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On the nonzero $x$ axis the function equals one, whereas its assigned value at the origin is zero. Therefore $f(x,0)$ does not tend to $f(0,0)$ as $x\to 0$. It is not a continuous function there, and differentiability implies continuity shows

Put $r=(x^2+y^2{)}^{1/2}$. Since $x^2{y}^2\le (x^2+y^2{)}^2/4$, the nonzero-point expression satisfiesThus the defining remainder condition for differentiability holds with the zero linear map:The estimate controls every approach direction at once, which is stronger than merely finding the two zero partial derivatives.

The dual constraints and objective coefficients on $x$ are unchanged, so $y^{\ast }$ remains dual feasible. Since every component of $y^{\ast }$ is positive, complementary slackness forces every constraint of any primal optimum paired with it to bind. Let $A$ be the original three-by-three constraint matrix and $b^{\prime }=b+ϵ$. It is invertible, and the only possible paired primal point isIf this point is nonnegative, it is primal feasible with zero slacks and $c^T{x}^{\prime }=y^{\ast T}{b}^{\prime }$, proving $y^{\ast }$ optimal. Conversely, if $y^{\ast }$ is optimal, linear programming duality supplies a paired primal optimum, and complementary slackness forces it to be exactly this point. This proves the strictly positive dual certificate and right-hand-side sensitivity criterion:Non-strict inequalities are essential: boundary cases can be degenerate but retain the same optimal dual vector. This is the complete range, not just a small-perturbation sufficient condition for linear programming sensitivity within a fixed optimal basis.

The dual linear program isThe negatives of the final slack payoff coefficients give $\boxed{y^{\ast }=(1,2/3,2/3)}$. Its three dual constraints are equalities, and its objective is $11+32/3+58/3=41$. Equality with the primal objective proves optimality by weak duality, independently of tableau sign conventions. It is also the complementary slackness certificate, with all primal variables and all dual variables strictly positive.

Continue the simplex algorithm using dictionaries, with nonbasic variables zero at the current vertex. The given basis has payoffThus $x_2$ enters. Its positive row coefficients give ratios $11$ for $z_1$ and $77/4$ for $z_2$; the $x_1$ row does not restrict this increase. Therefore $z_1$ leaves. After that pivot,Now $x_3$ enters. Only the $z_2$ row decreases along this move, and its ratio is $11/11=1$, so $z_2$ leaves. The final dictionary isAll reduced payoff coefficients are nonpositive. Since the slacks must be nonnegative, $f\le 41$, attained with $z=0$. Hence $\boxed{x^{\ast }=(7,22,1),f^{\ast }=41}$. Direct substitution makes all three original constraints equalities. The final dictionary is the optimal tableau expressed with basic variables on the left, not just a proposed feasible vertex.

The polynomial $p=x^5-4x+2$ satisfies the Eisenstein criterion at the prime two: every nonleading coefficient is even, the leading coefficient is not, and the constant term is not divisible by four. Hence it is irreducible over $\mathbb{Q}$. Its ideal is maximal in $\mathbb{Q}[x]$; explicitly, for any polynomial not divisible by $p$, the Euclidean algorithm gives $ag+bp=1$, furnishing the inverse of its residue class. ThusFor the integer quotient, division by the monic polynomial $p$ gives a unique remainder of degree below five with integer coefficients. If an integer polynomial becomes zero in the rational quotient, its integer remainder is still divisible by $p$ over $\mathbb{Q}$, so must be zero. This gives an injective ring map $\mathbb{Z}[x]/(p)\to \mathbb{Q}[x]/(p)$, proving the monic irreducible integer-polynomial quotient is a domain.

The same unique-remainder argument makes the integer quotient a free abelian group with basis $1,x,x^2,x^3,x^4$. The class of two is nonzero and cannot have an inverse: twice an integer remainder cannot equal the remainder one. Consequently

A nonzero primitive polynomial in $\mathbb{Z}[x]$ has greatest common divisor of its coefficients equal to one. Suppose primitive polynomials $f,g$ had a product whose coefficients were all divisible by some prime $p$. Their reductions in ${\mathbb{F}}_p[x]$ are both nonzero, because neither polynomial has every coefficient divisible by $p$. A polynomial ring over a field is an integral domain, so their product is nonzero modulo $p$, a contradiction. Hence the product of primitive polynomials is primitive.

