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About sums similar to gauss sums
Gauss-type sums for cube rootsA Gauss sum like summationRelation that holds for the Legendre symbol of an integer but not for the Jacobi symbol?A Trigonometric Sum Related to Gauss SumsPrime power Gauss sums are zeroA Gauss sum over a field.Determination of quartic Gauss sumsGauss Sum calculationQuestion About Primitive Root of UnityInfinite quadratic gauss sum.
$begingroup$
It is well known the case for sums like:
$$
sum_i=0^p^n -1zeta^-ai,
$$ where zeta is a primitive $p^n$-rooth of $1$.
But, is there a standard formula for sums like:
$$
sum_i=0^p^n -1i^Nzeta^-ai
$$
where $N$ is a fixed integer?
Do you know references? Thanks for suggestions!
algebraic-number-theory arithmetic roots-of-unity gauss-sums
$endgroup$
|
show 1 more comment
$begingroup$
It is well known the case for sums like:
$$
sum_i=0^p^n -1zeta^-ai,
$$ where zeta is a primitive $p^n$-rooth of $1$.
But, is there a standard formula for sums like:
$$
sum_i=0^p^n -1i^Nzeta^-ai
$$
where $N$ is a fixed integer?
Do you know references? Thanks for suggestions!
algebraic-number-theory arithmetic roots-of-unity gauss-sums
$endgroup$
$begingroup$
Look up the formula for $sum_i = 0^m-1 i^Nx^i$ and then set $m=p^n-1$ and $x=zeta^-a$.
$endgroup$
– KCd
Mar 12 at 10:35
2
$begingroup$
Start with a finite geometric series $1+x+ldots+x^m-1$ and repeatedly differentiate and multiply by $x$ to make $x^i$ have coefficients that are powers of $i$.
$endgroup$
– KCd
Mar 12 at 10:40
$begingroup$
@KCd I think to have found the formula for when $i^N $ is costant but the general formula seems quite complicated
$endgroup$
– andres
Mar 13 at 9:41
$begingroup$
I have no idea what "when $i^N$ is constant" means ($i$ is changing, so it's not constant), but in any case sure, the formula is quite complicated. There's no reason to expect tidy formulas for such things as $N$ grows.
$endgroup$
– KCd
Mar 13 at 10:46
$begingroup$
@KCd I mean in the case when $i^N =c$ it's a simpler case. In general i think that it is not really possibile give a formula. In fact i think that it does not exists formulas for $sum_i i^N$ too.
$endgroup$
– andres
Mar 13 at 11:06
|
show 1 more comment
$begingroup$
It is well known the case for sums like:
$$
sum_i=0^p^n -1zeta^-ai,
$$ where zeta is a primitive $p^n$-rooth of $1$.
But, is there a standard formula for sums like:
$$
sum_i=0^p^n -1i^Nzeta^-ai
$$
where $N$ is a fixed integer?
Do you know references? Thanks for suggestions!
algebraic-number-theory arithmetic roots-of-unity gauss-sums
$endgroup$
It is well known the case for sums like:
$$
sum_i=0^p^n -1zeta^-ai,
$$ where zeta is a primitive $p^n$-rooth of $1$.
But, is there a standard formula for sums like:
$$
sum_i=0^p^n -1i^Nzeta^-ai
$$
where $N$ is a fixed integer?
Do you know references? Thanks for suggestions!
algebraic-number-theory arithmetic roots-of-unity gauss-sums
algebraic-number-theory arithmetic roots-of-unity gauss-sums
edited Mar 12 at 10:07
Gurjinder
552417
552417
asked Mar 12 at 9:57
andresandres
2439
2439
$begingroup$
Look up the formula for $sum_i = 0^m-1 i^Nx^i$ and then set $m=p^n-1$ and $x=zeta^-a$.
$endgroup$
– KCd
Mar 12 at 10:35
2
$begingroup$
Start with a finite geometric series $1+x+ldots+x^m-1$ and repeatedly differentiate and multiply by $x$ to make $x^i$ have coefficients that are powers of $i$.
$endgroup$
– KCd
Mar 12 at 10:40
$begingroup$
@KCd I think to have found the formula for when $i^N $ is costant but the general formula seems quite complicated
$endgroup$
– andres
Mar 13 at 9:41
$begingroup$
I have no idea what "when $i^N$ is constant" means ($i$ is changing, so it's not constant), but in any case sure, the formula is quite complicated. There's no reason to expect tidy formulas for such things as $N$ grows.
$endgroup$
– KCd
Mar 13 at 10:46
$begingroup$
@KCd I mean in the case when $i^N =c$ it's a simpler case. In general i think that it is not really possibile give a formula. In fact i think that it does not exists formulas for $sum_i i^N$ too.
