OpenLLM-France/wikimedia huggingface dataset:
# Numerical Analysis/Newton form exercise

1. Using the Newton form, find the interpolating polynomial passing through the points $(x_{0},y_{0})=(9,9)$, $(x_{1},y_{1})=(5,7)$, $(x_{2},y_{2})=(3,7)$, and $(x_{3},y_{3})=(7,1)$.
Solution:
To find the four coefficients, we can use the divided differences.
    ${\begin{matrix}x_{i}&y_{i}&[y_{0},y_{1}]&[y_{0},y_{1},y_{2}]&[y_{0},y_{1},y_{2},y_{3}]\\9&9&&\\&&{7-9 \over 5-9}={1 \over 2}&\\5&7&&{{0-1 \over 2} \over {3-9}}={1 \over 12}\\&&{7-7 \over 3-5}={0}&&{{{-3 \over 4}-{1 \over 12}} \over {7-9}}={5 \over 12}\\3&7&&{{-3 \over 2}-0 \over {7-5}}={-3 \over 4}\\&&{1-7 \over 7-3}={-3 \over 2}&\\7&1&&\\\end{matrix}}$
Thus, the four coefficients are $x_{0}$ and the upper diagonal of the calculated divided differences.
Using the Newton Forward Divided Difference formula, we find the polynomial is
    $p_{3}(x)=[y_{0}]+[y_{0},y_{1}](x-x_{0})+[y_{0},y_{1},y_{2}](x-x_{0})(x-x_{1})+[y_{0},y_{1},y_{2},y_{3}](x-x_{0})(x-x_{1})(x-x_{2}).$
    $p_{3}(x)=9+{\frac {1}{2}}(x-9)+{\frac {1}{12}}(x-9)(x-5)+{\frac {5}{12}}(x-9)(x-5)(x-3).$
    $p_{3}(x)={\frac {5}{12}}x^{3}-7x^{2}+{\frac {427}{12}}x-48$.
2. Find the interpolating polynomial passing through the four points given in the exercise above as well as through the point $(x_{4},y_{4})=(4,2)$
Solution:
To find the new coefficient, we can simply add the new point to the end and calculate $[y_{0},y_{1},y_{2},y_{3},y_{4}]$.
    ${\begin{matrix}x_{i}&y_{i}&[y_{0},y_{1}]&[y_{0},y_{1},y_{2}]&[y_{0},y_{1},y_{2},y_{3}]&[y_{0},y_{1},y_{2},y_{3},y_{4}]\\9&9&&\\&&{1 \over 2}&\\5&7&&{1 \over 12}\\&&{0}&&{5 \over 12}\\3&7&&{-3 \over 4}&&{{{-23 \over 12}-{5 \over 12}} \over {4-9}}={17 \over 15}\\&&{-3 \over 2}&&{{{7 \over 6}-{-3 \over 4}} \over {4-5}}={-23 \over 12}\\7&1&&{{-1 \over 3}-0 \over {-3 \over 2}}={7 \over 6}\\&&{2-1 \over 4-7}={-1 \over 3}&\\4&2&&\\\end{matrix}}$
Thus, we find the polynomial $p_{4}$ is
    $p_{4}(x)=p_{3}+[y_{0},y_{1},y_{2},y_{3},y_{4}](x-x_{0})(x-x_{1})(x-x_{2})(x-x_{3}).$
    $p_{3}(x)={\frac {5}{12}}x^{3}-7x^{2}+{\frac {427}{12}}x-48+{\frac {17}{15}}(x-9)(x-5)(x-3)(x-7).$
    $p_{3}(x)={\frac {7}{15}}x^{4}-{\frac {647}{60}}x^{3}+{\frac {1337}{15}}x^{2}-{\frac {18697}{60}}x+393$.
# Numerical Analysis/Newton form example

