text:
# Which of the following is not an invalid BCD Code?

1. 1011
2. 1010
3. 1001
4. 1100

Option 3 : 1001

## Binary Codes MCQ Question 1 Detailed Solution

Binary Coded Decimal (BCD) code:

• BCD is a way to express each of the decimal digits with a binary code.
• In this code, each decimal digit is represented by its 4-bit binary equivalent.
• Also, with four bits we can represent sixteen numbers (0000 to 1111)
• But as there are 10 decimal digits from 0 to 9, BCD code uses only the first ten of these (0000 to 1001). The remaining six code combinations i.e. 1010 to 1111 are invalid in BCD.

 Decimal Number Binary Number Binary Coded Decimal (BCD) 0 0000 0000 1 0001 0001 2 0010 0010 3 0011 0011 4 0100 0100 5 0101 0101 6 0110 0110 7 0111 0111 8 1000 1000 9 1001 1001 10 1010 0001 0000 11 1011 0001 0001 12 1100 0001 0010 13 1101 0001 0011 14 1110 0001 0100 15 1111 0001 0101

# The 4-bit Gray code for decimal number 5 is:

1. 0111
2. 1010
3. 1110
4. 1000

Option 1 : 0111

## Binary Codes MCQ Question 2 Detailed Solution

Binary code for 5 → 0101

The procedure to convert a gray code to a binary one is as shown:

A3 → Copy MSB → 0

A2 → 0 ⊕ 1 → 1

A1 → 1 ⊕ 0 → 1

A0 → 0 ⊕ 1 → 1

Grey code → 0111

# In Binary-coded Decimal (BCD) systems, the decimal number 81 is represented as

1. 10000001
2. 10100010
3. 01010001
4. 00011000

Option 1 : 10000001

## Binary Codes MCQ Question 3 Detailed Solution

• In BCD each decimal digit is represented by a 4-bit binary number.
• The binary representation of 8 → 1000
• The binary representation of 1 → 0001
• (81)10 = 10000001

# The given logic circuit represents:

1. 4 bit binary to decimal converter
2. 4 bit decimal to excess-3 code converter
3. 4 bit binary to Gray code converter
4. 4 bit decimal to binary converter

Option 3 : 4 bit binary to Gray code converter

## Binary Codes MCQ Question 4 Detailed Solution

The circuit is redrawn as:

y= x1

y2 = x1 ⊕ x2

y3 = x⊕ x3

y4 = x3 ⊕ x4

Example:

Let the input to the circuit be 1010

∴ For an input 1010, we get the output as 1111.

Similarly, let the input be 0110.

The output for 0110 input is 0101

Observations:

$$1010\xrightarrow{Binary\;to\;Gray} 1111$$

$$0110\xrightarrow{Binary\;to\;Gray} 0101$$

∴ The given circuit converts 4 bit binary to Grey code converter.

# The gray code for decimal number 6 is equivalent to

1. 1100
2. 1001
3. 0101
4. 0111

Option 3 : 0101

## Binary Codes MCQ Question 5 Detailed Solution

6 can be written as 0110 in binary form.

The conversion of binary to gray as follows.

(G1 G2 G3 G4)2

After conversion, G1g = G1

G2g = G1⊗ G2

G3g = G2⊗ G3

G4g = G3⊗ G4

The gray code will be (G1g G2g G3g G4g) = 0101,

given the binary code is 0110

# Which of the following is an invalid state in 8-4-2-1 Binary Coded Decimal counter

1. 1000
2. 1001
3. 0011
4. 1100

Option 4 : 1100

## Binary Codes MCQ Question 6 Detailed Solution

Concept:

• 8421 is also known as BCD code
• BCD is a weighted code.
• In weighted codes, each successive digit from right to left represents weights equal to some specified value, and to get the equivalent decimal number to add the products of the weights by the corresponding binary digit.

The following represents the 4-bit binary representation of decimal values:

 Decimal Symbol BCD Digit 0 0000 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111 8 1000 9 1001

Explanation:

• A number with an 'n' decimal digit will require 4k bits in BCD.
• Ex- Decimal 396 is represented in BCD with 12 bits as 0011 1001 0110, with each group of 4 bits representing one decimal digit.
• As the decimal digits relating to 1010, 1011, 1100, 1101, 1110, and 1111 do not exist, these are invalid.

In BCD valid states are from 0 to 9. So 1100 is an invalid state.

# BCD code has always how many bits per number

1. 2
2. 4
3. 8
4. 16

Option 2 : 4

## Binary Codes MCQ Question 7 Detailed Solution

• BCD (Binary Coded Decimal) is simply the 4-bit binary code representation of a decimal digit.
• Each decimal digit replaced in the integer and fractional parts by its binary equivalent.
• BCD Code uses four bits to represent the 10 decimal digits of 0 to 9.
• BCD code is often called 8421 because the BCD is a weighted code and the weights used in binary coded decimal code are 8, 4, 2, 1.

The following represents the 4-bit binary representation of decimal values:

 Decimal Symbol BCD Digit 0 0000 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111 8 1000 9 1001

# BCD equivalent of (345)10 is:

1. 0011 1001 1010
2. 1001 1001 1111
3. 0011 0100 0101
4. 0101 1100 1001

Option 3 : 0011 0100 0101

## Binary Codes MCQ Question 8 Detailed Solution

Explanation:

• A Binary Coded Decimal (BCD) is a type of binary representation for decimal values where each digit is represented by its 4-bit binary equivalent.
• Binary-coded decimals are an easy way to represent decimal values, as each digit is represented by its own 4-bit binary sequence which only has 10 different combinations.

The following represents the 4-bit binary representation of decimal values:

 Decimal Symbol BCD Digit 0 0000 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111 8 1000 9 1001

The BCD equivalent of 345 is 0011 0100 0101

# In which digital code is decimal digit expressed by 4 binary digits.

1. BCD code
2. Binary code
3. ASCII code
4. Gray code

Option 1 : BCD code

## Binary Codes MCQ Question 9 Detailed Solution

BCD (Binary Coded Decimal):

1) BCD (Binary Coded Decimal) is simply the 4-bit binary code representation of a decimal digit.

2) Each decimal digit replaced in the integer and fractional parts by its binary equivalent.

3) BCD Code uses four bits to represent the 10 decimal digits of 0 to 9.

4) BCD code is often called 8421 because the BCD is a weighted code and the weights used in binary coded decimal code are 8, 4, 2, 1.

5) The first 8-bit Extended Binary Coded Decimal Interchange Code (EBCDIC) introduced by IBM.

The following represents the 4-bit binary representation of decimal values:

 Decimal Symbol BCD Digit 0 0000 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111 8 1000 9 1001

Important Points

ASCII code:

• The ASCII table has 128 characters, with values from 0 through 127.
• Thus, 7 bits are sufficient to represent a character in ASCII; however, most computers typically reserve 1 byte, (8 bits), for an ASCII character.
• One byte allows a numeric range from 0 through 255 which leaves room for growth in the size of the character set, or for a sign bit.
• Consequently, a character data type may optionally represent signed values;
• However, for now, we will assume that character data types are unsigned, i.e. positive integer values, in the range 0-127.

# A BCD decoder will have how many rows in truth table?

1. 3
2. 9
3. 8
4. 10

Option 4 : 10

## Binary Codes MCQ Question 10 Detailed Solution

A BCD decoder will have 10 rows in the truth table.

Explanation:

• In BCD to 7-segment Decoder, the outputs of a digital circuit are often displayed as decimal digits.
• BCD to 7-segment decoder is a combinational circuit that converts a BCD number into signals that are required for the display of the value of that number on a seven-segment display.
• The number of input lines is 4.

Notes:

• The decoder outputs are (a, b, c, d, e, f, g)
• For the display of the digit 0, segments a, b, c, d, e, f will be lit as shown above.

A truth table can be formed for all digits from 0 to 9.

