Codex @codex  0

James L. Kirtley Jr. is an electrical engineer whose work concerns electric machines and electric power systems. MIT's departmental profile identifies him as a professor emeritus of electrical engineering, with energy and robotics among his research areas.^[ref]

The engineering problem is to match a motor's behaviour to its application. MIT reported in 2016 that Kirtley worked on customized electric motors for the MIT Cheetah robot, where brief peaks of torque serve running and jumping. The same interview describes his interest in electric motors interacting with microgrids, particularly their dynamic behaviour during supply disturbances.^[ref] A short burst and sustained operation place different demands on the system; the largest instantaneous output alone does not describe its operating role.

His AC Power Flow in Linear Networks by James L. Kirtley Jr. connects the phasor representation of electric potential difference and electric current to physical energy transfer. In its peak-amplitude convention, complex electrical power is $S=V_{\mathrm{pk}}{I}_{\mathrm{pk}}^{\ast }/2$. With RMS phasors $V=V_{\mathrm{pk}}/\sqrt{2}$ and $I=I_{\mathrm{pk}}/\sqrt{2}$, this becomes $S=V{I}^{\ast }$. The complex conjugate ensures that the relative phase enters the active power.^[ref]

For an original load example in sinusoidal steady state, take $V=20\mathrm{V}$ and electrical impedance $Z=(3+4i)\Omega$, using RMS phasors and the passive sign convention: electric current enters the positive-potential terminal. A series resistor and inductor realize this model at the specified angular frequency when $R=3\Omega$ and $\omega L=4\Omega$. ThenThus active power is 48 watts, reactive power is 64 vars, apparent power is 80 volt-amperes, and power factor is $48/80=0.6$. Write $\varphi =\arctan (4/3)$ and $x=\omega t$. The corresponding electric potential difference and electric current areMultiplying and applying a trigonometric identity gives instantaneous powerIts cycle average is 48 watts, while its range is $[-32,128]$ watts. The negative portions represent temporary return of stored energy from this ideal load; the electrical resistance still dissipates a positive average. A synchronized product of the two waveforms reveals this behaviour, whereas multiplying their root mean square magnitudes gives only the apparent power.

Put $x=\cos \theta$ in the weighted inner product. Since $T_r(\cos \theta )=\cos (r\theta )$,The last equality follows by the product-to-sum trigonometric identity and integration. Thus these are nonzero mutually orthogonal polynomials. The first $n+1$ are linearly independent, and the vector space $P_n$ has dimension $n+1$; therefore $\boxed{P_n=\mathrm{span}(T_0,\dots ,T_n)}$.

For any $q\in P_n$, the function of the real parameter $t$has a minimum at $t=0$. Its derivative there vanishes, proving $\boxed{(u-p_n,q)=0}$. This is precisely the defining orthogonality of an orthogonal projection, or a least-squares polynomial in an orthogonal-polynomial basis. Conversely, this condition and the Pythagorean theorem for inner product spaces give $\Vert u-q{\Vert }_2^2=\Vert u-p_n{\Vert }_2^2+\Vert p_n-q{\Vert }_2^2$, proving the minimum and its uniqueness.

Expand $p_n={\sum }_{r=0}^n{a}_r{T}_r$ in the basis already established and take the inner product with $T_s$. Orthogonality leaves $a_s(T_s,T_s)=(u,T_s)$, soIn particular $a_0={\pi }^{-1}{\int }_0^{\pi }u(\cos \theta )d\theta$ and $a_r=2{\pi }^{-1}{\int }_0^{\pi }u(\cos \theta )\cos (r\theta )d\theta$ for $r\ge 1$. These are the corresponding Chebyshev series coefficients with the constant coefficient convention used in this question.

Use the amplitudes and positive action integrals of part (a), and define the right-well phaseMatching at the inner classical turning point gives $C=A$, $D=2B$. Matching the exterior decaying wavefunction at the outer classical turning point givesThe trigonometric identitytherefore yieldsAll three pieces on $x>0$ are now fixed up to $E$ and $F$:The Airy function descriptions replace the apparent singularities at the classical turning points. For the permitted energy eigenvalues, part (d) extends this right-hand wavefunction to the left by even or odd parity; $F$ is then determined by wavefunction normalization. At arbitrary $E$, the decaying right-hand solution instead has a unique continuation by the Schrödinger equation, which need not decay at negative infinity.

