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A cube root of a matrix $A$ is a square matrix $B$ of the same size over the specified field satisfying $B^3=A$. Every complex invertible matrix has one: put it in Jordan normal form, choose a scalar cube root of each nonzero eigenvalue, and apply the truncated binomial cube root of the identity plus a nilpotent matrix on each block. The Nilpotent Jordan block $J_3(0)$ has none, even though its square is nonzero. If $B^3=J_3(0)$, then $B^9=0$, making $B$ a nilpotent matrix of size three; the Cayley-Hamilton theorem then gives $B^3=0$, a contradiction.

Use the compatible open semicircles and set $r=\sqrt{x^2+y^2}$. The characteristic curve through $(\cos \theta ,\sin \theta ,-1)$ hasThe condition $s<1$ is essential: at $s=1$ there is finite-time blowup, so the opposite branch cannot be reached from the data. From $r=(1-s\sin \theta {)}^{-1}$ and $y=r\sin \theta$,Eliminating $s$ yieldsThe ratio inside the cube root is positive precisely on the two data-determined regionsOn $D_{+}$ each point traces to the upper semicircle, and on $D_{-}$ to the lower semicircle, with $s<1$. To check the parameter ranges explicitly, for $\eta =\sin \theta >0$ they are $-\infty <s<1$, whereas for $\eta <0$ they are $1/\eta <s<1$. At the latter lower endpoint the planar projection escapes to infinity. The inverse function theorem and uniqueness for the characteristic ordinary differential equations give a unique smooth solution on each region. Equivalently, each point has the unique foot $(x/r,y/r)$ and parameter $(r-1)/y$.

A direct check is particularly short. For $s=(r-1)/y$, let $\mathcal{D}=xy{\partial }_x+y^2{\partial }_y$. Then $\mathcal{D}r=yr$, $\mathcal{D}y=y^2$, andSince $dz/ds=-z^4/3$, the displayed formula solves the partial differential equation, and $r=1$ gives $z=-1$ on both open semicircles. The finite boundary where $r=1+y$, away from the axis, is the parabola $y=(x^2-1)/2$ and corresponds to blowup. Points on $y=0$ cannot be added to a classical continuation of these regions: where they are approached from a data-determined region, the height behaves like a nonzero multiple of $|y{|}^{1/3}$, or has incompatible limits at the endpoints. The formula is therefore the complete classical answer for the compatible arcs, while the full-circle problem as printed has no classical solution.

A characteristic curve in $(x,y,z)$ follows the vector field defining the first-order partial differential equation. Along such a curve the chain rule turns the equation intoIts characteristic projection is the curve $(x(s),y(s))$ in the independent-variable plane.

Writing $x(0)=\xi$, $y(0)=\eta$, $z(0)=\zeta$, separation of variables givesThese formulas hold on the maximal interval through $s=0$ where the denominators remain nonzero, using the real cube root. They include $\eta =0$ (a stationary characteristic projection) and $\zeta =0$ (the zero height). For $\zeta \ne 0$, the last formula follows from $d(z^{-3})/ds=1$. The alternative zero solution must be retained when deriving this by division by $z^4$.

We construct a cube root of a matrix on each Jordan block. Since $A$ is an invertible matrix over the complex numbers, its Jordan normal form isConsider one block $J_m(\lambda )=\lambda I+N$, where $N=J_m(0)$ is a Nilpotent Jordan block and $N^m=0$. Choose a complex cube root $\beta$ of $\lambda$, so ${\beta }^3=\lambda$.

We need a polynomial $q(z)$ with constant term one satisfyingThis can be constructed using only polynomial multiplication. Write $q(z)=1+{\sum }_{j=1}^{m-1}{c}_j{z}^j$. Once $c_1,\dots ,c_{j-1}$ have been chosen, the coefficient of $z^j$ in $q(z{)}^3$ is $3c_j$ plus an expression depending only on those earlier coefficients. Choose $c_j$ to make this coefficient one at $j=1$ and zero at $2\le j<m$. Division by three is possible over the complex numbers, so the recursion always succeeds. Its first coefficients areThis is the truncated binomial cube root of the identity plus a nilpotent matrix; equivalently $c_j=\left(\genfrac{}{}{0ex}{}{1/3}{j}\right)$, where the generalized binomial coefficient is $\alpha (\alpha -1)\cdots (\alpha -j+1)/j!$ and the empty product is one.