We'll find the interpolating polynomial passing through the points $(1,-6)=(x_{0},y_{0})$, $(2,2)=(x_{1},y_{1})$, $(4,12)=(x_{2},y_{2})$, using the Newton form of the interpolation polynomial.
The Newton form is given by the formula $p(x)=\sum _{j=0}^{k}a_{j}n_{j}(x)$, where $a_{j}=[y_{0},\ldots ,y_{j}]$ and $n_{j}(x)=\prod _{i=0}^{j-1}(x-x_{i})$, with $n_{0}(x)=1$. We start by finding each $n_{j}(x)$.
$n_{0}(x)=1$
$n_{1}(x)=x-1$
$n_{2}(x)=(x-1)(x-2)=x^{2}-3x+2$
Next, we find the necessary divided differences. First, $[y_{0}]=-6$, $[y_{1}]=2$, and $[y_{2}]=12$. For the next level, we have:
$[y_{0},y_{1}]={\frac {2+6}{2-1}}=8$
$[y_{1},y_{2}]={\frac {12-2}{4-2}}=5$
Finally, we can find:
$[y_{0},y_{1},y_{2}]={\frac {5-8}{4-1}}=-1$.
Now, we can find the coefficients $a_{j}$.
$a_{0}=[y_{0}]=-6$
$a_{1}=[y_{0},y_{1}]=8$
$a_{2}=[y_{0},y_{1},y_{2}]=-1$
Substituting and simplifying, we get our interpolating polynomial:
$p(x)=-6+8(x-1)-(x^{2}-3x+2)=-x^{2}+11x-16$.

## Adding a point

Now let's add the point $(3,-10)=(x_{3},y_{3})$ to our data set and find the new polynomial using the same method. Due to the formula for the Newton form, we only have to add the term $a_{3}n_{3}(x)$ to our previous interpolating polynomial.
First, we have
$n_{3}(x)=(x-1)(x-2)(x-4)=x^{3}-7x^{2}+14x-8$.
Now to find $a_{3}$ we calculate some more divided differences.
$[y_{3}]=-10$
$[y_{2},y_{3}]={\frac {-10-12}{3-4}}=22$
$[y_{1},y_{2},y_{3}]={\frac {22-5}{3-2}}=17$
$a_{3}=[y_{0},\ldots ,y_{3}]={\frac {17+1}{3-1}}=9$
So, our new interpolating polynomial is:
$p(x)=-x^{2}+11x-16+9(x^{3}-7x^{2}+14x-8)=9x^{3}-64x^{2}+137x-88$.
# Numerical Analysis/Neville's algorithm code

The basic outline of a Matlab program that evaluates an interpolating polynomial using Neville's Algorithm given inputs of a point at which to evaluate (x0), a matrix of the x terms of the ordered pairs (x), and a matrix of the y terms of the ordered pairs (y) is given below. The numbers in the parenthese at the end of the comments correspond to questions given below the code.

## Questions

(1) Is the degree of the polynomial equal to the length of x minus 1 or the length of x plus one?
Solution:
length of x minus 1
(2) How many rows and columns should matrix p have?
Solution:
n+1 rows and n+1 columns
(3) When should the loop end?
Solution:
when i is equal to the length of x, which is also when i is equal to n+1
(4) When should the loop end?
Solution:
when j is equal to n+1
(5) When should the loop end?
Solution:
when i is equal to n+1-j
(6) What term of the matrix is the output value?
Solution:
p = p(1,n+1)
# Numerical Analysis/Matrix norm exercises

