text:
// can't yet format YamlFrontmatter (["title: Contour plots"; "category: Distribution Charts"; "categoryindex: 5"; "index: 4"], Some { StartLine = 2 StartColumn = 0 EndLine = 6 EndColumn = 8 }) to pynb markdown

In [ ]:
#r "nuget: Plotly.NET, 2.0.0-preview.16"
#r "nuget: Plotly.NET.Interactive, 2.0.0-preview.16"

# Contour plots¶

Summary: This example shows how to create contour plot in F#.

let's first create some data for the purpose of creating example charts:

In [3]:
open System
open Plotly.NET

// Generate linearly spaced vector
let linspace (min,max,n) =
if n <= 2 then failwithf "n needs to be larger then 2"
let bw = float (max - min) / (float n - 1.)
[|min ..bw ..max|]

// Create example data
let size = 100
let x = linspace(-2. * Math.PI, 2. * Math.PI, size)
let y = linspace(-2. * Math.PI, 2. * Math.PI, size)

let f x y = - (5. * x / (x**2. + y**2. + 1.) )

let z =
Array.init size (fun i ->
Array.init size (fun j ->
f x.[j] y.[i]
)
)

A contour plot is a graphical technique for representing a 3-dimensional surface by plotting constant z slices, called contours, on a 2-dimensional format. That is, given a value for z, lines are drawn for connecting the (x,y) coordinates where that z value occurs.

The contour plot is an alternative to a 3-D surface plot.

In [4]:
let contour1 =
z
|> Chart.Contour
|> Chart.withSize(600.,600.)

In [ ]:
contour1

Out[ ]:

## Smooth Contour Coloring¶

to apply heatmap gradient coloring between each contour level, set the ContourColoring to heatmap:

In [6]:
let contour2 =
Chart.Contour(
z,
ContoursColoring = StyleParam.ContourColoring.Heatmap
)

In [ ]:
contour2

Out[ ]:

## Contour Line Labels¶

Use ContourLabelFont to set a contour label font, and display the labels with ShowContourLabels:

In [8]:
let contour3 =
Chart.Contour(
z,
ContoursColoring = StyleParam.ContourColoring.Heatmap,
ShowContourLabels = true,
ContourLabelFont = Font.init(Size = 12., Color = Color.fromKeyword White)
)

In [ ]:
contour3

Out[ ]:


text:
# Is this a comprehensive list of ruby hook methods?

I have been watching Dave Thomas' screencasts "The Ruby Object Model and Metaprogramming". He mentions 7 hook methods provided by Module and Class.

inherited
append_features
included
extend_object
extended
initialize_copy
const_missing

Is anyone aware of any further ruby hook methods?

-

I've now watched more of the screencasts' and Dave's list actually seems to be far more comprehensive than I first thought.

Method-related hooks

method_missing
method_removed
singleton_method_removed
method_undefined
singleton_method_undefined

Class & Module Hooks

inherited
append_features
included
extend_object
extended
initialize_copy
const_missing

Marshalling Hooks

marshal_dump

Coercion Hooks

coerce
induced_from
to_xxx

God willing, one day I'll read the ruby source and find the comprehensive list as I notice that Jörg has mentioned further methods which Dave has ommitted and I'm still interested to know them all.

-
method_missing
const_missing
respond_to_missing?
Kernel.trace_var
at_exit
BEGIN
END

-
örg, Thank you for your response. How would you hook BEGIN and END and have you found any use for them? –  dsjbirch Feb 28 '11 at 7:34
@dsjbirch: BEGIN, END and at_exit take blocks like so: BEGIN { puts "Hello from #{__FILE__}" }. I'm not sure that they are that useful. at_exit is the only one I have seen used, it's used by test/unit, RSpec, MiniTest et al. to automatically run the tests. –  Jörg W Mittag Feb 28 '11 at 10:27

Ruby Ahead has a list of callbacks:

http://www.nach-vorne.de/2007/3/18/list-of-callback-methods/

-

Check Ruby callbacks.

-
I'm going to put a list of the hooks into the question that I took from your post. Thank you. –  dsjbirch Feb 28 '11 at 7:38

format: plain


text:
If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Limits at infinity of quotients with square roots (even power)

AP.CALC:
LIM‑2 (EU)
,
LIM‑2.D (LO)
,
LIM‑2.D.3 (EK)
,
LIM‑2.D.4 (EK)
,
LIM‑2.D.5 (EK)

## Video transcript

- [Voiceover] Let's see if we can find the limit as x approaches negative infinity of the square root of 4x to the fourth minus x over 2x squared plus three. And like always, pause this video and see if you can figure it out. Well, whenever we're trying to find limits at either positive or negative infinity of rational expressions like this, it's useful to look at what is the highest degree term in the numerator or in the denominator, or, actually in the numerator and the denominator, and then divide the numerator and the denominator by that highest degree, by x to that degree. Because if we do that, then we're going to end up with some constants and some other things that will go to zero as we approach positive or negative infinity, and we should be able to find this limit. So what I'm talking about, let's divide the numerator by one over x squared and let's divide the denominator by one over x squared. Now, you might be saying, "Wait, wait, "I see an x to the fourth here. "That's a higher degree." But remember, it's under the radical here. So if you wanna look at it at a very high level, you're saying, okay, well x to the fourth, but it's under, you're gonna take the square root of this entire expression, so you can really view this as a second degree term. So the highest degree is really second degree, so let's divide the numerator and the denominator by x squared. And if we do that, dividing, so this is going to be the same thing as, so this is going to be the limit, the limit as x approaches negative infinity of, so let me just do a little bit of a side here. So if I have, if I have one over x squared, all right, let me write it. Let me just, one over x squared times the square root of 4x to the fourth minus x, like we have in the numerator here. This is equal to, this is the same thing as one over the square root of x to the fourth times the square root of 4x to the fourth minus x. And so this is equal to the square root of 4x to the fourth minus x over x to the fourth, which is equal to the square root of, and all I did is I brought the radical in here. You could view this as the square root of all this divided by the square root of this, which is equal to, just using our exponent rules, the square root of 4x to the fourth minus x over x to the fourth. And then this is the same thing as four minus, x over x to the fourth is one over x to the third. So this numerator is going to be, the numerator's going to be the square root of four minus one, x to the third power. And then the denominator is going to be equal to, well, you divide 2x squared by x squared. You're just going to be left with two. And then three divided by x squared is gonna be three over x squared. Now, let's think about the limit as we approach negative infinity. As we approach negative infinity, this is going to approach zero. One divided by things that are becoming more and more and more and more and more negative, their magnitude is getting larger, so this is going to approach zero. This over here is also going to be, this thing is also going to be approaching zero. We're dividing by larger and larger and larger values. And so what this is going to result in is the square root of four, the principal root of four, over two, which is the same thing as two over two, which is equal to one. And we are done.


text:
# A, B and C Are All Liquids. Liquid a Has a Comparatively Low Boiling Point. on Heating, Liquid a Vaporises Completely Without Leaving Behind Any Residue. Liquid a is Being Used - Science

Sum

A, B and C are all liquids. Liquid A has a comparatively low boiling point. On heating, liquid A vaporises completely without leaving behind any residue. Liquid A is being used increasingly as a fuel in motor vehicles either alone or by mixing with petrol. Liquid B has a very high boiling point. It also vaporises completely on heating, without leaving any residue. Liquid B is a conductor of electricity and used in making thermometers. Liquid C has a moderate boiling point. On heating, liquid C vaporises leaving behind a white solid D which is used in cooking vegetables. The condensation of vapours from C give a liquid E which turns anhydrous CuSO4 to blue.

(a) Which liquid could be an element ? Name this element.

(b) Which liquid could be a mixture ? Name this mixture.

(c) Which liquid could be a compound ? Name this compound.

(d) What could the solid D be ?

(e) What do you think is liquid E ?

#### Solution

1. Liquid B has a very high boiling point. It also vaporises on heating. It is a conductor of electricity and is used in making thermometers. So, B could be an element. Further, liquid B is mercury as mercury is a liquid metal, which expands on heating and is used extensively in thermometers.
2. Liquid C has a moderate boiling point. It could be a mixture. It is a salt solution because when C is heated, it vaporises and leaves behind a white solid (salt), which is used in the process of cooking.
3. Liquid A is used in motor vehicles as fuel, either alone or after mixing with petrol. It could be a compound because on heating, it does not leave any residue. It is alcohol as alcohol is being used in motor vehicles nowadays.
4. Solid D is sodium chloride. It is also known as common salt and is added in food.
5. Liquid E is water as, on heating, liquid C, which is a salt solution, leaves water vapours. On condensing these vapours, we obtain pure water. When water is mixed with anhydrous copper sulphate, we obtain a blue colour solution. This is because of the formation of hydrated copper sulphate solution.

$\ce{CuSO4 + 5H2O →\underset{\text{hydrated copper sulphate}}{CuSO4.5H2O}}$

Concept: Matter
Is there an error in this question or solution?

#### APPEARS IN

Lakhmir Singh Class 9 Chemistry - Science Part 2
Chapter 2 Is Matter Around Us Pure
Hots Questions | Q 75 | Page 60


text:
XmlPackage Exception: Exception=Last Trace Point=[]. An error occurred during a call to extension function 'LookupImage'. See InnerException for a complete description of the error.

Could not find a part of the path 'D:\Webstore8\Web\images\Product\icon\MID1'.

System.ArgumentException: Last Trace Point=[]. An error occurred during a call to extension function 'LookupImage'. See InnerException for a complete description of the error.

