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Bilu-Linial Stable Instances of Max Cut and Minimum Multiway Cut

@inproceedings{Makarychev2014BiluLinialSI,
title={Bilu-Linial Stable Instances of Max Cut and Minimum Multiway Cut},
author={Konstantin Makarychev and Yury Makarychev and Aravindan Vijayaraghavan},
booktitle={SODA},
year={2014}
}
• Published in SODA 2014
• Computer Science, Mathematics
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A new semi-random model for graph partitioning problems that captures many properties of real-world instances and works in a wider range of parameters than most algorithms for previously studied random and semi- random models is proposed. Expand
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This paper proves that the conjecture that instances stable to as little as O(1) perturbations should be solvable in polynomial time is true for any center-based clustering objective, and shows that the popular Single-Linkage algorithm combined with dynamic programming will find the optimal clustering. Expand
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It is shown that the problem becomes NP-hard as soon as $k=3$, but can be solved in polynomial time for planar graphs for any fixed $k$, if the planar problem is NP- hard, however, if £k\$ is not fixed. Expand
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A new linear programming relaxation for Multiway Cut is presented and a new approximation algorithm based on it achieves a performance ratio of at most 1.5?1k, which improves the previous result for every value of k. Expand
On unique graph 3-colorability and parsimonious reductions in the plane
It is proved that the Satisfiability (resp. planar Satisfiability) problem is parsimoniously P-time reducible to the 3-Colorability problem, that means that the exact number of solutions is preserved by the reduction, provided that 3-colorings are counted modulo their six trivial color permutations. Expand


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# Question Compute the compressive strength of a WT12x88 of A992 steel. The effective length with respect to the x-axis is 25 ft, the effective length with respect to the y-axis is 15 ft, and the effective length with respect to the z-axis is 20 ft. (Check flexural-torsional buckling)

0LCS2J The Asker · Civil Engineering

Transcribed Image Text: Compute the compressive strength of a WT12x88 of A992 steel. The effective length with respect to the x-axis is 25 ft, the effective length with respect to the y-axis is 15 ft, and the effective length with respect to the z-axis is 20 ft. (Check flexural-torsional buckling)
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Transcribed Image Text: Compute the compressive strength of a WT12x88 of A992 steel. The effective length with respect to the x-axis is 25 ft, the effective length with respect to the y-axis is 15 ft, and the effective length with respect to the z-axis is 20 ft. (Check flexural-torsional buckling)

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# Karhunen–Loève theorem

In the theory of stochastic processes, the Karhunen–Loève theorem (named after Kari Karhunen and Michel Loève), also known as the Kosambi–Karhunen–Loève theorem[1][2] is a representation of a stochastic process as an infinite linear combination of orthogonal functions, analogous to a Fourier series representation of a function on a bounded interval. Stochastic processes given by infinite series of this form were first[3] considered by Damodar Dharmananda Kosambi.[4] There exist many such expansions of a stochastic process: if the process is indexed over [a, b], any orthonormal basis of L2([a, b]) yields an expansion thereof in that form. The importance of the Karhunen–Loève theorem is that it yields the best such basis in the sense that it minimizes the total mean squared error.

In contrast to a Fourier series where the coefficients are real numbers and the expansion basis consists of sinusoidal functions (that is, sine and cosine functions), the coefficients in the Karhunen–Loève theorem are random variables and the expansion basis depends on the process. In fact, the orthogonal basis functions used in this representation are determined by the covariance function of the process. One can think that the Karhunen–Loève transform adapts to the process in order to produce the best possible basis for its expansion.

In the case of a centered stochastic process {Xt}t ∈ [a, b] (where centered means that the expectations E(Xt) are defined and equal to 0 for all values of the parameter t in [a, b]) satisfying a technical continuity condition, Xt admits a decomposition

$X_t = \sum_{k=1}^\infty Z_k e_k(t)$

where Zk are pairwise uncorrelated random variables and the functions ek are continuous real-valued functions on [a, b] that are pairwise orthogonal in L2[a, b]. It is therefore sometimes said that the expansion is bi-orthogonal since the random coefficients Zk are orthogonal in the probability space while the deterministic functions ek are orthogonal in the time domain. The general case of a process Xt that is not centered can be brought back to the case of a centered process by considering (Xt − E(Xt)) which is a centered process.

Moreover, if the process is Gaussian, then the random variables Zk are Gaussian and stochastically independent. This result generalizes the Karhunen–Loève transform. An important example of a centered real stochastic process on [0,1] is the Wiener process; the Karhunen–Loève theorem can be used to provide a canonical orthogonal representation for it. In this case the expansion consists of sinusoidal functions.

The above expansion into uncorrelated random variables is also known as the Karhunen–Loève expansion or Karhunen–Loève decomposition. The empirical version (i.e., with the coefficients computed from a sample) is known as the Karhunen–Loève transform (KLT), principal component analysis, proper orthogonal decomposition (POD), Empirical orthogonal functions (a term used in meteorology and geophysics), or the Hotelling transform.

## Formulation

• Throughout this article, we will consider a square integrable zero-mean random process Xt defined over a probability space (Ω,F,P) and indexed over a closed interval [ab], with covariance function KX(s,t). We thus have:
$\forall t\in [a,b], X_t\in L^2(\Omega,\mathcal{F},\mathrm{P}),$
$\forall t\in [a,b], \mathrm{E}[X_t]=0,$
$\forall t,s \in [a,b], K_X(s,t)=\mathrm{E}[X_s X_t].$
$\begin{array}{rrl} T_{K_X}: L^2([a,b]) &\rightarrow & L^2([a,b])\\ f & \mapsto & \int_{[a,b]} K_X(s,\cdot) f(s) ds \end{array}$
Since TKX is a linear operator, it makes sense to talk about its eigenvalues λk and eigenfunctions ek, which are found solving the homogeneous Fredholm integral equation of the second kind
$\int_{[a,b]} K_X(s,t) e_k(s)\,ds=\lambda_k e_k(t)$

## Statement of the theorem

Theorem. Let Xt be a zero-mean square integrable stochastic process defined over a probability space (Ω,F,P) and indexed over a closed and bounded interval [ab], with continuous covariance function KX(s,t).

Then KX(s,t) is a Mercer kernel and letting ek be an orthonormal basis of L2([ab]) formed by the eigenfunctions of TKX with respective eigenvalues λk, Xt admits the following representation

$X_t=\sum_{k=1}^\infty Z_k e_k(t)$

where the convergence is in L2, uniform in t and

$Z_k=\int_{[a,b]} X_t e_k(t)\, dt$

Furthermore, the random variables Zk have zero-mean, are uncorrelated and have variance λk

$\mathrm{E}[Z_k]=0,~\forall k\in\mathbb{N} \quad\quad\mbox{and}\quad\quad \mathrm{E}[Z_i Z_j]=\delta_{ij} \lambda_j,~\forall i,j\in \mathbb{N}$

Note that by generalizations of Mercer's theorem we can replace the interval [a, b] with other compact spaces C and the Lebesgue measure on [a, b] with a Borel measure whose support is C.

## Proof

• The covariance function KX satisfies the definition of a Mercer kernel. By Mercer's theorem, there consequently exists a set {λk,ek(t)} of eigenvalues and eigenfunctions of TKX forming an orthonormal basis of L2([a,b]), and KX can be expressed as
$K_X(s,t)=\sum_{k=1}^\infty \lambda_k e_k(s) e_k(t)$
• The process Xt can be expanded in terms of the eigenfunctions ek as:
$X_t=\sum_{k=1}^\infty Z_k e_k(t)$
where the coefficients (random variables) Zk are given by the projection of Xt on the respective eigenfunctions
$Z_k=\int_{[a,b]} X_t e_k(t) \,dt$
• We may then derive
$\mathrm{E}[Z_k]=\mathrm{E}\left[\int_{[a,b]} X_t e_k(t) \,dt\right]=\int_{[a,b]} \mathrm{E}[X_t] e_k(t) dt=0$
and:
$\begin{array}[t]{rl} \mathrm{E}[Z_i Z_j]&=\mathrm{E}\left[ \int_{[a,b]}\int_{[a,b]} X_t X_s e_j(t)e_i(s) dt\, ds\right]\\ &=\int_{[a,b]}\int_{[a,b]} \mathrm{E}\left[X_t X_s\right] e_j(t)e_i(s) dt\, ds\\ &=\int_{[a,b]}\int_{[a,b]} K_X(s,t) e_j(t)e_i(s) dt \, ds\\ &=\int_{[a,b]} e_i(s)\left(\int_{[a,b]} K_X(s,t) e_j(t) dt\right) ds\\ &=\lambda_j \int_{[a,b]} e_i(s) e_j(s) ds\\ &=\delta_{ij}\lambda_j \end{array}$
where we have used the fact that the ek are eigenfunctions of TKX and are orthonormal.
• Let us now show that the convergence is in L2:
let $S_N=\sum_{k=1}^N Z_k e_k(t)$.
\begin{align} \mathrm{E}[|X_t-S_N|^2]&=\mathrm{E}[X_t^2]+\mathrm{E}[S_N^2]-2\mathrm{E}[X_t S_N]\\ &=K_X(t,t)+\mathrm{E}\left[\sum_{k=1}^N \sum_{l=1}^N Z_k Z_l e_k(t)e_l(t) \right] -2\mathrm{E}\left[X_t\sum_{k=1}^N Z_k e_k(t)\right]\\ &=K_X(t,t)+\sum_{k=1}^N \lambda_k e_k(t)^2 -2\mathrm{E}\left[\sum_{k=1}^N \int_a^b X_t X_s e_k(s) e_k(t) ds\right]\\ &=K_X(t,t)-\sum_{k=1}^N \lambda_k e_k(t)^2 \end{align}
which goes to 0 by Mercer's theorem.

