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User:Marshallsumter/Radiation astronomy/Entities

Radiation astronomy entities, radiation entities, are any astronomical persons or things that have separate and distinct existences in empirical, objective or conceptual reality.

Some of them, like the astronomers of today, or at any time in the past, are relatively known. But there are many entities that are far less known or understood, such as the observers of ancient times who suggested that deities occupied the sky or the heavens. Likewise, these alleged deities may be entities, or perhaps something a whole lot less.

Radiation sources and objects are also entities but better understood as sources or objects.

Geology
Def. an "expanse of space that seems to be [overhead] like a dome" is called the sky, or the sometimes the heavens.

'Sky' is an astronomical entity.

While it is arguable that any entities are controlling the sky, entities may be assigned as a first approximation in the theory of cause and effect, for example. Three comet-like objects occur in lecture images. An entity that produces comet-like objects may exist. The Sun emits visual radiation that may reflect off a comet's tail. The coronal cloud in close proximity to the Sun also emits X-rays that produce visual fluorescence from gases in a comet's coma and tail.

The Sun appears to move across the sky during the daytime only. An entity or two may be responsible for this, or may influence it in some way.

A storm cloud blocks the daytime sky while releasing rain. Perhaps a different entity may be assigned for each of these effects, or one for all.

Many people like gravity as an entity, or spacetime.

The Moon also crosses the sky occasionally sometimes in the daylight other times at night. The Moon doesn't always reflect uniformly during its travels. A shadow often blocks some of the reflection. Which entity is the cause for this, or is it an object, or perhaps a source of shadow?

The entity Thor (also called Jupiter in some cultures) is an entity assigned to throwing lightning bolts.

Theoretical astronomy tries to assess which entities may be responsible for controlling the sky away from a pleasant fair weather day. And, in turn seeks to explore those that may allow us to control the sky for a few more fair weather days.

Synchrotron radiation
Particle radiation consists of a stream of charged or neutral particles, from the size of subatomic elementary particles upwards of rocky and gaseous objects to even larger more loosely bound entities.

In astrophysics, x is usually a ratio of frequencies, that is, the frequency over a critical frequency (critical frequency is the frequency at which most synchrotron radiation is radiated). This is needed when calculating the spectra for different types of synchrotron emission. It takes a spectrum of electrons (or any charged particle) generated by a separate process (such as a power law distribution of electrons and positrons from a constant injection spectrum) and converts this to the spectrum of photons generated by the input electrons/positrons.

Planetary sciences
There are many other natural objects, entities, bodies, or phenomena that occur in the sky. Some of these may occur frequently: the Sun passes overhead every day, so does the Moon either during the day or at night, a variety of clouds pass across the sky and sometimes completely fill the sky for days, occasionally a few go in the opposite direction across the sky or in different directions.

Usually when clouds fill the sky and associated with some of these clouds is lightning, a phenomenon that moves so quickly it’s difficult to think of it as an object or entity with a body.

Theoretical radiation entities
The overall theory of astronomy consists of three fundamental parts:
 * 1) the derivation of logical laws,
 * 2) the definitions of natural bodies (entities, sources, or objects), and
 * 3) the definition of the sky (and associated realms).

On the basis of dictionary definitions, what is the difference between a 'body', an 'entity', an 'object', a 'thing', and a 'phenomenon'?

The categories for synonymy and most common usage place 'body' in "3. SUBSTANTIALITY", 'entity' in the same, 'object' in "651. INTENTION" , 'thing' in "3. SUBSTANTIALITY" , and 'phenomenon' in "918. WONDER". The more common uses of the words 'object' and 'phenomenon' are not exactly the same as may be used in a specialized endeavor like a science such as astronomy. A slightly less common use of 'phenomenon' is in category "150. EVENTUALITY". For the word 'object' a slightly less common or popular meaning is in category "543. MEANING". The closest category of meaning or synonymy for 'object' to category 1. is category "375. MATERIALITY".

Of each of these words, 'entity' uses the word 'existence', category "1. EXISTENCE" in each definition. 'Entity' may be thought of as the most general of these terms because its meanings are the closest to category 1. The farthest from category 1. on the basis of conceptual meaning and synonymy is the word 'object' in category 375. A tentative order is 'entity' > 'phenomenon' > 'object' by generalness, or by preciseness (perhaps more description is needed beyond only existence) 'object' > 'phenomenon' > 'entity'.

'Thing' (category 3.) has the word 'entity' in three of four meanings and 'object' in the fourth. The second most popular meaning of 'thing' is in category 375.

'Body' (category 3.) has 'mass' and is closer to 'substantiality' in common usage than 'thing', and neither word has a synonym closer in meaning to 'existence'. The second most common meaning of 'body' is category "203. BREADTH, THICKNESS".

This suggests a hierarchy such as 'entity' > 'body' > 'thing' > 'phenomenon' > 'object' by generalness, where 'existence' is the most general word; or, 'object' > 'phenomenon' > 'thing' > 'body' > 'entity' by preciseness. An 'astronomical object' may be expected to require a more detailed description in its definition to indicate meaning than an 'astronomical entity'. In astronomy, the concept of an 'astronomical body' may suggest a meaning closer to category 203. rather than a 'thing' or 'entity'.

The choice of general order is 'entity' > 'source' > 'object' > 'phenomena'. The term 'astronomical body' has much less popularity per Google scholar than 'object'. The body of astronomers in the International Astronomical Union is auspicious and here is considered an astronomical entity.

Def. the theory of the science of the biological, chemical, physical, and logical laws (or principles) with respect to any natural being, body, thing, entity, source, object, or phenomena in the sky especially at night is called theoretical astronomy.

Physics deals with forces, fields, energy, kinetics, and radiation. Astronomy has its own laws with respect to entities or bodies in motion. Application of a field to an astronomical phenomenon may clarify what is happening. That's the focus of astrophysics. Theory is needed to bring the physics in line with the magnitude of the situation and its complexity.

Entity astronomy
Def.
 * 1.a: an "independent, separate, or self-contained [astronomical] existence",
 * 1.b: "the [astronomical] existence of a [person, place, or] thing as contrasted with its attributes", or
 * 2. "some [astronomical] thing that has separate and distinct existence and objective or conceptual reality",

is called an astronomical entity.

Def.
 * 1) that which has a distinct existence as an individual unit,
 * 2) an existent something that has the properties of being real, and having a real existence, or
 * 3) the state or quality of being or existence is called an entity.

"I shall introduce the themes with several quotations which are representative of a number of popular contemporary writings on cosmogony. ... First, they each personify an entity or concept-'the laws of physics', 'chance' and 'natural selection' respectively."

Two astronomical entities are shown in the images at left: the supernova remnant CTA 1 and the Natal Microcosm, the latter is an image from the Spitzer space telescope.

Semantics
Def.
 * 1.a: an "independent, separate, or self-contained existence",
 * 1.b: "the existence of a thing as contrasted with its attributes", or
 * 2. "something that has separate and distinct existence and objective or conceptual reality",

is called an entity.

Def.
 * 1.a: "something that is or is capable of being seen, touched, or otherwise sensed",
 * 1.b: "something physical or mental of which a subject is cognitively aware",
 * 2. "something that arouses an emotion in an observer", or
 * 3. "a thing that forms an element of or constitutes the subject matter of an investigation or science"

is called an object.

Def.
 * 1.a: "a mass of matter distinct from other masses" or
 * 2.b: "something that embodies or gives concrete reality to a thing; [specifically] : a sensible object in physical space"

is called a body.

Def.
 * 1.a: "a separate and distinct individual quality, fact, idea, or [usually] entity",
 * 1.b: "the concrete entity as distinguished from its appearances",
 * 1.c: "a spatial entity", or
 * 1.d: "an inanimate object distinguished from a living being"

is called a thing.

Def.
 * 1: "an observable fact or event",
 * 2.a: "an object or aspect known through the senses rather than by though or intuition",
 * 2.b: "an object of experience in space and time as distinguished from a thing-in-itself", or
 * 2.c: "a fact or event of scientific interest susceptible of scientific description and explanation"

is called a phenomenon.

Such words as "entity", "object", "thing", and perhaps "body", words "connoting universal properties, constitute the very highest genus or "summum genus"" of a classification of universals. To propose a definition for say a plant whose flowers open at dawn on a warm day to be pollinated during the day time using the word "thing", "entity", "object", or "body" seems too general and is. But, astronomy deals with the universe, sometimes only the very local universe just above the Earth's atmosphere. These "universals" may be just the words to use.

Astronomers
Observers and astronomers who make or made the records are entities.

Johannes Kepler (December 27, 1571 – November 15, 1630) was a German mathematician, astronomer and astrologer. A key figure in the 17th century scientific revolution, he is best known for his eponymous laws of planetary motion, codified by later astronomers, based on his works Astronomia nova, Harmonices Mundi, and Epitome of Copernican Astronomy.

Laurence E. Peterson is Emeritus Professor of Physics and Director of the Center for Astrophysics and Space Sciences at the University of California, San Diego, California. He was a pioneer in the field of X-ray astronomy. He led the high-energy astronomy group at UCSD for many years. In addition to carrying out numerous experiments using high-altitude balloons, he was principal investigator on several NASA satellite experiments, including one on the OSO 1, one on OSO 3, two on OSO 7, the A4 experiment on HEAO 1, and co-investigator on the The High Energy X-ray Timing Experiment (HEXTE) flown on the Rossi X-ray Timing Explorer.

Articles
Each of the articles in the literature focusing on any entity from astronomy is itself an astronomical entity. Where the entity is from radiation astronomy, it is a radiation astronomy entity.

Journals
Astronomical entities include some journals (such as The Astrophysical Journal, the Monthly Notices of the Royal Astronomical Society, and Astronomy & Astrophysics), articles in journals and magazines, books on astronomy that may be references or be cited for astronomy information or facts.

Books
At right is a copy of the title page of Johannes Kepler's Rudolphine Tables (1627). It is regarded as the most accurate and comprehensive star catalogue and planetary tables published up until that time. It contained the positions of over 1000 stars and directions for locating the planets within our solar system. Kepler finished the work in 1623 and dedicated it to his patron, the Emperor Rudolf II, but actually published it in 1627. The table's findings support Kepler's laws and the theory of a heliocentric astronomy.

Information processing
For information processing, astronomical 'being', 'body', 'something' or 'thing' are also astronomical entities.

Types of entities for Natural Language Processing (NLP):
 * 1) names - person, location, organization;
 * 2) temporal expressions - date, time;
 * 3) numeric expressions - money, percent;
 * 4) instrument name;
 * 5) source name;
 * 6) source type;
 * 7) spectral feature; and
 * 8) text and scientific databases.

Databases
"Astronomy is a broad scientific domain combining theoretical, observational and computational research, which all differ in conventions and jargon." "There is a major effort in astronomy to move towards integrated databases, software and telescopes." ("The Virtual Observatory" ).

Artifacts
The eyes of these statues contain rock-crystals. "These are probably the oldest portrait statues in the world. These people who sit before us side by colored to the life, fresh and glowing as the day they gave the artist his last sitting lived at a time when the great pyramids were not yet built and at a date which is variously calculated as from about 4,000 to 6,300 years from the present day. The princess wears her hair precisely as it is still worn in Nubia and her necklace is of a pattern still favored. The eyes of both statues are inserted. The eyeball which is set in an eyelid of bronze, is made of opaque white quartz with an iris of rock-crystal enclosing a pupil of some kind of brilliant metal. This treatment gives to the eyes a look of intelligence which is almost appalling." Amelia B. Edwards "These incomparable statues are most expressive and stand in vitality to the works of any later age in Egypt. They were found in the tomb chamber- Ra-Ho-tep is entitled a royal son [probably of Seneferu]- The signs carved in these tombs are among the earliest known. Instead of full-length burial with coffins, head rests, vases, and provision for a future life, the more usual method of burial at Medum is lying on the left side with the knees drawn up facing the east and without vases or other objects, showing a diversity of beliefs and probably of races." W.M. Flinders Petrie.

"The composition of these eyes is a lens of polished rock crystal (either alpha silica or fused silica, formerly known as cystalline quartz and fused quartz which had a convex front surface and a near hemispherical concave ground pupil surface in a flat iris plane (normally covered with resin) at the rear of the lens."

The middle image is a painting by artist Giorgio Vasari (1511–1574). Behind and above the main focus on Cronus (Saturn) castrating Uranus (the Greek sky god before Zeus) is what appears to be "some kind of Armillary sphere."

Even more than that is an object angled slightly upward (right to left) that appears to be pointed at or above the large metallic ring. This object seems to be painted as if it extends through the top hole of the armillary sphere. At the upper left end of the object is an embellished end piece that may have a lens (objective lens) or piece of material the diameter of which matches that of the tube behind the embellishment. The midsection appears to taper to a smaller diameter near the right end. The right end has an even smaller piece attached or inserted that looks like an eyepiece lens holder. This tubular object seems to be receiving hands-in-the-air praise from two people to its left. The armillary sphere may be tilted ~23 1/2°.

The tubular object bares a striking resemblance to a telescope. The image taken as a whole suggests an interaction between Cronus and Uranus as observed near the plane of the ecliptic by someone through a telescope.

The Castration of Uranus is apparently a fresco by Vasari & Cristofano Gherardi (c. 1560, Sala di Cosimo I, Palazzo Vecchio, Florence).

Usually a ball representing the Earth or, later, the Sun is placed in its center. It is used to demonstrate the motion of the stars around the Earth. Before the advent of the European telescope in the 17th century, the armillary sphere was the prime instrument of all astronomers in determining celestial positions.

