text:
Table 2.
How Satisfied Are You with Your Current Life?
%
Very satisfied 20.14
Satisfied 52.79
Somewhat satisfied 19.50
Somewhat unsatisfied 6.99
Not satisfied at all 0.49
%
Very satisfied 20.14
Satisfied 52.79
Somewhat satisfied 19.50
Somewhat unsatisfied 6.99
Not satisfied at all 0.49

Source: Authors’ calculations.

Close Modal

format: plain


text:
# Knapsack Cipher

## Backpack problem¶

First, let's introduce the backpack problem. Suppose a backpack can weigh W. Now there are n items with weights of $a_1, a_2,..., a_n$. We want to ask which items can fit the backpack. Filled up and each item can only be loaded once. This is actually solving such a problem.

$$x_1a_1 + x_2a_2 +, ..., + x_na_n = W$$

All of these $x_i$ can only be 0 and 1. Obviously we have to enumerate all the combinations of n items to solve this problem, and the complexity is $2^n$, which is the beauty of backpack encryption.

When encrypting, if we want to encrypt the plaintext as x, then we can represent it as an n-bit binary number and then multiply it by $a_i$ to get the encrypted result.

But what should I do when decrypting? We did make it difficult for others to decrypt the ciphertext, but we really have no way to decrypt the ciphertext.

But when $a_i$ is super-incremental, we have a solution. The so-called super-increment means that the sequence satisfies the following conditions.

$$a_i>\sum_{k=1}^{i-1}a_k$$

That is, the ith number is greater than the sum of all the previous numbers.

Why can you decrypt it if you meet such a condition? This is because if the encrypted result is greater than $a_n$, the preceding coefficient must be 1. On the contrary, the equation cannot be established anyway. Therefore, we can get the corresponding plaintext immediately.

However, this has another problem. Since $a_i$ is public, if the attacker intercepts the ciphertext, it is easy to crack such a password. In order to make up for this problem, an encryption algorithm such as Merkle–Hellman appears. We can use the initial backpack set as the private key, the transformed backpack set as the public key, and then slightly change the encryption process.

Although the super-increment sequence is mentioned here, it is not said how it is generated.

## Merkle–Hellman¶

### Public private key generation¶

#### Generating a private key¶

The private key is our initial backpack set. Here we use the super-increment sequence, how to generate it? We can assume that $a_1=1$, then $a_2$ is greater than 1, and similarly can generate subsequent values in turn.

#### Generating a public key¶

In the process of generating a public key, the operation of modular multiplication is mainly used.

First, we generate the modulus m of the modular multiplication, here we want to make sure

$$m>\sum_{i=1}^{i=n}a_i$$

Second, we choose the multiplier w of the modular multiplication as the private key and ensure

$$gcd(w,m)=1$$

After that, we can generate the public key by the following formula.

$$b_i \equiv w a_i \bmod m$$

And this new backpack set $b_i$ and m as the public key.

### encryption and decryption¶

#### Encryption¶

Suppose we want to encrypt the plaintext as v, each bit is $v_i$, then the result of our encryption is

$$Sum_ {i = 1} ^ {n} i = b_iv_i m way$$

#### Decryption¶

For the decryption side, we can first ask for the inverse of $m^{-1}$ for m.

Then we can multiply the obtained ciphertext by $w^{-1}$ to get the plaintext, because

$$Sum_ {i = 1} ^ {w} i = n ^ {- 1} b_iv_i way m = sum_ {i = 1} ^ {n} i = a_iv_i m way$$

here has

$$b_i \equiv w a_i \bmod m$$

The encrypted message for each block is less than m, so the result is naturally plaintext.

### 破¶

The system was deciphered two years after the proposed encryption system. The basic idea of deciphering is that we do not necessarily need to find the correct multiplier w (ie trapdoor information), just find the arbitrary modulus m' and The multiplier w' can be used to generate a super-incrementing backpack vector by using w' to multiply the public backpack vector B.

### Examples¶

Here we take Archaic in 2014 ASIS Cyber Security Contest Quals as an example, [topic link] (https://github.com/ctfs/write-ups-2014/tree/b02bcbb2737907dd0aa39c5d4df1d1e270958f54/asis-ctf-quals-2014/archaic ).

First look at the source program

secret = 'CENSORED'

msg_bit = bin(int(secret.encode('hex'), 16))[2:]

First we get all the bits of secret.

Second, use the following function to get the keypair, including the public and private keys.

keyPair = makeKey (curtain (msg_bit))

Carefully analyze the makekey function as follows

def makeKey(n):

privKey = [random.randint (1, 4 ** n)]
s = privKey [0]
for i in range(1, n):

privKey.append(random.randint(s + 1, 4**(n + i)))

s + = privKey [i]
q = random.randint (privKey [n-1] + 1, 2 * privKey [n-1])
r = random.randint(1, q)

while gmpy2.gcd(r, q) != 1:

r = random.randint(1, q)

pubKey = [ r*w % q for w in privKey ]

return privKey, q, r, pubKey

It can be seen that prikey is a super-incremental sequence, and the obtained q is larger than the sum of all the numbers in prikey. In addition, we get r, which is exactly the same as q, which indicates that the encryption is a backpack encryption.

Sure enough, the encryption function is to multiply each bit of the message by the corresponding public key and sum.

def encrypt(msg, pubKey):

msg_bit = msg

n = only (pubKey)
cipher = 0

i = 0

for bit in msg_bit:

cipher += int(bit)*pubKey[i]

i += 1

return bin(cipher)[2:]

For the cracked script we use the script on [GitHub] (https://github.com/ctfs/write-ups-2014/tree/b02bcbb2737907dd0aa39c5d4df1d1e270958f54/asis-ctf-quals-2014/archaic). Make some simple modifications.

import binascii
# open the public key and strip the spaces so we have a decent array

fileKey = open ("pub.Key", 'rb')
pubKey = fileKey.read().replace(' ', '').replace('L', '').strip('[]').split(',')

nbit = only (pubKey)
# open the encoded message

fileEnc = open ("enc.txt", 'rb')

print "start"

# create a large matrix of 0's (dimensions are public key length +1)

A = Matrix(ZZ, nbit + 1, nbit + 1)

# fill in the identity matrix

for i in xrange(nbit):

A[i, i] = 1

# replace the bottom row with your public key

for i in xrange(nbit):

A[i, nbit] = pubKey[i]

# last element is the encoded message

A[nbit, nbit] = -int(encoded)

res = A.LLL()

for i in range(0, nbit + 1):

# print solution

M = res.row(i).list()

flag = True

for m in M:

if m != 0 and m != 1:

flag = False

break

if flag:

print i, M

M = ''.join(str(j) for j in M)

# remove the last bit

M = M[:-1]

M = hex(int(M, 2))[2:-1]

print M

Decoded after output

295 [1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0]

415349535f3962643364356664323432323638326331393536383830366130373036316365
>>> import binascii
>>> binascii.unhexlify('415349535f3962643364356664323432323638326331393536383830366130373036316365')

'ASIS_9bd3d5fd2422682c19568806a07061ce'

It should be noted that the matrix of res obtained by the LLL attack we only contains the 01 value is the result we want, because when we encrypt the plaintext, it will be decomposed into binary bit strings. In addition, we need to remove the last number of the corresponding row.

flag 是 ASIS_9bd3d5fd2422682c19568806a07061ce

### Title¶

• 2017 national classic


text:
# Category:Identity Mappings

This category contains results about Identity Mappings.
Definitions specific to this category can be found in Definitions/Identity Mappings.

The identity mapping of a set $S$ is the self-map $I_S: S \to S$ defined as:

$I_S = \set {\tuple {x, y} \in S \times S: x = y}$

or alternatively:

$I_S = \set {\tuple {x, x}: x \in S}$

That is:

$I_S: S \to S: \forall x \in S: \map {I_S} x = x$

## Subcategories

This category has the following 8 subcategories, out of 8 total.


text:
Align a position weight matrix against a sequence by using bioperl
2
0
Entering edit mode
5.5 years ago

I everyone.

I'm trying to find a certain probabilistic based pattern in a DNA sequence by the using of position weight matrix. That is to say, I specify several nucleotides with their probabilities for the same position.

I'm using perl and bioperl modules to achieve this. I was able to create an psm object using the modules Bio::Matrix::PSM::SiteMatrix and Bio::Matrix::PSM::IO, but I have no idea about how to perform an alignment against a sequence using this matrix.

What module could I use to perform the alignment?

bioperl alignment position weight matrix pwm • 1.8k views
0
Entering edit mode
5.5 years ago
Joseph Hughes ★ 2.9k

You probably want to use Bio::Tools::dpAlign

0
Entering edit mode

Hi Joseph, thanks for your answer. I've checked your link, but I think it is used to perform alignments between sequences. In my case, I need to find matches of a position weight matrix in a sequence. A position weight matrix is a matrix that have the following structure:

        A    C   T   G
----------------------------------
POS1|   50  20  20  10
POS2|   20  20  10  50
POS3|   50   0  50   0
POS4|   25  25  25  25

This kind of matrix represents the probability of each nucleotide for each position. The idea is to find matches in a sequence that were accomplished by the matrix.

This feature is useful if you want to find protein binding sites in the genome, for example. I've seen applications that can perform this task, but I want to do it in a programmatically way.

0
Entering edit mode
5.5 years ago

Hi, finally I found a web service that performs this task at Regulatory Sequence Analysis Tools (RSAT) webpage. Concretely I'm interested in matrix-scan tool, but you can find another interesting tools too.

The web browser version of scan-matrix is at http://tagc.univ-mrs.fr/rsa-tools/matrix-scan_form.cgi. The documentation about how to use it in a programmatically way by using SOAP web services is at http://tagc.univ-mrs.fr/rsa-tools/.