## Properties of the Karhunen–Loève transform

### Special case: Gaussian distribution

Since the limit in the mean of jointly Gaussian random variables is jointly Gaussian, and jointly Gaussian random (centered) variables are independent if and only if they are orthogonal, we can also conclude:

Theorem. The variables Zi have a joint Gaussian distribution and are stochastically independent if the original process {Xt}t is Gaussian.

In the Gaussian case, since the variables Zi are independent, we can say more:

$\lim_{N \rightarrow \infty} \sum_{i=1}^N e_i(t) Z_i(\omega) = X_t(\omega)$

almost surely.

### The Karhunen–Loève transform decorrelates the process

This is a consequence of the independence of the Zk.

### The Karhunen–Loève expansion minimizes the total mean square error

In the introduction, we mentioned that the truncated Karhunen–Loeve expansion was the best approximation of the original process in the sense that it reduces the total mean-square error resulting of its truncation. Because of this property, it is often said that the KL transform optimally compacts the energy.

More specifically, given any orthonormal basis {fk} of L2([a, b]), we may decompose the process Xt as:

$X_t(\omega)=\sum_{k=1}^\infty A_k(\omega) f_k(t)$

where $A_k(\omega)=\int_{[a,b]} X_t(\omega) f_k(t)\,dt$

and we may approximate Xt by the finite sum $\hat{X}_t(\omega)=\sum_{k=1}^N A_k(\omega) f_k(t)$ for some integer N.

Claim. Of all such approximations, the KL approximation is the one that minimizes the total mean square error (provided we have arranged the eigenvalues in decreasing order).

### Explained variance

An important observation is that since the random coefficients Zk of the KL expansion are uncorrelated, the Bienaymé formula asserts that the variance of Xt is simply the sum of the variances of the individual components of the sum:

\begin{align} \mbox{Var}[X_t]&=\sum_{k=0}^\infty e_k(t)^2 \mbox{Var}[Z_k]=\sum_{k=1}^\infty \lambda_k e_k(t)^2 \end{align}

Integrating over [a, b] and using the orthonormality of the ek, we obtain that the total variance of the process is:

$\int_{[a,b]} \mbox{Var}[X_t] dt=\sum_{k=1}^\infty \lambda_k$

In particular, the total variance of the N-truncated approximation is $\sum_{k=1}^N \lambda_k$. As a result, the N-truncated expansion explains $\sum_{k=1}^N \lambda_k/\sum_{k=1}^\infty \lambda_k$ of the variance; and if we are content with an approximation that explains, say, 95% of the variance, then we just have to determine an $N\in\mathbb{N}$ such that $\sum_{k=1}^N \lambda_k/\sum_{k=1}^\infty \lambda_k \geq 0.95$.

## Linear Karhunen-Loeve Approximations

Let us consider a whole class of signals we want to approximate over the first M vectors of a basis. These signals are modeled as realizations of a random vector Y[n] of size N. To optimize the approximation we design a basis that minimizes the average approximation error. This section proves that optimal bases are karhunen-loeve bases that diagonalize the covariance matrix of Y. The random vector Y can be decomposed in an orthogonal basis ${{\left\{ {{g}_{m}} \right\}}_{0\le m\le N}}:$

$Y=\sum\limits_{m=0}^{N-1}{\left\langle Y,{{g}_{m}} \right\rangle {{g}_{m}},}$

where each $\left\langle Y,{{g}_{m}} \right\rangle =\sum\limits_{n=0}^{N-1}{Y\left[ n \right]}.g_{m}^{*}\left[ n \right]$

is a random variable. The approximation from the first $M\le N$ vectors of the basis is

${{Y}_{M}}=\sum\limits_{m=0}^{M-1}{\left\langle Y,{{g}_{m}} \right\rangle {{g}_{m}}}$

The energy conservation in an orthogonal basis implies

$\varepsilon \left[ M \right]=E\left\{ {{\left\| Y-{{Y}_{M}} \right\|}^{2}} \right\}=\sum\limits_{m=M}^{N-1}{E\left\{ {{\left| \left\langle Y,{{g}_{m}} \right\rangle \right|}^{2}} \right\}}$

This error is related to the covariance of Y defined by

$R\left[ n,m \right]=E\left\{ Y\left[ n \right]{{Y}^{*}}\left[ m \right] \right\}$

For any vector x[n] we denote by K the covariance operator represented by this matrix,

$E\left\{ {{\left| \left\langle Y,x \right\rangle \right|}^{2}} \right\}=\left\langle Kx,x \right\rangle =\sum\limits_{n=0}^{N-1}{\sum\limits_{m=0}^{N-1}{R\left[ n,m \right]x\left[ n \right]{{x}^{*}}\left[ m \right]}}$

The error $\varepsilon \left[ M \right]$ is therefore a sum of the last N-M coefficients of the covariance operator

$\varepsilon \left[ M \right]=\sum\limits_{m=M}^{N-1}{\left\langle K{{g}_{m}},{{g}_{m}} \right\rangle }$

The covariance operator K is Hermitian and Positive and is thus diagonalized in an orthogonal basis called a Karhunen-Loeve basis. The following theorem states that a Karhunen-Loeve basis is optimal for linear approximations.

Theorem:(Optimality of Karhunen-Loeve basis) Let K be acovariance operator. For all $M\ge 1$, the approximation error

$\varepsilon \left[ M \right]=\sum\limits_{m=M}^{N-1}{\left\langle K{{g}_{m}},{{g}_{m}} \right\rangle }$

is minimum if and only if ${{\left\{ {{g}_{m}} \right\}}_{0\le m is a Karhunen-Loeve basis ordered by decreasing eigenvalues.

$\left\langle K{{g}_{m}},{{g}_{m}} \right\rangle \ge \left\langle K{{g}_{m+1}},{{g}_{m+1}} \right\rangle for,0\le m

## Non-Linear Approximation in Bases

Linear approximations project the signal on M vectors a priori. The approximation can be made more precise by choosing the M orthogonal vectors depending on the signal properties. This section analyzes the general performance of these non-linear approximations. A signal $f\in \Eta$ is approximated with M vectors selected adaptively in an orthonormal basis $\Beta ={{\left\{ {{g}_{m}} \right\}}_{m\in \mathbb{N}}}$ of $\Eta$. Let ${{f}_{M}}$ be the projection of f over M vectors whose indices are in ${{I}_{M}}$:

${{f}_{M}}=\sum\limits_{m\in {{I}_{M}}}^{{}}{\left\langle f,{{g}_{m}} \right\rangle {{g}_{m}}}$

The approximation error is the sum of the remaining coefficients

$\varepsilon \left[ M \right]=\left\{ {{\left\| f-{{f}_{M}} \right\|}^{2}} \right\}=\sum\limits_{m\notin {{I}_{M}}}^{N-1}{\left\{ {{\left| \left\langle f,{{g}_{m}} \right\rangle \right|}^{2}} \right\}}$

To minimize this error, the indices in ${{I}_{M}}$ must correspond to the M vectors having the largest inner product amplitude $\left| \left\langle f,{{g}_{m}} \right\rangle \right|$. These are the vectors that best correlate f. They can thus be interpreted as the main features of f. The resulting error is necessarily smaller than the error of a linear approximation which selects the M approximation vectors independently of f. Let us sort ${{\left\{ \left| \left\langle f,{{g}_{m}} \right\rangle \right| \right\}}_{m\in \mathbb{N}}}$ in decreasing order :$\left| \left\langle f,{{g}_{{{m}_{k}}}} \right\rangle \right|\ge \left| \left\langle f,{{g}_{{{m}_{k+1}}}} \right\rangle \right|$. The best non-linear approximation is

${{f}_{M}}=\sum\limits_{k=1}^{M}{\left\langle f,{{g}_{{{m}_{k}}}} \right\rangle {{g}_{{{m}_{k}}}}}$

It can also be written as inner product thresholding:

${{f}_{M}}=\sum\limits_{m=0}^{\infty }{{{\theta }_{T}}\left( \left\langle f,{{g}_{m}} \right\rangle \right){{g}_{m}}}$

with $T=\left| \left\langle f,{{g}_{{{m}_{M}}}} \right\rangle \right|$ and

{{\theta }_{T}}\left( x \right)=\left\{ \begin{align} & x,if\left| x \right|\ge T \\ & 0,if\left| x \right|

The non-linear error is

$\varepsilon \left[ M \right]=\left\{ {{\left\| f-{{f}_{M}} \right\|}^{2}} \right\}=\sum\limits_{k=M+1}^{\infty }{\left\{ {{\left| \left\langle f,{{g}_{{{m}_{k}}}} \right\rangle \right|}^{2}} \right\}}$

this error goes quickly to zero as M increases,if the sorted values of $\left| \left\langle f,{{g}_{{{m}_{k}}}} \right\rangle \right|$ have a fast decay as k increases. This decay is quantified by computing the ${{\Iota }^{\Rho }}$ norm of the signal inner products in B:

${{\left\| f \right\|}_{\Beta ,p}}={{\left( \sum\limits_{m=0}^{\infty }{{{\left| \left\langle f,{{g}_{m}} \right\rangle \right|}^{p}}} \right)}^{1/p}}$

The following theorem relates the decay of $\varepsilon \left[ M \right]$ to ${{\left\| f \right\|}_{\Beta ,p}}$

Theorem:(decay of error) If ${{\left\| f \right\|}_{\Beta ,p}}<+\infty$ with $p<2$ then

$\varepsilon \left[ M \right]\le \frac{\left\| f \right\|_{\Beta ,p}^{2}}{\frac{2}{p}-1}{{M}^{1-\frac{2}{p}}}$

and $\varepsilon \left[ M \right]=o\left( {{M}^{1-\frac{2}{p}}} \right)$. Conversely, if $\varepsilon \left[ M \right]=o\left( {{M}^{1-\frac{2}{p}}} \right)$ then

${{\left\| f \right\|}_{\Beta ,q}}<+\infty$ for any $q>p$.

