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The given identity immediately yields $V_2=V_1-V_0\sin \omega t$. Thus only the fundamental sine coefficient changes: $V_0/2$ becomes $-V_0/2$. No new integral is needed. The first five nonzero terms areThis also follows from translation of a Fourier series: a half-period delay multiplies the $n$th harmonic by $(-1{)}^n$, leaving the constant and even harmonics unchanged and reversing the only odd harmonic present.

In a steady state, $\nu u_s^{\prime \prime }=-G\sin (n\pi x)$. Since the second derivative of the sine is $-(n\pi {)}^2$ times itself,The zero boundary values at $x=0,1$ force $A=B=0$. Therefore

Separate $\tilde{u}=f(t)g(x)$. On a nonzero product region,A decaying temporal factor requires $\lambda >0$, with $f=C{e}^{-\nu \lambda t}$. The zero spatial boundary values give $g=A\sin (\sqrt{\lambda }x)$ and $\sin \sqrt{\lambda }=0$, hence $\lambda =j^2{\pi }^2$ with $j=1,2,\dots$. Zero or negative separation parameters give no nonzero spatial mode with both boundary values zero.

The initial profile is already exactly the $j=n$ sine eigenfunction. Its coefficient therefore determines a single separated solution:The exponential tends to zero because $\nu >0$. More generally sine orthogonality would determine a sum of such modes, but all other coefficients vanish here. This is the transient of the heat equation with a single stationary sine forcing.

Adding the steady and transient parts givesIntegrating the sine across the channel gives ${\int }_0^1\sin (n\pi x)dx=[1-(-1{)}^n]/(n\pi )$. ThusFor even $n$ the signed flow rate is zero, because positive and negative velocity lobes cancel. For odd $n$ the bracket is two. Both expressions vanish initially and approach their steady values exponentially; $G=0$ gives the trivial zero solution throughout.

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.

In the homogeneous equation, substitute the exponential function $e^{kx}$. The characteristic equation isIts distinct roots are $0,1,-1,i,-i$. Combining the complex-conjugate pair into real sine and cosine gives the complementary functionFive linearly independent terms supply the required five homogeneous degrees of freedom.

In one period, sine equals $-\sqrt{3}/2$ at $4\pi /3$ and $5\pi /3$, and lies below that value between them. Since sine is a periodic function with period $2\pi$, all solutions are:The endpoints are included because the inequality is non-strict.

The required property is odd symmetry, $\boxed{f(-x)=-f(x)}$, up to irrelevant changes on a set of measure zero when discussing Fourier coefficients. An odd function has zero mean, and its products with cosine are odd, so every cosine coefficient vanishes. Its sine coefficients may remain nonzero.

For pointwise representation, the Fourier series takes the average of the one-sided limits at a jump. In particular a pure Fourier sine series has value zero at the origin and at the endpoints; arbitrary assigned values there need not equal the series sum.

If $\ell =mL$ with integer $m>0$, then $k_n\ell =2\pi nm$. All sine phase factors vanish and all cosine phase factors equal one. Thereforefor the periodic extension, and every Fourier coefficient is unchanged.

Use the periodic extension of $f$ when the translated argument falls outside the original interval. With $k_n=2\pi n/L$, the sine subtraction identity givesThis translation of a Fourier series generally introduces cosine terms: a translated odd function need not be odd about the original origin. Equality is understood wherever the original Fourier series represents its periodic extension, with midpoint values at jumps.

Apart from its endpoint convention, the ramp is odd, so its constant and cosine Fourier coefficients vanish. For the sine coefficient, evenness of the integrand and integration by parts giveHenceThis is the Fourier series of a periodic sawtooth. At the periodic jump, the series sums to zero rather than the assigned endpoint value $K/2$; changing a value at one endpoint does not change the coefficients.

This is the defining difference quotient for the derivative of sine at $a$. ThereforeThe result holds for every real constant $a$.

At $x=0$ and $x=L$, the fixed-end boundary conditions require $X(0)=X(L)=0$. In the oscillatory family, the first condition removes the cosine term and the second requires $\sin \lambda L=0$, so $\lambda =n\pi /L$ with $n\ge 1$. The linear and hyperbolic spatial families satisfy both boundary conditions only when identically zero.

Thus the nonzero normal modes and their superposition giveFor initial displacement $h(x)$ and velocity $j(x)$ compatible with the fixed ends, orthogonality of the sine functions determinesThese are all the string solutions in the usual class where the series has the regularity needed to satisfy the wave equation.

The initial displacement is already the first spatial normal mode, so sine orthogonality gives $A_1=1$ and every other $A_n=0$. Zero initial velocity gives every $B_n=0$. ConsequentlyBoth fixed-end values vanish for all time, and differentiation shows that the initial velocity is zero. This is the fundamental standing wave, with angular frequency $\pi c/L$.

The function is odd, so its constant and cosine coefficients are zero. Its sine coefficient isThe polynomial vanishes at both integration endpoints. Two integrations by parts therefore giveOne more integration yieldsConsequently $I_n=6\pi (-1{)}^n/n^3$ and $b_n=12(-1{)}^n/n^3$. HenceThe periodic continuation is continuous at $\pm \pi$, where its values are zero, so the series represents it there as well as in the open interval.

Matching the two parametric equations at an intersection gives $\cos \varphi =-t$ and $\sin \varphi =1+t$. The Pythagorean trigonometric identity therefore requiresFor $t=0$, the sine is one and the cosine is zero; for $t=-1$, the cosine is one and the sine is zero. Keeping the stated half-open angular interval givesThe corresponding points are $(0,1)$ and $(2,0)$; both satisfy the original parametric equations.

For $r>0$, the partial derivatives of radius are $r_x=x/r$, $r_y=y/r$ and $r_z=z/r$. Applying the chain rule to the outer sine and then the radial profile gives the Cartesian coordinates of the gradient:At the origin these coordinate formulas require a differentiable radial extension; differentiability of $f$ as a one-variable profile alone does not guarantee it.

The complex conjugate is $z^{\ast }=x-iy$. From the complex sine and cosine identities,Divide these expressions and multiply by the complex conjugate of the denominator. Its squared modulus is ${\cos }^{2}x{\cosh }^{2}y+{\sin }^{2}x{\sinh }^{2}y=(\cos 2x+\cosh 2y)/2$. The numerator simplifies using ${\cosh }^{2}y-{\sinh }^{2}y=1$ and the double-angle formulas. ThusThese expressions apply except at the poles $y=0$, $x=\pi /2+k\pi$ with integer $k$, where the original tangent function is undefined.

Write the sine argument as $\pi /2+\delta$, with $\delta =\frac{\pi }{2}(e^x-1)=\frac{\pi }{2}x+\frac{\pi }{4}{x}^2+O(x^3)$. The sine addition formula gives $\sin (\pi /2+\delta )=\cos \delta$. Its Maclaurin series is $1-{\delta }^2/2+O({\delta }^4)$, soThere is no linear term because the outer sine has zero derivative at $\pi /2$.

The double-angle formula gives $\cos 2x=1-2{\sin }^{2}x$. Setting $s=\sin x$ reduces the equation to $2s^2+s-1=(2s-1)(s+1)=0$. On $0\le x\le \pi$, the sine is nonnegative, so only $s=1/2$ is allowed. Therefore