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For a differentiable objective $f$ and regular constraint $g=r$, a constrained stationary point satisfieswhere $\lambda$ is a Lagrange multiplier. Solve these equations for both the coordinates and $\lambda$.

Geometrically, the gradient $\nabla g$ is normal to the constraint level set when $\nabla g\ne 0$. Every feasible infinitesimal displacement $\mathbf{v}$ is tangent to that level set, so $\nabla g\cdot \mathbf{v}=0$. At a constrained stationary point, the directional change $\nabla f\cdot \mathbf{v}$ must also vanish for every such tangent displacement. Thus $\nabla f$ is normal to the same tangent plane and is a scalar multiple of $\nabla g$. This is the geometric basis of the Lagrange multiplier method.

For a second constraint $h=s$, with linearly independent vectors $\nabla g$ and $\nabla h$, the feasible tangent directions are perpendicular to both. Their normal space is the linear span of these two gradients, so introduce a second Lagrange multiplier:These equations find candidates; deciding whether they are maxima, minima or saddle points requires further analysis. Constraint points where the required gradient independence fails must be examined separately.

L'Hôpital's rule applies to a quotient $f(x)/g(x)$ of differentiable functions near a limiting point, excluding that point itself, provided $g^{\prime }(x)\ne 0$ there. If the quotient has an indeterminate form $0/0$ or $\infty /\infty$, and the limit of $f^{\prime }(x)/g^{\prime }(x)$ exists as a finite or infinite value $L$, thenThe rule also applies to one-sided limits and to limits at infinity, under the corresponding hypotheses. Existence of the derivative-ratio limit and the indeterminate-form hypotheses are required; an arbitrary quotient cannot be replaced by a quotient of derivatives.

For a series $\sum u_k$ with eventually nonzero terms, supposeexists, allowing $L=+\infty$. The ratio test says that if $L<1$, the series converges absolutely; if $L>1$, it diverges; if $L=1$, the test is inconclusive. In the divergent case the terms fail to tend to zero. For example, the P-series with exponents $1$ and $2$ both have ratio limit $1$ but have different convergence behaviour.

Let $p(t)=\det (tI-A_{\alpha })$ be the real characteristic polynomial. It is a monic cubic, so $p(t)\to +\infty$ as $t\to +\infty$. At zero,The intermediate value theorem therefore gives some $t_0>0$ with $p(t_0)=0$. A root of the characteristic polynomial is an eigenvalue, hence $A_{\alpha }$ has a positive real eigenvalue for every real $\alpha$. This uses the general fact that an odd-dimensional real matrix with positive determinant has a positive eigenvalue.

Completing the square in the determinant givesfor every real $\alpha$. A square matrix is invertible exactly when its determinant is nonzero, so $A_{\alpha }$ is nonsingular for every real $\alpha$.

Expand the determinant along the first row:The cubic terms cancel in the expansion.

Substituting $\beta =1$ and $\gamma =1-\alpha$ into the matrix givesIts row sums are $4,4,6$, checking that $A_{\alpha }\mathbf{c}=\mathbf{b}$ for every real $\alpha$.

The matrix multiplication $A\mathbf{c}$ adds each row. Equating it to $\mathbf{b}$ gives the linear systemThe last equation is the sum of the first two, so it supplies no further restriction. Subtracting the first from the second gives $2\beta =2$, hence $\beta =1$, and then $\gamma =1-\alpha$. Thus all parameter choices are

Evaluate the Fourier sine series of part (d) at the interior point $x=1/2$. Only odd indices remain, with alternating signs:Thus $\boxed{A=4}$.

For the second identity, apply Parseval's identity to $h(x)=x^2$. From part (c), $a_0=2/3$, $a_n=4(-1{)}^n/(n^2{\pi }^2)$ and $b_n=0$, soThe left side is $2/5$, giving $\frac{16}{{\pi }^4}\sum n^{-4}=2/5-2/9=8/45$. ConsequentlyThe first series is conditionally convergent, while the second converges absolutely by the P-series criterion.

Termwise differentiation of a Fourier series, applied to the continuous periodic extension in part (c), gives for $-1<x<1$Hence the required Fourier sine series isThe endpoint deserves care: at $x=1$ every term is zero, so the Fourier series sums to $0$, the average of the periodic one-sided limits $1$ and $-1$. It therefore does not reproduce the specified endpoint value $g(1)=1$. Changing a value at one point does not change any Fourier coefficient, so this is still the Fourier series of the stated function on $(-1,1]$.

For $L=1$, the even function $h(x)=x^2$ has $b_n=0$, because $x^2\sin (n\pi x)$ is an odd function. Its constant coefficient is $a_0={\int }_{-1}^1{x}^{2}dx=2/3$. The remaining Fourier coefficients areTwo applications of integration by parts give the antiderivative, with $k=n\pi$,Its endpoint difference is $2(-1{)}^n/(n\pi {)}^2$, so $a_n=4(-1{)}^n/(n^2{\pi }^2)$. ThereforeThe periodic extension is continuous and piecewise smooth, so pointwise convergence of a piecewise smooth Fourier series gives equality, including at the endpoints.

The Fourier coefficients follow by multiplying the expansion by $1$, by the appropriate cosine, or by the appropriate sine, then using orthogonality. In the convention with constant term $a_0/2$,The factor $1/L$ comes from the squared norms of the nonconstant sine and cosine functions: the integral of each square over $[-L,L]$ is $L$. The constant function has squared norm $2L$.