Consider the general $3\times 3$ matrix $A={\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\\end{bmatrix}}$ and the specific example $B={\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\\\end{bmatrix}}$.
Find $\left\|A\right\|_{1}$
Solution:
given: $\left\|A\right\|_{1}=\max \limits _{1\leq j\leq n}\sum _{i=1}^{n}|a_{ij}|$
Solve for column 1: $\sum _{i=1}^{3}|a_{i1}|=|a_{11}|+|a_{21}|+|a_{31}|$
Solve for column 2: $\sum _{i=1}^{3}|a_{i2}|=|a_{12}|+|a_{22}|+|a_{32}|$
Solve for column 3: $\sum _{i=1}^{3}|a_{i3}|=|a_{13}|+|a_{23}|+|a_{33}|$
$\left\|A\right\|_{1}$ = the maximum absolute column sum of the matrix
Find $\left\|B\right\|_{1}$
Solution:
Solve for column 1: $\sum _{i=1}^{3}|b_{i1}|=|1|+|4|+|7|=12$
Solve for column 2: $\sum _{i=1}^{3}|b_{i2}|=|2|+|5|+|8|=15$
Solve for column 3: $\sum _{i=1}^{3}|b_{i3}|=|3|+|6|+|9|=18$
$\left\|B\right\|_{1}=\max[12,15,18]=18$
Find the infinite norm, $\left\|A\right\|_{\infty }$
Solution:
given: $\left\|A\right\|_{\infty }=\max \limits _{1\leq i\leq n}\sum _{j=1}^{n}|a_{ij}|$
Solve for row 1: $\sum _{j=1}^{3}|a_{1j}|=|a_{11}|+|a_{12}|+|a_{13}|$
Solve for row 2: $\sum _{j=1}^{3}|a_{2j}|=|a_{21}|+|a_{22}|+|a_{23}|$
Solve for row 3: $\sum _{j=1}^{3}|a_{3j}|=|a_{31}|+|a_{32}|+|a_{33}|$
$\left\|A\right\|_{\infty }$ = the maximum absolute row sum of the matrix
Find the infinite norm, $\left\|B\right\|_{\infty }$
Solution:
Solve for row 1: $\sum _{j=1}^{3}|b_{1j}|=|1|+|2|+|3|=6$
Solve for row 2: $\sum _{j=1}^{3}|b_{2j}|=|4|+|5|+|6|=15$
Solve for row 3: $\sum _{j=1}^{3}|b_{3j}|=|7|+|8|+|9|=24$
$\left\|B\right\|_{\infty }=\max[6,15,24]=24$
Find the Frobenius norm, $\left\|A\right\|_{F}$
Solution:
given: $\|A\|_{F}={\sqrt {\sum _{i=1}^{3}\sum _{j=1}^{3}|a_{ij}|^{2}}}$
$\|A\|_{F}={\sqrt {|a_{11}|^{2}+|a_{12}|^{2}+|a_{13}|^{2}+|a_{21}|^{2}+|a_{22}|^{2}+|a_{23}|^{2}+|a_{31}|^{2}+|a_{32}|^{2}+|a_{33}|^{2}}}$
Find the Frobenius norm, $\left\|B\right\|_{F}$
Solution:
$\|B\|_{F}={\sqrt {|1|^{2}+|2|^{2}+|3|^{2}+|4|^{2}+|5|^{2}+|6|^{2}+|7|^{2}+|8|^{2}+|9|^{2}}}={\sqrt {285}}\approx 16.9$
# Vancouver Flower Clock Project/Wall Lettuce 2

Back to Vancouver Flower Clock Project
This is a Vancouver Flower Clock Project Journal Page for Species Mycelis muralis, Common Names: Wall Lettuce.

## General Info

* Bloom Season
* Habitat

### Identification
#### Seeds
## Cultivation
## Notebook

Share a poem, story, memory or anecdote about this plant.

## Log a Bloom Observation

To log a bloom observation, sign your name by typing "~~~~". Please indicate if the time and date of observation are different from the time of signing. Be as detailed as possible in your description of the observation. Were the blooms open or closed? How long did you watch? Where were they situated? What were the weather conditions? Were there any surrounding plants, insects? What was your response?. Take a photograph, video or scan an image and upload it using Special:Upload.
# Variance

This statistics resource is a "stub". Learn how you can help Wikiversity to develop it.
In probability theory and statistics, the variance of a random variable, probability distribution, or sample is one measure of statistical dispersion, averaging the squared distance of its possible values from the expected value (mean). Whereas the mean is a way to describe the location of a distribution, the variance is a way to capture its scale or degree of being spread out. The unit of variance is the square of the unit of the original variable. The positive square root of the variance, called the standard deviation, has the same units as the original variable and can be easier to interpret for this reason.
# Variations of the Mandelbrot set

This short article by Dan Polansky explores variations of the Mandelbrot set. It asks: what happens if we change the exponent of the iterative function (away from 2) or if we let the orbits start from a fixed value different from 0?
The results are interesting, some of them bizarre. That a simple formula iterated could result in the curves of the Mandelbrot set is curious enough; the kind of deformations that result from changing these parameters perhaps even more so.
Let us recall: the function being iterated in the standard Mandelbrot set is f(x) = x ** 2 + c, where the iteration process starts at zero. We can instead take e.g. f(x) = x ** 2.1 + c, starting at 0.5 + 0.5i instead of 0.
Exercise: install PyPy and other requirements, copy the Python code below to a local file, and try your own hand in discovering interesting plots. Try different parameter values or even different iteration formulas.
On a subjective note, some people to whom I have shown the images asked whether it was something like Rorschach test. Indeed, one could perhaps produce a series of neo-Rorschach test using this technique.

