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For a sample of size $n\ge 2$, defineFor the zero-mean Gaussian sample, the standardized vector $(X_1/\sigma ,\dots ,X_n/\sigma )$ has independent standard normal coordinates. Apply an orthogonal transformation whose first row is $(1,\dots ,1)/\sqrt{n}$. The resulting coordinates $Z_1,\dots ,Z_n$ remain independent standard normals: their joint density depends only on squared Euclidean length and the transformation has unit absolute Jacobian. Its first coordinate is $Z_1=\sqrt{n}\overline{X}/\sigma$, while orthogonal decomposition givesThusThe ratio of the first independent standard normal and the square root of the normalized chi-square variable has a Student probability distribution, giving

For the paired height observations, form one cross-minus-self difference per pot. The sampling units for this test are the fifteen differences, not thirty separate plant heights. Under the usual independent normal model for differences, a null mean of zero gives a $t_{14}$ statistic. The observed statistic isFor the two-sided alternative, reject at significance level $\delta$ if $|t_{\mathrm{obs}}|>t_{14,1-\delta /2}$. Its two-sided $p$-value is $2[1-F_{t_{14}}(|t_{\mathrm{obs}}|)]$. This uses the recorded values without performing the numerical calculation forbidden by the question. The normal difference model is what makes the small-sample Student calibration exact.

Choose each factor with its argument between zero and $\pi$ in the upper half-plane. On an interval between consecutive real prevertices, each factor has a constant boundary argument, so $f^{\prime }$ has constant argument and the image is a straight side. Crossing $X_j$ from left to right decreases $\arg (Z-X_j)$ by $\pi$, hence increases $\arg f^{\prime }$ by ${\beta }_j\pi$, giving the prescribed exterior turn. Locally,so an upper half-disc maps to angle $(1-{\beta }_j)\pi$, the required interior angle. The overall constant rotates and scales the polygon; integration adds translation. For a general prescribed polygon, the prevertex positions must also satisfy the side-length conditions: the derivative formula fixes angles but arbitrary choices do not fix arbitrary side lengths.

For the slit domain, the slit tip has interior angle $2\pi$, so its exterior parameter is $-1$. The two vertices at infinity give the factors at $\pm 1$, leading to $f^{\prime }(Z)$ proportional to $Z(1-Z^2{)}^{-3/2}$. Integrating and imposing the tip value givesThe square root is the principal one: for $\Im Z>0$, $1-Z^2$ never lies on the nonpositive real axis. Thus $\Re f>0$. Boundary points $-1<X<1$ map to the two banks of the slit, with $X=0$ at the tip; the outside intervals map to the imaginary axis. The derivative is nonzero in the open half-plane. Conversely each point of the slit domain determines exactly one root $Z$ of $Z^2=1-z^{-2}$ with positive imaginary part, proving bijectivity.

The Möbius map $(Z-1)/(Z+1)$ takes the upper half-plane to itself, so its principal argument lies in $(0,\pi )$. The imaginary part of its analytic logarithm is harmonic, equals $\pi$ on $(-1,1)$ and zero on the other real intervals. ThusIt lies between zero and one and has the required boundary limits away from the two discontinuities. Pulling this harmonic function back by the inverse conformal map yieldsThis specifies all branches and remains bounded even as $z$ tends to either end at infinity.

To exhibit nonuniqueness without the growth restriction, putThis is harmonic in the domain. On its boundary the corresponding $Z$ is real and the rational expression is real, except at $\pm 1$, which represent infinity. At $Z=\infty$, representing $z=0$, the expression tends to zero. Thus $H$ has zero finite boundary data. It is nonzero, since for $Z=iY$ its value is $Y/(1+Y^2)>0$, and it is unbounded as $Z$ approaches either pole $\pm 1$ non-tangentially. Therefore $u+cH$ gives distinct solutions with the same finite boundary data when boundedness at infinity is removed.

