For $0<p<1$, independence of the remaining indicators givesBecause $0<k<n$, the latter probability is positive for interior $p$. The two events are independent events exactly when their joint probability equals $p\mathbb{P}(S_n=k)$, which is equivalent to $p=k/n$.

At $p=0$ or $p=1$, the event $\{S_n=k\}$ is null, and a null event is independent of every event. Therefore the full answer, including the degenerate endpoints, is

Put $h=(b-a)/n$, $g_i=g(x_i)$, and $z_0=0$. Under the endpoint-slope convention suggested by the requested connection, the recurrence is $z_{i+1}=z_i+\frac{h}{2}(g_i+g_{i+1})$. Summing telescopes, soSince $G(b)={\int }_a^{b}g(x)dx$ by the fundamental theorem of calculus, this recurrence gives exactly the composite trapezoidal rule. It replaces the average derivative over each strip by the arithmetic mean of its endpoint derivatives.

For the standard midpoint method, the literal endpoint-only request is not valid for general $g$. Both the explicit and implicit one-step midpoint formulas here give insteadthe composite midpoint quadrature rule. It requires half-grid samples. These samples cannot be recovered exactly from $g_0,\dots ,g_n$ for arbitrary smooth $g$: the zero function and $g(x)={\prod }_{j=0}^n(x-x_j{)}^2$ have identical grid values but different midpoint sums. For a smooth $g$, replacing each half-grid sample by $(g_i+g_{i+1})/2$ is an additional approximation, giving $T_n$ with an $O(h^2)$ total difference, rather than an identity. A precise useful relation, obtained by splitting the finer trapezoidal sum into old nodes and midpoints, isHere $T_{2n}$ uses the finer half-grid, so it does not eliminate the missing data.

If “Mid-Point Method” means the two-step explicit midpoint linear multistep method, the recurrence is $z_{i+1}=z_{i-1}+2h{g}_i$. It gives $z_{2r}=2h{\sum }_{j=1}^r{g}_{2j-1}$ and $z_{2r+1}=z_1+2h{\sum }_{j=1}^r{g}_{2j}$, so an odd final index additionally needs a starting value $z_1$. For even $n=2r$, the result is the midpoint quadrature on $r$ double-width strips; equivalently $z_n=2T_n-T_{n/2}$. The question specifies neither even $n$ nor a starter. Thus the trapezoidal expression above is the natural endpoint-average interpretation, while the other common midpoint meanings require the qualifications just given. The ambiguity is present in the original PDF.

Under the hypotheses of Taylor's theorem with Lagrange remainder, for example $f\in C^{n+1}$ on an interval containing $x$ and $x+h$, there is a point $\xi$ between them such thatThe Lagrange remainder makes this an exact equality, rather than just a truncated Taylor polynomial. For $h=0$ the equality is immediate.

Apply separation of variables to each mode of the damped wave equation $u_{tt}+2k^2{u}_t=c^2{u}_{xx}$, assuming $L,c>0$. The spatial ordinary differential equation is $X_n^{\prime \prime }+({\lambda }_n/c{)}^2{X}_n=0$. Its general solution is $C\sin ({\lambda }_{n}x/c)+D\cos ({\lambda }_{n}x/c)$. The boundary condition at $0$ gives $D=0$; a nonzero mode at $L$ requiresOnly odd $n=2m+1$ occur in the initial Fourier sine series. Put ${\omega }_m=(2m+1)\pi c/L$ and $d_m=\sqrt{k^4-{\omega }_m^2}$. The temporal equation $T_m^{\prime \prime }+2k^2{T}_m^{\prime }+{\omega }_m^2{T}_m=0$ has roots $-k^2\pm d_m$. DefineFor distinct roots, let the full amplitude be $q_m(t)=A_m{e}^{-{\alpha }_{m}t}+B_m{e}^{-{\beta }_{m}t}$. The initial displacement and zero initial velocity impose $A_m+B_m=F_m$ and ${\alpha }_m{A}_m+{\beta }_m{B}_m=0$. Solving givesThus the physically consistent superposition isFor ${\omega }_m<k^2$ there is overdamping, with two real decaying exponentials. For ${\omega }_m>k^2$ there is underdamping; set ${\nu }_m=\sqrt{{\omega }_m^2-k^4}$; the equivalent real expression isAt critical damping, the repeated-root case ${\omega }_m=k^2$, take the limit, giving $q_m(t)=F_m(1+k^{2}t)e^{-k^{2}t}$. These formulas all satisfy $q_m(0)=F_m$ and $q_m^{\prime }(0)=0$.

Three source qualifications matter. First, the sum must start at $m=0$, for the same reason as in (b). Second, the printed $X(x)=u_0(x)$ cannot hold for a single separated mode: the tent has $X^{\prime \prime }=0$ on each open linear segment but $X\ne 0$, contradicting $X^{\prime \prime }=-({\lambda }_n/c{)}^{2}X$ for positive ${\lambda }_n$, and its peak is not twice differentiable. Separation applies mode by mode, with the initial tent recovered by superposition. Third, no finite “large” $k$ makes every mode overdamped, because ${\omega }_m\to \infty$. The two-exponential expression permits complex conjugate roots, while a repeated root requires its displayed limiting form. The tent initial data are naturally interpreted through the convergent modal solution, rather than as a single smooth separated product.

For the Fourier sine series write $u_0(x)={\sum }_{n\ge 1}{b}_n\sin (n\pi x/L)$. Orthogonality of trigonometric functions givesIn the second half of the integral put $z=L-x$; then $u_0(L-z)=\rho z$ and $\sin [n\pi (L-z)/L]=(-1{)}^{n+1}\sin (n\pi z/L)$. HenceEvery even coefficient is zero. For odd $n$, part (a) or the supplied integral yieldsPutting $n=2m+1$ therefore givesThis is the Fourier sine series of a symmetric tent. The absolute values of its terms are uniformly bounded by a summable sequence, giving absolute convergence and uniform convergence; the piecewise smooth odd extension and the Dirichlet convergence theorem for Fourier series identify its value with the continuous $u_0$, including the zero endpoints and the peak.

The printed lower limit $m=1$ is an error. The original PDF visibly has this lower limit. It omits the nonzero fundamental coefficient $F_0=4L\rho /{\pi }^2$. In particular, the printed series equals $u_0(x)-F_0\sin (\pi x/L)$, rather than $u_0(x)$. At $x=L/2$ it gives $L\rho /2-4L\rho /{\pi }^2$ instead of $L\rho /2$. The corrected lower limit is displayed explicitly above.