 Digit D C B A a b c d c f g Hex equivalent 0 0 0 0 0 1 1 1 1 1 1 0 7E 1 0 0 0 1 0 1 1 0 0 0 0 30 2 0 0 1 0 1 1 0 1 1 0 1 6D 3 0 0 1 1 1 1 1 1 0 0 1 79 4 0 1 0 0 0 1 1 0 0 1 1 33 5 0 1 0 1 1 0 1 1 0 1 1 5B 6 0 1 1 0 1 0 1 1 1 1 1 5 F 7 0 1 1 1 1 1 1 0 0 0 0 70 8 1 0 0 0 1 1 1 1 1 1 1 7F 9 1 0 0 1 1 1 1 1 0 1 1 7B

# What would be the gray code equal to the number 14?

1. 1000
2. 1110
3. 1111
4. 1001

Option 4 : 1001

## Binary Codes MCQ Question 11 Detailed Solution

Concept:

The circuit for binary to gray code convertor is:

y= x1

y2 = x1 ⊕ x2

y3 = x⊕ x3

y4 = x3 ⊕ x4

Calculation:

Let the input to the circuit be 14

Binary value of 14 is 1110

∴ For an input 1110, we get the output as :

y= 1

y2 = 1 1 = 0

y3 = 1 ⊕ 1 = 0

y4 = 1 0 = 1

Hence the gray code for 14 is 1001

# If the constraint length of a (n, k, L) convolutional code is defined as the number of encoder output bits influenced by each message bit, then the constraint length is given by

1. L(n + 1)
2. n(L + 1)
3. n(L + k)
4. L(n + k)

Option 2 : n(L + 1)

## Binary Codes MCQ Question 12 Detailed Solution

• A convolutional code is a type of error-correcting code. For generating a convolutional code, the information is passed sequentially through a linear finite-state shift register.
• An (n, k, L) convolutional code can be implemented with a k-input, n-output linear sequential circuit with input memory L.
• The constraint length is defined as K = n(L + 1).
• Since each information bit remains in the encoder for up to (L + 1) time units, and during each time unit can affect any of the n encoder outputs.
• Therefore, K can be interpreted as the maximum number of encoder outputs that can be affected by a signal information bit.

# BCD equivalent of decimal number 43 is

1. 101 011
2. 1000 0011
3. 0100 0011
4. 0011 0100

Option 3 : 0100 0011

## Binary Codes MCQ Question 13 Detailed Solution

Concept:

• 8421 is also known as BCD code
• BCD is a weighted code.
• In weighted codes, each successive digit from right to left represents weights equal to some specified value, and to get the equivalent decimal number to add the products of the weights by the corresponding binary digit.

The following represents the 4-bit binary representation of decimal values:

 Decimal Symbol BCD Digit 0 0000 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111 8 1000 9 1001

Application:

From the above Table:

4 = 0100

3 = 0011

43 = 0100 0011 in BCD

• A number with 'n' decimal digit will require 4k bits in BCD.
• Ex- Decimal 396 is represented in BCD with 12 bits as 0011 1001 0110, with each group of 4 bits representing one decimal digit.
• As the decimal digits relating to 1010, 1011, 1100, 1101, 1110, and 1111 do not exist, these are invalid.

# In CRC if the data unit is 100111001 and the divisor is 1011 then what is dividend at the receiver?

1. 100111001101
2. 100111001011
3. 100111001
4. 100111001110

Option 2 : 100111001011

## Binary Codes MCQ Question 14 Detailed Solution

The correct answer is "option 2".

CONCEPT:

CRC stands for Cyclic Redundancy Check, is an error-detecting code used to detect errors in the frame received.

CRC is used to detect errors in the Data link layer.

Key Points

Steps to perform CRC on the sender side:

1. String of n 0’s is appended to the data unit to be transmitted.
2. Where, n -> number of bits in CRC divisor - 1
3. Binary division is performed of data unit with the CRC divisor.
4. The remainder obtained after division is called CRC.
5. Append CRC at last of data units.

CALCULATION:

STEP1:

No. of bits in divisor - 4 bits, append 3 0's to data bits

Data bits after appending 0's - 100111001000

STEP2:

Binary division of 100111001000 with 1011:

_____________

1011  ) 100111001000  ( 101000001

1011

----------

01011

1011

-----------------

1000

⊕  1011

------------

011

STEP 3:

Append CRC 011 at last of data unit → 100111001011.

Hence, the correct answer is "option 2".

# The circuit in the figure converts

1. BCD to binary code
2. Binary to Excess-3 code
3. Excess-3 to Gray code
4. Gray to Binary code

Option 4 : Gray to Binary code

## Binary Codes MCQ Question 15 Detailed Solution

Concept:

Gray code to Binary conversion:

b3 = g3 → MSB is same in both

b2 = b3 ⊕ g2

b1 = b2 ⊕ g1

b0 = b1 ⊕ g0

Analysis:

If we carefully observe the given circuit MSB is the same

AND and OR gates are used to butter the input of XOR gate {input in XOR is input → gray code and output of previous XOR} so it is gray code to binary converter.

Important Point:

Binary to gray conversion:

g3 = b3

g2 = b3 ⊕ b2

g1 = b2 ⊕ b1

g0 = b1 ⊕ b0

# The code used to reduce the error due to ambiguity in reading of a binary optical encoder is

1. Octal code
2. Excess-3 code
3. Gray code
4. BCD code

Option 3 : Gray code

## Binary Codes MCQ Question 16 Detailed Solution

Concept:

Gray codes are less error-prone because they only change in one-bit position at a time. Hence, they are considered as the minimum error code.

Explanation:

 Decimal Binary Gray code Octal Excess – 3 code 0 0000 0000 000 0011 1 0001 0001 001 0100 2 0010 0011 010 0101 3 0011 0010 011 0110 4 0100 0110 100 0111 5 0101 0111 101 1000 6 0110 0101 110 1001 7 0111 0100 111 1010 8 1000 1100 001 000 1011

From Table:

Distance between two successive values In Octal, Binary and Excess-3 code ≥ 1

Distance between two successive values in gray code is 1

Hence minimum error code is gray code.

# Match List-I with List-II: List-I List-II a) Baudot code i) Permits the correction of errors by receiver without retransmission b) Hamming code ii) Provides no error detection at all c) Error detecting code iii) Permits the detection of two errors or correction of only one error d) Forward error-correcting code iv) Uses an extra parity bit at the end of each word to detect errors in received data Choose the correct option from those given below:

1. a-ii; b-iii; c-iv; d-i
2. a-iii; b-ii; c-i; d-iv
3. a-iv; b-ii; c-i; d-iii
4. a-ii; b-iii; c-i; d-iv

Option 1 : a-ii; b-iii; c-iv; d-i

## Binary Codes MCQ Question 17 Detailed Solution

 List-I List-II a) Baudot code i) Provides no error detection at all b) Hamming code ii) Permits the detection of two errors or correction of only one error c) Error detecting code iii) Uses an extra parity bit at the end of each word to detect errors in received data d) Forward error-correcting code iv) Permits the correction of errors by the receiver without retransmission

Baudot Code:

• Baudot is a five-bit binary code that uses crosses and dots.
• Originally used in wireless telegraphy as a replacement for Morse Code.
• Binary code which uses crosses and dots.
• Used for teleprinter message and allowed to encode 25 = 32 characters.
• The Baudot code includes two 30-symbol character sets, and two Shift symbols, the shift symbols are used to shift between the two character sets, thereby allowing for 60 different
• It provides no error detection at all

Hamming code:

Hamming codes are linear block codes. The family of (n, k) hamming codes for q ≥ 3 is defined by the following expressions:

1. Number of Block diagrams (n) = 2q – 1

2. Number of message bits (k) = 2q – q – 1

3. Number of Parity bits (q) = (n – k)

Hamming code is a set of error-correction codes that can be used to detect and correct '1' bit errors that can occur when bit stream are sent through channel. It can be employed in both burst and signal error correction

Error detecting code:

• An error occurs when, it is altered between transmission and reception (1 is transmitted and 0 is received, and vice versa)
• For reliable communication, errors must be detected and corrected.
• Error detection means deciding whether the received data is correct or not without having a copy of the original message.
• Error detection and correction uses the concept of redundancy, which means adding extra bits for detecting errors at the destination.
• These redundant bits are added by the sender and removed by the receiver.
• The error detection methods include:

1). VRC (Vertical Redundancy Check): It is an error-detecting code commonly used in digital networks and storage devices to detect an error in data sent by the sender.

2). LRC (Longitudinal Redundancy Check): It is an error-checking method used on an eight-bit ASCII character

3). CRC (Cyclic Redundancy Check)

4). Checksum

# A Frequency Hopping Spread Spectrum system has its frequency synthesizers controlled by five stage shift-registers with feedback connections taken from the second and fifth stages. The number of slots available for frequency hopping is:

1. 32
2. 31
3. 24
4. 28

Option 1 : 32

## Binary Codes MCQ Question 18 Detailed Solution

Concept:

• In FH-SS, the spectrum of the transmitted signal is spread sequentially by randomly hopping the data modulated carrier from one frequency to the next.
• The synthesizer output at a given instant of time is the frequency hop.
• The output bits of the PN generator change randomly. Hence the synthesizer output frequency will also change randomly.
• If the number of successive bits at the output of the PN generator is n, then the total number of frequency hops will be 2n.