To deduce unique factorization, write any nonzero integer polynomial as its positive polynomial content times a primitive polynomial, with its sign treated as a unit. The preceding result gives multiplicativity of content. Factor the primitive part over the Euclidean polynomial ring $\mathbb{Q}[x]$, and rescale every irreducible rational factor to a primitive integer polynomial. If $f=q{\prod }_i{f}_i$ with all polynomials primitive and $q=a/b$ in lowest terms, the equality $bf=a{\prod }_i{f}_i$ has contents $b$ and $|a|$. Thus $b=|a|$, forcing $q=\pm 1$. The rational factorization therefore lifts to an integer factorization.

A primitive integer polynomial is irreducible over $\mathbb{Z}$ exactly when it is irreducible over $\mathbb{Q}$: an integer factorization gives a rational one, and a rational factorization lifts by the same primitive-content argument. Factor the content into integer primes. Existence and uniqueness over the integers and over $\mathbb{Q}[x]$ now give existence and uniqueness of the resulting factorization into constant prime factors and primitive irreducible polynomial factors, up to order and signs. Therefore $\boxed{\mathbb{Z}[x]\text{ is a unique factorization domain}}$. This is the full lifting step in Gauss lemma for polynomials.

Use first-kind Chebyshev polynomials $T_j$, characterized by $T_j(\cos \theta )=\cos (j\theta )$. Substituting $x=\cos \theta$ converts the weighted inner product into ${\int }_0^{\pi }g(\cos \theta )h(\cos \theta )d\theta$. Hence these polynomials are orthogonal, with squared norms $\pi$ for $j=0$ and $\pi /2$ for $j\ge 1$.

For the specified function, $f(\cos \theta )=\sin \theta$. Its constant projection coefficient is ${\pi }^{-1}{\int }_0^{\pi }\sin \theta d\theta =2/\pi$. For $j\ge 1$ the coefficient is $(2/\pi ){\int }_0^{\pi }\sin \theta \cos (j\theta )d\theta$. Symmetry about $\pi /2$ makes it zero for odd $j$. For $j=2m$, the product-to-sum identity givesTherefore the Chebyshev projection of a semicircle isIn particular the best degree-at-most-$n$ polynomial is even, so for odd $n$ its actual degree is at most $n-1$.

Take $Q_j$ nonzero of exact degree $j$, so $Q_0,\dots ,Q_n$ are an orthogonal basis of the polynomials of degree at most $n$. Their orthogonal projection isIndeed its residual is orthogonal to every such polynomial. For another polynomial $q=p_n+r$ of degree at most $n$, the squared error issince the cross term vanishes. It is uniquely minimized at $r=0$. The approximation need not have exact degree $n$ when the highest coefficient happens to vanish.

Suppose two distinct minimizers $x_1,x_2$ have the same minimum distance $d$ from $y$. Convexity puts their midpoint $z=(x_1+x_2)/2$ in $A$. Expanding the Euclidean norm givescontradicting minimality. Therefore the nearest point is unique. Together with the preceding existence proof, this defines the metric projection onto a nonempty closed convex set.

Choose $a_0\in A$ and put $R=\Vert a_0-y\Vert$. The intersection $K=A\cap \overline{B}(y,R)$ is nonempty, closed and bounded in ${\mathbb{R}}^n$, hence compact. The continuous distance function $a\mapsto \Vert a-y\Vert$ attains a minimum at some $x\in K$. For $a\in A\setminus K$ its value is strictly greater than $R$, while the value at $x$ is at most that at $a_0$, namely $R$. Thus this minimum on $K$ is a minimum on all of $A$:The set $A$ itself need not be bounded; only the compact portion needed to find the nearest point is bounded.

The cyclic matrix acts by $(v_1,v_2,v_3,v_4)\mapsto (v_2,v_3,v_4,v_1)$. For each fourth root of unity $\lambda$, the vector $v_{\lambda }=(1,\lambda ,{\lambda }^2,{\lambda }^3{)}^T$ satisfies $A{v}_{\lambda }=\lambda v_{\lambda }$, because ${\lambda }^4=1$. This supplies the four distinct eigenvalues and nonzero eigenvectors:Eigenvectors for distinct eigenvalues are linearly independent. Thus, with these columns in the indicated order,They form a basis over $\mathbb{C}$, so there are no other eigenvalues. Equivalently the characteristic polynomial is ${\lambda }^4-1$.

