Differentiation of the finite sum involves no convergence issue. The derivative has the Fourier cosine series

Use the double-angle identity $2\sin x\cos x=\sin 2x$. The Fourier series is already a finite trigonometric sum:In the usual Fourier sine series convention, $b_1=b_2=b_3=1$, all other $b_n$ vanish, and all cosine coefficients and the constant coefficient vanish.

The fundamental theorem for line integrals evaluates this conservative vector field integral by the change in potential. At the initial point $\varphi =0$. At the final point $z=0$, so the exponential factor is one, andThe detailed shape of the path does not affect this line integral.

The potential is smooth everywhere, so its mixed partial derivatives commute. The curl of a gradient therefore vanishes:

The characteristic polynomial is ${\lambda }^2-2\lambda \cos \theta +1$, with eigenvalues $\cos \theta \pm i\sin \theta$. A real diagonal form can have only real eigenvalues, so it requires $\sin \theta =0$. Conversely, when $\theta =k\pi$, the rotation matrix is already $(-1{)}^{k}I$. Thus

Yes. The matrix is a real symmetric matrix, so the spectral theorem for real symmetric matrices provides an orthonormal basis of real eigenvectors and hence a real diagonal form. Explicitly its characteristic polynomial is ${\lambda }^2-7\lambda -4$, with distinct real eigenvalues

Write $q=35/36$. Alice wins on her $(n+1)$st turn if $n$ complete rounds fail and then she succeeds. These disjoint events give the geometric seriesConsequentlyThe probability of never obtaining a winner is ${\lim }_{n\to \infty }{q}^{2n}=0$. The small advantage over $1/2$ comes from taking the first turn.

A double six has probability $p=1/36$ on each independent throw. For Alice to win on her second turn, her first throw and Bob's first throw must both fail, then her second throw must succeed. Therefore

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 $\pi c/L$.

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.

Use separation of variables, $f=X(x)T(t)$, and setFor $\kappa =-{\lambda }^2$, with $\lambda >0$, the separated solutions areFor $\kappa =0$, they areFor $\kappa =+{\lambda }^2$, they areArbitrary admissible superpositions of separated solutions are again solutions. An individual product is only a separated family, not the entire general solution. Without prescribed boundaries, the general twice-differentiable solution is equivalently D'Alembert's formula,For example, changing to $\xi =x-ct$, $\eta =x+ct$ makes the differential equation $f_{\xi \eta }=0$, whose two integrations give this form.

Yes. The wave equation is linear and homogeneous. For $\mathcal{W}={\partial }_x^2-c^{-2}{\partial }_t^2$ and constants $A,B$,If both terms on the right vanish, their combination is also a solution. Thus the superposition principle applies, and it also preserves homogeneous fixed-end boundary conditions.

Put $q=2.4048$ and let the fixed constraint value be $C>0$. For $V=\pi R^{2}H$, the Lagrange multiplier equations giveDividing removes $\lambda$ and yieldsTo establish a minimum, rather than merely a stationary ratio, set $u=q^2/R^2$ and $v={\pi }^2/H^2$. Then $u+v=C$ and $V={\pi }^2{q}^2/(u\sqrt{v})$. The denominator squared is $u^2(C-u)$; its derivative is $u(2C-3u)$, giving a unique maximum at $u=2C/3$, $v=C/3$. The volume tends to infinity at either endpoint. This proves the stated ratio is the unique global minimum, as in the optimal cylinder for a reciprocal-square constraint.

Let the fixed positive volume be $V=abc$. For an interior constrained extremum, Lagrange multipliers applied to the energy giveMultiplying these equations by $a,b,c$ shows that $2A/a^2=2A/b^2=2A/c^2=\lambda V$. Since the lengths are positive, $a=b=c=V^{1/3}$.

The arithmetic-geometric mean inequality proves this is the global minimum:with equality exactly for equal sides. Thus the optimal shape is a cube, and $\boxed{E_{\min }=3A{V}^{-2/3}}$.