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This is mid-edge subdivision, rather than a vertex-retaining refinement. Use the usual closed two-manifold mesh setting: each edge has two incident facets and each vertex has a cyclic fan. For boundary meshes, a boundary refinement rule has to be specified separately. The relevant generic-position condition excludes collapsed local tangent directions and singular regular patches.

Topology and the manifold property. Regard the abstract mesh as a cell decomposition of a surface. Join the edge midpoints in each facet. The inner polygon is the new face associated with that facet; the corner regions around an old vertex assemble into the face associated with that vertex. At an old edge midpoint, the four adjacent corner connections form a cyclic link. Thus every new interior vertex has valence four, every new edge has two incident faces, and the refined cell complex is homeomorphic to the original surface. If the original counts are $V,E,F$, the new counts are $E,2E,V+F$, preserving $V-E+F$. The cyclic-link argument, rather than Euler characteristic alone, verifies the local manifold condition. Convergence and local regularity, justified below, transfer this abstract surface structure to the limit.

Convex hull. Each newly created point is an average of two previous points. By induction it is a convex combination of original vertices. The same is true of points of any triangulation of the refined facets, and taking limits keeps them in the closed convex hull. Hence the entire limit stays in that hull, not necessarily inside the original nonconvex enclosed solid.

Facet centroids. Follow the descendant inner face of an original $n$-sided facet, with cyclic vertices $p_i$. Its vertices satisfy $p_i^{(r+1)}=(p_i^{(r)}+p_{i+1}^{(r)})/2$. Summing proves mid-edge facet centroid invariance:For a cyclic Fourier mode $e^{2\pi iji/n}$ the multiplier is $(1+e^{2\pi ij/n})/2$, whose modulus is $|\cos (\pi j/n)|<1$ for $j\ne 0$. Every mean-zero mode therefore decays, so all descendant face vertices converge to $\overline{p}$. Consequently the original vertex centroid belongs to the limit surface. No planarity assumption is needed.

Quadratic triangular pieces. On a regular quadrilateral mesh, apply two midpoint steps. Each new point combines three old neighboring controls with weights $1/2,1/4,1/4$. For example, around a face with cyclic corners $p_{00},p_{10},p_{11},p_{01}$, the four new points areThis is the binary refinement mask obtained by expandingwith the translation chosen to place the refined grid at quarter offsets. The limit is therefore represented by shifts of the quadratic four-direction box spline with directions $e_1,e_2,e_1+e_2,e_1-e_2$. Projection of a four-cube onto the plane has piecewise total degree $4-2=2$, and its knot grid is triangular. Coordinatewise linear combinations preserve that degree, giving parametric quadratic triangle patches. Around a nonregular center there are infinitely many shrinking rings of such patches; the isolated center is their limit, not generally part of a finite polynomial patch representation.

Only finitely many initially possible smoothness exceptions. Every vertex has valence four after the first step. Only the descendants of original facets and the faces created around original vertices can remain nonquadrilateral. All subsequently created faces around four-valent vertices are quadrilaterals. Thus every point except the limiting centers of those persistent nonquadrilateral faces eventually lies in a regular patch neighborhood. The number of possible exceptional centers is at most $V+F$. For the regular box spline, any three distinct directions still span the plane, so three directions must be removed to make the remainder nonspanning. The box spline continuity criterion gives $C^{3-2}=C^1$ across its triangle boundaries. This proves $C^1$ continuity away from at most those finitely many centers.

Removing the apparent exceptions. Center a finite neighborhood of an $n$-sided exceptional face, and analyze a full two-step subdivision matrix so the mesh orientation is restored. Cyclic Fourier basis modes diagonalize its sector symmetry. The face mode has eigenvalueThe local ring blocks also contribute eigenvalues $1/4$ and zero. For $n>3$, the constant mode is the single eigenvalue one, and the two first-frequency tangent modes dominate all nonconstant remaining modes. Construct their characteristic map of a subdivision surface; its quadratic derivative pieces must have a nonzero consistently oriented Jacobian, and its fundamental sector must be injective. These checks, not the eigenvalue ordering alone, give a regular chart and a continuously limiting tangent plane for generic data.

