MHB What are the values of α for any initial value of x(0) in a linear system?

[natalia]

Considering the following

View attachment 2409
find the values for α for any initial value of x⁽⁰⁾.

Any help will be useful, thank you!

 

[I like Serena]

Considering the following

View attachment 2409
find the values for α for any initial value of x⁽⁰⁾.

Any help will be useful, thank you!

Welcome to MHB natalia! :)

Your problem needs a little more context.

I'm going to venture out and guess that this is about numerical stability.
The question might then be which values of $\alpha$ will reduce a worst case error in $x^{(0)}$.
In that case we can say that $\alpha$ times the norm of the matrix must have an absolute value that is smaller than 1.

Which context can you provide for this problem?

 

[natalia]

I need to find the values of $\alpha$ for which the iteration converges.

 

[chisigma]

Considering the following

View attachment 2409
find the values for α for any initial value of x⁽⁰⁾.

Any help will be useful, thank you!

Wellcome on MHB natalia!...

In...

http://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/difference-equation-tutorial-draft-part-i-426-post2494.html

... it has been explained that the first order difference equation...$\displaystyle x_{n+1} = a\ x_{n} + b\ (1)$ ... has solution...

$\displaystyle x_{n} = x_{0}\ a^{n} + b\ \frac{1 - a^{n}}{1 - a}\ (2)$

In Your case I suppose that $\overrightarrow {x}_{n} $ and $\displaystyle \overrightarrow {b}$ are vectors of dimesion 2, $\alpha$ is a scalar and $\displaystyle A = \left | \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right |$ a 2 x 2 Matrix, so that is...

$\displaystyle \overrightarrow {x}_{n+1} = \overrightarrow {b} + \alpha\ A\ \overrightarrow {x}_{n}\ (3)$

... and the solution of (3) is...

$\displaystyle \overrightarrow {x}_{n} = \overrightarrow {x}_{0}\ \alpha^{n}\ A^{n} + \overrightarrow {b}\ (I- \alpha^{n}\ A^{n})\ (I- \alpha\ A)^{-1}\ (4)$

... where I is the 2 x 2 identity matrix...

Kind regards

$\chi$ $\sigma$

 

[I like Serena]

I need to find the values of $\alpha$ for which the iteration converges.

Aha!

Actually, that turns out the same: you need $\alpha$ such that
$$||\alpha A || < 1$$
where $|| \cdot ||$ represents the matrix norm.
This is the same as:
$$|\alpha| \cdot || A || < 1$$
$$|\alpha| < \frac 1 {|| A ||}$$To explain, suppose the sequence converges (or should converge) to $y$.

Let $\Delta y^{(k)} = x^{(k)} - y$, which represents how far you are from the limit.
And let's call your matrix $A$ for ease of notation.

Then:
\begin{aligned}
x^{(k+1)} &= b + \alpha A x^{(k)} \\

&= b + \alpha A (y + \Delta y^{(k)}) \\

&= (b + \alpha A y) + \alpha A \Delta y^{(k)} \\

&= y + \alpha A \Delta y^{(k)}
\end{aligned}

That means that:
$$\Delta y^{(k+1)} = \alpha A \Delta y^{(k)}$$

In a worst case scenario this will only converge if
$$||\alpha A || < 1$$
where $|| \cdot ||$ represents the matrix norm.

 

[natalia]

Thank you very much, I like Serena :), your explanations are really clear.