The integral construction requires differentiation under the integral and vanishing boundary terms. With those assumptions, integration by parts givesThe standard construction therefore sets $g^{\prime }={\zeta }^{2}g$, giving $g=C{e}^{{\zeta }^3/3}$. Convergent contours run between rays where $\Re {\zeta }^3<0$: convenient central rays have arguments $-\pi /3$, $\pi /3$ and $\pi$. Two choices are the oriented pathsThe cubic exponential dominates the linear one at both ends. To verify independence directly, put $A={\int }_0^{\infty }{e}^{-r^3/3}dr>0$ and $B={\int }_0^{\infty }r{e}^{-r^3/3}dr>0$. At $z=0$ the value and derivative of the integral along rays ${\theta }_1,{\theta }_2$ are $A(e^{i{\theta }_2}-e^{i{\theta }_1})$ and $B(e^{2i{\theta }_2}-e^{2i{\theta }_1})$. For the two contours above their derivative-to-value ratios are respectively $B/A$ and $(B/A)(e^{i\pi /3}-1)$, which differ. Hence their Wronskian is nonzero.

For the second equation multiply by $z$ before integrating by parts:Setting the residual weight to zero gives $g=C(1+{\zeta }^2{)}^{-1/2}$. One contour is the segment from $-i$ to $i$, with the square root positive at $\zeta =0$. Its integral is a nonzero multiple of $J_0(z)$: substituting $\zeta =it$ gives $i{\int }_{-1}^1{e}^{izt}(1-t^2{)}^{-1/2}dt$, an entire function nonzero at zero.

A second contour starts at $i$ and follows $\zeta =i-t$, $0\le t<\infty$, choosing a continuous branch along the path. For $\Re z>0$ it converges at infinity; at the branch endpoint its integrable singularity is $t^{-1/2}$. In both cases the boundary term contains $\sqrt{1+{\zeta }^2}$ and vanishes at branch endpoints. The second integral has a nonzero logarithmic divergence as $z\to 0$ through positive real values, because its large-$t$ integrand is a nonzero multiple of $e^{-zt}/t$. It is therefore independent of the entire first solution; analytic continuation supplies solutions elsewhere away from the equation's singular point.

Strictly, nonzero $w$ and an arbitrary contour alone do not prove the printed necessity for $g$. Necessity also needs injectivity of the chosen Laplace representation, while the vanishing-integrand argument above is the usual sufficient construction. For a concrete counterexample to unconditional necessity in the Bessel case, take a positively oriented circle $|\zeta |=R>1$, choose the branch $(1+{\zeta }^2{)}^{-1/2}\sim 1/\zeta$ outside the cut $[-i,i]$, and add one to that weight. The added entire integrand integrates to zero. Expanding the original weight on this circle gives integral $2\pi i{J}_0(z)$, which is nonzero and solves the equation, although the modified weight is not a constant multiple of the printed one. The standard weights and contours above still give the requested independent solutions.

Let $D$ be the separation of the Earth and Moon, and let $x$ be the spacecraft's distance from the Earth's centre towards the Moon. On $0<x<D$, the two gravitational forces oppose each other. Their equality, by Newton's law of universal gravitation, isTaking the positive square root givesThe centre of mass of the two bodies is at $x_G=D{M}_M/(M_E+M_M)$ from the Earth's centre. Thus the gravitational field null between two point masses lies a distancefrom their centre of mass, on the Moon's side. With $M_E/M_M=81$, $x=9D/10$ and $x_G=D/82$, soThe point is $3.84\times 10^7\mathrm{m}$ from the Moon. It is an instantaneous zero of the physical gravitational field, not a co-rotating Lagrange point: a spacecraft remaining fixed relative to the Earth-Moon line would need a nonzero centripetal acceleration. With zero net force it cannot follow that circular co-rotation.

For motion in a central potential, conservation of angular momentum gives $L=m{r}^2\dot{\theta }$. In polar coordinates, the conserved mechanical energy isConsequently, for $L\ne 0$,The sign in $\dot{r}$ distinguishes inward and outward motion. Choosing the orientation so $L>0$, it is also the sign of $dr/d\theta$. The expression under the square root must be nonnegative. If $L=0$, the motion is radial and $\theta$ is constant, so $r$ cannot be parametrized by $\theta$ this way.

Let $a={\mathbb{E}}_{\psi }(A)$, which is real. The dispersion, or quantum uncertainty, is the nonnegative standard deviationIndeed, self-adjointness and normalization giveA squared norm is nonnegative, so the square root is defined. It vanishes precisely when $(A-aI)\psi =0$, namely when $\psi$ is an eigenvector of $A$. Conversely, an eigenvector with eigenvalue $\lambda$ has $a=\lambda$ and hence zero dispersion. Thus zero dispersion is equivalent to being an eigenstate.

