Timeline for Closed form for sum of modulo remainders
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| May 28, 2022 at 5:28 | comment | added | userviswa | Basically I want to separate $N$ from $N_i$, so that all operations on $N_i$ can be pre-computed, instead of computing the mod every time I get a $N$. | |
| May 28, 2022 at 5:19 | comment | added | userviswa | Oh! i understand it now. May be my understanding of closed form is different, may be even wrong. What I searching for is to represent $K$ as a function of $n$, $N$ and some $f(N_i)$. An example would be the sum of natural number 1+2+3+ ... n = n(n+1)/2, a formula where I can avoid running through the numbers to compute the sum. Is something like this possible? | |
| May 28, 2022 at 4:28 | comment | added | Steven Clark | I don't see how either could have a closed form unless you consider a finite sum (which you already have) a closed form. | |
| May 28, 2022 at 4:28 | comment | added | Steven Clark | I understand your question, but I'll attempt to clarify my point using different variable names since reusing your variable names seems to have caused some confusion. The mod operation maps the values of $N_i$ to $0\le K_i<N$ where $K_i=N_i \% N$. I was just pointing out that finding a closed form for $J=\sum\limits_{i=1}^n K_i$ where each $K_i$ in an arbitrary set of integers $\{K_1, ... K_n\}$ is restricted to $0\le K_i<N$ doesn't seem to be any easier than finding a closed form for $M=\sum\limits_{i=1}^n N_i \% N$. | |
| May 28, 2022 at 3:59 | comment | added | userviswa | @StevenClark $M \neq \sum_{i=1}^n N_i$, but it is the sum of remainders $M = \sum_{i=1}^n N_i \% N$. Also, $N$ has no relation to $N_i$, so there is no such restriction as $0 < N_i < N$. | |
| May 27, 2022 at 14:38 | comment | added | Steven Clark | Ok, so arbitrary versus random. It seems to me finding a closed form for $M=\sum\limits_{i=1}^n N_i$ where each $N_i$ in the arbitrary set of integers $\{N_1, ... N_n\}$ is restricted to $0\le N_i<N$ isn't really any easier of a problem, is it? | |
| May 27, 2022 at 2:45 | comment | added | userviswa | @StevenClark I have modified the question to reduce confusions, wrt random. Please let me know if this clarifies your doubts. Distinct implied $N_i \neq N_j$, which is optional, just in case it helped to arrive at a solution quicker. $N$ can be any other number either inside or outside the set of {$N_i$} and $M$ is just the representation of the actual sum. | |
| May 27, 2022 at 2:34 | history | edited | userviswa | CC BY-SA 4.0 |
added 156 characters in body
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| May 26, 2022 at 12:51 | history | edited | userviswa | CC BY-SA 4.0 |
edited body
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| May 26, 2022 at 12:51 | history | edited | userviswa | CC BY-SA 4.0 |
added definition for $K$
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| May 26, 2022 at 12:46 | comment | added | userviswa | @NawafBou-Rabee K is the quotient when M is divided by N | |
| May 26, 2022 at 12:44 | comment | added | userviswa | @GerryMyerson Oops that right, it should start from $N_1$ not $N_0$. | |
| May 26, 2022 at 12:31 | comment | added | Gerry Myerson | Also: $\{N_0,\dots,N_n\}$ is $n+1$ numbers, not $n$ numbers. | |
| May 26, 2022 at 12:30 | comment | added | Gerry Myerson | What makes you think there is a closed form? | |
| S May 26, 2022 at 11:19 | review | First questions | |||
| May 26, 2022 at 11:35 | |||||
| S May 26, 2022 at 11:19 | history | asked | userviswa | CC BY-SA 4.0 |