"The Keplerian Telescope, invented by Johannes Kepler in 1611, is an improvement on Galileo's design. A diagram of one with planoconvex lenses is shown at right.

The lowest right image is a woodcut illustration of an aerial telescope. Johannes Hevelius built [an air telescope] with a 45 m (150 ft) focal length and even longer tubeless "aerial telescopes" were constructed.

Named entities
"Astronomical named entities":
 * 1) "Names of telescopes and other measurement instruments,"
 * 2) "Names of celestial objects,"
 * 3) "Types of objects," and
 * 4) "Features that can be pointed to on a spectrum".

At top right: An observation post, temporary or fixed, is any preselected position from which observations are to be made - this may include very temporary installations ... or even an airborne aircraft.

At right the Very Large Telescope (VLT) is a telescope operated by the European Southern Observatory on Cerro Paranal in the Atacama Desert of northern Chile. The VLT consists of four individual telescopes, each with a primary mirror 8.2m across, which are generally used separately but can be used together to achieve very high spatial resolution.

Named entity recognitions
"Gazetteers are useful for finding commonly referenced names of people, places or organisations" associated with astronomy. These are astronomical entities that can be used for information processing.

Named entity recognition (NER) for astronomy literature: NER "involves assigning broad semantic categories to entity references in text."

Dominant group/Astronomy
Notation: let the English capital letter Z stand for atomic number, as the number of protons in a chemical element's isotope atomic nucleus.

Notation: let various International System of Units, SI prefixes occur before electronvolt abbreviated as eV.

For example, PeV denotes 1015 eV.

"The most dominant group is the iron group (Z = 25 - 27), at energies around 70 PeV more than 50% of the all-particle [cosmic-ray] flux consists of these elements."

"The X-ray luminosity of the dominant group is an order of magnitude fainter than that of the X-ray jet."

The term "dominant group" is used in astronomy to identify other astronomical entities of importance. The genera differentia for possible definitions of "dominant group" fall into the following set of orderable pairs:

'Orderable' means that any synonym from within the first category can be ordered with any synonym from the second category to form an alternate term for "dominant group"; for example, "superior class", "influential sect", "master assembly", "most important group", and "dominant painting". "Dominant" falls into category 171. "Group" is in category 61. Further, any word which has its most or much more common usage within these categories may also form an alternate term, such as "ruling group", where "ruling" has its most common usage in category 739, or "dominant party", where "party" is in category 74.

dominant group

"A particular subject of interest is the cluster ion series (NH3)nNH4+, since it is the dominant group of ions over the whole investigated temperature range." For astrochemisty, "[t]hese studies are expected to throw light on the sputtering from planetary and interstellar ices and the possible formation of new organic molecules in CO--NH3–H2O ice by megaelectronvolt ion bombardment."

All alternate terms for "dominant group" used in astronomy are astronomical entities. Here are some examples from the literature:

master class

 * 1) "Once created, device class objects are registered with an instance of the master class."
 * 2) "For ATIC, a possible set of defined classes would be a master class event, and sub-classes header, silicon, scintillator, bgo and track."

superior size

 * 1) "The superior size and albedo of Venus completely turn the scale, with the result that Venus at her brightest is about 12 times brighter than Mercury at his brightest."
 * 2) "There is no reason to question but that they are simply ordinary meteors, which from their superior size and unusually slow speed have survived to reach the earth's surface."

influential set

 * 1) "Together with Leonard Searle, he wrote an influential set of papers which established that stellar disks are truncated at about four exponential scale-lengths, and that the vertical scale-height of disks is constant with radius."
 * 2) "Until now Themo has been best known for an influential set of questions on Aristotle's Meteorologica, which is closely related to similar sets by Nicole Oresme and, putatively, Simon Tunsted."

Astronomical sources
Def. an entity from which something comes or is acquired is called a source.

The visible light we see [from the outer surface of the photosphere] is produced as electrons react with hydrogen atoms to produce H− ions. This indicates that the outer surface of the Sun's photosphere is a primary source of visual radiation.

Source astronomy has its origins in the actions of intelligent life on Earth when they noticed things or entities falling from above and became aware of the sky. Sometimes what they noticed is an acorn or walnut being dropped on them or thrown at them by a squirrel in a tree. Other events coupled with keen intellect allowed these life forms to deduce that some entities falling from the sky are coming down from locations higher than the tops of local trees. It may have taken awhile to realize that the sky is penetrable.

Direct observation and tracking of the origination and trajectories of falling entities such as volcanic bombs presented early intelligent life with vital albeit sometimes dangerous opportunities to compose the science that led to source astronomy.

Def. an entity from which something comes or is acquired is called a source.

Constellations
The constellations, past and present, are astronomical entities. At right is a equirectangular plot of declination vs right ascension of the modern constellations, colour-coded by family and year established, with a dotted line denoting the ecliptic. Note that the right ascension axis increases from right to left to approximate the view of the night sky.

The ecliptic currently is in the following constellation: Aquarius, Capricornus, Sagittarius, Scorpius, Ophiuchus, Libra, Virgo, Leo, Cancer, Gemini, Taurus, Aries, and Pisces.

The Milky Way is shown as the fuzzy bitmap at higher, then lower declinations relative to the ecliptic.

The Galactic plane passes through Cepheus, Lacerta, Cygnus, Vulpecula, Sagitta, Aquila, Serpens Cauda, Scutum, Ophiuchus, Sagittarius, Scorpius, Ara, Norma, Circinus, Centaurus, Crux, Carina, Vela, Puppis, Canis Major, Monoceros, Orion, Gemini, Taurus, Auriga, Camelopardalis, Perseus, and Cassiopeia.

Objects
Included as astronomical entities are 'astronomical objects' and 'astronomical sources', even those with large error regions of whole degrees. Diffuse background radiations, from gamma ray to radio, are astronomical entities.

Entity categories include 'galaxy', 'nebula', 'star', 'star cluster', 'supernova', 'planet', 'frequency', 'duration', 'luminosity', 'position', 'telescope', 'ion', 'survey', and 'date'.

At top left is an apparent result of meteoritic radiation. This is an aerial view of the Barringer Meteor Crater about 69 km east of Flagstaff, Arizona. "Meteor Crater is a meteorite impact crater approximately 43 miles (69 km) east of Flagstaff, near Winslow in the northern Arizona desert of the United States. Because the US Department of the Interior Division of Names commonly recognizes names of natural features derived from the nearest post office, the feature acquired the name of "Meteor Crater" from the nearby post office named Meteor. The site was formerly known as the Canyon Diablo Crater, and fragments of the meteorite are officially called the Canyon Diablo Meteorite. Scientists refer to the crater as Barringer Crater in honor of Daniel Barringer, who was first to suggest that it was produced by meteorite impact.

The Holsinger meteorite is the largest discovered fragment of the meteorite that created Meteor Crater and it is exhibited in the crater visitor center.

The Canyon Diablo meteorite comprises many fragments of the asteroid that impacted at Barringer Crater (Meteor Crater), Arizona, USA. Meteorites have been found around the crater rim, and are named for nearby Canyon Diablo, which lies about three to four miles west of the crater. There are fragments in the collections of museums around the world including the Field Museum of Natural History in Chicago. The biggest fragment ever found is the Holsinger Meteorite, weighing 639 kg, now on display in the Meteor Crater Visitor Center on the rim of the crater.

Coronal clouds
Def. a cloud, or cloud-like, natural astronomical entity, composed of plasma is called a coronal cloud.

Historical evidence exists for the phrase coronal cloud with respect to "a cloud, or cloud-like, natural astronomical entity, composed of plasma". For now this may serve as a working definition.

Holes
Def. a hollow spot in a surface is called a hole.

Def. an opening in a solid, liquid, gas, or plasma is called a hole.

Def. a round or irregular patch on or apparently on the surface of an entity having a different color, texture etc. and generally round in shape is called a spot.

Def. an opening through which entities such as gases can pass is called a vent.

Sunspots are temporary phenomena on the photosphere of the Sun that appear visibly as dark spots compared to surrounding regions.

The Lockman Hole is an area of the sky in which minimal amounts of neutral hydrogen gas are observed. Clouds of neutral hydrogen glow faintly with infrared light and obscure distant views at extreme ultraviolet and soft x-ray wavelengths. They interfere with observations at those wavelengths in nearly all other directions since they are common in our galaxy. So the Lockman Hole serves as a relatively clear window on distant objects, which makes it an attractive area of the sky for observational astronomy surveys. It is located near the pointer stars of the Big Dipper in the constellation Ursa Major and is about 15 square degrees in size.

Voids
"[V]oids [are] now considered as regular astronomical entities in their own rights, [and] are clustered."

Variety
Each section in astronomy mentions or shows images of astronomical entities:
 * 1) Barringer Meteor Crater,
 * 2) Sun,
 * 3) tides,
 * 4) Moon,
 * 5) Earth (as a planet),
 * 6) Aurora Borealis,
 * 7) Lightning,
 * 8) Clouds,
 * 9) Meteors,
 * 10) Astronomical objects,
 * 11) Astronomical sources,
 * 12) astronomers and observers,
 * 13) Planets,
 * 14) Nebulas,
 * 15) Galaxies,
 * 16) calendars, star charts,
 * 17) Constellations,
 * 18) Telescopes and other technology,
 * 19) Observatories,
 * 20) physical astronomy and Astrophysics,
 * 21) Astrodesy, Astrognosy, and Astrometry,
 * 22) Dominant groups, Orbits and logical laws,
 * 23) Chemicals, Materials, and Ions, etc.

Generalness
In terms of meaning and generalness: 'being' > 'body' > 'something' or 'thing' > 'entity'.

Modern computing software has enabled new methods of creating astro art, and the internet has enabled new methods of sharing it. Digital photography has expanded and improved, increasing the number of devices that are able to capture astronomical beings with precision and clarity.

An astronomical body's Hill sphere is the region in which it dominates the attraction of satellites. To be retained by a planet, a moon must have an orbit that lies within the planet's Hill sphere. That moon would, in turn, have a Hill sphere of its own. Any object within that distance would tend to become a satellite of the moon, rather than of the planet itself.

The poles of astronomical bodies are determined based on their axis of rotation in relation to the celestial poles of the celestial sphere.

Agrippa (unkn-fl. 92 AD) was a Greek astronomer. The only thing that is known about him regards an astronomical observation that he made in 92 AD, which is cited by Ptolemy (Almagest, VII, 3). Ptolemy writes that in the twelfth year of the reign of Domitian, on the seventh day of the Bithynian month Metrous, Agrippa observed the occultation of a part of the Pleiades by the southernmost part of the Moon.

The deities worshipped are unknown, but were perhaps connected to water, or to astronomical entities (Sun, Moon, solstices).

In Western horoscopic astrology the interpretation of a horoscope is governed by:
 * The position of deduced astronomical entities, such as the Lunar nodes.

Relativistic jets
Relativistic jets are extremely powerful jets of plasma which emerge from presumed massive objects at the centers of some active galaxies, notably radio galaxies and quasars. Their lengths can reach several thousand or even hundreds of thousands of light years. The hypothesis is that the twisting of magnetic fields in the accretion disk collimates the outflow along the rotation axis of the central object, so that when conditions are suitable, a jet will emerge from each face of the accretion disk. If the jet is oriented along the line of sight to Earth, relativistic beaming will change its apparent brightness. The mechanics behind both the creation of the jets and the composition of the jets are still a matter of much debate in the scientific community; it is hypothesized that the jets are composed of an electrically neutral mixture of electrons, positrons, and protons in some proportion.

Meteors
At left is a radiated object and its associated phenomena.

Ultra-violet studies of Mira by NASA's Galaxy Evolution Explorer (Galex) space telescope have revealed that it sheds a trail of material from the outer envelope, leaving a tail 13 light-years in length, formed over tens of thousands of years. It is thought that a hot bow-wave of compressed plasma/gas is the cause of the tail; the bow-wave is a result of the interaction of the stellar wind from Mira A with gas in interstellar space, through which Mira is moving at an extremely high speed of 130 kilometres/second (291,000 miles per hour). The tail consists of material stripped from the head of the bow-wave, which is also visible in ultra-violet observations. Mira's bow-shock will eventually evolve into a planetary nebula, the form of which will be considerably affected by the motion through the interstellar medium (ISM).

Neutrons
Neutron stars are an entity of theoretical astrophysics. There does not appear to be any direct way using neutron astronomy to successfully detect neutron stars.

A "new type of neutron star model (Q stars) [is such that] high-density, electrically neutral baryonic matter is a coherent classical solution to an effective field theory of strong forces and is bound in the absence of gravity. [...] allows massive compact objects, [...] and has no macroscopic minimum mass."

"Compact objects in astronomy are usually analyzed in terms of theoretical characteristics of neutron stars or black holes that are based upon calculations of equations of state for matter at very high densities. At such high densities, the effects of strong forces cannot be neglected. There are several conventional approaches to describing nuclear forces, all of which find that for a baryon number greater than ~250, a nucleus will become energetically unbound. High-density hadronic matter is not stable in these theories until there are enough baryons for gravitational binding to form a neutron star, typically with a minimum mass ≳ 0.1 M⊙ and maximum mass ≲ 3 M⊙."

"[T]he huge number of neutrinos [a neutron star] emits carries away so much energy that the temperature falls within a few years [after formation] to around 106 kelvin. Even at 1 million kelvin, most of the light generated by a neutron star is in X-rays. In visible light, neutron stars probably radiate approximately the same energy in all parts of visible spectrum, and therefore appear white.

