"How to simplify f(x)=∑i=0∞ximod(k−1)i!" was answered by our community

Perla Benitez

Answered question

2022-04-08

How to simplify

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

Answer & Explanation

unduncjineei5r3

Beginner2022-04-09Added 19 answers

This is an expansion of the other answer. There's no point in working modulo $k - 1$ instead of k
$\sum_{m = 0}^{\infty} \frac{x^{(m mod k)}}{m!} = \sum_{m = 0}^{k - 1} (\frac{1}{m!} + \frac{1}{(m + k)!} + \frac{1}{(m + 2 k)!} + \dots) x^{m}$
Now if $ω$ is a primitive k-th root of unity we have the property
$\frac{1}{k} \sum_{l = 0}^{k - 1} ω^{j l} = (1 j \equiv 0 \mod k;), (0 j ≢ 0 \mod k) :$
Thus
$\frac{1}{k} \sum_{l = 0}^{k - 1} \exp (ω^{l}) ω^{- m l} = \frac{1}{k} \sum_{l = 0}^{k - 1} (\sum_{n = 0}^{\infty} \frac{{(ω^{l})}^{n}}{n!}) ω^{- m l} = \sum_{n = 0}^{\infty} \frac{1}{n!} (\frac{1}{k} \sum_{l = 0}^{k - 1} ω^{(n - m) l})$
The inner sum above is 1 when $n \equiv m mod k$ and 0 otherwise so this equals\
$\frac{1}{m!} + \frac{1}{(m + k)!} + \frac{1}{(m + 2 k)!} + \dots$
Therefore our original sum is
$(\cdot) = \frac{1}{k} \sum_{m = 0}^{k - 1} \sum_{l = 0}^{k - 1} \exp (ω^{l}) {(x ω^{- l})}^{m}$

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