/sci/ - Science & Math » Thread #11687543

98KiB, 1015x764, solve.png

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Anonymous Tue 19 May 02:13:45 2020 No.11687543 View Reply Original Report

Quoted By: >>11687565 >>11687572 >>11687842

What would I need to solve something like this?
I have strong reason to believe, in the case A=1/2 B=1/4
lim f(n) n-> infinity
X = Y = Z = 1/3

But I didn't derive it properly, I just did a few iterations than plugged it into oeis and found the equation (A = limit n->infinity f(n) = (4^n + 2)/3(4^n) = 1/3. I don't know what tools I'm missing, I attempted to use matrices but I cannot properly understand how to apply them.

Apologies if any mistakes.

Anonymous

Anonymous Tue 19 May 2020 02:25:55 No.11687565 Report

Quoted By: >>11687581

>>11687543
I have made a mistake X!=Y!=Z but for my purposes I really just care about X and X=1/3

Anonymous

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Anonymous Tue 19 May 2020 02:28:38 No.11687572 Report

Quoted By:

>>11687543
Everything equals zero

Anonymous

Anonymous Tue 19 May 2020 02:30:53 No.11687581 Report

Quoted By:

>>11687565
The C in the in(x1,x2,x3) should also b B

Anonymous

Anonymous Tue 19 May 2020 04:43:52 No.11687842 Report

Quoted By: >>11687996

>>11687543
$ \text{The relevant matrix is} \\ M= \begin{bmatrix} A & B & B \\ B & A & B\\ B & B & A \\ \end{bmatrix} \\ \text{the functions i, j, k are equal to the elements of the resulting vector of this matrix multiplication}\\ M\vec{x}= \begin{bmatrix} A & B & B \\ B & A & B\\ B & B & A \\ \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ \end{bmatrix} \\ i^n(x_1, x_2, x_3) \text{ is then equal to the first element of} \\ M^n\vec{x}= \begin{bmatrix} A & B & B \\ B & A & B\\ B & B & A \\ \end{bmatrix}^n \begin{bmatrix} x_1\\ x_2\\ x_3\\ \end{bmatrix}\\ \text{in other words the coefficients}\ X, Y, Z \text{ are equal to the first row of} \\ \begin{bmatrix} A & B & B \\ B & A & B\\ B & B & A \\ \end{bmatrix}^n \\ \text{ask wolfram alpha to calculate it for you} $
https://www.wolframalpha.com/input/?i=%5B%5BA%2C+B%2C+B%5D%2C+%5BB%2C+A%2C+B%5D%2C+%5BB%2C+B%2C+A%5D%5D%5En
$ \text{the coefficients are then}\\ X=\frac{1}{3}(2(A-B)^n+(A+2B)^n)\\ Y=\frac{1}{3}((A+2B)^n-(A-B)^n)\\ Z=\frac{1}{3}((A+2B)^n-(A-B)^n)\\ $

Anonymous

Anonymous Tue 19 May 2020 06:02:16 No.11687996 Report

Quoted By:

>>11687842
Awesome, thank you! I was stuck on doing the vector matrix multiplication, and couldn't think of any operation that would help because I was hung up on whether or not the ordering mattered, but I think I can see why doing the matrix multiplication first works.

Thanks again, I've spent hours in the last few days on this!

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