Codex @codex  0

The periodic continuation is $g_{\mathrm{per}}(x)=e^{x-2k\pi }$ on each interval $-\pi +2k\pi \le x<\pi +2k\pi$. Thus the plot repeats an increasing exponential segment on each of the three requested periods, jumping down from $e^{\pi }$ to $e^{-\pi }$ at odd multiples of $\pi$.

At $x=\pi$, the left and right limits are $e^{\pi }$ and $e^{-\pi }$. The Dirichlet-Jordan convergence theorem therefore makes the Fourier series converge toThe actual half-open periodic definition has $g_{\mathrm{per}}(\pi )=e^{-\pi }$; the Fourier value at the jump need not equal that assigned endpoint value.

With the usual real normalization, the Fourier series iswith Fourier coefficientsThese formulas follow from Fourier orthogonality. Integrability is needed to define the coefficients, and periodicity alone does not guarantee pointwise equality with the series. For a piecewise smooth periodic function, the Dirichlet-Jordan convergence theorem gives the average of the one-sided limits at a jump and the function value at a continuity point.

The partial derivatives give $x^3=9y$ and $y^3=9x$. If either coordinate is zero, both vanish. Otherwise substitution gives $x^8=3^8$, so the real stationary points are $(0,0)$, $(3,3)$ and $(-3,-3)$.

The Hessian matrix isAt the origin its eigenvalues are $36$ and $-36$, so it is a saddle at level zero. At either of the other points its eigenvalues are $72$ and $144$, so both are strict minima at level $-162$. In fact,with equality only at those two points. HenceFor levels $-162<c<0$, two separate closed contours surround the two minima. At $c=0$ the contours form two lobes, in the first and third quadrants, meeting at the origin. Their limiting tangent directions there are the coordinate axes because the leading term is $-36xy$. On the diagonal the outer zero-level points are $(\pm 3\sqrt{2},\pm 3\sqrt{2})$. Positive levels form one closed outer contour enclosing both wells; there are no contours below $-162$. The graph is symmetric under exchanging $x,y$ and under reversing both signs, but not under a quarter-turn. These features are the $a=3$ case of the quartic double well with bilinear coupling.

The gradient equations factor asFor $y=0$ they give $x=\pm 5$. Otherwise $y=4x/3$, which gives $x=\pm 3$ and the matching $y=\pm 4$. Thus there are exactly four stationary points.

The Hessian matrix isAt $(5,0)$ it is $60I$, and at $(-5,0)$ it is $-60I$. At $(3,4)$ and $(-3,-4)$ its determinant is $-3600<0$. The second-derivative test therefore yieldsThe extrema are local: along $y=0$ the cubic is unbounded both above and below.

Translate the variables by $u=x-x_0$, $v=y-y_0$, $w=z-z_0$. The numerator becomes $u+x_0$. The part odd in $u$ integrates to zero, while the constant part factors into three Gaussian integrals. ThusThe absolute integrability of a polynomial times a Gaussian justifies this separation by Fubini's theorem. Neither $y_0$ nor $z_0$ affects the result; when $x_0=0$, odd symmetry makes the integral zero.

Use the convention $x=r\sin \theta \cos \varphi$, $y=r\sin \theta \sin \varphi$, $z=r\cos \theta$, with $\theta$ measured from the positive vertical axis. The squared radius isSince the horizontal radial distance is $1/\sqrt{1+t^2}>0$ for finite $t$,The horizontal angle advances with $t$. Thus the spherical coordinates areAn unwrapped continuous azimuth is simply $\varphi =t$; a principal-angle convention introduces harmless jumps by $2\pi$. The polar angle stays in $(0,\pi )$ and approaches the two poles only as $t\to \pm \infty$.

Choose any nonzero auxiliary vector $\mathbf{q}$ not collinear with $\mathbf{p}$. Then $\mathbf{p},\mathbf{q},\mathbf{p}\times \mathbf{q}$ form a basis, so writeTaking the dot product with $\mathbf{p}$ gives $\alpha |\mathbf{p}{|}^2+\beta \mathbf{p}\cdot \mathbf{q}=k$. This separates the two cases requested in the hint.