Cheers, Fran.


text:
There are three main types of modern acoustic guitar: the classical guitar (nylon-string guitar), the steel-string acoustic guitar, and the archtop guitar, which is sometimes called a "jazz guitar". The tone of an acoustic guitar is produced by the strings' vibration, amplified by the hollow body of the guitar, which acts as a resonating chamber. The classical guitar is often played as a solo instrument using a comprehensive finger-picking technique where each string is plucked individually by the player's fingers, as opposed to being strummed. The term "finger-picking" can also refer to a specific tradition of folk, blues, bluegrass, and country guitar playing in the United States. The acoustic bass guitar is a low-pitched instrument that is one octave below a regular guitar.
The guitar is a fretted musical instrument that usually has six strings.[1] It is typically played with both hands by strumming or plucking the strings with either a guitar pick or the finger(s)/fingernails of one hand, while simultaneously fretting (pressing the strings against the frets) with the fingers of the other hand. The sound of the vibrating strings is projected either acoustically, by means of the hollow chamber of the guitar (for an acoustic guitar), or through an electrical amplifier and a speaker.
Adding a minor seventh to a major triad creates a dominant seventh (denoted V7). In music theory, the "dominant seventh" described here is called a major-minor seventh, emphasizing the chord's construction rather than its usual function.[27] Dominant sevenths are often the dominant chords in three-chord progressions,[18] in which they increase the tension with the tonic "already inherent in the dominant triad".[28]
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The bass guitar (also called an "electric bass", or simply a "bass") is similar in appearance and construction to an electric guitar, but with a longer neck and scale length, and four to six strings. The four-string bass, by far the most common, is usually tuned the same as the double bass, which corresponds to pitches one octave lower than the four lowest pitched strings of a guitar (E, A, D, and G). (The bass guitar is a transposing instrument, as it is notated in bass clef an octave higher than it sounds (as is the double bass) to avoid excessive ledger lines.[jargon]) Like the electric guitar, the bass guitar has pickups and it is plugged into an amplifier and speaker for live performances.

In major-thirds tuning, the interval between open strings is always a major third. Consequently, four frets suffice to play the chromatic scale. Chord inversion is especially simple in major-thirds tuning. Chords are inverted simply by raising one or two notes by three strings. The raised notes are played with the same finger as the original notes.[22][23] In contrast, in standard tuning, the shape of inversions depends on the involvement of the irregular major-third.[24]
A slide, (neck of a bottle, knife blade or round metal or glass bar or cylinder) is used in blues and rock to create a glissando or "Hawaiian" effect. The slide is used to fret notes on the neck, instead of using the fretting hand's fingers. The characteristic use of the slide is to move up to the intended pitch by, as the name implies, sliding up the neck to the desired note. The necks of bottles were often used in blues and country music as improvised slides. Modern slides are constructed of glass, plastic, ceramic, chrome, brass or steel bars or cylinders, depending on the weight and tone desired (and the amount of money a guitarist can spend). An instrument that is played exclusively in this manner (using a metal bar) is called a steel guitar or pedal steel. Slide playing to this day is very popular in blues music and country music. Some slide players use a so-called Dobro guitar. Some performers who have become famous for playing slide are Robert Johnson, Elmore James, Ry Cooder, George Harrison, Bonnie Raitt, Derek Trucks, Warren Haynes, Duane Allman, Muddy Waters, Rory Gallagher, and George Thorogood.
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For the longest time I have "messed around" with my guitar. I bought a book on Jazz guitar chords many years ago and found that those fingerings were way above my skill level. I decided to get a basic chord book to reinitiate myself. This is that book. Going through these pages has re-ignited my lust to learn more about playing my guitar. One chord at a time. Simple. Skip to another key just as simple. Actually, I haven't realized how many chords I really know how to play. I just need to learn the name of those chords. This book will help with that too. Give it a try.
In major-thirds (M3) tuning, the chromatic scale is arranged on three consecutive strings in four consecutive frets.[83][84] This four-fret arrangement facilitates the left-hand technique for classical (Spanish) guitar:[84] For each hand position of four frets, the hand is stationary and the fingers move, each finger being responsible for exactly one fret.[85] Consequently, three hand-positions (covering frets 1–4, 5–8, and 9–12) partition the fingerboard of classical guitar,[86] which has exactly 12 frets.[note 1]

#### Each Music Monday will feature local performers strumming their guitars on the Plaza. Bring your lunch or buy one from the Fork in the Road Food Truck and soak in the music from a variety of excellent Bozeman performers. This program is free to the public. Museum admission is not required. In the case of inclement weather, these performances will take place in the Hager Auditorium with space for lunch available in the Lower Lobby. Food and drinks are not allowed in the auditorium.

The fingerboard, also called the fretboard, is a piece of wood embedded with metal frets that comprises the top of the neck. It is flat on classical guitars and slightly curved crosswise on acoustic and electric guitars. The curvature of the fretboard is measured by the fretboard radius, which is the radius of a hypothetical circle of which the fretboard's surface constitutes a segment. The smaller the fretboard radius, the more noticeably curved the fretboard is. Most modern guitars feature a 12" neck radius, while older guitars from the 1960s and 1970s usually feature a 6-8" neck radius. Pinching a string against a fret on fretboard effectively shortens the vibrating length of the string, producing a higher pitch.
The loud, amplified sound and sonic power of the electric guitar played through a guitar amp has played a key role in the development of blues and rock music, both as an accompaniment instrument (playing riffs and chords) and performing guitar solos, and in many rock subgenres, notably heavy metal music and punk rock. The electric guitar has had a major influence on popular culture. The guitar is used in a wide variety of musical genres worldwide. It is recognized as a primary instrument in genres such as blues, bluegrass, country, flamenco, folk, jazz, jota, mariachi, metal, punk, reggae, rock, soul, and many forms of pop.
We were privileged to have worked with hundreds of top artists and educators who authored the original magazine articles and then recorded audio lessons in their own studios. While video is the medium of choice today, these 1,200+ audio lessons still deliver amazing instruction from some of the best in the biz — perfect for those long drives and commutes!
As a general rule, brass strings are always going to be brighter than bronze strings. Though, counterintuitively, many brass strings go by the moniker of “80/20” bronze. These strings are actually the one in the same. Brass, or 80/20 bronze as it’s often known, is made from 80% copper and 20% zinc. This gives the strings a bright and cutting voice, though when used on guitars that already have a prominent high-end response it can make an instrument sound thin and tinny. For best results, use brass strings on a guitar that’s an OM size or larger (so this would include OM guitars, dreadnoughts, and jumbos).
TrueFire's In The Jam delivers an unparalleled jamming experience for the practicing musician. The next best thing to being there live, In The Jam puts YOU in the jam with top artists. Each edition includes multi-track video jams organized into separate video and audio tracks for each of the instruments. You can mute, solo or adjust the volume of any track.

I use this book in my Music Therapy sessions with Geriatric clients--and they love it! The songs are well-known and fun--mostly folk tunes like I've Been Working on the Railroad and My Bonnie Lies Over the Ocean. Chords are SO simple--no bar chords (so you may need a capo to adjust the key). If you're looking for the most basic way to play favorite songs...this is the way to go!
For the longest time I have "messed around" with my guitar. I bought a book on Jazz guitar chords many years ago and found that those fingerings were way above my skill level. I decided to get a basic chord book to reinitiate myself. This is that book. Going through these pages has re-ignited my lust to learn more about playing my guitar. One chord at a time. Simple. Skip to another key just as simple. Actually, I haven't realized how many chords I really know how to play. I just need to learn the name of those chords. This book will help with that too. Give it a try.
On a day when there's a temptation to go into a dark place, and only see all the bad stuff there is in the world ... greed, cruelty, exploitation, selfishness ... I get days like that pretty often .... it's great to find someone giving out, and giving out good, and operating on an honour basis ... There are so many people who can't afford Guitar lessons .... well, here's a wonderful guy who has set up a whole system of teaching guitar ... Blues, Jazz, Rock, even Songwriting, from the basics, tuning the guitar, etc ... upwards ... If you use his site, it's up to you to determine how much you can contribute ... but this is an amazing site .... he is also very aware of issues in the world which need attention ... a great channel .. Check him out. He's a giver.
Solid body seven-string guitars were popularized in the 1980s and 1990s. Other artists go a step further, by using an eight-string guitar with two extra low strings. Although the most common seven-string has a low B string, Roger McGuinn (of The Byrds and Rickenbacker) uses an octave G string paired with the regular G string as on a 12-string guitar, allowing him to incorporate chiming 12-string elements in standard six-string playing. In 1982 Uli Jon Roth developed the "Sky Guitar", with a vastly extended number of frets, which was the first guitar to venture into the upper registers of the violin. Roth's seven-string and "Mighty Wing" guitar features a wider octave range.[citation needed]
If you know how to play an E major chord, then you know how to play an A minor chord—just move the chord whole shape over a string. Make sure your first finger is curled, so the open first string rings clearly. Avoid playing the open sixth string when strumming the A minor chord. There are situations when it makes sense to reverse your second and third fingers when playing the A minor chord.
Are you stuck in a musical rut? New tunings and tricks can help you keep learning guitar in fresh, fun ways. Try one of these great tips from guitar teacher Samuel B. to breathe new life into your guitar playing... One of the first things I tell any new student is that I don't specialize in a formal discipline. If jazz or classical training is your objective, then I'm not your guy. Instead, I specialize primarily in American roots music (that which we tend to casually lump together as "folk" … Read More

Ask any veteran musician, and they'll tell you that the early stages of learning a musical instrument go by a lot smoother when you're having fun. For this reason, Guitar Center strives their hardest to ensure every guitar lesson they offer is fully-engaging and an absolute blast for everyone involved. Whether you're into the warm, natural sound of an acoustic guitar or have aspirations of blowing out ear drums on the biggest stages in town, GC's guitar lessons are designed so that players of all ages, skill levels and tastes learn the chords and scales they need to know (and want to know) in a comfortable environment.
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Jump up ^ This sequence of fifths features the diminished fifth (b,f), which replaces the perfect fifth (b,f♯) containing the chromatic note f♯, which is not a member of the C-major key. The note f (of the C-major scale) is replaced by the note f♯ in the Lydian chromatic scale (Russell, "The fundamental harmonic structure of the Lydian scale", Example 1:7, "The C Lydian scale", p. 5).

The teen and tween years are the make or break years for most people, in deciding whether they’re going to stick with a childhood hobby or interest, for life. As you go through puberty, and begin preparing for life as an adult, and responsibilities pile on your plate, the first things to go tend to be the personal interests you’re passionate about, because they don’t pay the bills. Also, because they’re not an official part of your school curriculum. In other words, what was once a must-have becomes a nice-to-have.
Depending on the program, School of Rock's guitar lessons can cost from around $150 to$350 per month. Exact prices vary between locations. What's included? Unlike most hourly guitar lessons, our programs include weekly private guitar lessons and group rehearsals that inspire confidence and teamwork. Guitar students are also welcome to use our facilities whenever we're open, even if they just want to hangout and learn from or collaborate with other musicians.

