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False. The constant continuous function $f:\mathbb{R}\to \mathbb{R}$, $f(x)=0$, sends the open set $\mathbb{R}$ to $\{0\}$, which is not open. Therefore continuity alone does not imply an open map.

For the near-origin probability in a maximal-angular-momentum hydrogen state, integrate the normalized radial probability density:Throughout this small interval the exponential is $1+O(r_p/a)$ for fixed $l$. HenceUsing the normalization above,The high power of the small radius reflects the near-origin behavior $R\propto r^l$ of the hydrogen atom state.

The normalized spherical harmonics separate the angular variables. The Radial Schrodinger equation for the hydrogen atom isWrite $R=A{r}^l{e}^{-\kappa r}$. ThenSubstitution separates a constant term and a $1/r$ term. The latter vanishes precisely when ${\hslash }^2\kappa (l+1)/m=e^2/(4\pi {ϵ}_0)$. With $\kappa =1/[a(l+1)]$ this determines the Bohr radiusThe constant term determinesThese are the maximal-angular-momentum hydrogen states, with principal quantum number $n=l+1$.

Because the angular integral is one, normalization requiresThereforeThe phase of $A$ remains arbitrary.

The Duhamel principle superposes the response to source impulses:This expression uses the source at the origin, is causal, and has zero initial data in the distributional sense. Its integral over the plane is ${\int }_0^{t}p(s)ds$, consistent with the injected total amount.

For the finite-duration source and $t>t_0$, set $b=(x^2+y^2)/(4D)$. The causal diffusion from a finite-duration planar point source becomesWith $\eta =b/\tau$, or by differentiating the supplied exponential integral, this isIf $b/(t-t_0)\ll 1$, both arguments are small. Their logarithmic constants and $\ln b$ cancel, leavingAt the origin this logarithmic expression is exact for $t>t_0$, obtained as the limit $b\to 0$. The leading spatial correction is $-p_{0}b{t}_0/[4\pi Dt(t-t_0)]$. If also $t\gg t_0$, the leading value is $p_0{t}_0/(4\pi Dt)$.

Take the Fourier transform in $x$. The transform of the source is $\delta (t-t^{\prime })e^{-ik{x}^{\prime }}$, so the Green function equation becomesChoose the causal Green function, zero for $t<t^{\prime }$. Away from $t^{\prime }$ the solution is a decaying exponential, and integrating across $t^{\prime }$ gives the jump $\tilde{G}(t^{\prime }+)-\tilde{G}(t^{\prime }-)=e^{-ik{x}^{\prime }}$. ThereforeThe phase shifts the kernel's centre to $x^{\prime }$. Applying the inverse Fourier transform and the Fourier transform of a Gaussian with $a=D(t-t^{\prime })$ givesFor positive elapsed time this heat kernel solves the homogeneous diffusion equation and has unit integral; as elapsed time tends to zero it approaches the spatial delta distribution, supplying the required impulse.

True, and the closedness assumption on $Y$ is unnecessary. The Minkowski sum has the representationEach translate of the open set $X$ is open, and an arbitrary union of open sets is open. If $Y$ is empty the union is empty, which is also open.

False. In $\mathbb{R}$, letEach is closed: its distinct points escape to infinity, so it has no finite accumulation point. Their Minkowski sum contains $1/n$, tending to zero. However zero is not in the sum. If $m-n+1/n=0$ with positive integers $m,n$, then $n-m=1/n$. For $n>1$ the right side is strictly between zero and one, while the left side is an integer; for $n=1$ the equation would require $m=0$. Thus the sum of two closed sets need not be closed. The same counterexample embeds along a coordinate axis in ${\mathbb{R}}^n$.

True. The same subsequence argument applies to the sum of a compact set and a closed set. For a convergent sequence $z_j=x_j+y_j$, boundedness of $X$ provides a convergent subsequence of $x_j$, with limit $x\in X$. Then $y_j\to z-x\in Y$ along that subsequence. Consequently $z\in X+Y$. No boundedness of $Y$ is needed.

True. Boundedness follows from $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$. To prove closedness of the Minkowski sum, let $z_j=x_j+y_j\in X+Y$ converge to $z$. By the Bolzano-Weierstrass theorem, the bounded sequence $x_j$ has a subsequence converging to $x$. Closedness gives $x\in X$. Along that subsequence, $y_j=z_j-x_j\to z-x$, which belongs to $Y$ by closedness. Thus $z=x+(z-x)\in X+Y$. This proves that the sum is bounded and closed.

For the determinant of a complex block representation, putThe change-of-basis matrix $T$ is invertible, and direct block multiplication givesThus $T^{-1}MT$ is block diagonal. Using multiplicativity of the determinant and the preceding block formula,The identity follows from this constant change of basis for arbitrary complex $A,B$.

In the determinant expansion of the block upper triangular matrix, a nonzero product must assign each of the bottom $n$ rows to one of the last $n$ columns, because the lower-left block is zero. Those rows therefore occupy all the last $n$ columns, leaving the first $n$ columns for the top rows. The contributing permutations are precisely independent permutations within the two blocks. Their signs multiply, so the sum factors:The off-diagonal block contributes to no surviving term.

Induct on $n$. The one-dimensional assertion is immediate, and the zero-dimensional case is vacuous. If $T$ has a cyclic vector spanning all of $V$, the direct proof in part (ii) applies. Otherwise choose $x\ne 0$ and let $W$ be its cyclic subspace. Part (i) makes $W$ invariant; the absence of a cyclic vector makes it proper and nonzero.

