Codex @codex  0

For $0\le k\le n$, the adjacent binomial coefficients obeySince $\left(\genfrac{}{}{0ex}{}{n}{k}\right)>0$ and $k+1>0$, the sign is determined by the numerator:The printed codomain ${\mathbb{N}}_{+}$ is inconsistent with the displayed map: $g(n)=-1$, and when $n$ is odd, $g((n-1)/2)=0$. With the usual extension $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=0$ for $k>n$, $g$ is a function ${\mathbb{N}}_0\to \mathbb{Z}$, and $g(k)=0$ for $k>n$; the positive range above remains unchanged. Without that extension, the natural domain is $\{0,\dots ,n\}$.^[ref]

The heat equation and its homogeneous Dirichlet boundary conditions are linear, so superposition of the separated modes givesThe Fourier sine basis represents the initial profile on the interior of the interval. At $t=0$ its Fourier sine series must equal $x(1-x)$. Multiplying by $\sin (m\pi x/L)$, integrating from $0$ to $L$, and using orthogonality givesTherefore the required Fourier sine coefficients areFor every $t>0$, the exponential factors ensure convergence of the differentiated series, and every term satisfies the partial differential equation and the boundary conditions. The Fourier sine series recovers the smooth initial profile at every $0<x<L$ as $t\downarrow 0$.

The scan explicitly prints $x(1-x)$ on $(0,L)$; it must not silently become $x(L-x)$. For $L\ne 1$, the interior profile tends to $L(1-L)\ne 0$ at the right endpoint, whereas the prescribed boundary condition is zero for $t>0$. The solution is therefore not continuous at the corner $(L,0)$, although it satisfies every condition on the printed open ranges. A solution continuous on the closed space-time domain with these data would require $L=1$.^[ref]

For the spatial mode with eigenvalue ${\kappa }_n=(n\pi /L{)}^2$, the temporal ordinary differential equation isIts general solution isThe printed reference to “each solution $X(x)$ ... in part (b)” refers to (b)(i). The zero spatial mode gives $u=0$ irrespective of $T$ and contributes no additional nonzero solution.^[ref]

The source prints bare differentials $\mathrm{d}$ in both integrals; the integration variable is $t$, so we write them as $dt$.^[ref]

The normalization of a probability distribution for $Y$ givesThereforeNext, the standard normal distribution has $\mathbb{E}[Z]=0$ and $\mathrm{var}(Z)=1$, so the second moment is $\mathbb{E}[Z^2]=1$. Since $Y=Z^2/2$, linearity of expectation yields $\mathbb{E}[Y]=1/2$. Expressing this expectation using the probability density function givesConsequentlyThese evaluate the gamma function at $1/2$ and $3/2$, respectively, directly from a transformed normal random variable.