With $b=1$, nonnegativity of the probability density function at both $t$ and $-t$ requiresThis is equivalent to $|a|\le {\inf }_{0<t\le h}\cos t/t$. The derivativeshows that this infimum occurs at $t=h$. Therefore the necessary and sufficient restriction isSufficiency follows from the same decreasing-ratio argument at every interior point. The endpoint values of $a$ are allowed: the probability density function merely vanishes at one endpoint.

A probability density function must have total integral one. Put $h=\pi /6$. The term $ax$ is an odd function, soThus $\boxed{b=1}$.

Set $\ell =\ln 2$. Applying the exponential series and the Maclaurin series of the sine givesMultiply the two power series, retaining every contribution through degree four:Since $0<\ln 2<1$, the coefficient of $x^4$ is nonzero, so these are indeed the first two nonzero terms. The remainder is in Big O notation.

Use the exponential series to obtain the Maclaurin series of the hyperbolic sine and hyperbolic cosine:Dividing these power series gives the hyperbolic tangent expansionThe displayed remainder uses Big O notation as $x\to 0$.

For the stipulated linearly independent vectors ${\mathbf{v}}_1,{\mathbf{v}}_2$, the denominator in part (i) is nonzero. ThusEquivalently, ${\mathbf{a}}_2-{\mathbf{a}}_1$ belongs to the linear span of ${\mathbf{v}}_1,{\mathbf{v}}_2$. Geometrically the lines lie in one plane, and, since their direction vectors are not parallel, they intersect at a unique point. The linear combination expressing ${\mathbf{a}}_2-{\mathbf{a}}_1$ gives the two parameters of that intersection.

The nonzero direction vectors now have linear dependence, so the lines are parallel or coincident. Subtract from $\mathbf{d}={\mathbf{a}}_2-{\mathbf{a}}_1$ its orthogonal projection along ${\mathbf{v}}_1$; the remaining component is the shortest joining displacement. The distance between parallel lines isThe second expression follows from the relation between the cross product and the perpendicular component. It includes distance zero for coincident lines.

Let $\mathbf{d}={\mathbf{a}}_2-{\mathbf{a}}_1$ and $\mathbf{w}={\mathbf{v}}_1\times {\mathbf{v}}_2$. The assumed linear independence gives $\mathbf{w}\ne 0$. Every displacement joining one line to the other isIts dot product with $\mathbf{w}$ is always $\mathbf{d}\cdot \mathbf{w}$. The component in that direction is therefore fixed, so orthogonal decomposition gives $\Vert \mathbf{s}\Vert \ge |\mathbf{d}\cdot \mathbf{w}|/\Vert \mathbf{w}\Vert$.

The two linearly independent vectors ${\mathbf{v}}_1,{\mathbf{v}}_2$ span the plane perpendicular to $\mathbf{w}$. Suitable ${\lambda }_1,{\lambda }_2$ cancel the component of $\mathbf{d}$ in that plane, attaining the lower bound. Thus the distance between skew lines formula, also valid when the lines intersect, gives

Use the distance from a point to a plane formula with the unit normal $\hat{\mathbf{n}}$. Since the plane passes through the origin, the distance isFor a direct check, the closest point on a plane is $\mathbf{p}-(\mathbf{p}\cdot \hat{\mathbf{n}})\hat{\mathbf{n}}=(1,\sqrt{2},-1{)}^T$, whose displacement from $\mathbf{p}$ has Euclidean norm $\sqrt{2}$.

The plane equation $x+z=0$ has normal vector $(1,0,1{)}^T=\sqrt{2}\hat{\mathbf{n}}$. Thus its normal vector is parallel to the cylinder's axis and the plane is perpendicular to that axis. It passes through the origin, so the intersection is a circle of radius $2$ centred at the origin in that plane.