Could not find a part of the path 'D:\Webstore8\Web\images\Product\icon\MID1'.

at AspDotNetStorefrontCore.XmlPackage2.TransformString() at AspDotNetStorefrontCore.AppLogic.RunXmlPackage(XmlPackage2 p, Parser UseParser, Customer ThisCustomer, Int32 SkinID, Boolean ReplaceTokens, Boolean WriteExceptionMessage)


text:
If you like the ChemWiki, please "like" us on facebook, "share" us on Google+, or "tweet" about the project.

# EPR: Interpretation

Electron paramagnetic resonance spectroscopy (EPR), also called electron spin resonance (ESR), is a technique used to study chemical species with unpaired electrons.  EPR spectroscopy plays an important role in the understanding of organic and inorganic radicals, transition metal complexes, and some biomolecules. For theoretical background on EPR, please refer to EPR:Theory.

### Introduction

Like most spectroscopic techniques, EPR spectrometers measure the absorption of electromagnetic radiation. A simple absorption spectra will appear similar to the one on the top of Figure 1. However, a phase-sensitive detector is used in EPR spectrometers which converts the normal absorption signal to its first derivative. Then the absorption signal is presented as its first derivative in the spectrum, which is similar to the one on the bottom of Figure 1. Thus, the magnetic field is on the x-axis of EPR spectrum; dχ″/dB, the derivative of the imaginary part of the molecular magnetic susceptibility with respect to the external static magnetic field in arbitrary units is on the y-axis. In the EPR spectrum, where the spectrum passes through zero corresponds to the absorption peak of absorption spectrum. People can use this to determine the center of the signal. On the x-axis, sometimes people use the unit “gauss” (G), instead of tesla (T). One tesla is equal to 10000 gauss.

Figure 1. Comparison of absorption spectrum and EPR spectrum. http://en.wikipedia.org/wiki/File:EPR_lines.png

### Proportionality factor (g-factor)

As a result of the Zeeman Effect, the state energy difference of an electron with s=1/2 in magnetic field is

$\Delta E = gβB$ (1)

where β is the constant, Bohr magneton. Since the energy absorbed by the electron should be exactly the same with the state energy difference ΔE, ΔE=hv ( h is Planck’s constant), the equation (1) can be expressed as

$h \nu=gβB$ (2)

People can control the microwave frequency v and the magnetic field B. The other factor, g, is a constant of proportionality, whose value is the property of the electron in a certain environment. After plugging in the values of h and β in equation (2), g value can be given through equation (3):

$g = 71.4484v \text{(in GHz)/B (in mT)}$ (3)

A free electron in vacuum has a g value ge= 2.00232. For instance, at the magnetic field of 331.85 mT, a free electron absorbs the microwave with an X-band frequency of 9.300 GHz. However, when the electron is in a certain environment, for example, a transition metal-ion complex, the second magnetic field produced by the nuclei, ΔB, will also influence the electron. At this kind of circumstance, equation (2) becomes

$h\nu = gβ(B_e+ \Delta B)$ (4)

since we only know the spectrometer value of B, the equation is written as:

$h\nu = (g_e+ \Delta g)βB$ (5)

From the relationship shown above, we know that there are infinite pairs of v and B that fit this relationship. The magnetic field for resonance is not a unique “fingerprint” for the identification of a compound because spectra can be acquired at different microwave frequencies. Then what is the fingerprint of a molecule? It is Δg. This value contains the chemical information that lies in the interaction between the electron and the electronic structure of the molecule, one can simply take the value of g = ge+ Δg as a fingerprint of the molecule. For organic radicals, the g value is very close to ge with values ranging from 1.99-2.01. For example, the g value for •CH3 is 2.0026. For transition metal complexes, the g value varies a lot because of the spin-orbit coupling and zero-field splitting. Usually it ranges from 1.4-3.0, depending on the geometry of the complex. For instance, the g value of Cu(acac)2 is 2.13. To determine the g value, we use the center of the signal. By using equation (3), we can calculate the g factor of the absorption in the spectrum.  The value of g factor is not only related to the electronic environment, but also related to anisotropy. About this part, please refer to EPR:Theory, Parallel Mode EPR: Theory and ENDOR:Theory. An example from UC Davis is shown below[1] (Britt group, Published in J.A.C.S. ):

Figure 2. EPR spectra of some proteins (WT mitoNEET, H87C mitoNEET, and ferredoxin). Frequency is 30.89 GHz.1

### Hyperfine Interactions

Another very important factor in EPR is hyperfine interactions. Besides the applied magnetic field B0, the compound contains the unpaired electrons are sensitive to their local “micro” environment. Additional information can be obtained from the so-called hyperfine interaction. The nuclei of the atoms in a molecule or complex usually have their own fine magnetic moments. Such magnetic moments occurrence can produce a local magnetic field intense enough to affect the electron. Such interaction between the electron and the nuclei produced local magnetic field is called the hyperfine interaction. Then the energy level of the electron can be expressed as:

E = gmBB0MS + aMsmI                                                                                (6)

In which a is the hyperfine coupling constant, mI is the nuclear spin quantum number. Hyperfine interactions can be used to provide a wealth of information about the sample such as the number and identity of atoms in a molecule or compound, as well as their distance from the unpaired electron.

The rules for determining which nuclei will interact are the same as for NMR. For isotopes which have even atomic and even mass numbers, the ground state nuclear spin quantum number, I, is zero, and these isotopes have no EPR (or NMR) spectra.  For isotopes with odd atomic numbers and even mass numbers, the values of I are integers. For example the spin of 2H is 1. For isotopes with odd mass numbers, the values of I are fractions. For example the spin of 1H is 1/2 and the spin of 23Na is 7/2. Here are more examples from biological systems:

Table 1. Bio transition metal nuclear spins and EPR hyperfine patterns[3]

Table 2. Bio ligand atom nuclear spins and their EPR hyperfine patterns[3]

The number of lines from the hyperfine interaction can be determined by the formula: 2NI + 1. N is the number of equivalent nuclei and I is the spin. For example, an unpaired electron on a V4+ experiences I=7/2 from the vanadium nucleus. We can see 8 lines from the EPR spectrum. When coupling to a single nucleus, each line has the same intensity. When coupling to more than one nucleus, the relative intensity of each line is determined by the number of interacting nuclei. For the most common I=1/2 nuclei, the intensity of each line follows Pascal's triangle, which is shown below:

Figure 3. Pascal's triangle

For example, for •CH3, the radical’s signal is split to 2NI+1= 2*3*1/2+1=4 lines, the ratio of each line’s intensity is 1:3:3:1. The spectrum looks like this:

Figure 4. Simulated EPR spectrum of the •CH3 radical. http://en.wikipedia.org/wiki/File:EPR_methyl.png

If an electron couples to several sets of nuclei, first we apply the coupling rule to the nearest nuclei, then we split each of those lines by the coupling them to the next nearest nuclei, and so on. For the methoxymethyl radical, H2C(OCH3), there are (2*2*1/2+1)*(2*3*1/2+1)=12 lines in the spectrum, the spectrum looks like this:

Figure 5. Simulated EPR spectrum of the H2C(OCH3) radical. http://en.wikipedia.org/wiki/File:EP...hoxymethyl.png

For I=1, the relative intensities follow this triangle:

Figure 5. Relative Intensities of each line when I=1

### Simulations

The EPR spectra have very different line shapes and characteristics depending on many factors, such as the interactions in the spin Hamiltonian, physical phase of samples, dynamic properties of molecules. To gain the information on structure and dynamics from experimental data, spectral simulations are heavily relied. People use simulation to study the dependencies of spectral features on the magnetic parameters, to predict the information we may get from experiments, or to extract accurate parameter from experimental spectra.

Many methods were developed to simulate the EPR spectra. Dr. Stefan Stoll, who is now working at UC Davis, wrote EasySpin, a computational EPR package for spectral simulation. EasySpin is based on Matlab, which is a numerical computing environment and fourth-generation programming language. EasySpin is a powerful tool in EPR spectral simulation. It can simulate spectra under many different conditions. Some functions are shown below:

Spectral simulations and fitting functions:

• garlic: cw EPR (isotropic and fast motion)
• chili: cw EPR (slow motion)
• pepper: cw EPR (solid state)
• salt: ENDOR (solid state)
• saffron: pulse EPR/ENDOR (solid state)
• esfit: least-squares fitting

### References

1. Dicus, M.M; Conlan,A.; Nechushtai,R.; Jennings,P.A.;Paddock,M.L.; Britt,R.D.; Stoll S.  J. AM. CHEM. SOC. 2010, 132, 2037–2049
2. Hagen,W.R. 2009. Biomolecular EPR Spectroscopy. Boca Raton: CRC Press.
3. Hagen,W.R. Dalton Trans., 2006, 4415–4434
4. Stoll,S.;Schweiger,A.  Journal of Magnetic Resonance 178 (2006) 42–55

### Problems

1. If there is one unpaired electron in  Cu2+ (I=3/2) and the copper ion is coordinated by one nitrogen atom (I=1) and one OH- (I=1/2), how many lines can be expected in the EPR spectrum? ( (2*1*3/2+1)*(2*1*1+1)*(2*1*1/2+1)=24 )
2. For a radical, the magnetic field is 3810 G, the frequency of the microwave is 9600 MHz. What is the value of its g-factor?