The case $n=3$ needs a separate Jordan analysis: $\lambda =1/4$ is eightfold, so the usual semisimple double-eigenvalue test does not apply. Its first-frequency Jordan chains supply the leading tangent behavior, while the other contributions become relatively negligible. Regularity and injectivity of that generalized characteristic map complete the argument. These concrete spectral and characteristic-map checks are established for this scheme in Peters and Reif, Sections 3 and Appendix. Thus the generic limit is geometrically $C^1$ also at the exceptional centers; compatible local charts need not preserve the original uniform refinement parameterization.

The manifold statement concerns the abstract limit surface and its regular local parametrizations. It does not, from combinatorics and generic tangent data alone, prove global injectivity of the image in three-dimensional space. A particular model may require an additional global self-intersection test before claiming that its image is an embedded manifold.

The Markov interlacing lemma says that differentiation preserves interlacing roots of polynomials. More explicitly, if two degree-$d$ real polynomials have ordered real roots satisfying $a_1\le b_1\le a_2\le \cdots \le a_d\le b_d$, their $k$th derivatives have ordered roots $a_j^{(k)},b_j^{(k)}$ satisfyingThe analogous alternating order is preserved for polynomials of consecutive degrees. Common roots are allowed through weak inequalities; strictly interlacing simple roots give strict inequalities. When a derivative is constant, its root list is empty.

Work locally in $z$ where $u\ne 0$, and take the matrix entries literally as they appear in the original PDF. At fixed $z$, abbreviateThen $A=D{\lambda }^2+\left(\begin{array}{cc}0& U\\ V& 0\end{array}\right)\lambda +\left(\begin{array}{cc}d& P\\ R& -d\end{array}\right)$. DefineThe literal coefficients giveIn general neither $h=z/2$ nor $\theta =0$ is valid for the printed system.

The normalized formal fundamental matrix iswith a chosen logarithm branch. In particular the diagonal normal form has ${\Lambda }_1=0$, ${\Lambda }_2=hD$, ${\Lambda }_3=\theta D$. Its first two prefactor coefficients areThese formulas include the diagonal terms required by the derivative of the prefactor. Setting every diagonal prefactor coefficient to zero while retaining only the displayed exponent would generally fail the differential equation.

Here is the coefficient calculation and the full recursion. Write $A_1=\left(\begin{array}{cc}0& U\\ V& 0\end{array}\right)$, $A_2=\left(\begin{array}{cc}d& P\\ R& -d\end{array}\right)$ and $G_0=I$, with negative-index $G_j=0$. Substitution into the gauge transformation of a linear differential system gives, for $m\ge 1$,At $m=1$, the commutator determines $(G_1{)}_{12}=-U/2$, $(G_1{)}_{21}=V/2$ and the absence of a linear diagonal term. The diagonal part at $m=2$ gives $h=d+UV/2$. The next diagonal part gives $\theta =(UR+VP)/2$. At $m=4$ diagonal solvability gives $-\rho =PR/2-UV(d+h)/4$; the next order fixes the two diagonal entries of $G_2$ above. The off-diagonal entries are fixed at each stage because $[D,X{]}_{12}=2X_{12}$ and $[D,X{]}_{21}=-2X_{21}$. Later diagonal solvability fixes the remaining normalization coefficients. Thus this is a complete formal solution, not a convergence assertion. It is the formal cubic-exponential two-by-two system expansion. The printed hint's second prefactor term must have denominator ${\lambda }^2$, and the remainder after the $1/\lambda$ term of the diagonal expansion must start at $1/{\lambda }^2$.

