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[VA] SRC#003- New Year 2019 Special
02-09-2019, 01:55 AM (This post was last modified: 02-09-2019 02:41 AM by Albert Chan.)
Post: #27
RE: [VA] SRC#003- New Year 2019 Special
The idea of only doing only *last* row work for 3x3 matrix too. Smile

Let M = {{k, n, n}, {1, k, n}, {1, 1, k}}

-> M^p = {{c, n*a, n*b},
                 {b,     c, n*a},
                 {a,     b,     c}}, for some a, b, c

-> For M^(p+1), last row = {k*a+b+c, n*a+k*b+c, n*a+n*b+k*c}

Example: this is result of M^200 last row ratios:
lua> k, n = math.pi, 2019
lua> a, b, c = 1, 1, k
lua> for i=2,200 do
:       a, b, c = (k*a+b+c)/n, a+(k*b+c)/n, a+b+k*c/n
:    end
lua> =a, b, c
2.916424658351884e-212  3.686063969271537e-211  4.658809733574611e-210
lua> =b/a, c/b
12.638982319380704      12.638982319385288
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RE: [VA] SRC#003- New Year 2019 Special - Albert Chan - 02-09-2019 01:55 AM

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