$g = \dfrac{ (71.4484) (9600 \,\text{x} \,10^{-3})}{3810\, \text{x}\, 10^{-1}}=1.800$

### Contributors

• Pei Zhao (University of California, Davis)

09:44, 2 Oct 2013

## Tags

### Textbook Maps

An NSF funded Project


text:
How to calculate the radii of high orbits above the parallels of 45 and 80 degrees?

The new theories in the field demonstrate the possibility of non-Keplerian orbits.

https://www.pourlascience.fr/sd/spatial/de-nouvelles-orbites-geostationnaires-10636.php

I am interested in calculating orbits above the parallels of 45 and 80 degrees. A large disk placed on these orbits should always be above an area. Have geostationary behavior.

• Orbits like your top two are impossible. In simple terms, the satellite must orbit the center of mass of the Earth. Consider reading through this article: earthobservatory.nasa.gov/features/OrbitsCatalog – Organic Marble Oct 5 '19 at 1:34
• The new theories in the field demonstrate the possibility of non-Keplerian orbits. – Ion Corbu Oct 5 '19 at 2:27
• I edited the post and added the photo and link. – Ion Corbu Oct 5 '19 at 2:28
• The article says it might be possible to displace the satellites by 5 to 30 miles. At the distance of a geostationary satellite, that offset won't get you anywhere close to 45 or 80 degrees latitude. – JohnHoltz Oct 5 '19 at 3:27
• A satellite moves on its orbit without engine thrust. If thrust is neccessary for displacement, it is no orbital path. There is no thrust for the orbits of solar system planets and moons – Uwe Oct 5 '19 at 22:06

The cited article is talking about a 1984 idea by Robert Forward. (See this English summary). In it, he proposed that solar pressure could be used to move an geostationary satellite a few 10’s of kilometers north or south of the usual geostationary orbit. These are more generally called “statites”.

There’s really nothing here that points to much larger offsets from an equatorial orbit.

note: The question was modified while I was writing this answer, so I will have update(d) this (with a new) answer with more math shortly finally, here!

So let's think about what an orbit is. One way to think about it is a combination of moving forward while falling down. At each moment the object keeps moving forward, but it also accelerates towards the Earth because of the Earth's gravitational attraction.

In which direction is it always accelerating?

Since the Earth is very close to spherical, the gravitational attractions from each part of it add up to an attraction towards the center of the Earth. Even if an object started on one of those circles above or below the equatorial plane, Earth's gravity would pull accelerate it in a plane that intersects the spacecraft's position and the center of the Earth. That would be the orbital plane, and all orbital planes around spherical bodies pass through the center of the body.

There are no other free orbits than those. The only way to make circles like that is to constantly fight gravity with powerful propulsion, and there's no known technology that can sustain that right now for longer than several minutes.

That would be a spaceflight equivalent of Franky Zapata's Flyboard.

There might be some ideas of using pressure from sunlight on a reflective film only a few atoms thick so as to be extremely low mass, but it's not something you could really build today.

• Well said! Cool 'flyboard' too. – Organic Marble Oct 5 '19 at 2:39
• Have you seen the new post edit? With non-Keplerie orbits? The orbit remains high but less than the equatorial one. The attraction to the center of the Earth is counteracted by altitude, speed and trajectory in the form of a flattened figure 8. The altitude corrections do not require too many edges. – Ion Corbu Oct 5 '19 at 2:49
• @IonCorbu oh I see, that was added while I was writing. Okay I'll have a look now and then update my answer, thanks! – uhoh Oct 5 '19 at 2:54
• Here is apparently most of the same information in English: phys.org/news/2010-07-scots-space-year-old-theory.html . It includes cites to 2 papers: "-- R L Forward, Light-levitated geostationary cylindrical orbits using perforated light sails, can be found in the Journal of the Astronautical Sciences, Vol.32, Apr-June, pp.221-226, 1984. -- S Baig and C R McInnes, Light-Levitated Geostationary Cylindrical Orbits are Feasible, Journal of Guidance, Control and Dynamics, Vol. 33, No. 3, pp. 782-793, 2010. -- A pre-print of the new paper is available at: strathprints.strath.ac.uk/18865/" – Bit Chaser Oct 5 '19 at 21:43
• That preprint url doesn't resolve for me, but pureportal.strath.ac.uk/en/publications/… does and the full paper is at pure.strath.ac.uk/ws/portalfiles/portal/67305638/… – Bit Chaser Oct 5 '19 at 21:51

If the aim is to hover a solar powered ion satellite the basic numbers come out as:

Weight for 1KW of solar panels = 5.33kg

Force of gravity at at 400km up = 8.6 Newtons per Kg

Mass 5.33 kg so that = 46 Newtons of gravitational force

Thrust of a 0 mass Xenon engine per KW = 0.034 Newtons

So pretty clearly it will not be possible to make something that hovers over the poles on electric power, even with unlimited Xenon available and a magical 0 mass thruster. Suspect there may be better power per KG choices for solar panels than used above but they would need to be 1000 times better to make this math work even for a magical thruster.

The thrust force needed will be sine of the latitude you are hovering over so 80 degrees would need 45 Newtons and 45 degrees would need 32 Newtons per KW of power.

For non polar locations you also need to work with the Earth's shadow, cutting our available power by half for a single platform, by rather more for a fixed ring due shade from adjacent panels, less as orbit height increases.

Where the concept starts to work is at 0.04 degrees where the numbers above equal but only get a couple of Km of displacement from equatorial.

The other method where this might start to work for high latitudes is for very high orbits where the force of gravity from earth is reduced, but that will place your satellite at distances greater than the moons orbit above a pole, defeating most of the purpose of being geostationary, and still not doing much for 45 degrees latitude.

Okay, I had said that I would post another answer based on your clarification, but then I got temporarily distracted. Here it goes now.

Here is just what you've asked for; How to calculate the radii of high orbits above the parallels of 45 and 80 degrees and then some even higher ones!

Just fyi I have confirmed by a simple 3D orbit simulation that this works. I calculate the real acceleration, then zero-out the z-component in order to simulate the thrust. Since the magnitude of the thrust is not part of the question, there was no need to calculate it explicitly, but it will be sizable, and a realistic conventional spacecraft couldn't maintain it very long.

We will assume you have some "vertical" acceleration due to thrust $$a_T$$ produced by some unconventional thrust mechanism pushing your spacecraft "up and away" from the plane of the equator, that exactly cancels the "downward" component of Earth's gravitational acceleration. This leaves only a horizontal acceleration $$a_H$$.

Every unconventional orbit is defined by the axial distance $$R$$ and the height above the equatorial plane $$h$$. The distance to the center of the earth $$r=\sqrt{R^2 + h^2}$$ is used to calculate the radial acceleration

$$a = \frac{GM}{r^2} = \frac{GM}{R^2 + h^2}$$

and the horizontal acceleration is (by use of similar triangles)

$$a_H = \frac{GM}{R^2 + h^2} \times \frac{R}{\sqrt{R^2 + h^2}} = \frac{GM \ R}{(R^2+h^2)^{3/2}}.$$

For a circular orbit the acceleration is $$v^2/r$$, so we can write

$$v^2 = a_H R = \frac{GM \ R^2}{(R^2+h^2)^{3/2}}.$$

Also for a circular orbit, the relationship between the radius, velocity and the period is

$$T = \frac{2 \pi R}{v}$$

$$v^2 = \frac{4 \pi^2R^2}{T^2}$$

If we set the two expressions for v^2 equal, we get the surprisingly simple result:

$$R^2 + h^2 = \left( \frac{GM \ T^2}{4 \pi^2} \right)^{2/3}$$

Try it! The standard gravitational parameter $$GM$$ for the Earth is 3.986+14 m^3/s^2. Put in 86164 seconds for T (one sidereal day, 23h 56m 4s) and start with a normal GEO orbit with $$h=0$$. You should get 42,164 kilometers (42,164,000 meters).

Now, $$h$$ is given by

$$h = r_E \sin(lat)$$

and let's just use the average radius of the Earth of 6371 km.

Considering that the GEO orbit is so far from Earth, this is only a tiny variation. Let's do some crazy orbits way above the North Pole as well.

lat(degs)    h (km)       R (km)      v (km/s)
---------    ------      -------      --------
0            0       42,164        3.075
45        4,505       41,922        3.057
80        6,274       41,694        3.040
--       10,000       40,961        2.987
--       20,000       37,119        2.707
--       30,000       29,628        2.161
--       40,000       13,335        0.972

• Cool stuff! Now if we can just devise a bootstrap-lifter! – Organic Marble Oct 12 '19 at 22:24


text:
# Find the Compression!

Consider the diagram shown above where the two chambers separated by piston-spring arrangement contain equal amounts of same ideal gas. Initially when the temperatures of the gas in both the chambers are kept at $300 K$. The compression in the spring is $10 \text{ cm}$. The temperature of the left and the right chambers are now raised slowly to $400 K$ and $500 K$ respectively. The pistons are free to slide (i.e. no friction between piston and chamber). Final compression in the spring is found to be ($\frac{\alpha}{60}m$). Find $\alpha$.

Use $\sqrt{481} = 22$.

