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At zero, $g(0)=0$ andby continuity of $f$ at zero. Thus $g$ is always differentiable there, with $g^{\prime }(0)=0$, regardless of whether $f^{\prime }(0)$ exists.

At $x\ne 0$, if $f$ is differentiable, the product rule applied to $t^{2}f(t)$ proves that $g$ is differentiable. Conversely, if $g$ is differentiable at $x$, the function $t^{-2}$ is differentiable on a neighborhood of $x$. Apply the product rule to $f(t)=t^{-2}g(t)$ on that neighborhood. This proves the differentiability restored by a quadratic zero equivalence:At a nonzero differentiability point, $g^{\prime }(x)=2xf(x)+x^2{f}^{\prime }(x)$.

By differentiability implies continuity, $f(x+h)\to f(x)$. Add and subtract $f(x+h)g(x)$ in the difference quotient:The two quotient limits exist by the given differentiability. Taking the limit and using continuity yields the product rule:Thus the product is differentiable at the specified point; differentiability elsewhere is not needed.

Let $u,v\in f(I)$ with $u<y<v$. Choose preimages $x_u,x_v\in I$, and let $a=\min (x_u,x_v)$, $b=\max (x_u,x_v)$. The defining property of a real interval ensures $[a,b]\subseteq I$. The restriction of $f$ to this closed interval is continuous, and its endpoint values are $u,v$ in one order or the other. The intermediate value theorem supplies $c\in [a,b]\subseteq I$ with $f(c)=y$. Thus every value between two image values lies in the image, provingThe continuous image of a real interval result does not require $I$ itself to be closed or bounded; only the segment between the chosen preimages is used. Empty and single-point images also satisfy the interval property.

The intermediate value theorem states that, if $f$ is a continuous function on a closed interval $[a,b]$, every value between $f(a)$ and $f(b)$ equals $f(c)$ for some $c\in [a,b]$. If the value is strictly between the endpoint values, $c$ may be chosen in $(a,b)$.

Here is a proof from the supremum property of the real numbers. Suppose $f(a)<y<f(b)$; the reverse inequality is reduced to this case by replacing $f,y$ with $-f,-y$. LetThe set is nonempty because it contains $a$ and is bounded above by $b$. Continuity at $a$ supplies points of $E$ to the right of $a$, while continuity at $b$ excludes a neighborhood of $b$ from $E$. Hence $a<c<b$. By the definition of supremum, there are $t_n\in E$ with $t_n\to c$. Continuity gives $f(c)=\lim f(t_n)\le y$. If $f(c)<y$, continuity would give a point just to the right of $c$ still satisfying $f(t)<y$, contradicting that $c$ is an upper bound for $E$. Thus $f(c)=y$. Values equal to either endpoint are attained at that endpoint, completing the proof.

Put $I=I_1=I_2$. The third of the Euler equations for a torque-free rigid body gives ${\dot{\omega }}_3=0$. The first two become ${\dot{\omega }}_1=-\Omega {\omega }_2$, ${\dot{\omega }}_2=\Omega {\omega }_1$, whereTherefore ${\omega }_1+i{\omega }_2=C{e}^{i\Omega t}$. In body coordinates, $L_1+i{L}_2=IC{e}^{i\Omega t}$ and $L_3=I_3{\omega }_3$ is constant. The angular momentum precesses about the body symmetry axis, with periodFor positive ${\omega }_3$ this is the printed expression. In the inertial frame the torque-free angular momentum is constant; the precession here is its changing representation in the rotating body frame. If $e=0$ or ${\omega }_3=0$, this body-frame precession rate is zero.

Assume the spheroid is homogeneous, as implicit in the stated moments. Put $s=\sqrt{1-e^2}$ and map a uniform solid sphere by $(X_1,X_2,X_3)\mapsto (X_1,X_2,s{X}_3)$, preserving total mass by adjusting density. For the sphere, symmetry and its moment $2M{a}^2/5$ give $\int X_i^{2}dm=M{a}^2/5$ for each coordinate. The transformed squared-coordinate integrals are consequently $M{a}^2/5,M{a}^2/5,s^{2}M{a}^2/5$. Reflection symmetry makes the off-diagonal inertia entries vanish. ThusThe original PDF gives the correct solid-sphere factor $2/5$, rather than the converted TeX's $2/3$.

For $|z|<1$, differentiation under the convergent Euler integral giveswhere $(a{)}_n=a(a+1)\cdots (a+n-1)$ and $(a{)}_0=1$ are Pochhammer symbols. Thus its Taylor series iswhich is symmetric in $a,b$. In the common region of parameter convergence for the two Euler integrals, this proves equality near zero. The identity theorem extends it throughout their common principal slit domain, and parameter continuation gives the usual identity wherever both sides are defined:

For the Euler integral to converge directly, assume $\mathrm{Re}b>0$ and $\mathrm{Re}(c-b)>0$. At zero it is the beta function $B(b,c-b)=\Gamma (b)\Gamma (c-b)/\Gamma (c)$. The normalization is thereforeOutside these initial parameter conditions the hypergeometric function is defined by analytic continuation wherever the parameters permit it.