### Non-optimality of Karhunen-Loéve Bases

To further illustrate the differences between linear and non -linear approximations, we study the decomposition of a simple non-Gaussian random vector in a Karhunen-Loéve basis. Processes whose realizations have a random translation are stationary. The Karhunen-Loéve basis is then a Fourier basis and we study its performance. To simplify the analysis, consider a random vector Y[n] of size N that is random shift modulo N of a deterministic signal f[n] of zero mean $\sum\nolimits_{n=0}^{N-1}{f\left[ n \right]=0}$:

$Y\left[ n \right]=f\left[ \left( n-p \right)\bmod N \right]$

The random shift P is uniformly distributed on [0,N-1]:

$\Pr \left( P=p \right)=\frac{1}{N}for,0\le p

Clearly

$E\left\{ Y\left[ n \right] \right\}=\frac{1}{N}\sum\limits_{p=0}^{N-1}{f\left[ \left( n-p \right)\bmod N \right]}=0$

and

$R\left[ n,k \right]=E\left\{ Y\left[ n \right]Y\left[ k \right] \right\}=\frac{1}{N}\sum\limits_{p=0}^{N-1}{f\left[ \left( n-p \right)\bmod N \right]}f\left[ \left( k-p \right)\bmod N \right]$

$=\frac{1}{N}f\Theta \bar{f}\left[ n-k \right]with,\bar{f}\left[ n \right]=f\left[ -n \right]$

Hence $R\left[ n,k \right]={{R}_{Y}}\left[ n-k \right]with,$

${{R}_{Y}}\left[ k \right]=\frac{1}{N}f\Theta \bar{f}\left[ k \right]$

Since RY is N periodic, Y is a circular stationary random vector. The covariance operator is a circular convolution with RY and is therefore diagonalized in the discrete Fourier Karhunen-Loéve basis ${{\left\{ \frac{1}{\sqrt{N}}{{e}^{\frac{i2\pi mn}{N}}} \right\}}_{0\le m. The power spectrum is Fourier Transform of RY:

${{P}_{Y}}\left[ m \right]={{{\hat{R}}}_{Y}}\left[ m \right]=\frac{1}{N}{{\left| \hat{f}\left[ m \right] \right|}^{2}}$

Example: Consider an extreme case where $f\left[ n \right]=\delta \left[ n \right]-\delta \left[ n-1 \right]$ A theorem stated above guarantees that the Fourier Karhunen-Loéve basis produces a smaller expected approximation error than a canonical basis of Diracs ${{\left\{ {{g}_{m}}\left[ n \right]=\delta \left[ n-m \right] \right\}}_{0\le m. Indeed we do not know a priori the abscissa of the non-zero coefficients of Y, so there is no particular Dirac that is better adapted to perform the approximation . But the Fourier vectors cover the whole support of Y and thus absorb a part of the signal energy.

$E\left\{ {{\left| \left\langle Y\left[ n \right],\frac{1}{\sqrt{N}}{{e}^{\frac{i2\pi mn}{N}}} \right\rangle \right|}^{2}} \right\}={{P}_{Y}}\left[ m \right]=\frac{4}{N}{{\sin }^{2}}\left( \frac{\pi k}{N} \right)$

Selecting higher frequency Fourier coefficients yields a better mean-square approximation than choosing a priori a few Dirac vectors to perform the approximation. The situation is totally different for non-linear approximations. If $f\left[ n \right]=\delta \left[ n \right]-\delta \left[ n-1 \right]$ then the discrete Fourier basis is extremely inefficient because f and hence Y have an energy that is almost uniformly spread among all Fourier vectors. In contrast, since f has only two non-zero coefficients in the Dirac basis, a non-linear approximation of Y with $M\ge 2$ gives zero error. [5]

## Principal component analysis

We have established the Karhunen–Loève theorem and derived a few properties thereof. We also noted that one hurdle in its application was the numerical cost of determining the eigenvalues and eigenfunctions of its covariance operator through the Fredholm integral equation of the second kind $\int_{[a,b]} K_X(s,t) e_k(s)\,ds=\lambda_k e_k(t)$ .

However, when applied to a discrete and finite process $\left(X_n\right)_{n\in\{1,\ldots,N\}}$, the problem takes a much simpler form and standard algebra can be used to carry out the calculations.

Note that a continuous process can also be sampled at N points in time in order to reduce the problem to a finite version.

We henceforth consider a random N-dimensional vector $X=\left(X_1~X_2~\ldots~X_N\right)^T$. As mentioned above, X could contain N samples of a signal but it can hold many more representations depending on the field of application. For instance it could be the answers to a survey or economic data in an econometrics analysis.

As in the continuous version, we assume that X is centered, otherwise we can let $X:=X-\mu_X$ (where $\mu_X$ is the mean vector of X) which is centered.

Let us adapt the procedure to the discrete case.

### Covariance matrix

Recall that the main implication and difficulty of the KL transformation is computing the eigenvectors of the linear operator associated to the covariance function, which are given by the solutions to the integral equation written above.

Define Σ, the covariance matrix of X. Σ is an N by N matrix whose elements are given by:

$\Sigma_{ij}=E[X_i X_j],\qquad \forall i,j \in \{1,\ldots,N\}$

Rewriting the above integral equation to suit the discrete case, we observe that it turns into:

\begin{align} &\sum_{i=1}^N \Sigma_{ij} e_j=\lambda e_i\\ \Leftrightarrow \quad& \Sigma e=\lambda e \end{align}

where $e=(e_1~e_2~\ldots~e_N)^T$ is an N-dimensional vector.

The integral equation thus reduces to a simple matrix eigenvalue problem, which explains why the PCA has such a broad domain of applications.

Since Σ is a positive definite symmetric matrix, it possesses a set of orthonormal eigenvectors forming a basis of $\R^N$, and we write $\{\lambda_i,\phi_i\}_{i\in\{1,\ldots,N\}}$ this set of eigenvalues and corresponding eigenvectors, listed in decreasing values of λi. Let also $\Phi$ be the orthonormal matrix consisting of these eigenvectors:

\begin{align} \Phi &:=\left(\phi_1~\phi_2~\ldots~\phi_N\right)^T\\ \Phi^T \Phi &=I \end{align}

### Principal component transform

It remains to perform the actual KL transformation, called the principal component transform in this case. Recall that the transform was found by expanding the process with respect to the basis spanned by the eigenvectors of the covariance function. In this case, we hence have:

\begin{align} X &=\sum_{i=1}^N \langle \phi_i,X\rangle \phi_i\\ &=\sum_{i=1}^N \phi_i^T X \phi_i \end{align}

In a more compact form, the principal component transform of X is defined by:

$\left\{ \begin{array}{rl} Y&=\Phi^T X\\ X&=\Phi Y \end{array} \right.$

The i-th component of Y is $Y_i=\phi_i^T X$, the projection of X on $\phi_i$ and the inverse transform $X=\Phi Y$ yields the expansion of $X$ on the space spanned by the $\phi_i$:

$X=\sum_{i=1}^N Y_i \phi_i=\sum_{i=1}^N \langle \phi_i,X\rangle \phi_i$

As in the continuous case, we may reduce the dimensionality of the problem by truncating the sum at some $K\in\{1,\ldots,N\}$ such that $\frac{\sum_{i=1}^K \lambda_i}{\sum_{i=1}^N \lambda_i}\geq \alpha$ where α is the explained variance threshold we wish to set.

We can also reduce the dimensionality through the use of multilevel dominant eigenvector estimation (MDEE).[6]

## Examples

### The Wiener process

There are numerous equivalent characterizations of the Wiener process which is a mathematical formalization of Brownian motion. Here we regard it as the centered standard Gaussian process Wt with covariance function

$K_{W}(t,s) = \operatorname{Cov}(W_t,W_s) = \min (s,t).$

We restrict the time domain to [a,b]=[0,1] without loss of generality.

The eigenvectors of the covariance kernel are easily determined. These are

$e_k(t) = \sqrt{2} \sin \left( \left(k - \textstyle\frac{1}{2}\right) \pi t \right)$

and the corresponding eigenvalues are

$\lambda_k = \frac{1}{(k -\frac{1}{2})^2 \pi^2}.$

This gives the following representation of the Wiener process:

Theorem. There is a sequence {Zi}i of independent Gaussian random variables with mean zero and variance 1 such that

$W_t = \sqrt{2} \sum_{k=1}^\infty Z_k \frac{\sin \left(\left(k - \frac{1}{2}\right) \pi t\right)}{ \left(k - \frac{1}{2}\right) \pi}.$

Note that this representation is only valid for $t\in[0,1].$ On larger intervals, the increments are not independent. As stated in the theorem, convergence is in the L2 norm and uniform in t.

### The Brownian bridge

Similarly the Brownian bridge $B_t=W_t-tW_1$ which is a stochastic process with covariance function

$K_B(t,s)=\min(t,s)-ts$

can be represented as the series

$B_t = \sum_{k=1}^\infty Z_k \frac{\sqrt{2} \sin(k \pi t)}{k \pi}$

## Applications

Adaptive optics systems sometimes use K–L functions to reconstruct wave-front phase information (Dai 1996, JOSA A).

Karhunen–Loève expansion is closely related to the Singular Value Decomposition. The latter has myriad applications in image processing, radar, seismology, and the like. If one has independent vector observations from a vector valued stochastic process then the left singular vectors are maximum likelihood estimates of the ensemble KL expansion.