Calculation:

The number of register stages given is 5, i.e.

n = 5

∴ The number of slots available for frequency hopping will be:

Number of Slots = 2n = 25

= 32

# The number of bytes required to represent the decimal number 1856357 in packed BCD (Binary Coded Decimal) form is ________.

## Binary Codes MCQ Question 19 Detailed Solution

In BCD coding, every decimal bit is encoded into 4 bits of binary representation.

Also, 1 Byte = 8 bits.

The given decimal number is 1856357, which has a total of 7 digits.

Each digital will have a 4-bit binary equivalent.

∴ The total number of bits required to represent will be:

= 4 × 7 = 28 bits

Since 1 Byte = 8 bits, the total number of bytes required will be 4.

This is because a 3 Byte number has only 3 × 8 = 24 bits.

# The number of bits used to store a BCD digit is

1. 8
2. 4
3. 1
4. 2

Option 2 : 4

## Binary Codes MCQ Question 20 Detailed Solution

A Binary Coded Decimal (BCD) is a type of binary representation for decimal values where each digit is represented by its 4 bit binary equivalent.

Binary-coded decimals are an easy way to represent decimal values, as each digit is represented by its own 4-bit binary sequence which only has 10 different combinations.

The following represents the 4-bit binary representation of decimal values:

 Decimal Symbol BCD Digit 0 0000 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111 8 1000 9 1001

• A number with 'n' decimal digit will require 4k bits in BCD.
• Example: Decimal 396 is represented in BCD with 12 bits as 0011 1001 0110, with each group of 4 bits representing one decimal digit.
• As the decimal digits relating to 1010, 1011, 1100, 1101, 1110, and 1111 do not exist, these are invalid.


text:
Now showing items 1-20 of 31

• #### 2^n Bordered Constructions of Self-Dual codes from Group Rings

Self-dual codes, which are codes that are equal to their orthogonal, are a widely studied family of codes. Various techniques involving circulant matrices and matrices from group rings have been used to construct such codes. Moreover, families of rings have been used, together with a Gray map, to construct binary self-dual codes. In this paper, we introduce a new bordered construction over group rings for self-dual codes by combining many of the previously used techniques. The purpose of this is to construct self-dual codes that were missed using classical construction techniques by constructing self-dual codes with diﬀerent automorphism groups. We apply the technique to codes over ﬁnite commutative Frobenius rings of characteristic 2 and several group rings and use these to construct interesting binary self-dual codes. In particular, we construct some extremal self-dual codes length 64 and 68, constructing 30 new extremal self-dual codes of length 68.
• #### An Altered Four Circulant Construction for Self-Dual Codes from Group Rings and New Extremal Binary Self-dual Codes I

We introduce an altered version of the four circulant construction over group rings for self-dual codes. We consider this construction over the binary field, the rings F2 + uF2 and F4 + uF4; using groups of order 4 and 8. Through these constructions and their extensions, we find binary self-dual codes of lengths 16, 32, 48, 64 and 68, many of which are extremal. In particular, we find forty new extremal binary self-dual codes of length 68, including twelve new codes with \gamma=5 in W68,2, which is the first instance of such a value in the literature.
• #### Bordered Constructions of Self-Dual Codes from Group Rings and New Extremal Binary Self-Dual Codes

We introduce a bordered construction over group rings for self-dual codes. We apply the constructions over the binary field and the ring $\F_2+u\F_2$, using groups of orders 9, 15, 21, 25, 27, 33 and 35 to find extremal binary self-dual codes of lengths 20, 32, 40, 44, 52, 56, 64, 68, 88 and best known binary self-dual codes of length 72. In particular we obtain 41 new binary extremal self-dual codes of length 68 from groups of orders 15 and 33 using neighboring and extensions. All the numerical results are tabulated throughout the paper.
• #### Composite Constructions of Self-Dual Codes from Group Rings and New Extremal Self-Dual Binary Codes of Length 68

We describe eight composite constructions from group rings where the orders of the groups are 4 and 8, which are then applied to find self-dual codes of length 16 over F4. These codes have binary images with parameters [32, 16, 8] or [32, 16, 6]. These are lifted to codes over F4 + uF4, to obtain codes with Gray images extremal self-dual binary codes of length 64. Finally, we use a building-up method over F2 + uF2 to obtain new extremal binary self-dual codes of length 68. We construct 11 new codes via the building-up method and 2 new codes by considering possible neighbors.
• #### Composite Matrices from Group Rings, Composite G-Codes and Constructions of Self-Dual Codes

In this work, we define composite matrices which are derived from group rings. We extend the idea of G-codes to composite G-codes. We show that these codes are ideals in a group ring, where the ring is a finite commutative Frobenius ring and G is an arbitrary finite group. We prove that the dual of a composite G-code is also a composite G-code. We also define quasi-composite G-codes. Additionally, we study generator matrices, which consist of the identity matrices and the composite matrices. Together with the generator matrices, the well known extension method, the neighbour method and its generalization, we find extremal binary self-dual codes of length 68 with new weight enumerators for the rare parameters $\gamma$ = 7; 8 and 9: In particular, we find 49 new such codes. Moreover, we show that the codes we find are inaccessible from other constructions.
• #### Constructing Self-Dual Codes from Group Rings and Reverse Circulant Matrices

In this work, we describe a construction for self-dual codes in which we employ group rings and reverse circulant matrices. By applying the construction directly over different alphabets, and by employing the well known extension and neighbor methods we were able to obtain extremal binary self-dual codes of different lengths of which some have parameters that were not known in the literature before. In particular, we constructed three new codes of length 64, twenty-two new codes of length 68, twelve new codes of length 80 and four new codes of length 92.
• #### Constructions for Self-Dual Codes Induced from Group Rings

In this work, we establish a strong connection between group rings and self-dual codes. We prove that a group ring element corresponds to a self-dual code if and only if it is a unitary unit. We also show that the double-circulant and four-circulant constructions come from cyclic and dihedral groups, respectively. Using groups of order 8 and 16 we find many new construction methods, in addition to the well-known methods, for self-dual codes. We establish the relevance of these new constructions by finding many extremal binary self-dual codes using them, which we list in several tables. In particular, we construct 10 new extremal binary self-dual codes of length 68.
• #### Double Bordered Constructions of Self-Dual Codes from Group Rings over Frobenius Rings

In this work, we describe a double bordered construction of self-dual codes from group rings. We show that this construction is effective for groups of order 2p where p is odd, over the rings F2 + uF2 and F4 + uF4. We demonstrate the importance of this new construction by finding many new binary self-dual codes of lengths 64, 68 and 80; the new codes and their corresponding weight enumerators are listed in several tables
• #### Extending an Established Isomorphism between Group Rings and a Subring of the n × n Matrices

In this work, we extend an established isomorphism between group rings and a subring of the n × n matrices. This extension allows us to construct more complex matrices over the ring R. We present many interesting examples of complex matrices constructed directly from our extension. We also show that some of the matrices used in the literature before can be obtained by a direct application of our extended isomorphism.
• #### G-codes over Formal Power Series Rings and Finite Chain Rings

In this work, we define $G$-codes over the infinite ring $R_\infty$ as ideals in the group ring $R_\infty G$. We show that the dual of a $G$-code is again a $G$-code in this setting. We study the projections and lifts of $G$-codes over the finite chain rings and over the formal power series rings respectively. We extend known results of constructing $\gamma$-adic codes over $R_\infty$ to $\gamma$-adic $G$-codes over the same ring. We also study $G$-codes over principal ideal rings.
• #### G-Codes, self-dual G-Codes and reversible G-Codes over the Ring Bj,k

In this work, we study a new family of rings, Bj,k, whose base field is the finite field Fpr . We study the structure of this family of rings and show that each member of the family is a commutative Frobenius ring. We define a Gray map for the new family of rings, study G-codes, self-dual G-codes, and reversible G-codes over this family. In particular, we show that the projection of a G-code over Bj,k to a code over Bl,m is also a G-code and the image under the Gray map of a self-dual G-code is also a self-dual G-code when the characteristic of the base field is 2. Moreover, we show that the image of a reversible G-code under the Gray map is also a reversible G2j+k-code. The Gray images of these codes are shown to have a rich automorphism group which arises from the algebraic structure of the rings and the groups. Finally, we show that quasi-G codes, which are the images of G-codes under the Gray map, are also Gs-codes for some s.
• #### Group Rings, G-Codes and Constructions of Self-Dual and Formally Self-Dual Codes