Given a binary cylinder with finite prefix $b_1\dots b_N$, repeat a finite block beginning with that prefix indefinitely. The resulting binary sequence is periodic under the shift and represents a periodic point of an interval map of $F$ in the cylinder. If the given prefix consists entirely of ones, append a zero to the block so that its value is less than one; this still preserves the prefix. Since cylinders can be taken inside any nonempty open interval, periodic points are dense.

Every nonempty relatively open interval contains a binary cylinder whose points share a finite prefix. Given nonempty open sets $U,V$, choose such a cylinder $C\subset U$ of length $N$. After $N$ binary shifts, $F^N(C)$ contains $[0,1)$ with the usual half-open cylinder convention. In particular, some point of $C$ maps into $V$. Thus $F$ is topologically transitive: $F^N(U)\cap V\ne \varnothing$. Equivalently, concatenate a prefix placing the point in $U$ with a tail placing its $N$th iterate in $V$.

Choose binary expansions $x=0.b_1{b}_2\dots$ that do not end in an infinite string of ones. The doubling map deletes the first digit: $F(x)=0.b_2{b}_3\dots$. Given $x$ and any neighbourhood of it, choose $N$ sufficiently large that its binary cylinder of length $N$ is small enough, and choose a point $y$ sharing that prefix but with a prescribed tail. After $N$ shifts, the tail of $y$ may be chosen arbitrarily in $[0,1)$. Choose it close to zero or close to one, whichever is at distance more than $1/3$ from $F^N(x)$. Then $|x-y|\le 2^{-N}$ but $|F^N(x)-F^N(y)|>1/3$. Hence the map has sensitive dependence on initial conditions, with a uniform separation constant. At dyadic endpoints the cylinder sharing the chosen terminating expansion still contains points arbitrarily close to the endpoint.

The two independent columns of $V$ remain independent after multiplication by invertible $S$. Since $R=SV$ has zeros below the second row, its leading two-by-two triangular block is invertible. Hence the image under $S$ of the invariant subspace is exactly $\mathrm{span}(e_1,e_2)$. This coordinate plane is invariant under $\hat{A}=SA{S}^{-1}$, forcing the bottom-left block to vanish:The block determinant factors $\det (zI-\hat{A})=\det (zI-B)\det (zI-D)$. Therefore the eigenvalues of $A$ are exactly those of the indicated two-by-two and $(n-2)$-by-$(n-2)$ blocks, counted with algebraic multiplicity. No diagonalizability assumption is needed.

Because $Sw=c{e}_1$ with $c\ne 0$, the similarity transform satisfies $\hat{A}{e}_1={\lambda }_1{e}_1$. Thus its first column is $({\lambda }_1,0,\dots ,0{)}^T$ and it has block form $\left(\begin{array}{cc}{\lambda }_1& \ast \\ 0& B\end{array}\right)$. Taking the determinant gives $\det (zI-\hat{A})=(z-{\lambda }_1)\det (zI-B)$. Similarity preserves the characteristic polynomial. Hence the eigenvalues of $A$, with algebraic multiplicities, are ${\lambda }_1$ and those of $B$.

For side-by-side spheres, reflection in the vertical mid-plane exchanges the spheres while preserving their common velocity, so their axial drags are equal. For vertically aligned spheres, reflection in the horizontal mid-plane exchanges them and reverses velocity; applying linearity to reverse all velocities back restores the original problem and again gives equal axial drag. These are symmetries of the exact two-sphere Stokes problem.

At a distant side-by-side centre, the leading flow produced by one sphere is $3aU/(4b)$ in the direction of translation. On its axis it is $3aU/(2b)$. This ambient entrainment lowers the relative velocity of the other sphere. Using Stokes drag law with that relative velocity gives the mutual drag reduction of two distant translating spheres:The feedback of the corrected force into the other sphere's flow enters at the next order. If the coefficient $6\pi$ is needed directly from the preceding solution, its pressure anomaly is $p=3\mu Ua\cos \theta /(2R^2)$. At the surface the traction components are $t_R=-3\mu U\cos \theta /(2a)$ and $t_{\theta }=3\mu U\sin \theta /(2a)$; their axial projection is the constant $-3\mu U/(2a)$. Integrating over $4\pi a^2$ gives the single-sphere force $-6\pi \mu aU{\mathbf{e}}_z$, as used above.

Substitution givesRotational symmetry about the translation axis forbids a transverse force. Linearity of the Stokes equations and boundary conditions makes the force linear in $U$; dissipation makes the force on the translating sphere oppose its motion. Thus the drag is axial and proportional to $U$, without computing the stress integral. Dimensional analysis further gives scale $\mu aU$; the exact single-sphere coefficient is $6\pi$.