So for example the four roots of 1 can be represented as i, – 1, – i and 1, corresponding to 1

^{1/4}, 1

^{2/4}, 1

^{3/4 }and

^{ }1

^{4/4 }respectively.

I therefore wondered what would be the results when adding just the dimensional exponents of 1 in each case representing the degree of rotation with respect to the circular circumference with the simple restriction that values from 90 to 270 degrees be treated as negative. Alternatively we can state this as taking positive values for the fractional values in the 1st and 4th quadrants and negative in the 2nd and 3rd!

Now an interesting pattern applies.

If n (representing the fractional values associated with the n roots of 1) is divisible by 4 then the sum of corresponding values = 0.

So for example with respect to the fractional values associated with the four roots the 1st and 4th fall in the 1st and 4th quadrants respectively and are thereby positive, whereas the 2nd and 3rd fall in the 2nd and 3rd quadrants and thereby are treated as negative.

Therefore the corresponding sum of fractions = 1/4 – 2/4 – 3/4 + 4/4 = 0.

For the sum of even numbered fractions (where n is not divisible by 4) the corresponding sum = 1/2

So for example in the simplest case where n = 2, the 1st value will lie in the 2nd quadrant (as negative and the second in the 4th as positive).

Thus the sum of fractions = – 1/2 + 1 = 1/2.

For the sum of an odd number of fractions in n is one less than a number divisible by 4 then the total = 0.

For example 3 is one less than such a number.

With respect to the corresponding fractions 1/3, 2/3 and 3/3, the first two will be negative (lying in 2nd and 3rd quadrants respectively and the last positive (in the 4th quadrant) so the requisite sum =

**– 1/3 – 2/3 + 3/3 = 0.**

For the sum of an odd number of fractions where n is one more than a number divisible by 4, then the total = 1.

So for example when n = 5, the corresponding fractions are 1/5, 2/5, 3/5, 4/5 and 5/5. The 2nd, and 3rd of these values are located in the 2nd and 3rd quadrants respectively and are thereby negative.

Thus the requisite sum = 1/5 – 2/5 – 3/5 + 4/5 + 5/5 = 1.

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