We describe G-codes, which are codes that are ideals in a group ring, where the ring is a finite commutative Frobenius ring and G is an arbitrary finite group. We prove that the dual of a G-code is also a G-code. We give constructions of self-dual and formally self-dual codes in this setting and we improve the existing construction given in [13] by showing that one of the conditions given in the theorem is unnecessary and, moreover, it restricts the number of self-dual codes obtained by the construction. We show that several of the standard constructions of self-dual codes are found within our general framework. We prove that our constructed codes must have an automorphism group that contains G as a subgroup. We also prove that a common construction technique for producing self-dual codes cannot produce the putative [72, 36, 16] Type II code. Additionally, we show precisely which groups can be used to construct the extremal Type II codes over length 24 and 48. We define quasi-G codes and give a construction of these codes.
• #### A Modified Bordered Construction for Self-Dual Codes from Group Rings

We describe a bordered construction for self-dual codes coming from group rings. We apply the constructions coming from the cyclic and dihedral groups over several alphabets to obtain extremal binary self-dual codes of various lengths. In particular we find a new extremal binary self-dual code of length 78.
• #### Modified Quadratic Residue Constructions and New Exermal Binary Self-Dual Codes of Lengths 64, 66 and 68

In this work we consider modiﬁed versions of quadratic double circulant and quadratic bordered double circulant constructions over the binary ﬁeld and the rings F2 +uF2 and F4 +uF4 for diﬀerent prime values of p. Using these constructions with extensions and neighbors we are able to construct a number of extremal binary self-dual codes of diﬀerent lengths with new parameters in their weight enumerators. In particular we construct 2 new codes of length 64, 4 new codes of length 66 and 14 new codes of length 68. The binary generator matrices of the new codes are available online at [8].
• #### New binary self-dual codes of lengths 56, 58, 64, 80 and 92 from a modification of the four circulant construction.

In this work, we give a new technique for constructing self-dual codes over commutative Frobenius rings using $\lambda$-circulant matrices. The new construction was derived as a modification of the well-known four circulant construction of self-dual codes. Applying this technique together with the building-up construction, we construct singly-even binary self-dual codes of lengths 56, 58, 64, 80 and 92 that were not known in the literature before. Singly-even self-dual codes of length 80 with $\beta \in \{2,4,5,6,8\}$ in their weight enumerators are constructed for the first time in the literature.
• #### New binary self-dual codes via a generalization of the four circulant construction

In this work, we generalize the four circulant construction for self-dual codes. By applying the constructions over the alphabets $\mathbb{F}_2$, $\mathbb{F}_2+u\mathbb{F}_2$, $\mathbb{F}_4+u\mathbb{F}_4$, we were able to obtain extremal binary self-dual codes of lengths 40, 64 including new extremal binary self-dual codes of length 68. More precisely, 43 new extremal binary self-dual codes of length 68, with rare new parameters have been constructed.
• #### New Extremal Binary Self-dual Codes from block circulant matrices and block quadratic residue circulant matrices

In this paper, we construct self-dual codes from a construction that involves both block circulant matrices and block quadratic residue circulant matrices. We provide conditions when this construction can yield self-dual codes. We construct self-dual codes of various lengths over F2 and F2 + uF2. Using extensions, neighbours and sequences of neighbours, we construct many new self-dual codes. In particular, we construct one new self-dual code of length 66 and 51 new self-dual codes of length 68.
• #### New Extremal binary self-dual codes of length 68 from generalized neighbors

In this work, we use the concept of distance between self-dual codes, which generalizes the concept of a neighbor for self-dual codes. Using the $k$-neighbors, we are able to construct extremal binary self-dual codes of length 68 with new weight enumerators. We construct 143 extremal binary self-dual codes of length 68 with new weight enumerators including 42 codes with $\gamma=8$ in their $W_{68,2}$ and 40 with $\gamma=9$ in their $W_{68,2}$. These examples are the first in the literature for these $\gamma$ values. This completes the theoretical list of possible values for $\gamma$ in $W_{68,2}$.
• #### New Extremal Self-Dual Binary Codes of Length 68 via Composite Construction, F2 + uF2 Lifts, Extensions and Neighbors

We describe a composite construction from group rings where the groups have orders 16 and 8. This construction is then applied to find the extremal binary self-dual codes with parameters [32, 16, 8] or [32, 16, 6]. We also extend this composite construction by expanding the search field which enables us to find more extremal binary self-dual codes with the above parameters and with different orders of automorphism groups. These codes are then lifted to F2 + uF2, to obtain extremal binary images of codes of length 64. Finally, we use the extension method and neighbor construction to obtain new extremal binary self-dual codes of length 68. As a result, we obtain 28 new codes of length 68 which were not known in the literature before.
• #### New Self-Dual and Formally Self-Dual Codes from Group Ring Constructions

In this work, we study construction methods for self-dual and formally self-dual codes from group rings, arising from the cyclic group, the dihedral group, the dicyclic group and the semi-dihedral group. Using these constructions over the rings $_F2 +uF_2$ and $F_4 + uF_4$, we obtain 9 new extremal binary self-dual codes of length 68 and 25 even formally self-dual codes with parameters [72,36,14].

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--Calculating Discount Payback 1 answer below »

An investment project costs $10,000 and has annual cash flows of$2,600 for six years. What is the discounted payback period if the discount rate is 0%? What if the discount rate is 10%? If it is 15%?

Wahid A
Solution: If discount rate is 0, it will take 10,000/ 2,600 = 3.8 years If the discount rate is 10, the discounted cash flows are 2,600/ 1.10 = 2,363.64, 2,600/ 1.10^2 = 2,148.76, 2,600/ 1.10^3 = 1,953.42, 2,600/ 1.1^4 = 1,775.83, At the end of the 4th year, this adds up to 8,,241.65. 5th year cash flow = 2,600/ 1.1^5 5th year cash flow =...

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# Using the Riemann hypothesis to predict $p_n$, the $n^{th}$ prime number

In this video, Terence Tao states that if you assume the Riemann hypothesis to be true, you can predict the $$n^{th}$$ prime number $$p_n$$ using

$$\int_2^{p_n} \frac{dt}{\text{log}(t)} = n+O\left(\sqrt{n\cdot\text{log}^3 n}\right)$$

which will have an error of about $$\sqrt{n}$$. He goes on to say that the error $$O\left(\sqrt{n\cdot\text{log}^3 n}\right)$$ predicted by the Riemann hypothesis is essentially the same type of error one would have expected if the primes were distributed randomly.

$$\bullet$$ Does this mean for a given value $$n$$, $$O\left(\sqrt{n\cdot\text{log}^3 n}\right)$$ cannot be calculated and acts as a random variable?

$$\bullet$$ How then could you use the formula

$$\int_2^{p_n} \frac{dt}{\text{log}(t)} = n+O\left(\sqrt{n\text{log}^3 n}\right)$$

to get a prediction for $$p_n$$? Would you ignore $$O(\cdot)$$ and numerically find $$p_n$$ such that

$$\int_2^{p_n} \frac{dt}{\text{log}(t)} = n$$

is satisfied?

$$\bullet$$ He shows here that the $$10^{12}-th$$ prime number is $$29,996,224,275,833$$ and that the Riemann hypothesis predicts it to be $$29,996,219,470,277$$ which gives an error of about $$4,805,556$$. However, he says the error will be about $$\sqrt{n}=\sqrt{10^{12}}=10^6$$ which is nearly $$5$$ times less than our error. Am I interpreting this statement incorrectly?

• I think you need the actual error term. $O()$ only shows its asymptotic growth – Shine On You Crazy Diamond Mar 22 at 3:17

For your last point, he seems to be showing $$\rm{Li}^{-1}(n)$$. At this size we can get even closer using $$\rm{R}^{-1}(n)$$, though I believe asymptotically it doesn't matter.
Add: "[...] the error will be about [...]". Not really. The bounds for the error are asymptotically that amount (as BananaCats mentions in a comment). Especially with R(x) they will fluctuate around the actual value, so at any given point it could be quite close. We have bounds (e.g. Dusart, Axler, etc.) that tell us it won't get terribly far away from Li(x). If the Riemann Hypothesis is true then we know the bounds are much tighter even for values we haven't checked. Some researchers such as Büthe and Axler have gone over empirical Riemann zero data and shown we can correctly use the much tighter Schoenfeld bound up to $$5.5 \cdot 10^{25}$$.
• So in other words, he is saying that as $n$ gets large enough, the absolute error will be less than $\sqrt{n}$? $10^{12}$ is already quite large so I guess that bound doesn't occur until much later on? – Remy Mar 22 at 18:48


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# Classical car collision

I have a very confusing discussion with a friend of mine.