By induction, the restriction to $W$ is annihilated by ${\chi }_A$, and the induced map on $V/W$ is annihilated by ${\chi }_D$. The quotient assertion means ${\chi }_D(T)V\subset W$. Applying the restriction assertion next givesBy the characteristic factorization in part (iii), this is precisely ${\chi }_T(T)V=0$. HenceThis is the Cayley-Hamilton theorem from cyclic subspaces; using the quotient map is what makes the argument valid even when the block $B$ is nonzero.

Since the first $k$ basis vectors span the invariant subspace $W$, their images have no components in the last $n-k$ coordinates. ThusHere $A$ is the matrix of $T{|}_W$, $D$ is the matrix of the induced map on the quotient vector space $V/W$, and $B$ is unrestricted. The chosen complementary span need not itself be invariant, so $D$ should not in general be called a restriction to that complement. The block-triangular determinant gives

The $n$ spanning vectors form a basis. WriteIn this cyclic basis the matrix is the companion matrixExpanding $\det (tI-C)$ along its last column givesFor example, the term containing $a_j$ has minor $t^j$ and the signs from the subdiagonal cancel the cofactor sign; the diagonal term supplies $t^n$. The defining relation says ${\chi }_T(T)x=0$. Every polynomial in $T$ commutes with $T$, soIt therefore vanishes on a basis, provingThis direct argument for a cyclic vector does not assume the Cayley-Hamilton theorem.

If $x=0$, the generated subspace is zero and invariant. Otherwise, let $r$ be the first index for which $x,Tx,\dots ,T^{r}x$ are linearly dependent. Such an index exists with $1\le r\le n$, since there are $n+1$ vectors. The preceding vectors are independent, so the coefficient of $T^{r}x$ in this first relation cannot vanish. HenceThe linear operator $T$ sends each of the first $r-1$ generators to the next and sends the last into their span. This proves that their span is an invariant subspace. Every later iterate lies in the same subspace by induction, so it equals the span of the first $n$ iterates. Therefore the indicated subspace is $T$-invariant. This proof of a cyclic subspace uses only finite dimension and linear dependence.

By the bias-variance decomposition of mean squared error,For $\theta \ne 0$, comparison with the unbiased estimator reduces toIts roots are $(n-1)/(n+1)$ and $1$, soEquality at either endpoint gives the same risk as $\bar{X}$. At $\theta =0$, every risk is zero, so no strict improvement is possible. This illustrates shrinking a sample mean with variance proportional to mean squared: allowing a small bias can reduce variance enough to improve total risk. The best constant is $k=n/(n+1)$, whose mean squared error is ${\theta }^2/(n+1)$.

For an estimator with finite second moment, its bias and mean squared error areWrite $\hat{\theta }-\theta =(\hat{\theta }-{\mathbb{E}}_{\theta }\hat{\theta })+b_{\theta }$. On squaring and taking expectations, the cross term vanishes. This proves the bias-variance decomposition of mean squared error:For the sample mean, independence gives $\mathbb{E}\bar{X}=\theta$, $\mathrm{Var}(\bar{X})={\theta }^2/n$. Thus $k\bar{X}$ has bias $(k-1)\theta$. Unbiasedness throughout the parameter family requiresAt the single parameter value $\theta =0$, all these estimators are zero almost surely; that special value does not change the requirement of unbiasedness for all $\theta$.

The relevant one of the characteristic polynomials of a linear multistep method isThe coefficients on the right are chosen to give third order, so consistency of a numerical method is already available. By the Dahlquist equivalence theorem, convergence with convergent starting values is equivalent to zero-stability: every root of $\rho$ must satisfy $|w|\le 1$, and every root on the unit circle must be simple.

If $a<0$, the quadratic is negative at $w=-1$ and tends to $+\infty$ as $w\to -\infty$, so it has a root less than $-1$. If $a>1$, its root product is $a$, so at least one root has modulus greater than one. For $0\le a\le 1/4$, its real roots $(-1\pm \sqrt{1-4a})/2$ lie in $[-1,0]$. For $1/4<a\le 1$, its conjugate roots have modulus $\sqrt{a}\le 1$.

At $a=0$, the three roots are $1,0,-1$, with the unit-modulus roots simple. At $a=1$, they are $1,e^{2\pi i/3},e^{-2\pi i/3}$, again simple. The repeated quadratic root at $a=1/4$ is $-1/2$, strictly inside the unit circle, which the root condition for a multistep method permits. ThereforeThis is the root stability of a cubic multistep polynomial; oscillatory parasitic roots at the endpoints do not invalidate convergence when their starting amplitudes tend to zero.

Insert an exact smooth solution into the linear multistep method and define its unscaled local truncation error byIts Taylor expansion has coefficientsThe method has order at least $p$ exactly when $c_0=\cdots =c_p=0$, since then ${\mathcal{R}}_h[y]=O(h^{p+1})$ for every sufficiently smooth $y$. Necessity can be checked by inserting polynomials of successive degrees. This also corresponds to the usual truncation error ${\mathcal{R}}_h/h=O(h^p)$.

On the other hand, expansion of the characteristic polynomials of a linear multistep method givesThus the exponential-symbol order criterion for a multistep method isIf “order $p$” is meant to be the exact rather than guaranteed order, one additionally requires $c_{p+1}\ne 0$. The first two vanishing conditions are the usual consistency of a numerical method relations $\rho (1)=0$, ${\rho }^{\prime }(1)=\sigma (1)$.

The convergence theorem for an irreducible aperiodic positive recurrent Markov chain says ${\mathbb{P}}_i(X_n=j)\to {\pi }_j$ for every pair of states. Apply it to $i=j=0$ and take the complement:Although $p_{00}=0$, the chain is aperiodic because its possible return lengths have greatest common divisor one; a self-loop is sufficient but not necessary for aperiodicity.