## Gallery

A gallery follows.
Normal: exponent 2, start from 0:
Exponent 0.5:
Exponent 1.0:
Exponent 1.5:
Exponent 1.9:
Exponent 2.1:
Exponent 3:
Exponent 4:
Exponent 2, starting from 0.5:
Exponent 2, starting from 0.5i:
Exponent 2, starting from 0.5 + 0.5i:
Exponent 2.1, starting from 0.5 + 0.5i:
Exponent 2, starting from 0.3 + 0.3i:
Exponent 2, starting from 0.7 + 0.7i:
Exponent 2, starting from 0.9 + 0.9i:
Part of the above images are covered in the animation File:Mandelbrot Set Animation 1280x720.gif.

## Plotting code

The images above were plotted using the following Python code, run using PyPy JIT compiler rather than the standard CPython. For this code, the speed difference from using PyPy is huge; the speed with PyPy in plotting a 2000 x 2000 pixel image is pleasant, unlike with CPython. We make use of pure-Python PNG-creation library purepng. To be fair to CPython, the plotting could turn rather tolerable if it would use vectorization using numpy. The code is very straightforward, like code one might find written in Basic in a popular computing magazine decades ago.

## Plotting prerequisites

One option for installing prerequisites is this:
1) Install PyPy
2) Install pip for PyPy: pypy -m ensurepip
3) Use pip to install purepng: pypy -m pip install purepng
Links:
* python - Install pip on pypy, stackoverflow.com
* purepng, pypi.org
# Vector/Linear form/Zero test/Fact

Let $V$ denote a $K$-vector space and let $v\in V$ be a vector different from $0$.
linear form $f\colon V\rightarrow K$ such that
$f(v)\neq 0$
# Vancouver Flower Clock Project/Wall Lettuce

Back to Vancouver Flower Clock Project
This is a Vancouver Flower Clock Project Journal Page for Species Mycelis muralis, Common Names: Wall Lettuce.

## General Info

* Bloom Season
* Habitat

### Identification
#### Stems

Widely branched

#### Seeds
## Cultivation
## Notebook

Share a poem, story, memory or anecdote about this plant.

## Log a Bloom Observation

To log a bloom observation, sign your name by typing "~~~~". Please indicate if the time and date of observation are different from the time of signing. Be as detailed as possible in your description of the observation. Were the blooms open or closed? How long did you watch? Where were they situated? What were the weather conditions? Were there any surrounding plants, insects? What was your response?. Take a photograph, video or scan an image and upload it using Special:Upload.

#### 4:00pm-5:00pm

* June 21 2009 Petals closed
# Variable-Frequency Data Encoding and Transmission
## Abstract

Data is generally transmitted as a binary stream. This learning and research project aims to explore a system of encoding and transmitting data streams by using multi-frequency spectrum divisions (stations) as the basis for archetypal clusters of binary bits. An example mapping of the visual spectrum would be as follows:
* White -> 1
* Black -> 0
* Red -> 101
* Green -> 010
* Blue -> 111
Experiments and statistical analysis will offer insight as to the transmission speeds of various combinations of discreet frequency divisions and binary clusterings.

## Basic Algorithm

Before transmission, a binary stream can be passed through a clustering module. The clustering module will look for longer patterns of binary digits. If a pattern is found, e.g. '11111', the clustering module will replace that sequence with a single pulse of a specified frequency.
On the receiving end, the multi-frequency stream will be converted to binary data by replacing the agreed upon frequencies with binary patterns.
This process can speed up multi-hop transmission rates as well as reduce large scale energy consumption across regional or global networks.