Indeed, with the orthonormal basis $\mathbf{e}=(1,0,-1{)}^T/\sqrt{2}$ and $\mathbf{f}=(0,1,0{)}^T$ of the plane, write $\mathbf{r}=s\mathbf{e}+t\mathbf{f}$. The cylinder equation becomes $s^2+t^2=4$. Equivalently, the intersection has the parametric equation

The removed vector is the orthogonal projection of $\mathbf{r}$ onto the line through the origin in direction $\hat{\mathbf{n}}$. The remaining vector measures the perpendicular displacement from that line. Fixing its Euclidean norm at $2$ therefore describes an infinite circular cylinder of radius $2$, with axis $x=z$, $y=0$.

The dot product with the given unit vector is $\mathbf{r}\cdot \hat{\mathbf{n}}=(x+z)/\sqrt{2}$. Subtracting the corresponding orthogonal projection givesTaking its squared Euclidean norm yields the equivalent Cartesian equation

No. Every vector in the plane can be written asThe two linearly independent vectors in part (ii) therefore form a vector space basis for this vector subspace. Its dimension is $2$, so any three vectors in it have linear dependence.

For example, chooseBoth vectors satisfy the plane equation. They are linearly independent vectors: in $\alpha \mathbf{u}+\beta \mathbf{v}=0$, the second coordinate gives $\alpha =0$ and the third gives $\beta =0$.

This is a plane through the origin in three-dimensional Euclidean space, with nonzero normal vector $(1,2,-3{)}^T$:

The standing wave oscillates about the constant level $1/2$, with time period $2\pi /|v|$. Suitable snapshots areThe nonconstant amplitude shrinks to zero at a quarter-period, reverses sign at half a period, and returns after a full period. The nodes of a standing wave of the oscillatory part, $x=\pi /2+k\pi$, remain at height $1/2$.

For the two travelling profiles, the initial conditions implySince $v\ne 0$, $g-f$ is constant. One may therefore choosewhere the constants cancel in the sum. The D'Alembert formula then givesThis agrees exactly with the restricted-domain solution from part (b). The whole-line cosine profile already has zero spatial derivative at both $x=0$ and $x=\pi$ for every time, so its restriction satisfies the Neumann boundary conditions without further modification.

For a point in the first quadrant, $xy$ is the area of the axis-aligned rectangle with opposite vertices at the origin and $(x,y)$. Thus the maximum says that the largest such quarter-rectangle with its outer corner on the ellipse has area one.

Equivalently, reflect this rectangle in both axes. Its four vertices $(\pm |x|,\pm |y|)$ lie on the ellipse, and its full area is $4|xy|$. ThereforeIts width is $1/\sqrt{2}$ and its height is $4\sqrt{2}$.

Use the Lagrange multiplier condition $\nabla f=\lambda \nabla g$ for $g=4x^2+y^2/16$:Neither coordinate can vanish at a constrained stationary point: either would force both to vanish, contradicting the constraint. Substitution therefore gives ${\lambda }^2=1$, so $y=8x$ or $y=-8x$. In either case $8x^2=1$, givingEqual signs give the two points attaining the global maximum $f=1$; opposite signs give the two global minima with $f=-1$. These exhaust the extrema because the ellipse is compact, its constraint gradient never vanishes on it, and $f$ is smooth. The bound can also be checked by setting $X=2x$, $Y=y/4$: then $X^2+Y^2=1$ and $f=2XY$, so $|f|\le 1$.

The square is even, and direct integration givesUsing the nonzero Fourier coefficients in Parseval's identity,Hence

The function is an odd function, so $a_0=a_n=0$. Here $L=1$, and the Fourier coefficients areWrite $k=n\pi$. One integration by parts has zero endpoint terms because $x(1-x)$ vanishes at both endpoints:A second integration by parts givesTherefore $b_n=4(1-(-1{)}^n)/(n^3{\pi }^3)$, zero for even $n$ and $8/(n^3{\pi }^3)$ for odd $n$. The Fourier series isThe coefficients are absolutely summable, and the periodic function is continuous even at the joins $x=\pm 1$, so the displayed Fourier series represents it everywhere.