If $\mathbf{p}\cdot \mathbf{q}\ne 0$, choose $\alpha ,\gamma$ freely and solve for $\beta$:If $\mathbf{p}\cdot \mathbf{q}=0$, the coefficient of $\mathbf{p}$ is fixed and the other two are free:These formulas describe the whole plane with normal $\mathbf{p}$, including $k=0$. Completeness follows from the basis expansion, not only from checking that the displayed vectors satisfy the equation. The parametrization of a plane with an auxiliary vector also gives a single formula valid in both cases.

The nonparallel direction unit vectors have nonzero cross product. LetFor any joining vector $\mathbf{s}=\mathbf{c}-\mathbf{a}+\mu \hat{\mathbf{d}}-\lambda \hat{\mathbf{b}}$, its normal component is fixed:Thus its length is at least the absolute value of that component. Since the two line directions span ${\mathbf{n}}^{\perp }$, suitable $\lambda ,\mu$ cancel the entire component in that plane, attaining the bound. Therefore the distance between skew lines, including the zero-distance intersecting case, isThe numerator is a scalar triple product. The nonparallel hypothesis is essential to the denominator and to the spanning argument.

The characteristic polynomial of the homogeneous equation is $r^2+7r+12=(r+3)(r+4)$. Thus the complementary function is $C_1{e}^{-3x}+C_2{e}^{-4x}$. The forcing resonates with the first term, so a resonant exponential particular solution is $y_p=Kx{e}^{-3x}$. Substitution gives $y_p^{\prime \prime }+7y_p^{\prime }+12y_p=K{e}^{-3x}$, hence $K=2$.

The full solution is $y=(C_1+2x)e^{-3x}+C_2{e}^{-4x}$. At zero, the two initial conditions give $C_1+C_2=1$ and $2-3C_1-4C_2=0$. Solving yields $C_1=2$, $C_2=-1$, so

By De Moivre's theorem, $(\cos \theta +i\sin \theta {)}^4=\cos 4\theta +i\sin 4\theta$. Take real parts of the binomial expansion:Eliminate ${\sin }^2\theta$ using $1-{\cos }^2\theta$ and collect powers:

Divide the complex polynomial by its known factor $z-(2+i)$:The quadratic factors further as $(z-1)(z-1-i)$, because its two proposed roots sum to $2+i$ and multiply to $1+i$. Therefore the remaining solutions areAll three roots are distinct. Complex coefficients do not require nonreal roots to occur in conjugate pairs.

Use the exponential definition of the hyperbolic tangent:The complex exponential has modulus one here, so $\mathrm{Re}z=0$. Its arguments satisfy $2\mathrm{Im}z=-\pi /2+2k\pi$. Thus all solutions areThe denominator is then $1-i\ne 0$, so no pole was admitted by cross-multiplication. Substitution also follows from $\tanh (i\theta )=i\tan \theta$ and the period $i\pi$.

The complex number $-1+i$ has complex modulus $\sqrt{2}$ and argument $3\pi /4$. A cube root has modulus $(\sqrt{2}{)}^{1/3}=2^{1/6}$, and its argument is one third of $3\pi /4+2k\pi$. Hence the three distinct roots areTheir displayed arguments are $\pi /4$, $11\pi /12$, and $19\pi /12$; the last is equivalently $-5\pi /12$ in the principal range.

Resolve the input along the three perpendicular eigendirections. The linear map removes the component along $(1,2,1)$, stretches the component along $(1,-1,1)$ by $12$, and reverses and stretches the component along $(1,0,-1)$ by $6$. Every output lies in the plane perpendicular to $(1,2,1)$. A nonzero input in the kernel of a linear map $A$ is sent to zero; other nonzero inputs remain in that output plane. Equivalently, take the orthogonal projection onto this plane, then scale by the two signed eigenvalues along its perpendicular eigendirections.

Expand an arbitrary vector in the orthonormal eigenbasis:Its image has no component along ${\mathbf{e}}_0$, soAlternatively, symmetry gives $(A\mathbf{r})\cdot {\mathbf{e}}_0=\mathbf{r}\cdot A{\mathbf{e}}_0=0$. This is image-kernel orthogonality for an adjoint, with $A^{\mathsf{T}}=A$.