5-5 + 0.13 u
hero to zero in no time
can you kuroda?
using my new bases totals formula.....not sure how it will go, but it looks something like this:
The discriminant of the quadratic equation is
D = tan2 (t) - 4 tan2 (t) = -3 tan2 (t)
The roots x1 and x2 are
- tan(t) + i sqrt(3) tan (t)
x1 = -------------------------------
2 tan (t) tan (t)
-1 + sqrt(3)
= ------------- cot(t)
2
= (cos(2 pi/3) + i sin(2 pi/3)) cot(t)
- tan(t) - i sqrt(3) tan (t)
x2 = -------------------------------
2 tan (t) tan (t)
-1 - sqrt(3)
= ------------- cot(t)
2
= (cos(2 pi/3) - i sin(2 pi/3)) cot(t)
So,
x1n = (cos(2 pi/3) + i sin(2 pi/3))n cotn(t)
= (cos(2 n pi/3) + i sin(2 n pi/3)) cotn(t)
x2n = (cos(2 pi/3) - i sin(2 pi/3))n cotn(t)
= (cos(2 n pi/3) - i sin(2 n pi/3)) cotn(t)
and
(x1)n + (x2)n = 2 cos(2 n pi/3) cotn(t)
anyway
bos -1.5 + 123 x 4
mil/cin o 8.5 -115 x 2
atl/was u 7.5 + 102 x 2
az/col u 9.5 -104 x 2
gl
hero to zero in no time
can you kuroda?
using my new bases totals formula.....not sure how it will go, but it looks something like this:
The discriminant of the quadratic equation is
D = tan2 (t) - 4 tan2 (t) = -3 tan2 (t)
The roots x1 and x2 are
- tan(t) + i sqrt(3) tan (t)
x1 = -------------------------------
2 tan (t) tan (t)
-1 + sqrt(3)
= ------------- cot(t)
2
= (cos(2 pi/3) + i sin(2 pi/3)) cot(t)
- tan(t) - i sqrt(3) tan (t)
x2 = -------------------------------
2 tan (t) tan (t)
-1 - sqrt(3)
= ------------- cot(t)
2
= (cos(2 pi/3) - i sin(2 pi/3)) cot(t)
So,
x1n = (cos(2 pi/3) + i sin(2 pi/3))n cotn(t)
= (cos(2 n pi/3) + i sin(2 n pi/3)) cotn(t)
x2n = (cos(2 pi/3) - i sin(2 pi/3))n cotn(t)
= (cos(2 n pi/3) - i sin(2 n pi/3)) cotn(t)
and
(x1)n + (x2)n = 2 cos(2 n pi/3) cotn(t)
anyway
bos -1.5 + 123 x 4
mil/cin o 8.5 -115 x 2
atl/was u 7.5 + 102 x 2
az/col u 9.5 -104 x 2
gl