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Put $R=R_A+R_B$, $c=\sqrt{R^2-b^2}/R$, and $s=b/R$. At impact the unit vector from A to B is $n=(-c,s)$. The impulse is normal, so tangential velocity is unchanged, while the normal components undergo a one-dimensional elastic collision. Therefore

Newton's equation and its energy integral areThusLet $a=-V^{\prime \prime }(x_{+}^{\ast })>0$. Near the maximum,and the nearby turning point has distance $ϵ\sim \sqrt{2E^{\ast }\delta /a}$ from $x_{+}^{\ast }$. The singular part of the period is an $\mathrm{arcosh}$ integral:HencePhysically, the limiting orbit approaches the unstable equilibrium at the barrier top with vanishing speed and spends arbitrarily long there.

The series $\sum a_n$ converges when its partial sums tend to a finite real limit. Put $m=n+1$. Thenby strict concavity of the square root. Taylor expansion at $m$ givesso $m^2{a}_n\to 1/8$. Limit comparison with $\sum n^{-2}$ proves convergence.

With $\gamma =(1-v^2/c^2{)}^{-1/2}$,Differentiation with respect to proper time givesOrthogonality $U\cdot A=0$ implies that in the instantaneous rest frame $A^{\prime \mu }=(0,a^{\prime })$. For collinear motion a Lorentz boost givesFor constant positive proper acceleration $a^{\prime }$,This hyperbola has future null asymptote $x=ct-c^2/a^{\prime }$. A right-moving signal emitted from $x_0$ at $t=0$ follows $x=ct+x_0$; if $x_0<-c^2/a^{\prime }$, it lies beyond that acceleration horizon and never intersects the worldline.

For axis unit vector $\hat{\omega }$,For a body, replace the sum by $I=\int \rho (x)r_{\perp }^{2}dV$.

For the hollow ball,Energy conservation and $v=R_{+}\Omega$ giveThe factor $\beta$ increases with $\mu$, so if ${\mu }_A>{\mu }_B$, ball B accelerates more and arrives first. Moving the cavity off centre removes material with a larger lever arm and lowers the moment of inertia about the centre of mass; despite the resulting wobble, ball C therefore rolls down faster than A.

Binet's equation givesConsequently

Invertible matrices are closed under multiplication, contain $I$, have associative multiplication, and have inverses by the adjugate formula over the field ${\mathbb{F}}_p$. The first column can be any nonzero vector and the second any vector outside its span, so

For $p=2$, the matrix $\left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right)$ has order three. There is no element of order six: $G{L}_2({\mathbb{F}}_2)\cong S_3$, whose element orders are $1,2,3$.

For $p>2$, $S{L}_2({\mathbb{F}}_p)=\ker \det$ is a proper normal subgroup. For $p=2$, the order-three subgroup in the copy of $S_3$ is proper and normal.

In $G{L}_2({\mathbb{F}}_{11})$, takewhere $A\le {\mathbb{F}}_{11}^{\times }$ is the subgroup of order five. This semidirect product has order $55$ and is nonabelian because a nontrivial diagonal element does not commute with translations.

For $f(z)=(az+b)/(cz+d)$, finite fixed points obeywith the point at infinity included in the usual way when appropriate. The fundamental theorem of algebra on the Riemann sphere gives at least one fixed point. Unless $f$ is the identity, the equation is nonzero of degree at most two, so there are one or two distinct fixed points.

Let $\zeta$ be a primitive $m$th root of unity. For every $u\in \mathbb{C}$,has order $m$, and these transformations are distinct as $u$ varies.

Projectivising matrices is a homomorphism. Thus $B=CA{C}^{-1}$ in $S{L}_2(\mathbb{C})$ implies $g=[C]f[C{]}^{-1}$ in the Möbius group.

The converse as stated is false because matrix representatives may be rescaled: $A=I$ and $B=2I$ define the same Möbius transformation, while neither $B$ nor $-B$ is conjugate to $A$ (their traces differ).