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The dot product isBoth vectors have squared Euclidean norm $a^2+(a+1{)}^2=2a^2+2a+1$. Under the given condition $a^2+a=1/2$, their dot product is $1$ and each squared Euclidean norm is $2$. Hence

The chain rule givesFor this to vanish for every differentiable $f$, its coefficient must be zero:The resulting $f(x-t)$ is a profile transported to the right at unit speed by the transport equation.

Substitute $r=-x^2$ into the geometric series, without taking any derivatives:Thus the coefficient of $x^n$ in the Taylor series is $(-1{)}^{n/2}$ when $n$ is even and zero when $n$ is odd. If nonzero terms are numbered from one, the $n$th term is $(-1{)}^{n-1}{x}^{2n-2}$.

Normalization of the probability mass function and the geometric series giveThe tail is another geometric series:

Expanding the cross product givesTaking its divergence component by component,

Complete the square in the velocity variable:Since $m>0$, the final term is nonpositive and vanishes at exactly one velocity. ThereforeThis is the Legendre transform in mechanics that constructs the Hamiltonian mechanics energy from the kinetic-minus-potential Lagrangian.

For $n=0$, the change of variables $s=\sqrt{a}x$ reduces the integral to the given Gaussian integral:For $n=1$, the integrand is an integrable odd function, so its contributions from the two half-lines cancel:The condition $a>0$ ensures convergence in both cases.

For this first-order linear differential equation, the integrating factor is $e^x$. Thusand the general solution is

The boundary order fixes the orientation. Two edges issuing from $A$ are $\overrightarrow{AB}=(0,0,1)$ and $\overrightarrow{AD}=(1,1,0)$. Their cross product gives the oriented vector area:Its norm is $\sqrt{2}$, the ordinary rectangle area. The sign agrees with traversing $A\to B\to C\to D\to A$ by the right-hand orientation rule.