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The parametric equation of a straight line through $a$ with unit direction $\hat{t}$ isIf the two lines meet, then $a_1+s{\hat{t}}_1=a_2+t{\hat{t}}_2$ for some real $s,t$, so $a_1-a_2$ lies in the linear span of their direction vectors. Its dot product with their cross product is consequently zero:This is not sufficient without excluding parallel lines. For instance $a_1=(0,0,0)$, $a_2=(0,1,0)$ and ${\hat{t}}_1={\hat{t}}_2=(1,0,0)$ satisfy it, but the lines are distinct and never meet. For nonparallel lines the condition is sufficient, because the two directions span the plane perpendicular to their nonzero cross product.

For nonparallel skew lines, put $n=({\hat{t}}_1\times {\hat{t}}_2)/|{\hat{t}}_1\times {\hat{t}}_2|$. Every joining vector is $a_1-a_2+s{\hat{t}}_1-t{\hat{t}}_2$, with fixed component $(a_1-a_2)\cdot n$ along $n$. Its length is at least the absolute value of that component. Its perpendicular component can be canceled by a unique choice of $s,t$, since the two independent directions span $n^{\perp }$. This lower bound is attained, proving the distance between skew lines:

Adding the two expanded unitary matrix identities instead cancels the commutator and gives $2(A^2+B^2)=2I$. HenceTogether with part (i), this characterizes the commuting Hermitian parts of a unitary matrix.

By the unitary matrix identities,Subtracting the two equations yields $2i(AB-BA)=0$. ThereforeThe subtraction is important: entrywise real and imaginary parts cannot generally be used to separate these terms, since the Hermitian matrices need not have real entries.

For $|z|=r<1$, the inequality $n!\ge n$ for $n\ge 1$ givesThe first $r$ comes from $0!=1$; both initial terms have exponent one. Thus the factorial-gap power series converges absolutely inside the unit circle. For $|z|>1$, its terms have magnitudes $|z{|}^{n!}\to \infty$, so the term test for divergence proves divergence and $R=1$.

At every point with $|z|=1$, each term has magnitude one, so the terms cannot tend to zero, regardless of their phases. HenceThe large gaps between exponents make this a lacunary power series; cancellation cannot overcome the necessary condition that the terms of a convergent series tend to zero.

For $z\ne 0$, the ratio of successive absolute terms isThe ratio test therefore gives absolute convergence for $|z|<1$ and divergence for $|z|>1$, while at zero the series is trivially convergent. Hence $R=1$. On the boundary, $|z^n/n^2|=1/n^2$, and the convergent p-series proves

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.