A neutron star is a theoretical radiation source. It is a type of stellar remnant [(a compact star)] that can result from the gravitational collapse of a massive star during a Type II, Type Ib or Type Ic supernova event. Such stars are composed almost entirely of neutrons.

Neutron stars are theorized as the radiation source for anomalous X-ray pulsars (AXPs), binary pulsars, high-mass X-ray binaries, intermediate-mass X-ray binaries, low-mass X-ray binaries (LMXB), pulsars, and soft gamma-ray repeaters (SGRs).

Protons
"Deuterated isotopomers of methanol have been detected both in hot cores and in the protostellar source IRAS 16293-2422. [...] In studying the post-evaporative gas-phase chemistry of these isotopomers, it is important to know if pairs of isotopomers with D atoms in different places (eg CH3OD and CH2DOH) can be interconverted or whether they can be viewed as separate entities with depletion mechanisms that are independent of each other. Here we show that it is difficult to exchange protons and deuterons on the two different parts of the methanol backbone."

Electrons
"The electron is a subatomic particle with a negative charge, equal to -1.60217646x10-19 C. Current, or the rate of flow of charge, is defined such that one coulomb, so 1/-1.60217646x10-19, or 6.24150974x1018 electrons flowing past a point per second give a current of one ampere. The charge on an electron is often given as -e. note that charge is always considered positive, so the charge of an electron is always negative."

Positrons
There may be a "connection ... between the magnetic field strengths inside an electron, in newly-born pulsars, and the sun. ... the upper limit to the strength of magnetic field ... is that which would permit emission of a photon at the non-relativistic electron gyrofrequency, with the energy of the order of the electron rest mass."

A "basic process in the formation of pulsar magnetic fields [may be] a variant of electron-positron spin-zero annihilation, as follows
 * $$ e^-\uparrow + e^+\uparrow \, \rightarrow \, \uparrow \cup \uparrow + \gamma + \gamma,$$

where the [up] arrow represents the magnetic moment of an electron.

This relation "symbolises the formation of a magnetic entity, $$\uparrow \cup \uparrow$$, here called an M-particle, with twice the magnetic moment of an electron or a positron, and [γ] represents a photon."

Neutrinos
"For three quarters of a century, neutrinos have proven the most ghostly of all the quantum entities that make up the universe."

Gamma rays
Def. "the ratio of the area of causally connected regions that become active to the observable area of the shell" is called the surface filling factor.

"In the external shock model [for gamma-ray bursts] ... multiple peaks ... [may arise] because various patches (or emitting "entities") on the shell randomly become active. ... [S]o many entities [can] become simultaneously active that the overall envelope appears quite smooth [on the other hand when] fewer entities become active, ... random fluctuations in the number of simultaneously active entities cause the peak structure to be spiky. ... [E]ach observed peak is not necessarily caused by a single entity, but the peak structure is caused by random variations in the number of active entities".

X-rays
Serpens X-1 is an X-ray source with an error circle fixed for all time on the celestial sphere. It is also an X-ray entity in the sense that it has an independent, separate, or self-contained astronomical existence. It has a history, a spatial extent, and a spectral extent.

In 2001: "The first class is central-dominant group galaxies, which are very X-ray luminous and may be the focus of group cooling flows."

Ultraviolets
Charles Stuart Bowyer is generally given credit for starting this field ultraviolet astronomy.

Blues
In Egypt, blue was associated with the sky and with divinity. The Egyptian god Amun could make his skin blue so that he could fly, invisible, across the sky. Blue could also protect against evil; many people around the Mediterranean still wear a blue amulet, representing the eye of God, to protect them from misfortune.

Many of the gods in Hinduism are depicted as having blue-coloured skin, particularly those associated with Vishnu, who is said to be the Preserver of the world and thus intimately connected to water. Krishna and Ram, Vishnu's avatars, are usually blue. Shiva, the Destroyer, is also depicted in light blue tones and is called neela kantha, or blue-throated, for having swallowed poison in an attempt to turn the tide of a battle between the gods and demons in the gods' favour.

Huitzilopochtli at second right was the patron god of the Mexica tribe. Originally he was of little importance to the Nahuas, but after the rise of the Aztecs, Tlacaelel reformed their religion and put Huitzilopochtli at the same level as Quetzalcoatl, Tlaloc, and Tezcatlipoca, making him a solar god.

Over the South presides the Blue Tezcatlipoca, Huitzilopochtli, the god of war.

Rama imaged at left is the seventh avatar of the god Vishnu in Hinduism, and a king of Ayodhya in Hindu scriptures. In a few Rama-centric sects, Rama is considered the Supreme Being, rather than an avatar. Rama was born in Suryavansha (Ikshvaku Vansham) later known as Raghuvansha after king Raghu. When depicted with his brother Lakshman and consort Sita, with Hanuman kneeling in a state of prayer, this form is called Ram Parivar, and is the typical fixture depicting Rama in Hindu mandirs, or temples. The Hindi word parivar translates as "family."

Gaseous objects
"It is generally believed that stellar complexes appear to be the result of the evolution of gaseous superclouds which are the largest in size and mass (up to ~107 M⊙), entities of diffuse matter distributed in the galactic disks. The “top-down” mechanism of gravitational instability assumes that these [gaseous] superclouds are the first entities formed whereas the denser star-forming clouds are developing inside them [...] The "top-down" scenario implies the presence of a few fundamental scales of stellar groupings and the hierarchical arrangement of the developed structures."

Rocky objects
"There is no correlation between n (the exponent for the change in particle diameter) and any of the other entities, which shows that n-values cannot be used as an index of maturation of a soil."

 * $$ C_D = S (D/D_0)^{-n},$$

where $$C_D$$ is change in particle diameter, $$S$$ is the constant of surface area rate of change, $$D$$ is the measured particle diameter in the soil, $$D_0$$ is the initial particle diameter for all samples, and $$n$$ is the exponent.

Interplanetary mediums
A double layer is a structure in a plasma and consists of two parallel layers with opposite electrical charge. The sheets of charge cause a strong electric field and a correspondingly sharp change in voltage (electrical potential) across the double layer. Ions and electrons which enter the double layer are accelerated, decelerated, or reflected by the electric field. In general, double layers (which may be curved rather than flat) separate regions of plasma with quite different characteristics. Double layers are found in a wide variety of plasmas, from discharge tubes to space plasmas to the Birkeland currents supplying the Earth's aurora, and are especially common in current-carrying plasmas. Compared to the sizes of the plasmas which contain them, double layers are very thin (typically ten Debye lengths), with widths ranging from a few millimeters for laboratory plasmas to thousands of kilometres for astrophysical plasmas.

Double layers are formed in four main ways

Double layers are associated with (filamentary) currents,

Other names for a double layer are electrostatic double layer, electric double layer, plasma double layers, electrostatic shock (a type of double layer which is oriented at an oblique angle to the magnetic field in such a way that the perpendicular electric field is much stronger than the parallel electric field), space charge layer. In laser physics, a double layer is sometimes called an ambipolar electric field.

If there is a net current present, then the DL is oriented with the base of the L in line with direction of current.

The details of the formation mechanism depend on the environment of the plasma (e.g. double layers in the laboratory, ionosphere, solar wind, fusion, etc.).
 * 1) Between plasmas of different temperatures
 * 2) By pinches in cosmic plasma regions
 * 3) By an electrical discharge

"Since the double layer acts as a load, there has to be an external source maintaining the potential difference and driving the current. In the laboratory this source is usually an electrical power supply, whereas in space it may be the magnetic energy stored in an extended current system, which responds to a change in current with an inductive voltage".

Intergalactic medium
The production of a double layer requires regions with a significant excess of positive or negative charge, that is, where quasi-neutrality is violated. In general, quasi-neutrality can only be violated on scales of the order of the Debye length. The thickness of a double layer is of the order of ten Debye lengths, which is a few centimeters in the ionosphere, a few tens of meters in the interplanetary medium, and tens of kilometers in the intergalactic medium.

History
The Hominidae have apparently been on Earth for around seven million years, at least somewhere in Africa and possibly elsewhere. Fortunately and deliberately, many of these have worked out ways to record knowledge about the objects or entities in the sky observed by the radiation they produce.

Mathematics
An initial use of mathematics in astronomy is counting entities, sources, or objects in the sky.

While in astronomy most entities have names, demonstrating that one or more numbers, shapes, structures, or changes are associated with an entity is evidence of proof of concept that mathematics is applicable to astronomy.

Usually, in astronomy, a number is associated with a dimension or aspect of an entity. For example, the Earth is 1.50 x 108 km on average from the Sun. Kilometer (km) is a dimension and 1.50 x 108 is a number.

In theoretical astronomy, whether the Earth moves or not, serving as a fixed point with which to measure movements by objects or entities, or there is a solar system with the Sun near its center, is a matter of simplicity and calculational accuracy.

Copernicus's theory provided a strikingly simple explanation for the apparent retrograde motions of the planets—namely as parallactic displacements resulting from the Earth's motion around the Sun—an important consideration in Johannes Kepler's conviction that the theory was substantially correct.

The altitude of an entity in the sky is given by the angle of the arc from the local horizon to the entity.

The observations require precise measurement and adaptations to the movements of the Earth, especially when and where, for a time, an object or entity is available.

With the creation of a geographical grid, an observer needs to be able to fix a point in the sky. From many observations within a period of stability, an observer notices that patterns of visual objects or entities in the night sky repeat. Further, a choice is available: is the Earth moving or are the star patterns moving? Depending on latitude, the observer may have noticed that the days vary in length and the pattern of variation repeats after some number of days and nights. By choosing an equal day/night position among the fixed objects in the night sky, the observer can measure equatorial coordinates: declination (Dec) and right ascension (RA).

Once these can be determined, the apparent absolute positions of objects or entities are available in a communicable form. The repeat pattern of (day/night)s allows the observer to calculate the RA and Dec at any point during the cycle for a new object, or approximations are made using RA and Dec for recognized objects.

Sciences
Superfluid vacuum theory is an approach in theoretical physics and quantum mechanics where the physical vacuum is viewed as superfluid. The ultimate goal of the approach is to develop scientific models that unify quantum mechanics (describing three of the four known fundamental interactions) with gravity. This makes SVT a candidate for the theory of quantum gravity and an extension of the Standard Model. It is hoped that development of such theory would unify into a single consistent model of all fundamental interactions, and to describe all known interactions and elementary particles as different manifestations of the same entity, superfluid vacuum.

EDP Sciences is a publishing group gathering several entities:
 * EDP Sciences, the publishing partner of the scientific communities;
 * EDP Santé, the is the medical branch of the company;
 * EDP Open, the platform for open access journals.

Based on existing knowledge accumulated through previous missions, new science questions are articulated. Missions are developed in the same way an experiment would be developed using the scientific method. In this context, Goddard does not work as an independent entity but rather as one of the 10 NASA centers working together to find answers to these scientific questions.

Technology
Observatory geology has two forms: the geological study necessary to put an observatory on a solid foundation to maximize telescope function through minimizing ground-based vibration and imaging terrain so that geological study may be performed.

There are "a plethora of observations from heavenly bodies which did not agree with each other despite being from the same astronomical entities."

"In order that the [Virtual Astronomical Observatory] VAO would be seen as an entity that is of and for the research community, a dedicated not-for-profit company was established to manage the governance and business functions."

"The U.S. Virtual Astronomical Observatory (VAO; http://www.us-vao.org/) [...] goal is to enable new science through efficient integration of distributed multi-wavelength data."

Detectors
Def. "the fraction of photoelectric events which end up in the photopeak of the measured energy spectrum" is called the photopeak efficiency (ε).

Ending up in the photopeak means within ± 1 full-width at half maximum (FWHM) of the peak of the distribution.

"The peak to valley ratio is commonly used as a token for ε."

"Another common practice is to fit an exponential function to the “valley” and to extrapolate the fit to lower pulse heights to estimate the fraction of counts hidden in the Compton continuum."

"We have used a calibrated Cs137 source to determine the absolute photopeak efficiency at 662 keV. The source was placed at a sufficiently large distance from the detector so that the event rate was low and the dead time was less than 20%. Based on a log-histogram of the time intervals between events, the dead-time has been estimated to a fractional accuracy of better than 5%. We determine the photopeak efficiency by comparing the dead-time corrected event rate in the photopeak with the theoretical expectation assuming a perfect detector."

Def. the average energy loss of the particle per unit path length is called the stopping power.

Def. the slowing down of a projectile ion due to the inelastic collisions between bound electrons in the medium and the ion moving through it is called the electronic stopping power.

Def. the elastic collisions between the projectile ion and atoms in the sample involving the interaction of the ion with the nuclei in the target is called the nuclear stopping power.

Satellites
"Space is considered an environment — an extreme environment, filled with entities that can be harmful to spacecraft."

"In space, there are several environmental threats that can harm materials used to create spacecraft. These threats include ultraviolet rays and x-rays from the sun; solar wind particle radiation; thermal cycling (hot and cold cycles); space particles (micrometeoroids and debris); and atomic oxygen."

"Since 2001, NASA and its partners have operated a series of flight experiments called Materials International Space Station Experiment, or MISSE. The objective of MISSE is to test the stability and durability of materials and devices in the space environment."

"PECs [Passive Experiment Containers], which are attached to the exterior of the International Space Station, are about 2-feet by 2-feet and hold a variety of materials samples and devices whose reactions in space are of interest."