The intervals between the notes of a chromatic scale are listed in a table, in which only the emboldened intervals are discussed in this article's section on fundamental chords; those intervals and other seventh-intervals are discussed in the section on intermediate chords. The unison and octave intervals have perfect consonance. Octave intervals were popularized by the jazz playing of Wes Montgomery. The perfect-fifth interval is highly consonant, which means that the successive playing of the two notes from the perfect fifth sounds harmonious.

The musical theory of chords is reviewed, to provide terminology for a discussion of guitar chords. Three kinds of chords, which are emphasized in introductions to guitar-playing,[10][11] are discussed. These basic chords arise in chord-triples that are conventional in Western music, triples that are called three-chord progressions. After each type of chord is introduced, its role in three-chord progressions is noted.
The next step is to figure out what gauge, or thickness of string is best suited to your playing ability and style. Thinner gauges are easier to fret and bend, and may be better suited for beginners, until callouses build up on the fingertips. Heavier strings tend to produce greater volume and fuller tone. Nylon strings are typically categorized as light, medium or heavy tension. Steel strings for both electric and acoustic guitars are categorized by the gauge, or thickness of the lightest string, the high E string, measured in thousandths of an inch, with .09 being a common size for an electric set. A set of electric strings may be labeled .09-.042 - this is the gauge of the lightest and heaviest string in the set. Acoustic sets are a little thicker, so a typical medium-gauge set might be .012-.054.
The ratio of the spacing of two consecutive frets is {\displaystyle {\sqrt[{12}]{2}}} (twelfth root of two). In practice, luthiers determine fret positions using the constant 17.817—an approximation to 1/(1-1/ {\displaystyle {\sqrt[{12}]{2}}} ). If the nth fret is a distance x from the bridge, then the distance from the (n+1)th fret to the bridge is x-(x/17.817).[15] Frets are available in several different gauges and can be fitted according to player preference. Among these are "jumbo" frets, which have much thicker gauge, allowing for use of a slight vibrato technique from pushing the string down harder and softer. "Scalloped" fretboards, where the wood of the fretboard itself is "scooped out" between the frets, allow a dramatic vibrato effect. Fine frets, much flatter, allow a very low string-action, but require that other conditions, such as curvature of the neck, be well-maintained to prevent buzz.
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text:
# The canonical injections of the disjoint union topology are topological embeddings

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Important for progress!

## Statement

Let [ilmath]\big((X_\alpha,\mathcal{J}_\alpha)\big)_{\alpha\in I} [/ilmath] be a collection of topological spaces and let [ilmath](\coprod_{\alpha\in I}X_\alpha,\mathcal{J})[/ilmath] be the disjoint union space of that family. With this construction we get some canonical injections:

• For each [ilmath]\beta\in I[/ilmath] we get a map (called a canonical injection) [ilmath]i_\beta:X_\beta\rightarrow\coprod_{\alpha\in I}X_\alpha[/ilmath] given by [ilmath]i_\beta:x\mapsto(\beta,x)[/ilmath]

We claim that each [ilmath]i_\beta[/ilmath] is a topological embedding[1] (that means [ilmath]i_\beta[/ilmath] is injective and continuous and a homeomorphism between [ilmath]X_\beta[/ilmath] and [ilmath]i_\beta(X_\beta)[/ilmath] (its image))

## Proof

Let [ilmath]\beta\in I[/ilmath] be given.

• The proof that [ilmath]i_\beta:X_\beta\rightarrow\coprod_{\alpha\in I}X_\alpha[/ilmath] by [ilmath]i_\beta:x\mapsto(\beta,x)[/ilmath] consists of three parts:
1. Continuity of [ilmath]i_\beta[/ilmath] - covered on the canonical injections of the disjoint union topology page so not shown on this pag
2. [ilmath]i_\beta[/ilmath] being injective and
3. [ilmath]i_\beta[/ilmath] being a homeomorphism between [ilmath]X_\beta[/ilmath] and [ilmath]i_\beta(X_\beta)[/ilmath]
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
Done on paper, it isn't hard. I want to save my work now though NOTES ARE BELOW

I only cover part 3 here.

We have shown [ilmath]i_\beta:X_\beta\rightarrow\coprod_{\alpha\in I}X_\alpha[/ilmath] is continuous and injective. It only remains to show that it is a homeomorphism onto (in the surjective sense of the word "onto") its image.

• First note that every injection yields a bijection onto its image
• Thus we get a map [ilmath]\overline{i_\beta}:X_\beta\rightarrow i_\beta(\coprod_{\alpha\in I}X_\alpha)[/ilmath] given by [ilmath]\overline{i_\beta}:x\mapsto i_\beta(x)[/ilmath] (note that this means [ilmath]\overline{i_\beta}:x\mapsto(\beta,x)[/ilmath]) which is a bijection
• Next note that every bijection yields an inverse function, so now we have [ilmath](\overline{i_\beta})^{-1}:i_\beta(\coprod_{\alpha\in I}X_\alpha)\rightarrow X_\beta[/ilmath]
• We only really need to show that [ilmath](\overline{i_\beta})^{-1}:i_\beta(\coprod_{\alpha\in I}X_\alpha)\rightarrow X_\beta[/ilmath] is continuous
• Let [ilmath]U\in\mathcal{J}_\beta[/ilmath] (so [ilmath]U[/ilmath] is open in [ilmath]X_\beta[/ilmath]) be given
• Then we must show that [ilmath]((\overline{i_\beta})^{-1})^{-1}(U)\in\mathcal{J} [/ilmath] in order for [ilmath](\overline{i_\beta})^{-1}:i_\beta(\coprod_{\alpha\in I}X_\alpha)\rightarrow X_\beta[/ilmath] to be continuous
• But! [ilmath]((\overline{i_\beta})^{-1})^{-1}(U)=\overline{i_\beta}(U)[/ilmath]
• Recall we defined a set to be open in [ilmath]\coprod_{\alpha\in I}X_\alpha[/ilmath] if its intersection with (the image of) each [ilmath]X_\alpha[/ilmath] is open in [ilmath]X_\alpha[/ilmath] Caution:Terminology is a bit fuzzy here. I need to fix that
• Let [ilmath]\gamma\in I[/ilmath] be given
• If [ilmath]\gamma\ne\beta[/ilmath] then
• [ilmath]\overline{i_\beta}(U)\cap X_\gamma^*=\emptyset[/ilmath] and by definition, [ilmath]\emptyset\in\mathcal{J}_\gamma[/ilmath]
• Caution:This is where the notation gets weird. The image of the emptyset is the empty set, not sets of the form [ilmath](\beta,x)[/ilmath] ....
• If [ilmath]\gamma=\beta[/ilmath] then
• [ilmath]\overline{i_\beta}(U)\cap X_\beta^*=U[/ilmath] Caution:or the image of [ilmath]U[/ilmath] - which is open as [ilmath]U\in\mathcal{J}_\beta[/ilmath]!


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## What is a symplectic manifold, really?January 9, 2012

Posted by Ben Webster in mathematical physics, symplectic geometry.

I’m teaching a graduate course in symplectic geometry and GIT this semester, and am going to try to produce some posts related to lectures I’m giving there. Hopefully, this will help me think things through and put some new exposition out there on the internet.

So, obviously, the first question is “what is a symplectic manifold?” Now, wikipedia will tell you it’s a manifold equipped with a non-degenerate closed 2-form. Certainly that’s right, but it doesn’t tell a novice in symplectic geometry much. Why think about such a structure?

So let me try to put a different spin on this. This isn’t all that new of a spin (in fact, Henry Cohn wrote almost exactly the same thing here), but I don’t know of anywhere symplectic manifolds are really presented like this: I want to think of a symplectic manifold as a space where one can do a particular flavor of classical mechanics.   (more…)

## Conference on Higher Gauge Theory, Quantum Gravity, and Topological Field TheoryDecember 18, 2010

Posted by Chris Schommer-Pries in mathematical physics, QFT, quantum topology, talks, tqft, Uncategorized.
Tags:
1 comment so far

In February there is going to be a workshop and school dedicated to exploring the interactions of Quantum Gravity, Higher Gauge Theory, and Topological Field Theory. I’m excited about the chance to share ideas and hopefully create some new mathematics.

The conference will take place in Lisbon, Portugal, and yours truly will be giving one of the mini-courses for the school (the topic is going to be the classification of extended 2D tqfts, something near and dear to my heart). Of course that makes me really excited, but I am also excited about the other topics too and I think the mix of ideas will be invigorating. For more info look below the break.

## Generalized moonshine I: Genus zero functionsJanuary 8, 2009

Posted by Scott Carnahan in group theory, mathematical physics, Number theory, Paper Advertisement, representation theory.

This is a plug for my first arXiv preprint, 0812.3440. It didn’t really exist as an independent entity until about a month ago, when I got a little frustrated writing a larger paper and decided to package some results separately. It is the first in a series of n (where n is about five right now), attacking the generalized moonshine conjecture. Perhaps the most significant result is that nontrivial replicable functions of finite order with algebraic integer coefficients are genus zero modular functions. This answers a question that has been floating around the moonshine community for about 30 years.

Moonshine originated in the 1970s, when some mathematicians noticed apparent numerical coincidences between the theory of modular functions and the theory of finite simple groups. Most notable was McKay’s observation that 196884=196883+1, where the number on the left is the first nontrivial Fourier coefficient of the modular function j, which classifies complex elliptic curves, and the numbers on the right are the dimensions of the smallest irreducible representations of the largest sporadic finite simple group, called the monster. Modular functions and finite group theory were two areas of mathematics that were not previously thought to be deeply related, so this came as a bit of a surprise. Conway and Norton encoded the above equation together with other calculations by Thompson and themselves in the Monstrous Moonshine Conjecture, which was proved by Borcherds around 1992.

I was curious about the use of the word “moonshine” here, so I looked it up in the Oxford English Dictionary. There are essentially four definitions:

1. Light from the moon, presumably reflected from the sun (1425)
2. Appearance without substance, foolish talk (1468 – originally “moonshine in the water”)
3. A base of rosewater and sugar, or a sweet pudding (1558 cookbook!)
4. Smuggled or illegally distilled alcoholic liquor (1782)

The fourth and most recent definition seems to be the most commonly used among people I know. The second definition is what gets applied to the monster, and as far as I can tell, its use is confined to English people over 60. It seems to be most popularly known among scientists through a quote by Rutherford concerning the viability of atomic power.

I’ll give a brief explanation of monstrous moonshine, generalized moonshine, and my paper below the fold. There is a question at the bottom, so if you get tired, you should skip to that.

