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The Fibonacci recurrence givesThe map $x\mapsto 1+1/x$ is strictly decreasing for $x>0$. Since $a_3=2\ge a_1=1$, applying this decreasing map reverses each inequality and proves by mathematical induction thatTaking even $n$ shows $a_{2n+2}\le a_{2n}$, so $(a_{2n})$ is a decreasing sequence.

This can fail in the same example as in part (i): $g\circ f$ is injective on the singleton $X$, but $g(a)=g(b)$.

For the final count, $g\circ f={\mathrm{id}}_X$ forces $f$ to be injective. There aresuch functions $f$. Once $f$ is chosen, $g$ is forced on the $m$ elements of $f(X)$ and has $m$ independent choices at each of the other $n-m$ elements of $Y$. Hence the number of pairs is

This is always true. If $f(x_1)=f(x_2)$, thenThe assumed injectivity of $g\circ f$ gives $x_1=x_2$, so $f$ is injective.

This can fail. Let $X=\{1\}$ and $Y=\{a,b\}$, put $f(1)=a$, and let $g(a)=g(b)=1$. Then the function composition $g\circ f$ is the identity function on $X$, hence is a surjective function, although $f$ is not surjective.

The inertial acceleration obtained by differentiating the position used in part (i) isSince $N=m\ddot{r}+mg{e}_z$, substitution of the equation of motion gives the wire's force on the bead asThe first component is normal to the circle within its plane; the second is perpendicular to the rotating plane.

The downward and upward positions $\theta =0,\pi$ are equilibria. Linearization about $\theta =0$ givesso the downward equilibrium is stable for ${\omega }^2<g/R$ and unstable for ${\omega }^2>g/R$; at equality it is marginal at linear order. At $\theta =\pi$, the linear coefficient is ${\omega }^2+g/R>0$, so the upward equilibrium is unstable.

When ${\omega }^2\ge g/R$, there are also the two equilibriaAt either nonvertical equilibrium, differentiation of the right-hand side gives $-{\omega }^2{\sin }^2\theta <0$, so both are stable when distinct. This is a pitchfork bifurcation of a rotating hoop bead at ${\omega }^2=g/R$.

Let $e_i^{\prime }(t)$ be a Cartesian basis fixed in the rotating frame. A vector $A=A_i^{\prime }{e}_i^{\prime }$ then satisfiesbecause a derivative of a body-fixed basis vector is ${\dot{e}}_i^{\prime }=\omega \times e_i^{\prime }$. Applying this transport formula twice to $r$ gives, for constant $\omega$,The second and third terms correspond to Coriolis acceleration and centrifugal acceleration in the rotating description.

For the bead, use cylindrical unit vectors $(e_r,e_{\varphi },e_z)$ with $e_z$ upward andProjection of $m\ddot{r}=-mg{e}_z+N$ along the wire tangent $\cos \theta e_r+\sin \theta e_z$ eliminates the smooth-wire normal force $N$ and gives

Motion under a central force is planar. In plane polar coordinates, the radial component of the acceleration is $\ddot{r}-r{\dot{\theta }}^2$, soThe conserved specific angular momentum is $h=r^2\dot{\theta }$. Substitution of $\dot{\theta }=h/r^2$ yieldsThe last term is the centrifugal part of the radial effective potential.

The torque about the origin isSince $\dot{L}=\tau$, the angular momentum $L=mr\times \dot{r}$ is conserved. The force is the negative gradient of the inverse-square potentialso the conserved energy isDirectly, $\dot{E}=\dot{r}\cdot (m\ddot{r}+kmr/r^3)=0$.

For $n=(a,b,c)$ and $r=(x,y,z)$, the integrand is $\frac{1}{2}(n\times r)\cdot dr$. Sincethe Stokes theorem gives the signed planar area, and its absolute value gives the ordinary area.

The $xy$ projection of the given curve is the astroid $x=2{\cos }^{3}t$, $y={\sin }^{3}t$, whose area isThe plane $z=\sqrt{3}y$ stretches area from its $xy$ projection by $\sqrt{1+3}=2$. Hence

Differentiation givesand $|r^{\prime }|=6|\sin t\cos t|$. ThusThe plus sign applies on $(0,\pi /2)\cup (\pi ,3\pi /2)$ and the minus sign on $(\pi /2,\pi )\cup (3\pi /2,2\pi )$.

On each open quadrant,The constant binormal proves that the curve is planar; indeed $z=\sqrt{3}y$.

For a regular parametrized curve $r(t)$, the Frenet frame isEquivalently, wherever $dT/dt\ne 0$, one may write $N=(dT/dt)/|dT/dt|$ and $B=T\times N$.

The standard isomorphisms areand

For $z\in B$, let $R_z$ be the corresponding rotation. The identityshows thatrespects the multiplication of the generalized dihedral group. The decomposition in part (b) makes it bijective. Hence

If $Q\in O(2)$, then $Q^{T}Q=I$, soand $\det Q=\pm 1$. If $\det Q=1$, its first column is $(\cos \theta ,\sin \theta {)}^T$ for some $\theta$, and orthonormality plus positive orientation forceThus every element of $SO(2)$ is a rotation.

Let $J=\mathrm{diag}(1,-1)$. If $\det Q=-1$, then $QJ\in SO(2)$ and $Q=(QJ)J\in SO(2)J$. Conversely every matrix in $SO(2)J$ has determinant $-1$. Therefore the union is disjoint and

Choose $h\in H\setminus \mathrm{Stab}(x)$ and put $y=hx\ne x$. For any $z\ne x$, double transitivity supplies an element $s\in G$ taking the ordered pair $(x,y)$ to $(x,z)$. Since $sx=x$, one hasThen $sh\in H$ sends $x$ to $z$. Together with the identity for $z=x$, this proves that $H$ acts transitively on $X$.

Now let $g\in G$. Transitivity of $H$ gives $h_0\in H$ with $h_{0}x=gx$. Henceso $g\in H$. ThereforeThis is the point-stabilizer maximality property of a doubly transitive group action.