Now impose an isomonodromic deformation in the full irregular sense: its Stokes matrices and formal exponent are fixed in $z$, with the leading formal normalization fixed. Sectorial matrices satisfy $Y_{j+1}=Y_j{C}_j$ with $(C_j{)}_z=0$, so $B_j=(Y_j{)}_z{Y}_j^{-1}$ is the same on overlaps. Since the spectral system has no finite singularities, its fundamental matrix and inverse are entire in $\lambda$, making $B$ entire. Differentiating the formal expression, using ${\theta }_z=0$, givesBut $[G_1,D]=A_1$. The sectorial cover makes this a global growth bound, so Liouville's theorem applied to $B-h_z(\lambda D+A_1)$ gives the literal consequenceAn arbitrary parameter dependence of $u,w$ need not be isomonodromic; this analytic argument presupposes the stated constancy of generalized monodromy data.

The requested matrix would require $h_z=1/2$ from its diagonal and upper entry, and then its lower entry would require $(z-2)w=0$. On an open deformation domain this forces $w\equiv 0$. Thus that requested expression is not the general deformation matrix for the printed coefficients.

There is a consistent nearby correction: replace the lower entry of $A_1$ by $-2w/u$. Then $V=-2w/u$, so $h=z/2$ and $\theta =0$, and the same derivation proves exactlyFor this corrected system use the same prefactor formulas with $V=-2w/u$: $\rho =-w(w+z)/2-w(u^{\prime }/u{)}^2$ and both diagonal entries of $G_2$ are ${\rho }^2/2+w/4$. Compatibility determines the allowed $u,w$ in the next solution. This correction is stated explicitly rather than silently substituting it for the original PDF.

For an invertible fundamental matrix, differentiating $Y_{\lambda }=AY$ in $z$ and $Y_z=BY$ in $\lambda$ gives $Y_{\lambda z}=(A_z+AB)Y$ and $Y_{z\lambda }=(B_{\lambda }+BA)Y$. Thus the necessary and sufficient compatibility identity isHere both derivatives must use the actual deformation variable $z$; the printed $t$ has not been defined. The isomonodromic deformation identity can be checked as a polynomial in $\lambda$, without assuming the conclusion.

First retain the coefficients actually printed in the PDF. Write $p=u^{\prime }$, $d=w+z/2$, and introduce the two residualsDirect matrix multiplication and differentiation giveFor example, the quadratic lower entry is the difference between $[D,B_0{]}_{21}=2w/u$ and the contribution $(A_1{)}_{21}=-zw/u$; it does not cancel. The constant lower entry follows by differentiating $2u^{\prime }w/u^2$ and collecting it as $2wH/u^2+2u^{\prime }K/u^2$. These entries also provide independent checks of the two most important cancellations.

Work on an open connected deformation domain where $u\ne 0$, as the matrices require. The quadratic entry forces $(2-z)w=0$ throughout the domain, hence $w\equiv 0$ by continuity, including a possible point $z=2$. All remaining entries then vanish exactly when $u^{\prime \prime }+(z/2)u=0$. Thus the literal answer isThis is a rescaled Airy function equation: with $\xi =-2^{-1/3}z$, the equation is $u_{\xi \xi }-\xi u=0$, so $u=c_1\mathrm{Ai}(\xi )+c_2\mathrm{Bi}(\xi )$ on a domain avoiding its zeros. In particular, choose the local solution with $u(1)=u^{\prime }(1)=1$ and $w=0$. It gives compatible matrices but violates the printed $w=u^{\prime }$. This proves that the requested assertion is false for the original coefficients, independently of the undefined $t/z$ in the printed differential equation.

For completeness, use the explicitly corrected lower coefficient $(A_1{)}_{21}=-2w/u$ from the preceding solution. The same multiplication gives the much simpler residualConsequently compatibility is equivalent to $K=H=0$. The first equation integrates without dividing by $w$:The constant of integration $C$ therefore satisfies $w=C{u}^2$, and the second equation becomesConversely these two equations give $K=H=0$, and substitution makes every residual entry vanish, proving sufficiency as well as necessity. This is the quadratic spectral Lax pair for zero-parameter Painlevé II. For $C\ne 0$, choose constants $b,a$ with $b^3=-2$ and $C{a}^2{b}^2=-2$, set $z=bx$, $u=aq(x)$, and obtain $q^{\prime \prime }=2q^3+xq$, the zero-parameter Painlevé II equation. For $C=0$ it reduces to the Airy function equation. Even with this correction, the conclusions are $w=C{u}^2$ and the coefficient $z/2$, rather than the incompatible expressions printed in the question.

