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Use $d(\ln \sec x)/dx=\tan x$ and $\ln \sec 0=0$. To obtain enough terms of tangent, write $t(x)=x+a{x}^3+b{x}^5+O(x^7)$; it is an odd function. The identity $t^{\prime }=1+t^2$ givesso $a=1/3$, $b=2/15$. Integrate this Taylor expansion term by term from zero to $x$:There is no constant term, and only even powers occur because $\ln \sec x$ is an even function near zero. Thus reaching the first three nonzero terms requires expansion through degree six.

Let $g(x)=\tan (x+\pi /4)$. The derivative identity for tangent gives $g^{\prime }=1+g^2$, so $g(0)=1$, $g^{\prime }(0)=2$, and $g^{\prime \prime }(0)=2g(0)g^{\prime }(0)=4$. Its first three nonzero Taylor expansion terms are thereforeThe factor $1/2!$ on the second derivative is what turns $g^{\prime \prime }(0)=4$ into the coefficient $2$.

Use ${\tan }^{2}x={\sec }^{2}x-1$ and the derivative of the tangent:The antiderivative applies on each interval avoiding the poles of $\tan x$.

Use separation of variables on intervals avoiding zeros of $\cos x$:Taking the tangent gives the real general solutionEach solution is restricted to an interval where both the original equation and this expression are finite. There is no real constant solution because $1+y^2$ never vanishes for real $y$.

The tangent has poles at $\pi /2$ and $3\pi /2$, so these are vertical asymptotes of $x\tan x$. The zeros on the given interval are $0,\pi ,2\pi$. Near zero, $x\tan x=x^2+O(x^4)$, so the graph starts with a horizontal tangent line.

Each of its three branches is increasing. Indeedand the numerator starts at zero with derivative $2{\cos }^{2}x\ge 0$, so it is positive for $x>0$. The first branch rises from zero to $+\infty$, the middle branch rises from $-\infty$ through $(\pi ,0)$ to $+\infty$, and the last rises from $-\infty$ to $(2\pi ,0)$.

The tangent half-angle substitution gives $\sin \theta =2t/(1+t^2)$ and $\cos \theta =(1-t^2)/(1+t^2)$. Dividing and then taking the reciprocal square of the cosine givesThese formulas apply where $t$ is defined and $t\ne \pm 1$; at $t=\pm 1$ the original tangent and secant function are undefined because $\cos \theta =0$.

The initial condition gives $C=\sin (\pi /3)=\sqrt{3}/2$. Since $\pi /3$ belongs to the principal inverse sine range, choose $k=0$ in part (i). The particular solution isThe upper endpoint is excluded because $y/x\to \pi /2$ there and the original tangent ceases to be defined.

Use the substitution $v=y/x$ for a homogeneous first-order differential equation. Since $y^{\prime }=v+x{v}^{\prime }$, it givesWhere $\sin v\ne 0$, this separable differential equation integrates to $\ln |\sin v|=\ln |x|+C$, hence $\sin v=C_{1}x$. One can obtain the same relation without division by $\sin v$: the equation gives $(\sin v{)}^{\prime }=(\sin v)/x$, so $\sin v=C_{1}x$, including $\sin v\equiv 0$.

All real branches of the inverse sine are captured byEach branch is used on an interval with $x\ne 0$ and $|Cx|<1$, with fixed $k$. The strict inequality excludes zeros of $\cos (y/x)$, where the original tangent is undefined. When $C=0$, the formula includes all straight-line solutions $y=k\pi x$. Keeping both families of inverse sine branches is essential for the two different initial conditions below.

Put $s=\sin \theta$. By part (i), $\sin \varphi =\sqrt{2}s$. The acute-angle geometry requires $0<s<1/\sqrt{2}$. Expressing the tangents in terms of their sines in the geometric constraint gives the required implicit equationThe hint suggests $s=1/m$ for an integer $m\ge 2$. Testing $m=2$ givesThus $\boxed{\sin \theta =1/2}$. Each term on the left is strictly increasing in $s$ over the allowed interval, so this solution is unique.

