User blog comment:Ikosarakt1/Another research in comparisons between FGH and SGH/@comment-5529393-20131030210834

Proving $$g_{\theta(\alpha,m)}(n) \approx f_\alpha (n(m+1))$$ doesn't prove that $$g_{\theta(\alpha+1, 0)}(n) \approxf_{\alpha+1}(n)$$, because the former applies only to finite m and you are trying to apply it to infinite ordinals.

The analogy breaks down at \omega + 1; we have $$g_{\varphi(\alpha,0)}(n) \approx f_\alpha(n)$$ for $$\alpha \le \omega$$, but $$f_{\omega+1}(n) \approx g_{\Gamma_0}(n)$$. So you really need to focus on the area between $$f_\omega(n)$$ and $$f_{\omega+1}(n)$$.