The two derivatives are $2x$ and $2(x-1)$. At the intersection their slopes are $1$ and $-1$, so their direction vectors $(1,1)$ and $(1,-1)$ have zero dot product. Thus the tangent lines are perpendicular and their angle is

At an intersection, $x^2=(x-1{)}^2=x^2-2x+1$, so $x=1/2$. Substitution into either parabola givesThe resulting linear equation for $x$ has only one solution, so this is the unique intersection.

For $p\le 0$, the terms $k^{-p}$ do not tend to zero, so the series diverges. For $p>0$, group the positive terms into the dyadic blocks $2^j\le k<2^{j+1}$. Each block contains $2^j$ terms, and monotonicity givesWhen $p>1$, the upper bounds form a convergent geometric series with ratio $2^{1-p}<1$, proving convergence. When $0<p\le 1$, the lower bounds do not tend to zero, so infinitely many blocks each make a nonvanishing contribution and the sum diverges. Thus the p-series criterion isThis dyadic comparison is the mechanism of the Cauchy condensation test. In particular $p=1$ is the divergent harmonic series.

Apply the ratio test to the infinite series whose partial sums are $S_N$. Assuming the terms are eventually nonzero and the limit exists, letIf $\ell <1$, the series has absolute convergence, by eventual comparison with a decaying geometric series. If $\ell >1$, including an infinite limit, the terms fail to tend to zero and the series diverges. If $\ell =1$, the test is inconclusive: both $\sum 1/k$ and $\sum 1/k^2$ have that ratio limit, but only the second converges. A finite partial sum itself requires no convergence test; the issue is the limit as $N\to \infty$.

Use the Maclaurin series on the positive side:In the last expression the original square root covers $e^x-1$, whose leading term is $x>0$. Hence $g/f\to 0$ and $f/h\to 0$. All three are positive for sufficiently small positive $x$, soThus $h$ is greatest and $g$ least in the stated limit.

The Big O notation means that there exist constants $C>0$ and $\delta >0$ such thatwithin the domain of $f$. Thus $f(x)/x^2$ stays bounded in a punctured neighbourhood of zero. It need not have a limit: $f(x)=x^2\sin (1/x)$ is one example. The statement imposes no condition on a separately assigned value $f(0)$.

Let $A,B,C$ denote the mutually exclusive supplier events and $D$ the defective-bulb event. Their probabilities and conditional defect rates areThese supplier events partition the possible bulbs. The following calculations use the law of total probability and Bayes' theorem with that partition.

All size-six subsets are equally likely under uniform sampling without replacement. There are $\left(\genfrac{}{}{0ex}{}{12}{6}\right)$ subsets in total, and $\left(\genfrac{}{}{0ex}{}{6}{3}\right)\left(\genfrac{}{}{0ex}{}{6}{3}\right)$ contain the required three members from each group. Thus the hypergeometric distribution gives

Over one period, the graph rises linearly from the limiting point $(-\pi ,0)$ to $(0,\pi )$, then drops to zero and stays there up to $\pi$. Repeat this pattern every $2\pi$. The given open endpoints do not fix values at the jumps, but changing individual point values does not change any Fourier coefficient.

Only the negative half-interval contributes. The constant coefficient isFor $n\ge 1$, integration by parts givesandConsequently the requested Fourier series isIt equals the given function on each open continuity interval. By the Dirichlet-Jordan convergence theorem, at $x=2k\pi$ it takes the jump midpoint $\pi /2$, while at $x=(2k+1)\pi$ it takes zero because the periodic one-sided limits agree there.

With the standard real normalization, the Fourier series of a period-$2L$ function iswhereFor the partial sum $S_N$, Fourier orthogonality eliminates all cross terms. The constant term contributes $2L(a_0/2{)}^2$, while every squared sine or cosine integrates to $L$, givingFor a square-integrable function, completeness of the trigonometric orthogonal basis in the Hilbert space $L^2[-L,L]$ makes $S_N\to f$ in norm. Passing to the limit proves Parseval's identity:Square integrability is the needed hypothesis here; periodicity alone is not an integrability or convergence condition.

The gradient factors asIf a coordinate is zero, the solutions are $(0,0)$, $(\pm 1,0)$ and $(0,\pm 1)$. With both coordinates nonzero, subtracting the two bracket equations gives $x^2=y^2$, hence $x^2=y^2=1/4$. These give four more stationary points.

The Hessian matrix has equal diagonal entries $6xy$ and off-diagonal entry $3(x^2+y^2)-1$. At the origin its eigenvalues are $\pm 1$, and at the four nonzero axis points they are $\pm 2$, so all five are saddles. At the two same-sign interior points the eigenvalues are $1,2$, while at the opposite-sign points they are $-1,-2$. ThereforeThe extrema are not global, since the quartic takes arbitrarily large positive and negative values along suitable rays.

For the contour sketch, the zero level is exactly the two axes together with the unit circle. Inside the circle, $f$ is negative in the first and third quadrants and positive in the other two; outside, those signs reverse. Small negative levels contain closed loops around the two minima as well as unbounded branches in the second and fourth quadrants. Small positive levels similarly contain two loops around the maxima and unbounded branches in the first and third quadrants. The closed loops shrink to the extrema at levels $\mp 1/8$, and disappear beyond those levels. This is the unit-scale quartic surface with axes and a circle as its zero set.

For $h>0$, parametrize the curved surface by $(r\cos \theta ,r\sin \theta ,r^2/a)$, with $0\le r\le \sqrt{ah}$ and $0\le \theta \le 2\pi$. The magnitude of the cross product of its coordinate tangents gives the area element of a surface:Integrating givesThis is the curved surface area, not the area of a lid across the horizontal opening. The expression tends to zero at $h=0$ and has dimensions of area. The volume and centroid of a parabolic cup use the same geometry.

In cylindrical coordinates, $x=r\cos \theta$ and $y=r\sin \theta$, so $x^2+y^2=r^2$. Thus the circular elliptic paraboloid has equationAt height $h\ge 0$, the rim radius is $\sqrt{ah}$.