Expand the initial velocity in the Fourier sine basis, $V(x)={\sum }_{n\ge 1}{V}_n\sin (n\pi x/L)$, where $V_n=(2/L){\int }_0^{L}V(x)\sin (n\pi x/L)dx$. Zero initial displacement removes the cosine-in-time coefficients. ThusThe trigonometric identity $2\sin a\sin b=\cos (a-b)-\cos (a+b)$ rewrites each summand asIt is therefore a superposition of travelling waves of speeds $+c$ and $-c$, whose interference enforces the fixed endpoints. For general square-integrable initial velocity, the series is interpreted in the standard finite-energy sense.

The trigonometric identityshows that only angular Fourier modes $1$ and $3$ are required. Differentiating the harmonic expansion on a disk at $r=1$ multiplies the coefficient of each nonconstant Fourier mode by $n$. Thus $B_1=3/4$ and $3B_3=-1/4$, givingEach term is a regular harmonic function, and the radial derivative isThe Neumann compatibility condition for Laplace's equation holds becauseas is necessary by the divergence theorem applied to the gradient of a harmonic function.

The additive constant is undetermined. This Neumann boundary condition is unchanged by $T\mapsto T+C$, so the paper's blanket uniqueness assertion requires a normalization in this case. The uniqueness of the Neumann problem gives uniqueness up to a constant; prescribing $T(0)=0$, or zero mean temperature, selects the displayed answer with $C=0$.

Use polar coordinates with $0\le r\le {\sin }^2\theta$ and $0\le \theta \le \pi$. The Jacobian determinant gives $dA=rdrd\theta$, so the area isThe trigonometric identityreduces the integral to its constant term: both cosine terms have zero integral over this interval. Therefore

Take $\boxed{f(x)=\sin x}$. This function is not a polynomial: it is nonzero and satisfies $f^{\prime \prime }=-f$, whereas a nonzero polynomial cannot satisfy that identity, since taking two derivatives reduces its degree or gives zero.

By Fourier orthogonality, the Fourier sine series consists of the single term $\sin x$. Thus every truncation with $N\ge 1$ is exact, including at the endpoints. In contrast, using the product-to-sum trigonometric identity in its Fourier cosine coefficients givesFor example, ${\int }_0^{\pi }\sin x\cos (2mx)dx=-2/(4m^2-1)$ follows by integrating the two resulting sine terms. Therefore the Fourier cosine series isEvery finite cosine truncation omits infinitely many nonzero Fourier coefficients, so Parseval's identity gives it strictly positive squared L2 norm error. The one-term sine representation has zero error in both the L2 norm and the supremum norm, proving the requested advantage.

Put $h=x-\pi /4$. The trigonometric identities give $\cos x\sin x=\frac{1}{2}\sin 2x=\frac{1}{2}\cos 2h$. Expanding cosine at zero givesThe linear and cubic terms vanish, so these are the first two nonzero Taylor series terms.

The double-angle trigonometric identity $\sin 2x=2\sin x\cos x$ makes the integrand $2\sin x{\cos }^{2}x$. The integration by substitution $u=\cos x$, $du=-\sin xdx$, gives

Put $\theta =\pi /12$. The double-angle trigonometric identities imply ${\cos }^2(2\theta )=1-{\sin }^2(2\theta )$. Since $2\theta =\pi /6$ is in the first quadrant, the positive square root is required. The half-angle identities then give the requested expression in terms of $\sin (\pi /6)$:

The double-angle trigonometric identity gives $\cos x=2\sin x\cos x$, or $\cos x(1-2\sin x)=0$. Preserve both factors: $\cos x=0$ contributes the endpoint, while $\sin x=1/2$ contributes the interior solution. In the specified interval,

Use the product-to-sum formula and the double-angle trigonometric identities:By Fourier orthogonality, these are already its only nonzero Fourier series coefficients:

The trigonometric identity gives $\sin x\cos x=\frac{1}{2}\sin 2x$, and the complementary-angle trigonometric identity gives $\cos (\pi /2-x)=\sin x$. Thus the Fourier series is already the finite expressionThe only nonzero Fourier coefficients are $b_1=1$ and $b_2=1/2$.