For total nuclear density $n$, $n_e=n\zeta$ and $n_H=n(1-\zeta )$. Saha's equation becomes $n{\zeta }^2/(1-\zeta )=C{T}^{3/2}{e}^{-R/(k_{B}T)}$. At $\zeta \ll 1$,The factor two in the exponent follows from taking a square root; using the full ionization temperature would double the predicted logarithmic slope.

Homogeneity makes the inertial transformation linear. Align origins at $t=t^{\prime }=0$, let the second origin have velocity $v$, and impose isotropy and reciprocity. Then $x^{\prime }=\gamma (v)(x-vt)$, $y^{\prime }=y$, $z^{\prime }=z$. If the same limiting signal speed $c$ has trajectories $x=\pm ct$ in both systems, substituting these two trajectories fixes $t^{\prime }=\gamma (v)(t-vx/c^2)$. Requiring the inverse to have the same form with $-v$ fixes ${\gamma }^2(1-v^2/c^2)=1$. Continuity at rest selects the positive square root:Light enters this derivation through the empirical/in-principle operational identification of the invariant speed and Einstein synchronization. Group, homogeneity and isotropy arguments can instead derive a family with a universal parameter $K$, $t^{\prime }=\gamma (t-Kvx)$ and $\gamma =(1-K{v}^2{)}^{-1/2}$. They do not by themselves determine whether $K=0$ or a positive finite value. Empirical light propagation or other relativistic dynamics identifies $K=1/c^2$ for our world. Consequently a derivation need not assume light as its first premise, but it must acquire the physical invariant-speed structure somewhere rather than extract it from linearity alone.^[ref]

In this argument, homogeneity, isotropy, equivalence of inertial frames and the existence/identification of an invariant speed are physical structural posits. Coordinate origins, axis orientation and clock-unit choices are conventions. Einstein synchronization stipulates equal one-way signal times for outward and return paths; the empirical two-way light behavior and material-clock comparisons are not made true merely by that stipulation. Other synchronization coordinates can describe the same operational observations with a different displayed one-way light speed and transformed formula. Distinguishing those choices from the physical invariant structure is essential to the derivation.

Homogeneity makes the inertial transformation linear. Align origins at $t=t^{\prime }=0$, let the second origin have velocity $v$, and impose isotropy and reciprocity. Then $x^{\prime }=\gamma (v)(x-vt)$, $y^{\prime }=y$, $z^{\prime }=z$. If the same limiting signal speed $c$ has trajectories $x=\pm ct$ in both systems, substituting these two trajectories fixes $t^{\prime }=\gamma (v)(t-vx/c^2)$. Requiring the inverse to have the same form with $-v$ fixes ${\gamma }^2(1-v^2/c^2)=1$. Continuity at rest selects the positive square root:Light enters this derivation through the empirical/in-principle operational identification of the invariant speed and Einstein synchronization. Group, homogeneity and isotropy arguments can instead derive a family with a universal parameter $K$, $t^{\prime }=\gamma (t-Kvx)$ and $\gamma =(1-K{v}^2{)}^{-1/2}$. They do not by themselves determine whether $K=0$ or a positive finite value. Empirical light propagation or other relativistic dynamics identifies $K=1/c^2$ for our world. Consequently a derivation need not assume light as its first premise, but it must acquire the physical invariant-speed structure somewhere rather than extract it from linearity alone.^[ref]

A branch point is a point around which analytic continuation changes the value of a multivalued function. For $z^{1/2}=r^{1/2}{e}^{i\theta /2}$, one circuit around zero changes the sign. A branch cut removes paths producing this ambiguity; for example $-\pi <\arg z<\pi$ gives a single holomorphic square root off the nonpositive real axis. The position of a cut is a convention, while the branch point is intrinsic.