"The PECs are positioned in either a ram/wake orientation or in a zenith/nadir orientation. The ram orientation is the direction in which the space station is traveling, and the wake orientation faces the direction traveled. The zenith orientation faces away from Earth into space, while the nadir orientation faces straight down to Earth. Each orientation exposes the samples to different space environmental factors."

Hypotheses

 * 1) At least one radiation entity is or was an astronomical object or source that is unrecognized today for what it once was.

Early observations of origination, tracking, and recovery of entities are the proof of concept that entities falling from the sky have an origin or source.


text:
User:Egm4313.s12.team8.tclamb/R6-2

=R6.2 - Fourier Series Expansion=

Problem Statement
Find the Fourier series expansion for $$f(x)$$ on p.9-8 as follows:

Part 1
Develop the Fourier series expansion of $$f(\bar x)$$ Plot $$f(x)$$ and the truncated Fourier series $$f_n(\bar x)$$
 * $$f_n(\bar x) := \bar a_0 + \sum\limits_{k=1}^n \left(\bar a_k \cos{\frac{k \pi \bar x}{L}} + \bar b_k \sin{\frac{k \pi \bar x}{L}} \right)$$

for $$n\,=\,0,\,1,\,2,\,4,\,8$$. Observe the values of $$f_n(\bar x)$$ at the points of discontinuities, and the Gibbs phenomenon. Transform the variable so to obtain the Fourier series expansion of $$f(x)$$.

Part 2
Do the same as above, but using $$f(\tilde x)$$ to obtain the Fourier series expansion of $$f(x)$$; compare to the result obtained above. Plot the truncated Fourier series $$f_n(\tilde x)$$ for $$n\,=\,100,\,200$$:
 * $$f_n(\tilde x) := \tilde a_0 + \sum\limits_{k=1}^n \left(\tilde a_k \cos{\frac{k \pi \tilde x}{L}} + \tilde b_k \sin{\frac{k \pi \tilde x}{L}} \right)$$

Part 1
From inspection of the graph, we determine the value of L:
 * $$L=2$$

To find the coefficients of the Fourier expansion, we use the following formulas:
 * $$\begin{array}{rl}

\bar a_0 &= \frac1{2L}\int_{-L}^L f(\bar x)\, dx \\ \bar a_n &= \frac1L \int_{-L}^L f(\bar x) \cos{\frac{n \pi \bar x}L}\, dx \\ \bar b_n &= \frac1L \int_{-L}^L f(\bar x) \sin{\frac{n \pi \bar x}L}\, dx \end{array}$$

Solving for $$\bar a_0$$:
 * $$\begin{array}{rl}

\bar a_0 &= \frac1{2L}\int_{-L}^L f(\bar x)\, dx \\ &= \frac14 \int_{-2}^2 f(\bar x)\, dx \\ &= \frac14 \left( \int_{-2}^{-1} 0\, dx + \int_{-1}^1 A\, dx + \int_1^2 0\, dx \right) \\ &= \frac14 \int_{-1}^1 A\, dx \\ &= \frac12 A\end{array}$$

Solving for $$\bar a_n$$:
 * $$\begin{array}{rl}

\bar a_n &= \frac1L \int_{-L}^L f(\bar x) \cos{\frac{n \pi \bar x}{L}}\, dx \\ &= \frac12 \int_{-2}^2 f(\bar x) \cos{\frac{n \pi \bar x}2}\, dx \\ &= \frac12 \left( \int_{-2}^{-1} 0 \cos{\frac{n \pi \bar x}2}\, dx + \int_{-1}^1 A \cos{\frac{n \pi \bar x}2}\, dx + \int_1^2 0 \cos{\frac{n \pi \bar x}2}\, dx \right)\\ &= \frac12 A \int_{-1}^1 \cos{\frac{n \pi \bar x}2}\, dx \\ &= \frac{A}{n \pi} \left(\sin{\frac{n\pi}2}-\sin{-\frac{n\pi}2}\right) \\ &= \frac{2A}{n\pi} \sin{\frac{n\pi}2} \end{array}$$

From this, we can notice the following:
 * For even $$n$$, $$\bar a_n = 0$$.
 * For $$n \equiv 1 \mod 4$$, $$\bar a_n = \frac{2A}{n \pi}$$.
 * For $$n \equiv 3 \mod 4$$, $$\bar a_n = -\frac{2A}{n \pi}$$.

Instead of solving for $$\bar b_n$$, we note that $$f(\bar x)$$ is an even function. Sin is an odd function, therefore by orthogonality, $$\bar b_n = 0$$. Evaluating the integral yields the same result.

Taking our results, we insert them into the definition of $$f_n(\bar x)$$:
 * $$\begin{array}{rl}

f_n(\bar x) &= \bar a_0 + \sum\limits_{k=1}^n \left(\bar a_k \cos{\frac{k \pi \bar x}{L}} + \bar b_k \sin{\frac{k \pi \bar x}{L}} \right) \\ &= \frac12 A + \sum\limits_{k=1}^n \bar a_n \cos{\frac{n \pi \bar x}2} \end{array} $$

Using the substitution $$k = 2 m - 1$$ and our observations about $$\bar a_n$$, we obtain: $$f_n(\bar x) = \frac12 A - \sum\limits_{m=1}^{\lfloor \frac{n+1}2 \rfloor} (-1)^m \frac{2A}{(2m-1)\pi} \cos{\frac{(2m-1) \pi \bar x}2}$$

From this, we can generate the following solutions:
 * $$\begin{array}{rl}

f_0(\bar x) &= \frac{A}2 \\ f_1(\bar x) &= \frac{A}{2} + \frac{2 A\cos \left(\frac{\pi x}{2}\right)}{\pi } \\ f_2(\bar x) &= \frac{A}{2} + \frac{2 A\cos \left(\frac{\pi x}{2}\right)}{\pi } \\ f_4(\bar x) &= \frac{A}{2} + \frac{2 A\cos \left(\frac{\pi x}{2}\right)}{\pi }-\frac{2 A\cos \left(\frac{3 \pi    x}{2}\right)}{3 \pi } \\ f_8(\bar x) &= \frac{A}{2} + \frac{2 A\cos \left(\frac{\pi x}{2}\right)}{\pi }-\frac{2 A\cos \left(\frac{3 \pi    x}{2}\right)}{3 \pi }+\frac{2 A\cos \left(\frac{5 \pi  x}{2}\right)}{5 \pi }-\frac{2 A\cos \left(\frac{7 \pi x}{2}\right)}{7 \pi } \end{array}$$

To obtain a solution for $$f_n(x)$$, we observe the following relationship between $$x$$ and $$\bar x$$:
 * $$\bar x = x - \frac54$$

Substituting this into our above result for $$f_n(\bar x)$$, we obtain: $$f_n(x) = \frac12 A - \sum\limits_{m=1}^{\lfloor \frac{n+1}2 \rfloor} (-1)^m \frac{2A}{(2m-1)\pi} \cos{\frac{(2m-1) \pi \left(x - \frac54\right)}2}$$

Plots of $$f_n(\bar x)$$

Part 2
To obtain the Fourier expansion of $$f(\tilde x)$$, the same process as above is followed:

From inspection of the graph, we determine the value of L:
 * $$L=2$$

To find the coefficients of the Fourier expansion, we use the following formulas:
 * $$\begin{array}{rl}

\tilde a_0 &= \frac1{2L}\int_{-L}^L f(\tilde x)\, dx \\ \tilde a_n &= \frac1L \int_{-L}^L f(\tilde x) \cos{\frac{n \pi \tilde x}L}\, dx \\ \tilde b_n &= \frac1L \int_{-L}^L f(\tilde x) \sin{\frac{n \pi \tilde x}L}\, dx \end{array}$$

Solving for $$\tilde a_0$$:
 * $$\begin{array}{rl}

\tilde a_0 &= \frac1{2L}\int_{-L}^L f(\tilde x)\, dx \\ &= \frac14 \int_{-2}^2 f(\tilde x)\, dx \\ &= \frac14 \left( \int_{-2}^{-1} 0\, dx + \int_{-1}^1 A\, dx + \int_1^2 0\, dx \right) \\ &= \frac14 \int_{-1}^1 A\, dx \\ &= \frac12 A\end{array}$$

Instead of solving for $$\tilde a_n$$, we note that $$f(\tilde x)$$ is an odd function. Cos is an even function, therefore by orthogonality, $$\tilde a_n = 0$$. Evaluating the integral yields the same result.

Solving for $$\tilde b_n$$:
 * $$\begin{array}{rl}

\tilde b_n &= \frac1L \int_{-L}^L f(\tilde x) \sin{\frac{n \pi \tilde x}{L}}\, dx \\ &= \frac12 \int_{-2}^2 f(\tilde x) \sin{\frac{n \pi \tilde x}2}\, dx \\ &= \frac12 \left( \int_{-2}^0 0 \sin{\frac{n \pi \tilde x}2}\, dx + \int_0^2 A \sin{\frac{n \pi \tilde x}2}\, dx\right)\\ &= \frac12 A \int_0^2 \sin{\frac{n \pi \tilde x}2}\, dx \\ &= \frac{A}{n \pi} \left(-\cos{n \pi}+\cos0\right) \\ &= \frac{A}{n \pi} \left(1 - \cos{n \pi}\right) \end{array}$$

From this, we can notice the following:
 * For odd $$n$$, $$\tilde b_n = \frac{2A}{n \pi}$$.
 * For even $$n$$, $$\tilde b_n = 0$$.

Taking our results, we insert them into the definition of $$f_n(\tilde x)$$:
 * $$\begin{array}{rl}

f_n(\tilde x) &= \tilde a_0 + \sum\limits_{k=1}^n \left(\tilde a_k \cos{\frac{k \pi \tilde x}{L}} + \tilde b_k \sin{\frac{k \pi \tilde x}{L}} \right) \\ &= \frac12 A + \sum\limits_{k=1}^n \tilde b_n \sin{\frac{n \pi \tilde x}2} \end{array} $$

Using the substitution $$k = 2 m - 1$$ and our observations about $$\tilde b_n$$, we obtain: $$f_n(\tilde x) = \frac12 A + \sum\limits_{m=1}^{\lfloor \frac{n+1}2 \rfloor} \frac{2A}{(2m-1) \pi} \sin{\frac{(2m-1) \pi \tilde x}2}$$

From this, we can generate the following solutions:
 * $$\begin{array}{rl}

f_0(\tilde x) &= \frac{A}2 \\ f_1(\tilde x) &= \frac{A}{2} + \frac{2 A \sin \left(\frac{\pi x}{2}\right)}{\pi } \\ f_2(\tilde x) &= \frac{A}{2} + \frac{2 A \sin \left(\frac{\pi x}{2}\right)}{\pi } \\ f_4(\tilde x) &= \frac{A}{2} + \frac{2 A \sin \left(\frac{\pi x}{2}\right)}{\pi } + \frac{2 A \sin \left(\frac{3 \pi    x}{2}\right)}{3 \pi } \\ f_8(\tilde x) &= \frac{A}{2} + \frac{2 A \sin \left(\frac{\pi x}{2}\right)}{\pi } + \frac{2 A \sin \left(\frac{3 \pi    x}{2}\right)}{3 \pi } + \frac{2 A \sin \left(\frac{5 \pi  x}{2}\right)}{5 \pi } + \frac{2 A \sin \left(\frac{7 \pi x}{2}\right)}{7 \pi } \end{array}$$

To obtain a solution for $$f_n(x)$$, we observe the following relationship between $$x$$ and $$\tilde x$$:
 * $$\tilde x = x - \frac14$$

Substituting this into our above result for $$f_n(\tilde x)$$, we obtain: $$f_n(x) = \frac12 A + \sum\limits_{m=1}^{\lfloor \frac{n+1}2 \rfloor} \frac{2A}{(2m-1) \pi} \sin{\frac{(2m-1) \pi (x - \frac14)}2}$$

This is shown to be equivalent to the above solution by making the following substitution:
 * $$\sin\theta = \cos{\frac{\pi}2 - \theta}$$

Plots of $$f_n(\tilde x)$$

Plots of $$f_n(x)$$

Gibbs Phenomenon
Looking at the above plots, it is clear that the average error is decreasing as a function of $$n$$. However, the overshoot of the Fourier series remains, even as far as $$n = 255$$ as shown below. This overshoot decreases initially, but eventually converges to a limit. This behavior is known as the Gibbs phenomenon. The following table lists approximate average error and the maximum overshoot of the Fourier series as they vary with $$n$$, setting $$A=1$$:

=R6.4 - Solve an ODE with a Fourier Series Excitation=

Problem Statement
Consider the L2-ODE-CC with the window function from R6.2, $$f(x)$$, as excitation:
 * $$y'' - 3y' + 2y = r(x)$$
 * $$r(x) = f(x)$$

And the initial conditions:
 * $$y(0) = 1,\;y'(0) = 0$$

Part 1
Find $$y_n(x)$$ such that:
 * $$y_n'' + a y_n' + b y_n = r_n(x)$$

with the same initial conditions.

Plot $$y_n(x)$$ for $$n=2,\,4,\,8$$, for $$x \in \left[ 0, \, 10 \right]$$.

Part 2
Use the MATLAB command ode45 to integrate the L2-ODE-CC, and plot the numerical solution to compare with the analytical solution.

Solution
When plotting analytical solutions of y(x) over [0,10] the transient response to the excitation is dominated by the steady state response from the homogenous solution.

To more clearly illustrate the convergence of the transient response, we will also plot our solution over the range [-10,0].

