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Measure displacement downward from the release point and take downward speed $v\ge 0$. Newton's second law with quadratic drag givesWritingseparation of variables gives the terminal velocity solutionThe distance fallen by time $t$ is thereforeSetting $y(t)=h$ and solving for $t$ yields

During a short interval $dt$, let the rocket's mass change by $dm<0$ and its speed by $dv$. The expelled mass $-dm$ has inertial-frame speed $v-u$ to first order. Conservation of linear momentum, including the impulse $Fdt$, givesDiscarding the second-order product $dmdv$ leavesand hence the Tsiolkovsky rocket equation

In deep space $F=0$, so $dv=-udm/m$. Burning from initial mass $m_0$ to $m_0/2$ increases the speed by

The first series is telescoping, sinceIts $N$th partial sum is $1-1/(N+1)$, soa rational number.

The second series converges absolutely, for example by the ratio test, and is the odd part of the exponential series:This value is irrational. Indeed, if $a=\sinh 1$ were algebraic, then $e$ would satisfymaking $e$ algebraic over the algebraic numbers and hence algebraic, contrary to the Hermite theorem on the transcendence of e. Thus the second sum is in fact transcendental.

Since $4^{-1}\equiv 16(\mathrm{mod}21)$ and $2^{-1}\equiv 23(\mathrm{mod}45)$, the two linear congruences reduce toWrite $x=16+21k$. The second congruence becomesor $7k\equiv 3(\mathrm{mod}15)$ after division by three. Since $7^{-1}\equiv 13(\mathrm{mod}15)$, $k\equiv 9(\mathrm{mod}15)$. The Chinese remainder theorem therefore gives the complete familyThe modulus is $\mathrm{lcm}(21,45)=315$ because the compatibility condition modulo their greatest common divisor is satisfied.

For the one-to-one transformation $(u,v,w)\mapsto (x,y,z)$, the Jacobian determinant used in the stated formula isLocally, the derivative maps a small rectangular box of volume $dudvdw$ to a parallelepiped of volume $|J|dudvdw$. Partitioning the region into such boxes and passing to the Riemann integral gives the change of variables formula

In the plane $y=0$, the region is inside the unit circle and on the right-hand side of the hyperbolaIt is nonempty exactly when $\alpha \le 1$; it has positive area when $\alpha <1$. In $x>0$, its boundaries are $x=\sqrt{1-z^2}$ and $x=\alpha \sqrt{1+z^2/{\gamma }^2}$. The minimum and maximum $x$ values are $\alpha$ and $1$, both at $z=0$. The largest absolute $z$ occurs where the boundaries meet:

Now set $u=x$, $v=y/k$, and $w=z$. The inverse Jacobian determinant is $k$, and in cylindrical coordinates $r^2=u^2+v^2$ the transformed region isThus $|w|\le z_0$, and each horizontal cross-section is an annulus. Its volume isTherefore, for $\alpha <1$,For $\alpha =1$ the region has zero volume.

Represent a Möbius transformation by a matrix, with nonzero scalar multiples representing the same map. Matrix multiplication reproduces composition, so $\theta$ is a group homomorphism. Every Möbius map has a representing matrix $A\in G{L}_2(\mathbb{C})$; multiplying $A$ by either square root of $(\det A{)}^{-1}$ gives a matrix in $S{L}_2(\mathbb{C})$ representing the same map. Thus $\theta$ is surjective. A matrix in its kernel represents the identity and is therefore scalar; determinant one leaves

If a nonidentity Möbius map $T$ has two distinct fixed points, conjugate those points to $0$ and $\infty$. The conjugated map fixes both and hence has the form $S(z)=\mu z$ with $\mu \ne 0,1$. If $T$ has one fixed point, conjugate it to $\infty$. The resulting affine map has the form $az+b$; having no finite fixed point forces $a=1$ and $b\ne 0$, and conjugating by a scaling makes it $S(z)=z+1$.

Directly, the finite fixed points satisfyIncluding $\infty$ when appropriate, a nonidentity Möbius map consequently has one or two distinct fixed points.

Finally suppose $T$ has the unique fixed point $z_0$. With the conjugacy $R$ above,Every finite point tends to $\infty$ on the Riemann sphere, while $\infty$ itself is fixed. Applying the continuous map $R^{-1}$ shows thatfor every $z\in \mathbb{C}\cup \{\infty \}$.

The first isomorphism theorem for groups states that for a group homomorphism $f:G\to Q$,The map $g\ker f\mapsto f(g)$ is well defined because two representatives differ by an element of the kernel; it is a surjective homomorphism onto the image and is injective because $f(g)=e$ exactly when $g\in \ker f$.

For $n_i\in N$ and $h_i\in H$, normality of $N$ givesand $(nh{)}^{-1}=h^{-1}{n}^{-1}h{h}^{-1}\in NH$. Hence $NH$ is a subgroup. Also $N\cap H⊴H$, since conjugation by $H$ preserves both $N$ and $H$, and $N⊴NH$ because $NH\le G$.

Apply the first isomorphism theorem to $f:H\to NH/N$, $f(h)=hN$. Its kernel is $N\cap H$, and it is surjective because $nhN=hN$. Therefore

If $K,H⊴G$, then $KH$ is a normal subgroup: it is a subgroup by the calculation above, and $g(KH)g^{-1}=(gK{g}^{-1})(gH{g}^{-1})=KH$. Without normality a product need not be a subgroup. In $S_3$, take $H=⟨(12)⟩$ and $K=⟨(23)⟩$. The set $KH$ has four elements, so it cannot be a subgroup of the six-element group by Lagrange's theorem for finite groups.