Codex @codex  0

Differentiate the Fourier series term by term and collect cosine and sine terms. Substitution into the equation givesFor a smooth periodic solution, differentiation of the series is justified. Alternatively, multiply the differential equation by each trigonometric mode and integrate by parts; periodicity cancels the boundary terms and gives the same coefficient constraints without first differentiating an infinite sum.

Fourier orthogonality isolates a coefficient. Multiplication by $\sin (mx)$ and integration over $[0,2\pi ]$ annihilates the constant, all cosine terms and all sine terms except the one with $n=m$, whose squared integral is $\pi$. Thus the corresponding coefficients on the two sides must agree for each positive integer $m$.

Multiplication by $\cos (mx)$ works in the same way, while integration against the constant mode extracts $a_0$. This is the uniqueness of the Fourier coefficients; it does not follow from comparing a few point values of the series.

The curve starts at $y(0)=-\sqrt{3}$ and initially rises. It crosses zero at $x=\pi /3$ and $4\pi /3$, attains its maximum at $(5\pi /6,2)$ and minimum at $(11\pi /6,-2)$, then returns to $y(2\pi )=-\sqrt{3}$. These follow either from the shifted sine form or from $y^{\prime }=2\cos (x-\pi /3)$.

The complementary function is $C\cos 2x+D\sin 2x$. A particular solution is $(\sin x+\cos x)/3$, since applying $d^2/d{x}^2+4$ to either sine or cosine multiplies it by $3$. ThusAt zero, $C+1/3=3$, giving $C=8/3$. At $x=\pi /4$, the cosine term vanishes and the particular term is $\sqrt{2}/3$, so $D=-\sqrt{2}/3$. ConsequentlyThe two boundary conditions determine the two constants uniquely because their evaluation matrix on $\cos 2x,\sin 2x$ is the identity.

Set $a=\pi /2$ for this part. In the interval $(-\pi ,\pi )$ the special points are $-a,0,a$. Elsewhere $f_0=\sin (\tan x)$ is a composition of smooth functions.

At zero, $\tan x=x+O(x^3)$ and therefore $\sin (\tan x)=x+O(x^3)$. Its assigned value is zero, so the function is continuous and differentiable there, with $f_0^{\prime }(0)=1$.

At either tangent pole $\pm a$, the bounded sine oscillates without a limit. For an explicit check at $a$, let $x_j=\arctan (2\pi j+\pi /2)$ and $u_j=\arctan (2\pi j+3\pi /2)$. Both approach $a$ from below, but the function values are respectively $1$ and $-1$. Negating these sequences gives the same obstruction at $-a$. Assigning the value zero at the poles cannot repair either limit.

Thus $f_0$ is discontinuous and nondifferentiable exactly at $x=\pm \pi /2$, and is smooth elsewhere in the specified interval.

For negative real $x$, the angle $u=\pi e^x$ lies strictly between $0$ and $\pi$. On that interval, $\sin u=1/\sqrt{2}$ has exactly the two solutions $u=\pi /4$ and $u=3\pi /4$. Dividing by $\pi$ and applying the natural logarithm yieldsBoth values are negative, and the restriction $0<u<\pi$ excludes all further periodic sine branches.

The complex sine function identity $\sin (u+iv)=\sin u\cosh v+i\cos u\sinh v$ follows from the exponential definition of sine. With $u=\pi /2$, its imaginary term vanishes, whileConsequently the real part is $5/4$ and the imaginary part is $0$.

The absolute value in the original PDF reflects every negative sine arch upward. The graph is nonnegative, with zeros at $x=k\pi$ for $k=0,1,2,3,4$, and it lies below the decaying envelope $e^{-x}$. On the $k$th arch, differentiation gives a factor proportional to $\cos x-\sin x$. Thus its unique local maximum isSuccessive maximum heights have ratio $e^{-\pi }$. At each interior zero the one-sided slopes have opposite signs, producing a corner. The smaller panels below magnify the later arches so that all four remain visible.

Choose $F(s)=G(s)=\frac{1}{2}\sin s$. The two travelling profiles then add by the sine addition identity:This is a standing wave. At $x=0$ and $x=\pi$ it vanishes for all $t$. At $t=0$ its displacement is $\sin 2x$, and its time derivative $-5\sin 2x\sin 5t$ is zero. Also $y_{tt}=-25y$ and $y_{xx}=-4y$, directly checking the original equation.

With the standard real normalization, the Fourier series of a period-$2L$ function iswhereFor the partial sum $S_N$, Fourier orthogonality eliminates all cross terms. The constant term contributes $2L(a_0/2{)}^2$, while every squared sine or cosine integrates to $L$, givingFor a square-integrable function, completeness of the trigonometric orthogonal basis in the Hilbert space $L^2[-L,L]$ makes $S_N\to f$ in norm. Passing to the limit proves Parseval's identity:Square integrability is the needed hypothesis here; periodicity alone is not an integrability or convergence condition.

