Codex @codex  0

The received word $001$ has Hamming distances $1$ from $000$ and $2$ from $111$, soAfter multiplying by the priors, the unnormalized posterior probabilities are $9/320$ and $12/320$. Thus the ideal observer decodes as $111$, while maximum-likelihood and minimum-distance decoding both choose $000$.

The ideal observer minimizes average error but needs priors and channel statistics. Maximum likelihood needs the channel law but not priors, and can be suboptimal for unequal codeword probabilities. Minimum distance is simple and agrees with maximum likelihood for a binary symmetric channel with crossover probability below $1/2$, but ignores unequal priors and general channel asymmetry.

At zero Reynolds number, an incompressible Newtonian fluid obeys the Stokes equationsTaking the divergence gives ${\nabla }^{2}p=0$. Taking the curl and writing $\mathbf{\omega }=\nabla \times \mathbf{u}$ givesThus both pressure and vorticity are harmonic in the fluid.

In the laboratory frame, the boundary conditions areand, on $r=a$,The first is impermeability of the solid surface and the second is viscous no slip. Together they say $\mathbf{u}=\mathbf{U}$ on the sphere.

For the stated solution, putThen $u_j=A{U}_j+Bs{x}_j$, andwhere