For the first derivative, $f^{\prime }(x)=\frac{1}{2}(1+x{)}^{-1/2}$, which is the stated formula at $r=1$. If the formula holds at $r$, differentiating multiplies its coefficient by $\frac{1}{2}-r=-(2r-1)/2$. The factorial coefficients satisfyThe sign changes, proving by mathematical induction that

Apply the given Taylor formula with integral remainder at $x=1/2$. Its polynomial part is $1+2S_n$, where $S_n$ is the first $n$ terms of the requested series, sinceTo justify passage to the infinite series, for $0\le t\le 1/2$,Therefore the integral remainder obeysThe last inequality follows from $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)\le {\sum }_{j=0}^{2n}\left(\genfrac{}{}{0ex}{}{2n}{j}\right)=4^n$, a direct binomial theorem bound. Hence $\sqrt{3/2}=1+2\lim S_n$, and

Now impose the stated zero endpoint values and nonvanishing assumptions near $a$. For each $x\in (a,a+\delta )$, applying the proved identity on $[a,x]$ gives some $c_x\in (a,x)$ such thatAs $x\downarrow a$, every possible $c_x$ also tends to $a$. Explicitly, for any $\epsilon >0$, choose a right neighborhood of $a$ where $|f^{\prime }/g^{\prime }-L|<\epsilon$. If $x$ lies in it, so does $c_x$, and therefore $|f(x)/g(x)-L|<\epsilon$. This proves the endpoint zero-over-zero form of L'Hôpital's rule:No continuity of either derivative is required, only their ratio limit and the given nonvanishing conditions.

For the final limit, near zero $\cos x>0$, so the real logarithm is well defined. First apply the proved rule to $\tan x/x$ on the right: its derivative ratio is ${\sec }^{2}x\to 1$. Next apply it to $\log (\cos x)/x^2$; the derivative ratio isThe original quotient is even, so the left limit equals the right limit. Consequently

Define a candidate antiderivative by integration along the straight segment from zero:For $z$ in the open unit disc and sufficiently small $h$, the triangle with vertices $0,z,z+h$ lies within the disc, since the disc is convex. By the Cauchy integral theorem, its boundary contour integral vanishes. ConsequentlyFor $h\ne 0$, subtract $f(z)$ after division by $h$. The absolute value of the remainder is at most ${\sup }_{0\le t\le 1}|f(z+th)-f(z)|$, which tends to zero by continuity of the holomorphic function $f$ at $z$. Hence the complex derivative exists and $g^{\prime }(z)=f(z)$ for every point of the disc. This proves that $g$ is a holomorphic function and supplies the requested primitive directly from the Cauchy integral theorem.

Let $r=Ax-b$ and consider an arbitrary variation $x+th$. The objective for the linear least-squares problem isAt a minimizer the derivative at $t=0$ must vanish for every $h$, so $A^{T}r=0$. Conversely, when this holds, the same expansion at $t=1$ gives $\Vert A(x+h)-b{\Vert }^2=\Vert r{\Vert }^2+\Vert Ah{\Vert }^2\ge \Vert r{\Vert }^2$ for every $h$. Thus $x$ minimizes the residual norm if and only if $A^T(Ax-b)=0$. These least-squares normal equations do not require $A$ to have full rank; without full rank the minimizing vector need not be unique.

The original PDF has $A_{22}=1$, not the 2 in the converted TeX. Using the PDF's matrix givesThe least-squares normal equations consequently have the unique solutionAs a direct check, $Ax-b=(-279,-95,238,-61{)}^T/123$, whose inner product with either column of $A$ is zero. The positive determinant verifies full column rank for this particular problem.

Put ${\epsilon }_n={\sup }_{[-1,1]}|F_n-F|$. Uniform convergence means ${\epsilon }_n\to 0$, and continuity ensures that the functions have Riemann integrals. The integral triangle inequality givesThis proves that uniform convergence preserves integrals on a compact interval.

