For any $q\in {\Pi }_n$, write $f-q=(f-p)+(p-q)$. The stated orthogonality makes the two summands perpendicular in the given inner product, because $p-q\in {\Pi }_n$. The Pythagorean theorem for inner product spaces therefore givesEquality is attained at $q=p$, and positive definiteness makes it unique. Thus $p$ is the orthogonal projection of $f$ onto the polynomial subspace.

For the weighted example set $p(x)=A+Bx$. Orthogonality against the basis $1,x$ givesOdd terms vanish on the symmetric interval, so $A=3/5$ and $B=0$. ConsequentlyThe weight $x^2$ vanishes only at one point; its integral against $g^2$ is positive for every nonzero continuous $g$, so it really does define an inner product. For an optional error check, the squared minimum error is $2/7-12/25+6/25=8/175$.

The Gershgorin circle theorem states that every complex eigenvalue $\lambda$ of $A=(a_{ij})$ belongs to at least one closed discTo prove it, take an eigenvector $v\ne 0$ and choose $i$ with $|v_i|={\max }_j|v_j|>0$. The $i$th row of $Av=\lambda v$ givesDivision by $|v_i|$ proves the assertion. The proof works for real matrices even when their eigenvalues or eigenvectors are complex.

For $B+\mu I$, the three discs have centres $2+\mu ,3+\mu ,4+\mu$ and radii $2,3,4$. Every point in their union has real part at least $\mu$. Thus for $\mu >0$ no disc contains zero, and zero cannot be an eigenvalue. Hence $B+\mu C$ is nonsingular.

With immediate release and ideal independent uniform sampling of fish, each catch is a Bernoulli trial with tagged-success probability $p=k/N$. Stopping at the $r$th success makes the count of untagged catches a negative binomial random variable. A total of $x$ failures requires $r-1$ successes among the first $x+r-1$ trials and a success on the last trial, so the probability is $\left(\genfrac{}{}{0ex}{}{x+r-1}{r-1}\right)p^r(1-p{)}^x$. Separate days use disjoint independent sequences of trials, hence their counts form an independent and identically distributed sample. Equal catch probabilities alone do not mathematically imply independence; this is the usual repeated independent sampling assumption in the stated model.

By continuity of the cumulative distribution function and nonnegativity, $F(0)=0$; strict increase and $F(x)\to 1$ allow inversion for every $0<t<1$. The probability integral transform follows fromThus $F(X)$ is uniformly distributed on $(0,1)$.

The order statistics arrange the sample increasingly: $X_{(1)}\le \cdots \le X_{(n)}$. Strict monotonicity means $F(X_{(i)})$ is the $i$th order statistic of $n$ independent uniform variables. Its density isobtained by choosing the $i-1$ observations below $t$ and the $n-i$ above it. This is the beta distribution with parameters $i,n+1-i$, soFor the given log-logistic distribution, rearrangement givesConsequently plot $y_i=\log (i/(n+1-i))$ vertically against $x_i^{\ast }=\log X_{(i)}$ horizontally. Under a good fit, replacing $F(X_{(i)})$ by its plotting position $i/(n+1)$ gives an approximate straight line withThis is a diagnostic probability plot, rather than an assertion that the expectation of the nonlinear transformed order statistic equals that transformed plotting position. Positive observations occur almost surely under the model.

For the isolated line vortex, use polar coordinates $z=r{e}^{i\theta }$ to obtainThus for nonzero $\Gamma$ the streamlines are circles and their counterclockwise circulation is $2\pi r{u}_{\theta }=\Gamma$. If $\Gamma =0$ the fluid is stationary and the streamline description degenerates.

Place an opposite image vortex at $z=-1$. The method of images givesOn $z=iy$, the distances to the two vortex positions agree, so $\psi =-(\Gamma /2\pi )\log (|z-1|/|z+1|)=0$. Hence the wall is a streamline. Directly the wall velocity has $u=0$ andFor the stipulated steady flow, Bernoulli's equation in the irrotational fluid outside the vortex core givesThe force exerted by the fluid on this wall points in the $-x$ direction for positive pressure. Relative to uniform pressure $p_{\infty }$, however, its change is in the $+x$ direction. Using the supplied integral, the force change per unit span isThus the wall is attracted towards the vortex, with this magnitude if the other side remains at $p_{\infty }$. Subtracting the uniform pressure is essential for an infinite wall. A freely moving vortex would translate parallel to the wall under its image-induced velocity; the calculation here is for the stated steady vortex configuration, with its position held fixed.

For a planar incompressible flow, a stream function satisfies $u={\psi }_y$, $v=-{\psi }_x$. A velocity potential satisfies $u={\varphi }_x$, $v={\varphi }_y$ where the flow is irrotational; such a potential exists locally, and globally on a simply connected nonsingular region. The irrotational flow condition is $v_x-u_y=0$. Where both exist, the Cauchy-Riemann equations make the complex potential $w=\varphi +i\psi$ analytic with $w^{\prime }=u-iv$.

For the logarithmic potential, write $z-z_0=r{e}^{i\theta }$. ThenThe outward volume flux per unit span across a circle is $2\pi r{u}_r=m$. Hence it is a line source of strength $m$. The potential may use local logarithm branches; its velocity is single-valued away from the source.

By superposition, the full flow hasFor positive $m$ and $U$, the stagnation points areThey are real, one upstream of the source and one downstream of the sink. Away from the singularities, streamlines are level curves ofThe zero level through the two stagnation points contains an upper and a lower separating curve, symmetric about the real axis and about $x=1/2$. Together they bound a Rankine oval. Flow inside goes from source to sink; the external stream divides around this oval and tends to horizontal flow far away. This also specifies the sketch for every positive parameter ratio; the figure illustrates one ratio.

For a field point in the plane of the loop, both the directed line element $d\mathbf{s}$ and the displacement $\mathbf{r}-\mathbf{s}$ lie in that plane. Their cross product is normal to it, so every term of the Biot-Savart law is normal, and their integral is normal as well. This argument also allows the field to vanish at particular points.

Choose the loop in the $xy$-plane with centre at the origin and let positive current be counterclockwise as seen from $+z$. On its lower side, write $\mathbf{s}=(t,-a,0)$, $-a\le t\le a$, soThat side contributesAll four contributions have the same direction and size by rotation. Thuswhere $\mathbf{n}$ follows the right-hand rule for the current.