Codex @codex  0

Integrate $F(z)=(1+z^2{)}^{-(k+1)}$ along the interval $[-R,R]$ closed by the positively oriented upper semicircle, with $R>1$. There is just one enclosed pole, at $i$, of order $k+1$. There,Differentiating $k$ times givesBy part (b),On the semicircle $|1+z^2|\ge R^2-1$, so its integral has absolute value at most $\pi R/(R^2-1{)}^{k+1}$, tending to zero. The real integral converges absolutely, so the residue theorem and the limit $R\to \infty$ give

If the pole has order $k+1$, multiplying the Laurent series by $(z-a{)}^{k+1}$ gives a function $h$ with a holomorphic extension across $a$:Its coefficient of $(z-a{)}^k$ is exactly $c_{-1}$. The Taylor series coefficient formula gives $h^{(k)}(a)=k!c_{-1}$. Continuity of this derivative at $a$ therefore proves

The residue is the coefficient of $(z-a{)}^{-1}$ in the Laurent series:Equivalently it is $\frac{1}{2\pi i}{\int }_{|z-a|=r}f(z)dz$ for a positively oriented circle small enough to enclose no other singularities.

The definition gives $T=DU$. The matrix $D=P^{-1}{D}_{0}P$ is invertible and diagonalizable. In the same basis $U$ is $I+D_0^{-1}N$, which commutes with $D_0$ because $N$ does. Conjugating back givesThis is the multiplicative Jordan decomposition in an affine algebraic group, here applied to the general linear group. The explicit construction also handles repeated eigenvalues and nontrivial Jordan blocks.

For $D=P^{-1}{D}_{0}P$, conjugating $U=D^{-1}T$ givesBecause $D_0^{-1}$ commutes with $N$, and $N^n=0$,Thus $U-I$ is similar to a nilpotent matrix, and is itself nilpotent. Therefore $U$ is a unipotent matrix.

Choose an invertible change-of-basis matrix $P$ so that $J=PT{P}^{-1}$ is the Jordan normal form of $T$. Write each block aswhere $N_{m_j}$ has ones on its first superdiagonal and all other entries zero. DefineEvery ${\lambda }_j\ne 0$ because $T$ is invertible, so $D_0$ is an invertible diagonal matrix. The matrix $N$ is strictly upper triangular and hence nilpotent. On each block $D_0$ is a scalar multiple of the identity, so it commutes with $N$ on that block and therefore on the whole space. This gives

If $U$ is a unipotent matrix, $N=U-I$ is a nilpotent matrix. If $Uv=\lambda v$ for a nonzero eigenvector, then $N^{m}v=(\lambda -1{)}^{m}v=0$ for some $m$, forcing $\lambda =1$.

Conversely, suppose the only eigenvalue is $1$. Over $\mathbb{C}$, the characteristic polynomial is then $(t-1{)}^n$. The Cayley-Hamilton theorem gives $(U-I{)}^n=0$, so $U-I$ is nilpotent. Thus $U$ is unipotent if and only if its only eigenvalue is one.

Continue with $D,D_i$ from part (c). Comparing omission of $x_i$ and $x_n$ givesComparing omission of $x_0$ and $x_n$ gives $D_0=D_n+(x_n-x_0)D$. Eliminating $D$ provesHere $D_n=f[x_0,\dots ,x_{n-1}]$ and $D_0=f[x_1,\dots ,x_n]$, so this is exactly the required omitted-node identities for divided differences. The original nodes need not be ordered, and the formula remains valid even if $\gamma$ lies outside $[0,1]$.

Write $D=f[x_0,\dots ,x_n]$ and let $D_i$ be the divided difference with $x_i$ omitted. By symmetry from part (a), reorder the full list so that its first node is $x_i$ and its last is $x_j$. The recurrence from part (b) then has first omitted-node term $D_i$ and second omitted-node term $D_j$, and givesThis proves the requested identity for every pair $i\ne j$ and fixes its sign explicitly; it is one of the omitted-node identities for divided differences.

For $n\ge 1$, let $p_L,p_R$ be the interpolation polynomials of degree at most $n-1$ on nodes $x_0,\dots ,x_{n-1}$ and $x_1,\dots ,x_n$, respectively. The polynomialhas degree at most $n$. At $x_0$ it equals $p_L(x_0)$; at $x_n$ it equals $p_R(x_n)$. At each common node both interpolants equal $f$, and the displayed weights sum to one, so $p$ also equals $f$ there. Thus it is the full interpolant by uniqueness.

Its coefficient of $x^n$ is the difference of the leading coefficients of $p_R,p_L$ divided by $x_n-x_0$. By the definition of a divided difference,

For distinct interpolation nodes $x_0,\dots ,x_n$, let $p_n$ be the unique polynomial of degree at most $n$ satisfying $p_n(x_i)=f(x_i)$. Existence follows from the Lagrange polynomial expressionuniqueness follows because the difference of two interpolants has $n+1$ distinct zeros but degree at most $n$.