Part 1
From R6.2, we know that our excitation function is given by:
 * $$r_n(x) = f_n(x) = \frac12 A + \sum\limits_{m=1}^{\lfloor \frac{n+1}2 \rfloor} \frac{2A}{(2m-1) \pi} \sin{\frac{(2m-1) \pi (x - \frac14)}2}$$

By the Basic Rule and Sum Rule developed in Section 7 of the lecture notes (p.7-2), we know that our particular solution is of the form:
 * $$y_{p,n}(x) = \frac{A}2 + \sum\limits_{i=1}^{\lfloor \frac{n+1}2 \rfloor} \left( J_i \sin{\frac{(2i-1) \pi \left(x - \frac14\right)}{2}} + K_i \cos{\frac{(2i-1) \pi \left(x - \frac14\right)}{2}} \right)$$

And from our work in R3.3, we know that the homogeneous solution is of the form:
 * $$y_h(x) = C_1 e^x + C_2 e^{2x}$$

The cases $$n=0$$ and $$n=1$$ will be worked out below as demonstration, and further cases will be solved using computational software.

Case $$n = 0$$
Setting $$n=0$$ yields the following:
 * $$r_0(x) = \frac12 A$$
 * $$y_{p,0}(x) = \frac{A}4$$

Combining our solution for $$y_{p,0}(x)$$ with $$y_h(x)$$ yields:
 * $$y_0(x) = y_h(x) + y_{p,0}(x) = C_1 e^x + C_2 e^{2x} + \frac{A}4$$

Differentiating this yields:
 * $$y_0'(x) = C_1 e^x + 2 C_2 e^{2x}$$

Applying initial conditions allows us to solve for the homogeneous constants:
 * $$\begin{array}{rl}

y_0(0) &= C_1 + C_2 + \frac{A}4 = 1 \\ y_0'(0) &= C_1 + 2 C_2 = 0 \\ C_1 &= 2 - \frac{A}2 \\ C_2 &= \frac{A}4 - 1 \end{array}$$

Therefore, our final solution is: $$y_0(x) = \frac14 \left( A + (8 - 2 A) e^x + (A - 4) e^{2x} \right)$$

Case $$n = 1$$
Setting $$n=1$$ yields the following:
 * $$r_1(x) = \frac12 A + \frac{2A}{\pi} \sin{\frac{\pi \left(x - \frac14 \right)}2}$$
 * $$y_{p,1}(x) = \frac{A}4 + J \sin{\frac{\pi \left(x - \frac14 \right)}2} + K \cos{\frac{\pi \left(x - \frac14 \right)}2}$$

Differentiating the particular solution yields:
 * $$y_{p,1}'(x) = \frac{J\pi}2 \cos{\frac{\pi \left(x - \frac14 \right)}2} - \frac{K\pi}2 \sin{\frac{\pi \left(x - \frac14 \right)}2}$$
 * $$y_{p,1}''(x) = -\frac{J\pi^2}4 \sin{\frac{\pi \left(x - \frac14 \right)}2} - \frac{K\pi^2}4 \cos{\frac{\pi \left(x - \frac14 \right)}2}$$

Substituting these into the ODE:
 * $$\begin{array}{l}

- \frac{J\pi^2}4 \sin{\frac{\pi \left(x - \frac14 \right)}2} - \frac{K\pi^2}4 \cos{\frac{\pi \left(x - \frac14 \right)}2} \\ \;\;\;\; - \frac{3J\pi}2 \cos{\frac{\pi \left(x - \frac14 \right)}2} + \frac{3K\pi}2 \sin{\frac{\pi \left(x - \frac14 \right)}2} \\ \;\;\;\; + \frac{A}2 + 2J \sin{\frac{\pi \left(x - \frac14 \right)}2} + 2K \cos{\frac{\pi \left(x - \frac14 \right)}2} \\ \;\;\;\;\;\;\;\; = \frac12 A + \frac{2A}{\pi} \sin{\frac{\pi \left(x - \frac14 \right)}2} \end{array}$$

To solve for the coefficients $$J$$ and $$K$$, we set like terms' coefficients equal to each other:
 * $$\begin{array}{rll}

-\frac{J\pi^2}4 + \frac{3K\pi}2 + 2 J &= \left(2 - \frac{\pi^2}4 \right) J + \frac{3\pi}2 K &= \frac{2A}{\pi} \\ -\frac{K\pi^2}4 - \frac{3J\pi}2 + 2 K &= \left(2 - \frac{\pi^2}4 \right) K - \frac{3\pi}2 J &= 0 \\ J &= - \frac{8A\left(\pi^2 - 8\right)}{\pi\left(\pi^4 + 20 \pi^2 + 64\right)} &\\ K &= \frac{48A}{\pi^4 + 20 \pi^2 + 64} & \end{array}$$

As $$J$$ and $$K$$ are rather complex expressions, it is easier to leave them as such and substitute their true values in when necessary. Next, combining our solution for $$y_{p,0}(x)$$ with $$y_h(x)$$ yields:
 * $$y_1(x) = y_h(x) + y_{p,1}(x) = C_1 e^x + C_2 e^{2x} + \frac{A}4 + J \sin{\frac{\pi \left(x - \frac14 \right)}2} + K \cos{\frac{\pi \left(x - \frac14 \right)}2}$$

Differentiating this yields:
 * $$y_1'(x) = C_1 e^x + 2 C_2 e^{2x} + \frac{J\pi}2 \cos{\frac{\pi \left(x - \frac14 \right)}2} - \frac{K\pi}2 \sin{\frac{\pi \left(x - \frac14 \right)}2}$$

Applying initial conditions allows us to solve for the homogeneous constants:
 * $$\begin{array}{rl}

y_1(0) &= C_1 + C_2 + \frac{A}4 + J \sin{\frac{\pi}8} + K \cos{\frac{\pi}8} = 1 \\ y_1'(0) &= C_1 + 2 C_2 + \frac{J\pi}2 \cos{\frac{\pi}8} - \frac{K\pi}2 \sin{\frac{\pi}8} = 0 \\ C_1 &= -\left(\frac{A}2 - 2 + \frac{4 A \cos{\frac{\pi}8}}{\pi^2 + 4} + \frac{8 A \sin{\frac{\pi}8}}{\pi\left(\pi^2 + 4\right)} \right)\\ C_2 &= \frac{A}4 - 1 + \frac{4 A \cos{\frac{\pi}8}}{\pi^2 + 16} + \frac{16 A \sin{\frac{\pi}8}}{\pi\left(\pi^2 + 16\right)} \end{array}$$

Therefore, our final solution is: $$\begin{array}{rl} y_1(x) = & -\left(\frac{A}2 - 2 + \frac{4 A \cos{\frac{\pi}8}}{\pi^2 + 4} + \frac{8 A \sin{\frac{\pi}8}}{\pi\left(\pi^2 + 4\right)} \right) e^x \\ & + \left(\frac{A}4 - 1 + \frac{4 A \cos{\frac{\pi}8}}{\pi^2 + 16} + \frac{16 A \sin{\frac{\pi}8}}{\pi\left(\pi^2 + 16\right)}\right) e^{2x} \\ & + \frac{A}4 - \frac{8A\left(\pi^2 - 8\right)}{\pi\left(\pi^4 + 20 \pi^2 + 64\right)} \sin{\frac{\pi \left(x - \frac14 \right)}2} + \frac{48A}{\pi^4 + 20 \pi^2 + 64} \cos{\frac{\pi \left(x - \frac14 \right)}2} \end{array}$$

Cases $$n = 2, 4, 8$$
Finding the solution with exact coefficients becomes too tedious for cases of n=2 and greater. Instead, Mathematica is used to generate an analytical solution with approximate coefficients.

For n=2 we find the analytical solution to be:

$$\begin{array}{rl} y_2(x) = y_1(x) \approx & 1.30381 e^{1 x}-0.682487 e^{2 x} +0.25 \\ & + 0.038936 \sin (1.5708 x)+0.128673 \cos (1.5708 x) \end{array}$$

For n=4 we find the analytical solution to be:

$$\begin{array}{rl} y_4(x) \approx & 1.29577 e^{1. x} - 0.682847 e^{2. x} + 0.25 \\ & + 0.00705066 \sin (1.1781 -4.71239 x)+0.00493286 \cos (1.1781 -4.71239 x) \\ & + 0.038936 \sin (1.5708 x)+0.128673 \cos (1.5708 x) \end{array}$$

For n=8 we find the analytical solution to be:

$$\begin{array}{rl} y_8(x) \approx &1.31162 e^{1 x}-0.700209 e^{2 x} + 0.25 \\ &+0.0000985159 \sin (9.03208 -36.1283 x) \sin (6.28319 -25.1327 x) \\ &+0.000364302 \sin (8.24668 -32.9867 x) \sin (6.28319 -25.1327 x) \\ &+0.00246643 \sin (7.46128 -29.8451 x) \sin (6.28319 -25.1327 x) \\ &+0.0668894 \sin (6.67588 -26.7035 x) \sin (6.28319 -25.1327 x) \\ &+0.0668894 \sin (6.28319 -25.1327 x) \sin (5.89049 -23.5619 x) \\ &+0.00246643 \sin (6.28319 -25.1327 x) \sin (5.10509 -20.4204 x) \\ &+0.000364302 \sin (6.28319 -25.1327 x) \sin (4.31969 -17.2788 x) \\ &+0.0000985159 \sin (6.28319 -25.1327 x) \sin (3.53429 -14.1372 x) \\ &+0.0000985159 \cos (9.03208 -36.1283 x) \cos (6.28319 -25.1327 x) \\ &+0.000364302 \cos (8.24668 -32.9867 x) \cos (6.28319 -25.1327 x) \\ &+0.00246643 \cos (7.46128 -29.8451 x) \cos (6.28319 -25.1327 x) \\ &+0.0668894 \cos (6.67588 -26.7035 x) \cos (6.28319 -25.1327 x) \\ &+0.0668894 \cos (5.89049 -23.5619 x) \cos (6.28319 -25.1327 x) \\ &+0.00246643 \cos (5.10509 -20.4204 x) \cos (6.28319 -25.1327 x) \\ &+0.000364302 \cos (4.31969 -17.2788 x) \cos (6.28319 -25.1327 x) \\ &+0.0000985159 \cos (3.53429 -14.1372 x) \cos (6.28319 -25.1327 x) \\ &+0.000355107 \sin (9.03208 -36.1283 x) \cos (6.28319 -25.1327 x) \\ &+0.000922816 \sin (8.24668 -32.9867 x) \cos (6.28319 -25.1327 x) \\ &+0.00352533 \sin (7.46128 -29.8451 x) \cos (6.28319 -25.1327 x) \\ &+0.00663447 \sin (6.67588 -26.7035 x) \cos (6.28319 -25.1327 x) \\ &-0.00663447 \sin (5.89049 -23.5619 x) \cos (6.28319 -25.1327 x) \\ &-0.00352533 \sin (5.10509 -20.4204 x) \cos (6.28319 -25.1327 x) \\ &-0.000922816 \sin (4.31969 -17.2788 x) \cos (6.28319 -25.1327 x) \\ &-0.000355107 \sin (3.53429 -14.1372 x) \cos (6.28319 -25.1327 x) \\ &-0.000355107 \sin (6.28319 -25.1327 x) \cos (9.03208 -36.1283 x) \\ &-0.000922816 \sin (6.28319 -25.1327 x) \cos (8.24668 -32.9867 x) \\ &-0.00352533 \sin (6.28319 -25.1327 x) \cos (7.46128 -29.8451 x) \\ &-0.00663447 \sin (6.28319 -25.1327 x) \cos (6.67588 -26.7035 x) \\ &+0.00663447 \sin (6.28319 -25.1327 x) \cos (5.89049 -23.5619 x) \\ &+0.00352533 \sin (6.28319 -25.1327 x) \cos (5.10509 -20.4204 x) \\ &+0.000922816 \sin (6.28319 -25.1327 x) \cos (4.31969 -17.2788 x) \\ &+0.000355107 \sin (6.28319 -25.1327 x) \cos (3.53429 -14.1372 x) \\ \end{array}$$

Positive $$x$$ Plots

 * [[image:y0_pos.png]]
 * [[image:y1_pos.png]]
 * [[image:y2_pos.png]]
 * [[image:y4_pos.png]]
 * [[image:y8_pos.png]]

Negative $$x$$ Plots

 * [[image:y0_neg.png]]
 * [[image:y1_neg.png]]
 * [[image:y2_neg.png]]
 * [[image:y4_neg.png]]
 * [[image:y8_neg.png]]

Part 2
Using the following Matlab function and input allows us to compare the analytical and numerical solutions:

Positive $$x$$ Plots

 * [[image:y,_y2,_y4,_y8_ode45.png]]

 * [[image:y,_y2,_y4,_y8_ode45_zoom1.png]]

The zoomed plot clearly shows that the addition of higher order terms brings the analytical solution closer to y(x).

Negative $$x$$ Plots

 * [[image:Y,_y2,_y4,_y8_neg_NDSolve.png]]

 * [[image:Y,_y2,_y4,_y8_neg_NDSolve_zoom.png]]

Again, the addition of higher order terms brings the analytical solution closer to numerical y(x).