Set $q(z)=(z-1)/(z+1)$. For $z$ outside the closed segment joining $-1$ to $1$, $q(z)$ avoids the nonpositive real axis. Therefore the principal square root of $q$ defines a holomorphic function there, with the normalization $f(\infty )=1$. The endpoints are branch points. Expanding at infinity,For a positively oriented contour surrounding the cut, its integral equals the integral on a sufficiently large positively oriented circle, since the intervening annulus is holomorphic. ThusOn the upper and lower banks, $f_{+}(x)=+i\sqrt{(1-x)/(1+x)}$ and $f_{-}(x)=-i\sqrt{(1-x)/(1+x)}$. The upper bank is traversed from $1$ to $-1$, and the lower bank from $-1$ to $1$. Their sum is $-2i$ times the required real integral. Small endpoint arcs vanish because their singularities have exponent less than one. ConsequentlyFor $\alpha =1/n$, use $f_{\alpha }=q^{\alpha }$ with the same normalization. Its expansion is $1-2\alpha /z+O(z^{-2})$, and its bank values are $e^{\pm i\pi \alpha }[(1-x)/(1+x){]}^{\alpha }$. Equating the contour integral $-4\pi i\alpha$ with $-2i\sin (\pi \alpha )$ times the real integral givesThe condition $n\ge 2$ ensures endpoint integrability. Choosing the other square-root branch reverses $I_{\Gamma }$, but also reverses both bank values, leaving the real integral unchanged.

Write $\nu =m\gamma /{\hslash }^2$, so $t(k)\simeq k/(k+i\nu )$. Its nontrivial pole isFor negative $\gamma$ this is on the positive imaginary $k$ axis, the physical bound-state sheet. For positive $\gamma$ it lies on the negative imaginary axis and is a virtual-state pole reached by analytic continuation through the continuum cut. These different sheets can have the same negative numerical energy; specifying only $E_{\ast }$ would conceal the distinction. As a function of $E$, the momentum square root also introduces the continuum threshold branch point at zero. If $\gamma =0$, numerator and denominator cancel at this order and no pole is predicted by this leading approximation.

At the same temperature, momenta increase as the square root of mass, so wavelengths decrease as its inverse square root. For a nucleus of mass $M_N\simeq A{m}_p$,At the electron overlap threshold, ${\lambda }_e=d$, the nuclear wavelength is about $7.0\mathrm{pm}/\sqrt{A}$, far below the $0.3$ nm spacing. Nuclear exchange effects are consequently much less important than electron exchange effects. This statement about nuclear overlap does not mean lattice vibrations are always classical.

Take a singular value decomposition $F=Q_{1}D{Q}_2^T$, with $D=\mathrm{diag}({\lambda }_1,{\lambda }_2,{\lambda }_3)$ positive and $Q_1,Q_2\in SO(3)$. Such a choice is possible because $\det F>0$. Left material frame indifference removes $Q_1$, and right material isotropy removes $Q_2$, giving $W(F)=W(D)$.

Moreover $W(D)$ is unchanged when the positive diagonal entries are permuted. Every permutation can be implemented by a signed permutation matrix of determinant one; the signs disappear in conjugating a diagonal matrix. Thus the energy depends only on the unordered principal stretches.

The three invariants determine precisely this unordered triple: the roots of $t^3-I_B{t}^2+I{I}_{B}t-II{I}_B$ are ${\lambda }_1^2,{\lambda }_2^2,{\lambda }_3^2$, and their positive square roots give the stretches. Consequently, on the set of admissible invariant triples, define $h(I_B,I{I}_B,II{I}_B)$ to be $W(D)$. The preceding argument proves that this is well defined and thatNot every positive triple is realizable by three positive roots. Since the requested statement only asserts existence of a function on $(0,\infty {)}^3$, extend $h$ arbitrarily on the nonrealizable triples; no choice there affects the formula for a deformation in continuum mechanics. Smooth invariant representations, where needed for differentiating the constitutive equation, are understood locally on the realizable set.

Put $T=t-t^{\prime }$. momentum eigenstates evolve with energy $p^2/(2m)$, so the retarded momentum kernel is $\theta (T)\delta (p-p^{\prime })e^{-i{p}^{2}T/(2m\hslash )}$. Its Fourier transform reduces toCompleting the square and defining the oscillatory Gaussian integral by a convergent limiting prescription givesFor $T>0$ the square root has quantum phase $e^{-i\pi /4}$. As $T\downarrow 0$, this kernel tends to $\delta (x-x^{\prime })$ in the distributional sense. Differentiating the Heaviside step function consequently supplies the required $i\hslash \delta (T)\delta (x-x^{\prime })$ source; an extra $-i/\hslash$ factor would correspond to a different Green-function convention.