For the derivative assertion, take $F_n(x)=\sin (n^{2}x)/n$ and $F(x)=0$. These functions are differentiable, with $\sup |F_n|\le 1/n$, so their convergence is uniform. However, $F_n^{\prime }(0)=n$ whereas $F^{\prime }(0)=0$. The derivatives therefore need not converge even at one interior point.

In polar coordinates, the Laplace equation isFor a single-valued, $2\pi$-periodic harmonic function on an annulus, separation of variables and its angular Fourier series giveIndeed a nonzero angular mode has radial equation $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$, with solutions $r^{\pm n}$; the zero mode has solutions 1 and $\log r$. This expression describes the periodic annular problem here; angular terms not periodic, such as $\theta$, are excluded.

Decay at the origin leaves only positive powers in the interior, and decay at infinity leaves only negative powers in the exterior. Continuity on the unit circle matches the corresponding Fourier coefficients. If the common cosine coefficient is $K_n$, the interior and exterior terms are $K_n{r}^n\cos n\theta$ and $K_n{r}^{-n}\cos n\theta$. Their exterior-minus-interior radial derivative at the circle is $-2n{K}_n\cos n\theta$. Thus harmonic matching across a circle with a derivative jump gives $K_2=-1/4$, $K_4=-1/8$, and zero for every other sine or cosine mode. ConsequentlyEach branch satisfies the Laplace equation and its respective decay condition. Both give $-\cos 2\theta /4-\cos 4\theta /8$ at $r=1$, and direct differentiation gives the required jump $\cos 2\theta +\cos 4\theta$. The same mode calculation leaves no nonzero homogeneous solution in this class.

Finite second moment implies $\mathbb{E}X<\infty$. For $\lambda$ near any positive ${\lambda }_0$, the derivative of $e^{-\lambda X}$ is bounded in absolute value by $X$, provided the neighbourhood stays in $(0,\infty )$. The dominated convergence theorem therefore permits differentiation under the expectation:For the expansion, put $r(z)=e^{-z}-1+z-z^2/2$. Taylor expansion gives $r(z)/z^2\to 0$ as $z\downarrow 0$. Also, for $z\ge 0$,The inequalities follow by integrating $0\le 1-e^{-u}\le u$ from zero to $z$. Consequently $r(\lambda X)/{\lambda }^2\to 0$ pointwise and its absolute value is bounded by $X^2/2$. Another application of the dominated convergence theorem gives the second-moment expansion of a Laplace transform:No third moment is needed for the remainder estimate.

Use the riskless financial asset as numéraire, normalized to have initial value one and terminal value $R>0$. Write $S_0\in {\mathbb{R}}^s$ for the initial risky prices and $S_1$ for the finite, measurable terminal price random vector. Define the discounted gain vector $X=S_1/R-S_0$. A risky holding $h$ financed by the cash holding $-h\cdot S_0$ has zero initial cost and discounted terminal gain $h\cdot X$. An arbitrage is such a portfolio with $h\cdot X\ge 0$ almost surely and $\mathbb{P}(h\cdot X>0)>0$. Allowing nonpositive initial cost is equivalent: invest any initial surplus in the riskless financial asset to obtain a zero-cost arbitrage.

An equivalent martingale measure is a probability measure $Q$ with exactly the same null events as the physical probability measure $P$, under which discounted asset prices are integrable and satisfy ${\mathbb{E}}_Q[S_1/R]=S_0$. Equivalently, ${\mathbb{E}}_{Q}X=0$. Such a measure rules out arbitrage: every zero-cost portfolio has mean discounted gain zero, whereas a nonnegative gain positive with positive $P$-probability is also positive with positive $Q$-probability and has strictly positive $Q$-expectation.