2 cars ($car_a$ and $car_b$) of the same mass $m$ are on a collision course. Both cars travel at $50_\frac{km}{h}$ towards each other.

They collide. Ignoring any shreds and collateral damage, what is the speed of collision that the driver of $car_a$ felt?

What I mean is, if $car_a$ were to be driven into an infinite mass wall, what would the velocity be to replicate the damage caused by the initial collision?

-
I don't believe there is a formal definition of "the speed of collision" other than what you've defined here. A better assumption to start from is complete symmetry between the cars. The "imaginary" wall becomes impenetrable as well as frictionless, even for the shreds. The imaginary wall is stronger than any actual wall that could be built, in fact. Provided all this, I believe you answer will just be 50 kmph. –  Alan Rominger Nov 2 '11 at 19:03
Exactly what I meant, and that's what I argued as well. My friend argues it's 100kmph though. –  Madara Uchiha Nov 2 '11 at 19:09
This experiment was actually covered on tv (Mythbusters), sans the infinite wall and spherical cows. The damage in the two-car-same-speed collision was consistent with a single car and a wall (speed was not additive) ( Episode 143: Mythssion Control ) –  horatio Nov 2 '11 at 20:53
I need to see that, thanks @horatio :) –  Madara Uchiha Nov 3 '11 at 6:13

I think you are asking how much damage would be done to the driver in the two cases you described. If that is your question, then the single car that is driven at a speed of $50\frac{km}{hr}$ into an infinite mass wall would experience the same damage as two identical cars being driven exactly head on at a speed relative to the ground of $50\frac{km}{hr}$ into each other.

You can easily understand this if you imagine an infinitely thin wall between the two colliding cars at exactly the plane where they collide. Assuming no shards or other pieces breaking off and going through this imaginary wall you can see that this is exactly the same as having an infinitely massive wall in place of either car since everything is exactly symmetric about this infinitely thin imaginary wall.

To be absolutely correct, this answer would actually require that the cars be left-right symmetric so that the centers of mass exactly line up perpendicular to the plane of the collision. If they were asymmetric it would be similar to symmetric cars hitting slightly off center - so there would be some torque around the point on the collision plane where the line determined by centers of mass intersect the plane. Clear?

EDIT: A situation which is equivalent to 2 cars hitting head on at 50kmph is the following: one car sitting stationary (with brakes off) while the other car hits it head on at 100kmph. This assumes a perfectly inelastic collision so that the two cars will then proceed (joined together) in the original direction of the 100kmph car but they will both be going at 50kmph. In both of these cases the change in speed of each car is 50kmph in a short time so the damages will be equivalent (either 50->0, 100->50 or 0->-50). However, if a car that is traveling at 100kmph hits an infinite mass wall the change in speed in a short time will be 100kmph so it is not equivalent to the two cases.

-
Okay, and on a related note, if we were to drive at different lanes at 50kmph each, when I throw an object out of my window into your lane. Assuming the object does not decelerate when shattering your front window, that object will indeed hit you at 100kmph. How does that settle with the above answer? –  Madara Uchiha Nov 3 '11 at 6:12
Yes, the object will hit at 100kmph but that is irrelevant to the analysis in my answer. A situation which is equivalent is the following: one car sitting stationary (with brakes off) while the other car hits it head on at 100kmph - this would be equivalent to have 2 cars hitting head on at 50kmph. This assumes a perfectly inelastic collision so that the two cars will then proceed (joined together) in the original direction of the 100kmph car but they will both be going at 50kmph. –  FrankH Nov 3 '11 at 7:06
In all these cases the change in speed of each car is 50kmph in a short time so the damages will be equivalent. However, if a car traveling at 100kmph hits an infinite mass wall the change in speed in a short time will be 100kmph so it is not equivalent to the two cases in the previous comment... –  FrankH Nov 3 '11 at 7:11

There is no defined quantity called "speed of collision" and you cannot really measure "damage" in that sense either. I've heard this before when people at the drivers licence center talk about collisions and it is a bit strange.

If your question is with what force a car will be subject to when it comes to a halt it is the same as always: $F = ma$, where $a$ here will be the deceleration the car is subject to in the collision. So if you collide with an imaginary inelastic wall, as you imply, that brings you to an absolute stop in no time the force will be infinite. Of course this is impossible in reality since there is no such wall and the car itself will also compress and therefore slow the deceleration.

If you're interested in an inelastic crash it could give you the speed before and after the crash if you look at the momentum, $p=mv$. The total momentum must be the same before and after the crash, called the conservation of linear momentum, so $m_av_a + m_b+v_b = 0$ where $v_a = -v_b$ and $m_a = m_b$ as stated in the question. Here you can see that depending on the initial movement of the cars it is impossible to come to a halt if they do not move in opposite directions. You could simulate you problem by making the wall very heavy and making it move the opposite way with a small velocity. This is of course not what happens in reality either though but maybe closer to your question.

-
When speaking of damage in a car crash, there are several things that cause it. Mainly, the rapid deceleration, and the shreds and debris flying from the broken class and metal. The later is ignored. I'm asking in this question, would the deceleration be the same should the car hit a giant impenetrable wall instead of another symmetric car at the same velocity of 50kmph? –  Madara Uchiha Nov 2 '11 at 19:58
Yes exactly, it's the deceleration that results in the damage. If the car hits the wall in 50kmph and deforms the same way as when it hits the other car, which also deforms the same way, then the deceleration would be the same and the collisions would be equal, true. –  while Nov 2 '11 at 21:36
Okay, and on a related note, if we were to drive at different lanes at 50kmph each, when I throw an object out of my window into your lane. Assuming the object does not decelerate when shattering your front window, that object will indeed hit you at 100kmph. How does that settle with the above answer? –  Madara Uchiha Nov 3 '11 at 6:11


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Publications

2007

• Böttcher, A., Karlovich, A., Silbermann, B., "Generalized Krein algebras and asymptotics of Toeplitz determinants", Methods of Functional Analysis and Topology 13, no. 3 (2007), 236-261.
• Faria F, Oliveira J J, "Global attractivity for scalar differential equations with small delay", J. Math. Anal. Appl. 329 (2007) 1397-1420.
• Figueiredo, J.F. Rodrigues and L. Santos (Eds): / Free Boundary Problems: theory and applications, / Series: International Series of Numerical Mathematics Vol. 154, Birkhauser Verlag, Basel, 2007
• Karlovich, A., "Asymptotics of determinants and traces of Toeplitz matrices with symbols in weighted Wiener algebras", Zeitschrift für Analysis und ihre Anwendungen 26 (2007), 43-56.
• Karlovich, A., "Asymptotics of block Toeplitz determinants generated by factorable matrix functions with equal partial indices", Math. Nachr. 280, no. 9-10 (2007), 1118-1127.
• Karlovich, A., "Semi-Fredholm singular integral operators with piecewise continuous coefficients on weighted variable Lebesgue spaces are Fredholm", Operators and Matrices 1, no. 3 (2007), 427-444.
• Karlovich, A., "Asymptotics of Toeplitz determinants generated by functions with Fourier coefficients in weighted Orlicz sequence classes", Proceedings of the 5th Conference on Function Spaces. In: Function Spaces, Contemporary Mathematics 435 (2007), 229-243.
• Karlovich, A., "Higher order asymptotic formulas for traces of Toeplitz matrices with symbols in Hölder-Zygmund spaces", Proceedings of IWOTA 2005,In: Recent Advances in Matrix and Operator Theory, Operator Theory: Advances and Applications 179 (2007), 185-196.
• Kremer G, Pandolfi Bianchi M, Soares A J, The effect of the affinity on the equilibrium behavior of a reactive system, 25th International Symposium on Rarefied Gas Dynamics, pp 941-94
• Mena F C and Tod P, Lanczos potentials and a measure of gravitational entropy in perturbed FLRW models, Classical and Quantum Gravity, 24 (2007) 1733-1745
• Mena F C, Mulryne D and Tavakol R, Non-linear vector perturbations in a contracting universe, Classical and Quantum Gravity, 24 (2007)  2721-2734
• Mena F C and Tod P, "Lanczos potentials and gravitational entropy in perturbed FLRW models", in "Proceedings of the Spanish Relativity Meeting 2006", Journal of Physics: Conference Series, vol. 66 (2007) 012019 (4 pages)
• Mars M, Mena F C and Vera R, Linear perturbations of matched spacetimes: the gauge problem and background symmetries, Classical and Quantum Gravity, 24 (2007) 3673-368
• Soares, A.J., Oliveira, Filipe, "A non-trivial explicit solution for the 14-velocity Cabannes kinetic model", Nonlinear Analysis TMA, Nonlinear Analysis TMA 67 (2007), pp. 1167-1172
• Soares, A.J., Bianchi. M. Pandolfi, Kremer, G., "Analysis of the trend to equilibrium of a chemically reacting system", J. Phys. A: Math. Theor. 40 (2007), pp. 2553–2571.

