For distinct positive integers $m,n$, the product-to-sum identity givesBoth cosine integrals vanish over $[-\pi ,\pi ]$, soIf one index is zero, the corresponding function is identically zero and the same answer follows immediately.

For a positive integer $n$, the identity ${\sin }^2(nx)=[1-\cos (2nx)]/2$ givesThe stated nonnegative indexing includes $n=0$ as a special case: $f_0\equiv 0$, whose squared norm is zero. Thus $⟨f_0,f_0⟩=0$; it cannot belong to a basis.

A norm is nonnegative, so it cannot equal a negative $r$. The locus is therefore $\boxed{\varnothing }$, irrespective of $a,b,c$.

First assume $a\ne 0$. The first equation is a plane, and the second is a sphere centred at $c$. Define its signed normal separation and projected centre byThe intersection of a plane and a sphere isThe circle lies in the given plane. Its radius follows from the Pythagorean theorem, splitting $x-c=(x-q)+(q-c)$ into orthogonal components. If $a=0$, the first equation is the identity $0=0$, so the locus is the entire sphere, rather than a planar section.

For $r=0$, the norm equation requires $x=c$. Thus the locus isThis includes $a=0$, when the first condition imposes no restriction.

The orthogonal projection of the origin onto the line occurs at $t_{\ast }=-p\cdot u/|u{|}^2=-1/5$, giving closest point $(2,4/5,-2/5)$. Hence the distance from a point to a line in three dimensions is

Eliminating $2y-z$ from the two plane equations gives $x=2$, followed by $2y-z=2$. Taking $y=1$ gives the point $\boxed{p=(2,1,0)}$. The entire intersection line is $\mathbf{r}=p+tu$, where $u=(0,1,2)$.

A direction of the intersection line must be orthogonal to both normal vectors. Their cross product isThus a convenient parallel vector is $\boxed{u=(0,1,2)}$.

The product-to-sum formula givesWhen both indices are positive the two integer frequencies integrate to zero. If either index is zero the integrand already vanishes. This is Fourier orthogonality.

For $n\ge 1$, use ${\sin }^2(nx)=[1-\cos (2nx)]/2$. The inner product isThe stated nonnegative index range also includes $n=0$. Since $f_0$ is the zero function, $⟨f_0,f_0⟩=0$, so it must be removed from any proposed basis.

The logarithm requires $x>0$ for real-valued solutions. Put $v=xy$. Since $v^{\prime }=x{y}^{\prime }+y$, the equation becomes $v^{\prime }=v^2\log x$. For a nonzero branch, separation givesRenaming the constant and restoring $y=v/x$ givesDividing by $v^2$ would omit the equilibrium $v=0$, so the full solution set also includes $\boxed{y\equiv 0}$. The nonzero branches are defined on connected intervals with $x>0$ and $C+x-x\log x\ne 0$. A regular solution cannot switch from a nonzero branch to zero at an ordinary point, because the right-hand side is locally Lipschitz in $v$ there. These domain restrictions are part of the full real solution.

On an interval with $x\ne -1$, an integrating factor is $(x+1{)}^{-2}$. Multiplication givesIntegration therefore yieldsThe polynomial formula has a smooth continuation at $x=-1$, but the original coefficient $2/(x+1)$ is undefined there. Solution intervals for the stated equation must avoid that point, and the constants on its two sides need not agree.

Setting $\mathbf{b}=\mathbf{a}$ in i removes the cross product term and gives $\sigma (\mathbf{a}{)}^2=(\mathbf{a}\cdot \mathbf{a})I$. Repeated multiplication therefore yields, for nonnegative integers $m$,The zeroth power is $I$. If negative powers are intended, they require $\mathbf{a}\cdot \mathbf{a}\ne 0$, and the same parity formulas then follow from $\sigma (\mathbf{a}{)}^{-1}=\sigma (\mathbf{a})/(\mathbf{a}\cdot \mathbf{a})$. Every nonzero real vector meets that condition; a nonzero complex null vector need not. For example, $\mathbf{a}=(1,i,0)$ has $\sigma (\mathbf{a}{)}^2=0$ and no inverse.

For a real unit vector $\hat{\mathbf{n}}$, let $S=\sigma (\hat{\mathbf{n}})$. Then $S^2=I$, so even powers are $I$ and odd powers are $S$. Split the absolutely convergent matrix exponential into its even and odd series:The result holds for every real $\theta$, not merely small angles. Its two scalar series are the ordinary cosine and sine series.

Combine the diagonal and cyclic products from b into ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{\sum }_k{ϵ}_{ijk}{\sigma }_k$. Distribute the coefficients in the two Pauli-vector factors:The inner sums are precisely the components of $\mathbf{a}\times \mathbf{b}$, soInterchanging the two factors shows that their commutator is $2i\sigma (\mathbf{a}\times \mathbf{b})$. The three Pauli matrices are linearly independent: the entries of $\sigma (\mathbf{c})$ are $c_z,c_x-i{c}_y,c_x+i{c}_y,-c_z$, which vanish together only when $\mathbf{c}=0$. ThusFor nonzero real vectors, this means parallel or antiparallel directions. The dot and cross products here are bilinear; no complex conjugates are inserted if complex coefficients are considered.

