A tree is a connected graph with no cycles. Equivalently, between any two of its vertices there is exactly one graph path. A single vertex with no edges is also a tree.

For $y=\pm \sqrt{2/\pi }\sin x$, both the constraint and endpoint values hold, the extremal equation has $\lambda =1$, and direct integration gives $J[y]=1$.

For any admissible function, its derivative has Fourier cosine series coefficients $n{a}_n$ and zero constant coefficient. This follows by integration by parts: ${\int }_0^{\pi }{y}^{\prime }\cos (nx)dx=n{\int }_0^{\pi }y\sin (nx)dx$, and ${\int }_0^{\pi }{y}^{\prime }=0$. Thus the Parseval identity givesFor finite sums the same identity follows directly from the cosine product formula in the hint; the $L^2$ identity justifies the general case. Equality holds only if $a_n=0$ for all $n\ge 2$, and normalization then fixes $a_1=\pm \sqrt{2/\pi }$. ThereforeThis is the minimum Dirichlet energy at fixed L2 norm on an interval, equivalently the sharp Wirtinger inequality for this interval.

The sine functions obey ${\int }_0^{\pi }\sin (nx)\sin (mx)dx=(\pi /2){\delta }_{nm}$. The Parseval identity for the Fourier sine series givesConsequently

The extremal equation is $y^{\prime \prime }+\lambda y=0$. For $\lambda >0$, its general solution is $A\cos (\sqrt{\lambda }x)+B\sin (\sqrt{\lambda }x)$. The left endpoint forces $A=0$, while the right endpoint and $B\ne 0$ require $\sin (\sqrt{\lambda }\pi )=0$. ThereforeIn particular, each extremal has the requested Fourier sine series form $y={\sum }_{k\ge 1}{a}_k\sin (kx)$. More strongly, for an individual extremal only the coefficient at $k=n$ can be nonzero. An arbitrary combination of distinct modes is an admissible function, but does not solve this one-eigenvalue equation.

Part (a)(ii) now givesIf $\lambda =0$, continuity makes $y^{\prime }$ identically zero. Thus $y$ is constant, and the boundary condition forces $y=0$, contradicting ${\int }_0^{\pi }{y}^2=1$. Hence

Multiply the equation for $y_n$ by $y_m$ and that for $y_m$ by $y_n$, then subtract. Integrating the resulting identity yieldsAll endpoint terms vanish because both functions satisfy the zero Dirichlet boundary conditions. Since the multipliers differ,This is the weighted orthogonality of Sturm-Liouville eigenfunctions.

Multiply the Sturm-Liouville equation by $y$ and integrate. The endpoint values and integration by parts giveSince $I[y]=1$,Thus the multiplier equals the stationary value of $J$ at the extremal, as in stationary values of a quadratic functional with a quadratic constraint. It is also the corresponding Rayleigh quotient.

Apply a constant Lagrange multiplier to the isoperimetric constraint, using the augmented functional $J-\lambda (I-1)$. Its integrand isThe Euler-Lagrange equation gives $-2qy-2\lambda wy-2(p{y}^{\prime }{)}^{\prime }=0$, henceThe multiplier rule applies because the constraint derivative in direction $y$ is $D{I}_y[y]=2I[y]=2$, so the constraint derivative is not zero. The prescribed endpoint values remove boundary variations.

Complete the square: $x^2+2x+2=(x+1{)}^2+1$. In the Fourier transform, the substitution $u=x+1$ givesThe phase $e^{is}$ is important. Taking the real part at $s=a$ gives the cosine integral, since changing the sign in the exponential does not change its real part:

With this convention, the Fourier inversion theorem states that if $f,\hat{f}\in L^1(\mathbb{R})$, thenalmost everywhere and at every continuity point of $f$. Apply this to the continuous function in part (ii):The denominator is even. Replacing $x$ by $-t$ and renaming the integration variable givesBoth integrable members of this transform pair satisfy the inversion hypotheses.

Split the integral at the origin and reflect the negative half:Thus the Fourier transform of a two-sided exponential is

Use the angular-frequency Fourier transform conventionFor $f,f^{\prime }\in L^1(\mathbb{R})$ with the usual local absolute-continuity hypothesis, the Fourier transform of a derivative isFor example, this applies when $f$ is $C^1$ and both displayed functions are integrable. These assumptions make integration by parts valid and the endpoint terms vanish.

Since $A$ is invertible, the quadratic form gradient vanishes exactly at the origin. At every nonzero point of $C_{\alpha }$, the scalar derivative is surjective, so the regular level set theorem gives a smooth submanifold of dimension $n-1$.

If $\alpha \ne 0$, the origin is absent. Therefore the largest relatively open smooth subset is all of $C_{\alpha }$.