### Details

* take the first n letters of a string ( e.g. '0110100101001010')
* see of the first n letters match any of an number of patterns (e.g. '111', '110', '101')
* If so, return a single letter (e.g. 'R')
* If not, return the 0th letter of the string (e.g. '0')
* shorten the string by the length of the returned value.
This will repeat on the subsequent letters until the string has been shortened.
E.g. '111' = 'r'; '110' = 'g'; '101' = 'b'
string = "0 111 0 101 101 0 110 111 0 111 101 00 110"
output = '0 R 0 BB 0 GR 0 RB 00 G'

## Transmission Medium

The electromagnetic oscillations can be transmitted through a waveguide in order that they reach their destination while preserving signal strength. Additional methods of transmission can be LASER or optical transmissions, auditory oscillations, or various ranges of electromagnetic spectrum such as micro-waves.

## Modulation

Types of modulation include Quadrature amplitude modulation (QAM) and/or Multiple frequency-shift keying (MFSK).

## Possible Uses

* Parallel procesing, communicating data between multiple processors
* Fiber optic networking
* Wireless communications
# Vancouver Flower Clock Project/Wall Rocket 2

Back to Vancouver Flower Clock Project
This is a Vancouver Flower Clock Project Journal Page for Species Diplotaxis Tenuifolia, Common Names Perennial Wall Rocket.

## General Info

* Bloom Season
* Habitat

### Identification
#### Seeds
## Cultivation
## Notebook

Share a poem, story, memory or anecdote about this plant.

## Log a Bloom Observation

To log a bloom observation, sign your name by typing "~~~~". Please indicate if the time and date of observation are different from the time of signing. Be as detailed as possible in your description of the observation. Were the blooms open or closed? How long did you watch? Where were they situated? What were the weather conditions? Were there any surrounding plants, insects? What was your response?. Take a photograph, video or scan an image and upload it using Special:Upload.
# Vancouver Flower Clock Project/Wall Rocket

Back to Vancouver Flower Clock Project
This is a Vancouver Flower Clock Project Journal Page for Species Diplotaxis tenuifolia, Common Names: Wall Rocket.

## General Info

* Bloom Season
* Habitat

### Identification
#### Seeds
## Cultivation
## Notebook

Share a poem, story, memory or anecdote about this plant. Include sketches, photographs and videos as well.

## Log a Bloom Observation

To log a bloom observation, sign your name by typing "~~~~". Please indicate if the time and date of observation are different from the time of signing. Be as detailed as possible in your description of the observation. Were the blooms open or closed? How long did you watch? Where were they situated? What were the weather conditions? Were there any surrounding plants, insects? What was your response?. Take a photograph, video or scan an image and upload it using Special:Upload.
# PlanetPhysics/Riccati Equation 2

The nonlinear differential equation
   ${\begin{matrix}{\frac {dy}{dx}}=f(x)+g(x)y+h(x)y^{2}\end{matrix}}$
is called the Riccati equation .\, If\, $h(x)\equiv 0$,\, it becomes a linear differential equation; if\, $f(x)\equiv 0$,\, then it becomes a Bernoulli equation.\, There is no general method for integrating explicitely the equation (1), but via the substitution $y\,:=\,-{\frac {w'(x)}{h(x)w(x)}}$ one can convert it to a second order homogeneous linear differential equation with non-constant coefficients.\\
If one can find a particular solution \,$y_{0}(x)$,\, then one can easily verify that the substitution
   ${\begin{matrix}y\,:=\,y_{0}(x)+{\frac {1}{w(x)}}\end{matrix}}$
converts (1) to
   ${\begin{matrix}{\frac {dw}{dx}}+[g(x)\!+\!2h(x)y_{0}(x)]\,w+h(x)=0,\end{matrix}}$
which is a linear differential equation of first order with respect to the function \,$w=w(x)$.\\
Example. \, The Riccati equation
   ${\begin{matrix}{\frac {dy}{x}}=3+3x^{2}y-xy^{2}\end{matrix}}$
has the particular solution\, $y:=3x$.\, Solve the equation.
We substitute\, $y:=3x+{\frac {1}{w(x)}}$\, to (4), getting ${\frac {dw}{dx}}-3x^{2}w-x=0.$ For solving this first order equation we can put\, $w=uv$,\, $w'=uv'+u'v$,\, writing the equation as
   ${\begin{matrix}u\cdot (v'-3x^{3}v)+u'v=x,\end{matrix}}$
where we choose the value of the expression in parentheses equal to 0: ${\frac {dv}{dx}}-3x^{2}v=0$ After separation of variables and integrating, we obtain from here a solution\, $v=e^{x^{3}}$,\, which is set to the equation (5): ${\frac {du}{dx}}e^{x^{3}}=x$ Separating the variables yields $du={\frac {x}{e^{x^{3}}}}\,dx$ and integrating: $u=C+\int xe^{-x^{3}}\,dx.$ Thus we have $w=w(x)=uv=e^{x^{3}}\left[C+\int xe^{-x^{3}}\,dx\right],$ whence the general solution of the Riccati equation (4) is Failed to parse (syntax error): {\displaystyle y \,:=\, 3x+\frac{e^{-x^3}}{C+\int xe^{-x^3}\,dx}.\\}
It can be proved that if one knows three different solutions of Riccati equation (1), then any other solution may be expressed as a rational function of the three known solutions.
# PlanetPhysics/Rigged Hilbert Space

