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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.

For $z\ne 0$, rationalizing the denominators gives $1/z=(x-iy)/(x^2+y^2)$ and $1/\overline{z}=(x+iy)/(x^2+y^2)$. Their imaginary parts cancel, soThis also follows because the summands are complex conjugates.

For $z\ne 0$, write $z=r{e}^{i\theta }$ with $r=\sqrt{x^2+y^2}>0$. The complex logarithm has values $\ln r+i(\theta +2\pi k)$, $k\in \mathbb{Z}$. ThereforeFor a specified single-valued branch, choose the corresponding continuous determination of the argument. In particular, the principal value uses $\mathrm{Arg}z\in (-\pi ,\pi ]$; it is not represented correctly by $\arctan (y/x)$ in every quadrant. The complex logarithm is undefined at zero.

Using $\cos z=(e^{iz}+e^{-iz})/2$ and collecting real and imaginary terms gives the complex cosine identity $\cos (x+iy)=\cos x\cosh y-i\sin x\sinh y$. ThusThe sign follows from the factors $e^{-y}$ and $e^y$ in the two complex exponentials.

The complex exponential function factorizes as $e^{x+iy}=e^x{e}^{iy}=e^x(\cos y+i\sin y)$. HenceThe real exponential multiplies both components; it does not change the argument.

Interchange the two dummy indices $j,k$. The product $a_j{a}_k$ is symmetric under this interchange, while the Levi-Civita symbol changes sign. Thus the contraction equals its negative and is the zero vector, equivalently $\mathbf{a}\times \mathbf{a}=0$.

Compare with $(\mathbf{b}\times \mathbf{a}{)}_i={ϵ}_{ijk}{b}_j{a}_k$. The scalar factors commute, so the answer isThe order of the contracted indices matters because the Levi-Civita symbol is antisymmetric.

The contracted Kronecker delta is ${\delta }_{ll}=3$. The other scalar contraction is $a_i{C}_{ij}{b}_j={\mathbf{a}}^{T}C\mathbf{b}$, while $C_{ik}{d}_i$ is the $k$th component of the matrix transpose applied to $\mathbf{d}$. Hence the resulting vector isBoth terms have the same single free index $k$, as required by the Einstein summation convention.

This is the incorrectly formed expression. In the Einstein summation convention, an index in a term is either free and occurs once, or summed and occurs twice. Here $i$ occurs three times in the same product, so it cannot consistently be classified as either. Renaming a dummy index before multiplying would give a valid different expression, such as ${\delta }_{kk}{\delta }_{ij}=3{\delta }_{ij}$, but does not make the printed expression valid.

The two free indices are the row and column indices of a matrix. The Kronecker delta has value one on its diagonal and zero off it, so the answer is the $3\times 3$ identity matrix $I$.

Suppose an analytic extension $F$ existed on the punctured plane. The identity theorem would extend $F^2=z$ from $D$ to that connected domain, so $F$ would never vanish and differentiation would give $2F^{\prime }/F=1/z$. Integrate around the positively oriented unit circle. The argument principle says $\int F^{\prime }/F$ is $2\pi i$ times the integer winding number of the closed curve $F(e^{it})$ about zero, whereasThis would make that winding number $1/2$, a contradiction. Hence no such analytic extension exists. Equivalently, continuation of a square root once around zero reverses its sign, preventing a single-valued holomorphic function on the punctured plane.

The slit plane is connected and contains $D$. If two holomorphic functions on it extend the same $f$, their difference vanishes on the nonempty open disc $D$. The identity theorem forces their difference to vanish on the entire slit plane. Thus there is at most one analytic extension of the chosen $f$. Choosing the opposite square root initially changes that chosen function; it does not produce two extensions of the same function.

The disc lies in the upper-right half-plane and avoids zero and the negative real axis. On the slit plane take the holomorphic logarithm $\mathrm{Log}z=\log |z|+i\arg z$ with $-\pi <\arg z<\pi$. Its derivative is $1/z$, and $e^{\mathrm{Log}z}=z$. Therefore the holomorphic square rootis analytic and satisfies $f^2=z$. One can also construct it without a global branch theorem: fix either scalar square root $a$ of $3+3i$ and expand $a(1+(z-3-3i)/(3+3i){)}^{1/2}$ by its convergent binomial series. The series converges on the entire disc because $|z-3-3i|/|3+3i|<1/\sqrt{18}$. Either sign gives an admissible choice.

Let $P(x)$ denote the polynomial under consideration. The change $x\mapsto x+1$ is an automorphism of $\mathbb{Q}[x]$ and preserves degrees, so it preserves reducibility. Its transformed polynomial isIt is monic; every other coefficient is divisible by $p$, because $p$ is prime and $1\le j<p$; and its constant coefficient is $p$, not divisible by $p^2$. The Eisenstein criterion proves that $P(x+1)$ is irreducible, and therefore $P$ is irreducible. The argument also includes $p=2$, when the polynomial is linear. This is the prime-index case of an irreducible polynomial arising as a cyclotomic polynomial.