If $\alpha =0$, the nontriviality assumption forces the form to be an indefinite quadratic form: an invertible definite form would have zero set consisting only of the origin. Every nonzero point is regular. To show that the origin really fails to be a manifold point, use Sylvester's law of inertia to reduce to $|u{|}^2-|w{|}^2=0$. With positive and negative coordinate vectors $e_i,f_j$, the null vectors $e_i+f_j$ and $e_i-f_j$ span ${\mathbb{R}}^n$. Each line in one of these directions lies in the cone.

If a relatively open manifold neighborhood of the origin existed, its tangent space would contain all these line directions, so would have dimension $n$. Such an embedded manifold would be open in ${\mathbb{R}}^n$, impossible for the zero set of a nonzero quadratic form. ConsequentlyThis gives the smooth locus of a nondegenerate real quadratic level set, including maximality at the singular vertex.

Expand the increment of the quadratic form, using $A^T=A$:The last term obeys $|h^{T}Ah|\le \Vert A\Vert \Vert h{\Vert }^2=o(\Vert h\Vert )$. Thus $q$ is differentiable andThis is the quadratic form gradient. The derivative is a linear functional of the increment $h$, whereas the gradient is its representing vector under the Euclidean inner product.

A sufficient condition is that the derivative $D{f}_x:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is surjective, equivalently that its Jacobian matrix has rank $m$. Then the implicit function theorem makes the zero set near $x$ a $C^1$ embedded submanifold of dimensionThe rank condition persists nearby because the derivative is continuous. This is the local form of the regular level set theorem; only the rank at the specified point is needed for this local conclusion.

The map is differentiable at $x$ if there is a linear map $L:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatEquivalently, $f(x+h)=f(x)+Lh+o(\Vert h\Vert )$. This linear map is unique and is the Fréchet derivative $D{f}_x$. Its matrix in the standard bases is the Jacobian matrix, whose $(i,j)$ entry is $\partial f_i/\partial x_j$ at $x$. A function is differentiable on its domain if this condition holds at every point.

The corresponding side equations, computed using projective joins and meets by cross products, areSubtracting the two equations in each row and substituting back gives the three projective pointsFor example, the first pair forces $y=z$ and $x=3z$; the second pair forces $y=0$ and $x=-z$; the last pair forces $x=z=2y$. All three satisfysince $3-4+1=0$, $1-0-1=0$, and $2-4+2=0$. The coefficient vector is nonzero, so this is a projective line. Thus the corresponding-side intersections are collinear, explicitly confirming Desargues's theorem.

The diagram uses the affine chart $x+y+z\ne 0$ with coordinates $u=y/(x+y+z)$ and $v=z/(x+y+z)$. Here the center $O$ is finite, the axis is $u=1/5$, and $Y$ lies on the projective line at infinity. The apparent parallelism of $QR$ and $Q^{\prime }{R}^{\prime }$ records their common vertical point at infinity, which also belongs to the axis.

Apply projective joins and meets by cross products to each corresponding pair. The cross products of their representatives are proportional to $(0,1,0)$, $(0,0,1)$, and $(0,1,-1)$, respectively. HenceTheir common point isIt is a valid projective point, because its representative is nonzero. The three equations force $y=z=0$, so this common projective point is unique. Thus the two projective triangles are in perspectivity from a point $O$.

The dual projective space is $\mathbb{P}(({\mathbb{R}}^3{)}^{\ast })$. A projective line with coefficients $[\alpha :\beta :\gamma ]$ becomes a projective point in this dual plane; a projective point $[a:b:c]$ becomes the dual projective line $a\alpha +b\beta +c\gamma =0$.

The point-line duality reverses incidence:Thus the dual statement is that the dual point $l^{\ast }$ lies on the dual line $P^{\ast }$. A join of projective points becomes an intersection of dual projective lines, and the converse holds too. In particular, collinear projective points correspond to concurrent dual projective lines.

The dual of Desargues's theorem is its converse: if the intersections of the three pairs of corresponding sides are collinear, then the three lines joining corresponding vertices are concurrent. In the same nondegenerate setting, perspectivity from a line therefore implies perspectivity from a point.

Two projective triangles are in perspectivity from a point $O$ when their three corresponding-vertex projective lines $P{P}^{\prime }$, $Q{Q}^{\prime }$, and $R{R}^{\prime }$ all pass through $O$.

Desargues's theorem: perspectivity from a point implies perspectivity from a line. More explicitly, if $P{P}^{\prime },Q{Q}^{\prime },R{R}^{\prime }$ are concurrent, then the three corresponding-side intersectionsare collinear. The usual statement assumes nondegenerate projective triangles and well-defined, distinct corresponding-side intersections. The projective line containing $X,Y,Z$ is the axis of perspectivity from a line. All intersections are projective: parallel lines in an affine chart meet on its projective line at infinity.