Let $z=E-V_0+i0$. The outgoing analytic branch of Klein-Gordon step momentum iswhere each square root is continued from the upper half-plane. It behaves as $z/(\hslash c)$ at large positive or negative real $z$. The branch points are $E=V_0\pm M$; replacing this product by a principal square root of $z^2-M^2$ can choose the wrong sign below the lower threshold.

I. For $E-V_0>M$, $q>0$ and both transmitted kinetic energy and group velocity are positive. Thus $0\le \mathcal{R}<1$ and $\mathcal{T}=4kq/(k+q{)}^2>0$; the transmission decreases to zero on approaching the upper threshold from above.

II. For $|E-V_0|<M$, write $q=i\kappa$ with $\kappa =\sqrt{M^2-(E-V_0{)}^2}/(\hslash c)>0$. The right wave decays, $r=(k-i\kappa )/(k+i\kappa )$ has modulus one, and $\mathcal{T}=0$. There is an evanescent tail but no flux transmitted to infinity. Its reflection quantum phase changes continuously between the two thresholds.

III. For $E-V_0<-M$, the analytic continuation gives $q=-Q$, where $Q=\sqrt{(E-V_0{)}^2-M^2}/(\hslash c)>0$. The group velocity $c^2\hslash q/(E-V_0)$ is again rightward, while the charge density and transmitted charge current are negative. ThereforeChoosing $q=+Q$ would describe a wavepacket arriving from the right, inconsistent with the chosen incoming channel. As the lower threshold is approached, $q\to 0^{-}$ and $\mathcal{R}\to 1$ from above. This Klein-Gordon step amplification has a real denominator zero at $E=V_0/2$ when $V_0>2M$: there $Q=k$, and no finite stationary solution with unit incoming amplitude satisfies both matching equations. The resulting formal pole must not be described as an infinite single-particle probability. Finite switching, wavepackets and the treatment of the external source require a multiparticle analysis.^[ref]

For $r>0$, differentiate $r^2=x^2+y^2$ and use the two differential equations:Substituting $x=r\cos \theta$, $y=r\sin \theta$ gives the required equations in polar coordinates. The origin is handled in Cartesian coordinates, where it is an equilibrium point; an angular coordinate is not defined there.

A Lyapunov function near an equilibrium point is a continuously differentiable positive-definite scalar function $V$ whose derivative along solutions is nonpositive. This implies local Lyapunov stability under the usual comparison bounds; if its derivative is negative definite, the Second Lyapunov theorem gives local asymptotic stability. Here takeIf $a<f(0,0)$, continuity gives a small ball and $\delta >0$ such that $f(x,y)-a\ge \delta$ inside it. Consequently $\dot{V}\le -2\delta V$. A sufficiently small initial sublevel set stays inside the ball, and $V(t)\le V(0)e^{-2\delta t}$. The origin is asymptotically stable, in fact locally exponentially stable. If $a>f(0,0)$, continuity instead gives $\dot{V}\ge 2\delta V$ in a small ball. Every nonzero initial point there exits that ball in forward time, proving instability of the origin. These inequalities hold along all local solutions; continuity of $f$ alone need not supply uniqueness for a general vector field.

If $f=g(r)$, a nonzero constant radius $r_0$ satisfying $a=g(r_0)$ gives a periodic orbit of period $2\pi$, since $\dot{\theta }=1$. Linearizing the scalar radial equation there givesThus the periodic orbit is orbitally asymptotically stable for $g^{\prime }(r_0)>0$ and unstable for $g^{\prime }(r_0)<0$. The direction along the periodic orbit is neutral, as expected for an autonomous periodic orbit; orbital asymptotic stability means attraction in a dynamical system to the orbit with a possible phase shift.

For the specified radial response,The origin is therefore stable for $0<a<3/2$ and unstable for $a>3/2$. The nonzero-radius equation has roots of an equationretaining only positive radii. For $1<a<3/2$, the square root exceeds one, so only $r_{+}>1$ exists and $g^{\prime }(r_{+})<0$: the periodic orbit is unstable. For $3/2<a<2$, both radii are positive; $0<r_{-}<1$ gives $g^{\prime }(r_{-})>0$ and a stable periodic orbit, while $r_{+}>1$ remains unstable.