2006

• A. Karlovich, Higher order asymptotics of Toeplitz determinants with symbols in weighted Wiener algebras, Journal of Mathematical Analysis and Applications, 320 (2006) 944-963.
• Kremer Gilberto, Pandolfi Bianchi Miriam and Soares Ana Jacinta, A relaxation kinetic model for transport phenomena in a reactive flow, Physics of Fluids, 18 (2006) 037104 1-15
• G. Kremer, M. Pandolfi Bianchi, Ana Jacinta Soares,
Closure of the balance laws for gaseous mixtures near chemical equilibrium,
XIII International Conference WASCOM 2005, 304-309.
• R. Monaco, M. Pandolfi Bianchi, Ana Jacinta Soares,
Simulations at Kinetic Scale of Relaxation Models for Slow and Fast Chemical Reactions,
XIII International Conference WASCOM 2005, 378-389.
• Filipe Oliveira, Ana Jacinta Soares,
Global solutions of Boltzmann-type equations with three-body chemical interactions,
XIII International Conference WASCOM 2005, 414-419.
• Filipe C. Mena and Malcolm MacCallum, Locally discretely isotropic space-times, Pré-publicação CMAT 2006
• Hideo Iguchi, Tomohiro Harada and Filipe C. Mena, Spherically symmetric perfect fluids in area-radial coordinates, in "Albert Einstein Century International Conference", in American Institute of Physics Proceedings, 861, pgs. 6446-6453, Eds. J-M Alimi and A Fuzfa (New York: AIP, 2006)
• Filipe C. Mena, Non-equivalence of Newtonian and relativistic second order perturbations, in "A Century of Relativity Physics - Proceedings of the Spanish Relativity Meeting 2005 in American Institute of Physics Proceedings, 841, pgs. 639-640, Eds. L Mornas and J Diaz-Alonso (New York: AIP, 2006)
• Marc Mars, Filipe C. Mena and Raul Vera First order perturbations of the Einstein-Straus model, in "A Century of Relativity Physics - Proceedings of the Spanish Relativity Meeting 2005", in American Institute of Physics Proceedings, 841, pgs. 519-522, Eds. L Mornas and J Diaz-Alonso (New York: AIP, 2006)

2005

• Caiado, M.I. and Sarychev, A.V.,Remarks on stability of inverted pendula'', Rendiconti del Seminario Matematico dell'Universitá e del Politecnico di Torino 63 (2005), pp. 333-348.
• Caiado, M.I. and Sarychev, A., On Asymptotic Stabilization of Elastic Systems'', Second International Conference Physics and Control' (PhysCon 2005), St.Petersburg, Russia, August 2005.
• Faria, T., Liz, E., Oliveira, J.J., and Trofimchuk, S., On a generalized Yorke condition for scalar delayed population models'' Discrete Contin. Dynam. Systems 12 (2005), pp. 481-500.
• Iguchi, H., Harada, T., and Mena, F., Spherically symmetric perfect fluids in area-radial coordinates'', Classical and Quantum Gravity 22 (2005), pp. 841-854.
• Karlovich, A. and Lerner, A., Commutators of singular integrals on generalized $L^p$ spaces with variable exponent'', Publ. Matemátiques, Barcelona 49 (2005), pp. 111--125.
• Karlovich, A. and Santos, P.A., On asymptotics of Toeplitz determinants with symbols of nonstandard smoothness'', Journal of Fourier Analysis and Applications 11 (2005), pp. 43-72.
• Karlovich, A., Algebras of singular integral operators on Nakano spaces with Khvedelidze weights over Carleson curves with logarithmic whirl points'', Pseudodifferential Equations and Some Problems of  Mathematical Physics, Rostov-on-Don, Russia, 2005, pp. 135-142.
• Filipe C. Mena, Reza Tavakol and Raul Vera, Generalisations of the Einstein-Straus model to cylindrically symmetric settings, in "Proceedings of the Tenth Marcel Grossmann Meeting on General Relativity", edited by M. Novello, S. Perez-Bergliaffa and R. Ruffini, (Singapore: World Scientific, 2005)
• Monaco, R., Pandolfi Bianchi, M., and Soares, A.J., BGK-type models in strong reaction and kinetic chemical equilibrium regimes'',  J. Phys. A.: Math. and General. 38 (2005), pp. 10413-10431.
• Oliveira, F., A class of non-local operators for Vorticity waves'', Applicable Analysis 12 (2005), pp. 1287-1302.
• Oliveira, F. and Soares, A.J., `On global solutions to the discrete Boltzmann equations with chemical reactions'', Math. Meth. Appl. Sci. 28 (2005), pp. 1491-1506.

2004

• Mena, F.C., Nolan, B.C, Tavakol, R., The role of anisotropy and inhomogeneity in Lemaitre-Tolman-Bondi collapse, Physical Review D, vol 70, pp 84030-84039, 2004.
• Mena, F.C, Tod, P.; Matching non-stationary spatially homogeneous spacetimes with vacuum exteriors, Physical Review D, vol 70; pp 104028-104036, 2004.
• Oliveira, Filipe, Approximation of the DNLS equation by the cubic nonlinear Schrodinger equation, Proc. Roy. Soc. Edinburgh, Sect. A, vol 134, n. 3, pp 595-607, 2004.
• Soares,  A.J., Bianchi, M. Pandolfi, Kinetic approach to transport properties of a reactive gas, {\it World Scientific Proceedings of the XII International Conference Wascom}, Singapore, Eds. R. Monaco, S. Pennisi, S. Rionero, T. Ruggeri, pp. 380-385, 2004.

2003

• Oliveira, Filipe, Stability of the solitons for the one-dimensional Zakharov-Rubenchick equation, Physica D,  175:220-240, 2003.
• Mena, Filipe, Evolution of second order perturbations in flat dust FLRW, Proceedings of the Spanish Relativity Meeting 2002, pp. 263-266 (Publicaciones de la Universitat de Barcelona), 2003.
• Mena, Filipe, On modifications of the Einstein-Straus model to Bianchi metrics, Proceedings of the Spanish Relativity Meeting 2001, pg. 343-346 (Berlin: Springer-Verlag), 2003.

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# Limit of a function with e

Calculus Level 2

If $f(x)$ is a continuous function defined on the domain $x > -\frac{1}{2}$ that satisfies $(e^x-1)f(x)=\ln(1+2x),$ what is $f(0)?$

×


text:
# Why do tangent functions have asymptotes?

Nov 12, 2014

Since

$\tan x = \frac{\sin x}{\cos x}$,

it will have a vertical asymptote whenever the denominator $\cos x = 0$.

I hope that this was helpful.


text:
Homework Help: Questions on Complex Analysis

1. Sep 27, 2012

raopeng

1. The problem statement, all variables and given/known data
To prove that the sequence $a_{n}= \prod_{k}^\infty (1 + \frac{i}{k})$ when n is infinite constitutes points on a circle.

2. Relevant equations
Ehh no idea what equations shall be used.

3. The attempt at a solution
A friend asked me this, but I am usually engaged more with the physical aspects of Complex Analysis... so I have no idea how I should approach this question.

2. Sep 27, 2012

Studiot

Well points on a circle have a particular (finite) radius.

So you need to prove your infinite product has a finite limit,
Since we are in the complex plane this will be a radius, as every point with this modulus will be included.

3. Sep 28, 2012

raopeng

Thank you. Yes I noticed that too, as we can extract an infinite series from it...

4. Sep 28, 2012

Studiot

So what happens if you write a few of the terms of the series out and multiply them in pairs?

Further hint put 1/k = α.
So the terms take the form (1+αi)

5. Sep 28, 2012

raopeng

Thank you for the help. Also $arg a_{n} = \sum^{\inf}_{k}\frac{1}{k}$ covers the entire circle.


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Thứ Sáu, 8 tháng 5, 2015

Document type while transfer from fica to fico

What we have observed is that we post a payment using payment lot (FP05), The FI document generated at the end of the process (checked in FB03) is of document type MC. However when we post a payment using FPE1 t-code (using main and sub: 0600/0010), the FI document created at the end of the process is a MN.

