How to show that the method of steepest descent does not converge in a finite number of steps?

Question

I have a function, $$f(\mathbf{x})=x_1^2+4x_2^2-4x_1-8x_2,$$

which can also be expressed as $$f(\mathbf{x})=(x_1-2)^2+4(x_2-1)^2-8.$$ I've deduced the minimizer $\mathbf{x^*}$ as $(2,1)$ with $f^*=-8$ by finding the gradient $$\nabla f(\mathbf{x})=(2x_1-4,8x_2-8).$$

I understand that method of steepest descent is defining $$z(t)=\mathbf{x}-t\nabla f(\mathbf{x})$$ $$g(t)=f(z(t))$$ and then setting $g'(t)=0$ and solving for $t$ in terms of $x_1$ and $x_2$. I am asked to start from $\mathbf{x_0}=(0,0)^T$. I can iterate multiple times and I am assuming $f$ will not reach $f^*$. However I am lost in how to prove that this method diverges with a succinct proof.

Answer 1

I tried your example and found the following convergence:

0:  x1=0        x2=0        t=0        f(x)=0
1:  x1=0.588255 x2=1.17651  t=0.147064 f(x)=-5.88235
2:  x1=1.47049  x2=0.735289 t=0.312461 f(x)=-7.43933
3:  x1=1.62623  x2=1.04672  t=0.147064 f(x)=-7.85156
4:  x1=1.85982  x2=0.929922 t=0.312477 f(x)=-7.96071
5:  x1=1.90105  x2=1.01237  t=0.147064 f(x)=-7.9896
6:  x1=1.96289  x2=0.981448 t=0.312477 f(x)=-7.99725
7:  x1=1.9738   x2=1.00327  t=0.147064 f(x)=-7.99927
8:  x1=1.99018  x2=0.995088 t=0.312502 f(x)=-7.99981
9:  x1=1.99307  x2=1.00087  t=0.147064 f(x)=-7.99995
10: x1=1.9974   x2=0.9987   t=0.312436 f(x)=-7.99999
11: x1=1.99816  x2=1.00023  t=0.147079 f(x)=-8

I can't see any reason why the stepest-descent method should not converge in your case. The starting point is near enough to the global minimum.