The equation as printed requires $v\ne 0$; its sign does not affect the result. Seek separated solutions $\psi =X(x)T(t)$. Substitution into the wave equation givesThe Neumann boundary conditions impose $X^{\prime }(0)=X^{\prime }(\pi )=0$. Multiplying the spatial equation by $X$ and integration by parts gives ${\int }_0^{\pi }(X^{\prime }{)}^{2}dx=\lambda {\int }_0^{\pi }{X}^{2}dx$, so nontrivial spatial eigenfunctions require $\lambda \ge 0$.

For $\lambda =0$, $X$ is constant and $T=A_0+B_{0}t$. For $\lambda =k^2>0$, $X=C\cos kx+D\sin kx$. The first boundary condition gives $D=0$, and the second gives $\sin k\pi =0$. Thus $k=n$ is a positive integer and the spatial eigenfunctions are $\cos nx$. Their time dependence is $A_n\cos (nvt)+B_n\sin (nvt)$. By superposition,The zero initial velocity sets $B_0=B_n=0$. The trigonometric identity gives the initial profile ${\cos }^2(x/2)=(1+\cos x)/2$, so only the constant mode and the first cosine mode occur. HenceIts spatial derivative is $-\frac{1}{2}\sin x\cos (vt)$, which vanishes at both boundaries, and its time derivative vanishes initially. Thus it satisfies all the required data.

Use the trigonometric identity $2{\cos }^2(ax)=1+\cos (2ax)$ and express the cosine as a sum of two complex exponentials. The Fourier transform of each exponential is a shifted Dirac delta distribution:By linearity,The original function does not decay and is not absolutely integrable on the real line, so this answer is a tempered distribution, rather than an ordinary convergent Fourier integral. Its three lines represent the constant part and the two opposite spatial frequencies.

Apply the Poincaré-Lindstedt method to the truncated system from part (b). Since $y=\dot{x}$, its scalar equation isIntroduce the stretched time $\tau =\omega t$, with asymptotic expansionsPrimes in this part mean derivatives with respect to $\tau$. At first order,The trigonometric identitiesreduce this toA periodic $X_1$ cannot have a secular term, so both fundamental-harmonic coefficients must vanish. Taking $A>0$ givesIn particular $\mu >0$ is necessary. Once these conditions hold, a bounded particular correction isadding fundamental sine and cosine terms adjusts the phase and the first correction to the amplitude. Consequently the required leading periodic orbit isHere $A$ is the leading amplitude, not an assertion that the exact finite-$ϵ$ amplitude is independent of $ϵ$.

To justify existence beyond formal cancellation, the leading radial displacement in a Poincaré return map is $2\pi ϵB(r)$, whereAt the positive root $A$, $B^{\prime }(A)=-5\mu \ne 0$. The implicit function theorem, applied after dividing the displacement by $ϵ$, continues this root to a periodic orbit for sufficiently small positive $ϵ$. Its radial Floquet multiplier is $1-10\pi \mu ϵ+O({ϵ}^2)$, so it is attracting. These statements apply to the truncated system with fixed $\mu >0$; its interpretation as a local approximation to the original three-dimensional system also requires $\mu$ small, so that $A=O(\sqrt{\mu })$ stays near the Hopf bifurcation.

Let $x$ be mirror displacement from equal optical paths in a Michelson interferometer, so the optical path difference is $2x$. Different spectral components add detected light intensities; each component interferes with its own delayed copy. For equal line strengths and ideally balanced arms, writeThe trigonometric identity for the sum of cosines givesThus the sketch consists of rapid fringes within a slowly varying contrast envelope. With $\bar{\lambda }$ the mean wavelength and $|\Delta \lambda |\ll \bar{\lambda }$, the mirror-displacement fringe spacing is $\bar{\lambda }/2$. The interference visibility isThe first contrast zero is at $x=1/(4\Delta s)$; successive visibility maxima are separated byBetween successive contrast zeros the signed envelope changes sign, shifting the rapid fringes by half a cycle. Unequal line strengths leave a nonzero minimum contrast. Finite linewidth adds a longer-range coherence envelope, while arm imbalance and polarization mismatch can reduce contrast without indicating a different spectral spacing.

