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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.

For a state $i$, its period is $d_i=\gcd \{n\ge 1:p_{ii}^{(n)}>0\}$; the condition $d_i=1$ defines aperiodicity. It is a positive recurrent state if return occurs almost surely and its first strictly positive return time has finite expectation. An ergodic state of a Markov chain is positive recurrent and aperiodic.

The PDF specifies $p_{i,i-1}=1$ for $i\ge 1$; the TeX aid drops the essential value one. Consequently every positive state reaches zero deterministically. From zero, every positive state $j$ is reached in one step with probability $p{q}^{j-1}>0$. Paths through zero connect any pair of states, proving irreducibility.

Let $J$ be the jump out of zero. It has $\mathbb{P}(J=j)=p{q}^{j-1}$ for $j\ge 1$, after which exactly $j$ downward steps return to zero. Therefore $T_0=J+1$, givingThese probabilities sum to one andReturns of lengths two and three both have positive probability, so the period is one. Hence state zero is ergodic, and by irreducibility all states are positive recurrent and have the same period.

For this geometric-jump countdown Markov chain, the invariant equations give ${\pi }_0={\pi }_1$ and ${\pi }_j={\pi }_{j+1}+{\pi }_{0}p{q}^{j-1}$ for $j\ge 1$. Summing the tail, or counting visits in a regeneration cycle, yields ${\pi }_j={\pi }_0{q}^{j-1}$. Normalization givesThis is the stationary distribution of a regenerative countdown chain. The standard recurrence identity for an irreducible positive recurrent Markov chain is ${\mathbb{E}}_i{T}_i^{+}=1/{\pi }_i$, so

As the real charge descends to height $z(t)$, its instantaneous image charge is at $-z(t)$. The force is thereforeThe image is not kept at its initial position. Multiplying by $\dot{z}$ and using $z(0)=d$, $\dot{z}(0)=0$ givesFor the falling charge above a grounded conducting plane, the arrival time isPut $z=d{\sin }^2\theta$; the integral becomes $2d^{3/2}{\int }_0^{\pi /2}{\sin }^2\theta d\theta =\pi d^{3/2}/2$. HenceThis is the finite arrival time in the specified ideal force model, for $q\ne 0$. The field-force derivation also avoids incorrectly using the full real-image pair interaction energy: the conductor's effective energy is $-q^2/(16\pi {\epsilon }_{0}z)$, with the factor one-half appropriate to induced charge.

The method of images places an image charge $-q$ at the reflected position below the plane. The candidate potential in the physical upper half-space isAt $z=0$ the two terms cancel, so $\Phi =0$ and $E_x=E_y=0$. Direct differentiation givesThus the upper-side field satisfies exactly the conductor boundary conditions from part (i). The fictitious pair does have a field below the plane, but that part of its solution is not used for the physical conductor.

In $z>0$, the potential has the correct real point source, the correct zero Dirichlet boundary condition, and the required decay at infinity. By Uniqueness of the Dirichlet problem, it is the physical potential there. This is why the artificial charge can represent the induced conductor response without being a second physical charge.

An ideal conductor in electrostatic equilibrium has zero field in its interior and constant potential on its surface. Grounding fixes that potential to zero. The tangential field is continuous and therefore vanishes; the normal field has the jump fixed by the induced surface charge density $\sigma$:Here the normal points upward out of the conductor. The normal field is not generally zero; grounding prescribes the potential, not the induced charge density.

The Parseval identity for the orthonormal stationary-state expansion is ${\sum }_{n\ge 1}|c_n{|}^2=\Vert \psi {\Vert }^2=1$. Splitting the amplitudes into even and odd indices givesMultiply by ${\pi }^2/3$ to obtain the requested identity. Its integer constants areThe series converges absolutely, since both parity contributions are $O(p^{-2})$. This derivation uses the normalization of the prepared state and completeness of the well stationary states, rather than a separate summation trick.