What we want to find out is how is the system understanding whether to select MN or MC document type at the FI entry stage for the FICA postings that we are creating? And secondly, how can we control this, so that we can get MC document type being used for the FI entry for 0600/0010 payment that we are posting manually in FICA using FPE1.


text:
Capacitive reactance and capacitance

Unicorn Notreal

Joined Dec 24, 2015
5
I don't understand why is the capacitance of a capacitor is inversely proportional to the quantity of charges !
And also why the capacitive reactance is inversely proportional to the capacitance ?

crutschow

Joined Mar 14, 2008
25,659
I don't understand why is the capacitance of a capacitor is inversely proportional to the quantity of charges !
And also why the capacitive reactance is inversely proportional to the capacitance ?
You have it wrong. The value of the capacitance is proportional to the amount of charge stored: C = Q/V.

As the above equation shows, the larger the capacitance, the more charge that is moved for a given voltage (Q=CV), thus the "resistance" to moving that charge (the reactance) is reduced as the capacitance is increased.

Make sense?

Unicorn Notreal

Joined Dec 24, 2015
5
Yes, sorry I was probably asleep while reading the equation.
But what about the capacitive reactance (Xc=1/2πυC) why is it inversely proportional to the capacitance?
You have it wrong. The value of the capacitance is proportional to the amount of charge stored: C = Q/V.

As the above equation shows, the larger the capacitance, the more charge that is moved for a given voltage (Q=CV), thus the "resistance" to moving that charge (the reactance) is reduced as the capacitance is increased.

Make sense?

crutschow

Joined Mar 14, 2008
25,659
The larger the capacitance, the easier it is to move charge (Q=CV) through the capacitor, so it's reactance (resistance to charge movement) is inversely proportional to capacitance.

MrChips

Joined Oct 2, 2009
22,093
$$Reactance = \frac{1}{\omega C}$$

$$Susceptance =\omega C$$

If we ignore the j bit, we can say

$$Admittance =\omega C$$

Does that make you happy?

As C increases we admit more AC signal.

cmartinez

Joined Jan 17, 2007
7,174
$$Reactance = \frac{1}{\omega C}$$

$$Susceptance =\omega C$$

If we ignore the j bit, we can say

$$Admittance =\omega C$$

Does that make you happy?

As C increases we admit more AC signal.
Sorry for the dumb question, but what is "the j bit"?

Papabravo

Joined Feb 24, 2006
14,682
The "j bit" is a reference to the imaginary unit. Mathematicians use a lower case i, but Electrical Engineers might confuse that with current, so we use the letter j instead.
Reactance is is a real number, but impedance is a complex number.

Also:

$$\frac{1}{j \omega C}=-j \frac{1}{\omega C}$$

If the square root of -1 gives you a funny feeling, just think of j as a rotation operator.


text:
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solveda brass alloy is known to have a yield strength of

solveda brass alloy is known to have a yield strength of

Tel: 0086(371)86&15&18&27

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What is the tensile strength of brass?6.20 A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elastic modulus of 103 GPa (15.0 × 106psi). A cylindrical specimen of this alloy 12.7 mm (0.50 in.) in diameter and 250 mm (10.0 in.) long is stressed in tension and found to elongate 7.6 mm (0.30 in.).See all results for this question

brass 260 Product Guide from Online Metals

The most ductile of the yellow brasses, 260 brass also has a good strength, finishing capabilities and corrosion resistance. It is also easily machined and cold formed. 260 brass should not be used in projects where it will come into contact with acetic acid, hydrochloric acid, nitric acid, or ammonia.Some results are removed in response to a notice of local law requirement. For more information, please see here.[Solved] A light pressure vessel is made of 2024-T3 solveda brass alloy is known to have a yield strength ofA brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elastic modulus of 103 GPa (15.0 × 106 psi). A cylindrical specimen of this alloy 12.7 mm (0.50 in.) in diameter and 250 mm (10.0 in.) long is stressed in tension and found solveda brass alloy is known to have a yield strength of View Answer3.7/5(3)[Solved] A brass alloy to be used for a spring application solveda brass alloy is known to have a yield strength ofA brass alloy to be used for a spring application must have a modulus of resilience of at least 0.75 MPa (110 psi). What must be its minimum yield strength? Students also viewed these Thermodynamics questions A steel alloy to be used for a spring application must have a modulus of resilience of at least 2.07 MPa (300 psi).4.3/5(3)

[Solved] A brass alloy is known to have a yield strength solveda brass alloy is known to have a yield strength of

Answer to A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elastic modulus of 103 GPa (15.0 × | SolutionInn Toggle navigation Menu[Solved] A brass alloy is known to have a yield strength solveda brass alloy is known to have a yield strength ofA brass alloy is known to have a yield strength of 275MPa (40,000psi), a tensile strength of 380MPa (55,000psi), and an elastic modulus of 103GPa (15.0 × 10 6 psi). A cylindrical specimen of this alloy 12.7 mm (0.50 in.) in diameter and 250 mm (10.0 in.) long is stressed in 5/5(3)[Solved] (a) From the plot of yield strength versus (grain solveda brass alloy is known to have a yield strength ofA brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 solveda brass alloy is known to have a yield strength of.. solveda brass alloy is known to have a yield strength of to elongate 7.6 mm (0.30 in.). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length?5/5(1)

Which is the best source for metals and alloys?

The sources that have been most helpful are the metal and alloy producers, ALLOY DIGEST, WOLDMANS ENGI- NEERING ALLOYS, International Nickels publications and UNIFIED NUMBERING SYSTEM for METALS and ALLOYS. These data are presented to facilitate general compari- son and are not intended for specification or design purposes.See all results for this questionWhat is the tensile strength of brass?6.20 A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elastic modulus of 103 GPa (15.0 × 106psi). A cylindrical specimen of this alloy 12.7 mm (0.50 in.) in diameter and 250 mm (10.0 in.) long is stressed in tension and found to elongate 7.6 mm (0.30 in.).See all results for this questionWhat is the tensile strength of a plastic strain?It intersects the stress-strain curve at approximately 285 MPa (41,000 psi ). (d) The tensile strength is approximately 370 MPa (54,000 psi), corresponding to the maximum stress on the complete stress-strain plot. (e) The ductility, in percent elongation, is just the plastic strain at fracture, multiplied by one-hundred.See all results for this question

What is the stress of a bronze alloy?

6.7 For a bronze alloy, the stress at which plastic deformation begins is 275 MPa (40,000 psi), and the modulus of elasticity is 115 GPa (16.7 × 106 psi).See all results for this questionWhat To Know About Steel Yield StrengthSep 05, 2017 · What is yield strength? Yield strength is a measurement of how well a steel material can hold up to stress. Through mathematical formulas and testing, metal professionals calculate the amount of stress that is necessary to cause the metal to deform. In the case of steel, high yield steel will hold up to stress and pressure better than low yield solveda brass alloy is known to have a yield strength ofStress-Strain Behavior= 75.25 mm 6.20 A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elastic modulus of 103 GPa (15.0 × 106psi). A cylindrical specimen of this alloy 12.7 mm (0.50 in.) in diameter and 250 mm (10.0 in.) long is stressed in tension and found to elongate 7.6 mm (0.30 in.).

Solved: A Brass Alloy Is Known To Have A Yield Strength Of solveda brass alloy is known to have a yield strength of

A brass alloy is known to have a yield strength of 275 MPa, a tensile strength of 380 MPa, and an elastic modulus of 103 GPa. A cylindrical specimen of this alloy 12.7 mm in diameter and 250 mm long is stressed in tension and found to elongate 7.6 mm.Solved: A Brass Alloy Is Known To Have A Yield Strength Of solveda brass alloy is known to have a yield strength ofA brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elastic modulus of 103 GPa (15.0×106 psi). A cylindrical specimen of this alloy 9.0 mm (0.35 in.) in diameter and 245 mm (9.65 in.) long is stressed in Solved: A Brass Alloy Is Known To Have A Yield Strength Of solveda brass alloy is known to have a yield strength ofA brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elastic modulus of 103 GPa (15.0×10 6 psi). A cylindrical specimen of this alloy 7.4 mm (0.29 in.) in diameter and 293 mm (11.54 in.) long is

Solved: A Brass Alloy Is Known To Have A Yield Strength Of solveda brass alloy is known to have a yield strength of

A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elastic modulus of 103 GPa (15.0 x 100 psi). A cylindrical specimen of this alloy 12.7 mm (0.50 in.) in diameter and 250 mm (10.0 in.) long is stressed in Solved: A Brass Alloy Is Known To Have A Yield Strength Of solveda brass alloy is known to have a yield strength ofA brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elastic modulus of 103 GPa (15.0×10 6 psi). A cylindrical specimen of this alloy 7.4 mm (0.29 in.) in diameter and 293 mm (11.54 in.) long is stressed in tension and found to elongate 9.8 mm (0.39 in.).Solutions - 6.20 A brass alloy is known to have a yield solveda brass alloy is known to have a yield strength of6.20 A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elastic modulus of 103 GPa (15.0 10 6 psi). A cylindrical specimen of this alloy 12.7 mm (0.50 in.) in diameter and 250 mm (10.0 in.) long is stressed in tension and found to elongate 7.6 mm (0.30 in.).

