reinout-g

Yeah that does help!

How foolish of me not to think of it in that way. 

However I do find it suprising to see the $$\begin{pmatrix}
0\\
0\\
\end{pmatrix}$$ vector as an eigenvector since that eigenvector is basically applicable to any matrix. Nevertheless I wouldn't know of any other way to solve this.

Thanks Alan.

Hey Rosala,

I'm not, the question is to give a 'proof' that zn = 0 is an asymptotically stable steady state of the system for all n given that abs[r] < 1

Even though it seems obvious that $$z_{n+1} = Az_n \Rightarrow 0 = A 0$$. I don't know the conditions which must be checked before I can formally prove it to be an asymptotically stable steady state of the system

Hey Alan,

you're right, but they are not initial values.

They are constants which can be calculated for given initial values for x and y

Thank you Alan, that was really helpfull :)

Now that I come to think of it,

for the eigenvalues it is always true that

$$\lambda_1 + \lambda_2 + \lambda_3 = trace (A)$$

and

$$\lambda_1*\lambda_2*\lambda_3 = det(A)$$

In this case

$$det(A) = 12\\
trace(A) = 8\\$$

edit: Aaah I found it 

Hey Feitan,

Nice answer!

I haven't seen you before on the forum are you new here?

I was just wondering about the part of the signifant numbers...

Isn't it true that since 2g has only 1 significant number the final answer should also have only 1 significant number?

Reinout 

$$3x(2x^3-5x^2+3x+1)$$

$$3x \times 2x^3 - 3x \times 5x^2 + 3x \times 3x + 3x \times 1$$

$$3 \times 2 \times x \times x^3 - 3 \times 5 \times x \times x^2 + 3 \times 3 \times x \times x + 3 \times 1 \times x$$

$$3 \times 2 \times x^1 \times x^3 - 3 \times 5 \times x^1 \times x^2 + 3 \times 3 \times x^1 \times x^1 + 3 \times 1 \times x^1$$

$$6 \times x^{3+1} - 15 \times x^{1+2} + 9 \times x^{1+1} + 3 \times x^1$$

$$6x^4-15x^3+9x^2+3x$$

Reinout 

thank you Alan 

I know I'm nitpicking but if you belong to a nation, isn't the probability that you were born in it 100% (if we do not take emigration into account)

HEY!

THAT'S SUPPOSED TO BE A SECRET!!!