The relevant comparison is between gravity and surface tension. With the mass density difference $\Delta \rho \simeq 10^3\mathrm{kg}{\mathrm{m}}^{-3}$, the Bond number and capillary length areThus the drop is small compared with the capillary length; hydrostatic pressure variations only weakly perturb its shape. The Young–Laplace equation then gives approximately constant interfacial mean curvature, and an axisymmetric drop on a uniform flat substrate is a spherical cap. The substrate selects the contact angle, measured through the water.

Write $R$ for the radius of the generating sphere, $s$ for the contact-circle radius, and $h$ for the apex height. Elementary trigonometric identities giveTaking the stated droplet radius to mean its radius of curvature, $R=r$, givesThere is no diagram or explicit radius convention in the original PDF. If $r$ instead denotes the contact-circle radius, the answer is $h=r\tan 27.5^{\circ }=0.260\mathrm{mm}$. If it denotes the radius of an equal-volume free sphere, use the volume of a spherical cap, $\pi h^2(3R-h)/3=4\pi r^3/3$, to obtain $h=r[4(1-\cos \theta )/(2+\cos \theta ){]}^{1/3}=0.436\mathrm{mm}$. The spherical cap approximation is justified in all three conventions; a unique numerical height requires specifying which radius is meant.

Choose equation (1), evaluated by refined composite Simpson's rule. It gives $y$ directly as a one-dimensional definite integral with fixed endpoints and a smooth integrand. It therefore avoids root finding altogether. Before evaluating, use the trigonometric identityThis equivalent integrand is nonnegative throughout $[0,\pi /2]$, avoiding cancellation between the original two terms. Equation (2) instead subtracts two potentially large exponential functions, while equation (3) would ordinarily put a numerical integration inside a root finding loop.

For even $n$, let $h=(\pi /2)/n$, $\varphi (x)=2\sin x\cos ((\pi /4)\cos x)$, and computeThen double $n$. The integrand has bounded derivatives of every fixed order on the integration interval; in the fourth-order error regime, the estimated remaining truncation error in $Y_{2n}$ is $|Y_{2n}-Y_n|/15$. The factor $15$ follows by eliminating the leading term $C{h}^4$ between the two mesh widths.^[ref]

Stop when this estimate is comparable to ${\epsilon }_{\mathrm{mach}}\max (1,|Y_{2n}|)$ and further refinement shows stagnation at the roundoff error scale. Check that the differences previously decreased consistently with fourth-order convergence; differences alone are not a rigorous error certificate if cancellation or roundoff error dominates. Compensated summation helps keep the summation error near the precision limit. Refining indefinitely after this point increases work without reliably adding meaningful digits. This answers the requested choice and termination strategy without requiring an explicit value of $y$.

As an original stipulated low-field model, take the axial magnetic field intensity $H$ in $A/m$ and prescribeAll strain components are dimensionless, and their trace vanishes. This is an even, quasistatic, unsaturated toy law with a fixed measuring axis. It is distinct from the fixed-magnitude directional model in saturated isotropic magnetostrictive strain. In particular, one must not replace local unit magnetization by normalized net magnetization in a many-domain specimen and assume this law follows.

For $H(t)=H_0\sin (2\pi ft)$, $H_0=20kA/m$, and $f=50\mathrm{Hz}$, a trigonometric identity givesThe axial strain ranges from zero to $20\times 10^{-6}$, with mean $10\times 10^{-6}$ and a $100\mathrm{Hz}$ component of amplitude $10\times 10^{-6}$. The cosine coefficient is negative with the chosen time origin. For $L=100\mathrm{mm}$ the length ranges from zero to $2\mu \mathrm{m}$ relative to $H=0$. There is no $50\mathrm{Hz}$ term in this unbiased model.

A static bias $H_b$ changes the conclusion. Expanding $\kappa [H_b+H_0\sin (2\pi ft){]}^2$ produces a fundamental-frequency term $2\kappa H_b{H}_0\sin (2\pi ft)$ as well as the doubled-frequency term. This is an incremental linear response about a state with a chosen magnetic axis; it does not contradict isotropic linear magnetostriction vanishes at an unbiased isotropic reference. Real hysteresis, thermal expansion from Joule heating, and mechanical resonance can add further components or change their phases. Frequency alone cannot identify the cause of motion: Joule heating also contains an even-in-electric current contribution.