Taking real parts of the geometric sum givesso $(a_n)$ is bounded when $\cos y\ne 1$. It cannot converge: convergence would imply $\cos (ny)=a_n-a_{n-1}\to 0$, but thenwhereas the same necessary condition applied to the subsequence $2n$ would give a limit of zero.

Every term is nonnegative. Since $1-\cos x\le x^2/2$,Comparison with $\sum k^{-2}$ shows that the partial sums are bounded. Being increasing, they converge to the finite limit

Draw the even $\dot{y}=0$ nullcline through $(\pm 1,0)$ and use the arrows from part (iii). Trajectories near $(-1,0)$ spiral clockwise inward because, immediately to its right on the $x$-axis, $\dot{y}<0$. At $(1,0)$ the stable and unstable eigendirections have slopes $-4$ and $1$. Their four separatrix branches divide the portrait; the only trajectory actually reaching $(1,0)$ at finite time is the constant one, while these branches approach it as $t\to \pm \infty$. All remaining curves follow the sign field, with those in its basin spiralling into $(-1,0)$.

The sign of $\dot{x}$ is the sign of $y$. The $\dot{y}=0$ nullcline is$\dot{y}$ is positive below this curve and negative above it. Combining the upper and lower half-planes with the regions above and below this even curve labels the four requested sign combinations.

The Jacobian isAt $(1,0)$ its eigenvalues are $1,-4$, so this point is a saddle. At $(-1,0)$ they are $(-3\pm i\sqrt{7})/2$, so this point is a stable spiral.

With $y=\dot{x}$, the system isAt equilibrium $y=0$ and $x^2=1$, so the equilibrium points are $(1,0)$ and $(-1,0)$.

For a compactly supported test function $f$,Thus this is the distributional derivative of the Heaviside step function: $H^{\prime }=\delta$.

The derivative of a distribution is defined by integration by parts. Therefore

For every test function $\phi$ and nonzero $p$, substitution $u=pt$ with the orientation handled by $|p|$ givesHence the Dirac delta scaling identity is $\delta (pt)=\delta (t)/|p|$.

This is the sphere with centre $c$ and radius $\rho >0$.

This is the affine plane through $b,d,f$, provided $d-b$ and $f-b$ are independent. The scalars $\lambda ,\mu$ are affine coordinates in those two directions.

This is the plane through $a$ perpendicular to the nonzero normal vector $n$.

Let $L_h$ be the spatial matrix, so the Forward Euler method isThe conservative use of shared edge coefficients makes $L_h$ a real symmetric matrix. In an interior row, put $s=\alpha +\beta +\gamma +\delta <4a_{\max }$. Its diagonal entry is $-s/h^2$, and the sum of the absolute off-diagonal entries is $s/h^2$. Boundary rows have no larger disk because homogeneous Dirichlet data remove some unknown neighbors.

The Gershgorin circle theorem and symmetry therefore place every eigenvalue of $L_h$ inIf $\mu =k/h^2$, each amplification eigenvalue is $1+k\lambda$. It lies in $[-1,1]$ wheneverorSince the amplification matrix is symmetric, bounding all its eigenvalues in modulus by one bounds its discrete Euclidean operator norm by one. This proves the stated stability condition and is the Gershgorin stability bound for a variable-coefficient diffusion stencil.

Sample the diffusion coefficient at edge midpoints and writeCentered differences of the two fluxes give the conservative finite-difference stencilFor example, the $x$ part isTaylor expansion about $(ih,jh)$ shows that the odd powers cancel between the two face fluxes and that the result equals ${\partial }_x(a{u}_x)+O(h^2)$. The same calculation in $y$ gives total truncation error

The fluid velocity at the piston equals its velocity,Using $c-c_0=(\gamma -1)u/2$ in part (ii) givesOver $0\le t\le 1$,ThusThe linear pressure oscillation has zero mean, while the quadratic compressibility produces a positive mean pressure, the finite-amplitude acoustic analogue of radiation pressure.

The motion is isentropic before a shock forms, and the perfect-gas relations giveWriting $\delta =\delta c/c_0$ and applying the generalized binomial expansion,

Before shock formation, the piston launches a right-moving simple wave into the undisturbed gas. The opposite invariant is fixed at its ambient value:Therefore the simple-wave relation is

The covariant derivative adds one covariant index, so $\nabla S$ has type $(2,2)$. The outer product of two tensors adds their index counts. Hence

The map takes one covector argument and returns two vector slots, so $S$ has two contravariant and one covariant index. Its tensor type isor type $(2,1)$, with components $S^{ab}{}_c$.

Removing the closed unit ball leaveswhich is an open subset of ${\mathbb{R}}^3$. Every point therefore has an ordinary Euclidean coordinate neighborhood, soIt has no boundary; the unit sphere was removed together with the ball.