Is there any expression for numbers running in cyclic fashion? For example, let us assume a sequence $1, 2, 3, 4, 5, 6, 7, 8$. For any variable $n = 1$ it should be $1, 2, 3, 4, 5, 6, 7, 8$; if $n =2$ it should be $8, 1, 2, 3, 4, 5, 6, 7$; for $n =3$ it should be $7, 8, 1, 2, 3, 4, 5, 6,$; if $n = 4$ it should be $6, 7, 8, 1, 2, 3, 4, 5,...$ so on....
expression for a cyclic series
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0what do you mean? – 2017-01-07
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1Hint: $1 + (k+7) \bmod 8$ gives the sequence $8,1,2,3,4,5,6,7$ for $k=0 \dots 7$. – 2017-01-07
1 Answers
Yes, there is.
What you are asking about belongs to finite groups theory, a relatively advanced area in maths.
Focusing in your example, all the rearrangements that you can do with these eight numbers (or any eight distinguishable objects) form a set called $S_8$. Don't understand the word rearrangement as the order in which the objects are, but as the act of change the order of all, some of none of them. There is a special notation and a name for the rearrangement that you have described. It is called a cycle and its notation is $(12345678)$. That means that the first element go to the second place, the second to the third, ..., and the eighth to the first. The cycle $(153)$ moves the first object to the fifth place, the fifth to the third and the third to the first (the other objects remain in their place).
If you call $\sigma$ (a common name for these rearrangements, that, by the way, are called permutations) to the cycle $(12345678)$, if you apply once, you just write $\sigma$. If you apply twice, write $\sigma^2$, ..., and if you apply $n$ times, you write $\sigma^n$.
If you apply eight times, you leave the eight objects where they were at the beginning. You can write this fact $\sigma^8=\sigma^0$ or $\sigma^8=1$, where here $1$ is not the number known as $1$, but the permutation that consists on moving nothing. Also $\sigma^8=e$ is often used.