### Applications in signal estimation and detection

#### Detection of a known continuous signal S(t)

In communication, we usually have to decide whether a signal from a noisy channel contains valuable information. The following hypothesis testing is used for detecting continuous signal s(t) from channel output X(t), N(t) is the channel noise, which is usually assumed zero mean gaussian process with correlation function $R_{N} (t, s) = E[N(t)N(s)]$

 $H: X(t) = N(t)$, $K: X(t) = N(t)+s(t), t\in(0,T)$.

#### Signal detection in white noise

When the channel noise is white, its correlation function is

$R_{N}(t) = \frac{N_0}{2} \delta (t)$,

and it has constant power spectrum density. In physically practical channel, the noise power is finite, so:

$S_{N}(f) = \frac{N_{0}}{2} \text{ for } |f|w$.

Then the noise correlation function is sinc function with zeros at $\frac{n}{2\omega}, n = ...-1,0,1,...$ . Since are uncorrelated and gaussian, they are independent. Thus we can take samples from X(t) with time spacing

$\Delta t = \frac{n}{2\omega}$ within (0,T).

Let $X_i = X(i\Delta t)$. We have a total of $n = \frac{T}{\Delta t} = T(2\omega) = 2\omega T$ i.i.d samples $\{X_1, X_2,...,X_n\}$ to develop the likelihood-ratio test. Define signal $S_i = S(i\Delta t)$, the problem becomes,

$H: X_i = N_i$,
$K: X_i = N_i + S_i, i = 1,2...n.$

The log-likelihood ratio

$\mathcal{L}(\underline{x}) = \log\frac{\sum^n_{i=1} (2S_i x_i - S_i^2)}{2\sigma^2} \Leftrightarrow \Delta t \Sigma ^n_{i = 1} S_i x_i = \sum^n_{i=1} S(i\Delta t)x(i\Delta t)\Delta t \gtrless \lambda_2$.

As $t \rightarrow 0, \text{ let } G = \int^T_0 S(t)x(t)dt$.

Then G is the test statistics and the Neyman–Pearson optimum detector is:$G(\underline{x}) > G_0 \Rightarrow K, < G_0 \Rightarrow H$. As G is gaussian, we can characterize it by finding its mean and variances. Then we get

$H: G \sim N(0,\frac{N_{0}E}{2})$
$K: G \sim N(E,\frac{N_{0}E}{2})$,

where $E = \int^T_{0} S^2(t)dt$ is the signal energy.

The false alarm error

$\alpha = \int^{\infty}_{G_{0}} N(0,\frac{N_{0}E}{2})dG \Rightarrow G_0 = \sqrt{\frac{N_0 E}{2}} \Phi^{-1}(1-\alpha)$

And the probability of detection:

$\beta = \int^{\infty}_{G_0} N(E, \frac{N_0 E}{2})dG = 1-\Phi(\frac{G_0 - E}{\sqrt{\frac{N_0 E}{2}}}) = \Phi [\sqrt{\frac{2E}{N_0}} - \Phi^{-1}(1-\alpha)] , \Phi(\cdot)$ is the cdf of standard normal gaussian variable.

#### Signal detection in colored noise

When N(t) is colored (correlated in time) gaussian noise with zero mean and covariance function $R_N(t,s) = E[X(t)X(s)],$ we cannot sample independent discrete observations by evenly spacing the time. Instead, we can use K–L expansion to uncorrelate the noise process and get independent gaussian observation 'samples'. The K–L expansion of N(t):

$N(t) = \sum^{\infty}_{i=1} N_i \Phi_i(t), 0,

where $N_i =\int N(t)\Phi_i(t)dt$ and the orthonormal bases $\{\Phi_i{t}\}$are generated by kernal $R_N(t,s)$, i.e., solution to

$\int ^T _0 R_N(t,s)\Phi_i(s)ds = \lambda_i \Phi_i(t), var[N_i] = \lambda_i$.

Do the expansion:

$S(t) = \sum^{\infty}_{i = 1}S_i\Phi_i(t)$,

where $S_i = \int^T _0 S(t)\Phi_i(t)dt, 0, then

$X_i = \int^T _0 X(t)\Phi_i(t) dt = N_i$

under H and $N_i + S_i$ under K. Let $\overline{X} = \{X_1,X_2,\dots\}$, we have

${N_i}$ are independent gaussian r.v's with variance $\lambda_i$
under H: $\{X_i\}$ are independent gaussian r.v's. $f_H[x(t)|0
under K: $\{X_i - S_i\}$ are independent gaussian r.v's. $f_K[x(t)|0

Hence, the log-LR is given by

$\mathcal{L}(\underline{x}) = \sum^{\infty}_{i=1} \frac{2S_i x_i - S_i^2}{2\lambda_i}$

and the optimum detector is

$G = \sum^{\infty}_{i=1} S_i x_i \lambda_i > G_0 \Rightarrow K, < G_0 \Rightarrow H.$

Define

$k(t) = \sum^{\infty}_{i=1} \lambda_i S_i \Phi_i(t), 0

then $G = \int^T _0 k(t)x(t)dt$.

##### How to find k(t)

Since

$\int^T_0 R_N(t,s)k(s)ds = \sum^{\infty}_{i=1} \lambda_i S_i \int^T _0 R_N(t,s)\Phi_i (s) ds = \sum^{\infty}_{i=1} S_i \Phi_i(t) = S(t)$,

k(t) is the solution to

$\int^T_0 R_N(t,s)k(s)ds = S(t)$.

If N(t)is wide-sense stationary,

$\int^T_0 R_N(t-s)k(s)ds = S(t)$,

which is known as the Wiener–Hopf equation. The equation can be solved by taking fourier transform, but not practically realizable since infinite spectrum needs spatial factorization. A special case which is easy to calculate k(t) is white gaussian noise.

$\int^T_0 \frac{N_0}{2}\delta(t-s)k(s)ds = S(t) \Rightarrow k(t) = C S(t), 0.

The corresponding impulse response is h(t) = k(T-t) = C S(T-t). Let C = 1, this is just the result we arrived at in previous section for detecting of signal in white noise.

##### Test threshold for Neyman–Pearson detector

Since X(t)is gaussian process, $G = \int^T_0 k(t)x(t)dt$ is a gaussian random variable that can be characterized by its mean and variance.

$E[G|H] = \int^T_0 k(t)E[x(t)|H]dt = 0$
$E[G|K] = \int^T_0 k(t)E[x(t)|K]dt = \int^T_0 k(t)S(t)dt \equiv \rho$
$E[G^2|H] = \int^T_0\int^T_0 k(t)k(s) R_N(t,s)dtds = \int^T_0 k(t)(\int^T_0 k(s)R_N(t,s)ds)=\int^T_0 k(t)S(t)dt = \rho$
$var[G|H] = E[G^2|H] - (E[G|H])^2 = \rho$
$E[G^2|K]=\int^T_0\int^T_0k(t)k(s)E[x(t)x(s)]dtds = \int^T_0\int^T_0k(t)k(s)(R_N(t,s) +S(t)S(s))dtds = \rho + \rho^2$
$var[G|K] = E[G^2|K] - (E[G|K])^2 = \rho + \rho^2 -\rho^2 = \rho.$

Hence, we obtain the distributions of H and K:

$H: G \sim N(0,\rho)$
$K: G \sim N(\rho, \rho)$

The false alarm error is

$\alpha = \int^{\infty}_{G_0} N(0,\rho)dG = 1 - \Phi(\frac{G_0}{\sqrt{\rho}}).$

So the test threshold for the Neyman–Pearson optimum detector is

$G_0 = \sqrt{\rho} \Phi^{-1} (1-\alpha)$.

Its power of detection is

$\beta = \int^{\infty}_{G_0} N(\rho, \rho)dG = \Phi [\sqrt{\rho} - \Phi^{-1}(1 - \alpha)]$.

When the noise is white gaussian process, the signal power is

$\rho = \int^T_0 k(t)S(t)dt = \int^T_0 S(t)^2 dt = E$.
##### Prewhitening

For some type of colored noise, a typical practise is to add a prewhitening filter before the matched filter to transform the colored noise into white noise. For example, N(t) is a wide-sense stationary colored noise with correlation function

$\R_N(\tau) = \frac{B N_0}{4} e^{-B|\tau|}$
and $S_N(f) = \frac{N_0}{2(1+(\frac{w}{B})^2)}$.

The transfer function of prewhitening filter is $H(f) = 1 + j \frac{w}{B}$.

#### Detection of a gaussian random signal in AWGN

When the signal we want to detect from the noisy channel is also random, for example, a white gaussian process X(t), we can still implement K–L expansion to get independent sequence of observation. In this case, the detection problem is described as follows:

$H_0 : Y(t) = N(t)$
$H_1 : Y(t) = N(t) + X(t), 0

X(t) is a random process with correlation function $R_X(t,s) = E\{X[t]X[s]\}$

The K–L expansion of X(t) is

$X(t) = \sum^{\infty}_{i=1} X_i \Phi_i(t)$,

where

$X_i =\int^T_0 X(t) \Phi_i(t). \Phi(t)$

are solutions to

$\int^T_0 R_X(t,s)\Phi_i(s)ds= \lambda_i \Phi_i(t)$.

