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A triangular Bézier patch is a non-tensor-product example. On barycentric parameters $u,v,w\ge 0$ with $u+v+w=1$, defineThe indices satisfy a total-degree constraint, and $w=1-u-v$ couples the two independent parameters. The control net is triangular with $(n+1)(n+2)/2$ points, rather than a rectangular product index set. The multinomial theorem gives nonnegative basis functions summing to one, but the entire basis is not a product of two independent univariate bases.

A Poisson distribution with parameter $\lambda >0$ assigns probability mass functionThe exponential series makes these probabilities sum to one. Shifting the indices in the absolutely convergent sums givesHence $E{Z}^2={\lambda }^2+\lambda$ andFor independent random variables $Z_1,\dots ,Z_n$ of parameter one, let $S_n={\sum }_j{Z}_j$. Summing their product probability mass function over all nonnegative $k_1+\cdots +k_n=k$ givesThe final equality follows from the multinomial theorem applied to $(1+\cdots +1{)}^k$. Thus $S_n$ has the Poisson distribution of parameter $n$.

Each $Z_j$ has expectation and variance one, so the central limit theorem gives convergence in distributionZero is a continuity point of the standard normal cumulative distribution function. ThereforeThe discrete atom at $S_n=n$ causes no difficulty: the central limit theorem applies to the entire event with threshold zero.

By Fermat's little theorem,Thus the sum of $g$ over ${\mathbb{F}}_p^3$ counts, modulo $p$, the triples on which the ternary quadratic form $f$ vanishes.

Expand $f^{p-1}$ with the multinomial theorem. A term indexed by $r+s+t=p-1$ contributes a constant multiple ofAt least one of $r,s,t$ is strictly less than $(p-1)/2$. Its corresponding exponent is therefore in $\{0,1,\dots ,p-2\}$, so part (b), including its $k=0$ case, makes that power sum zero modulo $p$. Every term in the expansion consequently vanishes after summation. The constant term contributes $p^3\equiv 0$, and therefore

The zero triple is one solution of $f=0$. If it were the only solution, the displayed sum would be congruent to $1$ rather than $0$. Hence there is another triple $(x,y,z)$, not all divisible by $p$, satisfying

Write $S=x_1+\cdots +x_n$. The Taylor series of the cosine isAfter expanding each power by the multinomial theorem, the coefficient of $x^{\alpha }$ has absolute value $1/\alpha !$ when $|\alpha |$ is even and is zero when $|\alpha |$ is odd. The hyperbolic cosinetherefore has exactly the absolute values of the coefficients of $f$. Thus $g$ is an entire majorant series for $f$.

For nonnegative barycentric parameters, these total-degree Bernstein basis functions define a surface over a triangle. The triangular control net has $(n+1)(n+2)/2$ points. Its index constraint couples the three exponents, so it is not a tensor product of two independent univariate bases. The multinomial theorem gives nonnegativity and partition of unity.