×


text:
# Acronym/Abbreviation meaning of 1"'-- (2)

The acronym 1"'--(2) means : 1

Acronyms, abbreviations and meanings similar to 1"'--

## Famous / Popular results for 1"'--

2018–19 UEFA Champions League

determine the number of participating teams for each association: Associations 1–4 each have four teams qualify. Associations 5–6 each have three teams qualify

2017–18 UEFA Champions League

in Kiev, Ukraine. Real Madrid, the defending champions, beat Liverpool 3–1 to win a record-extending 13th title and their third title in a row. As winners

2018–19 Premier League

Premier League history was scored this season on 23 April by Shane Long in a 11 draw between his side Southampton and Watford after 7.69 seconds. On 4 May

List of NBA champions

court in Games 1, 2, 6 and 7) during 1949, 1953–1955 and 1985–2013, in a 1111111 format during 1956 and 1971 and in a 1–2–2–11 format during 1975

2018–19 UEFA Europa League

berth reserved will be given to the third-placed team of the 2018–19 Ligue 1, the 5th-ranked association according to next season's access list. For the

List of countries and dependencies by population

inclusion within the list being primarily based on the ISO standard ISO 3166-1. For instance, the United Kingdom is considered as a single entity, while

List of constituencies of the Lok Sabha

2008-10-14. Archived from the original (PDF) on 2008-10-31. Retrieved 2008-11-27. [1][dead link] "THE REPRESENTATION OF THE PEOPLE (AMENDMENT) BILL, 2008" (PDF)

Standard deviation

1%. A more accurate approximation is to replace N − 1.5 {\displaystyle N-1.5} above with N − 1.5 + 1 / ( 8 ( N − 1 ) ) {\displaystyle N-1.5+1/(8(N-1))}

Normal distribution

a + b = 1 1 a + 1 b = ( a − 1 + b − 1 ) − 1 . {\displaystyle {\frac {ab}{a+b}}={\frac {1}{{\frac {1}{a}}+{\frac {1}{b}}}}=(a^{-1}+b^{-1})^{-1}.} This

1

1 (one, also called unit, unity, and (multiplicative) identity) is a number, numeral, and glyph. It represents a single entity, the unit of counting or

2018–19 UEFA Champions League

determine the number of participating teams for each association: Associations 1–4 each have four teams qualify. Associations 5–6 each have three teams qualify

2017–18 UEFA Champions League

in Kiev, Ukraine. Real Madrid, the defending champions, beat Liverpool 3–1 to win a record-extending 13th title and their third title in a row. As winners

List of NBA champions

court in Games 1, 2, 6 and 7) during 1949, 1953–1955 and 1985–2013, in a 1111111 format during 1956 and 1971 and in a 1–2–2–11 format during 1975

2018–19 UEFA Europa League

berth reserved will be given to the third-placed team of the 2018–19 Ligue 1, the 5th-ranked association according to next season's access list. For the

2018–19 Premier League

Premier League history was scored this season on 23 April by Shane Long in a 11 draw between his side Southampton and Watford after 7.69 seconds. On 4 May

List of constituencies of the Lok Sabha

2008-10-14. Archived from the original (PDF) on 2008-10-31. Retrieved 2008-11-27. [1][dead link] "THE REPRESENTATION OF THE PEOPLE (AMENDMENT) BILL, 2008" (PDF)

List of countries and dependencies by population

inclusion within the list being primarily based on the ISO standard ISO 3166-1. For instance, the United Kingdom is considered as a single entity, while

Standard deviation

1%. A more accurate approximation is to replace N − 1.5 {\displaystyle N-1.5} above with N − 1.5 + 1 / ( 8 ( N − 1 ) ) {\displaystyle N-1.5+1/(8(N-1))}

Normal distribution

a + b = 1 1 a + 1 b = ( a − 1 + b − 1 ) − 1 . {\displaystyle {\frac {ab}{a+b}}={\frac {1}{{\frac {1}{a}}+{\frac {1}{b}}}}=(a^{-1}+b^{-1})^{-1}.} This

1

1 (one, also called unit, unity, and (multiplicative) identity) is a number, numeral, and glyph. It represents a single entity, the unit of counting or

## Meaning of CARES

CARES means: Ceramic Analysis And Reliability Evaluation Of Structures

## Meaning of HETA

HETA means: Hazard Evaluation Technical Assistance Branch (NIOSH)

## Meaning of SBD

SBD means: Schottky-barrier Diode

## Meaning of FAL

FAL means: First Approach And Landing (Test)

## Meaning of CSCG

CSCG means: Communications Security Control Group


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# Math Help - Two definite integrals

1. ## Two definite integrals

I'm a bit stuck on these. A nudge in the right direction would be much appreciated.

$\int_{- \frac {\pi}{2}}^{\frac {\pi}{2}} \frac {x^{2} sin{x}}{1+x^{6}} dx
$

I can't seem to figure out what to do with that sin(x) function. If I let $u = x^{3}$, I can get the whole thing into the form of the derivative of arctan(x), except for the sin(x).

The other one that's bugging me is:
$\int_{0}^{4} \frac {x}{\sqrt{1+2x}} dx$

No excuse here. I've tried quite a few different substitutions and algebraic manipulations without success. I can't shake the feeling that I'm over-looking something glaring with this problem. I think I found a method that would involve using partial fractions, but I was told that none of these problems required that method.

2. Originally Posted by DarrenM
I'm a bit stuck on these. A nudge in the right direction would be much appreciated.

$\int_{- \frac {\pi}{2}}^{\frac {\pi}{2}} \frac {x^{2} sin{x}}{1+x^{6}} dx
$

I can't seem to figure out what to do with that sin(x) function. If I let $u = x^{3}$, I can get the whole thing into the form of the derivative of arctan(x), except for the sin(x).

The other one that's bugging me is:
$\int_{0}^{4} \frac {x}{\sqrt{1+2x}} dx$

No excuse here. I've tried quite a few different substitutions and algebraic manipulations without success. I can't shake the feeling that I'm over-looking something glaring with this problem. I think I found a method that would involve using partial fractions, but I was told that none of these problems required that method.
1.
The integrated function is odd and the integral is from $\frac{-\pi}{2}$ to $\frac{\pi}{2}$ , What do you think ?

2. The substitution $u=1+2x$ will make it looks like :
$\frac{1}{4}\int_1^9 \frac{u-1}{\sqrt{u}} du$.
Which is easier.

The substitution $u=\sqrt{1+2x}$ works also.

3. Thanks!

1. Zero. I wasn't certain, looking at the function, whether it was even or odd. I know that sin(x) is an odd function, but I wasn't sure about the rest of it. $x^{2}$ is an even function, and $\frac {1}{1+x^{6}}$ is an even function.

I just did a quick search, however, and I think I may have found the justification. The product of an even function and an odd function is odd. The quotient of an odd function and an even function is an odd function. So, despite the fact that both of those were even, the overall result is still odd... is that correct?

2. Argh. I got to that point but didn't actually factor out the 2, hence the "partial fractions" thing. Always a bit embarassed when I make a dumb mistake like that.

Thank you very much for your help.

4. It's worth to say that $\int_{-a}^a f=0$ for $a>0$ when $f$ is odd, as long as this one is integrable.

For example, one could say $\int_{-1}^1\frac{dx}x=0,$ but this function is not bounded on $[-1,1]$ so it's not integrable there and the result does not apply.


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## anonymous 4 years ago $\sqrt{x}(2x ^{3}-1)$ Find the derivative using the product rule.

1. anonymous

i have gotten down to: $\sqrt{x}(6x ^{2})+(2x ^{3}-1)+(x ^{2})$

2. anonymous

just not sure how to simplify from here.

3. cwrw238

sqrtx * 6x^2 + (1/2) x^(-1/2) * (2x^3 - 1)

4. anonymous

√x(6x^2)+(2x^3−1)+(x^2) is what i got.

5. cwrw238

derivative of sqrtx or x^(1/2) is (1/2) x^(-1/2)

6. anonymous

the deriviate of $\sqrt{x} = x ^{2}$

7. anonymous

i think? lol, that's what my prof wrote down.

8. cwrw238

the rule is if f(x) = ax^n , f'(x) = an x^(n-1)

9. cwrw238

if n = 1/2, n-1 = -1/2

10. anonymous

so it's $\frac{ 1 }{ 2 }x ^{\frac{ -1 }{ 2 }}$ ??

11. cwrw238

yes

12. anonymous

okay im still not sure how to further simplfy from there.

13. anonymous

wait, i don't even need the deriviate of the square root of x ..

14. anonymous

nevermind i do lol anyways,

15. cwrw238

which can be written as 1 / 2x^(1/2) sorry can't help anymore - gotta go out

16. anonymous

okay.

17. anonymous

so im left with $x ^{\frac{ 1 }{ 2 }}(6x ^{2})+(2x-1)(\frac{ 1 }{ 2 }x ^{\frac{ -1 }{ 2 }}$ i need help further simplifying.

18. anonymous

$\huge 6x^2\sqrt x + \frac{2x^3-1}{2\sqrt x}$add them ~

19. anonymous

why 2 root x?

20. anonymous

because$\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt x}$

21. anonymous

$(fg)'=f'g+g'f$ with $f(x)=\sqrt{x},f'(x)=\frac{1}{2\sqrt{x}},g(x)=2x^3-1,g'(x)=6x^2$

22. anonymous

yes i got all that, im just not sure how to further simply once im at the last step with all deriviates completed.

23. anonymous

i have everything over 2 root x

24. anonymous

plug and play (algebra) not to butt in, but $$\sqrt{x}$$ is a very very common function, plus math teachers like to put it on tests and quizzes, so rather than use the power rule each time it would help your cause out to memorize $\frac{d}{dx}[\sqrt{x}]=\frac{1}{2\sqrt{x}}$ like you know $$8\times 9=72$$ for example it will save you time on exams if nothing else

25. anonymous

i know all that but what im having trouble with is simplyfying $\sqrt{x}(6x ^{2})+2x ^{3}-1 /2\sqrt{x}$

26. anonymous

$\huge 6x^2 \sqrt x + \frac{2x^3-1}{2 \sqrt x}$ you got to here, now just make the denominators common and add them

27. anonymous

i have $(\sqrt{x})(6x ^{2})(2\sqrt{x})+2x ^{3}-1 all \over 2\sqrt{x}$

28. anonymous

i cant get any further than that.