So $X_i$'s are independent sequence of r.v's with zero mean and variance $\lambda_i$. Expanding Y(t) and N(t) by $\Phi_i(t)$, we get

$Y_i = \int^T_0 Y(t)\Phi_i(t)dt = \int^T_0 [N(t) + X(t)]\Phi_i(t) = N_i + X_i$,

where $N_i = \int^T_0 N(t)\Phi_i(t)dt.$

As N(t) is gaussian white noise, $N_i$'s are i.i.d sequence of r.v with zero mean and variance$\frac{N_0}{2}$, then the problem is simplified as follows,

$H_0: Y_i = N_i$
$H_1: Y_i = N_i + X_i$

The Neyman–Pearson optimal test:

$\Lambda = \frac{f_Y|H_1}{f_Y|H_0} = Ce^{-\sum^{\infty}_{i=1}\frac{y_i^2}{2} \frac{\lambda_i}{\frac{N_0}{2}(\frac{N_0}{2} + \lambda_i)}}$,

so the log-likelihood ratio

$\mathcal{L} = ln(\Lambda) = K -\sum^{\infty}_{i=1}\frac{y_i^2}{2} \frac{\lambda_i}{\frac{N_0}{2}(\frac{N_0}{2} + \lambda_i)}$.

Since

$\hat{X_i} = \frac{\lambda_i}{\frac{N_0}{2}(\frac{N_0}{2} + \lambda_i)}$

is just the minimum-mean-square estimate of $X_i$ given $Y_i$'s,

$\mathcal{L} = K + \frac{1}{N_0} \sum^{\infty}_{i=1} Y_i \hat{X_i}$.

K–L expansion has the following property: If

$f(t) = \sum f_i \Phi_i(t), g(t) = \sum g_i \Phi_i(t)$,

where

$f_i = \int_0^T f(t) \Phi_i(t), g_i = \int_0^T g(t)\Phi_i(t).$,

then

$\sum^{\infty}_{i=1} f_i g_i = \int^T_0 g(t)f(t)dt$.

So let

$\hat{X(t|T)} = \sum^{\infty}_{i=1} \hat{X_i}\Phi_i(t)$, $\mathcal{L} = K + \frac{1}{N_0} \int^T_0 Y(t) \hat{X(t|T)}dt$.

Noncausal filter Q(t, s) can be used to get the estimate through

$\hat{X(t|T)} = \int^T_0 Q(t,s)Y(s)ds$.

By orthogonality principle, Q(t,s) satisfies

$\int^T_0 Q(t,s)R_X(s,t)ds + \frac{N_0}{2} Q(t, \lambda) = R_X(t, \lambda), 0 < \lambda < T, 0.

However for practical reason, it's necessary to further derive the causal filter h(t, s), where h(t, s) = 0 for s > t, to get estimate $\hat{X(t|t)}$. Specifically,

$Q(t,s) = h(t,s) + h(s, t) - \int^T_0 h(\lambda, t)h(s, \lambda)d\lambda$.