Substitute $z=N/\lambda$. The polynomial remainder is a multiple of $N^m$, hence zero. It follows thatFor $m=1$, take $q=1$, so the same construction covers single-entry blocks. Assemble the block diagonal matrix $C={⨁}_r{C}_{m_r}({\lambda }_r)$. Taking matrix powers respects these blocks, and conjugating back givesThus every invertible matrix over the complex numbers has at least one cube root of a matrix. The construction uses only finitely many matrix powers.

Write $27i=27e^{i(\pi /2+2\pi k)}$. Every cube root has complex modulus $3$ and an angle one third of one of these complex arguments:Equivalently the three complex roots areThey lie on the circle of radius $3$ centred at the origin in the Argand plane, spaced by $2\pi /3$. Further integer values of $k$ repeat these three points. Cubing any of them multiplies its angle by three and cubes its complex modulus from $3$ to $27$, giving $27i$.

A stationary point is a point where the function is differentiable and its first derivative vanishes. Interpret the cube root as the real cube root, and put $q(x)=x^2-8x+7$. Then $f(x)=15+(x-3)\sqrt[3]{q(x)}$. For $x\ne 1,7$, the product rule and chain rule giveThe denominator is positive wherever it is defined. Consequently the two stationary abscissae areUsing the stationary equation to simplify the value of $q$ gives $q(x_{\pm })=-2(x_{\pm }+5)/5$. Thus the complete stationary points areNumerically they are approximately $(1.46762,17.10362)$ and $(6.13238,9.84664)$. The numerator of $f^{\prime }$ is positive outside $[x_{-},x_{+}]$ and negative inside, so the first is a local maximum and the second a local minimum.

At $x=1,7$, the real function is continuous but has a vertical tangent; the factor $x-3$ does not cancel either cube-root singularity. Those points are not stationary points. This is the distinction required when finding stationary points with real cube-root factors.

If $L<1$, set $\epsilon =(1-L)/2>0$. The definition of limit of a sequence gives $a_n<L+\epsilon =(1+L)/2<1$ for all sufficiently large $n$, contradicting $a_n\ge 1$. Hence $\boxed{L\ge 1}$.

Factor the difference of cubes. Since both cube roots are at least one,Given $\epsilon >0$, the definition of limit of a sequence supplies $N$ with $|a_n-L|<3\epsilon$ for $n\ge N$. The displayed inequality then gives $|a_n^{1/3}-L^{1/3}|<\epsilon$, proving directly that $\boxed{a_n^{1/3}\to L^{1/3}}$.

Where $q(x)\ne 0$, the product rule and chain rule give the displayed real derivative. Its stationary points therefore satisfy $3q+(x-a)q^{\prime }=0$. The denominator uses the square of the real cube root, which stays positive for negative $q$. Points with $q=0$ must be checked in the original function: a simple zero of $q$ with $x\ne a$ creates a vertical tangent, not a stationary point. If the prefactor also vanishes, a cancellation can change differentiability and must be analyzed separately.

Over the complex numbers, if $N^m=0$, thenHere the generalized binomial coefficient is $\left(\genfrac{}{}{0ex}{}{\alpha }{j}\right)=\alpha (\alpha -1)\cdots (\alpha -j+1)/j!$, with $\left(\genfrac{}{}{0ex}{}{\alpha }{0}\right)=1$. This is an exact finite identity, even when an infinite binomial series would not converge.

One can construct the required polynomial $q(z)=1+{\sum }_{j=1}^{m-1}{c}_j{z}^j$ without any infinite series. At polynomial degree $j$, the coefficient of $q(z{)}^3$ is $3c_j$ plus terms involving only earlier coefficients. Recursively choose it to agree with $1+z$ through polynomial degree $m-1$; the first coefficients are $1/3,-1/9,5/81$. The stated generalized binomial coefficients solve this recursion: near $z=0$, the binomial series for $(1+z{)}^{1/3}$ cubes to $1+z$, and truncating it changes no coefficient below polynomial degree $m$. Thus $q(z{)}^3-(1+z)$ is divisible by $z^m$ as a polynomial identity. Substituting the nilpotent matrix $N$ kills the remaining terms, all divisible by $N^m$. For a Jordan block $\lambda I+N$ with $\lambda \ne 0$, replace $N$ by $N/\lambda$ and multiply $Q$ by a scalar cube root of $\lambda$.