For the converse, the probability space need not be finite, and the payoffs need not be integrable under $P$. Here is a direct Gaussian-damped martingale density construction. SetThe set $N$ is a linear subspace. Choosing a finite basis of $N$ shows that $X\in L$ almost surely: intersect the probability-one events on which each basis vector is orthogonal to $X$. If $h\in L\setminus \{0\}$, then $h\cdot X$ is not almost surely zero. Absence of arbitrage implies both $P(h\cdot X<0)>0$ and $P(h\cdot X>0)>0$, by applying the definition to $h$ and $-h$.

All the analytical steps in minimizing $F$ can be checked directly. On $|h|\le H$ the integrand is bounded by $e^{H^2/4}$, since $-|X{|}^2+H|X|\le H^2/4$. Its first two directional derivatives are bounded uniformly in $X$ by constants depending only on $H$ and the direction, because the functions $r^j{e}^{-r^2+(H+1)r}$ are bounded for $j=0,1,2$. The mean-value bound therefore proves continuity of $F$, and the Taylor theorem remainder bound, integrated over $P$, provesThus differentiation under the integral does not require an unproved physical moment assumption.

To prove coercivity, suppose $|h_k|\to \infty$ in $L$. A subsequence of $h_k/|h_k|$ converges to a unit vector $v\in L$. This elementary finite-dimensional compactness follows by successively taking convergent subsequences of the bounded coordinates. Since $P(v\cdot X<0)>0$, some event $A=\{v\cdot X\le -\epsilon ,|X|\le K\}$ has positive probability: the negative-gain event is the countable union of such events with $\epsilon =1/j$ and integer $K$. Eventually $(h_k/|h_k|)\cdot X\le -\epsilon /2$ on $A$, and henceIf coercivity failed, an unbounded sequence with bounded $F$ would have just such a contradictory subsequence. Consequently a minimizing sequence for $F$ is bounded and has a convergent subsequence; continuity makes its limit $h_{\ast }$ a minimizer. If $L=\{0\}$, simply take $h_{\ast }=0$. In every case the directional derivatives vanish at $h_{\ast }$. The vector ${\mathbb{E}}_P[X{e}^{-|X{|}^2-h_{\ast }\cdot X}]$ lies in $L$ and is orthogonal to $L$, so it is zero. Thereforeis a positive mean-one density defining an equivalent martingale measure. Integrability of $X$ under $Q$ follows from the same Gaussian bound with $j=1$. This establishes absence of arbitrage if and only if an equivalent martingale measure exists, without quoting a financial duality theorem or assuming a finite state space.

Now suppose the probability space is finite. Discard null states and enumerate its positive-probability atoms by $1,\dots ,m$. Every measurable terminal payoff is a vector in ${\mathbb{R}}^m$. Let $A$ be the $m\times (s+1)$ matrix whose columns are the discounted terminal payoffs of the traded financial assets, with first column $1$, and let $p=(1,S_0)$ be their initial price vector. A portfolio with holding vector $z$ replicates the discounted payoff $Az$, and an equivalent martingale measure is exactly a vector $q$ satisfyingThe first pricing equation includes ${\sum }_j{q}_j=1$. A complete market means $\mathrm{im}A={\mathbb{R}}^m$. If $q,q^{\prime }$ both price the financial assets, then $(q-q^{\prime })\cdot Az=0$ for every $z$. Completeness allows $Az$ to be each coordinate vector, so $q=q^{\prime }$.

Conversely, if the financial market is incomplete, there is a nonzero vector $v$ with $A^{T}v=0$. To see this without invoking a duality theorem, perform row elimination on the homogeneous system: its number of pivots is $\mathrm{rank}A<m$, leaving a free variable and hence a nonzero solution. The constant payoff column gives ${\sum }_j{v}_j=0$. Existence of an equivalent martingale measure was already proved, so choose one strictly positive vector $q$. Forboth $q+\epsilon v$ and $q-\epsilon v$ are distinct strictly positive probability vectors satisfying $A^T(q\pm \epsilon v)=p$. Thus uniqueness fails. On a finite arbitrage-free market, completeness is equivalent to uniqueness of the equivalent martingale measure.

