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The inclusion-exclusion principle for probabilities givesEach outcome belonging to two events is initially counted twice and must lose one count. An outcome in all three is counted three times and then subtracted three times, so the last term restores its single count.

The Hessian matrix isIts eigenvalues have opposite signs, so the stationary point is a saddle point. More directly, putting $x=1+u$ and $y=-1+v$ gives $f=u^2+uv-v^2$: it increases along $v=0$ and decreases along $u=0$.

A stationary point has zero gradient. The two partial derivatives giveSolving this system of linear equations gives

Here $I_0=x$ up to a constant. Substitution of $n=0$ in the integration by parts recurrence givesIndeed, its derivative is $\ln x+1-1=\ln x$.

Use integration by parts with $u=(\ln x{)}^{n+1}$ and $dv=dx$. Since $du=(n+1)(\ln x{)}^{n}dx/x$, this yieldsAs an identity between indefinite integrals, the formula is understood up to an additive constant.

For the Gaussian ansatz, partial differentiation givesCancel the positive exponential and compare the constant and quadratic powers of $x$ in the diffusion equation. The two coupled ordinary differential equations areThe initial profile requires $F(0)=T_0$ and $H(0)=L^{-2}$. The second equation is a separable differential equation:Substitute this into the first equation and integrate its logarithmic derivative:Thusand the solution isThis is Gaussian diffusion of a smooth initial profile: its width grows while its peak decreases. The Gaussian integral gives ${\int }_{-\infty }^{\infty }T(x,t)dx=\sqrt{\pi }{T}_{0}L$, independent of time, which checks conservation of the total amount under the diffusion equation.

The power series for $e^{y^3}$ has uniform convergence on every compact interval, so termwise integration from zero to $x$ is valid. ThusThis power series converges for every real $x$, including negative $x$ with the usual orientation of a definite integral.

Substitute $x^3$ into the Taylor series for the exponential function:The general nonzero term is $x^{3n}/n!$; the power series coefficients of powers not divisible by three are zero. The radius of convergence is infinite.

Because all variables are positive, the product constraint is equivalent to ${\sum }_i\ln x_i=\ln a$. The Lagrange multiplier equations forare $1-\lambda /x_i=0$. Thus every $x_i$ equals $\lambda$, and the constraint requires $\lambda =a^{1/n}$. The candidate minimum is thereforeFor a global justification, put $y_i=\ln x_i$. The objective becomes the strictly convex function ${\sum }_i{e}^{y_i}$ on the affine constraint ${\sum }_i{y}_i=\ln a$. A stationary point of this restricted convex function is its unique global minimum, establishing the claim without presupposing the desired inequality.

Finally take $a={\prod }_i{x}_i$ and divide the minimum bound by $n$. This is the arithmetic-geometric mean inequality:with equality exactly when all the positive variables are equal.

The base lies in the plane $z=R-h$. Its circular rim lies on the sphere, soMaximize the cone volume $V_c=\pi r^{2}h/3$ using the Lagrange multiplier expressionFor a nondegenerate cone, the $r$ equation gives $\lambda =\pi h/3$. The $h$ equation is $\pi r^2/3+2\lambda (R-h)=0$, hence $r^2=2h(h-R)$. Combining this with $r^2=2Rh-h^2$ givesIndeed, eliminating $r$ gives $V_c(h)=\pi h^2(2R-h)/3$, which vanishes at the endpoints and has a unique positive interior maximum at $4R/3$. Thus

For a closed circular cylinder, $S=2\pi r^2+2\pi rh$ and the volume constraint is $\pi r^{2}h=V$. Use a Lagrange multiplier:The stationary equations areFor $r>0$, the second gives $\lambda =2/r$; the first then gives $h=2r$. The constraint yieldsandTo verify a global minimum, eliminate $h$ to obtain $S(r)=2\pi r^2+2V/r$. It tends to infinity at both ends of $r>0$ and has $S^{\prime \prime }(r)=4\pi +4V/r^3>0$, so its unique stationary point is the global minimum.

For the period-two Fourier series of a square-integrable function $w$, Parseval's identity statesFor $w=f-g$, write $s=\sinh 1$. Its coefficients are $a_0=2s$, $a_n=2(-1{)}^{n}s/(1+n^2{\pi }^2)$ and $b_n=n\pi a_n$, because $g$ is subtracted. ThereforeThe sum was obtained in part (a). As a separate check, $f-g=e^{-x}$ in the interval, whose square has integral ${\int }_{-1}^1{e}^{-2x}dx=\sinh 2$. Endpoint jumps have no effect on this integral.

On the open interval, $g=f^{\prime }$. It is an odd function there, so its constant and cosine coefficients vanish; assigned values at the endpoints do not affect these integrals. For its sine coefficient, integration by parts givesConsequentlyThe endpoint qualification matters. The periodic $g$ has a jump discontinuity at every odd integer, with one-sided limits $\pm \sinh 1$. By the Dirichlet-Jordan convergence theorem, the Fourier series converges to their average, zero, there, not to the assigned value $g(1)=\sinh 1$. The integration by parts argument also avoids assuming unrestricted termwise differentiation of the first series.

For period two, the Fourier series convention isSince $f$ is an even function, every $b_n$ is zero. Also $a_0={\int }_{-1}^1\cosh xdx=2\sinh 1$. For $n\ge 1$, an antiderivative useful for the cosine coefficient isEvaluation at $\pm 1$ gives $a_n=2(-1{)}^n\sinh 1/(1+n^2{\pi }^2)$. ThusThe periodic extension is continuous at $x=1$, so its Fourier series converges to $\cosh 1$ there. Since $\cos (n\pi )=(-1{)}^n$,Therefore $\boxed{p=2,q=\pi }$, with $q=-\pi$ giving the same expression because only $q^2$ occurs. This is the $\beta =1$ case of Fourier series of periodic hyperbolic sine and cosine.