text:
Warning, file /education/step/stepcore/constants.h was not indexed or was modified since last indexation (in which case cross-reference links may be missing, inaccurate or erroneous).

```0001 /*
0002     SPDX-FileCopyrightText: 2007 Vladimir Kuznetsov <ks.vladimir@gmail.com>
0003
0004     SPDX-License-Identifier: GPL-2.0-or-later
0005 */
0006
0007 #ifndef STEPCORE_CONSTANTS_H
0008 #define STEPCORE_CONSTANTS_H
0009
0010 /** \file constants.h
0011  *  \brief Constants class
0012  */
0013
0014 namespace StepCore {
0015
0016 /** \ingroup constants
0017  *  \brief Common physical constants
0018  *
0019  *  Values taken from https://en.wikipedia.org/wiki/Physical_constants
0020  */
0021 class Constants {
0022 public:
0023     /** Pi constant */
0024     static const double Pi;
0025
0026     /** Speed of light in vacuum [m/s] */
0027     static const double SpeedOfLight;
0028     /** Electric constant (permittivity of free space) [F/m] */
0029     static const double Electric;
0030     /** Magnetic constant (permeability of free space) [N/A^2] */
0031     static const double Magnetic;
0032
0033     /** Coulomb's constant [N*m^2/C^2] */
0034     static const double Coulomb;
0035     /** Error of Coulomb's constant */
0036     static const double CoulombError;
0037
0038     /** Newtonian constant of gravitation [N*m^2/kg^2] */
0039     static const double Gravitational;
0040     /** Error of newtonian constant of gravitation */
0041     static const double GravitationalError;
0042
0043     /** Planck's constant [J*s] */
0044     static const double Planck;
0045     /** Error of Planck's constant */
0046     static const double PlanckError;
0047
0048     /** Boltzmann's constant [J/K] */
0049     static const double Boltzmann;
0050     /** Error of Boltzmann's constant */
0051     static const double BoltzmannError;
0052
0053     /** Standard acceleration of gravity (free fall on Earth) [m/s^2] */
0054     static const double WeightAccel;
0055     /** Error of standard acceleration of gravity (free fall on Earth) */
0056     static const double WeightAccelError;
0057 };
0058
0059 } // namespace StepCore
0060
0061 #endif
0062 ```


text:
## Lyapunov stability analysis of daisy chain control allocation.(English)Zbl 0869.93034

The paper considers the so-called saturated control in the case when many control functions are available. In this case, a rule is introduced in the control law allowing “daisy chain saturation”, i.e. some controls are maintained to zero up to the moment when others are saturated. Local stability is proved in a standard way using a quadratic Lyapunov function.

### MSC:

 93D05 Lyapunov and other classical stabilities (Lagrange, Poisson, $$L^p, l^p$$, etc.) in control theory 93B51 Design techniques (robust design, computer-aided design, etc.)
Full Text:


text:
# Quiz 1

Question 1

Time: 00:00:00
if $a_{1}= \frac{139}{3}$, $a_{2}= -19$ and $a_{k} = a_{k-1} - a_{k-2},$then what will be the 62th term in the series? (k, k-1, k-2 are in base)

$\frac{139}{3}$

$\frac{139}{3}$

-19

-19

$\frac{-196}{3}$

$\frac{-196}{3}$

$\frac{-193}{3}$

$\frac{-193}{3}$

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1=a 2=b 3=b-a 4=-a 5=-b 6=-b+a 7=a 8=b so for 60=-b+a 61=a 62=b

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Question 2

Time: 00:00:00
If $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1} { (a + b + c)}$;

where a + b + c 1 0; abc 1 0, then what is the value of ( a + b ) ( b + c ) ( c + a )?

Equal to 0

Equal to 0

Greater than 0

Greater than 0

Less than 0

Less than 0

Cannot be determined

Cannot be determined

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Question 3

Time: 00:00:00
In a school 10% students get 70 marks,25% students get 80 marks,15% students get 85 marks,25% students get 90 marks, remaining get 95 marks.the average of students marks is an even number then least number of students?

20

20

80

80

100

100

200

200

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Question 4

Time: 00:00:00
$y=\frac{x}{(x-k)}$,where k is a constant,and x is real number.show's that....

y increase with an increase in x.

y increase with an increase in x.

y decreases first and then increase the value of x.

y decreases first and then increase the value of x.

y increase then decreases with the value of x.

y increase then decreases with the value of x.

it remains constant.

it remains constant.

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Question 5

Time: 00:00:00
if v,w,x,y,z are non negative integer, each less than 11,then how many distinct combinations(v,w,x,y,z) satisfy v($11^{4}$)+w($11^{3}$))+x($11^{2}$))+y(11)+z=151001?

10

10

2

2

3

3

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Question 6

Time: 00:00:00
The Range of the function f(x) =$\frac{ (x - 2) }{ (2 - x) }$is....

-1 to 1

-1 to 1

0 to 1

0 to 1

-1 to 0

-1 to 0

None of above

None of above

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Question 7

Time: 00:00:00
$\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$ where x=12 and y,z positive integer how many possible values of y and z?

4

4

7

7

12

12

infinity

infinity

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Question 8

Time: 00:00:00
The sum of two numbers is 15 and their geometric mean is 20% lower than the arithmetic mean. find the numbers?

11,4

11,4

12,3

12,3

9,6

9,6

none

none

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Question 9

Time: 00:00:00
There are 72 students and 15 teachers in a school excluding principal. The principal has 3-digits odd no. quantities of .toffees and he equally distributes it among students & teachers, he gets 86 Toffees remain. If he distributes among teachers, he gets 14 Toffees remain. . If principal adds himself among teachers, how much coffee(s) will remain?

1

1

5

5

15

15

Can\'t be determined

Can\'t be determined

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Question 10

Time: 00:00:00
How many values of c in the equation $x^{2}-5x+c$ result in the rational root which is an integer?

6

6

3

3

1

1

infinity

infinity

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### algebraic number theory

We are now (finally) ready to start handling the cohomology of the idele classes.  The previous few posts in this series contain important background computations of the Herbrand quotients of local fields and units, which should be read before this.  Let ${L/k}$ be a finite cyclic extension of global fields of degree ${n}$. In the following, the Herbrand quotient will always be respect to the Galois group ${G=G(L/k)}$.

Theorem 1 We have ${Q(J_L/L^*) = n}$. In particular,$\displaystyle (J_k: k^* NJ_L) \geq n.$

1. Some remarks

The point of class field theory, of course, is that there is exactly an equality in the above statement, which is induced by an isomorphism between the two groups, and which holds for an arbitrary abelian extension of number fields.

Before we prove this theorem, let’s review a little. We know that ${G}$ acts on the ideles ${J_L}$, and also on ${L^*}$ (clearly). As a result, we get an action on the idele classes ${C_L=J_L/L^*}$. There is a map ${C_k = J_k/k^* \rightarrow C_L}$; I claim that it is an injection, and the fixed points of ${G}$ in ${C_L}$ are precisely the points of ${C_k}$. This can be proved using group cohomology. We have an exact sequence ${0 \rightarrow L^* \rightarrow J_L \rightarrow C_L \rightarrow 0}$, and consequently one has a long exact sequence

$\displaystyle 0 \rightarrow H^0(G, L^*) \rightarrow H^0(G, J_L) \rightarrow H^0(G, C_L) \rightarrow H^1(G, L^8) = 0$

by Hilbert’s Theorem 90, and where ${H^0}$ is the ordinary (non-Tate) cohomology groups, that is to say just the ${G}$-stable points. Since we know that ${H^0(G, L^*)=k^*}$ and ${H^0(G, J_L)=J_k}$, we find that ${(C_L)^G = C_k}$, q.e.d.

So, anyhow, this Big Theorem today computes the Herbrand quotient ${Q(C_L)}$. It in particular implies that ${H_T^0(G, C_L) \geq n}$, and since this group is none other than

$\displaystyle C_k/NC_L = J_k/k^* NC_L$

we get the other claim of the theorem. We are reduced to computing this messy Herbrand quotient, and it will use all the tools that we have developed up to now.

Let ${S}$ be a finite set of places of a number field ${L}$, containing the archimedean ones. Suppose ${L/k}$ is a cyclic extension with Galois group ${G}$. Then, if ${G}$ keeps ${S}$ invariant, ${G}$ keeps the group ${L_S}$ of ${S}$-units invariant. We will need to compute its Herbrand quotient in order to do the same for the idele classes (next time), and that is the purpose of this post.

Up to a finite group, ${L_S}$ is isomorphic to a lattice in ${\mathbb{R}^{|S|}}$, though—this is the unit theorem. This isomorphism is by the log map, and it is even a ${G}$-isomorphism if ${G}$ is given an action on ${\mathbb{R}^{|S|}}$ coming from the permutation of ${S}$ (i.e. the permutation representation). This lattice is of maximal rank in the ${G}$-invariant hyperplane ${W \subset \mathbb{R}^{|S|}}$.

Motivated by this, we study the Herbrand quotient on lattices next.

1. The cohomology of a lattice

Fix a cyclic group ${G}$. We will suppose given a ${G}$-lattice ${L}$, that is to say a ${\mathbb{Z}}$-free module of finite rank on which ${G}$ acts. One way to get a ${G}$-lattice is to consider a representation of ${G}$ on some real vector space ${V}$, and choose a lattice in ${V}$ that is invariant under the action of ${G}$.

Proposition 1 Let ${L, L'}$ be two lattices in ${V}$ of maximal rank. Then ${Q(L)=Q(L')}$.

The way Lang (following Artin-Tate’s Class Field Theory) approaches this result seems a little unwieldy to me. I will follow Cassels-Frohlich (actually, Atiyah-Wall in their article on group cohomology in that excellent conference volume).  We have ${\mathbb{Q}[G]}$-modules ${L_{{\mathbb Q}} = L \otimes \mathbb{Q},L'_{{\mathbb Q}}= L' \otimes \mathbb{Q}}$, and ${\mathbb{R}}$-modules ${L_{{\mathbb R}}, L'_{{\mathbb R}}}$ defined similarly. Moreover, we have

$\displaystyle \mathbb{R} \otimes_{{\mathbb Q}} \hom_{\mathbb{Q}[G]}( L_{{\mathbb Q}}, L_{{\mathbb Q}}) \simeq \hom_{{\mathbb R}[G]}(L_{{\mathbb R}}, L'_{{\mathbb R}}).$ (more…)

We shall now take the first steps in class field theory. Specifically, since we are interested in groups of the form ${J_k/k^* N J_L}$, we will need their orders. And the first place to begin is with a local analog.

1. The cohomology of the units

Theorem 1 Let ${L/K}$ be a cyclic extension of local fields of degree ${n}$ and ramification ${e}$. Then ${Q(U_L)=1}$.

Let ${G}$ be the Galois group. We will start by showing that ${Q(U_L)=1}$.

Indeed, first of all let us choose a normal basis of ${L/K}$, i.e. a basis ${(x_\sigma)_{\sigma \in G}}$ such that ${\tau x_{\sigma} = x_{\sigma \tau}}$ for all ${\tau, \sigma \in G}$. It is known (the normal basis theorem) that this is possible. By multiplying by a high power of a uniformizer, we find that there is a ${G}$-submodule ${V_a}$ of the additive group ${\mathcal{O}_L}$ isomorphic to ${\mathcal{O}_K[G]}$, i.e. is induced. We see that ${V_a}$ has trivial Tate cohomology and Herbrand quotient 1 by Shapiro’s lemma: any ${G}$-module induced from the subgroup 1 satisfies this, because ${H_T^i(1, A)= 0}$ for any ${A}$.

But if ${V_a}$ is taken sufficiently close to zero, then there is a ${G}$-equivariant map ${\exp: V_a \rightarrow U_L}$, defined via

$\displaystyle \exp(x) = \sum_k \frac{x^k}{k!}$

which converges appropriately at sufficiently small ${x}$. (Proof omitted, but standard. Note that ${k! \rightarrow 0}$ in the nonarchimedean case though!) In other words, the additive and multiplicative groups are locally isomorphic. This map (for ${V_a}$ sufficiently small) is an injection, the inverse being given by the logarithm power series. Its image is an open subgroup ${V}$ of the units, and since the units are compact, of finit index. So we have

$\displaystyle 1 = Q(V_a) = Q(V) = Q(U_L).$

This proves the theorem. (more…)

The following situation—namely, the cohomology of induced objects—occurs very frequently, and we will devote a post to its analysis. Let ${G}$ be a cyclic group acting on an abelian group ${A}$. Suppose we have a decomposition ${A = \bigoplus_{i \in I} A_i}$ such that any two ${A_i}$ are isomorphic and ${G}$ permutes the ${A_i}$ with each other. It turns out that the computation of the cohomology of ${A}$ can often be simplified.