As $a$ increases through $3/2$, the stable small-radius periodic orbit emerges from the origin while the origin loses Lyapunov stability. This is the supercritical Hopf-type transition with a nonsmooth radial response requested. It follows directly from the explicit radial dynamics. There is a regularity qualification: $g(r)=3/2+r/2+O(r^2)$ near zero, so the vector field in Cartesian coordinates contains $x\sqrt{x^2+y^2}$ and $y\sqrt{x^2+y^2}$ terms. It is continuously differentiable but not generally twice differentiable at the origin, and the periodic orbit radius grows as $r_{-}\sim 2(a-3/2)$ rather than the square-root law of a generic smooth Hopf normal form. One should therefore use the explicit computation here rather than invoke a smooth Hopf bifurcation theorem with unmet hypotheses. At $a=3/2$ itself the origin is still weakly attracting, because $a-g(r)<0$ for small positive $r$.

The update in Newton's method is exactly $x_n=g(x_{n-1})$. Positivity is preserved, and the arithmetic-geometric mean inequality from part (a) applies to every update. ConsequentlyThis holds even when $0<x_0<\sqrt{a}$: the first update moves to or above the positive square root.

Use the iterator adapter from part (b) with a primality test by trial division. A composite positive integer has a divisor no greater than its square root.Before each trial division iteration, no integer in $[2,d)$ divides n. If division succeeds, the condition becomes false with $d\le n/d$, and isPrime returns false. Otherwise the loop ends with $d>n/d$, so every possible small divisor has been excluded. The quotient comparison is equivalent to $d^2\le n$ for positive values and avoids integer overflow in d*d.Bounds at most two give an empty iterator. This method tests candidates lazily and uses $O(\sqrt{n})$ trial divisions for candidate $n$; no efficiency requirement in the question forces a sieve of Eratosthenes implementation.

Use trial division on a shrinking residual. Once the trial divisor exceeds its square root, any residual larger than one is itself prime.For the outer loop invariant, let $P$ be the set of primes below trial and $e_p$ their exponents in the original $n$. Thenand residual has no prime factor below trial. If trial divides the residual it must therefore be prime: a composite trial value would have a smaller prime factor, contradicting this clause. If it does not divide, the zero exponent contributes the factor $1$.

For the inner loop invariant, let $r_0$ be the residual on entry. Then $r_0={\mathrm{trial}}^{\mathrm{exponent}}\mathrm{residual}$. Each exact division reduces the positive residual by at least a factor of two, so this loop terminates. At exit the exponent is the full multiplicity of trial; multiplying by exponent+1 preserves the outer loop invariant. The outer loop terminates because trial increases and residual never increases. When it exits, a composite residual would have a prime factor at most its square root, yet every smaller trial factor has already been excluded. Hence the residual is $1$ or a single remaining prime of exponent $1$.

The procedure computes $\tau (n)$ in $O(\sqrt{n})$ unit-cost time and $O(1)$ auxiliary space, with the separate successful divisions bounded by $\lfloor {\log }_{2}n\rfloor$. No array or table of primes is needed, and $n=1$ correctly returns $1$.

The second procedure should reach large outputs faster, because the first makes exactly $n$ expensive MOD operations when processing argument $n$, whereas the second tries only up to the square root of a shrinking residual, with at most $\lfloor {\log }_{2}n\rfloor$ successful DIV operations. Each successful division also requires a successful MOD test; each outer iteration adds one final unsuccessful MOD test. Thus its expensive-operation count is $O(\sqrt{n}+\log n)$ per argument, and at most $O(n^{3/2})$ cumulatively through argument $n$, compared with $\Theta (n^2)$ for direct counting.

The speedup is not an assertion that every single call uses fewer divisions: for $n=4$, the second code makes three MOD tests and two DIV operations, while the first makes four MOD tests. Small constant costs can go either way. The many small prime factors in highly composite numbers also shrink the residual early, which often helps further; the cumulative comparison includes every preceding argument, not only the printed records.