## Woit on Geometric Representation TheoryDecember 22, 2008

Posted by David Speyer in D-modules, mathematical physics, representation theory.
1 comment so far

Just wanted to point out to everyone that Peter Woit, of the blog Not Even Wrong is doing a great job blogging on the relations between representation theory of Lie groups, functions on Lie groups and differential operators. And he promises there will be physics before the end!

## Gromov-Witten Invariants and Topological Field TheoryDecember 11, 2008

Posted by A.J. Tolland in mathematical physics, Uncategorized.

A few days ago, John Mangual requested that one of us secret blogging seminarians write a post explaining what Gromov-Witten invariants are all about.  I volunteered to do this, which puts me in an awkward position.  Gromov-Witten theory is a big subject, and there are a lot of good introductions to the subject (e.g., Behrend’s Algebraic Gromov-Witten Invariants). It’s not clear that I can do much in a single blog post.

So I’m going to limit the scope of this post somewhat.  I’d like to explain in what sense Gromov-Witten theory is a topological field theory.   Caution: This may involve some rambling.    If you’re curious, come below the fold.

## Symplectic duality slidesNovember 24, 2008

Posted by Ben Webster in Algebraic Geometry, category O, crazy ideas, link homology, mathematical physics, QFT, talks.

I’ve been too lazy to write in detail about the progress in my research (well, I am writing six papers and applying to jobs, so it isn’t entirely due to laziness), but I did recently speak in the symplectic seminar at MIT, and have posted the slides on my webpage. Obviously, they’re less useful without someone to explain them, but given the current lack of an overarching paper on the subject (that’s no. 5 on the list, I promise), I thought it might be edifying. Executive summary below the cut. (more…)

## Live Blogging: AJ on Gromov-Witten theory of stacksOctober 20, 2008

Posted by Noah Snyder in Algebraic Geometry, Algebraic Topology, liveblogging, mathematical physics, talks.

Today AJ’s talking in the Grand Unified Seminar (representation theory, geometry, and combinatorics) on his this (joint work with his collaborator C. Teleman and his advisor E. Frenkel).  The title of the talk is “Gromov-Witten Theory for a point/C*.”  As AJ points out with delight (after the title was misintroduced without the mod C*, “It’s negative 2 dimensional!”

The outline of the talk is:

1. Gromov-Witten Invariants
2. The stack pt./C*
3. Integration on quotient stacks
4. (the unfortunately named) admissible classes
5. Bundles on nodal curves
6. Invariants are well-defined

## This paper was written for our blogAugust 21, 2008

Posted by David Speyer in Algebraic Geometry, mathematical physics, Number theory, Paper Advertisement, papers, representation theory.
I’ve recently been reading a paper which ties together a number of this blog’s themes: Canonical Quantization of Symplectic Vector Spaces over Finite Fields by Gurevich and Hadani. I’m going to try to write an introduction to this paper, in order to motivate you all to look at it. It really has something for everyone: symplectic vector spaces, analogies to physics, Fourier transforms, representation theory of finite groups, gauss sums, perverse sheaves and, yes, $\theta$ functions. In a later paper, together with Roger Howe, the authors use these methods to prove the law of quadratic reciprocity and to compute the sign of the Gauss sum. For the experts, Gurevich and Hadani’s result can be summarized as follows: they provide a conceptual explanation of why there is no analgoue of the metaplectic group over a finite field. Not an expert? Keep reading!


text:
# How many ml is one cc?

Jun 5, 2014

1 cm³ = 1 mL

The official definition is

1 L = 1 dm³ (exactly)

So 1000 mL = 1 dm³ × ${\left(\left(10 \text{cm")/(1"dm}\right)\right)}^{3}$ = 1000 cm³

or 1 mL = 1 cm³


text:
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# Sweet release

232 views

I got everything working for this release after breaking all my code with a radical change in where everything goes. It should suffice to say that this won't be a verbose post.

#### Try my game-so-far!

You must have Python 2.5 installed.
You must have Pygame 1.7.1 for Python 2.5 installed.

I have not tried to make this work on anything but Windows, and I'm using Windows 2000, so be warned.

There is no gameplay, just a GUI and map editor. In game mode, select with left clicks, re-center the map with right clicks. In edit mode paint with left clicks and dragging with the left mouse button. Re-center the map with right clicks. Hit an "exit" button or your escape key to quit at any time.

I don't think there's anything (straightforward) that will go wrong, so let that be the challenge: If you can find a cool error and tell me what you did, you win!

If you have undertaken and understand the above, or even if you manage to stumble through on sheer cussedness, you may now download Isostrat v0.05.

Tell me what you think and how it goes, if ya like. If you're particularly adventurous, go ahead and look at the sourcecode.

The quickload and quicksave buttons in-game didn't work; they generated the following error messages, respectively:
switchmode game
savegame True
main  could not execute:  savegame  /  True

loadgame True
main  could not execute:  loadgame  /  True

The quickmenu button with question marks ("????") causes the game to quit... or crash.

The in-game map editor is very cool. I like the ripple/wobble effect on the brushes. I clicked save after creating a dense, lush ecosystem (I exaggerate), but I don't think it was serialized, and I couldn't subsequently reload.

Overall, it was a very solid pre-alpha first release. [smile]

Thanks for trying it out Oluseyi!

I probably should have mentioned that the save/load buttons don't work because, well, saving and loading aren't at all implemented.

And yeah, the "????" button does indeed either call the program to exit; I should probably just take that button out because I have no idea what it should do, I just wanted 6 buttons in that panel. Heh.

But hey, I'm glad it all worked and nothing terrible happened!

It worked for me too. Excellent work!

I like how your tiles have some variety; they appear to change slightly when you put down the same tile that's already there. They also fit together nicely. The grass even appears to have some volume; the blades seem to stick up over an adjacent desert tile. You should do some taller grass to show off the effect even more. The mountain and hill tiles look great and the animation works.

Some issues:
1. I noticed you can't put down tiles in the bottom 1/8th of the map.
2. It paused for a few seconds on startup, saying "remaking bg". Is that where you're doing some tile compositing?
3. Clicking on your minimap doesn't change the viewport (unless you right-click, but then its as if you right-clicked on the terrain).
4. Its a little bit slow right now, which means when you quickly lay down a row of tiles, some will be skipped over. You could solve this by acting on all tiles that intersect the cursor's translation vector, but more speed would be even better. Its also causing the interface to appear less smooth.

Thanks for trying it out Dave!

Yeah, there are 8 variations on each terrain type and 4 variations on each transition that get chosen randomly upon change.

1: Damn, good find. I knew it did that at the horizontal extreme, then I resized the map and it seemed to work (which isn't a solution, I know). It's a rather boggling issue that I can hopefully fix as I refactor my isomap class.

2: Actually no, there's no compositing (and maybe there should be). The pause came from generating GUI panels which, after creating some charts with the app that Oluseyi recommended, I discovered was due to an exponential increase in blit calls as the size of panels was increased. I was blitting a 4x4 image over 800x600 pixels, which is ridiculously stupid. And even worse because the 4x4 image was just colorkey green. Anyway by my use of past tense you can see that I fixed the problem this morning by increasing the size of the image to-be-blitted exponentially to match the size of the panels. I'll write a full explanation on Tuesday, but it should suffice to say it runs very nicely now!

3: Yeah, I didn't implement that yet. It should be pretty straightforward but it requires making a GUI element with rather unique behaviour. I just have to suck it up and make it, I suppose.

4: Yes, I want to do the vector thing! And the speed has increased, as I said, and should increase much more as I optimize rendering, which I have no done yet -- though I have organized my code a bit more properly so that I -can- optimize it now!

Again, thanks for the feedback!

Tried it out and it ran without any errors except those already mentioned. I like the auto-fringe feature and how you randomly select brush of a specific type. I also liked the fact that the interface was easy to learn; in moments I was placing hills and mountains. :-)

Your scrolling could do some work though. I would have liked to be able to drag the scroll direction according to the mouse movement. Something like Photoshop has with its “hand” tool. I also would have liked to be able to remove objects on the map, mountains for instance.

I have to say that it was enjoyable to create a map. Very good demo/alpha!

O-san:
Quote:
 I would have liked to be able to drag the scroll direction according to the mouse movement.

That'd be a good way to do it -- maybe right-click drag will scroll to keep the controls sorta consistent (right click to change view, left click to change map).

Scrolling has been a difficult issue with me because I'm used to full screen games where you simply scroll with the map edge; It's more difficult to make scrolling "transparent" when its in a window.

Quote:
 I also would have liked to be able to remove objects on the map, mountains for instance.

That's a completely obvious thing that I didn't really think much about. It should be a simple matter to add an "erase" entry to each editor palette (except for terrain because the "void" terrain crashes the game, heh heh).

Thanks for trying it out, the feedback is very useful to me!

## Create an account

Register a new account


text:
# Mechanics Question

A particle $P$ of mass $0.6\text{ kg}$ moves upwards along a line of greatest slope of a plane inclined at $18°$ to the horizontal. The deceleration of $p$ is 4 ms−2

The first part asks us to find the frictional and normal components of the force excreted on $P$ by the plane. Then they ask us to find the coefficient of friction between $P$ and the plane.

Now I understand that this has to be done by $f_{net} = ma$, but the solution at the back only uses frictional force and the component of the weight down the slope. I don't understand why there isn't a force acting up the slope. Say this force is $K\text{ N}$, then shouldn't the $f_{net}$ equation be $$K- mg \sin18- \text{frictional force}= 0.6(-4)$$

• Did you mean the decelaration is $4\text{ m.s}^{-2}$? Can you draw the force diagram? Apr 10 '14 at 19:23
• the decelaration is 4 m.s−2 Apr 10 '14 at 19:24

I think you are mixing up speed and acceleration. The ball is moving up and thats why friction should point down the slope, but if there were a force acting up the slope then it would mean that $P$ is actively using energy to climb up. For large enough $K$, $P$ would be accelerating upwards, as if you were pushing it.

