prove point series generation using exact line search method

Question

I have a function as:

min $f(x_1,x_2) = x_1^2 +2x_2^2$ where $x_0 = (2,1)^T$

Using steepest descent we compute $x_{k+1}$ as:

$x_{k+1} = x_k + \alpha*p_k$ where:

$p_k = -\nabla f(x_k)$ and $\alpha = - \frac{\nabla f(x_k)^Tp_k}{p_k^TQp_k}$

Here Q is simply the hessian given by:

$Q = (\begin{array}{cc} 2&0 \\ 0 & 4\end{array})$

Now the book states the points $x_k$ are given by the following sequence given the starting point $x_0$ as $x_k = \frac{1}{3}^k (\frac{2}{(-1)^k})$

Clearly this is correct given at k=0 $x_0 = (2,1)^T$.

However I have no idea how they derive this geometric seq

Answer 1

Thinking about this some more the answer is very simple, I'll prove for $x_1=2$

based on the starting point $\alpha$ is always $1/3$ and thus:

$x_1 = x_0 - 2\alpha x_0 \ni \alpha = 1/3$

$\implies x_1 = 1/3 *x_0$

$\implies x_2 = 1/3*x_1 \implies 1/3*1/3*x_0$

This implies the variable $(x_1)_k = 2*\frac{1}{3}^k$