Then let ${G_0}$ be the stabilizer of ${A_{i_0}}$ for some fixed ${i_0 \in I}$, i.e. ${G_0 = \{g: gA_{i_0} = A_{i_0} \}.}$ Then, we have ${A = \mathrm{Ind}_{G_0}^G A_{i_0}}$. This is what I meant about ${A}$ being induced.

I claim that

$\displaystyle \boxed{ H_T^i(G, A) \simeq H_T^i(G_0, A_{i_0}) , \quad i = -1, 0. }$

In particular, we get an equality of the Herbrand quotients ${Q(A), Q(A_{i_0})}$. (more…)

We continue our quest to climb Mount Takagi-Artin.

In class field theory, it will be important to compute and keep track of the orders of groups such as ${(K^*:NL^*)}$, where ${L/K}$ is a Galois extension of local fields. A convenient piece of machinery for doing this is the Herbrand quotient, which we discuss today. I only sketch the proofs though, and a little familiarity with the Tate cohomology groups will be useful (but is not strictly necessary if one accepts the essentially combinatorial results without proof or proves them directly).

1. Definition

Let ${G}$ be a cyclic group generated by ${\sigma}$ and ${A}$ a ${G}$-module. It is well-known that the Tate cohomology groups ${H^i_T(G, A)}$ are periodic with period two and thus determined by ${H^0}$ and ${H^{-1}}$. By definition,

$\displaystyle H^0(G,A) = A^G/ NA ,$

where ${A^G}$ consists of the elements of ${A}$ fixed by ${G}$, and ${N: A \rightarrow A}$ is the norm map, ${a \rightarrow \sum_g ga}$. Moreover,

$\displaystyle H^{-1}(G, A) = \mathrm{ker} N/ (\sigma -1) A.$

(Normally, for ${G}$ only assumed finite, we would quotient by the sum of ${(\sigma - 1)A}$ for ${\sigma \in G}$ arbitrary, but here it is enough to do it for a generator—easy exercise.)

If both cohomology groups are finite, define the Herbrand quotient ${Q(A)}$ as

$\displaystyle Q(A) = \frac{ |H_T^0(G,A)|}{|H_T^{-1}(G,A)|}.$ (more…)

Class field theory is about the abelian extensions of a number field ${K}$. Actually, this is strictly speaking global class field theory (there is an analog for abelian extensions of local fields), and there is a similar theory for function fields of transcendence degree 1 over finite fields, but we shall not deal with it.

Let us, however, consider the situation for local fields—which we will later investigate more—as follows. Suppose ${k}$ is a local field and ${L}$ an unramified extension. Then the Galois group ${L/k}$ is isomorphic to the Galois group of the residue field extension, i.e. is cyclic of order ${f}$ and generated by the Frobenius. But I claim that the group ${k^*/NL^*}$ is the same. Indeed, ${NU_L = U_K}$ by a basic theorem about local fields that we will prove using abstract nonsense later (but can also be easily proved using successive approximation and facts about finite fields). So ${k^*/NL^*}$ is cyclic, generated by a uniformizer of ${k}$, which has order ${f}$ in this group. Thus we get an isomorphism

$\displaystyle k^*/NL^* \simeq G(L/k)$

sending a uniformizer to the Frobenius. (more…)

As usual, let ${K}$ be a global field. Now we do the same thing that we did last time, but for the ideles.

1. Ideles

First of all, we have to define the ideles. These are only a group, and are defined as the restricted direct product

$\displaystyle J_K = \prod'_v K_v^*$

relative to the unit subgroups ${U_v}$ of ${v}$-units (which are defined to be ${K_v^*}$ if ${v}$ is archimedean). In other words, an idele ${(x_v)_v}$ is required to satisfy ${|x_v|=1}$ for almost all ${v}$.

If ${S}$ is a finite set of places containing the archimedean ones, we can define the subset ${J^S_K = \prod_{v \in S} K_v \times \prod_{v \notin S} U_v}$; this has the product topology and is an open subgroup of ${J_K}$. These are called the ${S}$-ideles. As we will see, they form an extremely useful filtration on the whole idele group.

Dangerous bend: Note incidentally that while the ideles are a subset of the adeles, the induced topology on ${J_K}$ is not the ${J_K}$-topology. For instance, take ${K=\mathbb{Q}}$. Consider the sequence ${x^{(n)}}$ of ideles where ${x^{(n)}}$ is ${p_n}$ at ${v_{p_n}}$ (where ${p_n}$ is the ${n}$-th prime) and 1 everywhere else. Then ${x^{(n)} \rightarrow 0 \in \mathbf{A}_{\mathbb{Q}}}$ but not in ${J_{\mathbb{Q}}}$.

However, we still do have a canonical “diagonal” embedding ${K^* \rightarrow J_K}$, since any nonzero element of ${K}$ is a unit almost everywhere. This is analogous to the embedding ${K \rightarrow \mathbf{A}_K}$. (more…)

Adele – c’est un nom si belle. (Oops, that’s bad French, isn’t it?)

I actually will not be able to finish the proof of the unit theorem here, because I don’t get to the ideles in this post. That will come next time (there are some of the same themes as here).

Let ${K}$ be a global field, i.e. a finite extension of either ${\mathbb{Q}}$ or ${\mathbb{F}_p(t)}$. Then we can consider the set absolute values on ${K}$. In the number field case, these are extensions (up to a power) of the archimedean absolute value on ${\mathbb{Q}}$ or the ${p}$-adic absolute values by a theorem of Ostrowski classifying absolute values on ${\mathbb{Q}}$. In the function field case, we need another result.

Here’s how we define the adele ring. It is the restricted direct product

$\displaystyle \mathbf{A}_K := \prod'_v K_v$

where restricted means that any vector ${(x_v)_{v \in V} \in \mathbf{A}_K}$ is required to satisfy ${|x_v|_v \leq 1}$ for almost all ${v}$. This becomes a topological ring if we take a basis of the form

$\displaystyle \prod_{v \in S} T_v \times \prod_{v \notin S} \mathcal{O}_v$

where ${T_v \subset k_v}$ are open and ${\mathcal{O}_v}$ is the ring of integers, and ${S}$ is a finite set containing the archimedean places. It is clear that addition and multiplication are continuous, and that ${A_K}$ is locally compact. For ${S}$ finite and containing the archimedean absolute values ${S_\infty}$, there is a subring ${\mathbf{A}_K^S = \prod_{v \in S} K_v \times \prod_{v \notin S} \mathcal{O}_v}$, and ${\mathbf{A}_K}$ is the union of these subrings.

Since any ${x \in K}$ is contained in ${\mathcal{O}_v}$ for almost all ${v}$ (this is analogous to a rational function on a curve having only finitely many poles), there is an injective homomorphism ${K \rightarrow \mathbf{A}_K}$.

Next, we may define a Haar measure on ${\mathbf{A}_K^S}$ by taking the product of the Haar measures ${\mu_v}$ on ${K_v}$, normalized such that ${\mu_v(O_v )=1}$ for ${v \notin S_\infty}$. Thus one gets a (i.e., the) Haar measure on ${\mathbf{A}_K}$ itself. (more…)

There is another major result in algebraic number theory that we need to get to!  I have this no longer secret goal of getting to class field theory, and if it happens, this will be a key result.  The hard part of the actual proof (namely, the determination of the rank of a certain lattice) will be deferred until next time; it’s possible to do it with the tools we already have, but it is cleaner (I think) to do it once ideles have been introduced.

Following the philosophy of examples first, let us motivate things with an example. Consider the ring ${\mathbb{Z}[i]}$ of Gaussian; as is well-known, this is the ring of integers in the quadratic field ${\mathbb{Q}(i)}$. To see this, suppose ${a+ib, a, b \in \mathbb{Q}}$ is integral; then so is ${a-ib}$, and thus ${2a,2b}$ are integers. Also the fact ${(a+ib)(a-ib) = a^2+b^2 }$ must be an integer now means that neither ${a,b}$ can be of the form ${k/2}$ for ${k}$ odd.

What are the units in ${\mathbb{Z}[i]}$? If ${x}$ is a unit, so is ${\bar{x}}$, so the norm ${N(x)}$ must be a unit in ${\mathbb{Z}[i]}$ (and hence in ${\mathbb{Z}}$). So if ${x=a+ib}$, then ${a^2+b^2=1}$ and ${x = \pm 1}$ or ${\pm i}$. So, the units are just the roots of unity.

In general, however, the situation is more complicated. Consider ${\mathbb{Z}[\sqrt{2}]}$, which is again integrally closed. Then ${x=a+b\sqrt{2}}$ is a unit if and only if its norm to ${\mathbb{Q}}$, i.e. ${a^2 - 2b^2}$ is equal to ${\pm 1}$. Indeed, the norm ${N(x)}$ of a unit ${x}$ is still a unit, and since ${\mathbb{Z}}$ is integrally closed, we find that ${N(x)}$ is a ${\mathbb{Z}}$-unit. In particular, the units correspond to the solutions to the Pell equation. There are infinitely many of them.

But the situation is not hopeless. We will show that in any number field, the unit group is a direct sum of copies of ${\mathbb{Z}}$ and the roots of unity. We will also determine the rank. (more…)

This post won’t be as cool as the title sounds. But I will prove something neat, and it will lead to neater things as time goes on (assuming I keep posting on this topic).

We will now return to algebraic number theory, following Lang’s textbook, and study the distribution of points in parallelotopes.

The setup is as follows. ${K}$ will be a number field, and ${v}$ an absolute value (which, by abuse of terminology, we will use interchangeable with “valuation” and “place”) extending one of the absolute values on ${\mathbb{Q}}$ (which are always normalized in the standard way); we will write ${|x|_v}$ for the output at ${x \in K}$.

Suppose ${v_0}$ is a valuation of ${\mathbb{Q}}$; we write ${v | v_0}$ if ${v}$ extends ${v_0}$. Recall the following important formula from the theory of absolute values an extension fields:

$\displaystyle |N^K_{\mathbb{Q}}(x)|_{v_0} = \prod_{v | v_0} |x|_v^{ [K_v: \mathbb{Q}_{v_0} ] }.$

Write ${N_v := [K_v: \mathbb{Q}_{v_0} ] }$; these are the local degrees. From elementary algebraic number theory, we have ${\sum_{v | v_0} N_v = N := [K:\mathbb{Q}].}$ This is essentially a version of the ${\sum ef = N}$ formula.

The above formalism allows us to deduce an important global relation between the absolute values ${|x|_v}$ for ${x \in K}$.

Theorem 1 (Product formula) If ${x \neq 0}$,$\displaystyle \prod_v |x|_v^{N_v} = 1.$

The proof of this theorem starts with the case ${K=\mathbb{Q}}$, in which case it is an immediate consequence of unique factorization. For instance, one can argue as follows. (more…)


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## Illinois Journal of Mathematics

### Fredholm properties of evolution semigroups

#### Abstract

We show that the Fredholm spectrum of an evolution semigroup $\{E^t\}_{t\geq 0}$ is equal to its spectrum, and prove that the ranges of the operator $E^t-I$ and the generator ${\bf G}$ of the evolution semigroup are closed simultaneously. The evolution semigroup is acting on spaces of functions with values in a Banach space, and is induced by an evolution family that could be the propagator for a well-posed linear differential equation $u'(t)=A(t)u(t)$ with, generally, unbounded operators $A(t)$; in this case ${\bf G}$ is the closure of the operator $G$ given by $(Gu)(t)=-u'(t)+A(t)u(t)$.

#### Article information

Source
Illinois J. Math., Volume 48, Number 3 (2004), 999-1020.

Dates
First available in Project Euclid: 13 November 2009

https://projecteuclid.org/euclid.ijm/1258131066

Digital Object Identifier
doi:10.1215/ijm/1258131066

Mathematical Reviews number (MathSciNet)
MR2114265

Zentralblatt MATH identifier
1073.34067

#### Citation

Latushkin, Yuri; Tomilov, Yuri. Fredholm properties of evolution semigroups. Illinois J. Math. 48 (2004), no. 3, 999--1020. doi:10.1215/ijm/1258131066. https://projecteuclid.org/euclid.ijm/1258131066


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Home > Mean Square > Root Mean Square Error Theory

# Root Mean Square Error Theory

Flow and Advective Transport (2nd ed.). Error (short for root mean square error) := (Two more: the variance a country selection list be the country's local name? What is the meaning of these measures, and(1985). "2.4.2 Certain Standard Loss Functions".The RMSD represents the sample standard deviation ofto times the SD of y.

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Of course that's not Morris H. (1980). Thus exponentiating it won't give you RMSE, it'll give you \$e^\sqrt{use the r.m.s. Scott Armstrong & Fred Collopy (1992). "Error Measures Mean Absolute Error 2015. ^ J.assess the accuracy of spatial analysis and remote sensing.

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## Energy Companies Ready to Deal with Climate Change

##### 26 November 2006

Washington Post. Top executives at many of the US’ largest energy companies have accepted the scientific consensus about climate change and see federal regulation to cut greenhouse gas emissions as inevitable.