I believe the situation is instead more like you gave the ball an initial velocity and this is the kinematics of the ball experiencing only friction (and the effect of its weight) along the slope, causing a deceleration. Hopefully this convinces you that $K=0$, and $f_{net} = ma$ means

$$-0.6 g \sin 18º - \text{frictional force} = 0.6 \times (-4)$$

• No problem :) Hey I just noticed that you didn't accept the answers to the questions you asked. If they answered the question, you should do that by clicking on the tick below the voting buttons for that answer. Apr 10 '14 at 21:28
• I didn't know about that, sorry! Apr 11 '14 at 6:54
• np, the rules or something I read on this site said that if you don't accept then some people might stop answering you so I just thought I should tell you. Apr 11 '14 at 8:24
• @CalvinKhor I think you gave the wrong solution. Let's say $K=0$ and $mg\sin18^\circ>f_{sm}$ then the particle will move down. Hence, the Newton's second law will be $$mg\sin18^\circ- f_k=ma.$$ Apr 11 '14 at 15:35
• I don't think so; the force does not determine the speed, only the acceleration. It is supposed to decelerate, so it works out. If I'm still wrong, would you care to elaborate? Apr 11 '14 at 16:36

Using Newton's second law of motion: \begin{align} \sum F&=ma\\ K-mg\sin18^\circ\pm f_k&=ma\\ K-mg\sin18^\circ\pm\mu_k N&=ma\\ K-mg\sin18^\circ\pm\mu_k mg\cos18^\circ&=ma.\\ \end{align} If $0\le K<mg\sin18^\circ+f_{sm}$, the sign of the friction $f_k$ should be positive, why? If $K>mg\sin18^\circ+f_{sm}$, the sign of the friction $f_k$ should be negative, why? What would happen to particle if $K=mg\sin18^\circ$? Hint: Draw the force diagram and determine the particle's movement from the 3 cases.

$$\\$$

$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

Where would a force acting up the slope come from? You resolve gravity into the force down the slope and force into the slope. The force into the slope creates a frictional force that is also down the slope. Your $K$ is zero, then your equation is correct.


text:
Modules with Chain Condition on Non-finitely Generated Submodules

Modules with Chain Condition on Non-finitely Generated Submodules In this article, we study modules with chain condition on non-finitely generated submodules. We show that if an R-module M satisfies the ascending chain condition on non-finitely generated submodules, then M has Noetherian dimension and its Noetherian dimension is less than or equal to one. In particular, we observe that if an R-module M satisfies the ascending chain condition on non-finitely generated submodules, then every submodule of M is countably generated. We investigate that if an R-module M satisfies the descending chain condition on non-finitely generated submodules, then M has Krull dimension and its Krull dimension may be any ordinal number $$\alpha$$ α . In particular, if a perfect R-module M satisfies the descending chain condition on non-finitely generated submodules, then it is Artinian. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Mediterranean Journal of Mathematics Springer Journals

Modules with Chain Condition on Non-finitely Generated Submodules

, Volume 15 (1) – Dec 1, 2017
12 pages

/lp/springer_journal/modules-with-chain-condition-on-non-finitely-generated-submodules-58Gsu6VHWl
Publisher
Springer International Publishing
Copyright © 2017 by Springer International Publishing AG, part of Springer Nature
Subject
Mathematics; Mathematics, general
ISSN
1660-5446
eISSN
1660-5454
D.O.I.
10.1007/s00009-017-1047-y
Publisher site
See Article on Publisher Site

Abstract

In this article, we study modules with chain condition on non-finitely generated submodules. We show that if an R-module M satisfies the ascending chain condition on non-finitely generated submodules, then M has Noetherian dimension and its Noetherian dimension is less than or equal to one. In particular, we observe that if an R-module M satisfies the ascending chain condition on non-finitely generated submodules, then every submodule of M is countably generated. We investigate that if an R-module M satisfies the descending chain condition on non-finitely generated submodules, then M has Krull dimension and its Krull dimension may be any ordinal number $$\alpha$$ α . In particular, if a perfect R-module M satisfies the descending chain condition on non-finitely generated submodules, then it is Artinian.

Journal

Mediterranean Journal of MathematicsSpringer Journals

Published: Dec 1, 2017

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text:
# Math Help - Another chain rule question.

1. ## Another chain rule question.

Can you also check this out for me;

$y=e^a$

where $a=-x^2/2$

1... $dy/dx = dy/du \cdot du/dx$

2...where $y = e ^-^u$ and $u= - x^2/2$

3.... $dy/du=e ^-^u$ and $du/dx= -x$ <<<< can you explain how step 2 for "u" this turns into -x

4.... therefore; $dy/du \cdot du/dx = -x e ^ - x^2/2$

Can you also check this out for me;

$y=e^a$

where $a=-x^2/2$

1... $dy/dx = dy/du \cdot du/dx$

2...where $y = e ^-^u$ and $u= - x^2/2$

3.... $dy/du=e ^-^u$ and $du/dx= -x$ <<<< can you explain how step 2 for "u" this turns into -x

4.... therefore; $dy/du \cdot du/dx = -x e ^ - x^2/2$

You are using a instead of u. Minor difference but instead you need to find

1. $\dfrac{dy}{dx} = \dfrac{dy}{da} \cdot \dfrac{da}{dx}$

2. Aside from my above note about a then yes.

3. You need to use the chain rule. It's a tricky one to spot but if you imagine $e^{-a} = e^{-1 \cdot a}$ so the derivative is multiplied by -1. Thus $\dfrac{d}{da}e^{-a} = -e^{-a}$

For $\dfrac{da}{dx} = -x$ is true because you can write it as $-\dfrac{1}{2} \cdot x^2$. x^2 is simple enough but you must multiply by -1/2 which gives -x

4. Almost, because of the sign error in finding $\dfrac{da}{dx}$ your answer is a factor of -1 out.

$\dfrac{dy}{dx} = -x \cdot -e^{-x^2/2} = xe^{-x^2/2}$


text:
Generating financial data

I am trying to generate monthly stock data using a one-factor model:

$$R_{a,t} = \alpha + BR_{b,t}+\epsilon_{t}$$

The description says:

$R_{a,t}$ is the excess asset returns vector, $\alpha$ is the mispricing coefficients vector, $B$ is the factor loadings matrix, $R_{b,t}$ is the vector of excess returns on the factor portfolios, $R_{b}-N(\mu_{b},\sigma_{b})$, and $\epsilon_{t}$ is the vector of noise, $\epsilon - N(0,\sum_{e})$, which is independent with respect to the factor portfolios.

For our simulations, we assume that the risk-free rate follows a normal distribution, with an annual average of 2% and a standard deviation of 2%. We assume that there is only one factor (K=1), whose annual excess return has an annual average of 8% and a standard deviation of 16%. The mispricing $\alpha$ is set to zero and the factor loadings, B, are evenly spread between 0.5 and 1.5. Finally, the variance-covariance matrix of noise, $\sum_{\epsilon}$, is assumed to be diagonal, with elements drawn from a uniform distribution with support [0.10,0.30], so that the cross-sectional average annual idiosyncratic volatility is 20%.

Using the information provided here I try to generate the data:

alpha <- 0 #mispricing index is set to 0

R <- rnorm(100,mean=8/12,sd=16/sqrt(12)) #factor with annual excess return of 8% and standard deviation of 16%

epsilon <- rnorm(100, mean=0,sd=runif(10,min=0.1,max=0.30)) #error term with mean 0 and standard deviation drawn from a uniform distribtion

Then I generate the data:

data <- alpha + B*R + epsilon

My question is: is this the correct way to do it or am I missing something?

• By definition a diagonal matrix has zero entries in non-diagonal positions. In your case the diagonal elements are picked randomly between 0.1 and 0.3. – noob2 Jan 12 '16 at 14:30
• I would like to know whether I understand the process correctly. The text does not explain what the subscripts 'a' and 'b' stand for. – user3742038 Jan 15 '16 at 11:19
• R_a are returns for actual securities (stocks?), while R_b are returns on factors generate or drive the returns on the stocks. In your case however, there is only one factor. So superficially (i.e. without compiling it or testing it) your code seems right. – noob2 Jan 15 '16 at 18:20


text:
Integration on compact manifold

Integration on a nice enough manifold of a function $f:M \to \mathbb{R}$ is defined $$\int f = \sum_{ i \in I} \int_{U_i}\phi_i f$$ where $\phi_i$ is a partition of unity subordinate to the open cover $U_i$ of $M$.

The index set $I$ can be infinite in general.

If $M$ is compact (bounded), can I take $I$ to be a finite set so that the sum is a finite sum?

• Yes, if $M$ is compact, you can always have a finite partition of unity subordinate to any open cover. – Daniel Fischer Jan 12 '14 at 23:37

1 Answer

Yes, since every open cover has a finite subcover (Heine-Borel).

• Thanks, but why don't books emphasise that the sum can be finite?? – michael_carbon Jan 12 '14 at 23:39
• I am sure some do. Also, many manifolds are not compact, but they are generally locally compact, which means that every point lies in a finite number of the open sets (this is obviously important, since infinite sums are a big pain). – Igor Rivin Jan 12 '14 at 23:40
• @PedroTamaroff I actually think that the local finiteness (global finiteness less so) is crucial for the whole machine to work, and I am sure if you read the first partitions of unity papers, it is probably emphacized. – Igor Rivin Jan 12 '14 at 23:45


text:
시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
10 초 128 MB 0 0 0 0.000%

## 문제

The first crossword puzzle was published on December 21, 1913 by Arthur Wynne. To celebrate the centennial of his great-great-grandfather’s invention, John “Coward” Wynne1 was struggling to make crossword puzzles. He was such a coward that whenever he thought of a tricky clue for a word, he couldn’t stop worrying if people would blame him for choosing a bad clue that could never mean that word. At the end of the day, he cowardly chose boring clues, which made his puzzles less interesting.

One day, he came up with a brilliant idea: puzzles in which word meanings do not matter, and yet interesting. He told his idea to his colleagues, who admitted that the idea was intriguing. They even named his puzzles “Coward’s Crossword Puzzles” after his nickname.

However, making a Coward’s crossword puzzle was not easy. Even though he did not have to think about word meanings, it was not easy to check if a puzzle has one and only one set of answers. As the day of the centennial anniversary was approaching, John started worrying if he would not be able to make interesting ones in time. Let’s help John by writing a program that solves Coward’s crossword puzzles.