29. anonymous

$\huge 6x^2 \sqrt x \times 2 \sqrt x = ?$

30. anonymous

no idea

31. anonymous

12x^2 root x ?

32. anonymous

???

33. anonymous

$\huge \sqrt x \times \sqrt x = x$$\huge 6x^2 \sqrt x \times 2 \sqrt x = 12 x ^3$

34. anonymous

why x cubed and not x squared?

35. anonymous

because~$\huge 6x^2 \times 2 = 12x^2$$\huge \sqrt x \times \sqrt x = x$$\huge 12x^2 \times x = 12 x^3$ then $\huge \frac{12x^3 + 2x^3 + 1}{2 \sqrt x}=\frac{ 14x^3 + 1}{2\sqrt x}$

36. anonymous

thank-you! then i can just rationalize the denominator and ill be done.


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# In an examination, one of the questions requires Shweta to calculate the distance between the sun and Jupiter. It is given that the time taken by light to cover this distance is about 43 minutes. To reduce calculation, the speed of light is to be taken as 10 units. However, the value used by Shweta is such that her answer comes out to be 40% less than the desired answer.  What is the value of the speed of light used by Shweta in her calculations?

Can you answer this question?

## 1 Answer

0 votes
6 units
Hence (B) is the correct answer.
answered Jun 1, 2014


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Now showing items 1-1 of 1

• #### Neutron electric dipole moment from QCD sum rules

(1996)
The electric dipole moments of nucleons (NEDM, $d\sb{N}$) are calculated using the method of QCD sum rules. Our calculations are based on the parity ($/\!\!\!P$) and time reversal ($/\!\!\!T$) violating parameter $\bar\Theta$ ...


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# multivariate hypergeometric distribution calculator

The hypergeometric distribution deals with successes and failures and is useful for statistical analysis with Excel. Population size # Successes in population. Observations: Let p = k/m. In a set of 16 light bulbs, 9 are good and 7 are defective. The test is often used to identify which sub-populations are over- or under-represented in a sample. Right? In probability theory and statistics, Fisher's noncentral hypergeometric distribution is a generalization of the hypergeometric distribution where sampling probabilities are modified by weight factors. Example 4.25. The hypergeometric distribution calculator finds the probability of success in a population. Various generalizations to this distribution exist for cases where the picking of colored balls is biased so that balls of one color are more likely to be picked than balls of another color. If I just wanted to calculate the probability for a single class (say 1 or more red marble), I could use the upper tail of the hypergeometric cumulative distribution function, in other words calculate 1 - the chance of not drawing a single red marble. The probability mass function (pmf) of the distribution is given by: Where: N is the size of the population (the size of the deck for our case) m is how many successes are possible within the population (if youâ€™re looking to draw lands, this would be the number of lands in the deck) n is the size of the sample (how many cards weâ€™re drawing) k is how many successes we desire (if weâ€™re looking to draw three lands, k=3) For the rest of this article, â€œpmf(x, n)â€, will be the pmf of the scenario weâ€… distributions are reduced to the (multivariate) binomial distribution when n = 1, or to the (multivariate) hypergeometric distribution when all wi’s are equal. Suppose a shipment of 100 DVD players is known to have 10 defective players. The Weibull Distribution¶ double gsl_ran_weibull (const … If I just wanted to calculate the probability for a single class (say 1 or more red marble), I could use the upper tail of the hypergeometric cumulative distribution function, in other words calculate 1 - the chance of not drawing a single red marble. A revised version of this article will appear in Communications in Statistics, Simulation and Computation, vol. Question 5.13 A sample of 100 people is drawn from a population of 600,000. Hypergeometric Distribution Calculator; Hypergeometric Distribution Calculator with source (Ruby, C++) The Hypergeometric Distribution and Binomial Approximation to a Hypergeometric Random Variable by Chris Boucher, Wolfram Demonstrations Project. Let Wj = ∑i ∈ AjYi and rj = ∑i ∈ Ajmi for j ∈ {1, 2, …, l} However, you can skip this section and go to the explanation of how the calculator itself works. So I can not use it. The Multivariate Hypergeometric Distribution Basic Theory As in the basic sampling model, we start with a finite population D consisting of m objects. So I can not use it. The cumulative multivariate hypergeometric distribution is more complex, as it calculates the chances of drawing both the card we wish to play, and then not having enough coloured mana to play that card. Let's consider the following setup: We take a set having N number of elements. The following conditions characterize the hypergeometric distribution: 1. This test has a wide range of applications. of successes in population, sample size and no. Gentle, J.E. The most common use of the hypergeometric distribution, which we have seen above in the examples, is calculating the probability of samples when drawn from a set without replacement. X ~ H(r, b, n) Read this as "X is a random variable with a hypergeometric distribution." The hypergeometric distribution is used for sampling without replacement. The method is used if the probability of success is not equal to the fixed number of trials. The Hypergeometric Calculator makes it easy to compute individual and cumulative hypergeometric probabilities. Multivariate Polya distribution: functions d, r of the Dirichlet Multinomial (also known as multivariate Polya) distribution are provided in extraDistr, LaplacesDemon and Compositional. Multivariate hypergeometric distribution accounts for the case that I got additional features of interest more than ns and ni in my mount, as far as I understand it. Calculation Methods for Wallenius’ Noncentral Hypergeometric Distribution Agner Fog, 2007-06-16. It is applied in number theory, partitions, physics, etc. Where k=sum(x), N=sum(n) and k<=N. In statistics, the hypergeometric test uses the hypergeometric distribution to calculate the statistical significance of having drawn a specific successes (out of total draws) from the aforementioned population. multivariate hypergeometric distribution. E.g. ; We categorize these elements along some arbitrary requirement or requirements into m number of categories. It is used for sampling without replacement $$k$$ out of $$N$$ marbles in $$m$$ colors, where each of the colors appears $$n_i$$ times. In probability theory and statistics, the hypergeometric distribution is a discrete probability distribution that describes the probability of successes (random draws for which the object drawn has a specified feature) in draws, without replacement, from a finite population of size that contains exactly objects with that feature, wherein each draw is either a success or a failure. This is a little digression from Chapter 5 of Using R for Introductory Statistics that led me to the hypergeometric distribution. Let x be a random variable whose value is the number of successes in the sample. 3 Homogeneity Testing for the Multivariate Hypergeometric Distribution 8 3.1 Introduction 8 3.2 Procedure 1 9 3.3 Procedure 2 12 3.4 Approximation Algorithm for P H 0 (X (k)t X (k 1)t D 2) 20 3.5 Simulation of Multivariate Hypergeometric Random Variables 23 4 Powers of Procedures and Sample Size in Multivariate Hypergeometric Distribution 24 I want to calculate the probability that I will draw at least 1 red and at least 1 green marble. p = hygecdf(2,100,20,10) p = 0.6812 Extended Capabilities . If so Start off with the fact that each group must contain at least 1 ball, that leaves you with 10 balls to place among the sets. It is used for sampling without replacement k out of N marbles in m colors, where each of the colors appears n[i] times. The parameters are r, b, and n; r = the size of the group of interest (first group), b = the size of the second group, n = the size of the chosen sample. of successes in sample. Suppose you have a lot of 100 floppy disks and you know that 20 of them are defective. Relevance and Uses of Hypergeometric Distribution Formula. In terms of the formula used. Multivariate hypergeometric distribution accounts for the case that I got additional features of interest more than ns and ni in my mount, as far as I understand it. card combination calculator multivariate hypergeometric distribution. This calculator finds probabilities associated with the hypergeometric distribution based on user provided input. Enter the number of size and success of the population and sample in the hypergeometric distribution calculator to find the cumulative and hypergeometric distribution. (2006). Hypergeometric Distribution Calculator; Hypergeometric Distribution Calculator with source (Ruby, C++) The Hypergeometric Distribution and Binomial Approximation to a Hypergeometric Random Variable by Chris Boucher, Wolfram Demonstrations Project. [Archive] Multivariate hypergeometric distribution General Questions. The multivariate hypergeometric distribution is generalization of hypergeometric distribution. In probability theory and statistics, Wallenius' noncentral hypergeometric distribution (named after Kenneth Ted Wallenius) is a generalization of the hypergeometric distribution where items are sampled with bias.. card combination calculator multivariate hypergeometric distribution. Along with that, “N” is the total number of draws which have to be done. In this section, we suppose in addition that each object is one of k types; that is, we have a multi-type population. Using a Hypergeometric Calculator The hypergeometric distribution can describe the likelihood of any number of successes when drawing from a deck of Magic cards. 37, no. The test is often used to identify which sub-populations are over- or under-represented in a sample. Thank you for your questionnaire.Sending completion. He is interested in determining the probability that, C/C++ Code Generation Generate C and C++ code using MATLAB® Coder™. 37, no. If you randomly select 6 light bulbs out of these 16, what’s the probability that 3 of the 6 are […] In probability theory and statistics, the hypergeometric distribution is a discrete probability distribution that describes the probability of successes in draws, without replacement, from a finite population of size that contains exactly successes, wherein each draw is either a success or a failure. Jump to navigation Jump to search. Let’s start with an example. Multivariate hypergeometric distribution: provided in extraDistr. 2, 2008. In the next section, I’ll explain the actual math, like I did with the single variable hypergeometric distribution. (2006). The ordinary hypergeometric distribution corresponds to k=2. I want to calculate the probability that I will draw at least 1 red and at least 1 green marble. The method uses the fact that a multivariate Gaussian distribution is spherically symmetric. Best How To : phyper(5, 8, 92, 30) gives the probability of drawing five or fewer red marbles. The probability density function (pdf) for x, called the hypergeometric distribution, is given by. Hypergeometric distribution has many uses in statistics and in practical life. Each component is generated to have a Gaussian distribution, and then the components are normalized. Your feedback and comments may be posted as customer voice. [1]  2020/05/20 18:37   Male / Under 20 years old / High-school/ University/ Grad student / Very /, [2]  2019/11/13 09:46   Male / 20 years old level / An office worker / A public employee / Very /, [3]  2018/02/12 09:10   Male / 50 years old level / A teacher / A researcher / A little /, [4]  2017/11/25 22:50   Male / 60 years old level or over / A retired people / Very /, [5]  2017/09/29 07:06   Male / 30 years old level / High-school/ University/ Grad student / Useful /, [6]  2017/03/14 10:42   Male / Under 20 years old / Elementary school/ Junior high-school student / Very /, [7]  2015/04/20 20:09   Male / 40 years old level / A teacher / A researcher / Very /, [8]  2015/03/08 03:47   Female / 60 years old level or over / A teacher / A researcher / Useful /, [9]  2014/11/23 06:00   Female / Under 20 years old / High-school/ University/ Grad student / A little /, [10]  2014/11/17 02:15   Male / 30 years old level / Self-employed people / A little /. Examples of how to use “hypergeometric” in a sentence from the Cambridge Dictionary Labs It takes into account the fact that each draw decreases the size of your library by one, and therefore the probability of success changes on each draw. For help, read the Frequently-Asked Questions or review the Sample Problems. What is the probability of drawing zero to two defective floppies if you select 10 at random? If you have a look at the concept of hypergeometric distribution, it is very similar to the binomial theorem. Example of a hypergeometric distribution problem. The density of this distribution with parameters m, n and k (named N p, N − N p, and n, respectively in the reference below) is given by p (x) = (m x) (n k − x) / (m + n k) for x = 0, …, k. It takes into account the fact that each draw decreases the size of your library by one, and therefore the probability of success changes on each draw. I wasn’t even aware that an online tool existed until two readers pointed it out to me last week. Multivariate Ewens distribution: not yet implemented? Examples of how to use “hypergeometric” in a sentence from the Cambridge Dictionary Labs $$\normalsize Hypergeometric\ distribution\\. After withdrawals, replacements are not made. A random variable X{\displaystyle X} follows the hypergeometric distribution if its probability mass functi… That is, a population that consists of two types of objects, which we will refer to as type 1 and type 0. This technique can be used by a marketing company to know the customers or public views. Hence, it is not surprising that the two distributions approximate each other when n á N and when the odds ratios are all close to 1. Definition 1: Under the same assumptions as for the binomial distribution, from a population of size m of which k are successes, a sample of size n is drawn. The Hypergeometric Distribution Basic Theory Dichotomous Populations. How to use Excel as a card probability calculator. In probability theory and statistics, Wallenius' noncentral hypergeometric distribution (named after Kenneth Ted Wallenius) is a generalization of the hypergeometric distribution where items are sampled with bias.. This distribution can be illustrated as an urn model with bias. Random number generation and Monte Carlo methods. Let x be a random variable whose value is the number of successes in the sample. In statistics, the hypergeometric distribution is the discrete probability distribution generated by picking colored balls at random from an urn without replacement.. Where \(k=\sum_{i=1}^m x_i$$, $$N=\sum_{i=1}^m n_i$$ and $$k \le N$$. References. The multivariate hypergeometric distribution is parametrized by a positive integer n and by a vector {m 1, m 2, …, m k} of non-negative integers that together define the associated mean, variance, and covariance of the distribution. “K” is the number of successes that have to be attained. Eric W. Weisstein, Hypergeometric Distribution at … Definition 1: Under the same assumptions as for the binomial distribution, from a population of size m of which k are successes, a sample of size n is drawn. The method is described by Knuth, v2, 3rd ed, p135–136, and attributed to G. W. Brown, Modern Mathematics for the Engineer (1956). For example, if you have an urn with 2 red marbles, 4 white marbles, 8 blue marbles, and 12 orange marbles, the probability of drawing 5 marbles and getting 1 red marble and 2 white marbles is as follows: Using an Online Multivariate Hypergeometric Calculator. Some functions are limited now because setting of JAVASCRIPT of the browser is OFF. Show the following alternate from of the multivariate hypergeometric probability density function in two ways: combinatorially, by considering the ordered sample uniformly distributed over the permutations Density, distribution function, quantile function and random generation for the hypergeometric distribution. The hypergeometric distribution can describe the likelihood of any number of successes when drawing from a deck of Magic cards. Hypergeometric Distribution Calculator References: Hypergeometric Distribution (on Wikipedia) Hypergeometric Calculator; Probability: Drawing Cards from Decks (in "The Mathematics of Magic The Gathering") Footnotes: (1) cf. Specifically, there are K_1 cards of type 1, K_2 cards of type 2, and so on, up to K_c cards of type c. (The hypergeometric distribution is … Calculation Methods for Wallenius’ Noncentral Hypergeometric Distribution Agner Fog, 2007-06-16. Notation for the Hypergeometric: H = Hypergeometric Probability Distribution Function. In statistics, the hypergeometric distribution is a function to predict the probability of success in a random 'n' draws of elements from the sample without repetition. In statistics, the hypergeometric test uses the hypergeometric distribution to calculate the statistical significance of having drawn a specific successes (out of total draws) from the aforementioned population. What is Cumulative hypergeometric distribution. For example, calculate your odds of getting a run of aces from a standard deck. SUMMARY.Two different probability distributions are both known in the literature as 2, 2008. Suppose that we have a dichotomous population $$D$$. Specifically, suppose that (A1, A2, …, Al) is a partition of the index set {1, 2, …, k} into nonempty, disjoint subsets. From formulasearchengine. The algorithm behind this hypergeometric calculator is based on the formulas explained below: 1) Individual probability equation: H(x=x given; N, n, s) = [ s C x] [ N-s C n-x] / [ N C n] 2) H(x