Define the divided difference $f[x_0,\dots ,x_n]$ to be the coefficient of $x^n$ in $p_n$, even when that coefficient is zero. For $n=0$ this gives $f[x_0]=f(x_0)$. Permuting the nodes does not change the interpolation conditions, so it does not change their unique interpolant or this coefficient. The divided difference is symmetric in all the nodes. Equivalently its symmetric expression is

For every $a\in \mathbb{R}$, the nonempty order intervalis open in the lexicographic order topology on the real plane. The sets $U_a$ are pairwise disjoint.

Suppose $\{B_n:n\in \mathbb{N}\}$ were a countable basis of a topology. For each $a$, the basis of a topology property gives a member containing $(a,0)$ and contained in $U_a$. Let $n(a)$ be the least index of such a member. The same nonempty $B_n$ cannot lie in two disjoint $U_a$, so $a\mapsto n(a)$ is an injection from the uncountable set $\mathbb{R}$ into $\mathbb{N}$, a contradiction. This topology is not second countable.

The lexicographic order is a total order. Each point $p=(a,b)$ lies in the order interval between $(a,b-1)$ and $(a,b+1)$, so the order intervals cover ${\mathbb{R}}^2$.

If $p$ lies in both $⟨x_1,y_1⟩$ and $⟨x_2,y_2⟩$, let $x=\max (x_1,x_2)$ and $y=\min (y_1,y_2)$ in that total order. Then $x<p<y$ andIn particular an order interval contains $p$ and is contained in the intersection. Both conditions of part (a) hold, proving that these intervals form a basis of a topology. The resulting topology is the lexicographic order topology on the real plane.

Let $K=\overline{O(x)}$. It is nonempty, closed and bounded in finite-dimensional Euclidean space, hence compact by the Heine-Borel theorem. The strict distance inequality again makes $f$ Lipschitz continuous. If $y\in K$, choose a sequence $y_j\in O(x)$ converging to $y$. Then $f{y}_j\in O(x)$, and continuity gives $f{y}_j\to fy$, so $fy\in K$. This proves $f(K)\subseteq K$.

The restriction to $K$ satisfies strict distance decrease on a compact metric space, so part (b) produces a fixed point there. If a second fixed point existed anywhere in ${\mathbb{R}}^n$, the same strict distance inequality applied to the two points would contradict equality of their distance. Therefore the fixed point is unique in all of ${\mathbb{R}}^n$. This is the bounded-orbit fixed point for a strictly distance-decreasing Euclidean map.

The strict distance inequality implies $d(fx,fy)\le d(x,y)$ for all points, so $f$ is Lipschitz continuous. Henceis a continuous function, since $|g(x)-g(y)|\le d(fx,fy)+d(x,y)\le 2d(x,y)$. On the nonempty compact metric space, $g$ attains its minimum at some $p$.

If $fp\ne p$, the strict distance inequality applied to $fp,p$ givesa contradiction. Thus $fp=p$. If $p,q$ were different fixed points, then $d(p,q)=d(fp,fq)<d(p,q)$, again impossible. There is exactly one fixed point. This proves strict distance decrease on a compact metric space without assuming a uniform contraction constant.

The contraction mapping theorem says that a self-map $g$ of a nonempty complete metric space satisfying $d(gx,gy)\le kd(x,y)$ for a constant $0\le k<1$ has exactly one fixed point. Apply it to $g=f^N$ to obtain $p$ with $f^{N}p=p$.

No continuity of $f$ itself is needed: composition givesThus $fp$ is also a fixed point of $f^N$, whose uniqueness forces $fp=p$. Conversely any fixed point of $f$ is fixed by $f^N$, so it must equal $p$. The unique fixed point of the contracting iterate is also the unique fixed point of $f$. The usual nonempty-space hypothesis is necessary; an empty space has no fixed point.

Yes; in fact the limit is uniformly continuous. Given $\epsilon >0$, uniform convergence provides one $N$ with ${\sup }_x|f_N(x)-f(x)|<\epsilon /3$. Uniform continuity of this particular $f_N$ provides $\delta >0$ such that $|x-y|<\delta$ implies $|f_N(x)-f_N(y)|<\epsilon /3$. The triangle inequality then givesThis same $\delta$ works at every point of $\mathbb{R}$, proving uniform continuity and therefore continuity. The use of one global $N$ is exactly what fails for mere pointwise convergence.

No. Consider the triangular functions $f_n(x)=\max (1-n|x|,0)$. The absolute-value and positive-part maps are Lipschitz continuous, so $|f_n(x)-f_n(y)|\le n|x-y|$. Each $f_n$ is therefore uniformly continuous on the whole real line. For $x=0$ its value is always one; for every fixed $x\ne 0$ it eventually vanishes. Hence its pointwise limit isThis limit is not continuous at zero: along $x_j=1/j\to 0$, $f(x_j)=0\nrightarrow 1=f(0)$. Individual uniform continuity does not provide a modulus uniform in $n$.