Let $S_N$ be the Fourier partial sum through frequency $N$. By orthogonality and the coefficient formulas,Expanding the nonnegative squared L2 norm of the error gives the exact finite identityFor $f$ in the L2 space, completeness of the Fourier basis makes the left-hand side tend to zero. Dividing the resulting equality by $2L$ proves Parseval's identity:This argument uses convergence in L2 rather than an unqualified exchange of an infinite sum with the integral of its square. The function in part (c) is continuous and piecewise polynomial, so these hypotheses hold.

The analogous orthogonal projection onto the cosine terms givesFor $n=0$, this means $a_0=L^{-1}{\int }_{-L}^{L}f(x)dx$; the actual constant in the Fourier series is $a_0/2$ because the constant function has squared norm $2L$.

Take the inner product of the Fourier series with $\sin (n\pi x/L)$. Fourier orthogonality eliminates the constant, all cosine terms, and all sine terms except the $n$th. The remaining squared norm issoFor a square-integrable function, the calculation is justified by taking inner products of finite Fourier partial sums and passing to their mean-square limit.

Here $\alpha \ne 3$, so the same elimination forces $z=0$. The first two linear equations would then require $x-y=1$ and $x-y=-1$ simultaneously. Thus

When $\alpha =3$, the third linear equation is $-3$ times the second, so it adds no constraint. Take $z=t$ freely. The remaining linear equations are $x-y=-1-3t$ and $x+6y=1-\beta t$. Solving gives the straight line of solutionsThis exists for every real $\beta$.

Adding three times the second linear equation to the third gives $3(3-\alpha )z=0$. Since $\alpha \ne 3$, it follows that $z=0$. The second linear equation then gives $y=x+1$, and the first becomes $(1+2\alpha )x=1-2\alpha$. Thus the unique solution isThe assumption $\alpha \ne -1/2$ makes the denominator nonzero; $\beta$ is unrestricted.

Add the four outward fluxes:For a check, $\nabla \cdot \mathbf{F}=2(x+y+z)$. The divergence theorem gives the same result from

On $OAC$, $y=0$ and the outward unit normal is $-{\mathbf{e}}_y$. Its flux isOn $OBC$, $x=0$ and the outward unit normal is $-{\mathbf{e}}_x$. HenceOn $ABC$, write $z=1-x-y$ over $x\ge 0$, $y\ge 0$, $x+y\le 1$. The oriented surface element is $(1,1,1{)}^{T}dxdy$, pointing away from the origin. Here $x+y+z=1$, so $\mathbf{F}=(1,1,0{)}^T$. ThusThe unnormalized normal vector in the oriented surface element already incorporates the factor $\sqrt{3}$ from the surface area; no additional factor is needed.

On the face $OBA$, $z=0$ and the outward unit normal is $-{\mathbf{e}}_z$. The flux integral is thereforeThe negative sign comes from the outward orientation, which points below the tetrahedron on this face.

The characteristic roots areTaking real linear combinations of the two complex exponential solutions yieldsThe coefficient $\beta$ is three here, as confirmed in the PDF; the OCR replaces it by an undefined symbol.

The characteristic equation factors asThe distinct characteristic roots therefore give the general solution

The characteristic equation has the repeated polynomial root $m=-2$, because both $e^{-2x}$ and $x{e}^{-2x}$ occur as independent solutions. Hence its characteristic polynomial isComparing constant terms gives

An excess of thirty seconds is $T>1/2$ in minute units. ThereforeThis is the shaded triangular area under the probability density function.

Since $q$ is an even function, $tq(t)$ is an odd function and its integral over the symmetric interval vanishes. Thus $\mathbb{E}[T]=0$. Its variance isConsequentlyThe standard deviation is approximately $0.40825$ minute, or $24.49$ seconds.

The probability density function graph joins $(-1,0)$ to $(0,1)$ to $(1,0)$ and is zero elsewhere, as shown in the right panel of the timing figure. Its area isThus the probability distribution is normalized.

Let $E$ be the finish-time offset from the start of its displayed minute. The elapsed time is $25+E-S$, so $T=E-S$. To obtain the stated density, use the intended model that $E$ and $S$ are independent random variables, both with the uniform distribution on $[0,1)$.

Their joint probability density function is one on the unit square. Equivalently, the convolution of probability densities givesThe integrand is one exactly where $s\in [0,1]$ and $s+t\in [0,1]$. The overlap has length $1+t$ for $-1\le t<0$, length $1-t$ for $0\le t<1$, and zero otherwise. ThereforeThis is a symmetric triangular distribution. The independence and finish-time uniformity are additional modelling assumptions: the two truncated clock readings alone do not imply them. For example, if $E=S$ then the elapsed time is exactly twenty-five minutes and $T=0$, even though each endpoint offset is individually uniform.

For the uniform distribution, the expected value and second moment areHence the variance is

The graph is a rectangle of height one over $[0,1)$, with zero height elsewhere. Its unit area is the required normalization of the probability density function.

A displayed minute contains one minute of possible start times, so the specified uniform distribution has probability density functionEndpoint choices for a probability density function do not affect any probabilities.

On the circular arc, $z=0$ and $dz=0$, so the line integral reduces toThis exact differential evaluation applies to the specified arc; for example its parametric equation is $(x,y,z)=(1-\cos \theta ,\sin \theta ,0)$, $0\le \theta \le \pi /2$. On the vertical segment $x=y=1$, only $dz$ contributes, giving ${\int }_0^1(\lambda +1)dz=\lambda +1$. The whole path therefore givesTo answer the final conservativeness requirement, compute the curl:It vanishes identically exactly whenOn the simply connected whole space this condition is sufficient for a conservative vector field. Explicitly a vector-field potential isBoth path calculations then give $13/6$, equal to $\varphi (1,1,1)-\varphi (0,0,0)$ by the fundamental theorem for line integrals. Equality of just these two path values would only force $\mu =2\lambda +1$; it is the curl calculation that establishes full path independence.