=R6.5 - Explore the Sensitivity of a Truncated Taylor Series Excitation and its Solution=

R4.3, p.7c-28
Consider the L2-ODE-CC (5) p.7b-7 with $$log(1+x)$$ as excitation:

$$y''-3y'+2y=r(x)$$

$$r(x)=log(1+x)$$

and the initial conditions

$$y(-\frac34)=1, y'(-\frac34)=0$$

Let $$r_{n}(x)$$ be the truncated Taylor series about $$\hat{x}=0$$, with n terms--which is also the highest degree of the Taylor (power) series--of $$log(1+x)$$.

Explore the sensitivity of the excitation and of the solution for n=11 by adding a small perturbation to the coefficients of the excitation $$r_{n}(x)$$, and propose and explanation.

R4.4, p.7c-29
Extend the accuracy of the solution beyond $$\hat{x}=1$$

Consider the point $$x_{1}=0.9$$ near the brink of non-convergence for the Taylor series of $$log(1+x)$$.

Find the values of $$y_{n}(x_1), y'_{n}(x_1)$$ that will serve as initial conditions for the next iteration to extend the domain accuracy of the analytical solution. Find n sufficiently high so that $$y_{n}(x_1), y'_{n}(x_1)$$ do not differ from the numerical solution by more than $$10^{-5}$$.

Develop $$log(1+x)$$ in Taylor series about $$\hat{x}=1$$.

With $$y_{n}(x)$$ such that:

$$y''_{n}+ay'_{n}+by_{n}=r_{n}(x)$$

For $$x$$ in $$[0.9, 3]$$ with the initial conditions $$y_{n}(x_1), y'_{n}(x_1)$$

For n=11, explore and find the reason why the solution $$y_{n}(x)$$ did not match the numerical solution $$y(x)$$ that serves as the baseline for the comparison.

R4.3
The following MATLAB code generates the numerical solution to the problem:

The following MATLAB function will generate the output, as well as the coefficients, of a series solution to the ODE:

Using the above code, we obtain the following solution and Taylor polynomial for the excitation.
 * $$\begin{array}{rl}y_{11}(x) = &0.045455 x^{11} + 0.70000 x^{10} + 8.0556 x^9 + 77.188 x^8 + 636.32 x^7 + 4520.0 x^6 + 27318 x^5 \\

&+ 137080 x^4 + 549320 x^3 + 164900 x^2 + 3300300 x + 3301100 - 3301800 e^x + 734.89 e^{2x}\end{array}$$
 * $$r_{11}(x) = \frac1{11} x^{11} - \frac1{10} x^{10} + \frac19 x^9 - \frac18 x^8 + \frac17 x^7 - \frac16 x^6 + \frac15 x^5 - \frac14 x^4 + \frac13 x^3 - \frac12 x^2 + x$$

As we know, the excitation blows up beyond the radius of convergence of the Taylor polynomial ($$-1 < x < 1$$). Similarly, the solution blows up at this scale for $$x > 2$$.

Perturbation of 0.01
First, we attempted perturbing the coefficients of $$r_{11}(x)$$ by adding 0.01 to each.
 * $$\begin{array}{rl}

r_{11,pert}(x) = &0.10091 x^{11} - 0.09 x^{10} + 0.12111 x^9 - 0.115 x^8 + 0.15286 x^7 - 0.15667 x^6 \\ &+ 0.21 x^5 - 0.24 x^4 + 0.34333 x^3 - 0.49 x^2 + 1.01 x + 0\end{array} $$

Solving the ODE with this perturbation yields:
 * $$\begin{array}{rl}

y_{11}(x) = &0.050455 x^{11} + 0.7875 x^{10} + 9.0981 x^9 + 87.329 x^8 + 720.49 x^7 + 5119.9 x^6 + 30949 x^5 \\ &+ 155320 x^4 + 622410 x^3 + 1868900 x^2 + 3739600 x + 3740500 - 3741360 e^x + 854.68 e^{2x} \end{array}$$

Adding a positive perturbation to the coefficients of the excitation resulted in a greater divergence of the solution. Noting this, we attempted a negative perturbation.

Perturbation of -0.01
We next attempted perturbing the coefficients of $$r_{11}(x)$$ by adding -0.01 to each.
 * $$\begin{array}{rl}

r_{11,pert}(x) = &0.08091 x^{11} - 0.11 x^{10} + 0.10111 x^9 - 0.135 x^8 + 0.13286 x^7 - 0.17667 x^6 \\ &+ 0.19 x^5 - 0.26 x^4 + 0.32333 x^3 - 0.51 x^2 + 0.99 x + 0\end{array} $$

Solving the ODE with this perturbation yields:
 * $$\begin{array}{rl}

y_{11}(x) = &0.040455 x^{11} + 0.6125 x^{10} + 7.0131 x^9 + 67.046 x^8 + 552.15 x^7 + 3920.2 x^6 + 23687 x^5 \\ &+ 118850 x^4 + 476220 x^3 + 1429900 x^2 + 2861000 x + 2861700 - 2862300 e^x + 615.11 e^{2x} \end{array}$$

As predicted, the accuracy of the solution improved through the addition of a negative perturbation. Noting this, we proceeded to vary the perturbation in an attempt to best match the numerical solution.

Perturbation of -0.035
We finally perturbed the coefficients of $$r_{11}(x)$$ by adding -0.035 to each, as we found through trial and error that this most closely matched the numerical ODE solution over our range.
 * $$\begin{array}{rl}

r_{11,pert}(x) = &0.055909 x^{11} - 0.135 x^{10} + 0.076111 x^9 - 0.16 x^8 + 0.10786 x^7 - 0.20167 x^6 \\ &+ 0.165 x^5 - 0.285 x^4 + 0.29833 x^3 - 0.535 x^2 + 0.965 x + 0\end{array} $$

Solving the ODE with this perturbation yields:
 * $$\begin{array}{rl}

y_{11}(x) = &0.027955 x^{11} + 0.39375 x^{10} + 4.4068 x^9 + 41.693 x^8 + 341.73 x^7 + 2420.6 x^6 + 14609 x^5 \\ &+ 73261 x^4 + 293470 x^3 + 8810600 x^2 + 1762800 x + 1763100 - 1763400 e^x + 315.64 e^{2x} \end{array}$$

This perturbation of -0.035 results in a solution closely matching that of the numerical solution. As shown above, very small perturbations in the coefficients of the excitation resulted in great variety of our solution. In other words, there is great sensitivity in the coefficients of the Taylor polynomial of the excitation.

R4.4

In general, $$y_n(x)$$ will not match the numerical solution, $$y(x)$$, beyond a certain radius of convergence. The solution depends on the excitation, which is a logarithm that we are trying to approximate by a Taylor polynomial of order $$n$$. The shape of a logarithm and a polynomial are fundamentally different. No matter how high we choose $$n$$, it will always diverge. Therefore, the solution must always diverge.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Egm4313.s12.team8.tclamb/R6-2


title: File:Report6.5n=3.png


text:
User:Egm4313.s12.team8/R6

=R6.1 - Prove Periodicity=

Problem Statement
Given $$\displaystyle f(x)=a_{0}+\sum_{n=1}^{\infty }[a_{n}cos(n\omega x)+b_{n}sin(n\omega x)]$$: (1) Find the (smallest) period of $$\displaystyle cos(n\omega x) $$ and $$\displaystyle sin(n\omega x) $$. (2) Show that these functions also have period $$\displaystyle p $$. (3) Show that the constant $$\displaystyle a_{0} $$ is also a periodic function with period $$\displaystyle p $$.

Solution (1)
We know that the period of a normal $$\displaystyle sin(x) $$ or $$\displaystyle cos(x) $$ is $$\displaystyle 2\pi $$. When there are values or variables being multiplied by this $$\displaystyle x $$ variable, the period becomes $$\displaystyle 2\pi $$ divided by the values or variables. We know that: $$\displaystyle \omega=\frac{\pi }{L}=\frac{2 \pi}{p}$$    (6.1.1) Where $$\displaystyle p $$ is the period. Using this relation, along with our variable $$\displaystyle n $$, we can solve for the period of $$\displaystyle cos(n\omega x) $$ and $$\displaystyle sin(n\omega x) $$ as follows: $$\displaystyle p=\frac{2\pi }{\tfrac{n\pi }{L}}$$ $$\displaystyle p=\frac{L\pi}{n} $$    (6.1.2) Since the period will be smallest at $$\displaystyle n=1$$, plugging into equation (6.1.2) shows that the smallest period of $$\displaystyle cos(n\omega x) $$ and $$\displaystyle sin(n\omega x) $$ is: $$\displaystyle p=L\pi $$    (6.1.3)

Solution (2)
We are also given that: $$\displaystyle \omega =\frac{2\pi}{p}$$    (6.1.4) Using this relation, we can solve for the period again as follows: $$\displaystyle p=\frac{2\pi}{\frac{n2\pi}{p}}$$ $$\displaystyle p=\frac{p}{n}$$    (6.1.5) We know that the period is smallest at $$\displaystyle n=1$$, and plugging this value into (6.1.5) proves that: $$\displaystyle p=p $$    (6.1.6)

Solution (3)
We know that, starting at 0: $$\displaystyle \widetilde{a}_{0}=\frac{1}{2L}\int_{0}^{2L}f(\widetilde{x})d\widetilde{x} $$    (6.1.7) Where the period is represented by 0 to 2L. We are also given that: $$\displaystyle \frac{\pi}{L}=\frac{2\pi}{p} $$    (6.1.8) Rearranging (6.1.8) allows us to solve for L: $$\displaystyle L=\frac{p}{2}$$     (6.1.9) Multiplying (6.1.9) by 2 allows us to find the period of $$\displaystyle a_{0}$$, as follows: $$\displaystyle 2L=p$$    (6.1.10) This shows that $$\displaystyle a_{0}$$ is indeed a periodic function with a period of $$\displaystyle p$$. This also shows that at any given $$\displaystyle x$$ value or period throughout the periodic function, $$\displaystyle a_{0}$$ holds its constant value.

=R6.2 - Fourier Series Expansion=

Problem Statement
Find the Fourier Series Expansion for f(x) as follows:

1. Determine if f(x) is even, odd, or neither. Develop the Fourier Series Expansion of $$f(\bar{x})$$. Plot $$f(\bar{x})$$ and the truncated Fourier Series $$f_{n}(\bar{x})$$.

$$f_{n}(\bar{x}):= \bar{a}_0 + \sum_{k=1}^n[\bar{a}_k \cos k \omega \bar{x} + \bar{b}_k \sin k \omega \bar{x}]$$ for n = 0,1.

Observe the values of $$f_{n}(\bar{x})$$ at the points of discontinuities and the Gibbs phenomenon. Transform the variable so to obtain the Fourier Series Expansion for f(x).

2. Repeat 1 but using $$f_{n}(\tilde{x})$$ to obtain the Fourier Series Expansion of f(x). Compare results for n=0,1.

$$f(x)=x^2 ,(-1<x<1), p=2 $$ $$f(x)=1-\frac{x^2}{4} ,(-2<x<2), p=4$$

Part 1
To determine if f(x) is even or odd, substitute -x in for x. If f(x)=f(-x) then the function is even.

$$f(-x)=(-x)^2=x^2=f(x)$$ $$f(-x)=1-\frac{(-x)^2}{4}=1-\frac{x^2}{4}=f(x)$$

Therefore both functions are even.

According to the book, the Fourier Series for even functions reduces to a Fourier cosine series.

$$f(x)= a_0 + \sum_{n=1}^\infty[a_n \cos \frac{n \pi}{L} x]$$

$$a_0= \frac{1}{L} \int_{0}^{L} f(x)dx$$

$$a_n= \frac{2}{L} \int_{0}^{L} f(x)\cos \frac{n \pi x}{L} dx$$

From the Figure shown in the Problem Statement, $$f(\bar{x})$$ has a period of p=4. Therefore L=2, $$\omega = \frac{2 \pi}{p}$$.

$$a_0= \frac{1}{2} \int_{0}^{2} f(\bar{x})d \bar{x}$$

$$a_0= \frac{A}{2}$$

$$a_n= \frac{2}{2} \int_{0}^{2} f(\bar{x})\cos \frac{n \pi \bar{x}}{2} d \bar{x}$$

$$a_n= \frac{2A}{k \pi} \sin \frac{k \pi}{2}$$

$$f_k(\bar{x})= \frac{A}{2} + \sum_{k=1}^n[(\frac{2A}{k \pi} \sin \frac{k \pi}{2}) \cos \frac{k \pi}{2} \bar{x}]$$

For n=0:

$$f_0(\bar{x})= \frac{A}{2}$$

For n=1:

$$f_1(\bar{x})= \frac{A}{2} + \frac{2A}{\pi} \cos \frac{\pi}{2} \bar{x}$$

$$\bar{x} = x - 1.25$$

$$f_k(x)= \frac{A}{2} + \frac{A}{\pi}\sin (\frac{\pi (x-1.25)}{2})$$

Let A=1

Part 2
From the Figure shown in the Problem Statement, $$f(\tilde{x})$$ has a period of p=4. Therefore L=2, $$\omega = \frac{2 \pi}{p}$$.

