= Solution
Choose a fixed positive <Landweber relaxation parameter> $\tau$ and rearrange the <normal equation for a linear inverse problem> as the <fixed point> equation
$$
x=(I-\tau A^*A)x+\tau A^*y.
$$
Successive substitution gives <Landweber iteration>
$$
\boxed{x_{n+1}=x_n+\tau A^*(y-Ax_n),\qquad x_0=0.}
$$
Set $B=A^*A$ and $P=I-\tau B$. Induction yields the closed polynomial-operator expression
$$
\boxed{x_n=\tau\sum_{j=0}^{n-1}P^jA^*y
=h_n(B)A^*y,\qquad
h_n(t)=\begin{cases}\dfrac{1-(1-\tau t)^n}{t},&t\ne0,\\n\tau,&t=0.\end{cases}}
$$
For $n=0$ the sum is empty, giving zero. The expression is well defined without an inverse of $A^*A$, which can have a <kernel of a linear map> and an unbounded inverse on its range. Each iterate lies in the closure of $\operatorname{Ran}A^*$, so initialization at zero eliminates the arbitrary null-space component of a solution. The unrelaxed choice $\tau=1$ gives $x_{n+1}=(I-A^*A)x_n+A^*y$ and the corresponding sum with $\tau=1$; it needs the convergence restriction in part (d).
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