For a complex square matrix, define its determinant byThe cofactor is $C_{ij}=(-1{)}^{i+j}\det A^{(ij)}$, where $A^{(ij)}$ deletes row $i$ and column $j$. The adjugate is the transposed cofactor matrix, $\mathrm{adj}(A{)}_{ij}=C_{ji}$.

The $(i,j)$ entry of $\mathrm{adj}(A)A$ is ${\sum }_k{C}_{ki}{a}_{kj}$. For $j=i$, expansion along column $i$ gives $\det A$. For $j\ne i$, this same sum is the determinant of the matrix obtained by replacing column $i$ by column $j$: deleting the replaced column leaves all its cofactors unchanged. The new matrix has two equal columns, so its determinant is zero. ThusThis proof does not assume invertibility and therefore also covers singular matrices.

A particle pathline follows the time-dependent velocity field, so its coordinates satisfy $\dot{x}-x=\sin t$ and $\dot{y}=-y$. The second equation and the initial data give $y=e^{-t}$. Using the integrating factor $e^{-t}$ in the first,Therefore the complete pathline isIf $x_0+\frac{1}{2}\ne 0$, its first coordinate grows without bound in magnitude. If $x_0=-\frac{1}{2}$, the first coordinate remains bounded and periodic while $y\to 0$. The unique release coordinate giving a bounded particle path is $x_0=-1/2$.

Integrating ${\psi }_y=x+\sin t$ and imposing $-{\psi }_x=-y$ gives the streamfunctionup to an arbitrary function of time. At the specified time, the streamlines areFor $C\ne 0$, the connected branches are rectangular hyperbolas. The zero level consists of the lines $y=0$ and $x=-1$, with the stagnation point $(-1,0)$ at their intersection; their half-lines and the stationary point are the distinct streamlines.

The PDF has the doubly exponential denominator $e^{e^n}$. For its positive summands, the successive-term ratio isThus the ratio test proves absolute convergence. Concretely, the tail is eventually dominated by a geometric series of ratio $1/2$; the polynomial numerator cannot offset the doubly exponential decay.

Let $a_n$ denote the summand. Its denominator is positive for every $n\ge 1$: odd $n$ gives $n^4+n^2$, while even $n\ge 2$ gives $n^4-n^2$. Dividing numerator and denominator by their highest powers givesThe terms are positive for sufficiently large $n$, so the limit comparison test with the divergent harmonic series gives divergence to $+\infty$. More explicitly, for $n\ge 16$ the numerator is at least $n^3/2$ and the denominator is at most $2n^4$, giving $a_n\ge 1/(4n)$. Any finitely many early negative terms change partial sums by only a finite constant and cannot prevent divergence of this positive tail.

For odd $n$, the denominator is $(n+1{)}^2$; for even $n$, it is $(n-1{)}^2$. There is no zero denominator in the printed index range. For all $n\ge 2$,The comparison test for series with the convergent p-series gives absolute convergence. In fact the consecutive pair $n=2k-1,2k$ contributes $1/(2k{)}^2+1/(2k-1{)}^2$, so the sum is the same as ${\sum }_{m=1}^{\infty }{m}^{-2}$. Positivity justifies this paired reordering; an alternating-sign appearance in the denominator does not make this an alternating series.

For $n\ge 2$, $0<n^{-2}\le [n(n-1){]}^{-1}=1/(n-1)-1/n$. The upper bound has telescoping partial sums, bounded by one. The positive partial sums of the original series are consequently bounded by two and increasing, so they converge as a monotone bounded sequence. The series converges absolutely. This also supplies the $p=2$ case of a p-series without requiring its exact sum.

The cubic terms are the expansion of $(z+\overline{z}{)}^3=(2x{)}^3$. Subtracting $\overline{z}=x-iy$ therefore givesHere the expression is a polynomial in $z$ and its complex conjugate, rather than a holomorphic polynomial in $z$ alone.