Beginning a Veblen heirarchy with $1 + x$? If I define:
$$
φ'_0(x)=1+x
$$
Enumerating the fixed points of $φ'_0(x)$, one would get:
$$
φ_1'(0)=ω φ_1'(1)=ω+1 \cdots φ_1'(ω)=ω+ω \cdots
$$
Which is identical to:
$$
φ_1'(x)=ω+x
$$
The next several levels seem to give us:
$$
φ'_2(x)=ω⋅x φ'_3(x)=\mathbf{φ_0(x)=ω^x} φ_4'(x)=ε_x \cdots
$$
So basically, $φ'_3(x) = φ_0(x)$. 
Is this version of the Veblen hierarchy correct? If so, why does the Veblen hierarchy start with $ω^x$? It seems that starting it with $1+x$ is a lot more meaningful.
 A: This is essentially a longer comment. Main reason for posting it here is that posting anything even slightly complicated becomes just too confusing in comments (usually a minor error just carries along without anyway to edit it). 
There is typo-issue with what you have written ..... just taking it as it is. Suppose we set  $\varphi_{0}(x)=1+x$ as you stated. Then we get $\varphi_{1}(x)=\omega+x$.
However $\varphi_{2}(x)$ won't be equal to $\omega \cdot x$. We will have $\varphi_{2}(x)=\omega^2+x$. Continuing this we should get $\varphi_{3}(x)=\omega^3+x$. And for any $n<\omega$, $\varphi_{n}(x)=\omega^n+x$.
Similarly for $\varphi_{\omega}(x)$ we get $\varphi_{\omega}(x)=\omega^\omega+x$. This simply enumerates the elements that are common to every $\varphi_{n}(x)$ (for $n<\omega$), in the image-set of these functions. Continuing this, it seems we would get $\varphi_{\omega+1}(x)=\omega^{\omega+1}+x$.
The previous values are highly suggestive for the function $x \mapsto \varphi_{x}(0)$ being equal to $\omega^x$. And the fixed point of this will be $\epsilon_0$ as we already know (roughly speaking, this is analogous to $\Gamma_0$ in the original hierarchy). 
