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PutThe log-likelihood, apart from an additive constant, isOver the full parameter space its maximum-likelihood estimates are $(\hat{\mu },{\hat{\sigma }}^2)=(\overline{X},S/n)$. Under $H_0$, maximizing over the nuisance parameter $\mu$ gives $({\hat{\mu }}_0,{\sigma }_0^2)=(\overline{X},1)$. Thus the generalized likelihood-ratio test statistic, in the convention $2({\sup }_{\Theta }{\ell }_n-{\sup }_{{\Theta }_0}{\ell }_n)$, is

Under $H_0$, the normal-sample decomposition givesWriting $Z_n=(S-(n-1))/\sqrt{2(n-1)}$, the central limit theorem for a sum of squared independent standard normal variables yields $Z_n\stackrel{d}{\to }N(0,1)$. Hence Slutsky theorem givesand in particular ${\hat{\sigma }}^2\stackrel{P}{\to }1$.

For $h(x)=x-1-\log x$, Taylor theorem at $x=1$ givesConsequentlyby the continuous mapping theorem and Slutsky's theorem. Since the square of a standard normal variable has the chi-squared distribution with one degree of freedom,

Writing $f,f^{\prime },f^{\prime \prime }$ at $y_n$, the implicit stage has the expansionThus one step isThe exact Taylor expansion has third-order termThe coefficients agree through order two for every real $a$, while the $f^{\prime \prime }{f}^2$ coefficient prevents order three for any $a$. Hence the order of a Runge-Kutta method is(The linear special case has one extra matched term when $a=1/3$, but the order for general nonlinear equations remains two.)

The integral form of the Taylor theorem about the left endpoint iswhere $q_{k-1}\in {\mathcal{P}}_{k-1}$. Apply the order-$k$ divided difference at $x_i,\dots ,x_{i+k}$. The polynomial term vanishes, and linearity permits interchange with the integral:By the definition $M_i(t)=k[x_i,\dots ,x_{i+k}](\cdot -t{)}_{+}^{k-1}$,This is the Peano kernel theorem for the divided-difference functional.

Write the pair energy asAt $A_H=0$ the interaction is repulsive, so the Mayer integrand and hence $B_2$ are positive. Increasing $A_H$ strengthens attraction andFor sufficiently strong attraction, negative configurations dominate and $B_2<0$. Continuity therefore gives a critical $A_H^{\ast }$ with $B_2(A_H^{\ast })=0$. The Taylor theorem giveswhere $r=-B_2^{\prime }(A_H^{\ast })>0$.

Iteration gives the interior derivative estimate for a harmonic functionThe growth hypothesis bounds the right side by $C^{\prime }{R}^{\alpha -1}+o(1)$, which tends to zero as $R\to \infty$. Thus every derivative of order $k+1$ vanishes everywhere. The Taylor theorem makes $u$ a polynomial of degree at most $k$, proving the polynomial-growth Liouville theorem for harmonic functions.

The relevant invariance principle for a low-degree multilinear polynomial is the following. Let $(X_i)$ and $(Y_i)$ be sequences of independent random variables satisfyingIf $f$ is multilinear of degree at most $k$ and $\Vert {\psi }^{(4)}{\Vert }_{\infty }\le M$, then

For the proof, use the Lindeberg replacement method. Replace $X_i$ by $Y_i$ one coordinate at a time and write $f=u_i+z_i{v}_i$, where $u_i=E_{i}f$ and $v_i=D_{i}f$ depend only on the other coordinates. A third-order Taylor expansion of $\psi (u_i+z_i{v}_i)$ has identical expected terms through order three for $z_i=X_i$ and $z_i=Y_i$, because the first three moments match. Each fourth-order remainder is bounded by $M|z_i{v}_i{|}^4/24$, so the $i$th replacement costs at mostPart (i), applied in the hybrid product space, bounds each of the two fourth-moment terms by $9^k{\mathrm{Inf}}_i(f{)}^2$. Thus the cost is at most $(M/12)9^k{\mathrm{Inf}}_i(f{)}^2$. The triangle inequality and summation over $i$ prove the result.