With $\mathbf{d}={\mathbf{q}}_1-{\mathbf{q}}_2$ and $r=|\mathbf{d}|$, the Lagrangian isAt configurations with $r>0$, the Euler-Lagrange equations giveThese formulas apply on the differentiable part of the potential; collision configurations require any additional smooth extension of $f(|\mathbf{d}|)$ to be specified. Summing the equations givesso the total linear momentum is conserved. The total energy isDifferentiating, the two interacting forces do work $-f^{\prime }(r)\dot{r}$, while the potential derivative is $f^{\prime }(r)\dot{r}$ and particle three contributes no force term. Hence $\dot{E}=0$. Both total momentum and total energy are conserved.

Take a smooth variation $u+\epsilon \eta$ with $\eta$ vanishing on the boundary; compactly supported test functions suffice for the interior equation. At an extremum, the first variation is zero:Integration by parts removes the derivatives from $\eta$. The boundary terms vanish, givingThe fundamental lemma of the calculus of variations gives the Euler-Lagrange equationThis is the classical equation for sufficiently smooth admissible extrema; for weaker regularity the first-variation identity supplies its weak form.

For the squared-gradient integrand, the equation is $2(u_{xx}+u_{yy}+u_{zz})=0$, namely the Laplace equation. Put $u_0=ax+by+cz$ and write any admissible function as $u=u_0+v$, with $v=0$ on the boundary. The energy identity isBy the divergence theorem, the middle integral is ${\int }_{\partial V}v(a,b,c)\cdot ndS=0$; equivalently it vanishes by weak integration by parts for zero-trace functions. The final term is nonnegative and vanishes at $v=0$. Thus this is a global minimum, not merely a stationary value:

One version of dominated differentiation under the integral sign is the following. Let the parameter range be open. Suppose $f(\cdot ,y)$ is integrable, ${\partial }_{y}f(x,y)$ exists for almost every $x$, and near each parameter value its absolute value is bounded by an integrable function of $x$ independent of $y$. ThenThis follows by applying the dominated convergence theorem to difference quotients, bounded by that same integrable function through the mean value theorem.

On any compact parameter interval $0<y\le B$, the original integrand is bounded by $B/(1+x^4)$, using $|\sin (x^{2}y)|\le B{x}^2$. Its first and second parameter derivatives areTheir absolute values are bounded respectively by $(1+x^4{)}^{-1}$ and $x^2/(1+x^4)$, both integrable on $(0,\infty )$. Applying the theorem twice proves that $F$ is twice differentiable and givesThis last integral is absolutely convergent: its integrand is bounded near zero and bounded by $x^{-2}$ at infinity. Substituting $t=x\sqrt{y}$ yieldsThe expression for $c$ is finite and independent of $y$, which is all that is required.

The Lebesgue outer measure can be defined byA null set has $m^{\ast }(E)=0$, equivalently it admits such a cover of total length below every positive $\epsilon$.

The usual Cantor set is the intersection of the sets obtained from $[0,1]$ by repeatedly removing every open middle third. At stage $n$ there are $2^n$ closed intervals of length $3^{-n}$. Enlarge them slightly to open intervals whose total added length is below $\epsilon$. Thus $m^{\ast }(C)\le (2/3{)}^n+\epsilon$ for every $n$ and $\epsilon >0$, proving $m^{\ast }(C)=0$.

For the different deletion rule in this problem, let $L_n$ be the total length of the stage-$n$ intervals. Removing the stated fraction from each component givesFor $n\ge 1$ the telescoping product simplifies toBecause $K\subseteq K_n$, the same finite interval-enlargement argument gives $m^{\ast }(K)\le L_n$. Taking the limit provesOnly interval covers and their lengths were used; no continuity theorem for integrals is needed.

A solvable group has a finite subnormal series from the identity to the group with abelian group factors. An equivalent definition uses the derived series: set $G^{(0)}=G$ and $G^{(j+1)}=[G^{(j)},G^{(j)}]$; the series must reach the identity. For completeness, an abelian-factor series forces successive derived subgroups into its successive terms, while a terminating derived series itself supplies such a series.

If $H\le G$, induction gives $H^{(j)}\le G^{(j)}$: every commutator of elements of $H^{(j)}$ is also a commutator in $G^{(j)}$. Thus termination for $G$ implies termination for $H$, proving that every subgroup of a solvable group is solvable.

In $S_4$, useThe Klein four-group is invariant under conjugation in $S_4$; the other normality assertions follow from the alternating group being the kernel of the sign map. The quotient group factors are $V_4,C_3,C_2$, all abelian. Hence $S_4$ is solvable.

For the last example take $A_5\times C_2$. Its subgroup $\{1\}\times C_2$ is nontrivial, proper and normal, so the product is not a simple group. Projection onto $A_5$ is surjective, and every quotient of a solvable group is solvable, because a homomorphism carries each derived subgroup onto the derived subgroup of its image. But the nonabelian simple group $A_5$ has $A_5^{\prime }=A_5$: its derived subgroup is nontrivial and normal. Thus the product is not solvable either. One standard justification of the simplicity used here is that the nonidentity conjugacy class sizes of $A_5$ are $15,20,12,12$; no proper nontrivial union with the identity has size dividing $60$, as a normal subgroup would require by Lagrange's theorem for finite groups.

The splitting field is the smallest subfield of $\mathbb{C}$ containing $\mathbb{Q}$ and all the roots of a polynomial under consideration. Equivalently it is generated over $\mathbb{Q}$ by those roots, and the polynomial splits there into linear factors.

Put ${\zeta }_6=e^{\pi i/3}$. All sixth roots of unity are powers of ${\zeta }_6$, so the required splitting field isIndeed ${\zeta }_6=(1+\sqrt{-3})/2$, and conversely $\sqrt{-3}=2{\zeta }_6-1$. Its field extension degree is two.

A natural cubic spline is piecewise cubic, interpolates the data, has continuous first and second derivatives at the knot, and has zero second derivative at the outer endpoints. Its nodal second derivatives here satisfy $M_0=M_2=0$ and the usual matching equationso $M_1=6$. The resulting pieces areDirectly, the values are $0,0,4$, both first derivatives at one equal two, both second derivatives there equal six, and $S^{\prime \prime }(0)=S^{\prime \prime }(2)=0$. Thus all spline conditions hold.

The second data set adds the values of the affine function $x$ to the first. Adding $x$ to the spline preserves derivative matching and the zero endpoint second derivatives, so uniqueness of the natural interpolant gives the new spline $S(x)+x$. Its first piece is $x^3$, and its second is $1+3t+3t^2-t^3$. This is the affine covariance of natural cubic spline interpolation.