\newcommand{\sqdiagram}[9]{Failed to parse (unknown function "\diagram"): {\displaystyle \diagram #1 \rto^{#2} \dto_{#4}& \eqno{\mbox{#9}}} }
A rigged Hilbert space is a pair $(\mathbb {H} ,\phi )$ with $\mathbb {H}$ a Hilbert space and $\phi$ is a dense subspace with a topological vector space structure for which the inclusion map {\mathbf $i$} is continuous. Between $\mathbb {H}$ and its dual space $\mathbb {H} ^{*}$ there is defined the adjoint map $i^{*}:\mathbb {H} ^{*}\to \phi ^{*}$ of the continuous inclusion map $i$. The duality pairing between $\phi$ and $\phi ^{*}$ also needs to be compatible with the inner product on $\mathbb {H}$: $\langle u,v\rangle _{\phi \times \phi ^{*}}=(u,v)_{\mathbb {H} }$ whenever $u\in \phi \subset \mathbb {H}$ and $v\in \mathbb {H} =\mathbb {H} ^{*}\subset \phi ^{*}$.
# PlanetPhysics/Rigid Body

It is often stated that a rigid body is any body in which the distances between all particles of the body (selected in pairs) remain constant. This definition might seem completely satisfactory to a theoretician who does not concern himself much with the meaning of distance. The practical person, however, has difficulty even imagining that he could determine the magnitude of a distance or its constancy without the employment of some device such as a yardstick which he was confident would be rigid at least while he was using it. The difficulty can not be resolved by substituting for a simple yardstick, some more complicated precision instrument such as an interferometer. One would have little confidence in the reading of an interferometer if he were not confident that its component parts such as screws and mirrors were behaving satisfactorily as rigid bodies. Moreover, an expert on the use of an interferometer would emphasize that the component parts of his instrument were actually elastic rather than rigid, and that their approximation to rigidity would be satisfactory only provided great care were taken to avoid all unnecessary stresses. All this ignores thermal agitation of molecules and atoms and motions of electrons, of course.
From this point of view, it appears that the concept of a rigid body is an idealization not only in the sense that nobody seems likely to find or construct one, but also in the sense that he could not be sure that the body was rigid even if he did acquire one. The difficulty in even defining satisfactorily the meaning of a rigid body appears to be associated intimately wit the difficulty in defining distance. While it is worth giving some serious attention to our definitions or specifications of elementary concepts, the procedure should not be followed to the point of discouragement. We do not give up scientific work as hopeless because our definitions of the fundamentals are somewhat hazy and we can not measure anything exactly. We are always approximating, it is true, but often our approximations are so good as to astonish others if not ourselves.
The concept of a rigid body is not merely fanciful, but extremely useful, and the extensive theoretical work which has been done on the statics and dynamics of rigid bodies has a multitude of practical applications. Practically, any body which satisfies the condition of rigidity to the extent that we are unable to detect any change in the distance between any pair of the particles constituting the body may be regarded as rigid. For some purposes we may regard a body as rigid even when departures from rigidity can be detected, provided the displacements associated with such observable departures from rigidity are sufficiently small in comparison with certain other displacements or distances involved in a particular problem.