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ADGA is a one-day workshop that focuses on the theoretical foundations of distributed graph algorithms. The program consists of several invited talks, which are aimed at the general DISC audience.

ADGA 2020 will be held as a virtual workshop on Friday, 16 October 2020, and will be co-located with DISC 2020. The workshop will be chaired by Jara Uitto.

For information on past and future ADGA workshops, see the web site of the ADGA workshop series. For information about the registration, please go to the DISC 2020 website.

# Schedule

15:00 talk 1 Krzysztof Onak: The Current Landscape of Massively Parallel Algorithms for Graphs · video
15:30 talk 2 Slobodan Mitrović: Recent LCA Techniques through the Lens of Matchings · video
16:00 coffee break
16:20 talk 3 Faith Ellen: Constant-Length Labelling Schemes for Deterministic Radio Broadcast · video
16:50 talk 4 Yannic Maus: Distributed vertex coloring: classic meets modern · video
17:20 coffee break
17:40 talk 5 Dennis Olivetti: Distributed Edge Coloring in Time Quasi-Polylogarithmic in Δ · video
18:10 talk 6 Michael Dinitz: New Graph Spanners and the Greedy Algorithm · video
18:40 Workshop ends

# Invited Speakers

Michael Dinitz

Johns Hopkins University

Michael Dinitz is an assistant professor in the Department of Computer Science at Johns Hopkins University. He received his PhD from Carnegie Mellon University in 2010 under Anupam Gupta, and was a postdoctoral fellow at the Weizmann Institute of Science with Robert Krauthgamer and David Peleg. He works broadly in algorithms, with a focus on approximation algorithms, graph algorithms, distributed computing, and theory of networking.

New Graph Spanners and the Greedy Algorithm

Graph spanners (subgraphs which approximately preserve distances) originated in the distributed computing community and have had a huge impact in distributed algorithms and in algorithms more broadly. In this talk we will discuss recent progress on two relatively new variants of graph spanners: $\ell_p$-norm and fault-tolerant spanners. For $\ell_p$-norm spanners this includes tight existential bounds and good approximation algorithms (at least for some parameter settings), and for fault-tolerant spanners this includes tight existential bounds with surprisingly efficient algorithms.

Both of these variants are particularly well-motivated in distributed settings. Yet for both of them the recent progress in the centralized setting is based on some variant of the greedy algorithm, which is not easily implementable in distributed settings. We hope that this talk will help spur research on distributed algorithms for these important objects.

Faith Ellen

University of Toronto

Faith Ellen received her Ph.D. from the University of California, Berkeley, in 1982, became a faculty member at the University of Washington in 1983, and moved to the University of Toronto in 1986, where she is a Professor of Computer Science. She was the program committee chair for DISC 2002, OPODIS 2018, and PODC 2019, served as the chair of the PODC steering committee from 2006 to 2009, and was vice-president of SIGACT from 1997 to 2001. Her research interests include the theory of distributed computing, data structures, and complexity and she is a co-author of the book “Impossibility Results for Distributed Computing”. Faith became a Fellow of the Association for Computing Machinery in 2014.