With the same active rotation convention, the $x$ axis remains fixed, while ${\mathbf{e}}_y$ moves toward ${\mathbf{e}}_z$. ThereforeFor instance, a positive quarter-turn maps ${\mathbf{e}}_y$ to ${\mathbf{e}}_z$ and ${\mathbf{e}}_z$ to $-{\mathbf{e}}_y$, checking both signs. This is again an orthogonal matrix of determinant one.

Use active rotations on column vectors and positive angles counter-clockwise when viewed from the positive axis toward the origin. Rotation about $z$ takes ${\mathbf{e}}_x$ to $(\cos \alpha ,\sin \alpha ,0)$, ${\mathbf{e}}_y$ to $(-\sin \alpha ,\cos \alpha ,0)$, and leaves ${\mathbf{e}}_z$ fixed. These images are the columns of the rotation matrix:It satisfies $R_z^T{R}_z=I$ and $\det R_z=1$. Stating the viewing convention avoids confusing this active rotation with a passive change of axes.

For the neo-Hookean solid, the axial Cauchy stress tensor under the unloaded lateral boundary conditions isWith $K,\mu >0$, the spring balance is equivalent, after multiplication by $\alpha >0$, to the cubic equationFor a fixed $\gamma \ge 1$, write its left side as $h(\alpha )$. It obeys $h(0)=-\mu <0$ and tends to $+\infty$ as $\alpha \to \infty$. MoreoverThus $h$ first decreases from a negative value and then increases strictly, crossing zero exactly once on $\alpha >0$. For $\gamma >1$,so the unique equilibrium has $1/\gamma <\alpha <1$. The tissue is elastically compressed, while its actual length $\alpha \gamma$ exceeds one and compresses the spring.

To determine the large-growth limit without presuming a power law, solve for $\gamma$:This is a strictly decreasing function of $\alpha >0$, since its derivative isIt tends to $+\infty$ only as $\alpha \to 0^{+}$. Consequently unlimited growth forces $\alpha \to 0^{+}$, andMore precisely, the dominant balance is $K\gamma {\alpha }^3\sim \mu$, givingThe surrounding spring softens the divergence relative to rigid confinement but does not bound the compressive stress. Both axial extension and lateral expansion help accommodate growth:whose product is the total volume $\gamma$, as required by elastic incompressibility and the imposed growth tensor.

The end displacement relative to the initial unit axial length isBy Hooke's law, the total resisting force supplied by the spring is $F_3=-K\delta$. To convert this resultant into the axial Cauchy stress tensor, divide by the current end cross-sectional area. The transverse total stretches are both ${\alpha }^{-1/2}$, so that area isUniform traction over that face consequently givesEquate this to the expression from part (ii):The extra factor $\alpha$ on the left is essential: the spring law specifies a total force, whereas the material constitutive equation specifies force per current area. The reference end area is one, so the same spring resultant can also be read as a nominal end traction.

Write $W_i=\partial W/\partial {\alpha }_i$, evaluated at $({\alpha }^{-1/2},{\alpha }^{-1/2},\alpha )$. Applying the chain rule to the reduced strain energy density givesMultiply by $\alpha$ and use ${\alpha }_1={\alpha }_2={\alpha }^{-1/2}$:The transverse symmetry specified in the hint has $W_1=W_2$. ThusThe lateral faces have zero traction, so their diagonal Cauchy stress tensor components satisfy $t_1=t_2=0$. The hydrostatic pressure enforcing elastic incompressibility is therefore $p={\alpha }_1{W}_1={\alpha }_2{W}_2$, and the axial component simplifies toThis elimination of the Lagrange multiplier $p$ is what makes the reduced strain energy density useful for the spring balance.

In morphoelasticity, the deformation gradient factors as $\mathbf{F}=\mathbf{AG}$, where $\mathbf{G}$ accounts for local stress-free growth and $\mathbf{A}$ accounts for the subsequent elastic accommodation. Elastic incompressibility means that the latter preserves volume:It does not forbid volume increase by growth: here $\det \mathbf{F}=\det \mathbf{G}=\gamma$.

For the homogeneous diagonal deformation, the two unloaded transverse directions are equivalent. Taking the transversely symmetric branch implicit in the paper's hint gives ${\alpha }_1={\alpha }_2>0$. Set ${\alpha }_3=\alpha >0$. The determinant constraint is thensoEquality of the transverse stretches uses that symmetry assumption; elastic incompressibility alone would only give ${\alpha }_1{\alpha }_2{\alpha }_3=1$. The total deformation gradient and the actual axial length arebecause the initial cuboid is a unit cube.