Codex @codex  0

The inequality is equivalent to $|\cos \theta |>1/\sqrt{2}$. The cosine exceeds $1/\sqrt{2}$ close to zero and $2\pi$, and is less than $-1/\sqrt{2}$ close to $\pi$. Within the specified interval this givesThe four equality angles are excluded because the inequality is strict; zero and $2\pi$ satisfy it and are included.

The increasing substitution $x={\sin }^2\theta$ maps $0\le \theta \le \pi /2$ onto $0\le x\le 1$, with $dx=2\sin \theta \cos \theta d\theta$. Since sine and cosine are nonnegative there, powers combine without a sign ambiguity:This is the trigonometric form of the beta function. The powers at the endpoints are integrable because $p,q>0$.

The cosine-difference identity gives $\cos x-\cos 3x=2\sin 2x\sin x$. HenceEach sine ratio tends to one, soEquivalently, the quadratic terms of the two cosine expansions differ by $4x^2$.

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 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.

Differentiate $x=r\cos \theta$, $y=r\sin \theta$:Squaring and adding cancels the mixed terms and uses ${\cos }^2\theta +{\sin }^2\theta =1$, givingFor this cardioid, the polar arc length integrand isThe cosine inside the absolute value changes sign at $\theta =\pi$. Therefore the perimeter isin coordinate length units. The isolated cusp does not obstruct this piecewise smooth arc length integral.

Assume $L>0$. For any nonzero integer $k$,For $m\ne n$, the product-to-sum identity expresses the integrand as one half the sum of cosines with nonzero frequencies $m-n$ and $m+n$, so the integral is zero. For $m=n>0$, use ${\cos }^{2}u=(1+\cos 2u)/2$ to get $L$. Finally, for $m=n=0$ the integrand is one. The complete Fourier orthogonality formula isThe constant mode has twice the squared norm of a positive-frequency cosine mode.

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

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

Put $q(x)=\sqrt{{\pi }^2/16+x}$. The original PDF confirms that the $+x$ is inside the radical. For the composite cosine,At $x=0$, $q=\pi /4$ and $\sin q=\cos q=\sqrt{2}/2$. Substituting into $f(0)+f^{\prime }(0)x+f^{\prime \prime }(0)x^2/2$ givesAll three displayed coefficients are nonzero.

For a short string element of length $\Delta x$, the small-slope approximation makes the upward components of tension at its two ends $T{y}_x(x+\Delta x,t)$ and $-T{y}_x(x,t)$. Its mass is $\rho \Delta x$. Newton's second law givesDivide by $\Delta x$ and take the limit to obtain the wave equation on a string,Here $\rho$ is linear mass density, and the derivation assumes small transverse slopes, uniform tension, and a flexible string.

At the right endpoint the string pulls the ring toward the interior. Its vertical force is therefore $-T{y}_x(L,t)$ to first order; the frictionless wire supplies no vertical force. The ring obeysThis is the dynamic endpoint law for a string with an endpoint mass.

For a separated oscillatory solution, write $y=X(x)q(t)$ with $q^{\prime \prime }=-{\omega }^{2}q$ and $X^{\prime \prime }+(\omega /c{)}^{2}X=0$. The fixed left end removes the spatial cosine, so the nonzero normal modes haveThe endpoint law gives $M{\omega }_n^2\sin ({\omega }_{n}L/c)=T({\omega }_n/c)\cos ({\omega }_{n}L/c)$. Since $\sin ({\omega }_{n}L/c)=0$ would also force the cosine to vanish, it can be divided out for a nonzero mode. Using $T=\rho c^2$ givesFor physical $M\ge 0$, these give all positive modal frequencies. There is no nonzero static linear mode: the left value and right slope conditions would force it to vanish. Positivity also follows by integration by parts from

For $M=0$, the right endpoint is free, with the Neumann boundary condition $y_x(L,t)=0$. At $L=\pi$, the allowed wave numbers are $k_n=n+1/2$, $n\ge 0$, and ${\omega }_n=c{k}_n$. Zero initial displacement removes every cosine-in-time coefficient. The initial velocity contains only the modes $k=1/2$ and $k=5/2$, each with coefficient one. Dividing those velocity coefficients by their angular frequencies givesAt $x=0$ each term vanishes, at $x=\pi$ each spatial derivative vanishes, and at $t=0$ the displacement and its time derivative reproduce the two initial conditions.

Here $P=y^2$ and $Q={\cos }^{2}x+y\sin 2x$. Their derivatives giveThus on intervals with $\cos x\ne 0$, choose the integrating factor $\mu =e^{\int 2\tan xdx}={\sec }^{2}x$. The multiplied form isConsequentlyThis implicit family includes $y=0$ and the branch $y=-\cot x$ when $C=0$. Where the coefficient of $y^{\prime }$ is nonzero, its differentiated form directly recovers the original equation. A useful equivalent relation is $y\cos x+y^2\sin x=C\cos x$. At $\cos x=0$, the original equation requires $y=0$; continuation there must be checked in that original equation rather than justified by the singular integrating factor. In particular the zero solution extends everywhere, and $y=-\cot x$ extends smoothly across the cosine zeros where it is defined.

Each proposed displacement has second derivative ${\ddot{x}}_i=-{\omega }^2{a}_i\cos \omega t$. Matching the coefficients of the common time factor in the three equations givesThese algebraic relations hold for the entire oscillation, including the isolated times when the common cosine vanishes.

Expanding the determinant along the first row givesThe eigenvalues and normalized spatial eigenvectors, in increasing frequency order, areFor each mode both signs of ${\omega }_j$ give the same cosine displacement. An arbitrary overall mode amplitude remains free. These vectors are the $N=3$ instances of the sine eigenvectors of the fixed-end spring chain.

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.

The original PDF contains the product $\sin x(1+4\cos x)$; the transcription drops the $x$ after the first sine. Using $2\sin x\cos x=\sin 2x$ gives the finite Fourier sine seriesThus $b_1=1$, $b_2=2$, and all other sine coefficients vanish. No cosine or constant terms are present.