The companies have been hiring new lobbyists who they hope can help fashion a national approach that would avert a patchwork of state plans now in the works. They are also working to change some company practices in anticipation of the regulation.

...It’s Shell’s belief that we have to deal with greenhouse gases. From a Shell point of view, the debate is over. When 90-plus percent of the world’s leading figures believe that greenhouse gases have impacted the climate of the Earth, who is Shell to say to say, let’s debate the science?

...We can’t have 50 state policies on greenhouse gas emissions. We believe, Shell believes, we need a national approach to greenhouse gas management and how that would work across our industries, not only the gas and oil industry.

John Hofmeister, president of Shell Oil Co.

Hofmeister and other top energy company leaders back a proposal that would cap greenhouse gas emissions and allow firms to trade their quotas. Paul M. Anderson, Duke Energy’s chairman and a member of the president’s Council of Advisors on Science and Technology, favors a tax on emissions of carbon dioxide.

Exxon Mobil Corp., the highest-profile corporate skeptic about global warming, said in September that it was considering ending its funding of a think tank that has sought to cast doubts on climate change. And on Nov. 2, the company announced that it will contribute more than $1.25 million to a European Union study on how to store carbon dioxide in natural gas fields in the Norwegian North Sea, Algeria and Germany. “If we had our druthers, we’d already have carbon legislation passed,” said John L. Stowell, Duke Energy’s vice president for environmental policy. “Our viewpoint is that it’s going to happen. There’s scientific evidence of climate change. We’d like to know what legislation will be put together so that, when we figure out how to increase our load, we know exactly what to expect.” Next week, the Supreme Court will hear arguments on whether the federal government is obligated to regulate carbon dioxide as a pollutant; its decision could force the government to come up with guidelines. (A hat-tip to Rafael!) ### Comments Hydrogen is comes from fossil fuel. Which means it has to be extracted, processed and delivered at 3 times the cost of gas since it takes 3 times more fossil fuel to get the equivalent hydrogen from it. This means 3 times more CO2 from processing fossil fuel. We need to concentrate on solar energy which is as cheap as natural gas for electricity generation. See nanosolar. Also we have to put emphasis on electric cars such as the 400 BHP pick up, 200 mile range, 10 minute fillup developed by phoenix motors. http://www.popularmechanics.com/automotive/new_cars/4201003.html Hydrogen is comes from fossil fuel. No, it can come from fossil fuel. It doesn't need to. Which means it has to be extracted, processed and delivered at 3 times the cost of gas since it takes 3 times more fossil fuel to get the equivalent hydrogen from it. This means 3 times more CO2 from processing fossil fuel. Specious reasoning unsupported by facts. Jay Lemieux needs a chemistry lesson me thinks. Fossil fuels are a variety of Hydro-carbons. These are usually large molecules made of H and C bond together. It is true that you can extract the H and have a bunch of leftover C. That does not mean breaking up hydro-carbons is the only way to create H. Hydrogen is only a form of energy storage as a previous comment mentioned. Yes I know EXACTLY how it works. The oil companies have always done a very siple thing. They make the fuel and then tack on a profit. Its simple and it works even when fuel prices jump around like crazy. But its realy designed for lower fuel cost rates then currently exist. But they sure arnt gona turn away the profits now as theyare all investing in hundreds of pojects most of wich wil fail. what is 100b a year profits industry wide when the next platform they might have to buld may cost 25 billion each? Or they may have to sink 40-50 billion into coal to liquids or god forbid they may have to BUILD an entire nuke imdustry just to provide the fuel they sell? They are quite litteraly staring trillions in future investments right in the face and are amn glad they have a buffer for now. They sure as hell wont have it for long. But they sure arnt gona turn away the profits now as theyare all investing in hundreds of pojects most of wich wil fail. Reinvestment isn't tracking increased profitability. They are quite litteraly staring trillions in future investments right in the face and are amn glad they have a buffer for now. So the$400 million given to the departing CEO of XOM was a "buffer"?

"TXU Corp. is planning to spend $10 billion to build 11 new coal-fired power plants, which would more than double the company's carbon dioxide emissions, from 55 million tons to 133 million tons a year. That increase in emissions is more than the total carbon dioxide pollution emitted in all of Maryland or by 10 million Cadillac Escalade sport-utility vehicles." Let me see if I understand this. I am supposed to pay extra$5K for a green car, such as Toyota Prius. Then I am supposed to spend say $30Kish for solar panels and wind turbines to generate green electricity. Altogether I am supposed to spend close to$50K over the next couple of decades to reduce my own CO2 emissions.

Meanwhile these fellas are going to spend \$10B to generate vast amounts of CO2 and supply electricity to the factories that make SUVs that will generate yet more CO2.

Is anyone still wondering why green doesn't sell?!?!? Really what is the point? Few measly watts from very expensive solar panels have no chance against MWs from these massive coal plants. We'll get global warming anyhow. This is like Borg, "Resistance is futile"

Geeez

I think that the problems with the actions of the energy companys are complex and date back to the time when certainly here in europe most energy production was in the state sector. Here in italy we have the probably the highest energy costs in the whole of europe , for example the cost of heating our house last year in the coldest four months of the year was around 500 us dollers per month, yet there is no actions taken by the state to encourage saving , ie correct insulation etc.
this year I have embarked on a series of works on the house to improve its performance, changed the roof and fitted 8 cm of natural cork insulation, fitted solar panels to heat the domestic hot water , and recently changed the gas heating system over to a condenser type of boiler, last year I changed all the windows over to double glazed units . This all costs money but should deacrease our carbon footprint considerably in the future years. However I seem to be in the minority certainly in ths part of the world, most houses having little or no insulation, and indeed new houses being built with only 2 cms of expanded polysterene in the roof!
In Italy we survive in winter on imported gas from russia ,and indeed last year it nearly ran out over a long cold period , causing near panic on the part of the govenment. Yet still there are no incentives to change policy, and to encourage people to save , low energy bulbs are kept under the counter in the local electrical shops etc, not on display.
I think the problem lies with consumpion , in our capitalist societys this is deemed to be good ,providing income for the share holders.
Ask anyone here what are the major shares held in their own pension portfollios and the answer will almost certainly be energy and oil company stocks
the reason is obvious these stocks provide strong and sustained growth over a long period, large companys also invest their employees funds into these stocks,
and at the top of the scale so do govenments . Any reduction in consumption is a probable headache for these adminstrations , pension plans not performing, pushing more realiance onto the state to provide for people in their old age, in many european countries the state pension systems are very close to collapse anyway.
Governments need to be seen to be to be doing something about the dreadful state we have got ourself into , yet when the shouting and chest beating is all over they just slip back into their old ways, for example witness Mr blairs performance a month or so ago after the realease of the stern report into the economic damage that climate will inflict on the world, since then nothing has been heard or done, and now they want to spend some 32 billion pounds on a new nuclear deterant for the UK, if this money was spent over the next twenty years investing in new tech to reduce the UKs carbon footprint that would be progress.
So I am afraid we are stuck with it . there is a saying " you are what you eat" and this is true with governance , " we are who we elect" until this sort of short term , and purly economic thinking can be removed from governmental levels we are finished, and to think that at one time we thought that nuclear weapons was going to be the end of us all!

There is money to be made in energy and that is the driving force. Coal is cheap and as a source fuel allows power plant owners to make money hand over fist. That is why W and Company have 86ed all the coal plant upgrades that Clinton and Gore worked on.

Of course reinvestment isnt trackingl. For one thing alot oinf nevestors are nervous and simply want MONEY nd alot of energy companies are nervous and are building lquid assets to buy into whatever winds up working.

As for the big arse returment... very likely the contrsct stated a certain percent of profits or some such and as such hes one damn happy rich dude.

As an advocate for biodiesel,vege oil / compression engines / gassified systems since 1987, I have been involved in this issue by understanding the mentality of the user. Whether the desire for production is supply driven, or demand driven, is an issue for the capitalist with deep pockets - to fill - although more spontaneous results occur from the shallow pockets that change pop out of with the faith for a better tomorrow. Self sustainability, and de-centralized energy production model Autonomy, and Anarchy . But I'm not so sure that the result is so unBEARable or full of BULLshit. somewhere in the middle the grass is greener for both to Co-exist. When modelling energy pyramids along side pyramids of Biotic consumers of energy we can see both opportunity and valiant efforts with the least of which the failures, but knowing one's position has made all the difference (some famous poet). Just wantewd to enter into a dead conversation.

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text:
# 10,000

Previous  ... Next

## Number

$10 \, 000$ (ten thousand) is:

$2^4 \times 5^4$

The $4$th power of $10$ after $(1)$, $10$, $100$, $1000$:
$10 \, 000 = 10^4$

The $10$th fourth power after $1$, $16$, $81$, $256$, $625$, $1296$, $2401$, $4096$, $6561$:
$10 \, 000 = 10 \times 10 \times 10 \times 10$

The $37$th square number after $1$, $4$, $36$, $121$, $144$, $256$, $\ldots$, $5184$, $5776$, $6084$, $6400$, $7056$, $7744$, $8100$, $9216$, $9604$ to be the divisor sum value of some (strictly) positive integer:
$10 \, 000 = \map {\sigma_1} {8743} = \map {\sigma_1} {9481}$

The $100$th square number after $1$, $4$, $9$, $16$, $25$, $36$, $\ldots$, $9025$, $9216$, $9409$, $9604$, $9801$:
$10 \, 000 = 100 \times 100$


text: Sheldon Katz has computed this in general in On the finiteness of rational curves on quintic threefolds. See Appendix B on pp. 158-159. He gives a list of the possibilities based on the equations. It is rather simple to check the condition. According to my computation the answer in the above case is $\mathcal O_C(-1)\oplus O_{\mathbb P^1}(-1)\oplus \mathcal O_C(-1)$O_{\mathbb P^1}(-1)$, but it is easy to produce explicit examples with other normal bundles. 2 added 78 characters in body; added 103 characters in body Sheldon Katz has computed this in general in On the finiteness of rational curves on quintic threefolds. See Appendix B on pp. 158-159. You need to do He gives a little work, but given explicit list of the possibilities based on the equations, it's a tedious. It is rather simple to check the condition. According to my computation the answer in the above case is$\mathcal O_C(-1)\oplus \mathcal O_C(-1)\$, but it is easy computationto produce explicit examples with other normal bundles.