I was going over a question and need your opinion about solution of the sum of the absolute values as
$$S_n = |0-a|+|1-a|+|2-a|+ \dots + |(n-1)-a|+|n-a|$$
where a is a constant term. What could be the general sum of this series?
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
The general sum of the series, assuming $a$ is real, is easily expressed in terms of the auxiliary function $s(n,a) := \sum_{k=0}^n (n-a) = (n+1)(\frac n2 - a)$. Then $$ S_n = \begin{cases} s(n,a) - 2s(\lfloor a \rfloor, a), & 0 < a< n; \\ s(n,a), & 0 \ge a; \\ -s(n,a), & a \ge n. \end{cases}$$
answered Oct 14, 2019 at 6:27
-
$\begingroup$ Thanks for the answer. Can you clarify how to get or prove that $S_n = s(n,a) - 2s(\lfloor a \rfloor, a)$. Because in my case $0 < a < n$. $\endgroup$– Zizu006Commented Oct 14, 2019 at 7:33
-
$\begingroup$ @Zizu006 The terms $|k-a|$ where $k\le \lfloor a \rfloor$ need to be treated as $a-k$ rather than $k-a$. We already have a notation to capture the sum of all such terms. To flip a $+B$ into a $-B$ you subtract $2B$. $\endgroup$ Commented Oct 14, 2019 at 15:12