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If i define $f(m,n)=$ $\sum_{1\leq k\leq mn}\left\{ \frac{k}{m}\right\} \left\{ \frac{k}{n}\right\} .$

Then prove $f(m+n,n) - f(m,n) =\frac{n^2-n}{4}$ for all $m$ and $n$.

This question came from part of answer from this question: A sum of fractional parts.

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Since $k$ goes from $1$ to $mn$, then the pairs $((k \bmod m),(k \bmod n))$ will meet all cases once.

thus the product is

$\sum_{1\le k\le m}\left\{\frac{k}{m}\right\}\sum_{1\le t\le n}\left\{\frac{t}{n}\right\}=\frac{m-1}{2}\frac{n-1}{2}\;.$

Done.

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    $\{a\}=a-[a]$, where [a] is the largest integer less than or equal to a.2012-05-04