The derivative of the tangent is ${\sec }^{2}x$, and the order of the endpoints must be retained. Thus the definite integral is

Differentiate the flux:On $-1<z_0<1$, $\Phi$ is strictly increasing. Its limiting endpoint values are $\Phi (-1)=0$ and $\Phi (1)=4\pi$, and $\Phi (0)=2\pi$. It has horizontal limiting tangents at the endpoints, curves upward for $z_0<0$, curves downward for $z_0>0$, and has an inflection point at zero.

The derivative graph is a downward-opening parabola, approaching zero at both endpoints and attaining its maximum $3\pi$ at $z_0=0$.

Consider an isolated intersection of the two curves. For two distinct lines, solving their linear equations by elimination gives coordinates in $F$ whenever there is an intersection.

For a line and a circle, solve the line for one coordinate with a nonzero coefficient and substitute into the circle equation. The resulting quadratic has coefficients in $F$. A real root lies in $F$ if its discriminant is a square in $F$, and otherwise in $F(\sqrt{D})$ for its positive nonsquare discriminant $D\in F$. The other coordinate is an $F$-linear expression in that root. Thus both coordinates lie in the same real quadratic extension. A tangent intersection has discriminant zero and already lies in $F$.

Write two circle equations in normalized form $x^2+y^2+u_{i}x+v_{i}y+w_i=0$. Subtracting eliminates the quadratic terms. If the circles are distinct and meet, this gives a nontrivial line, reducing the problem to the preceding case. Concentric distinct circles do not intersect. ThereforeThe isolated-intersection qualification is implicit in the construction operation. If coincident curves were allowed and every point on their common curve counted as an intersection point, the printed assertion would be false: a rational line contains points with transcendental coordinates.

Orient the body boundary counterclockwise and write its tangent displacement as $(dx,dy)$. The outward normal from the body has displacement-weighted components $(dy,-dx)$. Pressure therefore exerts the force components $F_x=-\oint pdy$ and $F_y=\oint pdx$, orFor steady irrotational incompressible flow at constant density, Bernoulli gives $p=p_0-\rho |w^{\prime }(z){|}^2/2$, with any constant far-field kinetic term absorbed into $p_0$. Its constant contribution integrates to zero. The boundary is a streamline, so $\mathrm{Im}w$ is constant there and $w^{\prime }(z)dz$ is real. Thus $w^{\prime }dz=\overline{w^{\prime }}d\bar{z}$, giving $|w^{\prime }{|}^{2}d\bar{z}=(w^{\prime }{)}^{2}dz$. This identity element remains valid at a stagnation point as both sides vanish. Substitution yields Blasius force theorem:

Choose the heavier particle's outgoing velocity above the initial direction. Let its speed be $V$, and let the lighter particle have components $(v\cos \theta ,-v\sin \theta )$. Thus $\theta$ and $\varphi$ are nonnegative deflections on opposite sides. Conservation of momentum givesThe elastic collision conserves kinetic energy:Taking the squared magnitude of the vector momentum equation yieldsEliminate $v^2$ with the kinetic energy equation. For a nontrivial collision, $V\ne 0$, soSubstituting back into the two momentum components givesThereforeThe ratio alone does not determine the quadrant: for $M>m$, the lighter particle can move backwards. The two component equations select the correct $0\le \theta \le \pi$. At $m/M=\cos 2\varphi$, the outgoing direction is perpendicular to the initial direction; the displayed tangent has a pole there.

Single-slit diffraction is the spreading of light transmitted by one slit. Consider an ideal rectangular slit of width $a$, illuminated normally by a uniform plane wave with a single wavelength $\lambda$ in the surrounding medium. This monochromatic wave has uniform wave amplitude and equal incident wave phases across the slit, providing the required optical coherence. In the scalar Fraunhofer diffraction model, the normalized angular wave intensity isHere $\theta$ is measured from the forward normal in the plane across the slit width. The sinc function uses the unnormalized convention $\mathrm{sinc}u=\sin u/u$. The limit makes the central value one. This ideal envelope neglects additional direction-dependent propagation and detector factors; near the forward direction it describes the normalized screen irradiance. ^[ref].

The noncentral zeros satisfyThe restriction follows from the range of the sine of a real angle. The first minima have $\sin \theta =\pm \lambda /a$ when $\lambda /a\le 1$. If $a<\lambda$, this model has no noncentral zero in the forward hemisphere; equality puts the first zeros at grazing directions, outside the small-angle approximation and off any finite perpendicular screen. ^[ref].