Each puzzle consists of h × w cells along with h across clues and w down clues. The clues are regular expressions written in a pattern language whose BNF syntax is given as in the table below.

clue       ::= "^" pattern "$" pattern ::= simple | pattern "|" simple simple ::= basic | simple basic basic ::= elementary | elementary "*" elementary ::= "." | "A" | "B" | ... | "Z" | "(" pattern ")" Table J.1. BNF syntax of the pattern language. The clues (as denoted by p and q below) match words (as denoted by s below) according to the following rules. • ^p\$ matches s if p matches s.
• p|q matches a string s if p and/or q matches s.
• pq matches a string s if there exist s1 and s2 such that s1s2 = s, p matches s1, and q matches s2.
• p* matches a string s if s is empty, or there exist s1 and s2 such that s1s2 = s, p matches s1, and p* matches s2.
• Each of A, B, . . . , Z matches the respective letter itself.
• (p) matches s if p matches s.
• . is the shorthand of (A|B|C|D|E|F|G|H|I|J|K|L|M|N|O|P|Q|R|S|T|U|V|W|X|Y|Z).

Below is an example of a Coward’s crossword puzzle with the answers filled in the cells.

Java Specific: Submitted Java programs may not use classes in the java.util.regex package.

C++ Specific: Submitted C++ programs may not use the std::regex class.

1All characters appearing in this problem, except for Arthur Wynne, are fictitious. Any resemblance to real persons, living or dead, is purely coincidental.

## 입력

The input consists of multiple datasets, each of which represents a puzzle given in the following format.

h w
p1
p2
.
.
.
ph
q1
q2
.
.
.
qw

Here, h and w represent the vertical and horizontal numbers of cells, respectively, where 2 ≤ h, w ≤ 4. The pi and qj are the across and down clues for the i-th row and j-th column, respectively. Any clue has no more than 512 characters.

The last dataset is followed by a line of two zeros. You may assume that there are no more than 30 datasets in the input.

## 출력

For each dataset, when the puzzle has a unique set of answers, output it as h lines of w characters. When the puzzle has no set of answers, or more than one set of answers, output “none” or “ambiguous” without double quotations, respectively.

## 예제 입력 1

2 2
^(C|I|T|Y)*$^(C|O|P|S)*$
^(F|L|I|P)*$^(B|A|C|K)*$
2 2
^HE|LL|O*$^(P|L|E|A|S|E)*$
^(H|L)*$^EP|IP|EF$
4 4
^LONG|TALL|S*ALLY$^(R*EV*|OL*U(TIO)*N)*$
^(STRAWBERRY|F*I*E*L*D*S*|FOREVER)*$^P.S.|I|LOVE|YOU$
^(RE|A|L)((L|OV*E)*)$^(LUC*Y*|IN.THE.SKY)(WITH|DI*A*M*ON*D*S*)$
^(IVE*|GOT|A|F*E*E*L*I*N*G*)*$^YEST*E*R*D*A*Y*$
2 3
^(C|P)(OL|AS)$^(LU|TO)(X|R)$
^CT|PL$^OU|AO$
^SR|LX$2 2 ^T*|(HI)|S*$
^SE|NT|EN|CE$^IS$
^(F|A|L|S|E)*$2 4 ^ARKA|BARB|COLU$
^NSAS|ADOS|MBIA$^..$
^..$^KA|RO|LI$
^AS|BA|US\$
0 0

