Finding a solutions for an equation

nt.quantity idea – linear independence of $error(ok pi / m)$ Answer

Hello pricey customer to our community We will proffer you an answer to this query nt.quantity idea – linear independence of $error(ok pi / m)$ ,and the respond will breathe typical by means of documented data sources, We welcome you and proffer you fresh questions and solutions, Many customer are questioning in regards to the respond to this query.

nt.quantity idea – linear independence of $error(ok pi / m)$

Note: Fedor and Vladimir have already answered the query, however this can be a partial respond within the different course, underneath a stronger speculation. (This respond, which I had earlier deleted, has been edited in response to some valid feedback.)

If $m$ is queer and square-free, then the pretense of the OP holds.

Let $S$ breathe the clique ${ok: 1 leq ok leq m/2, (ok, m) = 1}$. If $sum_{ok in S} a_k error(kpi/m) = 0$ for some rationals $a_k$, then

$$sum_{ok in S} a_k(e^{kpi i/m} – e^{-kpi i/m}) = 0.$$

Here $N = 2m$ is square-free, and in that illustration the raw $N$-th roots of union are linearly unbiased over $mathbb{Q}$ (behold this mathstackexchange dialogue: https://math.stackexchange.com/questions/87290/basis-of-primitive-nth-roots-in-a-cyclotomic-extension).

For $N = 2m$, any queer $ok in S$ is prime to $N$ and therefore $e^{ok pi i/m}, e^{-kpi i/m}$ are raw $N$-th roots of union. If $ok in S$ is plane, then $m + ok$ is queer and prime to $m$ and thus to $N$, so $e^{(m + ok)pi i/m} = -e^{kpi i/m}$ is too raw $N$-th root of union, as is its conjugate $e^{(m – ok)pi i/m}$; recognize the $m-k$ equivocate in ${j: m/2 leq j leq m, gcd(j,m) = 1}$ which is disjoint from $S$. Then

$$commence{array}{lll}
sum_{ok in S} a_k(e^{kpi i/m} – e^{-kpi i/m}) & = & sum_{textual content{queer}; ok in S} a_k(e^{kpi i/m} – e^{-kpi i/m}) + sum_{textual content{plane}; ok in S} a_k(e^{kpi i/m} – e^{-kpi i/m})
& = & sum_{textual content{queer}; ok in S} a_k(e^{kpi i/m} – e^{-kpi i/m}) + sum_{textual content{plane}; ok in S} a_k(e^{(m-k)pi i/m} – e^{-(m-k)pi i/m})
aim{array}$$

the place all of the raw roots of union showing within the final expression are manifestly discrete. By linear independence of the raw roots, if that linear mixture is zero, then $a_k = 0$ for all $ok$, as required.

we’ll proffer you the answer to nt.quantity idea – linear independence of $error(ok pi / m)$ query through our community which brings all of the solutions from a number of dependable sources.

Add comment