Matematické Fórum

stuart clark · 14. 04. 2019 10:23

If $\begin{pmatrix} 2 & 1\\ 1 & 0 \end{pmatrix}^n = (a_{ij}(n)).$ Then $\lim_{n\rightarrow \infty}\frac{a_{12}(n)}{a_{22}(n)}$ is

krakonoš

↑ stuart clark:
Hi Stuart.
a12 in step n is the same as a11 in step n-1 & in step n-1 is a11=a12+a21+a22.
If $L=\lim_{n\to\infty }\frac{a_{12}}{a_{22}}$
then you will get $L=1+1+1/L$
$L=1+\sqrt{2}$

laszky · 14. 04. 2019 12:56 — Editoval laszky (14. 04. 2019 13:11)

↑ stuart clark:

Hi, if it is a guessing, then I guess that $\lim_{n\rightarrow \infty}\frac{a_{12}(n)}{a_{22}(n)}=1+\sqrt{2}$ :-)

Autor skryl část textu. Zobrazit/skrýt!

krakonoš

↑ laszky:
Ahoj laszky.
Ja jsem si zkusila spocitat postupne mocniny az do radu 6,a vychazi mi,ze a12 v ntem kroku je souctem a21 plusa12 plus a22 v n-1 tem kroku a tento soucet je zaroven roven a11 v n-1 vem kroku .
A ty 2/5 jako limita (2n plus konst)/(5n plus konst)..
Uz na par tech prikladech ten pomer je docela stabilni uz po par krocich,a vychazi mi opravdu kolem2.4.Ten rozdil bude zrejme v tom ,ze konst ale taky je jeste zavisla na n (roste s n),takze tento vypocet nebude tak presny.
Tam bude lepsi uvazovat,ze pri dosazeni stability nastane rovnost L rovno 1 plus 1 plus 1/L.Takze ta limita bude presne opravdu jak pises ty.

stuart clark · 17. 04. 2019 11:08

Thanks ↑ krakonoš:.

stuart clark · 17. 04. 2019 11:09

↑ laszky:

I did not understand your solution. please explain me. Thanks

laszky · 17. 04. 2019 12:44 — Editoval laszky (17. 04. 2019 12:45)

↑ stuart clark:

If you decompose the matrix into the form $A=P\Lambda P^{-1}$ , then $A^n=P\Lambda^n P^{-1}$ .
If the matrix $\Lambda$ is diagonal, then it is easy to compute its power.
The decomposition with diagonal matrix $\Lambda$ can be (for diagonalizable matrices) obtained via Jordan decomposition

In our case

$\begin{pmatrix}2 & 1 \\ 1 & 0 \end{pmatrix}^n = P\Lambda^n P^{-1} = \begin{pmatrix}1 & 1 \\ -1-\sqrt{2} & -1+\sqrt{2} \end{pmatrix}\begin{pmatrix} 1-\sqrt{2} & 0 \\ 0 & 1+\sqrt{2} \end{pmatrix}^n\begin{pmatrix}\frac{\sqrt{2}-1}{2\sqrt{2}} & \frac{-1}{2\sqrt{2}} \\ \frac{\sqrt{2}+1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} \end{pmatrix} =$
$=\begin{pmatrix}1 & 1 \\ -1-\sqrt{2} & -1+\sqrt{2} \end{pmatrix}\begin{pmatrix} (1-\sqrt{2})^n & 0 \\ 0 & (1+\sqrt{2})^n \end{pmatrix}\begin{pmatrix}\frac{\sqrt{2}-1}{2\sqrt{2}} & \frac{-1}{2\sqrt{2}} \\ \frac{\sqrt{2}+1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} \end{pmatrix} =$
$= \frac{1}{2\sqrt{2}}\begin{pmatrix} (1-\sqrt{2})^n(\sqrt{2}-1)+(1+\sqrt{2})^n(\sqrt{2}+1) & (1+\sqrt{2})^n-(1-\sqrt{2})^n \\ (1+\sqrt{2})^n-(1-\sqrt{2})^n & (1-\sqrt{2})^n(1+\sqrt{2})+(1+\sqrt{2})^n(\sqrt{2}-1) \end{pmatrix}$

since $|1-\sqrt{2}|<|1+\sqrt{2}|$ we have $\lim_{n\rightarrow \infty}\frac{a_{12}(n)}{a_{22}(n)}= \frac{1}{\sqrt{2}-1} = 1+\sqrt{2}$