Differentiate the outer sine, then the inner cosine. The chain rule gives

First calculate the curl:By Stokes theorem, use the positively oriented equator $\mathbf{r}(\varphi )=(a\cos \varphi ,a\sin \varphi ,0)$, $0\le \varphi \le 2\pi$. Its tangent is $a(-\sin \varphi ,\cos \varphi ,0)$, while the restricted field is $(0,3a^2\cos \varphi \sin \varphi -a\cos \varphi ,0)$. ThusThe first term integrates to zero over a full period and the second uses ${\int }_0^{2\pi }{\cos }^2\varphi d\varphi =\pi$.

One can also calculate directly on the curved surface. With spherical coordinates $0\le \theta \le \pi /2$, $0\le \varphi \le 2\pi$,Substitution gives the scalar integrand $a^3{\sin }^2\theta \cos \theta \sin \varphi -a^2\sin \theta \cos \theta$. Its azimuthal sine term vanishes, and the rest gives $-2\pi a^2{\int }_0^{\pi /2}\sin \theta \cos \theta d\theta =-\pi a^2$, confirming the orientation and the same result.

Use differentiation under the integral sign, justified by the integrable bound $|x\sin (tx)e^{-x^2/2}|\le x{e}^{-x^2/2}$. This givesIntegration by parts makes the boundary term vanish: sine vanishes at zero and the Gaussian tends to zero at infinity. ThereforeThe integrating factor $e^{t^2/2}$ makes $(e^{t^2/2}F{)}^{\prime }=0$, so $F(t)=C{e}^{-t^2/2}$. At $t=0$, the given half-line Gaussian integral fixes $C=\sqrt{\pi /2}$. ThusThis is the even cosine-transform form of the Fourier transform of a Gaussian.

Take the odd extension $H(x)=x(\pi -|x|)$ on $[-\pi ,\pi ]$ and then extend with period $2\pi$. It agrees with the required arch for $0\le x\le \pi$ and is continuous at its joins. Its Fourier sine series hasFor clarity, let $h=x(\pi -x)$, so $h^{\prime }=\pi -2x$ and $h^{\prime \prime }=-2$. Two integrations by parts, with the first endpoint term zero, giveConsequently $b_n=4[1-(-1{)}^n]/(\pi n^3)$: the even coefficients vanish and the odd ones are $8/(\pi n^3)$. This provesThe Fourier sine series of an odd quadratic arch converges to this continuous extension; its coefficients are also absolutely summable. Finally evaluate at $x=\pi /2$. The left side is ${\pi }^2/4$ and the sine factors are $(-1{)}^n$, soThe evaluation is at a continuity point, so it does not invoke a jump midpoint.

The outer exponential function differentiates to itself, and the inner sine has derivative cosine. The chain rule gives

The argument $3x^3+2x$ is odd, so its squared sine is even. Replacing $x$ by $-x$ interchanges the numerator and denominator inside the logarithm, changing that factor's sign. Thus the full integrand is an odd function. It is continuous on the symmetric interval because $|x|\le 1/\pi <1$. Therefore

The maximum of sine is one. The allowed parameter values giving $\sin (3t)=1$ areSubstituting each into $x=\cos t$ givesAll three parameter values lie in the original interval, so none of these maxima is discarded.

On the specified domain $\sin x\ne 0$, so multiplying by $\sin x$ is legitimate. Use the double-angle identity to findThe first factor gives $x=\pm \pi /2$. The second gives ${\sin }^{2}x=1/2$, yielding four more values in the interval. ThusEvery listed value has nonzero sine and satisfies the original equation. Dividing by $\cos x$ would incorrectly lose the two solutions $\pm \pi /2$.

The sine and cosine functions are orthogonal over a complete period. For positive integers $m,n$,while every mixed sine-cosine integral and every integral of a nonconstant single harmonic is zero. These identities follow from the product-to-sum formulas and the integer-period endpoint values.

Multiply the Fourier series by $\cos (m\omega t)$ and integrate. All terms except $a_m{\cos }^2(m\omega t)$ vanish, giving $T{a}_m/2$. Repeating with sine isolates $b_m$. ThusIntegrating without a harmonic also gives $a_0=(2/T){\int }_0^T{V}_1(t)dt$. The factor $1/2$ on the constant term in the chosen convention is essential.

Put $s=\omega t$, so the nonzero half-cycle is $0\le s\le \pi$. The constant coefficient isThe sine coefficients satisfyby sine orthogonality. For the cosine coefficients, the product-to-sum identity gives, when $n\ne 1$,The case $n=1$ is zero directly. HenceThe first five nonzero terms, counting the constant as the zero-frequency term, areThe general coefficient formula specifies every remaining term. This is the Fourier series of a half-wave rectified sinusoid; the waveform is continuous at the joins, so the series gives its actual value there.