The original PDF threshold contains $\hslash$, not $h$. It is $8E_1$, so the energies strictly above it are exactly those with $n\ge 3$. By the Born rule, the required probability is $1-|c_1{|}^2-|c_2{|}^2$. The two excluded amplitudes areConsequentlyThus, in increasing powers of $1/\pi$,The threshold lies strictly between $E_2$ and $E_3$, so no equality-at-threshold convention affects the answer.

Choose the prepared wavefunction real and positive on its support. Its normalization isIn the ramp-state expansion in a symmetric infinite well, write $\psi ={\sum }_{n\ge 1}{c}_n{\psi }_n$. Orthogonality givesFor $\kappa =n\pi /(2a)$, a primitive is $-x\cos (\kappa (x+a))/\kappa +\sin (\kappa (x+a))/{\kappa }^2$. Evaluating both endpoints givesAn equivalent parity-separated form isThese formulas with ${\sum }_n{c}_n{\psi }_n$ are the requested explicit expansion. The convergence is in the square-integrable norm; at the idealized discontinuity at the right wall, a pointwise boundary value does not change the quantum state or its probability amplitudes.

The squared sine has integral $a$ over the interval of length $2a$, since it contains an integer number of half-wavelengths. A real positive phase convention therefore givesInside the infinite square well, the Time-independent Schrödinger equation is $-{\hslash }^2{\psi }_n^{\prime \prime }/(2m)=E_n{\psi }_n$. The wavenumber is $\pi n/(2a)$, soThe stationary states vanish at both walls and form an orthonormal basis of the square-integrable state space. A common arbitrary phase on each basis function does not alter any energy or measurement probability.

Set $U=R(r)\Theta (\theta )$ in the Laplace equation, multiply by $r^2/U$, and separate:Putting $x=\cos \theta$ turns the angular equation into the Legendre differential equation. For a solution smooth on the whole spherical axis, the allowed eigenvalues are $\lambda =n(n+1)$ with angular functions $P_n(\cos \theta )$. The radial equation has the two solutions $r^n,r^{-n-1}$. Superposition yields the regular axisymmetric expansionThe coefficients and convergence region are set by the domain and boundary data. Without regularity at the poles, singular Legendre functions and more general separated degrees would also be possible; they are excluded for this exterior spherical field.

The far-field matching fixes $A_1=v_0$ and all other growing coefficients to zero. The radial Neumann boundary condition at $r_0$ isIndependence of the angular modes gives $B_1=v_0{r}_0^3/2$ and all other decaying coefficients zero. Thus the exterior harmonic potential with no flux through a sphere isIts derivative vanishes on the sphere and its difference from $v_{0}r\cos \theta$ tends to zero at infinity. A spatial constant may be added if the far-field condition is intended only to specify the gradient or the leading growing term; matching the potential difference to zero fixes that constant to zero.

Substitute $y(x)={\sum }_{j\ge 0}{c}_j{x}^j$ into the Legendre differential equation. Matching the coefficient of $x^j$ givesChoose the parity of $n$: start with nonzero $c_0$ for even $n$, or nonzero $c_1$ for odd $n$, and set the other parity to zero. The numerator is nonzero for every preceding index of that parity below $n$, so $c_n\ne 0$, but at $j=n$ it vanishes. All higher coefficients are zero. This constructs a nonzero polynomial solution of degree exactly $n$.

Its value at $x=1$ cannot be zero. If it had a zero there of multiplicity $k\ge 1$, write $y=(x-1{)}^{k}R$ with $R(1)\ne 0$. The lowest-order term of $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }$ would be $-2k^{2}R(1)(x-1{)}^{k-1}$, whereas $n(n+1)y$ starts at order $k$. The differential equation could not hold. Hence the polynomial can be scaled to satisfy $P_n(1)=1$.

The resulting Legendre polynomials beginEach satisfies both the differential equation and the specified normalization.