A brass alloy is known to have a yield strength of 275 MPa solveda brass alloy is known to have a yield strength of

A brass alloy is known to have a yield strength of 275 MPa, a tensile strength of 380 MPa, and an elastic modulus of 103 GPa. A cylindrical specimen of this alloy 12.7 mm in diameter and 250 mm solveda brass alloy is known to have a yield strength ofA brass alloy is known to have a yield strength of 275 MPa solveda brass alloy is known to have a yield strength ofA brass alloy is known to have a yield strength of 275 MPa, a tensile strength of 380 MPa, and an elastic modulus of 103 GPa.A cylindrical specimen of this alloy 12.7 mm in is diameter and 250 mm solveda brass alloy is known to have a yield strength ofA brass alloy is known to have a yield strength of 275 MPa solveda brass alloy is known to have a yield strength ofA brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elastic modulus of 103 GPa (15.0 106 psi).

A brass alloy is known to have a yield strength of 275 MPa solveda brass alloy is known to have a yield strength of

6.20 A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elastic modulus of 103 GPa (15.0 10 6 psi). A cylindrical specimen of this alloy 12.7 mm (0.50 in.) in diameter and 250 mm (10.0 in.) long is stressed in tension and found to elongate 7.6 mm (0.30 in.).A brass alloy is known to have a yield strength of 240 MPa solveda brass alloy is known to have a yield strength of6.20 A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elastic modulus of 110 GPa (16.0 × 10 6 psi). A cylindrical specimen of this alloy 15.2 mm (0.60 in.) in diameter and 380 mm (15.0 in.) long is stressed in tension and found to elongate 1.9 mm (0.075 in.). On the basis of the information given, is it possible to solveda brass alloy is known to have a yield strength of(Solved) - A brass alloy is known to have a yield strength solveda brass alloy is known to have a yield strength ofOct 18, 2013 · A brass alloy is known to have a yield strength of 275MPa (40,000psi), a tensile strength of 380MPa (55,000psi), and an elastic modulus of 103GPa (15.0 (106psi). A cylindrical specimen of this alloy 12.7 mm (0.50 in.) in diameter and 250 mm (10.0 in.) long is stressed in tension and found to elongate 7.6 mm (0.30 in.).

(Solved) - A brass alloy is known to have a yield strength solveda brass alloy is known to have a yield strength of

Jan 13, 2016 · A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elastic modulus of 110 GPa (16.0 × 10 6 psi). A cylindrical specimen of this alloy 15.2 mm (0.60 in.) in diameter and 380 mm (15.0 in.) long is stressed in tension and found to elongate 1.9 mm (0.075 in.).

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(3 points) Label the following dienes as either a cumulated, conjugated or an isolated diene. A) B)

Question:

(3 points) Label the following dienes as either a cumulated, conjugated or an isolated diene. A) B)

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Quasicategories for non-simplicial model categories

If I have a simplicially enriched model category, then I can take the coherent nerve of the full subcategory of bifibrant opjects to obtain a quasicategory. If I have a model category that is not simplicially enriched, then I could take the hammock localisation first and then take the coherent nerve. I have no technical objection to this, but I find it aesthetically displeasing. I would prefer to have a construction that accepts an arbitrary model category $\mathcal{C}$ and produces a quasicategory $N'(\mathcal{C})$ with the same homotopy theory in a single step. Ideally, all objects of $\mathcal{C}$ would appear as $0$-simplices in $N'(\mathcal{C})$, not just the bifibrant ones. Does such a thing exist in the literature?

One idea (in the spirit of the thesis of James Cranch) is as follows. Given a left proper model category $\mathcal{C}$, an $n$-simplex of $N'(\mathcal{C})$ would be an $n$-fold cospan diagram consisting of objects $X_{ij}$ for $0\leq i\leq j\leq n$. There would be cofibrations $X_{ij}\to X_{i,j+1}$ and weak equivalences $X_{i+1,j}\to X_{ij}$ and the resulting squares would be cocartesian. I have not made any serious attempt to check whether this works.

Alternatively, it may be that an answer is essentially contained in the paper of Barwick and Kan on Relative Categories, but I have not properly digested that.

• The construction you outline, although very appealing, unfortunately does not work - it does not actually invert the weak equivalences. (For instance, the mapping spaces are just groupoids.) – Rune Haugseng Aug 11 '15 at 18:00
• Just as a remark: After taking hammock localization, you first would have to fibrantly replace your simplicial category (for example, by applying $Ex^\infty$ to all mapping spaces) and then apply the coherent nerve. – Lennart Meier Aug 12 '15 at 7:02

It's not quite in the literature, but there is a fully explicit construction that avoids hammock localisation or any kind of fibrant replacement: by a recent result of Lennart Meier, a certain "double cosubdivision" of the Rezk classification diagram of a model category is a complete Segal space, so (by a result of Joyal and Tierney) we can take degreewise 0-simplices to get a quasicategory. The vertices of the quasicategory constructed above are indeed all the objects of the model category we start with, and the edges are diagrams of the form $$\bullet \rightarrow \bullet \leftarrow \bullet \rightarrow \bullet \leftarrow \bullet$$ where every arrow except possibly the interior $\rightarrow$ is a weak equivalence. Perhaps this is more complicated than you hoped for, but it is unavoidable in the general case because we do not always have a two-arrow calculus.
You can consider the marked simplicial set $(N(\mathcal{C}),\mathcal{W})$, where $N$ is the usual nerve functor and $\mathcal{W}$ is the class of weak equivalences in your model category $\mathcal{C}$. Then take $N'(\mathcal{C})$ to be the underlying simplicial set of any fibrant replacement of $(N(\mathcal{C}),\mathcal{W})$ in the Cartesian model structure. This paper by Hinich shows that this procedure gives the "correct" $\infty$-category associated to $\mathcal{C}$. Perhaps some unwinding of definitions will give you a more explicit construction.


text:
• Describe the effect of the transformation of the form $y = a\cdot f(x)$ given the graph of $y = f(x)\text{,}$ and sketch a graph.
• Describe the effects on the graph of a function $f(x)$ when comparing $f(x)$ to $f(-x)$ or $-f(x)\text{.}$


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# Tikz-Feynman: Particles overlap and appear at only one external leg

# I am facing trouble with the tikz-feynman package that I have recently installed. I am currently using LuaTex as the engine in TexLive. Also, I am using a non-English language, so I am using ltjsarticle as my documentclass (this might not be the main problem but I will state it anyway).
As is said in the title, the main problem is that despite using a sample code from CTAN (or maybe some other site), I get a Feynman diagram that looks like the one posted in this page.
Here is the code that I have used:

\documentclass{ltjsarticle}
\usepackage[dvipdfmx]{graphicx}
\usepackage{gensymb}
\usepackage{here}
\usepackage{listings}
\usepackage{url}
\usepackage{cases}
\usepackage[compat=1.1.0]{tikz-feynman}

\begin{document}
\feynmandiagram [small, horizontal=a to t1]
{ a [particle=$$\pi^{0}$$] -- [scalar] t1 -- t2 -- t3 -- t1, t2 --
[photon] p1 [particle=$$\gamma$$], t3 -- [photon] p2 [particle=$$\gamma$$], };

\feynmandiagram [inline=(b), horizontal=a to b, red]
{a -- b -- {c [particle={$$\alpha$$}], d [particle=$$\beta$$]}
};

\end{document}

• If you remove \usepackage[dvipdfmx]{graphicx}, it works fine. (I dunno why, I'm a newcomer to lualatex.) It also works if you put \usepackage{graphicx}. – marmot Jan 4 '18 at 17:53
• There is at least one other place where lualatex and dvipdfmx appear to interfere, see p.8 of the microtype documentation. – marmot Jan 4 '18 at 18:15
• One more thing: if you put \usepackage[dvipdfmx]{graphicx} after \usepackage[compat=1.1.0]{tikz-feynman}, this produces an error. If you put \usepackage{graphicx}, it again works fine. – marmot Jan 4 '18 at 23:12