= Solution
Let $T(x)$ be the weighted average on the right, and define $F$ and $\Phi=\log F$ as in part (i). Their explicit <gradient> identity is
$$
\nabla\Phi(x)=2(x-T(x)).
$$
Therefore the displayed equation is exactly the <stationary point> equation for the strongly convex $\Phi$. It has one solution, which is the unique minimizer from part (i). \b[Reuse the BFGS iteration], with
$$
x_{n+1}=x_n-2\alpha_nH_n(x_n-T(x_n)).
$$
The <gradient> is explicit, the weights can again be scaled by their largest exponent, and the usual full-memory <BFGS> assumptions give \b[local <superlinear convergence>].
Simply setting $x_{n+1}=T(x_n)$ is not automatically a convergent <fixed-point iteration>. Differentiating the normalized weights gives
$$
J_T(x)=-2\sum_iw_i(a_i-T(x))(a_i-T(x))^T,
$$
whose <operator norm> need not be below $1$. For example, in one dimension with centres $-1,1$, $T(x)=-\tanh(2x)$ and $T'(0)=-2$: the desired fixed point is locally repelling for that naive iteration. If exact second <derivatives> are affordable, <multivariable Newton iteration> on $x-T(x)=0$ is another valid choice, with <Jacobian matrix> $I-J_T$ as given above and local <quadratic convergence>.
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