Broadcast is a fundamental network communication primitive, in which a source node has a message that has to be received by all other nodes. In synchronous radio networks, this problem is non-trivial, since if two or more neighbours of a node transmit at the same time, it hears nothing. In fact, if nodes do not store any information, broadcast is impossible deterministically, even in a four-cycle. If the nodes have distinct identifiers from a small name space, then a round-robin strategy suffices, but it takes a long time.

This talk will show that every radio network can be labelled using a small constant number of bits so that broadcast can be accomplished by a fixed deterministic algorithm that does not know the network topology nor any bound on its size. Specifically, there is a labelling scheme that stores 2 (carefully chosen) bits per nodes that allows broadcast to be performed in $O(n)$ rounds, where $n$ is the size of the network. There is a variant of this algorithm using 4 bits per node that completes broadcast in $O(\sqrt{Dn})$ rounds, where $D$ is the source eccentricity of the network. This number of rounds is shown to be optimal for a class of algorithms that includes both.

Then, using some ideas from some old algorithms, which assume nodes have distinct identifiers and a bound on the network size, a deterministic algorithm is constructed that uses 3 bits per node and completes in $O(D \log^2 n)$ rounds. A randomized construction of a labelling scheme with 3 bits per node for a broadcast algorithm that completes in $O(D \log n + \log^2 n)$ rounds is also presented.

This is joint work with Seth Gilbert and Barun Gorain, Avery Miller, and Andrzej Pelc.

Yannic Maus

Technion

Yannic Maus is a postdoc at the Technion in the group of Keren Censor-Hillel. His research focused on distributed algorithms, in particular on the limits of randomization and the power of locality. He received his PhD from University of Freiburg, supervised by Fabian Kuhn and, this year, he will be awarded the ACM principles of distributed computing dissertation award.

Distributed vertex coloring: classic meets modern

The last years have seen several new distributed graph coloring algorithms in the LOCAL and the CONGEST model. In this talk we will survey modern techniques and relate (some of) them to classic ones. In particular, we see how Linial’s seminal $O(\Delta^2)$-coloring algorithm for graphs with maximum degree $\Delta$ extends to list coloring and how this implies one of the current state of the art distributed graph coloring algorithms.

Slobodan Mitrović

MIT

Slobodan Mitrović is a postdoc at MIT, in a group led by Ronitt Rubinfeld. His main research interests are in the area of algorithms in memory-constrained settings, in particular parallel, distributed and streaming. He completed his PhD at EPFL in 2018 under the supervision of Aleksander Mądry.

Recent LCA Techniques through the Lens of Matchings

Local computation algorithms (LCA) formalize concepts that arise in various algorithmic fields. The main task of LCA is to solve a problem on a part of an instance by accessing only "a small" part of it, e.g., deciding whether a given edge is in a maximal matching without looking at the entire graph. In this talk, in the context of matchings, we will discuss two recent LCA techniques. These techniques start with a generic transformation of a LOCAL algorithm to an LCA, and then provide a way to reduce the amount of an input needed to be inspected by the generic approach. One of these techniques has implications for MIS, set cover and coloring as well.

Dennis Olivetti

University of Freiburg

Dennis Olivetti is a postdoc in the Algorithms and Complexity group of Prof. Fabian Kuhn at University of Freiburg, Germany. In 2018 and 2019 he was a postdoc in the Distributed Algorithms group of Prof. Jukka Suomela at Aalto University, Finland. He received his PhD from GSSI, Italy, in December 2017, supervised by Prof. Pierre Fraigniaud. His main research interests are on distributed graph algorithms, and in particular in understanding what can be computed locally.

Distributed Edge Coloring in Time Quasi-Polylogarithmic in $\Delta$

The problem of coloring the edges of an $n$-node graph of maximum degree $\Delta$ with $2\Delta-1$ colors is one of the key symmetry breaking problems in the area of distributed graph algorithms. While there has been a lot of progress towards the understanding of this problem, the dependency of the running time on $\Delta$ has been a long-standing open question. Very recently, Kuhn [SODA ’20] showed that the problem can be solved in time $2^{O(\sqrt{\log \Delta})}+O(\log^* n)$.

In this talk, we discuss the edge coloring problem in the distributed LOCAL model. We show that the (degree+1)-list edge coloring problem, and thus also the $(2\Delta-1)$-edge coloring problem, can be solved deterministically in time $2^{O(\log^2 \log \Delta)} + O(\log^* n)$.

Krzysztof Onak

IBM T.J. Watson Research Center

Krzysztof Onak is a research scientist in the Mathematics of AI group at the IBM T.J. Watson Research Center. His main research interests concern big data computation with limited resources, including algorithms for modern parallel and distributed systems, sublinear-time algorithms, and streaming. Krzysztof received his Master’s degree from the University of Warsaw and his PhD from the Massachusetts Institute of Technology. Before joining IBM, he was a Simons Postdoctoral Fellow at Carnegie Mellon University.

The Current Landscape of Massively Parallel Algorithms for Graphs

One of the most interesting settings for massively parallel algorithms on graphs is when the local space of each machine is sublinear in the number of vertices. This allows for distributing computation, even for sparse graphs, across a large number of machines that do not have to be very powerful. In recent years, we have seen several developments in this area, including both algorithms and conditional lower bounds. In many cases these results leverage connections to other areas such as distributed algorithms and property testing. I will survey some of them and talk in more detail about generating random walks and estimating PageRank, for which we were able to obtain nearly-exponential improvements in round complexity over direct implementations of known algorithms. Finally, I will share some of my favorite open questions in the area.


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# 65 (number)

65 (sixty-five) is the natural number following 64 and preceding 66.

 ← 64 65 66 →
Cardinalsixty-five
Ordinal65th
(sixty-fifth)
Factorization5 × 13
Divisors1, 5, 13, 65
Greek numeralΞΕ´
Roman numeralLXV
Binary10000012
Ternary21023
Senary1456
Octal1018
Duodecimal5512

## In mathematics

65 is the 23rd semiprime,[1] and the third of the form (5.q). It is an octagonal number.[2] It is also a Cullen number.[3] Given 65, the Mertens function returns 0.[4]

This number is the magic constant of a 5x5 normal magic square:

${\displaystyle {\begin{bmatrix}17&24&1&8&15\\23&5&7&14&16\\4&6&13&20&22\\10&12&19&21&3\\11&18&25&2&9\end{bmatrix}}.}$

This number is also the magic constant of n-Queens Problem for n = 5.[5]

65 is the smallest integer that can be expressed as a sum of two distinct positive squares in two ways, 65 = 82 + 12 = 72 + 42.[6]

It appears in the Padovan sequence, preceded by the terms 28, 37, 49 (it is the sum of the first two of these).[7]

65 is a Stirling number of the second kind, the number of ways of dividing a set of six objects into four non-empty subsets.[8]

65 = 15 + 24 + 33 + 42 + 51.[9]

65 is the length of the hypotenuse of 4 different Pythagorean triangles, the lowest number to have more than 2: 652 = 162 + 632 = 332 + 562 = 392 + 522 = 252 + 602.[10] The first two are "primitive", and 65 is the lowest number to be the largest side of more than one such triple.[11]

65 is the number of compositions of 11 into distinct parts.[12]

## In music

• "65 Love Affair", singer Paul Davis' hit song in 1982
• Sammy Hagar re-recorded his hit "I Can't Drive 55", with the 55 changed to 65, in 2001 for NBC's NASCAR broadcasts to reflect higher speed limits; the song was used from 2001 until 2004 to introduce Budweiser Pole Award winners on NBC and TNT broadcasts
• 65 is a commonly used abbreviation for the Sheffield, UK, post-rock band 65daysofstatic
• Referenced in "Heroes and Villains" by the Beach Boys: "At 60 and 5 / I'm very much alive / I've still got the jive / to survive with the Heroes and Villains"
• Odd Future group MellowHype has performed a song entitled "65"

## In other fields

• 65 miles per hour is a common speed limit on expressways in many U.S. states, primarily in the eastern and central United States. (In the western United States, a common speed limit is 70 m.p.h., and in some places it is 75 m.p.h.).
• +65 is the code for international direct dial telephone calls to Singapore.
• the traditional age for retirement in the United Kingdom, Germany, the United States, Canada, and several other countries.
• in the U.S., the age at which a person is eligible to obtain Medicare.
• CVN-65 is the designation of the U.S. Navy's first nuclear-powered aircraft carrier, the USS Enterprise (CVN-65).
• 65 is the minimum grade or average required to pass an exam, or a class, in some schools.
• The setting of the American classic TV series Naked City (1958–1963) was the 65th Precinct in New York City (sometimes referred to as simply "the 65").
• 65 is the number of the French department Hautes-Pyrénées.
• 65 is commonly used in names of many dishes of South India cuisine, for instance Chicken 65.
• The M-65 field jacket was commonly worn by American troops during the War in Vietnam.
• A 65th anniversary is sometimes referred to as a sapphire jubilee.
• The AGM-65 Maverick is a mass-produced air-to-ground tactical missile (AGM) designed for close air support against a wide range of targets.

## References

1. ^ Sloane, N. J. A. (ed.). "Sequence A001358". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
2. ^ "Sloane's A000567 : Octagonal numbers". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2016-05-30.
3. ^ "Sloane's A002064 : Cullen numbers". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2016-05-30.
4. ^ "Sloane's A028442 : Numbers n such that Mertens' function is zero". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2016-05-30.
5. ^ Sloane, N. J. A. (ed.). "Sequence A006003". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
6. ^
7. ^ "Sloane's A000931 : Padovan sequence". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2016-05-30.
8. ^ "Sloane's A008277 : Triangle of Stirling numbers of the second kind". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2021-12-24.
9. ^
10. ^
11. ^
12. ^ Sloane, N. J. A. (ed.). "Sequence A032020 (Number of compositions (ordered partitions) of n into distinct parts)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2022-05-24.