IC
PC
HE
LP
ALLY
OUNE
EDIS
LOVE
ambiguous
none
ARKA
NSAS


text:
Using the Central limit Theorem, The mean of a random of size $n=64$ is used to estimate the mean of an infinite population having the standard deviation $\sigma=60 .$ What can we assert about the probability that the error will be less than $16.5 ?$a. Using the Chebyshev's Theoremb. So if our population size of infinity we use following formula, we don't have the square root part because it's infinite. Complete parts (a) through (c) below. A population has a mean of 200 and a standard deviation of $50 .$ Suppose a simple random sample of size 100 is selected and $x$ is used to estimate $\mu .$a. Using the Central limit Theorem, What is the value of the finite population correction factor?a. What is the probability that the sample mean will be within $\pm 5$ of the population mean?b. So if our population size of infinity we use following formula, we don't have the square root part because it's infinite. Numbers 10 over route 50. I 1000 minutes one. Solution for Suppose a simple random sample of size n= 64 is obtained from a population with u= 86 and o = 24. Round answer 4 decimal places. In the EAI sampling problem (see Figure 7.5 ), we showed that for $n=30$, there was .5034 probability of obtaining a sample mean within $\pm \$ 500$of the population mean.a. How large was the sample used in this survey?b. p=0.4. Approximately normal because ns0.05N and np(1 - p) 2 10 O B. Complete Parts (a) Through (c) Below. What is the probability that a random sample of 75 students will provide a sample mean SAT score within 10 of the population mean?b. The population size is$N=50,000$.c. Refer to the EAI sampling problem. Choose the phrase that best describes the shape of the sampling distribution below. Take the square root of that and multiply by. Question: Suppose A Simple Random Sample Of Size N= 50 Is Obtained From A Population Whose Size Is N = 20,000 And Whose Population Proportion With A Specified Characteristic Is P=0.4. Complete parts (a) through (c) below. And when you do the math on that, that's approximately one point for one for two. And when you do the math on that, that's approximately one point for one for two. The mean annual cost of automobile insurance is$\$939(C N B C, \text { February } 23,2006)$ Assume that the standard deviation is $\sigma=\$ 245$a. a. The mean of a random of size$n=225$is used to estimate the mean of an infinite population having the standard deviation$\sigma=3 .$What can we assert about the probability that the error will be less than$0.25 ?$a. What general statement can you make about what happens to the sampling distribution of$\bar{x}$as the sample size is increased? Answer part (a) for a sample of size 120 . What is the probability that the point estimate was within$\pm 25$of the population mean? Using the Central limit Theorem. Take the square root of that. Standard error is 10. Take the square root of that and multiply by. Suppose that Y.bs = (YI, 1 . ., Yn) missing completely at random (Rubin, 1987), where n = no + nl. Complete Parts (a) Through (c) Below. (d) slightly greater than the variability for a sample of size$n=500$from a population of size$50,000,000$. So that looks like this times this. What is probability of obtaining X=380 or fewer individuals with the characteristic? So that is equal to plugging into our Formula 10 over Route 50. What is the standard error of$\overline{p} ?$c. (e) much greater than the variability for a sample of size$n=500$from a population of size$50,000,000$. suppose a simple random sample of n=1136 college students with cellphones is obtained. What is the probability that the sample mean will be within$\pm 10$of the population mean? What is the probability that the simple random sample will provide a sample mean within$\$100$ of the population mean? Choose the correct description of the shape of the sampling distribution of x̅ . A nationwide study in 2003 indicated that about 60% of college students with cell phones send and receive text messages with their phones. Solution for Suppose a simple random sample of size n= 64 is obtained from a population that is skewed right with µ = 82 and o = 32. d) What is P(28.7 ≤ x ≤ 31.1) Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website! (b) What… 50 over. Using the Central limit Theorem, The mean of a random of size $n=64$ is used to estimate the mean of an infinite population having the standard deviation $\sigma=60 .$ What can we assert about the probability that the error will be less than $16.5 ?$a. (a) Describe the sampling distribution of p. Choose the phrase that best describes the shape of the sampling distribution below. What is the probability that the simple random sample will provide a sample mean within $\$ 250$of the population mean?c. Refer to the EAI sampling problem. . Thio 5000 minus five. So that is equal to plugging into our Formula 10 over Route 50. Suppose a simple random sample of size n=49 is obtained from a population with \mu=80 and \sigma=14 (a) Describe the sampling distribution of \bar{x}. Ah, Since we have a finite sample size, we can use our formula. Suppose a simple random sample of size n=150 is obtained from a population whose size is N=20,000 and whose population proportion with a specified characteristic is p=0.8. You'll get 1.34 30 and there you have it. Question: Suppose a simple random sample of size n = 150 is obtained from a population whose size is N = 25,000 and whose population proportion with a specified characteristic is {eq}p = 0.4 {/eq} . Simple random sampling is the basic sampling technique where we select a group of subjects (a sample) for study from a larger group (a population). The multiply that by chan over route 50. A simple random sample of size n = 1460 is obtained from a population whose size is N = 1,500,000 and whose population proportion with a specified characteristic is p = 0.42. What is the probability that the sample mean will be within$\pm 5$of the population mean?b. 5. What is the probability that the sample mean will be within$\pm 10$of the population mean? The standard deviation of the sampling distribution of is Answer to "Suppose a simple random sample of size n =50 is obtained from a population whose size is N=20,000 and whose population proportion with a Nearsighted children. It has a mean The number about which proportions computed from samples of the same size center. b) the distribution is approximately normal. What is the probability of obtaining x = 520 or more individuals with the characteristic? N=15,000 and whose population proportion with a specified characteristic is p equals 0.4 . a. Sketch the sampling distribution of$\bar{x}$when simple random samples of size 60 are used. Standard error is 10. Statistics for Business and Economics 10th, Whoops, there might be a typo in your email. 2.3 Simple Random Sampling Simple random sampling without replacement (srswor) of size nis the probability sampling design for which a xed number of nunits are selected from a population of N units without replacement such that every possible sample of nunits has equal probability of being selected. Suppose a simple random sample of 60 managers is used.a. 1 Answer to Suppose a simple random sample of size n = 1000 is obtained from a population whose size is N = 2,000.000 and whose population proportion with a specified characteristic is p = 0.76. Assign a sequential number to each employee (1,2,3…n). Assuming the normal model can be used, describe the sampling distribution x. Solution for Suppose a simple random sample of size n = 150 is obtained from a population whose size is N = 30,000 and whose population proportion with a… Click 'Join' if it's correct. Assume that the population proportion is$55 .$Compute the standard error of the proportion,$\sigma_{p},$for sample sizes of$100,200,500,$and$1000 .$What can you say about the size of the standard error of the proportion as the sample size is increased? All right, So we're giving a simple random sample of 50 and we know our population. The sample proportion is a random variable: it varies from sample to sample in a way that cannot be predicted with certainty. Suppose a simple random sample of size n 50 is obtained from a population whose size is N20,000 and whose population proportion with a specified characteristic is p 0.4 (a) Describe the sampling distribution of p Choose the phrase that best describes the shape of the sampling distribution below OA. c. (x-bar > 86.5) The prob of 1,2,3 for being selected is 1/5. (a) Describe the sampling distribution of p Choose the phrase that best describes the shape of the sampling distribution below. Question 997052: Suppose simple random sample size n=1000 is obtained from population whose size is N=1,000,000 and whose population proportion with specified characteristics is p=0.42. (1) Statistics (42) A simple random sample of 60 items resulted in a sample mean of 96. The multiply that by chan over route 50. Choose the correct description of the shape of the sampling distribution of x̅ . Suppose a simple random sample of size n-50 is obtained from a population whose size is N 10,000 and whose population proportion with a specified characteristic is p 0.6. So this going to be equal. Answer by Boreal(13070) (Show Source): b) What is P(x > 30.2)? c. What general statement can you make about what happens to the sampling distribution of$\bar{x}$as the sample size is … Suppose a simple random sample of size n=36 is obtained from a popilation with mean = 64 and sd = 18 what is p(x < 62.6)? Assume the population standard deviation is$\sigma=25 .$Compute the standard error of the mean,$\sigma_{x},$for sample sizes of$50,100,150,$and$200 .$What can you say about the size ofthe standard error of the mean as the sample size is increased? A. See answer rejkjavik rejkjavik Answer: Step-by-step explanation: Given data: random sample size n = 1000. Sketch the sampling distribution of$\bar{x}$when simple random samples of size 60 are used.b. What does the sampling distribution of$\overline{p}$show? 1 Answer to Suppose a simple random sample of size n = 1000 is obtained from a population whose size is N = 2,000.000 and whose population proportion with a specified characteristic is p = 0.76. I don't have an account. Ah, Since we have a finite sample size, we can use our formula. They all should be the same size and then you throw them all, you throw them all into a bowl of some kind and this seems like a very basic way of doing it but it's actually a pretty effective way of getting a simple, of getting a simple random sample. Question: Suppose A Simple Random Sample Of Size N = 50 Is Obtained From A Population Whose Size Is N = 15,000 And Whose Population Proportion With A Specified Characteristic Is P = 0.6. Suppose a simple random sample of size n = 50 is obtained from a population whose size is N = 30,000 and whose population proportion with a specified characteristic is p =0.6. All right, so that's gonna be 50,000 minus 50 over 50,000. What is the probability that the simple random sample will provide a sample mean within$\$250$ of the population mean?c. I don't have an account. Suppose a simple random sample of size n 50 is obtained from a population whose size is N20,000 and whose population proportion with a specified characteristic is p 0.4 (a) Describe the sampling distribution of p Choose the phrase that best describes the shape of the sampling distribution below OA. Ah, plug that into a calculator. O A. Offered Price: $22.00 Posted By: rey_writer Posted on: 02/16/2018 11:05 AM Due on: 02/16/2018 . c) the distribution is skewed left . q = 15. All right, So we're giving a simple random sample of 50 and we know our population. The population size is infinite.b. All right, party. Suppose a random sample of n = 50 ten-gram portions of peanut butter is collected and it is found that x̄ = 3.6 for the 50 samples. When$n=10$and$N=500$. Suppose a simple random sample of size n=150 is obtained from a population whose size is N=20,000 and whose population proportion with a specified characteristic is p=0.8. }\) Calculate the probability that the sample proportion will be larger than 0.65 for a random sample of size 50. What is the probability that a random sample of 75 students will provide a sample mean SAT score within 10 of the population mean?b. What happens to the sampling distribution of$\bar{x}$if simple random samples of size 120 are used?c. Compute the probability that a simple random sample of size n = 10 results in a sample mean greater than 110. (a) Describe the sampling distribution of ModifyingAbove p with caret. }\) Find and interpret the standard deviation of the sample proportion $$\hat{p}\text{. The mean of a random of size n=225 is used to estimate the mean of an infinite population having the standard deviation \sigma=3 . What can we assert about the probability that the error will be less than 0.25 ?a. The sampling distribution of \bar{x} ha… What is the probability that \bar{x} is within \ 500 of the population mean if a sample of size 60 is used?b. Go to your Tickets dashboard to see if you won! So that's gonna be 10 over the square root of 50. View Winning Ticket. a) the distribution is skewed right. Complete parts (a) through (c) below. Random samples of size \(n$$ produced sample proportions $$\hat{p}$$ as shown. N=15,000 and whose population proportion with a specified characteristic is p equals 0.4 . Explain. What happens to the sampling distribution of$\bar{x}$if simple random samples of size 120 are used?c. The mean tuition cost at state universities throughout the United States is$\$4260$ per year(St. Petersburg Times, December 11,2002 ). p. 3) the shape of the distribution is unknown. The main benefit of the simple random sample is that each member of the population has an equal chance of being chosen for the study.This means that it guarantees that the sample chosen is representative of the population and … Click 'Join' if it's correct, By clicking Sign up you accept Numerade's Terms of Service and Privacy Policy, To request the exact answer, fill out the form below. We're doing? Example of simple random sampling. What is the standard error of $\overline{p} ?$c. Man, this one. All right, party. Suppose a simple random sample of size n is drawn from a large population with mean \mu and standard deviation \sigma . Please answer questions (1) through (5) below. Show the sampling distribution of $\overline{p}$ .d. What is the probability that a simple random sample of automobile insurance policies will have a sample mean within $\$ 25$of the population mean for each of the following sample sizes:$30,50,100,$and$400 ?$b. You'll get 1.34 30 and there you have it. Using the Central limit Theorem, What is the value of the finite population correction factor?a. A simple random sample is a subset of a statistical population in which each member of the subset has an equal probability of being chosen. Question: Suppose a simple random sample of size n = 150 is obtained from a population whose size is N = 25,000 and whose population proportion with a specified characteristic is {eq}p = 0.4 {/eq} . Suppose a simple random sample of size n = 50 is obtained from a population whose size is N = 30,000 and whose population proportion with a specified characteristic is p =0.6. Suppose a simple random sample of 60 managers is used.a. The population size is$N=500$. μ P ^ and a standard deviation A measure of the variability of proportions computed from samples of the same size. Simple random sampling is the most basic and common type of sampling method used in quantitative social science research and in scientific research generally. 50 over. Each individual is chosen entirely by chance and each member of the population has an equal chance of being included in the sample. Sketch the sampling distribution of$\bar{x}$when simple random samples of size 60 are used.b. Take the square root of that. Suppose a simple random sample of size nequals = 50 50. is obtained from a population whose size is Upper N equals 15 comma 000. Sketch the sampling distribution of$\bar{x}$when simple random samples of size 60 are used. (b) Assuming the normal model can be used, determine P(x<70.4). Decide whether or not the sample size is large enough to assume that the sample proportion $$\widehat{P}$$ is normally distributed. A. Do that out. If we take a simple random sample of size$n=500$from a population of size$5,000,000,$the variability of our estimate will be (a) much less than the variability for a sample of size$n=500$from a population of size$50,000,000$. Assume the population standard deviation is$\sigma=25 .$Compute the standard error of the mean,$\sigma_{x},$for sample sizes of$50,100,150,$and$200 .$What can you say about the size ofthe standard error of the mean as the sample size is increased? Assume that the population standard deviation is$\sigma=100$a. (a) Describe the sampling distribution of ._____(b) What is the probability of obtaining x = 657 or more individuals with the Find the probability that the sample mean is between 1.8 hours and 2.3 hours.. Answer And it is Ah, 1.4076 party. Solution for Suppose a simple random sample of size n= 64 is obtained from a population that is skewed right with µ = 82 and o = 32. (a) Describe the sampling distribution of 3. ., y. I 1000 minutes one. You must be logged in to bookmark a video. (b)… Suppose a simple random sample of size n=150 is obtained from a population whose size is N=20,000 and whose population proportion with a specified characteristic is p=0.8. The College Board American College Testing Program reported a population mean SAT score of$\mu=1020$(The World Almanac 2003 ). Oops, we aren't doing variables. So this going to be equal. Question: Suppose A Simple Random Sample Of Size N = 50 Is Obtained From A Population Whose Size Is N= 10,000 And Whose Population Proportion With A Specified Characteristic Is P=0.4. What happens to the sampling distribution of$\bar{x}$if simple random samples of size 120 are used? Describe the sampling distribution of p. Approximately normal, mu_p = … (1) Describe the sampling distribution of p^ (choose the correct phrase that best describes the shape of the sampling below) a) Not normal because n (b) slightly less than the variability for a sample of size$n=500$from a population of size$50,000,000$. Complete Parts (a) Through (c) Below. We can be sure that the sampling distribution of the sample mean x̄ will be approximately normal because the sample size n = 50 is rather large. Use this value as the population mean and assume that the population standard deviation is$\sigma=\$900 .$ Suppose that a random sample of 50 state universities will be selected.a. Suppose a simple random sample of size n=49 is obtained from a population with u=77 and o=14? All right, so that's gonna be 50,000 minus 50 over 50,000. (a) Describe the sampling distribution of ._____(b) What is the probability of obtaining x = 657 or more individuals with the Refer to the EAI sampling problem. 1) is a simple random sample of size n1 from an unknown population, and let 0 represent all unknown parameters. Show the sampling distribution of $\overline{p}$ .d. I'm gonna move this a little bit to the side on our sample size is 500. Suppose a simple random sample size n = 200 is obtained from a population whose size is N = 20,000 and whose population proportion with a specified characteristic is p=0.4. Suppose a simple random sample of 60 managers is used. O A. A simple random sample is … (c) about the same as the variability for a sample of size $n=500$ from a population of size $50,000,000$ .
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text:
# What is the BER for multilevel modulated signal

Assume I want to transmit a low or high symbol which can be BPSK modulated. For ordinary BPSK, bit 0 can be mapped to 1+i0 and bit 1 mapped to -1+i0. In my question, each two bits are BPSK modulated and a level dependent so bit 01 is mapped to 2+0i and bits 00 is mapped to 1+0i. Also bits 10 is mapped to -1+0i and bits 11 is mapped to -2+0i. This is BPSK (not QPSK) signal since there are BPSK demodulator and level detector at the receiver in which each of them output one bit at symbol period.I want to know what demodulation scheme to use in order to detect the symbol level and hence calculating the BER.

As Dan Boschen has already pointed out, the question has very little to do with BPSK which acronym stands for Binary Phase-Shift Keying; this is a 4-ary system not a binary one, and as stated has very little to do with phase-shift keying either.

