For distinct nodes, define the divided difference byThis is the coefficient of $x^k$ in the Lagrange interpolation polynomial through the first $k+1$ data points.

Let $p_k$ denote that interpolating polynomial. The difference $p_k-p_{k-1}$ vanishes at $x_0,\dots ,x_{k-1}$, soThe Lagrange formula shows that the leading coefficient of $p_k$ is $f[x_0,\dots ,x_k]$, whereas $p_{k-1}$ has degree at most $k-1$. Hence $c_k=f[x_0,\dots ,x_k]$. Starting with $p_0=f(x_0)$ and iterating gives the Newton interpolation polynomial

The divided-difference recurrence isFor three nodes, the triangular table iswhere each entry in a new column uses the two adjacent entries to its left. There are $n$ first differences, $n-1$ second differences, and so on, each requiring one division. The exact total is

At a regular constrained extremum of $f$ on $g=0$, the tangent derivatives of $f$ vanish. Since $\nabla g$ is normal to the constraint surface, the Lagrange multiplier condition isOne solves these equations and then compares the resulting candidates, including any boundary or singular cases.

Let the base have dimensions $x$ and $y$, and let the height be $z$. Measure cardboard relative to the thickness of the front and back. The weighted amount used isbecause the bottom has triple thickness, the two front and back faces have ordinary thickness, and the two side faces have double thickness. The constraint is $xyz=3$.

The multiplier equations areThey implyThus $x=2y$ and $z=3y/2$; imposing $xyz=3$ gives $y=1$. ThereforeThis is the global minimum: by the arithmetic-geometric mean inequality,and equality holds at these dimensions. This is an instance of weighted open-box minimization.

Sincethe critical points are exactlyThe holomorphic inverse function theorem shows that $f$ is conformal locally everywhere else.

For $z=x+iy$,Its imaginary part is positive throughout $S$. On the three boundary pieces,These intervals traverse the boundary of the upper half-plane. The inverse branchexists for $\Im \zeta >0$, proving that the image is preciselyThis is the hyperbolic-cosine half-strip map.

After the $2n$ sides are identified in pairs, the polygonal-schema Euler count gives one face, $n$ edges and some number $V\ge 1$ of vertices. The Euler characteristic of a closed orientable genus-$g$ surface is $2-2g$, so

For genus two, take a regular octagon in the Poincare disc model and identify each side with its opposite side, with the orientation reversed along the boundary. One cyclic labelling isThe quotient has $V=1$, $E=4$, and $F=1$, hence Euler characteristic $-2$ and genus two. All eight vertices become one point, so smoothness requires their angles to sum to $2\pi$. Each interior angle is consequentlyEquivalently, the hyperbolic polygon area formula shows that the regular hyperbolic octagon fundamental polygon has areaas required by the Gauss-Bonnet theorem for a genus-two surface of curvature $-1$.

For a linear map $T:V\to W$, its rank is $\dim \mathrm{im}T$ and its nullity is $\dim \ker T$. The rank-nullity theorem states that, when $V$ is finite-dimensional,To prove it, take a basis $v_1,\dots ,v_k$ of $\ker T$ and extend it to a basisof $V$. The vectors $T{v}_{k+1},\dots ,T{v}_n$ span $\mathrm{im}T$. They are also linearly independent: if ${\sum }_{j=k+1}^n{a}_{j}T{v}_j=0$, then ${\sum }_{j=k+1}^n{a}_j{v}_j\in \ker T$, and independence of the chosen basis forces every $a_j$ to vanish. Thus the rank is $n-k$ and the nullity is $k$.

For the given subspace, row reduction of the coefficient matrix givesTaking $x_3=s$ and $x_5=t$ therefore givesHenceis a basis of $W$.

A Givens rotation ${\Omega }^{[p,q]}$ is the identity except in rows and columns $p,q$, where it has the blockFor $a=A_{pj}$ and $b=A_{qj}$, choosewhen $(a,b)\ne (0,0)$. ThenIf both entries vanish, any angle works.

For the given matrix, first useIt givesThen useThus $R={\Omega }^{[2,3]}{\Omega }^{[1,3]}A$ isAll leading row entries are positive. The resulting QR decomposition by Givens rotations iswhereand $R$ is the matrix above. Since it is a product of transposed rotations, $Q$ is orthogonal.

Let $x$ point downslope and $z$ point normally away from the plane, with $0\le z\le h$. Write the parallel velocity as $u(z)e_x$ and take $\tau >0$ to be the magnitude of the air's upslope stress. The steady equations arewith boundary conditionsThe free-surface condition makes $p_x=0$, and integration gives

The surface velocity, downslope shear exerted by the fluid on the plane, and volume flux per unit width areThus the inclined viscous film with opposing surface shear reverses in the three senses whenThe order of increasing required air stress is therefore

Differentiate under the integral and use ${\nabla }_x|x-x^{\prime }{|}^{-1}=-{\nabla }_{x^{\prime }}|x-x^{\prime }{|}^{-1}$:The omitted boundary term vanishes if the current is localized sufficiently rapidly. A steady current obeys charge conservation ${\nabla }^{\prime }\cdot J=0$, hence

For $r=|x|$ much larger than the source size,Localization and ${\nabla }^{\prime }\cdot J=0$ imply $\int J{d}^3{x}^{\prime }=0$. They also implyby integrating ${\partial }_k^{\prime }(x_i^{\prime }{x}_j^{\prime }{J}_k)$. Thus the first nonzero moment is antisymmetric and can be written using the magnetic dipole momentConsequently the Coulomb-gauge vector potential of a localized steady current has far fieldThe dimensions are