$$a_0= \frac{1}{2L} \int_{0}^{2L} f(\tilde{x})d \tilde{x}$$

$$a_0= \frac{1}{4} \int_{0}^{4} f(\tilde{x})d \tilde{x}$$

$$a_0= \frac{A}{2}$$

$$a_n= \frac{1}{L} \int_{0}^{2L} f(\tilde{x})\cos (n \omega \tilde{x}) d \tilde{x}$$

$$a_n= \frac{1}{2} \int_{0}^{4} f(\tilde{x})\cos (\frac{n \pi \tilde{x}}{2}) d \tilde{x}$$

$$a_n= \frac{1}{2} f(\tilde{x})\cos (\frac{n \pi \tilde{x}}{2}) \mid _0^4 = \frac{A}{n \pi} \sin (\pi k)$$

$$b_n= \frac{1}{L} \int_{0}^{2L} f(\tilde{x})\sin (n \omega \tilde{x}) d \tilde{x}$$

$$b_n= \frac{1}{2} \int_{0}^{4} f(\tilde{x})\sin (\frac{n \pi \tilde{x}}{2}) d \tilde{x}$$

$$b_n= \frac{1}{2} f(\tilde{x})\sin (\frac{n \pi \tilde{x}}{2}) \mid _0^4 = \frac{A}{n \pi}(1 - \cos (\pi k))$$

Since $$f(\tilde{x})$$ is an odd function the Fourier series reduces to a Fourier sine series.

$$f_k(\tilde{x})= \frac{A}{2} + \sum_{k=1}^n ([\frac{A}{k \pi}(1 - \cos (\pi k)] \sin [\frac{k \pi \tilde{x}}{2}])$$

For n=0:

$$f_0(\tilde{x})= \frac{A}{2}$$

For n=1:

$$f_1(\tilde{x})= \frac{A}{2} + \frac{A}{\pi}\sin (\frac{\pi \tilde{x}}{2})$$

$$\tilde{x} = x - 0.25$$

$$f_1(x)= \frac{A}{2} + \frac{A}{\pi}\sin (\frac{\pi (x-.25)}{2})$$

Let A=1

=R6.3 - Two Fourier Series=

Problem Statement
K 2011 p. 491 numbers 15 and 17

-State whether the function is even, odd, or neither

-Find it's Fourier series

-Plot the truncated Fourier series for n = 2, 4, and 8

Problem #15
$$ \displaystyle f(x)=-x-\pi $$ on $$ \displaystyle (-\pi,-\pi/2) $$

$$ \displaystyle f(x)=x $$ on $$ \displaystyle (-\pi/2,\pi/2) $$

$$ \displaystyle f(x)=-x+\pi $$ on $$ \displaystyle (\pi/2,\pi) $$

This function is an odd function.

This is proven by the fact that $$ \displaystyle f(-x)=-f(x) $$

Since this is an odd function and of period $$ \displaystyle 2\pi $$ its Fourier series reduces to the Fourier sine series shown here:

$$ \displaystyle f(x)=\sum_{n = 1}^{\infty} b_n\sin(nx) $$

Where the coefficient equals:

$$ \displaystyle b_n=\frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx)dx $$

$$ \displaystyle b_n=\frac{2}{\pi}(\int_{0}^{\pi/2}x\sin(nx)dx + \int_{\pi/2}^{\pi}(-x+\pi)\sin(nx)dx) $$

$$ \displaystyle b_n=\frac{2}{\pi}(\int_{0}^{\pi/2}x\sin(nx)dx - \int_{\pi/2}^{\pi}x\sin(nx)dx + \int_{\pi/2}^{\pi}\pi\sin(nx)dx) $$

Using integration by parts and then cancelling like terms:

$$ \displaystyle b_n=\frac{2}{\pi}(\frac{2\sin(\frac{\pi}{2}n)}{n^2}) $$

Plugging this back into the function equation:

$$ \displaystyle f(x)=\sum_{n = 1}^{\infty}(\frac{4\sin(\frac{\pi}{2}n)}{\pi n^2})\sin(nx) $$

Expanding this, one obtains:

$$ \displaystyle f(x)=\frac{4}{\pi}(\sin x-\frac{1}{9}\sin3x+\frac{1}{25}\sin5x-...) $$

If n=2

$$ \displaystyle f(x)=\frac{4}{\pi}(\sin x) $$

If n=4

$$ \displaystyle f(x)=\frac{4}{\pi}(\sin x-\frac{1}{9}\sin3x) $$

If n=8

$$ \displaystyle f(x)=\frac{4}{\pi}(\sin x-\frac{1}{9}\sin3x+\frac{1}{25}\sin5x-\frac{1}{49}\sin7x) $$

Problem #17
$$ \displaystyle f(x)=x+1 $$ on $$ \displaystyle (-1,0) $$

$$ \displaystyle f(x)=-x+1 $$ on $$ \displaystyle (0,1) $$

This function is an even function.

This is proven by the fact that $$ \displaystyle f(-x)=f(x) $$

Since this is an even function and of period 2 its Fourier series reduces to the Fourier cosine series shown here:

$$ \displaystyle f(x)=a_0 + \sum_{n = 1}^{\infty} a_n\cos(n\pi x) $$

Where the coefficients are equal to:

$$ \displaystyle a_0=\int_{0}^{1}f(x)dx $$

$$ \displaystyle a_n=2\int_{0}^{1}f(x)\cos(n\pi x)dx $$

$$ \displaystyle a_0=\int_{0}^{1}(1-x) $$

$$ \displaystyle a_0=\frac{1}{2} $$

$$ \displaystyle a_n=2\int_{0}^{1}(1-x)\cos(n\pi x)dx $$

$$ \displaystyle a_n=2(\int_{0}^{1}\cos(n\pi x)dx - \int_{0}^{1}x\cos(n\pi x)dx ) $$

Using integration by parts and evaluating on the interval:

$$ \displaystyle a_n=2(\frac{1-\cos(n\pi)}{n^{2} \pi^{2}}) $$

Plugging this back into the function equation:

$$ \displaystyle f(x)=\frac{1}{2} + \sum_{n = 1}^{\infty} \frac{2-2\cos(n\pi)}{n^{2} \pi^{2}} \cos(n\pi x) $$

Expanding this, one obtains:

$$ \displaystyle f(x)=\frac{1}{2} + \frac{4}{\pi^2}(\cos \pi x+ \frac{1}{9}\cos3 \pi x+ \frac{1}{25}\cos5\pi x+...) $$

If n=2

$$ \displaystyle f(x)=\frac{1}{2} + \frac{4}{\pi^2}(\cos \pi x) $$

If n=4

$$ \displaystyle f(x)=\frac{1}{2} + \frac{4}{\pi^2}(\cos \pi x+ \frac{1}{9}\cos3 \pi x) $$

If n=8

$$ \displaystyle f(x)=\frac{1}{2} + \frac{4}{\pi^2}(\cos \pi x+ \frac{1}{9}\cos3 \pi x+ \frac{1}{25}\cos5\pi x+ \frac{1}{49}\cos7\pi x) $$

Problem #15
>> x=linspace(-pi,pi)

>> y2=(4/pi)*sin(x)

>> y4=(4/pi)*(sin(x)-(1/9)*sin(3*x))

>> y8=(4/pi)*(sin(x)-(1/9)*sin(3*x)+(1/25)*sin(5*x)-(1/49)*sin(7*x))

>> plot(x,y2,x,y4,x,y8)

Where y2 is blue, y4 is green and y8 is red

Problem #17
>> x=linspace(-1,1)

>> f2=.5+(4/pi^2)*cos(pi*x)

>> f4=.5+(4/pi^2)*(cos(pi*x)+(1/9)*cos(3*pi*x))

>> f8=.5+(4/pi^2)*(cos(pi*x)+(1/9)*cos(3*pi*x)+(1/25)*cos(5*pi*x)+(1/49)*cos(7*pi*x))

>> plot(x,f2,x,f4,x,f8)

Where f2 is blue, f4 is green and f8 is red

=R6.4 - Solve an ODE with a Fourier Series Excitation=

Problem Statement
Consider the L2-ODE-CC with the window function from R6.2, $$f(x)$$, as excitation:
 * $$y'' - 3y' + 2y = r(x)$$
 * $$r(x) = f(x)$$

And the initial conditions:
 * $$y(0) = 1,\;y'(0) = 0$$

Part 1
Find $$y_n(x)$$ such that:
 * $$y_n'' + a y_n' + b y_n = r_n(x)$$

with the same initial conditions.

Plot $$y_n(x)$$ for $$n=0,1$$, for $$x \in \left[ 0, \, 10 \right]$$.

Part 2
Use the MATLAB command ode45 to integrate the L2-ODE-CC, and plot the numerical solution to compare with the analytical solution.

Part 1
From R6.2, we know that our excitation functions are given by:
 * $$r_0(x) = f_0(x) = \frac{A}{2}$$
 * $$r_1(x) = f_1(x)= \frac{A}{2} + \frac{A}{\pi}\sin (\frac{\pi (x-.25)}{2})$$

Case $$n = 0$$
Setting $$n=0$$ yields the following:

$$r_0(x) = \frac12 A$$

So the ODE takes the form:

$$y''(x)-3y'(x)+2y(x)=A/2$$

Inputting into Wolfram Alpha with the initial conditions:

$$y(0)=1$$ and $$y'(0)=0$$

Yields the solution:

$$ y_{0}(x) = \frac14 (A (e^x-1)^2-4 e^x (e^x-2))$$

Case $$n = 1$$
Setting $$n=1$$ yields the following:

$$r_1(x) = \frac12 A + \frac{A}{\pi} \sin{\frac{\pi \left(x - \frac14 \right)}2}$$

So the ODE takes the form:

$$y''(x)-3y'(x)+2y(x)=\frac12 A + \frac{A}{\pi} \sin{\frac{\pi \left(x - \frac14 \right)}2}$$

Inputting into Wolfram Alpha with the initial conditions:

$$y(0)=1$$ and $$y'(0)=0$$

Yields the solution:

$$ y_{1}(x) = -0.598093 A e^x+0.283756 A e^{2x}+0.019468 A sin(1.5708 x)+0.0643367 A cos(1.5708 x)+0.25 A+2e^x-e^{2 x}$$

Part 2
Using the following Matlab function and inputs allows us to compare the analytical and numerical solutions:

 * [[Image:Egm4313.s12.team8.r6.4.y0.png]]

 * [[Image:Egm4313.s12.team8.r6.4.y1.png]]

 * [[Image:Egm4313.s12.team8.r6.4.y+y0+y1.png]]

 * [[Image:Egm4313.s12.team8.r6.4.y+y0+y1.zoomed.png]]

The zoomed plot shows, as expected, an increase in accuracy as n increases from 0 to 1.

=R6.5 - Analyzing Accuracy of Different Approximations=

Problem Statement
$$ \displaystyle \text{Part 1. R4.2, p7c-26} $$ $$ \displaystyle y_n(x)=y_{h,n}(x)+y_{p,n}(x) $$ $$ \displaystyle \text{For each of n=3,5,9, redisplay the expressions for } $$ $$ \displaystyle \text{the 3 functions } y_{p,n}(x), y_{h,n}(x), y_n(x), \text{ and plot these } $$ $$ \displaystyle \text{3 functions separately over the interval } \left[ 0,20 \pi \right]. $$

$$ \displaystyle \text{Exact solution: } y(x) = y_h(x)+y_p(x) $$ $$ \displaystyle \text{Redisplay the expressions for } y_p(x), y_h(x), y(x) $$

$$ \displaystyle \text{Superpose each of the above plot with that of the}$$ $$ \displaystyle \text{exact solution.} $$

$$ \displaystyle \text{Part 2. R4.3, p7c-28} $$ $$ \displaystyle \text{Understand and run the TA's code to produce} $$ $$ \displaystyle \text{a similar plot, but over a larger interval } \left[ 0, 10 \right]. \text{ Do } $$ $$ \displaystyle \text{zoom-in plots about points } x = -0.5, 0, +0.5 \text{ and } $$ $$ \displaystyle \text{comment on the accuracy of different approximations.} $$

$$ \displaystyle \text{Part 3. R4.4, p7c-29} $$ $$ \displaystyle \text{Understand and run the TA's code to produce} $$ $$ \displaystyle \text{a similar plot, but over a larger interval } \left[ 0.9, 10 \right], \text{ and } $$ $$ \displaystyle \text{for } n=4,7. \text{ Do zoom-in plots about } x = 1, 1.5, 2, 2.5 $$ $$ \displaystyle \text{and comment on the accurace of the approximations.} $$