$endgroup$
– andres
Mar 13 at 11:06
|
show 1 more comment
$begingroup$
Look up the formula for $sum_i = 0^m-1 i^Nx^i$ and then set $m=p^n-1$ and $x=zeta^-a$.
$endgroup$
– KCd
Mar 12 at 10:35
2
$begingroup$
Start with a finite geometric series $1+x+ldots+x^m-1$ and repeatedly differentiate and multiply by $x$ to make $x^i$ have coefficients that are powers of $i$.
$endgroup$
– KCd
Mar 12 at 10:40
$begingroup$
@KCd I think to have found the formula for when $i^N $ is costant but the general formula seems quite complicated
$endgroup$
– andres
Mar 13 at 9:41
$begingroup$
I have no idea what "when $i^N$ is constant" means ($i$ is changing, so it's not constant), but in any case sure, the formula is quite complicated. There's no reason to expect tidy formulas for such things as $N$ grows.
$endgroup$
– KCd
Mar 13 at 10:46
$begingroup$
@KCd I mean in the case when $i^N =c$ it's a simpler case. In general i think that it is not really possibile give a formula. In fact i think that it does not exists formulas for $sum_i i^N$ too.
$endgroup$
– andres
Mar 13 at 11:06
$begingroup$
Look up the formula for $sum_i = 0^m-1 i^Nx^i$ and then set $m=p^n-1$ and $x=zeta^-a$.
$endgroup$
– KCd
Mar 12 at 10:35
$begingroup$
Look up the formula for $sum_i = 0^m-1 i^Nx^i$ and then set $m=p^n-1$ and $x=zeta^-a$.
$endgroup$
– KCd
Mar 12 at 10:35
2
2
$begingroup$
Start with a finite geometric series $1+x+ldots+x^m-1$ and repeatedly differentiate and multiply by $x$ to make $x^i$ have coefficients that are powers of $i$.
$endgroup$
– KCd
Mar 12 at 10:40
$begingroup$
Start with a finite geometric series $1+x+ldots+x^m-1$ and repeatedly differentiate and multiply by $x$ to make $x^i$ have coefficients that are powers of $i$.
$endgroup$
– KCd
Mar 12 at 10:40
$begingroup$
@KCd I think to have found the formula for when $i^N $ is costant but the general formula seems quite complicated
$endgroup$
– andres
Mar 13 at 9:41
$begingroup$
@KCd I think to have found the formula for when $i^N $ is costant but the general formula seems quite complicated
$endgroup$
– andres
Mar 13 at 9:41
$begingroup$
I have no idea what "when $i^N$ is constant" means ($i$ is changing, so it's not constant), but in any case sure, the formula is quite complicated. There's no reason to expect tidy formulas for such things as $N$ grows.
$endgroup$
– KCd
Mar 13 at 10:46
$begingroup$
I have no idea what "when $i^N$ is constant" means ($i$ is changing, so it's not constant), but in any case sure, the formula is quite complicated. There's no reason to expect tidy formulas for such things as $N$ grows.
$endgroup$
– KCd
Mar 13 at 10:46
$begingroup$
@KCd I mean in the case when $i^N =c$ it's a simpler case. In general i think that it is not really possibile give a formula. In fact i think that it does not exists formulas for $sum_i i^N$ too.
$endgroup$
– andres
Mar 13 at 11:06
$begingroup$
@KCd I mean in the case when $i^N =c$ it's a simpler case. In general i think that it is not really possibile give a formula. In fact i think that it does not exists formulas for $sum_i i^N$ too.
$endgroup$
– andres
Mar 13 at 11:06
|
show 1 more comment
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active
oldest
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$begingroup$
Look up the formula for $sum_i = 0^m-1 i^Nx^i$ and then set $m=p^n-1$ and $x=zeta^-a$.
$endgroup$
– KCd
Mar 12 at 10:35
2
$begingroup$
Start with a finite geometric series $1+x+ldots+x^m-1$ and repeatedly differentiate and multiply by $x$ to make $x^i$ have coefficients that are powers of $i$.
$endgroup$
– KCd
Mar 12 at 10:40
$begingroup$
@KCd I think to have found the formula for when $i^N $ is costant but the general formula seems quite complicated
$endgroup$
– andres
Mar 13 at 9:41
$begingroup$
I have no idea what "when $i^N$ is constant" means ($i$ is changing, so it's not constant), but in any case sure, the formula is quite complicated. There's no reason to expect tidy formulas for such things as $N$ grows.
$endgroup$
– KCd
Mar 13 at 10:46
$begingroup$
@KCd I mean in the case when $i^N =c$ it's a simpler case. In general i think that it is not really possibile give a formula. In fact i think that it does not exists formulas for $sum_i i^N$ too.
$endgroup$
– andres
Mar 13 at 11:06