A (probably easy) question about exchanging the order of this double sum

Question

I would like to exchange the order of the following:

$\sum_{k=1}^{i-1}\sum_{r=1}^{2k}$ (stuff).

I feel like it should be easy, but so far I am only able to produce

$\sum_{r=1}^{2(i-1)}\sum_{k=??}^{??}$ (stuff).

In particular, I don't see how to reconcile the inequalities $1 \leq r\leq 2k\,$ and $1 \leq k \leq i-1$\, without introducing half integers into the sum.

Thanks,

score 2 · Accepted Answer · answered Jan 08 '19 at 06:12

2

The second sum should be $$\sum_{k=\lceil \frac r2\rceil}^{i-1}$$ $k$ is always at least $\frac r2$ and ranges up to $i-1$

answered Jan 08 '19 at 06:12

Ross Millikan

374,822

I think $k$ starts from $[\frac r 2]$ if $r$ is even and $[\frac r 2]+1$ if $r$ is odd. – Kavi Rama Murthy Jan 08 '19 at 06:17
1

@Kavi That's exactly what $\lceil{r\over2}\rceil$ means. – Ivan Neretin Jan 08 '19 at 06:23
I see. I didn't know this. Thanks. – Kavi Rama Murthy Jan 08 '19 at 06:29
Your answer is clear, thanks. – MathIsArt Jan 08 '19 at 07:06
@KaviRamaMurthy: those brackets represent the ceiling function-you round up to the next integer. Upside down ones are the floor, where you round down. – Ross Millikan Jan 08 '19 at 15:02
Sorry about asking a new question here, but it's so related that I can't justify making a seperate topic:
Did I compute this exchange correctly?

$$\sum_{k=1}^{i-1}\sum_{n=1}^{i-1-k}\sum_{s=1}^{2n} a_{kns} = \sum_{s=1}^{2(i-2)}\sum_{k=1}^{i-1-\lceil s/2 \rceil}\sum_{n=\lceil s/2 \rceil}^{i-1-k} a_{kns}$$
– MathIsArt Jan 08 '19 at 19:07
It looks good, but it can be confusing – Ross Millikan Jan 08 '19 at 19:39
Thank you for checking! Out of curiosity, what would you do to make it less confusing? – MathIsArt Jan 08 '19 at 19:51

A (probably easy) question about exchanging the order of this double sum

1 Answers1