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The implicit function theorem states that if $F(x_0,y_0)=0$ and the partial derivative $D_{y}F(x_0,y_0)$ is an invertible linear map, then near $(x_0,y_0)$ the zero set of $F$ is uniquely the graph $y=g(x)$ of a continuously differentiable function.

If $f$ is a differentiable bijection with differentiable inverse $g$, the chain rule applied to $g\circ f$ and $f\circ g$ givesThus $D{f}_x$ is an isomorphism with inverse $D{g}_{f(x)}$.

If a continuously differentiable map $F:{\mathbb{R}}^n\to {\mathbb{R}}^n$ has invertible derivative everywhere, the inverse function theorem makes it a local diffeomorphism. In particular it is an open map, so its image is open. Its image need not be closed: $x\mapsto \arctan x$ has nonzero derivative everywhere and image $(-\pi /2,\pi /2)$.

For the given map of elementary symmetric polynomials,and direct evaluation of the determinant givesHence the critical set isIts complement is the Zariski-open set on which the three coordinates are pairwise distinct. Each point has one of the six possible strict coordinate orderings, and each ordering defines a nonempty convex open region. A continuous path cannot change an ordering without crossing $C$. Therefore ${\mathbb{R}}^3\setminus C$ has exactlyconnected components.

A Euclidean domain is an integral domain $R$ equipped with a function $\delta :R\setminus \{0\}\to \mathbb{N}$ such that for $a,b\in R$, $b\ne 0$, there are $q,r\in R$ withFor the Gaussian integers $\mathbb{Z}[i]$, take $\delta (z)=N(z)=|z{|}^2$. Choosing a Gaussian integer nearest to $a/b$ makes the remainder norm smaller than $N(b)$.

The units are precisely the elements of norm one:Unique factorization in this Euclidean domain givesThe displayed factors have prime norms $2$ or $5$, so they are irreducible; factors appearing together are nonassociate.

Now suppose $x^2+4=y^3$. Necessarily $y>0$. First let $y$ be odd. Then $x$ is odd, and $x+2i$ and $x-2i$ are coprime in $\mathbb{Z}[i]$: a common Gaussian prime would divide $4i$, while their product has odd norm. Henceup to a unit, which can be absorbed into the cube. Comparing imaginary parts givesChecking $b\mid 2$ yields only $b=-2$, $a=\pm 1$, and therefore

If $y$ is even, congruence modulo $8$ gives $x=2X$ with $X$ odd and $y=2Y$, whereEach of $X\pm i$ contains exactly one factor $1+i$, so the coprime quotients are cubes up to units. ThusComparing the imaginary part after the four possible units reduces toEach integer factor must have absolute value one. Substitution then gives $a^2+b^2=1$, so $Y=1$ and $X=\pm 1$. HenceAll four pairs satisfy the equation, so the complete answer is

Starting from $x_0\in {\mathbb{R}}^n$, gradient descent repeatedly computes the gradient and updatesstopping when the gradient norm, step, or objective decrease is sufficiently small.

The Hessian bounds say that $f$ is $\alpha$-strongly convex and has $\beta$-smooth gradient. With $\eta =1/\beta$,Thus the iteration count isconvergence becomes slower linearly with the condition number $\kappa =\beta /\alpha$.

Forthe Hessian matrix is $\mathrm{diag}(2,200,20000)$, soTakeThenwhose Hessian is $2I$ and whose condition number is $1$.

The Rao-Blackwell theorem says that if $U$ is an estimator with finite second moment and $T$ is a sufficient statistic, thenhas the same expectation as $U$ and satisfieswith equality only when $U$ is already a function of $T$ almost surely.

Here $\hat{\theta }=1_{\{X_1=1\}}$ is unbiased becauseConditional on $T=t$, every weak composition $(x_1,\dots ,x_n)$ of $t$ has the same probability $p^n(1-p{)}^t$. There aresuch compositions. For $t\ge 1$, those with $x_1=1$ correspond to weak compositions of $t-1$ into $n-1$ parts, of which there areThe Rao-Blackwell estimator is thereforeFor $n\ge 2$ and $0<p<1$, $\hat{\theta }$ is not determined by $T$; for example, conditional on $T=1$, the sole failure can occur in any coordinate. The variance inequality is consequently strict:

Let the columns of $A$ be $a_1,a_2,a_3$. The Gram-Schmidt process givesNext,soFor the final column,andThus $q_3=\frac{1}{2}(-1,-1,1,-1{)}^T$ and $r_{33}=2$. The reduced QR decomposition isThe columns of $Q$ are orthonormal, and direct multiplication gives $QR=A$.

Forthe Euler-Lagrange equation isHere this becomesBecause the forcing is resonant with the complementary solution, a particular integral is $(x/2)\cos x$. HenceandThe Neumann boundary conditions giveThus

The Taylor series of the complex sine isTherefore, for $z\ne 0$,and the right side defines a holomorphic function at $z=0$ with value $1$. Thus the apparent singularity is a removable singularity.

The resulting power series isThe ratio test shows convergence for every $z\in \mathbb{C}$, so its radius of convergence is

A unit-speed curve on a smooth embedded surface is a geodesic exactly when its acceleration is normal to the surface, equivalently when its tangential acceleration vanishes.

Every unit-speed geodesic on the cylinder through $(1,0,0)$ has the formIndeed, unrolling the cylinder to its universal cover ${\mathbb{R}}^2$ turns these curves into straight lines. Directly,which is parallel to the cylinder's radial normal vector, verifying the geodesic characterization. Such a geodesic is closed exactly when $b=0$; its image is then the horizontal circle $z=0$.

Yes. In polar coordinates $(r,\theta )$ on ${\mathbb{R}}^2\setminus \{0\}$, use the Riemannian metricThe coordinate $u=\log r$ identifies this surface isometrically with the flat cylinder $\mathbb{R}\times S^1$. Through each point, the circle $u=\text{constant}$ is a closed geodesic, and every other geodesic has nonzero linear motion in $u$ and is not closed. Thus every point lies on a unique closed geodesic.