Do "Type-2" Turing machines with infinite length inputs have more computational power?
Type-2 Turing machines are not more powerful than ordinary Turing machines in the sense that any map $\mathbb{N} \to \mathbb{N}$ that can be computed by a type-2 machine can also be computed by an ordinary machine. To see this, suppose a type-2 Turing machine $T$ computes a function $f : \mathbb{N} \to \mathbb{N}$. We can convert $T$ to an ordinary machine by modifying it so that it stops as soon as it writes $f(n)$ on the output (and we can tell when this happens because the output is a single finite number, so it cannot take forever for the machine to produce it).

It is far more interesting to go up one level and compare computability of functionals, which are maps
$$F : \mathbb{N}^\mathbb{N} \to \mathbb{N}$$
that take an infinite sequence of numbers as an input and output a number. Now there is a difference between type 1 and type 2.

Say that a functional $F$ is type-2 computable if there exists a type-2 Turing machine $T$ such that, for any sequence $\alpha \in \mathbb{N}^\mathbb{N}$, when $\alpha$ is written on the input tape and machine $T$ is run, it writes $F(\alpha)$ on the output tape (and terminates).

Say that a functional $F$ is type-1 computable if there exists a type-1 Turing machine $T$ (i.e., an ordinary one), such that, for any computable sequence $\alpha \in \mathbb{N}^\mathbb{N}$, if we write onto the input tape the encoding $n$ of a machine that computes $\alpha$, then $T(n)$ writes $F(\alpha)$ on the output tape (and terminates).

Notice that a type-1 computable functional only accepts computable inputs, whereas a type-2 computable functional accepts all inputs. If we write $\mathcal{R}$ for the set of all computable sequences of numbers (i.e., the total computable functions), then a type-1 computable functional is a map $F : \mathcal{R} \to \mathbb{N}$ computed by some type-1 machine, whereas a type-2 computable functional is a map $F : \mathbb{N}^\mathbb{N} \to \mathbb{N}$ computed by some type-2 machine.

Because $\mathcal{R} \subseteq \mathbb{N}^\mathbb{N}$, we can convert any type-2 functional to a type-1 functional by simply feeding it only computable inputs. We have:


  Theorem: When a type-2 computable functional is restricted to computable inputs, it becomes a type-1 computable functional.


Proof. The theorem is not a complete triviality. The input is presented to a type-2 functional as a sequence, whereas to a type-1 functional it is presented as a description of a Turing machine computing the sequence. So, suppose $F$ is a type-2 functional computed by a type-2 machine~$T$. We convert $T$ to a type-1 machine as follows: when given an input $n$ which described a machine $M$ that computes some $\alpha \in \mathcal{R}$, we simulate an input type for $T$ so that whenever $T$ wants the $k$-th cell of the input, we run $M(k)$ to compute $\alpha(k)$ and give $T$ the answer. $\Box$

Can we go in the other direction? Suppose $F : \mathcal{R} \to \mathbb{N}$ is computed by a type-1 machine $T$. Perhaps we can extend $F$ to all of $\mathbb{N}^\mathbb{N}$ to get a computable $\overline{F} : \mathbb{N}^\mathbb{N} \to \mathbb{N}$ which agrees with $F$ on $\mathcal{R}$? The anwser is negative! (See Exercise 15-40 in H. Rogers Jr. "Theory of Recursive Functions and Effective Computability".)

The moral of the story is this (as I have often repeated on these forums): all known notions of computability of functions from numbers to numbers agree, but notions of computability of functionals do not, and they are incomparable in general.