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# Factors Affecting Interest

We will discuss here what are the different factors affecting interest.

Interest depends on three factors, namely they are

1. Principal

2. Rate of Interest

3. Period of Time

‘P’ is the principal. ‘T’ is the time in years, ‘R’ is the rate of interest per annum and S.I. is simple interest, then we can calculate simple interest by using the following formulae.

S.I. = P × R × T/100

Amount: The sum which is borrowed, repays to the lender along with the interest after using the principal for a certain period is called the Amount.

Therefore, Amount = Principal + Interest

Interest = Amount - Principal

Principal = Amount - Interest

Note: (i) If the time is in months, change it into years by dividing the number of months by 12.

(ii) The present rate has to be changed into a fraction. For example, if the percent rate is 5%, write 5/100 for rate. Do not write 5 for the rate.

Solved examples on calculating interest:

1. Find the interest on $2500 at 8 ½ per annum for 4 ½ years? Principal =$ 2500

Rate = 8 ½ % per annum

= 17/2 % per annum

Time = 4 ½ years = 9/2 years

Interest = P × R × T/100

= $(2500 × 9/2 × 17/2)/100 =$ (2500 × 9 × 17)/(2 × 2 × 100)

= $956.25 2. Find the interest to be paid at the end of one year on$ 500 at 5% per annum.

I = P × R × T/100

= (500 × 5 × 1)/100

= $25 3. Find the interest to be paid at the end of one year on$ 900 at 6% per annum

I = P × R × T/100

= (900 × 6 × 1)/100

= $54 4. Find the interest to be paid at the end of one year on$ 1200 at 8 ½ % per annum

I = P × R × T/100

= (2000 × 17/2 × 1)/100

= (2000 × 17 × 1)/(2 × 100)

= \$ 170

Word Problems on Simple Interest.

In Simple Interest when the Time is given in Months and Days.

To find Principal when Time Interest and Rate are given.

To find Rate when Principal Interest and Time are given .

To find Time when Principal Interest and Rate are given.

Worksheet on Simple Interest.

Worksheet on Factors affecting Interest


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# Asymptotic behavior of the solution of the high degree differential equation $(x^{2n}y^{(n)})^{(n)}-x^2y=\lambda \; y$

The following differential equation has two independent solutions, one of the two is decreasing exponentially at infinity (k-Bessel function).

$$(x^2y')'-x^2y=\lambda \;y$$

Now for a higher-degree differential equation like:

$$(x^{2n}y^{(n)})^{(n)}-x^2y=\lambda \; y$$

We have $2n$ independent solutions. How can we find information on their asymptotic behavior near infinity? How many linearly independent solutions can we find that will decrease exponentially at infinity? (I guess $n$?) (any book of reference on this subject?)

If you want a quick and dirty way to find the asymptotics of $y(x)$ as $x \to \infty$, you can use the WKB ansatz $y(x) = e^{S(x)}$, with the hypothesis that $S^{(k)}/S' \to 0$ as $x\to \infty$ for all $k>1$. Substituting this form into your equation and keeping only the leading terms at infinity, you find $$x^{2n} (S')^{2n} + x^2 = 0 .$$ The solutions are $S'(x) = (-x^{2-2n})^{1/2n} = (-1)^{1/2n} x^{1/n-1}$ and hence the expected asymptotics for the $2n$ independent solutions are $y_k(x) \sim \exp(e^{i\frac{\pi}{2n}(1+2k)} n x^{1/n})$, $k=0,\ldots,2n-1$. In the case $n=1$, you get the asymptotics $y(x) \sim e^{\pm i x}$, and not $e^{\pm x}$ as your question suggested. You get the $e^{\pm x}$ asymptotics by flipping the sign of the $x^2 y$ term in the equation.
If you want to be more systematic, it helps to apply some transformations to the equation first. If you introduce $z = x^n y^{(n)}$, then the equation is equivalent to $$\begin{bmatrix} x^n y^{(n)} \\ (x^n z)^{(n)} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -x^2+\lambda & 0 \end{bmatrix} \begin{bmatrix} y \\ z \end{bmatrix} .$$ Next, it is convenient to rescale the dependent variables such that the leading power of $x$ on the right-hand side has a diagonalizable matrix as coefficient (if it's actually possible). Here, this looks like $$\begin{bmatrix} x^n (x y)^{(n)} + P(x\frac{d}{dx}) y \\ (x^n z)^{(n)} \end{bmatrix} = x \begin{bmatrix} 0 & 1 \\ -1+\frac{\lambda}{x} & 0 \end{bmatrix} \begin{bmatrix} x y \\ z \end{bmatrix} ,$$ where now $P(x\frac{d}{dx})$ is a constant coefficient polynomial in its argument, of order less than $n$. The last step is to substitute $x=t^{n}$, where the power of $t$ is chosen so that the resulting equation has asymptotically constant coefficients as $t\to \infty$. Since $x\frac{d}{dx} = \frac{1}{n} t\frac{d}{dt}$, the result is $$\begin{bmatrix} (t^n y)^{(n)} + t^{-n} Q(t\frac{d}{dt}) y \\ z^{(n)} + t^{-n} R(t\frac{d}{dt}) z \end{bmatrix} = n^n \begin{bmatrix} 0 & 1 \\ -1+\frac{\lambda}{t^n} & 0 \end{bmatrix} \begin{bmatrix} t^n y \\ z \end{bmatrix} ,$$ where again the constant coefficient polynomials $Q$ and $R$ are of order less than $n$. From here the theory of asymptotics of ODEs at irregular singular points tells us that the independent solutions of your equation will have the asymptotic expansions $y_k(t) = t^{-n} Y_k(t) (1 + O(1/t))$, where the last factor stands for an asyptotic series in inverse powers of $t$ and $Y_k(t)$ solves the asymptotic constant coefficient equation. In this case corresponds to $Y_k(t) = \exp(\alpha_k t)$, where the $\alpha_k$ are the different $n$-th roots of the eigenvalues of matrix $n^n [\begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix}]$, namely $\alpha_k = e^{i\frac{\pi}{2n}(1+2k)} n$, $k=0,\ldots,2n-1$, which reproduces the form of the $S(x)$ phase function from the WKB solution. But now we have more detailed information about the asymptotics.
• Exact there is a mistake in the sign in front of $x^2y$ to have solutions with an exponential decrease, I will correct it. Thanks for you answer. – Bertrand Oct 1 '17 at 14:24
A good reference on the subject is W. Wasow, Asymptotic expansions for ordinary differential equations. John Wiley & Sons, Inc., New York-London-Sydney 1965. It is not difficult to obtain the answer for small $n$, but computation for arbitrary $n$ may be complicated. Anyway, there is an algorithm of obtaining these asymptotics.


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# Fight Finance

#### CoursesTagsRandomAllRecentScores

A European bond paying annual coupons of 6% offers a yield of 10% pa.

Convert the yield into an effective monthly rate, an effective annual rate and an effective daily rate. Assume that there are 365 days in a year.

All answers are given in the same order:

$$r_\text{eff, monthly} , r_\text{eff, yearly} , r_\text{eff, daily}$$


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# Polynomial Properties

Algebra Level 3

Given two polynomials $p(x)$ and $q(x)$ such that they are cubic, which of the following statements doesn't hold true?

Note:-

$\bullet$ $d(P(x))$ stands for degree of the polynomial $p(x)$

$\bullet$This question is taken from TamilNadu-NTSE 2014 paper.

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# Solve integral equation for unknown constant

Consider the equations

$$\int_0^L \mathbf W(\mathbf u, s) \, \mathrm ds = \mathbf 0$$

where $0 \leq s \leq L$ and $\mathbf u$ is a vector of constants.

Numerically, what is the best way to determine $\mathbf u$ that satisfy the equations?

• I think this isn't what's usually called an integral equation: en.wikipedia.org/wiki/Integral_equation, it's just a regular equation that happens to involve an integral. In other words, define the function $f(u) = \text{l.h.s.}$, and just apply a regular multidimensional root finder to $f(u)=0$. – Kirill Oct 8 '16 at 1:22

Like what @Kirill says, write a script that defines $f(u)$ to be a function that approximates the integral. Thus all that you need to is solve $f(u)=0$ which can use standard root finding tools like Newton's method.

You're best choice for this is probably Julia. Using the built in quadgk function for Gauss-Kronrad quadrature, you can define the f function via:

f(u) = quadgk((x)->W(u,x),0,L)[1]

(the [1] is because quadgk returns a tuple where the first part is the integral and the second part is the error estimate, so this function ignores the error estimate and just returns the integral approximation). Note that (x)->W(u,x) is using what's known as a closure: it makes an anonymous function g(x) = W(u,x) where it sticks in the value of u, allowing you to easily integrate among the x. Since Julia v0.5 compiles the anonymous functions as generic functions, this will be very fast.

Then you just have to solve the root finding problem. I particularly like the library NLSolve. Since this function is not in place, we would use the command

using NLsolve
nlsolve(not_in_place(f),initial_u)

where initial_x is a guess for the final value of u. This will use finite differencing. We can many times make it faster and more accurate by enabling auto-differentiation:

nlsolve(not_in_place(f),initial_u,autodiff = true)