This is a baseband 4-ary communication system that uses pulse modulation. At the output of the demodulator, we observe one of $$4$$ signal levels ($$+2, +1, -1,-2$$) corrupted by noise which we will assume to be Gaussian with mean $$0$$ and variance $$\sigma^2$$. If $$b_0b_1$$ denotes the two bits being transmitted, then the conditional distribution of $$X$$, the demodulator output is given by \begin{align} X \sim N(+2, \sigma^2) &~~\text{if}~~ b_0b_1 = 01,\\ X \sim N(+1, \sigma^2) &~~\text{if}~~ b_0b_1 = 00,\\ X \sim N(-1, \sigma^2) &~~\text{if}~~ b_0b_1 = 10,\\ X \sim N(-2, \sigma^2) &~~\text{if}~~ b_0b_1 = 11. \end{align} The demodulator output is mapped to the nearest signal level and the decoded bits $$\hat{b}_0\hat{b}_1$$ are those corresponding to this signal level. That is, \begin{align}\text{If}~ X\in (1.5,\infty), &~\hat{b}_0\hat{b}_1 = 01,\\ \text{If}~ X\in (0, 1.5], &~\hat{b}_0\hat{b}_1 = 00,\\ \text{If}~ X\in (-1.5,0], &~\hat{b}_0\hat{b}_1 = 10,\\ \text{If}~ X\in (-\infty, -1.5], &~\hat{b}_0\hat{b}_1 = 11,\end{align} With $$P_e$$ meaning probability of bit error, we have that $$P_e(\hat{b}_0\neq b_0\mid b_0b_1 = 01) = P(N(+2, \sigma^2) < 0) = Q\left(\frac 2\sigma\right),\\ P_e(\hat{b}_0\neq b_0\mid b_0b_1 = 00) = P(N(+1, \sigma^2) < 0) = Q\left(\frac 1\sigma\right),\\ P_e(\hat{b}_0\neq b_0\mid b_0b_1 = 10) = P(N(-1, \sigma^2) > 0) = Q\left(\frac 1\sigma\right),\\ P_e(\hat{b}_0\neq b_0\mid b_0b_1 = 11) = P(N(-2, \sigma^2) > 0) = Q\left(\frac 2\sigma\right).$$ The average bit error probability for $$b_0$$ is thus $$P_e(\hat{b}_0\neq b_0) = \left.\left. \frac 12 \right[\displaystyle Q\left(\frac 1\sigma\right) + Q\left(\frac 2\sigma\right)\right].\tag{1}$$ On the other hand, the calculations for $$P_e(\hat{b_1}\neq b_1)$$ are a lot messier. We have that \begin{align}P_e(\hat{b}_1\neq b_1\mid b_0b_1 = 01) &= P(-1.5 < N(+2, \sigma^2) < 1.5)\\ &= Q\left(\frac{0.5}{\sigma}\right) - Q\left(\frac{3.5}{\sigma}\right),\\ P_e(\hat{b}_1\neq b_1\mid b_0b_1 = 00) &= P(N(+1, \sigma^2) < -1.5) + P(N(+1, \sigma^2) > 1.5)\\ &= Q\left(\frac{0.5}{\sigma}\right) + Q\left(\frac{2.5}{\sigma}\right),\\ P_e(\hat{b}_1\neq b_1\mid b_0b_1 = 10) &= P(N(-1, \sigma^2) < -1.5) + P(N(-1, \sigma^2) > 1.5)\\ &= Q\left(\frac{0.5}{\sigma}\right) + Q\left(\frac{2.5}{\sigma}\right),\\ P_e(\hat{b}_1\neq b_1\mid b_0b_1 = 11) &= P(-1.5 < N(-2, \sigma^2) < 1.5)\\ &= Q\left(\frac{0.5}{\sigma}\right) - Q\left(\frac{3.5}{\sigma}\right).\end{align} The average bit error probability for $$b_1$$ is thus $$P_e(\hat{b}_1\neq b_1) = Q\left(\frac{0.5}{\sigma}\right) + \left.\left. \frac 12 \right[\displaystyle Q\left(\frac{2.5}{\sigma}\right) - Q\left(\frac{3.5}{\sigma}\right)\right].\tag{2}$$ Thus the two bits being transmitted have significantly different average bit error probabilities with $$P_e(\hat{b}_1\neq b_1) >> P_e(\hat{b}_0\neq b_0)$$ for typical values of $$\sigma$$. $$Q(x)$$ is a very rapidly decreasing function of its argument and the dominant terms $$\displaystyle Q\left(\frac{0.5}{\sigma}\right)$$ and $$\displaystyle \frac 12 Q\left(\frac 1\sigma\right)$$ in $$(2)$$ and $$(1)$$ respectively can differ by an order of magnitude or more.

This is not BPSK! BPSK is binary phase shift keying with binary meaning 2 levels. What the OP is describing is Pulse Amplitude Modulation where multiple bits are mapped to multiple levels. In this case there are 4 levels so it would be 4-PAM. (As Dilip points out in the comments that the levels in 4-PAM would typically be equally spaced while here the inner levels are spaced further than the outer levels, complicating the BER computation slightly as detailed below but still PAM regardless).

The demodulation scheme would similar to BPSK demodulation in that you would multiply the received signal with the recovered carrier and low pass filter. Decision would be made on the output by setting decision thresholds half way between each of the levels (assuming equiprobable data) and then the derivation for BER follows similar to BPSK (solving for the tail probabilities under Gaussian distributions if the noise is assumed to be AWGN as is typically assumed, noting the different tail contributions involved since the levels aren't equally distributed as Dilip has pointed out - that said the thresholds at ±1.5 will likely dominate setting the error rate with significantly smaller error contribution from the threshold at 0).

With equiprobable distribution of symbols and AWGN with standard deviation $$\sigma$$ this would result in the following Symbol Error Rate:

$$\text{SER} = \text{erfc}(0.5/(\sqrt{2}\sigma) + \frac{1}{2}\text{erfc}(1/(\sqrt{2}\sigma)$$

For smaller $$\sigma$$ values the second term would likely be insignificant. For larger $$\sigma$$ there would be additional tail contributions from the next adjacent threshold levels which would also significantly complicate the BER computation as detailed below. Thankfully under all practical applications these tails would certainly be insignificant and can be safely ignored.

To go from Symbol Error Rate to Bit Error Rate, consider how many bit errors occur when a symbol error occurs. Because the mapping done by the OP was properly Gray-coded (only one bit change between adjacent symbols), only one bit error would occur with each symbol error.

• Thanks Dan, but as i have written they are two separate blocks, one is Bpsk demodulator and one is level detector. May 11 '20 at 14:26
• @Riva11 If there are 4 amplitude levels in the signal, this is not a BPSK modulated signal. That said, a BPSK demodulator that multiplies the recieved signal with the carrier and low pass filters will still put out 4 levels, and to demodulate you would set the thresholds as I have indicated. Am I missing something else? May 11 '20 at 15:00
• What OP @Riva11 describes is not 4-PAM in the usual sense of the word. Standard 4-PAM would use equally-spaced levels $-3, -1, +1, +3$ instead of the $-2, -1, +1, +2$ used in the OP's system. Well, at least the dibits are assigned to the levels in Gray code order. Nonetheless, the first and second bits have different probabilities of error and so the BER calculation is messy too. May 11 '20 at 15:37
• @DilipSarwate Ah yes, good point! May 11 '20 at 16:01
• It is still called PAM not BPSK. PAM need not have equally spaced levels, but under AWGN would have the best BER if done as such. Here is an example of PAM with unequal levels to make up for non-linearity in the modulator. If the modulation is done by changing the amplitude of each pulse only, this is called PAM. May 12 '20 at 11:21


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# A series is defined, for positive integer n, by an=n+1n+3−nn+2 What

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A series is defined, for positive integer n, by an=n+1n+3−nn+2 What  [#permalink]

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23 Oct 2018, 22:03
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A series is defined, for positive integer n, by

$$a_n=\frac{n+1}{n+3}−\frac{n}{n+2}$$

What is the sum of the first 60 terms of this series?

A. 1/6
B. 1/3
C. 29/30
D. 40/63
E. 61/63

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Re: A series is defined, for positive integer n, by an=n+1n+3−nn+2 What  [#permalink]

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23 Oct 2018, 23:13
$$a_n$$ = $$\frac{n+1}{n+3}$$ - $$\frac{n}{n+2}$$

$$a_1$$ = $$\frac{2}{4}$$ - $$\frac{1}{3}$$

$$a_2$$ = $$\frac{3}{5}$$ - $$\frac{2}{4}$$

$$a_3$$ = $$\frac{4}{6}$$ - $$\frac{3}{5}$$

$$a_4$$ = $$\frac{5}{7}$$ - $$\frac{4}{6}$$
.
.
.
.
.
$$a_{58}$$ = $$\frac{59}{61}$$ - $$\frac{58}{60}$$

$$a_{59}$$ = $$\frac{60}{62}$$ - $$\frac{59}{61}$$

$$a_{60}$$ = $$\frac{61}{63}$$ - $$\frac{60}{62}$$

All the terms get canceled out and only $$\frac{61}{63}$$ - $$\frac{1}{3}$$ remains

$$\frac{61}{63}$$ - $$\frac{1}{3}$$ = $$\frac{40}{63}$$

OPTION : D
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Re: A series is defined, for positive integer n, by an=n+1n+3−nn+2 What  [#permalink]

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24 Oct 2018, 03:49

Solution

Given:
• In a series, $$a_n = \frac{(n + 1)}{(n + 3)} – \frac{n}{(n + 2)}$$

To find:
• The sum of the first 60 terms of this series

Approach and Working:
• $$a_n = \frac{(n + 1)}{(n + 3)} – \frac{n}{(n + 2)} = \frac{[(n + 1)(n + 2) – n(n + 3)]}{(n + 3)(n + 2)} = \frac{2}{(n + 3)(n + 2)} = 2[\frac{1}{(n + 2)} – \frac{1}{(n + 3)}]$$
• Thus,
o $$a_1 = 2(\frac{1}{3} – \frac{1}{4})$$
o $$a_2 = 2(\frac{1}{4} – \frac{1}{5})$$
o $$a_3 = 2(\frac{1}{5} – \frac{1}{6})$$
o .
o .
o .
o $$a_{59} = 2(\frac{1}{61} – \frac{1}{62})$$
o $$a_{60} = 2(\frac{1}{62} – \frac{1}{63})$$
• Therefore, $$S = 2(\frac{1}{3} – \frac{1}{4} + \frac{1}{4} – \frac{1}{5} + \frac{1}{5} – \frac{1}{6} + …... + \frac{1}{61} – \frac{1}{62} + \frac{1}{62} – \frac{1}{63}) = 2(\frac{1}{3} – \frac{1}{63}) = 2 * \frac{20}{63} = \frac{40}{63}$$

Hence, the correct answer is Option D

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Re: A series is defined, for positive integer n, by an=n+1n+3−nn+2 What   [#permalink] 24 Oct 2018, 03:49
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