For $0<\lambda /a<1$, define the central width as the separation between the two first minima. Its angular width and its width on a perpendicular screen areThe angular approximation uses radians and requires $\lambda /a\ll 1$; the screen approximation uses $y\approx L\theta$. The tangent relation is exact straight-line geometry, even though the underlying Fraunhofer diffraction field is approximate. This width is distinct from the full width at half maximum.

For $\lambda =500\mathrm{nm}$, $a=0.1\mathrm{mm}$ and $L=1\mathrm{m}$, the positive first minimum isThe dimensionless far-field parameter is $a^2/(\lambda L)=0.02$. Doubling the slit width to $a=0.2\mathrm{mm}$ gives $W\approx 5.00002\mathrm{mm}$ and parameter $0.08$; both curves below use the Fraunhofer diffraction approximation. Choosing $L=2\mathrm{m}$ halves those parameters and doubles both screen widths.

Each curve is normalized to its own $I_0$; the arrows measure between first minima. The calculation uses $\theta =\arctan (y/L)$ before evaluating the sinc function, rather than replacing $\sin \theta$ by $y/L$. The lower diagram shows the exact screen geometry with the vertical scale enlarged. Normalization removes the change in peak brightness when the slit width changes.

The wave interference explanation is especially simple: divide the slit into equal narrow strips and add their complex amplitudes, rather than their wave intensities. With wavenumber $k=2\pi /\lambda$, the continuous slit givesIntegrating the complex exponential gives the endpoint difference; its continuous value at zero is one. Taking the squared complex modulus gives the stated wave intensity. This is the Fourier transform of an interval indicator expressed in angular coordinates. Feynman's continuous-source argument obtains the same pattern by letting the number of sources tend to infinity. ^[ref].

At the positive first minimum, the wave phases span exactly $2\pi$ across the slit. Every strip in one half has a partner $a/2$ away, with equal wave amplitude and phase difference $\pi$, so the pairs cancel. At $\theta =0$, all the wave phases align instead. ^[ref].

The drawing uses $N=8$ equally weighted midpoint samples $x_j/a=-1/2+(j+1/2)/8$, with $j=0,\dots ,7$. At the first minimum their complex amplitudes are a common rotation of the eight roots of unity, so their head-to-tail polygon closes exactly. This finite sum illustrates cancellation; the continuous integral, rather than eight point sources, defines the slit pattern at general angles. The circle shows the continuous range of wave phases; coloured arrows show the eight samples.

A narrower aperture requires a larger angle to accumulate the cancelling phase difference, so narrowing the slit widens the beam. This is a useful limit on geometrical optics: confining light more tightly at an opening increases its subsequent angular spread. At fixed incident irradiance and slit height, reducing $a$ also reduces transmitted power, exposing a practical balance between diffraction-limited resolution and light throughput. In a diffraction grating, this single-slit diffraction envelope modulates the wave interference between different slits.

For $A,ϵ>0$, consider $\tan \theta =-ϵ\theta /\sqrt{A^2-{\theta }^2}$ on $0<\theta <A$. On each negative branch of the tangent, the left side increases strictly, while the right side decreases strictly. A unique intersection exists precisely when the branch starts at $p_n=(n-1/2)\pi <A$, since the difference increases from negative infinity to a positive value or positive infinity. Consequently the number of trapped normal modes is $\lceil A/\pi -1/2\rceil$. The floor function expression $\lfloor A/\pi +1/2\rfloor$ agrees except at $A=p_n$, where it counts an inadmissible cutoff state with zero exterior exponential decay rate.

For a point $B$ of a curve, let its ordinate meet the chosen coordinate axis at $C$, and its tangent meet that axis at $E$. The segment $CE$ is the subtangent. It depends on the chosen axis. For the standard parabola, $CE=2CD$, where $D$ is its vertex of a parabola.

The tangent is the trigonometric function $\tan x=\sin x/\cos x$ where $\cos x\ne 0$.

Posthumous Toulouse edition of mathematical works of Pierre de Fermat. The adequality and tangent discussions belong to an earlier period than this printing. The ETH-Bibliothek record gives the original title, date and publisher.

Online source: www.e-rara.ch/zut/content/titleinfo/6491216