    
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
2  
Look up multisection of series. –  marty cohen Oct 29 '11 at 6:20

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.