The parameter range is $0\le \theta \le 1$. Omitting data-only constants from the multinomial distribution likelihood, its log isFor an interior maximizer, the likelihood score must vanish:Multiply by $\theta (1-\theta )(7+4\theta )$ and collect powers to obtainThe likelihood is concave: its second derivative in the interior isTherefore an admissible interior root is the unique maximum. Endpoints must also be handled if some counts vanish. If $x_3>0$ and $x_1+x_2>0$, both endpoints have zero likelihood and the root is interior. If the maximum is at zero then $x_3=0$, and zero also satisfies the displayed polynomial; a maximum at one requires $x_1+x_2=0$, and one likewise satisfies it. Multiplying the score can introduce other endpoint roots, so not every polynomial root is an MLE: evaluate the actual likelihood on the feasible candidates.

Use a Minkowski external source term $+\int J\chi$, where $\chi$ is the integration variable, and normalize the vacuum functional:The usual vacuum boundary prescription is implicit. This is the logarithmic convention $Z=e^W$ for the connected generating functional. It makes the given quadratic expression consistent with ${\Delta }_F=⟨T\chi \chi ⟩$. Another common convention is $Z=e^{i{W}_c}$, with $W_c=-iW$; mixing the two conventions would change the factors of $i$ below.

Functional derivative generates the full and connected correlation functions:These are time-ordered products in the vacuum, and $G_1=G_1^c=0$ by the stated assumption. Differentiating $e^W$ four times partitions the four insertions into connected blocks. All partitions with a singleton vanish, leavingThe shorthand arguments stand for spacetime points; this is a functional cumulant identity, not an assumption of a free field.

To construct the quantum effective action, define the source-dependent classical fieldLocally invert this relation to obtain $J[\phi ]$, assuming the response integral kernel is invertible with the chosen regulator and boundary prescription. Take its Legendre transformThe chain-rule terms proportional to $\delta J$ cancel, so $\delta \Gamma /\delta \phi =-J$. A field-independent constant is fixed separately and is immaterial to the vertices. At zero external source the classical field is zero, and there is no linear term.

For the two-point vertex, differentiate the classical field once. If $G=G_2^c$, thenThe vertex coefficients in this problem are derivatives of $i\Gamma$, not of $\Gamma$ alone. ThereforeFor the three-point vertex, the derivative of the connected two-point integral kernel at a general external source isCombine this with $\delta J/\delta \phi =-i{G}^{-1}$ and differentiate the inverse identity, $\delta G^{-1}=-G^{-1}(\delta G)G^{-1}$. This gives the logarithmic-source inverse-kernel vertices relationAll kernels here are finally evaluated at zero external source. The connected three-point function is amputated by its three inverse quantum field theory propagators.

For the specified quadratic $W$, the external source response is $\phi =i{\Delta }_{F}J$, hence $J=-i{\Delta }_F^{-1}\phi$. Substitution in the transform givesFor the standard free quantum field theory propagator ${\tilde{\Delta }}_F(p)=i/(p^2-m^2+i0)$, this is the usual quadratic free action, up to the vacuum prescription. The generating functional is Gaussian, so connected functions beyond second order vanish and there are no interaction vertices or nontrivial scattering amplitudes. Full higher even-point functions can still be nonzero: they are products of two-point functions, as the four-point formula shows.

The kinetic normalization fixes $[\Phi ]=(d-2)/2$, since a derivative has mass dimension one and the Lagrangian density has dimension $d$. Every differentiated field is still a field insertion. Hence the interaction has $n+r$ legs, not just $n$, and its vertex contains a homogeneous polynomial of degree $r$ in the attached momenta. Its coupling dimension isFor a connected Feynman diagram of this vertex type,The superficial momentum degree is now $D=dL-2I+rV$. Eliminating $I$ and $L$ givesso derivative-interaction scalar power counting yieldsThe same counting applies to tensor components with quantum field theory propagators scaling as $p^{-2}$; contractions, identities and cancellations can lower the actual degree. A coupling of negative mass dimension lets the superficial degree increase with the number of vertices, explaining the need for progressively higher-dimensional counterterms in a perturbative effective field theory.