sum of fractions $(modp)$
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Compute $\sum\lfloor\frac{k^3}{p}\rfloor$ for nonnegative integers $r<p$ and an odd prime $p$.
Re: sum of fractions $(modp)$
ok, but where is $r$?
One one thing is neutral in the universe, that is $0$.
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Re: sum of fractions $(modp)$
Sorry, that would be $ k $ instead of $ r $.
Re: sum of fractions $(modp)$
Let \(0\le \{x\}<1\) be the fractional part of \(x\) so that \(x=\lfloor x\rfloor+\{x\}\). Now for \(1\le k\le p-1\) we have
\[\left\{\dfrac{k^3}{p}\right\}+\left\{\dfrac{\left(p-k\right)^3}{p}\right\}=\left\{\dfrac{k^3}{p}\right\}+\left\{p^2+3k^2-3kp-\dfrac{k^3}{p}\right\}=1.\] Therefore \[\begin{eqnarray}
\sum_{k=1}^{p-1}\left\lfloor\dfrac{k^3}{p}\right\rfloor &=& \sum_{k=1}^{p-1}\dfrac{k^3}{p} - \sum_{k=1}^{p-1}\left\{\dfrac{k^3}{p}\right\} \\ \\ &=& \sum_{k=1}^{p-1}\dfrac{k^3}{p} - \dfrac{1}{2}\sum_{k=1}^{p-1} \left(\left\{\dfrac{k^3}{p}\right\}+\left\{\dfrac{\left(p-k\right)^3}{p}\right\}\right) \\ \\ &=& \sum_{k=1}^{p-1}\dfrac{k^3}{p} - \sum_{k=1}^{p-1} 1\\ \\ &=& \dfrac{p(p-1)^2}{4} -\dfrac{p-1}{2} \\ \\ &=& \boxed{\dfrac{\left(p^2-1\right)\left(p-2\right)}{4}}
\end{eqnarray}\]
What does \(\bmod~p\) have to do with it?
\[\left\{\dfrac{k^3}{p}\right\}+\left\{\dfrac{\left(p-k\right)^3}{p}\right\}=\left\{\dfrac{k^3}{p}\right\}+\left\{p^2+3k^2-3kp-\dfrac{k^3}{p}\right\}=1.\] Therefore \[\begin{eqnarray}
\sum_{k=1}^{p-1}\left\lfloor\dfrac{k^3}{p}\right\rfloor &=& \sum_{k=1}^{p-1}\dfrac{k^3}{p} - \sum_{k=1}^{p-1}\left\{\dfrac{k^3}{p}\right\} \\ \\ &=& \sum_{k=1}^{p-1}\dfrac{k^3}{p} - \dfrac{1}{2}\sum_{k=1}^{p-1} \left(\left\{\dfrac{k^3}{p}\right\}+\left\{\dfrac{\left(p-k\right)^3}{p}\right\}\right) \\ \\ &=& \sum_{k=1}^{p-1}\dfrac{k^3}{p} - \sum_{k=1}^{p-1} 1\\ \\ &=& \dfrac{p(p-1)^2}{4} -\dfrac{p-1}{2} \\ \\ &=& \boxed{\dfrac{\left(p^2-1\right)\left(p-2\right)}{4}}
\end{eqnarray}\]
What does \(\bmod~p\) have to do with it?
- What is the value of the contour integral around Western Europe?
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
Re: sum of fractions $(modp)$
If possible, please use \dfrac instead of \frac if you are writing an inline fraction with even smaller stuff in the numerator/denominator. I though the $k^3$ to be $k^2$mutasimmim wrote:Compute $\sum\left \lfloor\dfrac{k^3}{p} \right \rfloor$ for nonnegative integers $r<p$ and an odd prime $p$.
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Re: sum of fractions $(modp)$
${\dfrac{\left(p^2-1\right)\left(p-2\right)}{4}}\equiv \dfrac{p+1}{2}$ (mod $p$)Nirjhor wrote:Let \(0\le \{x\}<1\) be the fractional part of \(x\) so that \(x=\lfloor x\rfloor+\{x\}\). Now for \(1\le k\le p-1\) we have
\[\left\{\dfrac{k^3}{p}\right\}+\left\{\dfrac{\left(p-k\right)^3}{p}\right\}=\left\{\dfrac{k^3}{p}\right\}+\left\{p^2+3k^2-3kp-\dfrac{k^3}{p}\right\}=1.\] Therefore \[\begin{eqnarray}
\sum_{k=1}^{p-1}\left\lfloor\dfrac{k^3}{p}\right\rfloor &=& \sum_{k=1}^{p-1}\dfrac{k^3}{p} - \sum_{k=1}^{p-1}\left\{\dfrac{k^3}{p}\right\} \\ \\ &=& \sum_{k=1}^{p-1}\dfrac{k^3}{p} - \dfrac{1}{2}\sum_{k=1}^{p-1} \left(\left\{\dfrac{k^3}{p}\right\}+\left\{\dfrac{\left(p-k\right)^3}{p}\right\}\right) \\ \\ &=& \sum_{k=1}^{p-1}\dfrac{k^3}{p} - \sum_{k=1}^{p-1} 1\\ \\ &=& \dfrac{p(p-1)^2}{4} -\dfrac{p-1}{2} \\ \\ &=& \boxed{\dfrac{\left(p^2-1\right)\left(p-2\right)}{4}}
\end{eqnarray}\]
What does \(\bmod~p\) have to do with it?
Re: sum of fractions $(modp)$
The question asks to find the sum only, not \(\bmod~p\).
- What is the value of the contour integral around Western Europe?
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
-
- Posts:107
- Joined:Sun Dec 12, 2010 10:46 am
Re: sum of fractions $(modp)$
Sorry, I wrote $(mod p)$ by mistake, I only meant to determine the sum.