Solution
$$ \displaystyle \text{Part 1.}$$

$$ \displaystyle \text{n = 3} $$ $$ \displaystyle \text{Particular solution: } $$ $$ \displaystyle y_{p,3}(x)=-9.9206*10^{-5}x^7-0.0010x^6-0.0031x^5-0.0078x^4 \cdots $$ $$ \displaystyle -0.0990x^3-0.3984x^2-0.3984x-0.1992 $$

$$ \displaystyle \text{Homogeneous solution: } $$ $$ \displaystyle y_{h,3}(x)=2e^x-0.8008e^{2x} $$

$$ \displaystyle \text{General solution: } $$ $$ \displaystyle y_{3}(x)=2e^x-0.8008e^{2x}-9.9206*10^{-5}x^7-0.0010x^6-0.0031x^5 \cdots $$ $$ \displaystyle -0.0078x^4-0.0990x^3-0.3984x^2-0.3984x-0.1992 $$

$$ \displaystyle \text{n = 5} $$ $$ \displaystyle \text{Particular solution: } $$ $$ \displaystyle y_{p,5}(x)=-1.2526*10^{-8}x^{11}-2.0668*10^{-7}x^{10}-1.0334*10^{-6}x^9 \cdots $$ $$ \displaystyle -4.6503*10^{-6}x^8-1.1781*10^{-4}x^7-0.0011x^6-0.0033x^5 \cdots $$ $$ \displaystyle -0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.2000 $$

$$ \displaystyle \text{Homogeneous solution: } $$ $$ \displaystyle y_{h,5}(x)=2.0001e^x-0.8001e^{2x} $$

$$ \displaystyle \text{General solution: } $$ $$ \displaystyle y_{5}(x)=2.0001e^x-0.8001e^{2x}-1.2526*10^{-8}x^{11}-2.0668*10^{-7}x^{10} \cdots $$ $$ \displaystyle -1.0334*10^{-6}x^9-4.6503*10^{-6}x^8-1.1781*10^{-4}x^7-0.0011x^6 \cdots $$ $$ \displaystyle -0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.2000 $$

$$ \displaystyle \text{n = 9} $$ $$ \displaystyle \text{Particular solution: } $$ $$ \displaystyle y_{p,9}(x)=-4.1103*10^{-18}x^{19}-1.1714*10^{-16}x^{18}-1.0543*10^{-15}x^{17} \cdots $$ $$ \displaystyle -8.9615*10^{-15}x^{16}-4.5405*10^{-13}x^{15}-9.1408*10^{-12}x^{14} \cdots $$ $$ \displaystyle -6.3985*10^{-11}x^{13}-4.1590*10^{-10}x^{12}-1.5021*10^{-8}x^{11} \cdots $$ $$ \displaystyle -2.2040*10^{-7}x^{10}-1.1020*10^{-6}x^{9}-4.9591*10^{-6}x^{8} \cdots $$ $$ \displaystyle -1.1904*10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4 \cdots $$ $$ \displaystyle -0.1000x^3-0.4000x^2-0.4000x-0.2000 $$

$$ \displaystyle \text{Homogeneous solution: } $$ $$ \displaystyle y_{h,9}(x)=2e^x-0.8e^{2x} $$

$$ \displaystyle \text{General solution: } $$ $$ \displaystyle y_{9}(x)=2e^x-0.8e^{2x}-4.1103*10^{-18}x^{19}-1.1714*10^{-16}x^{18} \cdots $$ $$ \displaystyle -1.0543*10^{-15}x^{17}-8.9615*10^{-15}x^{16}-4.5405*10^{-13}x^{15} \cdots $$ $$ \displaystyle -9.1408*10^{-12}x^{14}-6.3985*10^{-11}x^{13}-4.1590*10^{-10}x^{12} \cdots $$ $$ \displaystyle -1.5021*10^{-8}x^{11}-2.2040*10^{-7}x^{10}-1.1020*10^{-6}x^{9} \cdots $$ $$ \displaystyle -4.9591*10^{-6}x^{8}-1.1904*10^{-4}x^7-0.0011x^6 \cdots $$ $$ \displaystyle -0.0033x^5-0.0083x^4-0.1000x^3-0.4000x^2-0.4000x-0.2000 \cdots $$

$$ \displaystyle \text{Exact Overall solution: } $$ $$ \displaystyle y(x)=1.5e^x-0.8e^{2x}+0.3 \cos x+0.1 \sin x $$

$$ \displaystyle \text{Part 2.}$$ $$ \displaystyle \text{The following plot displays the Taylor series of } log(1+x) $$ $$ \displaystyle \text{ about } \hat{x} = 0.$$

$$ \displaystyle \text{The zoom-in about } x = -0.5 \text{ is displayed below. Here, we can}$$ $$ \displaystyle \text{see the plot broken, showing the inaccuracy beyond this point. }$$

$$ \displaystyle \text{The zoom-in about } x = 0 \text{ is displayed below. Here, we can}$$ $$ \displaystyle \text{see that the plot is very exact, displaying the accuracy at x = 0.}$$

$$ \displaystyle \text{The zoom-in about } x = 0.5 \text{ is displayed below. Here, we can}$$ $$ \displaystyle \text{see that the plot has begun to veer off the exact solution.}$$

$$ \displaystyle \text{Part 3.}$$ $$ \displaystyle \text{The following plot displays the Taylor series of } log(1+x) $$ $$ \displaystyle \text{ about } \hat{x} = 1.$$

$$ \displaystyle \text{The zoom-in about } x = 1 \text{ is displayed below. }$$

$$ \displaystyle \text{The zoom-in about } x = 1.5 \text{ is displayed below. }$$

$$ \displaystyle \text{The zoom-in about } x = 2 \text{ is displayed below. }$$

$$ \displaystyle \text{The zoom-in about } x = 2.5 \text{ is displayed below. }$$

$$ \displaystyle \text{It is noted that the plots begin to diverge from the exact}$$ $$ \displaystyle \text{solution beginning (ever so slightly) at x = 1. }$$

=R6.6 - Verify Particular Solution to ODE=

Problem Statement
Verify the following:
 * $$6 e^{-2x} ( n \cos{3x} - m \sin{3x} )$$

As the left hand side to the following ODE:
 * $$y_p'' + 4 y_p' + 13y_p = 2 e^{-2x} \cos{3x}$$

Where:
 * $$y_p(x) = x e^{-2x} ( m \cos{3x} + n \sin{3x} )$$

Then, solve for m and n.

Solution
First, we find and simplify each term of the left hand side:
 * $$\begin{array}{rl}

y_p &= x e^{-2x} ( m \cos{3x} + n \sin{3x} ) \\ &= e^{-2x} ( m x \cos{3x} + n x \sin{3x} ) \\ 13 y_p &= e^{-2x} ( 13 m x \cos{3x} + 13 n x \sin{3x} ) \\ &\\ y_p' &= e^{-2 x} x (3 n \cos{3 x}-3 m \sin{3 x})+e^{-2 x} (m \cos{3 x}+n \sin{3 x})-2 e^{-2 x} x (m \cos{3 x}+n \sin{3 x}) \\ &= e^{-2 x} ((-2 m x+m+3 n x) \cos{3 x}+ (-3 m x-2 n x+n)\sin{3 x}) \\ 4 y_p' &= e^{-2 x} ((-8 m x+4m+12 n x) \cos{3 x}+ (-12 m x-8 n x+4n)\sin{3 x}) \\ &\\ y_p'' &= e^{-2 x} (3 n \cos{3 x}-3 m \sin{3 x}) \\ &\;\;+\left(e^{-2 x}-2 e^{-2 x} x\right) (3 n \cos  (3 x)-3 m \sin{3 x})\\ &\;\;+e^{-2 x} x (-9 m \cos{3 x}-9 n \sin{3 x})\\ &\;\;-2 e^{-2 x} (m  \cos{3 x}+n \sin{3 x})\\ &\;\;+4 e^{-2 x} x (m \cos{3 x}+n \sin{3 x})\\ &\;\;-2 e^{-2 x} (x (3 n \cos{3 x}-3 m \sin{3 x})+m \cos{3 x}+n \sin{3 x}) \\ &= e^{-2 x} ( (-5 x m - 4 m - 12 x n + 6 n) \cos{3 x} + (12 m x - 6 m - 5 x n - 4 n) \sin{3 x}) \end{array}$$

Then, we combine the three terms and simplify:
 * $$\begin{array}{rl}

y_p'' + 4 y_p' + 13 y_p &= e^{-2 x} ( (-5 x m - 4 m - 12 x n + 6 n) \cos{3 x} + (12 m x - 6 m - 5 x n - 4 n) \sin{3 x}) \\ &\;\;+e^{-2 x} ((-8 m x+4m+12 n x) \cos{3 x}+ (-12 m x-8 n x+4n)\sin{3 x}) + e^{-2x} ( 13 m x \cos{3x} + 13 n x \sin{3x} ) \\ &= 6 e^{-2x} ( n \cos{3x} - m \sin{3x} )\end{array}$$

This verifies the given left hand side. To solve for m and n, we set this equal to the right hand side of the ODE:
 * $$\begin{array}{rcl}

6 e^{-2x} ( n \cos{3x} - m \sin{3x} ) &=& 2 e^{-2x} \cos{3x} \\ 6 ( n \cos{3x} - m \sin{3x} ) &=& 2 \cos{3x} \\ n \cos{3x} - m \sin{3x} &=& \frac13 \cos{3x} \end{array}$$

Therefore:
 * $$\begin{array}{rl}

n &= \frac13 \\ m &= 0 \\ y_p(x) &= \frac13 x e^{-2x} \sin{3x} \end{array}$$

=R6.7 - Seperate ODE for Different Variables=

Problem Statement
Find the seperated ODEs for the heat equation

$$\ \frac{\partial u}{\partial t}=\kappa \frac{\partial ^2 u}{\partial x^2}$$

Solution
Assume $$\ u(x,t)=F(x)\cdot G(t)$$

where:

$$\ F(x)=f(x)$$, a function of only x

$$\ G(t)=f(t)$$, a function of only t

Finding first and second partial derivatives of u with restpect to x gives:

$$\ \frac{\partial u}{\partial x}=F'(x) \cdot G(t)$$

$$\ \frac{\partial^2 u}{\partial x^2}=F''(x) \cdot G(t)$$

Finding first and second partial derivates of u with respect to t gives:

$$\ \frac{\partial u}{\partial t}=F(x) \cdot \dot{G}(t)$$

$$\ \frac{\partial^2 u}{\partial ^2t}=F(x) \cdot \ddot{G}(t)$$

Plugging these into the heat equation gives:

$$\ \frac{\partial u}{\partial t}=\kappa \frac{\partial ^2u}{\partial x^2}  =>  F(x) \cdot \dot{G}(t)=\kappa F''(x) \cdot G(t)$$

Combining like terms gives:

$$\ \frac{\dot G(t)}{\kappa G(t)}=\frac{F''(x)}{F(x)}=c$$

where c is a constant

$$\ \frac{\dot G(t)}{\kappa G(t)}=c   => \dot G(t)=c \kappa G(t)  =>  \dot G(t)-c \kappa G(t)=0$$

$$\ \frac{F(x)}{F(x)}=c    => F(x)=cF(x)     =>    F''(x)-cF(x)=0$$

$$\ \dot G(t)-c \kappa G(t)=0$$

$$\ F''(x)-cF(x)=0$$

=Team Contributions=

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text:
User:Viriditas

Do what you love and love what you do.

title: User:Viriditas


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text:
Talk:Controversies in Science/What killed the dinosaurs/A critique of More evidence that the Chicxulub impact predates the K/T mass extinction. Meteoritics & Planetary Science

Interesting point by Gerta Keller, yet it is evident that she chooses to ignore certain critical elements. Namely, if she acknowledged a meteor did in fact impact the earth and left a massive crater, what impacts would that have on life in the surrounding region? It is extremely likely that such an event would literally obliterate any land-based dinosaurs. Evidence of remaining animals or organisms after the impact may be true, but not enough is known of the survival capabilities of many organisms or species in the cretaceous period. In addition, how many samples did her study sample in order to arrive at a very specific 300 thousand year discrepancy? (a 0.002% difference). Dlodh574 (talk) 00:52, 9 April 2012 (UTC)

 * Yes, there was life after the asteroid hit in the Yucatan Peninsula in northestern Mexico. However, because the asteroid produced large amounts of NOGH-related CH4, it slowly killed off the dinosaurs.--Nzurg (talk) 00:07, 11 April 2012 (UTC)
 * I agree with Nzurg, a common misconception of the asteroid theory is that the impact of the asteroid hitting the Earth caused instantanious extinction, when in actuality, studies have shown that the extinction was a lengthy process starting with phytoplankton, moving onto herbivores and eventually, when all other food sources were depleated or gone entirely, the carnivores also starved to death until extinction. Jstal531 (talk) 01:26, 11 April 2012 (UTC)

Just because other life forms survived the meteorite does not mean that the dinosaurs did. Perhaps the life forms that surivived were able to survive in environments that did not require the sun to live, such as the bottom of the ocean. Dinosaurs did require the sun to live, and when the dust particles blocked the sun after the meteorite hit, they died. Bmina836 (talk) 00:02, 13 April 2012 (UTC)

I agree with Bmina836. Every specie requires different envrionments to survive and some can adapt to change quicker and better than others. The dinosaurs could not adapt to the major changes in their environment and slowly died off after the impact. Just because some species were able to survive and there was life afterwards does not make any conslusion about the survival of dinosaurs.Jmay3355 (talk) 22:25, 13 April 2012 (UTC)

 * The trajectory model suggests that the photochemical reactions formed ground-level O3 whose concentration was low (1.0 ppm) but much greater than the current level of 0.04 ppm. Because of this, creatures such as amphibians and reptiles survived because they were able to retreat to areas where O3 wouldn't affect them. Creatures that lived above the water were more affected because the NOx and CH4 reacted with the sunlight producing a deadly level of O3. .--Randana (talk) 05:11, 4 April 2012 (UTC)

 * As said in the critique above there was still life after the Chicxulub. But the main animal many scientists are focusing on is the dinosaurs. Many mammals after the extinction were also able to dig under ground where they were lower levels of O3. Because of this humans were later able to evolve which was on Dinosaur week on Discovery. Without the extinction of the dinosaurs we would not have been able to exist because they ate mammals at a accelerated rate. Therefore evolving would have been out of the question.

And as said in my previous critique many animals that lived above ground would have starved to death do to the lack of food. If the sun was not able to get through the dust many plants would die; therefore the animals eating the plants would die and the food chain would continue to to be affected from then on.Sambo1428 (talk)

 * This study does not prove that an asteroid didn't kill of the dinosaurs but that selected species were able to survive longer than others after the asteroid did hit. This is true for most disasters however, as those that are better adopted to the changing conditions will survive. So while this study can provide a partial objection it does not provide an absolute refutation.Alaxative (talk) 05:22, 12 April 2012 (UTC)